USA IMO, 2003

First Solution. (By Po-Ru Loh) We begin by showing that points $O_3, H_2, T_2$, and $O_3$ lie on a cyclic. We will prove this by establishing $\angle O_3O_1H_2 = \angle O_3T_2C = \angle O_3T_2H_2$. To find $\angle O_3O_1H_2$, observe that triangles $H_2AH_3$ and $H_2H_1C$ are similar. Indeed, quadrilateral $BH_3H_2C$

is cyclic so $\angle H_2H_3A = \angle C$, and likewise $\angle CH_1H_2 = \angle A$. Now, $O_1$ and $O_3$ are corresponding incenters of similar triangles, so it follows that triangles $H_2AO_1$ and $H_2H_1O_3$ are also similar, and hence are related by a spiral similarity about $H_2$. Thus,
$$\frac{AH_2}{H_1H_2} = \frac{O_1H_2}{O_3H_2}$$
and
$$\begin{aligned} \angle AH_2H_1 &= \angle AH_2O_1 + \angle O_1H_2H_1 \\ &= \angle O_1H_2H_1 + \angle H_1H_2O_3 = \angle O_1H_2O_3. \end{aligned}$$
It follows that another spiral similarity about $H_2$ takes triangle $H_2AH_1$ to triangle $H_2O_1O_3$. Hence $\angle O_3O_1H_2 = \angle H_1AH_2 = 90^\circ - \angle C$.
We wish to show that $\angle O_3T_2C = 90^\circ - \angle C$ as well, or in other words, $T_2O_3 \perp BC$. To do this, drop the altitude from $O_3$ to $BC$ and let it intersect $BC$ at $D$. Triangles $ABC$ and $H_1H_2C$ are similar as before, with corresponding incenters $I$ and $O_3$. Furthermore, $IT_2$ and $O_3D$ also correspond. Hence, $CT_2/T_2A = CD/DH_1$, and so $T_2D \parallel AH_1$. Thus, $T_2D \perp BC$, and it follows that $T_2O_3 \perp BC$.

Having shown that $O_1H_2T_2O_3$ is cyclic, we may now write $\angle O_1T_2O_3 = \angle O_1H_2O_3$. Since triangles $H_2AO_1$ and $H_2H_1O_3$ are related by a spiral similarity about $H_2$, we have
$$\angle O_1H_2O_3 = \angle AH_2H_1 = 180^\circ - \angle B,$$
by noting that $ABH_2H_1$ is cyclic. Likewise,
$$\angle O_2T_3O_1 = 180^\circ - \angle C \quad \text{and} \quad \angle O_3T_1O_2 = 180^\circ - \angle A,$$
and so $\angle O_1T_2O_3 + \angle O_2T_3O_1 + \angle O_3T_1O_2 = 360^\circ$. Therefore, $\angle T_3O_1T_2$, $\angle T_1O_2T_3$, and $\angle T_2O_3T_1$ of hexagon $O_1T_2O_3T_1O_2T_3$ also sum to $360^\circ$. Now let $H$ be the intersection of circles $\omega_1$ and $\omega_2$. Then $\angle T_2HT_3 = 180^\circ - \frac{1}{2}\angle T_3O_1T_2$ and $\angle T_3HT_1 = 180^\circ - \frac{1}{2}\angle T_1O_2T_3$. Therefore,
$$\begin{aligned} \angle T_1HT_2 &= 360^\circ - \angle T_2HT_3 - \angle T_3HT_1 \\ &= \frac{1}{2}\angle T_3O_1T_2 + \frac{1}{2}\angle T_1O_2T_3 = 180^\circ - \frac{1}{2}\angle T_1O_3T_2, \end{aligned}$$
and so $H$ lies on the circle $\omega_3$ as well. Hence, circles $\omega_1$, $\omega_2$, and $\omega_3$ share a common point, as wanted.

