Compiled and Solved Problems in Geometry and Trigonometry 255 Compiled and Solved Problems in Geometry and Trigonometry

Florentin Smarandache Compiled and Solved Problems in Geometry and Trigonometry 255 Compiled and Solved Problems in Geometry and Trigonometry

FLORENTIN SMARANDACHE

255 Compiled and Solved Problems in Geometry and Trigonometry

(from Romanian Textbooks)

Educational Publisher 2015

Florentin Smarandache

Peer reviewers: Prof. Rajesh Singh, School of Statistics, DAVV, Indore (M.P.), India. Dr. Linfan Mao, Academy of Mathematics and Systems, Chinese Academy of Sciences, Beijing 100190, P. R. China. Mumtaz Ali, Department of Mathematics, Quaid-i-Azam University, Islamabad, 44000, Pakistan Prof. Stefan Vladutescu, University of Craiova, Romania. Said Broumi, University of Hassan II Mohammedia, Hay El Baraka Ben M'sik, Casablanca B. P. 7951, Morocco.

E-publishing, Translation & Editing: Dana Petras, Nikos Vasiliou AdSumus Scientific and Cultural Society, Cantemir 13, Oradea, Romania

Educational Publisher, Columbus, USA ISBN: 978-1-59973-299-2

2 255 Compiled and Solved Problems in Geometry and Trigonometry

Table of Content

Explanatory Note ...... 4

Problems in Geometry (9th grade) ...... 5

Solutions ...... 11

Problems in Geometry and Trigonometry ...... 38

Solutions ...... 42

Other Problems in Geometry and Trigonometry (10th grade) ...... 60

Solutions ...... 67

Various Problems ...... 96

Solutions ...... 99

Problems in Spatial Geometry ...... 108

Solutions ...... 114

Lines and Planes ...... 140

Solutions ...... 143

Projections ...... 155

Solutions ...... 159

Review Problems ...... 174

Solutions ...... 182

Florentin Smarandache

Explanatory Note

College in Balcesti, Valcea (Romania), Lycée Sidi El Hassan Lyoussi in Sefrou (Morocco), then at the "Nicolae Balcescu" National College in Craiova and Dragotesti General School (Romania), but also I did intensive private tutoring for students preparing their university entrance examination. After that, I have escaped in Turkey in September 1988 and lived in a political refugee camp in Istanbul and Ankara, and in March 1990 I immigrated to United States. The degree of difficulties of the problems is from easy and medium to hard. The solutions of the problems are at the end of each chapter. One can navigate back and forth from the text of the problem to its solution using bookmarks. The book is especially a didactical material for the mathematical students and instructors.

The Author

4 255 Compiled and Solved Problems in Geometry and Trigonometry

Problems in Geometry (9th grade)

1. The measure of a regular polygon’s interior angle is four times bigger than

the measure of its external angle. How many sides does the polygon have?

Solution to Problem 1

2. How many sides does a convex polygon have if all its external angles are obtuse?

Solution to Problem 2

3. Show that in a convex quadrilateral the bisector of two consecutive angles forms an angle whose measure is equal to half the sum of the measures of the other two angles. Solution to Problem 3

4. Show that the surface of a convex pentagon can be decomposed into two quadrilateral surfaces. Solution to Problem 4

5. What is the minimum number of quadrilateral surfaces in which a convex polygon with 9, 10, 11 vertices can be decomposed? Solution to Problem 5

6. If (퐴퐵퐶)̂ ≡ (퐴̂′퐵′퐶′), then ∃ bijective function 푓 = (퐴퐵퐶)̂ → (퐴̂′퐵′퐶′) such that for ∀ 2 points 푃, 푄 ∈ (퐴퐵퐶)̂ , ‖푃푄‖ = ‖푓(푃)‖, ‖푓(푄)‖, and vice versa. Solution to Problem 6

Florentin Smarandache

7. If ∆퐴퐵퐶 ≡ ∆퐴′퐵′퐶′ then ∃ bijective function 푓 = 퐴퐵퐶 → 퐴′퐵′퐶′ such that (∀) 2 points 푃, 푄 ∈ 퐴퐵퐶, ‖푃푄‖ = ‖푓(푃)‖, ‖푓(푄)‖, and vice versa. Solution to Problem 7

8. Show that if ∆퐴퐵퐶~∆퐴′퐵′퐶′, then [퐴퐵퐶]~[퐴′퐵′퐶′]. Solution to Problem 8

9. Show that any two rays are congruent sets. The same property for lines. Solution to Problem 9

10. Show that two disks with the same radius are congruent sets. Solution to Problem 10

11. If the function 푓: 푀 → 푀′ is isometric, then the inverse function 푓−1: 푀 → 푀′ is as well isometric. Solution to Problem 11

′ ′ ′ ′ 12. If the convex polygons 퐿 = 푃1, 푃2, … , 푃푛 and 퐿 = 푃1, 푃2, … , 푃푛 have |푃푖, 푃푖+1| ≡ ′ ′ ̂ ′ ′̂′ |푃푖 , 푃푖+1| for 푖=1,2,…,푛−1, and 푃푖푃푖+1푃푖+2 ≡ 푃푖 푃푖+1푃푖+2, (∀) 푖 = 1, 2, … , 푛 − 2, then 퐿 ≡ 퐿′ and [퐿] ≡ [퐿′]. Solution to Problem 12

13. Prove that the ratio of the perimeters of two similar polygons is equal to their similarity ratio. Solution to Problem 13

14. The parallelogram 퐴퐵퐶퐷 has ‖퐴퐵‖ = 6, ‖퐴퐶‖ = 7 and 푑(퐴퐶) = 2. Find 푑(퐷, 퐴퐵). Solution to Problem 14

255 Compiled and Solved Problems in Geometry and Trigonometry

15. Of triangles 퐴퐵퐶 with ‖퐵퐶‖ = 푎 and ‖퐶퐴‖ = 푏, 푎 and 푏 being given numbers, find a triangle with maximum area. Solution to Problem 15

16. Consider a square 퐴퐵퐶퐷 and points 퐸,퐹,퐺,퐻,퐼,퐾,퐿,푀 that divide each side in three congruent segments. Show that 푃푄푅푆 is a square and its area is 2 equal to 휎[퐴퐵퐶퐷]. 9 Solution to Problem 16

17. The diagonals of the trapezoid 퐴퐵퐶퐷 (퐴퐵||퐷퐶) cut at 푂. a. Show that the triangles 퐴푂퐷 and 퐵푂퐶 have the same area; b. The parallel through 푂 to 퐴퐵 cuts 퐴퐷 and 퐵퐶 in 푀 and 푁. Show that ||푀푂|| = ||푂푁||. Solution to Problem 17

18. 퐸 being the midpoint of the non-parallel side [퐴퐷] of the trapezoid 퐴퐵퐶퐷, show that 휎[퐴퐵퐶퐷] = 2휎[퐵퐶퐸]. Solution to Problem 18

19. There are given an angle (퐵퐴퐶)̂ and a point 퐷 inside the angle. A line through 퐷 cuts the sides of the angle in 푀 and 푁. Determine the line 푀푁 such that the area ∆퐴푀푁 to be minimal. Solution to Problem 19

20. Construct a point 푃 inside the triangle 퐴퐵퐶, such that the triangles 푃퐴퐵, 푃퐵퐶, 푃퐶퐴 have equal areas. Solution to Problem 20

21. Decompose a triangular surface in three surfaces with the same area by parallels to one side of the triangle. Solution to Problem 21

Florentin Smarandache

22. Solve the analogous problem for a trapezoid. Solution to Problem 22

23. We extend the radii drawn to the peaks of an equilateral triangle inscribed in a circle 퐿(푂, 푟), until the intersection with the circle passing through the peaks of a square circumscribed to the circle 퐿(푂, 푟). Show that the points thus obtained are the peaks of a triangle with the same area as the hexagon inscribed in 퐿(푂, 푟). Solution to Problem 23

24. Prove the leg theorem with the help of areas. Solution to Problem 24

25. Consider an equilateral ∆퐴퐵퐶 with ‖퐴퐵‖ = 2푎. The area of the shaded surface determined by circles 퐿(퐴,푎),퐿(퐵,푎),퐿(퐴,3푎) is equal to the area of the circle sector determined by the minor arc (̂퐸퐹) of the circle 퐿(퐶, 푎). Solution to Problem 25

26. Show that the area of the annulus between circles 퐿(푂, 푟2) and 퐿(푂, 푟2) is equal to the area of a disk having as diameter the tangent segment to

circle 퐿(푂, 푟1) with endpoints on the circle 퐿(푂, 푟2). Solution to Problem 26

27. Let [푂퐴],[푂퐵] two ⊥ radii of a circle centered at [푂]. Take the points 퐶 and 퐷 on the minor arc 퐴퐵퐹̂ such that 퐴퐶̂ ≡퐵퐷̂ and let 퐸, 퐹 be the projections of 퐶퐷 onto 푂퐵. Show that the area of the surface bounded by [퐷퐹], [퐹퐸[퐸퐶]] and arc 퐶퐷̂ is equal to the area of the sector determined by arc 퐶퐷̂ of the circle 퐶(푂, ‖푂퐴‖). Solution to Problem 27

255 Compiled and Solved Problems in Geometry and Trigonometry

28. Find the area of the regular octagon inscribed in a circle of radius 푟. Solution to Problem 28

29. Using areas, show that the sum of the distances of a variable point inside the equilateral triangle 퐴퐵퐶 to its sides is constant. Solution to Problem 29

30. Consider a given triangle 퐴퐵퐶 and a variable point 푀 ∈ |퐵퐶|. Prove that between the distances 푥 = 푑(푀, 퐴퐵) and 푦 = 푑(푀, 퐴퐶) is a relation of 푘푥 + 푙푦 = 1 type, where 푘 and 푙 are constant. Solution to Problem 30

31. Let 푀 and 푁 be the midpoints of sides [퐵퐶] and [퐴퐷] of the convex quadrilateral 퐴퐵퐶퐷 and {푃} = 퐴푀 ∩ 퐵푁 and {푄} = 퐶푁 ∩ 푁퐷. Prove that the area of the quadrilateral 푃푀푄푁 is equal to the sum of the areas of triangles 퐴퐵푃 and 퐶퐷푄. Solution to Problem 31

32. Construct a triangle having the same area as a given pentagon. Solution to Problem 32

33. Construct a line that divides a convex quadrilateral surface in two parts with equal areas. Solution to Problem 33

34. In a square of side 푙, the middle of each side is connected with the ends of the opposite side. Find the area of the interior convex octagon formed in this way. Solution to Problem 34

Florentin Smarandache

35. The diagonal [퐵퐷] of parallelogram 퐴퐵퐶퐷 is divided by points 푀, 푁, in 3 segments. Prove that 퐴푀퐶푁 is a parallelogram and find the ratio between 휎[퐴푀퐶푁] and 휎[퐴퐵퐶퐷]. Solution to Problem 35

36. There are given the points 퐴, 퐵, 퐶, 퐷, such that 퐴퐵 ∩ 퐶퐷 = {푝}. Find the locus of point 푀 such that 휎[퐴퐵푀] = 휎[퐶퐷푀]. Solution to Problem 36

37. Analogous problem for 퐴퐵||퐶퐷. Solution to Problem 37

38. Let 퐴퐵퐶퐷 be a convex quadrilateral. Find the locus of point 푥1 inside 퐴퐵퐶퐷 such that 휎[퐴퐵푀] + 휎[퐶퐷푀] = 푘, 푘 – a constant. For which values of 푘 the desired geometrical locus is not the empty set? Solution to Problem 38

255 Compiled and Solved Problems in Geometry and Trigonometry

Solutions

Solution to Problem 1.

180 (푛 − 2) 180 = 4  푛 = 10 푛 5

Solution to Problem 2.

푥 > 900 Let 푛 = 3 1 ⟹ 푥 > 900} ⟹ 푥 + 푥 + 푥 > 2700, so 푛 = 3 is possible. 푥 , 푥 , 푥 ∢ ext 2 1 2 3 1 2 3 0 푥3 > 90

0 푥1 > 90 Let 푛 = 4 0 ⟹ ⋮ } ⟹ 푥1 + 푥2 + 푥3 + 푥4 > 360 , so 푛 = 4 is impossible. 푥1, 푥2, 푥3, 푥4∢ ext 0 푥3 > 90

Therefore, 푛 = 3.

Solution to Problem 3.

Florentin Smarandache

m(퐷̂) + m(퐶̂) m(퐴퐸̂퐵) = 2 m(퐴̂) + m(퐵̂) + m(퐶̂) + m(퐷̂) = 360° m(퐴̂) + m(퐵̂) m(퐶̂) + m(퐷̂) = 180° − 2 2 m(퐴̂) m(퐵̂) m(퐴퐸퐵̂) = 180° − − = 2 2 m(퐶̂) + m(퐷̂) m(퐶̂) + (퐷̂) = 180° − 180° + = 2 2

Solution to Problem 4.

Let 퐸퐷퐶̂ ⇒ 퐴,퐵 ∈ int.퐸퐷퐶̂. Let 푀 ∈ |퐴퐵| ⇒ 푀 ∈ int. 퐸퐷퐶̂ ⇒ |퐷푀 ⊂ int. 퐸퐷퐶̂, |퐸퐴| ∩ |퐷푀 = ∅ ⇒ 퐷퐸퐴푀 quadrilateral. The same for 퐷퐶퐵푀.

Solution to Problem 5.

9 vertices; 10 vertices; 11 vertices; 4 quadrilaterals. 4 quadrilaterals. 5 quadrilaterals.

255 Compiled and Solved Problems in Geometry and Trigonometry

Solution to Problem 6.

We assume that ABĈ ≡ A’B’C’̂ . We construct a function 푓:̂ 퐴퐵퐶 → 퐴′퐵′퐶′̂ such that 푓 (B) = B′ { if P ∈ |BA, 푓(P) ∈ B′A′ 푃 ∈ |퐵퐶, 푓(푃) ∈ 퐵′퐶′ such that ‖퐵푃‖ = ‖퐵′푃′‖ where 푃′ = 푓(퐹). The so constructed function is bijective, since for different arguments there are different corresponding values and ∀ point from 퐴′퐵′퐶′ is the image of a single point from 퐴퐵퐶̂ (from the axiom of segment construction).

If 푃, 푄 ∈ this ray, ‖BP‖ = ‖B′P′‖ } ⟹ ‖PQ‖ = ‖BQ‖ − ‖BP‖ = ‖B′Q′‖ − ‖B′P′‖ = ‖P′Q′‖ = ‖푓(P), 푓(Q)‖. ‖BQ‖ = ‖B′Q′‖

If 푃, 푄 ∈ a different ray, ‖BP‖ = ‖B′P′‖ ‖BQ‖ = ‖B′Q′‖} ⟹ ∆PBQ = ∆P′B′Q′ ‖PQ‖ = ‖P′Q′‖ = ‖푓 (P), 푓 (Q)‖. PBQ̂ ≡ P̂′B′Q′

Vice versa. Let 푓 ∶ ABC → A′B′C′ such that 푓 bijective and ‖PQ‖ = ‖푓(P), 푓(Q)‖.

Let 푃, 푄 ∈ |퐵퐴 and 푅푆 ∈ |퐵퐶.

Florentin Smarandache

‖PQ‖ = ‖P′Q′‖ ‖PS‖ = ‖P′S′‖ } ⟹ ∆PQS ≡ ∆P′Q′S′ ⟹ ̂QPS ≡ Q̂′P′S′ ⟹ BPŜ ≡ B̂′P′S′ (1); ‖QS‖ = ‖Q′S′‖ ‖푃푆‖ = ‖푃′푆′‖ ‖푅푃‖ = ‖푅′푃′‖} ⟹ ∆PRS ≡ ∆P′R′S′′ ⟹ PSB̂ ≡ P̂′S′B′ (2). ‖푃푆‖ = ‖푃′푆′‖ From (1) and (2) ⟹ PBĈ ≡ P′B′S′̂ (as diff. at 180°) i.e. ABĈ ≡ A′B′C′̂ .

Solution to Problem 7.

Let ∆퐴퐵퐶 ≡ ∆퐴′퐵′퐶′. We construct a function 푓 ∶ 퐴퐵퐶 → 퐴′퐵′퐶′ such that 푓(퐴) = 퐴′,푓(퐵) = 퐵′,푓(퐶) = 퐶′ and so 푃 ∈ |퐴퐵| → 푃′ = 푓(푃) ∈ |퐴′퐵′| such that ||퐴푃|| = ||퐴′푃′||; 푃 ∈ |퐵퐶| → 푃′ = 푓(푃) ∈ |퐵′퐶′| such that ||퐵푃|| = ||퐵′푃′||; 푃 ∈ |퐶퐴| → 푃′ = 푓(푃) ∈ |퐶′퐴′| such that ||퐶푃|| = ||퐶′푃′||. The so constructed function is bijective.

Let 푃 ∈ |퐴퐵| and 푎 ∈ |퐶퐴| ⟹ 푃′ ∈ |퐴′퐵′| and 푄′ ∈ |퐶′퐴′|. ‖AP‖ = ‖A′P′‖ ‖CQ‖ = ‖C′Q′‖} ⟹ ‖AQ‖ = ‖A′Q′‖; A ≡ A′ ⟹ ∆APQ ≡ ∆A′P′Q′ ⟹ ‖PQ‖ = ‖P′Q′‖. ‖CA‖ = ‖C′A′‖ Similar reasoning for (∀) point 푃 and 푄.

Vice versa.

We assume that ∃ a bijective function 푓 ∶ 퐴퐵퐶 → 퐴′퐵′퐶′ with the stated properties.

We denote 푓(A) = A′′, 푓(B) = B′′, 푓(C) = C′′ ⟹ ‖AB‖ = ‖A′′B′′‖, ‖BC‖ = ‖B′′C′′‖, ‖AC‖ = ‖A′′C′′‖∆ABC = ∆A′′B′′C′′. Because 푓(ABC) = 푓([AB] ∪ [BC] ∪ [CA]) = 푓([AB]) ∪ 푓([BC]) ∪ 푓([CA]) = [A′′B′′] ∪ [B′′C′′][C′′A′′] = A′′B′′C′′. But by the hypothesis 푓(퐴퐵퐶) = 푓(퐴’퐵’퐶’), therefore A′′B′′C′′ = ∆A′B′C′ ⟹ ∆ABC ≡ ∆A′B′C′. 14

255 Compiled and Solved Problems in Geometry and Trigonometry

Solution to Problem 8.

If ∆퐴퐵퐶~∆퐴’퐵’퐶’ then (∀) 푓: 퐴퐵퐶 → 퐴’퐵’퐶’ and 푘 > 0 such that: ||푃푄|| = 푘 ||푓(푃), 푓(푄)||, 푃, 푄 ∈ 퐴퐵퐶; ‖AB‖ ‖BC‖ ‖CA‖ ‖AB‖ = 푘‖A′B′‖ = = = 푘 ∆퐴퐵퐶~∆퐴′퐵′퐶′ ⟹ ‖A′B′‖ ‖B′C′‖ ‖C′A′‖ } ⟹ ‖BC‖ = 푘‖B′C′‖ . Â ≡ Â′;B̂ ≡ B̂′;Ĉ ≡ C′̂ ‖CA‖ = 푘‖C′A′‖ We construct a function 푓: 퐴퐵퐶 → 퐴′퐵′퐶′ such that 푓(퐴) = 퐴′, 푓(퐵) = 퐵′, 푓(퐶) = 퐶′; if 푃 ∈ |퐵퐶| → 푃 ∈ |퐵′퐶′| such that ||퐵푃|| = 푘||퐵′푃′||; if 푃 ∈ |퐶퐴| → 푃 ∈ |퐶′퐴′| such that ||퐶푃|| = 푘||퐶′푃′||; 푘 – similarity constant.

Let 푃, 푄 ∈ 퐴퐵 such that 푃 ∈ |퐵퐶|, 푄 ∈ |퐴퐶| ⟹ 푃′ ∈ |퐵′퐶′| and ||퐵푃|| = 푘||퐵′푃′|| Q′ ∈ |A′C′| and ‖CQ‖ = 푘‖C′Q′‖ (1); As ‖BC‖ = 푘‖B′C′‖ ⟹ ‖PC‖ = ‖BC‖ − ‖BP‖ = 푘‖B′C′‖ − 푘‖B′P′‖ = = 푘(‖B′C′‖ − ‖B′P′‖) = 푘‖P′C′‖ (2); Ĉ ≡ Ĉ′ (3). From (1), (2), and (3) ⟹ ∆PCQ~∆P′C′Q′ ⟹ ‖PQ‖ = 푘‖P′Q′‖ . Similar reasoning for 푃, 푄 ∈ 퐴퐵퐶. We also extend the bijective function previously constructed to the interiors of the two triangles in the following way:

Let 푃 ∈ int. 퐴퐵퐶 and we construct 푃’ ∈ int. 퐴’퐵’퐶’ such that ||퐴푃|| = 푘||퐴’푃’|| (1). Let 푄 ∈ int. 퐴퐵퐶 → 푄′ ∈ int. 퐴’퐵’퐶’ such that BAQ̂ ≡ B’A’Q’̂ and ||퐴푄|| = 푘||퐴’푄’|| (2).

Florentin Smarandache

From (1) and (2), AP AQ = = 푘, PAQ̂ ≡ P̂′A′Q′ ⟹ ∆APQ ~ ∆A′P′Q′ ⟹ ‖PQ‖ = 푘‖P′Q′‖, A′P′ A′Q′ but 푃, 푄 ∈ [퐴퐵퐶], so [퐴퐵퐶] ~[퐴′퐵′퐶′].

Solution to Problem 9.

a. Let |푂퐴 and |푂′퐴′ be two rays:

Let 푓: |푂퐴 → |푂′퐴′ such that 푓(푂) = 푂′ and 푓(푃) = 푃′ with ||푂푃|| = ||푂′푃′|. The so constructed point 푃′ is unique and so if 푃 ≠ 푄 ⟹ ||푂푃|| ≠ ||푂푄|| ⟹ ||푂′푃′|| ≠ ||푂′푄′|| ⟹ 푃′ ≠ 푄′ and (∀)푃′ ∈ |푂′퐴′ (∃) a single point 푃 ∈ |푂퐴 such that ||푂푃|| = ||푂′푃′||. The constructed function is bijective.

If 푃, 푄 ∈ |푂퐴, 푃 ∈ |푂푄| → 푃′푄′ ∈ |푂′퐴′ such that ‖OP‖ = ‖O′P′‖; ‖OQ‖ = ‖O′Q′‖ ⟹ ‖PQ‖ = ‖OQ‖ − ‖OP‖ = ‖O′Q′‖ − ‖O′P′‖ = ‖P′Q′‖(∀)P; 푄 ∈ |OA ⟹ the two rays are congruent.

b. Let 푑 and 푑′ be two lines.

Let 푂 ∈ 푑 and 푂′ ∈ 푑′. We construct a function 푓: 푑 → 푑′ such that 푓(푂) = 푂′ and 푓 (|푂퐴) = |푂′퐴′ and 푓 (|푂퐵) = |푂′퐵′ as at the previous point. It is proved in the same way that 푓 is bijective and that ||푃푄|| = ||푃′푄′|| when 푃 and 푄 belong to the same ray.

255 Compiled and Solved Problems in Geometry and Trigonometry

If 푃, 푄 belong to different rays:

‖푂푃‖ = ‖푂′푃′‖ } ⟹ ‖푃푄‖ = ‖푂푃‖ + ‖푂푄‖ = ‖푂′푃′‖ + ‖푂′푄′‖ = ‖푃′푄′‖ ‖푂푄‖ = ‖푂′푄′‖

and so the two rays are congruent.

Solution to Problem 10.

We construct a function 푓: 퐷 → 퐷′ such that 푓(푂) = 푂′,푓(퐴) = 퐴′ and a point (∀) 푃 ∈ 퐷 → 푃′ ∈ 퐷′ which are considered to be positive. From the axiom of segment and angle construction ⟹ that the so constructed function is bijective, establishing a biunivocal correspondence between the elements of the two sets. Let 푄 ∈ 퐷 → 푄′ ∈ 퐷′ such that ||푂푄′|| = ||푂푄||; 퐴푂푄̂ ≡ 퐴̂′푂′푄′. As: ‖푂푃‖ = ‖푂′푃′‖ ‖푂푄‖ = ‖푂′푄′‖} ⟹ ∆푂푃푄 ≡ 푃̂′푂′푄′ ⟹ ‖푃푄‖ = ‖푃′푄′‖, (∀) 푃, 푄 ∈ 퐷 ⟹ 퐷 ≡ 퐷′. 푃푂푄̂ ≡ 푃̂′푂′푄′

Solution to Problem 11.

Florentin Smarandache

푓: 푀 → 푀′ is an isometry ⟹ 푓 is bijective and (∀) 푃, 푄 ∈ 푀 we have ||푃푄|| = ||푓(푃), 푓(푄)||, 푓 – bijective ⟹ 푓 – invertible and 푓−1 – bijective. ‖푃′푄′‖ = ‖푓(푃); 푓(푄)‖ = ‖푃푄‖ } ⟹ ‖푓−1(푃′); 푓−1(푄′)‖ = ‖푓−1(푓(푃)), 푓−1(푓(푄))‖ = ‖푃푄‖ ‖푃′푄′‖ = ‖푓−1(푓(푃′)), 푓−1(푓(푄′))‖, (∀)푃′, 푄′ ∈ 푀, therefore 푓−1: 푀′ → 푀 is an isometry.

Solution to Problem 12.

′ We construct a function 푓 such that 푓(푃푖) = 푃푖 , 푖 = 1, 2, … , 푛, and if 푃 ∈

|푃푖, 푃푖+1|.

The previously constructed function is also extended inside the polygon as ̂ follows: Let 푂 ∈ int. 퐿 → 푂′ ∈ int. 퐿′ such that 푂푃̂푖푃푖+1 ≡ 푂′푃′푖푃′푖+1 and ‖푂푃푖‖ =

‖푂′푃′푖‖. We connect these points with the vertices of the polygon. It can be easily proved that the triangles thus obtained are congruent.

We construct the function 푔: [퐿] → [퐿′] such that 푓(푃), if 푃 ∈ 퐿 푔(푃) = { 푂′, if 푃 = 푂 ′ 푃 , if 푃 ∈ [푃푖푂푃푖+1] such that ′ ′ ′ 푃̂푖푂푃 ≡ 푃푖̂푂 푃 (∀)푖=1,2,…,푛−1

The so constructed function is bijective (∀) 푃, 푄 ∈ [퐿]. It can be proved by the congruence of the triangles 푃푂푄 and 푃′푂′푄′ that ||푃푄|| = ||푃′푄′||, so [퐿] = [퐿′]

⟹ if two convex polygons are decomposed in the same number of triangles respectively congruent, they are congruent.

255 Compiled and Solved Problems in Geometry and Trigonometry

Solution to Problem 13.

′ ′ ′ ′ 퐿 = 푃1푃2 … , 푃푛; 퐿 = 푃1푃2 … , 푃푛

퐿~퐿′ ⟹ (∃)퐾 > 0 and 푓: 퐿 → 퐿′ such that ‖푃푄‖ = 푘‖푓(푃)푓(푄)‖ (∀)푃, 푄 ∈ 퐿,

′ and 푃퐼 = 푓(푃푖).

Taking consecutively the peaks in the role of 푃 and 푄, we obtain:

‖ ‖ ′ ′ 푃1푃2 ‖푃1푃2‖ = 푘‖푃1푃2‖ ⇒ ′ ′ = 푘 ‖푃1 푃2 ‖ ‖ ‖ ′ ′ 푃2푃3 ‖푃2푃3‖ = 푘‖푃2푃3‖ ⇒ ′ ′ = 푘 ‖푃2 푃3 ‖ ⋮ ⟹

‖푃푛−1푃푛‖ ‖푃 푃 ‖ = 푘‖푃′ 푃′‖ ⇒ = 푘 푛−1 푛 푛−1 푛 ‖푃′ 푃′‖ 푛−1 푛 ‖ ‖ ′ ′ 푃푛푃1 ‖푃푛푃1‖ = 푘‖푃푛푃1‖ ⇒ ′ ′ = 푘 ‖푃푛푃1 ‖ }

‖푃1푃2‖ ‖푃2푃3‖ ‖푃1푃2‖ + ‖푃2푃3‖ + ⋯ + ‖푃푛−1푃푛‖ + ‖푃푛푃1‖ 푃 ⟹ 푘 = ′ ′ = ′ ′ = ⋯ = ′ ′ ′ ′ ′ ′ ′ ′ = ′ . ‖푃1 푃2 ‖ ‖푃2 푃3 ‖ ‖푃1 푃2 ‖ + ‖푃2 푃3 ‖ + ⋯ + ‖푃푛−1푃푛‖ + ‖푃푛푃1 ‖ 푃

Solution to Problem 14.

2∙7 14 7 휎[퐴퐷퐶] = = 7; 휎[퐴퐵퐶퐷] = 2 ∙ 7 = 14 = 6‖퐷퐹‖ ⟹ ‖퐷퐹‖ = = . 2 6 3

Solution to Problem 15.

Florentin Smarandache

ℎ = 푏 ∙ sin 퐶 ≤ 푏; 푎∙ℎ 휎[퐴퐵퐶] = is max. when ℎ is max. 2 max. ℎ = 푏 when sin퐶 = 1 ⇒ 푚(퐶) = 90 ⇒ 퐴퐵퐶 has a right angle at 퐶.

Solution to Problem 16.

‖푀퐷‖ = ‖퐷퐼‖ ⟹ 푀퐷퐼 – an isosceles triangle. ⟹ 푚(퐷푀퐼̂) = 푚(푀퐼퐷̂) = 450 ; The same way, 푚(퐹퐿퐴̂ ) = 푚(퐴퐹퐿̂ ) = 푚(퐵퐸퐻̂ ) = 푚(퐸퐻퐵̂ ). ‖푅퐾‖ ⟹ ‖푆푃‖ = ‖푃푄‖ = ‖푄푅‖ = ‖푅푆‖ ⟹ 푆푅푄푃 is a square.

2푎 4푎2 4푎2 2푎 2 ‖퐴퐵‖ = 푎, ‖퐴퐸‖ = , ‖푀퐼‖ = √ + = √ ; 3 9 9 3 푎2 푎2 푎 푎 2 2‖푅퐼‖2 = ⟹ ‖푅퐼‖2 = ⟹ ‖푅퐼‖ = = √ ; 9 18 3√2 6 2푎 2 푎 2 푎 2 ‖푆푅‖ = √ − 2 √ = √ ; 3 6 3 2푎2 2 휎[푆푅푄푃] = = 휎[퐴퐵퐶퐷]. 9 9

Solution to Problem 17.

‖퐷퐶‖ ∙ ‖퐴퐸‖ 휎[퐴퐶퐷] = 2 ‖퐷퐶‖ ∙ ‖퐵퐹‖ ⟹ 휎[퐴퐶퐷] = 휎[퐵퐶퐷] 휎[퐵퐶퐷] = 2 ‖퐴퐸‖ = ‖퐵퐹‖ }

255 Compiled and Solved Problems in Geometry and Trigonometry

휎[퐴푂퐷] = 휎[퐴푀푂] + 휎[푀푂퐷]

‖푀푂‖ ∙ ‖푂푃‖ [ ] [ ] ‖푂푀‖(‖푂푃‖ + ‖푂푄‖) ‖푂푀‖ ∙ ℎ 휎 퐴푀푂 = 휎 푀푃푂 = ⟹ 휎[퐴푂퐷] = = 2 2 2 ‖푂푀‖ ∙ ‖푂푄‖ 휎[푀푂퐷] = 휎[푀푂푄] = 2 }

The same way, ‖푂푁‖ ∙ ℎ 휎[퐵푂퐶] = . 2 Therefore, ‖푂푀‖ ∙ ℎ ‖푂푁‖ ∙ ℎ 휎[퐴푂퐷] = 휎[퐵푂퐶] ⟹ = ⟹ ‖푂푀‖ = ‖푂푁‖. 2 2

Solution to Problem 18.

