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

Revision as of 18:28, 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 P 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 $PA=16$ . However, we scaled down everything by 12 so our answer is $16*12=\boxed{192}$ . ~JHawk0224

@@ Line 14: / Line 14: @@
 <cmath>3s^2-2s(x+y+z)+x^2+y^2+z^2=63</cmath>
 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>PA=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>PA=16</math>. However, we scaled down everything by 12 so our answer is <math>16*12=\boxed{192}</math>.
 ~JHawk0224

2021 AIME I (Problems • Answer Key • Resources)
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Difference between revisions of "2021 AIME I Problems/Problem 6"

Revision as of 18:28, 11 March 2021

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Solution

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