Geometry Problem

Click the figure below to see the complete problem 449 about Congruent Tangent Circles, Diameter, Sagitta, Arc, Perpendicular, Chord.

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Complete Problem 449

Level: High School, SAT Prep, College geometry

## Monday, May 10, 2010

### Problem 449: Congruent Tangent Circles, Diameter, Sagitta, Arc, Perpendicular, Chord

Labels:
arc,
chord,
circle,
right triangle,
sagitta,
semicircle,
tangent

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Let O,I,R,r be center and radius of circumcircle and incircle of ABC respectively

ReplyDelete2[(R-a)+(R-b)]=2(OF+OD)=2(DB+FB)=AB+BC ⇒

r=(AB+BC-AC)/2={2[(R-a)+(R-b)]-2R}/2=R-a-b ...(1)

(R-a)²+(R-b)²=OF²+OD²=FD²=(AC/2)²=R² ⇒

(r+a)²+(r+b)²=(r+a+b)² ⇔

r=√(2ab) ...(2)

consider similar triangles about incenter

4x/(2R)=4x/AC=(r-x)/r ⇔

x=rR/(R+2r)

=r(a+b+r)/(a+b+3r) (by (1))

=√(2ab)(a+b+√(2ab))/(a+b+3√(2ab)) (by (2))

Please send diagram of this solution

DeleteRefer problems # 275 and 435 -- We've r^2=2ab, BC + 2a = AC, AB + 2b = AC & AC + 2r = AB + BC which leads to AC = 2(a+b+r). Further, 4x/AC = (r-x)/r or x = (r*AC)/(AC+4r)= 2r(a++b+r)/(2a+2b+6r) =(√(2ab)(a+b+√(2ab))/(a+b+3√(2ab)) QED.

ReplyDeleteAjit