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In the image below taken from article title "Trisection using Bisection", here; have made a graph using desmos here for the simplest case of trisection. On the last page (as in the image below), n-sect is dealt with & this post is related to that part.

Sorry if naive but found it interesting, irrespective of no other source, to follow the article.

enter image description here

Please vet my ideas below for the general case of $n$-sect, as shown on the below page:

  1. There are two cases shown, which as per my understanding deal with two separate cases:
    (a) of which the first case deals with $\exists m\in \mathbb{Z}, n = 2^m$, say for $n=4, m=2$. Here, there is bisection done twice, i.e. the interval $AB$ is divided twice in two parts, and the division is done from the left end. This can be done by compass purely, each time.



    (b) For the other case, my elaboration of the page's contents is stated below, for small groups of sentences:

    (i) "In other cases, we start by choosing m such that $2^{m-1} \lt n \lt 2^m$ and use repeated bisection to find a segment $\frac{1}{2^m}$ of the original one. This corresponds to $AD$ in the diagram, so label that point $D$."

    --> Say, $n= 3$ & $2^1\lt n \lt 2^2$, with $m-1=1 => m=2$. So, there are two (as $m=2$) bisections to find a segment $AD = 1/2^2(AB)$.

    (ii) "Copy that segment $AD \,\,\,n$ times starting at $D$ and in the direction of $B$. Label the end of these copies $K$, such that $D$ is between $A$ and $K$ and $AD/DK = 1/n$. Note that in the case of trisection, $K$ and $B$ coincided, but in general this won’t happen. Bisect $DK$ giving $E$. We now have the centers of the two circles ($D$ and $E$) and the construction proceeds as above to obtain $H$. The proof that $DH$ is $1/n$ of $DA$ is also as above."

    --> There would be $n+1$ copies of $AD$ starting from $A$ and continuing towards $B$. For the simplest case of $n=3$ need copy the segment $AD\,\,3$ times to the right, starting from the point $D$. For $n=3$, there will be a total of $1=3=4$ occurrences of $AD$, which coincides with $AD = AB/4$.
    But, taking the case of $n=5, m= 3$. So, $AE = AB/2^3 = AD/2$, with the first bisection finding $C$, second finding $D$, third finding $E$. Now, copying $AE\,\, 5$ times, will lead to $1+5=6$ copies of $AE$ forming the length of $AK$. This also means that the length of $AK \lt $ length of $AB$; and here for $n=5, |AK| = |AB| -2|AE|$.


    (iii) "At this point in the original proof, we moved from $DH$ being $1/n$ of $DA$ to $AH$ being $1/n$ of $AB$. However, this will not be the case in general. To finish the general case, we need to copy segment $DH, 2^m-n$ times in the direction of $B$. This gives a new point $L$, such that $H$ is between $D$ and $L$ and such that $DH/DL = 1/(2^m-n)$. Then, using $DA$ as the unit (so that $DH = 1/n$ and $AB= 2^m$), we have $AL = AD + DL = 1 + (2^m-n)DH = 1 + (2^m-n)/n = 2^m/n$.
    Thus, $AL/AB =(2^m/n)/(2^m) = 1/n$ as desired."

    --> In the lune formed between $c1, c2$ covering from center of $c1$ to end of $c1$, have $n$ intervals of the length $\frac{1}{n.2^m}$. Considering few specific cases, where $n\ne 2^m$ (i.e., general cases):
    (1) for the case of trisection, need add $\frac{1}{3.2^2}$ to length $AD=\frac{1}{2^2}$ to get $\frac{1}{3}$. This means only a single interval of $DH$ is being added to $AD$ to get the trisection. This value of $1$ interval is given by $2^2-3$ also.
    (2) for the case of $n=5$, the lune is formed between the two ends at $E: \frac{1}{2^3}$ and $D: \frac{1}{2^2}$, & want to get to the value $\frac{1}{5}$ from $E$. The difference is $\frac{1}{5} - \frac{1}{2^3} = \frac{3}{40}$. Each of the $n$ intervals in the lune is of length $\frac{1}{5.2^3}$, so a total of $3$ intervals are being used up. This means that after the intersection point of circles $c1, c2$ gives the length for a single interval ($DH$), need take $2$ more intervals to the right to form $DL$. This value of $3$ intervals is given by $2^3-5$.

