luong@tonto.DEC (Van Luong Nguyen UHO DTN 264-6560) (07/13/84)
There must be many possible ways to solve this problem. The solution below gives the radius of not only the third circle, but of any circle in the series. Terminology: ----------- By symmetry, the tangents L1, L2, and the line L3 joining the centers of the five circles are concurrent, meeting at the same point A. Call the centers of the circles C1,C2,C3,C4,C5. Let line L3 meet the smallest circle 1 at B1, the common point to circles 1 and 2 at B2, the common point to circles 2 and 3 at B3, etc... Call the distance ABn = X(n), and call by R(n) the radius of circle n. Thus, for example: AB1 = X(1) AC1 = AB1 + B1C1 = X(1) + R(1) AB2 = X(2) = AB1 + 2*R(1) = X(1) + 2R(1) etc... Proof: ----- From centers C1, C2, C3, ... draw perpendiculars C1H1, C2H2, C3H3, ... to the tangent L1. Thus C1H1 = R(1), C2H2 = R(2) ... Let X(1) = AB1 = d . Since C1H1 and C2H2 are parallel, C2H2 = AC2 ie. R(2) = d + 2R(1) + R(2) ---- --- ---- ---------------- C1H1 AC1 R(1) d + R(1) Using R(1) = 8 for solving: R(2) = 8 [(d + 16)/d] Also, X(2) = AB2 = d + 2R(1) therefore: X(2) = (d + 16) By recursion, it is easy to prove that, for n >= 2 : R(n) = 8 [(d+16)/d]**(n-1) (to the (n-1)th power) and: X(n) = [(d + 16)**(n-1)] / [d**(n-2)] (A recursion proof shows that if a statement or formula is true for n, it will also be true for (n+1). Since the statement is true for n=2, it follows that it is also true for all n>=2 . I will leave the recursion proof to the reader.) For n=5, R(5) = 8 [(d+16)/d]**4 = 18 therefore [(d+16)/d] = (18/8)**(0.25) For n=3, R(3) = 8 [(d+16)/d]**2 = 8 [(18/8)**0.25]**2 = 8 [(18/8)**0.5] Thus: R(3) = 12 ========= For any general value n>=2 : R(n) = 8 [(18/8)**(0.25*(n-1))] For example, if we draw four more larger circles 6,7,8,9, then: R(9) = 8[(18/8)**2] = 40.5 Van Luong Nguyen, Digital Equipment Corp.