Let $(1,1),(-1,1),(-1,-1),(1,-1)$ be a vertices of the square and $S$ the region of all the points i

Question

Let $(1,1),(-1,1),(-1,-1),(1,-1)$ be a vertices of the square and $S$ the region of all the points insides the square which are nearer to the origin than to any edge. If the area of $S=\frac{m\sqrt{2}-n}{3}$, find $m+n$

Professor PC Thomas Classes · Answer

Let $A(1,1), B(-1,1), C(-1,-1)$, $D(1,-1)$\ Equation of $A B$ is $y=1$\ Let $S$ be the shaded portion of the region formed by four PQRS\ Now four cases arise\ (i) $\quad y^{2} \leq 1-2 x$ by squaring\ $\sqrt{x^{2}+y^{2}} \leq|1-x|$\ (ii) $y^{2} \leq 1+2 x$ by squaring\ $\sqrt{x^{2}+y^{2}} \leq|1+x|$\ (iii) $x^{2} \leq 1-2 y$ by squaring\ $\sqrt{x^{2}+y^{2}} \leq|1-y|$\ (iv) $x^{2} \leq 1+2 y$ by squaring\ $\sqrt{x^{2}+y^{2}} \leq|1+y|$\ Solving curve (i) and curve (iii), we get\ $P(\sqrt{2}-1, \sqrt{2}-1)$\ Area bounded by the square $P Q R S=$\ $[2(\sqrt{2}-1)]^{2}$\ Area between the curve (i) and line\ $x=\sqrt{2}-1$ is\ $2 \int_{\sqrt{2}-1}^{1 / 2} y d x=2 \int_{\sqrt{2}-1}^{1 / 2} \sqrt{1-2 x} d x=\frac{2}{3}(\sqrt{2}-1)^{3}$\ Total area $=$ Area of the square $P Q R S+4$\ (Area of the curve (i)) & the line $x=\sqrt{2}-1$\ $=4(\sqrt{2}-1)^{2}+\frac{2}{3}(\sqrt{2}-1)^{3} 	imes 4$\ $=4(\sqrt{2}-1)^{2}\left[1+\frac{2(\sqrt{2}-1)}{3}ight]$\ $=\frac{4(3-2 \sqrt{2})(1+2 \sqrt{2})}{3}=\frac{16 \sqrt{2}-20}{3}$\