Your three circumcircles intersect at the orthocenter of triangle $ T_1 T_2 T_3$. Here is a proof using some triangle geometry facts. At first, for reasons of laziness, I rename everything: Suppose ABC is an acute, scalene triangle. Let $ H_a$, $ H_b$, $ H_c$ be the feet of the altitudes from A, B, and C, respectively. Also let X, Y, Z be the points where the incircle of triangle ABC touches BC, CA, and AB, respectively. Now, let X' be the point on the line $ H_a H_b$ such that $ H_a X' = H_a X$ and the angle $ \angle X' H_a X$ is acute. Define Y' and Z' by cyclic permutation. Prove that the circumcircles of triangles X'XY, Y'YZ and Z'ZX intersect at a common point. Proof. Let H' be the orthocenter of triangle XYZ. We will prove that the circumcircles of triangles X'XY, Y'YZ and Z'ZX pass through H'. At first, we show a classical result: Lemma 1. For every triangle ABC with orthocenter H, the circumcenter of triangle BCH is the reflection of the circumcenter of triangle ABC in the line BC. Proof. It is well-known ([1], Lemma 1) that the reflection A' of the orthocenter H in the line BC lies on the circumcircle of triangle ABC. Hence, the circumcenter of triangle BCA' coincides with the circumcenter of triangle ABC. But since A' is the reflection of H in the line BC, the triangles BCH and BCA' are symmetric with respect to BC, and the circumcenter of triangle BCH must be the reflection of the circumcenter of triangle BCA' in the line BC, i. e. the reflection of the circumcenter of triangle ABC in the line BC. Lemma 1 is proven. Now, let I be the incenter of triangle ABC. Then, I is the circumcenter of triangle XYZ. Applied to triangle XYZ, Lemma 1 yields that the circumcenter F of triangle XYH' is the reflection of I in the line XY. Thus, XF = XI and YF = YI. But obviously, XI = YI; so, XF = XI = YI = YF, and the quadrilateral XFYI is a rhombus, so that XF is parallel to YI, or, in other words, XF is perpendicular to CY. Similarly, YF is perpendicular to CX. Thus, F is the orthocenter of triangle CXY. But according to [2] Theorem 3.6, the following result holds: Lemma 2. The orthocenter of triangle CXY coincides with the incenter of triangle $ H_a H_b C$. In other words, F is the incenter of triangle $ H_a H_b C$. Hence, the line $ H_a F$ is the angle bisector of the angle $ \angle C H_a H_b = \angle X' H_a X$. Now, the isosceles triangle $ X' H_a X$ yields that the angle bisector of the angle $ X' H_a X$, i. e. in the line $ H_a F$, is the perpendicular bisector of the segment XX', i. e. the point X' is the reflection of the point X in the line $ H_a F$. But since F is the circumcenter of triangle XYH', the circumcircle of triangle XYH' has center F. The line $ H_a F$ is a diameter of this circumcircle, and since the reflection of a point on a circle in a diameter of the circle lies on the same circle again, the reflection X' of X in the line $ H_a F$ lies on the circumcircle of triangle XYH' again. In other words, the circumcircle of triangle X'XY passes through H. Similarly, the circumcircles of triangles Y'YZ and Z'ZX pass through H, and we are done. References (both avaliable at my geometry website; I was too lazy to browse other sites) [1] Darij Grinberg, Anti-Steiner points with respect to a triangle. [2] Darij Grinberg, Generalization of the Feuerbach point. Darij

Here I give my solution: Call the three circumcircles $k_1$, $k_2$ and $k_3$ Firstly we can see that the circumcenter of $T_1P_1T_2$ is the incenter of $CH_1H_2$. This is true because $S_{T_1T_2}$ is $l_{\angle H_1CH_2}$ and $S_{T_1P_2}$ is $l_{\angle CH_1H_2}$. Call that center $I_C$. So $\frac{CI_C}{CI} =\frac{CH_1}{CA} =cos{\gamma}$ because $\triangle{H_1H_2C}$ is similar to $\triangle{ABC}$ ! Consider $\triangle{CIT_1}$ and point $I_C$ on $CI$ so that $\frac{CI_C}{CI} =cos{2\angle{ICT_1}}$. Let $K$ be the reflection of $I_C$ over $CT_1$. Then $\angle{IKC}= 90'$. ' means degree sign(sorry I didn't found it ). Let $IK$ intersects $CT_1$ in point $L$. $\angle{CI_CK}= 90' - \frac{\gamma}{2} = \angle {CLK}$ So $C$, $I_C$, $L$ and $K$ lie on circle -> $\angle {CI_CL}=90$ -> $I_C$, $I$, $T_1$ and $L$ lie on circle -> $\angle{II_CT_1}=\angle{ILT_1}= 90' - \frac{\gamma}{2}$ Then $\angle{T_1I_CT_2}= 180' - \gamma$. We are almost ready! Let circle $k_1(I_C) \cap k_2(I_B)=U$. By simple calculation of angles we can see that point $U$ lies also on $k_3$. The problem is solved! Also knowing the angles of $\triangle{T_1T_2T_3}$ we can easily see also that $U$ is its orthocenter!