‖퐴퐸‖ = ‖퐸퐷‖ ; We draw 푀푁 ⊥ 퐴퐵; 퐷퐶; ℎ ‖퐸푁‖ = ‖퐸푀‖ = ; 2 (‖퐴퐵‖ + ‖퐷퐶‖) ∙ ℎ ‖퐴퐵‖ ∙ ℎ ‖퐷퐶‖ ∙ ℎ (‖퐴퐵‖ + ‖퐷퐶‖) ∙ ℎ 1 휎[퐵퐸퐶] = − − = = 휎[퐴퐵퐶퐷]; 2 4 4 4 2 Therefore, [퐴퐵퐶퐷] = 2휎[퐵퐸퐶] .

Solution to Problem 19.

휎[퐴퐸퐷푁′] is ct. because 퐴, 퐸, 퐷, 푁′ are fixed points. Let a line through 퐷, and we draw ∥ to sides 푁퐷 and 퐷퐸. No matter how we draw a line through 퐷, 휎[푄푃퐴] is formed of: 휎[퐴퐸퐷푁] + 휎[푁푃푂] + 휎[퐷퐸푄]. We have 휎[퐴퐸퐷푁] constant in all triangles 푃퐴푄.

Florentin Smarandache

Let’s analyse: ‖푁′퐷‖ ∙ ℎ ‖퐸푄′‖ ∙ ℎ ‖푁′퐷‖ ‖퐸푄‖ 휎[푃푁′퐷] + 휎[퐷퐸푄] = 1 + 2 = (ℎ + ∙ ℎ ) 2 2 2 1 ‖푁퐷‖ 2 ′ ‖푁′퐷‖ ℎ2 ‖푁 퐷‖ 2 = ∙ (ℎ1 + ∙ ℎ2) = [(ℎ1 − ℎ2) + 2ℎ1ℎ2]. 2 ℎ1 2ℎ1

∆퐴푀푁 is minimal when ℎ1 = ℎ2 ⟹ 퐷 is in the middle of |푃푄|. The construction is thus: ∆퐴푁푀 where 푁푀 || 퐸푁′. In this case we have |푁퐷| ≡ |퐷푀|.

Solution to Problem 20.

‖퐵퐶‖ ∙ ‖퐴퐴′‖ 휎[퐴퐵퐶] = 2

Let the median be |퐴퐸|, and 푃 be the centroid of the triangle. ‖퐵퐶‖∙‖푃퐷‖ Let 푃퐷 ⊥ 퐵퐶. 휎[퐵푃퐶] = . 2 ′ 퐴퐴′ ⊥ 퐵퐶 ‖푃퐷‖ ‖푃퐸‖ 1 ‖퐴퐴 ‖ } ⇒ 퐴퐴′ ∥ 푃퐷 ⇒ ∆푃퐷퐸~∆퐴퐴′퐸 ⇒ = = ⇒ ‖푃퐷‖ = ⇒ 휎[퐵푃퐶] 푃퐷 ⊥ 퐵퐶 ‖퐴퐴′‖ ‖퐴퐸‖ 3 3 ‖퐴퐴′‖ ‖퐵퐶‖ ∙ 1 ‖퐵퐶‖ ∙ ‖퐴퐴′‖ 1 = 3 = = 휎[퐴퐵퐶]. 2 3 2 3 1 We prove in the same way that 휎[푃퐴퐶] = 휎[푃퐴퐵] = 휎[퐴퐵퐶], so the specific 3 point is the centroid.

255 Compiled and Solved Problems in Geometry and Trigonometry

Solution to Problem 21.

Let 푀, 푁 ∈ 퐴퐵 such that 푀 ∈ |퐴푁|. We take 푀푀′ ∥ 퐵퐶, 푀푁′ ∥ 퐵퐶.

휎[퐴푀푀′] 퐴푀 ∆퐴푀푀′~∆퐴퐵퐶 ⇒ = ( ) 휎[퐴퐵퐶] 퐴퐵

1 ‖퐴푀‖ 2 1 ‖퐴퐵‖ 휎[퐴푀푀′] = 휎, ( ) = , ‖퐴푀‖ = ; ∆퐴푁푁′~∆퐴퐵퐶 ⟹ 3 ‖퐴퐵‖ 3 √3

2 휎[퐴푁푁′] ‖퐴푁‖ = ( ) ‖퐴푁‖ 2 2 2 휎[퐴퐵퐶] ‖퐴퐵‖ ⟹ ( ) = ⟹ ‖퐴푁‖ = √ ‖퐴퐵‖ . 2 ‖퐴퐵‖ 3 3 휎[퐴푁푁′] = 휎[퐴퐵퐶] 3 }

Solution to Problem 22.

‖푂퐷‖ = 푎, ‖푂퐴‖ = 푏 ; 1 휎[∆퐶푀′푀] = 휎[푀푀′푁′푁] = 휎[푁푁′퐵퐴] = 휎[퐴퐵퐶퐷] ; 3

휎[푂퐷퐶] ‖푂퐷‖2 푎 휎[푂퐷퐶] 푎2 휎[푂퐷퐶] ∆푂퐷퐶~∆푂퐴퐵 ⟹ = = ⟹ = ⟹ 휎[푂퐴퐵] ‖푂퐴‖2 푏 휎[푂퐴퐵] − 휎[푂퐷퐶] 푏2 − 푎2 휎[퐴퐵퐶퐷] 푎2 = (1) 푏2 − 푎2 23

Florentin Smarandache

휎[푂퐷퐶] ‖푂퐷‖ 2 휎[푂퐷퐶] 푎2 ∆푂퐷퐶~∆푂푀푀′ ⟹ = ( ) ⟹ = 휎[푂푀푀′] ‖푂푀‖ 휎[푂푀푀′] − 휎[푂퐷퐶] ‖푂푀‖2 휎[푂퐷퐶] 푎2 휎[푂퐷퐶] 푎2 ⟹ = ⟹ = (2) [ ′] ‖ ‖2 2 1 ‖ ‖2 2 휎 퐷퐶푀푀 푂푀 − 푎 휎[퐴퐵퐶퐷] 푂푀 − 푎 3 휎[푂퐷퐶] ‖푂퐷‖ 푎2 휎[퐼퐷퐶] 푎2 ∆푂푁푁′~∆푂퐷퐶 ⟹ = = ⟹ = 휎[퐷푁푁′] ‖푂푁‖ ‖푂푁‖ 휎[푂푁푁′] − 휎[푂퐷퐶] ‖푂푁‖2 휎[푂퐷퐶] 푎2 휎[푂퐷퐶] 푎2 ⟹ = ⟹ = (3) [ ] ‖ ‖2 2 2 ‖ ‖2 2 휎 퐷퐶푁푁′ 푂푁 − 푎 휎[퐴퐵퐶퐷] 푂푁 − 푎 3

We divide (1) by (3): 2 ‖푂푁‖2 − 푎2 푎2 + 2푏2 = ⟹ 3‖푂푁‖2 − 3푎2 = 2푏2 − 2푎2 ⟹ ‖푂푁‖2 = . 3 푏2 − 푎2 3

Solution to Problem 23.

3푟2 2 ‖푂퐴‖ = 푟 → ‖퐷퐸‖ = 2푟; 휎 = √ (1) hexagon 2

퐷퐸퐹푂 a square inscribed in the circle with radius 푅 ⟹

⟹ 푙4 = 푅√2 = ‖퐷퐸‖ ⟹ 푃√2 = 2푟 ⟹ 푅 = 푟√2

‖푂푀‖ = 푅 = 푟√2

√3 ‖푂푀‖ ∙ ‖푂푁‖ sin 120 푟√2 ∙ 푟√2 ∙ 푟2√3 휎[푂푀푁] = = 2 = 2 2 2

푟2√3 3푟2√2 휎[푀푁푃] = 3휎[푂푀푁] = 3 = (2) 2 2

From (1) and (2) ⟹ 휎[푀푁푃] = 휎hexagon.

255 Compiled and Solved Problems in Geometry and Trigonometry

Solution to Problem 24.

‖퐴퐵‖2 = ‖퐵퐶‖ ∙ ‖퐵퐴′‖ We construct the squares 퐵퐶퐸퐷 on the hypotenuse and 퐴퐵퐹퐺 on the leg. We draw 퐴퐴′ ⊥ 퐵퐶. 휎[퐴퐵퐹퐺] = ‖퐴퐵‖2 휎[퐴′퐵퐷퐻] = ‖퐵퐷‖ ∙ ‖퐵퐴′‖ = ‖퐵퐶‖ ∙ ‖퐵퐴′‖ …

Solution to Problem 25.

휎(푠1) = 휎[퐴퐵퐶] − 3휎[sect. 퐴퐷퐻] 푙2√3 (2푎)2√3 휎[퐴퐵퐶] = = = 푎2√3 4 4 푟2 휎[sect. 퐴퐷퐻] = 푚(퐷퐻̂ ) 2 휋 휋 휋 푚(퐷퐻̂ ) = 푚(퐷퐻̂ ) = ∙ 600 = 180 180 3 푎2 휋 휋푎2 휎[sect. 퐴퐷퐻] = ∙ = 2 3 2 휋푎2 휋푎2 휎(푠 ) = 푎2√3 − 3 = 푎2√3 − (1) 1 6 2

Florentin Smarandache

휎(푠2) = 휎[sect. 퐴퐸퐺] − 휎[sect. 퐴퐵퐶] − 휎[sect. 퐸퐶퐹] − 휎[sect. 퐺퐵퐹] (3푎)2 휋 푎2 휋 9푎2 휋 휋푎2 휋푎2 = ∙ ∙ 60 − 푎2√3 − ∙ ∙ 120 = ∙ − 푎2√3 − − 2 180 2 180 2 3 3 3 3휋푎2 2휋푎2 = − − 푎2√3 (2) 2 3

From (1) and (2) 3휋푎2 2휋푎2 휋푎2 2휋푎2 휋푎2 ⟹ 휎(푠 ) + 휎(푠 ) = − − = = . 1 2 2 3 2 6 3

Solution to Problem 26.

2 2 2 ‖퐴퐷‖ = 푟2 − 푟1

2 휎[퐿(푂, 푟1)] = 휋푟1

2 휎[퐿(푂, 푟2)] = 휋푟2

2 2 2 2 휎[annulus] = 휋푟2 − 휋푟1 = 휋(푟2 − 푟1 ) (1)

2 2 2 휎[disk diameter‖퐴퐵‖] = 휋‖퐴퐷‖ = 휋(푟2 − 푟1 ) (2)

From (1) and (2) ⟹ 휎[annulus] = 휎[disk diameter].

Solution to Problem 27.

255 Compiled and Solved Problems in Geometry and Trigonometry

휎[퐶퐷퐷′퐶′] 휎[퐶퐷퐸퐹] = 2 휎[퐶퐷퐷′퐶′] = 휎seg[퐶퐷퐵퐷′퐶′] − 휎seg[퐷퐵퐷′]

휋 We denote 푚(퐴퐶̂ ) = 푚(퐵퐷̂ ) = 푎 ⟹ 푚(퐶퐷̂ ) = − 2푎 2 푟2 휎 = (훼 − sin 훼) sect. 2 푟2 푟2 휎[퐶퐷퐵퐷′퐶′] = [휋 − 2훼 − sin(휋 − 2훼)] = 푞 [휋 − 2훼 − sin(휋 − 2훼)] 2 2 푟2 휎 = (2훼 − sin 2훼) 퐴 2 푟2 푟2 휎[퐶퐷퐷′퐶′] = 휎[퐶퐷퐵퐷′퐶′] − 휎[퐷퐵퐷′] = (휋 − 2훼 − sin 2훼) = (2훼 − sin 2훼) 2 2 푟2 푟2 = (휋 − 2훼 − sin 2훼 − 2훼 + sin 2훼) = (휋 − 4훼) 2 2 휎[퐶퐷퐷′퐶′] 푟2 푟2 휋 ⟹ 휎[퐶퐷퐸퐹] = = (휋 − 4훼) = ( − 2훼) (1) 2 4 2 2 푟2 푟2 휋 휎[sect. 퐶푂퐷] = 푚(퐶퐷̂ ) = ( − 2훼) (2) 2 2 2

From (1) and (2) ⟹ 휎[퐶퐷퐸퐹] = 휎[sect. 퐶푂퐷].

‖푂1퐹‖ = ‖푂퐸‖ 푏푐 휎[square] = ‖퐷퐸‖2 = ‖푂퐴‖2 = = 푉[퐴퐵퐶] 2

Solution to Problem 28.

휋 휇(퐴푂퐵̂) = 4

2 휋 2 √2 푟 sin 푟 푟2√2 휎[퐴푂퐵] = 4 = 2 = 2 2 4 푟2√2 휎[orthogon] = 8 ∙ = 2√2푟2 4

Florentin Smarandache

Solution to Problem 29.

휎[퐴퐵퐶] = 휎[퐴푀퐵] + 휎[퐴푀퐶] + 휎[푀퐵퐶] ⟹ 푎ℎ푎 = 푎푑3 + 푎푑2 + 푎푑1

푑1 + 푑2 + 푑3 = ℎ푎 (푎 is the side of equilateral triangle) 푎 3 푎 3 ⟹ 푑 + 푑 + 푑 = √ (because ℎ = √ ). 1 2 3 2 푎 2

Solution to Problem 30.

퐴퐵퐶 − given ∆ ⟹ 푎, 푏, 푐, ℎ − constant 푎ℎ 휎[퐴퐵퐶] = 2 휎[퐴퐵퐶] = 휎[퐴푀퐵] + 휎[퐴푀퐶] 푎ℎ 푐푥 푏푦 푐 푏 ⟹ = + ⟹ 푐푥 + 푏푦 = 푎푏 ⟹ 푥 + 푦 = 1 ⟹ 푘푥 + 푙푦 = 1, 2 2 2 푎ℎ 푎ℎ 푐 푏 where 푘 = and 푙 = . 푎ℎ 푎ℎ

Solution to Problem 31.

255 Compiled and Solved Problems in Geometry and Trigonometry

퐴퐴′ ∥ 푁푁′ ∥ 퐷퐷′ We draw 퐴퐴′ ⊥ 퐵퐶; 푁푁′ ⊥ 퐵퐶; 퐷퐷′ ⊥ 퐵퐶 ⇒ } ⟹ 푀푁′median line in ‖퐴푁‖ = ‖푁퐷‖ ‖퐴퐴′‖+‖퐷퐷′‖ ‖퐵퐶‖+‖푁푁′‖ the trapezoid 퐴퐴′퐷′퐷 ⟹ ‖푁푁′‖ = , 휎[퐵퐶푁] = . 2 2

Solution to Problem 32.

First, we construct a quadrilateral with the same area as the given pentagon. We draw through C a parallel to BD and extend |AB| until it intersects the parallel at M.

휎[퐴퐵퐶퐷퐸] = 휎[퐴퐵퐷퐸] + 휎[퐵퐶퐷], 휎[퐵퐶퐷] = 휎[퐵퐷푀] (have the vertices on a parallel at the base). Therefore, 휎[퐴퐵퐶퐷퐸] = 휎[퐴푀퐷퐸].

Then, we consider a triangle with the same area as the quadrilateral 퐴푀퐷퐸. We draw a parallel to 퐴퐷, 푁 is an element of the intersection with the same parallel.

휎[퐴푀퐷퐸] = 휎[퐴퐷퐸] + 휎[퐴퐷퐸] = 휎[퐴퐷퐸] + 휎[퐴퐷푁] = 휎[퐸퐷푁].

Florentin Smarandache

Solution to Problem 33.

‖퐴퐸‖ = ‖퐸퐶‖

‖퐸퐹‖퐵퐷 ⟹ 휎[퐵퐷퐹] = 휎[퐵퐷퐸]

휎[퐴퐵퐹퐷] = 휎[퐴퐵퐸퐷] (1)

휎[퐴퐷퐸] = 휎[퐷퐸퐶] equal bases and the same height;

휎[퐴퐷퐸] = 휎[퐷퐸퐶]

휎[퐴퐵퐸퐷] = 휎[퐵퐸퐷퐶] (2)

휎[퐷퐸퐹] = 휎[퐵퐸퐹] the same base and the vertices on parallel lines at the base;

255 Compiled and Solved Problems in Geometry and Trigonometry

Solution to Problem 34.

It is proved in the same way that:

⟹ 푀푁푃푄 rhombus with right angle⟹ 푀푁푃푄 is a square.

It is proved in the same way that all the peaks of the octagon are elements of the axis of symmetry of the square, thus the octagon is regular.

Florentin Smarandache

Consider the square separately.

Solution to Problem 35.

‖푂푀‖ = ‖푁푀‖ = ‖푁퐵‖ ‖퐷퐶‖? ⟹ ∆푀푂퐶 = ∆푁퐵퐴 ⟹ ‖푀퐶‖ = ‖퐴푁‖ It is proved in the same way that ∆퐷퐴푀 = ∆퐵퐶푁 ⟹ ‖푀퐶‖ = ‖푁퐶‖. Thus 퐴푁퐶푀 is a parallelogram.

255 Compiled and Solved Problems in Geometry and Trigonometry

Solution to Problem 36.

To determine the angle α:

We write

thus we have established the positions of the lines of the locus.

constant for 퐴, 퐵, 퐶, 퐷 – fixed points.

We must find the geometrical locus of points 푀 such that the ratio of the distances from this point to two concurrent lines to be constant.

Florentin Smarandache

‖푀퐸‖ ‖푀′퐸′‖ = 푘. Let M' be another point with the same property, namely = 푘. ‖푀퐹‖ ‖푀′퐹′‖

푃, 푀, 푀′ collinear ⟹ the locus is a line that passes through 푃.

When the points are in ∢퐶푃퐵 we obtain one more line that passes through 푃. Thus the locus is formed by two concurrent lines through 푃, from which we eliminate point 푃, because the distances from 푃 to both lines are 0 and their ratio is indefinite.

Vice versa, if points 푁 and 푁′ are on the same line passing through 푃, the ratio of their distances to lines 퐴퐵 and 퐶퐷 is constant.

Solution to Problem 37.

We show in the same way as in the previous problem that:

‖푀퐸‖ ‖푀퐸‖ 푘 푘푑 = 푘 ⟹ = ⟹ ‖푀퐸‖ = , ‖푀퐹‖ ‖푀퐹‖ + ‖푀퐸‖ 1 − 푘 1 + 푘

and the locus of the points which are located at a constant distance from a given line is a parallel to the respective line, located between the two parallels.

If ||퐴퐵|| > ||퐶퐷|| ⟹ 푑(푀퐴퐸) < 푑(푀퐶퐷).

255 Compiled and Solved Problems in Geometry and Trigonometry

Then, if

푀퐸 푀퐸 푘 푀퐸 푘 푘푑 = 푘 ⟹ = ⟹ = ⟹ 푀퐸 = , 푀퐹 푀퐹 − 푀퐸 1 − 푘 푑 1 − 푘 1 − 푘

thus we obtain one more parallel to 퐴퐵.

Solution to Problem 38.

Solution no. 1

We suppose that ABCD is not a parallelogram. Let {퐼} = 퐴퐵 ∩ 퐶퐷. We build 퐸 ∈ (퐼퐴 such that 퐼퐸 = 퐴퐵 and 퐹 ∈ (퐼퐶 such that 퐼퐹 = 퐶퐷. If 푀 a point that verifies 휎[퐴퐵푀] + 휎[퐶퐷푀] = 1 (1), then, because 휎[퐴퐵푀] = 휎[푀퐼퐸] and 휎[퐶퐷푀] = 휎[푀퐼퐹], it results that 휎[푀퐼퐸] + 휎[푀퐼퐹] = 푘 (2).

We obtain that 휎[푀퐸퐼퐹] = 푘.

On the other hand, the points 퐸, 퐹 are fixed, therefore 휎[퐼퐸퐹] = 푘′ = const. That is, 휎[푀퐸퐹] = 푘 − 푘′ = const.

2(푘−푘′) Because 퐸퐹 = const., we have 푑(푀, 퐸퐹) = = const., which shows that 푀 퐸퐹 2(푘−푘′) belongs to a line that is parallel to 퐸퐹, taken at the distance . 퐸퐹

Therefore, the locus points are those on the line parallel to 퐸퐹, located inside the quadrilateral 퐴퐵퐶퐷. They belong to the segment [퐸′퐹′] in Fig. 1.

Reciprocally, if 푀 ∈ [퐸′퐹′], then 휎[푀퐴퐵] + 휎[푀퐶퐷] = 휎[푀퐼퐸] + 휎[푀퐼퐹] = 퐸퐹∙2(푘−푘′) 휎[푀퐸퐼퐹] = 휎[퐼퐸퐹] + 휎[푀퐸푃] = 푘′ + = 푘. 2∙퐸퐹

Florentin Smarandache

In conclusion, the locus of points 푀 inside the quadrilateral 퐴퐵퐶퐷 which occurs for relation (1) where 푘 is a positive constant smaller than 푆 = 휎[퐴퐵퐶퐷] is a line segment.

If 퐴퐵퐶퐷 is a trapeze having 퐴퐵 and 퐶퐷 as bases, then we reconstruct the reasoning as 퐴퐷 ∩ 퐵퐶 = {퐼} and 휎[푀퐴퐷] + 휎[푀퐵퐶] = 푠 − 푘 = const.

If 퐴퐵퐶퐷 is a parallelogram, one shows without difficulty that the locus is a segment parallel to 퐴퐵.

Solution no. 2 (Ion Patrascu)

We prove that the locus of points 푀 which verify the relationship 휎[푀퐴퐵] + 휎[푀퐶퐷] = 푘 (1) from inside the convex quadrilateral 퐴퐵퐶퐷 of area 푠 (푘 ⊂ 푠) is a line segment.

Let’s suppose that 퐴퐵 ∩ 퐶퐷 = {퐼}, see Fig. 2. There is a point 푃 of the locus which 2푘 belongs to the line 퐶퐷. Therefore, we have (푃; 퐴퐵) = . Also, there is the point 퐴퐵 2푘 푄 ∈ 퐴퐵 such that 푑(푄; 퐶퐷) = . 퐶퐷

Now, we prove that the points from inside the quadrilateral 퐴퐵퐶퐷 that are on the segment [푃푄] belong to the locus.

Let 푀 ∈ int[퐴퐵퐶퐷] ∩ [푃푄]. We denote 푀1 and 푀2 the projections of 푀 on 퐴퐵 and

퐶퐷 respectively. Also, let 푃1 be the projection of 푃 on 퐴퐵 and 푄1 the projection of

푄 on 퐶퐷. The triangles 푃푄푄1 and 푃푀푀2 are alike, which means that

푀푀 푀푃 2 = (2), 푄푄1 푃푄

255 Compiled and Solved Problems in Geometry and Trigonometry

and the triangles 푀푀1푄 and 푃푃1푄 are alike, which means that

푀푀 푀푄 1 = (3). 푃푃1 푃푄

By adding member by member the relations (2) and (3), we obtain

푀푀 푀푀 푀푃 + 푀푄 2 + 1 = = 1 (4). 푄푄1 푃푃1 푃푄

2푘 2푘 Substituting in (4), 푄푄 = and 푃 = , we get 퐴퐵 ∙ 푀푀 + 퐶퐷 ∙ 푀푀 = 2푘, that 1 퐶퐷 1 퐴퐵 1 2 is 휎[푀퐴퐵] + 휎[푀퐶퐷] = 푘.

We prove now by reductio ad absurdum that there is no point inside the quadrilateral 퐴퐵퐶퐷 that is not situated on the segment [푃푄], built as shown, to verify the relation (1).

Let a point 푀′ inside the quadrilateral 퐴퐵퐶퐷 that verifies the relation (1), 푀′ ∉ [푃푄]. We build 푀′푇 ∩ 퐴퐵, 푀′푈 ∥ 퐶퐷, where 푇 and 푈 are situated on [푃푄], see Fig. 3.

′ ′ We denote 푀1, 푇1, 푈1 the projections of 푀1, 푇, 푈 on 퐴퐵 and 푀2, 푇2, 푈2 the projections of the same points on 퐶퐷.

We have the relations:

′ ′ ′ ′ 푀 푀1 ∙ 퐴퐵 + 푀 푀2 ∙ 퐶퐷 = 2푘 (5),

푇푇1 ∙ 퐴퐵 + 푇푇2 ∙ 퐶퐷 = 2푘 (6).

′ ′ ′ ′ Because 푀 푀1 = 푇푇1 and 푀 푀2 = 푈푈2, substituting in (5), we get:

푇푇1 ∙ 퐴퐵 + 푈푈2 ∙ 퐶퐷 = 2푘 (7).

From (6) and (7), we get that 푇푇2 = 푈푈2, which drives us to 푃푄 ∥ 퐶퐷, false!

Florentin Smarandache

Problems in Geometry and Trigonometry

39. Find the locus of the points such that the sum of the distances to two concurrent lines to be constant and equal to 푙. Solution to Problem 39

40. Show that in any triangle 퐴퐵퐶 we have: a. 푏 푐표푠퐶 + 푐 푐표푠퐵 = 푎; b. 푏 푐표푠퐵 + 푐 푐표푠퐶 = 푎 푐표푠(퐵 − 퐶). Solution to Problem 40

41. Show that among the angles of the triangle 퐴퐵퐶 we have: 푏2−푎2 a. 푏 푐표푠퐶 − 푐 푐표푠퐵 = ; 푎 b. 2(푏푐 푐표푠퐴 + 푎푐 푐표푠퐵 + 푎푏 푐표푠퐶 = 푎2 + 푏2 + 푐2. Solution to Problem 41

2 42. Using the law of cosines prove that 4푚 = 2(푏2 + 푐2) − 푎2, where 푚 is the 푎 푎 length of the median corresponding to the side of 푎 length. Solution to Problem 42

푎+푐 퐵 43. Show that the triangle 퐴퐵퐶 where = cot is right-angled. 푏 2 Solution to Problem 43

44. Show that, if in the triangle 퐴퐵퐶 we have cot 퐴 + cot 퐵 = 2 cot 퐶 ⇒ 푎2 + 푏2 = 2푐2. Solution to Problem 44

45. Determine the unknown elements of the triangle 퐴퐵퐶, given: a. 퐴, 퐵 and 푝; b. 푎 + 푏 = 푚, 퐴 and 퐵; c. 푎, 퐴; 푏 − 푐 = 푎. Solution to Problem 45

255 Compiled and Solved Problems in Geometry and Trigonometry

퐴−퐵 퐶 푎−푏 46. Show that in any triangle 퐴퐵퐶 we have tan tan = (tangents 2 2 푎+푏 theorem). Solution to Problem 46

푏 퐵−퐶 47. In triangle 퐴퐵퐶 it is given 퐴̂ = 60° and = 2 + √3. Find tan and angles 푐 2 퐵 and 퐶. Solution to Problem 47

48. In a convex quadrilateral 퐴퐵퐶퐷, there are given ‖퐴퐷‖ = 7(√6 − √2), ‖퐶퐷‖ = 33 휋 5 13, ‖퐵퐶‖ = 15, 퐶 = arccos , and 퐷 = + arccos . The other angles of the 65 4 13 quadrilateral and ‖퐴퐵‖ are required. Solution to Problem 48

49. Find the area of ∆퐴퐵퐶 when: 24 12 a. 푎 = 17, 퐵 = arcsin , 퐶 = arcsin ; 25 13 b. 푏 = 2, 퐴̂ ∈ 135°, 퐶̂ ∈ 30°; c. 푎=7,푏=5,푐=6; d. 퐴̂ ∈ 18°, 푏 = 4, 푐 = 6. Solution to Problem 49

50. How many distinct triangles from the point of view of symmetry are there such that 푎 = 15, 푐 = 13, 푠 = 24?

Solution to Problem 50

51. Find the area of ∆퐴퐵퐶 if 푎 = √6, 퐴̂ ∈ 60°, 푏 + 푐 = 3 + √3. Solution to Problem 51

52. Find the area of the quadrilateral from problem 48. Solution to Problem 52

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53. If 푆푛 is the area of the regular polygon with 푛 sides, find:

푆3, 푆4, 푆6, 푆8, 푆12, 푆20 in relation to 푅, the radius of the circle inscribed in the polygon. Solution to Problem 53

54. Find the area of the regular polygon 퐴퐵퐶퐷 … 푀 inscribed in the circle with 1 1 1 radius 푅, knowing that: = + . ‖퐴퐵‖ ‖퐴퐶‖ ‖퐴퐷‖ Solution to Problem 54

55. Prove that in any triangle ABC we have: 퐴 a. 푟 = (푝 − 푎) tan ; 2 퐴 b. 푆 = (푝 − 푎) tan ; 2 퐴 퐵 퐶 c. 푝 = 4푅 cos cos cos ; 2 2 2 퐴 퐵 퐶 d. 푝 − 푎 = 4푅 cos cos cos ; 2 2 2 2 2 2 e. 푚푎 = 푅 (sin 퐴 + 4 cos 퐴 sin 퐵 sin 퐶;

f. ℎ푎 = 2푅 sin 퐵 sin 퐶. Solution to Problem 55

56. If 푙 is the center of the circle inscribed in triangle 퐴퐵퐶 show that ‖퐴퐼‖ = 퐵 퐶 4푅 sin sin . 2 2 Solution to Problem 56

57. Prove the law of sine using the analytic method. Solution to Problem 57

58. Using the law of sine, show that in a triangle the larger side lies opposite to the larger angle. Solution to Problem 58

59. Show that in any triangle 퐴퐵퐶 we have: 푎 cos 퐶 − 푏 cos 퐵 a. + cos 퐶 = 0, 푎 ≠ 푏; 푎 cos 퐵 − 푏 cos 퐴

255 Compiled and Solved Problems in Geometry and Trigonometry

sin(퐴 − 퐵) sin 퐶 푎2 − 푏2 b. = ; 1 + cos(퐴 − 퐵) cos 퐶 푎2 + 푏2 퐵 3퐵 퐵 퐵 c. (푎 + 푐) cos + 푎 cos(퐴 + ) = 2푐 cos cos . 4 4 2 4 Solution to Problem 59

2 2 60. In a triangle 퐴퐵퐶, 퐴 ∈ 45°, ‖퐴퐵‖ = 푎, ‖퐴퐶‖ = √ 푎. Show that tan 퐵 = 2. 3 Solution to Problem 60

61. Let 퐴′, 퐵′, 퐶′ be tangent points of the circle inscribed in a triangle 퐴퐵퐶 with 휎[퐴′퐵′퐶′] 푟 its sides. Show that = . 휎[퐴퐵퐶] 2푅 Solution to Problem 61

퐴 푎 62. Show that in any triangle 퐴퐵퐶 sin ≤ . 2 2√푏푐 Solution to Problem 62

63. Solve the triangle 퐴퐵퐶, knowing its elements 퐴, 퐵 and area 푆. Solution to Problem 63

4 64. Solve the triangle 퐴퐵퐶, knowing 푎 = 13, arc cos , and the corresponding 5 1 median for side 푎, 푚 = √15√3 . 푎 2 Solution to Problem 64

2휋 65. Find the angles of the triangle 퐴퐵퐶, knowing that 퐵 − 퐶 = and 푅 = 8푟, 3 where 푅 and 푟 are the radii of the circles circumscribed and inscribed in the triangle.

Solution to Problem 65

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Solutions

Solution to Problem 39.

Let 푑1 and 푑2 be the two concurrent lines. We draw 2 parallel lines to 푑1 located

on its both sides at distance 푙. These intersect on 푑2 at 퐷 and 퐵, which will be

points of the locus to be found, because the sum of the distances 푑(퐵, 푑1) +

푑(퐵, 푑2) = 푙 + 0 verifies the condition from the statement.

We draw two parallel lines with 푑2 located at distance 푙 from it, which cut 푑1 in 퐴 and 퐶, which are as well points of the locus to be found. The equidistant parallel |퐷푂| ≡ |푂퐵| lines determine on 푑 congruent segments ⟹ , in the same way 퐴퐵퐶퐷 2 |퐴푂| ≡ |푂퐶| is a parallelogram.

‖퐶퐶′‖ = 푑(퐶, 푑 ) ∆퐵푂퐶, 2 } ⟹ ‖퐶퐶′‖ = ‖퐵퐵′‖ ‖퐵퐵′‖ = 푑(퐵, 푑1)

⟹ ∆퐵푂퐶 is isosceles.

