Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 807.
Saturday, September 29, 2012
Problem 807: Right Triangle Area, Incircle, Tangency Points, Semicircle Area
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Let AB = a, BC = b and AC = c.
ReplyDeleteThen
S = 1/2 ab
AD = 1/2 (a+c-b)
CE = 1/2 (b+c-a)
S1 = 1/32 π (a+c-b)^2
S2 = 1/32 π (b+c-a)^2
S1*S2
= 1/1024 π^2 [(a+c-b)(b+c-a)]^2
= 1/1024 π^2 [c^2 - (a-b)^2]^2
= 1/1024 π^2 [2ab]^2
= 1/256 π^2 (ab)^2
√[S1*S2]
= 1/16 π ab
= π/8 S
S = 8/π √[S1*S2]
ΑD=AF=x ,FC=CE=y,BE=BD=z .if a+b+c=2t We know that x=t-a,y=t-c,z=t-b
ReplyDeleteS1=π(t-a)^2/8 so √[8S1/π]=t-a similarly √[8S2/π]=t-c so (8/π)√(S1S2)=(t-a)t-c)
S=√t(t-a)(t-c)(t-b) but easily seen (t-a)t-c)=√t(t-a)(t-c)(t-b)because a^2+c^2=b^2
solution by Michael Tsourakakis from Greece
http://alnasiry.net/forums/uploaded/9_01349036799.png
ReplyDeletelet |EC|=x ,|EI|=r=|BD|,|AD|=√y
s=A(CEIF)+A(BEID)+A(ADIF)
s=x.r +r^2+y.r
but s=1/2 (x+r)(y+r)
=1/2 (xy+xr+yr+r^2 )=1/2 (xy+s)
s=xy
s_2=1/2 (x^2/4)π x=√((8s_2)/π)
s_1=1/2 (y^2/4)π y=√((8s_1)/π)
s=xy=8/π √(s_1.s_2 )