# algebra and geometry question

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.

• Keep everything in symbol with $\pi$ and $\sqrt{}$'s, then you can read off $M,N,P$ directly,. – user10354138 Feb 24 at 6:13

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. $$M+N+P = 18$$.