Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 765.
http://img43.imageshack.us/img43/9397/problem765.pngDenote S(XYZ)= area of triangle XYZDraw additional lines per attached sketch.We have ∆AA’F similar to ∆AIESo IE/A’F=r/rA =AI/AA’= S(IC’B’)/S(A’B’C’)Similarly r/rB=S(IA’C’)/S(A’B’C’) and r/rC=S(IA’B’)/S(A’B’C’)But S(A’B’C’)=S(IA’C’)+S(IA’B’)+S(IC’B’)So r/rA+r/rB+r/rC= 1 or 1/r=1/rA+1/rB+1/rC
Let S be the area of the triangle and X = (1/2)(a+b+c), where a,b,c are the side lengths of the triangle.By the properties of the exradii, S=(Ra)(X-a) ; S=(Rb)(X-b) ; S=(Rc)(X-c).Therefore,S/(Ra)=(X-a)S/(Rb)=(X-b)S/(Rc)=(X-c)By summing up, S[(1/Ra)+(1/Rb)+(1/Rc)]=3X-(a+b+c) = X.By the properties of the inradii, S=RX => X =S/RSo, S/R = S[(1/Ra)+(1/Rb)+(1/Rc)]1/R = (1/Ra)+(1/Rb)+(1/Rc)q.e.d.Now,
Let BC touch incircle at X and excircle opp to B at YTriangles BIX and BI₂Y are similar r/r₂ = AX/AY = (s - b)/sSimilarly r/r₃ = (s - c)/s and r/r₁= (s - a)/sr/r₁ + r/r₂ + r/r₃ =(s - a)/s + (s - b)/s + (s - c)/s= [3s - (a + b + c)] = (3s - 2s) /s = 1Hence 1/r₁ + 1/r₂ + 1/r₃ = 1/r