Choose any four points in a plane, such that no three are collinear and the four do not lie on a circle.

Show that one of the points must lie within the circle formed by the other three.

(In reply to a proof by Charlie)

Charlie's proof is convincing, but it is hard to follow without a diagram. This is a variant on his proof that, hopefully, it is a little easier to follow.

For any four points meeting the conditions of the problem two things are true. You can construct 4 circles, each containing three of the points, and you can construct a quadrilateral whose vertices are the given points.

If the quadrilateral is concave, then one of the points is within the triangle formed by the other three, and thereby also within the circumscribing circle. This is a trivial solution

If the quadrilateral is convex, then you can select any two adjacent vertices. Call them A and B. The other two points (Call them C and D) are both on the same side of the line which includes AB. (This is easy to prove using analytic algebra -- and even easier to see intuitively.)

Construct the circles that contain ABC and ABD. You have two circles that intersect at A and B. On each side of line AB all the points of one circle are contained within the other circle. Specifically, on the side in which both C an D lie, all of the points of the inner circle [for example, ABD] include the third point on which it was constructed [D]. This point is therefore within the other circle [ABC] The argument is the same with the labels reversed if the inner circle is the one that contains C.

When you cross the common chord AB, which circle is within the other switches, which is why it was important to show that you can select C and D such that they are on the same side.

Posted by TomM on 2004-02-20 20:24:59