Four Circle Theorem

## **Problem**

The following geometry problem was posed independently in Europe and Japan. Rene Descartes found the theorem without proof and sent the result in a letter to Princess Elizabeth of Bohemia in 1643. In Europe, the four circle theorem was first proven by Jakob Steiner 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* (筭法點竄初學抄) in 1830.

Suppose there are four circles

A
,
B
,
C
,
D

{\displaystyle A, B, C, D}

 lying in a plane, such that each circle is mutually tangent to the other three circles. Let the radius of each circle be

a
,
b
,
c
,
d

{\displaystyle a,b,c,d}

 (as shown in Figure 1), prove the identity

2

(

1

a
2

1

b
2

1

c
2

1

d
2

)

=

(

1
a

1
b

1
c

1
d

)

2

.

{\displaystyle 2\left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} + \frac{1}{d^2} \right)= {\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \right)}^{2}. }

## **Hints**

(1) Given three angles

θ
,
ϕ
,
ψ

{\displaystyle \theta, \phi, \psi }

, where

θ
ϕ
# ψ
2
π

{\displaystyle \theta + \phi + \psi = 2\pi }

, then

cos
2

⁡

θ

cos
2

⁡

ϕ

cos
2

⁡

ψ

=
2
cos
⁡

θ

cos
⁡

ϕ

cos
⁡

ψ

{\displaystyle \cos^2{\theta} + \cos^2{\phi} + \cos^2{\psi} = 2\cos{\theta}\cos{\phi}\cos{\psi} + 1. }

## **Solution**

Let

A
# B
a
b

{\displaystyle AB=a+b}

,

A
# C
a
c

{\displaystyle AC=a+c}

,

A
# D
a
d

{\displaystyle AD = a + d }

,

B
# C
b
c

{\displaystyle BC = b + c }

,

B
# D
b
d

{\displaystyle BD=b+d}

, and

C
# D
c
d

{\displaystyle CD=c+d}

. Furthermore, let there be three angles

# θ
∠
A
D
B

{\displaystyle \theta=\angle ADB}

,

# ϕ
∠
B
D
C

{\displaystyle \phi = \angle BDC }

, and

# ψ
∠
C
D
A

{\displaystyle \psi = \angle CDA }

. By applying the cosine law on triangle ADB, we get

A
B

2

=

A
D

2

B
D

2

−
2
A
B
⋅
B
D
cos
⁡

θ

{\displaystyle {AB}^{2}= {AD}^{2} + {BD}^{2} - 2 AB \cdot BD \cos{\theta} }

(

a
b

)

2

=

(

a
d

)

2

(

b
d

)

2

−
2

(

a
d

)

(

b
d

)

cos
⁡

θ

{\displaystyle {\left(a + b \right)}^{2} = {\left(a + d \right)}^{2} + {\left(b + d \right)}^{2} - 2 \left(a + d \right) \left(b + d \right) \cos{\theta} }

Expanding each bracket and rearranging yields

a
# b

d
2

a
d
b
d
−

(

a
d

)

(

b
d

)

cos
⁡

θ

{\displaystyle ab = d^2 + ad + bd - \left(a + d \right) \left(b + d \right) \cos{\theta} }

(

a
d

)

(

b
d

)

cos
⁡

θ

=

d
2

a
d
b
d
−
a
b

{\displaystyle \left(a + d \right) \left(b + d \right) \cos{\theta} = d^2 + ad + bd - ab }

(

a
d

)

(

b
d

)

cos
⁡

θ

=

d
2

a
d
b
d

(

a
b
−
a
b

)

−
a
b

{\displaystyle \left(a + d \right) \left(b + d \right) \cos{\theta} = d^2 + ad + bd + \left(ab - ab \right) - ab }

(

a
d

)

(

b
d

)

cos
⁡

θ

=

(

a
d

)

(

b
d

)

−
2
a
b

{\displaystyle \left(a + d \right) \left(b + d \right) \cos{\theta} = \left(a + d \right) \left(b + d \right) - 2ab }

Solving for

cos
⁡

θ

{\displaystyle \cos{\theta} }

 yields

cos
⁡

θ

=
1
−

2
a
b

(

a
d

)

