The Arbelos — A Curved Sliver Equal to a Circle
Carve two small semicircles out of a big one and a knife-shaped “arbelos” remains. Bounded by three arcs, its area equals exactly one round circle — a figure Archimedes loved.
Problem
Find the area of the arbelos (big semicircle minus two small ones on AC and CB) for a point C on diameter AB.
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Diameter AB, point C, two small semicircles on AC and CB. Big minus the two small = arbelos. Area?
Why this approach works
Three arcs, but the area is just add-and-subtract of semicircles: arbelos = ½π(R² − r₁² − r₂²). Since R = r₁+r₂, that is πr₁r₂. Now erect a perpendicular at C meeting the big arc at D. By Thales ∠ADB = 90°, so CD is the altitude of a right triangle and CD² = AC·CB = 4r₁r₂. The circle on diameter CD has area π(CD/2)² = πr₁r₂ — exactly the arbelos.
Proof
Arbelos = ½π(R²−r₁²−r₂²) = πr₁r₂ since R=r₁+r₂. With D where the perpendicular at C meets the big arc, ∠ADB=90° (Thales) gives CD²=AC·CB=4r₁r₂, so the circle on CD has area π(CD/2)²=πr₁r₂ — the arbelos.
Try it yourself
Slide C to the middle (r₁=r₂=R/2): the arbelos is largest, πr₁r₂ = π(R/2)², and CD becomes the big radius.
Related mathematics
Beyond MathVoyage
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