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 764.
Let S be the area of triangle ABC,and a,b,c be the lengths of the sidesBy the properties of incenterS = (1/2)(r)(a+b+c)Now we consider the area of the triangle and the altitudesS = (Ha)(a)/2 = (Hb)(b)/2 = (Hc)(c)/2Rearranging, a = 2S/(Ha) , b = 2S/(Hb), c = 2S/(Hc)Hence, S = (1/2)(r)(2S)[1/(Ha)+1/(Hb)+1/(Hc)]1 = r [1/(Ha)+1/(Hb)+1/(Hc)]q.e.d.
http://img850.imageshack.us/img850/6278/problem764x.pngDenote S(XYZ)= area of triangle XYZDraw additional lines per attached sketch.We have ∆BEB’ similar to ∆IFB’So IF/BE=r/hB =B’I/B’B= S(IAC)/S(ABC)Similarly r/hA=A’I/A’A=S(IBC)/S(ABC) and r/hC=C’I/C’C=S(IAB)/S(ABC)But S(ABC)=S(IAC)+S(IAB)+S(IBC)So r/hA+r/hB+r/hC= 1 or 1/r=1/hA+1/hB+1/hC
Let BI extended meet AC at E.BI:IE = (a + c): bSo r/h_b = IE/BE = b/(a + b +c)Similarly r/h_c = c/(a + b +c) and r/h_a = a/(a + b +c)Hence by adding r/h_b + r/h_c + r/h_a = 1, 1/h_b + 1/h_c + 1/h_a = 1/r
Result follows easily by noting that double the area of Tr. ABC = r(a+b+c) = a(ha) = b(hb) = c(hc) and eliminating a, b, c Sumith Peiris MoratuwaSri Lanka