What is your question?

- Jul 18th 2008, 11:08 PMfoomanchyu.
What is your question?

- Jul 19th 2008, 01:58 AMflyingsquirrel
Hello

We can do it using the intercept theorem. In the rectangle triangle $\displaystyle ABD$ (see attachment) $\displaystyle \frac{AB}{AG}=\frac{DB}{GF}$

Say $\displaystyle B$ is at 1, $\displaystyle A$ is at 0 ($\displaystyle \implies AB=1$) and we want $\displaystyle G$ to be at $\displaystyle \frac{1}{x}$. (hence we want $\displaystyle AG$ to be $\displaystyle \frac{1}{x}$) Using the intercept theorem we get $\displaystyle AG=\underbrace{AB}_1\times \frac{GF}{DB}=\frac{GF}{DB}$. If $\displaystyle FG=1$ and $\displaystyle DB=x$ then $\displaystyle AF=\frac{1}{x}$ !

To construct $\displaystyle G$, one can proceed as follows :

- Draw segment line $\displaystyle [AB]$ with $\displaystyle A(0,0),\,B(1,0)$.
- Draw segment line $\displaystyle [BD]$ perpendicular to the x-axis and such that $\displaystyle BD=AC=x$
- Draw segment line $\displaystyle [AD]$.
- Draw straight line $\displaystyle (c)$ perpendicular to the y-axis which passes through the point $\displaystyle (0,1)$. Let $\displaystyle F$ be the intersection of this straight line with $\displaystyle [AD]$.
- Draw straight line $\displaystyle (e)$ perpendicular to $\displaystyle (c)$ and passing through F.
- G is the intersection of $\displaystyle (e)$ with the x-axis.

- Jul 19th 2008, 01:58 AMticbol
All these questions about constructions of geometric figures...

Is there a subject, or branch or part of Geometry that deals about these kind of constructions? It was not taught to us in my school days. Is it something new?

Anyway, as is my way of solving things not taught to us in school, I will use "common sense", or is it intuitive reasoning, here.

From (0,0), or from a, as the center, strike a quarter of a circle whose radius is x units long.

Then draw a line from (1,0) or b to the intersection of the circular curve with the vertical axis at (0,0) or a.

Then subdivide that line into x number of equal parts.

Then from the 1st subdivision, draw a vertical line up or down to the horizontal axis at (0,0) or a.

The horizontal distance from (0,0) or a to the foot of that vertical line is 1/x. - Jul 19th 2008, 02:53 AMMoo
Hello ticbol and flyingsquirrel,

Actually, this guy edited his post because we asked him to post it instead of the pdf... But the true question is that you can only use a compass and draw circles...

http://www.misterhoffman.com/downloa...roblemSets.pdf

#7-4 - Jul 20th 2008, 03:53 AMflyingsquirrel
Let's do it using *only* circles :

(See the first picture) Let $\displaystyle A(0,0)\,B(1,0)$ and $\displaystyle C(y,0)$ for $\displaystyle y>1$. Construct the two following circles : $\displaystyle \begin{cases} (c)\,:\,\text{center = } \, C(y,0),\text{ radius = }y\\

(d)\,:\,\text{center = }\,A(0,0),\text{ radius = }1\\

\end{cases}$ Let $\displaystyle G$ be the intersection point of $\displaystyle (d)$ with the x-axis such that $\displaystyle G\neq A$. Let $\displaystyle D$ be an intersection point of the two circles. (it exists since $\displaystyle AG>\frac12$) As $\displaystyle AG$ is a diameter of $\displaystyle (c)$, the triangle $\displaystyle ACD$ is rectangle. Now let $\displaystyle H$ be the orthogonal projection of $\displaystyle D$ on the x-axis. $\displaystyle AHD$ and $\displaystyle ACD$ are both rectangle triangles. Using the Pythagorean theorem for the three triangles, we get

