HMMT 二月 2010 · 几何 · 第 4 题

HMMT February 2010 — Geometry — Problem 4

[ 4 ] Let ABCD be an isosceles trapezoid such that AB = 10, BC = 15, CD = 28, and DA = 15. There is a point E such that 4 AED and 4 AEB have the same area and such that EC is minimal. Find EC .

解析

[ 4 ] Let ABCD be an isosceles trapezoid such that AB = 10, BC = 15, CD = 28, and DA = 15. There is a point E such that 4 AED and 4 AEB have the same area and such that EC is minimal. Find EC . 216 √ Answer: 145 B A M X D C Y E Geometry Subject Test The locus of points E such that [ AED ] = [ AEB ] forms a line, since area is a linear function of the 2 coordinates of E ; setting the areas equal gives a linear equation in the coordinates E . Note that A and M , the midpoint of DB , are on this line; A because both areas are 0, and M because the triangles share an altitude, and bases DM and M B are equal in length. Then AM is the set of points satisfying the area condition. The point E , then, is such that 4 AEC is a right angle (to make the distance minimal) and E lies on AM . Let X be the point of intersection of AM and CD . Then 4 AM B ∼ 4 XM D , and since M D = BM , they are in fact congruent. Thus DX = AB = 10, and XC = 18. Similarly, BX = 15, so ABXD is a DC − AB parallelogram. Let Y be the foot of the perpendicular from A to DC , so that DY = = 9. Then 2 √ √ √ 2 2 2 2 AY = AD − DY = 225 − 81 = 12. Then Y X = DX − DY = 1 and AX = AY + Y X = √ √ 144 + 1 = 145. Since both 4 AXY and 4 CXE have a right angle, and ∠ EXC and ∠ Y XA are CE CX 18 √ congruent because they are vertical angles, 4 AXY ∼ 4 CXE . Then = , so CE = 12 · = AY AX 145 216 √ . 145