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2005 AIME I Problems/Problem 15

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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 = \boxed{038}.

See also

2005 AIME I (ProblemsResources)
Preceded by
Problem 14
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