Points O, A, B, C… are shown in figure where $OA=2AB=4BC=\mathrm{..}$ so on. Let A be the centroid of a triangle whose orthocenter and circum center 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 find the value of $(\alpha +\beta ).$

 

$\because$ centroid divides line joining orthocentre and circumcentre in the ratio 2:1

$\therefore A=(\frac{2\times \frac{7}{2}+1\times 2}{2+1},\frac{2\times \frac{5}{2}+1\times 4}{2+1})=(3,3)\Rightarrow OA=3\sqrt{2}$

x-coordinate of point $P=OA\cos {45}^0+AB\cos {45}^0+BC\cos {45}^0\mathrm{....}$

$=\cos {45}^0(OA+\frac{OA}{2}+\frac{OA}{4}+\mathrm{....})=\frac{1}{\sqrt{2}}\times \frac{OA}{1-\frac{1}{2}}=\frac{2}{\sqrt{2}}\times 3\sqrt{2}=6$

y-coordinate of point $P=OA\sin {45}^0-AB\sin {45}^0+BC\sin {45}^0\mathrm{....}=\sin {45}^0(OA-AB+BC\mathrm{.....})$

$=\frac{1}{\sqrt{2}}(OA-\frac{OA}{2}+\frac{OA}{4}\mathrm{...})=\frac{1}{\sqrt{2}}\frac{OA}{1-(1-\frac{1}{2})}=\frac{2}{3\sqrt{2}}OA=\frac{2\times 3\sqrt{2}}{3\sqrt{1}}=2$

$\therefore P\equiv (6,2)\Rightarrow \alpha +\beta =8$