Problem: Find a nice simple closed curve other than circle which passes through the points $(0,0)$ and $(1,0)$ on the Cartesian plane and whose length is $L\pi$.
If the given condition is not the loop length but the loop area, it is easy to find a nice curve. The ellipse $$(2x-1)^2+\frac{y^2}{4A^2}=1$$ has the area $A\pi .$
Can someone solve the problem above? Thanks in advance!
If one desires to fit a circular arc cut from a circle of radius $r$ and azimuth $\alpha$ (in radians), the arc length is $$\alpha r=L\pi.$$ To have a nice symmetry one could place the circle center at $(x_0,y_0)$ with $x_0=1/2$ (there may be other more asymmetric solutions). The circle equation is $$(x-x_0)^2+(y-y_0)^2=r^2$$. To let the circle pass through $(0,0)$ and $(1,0)$ we need $x_0^2+y_0^2=r^2$ and $(1-x_0)^2+y_0^2=r^2$, which requires $y_0=\pm \sqrt(r^2-x_0^2)$. The azimuth angle is $\tan(\alpha/2)=x_0/|y_0|=x_0/\sqrt{r^2-x_0^2}$. $$\alpha=2\arctan\frac{x_0}{\sqrt{r^2-x_0^2}}.$$ $$\alpha r=2r\arctan\frac{x_0}{\sqrt{r^2-x_0^2}}=L\pi.$$ and with $x_0=1/2$ $$2r\arctan\frac{1}{\sqrt{4r^2-1}}=L\pi.$$ This usually requires numerical solution. For small $L\pi-1$, therefore large $r$, one can get a first estimate from the Taylor series, $$2r\arctan\frac{1}{\sqrt{4r^2-1}}\approx 1+\frac{1}{24}\frac{1}{r^2}+\frac{3}{640}\frac{1}{r^4}=L\pi,$$ which is a biquadratic equation in $r$.