Singapore Mathematical Olympiad (SMO)

Consider the sum of the angles of the triangles at each type of the vertices. At type 0, the sum is $180^{\circ }$. At type 1, the sum is $180t_1$. At type 2, the sum is $360t_2$. Therefore $$180+180t_1+360t_2=180n\Rightarrow 1+t_1+2t_2=n.$$ Now each Type 1 vertex has at most 1 angle of $x^{\circ }$ and each Type 2 vertex has at most 2 which each Type 0 vertex has, none. So the number of angles of $x^{\circ }$ is $\le t_1+2t_2<n$, a contradiction.

Now we construct for each $x\in (0,120]$. It is trivial for $x=120$. Take the centre of the triangle and join it to the vertices to obtain a partition into 3 congruent triangles. Now assume $x\in (0,120)$.

Define an $(a,b)$-trapezium to be the trapezium $ABCD$ with $\angle A=\angle D=60^{\circ }$, $AB=CD=a$, $BC=b$. Then it follows that $AD=a+b$. We call a polygon $x$-good if it can be partitioned into triangles each with one $x^{\circ }$ angle.

Lemma 1: There exists $R(a)$ such that an $(a,b)$-trapezium is $x$-good when $b>R(a)$. Proof: The proof is clear from Fig. A. The second, third and fourth triangles are isosceles. The fifth and sixth triangles form a parallelogram with arbitrarily long horizontal side.



Fig.A

Fig.B

Lemma 2: An $(a,b)$-trapezium is $x$-good. Proof: It follows from the fact that the trapezium can be sliced into arbitrarily thin $(a^{\prime },b^{\prime })$-trapezium so that $b^{\prime }>R(a^{\prime })$. (See Fig. B.) Since an equilateral triangle can be partitioned in $(a,b)$-trapeziums, the proof is complete.