Triangle inscribed in a circle - maximize $a^{2}+b^{2}+c^{2}$.

Question

I am interested in the following problem:

What is maximum value of $a^2 +b^2+c^2$, where $a,b,c$ are sides of a triangle inscribed in a unit circle?

I know that this problem can be solved using multivariate analysis, e.g by Lagrange multipliers, etc. But this is a problem for high-school level math circle. That is why I would like to find out how to use elementary plane geometry ideas to solve it efficiently.

I believe that correct answer is equilateral triangle. I will be glad for solutions or hints.

Answer 1

This is a solution using complex numbers. The unit circle is defined by equation $z\bar{z}=1$ and the three vertices of the triangle is $z_{1},z_{2},z_{3}$.

$$ \begin{aligned} a^{2}+b^{2}+c^{2}&=(z_{1}-z_{2})(\bar{z_{1}}-\bar{z_{2}})+(z_{2}-z_{3})(\bar{z_{2}}-\bar{z_{3}})+(z_{3}-z_{1})(\bar{z_{3}}-\bar{z_{1}})\\ &=9-(z_{1}+z_{2}+z_{3})(\bar{z_{1}}+\bar{z_{2}}+\bar{z_{3}})\\ &\leq 9 \end{aligned} $$

$a^{2}+b^{2}+c^{2}$ is maximum when the triangle’s circumcenter coincides with its centroid i.e. when the triangle is equilateral.

Answer 2

The hint:

Use $$S=\frac{abc}{4R}=\frac{1}{4}\sqrt{\sum_{cyc}(2a^2b^2-a^4)}$$ and prove that $$a^2+b^2+c^2\leq9R^2.$$ Finally, we need to prove that: $$\sum_{cyc}(a^6-a^4b^2-a^4c^2+a^2b^2c^2)\geq0,$$ which is true by Schur.

Answer 3

There is also the following way.

In the standard notation we'll prove that: $$a^2+b^2+c^2\leq9R^2$$ or $$\sum_{cyc}4R^2\sin^2\alpha\leq9R^2$$ or $$2\sum_{cyc}(1-\cos2\alpha)\leq9$$ or $$2\sum_{cyc}\cos2\alpha\geq-3$$ or $$2(2\cos^2\alpha-1)+4\cos(\beta+\gamma)\cos(\beta-\gamma)\geq-3$$ or $$4\cos^2\alpha-4\cos(\beta-\gamma)\cos\alpha+1\geq0$$ or $$(2\cos\alpha-\cos(\beta-\gamma))^2+\sin^2(\beta-\gamma)\geq0,$$ the equality occurs for the equilateral triangle, which says that $9$ is a maximal value.