Connect the Dots

[TimeSpaceLightForce]

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Build a road system that access all the 9 towns shown on the map.
How long is the total length of shortest pavement that can be constructed?

[Buddyboy3000]

I finally found a way to make the road system less than 7.5 miles.

Spoiler

 

The idea I got was an edit of the idea that tojo928 gave. Instead of doing X's across four towns, it will be shorter to do something similar to the shape of two trapezoids. The left and right side of the X's are pushed inward to make a 0.5 mile road in the middle. Instead of 2.828 miles, this idea will be about 2.736 miles. Use two of these to replace the X's to make it shorter. The improved road system will be about 7.472 miles long.

 

[bonanova]

First thoughts

Spoiler

Obviously 8 miles will suffice, but the answer must be less because a good puzzle does not have an obvious answer.

I'll jump way out of the box and observe that if the pavement can be 2 miles wide, then it need be only 2 miles long.

 

[TimeSpaceLightForce]

Hi there bonanova. .

You may consider curved and angled paths. but assume paving a narrower road on a flat open area. about 10 ft wide.

[tojo928]

A little shorter then the obvious

Spoiler

2+4*sqrt(2) or 7.65.

Villages numbered like a phone keypad. Road from 1->5->9. Road from 2->4. Road from 6->8. Road from 2->3. Road from 7->8.

 

[TimeSpaceLightForce]

@tojo928  

Your 7.65 mile road  design  is a good  proposal.. but the budget  is good only  for road system  shorter than 7.50 miles. Can you revise it?

Thanks you for submitting your work.

 

[DejMar]

tojo928's proposal is better than the simple design of connecting the nine towns with orthogonal roadways. An improvement to his solution is

Spoiler

to recognize that one can branch off at points anywhere. (It should be evident that variations to the roadway system of the same length does exist). The change made to tojo928's proposal is replacing the √2, or approximately 1.4142 mile connection between the midpoint of the town-2-town-4 connection to the midpoint of the town-6-town-8 connection with the 1 mile town-4-town-7 connection.  The total length of this system of roads is  3+3√2, or approximately 7.26241, miles.

 

[Buddyboy3000]

21 hours ago, DejMar said:

tojo928's proposal is better than the simple design of connecting the nine towns with orthogonal roadways. An improvement to his solution is

  Hide contents

to recognize that one can branch off at points anywhere. (It should be evident that variations to the roadway system of the same length does exist). The change made to tojo928's proposal is replacing the √2, or approximately 1.4142 mile connection between the midpoint of the town-2-town-4 connection to the midpoint of the town-6-town-8 connection with the 1 mile town-4-town-7 connection.  The total length of this system of roads is  3+3√2, or approximately 7.26241, miles.

 

 This will work to make the road system shorter, but you will no longer have connection to town #5. 

 To find the shortest road system, I found it useful to use Desmos Calculator, if anyone else likes the idea. Anyways, this is my proposal...

Spoiler

Instead of going from towns 6 to 8, then to 7, I went a more direct route from 6 to 7. Keep the road from town 6 to the the area in the middle of town 6 and 8. Now go from there straight to 7. Now, connect town 8 to the nearest part of the new road. I did the same to towns 2, 3, and 4. With the new long roads made, make a road perpendicular to them going through town 5. Last, connect town 9 to the area between towns 6 and 8, and town 1 to the area between towns 2 and 4.

This should make it look like the road system below. The total length of it is 2*sqrt(2.6)+sqrt(8)+sqrt(1.6), which is about 7.318 miles long.

(Note: The image got stretched out a little bit when uploaded, so it is not completely at the right angle.)

 

 

[TimeSpaceLightForce]

@Buddyboy3000

Nice strategy ..

Spoiler

..but looks longer.

 

[Buddyboy3000]

Oops...

I did the math wrong. My idea is actually about 7.88 miles long. Instead ls 2*sqrt(2.6)+sqrt(8)+sqrt(1.6), it is 2*sqrt(2.5)+sqrt(8)+sqrt(1.6)+2*sqrt(0.1) I made the yellow sides a tiny bit longer, and forget the short roads to towns 2 and 8.

I am going to continue and try to find a shorter path.

[jasen]

Total length = 6 * sqrt 2 = 7.414

Spoiler

 

 

[jasen]

sorry I'm wrong, ignore it

[TimeSpaceLightForce]

17 hours ago, Buddyboy3000 said:

I finally found a way to make the road system less than 7.5 miles.

  Reveal hidden contents

 

The idea I got was an edit of the idea that tojo928 gave. Instead of doing X's across four towns, it will be shorter to do something similar to the shape of two trapezoids. The left and right side of the X's are pushed inward to make a 0.5 mile road in the middle. Instead of 2.828 miles, this idea will be about 2.736 miles. Use two of these to replace the X's to make it shorter. The improved road system will be about 7.472 miles long.

 

The construction contract is yours..nice job.

[DejMar]

Euler? Euler? Euler?
 

Spoiler

It appears Euler's constant comes into play for the shortest built roadway system.
TimeSpaceLight's improvement on tojo982's idea uses a point 0.25 miles (if not an approximate) east and 0.5 miles
north of town-1 for a NNE road of length of approxiately 0.559017 miles before it branches East for 0.5 miles, eventually forming the 7.472 mile long roadway system.
Yet, if the point is e/2 miles east and 0.5 miles north of town-1, the first dogleg is a length of approximately
0.577317 miles with that first East road being only approximately 0.422784 miles. Following this design, a roadway system of
only approximately 7.464102 miles can be built, with a savings of approximately 0.0079 miles more of pavement than TimeSpaceLight's solution.

 

[plainglazed]

and a Spoiler for some additional thoughts:

 

I think 1/sqrt3 is more likely than e.  Bb3k's described improvement to tojo982's offering: "The left and right side of the X's are pushed inward to make a 0.5 mile road in the middle."  If we vary the .5 mile connection, we can generalize the length of this path: 

4sqrt(.5² + (.5tan(a))²) + 1-tan(a)

where (a) is the angle formed by the y-axis and the X in Bb3k's drawing above.  This expression minimizes when (a) is 30 degrees.

EDIT:  although now I'm not so sure.  don't know if the rounding in Excel or my calculator is more accurate and the kind of calculus needed here i've long since forgotten.  really quite amazing that the two are so close. in any event, Bb3k's answer I thought was quite clever and satisfied the OP. 

 

Edited May 28, 2016 by plainglazed

[TimeSpaceLightForce]

Spoiler

The cad solution I made give the  values calculated by last 2 posts. No need for math w/ CAD. Just draw the lines ,click property , to get precise lengths , angles, etc. Best for geometric puzzles. While Bb3k's drawing got the reduced OP requirement by using Desmos Calculator.