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Offline Uranium235
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#661
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Let's assume something else. If the spider moves faster than the ant, then it just needs to be on the surface, and then it will eventually catch up. Apparently, the task isn't about that. Let's assume that due to the freedom of movement, the speed V should be sufficient. The route is to move in a straight line towards the ant, no matter where it is.
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Offline Hermit
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#662
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Uranium235
Let's assume something else. If the spider moves faster than the ant, then it just needs to be on the surface, and then it will eventually catch up. Apparently, the task isn't about that.
Let's assume that due to freedom of movement, a speed of V is sufficient. The route is to move in a straight line towards the ant, no matter where it is.
So, what is the minimum speed the spider needs to move at to catch the ant? If the answer is that the minimum speed at which it needs to move to catch the ant is v, then that is incorrect. (I'm not claiming that it's insufficient.)
Всё не так плохо как Вы думаете. Всё намного хуже!
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Offline Dybinkin
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#663
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Ant speed +1?
С уважением, Дубинкин.
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Offline Hermit
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#664
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No. This is not the minimum speed at which it needs to crawl to catch an ant.
Всё не так плохо как Вы думаете. Всё намного хуже!
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Offline Uranium235
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#665
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No, I won't derive the formula. I don't even remember what the height of a tetrahedron is, although it might not even be necessary. Also, we can assume that since the spider's speed will be lower, the ant needs to be cornered or kept near a corner. We need to find a point from which we can catch the ant regardless of one of the three directions it can choose, taking into account the possibility that it can stand still until the last moment.
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Offline Hermit
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#666
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Uranium235
We can also assume that since the spider's speed will be lower, the ant needs to be cornered or kept near a corner. Find a point from which you can catch the ant regardless of one of the three directions it can choose, taking into account the possibility that it can stand still until the very end.
Yes! However, I'd like to point out that in one of the three directions, the ant can move when it's at a vertex, but when it's just on an edge, it can only move in two directions.

Added after 13 hours and 52 minutes
Hint.

Solve the riddle! An ant runs along the sides of an equilateral triangle at a speed of v, and a spider runs across its entire area. What should be the minimum speed u of the spider so that it can catch it?

How do you like my hint? :)
Всё не так плохо как Вы думаете. Всё намного хуже!
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Offline Ment
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#667
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Uran is right, I think. Trap the ant in the corner of a tetrahedron (this is possible, as the ant needs to change edges. If the ant only crawls along one edge, the task of catching it becomes much simpler). Accordingly, the spider at that moment should be on the bisector to the corner, at the vertex where the ant is. If the ant moves, the spider should move along the perpendicular to the ant's edge, and it should have enough speed to cover this segment while the ant crawls to its base. At the same time, the ant may try to play with the spider, making continuous false sorties, without intending to finally escape from the trap. This will essentially ensure its safety with the minimum speed of the spider (it will not be able to get closer), so the spider's speed should be even infinitesimally greater.
The perpendicular to the edge from the bisector of the tetrahedron... After long calculations, I realized that it is apparently sqrt(2) times smaller than the distance to its base from the vertex. Then, if the ant's speed is v, the spider's speed is
v/sqrt(2) + o(v)
 

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Offline Ment
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#668
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Actually, no, o(v) doesn't seem to be necessary. It's better to return to the bisector; if the ant changes direction, it will be easier for the spider, as the path will be shorter compared to the one it took when deviating from the bisector. This is how it will gain an advantage.
 

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Offline Hermit
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#669
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Ment
Uran is right.
Yes, Uran is right, that the ant needs to be driven into a corner.
Ment
v/sqrt(2) + o(v)
No, that's incorrect.
Ment
If the ant moves, the spider should move perpendicularly to the ant's edge.
But why exactly perpendicularly?

