Archimedes' quadruplets

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Each of the Archimedes' quadruplets (green) have equal area to each other and to Archimedes' twin circles
Each of the Archimedes' quadruplets (green) have equal area to each other and to Archimedes' twin circles

In geometry, Archimedes' quadruplets, introduced by Frank Power in the summer of 1998, each have the same area as Archimedes' twin circles.[1]

[edit] Construction

Three semicircles are created with the diameters of AB, AC, and BC. The two smaller circles have radii of r1 and r2, meaning the larger semicircle has a radius that satisfies the equation: 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.

[edit] Proof of congruency

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

\frac{r_1\cdot r_2}{r}.

By the Pythagorean theorem:

\left(HE\right)^2=\left(r_1\right)^2+\left(r_2\right)^2.

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

\left(HJ_i\right)^2=\left(HE\right)^2+x^2=\left(r_1\right)^2+\left(r_2\right)^2+x^2

Also:

HJ_i=HL_i-x=r-x=r_1+r_2-x~

Combinding these gives:

\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:

2r_1r_2-2x\left(r_1+r_2\right)=0

Solving for x:

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.[2]

[edit] References

  1. ^ Power, Frank (2005), “Some More Archimedean Circles in the Arbelos”, in Yiu, Paul, Forum Geometricorum, vol. 5, 2005-11-02, pp. 133-134, ISBN 1534-1178, <http://forumgeom.fau.edu/FG2005volume5/FG200517.ps>. Retrieved on 13 April 2008 
  2. ^ Bogomolny, Alexander. Archimedes' Quadruplets. Retrieved on 2008-04-13.
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