Geometry Problem

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

Click the figure below to see the complete problem 764.

## Thursday, June 14, 2012

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

Labels:
altitude,
harmonic mean,
inradius,
triangle

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Let S be the area of triangle ABC,and a,b,c be the lengths of the sides

ReplyDeleteBy 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.

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

ReplyDeleteDenote 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

Let BI extended meet AC at E.

ReplyDeleteBI: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)

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

ReplyDeleteSumith Peiris

Moratuwa

Sri Lanka