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9 members have voted

Canyonero
Canyonero
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Thanks for this post from:
OnceDearChuckleberry
March 1st, 2019 at 12:00:52 PM permalink
Since I have known of this problem before, I must recuse myself. However, once you guys are done working on it, reward yourself with this clip (spoiler!):

https://youtu.be/dAyDi1aa40E?t=250
beachbumbabs
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beachbumbabs
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March 1st, 2019 at 12:05:29 PM permalink
Quote: Wizard

Thanks for the responses so far. I'd like to say that to get full credit I want to see a solution that for the given type of road shape, the details are optimal. Not just using goalseek in Excel, or some similar product.



Which means that, once again, Babs has the right answer, but not the definitive proof, correct Wizard? And once again, the goalposts move away from the Babs beer. (Edit: I'm wrong about the goalposts moving, perhaps. I don't know if my answer equals or beats the Wizard's. The POLL asked the best shape, but did not qualify for the beer.)

I would submit that the 30° angle is the correct answer for all squares. For other rectangles, the angle could change, but not exceed 45° nor fall to 0 (all angles measured from the shorter side). The crossbar of the H should always parallel the longer side of the rectangle. These rules are all reflective of the ratio of the hypotenuse to the shorter side of a right triangle.
If the House lost every hand, they wouldn't deal the game.
charliepatrick
charliepatrick
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March 1st, 2019 at 1:37:28 PM permalink
Here is an idea for the proof.

First consider the shortest roads between three points. It transpires (ignoring weird cases where going AB and BC are shorter) the point is such that the angles of the three roads AF, BF, CF all meet at 120 degrees. (I can prove it if needs be but the rough proof is to consider strings where they all pull a ring at point F; then by symmetry the net force has to be 0; so the angles have to be equal.)

Now consider the square ABCD and roads directly to the centre (E). Thus the initial idea is roads AE BE CE and DE. Then consider the three points A B and E. The set of roads at 120 degrees is shorter than AE+BE. So construct a point F so the roads are AF BF and EF where the angles at F are 120 degrees. Similarly construct G for C and D.

This leaves the roads AE BE EF FC FD as a local minimum. However this set of roads goes through the centre of the square so symmetry suggest this is one of two solutions (the other solution joins BCE and ADF instead).
Joeman
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Thanks for this post from:
ChuckleberryunJonbeachbumbabsBleedingChipsSlowly
March 1st, 2019 at 1:48:32 PM permalink
So after reading Babs' answer and Wiz's response above, I figured that I should be able to dust off my old Calc and Trig books and solve this. This solution crossed my mind before I gave my first answer, but I didn't have the time or resources then to tackle it at that point.

So, assuming the "slant H" shape, the the road is the total of the four angled lines plus the horizontal line. Since the state is symmetrical, we can just focus on half of the state, and double our answer at the end. So, all we need to do now is come up with a function that defines the length of road. Once we have that, we can take the first derivative of that function and set the result equal zero. Then by solving that equation, we can determine the minimum road length. Here's a crude drawing of this



From this, we see that the length of road is equal to the total length of the two angled lines plus the left segment of the horizontal line.
In terms of angle 'a,' the total length L = 2(100*sec a) + (100 - 100*tan a). Let's divide out the 100 to get a simplified:

L = 2(sec a) + 1 - tan a

Taking the first deravitive, we get

d/da = 2(sec a * tan a) - sec2a

Now, set d/da = 0 and solve for a:

sec2a = 2(sec a * tan a)

divide out sec a from both sides to get:

sec a = 2(tan a)

substitute the right angle triangle sides (Opp, Adj, Hyp) for the trig ratios, and you get:

H/A = 2 O/A

multiply both sides by A and rearrange, and you get:

O/H = 1/2 = sin a

So, a = 30°, just like Babs said! :)

Thus, L = 200*sec 30° + 100 - 100* tan 30°, yielding L = 273.2

Double that and you get the total paved km to be: 546.4

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Wizard
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March 1st, 2019 at 3:54:04 PM permalink
Quote: beachbumbabs

Which means that, once again, Babs has the right answer, but not the definitive proof, correct Wizard? And once again, the goalposts move away from the Babs beer.



Yes, that is pretty much the case. Lest, you feel I'm being unfair, I think most here can vouch that I like to see winners show their work to get full credit. I know it is difficult conveying a full solution with a keyboard, so I don't need to see every step. Glorified trial and error is not the kind of "work" I like to see.

Hint: Calculus is the perfect tool for problems like this, and we all know much I love calculus.
It's not whether you win or lose; it's whether or not you had a good bet.
charliepatrick
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March 1st, 2019 at 5:17:24 PM permalink
Quote: Wizard

...Hint: Calculus is the perfect tool for problems like this...

I agree with Babs answer (although I hadn't read it before answering). I'm not sure you need calculus to prove it, except you can think of little triangles...and prove the derivitave needs to be zero.
Just consider the base AB along the x-asis so A=(0,0) B=(200,0) and the center is at O=(100,100). The roads meet at Y=(100,y).

It is easier to think of triangles and how various lengths change as y changes.

As y increases the length of the road to the centre merely decreases by y (actually YO = 100-y).

Consider similar triangles and the angle BAY. (You actually need to look at a small triangle Y Y' and Z=a point on AY extended where Y Z Y' is a right angles.)

Angle(Y Y' Z) = Angle(BAY), and the length Y Y' is y (i.e. how long the road to the center changed) and the length Y Z is (for small triangles) how long the road AY got longer.

Since Y Z = Y Y' * Sin ( Y Y' Z )= Y Y' * Sin (BAY)

There are two roads (AY and BY) so the total length of these increase by 2 * { Y Y' * Sin (BAY) }.

Thus the total roads change by 2 y Sin(BAY) - y.

This is zero when Sin(BAY)= 1/2 i.e. when BAY = 30 degrees.

Thus AYB = 120 degrees.

Thus AY = 2 y and 100 = SQRT (2y*2y - y*y) = SQRT(3 y^2). So y = 100 / SQRT(3).

Total length of all roads = 4 * 2y + 2 * (100-y) = 200+6y = 200 + 600/SQRT(3) = 200 * (SQRT(3)+1) = 546.102.
sodawater
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March 1st, 2019 at 8:46:11 PM permalink
I don't understand how the soap KNOWS how to solve this problem.
DJTeddyBear
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March 1st, 2019 at 8:52:59 PM permalink
Quote: Wizard

... Glorified trial and error is not the kind of "work" I like to see.

Sounds like Mike is talking about the Excel spreadsheet that I discribed in my second post on page 1. But that’s OK. I kinda new that he would not like that kind of “proof.”
I invented a few casino games. Info: http://www.DaveMillerGaming.com/ ————————————————————————————————————— Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
Wizard
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March 1st, 2019 at 9:39:56 PM permalink
I have to give the beer to Charlie on this one. If he is claiming his solution didn't use calculus, I would claim it did, but avoiding calculus terminology. He can transfer his beer to Barbara, if he wishes.

Can anyone find a web page that gives solutions for higher order regular n-gons?
It's not whether you win or lose; it's whether or not you had a good bet.
sodawater
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March 2nd, 2019 at 12:08:24 AM permalink
apparently for an octagon it's just the perimeter. I would guess anything with more sides would be the same.

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