July 17, 2020. Math 9+. Geometry Revisited and more. Collinearity and Concurrence. Morley theorem. Morley’s theorem. Definition. Three lines trisecting angle ∠퐴퐵퐶 into three equal angles are called trisectors of ∠퐴퐵퐶. Theorem. The points of intersection of the adjacent trisectors of the angles of any triangle are the vertices of an equilateral triangle. Proof. Let the adjacent trisectors of B angles 퐵퐴퐶 and 퐵퐶퐴 meet at a point 퐵1 and the other two, non-adjacent trisectors meet at point 퐵2 (see figure). Y Then, 퐵 is the incenter of the triangle B 1 X 2 퐴퐵 퐶 and 퐵 퐵 is the bisector of 2 2 1 A1 C1 ∠퐴퐵2퐶. Let us now construct an 30o 30o equilateral triangle 퐴1퐵1퐶1 where 퐴1 B1 and 퐶 belong to the non-adjacent 1 trisectors, 퐶퐵2 and 퐴퐵2, respectively. A C In order to prove the theorem, we must prove that 퐵퐶1 and 퐵퐴1 are the trisectors of the angle 퐴퐵퐶. Exercise. Complete the above proof. C Q Exercise. Prove the following theorem B YD (butterfly). M Y XA Theorem (butterfly). Through the midpoint YC X 푀 of a chord PQ of a circle, any other two P X O chords, 퐴퐶 and 퐵퐷 are drawn. If chords 퐴퐵 B and 퐶퐷 meet 푃푄 at points 푋 and 푌, then M is the midpoint of 푋푌. A D Proof. Consider the figure. Napoleon triangles. Theorem (butterfly). If triangles are erected externally on the sides of an arbitrary triangle so that sum of the “remote” angles of these three triangles is 180°, then the circumcircles of these three triangles have a common point. A’ B B’ C O1 O O2 1 F B O3 O A C A O3 C’ O2 Proof. Consider the figure. Theorem. Three equilateral triangles are erected externally on the sides of an arbitrary triangle 퐴퐵퐶. Then, the triangle 푂1푂2푂3 obtained by connecting the centers of these equilateral triangles is also an equilateral triangle (Napoleon’s triangle, see Figure). Solution. Denote |AB| = c, |BC| = a, |AC| = b. Let us find the side |푂2푂3|. 1 1 Express 푂⃗⃗⃗⃗⃗⃗⃗푂⃗⃗⃗ = 퐴⃗⃗⃗⃗푂⃗⃗⃗ − 퐴푂⃗⃗⃗⃗⃗⃗⃗ , or, 푂⃗⃗⃗⃗⃗⃗⃗푂⃗⃗⃗ = 퐴⃗⃗⃗⃗퐵⃗ + 퐶⃗⃗⃗⃗′⃗푂⃗⃗⃗⃗ − 퐴퐶⃗⃗⃗⃗⃗ − 퐵⃗⃗⃗⃗′⃗⃗푂⃗⃗⃗ . 