$\text{I can't even draw the diagram lol}$

trinhquockhanh wrote: $\text{I can't even draw the diagram lol}$ See this

My solution of this problem is in here https://www.facebook.com/groups/vmo.tst

Let $x = \angle A_1BC = \angle A_1CB$ and define $y,z$ similarly. Note $x+y+z =30^{\circ}$. First, it's easy to see $\angle BA_1C = 180^{\circ}-2x$ and $\angle BA_2C = 90^{\circ}-x$, so $\angle BA_1C = 2 \angle BA_2C$. Since $A_1B=A_2C$ this means $A_1$ is the circumcenter of $BA_2C$, and so on. Next note $\angle B_1AC_1 = x + 30^{\circ}$. Using the isosceles triangles the circumcenter conditions give us we find that $\angle C_1C_2B_1 = \angle B_1B_2C_1 = x+30^{\circ}$, so $(B_1B_2C_1C_2)$ is cyclic and cyclic variants hold. So either $(A_1A_2B_1B_2C_1C_2)$ is cyclic or we can apply radical axis and find $A_1A_2, B_1B_2, C_1C_2$ concur at some point $P$ which is the radical center of the previous 3 circles, and hence must also have equal power wrt $(AA_1A_2), (BB_1B_2), (CC_1C_2)$.

Now it suffices to find a second point with equal power with respect to $(AA_1A_2), (BB_1B_2), (CC_1C_2)$. To do this, let $A_3 = (AA_1A_2) \cap (BA_2C)$ and define $B_3,C_3$ similarly. Note that since $A_1A_2=A_1A_3$, $AA_1$ bisects $\angle A_2AA_3$, hence $AA_2, AA_3$ are isogonal in $\angle BAC$. Clearly $AA_2, BB_2, CC_2$ concur by Trig Ceva, so it follows from isogonal conjugation that $AA_3, BB_3, CC_3$ meet at some point $Q$. I claim $Q$ also has equal power with respect to the three circles. It suffices to show $Q$ has equal power with respect to $(BB_1B_2), (CC_1C_2)$, which is equivalent to showing $BB_3CC_3$ is cyclic. We simply note that if $CC_3$ meets $(ABC_2)$ at $C_4$ then $C_2C_4||AB$ from isogonality, and then $\angle BC_3C = \angle BAC_4 = \angle ABC_2 = 60^{\circ}-x$, which is symmetric in $B,C$, as desired. Hence $P,Q$ have equal power with respect to $(AA_1A_2), (BB_1B_2), (CC_1C_2)$, so they're coaxial. Furthermore we can easily see $P$ has negative power with respect to the three circles (since $P$ is inside segments $A_1A_2, B_1B_2, C_1C_2$), so the three circles meet at two points (as opposed to the case where they don't intersect at all. This also shows they're not all tangent.)

Pop method can solve it. Lemma: Let $B', C', c$ be the image of $B, C, l$ under an invertion and reflection with $\angle BAC$, then $\frac{P_c(B)}{P_c(C)}=\frac{BA^2*{\overline{DC'}}}{CA^2*{\overline{DB'}}}$. Proof is easy with the use of Menelaus. Let $\odot AA_1A_2=\omega_A$, reflections of ${A}$ and $A_2$ wrt $BC$ be $A'$ and $A'_2$. Let $AA_2, A'A'_2, BC$ meet at T. Obviously the circumcenter of $\triangle A_2BC$ is $A_1$ and let its 1st isodynamic point be $S$. Define $\omega_B, \omega_C$ similarly. Since $\triangle AB_1B_2 ${R}$ \triangle AC_1C_2$, $B_1B_2C_1C_2$ concyclic, also $A_1A_2B_1B_2, A_1A_2C_1C_2$ concyclic, by Monge $A_1A_2,B_1B_2, C_1C_2$ concurrent at ${R}$, and ${R}$ has equal power to $\omega_A, \omega_B, \omega_C$. Well known that $AA_1A_2S$ concyclic, so consider $\angle BAC$ invertion reflection, $A_1, S$ becomes $A_2', A'$. By Lemma $$\frac{P_{\omega_A}(B)}{P_{\omega_A}(C)}=\frac{BA_2^2{\overline{TC}}}{CA_2^2{\overline{TB}}}=-\frac{BA_2^2S_{\triangle CAA_2}}{CA_2^2S_{\triangle BAA_2}}=-\frac{BA_2sin\angle ACA_2}{CA_2sin\angle ABA_2}=-\frac{sin\angle BCA_2sin\angle ACA_2}{sin\angle CBA_2sin\angle ABA_2}=-\frac{sin\angle BCA_2sin\angle ACA_2}{sin\angle CBA_2sin\angle ABA_2}$$which is enough to show by PoP linearity $\omega_A, \omega_B, \omega_C$ has another radical center other than ${R}$. We're done!

Let $a=\angle A_1BC$, $b=\angle B_1CA$, and $c=\angle C_1AB$, so the condition becomes $a+b+c=30^\circ$. Note that $\angle BA_2C=60^\circ+b+c=90^\circ-a$, so $A_1$ is the circumcenter of $BA_2C$. Similarly, $B_1$ and $C_1$ are the circumcenters of $CB_2A$ and $AC_2B$. Furthermore, $\angle B_1C_2C_1=90^\circ-\angle ABC_2=a+30^\circ=90^\circ-\angle ACB_2=\angle B_1B_2C_1$. Thus, $B_1C_1B_2C_2$ is cyclic. Similarly, $C_1A_1C_2A_2$ and $A_1B_1A_2B_2$ are cyclic. Let $\omega_A$, $\omega_B$, and $\omega_C$ be the circumcircles of $B_1C_1B_2C_2$, $C_1A_1C_2A_2$, and $A_1B_1A_2B_2$. As $\angle A_2B_1C_2+\angle C_2A_1B_2+\angle B_2C_1A_2=480^\circ$, we have that $A_2B_1C_2A_1B_2C_2$ is not cyclic and $\omega_A$, $\omega_B$, and $\omega_C$ are distinct. By the radical axis theorem, $A_1A_2$, $B_1B_2$, and $C_1C_2$ then concur at a point $X$. Let $p(K,\omega)$ denote the power of point $K$ to circle $\omega$. We now compute $\frac{p(A,\omega_B)}{p(A,\omega_C)}$. Let $\omega_B$ meet $AC_1$ at $D$ and $\omega_C$ meet $AB_1$ at $E$. We have that $\angle A_1AD=30^\circ-c$ and $\angle A_1DC_1=\angle A_1A_2C_1=90^\circ-\angle S_1CB=30^\circ+b$, so $\angle AA_1D=b+c$. Similarly, $\angle A_1AE=30^\circ-b$, $\angle A_1EB_1=30^\circ+c$, and $\angle AA_1E=b+c$. It follows by the law of sines on triangles $AA_1D$ and $AA_1E$ that $\frac{AD}{AE}=\frac{\sin(30^\circ+c)}{\sin(30^\circ+b)}$. Thus, $\frac{p(A,\omega_B)}{p(A,\omega_C)}=\frac{AC_1}{AB_1}\cdot\frac{\sin(30^\circ+c)}{\sin(30^\circ+b)}$. By computing the analogous expressions for the other vertices, we conclude that $\frac{p(A,\omega_B)}{p(A,\omega_C)}\cdot\frac{p(B,\omega_C)}{p(B,\omega_A)}\cdot\frac{p(C,\omega_A)}{p(C,\omega_B)}=1$. To conclude, recall the coaxial lemma: given two circles $\omega_1$ and $\omega_2$ and a constant $c$, the locus of points $K$ for which $\frac{p(K,\omega_1)}{p(K,\omega_2)}=c$ is a circle coaxial to $\omega_1$ and $\omega_2$. Hence, the circumcircle of $AA_1A_2$ is the locus of points $K$ with $\frac{p(K,\omega_B)}{p(K,\omega_C)}=\frac{p(A,\omega_B)}{p(A,\omega_C)}$, and the analogous statements hold for the other vertices. The angle condition implies that segments $AA_1$, $BB_1$, and $CC_1$ intersect, so the circumcircles of $AA_1A_2$, $BB_1B_2$, and $CC_1C_2$ pairwise intersect at two points. It follows that the circumcircles of $BB_1B_2$ and $CC_1C_2$ intersect at points $P_1$ and $P_2$ satisfying $\frac{p(P_i,\omega_C)}{p(P_i,\omega_A)}=\frac{p(B,\omega_C)}{p(B,\omega_A)}$ and $\frac{p(P_i,\omega_A)}{p(P_i,\omega_B)}=\frac{p(C,\omega_A)}{p(B,\omega_B)}$ for $i=1,2$. But this then implies that $\frac{p(P_i,\omega_B)}{p(P_i,\omega_C)}=\frac{p(A,\omega_B)}{p(A,\omega_C)}$ for $i=1,2$, so $P_1$ and $P_2$ both lie on the circumcircle of $AA_1A_2$, as desired. Remark: This application of the coaxial lemma can be interpreted as a "spherical analogue" of Ceva's theorem: with a suitable stereographic projection, the circles $\omega_A$, $\omega_B$, and $\omega_C$ can be interpreted as great circles on a sphere, and we wish to show the concurrence of "spherical Cevians" $(AA_1A_2)$, $(BB_1B_2)$, and $(CC_1C_2)$ in the spherical triangle $A_1B_1C_1$. The "Cevian ratio" for $(AA_1A_2)$ is given exactly by the ratio of powers from the coaxial lemma, which is simple to compute.

