Invert through $A$ with radius $\sqrt{AB \cdot AC} $ and then reflect everything over the bisector of angle $A$. Doing so, we can see that $B’ \longmapsto C’$ and $B \longmapsto C$. So the intersection of $BB’$ and $CC’$, which is the orthocenter of triangle $ABC$, will become $ \odot(ACC’) \cap \odot(ABB’)=\{A_1\}$. So $AH$ and $AA_1$ are isogonal and we’re done.

$\angle AA_1C=\angle AC'C=\angle ACC'=90-\alpha$. Similarly, $\angle AA_1B=90-\alpha$ Let $A_1C \cap (ABB'A_1)=X$, then $\angle AA_1X=90-\alpha=\angle AA_1B$. Therefore, $AB=AB'=AX$, i.e. $B'\equiv X$, i.e. $A_1-C-B'$ are collinear. Now, $\angle CAH=90-\gamma=\angle B'BC=\angle BB'C=\angle BB'A_1=\angle BAA_1$ and since the circumcenter $O$ and the orthocenter $H$ are isogonal conjugates, it follows that $O$ lies on $AA_1$. Similarly, $O$ lies on $BB_1$ and $CC_1$.

$\sqrt{bc}$ inversion FTW! Let $\Psi_A$ refer to the inversion around $A$ with radius $\sqrt{AB\cdot AC}$ followed by reflection across the angle bisector of $A$. Under $\Psi_A$, $A_1$ gets sent to the $H$, the orthocenter of $\triangle ABC$. Thus, $AA_1$ is isogonal to the $A$-altitude. Similarly for $BB_1$ and $CC_1$. Thus, they are concurrent at the isogonal conjugate of the orthocenter.

ok ok, daca iti pun like, imi pui inapoi likeurile?

Vrangr wrote: $\sqrt{bc}$ inversion FTW! Let $\Psi_A$ refer to the inversion around $A$ with radius $\sqrt{AB\cdot AC}$ followed by reflection across the angle bisector of $A$. Under $\Psi_A$, $A_1$ gets sent to the $H$, the orthocenter of $\triangle ABC$. Thus, $AA_1$ is isogonal to the $A$-altitude. Similarly for $BB_1$ and $CC_1$. Thus, they are concurrent at the isogonal conjugate of the orthocenter. Isn't this the exact same solution as #$\text{2}$?

@above This is just your opinion

rmtf1111 wrote: ok ok, daca iti pun like, imi pui inapoi likeurile? Da

Ok, tiam pus like

Electron_Madnesss wrote: Isn't this the exact same solution as #$\text{2}$? Yep. This problem was actually posted on Math LAB group on Telegram a few hours earlier (after JBMO ended of course). I posted this solution since that's how I solved it then.

Let $O= \odot A_1 BC \cap AA_1$ different from $A_1$. We have $\angle AA_1C=\angle AC'C$ and $\angle AA_1B=\angle AB'B$. But $\angle AB'B=\angle AC'C=\angle ABH$($H$-orthocenter), therefore $A_1A$ bisects $\angle BA_1C$. As a conclusion $\triangle BOC$ is isosceles, with $\angle OCB=\angle OBC=\angle AA_1C=\angle ABH$. Hence $O$ is the circumcenter and $O \in AA_1$. Similarly $O \in BB_1$, $O \in CC_1$, which concludes $AA_1\cap BB_1 \cap CC_1=O$.

Nice solution @Cezar and @Vragnar.

@above fastlikearabbit was mad at me because I have given a like to Vrangr's solution but I did not give one to his, even though they are exactly the same and he posted it first, consequently, in order to revenge, fastlikearabbit had unliked some of my posts, and after realizing this, I made him an offer he can't refuse.

But, you could have told in english. If I start writing my language here no aopsers will understand. Thank you for your response.

As $\Delta ACC'$ is an isoscel triangle, with $AC'=AC$, the antipode of $A$ in its circumcircle lies on the line $AB$. Denote this antipode by $M$. Also, denote the antipode of $A$ in the circumcircle of $\Delta ABB'$ by $N$ and a similar argument provides us that $N\in AC$. Note that $m(\angle ACM)=m(\angle ABN)=90^o$ by obvious reasons, hence if we denote by $A_2$ the intersection of $CM$ and $BN$, we get that $A_2$ is the antipode of $A$ in the circumcircle of $\Delta ABC$. However, $m(\angle MBN)=m(\angle MCN)=90^o$, therefore $MBCN$ is cyclic and the Power of Point Theorem tells us that $A_2C\cdot A_2M=A_2B\cdot A_2N$. The latter gives us that $A_2$ lies on the radical axis of the circumcircles of $\Delta ACC'$ and $\Delta ABB'$ i.e. on $AA_1$. But then the line $AA_1$ is the $A-diameter$ in the circumcircle of $\Delta ABC$. Similar arguments for $B$ and $C$ provides us that the three lines $AA_1,BB_1,CC_1$ are indeed concurrent (at the circumcenter of $\Delta ABC$).

