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Topic: BMAD Polygons Area=Perimeter (Read 939 times) |
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BMAD
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BMAD Polygons Area=Perimeter
« on: May 22nd, 2014, 3:27pm » |
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Let's define BMAD polygons as polygons where the perimeter and area values are identical. Does there exist two such polygons (p1, p2) where the number of sides of p1 > the number of sides of p2 but the perimeters are identical?
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dudiobugtron
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Re: BMAD Polygons Area=Perimeter
« Reply #1 on: May 22nd, 2014, 6:46pm » |
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For a given perimeter p, you can construct a corresponding BMAD rectangle by solving:
x * (p/2 - x) = p
where x is the length of one of the sides.
Rearranging gives x^2 - (p/2)*x + p = 0
This has a (positive real) solution as long as (-p/2)^2 - 4*1*p > 0 , ie: p > 16
It also has one at p= 16, which is the 4x4 square.
And here's my start on the triangle:
You can construct an equilateral BMAD triangle by using:
(x^2 - (x/2)^2)*x/2 = 3x
3/8 * x^3 = 3x
x^3 = 8x
Which has positive real solution x = sqrt(
meaning p = 6 * sqrt(2)
I conjecture that for each possible number of sides n, there is a minimum (regular) BMAD polygon with perimeter pn, and for any perimeter p > pn you can find a corresponding BMAD n-gon.
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pex
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Re: BMAD Polygons Area=Perimeter
« Reply #2 on: May 23rd, 2014, 2:06am » |
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To give a specific example using dudiobugtron's work: the triangle with sides (8, 20/3, 20/3) and the rectangle with sides 8 and 8/3 both have area and perimeter 64/3.
More on triangles:| hidden: | Create an isosceles triangle by gluing together two right triangles with sides (a,b,c), with the length-b sides together. This isosceles triangle has
area ab and perimeter 2(a + c) = 2(a + sqrt(a2 + b2)).
Setting these equal yields
b = 4a2 / (a2 - 4) and hence, c = a (a2 + 4) / (a2 - 4),
with area and perimeter 4a3 / (a2 - 4).
The example I give above is for a = 4, where the other variables also work out to reasonably nice numbers. |
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BMAD
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Re: BMAD Polygons Area=Perimeter
« Reply #3 on: May 23rd, 2014, 6:01am » |
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Can a BMAD polygon ever be regular?
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towr
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Re: BMAD Polygons Area=Perimeter
« Reply #4 on: May 23rd, 2014, 6:43am » |
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For any polygon you can make the area and perimeter equal just by scaling. So regular polygons can just as easily have equal perimeter and area as any other.
But I don't think you can get two different regular polygons that share the same area and perimeter.
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Grimbal
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Re: BMAD Polygons Area=Perimeter
« Reply #5 on: May 23rd, 2014, 9:15am » |
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To make the perimeter equal the area for a regular polygon, you just need to build it around an inscribed circle of radius 2.
Any polygon that has an inscribed circle with radius 2 satisfies the equation.
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