In the symmetric trapezoid $A B C D$ ($A B \| C D$ and $A B < C D$), the height is $m$. Furthermore, it is known that the diagonal $B D$ is perpendicular to the side $B C$, and that the circles constructed over the sides $B C$ and $A D$ as diameters (externally) touch each other. The task is to calculate $A B$ and $C D$!
Let $B H=m, A B=x, C D=y$.

The circles corresponding to the diameters $A D$ and $B C$ are symmetric with respect to the axis of symmetry $E F$ of the trapezoid; therefore, their point of tangency, $I$, lies on the line $E F$. ( $E$ is the midpoint of $A B$, and $F$ is the midpoint of $C D$.) The foot $H$ of the altitude $B H$ lies on the circle with diameter $B C$. (By Thales' theorem.) With respect to this circle,

$$
\overline{F I}^{2}=\overline{F H} \cdot \overline{F C}
$$

However, $\sqrt{1}$

$$
F I=\frac{m}{2}, \quad F H=E B=\frac{x}{2}, \quad F C=\frac{y}{2}
$$

Accordingly,

$$
m^{2}=x y \ldots
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_03c62673134b48d5de57g-1.jpg?height=306&width=526&top_left_y=725&top_left_x=797)

In the right triangle $C B D$, $\overline{B H^{2}}=\overline{D H} \cdot \overline{H C}$. Since

$$
D H=\frac{x+y}{2}, \quad H C=\frac{y-x}{2}, \quad 4 m^{2}=y^{2}-x^{2} \ldots
$$

If we write 1) in the form $-x^{2} y^{2}=-m^{4}$, then based on 1) and 2), we can say that $y^{2}$ and $-x^{2}$ are the roots of the equation

$$
z^{2}-4 m^{2} z-m^{4}=0 \ldots
$$

and specifically,

$$
y^{2}=m^{2}(2+\sqrt{5}) \quad \text { and } \quad-x^{2}=m^{2}(2-\sqrt{5}) \text {. }
$$

Since $y$ and $x$ can only be positive,

$$
y=m \sqrt{2+\sqrt{5}}, \quad x=m \sqrt{\sqrt{5}-2}
$$

György Mendelsohn (Izr. g. VI. o. Bp. )[^0]


[^0]:    ${ }^{1}$ If a perpendicular is drawn from $I$ to $E F$, it passes through the center of the circle, i.e., the midpoint of $B C$ (or $A D$). This implies: 1) $I$ is the midpoint of $E F$; 2) $B C=2 I O=\frac{x+y}{2}$.