Difference between revisions of "2021 AIME I Problems/Problem 6"
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==Solution== | ==Solution== | ||
− | <math> | + | First scale down the whole cube by 12. Let point M have coordinates <math>(x, y, z)</math>, A have coordinates <math>(0, 0, 0)</math>, and <math>s</math> be the side length. Then we have the equations |
+ | \begin{align*} | ||
+ | (s-x)^2+y^2+z^2&=250\ | ||
+ | x^2+(s-y)^2+z^2&=125\ | ||
+ | x^2+y^2+(s-z)^2&=200\ | ||
+ | (s-x)^2+(s-y)^2+(s-z)^2&=63 | ||
+ | \end{align*} | ||
+ | These simplify into | ||
+ | \begin{align*} | ||
+ | s^2+x^2+y^2+z^2-2sx&=250\ | ||
+ | s^2+x^2+y^2+z^2-2sy&=125\ | ||
+ | s^2+x^2+y^2+z^2-2sz&=200\ | ||
+ | 3s^2-2s(x+y+z)+x^2+y^2+z^2&=63 | ||
+ | \end{align*} | ||
+ | Adding the first three equations together, we get <math>3s^2-2s(x+y+z)+3(x^2+y^2+z^2)=575</math>. | ||
+ | Subtracting these, we get <math>2(x^2+y^2+z^2)=512</math>, so <math>x^2+y^2+z^2=256</math>. This means <math>AM=16</math>. However, we scaled down everything by 12 so our answer is <math>16*12=\boxed{196}</math>. | ||
+ | ~JHawk0224 | ||
==See also== | ==See also== | ||
{{AIME box|year=2021|n=I|num-b=5|num-a=7}} | {{AIME box|year=2021|n=I|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 16:20, 11 March 2021
Problem
Segments and are edges of a cube and is a diagonal through the center of the cube. Point satisfies and . What is ?
Solution
First scale down the whole cube by 12. Let point M have coordinates , A have coordinates , and be the side length. Then we have the equations
See also
2021 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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.