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My attempt to prove the angle trisection identity of limaçons.

In the diagram below, I drew a limaçon trisectrix having polar equation $r=a+2a\cos\theta$ (In this case, $a=1$). Suppose $O(0,0), A(a,0), B(3a,0)$. Then, draw a line which has cartesian equation $y=kx\; (k\in \mathbb{R}^+)$ intersecting the limaçon at point $C$. Connect line $AC$.

Now, I'm supposed to prove $3\angle ACO=\angle BAC$. However, I got stuck. Is there a method without using analytic geometry?
Limaçon

E. Huang
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  • What exactly have you tried to do? – heropup Jun 18 '21 at 04:37
  • What method can be used which avoids analytic geometry? For example what is the definition of an angle? I know of one involving the dot product and use of inverse cosine, but that would involve analytic geometry. Once a definition of angle is settled, one would of course need also to define the size of an angle, in order to show one angle is three times another. – coffeemath Jun 18 '21 at 06:25
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    I think that it would be very ugly if this problem is solved by analytic geometry. First, I would solve for point $C$, which is probably very ugly. Then, I would solve for line $AC$. Last, proving the tangent of the slope of line $AC$ is 1.5 times the tangent of the slope of line $OC$ should be necessary. I just don't like this solution. – E. Huang Jun 18 '21 at 07:46
  • It looks like there's a proof in the wiki article here: https://en.wikipedia.org/wiki/Lima%C3%A7on_trisectrix – coffeemath Jun 18 '21 at 16:35

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