Difference between revisions of "2021 AIME I Problems/Problem 6"
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==Solution 3== | ==Solution 3== | ||
− | By | + | Let E be the vertex of the cube such that ABED is a square. |
+ | By the British Flag Theorem, we can easily we can show that | ||
<cmath>PA^2 + PE^2 = PB^2 + PD^2</cmath> | <cmath>PA^2 + PE^2 = PB^2 + PD^2</cmath> | ||
+ | and | ||
<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, adding the two equations together, we get <math>2PA^2 + PG^2 = PB^2 + PC^2 + PD^2</math>. Substituting in the values we know, we get <math>2PA^2 + 7\cdot 36^2 =10\cdot60^2 + 5\cdot 60^2 + 2\cdot 120^2 </math>. |
− | Thus <math>PA</math> | + | Thus, we can solve for <math>PA</math>, which ends up being <math>\boxed{192}</math>. |
(Lokman GÖKÇE) | (Lokman GÖKÇE) |
Revision as of 13:00, 14 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 ?
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=vaRfI0l4s_8
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
Let E be the vertex of the cube such that ABED is a square. By the British Flag Theorem, we can easily we can show that and Hence, adding the two equations together, we get . Substituting in the values we know, we get .
Thus, we can solve for , which ends up being .
(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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.