Question

Points $O,A,B,C,\dots$ are shown in figure where $OA=2AB=4BC=\dots$ so on. Let $A$ be the centroid of a triangle whose orthocentre and circumcentre are $(2,4)$ and $(\frac{7}{2},\frac{5}{2})$ respectively. If an insect starts moving from the point $O(0,0)$ along the straight line in zig-zag fashion and terminates ultimately at point $P(\alpha ,\beta ),$ then the value of $\alpha +\beta$ is

Answer 1

Since, centroid divides line joining orthocentre and circumcentre in the ratio $2:1,$
$\therefore A\equiv (\frac{2\times \frac{7}{2}+1\times 2}{2+1},\frac{2\times \frac{5}{2}+1\times 4}{2+1})=(3,3)$
$\Rightarrow OA=3\sqrt{2}$
So, $x$-coordinate of point $P$ is $OA\cos {45}^{\circ }+AB\cos {45}^{\circ }+BC\cos {45}^{\circ }+\cdots$
$=\cos {45}^{\circ }(OA+\frac{OA}{2}+\frac{OA}{4}+\cdots )$
$=\frac{1}{\sqrt{2}}\times \frac{OA}{1-\frac{1}{2}}$
$=\frac{1}{\sqrt{2}}\times \frac{3\sqrt{2}\times 2}{1}=6$

Similarly, $y$-coordinate of point $P$ is $OA\sin {45}^{\circ }-AB\sin {45}^{\circ }+BC\sin {45}^{\circ }-\cdots$
$=\sin {45}^{\circ }(OA-\frac{OA}{2}+\frac{OA}{4}-\cdots )$
$=\frac{1}{\sqrt{2}}\frac{OA}{1-\left(\frac{-1}{2}\right)}$
$=\frac{2}{3\sqrt{2}}OA=2$
$\therefore P\equiv (6,2)$
$\Rightarrow \alpha +\beta =8$