# Contstructing Segments

• Dec 9th 2009, 06:05 AM
ReneePatt
Contstructing Segments
Given three segments whose lengths are 1, a, and b, construct segments of length a + b, |ab|, ab, $\frac{a}{b}$, and $\sqrt{ab}$. These are Euclidean constructions.
• Dec 9th 2009, 06:53 AM
Soroban
Hello, ReneePatt!

Surely, you can do the first two . . .

Quote:

Given three segments whose lengths are: $1,\;a,\text{ and }b,$
construct segments of lengths:

$(1)\;a + b \qquad (2)\;|a - b| \qquad (3)\;ab \qquad (4)\;\frac{a}{b} \qquad (5)\;\sqrt{ab}$

$(3)\;ab$

On a horizontal line, measure off: . $OP = 1,\;PQ = a$
Code:

       o - - - o - - - o - - -       O  1  P  a  Q

Through $O$ draw a diagonal line.
On the diagonal, measure off $OR = b$
Draw $PR.$
Code:

                              *                           *                         *                   R  *                   o           b  * /             *  /         *    /       o - - - o - - - o - - -       O  1  P  a  Q

Through $Q$ construct a line parallel to $PR$, cutting the diagonal at $S.$
Code:

                              S                               o                       x  * /                         *  /                   R  *    /                   o      /           b  * /      /             *  /      /         *    /      /       o - - - o - - - o - - -       O  1  P  a  Q

Then: . $x \,=\,RS \,=\,ab$

Proof
From similar triangles: . $\frac{b+x}{1+a} \:=\:\frac{b}{1}$

And we have: . $b + x \:=\:b + ab \quad\Rightarrow\quad x \:=\:ab$

• Dec 9th 2009, 08:51 AM
earboth
Quote:

Originally Posted by ReneePatt
Given three segments whose lengths are 1, a, and b, construct segments of length ... $\sqrt{ab}$. These are Euclidean constructions.

1. Draw a line with the length (a + b)

2. This line is the diameter of a circle.

3. Construct a perpendicular line at the end of a = begin of b. This line intersect the circle line. Construct a right triangle with (a + b) as hypotenuse .

4. According to Euclid's theorem you have in a right triangle:

$a \cdot b = h^2~\implies~\boxed{h = \sqrt{a \cdot b}}$
• Dec 9th 2009, 11:11 AM
Soroban
Hello, Renee!

Quote:

$(4)\;\frac{a}{b}$

On a horizontal line, measure off: . $OP = b,\;PQ = 1$
Code:

       o - - - o - - - o - - -       O  b  P  1  Q

Through $O$ draw a diagonal line.
On the diagonal, measure off $OR = a$
Draw $PR.$
Code:

                              *                           *                         *                   R  *                   o           a  * /             *  /         *    /       o - - - o - - - o - - -       O  b  P  1  Q

Through $Q$ construct a line parallel to $PR$, cutting the diagonal at $S.$
Code:

                              S                               o                       x  * /                         *  /                   R  *    /                   o      /           a  * /      /             *  /      /         *    /      /       o - - - o - - - o - - -       O  b  P  1  Q

Then: . $x \,=\,RS \,=\,\frac{a}{b}$

Proof

From similar triangles: . $\frac{a+x}{b+1} \:=\:\frac{a}{b}$

And we have: . $ab+bx \:=\:ab +a \quad\Rightarrow\quad bx \:=\:a \quad\Rightarrow\quad x \:=\:\frac{a}{b}$