The problem is from Kiselev's Geometry Book I. Exercise 591:

Three congruent circles intersect at one point. Prove that the three lines, each passing through the center of one of the circles and the second intersection point of the other two circles, are concurrent.

Here is a picture of the problem:

enter image description here

I am not sure how to approach this problem. I think algebraic method is the way to go, but my attempt was too messy; basically, I used the equation of the formula, picked up a point they all intersect, and tried to simply manipulate all the formulas to prove the statement. Not only was it arduous and fruitless but I also wanted to have (possibly) a more clean and intuitive solution.

[Edited] I was informed that this is the Johnson circles configuration. I have followed all the properties listed in the Wikipedia, but still I was not able to prove this. I suspected that the intersection of the $3$-intersection and $2$-intersection will cross each circle's center, but it was wrong. I also attempted to prove that the resulting point from the concurrent lines is the excircle or incircle of one of the triangles (Johnson or reference), but it does not seem like it.

I would really appreciate if someone could help me on this problem. I think there should be a solution using a basic geometric method.

  • $\begingroup$ You should include the work you did with the algebraic approach. This will help people avoid wasting time duplicating your effort; and it's possible that someone will see a way to streamline your argument. $\endgroup$ – Blue Jul 31 at 12:25
  • $\begingroup$ @Blue I know that and I have always provided all the detail if I could do something. In this case I could not find any elementary method and as I have said, algebraic method was just a bunch of manipulating the formula. I will make it more clear in the problem. $\endgroup$ – Taxxi Jul 31 at 13:02
  • 1
    $\begingroup$ This configuration is called "Johnson circles configuration". It has a lot of properties. $\endgroup$ – Jean Marie Jul 31 at 13:14
  • $\begingroup$ @JeanMarie Thank you so much for the reference. At my first glance, I still cannot figure what which property will help to prove the theorem. I will have a close look tomorrow (it is night time here.) $\endgroup$ – Taxxi Jul 31 at 13:21
  • $\begingroup$ @JeanMarie I was able to follow all the properties in the Wikipedia, but I was still unable to prove it. Maybe I am not yet fully familiar with working on it. Do you have any hint or advice about this problem? $\endgroup$ – Taxxi Aug 1 at 3:02

To make the question answered, here is the solution; each such line passes from one vertex of the Johnson triangle to the other homothetic counterpart in the reference triangle. The homothetic center should be on each line, hence the lines are concurrent.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.