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

Video by Art of Problem Solving.  Problem from the MAA American Mathematics Competitions. Created by Art of Problem Solving.

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Video transcript

- All right, we've got some 3D geometry here, so we're gonna have to read carefully and visualize what's going on 'cause it's kind of hard to draw in 3D. We've got six spheres with radius one. Their centers are at the vertices of a regular hexagon that has side length two. So we're starting with a regular hexagon, and we're gonna put spheres centered at each of the vertices, and since the radius of each sphere is one, the side length is two, that means each of these spheres is gonna 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 gonna 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've got our six little ones down here around the hexagon, and we're gonna take this new sphere and just set it right on top of those six, and it's gonna 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. One thing I like to do with these 3D problems is I like to take 2D cross-sections, turn 2D problems into 3D problems, and when I have a problem with the bunch of spheres, I like to throw my cross-sections through the 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 'cause that's gonna go through the centers of seven of these spheres and all kinds of points of tangency. So to start off, we'll draw a regular hexagon, and you're gonna have to bear with me. On the test, of course, you've got your, you've got your ruler, you've got your protractor, you've got your compass, so you can draw a perfect diagram. You can 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 gonna 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 one. And one thing that's really nice about regular hexagons is you can break 'em up into equilateral triangles. So this is an equilateral triangle. Here's the center of the hexagon, the center of the big circle. I can extend this out to the point of tangency of this small sphere and the big one. So this is one 'cause it's the radius of the small sphere, and this is one. This is an equilateral triangle, so this side is the same as this side. So it tells us that this is one. And now we know that the radius of the giant sphere is three. So we've got the radius of the giant sphere. We've got the radii of all of these little spheres. All we have left is that eighth sphere we sat on top. Now of course, that sphere is not in this diagram. It's sitting right up here. So we're gonna need a different cross-section to go after this sphere. Of course, we're gonna choose one that's gonna go right through the center of that sphere. We want to hit some points of tangency. We're gonna go through the centers of some of these little spheres, and of course, the center of our whole diagram, the center of the big sphere. So we're gonna take that cut right here, and that cut's gonna look like this. We've got our little spheres out here still. And then we've got our big sphere that we just found. It has radius three. And I'm not gonna worry about what's going on down here 'cause 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 gonna continue this down to the center of that big sphere. And we know that this length, this is r. We know that this length is three minus r. And well, we can build a right triangle right here. And you're gonna 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'd like to connect the centers. We know that this is one 'cause this is one of our small spheres, this is r, and then this right here, well, this is one, and we found before that this is one 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 three minus r squared, plus two squared, equals one plus r squared. I'll go ahead and square this out. We have r squared minus 6r plus nine plus four, equals r squared plus 2r plus one. Bring the one over here, and it's 13 minus one is 12. Add the 6r over there. We have 12 equals 8r, and that gives us r equals three halves. If we go back to the problem, we find the 3/2, and we're done.