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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.

jiten
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    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. – jiten Apr 14 '18 at 18:02

2 Answers2

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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 $$ L=D+(2^m-n)x={1\over 2^m}+{2^m-n\over n2^m}={1\over n}. $$

Intelligenti pauca
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  • Not clear why need two bisections for $n=3$ (trisection) case. I have graph at desmos, and that uses only one bisection by dividing once the interval $AB$. There is no second bisection, just the radius of second circle $c_2$ is extended. If a second bisection were there, then the interval $AD$ would have been used as diameter. I hope that either my above analysis is correct, or the graph drawn at 'desmos' is wrong too. The interpretation made for $n=4$ case, by me, is that there is exactly double bisection there. But, for $n=3$, need the general approach dealt in the post. – jiten Apr 14 '18 at 20:31
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    Sorry but I don't understand: in the diagrams contained in that article $AD/AB=1/4$. – Intelligenti pauca Apr 14 '18 at 20:38
  • Sorry, I am wrong. The text states the same as in b(i) above. However, there is confusion, that need only one bisection for the case of trisection; and can find $AD$ by that directly (as also by your comment). Please help resolve it, as will one or two bisections be needed for the trisection case. – jiten Apr 14 '18 at 20:44
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    Where did you read that one can get point $D$ with only one bisection? – Intelligenti pauca Apr 14 '18 at 21:11
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    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 $AC$). – Intelligenti pauca Apr 14 '18 at 21:21
  • Thanks a lot. I request the reason for Q.#1. It is stating that $K,B$ may not coincide. But, this is not clear as $AD$ is formed by repeated bisection. And so, $n.|AD|=|AB|$ for any value of $n$. Also, need a counter-example, if possible. – jiten Apr 14 '18 at 21:42
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    $DK=nAD$ and $AK=AD+DK=(n+1)AD$. But $n+1$ can be different from $2^m$ (e.g. for $n=5$): in that case $AB=2^mAD>AK$. – Intelligenti pauca Apr 14 '18 at 21:53
  • Thanks a lot. Please address the last part, and the failure for the case $n=3$, as shown at the end of the post. – jiten Apr 14 '18 at 22:28
  • Please help with the highly edited post, that is differing in the fact that for the general case of ($n=5$, as suggested by you) $AB=AK$. Please help as intend to code it, and any error would be catastrophic for larger values of $n$. – jiten Apr 15 '18 at 05:28
  • Sorry but I give up: you seem not to understand even the simplest examples. Good luck with your coding. – Intelligenti pauca Apr 15 '18 at 08:58
  • I hope my latest edit is having the level expected by you. I hope that the reason for part b(iii) will be made clear by you, which concerns finding the reason why $2^m -n$ times the length $AB\cdot \frac{1}{n\cdot 2^m}$ need be added to the length $AB. \frac{1}{2^m}$. – jiten Apr 15 '18 at 12:37
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    See my edited answer. – Intelligenti pauca Apr 15 '18 at 14:14
  • Please share your construction in GeoGebra. I would be highly thankful for that. Also, I am trying (as shown in the last part of the edited post) to find the value of $x$ for $n=5$ case, but am unsuccessful, although I feel the 'interpretation' made by me is the same as that detected by you. I realized this fact earlier, and made the 'total' number of copies of $DH=2^m-n$ starting from $D$. Still unable to get the coordinates correctly for $x$, as stated above. – jiten Apr 15 '18 at 14:45
  • I would also like to add that by just running, the problem of finding the coordinates for getting $x=AB/5$ is not solved. May be I am wrong. I have also run at :https://www.desmos.com/calculator/e6dhp99eiu, but need help on finding why am not getting the coordinates for $x$ correctly in the post (at the end). – jiten Apr 15 '18 at 14:55
  • Also, please answer my earlier question raised in comment regarding the derivation of the formula $2^m-n$ (or, $2^m-n-1$). – jiten Apr 15 '18 at 15:04
  • Please check the summary (in the edited post) for the logic for the general case ($n \ne 2^m$) having lune's two intersection points having the $x$ coordinate as the first intersection point at $\frac{b}{2^m\cdot n}$. – jiten Apr 16 '18 at 09:07
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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.

