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

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==Problem==
 
==Problem==
Segments <math>\overline{AB}, \overline{AC},</math> and <math>\overline{AD}</math> are edges of a cube and <math>\overline{AG}</math> is a diagonal through the center of the cube. Point <math>P</math> satisfies <math>BP=60\sqrt{10}, CP=60\sqrt{5}, DP=120\sqrt{2},</math> and <math>GP=36\sqrt{7}</math>. Find <math>AP.</math>
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Segments <math>\overline{AB}, \overline{AC},</math> and <math>\overline{AD}</math> are edges of a cube and <math>\overline{AG}</math> is a diagonal through the center of the cube. Point <math>P</math> satisfies <math>PB=60\sqrt{10}, PC=60\sqrt{5}, PD=120\sqrt{2},</math> and <math>PG=36\sqrt{7}</math>. What is <math>PA</math>?
  
 
==Solution==
 
==Solution==
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<math>PA=\text{Pennsylvania}</math> lol
  
 
==See also==
 
==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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:49, 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

$PA=\text{Pennsylvania}$ lol

See also

2021 AIME I (ProblemsAnswer KeyResources)
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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