# Maths and mountains, and Issac Newton

Below is a nice puzzle, nice being defined if you like maths or logic puzzles, like readers of my blog know that I do. And my family, to their expense, know that I do.

**Suppose a man sets out to climb a mountain at sunrise, arriving at the top at sunset. He sleeps at the top and descends the following day, also from sunrise, travelling more quickly downhill. Prove that there's a point on the path that he will pass at the same time on both days.**

I gave this problem to my (long suffering) daughter #2 and (even longer suffering) wife during a recent hike up the wonderful **Walla Crag**, a walk from Keswick into Borrowdale, near our home in Rosthwaite.

After the initial usual (I) moan (II) clarification/ recalibration of the question, and after some thought my wife said 'course not', and when I said there was, a brief discussion ensured (basically, said wife saying 'rubbish') until daughter #2 agreed to listen to the explanation. And then once explained, one final exchange (said wife saying 'no more puzzles today‘) and on with our walk.

I solved the puzzle in two ways, but there is a third, pretty way which is the solution given in the book where I found it, Futliity Closet by Greg Ross. My son gave me the book last Christmas and it has been a joy to read. As the hyperlink shows, it stems from a website whose webmaster turned some of his best contributions into a book published at the start of 2014, and my son guessed that I would like it: I do, it is a perfect last thing at night read, short snippets, some funny, some quirky, some history, some maths, some word plays…all the that I like.

**Solution**

**Newton's Laws of Motion approach**

Having tried to solve it intuitively, and failed, I first used algebra. Remembering 'v equals u plus a t' and the other equations of motion, I realised the way to solve this puzzle was 's equals u t plus half a t squared'.

Here, the acceleration is nil, so the equation simplifies considerably to s equal u t. If you assume Walla Crag has a height of h, then the height at time, the meeting point m, is m=ut, with u being the upward speed. On the way down, with speed d, the height at time t is also m, where m= h-dt.

So, if there is a meeting point, the equations are solvable, so that ut=h-dt. A simple rearrangement shows that the meeting time t= h/(u+d) which says that the meeting time is dependent on the sum of the up and down speeds. This led me on to my next solution, discussed below.

Substituting for t in the initial s=ut equation gives the final result that the meeting point **m=h * u/(u+d). **Again, this solution is revealing. The faster the upward speed, the higher up the mountain where the meeting point is, which in hindsight is obvious; and the faster the downward speed, the lower down; and finally the taller the mountain, the higher up the meeting point. All sensible, once you think about it.

**Rationalisation**

Armed with the solution, and there was no point in trying Newton's laws on either GCSE-result-expectant daughter or wife, I found this way to explain it.

Imagine a zip wire was installed, taking climbers instantly down from the summit to the starting point. Then at sunrise the next day, the descender would be at the starting point. When the response was that zip wires aren't instantaneous, I almost gave up, but persevered. Is is not easy being a mathematician.

Or, imagine that the fell walker died at the summit (perhaps his wife murdered him for asking too many maths questions) then the two walkers meet at the summit (speed of descent is zero, d standing for death).

So, at the two extremes of speed, there is a meeting point. From this the inference can be drawn that there is always a meeting point, the place depending on the ratio of upwards and downwards speed: descend twice as fast, meet 1/3 the way up, for instance.

Extending the logic, it can now be seen that the simplification of assuming no acceleration was unnecessary: the walkers still meet if there are sandwich and catching breath breaks.

**Logical solution**

The book has a prettier solution.

Imagine rather than there being one walker, there are two twins, Flora and Iona (twins we know). Iona walks up the hill whilst Flora sets off from the summit at the same moment, same day. Clearly they must meet somewhere: twins reunited.

Of course, my wife would say 'they might descend by different routes' and, by doing so, she would, as ever, win the argument. She also wins, of course, by just admiring the view.

(The two pictures are from googling, not of our family; but when we last walked up Walla Crag, a couple of weeks ago, the weather was every bit as good).