Geometry Problem

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

Click the figure below to see the complete problem 815.

## Wednesday, October 17, 2012

### Problem 815: Triangle, Incircle, Semicircles, Arbelos, Area

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Let BC=a, AC=b, AB=c, s=1/2(a+b+c).

ReplyDeleteThen

AD=AF=s−a, BD=BE=s−b, CE=CF=s−c

Sa

= 1/8 π[a^2 − (s−b)^2 − (s−c)^2]

= 1/8 π[a^2 − 1/4 (a+c−b)^2 − 1/4 (a+b−c)^2]

= 1/8 π[1/2 a^2 − 1/2 (b−c)^2]

= 1/16 π[(a+b−c)(a−b+c)]

= 1/4 π (s−b)(s−c)

1/Sa = 4/π 1/[(s−b)(s−c)]

1/Sb = 4/π 1/[(s−a)(s−c)]

1/Sc = 4/π 1/[(s−a)(s−b)]

1/Sa + 1/Sb + 1/Sc

= 4/π [1/[(s−b)(s−c)]+1/[(s−a)(s−c)]+1/[(s−a)(s−b)]]

= 4/π s/[(s−a)(s−b)(s−c)]

= 4/S

Let AB = c, BC = a, CA = b, and s = (a+b+c)/2

ReplyDeleteThen BD = BE = (s-b), EC = FC = (s-c), FA = AD = (s-a)

The area of arbelo

Sa = π((BE/2)(EC/2)) = π(s-b)(s-c)/4

Similarly,

Sb = π(s-a)(s-c)/4

Sc = π(s-a)(s-b)/4

1/Sa + 1/Sb + 1/Sc

= 4[(s-a)+(s-b)+(s-c)] / π[(s-a)(s-b)(s-c)]

= 4s/π[(s-a)(s-b)(s-c)]

= {4s/π[(s-a)(s-b)(s-c)]}

= (1/π)(4s/[(s-a)(s-b)(s-c)])

= (1/π)(s/Δ)^2, where Δ is the area of triangle ABC

= (1/π)(1/r)^2, where r is the inradius

= 1/S

*Typo :

DeleteFor the last 3 lines of my proof:

= 4(1/π)(s/Δ)^2, where Δ is the area of triangle ABC

= 4(1/π)(1/r)^2, where r is the inradius

= 4/S