tag:blogger.com,1999:blog-6933544261975483399.post6422674010089714783..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 815: Triangle, Incircle, Semicircles, Arbelos, AreaAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-83883983385782665142012-10-17T22:05:05.122-07:002012-10-17T22:05:05.122-07:00*Typo :
For the last 3 lines of my proof:
= 4(1/π...*Typo :<br /><br />For the last 3 lines of my proof:<br />= 4(1/π)(s/Δ)^2, where Δ is the area of triangle ABC<br />= 4(1/π)(1/r)^2, where r is the inradius<br />= 4/SW Fungnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-70768670596351427612012-10-17T17:46:23.388-07:002012-10-17T17:46:23.388-07:00Let AB = c, BC = a, CA = b, and s = (a+b+c)/2
Then...Let AB = c, BC = a, CA = b, and s = (a+b+c)/2<br />Then BD = BE = (s-b), EC = FC = (s-c), FA = AD = (s-a)<br />The area of arbelo <br />Sa = π((BE/2)(EC/2)) = π(s-b)(s-c)/4<br />Similarly, <br />Sb = π(s-a)(s-c)/4<br />Sc = π(s-a)(s-b)/4<br /><br />1/Sa + 1/Sb + 1/Sc <br />= 4[(s-a)+(s-b)+(s-c)] / π[(s-a)(s-b)(s-c)]<br />= 4s/π[(s-a)(s-b)(s-c)]<br />= {4s/π[(s-a)(s-b)(s-c)]}<br />= (1/π)(4s/[(s-a)(s-b)(s-c)])<br />= (1/π)(s/Δ)^2, where Δ is the area of triangle ABC<br />= (1/π)(1/r)^2, where r is the inradius<br />= 1/S<br /><br />W Fungnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-89547727589301638732012-10-17T17:19:23.117-07:002012-10-17T17:19:23.117-07:00Let BC=a, AC=b, AB=c, s=1/2(a+b+c).
Then
AD=AF=s...Let BC=a, AC=b, AB=c, s=1/2(a+b+c). <br /><br />Then<br />AD=AF=s−a, BD=BE=s−b, CE=CF=s−c<br /><br />Sa<br />= 1/8 π[a^2 − (s−b)^2 − (s−c)^2]<br />= 1/8 π[a^2 − 1/4 (a+c−b)^2 − 1/4 (a+b−c)^2]<br />= 1/8 π[1/2 a^2 − 1/2 (b−c)^2]<br />= 1/16 π[(a+b−c)(a−b+c)]<br />= 1/4 π (s−b)(s−c)<br /><br />1/Sa = 4/π 1/[(s−b)(s−c)]<br />1/Sb = 4/π 1/[(s−a)(s−c)]<br />1/Sc = 4/π 1/[(s−a)(s−b)]<br /><br />1/Sa + 1/Sb + 1/Sc<br />= 4/π [1/[(s−b)(s−c)]+1/[(s−a)(s−c)]+1/[(s−a)(s−b)]]<br />= 4/π s/[(s−a)(s−b)(s−c)]<br />= 4/SJacob HA (EMK2000)https://www.blogger.com/profile/17238561555526381028noreply@blogger.com