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Fox 330
Henk Reuling
SOLUTION
See figure for naming of the points and
edges.
ACP = 180º – α and BDQ = 180º – β.
Because PQ is tangent to both circles,
PQ is perpendicular to both radiuses AP
and BQ.
Therefore PAC = α – 90º and
QBD = β – 90º.
sin(90  φ)
cos(φ)
1


.
cos(90  φ)
sin(φ)
tan φ
1
x
r
Therefore: tan(PAC )  tan(α  90)  

→ tan α  
tan α r
x
1
x
R

and tan( QBD)  tan(β  90)  
→ tan β   .
tan β R
x
We know the identities: tan(φ  90)   tan(90   φ)  
From point A to radius BQ gives point S.
2
2
2
Pythagoras in triangle ABS: AS = AB – BS → (3 x) 2  ( R  r ) 2  ( R  r ) 2  ...  4rR .
2
9
2
This gives 9x = 4rR → rR = /4·x .
tan α  tan β  
r
R rR 9 x 2
  2  4 2  9 .
4
x
x x
x
The answer is: A