Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2002 AIME II Problems/Problem 2

Revision as of 12:31, 28 December 2007 by Johan.gunardi (talk | contribs)

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

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

See also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2002_AIME_II_Problems/Problem_2&oldid=21637"