# Archimedes's Twins and a Cobbler's Knife

There is a beautiful and interesting shape called the Arbelos - or cobble's knife. It consists of semicircle whose diameter is shared by the diameters of two other semicircles as shown.

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Among the many interesting properties this beautiful shape possesses, one is a pair of circles called Archimedes's twins. They are two circles tucked tangentially between two semicircles and a common tangent to the smaller two as shown in \red.

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Let O, C, D be the centers of the semicircles of radii $R = OA, r_{1} = CT, r_{2} = TD$ respectively.

Let $C_{1}$, r = center & radius of the Archimedes circle on the left. and $C_{1}M$ be perpendicular to AB.

Then in triangle $CC_{1}O$ we can mark some distances as follows -

$CC_{1} = r_{1}+r;~~~ OC_{1} = R-r; CM = r_{1}-r; ~~~MO = R-2r_{1}+r$

$MC_{1}^2 = CC_{1}^2 - CM^2$

$~~~~~~~~~= (r_{1}+r)^2-(r_{1}-r)^2$

$~~~~~~~~~= 4rr_{1} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)$

But we can also find MC as,

$MC_{1}^2 = OC_{1}^2 - OM^2 = (R-r)^2-(R-2r_{1}+r)^2$

$~~~~~~~~~=(R^2-2Rr+r^2)-(R^2+4r_{1}^2+r^2-4Rr_{1}+2Rr-4rr_{1})$

$~~~~~~~~~=-4Rr-4r_{1}^2+4Rr_{1}+4rr_{1} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)$

Equating (1) and (2) we get $4rr_{1}=-4Rr-4r_{1}^2+4Rr_{1}+4rr_{1}$ $Rr=-r_{1}^2+Rr_{1}$ $Rr=r_{1}(R-r_{1})$ but $R-r_{1}=r_{2}$ $r=\frac{r_{1}r_{2}}{R}$ Now we may derive the whole expression again for the Archimedes circle on the right or use symmetry of this expression to conclude it must hold for that circle too. Since the two circles have identical radii, they are called twins!

Note by Ujjwal Rane
5 years, 7 months ago

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Interesting !!!

- 5 years, 7 months ago

Another cool (but obvious) thing about this shape that I remember my 5th grade teacher pointing out is that the perimeter of the Cobbler's Knife is the circumference of the original circle, or equivalently, the sum of the lengths of the two arcs $AT$ and $TB$ is equal to the arc length of $AB$.

- 5 years, 7 months ago

That is a nice property! Thanks for sharing it Raj!

- 5 years, 7 months ago