The sum of the angles of two polygons, expressed in degrees, is 21 times less 39 than the sum of all sides and all diagonals. The number of all diagonals is 99 more than the number of all sides. How many sides do the polygons have?
Let the number of sides of the polygons be denoted by $x$ and $y$. The sum of the angles in degrees is $(x-2) 180$ and $(y-2) 180$, and the total number of diagonals is $\frac{x(x-3)}{2}$ and $\frac{y(y-3)}{2}$. According to the problem,

\[
\begin{gathered}
(x-2) 180 + (y-2) 180 = 21 \left[ x + y + \frac{x(x-3)}{2} + \frac{y(y-3)}{2} \right] - 39 \\
\frac{x(x-3)}{2} + \frac{y(y-3)}{2} - (x + y) = 99
\end{gathered}
\]

From this,

\[
\begin{gathered}
7 x^{2} + 7 y^{2} - 127 x - 127 y + 454 = 0, \\
x^{2} + y^{2} - 5 x - 5 y - 198 = 0 .
\end{gathered}
\]

Subtracting the 7 times of \((2')\) from \((1')\), we get

\[
92 x + 92 y - 1840 = 0,
\]

which simplifies to

\[
x + y = \frac{1840}{92} = 20
\]

and thus

\[
x^{2} + y^{2} = (x + y)^{2} - 2 x y = 20^{2} - 2 x y - 400 - 2 x y
\]

Substituting the values of (3) and (4) into \((2')\),

\[
400 - 2 x y - 5 \cdot 20 - 198 = 0,
\]

from which we get

\[
x y = 51.
\]

From (3) and (5), it follows that

\[
x_{1} = y_{2} = 17, \quad x_{2} = y_{1} = 3
\]

Bayer Márta (Bp. XX., Bagi Ilona lg. I. o. t.)