This solution was proposed by Yulhee Nam in Korean Team Intensive Training. Define a point $ X$ s.t. a quadrilateral $ CA_{1}XB_{3}$ be a parallelogram. Then $ BXB_{3}A_{3}$ and $ AXA_{1}B_{1}$ be a parallelogram, too. $ \Rightarrow \angle CC_{2}A_{1}=\angle CB_{1}A_{1}=\angle B_{3}AX$ $ \Rightarrow$ Lines $ AX$ and $ C_{2}A_{2}$ meet on the circumcircle of triangle $ ABC$ at $ Q$. $ \Rightarrow \angle CBQ=\angle CC_{2}Q=\angle CAQ=\angle A_{1}XQ$ $ \Rightarrow$ $ BXA_{1}Q$ is cyclic. $ \Rightarrow \angle B_{3}A_{3}C=\angle XBC=\angle XQA_{1}=\angle C_{2}CA$. Therefore, we get following amazing result : $ \angle B_{3}A_{3}C=\angle C_{2}CA$. We can easily prove that $ A_{2}B_{2}C_{2}$ and $ A_{3}B_{3}C_{3}$ are similar by this result. We can also use complex numbers. Because $ C_{2}AB_{1}\sim C_{2}BA_{1}$, $ \frac{c_{2}-b_{1}}{c_{2}-a}=\frac{c_{2}-a_{1}}{c_{2}-b}$. So we can easily express $ a_{2}, b_{2}, c_{2}$ by $ a,b,c,a_{1},b_{1},c_{1}$. Now, with some calculation, we get the result.

This one was proposed by Bodo Lass, a mathematician from Lyon in France : $ A_{2}$ is the center of the spiral similarity which maps $ BC_{1}$ on $ CB_{1}$. So we have $ \frac{A_{2}B}{A_{2}C}= \frac{BC_{1}}{CB_{1}}= \frac{AC_{3}}{AB_{3}}$, that is the triangles $ A_{2}BC$ and $ AC_{3}B_{3}$ are similar. Working with angles mod 180° we get $ (A_{2}B_{2},C_{2}B_{2}) = (A_{2}B_{2},BB_{2})+(BB_{2},C_{2}B_{2}) = (A_{2}C,BC)+(BA,C_{2}A) = (AC,C_{3}B_{3})+(A_{3}B_{3},CA) = (A_{3}B_{3},C_{3}B_{3})$, Quite Easily Done Benjamin

benjamin wrote: $ A_{2}$ is the center of the spiral similarity which maps $ BC_{1}$ on $ CB_{1}$. So we have $ \frac{A_{2}B}{A_{2}C}= \frac{BC_{1}}{CB_{1}}= \frac{AC_{3}}{AB_{3}}$, that is the triangles $ A_{2}BC$ and $ AC_{3}B_{3}$ are similar. Working with angles mod 180° we get $ (A_{2}B_{2},C_{2}B_{2}) = (A_{2}B_{2},BB_{2})+(BB_{2},C_{2}B_{2}) = (A_{2}C,BC)+(BA,C_{2}A) = (AC,C_{3}B_{3})+(A_{3}B_{3},CA) = (A_{3}B_{3},C_{3}B_{3})$, I think this is more likely what I did too, in fact a "polish" of the official solution. Little Gauss wrote: Define a point $ X$ s.t. a quadrilateral $ CA_{1}XB_{3}$ be a parallelogram. Then $ BXB_{3}A_{3}$ and $ AXA_{1}B_{1}$ be a parallelogram, too. $ \Rightarrow \angle CC_{2}A_{1}=\angle CB_{1}A_{1}=\angle B_{3}AX$ $ \Rightarrow$ Lines $ AX$ and $ C_{2}A_{2}$ meet on the circumcircle of triangle $ ABC$ at $ Q$. $ \Rightarrow \angle CBQ=\angle CC_{2}Q=\angle CAQ=\angle A_{1}XQ$ $ \Rightarrow$ $ BXA_{1}Q$ is cyclic. $ \Rightarrow \angle B_{3}A_{3}C=\angle XBC=\angle XQA_{1}=\angle C_{2}CA$. Therefore, we get following amazing result : $ \angle B_{3}A_{3}C=\angle C_{2}CA$. We can easily prove that $ A_{2}B_{2}C_{2}$ and $ A_{3}B_{3}C_{3}$ are similar by this result. And this solution is really amazing. Congratulations, Yulhee Nam!

Little Gauss wrote: This solution was proposed by Yulhee Nam in Korean Team Intensive Training. Define a point $ X$ s.t. a quadrilateral $ CA_{1}XB_{3}$ be a parallelogram. Then $ BXB_{3}A_{3}$ and $ AXA_{1}B_{1}$ be a parallelogram, too. $ \Rightarrow \angle CC_{2}A_{1}=\angle CB_{1}A_{1}=\angle B_{3}AX$ $ \Rightarrow$ Lines $ AX$ and $ C_{2}A_{2}$ meet on the circumcircle of triangle $ ABC$ at $ Q$. $ \Rightarrow \angle CBQ=\angle CC_{2}Q=\angle CAQ=\angle A_{1}XQ$ $ \Rightarrow$ $ BXA_{1}Q$ is cyclic. $ \Rightarrow \angle B_{3}A_{3}C=\angle XBC=\angle XQA_{1}=\angle C_{2}CA$. Therefore, we get following amazing result : $ \angle B_{3}A_{3}C=\angle C_{2}CA$. We can easily prove that $ A_{2}B_{2}C_{2}$ and $ A_{3}B_{3}C_{3}$ are similar by this result. I understand nothing. I don't see $ Q\in (O)$ because $ Q=AX\cap A_{2}C_{2}$ so if $ Q\in (O)$ then $ Q\equiv A_{2}?$ Or I missed some thing?

