How many "marvelous" points inside a triangle.

Question

We call a point $P$ inside a triangle $\triangle ABC$ marvellous if exactly $27$ rays can be drawn from it, intersecting the sides of $\triangle ABC$ such that the triangle is divided into $27$ smaller triangles of equal areas.

Determine with proof the total number of marvellous points inside a given triangle $\triangle ABC$.

Answer 1

Assuming "intersecting the sides or vertices of $\triangle ABC$, the three rays must be drawn through the vertices, otherwise a quadrilateral will be formed on the vertex. Then from the point $P$ there must be drawn $24$ rays to the sides of $\triangle ABC$. Let the areas of the triangles $\triangle PBC$, $\triangle PAC$, $\triangle PAB$ be in ratio $a:b:c$, respectively. Refer to the figure:

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Then: $$S_{\triangle PBC}=\frac{a}{27}S_{\triangle ABC}; \qquad S_{\triangle PAC}=\frac{b}{27}S_{\triangle ABC}; \qquad S_{\triangle PAB}=\frac{c}{27}S_{\triangle ABC}.$$

We will show that the point $P$ is unique for each ratio $a:b:c$. Note: $$a+b+c=27 \Rightarrow c=27-a-b; \\ PF=\frac{2S_{\triangle PBC}}{BC}=\frac{2aS_{\triangle ABC}}{27}; \ \ PD=\frac{2S_{\triangle PAC}}{AC}=\frac{2bS_{\triangle ABC}}{27}; \\ PE=\frac{2S_{\triangle PAB}}{AB}=\frac{2cS_{\triangle ABC}}{27}=\frac{2(27-a-b)S_{\triangle ABC}}{27}.$$ For each set of $a,b,c$, the altitudes $PF, PD, PE$ are uniquely determined.

Now we want the areas of $\triangle PBC$, $\triangle PAC$, $\triangle PAB$ to be in the ratio $a:b:c$ such that: $$a,b,c\in \mathbb{N^+};\\ 1\le a,b,c\le 25; \\ a+b+c=27.$$ Then we can have $a,b,c$ small triangles.

Using stars and bars method we find: $${27-1\choose 3-1}={26\choose 2}=325.$$

Answer 2

As you insist the regions are all triangles, the $27$ rays must include $PA$, $PB$ and $PC$. Then the triangles $PAB$, $PBC$ and $PAC$ must be divided into $r$, $s$ and $t$ little triangles where $r$, $s$ and $t$ are positive numbers adding to $27$. I claim that for each triple $(r,s,t)$ there is exactly one $P$ that works. To see this, the locus of $P$ with area$(PAB)=(r/27)$ area$(ABC)$ is a line parallel to $AB$. Likewise the locus of $P$ with area$(PBC)=(s/27)$ area$(ABC)$ is a line parallel to $BC$. These do meet at a unique point $P$, and then automatically area$(PCA)=(t/27)$ area$(ABC)$.

We now need to count the triples $(r,s,t)$. The map $(r,s,t)\to\{r,r+s\}$ bijects them to the two-element subsets of $\{1,\ldots,26\}$ and there are $\binom{26}2$ of these.

This answer is partly based on String's comments.