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

m (Solution in progress -- all are welcome to finish.)
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== 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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Let <math>E</math>, <math>F</math> and <math>G</math> be the points of tangency of the incircle with <math>BC</math>, <math>AC</math> and <math>AB</math>, respectively.  Without loss of generality, let <math>AC < AB</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 = c</math> and thus <math>AC = AF + CF = DE + CE = CD = 10</math>.  Then <math>DE = AF = AG = 10 - c</math> so <math>BG = BE = BD + DE = 20 - c</math> and thus <math>AB = AG + BG = 30 - 2c</math>.
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{{solution}}
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Now, by [[Stewart's Theorem]] in triangle <math>\triangle ABC</math> with [[cevian]] <math>\overline AD</math>, we have
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 +
<math>(3m)^2\cdot 20 + 20\cdot10\cdot10 = 10^2\cdot10 + (30 - 2c)^2\cdot 10</math>.  Our earlier result from Power of a Point was that <math>2m^2 = (10 - c)^2</math>, so we combine these two results to solve for <math>c</math> and we get
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 +
<math>9(10 - c)^2 + 200 = 100 + (30 - 2c)^2</math>
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 +
or equivalently
 +
 
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<math>c^2 - 12c + 20 = 0</math>.
 +
 
 +
Thus <math>c = 2</math> or <math>c = 10</math>.  We discard the value <math>c = 10</math> as extraneous (it gives us an equilateral triangle) and are left with <math>c = 2</math>, so our triangle has sides of length 10, 20 and 26.  Applying [[Heron's formula]] or the equivalent gives that the area is <math>A = \sqrt{28 \cdot 18 \cdot 8 \cdot 2} = 24\sqrt{14}</math> and so the answer is <math>24 + 14 = 038</math>.
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== See also ==
 
== See also ==

Revision as of 17:49, 17 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 $E$, $F$ and $G$ be the points of tangency of the incircle with $BC$, $AC$ and $AB$, respectively. Without loss of generality, let $AC < AB$, 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 $CE = CF = c$ and thus $AC = AF + CF = DE + CE = CD = 10$. Then $DE = AF = AG = 10 - c$ so $BG = BE = BD + DE = 20 - c$ and thus $AB = AG + BG = 30 - 2c$.

Now, by Stewart's Theorem in triangle $\triangle ABC$ with cevian $\overline AD$, we have

$(3m)^2\cdot 20 + 20\cdot10\cdot10 = 10^2\cdot10 + (30 - 2c)^2\cdot 10$. Our earlier result from Power of a Point was that $2m^2 = (10 - c)^2$, so we combine these two results to solve for $c$ and we get

$9(10 - c)^2 + 200 = 100 + (30 - 2c)^2$

or equivalently

$c^2 - 12c + 20 = 0$.

Thus $c = 2$ or $c = 10$. We discard the value $c = 10$ as extraneous (it gives us an equilateral triangle) and are left with $c = 2$, so our triangle has sides of length 10, 20 and 26. Applying Heron's formula or the equivalent gives that the area is $A = \sqrt{28 \cdot 18 \cdot 8 \cdot 2} = 24\sqrt{14}$ and so the answer is $24 + 14 = 038$.


See also

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