BrainDen.com - Brain Teasers

## Question

I once stood on a Cartesian plane at (0, 0) facing north (along the positive y-axis). I pulled out a rug in the shape of a regular triangle (which I call a 3-gon) and set it on the plane with one vertex at my feet at (0, 0) and with the center in the direction I was facing along the y-axis. I then started walking forward on the rug until I got to the center of the 3-gon, at which point I stopped and turned clockwise until I was facing a vertex, and I walked to that vertex of the 3-gon. Then I pulled out a rug in the shape of a square (which I call a 4-gon), put it with one vertex at my feet and with the center straight ahead of my current view (after that previous clockwise turn). I walked to the center of the 4-gon and then turned counterclockwise (instead of clockwise like on the odd-numbered N-gon rug), and started walking again as soon as I was facing a new vertex of the 4-gon. I kept repeating that for every odd numbered N-gon and every multiple of 4-gon.

Which way was I facing after I did that forever?

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• 1

Looks like

Spoiler

The angles now are

[ - 90o/1 - 90o/2 + 180o/3 ] + [ - 90o/3 - 90o/4 + 180o/5 ] + ... + [ - 90o/(2n-1) - 90o/2n  + 180o/(2n+1) ] + ....

= (90o) Sum(n=1, inf) { - 1/(2n-1) - 1/2n  + 2/(2n+1) }

= 90o { ln(2) - 2 } (Wolfram)

= -117.6167...o

Twice as much as the first answer. Showing that rearrangement of terms in an infinite series that is not absolutely convergent (both answers were forms of alternating harmonic series) can change the result. If the terms of a convergent series are all positive, the order does not affect the result.

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• 0

Just to be clear, n = 3 4 5 7 8 9 11 12 13 ..., and CW and CCW alternate untethered (my new favorite word) to parity?

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Assuming the above, we note that

Spoiler

Each n-gon bends the path uniquely. For odd n, the bend is given by 180o/n.
For each 4
p-gon, where p=1 2 3 ..., the bend is 90o/p.
Since the bends are all in the plane, they can be added in arbitrary order.

4p-gons (square, octagon, etc., 90/p, - + - + - + ...)

Bend4p = -90o, + 45o, -22.5o , = (-90o) Sum(i=1, inf) (-1)i+1(1/i) = (-90) ln 2 = -62.383246250395077847550890931236o

n-gons (n odd, 180/n, + + - - + + - - ...)

Bendnodd = 180o { 1/3 + 1/5 - 1/7 - 1/9 + 1/11 + 1/13 - 1/15 + ... }

Off the top of my head, I don't know this sum. Here are some thoughts ...

Alternating series are generally easier to deal with. We can sum pairs of terms to do this in two ways. In the first approach, we sum adjacent terms of like sign:

Bendnodd = 180o { (1/3+1/5) - (1/7+1/9) + (1/11+1/13) - (1/15+1/17) + ...}
= 180o ( 8/15 - 16/63 + 24/143 - 32/255 + ... (2
n+1/22n-1) ...

In the second approach, we sum adjacent odd terms and separately adjacent even terms to obtain two series:

Bendnodd1 = 180o (( 1/3 - 1/7) + (1/11 - 1/15) + (1/19 - 1/23) ...)
= 720o ( 1/21         + 1/165            + 1/437 + ... )

And then the even terms pairwise to get

Bendnodd2 = 180o ( (1/5 - 1/9) + (1/13 - 1/17) + (1/21 - 1/25) ...)
= 720o ( 1/45          + 1/221           + 1/525 + ... )

Final angle that you're facing after doing this forever is  { (-90)o ln 2 } + Bendnodd (CW is positive)

To be continued ...

I'm leaving now, to plow through some old texts on evaluating alternating series ... ugh.

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Clarification: when I walked the odd numbered N-gons I always turned clockwise. When I walked the multiple of 4-gons I always turned counterclockwise.

For now, suppose I always alternated between an odd-gon and a 4x-gon without bothering to keep edge number monotonically increasing. So 3-gon, 4-gon, 5-gon, 8-gon, 7-gon, 12-gon, 9-gon, etc with clockwise turns on blue and counterclockwise on red.

(Otherwise, yeah, it would be a mess.)

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With that interpretation,

Spoiler

The successive angles are 180/3 - 90/1 + 180/5 - 90/2 + ... so that

Final angle = 180o {  1/3 - 1/2  +  1/5 - 1/4  +  1/7 - 1/6  + ... }

= 180o { (1/3 - 1/2) + (1/5 - 1/4) + (1/7 - 1/6) + ... }

= 180o { -1/6 - 1/20 - 1/42 - ... - (1/(2
n+1) - 1/2n) - .... }

= -90o { 1/3 + 1/10 + 1/21 + ... + 1/(
n(2n+1)) + .... }

= -90o Sum(
n=1, inf) 1/(n(2n+1))

= -90o { 2 - 2 ln(2) } (Wolfram)

= -55.2335 ...o (CCW from
y-axis)

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Good, but now for the kicker. There's a reason I didn't specify the order in which I pulled out the rugs in the OP.

After I finished pulling out every rug and walking the pattern forever, I went back and picked up all my rugs. Then instead of alternating odd-gons and 4x-gons, I placed two 4x-gons in a row before placing the next odd-gon. I still turned the same direction on every rug that I did before: still clockwise on every odd-gon and counterclockwise on every 4x-gon, so in the pattern 4-gon, 8-gon, 3-gon, 12-gon, 16-gon, 5-gon, 20-gon, 24-gon, 7-gon, 28-gon, 32-gon with clockwise turns on blue and counterclockwise on red.

After that second forever, which way was I facing?

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1 hour ago, bonanova said:

Looks like

Hide contents

The angles now are

[ - 90o/1 - 90o/2 + 180o/3 ] + [ - 90o/3 - 90o/4 + 180o/5 ] + ... + [ - 90o/(2n-1) - 90o/2n  + 180o/(2n+1) ] + ....

= (90o) Sum(n=1, inf) { - 1/(2n-1) - 1/2n  + 2/(2n+1) }

= 90o { ln(2) - 2 } (Wolfram)

= -117.6167...o

Twice as much as the first answer. Showing that rearrangement of terms in an infinite series that is not absolutely convergent (both answers were forms of alternating harmonic series) can change the result. If the terms of a convergent series are all positive, the order does not affect the result.

Yes, and this has got to be among the most counter-intuitive properties of infinity that I know of.

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