Geometry Problem
Click the figure below to see the complete problem 461 about Three Circles, Tangent, Right Angle, Center, Distance, Measurement.
See also:
Complete Problem 461
Level: High School, SAT Prep, College geometry
Saturday, May 22, 2010
Problem 461: Three Circles, Tangent, Right Angle, Center, Distance, Measurement
Subscribe to:
Post Comments (Atom)
From C draw a perpendicular to OF and from B draw a perpendicular to OG, Let the two intersect in P. Now CP = GM + OM - b. GM is the common tangent between A & C and hence = 2√(ac)while OM=a and thus, CP=(2√(ac)+a-b). Similarly, BP = HF + OH – c = (2√(ab)+a-c) while x^2 = BP^2 + CP^2. Hence etc.
ReplyDeleteVihaan
BC = x; proof:
ReplyDeletex²=CN²+BC²=(2√(ac)+a-b)²+(2√(ab)+a-c)²
Draw a line from B to OG and a line from C to OF.
The intersection of these two lines is N.
Draw a line from A to OG, the intersection with CN is P. Draw a line from A to OF, the intersection with BN is Q.
Now BN = BQ + NQ and CN = CP + NP.
BQ²=AB²-AQ²=(a+b)²-(a-b)²=a²+2ab+b²-(a²-2ab+b²)=4ab
BQ=2√(ab)
CP²=AC²-AP²=(a+c)²-(a-c)²=a²+2ac+c²-(a²-2ac+c²)=4ac
CP=2√(ac)
NQ=a-c
NP=a-b
BN=2√(ab)+a-c
CN=2√(ac)+a-b
x²=CN²+BC²=(2√(ac)+a-b)²+(2√(ab)+a-c)²
QED