## 12 February 2012

### A faulty geometry problem

If you stand on level ground and look at The Pentagon, you will see either two sides of the building, or three (assuming you're far enough away not to see just one side).

Which is more likely?  Seeing two sides, or seeing three sides?

It's very simple, and can be deduced by logic rather than geometry per se.  But I got it wrong.

To find out whether your answer is right or not, you'll need to look in the Futility Closet.*

*Addendum - but it won't do you any good, because the answer at Futility Closet might be wrong, as three readers here pointed out within an hour.

**Second addendum - the errors claimed by the first three readers have been challenged by a fourth.

I'll leave the post up until we sort out the correct answer (or agree on how to word the proposition accurately).

1. The lack of comments on Futility Closet is very annoying when Greg posts things that are very clearly wrong.

If you draw out the geometry: a pentagon with sides extrapolated until they cross, you can see that the second sentence in the given answer is not actually true, and (with a bit of knowledge of similar triangles), what the correct answer is.

2. "If you can see two sides, she can see three, and vice versa."

This premise is untrue, even assuming that you and her are equidistant. It is untrue even if you are both infinitely far away.

While it is true that you can see three sides, she can see two, it is not true that if you can see two sides, she can see three. She may be seeing one, two or three.

This can be seen by extending the sides of a pentagon beyond where they intersect, marking the center of the pentagon, and then placing a straight edge at various angles through the center.

3. As others have pointed out, this is not true, which can clearly be seen in the magnificent (well ...) drawing I made:
http://shrani.si/f/q/YL/2Hahjdvc/5gon.jpg
Yellow, red and green mean 1, 2 and three sides seen, in that order.

4. Wow. Thank you all. I've modified the post, but will leave it up so others can note the error in the original.

5. OK, my turn to be contrarian.

In the limit of infinite distance, it is 50%, and the reason given applies, with one additional fact not mentioned in the Futility Closet solution, but implicit int he problem.

First, the case for 50% at infinite distance:
Looking at Jure B's excellent figure, it's clear that the "opening angle" of the red regions and the green regions are the same -- indeed, they are formed by the cross-over of the "same" two lines. Observe the cross-over directly below the "P" in the figure -- the green triangle spreads out below with some opening angle, and as you follow the log-side lines of the yellow triangle upwards from the bottom vertex, they become sides of the pentagon, then sides of other yellow triangles, and finally, for the long haul, become the sides of the red triangle opposite the initial green one. At large distances, the only thing that matters is the opening angle. Take a circle centered on the pentagon, and examine the limit of large radius -- as the radius increases, the fraction of red and green will tend to 50%.

Second, the case for FC's explanation:
In the large-distance limit, two and three sides are the only possibilities, *and* *the* *figure* *has* *odd* *parity*, so that if you are opposite a vertex, your 180-degree counterpart is opposite a face. Again in the large distance limit, the locus of angles where the FC assertion fails will diminish to zero.

I concur that the description is flawed, mostly because they didn't specify the large-distance limit in which their explanation makes sense.

1. Thank you, Urban. I'll modify the post to say that the answer at Futility Closet *might be* wrong. And wait for the math-minded readers to sort this out.

2. I actually did a much more laborious computation to come up with 50% in the long-distance limit, before I saw Jure B's figure. It's hard to describe, but I could share that, too, if by chance the discussion should heat up.

3. Urban Garlic, you are right, and just as you pointed out below, FC is still wrong. I should have put more thought into it, being a math student.

BTW, Minnesotastan, I love your blog. I've been reading it daily since mid-2010 and I learn something new almost every time. Great work! :)

4. Thanks, Jure B. In mid-2010, the visits here were about 1/4 what they are now, so that makes you one of the old-timers. Greetings.

5. Here a modification of Jure's drawing that I think definitively answers the question.
http://img59.imageshack.us/img59/3640/90917126.png

I've drawn a yellow line on the drawing passing through the center of the pentagon. Clearly the line intersects red on one side and red on the other for any observer distance, infinite or not. However, that line is unique . Even the slightest tilt in that line and at infinity one side will be in the red and one side will be in the green. Other than the 5 symmetrically equivalent lines there are no other lines that pass through the center and never cross into the green.

Thus the original answer is correct. For large distance two observers on opposite sides one will see 3 sides and one will see two. That special line is a vanishingly improbable arrangement of the two observers and thus can be ignored when computing the probability of seeing 3 or 2.

So the answer is in the limit of large distance the chance of seeing 3 sides is 50%

6. Great. Now I'm confused again. I'll concede your conclusions about people standing opposite one another around the pentagon. But...

