Hippocrates' Lune: Prove the Lune Equals the Isosceles Right Triangle (Age 13)

Step-by-step proof (age 13) that the lune formed from a circle and a semicircle on the hypotenuse of an isosceles right triangle has the same area as that triangle. Uses circle sectors, triangle area, and simple algebra — suitable for AoPS Prealgebra/Intro Geometry students.

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Goal

Show that the lune (the part of the semicircle on the triangle's hypotenuse lying outside the original circle) has the same area as the isosceles right triangle whose right angle is at the circle's centre.

Setup and diagram instructions

  1. Draw a circle with centre O and radius r.
  2. Choose points A and B on the circle so that triangle OAB is isosceles and right-angled at O (so OA = OB = r and angle AOB = 90°).
  3. Draw the semicircle with diameter AB on the side opposite O. The part of that semicircle outside the original circle is the lune.

Step-by-step proof

  1. Area of triangle OAB. Because OA and OB are perpendicular legs of length r, the triangle area is
    Area_triangle = (1/2)·OA·OB = (1/2)·r·r = r^2/2.
  2. Area of the quarter-sector of the original circle cut off by chord AB. The central angle AOB = 90°, so this sector is one quarter of the whole circle:
    Area_sector = (1/4)·π·r^2 = π r^2 / 4.
  3. Area of the circular segment inside the circle and bounded by chord AB. The segment = sector − triangle (the triangular part under the chord):
    Area_segment = Area_sector − Area_triangle = (π r^2 / 4) − (r^2 / 2).
  4. Area of the semicircle on AB. First find AB. Since OA and OB are legs of the right triangle, AB is the hypotenuse:
    AB = r·√2. The radius of the semicircle is (AB)/2 = r·√2 / 2 = r / √2.
    Area_semicircle = (1/2)·π·(radius)^2 = (1/2)·π·(r^2/2) = π r^2 / 4.
  5. Area of the lune. The lune is the part of the semicircle that lies outside the original circle, so
    Area_lune = Area_semicircle − Area_segment. Substitute the expressions:
    Area_lune = (π r^2 / 4) − [ (π r^2 / 4) − (r^2 / 2) ] = r^2 / 2.
  6. Compare: Area_lune = r^2 / 2 = Area_triangle. Therefore the lune and the isosceles right triangle have exactly the same area.

Conclusion

By calculating sector, segment and semicircle areas and subtracting appropriately, we find the lune's area equals the triangle's area: r^2 / 2. This is a classical result (one of Hippocrates' lunes).

ACARA v9 mapping (Mathematics — Measurement & Geometry)

  • Investigate properties of circles: sectors, segments, chords and arc relationships.
  • Apply area formulas for triangles and circles and reason algebraically to compare areas.
  • Develop geometric proof skills: decompose shapes, compute areas, and justify equalities.
  • Suggested Year level: ~Year 8–9 students (age ~13), extension for able students in Year 9–10.

Teacher comments and rubric (100 words, Tiger-Mother cadence)

You must do this perfectly. I expect flawless reasoning, precise diagrams, and exact arithmetic. Construct the circle with centre O and radius r, draw isosceles right triangle OAB with right angle at O and A,B on circle, then semicircle on AB. Prove area of lune equals area of triangle by computing triangle area (r^2/2), sector area (π r^2/4), semicircle area (π r^2/4) so difference gives r^2/2. Rubric: 3 Excellent—complete proof, neat diagram, correct algebra; 2 Good—correct idea but minor slips; 1 Needs Work—incomplete, unclear. Redo until perfect. No excuses; practise until it is memorised and flawless under test conditions. Now.

Additional enrichment and extensions

  • Explore other Hippocrates lunes: which combinations of semicircles and circles give lunes equal to simple polygon areas?
  • Generalise: try replacing the isosceles right triangle with other isosceles triangles — when (if ever) do analogous equalities hold?
  • Coordinate proof: place O at (0,0) and A,B on unit circle at angles ±45°, then use integrals or area formulas to verify the result algebraically.
  • Challenge: find all right triangles with vertex at the circle centre for which the semicircle on the hypotenuse produces a lune whose area equals the triangle area (investigate angles other than 90°).

Quick checklist for students

  1. Draw clearly and label O, A, B, r and AB.
  2. Compute triangle area correctly: (1/2)·r·r = r^2/2.
  3. Compute sector and semicircle areas and perform the subtractions step-by-step.
  4. Write a concluding sentence equating the two areas.

If you want, I can create a printable worksheet with the figure and guided steps, or give graded practice problems that build to this theorem.

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