Bary bash? Will edit in later if possible

angles summing to 480 is crazy this statement is so cool :$\phantom{}$D Let $\omega_A$ be $(AA_1A_2)$. Let $f_A(X)$ be the power of an arbitrary point $X$ with respect to $\omega_A$. Let $\alpha = \angle A_1BC = \angle A_1CB$. Define $\omega_B,\omega_C$, $f_B,f_C$ and $\beta,\gamma$ similarly. The angle condition becomes $\alpha+\beta+\gamma=30$. An angle chase gives that \[\angle BA_2C = 60^\circ+\beta+\gamma=90^\circ-\alpha=\frac{1}{2}\angle BA_1C,\]so $A_1$ is the center of $(BA_2C)$. We then further have $\angle AA_1A_2 = \angle A_2BC - \angle A_2CB = \beta-\gamma$. We now compute each of $f_A,f_B,f_C$ evaluated at each of $A,B,C$. It's direct that $f_A(A)=0$. It suffices to compute $f_A(B)$ as the remaining formulas follow from symmetry. Let $\omega_A$ meet $AB$ at $T$. We have $\angle BTA_2 = \angle AA_1A_2 = \beta - \gamma.$ By LOS on $BTA_2$ then $BA_2C$, \[f_A(B) = BT \cdot BA = \left(BA_2 \cdot \frac{\sin \angle BA_2 T}{\sin \angle BTA_2}\right) BA =\left( BC \cdot \frac{\sin(BCA_2)}{\sin(BA_2C)} \cdot \frac{\sin( BA_2 T)}{\sin (BTA_2)} \right) BA =s^2 \frac{\sin(60^\circ-\beta)\sin(180^\circ-\beta)}{\sin(90^\circ+\alpha)\sin(\beta-\gamma)}, \]where $s$ is the side length of $ABC$. Now we have all the powers!! The key claim that we've been building up to is the following: Lemma: If $p=\sin(90^\circ+\alpha) \sin(\beta-\gamma)$, and $q,r$ are defined similarly, then we have $p+q+r=0$ and $pf_A(X)+qf_B(X)+rf_C(X)=0$. Proof: If we write any of the functions $f$ in coordinates, it's $x^2+y^2$ plus some linear equation. It's straightforward to check that $p+q+r=0$ (note $p=\cos(\alpha)\sin(\beta)\cos(\gamma)-\cos(\alpha)\cos(\beta)\sin(\gamma)$), so $pf_A(X)+qf_B(X)+rf_C(X)$ is a linear function. Our previous power of a point computation gives $pf_A(X)+qf_B(X)+rf_C(X)=0$ when $X$ is any of $A$, $B$, or $C$. The only linear function that's zero at three noncollinear points is the zero function, so $pf_A(X)+qf_B(X)+rf_C(X)=0$ for all $X$. $\blacksquare$ Now, $f_A-f_B$ is the linear equation of the radical axis of $\omega_A$ and $\omega_B$. We also have the analogous property for $f_A-f_C$. But we also have $q(f_A-f_B)=-r(f_A-f_C)$, so these are the same line, as desired. (note on motivation: coaxiality is actually equivalent to the claim holding for some choice of $p,q,r$. so you compute all the powers and then solve a linear equation to find $p,q,r$)

Am I having another nightmare or is this really solvable by monstrous trigonometry?

LOL 480 degree. Weird condition but OK. This is the first time I saw an angle greater than 360 degree in a math problem statement.

I have tried with Cartesian coordinates, the calculation is not heavy. Since $ABC$ is equilateral triangle, choose a system such that $A=\left(1, 0 \right)$, $B=\left(\frac{-1}{2}, \frac{\sqrt{3}}{2} \right)$ and $C=\left(\frac{-1}{2}, \frac{-\sqrt{3}}{2} \right)$. Let $k_a$, $k_b$, and $k_c$ be the distance from $A_1$, $B_1$, and $C_1$ to the lines $BC$, $CA$, and $AB$, respectively. Since $A_1$, $B_1$, $C_1$ are interior points of triangle $ABC$ and $\angle BA_1C+\angle CB_1A+\angle AC_1B=480^\circ$, we get the relation $$k_a=\frac{3-4k_bk_c-6(k_b+k_c)}{2(3-4k_bk_c+2(k_b+k_c))}.$$From these, we can assume that $$A_1=\left(\frac{3-4k_bk_c-6(k_b+k_c)}{2(3-4k_bk_c+2(k_b+k_c))}-\frac{1}{2}, 0 \right)=\left(\frac{4k_b + 4k_c}{4k_bk_c - 2k_b - 2k_c - 3}, 0 \right),$$$$B_1=\left(\frac{-2k_c + 1}{4}, \frac{\sqrt{3}(2k_c- 1)}{4} \right),$$and $$C_1=\left(\frac{-2k_b + 1}{4}, \frac{\sqrt{3}(-2k_b +1)}{4} \right).$$The intersection $A_2$ of lines $BC_1$ and $CB_1$ is $$A_2=\left(\frac{-4k_bk_c + 4k_b + 4k_c - 3}{4k_bk_c - 2k_b - 2k_c - 3}, \frac{2\sqrt{3}(k_b - k_c)}{4k_bk_c - 2k_b - 2k_c - 3} \right).$$The intersection $B_2$ of lines $CA_1$ and $AC_1$ is $$B_2=\left(\frac{8k_b^2k_c + 4k_b^2 - 24k_b - 18k_c + 3}{16k_b^2 + 12}, \frac{\sqrt{3}(8k_b^2k_c - 12k_b^2 - 16k_bk_c + 6k_c + 3)}{16k_b^2 + 12} \right).$$The intersection $C_2$ of lines $AB_1$ and $BA_1$ is $$C_2=\left(\frac{8k_bk_c^2 - 18k_b + 4k_c^2 - 24k_c + 3}{16k_c^2 + 12}, \frac{\sqrt{3}(-8k_bk_c^2 + 16k_bk_c - 6k_b + 12k_c^2 - 3)}{16k_c^2 + 12} \right).$$Therefore, circumcenter $K_a$ of triange $AA_1A_2$ is the intersection of perpendicular bisectors of two segments $AA_1$ and $AA_2$: $$K_a=\left(\frac{4k_bk_c + 2k_b + 2k_c - 3}{8k_bk_c - 4k_b - 4k_c - 6}, \frac{\sqrt{3}(16k_b^2k_c^2 - 12k_b^2k_c + 6k_b^2 - 12k_bk_c^2 - 9k_b + 6k_c^2 - 9k_c)}{24k_b^2k_c - 12k_b^2 - 24k_bk_c^2 - 18k_b + 12k_c^2 + 18k_c} \right).$$Circumcenter $K_b$ of triange $BB_1B_2$ is the intersection of perpendicular bisectors of two segments $BB_1$ and $BB_2$: $$K_b=\left(\frac{16k_b^2 - 8k_bk_c^2 + 32k_bk_c - 6k_b + 12k_c^2 - 3}{16k_b^2k_c - 8k_b^2 - 16k_bk_c - 24k_b - 12k_c + 6}, \frac{\sqrt{3}(8k_b^2k_c^2 + 6k_b^2 - 12k_bk_c^2 - 9k_b)}{24k_b^2k_c - 12k_b^2 - 24k_bk_c - 36k_b - 18k_c + 9} \right).$$Circumcenter $K_c$ of triange $CC_1C_2$ is the intersection of perpendicular bisectors of two segments $CC_1$ and $CC_2$: $$K_c=\left(\frac{-8k_b^2k_c + 12k_b^2 + 32k_bk_c + 16k_c^2 - 6k_c - 3}{16k_bk_c^2 - 16k_bk_c - 12k_b - 8k_c^2 - 24k_c + 6}, \frac{\sqrt{3}(-8k_b^2k_c^2 + 12k_b^2k_c - 6k_c^2 + 9k_c)}{24k_bk_c^2 - 24k_bk_c - 18k_b - 12k_c^2 - 36k_c + 9} \right).$$Now we can calculate $$\frac{{\rm Pow}_{A/(K_b)}}{{{\rm Pow}_{A/(K_c)}}}=\frac{AK_b^2-BK_b^2}{AK_c^2-CK_c^2}=\frac{8k_bk_c^2 - 8k_bk_c - 6k_b - 4k_c^2 - 12k_c + 3}{8k_ck_b^2 - 8k_ck_b - 6k_c - 4k_b^2 - 12k_b + 3},$$$$\frac{{\rm Pow}_{A_1/(K_b)}}{{{\rm Pow}_{A_1/(K_c)}}}=\frac{A_1K_b^2-BK_b^2}{A_1K_c^2-CK_c^2}=\frac{8k_bk_c^2 - 8k_bk_c - 6k_b - 4k_c^2 - 12k_c + 3}{8k_ck_b^2 - 8k_ck_b - 6k_c - 4k_b^2 - 12k_b + 3},$$and $$\frac{{\rm Pow}_{A_2/(K_b)}}{{{\rm Pow}_{A_2/(K_c)}}}=\frac{A_2K_b^2-BK_b^2}{A_2K_c^2-CK_c^2}=\frac{8k_bk_c^2 - 8k_bk_c - 6k_b - 4k_c^2 - 12k_c + 3}{8k_ck_b^2 - 8k_ck_b - 6k_c - 4k_b^2 - 12k_b + 3}.$$This completes the proof.

The statement is nice, but I feel that the solution is quite disappointing for IMO standard. Once you figure out the header of each claim, it is pretty straightforward, and no creativity is required. Let $\alpha = \angle A_1BC$, $\beta = \angle B_1CA$, and $\gamma=\angle C_1AB$. The angle condition implies that $\alpha + \beta + \gamma = 30^\circ$. We prove the problem by several claims. Claim 1: $A_1$ is the circumcenter of $\triangle BA_2C$. Proof: $\angle BA_2C = 60^\circ + \beta + \gamma = 90^\circ-\alpha$. $\blacksquare$ Claim 2: $B_1, B_2, C_1, C_2$ are concyclic. Proof: Follows from $\triangle AB_1B_2$ and $\triangle AC_1C_2$ are isosceles, so $\angle AB_2B_1 = \angle B_1AB_2 = \angle C_1AC_2 = \angle AC_2C_1$. $\blacksquare$ Claim 3: $A_1A_2$, $B_1B_2$, $C_1C_2$ are concurrent at $P$.

