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

Revision as of 11:31, 28 December 2007

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

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

@@ Line 3: / Line 3: @@
 == 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, PQR is an equilateral triangle. Let the side of the cube is <math>a</math>.
+<math>a\sqrt{2}=\sqrt{98}</math>
+So, <math>a=7</math>, and hence the surface area=<math>6a^2=294</math>.
 == See also ==
 * [[2002 AIME II Problems/Problem 1 | Previous problem]]
 * [[2002 AIME II Problems/Problem 3 | Next problem]]
 * [[2002 AIME II Problems]]