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

Suppose I have a circumcircle $C$ about the three points $x_1, x_2, x_3$ in the plane. Now I have a fourth point $x_4$ that lies in the interior of $C$, and is on the side of the segment $x_2 x_3$ opposite from $x_1$. Does it necessarily follow that the circumcircle about $x_1, x_2, x_4$ must have a smaller radius than $C$?

share|cite|improve this question
up vote 0 down vote accepted

If $\widehat{x_2 x_1 x_3}\leq\frac{\pi}{2}$ you have a relation between the two circumradii, and the result follows from the law of sines. In any triangle you have: $$ 2R = \frac{a}{\sin \widehat{A}}, $$ so, if you call $a=\overline{x_2 x_3},\; \widehat{A}=\widehat{x_2 x_1 x_3},\; \widehat{C}=\widehat{x_3 x_4 x_2}\;$, you have: $$ \pi-\widehat{A}<\widehat{C}\leq\pi, $$ so: $$ \sin\widehat{C}<\sin\widehat{A}, $$ then the circumradius of $x_2 x_3 x_4$ is bigger than the circumradius of $x_1 x_2 x_3$.

If $\widehat{x_2 x_1 x_3}\leq\frac{\pi}{2}$ does not hold, you cannot say anything: consider the case in which $x_1,x_4$ are symmetric with respect to the line $x_2 x_3$.

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.