Let ABCD be an inscribed quadrilateral in a circumference with center O such that it lies inside ABCD and ∠BAC=∠ODA. Let E be the intersection of AC with BD. Lines r and s are drawn through E such that r is perpendicular to BC, and s is perpendicular to AD. Let P be the intersection of r with AD, and M the intersection of s with BC. Let N be the midpoint of EO. Prove that M, N, and P lie on a line.
Problem
Source:
Tags: geometry, circumcircle, parallelogram
23.08.2014 02:11
23.08.2014 22:30
Well, if you change D and E and angle-chase a little, than it's http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3109660&sid=64df69e2c903958785a9c858b780a3de#p3109660
24.08.2014 01:31
Let be <ODA= a and <ODB=<OBD=b => <DAB= 90-b so <CAD=<DBD=90-a-b so <BEC=90 by brahmagupta teorem, PA=PD and MB=MC, so PO// EM and OM//EP, so POME is a parallelogram so M, N and P are collinear.
26.08.2014 12:19
This is far from being an original problem. A very similar configuration (and the same idea) was used in Iberoamerican 2010 Problem 5 (for part b of course) http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2029577&sid=6d200e14ae41b691506c45f0c1a74c44#p2029577 All of the countries that participate in the Cono Sur Olympiad also participate in the Iberoamerican Olympiad. I find it weird that no one noticed the similarity.
18.08.2023 09:58
First we have AC⊥BD. Then let T be the foot from E to AD, we want to prove M,E,T are colinear. we have ∡TED=∡DAC=∡DBC=∡DBM=∡MEB=∡MED yielding the result. So M is the midpoint of BC, similarly P is the midpoint of AD. We have POME is a parallelogram and we're done.