Add an equation for a line that bisects both the perimeter and the area of the "Egyptian" triangle (sides: 3, 4, 5).
Solution. Let's denote the vertices, sides, and angles of the triangle in the usual way (the right angle is at vertex $C$). Let $AC=4$ and $BC=3$. We claim that the desired line does not pass through any vertex of the triangle. A line passing through a vertex divides the triangle into two triangles with equal heights. If their areas are also equal, then their bases are equal, but then the perimeters are not equal (Figure 1).
![](https://cdn.mathpix.com/cropped/2024_05_02_287b9a59460e2b45fc4dg-1.jpg?height=294&width=1134&top_left_y=354&top_left_x=472)

$k=4+1.5 \neq 5+1.5$

$k=4+2.5 \neq 3+2.5$

Figure 1

The desired line either intersects one leg and the hypotenuse (Figure 2), or two legs (Figure 3), in both cases dividing the right triangle into one triangle and one quadrilateral.
![](https://cdn.mathpix.com/cropped/2024_05_02_287b9a59460e2b45fc4dg-1.jpg?height=798&width=890&top_left_y=862&top_left_x=606)

In the first case, let the segments cut off be $x$ and $y$ as shown in Figure 2. The area of the right triangle is: $t=\frac{3 \cdot 4}{2}=6$, and the perimeter is: $k=3+4+5=12$. The area of the cut-off triangle, according to the condition, is:

$$
t_{1}=\frac{x y \sin \beta}{2}=3, \quad \text { or } \quad \frac{x y \sin \alpha}{2}=3
$$

Substituting $\sin \beta=\frac{4}{5}$ and $\sin \alpha=\frac{3}{5}$:

$$
t_{1}=\frac{x \cdot y \cdot \frac{4}{5}}{2}=\frac{4 x y}{10}=3, \quad \text { or } \quad \frac{x \cdot y \cdot \frac{3}{5}}{2}=\frac{3 x y}{10}=3
$$

From this,

$$
x y=\frac{15}{2}
$$

and

$$
x y=10
$$

Due to the equality of the perimeters: $x+y=6$. Substituting this into equation (1) gives the quadratic equation $2 x^{2}-12 x+15=0$. From this,

$$
x=\frac{12 \pm \sqrt{144-120}}{4}=\left\{\begin{array}{l}
\frac{6+\sqrt{6}}{2} \approx 4.22>3: \text { not a solution, } \\
\frac{6-\sqrt{6}}{2} \approx 1.78 \text { and } y=\frac{6+\sqrt{6}}{2} \approx 4.22: \text { solution. }
\end{array}\right.
$$

From equation (2), $x^{2}-6 x+10=0$. This equation has a negative discriminant, so it has no solution.

If the line intersects the two legs, then

$$
\frac{x y}{2}=3, \quad x+y=6
$$

The solution to this system of equations is: $\{x ; y\}=\{3-\sqrt{3} ; 3+\sqrt{3}\}$. The larger of these two numbers is greater than 4, so we do not get a solution.

Therefore, the problem has one solution; in this case, the line intersects the shorter leg and the hypotenuse, as calculated.