Open main menu
Home
Random
Recent changes
Special pages
Community portal
Preferences
About Wikipedia
Disclaimers
Incubator escapee wiki
Search
User menu
Talk
Dark mode
Contributions
Create account
Log in
Editing
Regular polygon
(section)
Warning:
You are not logged in. Your IP address will be publicly visible if you make any edits. If you
log in
or
create an account
, your edits will be attributed to your username, along with other benefits.
Anti-spam check. Do
not
fill this in!
===Circumradius=== [[Image:PolygonParameters.png|class=skin-invert-image|thumb|left|180px|Regular [[pentagon]] (''n'' = 5) with [[edge (geometry)|side]] ''s'', [[circumradius]] ''R'' and [[apothem]] ''a'']] <div class="skin-invert-image">{{regular polygon side count graph.svg}}</div> The [[circumradius]] ''R'' from the center of a regular polygon to one of the vertices is related to the side length ''s'' or to the [[apothem]] ''a'' by :<math>R = \frac{s}{2 \sin\left(\frac{\pi}{n}\right)} = \frac{a}{\cos\left(\frac{\pi}{n} \right)} \quad_,\quad a = \frac{s}{2 \tan\left(\frac{\pi}{n}\right)}</math> For [[constructible polygon]]s, [[algebraic expression]]s for these relationships exist {{xref|(see {{slink|Bicentric polygon|Regular polygons}})}}. The sum of the perpendiculars from a regular ''n''-gon's vertices to any line tangent to the circumcircle equals ''n'' times the circumradius.<ref name=Johnson/>{{rp|p. 73}} The sum of the squared distances from the vertices of a regular ''n''-gon to any point on its circumcircle equals 2''nR''<sup>2</sup> where ''R'' is the circumradius.<ref name=Johnson/>{{rp|p. 73}} The sum of the squared distances from the midpoints of the sides of a regular ''n''-gon to any point on the circumcircle is 2''nR''<sup>2</sup> β {{sfrac|1|4}}''ns''<sup>2</sup>, where ''s'' is the side length and ''R'' is the circumradius.<ref name=Johnson/>{{rp|p. 73}} {{Clear|left}} If <math>d_i</math> are the distances from the vertices of a regular <math>n</math>-gon to any point on its circumcircle, then <ref name= Mamuka /> :<math>3\biggl(\sum_{i=1}^n d_i^2\biggr)^2 = 2n \sum_{i=1}^n d_i^4 </math>.
Edit summary
(Briefly describe your changes)
By publishing changes, you agree to the
Terms of Use
, and you irrevocably agree to release your contribution under the
CC BY-SA 4.0 License
and the
GFDL
. You agree that a hyperlink or URL is sufficient attribution under the Creative Commons license.
Cancel
Editing help
(opens in new window)