# Describing the plane curve $α(θ)$ that has the following property: the area of the triangle given by $cQT$ is constant (details below)

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.

• 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. – Ted Shifrin Dec 21 '17 at 22:33
• How did you come to those results? – Matheus Andrade Dec 21 '17 at 22:35
• 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.) – Ted Shifrin Dec 21 '17 at 22:37
• 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 :) – Ted Shifrin Dec 21 '17 at 22:57
• @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. – Matheus Andrade Mar 31 at 1:15