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 17:20, 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
\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 
.
Subtracting these, we get 
, so 
. This means 
. However, we scaled down everything by 12 so our answer is 
.
~JHawk0224
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. 
