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# 2013 AMC 10 A #25

Video transcript

In this problem,
we're starting off with all 20 diagonals
of a regular octagon. Then we're going to count up all
the points inside that octagon where at least two of
these diagonals meet. Now, if you're a lot better at
drawing and a lot more patient than I am, you could draw
the octagon very carefully, draw all the diagonals, and then
count up all the intersection points inside. Good luck with that. That doesn't sound like any fun,
so we're not going to do that. But we are going
to at least start with a quick sketch
of an octagon that you'll have to
pretend is regular. And now let's think about how
we can organize our counting and where all these
diagonals intersect. There are 20 diagonals,
but there are really only three types of diagonals. Let's organize our counting by
the lengths of the diagonals. We have short
diagonals like that. I'll paint those
red, s for short. And we have these
medium length diagonals. I'm going to color those green. Bring colored
pencils to the test. And then we have the
really long diagonal that goes right through
the middle of the octagon. And I forgot to give a label
to my mediums. l will be long and my mediums will be m. And now we're going to organize
our counting by thinking about each of these
different types of diagonals and what sort of
intersections they generate. Now, I'm going to start
with the short ones. So I need my red pen. So here's one short diagonal. And I'm going to look at
where two short diagonals can intersect. I can draw that one. I can draw this one. And we see that, well, these
two diagonals intersect right in front of that side. These two diagonals intersect
right in front of that side. I can draw another
one here, and it'll intersect right
in front of that. For each side of
the octagon, there's going to be one point
right out in front of it, where two of these short
diagonals intersect. So we have eight intersections
between two short sides. Now let's look at
where a short side can intersect a medium side. And again, we'll still look
at this diagonal right there. It can intersect with
this medium diagonal or this medium diagonal. Those are the only
two that can hit it. And clearly, these
two points are different from those two points. So we've got two
new intersection points between two diagonals. So for each short diagonal
in our octagon here, I could find two new
intersection points with medium diagonals. There are eight short
diagonals total, so that gives us 8 times 2. 16 intersections between a short
diagonal and a medium diagonal. And we're on to looking at short
diagonals and long diagonals. You can see that for
each short diagonal, there is one long diagonal
that cuts that right in half. And clearly, that
point is not the same as any of the other four. So we have found eight
new intersection points, one for each of these
short diagonals. So that takes care of
the short diagonals. Let's move on to the
medium diagonals. We've already looked at
the medium intersections with the short. So let's look at mediums
intersecting with each other. Here's the medium diagonal
I'm going to focus on. We've got one hitting here,
another one hitting it there. And we've got-- this medium
diagonal hits it, and so does this medium diagonal. And those are the only four
that hit this medium diagonal. But we have to be careful here. We have to make sure
these four intersection points are different. Clearly these two are different. They're coming out
of the same vertex and go in different directions. These two are different. But what about these
two right here? We can think about
whether or not these two are
different by looking at how far they are
from this vertex. This isosceles
right triangle here tells me that this distance
is equal to the side length of the octagon. But then I look at this
isosceles right triangle, and I see that this
distance is less than the side length
of an octagon. So this is not the same as that. These two points are different. We also have to worry about
this point and this point. Maybe these two diagonals
actually intersect down here. And I'm a really terrible,
terrible, artist, terrible at drawing
these things. Well, to see that
this can't possibly be all the way down here, we
look at this isosceles right triangle. This is the side length
of the octagon, obviously. So the distance
from here to here is half the side
length of the octagon. You know, I'll just cut it
into two little isosceles right triangles. So this distance is half the
side length of the octagon. This distance here
is the side length divided by the square root of 2. That's a lot farther. This is not on here. These two points
are not the same. All right. That tells me I've got
four intersection points along this medium diagonal. And then I have eight of
these medium diagonals, so it looks like
I have 8 times 4 is 32 total intersections
between two medium diagonals. But we have to be careful
here, because that 32 counts each intersection point twice. First along this diagonal, we
count this intersection point, once along this
diagonal and once more counting intersections
along that diagonal. So that 32, we have
to divide it by 2. We have 16 distinct
points because that 32 counted each one of them twice. And now we're ready to move
on to the intersections between medium and long. And while we see that this long
diagonal intersects this medium diagonal. But maybe it goes through
that point right there. You have to check out
what's going on right there. Is this long diagonal going
through the intersection point of these two mediums? Because then we don't have
a new intersection point. One way to look at
this-- so I like to look at it right
along that diagonal. I'll stand here. Maybe tilt your head and see. Look at what's going
along this diagonal. If we flip this vertex
over that diagonal, we'll get that vertex. If we flip this vertex
over the diagonal, we'll get that vertex. In other words, if
I take this diagonal and I flip it over
this long diagonal, I'm going to get this diagonal. These two medium
diagonals are symmetric about this long diagonal. And that tells me that
these two intersect right along that diagonal. So this is not a new
intersection point. This line goes directly
between these two points, directly between
these two points. It's going to run right through
the intersection of these two diagonals. We don't get a new
intersection point there. And that's going to
be true every time we draw a long diagonal. You're just going to have
to use your imagination with my diagram. But I'm going to curve
that line a little bit. If we drew this
long diagonal, it's got to go through the
intersection point of these two diagonals, which are
symmetric about that diagonal. So this is going to give
us no new intersections. And now we're on
to just thinking about all the long
diagonals, which is easy. All four of the long
diagonals intersect in the center of the octagon. And just from that
observation alone, you could have looked
at this problem and been like, OK,
the answer is A or B, because it's got to be odd. You're going to get that
one point in the middle, and everything else is going
to come in groups of eight. So now we can total
everything up. We've got 8 and 16 gives us 24. 8 and 16 gives us another 24. That brings up to 48. We add on the 1. We got 49, and we're done.