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A plane curve, $α(θ)$, has the following property: if $c(θ)$ is the center of curvature of $α$ in $θ$, $Q(θ)$ is the projection of $α(θ)$ on the x axis and $T(θ)$ is the intersection point of the tangent line to $α$ in $θ$ with the x axis, then the area of the triangle $cQT$ is constant. Give the parametrization for the curve $α(θ)$, where the parameter $θ$ is the angle between the tangent to $α(θ)$ and the x axis.

I know $c(θ) = α(θ) + \frac{n(θ)}{k(θ)}$, and $Q(θ) = (x(θ), 0)$ (given that $α(θ) = (x(θ),y(θ))$, but I haven't been able to put all the other information together. I know I should find an expression for $T(θ)$ as well, but I'm having trouble with that too. I've been stuck on this for a while, so I would appreciate being pointed in the right direction.

EDIT: After over a year with no progress to this question, I'm giving it a bounty. It's eating away at me that I haven't been able to solve such a basic problem.

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    $\begingroup$ I've thought about this for ten minutes or so. It does not look like an appealing problem to me. But I will tell you that $n(\theta) = (-\sin\theta,\cos\theta)$. The distance from $T$ to $Q$ is not so bad — it's $y(\theta)\cot\theta$. You don't need $T$ and $Q$ individually. $\endgroup$ – Ted Shifrin Dec 21 '17 at 22:33
  • $\begingroup$ How did you come to those results? $\endgroup$ – Matheus Andrade Dec 21 '17 at 22:35
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    $\begingroup$ The last sentence in the first paragraph :) The unit tangent vector to the curve is $(\cos\theta,\sin\theta)$. And, for the distance, just look at the right triangle with vertices $T$, $Q$, and $\alpha(\theta)$. But the problem appears totally yucky. Where did you find it? (The other differential geometry problems you've asked on this site are totally standard ones.) $\endgroup$ – Ted Shifrin Dec 21 '17 at 22:37
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    $\begingroup$ Thanks, but I haven't answered it much :) I'm not yet convinced that it's an interesting question, either. But it may turn out to be. ... Plenty of good problems in my differential geometry text, too :P At least I know how to do (most of) those :) $\endgroup$ – Ted Shifrin Dec 21 '17 at 22:57
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    $\begingroup$ @AlexRavsky I've tried to do that, but believe me, it really doesn't... the resulting systems of differential equations get pretty ugly pretty fast. $\endgroup$ – Matheus Andrade Mar 31 at 1:15

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