Given a circle of radius r, and two points ('X' and 'Z') on that circle, can some circumcircular arc "XYZ" be constructed of length r? I am strictly an amateur, not a professional mathematician or some such.
This question occurred to me while considering the fact that an angle of 1 radian centered on the center of a circle will produce a circumcircular arc on the circle of the same length as the circle's radius, whereas two segments (AB and BC) of equal length intersecting at 60 degrees will, of course, define a third segment of equal length (between A and C).

To elaborate:


*

*Define a circle C with centerpoint O and radius 'r'.  

*Define two points, X and Z, on circle C.

*Define the lines OX and OZ.

*Define angle XOZ.

*Define the line OA bisecting XOZ.

*Define some point Y on OA such that the circumcircular arc XYZ is of
length r.



Point 6, of course, is the one which I do not know how to do.  It has occurred to me that this problem is, of course, restricted to cases in which angle XOZ is of fewer than 60 degrees (as at 60 degrees, XYZ becomes a line segment, and above 60 degrees, no arc XYZ of length <= r can exist.
It has also occurred to me that this problem could also be solved by defining some point P on OA such that a circle D with centerpoint P and which runs through X (and Z) exists, where the length of the arc XZ on circle D is r, but I also have no idea how to do that.
Diagram

EDIT: A related question of interest would be to define the function which describes the length of the line OY relative to the length of OX or OZ (ie: 'r'), and the angle of XOZ.  There would, naturally, be two valid values of OY, as indicated by @RossMillikan, one for a Y inside the circle, and one for a Y outside the circle.
 A: Point $Y$ will exist.  There will be two points that satisfy the requirement,  one on each side of $XZ$.  You can prove that using the intermediate value theorem.  The segment $XZ$ is too short and the arc with a proposed $Y$ far away will be too long, so there is a point in between that is just right.  I am sure it will generally not be constructible with compass and straightedge but I don't know an easy way to show that.
A: Here a way to solve your problem: in the attached figure  you can calculate the theta angle in the right triangle $\triangle{XX'O}$ which is the half of the isosceles triangle $\triangle{XOY}$ with angle $\angle{XOY}=2\alpha\lt 60^{\circ}$.
Each value of the segment $\overline{OA}=a$ where $0\lt \overline{OA}\lt R\cos(\alpha)$ determines both the angle $\theta$ and the "little" radius $r$.
You have $$\begin{cases}r=\sqrt{R^2+a^2-2aR\cos(\alpha)}\\\theta=\arcsin\left(\dfrac{R\sin(\alpha)}{r}\right)\end{cases}\qquad(*)$$
Finally your equation is $$2r\theta=R$$ that is
$$2\sqrt{R^2+a^2-2aR\cos(\alpha)}\arcsin\left(\dfrac{R\sin(\alpha)}{r}\right)=R\qquad(**)$$ You must put in $(**)$ the value of $r$ given in $(*)$, of course. You finally have an equation in one variable $a$ since $R$ and $\alpha$ are data of the problem.

A: AS noted by Ross Millikan the searched point will exist. 
I suggest a way to find it. I use figure with different names:

where we know: 
$\overline{AC}=\overline{AB}=r$
$\angle CAE=\angle CAD=\gamma$
and we search:
$\angle CED=\theta $
$\angle ACE=\varphi$
so that we have 
$$
\theta=\gamma+ \varphi \qquad (1)
$$
with:
$\overline{CE}=x\quad$
and
$\quad \overline{AE}=y$
using the low of cosines for the triangle $ACE$ we have
$$
y^2=r^2+x^2-2rx\cos \varphi \qquad (2)
$$
and, using the triangles $ACD$ and $CED$:
$$
r^2+(y+x)^2-2r(y+x)\cos \gamma = 2x^2-2x^2\cos \theta \qquad (3)
$$
and, with the equation that gives the arc $CDB=r$:
$$
2x\theta=r \qquad (4)
$$
we have four equations for the four unknowns.  
The system is not simple to solve, and maybe that it can be simplified with a different choice of the unknowns.
A: Preface. It took me some time to understand the question, so pardon my initial idiocy. I find that it is somewhat interesting. Here's a report of what I did.
Refer to the figure in OP. We want the arc $XYZ$ whose length is equal to $r$. As OP said, it is meaningless to seek this for $X\hat O Z=\theta\ge π/3$. Now it is easy to see that it is sufficient to find the radius $s$ of the sought arc with length $x$, for once we know this we can always find the centre from the constraints that the arc must pass through $X$ and $Z$ and centered somewhere on the line $OA$. Since we always have two such arcs (forming a complete circle), we choose the smaller one, and it is the one whose length we denote by $x$. The possible radii $s$ of these arcs must satisfy $s\ge c$, where $c$ is half of the chord $XZ$. Now for each such $s$ we always have two such arcs (counting multiplicities), but we consider them to be one (that is, we consider the congruence class of the arcs instead of individuals, with exactly two per class) since we are only interested in their lengths, which must be equal since they are congruent. Furthermore, we seek such $s$ as make $x=1$ for fixed $\theta\in(0,π/3)$.
Since for each $r\ge c$, we have one and only one value for $x$, we can form a function $x(s)$ and simply investigate the $s$ that solve $x=1$. This dissolves the problem for some positive $\theta<π/3$.
By considering the figure we can see immediately that $x$ must satisfy the following:


*

*The function $x(s)$ is continuous.

*It decreases strictly with increasing $r$.

*At $s=c=r\sin{\theta/2}$, the minimum in the domain, we have $x(c)=πc$.

*When $s=r$, we have that $x(r)=\theta$ whenever $\theta$ satisfies the constraints imposed on it.

*The graph of $x$ intercepts both axes nowhere.

*As $r\to\infty$, $x\to 2c$.

*Consequently (specifically by 2 and 5), $x$ is bounded.
We can now imagine, for $\theta=1$ and $r=1$, for example, that the graph of $x$ looks like a segment (namely from about $r=1/2$ to $\infty$) of the graph of $e^{-s}$ translated upwards.
From now on, we shall take $r=1$ since we must needs have a unit. If we consider the diagram and make some simple constructions, we will see that $x=s\phi$, where $\phi$ is the angle of the arc which we seek, subtended at its centre $P$. Note that $\phi$ decreases from $π$ to $0$ with increasing $r$. To find a relationship between $s$ and $\phi$ is easy. Let $P$ be the centre of the arc $XYZ$, then we consider the triangle $PXQ$, where $Q$ is the point where the chord $XZ$ and the line $OA$ intersect. From this we see that $s\sin\phi/2=c=r\sin\theta/2=\sin\theta/2$. Substituting in the previous relation and doing a little rearrangement gives $$x(s)=2s\arcsin\left(\frac 1s \sin\frac \theta2\right)$$ for $s\ge\sin(\theta/2)$ and fixed $\theta$. Indeed if we choose $\theta=1$ and set $x=1$, we find the only solution to be $s=1$, as expected. For any fixed $\theta$, it is easy to see that there is exactly one solution since the constant function $1$ must intersect $x$ at most once, being strictly decreasing.
