Hippocrates of Chios


  1.  A trapezoid is a four-sided figure with one pair of parallel opposite sides, called its bases. (In a parallelogram, both pairs of opposite sides are parallel.) We typically orient a trapezoid so that its bases are horizontal with the longer base below the shorter one, whence the height of the trapezoid is the distance between the bases. Draw a trapezoid, label its longer base B, its shorter base b, and its height h. By dissecting the trapezoid into smaller figures, derive the formula
  1. Evaluate these integrals by sketching graphs of the corresponding regions:




  3. The "lune on the triangle" has an area calculation which is revealed by an argument similar to Hippocrates' for the "lune on the square". Inscribe an equilateral triangle ABC in a circle with center O and build semicircles on the sides of the triangles. This creates three lunes, numbered 1, 2, 3 in the diagram below. Clearly, the three lunes have the same area. Also, let D be opposite end of the diameter of the large circle through B.
    (a) A fundamental result in geometry says that an angle of measure a whose vertex lies on a circle cuts off an arc of the circle whose measure from the center is 2a. For instance, since angle BAD cuts off an arc whose measure from the center of the circle is 180 degrees (namely arc BD), angle BAD measures 90 degrees. What are the measures of the other two angles in triangle BAD? What are the measures of the angles in triangle OAD?
    (b) It follows that AD = (1/2)BD. Why?
    (c) Let x = AB be the length of the side of the equilateral triangle ABC, and let d = BD be the diameter of the large circle. Then part (b) says that AD = d/2. We can apply the Pythagorean Theorem to the right triangle BAD. Show from this that
    (d) It follows that the ratio of the areas of the semicircle on diameter AB to the semicircle on diameter BD is 3/4. Why?
    (e) Use the diagram to argue that, in terms of area,  3(semicircle on AB) + (triangle ABC) = 3(lune) + 2(semicircle on BD)
    and from this show that the area of any one of the numbered lunes = (1/3)(triangle ABC) + (1/12)(semicircle BD) .
  1. Hippocrates is said to have attempted the quadrature of another lune, the "lune on the hexagon", which we repeat here. Inscribe a regular hexagon (regular means that all the sides and interior angles are equal) in a circle centered at O. Drawing in the diagonals that join opposite vertices of the hexagon dissects the hexagon into six triangles at the center C of the circle. On each side of the hexagon build a semicircle. This constructs six lunes, marked 1 through 6 in the diagram.

  2. (a) In triangle ABC, the angle at C is 60 degrees and sides AC = BC. Why?
    (b) It follows that the angles at A and B are equal. What do they measure? Use this to explain why all six triangles are equilateral.
    (c) The semicircles containing lunes 1, 2, 3, and 4 have area equal to the upper half of the large circle. Why?
    (d) The three lunes 1, 2, and 3, together with the semicircle containing lune 4 have total area equal to the upper half of the hexagon (which is a trapezoid). Why?
    (e) It follows that the area of one of these lunes equals one-third the difference between the area of the trapezoid and the area of the semicircle that contains the lune. Why?
Return to the calendar

last modified 8/28/02
Copyright (c) 2000. Daniel E. Otero