If you're wondering how hard this is, a Thai IMO team member who is good at combi couldn't solve this in 3 hours

WHY IS MOVED HERE IT NEEDS TO BE MOVED BACK TO COLLEGE MATH

i'm actually interested to know the solution for this, anyone has it ...?

I made this problem, and here is my solution. The answer is all $x\in (0,120]$. We first show that $x>120$ is impossible. Suppose $x>120$ and such a partition into small triangles. Consider the set of vertices of the small triangles, which can be partitioned into 3 types. Let $A$ be the 3 vertices of the original equilateral triangle. Let $B$ be the set of vertices that lies on an edge of a small triangle, or on an edge of the original equilateral triangle. Let $C$ be the remaining vertices. Count the total sum of all the angles of the small triangles. If there are $n$ small triangles, this sum is $180n$. But if we were to count this sum at each vertex, each vertex of $A$ contribute 60; each vertex of $B$ contribute 180; and each vertex of $C$ contribute 360. Thus $180+180|B|+360|C|=180n$, so $|B|+2|C|=n-1$. Now count the number of angles $x$, which is $n$ by assumption. But each vertex in $A$ cannot have an angle $x$, each vertex in $B$ can have at most one angle $x$, and each vertex in $C$ can have at most 2 angles $x$, since $x>120$. Thus $n\leq |B|+2|C|=n-1$, a contradiction. Now we construct for each $x\in (0,120]$. The case $x=120$ is trivial: partition the equilateral triangle $ABC$ into $AOB,BOC,COA$, where $O$ is the center of $ABC$. So assume that $0<x<120$. For $R>0$, define an $R$-trapezoid to be a trapezoid similar to $ABCD$, where $\angle A=\angle D=60^\circ$, $\angle B=\angle C=120^\circ$, $AB=CD=1$ and $BC=R$. We say that a shape is constructible if it can be partitioned into triangles, each of which has an angle $x^\circ$. Lemma 1: For all sufficiently large $R$, the $R$-trapezoid is constructible. Proof: See figure below. $\square$ Lemma 2: The 1-trapezoid is constructible. Proof: Slice the 1-trapezoid horizontally into sufficiently thin $R$-trapezoids (for $R$ sufficiently large), each of which is constructible by Lemma 1. $\square$ Finally, the equilateral triangle is constructible, since you can partition it into 3 1-trapezoids.

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beansenthusiast505 wrote: If you're wondering how hard this is, a Thai IMO team member who is good at combi couldn't solve this in 3 hours Yeah that was me. I livesolved this problem in MODS and it took me 8:30h.