An arbelos is formed from three collinear points A, B, and C, by the three semicircles with diameters AB, AC, and BC. Let the two smaller circles have radii r₁ and r₂, from which it follows that the larger semicircle has radius r = r₁+r₂. Let the points D and E be the center and midpoint, respectively, of the semicircle with the radius r₁. Let H be the midpoint of line AC. Then two of the four quadruplet circles are tangent to line HE at the point E, and are also tangent to the outer semicircle. The other two quadruplet circles are formed in a symmetric way from the semicircle with radius r₂.

According to Proposition 5 of Archimedes' Book of Lemmas, the common radius of Archimedes' twin circles is:

${\displaystyle {\frac {r_{1}\cdot r_{2}}{r}}.}$ 

By the Pythagorean theorem:

${\displaystyle \left(HE\right)^{2}=\left(r_{1}\right)^{2}+\left(r_{2}\right)^{2}.}$ 

Then, create two circles with centers J_i perpendicular to HE, tangent to the large semicircle at point L_i, tangent to point E, and with equal radii x. Using the Pythagorean theorem:

${\displaystyle \left(HJ_{i}\right)^{2}=\left(HE\right)^{2}+x^{2}=\left(r_{1}\right)^{2}+\left(r_{2}\right)^{2}+x^{2}}$ 

Also:

${\displaystyle HJ_{i}=HL_{i}-x=r-x=r_{1}+r_{2}-x~}$ 

Combining these gives:

${\displaystyle \left(r_{1}\right)^{2}+\left(r_{2}\right)^{2}+x^{2}=\left(r_{1}+r_{2}-x\right)^{2}}$ 

Expanding, collecting to one side, and factoring:

${\displaystyle 2r_{1}r_{2}-2x\left(r_{1}+r_{2}\right)=0}$ 

Solving for x:

${\displaystyle x={\frac {r_{1}\cdot r_{2}}{r_{1}+r_{2}}}={\frac {r_{1}\cdot r_{2}}{r}}}$ 

Proving that each of the Archimedes' quadruplets' areas is equal to each of Archimedes' twin circles' areas.^[4]

• Arbelos: Book of Lemmas, Pappus Chain, Archimedean Circle, Archimedes' Quadruplets, Archimedes' Twin Circles, Bankoff Circle, S. ISBN 1156885493