In triangle $ABC$, the points $X$ and $Y$ are chosen on the arc $BC$ of the circumscribed circle of $ABC$ that doesn't contain $A$ so that $\measuredangle BAX = \measuredangle CAY$. Let $M$ be the midpoint of the segment $AX$. Show that $$BM + CM > AY.$$
Problem
Source: Macedonia JBMO TST 2017, Problem 4
Tags: circumcircle, geometry, Geometric Inequalities
27.06.2018 20:18
Let $O$ be the circumcenter of $ABC$ , by the conditions we have $\measuredangle BAX=\measuredangle BOX$ and $\measuredangle CAY=\measuredangle COY$ and by $\measuredangle BAX = \measuredangle CAY$ we have that $BX=CY$ ( We can do this directly from Law of Sines $BX=2R \sin(BAX)=2R \sin(CAY)=CY$ ) Now , from triangle inequality we have $AY-MB<AY-(BX+MX)=AY-BX-MX<AC-MX=AC-MA<MC$ $\Longleftrightarrow AY<MB+MC$
04.08.2018 03:15
XbenX wrote: Let $O$ be the circumcenter of $ABC$ , by the conditions we have $\measuredangle BAX=\measuredangle BOX$ and $\measuredangle CAY=\measuredangle COY$ and by $\measuredangle BAX = \measuredangle CAY$ we have that $BX=CY$ ( We can do this directly from Law of Sines $BX=2R \sin(BAX)=2R \sin(CAY)=CY$ ) Now , from triangle inequality we have $AY-MB<AY-(BX+MX)=AY-BX-MX<AC-MX=AC-MA<MC$ $\Longleftrightarrow AY<MB+MC$ I think the first inequality should be the other way around, because $MB<BX+MX \implies -BX-MX<-MB \implies AY-BX-MX<AY-MB$.
25.09.2023 16:30
See Iranian Geometry Olympiad 2014 Problem J5.