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Let be a circle of radius $1$ and center $O$. Consider the points $A$ and $B$ such that $OA=OB$. Find $M$ on the circle such that $ MA + MB$ to be minimum. Actually I am looking for a synthetic solutions, using a construction. I tried with analytic methods but it isn't the best way and to make some inversions too.

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  • $\begingroup$ isn't it the point where perpendicular bisector of points A and B meet the circle? $\endgroup$ – emil Jul 31 '18 at 14:03
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  • $MA+MB=k$ where $k$ is a constant is an ellipse of foci $A,B$
  • The big axis of the ellipse is supported by line $(AB)$
  • Since $OA=OB$ then the small axis is also the perpendicular bissector of $[AB]$

For now let just consider the easy case when $[AB]$ is outside the circle (like in emil's figure).

For $k$ small the ellipse and the circle of centre $O$ have no intersection.

  • When $k$ is growing the ellipse and circle will intersect, for the minimal $k$, they will be tangent to each other at a point $\color{green}{M_1}$ on the small axis.

enter image description here

You can also visualize a similar problem here:

minimal distance between two points and point on a plane

The cases where the segment is inside or intersect the circle seems more complicated.

  • For instance when $O$ is the middle of the segment and it is inside the circle (case $[CD]$) then when $k$ grows the tangency will happen in $\color{purple}{M_2}$ on the big axis this time.

  • But if the points are outside the circle (case $[EF]$) then the minimal $k$ is reached for the degenerated flat ellipse $k=EF$ and $\mathbf{M_4}$ is just one of the two intersection points of the segment with the circle.

  • The remaining cases seem even more complicated and I feel that analytic solution is the only way... For instance the point $\color{orange}{M_3}$ does not seem to have any particular property.

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Say, $OA=OB=x$, also $A,B$ subtend an angle of $2\theta$ on the center of the circle. Construct a perependicular bisector for points $A,B$. Point $M$ should be on the constructed perpendicular. Let $P$ be the point where $AB$ meet perpendicular bisector. So$$ AP=x\sin\theta$$ $$PM =|x\cos\theta-1| $$ $$AM=\sqrt{x^2-2x\cos\theta +1}$$ $$AM+BM=2\sqrt{x^2-2x\cos\theta +1}$$

P.S. When points $A, B$ lie inside of the circle, same proof can be used. Only difference is,$PM =1-x\cos\theta $. This is why I put modulus value for $PM.$

enter image description here

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  • $\begingroup$ $A$ and $B$ are fixed $\endgroup$ – rafa Jul 31 '18 at 14:20
  • $\begingroup$ Fixed it. ( Previously I took points A and B as points on the circle) $\endgroup$ – emil Jul 31 '18 at 14:45
  • $\begingroup$ If $OA \gt 1$ and line $AB$ intersects the circle with center $O$, the minimum is the straight line connecting $A$ and $B$. $\endgroup$ – Jens Jul 31 '18 at 15:04
  • $\begingroup$ That happens when $x\cos\theta=1$ in which if we substitute to the result, $AM+BM=2\sqrt{x^2-1}$ $\endgroup$ – emil Jul 31 '18 at 15:16
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This solution is valid only if $A$ and $B$ lie outside the circle. If segment $AB$ intersects the circle, $M$ is obviously either intersection point.

Suppose then $AB$ doesn't intersect the circle: in that case $M$ is the nearest intersection between the circle and the perpendicular bisector of $AB$, such that the circle and $AB$ are on opposite sides with respect to the tangent $t$ at $M$.

Proof.

Let $A'$ be the reflection of $A$ about $t$: then $M$ is the midpoint of $A'B$. If $N$ is any other point on $t$, different from $M$, then $AN+BN>AM+BM$, because: $$ AN+BN=A'N+BN>A'B=A'M+BM=AM+BM, $$ where I applied the triangular inequality to triangle $A'BN$. Let now $P$ be a point on the circle, different from $M$. Segment $PB$ will intersect $t$ at some point $Q$ and: $$ AP+BP=AP+PQ+BQ>AQ+BQ\ge AM+BM, $$ where I applied the triangular inequality to triangle $APQ$ and used the result proved above.

enter image description here

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