IMO shortlist 1997/7 solution

38th IMO 1997 shortlist

Problem 7

ABCDEF is a convex hexagon with AB = BC, CD = DE, EF = FA. Show that BC/BE + DE/DA + FA/FC ≥ 3/2. When does equality occur?

Solution

Many thanks to David Krumm for the following solution

We need Ptolemy's inequality - WX·YZ + XY·ZW ≥ WY·XZ with equality iff the quadrilateral WXYZ is cyclic (or a straight line) - and Nesbitt's inequality: a/(b+c) + b/(a+c) + c/(a+b) >= 3/2 for any positive numbers a, b, c. Both should be well-known. For example, taking a/(b+c) + 1 = (a+b+c)/(b+c) etc and (a+b+c) = (A + B + C)/2, where A = b+c, B = a+c, C = a+b, Nesbitt is equivalent to (A+B+C)/(1/A + 1/B + 1/C) ≥ 9, which is easily seen to be true since x + 1/x ≥ 2. The proof also shows that we have equality in Nesbitt iff a = b = c.

Applying Ptolemy to ACDE we get DE·AC + DC·AE ≥ DA·CE. Using DE = DC, this becomes DE(AC + AE) ≥ DA·CE or DE/DA ≥ CE/(AC + AE). Similarly (move on two letters) we get FA/FC ≥ EA/(CE + CA), and BC/BE ≥ EA/(EA + EC). Adding, and using Nesbitt we get the result.

For equality we must have equality in each of the three Ptolemy relations and in the Nesbitt relation. Nesbitt equality implies ACE must be equilateral, so angle CAE = 60^o. Then ACDE is cyclic, so angle D must be 120^o. Similarly angles B and F must be 120^o. Now each of the triangles ABC, CDE, EFA must be congruent (ABC is isosceles, so its angles are 30^o, 120^o, 30^o and the long side AC is a side of the equilateral triangle). Thus the hexagon has all sides equal and all angles 120^o, so it is regular. Conversely, it is obvious that if the hexagon is regular, then we have equality (because the long diagonal of a regular hexagon is twice as long as a side).

38th IMO shortlist 1997