What is the maximum value of the product of the length of a side and the opposite the angle of a triangle?

Question

WLOG, let $a$ be the side of a triangle inscribed in a unit circle and $A$ be the opposite angle. Experimental data shows that there is a constant $C \approx 3.6394$ such that $aA \le C$.

Question: What is the maximum value of $aA$?

Progress: Trivially, $\max \left (aA\right) \ge \pi$. This follows from the case when the triangle is a right triangle in a semi-circle so that $a = 2$ and $A = \frac{\pi}{2}$. A slightly less trivial bound is $\max \left (aA\right) \ge \left( \frac{2\pi}{\sqrt{3}}\right) \approx 3.6276$ which very close to the experimentally observed maxima. This corresponds to the case when we have a triangle whose angles are $\left(\frac{2\pi}{3}, \frac{\pi}{6}, \frac{\pi}{6}\right)$.

Answer 1

Let us take a Point $D$ on the Unit Circle , such that the Diametrically Opposite Point is $A$.

Let a Point $P$ move on $DA$ , whereby we get the triangle with Altitude $AP$.

When Angle at $A$ is $a$ , then the angle at the Center will be $2a$ , hence the base will be $2\sin(a)$.

The value we want to maximize is $P = 2a \sin(a)$

$dP/da = 2 \sin(a) + 2a \cos(a)$

This has a Zero at $\tan(a) = -a$ which gives a value a ≈ 2.02875783811043... (according to wolfram)

That gives $P = 3.639411482319306092413915357584$

This Maximum value is the Constant $C$ we want.

Answer 2

In a (unit) circle of radius $r=1$, the product angle A $\times$ cord a is

$$Aa=2\cdot \frac {Aa}{2\cdot \underbrace{r}_{=1}}=2A\cdot \frac a{2r}=2A\sin A$$

Therefore the problem resumes to finding the maximum of $$f:(0, \pi)\to \Bbb R, f(x)=2x\sin x$$

The exact value of the maximum is attained for $x=-\tan x\Rightarrow x\approx 2.02876$ $$f(2.02876)\approx 3.639$$