Intrigued by the thread http://boards.straightdope.com/sdmb/showthread.php?t=463314&highlight=area+triangle, I got curious first where the number 7 came from (it seemed unlikely) and also what would happen if you replaced 1/3 by 1/4, 15/, …, 1/n. The ratio would obviously have to approach 1 as n gets large. So here is the challenge. Find the area of the inner triangle if you divide each side in ration 1/n, (n-1)/n. Solution follows:
The answer is (n-1)^2/(n^2-n+1). To get the solution, begin with Ultrafilter’s elegant observation that since any two triangles can be related by an affine transformation and all such preserve both length ratios along any line as area ratios. It simplifies the arithmetic somewhat to let k = n^2-n+1 and use the triangle with vertices (0,0), (k,0), and (0,k). Then the 1/n points will be at (k/n,0), (k-k/n,k/n), and (0,k-k/n). The three lines from the vertices to the opposite division points have slopes 1/(n-1), -(n-1)/n, and -n respectively. From this, it is easy to see that the vertices of the inner triangle are (x_1,y_1) = (n-1,n), (x_2,y_2) = ((n-1)^2,(n-1)) and (x_3,y_3) = (1,(n-1)^2). The area of such a triangle is 1/2 the absolute value of the determinant of the matrix (x_2-x_1,y_2-y_1;x_3-x_1,y_3-y_1) which simplifies to (n-1)^2/k. Now divide by k^2 (twice the area of the original triangle), you get (n-2)^2/k = (n-2)^2/(n^2-n+1). When n = 3, you get 1/7, the original problem. n = 4 and 5 give 4/13 and 9/21.