Here's another approach: As in the two previous solutions, we shall prove that the circumcircle of $ T_{1}P_{1}T_{2}$ passes through the orthocentre $ X$ of $ T_{1}T_{2}T_{3}$, which is clearly sufficient. Assume $ AB$ is the shortest side of the triangle (we can't WLOG this, but in all other cases, the solution will be the same with slight modifications. The difference is, that when $ AB$ is shortest side. $ H_{1}$ is closer to $ B$ than $ T_{1}$ is, and $ H_{2}$ is closer to $ A$ than $ T_{2}$ is.) Let $ Q_{2}$ be the point on $ H_{1}H_{2}$ such that $ H_{2}Q_{2} = H_{2}T_{2}$. Since $ \angle H_{1}H_{2}C = B$, $ \angle H_{2}Q_{2}T_{2} = 90 - \frac{B}{2}$. Since $ \angle H_{2}H_{1}C = A$, $ \angle P_{1}T_{1}H_{1} = 90 - \frac{A}{2}$, and $ \angle T_{2}T_{1}C = 90 - \frac{C}{2}$, $ \angle P_{1}T_{1}T_{2} = 180 - ( \angle H_{1}T_{1}P_{1} + \angle CT_{1}T_{2}) = 90 - \frac{B}{2} = \angle H_{2}Q_{2}T_{2}$. Thus quadrilateral $ T_{1}P_{1}Q_{2}T_{2}$ is concyclic. A little more angle chasing gives $ \angle Q_{2}T_{2}T_{1} = 180 - \angle H_{2}T_{2}Q_{2} - \angle CT_{2}T_{1} = 90 - \frac {A}{2}$, and since $ \angle P_{1}T_{1}T_{2} = 90 - \frac{B}{2}$, if we let $ T_{1}P_{1}$ intersect $ T_{2}Q_{2}$ at a point $ T'_{3}$, we have $ T_{1}T_{2}T_{3}$ and $ T_{1}T_{2}T'_{3}$ are congruent triangles. Thus, the quadrilateral $ T_{1}T_{2}T_{3}T'_{3}$ is a concyclic, iscoceles trapezium, so $ T'_{3}$ lies on the incircle. Computations give us $ H_{2}T_{2} = AT_{2} - AH_{2} = (s-a) - c*cos B = \frac{(s-b)(a-c)}{b}$. Also, $ Q_{2}H_{2}T_{2}$ is similar to $ BT_{3}T_{1}$, so $ \frac {Q_{2}T_{2}}{T_{2}T'_{3}} = \frac {Q_{2}T_{2}}{T_{3}T_{1}} = \frac {T_{2}H_{2}}{BT_{1}} = \frac { \frac {(s-b)(a-c)}{b} }{s-b} = \frac {a-c}{b}$. Choose a point $ S$, distinct from $ T_{3}$ on segment $ T_{1}T_{3}$ so that $ T_{2}S =T_{2}T_{3}$, so $ \angle T_{2}ST_{1} = 90 + \frac {C}{2}$. Sin Rule gives: $ \frac {ST_{1}}{T_{1}T_{3}} = \frac {ST_{1}}{T_{1}T_{2}} * \frac {T_{1}T_{2}}{T_{1}T_{3}} = \frac {sin ( \frac {A}{2} - \frac {C}{2} )}{cos ( \frac {B}{2} ) }$ $ = \frac {2 cos (\frac{A}{2} + \frac{C}{2}) * sin ( \frac {A}{2} - \frac {C}{2} )} {2 sin (\frac {B}{2}) * cos (\frac {B}{2})} = \frac {sin A - sin C}{sin B} = \frac {a-c}{b}$. Thus $ \frac {ST_{1}}{T_{1}T_{3}} = \frac {Q_{2}T_{2}}{T_{2}T'_{3}}$. Since $ T_{1}T_{2}T_{3}T'_{3}$ is an iscoceles trapezium, this shows us that $ T_{2}T_{1}Q_{2}S$ is also an iscoceles trapezium, and more importantly, is cyclic. But a quick angle chase shows us that $ \angle T_{2}ST_{1} = 90 + \frac{C}{2} = \angle T_{1}XT_{2}$, so $ T_{1}T_{2}XS$ is cyclic. This shows that $ T_{1}, T_{2}, X, S, Q_{2}, P_{1}$ all lie on a circle, as required.

