From a rectangular sheet, we want to cut out two circular discs of equal radius, one at a time. What is the maximum possible radius of the circles? - What would it be if we wanted to cut out three circular discs of equal radius from the same sheet?
I. Let the sides of the original rectangle $T$ be $a$ and $b$, where $a \leq b$. Then the diameter of the largest circle that can be cut out from it in one piece is obviously $a$. Two such circles can be cut out if $b \geq 2a$ (Figure 1a), otherwise not.

If $a \leq b < 2a$, and the diameter of the two largest circles that can be cut out is $2r$, then

$$
\begin{gathered}
(a-2 r)^{2}+(b-2 r)^{2}=(2 r)^{2} \\
4 r^{2}-4(a+b) r+\left(a^{2}+b^{2}\right)=0
\end{gathered}
$$

and thus the diameter of the largest pair of circles is

$$
2 r=a+b-\sqrt{2 a b}
$$

Taking the square root with the + sign gives $2 r > a$, which does not fit, otherwise the roots are clearly real.
![](https://cdn.mathpix.com/cropped/2024_05_02_8dc9a3aca1b8c33dc258g-1.jpg?height=720&width=572&top_left_y=1584&top_left_x=343)

a) $\quad$ c)

![](https://cdn.mathpix.com/cropped/2024_05_02_8dc9a3aca1b8c33dc258g-1.jpg?height=694&width=658&top_left_y=1602&top_left_x=972)

Figure 2

II. Similarly, it follows that if $b \geq 3a$, three circles with diameter $a$ can be cut out from $T$, and if $b < 3a$, the diameter of the largest circle that can be cut out is $2r$, where

$$
2 r=a+b-\sqrt{2 a b}
$$

The product of the roots is positive, hence the written $r$ is also positive, since the term of the equation not containing $r$ is positive. Otherwise, the roots are clearly real.

For a constant $a$ and a decreasing $b$, the two circles on the base do not overlap as long as $b-4r \geq 0$. The condition for this, based on (2), is

$$
\begin{gathered}
4 \sqrt{3 a^{2}+2 a b} \geq 8 a+b \\
48 a^{2}+32 a b \geq 64 a^{2}+16 a b+b^{2} \\
\left(\frac{b}{a}\right)^{2}-16\left(\frac{b}{a}\right)+16 \leq 0 \\
8-4 \sqrt{3} \leq \frac{b}{a} \leq 8+4 \sqrt{3}
\end{gathered}
$$

and considering the previous constraint $b / a < 3$,

$$
(1.072 \approx) \quad 8-4 \sqrt{3} \leq b / a < 3
$$

For the case $1 \leq b / a < 8-4 \sqrt{3} (\approx 1.072)$, all three circles touch each other pairwise, but the $k_{3}$ circle only touches the side of length $a$. With the notation of Figure 2c,

$$
r+E F+G O_{2}+r=a, \quad r+O_{1} D+r=b
$$

and using the similarity of the triangles $O_{2} F G, O_{1} O_{3} D$, and $O_{1} F E$, and the ratio of the hypotenuses $O_{2} F: O_{1} O_{3}: O_{1} F = \sqrt{3}: 2: 1$:

$$
\begin{gathered}
2 r+x+\sqrt{3} \cdot \sqrt{r^{2}-x^{2}}=a \\
2 r+2 \sqrt{r^{2}-x^{2}}=b
\end{gathered}
$$

Expressing the square root from (4) and substituting it into (3),

$$
x=a-2 r-\sqrt{3}\left(\frac{b}{2}-r\right)=a-\frac{\sqrt{3}}{2} b-(2-\sqrt{3}) r
$$

and substituting this into the equation obtained by rearranging and squaring (4),

$$
\begin{gathered}
r^{2}-x^{2}=r^{2}-\left(a-\frac{\sqrt{3}}{2} b\right)^{2}+(2 a-\sqrt{3} b)(2-\sqrt{3}) r-(2-\sqrt{3})^{2} r^{2}= \\
=\left(\frac{b}{2}-r\right)^{2}=\frac{b^{2}}{4}-b r+r^{2} \\
(2-\sqrt{3})^{2} r^{2}-2(2-\sqrt{3})(a+b) r+\left(a^{2}-\sqrt{3} a b+b^{2}\right)=0
\end{gathered}
$$

The discriminant, and finally the radius of the circles,

$$
\begin{gathered}
4(2-\sqrt{3})^{2}(a+b)^{2}-4(2-\sqrt{3})^{2}\left(a^{2}-\sqrt{3} a b+b^{2}\right)=4(2-\sqrt{3})^{3} a b \\
r=(2+\sqrt{3})(a+b-\sqrt{(2+\sqrt{3}) a b})
\end{gathered}
$$

Specifically, for $a=b$, considering that

$$
\begin{gathered}
2+\sqrt{3}=(\sqrt{3}+1)^{2} / 2 \\
2 r=(2+\sqrt{3})(4-\sqrt{6}-\sqrt{2}) a \approx 0.509 a
\end{gathered}
$$