169 points are chosen at random inside an equilateral triangle of perimeter 300.

Question

I have this excercise:

169 points are chosen at random inside an equilateral triangle of perimeter 300. Prove that there are 3 of these points that determine a triangle of area at most 68.

I think that i can use the generalized pigeonhole principle which i know as:

Let $n,m,r$ be integers such that $n>rm$. If $n$ objects are placed into $m$ boxes, then there is at least one box containing at least $r+1:=\lceil n/m\rceil$ objects.

Since $n=169$ and I would like 3 points to fall in the same hole to form a triangle with an area smaller than that of the hole, I take $r=2$, then $m$ is the largest square less than $\lfloor n/r\rfloor=84$ (This is because when dividing an equilateral triangle into triangles that are also equilateral, I can do it in a square number of them) so $m=81$.

Now by Heron's formula the area of triangle $XYZ$ is $$A=\sqrt{s(s-x)(s-y)(s-z)}$$Where $s=\frac{x+y+z}{2}$ is the semiperimeter, in this case $x=y=z:=l=100$, so since $m=81=9^2$, we are dividing each side into $9$ equal segments, so the side of each small triangle measures $100/9:=b$. The semiperimeter of the small triangle is $c=\frac{3b}{2}=\frac{50}{3}$, apply now the Heron's formula $$a=\sqrt{c(c-b)^3}=(c-b)\sqrt{c(c-b)}=\frac{2500}{27\sqrt{3}}\approx53.4583\leq68$$But i see that it's a bad approximation, so I thought that I could divide the triangle into fewer holes, that is, divide each side into one less space, where $m=64$ and in this way the calculations would be: $b=\frac{25}{2}$, $c=\frac{75}{4}$,$$a=\sqrt{c(c-b)^3}=\frac{25^2}{4^2}\sqrt{3}\approx67.6582\leq68$$Which is a much better approximation, but I don't know how to justify this reasoning, or if I'm thinking it right, I need help.

Answer 1

Start by subdividing the original equilateral triangle into $4$ small equilateral triangles, as shown in the figure. Repeat this process on each of the $4$ smaller triangles, producing $16$ triangles; then further subdivide each of these $16$ triangles, producing $64$ congruent equilateral subtriangles. The original triangle has an area of 4330.13 square units, so each of the $64$ smaller triangles has area $67.66$ square units. By the generalized pigeonhole principle, at least one of the $64$ smaller triangles contains at least $3$ of the $169$ points. The triangle formed by these $3$ points must be $67.66$ square units or less.