I have a solution using $ \sin, \cos$ calculation, but it is ugly. I proved that $ p_{a}\parallel BC$ by $ \sin, \cos$ calculation. and the answer is $ \frac{1}{4}$.

lemma. let $ E,F$ be the points on side of $ AB,AC$ of triangle $ ABC$ respectively that $ EFCB$ be cyclic and $ O$ be cirumcenter of $ ABC$ . prove that $ AO$ is perpendicular to $ EF$. proof. we know that $ AEF \sim ABC$ so if we reflect triangle $ ABC$ with respect to angle bisector of $ A$ there is homothecy that brings the reflected triangle to $ AEF$ also we know that angle bisector of $ A$ is also angle bisector of $ AO$ and $ A$-altitude hence $ AO\perp EF$ back to the problem. let $ T_{c}T_{b}$ intersect with $ AB,AC$ at $ E,F$ in order and $ P,Q$ be tangency point of $ p_{a}$ with $ \omega_{c}$ and $ \omega_{b}$ respectively.and let $ T$ be the midpoint of arc that created by $ p_{a}$ and not containing $ A$. we have $ \angle AEF=\frac{\angle B+\angle C}{2}\ ,\angle AFE=\frac{\angle B+\angle C}{2}$ hence $ AE=AF$ so angle bisector of $ \angle A$ is perpendicular to $ T_{c}T_{b}$ we know that $ T_{c},P,T$ and $ T_{b},Q,T$ are collinear and also $ TP\times TT_{c}=TQ\times TT_{b}$ so quarilateral $ PQT_{b}T_{c}$ is cyclic and cause $ TO\perp p_{a}$ hence according to the lemma $ TA$ is perpendicular to $ T_{c}T_{b}$ hence $ AT$ is angle bisector of $ \angle BAC$ hence $ T\equiv T_{a}$ hence $ p_{a}\parallel BC$. like this $ p_{b}\parallel AC\ ,\ p_{c}\parallel AB$. I didn't think on second part yet.4 days later I will send the rest of proof.

Little Gaus please can you post a short idea to your solution.

