Hello,

A diagram would be most helpful, but I will try to convey the basic idea in words. Imagine the triangle arranged with one side horizontal (call it Side 1). From the top vertex, mark off points along Sides 2 and 3 at 1-inch intervals until you reach the base (Side 1). Connecting "opposite" points on Sides 2 and 3 with horizontal line segments (segments parallel to Side 1), you will have divided the triangle into 27 "strips". Note that each strip (except the top one) is an isosceles trapezoid. Now starting from the top vertex, we see that the first strip consists of just one equilateral triangle (of side-length one). It is relatively easy to see that the second strip can be divided into 3 such equilateral triangles. Similarly, the third strip has 5 triangles, the fourth has 7, etc. The number of triangles in each strip thus follows the sequence 1,3,5,7,9,....,53 (the sequence of odd positive integers, up to the 27th odd number.) Therefore the number of triangles is

n = 1 + 3 + 5 + 7 + .... + 51 + 53

(You can compute the sum by brute force, or better, note that the sequence of terms in this sum is an **arithmetic sequence.)** One intuitive way to get the sum of such a series is

**Sum = (Number of terms)[(First term + Last Term)/2]**

Therefore,

n = 27[(1 + 53)/2]

n = 729

Hope that helps! If you need any further explanation (especially about arithmetic series and how to find their sums), let me know.

William