What relationship exists between the sides of a triangle if it can be divided into two parts of equal area and equal perimeter by a line parallel to one of its sides?
Let $ABC$ be a triangle with a desired property, and let the line $PQ$ parallel to side $AB$ bisect both the area and the perimeter of the triangle, intersecting sides $CA$ and $CB$ at points $P$ and $Q$, respectively. (The bisection of the perimeter is derived from the equality of the perimeters of the resulting triangle $PQC$ and the trapezoid $PQBA$, excluding their common side $PQ$.)

![](https://cdn.mathpix.com/cropped/2024_05_02_34d383e7e0d6431ba3e0g-1.jpg?height=423&width=545&top_left_y=274&top_left_x=779)

Triangle $PQC$ is similar to $ABC$ and has half the area. Denoting the ratio of corresponding sides by $\lambda$, the ratio of the areas is $\lambda^2$, so $\lambda^2 = 1/2$, and $\lambda = 1/\sqrt{2}$. Thus, from the bisection of the original perimeter: $PC + CQ = PA + AB + BQ$, or using the usual notation: $\lambda b + \lambda a = (1 - \lambda) b + c + (1 - \lambda) a$, hence the desired relationship is:

$$
c = (2\lambda - 1)(a + b) = (\sqrt{2} - 1)(a + b)
$$

Korenchy Emóke (Bp. VIII., Ságvári E. gyak. lg. II. o. t.)

Note. The derived relationship—an equality among the three sides—does not uniquely determine the shape of the triangle. It is therefore natural to ask whether there is a triangle that can be divided in this way by lines parallel to two or all three sides.

Suppose, in addition to (1), we also require the analogous relationship for side $b$: $b = (\sqrt{2} - 1)(a + c)$. Eliminating $a$ first, we get $c = b$, and then from (1) we get $\sqrt{2} b = a$, meaning that both conditions are only satisfied in the isosceles right triangle, with the legs parallel to the cutting lines.

This way, on the one hand, we have found a special shape of a triangle that meets the original requirement (the construction of $PQ$ is evident from the diagram), and on the other hand, we see that no triangle can have this property for all three sides.

Choosing $a = b$ from (1) yields another isosceles and "bisectable" shape, in which the base $c$ is $(2\sqrt{2} - 2)$ times the length of the legs (in the diagram, $A'Q = QB$, $A'C' = C'B = AB$, and $C'P' = C'Q' = CP$).