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NickH
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Farmer's enclosure
« on: Oct 2nd, 2002, 12:22am » |
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A farmer has four straight pieces of fencing: 1, 2, 3, and 4 yards in length. What is the maximum area he can enclose by connecting the pieces?
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| « Last Edit: Oct 23rd, 2003, 7:41pm by Icarus » |
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Pietro K.C.
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Re: NEW PROBLEM: Farmer's enclosure
« Reply #1 on: Oct 2nd, 2002, 7:33am » |
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Well, he could make a quadrilateral of very small area around him (it can approach zero, as 4+1=2+3, so it is limited below only by the farmer's size), and define himself to be on the outside. 
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NickH
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Re: NEW PROBLEM: Farmer's enclosure
« Reply #2 on: Oct 2nd, 2002, 2:43pm » |
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That's an intriguing approach, but, strange to say, it's not the solution I'm looking for!
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James Fingas
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Re: NEW PROBLEM: Farmer's enclosure
« Reply #3 on: Oct 3rd, 2002, 10:12am » |
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Pietro, 
maybe you should look up "enclose" in the dictionary...
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Pietro K.C.
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Re: NEW PROBLEM: Farmer's enclosure
« Reply #4 on: Oct 3rd, 2002, 12:29pm » |
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I'm assuming the 'enclosure' to be situated on a spherical Earth... 
Hint for the 'real' solution: think right angle.
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James Fingas
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Re: NEW PROBLEM: Farmer's enclosure
« Reply #5 on: Oct 3rd, 2002, 1:45pm » |
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Pietro,
I didn't find any right angles in my solution ... my answer:
The area is exactly 1/4*sqrt(9.6)*(sqrt(6.4)+sqrt(14.4)) square yards
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NickH
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Re: NEW PROBLEM: Farmer's enclosure
« Reply #6 on: Oct 3rd, 2002, 2:05pm » |
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James,
That's the solution I got. It simplifies because 14.4 = 1.5^2 * 6.4.
I'd be interested in seeing your method. I originally used Heron's formula on two triangles, which I'd guess was your approach. I've also seen a Cartesian coordinate solution. Then there is a very simple solution which relies upon a result analogous to Heron's formula.
Pietro may possibly be referring to TWO right angles...
Nick
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James Fingas
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Re: NEW PROBLEM: Farmer's enclosure
« Reply #7 on: Oct 4th, 2002, 5:39am » |
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I am quite sure that there are no right angled corners in the largest-size enclosure. I have to say, my solution wasn't particularly pretty ... I parameterized the enclosures using a diagonal, and, as you say, used Heron's formula for the area. Direct, but not pretty. I thought of this simplification, as well:
A = 1/20 * sqrt(48)*(sqrt(32) + sqrt(72))
with your simplification too,
A = sqrt(24) (with an answer like this, there's gotta be a simple way to find it ... but darned if I know what it is)
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| « Last Edit: Oct 4th, 2002, 5:39am by James Fingas » |
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Brett Danaher
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Re: NEW PROBLEM: Farmer's enclosure
« Reply #8 on: Oct 4th, 2002, 7:12am » |
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Care to enlighten?? What is Heron's Formula? What is it's significance?
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James Fingas
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Re: NEW PROBLEM: Farmer's enclosure
« Reply #9 on: Oct 4th, 2002, 8:04am » |
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Heron's formula is a mysterious method of determining the area of a triangle, using only the lengths of its sides (you can see why that would be important here!)
A = sqrt( s (s - a) (s - b) (s - c) )
where s is the "semiperimeter", (a+b+c)/2, and a, b, and c are the side lengths.
I have no idea why it works ... but it's out there!
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S. Owen
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Re: NEW PROBLEM: Farmer's enclosure
« Reply #10 on: Oct 4th, 2002, 12:06pm » |
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I learned something today while looking for info related to this problem - "Bretschneider's theorem". Is this what you're talking about Nick? The answer follows, well, pretty much immediately from it. The proof is probably interesting.
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Jeremy
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Re: NEW PROBLEM: Farmer's enclosure
« Reply #11 on: Oct 4th, 2002, 1:08pm » |
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ok, wait wait wait... i punched that equation into ol' trusty
"1/4*sqr(9.6)*(sqr(6.4)+sqr(14.4)"
and i got 4.899, or 5 square yards.
I found an area of 7.032 square yards... umm... and i just put the 4 pieces together and moved them around.
I started with a shape with an area of 0 (1 and 4 segment for one side, and 3 and 2 for the other side) and them rotated the 1 segment around where it touched the 4 segment until i got some shapes that that i felt like calculating the area of. I ended up calculating the area of 3 shapes (2 triangles and a polygon) and figured the one that closest resembled a circle (the polygon) would be the maximum... i wasn't sure if this was the best way to go about solving this, but yeah it seemed to work.
Ok, with the polygon i split it into 2 triangles, and solved for the area of each separately. one was a right triangle with a side of 1 and 4 (unknown hypotenuse, but easily solvable ( root(1^2+4^2) = root(17)) and that has an area of 2 (4*1/2)
The other triangle was some weird ass triangle with a side of 2 and 3, and shared a hypotenuse with the 1 by 4 triangle. So i dropped a line from the vertex that the 2 and 3 segments made, to make 2 right triangles along the root(17) base. so now i had split the base into 2 segments (x and y) such that x+y = root(17) ===> y=root(17)-x
so, 3^2-x^2=h and 2^2-y^2=h
or, 3^2-x^2=2^2-y^2
i substituted y and now had the equation:
3^2-x^2=2^2-(root(17)-x)^2
simplified
2x^2-2root(17)x+8 = 0
plugging that into the quadratic equation gave me x = 2.562, and 1.562 (basically x and y are interchangeable, depending on which 2 and 3 fence segment is touch which 1 and 4 segment.
