Revival of Sangaku 3

**Problem created by Tao Steven Zheng (郑涛)**

## **Problem**

Consider the configuration below. There are many circles nested between two circle quadrants (blue).

**Part 1:** Determine the radius of the red circle.

**Part 2:** Determine the recurrence relation for computing the radius of the circles above the red circle.

**Part 3:** Determine the recurrence relation for computing the radius of the circles extending left and right of the red circle.

**Part 4:** Suppose the nested circles extend ad infinitum between the two circle quadrants of radius 1. Question: What is the total area of the nested circles?

**Hint:**

ζ
(
2

# n
1

∞

1

n
2

=

π
2

6

{\displaystyle \zeta(2) = \sum\_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} }

 and

ζ
(
4

# n
1

∞

1

n
4

=

π
4

90

{\displaystyle \zeta(4) = \sum\_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90} }

 and

∑

# n
1

∞

(

1
n

−

1

n
1

)

=
1

{\displaystyle \sum\_{n=1}^{\infty} \left(\frac{1}{n} - \frac{1}{n+1}\right) = 1 }

## **Solution**

**Part 1**

From Figure 1, let

A
# G

r
1

{\displaystyle AG = r\_1 }

,

A
# F

r
2

{\displaystyle AF = r\_2 }

,

A
# B

r
1

r
2

{\displaystyle AB = r\_1 + r\_2 }

, and

B
# G

r
1

−

r
2

{\displaystyle BG = r\_1 - r\_2 }

. Applying the Pythagorean theorem, we get

(

r
1

−

r
2

)
2

(

r
1

)
2

=
(

r
1

r
2

)
2

{\displaystyle (r\_1 - r\_2)^2 + (r\_1)^2 = (r\_1 + r\_2)^2 }

r
1
2

−
2

r
1

r
2

r
2
2

r
1
2

=

r
1
2

2

r
1

r
2

r
2
2

{\displaystyle r\_1^2 -2r\_1 r\_2 + r\_2^2 + r\_1^2 = r\_1^2 + 2r\_1 r\_2 + r\_2^2 }

r
1
2

=
4

r
1

r
2

{\displaystyle r\_1^2 = 4 r\_1 r\_2 }

r
1
2

−
4

r
1

r
2

=
0

{\displaystyle r\_1^2 - 4 r\_1 r\_2 = 0 }

Solving for

r
1

{\displaystyle r\_1 }

 gives either

r
1

=
0

{\displaystyle r\_1 = 0 }

 and

r
1

=
4

r
2

{\displaystyle r\_1 = 4r\_2 }

. The solution

r
1

=
0

{\displaystyle r\_1 = 0 }

 does not reflect the geometry; therefore,

r
2

=

r
1

4

.

{\displaystyle r\_2 = \frac{r\_1}{4}.}

**Part 2**

Let us call the nested circles above the red circle **Series 1**. From Figure 3, let

B
# I
h

{\displaystyle BI = h }

,

H
# I

r
1

{\displaystyle HI = r\_1 }

, and

H
# B

r
1

r
n

{\displaystyle HB = r\_1 + r\_n }

. It is known from Part 1 that

r
2

=

r
1

4

{\displaystyle r\_2 = \frac{r\_1}{4}}

. Here,

h

{\displaystyle h}

 denotes the gap between the uppermost nested circle and the top of the rectangle

C
B
D
E

{\displaystyle CBDE }

and

r
n

{\displaystyle r\_n }

denotes the radius of that circle. As the circles extend upwards, we discover two relations:

r
1
2

h
2

=
(

r
1

r
n

)
2

{\displaystyle r\_1^2 + h^2 = (r\_1 + r\_n)^2 }

r
1

=
2

(

∑

# i
2

n
−
1

r
i

)

r
n

h

{\displaystyle r\_1 = 2\left(\sum\_{i=2}^{n-1} r\_i\right) + r\_n + h }

(1) Expand the first expression:

r
1
2

h
2

=

r
1
2

2

r
1

r
n

r
n
2

[
5
p
t
]

h
2

=
2

r
1

r
n

r
n
2

.

