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Imagine four bugs situated at each vertex of a unit square.

Suddenly, each bug begins to chase its counterclockwise neighbour.

If the bugs travel at 1 unit per minute, how long will it take for the four bugs to crash into one another?

enter image description here

First, the author explains that the bug is travelling towards its neighbor and maintaining a 45 degree angle, with respect to the center of the square.

enter image description here

Second, the author explains that why are trying to find out the radial component of the bug to the center of the square.

enter image description here

Third, the author explains the radial speed was r, then r is the same as the length radial line of a vertex of the starting square, which has length 1 to the center. The actual value of r is square root of 2/2.

Although this puzzle is popularly solved using equations, this is the first time I am seeing this solved using graphical approach.

I do not understand why did he chose to use the diagonal line as a reference to the center of the square to calculate the radial component.

ilovetolearn
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1 Answers1

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By symmetry, the four bugs always occupy the vertices of a square. The square shrinks and rotates as the bugs chase one another, but symmetry implies that its center does not move. Therefore the diagonals of the square always meet at the center of the unit square and the bug's radial velocity is the projection of the bug's velocity vector along the diagonal that passes through the bug's position.

The velocity vector of a bug always points to the next bug, and is therefore along a side of the square. The angle between that side and the diagonal is $45\unicode{xb0}$. The radial velocity is therefore $\frac{1}{\sqrt{2}}$ units per minute; the initial distance of the bug from the center of the unit square is also $\frac{1}{\sqrt{2}}$ units. Hence the four bugs meet at the center after $1$ minute.

Fabio Somenzi
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    If the radial velocity is constant in magnitude, and always point to the center of the unit square, the time a bug takes to reach the center is the radial distance it covers over its velocity. – Fabio Somenzi Nov 04 '18 at 15:17
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    The radial vector (of a bug) is the vector pointing from the bug to the (fixed) center. This vector rotates as the bug spirals around the center, but its magnitude remains constant. – Fabio Somenzi Nov 06 '18 at 15:06
  • what is the author's intention to illustrate the radial component and velocity vector, where he could have just explained that it could be calculated from the vertex to the center of the square.

    The other explanation I found here: http://www.puzzles.com/puzzleplayground/BugsTraffic/BugsTrafficPrintPlay.pdf is much simpler to understand.

    Sorry, I really can't understand the author's intention of highlighting the smaller triangle. It confuses me.

     – ilovetolearn Nov 06 '18 at 15:34
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    The explanation you linked is clear and well-written. It shows that each bug will reach the next bug in 1 minute. From that you can infer that each bug reaches the center in 1 minute, because symmetry dictates that. The explanation in your post follows the opposite path. It shows that all bugs reach the center in one minute, which is therefore also when they all catch up with each other. I don't disagree that the other explanation is a bit more direct, but it's nice to see them both. – Fabio Somenzi Nov 06 '18 at 16:45
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    I need some help in understanding where you get stumped. Perhaps what's confusing is that the radial vector is not perpendicular to the velocity vector as in circular motion? (Just a wild guess on my part.) – Fabio Somenzi Nov 07 '18 at 19:32
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    That's because the velocity vector of a bug always points to the next bug at the other end of a side of the square, and the angle between that side and the diagonal is 45°. – Fabio Somenzi Nov 10 '18 at 15:45
  • The ratio of the side to the hypotenuse is $1/\sqrt{2}$. The ratio of the magnitude of the radial velocity to the magnitude of the total velocity is the same. Since the total velocity has constant magnitude $1$, the radial velocity has magnitude $1/\sqrt{2}$. – Fabio Somenzi Nov 10 '18 at 16:03
  • The $1/\sqrt{2}$ is actually the side of the isosceles right triangle whose hypotenuse is $1$. You can prove it with the Pythagorean theorem: $1 = 2\ell^2$. – Fabio Somenzi Nov 10 '18 at 16:41
  • how did the author get radial speed and radial distance as r/r ? – ilovetolearn Jan 13 '19 at 01:46