Archimedes' quadruplets

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