A circle passes through vertex $B$ of the triangle $ABC$, intersects its sides $ AB $and $BC$ at points $K$ and $L$, respectively, and touches the side $ AC$ at its midpoint $M$. The point $N$ on the arc $BL$ (which does not contain $K$) is such that $\angle LKN = \angle ACB$. Find $\angle BAC $ given that the triangle $CKN$ is equilateral.
Problem
Source: Baltic Way 2015
Tags: geometry
serquastic
10.11.2015 08:25
It is easy to see that $BNCA$ is an isosceles trapezoid. $\angle{NMC} = \angle{NKM}=\angle {BNM}=\angle {AKM}$ So $\angle{BAM}=\angle{NCA}=\angle{NMK}.$ $\implies \angle{KNM}=\angle{MNC}=30$ By triangle congruences $NK=NM=NC=CK$, and $KM=AM=CM$. $\implies \angle{BKN}=30$ , $BKMN$= isosceles trapezoid. $\implies BM=KN$ So, finally $\angle{MBN}=\angle{BAC}=75$
MouN
10.11.2015 10:30
From the angle condition we have $\angle ACB = \angle LKN = \angle CBN$ so $AC\parallel BN$. Now there's a certain symmetry about the perpendiular bisector of $AC$. Specifically $ACNB$ is an isosceles trapezoid, and we have $\angle MNC = \angle ABM = \angle KNM$, so $MN$ is the internal angle bisector of $\angle KNC$, hence $\angle AMK = \angle MNK = 30^{\circ}$. Also $AB=CB=KN$, $MB=MN$ and $\angle ABM = \angle KNM$ so $\triangle ABM$ and $\triangle KNM$ are congruent. Hence $\triangle AKM$ is an isosceles triangle so $\angle BAC=75^{\circ}$
A-Thought-Of-God
09.11.2020 14:39
Is this unnecessarily long?
Notice that $$\angle ACB=\angle LKN=\angle LKB=\angle CBN$$so $\overline{AC}\parallel\overline{BN}$. Moreover, $$-1=(A,C;M,P_{\infty})\overset{B}{=} (K,L;M,N)$$so $\odot KMLN$ is Harmonic Quad. Let $\overline{KL}\cap\overline{AC}=J$. Notice, $$\angle JNL=\angle LKN=\angle ACB=\angle ACL\implies \{L,C,J,N\}~\text{are cyclic points}$$Also, $N$ is Miquel Point of $ACLK$. Consequently, using cyclicity of $ACNB$ we conclude, that it's Isosceles Trapezoid, and $\angle BCN=\angle A -\angle C$. Also, $$\angle LJN=\angle LCN=\angle BCN=\angle A - \angle B~\text{with}~\angle CJL=\angle CNL=60^\circ-\angle B \implies \angle JMN=\angle JNM=150^\circ-\angle A \implies \angle MNL=150^\circ-\angle A -\angle C=\angle B -30^\circ$$But then, $$\angle ABM=\angle AMK=\angle MNK=\angle LNK-\angle LNM=30^\circ$$Previously, $\angle MNC=30^\circ$. Since $NK=NC$ we have $\triangle KNM\overset{\sim}{=}\triangle CNM,$ by which we have $AM=MC=MK,$ we conclude $\angle BAC=\angle KAM=75^\circ$. $\blacksquare$
rafaello
14.08.2021 00:30
Note that given angle condition implies that $BN\parallel AC$. Hence, $\measuredangle MBN=\measuredangle CMN=\measuredangle BNM\implies MB=MN$. As $BN\parallel AC$ and $BN,AC$ share a perpendicular bisector, we get that $AB=CN$, after reflection over the perpendicular bisector. Now, because $AB=CN=KN$, $BM=NM$ and $\measuredangle ABM=\measuredangle KNM$, hence $\triangle ABM\cong\triangle KNM$, thus $KM=AM=MC$. We obtain that $\triangle KNM\cong \triangle CNM$, hence $\measuredangle KNM=30^\circ$. Furthermore, $\measuredangle KMA=30^\circ\implies \measuredangle CAB=\boxed{75^\circ}$.