2 3 3 2 2 3 2 3 2 2 3 3 Note, that |퐵⃗⃗⃗⃗′⃗⃗푂⃗⃗⃗ | = 푏 √ , and |퐶⃗⃗⃗⃗′⃗푂⃗⃗⃗⃗ | = 푐 √ . Also, (퐴퐵⃗⃗⃗⃗⃗ ∙ 퐴퐶⃗⃗⃗⃗⃗ ) = 푏푐 cos 훼, 2 6 3 6 3 1 (퐴퐵⃗⃗⃗⃗⃗ ∙ 퐵⃗⃗⃗⃗′⃗⃗푂⃗⃗⃗ ) = (퐴퐶⃗⃗⃗⃗⃗ ∙ 퐶⃗⃗⃗⃗′⃗푂⃗⃗⃗⃗ ) = 푏푐 √ cos(90˚ + 훼) = − 푏푐 sin 훼, and 2 3 6 2√3 1 1 (퐶⃗⃗⃗⃗′⃗푂⃗⃗⃗⃗ ∙ 퐵⃗⃗⃗⃗′⃗⃗푂⃗⃗⃗ ) = 푏푐 cos(180˚ − 훼) = − 푏푐 cos 훼, where 훼 = 퐵퐴퐶̂. Then, 3 2 12 12 2 1 2 2 1 2 2 1 |푂⃗⃗⃗⃗⃗⃗⃗푂⃗⃗⃗ | = |퐴퐵⃗⃗⃗⃗⃗ | + |퐶⃗⃗⃗⃗′⃗푂⃗⃗⃗⃗ | + |퐴퐶⃗⃗⃗⃗⃗ | + |퐵⃗⃗⃗⃗′⃗⃗푂⃗⃗⃗ | − (퐴퐵⃗⃗⃗⃗⃗ ∙ 퐴퐶⃗⃗⃗⃗⃗ ) 2 3 4 3 4 2 2 ⃗⃗⃗⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗⃗⃗⃗ − (퐴퐵⃗⃗⃗⃗⃗ ∙ 퐵′푂2) − (퐴퐶⃗⃗⃗⃗⃗ ∙ 퐶′푂3) − 2 (퐶′푂3 ∙ 퐵′푂2), or, 2 1 1 1 4 2 |푂⃗⃗⃗⃗⃗⃗⃗푂⃗⃗⃗ | = (푐2 + 푐2 + 푏2 + 푏2 − 2푏푐 cos 훼 + 푏푐 sin 훼 + 푏푐 cos 훼), 2 3 4 3 3 √3 3 2 1 1 1 1 |푂⃗⃗⃗⃗⃗⃗⃗푂⃗⃗⃗ | = 푐2 + 푏2 − 푏푐 cos 훼 + 푏푐 sin 훼. 2 3 3 3 3 √3 Now, using the Law of cosines, 2푏푐 cos 훼 = 푏2 + 푐2 − 푎2, and the Law of sines, 푎 2 sin 훼 = , where R is the radius of the circumcircle, we obtain |푂⃗⃗⃗⃗⃗⃗⃗푂⃗⃗⃗ | = 2푅 2 3 1 1 1 푎푏푐 푎2 + 푏2 + 푐2 + . Obviously, the same expression holds for the sides 6 6 6 2√3푅 |푂1푂3| and |푂1푂2|. Hence, triangle 푂1푂2푂3 is equilateral. Problem. Let 퐴, 퐵 and 퐶 be angles of a triangle 퐴퐵퐶. B 3 a. Prove that cos 퐴 + cos 퐵 + cos 퐶 ≤ . -B 2 n a b. *Prove that for any three numbers, 푚,푛,푝, 2 c -C 2푚푛cos 퐴 + 2푛푝cos 퐵 + 2푝푚cos 퐶 ≤ 푚 + p 2 2 m 푛 + 푝 C A b Solution. Let vectors 푚⃗⃗ , 푛⃗ , 푝 be parallel to 퐴퐶⃗⃗⃗⃗⃗ , 퐵퐴⃗⃗⃗⃗⃗ and -A 퐶퐵⃗⃗⃗⃗⃗ , respectively, as in the Figure. Then, (푚⃗⃗ + 푛⃗ + 푝 )2 = 푚2 + 푛2 + 푝2 − 2푚푛 cos 퐴 − 2푛푝 cos 퐵 − 2푚푝 cos 퐶 wherefrom immediately follows that, 2푚푛cos 퐴 + 2푛푝cos 퐵 + 2푝푚cos 퐶 ≤ 푚2 + 푛2 + 푝2. The statement in part (a) follows from the above for 푚 = 푛 = 푝 = 1. Problem. Point 퐴’ divides the side 퐵퐶 of the triangle 퐴퐵퐶 into two segments, 퐵퐴′ and 퐴′퐶, whose lengths have the ratio |퐵퐴′|: |퐴′퐶| = 푚: 푛. Express vector 퐴퐴⃗⃗⃗⃗⃗⃗ ′ via vectors 퐴퐵⃗⃗⃗⃗⃗ and 퐴퐶⃗⃗⃗⃗⃗ . Find the length of the Cevian 퐴퐴’ if the sides of the triangle are |퐴퐵| = 푐, |퐵퐶| = 푎, and |퐴퐶| = 푏. 푚 푚 Solution. It is clear from the Figure, that 퐵퐴⃗⃗⃗⃗⃗⃗⃗ ′ = 퐴⃗⃗⃗⃗′⃗퐶⃗ = 퐵퐶⃗⃗⃗⃗⃗ , and 퐶퐴⃗⃗⃗⃗⃗⃗ ′ = 푛 푚+푛 푛 푛 퐶퐵⃗⃗⃗⃗⃗ = (퐴퐵⃗⃗⃗⃗⃗ − 퐴퐶⃗⃗⃗⃗⃗ ). Therefore, 푚+푛 푚+푛 B 푛 푛 a 퐴퐴⃗⃗⃗⃗⃗⃗ ′ = 퐴퐶⃗⃗⃗⃗⃗ + 퐶퐴⃗⃗⃗⃗⃗⃗ ′ = 퐴퐶⃗⃗⃗⃗⃗ + (퐴퐵⃗⃗⃗⃗⃗ − 퐴퐶⃗⃗⃗⃗⃗ ) = 퐴퐵⃗⃗⃗⃗⃗ + m A’ 푚+푛 푚+푛 c 푚 퐴퐶⃗⃗⃗⃗⃗ . n 푚+푛 C Or, we can obtain the same result as A b 푚 푛 푚 퐴퐴⃗⃗⃗⃗⃗⃗ ′ = 퐴퐵⃗⃗⃗⃗⃗ + 퐵퐴⃗⃗⃗⃗⃗⃗⃗ ′ = 퐴퐵⃗⃗⃗⃗⃗ + (퐴퐶⃗⃗⃗⃗⃗ − 퐴퐵⃗⃗⃗⃗⃗ ) = 퐴퐵⃗⃗⃗⃗⃗ + 퐴퐶⃗⃗⃗⃗⃗ . 