⟹ ||푂퐶|| = ||푂퐵|| ⟹ 퐴퐵퐶퐷 is a rectangle. Any point 푀 we take on the sides of this

rectangle, we have ||푅1, 푑1|| + ||푀, 푑2|| = 푙, using the propriety according to which the sum of the distances from a point on the base of an isosceles triangle at the sides is constant and equal to the height that starts from one vertex of the base, namely 푙. Thus the desired locus is rectangle 퐴퐵퐶퐷.

Solution to Problem 40.

255 Compiled and Solved Problems in Geometry and Trigonometry

Solution to Problem 41.

Solution to Problem 42.

푎2 푎 푚2 = 푐2 + − 2 푐 cos 퐵 ; 푎 4 2 푎2 + 푐2 − 푏2 4푚2 = 4푐2 + 푎2 − 4푎푐 cos 퐵 = 4푐2 + 푎2 − 4푎푐 = 4푐2 + 푎2 − 2푎 − 2푐2 + 2푏2 푎 2푎푐 = 2푐2 + 2푏2 − 푎2 = 2(푏2 + 푐2) − 푎2 .

Florentin Smarandache

Solution to Problem 43.

Using the sine theorem, 푎 = 푚 sin 퐴.

0 0 퐴 − 퐶 퐵 퐴 = 퐵 + 퐶 2퐴 = 180 퐴 = 90 = − ⟹ 퐴 − 퐵 = 퐶 or 퐴 − 퐶 = 퐵 ⟹ or ⟹ or ⟹ or 2 2 퐴 + 퐵 = 퐶 2퐶 = 1800 퐶 = 900

Solution to Problem 44.

By substitution:

255 Compiled and Solved Problems in Geometry and Trigonometry

Solution to Problem 45.

a. Using the law of sine,

Therefore,

푎 sin 퐵 We solve the system, and find 퐵 and 퐶. Then we find 푏 = and 푐 = 푏 − 푑. sin 퐴

Solution to Problem 46.

푎 푏 푐 푎 = 푚 sin 퐴 = = = 푚 ⟹ ; sin 퐴 sin 퐵 sin 퐶 푏 = 푚 sin 퐵 퐴 − 퐵 퐴 + 퐵 퐶 푎 − 푏 푚 sin 퐴 − 푚 sin 퐵 sin 퐴 − sin 퐵 2 sin cos 퐴 − 퐵 sin = = = 2 2 = tan 2 퐴 + 퐵 퐴 − 퐵 퐶 푎 + 푏 푚 sin 퐴 + 푚 sin 퐵 sin 퐴 + sin 퐵 2 sin cos 2 cos 2 2 2 퐴 − 퐵 퐶 = tan tan . 2 2

Florentin Smarandache

Solution to Problem 47.

Using tangents’ theorem,

Solution to Problem 48.

In ∆퐵퐷퐶 we have

255 Compiled and Solved Problems in Geometry and Trigonometry

In ∆퐴퐷퐵,

In ∆퐴퐷퐵 we apply sine’s theorem:

휋 Or we find 휇(퐷퐵퐶̂) and we add it to . 6

Solution to Problem 49.

Florentin Smarandache

−1+ 5 sin 퐴 = √ , because 푚(퐴) < 1800 and sin 퐴 > 0. 4

Solution to Problem 50.

255 Compiled and Solved Problems in Geometry and Trigonometry

Solution to Problem 51.

Florentin Smarandache

Solution to Problem 52.

At problem 9 we’ve found that

With Heron’s formula, we find the area of each triangle and we add them up.

Solution to Problem 53.

The formula for the area of a regular polygon:

255 Compiled and Solved Problems in Geometry and Trigonometry

Solution to Problem 54.

In ∆퐵푂푀:

In ∆푁푂퐶:

In ∆푃푂퐷:

Substituting (1), (2), (3) in the given relation:

which is impossible.

푚(complete circle) 2휋 푛 = = = 7. 푚(퐴퐵̂ ) 2휋 7 Thus the polygon has 7 sides.

Florentin Smarandache

Solution to Problem 55.

Solution to Problem 56.

255 Compiled and Solved Problems in Geometry and Trigonometry

We apply the law of sine in ∆퐴퐵퐼:

The law of sine applied in ∆퐴퐵퐶:

Solution to Problem 57.

‖퐶퐶′‖ In ∆퐴퐶퐶′: sin(1800 − 퐴) = ⟹ ‖퐶퐶′‖ = 푏 sin 퐴 ; cos(1800 − 퐴) = 푏 cos 퐴. 푏 So the coordinates of 퐶 are (−푏 cos 퐴 , 푏 sin 퐴). The center of the inscribed circle is at the intersection of the perpendicular lines drawn through the midpoints of sides 퐴퐵 and 퐴퐶.

The equation of the line 퐸푂:

Florentin Smarandache

If we redo the calculus for the same draw, we have the following result: (푏 cos 퐴 , 푏 sin 퐴).

using the law of cosine.

Solution to Problem 58.

푎 푏 푐 = = = 2푅. sin 퐴 sin 퐵 sin 퐶

We suppose that 푎 > 푏. Let’s prove that 퐴 > 퐵.

푎 푏 푎 sin 퐴 = ⇒ = sin 퐴 sin 퐴 sin 퐵 푏 sin 퐵} ⟹ > 1 ⟹ 퐴, 퐵, 퐶 ∈ (0, 휋) ⟹ sin 퐵 > 0 ⟹ sin 퐴 > sin 퐵 푎 sin 퐵 푎 > 푏 ⇒ > 1 푏 퐴 − 퐵 퐴 + 퐵 퐴 + 퐵 1800 − 퐶 ⟹ sin 퐴 − sin 퐵 > 0 ⟹ 2 sin cos > 0 ⟹ = 2 2 2 2 퐶 = 900 − . 2 54

255 Compiled and Solved Problems in Geometry and Trigonometry

퐴+퐵 퐶 퐶 퐴−퐵 cos = cos (90circ − ) = sin > 0, therefore > 0 ⟹ 퐴 > 퐵; 2 2 2 2 휋 퐴 − 퐵 휋 (− < < ). 2 2 2

Solution to Problem 59.

b. We transform the product into a sum:

From (1) and (2) ⟹

We consider the last two terms:

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Solution to Problem 60.

We apply the law of cosines in triangle ABC:

Solution to Problem 61.

‖퐼퐴‖ = ‖퐼퐵‖ = ‖퐼퐶‖ = 푟 퐼퐶′ ⊥ 퐴퐵 } ⇒ 퐼퐴′퐵′퐶′ inscribable quadrilateral 퐼퐴′ ⊥ 퐵퐶 푚(퐴′퐼퐶′) = 180 − 퐵̂ ⟹ sin(퐴′̂퐼퐶′) = sin 퐵̂ Similarly, 퐴′̂퐼퐵′ = sin 퐶 and 퐶′̂퐼퐵′ = sin 퐴.

255 Compiled and Solved Problems in Geometry and Trigonometry

In the same way,

Solution to Problem 62.

퐴 휋 퐴 0 < 퐴 ⇒< < ⇒ sin > 0 ; 2 2 2 퐴 1 − cos 퐴 sin2 = 퐴 1 − cos 퐴 2 2 sin = √ ⟹ } 2 2 푏2 + 푐2 − 푎2 cos 퐴 = 2푏푐 푏2 − 푎2 + 푐2 퐴 1 − 푎2 − (푏 − 푐)2 푎2 퐴 푎 ⟹ sin2 = 2푏푐 = ≤ ⟹ sin ≤ . 2 2 4푏푐 4푏푐 2 2√푏푐

Solution to Problem 63.

퐶 = 휋 − (퐴 + 퐵)

푎2 sin 퐵 sin 퐶 푆 = 2 sin 퐴 } ⟹ We find 푎. 퐴, 퐵, 퐶 are known

푎 푏 푎 sin 퐵 = ⟹ 푏 = . In the same way, we find 푐. sin 푎 sin 푏 sin 퐴

Solution to Problem 64.

Florentin Smarandache

2 2 푏 = 21 푏 = 20 푏 + 푐 = 841} ⟹ { or { 푏푐 = 420 푐 = 20 푐 = 21

We find 퐵.

We find the sum. Or

Solution to Problem 65.

We already know that 푟 퐴 퐵 퐶 = 4 sin sin sin 푅 2 2 2

퐴 We write sin = 푡. We have 2

255 Compiled and Solved Problems in Geometry and Trigonometry

From this system we find 퐵 and 퐶.

Florentin Smarandache

Other Problems in Geometry and Trigonometry (10th grade)

66. Show that a convex polygon can’t have more than three acute angles. Solution to Problem 66

67. Let 퐴퐵퐶 be a triangle. Find the locus of points 푀 ∈ (퐴퐵퐶), for which 휎[퐴퐵푀] = 휎[퐴퐶푀]. Solution to Problem 67

68. A convex quadrilateral 퐴퐵퐶퐷 is given. Find the locus of points 푀 ∈ 푖푛푡. 퐴퐵퐶퐷, for which 휎[푀퐵퐶퐷] = 휎[푀퐵퐴퐷]. Solution to Problem 68

69. Determine a line 푀푁, parallel to the bases of a trapezoid 퐴퐵퐶퐷 (푀 ∈ |퐴퐷|, 푁 ∈ |퐵퐶|) such that the difference of the areas of [퐴퐵푁푀] and [푀푁퐶퐷] to be equal to a given number. Solution to Problem 69

퐵퐷 퐶퐸 퐴퐹 70. On the sides of ∆퐴퐵퐶 we take the points 퐷, 퐸, 퐹 such that = = = 2. 퐷퐶 퐸퐴 퐹퐵 Find the ratio of the areas of triangles 퐷퐸퐹 and 퐴퐵퐶. Solution to Problem 70

푎 71. Consider the equilateral triangle 퐴퐵퐶 and the disk [퐶 (푂, )], where 푂 is the 3 orthocenter of the triangle and 푎 = ‖퐴퐵‖. Determine the area [퐴퐵퐶] − 푎 [퐶 (푂, )]. 3 Solution to Problem 71

255 Compiled and Solved Problems in Geometry and Trigonometry

72. Show that in any triangle 퐴퐵퐶 we have: 푏2+푐2 a. 1 + cos 퐴 cos(퐵 − 퐶) = ; 4푅2 b. (푏2 + 푐2 = 푎2) tan 퐴 = 4푆; 퐴 sin( +퐶) 푏+푐 2 c. 퐴 = ; 2푐 cos sin(퐴+퐵) 2 퐴 퐵 d. 푝 = 푟 (cot + cot + cot 퐶2); 2 2 퐴 퐵 퐶 푝 e. cot + cot + cot = . 2 2 2 푟 Solution to Problem 72

73. If 퐻 is the orthocenter of triangle 퐴퐵퐶, show that: a. ‖퐴퐻‖ = 2푅 cos 퐴; b. 푎‖퐴퐻‖ + 푏‖퐵퐻‖ + 푐‖퐶퐻‖ = 4푆. Solution to Problem 73

74. If 푂 is the orthocenter of the circumscribed circle of triangle 퐴퐵퐶 and 퐼 is the center of the inscribed circle, show that ‖푂퐼‖2 = 푅(푅 – 2푟). Solution to Problem 74

퐵−퐶 2푟 75. Show that in any triangle 퐴퐵퐶 we have: cos2 ≥ . 2 푅 Solution to Problem 75

1 1 76. Find 푧푛 + knowing that 푧 + = 2 sin 훼. 푧푛 푧 Solution to Problem 76

77. Solve the equation: (푧 + 1)푛 − (푧 − 1)푛 = 0. Solution to Problem 77

1 3 78. Prove that if 푧 < then |(1 + 푖)푧3 + 푖푧| ≤ . 2 4 Solution to Problem 78

Florentin Smarandache

79. One gives the lines 푑 and 푑′. Show that through each point in the space passes a perpendicular line to 푑 and 푑′. Solution to Problem 79

80. There are given the lines 푑 and 푑′, which are not in the same plane, and

the points 퐴 ∈ 푑, 퐵 ∈ 푑′. Find the locus of points 푀 for which pr푑푀 = 퐴 and

pr푑′푀 = 퐵. Solution to Problem 80

81. Find the locus of the points inside a trihedral angle 푎푏푐̂ equally distant from the edges of 푎, 푏, 푐. Solution to Problem 81

82. Construct a line which intersects two given lines and which is perpendicular to another given line. Solution to Problem 82

83. One gives the points 퐴 and 퐵 located on the same side of a plane; find in this plane the point for which the sum of its distances to 퐴 and 퐵 is minimal. Solution to Problem 83

84. Through a line draw a plane onto which the projections of two lines to be parallel. Solution to Problem 84

85. Consider a tetrahedron [퐴퐵퐶퐷] and centroids 퐿, 푀, 푁 of triangles 퐵퐶퐷,퐶퐴퐷,퐴퐵퐷. a. Show that (퐴퐵퐶) ∥ (퐿푀푁); [ ] b. Find the ratio 휎 퐴퐵퐶 . 휎[퐿푀푁] Solution to Problem 85

255 Compiled and Solved Problems in Geometry and Trigonometry

86. Consider a cube [퐴퐵퐶퐷퐴′퐵′퐶′퐷′]. The point 퐴 is projected onto 퐴′퐵,퐴′퐶,퐴′퐷

respectively in 퐴1, 퐴2, 퐴3. Show that: ′ a. 퐴 퐶 ⊥ (퐴1퐴2퐴3);

b. 퐴퐴1 ⊥ 퐴1퐴2, 퐴퐴3 ⊥ 퐴3퐴2;

c. 퐴퐴1퐴2퐴3 is an inscribable quadrilateral. Solution to Problem 86

87. Consider the right triangles 퐵퐴퐶 and 퐴퐵퐷 (푚(퐵퐴퐶̂)) = 푚((퐴퐵퐷̂) = 900) located on perpendicular planes 푀 and 푁, being midpoints of segments [퐴퐵],[퐶퐷]. Show that 푀푁 ⊥ 퐶퐷. Solution to Problem 87

88. Prove that the bisector half-plane of a dihedral angle inside a tetrahedron divides the opposite edge in proportional segments with the areas of the adjacent faces. Solution to Problem 88

89. Let 퐴 be a vertex of a regular tetrahedron and 푃, 푄 two points on its surface. Show that 푚(푃퐴푄̂) ≤ 600. Solution to Problem 89

90. Show that the sum of the measures of the dihedral angles of a tetrahedron is bigger than 360°. Solution to Problem 90

91. Consider lines 푑1, 푑2 contained in a plane 훼 and a line 퐴퐵 which intersects plane 훼 at point 퐶. A variable line, included in 훼 and passing through 퐶 all

푑1, 푑2 respectively at 푀푁. Find the locus of the intersection 퐴푀 ∩ 퐵푁. In which case is the locus an empty set? Solution to Problem 91

Florentin Smarandache

92. A plane 훼 intersects sides [퐴퐵], [퐵퐶], [퐶퐷], [퐷퐴] of a tetrahedron [퐴퐵퐶퐷] at points 퐿, 푀, 푁, 푃. Prove that ‖퐴퐿‖ ∙ ‖퐵푀‖ ∙ ‖퐶푁‖ ∙ ‖푃퐷‖ = ‖퐵퐿‖ ∙ ‖퐶푀‖ ∙ ‖퐷푁‖ ∙ ‖퐴푃‖. Solution to Problem 92

93. From a point 퐴 located outside a plane 훼, we draw the perpendicular line

퐴푂, 푂 ∈ 훼, and we take 퐵, 퐶 ∈ 훼. Let 퐻, 퐻1 be the orthocenters of triangles

퐴퐵퐶, 푂퐵퐶; 퐴퐷 and 퐵퐸 heights in triangle 퐴퐵퐶; and 퐵퐸1 height in triangle 푂퐵퐶. Show that:

a. 퐻퐻1 ⊥ (퐴퐵퐶); 푂퐴 퐷퐻 퐵퐸 b. ‖ ‖ ∙ ‖ 1‖ ∙ ‖ ‖ = 1. 퐴퐷 퐻1퐵 퐸퐸1 Solution to Problem 93

94. Being given a tetrahedron [퐴퐵퐶퐷] where 퐴퐵 ⊥ 퐶퐷 and 퐴퐶 ⊥ 퐵퐷, show that: a. ‖퐴퐵‖2 + ‖퐶퐷‖2 = ‖퐵퐶‖2 + ‖퐴퐷‖2 = ‖퐶퐴‖2 + ‖퐵퐷‖2; b. The midpoints of the 6 edges are located on a sphere. Solution to Problem 94

95. It is given a triangular prism [퐴퐵퐶퐴′퐵′퐶′] which has square lateral faces. Let 푀 be a mobile point [퐴퐵′], 푁 the projection of 푀 onto (퐵퐶퐶′) and 퐴ʺ the midpoint of [퐵′퐶ʺ]. Show that 퐴′푁 and 푀퐴ʺ intersect in a point 푃 and find the locus of 푃. Solution to Problem 95

96. We have the tetrahedron [퐴퐵퐶퐷] and let 퐺 be the centroid of triangle 퐵퐶퐷. Show that if 푀 ∈ 퐴퐺 then 휐[푀퐺퐵퐶] = 휐[푀퐺퐶퐷] = 휐[푀퐺퐷퐵]. Solution to Problem 96

97. Consider point 푀 ∈ the interior of a trirectangular tetrahedron with its vertex in 푂. Draw through 푀 a plane which intersects the edges of the

255 Compiled and Solved Problems in Geometry and Trigonometry

respective tetrahedron in points 퐴, 퐵, 퐶 so that 푀 is the orthocenter of ∆퐴퐵퐶. Solution to Problem 97

98. A pile of sand has as bases two rectangles located in parallel planes and trapezoid side faces. Find the volume of the pile, knowing the dimensions 푎′, 푏′ of the small base, 푎, 푏 of the larger base, and ℎ the distance between the two bases. Solution to Problem 98

99. A pyramid frustum is given, with its height ℎ and the areas of the bases 퐵 and 푏. Unite any point 휎 of the larger base with the vertices 퐴, 퐵, 퐴′, 퐵′ of a 6 side face. Show that 휐[푂퐴′퐵′퐴] = √ 휐[푂퐴퐵퐵′]. √퐵 Solution to Problem 99

100. A triangular prism is circumscribed to a circle of radius 푅. Find the area and the volume of the prism. Solution to Problem 100

101. A right triangle, with its legs 푏 and 푐 and the hypotenuse 푎, revolves by

turns around the hypotenuse and the two legs, 푉1, 푉2, 푉3; 푆1, 푆2, 푆3 being the volumes, respectively the lateral areas of the three formed shapes, show that: 1 1 1 a. 2 = 2 = 2; 푉1 푉2 푉3 푆 푆 푆 +푆 b. 2 + 3 = 2 3 . 푆3 푆2 푆1 Solution to Problem 101

102. A factory chimney has the shape of a cone frustum and 10m height, the bases of the cone frustum have external lengths of 3,14m and 1,57m, and the wall is 18cm thick. Calculate the volume of the chimney. Solution to Problem 102

Florentin Smarandache

103. A regular pyramid, with its base a square and the angle from the peak of a side face of measure 훼 is inscribed in a sphere of radius 푅. Find: a. the volume of the inscribed pyramid; b. the lateral and total area of the pyramid; c. the value 훼 when the height of the pyramid is equal to the radius of the sphere.

Solution to Problem 103

255 Compiled and Solved Problems in Geometry and Trigonometry

Solutions

Solution to Problem 66.

Let 퐴1, 퐴2 … 퐴푛 the vertices of the convex polygon. Let’s assume that it has four

acute angles. The vertices of these angles form a convex quadrilateral 퐴푙퐴푘퐴푚퐴푛.

Due to the fact that the polygon is convex, the segments |퐴푙퐴푘|, |퐴푘퐴푚|, |퐴푚퐴푛|,

|퐴푛퐴푙| are inside the initial polygon. We find that the angles of the quadrilateral are acute, which is absurd, because their sum is 360°.

Another solution: We assume that 퐴푙퐴푘퐴푚퐴푛 is a convex polygon with all its angles acute ⟹ the sum of the external angles is bigger than 360°, which is absurd (the sum of the measures of the external angles of a convex polygon is 360°).

Solution to Problem 67.

Florentin Smarandache

Let |퐴퐴′| be the median from 퐴 and 퐶푄 ⊥ 퐴퐴′, 퐵푃 ⊥ 퐴퐴′. ∆퐵퐴′푃 ≡ 퐶퐴′푄 because: 푃퐵퐶̂ ≡ 퐵퐶푄̂ alternate interior {푃퐴̂′퐵 ≡ 퐶퐴̂′푄 vertical angles 퐵퐴′ ≡ 퐴′퐶 ⟹ ||퐵푃|| = ||푄퐶|| and by its construction 퐵푃 ⊥ 퐴퐴′, 퐶푄 ⊥ 퐴퐴′. The desired locus is median |퐴퐴′|. Indeed, for any 푀 ∈ |퐴퐴′| we have 휎[퐴퐵푀] = 휎[퐴퐶푀], because triangles 퐴퐵푀 and 퐴퐶푀 have a common side |퐴푀| and its corresponding height equal ||퐵푃|| = ||푄퐶||.

Vice-versa. If 휎[퐴퐵푀] = 휎[퐴퐶′푀], let’s prove that 푀 ∈ |퐴퐴′|. Indeed: 휎[퐴퐵푀] 휎[퐴퐶푀] ⇒ 푑 (퐵, 퐴푀) = 푑(퐶, 퐴푀), because |퐴푀| is a common side, 푑 (퐵, 퐴푀) = ||퐵푃|| and 푑(퐶, 퐴푀) = ||퐶푄|| and both are perpendicular to 퐴푀 ⟹ 푃퐵푄퐶 is a parallelogram, the points 푃, 푀, 푄 are collinear (푃, 푄 the feet of the perpendicular lines from 퐵 and 퐶 to 퐴푀). In parallelogram 푃퐵푄퐶 we have |푃푄| and |퐵퐶| diagonals ⟹ 퐴푀 passes through the middle of |퐵퐶|, so 푀 ∈ |퐴퐴′|, the median from 퐴.

Solution to Problem 68.

Let 푂 be the midpoint of diagonal |퐴퐶| ⟹ ‖퐴푂‖ = ‖푂퐶‖. 휎[퐴푂퐷] = 휎[퐶푂퐷] (1) ‖퐴푂‖ = ‖푂퐶‖ Because { ‖푂퐷′‖ common height 휎[퐴푂퐵] = 휎[퐶푂퐵] (2)

255 Compiled and Solved Problems in Geometry and Trigonometry

the same reasons; we add up (1) and (2) ⟹ 휎[퐴퐷푂퐵] = 휎[퐷퐶퐵푂] (3), so 푂 is a point of the desired locus. We construct through 푂 a parallel to 퐵퐷 until it cuts sides |퐵퐶| and |퐷퐶| at 푃 respectively 푄. The desired locus is |푃푄|. Indeed (∀)푀 ∈ |푃푄| we have:

휎[퐵퐷푂] = 휎[퐵퐷푀] because 푀 and 푄 belongs to a parallel to 퐵퐷.

퐵, 퐷 ∈ a parallel to 푂푀. 휎[푀퐵퐴퐷] = 휎[퐴퐵푂퐷] So and 휎[퐵퐶퐷푀] = 휎[푂퐵퐶퐷]} ⟹ 휎[푀퐵퐴퐷] = 휎[퐵퐶퐷푀]. and from (3) Vice-versa: If 휎[푀퐵퐶퐷] = 휎[푀퐵퐴퐷], let’s prove that 푀 ∈ parallel line through 푂 to 퐵퐷. Indeed:

휎[퐵퐶퐷푀] = 휎[푀퐵퐴퐷] 휎[퐴퐵퐶퐷] } ⟹ 휎[푀퐵퐶퐷] = 휎[푀퐵퐴퐷] = (2). and because 휎[퐵퐶퐷푀] + 휎[푀퐵퐴퐷] = 휎[퐴퐵퐶퐷] 2 So, from (1) and (2) ⟹ 휎[푀퐵퐴퐷] = 휎[퐴퐵푂퐷] ⟹ 휎[퐴퐵퐷] + 휎[퐵퐷푂] = 휎[퐴퐵퐷] + 휎[퐵퐷푀] ⟹ 휎[퐵퐷푀] ⟹ 푀 and 푂 are on a parallel to 퐵퐷.

Solution to Problem 69.

We write ||퐸퐴|| = 푎 and ||퐸퐷|| = 푏, ||퐸푀|| = 푥.

Florentin Smarandache

We subtract (2) from (3)

We subtract (2) from (4)

from the hypothesis

푆푏2 From the relation (3), by writing [퐴퐵퐶퐷] − 푆 ⟹ 휎[퐸퐶퐷] = . 푎2−푏2 We substitute this in the relation of 푥² and we obtain:

and taking into consideration that ||퐸푀|| = ||퐷푀|| + 푏, we have

so we have the position of point 푀 on the segment |퐷퐴| (but it was sufficient to find the distance ||퐸푀||).

Solution to Problem 70.

We remark from its construction that 퐸푄||퐴퐵||푅퐷, more than that, they are equidistant parallel lines. Similarly, 퐸푄, 푃퐷, 퐴퐶 and 퐴퐵, 퐸푄, 푅퐷 are also equidistant parallel lines.

255 Compiled and Solved Problems in Geometry and Trigonometry

We write 휎[퐵퐹푄] = 푆. Based on the following properties:  two triangles have equal areas if they have equal bases and the same height;  two triangles have equal areas if they have the same base and the third peak on a parallel line to the base, we have:

by addition ⇒

휎[퐷퐸퐹] 3푆 1 So = = . 휎[퐴퐵퐶] 9푆 3 (If necessary the areas 푆 can be arranged).

Solution to Problem 71.

푎 3 ‖푂퐵‖ = √ (퐵퐵′ median) 6 In ∆푀푂퐵′:

휋 So (푀푂푁̂ ) = . 3 We mark with 훴 the disk surface bordered by a side of the triangle outside the triangle.

Florentin Smarandache

휎[훴] = 휎[circle sector 푀푂푁] − 휎[푀푂푁] 휋푎2 푎2 휋푎2 푎2√3 푎2 휋 √3 = − sin 600 = − = ∙ ( − ). 9 ∙ 6 9 ∙ 2 9 ∙ 6 4 ∙ 9 18 3 2 If through the disk area we subtract three times 휎[훴], we will find the area of the disk fraction from the interior of 퐴퐵퐶. So the area of the disk surface inside 퐴퐵퐶 is:

The desired area is obtained by subtracting the calculated area form 휎[퐴퐵퐶]. So:

Solution to Problem 72.

푏2+푐2 a. 1 + cos 퐴 ∙ cos(퐵 − 퐶) = 4푅2

4푆 b. We prove that tan 퐴 = . 푏2+푐2−푎2

255 Compiled and Solved Problems in Geometry and Trigonometry

⟹ All terms reduce.

퐴 퐵 푝 d. Let’s prove that cot + cot + cot 퐶2 = . 2 2 푟 Indeed

We now have to prove that:

Solution to Problem 73.

Florentin Smarandache

a. In triangle 퐴퐵퐵′: ‖퐴퐵′‖ = 푐 cos 퐴 In triangle 퐴퐻퐵′:

We used:

Solution to Problem 74.

Using the power of point 퐼 in relation to circle 퐶(푂, 푅)

Taking into consideration (1), we have ‖퐼퐴‖ ∙ ‖퐼퐷‖ = 푅2 − ‖푂퐼‖². We now find the distances ||IA|| and ||ID|| In triangle ∆퐼퐴푃,

We also find ‖퐼퐷‖: 휇(퐵퐼퐷̂ ) = 휇(퐷퐵퐼̂ ) have the same measure, more exactly:

255 Compiled and Solved Problems in Geometry and Trigonometry

In ∆퐴퐵퐷 according to the law of sine, we have:

So taking into consideration (3),

Returning to the relation ‖퐼퐴‖ ∙ ‖퐼퐷‖ = 푅2 − ‖푂퐼‖². with (2) and (4) we have:

Solution to Problem 75.

Note. We will have to show that

Indeed:

(by Heron’s formula).

Florentin Smarandache

Solution to Problem 76.

So:

We calculate for 푧1 and 푧2:

1 so 푧푛 + takes the same value for 푧 and for 푧 and it is enough if we 푧푛 1 2

calculate it for 푧1.

Analogously:

Solution to Problem 77.

255 Compiled and Solved Problems in Geometry and Trigonometry

(we substitute −1 with 푖² at denominator)

Solution to Problem 78.

Solution to Problem 79.

We construct 훼 ⊥ 푑 and 퐴 ∈ 훼. The so constructed plane is unique. Similarly we construct 훽 ⊥ 푑′ and 퐴 ⊥ 훽, 훼 ∩ 훽 = 푎 ∋ 퐴.

훼 ⊥ 푑 ⇒ 푑 ⊥ 푎 From } ⟹ 푎 is a line which passes through 퐴 and is perpendicular 훽 ⊥ 푑′ ⇒ 푑′ ⊥ 푎 to 푑 and 푑′. The line 푎 is unique, because 훼 and 훽 constructed as above are unique.

Solution to Problem 80.

We construct plane 훼 such that 퐴 ∈ 훼 and 푑 ⊥ 훼. We construct plane 훽 such that 퐵 ∈ 훽 and 푑′ ⊥ 훽.

Florentin Smarandache

The so constructed planes 훼 and 훽 are unique.

Let 푎 = 훼 ∩ 훽 ⟹ 푎 ⊂ 훼 so (∀) 푀 ∈ 푎 has the property pr푑푀 = 퐴.

훼 ⊂ 훽 ⟹ (∀) 푀 ∈ 푎 has the property 푝푟푑′푀 = 퐵.

Vice-versa. If there is a point 푀 in space such that pr푑푀 = 퐴 and 푝푟푑′ 푀 = 퐵 ⟹ 푀 ∈ 푎 and 푀 ∈ 훽 ⟹푀∈훼∩훽⟹푀∈푎 (훼 and 훽 previously constructed).

Solution to Problem 81.

Let 퐴 ∈ 푎, 퐵 ∈ 푏, 퐶 ∈ 푐 such that ‖푂퐴‖ = ‖푂퐵‖ = ‖푂퐶‖. Triangles 푂퐴퐵, 푂퐵퐶, 푂퐴퐶 are isosceles. The mediator planes of segments ‖퐴퐵‖, ‖퐴퐶‖, ‖퐵퐶 ‖ pass through 푂 and 푂′ (the center of the circumscribed circle of triangle 퐴퐵퐶). Ray |푂푂′| is the desired locus. Indeed (∀) 푀 ∈ |푂푂′| ⟹ 푀 ∈ mediator plane of segments |퐴퐵|, |퐴퐶| and |퐵퐶| ⟹ 푀 is equally distant from 푎, 푏 and 푐.

Vice-versa: (∀) 푀 with the property: 푑(푀,푎) = 푑(푀,푏) = 푑(푀,푐) ⟹ 푀 ∈ mediator plan, mediator planes of segments |퐴퐵|, |퐴퐶| and |퐵퐶| ⟹ 푀 ∈ the intersection of these planes ⟹ 푀 ∈ |푂푂′|.

255 Compiled and Solved Problems in Geometry and Trigonometry

Solution to Problem 82.

Let 푎, 푏, 푐 be the 3 lines in space. I. We assume 푎 ⊥ 푐 and 푏 ⊥ 푐. Let 훼 be a plane such that:

The construction is possible because ⊥ 푐 and 푏 ⊥ 푐. Line 퐴퐵 meets 푎 on 푝 and it is perpendicular to 푐, because 퐴퐵 ⊂ 훼 and 푐 ⊥ 훼. II. If 푎 ⊥ 푐 or 푏 ⊥ 푐, the construction is not always possible, only if plane 푝(푎, 푏) is perpendicular to 푐.

III. If 푎 ⊥ 푐 and 푏 ⊥ 푐, we construct plane 푎 ⊥ 푐 so that 푎 ⊂ 훼 and 푏 ⊂ 훼 ≠ ∅. Any point on line a connected with point 푏 ∩ 훼 is a desired line.

Solution to Problem 83.

We construct 퐴′ the symmetrical point of 퐴 in relation to 훼. 퐴′ and 퐵 are on different half-spaces, 훼 ∩ |퐴′퐵| = 푂.

