A triangle $ABC$ is given. The point $K$ is the base of the external bisector of angle $A$. The point $M$ is the midpoint of the arc $AC$ of the circumcircle. The point $N$ on the bisector of angle $C$ is such that $AN \parallel BM$. Prove that the points $M,N,K$ are collinear. (Proposed by Ilya Bogdanov)
Problem
Source: Sharygin geometry olympiad 2016, grade 10, Final Round, Problem 6.
Tags: geometry, collinearity, circumcircle
navi_09220114
05.08.2016 16:38
I am going to type this before Anant does :p xD Let $I, I_b, I_c$ b the incenter, B-excenter, C-excenter of $ABC$, and let $KM\cap AI=X, KM\cap BI=N'$, then it suffices to prove $N'=N$ by showing $AN'\parallel BM$. Clearly we have $-1=(K,A;I_c,I_b)=I(K,X;N',M)=A(AI_b,AI;AN',AM)$, but $M$ is the midpoint of $II_b$, thus $AN'\parallel II_b$ must hold, as desired.
anantmudgal09
05.08.2016 16:49
navi_09220114 wrote: I am going to type this before Anant does :p xD Darn,... you win! Essentially the same.
Let the line $KM$ meet the side $AC$ at point $T$. It suffices to showing that $\{AK,AT;AN,AM\}=\{CK,CT;CN,CM\}$. Using the trigonometric form of cross ratio (namely, that $(PA,PB;PC,PD)=\frac{\sin APC \cdot \sin BPD}{\sin APD\cdot \sin BPC}$) we get that both these are equal to $\frac{\sin(B/2)}{\sin((C-A)/2)}$ and the result follows.
k.vasilev
24.07.2017 01:08
WLOG $AB>AC$ and let $\measuredangle ABC=\beta , \measuredangle ACB=\gamma , \measuredangle BAC=\alpha$ . We get $\measuredangle MAC=\measuredangle MCA=\frac{\beta}{2} , \measuredangle MAK=\frac{\gamma}{2} , \measuredangle MCK=\alpha + \frac{\beta}{2} \Rightarrow $ from the trigonometric form of Ceva's theorem applied for $\triangle ACK$ and the point $M$ we get $$\frac{\sin \measuredangle CKM}{\sin \measuredangle MKA}=\frac{\sin \alpha+\frac{\beta}{2}}{\sin \frac{\gamma}{2}}$$, analogously applying the same theorem for the same triangle and point $N$ we get $$\frac{\sin \measuredangle CKN}{\sin \measuredangle NKA}=\frac{\sin \alpha+\frac{\beta}{2}}{\sin \frac{\gamma}{2}}$$, but $\measuredangle NKA+\measuredangle NKC=\measuredangle MKA + \measuredangle MKC=\measuredangle AKC$. Therefore $$\frac{\sin(\measuredangle AKC - \measuredangle AKN)}{\sin \measuredangle AKN}=\frac{\sin(\measuredangle AKC - \measuredangle AKM)}{\sin \measuredangle AKM} $$$$\Rightarrow \sin {\measuredangle AKC}.\cot {\measuredangle AKN}-\cos \measuredangle AKC=\sin {\measuredangle AKC}.\cot {\measuredangle AKM}-\cos \measuredangle AKC \Rightarrow \cot \measuredangle NKA=\cot \measuredangle MKA \Rightarrow \measuredangle MKA= \measuredangle NKA $$and the conclusion follows.
math_pi_rate
25.07.2018 17:07
My solution: Redefine $K$ as the point where $MN$ and $BC$ meet. Let $I$ and $I_B$ be the incenter and the $B$-excenter of $\triangle ABC$. Let $AM \cap CI = T$. Now, $\angle NAB = \angle ABM = \angle CBM = \angle MAC \Rightarrow AM$ and $AN$ are isogonal in $\angle BAC$. Thus, by the Isogonal Line Lemma on the pairs of isogonal lines $AM,AN$ and $AB,AC$, we get that $BN$ and $CM$ meet on $AI$. And, By Desargues' Theorem of Two Triangles for $\triangle AMN$ and $\triangle ICB$, we get that $KT$ is parallel to $BI$, and so by the converse of Reim's Theorem, we get that $A,C,T,K$ are concyclic $\Rightarrow \angle AI_BI = \angle ACI = \angle AKT \Rightarrow I_B$ lies on $AK$ $\Rightarrow K$ is the foot of external angle bisector of $\angle BAC$. $\blacksquare$ EDIT: My 100th post, so yay!!
edfearay123
29.07.2021 06:55
math_pi_rate wrote: And, By Desargues' Theorem of Two Triangles for $\triangle AMN$ and $\triangle ICB$, we get that $KT$ is parallel to $BI$, and so by the converse of Reim's Theorem, we get that $A,C,T,K$ are concyclic $\Rightarrow \angle AI_BI = \angle ACI = \angle AKT \Rightarrow I_B$ lies on $AK$ $\Rightarrow K$ is the foot of external angle bisector of $\angle BAC$. $\blacksquare$ What is the Reim's theorem? I see the cyclic with angle chasing, and do you have a prove for the Isogonal line lemma? I think that would make it with trig.