# Rectangle problem (RMO 2015 Chennai Solution)

Problem: Two circles $$\Sigma_1$$ and $$\Sigma_2$$ having centers at $$C_1$$ and $$C_2$$ intersect at A and B. Let P be a point on the segment AB and let $$AP \neq PB$$. The line through P perpendicular to $$C_1 P$$ meets $$\Sigma_1$$ at C and D. The line through P perpendicular to $$C_2P$$ meets $$\Sigma_2$$ at E and F. prove that C,D, E and F form a rectangle.

Discussion:

Note that $$C_1 P \perp CD$$ at P . Since $$C_1$$ is center and and CD is chord of the circle $$\Sigma_1$$, perpendicular drawn from center to chord bisects the chord. Therefore $$PC = PD$$. Similarly since $$C_2 P \perp EF$$ at P, hence PE = EF.
Since diagonals bisects each other, clearly $$CEDF$$ is a parallelogram.
The power of the point P with respect to the circle $$\Sigma_2$$ is $$PA \times PB = PE \times PF = PF^2$$ since we previously showed PE = PF.
Similarly power of the point P with respect to the circle $$\Sigma_1$$ is $$PA \times PB = PD \times PC = PD^2$$ since we previously showed PC = PD.
Hence we have $$PA \times PB = PF^2 = PD^2 \Rightarrow PF = PD$$.
Similarly we can show PE = PC. Thus we have PE = PC = PF = PD.
Now it is trivial to show CEDF is rectangle. (Consider $$\Delta PEC$$. Since PE = PC we have $$\angle PEC = \angle PCE = x$$ (say). Similarly in $$\Delta PFC$$, since PF = PC, we have $$\angle PFC = \angle PCF = y$$ (say). Thus $$2x + 2y = 180^o$$ or $$\angle PCF + \angle PCE = x + y = 90^o$$. Similarly one can show that the remaining angle of the quadrilateral is right angle. )