## Saturday, May 22, 2010

### Problem 461: Three Circles, Tangent, Right Angle, Center, Distance, Measurement

Geometry Problem
Click the figure below to see the complete problem 461 about Three Circles, Tangent, Right Angle, Center, Distance, Measurement.

Complete Problem 461

Level: High School, SAT Prep, College geometry

1. 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.
Vihaan

2. BC = x; proof:

x²=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