Difference between revisions of "Mock Geometry AIME 2011 Problems/Problem 6"

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'''Solution'''
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==Problem==
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Three points <math>A,B,C</math> are chosen at random on a circle. The probability that there exists a point <math>P</math> inside an equilateral triangle <math>A_1B_1C_1</math> such that <math>PA_1=BC,PB_1=AC,PC_1=AB</math> can be expressed in the form <math>\frac{m} {n},</math> where <math>m,n</math> are relatively prime positive integers. Find <math>m+n.</math>
  
We claim that all angles in triangle ABC have to be less than 120 in order for P to be inside the equilateral triangle.
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==Solution==
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The problem asks for the probability that point <math>P</math> is inside an equilateral triangle <math>A_1B_1C_1</math>. Let <math>x</math>, <math>y</math>, and <math>z</math> be the three distances from point <math>P</math> to each of the vertices, with <math>x</math> being the longest distance. Let's consider the case in which point <math>P</math> is actually on the line:
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<asy>
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unitsize(1cm);
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draw((0, 18*sqrt(3))--(18, 0));
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draw((18, 0)--(36, 18*sqrt(3)));
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<\asy>

Revision as of 18:01, 7 July 2019

Problem

Three points $A,B,C$ are chosen at random on a circle. The probability that there exists a point $P$ inside an equilateral triangle $A_1B_1C_1$ such that $PA_1=BC,PB_1=AC,PC_1=AB$ can be expressed in the form $\frac{m} {n},$ where $m,n$ are relatively prime positive integers. Find $m+n.$

Solution

The problem asks for the probability that point $P$ is inside an equilateral triangle $A_1B_1C_1$. Let $x$, $y$, and $z$ be the three distances from point $P$ to each of the vertices, with $x$ being the longest distance. Let's consider the case in which point $P$ is actually on the line: <asy> unitsize(1cm); draw((0, 18*sqrt(3))--(18, 0)); draw((18, 0)--(36, 18*sqrt(3))); <\asy>