2.561+1.516 = 4.124 (root(17) = 4.1231)
now i can solve for h root(3^2-2.561^2) = 2.441
2.441 * root(17) / 2 = 5.032
5.032 + 2(the first triangle) = 7.032
umm... right?
there's another polygon of sides 1 and 4 across from each other, and 2 and 3 across from each other, but i haven't solved for that yet... and i need to use the bathroom.
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James Fingas
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Re: NEW PROBLEM: Farmer's enclosure
« Reply #12 on: Oct 4th, 2002, 1:33pm » |
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on Oct 4th, 2002, 1:08pm, Jeremy wrote:
This is where your error is. The -x2 on both sides cancel out, rather than adding. Also, the constant should be 12 (= 17-9+4). This line should read:
-2root(17)x + 12 = 0
This is easier to solve, giving:
x = 6/root(17) = 1.4552
so h = root(22 - x2) = 1.372
the area is therefore 2.828 square yards. Your total area is 4.828 yards, which is pretty close, but it just goes to show that when you gotta go, just go--don't finish your math problem first.
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NickH
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Re: NEW PROBLEM: Farmer's enclosure
« Reply #13 on: Oct 4th, 2002, 1:58pm » |
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Here's the solution I originally found, using Heron's formula.
Consider a non-concave (obviously) quadrilateral with respective sides 1,2,3,4. Connect the 1,2 and 3,4 vertices, giving two triangles, 2,3,x and 1,4,x. Use Heron's formula to represent the area of each triangle in terms of x. (For an a,b,c triangle, with s = semi-perimeter = (a+b+c)/2, Heron's formula gives area^2 = s(s-a)(s-b)(s-c).)
For 2,3,x, area^2 = (25-x^2)(x^2-1)/16.
For 1,4,x, area^2 = (25-x^2)(x^2-9)/16.
Let u = x^2, and A = area of quadriliteral. (u increases monotonically with x.)
Then 4A = sqrt[(25-u)(u-1)] + sqrt[(25-u)(u-9)].
Differentiating, 4 * dA/du = (-2u+26)/sqrt[(25-u)(u-1)] + (-2u+34)/sqrt[(25-u)(u-9)].
dA/du = 0 => (u-13)/sqrt(u-1) = (17-u)/sqrt(u-9).
Squaring => (u-9)(u-13)^2 = (u-1)(17-u)^2.
Therefore u^3 - 35u^2 + 403u - 1521 = u^3 - 35u^2 + 323u - 289, giving u = 15.4.
Hence 4A = sqrt(9.6)[sqrt(14.4) + sqrt(6.4)] = sqrt(9.6)*2.5*sqrt(6.4). (14.4 = 1.5^2 * 6.4.)
Hence 4A = 2.5*6.4*sqrt(3/2) = 16*sqrt(3/2).
Therefore A = 4*sqrt(3/2) = 2*sqrt(6).
Geometrical considerations show this to be a maximum.
Angles can be calculated using the rule of cosines.
Finally, given a quadrilateral with sides 1,2,3,4, we have three non-congruent forms:
1,2,3,4 = 1,4,3,2; 1,2,4,3 = 1,3,4,2; 1,3,2,4 = 1,4,2,3.
However, if we are interested only in areas, the three forms are equivalent, by considering the reflection perpendicular to the longest side of triangles formed by joining opposite vertices. (i.e., we can take each triangle and turn it over!)
Having found this relatively laborious solution, I then stumbled across a truly remarkable formula for the area of an arbitrary quadrilateral. This is known as "Bretschneider's theorem," as S. Owen points out.
For a quadrilateral with sides a, b, c, d, and for which q is half the sum of two opposite angles (it doesn't matter which pair), the area is given by:
A = sqrt[(s-a)(s-b)(s-c)(s-d) - abcd cos(q)^2].
For a cyclic quadrilateral, i.e., a quadrilateral that can be inscribed in a circle, and for which the sum of opposite angles is 180 degrees, cos(q) = 0, thereby maximizing the area.
From this formula, the answer of 2*sqrt(6) drops straight out, as S. Owen observed. Of course, a lot of work is embodied in that formula!
Nick
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| « Last Edit: Oct 4th, 2002, 2:00pm by NickH » |
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Jeremy
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Re: NEW PROBLEM: Farmer's enclosure
« Reply #14 on: Oct 4th, 2002, 3:02pm » |
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ahh... oops...
stupid math
Well how about this, Mr. Farmer can call up his carpenter neighbor, have him cut up the fence pieces into 10 one yard segments, make a decahedron, and he's got an enclosure of 7.69 yrds^2... or just have the carpenter cut them up into splinters, make a circle, and then he's got 7.95 yrds^2... or call up his buddy from a higher dimension and then he's got an enclosure of 8.34 yrds^4.
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Pyotr
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Re: Farmer's enclosure
« Reply #15 on: Jan 24th, 2011, 12:38pm » |
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I imagined that if you extended the problem to a very large number of sides, the maximum area would be contained by a polygon whose vertices described a circle. That makes sense, doesn't it?
Then I went backwards and decided that this polygon must be a cyclic quadrilateral.
I then applied Brahmaguptra's formula (much like Bretschneider's Theorem) (searching Wikipedia for "Quadrilateral"):
Area = sqrt ((s-n)(s-b)(s-c)(s-d)) where
s = semiperimeter
and got 2sqrt6, or about 4.899
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