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2003 AIME II problem 4 (part 1)
In a regular tetrahedron-- and that's just a four-sided polyhedron, and its regular, so all of the sides and all of the faces will be the same-- the centers of the four faces are the vertices of a smaller tetrahedron. The ratio of the volume of the smaller tetrahedron to that of the larger is m over n, where m and n are relatively prime positive integers. Find m plus n. So let's just draw this regular tetrahedron first. So this is a regular tetrahedron. So in order to have four equal sides, or four equals faces, I should say, in three dimensions, it needs to look like this. It essentially needs to be four triangles. So we have one triangle, just like that. Then we have another triangle like that. Then we have another triangle in the back. So this is another triangle in the back. And then we have a triangle as a base. This is a regular tetrahedron, where each of these faces are going to be equilateral triangles. Each of these faces are equilateral triangles. So if you looked, this length is the same as that length, this is the same as that length. Fair enough. So that's our regular tetrahedron. Now, they say the centers of the four faces, so let's look at the centers of the four faces. Let's say that this is the center of one face. This is the center of another face. Back here is the center of the face that we can't see that's behind these two faces. And then this is the center, this right here, let's say, is the center of the base. The center of the four faces are the vertices of a smaller tetrahedron, so we could connect them and get a smaller tetrahedron. So along, so we have this one going from the center of this face to the center of the base. Going from the center of the base to the center of this face over here, we have this line. You connect those, we have one of the faces of the smaller tetrahedron. You go like that, you have kind of the top. If you flipped it over, it would be the base, but you'd have the top of the smaller tetrahedron. And then if you connected those in the back, you would have the two faces that you can't see-- that face and that face. Now, they want us to essentially figure out the volume of the smaller tetrahedron, this magenta tetrahedron, to this yellow one over here. And you might already be wrecking your brain, how do you figure out the volume of a tetrahedron and whatnot, and all of that. But the important thing to realize if you take two similar objects and if you just know the ratio of one of their dimensions, and assuming that all of their dimensions have the same ratio-- and it is the case in this right over here-- then the volume will be the cube of the ratios of one of their sides. So let me show this to you right over here. Let me depict it. So let's say that this length right over here, let's call that Length Big for the big tetrahedron. And let's find the corresponding edge of the small tetrahedron. So let's say that we have another side of the small tetrahedron, let's call that Length Small. If we know the ratio of Length Big to Length Small, we can figure out the ratio of the big tetrahedron to the small tetrahedron, the volume of the big tetrahedron to the volume of the small tetrahedron by just cubing this ratio. And we can do that because these are completely similar in every direction, in every dimension. Whatever the ratio of this is to this is the same as the ratio of that is to that. And actually, all of these sides are the same, so all of these sides are going to be the same as well. So they're similar in every dimension. So with that said, we really just need to figure out the ratio of one of the sides of the big tetrahedron to the length of the side of a small tetrahedron. And to do that, let's just put ourselves-- And there are more elegant ways of doing it, but those get a little bit more abstract. What I'm going to do is just try to figure out some coordinates of some of these vertices of the larger tetrahedron. And from that we can take the average of the coordinates to find the coordinates of these vertices, the centres of the faces. So let me draw some coordinate axes here. So let's say that this is my y-axis. I'm going to try to draw it so that we can visualize it in three dimensions. So let's say that is my y-axis. Let's make my x-axis do something like this. So let's say that is the x-axis. And then you're going to have your z-axis go straight up and down. And actually, I won't draw that yet. What I want to just draw right now is the base of the larger tetrahedron. And I'm going to do that. I'm just going to pick arbitrary coordinates here. We really just care about the ratios, so we can pick the larger tetrahedron and have any size. So let's say that that vertex right over here is sitting right over here. This vertex sits right over here. And then this vertex right over here sit right over there. And then let's assign-- we'll make it so that looks a little bit better. And so this is the base. of our tetrahedron, of our big one. It's an equilateral triangle-- all the sides are the same. And just for simplicity of the coordinates, let's make this coordinate right over here, let's say that x is equal to 1, y is clearly equal to 0, because we are on the x-axis, we haven't moved in the y direction. And then we haven't even paid any attention to z, but it's clearly sitting in the xy-plane, so z is also going to be equal to 0. If we go over here, let's make this negative 1, comma 0, comma 0. So it's still on the x-axis. So this distance right over here is 1, this distance over here is 1, this entire side is 2. So we also know that this entire side over here is going to be 2. And you could use 30-60-90 triangles to figure out this distance, which we need to figure out the y-coordinate here. Let me be very clear. This, the coordinate of this point over here, x is clearly equal to 0 over here. X is 0. z is clearly equal to 0, we're still in the xy-plane. But we don't know what the y-coordinate is. So we're going to have to do a little bit of-- well, you could use your knowledge of 30-60-90 triangles or if you don't even want to do that, you can just go straight to the Pythagorean theorem. We're looking for the