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# 2013 AMC 10 A #25

Video by Art of Problem Solving. Problem from the MAA American Mathematics Competitions. Created by Art of Problem Solving.

## Want to join the conversation?

- In comparison to the AMC8, how much harder are the AMC10? I am an 8th grader taking geometry and these problems seem quite challenging so I was wondering how much I need to study and learn.(5 votes)
- The AMC 10 is more about analysis and "abuse" of the various laws and properties of any number of things, which is seemingly unrelated. The AMC 10 has a bit more algebra than the AMC 8, would, but it's otherwise pretty similar: lot of analysis.(5 votes)

- Thankyou sir.I think the methods of solving these problems seem to be based on assumptions.Can you guide me with better shortcut methods?(2 votes)
- Thankyou sir.I think the methods of solving these problems seem to be based on assumptions.Can you guide me with better shortcut methods?(2 votes)

- I am in fifth grade and trying to learn about factorials. Can someone help?(4 votes)
- Exactly but the thing is by definition 0!=1 which has a lot of math to prove it. Now n!=n(n-1).... goes on till zero but not zero because if you multiply anything by zero the answer is zero.

So 1!=1, 2!=2(2-1)=2(1)=2, 3!=3*2*1=6 and it goes on. Hope that helped.(1 vote)

- here's a problem.

cos 1+cos 2+cos 3+.....cos 178+cos 179=?(2 votes) - At5:53, wouldn't finding out if the three lines really did intersect be useless because in real tests you would use rulers and other tools?(2 votes)
- How do you do this please? I'm good at maths too but this is way too complicated SO MUCH WORKING OUT!! Can someone help me please??(1 vote)
- on what question are you talking about m8(1 vote)

- How did you get that

This is helping me a lot and but can help me(1 vote) - This only saves about 50 seconds as opposed to the drawing out and counting method, I would recommened you do that if you are good at art and can do it quickly. It's a good strategy though.(1 vote)
- Simple factorial issue. Takes under 1 min to solve.(1 vote)
- it was kinda confusing to me but know i see how many triangles can go into a octagon and add up how many isosceles triangles there to get 49 as your answer. Tricky question. Easy answer(1 vote)
- Yes, after you do it for the first time it's much easier. Good luck(1 vote)

## Video transcript

- In this problem, we're starting
off with all 20 diagonals of a regular octagon then we're gonna 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 can 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 gonna do that but we are gonna 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 of 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 kind of
medium length diagonals. I'm gonna color those green. Bring colored pencils to the test and then we have the really long diagonal goes right through the
middle of the octagon and while I forgot to give
a label to my mediums, L will be long and my mediums will be M and now we're gonna organize our counting by thinking about each of these
different types of diagonals and what sort of
intersections they generate. Now, I'm gonna start with the short ones. So I need my red pen. So here's one short diagonal. I'm gonna look at where two
short diagonals can intersect. Draw that one. You can draw this one and we see, 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 gonna 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 start look
at this diagonal right there can intersect 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 can find two new intersection points with medium diagonals. There are eight short diagonals total so that gives us eight
times two, 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's 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 that
the medium intersections with the short. So let's look at mediums
intersecting with each other. Here's the medium diagonal
I'm gonna focus on. We've got one hitting it 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. Now, 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? Well, 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. Now, 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 and 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 two. That's a lot farther. This is not going here. These two points are not the same. Alright, 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 eight times four is 32 total intersections
between two medium diagonals but we have to be careful here 'cause that 32 counts each
intersection point twice. First, along this diagonal, we
count this intersection point once along this diagonal and once when we're counting intersections along that diagonal. So that 32, we have to divide it by two. We have 16 distinct points 'cause 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 that maybe it goes through
that point right there. We have to check out what's
going on right there. Is this long diagonal going
through the intersection point of these two mediums 'cause then we don't have
a new intersection point. One way to look at this, I like to look at it
right along that diagonal, I'll stand here, maybe you tilt your head and see, look at what's
going on 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 get that vertex. In other words, if I take this diagonal and I flip it over this long diagonal, I'm gonna 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 gonna run right
through the intersection of these two diagonals. We don't get a new
intersection point there and that's gonna be true every
time we draw a long diagonal. Just gonna have to use your
imagination with my diagram but I'm gonna curve
that line a little bit. We drew this long diagonal. It's gotta 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 look at this
problem and been like, okay, the answer's A or B
'cause it's gotta be odd. You're gonna get that
one point in the middle and everything else is gonna
come in groups of eight. So now we can total everything up. We've got eight and 16 gives us 24. Eight and 16 gives us another 24. That brings us up to 48. We add on the one. We've got 49 and we're done.