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2002 AIME II Problems/Problem 2

Revision as of 11:31, 28 December 2007 by Johan.gunardi (talk | contribs)

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?

Solution

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

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

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

So, PQR is an equilateral triangle. Let the side of the cube is $a$. $a\sqrt{2}=\sqrt{98}$

So, $a=7$, and hence the surface area=$6a^2=294$.

See also

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