HMMT 二月 2005 · 几何 · 第 9 题

HMMT February 2005 — Geometry — Problem 9

Let AC be a diameter of a circle ω of radius 1, and let D be the point on AC such that CD = 1 / 5. Let B be the point on ω such that DB is perpendicular to AC , and let E be the midpoint of DB . The line tangent to ω at B intersects line CE at the point X . Compute AX .

解析

Let AC be a diameter of a circle ω of radius 1, and let D be the point on AC such that CD = 1 / 5. Let B be the point on ω such that DB is perpendicular to AC , and let E be the midpoint of DB . The line tangent to ω at B intersects line CE at the point X . Compute AX . Solution: 3 We first show that AX is perpendicular to AC . Let the tangent to ω at A intersect ′ ′ ′ CB at Z and CE at X . Since ZA is parallel to BD and BE = ED , ZX = X A . ′ Therefore, X is the midpoint of the hypotenuse of the right triangle ABZ , so it is also ′ ′ ′ its circumcenter. Thus X A = X B , and since X A is tangent to ω and B lies on ω , ′ ′ we must have that X B is tangent to ω , so X = X . 4 3 3 Let O be the center of ω . Then OD = , so BD = and DE = . Then AX = 5 5 10 AC 3 2 DE · = · = 3. DC 10 1 / 5 Z X’ B E A C O D 3