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# 2013 AMC 10 A #22 / AMC 12 A #18

Video transcript

We've got some 3D
geometry here so we're going to have read carefully,
visualize what's going on because it's kind of
hard to draw in 3D. We got six spheres
with a radius 1. Their centers are
at the vertices of a regular hexagon
that has side length 2. So we're starting with
a regular hexagon, and we're going to
put spheres centered at each of the vertices. And since the radius
of each sphere is 1, side length is 2, that
means each of these spheres is going to be tangent
to its two neighbors. So we start off with a hexagon,
six spheres, each one tangent to each of its two neighbors. And then we're going to have
a larger sphere centered at the center of
the hexagon such that it's tangent to each
of the little spheres. Now, each of the
little spheres will touch the inside of
this giant sphere. And then we bring
out an eighth sphere that's externally tangent
to the six little ones. So we got out six little ones
down here around the hexagon, and we're going to
take this new sphere and just set it right
on top of those six. And it's going to touch--
right at the top of it, it's going to touch
this larger sphere. So we have at least
somewhat of a picture of what's going on here,
and we want the radius of this last sphere that we
dropped in at the top there. And now one thing I like to
do with these 3D problems is I like to take
2D cross sections, turn 3D problems
into 2D problems. So when I have a problem with
a whole bunch of spheres, I like to throw
my cross sections through centers of those spheres
and through points of tangency whenever I have tangent spheres. Now, a natural place to
start here, of course, is the hexagon. We take the cross
section with the hexagon, because that's going to go
through the centers of seven of these spheres and all
kinds of points tangency. So to start off, we'll
draw a regular hexagon. And you're going to
have to bear with me. On the test, of course,
you've got your ruler, you got your protractor,
you got your compass so you can draw a
perfect diagram. You could probably draw
a freehand better diagram better than I can, too. But when we take cross
sections of our spheres, we make circles. And we include the points of
tangency in this cross section. Of course, we're also going
to get the big sphere. A cross section of
that is a circle that touches each of
these little circles. All right, and there we go. This is the cross section
through the hexagon. Now we can label some lengths. We know that the radii of
the little spheres is 1, and one thing that's really
nice about regular hexagons is you can break them up
into equilateral triangles. So this is an
equilateral triangle. Here's the center of the hexagon
center and the big circle, and I can extend this out
to the point of tangency of small sphere and the big one. So we know this
is 1 because it's a radius of the small sphere. This is 1. This is an equilateral
triangle so this side is the same as this side. So it tells us that
this is 1, and now we know that the radius of
the giant sphere is 3. So we've got the radius
of the giant sphere. We got the radii of all
these little spheres. All we have left is that
eighth sphere we sat on top. And, of course, that
sphere's not in this diagram. It's sitting right up here. So we're going to
need a different cross section to go after this sphere. Of course, we're
going to choose one that's it's going to
right through the center of that sphere. We want to hit some
points of tangency. We want to go through
the centers of some of these little
spheres and, of course, the center of our whole
diagram center, the big sphere. So we're going to take
that cut right here, and that cut's going
to look like this. We got our little
spheres out here still, and then we've
got our big sphere that we just found
has a radius 3. I'm not going to worry about
what's going on down here because then it won't be as
embarrassing how badly I draw circles, and then I've
got another circle. This is my eighth sphere,
this egg-shaped thing. That's a circle. You have to use
your imagination. This is the circle whose
radius we're trying to find. Right there, that's
the radius we want. Of course, we're going
to continue this down to the center of the big sphere. And we know that this
length-- this is r. We know that this
length is 3 minus r. And, well, we can build a
right triangle right here. And you're going to have
to use your imagination, but this cross section
goes through the point of tangency and
these two centers. When we connect
these centers, we go through the
point of tangency. That's why whenever we
have two tangent circles, we like to connect the centers. We know that this
is 1 because this is one of our small spheres. This is r, and then
this right here? Well, this is 1. And we found before
that this is 1 as well. So we've got our right
triangle, and we've got our classic, geometry
problem-solving strategy-- build a right triangle, use
the Pythagorean theorem, and that's what
we're going to do right here to
finish the problem. We have 3 minus r
squared plus 2 squared equals 1 plus r squared. Go ahead and square this out. We have r squared minus
6r plus 9 plus 4 equals r squared plus 2r plus 1. Bring the 1 over here where
it's 13 minus 1 is 12. Add the 6r over there. We have 12 equals 8r, and
that gives us r equals 3/2. We go back to the problem. We find the 3/2, and we're done.