Florentin Smarandache

푂 is the desired point, because ||푂퐴|| + ||푂퐵|| = ||푂퐴′|| + ||푂퐵|| is minimal when 푂 ∈ |퐴′퐵|, thus the desired point is 푂 = |퐴′퐵| ∩ 훼.

Solution to Problem 84.

Let 푎, 푏, 푑 be the 3 given lines and through 푑 we construct a plane in which 푎 and 푏 to be projected after parallel lines. Let 퐴 be an arbitrary point on 푎. Through 퐴 we construct line 푏′||푏. It results from the figure 푏||훼,훼 = 푝(푎,푏′). Let 훽 such that 푑 ⊂ 훽 and 훽 ⊥ 훼. Lines 푎 and 푏′ are projected onto 훽 after the same line 푐. Line 푏 is projected onto

훽 after 푏1 and 푏1 ∥ 푐.

If 푏1 ∦ 푐,

absurd because 푏||훼(푏||푏′).

Solution to Problem 85.

푀 is the centroid in ∆퐴퐶퐷 ⟹ |푀퐷| ⟹ = 2 (1) |푀푃|

255 Compiled and Solved Problems in Geometry and Trigonometry

푁 is the centroid in ∆퐴퐵퐷 ⟹ |푁퐷| ⟹ = 2 (2) |푁푄| 퐿 is the centroid in ∆퐵퐶퐷 ⟹ |퐿퐷| ⟹ = 2 (3) |퐿푆| From 1 and 2, and from 2 and 3

because:

Solution to Problem 86.

퐵퐷 ⊥ (퐴퐴′퐶) from the hypothesis 퐴퐵퐶퐷퐴′퐵′퐶′퐷′ cube (1).

퐴1 midpoint of segment |퐵퐴′| (퐴퐵퐴′) isosceles and 퐴퐴1 ⊥ 퐵퐴′} |퐴1퐴3| mid-side in ∆퐴′퐵퐷 ⟹ 퐴1퐴3 ∥ 퐵퐷 (2) 퐴3 midpoint of |퐴′퐷|

From (1) and (2) ⟹ 퐴1퐴3 ∥ (퐴퐴′퐶) ⟹ 퐴′퐶 ⊥ 퐴1퐴3 (3)

Florentin Smarandache

From ∆퐴퐶퐴′:

From ∆퐴퐵퐴′:

Similarly

In ∆퐴퐶퐴′:

and

퐴1퐴2퐴′ right with 푚(퐴′퐴2퐴1) = 90 because

From (4) and (3) ⟹ 퐴′퐶 ⊥ (퐴1퐴2퐴3).

퐴′퐶 ⊥ (퐴1퐴2퐴3) As ′ } ⟹ 퐴1퐴2퐴3퐴 coplanar ⟹ 퐴1퐴2퐴3퐴 quadrilateral with 퐴 퐶 ⊥ 퐴2퐴 (by construction)

opposite angles 퐴1 and 퐴3 right ⟹ 퐴1퐴2퐴3퐴 inscribable quadrilateral.

Solution to Problem 87. The conclusion is true only if ||퐵퐷|| = ||퐴퐶|| that is 푏 = 푐.

푀푁 ⊥ 퐷퐶 if

255 Compiled and Solved Problems in Geometry and Trigonometry

Solution to Problem 88.

bisector plane

(푏 bisector half-plane)

In triangle 퐷퐷1퐷′:

But

Florentin Smarandache

휎[퐴퐵퐷] ‖퐷퐸‖ From 1, 2, 3 ⟹ = q.e.d. 휎[퐴퐵퐶] ‖퐸퐶‖

Solution to Problem 89.

Because the tetrahedron is regular 퐴퐵 = … =

we increase the denominator

If one of the points 푃 or 푄 is on face 퐶퐵퐷 the problem is explicit.

Solution to Problem 90.

We consider tetrahedron 푂푥푦푧, and prove that the sum of the measures of the dihedral angles of this trihedron is bigger than 360°. Indeed: let 100′ be the internal bisector of trihedron 푂푥푦푧 (1000′ the intersection of the bisector planes of the 3 dihedral angles) of the trihedron in 퐴, 퐵, 퐶.

255 Compiled and Solved Problems in Geometry and Trigonometry

The size of each dihedron with edges 표푥, 표푦, 표푧 is bigger than the size of the corresponding angles of 퐴퐵퐶, the sum of the measures of the dihedral angles of trihedron 푂푥푦푧 is bigger than 180°. Let (푎, 푏) be a plane ⊥ to 표푧 at 퐶; 푎 ⊥ 표푧, 푏 ⊥ 표푧, but |퐶퐴 and |퐶퐵 are on the same half-space in relation to (푎푏) ⇒ 푚(퐶̂) < 푚(푎푏̂).

In tetrahedron 퐴퐵퐶퐷, let 푎1, 푎2, 푎3, 푎4, 푎5 and 푎6 be the 6 dihedral angles formed by the faces of the tetrahedron.

according to the inequality previously established.

Solution to Problem 91.

We mark with a the intersection of planes (퐴, 푑1) and (퐵, 푑2). So

Let 푏 be a variable line that passes through 퐶 and contained in 훼, which cuts 푑1

and 푑2 at 푀 respectively 푁. We have: 푀퐴 ⊂ (퐴, 푑1), 푀퐴 ∩ 푁퐵 = 푃(푀퐴 and 푁퐵 intersect because they are contained in the plane determined by (퐴푀, 푏)).

Thus 푃 ∈ (퐴, 푑1) and 푃 ∈ (퐵, 푑2), ⟹ 푃 ∈ 푎, so 푃 describes line a the intersection of

planes (퐴, 푑1) and (퐵, 푑2). Vice-versa: let 푄 ∈ 푎.

In the plane (퐴, 푑1): 푄퐴 ∩ 푑1 = 푀′

In the plane (퐵, 푑2): 푄퐵 ∩ 푑2 = 푁′

Florentin Smarandache

Lines 푁′푀′ and 퐴퐵 are coplanar (both are on plane (푄, 퐴, 퐵)). But because 푁′푀′ ⊂ 훼 and 퐴퐵 has only point 퐶 in common with 훼 ⟹ 푀′푁′ ∩ 퐴퐵 = 퐶.

So 푀′푁′ passes through 퐶. If planes (퐴, 푑1) and (퐵, 푑2) are parallel, the locus is the empty set.

Solution to Problem 92.

Remember the theorem: If a plane 훾 intersects two planes 훼 and 훽 such that 휎||훼 ⟹ (훾 ∩ 훼)||(훾 ∩ 훽). If plane (퐿푀푁푃)||퐵퐷 we have:

If (퐿푀푁푃)||퐴퐶 we have:

⟹ relation 푎. Solution: Let 퐴′,퐵′,퐶′,퐷′ the projections of points 퐴, 퐵, 퐶, 퐷 onto plane (푀푁푃퐿). For ex. points 퐵′, 퐿, 퐴′ are collinear on plane (퐿푃푀푁) because they are on the projection of line 퐴퐵 onto this plane.

Similarly we obtain:

255 Compiled and Solved Problems in Geometry and Trigonometry

By multiplying the 4 relations,

⟹ relation (푎) from 푑.

Solution to Problem 93.

푀 – Midpoint of |퐵퐶|.

Solution to Problem 94.

(1) 퐴퐵 ⊥ 퐶퐷 (hypothesis)

From 1 and 2

height in ∆퐴퐵퐶 a

From 3 and 4 퐴퐶 ⊥ (퐵퐷퐻) ⟹ 퐴퐶 ⊥ 퐵퐻 ⟹ 퐵퐻 height in ∆퐴퐵퐶 b

From a and b ⟹ 퐻 orthocenter ∆퐴퐵퐶. Let 퐶1 be the diametrical opposite point to 0 퐶 in circle 퐶(퐴퐵퐶)퐶퐶1 diameter 푚(퐶̂1퐵퐶) = 90 but 퐴퐻 ⊥ 퐵퐶 ⟹ 퐴퐻 ∥ 퐵퐶1. Similarly

퐵퐻||퐶1퐴, so 퐴퐻퐵퐶1 parallelogram, we have:

Florentin Smarandache

similarly

diametrical opposite to B

but

by substituting above, we have:

Let 푁,푀,푄,푃,푆,푅 midpoints of the edges of the quadrilateral 푁푀푃푄 because: 푁푀||퐶퐷||푃푄 (median lines), 푄푀||퐴퐵||푃푁 (median lines), but 퐶퐷 ⊥ 퐴퐵 ⇒ 푀푁푄푃 rectangle |푁푄| ∩ |푃푀| = {0}. Similarly 푀푆푃푅 rectangle with |푀푃| common diagonal with a, the first rectangle, so the 6 points are equally distant from “푂” the midpoint of diagonals in the two rectangles ⟹ the 6 points are on a sphere.

Solution to Problem 95.

푀 arbitrary point on |퐴퐵′|

퐴ʺ midpoint of segment [퐵′퐶′] When 푀 = 퐵′, point 푃 is in the position 퐵′.

255 Compiled and Solved Problems in Geometry and Trigonometry

When 푀 = 퐴, point 푃 is in the position {푃1} = [퐴′퐴1] ∩ [퐴퐴′′]

(퐴′퐴′′퐴1퐴 rectangle, so 푃1 is the intersection of the diagonals of the rectangle)

[The locus is [퐵′푃1]]. Let 푀 be arbitrary point 푀 ∈ |퐴퐵′|.

because:

By the way it was constructed

⟹ (∀) plane that contains 퐴퐴1 is perpendicular to (퐵′퐶퐶′′), particularly to (퐵′퐴퐴1) ⊥ (퐵′퐶′퐶).

because 퐵′, 푃 ∈ (퐵′퐴퐴′′)

from this reason 퐵′, 푃1 ∈ (퐴′퐵′퐴1). From 1 and 2

Let

So (∀) 푀 ∈ |퐵′퐴| and we have

Vice-versa. Let 푃 arbitrary point, 푃 ∈ |퐵′푃1| and In plane

In plane

Indeed: 퐴′퐴′′ || (퐵′퐴퐴1) thus any plane which passes through 퐴′퐴′′ will intersect

(퐵′퐴퐴1) after a parallel line to 퐴′퐴′′. Deci 푀푁||퐴′퐴′′ or 푀푁||퐴퐴1 as 푀 ∈ (퐵′퐴퐴1) ⟹ 푀푁 ⊥ (퐵′퐶퐶′′). We’ve proved

Florentin Smarandache

and

we have

describes |퐵′푃1| and vice-versa,

there is 푀|퐵′퐴| and 푁|퐵′퐴1| such that

and 푃 is the intersection of the diagonals of the quadrilateral 퐴′푁푀퐴′′.

Solution to Problem 96.

known result

From 1 and 3

Solution to Problem 97.

From the hypothesis: 푂퐴 ⊥ 푂퐵 ⊥ 푂퐶 ⊥ 푂퐴 We assume the problem is solved. Let 푀 be the orthocenter of triangle 퐴퐵퐶.

255 Compiled and Solved Problems in Geometry and Trigonometry

퐴퐵 ⊥ (푂퐶퐶′) But 퐶퐶′ ⊥ 퐴퐵 ⟹ } ⟹ 퐴퐵 ⊥ 푂푀 ⟹ 푀푂 ⊥ 퐴퐵 (1) 푀푂 ⊥ (퐶푂퐶′) 퐵퐶 ⊥ (퐴푂퐴′) 퐴푂 ⊥ (퐶푂퐵) ⟹ 퐴푂 ⊥ 퐵퐶, but 퐴퐴′ ⊥ 퐵퐶 ⟹ } ⟹ 퐵퐶 ⊥ 푀푂 ⟹ 푀푂 ⊥ 퐵퐶 (2) 푀푂 ⊂ (퐴푂퐴′) From (1) and (2) ⟹ 푀푂 ⊥ (퐴퐵퐶) So the plane (퐴퐵퐶) that needs to be drawn must be perpendicular to 푂푀 at 푀.

Solution to Problem 98.

퐴′푁 ⊥ 퐴퐷, 퐵′푀 ⊥ 퐵퐶 푎−푎′ 푏−푏′ ‖퐵푀‖ = , ‖푃푀‖ = 2 2 푎 − 푎′ 푏 − 푏′ ℎ 푣[퐵푀푃푆퐵′] = ∙ ∙ 2 2 3 휎[푆푃퐵′] ∙ ‖퐵′퐴′‖ 푎 − 푎′ ℎ 푣[푆푃푊푅퐴′퐵′] = = ∙ ∙ 푏′ 3 2 2 (푏 + 푏′)ℎ 푣[퐵′퐴′푁푀퐶′퐷′퐷 퐶 ] = ∙ 푎′ 1 1 2

푎−푎′ 푏−푏′ ℎ 푎−푎′ ℎ (푏+푏′)ℎ 푣[퐴퐵퐴′퐵′퐶퐷퐶′퐷′] = 2 [2 ∙ ∙ ∙ + ∙ ∙ 푏′] + ( ) ∙ 푎′ = 2 2 3 2 2 2 ℎ ℎ (2푎푏 − 2푎푏′ − 2푎′푏′ + 3푎푏′ − 3푎′푏′ + 3푎′푏 + 3푎′푏′) = [푎푏 + 푎′푏′ + 6 6 (푎 + 푎′)(푏 + 푏′)].

Florentin Smarandache

Solution to Problem 99.

퐵 ∙ ℎ 푣[푂퐴퐵퐵′] = 2 ℎ 퐵ℎ 푏ℎ 푣[푂퐴′퐵′퐴] = 푣[퐴퐵푂푂′퐴′퐵′] − 푣[퐴퐵퐵′푂] − 푣[퐴′퐵′푂′푂] = (퐵 + 푏 + √퐵푏) − − = 3 3 3 ℎ √푏퐵. 3 So: ℎ 푣[푂퐴′퐵′퐴] √퐵푏 ∙ √퐵푏 √푏 √푏 = 3 = ⟹ 푣[푂퐴′퐵′퐴] = ∙ 푣[푂퐴퐵퐵′] [ ] 퐵ℎ 푣 퐷퐴퐵퐵′ ∙ 퐵 √퐵 √퐵 3 For the relation above, determine the formula of the volume of the pyramid frustum.

Solution to Problem 100.

푑(퐺퐺′) = ℎ = 2푅

푙 3 푙 3 Let 푙 = ‖퐴퐶‖ ⟹ ‖퐴퐷‖ = √ ⟹ ‖퐺퐷‖ = √ 2 6

푙√3 Figure 퐺퐷푀푂 rectangle ⟹ ‖퐺퐷‖ = ‖푂푀‖ ⟹ = 푅 = 2√3푅 6

255 Compiled and Solved Problems in Geometry and Trigonometry

2 So, the lateral area is 푆푙 = 3 ∙ 2√3푅 ∙ 푅 = 12√3푅 .

2√3푅√3 푣[퐴퐵퐶퐴′퐵′퐶′] = 휎[퐴퐵퐶] ∙ 2푅 = 2√3푅 ∙ ∙ 2푅 = 6√3푅2. 4

The total area:

2 2 2 푆푡 = 푆푙 + 2휎[퐴퐵퐶] = 12√3푅 + 2 ∙ 3푅√3푅 = 18√3푅

Solution to Problem 101.

Let 푉1 and 푆1 be the volume, respectively the area obtained revolving around 푎.

푉2 and 푆2 be the volume, respectively the area obtained after revolving around 푏.

푉3 and 푆3 be the volume, respectively the area obtained after revolving around c. So: 휋 ∙ 푖2(‖퐶퐷‖ + ‖퐷퐵‖) 휋 ∙ 푖2 ∙ 푎 푉 = = 1 3 3

푆1 =휋∙푖∙푐+휋∙푖∙푏=휋∙푖∙ (푏 + 푐) 휋푐2푏 휋푐2푏2 휋푏2푐2푎 푉 = = = 2 3 3푏 3푎2

푆2 = 휋 ∙ 푐 ∙ 푎 휋푏2푐 휋푏2푐2 휋푏2푐2 푉 = = = 3 3 3푐 3푎

푆3 = 휋 ∙ 푏 ∙ 푎

Therefore: 1 1 1 9푎2 9푐2 9푐2 2 = 2 + 2 ⟺ 2 2 2 = 2 2 2 + 2 2 2 푉1 푉2 푉3 (휋푏 푐 ) (휋푏 푐 ) (휋푏 푐 ) 푆 푆 푆 + 푆 푐 푏 휋푎(푏 + 푐) 푐2 + 푏2 푎 2 + 3 = 2 3 ⟺ + = ⟺ = 푆3 푆2 푆1 푏 푐 휋푖(푏 + 푐) 푏 ∙ 푐 푖

푐2+푏2 푎2 But 푖 ∙ 푎 = 푏 ∙ 푐 ⟹ = , ‖퐴퐷‖ = 푖. 푏푐 푏푐

Florentin Smarandache

Solution to Problem 102.

푟 = ‖푂퐴‖ = 25 푐푚 푅 = ‖푂′퐵‖ = 50 푐푚 2휋푟 = 1,57 ⟹ 푟 = 0,25 푚 2휋푅 = 3,14 ⟹ 푅 = 0,50 푚 ‖퐶푁‖ = 18 푐푚 = 0,18 푚 ‖퐴′퐵‖ = 25 푐푚

‖퐴퐵‖ = √100 + 0,0625 = 10,003125 ‖퐴퐴′‖ ∙ ‖퐴′퐵‖ 10 ∙ 0,25 ‖퐴′푀‖ = = ≈ 0,25 ‖퐴퐵‖ 10,003125 ‖퐶푁‖ ‖퐶퐵‖ 0,18 ‖퐶퐵‖ = ⟹ = ⟹ ‖퐶퐵‖ = 0,18 ‖퐴′푀‖ ‖퐵퐴′‖ 0,25 0,25 ‖푂′퐶‖ = 푅′ = 0,50 − 0,18 = 0,32 ‖푂푃‖ = 푟′ = 0,25 − 0,18 = 0,07 휋1 푉 = (푅2 + 푟2 + 푅푟) 3 휋10 푉 = (0,502 + 0,252 + 0,50 ∙ 0,25 − 0,322 − 0,072 − 0,32 ∙ 0,07) 3 휋10 = (0,4375 − 0,1297) = 1,026휋푚3 3

Solution to Problem 103. 푎 훼 ‖푉푃‖ = 훼 cos 2 sin 2 2 퐴 In ∆푉퐴푃: ‖푉퐴‖ = 훼 . 2 sin 2

255 Compiled and Solved Problems in Geometry and Trigonometry

2 2 2 푎 푎 In ∆푉퐴푂′: ‖푉푂′‖ = 훼 − . 4 sin2 2 2 푎√cos 훼 ‖푉푂′‖ = 훼 2 sin 2

푎√cos 훼 In ∆푉푂푂′: ‖푂푂′‖ = 훼 − 푅. 2 sin 2 2 훼 2 2 2 4푅√cos 훼 sin2 2 푎 푎√cos 훼 푎 푎 cos 훼 2푎푅√cos 훼 2 푅 = + ( 훼 − 푅) ⟹ + 훼 − 훼 ⟹ 푎 = 훼 훼 2 2 sin 2 4 sin2 2 sin sin (2 cos2 + cos 훼) 2 2 2 2 2 훼 ⟹ 푎 = 4푅√cos 훼 ∙ sin 2

훼 훼 훼 푎2 cos 푎2 cos cos 2 2 2 2 훼 2 2 2 퐴푙 = 4 훼 = 훼 = 16푅 cos 훼 sin ∙ 훼 = 8푅 cos 훼 sin 훼 = 4푅 sin 2훼 2 ∙ 2 sin sin 2 sin 2 2 2 훼 퐴 = 퐴 + 푎2 = 4푅2 sin 2훼 + 16푅2 cos 훼 sin2 푡 푙 2

푎√cos 훼 훼 √cos 훼 0 ‖푉푂′‖ = 푅 ⟹ 푅 = 푎 ⟹ 푅 = 4푅√cos 훼 sin ∙ 훼 ⟹ 2 cos 훼 = 1 ⟹ 훼 = 60 . 2 sin 2 2 sin 2 2

Florentin Smarandache

Various Problems

104. Determine the set of points in the plane, with affine coordinates 푧 that satisfy: a. |푧| = 1; 3휋 b. 휋 < arg 푧 ≤ ; 푧 ≠ 0; 2 4휋 c. arg 푧 > , 푧 ≠ 0; 3 d. |푧 + 푖| ≤ 2 . Solution to Problem 104

105. Prove that the 푛 roots of the unit are equal to the power of the particular

root 휀1. Solution to Problem 105

106. Knowing that complex number 푧 verifies the equation 푧푛 = 푛, show that numbers 2, −푖푧 and 푖푧 verify this equation. Application: Find (1 − 2푖)4 and deduct the roots of order 4 of the number −7 + 24푖. Solution to Problem 106

107. Show that if natural numbers 푚 and 푛 are coprime, then the equations 푧푚 − 1 = 0 and 푧푛 − 1 = 0 have a single common root. Solution to Problem 107

108. Solve the following binomial equation: (2 − 3푖)푧6 + 1 + 5푖 = 0. Solution to Problem 108

109. Solve the equations:

255 Compiled and Solved Problems in Geometry and Trigonometry

Solution to Problem 109

110. Solve the equation 푧̅ = 푧푛−1, 푛 ∈ 푁, where 푧̅ the conjugate of 푧. Solution to Problem 110

111. The midpoints of the sides of a quadrilateral are the vertices of a parallelogram. Solution to Problem 111

112. Let 푀1푀2푀3푀4 and 푁1푁2푁3푁4 two parallelograms and 푃푖 the midpoints of

segments [푀푖푁푖], 푖 ∈ {1, 2, 3, 4}. Show that 푃1푃2푃3푃4 is a parallelogram or a degenerate parallelogram. Solution to Problem 112

113. Let the function 푓: 퐶 → 퐶, 푓(푧) = 푎푧 + 푏; (푎,푏,푐∈퐶,푎≠0). If 푀1 and ′ ′ 푀2 are of affixes 푧1 and 푧2, and 푀1 and 푀2 are of affixes 푓(푧1), 푓(푧2), show ′ ′ ′ ′ that ‖푀1푀2‖ = |푎| ∙ ‖푀1푀2‖. We have ‖푀1푀2‖ = ‖푀1푀2‖ ⇔ |푎| = 1. Solution to Problem 113

114. Prove that the function z → z̅, z ∈ C defines an isometry.

Solution to Problem 114

115. Let M1M2 be of affixes 푧1, 푧2 ≠ 0 and z2 = αz1. Show that rays |OM1, |OM2 coincide (respectively are opposed) ⟺ α > 0 (respectively α < 0).

Solution to Problem 115

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116. Consider the points M1M2M3 of affixes 푧1푧2푧3 and 푀1 ≠ 푀2. Show that:

z3−z1 a. M3 ∈ |M1M2 ⟺ > 0; z2−z1 z3−z1 b. M3 ∈ M1M2 ⟺ ∈ R . z2−z1 Solution to Problem 116

117. Prove Pompeiu’s theorem. If the point 푀 from the plane of the equilateral

triangle 푀1푀2푀3 ∉ the circumscribed circle ∆ 푀1푀2푀3  there exists a

triangle having sides of length ‖푀푀1‖, ‖푀푀2‖, ‖푀푀3‖.

Solution to Problem 117

255 Compiled and Solved Problems in Geometry and Trigonometry

Solutions

Solution to Problem 104.

|푧| = 1 a. } ⟹ 푥2 + 푦2 = 1, so the desired set is the circle 퐶 . |푧| = √푥2 + 푦2 (0,1) 3휋 b. 휋 < arg푧 ≤ . 2

The desired set is given by all the points of quadrant III, to which ray |푂푦 is added, so all the points with 푥 < 0, 푦 < 0.

4휋 arg푧 > , 푧 ≠ 0 4휋 c. 3 } ⟹ < arg푧 < 2휋 arg푧 ∈ [0, 2휋] 2

The desired set is that of the internal points of the angle with its sides positive semi-axis and ray |푂퐵.

d. |푧 + 푖| ≤ 2; 푧 = 푥 + 푦푖, its geometric image 푀.

where 푂′(0, −1).

′ Thus, the desired set is the disk centered at 푂(0,−1) and radius 2.

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Solution to Problem 105.

Solution to Problem 106.

Let the equation 푧4 = 푛. If 푧4 = 4 (푧 is the solution) then: (−푧)4 = (−1)4푧4 = 1 ∙ 푛 = 푛, so – 푧 is also a solution.

⟹ 푖푧 is the solution;

⟹ −푖푧 is the solution;

⟹ is the solution of the equation 푧4 = −7 + 24푖.

The solutions of this equation are:

but based on the first part, if 푧 − 1 − 2푖 is a root, then

are solutions of the given equation.

Solution to Problem 107.

If there exist 푘 and 푘′ with 푧푘 = 푧푘′, then

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255 Compiled and Solved Problems in Geometry and Trigonometry

because (푚, 푛) = 1. Because 푘′ < 푛, 푘 < 푚, we have 푘′ = 0, 푘 = 0.

Thus the common root is 푧0.

Solution to Problem 108.

Solution to Problem 109.

Solution to Problem 110.

As:

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From:

positive

The given equation becomes

Solution to Problem 111.

We find the sum of the abscissa of the opposite points:

⟹ 푀푁푃푄 a parallelogram.

Solution to Problem 112.

In the quadrilateral 푀1푀3푁3푁1 by connecting the midpoints we obtain the ′ ′′ parallelogram 푂 푃1푂 푃3, with its diagonals intersecting at 푂, the midpoint of |푂′푂′′|

and thus |푃1푂| ≡ |푂푃3|. (1)

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255 Compiled and Solved Problems in Geometry and Trigonometry

In the quadrilateral 푀4푀2푁2푁4 by connecting the midpoints of the sides we obtain ′ ′′ the parallelogram 푂 푃2푂 푃4 with its diagonals intersecting in 푂, the midpoint of

|푂′푂′′| and thus |푃2푂| ≡ |푂푃4|. (2)

From (1) and (2) 푃1푃2푃3푃4 a parallelogram.

Solution to Problem 113.

If:

Solution to Problem 114.

Let 푀1 and 푀2 be of affixes 푧1 and 푧2. Their images through the given function ′ ′ 푀1 and 푀2 with affixes 푧̅1 and 푧̅2, so

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′ ′ ′ ′ From (1) and (2) ⟹ ‖푀1푀2‖ = ‖푀1푀2‖ or ‖푀1푀2‖ = |푧̅2푧̅1| = |√푧2 − 푧1| =

|푧2 − 푧1| = ‖푀1푀2‖.

So 푓: 퐶 → 퐶, 푓(푧) = 푧̅ defines an isometry because it preserves the distance between the points.

Solution to Problem 115.

We know that the argument (푎푧1) = arg푧1 + arg푧훼 − 2푘휋, where 푘 = 0 or 푘 = 1.

Because arg푧2 = arg(푎푧1), arg푧2 = arg푧1 + arg푧훼 − 2푘휋.

a. We assume that

Vice versa,

b. Let |푂푀1 and |푂푀2 be opposed ⟹ arg푧2 = arg푧1 + 휋

∈ to the negative ray |푂푥′ ⟹ 훼 < 0. Vice versa,

arg푧2 = arg푧1 + 휋 푘 = 0 or 푘 = 1 ⟹ or } ⟹ |푂푀1 and |푂푀2 are opposed. arg푧2 = arg푧1 − 휋

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255 Compiled and Solved Problems in Geometry and Trigonometry

Solution to Problem 116.

If n and 푛′ are the geometric images of complex numbers 푧 and 푧′, then the image

of the difference 푧– 푧′ is constructed on |OM1| and |푀′푀| as sides.

We assume that 푀3 ∈ |푀1푀2

We construct the geometric image of 푧2– 푧1 . It is the fourth vertex of the

parallelogram 푂푀1푀2푄1. The geometric image of 푧3– 푧1 is 푄2, the fourth vertex of the

parallelogram 푂푀1푀3푄2.

푂푄1 ∥ 푀1푀2 푂푄2 ∥ 푀1푀3 } ⟹ 푄1, 푄2, 푄3 collinear ⟹ 푀1푀2푀3collinear

Vice versa, we assume that

If 푀3 and 푀2 ∈ the opposite ray to 푂, then 푧3– 푧1 = 훼(푧2– 푧1) with 훼 < 0. We repeat the reasoning from the previous point for the same case.

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Thus, when 푀3 ∈ 푀1푀2푀3 + 푀2 we obtain for the respective ratio positive,

z3−z1 negative or having 푀3 = 푀1, so ∈ R. z2−z1

Solution to Problem 117.

The images of the roots of order 3 of the unit are the peaks of the equilateral triangle.

2 But ɛ1 = ɛ2, so if we write ɛ2 = 휀, then ɛ1 = ɛ2. 2 Thus 푀1(1), 푀2(휀), 푀3(휀 ). We use the equality:

adequate (∀)푧 ∈ ℂ.

But

Therefore,

By substitution:

but

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255 Compiled and Solved Problems in Geometry and Trigonometry thus

Therefore ‖푀푀1‖, ‖푀푀2‖, ‖푀푀3‖ sides of a ∆. Then we use ‖푥| − |푦‖ ≤ |푥 − 푦| and obtain the other inequality.

