Inside the circle with center $O$, points $A$ and $B$ are marked so that $OA = OB$. Draw a point $M$ on the circle from which the sum of the distances to points $A$ and $B$ is the smallest among all possible.
Problem
Source: 2012 Oral Moscow Geometry Olympiad grades 10-11 p5
Tags: minimum, Sum, distance, geometry, circle
mathlomaniac
25.09.2019 15:29
That would be the midpoint of chord AB and u can easily find it by drawing perpendicular on AB from centre.
natmath
25.09.2019 15:52
mathlomaniac wrote: That would be the midpoint of chord AB and u can easily find it by drawing perpendicular on AB from centre. I don't think you saw this, but A and B are not necessarily on the edge of the circle (they can be inside), also M must be on the circle, not on the inside. The mid point of a chord is not on the circle.
zuss77
25.09.2019 16:38
It would be the tangency point of ellipse with focii $A,B$ with the circle. To construct it we may use optical property of ellipses (normal bisects angle between focii). So $MO$ bisects $\angle AMB$. With $AO=BO$ it means that $AOBM$ - cyclic. So $M$ lays on intersection of $(AOB)$ with original circle. If $(AOB)$ lays inside circle, closest point would be where ray $ON$ cut the circle ($N$ - midpoint of $AB$).
mathlomaniac
25.09.2019 19:10
natmath wrote: mathlomaniac wrote: That would be the midpoint of chord AB and u can easily find it by drawing perpendicular on AB from centre. I don't think you saw this, but A and B are not necessarily on the edge of the circle (they can be inside), also M must be on the circle, not on the inside. The mid point of a chord is not on the circle. Oh i didn't notice