# 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:

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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### One comment on “Rectangle problem from RMO 2015 | Chennai Region”

1. […] Two circles and having centers at and intersect at A and B. Let P be a point on the segment AB and let . The line through P perpendicular to meets at C and D. The line through P perpendicular to meets at E and F. prove that C,D, E and F form a rectangle. SOLUTION: here […]

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