HMMT 二月 2007 · TEAM2 赛 · 第 6 题
HMMT February 2007 — TEAM2 Round — Problem 6
题目详情
- [ 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. A brief review of cyclic Quadrilaterals. The following discussion of cyclic quadrilaterals is included for reference. Any of the results given here may 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 circumcircle is called the circumcenter of the quadrilateral. For a convex quadrilateral ABCD , the following are equivalent: • 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 that given 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. 2
解析
- [ 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 P and 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 is cyclic if and only if π = ∠ BAD + ∠ DCB = 2 π − 2 α − 2 β, or simply π/ 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 here may 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 circumcircle is called the circumcenter of the quadrilateral. For a convex quadrilateral ABCD , the following are equivalent: • 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 that given 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