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 ==
-The diagram shows twenty congruent circles arranged in three rows and enclosed in a rectangle.  The circles are tangent to one another and to the sides of the rectangle as shown in the diagram.  The ratio of the longer dimension of the rectangle to the shorter dimension can be written as <math>\frac{1}{2}\left(\sqrt{p}-q\right),</math> where <math>p</math> and <math>q</math> are positive integers.  Find <math>p+q.</math>
+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?
-{{image}} (the image needed is image 6601 in the Album.)
 == 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}}

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