Subfield: Geometry


Question: Let $A B C$ be a triangle with circumcircle $\omega$ and incentre $I$. A line $\ell$ intersects the lines $A I, B I$, and $C I$ at points $D, E$, and $F$, respectively, distinct from the points $A, B, C$, and $I$. The perpendicular bisectors $x, y$, and $z$ of the segments $A D, B E$, and $C F$, respectively determine a triangle $\Theta$. Show that the circumcircle of the triangle $\Theta$ is tangent to $\omega$.

(Denmark)

Preamble. Let $X=y \cap z, Y=x \cap z, Z=x \cap y$ and let $\Omega$ denote the circumcircle of the triangle $X Y Z$. Denote by $X_{0}, Y_{0}$, and $Z_{0}$ the second intersection points of $A I, B I$ and $C I$, respectively, with $\omega$. It is known that $Y_{0} Z_{0}$ is the perpendicular bisector of $A I, Z_{0} X_{0}$ is the perpendicular bisector of $B I$, and $X_{0} Y_{0}$ is the perpendicular bisector of $C I$. In particular, the triangles $X Y Z$ and $X_{0} Y_{0} Z_{0}$ are homothetic, because their corresponding sides are parallel.

The solutions below mostly exploit the following approach. Consider the triangles $X Y Z$ and $X_{0} Y_{0} Z_{0}$, or some other pair of homothetic triangles $\Delta$ and $\delta$ inscribed into $\Omega$ and $\omega$, respectively. In order to prove that $\Omega$ and $\omega$ are tangent, it suffices to show that the centre $T$ of the homothety taking $\Delta$ to $\delta$ lies on $\omega$ (or $\Omega$ ), or, in other words, to show that $\Delta$ and $\delta$ are perspective (i.e., the lines joining corresponding vertices are concurrent), with their perspector lying on $\omega$ (or $\Omega$ ).

We use directed angles throughout all the solutions.

Solution:
Claim 1. The reflections $\ell_{a}, \ell_{b}$ and $\ell_{c}$ of the line $\ell$ in the lines $x, y$, and $z$, respectively, are concurrent at a point $T$ which belongs to $\omega$.

<img_3601>

Proof. Notice that $\sphericalangle\left(\ell_{b}, \ell_{c}\right)=\sphericalangle\left(\ell_{b}, \ell\right)+\sphericalangle\left(\ell, \ell_{c}\right)=2 \sphericalangle(y, \ell)+2 \sphericalangle(\ell, z)=2 \sphericalangle(y, z)$. But $y \perp B I$ and $z \perp C I$ implies $\sphericalangle(y, z)=\sphericalangle(B I, I C)$, so, since $2 \sphericalangle(B I, I C)=\sphericalangle(B A, A C)$, we obtain

$$
\sphericalangle\left(\ell_{b}, \ell_{c}\right)=\sphericalangle(B A, A C) .
\tag{1}
$$

Since $A$ is the reflection of $D$ in $x, A$ belongs to $\ell_{a}$; similarly, $B$ belongs to $\ell_{b}$. Then (1) shows that the common point $T^{\prime}$ of $\ell_{a}$ and $\ell_{b}$ lies on $\omega$; similarly, the common point $T^{\prime \prime}$ of $\ell_{c}$ and $\ell_{b}$ lies on $\omega$.

If $B \notin \ell_{a}$ and $B \notin \ell_{c}$, then $T^{\prime}$ and $T^{\prime \prime}$ are the second point of intersection of $\ell_{b}$ and $\omega$, hence they coincide. Otherwise, if, say, $B \in \ell_{c}$, then $\ell_{c}=B C$, so $\sphericalangle(B A, A C)=\sphericalangle\left(\ell_{b}, \ell_{c}\right)=\sphericalangle\left(\ell_{b}, B C\right)$, which shows that $\ell_{b}$ is tangent at $B$ to $\omega$ and $T^{\prime}=T^{\prime \prime}=B$. So $T^{\prime}$ and $T^{\prime \prime}$ coincide in all the cases, and the conclusion of the claim follows.



Now we prove that $X, X_{0}, T$ are collinear. Denote by $D_{b}$ and $D_{c}$ the reflections of the point $D$ in the lines $y$ and $z$, respectively. Then $D_{b}$ lies on $\ell_{b}, D_{c}$ lies on $\ell_{c}$, and

$$
\begin{aligned}
\sphericalangle\left(D_{b} X, X D_{c}\right) & =\sphericalangle\left(D_{b} X, D X\right)+\sphericalangle\left(D X, X D_{c}\right)=2 \sphericalangle(y, D X)+2 \sphericalangle(D X, z)=2 \sphericalangle(y, z) \\
& =\sphericalangle(B A, A C)=\sphericalangle(B T, T C),
\end{aligned}
$$

hence the quadrilateral $X D_{b} T D_{c}$ is cyclic. Notice also that since $X D_{b}=X D=X D_{c}$, the points $D, D_{b}, D_{c}$ lie on a circle with centre $X$. Using in this circle the diameter $D_{c} D_{c}^{\prime}$ yields $\sphericalangle\left(D_{b} D_{c}, D_{c} X\right)=90^{\circ}+\sphericalangle\left(D_{b} D_{c}^{\prime}, D_{c}^{\prime} X\right)=90^{\circ}+\sphericalangle\left(D_{b} D, D D_{c}\right)$. Therefore,

