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

Labels:
area,
diameter,
incircle,
right triangle,
semicircle,
tangency point

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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 )