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Difference between revisions of "2021 AMC 12A Problems/Problem 9"

(Created page with "==Problem== These problems will not be posted until the 2021 AMC12A is released on Thursday, February 4, 2021. ==Solution== The solutions will be posted once the problems are...")
 
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==Problem==
 
==Problem==
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These problems will not be posted until the 2021 AMC12A is released on Thursday, February 4, 2021.
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Triangle <math>ABC</math> lies in a plane with <math>AB=13</math>, <math>AC=14</math>, and <math>BC=15</math>. For any point <math>X</math> in the plane of <math>\triangle ABC</math>, let <math>f(X)</math> denote the sum of the three distances from <math>X</math> to the three vertices of <math>\triangle ABC</math>. Let <math>P</math> be the unique point in the plane of <math>\triangle ABC</math> where <math>f(X)</math> is minimized. Then <math>AP^2+BP^2+CP^2=\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is the value of <math>m+n</math>?
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==Solution==
 
==Solution==
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The solutions will be posted once the problems are posted.
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Point <math>P</math> is the [[Brocard point|Brocard point]] of <math>\triangle ABC</math>, where <math>\angle APB = \angle BPC = \angle APC = 120^\circ</math> and <math>\triangle APB, \triangle BPC, \triangle APC</math> are all <math>30-30-120</math> triangles, and the squares of the side lengths are in the ratio <math>\frac{\frac{1}{1}}{3}</math> which can easily be seen by dividing this triangle into two smaller <math>30-60-90</math> triangles. It follows <math>AP^2+BP^2=\frac{2}{3}AB^2</math>, <math>AP^2+CP^2=\frac{2}{3}AC^2</math>, and <math>BP^2+CP^2=\frac{2}{3}BC^2</math>. Now <math>2AP^2+2BP^2+2CP^2=\frac{2}{3}(AB^2+AC^2+BC^2)</math> because each got counted twice, so <math>AP^2+BP^2+CP^2=\frac{1}{3}(13^2+14^2+15^2)=\frac{590}{3}</math>, and <math>m+n=\boxed{593}</math>.
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==Note==
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~icematrix2
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See [[2021 AMC 12A Problems/Problem 1|problem 1]].
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==See also==
 
==See also==
 
{{AMC12 box|year=2021|ab=A|num-b=8|num-a=10}}
 
{{AMC12 box|year=2021|ab=A|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:30, 13 October 2020

Problem

Triangle $ABC$ lies in a plane with $AB=13$, $AC=14$, and $BC=15$. For any point $X$ in the plane of $\triangle ABC$, let $f(X)$ denote the sum of the three distances from $X$ to the three vertices of $\triangle ABC$. Let $P$ be the unique point in the plane of $\triangle ABC$ where $f(X)$ is minimized. Then $AP^2+BP^2+CP^2=\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is the value of $m+n$?

Solution

Point $P$ is the Brocard point of $\triangle ABC$, where $\angle APB = \angle BPC = \angle APC = 120^\circ$ and $\triangle APB, \triangle BPC, \triangle APC$ are all $30-30-120$ triangles, and the squares of the side lengths are in the ratio $\frac{\frac{1}{1}}{3}$ which can easily be seen by dividing this triangle into two smaller $30-60-90$ triangles. It follows $AP^2+BP^2=\frac{2}{3}AB^2$, $AP^2+CP^2=\frac{2}{3}AC^2$, and $BP^2+CP^2=\frac{2}{3}BC^2$. Now $2AP^2+2BP^2+2CP^2=\frac{2}{3}(AB^2+AC^2+BC^2)$ because each got counted twice, so $AP^2+BP^2+CP^2=\frac{1}{3}(13^2+14^2+15^2)=\frac{590}{3}$, and $m+n=\boxed{593}$. ~icematrix2

See also

2021 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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