The line $y=-\frac{1}{2} x+4$ is intersected by the line $y=m x$ at point $M$. The projection of $M$ on the $X$-axis is $N$. Let's examine and graph the variation of the area of $\Delta O M N$ as $m$ varies from 0 to $+\infty$.


I. Solution. The line $y=-\frac{1}{2} x+4$ intersects the $X$-axis at point $A(x=8, y=0)$ and the $Y$-axis at point $B(x=0, y=4)$. These two points determine the line.

$y=m x$ represents a line passing through the origin. If $m=0$, it coincides with the $X$-axis. As $m$ varies from 0 to $+\infty$, the line $y=m x$ rotates around $O$, starting from the $X$-axis in the positive direction until it coincides with the $Y$-axis. The intersection point with the solid line moves from $A$ to $B$, so the abscissa of $M$ decreases from 8 to 0.

![](https://cdn.mathpix.com/cropped/2024_05_02_41672434f72550a52512g-1.jpg?height=320&width=583&top_left_y=409&top_left_x=760)

The area of $\triangle O M N$, if the coordinates of $M$ are $(x, y)$, is $t=\frac{x y}{2}$. For $M$,

$$
\begin{gathered}
y=-\frac{1}{2} x+4 \\
t=-\frac{1}{4} x^{2}+2 x
\end{gathered}
$$

where $x$ decreases from 8 to 0. As we can see, the area of $\triangle O M N$ is a quadratic function of $x$, which is zero at $x=8$ and $x=0$. This function has a maximum at $x=\frac{1}{2} O A=4$. Here, $t_{\max }=4$.

Accordingly, the area of $\triangle O M N$ increases from 0 to 4, then decreases back to 0 (symmetrically with respect to $x=4$). The variation in area is represented by a parabolic arc, whose vertex is the highest point of the curve, and whose axis of symmetry is the line $x=4$.

![](https://cdn.mathpix.com/cropped/2024_05_02_41672434f72550a52512g-1.jpg?height=303&width=569&top_left_y=1377&top_left_x=775)

Detailed value table:

| $x$ | 0 |  | 1 |  | 2 |  | 3 |  | 4 |  | 5 |  | 6 |  | 7 |  | 8 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $t$ | 0 | $\nearrow$ | $\frac{7}{4}$ | $\nearrow$ | 3 | $\nearrow$ | $\frac{15}{4}$ | $\nearrow$ | 4 | $\searrow$ | $\frac{15}{4}$ | $\searrow$ | 3 | $\searrow$ | $\frac{7}{4}$ | $\searrow$ | 0 |

Jesch Aladár (Kegyesrendi g. VI. o. Bp.)

II. Solution. The area of $\triangle O M N$ can be expressed as a function of the slope $m$. Specifically, for the coordinates of point $M$,

$$
y=-\frac{1}{2} x+4 \text { and } y=m x
$$

From this,

$$
x=\frac{8}{2 m+1}, \quad y=\frac{8 m}{2 m+1}
$$

and

$$
t=\frac{x y}{2}=\frac{32 m}{(2 m+1)^{2}} \cdots
$$

For $m=0$ and $m=+\infty$, $t=0$. As $m$ varies from 0 to $+\infty$, $t$ is always positive. Since $t$ is a continuous function of $m$ for the values of $m$ in question, it must have at least one maximum value. To determine this, we need to examine when the equation

$$
t=\frac{32 m}{(2 m+1)^{2}}, \text { or } 4 t m^{2}+(4 t-32) m+t=0 \ldots
$$

has real solutions.

Clearly, this happens when the discriminant of the equation is:

$$
(4 t-32)^{2}-16 t^{2} \geqq 0 \text { or } t \leqq 4
$$

Thus, the maximum value of the area of $\triangle O M N$ is 4.

The sum of the roots of equation (2) is $\frac{32-4 t}{4 t}$, and the product is $\frac{t}{4 t}=\frac{1}{4}$.

If $0<t<4$, both the sum and the product of the roots are positive; hence, both roots are positive.

If $t=4$, then $m=\frac{1}{2}$. (For point $M$: $x=4, y=2$).

Vizhányó Ferenc (Áll. Szent-István g. VI. o. Bp. XIV.)

Note. The curve representing the function $t=\frac{32 m}{(2 m+1)^{2}}$ asymptotically approaches the $X$-axis.