Subfield: Geometry


Question: Let $A B C D$ be a cyclic quadrilateral with no two sides parallel. Let $K, L, M$, and $N$ be points lying on sides $A B, B C, C D$, and $D A$, respectively, such that $K L M N$ is a rhombus with $K L \| A C$ and $L M \| B D$. Let $\omega_{1}, \omega_{2}, \omega_{3}$, and $\omega_{4}$ be the incircles of triangles $A N K$, $B K L, C L M$, and $D M N$, respectively. Prove that the internal common tangents to $\omega_{1}$ and $\omega_{3}$ and the internal common tangents to $\omega_{2}$ and $\omega_{4}$ are concurrent.

Solution:
Let $I_{i}$ be the center of $\omega_{i}$, and let $r_{i}$ be its radius for $i=1,2,3,4$. Denote by $T_{1}$ and $T_{3}$ the points of tangency of $\omega_{1}$ and $\omega_{3}$ with $N K$ and $L M$, respectively. Suppose that the internal common tangents to $\omega_{1}$ and $\omega_{3}$ meet at point $S$, which is the center of homothety $h$ with negative ratio (namely, with ratio $-\frac{r_{3}}{r_{1}}$ ) mapping $\omega_{1}$ to $\omega_{3}$. This homothety takes $T_{1}$ to $T_{3}$ (since the tangents to $\omega_{1}$ and $\omega_{3}$ at $T_{1}$ to $T_{3}$ are parallel), hence $S$ is a point on the segment $T_{1} T_{3}$ with $T_{1} S: S T_{3}=r_{1}: r_{3}$.

Construct segments $S_{1} S_{3} \| K L$ and $S_{2} S_{4} \| L M$ through $S$ with $S_{1} \in N K, S_{2} \in K L$, $S_{3} \in L M$, and $S_{4} \in M N$. Note that $h$ takes $S_{1}$ to $S_{3}$, hence $I_{1} S_{1} \| I_{3} S_{3}$, and $S_{1} S: S S_{3}=r_{1}: r_{3}$. We will prove that $S_{2} S: S S_{4}=r_{2}: r_{4}$ or, equivalently, $K S_{1}: S_{1} N=r_{2}: r_{4}$. This will yield the problem statement; indeed, applying similar arguments to the intersection point $S^{\prime}$ of the internal common tangents to $\omega_{2}$ and $\omega_{4}$, we see that $S^{\prime}$ satisfies similar relations, and there is a unique point inside $K L M N$ satisfying them. Therefore, $S^{\prime}=S$.
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Further, denote by $I_{A}, I_{B}, I_{C}, I_{D}$ and $r_{A}, r_{B}, r_{C}, r_{D}$ the incenters and inradii of triangles $D A B, A B C, B C D$, and $C D A$, respectively. One can shift triangle $C L M$ by $\overrightarrow{L K}$ to glue it with triangle $A K N$ into a quadrilateral $A K C^{\prime} N$ similar to $A B C D$. In particular, this shows that $r_{1}: r_{3}=r_{A}: r_{C}$; similarly, $r_{2}: r_{4}=r_{B}: r_{D}$. Moreover, the same shift takes $S_{3}$ to $S_{1}$, and it also takes $I_{3}$ to the incenter $I_{3}^{\prime}$ of triangle $K C^{\prime} N$. Since $I_{1} S_{1} \| I_{3} S_{3}$, the points $I_{1}, S_{1}, I_{3}^{\prime}$ are collinear. Thus, in order to complete the solution, it suffices to apply the following Lemma to quadrilateral $A K C^{\prime} N$.

Lemma 1. Let $A B C D$ be a cyclic quadrilateral, and define $I_{A}, I_{C}, r_{B}$, and $r_{D}$ as above. Let $I_{A} I_{C}$ meet $B D$ at $X$; then $B X: X D=r_{B}: r_{D}$.

Proof. Consider an inversion centered at $X$; the images under that inversion will be denoted by primes, e.g., $A^{\prime}$ is the image of $A$.

