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A cylinder has a circumference of $4$, height of $3$, A is directly above B, and the distance from B to C along the circumference is $1$. The shortest distance through the cylinder from point A to C is same as $\displaystyle \sqrt {\frac{M+Nπ^2}{Pπ^2}}$ where M, N and P are all positive integers.I need to find the smallest value of the sum of M, N, P.

I drew the bottom of the cylinder as a circle and drew a square inside it, I also drew a line through the square representing the diameter of the circle. Then I calculated the diameter so I can find the distance of the straight line BC.
$4/π = 1.273 = \sqrt{2x^2}, x = 0.798$. So that means $AC = \sqrt{3^2 + 0.798^2} = 3.104$

Now the part I'm stuck on is where I have to find the value of M,N, and P. I have no clue what I should do here. Any help would be much appreciated.

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    $\begingroup$ Keep everything in symbol with $\pi$ and $\sqrt{}$'s, then you can read off $M,N,P$ directly,. $\endgroup$ – user10354138 Feb 24 at 6:13
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Given the circumference is $4$ and the distance from $B$ to $C$ along circumference is $1$, $\angle BOC = 90^0$.

$\displaystyle OB = OC = \frac{2}{\pi}$. So, $\displaystyle BC = \frac{2 \sqrt2}{\pi}$.

We also know that $AB = 3$ and $AB \perp BC$ (the question could have been written more clearly that $B$ and $C$ are on the circumference at the base of the cylinder and $A$ is above $B$ on the circumference at the top of the cylinder).

$AC^2 = AB^2 + BC^2 = 9 + \displaystyle \frac{8}{\pi^2}$

$AC = \displaystyle \sqrt{\frac{8 + 9 \pi^2}{\pi^2}} $

Hence $M = 8, N = 9, P = 1$ i.e. $M+N+P = 18$.

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