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
Owner confidence rating: Very high Entry average rating: No information on entry rating
[parent] compass and straightedge construction of geometric mean (Algorithm)

Given line segments of lengths $ a$ and $ b$, one can construct a line segment of length $ \sqrt{ab}$ using compass and straightedge as follows:

  1. Draw a line segment of length $ a$. Label its endpoints $ A$ and $ C$.

    \begin{pspicture}(-3,-2)(3,3) \rput[r](-0.4,0){.} \psline[linecolor=blue](-2.5,0... ...-2.5,0)(-0.5,0) \rput[a](-2.5,-0.3){$A$} \rput[a](-0.8,0.2){$C$} \end{pspicture}
  2. Extend the line segment past $ C$.

    \begin{pspicture}(-3,-2)(3,3) \rput[r](3,0){.} \psline(-2.5,0)(-0.5,0) \psline[l... ...-2.5,0)(-0.5,0) \rput[a](-2.5,-0.3){$A$} \rput[a](-0.8,0.2){$C$} \end{pspicture}
  3. Mark off a line segment of length $ b$ such that one of its endpoints is $ C$. Label its other endpoint as $ B$.

    \begin{pspicture}(-3,-2)(3,3) \rput[r](3,0){.} \psline(-2.5,0)(-0.5,0) \psline{-... ...(-2.5,-0.3){$A$} \rput[a](-0.8,0.2){$C$} \rput[a](2.5,-0.3){$B$} \end{pspicture}
  4. Construct the perpendicular bisector of $ \overline{AB}$ in order to find its midpoint $ M$.

    \begin{pspicture}(-3,-3)(3,3) \rput[r](3,0){.} \rput[a](0,3){.} \rput[b](0,-3){.... ...](-0.8,0.2){$C$} \rput[a](0.3,-0.3){$M$} \rput[a](2.5,-0.3){$B$} \end{pspicture}
  5. Construct a semicircle with center $ M$ and radii $ \overline{AM}$ and $ \overline{BM}$.

    \begin{pspicture}(-3,-3)(3,3) \rput[r](3,0){.} \rput[a](0,3){.} \rput[b](0,-3){.... ...](-0.8,0.2){$C$} \rput[a](0.3,-0.3){$M$} \rput[a](2.5,-0.3){$B$} \end{pspicture}
  6. Erect the perpendicular to $ \overline{AB}$ at $ C$ to find the point $ D$ where it intersects the semicircle. The line segment $ \overline{DC}$ is of the desired length.

    \begin{pspicture}(-3,-3)(3,3) \rput[r](3,0){.} \rput[a](0,3){.} \rput[b](0,-3){.... ...(0.3,-0.3){$M$} \rput[a](2.5,-0.3){$B$} \rput[b](-0.5,2.65){$D$} \end{pspicture}

    This construction is justified because, if $ \overline{AD}$ and $ \overline{BD}$ were drawn, then the two smaller triangles would be similar, yielding

    $\displaystyle \frac{AC}{DC}=\frac{DC}{BC}.$

    Plugging in $ AC=a$ and $ BC=b$ gives that $ DC=\sqrt{ab}$ as desired.

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



"compass and straightedge construction of geometric mean" is owned by Wkbj79.
(view preamble)

View style:

See Also: construction of central proportional


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

Cross-references: compass and straightedge constructions, similar, triangles, intersects, point, erect the perpendicular, radii, center, midpoint, perpendicular bisector, endpoints, straightedge, compass, lengths, line segments
There are 4 references to this entry.

This is version 7 of compass and straightedge construction of geometric mean, born on 2007-06-13, modified 2007-06-24.
Object id is 9581, canonical name is CompassAndStraightedgeConstructionOfGeometricMean.
Accessed 1438 times total.

Classification:
AMS MSC51-00 (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)