# Are there curves other than circles such that the line through two points on the curve is parallel the line tangent to the curve at midpoint?

In a circle, if we pick any two distinct points $$p_1$$ and $$p_2$$ and draw a line passing through $$p_1$$ and $$p_2$$ that line is parallel to the line tangent to the circle at the midpoint of the arc with endpoints $$p_1$$ and $$p_2$$.

Are there curves other than cicles such that is true? What about parabolas?

• parallel to "the line tangent to the circle at the midpoint of $p_1$ and $p_2$", what do you mean? which circle? Commented Aug 30, 2022 at 3:55
• "In a circle..." That circle, @MathFail What circle could OP mean? Commented Aug 30, 2022 at 3:57
• Well, a line satisfies this condition, of course. But a line can be thought of as a degenerate circle. Commented Aug 30, 2022 at 3:59
• How are you defining the midpoint of the curve between the two points? Commented Aug 30, 2022 at 4:34
• The parabola $f(x)=x^2$ satisfies that for $p_1=(a,a^2)$ and $p_2=(b,b^2)$ the tangent parallel to the secant through $p_1$ and $p_2$ is the tangent at x$x = (a+b)/2$. Commented Aug 30, 2022 at 18:08

In my opinion, this is a great question with a clear interpretation. I don't have an answer yet, but I can at least explain how to formulate the question precisely, which may help others find a solution.

Let's suppose the curve is a regular smooth plane curve parametrized by $$\mathbb{R}$$. Then without loss of generality it may be parametrized by arc-length, i.e. we have a smooth function $$\gamma : \mathbb{R} \to \mathbb{R}^2$$ such that $$\lVert \gamma'(t) \rVert = 1$$ for all $$t \in \mathbb{R}$$.

The condition that the secant line be parallel to the tangent at the midpoint then says that there exists a function $$c : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$$ such that $$\gamma(b) - \gamma(a) = c(a,b) \gamma'\left(\frac{a+b}{2}\right)$$ for all $$a,b \in \mathbb{R}$$. Using the fact that $$\gamma'$$ is a unit vector, we have $$c(a,b) = c(a,b) \left\lVert \gamma'\left(\frac{a+b}{2}\right) \right\rVert^2 = c(a,b) \gamma'\left(\frac{a+b}{2}\right) \cdot \gamma'\left(\frac{a+b}{2}\right) = (\gamma(b) - \gamma(a)) \cdot \gamma'\left(\frac{a+b}{2}\right).$$ We wish to know if this forces $$\gamma$$ to be a circle or a line, i.e. a curve of constant curvature. So the question can be posed as:

Let $$\gamma : \mathbb{R} \to \mathbb{R}^2$$ be a smooth function such that $$\lVert \gamma'(a) \rVert = 1$$ and $$\gamma(b) - \gamma(a) = \left((\gamma(b) - \gamma(a)) \cdot \gamma'\left(\frac{a+b}{2}\right)\right) \gamma'\left(\frac{a+b}{2}\right)$$ for all $$a, b \in \mathbb{R}$$. Must $$t \mapsto \lVert \gamma''(t) \rVert : \mathbb{R} \to \mathbb{R}$$ be constant?

Let $$\gamma$$ be immersive. As $$\gamma(s+h)-\gamma(s-h)$$ is parallel to $$\gamma'(s)$$, we have $$\gamma(s+h)-\gamma(s-h)=\langle\gamma(s+h)-\gamma(s-h),\gamma'(s)/\|\gamma'(s)\|\rangle \frac{\gamma'(s)}{\|\gamma'(s)\|}.$$ Now take the second derivative in respect to $$h$$ to get $$\gamma''(s+h)-\gamma''(s-h)=\langle\gamma''(s+h)-\gamma''(s-h),\gamma'(s)/\|\gamma'(s)\|\rangle \frac{\gamma'(s)}{\|\gamma'(s)\|}.$$ Divide by $$2h$$ and let $$h\to0$$: $$\gamma'''(s)=\langle\gamma'''(s),\gamma'(s)/\|\gamma'(s)\|\rangle\gamma'(s)\frac{\gamma'(s)}{\|\gamma'(s)\|},$$ that is, $$\gamma'''(s)$$ is parallel to $$\gamma'(s)$$,
In this case $$\left( \det(\gamma',\gamma'') \right)'=0,$$ hence $$\kappa(s)=\frac{\det\bigl(\gamma'(s),\gamma''(s)\bigr)}{\|\gamma'(s)\|^3} =\frac{\text{constant}}{\|\gamma'(s)\|^3},$$ that is, the curvature solely depends on the length of the velocity vector. For an arc-length parametrisation it follows that the curvature is constant.
As example would serve $$\gamma(s)=(a\cdot e^s,b\cdot e^{-s})$$ for constants $$a$$ and $$b$$.
As $$\gamma'(s)$$ is a unit vector and $$\gamma(s+h)-\gamma(s-h)$$ is parallel to $$\gamma'(s)$$, we have $$\gamma(s+h)-\gamma(s-h)=\langle\gamma(s+h)-\gamma(s-h),\gamma'(s)\rangle\gamma'(s),$$ as @diracdeltafunk remarked.
Now take the second derivative in respect to $$h$$ to get $$\gamma''(s+h)-\gamma''(s-h)= \langle\gamma''(s+h)-\gamma''(s-h),\gamma'(s)\rangle\gamma'(s).$$ Divide by $$2h$$ and let $$h\to0$$: $$\gamma'''(s)=\langle\gamma'''(s),\gamma'(s)\rangle\gamma'(s),$$ that is, $$\gamma'''(s)$$ is parallel to $$\gamma'(s)$$, hence the derivative of $$\|\gamma''(s)\|$$ vanishes: \begin{align} \frac{d}{ds}\|\gamma''(s)\|^2&=2\langle\gamma''(s),\gamma'''(s)\rangle\\ &=2\langle\gamma''(s),\langle\gamma'''(s),\gamma'(s)\rangle\gamma'(s)\rangle\\ &=2\langle\gamma'''(s),\gamma'(s)\rangle\langle\gamma''(s),\gamma'(s)\rangle\\ &=\langle\gamma'''(s),\gamma'(s)\rangle \frac{d}{ds} \lVert \gamma'(s) \rVert^2\\ &=0. \end{align}