IMO Shortlist 2010 problem G5


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23. lipnja 2013.
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Let ABCDE be a convex pentagon such that BC \parallel AE, AB = BC +  AE, and \angle ABC = \angle CDE. Let M be the midpoint of CE, and let O be the circumcenter of triangle BCD. Given that \angle DMO = 90^{\circ}, prove that 2 \angle BDA = \angle CDE.

Proposed by Nazar Serdyuk, Ukraine
Izvor: Međunarodna matematička olimpijada, shortlist 2010