Subfield: Geometry


Question: Three circular arcs $\gamma_{1}, \gamma_{2}$, and $\gamma_{3}$ connect the points $A$ and $C$. These arcs lie in the same half-plane defined by line $A C$ in such a way that $\operatorname{arc} \gamma_{2}$ lies between the $\operatorname{arcs} \gamma_{1}$ and $\gamma_{3}$. Point $B$ lies on the segment $A C$. Let $h_{1}, h_{2}$, and $h_{3}$ be three rays starting at $B$, lying in the same half-plane, $h_{2}$ being between $h_{1}$ and $h_{3}$. For $i, j=1,2,3$, denote by $V_{i j}$ the point of intersection of $h_{i}$ and $\gamma_{j}$ (see the Figure below).

Denote by $\overparen{V_{i j} V_{k j}} \overparen{k_{k \ell} V_{i \ell}}$ the curved quadrilateral, whose sides are the segments $V_{i j} V_{i \ell}, V_{k j} V_{k \ell}$ and $\operatorname{arcs} V_{i j} V_{k j}$ and $V_{i \ell} V_{k \ell}$. We say that this quadrilateral is circumscribed if there exists a circle touching these two segments and two arcs.

Prove that if the curved quadrilaterals $\sqrt{V_{11} V_{21}} \sqrt{V_{22} V_{12}}, \sqrt{V_{12} V_{22}} \sqrt{V_{23} V_{13}}, \sqrt{V_{21} V_{31}} \sqrt{V_{32} V_{22}}$ are circumscribed, then the curved quadrilateral $\overparen{V_{22} V_{32}} \overparen{V_{33} V_{23}}$ is circumscribed, too.

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Fig. 1

Solution:
Denote by $O_{i}$ and $R_{i}$ the center and the radius of $\gamma_{i}$, respectively. Denote also by $H$ the half-plane defined by $A C$ which contains the whole configuration. For every point $P$ in the half-plane $H$, denote by $d(P)$ the distance between $P$ and line $A C$. Furthermore, for any $r>0$, denote by $\Omega(P, r)$ the circle with center $P$ and radius $r$.

Lemma 1. For every $1 \leq i<j \leq 3$, consider those circles $\Omega(P, r)$ in the half-plane $H$ which are tangent to $h_{i}$ and $h_{j}$.

(a) The locus of the centers of these circles is the angle bisector $\beta_{i j}$ between $h_{i}$ and $h_{j}$.

(b) There is a constant $u_{i j}$ such that $r=u_{i j} \cdot d(P)$ for all such circles.

Proof. Part (a) is obvious. To prove part (b), notice that the circles which are tangent to $h_{i}$ and $h_{j}$ are homothetic with the common homothety center $B$ (see Fig. 2). Then part (b) also becomes trivial.

Lemma 2. For every $1 \leq i<j \leq 3$, consider those circles $\Omega(P, r)$ in the half-plane $H$ which are externally tangent to $\gamma_{i}$ and internally tangent to $\gamma_{j}$.

(a) The locus of the centers of these circles is an ellipse arc $\varepsilon_{i j}$ with end-points $A$ and $C$.

(b) There is a constant $v_{i j}$ such that $r=v_{i j} \cdot d(P)$ for all such circles.

Proof. (a) Notice that the circle $\Omega(P, r)$ is externally tangent to $\gamma_{i}$ and internally tangent to $\gamma_{j}$ if and only if $O_{i} P=R_{i}+r$ and $O_{j}=R_{j}-r$. Therefore, for each such circle we have

$$
O_{i} P+O_{j} P=O_{i} A+O_{j} A=O_{i} C+O_{j} C=R_{i}+R_{j} .
$$

Such points lie on an ellipse with foci $O_{i}$ and $O_{j}$; the diameter of this ellipse is $R_{i}+R_{j}$, and it passes through the points $A$ and $C$. Let $\varepsilon_{i j}$ be that arc $A C$ of the ellipse which runs inside the half plane $H$ (see Fig. 3.)

