# 2021 AIME I Problems/Problem 6

## 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 1

First scale down the whole cube by 12. Let point P have coordinates $(x, y, z)$, A have coordinates $(0, 0, 0)$, and $s$ be the side length. Then we have the equations $$(s-x)^2+y^2+z^2=(5\sqrt{10})^2$$ $$x^2+(s-y)^2+z^2=(5\sqrt{5})^2$$ $$x^2+y^2+(s-z)^2=(10\sqrt{2})^2$$ $$(s-x)^2+(s-y)^2+(s-z)^2=(3\sqrt{7})^2$$ These simplify into $$s^2+x^2+y^2+z^2-2sx=250$$ $$s^2+x^2+y^2+z^2-2sy=125$$ $$s^2+x^2+y^2+z^2-2sz=200$$ $$3s^2-2s(x+y+z)+x^2+y^2+z^2=63$$ Adding the first three equations together, we get $3s^2-2s(x+y+z)+3(x^2+y^2+z^2)=575$. Subtracting this from the fourth equation, we get $2(x^2+y^2+z^2)=512$, so $x^2+y^2+z^2=256$. This means $PA=16$. However, we scaled down everything by 12 so our answer is $16*12=\boxed{192}$. ~JHawk0224

## Solution 2 (Solution 1 with Slight Simplification)

Once the equations for the distance between point P and the vertices of the cube have been written. We can add the first, second, and third to receive, $$2(x^2 + y^2 + z^2) + (s-x)^2 + (s-y)^2 + (s-z)^2 = 250 + 125 + 200.$$ Subtracting the fourth equation gives, $$2(x^2 + y^2 + z^2) = 575 - 63$$ $$x^2 + y^2 + z^2 = 256$$ $$\sqrt{x^2 + y^2 + z^2} = 16.$$ Since point $A = (0,0,0), PA = 16$, and since we scaled the answer is $16 \cdot 12 = \boxed{192}$ ~Aaryabhatta1

## Solution 3

Let E be the vertex of the cube such that ABED is a square. By the British Flag Theorem, we can easily we can show that $$PA^2 + PE^2 = PB^2 + PD^2$$ and $$PA^2 + PG^2 = PC^2 + PE^2$$ Hence, adding the two equations together, we get $2PA^2 + PG^2 = PB^2 + PC^2 + PD^2$. Substituting in the values we know, we get $2PA^2 + 7\cdot 36^2 =10\cdot60^2 + 5\cdot 60^2 + 2\cdot 120^2$.

Thus, we can solve for $PA$, which ends up being $\boxed{192}$.

(Lokman GÖKÇE)

## See Also

 2021 AIME I (Problems • Answer Key • Resources) Preceded byProblem 5 Followed byProblem 7 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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