Little Gauss wrote: $ \Rightarrow$ Lines $ AX$ and $ C_{2}A_{2}$ meet on the circumcircle of triangle $ ABC$ at $ Q$. It is only a typo . It should be <<the lines $ AX$ and $ C_{2}A_{1}$ BTW , great solution Yulhee Nam

Lemma: Consider a triangle $ ABC$ and points $ A',B',C'$ on $ BC, CA, AB$. The perpendiculars from $ A'$ to $ BC$, $ B'$ to $ AC$ and $ C'$ to $ AB$ intersect in points $ A_{1}, B_{1}, C_{1}$ (correspondingly). The lines $ AA_{1},BB_{1},CC_{1}$ meet the circumcircle of $ ABC$ in $ A_{2},B_{2},C_{2}$. Then $ A_{2}B_{2}C_{2}$ is similar to $ A'B'C'$. Proof: $ m(\angle BAA_{2})=m(\angle C'AA_{1})=m(\angle C'B'A_{1})$ as $ A_{1}B'C'A$ is cyclic. Analogously $ m(\angle BCC_{2})=m(\angle A'CA_{1})=m(\angle A'B'A_{1})$ as $ A_{1}B'C'C$ is cyclic. Therefore $ m(\angle C_{2}B_{2}A_{2})=m(\angle BCC_{2})+m(\angle BAA_{2})=$ $ m(\angle C'B'A_{1})+m(\angle A'B'A_{1})=m(\angle A'B'C')$. So $ m(\angle A_{2}B_{2}C_{2})=m(\angle A'B'C')$. Together with the analogously deduced equalities this means that $ A_{2}B_{2}C_{2}$ and $ ABC$ are similar. Let's return to the problem. Let $ O$ be the circumcircle of $ ABC$. Let $ A', B', C'$ be diametrally opposite to $ A,B,C$. Also pick up $ A_{4}$ be the intersection of the perpendicular from $ B_{2}$ to $ AC$ and $ C_{2}$ to $ AB$, and analogously define $ B_{4}, C_{4}$. $ A_{2}$ is symmetric to $ A$ wrt $ OS$ where $ S$ is the midpoint of $ AA_{4}$ (the circumcenter of $ AB_{2}C_{2}$(, therefore $ A_{2}$ is the intersection of $ A'A_{4}$ with the circumcircle of $ ABC$. If we let the perpendicular from $ B_{3}$ to $ AC$ intersect the perpendicular from $ C_{3}$ to $ AB$ in $ A_{4}'$ then clearly $ A_{4}$ and $ A_{4}'$ are symmetric with respect to $ O$. Finally if we let $ A_{5},B_{5}, C_{5}$ be symmetric to $ A_{3}B_{3}C_{3}$ with respect to $ O$ then the perpendicular from $ B_{5}$ to $ A'C'$ meets the perpendicular from $ C_{5}$ to $ A'B'$ in a point which is symmetric to $ A_{4}'$ with respect to $ O$ (by symmetry), therefore in $ A_{4}$. It remains to apply the lemma for triangle $ A'B'C'$ and points $ A_{5},B_{5},C_{5}$ on its sides, as $ A_{5}B_{5}C_{5}$ is congruent to $ A_{3}B_{3}C_{3}$ by symmetry.

We have $ \left\{\begin{array}{c}\widehat{A_{2}BC_{1}}=\widehat{A_{2}CB_{1}}\\ \widehat{A_{2}C_{1}B}=180^\circ-\widehat{A_{2}A_{1}A}=180^\circ-\widehat{A_{2}B_{1}A}=\widehat{A_{2}B_{1}C}\end{array}\right\|$ $ \Rightarrow\triangle A_{2}BC_{1}\sim\triangle A_{2}CB_{1}\Rightarrow\frac{C_{1}B}{B_{1}C}=\frac{A_{2}B}{A_{2}C}\Rightarrow\frac{AC_{3}}{AB_{3}}=\frac{A_{2}B}{A_{2}C}$ $ \Rightarrow\triangle AC_{3}B_{3}\sim\triangle A_{2}BC\Rightarrow\widehat{AC_{3}B_{3}}=\widehat{A_{2}BC}$ Similarly, we also have $ \widehat{BC_{3}A_{3}}=\widehat{B_{2}AC}$ Thus, $ \widehat{A_{3}C_{3}B_{3}}=180^\circ-\widehat{AC_{3}B_{3}}-\widehat{BC_{3}A_{3}}=180^\circ-\widehat{A_{2}BC}-\widehat{B_{2}AC}$ On the other hand, \[ \begin{eqnarray*}\widehat{A_{2}C_{2}B_{2}}&=&\widehat{AC_{2}C}-\widehat{AC_{2}A_{2}}-\widehat{B_{2}C_{2}C}\\ &=&180^\circ-\widehat{ABC}-\widehat{ABA_{2}}-\widehat{B_{2}BC}\\ &=&180^\circ-\widehat{A_{2}BC}-\widehat{B_{2}AC}\] Therefore $ \widehat{A_{3}C_{3}B_{3}}=\widehat{A_{2}C_{2}B_{2}}$ And similarly for the other angles, we get $ \triangle A_{3}B_{3}C_{3}\sim\triangle A_{2}B_{2}C_{2}$ Source: http://imocompendium.com/phpBB1/viewtopic.php?t=113

