Question

A triangle has corners at `(1 , 5 )`, `(9 ,4 )`, and `(1 ,8 )`. What is the radius of the triangle's inscribed circle?

Answer 1

`1.200`

Explanation:

Refer to figure below. (Obs.: D is the circle's center)

With `A(1,5), B(9,4) and C(1,8)`
`AB=sqrt((9-1)^2+(4-5)^2)=sqrt(64+1)=sqrt(65)~=8.1`
`BC=sqrt((1-9)^2+(8-4)^2)=sqrt(64+16)=sqrt(80)=4sqrt(5)~=8.9`
`CA=sqrt((1-1)^2+(5-8)^2)=3`

Using the variables `m, n and o`
`m+n=4sqrt(5)`
`m+o=sqrt(65)`
`n+o=3` => `o=3-n`
=> `m+3-n=sqrt(65)` => `m-n=sqrt(65)-3`
Adding the last with the first equation, we get
`2m=4sqrt(5)+sqrt(65)-3` => `m=2sqrt(5)+sqrt(65)/2-1.5`

In the figure we can see that
`tan alpha=r/m` => `r=m*tan alpha`

We only need to know `alpha`

Using vectors
`BvecA=(1-9)hat i+(5-4)hat j=-8hat i+hat j`
`BvecC=(1-9)hat i+(8-4)hat j=-8hat i+4hat j`

`cos (2alpha)=(BvecA*BvecC)/(|BA|*|BC|)=(8*8+1*4)/(sqrt(65)*4sqrt(5))=68/(20sqrt(13))=17/(5sqrt(13))`
=> `2alpha=19.440^@` => `alpha=9.720^@`

Using the Law of Sines
`3/sin(180^@-2alpha)=sqrt(65)/sin(180^@-2beta)=(4sqrt(5))/sin(180^@-2gamma)`
Since sines of supplementary angles are equal we can rewrite the previous equation as
`3/(sin 2alpha)=sqrt(65)/(sin 2beta)=(4sqrt(5))/(sin 2gamma)`
`3/(sin 2alpha)=sqrt(65)/(sin 2beta)` => `sin 2beta=sqrt(65)/3*sin 2alpha`

We know also that `2alpha+2beta+2gamma=180^@` => `2gamma=180^@-2alpha-2beta`
So
`3/sin(2alpha)=(4sqrt(5))/sin(180^@-2alpha-2beta)=(4sqrt(5))/sin (2alpha+2beta)=(4sqrt(5))/(sin 2alpha *cos 2beta+sin 2beta *cos 2alpha)`

`-> 3/cancel (sin2alpha)=` `(4sqrt(5))/(cancel(sin 2alpha)*sqrt(1-65/9*sin^2 2alpha)+sqrt(65)/3*cancel(sin 2alpha) *sqrt(1-sin^2 2alpha)`

`-> cancel(3)*sqrt(9-65sin^2 2alpha)/cancel(3)+cancel(3)*sqrt(65)/cancel(3)*sqrt(1-sin^2 2alpha)=4sqrt(5)`
`-> (sqrt(9-65sin^2 2alpha))^2=(4sqrt(5)-sqrt(65)*sqrt(1-sin^2 2alpha))^2`
`-> 9-cancel(65sin^2 2alpha)=80-8*5*sqrt(13)*sqrt(1-sin^2 2alpha)+65-cancel(65sin^2 2alpha)`

`40sqrt(13)*sqrt(1-sin^2 2alpha)=136`
`1300-1300*sin^2 2alpha=1156` => `sin^2 2alpha=144/1300` => `sin 2alpha=6/(5sqrt(13))` => `2alpha=19.440^@` => `alpha=9.720^@`

Finally,
`r=m*tan alpha=(2sqrt(5)+sqrt(65)/2-1.5)*tan 9.720^@` => `r=1.220`