1988 IMO Problem #5

[cr]

I found a simpler solution. Triangle I1DI2 has side length proportional to AB and AC and has an right angle in between so it's similar to triangle ABC. So angle I1I2D is same as angle C. So I2NCD is co-cyclic therefore angle ANM = I2DC = I2DA. Therefore triangles ANI2 and ADI2 are congruent so AN = AD. Similarly AM = AD. So AM*AN = AD^2 = AB * cosC * AC * sin C = AB*AC*0.5sin(2) <= AB*AC*0.5.