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I know how to find the sides of the triangle when seeking to maximize the area of ​​the triangle, since the following relationship exists $Area= \frac{abc}{4R}$ where $a, b,c$ are the sides and $R$ the radius. But I don't know what to occupy to maximize the perimeter.

Jessica Flores Robles
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2 Answers2

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Recall that the Law of Sines tells you that $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R,$$ where $R$ is the circumradius. Then the perimeter is $$2R(\sin A + \sin B + \sin C).$$ Note that $$\sin A + \sin B = 2\cos(\tfrac{A - B}{2})\sin(\tfrac{A + B}{2}) \le 2\sin(\tfrac{A + B}{2})$$ with equality only when $A = B$. Thus, the perimeter is maximized when $A = B$. Using similar logic, you can show that $B = C$ as well when the perimeter is maximized. So, the perimeter is maximized when $A = B = C = 60^\circ$ with a value of $$2R(\sin 60^\circ + \sin 60^\circ + \sin 60^\circ) = 3\sqrt3 R.$$

Joseph Camacho
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  • You can avoid the final bounding argument by jumping to Jensen's inequality like in my solution, but keep your argument because it's different. Also, might be worth mentioning how the right-angle triangle from the circumcentre on each side gives the $a/2R=\sin A$ at the beginning. – Rod Laver Jan 21 '22 at 13:26
  • I know I can use Jensen's inequality. But I chose to do it this way because the original asker probably doesn't know Jensen's inequality or the math needed to prove that $\sin$ is concave on $[0, \pi]$. – Joseph Camacho Jan 21 '22 at 17:28
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Let $O$ be the circumcentre, and let $\alpha,\beta,\gamma$ be the angles at $O$ opposite the sides of the triangle of lengths $a$, $b$ and $c$ respectively. By the cosine rule, we have $$ a+b+c=\sqrt{2}R\left(\sqrt{1-\cos \alpha}+\sqrt{1-\cos \beta}+\sqrt{1-\cos \gamma}\right). $$ Now $\frac{d^2}{dx^2}\left(\sqrt{1-\cos x}\right)=\frac{-1}{4}\sqrt{1-\cos x}\leq 0$, so $\sqrt{1-\cos x}$ is concave. Thus by Jensen's inequality, $$ \sqrt{1-\cos \alpha}+\sqrt{1-\cos \beta}+\sqrt{1-\cos \gamma}\leq 3\cdot \sqrt{1-\cos \frac{\alpha+\beta+\gamma }{3}}=3\sqrt{1-\cos\frac{2\pi }{3}} $$ with equality iff $\alpha=\beta=\gamma=\frac{2\pi }{3}\Longleftrightarrow a=b=c$.

Rod Laver
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    @N.F.Taussig No. My $A$, $B$, $C$ are the angles at $O$, not the vertices of the triangle. I'll relabel them to $\alpha,\beta,\gamma$ to avoid confusion as I just realised Joseph used them as vertices of the triangle. – Rod Laver Jan 21 '22 at 12:58