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 | + | 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*} | <cmath>\begin{align*} | ||
2(x^2 + y^2 + z^2) &= 575 - 63 \\ | 2(x^2 + y^2 + z^2) &= 575 - 63 \\ | ||
Line 36: | Line 36: | ||
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 | ||
<cmath>PA^2 + PG^2 = PC^2 + PE^2</cmath> | <cmath>PA^2 + PG^2 = PC^2 + PE^2</cmath> | ||
− | 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>. | + | Hence, by 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, we can solve for <math>PA</math>, which ends up being <math>\boxed{192}</math>. | Thus, we can solve for <math>PA</math>, which ends up being <math>\boxed{192}</math>. | ||
(Lokman GÖKÇE) | (Lokman GÖKÇE) | ||
− | [[File:2021 AIME I 6b.jpg]] | + | [[File:2021 AIME I 6b.jpg|size:100px]] |
==Solution 4== | ==Solution 4== | ||
Line 63: | Line 63: | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
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==See Also== | ==See Also== |
Latest revision as of 23:50, 11 January 2024
Contents
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.