# Equilateral pentagon with increasing/equal angles

Suppose all sides of a convex pentagon ABCDE have the same size, and $$\angle A \ge \angle B \ge \angle C \ge \angle D \ge \angle E$$. Prove that this pentagon is a regular pentagon.

I know this should be proved using contradiction, but I'm not sure how to reach that point. Can anyone guide me to the right solution?

• Calculate the sum of the angles then divide by $5$. DONE! – David G. Stork Oct 21 '19 at 4:49
• @DavidG.Stork Could you please elaborate? I'm not sure how that would help solve the problem. – Taiga Agiat Oct 21 '19 at 4:50

WOLOG, we only need to consider the case where the side of the equilateral pentagon is $$1$$ and the vertices $$A,B,C,D,E$$ are ordered counterclockiwisely on circumference of the pentagon.

Let $$\alpha,\beta,\gamma,\delta,\epsilon$$ be the external angles at $$A, B, C, D, E$$. We have

$$\angle A \ge \angle B \ge \angle C \ge \angle D \ge \angle E \quad\implies\quad \alpha \le \beta \le \gamma \le \delta \le \epsilon$$ Since $$2\pi = \alpha + \beta + \gamma + \delta + \epsilon \le 5 \epsilon$$, we have $$\epsilon \ge \frac{2\pi}{5}$$. This leads to $$\alpha + \beta + \gamma + \delta \le \frac{8\pi}{5} \implies \alpha + \beta \le \frac{4\pi}{5} \implies \alpha \le \frac{2\pi}{5}$$

Choose a coordinate system so that $$E = (-\frac12,0), A = (\frac12,0)$$.
In this coordinate system, it is not hard to see the $$x$$-coordinate of $$C$$ is $$\frac12 + \cos\alpha + \cos(\alpha+\beta) \ge \frac12 + \cos\frac{2\pi}{5} + \cos\frac{4\pi}{5} = 0\tag{*1}$$ THis means $$C$$ is lying in the left-half plane and hence $$|AC| \le |EC|$$.

However \begin{align} &|AC|^2 = |AB|^2 + |BC|^2 + 2|AB||BC|\cos\beta = 2(1+\cos\beta)\\ & |CE|^2 = |CD|^2 + |DE|^2 + 2|CD||DE|\cos\delta = 2(1+\cos\delta) \end{align} Together with $$\beta \le \delta$$, we obtain $$|AC| \ge |EC|$$.

Combine with above, we get $$|AC| = |EC|$$. This forces $$C$$ to lies on the $$y$$-axis.

Notice if either $$\alpha < \frac{2\pi}{5}$$ or $$\alpha + \beta < \frac{4\pi}{5}$$, the inequality in $$(*1)$$ becomes strict. For $$C$$ to lies on the $$y$$-axis, we need $$\alpha = \frac{2\pi}{5}$$. This leads to

$$2\pi = 5\alpha \le \alpha + \beta + \gamma + \delta + \epsilon = 2\pi$$ Since $$\alpha \le \beta \le \gamma \le \delta \le \epsilon$$, this forces $$\alpha = \beta = \gamma = \delta = \epsilon$$ and $$ABCDE$$ is a regular pentagon.