# Amorous Bugs - a classic Martin Gardner puzzle

- Author:
- jeromeawhite

This construction models a generalized case of the following Martin Gardner puzzle:

Four bugs –A, B, C, and D– occupy the corners of a square 10 inches on a side. Each bug is attracted to one of its immediate neighbors. Simultaneously A crawls directly toward B, B toward C, C toward D, and D toward A. If all four bugs crawl at the same constant rate, they will describe four congruent logarithmic spirals which meet at the corner of the square. How far does each bug travel before they meet? The problem can be solved without calculus.

**SPOILERS AHEAD:**The solution is provided below the GeoGebra construction. Furthermore, checking the "show paths" box in the construction will reveal the logarithmic spirals and the general-case ratio of path length to polygon side length. If you're not already familiar with this puzzle, consider attempting it first before proceeding. Most of the sliders and other controls are best understood by playing with them. However, it may not be obvious that the "R_{0}" slider determines the radius of the circle that passes through all vertices of the starting polygon. For a general case, this seemed more sensible to me than having a slider that directly controls the polygon side lengths. Note: When animating the θ slider, the bugs do NOT move at constant linear speeds along their paths. Instead, the bugs experience an exponential decay in their speeds, but this may not be evident since the construction zooms in exponentially. Although this doesn't capture the true premise of the puzzle, it makes for a better animation. And yes, apparently these magical bugs also scale down in size exponentially as we zoom in :-)Martin Gardner solution:
the length of the original polygon's side.
Be sure to notice that logarithmic spirals are "self-similar." No matter how far you zoom in, the size and shape of each path appears identical to how it did before you started the zoom. If there were no scaling on the coordinate axes, you'd have no clue how far you had zoomed in! Also, theoretically, these paths can only reach the center of the original polygon (i.e. the origin) upon spiraling infinitely-many rotations inward.

At any given instant the four bugs form the corners of a square which shrinks and rotates as the bugs move closer together. The path of each pursuer will therefore at all times be perpendicular to the path of the pursued. This tells us that as A, for example, approaches B, there is no component in B's motion which carries B toward or away from A. Consequently A will capture B in the same time that it would take if B had remained stationary. The length of each spiral path will be the same as the side of the square: 10 inches. If three bugs start from the corners of an equilateral triangle, each bug's motion will have a component of 1/2 (the cosine of a 60° angle is 1/2) its velocity that will carry it toward its pursuer. Two bugs will therefore have a mutual approach speed of 3/2 velocity. The bugs meet at the center of the triangle after a time interval equal to twice the side of the triangle divided by three times the velocity, each tracing a path that is 2/3 the length of the triangle's side.Generally,

*n*bugs at the corners of an*n*-sided polygon will lead to each bug tracing a path that is