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?

  • $\begingroup$ Calculate the sum of the angles then divide by $5$. DONE! $\endgroup$ – David G. Stork Oct 21 '19 at 4:49
  • 2
    $\begingroup$ @DavidG.Stork Could you please elaborate? I'm not sure how that would help solve the problem. $\endgroup$ – 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.


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