HMMT 二月 2016 · 冲刺赛 · 第 36 题
HMMT February 2016 — Guts Round — Problem 36
题目详情
- [ 20 ] (Self-Isogonal Cubics) Let ABC be a triangle with AB = 2, AC = 3, BC = 4. The isogonal ∗ conjugate of a point P , denoted P , is the point obtained by intersecting the reflection of lines P A , P B , P C across the angle bisectors of ∠ A , ∠ B , and ∠ C , respectively. Given a point Q , let K ( Q ) denote the unique cubic plane curve which passes through all points P such ∗ that line P P contains Q . Consider: (a) the M’Cay cubic K ( O ), where O is the circumcenter of 4 ABC , (b) the Thomson cubic K ( G ), where G is the centroid of 4 ABC , (c) the Napoleon-Feurerbach cubic K ( N ), where N is the nine-point center of 4 ABC , (d) the Darboux cubic K ( L ), where L is the de Longchamps point (the reflection of the orthocenter across point O ), (e) the Neuberg cubic K ( X ), where X is the point at infinity along line OG , 30 30 (f) the nine-point circle of 4 ABC , (g) the incircle of 4 ABC , and (h) the circumcircle of 4 ABC . Estimate N , the number of points lying on at least two of these eight curves. An estimate of E earns ⌊ ⌋ −| N − E | / 6 20 · 2 points.
解析
- [ 20 ] (Self-Isogonal Cubics) Let ABC be a triangle with AB = 2, AC = 3, BC = 4. The isogonal ∗ conjugate of a point P , denoted P , is the point obtained by intersecting the reflection of lines P A , P B , P C across the angle bisectors of ∠ A , ∠ B , and ∠ C , respectively. Given a point Q , let K ( Q ) denote the unique cubic plane curve which passes through all points P such ∗ that line P P contains Q . Consider: (a) the M’Cay cubic K ( O ), where O is the circumcenter of 4 ABC , (b) the Thomson cubic K ( G ), where G is the centroid of 4 ABC , (c) the Napoleon-Feurerbach cubic K ( N ), where N is the nine-point center of 4 ABC , (d) the Darboux cubic K ( L ), where L is the de Longchamps point (the reflection of the orthocenter across point O ), (e) the Neuberg cubic K ( X ), where X is the point at infinity along line OG , 30 30 (f) the nine-point circle of 4 ABC , (g) the incircle of 4 ABC , and (h) the circumcircle of 4 ABC . Estimate N , the number of points lying on at least two of these eight curves. An estimate of E earns ⌊ ⌋ −| N − E | / 6 20 · 2 points. Proposed by: Evan Chen Answer: 49 The first main insight is that all the cubics pass through the points A , B , C , H (orthocenter), O , and the incenter and three excenters. Since two cubics intersect in at most nine points, this is all the intersections of a cubic with a cubic. On the other hand, it is easy to see that among intersections of circles with circles, there are exactly 3 points; the incircle is tangent to the nine-point circle at the Feurerbach point while being contained completely in the circumcircle; on the other hand for this obtuse triangle the nine-point circle and the circumcircle intersect exactly twice. All computations up until now are exact, so it remains to estimate: • Intersection of the circumcircle with cubics. Each cubic intersects the circumcircle at an even number of points, and moreover we already know that A , B , C are among these, so the number of additional intersections contributed is either 1 or 3; it is the former only for the Neuberg cubic which has a “loop”. Hence the actual answer in this case is 1 + 3 + 3 + 3 + 3 = 13 (but an estimate of 3 · 5 = 15 is very reasonable). • Intersection of the incircle with cubics. Since ∠ A is large the incircle is small, but on the other hand we know I lies on each cubic. Hence it’s very likely that each cubic intersects the incircle twice (once “coming in” and once “coming out”). This is the case, giving 2 · 5 = 10 new points. • Intersection of the nine-point with cubics. We guess this is close to the 10 points of the incircle, as we know the nine-point circle and the incircle are tangent to each other. In fact, the exact count is 14 points; just two additional branches appear. In total, N = 9 + 3 + 13 + 10 + 14 = 49. I A H I B A I C N I F e B C O