Subfield: Geometry


Question: Let $A B C$ be an acute triangle with circumcircle $\omega$. Let $t$ be a tangent line to $\omega$. Let $t_{a}, t_{b}$, and $t_{c}$ be the lines obtained by reflecting $t$ in the lines $B C, C A$, and $A B$, respectively. Show that the circumcircle of the triangle determined by the lines $t_{a}, t_{b}$, and $t_{c}$ is tangent to the circle $\omega$.

To avoid a large case distinction, we will use the notion of oriented angles. Namely, for two lines $\ell$ and $m$, we denote by $\angle(\ell, m)$ the angle by which one may rotate $\ell$ anticlockwise to obtain a line parallel to $m$. Thus, all oriented angles are considered modulo $180^{\circ}$.

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Solution:
Denote by $T$ the point of tangency of $t$ and $\omega$. Let $A^{\prime}=t_{b} \cap t_{c}, B^{\prime}=t_{a} \cap t_{c}$, $C^{\prime}=t_{a} \cap t_{b}$. Introduce the point $A^{\prime \prime}$ on $\omega$ such that $T A=A A^{\prime \prime}\left(A^{\prime \prime} \neq T\right.$ unless $T A$ is a diameter). Define the points $B^{\prime \prime}$ and $C^{\prime \prime}$ in a similar way.

Since the points $C$ and $B$ are the midpoints of arcs $T C^{\prime \prime}$ and $T B^{\prime \prime}$, respectively, we have

$$
\begin{aligned}
\angle\left(t, B^{\prime \prime} C^{\prime \prime}\right) & =\angle\left(t, T C^{\prime \prime}\right)+\angle\left(T C^{\prime \prime}, B^{\prime \prime} C^{\prime \prime}\right)=2 \angle(t, T C)+2 \angle\left(T C^{\prime \prime}, B C^{\prime \prime}\right) \\
& =2(\angle(t, T C)+\angle(T C, B C))=2 \angle(t, B C)=\angle\left(t, t_{a}\right) .
\end{aligned}
$$

It follows that $t_{a}$ and $B^{\prime \prime} C^{\prime \prime}$ are parallel. Similarly, $t_{b} \| A^{\prime \prime} C^{\prime \prime}$ and $t_{c} \| A^{\prime \prime} B^{\prime \prime}$. Thus, either the triangles $A^{\prime} B^{\prime} C^{\prime}$ and $A^{\prime \prime} B^{\prime \prime} C^{\prime \prime}$ are homothetic, or they are translates of each other. Now we will prove that they are in fact homothetic, and that the center $K$ of the homothety belongs



to $\omega$. It would then follow that their circumcircles are also homothetic with respect to $K$ and are therefore tangent at this point, as desired.

We need the two following claims.

Claim 1. The point of intersection $X$ of the lines $B^{\prime \prime} C$ and $B C^{\prime \prime}$ lies on $t_{a}$.

Proof. Actually, the points $X$ and $T$ are symmetric about the line $B C$, since the lines $C T$ and $C B^{\prime \prime}$ are symmetric about this line, as are the lines $B T$ and $B C^{\prime \prime}$.

Claim 2. The point of intersection $I$ of the lines $B B^{\prime}$ and $C C^{\prime}$ lies on the circle $\omega$.

Proof. We consider the case that $t$ is not parallel to the sides of $A B C$; the other cases may be regarded as limit cases. Let $D=t \cap B C, E=t \cap A C$, and $F=t \cap A B$.

Due to symmetry, the line $D B$ is one of the angle bisectors of the lines $B^{\prime} D$ and $F D$; analogously, the line $F B$ is one of the angle bisectors of the lines $B^{\prime} F$ and $D F$. So $B$ is either the incenter or one of the excenters of the triangle $B^{\prime} D F$. In any case we have $\angle(B D, D F)+\angle(D F, F B)+$ $\angle\left(B^{\prime} B, B^{\prime} D\right)=90^{\circ}$, so

$$
\angle\left(B^{\prime} B, B^{\prime} C^{\prime}\right)=\angle\left(B^{\prime} B, B^{\prime} D\right)=90^{\circ}-\angle(B C, D F)-\angle(D F, B A)=90^{\circ}-\angle(B C, A B) .
$$

Analogously, we get $\angle\left(C^{\prime} C, B^{\prime} C^{\prime}\right)=90^{\circ}-\angle(B C, A C)$. Hence,

$$
\angle(B I, C I)=\angle\left(B^{\prime} B, B^{\prime} C^{\prime}\right)+\angle\left(B^{\prime} C^{\prime}, C^{\prime} C\right)=\angle(B C, A C)-\angle(B C, A B)=\angle(A B, A C),
$$

which means exactly that the points $A, B, I, C$ are concyclic.

