Difference between revisions of "2021 AIME I Problems/Problem 6"
MRENTHUSIASM (talk | contribs) m (Changed the title of the solutions.) |
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<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 | + | Subtracting this from the fourth equation, 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>PA=16</math>. However, we scaled down everything by 12 so our answer is <math>16*12=\boxed{192}</math>. |
~JHawk0224 | ~JHawk0224 | ||
Revision as of 00:05, 13 March 2021
Contents
[hide]Problem
Segments and are edges of a cube and is a diagonal through the center of the cube. Point satisfies and . What is ?
Solution 1
First scale down the whole cube by 12. Let point P 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 this from the fourth equation, we get , so . This means . However, we scaled down everything by 12 so our answer is . ~JHawk0224
Solution 2 (Solution 1 with slight simplification)
Once the equations for the distance between point P and the vertices of the cube have been written. We can add the first, second, and third to receive, Subtracting the fourth equation gives, Since point , and since we scaled the answer is ~Aaryabhatta1
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.