## The Hermit

Things are getting interesting!. Here is the **2nd logic puzzle**:

One morning at 7 am a hermit began to climb a mountain. The narrow path, no more than a foot or two wide, spiraled around the mountain to the summit. The hermit ascended the path at varying rates of speed, stopping along the way to rest and eat. He reached the summit shortly before sunset. He spent the night and began his journey back along the same path, starting at 7 am the next day.

Can you prove that there is a spot along the path that the hermit has occupied on both trips at precisely the same time of day?

*Responses can be submitted in the comment section below. Will unmask later on this week, thanks!*

jmart574| Profile September 22nd, 2009 - 1:06 amIf you had 2 hermits, one going up and one going down, at some point they would meet during the day. Each hermit would represent one of the trips.

pgzzz| Profile September 22nd, 2009 - 4:23 amCreate a graph showing time of day versus distance, running bottom to top on the up leg and top – bottom on the down leg. At some point on the graph (no matter how you vary the speeds)the two lines cross each other and at that point the time of day and the position on the mountain are the same.

Shawn| PUZZLE GRANDMASTER | Profile September 22nd, 2009 - 8:29 amIntuitively, it makes sense.

If you imagine that sunset is 7pm, and that the hermit walked at a constant speed both up and down, then the crossing point would occur at 1pm. If he walks more slowly on the way up, the crossing point will be farther down the mountain, and farther up the mountain if he takes more time on the way down.

Any attempt by the hermit to avoid the crossover will unsuccessful, because the pace that he set the previous day will march onward to meet him.

How about (limit, as A approaches B, of A/B) = 1

where A is the progress up the mountain, and B is the progress down, and A/B is the location at any given time of the hermit on one trip as compared to the other?

Ari| Profile September 22nd, 2009 - 10:05 amSay there were two hermits, one at the bottom and the other at the summit. Both begin their journey at the same time, the other is ascending, the other descending. Both are taking the same trail so they are bound to meet at some point.

The hermits’ speed doesent matter or the time they take to rest, eat, pee etc. They always meet at some point.

This also applies to the hermit in question. If you plot his position/time, the graphs are bound to intersect at one point.

hex| PUZZLE MASTER | Profile September 22nd, 2009 - 10:10 amDraw a graph x=f1(t) and x=f2(t) where x is the linear distance from the bottom to the top of the mountain, t is the time, f1 is the function during ascent, f2 during descent.

f1 will start at (t,x)=(7,0) and end at (T1, L)

f2 will start at (t,x)=(7,L) and end at (T2, 0)

T1=~sunset

T2=unknown

L=x at summit

The 2 curves have to intersect as the functions are continuous (think of it as a quadrilateral where the functions have to join opposite corners each).

suineg| PUZZLE MASTER | Profile September 22nd, 2009 - 6:03 pmok this problem is cool, to me the more easy proof is the visual one, lets put it this way:

A————B thats the direct path (jaja cool)

lets assume that at the same time the hermit began to climb the mountain there is a clone of the hermit that start his journey back along the same path there is a point P where they are going to meet at the same time of the day that does not depend on the velocity of each one:

A—–—B in this point the hermit is in the same spot in the same hour of the day cool, I know you expect difficult equations and all that but i find this example super easy, hope is enough, cool.

Mashplum| PUZZLE MASTER | Profile September 22nd, 2009 - 8:35 pmImagine that there is a second hermit who, on the second day, undertakes a faithful replica o f the first hermit’s journey. The original hermit can’t help but encounter his doppelganger on the way down. That meeting spot is where the original was at that time the previous day.

BillyPilgrim| Profile September 23rd, 2009 - 12:32 amI’m not sure how you would “prove” it exactly, but the way I like to think about it is this… imagine climbing day hermit and descending day hermit start their climb/descent on the same day at the same time. They have to pass each other at some point on their journey, meaning they’d be in the same spot at the same time.

aaronlau| Profile September 23rd, 2009 - 5:03 amI do not know if there is mathematical equation proof.

But graphically, if you were to draw two distance time graphs, beginning at either end and ending at the opposite side.

They WILL intersect at a point, representing the spot and time of occurrence.

Allen| Profile September 23rd, 2009 - 7:04 amHis ascend will have an increasing displacement from start and his descend will have a decreasing displacement from start.

Take the two as functions of time and displacement from start.

No matter what form each function takes, because one increases and the other decreases they will intercept (once). At this point, the Hermit will be the same distance from start at the same time during both descent and ascend.

Blusummers13| Profile September 23rd, 2009 - 2:43 pmAssuming that he never leaves the path, yes.

In trying to answer this question for myself, I had to think of it in terms of a similar example. Two people start along the narrow mountain path at 7 AM. One from the bottom and one from the summit. They walk at varying rates of speed and take several stops along the way to eat and rest. The one starting at the bottom reaches the summit just before sunset. Do they ever cross paths?

Again, the answer is yes. As long as their paths overlap, they have to cross at some point no matter what speed each one is going.

michaelc| Profile September 23rd, 2009 - 5:49 pmSeems complicated at first, but just picture the hermit starting up the mountain and down the mountain on the same day at the same time. Eventually they meet at some point at the same time. qed

the_god_dellusion| Profile September 24th, 2009 - 3:41 ammmm i’m not too sure if this is gd enough explanation let alone proof, lol, but here goes:

Lets place 2 hermits on this path, one at top and one at the bottom so one will be ascending the mountain and the other descending. then if they start at same time of day there will be an instance when they cross each other somewhere along the path. hence at this instant they will be occupying the same spot at precisely the same time…

i realise i have changed the puzzle slightly by placing 2 hermits instanteniously ascending and descending on same day, but the idea is the same as you can visualise combining the 2 days to get the effect that i have given.

As they start at same time in each day i think it is possible to use the example of having 2 hermits instantaniously ascending and descending the mountain.

i hope this is clear enough…

bilbao| Profile September 25th, 2009 - 2:21 amAwesome!! Not one but two different ways of ‘proving’ it (graphically and 2 hermits). Congrats to all of you!

Original source:

Monograph entitled “On Problem-Solving,” by the German Gestalt psychologist Karl Duncker. (1903-1940)

Tsopi| Profile January 19th, 2010 - 3:25 pmI think this is easy…noone can eat and rest and eat at a place that is 1 or 2 feet wide.So there must have been a place that was suitable for resting and eating.Now getting to the specific time…we all eat around a specific time for launch(1.30 or 2.00 pm here in greece) So he stopped again at the only spot he could rest and eat(along the path)obviously at the same time of the day