From point $A$ on a horizontal plane, we launch a body at an angle $\alpha$ with a speed of $c \mathrm{~m} /$ sec. On the orthogonal projection of the parabolic trajectory, from point $B$ which is $a \mathrm{~m}$ away from point $A$, we simultaneously launch another body vertically upwards with the same speed. After how many seconds will the distance between the two bodies be the smallest? What is this minimum distance if $\alpha=30^{\circ}, c=50 \mathrm{~m}$ sec and $a=50 \sqrt{3} \mathrm{~m}$?
I. Solution: It is obvious that the two motions occur in one and the same vertical plane. If in this plane we take the $x$-axis as the projection of the obliquely thrown body on the horizontal plane, with the starting point at point $A$, then it is known from physics that $t$ seconds after the common launch time, the coordinates of the first body are

$$
x_{1}=c t \cos \alpha, \quad y_{1}=c t \sin \alpha-\frac{g}{2} t^{2}
$$

and the coordinates of the second body are

$$
x_{2}=a \quad y_{2}=c t-\frac{g}{2} t^{2}
$$

and thus the square of the distance between the two bodies is

$$
\begin{gathered}
d^{2}=\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}=(c t \cos \alpha-a)^{2}+(c t \sin \alpha-c t)^{2}= \\
=c^{2} t^{2} \cos ^{2} \alpha-2 a c t \cos \alpha+a^{2}+c^{2} t^{2} \sin ^{2} \alpha-2 c^{2} t^{2} \sin \alpha+c^{2} t^{2}= \\
=2 c^{2} t^{2}(1-\sin \alpha)-2 a c t \cos \alpha+a^{2}
\end{gathered}
$$

This is a quadratic function of $t$, which has a minimum at

$$
t_{0}=-\frac{-2 a c \cos \alpha}{2 \cdot 2 c^{2}(1-\sin \alpha)}=\frac{a \cos \alpha}{2 c(1-\sin \alpha)}
$$

since the coefficient of $t^{2}$ is $2 c^{2}(1-\sin \alpha)>0$. 

The minimum value of $d^{2}$ is obtained by substituting the value of $t_{0}$ into the function under (1):

$$
\begin{aligned}
& d_{\min }^{2}=\frac{a^{2} \cos ^{2} \alpha}{2(1-\sin \alpha)}-\frac{a^{2} \cos ^{2} \alpha}{1-\sin \alpha}+a^{2}=a^{2}-\frac{a^{2} \cos ^{2} \alpha}{2(1-\sin \alpha)}= \\
= & a^{2}\left(1-\frac{1-\sin ^{2} \alpha}{2(1-\sin \alpha)}\right)=a^{2}\left(1-\frac{1+\sin \alpha}{2}\right)=a^{2} \frac{1-\sin \alpha}{2}
\end{aligned}
$$

thus

$$
d_{\min }=a \sqrt{\frac{1-\sin \alpha}{2}}
$$

Note that while the value of $t_{0}$ still depends on the common initial speed $c$, $d_{\text {min }}$ is independent of $c$.

With the given values,

$$
d_{\min }=50 \sqrt{3} \sqrt{\frac{1-\frac{1}{2}}{2}}=25 \sqrt{3} \mathrm{~m}
$$

Plichta Jenö (Mezókövesd, I. László g. IV. o. t.)

II. Solution: The gravitational force gives both bodies an equal vertical acceleration, and thus an equal vertical displacement, which does not affect the distance between the two bodies. Therefore, in our problem, we can neglect the gravitational force. Thus, we are examining the minimum distance between two bodies moving with speed $c$ along the hypotenuse $A H=p$ and the leg $B H=q$ of the right triangle $A B H$ (see figure). The motion starts simultaneously from $A$ and $B$. After time $t$, the body moving along the hypotenuse reaches $A^{\prime}$, and the body moving along the leg reaches $B^{\prime}$. $A A^{\prime}=B B^{\prime}=c t$.

![](https://cdn.mathpix.com/cropped/2024_05_02_031ec6c3041fd1545112g-1.jpg?height=464&width=599&top_left_y=1947&top_left_x=752)

Consider the quantities

$$
x=p-c t, \quad y=c t-q
$$

as variables. In this case, $x+y=p-q$ is constant.

It is obvious that as long as $x>0$ and $y>0$, $d$ decreases as $t$ increases. When $x \leq 0$ or $y \leq 0$, $d$ increases as $t$ increases (beyond any limit), so the minimum can only occur when $x>0$ and $y>0$.

In this case, by the cosine rule,

$$
\begin{gathered}
d^{2}=x^{2}+y^{2}-2 x y \cos \left(90^{\circ}+\alpha\right)=x^{2}+2 x y+y^{2}-2 x y+2 x y \sin \alpha= \\
=(x+y)^{2}-2 x y(1-\sin \alpha)
\end{gathered}
$$

Since $x+y$ is constant, $d^{2}$ will be minimal (considering that $1-\sin \alpha>0$) when $x y$ is maximal. If the sum of two quantities is constant, their geometric mean is maximal when the two quantities are equal. Thus, $x y$ is maximal when $x=y$, or $x+y=2 x=p-q$, i.e., $x=y=\frac{p-q}{2}$.

Now we can calculate the value of $t_{0}$, based on the fact that $p-c t_{0}=c t_{0}-q$, i.e., $t_{0}=\frac{p+q}{2 c}$, but we will show that we can determine the value of $d_{\text {min }}$ without calculating $t_{0}$, which is different from the first solution.

Indeed, from (1),

$$
\begin{gathered}
d_{\min }^{2}=(p-q)^{2}-\frac{(p-q)^{2}}{2}(1-\sin \alpha)=\frac{(p-q)^{2}}{2}+\frac{(p-q)^{2}}{2} \sin \alpha= \\
=\frac{(p-q)^{2}}{2}(1+\sin \alpha)
\end{gathered}
$$

Since $p=\frac{a}{\cos \alpha}$ and $q=a \operatorname{tg} \alpha$, we have

$$
p-q=\frac{a}{\cos \alpha}-\frac{a \sin \alpha}{\cos \alpha}=\frac{a(1-\sin \alpha)}{\cos \alpha}
$$

and thus from (2),

$$
d_{\min }^{2}=\frac{a^{2}(1-\sin \alpha)^{2}}{2 \cos ^{2} \alpha}(1+\sin \alpha)=\frac{a^{2}\left(1-\sin ^{2} \alpha\right)(1-\sin \alpha)}{2 \cos ^{2} \alpha}=\frac{a^{2}(1-\sin \alpha)}{2}
$$

hence

$$
d_{\min }=a \sqrt{\frac{1-\sin \alpha}{2}}
$$

Bártfai Pál (Bp., I., Petőfi g. III. o. t.)