Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.8 AIME Prove it Austria

The two equilateral triangles $ABC$ and $ADB$ (with $C \neq D$ ) share the common side $AB$ . The midpoints of $AC$ and $BC$ are denoted by $E$ and $F$ , respectively. Show that $DE$ and $DF$ divide $AB$ into three parts of equal length.

G. Kirchner, Innsbruck

Figure 1: Problem 4.

Solution

The intersections of $DE$ , $DF$ and $DC$ with $AB$ are denoted by $S_1$ , $S_2$ and $M$ , respectively.

We consider the triangle $ACD$ . In this triangle, $DE$ and $AM$ are medians. Therefore, their intersection $S_1$ is the centroid of this triangle and we have $\overline{AS_1} = 2 \cdot \overline{S_1M}$ . An analogous result follows for $S_2$ , which proves the assertion. $\square$

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