2021 AIME I Problems/Problem 6

Revision as of 16:20, 11 March 2021 by Jhawk0224 (talk | contribs) (Solution)

Problem

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 $PB=60\sqrt{10}, PC=60\sqrt{5}, PD=120\sqrt{2},$ and $PG=36\sqrt{7}$. What is $PA$?

Solution

First scale down the whole cube by 12. Let point M have coordinates $(x, y, z)$, A have coordinates $(0, 0, 0)$, and $s$ be the side length. Then we have the equations (sx)2+y2+z2=250x2+(sy)2+z2=125x2+y2+(sz)2=200(sx)2+(sy)2+(sz)2=63 These simplify into s2+x2+y2+z22sx=250s2+x2+y2+z22sy=125s2+x2+y2+z22sz=2003s22s(x+y+z)+x2+y2+z2=63 Adding the first three equations together, we get $3s^2-2s(x+y+z)+3(x^2+y^2+z^2)=575$. Subtracting these, we get $2(x^2+y^2+z^2)=512$, so $x^2+y^2+z^2=256$. This means $AM=16$. However, we scaled down everything by 12 so our answer is $16*12=\boxed{196}$. ~JHawk0224

See also

2021 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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