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[Geometry] Descartes Four Circle Theorem

2021-08-21 13:58 作者:AoiSTZ23  | 我要投稿

By: Tao Steven Zheng(鄭濤)

Rene Descartes (1596 - 1650) found the theorem without proof and sent it to Princess Elizabeth of Bohemia in 1643. The theorem was first proven by Jakob Steiner (1796-1863) nearly 200 years later in 1826. In Edo Japan, this theorem first appeared in a sangaku problem from 1796. The theorem was explained in Hashimoto Masataka's book ''Sanpo Tenzan Shogakusho'' (筭法點(diǎn)竄初學(xué)抄) in 1830.

【Problem】

Suppose there are four circles?A%2C%20B%2C%20C%2C%20D lying in a plane, such that each circle is mutually tangent to the other three circles. Let the radius of each circle be a%2C%20b%2C%20c%2C%20d. Prove that

%202%5Cleft(%5Cfrac%7B1%7D%7Ba%5E2%7D%20%2B%20%5Cfrac%7B1%7D%7Bb%5E2%7D%20%2B%20%5Cfrac%7B1%7D%7Bc%5E2%7D%20%2B%20%5Cfrac%7B1%7D%7Bd%5E2%7D%20%5Cright)%3D%20%7B%5Cleft(%5Cfrac%7B1%7D%7Ba%7D%20%2B%20%5Cfrac%7B1%7D%7Bb%7D%20%2B%20%5Cfrac%7B1%7D%7Bc%7D%20%2B%20%5Cfrac%7B1%7D%7Bd%7D%20%5Cright)%7D%5E%7B2%7D


Hint: Given three angles?%5Ctheta%2C%20%5Cphi%2C%20%5Cpsi where %20%5Ctheta%20%2B%20%5Cphi%20%2B%20%5Cpsi%20%3D%202%5Cpi, then %5Ccos%5E2%7B%5Ctheta%7D%20%2B%20%5Ccos%5E2%7B%5Cphi%7D%20%2B%20%5Ccos%5E2%7B%5Cpsi%7D%20%3D%202%5Ccos%7B%5Ctheta%7D%5Ccos%7B%5Cphi%7D%5Ccos%7B%5Cpsi%7D%20%2B%201.


【Solution】


Let AB%20%3D%20a%20%2B%20b, AC%20%3D%20a%20%2B%20c%20, AD%20%3D%20a%20%2B%20d,? BC%20%3D%20b%20%2B%20c%20,? BD%20%3D%20b%20%2B%20d, and CD%20%3D%20c%20%2B%20d.
?
Furthermore, let there be three angles %5Ctheta%20%3D%20%5Cangle%20ADB, %5Cphi%20%3D%20%5Cangle%20BDC, and %5Cpsi%20%3D%20%5Cangle%20CDA.

By applying the cosine law on triangle ADB, we get

%7BAB%7D%5E%7B2%7D%3D%20%7BAD%7D%5E%7B2%7D%20%2B%20%7BBD%7D%5E%7B2%7D%20-%202%20AB%20%5Ccdot%20BD%20%5Ccos%7B%5Ctheta%7D%20


%7B%5Cleft(a%20%2B%20b%20%5Cright)%7D%5E%7B2%7D%20%3D%20%7B%5Cleft(a%20%2B%20d%20%5Cright)%7D%5E%7B2%7D%20%2B%20%7B%5Cleft(b%20%2B%20d%20%5Cright)%7D%5E%7B2%7D%20-%202%20%5Cleft(a%20%2B%20d%20%5Cright)%20%5Cleft(b%20%2B%20d%20%5Cright)%20%5Ccos%7B%5Ctheta%7D


Expanding each bracket and rearranging yields

ab%20%3D%20d%5E2%20%2B%20ad%20%2B%20bd%20-%20%5Cleft(a%20%2B%20d%20%5Cright)%20%5Cleft(b%20%2B%20d%20%5Cright)%20%5Ccos%7B%5Ctheta%7D

%5Cleft(a%20%2B%20d%20%5Cright)%20%5Cleft(b%20%2B%20d%20%5Cright)%20%5Ccos%7B%5Ctheta%7D%20%3D%20d%5E2%20%2B%20ad%20%2B%20bd%20-%20ab

