2002 AIME II Problems/Problem 2

Problem

Three vertices of a cube are $P=(7,12,10)$ , $Q=(8,8,1)$ , and $R=(11,3,9)$ . What is the surface area of the cube?

$PQ=\sqrt{(8-7)^2+(8-12)^2+(1-10)^2}=\sqrt{98}$

$PR=\sqrt{(11-7)^2+(3-12)^2+(9-10)^2}=\sqrt{98}$

$QR=\sqrt{(11-8)^2+(3-8)^2+(9-1)^2}=\sqrt{98}$

So, $PQR$ is an equilateral triangle. Let the side of the cube be $a$ .

$a\sqrt{2}=\sqrt{98}$

So, $a=7$ , and hence the surface area is $6a^2=\framebox{294}$ .

2002 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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