Three spheres of radius 10 are placed on a table all touching each other. A fourth sphere of radius 10 is placed so that it lies on top of the other three. The distance from the bottom of the fourth sphere to the table is h, and h^2=a/b, where a and b are coprime positive integers. What is the value of a+b?
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Hmmm. As I recall, the vertical distance between sphere centers in the stack is
√6/3 * 20, so
h^2 = 400*6/9 = 800/3
To find the value of a+b, we need to calculate the value of h^2, given that the radius of each sphere is 10.
Let's consider the center of the bottom three spheres. It forms an equilateral triangle with sides of length 20 (twice the radius of each sphere). The height of this equilateral triangle can be found using the formula:
height = side * √3 / 2
So, the height of the equilateral triangle formed by the centers of the bottom three spheres is 20 * √3 / 2 = 10√3.
Now, let's consider the fourth sphere that is placed on top of the other three spheres. The center of this sphere will be height h (distance from the center of the equilateral triangle) above the topmost point of the bottom three spheres.
The center of the fourth sphere forms a right-angled triangle with the centers of two adjacent spheres from the bottom three. The hypotenuse of this right-angled triangle will be the sum of the radius of each sphere, which is 10 + 10 = 20.
Now, we can find the height h of the fourth sphere by applying the Pythagorean theorem:
(20)^2 = h^2 + (10√3)^2
400 = h^2 + 300
h^2 = 400 - 300
h^2 = 100
Therefore, we have h^2 = 100 = 100/1.
The value of a is 100, and the value of b is 1.
Finally, the value of a+b is 100 + 1 = 101.
Hence, the value of a+b is 101.