Problem: A circle is divided into congruent arcs by $432$ points.
The points are colored in four colors such that
some $108$ points are colored red,
some $108$ points are colored green,
some $108$ points are colored blue,
and the remaining $108$ points are colored yellow.
Prove that one can choose three points of each color in such a way
that the four triangles formed by the chosen points
of the same color are congruent.

Solution: First, consider the $431$ possible non-identity rotations
of the red points, and count overlaps with green points.
If we select a rotation randomly,
then each red point lies over a green point
with probability $\frac{108}{431}$;
hence the expected number of red-green incidences is
\[ \frac{108}{431} \cdot 108 > 27 \]
and so by pigeonhole, we can find a red $28$-gon
and a green $28$-gon which are rotations of each other.

Now, look at the $430$ rotations of this $28$-gon
(that do not give the all-red or all-green configuration)
and compare it with the blue points.
The same approach gives
\[ \frac{108}{430} \cdot 28 > 7 \]
incidences, so we can find red, green, blue $8$-gons
which are similar under rotation.

Finally, the $429$ nontrivial rotations of this $8$-gon expect
\[ \frac{108}{429} \cdot 8 > 2 \]
incidences with yellow.
So finally we have four monochromatic $3$-gons,
one of each color, which are rotations of each other.