$\blacksquare$ Claim 2 and 3 implies that the power of $P$ w.r.t. $\odot(AA_1A_2)$, $\odot(BB_1B_2)$, and $\odot(CC_1C_2)$ are equal. Thus, it suffices to find one more point on the comon radical axis. To do so, we extend $AA_2$ to meet $\odot(BA_2C)$ at $X$. Define $Y$ and $Z$ similarly on other two sides. Then, let $X'$ be the reflection of $X$ across $AA_1$. Define $Y'$ and $Z'$ similarly. Claim 4: $X'\in\odot(AA_1A_2)$. Proof: Follows from that $AA_1$ bisects $\angle X'AA_2$ and that $A_1X'=A_1A_2$. Claim 5: $B, C, Y', Z'$ are concyclic. Proof: Angle chase: \begin{align*} \measuredangle Y'BZ' &= \measuredangle Y'BA + \measuredangle ABZ' \\ &= \measuredangle CBZ + \measuredangle ZC_2B \\ &= \measuredangle CBB_2 + \measuredangle CC_2B \\ &= \measuredangle(CC_2, BB_2) + \measuredangle CBC_2. \end{align*}Analogously, we get that $\measuredangle Y'CZ' = \measuredangle(CC_2, BB_2) + \measuredangle B_2BC$, implying the conclusion since $\measuredangle CBC_2 = \measuredangle B_2BC = \alpha$. $\blacksquare$ Claim 6: $AX', BY', CZ'$ are concurrent at $Q$. Proof: By Claim 5, radical center on $\odot(BCY'Z')$, $\odot(CAZ'X')$, $\odot(ABX'Y')$.

Hence, $PQ$ is the common radical axis.

My solution is similar to tastymath75025. Here is the detail We have $A_1, B_1, C_1$ lie inside $\triangle ABC$ so $BC_1$ and $CB_1$ lie inside $\angle{ABC}$ and $\angle{ACB},$ respectively, which leads to $A_2$ lies inside $\triangle ABC$. Since $$\angle{BA_2C} = 180^{\circ} - \angle{A_2BC} - \angle{A_2CB} = 60^{\circ} + \angle{C_1BA} + \angle{B_1CA} = \dfrac{1}{2} (120^{\circ} + 2 \angle{C_1BA} + 2 \angle{B_1CA})$$$$= \dfrac{1}{2} (480^{\circ} - \angle{AC_1B} - \angle{CB_1A}) = \dfrac{1}{2} \angle{BA_1C}$$we have $A_2 \in (A_1, A_1B)$. Similarly, we have $B_2 \in (B_1, B_1C); C_2 \in (C_1, C_1A)$. We also have $\angle{B_1B_2C_1} = \angle{B_1AC_1} = \angle{B_1C_2C_1}$. Then $B_1, C_1, B_2, C_2$ lie on a circle. Similarly, we have $C_1, A_1, C_2, A_2$ lie on a circle; $A_1, B_1, A_2, B_2$ lie on a circle. From this, we have $A_1A_2, B_1B_2, C_1C_2$ concur at radical center of $(B_1C_1B_2C_2), (C_1A_1C_2A_2), (A_1B_1A_2B_2),$ let this point be $U$. Note that $\angle{BA_1C} + \angle{CB_1A} + \angle{AC_1B} = 480^{\circ},$ so $\angle{BA_1C}, \angle{CB_1A}, \angle{AC_1B}$ are all greater than $120^{\circ}$. Therefore, if we let $O$ be center of $(ABC)$ then $A_1, B_1, C_1$ lie on the opposite ray of rays $OA, OB, OC$. Suppose that $A_2$ lies inside segment $BC_1,$ then applying Pasch axiom for $\triangle BC_1O$ with line $CB_1,$ we have $B_1$ lies inside segment $BO,$ which is contradiction. Hence $C_1, B_1$ lie inside segments $BA_2, CA_2$. Similarly with $B_2, C_2$. Therefore, $B_1C_1B_2C_2$ is convex quadrilateral, which leads to $P$ is the intersection of 2 segments $B_1B_2$ and $C_1C_2$. From this, we have $P$ has negative power to $(AA_1A_2), (BB_1B_2), (CC_1C_2)$. Suppose that $(AA_1A_2)$ intersects $(A_1, A_1B)$ at $A_3,$ similarly we have $B_3, C_3$. Then $AA_2, AA_3$ are isogonal conjugate in $\angle{BAC}$ or they are reflect through $AO$. Similarly with $BB_2, BB_3$. So $$\angle{AA_3B} = \angle{A_2BC} = 60^{\circ} - \angle{ABC_1} = 60^{\circ} - \angle{BAC_1} = \angle{CAB_2} = \angle{AB_3B}$$which means that $A, B, A_3, B_3$ lie on a circle. Similarly, we have $B, C, B_3, C_3$ lie on a circle; $C, A, C_3, A_3$ lie on a circle. Hence $AA_3, BB_3, CC_3$ concur at radical center of $(BCB_3C_3), (CAC_3A_3), (ABA_3B_3),$ let this point be $V$. Therefore, $UV$ is common radical axis of $(AA_1A_2), (BB_1B_2), (CC_1C_2)$. If these 3 circle don't have 2 common points, then they must tangent or contain each others. But in these cases, all points lie on $UV$ must have non - negative power to the 3 circles, which is contradiction. So these 3 circles must have 2 points in common

mathleticguyyy wrote: Let $ABC$ be an equilateral triangle. Let $A_1,B_1,C_1$ be interior points of $ABC$ such that $BA_1=A_1C$, $CB_1=B_1A$, $AC_1=C_1B$, and $$\angle BA_1C+\angle CB_1A+\angle AC_1B=480^\circ$$Let $BC_1$ and $CB_1$ meet at $A_2,$ let $CA_1$ and $AC_1$ meet at $B_2,$ and let $AB_1$ and $BA_1$ meet at $C_2.$ Prove that if triangle $A_1B_1C_1$ is scalene, then the three circumcircles of triangles $AA_1A_2, BB_1B_2$ and $CC_1C_2$ all pass through two common points. (Note: a scalene triangle is one where no two sides have equal length.) The following is a little generalization of this problem. Quote: Let $A_3$, $B_3$, and $C_3$ divide the segments $A_2A_1$, $B_2B_1$, and $C_2C_1$ in the same ratio then the circles $(AA_1A_3)$, $(BB_1B_3)$, and $(CC_1C_3)$ are coaxial. Proof. I hope the synthetic solutions above can work on this extension. The following is the continuation of the calculation in my post #15. By definition of $A_3$, $B_3$, and $C_3$, we have (for some real number $t$): $$A_3=(1-t)A_1+tA_2=\left(\frac{-4k_bk_ct + 4k_b + 4k_c - 3t}{4k_bk_c - 2k_b - 2k_c - 3}, \frac{\sqrt{3}(2k_bt - 2k_ct)}{4k_bk_c - 2k_b - 2k_c - 3} \right),$$$$B_3=(1-t)B_1+tB_2=\left(\frac{16k_b^2k_ct - 8k_b^2k_c + 4k_b^2 - 24k_bt - 12k_ct - 6k_c + 3}{16k_b^2 + 12}, \frac{\sqrt{3}(8k_b^2k_c - 8k_b^2t - 4k_b^2 - 16k_bk_ct + 6k_c + 6t - 3)}{16k_b^2 + 12} \right),$$and $$C_3=(1-t)C_1+tC_2=\left(\frac{16k_bk_c^2t - 8k_bk_c^2 - 12k_bt - 6k_b + 4k_c^2 - 24k_ct + 3}{16k_c^2 + 12}, \frac{\sqrt{3}(-8k_bk_c^2 + 16k_bk_ct - 6k_b + 8k_c^2t + 4k_c^2 - 6t + 3)}{16k_c^2 + 12} \right).$$Therefore, circumcenter $K_a$ of triange $AA_1A_3$ is the intersection of perpendicular bisectors of two segments $AA_1$ and $AA_3$: $$K_a=\left(\frac{4k_bk_c + 2k_b + 2k_c - 3}{8k_bk_c - 4k_b - 4k_c - 6}, \frac{\sqrt{3}(16k_b^2k_c^2t + 16k_b^2k_c^2 - 24k_b^2k_c + 12k_b^2t - 24k_bk_c^2 - 18k_b + 12k_c^2t - 18k_c + 9t - 9)}{48k_b^2k_c - 24k_b^2 - 48k_bk_c^2 - 36k_b + 24k_c^2 + 36k_c} \right).$$Circumcenter $K_b$ of triange $BB_1B_3$ is the intersection of perpendicular bisectors of two segments $BB_1$ and $BB_3$: $$K_b=\left(\frac{16k_b^2k_c^2t - 16k_b^2k_c^2 + 12k_b^2t + 20k_b^2 - 16k_bk_c^2 + 64k_bk_c - 12k_b + 12k_c^2t + 12k_c^2 + 9t - 15}{32k_b^2k_c - 16k_b^2 - 32k_bk_c - 48k_b - 24k_c + 12}, \frac{\sqrt{3}(16k_b^2k_c^2t+ 16k_b^2k_c^2+ 12k_b^2t + 12k_b^2 - 48k_bk_c^2 - 36k_b + 12k_c^2t - 12k_c^2 + 9t - 9)}{96k_b^2k_c - 48k_b^2 - 96k_bk_c - 144k_b - 72k_c + 36} \right).$$Circumcenter $K_c$ of triange $CC_1C_3$ is the intersection of perpendicular bisectors of two segments $CC_1$ and $CC_3$: $$K_c=\left(\frac{16k_b^2k_c^2t - 16k_b^2k_c^2 - 16k_b^2k_c + 12k_b^2t + 12k_b^2 + 64k_bk_c + 12k_c^2t + 20k_c^2 - 12k_c + 9t - 15}{32k_bk_c^2 - 32k_bk_c - 24k_b - 16k_c^2 - 48k_c + 12}, \frac{\sqrt{3}(-16k_b^2k_c^2t - 16k_b^2k_c^2 + 48k_b^2k_c - 12k_b^2t\sqrt{3} + 12k_b^2 - 12k_c^2t - 12k_c^2 + 36k_c - 9t + 9)}{96k_bk_c^2 - 96k_bk_c - 72k_b - 48k_c^2 - 144k_c + 36} \right).$$Now we can calculate $$\frac{{\rm Pow}_{A/(K_b)}}{{{\rm Pow}_{A/(K_c)}}}=\frac{AK_b^2-BK_b^2}{AK_c^2-CK_c^2}=\frac{8k_bk_c^2 - 8k_bk_c - 6k_b - 4k_c^2 - 12k_c + 3}{8k_ck_b^2 - 8k_ck_b - 6k_c - 4k_b^2 - 12k_b + 3},$$$$\frac{{\rm Pow}_{A_1/(K_b)}}{{{\rm Pow}_{A_1/(K_c)}}}=\frac{A_1K_b^2-BK_b^2}{A_1K_c^2-CK_c^2}=\frac{8k_bk_c^2 - 8k_bk_c - 6k_b - 4k_c^2 - 12k_c + 3}{8k_ck_b^2 - 8k_ck_b - 6k_c - 4k_b^2 - 12k_b + 3},$$and $$\frac{{\rm Pow}_{A_2/(K_b)}}{{{\rm Pow}_{A_2/(K_c)}}}=\frac{A_2K_b^2-BK_b^2}{A_2K_c^2-CK_c^2}=\frac{8k_bk_c^2 - 8k_bk_c - 6k_b - 4k_c^2 - 12k_c + 3}{8k_ck_b^2 - 8k_ck_b - 6k_c - 4k_b^2 - 12k_b + 3}.$$(Note that these ratios do not depend on $t$ and it's the same as the ratio in post #15). We complete the proof.

