I found an article in a book about trisecting an angle equally. It was written there that Archimedes tried to solve that process by applying pure geometry (using only compass and scale without its reading). But he failed to do that. However, there was no other method by which the angle can be trisected equally. However, out of curiosity, I wanted to solve the problem with some another process.

It was also narrated in that text that $\angle BCF$ = $\phi$ = $\frac{\theta}{3}$, when $CD$ will be equal to $r$. And $C$ is to be a surmised point so that $CD$ can be equal to $r$. But I wanted to calculate the specific location of $C$ from the circumferential point $I$ and measure the length of $IC$ (the diagram is given below) through mathematical term.

My Attempt:

At first I made a reverse approach by drawing $\angle BCF$ = $\phi$ = $\frac{\theta}{3}$.

Than, I drew two altitude line $BF$ and $DG$, where $BF$ $\perp$ $CE$ and $DG$ $\perp$ $CA$. And than, I connected $AD$ and $DI$.

From right angled $\triangle DGC$,

$GC$ = $r\cos\phi$ and $DG$ = $r\sin\phi$, because $DC$ = $r$.

Again, we get an isosceles $\triangle ABD$ with having $AB$ = $AD$ = $r$. From the $\triangle ABC$, we get

$180^\circ-\theta$ + $\frac{\theta}{3}$ + $\angle ABD$ = $180^\circ$

Than calucating the value, I got $\angle ABD$ = $\frac{2\theta}{3}$ and so, from isosceles $\triangle ABD$,

$\angle BAD$ = $180^\circ$ - $\frac{4\theta}{3}$ and from here we can write that $\angle DAI$ = $\frac{\theta}{3}$.

In $\triangle ADI$, $AD$ = $r$ and $AI$ = $r$. So, $\triangle ADI$ is an isosceles and $\angle DIA$ = $\frac{540-\theta}{6}$.

And now from right angled $\triangle DIG$, we get

$\frac{DG}{IG}$ = $\tan(\frac{540-\theta}{6}$) $\implies$ $IG$ = $\frac{r\sin\phi}{\tan(\frac{540-\theta}{6})}$

Subtracting |$IG$| from |$CG$|,

$|IC|$ = $r\cos\phi$ - $\frac{r\sin\phi}{\tan(\frac{540-\theta}{6})}$ $\implies$ $|IC|$ = $r\cos(\frac{\theta}{3})$ - $\frac{r\sin(\frac{\theta}{3})}{\tan(\frac{540-\theta}{6})}$

Here we get the value of $IC$, and so the point $C$ has an exact and individual location depending on the radius of the circle and the value of the $\theta$.

Is my formula\answer correct? If it so, then I would like to see another method except trigonometry. And my query is that how can we find the length of $IC$ without applying trigonometry or in how many ways the angle $\theta$ can be trisected equally?

If I have a mistake, then please forgive me and help me correct the error. Thanks in advance.


I'm not sure what you mean by "reverse proof". Assuming $\phi=\theta/3$ and then reaching conclusions from there can't convince me that $\phi$ must be $\theta/3$, since I don't know we're not going to reach a contradiction later, or if $\theta/3$ is just one of multiple possibilities. This is the logical fallacy of "circular logic" or the classical meaning of "begging the question".

For Archimedes' argument using the same diagram but without trigonometry, see this subsection of the Wikipedia article on angle trisection. For other methods, see the rest of that entire section.

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    $\begingroup$ Yeah. You are right. I wanted to mean reverse approach instead of proof. That happened for my deep lack of knowledge. Thanks for giving me link of logical fallacy and I hope that my proof and proceeding towards something aren't gonna be contradicted anymore specially in the case of circle. $\endgroup$ – Anirban Niloy Feb 18 '19 at 13:05

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