Question

Points O, A, B, C ...... are shown in figure where $OA=2AB=4BC=\dots ..$. so on. Let $A$ is the centroid of a triangle whose orthocentre and circumcenter are $(2,4)$ and $\left(\frac{7}{2},\frac{5}{2}\right)$ 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 $(\alpha +\beta )$

Answer 1

Answer: 8

$\because$ Centroid divides line joining orthocentre & circumcentre in the ratio $2:1$
$\because A=\left(\frac{2\times \frac{7}{2}+1\times 2}{2+1},\frac{2\times \frac{5}{2}+1\times 4}{2+1}\right)=(3,3)\Rightarrow OA=3\sqrt{2}$
$x$ coordinate of point $P=OA\cos 45^{\circ }+AB\cos 45^{\circ }+BC\cos 45^{\circ }\dots \dots ..$.
$=\cos 45^{\circ }\left(OA+\frac{OA}{2}+\frac{OA}{4}+\dots \dots ..\right)=\frac{1}{\sqrt{2}}\times \frac{OA}{1-\frac{1}{2}}=\frac{2}{\sqrt{2}}\times 3\sqrt{2}=6$
$y\text{-coordinate of }P=OA\sin 45^{\circ }-AB\sin 45^{\circ }+BC\sin 45^{\circ }\dots \dots \dots ..$
$=\sin 45^{\circ }(OA-AB+BC\dots \dots ..)$
$=\frac{1}{\sqrt{2}}\left(OA-\frac{OA}{2}+\frac{OA}{4}\dots \dots .\right)=\frac{1}{\sqrt{2}}\frac{OA}{1-\left(-\frac{1}{2}\right)}=\frac{2}{3\sqrt{2}}OA=\frac{2\times 3\sqrt{2}}{3\sqrt{2}}=2$
$\therefore P\equiv (6,2)\Rightarrow \alpha +\beta =8$