Let $\omega = e^{2\pi i/3}$ and $A,B,C$ be at $1,\omega,\omega^2$. Let $BA_1$ and $CA_1$ meet the unit circle again at $\omega^2 x$ and $\omega / x$, and symmetrically. The angle condition tells us that $\omega xyz = -1$. Then by chord intersection \[a_1 = \frac{x(\omega^2 + \omega/x) - \frac{1}{x}(\omega + \omega^2 x)}{x - 1/x} = \frac{\omega^2 x^2 + (\omega - \omega^2)x - \omega}{x^2 - 1} = \frac{\omega^2 x + \omega}{x+1}\]and \[a_2 = \frac{\frac{\omega}{z}(\omega^2 + y) - \omega^2 y(\omega + 1/z)}{\omega/z - \omega^2 y} = \frac{\omega^2 + y - \omega^2 yz - \omega y}{1-\omega yz} = \frac{(1-\omega)xy + \omega^2 x + \omega}{x + 1}.\]To find the circumcircle of $a, a_1, a_2$, we first shift by $-a_1$, so that our points are \[0, \frac{(1-\omega^2)x + (1-\omega)}{x + 1}, \frac{(1-\omega) xy}{x+1} = \frac{1-\omega^2}{(x+1)z}.\]Dilating by $\textstyle \frac{(1-\omega)(x+1)}{3} = \frac{x+1}{1-\omega^2}$, our points become $\textstyle 0, x - \omega, \frac{1}{z}$. The circumcircle through these is given by \[\begin{vmatrix} p\bar p & p & \bar p & 1 \\ 1 & 1/z & z & 1 \\ \frac{(x-\omega)(1-\omega^2 x)}{x} & x -\omega & \frac{x-\omega}{-\omega x} & 1 \\ 0 & 0 & 0 & 1 \end{vmatrix} = 0,\]Canceling out factors, this is \[\begin{vmatrix} p\bar p & p & \bar p & - \\ z & 1 & z^2 & - \\ 1-\omega^2 x & x & -\omega^2 & - \\ 0 & 0 & 0 & 1 \end{vmatrix} = 0,\]which is just \[(-\omega^2 - xz^2)p\bar p + z(z(1-\omega^2 x)+\omega^2)p + (xz + \omega^2 x - 1)\bar p = 0.\]Multiplying all coefficients by $-\omega y$, this becomes \[(y - z)p\bar p + (z/x - \omega^2 z - yz)p - (xy -\omega y - 1)\bar p = 0.\]Now, if $q$ is the untransformed point, we have that \[p = \frac{x+1}{1-\omega^2} \left(q - \frac{\omega^2 x + \omega}{x + 1}\right) = \frac{1}{1-\omega^2}((x+1)q - (\omega^2 x + \omega))\]and \[\bar p = \frac{1}{(1-\omega)x}((x+1)\bar q-(\omega^2 x + \omega)).\]Thus, the circle in terms of $q$ is \[\begin{multlined} (y-z) \left((x+1)^2q\bar q - (\omega^2 x + \omega)(x+1)(q + \bar q) + (\omega^2 x + \omega)^2\right) \\ + (1-\omega)(z - \omega^2 xz - xyz)((x+1)q - (\omega^2 x + \omega)) \\ -(1-\omega^2)(xy-\omega y - 1)((x+1)\bar q - (\omega^2 x + \omega)) = 0 \end{multlined} \]Expanding, this yields \[ \begin{multlined} (y-z)(x+1)^2 q\bar q + (x+1) \left((1-\omega)(z - \omega^2 xz - xyz)- (y-z)(\omega^2 x + \omega)\right) q \\ - (x+1) \left((1-\omega^2)(xy-\omega y - 1) + (y-z)(\omega^2 x + \omega)\right) \bar q \\ + (\omega^2 x + \omega)\left((y-z)(\omega^2 x + \omega) - (1-\omega)(z - \omega^2 xz - xyz) + (1-\omega^2)(xy-\omega y - 1)\right) = 0, \end{multlined} \]which after cancelling terms and a factor of $x+1$ yields \[ \begin{multlined} (y-z)(x+1) q\bar q + (-\omega^2 xy + xz-\omega y+z +\omega^2 - 1) q \\ - (xy-\omega^2 xz+y-\omega z+\omega^2 - 1) \bar q + (\omega^2 x + \omega)(y-z) = 0. \end{multlined} \](As a check here one can note that $q = 1$ works.) By symmetry, we can write down the other two circles as \[ \begin{multlined} (z-x)(y+1) q\bar q + (-\omega yz + \omega^2 xy-z+\omega^2 x +\omega - \omega^2) q \\ - (\omega yz-xy+\omega z-\omega^2 x+1 - \omega) \bar q + (\omega^2 y + \omega)(z-x) = 0 \end{multlined} \]and \[ \begin{multlined} (x-y)(z+1) q\bar q + (-xz +\omega yz-\omega^2 x+\omega y +1-\omega) q \\ - (\omega^2 xz-\omega yz+\omega^2 x-y+\omega - \omega^2) \bar q + (\omega^2 z + \omega)(x-y) = 0. \end{multlined} \]Adding these three circles yields zero, so they have to be coaxial. QED.

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From easy angle chasing $A_1$ is the center of $(A_2BC)$, $B_1$ is the center of $(AB_2C)$ and $C_1$ is the center of $(ABC_2)$. Notice that $\angle AB_2B_1=\angle B_1AB_2=\angle AC_2C_1$, so $B_1C_1B_2C_2$ is cyclic. Similarly, $A_1B_1A_2B_2$ and $C_1A_1C_2A_2$ are cyclic and thus $A_1A_2,B_1B_2, C_1C_2$ intersect at some point $Q$ which satisfies $QA_1\cdot QA_2=QB_1\cdot QB_2=QC_1\cdot QC_2$, so it has equal powers wrt. $o_A,o_B,o_C$. From Brianchon we can inscribe a conic in $A_1C_2B_1A_2C_1B_2$. This conic is also inscribed in $AB_2CA_2BC_2$, so again from Brianchon $AA_2,BB_2,CC_2$ intersect at some point $P$. Call $P_1$ the isogonal conjugate of $P$ wrt. $ABC$. We will show that this point also has equal powers wrt. $o_A,o_B,o_C$. Let $A_3$ be the intersection of $AP_1$ and $o_A$. Then $AA_1$ is the angle bisector of $A_2AA_3$ and thus $A_1A_2=A_1A_3$, so $A_3$ also lies on $(A_2BC)$. Define $B_3,C_3$ similarly. We want to show that $AP_1\cdot P_1A_3=CP_1\cdot P_1C_3$. But notice that $\angle AA_3C= \angle A_2CB= \angle C_2AB=\angle CC_3A$, so $CAC_3A_3$ is cyclic and $AP_1\cdot P_1A_3=CP_1\cdot P_1C_3$. We do the same thing for $B_3$ and get that $P_1$ has equal powers wrt. $o_A,o_B,o_C$. Obviously $P_1 \not=Q$, so $o_A,o_B,o_C$ have a common radical axis which finishes the poof.

Most of the solutions in this thread don’t show that the circles actually intersect.

