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:

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

where $A$ is the area of a triangle with sides $a$, $b$, $c$ and semi-perimeter$s=\frac{1}{2}\left(a+b+c\right)$ .

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

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

Qin Jiushao’s Formula

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

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

where $A$ is the area of a triangle, $a$ is the short side,  $b$ is the middle side, and $c$ is the long side. 

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

Figure 1

Figure 1 shows that $\bigtriangleup ABC$ can be split into three triangles: $\bigtriangleup OAB$, $\bigtriangleup OAC$ and $\bigtriangleup OBC$. Hence, the area of  is the sum of the areas

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

where

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

Since the line segments $OD$,$OE$, and $OF$ are radii of the inscribed circle, let $OD=OE=OF=r$. Furthermore, let $AB=a$, $AC=b$, and $BC=c$.

Hence,

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

Since the radius is half the diameter,

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

Therefore, the diameter of the inscribed circle is

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