Revival of Sangaku 2

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

## **Problem**

Suppose there is a blue circle and a red square. The bottom side of the red square is tangent to the blue circle. The top side of the red square is a chord of the blue circle. Let the

A
S

,

A
C

{\displaystyle A\_S, A\_C}

 denote the areas of the square and circle respectively. Determine the ratio of the areas

A
S

:

A
C

{\displaystyle A\_S:A\_C}

.

## **Solution**

### **Method 1 : Pythagorean theorem**

Let

r

{\displaystyle r}

 be the radius of the circle, and

s

{\displaystyle s}

 be the side length of the square. In Figure 2, three radii of the circle are drawn in black. Furthermore, an orange line is drawn from the center of the circle to the top side of the square. Let the length of the orange line be

m

{\displaystyle m}

. From the diagram, we get

{

# s
r
m

# m

r
2

−

(

s
2

)

2

{\displaystyle \begin{cases}
s = r + m \\[5 pt]
m = \sqrt{r^2 - \left(\frac{s}{2}\right)^2}
\end{cases}}

and

# s
r

r
2

−

(

s
2

)

2

.

{\displaystyle s = r + \sqrt{r^2 - \left(\frac{s}{2}\right)^2}. }

Working out the algebra, we get

s
−
# r

r
2

−

(

s
2

)

2

{\displaystyle s - r = \sqrt{r^2 - \left(\frac{s}{2}\right)^2}}

(
s
−
r

)
2

=

r
2

−

(

s
2

)

2

{\displaystyle (s-r)^2 = r^2 - \left(\frac{s}{2}\right)^2}

s
2

−
2
s
r

r
2

=

r
2

−

s
2

4

{\displaystyle s^2 -2sr +r^2 = r^2 - \frac{s^2}{4} }

s
2

−
2
s
# r
−

s
2

4

{\displaystyle s^2 -2sr = - \frac{s^2}{4} }

5

s
2

−
8
s
# r

{\displaystyle 5s^2 - 8sr = 0. }

Factorize the quadratic equation and solve for

s

{\displaystyle s}

:

s
(
5
s
−
8
r
# )
0

{\displaystyle s(5s - 8r) = 0 }

{

# s
0

[
5
p
t
]
# s

8
r

5

{\displaystyle \begin{cases}
s=0\\ [5 pt]
s = \frac{8r}{5}
\end{cases} }

Since

# s
0

{\displaystyle s=0}

 gives no square, the answer must be

# s

8
r

5

{\displaystyle s = \frac{8r}{5} }

.

### **Method 2 : Power of a point theorem**

From Figure 3, let

r

{\displaystyle r}

 be the radius of the circle,

s

{\displaystyle s}

 be the side length of the square, and

x

{\displaystyle x}

be the length of line

A
F

{\displaystyle AF}

. Apply the power of a point theorem for the case where one of the lines is tangent to the circle while the other is a secant. In this problem, we have

A
C
⋅
C
# F
C

H
2

.

{\displaystyle AC \cdot CF = CH^2.}

We have

A
# C
s

{\displaystyle AC = s}

,

C
# F
s
−
x

{\displaystyle CF = s-x}

, and

C
# H

s
2

{\displaystyle CH = \frac{s}{2}}

; thus, the above formula becomes

s
⋅
(
s
−
x
)

=

(

s
2

)

2

[
5
p
t
]
s
x

=

3

s
2

4

[
5
p
t
]
x

=

3
s

4

{\displaystyle \begin{align}
s \cdot (s-x) &= \left(\frac{s}{2}\right)^2 \\ [5 pt]
sx &= \frac{3s^2}{4} \\ [5 pt]
x &= \frac{3s}{4}
\end{align}}

Applying the Pythagorean theorem, we get

A

F
2

A

B
2

=
B

F
2

[
5
p
t
]

(

3
s

4

)

2

s
2

=
(
2
r

)
2

[
5
p
t
]

9

s
2

16

s
2

=
4

r
2

[
5
p
t
]

25

s
2

16

=
4

r
2

[
5
p
t
]

s
2

=

64
25

r
2

[
5
p
t
]

{\displaystyle \begin{align}
AF^2 + AB^2 &= BF^2\\ [5 pt]
\left(\frac{3s}{4}\right)^2 + s^2 &= (2r)^2\\ [5 pt]
\frac{9s^2}{16} + s^2 &= 4r^2\\ [5 pt]
\frac{25s^2}{16} &= 4r^2\\ [5 pt]
s^2 &= \frac{64}{25}r^2\\ [5 pt]
\end{align}}

Hence,

# s

8
5

r

{\displaystyle s = \frac{8}{5}r}

.

**Area of the circle**

A
C

=
π

r
2

{\displaystyle A\_C = \pi r^2}

**Area of the square**

A
S

=

(

8
r

5

)

2

=

64

r
2

25

{\displaystyle A\_S = \left(\frac{8r}{5} \right)^2= \frac{64r^2}{25}}

Therefore, the desired area ratio is

A
S

:

A
C

=
64
:
25
π
.

{\displaystyle A\_S:A\_C = 64:25\pi. }