Question

Segments $\overline{AB},$ $\overline{AC},$ and $\overline{AD}$ are edges of a cube and $\overline{AG}$ is a diagonal through the center of the cube. Point $P$ satisfies $BP = 60\sqrt{10},$ $CP = 60\sqrt{5},$ $DP = 120\sqrt{2},$ and $GP = 36\sqrt{7}.$ Find $AP.$

答案

Solution:

Let $A$ be the origin with $B = (s, 0, 0),$ $C = (0, s, 0),$ $D = (0, 0, s),$ $G = (s, s, s),$ and $P = (x, y, z).$ Expanding, $BP^2 = AP^2 - 2sx + s^2, \quad CP^2 = AP^2 - 2sy + s^2, \quad DP^2 = AP^2 - 2sz + s^2,$ while $GP^2 = AP^2 - 2s(x + y + z) + 3s^2.$ Therefore $BP^2 + CP^2 + DP^2 - GP^2 = 2\,AP^2,$ with every term involving $s$ or the coordinates of $P$ cancelling.

The given lengths yield $BP^2 = 36000,$ $CP^2 = 18000,$ $DP^2 = 28800,$ and $GP^2 = 9072,$ so $2\,AP^2 = 36000 + 18000 + 28800 - 9072 = 73728,$ giving $AP^2 = 36864$ and $AP = 192.$

2021 AIME I Problem 6

Solution:

Problem 6 in Other Years