Klamkin's Quickie via Peru Geometrico

Source

Klamkin's Quickie via Peru Geometrico, problem

Solution

By the Law of Cosines, $a^2=b^2-bc+c^2\,$ iff $\angle A=60^{\circ}.\,$ That identity is equivalent to $a^2(b+c)=b^3+c^3,\,$ or $a^2(a+b+c)=a^3+b^3+c^3.$

Acknowledgment

I borrowed this problem from the Peru Geometrico facebook group. The problem is attributed to the late Murray S. Kamkin (probably one of his quickies.) There are several additional solutions available.

 

[an error occurred while processing this directive]

|Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny
[an error occurred while processing this directive]
[an error occurred while processing this directive]