Math 531, Exam 2. 10/26/11. Name: • Read Problems Carefully

Home , Altitude (triangle), Euler line, Median (geometry)

Math 531, Exam 2. 10/26/11. Name: • Read problems carefully. Show all work. No notes, calculator, or text. • The exam is roughly 15 percent of the total grade. There are 100 points total. • Write your full name in the upper right corner of page 1. • Do problem 1 on p. 1, problem 2 on p. 2,...(or at least present your solutions in numerical order). • Circle or otherwise clearly identify your ﬁnal answer. 1. (40 points): Short answer/calculation: (a) (12 points): Suppose that UV and XY are chords of circle S that meet at P inside S. Suppose that PU = x, P V = x − 1,PX = 3,PY = 1. Find x. You can do this quickly by using the appropriate fact from §1H. You do not need to prove that any triangles are similar. Solution: We have (Thm. 1.35) PU · PV = x(x − 1) = 3 · 1 = PX · PY ⇐⇒ x2 − x − 3 = 0.

Since x > 0, it follows that √ 1 + 13 x = . 2 (b) (18 points): i. (3 points): Define what it means for quadrilateral ABCD to be cyclic. Solution: A quadrilateral is cyclic if and only if it can be inscribed in a circle. ii. (3 points): Give a simple condition (as in the homework) equivalent to the cyclic- ity of quadrilateral ABCD. No justification required. Solution: A quadrilateral is cyclic if and only if one pair of opposite interior angles are supplementary. iii. (12 points): Suppose that ABCD is cyclic with sides AB = 3,BC = 2,CD = 4, AD = 1, and suppose that the ratio of the diagonals AC and BD is 7 BD = · AC. 5 Use Ptolemy’s Theorem to find AC. (Set x = AC; write down and solve a quadratic equation for x.) Solution: We have BD = 7x/5. It follows from Ptolemy’s Theorem that 7x2 AB · CD + BC · AD = 3 · 4 + 2 · 1 = 14 = = AC · BD ⇐⇒ 70 = 7x2 √ 5 ⇐⇒ 10 = x2 ⇐⇒ x = 10 > 0. (c) (10 points): Suppose that 4ABC ∼ 4DEF , and that AB = 3. DE Suppose that 4ABC has area 63. What is the area of 4DEF ? Briefly explain. (How do areas scale with similar triangles?) Solution: We have

Area(4ABC) 63 Area(4ABC) = 32Area(4DEF ) ⇐⇒ Area(4DEF ) = = = 7. 9 9 (The hypothesis implies that base(4ABC) = 3 · base(4DEF ) and that height(4ABC) = 3 · height(4DEF ).)

2. (15 points): In 4ABC suppose that D, E, F satisfy

DE k BC,FE k DC, AF = 4, AD = 10.

Find the value of BD. Carefully justify your reasoning. (If you use the appropriate fact from §1H, you can do this quickly without the need to prove that any triangles are similar.)

Solution: We apply Lemma 1.29 ﬁrst to 4ADC: since FE k DC, we have AE AF 4 2 = = = . AC AD 10 5 We apply Lemma 1.29 to 4ABC: since DE k BC, we have AE AD 10 = = . AC AB AB Combining these facts, we see that 2 AE 10 = = ⇐⇒ AB = 25. 5 AC AB It follows that BD = AB − AD = 25 − 10 = 15.

3. (23 points): Let 4ABC be given.

(a) (13 points): Let G, H, N, and O have their usual meaning associated to 4ABC. i. (4 points): Give the name and a definition for each point. Solution: G : centroid. The centroid is the concurrence of the medians. H : orthocenter. The orthocenter is the concurrence of the altitudes. 2 N : nine-point center. The nine-point center is the center of the nine-point circle. O : circumcenter. The circumcenter is the concurrence of the perpendicular bisectors. ii. (3 points): Give the value of GH/GO. No justification required. Solution: We have GH 2GO = = 2. GO GO iii. (3 points): Give the value of NH/NO. No justification required. Solution: Since N is the midpoint of HO on the Euler line, we have NH = 1. NO

iv. (3 points): Suppose that Q is the midpoint of side AC opposite vertex B. What is the value of GQ/GB? No justiﬁcation required. Solution: We have GQ (1/3) · BQ 1 = = . GB (2/3) · BQ 2

(b) (10 points): Suppose that 4ABC is not a right triangle, and suppose that the Euler line of 4ABC passes through the vertex A. Brieﬂy explain why 4ABC must be isosceles. (Which important point(s) lie on the Euler line?) Solution: Since H is on the Euler line, the Euler line is the altitude from A; hence, it is perpendicular to BC. Since G is on the Euler line, the Euler line is a median from A; hence it bisects BC. It follows that the Euler line is the perpendicular bisector of BC. Since it passes through A, we must have AB = AC. This is essentially homework problem 2C.1.

4. (9 points): Let S be the nine-point circle of 4ABC. No justiﬁcation required. (a) (3 points): Give nine points on S. Solution: The nine points are: • Three Euler points. The Euler points are midpoints between vertices and the orthocenter. • Three midpoints of the sides of the triangles. • Three feet of the altitudes.

(b) (3 points): Give a diameter of S. Solution: Three diameters are obtained by joining an Euler point to the midpoint of the opposite side of the triangle. These are: PX, QY , RZ. 3 (c) (3 points): Express the radius of S in terms of R, the circumradius of 4ABC. Solution: The radius of the nine-point circle is R/2.

5. (13 points): Let 4ABC be isosceles with base BC, and let P be a point on BC. Suppose that

R1 = circumradius(4ABP ),R2 = circumradius(4ACP ).

Apply the extended law of sines to triangles 4ABP and 4ACP to show that R1 = R2. What do you know about the base angles of 4ABC?

Solution: Since 4ABC is isosceles, we note that ∠B = ∠C. We apply the extended law of sines to obtain: AP AP 2R = = = 2R . 1 sin B sin C 2

It follows that R1 = R2.