푚+푛 푚+푛 푚+푛 For the length of the segment 퐴퐴’ we have, 푛 푚 2 푛2푐2+푚2푏2+(푛푚)2푏푐 cos 퐵퐴퐶̂ |퐴퐴′|2 = 퐴퐴⃗⃗⃗⃗⃗⃗ ′2 = ( 퐴퐵⃗⃗⃗⃗⃗ + 퐴퐶⃗⃗⃗⃗⃗ ) = . Using the 푚+푛 푚+푛 (푚+푛)2 Law of cosines, we write 2푏푐 cos 퐵퐴퐶̂ = 푏2 + 푐2 − 푎2, and obtain the final result, (푛2+푛푚)푐2+(푚2+푛푚)푏2−(푚푛)푎2 푚푏2+푛푐2 푚푛푎2 |퐴퐴′|2 = = − . (푚+푛)2 푚+푛 (푚+푛)2 푚푛푎2 Or, equivalently, (푚 + 푛)|퐵퐵′|2 = 푚푏2 + 푛푐2 − . 푚+푛 Substituting 푚 + 푛 = 푎, we obtain the Stewart’s theorem (Coxeter, Greitzer, exercise 4 on p. 6). If 퐴퐴’ is a median, then |퐵퐴′|: |퐴′퐶| = 1: 1, i.e. 푚 = 푛 = 1, and we have, 1 1 1 1 1 퐴퐴⃗⃗⃗⃗⃗⃗ ′ = 퐴퐵⃗⃗⃗⃗⃗ + 퐴⃗⃗⃗⃗퐶⃗ , |퐴퐴′|2 = 푏2 + 푐2 − 푎2 (퐴퐴’ is a median). 2 2 2 2 4 If 퐴퐴’ is a bisector, |퐵퐴′|: |퐴′퐶| = 푐: 푏, i.e. 푚 = 푐, 푛 = 푏, and we obtain 푏 푐 푏2푐+푐2푏 푏푐푎2 푎2 퐴퐴⃗⃗⃗⃗⃗⃗ ′ = 퐴퐵⃗⃗⃗⃗⃗ + 퐴퐶⃗⃗⃗⃗⃗ , as well as |퐴퐴′|2 = − = 푏푐 (1 − ) 푏+푐 푏+푐 푏+푐 (푏+푐)2 (푏+푐)2 (퐴퐴’ is a bisector). The nine-points circle problem. Theorem. The feet of the three altitudes of any triangle, the midpoints of the three sides, and the midpoints of the segments from the three vertices lo the orthocenter, all lie on the same circle, of radius ½푅. This theorem is usually credited to a German geometer Karl Wilhelm von Feuerbach, who actually rediscovered the theorem. The first complete proof appears to be that of Jean-Victor Poncelet, published in 1821, and Charles Brianson also claimed proving the same theorem prior to Feuerbach. The theorem also sometimes mistakenly attributed to Euler, who proved, as early as 1765, that the orthic triangle and the medial triangle have the same circumcircle, which is why this B circle is sometimes called "the H C H Euler circle". Feuerbach A H rediscovered Euler's partial result MC even later, and added a further MA property which is so remarkable N that it has induced many authors A O to call the nine-point circle "the Feuerbach circle". HB MB C Proof. Consider rectangles formed by the mid-lines of triangle ABC and of