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