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Four marching bugs


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Four bugs sit in a perfect 10x10 square, ABCD. Simultaneously, they start marching: A toward B, B toward C, C toward D, and D toward A. They march, spiraling, until they meet at the center. What distance does each bug cover?

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It is important to specify the condition that the bugs always move directly towards their target, not just at the starting gun.

This leads to the spiral path you mentioned.

It also leads to an easy solution without adevanced maths.

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16 minutes ago, studiot said:

It is important to specify the condition that the bugs always move directly towards their target, not just at the starting gun.

This leads to the spiral path you mentioned.

It also leads to an easy solution without adevanced maths.

Yes, to all three points above.

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40 minutes ago, TheVat said:
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Might help to switch to polar coordinates?  So any bug's path is a logarithmic spiral.  I get 10.  Can that be right?  Same as side of the square.

 

 

Spoiler

Yes, it is right. +1. However, it can be achieved without calculating the spiral. Change reference frame, again. 

At any moment, the bugs are in the corners of a square, which rotates and shrinks. What does the bug A see? It sees the bug B always in a position perpendicular to the line between them. The bug B moves its legs, but being perpendicular to the line between them, it doesn't march toward or away from the bug A. So, to get to the bug B, the bug A should cover the distance between them, no more, no less. Which is 10.

 

Edited by Genady
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12 hours ago, Genady said:
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Yes, it is right. +1. However, it can be achieved without calculating the spiral. Change reference frame, again. 

At any moment, the bugs are in the corners of a square, which rotates and shrinks. What does the bug A see? It sees the bug B always in a position perpendicular to the line between them. The bug B moves its legs, but being perpendicular to the line between them, it doesn't march toward or away from the bug A. So, to get to the bug B, the bug A should cover the distance between them, no more, no less. Which is 10.

Yes, I see the FoR aspect  now.  So simple, and there I was doing it the hard way.  Haha.

 

Edited by TheVat
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26 minutes ago, TheVat said:

 

OK!

Do you want to try the same problem but with three bugs in the corners of equilateral triangle? Six bugs in the corners of equilateral hexagon? Just for practice :) 

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