Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let A, B, and C be three points on a circle of radius 1.

Suppose that the magnitude of $\angle ABC$ is fixed. Then show that the area of the triangle ABC is maximized when $\angle BCA$ = $\angle CAB$.

share|cite|improve this question
up vote 1 down vote accepted

Let $\angle BAC=\alpha, \angle ABC=\beta,\angle BCA=\gamma$. We know that $\beta$ is fixed. Let $O$ the be center of the circle, then $\angle BOC=2\alpha,\angle AOC=2\beta,\angle BOA=2\gamma$. So, \begin{align} [ABC]&=\frac12\sin2\beta+\frac12(\sin2\alpha+\sin2\gamma)\\ &=\frac12\sin2\beta+\sin(\alpha+\gamma)\cos(\alpha-\gamma)\\ &=\frac12\sin2\beta+\sin(\pi-\beta)\cos(\alpha-\gamma)\\ &=\frac12\sin2\beta+\sin\beta\cos(\alpha-\gamma).\\ \end{align} Since $\beta$ is fixed and $\sin\beta>0$ then $[ABC]$ is maximum whenever $\cos(\alpha-\gamma)$ takes its maximum, i.e., $\cos(\alpha-\gamma)=1=\cos0\iff \alpha=\gamma$ as you wanted.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.