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

(→‎Solution 3)
(99 percent of the time, we use O for the center)
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==Solution 4==
 
==Solution 4==
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For all points <math>X</math> in space, define the function <math>f:\mathbb{R}^{3}\rightarrow\mathbb{R}</math> by <math>f(X)=PX^{2}-GX^{2}</math>. Then <math>f</math> is linear. Additionally, let <math>H=\frac{2A+G}{3}</math> be the centroid of <math>\triangle BCD</math>. Then since <math>f</math> is linear,
+
For all points <math>X</math> in space, define the function <math>f:\mathbb{R}^{3}\rightarrow\mathbb{R}</math> by <math>f(X)=PX^{2}-GX^{2}</math>. Then <math>f</math> is linear; let <math>O=\frac{2A+G}{3}</math> be the center of <math>\triangle BCD</math>. Then since <math>f</math> is linear,
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
−
3f(H)=f(B)+f(C)+f(D)&=2f(A)+f(G) \\  
+
3f(O)=f(B)+f(C)+f(D)&=2f(A)+f(G) \\  
 
\left(PB^{2}-GB^{2}\right)+\left(PC^{2}-GC^{2}\right)+\left(PD^{2}-GD^{2}\right)&=2\left(PA^{2}-GA^{2}\right)+PG^{2} \\
 
\left(PB^{2}-GB^{2}\right)+\left(PC^{2}-GC^{2}\right)+\left(PD^{2}-GD^{2}\right)&=2\left(PA^{2}-GA^{2}\right)+PG^{2} \\
 
\left(60\sqrt{10}\right)^{2}-2x^{2}+\left(60\sqrt{5}\right)^{2}-2x^{2}+\left(120\sqrt{2}\right)^{2}-2x^{2}&=2PA^{2}-2\cdot 3x^{2}+\left(36\sqrt{7}\right)^{2},
 
\left(60\sqrt{10}\right)^{2}-2x^{2}+\left(60\sqrt{5}\right)^{2}-2x^{2}+\left(120\sqrt{2}\right)^{2}-2x^{2}&=2PA^{2}-2\cdot 3x^{2}+\left(36\sqrt{7}\right)^{2},

Revision as of 18:06, 9 June 2022

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) 2021 AIME I 6b.jpg

Solution 4

For all points $X$ in space, define the function $f:\mathbb{R}^{3}\rightarrow\mathbb{R}$ by $f(X)=PX^{2}-GX^{2}$. Then $f$ is linear; let $O=\frac{2A+G}{3}$ be the center of $\triangle BCD$. Then since $f$ is linear, \begin{align*} 3f(O)=f(B)+f(C)+f(D)&=2f(A)+f(G) \\ \left(PB^{2}-GB^{2}\right)+\left(PC^{2}-GC^{2}\right)+\left(PD^{2}-GD^{2}\right)&=2\left(PA^{2}-GA^{2}\right)+PG^{2} \\ \left(60\sqrt{10}\right)^{2}-2x^{2}+\left(60\sqrt{5}\right)^{2}-2x^{2}+\left(120\sqrt{2}\right)^{2}-2x^{2}&=2PA^{2}-2\cdot 3x^{2}+\left(36\sqrt{7}\right)^{2}, \end{align*} where $x$ denotes the side length of the cube. Thus \begin{align*} 36\text{,}000+18\text{,}000+28\text{,}800-6x^{2}&=2PA^{2}-6x^{2}+9072 \\ 82\text{,}800-6x^{2}&=2PA^{2}-6x^{2}+9072 \\ 73\text{,}728&=2PA^{2} \\ 36\text{,}864&=PA^{2} \\ PA&=\boxed{192}. \end{align*}

Video Solution by Punxsutawney Phil

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

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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