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Is this statement true?

A rectangle is a subset of a circle if and only if all its vertices are in the circle (or in its boundary).

Intuitively it's true, but can someone give me proof?

Thank you in advance!

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  • $\begingroup$ If any of the 4 vertices are outside the circle then its obviously not contained within the circle. For all 4 vertices being on or inside the circle use convexity - both a disk and a rectangle are convex shapes. Proving convexity for either shape is trivial. $\endgroup$ – AlphaNumeric Sep 21 '16 at 14:33
  • $\begingroup$ If you can prove that all the points in a line between two points in a circle are also in the circle that should be sufficient. As AlphaNumeric points out that's the cocept of convexity. I'm not sure I agree that it is trivial as we can't just take "A circle is convex as an act of god" as a given. We need some context. For example are we allowed the analytic definitions of Circle ={(x,y)|(x - a)^2 + (y-b)^2 le r^2} and Rectagle = ... whatever. Then it is easy. Not trivial but easy. $\endgroup$ – fleablood Sep 21 '16 at 15:50
  • $\begingroup$ I solved it. Yes we are allowed to use the analytic definition of the circle $\endgroup$ – Marios Gretsas Sep 22 '16 at 13:50
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Your statement is true. A proof relies on the fact that the points in a circle form a convex set. Suppose the four vertices of the rectangle are in the circle. Then because the circle is convex, the four edges of rectangle must also be in the circle. Then if you take any point on the interior of the rectangle, it must lie on a line-segment connecting a vertex of the rectangle to some point on an edge of the rectangle, so it too is in the circle.


Just a note on terminology, many people use the words circle and rectangle to refer to just the boundary of those shapes.

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The statement is true, but I don't know of an elementary proof. The proof that comes to mind uses the Krein-Milman Theorem: a compact set is equal to the closed convex hull of its extreme points.

Suppose all the vertices of a rectangle are inside the circle. Then every convex combination of the vertices is also inside the circle (since the circle is also convex). Since the set of all convex combinations of the vertices generates the rectangle, the rectangle is inside the circle.

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Circle is convex. Thus, for any 2 points in a circle, there is a straight line entirely contained in a circle that connects our 2 points. Thus, if 4 vertices of a rectangle are in a circle, so are the boundaries of a rectangle. Can you think about an argument explaining why every interior point of a rectangle is in your circle as well?

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