Difference between revisions of "2013 AMC 10A Problems/Problem 22"
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<math>(3\sqrt{2})^2=3^2+(h+r)^2</math> | <math>(3\sqrt{2})^2=3^2+(h+r)^2</math> | ||
| − | r = 3/2 (B) | + | r = 3/2 \qquad\textbf{(B)} |
However, the answer key page says that the answer is E, so somebody needs to check this | However, the answer key page says that the answer is E, so somebody needs to check this | ||
Revision as of 22:18, 7 February 2013
Problem
Six spheres of radius 
are positioned so that their centers are at the vertices of a regular hexagon of side length 
. The six spheres are internally tangent to a larger sphere whose center is the center of the hexagon. An eighth sphere is externally tangent to the six smaller spheres and internally tangent to the larger sphere. What is the radius of this eighth sphere?
Solution
Set up an isosceles triangle between the center of the 8th sphere and two opposite ends of the hexagon. Then set up another triangle between the point of tangency of the 7th and 8th spheres, and the points of tangency between the 7th sphere and 2 of the original spheres on opposite sides of the hexagon. Express each side length of the triangles in terms of r (the radius of sphere 8) and h (the height of the first triangle). You can then use pythagorean theorem to set up two equations for the two triangles, and find the values of h and r.
r = 3/2 \qquad\textbf{(B)}
However, the answer key page says that the answer is E, so somebody needs to check this
See Also
| 2013 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
| 2013 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 17 |
Followed by Problem 19 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
