A Problem on a Pentagonal Star

Every one with even the slightest familiarity of the most basic geometric ideas must be able to solve the problem I am going to offer shortly. A proof is therefore supplied only for the completeness' sake. For me, the main attraction of the problem is in that it's formulated for a pentagon, or, more accurately, for a pentagonal star. The greatest part of Planimetry - Plane Geometry - deals with properties of triangle, quadrilateral and sometimes hexagons, like in Pascal's theorem (In a hexagon inscribed in a conic, the three points of intersection of opposite pairs of sides are collinear.) Very few problems deal with other polygons.

Here is then the problem. Let AEICG be a pentagonal star as depicted below.

Is it possible that all five inequalities AJ > AB, CB > CD, ED > EF, GF > GH, IH > IJ hold simultaneously?

Reference

1. D. Fomin, S. Genkin, I. Itenberg, Mathematical Circles (Russian Experience), AMS, 1996

Proof

Assume all the inequalities in the formulation of the problem hold. As is well known, in a triangle, a greater angle lies opposite a greater side. Therefore, for angles b, d, f, h, j we have the following chain of inequalities:

b > d > f > h > j > b

Contradiction.

The problem and its proof can be adapted to other polygons, but not all. Which?

Since the original formulation that included 10 different segments has been reduced to a question about 5 pairs of equal angles, it appears possible to obtain a stronger result. Indeed, let's apply the Law of Sines in each of the five triangles ABJ, BCD, DEF, FGH and HIJ (in all cases the angles are denoted with a single letter with no ambiguity):

Multiplying the five identities we get AJ·CB·ED·GF·IH = AB·CD·EF·GH·IJ, which also answers the original question.

The latter identity was first proved by Larry Hoehn (Mathematics Magazine, 66:2 121-123). The above proof is due to Cheng-Shyong Lee (The College Mathematics Journal, 29:2, 1998, 144-145.)

Copyright © 1996-2009 Alexander Bogomolny