KVPY SX 2017

Geometry Level 3

Consider the disc $x^2+y^2≤1$ . The area of the largest circle that can be inscribed in the $1\text{st}$ quadrant of the disc is

\frac{\pi}{2\sqrt{3}}

\frac{\pi}{8}

\pi\left(3-2\sqrt{2}\right)

\pi\left(4-2\sqrt{3}\right)

1 solution

Rishu Jaar
Nov 7, 2017

Consider the unit circle \ disc $\color{#CEBB00}{(yellow)}$ and the largest circle $\color{#20A900}{(green)}$ that can be inscribed in the unit circle in the first quadrant.
Let it have a radius r . Then by geometry in figure above $OA = r\sqrt{2}$ and $AP = r$ .
Now , here $OP = 1~ unit = OA + AP$ $\Large \implies r\sqrt{2} + r = 1 \\ \Large \implies r = \dfrac{1}{ \sqrt{2} +1 } \\ \Large \implies \color{#3D99F6}{\boxed{r = \sqrt{2} - 1}} \\ \LARGE \therefore ~ Area = \pi \cdot r^2 = \color{#69047E}{\boxed{\pi(3 - 2 \sqrt{2}) }}$