triangles 퐴퐵퐻, 퐵퐶퐻 and 퐴퐶퐻. Theorem. The orthocenter, 퐻, centroid, 푀, and the circumcenter, 푂, of any triangle are collinear: all these three points lie on the same line, 푂퐻, which is called the Euler line of the triangle. The orthocenter divides the distance from the centroid to the circumcenter in 2: 1 ratio. B HC Proof. Note that the altitudes of the medial HA H triangle푀퐴푀퐵푀퐶 are the perpendicular MC bisectors of the triangle 퐴퐵퐶, so the M orthocenter of Δ 푀 푀 푀 is the M A 퐴 퐵 퐶 A circumcenter, 푂, of Δ 퐴퐵퐶. Now, using the O property that centroid divides medians of a HB MB triangle in a 2:1 ratio, we note that triangles C 퐵푀퐻 and 푀퐵푀푂 are similar, and homothetic with respect to point 푀, with the homothety coefficient 2. Theorem. The center of the nine-point-circle lies on the (Euler’s) line passing through orthocenter, centroid, and circumcenter, midway between the orthocenter and the circumcenter. B HC B1 Proof. Consider the figure. Note the colored HA triangle 퐴1퐵1퐶1, which is formed by MC H medians of triangles 퐴퐵퐻, 퐵퐻퐶 and 퐶퐻퐴, A O9 1 M and is therefore congruent to the medial M A A C1 triangle 푀퐴푀퐵푀퐶, but rotated 180 degrees. O The 9 points circle is the circumcircle for HB MB both triangles, which means that rotation C by 180 degrees about the center 푂9 of the 9 point circle moves Δ 푀퐴푀퐵푀퐶 onto Δ 퐴1퐵1퐶1, and the orthocenter, 푂, of the Δ 푀퐴푀퐵푀퐶 onto the orthocenter, 퐻, of the Δ 퐴1퐵1퐶1. More problems. Problem. Rectangle DEFG is inscribed in triangle ABC such that the side DE belongs to the base AB of the triangle, while points F and G belong to sides BC abd CA, respectively. What is the largest area of rectangle DEFG? C Solution. Notice similar triangles, 퐶퐷퐸~퐴퐵퐶, (a) wherefrom the vertical side of the rectangle is, x|CH| |퐷퐸| D H’ E |퐷퐺| = |퐸퐹| = |퐶퐻| − |퐶퐻′| = (1 − ) |퐶퐻|, |퐴퐵| (1-x)|CH| A B so that the area of the rectangle is, 푆퐷퐸퐹퐺 = G G’ H F |퐷퐸| S +S +S = S +S = x2 S +(1-x)2S >= ½ S |퐷퐸||퐷퐺| = |퐷퐸| (1 − ) |퐶퐻| = AGD EFB DEC AG’D DEC AB C ABC ABC |퐴퐵| x22 +(1-x) = 1-x+2x22 = ½+2(x-½) ) >= ½ |퐷퐸| |퐷퐸| |퐷퐸| |퐷퐸| (1 − ) |퐴퐵||퐶퐻| = (1 − ) 2푆퐴퐵퐶.