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
<cmath>(s-x)^2+y^2+z^2=(5\sqrt{10})^2=250</cmath>
+
<cmath>(s-x)^2+y^2+z^2=(5\sqrt{10})^2</cmath>
<cmath>x^2+(s-y)^2+z^2=(5\sqrt{5})^2=125</cmath>
+
<cmath>x^2+(s-y)^2+z^2=(5\sqrt{5})^2</cmath>
<cmath>x^2+y^2+(s-z)^2=(10\sqrt{2})^2=200</cmath>
+
<cmath>x^2+y^2+(s-z)^2=(10\sqrt{2})^2</cmath>
<cmath>(s-x)^2+(s-y)^2+(s-z)^2=(3\sqrt{7})^2=63</cmath>
+
<cmath>(s-x)^2+(s-y)^2+(s-z)^2=(3\sqrt{7})^2</cmath>
 
These simplify into
 
These simplify into
 
<cmath>s^2+x^2+y^2+z^2-2sx=250</cmath>
 
<cmath>s^2+x^2+y^2+z^2-2sx=250</cmath>

Revision as of 17:25, 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 \[(s-x)^2+y^2+z^2=(5\sqrt{10})^2\] \[x^2+(s-y)^2+z^2=(5\sqrt{5})^2\] \[x^2+y^2+(s-z)^2=(10\sqrt{2})^2\] \[(s-x)^2+(s-y)^2+(s-z)^2=(3\sqrt{7})^2\] These simplify into \[s^2+x^2+y^2+z^2-2sx=250\] \[s^2+x^2+y^2+z^2-2sy=125\] \[s^2+x^2+y^2+z^2-2sz=200\] \[3s^2-2s(x+y+z)+x^2+y^2+z^2=63\] 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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