Challenge question about triangle areas

Intrigued by the thread, 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.

If I plug n = 3 into that, I get 4/7. You state what looks like the correct answer further down, where you have n - 2 all squared in the numerator

Routh’s theorem was linked in the previous thread.

Letting x = y = z = n - 1 in Routh’s theorem yields:

(n - 2)^2 / (n^2 - n + 1)

I looked at the triangle problem and went on.

My father in law. maybe a reliable source or not once explained our numerals were once based on how many angles the figure had. A vertical line with a little line at the top had one angle. A European crossed 7 would have 7 angles. 6, 8, and 9 needed to have polygons, not loops. A 0 had no angles.

A) This isn’t true. The history of the digit glyphs is well-known, and while there was originally a straightforward “n strokes (NOT n angles) for the digit n” pattern for the digits up through 3 (and, in a looser sense, 4), this does not extend to the larger ones, whose glyphs are quite arbitrary. See Wikipedia.

B) The story you were told isn’t even coherent; you have to be extremely contrived with the drawings of 4, 5, 7, and 9 to make it work. No one writes these digits the way this story demands you write these digits.

Since OP has been answered (in OP itself :cool:), here are two more very challenging puzzles about triangle areas (possibly due to J.J. Sylvester).

(1) Choose three points randomly inside a square of unit area. What is the expected area of the triangle formed by those 3 points?

(2) Choose three points randomly inside a triangle of unit area. What is the expected area of the triangle formed by those 3 points?

This fun challenge may have attracted no interest, but I’ll add a comment if only for the archive. Although difficult, the puzzle isn’t too difficult if approached in steps.

Here is the most difficult subproblem in those individual steps:
Given the following three points defined by uniform variates x,y
[ul][li] (0,0)[/li][li] (x,1) , 0 < x < 1[/li][li] (1,y) , 0 < y < 1[/li][/ul]
which divide the unit square into four triangles, determine the expected areas of those four triangles. (With two variable coordinates instead of six, this seems much easier than the quoted puzzle yet, as I say, is the hardest single step in solving that puzzle.)

OK, I did the challenge by doing a sextuple integral of a piecewise function. Mathematica helped a lot, but even so I’m not sure I didn’t make an error. Although both answers agree fairly closely with the average of a million randomly generated trials.

(1) 11/144

(2) 1/12

Interesting that the numbers are closely related.
11/144 = 1/12 (1-1/12)

The sextuple piecewise integral seems almost masochistic, but I guess Mathematica makes it a breeze. In another solution method, Integral 2x = x^2 is the “worst” calculus encountered.