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Travel Diary of Mathematician Yamaguchi Kanzan
a = 2r
B
A
r
r
b q
p h
r
t
t C
x t p
Figure 7.9. For Katori Zentaro¯’s problem.
D
Squaring this equation and solving for x gives x = 2 r t , and therefore from equation (1) t r = . p 2 rt + p Solving for p yields p=
2t rt . r −t
(2)
Convince yourself that the marked angles are the same. [Hint: It may help to drop a perpendicular from the center of the large circle to the hypotenuse of the triangle.] Then once more, by similar triangles, q rt t (3) = or q = . r p p Since the two sides BC and AD are equal, we have t + p + q + r = p + 2 r t + r , or, from the above expression for q, 2p r t = pt + rt .
Eliminating p with equation (2) gives after a few steps 5rt − r 2 = 2t r t .
Squaring both sides yields the cubic equation r 3 − 10r 2t + 25rt 2 − 4t 3 = 0,