You are given some equilateral triangles and squares, all with side length 1, and asked to form convex $n$ sided polygons using these pieces. If both types must be used, what are the possible values of $n$, assuming that there is sufficient supply of the pieces?
Problem
Source: 2018 Singapore Mathematical Olympiad Senior Q1
Tags: combinatorics, geometry, combinatorial geometry
ACGNmath
27.07.2023 07:45
The possible internal angles are $60^\circ, 90^\circ, 120^\circ$ and $150^\circ$. Let their respective numbers be $a, b, c, d$. Then we have \[a+b+c+d=n\]\[60a+90b+120c+150d=(n-2)180\]Eliminating $n$, we obtain \[4a+3b+2c+d=12\]Therefore, $n\leq 12$. First, we construct polygons with $5 \leq n \leq 12$. [asy][asy] unitsize(24); pair A = (0,0); pair B = dir(30); pair C = dir(330); pair D = B+(1,0); pair E = C+(1,0); draw(A--B--C--A--cycle); draw(B--D--E--C); [/asy][/asy] [asy][asy] unitsize(24); pair A = (0,0); pair B = dir(30); pair C = dir(330); pair D = B+(1,0); pair E = C+(1,0); pair G = E+dir(30); draw(A--B--C--A--cycle); draw(B--D--E--C); draw(D--G--E); [/asy][/asy] [asy][asy] unitsize(24); draw((0,0)--(1,0)--(1,1)--(0,1)--(0,0)--cycle); draw((0,1)--(0,1)+dir(60)--(1,1)); draw((1,0)--(1,0)+dir(30)--(1,1)); draw((1,1)--(1,1)+dir(60)--(1,1)+dir(120)); draw((1,1)+dir(60)--(1,1)+dir(60)+dir(330)--(1,1)+dir(330)); [/asy][/asy] [asy][asy] unitsize(24); draw((0,0)--(2,0)--(2,1)--(0,1)--(0,0)--cycle); draw((1,0)--(1,1)); draw((0,1)--(0,1)+dir(60)--(1,1)+dir(60)--(2,1)); draw((0,1)+dir(60)--(1,1)--(1,1)+dir(60)); draw((0,0)--(0,0)+dir(300)--(1,0)--(1,0)+dir(300)--(2,0)); draw((0,0)+dir(300)--(1,0)+dir(300)); [/asy][/asy] [asy][asy] unitsize(24); draw((0,0)--(1,0)--(1,1)--(0,1)--(0,0)--cycle); draw((0,1)--(0,1)+dir(60)--(1,1)); draw((1,0)--(1,0)+dir(30)--(1,1)); draw((1,1)--(1,1)+dir(60)--(1,1)+dir(120)); draw((1,1)+dir(60)--(1,1)+dir(60)+dir(330)--(1,1)+dir(330)); draw((0,0)--dir(300)--(1,0)--(1,0)+dir(300)--dir(300)); draw((1,0)+dir(300)--(1,0)+dir(300)+dir(30)--(1,0)+dir(30)--(1,0)+dir(30)+dir(0)); draw((1,0)+dir(30)+dir(0)+dir(120)--(1,0)+dir(30)+dir(0)--(1,0)+dir(30)+dir(0)+dir(240)); [/asy][/asy] [asy][asy] unitsize(24); draw((0,0)--dir(0)--dir(60)--(0,0)--dir(300)--dir(0)); draw(dir(60)--dir(60)+dir(30)--dir(60)+dir(30)+dir(-60)--dir(0)--dir(0)+dir(330)--dir(0)+dir(330)+dir(240)--dir(300)--dir(300)+dir(210)--dir(300)+dir(210)+dir(120)--dir(210)--(0,0)--dir(150)--dir(150)+dir(60)--dir(60)--dir(60)+dir(90)--dir(60)+dir(90)+dir(210)); draw(dir(60)+dir(90)--dir(60)+dir(90)+dir(330)); draw(dir(0)+dir(30)--dir(0)+dir(330)); draw(dir(150)--dir(210)); draw(dir(300)--dir(300)+dir(270)--dir(300)+dir(270)+dir(150)); draw(dir(300)+dir(270)--dir(300)+dir(270)+dir(30)); [/asy][/asy] [asy][asy] unitsize(24); draw(dir(90)--dir(270)); draw(dir(30)--dir(210)); draw(dir(150)--dir(330)); draw(dir(90)--dir(30)--dir(330)--dir(270)--dir(210)--dir(150)--dir(90)--cycle); draw(dir(90)--dir(90)+dir(120)--dir(90)+dir(60)--dir(90)--cycle); draw(dir(90)+dir(60)--dir(90)+dir(60)+dir(330)--dir(30)); draw(dir(30)+dir(60)--dir(30)+dir(0)--dir(30)--cycle); draw(dir(30)+dir(0)--dir(30)+dir(0)+dir(270)--dir(330)); draw(dir(330)+dir(0)--dir(330)+dir(0)+dir(240)--dir(330)--cycle); draw(dir(330)+dir(300)--dir(330)+dir(300)+dir(210)--dir(270)--dir(270)+dir(240)--dir(270)+dir(300)); draw(dir(270)+dir(240)--dir(270)+dir(240)+dir(150)--dir(210)--dir(210)+dir(150)--dir(210)+dir(150)+dir(240)--dir(210)+dir(240)); draw(dir(90)+dir(120)--dir(90)+dir(120)+dir(210)--dir(150)--dir(150)+dir(210)--dir(150)+dir(210)--dir(150)+dir(210)+dir(120)--dir(150)+dir(120)); draw(dir(150)+dir(210)+dir(120)--dir(150)+dir(210)+dir(120)+dir(240)--dir(150)+dir(210)+dir(120)+dir(240)+dir(300)); draw(dir(150)+dir(210)--dir(150)+dir(210)+dir(180)); [/asy][/asy] [asy][asy] unitsize(24); draw(dir(90)--dir(270)); draw(dir(30)--dir(210)); draw(dir(150)--dir(330)); draw(dir(90)--dir(30)--dir(330)--dir(270)--dir(210)--dir(150)--dir(90)--cycle); draw(dir(90)--dir(90)+dir(120)--dir(90)+dir(60)--dir(90)--cycle); draw(dir(90)+dir(60)--dir(90)+dir(60)+dir(330)--dir(30)); draw(dir(30)+dir(60)--dir(30)+dir(0)--dir(30)--cycle); draw(dir(30)+dir(0)--dir(30)+dir(0)+dir(270)--dir(330)); draw(dir(330)+dir(0)--dir(330)+dir(0)+dir(240)--dir(330)--cycle); draw(dir(330)+dir(300)--dir(330)+dir(300)+dir(210)--dir(270)--dir(270)+dir(240)--dir(270)+dir(300)); draw(dir(270)+dir(240)--dir(270)+dir(240)+dir(150)--dir(210)--dir(210)+dir(180)--dir(210)+dir(180)+dir(300)); draw(dir(150)--dir(150)+dir(180)--dir(150)+dir(180)+dir(270)); draw(dir(150)+dir(180)--dir(150)+dir(180)+dir(60)--dir(150)--cycle); draw(dir(150)+dir(180)+dir(60)--dir(150)+dir(180)+dir(60)+dir(30)); [/asy][/asy] To complete the proof, we now show that $n$ cannot be 3 or 4. First note that when a triangle and a square meet at the boundary, they meet at a vertex of the polygon and the angle at the vertex is $150^\circ$. Thus when $n=4$, the boundary of the polygon formed consists entirely of squares, or entirely of triangles (i.e. a rectangle or a parallelogram). Removing this outermost layer will yield a smaller rectangle or parallelogram respectively. As for $n=3$, the boundary of the triangle formed contains only triangles. Removing the bottom row of small triangles yields a smaller triangle, and we can conclude that no square is used. Therefore, $n$ cannot be 3 or 4.