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

**Problem: Rectangle problem from RMO 2015**

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.

** Discussion: **

Note that at P . Since is center and and CD is chord of the circle , perpendicular drawn from center to chord bisects the chord. Therefore . Similarly since at P, hence PE = EF.

Since diagonals bisects each other, clearly is a parallelogram.

The power of the point P with respect to the circle is since we previously showed PE = PF.

Similarly power of the point P with respect to the circle is since we previously showed PC = PD.

Hence we have .

Similarly we can show PE = PC. Thus we have PE = PC = PF = PD.

Now it is trivial to show CEDF is rectangle. (Consider . Since PE = PC we have (say). Similarly in , since PF = PC, we have (say). Thus or . Similarly one can show that the remaining angle of the quadrilateral is right angle. )

## Chatuspathi:

**Paper:**RMO 2015 (Chennai)**What is this topic:**Geometry**What are some of the associated concepts:**Power of a point**Where can learn these topics:**Cheenta**Book Suggestions:**Challenges and Thrills of Pre-College Mathematics by Venkatchala

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