$T R N$ and $H A M$ are congruent equilateral triangles. Moreover, point $T$ is the centroid of triangle $H A M$ and point $R$ lies on the ray $T A$.

What is the ratio of the areas of the parts of triangle $T R N$ that are inside and outside triangle $H A M$?

(E. Semerádová)

Hint. The centroid is the intersection of the medians.
The centroid is the intersection of the medians, and in an equilateral triangle, these are simultaneously altitudes and angle bisectors. These lines divide the triangle $H A M$ into six congruent triangles.

In particular, the internal angles of each of these triangles are $90^{\circ}$ at the vertices on the sides of $H A M$ (the foot of the altitude), $30^{\circ}$ at the vertices $H A M$ (half of the internal angle of the equilateral triangle), and $60^{\circ}$ at vertex $T$ (so that the sum of all angles is $180^{\circ}$).

Let $P$ be the intersection of the sides of triangles $H A M$ and $T R N$. Triangle $A T P$ is the common part of triangles $H A M$ and $T R N$. By comparing the internal angles at vertex $T$, we find that triangle $A T P$ is one of the six congruent triangles mentioned above.

![](https://cdn.mathpix.com/cropped/2024_04_17_e57040b3fafc3d67d9eag-2.jpg?height=591&width=671&top_left_y=1047&top_left_x=698)

Triangles $H A M$ and $T R N$ are congruent, and their common part represents $\frac{1}{6}$ of the area of each. Therefore, the part of triangle $T R N$ that lies outside triangle $H A M$ represents $\frac{5}{6}$ of its area. The ratio of the areas of these parts is $1: 5$.

Note. From the problem statement, we do not know whether point $P$ lies on side $A M$ or $A H$. The choice in the diagram is not significant; in both cases, the conclusions and their justifications are the same.