%5Cleft(a%20%2B%20d%20%5Cright)%20%5Cleft(b%20%2B%20d%20%5Cright)%20%5Ccos%7B%5Ctheta%7D%20%3D%20d%5E2%20%2B%20ad%20%2B%20bd%20%2B%20%5Cleft(ab%20-%20ab%20%5Cright)%20-%20ab

%5Cleft(a%20%2B%20d%20%5Cright)%20%5Cleft(b%20%2B%20d%20%5Cright)%20%5Ccos%7B%5Ctheta%7D%20%3D%20%5Cleft(a%20%2B%20d%20%5Cright)%20%5Cleft(b%20%2B%20d%20%5Cright)%20-%202ab


Solving for?%5Ccos%7B%5Ctheta%7D yields

%5Ccos%7B%5Ctheta%7D%20%3D%201%20-%20%5Cfrac%7B2ab%7D%7B%5Cleft(a%20%2B%20d%20%5Cright)%20%5Cleft(b%20%2B%20d%20%5Cright)%7D


Using the same method, we can also obtain

%5Ccos%7B%5Cphi%7D%20%3D%201%20-%20%5Cfrac%7B2bc%7D%7B%5Cleft(b%20%2B%20d%20%5Cright)%20%5Cleft(c%20%2B%20d%20%5Cright)%7D

%5Ccos%7B%5Cpsi%7D%20%3D%201%20-%20%5Cfrac%7B2ac%7D%7B%5Cleft(a%20%2B%20d%20%5Cright)%20%5Cleft(c%20%2B%20d%20%5Cright)%7D%20

Let

x%20%3D%5Cfrac%7B2ab%7D%7B%5Cleft(a%20%2B%20d%20%5Cright)%20%5Cleft(b%20%2B%20d%20%5Cright)%7D%20

y%20%3D%20%5Cfrac%7B2bc%7D%7B%5Cleft(b%20%2B%20d%20%5Cright)%20%5Cleft(c%20%2B%20d%20%5Cright)%7D%20

z%20%3D%20%5Cfrac%7B2ac%7D%7B%5Cleft(a%20%2B%20d%20%5Cright)%20%5Cleft(c%20%2B%20d%20%5Cright)%7D


Using the trigonometric identity from the hint gives

%7B%5Cleft(1-x%20%5Cright)%7D%5E%7B2%7D%20%2B%20%7B%5Cleft(1-y%20%5Cright)%7D%5E%7B2%7D%20%2B%20%7B%5Cleft(1-z%20%5Cright)%7D%5E%7B2%7D%20%3D%202%5Cleft(1-x%20%5Cright)%20%5Cleft(1-y%20%5Cright)%20%5Cleft(1-z%20%5Cright)%20%2B%201

Expanding and rearranging the above equation yields

3%20-2%5Cleft(x%2By%2Bz%5Cright)%20%2B%20x%5E2%20%2B%20y%5E2%20%2B%20z%5E2%20%3D%203%20%2B%202%5Cleft(xy%2Bxz%2Byx%20-(x%2By%2Bx)%20-%20xyz%20%5Cright)%20


x%5E2%20%2B%20y%5E2%20%2B%20z%5E2%20%2B%202xyz%20%3D%202%5Cleft(xy%2Bxz%2Byz%5Cright)

Divide both sides of the equation by xyz:

%20%5Cfrac%7Bx%7D%7Byz%7D%20%2B%20%5Cfrac%7By%7D%7Bxz%7D%20%2B%5Cfrac%7Bz%7D%7Bxy%7D%20%2B%202%20%3D%202%5Cleft(%5Cfrac%7B1%7D%7Bx%7D%20%2B%20%5Cfrac%7B1%7D%7By%7D%20%2B%20%5Cfrac%7B1%7D%7Bz%7D%5Cright)%20

Replace the variables for?%20x%2C%20y%2C%20z? in terms of?%20a%2C%20b%2C%20c%2C%20d:


%20%5Cfrac%7B%7B(c%2Bd)%7D%5E%7B2%7D%7D%7B2%7Bc%7D%5E%7B2%7D%7D%20%2B%5Cfrac%7B%7B(a%2Bd)%7D%5E%7B2%7D%7D%7B2%7Ba%7D%5E%7B2%7D%7D%20%2B%5Cfrac%7B%7B(b%2Bd)%7D%5E%7B2%7D%7D%7B2%7Bb%7D%5E%7B2%7D%7D%20%2B%202%20%3D%202%5Cleft%5B%5Cfrac%7B(a%2Bd)(b%2Bd)%7D%7B2ab%7D%20%2B%20%5Cfrac%7B(b%2Bd)(c%2Bd)%7D%7B2bc%7D%20%2B%20%5Cfrac%7B(a%2Bd)(c%2Bd)%7D%7B2ac%7D%5Cright%5D%20


