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?
 A: 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 $\frac{3\sqrt{3}}{4}$.
A: 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".
A: 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
A: 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.)
A: 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).
A: For the maximum area the normal height of the triangle must be maximum for a given base Hence, the triangle ABC inscribed in unit circle should be an isosceles triangle having, $\angle B=\angle C$,  base BC & angle between equal sides AB & AC be $\theta$ ($=\angle A$). Now, from right triangle, we get $$\sin\theta=\frac{\frac{BC}{2}}{1}\implies BC=2\sin\theta$$ Length of equal sides of triangle is determined as $$AB=AC=\frac{BC}{2\sin\frac{\theta}{2}}=\frac{2\sin\theta}{2\sin\frac{\theta}{2}}=2\cos\frac{\theta}{2}$$ Now, the area of isosceles $\Delta ABC$ is given as
$$A=\frac{1}{2}(AB)(AC)\sin \theta=\frac{1}{2}\left(2\cos\frac{\theta}{2}\right)^2\sin \theta=2\sin\theta\cos^2\frac{\theta}{2}=\sin\theta(1+\cos\theta)$$$$\implies A=\sin\theta+\frac{1}{2}\sin2\theta$$ Now differentiating the area (A) w.r.t. $\theta$ & equating it to zero, we get $$\frac{dA}{d\theta}=\cos\theta+\cos2\theta=0$$$$ \implies\cos\theta=-\cos2\theta=\cos(\pi-2\theta)$$ $$\implies \theta=\pi-2\theta \quad \text{or} \quad 3\theta=\pi \quad \text{or} \quad \theta=\frac{\pi}{3}=60^o=\angle A$$ $$\implies \angle B=\angle C=\frac{180^o-\angle A}{2}=60^o$$
Hence, the triangle having maximum area in a unit circle must be an equilateral triangle
($\color{#0ae}{\angle A=\angle B=\angle C=60^o}$) 
A: To add to the answers already here, the area of the largest (i.e. equilateral) triangle that can be drawn inside a circle, will have side-length: $$\mathcal{a} = r\sqrt{3}$$
This can be proved by using a property: the centroid of the triangle (i.e. the intersection point of the medians) will be the same as the center of the circle. Thus, the length from the centroid to any vertex will be the radius. Considering that the centroid divides the median in a 2:1 ratio, the line from the centroid to any side will have length r/2. This line will also be perperndicular to the side, since it's an equilateral triangle. Using simple Pythogyras' theorum, you can prove the above result.
So, the area of the largest (equilateral) triangle that can be inscribed in a circle would be:
$$\mathcal{Area} = \frac{\sqrt{3}}{4}.a^{2} = \frac{\sqrt{3}}{4}.{(r\sqrt{3})}^{2} = \frac{3\sqrt{3}}{4}.r^{2} $$
Where 'r' is the radius of the circle and 'a' is the side of the triangle. For a unit circle r=1 obviously.
A: 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
asinA^=bsinB^=2R
where R is the radius of the circumscribed circle, that is, 1. I hace omitted the c/sinC^ fraction because we have already shown that b=c. The area of the triangle is
12absinB^=2R2sinA^sinB^=2R2sin2B^sinB^=4R2sin2B^cosB^=4R2(cosB^−cos3B^)
Now define
f(x)=cosx−cos3x
take the derivative
f′(x)=−sinx+3sinxcos2x=2sinx−3sin3x
which vanishes at x=0 and x=π/3. This latter value will give the maximum area (this is indeed the equilatheral triangle).
