Problem:
In △ABC,AC=BC, and point D is on BC so that CD=3⋅BD. Let E be the midpoint of AD. Given that CE=7 and BE=3, the area of △ABC can be expressed in the form mn, where m and n are positive integers and n is not divisible by the square of any prime. Find m+n.
Solution:
Let AB=2x, and AC=BC=y. Then cos∠BAC=cos∠ABC=yx. Applying the Law of Cosines to △ABD yields
Because CE is a median in △ADC, Stewart's Theorem shows that 4CE2=2CD2+2AC2−AD2, or 28=1618y2+2y2−3x2−16y2. Hence 1649y2−3x2=28. Similarly, in △ABD,4BE2=2BD2+2AB2−AD2=162y2+8x2−3x2−16y2. Hence 16y2+5x2=36. Solving the system
1649y2−3x2=2816y2+5x2=36
yields y2=16,x2=7. Thus the area of △ABC is xy2−x2=37, and the requested sum is 3+7=10.
