HMMT-Nov 2019 gen, Problem 4

In $\triangle A B C, A B=2019, B C=2020$, and $C A=2021$. Yannick draws three regular $n$-gons in the plane of $\triangle A B C$ so that each $n$-gon shares a side with a distinct side of $\triangle A B C$ and no two of the $n$-gons overlap. What is the maximum possible value of $n$ ?

Solution: If any $n$-gon is drawn on the same side of one side of $\triangle A B C$ as $\triangle A B C$ itself, it will necessarily overlap with another triangle whenever $n>3$. Thus either $n=3$ or the triangles are all outside $A B C$. The interior angle of a regular $n$-gon is $180^{\circ} \cdot \frac{n-2}{n}$, so we require \[ 360^{\circ} \cdot \frac{n-2}{n}+\max (\angle A, \angle B, \angle C)<360^{\circ} \] As $\triangle A B C$ is almost equilateral (in fact the largest angle is less than $60.1^{\circ}$ ), each angle is approximately $60^{\circ}$, so we require \[ 360 \cdot \frac{n-2}{n}<300 \Longrightarrow n<12 \] Hence the answer is $n=11$. $\fbox{11}$.

Answer: 11