Difference between revisions of "2021 AIME I Problems/Problem 6"
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==Solution 2 (Solution 1 with Slight Simplification)== | ==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, <cmath>2(x^2 + y^2 + z^2) + (s-x)^2 + (s-y)^2 + (s-z)^2 = 250 + 125 + 200.</cmath> Subtracting the fourth equation gives, <cmath>2(x^2 + y^2 + z^2) = 575 - 63 | + | 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, <cmath>2(x^2 + y^2 + z^2) + (s-x)^2 + (s-y)^2 + (s-z)^2 = 250 + 125 + 200.</cmath> Subtracting the fourth equation gives, |
+ | <cmath>\begin{align*} | ||
+ | 2(x^2 + y^2 + z^2) &= 575 - 63, \ | ||
+ | x^2 + y^2 + z^2 &= 256, \ | ||
+ | \sqrt{x^2 + y^2 + z^2} &= 16. | ||
+ | \end{align*}</cmath> | ||
+ | Since point <math>A = (0,0,0), PA = 16</math>, and since we scaled the answer is <math>16 \cdot 12 = \boxed{192}</math>. | ||
+ | |||
~Aaryabhatta1 | ~Aaryabhatta1 | ||
Revision as of 18:39, 12 December 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 . 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 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)
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=vaRfI0l4s_8
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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