The following fact is the heart of my solution (with changed notations): Theorem 1. Let ABC be a triangle, and let A', B', C' be the midpoints of its sides BC, CA, AB. Let X, Y, Z be the midpoints of the arcs BC, CA, AB (not containing the points A, B, C, respectively) on the circumcircle of triangle ABC. Let x, y, z be the circles with diameters A'X, B'Y, C'Z. Let I be the incenter of triangle ABC. Then, the image of the line C'A' under the homothety $ \left(I,\frac12\right)$ is the external common tangent of the circles z and x such that y lies on the opposite side of this tangent than z and x do. Here, the "homothety $ \left(P,q\right)$" means the homothety with center P and ratio q. Proof of Theorem 1. Let $ I_{a}$ be the A-excenter of triangle ABC, and let $ H_{a}$ be the orthocenter of triangle $ I_{a}BC$. Let x' be the circle with diameter $ H_{a}I_{a}$. Similarly define two circles y' and z'. We have $ BH_{a}\perp CI_{a}$ (since $ H_{a}$ is the orthocenter of triangle $ I_{a}BC$) and $ CI\perp CI_{a}$ (since the interior and the exterior angle bisector of an angle are always perpendicular). Thus, $ BH_{a}\parallel CI$. Similarly, $ CH_{a}\parallel BI$. Thus, the quadrilateral $ BH_{a}CI$ is a parallelogram. Therefore, its diagonals BC and $ IH_{a}$ bisect each other. This means that the midpoint A' of BC is also the midpoint of $ IH_{a}$. In other words, the point A' is the image of the point $ H_{a}$ under the homothety $ \left(I,\frac12\right)$. We have $ CI\perp CI_{a}$, so that $ \measuredangle ICI_{a}=90^{\circ}$. Similarly, $ \measuredangle IBI_{a}=90^{\circ}$. Thus, the points C and B lie on the circle with diameter $ II_{a}$. In other words, the circle with diameter $ II_{a}$ is the circumcircle of triangle BIC. Thus, the midpoint of the segment $ II_{a}$ (being the center of the circle with diameter $ II_{a}$) is the circumcenter of triangle BIC. But the circumcenter of triangle BIC is the point X (in fact, it is well-known that the circumcenter of triangle BIC is the midpoint of the arc BC (not containing A) of the circumcircle of triangle ABC - but this midpoint is X). Hence, the point X is the midpoint of the segment $ II_{a}$. In other words, the point X is the image of the point $ I_{a}$ under the homothety $ \left(I,\frac12\right)$. Since the points A' and X are the images of the points $ H_{a}$ and $ I_{a}$ under the homothety $ \left(I,\frac12\right)$, we can conclude that the circle with diameter A'X is the image of the circle with diameter $ H_{a}I_{a}$ under the homothety $ \left(I,\frac12\right)$. In other words, the circle x is the image of the circle x' under the homothety $ \left(I,\frac12\right)$. Similarly, the circles y and z are the images of the circles y' and z' under this homothety. According to part (c) of http://www.mathlinks.ro/Forum/viewtopic.php?t=157492 , the line C'A' touches the circle with diameter $ H_{a}I_{a}$. In other words, the line C'A' touches the circle x'. Similarly, the line C'A' touches the circle z'. Hence, the line C'A' is a common tangent of the circles z' and x'. Thus, the image of the line C'A' under the homothety $ \left(I,\frac12\right)$ is a common tangent of the images of the circles z' and x' under this homothety. But the images of the circles z' and x' under this homothety are the circles z and x. Hence, it follows that the image of the line C'A' under the homothety $ \left(I,\frac12\right)$ is a common tangent of the circles z and x. Adding in some arrangement considerations which you are hardly interested in, we can extend this to: The image of the line C'A' under the homothety $ \left(I,\frac12\right)$ is an external common tangent of the circles z and x, namely the one satisfying the property that y lies on the opposite side of this tangent than z and x do. This proves Theorem 1. Now, the following result obviously solves the problem and even strengthens it: Theorem 2. Let ABC be a triangle, and let A', B', C' be the midpoints of its sides BC, CA, AB. Let X, Y, Z be the midpoints of the arcs BC, CA, AB (not containing the points A, B, C, respectively) on the circumcircle of triangle ABC. Let x, y, z be the circles with diameters A'X, B'Y, C'Z. Let $ p_{b}$ be the external common tangent of the circles z and x such that y lies on the opposite side of this tangent than z and x do. Similarly we define two lines $ p_{c}$ and $ p_{a}$. Let $ P$ be the triangle formed by the lines $ p_{a}$, $ p_{b}$, $ p_{c}$. Then: - This triangle $ P$ is the image of the triangle A'B'C' under the homothety $ \left(I,\frac12\right)$. - This triangle $ P$ is homothetic to triangle ABC with homothetic factor $ -\frac14$. Proof of Theorem 2. From Theorem 1, we know that the image of the line C'A' under the homothety $ \left(I,\frac12\right)$ is the external common tangent of the circles z and x such that y lies on the opposite side of this tangent than z and x do. In other words, the image of the line C'A' under the homothety $ \left(I,\frac12\right)$ is the line $ p_{b}$. Similarly, the images of the lines A'B' and B'C' under this homothety are the lines $ p_{c}$ and $ p_{a}$. Thus, the triangle $ P$, being defined as the triangle formed by the lines $ p_{a}$, $ p_{b}$, $ p_{c}$, must be the triangle formed by the images of the lines B'C', C'A', A'B' under the homothety $ \left(I,\frac12\right)$. Now, since the triangle formed by the lines B'C', C'A', A'B' is the triangle A'B'C', the triangle formed by the images of these lines under the homothety $ \left(I,\frac12\right)$ is the image of the triangle A'B'C' under the homothety $ \left(I,\frac12\right)$. In other words, the triangle $ P$ is the image of the triangle A'B'C' under the homothety $ \left(I,\frac12\right)$. In order to prove Theorem 2, it thus only remains to show that this triangle $ P$ is homothetic to triangle ABC with homothetic factor $ -\frac14$. But this is trivial: This triangle $ P$ is homothetic to triangle A'B'C' with homothetic factor $ \frac12$ (because it is the image of triangle A'B'C' under the homothety $ \left(I,\frac12\right)$), while the triangle A'B'C' is homothetic to triangle ABC with homothetic factor $ -\frac12$ (in fact, triangle A'B'C' is the medial triangle of triangle ABC). Hence, the triangle $ P$ is homothetic to triangle ABC with homothetic factor $ \frac12\cdot\left(-\frac12\right)=-\frac14$. This completes the proof of Theorem 2. Darij

Your Solution is very beautiful!