    Similarly, the author has given a formula of $2^m -n$ , but the reasoning is not clear apart from checking value for a few cases, and (implictly) using induction for all higher values of $n$.

    Also, it is not clear why the intersection points of circles $c1, c2$ will not give the correct length needed to form the $n$-th sect in the lune. The article shows the case of trisection when $DH = DL$, but for higher general case values of $n$ (say, $n=5$) $DH\ne DL$, leading to need for reason being asked.
    Have taken the case for $n=5$ below algebraically, that gives $x= 3b/20$, with $b= |AB|$. I have no way to check except that on adding $DH$ twice, get the coordinates of $AL$ correctly, i.e. $3b/20+ 2\cdot \frac{b}{8\cdot 5}$ would lead to $x=b/5$. Please vet it:
    $$circle \,\,c1 :> (x-\frac{b}{8})^2 + y^2 = (\frac{b}{8})^2$$ $$circle \,\,c2 :> (x-\frac{7b}{16})^2 + y^2 = (\frac{5b}{16})^2$$ For forming circle $c1$, there is done $3$ times bisection, followed by adding $5\cdot AE$ to the right of $AE$, leading to :

A----E----D-----------C--------------------K-------B

The first interval $AE$ will form radius of circle $c1$ with length $=\frac{AB}{2^3}$; while the second circle $c2$ will have center at $AE+EK/2 =\frac{AB}{8}+\frac{5AB}{16}= \frac{7AB}{16}$, and radius as $EK=\frac{5AB}{16}$.
So, the core issue left is how the factor $2^m -n$ is derived in the first place, is it by induction or some better real logic.


Summary --- The logic for the general case, $n\ne 2^m$ case arrives by taking two divisions, one of length $1$ unit (with unit = $\frac{1}{2^m}$, with $2^{m-1} \lt n \lt 2^{m}$). Hence, the lune will be of the size $\frac{1}{2^m}$, with the intersection points of two circles yielding single addition of the interval $\frac{1}{n\cdot 2^m}$. The reason for the single interval addition is the ratio $1:n$ of the two radius over which the circles are being formed.
The reason for adding a total of $2^m-n$ intervals is given by both answers below.

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    $\begingroup$ Wish got at least a reason for down-voting, for at least pursuing this much. Also, got a 'close' vote, so for this level of effort should get some reason too. $\endgroup$ – jiten Apr 14 '18 at 18:02
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  • Minor mistake in $(b)(i)$To get $\frac14$ of the original length, you need two bisections as you have stated in part $(a)$.

  • For part $(b)(ii)$, copy $AD$ $n$ times. if $n=3$, copy it $3$ times.

  • The statement in general $K \ne B$ means that thet statement $\forall n, K=B$ is false. We can still have $K=B$ sometimes. It means there exists $n$ such that $K \ne B$.

  • For the case of $n=3$, if we copy one time, then $H=L$, $DH=DL$

Edit $1$:

So, after copying one time $DL= DH+HL= DH=\frac{AD}{3}=\frac{AB}{\color{red}{12}}$.

$AL = AD+DL = AB(\frac{1}{4}+\frac{1}{\color{red}{12}})=\frac{\color{red}4}{12}AB$.

so $\frac{AL}{AB}=\frac13$.

Edit $2$:

\begin{align} AK &= AD + DK \\ &= AD + nAD\\ &= (n+1)AD \end{align}

Edit $3$:

You goal is to construct $L$ such that $\frac{AL}{AB}=\frac1n.$

By defining $m$, we have $2^{m-1} < n < 2^m$.

By definining $D$, we have $\frac{AD}{AB} = \frac1{2^m}$.

After which, we define $ K, E$ so that we can construct $H$ such that $\frac{DH}{AD}=\frac1n$.

That is we have $DH = \frac{AD}{n}=\frac{AB}{n2^m}.$

Let's construct $L$ by copying the length of $DH$ starting from $D$ $p$ times where $p$ is a quantity to decide.

