Question

For a point $M$ in the plane, let $d_1(M)$ and $d_2(M)$ be the distances of the point $M$ from the lines $x-3y=0$ and $x+3y=0$ respectively. If area of the region $R$ consisting of all points $M$ lying in the first quadrant of the plane and satisfying $6\le d_1(M)+d_2(M)\le 12$ is $A$, then find the sum of digits of [A].
[Note: [k] denotes greatest integer function less than or equal to k.]

Answer 1

Answer: 9

$\Theta 6\le \left|\frac{x-3y}{\sqrt{10}}\right|+\left|\frac{x+3y}{\sqrt{10}}\right|\le 12$
$\Theta$ In first quadrant $x,y>0$
$\therefore 6\sqrt{10}\le |x-3y|+x+3y\le 12\sqrt{10}$
(i) If $x\ge 3y$ $6\sqrt{10}\le 2x\le 12\sqrt{10}$
$\Rightarrow 3\sqrt{10}\le x\le 6\sqrt{10}$
(ii)$6\sqrt{10}\le 6y\le 12\sqrt{10}$
$\Rightarrow \sqrt{10}\le y\le 2\sqrt{10}$
$\therefore$ Required area $=$ shaded area
$=6\sqrt{10}\times 2\sqrt{10}-3\sqrt{10}\times \sqrt{10}$
$\therefore A=90$
$\therefore \text{ Sum of digits }=9$