Problem:
In △ABC,AB=20. The incircle of the triangle divides the median containing C into three segments of equal length. Given that the area of △ABC is mn, where m and n are integers and n is not divisible by the square of any prime, find m+n.
Solution:
Let M be the midpoint of AB, and let S and N be the points where median CM meets the incircle, with S between C and N. Let AC and AB touch the incircle at R and T, respectively. Assume, without loss of generality, that T is between A and M. Then AR=AT. Use the Power-of-a-Point Theorem to conclude that
MT2=MN⋅MS and CR2=CS⋅CN
Because CS=SN=MN, conclude that CR=MT, and
AC=AR+CR=AT+MT=AM=21AB=10
Let s=(1/2)(AB+BC+CA). Then AT=s−BC, and
MT=MA−AT=21AB−s+BC=2BC−AC=2BC−10
But MT2=MN⋅MS=(2/9)CM2, so 2BC−10=CM⋅32. Hence
CM=223⋅(BC−10)
Apply the Law of Cosines to triangles AMC and ABC to obtain
2⋅10⋅10102+102−CM2=cosA=2⋅10⋅20102+202−BC2
Then BC2=100+2⋅CM2, so BC2=100+(9/4)(BC−10)2. The solutions of this equation are 26 and 10, but BC>AB−AC=10. It follows that BC=26, and then that CM=122. The length of the altitude from A in isosceles △AMC is therefore 27. Thus [ABC]=2[AMC]=2414, and m+n=38.