We want $$\frac{AD+DL}{AB}=\frac{AD+pDH}{AB}=\frac1n$$

$$\frac{AD}{AB}+\frac{pDH}{AB}=\frac1n$$

$$\frac{1}{2^m}+\frac{p}{n2^m}=\frac1n$$

$$\frac{p}{n2^m}=\frac1n-\frac{1}{2^m}$$

$$\frac{p}{n2^m}=\frac{2^m-n}{n2^m}$$

Hence $p=2^m-n$.

Edit $4$:

$\triangle DJK$ is similar to $\triangle DHJ$ are similar right angle triangle.

$$\frac{DH}{DA}=\frac{DH}{DJ}=\frac{DJ}{DK}=\frac{DA}{DK}=\frac1n$$

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    $\begingroup$ edited, it is $\frac13$. $\endgroup$ – Siong Thye Goh Apr 14 '18 at 23:07
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    $\begingroup$ edited, made a typo earlier. $\endgroup$ – Siong Thye Goh Apr 14 '18 at 23:17
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    $\begingroup$ see edit. I don't understand why do u say $AL/AB = \frac15$ is incorrect. Also, try to avoid writing a number followed by braces with some numebrs in it. Braces are interpreted as multiplication usually. $\endgroup$ – Siong Thye Goh Apr 15 '18 at 6:40
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    $\begingroup$ you mean $2^{m-1} < n < 2^m$ $\endgroup$ – Siong Thye Goh Apr 16 '18 at 7:00
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    $\begingroup$ I don't think I mention trisection anywhere in my last edit. The main point is to construct point $H$. the rest are just details and distractions. btw, it seems that you are not following the label/notaton of the paper. $E$ is the center of the second circle, $\endgroup$ – Siong Thye Goh Apr 16 '18 at 17:36
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You wrote: "Say, $n=3$ & $2^1<n<2^2$, with $m−1=1$, $m=2$. So, there is a single (as $m−1=1$) bisection to find a segment $1/2$ of the original one."

You didn't read carefully: the article says "to find a segment $1/2^m$ of the original one" and $1/2^m=1/4$. Thus you need $2$ bisections.

EDIT.

There is in fact an error in the final part of the general algorithm: the sentence

To finish the general case, we need to copy segment $DH$ $2^m-n$ times in the direction of $B$.

should be replaced by

To finish the general case, we need to copy segment $DH$ $2^m-n-1$ times in the direction of $B$.

This is probably a misprint, because immediately after it they give the correct statement $DH/DL = 1/(2^m-n)$. In other words:

you must construct a point $L$ between $H$ and $B$ such that $HL = (2^m-n-1)DH$.

With this correction everything should be OK: I tested the construction with GeoGebra for the case $n=5$ and it works like a charm.

EDIT 2.

Just for the record, there is no need for induction. If $A=0$ and $B=1$ then $D=1/2^m$ and $K=(n+1)/2^m$. It follows that $$ x=DH={AD^2\over DK}={1/2^{2m}\over n/2^m}={1\over n2^m} $$ and $$ L=D+(2^m-n)x={1\over 2^m}+{2^m-n\over n2^m}={1\over n}. $$

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    $\begingroup$ Sorry but I don't understand: in the diagrams contained in that article $AD/AB=1/4$. $\endgroup$ – Aretino Apr 14 '18 at 20:38
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    $\begingroup$ Where did you read that one can get point $D$ with only one bisection? $\endgroup$ – Aretino Apr 14 '18 at 21:11
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    $\begingroup$ But you don't have point $C$, at the beginning, just $A$ and $B$. So you must construct $C$ (first bisection, of $AB$) and only after that you can construct $D$ (second bisection, of $AC$). $\endgroup$ – Aretino Apr 14 '18 at 21:21
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    $\begingroup$ $DK=nAD$ and $AK=AD+DK=(n+1)AD$. But $n+1$ can be different from $2^m$ (e.g. for $n=5$): in that case $AB=2^mAD>AK$. $\endgroup$ – Aretino Apr 14 '18 at 21:53
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    $\begingroup$ See my edited answer. $\endgroup$ – Aretino Apr 15 '18 at 14:14

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