Olympiad Geometry Problem #90: Incircle, Midpoint, Equal Angles

Here is a very enjoyable problem from the 2009 Oral Moscow Math Olympiad, posted by the user parmenides51 on the Art of Problem Solving Forum. A little bit easier than some of my latest videos, but still fun to solve. Enjoy! Link below.

[mishania]

can also be easily solved with sparrow lemmas 1 and 2

[draj]

Hi there i m frm india and first of all i wanna(thank u) bcz my all searchings ends here as i was looking for a dedicated channel of geometry consisting of questions from various national and international levels.Thank god ! Finally i found a gem.Its my humble request to u that in future also, u pls keep posting videos of geometry (theory+olympiad questions)

[MarinHristov-nm]

Quite an easy problem. My solution introduces just one point though. Let MN intersect BI at S(its point G in your solutiom). Now we know that S is the midpoint of the arc AC on the circumcircle of triangle ABC. Furthermore now NS is a diameter of this circle. Angle SCN=90deg=SMC so triangles SCM and SNC are similar triamgles. Now SC^2=SM.SN, but also SA=SI=SC by a very well known and easy to prove lemma. So SI^2=SM.SN and thus triangles SIM and SNI are similar. It follows that angle SIM= angle SNI. We are almost ready, just have to notice that 90deg-angle IMA= angle IMN= angle SIM + angle MSI= angle SNI+ angle ISN=angle BIN = 90deg - angle BNI (because MN is a diameter) and it follows that angle IMA= angle INB.

Although an easy problem I really liked it and overall want to support yt channels solving olympiad problems. Thank you very much, you have earned my subscription!

[yashvardhan]

Thank you Sir
I am solving EGMO by Evan Chen .It contains Many of the good Lemmas that u also use .Should I continue practising it.U can consider me as a beginner but I am very familiar with the usual theorems and it takes time for me to solve the EGMO problems.Can u plz give any advice.