PlanetMath (more info)
 Math for the people, by the people.
Encyclopedia | Requests | Forums | Docs | Wiki | Random | RSS  
Login
create new user
name:
pass:
forget your password?
Main Menu
sections
Encyclopædia
Papers
Books
Expositions

meta
Requests (211)
Orphanage
Unclass'd (1)
Unproven (472)
Corrections (36)
Classification

talkback
Polls
Forums
Feedback
Bug Reports

downloads
Snapshots
PM Book

information
News
Docs
Wiki
ChangeLog
TODO List
Legalese
About
Owner confidence rating: Very high Entry average rating: Very high
[parent] compass and straightedge construction of regular triangle (Algorithm)

One can construct a regular triangle with sides of a given length $ s$ using compass and straightedge as follows:

  1. Draw a line segment of length $ s$. Label its endpoints $ P$ and $ Q$.

    \begin{pspicture}(-3,-1)(3,1) \rput[a](-2,0.04){.} \psline[linecolor=blue](-2,0)... ...dots(-2,0)(2,0) \rput[a](-2.2,-0.2){$P$} \rput[a](2.2,-0.2){$Q$} \end{pspicture}
  2. Draw an arc of the circle with center $ P$ and radius $ \overline{PQ}$.

    \begin{pspicture}(-3,-1)(3,4) \rput[l](-0.63192,3.75877){.} \psline(-2,0)(2,0) \... ...dots(-2,0)(2,0) \rput[a](-2.2,-0.2){$P$} \rput[a](2.2,-0.2){$Q$} \end{pspicture}
  3. Draw an arc of the circle with center $ Q$ and radius $ \overline{PQ}$ to find a point $ R$ where it intersects the arc from the previous step.

    \begin{pspicture}(-3,-1)(3,4) \psline(-2,0)(2,0) \psarc(-2,0){4}{-10}{70} \psarc... ...[a](-2.2,-0.2){$P$} \rput[a](2.2,-0.2){$Q$} \rput[b](0,3.7){$R$} \end{pspicture}
  4. Draw the regular triangle $ \triangle PQR$.

    \begin{pspicture}(-3,-1)(3,4) \psarc(-2,0){4}{-10}{70} \psarc(2,0){4}{110}{190} ... ...[a](-2.2,-0.2){$P$} \rput[a](2.2,-0.2){$Q$} \rput[b](0,3.7){$R$} \end{pspicture}

This construction is justified by the following:

  • $ \overline{PQ}\cong\overline{PR}$ since they are both radii of the circle from step 2;
  • $ \overline{PQ}\cong\overline{QR}$ since they are both radii of the circle from step 3;
  • Thus, $ \triangle PQR$ is an equilateral triangle;
  • In Euclidean geometry, any equilateral triangle is automatically a regular triangle. Therefore, $ \triangle PQR$ is a regular triangle.

This construction is based off of the one that Euclid provides in The Elements as the first proposition of the first book. Please see this post for more details.

This construction also yields a method for constructing a $ 60^{\circ}$ angle using compass and straightedge.

Note that, with the exception of actually drawing the sides of the triangle, only the compass was used in this construction. Since regular triangles tessellate, repeated use of this construction provides a way to find infinitely many points on a line given two points on a line using just a compass.

If you are interested in seeing the rules for compass and straightedge constructions, click on the link provided.



"compass and straightedge construction of regular triangle" is owned by Wkbj79.
(view preamble)

View style:

Other names:  compass and straightedge construction of equilateral triangle, compass and straightedge construction of equiangular triangle
Also defines:  compass and straightedge construction of $60^{\circ}$ angle

This object's parent.
Log in to rate this entry.
(view current ratings)

Cross-references: compass and straightedge constructions, line, triangle, angle, Euclidean geometry, equilateral triangle, radii, intersects, point, radius, center, circle, arc, endpoints, line segment, straightedge, compass, length, sides, regular triangle
There are 2 references to this entry.

This is version 7 of compass and straightedge construction of regular triangle, born on 2007-06-24, modified 2007-06-25.
Object id is 9670, canonical name is CompassAndStraightedgeConstructionOfRegularTriangle.
Accessed 1935 times total.

Classification:
AMS MSC: 51-00 (Geometry :: General reference works )
 51M15 (Geometry :: Real and complex geometry :: Geometric constructions)
Pending Errata and Addenda
None.
Discussion
Style: Expand: Order:
forum policy

No messages.

Interact
post | correct | update request | add example | add (any)