Two points X,Y lie on the arc BC of the circumcircle of △ABC (this arc does not contain A) such that ∠BAX=∠CAY . Let M denotes the midpoint of the chord AX . Show that BM+CM>AY . by Mahan Tajrobekar
Problem
Source: IGO 2014 Junior 5
Tags: geometry, circumcircle, arc midpoint, geometric inequality
22.07.2018 17:40
Double post
22.07.2018 18:23
because it is 3 years before, we can say that this one is that problem's source
30.11.2020 09:38
Let O be the circumcenter of the circumcircle of triangle ABC . Then OM⊥AX. We draw a normal line from the point B at OM and let it intersect the circumcircle in the point Z . Since BZ⊥OM we have that OM is a line of symmetry of BZ. According to this, MZ=MB. Now, from the triangle inequality we have that BM+MC=ZM+MC>CZ. But, BZ||AX , so ^AZ=^BX=^CY where from we get ^ZAC=^YCA i.e. CZ=AY. That is why BM+CM>AY.
25.09.2023 16:29
Hmm, why do I think that equality actually can hold (particularly checking Gold_man's approach, in which the triangle inequality application the equality case has not been ruled out)?
Can anybody give an even more natural complex numbers approach?
25.09.2023 20:42
The equality holds when X and Y are the same point.