{
  "index": "1967-A-5",
  "type": "GEO",
  "tag": [
    "GEO",
    "ANA"
  ],
  "difficulty": "",
  "question": "A-5. Show that in a convex region in the plane whose boundary contains at most a finite number of straight line segments and whose area is greater than \\( \\pi / 4 \\) there is at least one pair of points a unit distance apart.",
  "solution": "A-5 Let the maximum diameter be \\( 2 d \\) and assume \\( d<\\frac{1}{2} \\). Take such a diameter as the \\( x \\)-axis with the origin at the mid-point. Since this is a maximum diameter the region is bounded between the lines \\( x=-d \\) and \\( x=d \\). The upper and lower boundaries of the region are functions, because of convexity. Denote them by \\( f(x) \\) and \\( -g(x) \\), where \\( f \\) and \\( g \\) are nonnegative for \\( -d \\leqq x \\leqq d \\). Calculating the distance between \\( (x, f(x)) \\) and \\( (-x,-g(x)) \\) shows that \\( f(x)+g(-x) \\) \\( <\\sqrt{ }\\left(1+4 x^{2}\\right) \\), for \\( -d \\leqq x \\leqq d \\). Area \\( =\\int_{-d}^{d}\\{f(x)+g(-x)\\} d x<\\int_{-d}^{d} \\sqrt{ }\\left(1+4 x^{2}\\right) d x \\) \\( <\\int_{-1 / 2}^{1 / 2} \\sqrt{ }\\left(1+4 x^{2}\\right) d x=\\frac{1}{2} \\pi \\). This contradiction proves that \\( d \\geqq \\frac{1}{2} \\) and so there must be at least two points a unit distance apart.\n\nComment: The requirement that the boundary contain at most a finite number of straight line segments was extraneous.",
  "vars": [
    "x"
  ],
  "params": [
    "d",
    "f",
    "g"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "abscissa",
        "d": "halfdiam",
        "f": "upperfnc",
        "g": "lowerfnc"
      },
      "question": "A-5. Show that in a convex region in the plane whose boundary contains at most a finite number of straight line segments and whose area is greater than \\( \\pi / 4 \\) there is at least one pair of points a unit distance apart.",
      "solution": "A-5 Let the maximum diameter be \\( 2 \\halfdiam \\) and assume \\( \\halfdiam<\\frac{1}{2} \\). Take such a diameter as the \\( \\abscissa \\)-axis with the origin at the mid-point. Since this is a maximum diameter the region is bounded between the lines \\( \\abscissa=-\\halfdiam \\) and \\( \\abscissa=\\halfdiam \\). The upper and lower boundaries of the region are functions, because of convexity. Denote them by \\( \\upperfnc(\\abscissa) \\) and \\( -\\lowerfnc(\\abscissa) \\), where \\( \\upperfnc \\) and \\lowerfnc \\) are nonnegative for \\( -\\halfdiam \\leqq \\abscissa \\leqq \\halfdiam \\). Calculating the distance between \\( (\\abscissa, \\upperfnc(\\abscissa)) \\) and \\( (-\\abscissa,-\\lowerfnc(\\abscissa)) \\) shows that \\( \\upperfnc(\\abscissa)+\\lowerfnc(-\\abscissa) \\)<\\sqrt{ }\\left(1+4 \\abscissa^{2}\\right) , for \\( -\\halfdiam \\leqq \\abscissa \\leqq \\halfdiam \\). Area \\( =\\int_{-\\halfdiam}^{\\halfdiam}\\{\\upperfnc(\\abscissa)+\\lowerfnc(-\\abscissa)\\} \\halfdiam \\abscissa<\\int_{-\\halfdiam}^{\\halfdiam} \\sqrt{ }\\left(1+4 \\abscissa^{2}\\right) \\halfdiam \\abscissa \\) \\( <\\int_{-1 / 2}^{1 / 2} \\sqrt{ }\\left(1+4 \\abscissa^{2}\\right) \\halfdiam \\abscissa=\\frac{1}{2} \\pi \\). This contradiction proves that \\( \\halfdiam \\geqq \\frac{1}{2} \\) and so there must be at least two points a unit distance apart.\n\nComment: The requirement that the boundary contain at most a finite number of straight line segments was extraneous."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "lemonadecup",
        "d": "marshmallow",
        "f": "broccolihead",
        "g": "waterfallmist"
      },
      "question": "A-5. Show that in a convex region in the plane whose boundary contains at most a finite number of straight line segments and whose area is greater than \\( \\pi / 4 \\) there is at least one pair of points a unit distance apart.",
      "solution": "A-5 Let the maximum diameter be \\( 2 marshmallow \\) and assume \\( marshmallow<\\frac{1}{2} \\). Take such a diameter as the \\( lemonadecup \\)-axis with the origin at the mid-point. Since this is a maximum diameter the region is bounded between the lines \\( lemonadecup=-marshmallow \\) and \\( lemonadecup=marshmallow \\). The upper and lower boundaries of the region are functions, because of convexity. Denote them by \\( broccolihead(lemonadecup) \\) and \\( -waterfallmist(lemonadecup) \\), where \\( broccolihead \\) and \\( waterfallmist \\) are nonnegative for \\( -marshmallow \\leqq lemonadecup \\leqq marshmallow \\). Calculating the distance between \\( (lemonadecup, broccolihead(lemonadecup)) \\) and \\( (-lemonadecup,-waterfallmist(lemonadecup)) \\) shows that \\( broccolihead(lemonadecup)+waterfallmist(-lemonadecup) \\) \\( <\\sqrt{ }\\left(1+4 lemonadecup^{2}\\right) \\), for \\( -marshmallow \\leqq lemonadecup \\leqq marshmallow \\). Area \\( =\\int_{-marshmallow}^{marshmallow}\\{broccolihead(lemonadecup)+waterfallmist(-lemonadecup)\\} d lemonadecup<\\int_{-marshmallow}^{marshmallow} \\sqrt{ }\\left(1+4 lemonadecup^{2}\\right) d lemonadecup \\) \\( <\\int_{-1 / 2}^{1 / 2} \\sqrt{ }\\left(1+4 lemonadecup^{2}\\right) d lemonadecup=\\frac{1}{2} \\pi \\). This contradiction proves that \\( marshmallow \\geqq \\frac{1}{2} \\) and so there must be at least two points a unit distance apart.\n\nComment: The requirement that the boundary contain at most a finite number of straight line segments was extraneous."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "stillpoint",
        "d": "minimumgap",
        "f": "steadyvalue",
        "g": "uniformvalue"
      },
      "question": "A-5. Show that in a convex region in the plane whose boundary contains at most a finite number of straight line segments and whose area is greater than \\( \\pi / 4 \\) there is at least one pair of points a unit distance apart.",
      "solution": "A-5 Let the maximum diameter be \\( 2 minimumgap \\) and assume \\( minimumgap<\\frac{1}{2} \\). Take such a diameter as the \\( stillpoint \\)-axis with the origin at the mid-point. Since this is a maximum diameter the region is bounded between the lines \\( stillpoint=-minimumgap \\) and \\( stillpoint=minimumgap \\). The upper and lower boundaries of the region are functions, because of convexity. Denote them by \\( steadyvalue(stillpoint) \\) and \\( -uniformvalue(stillpoint) \\), where \\( steadyvalue \\) and \\( uniformvalue \\) are nonnegative for \\( -minimumgap \\leqq stillpoint \\leqq minimumgap \\). Calculating the distance between \\( (stillpoint, steadyvalue(stillpoint)) \\) and \\( (-stillpoint,-uniformvalue(stillpoint)) \\) shows that \\( steadyvalue(stillpoint)+uniformvalue(-stillpoint) \\) \\( <\\sqrt{ }\\left(1+4 stillpoint^{2}\\right) \\), for \\( -minimumgap \\leqq stillpoint \\leqq minimumgap \\). Area \\( =\\int_{-minimumgap}^{minimumgap}\\{steadyvalue(stillpoint)+uniformvalue(-stillpoint)\\} minimumgap stillpoint<\\int_{-minimumgap}^{minimumgap} \\sqrt{ }\\left(1+4 stillpoint^{2}\\right) minimumgap stillpoint \\) \\( <\\int_{-1 / 2}^{1 / 2} \\sqrt{ }\\left(1+4 stillpoint^{2}\\right) minimumgap stillpoint=\\frac{1}{2} \\pi \\). This contradiction proves that \\( minimumgap \\geqq \\frac{1}{2} \\) and so there must be at least two points a unit distance apart.\n\nComment: The requirement that the boundary contain at most a finite number of straight line segments was extraneous."
    },
    "garbled_string": {
      "map": {
        "x": "pflqwztu",
        "d": "kxjrmval",
        "f": "vdmyhezo",
        "g": "zrnfqiwp"
      },
      "question": "A-5. Show that in a convex region in the plane whose boundary contains at most a finite number of straight line segments and whose area is greater than \\( \\pi / 4 \\) there is at least one pair of points a unit distance apart.",
      "solution": "A-5 Let the maximum diameter be \\( 2 kxjrmval \\) and assume \\( kxjrmval<\\frac{1}{2} \\). Take such a diameter as the \\( x \\)-axis with the origin at the mid-point. Since this is a maximum diameter the region is bounded between the lines \\( pflqwztu=-kxjrmval \\) and \\( pflqwztu=kxjrmval \\). The upper and lower boundaries of the region are functions, because of convexity. Denote them by \\( vdmyhezo(pflqwztu) \\) and \\( -zrnfqiwp(pflqwztu) \\), where \\( vdmyhezo \\) and \\( zrnfqiwp \\) are nonnegative for \\( -kxjrmval \\leqq pflqwztu \\leqq kxjrmval \\). Calculating the distance between \\( (pflqwztu, vdmyhezo(pflqwztu)) \\) and \\( (-pflqwztu,-zrnfqiwp(pflqwztu)) \\) shows that \\( vdmyhezo(pflqwztu)+zrnfqiwp(-pflqwztu) <\\sqrt{}\\left(1+4 pflqwztu^{2}\\right) \\), for \\( -kxjrmval \\leqq pflqwztu \\leqq kxjrmval \\). Area \\( =\\int_{-kxjrmval}^{kxjrmval}\\{vdmyhezo(pflqwztu)+zrnfqiwp(-pflqwztu)\\} d pflqwztu<\\int_{-kxjrmval}^{kxjrmval} \\sqrt{}\\left(1+4 pflqwztu^{2}\\right) d pflqwztu <\\int_{-1 / 2}^{1 / 2} \\sqrt{}\\left(1+4 pflqwztu^{2}\\right) d pflqwztu=\\frac{1}{2} \\pi \\). This contradiction proves that \\( kxjrmval \\geqq \\frac{1}{2} \\) and so there must be at least two points a unit distance apart.\n\nComment: The requirement that the boundary contain at most a finite number of straight line segments was extraneous."
