Problem 6？ Thanks.

Let $ \tau_d $ be the tangent of $ \Omega $ such that $ \tau_d $ $ \parallel $ $ A_2B_3 $ and $ \tau_d, $ $ \omega_1 $ are at the same side of $ A_2B_3 $ (define $ \tau_e, $ $ \tau_f $ similarly). Let $ \triangle_T, $ $ \triangle_D, $ $ \triangle_E, $ $ \triangle_F $ be the triangle formed by $(\tau_d,\tau_e,\tau_f), $ $(\tau_d,A_1X,B_1X), $ $(\tau_e,A_2X,B_2X), $ $(\tau_f,A_3X,B_3X) $, respectively. Clearly, the radius of the incircle of $ \triangle_D $ is greater than or equal to $ r_1 $, so $ \tfrac{r_1}{R} $ $ \leq $ $ k_d $ ($ k_d $ is the ratio of similitude of $ \triangle_D, $ $ \triangle_T $). Similarly, we can prove $ \tfrac{r_2}{R} $ $ \leq $ $ k_e $ and $ \tfrac{r_3}{R} $ $ \leq $ $ k_f $ ($ k_e, $ $ k_f $ is the ratio of the similitude of $ (\triangle_E,\triangle_T), $ $ (\triangle_F,\triangle_T) $, respectively), so $$ \frac{r_1}{R} + \frac{r_2}{R} + \frac{r_3}{R} \leq k_d + k_e + k_f =1 \Longrightarrow r_1+r_2+r_3 \leq R. $$____________________________________________________________ Let $ D, $ $ E, $ $ F $ be the tangency point of $ \Omega $ with $ \tau_d, $ $ \tau_e, $ $ \tau_f $, respectively. If $ R $ $ = $ $ r_1 $ $ + $ $ r_2 $ $ + $ $ r_3 $, then $ \omega_1, $ $ \omega_2, $ $ \omega_3 $ is the incircle of $ \triangle_D, $ $ \triangle_E, $ $ \triangle_F $, respectively $ \Longrightarrow $ $ D $ $ \in $ $ \omega_1, $ $ E $ $ \in $ $ \omega_2, $ $ F $ $ \in $ $ \omega_3 $. We will prove that the distance between $ X $ and the tangency point of $ \omega_i $ with $ XA_i, $ $ XB_i $ are equal. From symmetry, it suffices to prove $ XY $ $ = $ $ XZ $ where $ Y $ $ \equiv $ $ \omega_1 $ $ \cap $ $ XB_1, $ $ Z $ $ \equiv $ $ \omega_2 $ $ \cap $ $ XA_2 $. Since $ D $ is the midpoint of arc $ A_2B_3 $ in $ \Omega $, so $ Z $ lies on $ DE $. Similarly, we can prove $ Y $ $ \in $ $ DE $, so notice $ XY $ $ \parallel $ $ \tau_e, $ $ XZ $ $ \parallel $ $ \tau_d $ we conclude that $ XY $ $ = $ $ XZ $.

Sorry for posting in this thread, but does anyone have problem 6 on the RMM 2016? Thanks!

The OP has a small typo, it should be "convex hexagon", not "hexagon convex"

