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

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<cmath>(10 - c)^2 + 200 = 100 + (30 - 2c)^2 \quad \Longrightarrow \quad c^2 - 12c + 20 = 0.</cmath>
 
<cmath>(10 - c)^2 + 200 = 100 + (30 - 2c)^2 \quad \Longrightarrow \quad c^2 - 12c + 20 = 0.</cmath>
  
−
Thus <math>c = 2</math> or <math> = 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 <math>\sqrt{28 \cdot 18 \cdot 8 \cdot 2} = 24\sqrt{14}</math> and so the answer is <math>24 + 14 = \boxed{038}</math>.
+
Thus <math>c = 2</math> or <math> = 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 area <math>\sqrt{28 \cdot 18 \cdot 8 \cdot 2} = 24\sqrt{14}</math> and so the answer is <math>24 + 14 = \boxed{038}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 13:58, 28 August 2016

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

\[(3m)^2\cdot 20 + 20\cdot10\cdot10 = 1^2\cdot0 + (30 - 2c)^2\cdot 10.\]

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

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

Thus $c = 2$ or $= 10$. We discard the value $c = 10$ as extraneous (it gives us an equilateral triangle) and are left with $c = 2$, so our triangle has area $\sqrt{28 \cdot 18 \cdot 8 \cdot 2} = 24\sqrt{14}$ and so the answer is $24 + 14 = \boxed{038}$.

See also

2005 AIME I (Problems • Answer Key • Resources)
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