# “push” a line segment to a arc

circlesvectors

Graph:

There is a circle with a radius of 5, the center is O

There are two points A and B on the circle. OA and OB are vertical.

Connect A and B to form a line segment AB.

The arc between A and B is called arcAB.

What I want:

Imagine that there are countless points on the line AB (only 5 are marked in the graph), and I want to "push" these points to arcAB.

If you consider these 5 points as vectors, you can make them overlap with arcAB by scaling up the different values.

For example [1, 1.2, 1.414, 1.2, 1].

Question:

Is there any way to confirm this ratio?

In $$\bigcirc O$$, let $$\angle AOB=2\theta$$. OP asks how to "push" chord $$\overline{AB}$$ to arc $$\stackrel{\frown}{AB}$$, and proposes specifically the strategy of projecting from the center $$O$$, as shown with points $$P$$ and $$P'$$ in these figures:

Formulating of the projection is simple: divide $$\overrightarrow{OP}$$ by its length, then scale it by the radius of the circle; ie, $$\overrightarrow{OP'} = \frac{r}{|OP|} \overrightarrow{OP} \tag1$$ Less elegantly (but more-instructively), if we define $$p := \dfrac{|AP|}{|PB|}$$ (so that $$P=\dfrac{A+pB}{1+p}$$) as a parameter that ranges from $$0$$ (when $$P=A$$) to $$\infty$$ (when $$P=B$$), we can write $$P' = \left(\;\frac{r (1 + p \cos2\theta)}{\sqrt{1 + 2 p \cos2\theta+p^2}}\;,\; \frac{r p\sin2\theta}{\sqrt{1+ 2 p \cos2 \theta+p^2}}\;\right) \tag2$$

Unfortunately, such a projection is problematic when $$\angle AOB=180^\circ$$, as shown in the second figure; all the vectors $$\overline{OP}$$ are "flat", so their projections will only land at $$A$$ or $$B$$ (point $$O$$ projects to itself). This is reflected in how $$(2)$$ reduces when $$\theta=90^\circ$$: $$P' = \left(\;\frac{r (1-p)}{|1-p|}\;,\; 0\;\right) \tag{2'}$$

In a comment, I suggested an alternative.

Instead of projecting from $$O$$, project from the midpoint of the "other" arc of the circle ($$M$$ in the figures), as shown with points $$Q$$ and $$Q'$$. Here, defining $$q := \dfrac{|AQ|}{|AB|}$$, we can write $$Q' =\left(\;\frac{r (1 + 2 q \cos\theta + q^2 \cos2\theta)}{1 + 2q\cos\theta + q^2}, \frac{2r q \sin\theta(1 + q\cos\theta)}{1 + 2 q\cos\theta+q^2}\;\right) \tag3$$ This strategy works even when $$\angle AOB = 180^\circ$$, as the $$\theta=90^\circ$$ form of $$(3)$$ is $$Q' =\left(\;\frac{r (1-q^2)}{1+q^2}, \frac{2rq}{1+q^2}\;\right) \tag{3'}$$ which, by interesting coincidence, corresponds to the Weierstrauss tangent-half-angle parameterization of the (upper semi)circle. (Fun Fact: It's not a coincidence!)

OP has commented that the second projection (via $$M$$) is "flawed" because it doesn't divide the arc evenly. This is true: except when $$Q$$ is the midpoint (or an endpoint) of the chord, the projection doesn't transfer the chord ratio to the arc; ie, $$|AQ|:|QB|\neq|\stackrel{\frown}{AQ'}|:|\stackrel{\frown}{Q'B}|$$. But this is also true of OP's projection (via $$O$$).

Indeed, there is no "geometric construction" (in the classical sense) to get from every $$P$$ on the chord to the corresponding $$P'$$ on the arc such that $$|AP|:|PB|\neq|\stackrel{\frown}{AP'}|:|\stackrel{\frown}{P'B}|$$. In particular, if $$P$$ "trisects" the chord then $$P'$$ would have to trisect the arc, giving a trisection of $$\angle AOB$$ which is a famously impossible thing to accomplish via straightedge-and-compass techniques. (So, it's certainly impossible to accomplish via a simple projection from some given point. See this answer for a quick proof when the point is the center of the circle.)

If $$P'$$ is required to divide the arc the same way $$P$$ divides the chord, for every $$P$$ on that chord, then essentially the best one can do is to write $$P' = \left(r \cos\frac{2p\,\theta}{1+p}, r\sin\frac{2p\,\theta}{1+p}\right) \tag4$$ although this lacks the sense of geometrically "pushing" the chord to the arc.