Construct points $X\in\omega_c$ and $Y\in\omega_b$ such that $X,Y$ lie in between $O_bO_c,BC$ and $O_cX\| O_bY\perp BC$. By simple trigonometric calculations, $r_b=R\sin^2\beta$, etc., so \[d(X,BC)=d(Y,BC)=2R(\cos^2\beta\cos^2\gamma - \sin^2\alpha),\]where $\alpha=A/2$, etc., so $XY\| BC$, etc. Now it's easy to check that \[\frac{d(A,XY)}{d(A,BC)} = \frac{2+\tan\beta\tan\gamma}{4}=\frac{1}{2}+\frac{r^2}{4vw}=\frac{1}{2}+\frac{u}{4(u+v+w)}=\frac{3u+2v+2w}{4(u+v+w)},\]where $u,v,w$ are the lengths of the tangents from $A,B,C$ to the incircle. By a simple PIE computation the desired ratio is \[\frac{1}{2}+\frac{u}{4(u+v+w)}-\frac{u+v}{4(u+v+w)}-\frac{u+w}{4(u+v+w)}=\frac{1}{4}.\]

Let line thru O_b and || to BT_B intersect P_C at point X Easy to see that P_C || BA and angle O_BQX = 90 Let make simmetry of point O_B wrt point X and get point Y Easy to see that Y is on BT_B Let make simmetry of point M_B wrt Q and get point Z Easy to see that YZ || XQ || BA Let line T_CT_B intersect AC at point F Easy to see that M_BT_B = ZT_B and FM_B = FZ and FZ || BA , so F is on YZ Angle FYT_B = ABT_B = ACT_B , so T_BFYC is cyclic after Reim's theorem we get that Y is on T_CC , so Y is incenter of ABC And let make Homotety of triangle N_AM_BM_C wrt point Y and k = -1/2 and we get our triangle . done

darij grinberg wrote: ... Thus, $ BH_{a}\parallel CI$. Similarly, $\boxed{\boxed{CH_{a}\parallel BI}}$. Thus, the quadrilateral $ BH_{a}CI$ is a parallelogram. Therefore, its diagonals BC and $ IH_{a}$ bisect each other... Sorry to bring you back Darij, but I don't understand the boxed part. How did you get $CH_a \parallel CI$ ? I think it is $CH_a \perp CI$, and in that case $BH_a CI$ is not a parallelogram. I have attached a picture to make it easier.

pohoatza wrote: In a triangle $ ABC$, let $ M_{a}$, $ M_{b}$, $ M_{c}$ be the midpoints of the sides $ BC$, $ CA$, $ AB$, respectively, and $ T_{a}$, $ T_{b}$, $ T_{c}$ be the midpoints of the arcs $ BC$, $ CA$, $ AB$ of the circumcircle of $ ABC$, not containing the vertices $ A$, $ B$, $ C$, respectively. For $ i \in \left\{a, b, c\right\}$, let $ w_{i}$ be the circle with $ M_{i}T_{i}$ as diameter. Let $ p_{i}$ be the common external common tangent to the circles $ w_{j}$ and $ w_{k}$ (for all $ \left\{i, j, k\right\}= \left\{a, b, c\right\}$) such that $ w_{i}$ lies on the opposite side of $ p_{i}$ than $ w_{j}$ and $ w_{k}$ do. Prove that the lines $ p_{a}$, $ p_{b}$, $ p_{c}$ form a triangle similar to $ ABC$ and find the ratio of similitude. Proposed by Tomas Jurik, Slovakia Let $(M_AT_A) \cap \{T_AT_C, T_AT_B\}=\{X_A,Y_A\},$ and define $\{X_B,Y_B\},\{X_C,Y_C\}$ analogously. Then the first key claim is the following: Claim 1: We have $p_a \equiv X_BY_C, p_b \equiv X_CY_A$ and $p_c \equiv X_AY_B.$ Further, $p_a, p_b,p_c$ are respectievly parallel to $BC,CA,AB.$ Proof: Let $Y_CM_C \cap X_BM_B=T.$ Then $\angle TM_CA=\angle Y_CMB=\angle Y_CT_CM_B=B/2$ and $M_CT$ bisects $\angle AM_CM_B.$ Similarly $M_BT$ bisects its angle and so $T$ is the incenter of $AM_BM_C.$ Hence $AT$ passes through $T_A.$ Now, $TY_CT_AX_B$ is cyclic because of the opposite $90^\circ$ angles. Hence, $$\angle B/2=\angle TT_AX_B=\angle TY_CX_B,$$which gives $Y_CX_B$ is tangent to $(M_CT_C)$ and parallel to $BC,$ as desired. $\square$ Let $p_a,p_b,p_c$ form the triangle $DEF.$ Then the above claim implies $\triangle DEF \sim \triangle ABC$ (negatively) as desired. Now, to find the ratio, we have another claim: Claim 2: Let $I$ be the incenter of $\triangle ABC.$ Then $I$ is the center of homothety taking $DEF$ to $M_AM_BM_C$ and the ratio of homothety is $2.$ Proof: Let $\{DX_A, DY_A\} \cap BC=\{U,V\}.$ Then $DUV \sim ABC$ positively, and $(M_AT_A)$ is the corresponding excircle of $DUV.$ Thus $DM_A$ is parallel to the line joining $A$ to the $A-$excircle touch point (on $BC),$ which is well known to be parallel to $IM_A.$ Hence, $I,D,M_A$ are collinear. Analogous results imply that $I$ is the desired center. 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Better solution that actually lets you see what's going on with this problem: Let $\Omega_A$, $\Omega_B$, and $\Omega_C$ be the circles with center $T_A$, $T_B$, and $T_C$ passing through $M_A$, $M_B$, and $M_C$, respectively. We claim that the corresponding common tangents of these circles concur at the incenter $I$ of $\triangle ABC$ and are parallel to the sides of $\triangle ABC$. Let $\ell$ be the line through $I$ parallel to $\overline{BC}$. Since $\overline{T_AT_C}$ is perpendicular to the angle bisector of $\angle ABC$ and $\overline{AB}$ and $\overline{BC}$ are isogonal with respect to $\angle ABC$, the reflection of $\ell$ over $\overline{T_A}{T_C}$ is parallel to $\overline{AC}$. We also have $T_AA=T_AI$ and $T_CA=T_CI$, so $AT_AIT_C$ is a kite. Thus, the reflection of $I$ over $\overline{T_AT_C}$ is $B$, so the image of $\ell$ is $\overline{AC}$. Since $\overline{AC}$ is tangent to $\Omega_C$, $\ell$ is tangent to $\Omega_C$. The other tangencies follow symmetrically, done. Now, notice that the common tangent to $\omega_B$ and $\omega_C$ is the midpoint of $\overline{M_BM_C}$ and $\overline{WZ}$. Thus, the triangle created is the image of $M_AM_BM_C$ after a homothety of scale factor $\frac{1}{2}$ at $I$. Thus, the ratio of similitude is $\frac{1}{4}$.

