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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.

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Solution

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 h1 be the distance of P from BC. Then h1 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 h2andh3 be the distances of P from CAandAB respectively. Then
h1x=h2y=h3z=2Δ27,
where Δ denotes the area of the triangle ABC. These lead to
h1=2Δ27la,h1=2Δ27mb,h1=2Δ27nc.
But
2Δa=ha,2Δb=hb,2Δc=hc.
h1ha=l27,h2hb=m27,h3hc=n27.
Also, we have
h1ha=[PBC]Δ,[PCA]Δ,[PAB]Δ.
Adding these three relations,
h1ha+h2hb+h3hc=1.
Thus
l27+m27+n27=h1ha+h2hb+h3hc=1.
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
h1=2lΔ27a,h2=2mΔ27b.
Draw a line parallel to BC at a distance h2 from CA. Both lines are drawn such that they intersect at a point P inside the triangle ABC. Then
[PBC]=12ah1=lΔ27,[PCA]=mΔ27.
Hence
[PAB]=nΔ27.
This shows that the distance of P from AB is h3=2nΔ27c.
Therefore each triangle with base on CA has area Δ27. 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
(262)=325.

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