At time t=0, point A is at the origin (0,0) and point B is distance 1 appart at (0,1). A moves to the right (on the x-axis) with velocity 1 and B moves with velocity 1 towards A. What will be the distance between A and B when t goes to infinity?

Not sure what you mean by “towards A”. what happens if it reaches A as A approaches B. Since A continues to move right I understand that B will “follow” A with the same speed (both 1) and thus distance is 0.

In the X-Y plain A is directly below B and A starts moving right so at time t it is in coordinate (t, 0). Since B goes towards A it always goes towards the x axes and to the right. They can never meet because they have the same speed and A took the shortest path to get where he is. Eventually B is going to be almost on the x axes, so there will be a limiting distance between them. The question is how much 🙂

Very very nice! I got it wrong, although I realized the flaw in my intuition shortly after guessing.
After some though, I came up with an elementary way to get the right answer without solving the ODE (which I didn’t do mostly out of laziness), but I’m not sure if it counts as “intuition”.

A follow up: Suppose that the same process actually started before t=0 (and the conditions given are what was observed in medias res). At the negative starting time, B was actually on the line y = 3. What was the starting distance between A and B?

I also think I got it with an elegant method. I can give you hint if you send me an e-mail at parity@gmail.com, as there might be people here who don’t want to see any hint.

I’m proud of my solution, but perhaps others do not want the spoiler, so here it is in cipher (enter text at https://www.rot13.com/ to decode).

N irpgbe qvntenz znxrf vg dhvgr rivqrag gung N’f pbzcbarag bs irybpvgl va gur qverpgvba njnl sebz O vf ng rirel vafgnag rdhny va zntavghqr gb O’f pbzcbarag bs irybpvgl va gur evtugjneq qverpgvba. Gurersber, nf gur qvfgnapr orgjrra N naq O inevrf, gur ubevmbagny pbzcbarag bs qvfcynprzrag sebz O gb N inevrf ng na rdhny naq bccbfvgr engr, fb gung gur gbgny bs qvfgnapr naq ubevmbagny qvfcynprzrag vf pbafgnag.

Gur qvfgnapr vf vavgvnyyl 1 naq gur ubevmbagny qvfcynprzrag vf vavgvnyyl 0, fb gur pbafgnag inyhr bs gur fhz vf 1. Nf gur gvzr tbrf gb vasvavgl, gur qvfgnapr naq gur ubevmbagny qvfcynprzrag graq gb gur fnzr yvzvg, juvpu zhfg gurersber rdhny bar-unys.

Not sure what you mean by “towards A”. what happens if it reaches A as A approaches B. Since A continues to move right I understand that B will “follow” A with the same speed (both 1) and thus distance is 0.

But, can B reach A?

Ah, I misread B as starting at (1,0). This makes more sense.

I agree.

In the X-Y plain A is directly below B and A starts moving right so at time t it is in coordinate (t, 0). Since B goes towards A it always goes towards the x axes and to the right. They can never meet because they have the same speed and A took the shortest path to get where he is. Eventually B is going to be almost on the x axes, so there will be a limiting distance between them. The question is how much 🙂

I couldn’t see how one can get intuition about this problem, so I cheated and solved the ODE.

Conclusion: as long as B does not start exactly to the right of A it will never catch up: there will be a positive limiting distance.

Very very nice! I got it wrong, although I realized the flaw in my intuition shortly after guessing.

After some though, I came up with an elementary way to get the right answer without solving the ODE (which I didn’t do mostly out of laziness), but I’m not sure if it counts as “intuition”.

A follow up: Suppose that the same process actually started before t=0 (and the conditions given are what was observed in medias res). At the negative starting time, B was actually on the line y = 3. What was the starting distance between A and B?

can you give a hint on your method?

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Can you give a hint on your method?

I also think I got it with an elegant method. I can give you hint if you send me an e-mail at parity@gmail.com, as there might be people here who don’t want to see any hint.

Can anyone provide a hint for this? I believe the limiting distance is smaller than 1/2, but I don’t know how to find the precise one.

I’m curious why you think the limiting distance is less than 1/2.

(Read this only if you want some hints!)

Write $|AB|=:s$. Then $t-x=s \sin\theta= s\dot x$, y=s\cos\theta=-s\dot y$. Differentiate $(t-x)^2+y^2=s^2$ and simplify. You obtain $\dot s=\dot x-1=-(s\dot x)^\dot$. Draw conclusions.

I’m proud of my solution, but perhaps others do not want the spoiler, so here it is in cipher (enter text at https://www.rot13.com/ to decode).

N irpgbe qvntenz znxrf vg dhvgr rivqrag gung N’f pbzcbarag bs irybpvgl va gur qverpgvba njnl sebz O vf ng rirel vafgnag rdhny va zntavghqr gb O’f pbzcbarag bs irybpvgl va gur evtugjneq qverpgvba. Gurersber, nf gur qvfgnapr orgjrra N naq O inevrf, gur ubevmbagny pbzcbarag bs qvfcynprzrag sebz O gb N inevrf ng na rdhny naq bccbfvgr engr, fb gung gur gbgny bs qvfgnapr naq ubevmbagny qvfcynprzrag vf pbafgnag.

Gur qvfgnapr vf vavgvnyyl 1 naq gur ubevmbagny qvfcynprzrag vf vavgvnyyl 0, fb gur pbafgnag inyhr bs gur fhz vf 1. Nf gur gvzr tbrf gb vasvavgl, gur qvfgnapr naq gur ubevmbagny qvfcynprzrag graq gb gur fnzr yvzvg, juvpu zhfg gurersber rdhny bar-unys.

It seems Mr. Gil Kalai totally forgot about this test of intuition. Not sure what interesting thing he had in mind when posting this problem…

This is a lovely problem and with some help from the comments above I was able to get a closed form solution for the path of B. It is:

{1/2 (W(E^(1-4 t))+2 t-1),Sqrt[W(E^(1-4 t))]}

where W is the LambertW function.