geometryoptimizationtriangles
How to prove that isosceles triangle has maximum perimeter from all trangles inscribed in circle?
I found that from all isosceles trinagles – equilateral has maximum perimeter:
Maximum perimeter of an isosceles triangle inscribed in the unit circle?, but I wonder how to prove that a triangle with maximum perimeter should be isosceles.
I thought an image could help:
The perimeter is $$\overline{BC} + 2\overline{AB} = 2\sin \alpha + 2\cdot 2\sin \beta = 2\sin \alpha + 2\cdot 2\sin \left(\frac{\pi}{2} - \frac{\alpha}{2}\right) = 2\left(\sin \alpha+2\cos\left(\frac{\alpha}{2} \right) \right)$$ For further simplifications let $\alpha = 2t$, then you have to maximize $$f(t) = 2\left(\sin 2t+2\cos t) \right) = 2(2\sin t \cos t + 2\cos t ) = 4(\sin t \cos t + \cos t) $$ Finding $f'(t)$ is easy: $f'(t) = 4(\cos t \cdot \cos t + \sin t \cdot -\sin t -\sin t) = 4(\cos^2 t - \sin^2 t -\sin t )=4(-2\sin^2 t -\sin t +1)$ and it equals $0$ when $\sin t = \frac12 \Rightarrow t = \frac{\pi}{6} \Rightarrow \alpha = 2t = \frac{\pi}{3} $ so the triangle is equilater!
Computing the area and perimeter, we get $A=x\sqrt{1-x^2}$ and $P=2+2x$.
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Therefore, $$ \begin{align} r^2 &=\frac{4A^2}{P^2}\\ &=x^2\frac{1-x}{1+x}\tag{1} \end{align} $$ This reaches maximum when the derivative of $(1)$ is $0$; that is, when $$ x^3+x^2-x=0\require{cancel}\tag{2} $$ The only root of $(2)$ within the allowable range of $x\in(0,1)$ is $x=\frac{-1+\sqrt5}2=\frac1\phi$. Thus, the maximum area of the inscribed circle is $$ \begin{align} \pi r^2 &=\frac\pi{\phi^5}\\ &\doteq0.283277232857953\tag{3} \end{align} $$ Divide the area by $\pi$ and take the square root to get the radius of the circle.
Best Answer
Let a triangle be inscribed in a unit circle, and let $A$ and $B$ mark two vertices. Let $\theta$ be half the length of the arc connecting $A$ and $B$, and let $\ell$ be the length of the chord, you have that from elementary trigonometry
$$ \ell / 2 = \sin (\theta) $$
Now let $\theta_1, \theta_2, \theta_3$ be half the lengths of the three arcs demarcated by the vertices of the triangle, from the above we have that the perimeter is equal to
$$ 2 \left[ \sin (\theta_1) + \sin(\theta_2) + \sin(\theta_3) \right] $$
and we know that
$$ \theta_1 + \theta_2 + \theta_3 = \pi $$
while
$$ \theta_1, \theta_2, \theta_3 \in [0,\pi] $$
Now from the fact that $\sin$ is a concave function on $[0,\pi]$, we have that the perimeter is equal to
$$ 6 * \frac13 \left[ \sin(\theta_1) + \sin(\theta_2) + \sin(\theta_3) \right] \leq 6 * \sin (\pi / 3) $$
by Jensen's inequality, with the optimum (being the case of the equality) holding when $\theta_1 = \theta_2 = \theta_3$.