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

(Solution)
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==Solution==
 
==Solution==
<math>PA=\text{Pennsylvania}</math> lol
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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
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\begin{align*}
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(s-x)^2+y^2+z^2&=250\
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x^2+(s-y)^2+z^2&=125\
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x^2+y^2+(s-z)^2&=200\
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(s-x)^2+(s-y)^2+(s-z)^2&=63
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\end{align*}
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These simplify into
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\begin{align*}
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s^2+x^2+y^2+z^2-2sx&=250\
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s^2+x^2+y^2+z^2-2sy&=125\
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s^2+x^2+y^2+z^2-2sz&=200\
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3s^2-2s(x+y+z)+x^2+y^2+z^2&=63
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\end{align*}
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Adding the first three equations together, we get <math>3s^2-2s(x+y+z)+3(x^2+y^2+z^2)=575</math>.
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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>.
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~JHawk0224
  
 
==See also==
 
==See also==
 
{{AIME box|year=2021|n=I|num-b=5|num-a=7}}
 
{{AIME box|year=2021|n=I|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:20, 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 (sx)2+y2+z2=250x2+(sy)2+z2=125x2+y2+(sz)2=200(sx)2+(sy)2+(sz)2=63 These simplify into s2+x2+y2+z22sx=250s2+x2+y2+z22sy=125s2+x2+y2+z22sz=2003s22s(x+y+z)+x2+y2+z2=63 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 (ProblemsAnswer KeyResources)
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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