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The following problem has been given to me and I have not been able to solve it at first glance completely (only that there is a unique solution).

Given two points A and B located on either side of a line MN determine with the ruler and compass a point of MN whose difference of the distances to points A and B is maximum.

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  • $\begingroup$ Can you explain how you know there is a unique solution? $\endgroup$ – David Quinn Nov 19 '16 at 11:54
  • $\begingroup$ @David: read the Cave's answer, please. $\endgroup$ – Piquito Nov 19 '16 at 12:46
  • $\begingroup$ For any point $P$ in the plane. The locus for point $Q$ satisfying $|BQ - AQ| = |BP - AP|$ is a hyperbola having foci $A, B$ passing through $P$. Furthermore, the tangent vector of that hyperbola at $P$ bisects the two rays $AP$, $BP$. For any point $K$ on $MN$ to maximize $|BK - AK|$, the line $MN$ will be tangent to the hyperbola through $K$. So $MN$ bisects rays $AK$ and $BK$. This in turns implies $K$ lies on $AB'$ as described in Cave's answer. $\endgroup$ – achille hui Nov 19 '16 at 18:49
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Suppose $A,B$ is not symmetric, otherwise the problem is meaningless. And when the distances of $A,B$ to the line $MN$ is the same, the maximum does not exist since the difference of distances will keep growing as the chosen point tends to infinity. Now we rule out those two cases. Reflect the point $B$ along the line $MN$ to get a point $B'$(this is possible with ruler and compass). Suppose the line $B'A$ intersect with $MN$ at $K$. Then $|BK-AK|$ attains its maximum.

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  • $\begingroup$ Actually A and B should be not symmetric as you say but also the distances of both points to the line MN should be distinct because if not your point K is not a point of MN. Your answer is correct so my upvote and if you explain better it I will put the mark of best answer. $\endgroup$ – Piquito Nov 19 '16 at 12:44
  • $\begingroup$ @Piquito Thanks for pointing out the flaw. I added the argument for that case. $\endgroup$ – Cave Johnson Nov 19 '16 at 13:16
  • $\begingroup$ I did not refer to the missing case of no solution but to explain why your $|BK-AK|$ is maximum. If you won't added this, I shall do after some waiting. $\endgroup$ – Piquito Nov 19 '16 at 15:16

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