Let $\Omega$ and $O$ be the circumcircle of acute triangle $ABC$ and its center, respectively. $M\ne O$ is an arbitrary point in the interior of $ABC$ such that $AM$, $BM$, and $CM$ intersect $\Omega$ at $A_{1}$, $B_{1}$, and $C_{1}$, respectiuvely. Let $A_{2}$, $B_{2}$, and $C_{2}$ be the circumcenters of $MBC$, $MCA$, and $MAB$, respectively. It is to be proven that $A_{1}A_{2}$, $B_{1}B_{2}$, $C_{1}C{2}$ concur.
Problem
Source: MDA TST 2016 P7
Tags: geometry, circumcircle
ABCCBA
29.03.2019 05:06
Solution 1 Let $D$ be the foot of $A_2$ to $BM,$ $A_2M \cap B_1C_1$ at $E$ $\angle EB_1M = \angle C_1CB = \frac{1}{2} \angle BA_2M=\angle DA_2E \Rightarrow DEB_1A_2$ is cyclic $\Rightarrow \angle A_2EB_1 = \angle A_2DB_1 = 90^{\circ}$ Anagously, we can prove $B_2M \perp C_1A_1, C_2M \perp A_1B_1, \Rightarrow \triangle A_1B_1C_1$ and $\triangle A_2B_2C_2$ are bilogic $($common orthology center $M),$ so they are perspective
ABCCBA
05.04.2019 13:22
Solution 2 Let $D = B_1C_1 \cap B_2C_2$ and define $E, F$ similarly We have $D, E, F$ are collinear according to the topic Prove A_1,B_1,C_1 is collinear,so by Desargues theorem, $A_1A_2, B_1B_2, C_1C_2$ are concurrent Solution 3 Let $X, Y, Z$ be the reflection of $M$ wrt $B_1C_1, C_1A_1, A_1B_1,$ then $(MXA), (MYB), (MZC)$ are coaxial at $M, M',$ with $M'$ the anti-steiner point of $M$ wrt $\triangle A_1B_1C_1$ (See China TST 2016 Test 2 Day 1 Q1) Let $MM'$ cuts $(O)$ again at $T$ Let $P$ be the anti-pode of $M$ wrt $(BMC)$, $Q$ is the projection of $M$ on $B_1C_1$ From post #2, we know that $MA_2 \perp B_1C_1,$ so $P, A_2, M, Q, X$ are collinear $\angle PCC_1 = \angle PQC_1 = 90^{\circ} \Rightarrow P, C, Q, C_1$ are concyclic $\Rightarrow MA_2 \cdot MX = MP \cdot MQ = MC \cdot MC_1 = MA \cdot MA_1 \Rightarrow A_1, A_2, X, A$ are concyclic $\Rightarrow \angle A_2A_1M = \angle A_2XA = \angle MM'A = \angle TM'A = \angle TA_1A, $ so $A_1A_2$ passes through $T$ Anagously, we can prove $B_1B_2, C_1C_2$ passes through $T$, so these lines are concurrent at $T$ $\blacksquare$
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mathsworm
05.04.2019 13:57
How to prove that the triangles having common orthology center are perspective?
Flash_Sloth
04.07.2019 22:25
By basic angle chasing, $\angle A_1B_1C = \angle A_1AC = \angle MAC = \frac{1}{2} \angle MB_2C = \angle A_2B_2C$. Similarly $\angle CA_1B_1 = \angle CA_2B_2$. Therefore $\triangle CA_1B_1 \sim \triangle CA_2B_2$. $C$ is the center spiral transform sending $A_1B_1$ to $A_2B_2$. Let $T = A_1A_2 \cap B_1B_2$, then by Miquel point $T \in \odot (CA_1B_1) \cap \odot (CA_2B_2)$. Therefore $T$ is the second intersection of $A_1A_2/ B_1B_2$ with $\odot O$. Similarly, $C_1C_2$ also pass through $T$. BTW, perform inversion on the point $M$ gives exactly Chinese TST 2016 Test 2 Day 1 Q1
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rmtf1111
04.07.2019 23:15
Such a nice problem, wouldn't surprise me if it appeared before in other contests or if it will appear in future contests.
tenplusten
05.07.2019 16:48
rmtf1111 wrote: Such a nice problem, wouldn't surprise me if it appeared before in other contests or if it will appear in future contests. Lol this is fourth source I know for this problem(including inverted version) 1.ELMO SL 2015 2.China TST 2016 4.Moldova TST 2016
Systematicworker
05.07.2019 17:26
tenplusten wrote: rmtf1111 wrote: Such a nice problem, wouldn't surprise me if it appeared before in other contests or if it will appear in future contests. Lol this is fourth source I know for this problem(including inverted version) 1.ELMO SL 2015 2.China TST 2016 4.Moldova TST 2016 Hmmm, What about third source?
AlastorMoody
05.07.2019 21:09
Denote $\Delta A'B'C'$ in place of $\Delta A_1B_1C_1$ and $\Delta A_1B_1C_1$ in place of $\Delta A_2B_2C_2$. Let $\odot (C_1BA_1)$ $\cap$ $\odot (ABC)$ $=$ $D$. WLOG, assume $D$ lies on arc $AC$ not containing $B$. $$\angle BC_1A_1= \angle BDA_1=\angle BAM=\angle BDA' \implies D \in A'A_1$$$$\angle C_1DB=\angle BA_1C_1=\angle C'CB=\angle C'DB \implies D \in C'C_1$$Similarly, $B'B_1$ also passes through $D$