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Difference between revisions of "2021 AIME I Problems/Problem 6"

(→‎Solution)
(→‎Solution)
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==Solution==
 
==Solution==
 
First scale down the whole cube by 12. Let point M have coordinates <math>(x, y, z)</math>, A have coordinates <math>(0, 0, 0)</math>, and <math>s</math> be the side length. Then we have the equations
 
First scale down the whole cube by 12. Let point M have coordinates <math>(x, y, z)</math>, A have coordinates <math>(0, 0, 0)</math>, and <math>s</math> be the side length. Then we have the equations
−
\[\begin{align*}
+
<cmath>(s-x)^2+y^2+z^2&=250\\
−
(s-x)^2+y^2+z^2&=250\\
 
 
x^2+(s-y)^2+z^2&=125\\
 
x^2+(s-y)^2+z^2&=125\\
 
x^2+y^2+(s-z)^2&=200\\
 
x^2+y^2+(s-z)^2&=200\\
−
(s-x)^2+(s-y)^2+(s-z)^2&=63
+
(s-x)^2+(s-y)^2+(s-z)^2&=63</cmath>
−
\end{align*}\]
 
 
These simplify into
 
These simplify into
−
\[\begin{align*}
+
<cmath>s^2+x^2+y^2+z^2-2sx&=250\\
−
s^2+x^2+y^2+z^2-2sx&=250\\
 
 
s^2+x^2+y^2+z^2-2sy&=125\\
 
s^2+x^2+y^2+z^2-2sy&=125\\
 
s^2+x^2+y^2+z^2-2sz&=200\\
 
s^2+x^2+y^2+z^2-2sz&=200\\
−
3s^2-2s(x+y+z)+x^2+y^2+z^2&=63
+
3s^2-2s(x+y+z)+x^2+y^2+z^2&=63</cmath>
−
\end{align*}\]
 
 
Adding the first three equations together, we get <math>3s^2-2s(x+y+z)+3(x^2+y^2+z^2)=575</math>.
 
Adding the first three equations together, we get <math>3s^2-2s(x+y+z)+3(x^2+y^2+z^2)=575</math>.
 
Subtracting these, we get <math>2(x^2+y^2+z^2)=512</math>, so <math>x^2+y^2+z^2=256</math>. This means <math>AM=16</math>. However, we scaled down everything by 12 so our answer is <math>16*12=\boxed{196}</math>.
 
Subtracting these, we get <math>2(x^2+y^2+z^2)=512</math>, so <math>x^2+y^2+z^2=256</math>. This means <math>AM=16</math>. However, we scaled down everything by 12 so our answer is <math>16*12=\boxed{196}</math>.

Revision as of 16:21, 11 March 2021

Problem

Segments $\overline{AB}, \overline{AC},$ and $\overline{AD}$ are edges of a cube and $\overline{AG}$ is a diagonal through the center of the cube. Point $P$ satisfies $PB=60\sqrt{10}, PC=60\sqrt{5}, PD=120\sqrt{2},$ and $PG=36\sqrt{7}$. What is $PA$?

Solution

First scale down the whole cube by 12. Let point M have coordinates $(x, y, z)$, A have coordinates $(0, 0, 0)$, and $s$ be the side length. Then we have the equations 

\[(s-x)^2+y^2+z^2&=250\\
x^2+(s-y)^2+z^2&=125\\
x^2+y^2+(s-z)^2&=200\\
(s-x)^2+(s-y)^2+(s-z)^2&=63\] (Error compiling LaTeX. Unknown error_msg)

These simplify into 

\[s^2+x^2+y^2+z^2-2sx&=250\\
s^2+x^2+y^2+z^2-2sy&=125\\
s^2+x^2+y^2+z^2-2sz&=200\\
3s^2-2s(x+y+z)+x^2+y^2+z^2&=63\] (Error compiling LaTeX. Unknown error_msg)

Adding the first three equations together, we get $3s^2-2s(x+y+z)+3(x^2+y^2+z^2)=575$. Subtracting these, we get $2(x^2+y^2+z^2)=512$, so $x^2+y^2+z^2=256$. This means $AM=16$. However, we scaled down everything by 12 so our answer is $16*12=\boxed{196}$. ~JHawk0224

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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