    },
    "kernel_variant": {
      "question": "Let  \n\\[\n\\omega_{n}\\;=\\;\\frac{\\pi^{\\,n/2}}{\\Gamma\\!\\bigl(\\tfrac n2+1\\bigr)},\\qquad n\\ge 2 ,\n\\]  \nbe the volume of the Euclidean unit ball  \n\\[\nB_{n}(1)=\\bigl\\{x\\in\\mathbb R^{n}:\\|x\\|\\le 1\\bigr\\}\\subset\\mathbb R^{n}.\n\\]\n\nFor a non-empty compact convex set \\(K\\subset\\mathbb R^{n}\\) write  \n\\[\n\\operatorname{Vol}_{n}(K)=\\text{its $n$–dimensional Lebesgue measure},\\qquad\n\\operatorname{diam}K=\\sup\\{\\|x-y\\|:x,y\\in K\\}.\n\\]\n\n1. (Existence of a prescribed distance)  \n   Prove that if  \n   \\[\n       \\operatorname{Vol}_{n}(K)\\;>\\;2^{-n}\\,\\omega_{n},\n   \\]\n   then \\(K\\) contains two points whose Euclidean distance is exactly \\(1\\).\n\n2. (Sharp isodiametric inequality)  \n   Show that for every compact convex \\(K\\subset\\mathbb R^{n}\\)\n   \\[\n       \\operatorname{Vol}_{n}(K)\\;\\le\\;2^{-n}\\,\\omega_{n}\\bigl(\\operatorname{diam}K\\bigr)^{n},\\tag{$\\star$}\n   \\]\n   and that equality in \\((\\star)\\) holds if and only if \\(K\\) is a Euclidean ball.  \n   Deduce, in particular, that among all convex bodies of diameter \\(1\\) the (unique) maximiser of the volume is the ball of radius \\(1/2\\).",
      "solution": "Throughout the proof put  \n\\[\nd=\\operatorname{diam}K\\qquad(d>0).\n\\]\n\n--------------------------------------------------------------------\nStep 0.  Steiner symmetrisation  \n\nFor a unit vector \\(v\\in\\mathbb S^{\\,n-1}\\) let  \n\\[\nH_{v}=\\{x\\in\\mathbb R^{n}:x\\cdot v=0\\},\n\\]\nand denote by \\(S_{v}(A)\\) the Steiner symmetral of a measurable set\n\\(A\\subset\\mathbb R^{n}\\) with respect to \\(H_{v}\\):\n\\[\nS_{v}(A)=\\Bigl\\{y+tv : y\\in H_{v},\\;\n                    |t|\\le \\tfrac12\\,\\lambda_{1}\\bigl(A\\cap(y+\\mathbb R v)\\bigr)\\Bigr\\},\n\\]\nwhere \\(\\lambda_{1}\\) is one–dimensional Lebesgue measure on the line\n\\(y+\\mathbb R v\\).\n\nClassical facts.\n\n(i) Volume is preserved: \\(\\operatorname{Vol}_{n}(S_{v}(A))=\\operatorname{Vol}_{n}(A)\\).\n\n(ii) Diameter never increases:  \n\\[\n\\operatorname{diam}(S_{v}(A))\\le\\operatorname{diam}(A).\\tag{1}\n\\]\n\n--------------------------------------------------------------------\nStep 1.  A limiting ball and the basic inequality  \n\nChoose a countable dense set \\(\\{v_{m}\\}_{m\\ge 1}\\subset\\mathbb S^{\\,n-1}\\) and define\n\\(K_{0}=K\\) and \\(K_{m}=S_{v_{m}}(K_{m-1})\\;(m\\ge 1)\\).\nBy (i)–(1)\n\\[\n\\operatorname{Vol}_{n}(K_{m})=\\operatorname{Vol}_{n}(K),\\qquad\n\\operatorname{diam}(K_{m})\\le d\\quad\\forall m.\n\\]\nAll \\(K_{m}\\) lie in the fixed ball \\(B_{n}(d)\\); hence, by\nBlaschke’s selection theorem, a subsequence converges in the Hausdorff metric\nto a compact convex set \\(L\\).\n\nSince every \\(K_{m}\\) is symmetric with respect to the first\n\\(m\\) hyperplanes of the dense family, the limit is symmetric with respect to\nall hyperplanes through the origin; consequently  \n\\[\nL=B_{n}(r)\\quad\\text{for some }r\\ge 0.\\tag{2}\n\\]\nPassing to the limit gives\n\\[\n\\operatorname{Vol}_{n}(K)=\\omega_{n} r^{\\,n},\\qquad r\\le\\frac d2.\\tag{3}\n\\]\nTherefore\n\\[\n\\operatorname{Vol}_{n}(K)\\;=\\;\\omega_{n} r^{\\,n}\\;\\le\\;\\omega_{n}\\Bigl(\\frac d2\\Bigr)^{n}\n      \\;=\\;2^{-n}\\omega_{n}d^{\\,n},\n\\]\nwhich is \\((\\star)\\).\n\n--------------------------------------------------------------------\nStep 2.  Equality in \\((\\star)\\): first consequences  \n\nAssume henceforth that  \n\\[\n\\operatorname{Vol}_{n}(K)=2^{-n}\\omega_{n}d^{\\,n}.