Question

Los segmentos $\overline{AB},$ $\overline{AC},$ y $\overline{AD}$ son aristas de un cubo y $\overline{AG}$ es una diagonal que pasa por el centro del cubo. El punto $P$ satisface $BP = 60\sqrt{10},$ $CP = 60\sqrt{5},$ $DP = 120\sqrt{2},$ y $GP = 36\sqrt{7}.$ Halle $AP.$

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 $AP.$

Respuesta

Solución:

Sea $A$ el origen con $B = (s, 0, 0),$ $C = (0, s, 0),$ $D = (0, 0, s),$ $G = (s, s, s),$ y $P = (x, y, z).$ Al expandir, $\begin{aligned} BP^2 &= AP^2 - 2sx + s^2, \\ CP^2 &= AP^2 - 2sy + s^2, \\ DP^2 &= AP^2 - 2sz + s^2, \end{aligned}$ mientras que $GP^2 = AP^2 - 2s(x + y + z)$ $+ 3s^2.$ Por lo tanto $\begin{aligned} &BP^2 + CP^2 \\ &\quad {}+ DP^2 - GP^2 \\ &= 2\,AP^2, \end{aligned}$ con la cancelación de todo término que contiene $s$ o las coordenadas de $P$ .

Las longitudes dadas dan $BP^2 = 36000,$ $CP^2 = 18000,$ $DP^2 = 28800,$ y $GP^2 = 9072,$ así que $\begin{aligned} 2\,AP^2 &= 36000 + 18000 \\ &\quad {}+ 28800 - 9072 \\ &= 73728, \end{aligned}$ lo que da $AP^2 = 36864$ y $AP = 192.$

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, $\begin{aligned} BP^2 &= AP^2 - 2sx + s^2, \\ CP^2 &= AP^2 - 2sy + s^2, \\ DP^2 &= AP^2 - 2sz + s^2, \end{aligned}$ while $GP^2 = AP^2 - 2s(x + y + z)$ $+ 3s^2.$ Therefore $\begin{aligned} &BP^2 + CP^2 \\ &\quad {}+ DP^2 - GP^2 \\ &= 2\,AP^2, \end{aligned}$ 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 $\begin{aligned} 2\,AP^2 &= 36000 + 18000 \\ &\quad {}+ 28800 - 9072 \\ &= 73728, \end{aligned}$ giving $AP^2 = 36864$ and $AP = 192.$

2021 AIME I Problema 6

Solución:

El Problema 6 en otros años