Circles and Tangents

Recently I came across a very nice puzzle related to geometry.

Problem : Take any three non-intersecting circles of distinct radii. Common tangents of any two will intersect at a point. Prove that the three intersection points due to three pairs of tangents are co-linear.

Solution : Ramprasad and I came up with a proof but we believe that there should be much simple and elegant proof. If you can throw any light, please comment. Here is the proof.

Let the center of circles A, B and C be denoted by C_A, C_B,C_C. Also the intersection points be denoted by X_{AB}, X_{BC}, X_{AC} and radii are r_A,r_B,r_C.

Given any two circles, say A and B. Following relation holds due to congruence of the triangles formed.

\frac{X_{AB}C_{A}}{X_{AB}C_B} = \frac { r_A} { r_B}. Now, X_{AB}C_{A} = X_{AB}C_B + C_B C_A

Simplifying, we will get X_{AB}C_A = \frac{r_A}{r_A-r_B}C_AC_B. Thus, we have got how far a point will be given the radii of two circles and the distance between their centers. Let \alpha_{AB} = \frac{r_A}{r_A-r_B}C_AC_B.

Now, Let C_A be the origin and with respect to that the vectors of C_B,C_C be denoted as v_B,v_C.

So, vector v_{XAB} for X_{AB} will be \alpha_{AB}v_B.

Similarly, v_{XAC} = \alpha_{AC}v_C and v_{XBC} = \alpha_{BC}(v_C-v_B) + v_B = ( 1 - \alpha_{BC})v_B + \alpha_{BC}v_C

For three points to be collinear, following must hold for some \lambda.

v_{XBC} = \lambda v_{XAB} + ( 1 - \lambda ) v_{XAC}

Therefore, we must prove that for some \lambda

( 1 - \alpha_{BC})v_B + \alpha_{BC}v_C = \lambda \alpha_{AB}v_B + ( 1 - \lambda) \alpha_{AC}v_C

\lambda=\frac{ 1- \alpha_{BC} }{ \alpha_{AB} } and \lambda = 1 - \frac{\alpha_{BC} }{\alpha_{AC} }


You can verify that it is indeed the case that

\frac{ 1- \alpha_{BC} }{ \alpha_{AB} } = 1 - \frac{\alpha_{BC} }{\alpha_{AC} }


Very ugly math, and with very high probability I must have made a typo in the calculation. Please try it yourself and convince that the property holds. I am looking for a better proof!


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6 Responses to “Circles and Tangents”

  1. Dinesh Says:
    Is this what you are looking for? 🙂
    Proof with pictures I guess.

  2. Saket Says:

    Hey, another similar problem I read was this,

    “Four roads on a plane, each a straight line, are in general position so that no two are parallel and no three pass through the same point. Along each road walks a traveler at a constant speed. Their speeds, however, may not be the same. It’s known that traveler #1 met with Travelers #2, #3, and #4. #2, in turn, met #3 and #4 and, of course, #1. Please show that #3 and #4 have also met.”

    Try to come up with a solution that could be explained very well without any formulae at all. There is a very pretty solution to it.

  3. Boris Yakubchik Says:

    I may be mistaken, but I think here’s a clever solution:

    Any picture you draw, one can imitate it on a screen of a camera pointing to a 3D scene of 3 identically-sized spheres sitting on a 2D plane (move spheres back to make them look smaller on the screen). It should not be a surprise that when you connect the edges of the spheres – the lines that recede to the distance meet at the ‘horizon’.


  4. Boris Yakubchik Says:

    I hope I remember the way to do it correctly: in projective geometry there’s “duality” which can convert the statement that you wanted to prove to the following claim:

    “If you draw 3 intersecting circles, connect the 3 pairs of intersecting points, the resulting 3 lines all intersect at one point”

    Do you think this statement is easier to prove?

    Geometer’s Sketchpad construction I made seems to confirm the claim I made. Math blows my mind.

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