Given a circle $(O)$ with radius $R$, we project the point $M$ on its circumference onto two perpendicular diameters, thus forming the rectangle $M P O Q$. Under what condition of the position of $M$ will $t=a k$, where $t$ is the area of the rectangle, $k$ is half the perimeter of the rectangle, and $a$ is a given length? Discuss!

Apply the results if $a=\frac{12}{35} R$.
Let $O P=Q M=x, O Q=P M=y$. The problem's requirement is:

$$
x y=a(x+y) \ldots
$$

$M$ is a point on the circle, so

$$
x^{2}+y^{2}=R^{2} \ldots
$$

To solve this system of equations, we write:

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

If we let $x+y=u$, then according to 2),

$$
u^{2}-2 a u-R^{2}=0 \ldots
$$

This equation always has two real roots with opposite signs; we can only consider the positive one, so

$$
x+y=u=a+\sqrt{a^{2}+R^{2}} \ldots
$$

From 1),

$$
x y=v=a u=a^{2}+a \sqrt{a^{2}+R^{2}} \ldots
$$

Accordingly, $x$ and $y$ are the roots of the equation

$$
z^{2}-u z+v=0 \ldots
$$

These are real if $u^{2}-4 v \geq 0$, i.e.,

$$
u^{2}-4 a u \geq 0, \quad \text { so } \quad u \geq 4 a
$$

This occurs if

$$
\sqrt{a^{2}+R^{2}} \geq 3 a, \quad \text { i.e., } \quad R^{2} \geq 8 a^{2}
$$

Thus, the problem has a solution if $2 \sqrt{2} a \leq R$.

Application. If $a=\frac{12}{35} R$, then

$$
2 \sqrt{2} \cdot \frac{12}{35}<1, \quad \text { because } 24 \sqrt{2}<35 . \quad \text { Indeed } 576 \cdot 2=1152<1225
$$

Now

$$
a^{2}+R^{2}=R^{2}\left(\frac{144}{1225}+1\right)=R^{2} \frac{1369}{1225}=\left(\frac{37}{35} R\right)^{2}
$$

Thus

$$
\begin{aligned}
& x+y=u=\frac{12}{35} R+\frac{37}{35} R=\frac{49}{35} R=\frac{7}{5} R . \\
& x y=v=a u=\frac{7}{5} \cdot \frac{12}{35} R^{2}=\frac{12}{25} R^{2}
\end{aligned}
$$

$x$ and $y$ are the roots of the equation

$$
z^{2}-\frac{7 R}{5} z+\frac{12}{25} R^{2}=0
$$

From this, $x=\frac{3}{5} R, y=\frac{4}{5} R$ or $x=\frac{4}{5} R$ and $y=\frac{3}{5} R$.