   Pentagon probability.. A man arrives at a random spot several miles from the Pentagon. He looks at the building through binoculars. What is the probability that he will see three sides of the building? There is a generalization of this puzzle. What is the probability that a person can see 3 sides of a n-sided regular polygon building (all sides are equal and all angles are same, eg hexagon, decagon etc.) if he/she is standing at a random point sufficiently far from the building. (i.e., perpendicular distance to the closest side > 3.078 x side) As you can tell, n > 4. shailesh kumar Friday, February 15, 2002 Consider the Pentagon to be the center of a five-pointed star (that is, the star formed from five lines, coincidental with each of the Pentagon's sides).  Extend those lines forever.  Beyond the star, they divide the plane into ten regions.  In five of them, three sides are visible; in the other five, only two are visible. I'm noticing that all ten regions are wedge-shaped and congruent (the apex of each wedge is 54 degrees).  I'm tempted to say that since all the wedges are infinite in size, the probability approaches 50% that three sides are visible.  But it doesn't feel right.  If you restrict the man to being no farther than N miles away, the 3-side regions are always smaller in area. About all I can say right now is that the probability can be calculated as a function of distance.  It approaches 50% as the distance increases without bound.  It starts at 0% at the star points.  It's roughly equal to the ratio between two quantities.  Both quantities are proportional to the length of the base of isoceles triangles, and the difference between the heights of those two triangles is a constant.  That constant is equal to the height of the isoceles triangle formed by any three points of the star.  I'm too lazy to work that out right now.  :-) In general, the formula seems - different.  For a hexagon, the chance of seeing three sides approaches 100% (there's lanes of constant width in which you can only see two).  For N-gons with 7 or more sides, the chance of seeing exactly three sides actually decreases, since it's now possible to see four sides (more for bigger Ns). Paul Brinkley Tuesday, February 19, 2002 this was just a puzzle, not a computational geometry !!! u were right....u just need to make the approx...it says that u r far away from the pentagon...and for hexagon or more, u will see all the 3 sides... shailesh kumar Tuesday, February 26, 2002 Recent Topics Fog Creek Home