Given a semicircle with radius $R$ and diameter $AB$, the point $C$ on $AB$ is to be determined such that if the projection of a point $D$ on the semicircle arc is $C$, then $AC + AD = l$, where $l$ is a given length. For what values of $l$ does a solution exist?
Let $A D=x$. Then $A C=\frac{x^{2}}{2 R}$. 1.

Accordingly, we need to solve the equation

$$
x+\frac{x^{2}}{2 R}=l \quad \text { or } \quad x^{2}+2 R x-2 R l=0
$$

This equation always has real roots, and they are of opposite signs. Only the positive one can be valid, i.e.,

$$
x=\sqrt{R^{2}+2 R l}-R
$$

However, this value of $x$ is only valid if it is not greater than $2 R$; thus, it must be that

$$
\sqrt{R^{2}+2 R l} \leq 3 R, \quad \text { or } \quad 2 R l \leq 8 R^{2}, \quad \text { hence } \quad l \leq 4 R
$$

If $l=4 R$, then $x=2 R$, i.e., $A D=A C=2 R$.

Iván Steiner (Toldy Ferenc r. V. o. Bp. II.)

Note. If $D$ runs along the semicircular arc $\widehat{A B}$ from $A$ to $B$, then both $A D$ and $A C$ increase from 0 to $4 R$, their sum increases from 0 to $4 R$, and in the process, they take on every value exactly once. (The reader should examine the variation and plot the function

$$
y=\frac{x^{2}}{2 R}+x
$$

from $x=0$ to $x=2 R$.)[^0]


[^0]:    ${ }^{1}$ Indeed - By Euclid's theorem $-\overline{A D^{2}}=\overline{A B} \cdot \overline{A C}$.