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Square Circles (Posted on 2004-05-27) Difficulty: 3 of 5
Given:

Three circles A, B and C.

Each circle is tangent to the other two.

The radius of A is 20.

The radius of B is 30.

Questions:

How many unique values of radius C exist where the centers of A, B and C form a right triangle? (Unique: Do not count triangles which are equal through flips and rotations. You may only count dissimilar triangles and similar triagles of differing sizes.)

What are the values?

See The Solution Submitted by Axorion    
Rating: 4.0000 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution re: are you from Russia? | Comment 12 of 24 |
(In reply to are you from Russia? by GOM)

Indeed you are right: there are only three size/shaped unique right triangles that can be formed.  That brings up a peculiarity in the way the question was asked: first it asks for the number of unique values for the radius of C (which is 6; the radii are all different); then it asks that the triangles themselves not be congruent, leaving a count of only three.  The summary is:

If A is internally tangent to B and the center of the small circle is at the right angle, the distance between their centers is 10, the longer leg is 20+r where r is the radius of C, and the hypotenuse is 30-r, so: 10^2+(20+r)^2 = (30-r)^2; this makes r = 4.  The sides of the triangle are 10,24,26.

If the center of the large circle is the right angle, then 10^2+(30-r)^2 = (20+r)^2; thus r = 6. The sides of the triangle are 10,24,26.

If  A and B are externally tangent and both are external to C, A and B are separated by 50 units.

The center of C will then be at a right angle of the formed triangle if (20+r)^2 + (30+r)^2 = 50^2, so r = 10. The sides of the triangle are 30, 40, 50.

The right angle will be at the center of A if (20+r)^2 + 50^2 = (30+r)^2; so r = 100. The sides of the triangle are 120,50,130.

If A and B are externally tangent to each other but internal to C, then if the right angle is at the center of C then, (r-30)^2 +(r-20)^2 = 50^2; so r = 60. The sides of the triangle are 30, 40, 50.

If A and B are externally tangent to each other but internal to C but this time the right angle is at the center of B, (r-30)^2 + 50^2 = (r-20)^2; r = 150. The sides of the triangle are 50, 120, 130.

So while there are 6 different values for the radius of C, there are only three unique size-shape of triangle formed.  I had interpreted the question to merely rule out flips of the entire diagram, but indeed it rules out any congruent triangles.  But then we're not really finding all the unique values for the radius of C.

Corrected one of the triangle sizes above.

 

Edited on May 28, 2004, 3:02 pm
  Posted by Charlie on 2004-05-27 18:05:50

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