Inscribed Circle of a Triangle

## **Problem**

The following geometry problem is taken from *Sanpo tensei-ho shinan* 《算法天生法指南》 (1811) by Aida Yasuaki (会田 安明, 1747 – 1817):

Suppose a circle is inscribed in a triangle. The short side is 13, the middle side is 14, and the long side is 15. Question: What is the diameter of the inscribed circle1?

1 An inscribed circle can also be called an incircle.

## **Hint**

(1) Use Heron’s formula or Qin Jiushao’s formula to calculate the area of a triangle.

## **Solution**

Calculate the area of the triangle.

**Heron’s formula**

Heron of Alexandria gave his formula in Book I of the *Metrica*:

# A

s
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)

{\displaystyle A = \sqrt{s(s-a)(s-b)(s-c)}}

where

A

{\displaystyle A}

 is the area of a triangle with sides

a

{\displaystyle a}

,

b

{\displaystyle b}

,

c

{\displaystyle c}

 and semi-perimeter

# s

1
2

(

a
b
c

)

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

 .

# s

1
2

(

13
14
15

)

=
21

{\displaystyle s=\frac{1}{2}\left(13+14+15\right) = 21}

# A

21
(
21
−
13
)
(
21
−
14
)
(
21
−
15
)

=

{\displaystyle A = \sqrt{21(21-13)(21-14)(21-15)} = 84. }

**Qin Jiushao’s Formula**

In the *Shushu Jiuzhang*《数书九章》, Qin Jiushao gave the formula:

# A

1
4

[

a

2

c

2

−

(

a

2

c

2

−

b

2

2

)

2

]

{\displaystyle A = \sqrt{\frac{1}{4} \left[{a}^{2}{c}^{2} - {\left(\frac{{a}^{2} + {c}^{2} - {b}^{2}}{2}\right)}^{2} \right]}}

where

A

{\displaystyle A}

 is the area of a triangle,

a

{\displaystyle a}

 is the short side,

b

{\displaystyle b}

 is the middle side, and

c

{\displaystyle c}

 is the long side.

# A

1
4

[

13

2

15

2

−

(

13

2

15

2

−

14

2

2

)

2

]

=

{\displaystyle A = \sqrt{\frac{1}{4} \left[{13}^{2}{15}^{2} - {\left(\frac{{13}^{2} + {15}^{2} - {14}^{2}}{2}\right)}^{2} \right]} = 84.}

Figure 1 shows that

△
A
B
C

{\displaystyle \bigtriangleup ABC}

 can be split into three triangles:

△
O
A
B

{\displaystyle \bigtriangleup OAB }

,

△
O
A
C

{\displaystyle \bigtriangleup OAC }

 and

△
O
B
C

{\displaystyle \bigtriangleup OBC }

. Hence, the area of  is the sum of the areas

A

△
A
B
C

=

A

△
O
A
B

A

△
O
A
C

A

△
O
B
C

{\displaystyle {A}\_{\bigtriangleup ABC} = {A}\_{\bigtriangleup OAB} + {A}\_{\bigtriangleup OAC} + {A}\_{\bigtriangleup OBC}}

where

A

△
O
A
B

=

1
2

O
D
⋅
A
B

A

△
O
A
C

=

1
2

O
E
⋅
A
C

A

△
O
B
C

=

1
2

O
F
⋅
B
C

{\displaystyle \begin{align}
{A}\_{\bigtriangleup OAB} &= \frac{1}{2}OD \cdot AB \\[5 pt]
{A}\_{\bigtriangleup OAC} &= \frac{1}{2}OE \cdot AC \\[5 pt]
{A}\_{\bigtriangleup OBC} &= \frac{1}{2}OF \cdot BC
\end{align}}

Since the line segments

O
D

{\displaystyle OD}

,

O
E

{\displaystyle OE}

, and

O
F

{\displaystyle OF}

 are radii of the inscribed circle, let

O
# F
r

{\displaystyle OD=OE=OF=r }

. Furthermore, let

A
# B
a

{\displaystyle AB=a}

,

A
# C
b

{\displaystyle AC=b}

, and

B
# C
c

{\displaystyle BC=c}

.

Hence,

A

△
A
B
C

=

1
2

r
a

1
2

r
b

1
2

r
c

=

1
2

r

(

a
b
c

)

.

{\displaystyle \begin{align}
{A}\_{\bigtriangleup ABC} &= \frac{1}{2}ra + \frac{1}{2}rb + \frac{1}{2}rc \\[5pt]
&= \frac{1}{2}r\left(a+b+c\right).
\end{align}
 }

Since the radius is half the diameter,

A

△
A
B
C

=

1
4

d

(

a
b
c

)

{\displaystyle {A}\_{\bigtriangleup ABC} = \frac{1}{4}d\left(a+b+c\right)}

Therefore, the diameter of the inscribed circle is

d

=

4
×

A

△
A
B
C

a
b
c

=

4
×
84

13
14
15

=

{\displaystyle \begin{align}
d &= \frac{4 \times {A}\_{\bigtriangleup ABC}}{a+b+c}\\[5 pt]
&= \frac{4 \times 84}{13+14+15} \\[5 pt]
&= 8.
\end{align}}