what is the minimal number of lines to ensure there is at least one angle less than 26 degrees (The applications of pigeonhole principle)

Question

Question: In the plane, there are n non-parallel straight lines. These lines will form several angles (Fig. Q11 a). To ensure there is at least one angle less than 26 degrees, what is the minimal number of lines? Why? (Tip: The translation of the straight line does not change the angle, shown in Fig. Q11 b)

Solution: 7 lines are required. The reason is:

Since the translation of the straight line does not change the angle, we can move these straight lines and let them have only one intersection point. Therefore, n straight lines will form 2n angles. To let there have at least one angle less than 26 degrees, we need $\lceil \frac{360}{26}\rceil +1=14$. Therefore, 14 angles are needed, and we need 7 lines.

Although I have read the solution, there are two things confusing me.

1. Given the fact that the length of a line is infinite and they are non-parallel lines, these straight lines will have two or above intersections with others' lines when there are three lines or above. Therefore, I don't understand this explanation "Since the translation of the straight line does not change the angle, we can move these straight lines and let them have only one intersection point."
   
2. I am wondering why the pigeons are 360 and the pigeonholes are 26. and why would I need to + 1?

Answer 1

With $6$ lines, obviously you can have all angles bigger than $26^\circ$, namely make $6$ lines go through the same point and intersect at angles of $30^\circ$ - for example the longest diagonals of a regular $12$-gon, which all go through its centre.

Now, with $7$ lines this is impossible. Any $7$ lines non-parallel lines can be translated to go through a fixed point $O$. This does not change the angles between them. Thus, it is sufficient to consider only $7$-tuplets of lines that all go through a single point.

Those $7$ lines will make up $14$ angles with the corner at $O$. If all those angles were at least $26^\circ$, then the total would be at least $14\times 26^\circ=364^\circ > 360^\circ$, which is impossible, as they all add up to $360^\circ$.

The formula the solution used: $\lceil\frac{360}{26}\rceil+1$ in fact gives the wrong result $15$ instead of $14$ (due to $1$ being added). The correct formula comes from solving $26x>360$ for integer $x$ (the number of angles). This boils down to $x>\frac{360}{26}$, which again boils down to $x>\lfloor\frac{360}{26}\rfloor$ (as $x$ - the number of angles - is an integer), which is the same as $x\ge\lfloor\frac{360}{26}\rfloor+1$ (as an integer is bigger than another integer $n$ if and only if it is at least $n+1$).

Answer 2

Too much to fit in a comment, so here we are.

I disagree that we can't apply the pigeonhole principle; the pigeonholes are degrees, and the pigeons are one degree each--you're clumping them in bunches. You let the pigeons fly into the holes $26$ at a time, letting go $13$ bundles $13\cdot26=338$. The next bundle of pigeons won't have enough holes, which is where the $+1$ appears. If you'd used larger bunches, the last bunch would have even fewer holes (a smaller angle). Smaller bunches would already be under $26$ degrees. So, we need $14$ bunches--$14$ angles.

For the first part, let's consider things backward. Imagine three lines that cross at one point. Two are perpendicular, vertical and horizontal. The third intersects at their intersection, at a $35$ degree angle. You now have six angles, two each of $90, 55, 35$ degrees.

Now move the slanted line upward without changing its slope. One intersection becomes three intersections and $12$ angles, but notice that all the angles have the same measures as before. We have four $90$ degree angles, four $55$ degree angles, and four $35$ degree angles.

In other words, translating the line--moving it without changing its angle--preserved all the angles. And this will be true for any number of lines. So wherever our non-parallell lines start, we can translate them to all cross at one point, and the angle measures won't change.

Seven lines make $14$ angles, and we've seen that you can't fit $14$ separate $26$ degree angles in $360$ degrees. Therefore seven lines is sufficient to make it so that somewhere in the plane is an angle smaller than $26$ degrees.