My solution at the actual contest: Let $T_i$ be the tangency point of $\omega_i$ and $\Omega$, let $M_1, M_2, M_3$ be the midpoints of arcs $A_2B_3, A_3B_1, A_1B_2$ of $\Omega$ respectively which contain exactly one of the points $T_1, T_2, T_3$, let $\ell_i$ be the tangent to $\Omega$ at $M_i$, let $K_1, K_2, K_3$ be the intersections of $\ell_2$ and $\ell_3$, $\ell_3$ and $\ell_1$, and $\ell_1$ and $\ell_2$, respectively, and for each $i \in \{1, 2, 3\}$ let $K_iT_i$ intersect the opposite side of triangle $K_1K_2K_3$ at $S_i$. Notice that $\ell_2$ and $\ell_3$ are parallel to $A_3B_1$ and $A_1B_2$ respectively, so the homothety with center $T_1$ with ratio $\frac{R}{r_1}$, which takes circle $\omega_1$ to circle $\Omega$, takes point $X$ to point $K_1$, since it takes tangent lines $A_1B_2$ and $A_3B_1$ to tangent lines $\ell_2$ and $\ell_3$. Therefore $\frac{T_1X}{T_1K} = \frac{r_1}{R}$, and analogous relations hold. Thus $$\frac{T_1X}{T_1K_1} + \frac{T_2X}{T_2K_2} + \frac{T_3X}{T_3K_3} = \frac{r_1 + r_2 + r_3}{R}$$ Now notice that $\frac{T_iX}{T_iK_i} \leq \frac{S_iX}{S_iK_i}$, with equality if and only if $T_i$ and $S_i$ coincide (just set $|K_iX| = x, |XT_i| = y, |T_iS_i| = z$ and see that the inequality above is equivalent to $xy \geq 0$). Since the three ratios $\frac{S_iX}{S_iK_i}$ evidently add up to $1$, this gives us that $\frac{r_1 + r_2 + r_3}{R} \leq 1$ and $r_1 + r_2 + r_3 \leq R$. Now, in order for $R = r_1 + r_2 + r_3$ to hold, the points $S_i$ and $T_i$ must coincide, and thus they must both be equal to $M_i$, since it is the only point which is both on $\Omega$ and on the side of $K_1K_2K_3$ which is opposite to $K_i$. Now let $P$ be the point where $\omega_1$ is tangent to $XA_1$. Then homothety gives $$\frac{XP}{K_1M_3} = \frac{T_1X}{T_1K_1}$$Now let $K_2K_3 = a, K_3K_1 = b, K_1K_2 = c, s = \frac{a + b + c}{2}$. Easy length calculations give us that $\frac{T_1X}{T_1K_1} = \frac{4(s-b)(s-c)}{2(ab + bc + ca) - (a^2 + b^2 + c^2)}$, so that $$XP = \frac{4(s-a)(s-b)(s-c)}{2(ab + bc + ca) - (a^2 + b^2 + c^2)}$$Since this expression is symmetric in $a, b$ and $c$, we obtain the same quantity for the distance from $X$ to the other five tangency points of the circles, so the six tangency points are concyclic with center $X$.

You don't want to see my handdrawn diagram(s) to this one. a) Let lines $\ell_1, \ell_2, \ell_3$ form a triangle with $\omega$ as its incircle and $\ell_1 \parallel A_2B_3, \ell_2 \parallel A_3B_1$, and $\ell_3 \parallel A_1B_2$. The lines $\ell_1, \ell_2, \ell_3$ are tangent to $\omega$ at $D, E, F$. Let $\Omega_1$ be the incircle of $\triangle (\ell_1, A_1B_2, A_3B_1)$, and analogously define $\Omega_2, \Omega_3$. Clearly for each $i$, the radius of $\Omega_i$, which we denote $s_i$, is at least $r_i$. with equality occurring if and only if $\Omega_i, \ell_i, \omega$ are all tangent at a single point. Note that the sum of the ratios of similarity of $\Omega_1$ to $\omega$, $\Omega_2$ to $\omega$, and $\Omega_3$ to $\omega$ is $1$ since chords concur at $X$. Hence,\[s_1 + s_2 + s_3 = R \implies r_1 + r_2 + r_3 \leq R\]as desired. $\square$ b) Equality holds when $D, E, F$ are the tangency points of the $\omega_i$ with $\omega$. So $D \in \omega_1, E \in \omega_2, F \in \omega_3$. I claim that the center is $X$. Let $\omega_3$ be tangent to $A_3B_1$ at $R$ and $\omega_2$ be tangent to $A_1B_2$ at $T$. I will prove that $XR = XT$, and the result will follow through analogous arguments. Note that $F, R, E$ are collinear by Shooting Lemma on $\omega_3$ tangent to $\omega$ and $A_3B_1$. Similarly, note that $E, T, F$ are collinear by Shooting Lemma on $\omega_2$ tangent to $\omega$ and $A_1B_2$. Hence $E, R, T, F$ are collinear. One can show through angle chasing that $EF$ is perpendicular to the angle bisector of $\angle B_2XA_3$ so thus $XRT$ is isosceles as desired. $\blacksquare$