\\tag{4}\n\\]\n\n(a)  Central symmetral.  \nDefine\n\\[\nK^{\\circ}=\\tfrac12\\bigl(K+(-K)\\bigr).\n\\]\nBy Brunn–Minkowski,\n\\(\\operatorname{Vol}_{n}(K^{\\circ})\\ge\\operatorname{Vol}_{n}(K)\\).\nOn the other hand,\n\\(\\operatorname{diam}K^{\\circ}\\le d\\).\nUsing \\((\\star)\\) we obtain\n\\[\n2^{-n}\\omega_{n}d^{\\,n}\\overset{(4)}{=}\\operatorname{Vol}_{n}(K)\n\\le\\operatorname{Vol}_{n}(K^{\\circ})\n\\le 2^{-n}\\omega_{n}d^{\\,n},\n\\]\nhence  \n\\[\n\\operatorname{Vol}_{n}(K^{\\circ})=\\operatorname{Vol}_{n}(K),\\qquad\n\\operatorname{diam}K^{\\circ}=d.\\tag{5}\n\\]\nConsequently equality occurs in the Brunn–Minkowski inequality for the\npair \\((K,-K)\\).  \nEquality for two convex bodies \\(A,B\\) implies that\nthey are translates of homothetic copies of each other.  Here the homothety\nratio is \\(1\\); thus there exists a vector \\(a\\) such that\n\\[\n-K = K + a.\\tag{6}\n\\]\nBecause \\(-K\\) is obtained from \\(K\\) by the point reflection in \\(a/2\\),\nrelation (6) means that \\(K\\) is centrally symmetric with centre\n\\(\\tfrac12 a\\).  By translating \\(K\\) we may (and do) assume that this centre\nis the origin; from now on\n\\[\nK=-K,\\qquad h_{K}(-u)=h_{K}(u)\\quad\\forall u\\in\\mathbb S^{\\,n-1},\\tag{7}\n\\]\nwhere \\(h_{K}\\) is the support function.\n\nBecause the diameter of a centrally symmetric body equals twice its maximal\nsupport value,\n\\[\n\\max_{u\\in\\mathbb S^{\\,n-1}} h_{K}(u)=\\frac d2.\\tag{8}\n\\]\n\n--------------------------------------------------------------------\nStep 3.  A variational argument: the support function must be constant  \n\nSuppose that \\(K\\) is not a ball.  \nThen by continuity of \\(h_{K}\\) there exists a direction\n\\(v\\in\\mathbb S^{\\,n-1}\\) and a number \\(\\delta>0\\) such that  \n\\[\nh_{K}(v)\\le\\frac d2-\\delta.\\tag{9}\n\\]\n\nChoose an even, continuous function\n\\(\\varphi:\\mathbb S^{\\,n-1}\\to[0,1]\\) satisfying  \n\\[\n\\varphi(u)=1\\text{ for }u\\text{ in a small neighbourhood of }v,\\quad\n\\varphi(u)=0\\text{ whenever }h_{K}(u)=\\frac d2.\n\\]\nFor \\(\\varepsilon>0\\) define a new support function\n\\[\nh_{\\varepsilon}(u)=h_{K}(u)+\\varepsilon\\varphi(u),\\qquad u\\in\\mathbb S^{\\,n-1}.\n\\]\nBecause \\(0\\le\\varphi\\le 1\\) and \\(\\varphi\\) vanishes on the set where\n\\(h_{K}\\) already attains the value \\(d/2\\), we can fix\n\\(\\varepsilon_{0}>0\\) so small that  \n\\[\nh_{\\varepsilon}(u)\\le\\frac d2\\quad\\text{for every }u\\in\\mathbb S^{\\,n-1}\n\\text{ and every }0<\\varepsilon\\le\\varepsilon_{0}.\\tag{10}\n\\]\nConsequently the convex body \\(K_{\\varepsilon}\\) corresponding to\n\\(h_{\\varepsilon}\\) satisfies \\(\\operatorname{diam}K_{\\varepsilon}\\le d\\).\n\nNext recall the first variation formula for the volume of a convex body in\nterms of its support function (see, e.g., Schneider, *Convex Bodies:\nThe Brunn–Minkowski Theory*, Th. 5.1.8):\nif \\(h_{t}=h_{K}+t\\psi\\) with an even continuous \\(\\psi\\), then  \n\\[\n\\left.\\frac{d}{dt}\\right|_{t=0^{+}}\\operatorname{Vol}_{n}(K_{t})\n   =\\frac1n\\int_{\\mathbb S^{\\,n-1}}\\psi(u)\\,S_{K}(du),\\tag{11}\n\\]\nwhere \\(S_{K}\\) is the surface area measure of \\(K\\) (an even finite Borel\nmeasure that charges every open subset of the sphere).\n\nBecause \\(\\varphi\\ge 0\\) and is not identically zero,\n\\(\\displaystyle\\int_{\\mathbb S^{\\,n-1}}\\varphi\\,dS_{K}>0\\).