In triangle ABC  P, Q and R are respectively on the sides BC, AC, and AB.
Given that AP, BQ , and CR are concurrent at the point O , and that:
AO/OP + BO/OQ + CO/OR = 15 , find (AO/OP)*(BO/OQ)*(CO/OR)
I made a sketch with GSP. It is very surprising to me that the product always exceeds the sum by 2. Once the relative positions of P and Q are set, the values of the sum and product are fixed, irrespective of the size or angles of the triangle. It appears the minimum sum is 6.
Anyways the answer to the problem is 15+2 = 17 but it will take some more work to figure out why.

Posted by Jer
on 20150609 12:06:23 