HMMT 二月 2023 · 团队赛 · 第 3 题
HMMT February 2023 — Team Round — Problem 3
题目详情
- [35] Let ABCD be a convex quadrilateral such that ∠ ABC = ∠ BCD = θ for some angle θ < 90 . ◦ Point X lies inside the quadrilateral such that ∠ XAD = ∠ XDA = 90 − θ . Prove that BX = XC .
解析
- [35] Let ABCD be a convex quadrilateral such that ∠ ABC = ∠ BCD = θ for some angle θ < 90 . ◦ Point X lies inside the quadrilateral such that ∠ XAD = ∠ XDA = 90 − θ . Prove that BX = XC . Proposed by: Pitchayut Saengrungkongka Solution 1: T D X A B C Let lines AB and CD meet at T . Notice that ◦ ◦ ∠ AT D = 180 − ∠ ABC − ∠ DBC = 180 − 2 θ ◦ ◦ ∠ AXD = 180 − 2(90 − θ ) = 2 θ. ◦ Therefore, A , T , X , and D are concyclic. In particular, this implies that ∠ XT A = 90 − θ = ∠ XT D . Thus, XT bisects ∠ BT C . However, notice that ∠ T BC is isosceles, so XT is actually the perpendicular bisector of BC , implying that BX = XC . Solution 2: D X A P B C Without loss of generality, let AB > CD . Draw the circle γ centered at X and passes through A and D . Let this circle intersects CD again at point P 6 = D . Then, notice that ∠ AXD ∠ AP D = = θ, 2 implying that AP ‖ BC . Combining with ∠ ABC = ∠ BCP , we get that quadrilateral AP CB is isosceles trapezoid. Since X ∈ γ , we have X lies on the perpendicular bisector of AP , which is the same as the perpendicular bisector of BC , so we are done. Solution 3: D A X B C K Let the perpendicular bisector of AD intersects BC at point K . Notice that ◦ ◦ ∠ AXK = 90 + ∠ XAD = 180 − θ = ∠ ABK = ⇒ A, B, X, K are concyclic. = ⇒ ∠ XBC = ∠ XAK Similarly, we get that ∠ XCB = ∠ XDK . However, since both X and K lies on the perpendicular bisector of AD , implying that ∠ XBC = ∠ XCB . Solution 4: Fix θ and points A , B , and C . Animate point D along the fixed line through C . Since 4 XAD as a fixed shape, it follows that X moves linearly along a fixed line. Since we want to show that X lies on the perpendicular bisector of BC , which is fixed, it suffices to prove this for only two locations of D . • When AD ‖ BC , it follows that ABCD is an isosceles trapezoid, implying the result. • When D = C , we notice that ∠ AXC = 2 θ = 2 ∠ ABC, implying that X is the circumcenter of 4 ABC , and the result follows.