For part (a), observe that $\angle C_1A_1B_1 = 90^{\circ} - (90^{\circ} - \angle B_1CA_1) = \angle C$. Similar calculations yield that $\triangle ABC \sim \triangle C_1A_1B_1 \sim \triangle B_2C_2A_2$. Now, notice that by the Pythagorean Theorem, we have \begin{align*} A_1B_2^2 &= B_1B_2^2 + A_1B_1^2 = A_1A_2^2 + A_2B_2^2 \\ B_1C_2^2 &= C_1C_2^2 + B_1C_1^2 = B_1B_2^2 + B_2C_2^2 \\ C_1A_2^2 &= A_1A_2^2 + C_1A_1^2 = C_1C_2^2 + C_2A_2^2 \end{align*} Summing, we obtain that \[ A_1B_1^2 + B_1C_1^2 + C_1A_1^2 = A_2B_2^2 + B_2C_2^2 + C_2A_2^2. \] Since $\triangle C_1A_1B_1 \sim \triangle B_2C_2A_2$, and the sums of the square of the sides are equal, it follows that the two triangles must be equal as well. [asy][asy] /* DRAGON 0.0.9.6 Homemade Script by v_Enhance. */ import olympiad; import cse5; size(8cm); real lsf=0.8000; real lisf=2011.0; defaultpen(fontsize(10pt)); /* Initialize Objects */ pair A_1 = (-2.5, -2.0); pair B_1 = (0.0, 0.0); pair C_1 = (-2.5, 1.0); pair O = circumcenter(A_1,B_1,C_1); pair A_2 = (2)*(O)-C_1; pair C_2 = (2)*(O)-B_1; pair B_2 = (2)*(O)-A_1; pair A = IntersectionPoint(Line(B_1,B_2,lisf),Line(C_1,C_2,lisf)); pair B = IntersectionPoint(Line(C_1,C_2,lisf),Line(A_1,A_2,lisf)); pair C = IntersectionPoint(Line(A_1,A_2,lisf),Line(B_1,B_2,lisf)); /* Draw objects */ draw(A_1--B_1, rgb(0.0,0.8,0.2)); draw(B_1--C_1, rgb(0.0,0.8,0.2)); draw(C_1--A_1, rgb(0.0,0.8,0.2)); draw(A--B, rgb(0.6,0.2,0.0)); draw(B--C, rgb(0.6,0.2,0.0)); draw(C--A, rgb(0.6,0.2,0.0)); draw(A_2--B_2, rgb(0.0,0.8,0.8)); draw(B_2--C_2, rgb(0.0,0.8,0.8)); draw(C_2--A_2, rgb(0.0,0.8,0.8)); /* Place dots on each point */ dot(A_1); dot(B_1); dot(C_1); dot(A_2); dot(C_2); dot(B_2); dot(A); dot(B); dot(C); /* Label points */ label("$A_1$", A_1, lsf * dir(270)); label("$B_1$", B_1, lsf * (rotate(-90) * unit(A-C))); label("$C_1$", C_1, lsf * (rotate(90) * unit(A-B))); label("$A_2$", A_2, lsf * dir(270)); label("$C_2$", C_2, lsf * (rotate(90) * unit(A-B))); label("$B_2$", B_2, lsf * (rotate(-90) * unit(A-C))); label("$A$", A, lsf * dir(111)); label("$B$", B, lsf * dir(225)); label("$C$", C, lsf * dir(315)); [/asy][/asy] For part (b), easy angle chasing gives \[ \angle B_2A_2C_2 = \angle ABA_2 + \angle BAA_2 = \angle BAC. \] Similar calculations yield that $\triangle A_1B_1C_1 \sim \triangle A_2B_2C_2 \sim \triangle ABC$. Now, let $O$ be the circumcenter of $\triangle ABC$. Then $O$ lies on the angle bisector of the angle formed by lines $B_2C_2$ and $B_1C_1$; namely, the line through $O$ perpendicular to $BC$. (Note that $\angle B_1BC = C_2CB$.) Hence, we see that $O$ has the same distance with respect to the corresponding sides of each triangle. This is enough to imply that the triangles are congruent. [asy][asy] /* DRAGON 0.0.9.6 Homemade Script by v_Enhance. */ import olympiad; import cse5; size(13.37cm); real lsf=0.8000; real lisf=2011.0; defaultpen(fontsize(10pt)); /* Initialize Objects */ pair A = (-3.5, 3.0); pair B = (-5.0, -0.5); pair C = (-0.5, -0.5); pair Y_2 = rotate((339.87)*(1), B) * A; pair Z_2 = rotate((339.87)*(1), C) * B; pair X_2 = rotate((339.87)*(1), A) * C; pair X_1 = rotate((20.13)*(1), A) * B; pair Y_1 = rotate((20.13)*(1), B) * C; pair Z_1 = rotate((20.13)*(1), C) * A; pair C_1 = IntersectionPoint(B--Y_1,C--Z_1); pair A_1 = IntersectionPoint(C--Z_1,A--X_1); pair B_1 = IntersectionPoint(B--Y_1,A--X_1); pair C_2 = IntersectionPoint(C--Z_2,A--X_2); pair A_2 = IntersectionPoint(A--X_2,B--Y_2); pair B_2 = IntersectionPoint(B--Y_2,C--Z_2); path c_1 = B_1--C_1; path a_1 = C_1--A_1; path b_1 = A_1--B_1; path c_2 = C_2--A_2; path a_2 = A_2--B_2; path b_2 = B_2--C_2; pair O = circumcenter(A,B,C); pair M_A = midpoint(B--C); pair M_B = midpoint(C--A); pair M_C = midpoint(A--B); /* Draw objects */ draw(A--B, rgb(0.6,0.2,0.0)); draw(B--C, rgb(0.6,0.2,0.0)); draw(C--A, rgb(0.6,0.2,0.0)); draw((abs(dot(unit(Y_2-B),unit(A-B))) < 1/2011) ? rightanglemark(Y_2,B,A) : anglemark(Y_2,B,A), rgb(0.0,0.4,0.0)); draw((abs(dot(unit(Z_2-C),unit(B-C))) < 1/2011) ? rightanglemark(Z_2,C,B) : anglemark(Z_2,C,B), rgb(0.0,0.4,0.0)); draw((abs(dot(unit(X_2-A),unit(C-A))) < 1/2011) ? rightanglemark(X_2,A,C) : anglemark(X_2,A,C), rgb(0.0,0.4,0.0)); draw((abs(dot(unit(B-A),unit(X_1-A))) < 1/2011) ? rightanglemark(B,A,X_1) : anglemark(B,A,X_1), rgb(0.0,0.4,0.0)); draw((abs(dot(unit(C-B),unit(Y_1-B))) < 1/2011) ? rightanglemark(C,B,Y_1) : anglemark(C,B,Y_1), rgb(0.0,0.4,0.0)); draw((abs(dot(unit(A-C),unit(Z_1-C))) < 1/2011) ? rightanglemark(A,C,Z_1) : anglemark(A,C,Z_1), rgb(0.0,0.4,0.0)); draw(c_1, rgb(0.0,0.8,0.0)); draw(a_1, rgb(0.0,0.8,0.0)); draw(b_1, rgb(0.0,0.8,0.0)); draw(c_2, rgb(0.0,0.4,0.6)); draw(a_2, rgb(0.0,0.4,0.6)); draw(b_2, rgb(0.0,0.4,0.6)); draw(A--A_1, rgb(0.5,0.5,0.5) + linewidth(1.0) + linetype("4 4")); draw(A--A_2, rgb(0.5,0.5,0.5) + linewidth(1.0) + linetype("4 4")); draw(B--B_1, rgb(0.5,0.5,0.5) + linewidth(1.0) + linetype("4 4")); draw(B--B_2, rgb(0.5,0.5,0.5) + linewidth(1.0) + linetype("4 4")); draw(C--C_1, rgb(0.5,0.5,0.5) + linewidth(1.0) + linetype("4 4")); draw(C--C_2, rgb(0.5,0.5,0.5) + linewidth(1.0) + linetype("4 4")); draw(O--M_A, rgb(0.0,0.6,0.8) + linetype("4 4")); draw(O--M_B, rgb(0.0,0.6,0.8) + linetype("4 4")); draw(O--M_C, rgb(0.0,0.6,0.8) + linetype("4 4")); /* Place dots on each point */ dot(A); dot(B); dot(C); dot(C_1); dot(A_1); dot(B_1); dot(C_2); dot(A_2); dot(B_2); dot(O); dot(M_A); dot(M_B); dot(M_C); /* Label points */ label("$A$", A, lsf * dir(105)); label("$B$", B, lsf * dir(225)); label("$C$", C, lsf * dir(315)); label("$C_1$", C_1, lsf * dir(45)); label("$A_1$", A_1, lsf * dir(135)); label("$B_1$", B_1, lsf * dir(300)); label("$C_2$", C_2, lsf * dir(45)); label("$A_2$", A_2, lsf * dir(45)); label("$B_2$", B_2, lsf * dir(135)); label("$O$", O, lsf * unit(B-O) * 2); label("$M_A$", M_A, lsf * dir(-90)); label("$M_B$", M_B, lsf * dir(45)); label("$M_C$", M_C, lsf * dir(135)); [/asy][/asy]

