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.
Do not write an equation yet. Shake the figure first.
Play the reversal slowly
Now move the same pieces step by step and prove that the pattern you touched was not an accident.
Diameter AB, point C, two small semicircles on AC and CB. Big minus the two small = arbelos. Area?
Return to the exact question
Find the area of the arbelos (big semicircle minus two small ones on AC and CB) for a point C on diameter AB.
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
Curated sources and problems. Bring one discovery back from OEIS, Project Euler, MathOverflow, or arXiv.
- Wikipedia
- Wolfram MathWorld