@above. Yes, but it is not difficult to fix tho

Haven't seen anyone write a solution inverting round one of the vertices. Can't be bothered to write a full solution, but I'll write out the steps. Each step is fairly straightforward to do. Invert around C with radius AB. Let P' denote the image of P upon this inversion. Let $D=A_1'A_2'\cap{}B_1'B_2'$, $X=A_2'C_1'\cap{}AB_2'$, $Y=B_2'C_1'\cap{}BA_2'$. The problem becomes equivalent to proving that $C_1'C_2'$ is the radical axis of $(AA_1'A_2')$ and $(BB_1'B_2')$. This results from the following claims: 1. $A_1'B_1'A_2'B_2'$ is cyclic. 2. $D\in{}C_1'C_2'$. 3. $X\in{}(AA_1'A_2')$, $Y\in{}(BB_1'B_2')$. 4. $A_2B_2YX$ cyclic. From 1. we have that $D$ is on the radical axis of $(AA_1'A_2')$ and $(BB_1'B_2')$, so because of 2 it suffices to prove that $C_1'$ as well. This follows directly from 3. and 4., and we're done.

Note that for a point \(P\) in the interior of \(\triangle ABC\) with \(\angle PCB=r\) and \(\angle PBC=s\), trig Ceva gives that the barycentric coordinates of \(P\) are given by \begin{align*} P&=\left( 1 :\frac{\sin(60^\circ-r)}{\sin r} :\frac{\sin(60^\circ-s)}{\sin s}\right)\\ &=\left( 1:\frac{\sqrt3}2\cot r-\frac12 :\frac{\sqrt3}2\cot s-\frac12\right). \end{align*}A circle through \(A\) has equation of the form \(xy+yz+zx=(x+y+z)(vy+wz)\), and \(P\) lying on this circle gives \begin{align*} \left( \frac{\sqrt3}2\cot r-\frac12 \right)v +\left( \frac{\sqrt3}2\cot s-\frac12 \right)w &=\frac{\left( \frac{\sqrt3}2\cot r+\frac12 \right) \left( \frac{\sqrt3}2\cot s+\frac12 \right)-1} {\frac{\sqrt3}2(\cot r+\cot s)}\\ &=\frac12+\frac{\sqrt3}2\cdot\frac{\cot r\cot s-1}{\cot r+\cot s}\\ &=\frac12+\frac{\sqrt3}2\cot(r+s). \end{align*} Let \(\alpha=\angle A_1BC\) and define \(\beta\) and \(\gamma\) similarly, so we are given \(\alpha+\beta+\gamma=30^\circ\). Now we have \begin{align*} A_1&=\left( 1:\frac{\sqrt3}2\cot\alpha-\frac12: \frac{\sqrt3}2\cot\alpha-\frac12\right)\\ \text{and}\quad A_2&=\left( 1:\frac{\sqrt3}2\cot(60^\circ-\beta)-\frac12: \frac{\sqrt3}2\cot(60^\circ-\gamma)-\frac12\right) \end{align*}If we write the equation of \((AA_1A_2)\) as \(xy+yz+zx=(x+y+z)(vy+wz)\), the condition \(A_1\in(AA_1A_2)\), \begin{align*} \left( \frac{\sqrt3}2\cot\alpha-\frac12 \right)(v+w) &=\frac12+\frac{\sqrt3}2\cot(2\alpha)\\ &=\frac12+\frac{\sqrt3}2\cdot\frac{\cot^2\alpha-1}{2\cot\alpha}\\ &=\frac{\left( \frac{\sqrt3}2\cot\alpha-\frac12 \right) (\frac12\cot\alpha+\frac{\sqrt3}2)}{\cot\alpha}\\ \implies v+w&=\frac12+\frac{\sqrt3}2\tan\alpha =\frac{\cos(60^\circ-\alpha)}{\cos\alpha}. \end{align*}The condition \(A_2\in(AA_1A_2)\) gives \begin{align*} \left( \frac{\sqrt3}2\cot(60^\circ-\beta)-\frac12 \right)v +\left( \frac{\sqrt3}2\cot(60^\circ-\gamma)-\frac12 \right)w &=\frac12+\frac{\sqrt3}2\cot(120^\circ-\beta-\gamma)\\ &=\frac12+\frac{\sqrt3}2\cot(90^\circ+\alpha)\\ &=\frac12-\frac{\sqrt3}2\tan\alpha\\ \implies \frac{\sqrt3}2\cot(60^\circ-\beta)v+\frac{\sqrt3}2\cot(60^\circ-\gamma)w &=\left( \frac12-\frac{\sqrt3}2\tan\alpha \right) +\frac12\left( \frac12+\frac{\sqrt3}2\tan\alpha \right)\\ &=\frac34-\frac{\sqrt3}4\tan\alpha\\ \implies \cot(60^\circ-\beta)v+\cot(60^\circ-\gamma)w &=\frac{\sqrt3}2-\frac12\tan\alpha =\frac{\cos(30^\circ+\alpha)}{\cos\alpha}. \end{align*} Solving, we have \begin{align*} v&=\frac{\frac{\cos(30^\circ+\alpha)}{\cos\alpha} -\cot(60^\circ-\gamma)\frac{\cos(60^\circ-\alpha)}{\cos\alpha}} {\cot(60^\circ-\beta)-\cot(60^\circ-\gamma)}\\ &=\frac{\sin(60^\circ-\beta)(\cos(30^\circ+\alpha)\sin(60^\circ-\gamma) -\cos(60^\circ-\gamma)\cos(60^\circ-\alpha))} {\cos\alpha(\cos(60^\circ-\beta)\sin(60^\circ-\gamma) -\cos(60^\circ-\gamma)\sin(60^\circ-\beta))}\\ &=\frac{-\sin(60^\circ-\beta)\cos(120^\circ-\alpha-\gamma)} {\cos\alpha\sin(\beta-\gamma)} =\frac{\sin(60^\circ-\beta)\sin\beta}{\cos\alpha\sin(\beta-\gamma)} \end{align*}and thus \[vy+wz=\frac{\sin(60^\circ-\beta)\sin\beta}{\cos\alpha\sin(\beta-\gamma)}y -\frac{\sin(60^\circ-\gamma)\sin\gamma}{\cos\alpha\sin(\beta-\gamma)}z.\] Now write the equation of \((AA_1A_2)\) as \(0=-(xy+yz+zx)+(x+y+z)(u_ax+v_ay+w_az)\), and similarly express \((BB_1B_2)\) and \((CC_1C_2)\). (Of course \(u_a=v_b=w_c=0\).) To show these three circles are coaxial, it will suffice to show the existence of \(\lambda_1\), \(\lambda_2\), \(\lambda_3\) such that \[\lambda_1\cdot(AA_1A_2)+\lambda_2\cdot(BB_1B_2)+\lambda_3\cdot(CC_1C_2) =0,\]or rather so that \begin{align*} \lambda_1+\lambda_2+\lambda_3&=0\\ \lambda_1u_a+\lambda_2u_b+\lambda_3u_c&=0\\ \lambda_1v_a+\lambda_2v_b+\lambda_3v_c&=0\\ \lambda_1w_a+\lambda_2w_b+\lambda_3w_c&=0. \end{align*}It is clear that the choice of \(\lambda_1=\cos\alpha\sin(\beta-\gamma)\), and \(\lambda_2\) and \(\lambda_3\) symetrically satisfies the last three equations, and we also have \[\lambda_1=\cos\alpha\sin\beta\cos\gamma-\cos\alpha\cos\beta\sin\gamma\]which cyclically sums to zero.

statistics show that this is the second hardest pure geometry IMO P3/6 since 2000 (by counting complete or almost complete solutions e.g. 6 or 7 points resp.) 1st is 2011/6 (0+6=6 had 6 or 7 points resp.) [MOHS 45] mean 0.318 2nd is this 2023/6 (4+6=10 had 6 or 7 points resp.) [MOHS ??] mean 0.275 3rd 2002/6 (0+12=12 had 6 or 7 points resp.) [MOHS 45] mean 0.409 4th is 2008/6 (1+12=13 had 6 or 7 points resp.) [MOHS 40] mean 0.260 5th is 2021/3 (0+15=15 had 6 or 7 points resp.) [MOHS 45] mean 0.372 6th is 2018/6 (5+18=23 had 6 or 7 points resp.) [MOHS 45] mean 0.638 7th 2003/3 (1+23=24 had 6 or 7 points resp.) [MOHS 40] mean 0.405 8th 2019/6 (3+27=30 had 6 or 7 points resp.) [MOHS 35] mean 0.403 9th 2015/3 (1+30=31 had 6 or 7 points resp.) [MOHS 25] mean 0.653 10th 2014/3 (4+26=33 had 6 or 7 points resp.) [MOHS 40] mean 0.505 11th 2000/6 (2+33=35 had 6 or 7 points resp.) [MOHS 35] mean 1.050 12th 2013/3 (4+41=45 had 6 or 7 points resp.) [MOHS 35] mean 0.786

Step 1$A_1A_2,B_1B_2,C_1C_2$ concur at a point lying on radax of $(AA_1A_2),(BB_1B_2),(CC_1C_2)$. Proof Let $\angle A_1BC=x,\angle B_1CA=y,\angle C_1AB=z$. Note that the angle condition in problem implies $x+y+z=30^\circ$.Since $C_1A=C_1B$ and $\angle AC_2B+\angle C_1AB=60^\circ+x+y+z=90^\circ \implies C_1$ is circumcenter of $C_2AB$ and similarly $A_1,B_1$ are circumcenters of $A_2BC,B_2CA$ respectively.So $\angle C_1C_2A_1=\angle C_1BA_1=\angle C_1A_2A_1 \implies A_1C_1A_2C_2$ is cyclic and similarly $B_1C_1B_2C_2$ and $A_1B_1A_2B_2$ are cyclic quads.So by radax theorem or pop we get that $A_1A_2,B_1B_2,C_1C_2$ concur at a point lying on radax of $(AA_1A_2),(BB_1B_2),(CC_1C_2)$. LemmaLet $BAC$ be an isosceles triangle with $BA=AC$ and $D$ be any point inside the triangle. $E$ be the isogonal conjugate of $D$. $O$ be the circumcenter of $(BDC)$.$F$ be the inverse of $E$ wrt the circle centered at $A$ with radius $AB=AC$. Then $F =(ADO) \cap (BDC)$. Proof:- $\angle BDC+\angle BFC=(\angle DBA+\angle BAC+\angle DCA)+(\angle ABE+\angle ACE)=\angle DBA+\angle BAC+\angle DCA+\angle DBC+\angle DCB=180^\circ \implies$ $BDCF$ is cyclic. Since $AO$ is bisector of triangle $BAC$ so $AO$ bisects $\angle DAF$ and since $OD=OF$ we get that $O$ is midpoint of arc in $(ADF)$. Step 2 Let $(CC_2C_1)$ meet $(BC_2A)$ again at $C_3$, define $A_3,B_3$ similarly. $AA_3$, $BB_3$,$CC_3$ concur at a point lying on radax of $(AA_1A_2),(BB_1B_2),(CC_1C_2)$. Proof Let $F$ be the isogonal conjugate of $C_2$, by above lemma we have $CF \cdot CC_3=CA^2 \implies \angle AC_3C=\angle CAF=60^\circ-y$ and similarly $\angle CA_3A=60^\circ-y$ so $ACA_3C_3$ is cyclic and similarly $BAB_3A_3$ and $CBC_3B_3$ are also cylic quads. By radax theorem or by pop we get that $AA_3$, $BB_3$,$CC_3$ concur at a point lying on radax of $(AA_1A_2),(BB_1B_2),(CC_1C_2)$. So the $3$ circles share a common radax.

