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[Geometry] Sangaku Three Circles Problem

2021-07-03 14:12 作者:AoiSTZ23  | 我要投稿

By: Tao Steven Zheng (郑涛)

【Problem】

Figure 1 shows a famous Japanese sangaku (算額) problem from Gunma prefecture in 1824. Suppose the radius of the left circle is %7Br%7D_%7B1%7D, the radius of the right circle is %7Br%7D_%7B2%7D , and the radius of the circle at the center is %7Br%7D_%7B3%7D, determine the formula for the radius of center circle.

Figure 1

【Solution】

Let  %7Br%7D_%7B1%7D%2C%20%7Br%7D_%7B2%7D%2C%20%7Br%7D_%7B3%7D represent the radius of the left circle, right circle, and center circle respectively. According to Figure 2, AG%3D%7Br%7D_%7B1%7DBH%3D%7Br%7D_%7B2%7D and OC%3D%7Br%7D_%7B3%7D. Furthermore, AE%3D%7Br%7D_%7B1%7D-%7Br%7D_%7B3%7D, BF%3D%7Br%7D_%7B2%7D-%7Br%7D_%7B3%7D, BD%3D%7Br%7D_%7B2%7D-%7Br%7D_%7B1%7D and AB%3D%7Br%7D_%7B1%7D%2B%7Br%7D_%7B2%7D. Let EO%3Dx and FO%3Dy, then AD%3Dx%2By.

 

Figure 2

Applying the Pythagorean theorem, we get

x%5E2%2B(%7Br%7D_%7B1%7D-%7Br%7D_%7B3%7D)%5E2%3D(%7Br%7D_%7B1%7D%2B%7Br%7D_%7B3%7D)%5E2

y%5E2%2B(%7Br%7D_%7B2%7D-%7Br%7D_%7B3%7D)%5E2%3D(%7Br%7D_%7B2%7D%2B%7Br%7D_%7B3%7D)%5E2%0A

(x%2By)%5E2%2B(%7Br%7D_%7B2%7D-%7Br%7D_%7B1%7D)%5E2%3D(%7Br%7D_%7B1%7D%2B%7Br%7D_%7B2%7D)%5E2%0A

Simplify the above three equations as

x%5E2%3D4%7Br%7D_%7B1%7D%20%7Br%7D_%7B3%7D%0A%0A

y%5E2%3D4%7Br%7D_%7B2%7D%20%7Br%7D_%7B3%7D%0A%0A

(x%2By)%5E2%3D4%7Br%7D_%7B1%7D%20%7Br%7D_%7B2%7D%0A

Taking the positive square roots give

x%3D2%5Csqrt%7B%7Br%7D_%7B1%7D%20%7Br%7D_%7B3%7D%7D%0A%0A

y%3D2%5Csqrt%7B%7Br%7D_%7B2%7D%20%7Br%7D_%7B3%7D%7D%0A%0A

x%2By%3D2%5Csqrt%7B%7Br%7D_%7B1%7D%20%7Br%7D_%7B2%7D%7D%0A


Hence,

%5Csqrt%7B%7Br%7D_%7B1%7D%20%7Br%7D_%7B3%7D%7D%2B%5Csqrt%7B%7Br%7D_%7B2%7D%20%7Br%7D_%7B3%7D%7D%3D%5Csqrt%7B%7Br%7D_%7B1%7D%20%7Br%7D_%7B2%7D%7D%0A%0A

%5Csqrt%7B%7Br%7D_%7B3%7D%7D%20%20(%5Csqrt%7B%7Br%7D_%7B1%7D%7D%2B%5Csqrt%7B%7Br%7D_%7B2%7D%7D)%3D%5Csqrt%7B%7Br%7D_%7B1%7D%20%7Br%7D_%7B2%7D%7D%0A%0A

%5Csqrt%7B%7Br%7D_%7B3%7D%7D%3D%5Cfrac%7B%5Csqrt%7B%7Br%7D_%7B1%7D%20%7Br%7D_%7B2%7D%7D%7D%7B%5Csqrt%7B%7Br%7D_%7B1%7D%7D%2B%5Csqrt%7B%7Br%7D_%7B2%7D%7D%7D%0A

 Therefore,

%7Br%7D_%7B3%7D%3D%5Cfrac%7B%7Br%7D_%7B1%7D%20%7Br%7D_%7B2%7D%7D%7B(%5Csqrt%7B%7Br%7D_%7B1%7D%7D%2B%5Csqrt%7B%7Br%7D_%7B2%7D%7D)%5E2%7D



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