(

b
d

)

{\displaystyle \cos{\theta} = 1 - \frac{2ab}{\left(a + d \right) \left(b + d \right)} }

Using the same method, we can also obtain

cos
⁡

ϕ

=
1
−

2
b
c

(

b
d

)

(

c
d

)

{\displaystyle \cos{\phi} = 1 - \frac{2bc}{\left(b + d \right) \left(c + d \right)} }

cos
⁡

ψ

=
1
−

2
a
c

(

a
d

)

(

c
d

)

{\displaystyle \cos{\psi} = 1 - \frac{2ac}{\left(a + d \right) \left(c + d \right)} }

Let

# x

2
a
b

(

a
d

)

(

b
d

)

{\displaystyle x =\frac{2ab}{\left(a + d \right) \left(b + d \right)} }

# y

2
b
c

(

b
d

)

(

c
d

)

{\displaystyle y = \frac{2bc}{\left(b + d \right) \left(c + d \right)} }

# z

2
a
c

(

a
d

)

(

c
d

)

{\displaystyle z = \frac{2ac}{\left(a + d \right) \left(c + d \right)} }

Using the trigonometric identity from the hint gives

(

1
−
x

)

2

(

1
−
y

)

2

(

1
−
z

)

2

=
2

(

1
−
x

)

(

1
−
y

)

(

1
−
z

)

1

{\displaystyle {\left(1-x \right)}^{2} + {\left(1-y \right)}^{2} + {\left(1-z \right)}^{2} = 2\left(1-x \right) \left(1-y \right) \left(1-z \right) + 1 }

Expanding and rearranging the above equation yields

3
−
2

(

x
y
z

)

x
2

y
2

z
2

=
3
2

(

x
y
x
z
y
x
−
(
x
y
x
)
−
x
y
z

)

{\displaystyle 3 -2\left(x+y+z\right) + x^2 + y^2 + z^2 = 3 + 2\left(xy+xz+yx -(x+y+x) - xyz \right) }

x
2

y
2

z
2

2
x
y
# z
2

(

x
y
x
z
y
z

)

{\displaystyle x^2 + y^2 + z^2 + 2xyz = 2\left(xy+xz+yz\right) }

Divide both sides of the equation by

x
y
z

{\displaystyle xyz}

:

x

y
z

y

x
z

z

x
y

# 2
2

(

1
x

1
y

1
z

)

{\displaystyle \frac{x}{yz} + \frac{y}{xz} +\frac{z}{xy} + 2 = 2\left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) }

Replacing the variables for

x
,
y
,
z

{\displaystyle x, y, z}

  in terms of

a
,
b
,
c
,
d

{\displaystyle a,b,c,d}

 gives

(
c
d
)

2

2

c

2

(
a
d
)

2

2

a

2

(
b
d
)

2

2

b

2

# 2
2

[

(
a
d
)
(
b
d
)

2
a
b

(
b
d
)
(
c
d
)

2
b
c

(
a
d
)
(
c
d
)

2
a
c

]

{\displaystyle \frac{{(c+d)}^{2}}{2{c}^{2}} +\frac{{(a+d)}^{2}}{2{a}^{2}} +\frac{{(b+d)}^{2}}{2{b}^{2}} + 2 = 2\left[\frac{(a+d)(b+d)}{2ab} + \frac{(b+d)(c+d)}{2bc} + \frac{(a+d)(c+d)}{2ac}\right] }

Expanding the left-hand side of the equation yields

1
2

[

7
2
d

(

1
a

1
b

1
c

)

d

2

(

1

a
2

1

b
2

1

c
2

)

]

{\displaystyle \frac{1}{2} \left[7 + 2d\left(\frac{1}{a}+\frac{1}{b} +\frac{1}{c} \right) + {d}^{2}\left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \right) \right] }

Expanding the right-hand side of the equation yields

3
2
d

(

1
a

1
b

1
c

)

d

2

(

1

a
b

1

a
c

1

b
c

)

{\displaystyle 3 + 2d\left(\frac{1}{a}+\frac{1}{b} +\frac{1}{c} \right) + {d}^{2}\left(\frac{1}{ab} + \frac{1}{ac} + \frac{1}{bc} \right) }

