Difference between revisions of "2021 AIME I Problems/Problem 6"
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==Solution== | ==Solution== | ||
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 | 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*} | + | \[\begin{align*} |
(s-x)^2+y^2+z^2&=250\\ | (s-x)^2+y^2+z^2&=250\\ | ||
x^2+(s-y)^2+z^2&=125\\ | x^2+(s-y)^2+z^2&=125\\ | ||
x^2+y^2+(s-z)^2&=200\\ | x^2+y^2+(s-z)^2&=200\\ | ||
(s-x)^2+(s-y)^2+(s-z)^2&=63 | (s-x)^2+(s-y)^2+(s-z)^2&=63 | ||
| − | \end{align*} | + | \end{align*}\] |
These simplify into | These simplify into | ||
| − | \begin{align*} | + | \[\begin{align*} |
s^2+x^2+y^2+z^2-2sx&=250\\ | s^2+x^2+y^2+z^2-2sx&=250\\ | ||
s^2+x^2+y^2+z^2-2sy&=125\\ | s^2+x^2+y^2+z^2-2sy&=125\\ | ||
s^2+x^2+y^2+z^2-2sz&=200\\ | s^2+x^2+y^2+z^2-2sz&=200\\ | ||
3s^2-2s(x+y+z)+x^2+y^2+z^2&=63 | 3s^2-2s(x+y+z)+x^2+y^2+z^2&=63 | ||
| − | \end{align*} | + | \end{align*}\] |
Adding the first three equations together, we get <math>3s^2-2s(x+y+z)+3(x^2+y^2+z^2)=575</math>. | 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>. | 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>. | ||
Revision as of 16:21, 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. 
