2005 AIME I Problems/Problem 15

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 $24 + 14 = 038$ .

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2005 AIME I Problems/Problem 15

Problem

Solution

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