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[Geometry] Inscribed Circle of a Triangle

2021-07-01 21:21 作者:AoiSTZ23  | 我要投稿

By: Tao Steven Zheng (鄭濤)

【Problem】

The following problem is selected from Aida Yasuaki’s 《算法天生法指南》 (1811):

Suppose a circle is inscribed in a triangle with side lengths 13, 14, and 15. What is the diameter of the inscribed circle[1]?

[1] An inscribed circle is also called an incircle.


【Solution】

?(1) Use either Heron’s formula or Qin Jiushao’s formula for computing the area of a triangle.

Heron’s formula

Heron of Alexandria gave his formula in Book I of “Metrica”:

A%3D%5Csqrt%7Bs(s-a)(s-b)(s-c)%20%7D%20

where A ?is the area of a triangle with sides a, b, c and semi-perimeter s%3D%5Cfrac%7B1%7D%7B2%7D%20(a%2Bb%2Bc) .

s%3D%5Cfrac%7B1%7D%7B2%7D%20%20(13%2B14%2B15)%20%3D%2021%0A%0A%0A

A%3D%5Csqrt%7B21(21-13)(21-14)(21-15)%7D%20%0A%0A%0A

A%20%3D%20%5Csqrt%7B21%C3%978%C3%977%C3%976%7D%20%3D%2084

Qin Jiushao’s Formula

In the “Shushu Jiuzhang”《數(shù)書(shū)九章》, Qin Jiushao gave the formula:

A%3D%5Csqrt%7B%5Cfrac%7B1%7D%7B4%7D%5Cleft%5Ba%5E2%20c%5E2-%20%5Cleft(%5Cfrac%7Ba%5E2%2Bc%5E2-b%5E2%7D%7B2%7D%20%5Cright)%5E2%20%5Cright%5D%7D%20

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%3D%5Csqrt%7B%5Cfrac%7B1%7D%7B4%7D%20%20%5B(13)%5E2%20(15)%5E2-%5Cleft(%5Cfrac%7B13%5E2%2B15%5E2-14%5E2%7D%7B2%7D%20%5Cright)%5E2%20%5D%7D%20%3D%2084

?

(2)

Figure 1


Figure 1 shows that %E2%96%B3ABC?can be split into three triangles: %E2%96%B3OAB, ?%E2%96%B3OAC%20 and %E2%96%B3OBC. Hence, the area of %E2%96%B3ABC?is the sum

%7BA%7D_%7B%E2%96%B3ABC%7D%20%3D%20%7BA%7D_%7B%E2%96%B3OAB%7D%20%2B%20%7BA%7D_%7B%E2%96%B3OAC%7D%20%2B%20%7BA%7D_%7B%E2%96%B3OBC%7D

where

%7BA%7D_%7B%E2%96%B3OAB%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20OD%5Ccdot%20AB

%7BA%7D_%7B%E2%96%B3OAC%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20OE%5Ccdot%20AC

%7BA%7D_%7B%E2%96%B3OBC%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20OF%5Ccdot%20BC

?Since the line segments OD, OE, and ?OF are radii of the inscribed circle, let OD%3DOE%3DOF%3Dr. Let AB%3Da, AC%3Db, and BC%3Dc.

?Hence,

?%7BA%7D_%7B%E2%96%B3ABC%7D%3D%20%5Cfrac%7B1%7D%7B2%7D%20ra%2B%5Cfrac%7B1%7D%7B2%7D%20rb%2B%5Cfrac%7B1%7D%7B2%7Drc

%7BA%7D_%7B%E2%96%B3ABC%7D%3D%20%5Cfrac%7B1%7D%7B2%7D%20r%EF%BC%88a%2Bb%2Bc%EF%BC%89

?Since the radius is half the diameter,

%7BA%7D_%7B%E2%96%B3ABC%7D%3D%20%5Cfrac%7B1%7D%7B4%7D%20d%EF%BC%88a%2Bb%2Bc%EF%BC%89

Therefore, the diameter of the inscribed circle is

d%3D%5Cfrac%7B4%5Ctimes%20%7BA%7D_%7B%E2%96%B3ABC%7D%7D%7Ba%2Bb%2Bc%7D%20

d%3D%5Cfrac%7B4%5Ctimes%2084%7D%7B13%2B14%2B15%7D%20

d%3D8




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