$\mathrm{Az} A B=2 R$ diameter semicircle has a tangent drawn at point $B$. A secant from point $\mathrm{A}$ intersects the semicircle at $C$ and the tangent at point $D$, such that

$$
\overline{A C}^{2}+\overline{C D}^{2}+\overline{B D}^{2}=4 a^{2}
$$

Calculate the distance $A E=x$; $E$ is the projection of $C$ onto $A B$. For which value of $a^{2}$ does a solution exist and how many are there?
$R x$

$A B C$ is a right-angled triangle at $C$. According to the theorem for the legs of a right-angled triangle: $\overline{A C}^{2}=\overline{A B} \cdot \overline{A E}=$

![](https://cdn.mathpix.com/cropped/2024_05_02_7b6d3d8cf10632a3928ag-1.jpg?height=255&width=423&top_left_y=263&top_left_x=840)

Since $C E \| D B$, we can write: $C D: E B=A C: A E$. Therefore,

$$
\overline{C D}^{2}=\frac{\overline{E B}^{2} \cdot \overline{A C}^{2}}{C E^{2}}=\frac{(2 R-x)^{2} 2 R x}{x^{2}}=\frac{2 R(2 R-x)^{2}}{x}
$$

Furthermore: $D B: C E=A B: A E$ so

$$
\overline{D B}^{2}=\frac{\overline{C E}^{2} \cdot \overline{A B}^{2}}{\overline{A E}^{2}}=\frac{x(2 R-x) 4 R^{2}}{x^{2}}=\frac{(2 R-x) 4 R^{2}}{x}
$$

Based on these, our equation is:

$$
2 R \cdot x+\frac{2 R(2 R-x)^{2}}{x}+\frac{(2 R-x) 4 R^{2}}{x}=4 a^{2}
$$

and after appropriate rearrangement:

$$
f(x) \equiv R x^{2}-\left(3 R^{2}+a^{2}\right) x+4 R^{2}=0
$$

The value of $x$ that corresponds to our problem must be real, positive, and less than $2 R$. The roots of the equation $f(x)=0$ are real if

$$
\left(3 R^{2}+a^{2}\right)^{2}-16 R^{4} \equiv\left(3 R^{2}+a^{2}+4 R^{2}\right)\left(3 R^{2}+a^{2}-4 R^{2}\right) \geq 0
$$

The first factor is always positive; the second cannot be negative. Therefore, it must be that $a^{2} \geq R^{2}$.

The product of the roots is $4 R^{2}$ and their sum is $\frac{3 R^{2}+a^{2}}{R}$, both positive, so both roots are positive. $\sqrt{4 R^{2}}=2 R$ is the geometric mean of the roots; this implies that one root is smaller and the other larger than $2 R$, or both equal to $2 R$ in the boundary case.

Accordingly, the problem has only one solution if $a^{2} \geq R^{2}$, which is

$$
x=\frac{3 R^{2}+a^{2}-\sqrt{\left(a^{2}+7 R^{2}\right)\left(a^{2}-R^{2}\right)}}{2 R}
$$

Sándor Horváth (Br. Kemény Zsigmond g. VI. o. Bp.)