lemma 1: Let $ P$ be a Miquel point of four line $ l_1,l_2.l_3,l_4$. Simson line of $ P$ wrt triangle that formed by $ l_1,l_2,l_3,l_4$(of course, there are four triangle and four Simson line are coincide) is perpendicular to the Gauss line of $ l_1.l_2,l_3,l_4$. proof) Since Aubert line is perpendicular to the Gauss line(see http://www.mathlinks.ro/Forum/viewtopic.php?t=842), it is suffice to show Simson line is parallel to Aubert line but it is true since Simson line pass midpoint of $ PH_i(i = 1,2,3,4)$ where $ H_i$ is orthocenter of four triangle that formed by four line $ l_1,l_2,l_3,l_4$. lemma 2: Let $ P,Q$ are points on circumcircle of triangle $ ABC$ and $ l_1,l_2$ be Simson line of $ P,Q$ wrt $ \triangle ABC$. Then $ \measuredangle (l_1;l_2) = - \measuredangle PAQ$(directed angle mod 180). proof) Let $ X,Y$ are on circumcircle of triangle $ ABC$ such $ PX\perp BC,QY\perp BC$. It is known that $ AX\parallel l_1,AY\parallel l_2$. Hence $ \measuredangle (l_1;l_2) = \measuredangle XAY = \measuredangle XPY = - \measuredangle PYQ = - \measuredangle PAQ$. In above problem, Let $ X,Y$ be midpoint of $ BB_1,CC_1$ and $ l_1$ be Simson line of $ A_2$ wrt $ \triangle ABC$. By lemma 1, $ l_1\perp XY$.Let $ M$ be midpoint of $ B_1C_1$. Then $ MX = \frac {1}{2}C_1B = \frac {1}{2}AC_3$ and $ MY = \frac {1}{2}AB_3$. Hence triangle $ AC_3B_3$ and $ MXY$ are similar and we obtain $ B_3C_3\parallel XY$ But $ l_1\perp XY$. So $ B_3C_3\perp l_1$. Let $ l_2$ be Simson line of $ B_2$ wrt $ \triangle ABC$. Then we also have, $ l_2\perp C_3A_3$. By lemma 2,$ \measuredangle (l_1,l_2) = - \measuredangle A_2C_2B_2$. But since $ l_1\perp B_3C_3,l_2\perp C_3A_3, \measuredangle (l_1,l_2) = \measuredangle B_3C_3A_3$. Hence $ \measuredangle A_3B_3C_3 = \measuredangle A_2B_2C_2$ and we are done.

For convenience, let's call A instead of " Yulhee Nam's solution " Can A be finished solution? I think that she's solution is just a cool Theorem, but It can't cover all of the solution, isn't it?

From here, we have two impacts: (1): Given triangle $ ABC$ with circumcenter $ O$. If $ A_1, B_1, C_1$ lie on $ BC, CA, AB$ and $ A_2$ is the reflection of $ A_1$ through midpoint of $ BC$, similar for $ B_2, C_2. M_1, M_2$ is Miquel points of triangle $ ABC$ wrt $ (A_1,B_1,C_1)$ and $ (A_2,B_2,C_2), \angle M_1A_1C = \alpha$ then $ OM_1 = OM_2$ and $ \angle M_1OM_2 = 180^o - \alpha$. (2): Given triangle $ ABC$. A circle $ (O) \cap BC = \{A_1, A_2\}, \cap AC = \{B_1,B_2\}, \cap AB = \{C_1,C_2\}$. Let $ M_1, M_2$ be Miquel points of triangle $ ABC$ wrt $ (A_1,B_1,C_1)$ and $ (A_2,B_2,C_2), \angle M_1A_1C = \alpha$ then $ OM_1 = OM_2$ and $ \angle M_1OM_2 = 180^o - \alpha$. Back to our problem: $ \angle B_2A_2C_2 = \angle B_2AC_2 = \angle B_2AB + \angle BCC_2$ $ = 180^o - \angle B_2BA - \angle BB_2A + \angle A_2M_1C_2 = C + \angle A_2M_1C_2 - \angle C_1M_1B_2$ $ = \angle C + \angle B_2M_1C_2 - \angle C_1M_1A_1 = \angle C + \angle B - 180^o + \angle B_2M_1C_2 = \angle B_2M_1C_2 - \angle A$ On the other side, $ \angle C_2A_3B_3 = \angle BM_2C - \angle A$. So we need to show that $ \angle B_2M_1C_2 = \angle BM_2C (*)$ Let $ AA_2, BB_2, CC_2$ intersect each other and make triangle $ A'B'C', M_1, M_2$ be Miquel points of triangle $ ABC$ wrt $ (A_1,B_1,C_1)$ and $ (A_3,B_3,C_3)$ Put $ \angle M_1A_1C = \alpha$ We have $ \angle M_1B_2B = \angle BC_1M_1 = \angle M_1A_1C = \angle M_1A_2C$ then $ B_2M_1C_2A'$ is cyclic quadrilateral. Similarly we get $ M_1$ is Miquel point of triangle $ A'B'C'$ wrt $ (A_2,B_2,C_2)$ From $ (1): OM_1 = OM_2$ and $ \angle M_1OM_2 = 180^o - \alpha (3)$ Let $ M'_2$ be Miquel point of triangle $ A'B'C'$ wrt $ (A,B,C)$. From $ (2)$ we obtain $ OM_1 = OM'_2$ and $ \angle M_1OM'_2 = 180^o - \angle M_1C_2C = 180^o - \alpha (4)$ From $ (3)$ and $ (4), M'_2\equiv M_2$ so $ M_2$ is Miquel point of triangle $ A'B'C'$ wrt $ (A,B,C)$ $ M_2BA'C$ is cyclic thus $ \angle BM_2C = 180^o - \angle C'A'B' = \angle B_2M_1C_2$. Therefore $ (*)$ is true. We are done!

