Given $n \geq 3$ positive reals $\alpha_1, \dots, \alpha_n$ such that $\alpha_i < 180$ and $\alpha_1 + \dots + \alpha_n = 180(n – 2)$, how do we show the existence of an $n$-sided simple (convex) polygon with interior angles of measure $\alpha_1, \dots, \alpha_n$? This is a simpler version of this question, which considers non-convex polygons as well (but unfortunately has no posted answer).
EDIT:
I have accepted Jim Ferry's answer, but for completeness I will fill in some details he left implicit.
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We can guarantee the tangency of each successive ray because we have the freedom to choose when the previous ray ends.
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If the point of tangency $P'$ determined by the final ray was not equal to our starting point $P$ (the point on the circle from where our first ray birthed), then the ray from $P$ in the opposite direction and the ray from $P'$ as a continuation of the previous ray would intersect at another point $M$. Now, the $n$-points on our failed polygon along with $M$ determine the vertices of an $n + 1$ sided polygon, and moreover, $n$ of its interior angles are equal to the $\alpha_i$. But then, the interior angle $\alpha_{n + 1}$ at $M$ is equal to $180(n – 1) – (\alpha_1 + \dots + \alpha_n) = 180$, a contradiction.
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The reason why the process doesn't "loop too far around" the circle is given by a similar argument as above.
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The motivation for centering the construction around a circle was to prevent our polygon from hitting itself.
Best Answer
You can construct such a polygon explicitly by adding the requirement that it have an incircle.
Consider a circle and place a ray tangent to it, pointing counterclockwise. Add $180 - \alpha_1$ to the orientation of this ray to get the orientation of the next ray, and place this ray tangent (and counterclockwise) to the circle. Continuing like this yields the unique polygon with the given angle sequence and inradius, up to congruence.