
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 nsided 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 fivepointed 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 wedgeshaped 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 3side 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 Ngons 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
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