If the problem speaks about a single person at a random distance (beyond the yellow), then that person's view is determined by whether he/she is staning on red or green. It seems to me that for any circle centered on the pentagon, the majority of the circumference is red (a 2-sides-view). No?

7. "If the problem speaks about a single person at a random distance (beyond the yellow), then that person's view is determined by whether he/she is staning on red or green. It seems to me that for any circle centered on the pentagon, the majority of the circumference is red (a 2-sides-view). No?"

No. But it does looks like that as drawn. Here's how to think about that aspect. Notice that the angle one of the red section is expanding is exactly the same angle as the green section. Thus they are both triangles with the same growth rate. The difference between them is that the apex of the green one is offset from the middle. This gives it a little head start.

that head start matters less and less as you move further out. since they grow at the same rate they are asymptotically the same area as your get towards infinitely far away.

8. Well, yes and no.

I'll grant you the part out towards infinity, where the red/green areas may achieve parity.

But the question speaks of a random person standing anywhere (beyond the points of the yellow). For a nearby person (in Arlington for example) or someone within the confines of your diagram, the probability of standing on red>>green.

Certainly if we take into consideration all points on the plane of the Pentagon in the universe, it virtually evens out [it's dangerous for an English major like me to argue anything involving infinity...), but it seems that red never can lose that initial advantage from the near points.

btw, I'm not really "arguing" with you - just playing mathematical mind games...

6. The question states that "as you walk into Arlington, Virginia", which puts a bound on the distance from the pentagon that the observer can be (something well-short of infinity), so we are back to FC being incorrect.

7. The answer approaches 50/50 as the distance away from the pentagon increases, because each area opens from a 36 degree angle, but within a circle extending approximately 1568 feet from the center the building (based on the actual size of the building) the only possibilities are seeing 1 or 2 sides. Jure B's diagram perfectly shows the options visible from "close" in (i.e. about 3 tenths of a mile). Just outside this radius, where the option of 3 sides is just beginning, the odds of seeing 2 sides is much greater than seeing 3 (and will always be slightly more likely at any finite distance).

So the Futility Closet got this one wrong.

8. @Len, excellent work actually reading the question, which does indeed set a distance scale. I retract my defense of FC.

In fact, I have walked into Arlington, VA, from time to time, and can state with some confidence that, owing to intervening buildings and terrain, you mostly can't see the Pentagon at all.

I also note, with some amusement (and residual contrariness...) that in this post's accompanying photograph, you can actually see *five* exterior walls, owing to the elevated vantage point and the building's courtyard.

1. I was going to make the same comment about seeing five sides :)

9. In response to the following question: If you stand on level ground and look at The Pentagon, you will see either two sides of the building, or three (assuming you're far enough away not to see just one side).

Which is more likely? Seeing two sides, or seeing three sides?

The answer is 3, no question.

10. If you're limiting yourself to standing on the surface of the Earth, then you can't really use the mathematical argument that the 2 vs. 3 side visual approaches 50/50 as you go to infinity. Because, as some scientists have observed, the Earth is not flat :-) So my vote goes towards 2 sides being more probable.

11. Seven years ago when I was stationed there, that long bit of parking lot at the bottom was a freeway, which has been re-routed closer to the Potomac River. They rerouted the freeway because they were afraid of a terrorist attack (like a bomb in a truck), but I never got to see the completed project(and hadn't thought of it in years)until just now.

Anyhow it was a really glaring security gap that is no more (don't even think about it) but you've inadvertently brought back some old memories, so thanks for that.

To answer the original question, if you stand on level ground (or one of the hills) stare at the Pentagon long enough, you are very likely to see two OR three large men with guns ask for your name and ID, and what you're doing. Trust me.

12. Kinda late to the party, but in any case:

Imagine someone standing a distance d away from the centre of a regular pentagonal building with side length r. Using Jure B's image (http://shrani.si/f/q/YL/2Hahjdvc/5gon.jpg) we can intuitively see that if d is less than some critical distance d0 (so standing within that yellow triangle area or a similar distance from the pentagon), it is impossible for that person to see three sides. Using some trigonometry, we see that d0 = (r/2)(tan(72deg) + tan(54deg)), tan(72) coming from the exterior triangle, tan(54)from the interior. If d is greater than this, the person now has a chance of seeing three sides. I'll do pictures/working if anyone wants it, but long story short, the percentage chance of seeing three sides of the pentagon = (5/360)*tan^-1((d-d0)tan(18)/d) * 100