## Tuesday, December 16, 2008

### Archimedes' Book of Lemmas, Proposition #4, Arbelos

Problem 643
Exercise your brain. Archimedes wrote the "Book of Lemmas" more than 2200 years ago. Solve the proposition #4 (high school level) and lift up your geometry skills.

gogeometry.com/ArchBooLem04.htm

1. bjhopper bjhvash44@sbcglobal.netMarch 25, 2010 at 4:26 PM

AB dia of largest semicircle = 1
AD dia of smallest semicircle = F fraction of AB
DB dia of intermediate semicircle = 1-F

Arbelo area pi /8 - pi/8 . F^2 -pi/8.(1-F)^2
This simplifies to pi/4(F-F^2)
call CD a
ACB is a right triangle and F/a= a/1-F
a^2=F-F^2
area of cirle dia a=pi/4 x( F-F^2) = to arbelo

2. AD^2 + DB^2 = AC^2 + BC^2 - 2 CD^2
= AB^2 - 2 CD^2
Area of arbelos = Area of semicircle ACBA
- Area of semicircle on AD as diameter
- Area of semicircle on DB as diameter
= (Pi/8) (AB^2 - AD^2 - DB^2)
= (Pi/4) CD^2
= Area of circle on CD as diameter

3. Area of Arbelos = piAB^2/8- piAD^2/8 - piDB^2/8 = piAD.DB/4 = piCD^2/4 = Area of circle CD

Sumith Peiris
Moratuwa
Sri Lanka