Draw a pentagon with any 2 angle sums being less than $216^o$

Question

I've come across challenge proof question in my discrete mathematics textbook that I'm trying to solve for practice but unfortunately it does not have a solution. Any help with a reasonable explanation or solution so that I can understand where to start and verify my work would be greatly appreciated:

Can you draw a pentagon (5-sided figure) in which the sum of any two angles – for every two angles you pick – is less than $216^o$? Explain how to do it, or prove that it can not be done.

Hints:

(1) the sum of the angles in a pentagon is always $540^o$.

(2) Look at two different ways of computing $4(\Theta1 + \Theta2 + \Theta3 + \Theta4 + \Theta5)$, where $\Theta i$ is the value of the $i$th angle of the pentagon.

Thank you!

Answer 1

Let $\theta_i$ be the angles and let's assume that it's possible to draw such a pentagon. So, $\theta_i + \theta_{j} < 216º$ for all $i$ and $j \neq i$.

Hence, $(\theta_1 + \theta_2) + (\theta_2 + \theta_3) + (\theta_3 + \theta_4) + (\theta_4 + \theta_5) + (\theta_5 + \theta_1) < 5 \times 216º$.

Rearranging the terms we get: $2(\theta_1 + \theta_2 + \theta_3 + \theta_4 + \theta_5) < 1080º$, i.e., $2 \times 540º < 1080º$, which is false.

Answer 2

We'll assume that it's possible to draw the pentagon you're describing.

Observe that $$(\theta_1+\theta_2)+(\theta_1+\theta_3)+(\theta_1+\theta_4)+(\theta_1+\theta_5)+(\theta_2+\theta_3)+(\theta_2+\theta_4)+(\theta_2+\theta_5)+(\theta_3+\theta_4)+(\theta_3+\theta_5)+(\theta_4+\theta_5)=4·540°=2160°$$

By the Pigeonhole principle, at least one of the summands is $$\geq \frac{2160°}{10°}=216°$$

Contradiction

Answer 3

This is impossible. Let the angles be $\theta_i$ , $i=1,2,3,4,5$. Therefore $$\theta_i+\theta_j<216^o\quad,\quad i\ne j$$and $$\sum\theta_i=540^o$$by summation over $\theta_i+\theta_j$ we obtain $$\sum_{i\ne j}\theta_i+\theta_j=4\sum_{i=1}^{5}\theta_i=4\times 540^o=2160$$while $$\sum_{i\ne j}\theta_i+\theta_j<\binom{5}{2}\times 216=2160^o$$which leads to impossible statement $2160^o<2160^o$. Therefore such a pentagon doesn't exist.