A plane passing through the axis of a right circular cylinder intersects the lateral surface of the cone along the generators $S A, S B$. On the diameter $A B$ of the base circle, we mark the point $M$, and from the point $M$, we draw a line parallel to $S A$, which intersects $S B$ at point $N$. We draw a plane through the line $M N$, which is perpendicular to the plane $A S B$ and intersects the base circle of the cone in $P M Q$. $A B=2 R$, $S A=a$.

$1^{0}$. Determine the point $M$ such that $M N+M P=l$, a given length. Investigate for which values of $l$ there is a solution? Investigate and plot $l$ as a function of $A M=x$!

$2^{0}$. Investigate and plot the variation of the area of $\triangle P N Q$!
![](https://cdn.mathpix.com/cropped/2024_05_02_b55c341e7695541c247dg-1.jpg?height=485&width=445&top_left_y=129&top_left_x=840)

$1^{0}$. Given that $\triangle BMN \sim \triangle BAS$, we have $MN: AS = BM: BA$, so $MN = \frac{AS \cdot BM}{BA} = \frac{a(2R - x)}{2R}$. Since $\triangle APB$ is a right triangle at $P$:

$$
\overline{MP^2} = AM \cdot MB = x(2R - x).
$$

Our condition is: $MN + MP = l$, thus

$$
\frac{a(2R - x)}{2R} + \sqrt{x(2R - x)} = l
$$

or

$$
2R \sqrt{x(2R - x)} - 2R(l - a) + ax \ldots
$$

1) The left side is positive (zero in the limit); therefore, the right side must also be positive, i.e., $x \geq \frac{2R(a - l)}{a}$. Of course, $x \leq 2R$.
2) Square both sides and then rearrange; we get:

$$
f(x) \equiv (a^2 + 4R^2)x^2 - 4R(2R^2 + a^2 - al)x + 4R^2(l^2 - 2al + a^2) = 0 \ldots
$$

From this,

$$
x = \frac{2R}{a^2 + 4R^2} \left(2R^2 + a^2 - al + 2R \sqrt{R^2 + al - l^2}\right) \ldots
$$

The roots are real if $l^2 - al - R^2 \leq 0$; this occurs if

$$
\frac{a - \sqrt{a^2 + 4R^2}}{2} < l < \frac{a + \sqrt{a^2 + 4R^2}}{2}.
$$

This is true. Finally,

$$
f\left[\frac{2R(a - l)}{a}\right] = \frac{16R^4 l}{a^2}(l - a)
$$

This is negative if $l < a$, and both roots are valid if

$$
\frac{2R(a - l)}{a} < x < 2R.
$$

If $l > a$, both roots are valid because

$$
\frac{2R(a - l)}{a} < \frac{a}{2},
$$

which is true because we considered the case $l > a$.

In summary: if $0 < l < \frac{a + \sqrt{a^2 + 4R^2}}{2}$, then $f(x) > 0$ if $x < x_1$ and $f(x) < 0$ if $x > x_1$, so $\frac{dl}{dx} > 0$ if $0 < x < x_1$ and $\frac{dl}{dx} < 0$ if $x > x_1$. Therefore, the condition is satisfied if

$$
\left(4R^2 + a^2\right)^2 > a^2 \left(a^2 + 4R^2\right).
$$

This is true. If $l$ satisfies the condition necessary for the roots to be real, then condition 5) is also satisfied.