Difference between revisions of "2021 AIME I Problems/Problem 6"
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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 \ \ \ These simplify into \ \ \ 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.