## 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?

June 4th, 2011 at 7:45 pm

D

June 4th, 2011 at 9:19 pm

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.

June 5th, 2011 at 4:06 am

Why is it D?

June 5th, 2011 at 11:41 am

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

June 5th, 2011 at 12:17 pm

Hi Ajai. Why won’t they meet?

June 5th, 2011 at 12:32 pm

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

June 5th, 2011 at 3:21 pm

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.

June 5th, 2011 at 4:32 pm

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.

June 5th, 2011 at 9:07 pm

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.

June 6th, 2011 at 4:43 am

Hi JG. What is a constant curve?

June 6th, 2011 at 9:04 am

why are the bugs on a curve?

in math why not staight lines

June 6th, 2011 at 10:18 am

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.

June 6th, 2011 at 10:32 am

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

i think

June 6th, 2011 at 11:04 am

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

June 6th, 2011 at 11:18 am

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.

June 6th, 2011 at 1:50 pm

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.

June 6th, 2011 at 2:26 pm

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.

June 6th, 2011 at 3:17 pm

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.

June 6th, 2011 at 6:49 pm

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.

June 7th, 2011 at 4:57 am

In the wolfram link, the equation r = a e

^{bθ}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

^{-θ}.June 8th, 2011 at 7:48 am

Any chance we are looking at the perfect ratio?

June 8th, 2011 at 8:08 am

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.

June 8th, 2011 at 5:12 pm

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.

June 8th, 2011 at 5:17 pm

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?