\nApplying (11) with \\(\\psi=\\varphi\\) yields\n\\[\n\\operatorname{Vol}_{n}(K_{\\varepsilon})\n   \\;=\\;\\operatorname{Vol}_{n}(K)+c\\,\\varepsilon+O(\\varepsilon^{2})\n   \\quad(c>0).\\tag{12}\n\\]\nFor \\(0<\\varepsilon\\le\\varepsilon_{0}\\) both (10) and (12) hold, hence\n\\[\n\\operatorname{diam}K_{\\varepsilon}\\le d,\\qquad\n\\operatorname{Vol}_{n}(K_{\\varepsilon})>\\operatorname{Vol}_{n}(K).\n\\]\nThis contradicts the assumed maximality of \\(K\\) in (4).\nTherefore the hypothesis (9) is impossible, and we conclude that\n\\[\nh_{K}(u)\\equiv\\frac d2\\quad\\forall u\\in\\mathbb S^{\\,n-1}.\\tag{13}\n\\]\n\n--------------------------------------------------------------------\nStep 4.  Completion of the equality case  \n\nEquation (13) says that the support function of \\(K\\) is constant.\nFor a centrally symmetric body this is equivalent to being a Euclidean\nball, indeed  \n\\[\nK=\\bigl\\{x\\in\\mathbb R^{n} : x\\cdot u\\le\\tfrac d2\n     \\text{ for every }u\\in\\mathbb S^{\\,n-1}\\bigr\\}=B_{n}\\Bigl(\\frac d2\\Bigr).\n\\]\nConversely, the ball of radius \\(d/2\\) obviously satisfies\n\\(\\operatorname{Vol}_{n}=2^{-n}\\omega_{n}d^{\\,n}\\), so equality in\n\\((\\star)\\) occurs if and only if \\(K\\) is a ball.\n\n--------------------------------------------------------------------\nStep 5.  Prescribed unit distance (Part 1)  \n\nAssume \\(\\operatorname{Vol}_{n}(K)>2^{-n}\\omega_{n}\\).\nBy \\((\\star)\\) we must have \\(\\operatorname{diam}K>1\\); pick\n\\(A,B\\in K\\) with \\(\\|A-B\\|=\\operatorname{diam}K=d>1\\) and set\n\\[\n\\theta=\\frac{B-A}{\\|B-A\\|}\\in\\mathbb S^{\\,n-1}.\n\\]\nConvexity implies that the entire segment\n\\(\\{A+t\\theta:0\\le t\\le d\\}\\subset K\\).\nBecause the continuous function \\(t\\mapsto t\\) covers the interval\n\\([0,d]\\), there is \\(t_{0}\\in(0,d)\\) such that \\(t_{0}=1\\).\nPut \\(P=A,\\;Q=A+\\theta\\); then \\(P,Q\\in K\\) and \\(\\|P-Q\\|=1\\), completing\nPart 1.\n\n--------------------------------------------------------------------\nStep 6.  Summary  \n\n(i)  Successive Steiner symmetrisations lead to a limiting ball of\nradius \\(d/2\\), giving the sharp bound \\((\\star)\\).\n\n(ii)  When equality is assumed, Brunn–Minkowski implies central\nsymmetry; a careful variational argument shows that every support value\nalready equals \\(d/2\\), hence the body is the ball \\(B_{n}(d/2)\\).\n\n(iii)  The contrapositive of \\((\\star)\\) immediately yields\n\\(\\operatorname{diam}K>1\\) under the hypothesis of Part 1, and convexity\nthen produces a pair of points at distance \\(1\\).\n\nThus both requested assertions are rigorously established.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.571933",
        "was_fixed": false,
        "difficulty_analysis": "The original A-5 deals only with two-dimensional area and merely shows the existence of a unit segment.  The enhanced variant\n\n• passes to arbitrary dimension n,  \n• demands an optimal volume–diameter inequality valid for all convex bodies (necessitating repeated Steiner symmetrisation and knowledge that the ball is the unique maximiser),  \n• requires a rigorous extremal characterisation (strictly harder than just an existence statement), and  \n• obliges the solver to assemble ideas from several advanced areas of convex geometry—Steiner symmetrisation, isodiametric inequalities, and uniqueness issues.\n\nCarrying out these steps is markedly more technical and conceptually deeper than anything needed for the planar case, thus making the new kernel variant significantly harder in both breadth and depth."