Let $O$ and $O'$ be two circles intersecting at two points, $A$ and $B$. Construct a line segment $l$ going through $A$ and such that if $l$ meet $O$ at $M$ and $O'$ at $M'$, the length of $AM$ and $AM'$ is the same.

You don't have to draw a picture; an instruction would suffice.

We can find the centers of two given circles, and we may assume one circle is strictly greater than the other. We can also consider that the perpendicular bisector of $AB$ makes things symmetric.

I am not sure what to do here. I remember seeing a solution before but cannot remember of it other than that it involved drawing two or more circles (there may be other ways to solve this problem)..


A hint: Reflect one circle across its tangent at $A$.

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  • $\begingroup$ Thank you, I got the solution. May I ask how you came up with the solution (i.e. motivation)? $\endgroup$ Apr 6 '19 at 13:34
  • $\begingroup$ Could you please explain the construction? $\endgroup$
    – user
    Apr 7 '19 at 14:59
  • $\begingroup$ @user: Draw a figure until you have three circles, two of them making a figure eight centered at $A$. $\endgroup$ Apr 7 '19 at 15:03
  • $\begingroup$ This part is clear. If I understood correctly you hint that the second intersection point of the reflected circle with $O$ is the point $M$ in question. What is the reason behind? $\endgroup$
    – user
    Apr 7 '19 at 15:11

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