HMMT 二月 2010 · GEN2 赛 · 第 6 题

英文解析

6. [ 5 ] Let ABCD be an isosceles trapezoid such that AB = 10, BC = 15, CD = 28, and DA = 15. Thereis a point E such that 4 AED and 4 AEB have the same area and such that EC is minimal. Find
   EC .
   √216
   Answer:
   B145
   DXMA
   The locus of points E such that [ AED ] = [ AEB ] forms a line, since area is a linear function of the eYCcoordinates of E ; setting the areas equal gives a linear equation in the coordinates E . Note that A2
   and M , the midpoint of DB , are on this line; A because both areas are 0, and M because the trianglesshare an altitude, and bases DM and M B are equal in length. Then AM is the set of points satisfyingthe area condition. The point E , then, is such that 4 AEC is a right angle (to make the distanceminimal) 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 − ABparallelogram. Let Y be the foot of the perpendicular from A to DC , so that DY = = 9. Then
   The notation [ A A . . . A ] means the area of polygon A A . . . A .22
   1 2 n 1 2 n
   General Test, Part 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
   216145
   √
   .
   20 20145