Arbelos  the Shoemaker's Knife
Problem
Two smaller circles are outside each other, but inside a third, larger circle. Each of the three circles is tangent to the other two and their centers are along the same straight line. Given r the radius of the larger circle, and $t,$ the piece of the tangent to the two smaller circles at their common point enclosed by the larger circle. Find the area that is inside the larger circle but outside the two smaller ones.
Solution
Let $r_{1}$ and $r_{2}$ denote the radii of the two small circles. Obviously, $r = r_{1} + r_{2}.$ Also, the chord being perpendicular to the diameter, $2r_{1} \cdot 2r_{2} = (t/2)^2.$ Putting everything together yields the sought area
$\begin{align} S &= \pi r^2  \pi r_{1}^2  \pi r_{2}^2\\ &=\pi ((r_{1} + r_{2})^2  r_{1}^2  r_{2}^2)\\ &=\pi(2r_{1}r_{2})\\ &=\pi (2t^2/16) \end{align}$
Finally, $S = \pi t^2/8$ independent of either $r,$ $r_{1},$ or $r_{2}.$
Archimedes called one half (upper or lower) of this shape arbelos which literally means "a shoemaker's knife." The property we just proved appears as Proposition 4 in his Book of Lemmas:
If $AB$ be the diameter of a semicircle and $N$ any point on $AB,\;$ and if semicircles be described within the first semicircle and having $AN,\;$ $BN$ as diameters respectively, the figure included between the circumferences of the three semicircles is "what Archimedes called arbelos"; and its area is equal to the circle on $PN$ as diameter, where $PN$ is perpendicular to $AB$ and meets the original semicircle in $P.$
The circle in question is circumscribed around a rectangle characteristically formed in $AN$ arbelos. (An attempt to present a dynamic proof without words can be found elsewhere.)
Archimedes devoted two more Propositions to the study of arbelos that kept attracting mathematicians through the ensuing centuries. L.Bankoff lists several wonderful properties of arbelos and pertinent bibliography.
The following must be obvious:
Of the three arcs that form the boundary of $AN$ arbelos, the larger one is the sum of the two small arcs.
(For a generalization of this statement see An Isoperimetric Theorem. See also Chain of Inscribed Circles. The said circles are inscribed in $AN$ arbelos.)
26 November 2015, Created with GeoGebra
Reference
 L. Bankoff, The Marvelous Arbelos, in The Lighter Side of Mathematics, R.K.Guy and R.E.Woodrow, eds, MAA, 1994
 G. Polya, Mathematical Discovery, John Wiley & Sons, 1981
 Great Books of The Western World, v. 11, Encyclopaedia Britannica, 1952
On the Web
 H. $P.\;$ Boas, Reflections on the Arbelos, a pdf file https://www.math.tamu.edu/~harold.boas/preprints/arbelos.pdf
 T. Schoch, Arbelos: Amaizing Properties
 E. W. Weisstein, Arbelos, From MathWorldA Wolfram Web Resource.
https://mathworld.wolfram.com/Arbelos.html

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Archimedes' Book of Lemmas
Proposition 1: If two circles touch at $A,$ and if $CD,$ $EF$ be parallel diameters in them, $ADF$ is a straight line.
Proposition 2: Let $AB$ be the diameter of a semicircle, and let the tangents to it at $B$ and at any other point $D$ on it meet in $T.$ If now $DE$ be drawn perpendicular to $AB,\;$ and if $AT,$ $DE$ meet in $F,$ then $DF = FE.$.
Proposition 3: Let $P$ be any point on a segment of a circle whose base is $AB,\;$ and let $PN$ be perpendicular to $AB.$ Take $D\;$ on $AB$ so that $AN = ND.$ If now PQ be $AN$ arc equal to the arc PA, and BQ be joined, then BQ, BD shall be equal.
Proposition 4: If $AB$ be the diameter of a semicircle and $N$ any point on $AB,\;$ and if semicircles be described within the first semicircle and having $AN,\;$ $BN$ as diameters respectively, the figure included between the circumferences of the three semicircles is "what Archimedes called arbelos"; and its area is equal to the circle on $PN$ as diameter, where $PN$ is perpendicular to $AB$ and meets the original semicircle in $P$.
Proposition 5: Let $AB$ be the diameter of a semicircle, $C\;$ any point on $AB,\;$ and $CD\;$ perpendicular to it, and let semicircles be described within the first semicircle and having $AC,\;$ CB as diameters. Then if two circles be drawn touching $CD\;$ on different sides and each touching two of the semicircles, the circles so drawn will be equal.
Proposition 6: Let $AB,\;$ the diameter of a semicircle, be divided at $C\;$ so that $AC = 3/2\cdot CB\;$ [or in any ratio]. Describe semicircles within the first semicircle and on $AC,\;$ $CB\;$ as diameters, and suppose a circle drawn touching the all three semicircles. If $GH$ be the diameter of this circle, to find relation between $GH$ and $AB$.
Proposition 7: If circles are circumscribed about and inscribed in a square, the circumscribed circle is double of the inscribed square..
Proposition 8: If $AB$ be any chord of a circle whose centre is $O,$ and if $AB$ be produced to $C\;$ so that $BC$ is equal to the radius; if further $CO$ meets the circle in $D\;$ and be produced to meet the circle the second time in $E,$ the arc $AE$ will be equal to three times the arc $BD$.
Proposition 9: If in a circle two chords $AB,\;$ $CD\;$ which do not pass through the centre intersect at right angles, then $(arc AD) + (arc CB) = (arc AC) + (arc DB)$.
Proposition 10: Suppose that $TA,$ $TB$ are two tangents to a circle, while $TC$ cuts it. Let $BD$ be the chord through $B$ parallel to $TC,$ and let $AD$ meet $TC$ in $E.$ Then, if $EH$ be drawn perpendicular to $BD,$ it will bisect it in $H.$
Proposition 11: If two chords $AB,\;$ $CD\;$ in a circle intersect at right angles in a point $O,$ not being the centre, then $AO^2 + BO^2 + CO^2 + DO^2 = (diameter)^2.$
Proposition 12: If $AB$ be the diameter of a semicircle, and $TP,$ $TQ$ the tangents to it from any point $T,$ and if $AQ,$ $BP$ be joined meeting in $R,$ then $TR$ is perpendicular to $AB.$
Proposition 13: If a diameter $AB$ of a circle meet any chord $CD,$ not a diameter, in $E,$ and if $AM,$ $BN$ be drawn perpendicular to $CD,$ then $CN = DM.$
Proposition 14: Let $ACB$ be a semicircle on $AB$ as diameter, and let $AD,$ $BE$ be equal lengths measured along $AB$ from $A,$ $B$ respectively. On $AD,$ $BE$ as diameters describe semicircles on the side towards $C,$ and on $DE$ as diameter a semicircle on the opposite side. Let the perpendicular to $AB$ through $O,$ the centre of the first semicircle, meet the opposite semicircles in $C,$ $F$ respectively. Then shall the area of the figure bounded by the circumferences of all the semicircles be equal to the area of the circle on $CF$ as diameter.
Proposition 15: Let $AB$ be the diameter of a circle, $AC$ a side of $AN$ inscribed regular pentagon, $D\;$ the middle point of the arc $AC.$ Join $CD\;$ and produce it to meet $BA$ produced in $E;$ join $AC,\;$ $DB$ meeting in $F,$ and draw $FM$ perpendicular to $AB.$ Then $EM = (radius of circle).$
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