Let $M$ be an arbitrary point on the semicircle with diameter $AB = 2R$, and let $H$ be the orthogonal projection of $M$ onto $AB$. Write the sum $y = 2AM + 3HB$ as a function of $AM = x$. For which $x$ is this sum equal to a given length $l$? (Discussion). Write down the variation of $y$ and plot the function. Show that the question of the common points of the curve and the line $y = l$ is a visual equivalent to the previous discussion. Let $M_1$ be the position of $M$ for which $y$ is maximum. Calculate the trigonometric functions of the angle $AOM_1$. ($O$ is the midpoint of segment $AB$).
![](https://cdn.mathpix.com/cropped/2024_05_02_42af817e3851c546a7acg-1.jpg?height=363&width=523&top_left_y=119&top_left_x=796)

The triangle $ABM$ is a right triangle, and $HM$ is its altitude. According to the known geometric mean theorem, $BM^2 = AB \cdot HB$, which means $4R^2 - x^2 = 2R \cdot HB$, so $HB = 2R - \frac{x^2}{2R}$, and the desired function is:

$$
\begin{gathered}
y = 2x + 3\left(2R - \frac{x^2}{2R}\right) = -\frac{3}{2R} x^2 + 2x + 6R \\
0 \leq x \leq 2R
\end{gathered}
$$

because the chord $x$ cannot be negative or greater than $AB = 2R$.

We need to solve the equation $y = l$, which is

$$
-\frac{3}{2R} x^2 + 2x + (6R - l) = 0
$$

The discriminant of this equation is

$$
D = 4 + \frac{6}{R}(6R - l) = \frac{2}{R}(20R - 3l)
$$

which is negative if $l > \frac{20R}{3}$, in which case there are no real roots. For $D \geq 0$, i.e., $l \leq \frac{20R}{3}$, the roots (with $x_1 \leq x_2$) are

$$
x_1 = \frac{1}{3}\left(2R - \sqrt{40R^2 - 6Rl}\right), \quad x_2 = \frac{1}{3}\left(2R + \sqrt{40R^2 - 6Rl}\right)
$$

When $D = 0$, the double root $x_1 = x_2$ satisfies (2).

For $D > 0$, the smaller root $x_1$ is positive as long as $0 \leq x_1$, which is true as long as

$$
2R \geq \sqrt{40R^2 - 6Rl}
$$

or, since both sides are positive,

$$
4R^2 \geq 40R^2 - 6Rl, \quad \text{which means} \quad 6Rl \geq 36R^2, \quad l \geq 6R
$$

For the larger root $x_2$, $x_2 \leq 2R$ as long as

$$
\sqrt{40R^2 - 6Rl} \leq 4R, \quad \text{which means} \quad 24R^2 \leq 6Rl, \quad l \geq 4R
$$

Therefore,

$$
\begin{array}{ccccccccc}
l > & \frac{20R}{3} & \text{implies} & \text{no} & \text{chord} & x & \text{such that} & y = l \\
l = & \frac{20R}{3} & " & \text{one} & \text{chord} & x & \text{such that} & y = l \\
6R \leq l < & \frac{20R}{3} & " & \text{two} & \text{chords} & x & \text{such that} & y = l \\
4R \leq l < & 6R & " & \text{one} & \text{chord} & x & \text{such that} & y = l \\
l < & 4R & " & \text{no} & \text{chord} & x & \text{such that} & y = l
\end{array}
$$

For $l > \frac{20R}{3}$, the arc does not intersect because it has no point with $l$-ordinate. For $4R \leq l < 6R$, it intersects at one point. For $6R \leq l < \frac{20R}{3}$, it intersects at two points. For $l = \frac{20R}{3}$, it is tangent, having one common point. This indeed matches the findings in the discussion of (3).

The value of $x_m = \frac{2R}{3}$ that maximizes $y$ corresponds to the acute angle $AOM_1 = \alpha$ because $x_m < R$ implies that $M_1$ lies on the first third of the semicircle from $A$, and $\alpha < 60^\circ$. From $\cos \alpha = \frac{H_1O}{M_1O} = \frac{H_1B - OB}{R}$, using the expression for $HB$, we get $H_1B - OB = R - \frac{x_m^2}{2R} = \frac{7R}{9}$; thus, $\cos \alpha = \frac{7}{9}$, $\sin \alpha = \frac{4\sqrt{2}}{9}$, $\tan \alpha = \frac{4\sqrt{2}}{7}$, and $\cot \alpha = \frac{7}{4\sqrt{2}}$.