23. Construct a Triangle Given Two Sides and the Inradius

a

7.

b

5

r

1

show incircle

96.-12.

2

x

+

3

x

= 0

-32.-48.y+

3

y

= 0

This Demonstration draws a triangle

ABC

given two side lengths

a

and

b

and the inradius

r

(the radius of the inscribed circle). This construction involves solving a cubic and is not possible with a ruler and compass.

Let the third side length be

c

and let the area of

ABC

be

Δ

. Define the semiperimeter

s=(a+b+c)/2

. Since

(ra+rb+rc)/2

equals the sum of the areas of the three triangles with apex at the incenter,

Δ=rs

.

Using that together with Heron's formula,

Δ=

s(s-a)(s-b)(s-c)

gives

(s-a)(s-b)(s-c)=

2

r

s

, which is a cubic equation for

c

. The equations for

x=c

and

y=c-4

are shown at the bottom of the graphic.

Details

The case

a=7

,

b=5

,

r=2

gives

3

x

-12x+96=0

, where

x=c

. Substitute

x=y+4

to get

3

y

-48y-32=0

.

Since the leading coefficient of the equation is

1

, a rational root would have to be an integer that divides 32. But no integer

-32

,

-16

,

-8

,

-4

,

-2

,

2

,

4

,

8

,

16

,

32

is a root of the equation. Therefore the last equation in

y

(and so also the first in

x

) has no rational solutions.

According to the theorem on p. 42 of [1], none of the roots can be constructed by ruler and compass, but the roots can be constructed using a marked ruler [1, p. 134].

References

[1] G. E. Martin, Geometric Constructions, New York: Springer, 1998.

External Links

Triangle Area (Wolfram MathWorld)

Heron's Formula (Wolfram MathWorld)

Constructing a Square with Sticks of Equal Length

Descartes's Method of Evaluating the Principal Cube Root

Bisecting a Line Segment with a Ruler Given a Parallel Line

Permanent Citation

Izidor Hafner

"23. Construct a Triangle Given Two Sides and the Inradius" from the Wolfram Demonstrations Project http://demonstrations.wolfram.com/23ConstructATriangleGivenTwoSidesAndTheInradius/
Published: November 3, 2017