$$
\begin{gathered}
\sphericalangle\left(\ell_{b}, X T\right)=\sphericalangle\left(D_{b} T, X T\right)=\sphericalangle\left(D_{b} D_{c}, D_{c} X\right)=90^{\circ}+\sphericalangle\left(D_{b} D, D D_{c}\right) \\
=90^{\circ}+\sphericalangle(B I, I C)=\sphericalangle(B A, A I)=\sphericalangle\left(B A, A X_{0}\right)=\sphericalangle\left(B T, T X_{0}\right)=\sphericalangle\left(\ell_{b}, X_{0} T\right),
\end{gathered}
$$

so the points $X, X_{0}, T$ are collinear. By a similar argument, $Y, Y_{0}, T$ and $Z, Z_{0}, T$ are collinear. As mentioned in the preamble, the statement of the problem follows.
As mentioned in the preamble, it is sufficient to prove that the centre $T$ of the homothety taking $X Y Z$ to $X_{0} Y_{0} Z_{0}$ belongs to $\omega$. Thus, it suffices to prove that $\sphericalangle\left(T X_{0}, T Y_{0}\right)=$ $\sphericalangle\left(Z_{0} X_{0}, Z_{0} Y_{0}\right)$, or, equivalently, $\sphericalangle\left(X X_{0}, Y Y_{0}\right)=\sphericalangle\left(Z_{0} X_{0}, Z_{0} Y_{0}\right)$.

Recall that $Y Z$ and $Y_{0} Z_{0}$ are the perpendicular bisectors of $A D$ and $A I$, respectively. Then, the vector $\vec{x}$ perpendicular to $Y Z$ and shifting the line $Y_{0} Z_{0}$ to $Y Z$ is equal to $\frac{1}{2} \overrightarrow{I D}$. Define the shifting vectors $\vec{y}=\frac{1}{2} \overrightarrow{I E}, \vec{z}=\frac{1}{2} \overrightarrow{I F}$ similarly. Consider now the triangle $U V W$ formed by the perpendiculars to $A I, B I$, and $C I$ through $D, E$, and $F$, respectively (see figure below). This is another triangle whose sides are parallel to the corresponding sides of $X Y Z$.

Claim 2. $\overrightarrow{I U}=2 \overrightarrow{X_{0} X}, \overrightarrow{I V}=2 \overrightarrow{Y_{0} Y}, \overrightarrow{I W}=2 \overrightarrow{Z_{0} Z}$.

Proof. We prove one of the relations, the other proofs being similar. To prove the equality of two vectors it suffices to project them onto two non-parallel axes and check that their projections are equal.

The projection of $\overrightarrow{X_{0} X}$ onto $I B$ equals $\vec{y}$, while the projection of $\overrightarrow{I U}$ onto $I B$ is $\overrightarrow{I E}=2 \vec{y}$. The projections onto the other axis $I C$ are $\vec{z}$ and $\overrightarrow{I F}=2 \vec{z}$. Then $\overrightarrow{I U}=2 \overrightarrow{X_{0} X}$ follows.

Notice that the line $\ell$ is the Simson line of the point $I$ with respect to the triangle $U V W$; thus $U, V, W$, and $I$ are concyclic. It follows from Claim 2 that $\sphericalangle\left(X X_{0}, Y Y_{0}\right)=\sphericalangle(I U, I V)=$ $\sphericalangle(W U, W V)=\sphericalangle\left(Z_{0} X_{0}, Z_{0} Y_{0}\right)$, and we are done.

<img_3588>
Let $I_{a}, I_{b}$, and $I_{c}$ be the excentres of triangle $A B C$ corresponding to $A, B$, and $C$, respectively. Also, let $u, v$, and $w$ be the lines through $D, E$, and $F$ which are perpendicular to $A I, B I$, and $C I$, respectively, and let $U V W$ be the triangle determined by these lines, where $u=V W, v=U W$ and $w=U V$ (see figure above).

Notice that the line $u$ is the reflection of $I_{b} I_{c}$ in the line $x$, because $u, x$, and $I_{b} I_{c}$ are perpendicular to $A D$ and $x$ is the perpendicular bisector of $A D$. Likewise, $v$ and $I_{a} I_{c}$ are reflections of each other in $y$, while $w$ and $I_{a} I_{b}$ are reflections of each other in $z$. It follows that $X, Y$, and $Z$ are the midpoints of $U I_{a}, V I_{b}$ and $W I_{c}$, respectively, and that the triangles $U V W$, $X Y Z$ and $I_{a} I_{b} I_{c}$ are either translates of each other or homothetic with a common homothety centre.

Construct the points $T$ and $S$ such that the quadrilaterals $U V I W, X Y T Z$ and $I_{a} I_{b} S I_{c}$ are homothetic. Then $T$ is the midpoint of $I S$. Moreover, note that $\ell$ is the Simson line of the point $I$ with respect to the triangle $U V W$, hence $I$ belongs to the circumcircle of the triangle $U V W$, therefore $T$ belongs to $\Omega$.



Consider now the homothety or translation $h_{1}$ that maps $X Y Z T$ to $I_{a} I_{b} I_{c} S$ and the homothety $h_{2}$ with centre $I$ and factor $\frac{1}{2}$. Furthermore, let $h=h_{2} \circ h_{1}$. The transform $h$ can be a homothety or a translation, and

$$
h(T)=h_{2}\left(h_{1}(T)\right)=h_{2}(S)=T
$$

hence $T$ is a fixed point of $h$. So, $h$ is a homothety with centre $T$. Note that $h_{2}$ maps the excentres $I_{a}, I_{b}, I_{c}$ to $X_{0}, Y_{0}, Z_{0}$ defined in the preamble. Thus the centre $T$ of the homothety taking $X Y Z$ to $X_{0} Y_{0} Z_{0}$ belongs to $\Omega$, and this completes the proof.