By properties of inversion, we have

$$
\angle I_{C}^{\prime} I_{A}^{\prime} D^{\prime}=\angle X I_{A}^{\prime} D^{\prime}=\angle X D I_{A}=\angle B D A / 2=\angle B C A / 2=\angle A C I_{B} .
$$

We obtain $\angle I_{A}^{\prime} I_{C}^{\prime} D^{\prime}=\angle C A I_{B}$ likewise; therefore, $\triangle I_{C}^{\prime} I_{A}^{\prime} D^{\prime} \sim \triangle A C I_{B}$. In the same manner, we get $\triangle I_{C}^{\prime} I_{A}^{\prime} B^{\prime} \sim \triangle A C I_{D}$, hence the quadrilaterals $I_{C}^{\prime} B^{\prime} I_{A}^{\prime} D^{\prime}$ and $A I_{D} C I_{B}$ are also similar. But the diagonals $A C$ and $I_{B} I_{D}$ of quadrilateral $A I_{D} C I_{B}$ meet at a point $Y$ such that $I_{B} Y$ :



$Y I_{D}=r_{B}: r_{D}$. By similarity, we get $D^{\prime} X: B^{\prime} X=r_{B}: r_{D}$ and hence $B X: X D=D^{\prime} X:$ $B^{\prime} X=r_{B}: r_{D}$.
This solution is based on the following general Lemma.

Lemma 2. Let $E$ and $F$ be distinct points, and let $\omega_{i}, i=1,2,3,4$, be circles lying in the same halfplane with respect to $E F$. For distinct indices $i, j \in\{1,2,3,4\}$, denote by $O_{i j}^{+}$ (respectively, $O_{i j}^{-}$) the center of homothety with positive (respectively, negative) ratio taking $\omega_{i}$ to $\omega_{j}$. Suppose that $E=O_{12}^{+}=O_{34}^{+}$and $F=O_{23}^{+}=O_{41}^{+}$. Then $O_{13}^{-}=O_{24}^{-}$.

Proof. Applying Monge's theorem to triples of circles $\omega_{1}, \omega_{2}, \omega_{4}$ and $\omega_{1}, \omega_{3}, \omega_{4}$, we get that both points $O_{24}^{-}$and $O_{13}^{-}$lie on line $E O_{14}^{-}$. Notice that this line is distinct from $E F$. Similarly we obtain that both points $O_{24}^{-}$and $O_{13}^{-}$lie on $\mathrm{FO}_{34}^{-}$. Since the lines $E O_{14}^{-}$and $\mathrm{FO}_{34}^{-}$are distinct, both points coincide with the meeting point of those lines.
<img_3961>



Turning back to the problem, let $A B$ intersect $C D$ at $E$ and let $B C$ intersect $D A$ at $F$. Assume, without loss of generality, that $B$ lies on segments $A E$ and $C F$. We will show that the points $E$ and $F$, and the circles $\omega_{i}$ satisfy the conditions of Lemma 2, so the problem statement follows. In the sequel, we use the notation of $O_{i j}^{ \pm}$from the statement of Lemma 2, applied to circles $\omega_{1}, \omega_{2}, \omega_{3}$, and $\omega_{4}$.

Using the relations $\triangle E C A \sim \triangle E B D, K N \| B D$, and $M N \| A C$. we get

$$
\frac{A N}{N D}=\frac{A N}{A D} \cdot \frac{A D}{N D}=\frac{K N}{B D} \cdot \frac{A C}{N M}=\frac{A C}{B D}=\frac{A E}{E D}
$$

Therefore, by the angle bisector theorem, point $N$ lies on the internal angle bisector of $\angle A E D$. We prove similarly that $L$ also lies on that bisector, and that the points $K$ and $M$ lie on the internal angle bisector of $\angle A F B$.

Since $K L M N$ is a rhombus, points $K$ and $M$ are symmetric in line $E L N$. Hence, the convex quadrilateral determined by the lines $E K, E M, K L$, and $M L$ is a kite, and therefore it has an incircle $\omega_{0}$. Applying Monge's theorem to $\omega_{0}, \omega_{2}$, and $\omega_{3}$, we get that $O_{23}^{+}$lies on $K M$. On the other hand, $O_{23}^{+}$lies on $B C$, as $B C$ is an external common tangent to $\omega_{2}$ and $\omega_{3}$. It follows that $F=O_{23}^{+}$. Similarly, $E=O_{12}^{+}=O_{34}^{+}$, and $F=O_{41}^{+}$.