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 | ||
| − | <cmath>(s-x)^2+y^2+z^2=250\\ | + | <cmath>(s-x)^2+y^2+z^2=250</cmath>\\ |
| − | x^2+(s-y)^2+z^2=125\\ | + | <cmath>x^2+(s-y)^2+z^2=125</cmath>\\ |
| − | x^2+y^2+(s-z)^2=200\\ | + | <cmath>x^2+y^2+(s-z)^2=200</cmath>\\ |
| − | (s-x)^2+(s-y)^2+(s-z)^2=63</cmath> | + | <cmath>(s-x)^2+(s-y)^2+(s-z)^2=63</cmath> |
These simplify into | These simplify into | ||
| − | <cmath>s^2+x^2+y^2+z^2-2sx=250\\ | + | <cmath>s^2+x^2+y^2+z^2-2sx=250</cmath>\\ |
| − | s^2+x^2+y^2+z^2-2sy=125\\ | + | <cmath>s^2+x^2+y^2+z^2-2sy=125</cmath>\\ |
| − | s^2+x^2+y^2+z^2-2sz=200\\ | + | <cmath>s^2+x^2+y^2+z^2-2sz=200</cmath>\\ |
| − | 3s^2-2s(x+y+z)+x^2+y^2+z^2=63</cmath> | + | <cmath>3s^2-2s(x+y+z)+x^2+y^2+z^2=63</cmath> |
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 16:22, 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
![\[(s-x)^2+y^2+z^2=250\]](http://latex.artofproblemsolving.com/d/8/5/d857d4aee3e6e4a3cc9d484310d223718a97e54f.png)
\\
![\[x^2+(s-y)^2+z^2=125\]](http://latex.artofproblemsolving.com/d/f/5/df5efcd8f97830602538ffc436b24bd144c5e875.png)
\\
![\[x^2+y^2+(s-z)^2=200\]](http://latex.artofproblemsolving.com/2/9/4/294becd38b99cc4844715682d872005ba3fd4fbe.png)
\\
![\[(s-x)^2+(s-y)^2+(s-z)^2=63\]](http://latex.artofproblemsolving.com/0/e/2/0e251df567906828220ba868e2956ff7a7b71ac2.png)
These simplify into
![\[s^2+x^2+y^2+z^2-2sx=250\]](http://latex.artofproblemsolving.com/9/4/9/949422295a962056bc265f2f2c0c510751e3349e.png)
\\
![\[s^2+x^2+y^2+z^2-2sy=125\]](http://latex.artofproblemsolving.com/a/2/b/a2b50eca5e8b87b679336c4d9d24f0fdaf30557f.png)
\\
![\[s^2+x^2+y^2+z^2-2sz=200\]](http://latex.artofproblemsolving.com/6/8/e/68e2a7e39da10022375680d715cf695819a2e3b7.png)
\\
![\[3s^2-2s(x+y+z)+x^2+y^2+z^2=63\]](http://latex.artofproblemsolving.com/0/f/c/0fc67e75216e609499f9cd8036c3d7398e8aff57.png)
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. 
