CONSTRUCTION
Construct a circle with center O and radius R. Construct a diameter EF of circle O. Construct the chord AC of circle O that is perpendicular to EF at point G (with |GE| = r). Construct a circle with center F that passes through point A. Construct a circle with center A and radius 2|AC|. Pick either of the points in the intersection of circles A and F and label it point D. Line AD intersects circle O at point B. The desired triangle is ABC.
PROOF
If a, b, and c are the side lengths of any triangle ABC, then the circumradius and inradius are given by
R = abc/sqrt[(a+b+c)(b+c-a)(c+a-b)(a+b-c)] (1)
r = abc/2R(a+b+c) (2)
If the side lengths of triangle ABC are in arithmetic progression and b is the middle length, then
2b = a + c (3)
Eliminating a and c from (1), (2), and (3) gives
b = 2 sqrt[r(2R - r)] (4)
From the construction,
b = 2 sqrt[R2 - (R - r)2]
which agrees with (4).
Now all we have to do is show that 2b = a + c in the construction.
<BCD = <ABC - <BDC = <AFC - <BDC = 2 <ADC - <BDC
= 2 <BDC - <BDC = <BDC
==> |BD| = |BC|
2b = 2|AC| = |AD| = |AB| + |BD| = |AB| + |BC| = c + a.
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