# What is the area of the largest isosceles triangle that can be inscribed in a circle of radius 4?

Feb 25, 2016

$12 \sqrt{3} \cong 20.784$

#### Explanation:

One could start by saying that the isosceles triangle with largest area inscribed in a triangle is also an equilateral triangle.

However if you need a formal demonstration of this statement read the first part of this explanation.

Suppose an isosceles ${\triangle}_{A B C}$ inscribed in a circle with center in $D$ and radius $r$, like the figure below.

We can obtain the side $a$ in function of $r$ and $\alpha$ in this way (Law of Sines applied to ${\triangle}_{B C D}$):

$\frac{a}{\sin} \left(2 \alpha\right) = \frac{r}{\sin} \beta$
Since $2 \alpha + \beta + \beta = {180}^{\circ}$ => $\beta = {90}^{\circ} - \alpha$
=> $a = r \sin \frac{2 \alpha}{\sin} \left({90}^{\circ} - \alpha\right) = 2 r \frac{\sin \alpha \cdot \cancel{\cos \alpha}}{\cancel{\cos \alpha}}$
=> $a = 2 r \sin \alpha = 4 r \sin \left(\frac{\alpha}{2}\right) \cos \left(\frac{\alpha}{2}\right)$

We can obtain the height $h$ in function of $r$ and $\alpha$ in this way:

$\tan \left(\frac{\alpha}{2}\right) = \frac{\frac{a}{2}}{h}$ => $h = \frac{a}{2 \tan \left(\frac{\alpha}{2}\right)}$
Replacing $a$ for its value in function of $r$ and $\alpha$:
$h = \frac{4 r \cancel{\sin \left(\frac{\alpha}{2}\right)} \cos \left(\frac{\alpha}{2}\right)}{2} \cdot \cos \frac{\frac{\alpha}{2}}{\cancel{\sin \left(\frac{\alpha}{2}\right)}}$
=> $h = 2 r {\cos}^{2} \left(\frac{\alpha}{2}\right)$

Then we can obtain the area of the triangle in function of $r$ and $\alpha$:

${S}_{{\triangle}_{A B C}} = \frac{\text{base"*"height}}{2}$
${S}_{{\triangle}_{A B C}} = \frac{\left(4 r \sin \left(\frac{\alpha}{2}\right) \cos \left(\frac{\alpha}{2}\right)\right) \left(\cancel{2} r {\cos}^{2} \left(\frac{\alpha}{2}\right)\right)}{\cancel{2}}$
${S}_{{\triangle}_{A B C}} = 4 {r}^{2} \sin \left(\frac{\alpha}{2}\right) {\cos}^{3} \left(\frac{\alpha}{2}\right)$

Since in this problem $r$ is constant, we need to find the derivative relatively to $\alpha$ of ${S}_{{\triangle}_{A B C}}$ and equal it to zero to find the maximum or minimum of the area of the triangle.

So

$\frac{d}{\mathrm{da} l p h a} \left({S}_{{\triangle}_{A B C}}\right) = 4 {r}^{2} \cos \left(\frac{\alpha}{2}\right) \cdot \left(\frac{1}{2}\right) {\cos}^{3} \left(\frac{\alpha}{2}\right) + 4 {r}^{2} \sin \left(\frac{\alpha}{2}\right) \cdot 3 {\cos}^{2} \left(\frac{\alpha}{2}\right) \left(- \sin \left(\frac{\alpha}{2}\right)\right) \left(\frac{1}{2}\right)$
$= 2 {r}^{2} {\cos}^{4} \left(\frac{\alpha}{2}\right) - 6 {r}^{2} {\sin}^{2} \left(\frac{\alpha}{2}\right) {\cos}^{2} \left(\frac{\alpha}{2}\right)$
$= 2 {r}^{2} {\cos}^{2} \left(\frac{\alpha}{2}\right) \left({\cos}^{2} \left(\frac{\alpha}{2}\right) - 3 {\sin}^{2} \left(\frac{\alpha}{2}\right)\right)$
Equating the derivative to zero we get:
${\cos}^{2} \left(\frac{\alpha}{2}\right) = 0$ => $\cos \left(\frac{\alpha}{2}\right) = 0$ => $\frac{\alpha}{2} = {90}^{\circ}$ => $\alpha = {180}^{\circ}$ (this is the point of minimum area)
And
${\cos}^{2} \left(\frac{\alpha}{2}\right) - 3 {\sin}^{2} \left(\frac{\alpha}{2}\right) = 0$ => $3 {\sin}^{2} \left(\frac{\alpha}{2}\right) = {\cos}^{2} \left(\frac{\alpha}{2}\right)$ => $\tan \left(\frac{\alpha}{2}\right) = \frac{1}{\sqrt{3}}$ => $\frac{\alpha}{2} = {30}^{\circ}$ => $\alpha = {60}^{\circ}$ (this is the point of maximum area)
$\to$ But if $\alpha = {60}^{\circ}$ this means that $A \hat{B} C = A \hat{C} B = {60}^{\circ}$ and the isosceles triangle of maximum area is also an equilateral triangle.

Area of the triangle of maximum area

${S}_{{\triangle}_{A B C}} = 4 \cdot {4}^{2} \sin {30}^{\circ} \cdot {\cos}^{3} {30}^{\circ} = 4 \cdot 16 \cdot \left(\frac{1}{2}\right) {\left(\frac{\sqrt{3}}{2}\right)}^{3} = 32 \cdot \frac{3 \sqrt{3}}{8} = 12 \sqrt{3} \cong 20.784$