HMMT 二月 2015 · 团队赛 · 第 5 题
HMMT February 2015 — Team Round — Problem 5
题目详情
- [ 20 ] For a convex quadrilateral P , let D denote the sum of the lengths of its diagonals and let S S denote its perimeter. Determine, with proof, all possible values of . D
解析
- [ 20 ] For a convex quadrilateral P , let D denote the sum of the lengths of its diagonals and let S S denote its perimeter. Determine, with proof, all possible values of . D S Answer: 1 < < 2 Suppose we have a convex quadrilateral ABCD with diagonals AC and BD D intersecting at E (convexity is equivalent to having E on the interiors of segments AC and BD ). To prove the lower bound, note that by the triangle inequality , AB + BC > AC and AD + DC > AC , so S = AB + BC + AD + DC > 2 AC . Similarly, S > 2 BD , so 2 S > 2 AC + 2 BD = 2 D gies S > D . To prove the upper bound, note that again by the triangle inequality , AE + EB > AB , CE + BE > BC , AE + ED > AD , CE + ED > CD . Adding these yields 2( AE + EC + BE + ED ) > AB + BC + AD + CD = S. Now since ABCD is convex, E is inside the quadrilateral, so AE + EC = AC and BE + ED = BD . Thus 2( AC + BD ) = D > S . √ S To achieve every real value in this range, first consider a square ABCD . This has = 2. Suppose D now that we have a rectangle with AB = CD = 1 and BC = AD = x , where 0 < x ≤ 1. As x 1 The cleanest way to see this is via directed angles, or more rigorously by checking that the problem is equivalent to a four-variable polynomial identity in complex numbers. 2 This is true in general; it doesn’t require CB = CD . It’s part of the spiral similarity configuration centered at B : Y X → ZA and B : ZY → AX , due to Y Z ∩ AX = C and B = ( CY X ) ∩ ( CZA ). More explicitly, this follows from the angle chase ∠ DBA = ∠ XBY = ∠ XCY = ∠ ACZ = ∠ ABZ. Team S approaches 0 (i.e. our rectangle gets thinner ), gets arbitrarily close to 1, so by intermediate D √ S value theorem, we hit every value ∈ (1 , 2]. D ◦ To achieve the other values, we let AB = BC = CD = DA = 1 and let θ = m ∠ ABE vary from 45 ◦ down to 0 (i.e. a rhombus that gets thinner ). This means AC = 2 sin θ and BD = 2 cos θ . We √ ◦ S ◦ S have S = 4 and D = 2(sin θ + cos θ ). When θ = 45 , = 2, and when θ = 0 , = 2. Thus by the D D √ intermediate value theorem, we are able to choose θ to obtain any value in the range [ 2 , 2). Putting this construction together with the strict upper and lower bounds, we find that all possible S values of are all real values in the open interval (1 , 2). D