# [SOLVED] Construction problem(geometric mean)

• Jan 15th 2010, 09:58 AM
Garas
[SOLVED] Construction problem(geometric mean)
Can someone help me with this:
Construct line segment , which is a geometric mean of two given line segments.
• Jan 15th 2010, 10:12 AM
earboth
Quote:

Originally Posted by Garas
Can someone help me with this:
Construct line segment , which is a geometric mean of two given line segments.

According to Euclid's theorem the length of the height in a right triangle is the geometric mean of the 2 hypotenuse segments.

See attachment.
• Jan 15th 2010, 10:17 AM
masters
Quote:

Originally Posted by Garas
Can someone help me with this:
Construct line segment , which is a geometric mean of two given line segments.

Hi Garas,

This is very simple.

First, draw your two line segments end to end horizontally on your paper.

A.......................B............C

Next, construct the perpendicular bisector of AC. Let the perpendicular bisector intersect AC at X.

Next, using X as the center of AC, construct a semicircle through A and C using your compass.

Finally, Construct a perpendicular through B intersecting the semicircle at Y.

The length BY is the geometric mean between AB and BC.
• Jan 15th 2010, 10:48 AM
Soroban
Hello, Garas!

Quote:

Construct the line segment which is a geometric mean of two given line segments.
Code:

          S  * * *           *          *         * |            *       *  | x            *           |       *  |    M        *       *---*-----+---------*       P a Q      b      R

Measure $a=PQ$ and $b=QR$ consecutively on a line.

Bisect the line and locate its midpoint $M$.

Use the $M$ as center and $PM$ as radius,
. . and draw a semicircle.

At $Q$, erect a perpendicular $x$ meeting the semicircle at $S.$

Then $x$ is the geometric mean of $a$ and $b\!:\;\;x \:=\:\sqrt{ab.}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Why would we need such a construction?

Suppose we have an $a\times b$ rectangle
. . and we need a square with the same area.

Code:

                        *---------*       *-----------*    |        |       |          |    |        | x     b |          |    |        |       |          |    |        |       *-----------*    *---------*             a                x

Then: . $x^2 \:=\:ab \quad\Rightarrow\quad x \:=\:\sqrt{ab}$

Edit: too slow (again) . . . *sigh*
.