Solution from Twitch Solves ISL: We let $M_1$ be the arc midpoint of the arc $A_2B_3$ containing $A_1B_1$, and so on. The tangents to $\Omega$ at $M_1$, $M_2$, $M_3$ determine a triangle $ABC$ with incircle $\Omega$. We also let $\omega_i$ be tangent to $\Omega$ at $T_i$, and let $U_i$ and $V_i$ be tangency points as in the figure below. [asy][asy] import graph; size(13cm); pen qqwuqq = rgb(0.,0.39215,0.); pair O = (0.,0.), X = (-0.09676,0.19152), A_1 = (-0.67013,0.74223), A_2 = (0.98123,0.19283), A_3 = (-0.6,-0.8), B_2 = (0.76864,-0.63967), B_3 = (-0.98169,0.19044), B_1 = (0.29152,0.95656), T_1 = (-0.25520,0.96688), T_2 = (0.98478,-0.17375), T_3 = (-0.97531,-0.22083), M_1 = (-0.00121,0.99999), M_2 = (0.89172,-0.45258), M_3 = (-0.69271,-0.72121), A = (0.28081,-1.65625), B = (1.62997,1.00198), C = (-2.48160,0.99697); fill((-0.93731,0.99885)--X--(0.31376,1.00038)--cycle, palegreen); fill((0.78754,-0.65783)--X--(1.21944,0.19313)--cycle, palegreen); fill((-0.60350,-0.80689)--X--(-1.64105,0.18964)--cycle, palegreen); draw(A--B--C--cycle, linewidth(0.6) + blue); draw(circle(O, 1.), linewidth(1.2)); draw(circle((-0.17976,0.68109), 0.29558), linewidth(0.6) + red); draw(circle((0.67752,-0.11954), 0.31201), linewidth(0.6) + red); draw(circle((-0.64641,-0.14636), 0.33721), linewidth(0.6) + red); draw(A--B, linewidth(0.6) + blue); draw(B--C, linewidth(0.6) + blue); draw(C--A, linewidth(0.6) + blue); draw((-0.93731,0.99885)--(0.78754,-0.65783), linewidth(0.6) + qqwuqq); draw((1.21944,0.19313)--(-1.64105,0.18964), linewidth(0.6) + qqwuqq); draw((0.31376,1.00038)--(-0.60350,-0.80689), linewidth(0.6) + qqwuqq); dot("$O$", O, dir(-90)); dot("$X$", X, dir(45)); dot("$A_1$", A_1, dir((1.312, 2.987))); dot("$A_2$", A_2, dir((1.325, 3.074))); dot("$A_3$", A_3, dir(10)); dot("$B_2$", B_2, dir(180)); dot("$B_3$", B_3, dir(135)); dot("$B_1$", B_1, dir(-90)); dot("$T_1$", T_1, dir(-T_1)); dot("$T_2$", T_2, dir(-T_2)); dot("$T_3$", T_3, dir(-T_3)); dot("$M_1$", M_1, dir(M_1)); dot("$M_2$", M_2, dir(M_2)); dot("$M_3$", M_3, dir(M_3)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$U_1$", (-0.38452,0.46790), dir(M_1-T_3)); dot("$V_1$", (0.08381,0.54731), dir(T_1-M_2)); dot("$U_2$", (0.67714,0.19246), dir(90)); dot("$V_2$", (0.46139,-0.34456), dir(225)); dot("$U_3$", (-0.34571,-0.29898), dir(M_2-T_3)); dot("$V_3$", (-0.64683,0.19085), dir(90)); [/asy][/asy] The three green segments drawn (the extension of $A_1B_2$, $A_2B_3$, $A_3B_1$) together with the sides of triangle $ABC$ determine three smaller triangles, shaded light green in the diagram above. Notice that The red circles have radius smaller than the inradius of the green triangles they are contained in The sum of the three inradii of the green triangles equals $R$, because the homothety factors between the green triangles sum to $1$. This immediately implies that $R \ge r_1 + r_2 + r_3$, so (a) is solved. Moreover, equality holds if and only if $M_i = T_i$ for each $i$. To finish (b), note that $T_3M_1$ should unconditionally pass through the tangency point of $\omega_1$ with $\overline{A_1B_2}$, etc. Hence we always have \[ M_3 M_1 \cdot M_3 U_1 = M_3 V_2 \cdot M_3 M_2. \]However, we also have \[ \frac{T_3V_3}{T_3M_1} = \frac{T_3U_3}{T_3M_2}. \]So if $M_i = T_i$ for all $i$, then by power of a point from $T_3=M_3$ we are able to conclude that $U_1$, $V_3$, $U_3$, $V_2$ are cyclic. Analogous statements imply that unless $U_1V_1$, $U_2V_2$, $U_3V_3$ are cyclic --- which they obviously are not, because they are perpendicular to angle bisectors of $\angle A_i X B_i$ --- then the six points must be cyclic, as desired.

