Problem 1265

Step 1. Let $P$ be the point of intersection of the common external tangents to $\omega_{A}$ and $\omega_{D}$ (point $D$ can be at infinity), and $R$ be the point of intersection of the common external tangents to $\omega_{A}$ and $\omega_{C}$ (left figure). Note that $P$ lies on the line $A D$; $R$ - on $B D$.

We will show that the condition $\angle B I_{A} A=180^{\circ}-\angle I_{C} I_{A} I_{D}\left(^{*}\right)$ is equivalent to the fact that the line $I_{A}$, passing through $P$ and $R$, is a common tangent to the circles $\omega_{A}, \omega_{C}$, and $\omega_{D}$. Let for definiteness point $P$ lie on the ray $D A$ (other cases are similar). Since $\omega_{A}$ is inscribed in triangle $A B D$, then

$\angle A I_{A} B=90^{\circ}+1 / 2 \angle A D B$; on the other hand, $180^{\circ}-\angle I_{C} I_{A} I_{D}=\angle P I_{A} R$. Thus, (*) is equivalent to the equality $\angle P I_{A} R=90^{\circ}+1 / 2 \angle P D R$.

Let $J$ be the center of the inscribed circle of triangle PDR. Points $I_{A}$ and $J$ lie on the bisector of this triangle from point $D$, and (*) is equivalent to the equality $\angle P J R=\angle P I_{A} R$ - that is, the coincidence of points $J$ and $I_{A}$. This, in turn, means that the lines $P R$ and $R D$ are symmetric with respect to the line $I_{A} I_{C}$, and the lines $P D$ and $P R$ are symmetric with respect to $I_{A} I_{D}$, that is, that PR is tangent to all three circles $\omega_{A}, \omega_{C}$, and $\omega_{D}$ ( $\omega_{C}$ lies on the other side of $P R$ compared to the other two).

Similarly, the condition $\angle B I_{B} A+\angle I_{C} I_{B} I_{D}=180^{\circ}$ is equivalent to the fact that the circles $\omega_{A}, \omega_{C}$, and $\omega_{B}$ have a common tangent $I_{B}$, relative to which $\omega_{C}$ and $\omega_{D}$ lie on one side, and $\omega_{A}$ - on the other. It remains to prove that these two facts are equivalent.
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Step 2. Let the point of tangency of the circle $\omega_{A}$ with $A D$ be denoted as $A_{A D}$; similarly, denote other points of tangency (right figure). Note that

$A_{A D} D_{A D}=\left|A A_{A D}-A D_{A D}\right|=|1 / 2(A B+A D-B D)-1 / 2(A C+A D-C D)|=1 / 2|A B+C D-B D-A C| ;$ similarly $B_{B C} C_{B C}=1 / 2|A B+C D-B D-A C|=A_{A D} D_{A D}, A_{B D} C_{B D}=1 / 2|A D+B C-A B-C D|=B_{A C} D_{A C}$.

Now suppose that $l_{A}$ is tangent to the circles $\omega_{A}, \omega_{C}$, and $\omega_{B}$ at points $L_{A}, L_{C}$, and $L_{D}$, respectively.

Then $L_{A} L_{D}=A_{A D} D_{A D}=B_{B C} C_{B C}$ and

$L_{A} L_{C}=A_{B D} C_{B D}=D_{A C} B_{A C}$

Consider the circles $\omega_{B}^{\prime}, \omega_{C}^{\prime}$, and $\omega_{D}^{\prime}$ with centers $I_{B}^{\prime}, I_{C}^{\prime}$, and $I_{D}^{\prime}$, having the same radii as $\omega_{B}, \omega_{C}$, and $\omega_{D}$, respectively, and tangent to $l_{A}$ at points $L_{A}, L_{D}$, and $L_{C}$, respectively (where $\omega_{B}^{\prime}$ and $\omega_{C}^{\prime}$ lie on one side of $l_{A}$, and $\omega_{D}^{\prime}$ - on the other). Then the corresponding segments of the common tangents to $\omega_{B}^{\prime}, \omega_{C}^{\prime}, \omega_{D}^{\prime}$ and to $\omega_{B}, \omega_{C}, \omega_{D}$ have the same lengths (for $\omega_{C}$ and $\omega_{D}$ this is obvious, for the other pairs it follows from the above). From this, it easily follows that the corresponding sides of the triangles $I_{B} I_{C} I_{D}$ and $I_{B}^{\prime} I^{\prime} C^{\prime} I_{D}^{\prime}$ are equal (for example, $I_{B}^{\prime} I^{\prime}{ }_{C}=I_{B} I_{C}$ from the equality of quadrilaterals $I_{B}^{\prime} L_{A} L_{D} I^{\prime}{ }_{C}$ and $I_{B} B_{B C} C_{B C} I_{C}$). Therefore, the configurations of circles ( $\omega_{B}, \omega_{C}, \omega_{D}$ ) and ( $\omega_{B}^{\prime}, \omega_{C}^{\prime}, \omega_{D}^{\prime}$ ) are also equal. Since the circles in one triplet touch one line $l_{A}$, the same is true for the other triplet. This is what needed to be proved.

## | Authors: Khurmi A., Sudharshan K.V.

Given a cyclic quadrilateral $ABCD$ inscribed in a circle $\Omega$, find the geometric locus of the centers of the circles $\omega_{P D Q}$, where $P$ and $Q$ are such that the rays $BP$ and $BQ$ are symmetric with respect to the bisector of angle $B$.

## Solution

Let the lines $BP$ and $BQ$ be symmetric with respect to the bisector of angle $B$. Since $\angle ABP = \angle CBQ$, the arcs $AR$ and $CS$ are equal, i.e., $AC \parallel RS$. Therefore, the homothety with center $B$ translates triangle $BPQ$ into triangle $BRS$. Let their common tangent intersect $AC$ at point $X$, and the line $DX$ intersects

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Since $BX$ is the radical axis of circles $\Omega$ and $\omega_{BPQ}$, and $AC$ is the radical axis of circles $\omega_{BPQ}$ and $\omega_{DPQ}$, the line $DEX$ is the radical axis of circles $\Omega$ and $\omega_{DPQ}$. Therefore, $O$ lies on the circle $\omega_{DPQ}$ for any positions of points $P$ and $Q$. Thus, the geometric locus of the centers of the circles $\omega_{DPQ}$ consists of points on the perpendicular bisector of $DE$ for which the circle with center $O$

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interval.