Let $ABC$ be a triangle with circumcircle $(O)$. Let $M,N$ be the midpoint of arc $AB,AC$ which does not contain $C,B$ and let $M',N'$ be the point of tangency of incircle of $\triangle ABC$ with $AB,AC$. Suppose that $X,Y$ are foot of perpendicular of $A$ to $MM',NN'$. If $I$ is the incenter of $\triangle ABC$ then prove that quadrilateral $AXIY$ is cyclic if and only if $b+c=2a$.
Problem
Source: Iran Third Round 2013 - Geometry Exam - Problem 2
Tags: geometry, circumcircle, incenter, geometric transformation, geometry unsolved
Luis González
08.09.2013 02:29
From the parallel radii $\overline{OM} \parallel \overline{IM'}$ and $\overline{ON} \parallel \overline{IN'},$ it follows that $V \equiv MM' \cap NN'$ is the exsimilicenter of $(I) \sim (O)$ $\Longrightarrow$ $V \in OI.$ Hence, $A,X,Y,I$ are concyclic $\Longleftrightarrow$ $\angle AIV=90^{\circ}$ $\Longleftrightarrow$ $\angle AIO=90^{\circ}.$ It suffices to prove that $\angle AIO=90^{\circ}$ $\Longleftrightarrow$ $b+c=2a,$ which has been posted before. For instance, see Incentre, circumcentre and midpoints of AC,BC are concyclic and elsewhere.
sayantanchakraborty
07.04.2014 10:52
What do you mean by exsimilicenter???From where do I get it??
wiseman
19.09.2014 21:43
sayantanchakraborty wrote: What do you mean by exsimilicenter???From where do I get it?? I think this can help.
IDMasterz
21.09.2014 12:51
Tangency and midpoints of arcs... If $MM' \cap NN' = E$, then $E$ is the external centre of similitude of $(I)$ and $(O)$. If $(E)$ is the circle with diametre $AE$ then $I \in (E) \implies \angle AIE \equiv \angle AIO = \dfrac{\pi}{2}$. Now, let the midpoint of arc $CAB$ be $X$, $P$ midpoint of arc $BC$, $I_A$ be the A-excentre. As $OI \parallel AX$, and $OX \cap AI = P$, we have $AI = IM \implies 3 = \dfrac{AI_A}{AI} = \dfrac{s}{s-a} \implies \dfrac{a}{s-a} = 2 \implies 2a=b+c$.