Ants on a cylinder

Actually if Billi were to walk up from $B$ to $A$ , then to $E$ and down to $M$ , it would only be $2 + 2 \times 5 \sin 75^\circ + 1 \approx 12.66 < 13$ . So Billy isn't that smart. This also tells Billi that it is worthwhile to climb up.

Let $\angle AOC = \alpha$ and $\angle DOE = \beta$ . Then the length of the path $B$ - $C$ - $D$ - $M$ is given by:

$\begin{aligned} \ell & = BC + CD + DM \\ & = \sqrt{2^2+(5\alpha)^2} + 2 \times 5 \sin \left(\frac {\frac 56\pi - \alpha -\beta}2 \right) + \sqrt{1^2 +(5\beta)^2} \\ & = \sqrt{4+25\alpha^2} + 10 \sin \left(\frac {5\pi}{12} - \frac \alpha 2 - \frac \beta 2 \right) + \sqrt{1 + 25\beta^2} \end{aligned}$

Let us check, if we fix the value of $\beta$ , what value of $\alpha$ minimizes $\ell$ . We apply

$\begin{aligned} \frac {\partial \ell}{\partial \alpha} & = \frac {25\alpha}{\sqrt{4+25\alpha^2}} - 5 \cos \left(\frac {5\pi}{12} - \frac \alpha 2 - \frac \beta 2 \right) & \small \blue{\text{Putting }\frac {\partial \ell}{\partial \alpha} = 0} \\ \implies \frac {25\alpha}{\sqrt{4+25\alpha^2}} & = 5 \cos \left(\frac {5\pi}{12} - \frac \alpha 2 - \frac \beta 2 \right) \end{aligned}$

This means that for any value of $\beta$ , $\alpha$ minimizes $\ell$ , when $\frac {5\alpha}{\sqrt{4+25\alpha^2}} = \cos \left(\frac {5\pi}{12} - \frac \alpha 2 - \frac \beta 2 \right)$ . Similarly, we find that for any value of $\alpha$ , $\beta$ minimizes $\ell$ , when $\frac {5\beta}{\sqrt{1+25\beta^2}} = \cos \left(\frac {5\pi}{12} - \frac \alpha 2 - \frac \beta 2 \right)$ . Therefore $\ell$ is minimized, when

$\begin{aligned} \frac {5\alpha}{\sqrt{4+25\alpha^2}} & = \frac {5\beta}{\sqrt{1+25\beta^2}} = \cos \left(\frac {5\pi}{12} - \frac \alpha 2 - \frac \beta 2 \right) \\ \frac {\alpha^2}{4+25\alpha^2} & = \frac {\beta^2}{1+25\beta^2} \\ \implies \alpha & = 2 \beta \\ \implies \frac {5\beta}{\sqrt{1+25\beta^2}} & = \cos \left(\frac {5\pi}{12} - \frac {3\beta} 2 \right) \\ \implies \sin \left(\frac {5\pi}{12} - \frac {3\beta} 2 \right) & = \frac 1{\sqrt{1+25\beta^2}} \end{aligned}$

Solving numerically, we have $\beta \approx 0.080460651$ . Substitute in $\ell_{\min} = 3\sqrt{1+25\beta^2} + \dfrac {10}{\sqrt{1+25\beta^2}} \approx 12.51105177$ . Therefore $\lfloor 100p \rfloor = \boxed {1251}$ .

Let's find the optimal path. If we won't come on top of the cylinder the path would take $\sqrt{{(5 * \frac{5}{6} \pi)}^{2} + {(2-1)}^{2}} > 13$ . So, let's call the entry point $C$ and the exit point $N$ , thus we have two right triangles $\bigtriangleup ABC$ and $\bigtriangleup LMN$ , where A and L - points on the top directly above B and M accordingly (they become right triangles when we place the side of the cylinder on a plane). O is the center of the top side of the cylinder, $\bigtriangleup CON$ is an isosceles triangle with legs $OC = ON = 5$ . Our path then is the polygonal chain $(B, C, N, M)$ .

Let's deconstruct the cylinder into a rectangle (the sides) and a circle (the top) and place them on a plane, so that the circle touches rectangle in point $N$ . Then points $C, N, M$ should be inline, otherwise the path isn't minimal: we can draw a segment $CM$ which would be shorter and moreover the part that isn't in the circle or rectangle is "non-existent", so new path is even shorter. $ON \perp LN$ and $LM \perp LN$ , so $ON \parallel LM$ . And since $CNM$ is one line, $\angle CNO = \angle NML$ .

Similarly, we can prove that $\angle NCO = \angle CBA$ . $\bigtriangleup CON$ is an isosceles triangle which means $\angle CBA = \angle NCO = \angle CNO = \angle NML = \alpha$ . Since $\bigtriangleup ABC$ and $\bigtriangleup LMN$ are right triangle, they are similar with coefficient 2 and $LN = \frac{1}{\cos{\alpha}}$ and $AB = \frac{2}{\cos{\alpha}}$ . $\bigtriangleup CON$ is an isosceles triangle with legs $OC = ON = 5$ , so $CN = 10 \cos{\alpha}$ . Thus the path length is $\frac{3}{\cos{\alpha}} + 10 \cos{\alpha}$ , we just need to find $\alpha$ .

That's the tricky part where I kinda cheated, so beware. Let's focus on arcs of the top circle. We know the lengths of arcs $\cap NL = \tan{\alpha}$ and $\cap AC = 2\tan{\alpha}$ because we know other sides of the right triangles. Arc $\cap CN$ has length $5 * \angle CON = 5 * (\pi - 2\alpha) = 5\pi - 10\alpha$ . But $\angle AOL = \frac{5}{6} \pi$ , so $\cap AL = \frac{25}{6} \pi$ . So we have:

$\cap AC + \cap CN + \cap NL = \cap AL$

$3\tan{\alpha} + 5\pi - 10\alpha = \frac{25}{6} \pi$

$\alpha = \frac{\pi}{12} + \frac{3}{10}\tan{\alpha}$

And this is the moment where I cheated: I typed the equation in WolframAlpha and got the root, $\alpha = 0.382490364764667...$ The equation doesn't seem to have an analytical answer but it would be nice if someone proved me wrong.

