A Stronger Triangle Inequality

In 1997, a retired engineer (H. R. Bailley) and a retired chemist (R. Bannister) published a curious result - a strengthened triangle inequality - that serendipitously incorporated a rather ubiquitous consumer services symbol (24/7). A short while later, the now late math problem solver extraordinaire, Professor Murray Klamkin, gave an elegant 1 page proof of that result. I'll start with the motivation that drove Bailley and Bannister and proceed with Klamkin's proof.

Bailley and Bannister began there investigation with a right triangle and an inscribed square. They compared two ways of inscribing a square into a right triangle. Which of the squares is bigger?

Denote the legs of the triangle as a and b, the hypotenuse c, the altitude to the hypotenuse h, and the sides of the two squares s and t.

In both cases, the presence of similar triangles leads to the proportions:

from which

and ultimately

Since, in a right triangle, both ab and hc equal double the area of the triangle, ab = hc. It is obvious then that the relative sizes of the squares inversely depend on the relative magnitudes of (a + b) and (c + h). To compare the two, consider

It follows that in a right triangle we always have

Which is equivalent to s > t. Thus addition of the altitude h led to the inversion of the triangle inequality: a + b > c. Naturally, Bailley and Bannister decided to proceed and compare (a + b) and (c + h) for triangles other than right. They came up with a surprising result.

In a triangle with side length a, b, c, altitude h and the angle C opposite side c

For (2) to hold we must have 2·cos(C/2) / (1 + sin C/2) > 1. The left side is a decreasing function of C and equals 1 when sin C/2 = 3/5, which corresponds to C = arctan(24/7). Thus (2) indeed holds for all C < arctan(24/7).

It remains only to prove (4). If R is the circumradius of the triangle,

so that

Using elementary trigonometric identities and C = π - (A + B), the latter ratio becomes

Using (A+B)/2 = π/2 - C/2, the ratio takes the form

which is (4).

References

1. H. R. Bailley, R. Bannister, A Stronger Triangle Inequality, The College Mathematics Journal, vol. 28, no. 3, May 1997, 182-186.
2. M. S. Klamkin, A Sharp Triangle Inequality, The College Mathematics Journal, vol. 29, no. 1, January 1998, 33.

Copyright © 1996-2009 Alexander Bogomolny