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

1 Answer
Jan 19, 2016

1.2001.200

Explanation:

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

I created this figure using MS ExcelI created this figure using MS Excel

With A(1,5), B(9,4) and C(1,8)A(1,5),B(9,4)andC(1,8)
AB=sqrt((9-1)^2+(4-5)^2)=sqrt(64+1)=sqrt(65)~=8.1AB=(91)2+(45)2=64+1=658.1
BC=sqrt((1-9)^2+(8-4)^2)=sqrt(64+16)=sqrt(80)=4sqrt(5)~=8.9BC=(19)2+(84)2=64+16=80=458.9
CA=sqrt((1-1)^2+(5-8)^2)=3CA=(11)2+(58)2=3

Using the variables m, n and om,nando
m+n=4sqrt(5)m+n=45
m+o=sqrt(65)m+o=65
n+o=3n+o=3 => o=3-no=3n
=> m+3-n=sqrt(65)m+3n=65 => m-n=sqrt(65)-3mn=653
Adding the last with the first equation, we get
2m=4sqrt(5)+sqrt(65)-32m=45+653 => m=2sqrt(5)+sqrt(65)/2-1.5m=25+6521.5

In the figure we can see that
tan alpha=r/mtanα=rm => r=m*tan alphar=mtanα

We only need to know alphaα

Using vectors
BvecA=(1-9)hat i+(5-4)hat j=-8hat i+hat jBA=(19)ˆi+(54)ˆj=8ˆi+ˆj
BvecC=(1-9)hat i+(8-4)hat j=-8hat i+4hat jBC=(19)ˆi+(84)ˆj=8ˆi+4ˆj

cos (2alpha)=(BvecA*BvecC)/(|BA|*|BC|)=(8*8+1*4)/(sqrt(65)*4sqrt(5))=68/(20sqrt(13))=17/(5sqrt(13))cos(2α)=BABC|BA||BC|=88+146545=682013=17513
=> 2alpha=19.440^@2α=19.440 => alpha=9.720^@α=9.720

Using the Law of Sines
3/sin(180^@-2alpha)=sqrt(65)/sin(180^@-2beta)=(4sqrt(5))/sin(180^@-2gamma)3sin(1802α)=65sin(1802β)=45sin(1802γ)
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)3sin2α=65sin2β=45sin2γ
3/(sin 2alpha)=sqrt(65)/(sin 2beta)3sin2α=65sin2β => sin 2beta=sqrt(65)/3*sin 2alphasin2β=653sin2α

We know also that 2alpha+2beta+2gamma=180^@2α+2β+2γ=180 => 2gamma=180^@-2alpha-2beta2γ=1802α2β
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)3sin(2α)=45sin(1802α2β)=45sin(2α+2β)=45sin2αcos2β+sin2βcos2α

-> 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