Solution:

EDIT: The argument I use to prove the first claim is wrong--it fails when the hexagon has acute angles. Many other solutions in this thread have similar flaws, see Darij's post later in the thread: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=44478&postorder=asc&start=26

Please tell me if this hunch is on the right track:

Using schulmannerism's claim, one can also easily finish Ceva's Theorem using ratio of areas (half of product of two sides and the sine of the angle in between.)

Strange why most participants I have met used inequalities to solve this problem. My solution, at least, doesn't explicitely make use of them (of course, it uses the arrangement of the points, which can be considered as a kind of inequalities, but I guess there is no way to avoid using it...). So here is, depending of how soon we will quit the internet cafe, my solution or just a rough outline of it: Since the hexagon $A_1A_2B_1B_2C_1C_2$ has equal sidelengths, we have $C_2A_1=A_1A_2$; thus, the triangle $C_2A_1A_2$ is isosceles, so its base angle is $\measuredangle A_1C_2A_2=\frac{180^{\circ}-\measuredangle C_2A_1A_2}{2}=\frac{\measuredangle C_2A_1B}{2}$. Similarly, $\measuredangle C_1A_1C_2=\frac{\measuredangle A_1C_2B}{2}$. But by the sum of angles in triangle $C_2BA_1$, we have $\measuredangle C_2A_1B+\measuredangle A_1C_2B=180^{\circ}-\measuredangle C_2BA_1=180^{\circ}-\measuredangle ABC$. Finally, < ABC = 60°, since the triangle ABC is equilateral. Hence, if the lines $C_1A_1$ and $C_2A_2$ intersect at a point K, the exterior angle theorem in triangle $C_2KA_1$ yields $\measuredangle A_2KA_2=\measuredangle A_1C_2K+\measuredangle KA_1C_2=\measuredangle A_1C_2A_2+\measuredangle C_1A_1C_2$ $=\frac{\measuredangle C_2A_1B}{2}+\frac{\measuredangle A_1C_2B}{2}=\frac{\measuredangle C_2A_1B+\measuredangle A_1C_2B}{2}$ $=\frac{180^{\circ}-\measuredangle ABC}{2}=\frac{180^{\circ}-60^{\circ}}{2}=60^{\circ}$. So the lines $C_1A_1$ and $C_2A_2$ intersect at an angle of 60°. Similarly, the lines $A_1B_1$ and $A_2B_2$ intersect at an angle of 60°, and so do the lines $B_1C_1$ and $B_2C_2$. Hence, the triangle $A_2B_2C_2$ is the image of the triangle $A_1B_1C_1$ under a spiral similitude with rotation angle 60°. [On the exam, I avoided speaking of "angles between lines" without properly defining the orientation of these angles, and thus I rewrote the above argument in a non-transformational way.] Now, let Z be the center of the spiral similitude mapping the triangle $A_1B_1C_1$ to the triangle $A_2B_2C_2$. Then, since the rotation angle of the similitude is 60°, the triangles $ZA_1A_2$, $ZB_1B_2$, $ZC_1C_2$ are all directly similar, and have the angles $\measuredangle A_1ZA_2=\measuredangle B_1ZB_2=\measuredangle C_1ZC_2=60^{\circ}$. Actually, since $A_1A_2=B_1B_2=C_1C_2$ (as the hexagon $A_1A_2B_1B_2C_1C_2$ has equal sidelengths), these triangles are even congruent. Thus, for instance, $ZA_1=ZB_1$. Together with $A_1A_2=B_1A_2$ (again since the hexagon $A_1A_2B_1B_2C_1C_2$ has equal sidelengths) and $ZA_2=ZA_2$, the triangles $ZA_1A_2$ and $ZB_1A_2$ are congruent, so that $\measuredangle B_1ZA_2=\measuredangle A_1ZA_2$. Since $\measuredangle A_1ZA_2=60^{\circ}$, we thus have $\measuredangle B_1ZA_2=60^{\circ}$. Hence, $\measuredangle A_1ZA_2+\measuredangle B_1ZA_2+\measuredangle B_1ZB_2=60^{\circ}+60^{\circ}+60^{\circ}=180^{\circ}$. And thus, the point Z lies on the line $A_1B_2$. Similarly, the point Z lies on the lines $B_1C_2$ and $C_1A_2$. And we are done. Oh my goodness, I guess 75% of my above argumentation were redundant. But actually, the whole thing took me 25 minutes, and I found it too much (I expected a geometry problem 1 to be doable in 10 minutes, like the one last year, but this one was harder), so I did not try to simplify but rather wrote it up as quickly as possible to have the time for the other two, especially since problem 2 seemed to be absolutely hard . Anyway, a good first day, although not for me, since I could have done the third one... Darij

The whole point of the problem is to prove that $A_1 B_1 C_1$ and $ A_2 B_2 C_2$ are equilateral. To prove that let angle $C_1 B_1 B_2 = b$, $C_2 C_1 A_1 = c$ and $A_1 B_1 A_2 = a$. Then we easily get : $ A_1 C_1 B_1 = 60 +b-c$, $ C_1 A_1 B_1 = 60 + c-a$ and $ A_1 B_1 C_1 = 60 + a-b$. We can also prove in an easy way that : $ |a-b| < 30 $... and $ a<60$ ... respectivly. Then by the cosine rule to the triangle $C_1 C_2 B_1$ we get that : $(C_1 B_1)^2 = (2x cos(b))^2 $ ... By the sine rule to the trianlge $A_1 B_1 C_1$ and some calulation, we get the following system of equaltions : (1) $ sin(a+b-c)\cdot cos(60 +b-a)=\frac{1}{2} \cdot sin(c) $ (2) $ sin(b+c-a)\cdot cos(60 +c-b)=\frac{1}{2} \cdot sin(a) $ (3) $ sin(a+c-b)\cdot cos(60 +a-c)=\frac{1}{2} \cdot sin(b) $ Now, if $a=b$ we easily get from (1) that $a=b=c$. Let , wlog, that $a>b$. then from (1) we get that $sin(c) < sin(a+b-c)$ or $ 2c<a+b$ so $c<b$. If $a>c$ then similarly from (3) $ 2b <a+c$ . Contradiction. If $b>c>a$ then from (2) $sin(b+c-a) > sin(c)>sin(a)$ so $\frac{1}{2} > cos(60+c-b)$ so $60+c-b > 60$ or $c>b$ . Contradiction. So $a=c$ but then from (3) we get $a=b=c$. Contradiction. At last, we proved that $a=b=c$. But then $A_1 B_1 C_1$ is equilateral and angle $C_1 C_2 A = B_1 A_2 C = A_1 C_2 B$. Simirly for the triangle $A_2 B_2 C_2$ and finally it is obvious that triangles $A C_1 B_2 = B A_1 C_2 = C A_2 B_1$. The rest follows easily.

