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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.

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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 $n+3=9.$

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  • $\begingroup$ Alright what the hell. I was adding a comment and it edited your answer instead... I'm sorry if that deleted any changes you've made, it was not my intention. EDIT: I see now, it popped up an 'edit summary box' right where my comment box was when I went to copy/paste some LaTeX and I happily added my 'comment' there. $\endgroup$ – orlp Jul 20 '18 at 6:33
  • $\begingroup$ Either way, I'm not too familiar with geometric notation, could you explain what $\overset{m} =$ and $\widehat{DE\cdots A}$ mean? $\endgroup$ – orlp Jul 20 '18 at 6:35
  • $\begingroup$ Yep. That is to say, $\angle ACD$ equals half of the central angle opposite to the arc $DE\cdots A$. $\endgroup$ – mengdie1982 Jul 20 '18 at 6:40
  • $\begingroup$ did you mean n=9? $\endgroup$ – ShadyAF Jul 20 '18 at 6:42
  • $\begingroup$ @ShadyAF yes,a typo. $\endgroup$ – mengdie1982 Jul 20 '18 at 6:43
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The interior angle of a regular polygon is $$\angle ABC=\angle BCD=\frac{180(n-2)}{n}$$ Now the triangle ABC is isosceles, so $$\angle BCA = \frac12(180-\frac{180(n-2)}{n}) \\ \angle BCA+\angle ACD = \angle BCD \\ \frac12(180-\frac{180(n-2)}{n})+120=\frac{180(n-2)}{n}$$ This gives $n=9$.

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