Hi!

Hi! Edit: sorry, my Ipad kinda glitched, it double sent. Is aops normally a bit glitchy on mobile?

First invert trough A to see that A1 C B' and A1 B C' are collinear.After that trigonometric form of ceva finishes problem .

$Lemma:$ $AA_1$ passes through circumcenter $O$ of $\triangle ABC$. $Proof of Lemma:$ Since length of $AB'$ is equal to that of $AB$, it follows that circumcenter of $\triangle ABB', O1$ lies on side $AC$ of $\triangle ABC$ . Similarly, circumcenter of $\triangle ACC', O2$ lies on side $AB$ of $\triangle ABC$. So the radical axis ($AC$) of circles $ABC$ and $ACC'$ passes through $O1$. Also note that Mid-point $E$ of $AC$ and center of circle ACC' ($O2$) passes through $O$ and is perpendicular to $AO1$. Similarly, radical axis ($AB$) of circles $ABC$ and $ABB'$ passes through $O2$. Also, Mid-point $D$ of $AB$ and center of circle ABB' ($O1$) passes through $O$ and is perpendicular to $AO2$. Thus, circumcenter, $O$ of $\triangle ABC$ is the orthocenter of $\triangle AO1O2$. Thus, $AO$ is perpendicular to $O1O2$. Thus, $AO$ passes through $A1$, implying $AA1$ passes through circumcenter of $\triangle ABC$. Now, using $Lemma$ above, it can be proven similarly that $BB1$ and $CC1$ also pass through circumcenter of $\triangle ABC$. Thus, lines $AA_1$,$BB_1$ and $CC_1$ are concurrent at circumcenter of $\triangle ABC$.

$LEMMA1$ Involution is uniquely defined by two reciprocal pairs. Since $C'$ is the reflection of $C$ over $AB$ (similarly with $B'$),we have $\angle C'AB=\angle BAC=\angle CAB'$.Let l be the angle bisector of $\angle BAC$ ,thus $C'$ is the reflection of $B'$ over l and $C$ is the isogonal of $B$ over l.since the isogonalss are invoulution$(C',B'),(B,C)$(are reciprocal pairs) and from Lemma 1,we have $(H,X)$ reciprocal pairs,we know that $H$ and $O$ are isogonal and $H$ and $X$ are isogonal ,which means $A-O-X$ collinear.

itslumi wrote: thus $C'$ is the reflection of $B'$ over l and $C$ is the reflection of $B$ over l This is true only if $\Delta ABC$ is isoceles.

Olympikus wrote: itslumi wrote: thus $C'$ is the reflection of $B'$ over l and $C$ is the reflection of $B$ over l This is true only if $\Delta ABC$ is isoceles. Sorry i mean the isogonal

Hi there! just made a video about it. Hope you guys like it https://youtu.be/40UgHZV34C4

Amazing problem! Solution.I'll present my solution which is merely angle chasing. $\angle ABB'=\angle AB'B=\angle ACC'=\angle AC'C=\angle AA_1C=\angle AA_1B \implies A_1-C-B'$ and $A_1-B-C'$. $\angle BAA'=\angle BCC'=\angle BC'C=\angle CAA_1, \angle CAA'=\angle CBB'=\angle CB'B=\angle BAA_1.$ So, $\angle BAA_1=\angle ABB_1, \angle CAA_1=\angle ACC_1$ and $\angle BCC_1=\angle CBB_1$. By Trig. Ceva $AA_1$,$BB_1$ and $CC_1$ are concurrent $\iff \frac{\sin BAA_1}{\sin A_1AC} \cdot \frac{\sin CBB_1}{\sin B_1BA} \cdot \frac{\sin ACC_1}{\sin C_1CB} = 1$, which is true.$\blacksquare$

a nice problem , Let $O$ the circumcenter of $\triangle ABC$ Claim : $AA_1$ passe trought $O$ : proof : first we have that $C' , B , A_1 $ collinear , because we have that $\angle AA_1B = \angle AB'B = \angle ABB' = \angle ACC' = \angle AA_1C$ $\angle CAA_1 = \angle CC'A_1 = \angle CC'B = \angle FCB = 90 - \angle ABC = \angle BAH $ $\blacksquare$ similarly $BB_1$ and $CC_1$ passe throught $O$ . $AA_1$ , $BB_1$ and $CC_1$are concurent.

