Three semicircles and a circle all touching each other. The diameters of the two smallest semicircles are given. The large triangle connects three tangency points. What is the area of the green quadrilateral?
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Solution
The green quadrilateral is a square with area 36/13.
2 replies on “The green quad”
The diagram is basically that in Archimedes’ “Book of Lemmas”, Proposition 6, where the author proves that the quadrilateral is a rectangle, but does not point out that it is really a square. The purpose of his Proposition 6 is to arrive at the formula for the radius r of the inscribed circle in terms of the radii a, b of the small semicircles. Archimedes finds r= ab(a+b)/(a^2+ab+b^2).
A modern proof of Archimedes’ result would use the law of cosines twice, as in Peter Gallin’s solution above, and would find that Cosine(alpha) =
(a^2-b^2)/(a^2+b^2), giving 2a(a+b)/sqrt(a^2+b^2) for the left side length of the big right triangle, and 2b(a+b)/sqrt(a^2+b^2) for the right side length. The ration a:b for these side lengths easily translates (using similarity of all the right triangles) to a ratio 1:1 for the side lengths of the rectangle/square in question.
A natural question to ask: is there a simple direct proof of the square property that avoids the algebra needed to determine the formula for the cosine of alpha?
Since the procedure to construct the diagram is left-right symmetrical, so must be the ratio of green rectangle sides. It can only be 1, otherwise one would be able discern a preferred side.