# Can you outrun the angry ram?

The Riddler puzzle this week appears simple at first glance, but I promise you it’s not!

You, a hard-driving sheep farmer, are tucked into the southeast corner of your square, fenced-in sheep paddock. There are two gates equidistant from you: one at the southwest corner and one at the northeast corner. An angry, recalcitrant ram enters the paddock from the southwest gate and charges directly at you at a constant speed. You run — obviously! — at a constant speed along the eastern fence toward the northeast gate in an attempt to escape. The ram keeps charging, always directly at you.

How much faster than you does the ram have to run so that he catches you just as you reach the gate?

Here is a very simple solution by Hector Pefo. Minimal calculus required!
[Show Solution]

And here is my solution, which finds an equation for the path of the ram but requires knowledge of calculus and differential equations.
[Show Solution]

## 10 thoughts on “Can you outrun the angry ram?”

1. Hector Pefo says:

There is, as you suggest, a simple way to obtain the length without SOLVING integrals, because you can eliminate integrals in equations for the total change in distance between farmer and ram (1 = integral[cos(a)]dt – vT) and the total y distance covered by the ram (1 = integral[v cos(a)]dt), where a is the angular difference in their paths at a given time, T the total time, and the integrals are from 0 to T. Then simple algebra and the quadratic formula gives v = (sqrt(5)+1)/2.

1. Laurent says:

Wonderful! I was looking for a solution like this but couldn’t find one. I think I may have a solution that doesn’t use any calculus at all, but I think it may end up actually being more complicated than yours, so probably not worth it. I updated my solution to include yours. Thanks!

2. Dmytro Taranovsky says:

Inspired by the solution from Hector Pefo, here is a solution that does not use calculus. The key observation is that at each point, the speed of the ram in the y direction (the direction of the farmer’s movement) is v times higher than the speed of the farmer in the direction away from the ram. This is because the ram is v times faster than the farmer and the angle between the bearing of the ram and the y direction equals the angle between the bearing of the farmer (the y direction) and the direction away from ram (which equals the bearing of the ram).
At time 1, the ram (initially at (0,0)) and the farmer (initially at (1,0)) meet at (1,1), so the average speed in the y direction is 1. Thus, the farmer moves away from the ram with the average speed 1/v, so the ram approaches the farmer with the average speed v-1/v. At time 1, the distance between the farmer and the ram is 1-(v-1/v) = 0, so v^2-v-1=0 and v=(1+sqrt(5))/2.

3. hypergeometric says:

Nice solutions!
What is the closed form equation of the trajectory of the ram when starting at $(0,0)$? The length of the trajectory of the ram is $v=\phi$. If the trajectory is an ellipse, this means the value of the elliptic integral is $\phi$ – how is this possible?
(NB – in your solution it might helpful to point out, for completeness, the assumption WLOG that the speed of the man is $1$. )

1. Laurent says:

If you choose some fixed distance, the set of all possible ram starting points such that the ram ends up traveling this fixed distance forms an ellipse. The ram trajectories themselves, however, are not ellipses! To find the equation for the ram’s trajectory when starting from (0,0), set $a=b=0$ in my equation for $y$. Then you obtain:
$y = \frac{1}{2}\left[ \frac{-1}{1-1/v} \left(\left(1-x\right)^{1-1/v} – 1\right)+\frac{ 1}{1+1/v} \left(\left(1-x\right)^{1+1/v} – 1\right)\right]$To find the appropriate value of $v$, use the formula for $v_\text{ram}$, again setting $a=b=0$, and you obtain
$v=\frac{1+\sqrt{5}}{2}$Here is a numerical approximation, rounding to 5 decimal digits:
$y = 1-1.30902\,(1-x)^{0.38197} + 0.30902\,(1-x)^{1.61803}$You can see that when $x=0$, $y=0$. And when $x=1$, $y=1$. This is a formula for $y$ in terms of $x$ for the ram’s trajectory, and it’s definitely not an ellipse.

1. hypergeometric says:

Thanks for the clarification on the closed form (I managed to get the same from your original solution) and also for confirming that the ellipse reference was not for the trajectory.
It’s interesting to note that whilst the closed form for $y$ appears messy it can be written as
$$y=1-\int \sin(\frac 1v \ln (1-x)) dx$$.

1. Laurent says:

I think you meant hyperbolic sin, correct?
$y=1-\int \sinh\left(\tfrac{1}{v}\log(1-x)\right)\,dx$

1. hypergeometric says:

$$y=1-\int \sinh\left( \ln (1-x)\right) dx$$

(It’s great that your comments section accepts TeX input 🙂 Would it be too much to ask to include a preview box, like on math.stackexchange.com, in order to minimize typesetting errors since the comment can’t be changed once it’s posted?)

4. hypergeometric says:

Aarrgh!
$$y=1-\int \sinh \left(\frac 1v \ln (1-x) \right) dx$$

(Sorry about that, it’s past midnight here)

1. Laurent says:

Couldn’t find an easy way to do previews, but I added a plugin that allows anonymous commenters to edit their posts for 5 min after posting.