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In my lecture notes (and elsewhere on the Internet), the proof starts like this:

Let's consider a $60^{\circ}$ angle and let's show that trisecting the angle (i.e. constructing 3 angles of $20^\circ$) would imply constructing $\cos(20^\circ)=\cos(\pi/9)$ which is not constructible with a straightedge and compass.

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Then they proceed to show that $\cos(\pi/9)$ is not constructible.

The problem is that I feel like this also shows that bisecting the angle is impossible because we could consider bisecting a $40^\circ$ angle (to construct 2 angles of $20^\circ$) and show, in a similar way, that $\cos(\pi/9)$ is not constructible. But, this is not true since we know that we can bisect any angle.

Where is the flaw in my reasoning ?

Stef
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IsaKEKW
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  • Can someone explain what I'm missing, because given the green and black lines I can split that angle into thirds, using only a straight edge and a compass. So . . . what am I missing? Method: With compass draw an arc from green to black line. Draw a straight line between the points the arc touches on green and black lines. Split the line into 3 segments (using this mothod)[https://www.mathsisfun.com/geometry/construct-segment3.html]. each point on the line to the center of the original arc is one third of the original angle, so here 20 degrees. Thanks folks. – Binary Worrier Oct 11 '23 at 11:27
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    @BinaryWorrier Trisecting the chord does not lead to trisecting the angle. Imagine the arc is close to 180 degrees, almost flat. Divide the chord in 3 parts. The central part is associated to an angle which is still very close to 180 degrees, while the other two parts correspond to angles of nearly zero degrees. – chi Oct 11 '23 at 11:46
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    @BinaryWorrier: See, for instance, this old answer of mine. – Blue Oct 11 '23 at 12:13
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    OK, I see it now, my method it looks like it works for small angles but actually doesn't. Thank you – Binary Worrier Oct 11 '23 at 13:08

1 Answers1

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You showed that an angle of $40°$ is not constructible. The reason we argue that trisecting an angle of specifically $60°$ is not possible is because we can construct an angle (of $60°$) which we cannot trisect.

If we had an angle of $40°$ available then we could construct a $20°$ angle, but if we start only with two points, we cannot construct this $40°$ angle.

Kovomaka
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    I'd add that there appears to be one point missing in their proof: they go from "constructing cos(pi/9) is impossible" to "trisecting a 60° angle is impossible" and the missing link to make that implication is that constructing a 60° angle is easy, so if you can construct cos(pi/9) when given a 60° angle, then you can also construct cos(pi/9) without being given a 60° angle, which is a contradiction. – Stef Oct 11 '23 at 10:17
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    Possibly related to the discussion: for example if you're given a $(540/7)^\circ$ angle, you can trisect that angle by doubling it then subtracting from a straight angle. On the other hand, neither a $(540/7)^\circ$ angle nor a $(180/7)^\circ$ angle is constructible from scratch. (Might be easier to think of it as trisecting a given angle of $\frac{3\pi}{7}$ radians.) – Daniel Schepler Oct 11 '23 at 18:33
  • Can you explain '… trisecting an angle of specifically 60° is not possible because we can construct an angle (of 60°) which we cannot trisect'?

    How is that not invalidated by relying on itself for any definition?

    How is it not equivalent to '… trisecting an angle of 60° is not possible because an angle of 60° cannot be trisected'?

     – Robbie Goodwin Oct 12 '23 at 19:04
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    @RobbieGoodwin We can construct an angle of $60°$ but we cannot construct an angle of $20°$ sharing a side with it in the interior. Here I used "trisecting" in the sense "constructing an angle and its third". Since it can be shown that this $60°$ angle is constructible and this $20°$ angle is not, we have a non-trisectable angle. – Kovomaka Oct 12 '23 at 19:32
  • Yes, and you seem miss both points.

    First and foremost, geometry recognises the basic arithmetic principle that no line can be trisected and if any angle can, that's purely due to its specific measure; never because of a general principle. The ability to construct an angle of 60° has nothing to do with 'trisecting' angles of 180° nor, for that matter, 90°.

    Quite separately how is '… trisecting an angle of 60° is not possible because an angle of 60° cannot be trisected' not tautological meaninglessness?

     – Robbie Goodwin Oct 12 '23 at 19:57
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    @RobbieGoodwin I don't see where you are going. I never said that "trisecting an angle of $60°$ is impossible because an angle of $60°$ cannot be trisected", I gave a real reason why angles of $60°$ cannot be trisected in general : this is because, with a ruler and a compass, we can construct $60°$ angles (such as in an equilateral triangle) and we cannot construct $20°$ angles (because of some Galois theory). This is not a tautology or some word play, this is precise mathematics. – Kovomaka Oct 12 '23 at 20:46
  • @Kovomaka You did say 'The reason we argue that trisecting an angle of specifically 60° is not possible is because we can construct an angle (of 60°) which we cannot trisect.' If you believe that has any usefully different meaning in either ordinary English or geometry than 'trisecting an angle of 60° is impossible because an angle of 60° cannot be trisected' then please, explain that difference. – Robbie Goodwin Oct 13 '23 at 21:59
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    @RobbieGoodwin This is not a tautology : there is a precise definition of trisecting, which we use to construct a precise proof of the fact that this $60°$ angle is not trisectable. This is a proof, not a tautology. – Kovomaka Oct 14 '23 at 19:53
  • @Kovomaka Thanks and can you explain how 'The reason we argue that trisecting an angle of specifically 60° is not possible is because we can construct an angle (of 60°) which we cannot trisect' has any different meaning than 'trisecting an angle of 60° is impossible because an angle of 60° cannot be trisected'? – Robbie Goodwin Oct 19 '23 at 21:47