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Problems in Spatial Geometry

118. Show that if a line 푑 is not contained in plane 훼, then 푑 ∩ 훼 is ∅ or it is formed of a single point. Solution to Problem 118

119. Show that (∀) 훼, (∃) at least one point which is not situated in 훼. Solution to Problem 119

120. The same; there are two lines with no point in common. Solution to Problem 120

121. Show that if there is a line 푑 (∃) at least two planes that contain line 푑. Solution to Problem 121

122. Consider lines 푑, 푑′, 푑′′, such that, taken two by two, to intersect. Show that, in this case, the 3 lines have a common point and are located on the same plane. Solution to Problem 122

123. Let 퐴, 퐵, 퐶 be three non-collinear points and 퐷 a point located on the plane (퐴퐵퐶). Show that: a. The points 퐷, 퐴, 퐵 are not collinear, and neither are 퐷, 퐵, 퐶; 퐷, 퐶, 퐴. b. The intersection of planes (퐷퐴퐵), (퐷퐵퐶), (퐷퐶퐴) is formed of a single point. Solution to Problem 123

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255 Compiled and Solved Problems in Geometry and Trigonometry

124. Using the notes from the previous exercise, take the points 퐸, 퐹, 퐺 distinct from 퐴, 퐵, 퐶, 퐷, such that 퐸 ∈ 퐴퐷, 퐹 ∈ 퐵퐷, 퐺 ∈ 퐶퐷. Let 퐵퐶 ∩ 퐹퐺 = {푃}, 퐺퐸 ∩ 퐶퐴 = {푄}, 퐸퐹 ∩ 퐴퐵 = {푅}. Show that 푃, 푄, 푅 are collinear (T. Desarques). Solution to Problem 124

125. Consider the lines 푑 and 푑′ which are not located on the same plane and the distinct points 퐴, 퐵, 퐶 ∈ 푑 and 퐷, 퐸 ∈ 푑′. How many planes can we draw such that each of them contains 3 non-collinear points of the given points? Generalization. Solution to Problem 125

126. Show that there exist infinite planes that contain a given line 푑. Solution to Problem 126

127. Consider points 퐴, 퐵, 퐶, 퐷 which are not located on the same plane. a. How many of the lines 퐴퐵, 퐴퐶, 퐴퐷, 퐵퐶, 퐵퐷, 퐶퐷 can be intersected by a line that doesn’t pass through 퐴, 퐵, 퐶, 퐷? b. Or by a plane that doesn’t pass through 퐴, 퐵, 퐶, 퐷? Solution to Problem 127

128. The points 훼 and 훽 are given, 퐴, 퐵 ∈ 훼. Construct a point 푀 ∈ 훼 at an equal distance from 퐴 and 퐵, that ∈ also to plan 훽. Solution to Problem 128

129. Determine the intersection of three distinct planes 훼, 훽, 훾. Solution to Problem 129

130. Given: plane 훼, lines 푑1, 푑2 and points 퐴, 퐵 ∉ 훼 ∪ 푑1 ∪ 푑2. Find a point 푀 ∈

훼 such that the lines 푀퐴, 푀퐵 intersect 푑1 and 푑2. Solution to Problem 130

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131. There are given the plane 훼, the line 푑 ∉ 훼, the points 퐴, 퐵 ∉ 훼 ∪ 푑, and 퐶 ∈ 훼. Let 푀 ∈ 푑 and 퐴′, 퐵′ the points of intersection of the lines 푀퐴, 푀퐵 with plane 훼 (if they exist). Determine the point 푀 such that the points 퐶, 퐴′, 퐵′ to be collinear. Solution to Problem 131

132. If points 퐴 and 퐵 of an open half-space 휎, then [퐴퐵] ⊂ 휎. The property is as well adherent for a closed half-space. Solution to Problem 132

133. If point 퐴 is not situated on plane 훼 and 퐵 ∈ 훼 then |퐵퐴 ⊂ |훼퐴. Solution to Problem 133

134. Show that the intersection of a line 푑 with a half-space is either line 푑 or a ray or an empty set. Solution to Problem 134

135. Show that if a plane 훼 and the margin of a half-space 휎 are secant planes, then the intersection 휎 ∩ 훼 is a half-plane. Solution to Problem 135

136. The intersection of a plane 훼 with a half-space is either the plane 훼 or a half-plane, or an empty set. Solution to Problem 136

137. Let 퐴, 퐵, 퐶, 퐷 four non coplanar points and 훼 a plane that doesn’t pass through one of the given points, but it passes trough a point of the line |퐴퐵|. How many of the segments |퐴퐵|, |퐴퐶|, |퐴퐷|, |퐵퐶|, |퐵퐷|, |퐶퐷| can be intersected by plane 훼? Solution to Problem 137

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255 Compiled and Solved Problems in Geometry and Trigonometry

138. Let 푑 be a line and 훼, 훽 two planes such that 푑 ∩ 훽 = ∅ and 훼 ∩ 훽 = ∅. Show that if 퐴 ∈ 푑 and 퐵 ∈ 훼, then 푑 ⊂ |훽퐴 and 훼 ⊂ |훽퐵. Solution to Problem 138

139. Let |훼퐴 and |훽퐵 be two half-spaces such that 훼 ≠ 훽 and |훼퐴 ⊂ |훽퐵 or |훼퐴 ∩ |훽퐵 = ∅. Show that 훼 ∩ 훽 = ∅. Solution to Problem 139

140. Show that the intersection of a dihedral angle with a plane 훼 can be: a right angle, the union of two lines, a line, an empty set or a closed half- plane and cannot be any other type of set. Solution to Problem 140

141. Let 푑 be the edge of a proper dihedron ∠훼′훽′, 퐴 ∈ 훼′– 푑, 푏 ∈ 훽′– 푑 and

푃 ∈ int. ∠훼′훽′. Show that:

a. (푃푑) ∩ int. ∠훼′훽′ = |푑푃;

b. If 푀 ∈ 푑, int. ∠퐴푀퐵 = int. 훼′훽′ ∩ (퐴푀퐵). Solution to Problem 141

142. Consider the notes from the previous problem. Show that: a. The points 퐴 and 퐵 are on different sides of the plane (푃푑); b. The segment |퐴퐵| and the half-plane |푑푃 have a common point. Solution to Problem 142

143. If ∠푎푏푐 is a trihedral angle, 푃 ∈ int. ∠푎푏푐 and 퐴, 퐵, 퐶 are points on edges 푎, 푏, 푐, different from 푂, then the ray |푂푃 and int. 퐴퐵퐶 have a common point. Solution to Problem 143

144. Show that any intersection of convex sets is a convex set. Solution to Problem 144

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145. Show that the following sets are convex planes, half-planes, any open or closed half-space and the interior of a dihedral angle. Solution to Problem 145

146. Can a dihedral angle be a convex set? Solution to Problem 146

147. Which of the following sets are convex: a. a trihedral angle; b. its interior; c. the union of its faces; d. the union of its interior with all its faces? Solution to Problem 147

148. Let 휎 be an open half-space bordered by plane 훼 and 푀 a closed convex set in plane 훼. Show that the set 푀 ∩ 휎 is convex. Solution to Problem 148

149. Show that the intersection of sphere 푆(푂, 푟) with a plane which passes through 푂, is a circle. Solution to Problem 149

150. Prove that the int. 푆(푂, 푟) is a convex set. Solution to Problem 150

151. Show that, by unifying the midpoints of the opposite edges of a tetrahedron, we obtain concurrent lines. Solution to Problem 151

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255 Compiled and Solved Problems in Geometry and Trigonometry

152. Show that the lines connecting the vertices of a tetrahedron with the centroids of the opposite sides are concurrent in the same point as the three lines from the previous example. Solution to Problem 152

153. Let 퐴퐵퐶퐷 be a tetrahedron. We consider the trihedral angles which have as edges [퐴퐵,[퐴퐸,[퐴퐷,[퐵퐴,[퐵퐶,[퐵퐷,[퐶퐴,[퐶퐵,[퐶퐷,[퐷퐴,[퐷퐵,[퐷퐶. Show that the intersection of the interiors of these 4 trihedral angles coincides with the interior of tetrahedron [퐴퐵퐶퐷]. Solution to Problem 153

154. Show that (∀) 푀 ∈ int. [퐴퐵퐶퐷] (∃) 푃 ∈ |퐴퐵| and 푄 ∈ |퐶퐷| such that 푀 ∈ ‖푃푄. Solution to Problem 154

155. The interior of tetrahedron [퐴퐵퐶퐷] coincides with the union of segments |푃푄| with 푃 ∈ |퐴퐵| and 푄 ∈ |퐶퐷|, and tetrahedron [퐴퐵퐶퐷] is equal to the union of the closed segments [푃푄], when 푃 ∈ [퐴퐵] and 푄 ∈ [퐶퐷]. Solution to Problem 155

156. The tetrahedron is a convex set. Solution to Problem 156

157. Let 푀1 and 푀2 convex sets. Show that by connecting segments [푃푄], for

which 푃 ∈ 푀1 and 푄 ∈ 푀2 we obtain a convex set. Solution to Problem 157

158. Show that the interior of a tetrahedron coincides with the intersection of the open half-spaces determined by the planes of the faces and the opposite peak. Define the tetrahedron as an intersection of half-spaces.

Solution to Problem 158

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Solutions

Solution to Problem 118.

We assume that 푑 ∩ 훼 = {퐴, 퐵} ⇒ 푑 ⊂ 훼. It contradicts the hypothesis ⟹ 푑 ∩ 훼 = {퐴} or 푑 ∩ 훼 = ∅.

Solution to Problem 119.

We assume that all the points belong to the plane 훼 ⟹ (∄) for the points that are not situated in the same plane. False!

Solution to Problem 120.

∃ 퐴, 퐵, 퐶, 퐷, which are not in the same plane. We assume that 퐴퐵 ∩ 퐶퐷 = {0} ⟹ 퐴퐵 and 퐶퐷 are contained in the same plane and thus 퐴, 퐵, 퐶, 퐷 are in the same plane. False, it contradicts the hypothesis ⟹ 퐴퐵 ∩ 퐶퐷 = ∅ ⟹ (∃) lines with no point in common.

Solution to Problem 121.

(∃) 퐴 ∉ 푑 (if all the points would ∈ 푑, the existence of the plane and space would be negated). Let 훼 = (푑퐴), (∃)퐵 ∉ 훼 (otherwise the space wouldn’t exist). Let 훽 = (퐵푑), 훼 ≠ 훽 and both contain line 푑.

Solution to Problem 122.

We show that 푑 ≠ 푑′ ≠ 푑′′ ≠ 푑.

푑 ⊂ 훼 Let 푑 ∩ 푑′ = {퐴} = (푑, 푑′) ⟹ { 푑′ ⊂ 훼 푑 ∩ 푑′ = {퐵} 퐵 ∈ 푑 ⟹ { ⟹ 퐵 ∈ 푎 , 퐵 ∈ 푑′ 퐵 ≠ 퐴 푑 ⊂∝

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255 Compiled and Solved Problems in Geometry and Trigonometry

푑′′ ∩ 푑′ = {퐶} 퐶 ∈ 푑′ 퐶 ≠ 퐵 ⟹ ⟹ 퐶 ∈ 훼, 퐶 ∈ 푑′′ 푑′ ⊂ ∝ 퐶 ≠ 퐴 푑 = 푑′ ⟹ or 푑 = 푑′′

⟹ 푑′′ ⊂ 훼, so the lines are located on the same plane α.

푑 ∩ 푑′ = {퐴} ⟹ 퐴 ∈ 푑′ If } ⟹ 푑′ ∩ 푑′′ = {퐴}, and the three lines have a point in 푑′′ ∩ 푑 = {퐴} ⟹ 퐴 ∈ 푑′′ common.

Solution to Problem 123.

a. 퐷 ∉ (퐴퐵퐶).

퐷∈푑,퐴∈푑,퐵∈푑 푇2 We assume that 퐷, 퐴, 퐵 collinear ⟹ (∃)푑 such that } ⇒ 푑 ⊂ 퐴 ∈ (퐴퐵퐶), 퐵 ∈ (퐴퐵퐶) (퐴퐵퐶) ⟹ 퐷 ∈ (퐴퐵퐶) – false. Therefore, the points 퐷, 퐴, 퐵 are not collinear.

b. Let (퐷퐴퐵) ∩ (퐵퐶퐷) ∩ (퐷퐶퐴) = 퐸.

As the planes are distinct, their intersections are:

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(퐷퐴퐵) ∩ (퐷퐵퐶) = 퐷퐵 ⟹ If (퐷퐴퐵) = (퐷퐵퐶) (퐷퐴퐵) ∩ (퐷퐶퐴) = 퐷퐴} ⟹ 퐴, 퐵, 퐶, 퐷 coplanar, contrary to the hypothesis. (퐷퐵퐶) ∩ (퐷퐶퐴) = 퐷퐶 푀 ∈ 퐷퐵 퐵 ∈ 푀퐷 We suppose that (∃)푀 ∈ 퐸, 푀 ≠ 퐷 ⟹ } ⟹ } ⟹ 퐴, 퐵, 퐷 are collinear 푀 ∈ 퐷퐴 퐴 ∈ 푀퐷 (false, contrary to point a.). Therefore, set 퐸 has a single point 퐸 = {퐷}.

Solution to Problem 124.

We showed at the previous exercise that if 퐷 ∉ (퐴퐵퐶), (퐷퐴퐵) ≠ (퐷퐵퐶). We show that 퐸, 퐹, 퐺 are not collinear. We assume the opposite. Then,

Having three common points 퐷, 퐵 and 퐺 ⟹ false. So 퐸, 퐹, 퐺 are not collinear and determine a plane (퐸퐹퐺).

⟹ 푃, 푄, 푅 are collinear because ∈ to the line of intersection of the two planes.

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255 Compiled and Solved Problems in Geometry and Trigonometry

Solution to Problem 125.

The planes are (퐴, 푑′); (퐵, 푑′); (퐶, 푑′).

Generalization: The number of planes corresponds to the number of points on line 푑 because 푑′ contains only 2 points.

Solution to Problem 126.

Let line 푑 be given, and 퐴 any point such that 퐴 ∉ 푑.

We obtain the plane 훼 = (퐴, 푑), and let 푀 ∉ 훼. The line 푑′ = 퐴푀, 푑′ ⊄ 훼 is not thus contained in the same plane with 푑. The desired planes are those of type (푀푑), 푀 ∈ 푑′, that is an infinity of planes.

Solution to Problem 127.

a. (∀) 3 points determine a plane. Let plane (퐴퐵퐷). We choose in this plane 푃 ∈ |퐴퐷| and 푄 ∈ |퐴퐵| such that 푃 ∈ |퐵푄|, then the line 푃푄 separates the points 퐴 and 퐷, but does not separate 퐴 and 퐵, so it separates 푃 and 퐷 ⇒ 푃푄 ∩ |퐵퐷| = 푅, where 푅 ∈ |퐵퐷|.

Thus, the line 푃푄 meets 3 of the given lines. Let’s see if it can meet more.

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We assume that

it has two points in common with the plane.

⟹ 퐴, 퐵, 퐶, 퐷 coplanar – false.

Thus,

false.

We show in the same way that 푃푄 does not cut 퐴퐶 or 퐷퐶, so a line meets at most three of the given lines.

b. We consider points 퐸, 퐹, 퐺 such that 퐸 ∈ |퐵퐶|, 퐴 ∈ |퐷퐹|, 퐷 ∈ |퐵퐺|. These points determine plane (퐸퐹퐺) which obviously cuts the lines 퐵퐶, 퐵퐷 and 퐵퐷. 퐹퐺 does not separate 퐴 and 퐷 or 퐵퐷 ⟹ it does not separate 퐴 or 퐵 ⟹ 퐴 ∈ |퐵푅|.

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255 Compiled and Solved Problems in Geometry and Trigonometry

Let’s show that (퐸퐹퐺) meets as well the lines 퐴퐵, 퐶퐷, 퐴퐶. In the plane (퐴퐵퐷) we consider the triangle 퐹퐷퐺 and the line 퐴퐵.

As this line cuts side |퐹퐷|, but it does not cut |퐷퐺|, it must cut side |퐹퐺|, so 퐴퐵 ∩ |퐹퐺| = {푅} ⟹ 푅 ∈ |퐹퐺| ⊂ (퐸퐹퐺), so 퐴퐵 ∩ (퐸퐹퐺) = {푅}. In the plane (퐵퐶퐷), the line 퐸퐺 cuts |퐵퐶| and does not cut |퐵퐷|, so 퐸퐺 cuts the side |CD|, 퐸퐺 ∩ |퐶퐷| = {푃} ⟹ 푃 ∈ 퐸퐺 ⊂ (퐸퐹퐺) ⟹ 퐶퐷 ∩ (퐸퐹퐺) = {푃}.

푅 ∈ (퐸퐹퐺), 푅 does not separate 퐴 and 퐵 } ⟹ 푅 ∈∩ |퐴퐶| = 푄 ⟹ 푄 ∈ 푅퐸 ⟹ 푄 퐸 separates 퐵 and 퐶 ∈ (퐸퐹퐺) ∩ 퐴퐶 = {푄}.

Solution to Problem 128.

푀 ∈ 훼 We assume problem is solved, if } ⟹훼∩훽≠∅,⟹훼∩훽=푑. 푀 ∈ 훽

As ||푀퐴|| = ||푀퐵|| ⟹ 푀 ∈ the bisecting line of the segment [퐴퐵].

So, to find 푀, we proceed as follows:

1. We look for the line of intersection of planes 훼 and 훽, d. If 훼 ∥ 훽, the problem hasn’t got any solution.

2. We construct the bisecting line 푑′ of the segment [퐴퐵] in the plane 훼. 3. We look for the point of intersection of lines 푑 and 푑′. If 푑 ∥ 푑′, the problem hasn’t got any solution.

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Solution to Problem 129.

If 훼∩훽=∅⟹훼∩훽∩훾=∅. If 훼 ∩ 훽 = 푑, the desired intersection is 푑 ∩ 훾, which can be a point (the 3 planes are concurrent), the empty set (the line of intersection

of two planes is || with the third) or line 푑 (the 3 planes which pass through 푑 are secant).

Solution to Problem 130.

To determine 푀, we proceed as follows:

1. We construct plane (퐴푑1) and we look for the line of intersection with 훼1, 푑1.

If 푑1 (/∃), ∄ neither does 푀.

2. We construct plane (퐵푑2) and we look for the line of intersection with 훼, 푑2′.

If 푑2′ does not exist, neither does 푀.

3. We look for the point of intersection of lines 푑1′ and 푑2′. The problem has only one solution if the lines are concurrent, an infinity if they are coinciding lines and no solution if they are parallel.

Solution to Problem 131.

We assume the problem is solved. a. First we assume that 퐴. 퐵, 퐶 are collinear. As 퐴퐴′ and 퐵퐵′ are concurrent lines, they determine a plane 훽, that intersects 훼 after line 퐴′퐵′. As

and points 퐶, 퐴, 퐵′ are collinear (∀)푀 ∈ 푑.

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255 Compiled and Solved Problems in Geometry and Trigonometry

b. We assume that 퐴, 퐵, 퐶 are not collinear. We notice that: (퐴퐴′, 퐵퐵′) = 훽 (plane determined by 2 concurrent lines). 훽 ∪ 훼 = 푑′ and 퐶 ∈ 푑′.

To determine 푀 we proceed as follows: 1) We determine plane (퐴퐵퐶); 2) We look for the point of intersection of this plane with line 푑, so 푑 ∩ (퐴퐵퐶) = {푀} is the desired point. Then (퐴퐵퐶) ∩ 훼 = 푑′.

⟹ 퐴′, 퐵′, 퐶′ are collinear.

Solution to Problem 132.

퐴 ∈ 휎 and 퐵 ∈ 휎 ⟹ [퐴퐵] ∩ 훼 ≠ ∅. Let 휎 = |훼퐴 = |훼퐵. Let 푀 ∈ |퐴퐵| and we must show that 푀 ∈ 휎(∀)푀 inside the segment. We assume the contrary that 푀 ∉ 휎 ⟹ (∃)푃 such that [퐴푀] ∩ 푑 = {푃} ⟹ 푃 ∈ [퐴푀] ⟹ 푃 ∈ [퐴퐵] ⟹ [퐴퐵] ∩ 훼 ≠ ∅ false. 푃 ∈ 훼, so 푀 ∈ 휎. The property is also maintained for the closed half-space.

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Compared to the previous case there can appear the situation when one of the points 퐴 and 퐵 ∈ 훼 or when both belong to 훼.

If 퐴 ∈ 훼, 퐵 ∈ 휎, |퐴퐵| ∩ 훼 ≠ ∅ and we show as we did above that:

If:

Solution to Problem 133.

Let

Solution to Problem 134.

Let 훼 be a plane and 휎1, 휎2 the two half-spaces that it determines. We consider

half-space 휎1.

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255 Compiled and Solved Problems in Geometry and Trigonometry

푃 determines on 푑 two rays, |푃퐴 and |푃퐵 where

퐴 and 퐵 are in different half-spaces. We assume

Solution to Problem 135.

Let 휎 be an open half-space and 푝 its margin and let 푑 = 훼 ∩ 훽. We choose points 퐴 and 퐵 ∈ 훼 – 푑, on both sides of line 푑 ⟹

⟹ 퐴, 퐵 are on one side and on the other side of 훽 and it means that only one of them is on 휎. We assume that 퐴 ∈ 휎 ⟹ 퐵 ∈ 휎. We now prove 훼 ∩ 휎 = |푑퐴. 훼 ∩ 휎 ⊂ |푑퐴

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Let

[푀퐵] ∈ 푑 ≠ ∅ ⟹ 푀 and 퐵 are on one side and on the other side of line 푑 ⟹ 푀 is on the same side of line 푑 with 퐴 ⟹ 푀 ∈ |푑퐴

⟹ 푀 ∈ 푎 ∩ 휎, so |푑퐴 ⊂ 푎 ∩ 휎.

Solution to Problem 136. Let σ be the considered half-space and β its margin. There are more possible cases:

In this case it is possible that:

Let

is a half-plane according to a previous problem.

Solution to Problem 137.

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255 Compiled and Solved Problems in Geometry and Trigonometry

The intersection of two planes is a line and it cuts only two sides of a triangle. There are more possible cases: 1. 푑 cuts |퐴퐵| and |퐵퐶| 푑′ cuts |퐴퐵| and |퐴퐷|, 훼 cuts |퐴퐷| so it has a point in common with (퐴퐷퐶) and let (퐴퐷퐶) ∩ 훼 = 푑′′. 푑′′ cuts |퐴퐷| and does not cut |퐴퐶| ⟹ 푑′′ cuts |퐷퐶| 훼 cuts |퐷퐶| and |퐵퐶| ⟹ it does not cut |퐵퐷|. In this case 훼 cuts 4 of the 6 segments (the underlined ones). 2. 푑 cuts |퐴퐵| and |퐴퐶|, it does not cut |퐵퐷| 푑′ cuts |퐴퐵| an |퐴퐷|, it does not cut |퐵퐷| 푑′′ cuts |퐴퐷| an |퐴퐶|, it does not cut |퐷퐶| ⟹ 훼 does not intersect plane (퐵퐶퐷). In this case 훼 intersects only 3 of the 6 segments. 3. 푑 cuts |퐴퐵| and |퐵퐶|, it does not cut |퐴퐶| 푑′ cuts |퐴퐵| an |퐵퐷|, it does not cut |퐷퐶| 훼 intersects |퐵퐷| and |퐵퐶|, so it does not cut |퐷퐶| In ∆퐵퐷퐶 ⟹ 훼 does not intersect plane (퐴퐷퐶) In this case 훼 intersects only three segments. 4. 푑 cuts |퐴퐵| and |퐴퐶|, it does not cut |퐵퐶| 푑′ cuts |퐴퐵| an |퐵퐷|, it does not cut |퐴퐷| 푑′′ cuts |퐴퐶| an |퐷퐶| 훼 does not cut |퐵퐶| in triangle 퐵퐷퐶. So 훼 intersects 4 or 3 segments.

Solution to Problem 138.

Let

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Solution to Problem 139. We first assume that 훼 ≠ 훽 and |훼퐴 ⊂ |훽퐵.

The hypothesis can then be written as 훼 ≠ 훽 and |훼퐴 ⊂ |훽퐵. Let’s show that 훼 ∩ 훽 = ∅. By reductio ad absurdum, we assume that 훼 ∩ 훽 ≠ ∅ ⟹ (∃)푑 = 훼 ∩ 훽 and let 푂 ∈ 푑, so 푂 ∈ 훼 and 푂 ∈ 훽. We draw through 퐴 and 푂 a plane 푟, such that 푑 ∈ 푟, so the three planes 훼, 훽 and 푟 do not pass through this line. As 푟 has the common point 푂 with 훼 and 훽, it is going to intersect these planes.

which is a common point of the 3 planes. Lines 훿 and 훿′ determine 4 angles in plane 푟, having 푂 as a common peak, 퐴 ∈ the interior of one of them, let 퐴 ∈ int. ℎ̂푘. We consider 퐶 ∈ int. ℎ̂푘. Then 퐶 is on the same side with 퐴 in relation to 훿′, so 퐶 is on the same side with 퐴 in relation to 훼 ⟹ 퐶 ∈ |훼퐴.

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255 Compiled and Solved Problems in Geometry and Trigonometry

But 퐶 is on the opposite side of 퐴 in relation to 훿, so 퐶 is on the opposite side of 퐴 in relation to 훽 ⟹ 퐶 ∉ |훽퐴. So |훼퐴 ⊄ |훽퐴 – false – it contradicts the hypothesis ⟹ So 훼 ∩ 훽 = ∅.

Solution to Problem 140.

Let 푑 be the edge of the given dihedral angle. Depending on the position of a line in relation to a plane, there can be identified the following situations:

The ray with its origin in 푂, so 훼 ⊂ 훽̂′훾′ = 푑̂′푑′′ thus an angle.

Indeed, if we assumed that 푑′ ∩ 푑′′ ≠ ∅ ⟹ (∃)푂 ∈ 푑′ ∩ 푑′′.

false – it contradicts the hypothesis.

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in this case 훼 ∩ 훽̂′훾′ = 푑′′ - a line.

Then 훼 ∩ 훽̂′훾′ = ∅.

푑 ∩ 훼 = 푑, but 훼 ≠ 훽, 훼 ≠ 훾 훼 ∩ 훽̂′훾′ = 푑 thus the intersection is a line.

In this case the intersection is a closed half-plane.

Solution to Problem 141.

is a half-plane

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255 Compiled and Solved Problems in Geometry and Trigonometry

is a half-plane

From (*) and (**),

so they are secant planes

Solution to Problem 142.

⟹ points 퐴 and 퐵 are on different sides of (푑푃).

Solution to Problem 143.

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Let rays As 푃 is interior to the dihedron formed by any half-plane passing through 푂 of the trihedral, so

푃 and 푄 in the same half-plane det. 푂퐴 ⇒ 푃 and 푄 on the same side of 푂퐴 (1)

⟹ 푃 and 퐴 are on the same side of (푂퐵퐶) ∩ 훾′ ⟹ 푃 and 퐴 are on the same side of 푂푄 (2). From (1) and (2) ⟹

Solution to Problem 144.

Let 푀 and 푀′′be two convex sets and 푀 ∩ 푀′ their intersection. Let

so the intersection is convex.

Solution to Problem 145.

a. Let 푃, 푄 ∈ 훼; 푃 ≠ 푄 ⟹ |푃푄 = 푃푄 (the line is a convex set) 푃푄 ⊂ 훼, so |푃푄| ⊂ 훼, so the plane is a convex set.

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255 Compiled and Solved Problems in Geometry and Trigonometry

b. Half-planes: Let 푆 = |푑퐴 and 푃, 푄 ∈ 푆 ⟹ |푃푄| ∩ 푑 = ∅. Let 푀 ∈ |푃푄| ⟹ |푃푀| ⊂ |푃푄| ⟹ |푃푀| ∩ 푑 = ∅ ⟹ 푃 and 푀 are in the same half-plane ⟹ 푀 ∈ 푆. So |푃푄| ⊂ 푆 and 푆 is a convex set.

Let 푆′ = [푑퐴. There are three situations: 1) 푃, 푄 ∈ |푑퐴 – previously discussed; 2) 푃, 푄 ∈ 푑 ⟹ |푃푄| ⊂ 푑 ⊂ 푆′; 3) 푃 ∈ 푑, 푄 ∉ 푑 ⟹ |푃푄| ⊂ |푑푄 ⟹ |푃푄| ⊂ |푑퐴 ⊂ [푑퐴 so [푑퐴 is a convex set.

c. Half-spaces: Let 휎 = |훼퐴 and let 푃, 푄 ∈ 휎 ⟹ |푃푄| ∩ 훼 = ∅. Let 푀 ∈ |푃푄| ⟹ |푃푀| ⊂ |푃푄| ⟹ |푃푀| ∩ 훼 = ∅. Let 휎′ = [훼퐴. There are three situations: 1) 푃, 푄 ∈ |훼퐴 previously discussed; 2) 푃, 푄 ∈ 훼 ⟹ |푃푄| ⊂ 훼 ⊂ 휎′; 3) 푃 ∈ 훼, 푄 ∉ 훼.

and so 휎′ is a convex set.

d. the interior of a dihedral angle: 푖푛푡. 훼′훽′ = |훼퐴 ∩ |훽퐵 and as each half-space is a convex set and their intersection is the convex set.

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Solution to Problem 146.

No. The dihedral angle is not a convex set, because if we consider it as in the previous figure 퐴 ∈ 훽′ and 퐵 ∈ 훼′.

Only in the case of the null or straight angle, when the dihedral angle becomes a plane or closed half-plane, is a convex set.

Solution to Problem 147.

a. No. The trihedral angle is not the convex set, because, if we take 퐴 ∈ 푎 and 푄 ∈ the int. 푏푐̂ determined by 푃 ∈ the int. 푎푏푐̂ , (∃)푅 such that |푂푃 ∩ the int. 퐴퐵퐶 = {푅}, 푅 ∈ |퐴푄|, 푅 ∉ 푎푏푐̂ . So 퐴, 푄 ∈ 푎푏푐̂ , but |퐴푄| ∉ 푎푏푐. b. 훣 = (푂퐶퐴), 훾 = (푂퐴퐵) is a convex set as an intersection of convex sets. C) It is the same set from a. and it is not convex. D) The respective set is [훼퐴 ∩ [훽퐵 ∩ [훾퐶, intersection of convex sets and, thus, it is convex.

Solution to Problem 148.

Let 휎 = |훼퐴 and 푀 ⊂ 훼. Let 푃, 푄 ∈ 푀 ∩ 휎. We have the following situations:

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255 Compiled and Solved Problems in Geometry and Trigonometry

Solution to Problem 149.

Solution to Problem 150.

Let

In plane (푂푃푄), let 푀 ∈ (푃푄).

Solution to Problem 151.

Let: 푃 midpoint of |퐴퐵| 푅 midpoint of |퐵퐶| 푄 midpoint of |퐷퐶| 푆 midpoint of |퐴퐷| 푇 midpoint of |퐵퐷| 푈 midpoint of |퐴퐶|

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In triangle ABC:

In triangle DAC:

⟹ parallelogram ⟹ |푃푄| and |푆푅| intersect at their midpoint 푂.

⟹ 푆푇푅푈 parallelogram. ⟹ |푇푈| passes through midpoint 푂 of |푆푅|. Thus the three lines 푃푅, 푆푅, 푇푈 are concurrent in 푂.

Solution to Problem 152.

Let tetrahedron 퐴퐵퐶퐷 and 퐸 be the midpoint of |퐶퐷|. The centroid 퐺 of the face 퐴퐶퐷 is on |퐴퐸| at a third from the base. The centroid 퐺′ of the face 퐵퐶퐷 is on |퐵퐸| at a third from the base |퐶퐷|. We separately consider ∆퐴퐸퐵. Let 퐹 be the midpoint of 퐴퐵, so 퐸퐹 is median in this triangle and, in the previous problem, it was one of the 3 concurrent segments in a point located in the middle of each.

Let 푂 be the midpoint of |퐸퐹|. We write 퐴푂 ∩ 퐸퐵 = {퐺′} and 퐵퐺 ∩ 퐸퐴 = {퐺}.

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255 Compiled and Solved Problems in Geometry and Trigonometry

From (1) and (2)

⟹ 퐺′ is exactly the centroid of face 퐵퐶퐷, because it is situated on median |퐸퐵| at a third from 퐸. We show in the same way that 퐺 is exactly the centroid of face 퐴퐶퐷. We’ve thus shown that 퐵퐺 and 퐴퐺′ pass through point 푂 from the previous problem. We choose faces 퐴퐶퐷 and 퐴퐶퐵 and mark by 퐺′′ the centroid of face 퐴퐶퐵, we show in the same way that 퐵퐺 and 퐷퐺′′ pass through the middle of the segment |푀푁| (|퐴푀| ≡ |푀퐶|, |퐵푁| ≡ |푁퐷|) thus also through point 푂, etc.

Solution to Problem 153.

We mark planes (퐴퐵퐶) = 훼, (퐴퐷퐶) = 훽, (퐵퐷퐶) = 훾, (퐴퐵푂) = 훿. Let 푀 be the intersection of the interiors of the 4 trihedral angles. We show that: 푀 = int. [퐴퐵퐶퐷], by double inclusion. 1. 푃 ∈ 푀 ⟹ 푃 ∈ int. 푎푏푐̂ ∩ int. 푎푓푑̂ ∩ int. 푑푒푐̂ ∩ int. 푏푓푐̂ ⟹ 푃 ∈ |훼퐷 ∩ |훾퐶 ∩ 훽퐵 and 푃 ∈ |훿퐴 ∩ |훾퐶 ∩ |훽퐵 ⟹ 푃 ∈ |훼퐷 and 푃 ∈ |훽퐵 and 푃 ∈ |훾퐶 and 푃 ∈ |훿퐶 ⟹ 푃 ∈ |훼퐷 ∩ |훾퐶 ∩ 훽퐵 ∩ 훿퐴 ⟹ 푃 ∈ int. [퐴퐵퐶퐷]. So 푀 ∈ [퐴퐵퐶퐷]. 2. Following the inverse reasoning we show that [퐴퐵퐶퐷] ⊂ 푀 from where the equality.

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Solution to Problem 154.

such that

such that 푁 ∈ |퐶푃|, (퐴퐷퐵) ∩ (퐷푃퐶) = 퐷푃. 푙푒푚푚푎 From 푁 ∈ |퐶푃| and ∈ |퐷푁| ⇒ int. 퐷푃퐶 ⟹ 푀 ∈ int. 퐷푃퐶 ⟹ (∃)푄 ∈ |퐷퐶|. So we showed that (∃)푃 ∈ |퐴퐵| and 푄 ∈ |퐷퐶| such that 푀 ∈ |푃푄|.

Solution to Problem 155.