$\displaystyle

\begin{cases}

AH^2+HD^2=AD^2\\

HG^2+HD^2=DG^2\\

AD^2+DG^2=AG^2\\

AH+HG=AG

\end{cases}$

which gives us $\displaystyle \boxed{AH=\frac{1}{AG}}$. (see (*)) As we want to find $\displaystyle \frac{1}{x}$ one can think that letting $\displaystyle AG=x$ is a good idea but it requires us to construct a point $\displaystyle C$ such that $\displaystyle AC=\frac{x}{2}$ which is not easy using only a compass. (at least, I don't know how to do it :D) A better idea is to let $\displaystyle AC=2x$. It 'll give us $\displaystyle AH=\frac{1}{2x}$ from which one can easily get $\displaystyle \frac{1}{x}$. Unfortunately, there is another problem : how can we construct $\displaystyle H$ using only a compass ? I don't know... Here is a solution I found to have not to construct this point :

(see the second picture) Let $\displaystyle A(0,0),\,B(1,0)$ and $\displaystyle C(x,0)$ for $\displaystyle x>1$. Construct the two circles $\displaystyle (c)$ and $\displaystyle (d)$ as before : $\displaystyle \begin{cases} (c)\,:\,\text{center = }C(x,0),\text{ radius = }x\\

(d)\,:\,\text{center = }A(0,0),\text{ radius = }1\\

\end{cases}$ (hence $\displaystyle AG=2x$ and $\displaystyle AH=\frac{1}{2x}$) and construct the circle $\displaystyle (e)$ of radius 1 and which center is $\displaystyle G$. Let $\displaystyle I$ be the intersection point of $\displaystyle (e)$ with $\displaystyle (c)$ which is in the same quadrant as $\displaystyle D$. Let $\displaystyle K$ be the orthogonal projection of $\displaystyle I$ on the x-axis then, using the three rectangle triangles $\displaystyle GKI,\,AGI,\,GKA$, one can show that $\displaystyle KG=\frac{1}{AG}=\frac{1}{2x}$. (same proof as before : see (*)) This is interesting because it gives us $\displaystyle DI=HK=AG-AH-AK=2x-\frac{1}{2x}-\frac{1}{2x}=2x-\frac{1}{x}$ : we only need to construct $\displaystyle D$ and $\displaystyle I$ to find $\displaystyle \frac{1}{x}$.

(see picture #3) Solution of the problem : Given $\displaystyle A(0,0),\,B(1,0)$ and $\displaystyle C(x,0)$ for $\displaystyle x>1$ construct the three circles $\displaystyle \begin{cases} (c)\,:\,\text{center = }C(x,0),\text{ radius = }x\\

(d)\,:\,\text{center = }A(0,0),\text{ radius = }1\\

(e)\,:\,\text{center = }G(2x,0),\text{ radius = }1

\end{cases}$ Let $\displaystyle D$ and $\displaystyle I$ be the intersection points of $\displaystyle (d)$ with $\displaystyle (c)$ and of $\displaystyle (e)$ with $\displaystyle (c)$, respectively. (choose these two points so that they lie in the same quadrant) Then $\displaystyle DI=2x-\frac{1}{x}\implies \boxed{\frac{1}{x}=AG-DI}$ which can be easily constructed using *only* a compass.

Don't know if it's the "best" solution but that was interesting anyway. (Nod)

___________________________________________

(*) Let's show that $\displaystyle AH=\frac{1}{AG}$ :

$\displaystyle \begin{cases}

AH^2+HD^2=AD^2\\

HG^2+HD^2=DG^2\\

AD^2+DG^2=AG^2\\

AH+HG=AG

\end{cases}\underset{AD=1}{\implies} \begin{cases}

HD^2=1-AH^2 & (1)\\

HG^2+HD^2=DG^2 & (2)\\

DG^2=AG^2-1 & (3)\\

HG=AG-AH & (4)\\

\end{cases}

$

$\displaystyle (1)$ and $\displaystyle (2)$ give $\displaystyle HG^2+1-AH^2=DG^2$. Substituting $\displaystyle HG=AG-AH$ and $\displaystyle DG=AG^2-1$ in this equation :

$\displaystyle

(AG-AH)^2+1-AH^2 =AG^2-1 $

$\displaystyle

AG^2-2\cdot AG\cdot AH+AH^2+1-AH^2 =AG^2-1$

$\displaystyle

-2\cdot AG\cdot AH+1=-1$

$\displaystyle \boxed{AH=\frac{1}{AG}}

$