HMMT 二月 2019 · 冲刺赛 · 第 21 题
HMMT February 2019 — Guts Round — Problem 21
题目详情
- [ 12 ] A regular hexagon ABCDEF has side length 1 and center O . Parabolas P , P , . . . , P are 1 2 6 constructed with common focus O and directrices AB, BC, CD, DE, EF, F A respectively. Let χ be the set of all distinct points on the plane that lie on at least two of the six parabolas. Compute ∑ | OX | . X ∈ χ (Recall that the focus is the point and the directrix is the line such that the parabola is the locus of points that are equidistant from the focus and the directrix.)
解析
- [ 12 ] A regular hexagon ABCDEF has side length 1 and center O . Parabolas P , P , . . . , P are 1 2 6 constructed with common focus O and directrices AB, BC, CD, DE, EF, F A respectively. Let χ be the set of all distinct points on the plane that lie on at least two of the six parabolas. Compute ∑ | OX | . X ∈ χ (Recall that the focus is the point and the directrix is the line such that the parabola is the locus of points that are equidistant from the focus and the directrix.) Proposed by: Yuan Yao √ Answer: 35 3 Recall the focus and the directrix are such that the parabola is the locus of points equidistant from the focus and the directrix. We will consider pairs of parabolas and find their points of intersections (we label counterclockwise): (1): P ∩ P , two parabolas with directrices adjacent edges on the hexagon (sharing vertex A ). The 1 2 intersection inside the hexagon can be found by using similar triangles: by symmetry this X must lie on OA and must have that its distance from AB and F A are equal to | OX | = x , which is to say √ √ 3 x x ◦ sin 60 = = = = ⇒ x = 2 3 − 3 . 2 | OA | − x 1 − x By symmetry also, the second intersection point, outside the hexagon, must lie on OD . Furthermore, X must have that its distance AB and F A are equal to | OX | . Then again by similar triangles √ √ 3 x x ◦ sin 60 = = = = ⇒ x = 2 3 + 3 . 2 | OA | + x 1 + x (2): P ∩ P , two parabolas with directrices edges one apart on the hexagon, say AB and CD . The 1 3 intersection inside the hexagon is clearly immediately the circumcenter of triangle BOC (equidistance condition), which gives √ 3 x = . 3 Again by symmetry the X outside the hexagon must lie on the lie through O and the midpoint of EF ; √ then one can either observe immediately that x = 3 or set up √ 1 x ◦ sin 30 = = √ = ⇒ x = 3 2 x + 3 √ where we notice 3 is the distance from O to the intersection of AB with the line through O and the midpoint of BC . (3): P ∩ P , two parabolas with directrices edges opposite on the hexagon, say AB and DE . Clearly 1 4 the two intersection points are both inside the hexagon and must lie on CF , which gives √ 3 x = . 2 These together give that the sum desired is ( ) ( ) √ √ √ √ √ √ 3 3 6(2 3 − 3) + 6(2 3 + 3) + 6 + 6( 3) + 6 = 35 3 . 3 2