Question

Show that the triangle of maximum area that can be inscribed in a given cricle is an equilateral triangle.

Answer 1

Let $\Delta ABC$ is inscribed in a circle of radius r. A perpendicular $E$ is drawn of $AC$ from the vertex $B$ of triangle which meets $AC$ at $D$ and circle at $E$.

Let base of triangle $AC=2x$

and height of triangle $BD=y$

Then, area of triangle $A=\frac{1}{2}\times 2x\times y=xy$ ....(i)

$\because AC$ and $BE$ are chords of circle.

So, $AD\times DC=BD\times DE$

$\because$ radius of circle $=r$

$DE=BE-DB=2r-y]$

$\Rightarrow x\times x=y\times (2r-y)$

$\Rightarrow x^2=2ry-y^2$

$\Rightarrow x=\sqrt{2ry-y^2}$ ....(ii)

From equation (i) and (ii),

$A=\sqrt{2ry-y^2}y$

$\Rightarrow A^2=(2ry-y^2)y^2=2r{y}^3-y^4$

$\Rightarrow A^2=2r{y}^3-y^4$

Diff. w.r.t. y,

$2A\frac{dA}{dy}=6r{y}^2-4y^3$ ....(iii)

For maximum value of $A,\frac{dA}{dy}=0$

$\therefore \frac{dA}{dy}=0$

$\Rightarrow 6r{y}^2-4y^3=0$ $[\because A\ne 0]$

$\Rightarrow 2h{y}^2(3r-2y)=0$

$3r=2y$ $[\because y\ne 0]$

$\Rightarrow y=\frac{3}{2}r$

Diff. (iii) again w.r.t. y.

$2\frac{dA}{dy}.\frac{dA}{dy}+2A.\frac{d^{2}A}{d{y}^2}=12ry-12y^2$

$\Rightarrow 2{\left(\frac{dA}{dy}\right)}^2+2A\frac{d^{2}A}{d{y}^2}=12ry-12y^2$ ...(iv)

Putting in equation (iv) $\frac{dA}{dy}=0$ and $y=\frac{3}{2}r$

$2A.\frac{d^{2}A}{d{y}^2}=12r\times \frac{3}{2}r-12\times \frac{9}{4}{r}^2$

$=18r^2-27r^2=--r^2<0$

$\therefore \frac{d^{2}A}{d{y}^2}<0$ $(\because 2A>0)$

So, area of triangle will be maximum

When $y=\frac{3}{2}r$ then $x=\sqrt{2r\times \frac{3}{2}r-\frac{9}{4}{r}^2}$

$\Rightarrow x=\sqrt{3r^2-\frac{9}{4}{r}^2}$

$=\sqrt{\frac{12r^2-9r^2}{4}}=\sqrt{\frac{3r^2}{4}}$

$x=\frac{\sqrt{3}r}{2}$

In right angle $\Delta ABC$,

$tanA=\frac{BD}{AD}=\frac{y}{x}=\frac{\frac{3}{2}r}{\frac{\sqrt{3}r}{2}}=\sqrt{3}$

$\Rightarrow tanA=tan{60}^{\circ }$

$\Rightarrow A={60}^{\circ }$

Similarly $tanC=\frac{BD}{AD}=\frac{y}{x}=\frac{\frac{3}{2}r}{\frac{\sqrt{3}r}{2}}=\sqrt{3}$

$\Rightarrow tanC=tan{60}^{\circ }$

$\Rightarrow C={60}^{\circ }$

$\therefore \angle ABC={60}^{\circ }$

Hence, $\Delta ABC$ is an equilateral triangle.