Olympiad Geometry Problem #57: Altitudes, Midpoint, Circumcircle

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Here is a very beautiful problem from the Cono Sur Math Olympiad in 2007, posted on the Art of Problem Solving Forum. There are many different approaches to it, hope you all enjoy! Link below.
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Very nice solution here is mine. Orthocenther (H) and X are on (AEF), DMFE and DHXM are cyclic so by radical axis theorem we got that DM, XH, FE are concurrent and let it be G, then GB*GC=GF*GE=GH*GX therefore BCXH is cyclic and rest is easy angle chase, have to say motivation from this came from your "H-M" video so thanks for this and lots of more idea that I saw on this chanell.
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Dule-fkrd
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Nice video and even more nice solution!!!

ordanguric
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