it seemed to me that this problem is harder than the 2nd, but only because I was looking for a more complicated solution denote |A(1)A(2)|=d, |AB|=a, |A(2)C|=x, |B(2)A|=y, |C(2)B|=z. let's assume x>y, then a-d-y>a-d-x, and from triangles BA(1)C(2) and CB(1)A(2) follows that z>x, then a-d-x>a-d-z and similarly from the 2 triangles BA(1)C(2) and AC(1)B(2) it follows that y>z. So we have x>y>z>x, a contradiction. Similarly for x<y. So x=y and by symmetry x=y=z. Hence A(1)B(1)C(1) and A(2)B(2)C(2) are equilateral and the lines from the problems are their heights (medians, bisectors.. ) and all meet in one point.

Some say it is not an IMO problem? Well, I agree, since even I am able to solve it. My solution to problem one: Lemma(Poland 1965--1966) If the diagnals AD, BE, CF half the area of the convex hexagon, then AD, BE and CF are concurrent. Proof: Assume the do not meet at one point. Let's say $O= AD \cap BE$, $P= CF \cap BE$, $Q= AD \cap CF$ Since AD, BE, CF half the area, so: \[\triangle{AOB}=\triangle{DOE}, \triangle{EFP}=\triangle{BCP}, \triangle{CQD}=\triangle{AQF}\] then: $1=\frac{\triangle{AOB}}{\triangle{DOE}} \cdot \frac{\triangle{EFP}}{\triangle{BCP}} \cdot \frac{\triangle{CQD}}{\triangle{AQF}}= \frac{AO \cdot BO}{DO \cdot EO} \cdot \frac{EP \cdot FP}{BP \cdot CP} \cdot \frac{CQ \cdot DQ}{AQ \cdot FQ}$ $= \frac{AO}{AQ} \cdot \frac{BO}{BP} \cdot \frac{CQ}{CP} \cdot \frac{DQ}{DO} \cdot \frac{EP}{EO} \cdot \frac{FP}{FQ} $ $ \neq 1$ Hence they must be concurrent. Now get back to this problem. Since $C_1C_2 = B_1B_2= A_1A_2$, and $A=C=B=60^o$, so we shall have: $R_{BA_1C_2}=R_{AC_1B_2}=R_{CA_2B_1}$ ($R$ is the circumradius.) Assume $A_2C > B_2A$. Since $BA_1+A_2C=CB_1+B_2A$, so $BA_1 < CB_1$. Since $2R=\frac{BA_1}{\sin{BC_2A_1}}=\frac{CB_1}{\sin{B_1A_2C}}$ So $\angle{BC_2A_1} < \angle{CA_2B_1}$ So $ \angle{BA_1C_2} > \angle{CB_1A_2}$ Use the law of sines again, we get $BC_2 > CA_2$. So $BC_2 > B_2A$. Hence $\angle{BA_1C_2} > \angle{AC_1B_2}$, Hence $\angle{BC_2A_1} < \angle{AB_2C_1}$ Hence $BA_1 < AC_1$. Because we already have $C_2B > CA_2$, so add them togather we have $AB-C_1C_2 > BC- A_1A_2$. It is impossible. By symmetry, we can know that $CA_2=BC_2=AB_2$, $A_1B=C_1A=B_1C$. The rest is easy. We can prove that $\triangle{AC_1B_2}=\triangle{CB_1A_2}=\triangle{BA_1C_2}$ $\triangle{A_1C_1C_2}=\triangle{B_1A_1A_2}=\triangle{C_1B_1B_2}$ $\cdots \cdots \cdots$ It is easy to prove that the diagnols half the area of the hexagon. Use our lemma now... Cheers! How much point can this proof recieve?

I shall also add a pic for the lemma. P.S. How do I add two pictures in one post?

This question is very easy. Step 1: prove AC1=BA1=CB1. Shift AC1 to A'C'', C2B to C''B', AB2 to A'B'', B1C to B''C'. Then it is easy to show that B'A'' and A''C' are also shifted BA1 and A2C. Since triangle A'B'C' and A''B''C'' are both equilateral triangles, A'C''=B'A''=C'B'' => AC1=BA1=CB1 Step 2: A2C1 is the middle line of C2B2 and B1A1 and so on. It is easy. Step 3: triangle B2C2A2 is an equilateral triangle so the three lines meet.

I note: $BA_1=a_1,CA_2=b_1;CB_1=a_2,AB_2=b_2;AC_1=a_3,BC_2=b_3.$ ...........$A_1A_2=B_1B_2=C_1C_2=x,a_1+a_2+a_3=l,$ $a_1+b_1=a_2+b_2=a_3+b_3=k.$ **************************************************************************** $a_1^2+b_3^2-a_1b_3=a_2^2+b_1^2-a_2b_1=a_3^2+b_2^2-a_3b_2=x^2.$ $a_1^2+(k-a_3)^2-a_1(k-a_3)=a_2^2+(k-a_1)^2-a_2(k-a_1)...iff...$ $a_3^2-2ka_3+ka_1+a_1a_3=a_2^2-ka_2+a_1a_2...iff...$ $(a_3-a_2)(a_1+a_2+a_3)+k(a_1+a_2-2a_3)=0\ a.s.o.$ $l(a_3-a_1)=k(a_2+a_3-2a_1);\ l(a_1-a_2)=k(a_3+a_1-$ $2a_2);\ l(a_2-a_3)=k(a_1+a_2-2a_3).$ (*) ***************************************************************************************** I pressupose that $a_1\ne a_2\ne a_3\ne a_1.$ Thus, $\frac{a_2+a_3-2a_1}{a_3-a_1}=\frac{a_3+a_1-2a_2}{a_1-a_2}=\frac{a_1+a_2-2a_3}{a_2-a_3}=\frac{l}{k}...iff...$ $1-\frac{l}{k}=\frac{a_1-a_2}{a_3-a_1}=\frac{a_2-a_3}{a_1-a_2}=\frac{a_3-a_1}{a_2-a_3}=$ $\sqrt [3]{\frac{(a_1-a_2)(a_2-a_3)(a_3-a_1)}{(a_3-a_1)(a_1-a_2)(a_2-a_3)}}=1...iff...l=0$, what is absurd. Thus, there are $i,j\in \{1,2,3\},\ i\ne j$, such that $a_i=a_j.$ From the relations (*) results $a_1=a_2=a_3,$ a.s.o. Thanks, Darij Grinberg, for correction of calculus !

I have beautiful solution: I is point inside convex hexagon A_1A_2B_1B_2C_1C_2 such that IA_1A_2 is regular triangle. It's easy to prove that IC_1B_2 is regular triangle. Let the lines $A_1B_2$ and $A_2C_1$ meet at $M$. So I, M, B_1 are collinear (because they lie on the perpendicular bisector of segment B_2A_2); and I, M, C_2 are collinear (because they lie on the perpendicular bisector of segment A_1C_1). So C_2, M, B_1 are conlinear or A_1B_2, B_1C_2, C_1A_2 are concurrent. [Moderator edit: Corrected some language mistakes and defined point M.]

I think only Darij 's proof and Anto 's proof are valuable . Check your proofs again ! Something is vague .

check out this one: as the three vectors A1A2, B1B2, C1C2 are similar (in the same scale) to BC, AC, AB respectively, we have A1A2+B1B2+C1C2=0, so A2B1+B2C1+C2A1 is also a zero vector, so we can build an equilateral triangle from vectors A2B1, B2C1, C2A1, which has its sides parallel to the sides of the triangle made by the prolongations of A2B1, B2C1, C2A1, so this triangle is also equilateral (denote its vertices as A',B',C'). as we have angle B1B'C1=60=B1AC1, B1B'AC1 is cyclic, so angle B'B1A=B'C1A, so the angles B1 and C1 of the hexagon are equal. analogically we can prove that the angles of the hexagon A1=B1=C1=alpha, A2=B2=C2=beta, so A1B2, B1C2, C1A2 are the axes of symmetry of the hexagon....

Steps of the proof: 1 - Construction of the hexagon. 2 - Show that A_1B_1C_1 and A_2B_2C_2 are equilateral triangles. 3 - Show that the point of concurrence is the circuncenter of the triangle ABC. 1 - It's easy to see that we can construct the hexagon described by the problem by the following manner: Draw the equilateral triangle and its circuncircle. Choose other three points on the circuncircle and draw another equilateral triangle, different from ABC, named \overline{A}\overline{B}\overline{C}. The points where the triangles concur are exactly the vertices of the hexagon. 2 - By marking some anlgles and with the known fact of congruence, it's very easy to see that the triangles A_1B_1C_1 and A_2B_2C_2 are equilateral. 3 - Call P the circuncenter of the triangle ABC. Draw AP,C_1P, B_1P and \angle{AC_1B_2}=\alpha. We know that \overline{C_1P} bissects \angle{B_2C_1C_2}, and then \angle{B_2C_1P}=\frac{\pi-\alpha}{2} and using the fact that C_1A\overline{A}B_1 are concyclic e conclude that \angle{AB_1P}=\frac{\pi-\alpha}{2}, since \angle{AC_1P}+\angle{AB_1P}=\pi then, this quadrilateral is cyclic to, which proves that \overline{PC_1}=\overline{PB_1}. By the same way, you prove the same for the other segment, which makes us conclude that P is the circuncenter of A_1B_1C_1. Analogously, you can prove the P is circuncenter of A_2B_2C_2. Since we know then that P belong to the mediatrice of the equilateral triangle, we conclude that the segments concur in P If you find any mistake, I'll be glad to be noticed =] thanks =]

Steps of the proof: 1 - Construction of the hexagon. 2 - Show that A_1B_1C_1 and A_2B_2C_2 are equilateral triangles. 3 - Show that the point of concurrence is the circuncenter of the triangle ABC. 1 - It's easy to see that we can construct the hexagon described by the problem by the following manner: Draw the equilateral triangle and its circuncircle. Choose other three points on the circuncircle and draw another equilateral triangle, different from ABC, named \overline{A}\overline{B}\overline{C}. The points where the triangles concur are exactly the vertices of the hexagon. 2 - By marking some anlgles and with the known fact of congruence, it's very easy to see that the triangles A_1B_1C_1 and A_2B_2C_2 are equilateral. 3 - Call P the cinrcuncenter of the triangle ABC, and draw \overline{P\overline{A}},\overline{PA},\overline{PB_2}. It's easy to check that triangles \overline{A}B_1P, AC_1P are congruents, and then \overline{B_1P}=\overline{C_1P}, and by the same way we prove that PA_1 is also equal to those segments. So P is the circuncenter of A_1B_1C_1. Using the same facts one can prove that P is circuncenter of A_2B_2C_2. Since we know that P belong to the mediatrice of the sides of the triangle, we conclude that P is the concurence point. If you find any mistake, I'll be glad to be noticed =] thanks =]

