Question

Let $ABC$ be a triangle. An interior point $P$ of $ABC$ is said to be good if we can find exactly $27$ rays emanating from $P$ intersecting the sides of the triangle $ABC$ 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 $ABC$.

Answer 1

Let $P$ be a good point. Let $l,m,n$ be respectively the number of parts the sides $BC,CA,AB$ are divided by the rays starting from $P$.
A ray must pass through each of the vertices the triangle $ABC$; otherwise we get some quadrilaterals.
Let $h_1$ be the distance of $P$ from $BC$. Then $h_1$ is the height for all the triangles with their bases on $BC$. Equality of areas implies that all these bases have equal length. If we denote this by $x$, we get $lx=a$.
Similarly, taking $yandz$ as the lengths of the bases of triangles on $CAandAB$ respectively, we get $my=bandnz=c$. Let $h_{2}and{h}_3$ be the distances of $P$ from $CAandAB$ respectively. Then
$h_{1}x=h_{2}y=h_{3}z=\frac{2\Delta }{27}$,
where $\Delta$ denotes the area of the triangle $ABC$. These lead to
$h_1=\frac{2\Delta }{27}\frac{l}{a},h_1=\frac{2\Delta }{27}\frac{m}{b},h_1=\frac{2\Delta }{27}\frac{n}{c}$.
But
$\frac{2\Delta }{a}=h_a,\frac{2\Delta }{b}=h_b,\frac{2\Delta }{c}=h_c$.
$\therefore \frac{h_1}{h_a}=\frac{l}{27},\frac{h_2}{h_b}=\frac{m}{27},\frac{h_3}{h_c}=\frac{n}{27}$.
Also, we have
$\frac{h_1}{h_a}=\frac{\left[PBC\right]}{\Delta },\frac{\left[PCA\right]}{\Delta },\frac{\left[PAB\right]}{\Delta }$.
Adding these three relations,
$\frac{h_1}{h_a}+\frac{h_2}{h_b}+\frac{h_3}{h_c}=1$.
Thus
$\frac{l}{27}+\frac{m}{27}+\frac{n}{27}=\frac{h_1}{h_a}+\frac{h_2}{h_b}+\frac{h_3}{h_c}=1$.
$\therefore l+m+n=27$.
Thus every good point $P$ determines a partition $(l,m,n)$ of $27$ such that there are $l,m,n$ equal segments respectively on $BC,CA,AB$.
Conversely, take any partition $(l,m,n)$ of $27$. Divide $BC,CA,AB$ respectively into $l,m,n$ equal parts. Define
$h_1=\frac{2l\Delta }{27a},h_2=\frac{2m\Delta }{27b}$.
Draw a line parallel to $BC$ at a distance $h_2$ from $CA$. Both lines are drawn such that they intersect at a point $P$ inside the triangle $ABC$. Then
$\left[PBC\right]=\frac{1}{2}a{h}_1=\frac{l\Delta }{27},\left[PCA\right]=\frac{m\Delta }{27}$.
Hence
$\left[PAB\right]=\frac{n\Delta }{27}$.
This shows that the distance of $P$ from $AB$ is $h_3=\frac{2n\Delta }{27c}$.
Therefore each triangle with base on $CA$ has area $\frac{\Delta }{27}$. We conclude that all the triangles which partitions $ABC$ 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
$\left(\begin{array}{c}26\\ 2\end{array}\right)=325$.