## Task A-4.3.

For a point $P$ inside triangle $A B C$, we say it is brilliant if exactly 27 rays can be drawn from it intersecting the sides of triangle $A B C$ such that the triangle is divided into 27 smaller triangles of equal area. Determine the number of all brilliant points of triangle $A B C$.
## Solution.

Let $P$ be the shining point of triangle $ABC$ and, without loss of generality, let the area of triangle $ABC$ be 27. First, note that among the 27 drawn rays, there must always be rays $PA$, $PB$, and $PC$; otherwise, not all regions into which triangle $ABC$ is divided will be triangles.

Each of the 27 smaller triangles is located in (or is equal to) one of the triangles $PAB$, $PBC$, or $PCA$. Let there be $m$ of them in triangle $PAB$, and $n$ in triangle $PBC$. The numbers $m$ and $n$ are in the set $\{1, 2, \ldots, 25\}$ and for them, $m+n \leqslant 26$, and in triangle $PCA$ there are $27-m-n$ smaller triangles.

Consider the $m$ triangles that are located within triangle $PAB$. All of them have an area of 1 and share a common height which is the distance from point $P$ to $\overline{AB}$, let this be $d_{AB}$. These $m$ triangles have congruent sides to which the observed height is perpendicular, so we see that all these congruent sides are of length $\frac{1}{m}|AB|$. We conclude that the length $d_{AB}$ is uniquely determined by the number $m$. More precisely, $d_{AB} = \frac{2m}{|AB|}$. If we denote the distance from point $P$ to $\overline{BC}$ as $d_{BC}$, we see in the same way that the length $d_{BC}$ is uniquely determined by the number $n$.

Note that $d_{AB}$ is less than the distance from point $C$ to $\overline{AB}$, otherwise the area of triangle $PAB$ would be at least equal to the area of triangle $ABC$, which is impossible. Similarly, we conclude that $d_{BC}$ is less than the distance from point $A$ to $\overline{BC}$. Let $p$ be a line parallel to $AB$ that lies between $AB$ and point $C$ and is at a distance $d_{AB}$ from $AB$. Similarly, let line $q$ be between $BC$ and $A$ and be at a distance $d_{BC}$ from $BC$. Point $P$ is located at the intersection of lines $p$ and $q$ and is thus uniquely determined. For such a point $P$, the area of triangle $PAB$ is $m$, the area of triangle $PBC$ is $n$, and the area of triangle $PCA$ is then $27-m-n$.

![](https://cdn.mathpix.com/cropped/2024_05_30_9bdde249022d274b7f19g-26.jpg?height=605&width=782&top_left_y=337&top_left_x=637)

The question of the problem is now equivalent to the question of how many ways we can choose the numbers $m$ and $n$. For each $m \in \{1, 2, \ldots, 25\}$, $n$ can be chosen in exactly $26-m$ ways. Therefore, the solution is

$$
\sum_{m=1}^{25}(26-m) = \frac{25 \cdot 26}{2} = 325
$$