I think this problem is not suitable for the 1st problem of IMO . I have a solution :(It's similar to Anto's) We'll show : triangle $ A_1B_1C_1 $ is equilateral . Put $ A_1A_2 =b ; \measuredangle C_2C_1A_1 = \measuredangle C_2A_1C_1 = \alpha ; \measuredangle B_2C_1B_1 = \measuredangle B_2B_1C_1 = \beta ; \measuredangle A_2B_1A_1 = \measuredangle A_2A_1B_1 = \gamma $ We have : $ \measuredangle A_1C_2B = 2\alpha ; \measuredangle AB_2C_1 = 2\beta ; \measuredangle B_1A_2C = 2\gamma ; $ So $ \measuredangle A_1C_1B_1 = 60 + \beta- \alpha ; \measuredangle A_1B_1C_1 = 60 + \gamma- \beta ; \measuredangle B_1A_1C_1 = 60 + \alpha- \gamma ; $ Apply Sine law in triangles $ B_1B_2C_1 , C_1C_2A_1 , A_1A_2B_1$ , we have : $ B_1C_1 = 2b\cos \beta ; B_1A_1 = 2b\cos \gamma ; A_1C_1 = 2b\cos \alpha $ Apply Sine law in triangle $ A_1B_1C_1 $ , we have : $ \frac{\cos\alpha}{\sin(60+\gamma -\beta)} = \frac{\cos\beta}{\sin(60+\alpha- \gamma)} = \frac{\cos\gamma}{\sin(60+\beta -\alpha)} (1) $ We have : $ \alpha + \frac12 \measuredangle C_2A_1B = 60 $ . So $ \alpha < 60 $ Similarly , $ \beta , \gamma < 60 $ From (1) , we get $ |\beta - \alpha| , |\alpha- \gamma| , |\gamma- \beta| < 30 $ WLOG , We can suppose $ \alpha = \max(\alpha ,\beta, \gamma) $ Case 1: $ \alpha > \beta > \gamma (2) $ This infers $ \cos \alpha < \cos \beta < \cos \gamma $ From (2) , we get $ \sin(60+\beta -\alpha) < \sin(60+\alpha- \gamma) $ Contradict with (1) Case 2: $ \alpha > \gamma > \beta (3) $ This infers $ \cos \alpha < \cos \gamma < \cos \beta $ From (3) , we get $ \sin(60+\beta -\alpha) < \sin(60+\gamma- \beta) $ Contradict with (1) Eventually , we get $ \alpha = \beta = \gamma $ So $ \triangle{AC_1B_2} = \triangle{BA_1C_2} = \triangle{CB_1A_2} $ This infers $ \triangle{AC_1B_1} = \triangle{BA_1C_1} = \triangle{CB_1A_1} $ So $ \triangle{A_1B_1C_1} $ is equilateral The rest is easy .

Quote: Show that the incircle of ABC is tangent to every vertex of the hexagon, then use Brianchon's Theorem? Yes, you are on right track! You want a hint?

Show that there is a circle is tangent to every side of the hexagon and rest it is clear........!

I found a nice solution Let I be a point such that $\vec{IA_1}=\vec{C_1C_2}$ It is easy to see that $IA_1A_2$ and $IB_2C_1$ are equilateral triangles Hence, $IA_1=IA_2=IC_1=IB_2$ $\rightarrow A_1A_2B_2C_1$ is inscribable $\rightarrow \angle A_1C_1A_2=\frac{1}{2}%Error. "ANGLE" is a bad command. A_1IA_2=30$ and $\angle C_1A_1B_2=\frac{1}{2}%Error. "ANGLE" is a bad command. C_1IB_2=30$ Similarly, $B_1B_2C_2A_1$is inscribable$\rightarrow \angle A_1B_1C_2=\angle B_1A_1B_2=30$ $C_1C_2A_2B_1$is inscribable$\rightarrow \angle C_1B_1C_2=\angle B_1C_1A_2=30$ Hence, $A_1B_2,B_1C_2,C_1A_2$ are 3 bisectors of triangle $A_1B_1C_1$ So that they are concurent

Valentin Vornicu wrote: Six points are chosen on the sides of an equilateral triangle $ABC$: $A_1$, $A_2$ on $BC$, $B_1$, $B_2$ on $CA$ and $C_1$, $C_2$ on $AB$, such that they are the vertices of a convex hexagon $A_1A_2B_1B_2C_1C_2$ with equal side lengths. Prove that the lines $A_1B_2$, $B_1C_2$ and $C_1A_2$ are concurrent. Bogdan Enescu, Romania I found this problem to be pretty interesting. Let $A_1, B_1, C_1$ to be the point closer to $B$, $C$ and $A$ respectively. Take $X$ to be a point inside the hexagon such that $\triangle XA_1 A_2$ is equilateral. We know that $XA_1 \parallel C_1 C_2$ and $XA_2 \parallel B_1 B_2$. Therefore by the condition of the problem, we get that $XA_1 C_2 C_1$ and $XB_2 B_1 A_2$ are both rhombus. This gives us $XB_2 = B_2 B_1 = C_1 B_2 = C_2 C_1 = C_1 X$, and therefore we conclude tat $\triangle XC_1 B_2$ is equilateral. Claim. $\triangle A_1 B_1 C_1$ and $\triangle A_2 B_2 C_2$ are equilateral. Proof. Let $A_1 C_2 \cap A_2 B_1 = A^*$. Since $A_1 C_2 \parallel XC_1$ and $A_2 B_1 \parallel XB_2$, we have \[ \angle(A_1 C_2, A_2 B_1) = \angle(XC_1, XB_2) = 60^{\circ}. \]Therefore we conclude that $\angle C_2 A^* B_1 = 60^{\circ} = \angle C_2 BA_2$, and therefore $C_2 BA^*A_2$ cyclic. Define $B^*, C^*$ similarly, we could get $A^* A_1 B_1C$, $CA_2 B_2 B^*, AC_1 B_1 B^*, BA_1 C_1 C^*, AB_2 C_2 C^*$ all being cyclic. This is enough to forces $\angle BA_1 C_2= \angle CB_1 A_2 = \angle AC_1 B_2$, and therefore, we conclude that $\triangle AC_1 B_2, \triangle BA_1 C_2, \triangle CB_1 A_2$ are all congruent. This is enough to forces $\triangle A_1 B_1 C_1, \triangle A_2 B_2 C_2$ are equilateral. Now, we conclude that $A_1 C_1 B_2 B_1$ is a kite, and therefore, $A_1 B_2$ bisects $B_1 C_1$. Similarly, $B_1 C_2$ bisects $C_1 A_1$ and $C_1 A_2$ bisects $A_1 B_1$. Hence, $A_1 B_2, B_1 C_2, C_1 A_2$ concur at the centroid of $\triangle A_1 B_1 C_1$.

