Problem:
Let P be a point in the plane of △ABC, and γ a line passing through P. Let A′,B′,C′ be the points where the reflections of lines PA,PB,PC with respect to γ intersect lines BC,AC,AB, respectively. Prove that A′,B′,C′ are collinear.
Solution:
Solution 1: The proof is split into two cases.
Case 1: P is on the circumcircle of ABC. Then P is the Miquel point of A′,B′,C′ with respect to ABC. Indeed, because ∠A′B′C′=∠CBA=∠CPA=∠A′PC′, points P,A′,B′,C′ are concyclic, and the same can be said for P,A,B′,C′ and P,A′,B′,C. Hence ∠CA′B′=∠CPB′=∠BPC′=∠BA′C′, so A′B′C′ are collinear.
Case 2: P is not on the circumcircle of ABC. Let Q be isogonal conjugate of P with respect to ABC (which is not degenerate).
Claim. Let Q′ be the isogonal conjugate of P with respect to AB′C′. Then Q=Q′. Proof of the claim. Note that
∠BQC=∠BAC+∠CPB (because P and Q are isogonal conjugates in ABC ) =∠C′AB′+∠B′PC′=∠C′Q′B′ (because P and Q are isogonal conjugates in AB′C′ ).
Let X,Y,Z denote the reflections of P in sides BC,CA,AB, respectively, and let X′ denote P 's reflection in side B′C′ of triangle AB′C′. Then ∠ZXY=∠BQC (because QC is orthogonal to XY and QB is orthogonal to XZ ), whereas ∠ZX′Y′=∠C′Q′B′ because Q′B′ is orthogonal to X′Y and Q′C′ is orthogonal to X′Z and Q′C′ is orthogonal to X′Z, so since ∠C′Q′B′=∠BQC, we get ∠ZXY=∠ZX′Y. It follows that X,Y,Z,X′ are concyclic. The center of the XYZ-circle is Q while the center of the X′Y′Z-circle is Q′. Thus Q=Q′.
Note. We have made use of the well-known fact that the circumcenter of the triangle determined by the reflections of a point across the sidelines of another given triangle is precisely the isogonal conjugate of the point with respect to that triangle. For a proof see R. A. Johnson, Advanced Euclidean Geometry, 1929 ed., reprinted by Dover, 2007.
Similar arguments show that Q is also the isogonal point of P with respect to triangles A′BC′ and A′B′C. Therefore,
∠BC′A′=∠AC′A′=∠AC′P+∠PC′Q+∠QC′A′=∠QC′B′+∠PC′Q+∠BC′P=∠BC′B′=∠AC′B′
This means that A′,B′,C′ are collinear.
This problem and solution were suggested by Titu Andreescu and Cosmin Pohoata.
Solution 2: It's easy to see (say, by law of sines) that
BC′AC′=BPsin∠BPC′APsin∠APC′,CA′BA′=CPsin∠CPA′BPsin∠BPA′,AB′CB′=APsin∠APB′CPsin∠CPB′
The construction of A′,B′,C′ by reflections implies that
sin∠APC′=sin∠CPA′,sin∠BPC′=sin∠CPB′,sin∠BPC′=sin∠CPB′
Hence,
BC′AC′⋅CA′BA′⋅AB′CB′=1
and the proof is complete by Menelaus' theorem.
This second solution was suggested by Li Zhou, Polk State College, Winter Haven FL.
The problems on this page are the property of the MAA's American Mathematics Competitions