# General division of an angle into an arbitrary ratio using compass and ruler

Given any acute angle Ө and a known line which divides it in the ratio Ө/n where n $\ge 2$ is a real number, can we construct a line, using compass and unmarked ruler, which divides the angle in the ratio Ө/n+1? In other words, if an angle bisector is given, we want to find an angle trisector and if a line that divides an angle in the ration 1:5.2 then we want to draw a line which divides the angle in the ration 1:6.2.

It is know the dividing an angle in the ratio 1:3 some other ratios is not possible in general because of the impossibility of constructing transcendental numbers.

The construction given below however seems to give an approximate division of the angle in the ratio Ө/n+1 for any n $\ge 2$,

Question 1: For a given Ө and n, can we quantify the error between the exact value of Ө/n+1 and the approximate division of the angle in the ratio Ө/n+1?

Construction

Let [COB] = Ө be any acute angle and OI be a line such that [IOB] = Ө/n, where n is a positive integer greater than or equal to 2.

1. With O as center and radius OC, draw an arc to cut OB at D.

2. Join CD and draw DJ perpendicular to OB.

3. Let OI intersect DJ at I and the arc CD at K.

4. Extend DK and OC to intersect at G.

5. Let OI intersect CD at E.

6. Extend GE to intersect OB at H.

7. Join IH to intersect GD at F.

8. Draw a line OA through F.

9. [AOB] is the required angle which is approximately equal to Ө/(n + 1).

• It is famously impossible for $n=2$ -- since it is easy to construct a bisector for your arbitrary angle, being able to do what you want would yield a general trisection procedure. – hmakholm left over Monica Oct 16 '16 at 9:26
• By the way, the impossibility of trisection is not directly related to transcendental numbers -- the quantities you need are algebraic, just not constructible. – hmakholm left over Monica Oct 16 '16 at 9:29
• Yes it division impossible in general for that is why this is an approximate division and so we want to estimate the error between the approximate and the actual division in terms of $'theta$ and $n$. – Nilotpal Sinha Oct 16 '16 at 9:33
• You can do an approximate division of any angle into any number of parts to any desired degree of accuracy – it's just a question of how complicated a construction you are willing to contemplate. – Gerry Myerson Oct 17 '16 at 6:35
• "In other words, if an angle bisector is given, we want to find an angle trisector" An angle bisector needn't be given. It can always be constructed and taken. So if this were true we could always trisect an angle. We know we can't. – fleablood Oct 18 '16 at 16:37

One can compute the coordinates of $F$ by letting $O=(0,0)$, $D=(1,0)$, $C=(\cos\theta,\sin\theta)$, $K=(\cos\theta/n,\sin\theta/n)$, $I=(1,\tan\theta/n)$. From there one can find the equations of lines $OC$ and $DK$ to obtain their intersection $G$, then the equations of lines $OI$ and $CD$ to obtain their intersection $E$, and so on. At the end one gets the result: $${F_y\over F_x}=\tan(\angle AOB)= {2\sin{\theta\over2}\sin{\theta\over2n}\over\sin{(n+1)\theta\over2n}}= {2\tan{\theta\over2}\tan{\theta\over2n}\over \tan{\theta\over2}+\tan{\theta\over2n}}.$$ On the right hand side you may discern the harmonic mean between $\tan{\theta\over2}$ and $\tan{\theta\over2n}$: that makes sense, because in the small angle approximation ($\tan\theta\approx\theta$) it becomes the harmonic mean between ${\theta\over2}$ and ${\theta\over2n}$, which is ${\theta\over n+1}$.
To compare $\angle AOB=\arctan(F_y/F_x)$ with $\theta/(n+1)$ we can compute the Taylor expansion: $$\arctan{F_y\over F_x}-{\theta\over n+1} =\frac{(n-1)^2 }{12 n (n+1)^3}\theta^3+O\left(\theta^5\right).$$ For not too large values of $\theta$ the approximation is indeed very good and improves as $n$ increases. Below is a plot of $\angle AOB$ as a function of $\theta$ for $n=2$ (blue) and of $\theta/3$ (red). For $\theta=\pi/2$ we have $\angle AOB=0.585786...$ and $\theta/3=0.523599...$
• Can you share you computation to derive $\tan(AOB)$? – Nilotpal Sinha Oct 17 '16 at 5:39