Excellent problem! Consider what happens when we construct the point $P_A$ such that $B_2B_1A_2P_A$ is a parallelogram. We know that $\overrightarrow{A_2B_1} = \overrightarrow{P_AB_2}$. Next, observe that $\overrightarrow{A_1A_2} + \overrightarrow{B_1B_2} + \overrightarrow{C_1C_2} = 0$ since they are the same magnitude and are rotations of $120^\circ$ from each other(they form an equilateral triangle). This implies that $\overrightarrow{C_1B_2}$, $\overrightarrow{B_2P_A}$, and $\overrightarrow{A_1C_2}$ should form an equilateral triangle too, implying that $P_A$ is also the point that makes $C_1C_2A_1P_A$ a parallelogram. Now since $P_AA_2 = B_2B_1 = A_1A_2$ and so forth, we know that $\triangle A_1A_2P_A$ and $\triangle P_AB_2C_1$ are equilateral. Consequently, $P_A$ is the center of a circle $(B_2C_1A_1A_2)$, and we finish by radical axis applied to $(B_2C_1A_1A_2)$, $(A_1C_2B_1B_2)$, and $(A_2B_1C_1C_2)$.

Notice that $$\overrightarrow{A_1A_2}+\overrightarrow{A_2B_1}+\overrightarrow{B_1B_2}+\overrightarrow{B_2C_1}+\overrightarrow{C_1C_2}+\overrightarrow{C_2A_1}=0$$as these vectors form hexagon $A_1A_2B_1B_2C_1C_2$. However, $$\overrightarrow{A_1A_2}+\overrightarrow{B_1B_2}+\overrightarrow{C_1C_2}=0$$as well due to these three vectors having the same magnitude and being rotated $120$ degrees from each other. Therefore $$\overrightarrow{A_2B_1}+\overrightarrow{B_2C_1}+\overrightarrow{C_2A_1}=0$$which means that these three vectors are also rotated $120$ degrees from each other. This means that $\overrightarrow{B_1C_1}=\overrightarrow{B_1B_2}+\overrightarrow{B_2C_1}$ is rotated $120$ degrees from $\overrightarrow{A_1B_1}=\overrightarrow{A_1A_2}+\overrightarrow{A_2B_1}$. A similar thing applies to $C_1A_1$, and therefore $A_1B_1C_1$ form an equilateral triangle. Then $A_1B_2,B_1C_2,C_1A_2$ are simply the altitudes of $A_1B_1C_1$, and thus concur at the center of $A_1B_1C_1$.

Let $s$ be the side length of hexagon $A_1A_2B_1B_2C_1C_2$. Let $D, E, F$ be the points such that $DC_1C_2$, $EB_1B_2$, and $FA_1A_2$ are equilateral triangles. There are many observations to be made. Claim: $DA_2B_1$, $EC_2A_1$, and $FB_2C_1$ are equilateral triangles. Proof: I will just prove $FB_2C_1$ is equilateral as the other cases are symmetrical. Notice that $A_1F \parallel C_2C_1$ and $C_1C_2 = FA_1$, so $FA_1C_2C_1$ is a parallelogram and $A_1C_2 = FC_1 = s$. Also, $A_2F \parallel B_1B_2$ and $A_2F = B_1B_2$, so $FA_2B_1B_2$ is a parallelogram and $A_2B_1 = FB_2 = s$. Therefore $FB_2C_1$ is equilateral as desired. With this claim, notice that $D, E, F$ are the circumcenters of $(C_1C_2A_2B_1), (B_1B_2C_2A_1), (A_1A_2B_2C_1)$. The result follows from the Radical Axis Theorem on these $3$ circles. Remarks: This problem is so beautiful and uses equilateral triangles in such a nice way. It's unfortunate that this is a IMO #1 and definitely should've been at least a 2 or 5.

I think this problem is underrated... anyway : Let A2B2 and A1B1 meet at X, B1C1 and B2C2 meet at Y, A2C2 and A1C1 meet at Z. Step1 : CB1XA2 and AB2YC1 and BA1ZC2 are cyclic. ∠CA2B1 + ∠CB1A2 = 120 so ∠A2A1B1 + ∠A2B2B1 = 60 so ∠CB1X + ∠CA2X = 180 so CB1XA2 is cyclic. we prove the rest with same approach. Step2 : A2A1C1B2 and A1C2B2B1 and A2C2C1B1 are cyclic. By simple angle chasing we have XC = XA1 = XB2 so X is center of CA1B2 so ∠XA1B2 = ∠XB2A1 = 30. with proving that Y and Z are also centers we'll have plenty of 30 angles which they give us what we wanted. Now by radical axis theorem we have A1B2 and B1C2 and C1A2 are concurrent. we're Done.

Feels easy for even $P1$. Let $\angle B_2B_1C_1=\angle B_2C_1B_1=\alpha , \angle C_1B_1C_2=\theta \implies \angle C_1B_2C_2=\angle C_1C_2B_2=60-\alpha, \angle B_2C_2B_1=60-\theta$. So $\frac{\sin\theta}{\sin(60+\alpha)}=\frac{C_1C_2}{B_1C_2}=\frac{B_1B_2}{B_1C_2}=\frac{\sin(60-\theta)}{\sin(120-\alpha)} \implies \sin(\theta)=\sin(60-\theta) \implies \theta=30$. So we get $\angle B_2C_2B_1=\angle B_2A_1B_1=30 \implies B_1B_2A_1C_2$ is cyclic. Similarly $A_1A_2B_2C_1$ and $C_1C_2B_1A_2$ are cyclic. Since $A_1B_2, B_1C_2, C_1A_2$ are radical axises of these $3$ cirlces we get the result.

