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

Revision as of 18:39, 12 December 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 1

First scale down the whole cube by $12$ . Let point $P$ have coordinates $(x, y, z)$ , point $A$ have coordinates $(0, 0, 0)$ , and $s$ be the side length. Then we have the equations $\begin{align*} (s-x)^2+y^2+z^2&=\left(5\sqrt{10}\right)^2, \\ x^2+(s-y)^2+z^2&=\left(5\sqrt{5}\right)^2, \\ x^2+y^2+(s-z)^2&=\left(10\sqrt{2}\right)^2, \\ (s-x)^2+(s-y)^2+(s-z)^2&=\left(3\sqrt{7}\right)^2. \end{align*}$ These simplify into $\begin{align*} s^2+x^2+y^2+z^2-2sx&=250, \\ s^2+x^2+y^2+z^2-2sy&=125, \\ s^2+x^2+y^2+z^2-2sz&=200, \\ 3s^2-2s(x+y+z)+x^2+y^2+z^2&=63. \end{align*}$ Adding the first three equations together, we get $3s^2-2s(x+y+z)+3(x^2+y^2+z^2)=575$ . Subtracting this from the fourth equation, 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

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, $2(x^2 + y^2 + z^2) + (s-x)^2 + (s-y)^2 + (s-z)^2 = 250 + 125 + 200.$ Subtracting the fourth equation gives $\begin{align*} 2(x^2 + y^2 + z^2) &= 575 - 63 \\ x^2 + y^2 + z^2 &= 256 \\ \sqrt{x^2 + y^2 + z^2} &= 16. \end{align*}$ Since point $A = (0,0,0), PA = 16$ , and since we scaled the answer is $16 \cdot 12 = \boxed{192}$ .

~Aaryabhatta1

Solution 3

Let E be the vertex of the cube such that ABED is a square. By the British Flag Theorem, we can easily we can show that $PA^2 + PE^2 = PB^2 + PD^2$ and $PA^2 + PG^2 = PC^2 + PE^2$ Hence, adding the two equations together, we get $2PA^2 + PG^2 = PB^2 + PC^2 + PD^2$ . Substituting in the values we know, we get $2PA^2 + 7\cdot 36^2 =10\cdot60^2 + 5\cdot 60^2 + 2\cdot 120^2$ .

Thus, we can solve for $PA$ , which ends up being $\boxed{192}$ .

(Lokman GÖKÇE)

Video Solution by Punxsutawney Phil

https://youtube.com/watch?v=vaRfI0l4s_8

@@ Line 23: / Line 23: @@
 ==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, <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,
+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*}
-(x^2 + y^2 + z^2) &= 575 - 63, \\
+(x^2 + y^2 + z^2) &= 575 - 63 \\
-x^2 + y^2 + z^2 &= 256, \\
+x^2 + y^2 + z^2 &= 256 \\
 \sqrt{x^2 + y^2 + z^2} &= 16.
 \end{align*}</cmath>

2021 AIME I (Problems • Answer Key • Resources)
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