Olympiad Geometry Problem #8: Perpendiculars, Midpoints, Equal Angles

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This is a simple but fun problem from SMO 2018. A couple of right angles are given, and you want to use that to show two angles are equal. Enjoy!
  1. Michael Greenberg
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Wow, that is really a nice solution :D
And here is mine :

Let OM intersect AB at E and let F be the reflection of C respect to BO, so we get F, O and C colliner and FA//OM. İt is enough to show that FAOB is cyclic. <OFA = <FOE = <EBO = <ABO and we get FAOB is cyclyc as desired.

kerimbendov
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neat insight on the parallelogram construction! here's mine:
-> Construct F to be the foot of the altitude from C to AB. Notice that BFOC is cyclic.
-> Extend OM such that it intersects AB at X. By some angle chasing (on quadrilateral XMNB), we find that MX must be perpendicular to AB, implying that MX//CF. Since M is the midpoint of AC, X must be the midpoint of AF. Thus, O lies on the perpendicular bisector of AF.
-> From this, <BAO = <XAO = 90 - <AOX = 90 - <FOX = 90 - <OFC = 90 - <OBC = <BCO

justinvillafuerte
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Let P be the point on the extension of OM such that MP=MO. Then AOCP is a parallelogram and NM is an altitude and a median in triangle ONP. Therefore NO=NP and P lies on the circumcircle of BOC and angle BPC = 90. AO=CP. Let`s take D such that DBPC is a rectangle. Then D lies on the circumcircle of BOC and BD=CP. Therefore BD=CP=AO and AO || CP || BD. It follows that AODB is a parallelogramm and angle BAO = angle BDO = angle BCO.

rustamnovikov