Second Solution. (By Anders Kaseorg) Note that $AH_2 = AB \cos \angle A$ and $AH_3 = AC \cos \angle A$, so triangles $AH_2H_3$ and $ABC$ are similar with ratio $\cos \angle A$. Thus, since $O_1$ is the incenter of triangle $AH_2H_3$, $AO_1 = AI \cos \angle A$. If $X_1$ is the intersection of segments $AI$ and $T_2T_3$,

we have $\angle IX_1T_2 = \angle AT_2I = 90^\circ$, and so
$$\begin{aligned} X_1I = T_2I \cos \angle T_2IA &= AI \cos^2 \angle T_2IA = AI \sin^2 \frac{\angle A}{2} \\ &= AI \cdot \frac{1 - \cos \angle A}{2} = \frac{AI - AO_1}{2} = \frac{O_1I}{2}. \end{aligned}$$
Hence $O_1X_1 = X_1I$, so $O_1$ is the reflection of $I$ across line $T_2T_3$, and $O_1T_2 = IT_2 = IT_3 = O_1T_3$. Therefore, $O_1T_2IT_3$, and similarly $O_2T_3IT_1$ and $O_3T_1IT_2$, are rhombi with the same side length $r$, implying that circles $\omega_1, \omega_2, \omega$ have the same radius $r$. We also conclude that $O_1T_2 = T_3I = O_2T_1$ and $O_1T_2 \parallel T_3I \parallel O_2T_1$, and so $O_1O_2T_1T_2$ is a parallelogram. Hence the midpoints of $O_1T_1$ and $O_2T_2$ (similarly $O_3T_3$) are the same point $P$, and $O_1O_2O_3$ is the reflection of $T_1T_2T_3$ across $P$. If $H$ is the reflection of $I$ across $P$, we have $O_1H = O_2H = O_3H = r$, that is, $H$ is a common point of the three circumcircles.

Third Solution. We use directed lengths (along line $BC$, with $C$ to $B$ as the positive direction) and directed angles modulo $180^\circ$ in this proof. (For segments not lying on line $BC$, we assume its direction as the direction of its projection on line $BC$.) We claim that $\omega_i$, $i = 1, 2, 3$, all pass through $H$, the orthocenter of triangle $T_1T_2T_3$. Without loss of generality, it suffices to prove that $T_1P_1T_2H$ is cyclic. If $AB = AC$, then $T_1 = H_1 = P_1$ and the case is trivial. Let $AB = c$, $BC = a$, $CA = b$, $\angle BAC = \alpha$, $\angle CBA = \beta$, and $\angle ACB = \gamma$.

Let $Q$ be the intersection of lines $HP_1$ and $BC$. Note that
$$\begin{align*} \angle HT_2T_1 &= 90^\circ - \angle T_2T_1T_3 \\ &= 90^\circ - [180^\circ - \angle T_3T_1B - \angle CT_1T_2] \\ &= 90^\circ - \left[180^\circ - \left(90^\circ - \frac{\beta}{2}\right) - \left(90^\circ - \frac{C}{2}\right)\right] \\ &= \frac{\alpha}{2}. \end{align*}$$
(Likewise, $\angle T_2T_1H = \beta/2$.) Thus to prove that $T_1P_1T_2H$ is cyclic is equivalent to prove that $\angle QP_1T_1 = \alpha/2$.
Let $Q_H$ and $Q_P$ be the respective feet of perpendiculars from $H$ and $P_1$ to line $BC$. Because $\angle AH_1B = \angle AH_2B = 90^\circ$, $ABH_1H_2$ is cyclic, and so $\angle T_1H_1P_1 = \angle BH_1P_1 = \alpha$. Thus triangles $AT_3T_2$ and $H_1T_1P_1$

