## Task B-2.5.

In an equilateral triangle $ABC$, the midpoint of side $\overline{AB}$ is point $P$. Points $G$ and $H$ are between points $A$ and $P$, and points $I$ and $J$ are between points $P$ and $B$. Points $D, E$, and $F$ divide the length of $\overline{AC}$ into four equal parts. Parallels to side $\overline{AB}$ are drawn through these points. We consider all triangles with one vertex at point $C$ and the other two on one of the constructed parallels to side $\overline{AB}$, including $\overline{AB}$ itself, at the points of intersection with the lines $AC, GC, HC, IC, JC$, or $BC$.

If the total number of such triangles is $x$, and the total number of such triangles that do not contain the centroid $T$ is $y$, determine the ratio

![](https://cdn.mathpix.com/cropped/2024_05_30_1ebddb66d71be8305afcg-09.jpg?height=536&width=585&top_left_y=246&top_left_x=1358)
$x: y$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_1ebddb66d71be8305afcg-09.jpg?height=751&width=825&top_left_y=932&top_left_x=204)

Let the side $\overline{A B}$ be colored black, the parallel through $D$ be red, the parallel through $E$ be blue, and the parallel through $F$ be green. The centroid $T$ is located on the median $\overline{C P}$ and is between the blue and red parallels because it divides the median in the ratio 2:1 from vertex $C$.

Let's count the triangles whose side opposite vertex $C$ is black.

On the segment $\overline{A B}$, we need to choose two points from the 6 points $A, G, H, I, J$, and $B$.

If we choose point $A$, the second point can be chosen in 5 ways.

If we choose point $G$, the second point can be chosen in 4 ways (excluding point $A$ because triangle $C G A$ is included in the previous case).

If we choose point $H$, the second point can be chosen in 3 ways.

If we choose point $I$, the second point can be chosen in 2 ways.

If we choose point $J$, the second point can be chosen in 1 way.

Thus, there are 15 such triangles.

We have the same number of triangles whose one side is red, blue, or green. Therefore, the total number of triangles is $4 \cdot 15 = 60$.

Now we will calculate how many triangles contain point $T$. Such triangles must have one side either black or red.

If one vertex is at point $A$, the other vertex must be $I, J$, or $B$.

Similarly, if one vertex is at point $G$ or $H$, there are 3 possibilities for the second vertex.

Thus, the number of triangles that contain point $T$ with one black side is $3 + 3 + 3 = 9$, and the same number of triangles have a red side.

In total, 18 triangles contain point $T$, and $60 - 18 = 42$ triangles do not contain point $T$.

The desired ratio is $60: 42 = 10: 7$.

## NATIONAL MATHEMATICS COMPETITION

## 3rd grade - high school - B variant

Poreč, March 29, 2019.