cauchyguy wrote: Let $ \ol{AH_1}, \ol{BH_2}$, and $ \ol{CH_3}$ be the altitudes of an acute scalene triangle $ ABC$. The incircle of triangle $ ABC$ is tangent to $ \ol{BC}, \ol{CA},$ and $ \ol{AB}$ at $ T_1, T_2,$ and $ T_3$, respectively. For $ k = 1, 2, 3$, let $ P_i$ be the point on line $ H_iH_{i + 1}$ (where $ H_4 = H_1$) such that $ H_iT_iP_i$ is an acute isosceles triangle with $ H_iT_i = H_iP_i$. Prove that the circumcircles of triangles $ T_1P_1T_2$, $ T_2P_2T_3$, $ T_3P_3T_1$ pass through a common point. Notice that the problem is vitrually the same with this one: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=351094#351094 Reflect $ H_1$ and $ H_2$ into $ T_1T_2$ to form $ X$ and $ Y$ respectively. One intersection, $ K$, of $ XY$ and the incircle will be such that $ T_1K\parallel T_2T_3$ by the link. We will now prove that $ K$ is the reflection of $ P_1$ into $ T_1T_2$. Since $ K$ lies on $ XY$ and $ P_1$ lies on $ H_1H_2$, it suffices to show that $ \angle P_1T_1T_2=\angle T_2T_1K$. Yet, \[ \angle KT_1T_2=\angle T_3T_2T_1=180-\angle T_1T_3T_2-\angle T_3T_1T_2\] We see that $ \angle T_1T_3T_2=\angle T_2T_1C$ and $ \angle PT_1H_1=\angle AT_3T_2=\angle T_3T_1T_2$ the first because $ \triangle H_1P_1T_1\sim \triangle AT_3T_2$ since $ AT_3=AT_2$, $ P_1H_1=T_1H_1$, and $ \angle T_3AT_2=\angle H_2H_1T_1=\angle P_1H_1T_2$. Thus, we have that \[ \angle T_2T_1K=180-\angle P_1T_1H_1-\angle T_2T_1C=\angle PT_1T_2\] implying that $ P_1$ and $ K$ are reflections of each other over $ T_1T_2$. Thus, the circumcircle of $ \triangle P_1T_1T_2$ is the reflection of the circumcircle of $ \triangle T_1T_3T_2$ into $ T_1T_2$. Since the reflection of the orthocenter of $ \triangle T_1T_2T_3$, $ H$, into $ T_1T_2$ lies on the circumcircle of $ \triangle T_1T_3T_2$, we see that $ H$ lies on the circumcircle of $ \triangle T_1T_2P_1$. Doing the same with the other circles, we see that they all pass through $ H$, and are hence concurrent.

See at http://www.mathlinks.ro/viewtopic.php?t=315623 that the three circles are equal with the incircle of triangle $ ABC$, hence they all pass through the orthocenter of triangle $ T_1T_2T_3$ (4 circles theorem?) Best regards, sunken rock

Notice that $\frac{sin T_1P_1T_2}{sin T_1T_2P_1} = \frac{T_1T_2}{T_1P_1} = \frac{CI sin C}{2(sin A/2)T_1H_1} = \frac{CI sin C}{2IA(sin(A/2+B-90))sin(A/2)} = \frac{sin(90+C/2)}{sin(A/2+B-90)}$ and that $90+C/2+A/2+B-90 = 90-B/2 = \angle T_1P_1T_2 + \angle T_1T_2P_1$. Thus, $ \angle T_1P_1T_2 = 90+C/2 = 180 - \angle T_1T_3T_2$. So the circumcircle of $T_1P_1T_2$ is the reflection across $T_1T_2$ of the circumcircle of $T_1T_2T_3$, and we finish easily.