where s and r denote its semiperimeter and its inradius, respectively. Prove that triangle ABC is similar to a triangle T whose side lengths are all positive integers with no common divisor and determine these integers.
Solution:
First Solution: For simplification, let u=cot2A,v=cot2B,w=cot2C. We stact with a few basic facts.
Fact 1. Let [R] denote the area region R. Then
[ABC]=s(s−a)(s−b)(s−c)=rs.
The first equality is the Heron's formula. The second equality follows from [ABC]=[AIB]+[BIC]+[CIA]=rs, where I is the incenter of triangle ABC.
Fact 2. We have
u=(s−b)(s−c)s(s−a)
Let w be the excircle of triangle ABC opposite A, and let IA be its center. (irclo w is tangent to side BC. rays AB and AC and X,Y,Z, respectively: By equal tangents, AY=AZ,BX=BY and CX=CZ. Hence AX=AY=s. Then
By (1), and by noticing that 22+32+62=72, we can rewrite the given relation ar
(62+32+22)[u2+(2v)2+(3w)2]=(6u+6v+6w)2
This means that we have equality in the Chauchy-Schwartz inequality. It follows that
6u=32v=23w
or,
u=36k,v=9k,u=4k,
for some positive real number k. Pluging these back into (1) gives k=317. and cons quently u=T⋅c=4T. and w=97. Hence by the Double angle formulas. sin 1=2Zˉ. sinB=6556. and sinC=6563. or,
sinA=732513,sinB=732540.sinC=73254.5
By the Extanded law of sines, triangle ABC is similar to triangle I with the side lengths 13, 40, and 45. (The circumradius of T is 7325.)
Second Solution: Let D be the point of tangency of the incircle of triangle ABC and side AB. Then AI=r and AE=s−a, where I is the incenter of triangle ABC. Hence u=AIAE=rs−a. Likewise, v=rs−b and w=rs−c. Since
rs=r(s−a)+(s−b)+(s−c)=u+v+w
we can rewrite the given relation as
49[u2+4v2+9w2]=36(u+v+w)2
Exapnding the last equality aud cancelling the like terms, we obtain