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Topic: Dividing a Ranch (Read 665 times) |
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ThudnBlunder
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Dividing a Ranch
« on: Feb 17th, 2009, 2:29am » |
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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?
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Grimbal
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Re: Dividing a Ranch
« Reply #1 on: Feb 17th, 2009, 2:38am » |
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Easy, just solve the polynomial of degree 16. 
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towr
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Re: Dividing a Ranch
« Reply #2 on: Feb 17th, 2009, 3:25am » |
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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 -> ! j
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Wikipedia, Google, Mathworld, Integer sequence DB
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Grimbal
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Re: Dividing a Ranch
« Reply #3 on: Feb 17th, 2009, 5:05am » |
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Well, I solved it numerically, then I googled the numerical value and found that
j = 2·cos(2pi/15) = 1.827090915
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| « Last Edit: Feb 17th, 2009, 5:05am by Grimbal » |
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pex
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Re: Dividing a Ranch
« Reply #4 on: Feb 17th, 2009, 5:58am » |
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on Feb 17th, 2009, 5:05am, Grimbal wrote:Well, I solved it numerically, then I googled the numerical value and found that
j = 2·cos(2pi/15) = 1.827090915 |
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Interesting! That implies that
t = 2*cos(1*pi/15)
j = 2*cos(2*pi/15)
h = 2*cos(4*pi/15)
b = 2*cos(7*pi/15)
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pex
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Re: Dividing a Ranch
« Reply #5 on: Feb 17th, 2009, 7:13am » |
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This solution actually suggests a solution method.
| hidden: | 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 |
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ThudnBlunder
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Re: Dividing a Ranch
« Reply #6 on: Feb 18th, 2009, 4:39am » |
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That's the solution I had in mind, pex. 
on Feb 17th, 2009, 5:05am, Grimbal wrote:Well, I solved it numerically, then I googled the numerical value and found that
j = 2·cos(2pi/15) = 1.827090915 |
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One can only marvel in awe at the many serpentine roads to mathematical truth.
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