MathLabs

Problem 1

Given a triangle ABC, let I be the incenter. The internal bisectors of angles A, B, C meet the opposite sides in A', B', C' respectively. Prove that 1/4<AI⋅BI⋅CI/(AA′⋅BB′⋅CC′)≤8/271/4<AI\cdot BI\cdot CI/(AA'\cdot BB'\cdot CC')\le8/27.
Step 1 of 4: Convert the three ratios by areas
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p=AB+BC+CA,AI/AA′=(CA+AB)/pp=AB+BC+CA,\qquad AI/AA'=(CA+AB)/p
Detailed analysis

Let p be the perimeter and r the inradius. Since the areas of ABI and CAI are AB·r/2 and CA·r/2, their sum divided by the area of ABC equals AI/AA'. Hence AI/AA'=(CA+AB)/p, and cyclically the other two ratios are (AB+BC)/p and (BC+CA)/p.