{\displaystyle \begin{align}
r\_1^2 + h^2 &= r\_1^2 + 2r\_1 r\_n + r\_n^2 \\ [5 pt]
h^2 &= 2r\_1 r\_n + r\_n^2.
\end{align} }

Let

# u

r
1

−
2

(

∑

# i
2

n
−
1

r
i

)

{\displaystyle u = r\_1 - 2\left(\sum\_{i=2}^{n-1} r\_i\right) }

. From the expression

r
1

=
2

(

∑

# i
2

n
−
1

r
i

)

r
n

h

{\displaystyle r\_1 = 2\left(\sum\_{i=2}^{n-1} r\_i\right) + r\_n + h }

, we discover that

# h
u
−

r
n

{\displaystyle h = u - r\_n }

; thus,

h
2

=

(

u
−

r
n

)

2

{\displaystyle h^2 = \left(u - r\_n\right)^2 }

. Equating both expressions for

h
2

{\displaystyle h^2 }

, we get

(

u
−

r
n

)

2

=
2

r
1

r
n

r
n
2

[
5
p
t
]

u
2

−
2
u

r
n

r
n
2

=
2

r
1

r
n

r
n
2

[
5
p
t
]

u
2

−
2
u

r
n

=
2

r
1

r
n

.

{\displaystyle \begin{align}
\left(u - r\_n\right)^2 &= 2r\_1 r\_n + r\_n^2 \\ [5 pt]
u^2 - 2u r\_n +r\_n^2 &= 2r\_1 r\_n + r\_n^2 \\ [5 pt]
u^2 - 2u r\_n &= 2r\_1 r\_n.
\end{align} }

Solving for

r
n

{\displaystyle r\_n }

 gives

r
n

=

u
2

2
(

r
1

u
)

{\displaystyle r\_n = \frac{u^2}{2(r\_1 + u) } }

for

# n
3
,
4
,
5
,
.
.
.

{\displaystyle n = 3, 4, 5, ... }

.

**Part 3**

Let us call the nested circles circles extending left and right of the red circle **Series 2**. Each side is symmetric, so we will focus on the right side. From Figure 4, let

A
# B

r
1

r
2

{\displaystyle AB = r\_1 + r\_2 }

,

A
# G

r
2

r
3

{\displaystyle AG = r\_2 + r\_3 }

,

B
# G

r
1

r
3

{\displaystyle BG = r\_1 + r\_3 }

, and

A
# I

r
2

−

r
3

{\displaystyle AI = r\_2 - r\_3 }

. Furthermore, let

I
# G
x

{\displaystyle IG = x }

 and

G
# J
y

{\displaystyle GJ = y }

. Applying the Pythagorean theorem, we get

x
2

(

r
2

−

r
3

)
2

=
(

r
2

r
3

)
2

{\displaystyle x^2 + (r\_2 - r\_3)^2 = (r\_2 + r\_3)^2 }

y
2

(

r
1

−

r
3

)
2

=
(

r
1

r
3

)
2

{\displaystyle y^2 + (r\_1 - r\_3)^2 = (r\_1 + r\_3)^2 }

(
x
y

)
2

(

r
1

−

r
2

)
2

=
(

r
1

r
2

)
2

.

{\displaystyle (x+y)^2 + (r\_1 - r\_2)^2 = (r\_1 + r\_2)^2. }

(1) Expand the first expression:

x
2

r
2
2

−
2

r
2

r
3

r
3
2

=

r
2
2

2

r
2

r
3

r
3
2

[
5
p
t
]

x
2

=
4

r
2

r
3

[
5
p
t
]
x

=
2

r
2

r
3

.

{\displaystyle \begin{align}
x^2 + r\_2^2 - 2r\_2 r\_3 + r\_3^2 &= r\_2^2 + 2r\_2 r\_3 + r\_3^2 \\ [5 pt]
x^2 &= 4r\_2 r\_3 \\ [5 pt]
x &= 2\sqrt{r\_2 r\_3}.
\end{align} }

(2) Expand the second expression:

y
2

r
1
2

−
2

r
1

r
3

r
3
2

=

r
1
2

2

r
1

r
3

r
3
2

[
5
p
t
]

y
2

=
4

r
1

r
3

[
5
p
t
]
y

=
2

r
1

r
3

.

{\displaystyle \begin{align}
y^2 + r\_1^2 - 2r\_1 r\_3 + r\_3^2 &= r\_1^2 + 2r\_1 r\_3 + r\_3^2 \\ [5 pt]
y^2 &= 4r\_1 r\_3 \\ [5 pt]
y &= 2\sqrt{r\_1 r\_3}.
\end{align} }

(3) Expand the third expression:

(
x
y

)
2

r
1
2

−
2

r
1

r
2

r
3
2

=

r
1
2

2

r
1

r
2

r
3
2

[
5
p
t
]
(
x
y

)
2

=
4

r
1

r
2

[
5
p
t
]
x
y

=
2

r
1

r
2

.