Hello, the result is very ugly. We use complex numbers. Let lowercase letters represent the points. We prove a useful lemma. Lemma 1: If $K$ is the center of spiral similarity sending $AB \rightarrow CD$, the $k = \frac{ad-bc}{a+d-b-c}$. Proof: Note that the conditions imply that $\frac{k-a}{k-b} = \frac{k-c}{k-d}$. Now solving for $k$ gives the desired. By the spiral similarity lemma, $A_2$ sends $B_1C_1 \rightarrow CB$. So $a_2 = \frac{bb_1-cc_1}{b+b_1-c-c_1}$. $b_2, c_2$ are similar. Also, $a_3 = b+c-a_1$, and $b_3, c_3$ are similar. Now, we find the center of spiral similarity (let's name it $K$) sending $A_2A_3 \rightarrow B_2B_3$. By the Lemma, \[ k = \frac{a_2b_3-a_3b_2}{a_2-b_2-(a_3-b_3)} = \frac{\frac{bb_1-cc_1}{b+b_1-c-c_1}(a+c-b_1)-\frac{cc_1-aa_1}{c+c_1-a-a_1}(b+c-a_1)}{\frac{bb_1-cc_1}{b+b_1-c-c_1}-\frac{cc_1-aa_1}{c+c_1-a-a_1}-(b+c-a_1)+(a+c-b_1)} = \frac{\frac{(bb_1-cc_1)(a+c-b_1)(c+c_1-a-a_1)-(cc_1-aa_1)(b+c-a_1)(b+b_1-c-c_1)}{(b+b_1-c-c_1)(c+c_1-a-a_1)}}{\frac{(bb_1-cc_1)(c+c_1-a-a_1)-(cc_1-aa_1)(b+b_1-c-c_1)-(a+a_1-b-b_1)(b+b_1-c-c_1)(c+c_1-a-a_1)}{(b+b_1-c-c_1)(c+c_1-a-a_1)}} = \frac{(bb_1-cc_1)(a+c-b_1)(c+c_1-a-a_1)-(cc_1-aa_1)(b+c-a_1)(b+b_1-c-c_1)}{(bb_1-cc_1)(c+c_1-a-a_1)-(cc_1-aa_1)(b+b_1-c-c_1)-(a+a_1-b-b_1)(b+b_1-c-c_1)(c+c_1-a-a_1)}\] Now we evaluate the numerator and denominator. The numerator is equal to \[ (bb_1-cc_1)(a+c-b_1)(c+c_1-a-a_1)-(cc_1-aa_1)(b+c-a_1)(b+b_1-c-c_1) = \] $(a^2cc_1+ab^2a_1+abb^2_1+abb_1c_1+aca^2_1+aca_1b_1+aa^2_1c_1+bc^2b_1+bca_1c_1+bcc^2_1+ba_1b^2_1+cb_1c^2_1-a^2bb_1-aba^2_1-aba_1c_1-ac^2a_1-acb_1c_1-acc^2_1-aa^2_1b_1-b^2cc_1-bca_1b_1-bcb^2_1-bb^2_1c_1-ca_1c^2_1)$ The denominator is equal to \[(bb_1-cc_1)(c+c_1-a-a_1)-(cc_1-aa_1)(b+b_1-c-c_1)-(a+a_1-b-b_1)(b+b_1-c-c_1)(c+c_1-a-a_1) =\] $(bcb_1+acc_1+aba_1+aa_1b_1+bb_1c_1+ca_1c_1-bcc_1-aca_1-abb_1-ba_1b_1-cb_1c_1-aa_1c_1)-(a+a_1-b-b_1)(b+b_1-c-c_1)(c+c_1-a-a_1)$. Now since the numerator and denominator are both equal under the cyclic rotation $a \rightarrow b \rightarrow c \rightarrow a$ and $a_1 \rightarrow b_1 \rightarrow c_1 \rightarrow a_1$, this same spiral similarity sends $B_2B_3 \rightarrow C_2C_3 \rightarrow A_2A_3 \implies \triangle A_2B_2C_2 \sim \triangle A_3B_3C_3$. Note 1: This looks horrible, but it took less that 30 minutes to bash out (especially because there were no conjugations required). Note 2: In fact, this proved a stronger statement. (I never used the fact that $A_1$ lied on $BC$ completely, only for the spiral similarity part.) The stronger statement follows. Let $A_1, B_1, C_1$ be points in the plane. Let $A_2, B_2, C_2$ be the centers of spiral similarity sending $B_1C_1 \rightarrow CB$, etc. Let $A_3, B_3, C_3$ be the reflections of $A_1, B_1, C_1$ over the midpoints of $BC, AC, AB$. Show that $\triangle A_2B_2C_2 \sim \triangle A_3B_3C_3$. It would be interesting to see a synthetic proof of this. Edit: I was being stupid. It suffices to show that \[\frac{a_3-b_3}{a_3-c_3} = \frac{a_2-b_2}{a_2-c_2}\], and this is much easier to compute than what I did.

Usually, I dont post on old problems, but I feel some sort of victory. At first, I read it as $A_3$ is the symmetric of $A_2$... $3$ hours later I then see its not that... 5 minutes later I come up with this solution. Note that by length relations we spiral similarity + equal angle at $A$ we get $AC_3B_3 \sim BA_2C \implies \angle B_3C_3A = \angle A_2BC$. Similarly, one gets $\angle A_3C_3B = \angle B_2AC$. Hence, $\angle A_3C_3B_3 = 180 - (\angle B_2AC + \angle A_2BC) = \angle A_2C_2B_2$.

