A right angle is formed by the rays $O A$ and $O B$; we draw a circle tangent to these rays within the angle; let us determine a point $M$ on the circumference of this circle such that the perimeter of the rectangle $M P O Q$ is equal to a given length $2 p$; $M P\|O B, \quad M Q\| O A$.
Let $M Q=x, M P=y, C M=r$. In the right-angled triangle $C D M$,

$$
r^{2}=\overline{C D}^{2}+\overline{D M}^{2}
$$

But $C D=x-r, D M=y-r$, so:

$$
r^{2}=(x-r)^{2}+(y-r)^{2}
$$

or

$$
x^{2}+y^{2}-2 r(x+p)+r^{2}=0
$$

According to the problem,

$$
x+p=p
$$

Substituting this into (1), we get:

$$
x^{2}+y^{2}+r^{2}-2 r p=0
$$

From the equations under (2) and (3), we get:

$$
\begin{aligned}
& x=\frac{1}{2}\left[p \pm \sqrt{p^{2}-2(p-r)^{2}}\right] \\
& x=\frac{1}{2}\left[p \mp \sqrt{p^{2}-2(p-r)^2}\right] .
\end{aligned}
$$

The problem is possible if

$$
p^{2}-2(p-r)^{2} \geqq 0
$$

or

$$
[p+(p-r) \sqrt{2}][p-(p-r) \sqrt{2}] \geqq 0
$$

This expression is $>0$ if the two factors have the same sign; the two factors cannot be negative, because if $p>r$, the first factor is positive, and if $p<r$, the second factor is positive.

The first factor $\geqq 0$ if $p \geqq r(2-\sqrt{2})$.

The second factor $\geqq 0$ if $p \leqq r(2+\sqrt{2})$.

Thus, the problem can be solved if:

$$
r(2-\sqrt{2}) \leqq p \leqq r(2+\sqrt{2})
$$

In (4), the left side $=0$ if

$$
p=r(2 \pm \sqrt{2})
$$

then

$$
x=y=\frac{p}{2}
$$

In this case, the problem has only one solution, and the quadrilateral is regular; otherwise, the problem has two solutions.

Construction. Measure the distances $O C=O D=p$ on the two legs of the right angle. The intersection of the line $C D$ with the circle is the sought point $M$. Indeed, both the triangles $M Q D$ and $M P C$ are similar to the isosceles triangle $C O D$, so $M Q=D Q$, and $M P=Q O$, hence $M Q+M P=D Q+Q O=D O=p$. The problem has two points that satisfy it if $D C$ is a secant of the circle, and one point if it is a tangent. These conditions are identical to those stated in the previous problem.

(Friedmann Bernát, S.-A.-Ujhely.)

The 296th problem was solved by: Bálint B., Devecis M., Fischer O., Kornis Ö., Lichtenberg S., Spitzer Ö., Weisz Á.

Both problems were solved by: Fekete J., Freund A., Geist E., Goldstein Zs., Grünhut B., Hofbauer E., Kántor N., Klein M., Preisz K., Riesz F., Roth M., Szabó I., Szabó K