Question

Points O, A, B, C ...... are shown in figure where $OA =2 AB =4 BC =\ldots . .$. 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

$\because$ 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 $P = OA \cos 45^{\circ}+ AB \cos 45^{\circ}+ BC \cos 45^{\circ} \ldots \ldots . .$.
$=\cos 45^{\circ}\left( OA +\frac{ OA }{2}+\frac{ OA }{4}+\ldots \ldots . .\right)=\frac{1}{\sqrt{2}} \times \frac{ OA }{1-\frac{1}{2}}=\frac{2}{\sqrt{2}} \times 3 \sqrt{2}=6$
$y \text {-coordinate of } P = OA \sin 45^{\circ}- AB \sin 45^{\circ}+ BC \sin 45^{\circ} \ldots \ldots \ldots . .$
$ =\sin 45^{\circ}( OA - AB + BC \ldots \ldots . .) $
$=\frac{1}{\sqrt{2}}\left( OA -\frac{ OA }{2}+\frac{ OA }{4} \ldots \ldots .\right)=\frac{1}{\sqrt{2}} \frac{ OA }{1-\left(-\frac{1}{2}\right)}=\frac{2}{3 \sqrt{2}} OA =\frac{2 \times 3 \sqrt{2}}{3 \sqrt{2}}=2$
$\therefore P \equiv(6,2) \Rightarrow \alpha+\beta=8$