4. Given is a triangle $A B C$. On the sides $\overline{A C}$ and $\overline{B C}$ of triangle $\triangle A B C$, points $M$ and $N$ are marked such that: $|A M|:|M C|=1: m$ and $|B N|:|N C|=1: n$. Determine the natural numbers $m$ and $n$ for which $P_{\triangle A B C}: P_{\triangle M N C}=2: 1$.
4. Let's denote the height $\overline{A D}$ of triangle $\triangle A B C$ to side $\overline{B C}$ and the height $\overline{M E}$ of triangle $\triangle M N C$ to side $\overline{N C}$.

The right triangles $\triangle A D C$ and $\triangle M E C$ share the angle at vertex $C$, so they are similar by the AA (Angle-Angle) similarity criterion, and we have $|A C|:|M C|=|A D|:|M E|$.

Since $|A M|:|M C|=1: m$, it follows that

$|A C|:|M C|=(1+m): m$ and thus the ratio of the heights of triangles $\triangle A D C$ and $\triangle M E C$ is

![](https://cdn.mathpix.com/cropped/2024_05_30_3b064f6bc407e279e59eg-12.jpg?height=434&width=851&top_left_y=697&top_left_x=1162)

$|A D|:|M E|=(1+m): m$, or $\frac{|A D|}{|M E|}=\frac{(1+m)}{m}$.

Furthermore, since $|B N|:|N C|=1: n$, the ratio of the lengths of sides $\overline{B C}$ and $\overline{N C}$ of triangles $\triangle A B C$ and $\triangle M N C$ is $|B C|:|N C|=(1+n): n$, or

$\frac{|B C|}{|N C|}=\frac{(1+n)}{n}$.

Given that $P_{\triangle A B C} = \frac{|B C| \cdot|A D|}{2}$, $P_{\triangle M N C}=\frac{|N C| \cdot|M E|}{2}$, and $P_{\triangle A B C}: P_{\triangle M N C}=2: 1$, we have

$$
(|B C| \cdot|A D|):(|N C| \cdot|M E|)=2: 1
$$

or

$$
\frac{|B C| \cdot|A D|}{|N C| \cdot|M E|}=\frac{|B C|}{|N C|} \cdot \frac{|A D|}{|M E|}=2
$$

Therefore, we have:

$$
\begin{gathered}
\frac{1+n}{n} \cdot \frac{1+m}{m}=2 \\
(1+n) \cdot(1+m)=2 n m \\
1+m+n+n m=2 n m \\
n m-n-m=1 \\
n m-n-m+1=1+1 \\
n(m-1)-(m-1)=2 \\
(m-1)(n-1)=2
\end{gathered}
$$

Since $m$ and $n$ are natural numbers, we have two possibilities:

$m-1=1$ and $n-1=2$ or $m-1=2$ and $n-1=1$.

Thus, the desired numbers are $m=2$ and $n=3$ or $m=3$ and $n=2$.

Note 1: The equation $n m-n-m=1$ can also be solved as follows:

$$
n m-m=1+n
$$

$$
\begin{gathered}
m(n-1)=1+n \\
m=\frac{1+n}{n-1}=\frac{n-1+2}{n-1}=1+\frac{2}{n-1}
\end{gathered}
$$

For $m$ to be a natural number, $n-1$ must be a natural number and a divisor of 2.

Therefore, $n-1$ can be 1 or 2, i.e., $n$ can be 2 or 3.

The number $m$ can be, respectively depending on $n$, 3 or 2.

Thus, the desired numbers are $m=2$ and $n=3$ or $m=3$ and $n=2$.

Note 2: The solution does not depend on the type of triangle, i.e., the type of the internal angle at vertex $A$.
![](https://cdn.mathpix.com/cropped/2024_05_30_3b064f6bc407e279e59eg-13.jpg?height=490&width=1814&top_left_y=748&top_left_x=130)

However, if the problem is solved by considering the height $\overline{B D}$ of triangle $\triangle A B C$ to side $\overline{A C}$ and the height $\overline{N E}$ of triangle $\triangle M N C$ to side $\overline{M C}$, the solution is obtained analogously with the appropriate labels.

![](https://cdn.mathpix.com/cropped/2024_05_30_3b064f6bc407e279e59eg-13.jpg?height=351&width=1327&top_left_y=1452&top_left_x=296)

Let's denote the height $\overline{B D}$ of triangle $\triangle A B C$ to side $\overline{A C}$ and the height $\overline{N E}$ of triangle $\triangle M N C$ to side $\overline{M C}$.

The right triangles $\triangle B C D$ and $\triangle N C E$ share the angle at vertex $C$, so they are similar by the AA (Angle-Angle) similarity criterion, and we have $|B C|:|N C|=|B D|:|N E|$.

Since $|B N|:|N C|=1: n$, it follows that $|B C|:|N C|=(1+n): n$, so the ratio of the heights of triangles $\triangle B D C$ and $\triangle N E C$ is $|B D|:|N E|=(1+n): n$, or $\frac{|B D|}{|N E|}=\frac{(1+n)}{n}$.

Furthermore, since $|A M|:|M C|=1: m$, the ratio of the lengths of sides $\overline{A C}$ and $\overline{M C}$ of triangles $\triangle A B C$ and $\triangle M N C$ is $|A C|:|M C|=(1+m): m$, or

$\frac{|A C|}{|M C|}=\frac{(1+m)}{m}$.

Given that $P_{\triangle A B C}=\frac{|A C| \cdot|B D|}{2}$, $P_{\triangle M N C}=\frac{|M C| \cdot|N E|}{2}$, and $P_{\triangle A B C}: P_{\triangle M N C}=2: 1$, we have

$$
(|A C| \cdot|B D|):(|M C| \cdot|N E|)=2: 1
$$

or

$$
\frac{|A C| \cdot|B D|}{|M C| \cdot|N E|}=\frac{|A C|}{|M C|} \cdot \frac{|B D|}{|N E|}=2
$$

Therefore, we have:

$$
\begin{gathered}
\frac{1+m}{m} \cdot \frac{1+n}{n}=2 \\
(1+m) \cdot(1+n)=2 m n \\
1+n+m+m n=2 m n \\
m n-n-m=1 \\
m n-n-m+1=1+1 \\
n(m-1)-(m-1)=2 \\
(m-1)(n-1)=2
\end{gathered}
$$

Since $m$ and $n$ are natural numbers, we have two possibilities:

$m-1=1$ and $n-1=2$ or $m-1=2$ and $n-1=1$.

Thus, the desired numbers are $m=2$ and $n=3$ or $m=3$ and $n=2$.