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\\[\n\\omega_{n}\\;=\\;\\frac{\\pi^{\\,n/2}}{\\Gamma\\!\\bigl(\\tfrac n2+1\\bigr)},\\qquad n\\ge 2 ,\n\\]  \nbe the volume of the Euclidean unit ball  \n\\[\nB_{n}(1)=\\bigl\\{x\\in\\mathbb R^{n}:\\|x\\|\\le 1\\bigr\\}\\subset\\mathbb R^{n}.\n\\]\n\nFor a non-empty compact convex set \\(K\\subset\\mathbb R^{n}\\) write  \n\\[\n\\operatorname{Vol}_{n}(K)=\\text{its $n$–dimensional Lebesgue measure},\\qquad\n\\operatorname{diam}K=\\sup\\{\\|x-y\\|:x,y\\in K\\}.\n\\]\n\n1. (Existence of a prescribed distance)  \n   Prove that if  \n   \\[\n       \\operatorname{Vol}_{n}(K)\\;>\\;2^{-n}\\,\\omega_{n},\n   \\]\n   then \\(K\\) contains two points whose Euclidean distance is exactly \\(1\\).\n\n2. (Sharp isodiametric inequality)  \n   Show that for every compact convex \\(K\\subset\\mathbb R^{n}\\)\n   \\[\n       \\operatorname{Vol}_{n}(K)\\;\\le\\;2^{-n}\\,\\omega_{n}\\bigl(\\operatorname{diam}K\\bigr)^{n},\\tag{$\\star$}\n   \\]\n   and that equality in \\((\\star)\\) holds if and only if \\(K\\) is a Euclidean ball.  \n   Deduce, in particular, that among all convex bodies of diameter \\(1\\) the (unique) maximiser of the volume is the ball of radius \\(1/2\\).",
      "solution": "Throughout the proof put  \n\\[\nd=\\operatorname{diam}K\\qquad(d>0).\n\\]\n\n--------------------------------------------------------------------\nStep 0.  Steiner symmetrisation  \n\nFor a unit vector \\(v\\in\\mathbb S^{\\,n-1}\\) let  \n\\[\nH_{v}=\\{x\\in\\mathbb R^{n}:x\\cdot v=0\\},\n\\]\nand denote by \\(S_{v}(A)\\) the Steiner symmetral of a measurable set\n\\(A\\subset\\mathbb R^{n}\\) with respect to \\(H_{v}\\):\n\\[\nS_{v}(A)=\\Bigl\\{y+tv : y\\in H_{v},\\;\n                    |t|\\le \\tfrac12\\,\\lambda_{1}\\bigl(A\\cap(y+\\mathbb R v)\\bigr)\\Bigr\\},\n\\]\nwhere \\(\\lambda_{1}\\) is one–dimensional Lebesgue measure on the line\n\\(y+\\mathbb R v\\).\n\nClassical facts.\n\n(i) Volume is preserved: \\(\\operatorname{Vol}_{n}(S_{v}(A))=\\operatorname{Vol}_{n}(A)\\).\n\n(ii) Diameter never increases:  \n\\[\n\\operatorname{diam}(S_{v}(A))\\le\\operatorname{diam}(A).\\tag{1}\n\\]\n\n--------------------------------------------------------------------\nStep 1.  A limiting ball and the basic inequality  \n\nChoose a countable dense set \\(\\{v_{m}\\}_{m\\ge 1}\\subset\\mathbb S^{\\,n-1}\\) and define\n\\(K_{0}=K\\) and \\(K_{m}=S_{v_{m}}(K_{m-1})\\;(m\\ge 1)\\).\nBy (i)–(1)\n\\[\n\\operatorname{Vol}_{n}(K_{m})=\\operatorname{Vol}_{n}(K),\\qquad\n\\operatorname{diam}(K_{m})\\le d\\quad\\forall m.\n\\]\nAll \\(K_{m}\\) lie in the fixed ball \\(B_{n}(d)\\); hence, by\nBlaschke’s selection theorem, a subsequence converges in the Hausdorff metric\nto a compact convex set \\(L\\).\n\nSince every \\(K_{m}\\) is symmetric with respect to the first\n\\(m\\) hyperplanes of the dense family, the limit is symmetric with respect to\nall hyperplanes through the origin; consequently  \n\\[\nL=B_{n}(r)\\quad\\text{for some }r\\ge 0.\\tag{2}\n\\]\nPassing to the limit gives\n\\[\n\\operatorname{Vol}_{n}(K)=\\omega_{n} r^{\\,n},\\qquad r\\le\\frac d2.\\tag{3}\n\\]\nTherefore\n\\[\n\\operatorname{Vol}_{n}(K)\\;=\\;\\omega_{n} r^{\\,n}\\;\\le\\;\\omega_{n}\\Bigl(\\frac d2\\Bigr)^{n}\n      \\;=\\;2^{-n}\\omega_{n}d^{\\,n},\n\\]\nwhich is \\((\\star)\\).\n\n--------------------------------------------------------------------\nStep 2.  Equality in \\((\\star)\\): first consequences  \n\nAssume henceforth that  \n\\[\n\\operatorname{Vol}_{n}(K)=2^{-n}\\omega_{n}d^{\\,n}.