I recently came up with a question. Consider $\triangle ABC$. How many distinct triangles $\triangle DEF$ are there such that
- Centroid of $\triangle DEF$ coincides with that $\triangle ABC$
- Each side of $\triangle ABC$ contains exactly one of the three points $D, E$ and $F$
I know as a property that the medial triangle of $\triangle ABC$ is one such triangle satisfying the two conditions.
However, is it the only triangle having such a property or, are there more (possibly infinite) such triangles?
Here's how I started off, let $D, E$ and $F$ lie on sides $a, b$ and $c$ respectively. Now divide $a$ in the ratio $m : 1$, $b$ in the ratio $n : 1$ and $c$ in the ratio, $p : 1$. The points $D, E$ and $F$ must satisfy the equation of sides $a, b$ and $c$, and using the ratios, their coordinates can be obtained.
From here, I'm unable to come to a conclusion. It would be appreciated if someone could help me with the proof or give some kind of hint. Thank you in advance.
Conclusion: Infinite such triangles exist when we divide all sides in equal ratios. More specifically when: $\frac{BD}{DC} = \frac{CE}{EA} = \frac{AF}{FB}$
Proof: Let side $A = (x_1, y_1)$, $B = (x_2, y_2)$, and $C = (x_3, y_3)$. Centroid of $\triangle ABC = G$
$G_x = \frac{x_1 + x_2 + x_3}{3}$
Now divide these sides in the ratio m : 1. Solving this using section formula gives
$D_x = \frac{m x_3 + x_2}{m + 1}$
$E_x = \frac{m x_1 + x_3}{m + 1}$
$F_x = \frac{m x_2 + x_1}{m + 1}$
Centroid of $\triangle DEF = G'$
$G'_x = \frac{m (x_1 + x_2 + x_3) + x_1 + x_2 + x_3}{3 (m + 1)}$
$G'_x = \frac{x_1 + x_2 + x_3}{3}$
Solving similarly for y coordinates, we can show that G' coincides with G, thus completing the proof.