@above a similar but slightly different approach would be

wow. Let $p_b$ and $p_c$ meet at $D$, and define $E$ and $F$ similarly. The main claim is that $EF\parallel BC$. To do this, we will show that $\omega_b$ and $\omega_c$ reach the same "height" at the bottom of the circle. Let $r_b=R\sin^2\beta/2$ denote the radius of $\omega_b$. Then, treating $M_bM_c$ as zero, the center of $\omega_b$ is $r_b\cos\gamma$ above zero, so the bottom is $$r_b-r_b\cos\gamma=r_b(1-\cos\gamma)=R\sin^2\beta/2\cdot 2\sin^2\gamma/2$$below zero. Since this is symmetric in $B$ and $C$, the corresponding point of $\omega_c$ is also at this height, and thus $EF\parallel BC$. This finishes the proof that $\triangle DEF\sim\triangle ABC$. We claim the similarity ratio is $\frac{1}{4}$. Let $T$ denote the intersection of $p_a$ with $AC$. Then, $TM_b=r_b\tan\gamma/2,$ so $$CT=\frac{b}{2}-r_b\tan\gamma/2=\frac b2-R\sin^2\beta/2\tan\gamma/2.$$Note that $$\frac{CT}{AC}=\frac{1}{2}-\frac{R\sin^2\beta/2\tan\gamma/2}{b}=\frac{1}{2}-\frac{\sin^2\beta/2\tan\gamma/2}{2\sin\beta}$$$$=\frac{1}{2}(1-\frac{1}{2}\tan\beta/2\tan\gamma/2)=\frac{2-\tan\beta/2\tan\gamma/2}{4}.$$Thus, since we know that $EF$ is parallel to $BC$ and contains $T$, the barycentric equation of line $EF$ is $$x=\frac{2-\tan\beta/2\tan\gamma/2}{4},$$and similarly for lines $DE$ and $DF$. Thus, we have $$E=(\frac{2-\tan\beta/2\tan\gamma/2}{4},\frac{\tan\alpha/2\tan\beta/2+\tan\gamma/2\tan\beta/2}{4},\frac{2-\tan\alpha/2\tan\beta/2}{4})$$and $$F=(\frac{2-\tan\beta/2\tan\gamma/2}{4},\frac{2-\tan\alpha/2\tan\gamma/2}{4},\frac{\tan\alpha/2\tan\gamma/2+\tan\gamma/2\tan\beta/2}{4}).$$ Taking the difference between the $y$ coordinates yields $$\frac{EF}{BC}=\frac{2-\sum_{cyc}\tan\alpha/2\tan\beta/2}{4},$$so our goal then becomes to show that $$\tan\alpha/2\tan\beta/2+\tan\beta/2\tan\gamma/2+\tan\gamma/2\tan\alpha/2=1,$$since that would show that $EF=\frac{1}{4}a$. Let $S=\tan\alpha/2\tan\beta/2+\tan\beta/2\tan\gamma/2+\tan\gamma/2\tan\alpha/2.$ Then, note that $$\tan\alpha/2\tan\beta/2=1-\frac{\tan\alpha/2+\tan\beta/2}{\tan(\alpha/2+\beta/2)}=1-(\tan \alpha/2+\tan\beta/2)\tan\gamma/2$$since $\alpha/2+\beta/2=90-\gamma/2$. Finally, cyclically summing both sides of $$\tan\alpha/2\tan\beta/2=1-(\tan\alpha/2+\tan\beta/2)\tan\gamma/2$$yields $$S=3-2S\implies S=1,$$as desired.

