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        riddles >> medium >> BMAD Polygons Area=Perimeter
        
(Message started by: BMAD on May 22^nd, 2014, 3:27pm)

Title: BMAD Polygons Area=Perimeter
Post by BMAD on May 22^nd, 2014, 3:27pm

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?

Title: Re: BMAD Polygons Area=Perimeter
Post by dudiobugtron on May 22^nd, 2014, 6:46pm

For a given perimeter p, you can construct a corresponding BMAD rectangle by solving:
[hide] 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.[/hide]

And here's my start on the triangle:
[hide]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(8)
meaning p = 6 * sqrt(2)[/hide]

I conjecture that for each possible number of sides n, there is a minimum (regular) BMAD polygon with perimeter p_n, and for any perimeter p > p_n you can find a corresponding BMAD n-gon.

Title: Re: BMAD Polygons Area=Perimeter
Post by pex on May 23^rd, 2014, 2:06am

To give a specific example using dudiobugtron's work: the triangle with sides [hide](8, 20/3, 20/3)[/hide] and the rectangle with sides [hide]8 and 8/3[/hide] both have area and perimeter [hide]64/3[/hide].

More on triangles:[hideb]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.[/hideb]

Title: Re: BMAD Polygons Area=Perimeter
Post by BMAD on May 23^rd, 2014, 6:01am

Can a BMAD polygon ever be regular?

Title: Re: BMAD Polygons Area=Perimeter
Post by towr on May 23^rd, 2014, 6:43am

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.

Title: Re: BMAD Polygons Area=Perimeter
Post by Grimbal on May 23^rd, 2014, 9:15am

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.