BDMO National Junior 2018/6

[nahin munkar]

Given $8$ lines on a plane and no two of them are parallel. Prove that, at least two of them form an angle less than 23°.

[samiul_samin]

Using $PHP$

$23×8=184$
$184>180$
Which is not possible as two lines cannot form angle more than $180$ degree.
So,at least two of them form an angle less than $23$ degree. [Proved]

Anyone please clarify my short solution.

[Enthurelxyz]

Let the $8$ lines are $l_1,l_2,\cdots ,l_8$. Take a point $O$ on the same plane. Now draw $8$ lines $L_i$ on point $O%$ such that $L_i \parallel l_i$.

Now, on point $O$, there are $8$ different angles created by $L_1,L_2,\cdots ,L_8$. So, there should be at least an angle less than $23$.

Let $\angle L_iOL_j < 23$ then $l_i , l_j$ create an angle less than $23$ as they create an equal angle with $\angle L_iOL_j$.

[Anindya Biswas]

Let the $8$ lines are $l_1,l_2,\cdots ,l_8$. Take a point $O$ on the same plane. Now draw $8$ lines $L_i$ on point $O%$ such that $L_i \parallel l_i$.

Now, on point $O$, there are $8$ different angles created by $L_1,L_2,\cdots ,L_8$. So, there should be at least an angle less than $23$.

Let $\angle L_iOL_j < 23$ then $l_i , l_j$ create an angle less than $23$ as they create an equal angle with $\angle L_iOL_j$.

I have almost same argument, since those angles add up to $180^{\circ}$, then on average they are $\frac{180^{\circ}}{8}=22.5^{\circ}$, so in order to get such average, we must need at least one angle $\leq 22.5^{\circ}$

[Mehrab4226]

since those angles add up to $180^{\circ}$, then on average they are $\frac{180^{\circ}}{8}=22.5^{\circ}$, so in order to get such average, we must need at least one angle $\leq 22.5^{\circ}$

This was a clever one! And easy to express too!