Subfield: Geometry


Question: A triangle $A B C$ is inscribed in a circle $\omega$. A variable line $\ell$ chosen parallel to $B C$ meets segments $A B$, $A C$ at points $D, E$ respectively, and meets $\omega$ at points $K, L$ (where $D$ lies between $K$ and $E$ ). Circle $\gamma_{1}$ is tangent to the segments $K D$ and $B D$ and also tangent to $\omega$, while circle $\gamma_{2}$ is tangent to the segments $L E$ and $C E$ and also tangent to $\omega$. Determine the locus, as $\ell$ varies, of the meeting point of the common inner tangents to $\gamma_{1}$ and $\gamma_{2}$.

Solution:
Let $P$ be the meeting point of the common inner tangents to $\gamma_{1}$ and $\gamma_{2}$. Also, let $b$ be the angle bisector of $\angle B A C$. Since $K L \| B C, b$ is also the angle bisector of $\angle K A L$.



Let $\mathfrak{H}$ be the composition of the symmetry $\mathfrak{S}$ with respect to $b$ and the inversion $\mathfrak{I}$ of centre $A$ and ratio $\sqrt{A K \cdot A L}$ (it is readily seen that $\mathfrak{S}$ and $\mathfrak{I}$ commute, so since $\mathfrak{S}^{2}=\mathfrak{I}^{2}=\mathrm{id}$, then also $\mathfrak{H}^{2}=\mathrm{id}$, the identical transformation). The elements of the configuration interchanged by $\mathfrak{H}$ are summarized in Table I.



Let $O_{1}$ and $O_{2}$ be the centres of circles $\gamma_{1}$ and $\gamma_{2}$. Since the circles $\gamma_{1}$ and $\gamma_{2}$ are determined by their construction (in a unique way), they are interchanged by $\mathfrak{H}$, therefore the rays $A O_{1}$ and $A O_{2}$ are symmetrical with respect to $b$. Denote by $\rho_{1}$ and $\rho_{2}$ the radii of $\gamma_{1}$ and $\gamma_{2}$. Since $\angle O_{1} A B=\angle O_{2} A C$, we have $\rho_{1} / \rho_{2}=A O_{1} / A O_{2}$. On the other hand, from the definition of $P$ we have $O_{1} P / O_{2} P=$ $\rho_{1} / \rho_{2}=A O_{1} / A O_{2}$; this means that $A P$ is the angle bisector of $\angle O_{1} A O_{2}$ and therefore of $\angle B A C$.



The limiting, degenerated, cases are when the parallel line passes through $A$ - when $P$ coincides with $A$; respectively when the parallel line is $B C$ - when $P$ coincides with the foot $A^{\prime} \in B C$ of the angle bisector of $\angle B A C$ (or any other point on $B C$ ). By continuity, any point $P$ on the open segment $A A^{\prime}$ is obtained for some position of the parallel, therefore the locus is the open segment $A A^{\prime}$ of the angle bisector $b$ of $\angle B A C$.

|         point $K$        | $\longleftrightarrow$ |    point $L$    |
|:------------------------:|:---------------------:|:---------------:|
|        line $K L$        | $\longleftrightarrow$ | circle $\omega$ |
|         ray $A B$        | $\longleftrightarrow$ |    ray $A C$    |
|         point $B$        | $\longleftrightarrow$ |    point $E$    |
|         point $C$        | $\longleftrightarrow$ |    point $D$    |
|       segment $B D$      | $\longleftrightarrow$ |  segment $E C$  |
|         arc $B K$        | $\longleftrightarrow$ |  segment $E L$  |
| $\operatorname{arc} C L$ | $\longleftrightarrow$ |  segment $D K$  |
TABLE I: Elements interchanged by $\mathfrak{H}$.



![](https://cdn.mathpix.com/cropped/2023_12_21_e0379f976a2878f867f0g-1.jpg?height=654&width=731&top_left_y=438&top_left_x=1231)

Final answer: the open segment $AA'$ of the angle bisector $b$ of $\angle B A C$