Let ℳ be the union of the open segments |푃푄|. We must prove that: int. [퐴퐵퐶퐷] = ℳ through double inclusion. 1. Let 푀 ∈ int. [퐴퐵퐶퐷] ⟹ (∀)푃 ∈ |퐴퐵 and 푄 ∈ |퐶퐷 such that 푀 ∈ |푃푄| ⟹ 푀 ∈ ℳ so int. [퐴퐵퐶퐷] ⊂ ℳ. 2. Let 푀 ∈ ℳ ⟹ (∃)푃 ∈ |퐴퐵| and 푄 ∈ |퐶퐷| such that 푀 ∈ |푃푄|. Points 퐷, 퐶 and 푃 determine plane (푃퐷퐶) and (푃퐷퐶) ∩ (퐴퐶퐵) = 푃퐶, (푃푈퐶) ∩ (퐴퐷퐵) = 푃퐷. As (∀)푄 ∈ |퐶퐷| such that 푀 ∈ |푃푄| ⟹ 푀 ∈ [푃퐶퐷] ⟹ |(∀)푅 ∈ |푃퐶| such that 푀 ∈ |퐷푅|. If 푃 ∈ |퐴퐵| and 푅 ∈ |푃퐶| ⟹ 푅 ∈ int. 퐴퐶퐵 such that 푀 ∈ |퐷푅| ⟹ 푀 int. [퐴퐵퐶퐷] ⟹ ℳ ⊂ int. [퐴퐵퐶퐷]. Working with closed segments we obtain that (∀)푅 ∈ [퐴퐶퐵] such that 푀 ∈ [퐷푅], thus obtaining tetrahedron [퐴퐵퐶퐷].

Solution to Problem 156.

Let 푀 ∈ [퐴퐵퐶퐷] ⟹ (∃) 푃 ∈ [퐴퐵퐶] such that 푀 ∈ [퐷푃]. Let 푁 ∈ [퐴퐵퐶퐷] ⟹ (∃) 푄 ∈ [퐴퐵퐶] such that 푁 ∈ [퐷푄]. The concurrent lines 퐷푀 and 퐷푁 determine angle 퐷푀푁. The surface of triangle 퐷푃푄 is a convex set.

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255 Compiled and Solved Problems in Geometry and Trigonometry

Let

such that 푂 ∈ [퐷푅]. But [푃푄] ⊂ [퐴퐵퐶] because 푃 ∈ [퐴퐵퐶] ∩ 푄 ∈ [퐴퐵퐶] and the surface of the triangle is convex. So (∃)푅 ∈ [퐴퐵퐶] such that

and the tetrahedron is a convex set.

Note: The tetrahedron can be regarded as the intersection of four closed half- spaces which are convex sets.

Solution to Problem 157.

Let ℳ be the union of the segments [푃푄] with 푃 ∈ ℳ1 and 푄 ∈ ℳ2.

Let 푥,푥’∈ℳ⟹(∀)푃∈ℳ1 and 푄 ∈ ℳ2 such that 푥 ∈ [푃푄];

(∃)푃′ ∈ ℳ1 and 푄′ ∈ ℳ2 such that 푥′ ∈ [푃′푄′]. ′ ′ From 푃, 푃′ ∈ ℳ1 ⟹ [푃푃 ] ∈ ℳ1 which is a convex set.

From 푄, 푄′ ∈ ℳ2 ⟹ [푄푄′] ∈ ℳ2 which is a convex set. The union of all the segments [푀푁] with 푀 ∈ [푃푃′] and 푁 ∈ [푄푄′] is tetrahedron [푃푃′푄푄′] ⊂ ℳ. So from 푥, 푥′ ∈ ℳ ⟹ |푥푥’| ⊂ ℳ, so set ℳ is convex.

Solution to Problem 158.

The interior of the tetrahedron coincides with the union of segments |푃푄|, 푃 ∈ |퐴퐵| and 푄 ∈ |퐶퐷|, that is int. [퐴퐵퐶퐷] = {|푃푄| ∖ 푃 ∈ |퐴퐵|, 푄 ∈ |퐶퐷|}. Let’s show that:

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1. Let

2. Let

If we assume 푁 ∈ |퐷푀| ⟹ |퐷푀| ∩ (퐴퐵퐶) ≠ ∅ ⟹ 푀 and 퐷 are in different half- spaces in relation to (퐴퐵퐶) ⟹ 푀 ∉ (퐴퐵퐶), 퐷, false (it contradicts the hypothesis). So

and the second inclusion is proved. As regarding the tetrahedron: [퐴퐵퐶퐷] = {[푃푄] ∖ 푃 ∈ [퐴퐵] and 푄 ∈ [퐶퐷]}.

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255 Compiled and Solved Problems in Geometry and Trigonometry

푃 = 퐴, 푄 ∈ [퐶퐷], [푃푄] describes face [퐴퐷퐶] If 푃 = 퐵, 푄 ∈ [퐶퐷], [푃푄] describes face [퐵퐷퐶] 푄 = 퐶, 푃 ∈ [퐴퐵], [푃푄] describes face [퐴퐵퐶]. Because the triangular surfaces are convex sets and along with their two points 푃, 푄, segment [푃푄] is included in the respective surface. So, if we add these two situations to the equality from the previous case, we obtain:

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Lines and Planes

159. Let 푑, 푑′ be two parallel lines. If the line 푑 is parallel to a plane 훼, show that 푑′||훼 or 푑′ ⊂ 훼. Solution to Problem 159

160. Consider a line 푑, parallel to the planes 훼 and 훽, which intersects after the line 푎. Show that 푑‖푎. Solution to Problem 160

161. Through a given line 푑, draw a parallel plane with another given line 푑′. Discuss the number of solutions. Solution to Problem 161

162. Determine the union of the lines intersecting a given line 푑 and parallel to another given line 푑′ (푑 ∦ 푑′). Solution to Problem 162

163. Construct a line that meets two given lines and that is parallel to a third given line. Discuss. Solution to Problem 163

164. If a plane 훼 intersects the secant planes after parallel lines, then 훼 is parallel to line 훽 ∩ 훾. Solution to Problem 164

165. A variable plane cuts two parallel lines in points 푀 and 푁. Find the geometrical locus of the middle of segment [푀푁]. Solution to Problem 165

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255 Compiled and Solved Problems in Geometry and Trigonometry

166. Two lines are given. Through a given point, draw a parallel plane with both lines. Discuss. Solution to Problem 166

167. Construct a line passing through a given point, which is parallel to a given plane and intersects a given line. Discuss. Solution to Problem 167

168. Show that if triangles 퐴퐵퐶 and 퐴′퐵′퐶′, located in different planes, have 퐴퐵 ∥ 퐴′퐵′, 퐴퐶 ∥ 퐴′퐶′ and 퐵퐶 ∥ 퐵′퐶′, then lines 퐴퐴′, 퐵퐵′, 퐶퐶′ are concurrent or parallel. Solution to Problem 168

169. Show that, if two planes are parallel, then a plane intersecting one of them after a line cuts the other one too. Solution to Problem 169

170. Through the parallel lines 푑 and 푑′ we draw the planes 훼 and 훼′ distinct from (푑, 푑′). Show that 훼 ∥ 훼′ or (훼 ∩ 훼′) ∥ 푑. Solution to Problem 170

171. Given a plane 훼, a point 퐴 ∈ 훼 and a line 푑 ⊂ 훼. a. Construct a line 푑′ such that 푑′ ⊂ 훼, 퐴 ∈ 푑′ and 푑′ ∥ 푑. b. Construct a line through 퐴 included in 훼, which forms with 푑 an angle of a given measure 푎. How many solutions are there? Solution to Problem 171

172. Show that relation 훼 ∥ 훽 defined on the set of planes is an equivalence relation. Define the equivalence classes. Solution to Problem 172

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173. Consider on the set of all lines and planes the relation “푥 ∥ 푦” or 푥 = 푦, where 푥 and 푦 are lines or planes. Have we defined an equivalence relation? Solution to Problem 173

174. Show that two parallel segments between parallel planes are concurrent. Solution to Problem 174

175. Show that through two lines that are not contained in the same plane, we can draw parallel planes in a unique way. Study also the situation when the two lines are coplanar. Solution to Problem 175

176. Let 훼 and 훽 be two parallel planes, 퐴, 퐵 ∈ 훼, and 퐶퐷 is a parallel line with 훼 and 훽. Lines 퐶퐴, 퐶퐵, 퐷퐵, 퐷퐴 cut plane 훽 respectively in 푀, 푁, 푃, 푄. Show that these points are the vertices of a parallelogram. Solution to Problem 176

177. Find the locus of the midpoints of the segments that have their extremities in two parallel planes. Solution to Problem 177

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255 Compiled and Solved Problems in Geometry and Trigonometry

Solutions

Solution to Problem 159.

퐴 ∈ 훼 ′′ 푑 ∥ 훼 } ⟹ 푑 ∥ 훼 Let 푑′′ ∥ 푑 ⟹ 푑′′ ∥ 훼 or 푑′ ⊂ 훼. 푑′ ∥ 푑 } ⟹ 푑′ ∥ 푑′′ 푑′′ ∥ 푑 }

Solution to Problem 160.

Let 퐴∈푎⇒퐴∈훼∩퐴∈훽. We draw through A, 푑′ ∥ 푑.

푑 ∥ 훼 퐴 ∈ 훼, } ⟹ 푑′ ⊂ 훼 푑′ ∥ 푑 푑′ ⊂ 훼 ⊂ 훽 푑′ = 푎 } ⟹ } ⟹ } ⟹ 푎 ∥ 푑 푑 ∥ 훽 훼 ∩ 훽 = 푎 푑′ ∥ 푑 퐴 ∈ 훽, } ⟹ 푑′ ⊂ 훽 푑′ ∥ 푑

Solution to Problem 161.

a. If 푑 ∦ 푑′ there is only one solution and it can be obtained as it follows:

Let 퐴 ∈ 푑. In the plane (퐴, 푑′′) we draw 푑′′ ∥ 푑′. The concurrent lines 푑 and 푑′′ determine plane 푎. As 푑′′ ∥ 푑 ⟹ 푑 ∥ 훼, in the case of the non-coplanar lines.

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b. If 푑 ∥ 푑′ || d'', (∃) infinite solutions. Any plane passing through 푑 is parallel to 푑′′, with the exception of plane (푑, 푑′).

c. 푑 ∦ 푑′, but they are coplanar (∄) solutions.

Solution to Problem 162.

Let 퐴 ∈ 푑, we draw through 퐴, 푑1 ∥ 푑′. We write 훼 = (푑, 푑1). As 푑1 ∥ 푑′ ⟹ 푑′ ∥ 훼.

Let 푀 ∈ 푑, arbitrary ⟹ 푀 ∈ 훼.

훿 ∥ 훿′, 푀 ∈ 훿 We draw } ⟹ 훿 ⊂ 훼, so all the parallel lines to 푑′ intersecting 푑 are 푑′ ∥ 푎 contained in plane 훼.

Let 훾 ⊂ 훼, 훾 ∥ 푑′ ⟹ 훾 ∩ 푑 = 퐵, so (∀) parallel to 푑′ from 훼 intersects 푑. Thus, the plane 훼 represents the required union.

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Solution to Problem 163.

푑 ∥ 푑′ We draw 푑 through 푀 such that } ⟹ 푑 ∥ 푑3. According to previous 푑′ ∥ 푑3

problem: 푑 ∩ 푑1 = {푁}. Therefore,

a. If 푑3 ∥ 푑1, the plane 훼 is unique, and if 푑2 ∩ 훼 ≠ ∅, the solution is unique.

푑 ∥ 푑1 b. If 푑1 ∥ 푑3, (∄) , because it would mean that we can draw through a 푑 ∩ 푑1 ≠ ∅

point two parallel lines 푑, 푑1 to the same line 푑3. So there is no solution.

c. If 푑1 ∦ 푑3 and 푑2 ∩ 훼 ≠ ∅, all the parallel lines to 푑2 cutting 푑1 are on the plane 훼

and none of them can intersect 푑2, so the problem has no solution.

d. If 푑2 ⊂ 훼, 푑1 ∩ 푑2 ≠ ∅, let 푑1 ∩ 푑2 = {푂}, and the required line is parallel to 푑3 drawn through 푂 ⟹ one solution.

e. If 푑2 ⊂ 훼, 푑1 ∥ 푑2. The problem has infinite solutions, (∀) || to 푑3 which cuts 푑1,

also cuts 푑2.

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Solution to Problem 164.

Solution to Problem 165.

The problem is reduced to the geometrical locus of the midpoints of the segments that have extremities on two parallel lines. 푃 is such a point |푀푃| = |푃푁|. We draw 퐴퐵 ⊥ 푑1 ⇒ 퐴퐵?푀푃퐴̂ = 퐵푃푁̂ ⟹ 훥푀퐴푃 = 훥푁퐵푃 ⟹ |푃퐴| ≡ |푃퐵| ⟹ ||퐴푃||? ⟹

the geometrical locus is the parallel to 푑1 and 푑2 drawn on the mid-distance between them. It can also be proved vice-versa.

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255 Compiled and Solved Problems in Geometry and Trigonometry

Solution to Problem 166.

′ ′ Let 푑1 ∦ 푑2. In plane (푑, 푀) we draw 푑1 ∥ 푑1, 푀 ∈ 푑1. In plane (푑2푀) we draw ′ ′ ′ ′ 푑2 ∥ 푑2, 푀 ∈ 푑2. We note 훼 = 푑1푑2 the plane determined by two concurrent lines.

푀 ∈ 훼 the only solution.

Let 푑1 ∥ 푑2, 푁 ∉ 푑1, 푀 ∉ 푑2. ′ ′ 푑1 = 푑1 = 푑2 ′ ′ In this case 푑1 = 푑2 = 푑 and infinite planes pass through 푑; 푑2 ∥ 푑 } ⟹ 푑1, 푑2 are parallel lines with (∀) of the planes passing through 푑. 푑1 ∥ 푑

The problem has infinite solutions. But 푀 ∈ 푑1 or 푀 ∈ 푑2, the problem has no solution because the plane can’t pass through a point of a line and be parallel to that line.

Solution to Problem 167.

Let 퐴 be the given point, 훼 the given plane and 푑 the given line. a. We assume that 푑 ∦ 훼, 푑 ∩ 훼 = {푀}. Let plane (푑퐴) which has a common point 푀 with 훼 ⇒ (푑퐴) ∩ 훼 = 푑′.

We draw in plane (푑퐴) through point 퐴 a parallel line to 푑′.

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⟹ 푎 is the required line. b. 푑 ∥ 훼, (푑퐴) ∩ 훼 ≠ ∅.

Let (dA) ∩ α = α’} ⇒ 푑′ ∥ 푑 푑 ∥ 훼

All the lines passing through 퐴 and intersecting 푑 are contained in plane (푑퐴). But all these lines also cut 푑′ ∥ 푑, so they can’t be parallel to 훼. There is no solution. c. 푑 ∥ 훼, (푑퐴) ∩ 훼 = ∅.

Let 푀 ∈ 푑 and line 퐴푀 ⊂ (푑퐴); (푑퐴) ∩ 훼 = ∅ ⟹ 퐴푀 ∩ 훼 = ∅ ⟹ 퐴푀 ∥ 훼, (∀)푀 ∈ 푑. The problem has infinite solutions.

Solution to Problem 168.

(퐴퐵퐶) and (퐴′퐵′퐶′) are distinct planes, thus the six points 퐴,퐵,퐶,퐴′,퐵′,퐶′ can’t be coplanar. 퐴퐵 ∥ 퐴′퐵’ ⟹ 퐴, 퐵, 퐴′, 퐵′ are coplanar.

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255 Compiled and Solved Problems in Geometry and Trigonometry

The points are coplanar four by four, that is (퐴퐵퐵′퐴), (퐴퐶퐶′퐴′), (퐵퐶퐶′퐵′), and determine four distinct planes. If we assumed that the planes coincide two by two, it would result other 6 coplanar points and this is false. In plane 퐴퐵퐵′퐴′, lines 퐴퐴′, 퐵퐵′ can be parallel or concurrent. First we assume that:

is a common point to the 3 distinct planes, but the intersection of 3 distinct planes can be only a point, a line or ∅. It can’t be a line because lines

⟹ are distinct if we assumed that two of them coincide, the 6 points would be coplanar, thus there is no common line to all the three planes. There is one possibility left, that is they have a common point 푆 and from

We assume 푑 ∩ 훽 = ∅ ⟹ 푑 ∥ 훽 ⟹ 푑 ∈ plane ∥ 훽 drawn through 퐴 ⟹ 푑 ⊂ 훼, false. So 푑 ∩ 훽 = {퐵}.

Solution to Problem 169.

Hypothesis: 훼 ∥ 훽, 훾 ∩ 훼 = 푑1.

Conclusion: 훾 ∩ 훽 = 푑2. We assume that 훾 ∩ 훽 = ∅ ⟹ 훾 ∥ 훽. Let

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because from a point we can draw only one parallel plane with the given plane.

But this result is false, it contradicts the hypothesis 훾 ∩ 훼 = 푑1 so 훾 ∩ 훽 = 푑2.

Solution to Problem 170.

Hypothesis: 푑 ∥ 푑′; 푑 ⊂ 훼; 푑′ ⊂ 훼′; 훼, 푎′ ≠ (푑푑′). Conclusion: 훼 ∥ 훼′ or 푑′′ ∥ 푑. As 훼, 푎′ ≠ (푑푑′) ⟹ 훼 ≠ 푎′. If 푎 ∩ 푎′ = ∅ ⟹ 푎 ∥ 푎′. If 푎 ∩ 푎′ = ∅ ⟹ 푎 ∩ 푎′ = 푑′′. 푑 ∥ 푑′′ 푑 ∥ 푎′ If ⟹ } ⟹ 푑′′‖푑. 푑′ ⊂ 푎′ 푑 ⊂ 푎

Solution to Problem 171.

a. If 퐴 ∈ 푑, then 푑′ = 푑. If 퐴 ∉ 푑, we draw through 퐴, 푑′ ∥ 푑.

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255 Compiled and Solved Problems in Geometry and Trigonometry

′ b. We draw 푑1 ⊂ 훼, 퐴 ∈ 푑, such that 푚(푑̂1푑 ) = 푎 and 푑 ⊂ 훼, 퐴 ∈ 푑2, such that ′ ′ 푚(푑̂2푑 ) = 푎, a line in each half-plane determined by 푑 . So (∃) 2 solutions excepting the situation 푎 = 0 or 푎 = 90 when (∃) only one solution.

Solution to Problem 172.

훼 ∥ 훽 or 훼 = 훽 ⟺ 훼~훽

1. 훼 = 훼 ⟹ 훼~훼, the relation is reflexive; 2. 훼~훽 ⟹ 훽~ ∝, the relation is symmetric. 훼 ∥ 훽 or 훼 = 훽 ⟹ 훽 ∥ 훼 or 훽 = 훼 ⟹ 훽~ ∝; 3. 훼~훽∩훽~훾 ⟹ 훼~훾.

If 훼=훽∩훽~훾⟹훼~훾.

훼 ≠ 훽 and 훼~훽 ⟹ 훼 ∥ 훽 If } ⟹ 훼 ∥ 훾 ⟹ 훼~훾. 훽~훾 ⟹ 훽 = 훾 or 훽 ∥ 훾

The equivalence class determined by plane 훼 is constructed of planes 훼′ with 훼′~훼, that is of 훼 and all the parallel planes with 훼.

Solution to Problem 173.

No, it is an equivalence relation, because the transitive property is not true. For example, 푥 is a line, 푦 a plane, 푧 a line. From 푥 || 푦 and || 푧 ⇏ 푥 || 푧, lines 푥 and 푧 could be coplanar and concurrent or non-coplanar.

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Solution to Problem 174.

⟹ 퐴퐵퐶퐷 parallelogram. So ||퐴퐶|| = ||퐵퐷||.

Solution to Problem 175.

′ ′ We consider 퐴 ∈ 푑 and draw through it 푑1 ∥ 푑 . We consider 퐵 ∈ 푑 and draw

푑2 ∥ 푑. Plane (푑1푑2) ∥ (푑푑1), because two concurrent lines from the first plane are parallel with two concurrent lines from the second plane.

When 푑 and 푑′ are coplanar, the four lines 푑, 푑1, 푑2 and 푑′ are coplanar and the two planes coincide with the plane of the lines 푑 and 푑′.

Solution to Problem 176.

Let planes:

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255 Compiled and Solved Problems in Geometry and Trigonometry

From (1), (2), (3), (4) ⟹ 푀푁푃푄 parallelogram.

Solution to Problem 177.

Let [퐴퐵] and [퐶퐷] be two segments, with 퐴, 퐶 ∈ 훼 and 퐵, 퐷 ∈ 훽 such that |퐴푀| = |푀퐵| and |퐶푁| = |푁퐷|.

In plane (푀퐶퐷) we draw through 푀, 퐸퐹||퐶퐷 ⟹ 퐸퐶||퐷퐹 ⟹ 퐸퐹퐷퐶 parallelogram ⟹ |퐸퐹| ≡ |퐶퐷|. Concurrent lines 퐴퐵 and 퐸퐹 determine a plane which cuts planes 훼 after 2 parallel lines ⟹ 퐸퐴||퐵퐹. In this plane, |퐴푀| ≡ |퐵푀|. 퐸푀퐴̂ ≡ 퐵푀퐹 (angles ̂opposed at peak) } 퐸퐴푀̂ ≡ 퐹퐵푀(alternatê interior angles)

In parallelogram 퐸퐶퐷퐹,

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So the segment connecting the midpoints of two of the segments with the extremity in 훼 and 훽 is parallel to these planes. We also consider [퐺퐻] with 퐺 ∈ 훼, 퐻 ∈ 훽 and |퐺푄| ≡ |푄퐻| and we show in the same way that 푂푀||훼 and 푂푀||훽. (2) From (1) and (2) ⟹ 푀, 푁, 푄 are elements of a parallel plane to 훼 and 훽, marked by 훾. Vice-versa, let’s show that any point from this plane is the midpoint of a segment, with its extremities in 훼 and 훽. Let segment [퐴퐵] with 퐴 ∈ 훼 and 퐵 ∈ 훽 and |퐴푀| = |퐵푀|. Through 푀, we draw the parallel plane with 훼 and 훽 and in this plane we consider an arbitrary point 푂 ∈ 훾. Through 푂 we draw a line such that 푑 ∩ 훼 = {퐼} and 푑 ∩ 훽 = {퐼}. In plane (푂퐴퐵) we draw 퐴′퐵′ || 퐴퐵. Plane (퐴퐴′퐵′퐵) cuts the three parallel planes after parallel lines ⟹

|퐴′푂| ≡ |푂퐵′| In plane (퐴′퐵′퐼) ⟹ |퐴′푂| ≡ |푂퐵′| ⟹ 퐼퐴′ || 퐵′퐼 and thus and 퐼퐴̂′푂 ≡ 퐼푂퐴′̂ ≡ 퐼푂퐵′̂ 퐼퐵̂′푂 ⟹

푂 is the midpoint of a segment with extremities in planes 훼 and 훽. Thus the geometrical locus is plane 훾, parallel to 훼 and 훽 and passing through the mid-distance between 훼 and 훽.

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255 Compiled and Solved Problems in Geometry and Trigonometry

Projections

′ ′ ′ 178. Show that if lines 푑 and 푑 are parallel, then pr훼푑 ∥ pr훼푑 or pr훼푑 = pr훼푑 . What can we say about the projective planes of 푑 and 푑′? Solution to Problem 178

179. Show that the projection of a parallelogram on a plane is a parallelogram or a segment. Solution to Problem 179

180. Knowing that side [푂퐴 of the right angle 퐴푂퐵 is parallel to a plane 훼, show that the projection of 퐴푂퐵̂ onto the plane 훼 is a right angle. Solution to Problem 180

181. Let 퐴′퐵′퐶′ be the projection of ∆퐴퐵퐶 onto a plane 훼. Show that the centroid of ∆퐴퐵퐶 is projected onto the centroid of ∆퐴’퐵’퐶’. Is an analogous result true for the orthocenter? Solution to Problem 181

182. Given the non-coplanar points 퐴, 퐵, 퐶, 퐷, determine a plane on which the points 퐴, 퐵, 퐶, 퐷 are projected onto the peaks of parallelogram. Solution to Problem 182

183. Consider all triangles in space that are projected onto a plane 훼 after the same triangle. Find the locus of the centroid. Solution to Problem 183

184. Let 퐴 be a point that is not on line 푑. Determine a plane 훼 such that pr훼푑

passes through pr훼퐴. Solution to Problem 184

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185. Determine a plane onto which three given lines to be projected after concurrent lines. Solution to Problem 185

186. Let 훼, 훽 be planes that cut each other after a line 푎 and let 푑 be a perpendicular line to 푎. Show that the projections of line 푑 onto 훼, 훽 are concurrent. Solution to Problem 186

187. Consider lines 푂퐴, 푂퐵, 푂퐶 ⊥ two by two. We know that ||푂퐴|| = 푎, ||푂퐵|| = 푏, ||푂퐶|| = 푐. Find the measure of the angle of planes (퐴퐵퐶) and (푂퐴퐵). Solution to Problem 187

188. A line cuts two perpendicular planes 훼 and 훽 at 퐴 and 퐵. Let 퐴′ and 퐵′ be the projections of points 퐴 and 퐵 onto line 훼 ∩ 훽. a. Show that ||퐴퐵||² = ||퐴퐴′||² + ||퐴′퐵′||² + ||퐵′퐵||²; b. If 푎, 푏, 푐 are the measures of the angles of line 퐴퐵 with planes 훼, 훽 and ‖퐴′퐵′‖ with 훼 ∩ 훽, then cos 푐 and sin2 푎 + sin2 푏 = sin2 푐. ‖퐴퐵‖ Solution to Problem 188

189. Let 퐴퐵퐶 be a triangle located in a plane 훼, 퐴′퐵′퐶′ the projection of ∆퐴′퐵′퐶′ onto plane 훼. We mark with 푆, 푆′, 푆′′ the areas of ∆퐴퐵퐶, ∆퐴′퐵′퐶′, ∆퐴ʺ퐵ʺ퐶ʺ, show that 푆′ is proportional mean between 푆 and 푆′′. Solution to Problem 189

190. A trihedral [퐴퐵퐶퐷] has |퐴퐶| ≡ |퐴퐷| ≡ |퐵퐶| ≡ |퐵퐷|. 푀, 푁 are the midpoints of edges [퐴퐵],[퐶퐷], show that: a. 푀푁 ⊥ 퐴퐵, 푀푁 ⊥ 퐶퐷, 퐴퐵 ⊥ 퐶퐷

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255 Compiled and Solved Problems in Geometry and Trigonometry

b. If 퐴′,퐵′,퐶′,퐷′ are the feet of the perpendicular lines drawn to the peaks 퐴, 퐵, 퐶, 퐷 on the opposite faces of the tetrahedron, points 퐵, 퐴′, 푁 are collinear and so are 퐴, 퐵′, 푁; 퐷,퐶′,푀; 퐶,퐷′,푀. c. 퐴퐴′, 퐵퐵′, 푀푁 and 퐶퐶′, 퐷퐷′, 푀푁 are groups of three concurrent lines. Solution to Problem 190

191. If rays [푂퐴 and [푂퐵 with their origin in plane 훼, 푂퐴 ⊥ 훼, then the two rays form an acute or an obtuse angle, depending if they are or are not on the same side of plane 훼. Solution to Problem 191

192. Show that the 6 mediator planes of the edges of a tetrahedron have a common point. Through this point pass the perpendicular lines to the faces of the tetrahedron, drawn through the centers of the circles of these faces. Solution to Problem 192

193. Let 푑 and 푑′ be two non-coplanar lines. Show that (∃) unique points 퐴 ∈ 푑, 퐴′ ∈ 푑′ such that 퐴퐴′ ⊥ 푑 and 퐴퐴′ ⊥ 푑′. The line 퐴퐴′ is called the common perpendicular of lines 푑 and 푑′. Solution to Problem 193

194. Consider the notations from the previous problem. Let 푀 ∈ 푑, 푀′ ∈ 푑′. Show that ‖퐴퐴′‖ ≤ ‖푀푀′‖. The equality is possible only if 푀 = 퐴, 푀′ = 퐴′. Solution to Problem 194

195. Let 퐴퐴′ be the common ⊥ of non-coplanar lines 푑, 푑′′ and 푀 ∈ 푑, 푀′ ∈ 푑′ such that |퐴푀| ≡ |퐴′푀′|. Find the locus of the midpoint of segment [푀푀′]. Solution to Problem 195

196. Consider a tetrahedron 푉퐴퐵퐶 with the following properties. 퐴퐵퐶 is an equilateral triangle of side 푎, (퐴퐵퐶) ⊥ (푉퐵퐶), the planes (푉퐴퐶) and (푉퐴퐵)

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form with plane (퐴퐵퐶) angles of 60°. Find the distance from point 푉 to plane (퐴퐵퐶). Solution to Problem 196

197. All the edges of a trihedral are of length a. Show that a peak is projected onto the opposite face in its centroid. Find the measure of the dihedral angles determined by two faces. Solution to Problem 197

198. Let 퐷퐸 be a perpendicular line to the plane of the square 퐴퐵퐶퐷. Knowing that ‖퐵퐸‖ = 푙 and that the measure of the angle formed by [퐵퐸 and (퐴퐵퐶) is 훽, determine the length of segment 퐴퐸 and the angle of [퐴퐸 with plane (퐴퐵퐶). Solution to Problem 198

199. Line 퐶퐷 ⊥ plane of the equilateral ∆퐴퐵퐶 of side 푎, and [퐴퐷 and [퐵퐷 form with plane (퐴퐵퐶) angles of measure 훽. Find the angle of planes (퐴퐵퐶) and 퐴퐵퐷. Solution to Problem 199

200. Given plane 훼 and ∆퐴퐵퐶, ∆퐴’퐵’퐶’ that are not on this plane. Determine a ∆퐷퐸퐹, located on 훼 such that on one side lines 퐴퐷, 퐵퐸, 퐶퐹 and on the other side lines 퐴′퐷,퐵′퐸,퐶′퐹 are concurrent.

Solution to Problem 200

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255 Compiled and Solved Problems in Geometry and Trigonometry

Solutions

Solution to Problem 178.

Let 푑 ∥ 푑′, 훽 the projective plane of 푑.

We assume that 푑′ ⊄ 훽, which means that is plane 푑, 푑′ ⊥ 훼, ⟹ the projective ′ ′ ′ plane of 푑 is 훽 . We want to show that pr푎푑 ∥ pr푎푑 . We assume that pr푎푑 ∩ ′ ′ ′ ′ pr푎푑 = {푃} ⇒ (∃) 푀 ∈ 푑 such that pr푎푀 = 푃 and (∃)푀 ∈ 푑 such that pr푎푀 = 푃. 푃푀 ⊥ 훼 ⟹ } ⟹ in the point 푃 on plane 훼 we can draw two distinct perpendicular 푃푀′ ⊥ 훼 lines. False.

If 훽 is the projective plane of 푑 and 훽 of 푑′, then 훽 ∥ 훽′, because if they had a ′ common point their projections should be elements of pr푎푑 and pr푎푑 , and thus they wouldn’t be anymore parallel lines.

If 푑′ ⊂ 훽 or 푑 ⊂ 훽′, that is (푑, 푑′) ⊥ 훼 ⟹ 푑 and 푑′ have the same projective plane ′ ⟹ pr푎푑 = pr푎푑 .

Solution to Problem 179.

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We assume that 퐴퐵퐶퐷 ⊥ 훼. Let 퐴′,퐵′,퐶′,퐷′ be the projections of points 퐴, 퐵, 퐶, 퐷.

푝푟1 ′ ′ ′ ′ 퐴퐵 ∥ 퐷퐶 ⇒ 퐴 퐵 ∥ 퐷 퐶 } ⟹ 퐴′,퐵′,퐶′,퐷′ parallelogram. 퐴퐷 ∥ 퐵퐶 ⟹ 퐴′퐷′ ∥ 퐵′퐷′

If (퐴퐵퐶퐷) ⊥ 훼 ⇒ the projection 퐴′, 퐵′, 퐶′, 퐷′ ∈ the line (퐴퐵퐶퐷) ∩ 훼 ⟹ the projection of the parallelogram is a segment.

Solution to Problem 180.

If 푂퐴||훼 ⟹ proj훼푂퐴||푂퐴 ⟹ 푂′퐴′||푂퐴 because (∀) a plane which passes through 푂퐴 cuts the plane 훼 after a parallel to 푂퐴.

⟹ 푂′퐴′ ⊥ 푂′퐵′ ⟹ 퐴′푂′퐵′̂ is a right angle.

