# Just finished trisecting...
**URL:** https://boards.straightdope.com/t/just-finished-trisecting/31307
**Category:** Factual Questions
**Created:** [August 30, 2000, 6:37am UTC](https://boards.straightdope.com/t/just-finished-trisecting/31307 "2000-08-30T06:37:17Z")
**Posts on this page:** 14
**Page:** 1
### Author:  [@RM\_Mentock](https://boards.straightdope.com/u/RM_Mentock)
#### Post date: [August 30, 2000, 6:37am UTC](https://boards.straightdope.com/t/just-finished-trisecting/31307/1 "2000-08-30T06:37:17Z")
Just finished [trisecting the angle](http://mentock.home.mindspring.com/trisect.htm). Why do you suppose everybody thinks it is so hard? I’ll post my short version of the proof of Fermat’s Last Theorem tomorrow–it won’t fit in this window.
---
### Author:  [@scr4](https://boards.straightdope.com/u/scr4)
#### Post date: [August 30, 2000, 7:16am UTC](https://boards.straightdope.com/t/just-finished-trisecting/31307/2 "2000-08-30T07:16:48Z")
I checked your page. The way you described it, length of OC is twice the length of OD. The triangle DOC is not an isosceles triangle, and the “angles DOC and OCD are equal” part is incorrect. If you set point C so that OD=DC, it would work, but I don’t think you can mark point C in such a way with just a straight edge and a compass.
---
### Author:  [@RM\_Mentock](https://boards.straightdope.com/u/RM_Mentock)
#### Post date: [August 30, 2000, 7:48am UTC](https://boards.straightdope.com/t/just-finished-trisecting/31307/3 "2000-08-30T07:48:21Z")
Good catch. The point of the compass is to slide along the line to C. I’ve changed the illustration.
---
### Author:  [@Lance\_Turbo](https://boards.straightdope.com/u/Lance_Turbo)
#### Post date: [August 30, 2000, 10:03am UTC](https://boards.straightdope.com/t/just-finished-trisecting/31307/4 "2000-08-30T10:03:25Z")
I’m not sure what you’re doing when you’re sliding the compass. Are points C, D, and E colinear? If that is the case, wht not just put the straig edge from C to E and mark D where the straight edge intersects the circle?
However, if C, D, and E are colinear and CF is 3r (where r is the radius of the circle), your construction doesn’t work. Let me try explain this my half assed attempt at proving this in plain text:
Draw a line down from D perpendicular to CF, label the intersection of this line and CF “G”. Draw a line down from E perpendicular to CF, label the intersection of this line and CF “H”. There are two important angles in this problem angle FOE and angle DOG. I’m going to call angle FOE “x” and angle DOG “y” from now on. Triangle DOG is a right triangle with hyponteneuse (segment OD) of length r, leg DG of length r_sin(y), and leg OG of length r_cos(y). Triangle ROE is a right triangle with hyponteneuse (segment OE) of length r, leg EH of length r_sin(x), and leg OH of length r_cos(x). The length of CG = OC - OG. OC was defined as 2r in the construction and we already know the length of OD. So CG = 2r-r_cos(y). CH = OC + OH = 2r + r_cos(x). Triangles DCG and ECH are similar. (Angle Angle) Therefore DG/CG = EH/CH. Substitute some crap into that and you have…
sin(y)/(2-cos(y)) = sin(x)/(2+cos(x))
that can be simplifeied to:
2sin(y) + sin(x+y) = 2 sin(x)
At this hour, I was unable to figure out what y was in terms of x. (There’s that half assed part I mentioned earlier.) But I did check a few values. If your construction trisected the angle then x would equal 3y. So I substituted 45 degrees for x and 15 degrees for y. The equation did not balance. I tried a couple other values and they too did not support your theory. I substituted 3y for x in the above equation to find out if there were any angles x for which this method did work. I found that the method only works if x is 0 degrees.
---
### Author:  [@JonF](https://boards.straightdope.com/u/JonF)
#### Post date: [August 30, 2000, 10:41am UTC](https://boards.straightdope.com/t/just-finished-trisecting/31307/5 "2000-08-30T10:41:27Z")
I’ll leave the validity of the trisection to Lance Turbo, who seems to have it in hand. The challenge that has been proved impossible is to trisect the angle with compass and straightedg alone. Your construction is not a “compass and straightedge” construction.
You are not allowed to leave the compass open. The compass must collapse to closed when not in the process of drawing a circle.
You are not allowed to use the straightedge to measure distances; you may use it only to draw straight lines. Laying the compass along the straightedge in step 3 is equivalent to putting marks on the straightedge and using it as a measuring device.
