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Chapter 6
From any of the figures, the relevant root is k = 4r, and thus we have the final result r=
CH . 4
Problem 4 ¯ mura Kazuhide (1824–1891), from his 1841 This solution is by O book Sanp¯o Tenzan Tebikigusa, or Algebraic Methods in Geometry. First we draw figure 6.27. The radii of circles O, O 1, O 2, and O 3 are r, r 1, r 2, and r 3, respectively. The basic approach will be to calculate a number of relevant areas as functions of the four radii, then eliminate the areas to get r in terms of r 1, r 2, and r 3. To do this, start by the same method employed in many of this book’s problems (e.g., chapter 3, problem 13) to find that the lengths are GD = 2 r1r3 ;
D ′E ′ = 2 r1r2 ;
EF = 2 r2r3 .
(1)
By using the equality of tangents from vertices A, B, and C to their respective inscribed circles, one also sees that E′C′ = EA′, GB′ = FA′, etc. With these one can establish GD + D′E′ − EF = 2B′D; D′E′ + EF − GD = 2C′E′; GD + EF − D′E′ = 2GB′. (2)
The area of the quadrilaterals formed by the auxiliary lines can easily be calculated by the trapezoid formula and equation (1): 1 (r1 + r3 )GD = (r1 + r3 ) r1r3 , 2 1 D ′O1O 2 E ′ = (r1 + r2 )D ′E ′ = (r1 + r2 ) r1r2 , 2 1 EO 2O 3 F = (r2 + r3 )EF = (r2 + r3 ) r2r3 . 2 GO 3O1D =
(3)
A D' C'
D O1
E'
B'
O G
Figure 6.27. Mark the important points and draw the auxiliary lines shown.
O2 B
E
O3 A'
F
C