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# 2003 AIME II problem 4 (part 1)

## Video transcript

in a regular tetrahedron and that's just a four-sided polyhedron and it's regular so all of the sides and all 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 + n so let's just draw this regular tetrahedron first so this is a regular tetrahedron so in order to have four equal sides why four equal faces I should say in three dimensions it needs to look like this it essentially needs to be three and this 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 phases are equilateral triangles so if you looked so this length is the same as that length it's 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 centre of one face this is the centre of another face back here is the centre of the face that we can't see that's behind these two faces and then this is the centre this right here let's say is the centre of the base the centre 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 centre of this face to the center of the base going from the center of the base to the centre 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 if you flipped it over be the base but you 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 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 racking your brain how do you figure out the volume of a tetrahedron what not and all of that but the important thing to realize the important thing to realize if you if you take two similar really 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 let me show this to you right over here let me depict it so if we let's say let's say that this this length right over here let's call that let's call that length big for the big tetrahedron and let's find a corresponding edge of the small tetrahedron so let's say that we have this correspondent we have another small side where we have another side of the small tattoo he drew let's call that let's call that length the small if we know the ratio of lengths big to length small if we know if we know the ratio length big to length small we can figure out the ratio of the big tetrahedron we can figure out the ratio of the big tetrahedron to the small tetrahedron to the the the volume of the big tetrahedron to the volume of the small tetrahedron by just cubing this ratio by just cubing this ratio and we can do that because these are completely similar in every direction in every dimension whatever the ratio whatever 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 Long's 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 gonna 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 Centers of the faces so let me just draw let me draw some coordinate axes here so let's say that this is my y-axis I'm gonna 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 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 gonna do that I'm just gonna pick arbitrary coordinates here we really just care about the ratios we care about the ratio so we can pick the larger tetrahedron have any any size so let's just say that this let's say that that vertex right over here is sitting is sitting right over here this vertex sits right over here and then this vertex this vertex right over here sits right over there and then let's assign let me make it so it looks a little bit more that looks a little bit better and so this is the base 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 just for simplicity of the coordinates let's make 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 if we go over here let's make this negative 1 comma 0 comma 0 so 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 this entire side over here is going to be too 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 gonna 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 this so this y-coordinate this y coordinate squared plus 1 squared is going to equal 2 squared so let me write that over here so this let me do the new color so this 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 its coordinate is square root of 3 in the Y Direction now while we're focusing on the base while we're focusing 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 all 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 its it has it does not move in the 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 three over three so it's one third of the way all the way to the top of the base and then the z-coordinate you can take the average but it's clearly still sitting on the XY plane so it's also going to be zero or you could say zero plus zero plus zero divided by three curly still zero so we figured out also one of the vertices of our smaller of our smaller tetrahedron what I want to do now is try to figure out is 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 two and then essentially take the cube of it and we'll have our we'll have the ratio we need and then we can figure out the rest of the problem the n plus n now to figure out 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 if we were to go straight up from this point so if we were to go straight up from that center point that we just figured out let's say we go right let me 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 saying it a little bit weird so now the tetrahedron you can imagine looks like this this is 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 having trouble drawing it making my hand go in an awkward position let me see 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 I've just drawn this magenta face let me I've just drawn this magenta face of the tetrahedron right over here right over here and I think and I could draw the backside 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 this point right here is going to right on top right on top of this Centrepointe over here so we already know a few we already know a few of its coordinates we know it's X&Y coordinate we just need to figure out its Z coordinate so its x-coordinate is going to be zero 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 three over three 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 draw 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 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's 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 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 we can figure out the height of this 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 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 alright it is I'll write the it's 3 minus 3 over 9 3 minus 1/3 which is the same thing as 9 thirds minus 1 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 square root of 8 is the 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 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 it in the next video but we now have the coordinate for this top point so we can take the average of these 3 coordinates find the coordinate for this point right over here then use 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 compare the ratio of that to that cubit to get the ratio of the volumes