Hmm wait, is it just me or is this a new solution... Let the perpendicular-bisector of $\overline{A_1C_1}$ and $\overline{A_2B_2}$ meet at $T$. Then notice that $TC_1 = TA_1$, and $TA_2 = TB_2$. Moreover notice that $\triangle TA_1C_2\equiv TC_1C_2$, $\triangle TA_1A_2\equiv \triangle TC_1B_2$, $\triangle TA_2B_1\equiv \triangle TB_2B_1$, all by SSS. Hence we have that $C_2,T,B_1$ are collinear. Since $C_1A_1, B_2A_2$ both perpendicular to this line, $A_1C_1\parallel A_2B_2$. Do this similar for the other two symmetries, then finish by Desargue's Theorem on perspective triangles.

A very bold solution, not sure if the second paragraph works. If not, simple congruent triangles does the trick. $A_1A_2B_1B_2C_1C_2$ is a closed polygon, and you can translate $A_1A_2,B_1B_2,C_1C_2$ to form a closed polygon, namely an equilateral triangle. Thus, by vectors you can translate $A_1C_2,C_1B_2,B_1A_2$ to form a closed polygon. Since the lengths are equal this closed polygon is equilateral. Extend $A_1C_2,C_1B_2,B_1,A_2$ to form another equilateral triangle, and we see that this one and $ABC$ are simply reflections of each other. Note that $A_1B_2,B_1C_2,C_1A_2$ are both possible symmetry lines of the union of the two equilateral triangles, and so they must be concurrent (at the center of gravity).

Solved with pi271828. We use vectors; note that $\overrightarrow{A_1A_2}+\overrightarrow{B_1B_2}+\overrightarrow{C_1C_2}=0$ which implies $\overrightarrow{A_1C_2}+\overrightarrow{B_1A_2}+\overrightarrow{C_1B_2}=0$ meaning that $A_2B_1,B_2C_1,C_2A_1$ form $60^{\circ}$ angles with each other. Extend $A_1C_2$ and $A_2B_1$ to intersect at $A_3$ and define $B_3$ and $C_3$ similarly. Then note that $\triangle BA_1C_2\cong \triangle A_3A_1A_2\cong \triangle CB_1A_2$ so we easily obtain that $A_1B_1C_1$ and $A_2B_2C_2$ are equilateral. It follows that they share a common circumcenter, implying that all six angle bisectors concur, so we are done by Brianchon's.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.5) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -3.222082806745295, xmax = 27.526893444717444, ymin = -9.045917501236078, ymax = 12.440700090960886; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); draw((0,0)--(10,0)--(5,8.660254037844387)--cycle,zzttqq); draw((8,3.4641016151377526)--(6.25,6.495190528383291)--(3,5.196152422706633)--(1.25,2.165063509461098)--(4,0)--(7.5,0)--cycle,zzttqq); /* draw figures */ draw((0,0)--(10,0),zzttqq); draw((10,0)--(5,8.660254037844387), zzttqq); draw((5,8.660254037844387)--(0,0), zzttqq); draw((8,3.4641016151377526)--(6.25,6.495190528383291), zzttqq); draw((6.25,6.495190528383291)--(3,5.196152422706633), zzttqq); draw((3,5.196152422706633)--(1.25,2.165063509461098), zzttqq); draw((1.25,2.165063509461098)--(4,0), zzttqq); draw((4,0)--(7.5,0), zzttqq); draw((7.5,0)--(8,3.4641016151377526), zzttqq); draw((3,5.196152422706633)--(7.5,0)); draw((6.25,6.495190528383291)--(4,0)); draw((1.25,2.165063509461098)--(8,3.4641016151377526)); draw((3,5.196152422706633)--(4,0)); draw((3,5.196152422706633)--(8,3.4641016151377526)); draw((4,0)--(8,3.4641016151377526)); draw((1.25,2.165063509461098)--(7.5,0)); draw((7.5,0)--(6.25,6.495190528383291)); draw((1.25,2.165063509461098)--(6.25,6.495190528383291)); draw(circle((2,0.28867513459481353), 2.0207259421636907), linetype("4 4")); draw(circle((4.25,6.783865662978102), 2.020725942163689), linetype("4 4")); draw(circle((8.75,1.5877132402714695), 2.0207259421636907), linetype("4 4")); /* dots and labels */ dot((0,0),dotstyle); label("$B$", (0.06546804874445067,0.15699472126169542), NE * labelscalefactor); dot((10,0),dotstyle); label("$C$", (10.056748546437696,0.15699472126169542), NE * labelscalefactor); dot((5,8.660254037844387),dotstyle); label("$A$", (5.064389635185645,8.828209005738927), NE * labelscalefactor); dot((4,0),dotstyle); label("$A_{1}$", (3.5748790929377274,-0.2971831668833692), NE * labelscalefactor); dot((3,5.196152422706633),dotstyle); label("$C_{1}$", (2.7188385514159363,5.472530082973504), NE * labelscalefactor); dot((1.25,2.165063509461098),dotstyle); label("$C_{2}$", (0.886911792559303,2.37366332266462), NE * labelscalefactor); dot((8,3.4641016151377526),dotstyle); label("$B_{1}$", (8.060531530511915,3.6063617024559997), NE * labelscalefactor); dot((7.5,0),dotstyle); label("$A_{2}$", (7.187370178159687,-0.2971831668833692), NE * labelscalefactor); dot((6.25,6.495190528383291),dotstyle); label("$B_{2}$", (6.31420882580746,6.636745219443141), NE * labelscalefactor); dot((4.25,4.763139720814413),dotstyle); label("$P$", (4.328194769476903,4.393919000656048), NE * labelscalefactor); dot((7,2.5980762113533142),dotstyle); label("$Q$", (6.536779366603126,2.6304754851211576), NE * labelscalefactor); dot((3.75,1.299038105676659),dotstyle); label("$R$", (3.8659328770551364,1.3977771053297778), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ /* end of picture */ [/asy][/asy] Let $\overline{C_1B_1} \cap \overline{C_2B_2} = P$ and cyclic variants. In addition, let $\angle B_1C_1B_2 = b$. Claim: $AB_2PC_1$ is cyclic. Proof. Angle chase; we have that \[\angle AC_1B_1 = 180 - 60 - b = 120 - b \implies \angle AC_1B_2 = 120 - 2b \implies \angle C_2C_1B_2 = 60 + 2b \implies \angle C_1C_2B_2 = 60 - b;\]this coupled with $\angle C_2C_1B_1 = 60+b$ implies $\angle C_1PC_2 = 60, \angle C_1PB_2 = 120$, so $AB_2PC_1$ is cyclic as desired. $\square$ Now, $\angle C_1AP = \angle C_1B_2P = \angle C_1C_2P$, so $PC_2 = PA$; similarly, $PB_1 = PA$, so $P$ is the circumcenter of triangle $AB_1C_2$; in particular, $\angle PC_2B_1 = 30$. Now analogously, $\angle QA_1B_2 = 30$, so in fact $A_1A_2B_2C_1$ is cyclic, and analogously $B_1B_2C_2A_1$, $C_1C_2A_2B_1$ are cyclic; now radical axis finishes (or $A_1A_2B_1B_2C_1C_2$ is cyclic, but then the diagonals trivially concur since the hexagon is regular). $\blacksquare$

