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

m (→‎Solution 1: LaTeX'ed the solution.)
m (→‎Solution 1)
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==Solution 1==
 
==Solution 1==
−
First scale down the whole cube by <math>12</math>. Let point <math>P</math> have coordinates <math>(x, y, z)</math>, <math>A</math> 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 <math>12</math>. Let point <math>P</math> have coordinates <math>(x, y, z)</math>, point <math>A</math> 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</cmath>
+
<cmath>\begin{align*}
−
<cmath>x^2+(s-y)^2+z^2=(5\sqrt{5})^2</cmath>
+
(s-x)^2+y^2+z^2&=\left(5\sqrt{10}\right)^2, \\
−
<cmath>x^2+y^2+(s-z)^2=(10\sqrt{2})^2</cmath>
+
x^2+(s-y)^2+z^2&=\left(5\sqrt{5}\right)^2, \\
−
<cmath>(s-x)^2+(s-y)^2+(s-z)^2=(3\sqrt{7})^2</cmath>
+
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*}</cmath>
 
These simplify into
 
These simplify into
−
<cmath>s^2+x^2+y^2+z^2-2sx=250</cmath>
+
<cmath>\begin{align*}
−
<cmath>s^2+x^2+y^2+z^2-2sy=125</cmath>
+
s^2+x^2+y^2+z^2-2sx&=250, \\
−
<cmath>s^2+x^2+y^2+z^2-2sz=200</cmath>
+
s^2+x^2+y^2+z^2-2sy&=125, \\
−
<cmath>3s^2-2s(x+y+z)+x^2+y^2+z^2=63</cmath>
+
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*}</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>.
 
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 this from the fourth equation, 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 <math>12</math> so our answer is <math>16*12=\boxed{192}</math>.
 
Subtracting this from the fourth equation, 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 <math>12</math> so our answer is <math>16*12=\boxed{192}</math>.

Revision as of 19:37, 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, \[2(x^2 + y^2 + z^2) = 575 - 63\] \[x^2 + y^2 + z^2 = 256\] \[\sqrt{x^2 + y^2 + z^2} = 16.\] 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

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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