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I have a geometry problem for an architectural application that I hope someone can help me with.

I have a rectangle $(a times b)$ which contains a reverse curve from bottom left to top right formed of two circular arcs of the same radius $R$ and angle $theta$ which are joined tangentially by a line of length $S$. For the various situations I wish to use this in, I will know $a$, $b$, and $S$ and I wish to solve for $R$ for which I think there is only one solution.

When $S = 0$, with the line disappearing and the two arcs joining tangentially in the centre of the rectangle, I am able to derive the simple formula

$$R = frac{0.25 a^2 + 0.25 b^2}{b}$$

but when $S > 0$ I can derive the following formulas of the relationship between the variables but I am struggling to rearrange them in terms of $R$ or $theta$

$$R=frac{b-Ssin(theta )}{2-2cos(theta)} = frac{a-Scos(theta)}{2sin(theta)}$$

Thank you.

I'll coordinatize the $square OACB$ with
$$O = (0,0) qquad A = (a,0) qquad B = (0,b) qquad C = (a,b)$$
The centers of the arcs will be at
$$A^prime = C - (0,r) qquad B^prime = O + (0,r)$$
Let $C^prime$ be the image of $C$ rotated about $A^prime$ by angle $2theta$, and $O^prime$ the image of $O$ rotated about $B^prime$:
$$C^prime = left(;a - r sin 2theta, b - r (1-cos 2theta);right)$$
$$O^prime = left(;r sin 2theta, r (1 - cos 2theta);right)$$

Then the S-curve conditions are that $|O^prime C^prime|=s$, and $overline{O^prime C^prime}perp overline{C^prime A^prime}$. That is,
$$begin{align}
(O^prime-C^prime)cdot(O^prime-C^prime) &= s^2 tag{1a} \
(O^prime-C^prime)cdot(C^prime-A^prime) &= 0 tag{1b}
end{align}$$
These imply
$$begin{align}
4 r left(; a sin 2theta - (b - 2 r) cos 2theta ;right) &= a^2 + b^2 - 4 b r + 8 r^2 - s^2 tag{2a} \
a sin 2theta - (b - 2 r) cos 2thetaphantom{left.;,right)} &= 2 r tag{2b}\
end{align}$$

Eliminating $theta$ is surprisingly easy, as is solving for $r$.

$$r = frac{1}{4b}left(a^2+b^2-s^2right) tag{$star$}$$

Back-substituting to get $theta$ is a little trickier, but we ultimately get
$$cos 2theta = frac{(a - b + s) (a + b + s)}{
(a+s)^2 + b^2} quadtext{or}quad frac{(a - b - s) (a + b - s)}{
(a-s)^2+b^2}$$
The latter of these turns out to be extraneous. The former yields this more-convenient form:

$$tan theta = frac{b}{a+s} tag{$starstar$}$$

The simplicity of $(starstar)$ suggests that there should be a way to see this solution; as it turns out, there's also a straightforward construction:

• We introduce $S$ and $S^prime$, at distance $s$ from $O$ and $C$, respectively, on extensions of $overline{AO}$ and $overline{BC}$. Note that $tan angle ASC = b/(a+s) = tantheta$, so that $angle ASC = theta$.




• Let $K$ be the center of $square OACB$ (constructed, say, as the intersection of the rectangle's two diagonals). If $bigcirc K$ has diameter $s$, then $overline{SC}$ and $overline{S^prime O}$ meet $bigcirc K$ at $C^prime$ and $O^prime$, respectively.




• Finally, $A^prime$ is where the perpendicular to $overline{C^prime O^prime}$ at $C^prime$ meets $overleftrightarrow{AC}$. Similarly for $B^prime$.

(Incidentally, the extraneous solution corresponds to taking $overline{OS}$ and $overline{CS^prime}$ "pointing the other way". That is, we can take $s$ to be negative in the above analysis. We see that the sign change has no effect on $(star)$, so that the radius $r$ ---and thus also centers $A^prime$ and $B^prime$--- remains unchanged. Therefore, the extraneous $C^prime$ and $O^prime$ are the "other" points where $bigcirc A^prime$ and $bigcirc B^prime$ meet $bigcirc K$.)