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(11.5cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -6.956539066076455, xmax = 5.865492584867444, ymin = -4.1805502508227566, ymax = 4.802184847325265; /* image dimensions */ pen ccqqqq = rgb(0.8,0,0); pen yqqqyq = rgb(0.5019607843137255,0,0.5019607843137255); pen qqwwzz = rgb(0,0.4,0.6); pen qqwuqq = rgb(0,0.39215686274509803,0); /* draw figures */ draw((-1.7922024232229385,0.6970303678733005)--(-1.2734696279486881,-2.236695965234121), linewidth(0.9) + ccqqqq); draw((-1.2734696279486881,-2.236695965234121)--(1.0078455066365823,-0.3205970201967967), linewidth(0.9) + yqqqyq); draw((1.0078455066365823,-0.3205970201967967)--(-1.7922024232229385,0.6970303678733005), linewidth(0.9) + yqqqyq); draw((0.04296523759040514,1.3855342963139519)--(-2.787315350990347,-0.9916461149864289), linewidth(0.9) + yqqqyq); draw((-2.787315350990347,-0.9916461149864289)--(0.6865235688648985,-2.254150798885142), linewidth(0.9) + yqqqyq); draw((0.6865235688648985,-2.254150798885142)--(0.04296523759040514,1.3855342963139519), linewidth(0.9) + ccqqqq); draw((-2.787315350990347,-0.9916461149864289)--(-1.2734696279486881,-2.236695965234121), linewidth(0.9) + yqqqyq); draw((0.6865235688648985,-2.254150798885142)--(1.0078455066365823,-0.3205970201967967), linewidth(0.9) + yqqqyq); draw((0.04296523759040514,1.3855342963139519)--(-1.7922024232229385,0.6970303678733005), linewidth(0.9) + ccqqqq); draw((0.04296523759040514,1.3855342963139519)--(-1.2734696279486881,-2.236695965234121), linewidth(0.9) + yqqqyq); draw(circle((1.4053534350599077,-1.380131563175476), 1.1316474720178729), linewidth(0.9) + qqwwzz); draw((0.4303956131066528,-0.8056028464548862)--(2.1085032670216415,-2.2668143219661436), linewidth(0.9) + yqqqyq); draw((-3.5092721327582757,-2.2167848957153953)--(-1.2734696279486881,-2.236695965234121), linewidth(0.9) + qqwuqq); draw((-1.2734696279486881,-2.236695965234121)--(0.6865235688648985,-2.254150798885142), linewidth(0.9) + ccqqqq); draw((0.6865235688648985,-2.254150798885142)--(2.1085032670216415,-2.2668143219661436), linewidth(0.9) + qqwuqq); draw((-0.6570576787984099,2.6233366001239204)--(0.04296523759040514,1.3855342963139519), linewidth(0.9) + qqwuqq); draw((0.04296523759040514,1.3855342963139519)--(1.0078455066365823,-0.3205970201967967), linewidth(0.9) + qqwuqq); draw((1.0078455066365823,-0.3205970201967967)--(2.1085032670216415,-2.2668143219661436), linewidth(0.9) + qqwuqq); draw((-0.6570576787984099,2.6233366001239204)--(-1.7922024232229385,0.6970303678733005), linewidth(0.9) + qqwuqq); draw((-1.7922024232229385,0.6970303678733005)--(-2.787315350990347,-0.9916461149864289), linewidth(0.9) + qqwuqq); draw((-2.787315350990347,-0.9916461149864289)--(-3.5092721327582757,-2.2167848957153953), linewidth(0.9) + qqwuqq); draw(circle((-0.27835670018127956,-0.5480194823743929), 1.960070917791835), linewidth(0.9) + qqwwzz); /* dots and labels */ dot((-0.6570576787984099,2.6233366001239204),dotstyle); label("$A$", (-0.7161586428033785,2.8033129937346604), NE * labelscalefactor); dot((-3.5092721327582757,-2.2167848957153953),dotstyle); label("$B$", (-3.7266546673120695,-2.4741962294649236), NE * labelscalefactor); dot((2.1085032670216415,-2.2668143219661436),dotstyle); label("$C$", (2.221207923620081,-2.4863844724746222), NE * labelscalefactor); dot((-1.7922024232229385,0.6970303678733005),dotstyle); label("$C_1$", (-2.1056183464227742,0.853194112182851), NE * labelscalefactor); dot((-1.2734696279486881,-2.236695965234121),dotstyle); label("$A_1$", (-1.3621355225562555,-2.5229492015037187), NE * labelscalefactor); dot((1.0078455066365823,-0.3205970201967967),dotstyle); label("$B_1$", (0.9780071361711481,-0.08530059956395718), NE * labelscalefactor); dot((-2.787315350990347,-0.9916461149864289),linewidth(4pt) + dotstyle); label("$C_2$", (-3.117242516601808,-0.8165951801458856), NE * labelscalefactor); dot((0.6865235688648985,-2.254150798885142),linewidth(4pt) + dotstyle); label("$A_2$", (0.6123598457449912,-2.5229492015037187), NE * labelscalefactor); dot((0.04296523759040514,1.3855342963139519),linewidth(4pt) + dotstyle); label("$B_2$", (0.06388891010575604,1.5601122067453819), NE * labelscalefactor); dot((0.4303956131066528,-0.8056028464548862),linewidth(4pt) + dotstyle); label("$T$", (0.1370183681909874,-0.8044069371361868), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] This took much longer than I would have preferred, mostly because I tried overcomplicating it. Like every other concurrency problem we use radical axes. Claim: $\angle A_1B_1C = \angle BA_2B_2$. Proof. Note that, \begin{align*} \angle BA_2B_2 &= 240 - \angle A_2B_2A\\ & = 60 + \angle A_2B_2B_1 \\ &= 60 + \angle B_2A_2B_1 \end{align*}Then $\angle B_2A_2B_1 = \angle BA_2B_2 - 60$ so we find $\angle B_1A_2A_1 = 2\angle BA_2B_2 - 60$. Clearly then $\angle A_2A_1B_1 = 120 - \angle BA_2B_2$. Hence we find, $\angle A_1B_1C = \angle BA_2B_2$. $\blacksquare$ Now letting $T = A_1B_1 \cap A_2B_2$ we find that $TA_2CB_1$ is cyclic. Claim: $\angle A_2B_2A_1 = 30$ Proof. Note that from cyclic $TA_2CB_1$ we have, \begin{align*} \angle B_1B_2A_2 &= \angle B_1A_2B_2\\ & = \angle B_1CT \end{align*}implying that $B_2T = CT$. Using a similar analysis we find that $A_1T = CT$, so we find $B_2T = A_1T$. However then it follows that $\angle A_2B_2A_1 = 30$. $\blacksquare$ Now cyclic permutations allows us to find that $\angle A_2C_1A_1 = 30$, hence we find that $A_1A_2B_2C_1$ is cyclic as desired. From here radical axes on $A_1A_2B_2C_1$, $B_1B_2C_2A_1$ and $C_1C_2A_2B_1$ finishes.

