Trisect each of the three angles of an arbitrary triangle. For each pair of adjacent vertices, take the intersection of the two trisectors nearest their common side.

The three intersection points appear to form an equilateral triangle.

Prove that they always do, no matter the shape of the original triangle.

Hints

Open one at a time

A direct angle chase is possible, but a constructed equilateral triangle is cleaner.

Write the three original angles as triples.

Solution

Best opened after a real attempt

Reverse the construction

Write the angles of the original triangle as $3\alpha$, $3\beta$, and $3\gamma$, so

$$\alpha +\beta +\gamma ={60}^{\circ }.$$

Instead of starting with the arbitrary triangle, begin with an equilateral triangle and build three surrounding triangles whose angles are chosen from $\alpha$, $\beta$, and $\gamma$. Repeated applications of the sine rule show that the outer boundary closes and has angles $3\alpha$, $3\beta$, and $3\gamma$.

The constructed lines are therefore exactly the trisectors of that outer triangle. Because the inner triangle was equilateral by construction, the three adjacent-trisector intersections in the original configuration must also form an equilateral triangle.

This is a case where synthesis is cleaner than analysis: constructing the desired figure first reveals why the apparently accidental ${60}^{\circ }$ is forced by $\alpha +\beta +\gamma$.