Expand the left-hand side of the equation:

%5Cfrac%7B1%7D%7B2%7D%20%5Cleft%5B7%20%2B%202d%5Cleft(%5Cfrac%7B1%7D%7Ba%7D%2B%5Cfrac%7B1%7D%7Bb%7D%20%2B%5Cfrac%7B1%7D%7Bc%7D%20%5Cright)%20%2B%20%7Bd%7D%5E%7B2%7D%5Cleft(%5Cfrac%7B1%7D%7Ba%5E2%7D%20%2B%20%5Cfrac%7B1%7D%7Bb%5E2%7D%20%2B%20%5Cfrac%7B1%7D%7Bc%5E2%7D%20%5Cright)%20%5Cright%5D


Expand the right-hand side of the equation:

3%20%2B%202d%5Cleft(%5Cfrac%7B1%7D%7Ba%7D%2B%5Cfrac%7B1%7D%7Bb%7D%20%2B%5Cfrac%7B1%7D%7Bc%7D%20%5Cright)%20%2B%20%7Bd%7D%5E%7B2%7D%5Cleft(%5Cfrac%7B1%7D%7Bab%7D%20%2B%20%5Cfrac%7B1%7D%7Bac%7D%20%2B%20%5Cfrac%7B1%7D%7Bbc%7D%20%5Cright)

Subsequently,

%5Cfrac%7B1%7D%7B2%7D%20%5Cleft%5B7%20%2B%202d%5Cleft(%5Cfrac%7B1%7D%7Ba%7D%2B%5Cfrac%7B1%7D%7Bb%7D%20%2B%5Cfrac%7B1%7D%7Bc%7D%20%5Cright)%20%2B%20%7Bd%7D%5E%7B2%7D%5Cleft(%5Cfrac%7B1%7D%7Ba%5E2%7D%20%2B%20%5Cfrac%7B1%7D%7Bb%5E2%7D%20%2B%20%5Cfrac%7B1%7D%7Bc%5E2%7D%20%5Cright)%20%5Cright%5D%20%3D%203%20%2B%202d%5Cleft(%5Cfrac%7B1%7D%7Ba%7D%2B%5Cfrac%7B1%7D%7Bb%7D%20%2B%5Cfrac%7B1%7D%7Bc%7D%20%5Cright)%20%2B%20%7Bd%7D%5E%7B2%7D%5Cleft(%5Cfrac%7B1%7D%7Bab%7D%20%2B%20%5Cfrac%7B1%7D%7Bac%7D%20%2B%20%5Cfrac%7B1%7D%7Bbc%7D%20%5Cright)%20


7%20%2B%202d%5Cleft(%5Cfrac%7B1%7D%7Ba%7D%2B%5Cfrac%7B1%7D%7Bb%7D%20%2B%5Cfrac%7B1%7D%7Bc%7D%20%5Cright)%20%2B%20%7Bd%7D%5E%7B2%7D%5Cleft(%5Cfrac%7B1%7D%7Ba%5E2%7D%20%2B%20%5Cfrac%7B1%7D%7Bb%5E2%7D%20%2B%20%5Cfrac%7B1%7D%7Bc%5E2%7D%20%5Cright)%20%3D%206%20%2B%204d%5Cleft(%5Cfrac%7B1%7D%7Ba%7D%2B%5Cfrac%7B1%7D%7Bb%7D%20%2B%5Cfrac%7B1%7D%7Bc%7D%20%5Cright)%20%2B%202%7Bd%7D%5E%7B2%7D%5Cleft(%5Cfrac%7B1%7D%7Bab%7D%20%2B%20%5Cfrac%7B1%7D%7Bac%7D%20%2B%20%5Cfrac%7B1%7D%7Bbc%7D%20%5Cright)