Subsequently,

1
2

[

7
2
d

(

1
a

1
b

1
c

)

d

2

(

1

a
2

1

b
2

1

c
2

)

]

=
3
2
d

(

1
a

1
b

1
c

)

d

2

(

1

a
b

1

a
c

1

b
c

)

{\displaystyle \frac{1}{2} \left[7 + 2d\left(\frac{1}{a}+\frac{1}{b} +\frac{1}{c} \right) + {d}^{2}\left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \right) \right] = 3 + 2d\left(\frac{1}{a}+\frac{1}{b} +\frac{1}{c} \right) + {d}^{2}\left(\frac{1}{ab} + \frac{1}{ac} + \frac{1}{bc} \right) }

7
2
d

(

1
a

1
b

1
c

)

d

2

(

1

a
2

1

b
2

1

c
2

)

=
6
4
d

(

1
a

1
b

1
c

)

2

d

2

(

1

a
b

1

a
c

1

b
c

)

{\displaystyle 7 + 2d\left(\frac{1}{a}+\frac{1}{b} +\frac{1}{c} \right) + {d}^{2}\left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \right) = 6 + 4d\left(\frac{1}{a}+\frac{1}{b} +\frac{1}{c} \right) + 2{d}^{2}\left(\frac{1}{ab} + \frac{1}{ac} + \frac{1}{bc} \right) }

1

d

2

(

1

a
2

1

b
2

1

c
2

)

=
2
d

(

1
a

1
b

1
c

)

2

d

2

(

1

a
b

1

a
c

1

b
c

)

{\displaystyle 1 + {d}^{2}\left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \right) = 2d\left(\frac{1}{a}+\frac{1}{b} +\frac{1}{c} \right) + 2{d}^{2}\left(\frac{1}{ab} + \frac{1}{ac} + \frac{1}{bc} \right) }

Divide both sides by

d
2

{\displaystyle d^2}

:

1

d
2

1

a
2

1

b
2

1

c
2

=
2

(

1

a
d

1

b
d

1

c
d

)

2

(

1

a
b

1

a
c

1

b
c

)

{\displaystyle \frac{1}{d^2} + \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = 2\left(\frac{1}{ad}+\frac{1}{bd} +\frac{1}{cd} \right) + 2\left(\frac{1}{ab} + \frac{1}{ac} + \frac{1}{bc} \right) }

Add

1

a
2

1

b
2

1

c
2

1

d
2

{\displaystyle \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} + \frac{1}{d^2} }

 on both sides:

2

(

1

a
2

1

b
2

1

c
2

1

d
2

)

=

1

a
2

1

b
2

1

c
2

1

d
2

2

(

1

a
b

1

a
c

1

a
d

1

b
c

1

b
d

1

c
d

)

{\displaystyle 2 \left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} + \frac{1}{d^2} \right) = \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} + \frac{1}{d^2} + 2\left(\frac{1}{ab} + \frac{1}{ac} + \frac{1}{ad}+\frac{1}{bc} +\frac{1}{bd} + \frac{1}{cd} \right) }

This can be compactly expressed as

2

(

1

a
2

1

b
2

1

c
2

1

d
2

)

=

(

1
a

1
b

1
c

1
d

)

2

{\displaystyle 2 \left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} + \frac{1}{d^2} \right) = {\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}+\frac{1}{d} \right)}^{2} }

### **Additional Remark**

Solving the result for

d

{\displaystyle d}

 gives two solutions: the radius of the incircle (inner circle), and the radius of the excircle (outer circle).

**Incircle radius**

# d

a
b
c

a
b
a
c
b
c
2

a
b
c

(

a
b
c

)

{\displaystyle d = \frac{abc}{ab + ac + bc + 2\sqrt{abc \left(a + b + c\right)}} }

**Excircle radius**

# d

−
a
b
c

a
b
a
c
b
c
−
2

a
b
c

(

a
b
c

)

{\displaystyle d = \frac{-abc}{ab + ac + bc - 2\sqrt{abc \left(a + b + c\right)}} }