   SEARCH HOME Math Central Quandaries & Queries  Question from omkar, a student: Any side of a triangle is smaller than half of its perimeter, prove this in short ? Hi Omkar.

In short, prove it by contradiction:

Hypothesis: Let P be the perimeter of a triangle ABC and let |AB| ≥ P/2.

The shortest distance between any two points is a straight line, so to get from A to B is at least of distance |AB|. To return from B to C to A therefore, we know that |BC| + |CA| ≥ |AB|. But the perimeter of a triangle is the sum of the lengths of the sides, so if we add |AB| to each side of this latest inequality, we have: |AB| + |CA| + |AB| ≥ 2 |AB|, which reduces to P/2 ≥ |AB|.

Thus P/2 ≥ |AB| ≥ P/2, which means that |AB| = P/2. So |AB| must be half the perimeter (it cannot be more).

However, this means |BC| + |CA| = |AB|, so B, C and A are colinear (three points in a straight line). This is not what we call a triangle, since in a triangle 0° < each angle < 180°.

This is a contraction, so the hypothesis is erroneous. Thus each side of a triangle must be less than half the triangle's perimeter.

Stephen.     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.