Can you explain your claim $\angle C_1 C_2B_1 = x + 30 ^{\circ}$ made in the 3rd paragraph?

Glancing through the thread, I realize there are no solutions which use barycentric coordinates without heavy use of trigonometry. For the purpose of completion, I will present such a solution (which is essentially the solution I found when first working on the problem), which also proves the following generalization. Generalized IMO 2023 P6 wrote: Let $ABC$ be an equilateral triangle, and let $A_1$, $B_1$, and $C_1$ be arbitrary points on the $A$-, $B$-, and $C$-medians of $\triangle ABC$, respectively. Let $A_1'$ be the point on the $A$-median satisfying $$\angle BA_1'C+\angle CB_1A+\angle AC_1B=480^\circ$$(where we define $\angle BA_1'C$ to be greater than $180^\circ$ if $A_1'$ lies on the opposite side of line $BC$ from $A$, et cetera), and define $B_1'$ and $C_1'$ similarly. Let $BC_1$ and $CB_1$ meet at $A_2$, and define $B_2$ and $C_2$ similarly. Then the circumcircles $(AA_1'A_2)$, $(BB_1'B_2)$, and $(CC_1'C_2)$ are coaxial. Note that it is not necessary to know the generalization to find this solution; I found it after realizing exactly when in the solution the angle condition was used. I would be very interested in seeing a synthetic proof of the generalization. It is possible that the standard synthetics solutions suffice; I have not thought very much about it. Solution. We let $A=(1:0:0)$, $B=(0:1:0)$, $C=(0:0:1)$, $A_1=(2\delta:1-\delta:1-\delta)$, $B_1=(1-\epsilon:2\epsilon:1-\epsilon)$, and $C_1=(1-\zeta:1-\zeta:2\zeta)$. Let $A_1'=(2\delta':1-\delta':1-\delta')$, and define $B_1'$ and $C_1'$ similarly with $\epsilon'$ and $\zeta'$. We compute $$A_2=\big((1-\epsilon)(1-\zeta):2\epsilon(1-\zeta):2\zeta(1-\epsilon)\big).$$Claim 1. We have $\delta'=\frac{1-3\epsilon-3\zeta-3\epsilon\zeta}{3(1+\epsilon+\zeta-3\epsilon\zeta)}$.

Now, define $(u_A,v_A,w_A)$ to be such that circle $(AA_1'A_2)$ is given by $xy+yz+zx=(x+y+z)(u_Ax+v_Ay+w_Az)$, and define $(u_B,v_B,w_B)$ and $(u_C,v_C,w_C)$ similarly. Note that $u_A=0$ since $A$ lies on $(AA_1'A_2)$; similarly, $v_B=w_C=0$. Claim 2. To solve the problem, it suffices to show $\frac{v_A}{v_C}+\frac{w_A}{w_B}=1$.

Claim 3. We have $v_A=\frac{\epsilon(1-\epsilon)(1+3\zeta^2)}{(1+\epsilon+\zeta-3\epsilon\zeta)(\epsilon-\zeta)}$.

From this expression, we can compute $v_C$, $w_A$, and $w_B$ by permuting the variables $(\delta,\epsilon,\zeta)$ appropriately. This gives $$\frac{v_A}{v_C}+\frac{w_A}{w_B}=\frac{(1+3\zeta^2)(1+\delta+\epsilon-3\delta\epsilon)(\epsilon-\delta)}{(1+3\delta^2)(1+\epsilon+\zeta-3\epsilon\zeta)(\epsilon-\zeta)}+\frac{(1+3\epsilon^2)(1+\zeta+\delta-3\zeta\delta)(\zeta-\delta)}{(1+3\delta^2)(1+\epsilon+\zeta-3\epsilon\zeta)(\zeta-\epsilon)},$$so the following claim allows us to finish the problem. Claim 4. $\sum_{\text{cyc}}(1+3\zeta^2)(1+\delta+\epsilon-3\delta\epsilon)(\epsilon-\delta)=0$.

I usually find it easier to understand a geometry proof with animation and diagram, so I did up a video that covers the solution to this monster problem: https://www.youtube.com/watch?v=jZNIpapyGJQ Hope it helps!

Here's a solution based on Ali Zamani's idea: For each point $Z$ on plane , define $f(Z) = f_A(Z)= \frac {ZB}{ZC}$ . Define $f_B,f_C$ similarly but we mostly don't need them.It's well-known that for cyclic quadlirateral $PQRS$ if $W = PS \cap QR$ , then $f(W)=f(P)f(R)$ where this time $f(X)=\frac {XQ}{XR}$ First , note that $A_1$ is the circumcenter of $(A_2BC)$ since $A_1B=A_1C$ and $\angle BA_1C = 2\angle BA_2C$.Which implies that $\angle C_1C_2B_1=\angle B_1AC_1=\angle B_1B_2C_1$ which lead us to the fact that $B_1B_2C_1C_2$ is cyclic and by radical center theorem on the circles $(B_1B_2C_1C_2)$ , the lines $A_1A_2$ are concurrent at a point which is on the radical axis of the circles $(AA_1A_2)$. So we need to show that there exist another point with this property. Now , let $(AA_1A_2) \cap (ABC) = A_3$ .We need to show that the lines $AA_3$ are concurrent which is proving that $\prod f_A(A_3) = 1$.We'll use the following lemmas: Lemma1. $f(A_3) = f(A_1^{*})f(A_2^{*}) \frac {sin \angle A_1^{*}BA_2^{*}}{sin \angle A_1^{*}CA_2^{*}}$ where $A_1^{*},A_2^{*}$ are the images of $A_1,A_2$ under an inversion wrt $(A,AB)$. Proof. Just note that after inversion , $A_3^{*} = A_1^{*}A_2^{*} \cap BC$ and it's well-known that if $XY \cap BC = Z$ then $f(Z)=f(X)f(Y)\frac {sin \angle XBY}{sin \angle XCY}$ Lemma2.the lines $AA_2$ are concurrent. Proof.Trivial by trig Ceva. Also note $\angle A_1^{*}BA_2^{*} = \angle AA_2B - \angle AA_1B$ where angles are directed.Finally , let $Y_i = AA_i \cap BC$ for $i=1,2$: $1 =? \prod {f_A(A_1^{*})f_A(A_2^{*}) \frac {sin \angle AA_2B - \angle AA_1B}{sin \angle AA_2C - \angle AA_1C}} = \prod {\frac {f(Y_2)}{f(A_2)} \prod \frac {f(Y_1)}{f(A_1)} \prod \frac {sin \angle AA_2B - \angle AA_1B}{sin \angle AA_2C - \angle AA_1C}} = \prod {\frac {sin \angle AA_2B - \angle AA_1B}{sin \angle AA_2C - \angle AA_1C}}$ For this , let $A_4 = (A_1,A_1A_2) \cap AA_2$ and assume that $A_4,A_2$ are different points(note that $B,C$ lie on $(A_1,A_1A_2)$). By an easy angle chasing and sine law in $(A_1,A_1A_2)$ we can see that the final product of $\frac {sin ...}{sin ...}$ is just $\prod f_A(A_4) = \prod {\frac {f(Y_2)}{f(A_2)} } = 1$ ; since those angles are from the arcs $A_4B,A_4C$ in$(A_1,A_1A_2)$ .So we're done.

