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| − | ==Problem==
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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==
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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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| − | ~icematrix2
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| − | ==See also==
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| − | {{AMC12 box|year=2021|ab=A|num-b=8|num-a=10}}
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| − | {{MAA Notice}}
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Revision as of 17:43, 13 October 2020
This page will open in February