The inversion solution is fine, but I like this one a little more. Let $(AC'C)$ intersect $AB$ again at $C''$, and let $(AB'B)$ intersect $AC$ again at $B''$. Since $(ACC')$ is symmetric about line $AB$, we have that $AC''$ is a diameter of $(ACC')$. Similarly, $AB''$ is a diameter of $(ABB')$. Thus, $\angle AA_1C''=\angle AA_1B''=90^\circ$, so $C''$, $A_1$, and $B''$ are collinear and $AA''$ is an altitude in $\triangle AB''C''$. Also, $AC''=\frac{AC}{\cos \angle A}$ and $AB''=\frac{AB}{\cos \angle A}$, so $AB\cdot AC''=AC\cdot AB''$, i.e. $BCB''C''$ is cyclic. Thus, $BC$ and $C''B''$ are antiparallel in $\angle A$, so $AA''$ is isogonal to the $A$-altitude in $\triangle ABC$. This finishes since the altitudes concur at the circumcenter.

Let $H$ be the orthocenter of $\triangle ABC$. Let $\angle ABH=\angle ACH=\alpha,$ $\angle BAH=\angle BCH=\beta$, and $\angle CAH=\angle CBH=\gamma.$ Let $D$ be the foot from $A$ to $BC$, $E$ be the foot from $B$ to $AC$, and $F$ be the foot from $C$ to $AB.$ Since $BE=B'E$, we have $\angle BB'C=\angle B'BC=\gamma$. Then, from cyclic $AB'A_1B$, $$\angle BAA_1=\angle BB'A_1=\gamma.$$Then, from $CF=C'F$, we have $$\angle CC'B=\angle C'CB=\beta,$$and from cyclic $AC'A_1C$, $$\angle CAA_1=\angle CC'A_1=\beta.$$Hence, $$\frac{\sin\angle BAA_1}{\sin \angle CAA_1}=\frac{\sin\gamma}{\sin\beta}.$$Clearly, $$\prod_{cyc}\frac{\sin\angle BAA_1}{\sin \angle CAA_1}=\frac{\sin\gamma}{\sin\beta}\frac{\sin\alpha}{\sin\gamma}\frac{\sin\beta}{\sin\alpha}=1,$$so by Trig Ceva we are done.

Trivial by Jacobi's Theorem

Let $H$ be the orthocenter of $\triangle ABC$ and $O$ its circumcenter. Claim: $A - O - A_1$ Proof: Note that there is a spiral similarity centered at $A$ that sends $C´C$ to $BB'$, this implies that $C'B \cap B'C = A_1$. Now, note that $AHBC'$ (and $AHB'C , A'HBC$) are cyclic so $\angle CAA_1 = \angle CC'A_1 = \angle BC'H = \angle BAH$ and thus $A'$ and $A_1$ are isogonal with respect to $\angle A$ which implies our claim. The same arguement can be also applied to $BB_1$ and $CC_1$ so we conclude that this lines concur at the circumcenter $O$.

Let $O$ denote the circumcenter of $\triangle ABC$. The claim is that this is the desired concurrence point. We start off with the following easy observation. Claim : Lines $\overline{B_1C'}$ and $\overline{C_1B'}$ intersect at $A$ (and similarly). Proof : It suffices to show that one of these lines passes through $A$ since then we shall be done by symmetry. Note that, \[\measuredangle B'C_1C = \measuredangle B'BC = \frac{\pi}{2} + \measuredangle ACB = \measuredangle AA'C = \measuredangle AC_1C\]which implies that points $A$ , $B'$ and $C_1$ are indeed collinear as claimed. Now we are essentially done since, it is well known that lines $CO$ and $CC'$ are isogonal with respect to $\angle A$ and, \[\measuredangle C_1CA = \measuredangle C_1A'A = \measuredangle BA'A = \measuredangle A'AB = \measuredangle CAO = \measuredangle OCA\]which implies that $O$ indeed lies on $\overline{CC_1}$. Similarly one can show that $O$ also lies on lines $\overline{AA_1}$ and $\overline{BB_1}$ which finishes the problem.

Solution purely by angle chasing. Let $O$ be the circumcenter of $\triangle{ABC}$ and $\angle{B}=\beta,\angle{C}=\gamma$. Claim: $B_1-C-A'$ and $B_1-O-B$. Proof: For the first part, note that $\triangle{BC'C}$ is isosceles and $AB \perp C'C$ so $\angle{BC'C}=\angle{BB_1C}=90^{\circ}-\beta$. Similarly, $\angle{BA'A}=\angle{BB_1A}=90^{\circ}-\beta$, but note that $\angle{AB_1A'}=180-2\beta$ since $\angle{ABA'}=2\beta$, but this implies that $\angle{BB_1A'}=\angle{BB_1C}=90^{\circ}-\beta$ so $B_1-C-A'$. Now note that because $\angle{AOC}=2\beta$, $AOCB_1$ is cyclic. By simple angle chasing we get $\angle{ACB_1}=\angle{AOB_1}=180-2\gamma$ and because $\angle{AOB}=2\gamma$, we finally get that $B_1-O-B$. Similarly, $A_1-O-A$ and $C_1-O-C$ so $AA_1,BB_1,CC_1$ are concurrent at $O$. $\square$.