\\tag{4}\n\\]\n\n(a)  Central symmetral.  \nDefine\n\\[\nK^{\\circ}=\\tfrac12\\bigl(K+(-K)\\bigr).\n\\]\nBy Brunn–Minkowski,\n\\(\\operatorname{Vol}_{n}(K^{\\circ})\\ge\\operatorname{Vol}_{n}(K)\\).\nOn the other hand,\n\\(\\operatorname{diam}K^{\\circ}\\le d\\).\nUsing \\((\\star)\\) we obtain\n\\[\n2^{-n}\\omega_{n}d^{\\,n}\\overset{(4)}{=}\\operatorname{Vol}_{n}(K)\n\\le\\operatorname{Vol}_{n}(K^{\\circ})\n\\le 2^{-n}\\omega_{n}d^{\\,n},\n\\]\nhence  \n\\[\n\\operatorname{Vol}_{n}(K^{\\circ})=\\operatorname{Vol}_{n}(K),\\qquad\n\\operatorname{diam}K^{\\circ}=d.\\tag{5}\n\\]\nConsequently equality occurs in the Brunn–Minkowski inequality for the\npair \\((K,-K)\\).  \nEquality for two convex bodies \\(A,B\\) implies that\nthey are translates of homothetic copies of each other.  Here the homothety\nratio is \\(1\\); thus there exists a vector \\(a\\) such that\n\\[\n-K = K + a.\\tag{6}\n\\]\nBecause \\(-K\\) is obtained from \\(K\\) by the point reflection in \\(a/2\\),\nrelation (6) means that \\(K\\) is centrally symmetric with centre\n\\(\\tfrac12 a\\).  By translating \\(K\\) we may (and do) assume that this centre\nis the origin; from now on\n\\[\nK=-K,\\qquad h_{K}(-u)=h_{K}(u)\\quad\\forall u\\in\\mathbb S^{\\,n-1},\\tag{7}\n\\]\nwhere \\(h_{K}\\) is the support function.\n\nBecause the diameter of a centrally symmetric body equals twice its maximal\nsupport value,\n\\[\n\\max_{u\\in\\mathbb S^{\\,n-1}} h_{K}(u)=\\frac d2.\\tag{8}\n\\]\n\n--------------------------------------------------------------------\nStep 3.  A variational argument: the support function must be constant  \n\nSuppose that \\(K\\) is not a ball.  \nThen by continuity of \\(h_{K}\\) there exists a direction\n\\(v\\in\\mathbb S^{\\,n-1}\\) and a number \\(\\delta>0\\) such that  \n\\[\nh_{K}(v)\\le\\frac d2-\\delta.\\tag{9}\n\\]\n\nChoose an even, continuous function\n\\(\\varphi:\\mathbb S^{\\,n-1}\\to[0,1]\\) satisfying  \n\\[\n\\varphi(u)=1\\text{ for }u\\text{ in a small neighbourhood of }v,\\quad\n\\varphi(u)=0\\text{ whenever }h_{K}(u)=\\frac d2.\n\\]\nFor \\(\\varepsilon>0\\) define a new support function\n\\[\nh_{\\varepsilon}(u)=h_{K}(u)+\\varepsilon\\varphi(u),\\qquad u\\in\\mathbb S^{\\,n-1}.\n\\]\nBecause \\(0\\le\\varphi\\le 1\\) and \\(\\varphi\\) vanishes on the set where\n\\(h_{K}\\) already attains the value \\(d/2\\), we can fix\n\\(\\varepsilon_{0}>0\\) so small that  \n\\[\nh_{\\varepsilon}(u)\\le\\frac d2\\quad\\text{for every }u\\in\\mathbb S^{\\,n-1}\n\\text{ and every }0<\\varepsilon\\le\\varepsilon_{0}.\\tag{10}\n\\]\nConsequently the convex body \\(K_{\\varepsilon}\\) corresponding to\n\\(h_{\\varepsilon}\\) satisfies \\(\\operatorname{diam}K_{\\varepsilon}\\le d\\).\n\nNext recall the first variation formula for the volume of a convex body in\nterms of its support function (see, e.g., Schneider, *Convex Bodies:\nThe Brunn–Minkowski Theory*, Th. 5.1.8):\nif \\(h_{t}=h_{K}+t\\psi\\) with an even continuous \\(\\psi\\), then  \n\\[\n\\left.\\frac{d}{dt}\\right|_{t=0^{+}}\\operatorname{Vol}_{n}(K_{t})\n   =\\frac1n\\int_{\\mathbb S^{\\,n-1}}\\psi(u)\\,S_{K}(du),\\tag{11}\n\\]\nwhere \\(S_{K}\\) is the surface area measure of \\(K\\) (an even finite Borel\nmeasure that charges every open subset of the sphere).\n\nBecause \\(\\varphi\\ge 0\\) and is not identically zero,\n\\(\\displaystyle\\int_{\\mathbb S^{\\,n-1}}\\varphi\\,dS_{K}>0\\).\nApplying (11) with \\(\\psi=\\varphi\\) yields\n\\[\n\\operatorname{Vol}_{n}(K_{\\varepsilon})\n   \\;=\\;\\operatorname{Vol}_{n}(K)+c\\,\\varepsilon+O(\\varepsilon^{2})\n   \\quad(c>0).