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

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== Solution ==
 
== Solution ==
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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>.
 
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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Now, by [[Stewart's Theorem]] in triangle <math>\triangle ABC</math> with [[cevian]] <math>\overline AD</math>, we have
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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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<cmath>(3m)^2\cdot 20 + 20\cdot10\cdot10 = 10^2\cdot10 + (30 - 2c)^2\cdot 10.</cmath>
  
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<math>9(10 - c)^2 + 200 = 100 + (30 - 2c)^2</math>
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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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or equivalently
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<cmath>9(10 - c)^2 + 200 = 100 + (30 - 2c)^2 \quad \Longrightarrow \quad c^2 - 12c + 20 = 0.</cmath>
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<math>c^2 - 12c + 20 = 0</math>.
 
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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>.
 
  
 +
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 <math>10, 20</math> and <math>26</math>.  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>.
  
 
== See also ==
 
== See also ==

Revision as of 12:44, 2 December 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

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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

\[9(10 - c)^2 + 200 = 100 + (30 - 2c)^2 \quad \Longrightarrow \quad c^2 - 12c + 20 = 0.\]

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

2005 AIME I (Problems • Answer Key • Resources)
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