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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: 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. )


December 26, 2015

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