Problem:
A pyramid has a triangular base with side lengths 20,20, and 24. The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length 25. The volume of the pyramid is mn​, where m and n are positive integers, and n is not divisible by the square of any prime. Find m+n.
Solution:
Let A,B, and C be the vertices of the base triangle so that AB=AC=20 and BC=24. Let D be the fourth vertex of the pyramid, P the foot of the altitude of the pyramid from D to its base, and M the midpoint of side BC. By symmetry, P lies on AM. By the Pythagorean Theorem, AM=202−122​=16 and DM2=252−122=481. Then AP2+DP2=252 and (16−AP)2+DP2=DM2=481. Thus
AP2+DP2−((16−AP)2+DP2)=625−481=144
and so 32(AP)−256=144. Solving for AP yields AP=225​. Therefore DP=625−(225​)2​=225​3​, and the volume of the pyramid is 31​⋅225​3​⋅21​⋅24⋅16=8003​. The requested sum is 800+3=803​.