It is not hard to see that $\triangle{BC_1A_2} \sim \triangle{CB_1A_2}$ so we get $\frac{A_2C_1}{A_2B_1}=\frac{BC_1}{CB_1}=\frac{AC_3}{AB_3}$.We also have $\angle{C_3AB_3}=\angle{C_1A_2B_1}$.So $\triangle{AB_3C_3} \sim \triangle{A_2B_1C_1}$.Likewise we also get $\triangle{BA_3C_3} \sim \triangle{B_2A_1C_1},\triangle{CA_3B_3} \sim \triangle{C_2A_3B_3}$.So now by easy angle chasing we get that $AA_2,BB_2,CC_2$ are tangent to the circumcircles of $\triangle{AB_3C_3},\triangle{BA_3C_3}$ and $\triangle{CA_3B_3}$ respectively.Let $\angle{AB_3C_3}=x,\angle{BC_3A_3}=y,\angle{CA_3B_3}=z$.Now simple angle chasing gives $\angle{A_3}=B+y-z,\angle{B_3}=C+z-x$.Also note that $\angle{B_2A_2C_2}=\angle{AA_2C_2}-\angle{AA_2B_2}=180-z-(180-B-y)=B+y-z$.Analogously $\angle{A_2B_2C_2}=C+z-x$.So $\triangle{A_3B_3C_3} \sim \triangle{A_2B_2C_2}$ as desired.

We use complex numbers. $A_2$ is the spiral similarity center mapping $B_1C_1$ to $CB$, so \begin{align*} \frac{a_2-b_1}{a_2-c_1}&=\frac{a_2-c}{a_2-b}\\ \implies a_2&=\frac{bb_1-cc_1}{b+b_1-c-c_1} \end{align*} Using the fact that directly similar triangles have a spiral similarity center common to each pair of sides, we get that two triangles $PQR$ and $XYZ$ are directly similar iff \[ p(y-z)+q(z-x)+r(x-y)=0 \] Note that \[ b_3-c_3=(a+c-b_1)-(a+b-c_1)=c+c_1-b-b_1 \] so plugging in $PQR=A_2B_2C_2$ and $XYZ=A_3B_3C_3$ yields \[ cc_1-bb_1+aa_1-cc_1+bb_1-aa_1=0 \] giving that the triangles are indeed similar.

:O Wow pi37, I that you said you don't bash!

sayantanchakraborty wrote: So now by easy angle chasing we get that $AA_2,BB_2,CC_2$ are tangent to the circumcircles of $\triangle{AB_3C_3},\triangle{BA_3C_3}$ and $\triangle{CA_3B_3}$ respectively. can u explain it. i cant see.

Very similar to pi37's solution, but with a less elegant finish. We use complex numbers. Let $a$, $b$ and $c$ correspond to points on $A$, $B$ and $C$ respectively. By considering the spiral similarity with center $A_2$ mapping $B_1C_1$ to $CB$, we calculate that $$a_2=\frac{bb_1-cc_1}{b+b_1-c-c_1}$$By exploiting symmetry, we find that $$a_2=\frac{bb_1-cc_1}{b+b_1-c-c_1}\;\;\;\;b_2=\frac{cc_1-aa_1}{c+c_1-a-a_1}\;\;\;\;c_2=\frac{aa_1-bb_1}{a+a_1-b-b_1}$$Since $a_3$, $b_3$ and $c_3$ are reflections of $a_1$, $b_1$ and $c_1$ over the midpoints of $BC$, $AC$ and $AB$ respectively, we compute $$a_3=b+c-a_1\;\;\;\;b_3=a+c-b_1\;\;\;\;c_3=a+b-c_1\;\;\;$$Now it suffices to show that $$\frac{a_2-b_2}{a_2-c_2}=\frac{a_3-b_3}{a_3-c_3}$$Expanding the right hand side, and simplifying, we obtain $$\frac{a_2-b_2}{a_2-c_2}=\frac{\frac{aa_1(b+b_1-c-c_1)+bb_1(c+c_1-a-a_1)+cc_1(a+a_1-b-b_1)}{(b+b_1-c-c_1)(c+c_1-a-a_1)}}{\frac{aa_1(c+c_1-b-b_1)+bb_1(a+a_1-c-c_1)+cc_1(b+b_1-a-a_1)}{(b+b_1-c-c_1)(a+a_1-b-b_1)}}=\frac{b+b_1-a-a_1}{c+c_1-a-a_1}$$Expanding the left hand side, we easily obtain $$\frac{a_3-b_3}{a_3-c_3}=\frac{b+b_1-a-a_1}{c+c_1-a-a_1}$$Implying that$ \triangle{A_2B_2C_2} \sim \triangle{A_3B_3C_3}$, as desired. $\square$

We use complex numbers. Easy way to look, $A_3$ is reflection of $A_1$ over midpoint of $BC$, so $$a_3=b+c-a_1$$and symmetrically we can find $b_3, c_3$. As $A_2$ is spiral similarity center mapping $\Delta B_1C_1A_2$ to $\Delta CBA_2$, $$a_2=\frac{bb_1-cc_1}{b+b_1-c-c_1}$$and symmetrically we can find $b_2, c_2$. We now compute determinant to verify similarity. $$\left \lvert \begin{array}{ccc} a_2 & a_3 &1 \\ b_2 &b_3&1 \\ c_2 &c_3 &1 \\ \end{array} \right \rvert =\left \lvert \begin{array}{ccc} \frac{bb_1-cc_1}{b+b_1-c-c_1}&b+c-a_1&1 \\ \frac{cc_1-aa_1}{c+c_1-a-a_1}&a+c-b_1&1 \\ \frac{aa_1-bb_1}{a+a_1-b-b_1}&a+b-c_1&1 \\ \end{array} \right \rvert =-\left \lvert \begin{array}{ccc}\frac{bb_1-cc_1}{b+b_1-c-c_1}&a+a_1&1 \\ \frac{cc_1-aa_1}{c+c_1-a-a_1}&b+b_1&1 \\ \frac{aa_1-bb_1}{a+a_1-b-b_1}&c+c_1&1 \\ \end{array} \right \rvert$$This determinant is equal to $$ -\sum_{cyc} \left(\frac{bb_1-cc_1}{b+b_1-c-c_1} \right)(b+b_1-c-c_1)=-\sum_{cyc} bb_1-cc_1=0$$Hence triangles are similar, as desired. $\blacksquare$

I made funny diagram!!

