Find the length of the shortest segment that bisects the area of a triangle with sides 3, 4, and 5.
Solution. First, we generally provide a lower estimate for the length of a segment that bisects the area of a triangle. If the segment $P Q$ bisects the area of triangle $A B C$, then at least one of the two parts of the original triangle is a triangle (both parts will be triangles if the segment $P Q$ contains one of the vertices of triangle $A B C$). Since we are providing a lower estimate, we can assume that $P$ and $Q$ fall on two sides of triangle $A B C$, and due to the symmetric role of the sides, we can also assume that $P$ is on segment $A B$ and $Q$ is on segment $A C$. Let $A P = x$ and $A Q = y$ (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_6f90d29b5d04c2c7da39g-1.jpg?height=360&width=588&top_left_y=373&top_left_x=758)

Figure 1

![](https://cdn.mathpix.com/cropped/2024_05_02_6f90d29b5d04c2c7da39g-1.jpg?height=371&width=472&top_left_y=785&top_left_x=813)

Figure 2

If the area of triangle $A B C$ is $T$ and the angle at vertex $A$ is denoted by $\alpha$, then the area of triangle $A P Q$ is

$$
\frac{T}{2} = \frac{A P \cdot A Q \cdot \sin \alpha}{2} = \frac{x y \sin \alpha}{2}
$$

From this, we get the relationship $x y = \frac{T}{\sin \alpha}$. Using this relationship, the cosine rule, and the inequality between the arithmetic and geometric means, we obtain

$$
P Q^{2} = x^{2} + y^{2} - 2 x y \cos \alpha \geq 2 x y (1 - \cos \alpha) = 2 T \frac{1 - \cos \alpha}{\sin \alpha}
$$

Since $1 - \cos \alpha = 2 \sin^2 \frac{\alpha}{2}$ and $\sin \alpha = 2 \sin \frac{\alpha}{2} \cdot \cos \frac{\alpha}{2}$, we have

$$
P Q^{2} \geq 2 T \frac{1 - \cos \alpha}{\sin \alpha} = 2 T \tan \frac{\alpha}{2}
$$

where equality holds if and only if $x = y = \sqrt{A B \cdot \frac{A C}{2}}$.

Returning to our original problem, the tangent function is strictly increasing for acute angles, so the shortest segment that bisects the area is obtained when the two points $P$ and $Q$ are taken on the sides of the triangle with lengths 4 and 5, such that $A P = A Q = \sqrt{5 \cdot \frac{4}{2}} = \sqrt{10}$. This is possible because $\sqrt{10} < 4$ (Figure 2). In this case, $\cos \alpha = \frac{4}{5}$, so from equation (1) we get

$$
P Q^{2} = 10 + 10 - 20 \cdot \frac{4}{5} = 4
$$

Thus, the length of the shortest segment that bisects the area of the triangle is 2.