2017 AMC 10 A

Complete problem set with solutions and individual problem pages

Problem 11 Easy

The region consisting of all points in three-dimensional space within 3 units of line segment \overline{AB} has volume 216\pi . What is the length AB? (2017 AMC 10A Problem, Question#11)

  • A.

    6

  • B.

    12

  • C.

    18

  • D.

    20

  • E.

    24

Answer:D

In order to solve this problem, we must first visualize what the region contained looks like. We know that, in a three dimensional plane, the region consisting of all points within 3 units of a point would be a sphere with radius 3. However, we need to find the region containing all points within 3 units of a segment. It can be seen that our region is a cylinder with two hemispheres on either end. We know the volume of our region, so we set up the following equation (the volume of our cylinder + the volume of our two hemispheres will equal 216\pi ): \dfrac{4 \pi }{3} \cdot 3^{3}+9 \pi x=216 \pi, where x is equal to the length of our line segment. Solving, we find that x= (\rm D)20.

Because this is just a cylinder and 2 "half spheres", and the radius is 3, the volume of the 2 half spheres is 4(3^3)/3\pi = 36\pi. Since we also know that the volume of this whole thing is 216\pi , we do 216 - 36 to get 180\pi as the area of the cylinder. Thus the height is 180\pi over the base, or 180\pi /9\pi = 20, D.