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

Labels:
circle,
distance,
perpendicular,
radius,
right angle,
tangent

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