HMMT 十一月 2012 · 冲刺赛 · 第 31 题
HMMT November 2012 — Guts Round — Problem 31
题目详情
英文原题
- [ 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 .
nn
解析
英文解析
- [ 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
√
√
through B and M is the line x = 0, y = − z 2. The plane through A , C , and D has equation 3
√3
√
z = 2 2 y + . The coordinates of Q are the coordinates of the intersection of this line and this 2
√3
1 6
√
plane. Equating the equations and solving for y and z , we see that y = − and z = , so the
5 35
√
1 6
√
coordinates of Q are (0 , − , ).
5 35
√ √
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