Since $A_2$ is the Miquel Point of $BCB_1C_1$, we know $A_2BC_1 \sim A_2CB_1$. Hence, we know $$\frac{AB_3}{AC_3} = \frac{CB_1}{BC_1} = \frac{A_2C}{A_2B}.$$But $$\angle B_3AC_3 = \angle CAB = \angle CA_2B$$so $AB_3C_3 \sim A_2CB$ by SAS. Analogously, we conclude $BC_3A_3 \sim B_2AC$ and $CA_3B_3 \sim C_2BA$. Now, it's easy to see $$\angle C_3A_3B_3 = 180^{\circ} - \angle B_3A_3C - \angle C_3A_3B = 180^{\circ} - \angle ABC_2 - \angle ACB_2$$$$= \angle AA_2C_2 - \angle AA_2B_2 = \angle C_2A_2B_2.$$By symmetry, we conclude $A_2B_2C_2 \sim A_3B_3C_3$ as desired. $\blacksquare$ Random Properties: Let $AK$ meet $(ABC)$ again at $K_a$. Then, $A_2KK_a \sim A_2C_1B \sim A_2B_1C$. Let $l_a$ denote the tangent to $(AB_1C_1)$ at $A$. If $l_a$ meets $(ABC)$ again at $A_4$, then $A_2AA_4 \sim A_2KK_a \sim A_2C_1B \sim A_2B_1C$. The main idea behind this problem is realizing $A_3, B_3, C_3$ are only attached to length conditions. This motivates us to compute $\frac{AB_3}{AC_3} = \frac{CB_1}{BC_1}$ in order to take advantage of symmetry. But the second ratio is connected to $A_2BC_1 \sim A_2CB_1$, and the rest follows easily.

First G9!yee ,literally took me no hints if I didn't stupidly rolled over the pages.

Claim: $\triangle AB_3C_3\stackrel{+}{\sim} \triangle A_2B_1C_1.$ We use composition of symmetry wrt midpoint of $AC,$ of spiral similarity $B_1C\mapsto C_1B$ and of symmetry wrt midpoint of $AB.$ Clearly it's a spiral similarity $AB_3\mapsto AC_3,$ and it has same rotation and ratio as $A_2B_1C\mapsto A_2C_1B,$ thus the conclusion $\Box$ Using claim and analogous assertions, we obtain $$\measuredangle A_2B_2C_2=\measuredangle A_2BC_2=\measuredangle A_2AC_1+\measuredangle A_1CC_2=\measuredangle A_2B_1C_1+\measuredangle A_1B_1C_2=\measuredangle AB_3C_3+\measuredangle A_3B_3C=\measuredangle A_3B_3C_3.$$This and analogous equalities yield the desired result.

6.18 is weird. spent about an hour trying to bash out a2 using cyclic cond and just found 6.18 and facepalmed so hard i fell out of my chair. incidentally 618 is an important number...

$\square$

All right, I see so many complex bash solutions on this, but where is @hukilau17? Someone please call bro up to do the heavenly task.

I don't know why you want to see this, but here you go.

The main claim is that $\triangle AB_3C_3 \sim \triangle A_2B_1C_1$; this is by SAS as $\angle BAC = \angle C_1A_2B_1$ and $$\frac{AB_2}{AC_3} = \frac{CB_1}{BC_1} = \frac{A_2C_1}{A_2B_1}$$by spiral similarity. Then the rest is a quick angle chase to find $$\angle B_3C_3C_3 = 180^\circ - \angle B_2CA - \angle C_2BA = 180^\circ - \frac 12 \widehat{AB_2} - \frac 12 \widehat{AC_2} = \frac 12 \widehat{B_2C_2} = \angle B_2A_2C_2,$$as requested.

We focus on $A_2$ and A, because the other cyclic variants follow. Note that there is a spiral similarity at $A_2$ mapping $B_1C$ to $BC_1$. Then $$\frac{A_2C}{AB_3}=\frac{A_2B}{AC_3}\iff \triangle C_3AB_3\sim\triangle BA_2C.$$Analogously, $\triangle A_3BC_3\sim\triangle CB_2A$, $\triangle B_3CA_3\sim\triangle AC_2B$, we get $$\angle A_3B_3C_3=180-C_3B_3A-A_3B_3C=180-BCA_2-BAC_2=\angle BAA_2-\angle BAC_2=\angle A_2AC_2=\angle A_2B_2C_2,$$and analogous results finishes it. $\blacksquare$ great problem, misplaced for g9 cuz imo that g4 this year was BRUTAL compared to g1-3 which were decently trivial (seems too complicated so im not doing it )