This ellipse arc lies between the $\operatorname{arcs} \gamma_{i}$ and $\gamma_{j}$. Therefore, if some point $P$ lies on $\varepsilon_{i j}$, then $O_{i} P>R_{i}$ and $O_{j} P<R_{j}$. Now, we choose $r=O_{i} P-R_{i}=R_{j}-O_{j} P>0$; then the



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Fig. 2

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Fig. 3

circle $\Omega(P, r)$ touches $\gamma_{i}$ externally and touches $\gamma_{j}$ internally, so $P$ belongs to the locus under investigation.

(b) Let $\vec{\rho}=\overrightarrow{A P}, \vec{\rho}_{i}=\overrightarrow{A O_{i}}$, and $\vec{\rho}_{j}=\overrightarrow{A O_{j}}$; let $d_{i j}=O_{i} O_{j}$, and let $\vec{v}$ be a unit vector orthogonal to $A C$ and directed toward $H$. Then we have $\left|\vec{\rho}_{i}\right|=R_{i},\left|\vec{\rho}_{j}\right|=R_{j},\left|\overrightarrow{O_{i} P}\right|=$ $\left|\vec{\rho}-\vec{\rho}_{i}\right|=R_{i}+r,\left|\overrightarrow{O_{j} P}\right|=\left|\vec{\rho}-\vec{\rho}_{j}\right|=R_{j}-r$, hence

$$
\begin{gathered}
\left(\vec{\rho}-\vec{\rho}_{i}\right)^{2}-\left(\vec{\rho}-\vec{\rho}_{j}\right)^{2}=\left(R_{i}+r\right)^{2}-\left(R_{j}-r\right)^{2} \\
\left(\vec{\rho}_{i}^{2}-\vec{\rho}_{j}^{2}\right)+2 \vec{\rho} \cdot\left(\vec{\rho}_{j}-\vec{\rho}_{i}\right)=\left(R_{i}^{2}-R_{j}^{2}\right)+2 r\left(R_{i}+R_{j}\right) \\
d_{i j} \cdot d(P)=d_{i j} \vec{v} \cdot \vec{\rho}=\left(\vec{\rho}_{j}-\vec{\rho}_{i}\right) \cdot \vec{\rho}=r\left(R_{i}+R_{j}\right)
\end{gathered}
$$

Therefore,

$$
r=\frac{d_{i j}}{R_{i}+R_{j}} \cdot d(P)
$$

and the value $v_{i j}=\frac{d_{i j}}{R_{i}+R_{j}}$ does not depend on $P$.

Lemma 3. The curved quadrilateral $\mathcal{Q}_{i j}=\overparen{V_{i, j} V_{i+1}, j} \overparen{V_{i+1, j+1} V_{i, j+1}}$ is circumscribed if and only if $u_{i, i+1}=v_{j, j+1}$.

Proof. First suppose that the curved quadrilateral $\mathcal{Q}_{i j}$ is circumscribed and $\Omega(P, r)$ is its inscribed circle. By Lemma 1 and Lemma 2 we have $r=u_{i, i+1} \cdot d(P)$ and $r=v_{j, j+1} \cdot d(P)$ as well. Hence, $u_{i, i+1}=v_{j, j+1}$.

To prove the opposite direction, suppose $u_{i, i+1}=v_{j, j+1}$. Let $P$ be the intersection of the angle bisector $\beta_{i, i+1}$ and the ellipse arc $\varepsilon_{j, j+1}$. Choose $r=u_{i, i+1} \cdot d(P)=v_{j, j+1} \cdot d(P)$. Then the circle $\Omega(P, r)$ is tangent to the half lines $h_{i}$ and $h_{i+1}$ by Lemma 1 , and it is tangent to the $\operatorname{arcs} \gamma_{j}$ and $\gamma_{j+1}$ by Lemma 2. Hence, the curved quadrilateral $\mathcal{Q}_{i j}$ is circumscribed.

By Lemma 3, the statement of the problem can be reformulated to an obvious fact: If the equalities $u_{12}=v_{12}, u_{12}=v_{23}$, and $u_{23}=v_{12}$ hold, then $u_{23}=v_{23}$ holds as well.