Let's construct a line parallel to each side of a given triangle so that after removing the smaller triangles cut off by these lines, the remaining (convex) hexagon has all 6 sides equal.
I. solution. Let's denote the vertices of the sought hexagon on the side $AB$ as $D, E$, on the side $BC$ as $F, G$, and on the side $CA$ as $H, J$, and the angles of the triangle as $\alpha, \beta, \gamma$ in the usual way.

![](https://cdn.mathpix.com/cropped/2024_05_02_7d20cbfe23761265d4d9g-1.jpg?height=612&width=1238&top_left_y=263&top_left_x=430)

The angles of the hexagon can be easily calculated, as its sides are parallel to the sides of the triangle:

$$
\begin{aligned}
& \angle DEF = \angle GHJ = 180^{\circ} - \alpha, \\
& \angle EFG = \angle HJD = 180^{\circ} - \gamma, \\
& \angle FGH = \angle JDE = 180^{\circ} - \beta.
\end{aligned}
$$

Since the sides of the hexagon are equal and we know its angles, we can construct a similar hexagon. By extending every second side of the constructed hexagon $D', E', F', G', H', J'$, we obtain a triangle $A' B' C'$ similar to the original triangle. By enlarging the resulting figure from an appropriate center by the desired scale, we obtain the hexagon to be constructed.

It is convenient to immediately place the sides $J' D', D' E', E' F'$ of the hexagon parallel to the sides $BC, AB, AC$, respectively, so that the triangle $A' B' C'$ will be in a similar position to $ABC$, making the enlargement process somewhat simpler.

II. solution. The triangles cut by the sides of the hexagon, $HGC, EBF, ADJ$, are similar to the triangle $ABC$ and thus to each other, with angles $\alpha, \beta$, and $\gamma$.

Each of these triangles contains one side of the hexagon - placed opposite different angles in sequence - and furthermore, $AJ + JH + HC = AB$.

Choose a segment of arbitrary length, consider this as one side of the hexagon, and construct triangles $H'G'C'$ and $A'D'J'$ similar to the triangles $HGC$ and $ADJ$ according to the conditions mentioned earlier. Then divide the side $AC$ of the triangle $ABC$ in the ratio $A'J' : J'H' : H'C'$ obtained.

The construction directly provides one side of the sought hexagon.

László Fodor (Vác, Sztáron S. Gym., II. grade)

Note. Denoting the side of the hexagon by $d$ and the sides of the triangle by $a, b, c$, $d$ can also be calculated after expressing the segments formed on the side $AC$ with it:

$$
\begin{gathered}
AC = AJ + JH + HC = \frac{d}{a} \cdot b + d + \frac{d}{c} \cdot b = b \\
d = \frac{abc}{ab + bc + ca}, \quad \frac{1}{d} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}
\end{gathered}
$$[^0]


[^0]:    ${ }^{1}$ Although the choice of extended sides can be done in two ways, the two resulting triangles are congruent, as the hexagon is centrally symmetric based on its prescribed properties and can be brought into coincidence with a $180^{\circ}$ rotation.