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Math Help - Construct the point on this number line which represents 1/x ..? [pdf]

  1. #1
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    What is your question?
    Last edited by foomanchyu; February 1st 2009 at 09:22 PM.
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hello
    Quote Originally Posted by foomanchyu View Post
    there is a line with points a b and c

    point a is at 0, b at 1, and c at x, some number greater than 1, how do you construct the point equivalent to 1/x on this line?
    We can do it using the intercept theorem. In the rectangle triangle ABD (see attachment) \frac{AB}{AG}=\frac{DB}{GF}

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

    To construct G, one can proceed as follows :
    • Draw segment line [AB] with A(0,0),\,B(1,0).
    • Draw segment line [BD] perpendicular to the x-axis and such that BD=AC=x
    • Draw segment line [AD].
    • Draw straight line (c) perpendicular to the y-axis which passes through the point (0,1). Let F be the intersection of this straight line with [AD].
    • Draw straight line (e) perpendicular to (c) and passing through F.
    • G is the intersection of (e) with the x-axis.
    Attached Thumbnails Attached Thumbnails Construct the point on this number line which represents 1/x ..? [pdf]-thales2.png  
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  3. #3
    MHF Contributor
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    Quote Originally Posted by foomanchyu View Post
    there is a line with points a b and c

    point a is at 0, b at 1, and c at x, some number greater than 1, how do you construct the point equivalent to 1/x on this line?
    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.
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  4. #4
    Moo
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    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
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  5. #5
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by Moo View Post
    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...
    Let's do it using *only* circles :

    Construct the point on this number line which represents 1/x ..? [pdf]-inverse1.png

    (See the first picture) Let A(0,0)\,B(1,0) and C(y,0) for y>1. Construct the two following circles : \begin{cases} (c)\,:\,\text{center = } \, C(y,0),\text{ radius = }y\\<br />
(d)\,:\,\text{center = }\,A(0,0),\text{ radius = }1\\<br />
\end{cases} Let G be the intersection point of (d) with the x-axis such that G\neq A. Let D be an intersection point of the two circles. (it exists since AG>\frac12) As AG is a diameter of (c), the triangle ACD is rectangle. Now let H be the orthogonal projection of D on the x-axis. AHD and ACD are both rectangle triangles. Using the Pythagorean theorem for the three triangles, we get

    <br />
\begin{cases}<br />
AH^2+HD^2=AD^2\\<br />
HG^2+HD^2=DG^2\\<br />
AD^2+DG^2=AG^2\\<br />
AH+HG=AG<br />
\end{cases}

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

    Construct the point on this number line which represents 1/x ..? [pdf]-inverse2.png

    (see the second picture) Let A(0,0),\,B(1,0) and C(x,0) for x>1. Construct the two circles (c) and (d) as before : \begin{cases} (c)\,:\,\text{center = }C(x,0),\text{ radius = }x\\<br />
(d)\,:\,\text{center = }A(0,0),\text{ radius = }1\\<br />
\end{cases} (hence AG=2x and AH=\frac{1}{2x}) and construct the circle (e) of radius 1 and which center is G. Let I be the intersection point of (e) with (c) which is in the same quadrant as D. Let K be the orthogonal projection of I on the x-axis then, using the three rectangle triangles GKI,\,AGI,\,GKA, one can show that KG=\frac{1}{AG}=\frac{1}{2x}. (same proof as before : see (*)) This is interesting because it gives us DI=HK=AG-AH-AK=2x-\frac{1}{2x}-\frac{1}{2x}=2x-\frac{1}{x} : we only need to construct D and I to find \frac{1}{x}.

    Construct the point on this number line which represents 1/x ..? [pdf]-inverse3.png

    (see picture #3) Solution of the problem : Given A(0,0),\,B(1,0) and C(x,0) for x>1 construct the three circles \begin{cases} (c)\,:\,\text{center = }C(x,0),\text{ radius = }x\\<br />
(d)\,:\,\text{center = }A(0,0),\text{ radius = }1\\<br />
(e)\,:\,\text{center = }G(2x,0),\text{ radius = }1<br />
\end{cases} Let D and I be the intersection points of (d) with (c) and of (e) with (c), respectively. (choose these two points so that they lie in the same quadrant) Then 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.


    ___________________________________________

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

    \begin{cases}<br />
AH^2+HD^2=AD^2\\<br />
HG^2+HD^2=DG^2\\<br />
AD^2+DG^2=AG^2\\<br />
AH+HG=AG<br />
\end{cases}\underset{AD=1}{\implies} \begin{cases}<br />
HD^2=1-AH^2 & (1)\\<br />
HG^2+HD^2=DG^2 & (2)\\<br />
DG^2=AG^2-1 & (3)\\<br />
HG=AG-AH & (4)\\<br />
\end{cases}<br />

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

    <br />
(AG-AH)^2+1-AH^2  =AG^2-1

    <br />
 AG^2-2\cdot AG\cdot AH+AH^2+1-AH^2 =AG^2-1

    <br />
  -2\cdot AG\cdot AH+1=-1

    \boxed{AH=\frac{1}{AG}}<br />
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