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Path Less Than 1 + sqrt(3)

Date: 03/18/2003 at 21:17:40
From: Dan Schellenberg
Subject: Geometry Question

Is there a way to connect the four vertices of a square (of side 
length 1) such that the path travelled is less than 1 + sqrt(3)?

My first hunch was that, by the triangle inequality, the shortest way 
between any two of the vertices of the square ABCD is a straight line 
connecting them. However, if I use two such lines, connecting AC and 
BD which intersect at their midpoints, the total combined distance is 
still over the allotted amount of 1 + sqrt(3). I then tried connecting 
them by using a straight line along AB, and then two straight lines 
from the midpoint of AB to C and D respectively. This is much too 
large. I'm running out of ideas, and any thoughts or advice would be 
really appreciated.

Date: 03/18/2003 at 23:55:36
From: Doctor Douglas
Subject: Re: Geometry Question

Hi, Dan

Thanks for submitting your question to the Math Forum.

I don't think it is possible to get a total path length LESS than
1+sqrt(3). However, it is possible to obtain 1+sqrt(3), exactly.

Here's a hint: consider the center point when the two diagonals AC and 
BD intersect (the center of the square). You can "split" it into two 
"knots" and separate them so that each diagonal doesn't have to come 
all the way into the center:

   A       B
    \     /
     *---*            here the knots are asterisks
    /     \
   D       C

See if you can use this hint to solve this problem.

If you're still stuck, you can either write back, or search the 
web for the term "Steiner tree."

- Doctor Douglas, The Math Forum 

Date: 03/19/2003 at 16:37:56
From: Dan Schellenberg
Subject: Geometry Question

Thanks for the idea. I am now able to get much closer to the required 
amount of 1+sqrt(3), but I still can't figure out how to make it 
EXACTLY that amount. Setting the length of the line segment between 
the two "knots" to 1/2, I am able to get a total path distance of 
4(sqrt(5/16))+(1/2). This is slightly larger (by  about .004) than 
sqrt(3)+1. Am I missing something?  

I tried a couple of other values for the length of the line segments 
between the knots, but can't seem to figure out the way to minimize 
the path. Any further help would be appreciated.

Date: 03/19/2003 at 23:29:26
From: Doctor Douglas
Subject: Re: Geometry Question

Hi again, Dan,

It is possible to prove that the strings that connect to the knot must 
come together with angles of 120 degrees between them.

Can you use this to determine how far the knots should be?

- Doctor Douglas, The Math Forum 

Date: 03/21/2003 at 13:10:50
From: Dan Schellenberg
Subject: Thank you (Geometry Question)

Well, after much brain stress trying to remember my trig functions :)  
I was able to solve the problem. 

Using sin(pi/3)=(1/2)(1/x) to find the length of the "connecting line" 
to be 1/sqrt(3), and then using the Pythagorean theorem to evaluate 
the "missing side" to be sqrt(1/12), thus the length of the line 
connecting the two knots is 1-2sqrt(1/12). Then to evaluate the total 
path distance, can add (1-2sqrt(1/12) + 4(1/sqrt(3)) and evaluate this 
to obtain a total path of 1+sqrt(3).

Thanks so much for your help! 
Associated Topics:
College Euclidean Geometry
High School Euclidean/Plane Geometry

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