PUMaC 2020 · 个人决赛（A 组） · 第 2 题
PUMaC 2020 — Individual Finals (Division A) — Problem 2
题目详情
- Helen has a wooden rectangle of unknown dimensions, a straightedge, and a pencil (no com- pass). Is it possible for her to construct a line segment on the rectangle connecting the mid- points of two opposite sides, where she cannot draw any lines or points outside the rectangle? Note: Helen is allowed to draw lines between two points she has already marked, and mark the intersection of any two lines she has already drawn, if the intersection lies on the rectangle. Further, Helen is allowed to mark arbitrary points either on the rectangle or on a segment she has previously drawn. Assume that only the four vertices of the rectangle have been marked prior to the beginning of this process. n
解析
- Helen has a wooden rectangle of unknown dimensions, a straightedge, and a pencil (no com- pass). Is it possible for her to construct a line segment on the rectangle connecting the mid- points of two opposite sides, where she cannot draw any lines or points outside the rectangle? Note: Helen is allowed to draw lines between two points she has already marked, and mark the intersection of any two lines she has already drawn, if the intersection lies on the rectangle. Further, Helen is allowed to mark arbitrary points either on the rectangle or on a segment she has previously drawn. Assume that only the four vertices of the rectangle have been marked prior to the beginning of this process. Solution : We will show that we can construct the midpoint of any edge of the rectangle. First, we draw the diagonals, and mark their intersection E . Then, we pick an arbitrary point X on AE , not being A or E . Then, we intersect DX with AB to get Y and EY with BX to get Z . The claim is that the pencil of lines ( AB, AE ; AZ, AD ) are harmonic. To see why, it suffices to let T be the intersection of AZ and BE . From Ceva’s − → − − → − − → BT EX AY theorem in triangle ABE we have that · · = 1. Similarly, Menelaus’ theorem in − → − − → − − → T E XA Y B − − → − − → − − → − − → − → BD EX AY BD BT triangle ABE gives · · = − 1. These equations imply = − , which gives the − − → − − → − − → − − → − → DE XA Y B DE T E above claim. This means that the line AZ passes through the midpoint of BC (because AD || BC , and projecting the harmonic pencil ( AB, AE ; AD, AZ ) onto BC gives respectively the B , C , the point at infinity and the midpoint of BC ). Other midpoints can be constructed similarly. Proposed by Daniel Carter. n