In a triangle $ABC$, $M$ is the midpoint of the $AB$. A point $B_1$ is marked on $AC$ such that $CB=CB_1$. Circle $\omega$ and $\omega_1$, the circumcircles of triangles $ABC$ and $BMB_1$, respectively, intersect again at $K$. Let $Q$ be the midpoint of the arc $ACB$ on $\omega$. Let $B_1Q$ and $BC$ intersect at $E$. Prove that $KC$ bisects $B_1E$. M. Kungozhin
Problem
Source: 2021 SRMC P3
Tags: geometry
gnoka
30.06.2021 16:39
here is a video solution in Chinese Language
SerdarBozdag
30.06.2021 19:50
BUMP....
hakN
30.06.2021 21:19
By projective, it suffices to show that $\angle BB_1M = \angle EB_1C$. So we can completely ignore point $K$ and $\omega_1$. But I couldn't show that these angles are equal... @2below You are right I felt so silly for not seeing that Ok so I will write my solution as well
As @Aryan-23 proved, we have $\angle BB_1M = \angle EB_1C$. Let $D = KC\cap B_1E$. Let the parallel to $B_1E$ through $C$ cut $(ABC)$ at $F$. Note that since $\angle EB_1C = \angle FCA$ and $\angle BB_1M = \angle BKM$, we have $\angle BKM = \angle FCA = \angle AKF$. So $KF$ is a symmedian in $\triangle AKB$, implying that $(B,A;K,F) = -1$. So, $-1 = (B,A;K,F) \stackrel{C} = (E,B_1;D, B_1E_{\infty})$, implying that $D$ is the midpoint of $B_1E$, as required.
SerdarBozdag
30.06.2021 22:19
Let $CK$ cut $BB_1$ at $F, QB_1$ cut $(ABC)$ at $I, CH \parallel IQ$ and $H\in (ABC)$, $G=IKFB_1\cap B_1M$. Because $CQ \parallel B_1F$, $IKFB_1$ is cyclic. Thus $H-I-F$. We will prove that $HK$ is symmedian in $AHB$. $\angle KFG=\angle KB_1M=\angle KBM=\angle KCA \implies GF\parallel CB_1$. Thus $\angle B_1GF=\angle B_1IF=\angle CQI$ and $\angle GFB_1=180- \angle CBB_1=180-\angle QBA=\angle QCB_1 \implies \angle CB_1Q=GB_1F \implies \angle HKA=\angle HCB_1= \angle CB_1Q=\angle MB_1B=\angle MKB \implies AKH \sim MKB \implies -1=(H,K;A,B)=(CH\cap B_1Q,KC\cap B_1E;B_1E)$ as desired. Note: I did not pay atteintion the crucial observation (which hakN said) when I realized it. I would spend much less time
Aryan-23
30.06.2021 22:21
hakN wrote: By projective, it suffices to show that $\angle BB_1M = \angle EB_1C$. So we can
By angle chasing you can prove that $QA,QB$ are tangents to $\odot (ABB_1)$, hence $B_1Q$ is a symmedian in $\odot (ABB_1)$. Hence $B_1Q$, $BM$ are isogonal in $\angle BB_1A$, and the result follows
GoodMorning
22.08.2023 08:47
Outline: Let the parallel of $B'Q$ through $C$ intersect the circle at $D$. Suffices to prove that $KADB$ is a harmonic quad, which is equivalent to $\angle DKA = \angle MKB$, which through angle chasing is equivalent to $\angle CB'Q = \angle BB'M$. We do this by proving that $B'Q$ is a symmedian of $AB'B$, which can be done by noticing that angle chasing yields that $BQ$ is tangent to $(ABB')$, and that $MQ$ is the perpendicular bisector of $AB$.
Attachments:
alinazarboland
17.10.2023 21:55
Fix $C,B_1,B$ and move $A$ projectively on $B_1C$.Since $\angle QBA = 90 - \frac{\angle C}{2}$ is constant , and $Q$ moves on a fixed line(since $\angle QCB$ is fix) , we get that the map $A \to Q$ is projective. So $A \to QB_1 \cap BC = E$ is also projective. So $A \to N$ where $N$ is the midpoint of $B_1E$ is a projective map.We also know that $A \to M$ is projective so by inverting through $B$ , $A \to K$ is projective. So the lines $CK$ rotate projectively in the pencil lines of $C$. But since $B_1E$ moved projectively , the lines from $C$ parallel to $B_1E$ also move projectively in the pencil lines of $C$. But we wanna prove that $(CB_1 , CB ; CK , C \infty_{B_1E}) = -1$ which is clearly a projective statement(what I mean by that , is that if $CK$ moves projectively , $C \infty_{B_1E}$ also moves projectively.) So By the moving point lemma , it's enough to check 3 cases for $A$. Which is ok : $\infty _{CB_1} , B_1 , C$ and we're done.