In regular polygon ABCDE…, we have ∠ACD = 120°. How many sides does the polygon have?
How would I start with solving this? AC would have different lengths depending on the polygon.
[Math] In regular polygon ABCDE…, we have ∠ACD = 120°. How many sides does the polygon have
geometrypolygons
Related Solutions
Consider a regular polygon with $n$ number of vertices $\mathrm{A_1, \ A_2,\ A_3, \ A_3, \ldots , A_{n-1}}$ & $\mathrm{A_{n}}$
Total number of triangles formed by joining the vertices of n-sided regular polygon $$N=\text{number of ways of selecting 3 vertices out of n}=\color{}{\binom{n}{3}}$$ $$N=\color{red}{\frac{n(n-1)(n-2)}{6}}$$ $\forall \ \ \color{blue}{n\geq 3}$
Consider a side $\mathrm{A_1A_2}$ of regular n-polygon. To get a triangle with only one side $A_1A_2$ common (As shown in figure-1 below)
(figure-1)
Join the vertices $A_1$ & $A_2$ to any of $(n-4)$ vertices i.e. $A_4, \ A_5,\ A_6, \ \ldots \ A_{n-1}$ to get triangles with only one side common. Thus there are $(n-4)$ different triangles with only one side $A_1A_2$ common. Similarly, there are $(n-4)$ different triangles with only one side $A_2A_3$ common & so on. Thus there are $(n-4)$ different triangles with each of $n$ sides common. Therefore, number of triangles $N_1$ having only one side common with that of the polygon $$N_1=\text{(No. of triangles corresponding to one side)}\text{(No. of sides)}=\color{blue}{(n-4)n}$$
(figure-2)
Now, join the alternate vertices $A_1$ & $A_3$ by a straight (blue) line to get a triangle $A_1A_2A_3$ with two sides $A_1A_2$ & $A_2A_3$ common. Similarly, join alternate vertices $A_2$ & $A_4$ to get another triangle $A_2A_3A_4$ with two sides $A_2A_3$ & $A_3A_4$ common & so on (as shown in above figure-2). Thus there are $n$ pairs of alternate & consecutive vertices to get $n$ different triangles with two sides common (Above fig-2 shows $n$ st. lines of different colors to join alternate & consecutive vertices). Therefore, number of triangles $N_2$ having two sides common with that of the polygon $$N_2=\color{blue}{n}$$ If $N_0$ is the number of triangles having no side common with that of the polygon then we have $$N=N_0+N_1+N_2$$ $$N_0=N-N_1-N_2$$ $$=\binom{n}{3}-(n-4)n-n$$ $$=\color{}{\frac{n(n-1)(n-2)}{6}-n^2+3n}$$ $$N_0=\color{red}{\frac{n(n-4)(n-5)}{6}}$$ The above formula $(N_0)$ is valid for polygon having $n$ no. of the sides such that $ \ \ \color{blue}{n\geq 6}$
I'm not saying you can't refine the proof you attempted, but the first thing that comes to mind when dealing with symmetries of polygons is a group action of the dihedral group on the polygon. In a sense, group theory is the study of symmetry: when a group acts on a set, each group element describes an operation that will rearrange the set in such a way that it looks like it did before, and only the labels have changed. A permutation is precisely a reordering of elements, and Cayley's theorem states that every group is isomorphic to a subgroup of a symmetric group (which is the group of all permutations on a certain number of elements).
I'll use $D_{2n}$ to denote the dihedral group of $2n$ elements. It is generated by two elements: a reflection $R$ (which you can think of as reflecting an $n$-gon across your favorite axis, say the vertical one), and a rotation $r$ (which you can think of as rotating the $n$-gon clockwise). It should be clear that if you reflect across the same axis twice, you're back to where you started (so $R^2 = 1$), and the same is true if you rotate $n$ times in the same direction (so $r^n = 1$). Another relation on these elements is that if you reflect your polygon, and then rotate it clockwise, and then reflect it again, it's the same as rotating it counterclockwise. In fact, the dihedral group can be presented as follows:
$$D_{2n} = <r,R : R^2 = r^n = 1, RrR = r^{-1}>.$$
This is more information than you asked for, but it means that there are only two kinds of symmetries of an $n$-gon: those made of just rotations, and those made of reflections too. Moreover, there are the same number of each, and the number of symmetries with reflections is just the number of lines of symmetry.
Best Answer
Assume that there exist totally $n$ sides along the vertexes $D,E,\cdots,A$. Notice that $$120^o=\angle ACD \overset{m}=\frac{1}{2}\widehat{DE\cdots A}=\frac{1}{2}\cdot 360^o\cdot \frac{n}{n+3}.$$
Thus, $n=6.$ As a result, the number of the sides of the polygon is $n+3=9.$