Yes; for anyone reading this thread in the future, I’ll re-write the proof:
The answer is also confirmed by Routh’s Theorem with r=s=t=2.
There was a time long ago when I wouldn’t make such elementary mathematical mistakes in a proof; I hope it’s just an off day…and appreciate the indulgence of the board. I think this was more valuable for me than the OP, as the exercise demonstrates that “fighting ignorance” isn’t always an external battle.
Damn you! I was just about to post the shocking revelation that I had figured out a better way to space my triangular grid which made it easy to precisely see the area of both triangles for the equilateral case, and that it proved you were all wrong and that the answer was 1/7 (one has an area of 63 and the other has an area of 441) 
That’s the idea I had in mind, though my solution is a little smaller. I like how it reduces the problem to one of counting triangles.
And about a million times better. Very nice.
Thx, I’m still digesting all this. FWIW I asked a math teacher at school who worked it out and got 1/7 of the area of the larger triangle.
I have a similar issue in that I have been give a triangle of any given size or type. I need to find the area a smaller triangle that is equally smaller than the larger on all sides (IE: 18" smaller). How is this calculated and what would it look like for an Excel spreadsheet? Ruthe’s theorem appears to be a bit difficult to translate for this situation.
That’ll depend on the exact dimensions (all three of them) of the larger triangle. For instance, if any of the larger triangle’s sides is 18", then the area of the smaller triangle vanishes. Once you have all three sides of the larger triangle, then you can easily get the sides of the smaller triangle, and then apply Hero’s Formula.
I’ve a hunch, though, that the problem you’re facing actually asks for the sides of the new triangle to be a certain proportion smaller (such as 18%), rather than a certain absolute amount smaller (like 18").
At this point, I don’t know if anyone cares, but here is my solution. Starting from Ultrafilter’s brilliant suggestion, I said to myself, why not take an isoscelese right triangle, specifically the one whose sides are the x-axis, the y-axis and the line x + y + 1, whose area is 1/2. Take the 1/3 points to have coordinates (1/3,0), (2/3,1/3), and (0,2/3). A little bit of work gives the triangle in question as having vertices (2/7,1/7), (4/7,2/7), and (1/7,4/7). The side lengths are then 1/7 times sqrt(5), sqrt(10), and sqrt(13), respectively. Now Heron’s formula for the area is not too useful in its usual form, but it is not hard to show that it is equivalent to (1/4)sqrt(2a^2b^2 + 2b^2c^2 + 2b^2c^2 - a^4 - b^4 -c^4) which gets rid of all but the outside radial. When evaluated you get (1/196)sqrt(196) which is 1/14, exactly 1/7 times the original area.
