Is it true that "Among all shapes with the same area, a circle has the shortest perimeter" ? and how to prove it ?
[Math] Among all shapes with the same area, a circle has the shortest perimeter
geometry
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Best Answer
The question can be asked such as "Find the plane curve which encloses given area with shortest perimeter". Then by parametric formulation of coordinates $x(t),y(t)$ and assuming non-selfintersecting curve the area can be written as $$A=\frac 12 \int_{t_1}^{t_2}\big(xy'-x'y\big)dt$$ and the perimeter $$I=\int_{t_1}^{t_2}\sqrt{x'^2+y'^2}dt$$ Then the Euler-Lagrange becomes $$-\frac{d}{dt}\bigg(\frac{x'}{\sqrt{x'^2+y'^2}}\bigg)-\frac{\lambda}2\bigg(y'+\frac{d}{dt}\bigg(y\bigg)\bigg)=0$$ $$-\frac{d}{dt}\bigg(\frac{y'}{\sqrt{x'^2+y'^2}}\bigg)-\frac{\lambda}2\bigg(-x'-\frac{d}{dt}\bigg(x\bigg)\bigg)=0$$ If $ds$ is the element of arc the equations can be reduced to $$\frac{d^2x}{ds^2}+\lambda \frac{dy}{ds}=0$$ $$\frac{d^2y}{ds^2}-\lambda \frac{dx}{ds}=0$$ which has the solution $$x(t)=K_1+C_1\cos(\lambda s)+C_2\sin(\lambda s)$$ $$y(t)=K_2-C_2\cos(\lambda s)+C_2\sin(\lambda s)$$ It can be further simplified by "Angle sum and difference identities" $$x(t)=K_1+D\sin(\lambda s+\alpha)$$ $$y(t)=K_2-D\cos(\lambda s+\alpha)$$ where $D\sin(\alpha)=C_1$ and $D\cos(\alpha)=C_2$; and this solution represents a circle of radius $D$ and centre $(K_1,K_2)$