a) Two heavy bodies start moving on two inclined planes without initial velocity. The inclined planes form angles $\alpha$ and $\beta$ with the horizontal plane, respectively. One body falls from point $A$, the other from point $B$. $O A=a, O B=b$. After how many seconds will the line $M N$, which connects the two moving bodies, be horizontal? Under what conditions is this problem possible?

![](https://cdn.mathpix.com/cropped/2024_05_02_cedbfe4ef8ea5fcedd1ag-1.jpg?height=262&width=409&top_left_y=281&top_left_x=847)

b) If $\alpha$ and $\beta$ are not given, but we know the angle $\varphi$ between the two planes, calculate the angles $\alpha$ and $\beta$, under the condition that the heavy bodies starting from points $A$ and $B$ without initial velocity arrive at $O$ simultaneously.
I.) The line $M N$ is horizontal if

$$
N O \sin \beta = M O \sin \alpha
$$

If $t$ denotes the time it takes for the bodies to reach $M$ and $N$, respectively, then:

$$
\left(b - \frac{1}{2} g t^{2} \sin \beta\right) \sin \beta = \left(a - \frac{1}{2} g t^{2} \sin \alpha\right) \sin \alpha
$$

From this equation:

$$
t = \sqrt{\frac{2(a \sin \alpha - b \sin \beta)}{g(\sin^2 \alpha - \sin^2 \beta)}}
$$

Furthermore,

$$
O M = \frac{b \sin \alpha - a \sin \beta}{\sin^2 \alpha - \sin^2 \beta} \sin \beta \text{ and } O N = \frac{b \sin \alpha - a \sin \beta}{\sin^2 \alpha - \sin^2 \beta} \sin \alpha
$$

The problem is only possible if $t$ has a real value and

$$
O M \geq 0, O N \geq 0
$$

These conditions are satisfied if:

$$
\frac{a \sin \alpha - b \sin \beta}{\sin^2 \alpha - \sin^2 \beta} \geq 0, \frac{b \sin \alpha - a \sin \beta}{\sin^2 \alpha - \sin^2 \beta} \geq 0
$$

First, assume that $\alpha > \beta$, then it must be:

$$
a \sin \alpha - b \sin \beta \geq 0, \text{ and } b \sin \alpha - a \sin \beta \geq 0
$$

From this,

$$
\frac{\sin \alpha}{\sin \beta} \geq \frac{a}{b} \geq \frac{\sin \beta}{\sin \alpha}
$$

If $\frac{a}{b} = \frac{\sin \alpha}{\sin \beta}$, then $O M = O N = 0$, meaning the two moving points arrive at $O$ simultaneously; if $\frac{a}{b} = \frac{\sin \beta}{\sin \alpha}$, then $t = 0$, meaning $A B$ is a horizontal line. If $\alpha = \beta$, then $a = b$ and $t = \frac{0}{0}$, which means $M N$ is horizontal throughout the entire motion.

II.) Since in this case

$$
a = \frac{1}{2} g t^{2} \sin \alpha \text{ and } b = \frac{1}{2} g t^{2} \sin \beta
$$

we get:

$$
\frac{1}{2} g t^{2} = \frac{a}{\sin \alpha} = \frac{b}{\sin \beta}
$$

From this,

$$
\frac{a + b}{a - b} = \frac{\tan \frac{\alpha + \beta}{2}}{\tan \frac{\alpha - \beta}{2}}
$$

From this equation,

$$
\tan \frac{\alpha - \beta}{2} = \frac{a - b}{a + b} \tan \frac{\alpha + \beta}{2}
$$

But $\alpha + \beta + \gamma = 180^\circ$ so

$$
\tan \frac{\alpha + \beta}{2} = \cot \frac{\gamma}{2}
$$

Thus,

$$
\tan \frac{\alpha - \beta}{2} = \frac{a - b}{a + b} \cot \frac{\gamma}{2}
$$

From this, we can calculate $\alpha - \beta$; we know $\alpha + \beta$, so we can determine $\alpha$ and $\beta$.

The problem was solved by: Friedmann Bernát, Goldstein Zsigmond, Goldziher Károly, Grünhut Béla, Hofbauer Ervin, Kántor Nándor, Klein Mór, Riesz Frigyes, Szabó István.