User:Lethal lammy/sandbox

Reim's theorem
Reim's theorem is a result of Euclidean geometry which may be stated as follows:

Let ω1, ω2 be two circles intersecting at M,N. Let a line through M intersect ω1, ω2 at A1, A2. Let B1, B2 be points on ω1, ω2 respectively. Then A1B1 || A2B2 if and only if B1, N and B2 are  collinear.

We may prove this result as follows:

Suppose that B1NB2 is a straight line, and P is the intersection point of A1A2 and B1B2 (can be at infinity). Then MN is antiparallel to A1B1, since


 * $$PA_1 \cdot PM = PB_1 \cdot PN$$

by the intersecting chords theorem, and similarly, MN is also antiparallel to A2B2, hence A1B1 || A2B2.

For the reverse implication, let the second intersection of B1B2 with ω1 be N*. Then MN* is antiparallel to A1B1, by similar reasoning as in the first part, so A2B2 is antiparallel to MN* by parallelism, so N* lies on ω2 and N = N*. Q.E.D.

Applications
Reim's theorem has many appliations, especially in Olympiad geometry, since despite its simplistic nature, it is actually very profound and is useful in dealing with difficult configurations involving many lines and circles. Here are some applications of the theorem at the level of the International Mathematical Olympiad.

International Mathematical Olympiad (IMO) 2013, Problem 4
Let ABC be an acute triangle with orthocenter H, and let W be a point on the side BC, lying strictly between B and C. The points M and N are the feet of the altitudes from B and C, respectively. Denote by ω1 the circumcircle of BWN, and let X be the point on ω1 such that WX is a diameter of ω1. Analogously, denote by ω2 the circumcircle of triangle CWM, and let Y be the point such that WY is a diameter of ω2. Prove that X, Y and H'' are collinear.

Proposed by Warut Suksompong and Potcharapol Suteparuk, Thailand

Solution: If K is the second intersection of those two circles, then it lies on circle ANMH. Now, it is easy to see that X, K, Y are collinear. Suppose KH intersects the circle BNW at X*. Then, by Reim's theorem, we have X*B || AH, so XB is perpendicular to BC, and hence X* is the antipode of W. It follows X = X*, and X, Y and H are collinear, as required.