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

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    $\begingroup$ By symmetry, should be equilateral, no ? $\endgroup$ – Yves Daoust May 25 '15 at 17:38
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    $\begingroup$ @YvesDaoust : There are logical arguments from symmetry, but yous is an intuitive argument from symmetry. ${}\qquad{}$ $\endgroup$ – Michael Hardy May 25 '15 at 17:39
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    $\begingroup$ I guess that it can be made rigorous by showing that when the triangle isn't equilateral there is a way to increase the area by moving a vertex. $\endgroup$ – Yves Daoust May 25 '15 at 17:41
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    $\begingroup$ @YvesDaoust How do you know equilateral triangles aren't the smallest, by symmetry? $\endgroup$ – PyRulez May 25 '15 at 21:01
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    $\begingroup$ @PyRulez: see my answer. (Also, flat triangles have no area and by continuity there are triangles as small as you want.) $\endgroup$ – Yves Daoust May 25 '15 at 21:13
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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.

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

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    $\begingroup$ I believe you mean "Take an arbitrary triangle..." :) $\endgroup$ – mathmandan May 25 '15 at 19:03
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    $\begingroup$ I believe it too ;-) $\endgroup$ – Yves Daoust May 25 '15 at 19:04
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    $\begingroup$ I guess we should at least mumble something about compactness and smoothness (or equivalent) to show that a largest triangle exists. $\endgroup$ – Drunix May 26 '15 at 6:03
  • $\begingroup$ @Drunix: quite right. We know that the area of any triangle lies in range $[0,A_E]$ and by continuity all values in that range can be achieved. Isn't that sufficient ? $\endgroup$ – Yves Daoust May 26 '15 at 7:02
  • $\begingroup$ @YvesDaoust Yes, that's what I actually meant. I just mixed up smoothness and continuity in my comment. I just wanted to point out that the "Make other triangles larger" argument for completeness also requires that a solution exists. $\endgroup$ – Drunix May 26 '15 at 9:25
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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 $\frac{3\sqrt{3}}{4}$.

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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

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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".

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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).

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  • $\begingroup$ My guts say me that there must be another much shorter way to show it, but I can't find it! $\endgroup$ – ajotatxe May 25 '15 at 18:07
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    $\begingroup$ You initially showed that the triangle must be isosceles with respect to any side. Isn't that enough? $\endgroup$ – Bill Kleinhans May 25 '15 at 18:08
  • $\begingroup$ Possibly, but not sure. I have shown that if the triangle is not isosceles, then there is another triangle which is isosceles and have a greater area.I can't see how to jump to the third side. $\endgroup$ – ajotatxe May 25 '15 at 18:09
  • $\begingroup$ @ajotatxe: that's it: if it is not isosceles taking some side as the basis, then you can increase the area. The maximum is reached when it is triply isosceles. $\endgroup$ – Yves Daoust May 25 '15 at 18:13
  • $\begingroup$ If the triangle is not equilateral, you can modify it to increase its area. So, if there is a triangle with maximum area, it must be equilateral. There are several optimization problems that have this subtlety. For example, the proof that the figure with maximum area for fixed perimeter is a circle. $\endgroup$ – Bill Kleinhans May 26 '15 at 23:33
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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
($\color{#0ae}{\angle A=\angle B=\angle C=60^o}$)

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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.

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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).

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protected by hardmath Jun 11 '18 at 2:17

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