Pentagon

[cinelli]

A man stands 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 of its sides?

CInelli

[UncleEbenezer]

When you say "several miles", I guess you mean "effectively infinite"?

Are we to assume the pentagon has five equal sides and equal angles? If so, that makes each angle 2*pi/5.

What's the orientation of the pentagon? If he views it into blinding winter sun, he'll see a whole lot less than in good light.

What's the distribution of spots where a man could stand with a sufficiently uninterrupted view of the pentagon? Won't he get shot or at best arrested, especially if he looks black or arabic or ...

Ok, stated purely as a puzzle in elementary geometry, to have sight of three sides is equivalent to having a line of sight to four vertices. The two ends of the nearest side[1], and two more. That implies the nearest side is viewed at an angle close to perpendicular.

By symmetry, we need only consider one of the five sides being nearest, and we need only consider half of viewpoints on it (i.e. one or other side of perpendicular). The angle from perpendicular will be between zero and pi/5, above which the adjacent side becomes the nearest. A third side will be visible if and only if the angle is less than pi/10, being the obtuseness of the pentagon's vertices. So the third side is visible from exactly half of all viewing angles.

[1] In the case of there being two equally-near sides, pick either. That zero-probability case is obviously one where only those two sides are visible.

[GoSeigen]

Agree with Uncle, 50%, calculated independently but using a similar method.


GS

[9873210]

An alternate method that arrives at the same answer.

Because the pentagon is convex any particular side can be seen from half the distant points. So the average number of sides in view is 2.5.

Since the only possible number of sides in view from a distant point for a regular pentagon are 2 and 3 each must be 50% to get the average.//

This also holds for irregular pentagons as long as each interior angle is greater than a right angle.

[cinelli]

I do have a geometrical argument but I’m not entirely happy with that. Neither am I entirely happy with this alternative approach.

Suppose the man has a companion and that she stands at a point diametrically opposite from his position relative to the centre of the Pentagon. Then by symmetry, if he sees 3 sides she will see 2 and if he sees 2 sides, she will see 3. So, as 9873210 writes, the average number of sides in view from any position is 2.5. Therefore the probability is one half.

But an interesting problem, I think.

And to answer one of UncleEbenezer’s questions, yes, the Pentagon does have five equal sides and angles. You can see it on Google Maps.

Cinelli