Create an isosceles triangle from only equilateral triangles. Possible or impossible?

Question

My friend recently asked me a question. And I think I have a nice solution but wanted to share this question here to see how you approach this problem and hopefully with a solid proof.

The question is:

Can you create an isosceles triangle with an infinite number of any sizes of only equilateral triangles?

Answer 1

Start from the 120-30-30 isosceles triangle. We aim to fill this triangle with equilateral triangles in infinite steps.

Draw an equilateral triangle at one-third of the longest side.

Now the triangle is $\frac13$ filled, with two smaller 120-30-30 isosceles triangles created. Repeat the process:

And now $\frac13 + \frac29$ of the triangle is filled. 3rd iteration below.

Why would the initial triangle be filled in infinite steps? The area filled in $n$ steps is equal to:

$$\sum_{k=1}^n \frac13(\frac23)^{k-1}$$

At infinity, this geometric sum evaluates to

$$\sum_{k=1}^\infty \frac13(\frac23)^{k-1} = \frac {1/3}{1-2/3}=1$$

so the whole triangle will be filled, given that we have infinite equilateral triangles of all sizes.

Or, equivalently, considering the fact that $1/3$ of the remaining area is filled in each iteration, we have the area remaining after the $n$th step is $\dfrac1{3^n}$, which $\to 0$ as $n \to \infty$.

Answer 2

Okay, I'll bite. I don't know if the triangles are allowed to overlap, but my argument would work in either case.

Assume that we can choose a triangle formed from equilateral triangles that is isosceles but not equilateral. Then the smallest angle of the isosceles triangle is strictly less than 60 degrees. Let A be the vertex of that angle. Among the infinite number of equilateral triangles that cover the triangle, choose one that includes point A. Let A' be the point on that equilateral triangle that covers A. Any neighborhood of A' would have to include points B' and C' such that m$\angle B'A'C'\ge60^\circ$. But then either B' or C' would be outside the isosceles triangle, which is a contradiction.

Needless to say, except that this is a puzzle: an equilateral triangle itself is isosceles, and you could construct that from an infinite number of equilateral triangle using the Sierpinski triangle as a guide.

Answer 3

You can fill the interior of any triangle (or more generally, any plane region) with non-overlapping equilateral triangles. Consider an equilateral triangular lattice $L$, and start with those lattice triangles that are contained in your region $R$.
Then refine the lattice (adding the midpoints of all edges), and take the triangles of the new lattice that are contained in $R$ but are not already covered. Repeat ad infinitum.

EDIT: You get something that looks like this

Answer 4

In any triangle, at least one angle is $\ge 60^{\circ}$.

So starting from the corresponding vertex you can draw an equilateral triangle, inscribed in the starting one , and which leaves out two triangles.

For each of these remaining triangles you can repeat the process and leave out four. And so on.