AoPS - Problem

Source: 2003 Romania JBMO TST4 p1

Tags: geometry, congruent, circumcircle, equal angles, incircle

parmenides51

23.05.2020 18:55

Let $ABC$ be a triangle inscribed in the circle $K$ and consider a point $M$ on the arc $BC$ that do not contain $A$. The tangents from $M$ to the incircle of $ABC$ intersect the circle $K$ at the points $N$ and $P$. Prove that if $\angle BAC = \angle NMP$, then triangles $ABC$ and $MNP$ are congruent. Valentin Vornicu

HIDE: about Romania JBMO TST 2004 in aops I found the Romania JBMO TST 2004 links here but they were inactive. So I am asking for solution for the only geo I couldn't find using search. The problems were found here.

parmenides51 wrote: Let $ABC$ be a triangle inscribed in the circle $K$ and consider a point $M$ on the arc $BC$ that do not contain $A$. The tangents from $M$ to the incircle of $ABC$ intersect the circle $K$ at the points $N$ and $P$. Prove that if $\angle BAC = \angle NMP$, then triangles $ABC$ and $MNP$ are congruent. Valentin Vornicu Let $L$ be the incircle with incenter $I$ and let $O$ be center of $K$ As $\angle BAC = \angle NMP$ we get that $AI=MI$ and we also know $OM=OA$ Consider reflection in the line $\overleftrightarrow{OI}$, we get that $K,L$ go to themselves and $A$ goes to $M$ Therefore $B,C$ go to $N,P$ and hence the triangle $ABC$ and $MNP$ are congruent. Hence proved.