Difference between revisions of "2002 AIME II Problems/Problem 2"

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== Problem ==
 
== Problem ==
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Three [[vertex|vertices]] of a [[cube]] are <math>P=(7,12,10)</math>, <math>Q=(8,8,1)</math>, and <math>R=(11,3,9)</math>. What is the [[surface area]] of the cube?
  
 
== Solution ==
 
== Solution ==
{{solution}}
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<math>PQ=\sqrt{(8-7)^2+(8-12)^2+(1-10)^2}=\sqrt{98}</math>
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<math>PR=\sqrt{(11-7)^2+(3-12)^2+(9-10)^2}=\sqrt{98}</math>
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<math>QR=\sqrt{(11-8)^2+(3-8)^2+(9-1)^2}=\sqrt{98}</math>
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So, <math>PQR</math> is an equilateral triangle. Let the side of the cube be <math>a</math>.
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<math>a\sqrt{2}=\sqrt{98}</math>
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So, <math>a=7</math>, and hence the surface area is <math>6a^2=\framebox{294}</math>.
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== See also ==
 
== See also ==
* [[2002 AIME II Problems/Problem 1 | Previous problem]]
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{{AIME box|year=2002|n=II|num-b=1|num-a=3}}
* [[2002 AIME II Problems/Problem 3 | Next problem]]
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* [[2002 AIME II Problems]]
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[[Category: Intermediate Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 03:32, 6 December 2019

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 be $a$.

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

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

See also

2002 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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