Determining Equable ShapesDate: 06/15/99 at 08:11:46 From: Jonathan Subject: Equable shapes Could you please list some of the equable shapes and show me how you determined them? Can you please write down the formula? Date: 06/15/99 at 08:18:12 From: Doctor Anthony Subject: Re: Equable shapes Consider figures that have perimeters and areas NUMERICALLY equal. I emphasize the words 'numerically equal' because one is a length and the other is an area so they cannot be 'equal' in the normal sense of the word. A general formula relating the number of sides to the length of side of an equable shape (equal area and perimeter)? The formula given below is for any regular polygon of 'n' sides, each side being of length 'x': x = 4 / [tan(90 - 180/n)] For example, Square: ------ x = 4 / [tan(90 - 180/4)] = 4 / tan(45) = 4 Area = x*x = 16 Perimeter = 4*x = 16 Hexagon: ------- x = 4 / [tan(90 - 180/6)] = 4 / tan(60) = 4/sqrt(3) so area = perimeter = 24/sqrt(3) The formula works for any regular polygon. For the case of a regular concave polygon - like a star shape - the Koch snowflake is an interesting example. The Koch Snowflake ------------------ To draw this you start with an equilateral triangle of side a. Now divide each side into three equal parts and on the middle third of each side construct an equilateral triangles pointing outwards from the original triangle. The total perimeter is now (4/3)(3a) = 4a. We now further subdivide each straight edge into 3 parts and construct equilateral triangles on the middle third of each side - again pointing outwards from the original figure. This process will enlarge the perimeter by a further factor of 4/3. There is no overlapping of the extra sides with those already present. The above process is repeated n times, at each stage the perimeter being increased by a factor of 4/3. So we have: perimeter = (3a)(4/3)(4/3)(4/3) ... to n terms. Clearly the perimeter will increase in this process, but the area of the figure will be less than the area of the circumcircle of the original equilateral triangle. So the perimeter and area will converge in numerical value. If A = area of original triangle then: The area of the snowflake follows the pattern: A[1 + 1/3 + 4/27 + 16/243 + 64/2187 + ...] where after the first term we have a GP with common ratio 4/9. = A + A[1/3 + 4/27 + .....to n terms] = A + A(1/3)[1-(4/9)^n]/(1-4/9) = A + A(1/3)[1-(4/9)^n]/(5/9) = A + (3A/5)[1-(4/9)^n] Now if area and perimeter are to be equal we need solutions to the equation (3a)(4/3)^n = A + (3A/5)[1-(4/9)^n] = A[1+(3/5)(1-(4/9)^n)] where A = (1/2)a^2.sin(60) = a^2.sqrt(3)/4 (3a)(4/3)^n = a^2.sqrt(3)/4[1+(3/5)(1-(4/9)^n)] (4/3)^n = a.sqrt(3)/12[1+(3/5)(1-(4/9)^n)] We can see that for different values of 'a' there will be different values of n. Solving for n with a given value of 'a' would be difficult, so choose a value of n and see what value of 'a' this requires. For example, n = 3 gives 2.3703 = 0.144337a[1+(3/5)(0.912208)] 2.3703 = 0.144337a[1.547325] 10.61314 = a So for any value of n there will be a value of 'a' that makes the perimeter and area numerically equal. This is just an example of the type of figure where perimeter and area can be numerically equal. However there will be any number of concave polygons where a similar relation could be established. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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