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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 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.