HMMT 二月 2007 · TEAM2 赛 · 第 6 题

英文解析

6. [ 40 ] Let the incircle of ABCD be tangent to sides AB, BC, CD, and AD at points P, Q, R, and S ,
   respectively. Show that ABCD is cyclic if and only if P R ⊥ QS.
   Solution. Let the diagonals of P QRS intersect at T. Because AP and AS are tangent to ω at Pand S, we may write α = ∠ ASP = ∠ SP A = ∠ SQP and β = ∠ CQR = ∠ QRC = ∠ QP R. Then
   ∠ P T Q = π − α − β. On the other hand, ∠ P AS = π − 2 α and ∠ RCQ = π − 2 β, so that ABCD iscyclic if and only ifπ = ∠ BAD + ∠ DCB = 2 π − 2 α − 2 β,
   or simplyπ/ 2 = π − α − β = ∠ P T Q,
   as desired.
   A brief review of cyclic Quadrilaterals.
   The following discussion of cyclic quadrilaterals is included for reference. Any of the results given heremay be cited without proof in your writeups.
   A cyclic quadrilateral is a quadrilateral whose four vertices lie on a circle called the circumcircle (the circle is unique if it exists.) If a quadrilateral has a circumcircle, then the center of this circumcircleis called the circumcenter of the quadrilateral. For a convex quadrilateral ABCD , the following areequivalent:
   • Quadrilateral ABCD is cyclic;
   • ∠ ABD = ∠ ACD (or ∠ BCA = ∠ BDA, etc.);
   °
   • Angles ∠ ABC and ∠ CDA are supplementary , that is, m ∠ ABC + m ∠ CDA = 180 (or angles
   ∠ BCD and ∠ BAD are supplementary);
   Cyclic quadrilaterals have a number of interesting properties. A cyclic quadrilateral ABCD satisfies
   AC · BD = AB · CD + AD · BC,
   a result known as Ptolemy’s theorem. Another result, typically called Power of a Point , asserts thatgiven a circle ω , a point P anywhere in the plane of ω , and a line ` through P intersecting ω at points
   ′
   A and B , the value of AP · BP is independent of ` ; i.e., if a second line ` through P intersects ω at
   ′ ′ ′ ′
   A and B , then AP · BP = A P · B P. This second theorem is proved via similar triangles. Say P lies
   ′ ′ ′
   outside of ω , that ` and ` are as before and that A and A lie on segments BP and B P respectively.
   ′ ′
   Then triangle AA P is similar to triangle B BP because the triangles share an angle at P and we have
   ′ ° ′ ′ ′ ′
   m ∠ AA P = 180 − m ∠ B A A = m ∠ ABB = m ∠ P BB .
   The case where A = B is valid and describes the tangents to ω. A similar proof works for P inside ω. 5