{\displaystyle \begin{align}
(x+y)^2 + r\_1^2 - 2r\_1 r\_2 + r\_3^2 &= r\_1^2 + 2r\_1 r\_2 + r\_3^2 \\ [5 pt]
(x+y)^2 &= 4r\_1 r\_2 \\ [5 pt]
x+y &= 2\sqrt{r\_1 r\_2}.
\end{align} }

Combining the three results yields

2

r
2

r
3

2

r
1

r
3

=
2

r
1

r
2

{\displaystyle 2\sqrt{r\_2 r\_3} + 2\sqrt{r\_1 r\_3} = 2\sqrt{r\_1 r\_2} }

or

r
2

r
3

r
1

r
3

=

r
1

r
2

.

{\displaystyle \sqrt{r\_2 r\_3} + \sqrt{r\_1 r\_3} = \sqrt{r\_1 r\_2}. }

Solving for

r
3

{\displaystyle r\_3}

 gives

r
3

=

r
1

r
2

(

r
1

r
2

)

2

.

{\displaystyle r\_3 = \frac{r\_1 r\_2}{\left(\sqrt{r\_1} +\sqrt{r\_2}\right)^2 }. }

The radius of subsequent circles are

r

n
1

=

r
1

r
n

(

r
1

r
n

)

2

.

{\displaystyle r\_{n+1} = \frac{r\_1 r\_n}{\left(\sqrt{r\_1} +\sqrt{r\_n}\right)^2 }. }

**Part 4**

Given that

r
1

=
1

{\displaystyle r\_1 = 1 }

 and thus

r
2

=

1
4

{\displaystyle r\_2 = \frac{1}{4} }

, we calculate each series of radii.

Series 1

r
3

=

1
12

,

r
4

=

1
24

,

r
5

=

1
40

,

⋯
,

r
n

=

1

2
n
(
n
−
1
)

{\displaystyle r\_3 = \frac{1}{12}, \quad r\_4 = \frac{1}{24}, \quad r\_5 = \frac{1}{40}, \; \cdots, \; r\_n = \frac{1}{2n(n-1)} }

Series 2

r
3

=

1
9

,

r
4

=

1
16

,

r
5

=

1
25

,

⋯
,

r
n

=

1

n
2

{\displaystyle r\_3 = \frac{1}{9}, \quad r\_4 = \frac{1}{16}, \quad r\_5 = \frac{1}{25}, \; \cdots, \; r\_n = \frac{1}{n^2} }

The total area of the nested circles is thus

π
16

S
e
r
i
e
s

1
2
×
S
e
r
i
e
s

2

{\displaystyle \frac{\pi}{16} + Series \; 1 + 2 \times Series \; 2 }

.

# A

π
16

2
π

(

1

9
2

1

16
2

1

25
2

⋯

)

π

(

1

12
2

1

24
2

1

40
2

⋯

)

{\displaystyle A = \frac{\pi}{16} + 2\pi \left(\frac{1}{9^2} + \frac{1}{16^2} + \frac{1}{25^2} + \cdots \right) + \pi \left(\frac{1}{12^2} + \frac{1}{24^2} + \frac{1}{40^2} + \cdots \right) }

# A
2
π

(

1

3
4

1

4
4

1

5
4

⋯

)

π

(

1

4
2

1

12
2

1

24
2

1

40
2

⋯

)

{\displaystyle A = 2\pi \left(\frac{1}{3^4} + \frac{1}{4^4} + \frac{1}{5^4} + \cdots \right) + \pi \left(\frac{1}{4^2} + \frac{1}{12^2} + \frac{1}{24^2} + \frac{1}{40^2} + \cdots \right) }

# A
2
π

∑

# n
3

∞

1

n
4

π

∑

# n
2

∞

(

1

2
n
(
n
−
1
)

)

2

{\displaystyle A = 2\pi \sum\_{n=3}^{\infty} \frac{1}{n^4} + \pi\sum\_{n=2}^{\infty} \left(\frac{1}{2n(n-1)} \right)^2 }

It is known that

ζ
(
2

# n
1

∞

1

n
2

=

π
2

6

{\displaystyle \zeta(2) = \sum\_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} }

 and

ζ
(
4

# n
1

∞

1

n
4

=

π
4

90

{\displaystyle \zeta(4) = \sum\_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90} }

; thus,

# A
2
π

∑

# n
3

∞

1

n
4

π
4

∑

# n
1

∞

(

1

n
1

−

1
n

)

2

{\displaystyle A = 2\pi \sum\_{n=3}^{\infty} \frac{1}{n^4} + \frac{\pi}{4} \sum\_{n=1}^{\infty} \left(\frac{1}{n+1} - \frac{1}{n}\right)^2 }