Let $\angle A_1BC=\alpha ,\angle B_1CA=\beta ,\angle C_1AB=\gamma ,$ so by angle condition $\alpha +\beta +\gamma =30^{\circ},$ and all three angles are different when $A_1B_1C_1$ is scalene. WLOG $\gamma$ is the most of them. Step 1. Observe that $\angle BA_2C=60^{\circ} +\beta +\gamma =90^{\circ} -\alpha,$ and analogously for other angles. By sine law in $ABC_2,$ $ACB_2$ $$\frac{|AC_1|}{|AC_2|}=\frac{\sin (60^{\circ} -\alpha)}{2\cos \beta \cos \gamma}=\frac{|AB_1|}{|AB_2|},$$hence $B_1C_1B_2C_2$ and all analogous quadrilaterals are cyclic. Step 2. Lines $AA_1,BB_1,CC_1$ concur at center of $ABC,$ so by converse of Brianchon for $AC_1BA_1CB_1$ this hexagon has an inconic. Then by Brianchon for $A_1C_2B_1A_2C_1B_2,$ $AB_2CA_2BC_2$ lines $A_1A_2,$ $B_1B_2,$ $C_1C_2$ concur at some point $P,$ and lines $AA_2,$ $BB_2,$ $CC_2$ concur at some point $Q.$ Step 3. Let $A_3$ be the second common point of $\odot (AA_1A_2), \odot (A_2BC),$ and similarly define $B_3,C_3.$ Since $\angle BA_1C=180^{\circ} -2\alpha =2\angle BA_2C_2,$ it follows that $A_1$ is the circumcenter of $\odot (A_2BC)$ and so $AA_1$ bisects angle $A_2AA_3.$ By the analogous reasoning lines $AA_3,$ $BB_3,$ $CC_3$ concur at the isogonal conjugate $R$ of $Q$ wrt $ABC.$ Step 4. Notice that $ABB_3A_3$ and all analogous quadrilaterals are cyclic since $$\angle AA_3B=\angle A_2A_3B- \angle AA_3A_2=\angle A_2CB-\angle AA_1A_2=\angle A_2CB-\angle A_2A_1B+\angle AA_1B=$$$$=\angle AA_1B-\angle A_2CB=30^{\circ} +\alpha +\beta =\ldots =\angle AB_3B.$$ Finish. Notice that points $P\neq R$ both have equal powers wrt $\odot (AA_1A_2),\odot (BB_1B_2),\odot (CC_1C_2),$ so they are coaxial $\blacksquare$

Solved with some help patching minor holes from DL. Below is a solution that addresses all possible traps of the problem. Let $\alpha=\angle ABC_1, \beta=BCA_1, \gamma=CAB_1.$ From the angle condition, we see $\alpha+\beta+\gamma=30^\circ.$ Observe \[\angle AC_2B = 180^\circ - (60^\circ -\gamma) - (60^\circ - \beta)=60^\circ+\beta + \gamma=90^\circ-\alpha,\]and $\angle AC_1B=180^\circ - 2\alpha,$ so $C_1$ is the center of $(AC_2B).$ Symmetrically, $A_1$ is the center of $(BA_2C)$ and $B_1$ is the center of $(CB_2A).$ Thus, $B_1$ lies on the perpendicular bisector of $B_2C,$ so $\triangle B_1B_2C$ is isosceles. Similarly, $\triangle A_1CA_2$ is also isosceles. Now $\angle B_1B_2 C = \angle B_1CB_2 = \angle A_1A_2C,$ so $A_1B_1A_2B_2$ is cyclic. Similarly $B_1C_1B_2C_2$ and $C_1A_1C_2A_2$ are cyclic. We now check $A_1B_2C_1A_2B_1C_2$ is not cyclic. But in order for a hexagon to be cyclic, the alternating angles must sum to $360^\circ,$ which cannot be true as $\angle A_1C_1B_2 + \angle B_2A_1C_2 + \angle C_2B_1A_2=480^\circ.$ Thus, by radical axes on $(A_1B_1A_2B_2), (B_1C_2B_2C_2), (C_1A_1C_2A_2)$ we find $A_1A_2, B_1B_2, C_1C_2$ concur. Let this concurrency point be $P.$ Now $PA_1 \cdot PA_2 = PB_1 \cdot PB_2 = PC_1 \cdot PC_2,$ so $P$ has equal powers with respect to $(AA_1A_2), (BB_1B_2), (CC_1C_2),$ and thus lies on their radical axis. Construct $A_3=(BA_2C) \cap (AA_1A_2) \neq A_2,$ and similarly define $B_3$ and $C_3.$ Note $A_1A_3 = A_1A_2,$ so we know $AA_1$ bisects $\angle A_3AA_2.$ Apply Brianchon on $AB_2CA_2BC_2$ to obtain $AA_2, BB_2, CC_2$ concur. Since $AA_2$ and $AA_3$ are isogonal in $\angle BAC,$ it follows $AA_3, BB_3, CC_3$ concur at the isogonal conjugate of $AA_2 \cap BB_2 \cap CC_2.$ Let this concurrency point be $Q.$ Reflect $C_2$ across $CC_1$ to a point $C_2'.$ Then we know $C, C_2', C_3$ are collinear. Observe $C_2'$ lies on $(ABC_2)$ since $\angle AC_2B = \angle AC_2'B,$ which holds as $CC_1$ is the perpendicular bisector of $AB.$ Note \[\angle BC_3C = \angle BC_3C_2' = \angle BAC_2' = \angle ABC_2 = 60^\circ - \beta.\]Similarly, reflecting $B_2$ over $BB_1$ to a point $B_2'$ gives us $\angle CB_3B = 60^\circ - \beta$ as well, implying $BCB_3C_3$ is cyclic. Now $QB \cdot QB_3 = QC \cdot QC_3,$ so $Q$ lies on the radical axis of $(BB_1B_2)$ and $(CC_1C_2).$ But $Q$ must also lie on the radical axis of $(AA_1A_2)$ due to the concurrency. Thus $Q$ has equal powers with respect to all three circles. We must check $P \neq Q.$ But this is clear since $P$ lies on $A_1A_2$ while $Q$ lies on $AA_3.$ Thus $PQ$ is the radical axis of $(AA_1A_2), (BB_1B_2),$ and $(CC_1C_2).$ It remains to verify that the circles intersect at two points. If otherwise, observe that in all other cases the radical axis of two circles will lie strictly outside both circles. But from definition of $P,$ we know it lies on chords $A_1A_2, B_1B_2, C_1C_2,$ so it must lie inside all three circles. Therefore the circumcircles of the three triangles pass through two common points.

Let $\angle{B_1AC}=\alpha, \angle{C_1AB}=\beta; \angle{A_1BC}=\theta$, by angle chasing, $\alpha+\beta+\theta=(180\cdot 3-480)/2=30$ degrees Realize $\angle{CA_2B}=180-(60-\alpha+60-\beta)=60+\alpha+\beta=60+(30-\alpha)=90-\alpha; \angle{BA_1C}=180-2\alpha$, we can conclude $A_1$ is the center of $(A_2CB)$. Similarly, we can deduce $B_1,C_1$ are centers of $(B_2AC);(C_2AB)$ respectively. As $A_1C=A_1A_2, B_1C=B_1B_2, \angle{A_1CB_1}=\angle{B_1B_2C}=\angle{A_1A_2C}, A_1,B_1,A_2,B_2$ are concyclic. Same reason, $C_1B_1C_2B_2,A_1C_1A_2C_2$ are cyclic quadrilaterals.Radical axes imply that $A_1A_2, B_1B_2, C_1C_2$ concur at point $M$. Then, we apply Brianchon to show that there is a conic section inscribed in $A_2B_1C_2A_1B_2C_1$. Then, we discover that the conic section is also tangent to the self-intersecting hexagon $AB_2CA_2BC_2$, which indicates $AA_2,BB_2,CC_2$ are concurrent as well, calling the intersection as $L$ Let $(AA_2A_1)$ meets $(A_2CB)$ at $J$, $(BB_2B_1)$ meets $(ACB_2)$ at $I$ and $(CC_2C_1)$ meets $(C_2AC)$ at $N$. As $A_1A_2=A_1J, \angle{A_1AA_2}=\angle{A_1AJ}$, $AA_1$ is bisecting $\angle{JAA_2}$, $BB_1,CC_2$ are bisecting $\angle{IBB_2}, \angle{NCC_2}$ respectively. Then we can have $AJ,BI,CN$ intersecting at point $K$, which is the isogonal conjugate of $L$ wrt $\triangle{ABC}$ Then, we can find $\angle{AJB}=\angle{AJA_2}+\angle{A_2JB}=\angle{A_2CB}+\angle{AA_1A_2}=\angle{A_2CB}+\angle{AA_1B}-\angle{A_2A_1B}=\angle{AA_1B}-\angle{A_2CB}=90+\theta-(60-\alpha)=30+\alpha+\theta=\angle{AIB}$ Thus $AIFB,ACJN,ICBN$ are all cyclic quadrilateral which radical center is $K$. Thus $L,K$ have the same power wrt $(AA_1A_2),(BB_1B_2),(CC_1C_2)$ which lies on the common radical axis of three circles which satisfies the statement.

Can someone please post a full solution using spherical triangles? I promise I will compile all the solutions to this beast of a problem once everybody has posted their solutions! This is encouraging me to make a whole set of geometry problems... I'm referring to this remark by ABCDE: ABCDE wrote: Remark: This application of the coaxial lemma can be interpreted as a "spherical analogue" of Ceva's theorem: with a suitable stereographic projection, the circles $\omega_A$, $\omega_B$, and $\omega_C$ can be interpreted as great circles on a sphere, and we wish to show the concurrence of "spherical Cevians" $(AA_1A_2)$, $(BB_1B_2)$, and $(CC_1C_2)$ in the spherical triangle $A_1B_1C_1$. The "Cevian ratio" for $(AA_1A_2)$ is given exactly by the ratio of powers from the coaxial lemma, which is simple to compute.

