HMMT 十一月 2012 · 冲刺赛 · 第 31 题

HMMT November 2012 — Guts Round — Problem 31

[ 17 ] Let ABCD be a regular tetrahedron with AB = 1, and let P be the center of face BCD . Let M be the midpoint of segment AP . Ray BM meets face ACD at Q . Determine the area of triangle QDC . n n

解析

[ 17 ] √ √ √ 3 3 6 3 Answer: We place the points in the coordinate plane. We let A = (0 , 0 , ), B = (0 , , 0), 20 3 3 √ √ √ 1 3 1 3 6 C = ( − , − , 0), and D = ( , , 0). The point P is the origin, while M is (0 , 0 , ). The line 2 6 2 6 6 √ √ 3 through B and M is the line x = 0, y = − z 2. The plane through A , C , and D has equation 3 √ √ 2 z = 2 2 y + . The coordinates of Q are the coordinates of the intersection of this line and this 3 √ 1 6 √ plane. Equating the equations and solving for y and z , we see that y = − and z = , so the 5 5 3 √ 1 6 √ coordinates of Q are (0 , − , ). 5 5 3 √ √ 3 3 3 Let N be the midpoint of CD , which has coordinates (0 , − , 0). By the distance formula, QN = . 6 10 √ QN · CD 3 3 Thus, the area of QCD is = . 2 20