Problem:
The inscribed circle of triangle ABC is tangent to AB at P, and its radius is 21. Given that AP=23 and PB=27, find the perimeter of the triangle.
Solution:
Let I be the center of the inscribed circle, and R be the point where the circle is tangent to CA. Because I is the intersection of the angle bisectors of ABC, it follows that ∠IAB+∠IBC+∠ICA=90∘. Notice that tan∠IAB=21/23,tan∠IBC=21/27, and tan∠ICA=21/CR. Thus
Therefore CR=245/2 and the perimeter of triangle ABC is 2(23+27+2245)=345.
OR
Use the same figure. Let a=BC,b=CA,c=AB,x=CR, and s=21(a+b+c). Then s=x+50,s−a=23,s−b=27, and s−c=x. The area of any triangle is the product of its inradius and its semiperimeter, so 21s=s(s−a)(s−b)(s−c), by Heron's formula. It follows that 212(x+50)=23⋅27⋅x. Solve this equation to obtain x=245/2 and 2s=2(x+50)=345.
OR
As in the preceding, let s be the semiperimeter of triangle ABC, and notice that BC=s−23. Because IB bisects ∠ABC, it follows that
sin∠ABC=2sin∠PBIcos∠PBI=2⋅1307⋅1309=6563
Hence the area of triangle ABC is 21AB⋅BCsin∠ABC=25(s−23)6563. The area of a triangle is also equal to the product of its inradius and its semiperimeter, so 25(s−23)6563=21s. Solve this equation to find that 2s=345, which is the perimeter.