Added 1 minute ago
Ment
Although, no, o(v) doesn't seem to be needed. It's better to return to the bisector if the ant changes direction; it will be easier for the spider, and the path will be shorter compared to the one it traveled while deviating from the bisector. It will gain an advantage because of this.
No, and the spider's speed u > v/21/2, and u = v/21/2 are also incorrect answers.

Added 6 minutes ago
I want to clarify that we have an ideal spider that has instant reaction to changes in the ant's velocity vector
Всё не так плохо как Вы думаете. Всё намного хуже!
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Offline Ment
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#670
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Why specifically along the perpendicular?
It's the shortest path to the edge. It's easy to show that in my scheme, it's enough to prevent the ant from escaping the trap. If the ant is attached to a corner, the spider will eventually be able to reduce the distance to the corner.
Let's say the ant crawls along the edge from the corner. Then the spider slides down from the bisector and goes to intercept. The ant hasn't changed direction => it's caught. The ant decided to crawl in the opposite direction (towards the vertex) => the spider returns to the bisector and still manages to get closer to the vertex in the process. Something like that.
 

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Offline Hermit
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#671
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Ment
The shortest path to the rib.
Why do you think that the shorter the path the spider has to crawl to the interception point, the less speed it will need to intercept the ant?
Всё не так плохо как Вы думаете. Всё намного хуже!
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Offline Ment
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#672
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Yes, that's exactly what I was thinking. If it's more convenient for the spider to return to the bisector along a line perpendicular to the bisector, then it might also be more convenient to crawl along that same perpendicular line from the bisector. In other words, let the spider intercept the ant along a line perpendicular to its bisector, specifically the one that intersects the edge of the tetrahedron. Then it will need a speed of v/sqrt(3) to intercept it.
But here's the catch: the ant can constantly return to the vertex and change edges. It won't be able to escape the trap that way, but the spider will also have to return to the same point, and it won't be able to get close to the vertex. So, in the end, I think the answer is v/sqrt(3) + o(v). In that case, the ant has no winning strategy.
 

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Offline Hermit
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#674
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Ment
Yes, that's exactly what I was thinking. If it's easier for the spider to return to the bisector along a line perpendicular to the bisector, then it might also be easier to crawl from the bisector along the same perpendicular. In other words, let the spider intercept the ant along a line perpendicular to its bisector, specifically the one that intersects the edge of the tetrahedron. Then it will need a speed of v/sqrt(3) to intercept.
But here's the catch: the ant can constantly return to the vertex and change edges. It won't be able to escape the trap that way, but the spider will also have to return to the same point, and it won't be able to get close to the vertex. So, in the end, I see the answer as v/sqrt(3)+o(v). In that case, the ant has no winning strategy.

Yes! The spider's speed must be greater than the ant's speed divided by the square root of three!!! u>v/31/2
This is the condition under which it can move its "three-pointed star of death" towards the ant. In the case of equality, it can only hold the ant in the corner, preventing it from escaping, but it also cannot bring the plane in which it crawls (the plane in which the three-pointed star is located) closer to the vertex of the corner in which the ant is trapped! But if it's greater, then, without letting the ant out of the corner, it can bring the plane closer to it!

Thank you all for your interest in the puzzle!

Go ahead, Ment, pose the next one!
Всё не так плохо как Вы думаете. Всё намного хуже!
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Offline Злобный Ящур
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#675
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Yes! The spider's speed must be greater than the ant's speed divided by the square root of three!!! u>v/31/2
In case of equality, just hold it.
But if it's greater...
So, what minimum speed should the spider have?:smile08:

Ничто не возникает из ничего и ничто не пропадает бесследно.
Если где-то чего-то убудет, то в другом месте добавится.
(Закон сохранения).

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Offline Hermit
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#676
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P. S. Here, the spider's speed will be minimal when the **ratio** of the distance the spider crawls to the interception point to the distance the ant crawls to the interception point is minimal, and not the distance the spider crawls to the interception point.
Всё не так плохо как Вы думаете. Всё намного хуже!
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