Two circles touch each other at point $A$, the point opposite to $A$ in the first circle is $B$, and in the second circle is $C$. A line rotating around point $A$ intersects the first circle at $D$ and the second circle at $E$. Determine the position of the line such that the area of the convex quadrilateral defined by points $B, C, D, E$ is maximized.
a) The two circles can touch each other in two different ways. In the case of external tangency, the order of the points on the line connecting the centers is $B, A, C$, and $D, E$ are on the two sides of this line, with the vertices of the convex quadrilateral in the order $B D C E$. By cutting this quadrilateral with the diagonal $B C$ into two triangles, the common base of these triangles is the sum of the diameters of the circles, $d_{1}$ and $d_{2}$, and the heights $m_{1}$ and $m_{2}$ are the perpendicular heights from $B C$ in the triangles $A B D$ and $A C E$, respectively. These two triangles are similar because the angles at $A$ are vertical angles, and the angles at $D$ and $E$ are right angles by Thales' theorem. Thus, $m_{1}: m_{2}=d_{1}: d_{2}$, $m_{2}=m_{1} d_{2} / d_{1}$, and the area in question is

$$
t_{a}=\left(d_{1}+d_{2}\right)\left(\frac{m_{1}}{2}+\frac{m_{2}}{2}\right)=\frac{m_{1}}{2} \cdot\left(d_{1}+d_{2}\right)\left(1+\frac{d_{2}}{d_{1}}\right)=m_{1} \cdot \frac{\left(d_{1}+d_{2}\right)^{2}}{2 d_{1}}
$$

By changing the angle $\alpha = \angle B A D$, only the factor $m_{1}$ changes in the expression for $t_{a}$, so $t_{a}$ is maximized when $D$ is farthest from $B C$. In this case, $D$ is the endpoint of the diameter of the first circle perpendicular to $A B$, making $A B D$ an isosceles right triangle, $\alpha=45^{\circ}, m_{1}=d_{1} / 2$, and the maximum area is $t_{a, \max }=\left(d_{1}+d_{2}\right)^{2} / 4$.
![](https://cdn.mathpix.com/cropped/2024_05_02_3170f78b0c2613f0ccd5g-1.jpg?height=454&width=1088&top_left_y=716&top_left_x=541)

b) In the case of internal tangency, $B$ and $C$ are on the same side of point $A$, and $D$ and $E$ are on the same side of the line $A B$. The similarity mentioned in case a) also holds here, and $B D$ and $C E$ are parallel because they are perpendicular to $A D$. The vertices of the convex quadrilateral are in the order $B C E D$. The area is equal to the difference in the areas of triangles $A C E$ and $A B D$ (the notation was chosen such that $d_{2} > d_{1}$):

$$
t_{b}=\frac{d_{2} m_{2}}{2}-\frac{d_{1} m_{1}}{2}=\frac{d_{2}^{2} m_{1}}{2 d_{1}}-\frac{d_{1} m_{1}}{2}=m_{1} \cdot \frac{d_{2}^{2}-d_{1}^{2}}{2 d_{1}}
$$

and its maximum value, similarly to case a), is achieved when $\alpha=45^{\circ}, m_{1}=d_{1} / 2$: $t_{b, \max }=\left(d_{2}^{2}-d_{1}^{2}\right) / 4$.

Schinagl Gábor (Budapest, Képző- és Iparművészeti Gimn., II. o. t.)