\\tag{12}\n\\]\nFor \\(0<\\varepsilon\\le\\varepsilon_{0}\\) both (10) and (12) hold, hence\n\\[\n\\operatorname{diam}K_{\\varepsilon}\\le d,\\qquad\n\\operatorname{Vol}_{n}(K_{\\varepsilon})>\\operatorname{Vol}_{n}(K).\n\\]\nThis contradicts the assumed maximality of \\(K\\) in (4).\nTherefore the hypothesis (9) is impossible, and we conclude that\n\\[\nh_{K}(u)\\equiv\\frac d2\\quad\\forall u\\in\\mathbb S^{\\,n-1}.\\tag{13}\n\\]\n\n--------------------------------------------------------------------\nStep 4.  Completion of the equality case  \n\nEquation (13) says that the support function of \\(K\\) is constant.\nFor a centrally symmetric body this is equivalent to being a Euclidean\nball, indeed  \n\\[\nK=\\bigl\\{x\\in\\mathbb R^{n} : x\\cdot u\\le\\tfrac d2\n     \\text{ for every }u\\in\\mathbb S^{\\,n-1}\\bigr\\}=B_{n}\\Bigl(\\frac d2\\Bigr).\n\\]\nConversely, the ball of radius \\(d/2\\) obviously satisfies\n\\(\\operatorname{Vol}_{n}=2^{-n}\\omega_{n}d^{\\,n}\\), so equality in\n\\((\\star)\\) occurs if and only if \\(K\\) is a ball.\n\n--------------------------------------------------------------------\nStep 5.  Prescribed unit distance (Part 1)  \n\nAssume \\(\\operatorname{Vol}_{n}(K)>2^{-n}\\omega_{n}\\).\nBy \\((\\star)\\) we must have \\(\\operatorname{diam}K>1\\); pick\n\\(A,B\\in K\\) with \\(\\|A-B\\|=\\operatorname{diam}K=d>1\\) and set\n\\[\n\\theta=\\frac{B-A}{\\|B-A\\|}\\in\\mathbb S^{\\,n-1}.\n\\]\nConvexity implies that the entire segment\n\\(\\{A+t\\theta:0\\le t\\le d\\}\\subset K\\).\nBecause the continuous function \\(t\\mapsto t\\) covers the interval\n\\([0,d]\\), there is \\(t_{0}\\in(0,d)\\) such that \\(t_{0}=1\\).\nPut \\(P=A,\\;Q=A+\\theta\\); then \\(P,Q\\in K\\) and \\(\\|P-Q\\|=1\\), completing\nPart 1.\n\n--------------------------------------------------------------------\nStep 6.  Summary  \n\n(i)  Successive Steiner symmetrisations lead to a limiting ball of\nradius \\(d/2\\), giving the sharp bound \\((\\star)\\).\n\n(ii)  When equality is assumed, Brunn–Minkowski implies central\nsymmetry; a careful variational argument shows that every support value\nalready equals \\(d/2\\), hence the body is the ball \\(B_{n}(d/2)\\).\n\n(iii)  The contrapositive of \\((\\star)\\) immediately yields\n\\(\\operatorname{diam}K>1\\) under the hypothesis of Part 1, and convexity\nthen produces a pair of points at distance \\(1\\).\n\nThus both requested assertions are rigorously established.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.464757",
        "was_fixed": false,
        "difficulty_analysis": "The original A-5 deals only with two-dimensional area and merely shows the existence of a unit segment.  The enhanced variant\n\n• passes to arbitrary dimension n,  \n• demands an optimal volume–diameter inequality valid for all convex bodies (necessitating repeated Steiner symmetrisation and knowledge that the ball is the unique maximiser),  \n• requires a rigorous extremal characterisation (strictly harder than just an existence statement), and  \n• obliges the solver to assemble ideas from several advanced areas of convex geometry—Steiner symmetrisation, isodiametric inequalities, and uniqueness issues.\n\nCarrying out these steps is markedly more technical and conceptually deeper than anything needed for the planar case, thus making the new kernel variant significantly harder in both breadth and depth."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}