ball and cylinder question Can you help me with this?
Thank you!
The surface area of the ball is two times larger than the side surface of the cylinder inscribed in the ball. Find the ratio of the height of the cylinder and the radius of its base.
 A: Let's assume the ball has a radius $R$, and we'll say the inscribed cylinder has radius $r$ and height $h$. Then the surface areas for each are:
\begin{align}
A_\textrm{ball} &= 4\pi R^2 \\
A_\textrm{cyl}  &= 2\pi r h
\end{align}
And we want $A_\textrm{ball} = 2A_\textrm{cyl}$. In other words:
\begin{align}
4\pi R^2 &= 4\pi r h \\
\iff R^2 &= r h
\end{align}
So far we haven't used the fact that the cylinder is inscribed within the ball. This means $r$ and $h$ cannot be independent of each other: For example if I make $r$ large, then $h$ must be small in order to make the cylinder fit inside the ball.
If you imagine a line connecting the center of the ball to a point (any point) where the cylinder touches the surface of the ball, then that line will (of course) have length $R$ because that line is a radius of the ball, but also, the length of that line is the hypoteneuse of the triangle whose base is the radius of the cylinder $r$, and whose height is half the height of the cylinder $h/2$. Thus by the Pythagorean Theorem:
$$ r^2 + (h/2)^2 = R^2 $$
And substituting for $R^2$ in the earlier equation:
$$ r^2 + (h/2)^2 = rh $$
From this point, you'll want to compute $h/r$, the ratio of the cylinder's height to its radius. Can you take it from here?
A: Slice the solid through the axis of the cylinder and draw a picture.  You should have a rectangle that touches a circle at the corners.  You might as well let the radius of the sphere be $1$ as everything will scale.  Let the cylinder have a base diameter of $d$ and height of $h$.  You should be able to find two equations in those two unknowns, one from the surface areas and one from the rectangle fitting in the circle.
A: Building on WB-man's answer, as I think that's the best approach, once you have that last equation
$$r^2+(h/2)^2=rh$$
just take the following steps
$$4r^2-4*rh+h^2= 0 = (2r-h)^2$$
$$2r-h=0$$ $$\Rightarrow 2r=h$$ $$\Rightarrow 2=h/r$$ $$\equiv$$ $$r/h=1/2$$
and there you have it (or you can try and solve via other methods for quadratic equations, but that will most likely be a pain).
