217
Still Harder Temple Geometry Problems
H1
r
O
O
H3
O
H4
O2
O3
O3
O1
r3
G
H2
r2
r1
B'
DD'
C'
E'E
r3
A'
Figure 6.28. The central figure in the triangle cut out, unfolded, and joined together such that the end points F and G match, as well as intermediate points DD′ and EE′. The points Hi are the endpoints (“perpendicular feet”) of the radii of each circle, e.g., GO 3, extended to the height of O.
F
We now, hopefully, simplify by drawing a second diagram, figure 6.28, which represents the central figure (EFO 3GDO 1D′E′O 2) cut out from the triangle and unfolded. (In other words, this is the figure that remains after cutting away the three quadrilaterals at the vertices of the triangle.) The points H1, . . . , H4 are the points at which extensions of the radii GO 3 etc. reach the height O From figure 6.28 and equations (1) and (2) we see that the areas of the missing triangles are 1 1 (r − r3 )GB ′ = (r − r3 )( r1r3 + r2r3 − r1r2 ), 2 2 1 1 ΔOO1H 2 = (r − r1)B ′D = (r − r1)( r1r3 + r1r2 − r2r3 ), 2 2 1 1 ΔO 2 H 2O = (r − r2 )C ′E = (r − r2 )( r1r2 + r2r3 − r1r3 ). 2 2 ΔO 3 H 1O =
(4)
It is now convenient to employ Heron’s formula, which gives the area of a triangle in terms of its sides and semiperimeter5 : Area Δ = s(s − a)(s − b )(s − c ) =
1 (a + b + c )(a + b − c )(a + c − b )(b + c − a ). 4
Focusing on the central triangle O 1O 2O 3, we have a = r 1 + r 2, b = r 1 + r 3, c = r 2 + r 3, and get for the area ΔO1O2O3 = (r1 + r2 + r3 )(r1 r2 r3 ) = ΔOO1O2 + ΔOO1O3 + ΔOO2O3 ,
(5)
where the last three triangles are shown on figure 6.28. Of course, the area of the big rectangle GH1H4 F in figure 6.28 is just rGF, which with the help of equation (1) is GH1 H4 F = 2 r ( r1 r3 + r1 r2 + r2 r3 ).
(6)
5 See problem 1, this chapter. Heron’s formula is discussed in most elementary geometry texts and on many websites.