2013 AMC 10 A #22 / AMC 12 A #18

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.