Now we can complete the proof. Let $K$ be the second intersection point of $B^{\prime} B^{\prime \prime}$ and $\omega$. Applying PASCAL's theorem to hexagon $K B^{\prime \prime} C I B C^{\prime \prime}$ we get that the points $B^{\prime}=K B^{\prime \prime} \cap I B$ and $X=B^{\prime \prime} C \cap B C^{\prime \prime}$ are collinear with the intersection point $S$ of $C I$ and $C^{\prime \prime} K$. So $S=$ $C I \cap B^{\prime} X=C^{\prime}$, and the points $C^{\prime}, C^{\prime \prime}, K$ are collinear. Thus $K$ is the intersection point of $B^{\prime} B^{\prime \prime}$ and $C^{\prime} C^{\prime \prime}$ which implies that $K$ is the center of the homothety mapping $A^{\prime} B^{\prime} C^{\prime}$ to $A^{\prime \prime} B^{\prime \prime} C^{\prime \prime}$, and it belongs to $\omega$.
Define the points $T, A^{\prime}, B^{\prime}$, and $C^{\prime}$ in the same way as in the previous solution. Let $X, Y$, and $Z$ be the symmetric images of $T$ about the lines $B C, C A$, and $A B$, respectively. Note that the projections of $T$ on these lines form a Simson line of $T$ with respect to $A B C$, therefore the points $X, Y, Z$ are also collinear. Moreover, we have $X \in B^{\prime} C^{\prime}, Y \in C^{\prime} A^{\prime}$, $Z \in A^{\prime} B^{\prime}$.

Denote $\alpha=\angle(t, T C)=\angle(B T, B C)$. Using the symmetry in the lines $A C$ and $B C$, we get

$$
\angle(B C, B X)=\angle(B T, B C)=\alpha \quad \text { and } \quad \angle\left(X C, X C^{\prime}\right)=\angle(t, T C)=\angle\left(Y C, Y C^{\prime}\right)=\alpha \text {. }
$$

Since $\angle\left(X C, X C^{\prime}\right)=\angle\left(Y C, Y C^{\prime}\right)$, the points $X, Y, C, C^{\prime}$ lie on some circle $\omega_{c}$. Define the circles $\omega_{a}$ and $\omega_{b}$ analogously. Let $\omega^{\prime}$ be the circumcircle of triangle $A^{\prime} B^{\prime} C^{\prime}$.



Now, applying MiqueL's theorem to the four lines $A^{\prime} B^{\prime}, A^{\prime} C^{\prime}, B^{\prime} C^{\prime}$, and $X Y$, we obtain that the circles $\omega^{\prime}, \omega_{a}, \omega_{b}, \omega_{c}$ intersect at some point $K$. We will show that $K$ lies on $\omega$, and that the tangent lines to $\omega$ and $\omega^{\prime}$ at this point coincide; this implies the problem statement.

Due to symmetry, we have $X B=T B=Z B$, so the point $B$ is the midpoint of one of the $\operatorname{arcs} X Z$ of circle $\omega_{b}$. Therefore $\angle(K B, K X)=\angle(X Z, X B)$. Analogously, $\angle(K X, K C)=$ $\angle(X C, X Y)$. Adding these equalities and using the symmetry in the line $B C$ we get

$$
\angle(K B, K C)=\angle(X Z, X B)+\angle(X C, X Z)=\angle(X C, X B)=\angle(T B, T C) .
$$

Therefore, $K$ lies on $\omega$.

Next, let $k$ be the tangent line to $\omega$ at $K$. We have

$$
\begin{aligned}
\angle\left(k, K C^{\prime}\right) & =\angle(k, K C)+\angle\left(K C, K C^{\prime}\right)=\angle(K B, B C)+\angle\left(X C, X C^{\prime}\right) \\
& =(\angle(K B, B X)-\angle(B C, B X))+\alpha=\angle\left(K B^{\prime}, B^{\prime} X\right)-\alpha+\alpha=\angle\left(K B^{\prime}, B^{\prime} C^{\prime}\right),
\end{aligned}
$$

which means exactly that $k$ is tangent to $\omega^{\prime}$.

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