Difference between revisions of "2002 AIME II Problems/Problem 2"
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== Problem == | == Problem == | ||
+ | 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 == | ||
− | {{ | + | <math>PQ=\sqrt{(8-7)^2+(8-12)^2+(1-10)^2}=\sqrt{98}</math> |
+ | |||
+ | <math>PR=\sqrt{(11-7)^2+(3-12)^2+(9-10)^2}=\sqrt{98}</math> | ||
+ | |||
+ | <math>QR=\sqrt{(11-8)^2+(3-8)^2+(9-1)^2}=\sqrt{98}</math> | ||
+ | |||
+ | So, <math>PQR</math> is an equilateral triangle. Let the side of the cube be <math>a</math>. | ||
+ | |||
+ | <math>a\sqrt{2}=\sqrt{98}</math> | ||
+ | |||
+ | 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 == | ||
− | + | {{AIME box|year=2002|n=II|num-b=1|num-a=3}} | |
− | + | ||
− | + | [[Category: Intermediate Geometry Problems]] | |
+ | {{MAA Notice}} |
Latest revision as of 03:32, 6 December 2019
Problem
Three vertices of a cube are ,
, and
. What is the surface area of the cube?
Solution
So, is an equilateral triangle. Let the side of the cube be
.
So, , and hence the surface area is
.
See also
2002 AIME II (Problems • Answer Key • Resources) | ||
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.