Notice that $\overrightarrow{A_1A_2} + \overrightarrow{B_1B_2} +\overrightarrow{C_1C_2} = 0$ is the sum of three unit vectors that are $120^\circ$ rotations of each other. It follows that $\overrightarrow{C_2A_1} + \overrightarrow{A_2B_1} + \overrightarrow{B_2C_1} = 0$ is the sum of three unit vectors; these three unit vectors then must form an equilateral triangle. In particular, the lines $\overline{C_2A_1}$, $\overline{C_1B_2}$, $\overline{A_2B_1}$ form an equilateral triangle $\Delta$, say $\delta_A \delta_B \delta_c$. We can show that $\triangle C_1\delta_C C_2 \cong \triangle A_1BC_2$ and cyclic permutations by AAS congruence. Then $B \to \delta_C$ and $C \to \delta_B$ under reflection about $\overline{B_1C_2}$, and it follows that $\overline{BC} \to \overline{\delta_B \delta_C}$ under this reflection. It then follows that this reflection sends $A_1C_1B_1A_2$ to $C_1C_2B_1B_2$, i.e. $\overline{B_1C_2}$ is an axis of symmetry of the hexagon. It thus passes through the center of mass $O$, and the other two lines are concurrent there similarly.

The main idea is that $\overrightarrow{A_1A_2}+\overrightarrow{A_2B_1}+\overrightarrow{B_1B_2}+\overrightarrow{B_2C_1}+\overrightarrow{C_1C_2}+\overrightarrow{C_2A_1}=0,$ but $\overrightarrow{A_1A_2}+\overrightarrow{B_1B_2}+\overrightarrow{C_1C_2}=0,$ since they are equal length and spaced at $120^\circ$ angles. Thus $\overrightarrow{A_2B_1}+\overrightarrow{B_2C_1}+\overrightarrow{C_2A_1}=0,$ and since they are also equal length they must be at $120^\circ$ angles. But now we claim that $A_1C_1\parallel A_2B_2.$ This is by writing $\overrightarrow{C_1A_1}=\overrightarrow{C_1C_2}+\overrightarrow{C_2A_1}$ and $\overrightarrow{A_2B_2}=\overrightarrow{A_2B_1}+\overrightarrow{B_1B_2},$ and then translating these four vectors to the origin and angle chasing on the circle centered at the origin they form. Thus $\triangle A_1B_1C_1$ and $\triangle B_2C_2A_2$ are homothetic, so the concurrency follows.

Consider vectors representing the sides of the triangle. Since the vectors going from $A_1$ to $A_2$ and cyclic variants are all of equal magnitude and spaced equally in orientation, they sum to zero. However, this forces the vector from $A_2$ to $B_1$ and cyclic variants all sum to zero, since they have equal magnitude this forces them to also be equally spaced. We then see that symmetry forces the vectors from $A_1$ to $B_1$ and cyclic variants to have equal magnitude and be equally spaced in orientation, so $A_1B_1C_1$ is equilateral. Consider the center of this equilateral triangle, then the translation to get from $A_1$ to $A_2$ can be interpreted at a spiral similarity at this center, which when applied to $A_1B_1C_1$ results in $A_2B_2C_2$, so this triangle is also equilateral with the same center. Lastly, the perpendicular bisector of $A_1B_1$ goes through $A_2$ and the center of both equilateral triangles, so the perpendicular bisector of $A_1B_1$ is the same as that of $B_2C_2$, so $A_1B_1$ is parallel to $B_2C_2$ symmetrically doing this forces that the triangles are homothetic, done.