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.
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.
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?
Return to the exact question
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.
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.
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