HMMT 二月 2019 · 团队赛 · 第 1 题

HMMT February 2019 — Team Round — Problem 1

[ 20 ] Let ABCD be a parallelogram. Points X and Y lie on segments AB and AD respectively, and AC intersects XY at point Z . Prove that AB AD AC

解析

[ 20 ] Let ABCD be a parallelogram. Points X and Y lie on segments AB and AD respectively, and AC intersects XY at point Z . Prove that AB AD AC

= . AX AY AZ Proposed by: Yuan Yao Solution 1. (Similar Triangles) ′ ′ ′ ′ Let X and Y lie on segments AB and AD respectively such that ZX ‖ AD and ZY ‖ AB . We note ′ ′ that triangles AXY and Y Y Z are similar, and that triangles AY Z and ADC are similar. Thus, we have ′ AC AD AY XZ = and = . ′ AZ AY AY XY This means that ′ AD AD AY XZ AC = · = · , ′ AY AY AY XY AZ and similarly, AB ZY AC = · . AX XY AZ Therefore we have ( ) AB AD XZ ZY AC AC
= + · = , AX AY XY XY AZ AZ as desired. Solution 2. (Affine Transformations) We recall that affine transformations preserve both parallel lines and ratios between distances of collinear points. It thus suffices to show the desired result when ABCD is a square. This can be done in a variety of ways. For instance, a coordinate bash can be applied by setting A to be the origin. Let the length of the square be 1 and set X and Y as ( a, 0) and (0 , b ) respectively, so the line XY has ab ab equation bx + ay = ab . Then, we note that Z is the point ( , ), so a + b a + b AB AD 1 1 a + b AC
= + = = . AX AY a b ab AZ