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.
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. )
- 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 I.S.I. & C.M.I. course,Cheenta Math Olympiad Program, discuss these topics in the ‘Geometry’ module.
- Book Suggestions: Challenges and Thrills of Pre-College Mathematics by Venkatchala