That was the most insane spiral problem I've ever done. (*cough* probably because I haven't done that many) Lemma: $\triangle AC_3B_3 \sim \triangle A_2BC$, and likewise for its cyclic variants. This is the critical claim. Proof: $A_2$ is the spiral center from $C_1B_1 \mapsto BC$, and $C_1B \mapsto B_1C$. We can also easily see that $\angle BA_2C = \angle BAC$. Then, $$ \frac{A_2B}{A_2C} = \frac{\frac{A_2C}{BC_2} \cdot BC_1}{\frac{A_2C}{CB_1} \cdot CB_1} = \frac{BC_1}{CB_1} = \frac{AC_3}{AB_3} $$By spiral similarity. We can also repeat this same process for the cyclic variants. By SAS, Lemma is proved. $\Box$ We finish with an angle chase, using Lemma to help us with the angles: \begin{align*} \angle B_2A_2C_2 &= \angle B_2A_2A - \angle AA_2C_2 \\ &= (180^\circ - \angle B_2CA) - \angle ABC_2 \\ &= 180^\circ - (\angle B_2CA + \angle ABC_2) \\ &= 180^\circ - (\angle BA_3C_3 + \angle B_3A_3C) \\ &= 180^\circ - (180^\circ - \angle C_3A_3B_3) \\ \angle B_2A_2C_2 &= \angle C_3A_3B_3 \end{align*}We can repeat the same process for the cyclic variants of $\angle B_2A_2C_2$, and we are done. $\blacksquare$

We focus on A_2 and A, because the other cyclic variants follow. Note that there is a spiral similarity at A_2 mapping B_1C to BC_1. Then \frac{A_2C}{AB_3}=\frac{A_2B}{AC_3}\iff \triangle C_3AB_3\sim\triangle BA_2C. Analogously, \triangle A_3BC_3\sim\triangle CB_2A,\triangle B_3CA_3\sim\triangle AC_2B, we get \angle A_3B_3C_3=180-C_3B_3A-A_3B_3C=180-BCA_2-BAC_2=\angle BAA_2-\angle BAC_2=\angle A_2AC_2=\angle A_2B_2C_2, and analogous results finishes it.

Imagine not complecks bash this tho. Sid03 sar motivated me not to :prayge: . (Bro headsolved the problem ) [asy][asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ /* A few re-additions are done using bubu-asy.py. This adds the dps, xmin, linewidth, fontsize and directions. */ pair A = (-91.53138,70.81505); pair B = (-110.35458,-31.32483); pair C = (7.06804,-31.32483); pair A_1 = (-79.20200,-31.32483); pair B_1 = (-54.54733,32.50300); pair C_1 = (-99.88684,25.47601); pair A_3 = (-24.08453,-31.32483); pair B_3 = (-29.91600,6.98722); pair C_3 = (-101.99912,14.01421); pair A_2 = (-102.39840,61.97874); pair B_2 = (-123.65121,5.70195); pair C_2 = (-31.45935,-58.63911); import graph; size(12cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); real xmin = -5, xmax = 5, ymin = -5, ymax = 5; draw(A_2--B_2--C_2--cycle, linewidth(0.5) + blue); draw(A_3--B_3--C_3--cycle, linewidth(0.5) + blue); draw(A--B, linewidth(0.5)); draw(B--C, linewidth(0.5)); draw(C--A, linewidth(0.5)); draw(circle((-51.64327,10.65974), 72.17841), linewidth(0.5)); draw(circle((-79.72959,45.20069), 28.20244), linewidth(0.5)); draw(circle((-94.77829,-4.83039), 30.73395), linewidth(0.5)); draw(circle((-36.06698,-11.31095), 47.55192), linewidth(0.5)); draw(A_2--B_2, linewidth(0.5) + blue); draw(B_2--C_2, linewidth(0.5) + blue); draw(C_2--A_2, linewidth(0.5) + blue); draw(A_3--B_3, linewidth(0.5) + blue); draw(B_3--C_3, linewidth(0.5) + blue); draw(C_3--A_3, linewidth(0.5) + blue); dot("$A$", A, NW); dot("$B$", B, SW); dot("$C$", C, SE); dot("$A_1$", A_1, S); dot("$B_1$", B_1, NE); dot("$C_1$", C_1, W); dot("$A_3$", A_3, SE); dot("$B_3$", B_3, NE); dot("$C_3$", C_3, NW); dot("$A_2$", A_2, NW); dot("$B_2$", B_2, W); dot("$C_2$", C_2, SE); [/asy][/asy] First note that clearly $A_2$ is the center of spiral similarity mapping $\overline{C_1B_1}\mapsto \overline{BC}$ and thus also the center of spiral similarity mapping $\overline{C_1B}\mapsto \overline{B_1C}$. Now note that $\measuredangle C_3AB_3=\measuredangle BAC=\measuredangle BA_2C$. Also note that $\dfrac{C_3A}{B_3A}=\dfrac{BC_1}{CB_1}\overset{\text{Spiral Sim}}{=}\dfrac{BA_2}{CA_2}$ which gives us that $\triangle AC_3B_3 \overset{+}{\sim}\triangle A_2BC$. This further gives us that $\measuredangle AC_3B_3=\measuredangle A_2BC$. In a similar logic, we also get that $\measuredangle BC_3A_3 = \measuredangle B_2AC$. Now to finish, note that, \begin{align*} \measuredangle B_3C_3A_3 &= \measuredangle B_3C_3A + \measuredangle BC_3A_3\\ &= \measuredangle CBA_2 + \measuredangle B_2AC\\ &= \measuredangle (CBA + \measuredangle ABA_2) + (\measuredangle B_2AB + \measuredangle BAC)\\ &= \measuredangle (CBA + \measuredangle BAC) + (\measuredangle ABA_2 + \measuredangle B_2AB)\\ &= \measuredangle BCA + (\measuredangle ACA_2 + \measuredangle B_2CB)\\ &= \measuredangle BCA - (\measuredangle A_2CA + \measuredangle BCB_2)\\ &= \measuredangle B_2CA_2\\ &= \measuredangle B_2C_2A_2 .\end{align*} This finally gives us $\measuredangle B_3C_3A_3 = \measuredangle B_2C_2A_2$ and similarly we also get that $\measuredangle C_3A_3B_3 = \measuredangle C_2A_2B_2$. Combining these two, we get $\triangle A_3B_3C_3 \overset{+}{\sim}\triangle A_2B_2C_2$ and we are done.

