We create a rectangular grid with steps of width $2c$. The grid openings are identical rectangles, with sides $a$ and $b$ (where $a > b$). We shoot a ball of radius $r$ perpendicularly to the plane of the grid. a) What is the probability that the ball will pass through the grid without touching it? b) With a constant grid opening, what step width should be chosen so that the above probability - for any ball - is always less than 1/2? Construct this step width.
It is sufficient to consider a rectangular aperture (grid aperture) with sides $a, b$ and the associated half-width $c$ of the slit. The possible area $T$ is the entire rectangle with sides $a+2c$ and $b+2c$.

![](https://cdn.mathpix.com/cropped/2024_05_02_9248375c5219ac0c69beg-1.jpg?height=453&width=585&top_left_y=275&top_left_x=778)

Figure 1

Favorable points for the center of the ball within the grid aperture are those points whose distance from the sides is at least $r$. (The shaded rectangle in the figure.) The favorable area is thus $t=(a-2r)(b-2r)$, and therefore the sought probability is

$$
v=\frac{t}{T}=\frac{(a-2r)(b-2r)}{(a+2c)(b+2c)}
$$

Since $v$ cannot be negative, it is necessary that $r<\frac{b}{2}<\frac{a}{2}$.

b) For $v$ to be less than $\frac{1}{2}$ for any small $r$ with fixed $a, b$ (i.e., at least $\frac{1}{2}$ when $r=0$), it is necessary that

$$
\frac{ab}{(a+2c)(b+2c)} \leq \frac{1}{2}
$$

The minimum value of $c$ is obtained from the equation

$$
\frac{ab}{(a+2c)(b+2c)}=\frac{1}{2}
$$

which simplifies to

$$
2ab=(a+2c)(b+2c)
$$

and further to

$$
(2c)^2 + (a+b) \cdot 2c - ab = 0
$$

from which

$$
2c = \frac{-(a+b) + \sqrt{(a+b)^2 + 4ab}}{2} = \sqrt{\left(\frac{a+b}{2}\right)^2 + ab} - \frac{a+b}{2}
$$

However, constructing $2c$ is simpler from (1) than from (2).

We construct the geometric mean of $a=AB$ and $2b=BD$: $BP$ using the Thales circle centered at $O_1$ over $AD$ (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_9248375c5219ac0c69beg-1.jpg?height=371&width=734&top_left_y=2254&top_left_x=671)

Figure 2

Let the circle with radius $QP$ centered at the midpoint $O_2$ of the distance $a+b=AC$ intersect the line $AD$ at points $Q$ and $R$. Then clearly

$$
\begin{gathered}
QA = CR = 2c \\
\text{since indeed } \sqrt{(a+2c)(b+2c)} = BP = \sqrt{a \cdot 2b}
\end{gathered}
$$

Otmár Solymoss (Kőszeg, Miklós Jurisich g. IV. o. t.)