Posted May 22, 2014 (edited) 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? Edited May 22, 2014 by BMAD 0 Share this post Link to post Share on other sites

0 Posted May 23, 2014 one such polygon exists where perimeter = area, then put an arbitrary point O on any one of its sides AB. introduce a infinitesimally small kink at O, such that AO and OB are almost almost almost colinear but are not. this new polygon has the same area and same perimeter as the old one so it satisfies the original conditions and also has greater number of sides. 0 Share this post Link to post Share on other sites

0 Posted May 23, 2014 Regular? no not necessarily. 0 Share this post Link to post Share on other sites

0 Posted May 23, 2014 one such polygon exists where perimeter = area, then put an arbitrary point O on any one of its sides AB. introduce a infinitesimally small kink at O, such that AO and OB are almost almost almost colinear but are not. this new polygon has the same area and same perimeter as the old one so it satisfies the original conditions and also has greater number of sides. wouldn't the area be infinitesimally off 0 Share this post Link to post Share on other sites

0 Posted May 23, 2014 (edited) For example: BMAD Quadrilateral Perimeter = 4 units Area = 4 units^2 (assuming the above is possible) Could I find a BMAD Triangle where Perimeter = 4 units Area = 4 units^2 assuming the same size unit in both cases Edited May 23, 2014 by BMAD 0 Share this post Link to post Share on other sites

0 Posted May 23, 2014 Let p2 be equilateral triangle with side of length 4*3^0.5. Perimeter and area of p2 are both equal to 12*3^0.5. Let p1 be rhombus with side of length 3*3^0.5. P1 and p2 have equal perimeters. Obviously we can make p1's area anything between 0 (when shorter diagonal of p1 tends to 0) and 27 (when p1 becomes a square) without changing it's perimeter, so we can make it equal to 12*3^0.5 (=~20.785) in particular. 0 Share this post Link to post Share on other sites

0 Posted May 23, 2014 Does this hold for shapes with sides > 3? >4? 0 Share this post Link to post Share on other sites

0 Posted May 23, 2014 one such polygon exists where perimeter = area, then put an arbitrary point O on any one of its sides AB. introduce a infinitesimally small kink at O, such that AO and OB are almost almost almost colinear but are not. this new polygon has the same area and same perimeter as the old one so it satisfies the original conditions and also has greater number of sides. wouldn't the area be infinitesimally off yup, sowwee 0 Share this post Link to post Share on other sites

Posted (edited)

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?

Edited by BMAD## Share this post

## Link to post

## Share on other sites