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

Latest revision as of 02:32, 6 December 2019

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 $6a^2=\framebox{294}$ .

@@ Line 1: / Line 1: @@
 == Problem ==
-Three 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?
+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>.
 == See also ==
-* [[2002 AIME II Problems/Problem 1 | Previous problem]]
+{{AIME box|year=2002|n=II|num-b=1|num-a=3}}
-* [[2002 AIME II Problems/Problem 3 | Next problem]]
-* [[2002 AIME II Problems]]
+[[Category: Intermediate Geometry Problems]]
+{{MAA Notice}}

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