Difference between revisions of "2021 AIME I Problems/Problem 6"
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<cmath>PA^2 + PG^2 = PC^2 + PE^2</cmath> | <cmath>PA^2 + PG^2 = PC^2 + PE^2</cmath> | ||
Hence, <math>2PA^2 + PG^2 = PB^2 + PC^2 + PD^2</math>. <math>2PA^2 + 7\cdot 36^2 =10\cdot60^2 + 5\cdot 60^2 + 2\cdot 120^2 </math>. | Hence, <math>2PA^2 + PG^2 = PB^2 + PC^2 + PD^2</math>. <math>2PA^2 + 7\cdot 36^2 =10\cdot60^2 + 5\cdot 60^2 + 2\cdot 120^2 </math>. | ||
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Thus <math>PA</math> is <math>\boxed{192}</math>. | Thus <math>PA</math> is <math>\boxed{192}</math>. | ||
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(Lokman GÖKÇE) | (Lokman GÖKÇE) | ||
Revision as of 06:01, 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
Solution 3
By Pythagorean Theorem, easly we can show that Hence, . .
Thus is .
(Lokman GÖKÇE)
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 |
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