Let the circumcircle of $ABC$ be the complex unit circle. Then if we let the lowercase letters represent their respective points translated onto the complex plane, we obtain that as $A_2, B_2, C_2$ are the second intersection points of circles $(AB_1C_1)$ and $(ABC)$ that \[a_2 = \frac{bb_1 - cc_1}{b + b_1 - c - c_1}; \phantom{e} b_2 = \frac{cc_1 - aa_1}{c + c_1 - a - a_1}; \phantom{e} c_2 = \frac{aa_1 - bb_1}{a + a_1 - b - b_1},\]and also \[a_3 = b + c - a_1; \phantom{e} b_3 = c + a - b_1; \phantom{e} c_3 = a + b - c_1.\]Now in order to show $\triangle A_2B_2C_2 \sim \triangle A_3B_3C_3$, it suffices to equivalently show $\frac{a_2 - b_2}{a_2 - c_2} = \frac{a_3 - b_3}{a_3 - c_3}$: \[\frac{a_2 - b_2}{a_2 - c_2} = \frac{\frac{bb_1 - cc_1}{b + b_1 - c - c_1} - \frac{cc_1 - aa_1}{c + c_1 - a - a_1}}{\frac{bb_1 - cc_1}{b + b_1 - c - c_1} - \frac{aa_1 - bb_1}{a + a_1 - b - b_1}} = \frac{\frac{aa_1(b + b_1 - c - c_1) + bb_1(c + c_1 - a - a_1) + cc_1(a + a_1 - b - b_1)}{(c + c_1 - a - a_1)}}{\frac{aa_1(c + c_1 - b - b_1) + bb_1(a + a_1 - c - c_1) + cc_1(b + b_1 - a - a_1)}{(a + a_1 - b - b_1)}} = \frac{b + b_1 - a - a_1}{c + c_1 - a - a_1}, \]which is also precisely $(a_3 - b_3)/(a_3 - c_3)$, as was needed to show. $\blacksquare$

We claim that $\triangle C_2AB \sim \triangle C_2A_1B_1 \sim \triangle CA_3B_3$. The first two just comes from spiral similarity, and we have the third from $\angle A_3CB_3 = \angle ACB$ and the similarity ratio \[\frac{CA_3}{CB_3} = \frac{BA_1}{AB_1} = \frac{C_2A}{C_2B}.\] Thus we get our desired similarity from \begin{align*} \measuredangle A_3C_3B_3 &= \measuredangle A_3C_3A + \measuredangle AC_3B_3 \\ &= \measuredangle A_1C_1B_2 + \measuredangle A_2C_1B_1 \\ &= \measuredangle A_1BB_2 + \measuredangle A_2AB_1 \\ &= \measuredangle CC_2B_2 + \measuredangle A_2C_2C \\ &= \measuredangle A_2C_2B_2. \quad \blacksquare \end{align*}

The key claim is the following. Claim: $$\triangle AC_3B_3\sim\triangle A_2BC$$and symmetric variants. Note that clearly $\angle BA_2C=\angle C_3AB_3$ as $ABCA_2$ is cyclic. Furthermore, since $A_2$ is the center of spiral similarity taking $BC_1$ to $CB_1$, we have $$\frac{A_2B}{A_2C}=\frac{BC_1}{CB_1}=\frac{AC_3}{AB_3},$$as desired. Thihs means that $\angle BA_3C_3=\angle B_2CA$ and $\angle CA_3B_3=\angle C_2BA$. Note that the former angle is inscribed in arc $B_2A$, and the latter in arc $C_2A$. Thus, $\angle C_3A_3B_3=180-\angle B_2CA-\angle C_2BA$ is inscribed in arc $B_2C_2$, and thus $\angle C_3A_3B_3=\angle C_2A_2B_2$, and symmetric variants, as desired. remark: this problem illustrates several things very well. First of all, the main idea here is really the usage of a spiral similarity to "move" length ratios. The spiral similarity allows us to move the ratio $\frac{A_2B}{A_2C}$ "onto the triangle". And "length flipping" shows up once again. After moving the ratio with a spiral similarity, it was then "flipped" using the isotomic condition in order to include vertex $A$, which is good since at this point we are only analyzing the $A$-centered version. This length ratio flipping directly induces the similar triangles there led to this solution. remark 2: oh wait, complex bash also works :flushed: as $A_2$ and $A_3$ can both be computed relatively easily in terms of $A_1,B_1,C_1$ (the former using spiral center and the latter just by reflecting).

Note that $A_2$ is the center of the spiral similarity taking $B_1C_1$ to $CB$. Therefore, $\angle BA_2C=\angle C_3AB_3$ and \[\frac{A_2C}{A_2B}=\frac{B_1C}{C_1B}=\frac{AB_3}{AC_3}\]which implies $\triangle BA_2C\sim \triangle C_3AB_3$. Thus, $\angle C_3B_3A=\angle BCA_2=\angle BC_2A_2$. Similarly, $\angle A_3B_3C=\angle BA_2C_2$ and so $\angle A_3B_3C_3=\angle A_2BC_2=\angle A_2B_2C_2$. Considering analogous finishes.