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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 [[perfect square | 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 integers 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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[[Image:2005_I_AIME-15.png]]
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<center><asy>
 +
size(300); pointpen=black;pathpen=black+linewidth(0.65); pen s = fontsize(10);
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pair A=(0,0),B=(26,0),C=IP(circle(A,10),circle(B,20)),D=(B+C)/2,I=incenter(A,B,C);
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path cir = incircle(A,B,C); pair E1=IP(cir,B--C),F=IP(cir,A--C),G=IP(cir,A--B),P=IP(A--D,cir),Q=OP(A--D,cir);
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D(MP("A",A,s)--MP("B",B,s)--MP("C",C,N,s)--cycle); D(cir);
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D(A--MP("D",D,NE,s)); D(MP("E",E1,NE,s)); D(MP("F",F,NW,s)); D(MP("G",G,s)); D(MP("P",P,SW,s)); D(MP("Q",Q,SE,s));
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MP("10",(B+D)/2,NE); MP("10",(C+D)/2,NE);
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</asy></center><!-- Asymptote replacement for Image:2005_I_AIME-15.png by azjps -->
  
 
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>.

Revision as of 17:09, 26 April 2008

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

[asy] size(300); pointpen=black;pathpen=black+linewidth(0.65); pen s = fontsize(10); pair A=(0,0),B=(26,0),C=IP(circle(A,10),circle(B,20)),D=(B+C)/2,I=incenter(A,B,C); path cir = incircle(A,B,C); pair E1=IP(cir,B--C),F=IP(cir,A--C),G=IP(cir,A--B),P=IP(A--D,cir),Q=OP(A--D,cir); D(MP("A",A,s)--MP("B",B,s)--MP("C",C,N,s)--cycle); D(cir); D(A--MP("D",D,NE,s)); D(MP("E",E1,NE,s)); D(MP("F",F,NW,s)); D(MP("G",G,s)); D(MP("P",P,SW,s)); D(MP("Q",Q,SE,s)); MP("10",(B+D)/2,NE); MP("10",(C+D)/2,NE); [/asy]

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 $24 + 14 = \boxed{038}$.

See also

2005 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Last Question
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All AIME Problems and Solutions
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