Finding the largest triangle inscribed in the unit circle

Among all triangles inscribed in the unit circle, how can the one with the largest area be found?

Solution 1:

Take an arbitrary triangle inscribed in the circle and let one of the sides subtend the central angle $\alpha$.

Keeping this side fixed and moving the opposite vertex to form an isoceles triangle, we get a larger triangle, and the two other sides will both subtend the central angle $\pi-\dfrac\alpha2$.

Repeating with one of the other sides, we establish the recurrence $\alpha_{k+1}=\pi-\dfrac{\alpha_k}2$. This sequence always converges to $\alpha=\dfrac{2\pi}3$, which yields the largest area.

(Actually it suffices to say that a non-equilateral triangle can always be enlarged.)

Solution 2:

Given any triangle $\triangle ABC$ of sides $a,b$ and $c$, let $R$ be its circumradius and $\mathcal{A}$ be its area. We have this interesting identity: $$4 R \mathcal{A} = abc$$ When $ABC$ is inscribed inside the unit circle, $R = 1$ and by $GM \le AM$, we have

$$\mathcal{A} = \frac14 abc \le \frac14 \left(\frac{a^2+b^2+c^2}{3}\right)^{3/2}$$

Notice $$\begin{align}a^2 + b^2 + c^2 &= |\vec{A} - \vec{B}|^2 + |\vec{B} - \vec{C}|^2 + |\vec{C}-\vec{A}|^2\\ &= 6 - 2\left(\vec{A}\cdot\vec{B} + \vec{B}\cdot\vec{C} + \vec{C}\cdot\vec{A}\right)\\ &= 9 - |\vec{A} + \vec{B} + \vec{C}|^2 \end{align} $$ This leads to an upper bound for the area

$$\mathcal{A} \le \frac14 \left(\frac{9}{3}\right)^{3/2} = \frac{3\sqrt{3}}{4}$$

Since this upper bound is attained by an equilateral triangle of side $\sqrt{3}$, the maximum area is $\frac{3\sqrt{3}}{4}$.

Solution 3:

Take a unit circle and take $A$ one of the vertices of the triangle to be on the $x$ axis so $A(1,0)$. Let $\theta$ and $\phi$ be the angles between $\vec{OA}$, $\vec{OB}$ and $\vec{OC}$ respectively. So one has

$$\vec{AB}=(\cos\theta-1,\sin\theta)$$ $$\vec{AC}=(\cos\phi-1,\sin\phi)$$

We want to maximize

$$\mathfrak{A}(\theta,\phi)=\begin{vmatrix} \cos\theta-1&\sin\theta\\ \cos\phi-1&\sin\phi \end{vmatrix}$$

The partial derivatives are

$$\sin\theta\sin\phi+\cos\theta(\cos\phi-1)=0$$ $$(\cos\theta-1)\cos\phi+\sin\phi\sin\theta=0$$

Which can be rewritten as

$$\cos(\theta-\phi)=\cos\theta$$ $$\cos(\theta-\phi)=\cos\phi$$

This means $\phi=2\pi-\theta$ and $3\theta=2\pi$ whence $\theta=\frac{2\pi}{3}$ and the triangle is equilateral. "The symmetry is in the cosine equations".

Solution 4:

HINT:

Like ajotatxe,

$$\dfrac a{\sin A}=\cdots=2R$$ where $R$ is the circum-radius

and $\triangle=\dfrac{abc}{4R}=2R^2\sin A\sin B\sin C$

Now follow this

Solution 5:

Fix WLOG a side $a$ to be parallel to $X$ axis. The maximum area with this side fixed is when $A$ is at the $Y$ axis, because the altitude is maximum. This way you can show that the maximum triangle is (at least) isosceles.

Now, the law of the sines states that $$\frac a{\sin \hat A}=\frac b{\sin\hat B}=2R$$ where $R$ is the radius of the circumscribed circle, that is, $1$. I hace omitted the $c/\sin \hat C$ fraction because we have already shown that $b=c$. The area of the triangle is $$\frac12ab\sin\hat B=2R^2\sin\hat A\sin\hat B=2R^2\sin2\hat B\sin\hat B=4R^2\sin^2\hat B\cos\hat B=4R^2(\cos\hat B-\cos^3\hat B)$$

Now define $$f(x)=\cos x-\cos^3 x$$ take the derivative $$f'(x)=-\sin x+3\sin x\cos^2x=2\sin x-3\sin^3 x$$ which vanishes at $x=0$ and $x=\pi/3$. This latter value will give the maximum area (this is indeed the equilatheral triangle).