wu :: forums - Print Page


    
      
        wu :: forums
        (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi)
      

        riddles >> medium >> Dividing a Ranch
        
(Message started by: ThudanBlunder on Feb 17^th, 2009, 2:29am)

Title: Dividing a Ranch
Post by ThudanBlunder on Feb 17^th, 2009, 2:29am

A rancher left a rather unusual will. Most of his estate went to his widow, but he stipulated that each of his four sons could take an exactly square plot of land if the dimensions fulfilled the following conditions. The area of Jeb's plot expressed in square miles must equal the length of Harry’s plot in miles, plus two. Similarly, Tony’s area must equal a side of Jeb’s area, plus two. Also, Harry’s area added to the length of Bob’s plot must equal two, and likewise Bob’s area plus the length of Tom’s plot must equal two.

What was the sons’ settlement?

Title: Re: Dividing a Ranch
Post by Grimbal on Feb 17^th, 2009, 2:38am

Easy, just solve the polynomial of degree 16. ::)

Title: Re: Dividing a Ranch
Post by towr on Feb 17^th, 2009, 3:25am

If all lengths have to be integer, then I don't see how there's a solution.

"Harry’s area added to the length of Bob’s plot must equal two"
h^2+b=2 -> h=b=1

"The area of Jeb's plot expressed in square miles must equal the length of Harry’s plot in miles, plus two"
j^2=h+2
j^2=3 -> !http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/exists.gifj

Title: Re: Dividing a Ranch
Post by Grimbal on Feb 17^th, 2009, 5:05am

Well, I solved it numerically, then I googled the numerical value and found that
j = [hide] 2·cos(2pi/15) = 1.827090915 [/hide]

Title: Re: Dividing a Ranch
Post by pex on Feb 17^th, 2009, 5:58am

on 02/17/09 at 05:05:28, Grimbal wrote:

Well, I solved it numerically, then I googled the numerical value and found that
j = [hide] 2·cos(2pi/15) = 1.827090915 [/hide]

Interesting! That implies that
t = [hide]2*cos(1*pi/15)[/hide]
j = [hide]2*cos(2*pi/15)[/hide]
h = [hide]2*cos(4*pi/15)[/hide]
b = [hide]2*cos(7*pi/15)[/hide]

Title: Re: Dividing a Ranch
Post by pex on Feb 17^th, 2009, 7:13am

This solution actually suggests a solution method. ::)
[hideb]The constraints can be summarized as follows:
h = j² - 2
b = 2 - h²
t = 2 - b²
j = t² - 2

If we make the substitution j = 2*cos(a), we find h = 4*cos²(a) - 2 = 2*[2*cos²(a) - 1] = 2*cos(2a). Continuing in this manner,
j = 2*cos(a)
h = 2*cos(2a)
b = -2*cos(4a)
t = -2*cos(8a)
j = 2*cos(16a)

Thus, we are looking for a solution to cos(a) = cos(16a). Restricting to 0 < a < pi (which is without loss of generality), the only interval in which j, h, b, and t are all positive is pi/8 < a < 3*pi/16.

The equation cos(a) = cos(16a) has only one solution in this interval. How to find it? Note that it is obvious that a = 2*pi/15 solves the equation, because then 16a = 15a + a = 2*pi + a. We happen to be lucky: pi/8 < 2*pi/15 < 3*pi/16.

Thus, the only solution is as follows:
j = 2*cos(2*pi/15) ~ 1.8271
h = 2*cos(4*pi/15) ~ 1.3383
b = -2*cos(8*pi/15) = 2*cos(7*pi/15) ~ 0.2091
t = -2*cos(16*pi/15) = 2*cos(pi/15) ~ 1.9563[/hideb]

Title: Re: Dividing a Ranch
Post by ThudanBlunder on Feb 18^th, 2009, 4:39am

That's the solution I had in mind, pex. ;)

on 02/17/09 at 05:05:28, Grimbal wrote:

Well, I solved it numerically, then I googled the numerical value and found that
j = [hide] 2·cos(2pi/15) = 1.827090915 [/hide]

One can only marvel in awe at the many serpentine roads to mathematical truth. ::)