Solution to Problem 181.

In the trapezoid 퐵퐶퐶′퐵′ (퐵퐵′||퐶퐶′),

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255 Compiled and Solved Problems in Geometry and Trigonometry

⟹ 퐴′푀′ is a median.

푀푀′ ∥ 퐴퐴′ ⟹ 푀푀′퐴′퐴 trapezoid ‖퐴′퐺‖ ‖퐴퐺‖ ‖퐴퐺‖ } = = 2 = 2, 퐺퐺′ ∥ 퐴퐴′ ‖퐺′푀′‖ ‖퐺푀‖ ‖퐺푀‖

⟹ 퐺′ is on median A'M' at 2/3 from the peak and 1/3 from the base.

Generally no, because the right angle 퐴푀퐶 should be projected after a right angle. The same thing is true for another height. This is achieved if the

sides of the ∆ are parallel to the plane.

Solution to Problem 182.

Let 퐴, 퐵, 퐶, 퐷 be the 4 non-coplanar points and 푀, 푁 midpoints of segments |퐴퐵| and |퐶퐷|.

푀 and 푁 determine a line and let a plane 훼 ⊥ 푀푁, 푀 and 푁 are projected in the same point 푂 onto 훼.

⟹ 퐴′퐵′퐶′퐷′ a parallelogram.

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Solution to Problem 183.

Let 퐴′퐵′퐶′, 퐴′′퐵′′퐶′′ two triangles of this type, with the following property:

⟹ 퐺′퐺 = 퐺퐺′′ ⟹ 퐺′′, 퐺′, 퐺 are collinear. Due to the fact that by projection the ratio is maintained, we show that 퐺′′ is the centroid of 퐴, 퐵, 퐶.

Solution to Problem 184.

Let 푀 ∈ 푑 and 퐴 ∉ 푑. The two points determine a line and let 훼 be a perpendicular plane to this line, 퐴푀 ⊥ 훼 ⟹ 퐴 and 푀 are projected onto 훼 in the

same point 퐴′ through which also passes projα푑 = projα퐴 ∈ projα푑.

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255 Compiled and Solved Problems in Geometry and Trigonometry

Solution to Problem 185.

We determine a line which meets the three lines in the following way. Let

Let now a plane

Solution to Problem 186.

Let 훼 ∩ 훽 = 푎 and 푀 ∈ 푑. We project this point onto 훼 and 훽:

⟹ 푎 ⊥ onto the projective plane of 푑 onto 훽.

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Let

⟹ 푂푀′ ∩ 푂푀′′ = {푂}, so the two projections are concurrent.

Solution to Problem 187.

the angle of planes (퐴퐵퐶) and (푂퐴퐵) is 푂푀퐶̂ = 훼.

Solution to Problem 188. Let 훼 ∩ 훽 = a and 퐴퐴′ ⊥ 푎, 퐵퐵′ ⊥ 푎.

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255 Compiled and Solved Problems in Geometry and Trigonometry

⟹ ∢ of line 퐴퐵 with 훼, 퐵퐴̂퐵′ = 푎

⟹ ∢ of line 퐴퐵 with 훽, 퐴퐵퐴′ = 푏 In the plane 훽 we draw through 퐵 a parallel line to 푎 and through 퐴′ a parallel line to 퐵퐵′. Their intersection is 퐶, and ||퐴′퐵′|| = ||퐵퐶||, ||퐵퐵′|| = ||퐴′퐶||. The angle of line 퐴퐵 with 훼 is 퐴퐵퐶̂ = 푐. As 퐴퐴’ ⊥ 훽 ⟹ 퐴퐴′ ⊥ 퐴′퐶 ⟹ ‖퐴퐶‖2 = ‖퐴퐴′‖2 + ‖퐴′퐶‖2 = ‖퐴퐴′‖2 + ‖퐵′퐵‖2(1) 퐵′퐵퐶퐴 rectangle

⟹ ∆퐴퐶퐵 is right in 퐶. We divide the relation (1) with ||퐴퐵||²:

Solution to Problem 189.

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Solution to Problem 190.

|퐴퐶| ≡ |퐵퐶| ⟹ ∆퐴퐶퐵 isosceles a. } ⟹ 퐶푀 ⊥ 퐴퐵 (1) 퐶푀 median |퐴퐷| ≡ |퐵퐶| ⟹ ∆퐴퐵퐷 isosceles } ⟹ 퐷푀 ⊥ 퐴퐵 (2) 퐷푀 median 퐴퐵 ⊥ 푀푁 From (1) and (2) ⟹ 퐴퐵 ⊥ (퐷푀퐶) = } 퐴퐵 ⊥ 퐷퐶 |퐵퐶| ≡ |퐵퐷| } ⟹ 퐵푁 ⊥ 퐷퐶 퐵푁 median } ⟹ 퐷퐶 ⊥ (퐴퐵푁) ⟹ 퐷퐶 ⊥ 푀푁 |퐴퐷| ≡ |퐴퐶| } ⟹ 퐴푁 ⊥ 퐷퐶 퐴푁 median

⟹ 퐴′ ∈ 퐵푁 ⟹ 퐵, 퐴′, 푁 are collinear. In the same way: From (퐴퐷퐶) ⊥ (퐴퐵푁) ⟹ 퐴, 퐵′, 푁 collinear (퐴퐵퐶) ⊥ (퐷푀퐶) ⟹ 푀, 퐷′, 퐶 collinear (퐴퐵퐷) ⊥ (퐷푀퐶) ⟹ 퐷, 퐶′, 푀 collinear C. At point a. we’ve shown that 푀푁 ⊥ 퐴퐵

퐴퐴′, 퐵퐵′ and 푀푁 are heights in ∆퐴퐵푁, so they are concurrent lines. In the same way, 퐷퐷′, 퐶퐶′, 푀푁 will be heights in ∆퐷푀퐶.

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255 Compiled and Solved Problems in Geometry and Trigonometry

Solution to Problem 191.

We assume that [푂퐴, [푂퐵 are on the same side of plane 훼. We draw

(∃) plane (퐴퐷, 퐵퐵′) = 훽 ⟹ |푂퐴, |푂퐵 are in the same half-plane.

In plane 훽 we have 푚(퐴푂퐵̂) = 900 = 푚(퐵푂퐵̂′) < 900 ⟹ 퐴푂퐵̂ acute. We assume that [푂퐴 and [푂퐵 are in different half-planes in relation to 훼 ⟹ 퐴 and 퐵 are in different half-planes in relation to 푂퐵′ in plane 훽 ⟹ |푂퐵’ ⊂ int. 퐵푂퐴̂ ⟹ 푚(퐴푂퐵̂) = 900 + 푚(퐵푂퐵̂’) > 900 ⟹ 퐴푂퐵 obtuse.

Solution to Problem 192.

We know the locus of the points in space equally distant from the peaks of ∆퐵퐶퐷 is the perpendicular line 푑 to the pl. ∆ in the center of the circumscribed circle of this ∆, marked with 푂. We draw the mediator plane of side |퐴퐶|, which intersects this ⊥ 푑 at point 푂. Then, point 푂 is equally distant from all the peaks of the tetrahedron ||푂퐴|| = ||푂퐵|| = ||푂퐶|| = ||푂퐷||. We connect 푂 with midpoint 퐸 of side |퐴퐵|. From |푂퐴| ≡ |푂퐵| ⟹ ∆푂퐴퐵 isosceles ⟹ 푂퐶 ⊥ 퐴퐵 (1).

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We project 푂 onto plane (퐴퐵퐷) in point 푂2. |푂퐴| ≡ |푂퐵| ≡ |푂퐷| As } ⟹ ∆푂퐴푂2 = ∆푂퐵푂2 = ∆푂퐷푂2 |푂)푂2 common side ⟹ |푂2퐴| ≡ |퐵푂2| ≡ |퐷푂2| ⟹

⟹ 푂2 is the center of the circumscribed circle of ∆퐴퐵퐷. We show in the same way that 푂 is also projected on the other faces onto the centers of the circumscribed circles, thus through 푂 pass all the perpendicular lines to the faces of the tetrahedron. These lines are drawn through the centers of the circumscribed circles. So b. is proved.

From |푂2퐴| ≡ |푂퐵2| ⟹ ∆푂2퐴퐵 isosceles 푂2퐸 ⊥ 퐴퐵 (2) 퐴퐵 ⊥ (퐸푂 푂) From (1) and (2) ⟹ 2 } ⟹ |퐴퐸| ≡ |퐸퐵|

⟹ (퐸푂2푂) is a mediator plane of side |퐴퐵| and passes through 푂 and the intersection of the 3 mediator planes of sides |퐵퐶|, |퐶퐷|, |퐵퐷| belongs to line 푑, thus O is the common point for the 6 mediator planes of the edges of a tetrahedron.

Solution to Problem 193.

Let 푀 ∈ 푑 and 훿||푑′, 푀 ∈ 훿. Let = (푑, 훿) ⟹ 푑′||훼. Let

= {퐴} otherwise 푑 and 푑′ would be parallel, thus coplanar. Let 훽 be the projective plane of line

In plane 훽 we construct a perpendicular to 푑′′ in point 퐴 and

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255 Compiled and Solved Problems in Geometry and Trigonometry

Solution to Problem 194.

We draw

We can obtain the equality only when 푀 = 퐴 and 푀′ = 퐴′.

Solution to Problem 195.

′′ Let 푀 ∈ 푑, 푀′ ∈ 푑′ such that |퐴푀| ≡ |퐴′푀′|. Let 푑 = pr훼푑′ and 푀′푀′′ ⊥ 푑′′ ⟹ 푀′푀′′ ⊥ 훼 ⟹ 푀′푀′′ ⊥ 푀′′푀.

⟹ ∆퐴푀푀′ isosceles.

Let 푃 be the midpoint of |푀푀′| and 푃′ = pr훼푃 ⟹ 푃푃′ ∥ 푀′푀′′ ⟹ 푃′ is the midpoint of 푀푀′′, ∆퐴푀푀′′ isosceles ⟹ [퐴푃′ the bisector of 푀′̂퐴푀. (푃푃′) is midline in 1 1 ∆푀′푀ʺ푀 ⟹ ‖푃푃′‖ = ‖푀′푀′′ = ‖퐴′퐴‖ = constant. 2 2 Thus, the point is at a constant distance from line 퐴푃′, thus on a parallel line to this line, located in the ⊥ plane 훼, which passes through 퐴푃′. When 푀 = 퐴 and 푀′ = 퐴′ ⟹ ||퐴푀|| = ||푁′퐴′ = 0 ⟹ 푃 = 푅, where 푅 is the midpoint of segment |퐴퐴′|. So the locus passes through 푅 and because

⟹ 푅푃 is contained in the mediator plane of segment |퐴퐴′|. So 푅푃 is the intersection of the mediator plane of segment |퐴퐴′| with the ⊥ plane to 훼, passing through one of the bisectors of the angles determined by 푑 and 푑′, we obtain one more line contained by the mediator plane of [퐴퐴′], the parallel line with the other bisector of the angles determined by 푑 and 푑′′. So the locus will be formed by two perpendicular lines.

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Vice-versa, let 푄 ∈ 푅푃 a (∀) point on this line and 푄’ = pr훼푄 ⟹ 푄′ ∈ |퐴푃′ bisector. We draw 푁푁′′ ⊥ 퐴푄′ and because 퐴푄′ is both bisector and height ⟹ ∆퐴푁푁′′ isosceles ⟹ |퐴푄′| median ⟹ |푁푄′| ≡ |푄′푁′′|. We draw

coplanar

⟹ |푄′푄| midline in ∆푁푁′푁′′ ⟹ 푄, 푁′, 푁 collinear and |푄푁′| ≡ |푄푁|.

Solution to Problem 196.

where α = (ABC).

VD common side

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255 Compiled and Solved Problems in Geometry and Trigonometry

Solution to Problem 197.

Let

||VO|| common

⟹ 푂 is the center of the circumscribed circle and as ∆퐴퐵퐶 is equilateral ⟹ 푂 is the centroid ⟹

Solution to Problem 198.

In ∆퐴퐸퐵, right in 퐴:

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Solution to Problem 199.

퐶퐸 ⊥ 퐵퐴 } ⟹ ∢pl. (퐴퐵퐶) and 퐴퐵퐷 are 푚(퐷퐸퐶̂). 퐷퐸 ⊥ 퐴퐵 푎 3 퐴퐵퐶 equilateral ⟹ ‖퐶퐸‖ = √ . 2

Solution to Problem 200.

We consider the problem solved and we take on plane 훼, ∆퐷퐸퐹, then points 푂 and 푂′ which are not located on 훼. We also construct lines |퐷푂, |퐹푂, |퐸푂 respectively |퐷푂′, |퐹푂′, |퐸푂′. On these rays we take ∆퐴퐵퐶 and ∆퐴′퐵′퐶′. Obviously, the way we have constructed the lines 퐴퐷, 퐵퐸, 퐶퐹 shows that they intersect at 푂. We extend lines 퐵퐴, 퐵퐶, 퐶퐴 until they intersect plane

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255 Compiled and Solved Problems in Geometry and Trigonometry

훼 at points 퐵, 퐶 respectively 퐴. Then, we extend lines 퐶′퐴′,퐶′퐵′,퐴′퐵′ until they intersect plane 훼 at points 퐴2, 퐶2 respectively 퐵2.

Obviously, points 퐴1, 퐵1, 퐶1 are collinear (because ∈ 훼 ∩ (퐴퐵퐶)) and 퐴2, 퐵2, 퐶2 are as well collinear (because ∈ 훼 ∩ (퐴′퐵′퐶′)).

On the other side, points 퐷, 퐹, 퐴1, 퐴2 are collinear because:

thus collinear (1)

⟹ 퐷, 퐹, 퐴2 collinear (2)

From (1) and (2) ⟹ 퐷, 퐹, 퐴1, 퐴2 collinear. Similarly 퐶, 퐸, 퐹, 퐶2 collinear and

퐵1, 퐸, 퐷, 퐷2 collinear.

Consequently, 퐷퐸퐹 is at the intersection of lines 퐴1퐴2, 퐶1퐶2 , 퐵1퐵2 on plane 훼, thus uniquely determined.

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Review Problems

201. Find the position of the third peak of the equilateral triangle, the affixes

of two peaks being 푧1 = 1, 푧2 = 2 + 푖. Solution to Problem 201

202. Let 푧1, 푧2, 푧3 be three complex numbers, not equal to 0, + two by 2, and

of equal moduli. Prove that if 푧1 + 푧2푧3, 푧2 + 푧3푧1, 푧2 + 푧 1푧3 ∈ 푅 ⇒ 푧1푧2푧3 = 1. Solution to Problem 202

203. We mark by 퐺 the set of 푛 roots of the unit, 퐺 = {휀0, 휀1, … , 휀푛−1}. Prove that:

a. 휀푖 ∙ 휀푗 ∈ 퐺, (∀) 푖, 푗 ∈ {0, 1, … , 푛 − 1}; −1 b. 휀푖 ∈ 퐺, (∀ ) 푖 ∈ {0, 1, … , 푛 − 1}. Solution to Problem 203

204. Let the equation 푎푧² + 푏푧² + 푐 = 0, 푎, 푏, 푐 ∈ 퐶 and arg푎 + arg푐 = 2arg푏, and |푎|+|푐| = |푏|. Show that the given equation has at list one root of unity. Solution to Problem 204

205. Let 푧1, 푧2, 푧3 be three complex numbers, not equal to 0, such that |푧1| =

|푧2| = |푧3|.

a. Prove that (∃) complex numbers 훼 and 훽 such that 푧2 = 훼푧1, 푧3 = 훽푧2 and |훼| = |훽| = 1; b. Solve the equation 훼² + 훽²– 훼 ∙ 훽– 훼– 훽 + 1 = 0 in relation to one of the unknowns.

2 2 2 c. Possibly using the results from 푎. and 푏., prove that if 푧1 + 푧2 + 푧3 =

푧1푧2 + 푧2푧3 + 푧1푧3 , then we have 푧1 = 푧2 = 푧3 or the numbers 푧1, 푧2, 푧3 are affixes of the peaks of an equilateral ∆. Solution to Problem 205

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255 Compiled and Solved Problems in Geometry and Trigonometry

206. Draw a plane through two given lines, such that their line of intersection to be contained in a given plane. Solution to Problem 206

207. Let 푎, 푏, 푐 be three lines with a common point and 푃 a point not located on any of them. Show that planes (푃푎), (푃푏), (푃푐) contain a common line. Solution to Problem 207

208. Let 퐴, 퐵, 퐶, 퐷 be points and 훼 a plane separating points 퐴 and 퐵, 퐴 and 퐶, 퐶 and 퐷. Show that 훼 ∩ |퐵퐷| ≠ ∅ and 훼 ∩ |퐴퐷| = ∅. Solution to Problem 208

210. Show that the following sets are convex: the interior of a trihedral angle, a tetrahedron without an edge (without a face). Solution to Problem 210

211. Let 퐴, 퐵, 퐶, 퐷 be four non-coplanar points and 퐸, 퐹, 퐺, 퐻 the midpoints of segments [퐴퐵],[퐵퐶],[퐶퐷], [퐷퐴]. Show that 퐸퐹 || (퐴퐶퐷) and points 퐸, 퐹, 퐺, 퐻 are coplanar. Solution to Problem 211

212. On lines 푑, 푑′ consider distinct points 퐴, 퐵, 퐶; 퐴′, 퐵′, 퐶′ respectively. Show that we can draw through lines 퐴퐴′, 퐵퐵′, 퐶퐶′ three parallel planes if and only ‖퐴퐵‖ ‖퐵퐶‖ if = . ‖퐴′퐵′‖ ‖퐵′퐶′‖ Solution to Problem 212

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213. Let 푀, 푀′ be each mobile points on the non-coplanar lines 푑, 푑′. Find the locus of points 푃 that divide segment |푀푀′| in a given ratio. Solution to Problem 213

214. Construct a line that meets three given lines, respectively in 푀, 푁, 푃 and ‖ ‖ for which 푀푁 to be given ratio. ‖푁푃‖ Solution to Problem 214

215. Find the locus of the peak 푃 of the triangle 푀, 푁, 푃 if its sides remain parallel to three fixed lines, the peak 푀 describes a given line 푑, and the peak 푁 ∈ a given plane 훼. Solution to Problem 215

216. On the edges [푂퐴,[푂퐵,[푂퐶 of a trihedral angle we consider points 푀, 푁, 푃 such that ‖푂푀‖ = 휆‖푂퐴‖, ‖푂푁‖ = 휆‖푂퐵‖, ‖푂푃‖ = 휆‖푂퐶‖, where 휆 is a positive variable number. Show the locus of the centroid of triangle 푀푁푃. Solution to Problem 216

217. 퐴퐵퐶퐷 and 퐴1퐵1퐶1퐷1 are two parallelograms in space. We take the points

퐴2, 퐵2, 퐶2, 퐷2 which divide segments [퐴퐴1], [퐵퐵1], [퐶퐶1], [퐷퐷1] in the same

ratio. Show that 퐴2퐵2퐶2퐷2 is a parallelogram. Solution to Problem 217

218. The lines 푑, 푑′ are given, which cut a given plane 훼 in 퐴 and 퐴′. Construct the points 푀, 푀′ on 푑, 푑′ such that 푀푀′ ∥ 훼 and segment [푀푀′] to have a given length 푙. Discuss. Solution to Problem 218

219. Construct a line which passes through a given point 퐴 and that is perpendicular to two given lines 푑 and 푑′. Solution to Problem 219

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255 Compiled and Solved Problems in Geometry and Trigonometry

220. Show that there exist three lines with a common point, perpendicular two by two. Solution to Problem 220

221. Let 푎 푏, 푐, 푑 four lines with a common point, 푑 is perpendicular to 푎 푏, 푐. Show that lines 푎, 푏, 푐 are coplanar. Solution to Problem 221

222. Show that there do not exist four lines with a common point that are perpendicular two by two. Solution to Problem 222

223. Let 푑 ⊥ 훼 and 푑′ ∥ 푑. Show that 푑′ ⊥ 훼. Solution to Problem 223

224. Show that two distinct perpendicular lines on a plane are parallel. Solution to Problem 224

225. Let 푑 ⊥ 훼 and 푑′[∥ 훼. Show that 푑′ ⊥ 푑. Solution to Problem 225

226. Show that two perpendicular planes on the same line are parallel with each other. Solution to Problem 226

227. Show that the locus of the points equally distant from two distinct points 퐴 and 퐵 is a perpendicular plane to 퐴퐵, passing through midpoint 푂 of the segment [퐴퐵] (called mediator plane of [퐴퐵]). Solution to Problem 227

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228. Find the locus of the points in space equally distant from the peaks of a triangle 퐴퐵퐶. Solution to Problem 228

229. The plane 훼 and the points 퐴 ∈ 훼, 퐵 ∉ 훼 are given. A variable line 푑 passes through 퐴 and it is contained in plane 훼. Find the locus of the ⊥ feet from 퐵 to 푑. Solution to Problem 229

230. A line 훼, and a point 퐴 ∉ 훼 are given. Find the locus of the feet of the perpendicular lines from 퐴 to planes passing through 훼. Solution to Problem 230

231. Consider a plane 훼 that passes through the midpoint of segment [퐴퐵]. Show that points 퐴 and 퐵 are equally distant from plane 훼. Solution to Problem 231

232. Through a given point, draw a line that intersects a given line and is ⊥ to another given line. Solution to Problem 232

233. Let 훼 and 훽 be two distinct planes and the line 푑 their intersection. Let 푀

be a point that is not located on 훼 ∪ 훽. We draw the lines 푀푀1 and 푀푀2 ⊥

on 훼 and 훽. Show that the line 푑 is ⊥ to (푀푀1푀푀2). Solution to Problem 233

234. A plane 훼 and a point 퐴, 퐴 ∉ 훼 are given. Find the locus of points 푀 ∈ 훼 such that segment |퐴푀| has a given length. Solution to Problem 234

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255 Compiled and Solved Problems in Geometry and Trigonometry

235. Let 푂, 퐴, 퐵, 퐶 be four points such that 푂퐴 ⊥ 푂퐵 ⊥ 푂퐶 ⊥ 퐷퐴 and we write 푎 = ‖푂퐴‖, 푏 = ‖푂퐵‖, 푐 = ‖푂퐶‖. a. Find the length of the sides of ∆퐴퐵퐶 in relation to 푎, 푏, 푐; b. Find 휎[퐴퐵퐶] and demonstrate the relation 휎[퐴퐵퐶]2 = 휎[퐷퐴퐵]2 + 휎[푂퐵퐶]2 + 휎[푂퐶퐴]2; c. Show that the orthogonal projection of point 푂 on plane (퐴퐵퐶) is the orthocenter 퐻 of ∆퐴퐵퐶; d. Find the distance ‖푂퐻‖. Solution to Problem 235

236. Consider non-coplanar points 퐴, 퐵, 퐶, 퐷 and lines 퐴퐴′, 퐵퐵′, 퐶퐶′, 퐷퐷′ perpendicular to (퐵퐶퐷), (퐴퐶퐷), (퐴퐵퐷). Show that if lines 퐴퐴′ and 퐵퐵′ are concurrent, then lines 퐶퐶′, 퐷퐷′ are coplanar. Solution to Problem 236

237. Let 퐴, 퐵, 퐶, 퐷 four non-coplanar points. Show that 퐴퐵 ⊥ 퐶퐷 and 퐴퐶 ⊥ 퐵퐷 ⟹ 퐴퐷 ⊥ 퐵퐶. Solution to Problem 237

238. On the edges of a triangle with its peak 푂, take the points 퐴, 퐵, 퐶 such that |푂퐴| ≡ |푂퐵| ≡ |푂퐶|. Show that the ⊥ foot in 푂 to the plane (퐴퐵퐶) coincides with the point of intersection of the bisectors ∆퐴퐵퐶. Solution to Problem 238

239. Let a peak 퐴 of the isosceles triangle 퐴퐵퐶 (|퐴퐵| ≡ |퐴퐶|) be the

orthogonal projection onto 퐴′ on a plane 훼 which passes through 퐵퐶. Show that 퐵̂퐴′퐶 > 퐵퐴퐶̂. Solution to Problem 239

240. With the notes of Theorem 1, let [퐴퐵′ be the opposite ray to [퐴퐵′′. Show that for any point 푀 ∈ 훼– [퐴퐵′′ we have 퐵̂′′퐴퐵 > 푀퐴퐵̂ . Solution to Problem 240

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241. Let 훼 be a plane, 퐴 ∈ 훼 and 퐵 and 퐶 two points on the same side of 훼 such that 퐴퐶 ⊥ 훼. Show that 퐶퐴퐵̂ is the complement of the angle formed by [퐴퐵 with 훼. Solution to Problem 241

242. Let 훼′훽′ be a trihedral angle with edge 푚 and 퐴 ∈ 푚. Show that of all the rays with origin at 퐴 and contained in half-plane 훽′, the one that forms with plane 훼 the biggest possible angle is that ⊥ 푝 ∈ 푚 (its support is called the line with the largest slope of 훽 in relation to 훼). Solution to Problem 242

243. Let 훼 be a plane, 휎 a closed half-plane, bordered by 훼, 훼′ a half-plane contained in 훼 and 푎 a real number between 00 and 1800. Show that there is only one half-space 훽′ that has common border with 훼′ such that 훽′ ⊂ 휎 and 푚(훼′훽′) = 푎. Solution to Problem 243

244. Let (훼̂′훽′) be a proper dihedral angle. Construct a half-plane 훾′ such that 푚(훼̂′훽′) = 푚(훾̂′훽′). Show that the problem has two solutions, one of which is located in the int. 훼̂′훽′ (called bisector half-plane of 훼̂′훽′). Solution to Problem 244

245. Show that the locus of the points equally distant from two secant planes 훼, 훽 is formed by two ⊥ planes, namely by the union of the bisector planes of the dihedral angles 훼 and 훽. Solution to Problem 245

246. If 훼 and 훽 are two planes, 푄 ∈ 훽 and 푑 ⊥ through 푄 on 훼. Show that 푑 ⊂ 훽. Solution to Problem 246

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255 Compiled and Solved Problems in Geometry and Trigonometry

247. Consider a line 푑 ⊂ 훼. Show that the union of the ⊥ lines to 훼, which intersect line 푑, is a plane ⊥ 훼. Solution to Problem 247

248. Find the locus of the points equally distant from two concurrent lines. Solution to Problem 248

249. Show that a plane 훼 ⊥ to two secant planes is ⊥ to their intersection. Solution to Problem 249

250. Let 퐴 be a point that is not on plane 훼. Find the intersection of all the planes that contain point 퐴 and are ⊥ to plane 훼. Solution to Problem 250

251. From a given point draw a ⊥ plane to two given planes. Solution to Problem 251

252. Intersect a dihedral angle with a plane as the angle of sections is right. Solution to Problem 252

253. Show that a line 푑 and a plane 훼, which are perpendicular to another plane, are parallel or line 푑 is contained in 훼. Solution to Problem 253

254. If three planes are ⊥ to a plane, they intersect two by two after lines 푎, 푏, 푐. Show that 푎 ∥ 푏 ∥ 푐. Solution to Problem 254

255. From a point 퐴 we draw perpendicular lines 퐴퐵 and 퐴퐶 to the planes of the faces of a dihedral angle 훼̂′훽′. Show that 푚(퐵퐴퐶̂) = 푚(훼̂′훽′) or 푚(퐵퐴퐶̂) = 1800 − 푚(훼̂′훽′).

Solution to Problem 255

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Solutions

Solution to Problem 201.

푀1 − 푧1 = 1 푀2 − 푧1 = 2 + 푖 푀1 − 푧1 = 푥 + 푦푖

∆푀1푀2푀3 equilateral ⟹ ‖푀1푀2‖ = ‖푀1푀3‖ = ‖푀2푀3‖ ⇒ |푧2 − 푧1| = |푧3 − 푧2| =

|푧1 − 푧3| (푥 − 2)2 + (푦 − 1)2 = 2 푥 + 푦 = 2 ⇒ √2 = √(푥 − 2)2 + (푦 − 1)2 ⇒ { ⇒ { (1 − 푥)2 + 푦2 = 2 푥2 + 푦2 − 2푥 = 1 ⇒ 푦 = 2 − 푥 1 − √3 푦1 = 2 2 3 ± √3 2 푥 + 4 + 푥 − 4푥 − 2푥 = 1 ⇒ 푥1,2 = ⇒ 2 1 + √3 푦 [ 2 2 3+ 3 1− 3 3− 3 1+ 3 Thus: 푀 ( √ , √ ) or 푀 ( √ , √ ). 3 2 2 3 2 2 There are two solutions!

Solution to Problem 202.

푧1 = 푟(cos 푡1 + 푖 sin 푡1) 푧2 = 푟(cos 푡2 + 푖 sin 푡2) 푧3 = 푟(cos 푡3 + 푖 sin 푡3) 푧1 ≠ 푧2 ≠ 푧3 ⇒ 푡1 ≠ 푡2 ≠ 푡3 푧1 + 푧2푧3 ∈ ℝ ⇒ sin 푡1 + 푟 sin(푡2 + 푡3) = 0 {푧2 + 푧3푧1 ∈ ℝ ⇒ sin 푡2 + 푟 sin(푡1 + 푡3) = 0 ⟹ 푧3 + 푧1푧2 ∈ ℝ ⇒ sin 푡3 + 푟 sin(푡1 + 푡2) = 0

sin 푡1(1 − 푟 cos 푡) + 푟 sin 푡 ∙ cos 푡1 = 0 {sin 푡2(1 − 푟 cos 푡) + 푟 sin 푡 ∙ cos 푡2 = 0 sin 푡3(1 − 푟 cos 푡) + 푟 sin 푡 ∙ cos 푡3 = 0

푡1 ≠ 푡2 ≠ 푡3 These equalities are simultaneously true only if 1 − 푟 ∙ cos 푡 = 0 and 푟 ∙ sin 푡 = 0, as 푟 ≠ 0 ⇒ sin 푡 = 0 ⇒ 푡 = 0 ⇒ cos 푡=1⇒1−푟=0⇒푟=1, so

푧1푧2푧3 = 1 ∙ (cos 0 + sin 0) = 1.

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255 Compiled and Solved Problems in Geometry and Trigonometry

Solution to Problem 203.

2푘휋 2푘휋 a. 휀 = + 푖 sin , 푘 ∈ {0, 1, … , 푛 − 1}. 푘 푛 푛

2푖휋 2푖휋 휀 = cos + 푖 sin 푖 푛 푛 2휋(푖+푗) 2휋(푖+푗) So 2푗휋 2푗휋} ⟹ 휀푖휀푗 = cos + 푖 sin , 푖, 푗 ∈ {0, 1, … , 푛 − 1}. 휀 = cos + 푖 sin 푛 푛 푗 푛 푛

1) 푖+푗<푛−1⟹푖+푗=푘∈ {0, 1, … , 푛 − 1} ⟹ 휀푖휀푗 = 휀푘 ∈ 퐺;

2) 푖 + 푗 = 푛 ⟹ 휀푖휀푗 = cos 2휋 + 푖 sin 2휋 = 1 = 휀표 ∈ 퐺; 2휋(푛∙푚+푟) 2휋(푛∙푚+푟) 3) 푖+푗>푛⟹푖+푗=푛∙푚+푟,0≤푟<푛,휀 휀 = cos + 푖 sin = 푖 푗 푛 푛 2휋푟 2휋푟 2휋푟 2휋푟 cos (2휋푚 + ) + 푖 sin (2휋푚 + ) = cos + 푖 sin = 휀 ∈ 퐺. 푛 푛 푛 푛 푟 2휋푖 2휋푖 b. 휀 = cos + 푖 sin 푖 푛 푛

−1 1 cos 0+푖 sin 0 2휋푖 2휋푖 2휋푖 휀푖 = = 2휋푖 2휋푖 = cos (− ) + 푖 sin (− ) = cos (2휋 − ) + 휀푖 cos +푖 sin 푛 푛 푛 푛 푛 2휋푖 2휋푛−2휋푖 2휋푛−2휋푖 2휋(푛−1) 2휋(푛−1) 푖 sin (2휋 − ) = cos + 푖 sin = cos + 푖 sin , 푛 푛 푛 푛 푛 푖 ∈ {0, 1, … , 푛 − 1}.

