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## Bug walk

Posted by Chris on June 4, 2011 – 5:05 pm

Four bugs are at the four corners of a square of side length D. They start walking at constant speed in an anticlockwise direction at all times directly towards the bug ahead of them.

How far does each bug walk before they meet with each other?

This post is under “Mathemagic, MathsChallenge” and has 24 respond so far.
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### 24 Responds so far- Add one»

1. 1. Tyler Salners Said：

D

2. 2. Your name Said：

If they walk along the side of the square then they would never touch as they are going at a constant speed. If they go straight to the other bug then they would travel exactly ‘D’ I think.

3. 3. Chris Said：

Why is it D?

4. 4. Ajai Said：

They would never meet each other as long as they walk at a constant speed.

5. 5. Chris Said：

Hi Ajai. Why won’t they meet?

6. 6. Ashkard Said：

They won’t meet since they are walking at a cst speed for a same distance!

7. 7. JG Said：

There are two answers:

1. For now, T represents the constant speed at which each bug can travel d. If each bug traveles d at a speed of T, then after the amount of time it takes to travel d, each bug would be at the next corner in an anticloclwise direction and this would continue forever. Therefore, the bugs would never meet.

2. Technically, the bugs would not be able to walk on a square because it is impossible to have a two dimensional object in our universe.

8. 8. Chris Said：

I’m not sure that the question is being interpreted correctly by some of you. For clarity, although the bugs start at the corners of the square they don’t walk along the edges of it, they always walk in the direction of the next (anti-clockwise) bug ahead of themselves – they are (or will be) walking on a curved path.

For bonus marks, describe the path that they walk.

—-

Off topic. I read an article on the James Webb Space Telescope http://en.wikipedia.org/wiki/James_Webb_Space_Telescope, (which will replace the Hubble ST) which was quite interesting in itself. It mentioned that it would sited be at the L2 point. The telescope will be considerably further away than the Moon, and will be going in an orbit parallel to the Earth’s. Normally an orbiting object further out than the Earth’s would take more than a year to do the orbit. But if it is placed in line with the Earth and the Sun, the extra gravitational attraction of the Earth causes it’s period to reduce (and is equivalent to making the Sun more massive) – a nice trick. There is also an L1 point. It’s inside the Earth’s orbit and also is associated with an orbit of 1 year period. This time the Earth is counteracting the Sun. There are 5 of these Lagrange points. See http://en.wikipedia.org/wiki/Lagrangian_point for more info.

In fact the telescope won’t really be at the L2 point, but near it and will actually have a wobbly orbit.

9. 9. JG Said：

The bugs would be walking in a constant curve, until they hit the center of the square, so the arc would have a radius of .5D. The bugs would walk 2*pi*(.5D)= D*pi but the arc is only 1/4 of a circle. This makes the distance .25pi*D-b, with b being the length of each bug’s body. Someone please correct me if I am wrong.

10. 10. Chris Said：

Hi JG. What is a constant curve?

11. 11. Lee Button Said：

why are the bugs on a curve?
in math why not staight lines

12. 12. Chris Said：

Hi Lee. That’s what I want people to show.

My bonus question is: what is the shape of the curve that the bugs walk on?

You don’t need to know the shape to answer the posted question. The correct answer (by the first two posters) is D. But they haven’t explained why that’s so, they may well have got the right answer for the wong reason.

13. 13. Danny Said：

take the diagonals and twist them anticlockwise that`s the shape of the movement
i think

14. 14. John24 Said：

Spiral since it is a curve and only ends at the center of the spiral.

15. 15. John24 Said：

All bugs start moving at same time. Each bug curves slightly towards the center of the square since this is the shortest path to the bug they are chasing. As each bug progresses the curve becomes sharper. The length of the path is the same as the length of one side of the square regardless of speed.

16. 16. Chris Said：

Hi John24. That’s more like it Thank you.

Consider two points, A and B, a distance D apart and point B then moves a very small distance, d, at 90° to the line joining A and B, then the increase in distance between A and B is about ½ d²/D. If point B moves a distance d directly towards point A, then the decrease in distance is d. If D >> d, then d >> ½ d²/D, so altogether the distance decreases from D to D-d. By choosing d to be as as small as we please, we can make the error in the approximation as small as we please. Hence the bugs meet after walking a distance D i.e. the length of the side of the original square.

If you make a sketch, you’ll realise that each bug is walking at 45° to a radial line from the centre of the square. So the bugs walk in a logarithmic spiral and will go round the centre an infinite number of times before meeting. Adding a touch of reality – if the bugs are such that when they meet they form a square of side b, then they will have walked a distance d – b before meeting.

See “Visiting Santa”L http://trickofmind.com/?p=138&trashed=1&ids=1145 posts 15 and on – the same 45° applies to that problem too.

17. 17. Chris Said：

Another way to see it: Initially the bugs will start walking along the edges of the original square. By symmetry, the bugs will remain in a square formation. The new square must be smaller than the old one as its corners are on the sides of the original square. But now the bugs start walking on the edges of the new square, etc.

If the bugs move a certain (not necessarily small) fraction of the distance to go, then the square becomes smaller by the same linear fraction.

18. 18. JG Said：

By constant curve I mean that the bugs will walk along the edges of a circle for exactly 90 degrees of the circle, minus the length of the bug, with the edge of the circle intersecting the center of the square and the corner that the bug starts in.

19. 19. Chris Said：

Hi JG. I had wondered if you meant a circle (a curve of constant curvature).

If you want to see the shape (and get more info) then http://mathworld.wolfram.com/LogarithmicSpiral.html is worth a glance.

20. 20. Chris Said：

In the wolfram link, the equation r = a e is given. If φ is the constant bearing with respect to the outward radial line through the bugs current position (such that φ = 0 => radially outward and φ = 90° => anticlockwise circular path) then b = 1/tanφ. a is the radial position when θ = 0.

In the posted problem φ = 135°, so 1/tanφ = -1. If look at one bug that we can suppose to start at r = D√2, θ = 0, then r = D√2 e.

21. 21. cazayoux Said：

Any chance we are looking at the perfect ratio?

22. 22. Chris Said：

Hi cazayoux. No. But there is a “Golden Ratio” version of the spiral, each rotation takes you one ratio nearer/further from the pevious position. I’ve no idea why or if it’s really special, but the literature does refer to it.

23. 23. lydia Said：

depends if the bugs are all going using the same speed then they will never reach each other but if one bug goes faster then the other he could reach the other bug but u would need to know what d is and the speed the bug is going at.

24. 24. Chris Said：

Hi lydia. The bugs are all going at the same speed and they do meet each other, and that’s after they have walked a distance d. You only need their speed if you wanted to calculate how long it would take for them to meet. Why do you say they won’t meet?

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