# A
2
π

[

∑

# n
1

∞

1

n
4

−

(

1

1

2
4

)

]

π
4

[

∑

# n
1

∞

1

n
2

∑

# n
1

∞

1

(
n
1

)
2

−
2

∑

# n
1

∞

(

1
n

−

1

n
1

)

]

{\displaystyle A = 2\pi \left[\sum\_{n=1}^{\infty} \frac{1}{n^4} - \left(1 + \frac{1}{2^4}\right)\right] + \frac{\pi}{4} \left[\sum\_{n=1}^{\infty} \frac{1}{n^2} + \sum\_{n=1}^{\infty} \frac{1}{(n+1)^2} - 2\sum\_{n=1}^{\infty} \left(\frac{1}{n} - \frac{1}{n+1}\right)\right] }

# A
2
π

[

π
4

90

−

17
16

]

π
4

[

π
2

6

(

π
2

6

−
1

)

−
2

]

{\displaystyle A = 2\pi \left[\frac{\pi^4}{90} - \frac{17}{16} \right] + \frac{\pi}{4} \left[\frac{\pi^2}{6} + \left(\frac{\pi^2}{6} - 1\right) - 2\right] }

# A
2
π

[

π
4

90

−

17
16

]

π
4

[

2
×

π
2

6

−
3

]

{\displaystyle A = 2\pi \left[\frac{\pi^4}{90} - \frac{17}{16} \right] + \frac{\pi}{4} \left[2\times \frac{\pi^2}{6} - 3 \right] }

# A

π
5

45

−

17
π

8

π
3

12

−

3
π

4

{\displaystyle A = \frac{\pi^5}{45} - \frac{17\pi}{8} + \frac{\pi^3}{12} - \frac{3\pi}{4} }

# A

π
5

45

π
3

12

−

23
π

8

{\displaystyle A = \frac{\pi^5}{45} + \frac{\pi^3}{12} - \frac{23\pi}{8} }

# A

8

π
5

30

π
3

−
1035
π

360

{\displaystyle A = \frac{8\pi^5 + 30\pi^3 - 1035 \pi}{360} }

The area of the nested circles to three decimal places is 0.352.

## **About the Diagram**

**Equations of the circles in Figure 1**

Red circle:

x
2

(

y
−

1
4

)

2

=

1
16

{\displaystyle x^2 + \left(y - \frac{1}{4}\right)^2 = \frac{1}{16} }

**Series 1**

Vermillion circle:

x
2

(

y
−

7
12

)

2

=

1
144

{\displaystyle x^2 + \left(y - \frac{7}{12}\right)^2 = \frac{1}{144} }

Green circle:

x
2

(

y
−

17
24

)

2

=

1
576

{\displaystyle x^2 + \left(y - \frac{17}{24}\right)^2 = \frac{1}{576} }

Grey circle:

x
2

(

y
−

31
40

)

2

=

1
1600

{\displaystyle x^2 + \left(y - \frac{31}{40}\right)^2 = \frac{1}{1600} }

**Series 2 (Right)**

Vermillion circle:

(

x
−

1
3

)

2

(

y
−

1
9

)

2

=

1
81

{\displaystyle \left(x - \frac{1}{3}\right)^2 + \left(y - \frac{1}{9}\right)^2 = \frac{1}{81} }

Green circle:

(

x
−

1
2

)

2

(

y
−

1
16

)

2

=

1
256

{\displaystyle \left(x - \frac{1}{2}\right)^2 + \left(y - \frac{1}{16}\right)^2 = \frac{1}{256} }

Grey circle:

(

x
−

3
5

)

2

(

y
−

1
25

)

2

=

1
625

{\displaystyle \left(x - \frac{3}{5}\right)^2 + \left(y - \frac{1}{25}\right)^2 = \frac{1}{625} }

**Series 2 (Left)**

Vermillion circle:

(

x

1
3

)

2

(

y
−

1
9

)

2

=

1
81

{\displaystyle \left(x + \frac{1}{3}\right)^2 + \left(y - \frac{1}{9}\right)^2 = \frac{1}{81} }

Green circle:

(

x

1
2

)

2

(

y
−

1
16

)

2

=

1
256

{\displaystyle \left(x + \frac{1}{2}\right)^2 + \left(y - \frac{1}{16}\right)^2 = \frac{1}{256} }

Grey circle:

(

x

3
5

)

2

(

y
−

1
25

)

2

=

1
625

{\displaystyle \left(x + \frac{3}{5}\right)^2 + \left(y - \frac{1}{25}\right)^2 = \frac{1}{625} }