1%20%2B%20%7Bd%7D%5E%7B2%7D%5Cleft(%5Cfrac%7B1%7D%7Ba%5E2%7D%20%2B%20%5Cfrac%7B1%7D%7Bb%5E2%7D%20%2B%20%5Cfrac%7B1%7D%7Bc%5E2%7D%20%5Cright)%20%3D%202d%5Cleft(%5Cfrac%7B1%7D%7Ba%7D%2B%5Cfrac%7B1%7D%7Bb%7D%20%2B%5Cfrac%7B1%7D%7Bc%7D%20%5Cright)%20%2B%202%7Bd%7D%5E%7B2%7D%5Cleft(%5Cfrac%7B1%7D%7Bab%7D%20%2B%20%5Cfrac%7B1%7D%7Bac%7D%20%2B%20%5Cfrac%7B1%7D%7Bbc%7D%20%5Cright)%20


Divide both sides by d%5E2:


%5Cfrac%7B1%7D%7Bd%5E2%7D%20%2B%20%5Cfrac%7B1%7D%7Ba%5E2%7D%20%2B%20%5Cfrac%7B1%7D%7Bb%5E2%7D%20%2B%20%5Cfrac%7B1%7D%7Bc%5E2%7D%20%3D%202%5Cleft(%5Cfrac%7B1%7D%7Bad%7D%2B%5Cfrac%7B1%7D%7Bbd%7D%20%2B%5Cfrac%7B1%7D%7Bcd%7D%20%5Cright)%20%2B%202%5Cleft(%5Cfrac%7B1%7D%7Bab%7D%20%2B%20%5Cfrac%7B1%7D%7Bac%7D%20%2B%20%5Cfrac%7B1%7D%7Bbc%7D%20%5Cright)%20


Add?%5Cfrac%7B1%7D%7Ba%5E2%7D%20%2B%20%5Cfrac%7B1%7D%7Bb%5E2%7D%20%2B%20%5Cfrac%7B1%7D%7Bc%5E2%7D%20%2B%20%5Cfrac%7B1%7D%7Bd%5E2%7D on both sides:


2%20%5Cleft(%5Cfrac%7B1%7D%7Ba%5E2%7D%20%2B%20%5Cfrac%7B1%7D%7Bb%5E2%7D%20%2B%20%5Cfrac%7B1%7D%7Bc%5E2%7D%20%2B%20%5Cfrac%7B1%7D%7Bd%5E2%7D%20%5Cright)%20%3D%20%5Cfrac%7B1%7D%7Ba%5E2%7D%20%2B%20%5Cfrac%7B1%7D%7Bb%5E2%7D%20%2B%20%5Cfrac%7B1%7D%7Bc%5E2%7D%20%2B%20%5Cfrac%7B1%7D%7Bd%5E2%7D%20%2B%202%5Cleft(%5Cfrac%7B1%7D%7Bab%7D%20%2B%20%5Cfrac%7B1%7D%7Bac%7D%20%2B%20%5Cfrac%7B1%7D%7Bad%7D%2B%5Cfrac%7B1%7D%7Bbc%7D%20%2B%5Cfrac%7B1%7D%7Bbd%7D%20%2B%20%5Cfrac%7B1%7D%7Bcd%7D%20%5Cright)


This can be compactly expressed as

2%20%5Cleft(%5Cfrac%7B1%7D%7Ba%5E2%7D%20%2B%20%5Cfrac%7B1%7D%7Bb%5E2%7D%20%2B%20%5Cfrac%7B1%7D%7Bc%5E2%7D%20%2B%20%5Cfrac%7B1%7D%7Bd%5E2%7D%20%5Cright)%20%3D%20%7B%5Cleft(%5Cfrac%7B1%7D%7Ba%7D%20%2B%20%5Cfrac%7B1%7D%7Bb%7D%20%2B%20%5Cfrac%7B1%7D%7Bc%7D%2B%5Cfrac%7B1%7D%7Bd%7D%20%5Cright)%7D%5E%7B2%7D

Additional Remark


Solving the result for?d gives two solutions, one of the incircle (inner circle) and one for the excircle (outer circle).

Incircle radius

d%20%3D%20%5Cfrac%7Babc%7D%7Bab%20%2B%20ac%20%2B%20bc%20%2B%202%5Csqrt%7Babc%20%5Cleft(a%20%2B%20b%20%2B%20c%5Cright)%7D%7D

Excircle radius


d%20%3D%20%5Cfrac%7B-abc%7D%7Bab%20%2B%20ac%20%2B%20bc%20-%202%5Csqrt%7Babc%20%5Cleft(a%20%2B%20b%20%2B%20c%5Cright)%7D%7D






[Geometry] Descartes Four Circle Theorem的評(píng)論 (共 條)

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