Difference between revisions of "2021 AIME I Problems/Problem 6"
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Let <math>E</math> be the vertex of the cube such that <math>ABED</math> is a square. | Let <math>E</math> be the vertex of the cube such that <math>ABED</math> is a square. | ||
− | + | Using the [[British Flag Theorem]], we can easily show that | |
<cmath>PA^2 + PE^2 = PB^2 + PD^2</cmath> | <cmath>PA^2 + PE^2 = PB^2 + PD^2</cmath> | ||
and | and | ||
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\end{align*}</cmath> | \end{align*}</cmath> | ||
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==See Also== | ==See Also== |
Latest revision as of 23:50, 11 January 2024
Contents
[hide]Problem
Segments and are edges of a cube and is a diagonal through the center of the cube. Point satisfies , , , and . Find
Solution 1
First scale down the whole cube by . Let point have coordinates , point 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 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 be the vertex of the cube such that is a square. Using the British Flag Theorem, we can easily show that and Hence, by adding the two equations together, we get . Substituting in the values we know, we get .
Thus, we can solve for , which ends up being .
Solution 4
For all points in space, define the function by . Then is linear; let be the center of . Then since is linear, where denotes the side length of the cube. Thus
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.