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Rectangle problem from RMO 2015 | Chennai Region

This is a Rectangle Problem from RMO (Regional Mathematics Olympiad) 2015 from Chennai Region.

Problem: Rectangle problem from RMO 2015

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: Screen Shot 2015-12-26 at 12.58.56 AM

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. )

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