Jump to content
BrainDen.com - Brain Teasers
  • 0

Polygons


BMAD
 Share

Question

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
Link to comment
Share on other sites

9 answers to this question

Recommended Posts

  • 0

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.

Link to comment
Share on other sites

  • 0

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.
Link to comment
Share on other sites

  • 0

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

Link to comment
Share on other sites

  • 0

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 by BMAD
Link to comment
Share on other sites

  • 0

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

Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Answer this question...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

Loading...
 Share

  • Recently Browsing   0 members

    • No registered users viewing this page.
×
×
  • Create New...