We have a paper sheet in the shape of an equilateral triangle, with a side length of $20 \mathrm{~cm}$. We want to cut out a regular 12-sided (dodecagon) sheet in one piece. What is the maximum possible length of its side, and what fraction of the sheet will be wasted in this case? (In addition to the calculation, the solution also requires proving that a regular 12-sided sheet with a side longer than the calculated one cannot be cut out in one piece!
1. We assert that the maximum area of a regular 12-sided polygon cut out from a given triangular disk $ABC = H$ has a vertex on each side of the triangle. If a regular 12-sided polygon $S_1$ is entirely within $H$, it can be enlarged from its center $O_1$ to a polygon $S_2$ such that at least one of its vertices (the farthest from $O_1$) lies on the perimeter of $H$, but no point of $S_2$ lies outside $H$. Let one such vertex be $D_2$ and the side of $H$ it lies on be $AB$. If $S_2$ has no vertices on the sides $AC$ or $BC$, it can be further enlarged from the center $D_2$ to a polygon $S_3$ such that $S_3$ has a vertex $E_3$ on $AC$. If $S_3$ is entirely in the half-plane containing $A$ and divided by $BC$, then by an appropriate enlargement from the center $A$, a regular 12-sided polygon $S_4$ can be obtained with a vertex $F$ on the side $BC$ of the disk; let the vertices obtained from $D_2$ and $E_3$ now be $D$ and $E$.

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For now, assume that none of the sides of $S_4$ starting from $D$ coincide with $AB$. Then consider the direction angles of the consecutive sides of $S_4$ from $D$ to $F$ relative to $AB$ as the base direction. We want to show that $F$ is the fourth vertex of $S_4$ from $D$. Since the exterior angle of a regular 12-sided polygon is $360^\circ / 12 = 30^\circ$, the direction angle of the first side $(D D')$ is between $0^\circ$ and $30^\circ$, the second side $(D' D'')$ is $30^\circ$ more, so between $30^\circ$ and $60^\circ$, and the $k$-th side is between $(k-1) \cdot 30^\circ$ and $k \cdot 30^\circ$. On the other hand, the direction angle of $BC$ is $120^\circ = 4 \cdot 30^\circ$, which falls between the direction angles of the 4th and 5th sides from $D$, thus confirming our statement. (The 5th side is $F F'$.) The line of the 5th side $F F'$ already intersects $AB$ beyond $B$. Similarly, the position of vertex $E$ from $F$ and from $D$ is also the 4th, so the diagonals $DF$, $FE$, and $ED$ are equal.

A more precise examination also reveals that the direction angle $\varepsilon$ of the side $F F'$ relative to $BC$ is equal to the angle $D' D B$, so $\angle FDB = \angle FDD' + \varepsilon = \angle EFF' + \varepsilon = \angle EFC = \angle DEA$, thus the triangles $DFB$, $FEC$, and $EDA$ (with the same orientation) are congruent and can be mapped to each other by $120^\circ$ rotations around the center $O_4$ of triangle $DFE$. However, the cyclic $120^\circ$ rotations of $A$, $B$, and $C$ are only possible around the center $O$ of $H$, so the center $O_4$ of $S_4$ coincides with $O$.

We can also assume that $\angle D' DB = \angle E''' DA$—this is just a matter of notation—so $DB \geq DA$.

It is now almost obvious that a regular 12-sided polygon larger than $S_4$ can also be cut out from $H$. We can achieve this by rotating $S_4$ around $O$ until $D D'$ is parallel to $AB$—during this process, $D$ and the entire $S_4$ move into the interior of $H$—and then enlarging this position from $O$. By increasing the ratio, the new positions of $D$ and $D'$ (or $F$, $F'$, $E$, $E'$) simultaneously reach the perimeter of the triangle, and a larger 12-sided polygon cannot be cut out, as the radius of its inscribed circle would equal the radius of the circle inscribed in $H$.

This situation would have immediately followed the enlargements if one side of $S_1$ were parallel to $AB$, so $D D'$ or $D E'''$ would coincide with $AB$.

2. In the obtained situation, half of the angle $ODA$ is equal to the interior angle of the 12-sided polygon, so $BOD$ is an isosceles triangle, and $D$ can be found by intersecting the circle centered at $B$ with radius $BO$. Thus, $D D' = 2(DB - GB) = 2 \cdot OB - AB = \frac{4}{3} CG - AB = \left(\frac{2\sqrt{3}}{3} - 1\right) AB = 30.9 \text{ mm}$ (where $G$ is the midpoint of $AB$).

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3. It is known that the area of a regular 12-sided polygon is three times the area of the square constructed on the radius of the circumscribed circle: $3 \cdot OD^2$.

$3 \cdot OD^2 = 3(O G^2 + D G^2) = (2 - \sqrt{3}) AB^2$. The area of the triangular disk is $(\sqrt{3} / 4) \cdot AB^2$, and the waste is $(5 \sqrt{3} - 8) / 4 \cdot AB^2$, which is $5 - \frac{8}{\sqrt{3}} = 0.381$ of the area of the given disk.