The Lune of Hippocrates — A Curved Area That Equals a Triangle
A crescent bounded by two arcs — yet its area exactly equals that of a triangle. It is among the earliest surviving exact quadratures of a curvilinear figure.
Problem
Inscribe a right isosceles triangle ACB (right angle at C) in a semicircle with hypotenuse AB as diameter. Draw a small semicircle outward on leg AC; a lune appears between it and the large arc AC. Find the area of the lune.
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Inscribe a right isosceles triangle ACB (right angle at C) in the semicircle on AB. A small semicircle on AC creates a lune. Its area?
Why this approach works
Two curved boundaries suggest calculus, but again it is add-and-subtract accounting. The lune = small semicircle − the segment the big circle cuts along chord AC. Small semicircle = ½πr² with r = AC/2; segment = sector − triangle = ¼πR² − ½R². Since AC subtends the right angle at C it cuts a 90° arc, and r = R/√2 makes the small semicircle ¼πR² as well. So lune = ¼πR² − (¼πR² − ½R²) = ½R². The π cancels entirely — the curved lune equals triangle AOC, exactly half of triangle ACB.
Proof
Subtract the segment cut by chord AC from the small semicircle. The right-isosceles condition makes the two π·R²/4 terms cancel exactly, so the curved lune equals the rectilinear triangle AOC — the first exact quadrature of a curved figure in history.
Try it yourself
Tilt the triangle flatter and it is no longer right-isosceles — the π no longer cancels cleanly. The magic hinges on that exact condition, which is why squaring a general lune stays hard.
Related mathematics
Beyond MathVoyage
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