
4. Let $A B C$ be a triangle. An interior point $P$ of $A B C$ is said to be good if we can find exactly 27 rays emanating from $P$ intersecting the sides of the triangle $A B C$ such that the triangle is divided by these rays into 27 smaller triangles of equal area. Determine the number of good points for a given triangle $A B C$.


Solution: Let $P$ be a good point. Let $l, m, n$ be respetively the number of parts the sides $B C, C A, A B$ are divided by the rays starting from $P$. Note that a ray must pass through each of the vertices the triangle $A B C$; otherwise we get some quadrilaterals.

Let $h_{1}$ be the distance of $P$ from $B C$. Then $h_{1}$ is the height for all the triangles with their bases on $B C$. Equality of areas implies that all these bases have equal length. If we denote this by $x$, we get $l x=a$. Similarly, taking $y$ and $z$ as the lengths of the bases of triangles on $C A$ and $A B$ respectively, we get $m y=b$ and $n z=c$. Let $h_{2}$ and $h_{3}$ be the distances of $P$ from $C A$ and $A B$ respectively. Then

$$
h_{1} x=h_{2} y=h_{3} z=\frac{2 \Delta}{27}
$$

where $\Delta$ denotes the area of the triangle $A B C$. These lead to

$$
h_{1}=\frac{2 \Delta}{27} \frac{l}{a}, \quad h_{1}=\frac{2 \Delta}{27} \frac{m}{b}, \quad h_{1}=\frac{2 \Delta}{27} \frac{n}{c}
$$

But

$$
\frac{2 \Delta}{a}=h_{a}, \quad \frac{2 \Delta}{b}=h_{b}, \quad \frac{2 \Delta}{c}=h_{c}
$$

Thus we get

$$
\frac{h_{1}}{h_{a}}=\frac{l}{27}, \quad \frac{h_{2}}{h_{b}}=\frac{m}{27}, \quad \frac{h_{3}}{h_{c}}=\frac{n}{27}
$$

However, we also have

$$
\frac{h_{1}}{h_{a}}=\frac{[P B C]}{\Delta}, \quad \frac{h_{2}}{h_{b}}=\frac{[P C A]}{\Delta}, \quad \frac{h_{3}}{h_{c}}=\frac{[P A B]}{\Delta}
$$

Adding these three relations,

$$
\frac{h_{1}}{h_{a}}+\frac{h_{2}}{h_{b}}+\frac{h_{3}}{h_{c}}=1
$$

Thus

$$
\frac{l}{27}+\frac{m}{27}+\frac{n}{27}=\frac{h_{1}}{h_{a}}+\frac{h_{2}}{h_{b}}+\frac{h_{3}}{h_{c}}=1
$$

We conclude that $l+m+n=27$. Thus every good point $P$ determines a partition $(l, m, n)$ of 27 such that there are $l, m, n$ equal segments respectively on $B C, C A, A B$.

Conversely, take any partition $(l, m, n)$ of 27 . Divide $B C, C A, A B$ respectively in to $l, m, n$ equal parts. Define

$$
h_{1}=\frac{2 l \Delta}{27 a}, \quad h_{2}=\frac{2 m \Delta}{27 b}
$$

Draw a line parallel to $B C$ at a distance $h_{1}$ from $B C$; draw another line parallel to $C A$ at a distance $h_{2}$ from $C A$. Both lines are drawn such that they intersect at a point $P$ inside the triangle $A B C$. Then

$$
[P B C]=\frac{1}{2} a h_{1}=\frac{l \Delta}{27}, \quad[P C A]=\frac{m \Delta}{27}
$$

Hence

$$
[P A B]=\frac{n \Delta}{27}
$$

This shows that the distance of $P$ from $A B$ is

$$
h_{3}=\frac{2 n \Delta}{27 c}
$$

Therefore each traingle with base on $C A$ has area $\frac{\Delta}{27}$. We conclude that all the triangles which partitions $A B C$ have equal areas. Hence $P$ is a good point.

Thus the number of good points is equal to the number of positive integral solutions of the equation $l+m+n=27$. This is equal to

$$
\binom{26}{2}=325
$$
