In the circumcircle of triange $\triangle ABC,$ $AA'$ is a diameter. We draw lines $l'$ and $l$ from $A'$ parallel with Internal and external bisector of the vertex $A$. $l'$ Cut out $AB , BC$ at $B_1$ and $B_2$. $l$ Cut out $AC , BC$ at $C_1$ and $C_2$. Prove that the circumcircles of $\triangle ABC$ $\triangle CC_1C_2$ and $\triangle BB_1B_2$ have a common point. (20 points)
Problem
Source: Iranian 3rd round Geometry exam P1
Tags: geometry, circumcircle, geometry proposed
Luis González
25.09.2014 23:51
Let $M$ and $N$ be the midpoints of the arcs $BC$ and $BAC$ of $\odot(ABC).$ $MB_2$ cuts $\odot(ABC)$ again at $P.$ $\angle BB_1B_2=\angle BAM=\angle BPM$ $\Longrightarrow$ $P \in \odot(BB_1B_2).$ But since $A'(B,C,N,M)=A'(B,C,B_2,C_2)=A(B,C,N,M)=-1,$ it follows that $PB_2,PC_2$ bisect $\angle BPC$ internally and externally, i.e. $N,P,C_2$ are collinear $\Longrightarrow$ $\angle CPC_2=\angle CAN=\angle CC_1M$ $\Longrightarrow$ $P \in \odot(CC_1C_2).$ Hence $\odot(BB_1B_2),$ $\odot(CC_1C_2)$ and $\odot(ABC)$ concur at $P,$ as desired.
AHZOLFAGHARI
27.09.2014 22:41
Here is my solution ! $M$ is the midpoint of the arc $BC$. $MB_2$ cuts the circle at $P$. $\angle BB_1B_2=\angle BAM=\angle BPM $ so $P$ is on the circle of triangle $BB_1B_2$. so we must prove that $P$ is on the circle of triangle $CC_1C_2$. $\angle AC_1M$ = $ \frac{180-CM}{2} = 90 - \angle \frac{A}{2}$ so $ C_1C_2C = \angle C - 90 + \angle \frac{A}{2}= \frac{C-B}{2} $ $ \angle B_2PA' = \frac{C-B}{2} $ so $ P , C_2 , A' , B_2 $ are on a circle . so $ \angle B_2PC_2 = 90 $ $ \angle CPC_2 = 90 + \angle \frac{A}{2} $ so $ \angle CC_1C_2 + \angle CPC_2 = 180 $ so $P$ is on the circle triangle $CC_1C_2$ so three circle have a common point . )$P$
jayme
22.10.2014 18:23
Dear Mathlinkers, 1. U, V the midpoints of the arcs BC which contain A, not A 2. P, Q second points of intersection of (ABC) wrt (CC1C2), (BB1B2) 3. according to the Reim's theorem, U, P and C2 collinear, V, Q, B2 collinear 4. B2 orthocenter of UVC2 and we are done. Sincerely Jean-Louis
GMOH
23.02.2018 15:33
let $AO$ cut $BC$ in $X$ . second intersection of $XT$ and $\odot(ABC)$ is the intersection point of three circles
Mahdi_Mashayekhi
28.05.2022 07:37
Let $T'$ and $T$ be midpoint of arcs $BAC$ and $BC$. Note that with some simple angle chasing we have $l'$ passes through $T'$ and $l$ passes through $T$. Let $TB_2$ meet circumcircle of $ABC$ at $S$. Claim $: C_2A'B_2S$ is cyclic. Proof $:$ Note that $TB_2 . TS = TC^2 = TA'.TC_2$. Now Note that $\angle SC_2A' = \angle SB_2T' = \angle STA = \angle SCA$ so $SCC_1C_2$ is cyclic. Note that $\angle BB_1B_2 = \angle BAT = \angle BST = \angle BSB_2$ so $SB_2BB_1$ is cyclic. so $S$ is the intersection point and we're Done.