pi37 wrote: Draw the lines $\ell_i$ tangent to $\omega_i$ and parallel to $A_{i+1}B_{i+2}$ with $\omega_i$ on the same side of $\ell_i$ as $X$. Consider incircle $\omega$ of the triangle formed by $\ell_1,\ell_2,\ell_3$, and check that its inradius $r$ satisfies $r=r_1+r_2+r_3$. Next, by applying the converse of Casey on $\omega_1,\omega_2,\omega_3,\omega$, we get that there is a circle tangent to all four, implying that $\Gamma$ is tangent to $\omega$. As $\Gamma$ intersects $\ell_i$, it contains $\omega$, so $R\ge r$. If equality holds, we need $\ell_i$ to be the common tangent of $\omega_i$ and $\Gamma=\omega$. Let $\delta_i$ be the length of the tangent from $X$ to $\omega_i$. In our computation for Casey, we check that the length of the common tangent between $\omega$ and $\omega_i$ is $|\delta_{i+1}-\delta{i+2}|$, so if these lengths are all $0$, then $\delta_1=\delta_2=\delta_2$, implying $X$ is equidistant from all $6$ points of tangency. I have a doubt in the above solution. I can't why the length of external tangent between $\omega_1,\omega$ is equal to $|\delta_2-\delta_3|$. This is equivalent to the following result Result: Let $ABC$ be a triangle with incircle $\omega$ and $X$ be a point inside it. Lines $\ell_A,\ell_B,\ell_C$ pass through $X$ and are parallel to $BC,CA,AB$, respectively. Let $\omega_A$ be incircle of triangle formed by lines $BC,\ell_B,\ell_C$. Then, $$ \delta(\omega,\omega_A) = |\delta(X,\omega_B) - \delta(X,\omega_C)| $$ Geogebra figure link: https://www.geogebra.org/geometry/jtxt75vx Can someone please tell how to prove this result.

Sketch: Let $M_{12},M_{23},M_{31}$ be relevant arc midpoints and let the tangential triangle formed be $L_1L_2L_3$. We can verify $T_1XL_1$ collinear where $T_1$ is the tangency point of $\omega_1$ and $X$ is the concurrency point. Extend $L_1T_1$ to intersect $L_2L_3$ again at $T_1'$ and note by area ratios that summing $\frac{T_1'X}{T_1'L_1}$ gives exactly $1$. This proves part (a). For part (b) note that $X$ is the Gergonne point of $\triangle L_1L_2L_3$ and we finish by ratios.

(a): Let $\ell_i$ denote the tangent line to $\Omega$ parallel to $\overline{A_iB_{i+1}}$. Let $\Delta_i$ denote the triangle induced by $\overline{XA_i}$, $\overline{XB_i}$, and $\ell_i$, and denote $R_i$ by the inradius of $\Delta_i$, and let $\Delta$ denote the triangle induced by the $\ell_i$. Clearly $r_i \le R_i$. By a well-known configuration (say, 1984 AIME #3), we see by similar triangles that $R_1+R_2+R_3=R$, so $r_1+r_2+r_3 \le R$, which completed part (a). (b): We are now required to look at the equality case for the inequality obtained in part (a). This itself sets $r_i=R_i$ for each $i$. Looking into further detail in this scenario, we figure that the midpoint of arc $A_iB_{i+1}$ that does not contain $A_{i+1}$ coincides with the point of tangency $T_i$. Let $M_i$ denote these arc midpoints. Now by Pascal on $M_3T_1M_2T_3M_1T_2$ and cyclic variants, we find that $\overline{R_iA} \parallel \overline{T_{i+1}T_{i+2}}$ for each $i$, where $R_i$ denotes the tangent to $XA_i$. A similar parallelism holds for the $S_i$, which is the tangent to $XB_i$. The desired concyclicity holds by similar triangles and PoP easily.