Question 703977
I thought hard too, and this is what I came up with.
At one crossing of 2 lines, there are four angles, and the wording hints that we should count all four.
 
With 3 lines, each line has at most {{{3-1=2}}} crossings
(less if one or both of the other lines are parallel).
If we multiply 3 (lines) times {{{2}}} crossings per line,
you get at most 6 crossings, but each crossing is counted twice,
one time from the point of view of one of the crossing lines,
and another time from the point of view of the other crossing line.
So there are really at most {{{6/2=3}}} crossings
(less if all 3 lines were to cross together at the same point).
That would mean at most {{{3*4=highlight(12)}}} angles.
{{{drawing(300,300,0,10,0,10,
line(3,0,3,10),line(0,7,10,7),line(0,0,10,10)
)}}}
 
With four lines there are at most {{{4*(4-1)/2=6}}} crossings.
That would give you at most {{{6*4=highlight(24)}}} angles.
{{{drawing(300,300,0,10,0,10,
line(3,0,3,10),line(0,7,10,7),
line(0,0,10,10),line(0,8,10,0)
)}}}
 
EXTRA:
{{{n}}} lines could have at most {{{n(n-1)/2}}} crossings and
that would make a maximum of {{{4n(n-1)/2=2n(n-1)}}} angles

NOTE:
Maybe you should try the artofproblemsolving.com forums.
Some of the people there are future Math Olympics competitors.