278
Chapter 7
O
r
A r
r
b
lk
Figure 7.8. For Matsumiya’s problem.
B
C a
r 2 = a 2 + (r − b)2, or r =(a 2 + b 2)/2b. If we label the lk from the left, then the Pythagorean Theorem also gives r 2 =(ak/8)2 + (r − lk)2, and hence the desired solution ⎛ ak ⎞ lk = r − r 2 − ⎜ ⎟ ⎝ 8⎠
2
(k = 1, 2, . . . , 8).
Problem 9: Katori Zentaro ¯’s Problem Katori’s answer is 280 ⎞ ⎛ S 1 = ⎜ 17 − ⎟ S2. ⎝ 28 − 3π ⎠
To get this result we follow another proof by Yoshida, which requires nothing more than elementary geometry and patience. The plan is to find expressions for t and b (see figure 7.9) in terms of r; this will allow us to compute the required areas. First, draw the auxiliary triangle shown in the figure. Then, by similar triangles, t r (1) = . p x+p But by Pythagoras x 2 + (r − t)2 = (r + t)2.