[Geometric Construction](http://mathworld.wolfram.com/GeometricConstruction.html).
---
### Author:  [@jcgmoi](https://boards.straightdope.com/u/jcgmoi)
#### Post date: [August 30, 2000, 11:57am UTC](https://boards.straightdope.com/t/just-finished-trisecting/31307/6 "2000-08-30T11:57:30Z")
**Lance** : CF = 3r after step two. Step three, where one point of the compass crawls up the circle to point D, requires point C to move toward O.
**JonF** is right. Not a classical construction.
I believe this means of trisection is due to Archimedes.
---
### Author:  [@AWB](https://boards.straightdope.com/u/AWB)
#### Post date: [August 30, 2000, 12:33pm UTC](https://boards.straightdope.com/t/just-finished-trisecting/31307/7 "2000-08-30T12:33:32Z")
Empirically, I’ve proven this. But I agree with JonF that this is not following the “compass and straightedge” construction.
---
### Author:  [@RM\_Mentock](https://boards.straightdope.com/u/RM_Mentock)
#### Post date: [August 30, 2000, 2:19pm UTC](https://boards.straightdope.com/t/just-finished-trisecting/31307/8 "2000-08-30T14:19:47Z")
> [@](#):
>
> \*Originally posted by JonF \*
> **You are not allowed to leave the compass open. The compass must collapse to closed when not in the process of drawing a circle.**
You are correct.
And jcgmoi is right that this is due to Archimedes.
You guys are sharp.
---
### Author:  [@hawthorne](https://boards.straightdope.com/u/hawthorne)
#### Post date: [August 30, 2000, 2:45pm UTC](https://boards.straightdope.com/t/just-finished-trisecting/31307/9 "2000-08-30T14:45:43Z")
It’s a lovely page for all that.
---
### Author:  [@Chronos](https://boards.straightdope.com/u/Chronos)
#### Post date: [August 30, 2000, 6:36pm UTC](https://boards.straightdope.com/t/just-finished-trisecting/31307/10 "2000-08-30T18:36:56Z")
Just for some further background: First, it is, of course, possible to trisect certain specific angles, notably a right angle. Secondly, it is possible to trisect a line segment, or indeed, to divide a line segment into an arbitrary number of segments of equal length. Thirdly, the proof that it is, in general, impossible to trisect the angle stems from the fact that the sine of 20[sup]o[/sup] has been proven to be a trancendental number, and hence inconstructible. This, in turn, implies that a 20[sup]o[/sup] angle is inconstructible, and hence that a 60[sup]o[/sup] angle cannot be trisected.
Just don’t ask me to prove the first step in that, that that number is trancendental.
---
### Author:  [@KneadToKnow](https://boards.straightdope.com/u/KneadToKnow)
#### Post date: [August 30, 2000, 8:56pm UTC](https://boards.straightdope.com/t/just-finished-trisecting/31307/11 "2000-08-30T20:56:34Z")
> [@](#):
>
> \*Originally posted by RM Mentock \*
> You guys are sharp.
I hope I can be this gracious if I am ever in this situation. Hats off, **RM Mentock**.
---
### Author:  [@Cabbage](https://boards.straightdope.com/u/Cabbage)
#### Post date: [August 30, 2000, 9:29pm UTC](https://boards.straightdope.com/t/just-finished-trisecting/31307/12 "2000-08-30T21:29:34Z")
> [@](#):
>
> …the sine of 20[sup]o[/sup] has been proven to be a trancendental number, and hence inconstructible.
Actually, sin(20[sup]o[/sup]) isn’t transcendental, it’s a root of the polynomial 64x[sup]6[/sup] - 96x[sup]4[/sup] + 36x[sup]2[/sup] - 3. You’re right about it being inconstructible, though.
---
### Author:  [@Achernar](https://boards.straightdope.com/u/Achernar)
#### Post date: [August 31, 2000, 4:44am UTC](https://boards.straightdope.com/t/just-finished-trisecting/31307/13 "2000-08-31T04:44:05Z")
Furthermore, sin(Q°), where Q is any rational number, can be easily shown to be Algebraic, that is, not Transcendental. **Chronos** , you even helped me prove this. How could you forget so easily? 🙂
---
### Author:  [@Chronos](https://boards.straightdope.com/u/Chronos)
#### Post date: [August 31, 2000, 8:05pm UTC](https://boards.straightdope.com/t/just-finished-trisecting/31307/14 "2000-08-31T20:05:00Z")
Open mouth.
Insert foot.
Repeat.
Yes, I meant inconstructible… Sorry for the confusion there.