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Two circles with centres $O_1$ and $O_2$ intersect in points $A$ and $B$. The line $MP$ goes through points $M$, $A$ and $P$ ($M$ and $P$ are two other intersection points).
How can I prove that MP will have a maximum length if $MP$ is parallel to the line $O_1O_2$?enter image description here

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  • $\begingroup$ This question is missing some information. It looks like the line has to go through one of the circle intersections. Is that true? $\endgroup$ – amd Apr 17 '17 at 17:43
  • $\begingroup$ The line MP goes through the point A. (it is the line MAP) @amd $\endgroup$ – idliketodothis Apr 17 '17 at 17:49
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The quadrilateral made by $O_1,O_2$, the midpoint of $AM$ and the midpoint of $AP$ is a right trapezoid, hence the length of $MP$ is at most twice the length of $O_1 O_2$, unless the previous trapezoid is indeed a rectangle, i.e. iff $MP\parallel O_1 O_2$.

enter image description here

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  • $\begingroup$ @idliketodothis: because in a circle the perpendicular bisector of a chord always goes through the center of the circle. I am updating with a picture. $\endgroup$ – Jack D'Aurizio Apr 17 '17 at 18:12
  • $\begingroup$ And how did you decide that the length of MP is the most twice as the length of O1O2? Have never heard of such a theorem. $\endgroup$ – idliketodothis Apr 17 '17 at 18:18
  • $\begingroup$ @idliketodothis: let $JK$ be the segments having as endpoints the midpoints of $AM$ and $AP$. Then $JK\leq O_1 O_2$ because we have a right trapezoid and $MP=2\,JK$ because that is obvious, hence $MP\leq 2\,O_1 O_2$. $\endgroup$ – Jack D'Aurizio Apr 17 '17 at 18:30

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