−1 If 푖=0⟹푛−푖=푛⟹휀0 = 휀0 ∈ 푔.

2휋ℎ If 푖≠0⟹푛−푖≤푛−1⟹ℎ=푛−푖∈ {0, 1, … , 푛 − 1} ⟹ 휖−1 = cos + 푛 2휋ℎ 푖 sin ∈ 퐺. 푛

Solution to Problem 204.

푎 = 푟1(cos 푡1 + 푖 sin 푡1) {푏 = 푟2(cos 푡2 + 푖 sin 푡2) 푐 = 푟3(cos 푡3 + 푖 sin 푡3)

arg푎 + arg푐 = 2arg푏 ⟹ 푡1 + 푡3 = 2푡2

and |푎| + |푐| = |푏| ⟹ 푟1 + 푟3 = 푟2

−푏 ± √푏2 − 4푎푐 푎푧2 + 푏푧 + 푐 = 0 ⟹ 푧 = 1,2 2푎 −푟 (cos 푡 + 푖 sin 푡 ) ± √푟2(cos 2푡 + 푖 sin 2푡 ) − 4푟 푟 (cos(푡 + 푡 ) + 푖 sin(푡 + 푡 )) = 2 2 2 2 2 2 1 3 1 3 1 3 2푟1(cos 푡1 + 푖 sin 푡1) −푟 (cos 푡 + 푖 sin 푡 ) ± √(cos 2푡 + 푖 sin 2푡 )(푟2 − 4푟 푟 ) = 2 2 2 2 2 2 1 3 2푟1(cos 푡1 + 푖 sin 푡1)

2 2 2 2 2 2 2 2 But 푟1 + 푟3 = 푟2 ⟹ 푟2 = 푟1 + 푟1 + 푟3 + 2푟1푟3 ⟹ 푟2 − 4푟1푟3 = 푟1 + 푟1 + 푟3 + 2푟1푟3 − 2 4푟1푟3 = (푟1 − 푟3) .

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Therefore:

−푟2(cos 푡2 + 푖 sin 푡2) ± (cos 푡2 + 푖 sin 푡2)(푟1 − 푟3) 푧1,2 = 2푟1(cos 푡1 + 푖 sin 푡1)

We observe that:

(cos 푡2+푖 sin 푡2)(−2푟1) 푧2 = = −[cos(푡2 − 푡1) + 푖 sin(푡2 − 푡1)] = cos[휋 + (푡2 − 푡1)] + 2푟1(cos 푡1+푖 sin 푡1)

푖 sin[휋 + 푡2 − 푡1] and 푡2 = 1.

Solution to Problem 205.

Let

So 훼 is determined.

So 훽 is determined.

If we work with reduced arguments, then 푡4 = 푡2 − 푡1 or 푡4 = 푡2 − 푡1 + 2휋, in the

same way 푡5.

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255 Compiled and Solved Problems in Geometry and Trigonometry

According to a. (∃) the complex numbers of modulus 1, 훼 and 훽 such that 푧2 = 훼푧1 and 푧3 = 훽푧1. In the given relation, by substitution we obtain:

훼 = 1 and 훽 = 1 verify this equality, so in this case 푧2 = 푧3 = 푧1. According to point b.,

where 훽 = 푥 + 푖푦, when

We construct the system:

The initial solution leads us to 푧1 = 푧2 = 푧3.

and gives

By substituting,

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|훼| = 2 does not comply with the condition |훼| = 1. But

then

and then

are on the circle with radius 푟 and the arguments are

they are the peaks of an equilateral triangle.

Solution to Problem 206.

a. We assume that 푑 ∩ 훼 ≠ ∅ and 푑′ ∩ 훼 = {퐵}.

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255 Compiled and Solved Problems in Geometry and Trigonometry

Let 푑 ∩ 훼 = {퐴} and 푑′ ∩ 훼 = {퐵} and the planes determined by pairs of concurrent lines (푑, 퐴퐵); (푑′, 퐴퐵). We remark that these are the required planes, because

b. We assume 푑 ∩ 훼 = {퐴} and 푑′||훼. We draw through 퐴, in plane 훼, line 푑′||푑 and we consider planes (푑, 푑′′) and (푑′, 푑′′) and we remark that

c. We assume 푑 ∩ 훼 = ∅ and 푑′ ∩ 훼 = ∅ and 푑′ ∈ direction 푑. Let 퐴 ∈ 훼 and 푑′′||푑 ⟹ 푑′′||푑′ and the planes are (푑, 푑′′) and (푑′, 푑′′). The reasoning is the same as above.

Solution to Problem 207.

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Solution to Problem 208.

If 훼 separates points 퐴 and 퐵, it means they are in different half-spaces and let 휎 = |훼퐴 and 휎′ = |훼퐵. Because 훼 separates 퐴 and 퐶 ⟹ 퐶 ∈ 휎′. Because 훼 separates 퐶 and 퐷 ⟹ 퐷 ∈ 휎. From 퐵 ∈ 휎′ and 퐷 ∈ 휎 ⟹ 훼 separates points 퐵 and 퐷

From 퐴 ∈ 휎 and 퐷 ∈ 휎 ⟹ |퐵퐷| ∩ 훼 = ∅.

Solution to Problem 209.

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255 Compiled and Solved Problems in Geometry and Trigonometry

From (1), (2), (3)

Solution to Problem 210.

a. int. (|푉퐴, |푉퐵̂ , |푉퐶) = |(푉퐴퐵), 퐶 ∩ |(푉퐵퐶), 퐴 ∩ |(푉퐴퐶), 퐵 is thus an intersection of convex set and thus the interior of a trihedron is a convex set.

b. Tetrahedron [푉퐴퐵퐶] without edge [퐴퐶]. We mark with ℳ1 = [퐴퐵퐶]– [퐴퐶] = [퐴퐵, 퐶 ∩ [퐵퐶, 퐴 ∩ |퐴퐶, 퐵 is thus a convex set, being intersection of convex sets.

is a convex set. In the same way

is a convex set, where

But [푉퐴퐵퐶] – [퐴퐶] =

and thus it is a convex set as intersection of convex sets. c. Tetrahedron [푉퐴퐵퐶] without face [퐴퐵퐶]

푉 is thus intersection of convex sets ⟹ is a convex set.

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Solution to Problem 211.

In plane (퐵퐴퐶) we have 퐸퐹||퐴퐶. In plane (퐷퐴퐶) we have 퐴퐶 ⊂ (퐷퐴퐶) ⟹ 퐸퐹||(퐷퐴퐶). In this plane we also have 퐻퐺||퐴퐶. So 퐸퐹||퐻퐺 ⟹ 퐸, 퐹, 퐺, 퐻 are coplanar and because ‖퐴퐶‖ ‖퐸퐹‖ = = ‖퐻퐺‖ ⟹ 퐸퐹퐺퐻 is a parallelogram. 2

Solution to Problem 212. We assume we have 훼||훽||훾 such that 퐴퐴’ ⊂ 훼, 퐵퐵’ ⊂ , 퐶퐶’ ⊂ 훾.

We draw through 퐴′ a parallel line with 푑: 푑′′||푑. As 푑 intersects all the 3 planes 퐴′ ⊂ 푑′′ at 퐴, 퐵, 퐶 ⟹ and its || 푑’’ cuts them at 퐴′, 퐵′′, 퐶′′. Because

Let plane (푑′, 푑′′ ). Because this plane has in common with planes 훼, 훽, 훾 the points 퐴′, 퐵′′, 퐶′′ and because 훼 || 훽 || 훾 ⟹ it intersects them after the parallel lines

Taking into consideration (1) and (2)

The vice-versa can be similarly proved.

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255 Compiled and Solved Problems in Geometry and Trigonometry

Solution to Problem 213.

Let

such that

and

such that

according to problem 7, three planes can be drawn || 훽 || 훼 || 훾 such that

and 푃푃′ ⊂ 훼. So by marking 푃 and letting 푃′ variable, 푃′ ∈ a parallel plane with the two lines, which passes through 푃. It is known that this plane is unique, because by drawing through 푃 parallel lines to 푑 and 푑’ in order to obtain this plane, it is well determined by 2 concurrent lines. Vice-versa: Let 푃 ∈ 훼, that is the plane passing through 푃 and it is parallel to 푑 and 푑’.

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(푃′′, 푑) determines a plane, and (푃′′, 푑′) determines a plane ⟹ the two planes, which have a common point, intersect after a line(푃′′, 푑)∩ (푃′′, 푑′) = 푄푄′ where 푄 ∈ 푑 and 푄′ ∈ 푑′. Because

such that 푀푄 ⊂ 훽, 훽||훼. Because

such that 푀′푄′ ⊂ 훾, 훾||훼. So the required locus is a parallel plane with 푑 and 푑’.

Solution to Problem 214.

We consider the plane, which according to a previous problem, represents the

locus of the points dividing the segments with extremities on lines 푑1 and 푑3 in a

given ratio 푘. To obtain this plane, we take a point 퐴 ∈ 푑1, 퐵 ∈ 푑3 and point 퐶 ∈ 퐴퐵 ‖퐴퐶‖ such that = 푘. Through this point 퐶 we draw two parallel lines 푑 and 푑 which ‖퐶퐵‖ 1 3 determine the above mentioned plane 훼.

Let 푑2 ∩ 훼 = {푁}. We must determine a segment that passes through 푁 and with

its extremities on 푑1 and 푑3, respectively at 푀 and 푃. As the required line passes through 푁 and 푀

The same line must pass through 푁 and 푃 and because

푀, 푁, 푃 collinear.

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255 Compiled and Solved Problems in Geometry and Trigonometry

From (1) and (2)

Then, according to previous problem 8:

and the required line is 푀푃.

Solution to Problem 215.

Let ∆푀푁푃 such that

Let ∆푀′푁′푃′ such that

Line 푀푃 generates a plane 훽, being parallel to a fixed direction 푑1 and it is based on a given line 푑. In the same way, the line 푀푁 generates a plane 훾, parallel to a

fixed direction 푑2, and based on a given line 푑. As 푑 is contained by 훾 ⟹ 푂 is a common point for 훼 and 훾 ⟹ 훼 ∩ 훾 ≠ ∅ ⟹ 훼 ∩ 훾 = 푑′, 푂 ∈ 푑′.

(∀) the considered ∆, so 푁 also describes a line 푑′ ⊂ 훼.

Because plane 훾 is well determined by line 푑 and direction 푑2, is fixed, so 푑′ = 훼 ∩ 훾 is fixed. In the same way, 푃푁 will generate a plane 훿, moving parallel to the fixed direction

푑3 and being based on the given line 푑’. As

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(∀)푃 variable peak, 푃 ∈ 푑′′. Thus, in the given conditions, for any ∆푀푁푃, peak 푃 ∈ 푑′′. Vice-versa, let 푃′ ∈ 푑′′. On plane (푑′, 푑′′) we draw 푃′, 푀′||푃푀 ⟹ (푀′푃′푁′)||(푃푀푁) ⟹ (푑푑′) the intersection of two parallel planes after parallel lines 푀′푁′||푀푁 and the so constructed ∆푀′푃′푁′ has its sides parallel to the three fixed lines, has 푀′ ∈ 푑 and 푁′ ∈ 훼, so it is one of the triangles given in the text. So the locus is line 푑′′. We’ve seen how it can be constructed and it passes through 푂. In the situation when 퐷||훼 we obtain

In this case the locus is a parallel line with 푑.

Let 푀푁푃 and 푀′푁′푃′ such that

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255 Compiled and Solved Problems in Geometry and Trigonometry

We assume 훼 ∩ 훽 = 푑 and let 푑 ∩ (푀푁푃) = {푂} and 푑 ∩ (푀′푁′푃′) = {푂′} ⟹

a plane cuts the parallel planes after parallel lines. In the same way, 푂푁||푂′푁′ and because

We use the property: Let 휋1 and 휋2 2 parallel planes and 퐴, 퐵, 퐶 ⊂ 휋1 and 퐴′퐵′퐶′ ⊂

휋2, 퐴퐵 ∥ 퐴′퐵′,

Let’s show that 퐵퐶||퐵′퐶′. Indeed (퐵퐵′퐶′) is a plane which intersects the 2 planes after parallel lines.

Applying in (1) this property ⟹ 푂푃||푂′푃′. Maintaining 푂푃 fixed and letting 푃’ variable, always 푂푃||푂′푃′.=, so 푂′푃′ generates a plane which passes through 푑. We assume 훽||훼.

푀푁푁′푀′ parallelogram

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Considering 푃′ fix and 푃 variable ⟹ 푃푃′||훼 and the set of parallel lines drawn 푃푃′||훽 to a plane through an exterior point is a parallel plane with the given plane. So the locus is a parallel plane with 훼 and 훽.

Solution to Problem 216.

In plane 퐷퐴퐶 we have:

In plane 퐷퐴퐵 we have:

In plane 푂퐵퐶 we have:

From 푃푀||퐴퐶 and 푃푁||퐵퐶 ⟹ (푀푁푅)||(퐴퐵퐶). Let 푄 and 퐷 be midpoints of sides |푀푁| and |퐴퐵|.

are collinear.

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255 Compiled and Solved Problems in Geometry and Trigonometry

Concurrent lines 푂퐷 and 푂퐶 determine a plane which cuts the parallel planes

are collinear. So 퐺’ ∈ |푂퐺 ⟹ the required locus is ray |푂퐺. Vice-versa: we take a point on |푂퐺, 퐺′′, and draw through it a parallel plane to (퐴퐵퐶), plane (푀′′, 푁′′, 푃′′), similar triangles are formed and the ratios from the hypothesis appear.

Solution to Problem 217.

Let 퐴2, 퐵2, 퐶2, 퐷2 such that

Mark on lines 퐴퐷1 and 퐵퐶1 points 푀 and 푁 such that

From

Next is

The same,

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we obtain

From

is a parallelogram.

is parallelogram.

is a parallelogram.

Solution to Problem 218.

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255 Compiled and Solved Problems in Geometry and Trigonometry

We draw through 퐴′ a line 푑′′||푑. We draw two parallel planes with 훼, which will intersect the three lines in 퐵′, 퐵∗, 퐵 and 퐶′, 퐶∗, 퐶. Plane (푑, 푑∗) intersects planes 훼, (퐵′퐵∗퐵), (퐶′퐶∗퐶) after parallel lines

= ‖퐵퐵∗‖ = ‖퐶′퐶∗‖. Plane (푑′, 푑∗) intersects parallel planes (퐵′퐵∗퐵), (퐶′퐶∗퐶) after parallel lines

So (∀) parallel plane with 훼 we construct, the newly obtained triangle has a side of 훼 length and the corresponding angle to 퐵′퐵̂∗퐵 is constant. We mark with a line that position of the plane, for which the opposite length of the required angle is 푙. With the compass spike at 퐶 and with a radius equal with 푙, we trace a circle arc that cuts segment | 퐶′퐶∗| at 푁 or line 퐶′퐶∗. Through 푁 we draw at (푑′, 푑∗) a parallel line to 푑∗ which precisely meets 푑′ in a point 푀′. Through 푀′, we draw the || plane to 훼, which will intersect the three lines in 푀, 푀′, 푀∗.

is a parallelogram.

⟹ 퐶푁푀′푀 is a parallelogram.

and line 푀푀′, located in a parallel plane to 훼, is parallel to 훼. Discussion: Assuming the plane (퐶′퐶∗퐶) is variable, as | 퐶퐶∗| and 퐶퐶̂′퐶∗ are constant, then 푑(퐶′퐶∗퐶) = 푏 = also constant If 푙 < 푑 we don’t have any solution. If 푙 = 푑 (∃) a solution, the circle of radius 푙, is tangent to 퐶′퐶∗. If 푙 > 푑 (∃ ) two solutions: circle of radius 푙, cuts 퐶′퐶∗ at two points 푁 and 푃.

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Solution to Problem 219.

We draw through 퐴 planes 훼 ⊥ 푑 and 훼′ ⊥ 푑′. As 퐴 is a common point

⟹ 훼 ∩ 훼′ = ∆⟹ 퐴 ∈ ∆.

⟹ ∆ the required line If 훼 ≠ 훼′ - we have only one solution. If 훼 = 훼′ (∀) line from 훼 which passes through 퐴 corresponds to the problem, so (∃) infinite solutions.

Solution to Problem 220.

Let 푑1 ⊥ 푑2 two concurrent perpendicular lines, 푑1 ∩ 푑2 = {푂}. They determine a

plane 훼 = (푑1, 푑2) and 푂 ∈ 훼. We construct on 훼 in 푂.

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255 Compiled and Solved Problems in Geometry and Trigonometry

Solution to Problem 221.

We use the reductio ad absurdum method. Let 푑⊥푎,푑⊥푎,푑⊥푐. We assume that these lines are not coplanar. Let 훼 = (푏, 푐), 훼′ = (푎, 푏), 훼 ≠ 훼′. Then 푑 ⊥ 훼, 푑 ⊥ 훼’. Thus through point 푂, 2 perpendicular planes to 푑 can be drawn. False ⟹ 푎, 푏, 푐 are coplanar.

Solution to Problem 222.

By reductio ad absurdum: Let 푎 ∩ 푏 ∩ 푐 ∩ 푑 = {푂} and they are perpendicular two by two. From 푑 ⊥ 푎, 푑 ⊥ 푎, 푑 ⊥ 푐 ⟹ 푎, 푏, 푐 are coplanar and 푏 ⊥ 푎, 푐 ⊥ 푎, so we can draw to point 푂 two distinct perpendicular lines. False. So the 4 lines cannot be perpendicular two by two.

Solution to Problem 223.

We assume that 푑 ⊥ 훼. In 푑′ ∩ 훼 = {푂} we draw line 푑′′ ⊥ 훼. Lines 푑′ and 푑′′ are concurrent and determine a plane 훽 = (푑′, 푑′′) and as 푂′ ∈ 훽, 푂′ ∈ 훼 ⟹

From (1) and (2) ⟹ in plane 훽, on line 푎, at point 푂′ two distinct perpendicular lines had been drawn. False. So 푑′||훼.

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Solution to Problem 224.

Reductio ad absurdum. Let 푑 ∦ 푑′. We draw 푑′′||푑 through 푂’.

⟹ at point 푂′ we can draw two perpendicular lines to plane 훼. False. So 푑||푑′.

Solution to Problem 225.

Let 푑 ⊥ 훼 and 푑 ∩ 훼 = {푂}. We draw through 푂 a parallel to 푑′, which will be contained in 훼, then 푑||훼.

Solution to Problem 226.

We assume 훽 ⟹ 훼 ∩ 훽 ≠ ∅ and let 퐴 ∈ 훼 ∩ 훽 ⟹ through a point 퐴 there can be drawn two distinct perpendicular planes on this line. False. ⟹ 훼 || 훽.

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255 Compiled and Solved Problems in Geometry and Trigonometry

Solution to Problem 227.

Let 푀 be a point in space with the property ||푀퐴|| = ||푀퐵||. We connect 푀 with the midpoint of segment [퐴퐵], point 푂.

So 푀 is on a line drawn through 푂, perpendicular to 퐴퐵. But the union of all perpendicular lines drawn through 푂 to 퐴퐵 is the perpendicular plane to 퐴퐵 at point 푂, marked with 훼, so 푀 ∈ 훼.

Vice-versa: let 푀 ∈ 훼,

common side

Solution to Problem 228.

Let 푀 be a point in space with this property:

Let 푂 be the center of the circumscribed circle ∆퐴퐵퐶 ⟹ ||푂퐴|| = 푂퐵|| = ||푂퐶||, so 푂 is also a point of the desired locus.

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According to the previous problem the locus of the points in space equally distant from 퐴 and 퐵 is in the mediator plane of segment [퐴퐵], which also contains 푀. We mark with 훼 this plane. The locus of the points in space equally distant from 퐵 and 퐶 is in the mediator plane of segment [퐵퐶], marked 훽, which contains both 푂 and 푀. So 훼 ∩ 훽 = 푂푀.

so 푀 ∈ the perpendicular line to plane (퐴퐵퐶) in the center of the circumscribed circle ∆퐴퐵퐶. Vice-versa, let 푀 ∈ this perpendicular line

= ||퐶푀||, so 푀 has the property from the statement.

Solution to Problem 229.

We draw ⊥ from 퐵 to the plane. Let 푂 be the foot of this perpendicular line. Let

the circle of radius 푂퐴. Vice-versa, let 푀 ∈ this circle

so 푀 represents the foot from 퐵 to 퐴푀.

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255 Compiled and Solved Problems in Geometry and Trigonometry

Solution to Problem 230.

Let 훼 be a plane that passes through 푎 and let 푀 be the ⊥ foot from 퐴 to 훼 ⟹ 퐴푀 ⊥ 훼. From

so 푀 ∈ a perpendicular line to 푎 in 퐴′, thus it is an element of the perpendicular plane to 푎 in 퐴′, which we mark as 휋 and which also contains 퐴.

푀 ∈ the circle of radius 퐴퐴′ from plane π.

*Vice-versa, let 푀 be a point on this circle of radius 퐴퐴′ from plane 휋.

⟹ 푀 is the foot of a ⊥ drawn from 퐴 to a plane that passes through 푎.

Solution to Problem 231.

Let 퐴′ and 퐵′ be the feet of the perpendicular lines from 퐴 and 퐵 to 훼

(∃) a plane 훽 = (퐴퐴′, 퐵퐵′) and 퐴퐵 ⊂ 훽

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are collinear. In plane 훽 we have

right

Solution to Problem 232.

Let 푑, 푑′ be given lines, 퐴 given point. We draw through 퐴 plane 훼 ⊥ to 푑′. If 푎 ∩ 훼 = {퐵}, then line 퐴퐵 is the desired one, because it passes through 퐴, meets 푑 and from 푑′훼 푑′퐴퐵. If 푑 ∩ 훼 = ∅ there is no solution. If 푑 ⊂ 훼, then any line determined by 퐴 and a point of 푑 represents solution to the problem, so there are infinite solutions.

Solution to Problem 233.

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255 Compiled and Solved Problems in Geometry and Trigonometry

Solution to Problem 234.

Let 푀 be a point such that ||퐴푀|| = 푘. We draw 퐴퐴′ ⊥ 훼 ⇒ 퐴′ fixed point and 퐴퐴′ ⊥ 퐴′푀. We write ||퐴퐴′|| = 푎. Then

푀 ∈ a circle centered at 퐴′ and of radius √푘2 − 푎2, for 푘 > 푎. For 푘 = 푎 we obtain 1 point. For 푘 < 푎 empty set. Vice-versa, let 푀 be a point on this circle ⟹

so 푀 has the property from the statement.

Solution to Problem 235.

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In ∆푂퐶푀:

But

3. Let 퐻 be the projection of 푂 lcp. plane 퐴퐵퐶, so

퐻 ∈ corresponding heights of side 퐴퐵. We show in the same way that 퐴퐶 ⊥ 퐵퐻 and thus 퐻 is the point of intersection of the heights, thus orthocenter.

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255 Compiled and Solved Problems in Geometry and Trigonometry

Solution to Problem 236.

First we prove that if a line is ⊥ to two concurrent planes ⟹ the planes coincide.

Let

∆퐴퐵푀 has two right angles. False. We return to the given problem.

being concurrent, they determine a plane ⟹ 퐶퐷 ⊥ 퐴퐵.

퐶, 퐷, 퐶′, 퐷′ are coplanar ⟹ 퐶퐶′ and 퐷퐷′ are coplanar.

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Solution to Problem 237.

We draw

⟹ 퐵퐴′ height in ∆퐵퐶퐷 (1)

⟹ 퐴′퐶 height in ∆퐴퐵퐶 (2) From (1) and (2) ⟹ 퐴′ is the orthocenter ∆퐴퐵퐶 ⟹ 푂퐴′ ⊥ 퐵퐶.

Solution to Problem 238.

Let

∆퐵푂푂′ and ∆퐶푂푂′ are right at 푂′.

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255 Compiled and Solved Problems in Geometry and Trigonometry

|푂퐴| ≡ |푂퐵| ≡ |푂퐶| As } |푂푂′| common side

is the center of the circumscribed circle ∆퐴퐵퐶.

Solution to Problem 239.

Let 퐷 be the midpoint of [퐵퐶] and 퐸 ∈ |퐷퐴′ such that ||퐷퐸|| = ||퐷퐴||. 퐴퐷 is median in the ∆ isosceles

common

being external for

Solution to Problem 240.

Let 푀 be a point in the plane and |퐴푀′ the opposite ray to 퐴푀. According to theorem 1

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Solution to Problem 241.

We construct 퐵 on the plane

⟹ 퐴퐶 and 퐵퐵′ determine a plane 훽 = (퐴퐶, 퐵퐵′) ⟹ 퐴퐵 ⊂ 훽 and on this plane 푀(퐶퐴퐵̂) = 900 − 푚(퐵퐴퐵′)̂ .

Solution to Problem 242.

Let ray |퐴퐵 ⊂ 훽′ such that 퐴퐵 ⊥ 푚. Let |퐴퐶 another ray such that |퐴퐶 ⊂ 훽’. We draw 퐵퐵′ ⊥ 훼 and 퐶퐶′ ⊥ 훼 to obtain the angle of the 2 rays with 훼, namely 퐵퐴퐵̂′ > 퐶퐴퐶̂′. We draw line |퐴퐴’ such that 퐴퐴′ ⊥ 훼 and is on the same side of plane 훼 as well as half-plane 훽’.

[퐴퐵 is the projection of ray [퐴퐴′ on plane 훽

Solution to Problem 243. Let 푑 be the border of 훼′ and 퐴 ∈ 푑. We draw a plane ⊥ on 푑 in 퐴, which we mark as 훾.

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255 Compiled and Solved Problems in Geometry and Trigonometry

In this plane, there is only one ray 푏, with its origin in 퐴, such that 푚(푐,̂ 푏) = 푎. The desired half-plane is determined by 푑 and ray 푏, because from

Solution to Problem 244.

Let 푑 be the edge of the dihedral angle and 퐴 ∈ 푑. We draw 푎 ⊥ 푑, 푎 ⊂ 훼′ and 푏 ⊥ 푑, 푏 ⊂ 훽’ two rays with origin in 퐴. It results 푑 ⊥ (푎푏). We draw on plane (푎, 푏) ray 푐 such that 푚(푎푐̂) = 푚(푐푏̂) (1). As 푑 ⊥ (푎푏) ⟹ 푑 ⊥ 푐. Half-plane 훾′ = (푑, 푐) is the desired one, because

If we consider the opposite ray to 푐, 푐′, half-plane 훾′′ = (푑, 푐′) also forms concurrent angles with the two half-planes, being supplementary to the others.

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Solution to Problem 245.

Let 푀 be a point in space equally distant from the half-planes 훼′, 훽′ ⟹ ||푀퐴|| = ||푀퐵||.

where 푑 = 훼 ∩ 훽. Let

|푀퐴| = |푀퐵| } ⟹ |푂푀| common side right triangle

⟹ 푀 ∈ bisector of the angle. 퐴퐷퐵̂ ⟹ 푀 ∈ bisector half-plane of the angle of half-planes 훼′, 훽′. If 푀′ is equally distant from half-planes 훽′ and 훼′′ we will show in the same way that 푀′ ∈ bisector half-plane of these half-planes. We assume that 푀 and 푀′ are on this plane ⊥ to 푑, we remark that 푚(푀푂푀′̂ ) = 900, so the two half-planes are ⊥. Considering the two other dihedral angles, we obtain 2 perpendicular planes, the 2 bisector planes.

Vice-versa: we can easily show that a point on these planes is equally distant form planes 훼 and 훽.

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255 Compiled and Solved Problems in Geometry and Trigonometry

Solution to Problem 246.

Let 훼 ∩ 훽 = 훼. In plane 훽 we draw

but

so from a point it can be drawn only one perpendicular line to a plane,

Solution to Problem 247.

Let

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the line with the same direction. We know that the union of the lines with the same direction and are based on a given line is a plane. As this plane contains a perpendicular line to 훼, it is perpendicular to 훼.

Solution to Problem 248.

Let 훼 = (푑1, 푑2) the plane of the two concurrent lines and 푀 is a point with the

property 푑(푀, 푑1) = 푑(푀, 푑2). We draw

Let

a bisector of the angle formed by the two lines, and 푀 is on a line 훼 which meets a bisector ⇒ 푀 ∈ a plane ⊥ 훼 and which intersects 훼 after a bisector. Thus the locus will be formed by two planes ⊥ α and which intersects 훼 after the two

bisectors of the angle formed by 푑1, 푑2. The two planes are ⊥. ‖푀′퐴‖ = ‖푀′퐵‖ ⟹ } ⟹ 푀퐴 ⊥ 푑 ||MM′|| common side 1

And in the same way 푀퐵 ⊥ 푑2 ⇒ 푀 has the property from the statement.

Solution to Problem 249.

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255 Compiled and Solved Problems in Geometry and Trigonometry

Let 훽 ∩ 훾 = 푑 and 푀 ∈ 푑 ⟹ 푀 ∈ 훽, 푀 ∈ 훾. We draw ⊥ from 푀 to 훼, line 푑’. According to a previous problem

Solution to Problem 250.

Let 훽 and 훾 be such planes, that is

From

are secant planes and ⊥ to 훼.

So their intersection is ⊥ through 퐴 to plane 훼.

Solution to Problem 251.

We construct the point on the two planes and the desired plane is determined by the two perpendicular lines.

Solution to Problem 252.

Let 훼 ∩ 훽 = 푑 and 푀 ∈ 푑. We consider a ray originating in 푀, 푎 ∈ 훼 and we construct a ⊥ plane to 푎 in 푀, plane 훾. Because

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and let a ray originating in 푀, 푏 ⊂ 훽 ∩ 훾 ⇒ 푏 ⊂ 훽 , 푏 ⊂.훾 As 푎 ⊥ 훾 ⟹ 푎 ⊥ 푏 and the desired plane is that determined by rays (푎, 푏).

Solution to Problem 253.

Let 훼 ∩ 훽 = 푎 and 푑 ∩ 훽 = {퐴}. We suppose that 퐴 ∉ 푎. Let 푀 ∈ 푎, we build 푏 ⊥ 훽, 푀 ∈ 푏 ⟹ 푏 ⊂ 훼.

Solution to Problem 254.

218 255 Compiled and Solved Problems in Geometry and Trigonometry

From (1), (2), (3) ⟹ 푎 ∥ 푏 ∥ 푐.

Solution to Problem 255.

Let 퐴 ∈ int. (훼̂′훽′), 훼 ∩ 훽 = 푑.

Let 퐴 ∈ int. (훼̂′′훽′). We show the same way that 푚(퐵̂퐴퐶) = 1800 − 푚(훼̂′′훽′) } ⟹ 푚(퐵퐴퐶̂) = 1800 − 1800 + 푚(훼̂′훽′) = 푚(훼̂′훽′). 푚(훼̂′′훽′) = 1800 − 푚(훼̂′훽′) If 퐴 ∈ int. (훼̂′′훽′′) ⟹ 푚(퐵퐴퐶̂) = 1800 − 푚(훼̂′′훽′). If 퐴 ∈ int. (훼̂′훽′′) ⟹ 푚(퐵̂퐴퐶) = 1800 − 푚(훼̂′훽′).

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This book is a translation from Romanian of "Probleme Compilate şi Rezolvate de Geometrie şi Trigonometrie" (University of Kishinev Press, Kishinev, 169 p., 1998), and includes problems of 2D and 3D Euclidean geometry plus trigonometry, compiled and solved from the Romanian Textbooks for 9th and 10th grade students, in the period 1981-1988, when I was a professor of mathematics at the "Petrache Poenaru" National College in Balcesti, Valcea (Romania), Lycée Sidi El Hassan Lyoussi in Sefrou (Morocco), then at the "Nicolae Balcescu" National College in Craiova and Dragotesti General School (Romania), but also I did intensive private tutoring for students preparing their university entrance examination. After that, I have escaped in Turkey in September 1988 and lived in a political refugee camp in Istanbul and Ankara, and in March 1990 I immigrated to United States. The degree of difficulties of the problems is from easy and medium to hard. The solutions of the problems are at the end of each chapter. One can navigate back and forth from the text of the problem to its solution using bookmarks. The book is especially a didactical material for the mathematical students and instructors. The Author

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