Difference between revisions of "2021 AIME I Problems/Problem 6"
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(Lokman GÖKÇE) | (Lokman GÖKÇE) | ||
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+ | ==Solution 4== | ||
+ | For all points <math>X</math> in space, define the function <math>f:\mathbb{R}^{3}\rightarrow\mathbb{R}</math> by <math>f(X)=PX^{2}-GX^{2}</math>. Then <math>f</math> is linear. Additionally, let <math>H=\frac{2A+G}{3}</math> be the centroid of <math>\triangle BCD</math>. Then since <math>f</math> is linear, | ||
+ | <cmath>\begin{align*} | ||
+ | 3f(H)=f(B)+f(C)+f(D)&=2f(A)+f(G) \ | ||
+ | \left(PB^{2}-GB^{2}\right)+\left(PC^{2}-GC^{2}\right)+\left(PD^{2}-GD^{2}\right)&=2\left(PA^{2}-GA^{2}\right)+PG^{2} \ | ||
+ | \left(60\sqrt{10}\right)^{2}-2x^{2}+\left(60\sqrt{5}\right)^{2}-2x^{2}+\left(120\sqrt{2}\right)^{2}-2x^{2}&=2PA^{2}-2\cdot 3x^{2}+\left(36\sqrt{7}\right)^{2}, | ||
+ | \end{align*}</cmath> | ||
+ | where <math>x</math> denotes the side length of the cube. Thus | ||
+ | <cmath>\begin{align*} | ||
+ | 36\text{,}000+18\text{,}000+28\text{,}800-6x^{2}&=2PA^{2}-6x^{2}+9072 \ | ||
+ | 82\text{,}800-6x^{2}&=2PA^{2}-6x^{2}+9072 \ | ||
+ | 73\text{,}728&=2PA^{2} \ | ||
+ | 36\text{,}864&=PA^{2} \ | ||
+ | PA&=\boxed{192}. | ||
+ | \end{align*}</cmath> | ||
==Video Solution by Punxsutawney Phil== | ==Video Solution by Punxsutawney Phil== |
Revision as of 18:44, 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 be the vertex of the cube such that 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)
Solution 4
For all points in space, define the function by . Then is linear. Additionally, let be the centroid of . Then since is linear, where denotes the side length of the cube. Thus
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.