are similar, implying that
$$\angle Q_P P_1 T_1 = 90^\circ - \angle P_1 T_1 H_1 = 90^\circ - \left( 90^\circ - \frac{\angle T_1 H_1 P_1}{2} \right) = \frac{\alpha}{2}.$$
Therefore, to prove that $\angle Q_P P_1 T_1 = \alpha/2$, we have now reduced to proving that $Q_P = Q_H$, or
$$\frac{T_1 Q_P}{T_1 H_1} = \frac{T_1 Q_H}{T_1 H_1}. \qquad (1)$$
Note that
$$T_1 H_1 = P_1 H_1 \quad \text{and} \quad \frac{T_1 Q_P}{T_1 H_1} = 1 - \frac{Q_P H_1}{T_1 H_1},$$
that is,
$$\frac{T_1 Q_P}{T_1 H_1} = 1 - \frac{Q_P H_1}{P_1 H_1} = 1 - \cos \angle T_1 H_1 P_1 = 1 - \cos \alpha. \quad (2)$$
On the other hand, applying the Law of Cosines to triangle $ABC$ gives
$$\begin{aligned} T_1 H_1 &= T_1 C - H_1 C = \frac{a+b-c}{2} - b \cos \gamma \\ &= \frac{a+b-c}{2} - \frac{a^2+b^2-c^2}{2a} = \frac{a(b-c) - (b^2-c^2)}{2a}, \end{aligned}$$
or
$$T_1 H_1 = \frac{(b-c)(a-b-c)}{2a} = \frac{(c-b)(b+c-a)}{2a}. \quad (3)$$
Now we calculate $T_1 Q_H$. Because $H$ is the orthocenter of triangle $T_1 T_2 T_3$,
$$\begin{aligned} \angle T_1 H T_2 &= 180^\circ - \angle H T_2 T_1 - \angle T_2 T_1 H \\ &= (90^\circ - \angle H T_2 T_1) + (90^\circ - \angle T_2 T_1 H) \\ &= \angle T_2 T_1 T_3 + \angle T_3 T_2 T_1 = 180^\circ - \angle T_1 T_3 T_2. \end{aligned}$$
Applying the Law of Sines to triangle $T_1 T_2 H$ and applying the Extended Law of Sines to triangle $T_1 T_2 T_3$ gives
$$\frac{T_1 H}{\sin \angle H T_2 T_1} = \frac{T_1 T_2}{\sin \angle T_1 H T_2} = \frac{T_1 T_2}{\sin \angle T_1 T_3 T_2} = 2r,$$
and consequently,
$$T_1 H = 2r \sin \frac{\alpha}{2}.$$
Because
$$\begin{aligned} \angle Q_H T_1 H &= \angle CT_1 T_2 + \angle T_2 T_1 H = (90^\circ - \frac{\gamma}{2}) + \frac{\beta}{2} \\ &= 90^\circ + \frac{\beta - \gamma}{2}, \end{aligned}$$
we obtain
$$T_1 Q_H = T_1 H \cos \angle H T_1 Q_H = 2r \sin \frac{\alpha}{2} \sin \frac{\gamma - \beta}{2}. \quad (4)$$
Combining equations (1), (2), (3), and (4), we conclude that it suffices to prove that
$$1 - \cos \alpha = \frac{4ar \sin \frac{\alpha}{2} \sin \frac{\gamma-\beta}{2}}{(c-b)(b+c-a)}. \quad (5)$$
Applying the fact
$$\frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}} = \tan \frac{\alpha}{2} = \frac{r}{AT_2} = \frac{2r}{b+c-a},$$
and applying the Law of Sines to triangle $ABC$, (5) becomes
$$1 - \cos \alpha = \frac{2 \sin \alpha \sin^2 \frac{\alpha}{2} \sin \frac{\gamma-\beta}{2}}{\cos \frac{\alpha}{2} (\sin \gamma - \sin \beta)}. \quad (6)$$
By the Double-angle formulas, $1 - \cos \alpha = 2 \sin^2 \frac{\alpha}{2}$ and $\sin \alpha = 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}$ and so (6) reads
$$\sin \gamma - \sin \beta = 2 \sin \frac{\alpha}{2} \sin \frac{\gamma - \beta}{2}.$$
By the Difference-to-product formulas, the last equation reduces to
$$2 \cos \frac{\beta + \gamma}{2} \sin \frac{\gamma - \beta}{2} = 2 \sin \frac{\alpha}{2} \sin \frac{\gamma - \beta}{2},$$
which is evident.