There are 27 small triangles of equal area to be constructed using a common point interior of a given triangle ABC. We can deduce that in this way, each side of the larger triangle ABC will have the base of some number of small triangles. LET side AB have the base of only one triangle on it, then sides BC and AC will take turns to share 26 triangles in between them. i.e. (25,1) triangle, and (24,2) triangles, and (23,3) triangles and so on, respectively. We can see that this can be done in 25 ways.

Paragraph 2 Now, if AB has 2 triangles on it, then AC and BC take turn to share 25 triangles in between them. i.e. (24,1); (13,2) ; and so on, in 24 ways...

If we go on this way, we will find that there are (25 + 24+ 23 + 22 + ....... + 2 + 1) ways to do it. That is, 325 ways!

The stars and bars method applies here. We need to partition 27 units into 3 sectors. Since each sector must have a minimum of one unit, we need to partition 24 units into 3 sectors. The answer is C(26,2) = 325.

Refer to INMO 2012 solutions Q.4. Link: http://olympiads.hbcse.tifr.res.in/uploads/inmo-2012.pdf

From the point P, there must be 3 rays that go through A, B, C. Then we have 3 new triangles: ABP, ACP, BCP. All the area must be equal for each small triangle, so it depends on how we split ABP, ACP and BCP into smaller ones. Denote x, y and z are the number of small triangles in ABP, CBP, ACP, we have x+y+z=27 , with x, y, z are positive integers. Follow by Euler formular, there are C(3-1)/(27-1) ways to do so. For each way that we split, the point P is unique due to the sectors AB, BC, AC. So, there are 325 good points

From point P length of altitudes of triangle formed on one side will be equal. For another side just we have to manage the length of base. So now we have to find out the no. Of solution of the equation a+b+c=27 I.e. 325