## Thursday, June 14, 2012

### Problem 764: Triangle, Inradius, Altitudes, Harmonic Mean

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 764. 1. Let S be the area of triangle ABC,and a,b,c be the lengths of the sides
By the properties of incenter
S = (1/2)(r)(a+b+c)
Now we consider the area of the triangle and the altitudes
S = (Ha)(a)/2 = (Hb)(b)/2 = (Hc)(c)/2
Rearranging, 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.

1. Thanks bro

2. http://img850.imageshack.us/img850/6278/problem764x.png

Denote S(XYZ)= area of triangle XYZ
Draw 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

3. Let BI extended meet AC at E.
BI:IE = (a + c): b
So 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)
4. 