2021 AIME I Problem 6

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.

Answer

Put

A

at the origin with the cube’s edges along the axes and side length

s;

expand each squared distance in terms of

AP^2,

s,

and the coordinates of

P

The combination

BP^2 + CP^2 + DP^2 - GP^2

collapses to

2\,AP^2:

all terms involving

s

cancel

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.

Problem 6 in other years