Beautiful problem. Without geogebra is very hard. Let $P,Q = (BB_1B_2)\cap (CC_1C_2)$, $X = B_1B_2\cap C_1C_2$ and $Y = (B_2C_1X)\cap (C_2B_1X)$. Let $\ell$ be the length of the triangle side and let $\angle A_1BC=\alpha, \angle B_1AC=\beta, \angle C_1AB=\gamma$. With easy angle working we get $\alpha+\beta+\gamma==30^{o}$. $\textbf{Claim 1.}$ $B_1C_2B_2C_1,\; A_1B_2A_2B_1,\; A_1C_2A_2C_1$ are cyclic. $\textbf{Proof.}$ We have $AB_1 = \frac{\ell}{2\cos\beta}$ and $AC_1 = \frac{\ell}{2\cos\gamma}$. Now by sine law on $\triangle AC_2B$ we have \[ \frac{AC_2}{\sin(60-\alpha)} = \frac{\ell}{\sin(\angle AC_2B)}= \frac{\ell}{\sin(60+\alpha+\beta)}= \frac{\ell}{\sin(60+\alpha+\beta+\gamma-\gamma)}= \frac{\ell}{\sin(90-\gamma)}= \frac{\ell}{\cos\gamma} \]then $AC_2 = \frac{\ell\sin(60-\alpha)}{\cos\gamma}$ and also $AB_2 = \frac{\ell\sin(60-\alpha)}{\cos\beta}$. Combining everything we have that $AB_1\cdot AC_2 = AC_1\cdot AB_2$ so $B_1C_2B_2C_1$ is cyclic, the rest is analogous.$\square$ $\textbf{Claim 2.}$ $A_1,X,A_2$ are collinear and $A_2X\cdot XA_1 = B_1X\cdot XB_2$. $\textbf{Proof.}$ Using previous claim and radical axis on $(A_1B_2A_2B_1), (A_1C_2A_2C_1),(B_1C_2B_2C_1)$ we get the collinearity, the rest is power of point.$\square$ $\textbf{Claim 3.}$ $P,X,Q$ are collinear and $PX\cdot XQ = B_1X\cdot XB_2$. $\textbf{Proof.}$ Radical axis on $(CC_1C_2), (BB_1B_2),(B_1C_2B_2C_1)$ and power of point.$\square$ $\textbf{Claim 4.}$ $AA_2PYA_1Q$ is cyclic. $\textbf{Proof.}$ By radical axis on $(B_1C_2B_2C_1),(B_2C_1X),(C_2B_1X)$ we have that $A,X,Y$ are collinear. Inversion at $A$ with radius $\sqrt{AX\cdot AY}$ shows that $AB_1YB_2$ is cyclic, so by power of point $AX\cdot XY = B_1X\cdot XB_2$, using claims $2,3$ we have \[ B_1X\cdot XB_2 = AX\cdot XY = A_2X\cdot XA_1 = PX\cdot XQ.\square \] Using last claim $P,Q\in (AA_1A_2)$ and we are done.

alinazarboland wrote: Here's a solution based on Ali Zamani's idea: For each point $Z$ on plane , define $f(Z) = f_A(Z)= \frac {ZB}{ZC}$ . Define $f_B,f_C$ similarly but we mostly don't need them.It's well-known that for cyclic quadlirateral $PQRS$ if $W = PS \cap QR$ , then $f(W)=f(P)f(R)$ where this time $f(X)=\frac {XQ}{XR}$ First , note that $A_1$ is the circumcenter of $(A_2BC)$ since $A_1B=A_1C$ and $\angle BA_1C = 2\angle BA_2C$.Which implies that $\angle C_1C_2B_1=\angle B_1AC_1=\angle B_1B_2C_1$ which lead us to the fact that $B_1B_2C_1C_2$ is cyclic and by radical center theorem on the circles $(B_1B_2C_1C_2)$ , the lines $A_1A_2$ are concurrent at a point which is on the radical axis of the circles $(AA_1A_2)$. So we need to show that there exist another point with this property. Now , let $(AA_1A_2) \cap (ABC) = A_3$ .We need to show that the lines $AA_3$ are concurrent which is proving that $\prod f_A(A_3) = 1$.We'll use the following lemmas: Lemma1. $f(A_3) = f(A_1^{*})f(A_2^{*}) \frac {sin \angle A_1^{*}BA_2^{*}}{sin \angle A_1^{*}CA_2^{*}}$ where $A_1^{*},A_2^{*}$ are the images of $A_1,A_2$ under an inversion wrt $(A,AB)$. Proof. Just note that after inversion , $A_3^{*} = A_1^{*}A_2^{*} \cap BC$ and it's well-known that if $XY \cap BC = Z$ then $f(Z)=f(X)f(Y)\frac {sin \angle XBY}{sin \angle XCY}$ Lemma2.the lines $AA_2$ are concurrent. Proof.Trivial by trig Ceva. Also note $\angle A_1^{*}BA_2^{*} = \angle AA_2B - \angle AA_1B$ where angles are directed.Finally , let $Y_i = AA_i \cap BC$ for $i=1,2$: $1 =? \prod {f_A(A_1^{*})f_A(A_2^{*}) \frac {sin \angle AA_2B - \angle AA_1B}{sin \angle AA_2C - \angle AA_1C}} = \prod {\frac {f(Y_2)}{f(A_2)} \prod \frac {f(Y_1)}{f(A_1)} \prod \frac {sin \angle AA_2B - \angle AA_1B}{sin \angle AA_2C - \angle AA_1C}} = \prod {\frac {sin \angle AA_2B - \angle AA_1B}{sin \angle AA_2C - \angle AA_1C}}$ For this , let $A_4 = (A_1,A_1A_2) \cap AA_2$ and assume that $A_4,A_2$ are different points(note that $B,C$ lie on $(A_1,A_1A_2)$). By an easy angle chasing and sine law in $(A_1,A_1A_2)$ we can see that the final product of $\frac {sin ...}{sin ...}$ is just $\prod f_A(A_4) = \prod {\frac {f(Y_2)}{f(A_2)} } = 1$ ; since those angles are from the arcs $A_4B,A_4C$ in$(A_1,A_1A_2)$ .So we're done. Nice solution bro, you show that Energy's guys are much much different than other

By the way who's the proposer ?

PNT wrote: By the way who's the proposer ? Ankan.

To prove that the three circumcircles of triangles $AA_1A_2, BB_1B_2,$ and $CC_1C_2$ all pass through two common points, we need to show that these circles are coaxial. Let's denote the circumcircles of triangles $AA_1A_2, BB_1B_2,$ and $CC_1C_2$ as $\Gamma_A, \Gamma_B,$ and $\Gamma_C,$ respectively. Consider an inversion with center $A$ and an arbitrary radius. For simplicity, let's denote the inverse of a point $X$ as $X'.$ Under this inversion: - $A_1$ is the inverse of $C.$ - $A_2$ is the inverse of $C_2.$ - $\Gamma_A$ is the image of line $C_1A_2$ after inversion. Let's call this image line $l_A.$ Similarly, we can define the inverses of points and lines with respect to centers $B$ and $C.$ Now, let's prove that $l_A, l_B,$ and $l_C$ (the images of lines $C_1A_2, A_1B_2,$ and $B_1C_2,$ respectively) are concurrent. Since $\angle BA_1C + \angle CB_1A + \angle AC_1B = 480^\circ,$ and the sum of the angles in an equilateral triangle is $180^\circ \times (3 - 2) = 60^\circ,$ we have: $$\angle BA_1C + \angle CB_1A + \angle AC_1B = \angle CAB + \angle ABC + \angle BCA = 180^\circ.$$ Therefore, $A_1, B_1, C_1$ are collinear points. Hence, under the inversion, $C_1$ is the inverse of line $l.$ Similarly, $A_1$ is the inverse of line $l_C,$ and $B_1$ is the inverse of line $l_B.$ We know that the inverses of three collinear points are concurrent. Therefore, $l_A, l_B,$ and $l_C$ are concurrent at a point $P.$ Now, let's prove that $\Gamma_A, \Gamma_B,$ and $\Gamma_C$ pass through $P.$ Since $l_A$ is the image of line $C_1A_2$ under inversion with center $A,$ we have $C_1A_2 \perp l_A$ (i.e., $l_A$ is perpendicular to $C_1A_2$ at the intersection point $P$). Similarly, $A_1B_2 \perp l_B$ and $B_1C_2 \perp l_C.$ Now, let's consider $\Gamma_A$ and prove that it passes through $P.$ Since $C_1A_2 \perp l_A,$ and $P$ is the intersection of $C_1A_2$ and $l_A,$ we have $\angle A_2PP' = 90^\circ.$ Therefore, $A_2P'P$ is a straight line, where $P'$ is the inverse of $P.$ But we also know that $A_1A_2P'C_1$ is cyclic (since $\Gamma_A$ is the circumcircle of triangle $AA_1A_2$). So, $\angle A_2P'A_1 = 180^\circ - \angle C_1A_1P' = 180^\circ - \angle C_1AA_1.$ Now, $\angle C_1A_1A = \angle C_1BA = \angle BCA$ (since $C_1A = AC, C_1B = BC$). Therefore, $\angle C_1A_1P' = 180^\circ - \angle BCA.$ Thus, $\angle A_2P'A_1 = \angle BCA.$ Since $\angle A_2PP' = 90^\circ,$ it implies that $\angle A_2PA_1 = 90^\circ + \angle BCA.$ Similarly, we can prove that $\angle B_2PB_1 = 90^\circ + \angle CAB$ and $\angle C_2PC_1 = 90^\circ + \angle ABC.$ Now, consider the angles of triangle $PAB$: $$\angle APB = 360^\circ - (\angle A_2PA_1 + \angle B_2PB_1) = 360^\circ - [(90^\circ + \angle BCA) + (90^\circ + \angle CAB)] = 180^\circ - \angle ABC.$$ Similarly, $\angle APC = 180^\circ - \angle BCA$ and $\angle BPC = 180^\circ - \angle CAB.$ These angle relations show that points $P, A, B, C$ are concyclic. Thus, $\Gamma_A, \Gamma_B,$ and $\Gamma_C$ all pass through $P.$ Hence, we have shown that $l_A, l_B,$ and $l_C$ are concurrent at $P,$ and $\Gamma_A, \Gamma_B,$ and $\Gamma_C$ all pass through $P.$ Thus, the three circumcircles are coaxial, and they share the two common points $P$ and the center of the equilateral triangle $ABC.$ This completes the proof.

My solution https://nttuan.org/2023/07/10/imo2023p6/.

how does one draw the diagram in an exam

PNT wrote: By the way who's the proposer ? Ankan and Luke

My $50^\text{th}$ post.