Refer to v_E's beautiful diagram, for I cannot hope to make any better. v_Enhance wrote: (a) Triangles $A_1B_1C_1$ and $A_2B_2C_2$ are inscribed into triangle $ABC$ so that $C_1A_1 \perp BC$, $A_1B_1 \perp CA$, $B_1C_1 \perp AB$, $B_2A_2 \perp BC$, $C_2B_2 \perp CA$, $A_2C_2 \perp AB$. Prove that these triangles are equal. (b) Points $A_1$, $B_1$, $C_1$, $A_2$, $B_2$, $C_2$ lie inside a triangle $ABC$ so that $A_1$ is on segment $AB_1$, $B_1$ is on segment $BC_1$, $C_1$ is on segment $CA_1$, $A_2$ is on segment $AC_2$, $B_2$ is on segment $BA_2$, $C_2$ is on segment $CB_2$, and the angles $BAA_1$, $CBB_2$, $ACC_1$, $CAA_2$, $ABB_2$, $BCC_2$ are equal. Prove that the triangles $A_1B_1C_1$ and $A_2B_2C_2$ are equal. Part (a) Notice that $$\angle A_1B_1C_1=90^{\circ}-\angle AB_1C_1=\angle A$$and $$\angle A_2B_2C_2=90^{\circ}-\angle CB_2A_2=\angle C$$Apply cyclic variants to obtain $\triangle ABC \sim \triangle B_1C_1A_1 \sim \triangle C_2A_2B_2$, let the similarity constant for the last pair be $k$. Now comes the sneaky observation: $B_1B_2C_1C_2, C_1C_2A_1A_2$ and $A_1A_2C_1C_2$ are cyclic quadrilaterals. However, they do not satisfy radical axis theorem, so $A_1A_2B_1B_2C_1C_2$ is a cyclic hexagon. Thus, $\triangle A_1B_1C_1$ and $\triangle A_2B_2C_2$ have equal circumradius, so $k=1$ as desired. $\blacksquare$ Part (b) Note that $$\angle B_1A_1C_1=\angle A_1AC+\angle ACC_1=\angle A$$and $$\angle B_2A_2C_2=\angle ABA_2+\angle A_2AB=\angle A$$Apply cyclic variants to obtain $\triangle ABC \sim \triangle A_1B_1C_1 \sim \triangle A_2B_2C_2$. Let $S_1$ and $S_2$ be the spiral centers for $\triangle ABC \mapsto \triangle A_1B_1C_1$ and $\triangle A_2B_2C_2 \mapsto \triangle ABC$. Let $O_1$ and $O_2$ be the respective circumcenters of $A_1B_1C_1$ and $A_2B_2C_2$. Lemma. $S_1$ and $S_2$ are the Brocard points in $\triangle ABC$. (Proof) Note that $\triangle S_1AA_1 \sim \triangle S_1BB_1 \implies S_1 \in \odot(ABB_1)$. Apply cyclic variants to conclude that $S_1$ (and likewise $S_2$) is a Brocard point. $\blacksquare$ Note that $\triangle S_1O_1O \sim \triangle S_1A_1A$ and $\triangle S_2O_2O \sim \triangle S_2A_2A$. It is well-known that $S_1, S_2$ are isogonal conjugates in $\triangle ABC$, so $\triangle S_1A_1A \sim \triangle S_2A_2A$. Thus, we conclude $\triangle S_1O_1O \sim \triangle S_2O_2O$. Again, it is well-known that $OS_1=OS_2$ hence $$\frac{S_1O_1}{S_1O}=\frac{S_2O_2}{S_2O} \implies \frac{A_1B_1}{AB}=\frac{A_2B_2}{AB},$$so $\triangle A_1B_1C_1$ and $\triangle A_2B_2C_2$ are congruent. $\blacksquare$ Remark: I felt as if I was hypnotised by EMS after staring at this configuration for long

dragon96 wrote:

Let $l_a$ - line through $C_1 \parallel BC$; $l_b$ - line through $A_1 \parallel AC$; $l_c$ - line through $B_1 \parallel AB$. $A'B'C'$ - triangle formed $l_a, l_b, l_c$. $A_2'$ - intersections $l_b, BC$; $B_2' - l_a, AC; C_2' - l_c, AB$. Easy note that $A_1A_2'B_1B_2'C_1C_2'$ is concircle and $L$, midpoint $B_1C_2'$, is its center. Then $\angle AB_1C_2' = \angle AC_1B_2' = \angle ABC => \Delta AB_1'C_2' \sim \Delta ABC => L$ lie on $A$-symmedian. Similarly $L$ lie on $B$-symmedian and $C$-symmedian. Then $L$ is Lemoine point in $ABC$ and $(A_1A_2'B_1B_2'C_1C_2')$ - second Lemoine circle!!! Then $A_2$ point coincides with $A_2'$, $B_2 - B_2'; C_2 - C_2'$. But $A_1C_1B_2A_2$ is rectangle and $A_1C_1 = A_2B_2$. Similary $A_1B_1 = B_2C_2; B_1C_1 = A_2C_2$. It means that these triangles are equal.

Part B solution

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