Let $O_A,O_B,O_C$ be midpoints of $M_AT_A,M_BT_B,M_CT_C$ respectively. Let $P_1$ and $P_2$ be the points on the circle, such that they are between $M_BM_C$ and $BC$ and $O_BP_1$ and $O_CP_2$ are perpendicular to $M_BM_C$. Note that \[d(O_B,M_BM_C)=\frac12 d(T_B,M_BM_C)=\frac12 M_BT_B\cos(\angle C)\]where the distance is signed. On the other hand, $O_BP_1=\frac12 M_BT_B$ so \begin{align*} d(P_1,M_BM_C)&=\frac12 M_BT_B (1-\cos(\angle C))\\ &=\frac14 AC \tan(\tfrac{\angle B}{2})(1-\cos(\angle C))\\ &=\frac14 \frac{AC}{\sin{\angle B}}(1-\cos(\angle B))(1-\cos(\angle C)) \end{align*}Similarly, \[d(P_2,M_BM_B)=\frac14 \frac{AB}{\sin{\angle C}}(1-\cos(\angle B))(1-\cos(\angle C))\]and these are clearly equal by the Law of Sines. Therefore, $P_1P_2$ is parallel to $M_BM_C$ and so $O_BP_1$ and $O_CP_2$ are both perpendicular to $P_1P_2$, meaning that $P_1P_2$ is $p_a$. We deduce similar things are the other two sides. This means that there is a homothety from $\triangle M_AM_BM_C$ to the triangle formed by $p_a,p_b,p_c$ which means they're similar. $~$ Now to find the ratio of similitude. We claim that it is $\tfrac14$ and we again compare with the medial triangle. We will prove that the triangle formed is half the size of the medial triangle. Consider the triangle that is formed by $M_AM_C,M_AM_B$, and $p_a$. Note that \[\frac{d(p_a,M_BM_C)}{d(M_A,M_BM_C)}=\frac{(1-\cos(\angle B))(1-\cos(\angle C))}{2\sin(\angle B)\sin(\angle C)}=\frac12 \tan\left(\frac{B}{2}\right)\tan\left(\frac{C}{2}\right)\]Let $a_1$ denote this ratio, and define $b_1$ and $c_1$ analogously. It is well-known that $a_1+b_1+c_1=\tfrac12$. Let $p_A$ intersect $M_AM_B,p_b,p_c,M_AM_C$ at $A_1,A_2,A_3,A_4$, respectively. Let $M_BM_C$ intersect $p_b$ and $p_c$ at $A_5$ and $A_6$, respectively. Note that $A_5A_6=(1-b_1-c_1)(M_BM_C)$, $A_1A_4=(1-a_1)M_BM_C$ and \[A_1A_2+A_3A_4=M_BA_5+A_6M_C=(b_1+c_1)M_BM_C\]Let $p_b$ and $p_C$ intersect at $A_7$. Then, $\triangle A_2A_3A_7$ and $\triangle A_5A_6M_7$ are similar with ratio of similitude \[\tfrac{A_2A_3}{A_5A_6}=\frac{(1-a_1-b_1-c_1)M_BM_C}{(1-b_1-c_1)M_BM_C}=\frac{1-a_1-b_1-c_1}{1-b_1-c_1}\]but $\triangle A_1A_4M_A$ and $\triangle M_AM_BM_C$ are similar with ratio of similitude $1-b_1-c_1$ which means that the ratio of similitude of the triangle formed by $p_a,p_b,p_c$ with $\triangle M_AM_BM_C$ is $\tfrac12$ as desired.