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Difference between revisions of "2005 AIME I Problems/Problem 15"

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== Problem ==
 
== Problem ==
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[[Triangle]] <math> ABC </math> has <math> BC=20. </math> The [[incircle]] of the triangle evenly [[trisect]]s the [[median of a triangle | median]] <math> AD. </math> If the [[area]] of the triangle is <math> m \sqrt{n} </math> where <math> m </math> and <math> n </math> are [[integer]]s and <math> n </math> is not [[divisor | divisible]] by the [[square]] of a [[prime]], find <math> m+n. </math>
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[[Triangle]] <math> ABC </math> has <math> BC=20. </math> The [[incircle]] of the triangle evenly [[trisect]]s the [[median of a triangle | median]] <math> AD. </math> If the [[area]] of the triangle is <math> m \sqrt{n} </math> where <math> m </math> and <math> n </math> are [[integer]]s and <math> n </math> is not [[divisor | divisible]] by the [[perfect square | square]] of a [[prime]], find <math> m+n. </math>
  
 
== Solution ==
 
== Solution ==
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Let <math>E</math> and <math>F</math> be the points of tangency of the incircle with <math>BC</math> and <math>AC</math>, respectively.  Without loss of generality, let <math>AB > AC</math>, so that <math>E</math> is between <math>D</math> and <math>C</math>.  Let the length of the median be <math>3m</math>.  Then by two applications of the [[Power of a Point Theorem]], <math>DE^2 = 2m \cdot m = AF^2</math>, so <math>DE = AF</math>.  Now, <math>CE</math> and <math>CF</math> are two tangents to a circle from the same point, so <math>CE = CF</math> and thus <math>AC = AF + CF = DE + CE = CD = 10</math>.  Then triangle <math>\triangle ACD</math> is [[isosceles triangle | isosceles]],
 
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== See also ==
 
== See also ==
 
* [[2005 AIME I Problems/Problem 14 | Previous problem]]
 
* [[2005 AIME I Problems/Problem 14 | Previous problem]]

Revision as of 21:06, 16 January 2007

Problem

Triangle $ABC$ has $BC=20.$ The incircle of the triangle evenly trisects the median $AD.$ If the area of the triangle is $m \sqrt{n}$ where $m$ and $n$ are integers and $n$ is not divisible by the square of a prime, find $m+n.$

Solution

Let $E$ and $F$ be the points of tangency of the incircle with $BC$ and $AC$, respectively. Without loss of generality, let $AB > AC$, so that $E$ is between $D$ and $C$. Let the length of the median be $3m$. Then by two applications of the Power of a Point Theorem, $DE^2 = 2m \cdot m = AF^2$, so $DE = AF$. Now, $CE$ and $CF$ are two tangents to a circle from the same point, so $CE = CF$ and thus $AC = AF + CF = DE + CE = CD = 10$. Then triangle $\triangle ACD$ is isosceles, This problem needs a solution. If you have a solution for it, please help us out by adding it.

See also

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