Construct a figure equal in area to the region inside a semicircle and outside two smaller semicircles...

geometry/construction/arbelos New

New entry, edited from material collected for the archive and not yet in a released version.

Problem

Construct a figure equal in area to the region inside a semicircle and
outside two smaller semicircles drawn inside it.

Draw a semicircle on the diameter AB.  Arbitrarily pick a point C on AB.
Inside the semicircle draw two semicircles on diameters AC and BC.  Find
a construction that will give the equivalent area of the internal area
that is contiguous to the three semicircles.

This problem is not in the FAQ, but it is 2440 years old.

-- Edward Cooper

Show solution

Area of a semicircle is proportional to the square of its diameter.
A(semicircle) = 1/2*pi*r^2 = 1/2*pi*(d/2)^2 = 1/2*pi*d^2 / 4.
Then A(semicircle) = pi/8*d^2

The desired area A satisfies:
A = pi/8*[(AB)^2 - (AC)^2 - (BC)^2]
A = pi/8*[(AC + BC)^2 - (AC)^2 - (BC)^2]
A = pi/8*2(AC)(BC)
A = pi/4*(AC)(BC)
A = area of a quarter circle of radius r = sqrt(AC*BC).
We just need to construct r (and draw the quarter circle).

Construct a perpendicular to AB at C that intersects the semicircle at
D.  The triangle ADC is a right triangle with right angle at D.  Then CD
is the mean proportional between AC and BC; that is,

  (CD)^2 = (AC)(BC)

[ Editor: the triangle with the right angle at D is ADB (angle in a
  semicircle); CD is its altitude, which gives (CD)^2 = (AC)(BC). ]

But then r = CD.  We set the width of our compass to CD, put the point
at C, draw an arc from D to AB, and behold the quarter circle with the
desired area has been constructed.

-- Nathaniel Silver <mathelp@worldnet.att.net>

[ Editor: line breaks of the original re-wrapped. ]