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

Siong Thye Goh
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  • Thanks a lot. Sorry, realized my folly for taking $DH\ne DL$ & edited that part by removal, just before your answer came. – jiten Apr 14 '18 at 22:41
  • Please help with the part (iii) ratio of $\frac{AL}{AB}=\frac{5}{12}$, rather than the stated one of $\frac{1}{3}$. – jiten Apr 14 '18 at 22:53
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    edited, it is $\frac13$. – Siong Thye Goh Apr 14 '18 at 23:07
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    edited, made a typo earlier. – Siong Thye Goh Apr 14 '18 at 23:17
  • Please see my highly edited post, and the last comment to @Aretino answer. – jiten Apr 15 '18 at 05:33
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    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. – Siong Thye Goh Apr 15 '18 at 06:40
  • Thanks, it is stated in both parts taken from the article in b(ii) , b(iii), that there would be a mismatch, in the general case, i.e. where $n \ne 2^m$. This is also asked in the Question#1, asked in the post. I have an additional question, as to how the factor $2^m -n$ was arrived at, for finding the no. of times to add $DH$. Can see that there are two facts: (i) $2^{m-1}$ values in the interval between $2^{m-1}$ & $2^m$. (ii) Also, the chosen number of bisections is $m = log_2n$ from the left end of the segment, and taking $2^m -n = 2^m - \frac{m}{log_2n}$. But, this does not lead anywhere. – jiten Apr 15 '18 at 07:25
  • Let us continue this discussion in chat. – Siong Thye Goh Apr 15 '18 at 08:48
  • Please see the newly edited post, and sorry the root question still is: how the formula $2^m- n$ is derived. Is it by induction, as I have hinted, or something better, for which also have given hints in the edited post. Also, would like to know the way to find the error introduced by not following the methodology for the general case ($n\ne 2^m$), i.e. still taking simple bisection of $AB$ $m$ times, with $2^{m-1} \lt n \lt 2^m$. Have run the graph for $n=5$ case at : https://www.desmos.com/calculator/e6dhp99eiu, and request vetting it. – jiten Apr 15 '18 at 15:42
  • Please check the summary added at the end of the post. – jiten Apr 16 '18 at 06:58
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    you mean $2^{m-1} < n < 2^m$ – Siong Thye Goh Apr 16 '18 at 07:00
  • Sorry for being too late. Was elsewhere. Yes, for the general case - that is the criteria for choosing $m$ for a given $n \ne 2^m$. Sorry for typo too, have edited it. – jiten Apr 16 '18 at 08:51
  • I hope that in your last edit, the line: "After which we construct $K, E$", means the same as : the ratio of $AD/AK = 1/(n+1)$ ensures that $DH/AD= 1/n$. This is also the same as stated in my post's summary. But, seemingly you misstated $E$, as have used the point $D$, as for trisection, rather than $E$ for $n=5$ in your derivation of the factor $2^m -n$. – jiten Apr 16 '18 at 10:48
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    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, – Siong Thye Goh Apr 16 '18 at 17:36
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    Note that bisect $DK$ gives $E$ in the original paper. – Siong Thye Goh Apr 16 '18 at 17:44
  • Sorry, actually I also used $E$ for $|AE| = \frac{AB}{2^3}$. Request help on :> I offered a simple reason for the general case having two circles with diameters in the ratio $1:n$ to have the lune's $2$ intersection points with common $x$ coordinate, dividing lune's length in first of the $n$ divisions. There are $n$ divisions of the lune, each with length $\frac{AB}{2^m\cdot n}$. I found till now no geometrical reason for this. The above stated logic is also to be found nowhere. Please help me by providing some source to look for a geometrical reason, or some keywords to google. – jiten Apr 17 '18 at 00:26
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    see edit, the proof is provided in the paper. – Siong Thye Goh Apr 17 '18 at 03:09
  • But, the proof to me is faulty. Although, shown for general case, it applies only for $n=3$. For $n=5$, the ratio $\frac{DH}{DA} \ne \frac{1}{n}$; hence need add $2^m-n$ times $DH$ to get $DL$. The only fact that works (for $n=5,6,7, 9,10,11, 12, 13, 14, 15,\cdots$) is that the ratio of diameters of circles $c1: c2$ is in the ratio of $1:n$. I am quite convinced of it, and hence asking of geometrical proof for that. Searched a lot, but seems like topic not worthy of mention even, may be it is too simple. Have my proof in lune formation, as can be seen by drawing differing ratios' $2$ circles. – jiten Apr 17 '18 at 04:41
  • Seems odd, but I was correct; as in answer to: https://math.stackexchange.com/q/2740911/424260, a different proof is provided. It uses a different mechanism to prove. – jiten Apr 17 '18 at 11:20
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    ok, good that you sort it out. – Siong Thye Goh Apr 17 '18 at 13:33