Now we can calculate the minimal path length, $\frac{3}{\cos{\alpha}} + 10 \cos{\alpha} = 12.511051767653990170...$ The answer is $\boxed{1251}$ .

First step: look for the optimal distance on the side of the fencepost. "Unrolling" the cylinder we have a right triangle with sides 1 and the arc formed by the 150° angle.

Nope! Longer than the full red distance passing by the center of the fencepost. It's shorter if we take the chord of the 150° angle, but we can do better, let's look how we can find the best angle to optimise his walk on the fencepost.

We know that the shortest path is a straight line. With the same angle in the center, the angle on the outside of the post change by a factor of two since Milly is twice the distance from the top than Brilli. The optimal straight line thus have angles that are x and x/2. This is the equation representing the distance that Brilli will walk, depending on the angle x from the center, and its graphic (from Pythagor and the calculation of a chord).

With the floor function as 100p, we thus have 1251! But anyway Brilli and Milly should be social distancing by now lol!

The key idea here is to find a way to "unfold" the cylinder into a flat net. Then the shortest path will be a straight line on that net. This gets a bit involved, so I've split the solution into a few parts.

Brilli has two options; stay on the curved surface, or cross over the flat top of the fencepost. It's difficult to find a single way to address these, so I'll do them separately. But first...

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Part 0: preliminaries

Let's start with notation. I want to solve the general problem, so let the radius of the fencepost be $r$ ; let Brilli's distance from the top be $a$ ; let Milly's be $b$ ; and let the angle separating them be $\alpha$ . So in this case $r=5$ , $a=2$ , $b=1$ , and $\alpha=150^\circ=\frac{5\pi}{6} \text{rad}$ (I'll be using radians for the rest of this solution).

For convenience, let's also define a cylindrical polar coordinate system $(\rho,\theta,z)$ : say the $z$ -axis is the axis of the cylinder, with the origin at the centre of the top of the of the fencepost, and take the positive $z$ direction to be downwards. Also, say Brilli starts at $\theta=0$ . The coordinates of the start and end points are then $B(r,0,a)$ and $M(r,\alpha,b)$ .

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Part 1: what if we stay on the curved surface?

Brilli has to travel a distance of $r\alpha$ in the $\theta$ direction, and $|b-a|$ in the $z$ direction. Since these are perpendicular, the straight-line distance (on the "unwrapped" curved surface) is, by Pythagoras, $D_\text{curved}=\sqrt{r^2 \alpha^2 + (b-a)^2}$

In this problem, this is about $13.128$ cm (this was Billy's suggestion, by the way).

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Part 2: crossing the top

Alternatively, Brilli can try going over the top of the fencepost. This path will take Brilli from his starting point $B$ to a point $B'$ where he moves from the curved surface to the flat top of the post, across the top to a point $M'$ , and down to Milly at $M$ .

To find our optimal straight line path, we need to unwrap the cylinder into its net. This net consists of a circle (the top of the fencepost) and a rectangle corresponding to the curved surface. The trick is to cut this rectangle into two smaller rectangles, one of which is tangent to the circle at $B'$ and the other at $M'$ :

The orange line represents the optimal path (which is a straight line on the net).

Point $B''$ has the same $z$ coordinate as $B$ and the same $\theta$ coordinate as $B'$ , so that $\Delta BB'B''$ has a right-angle at point $B''$ . Point $M''$ is defined likewise.

Let the angle $\angle B'OM'=2\phi$ . Then $\angle B'B''B=\angle M'M''M=\phi$ .

We can (finally) work out some distances. Considering triangle $\Delta BB'B''$ , we have $B'B''=a,\;\;\;BB''=a \cot \phi,\;\;\;BB'=\frac{a}{\sin \phi}$

Similarly, $M'M''=b,\;\;\;MM''=b \cot \phi,\;\;\;MM'=\frac{b}{\sin \phi}$

Crossing the circle, we have $B'M'=2r\sin\phi$ . So the length of the path is $D_\text{top}=BB'+B'M'+M'M=\frac{a+b}{\sin \phi}+2r \sin \phi$

Now, recall that the $\theta$ -component of the distance between $B$ and $M$ is $r\alpha$ . But we can also work this out on the net diagram; we have $r\alpha=BB''+2r\phi+MM''=(a+b)\cot \phi+2r\phi$

This gives us an equation we can solve (numerically) in $\phi$ ; in this case, we get $\phi \approx 1.188$ , which gives $D_\text{top}=12.51105\ldots$

Since this is the shorter of the two distances, it is optimal, giving the problem's answer $\boxed{1251}$ .

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Part 3: solving for $\phi$

To find $\phi$ , we had to solve the equation $(a+b)\cot \phi+2r\phi=r\alpha$

This is a transcendental equation, so has to be solved numerically. It's easy to plug into Wolfram|Alpha, or use Newton-Raphson; but a nice alternative is to simply iterate $\phi_{n+1}=\frac{r\alpha-(a+b)\cot \phi_n}{2r}$

for some sensible starting value of $\phi_0$ (eg $\frac{\pi}{4}$ ).

We in fact get more than one solution for $\phi$ , but it's easy to check among these for optimality. A nice property of the above iteration scheme is that only the optimal solution is stable (ie, for just about any starting value, it will converge to the correct $\phi$ ).