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

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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>2(x^2 + y^2 + z^2) = 575 - 63</cmath> <cmath>x^2 + y^2 + z^2 = 256</cmath> <cmath>\sqrt{x^2 + y^2 + z^2} = 16.</cmath> Since point <math>A = (0,0,0), PA = 16</math>, and since we scaled the answer is <math>16 \cdot 12 = \boxed{192}</math>
 
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>2(x^2 + y^2 + z^2) = 575 - 63</cmath> <cmath>x^2 + y^2 + z^2 = 256</cmath> <cmath>\sqrt{x^2 + y^2 + z^2} = 16.</cmath> Since point <math>A = (0,0,0), PA = 16</math>, and since we scaled the answer is <math>16 \cdot 12 = \boxed{192}</math>
 
~Aaryabhatta1
 
~Aaryabhatta1
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==See also==
+
==Solution 3==
 +
 
 +
By Pythagorean Theorem, easly we can show that
 +
<cmath>PA^2 + PE^2 = PB^2 + PD^2</cmath>
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<cmath>PA^2 + PG^2 = PC^2 + PE^2</cmath>
 +
Hence, <math>2PA^2 + PG^2 = PB^2 + PC^2 + PD^2</math>. <math>2PA^2 + 7\cdot 36^2 =10\cdot60^2 + 5\cdot 60^2 + 2\cdot 120^2 </math>.
 +
Thus <math>PA</math> is <math>\boxed{192}</math>.
 +
 
 +
==See Also==
 
{{AIME box|year=2021|n=I|num-b=5|num-a=7}}
 
{{AIME box|year=2021|n=I|num-b=5|num-a=7}}
  

Revision as of 05:59, 13 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 1

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 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

By Pythagorean Theorem, easly we can show that \[PA^2 + PE^2 = PB^2 + PD^2\] \[PA^2 + PG^2 = PC^2 + PE^2\] Hence, $2PA^2 + PG^2 = PB^2 + PC^2 + PD^2$. $2PA^2 + 7\cdot 36^2 =10\cdot60^2 + 5\cdot 60^2 + 2\cdot 120^2$. Thus $PA$ is $\boxed{192}$.

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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