y-coordinate over here. We're looking for this distance right over here. So this y-coordinate squared plus 1 squared is going to equal 2 squared. So let me write that over here. So this-- I will do a new color-- so the y-coordinate squared plus 1 squared-- so plus 1 is going to be equal to 2 squared, it's going to be equal to 4, where this is the y-coordinate. So y squared is going to be equal to 3. y is going to be equal to the square root of 3. So this distance right over here is the square root of 3. Or so I could write square root of 3 here-- you could see that this satisfies the Pythagorean theorem-- or I could just put its coordinate-- it's coordinate is square root of 3 in the y-direction. Now, while we're focusing on the base on the base of the large tetrahedron, we can right now figure out the center of that base, figure out the center of that face, which maybe might sit right about here. And the easy way to figure out the center of this face is literally just to average of the coordinates. So the coordinate for this character right over here is going to be the average of the x's. So 0 plus 1 minus 1 is 0 divided by 3 is just going to be 0, which is very clear. It's sitting on the y-axis. It does not move in the x x-direction, either to the right or to the left. The y-coordinate, we're going to have square root of 3 plus 0 plus 0 over 3. So that's just going to be square root of 3 over 3. So it's one third of the way all the way to the top of the base. And then the z-coordinate coordinate, you could take the average, but it's clearly still sitting on the xy-plane. So it's also going to be 0. You can take 0 plus 0 plus 0 divided by 3-- clearly still 0. So we figured out also one of the vertices of our smaller tetrahedron. What I want to do now to try to figure out the coordinates of this vertex right here. Then we can figure out this distance, compare it to a distance of 2, and then, essentially, take the cube of it, and we'll have the ratio we need. And then we can figure out the rest of the problem-- the m plus n. Now, to figure out the coordinates here, we need to take the average of this coordinate, this coordinate, and this coordinate up here. We know these two coordinates. We know these two coordinates. We just need to figure out this coordinate up here. And to do that, let's now extend into the z-direction. So if we were to go straight up from this point. So if we were to go straight up from that center point, we just figured out. Let's say we go right. I could draw a neater altitude than that. So let's say I were to go straight up from this point right here to the top of this tetrahedron-- so let's say tetrahedron, because I was saying it a little bit weird. So now the tetrahedron, you can imagine, looks like this. The drawing is the hard part here. So now this is a face of the tetrahedron. I don't even have to draw the whole thing. I think we can visualize it. I'm having trouble drawing it, making my hand go in an awkward position. Let me see if I can just draw it with dotted lines. So that side. So what I've just drawn here is this face of the tetrahedron. I've just drawn this magenta face of the tetrahedron right over here. And I could draw the back side too. I could draw the backside just like that. I could draw it over there. And I think it's pretty clear to you that this point right here is going to lie right on top of this center point over here. So we already know a few of its coordinates. We know its x- and y-coordinate. We just need to figure out its z-coordinate. So its x-coordinate is going to be 0. It's going to be the same x-coordinate here-- we haven't moved along the x-axis. Its y-coordinate is going to be the square root of 3 over 3. Same as this guy over here, but it's going to have some height in the z-direction. So we need to figure out what that is. And to do that, what I'm going to do is to draw, essentially, an altitude for the face of this front face. I'm going to go from this point up to here, or from this point up to here. So if you look at this blue line right over here, we already figured out what this distance is. All the faces-- it's a regular tetrahedron-- so all of the faces have the same length. So this blue distance right over here is going to be the same as this distance over here, because this was the altitude for that bottom face. And we already figured out that this distance over here is square root of 3. We did that, that was the first thing we did that this distance over here from this point to this point is square root of 3. So this distance is also the square root of 3. We also know that this distance right over here-- let me do this in a new color-- we also know that this distance over here is the square root of 3 over 3. This is a right angle. It might not be completely obvious the way I did it, but this was an altitude. So now we can figure out the height of the pyramid, or the tetrahedron. And that height to this point is going to be our z-coordinate. So let's call that height h. So we know from the Pythagorean theorem, h squared plus the base squared right over here-- which is going to be square root of 3 over 3 squared is going to be 3 over 3 over 9 is going to be equal to the square root of 3 squared-- it's going to be equal to 3. Or subtracting this from both sides, we have h squared is equal to-- I'll write the-- it's 3 minus 3 over 9, 3 minus one third, which is the same thing as 9/3 minus one third, which is equal to 8/3. And so if we want to take the square root of both sides, we'll get h is equal to the square root of 8 is a square root of 4 times square root of 2. So that's 2 square roots of 2 over the square root of 3. So that is h. And that's also the z-coordinate here. So this is 2 square roots of 2 over the square root of 3. I'm already close to 14 minutes here, so I'm going to continue in the next video. But we now have the coordinate for this top point. So we can take the average of these three coordinates, find the coordinate for this point right over here, then find the distance between this point and this point that we already got the coordinate for to figure out how long this side is. And then we can prepare the ratio of that to that, cube it to get the ratio of the volumes.