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Studying for a test? Prepare with this lesson on AMC 10.
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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.