{
  "index": "1979-B-5",
  "type": "GEO",
  "tag": [
    "GEO",
    "COMB"
  ],
  "difficulty": "",
  "question": "Problem B-5\nIn the plane, let \\( C \\) be a closed convex set that contains \\( (0,0) \\) but no other point with integer coordinates. Suppose that \\( A(C) \\), the area of \\( C \\), is equally distributed among the four quadrants. Prove that \\( A(C)<4 \\).",
  "solution": "B-5.\nA support line for \\( C \\) is a straight line touching \\( C \\) such that one side of the line has no points of \\( C \\). There is a support line containing \\( (0,1) \\); let its slope be \\( m \\). If \\( m>1 / 2 \\), the part of the area of \\( C \\) in the fourth quadrant is no more than 1 and we are done. Similarly, if \\( m<-1 / 2 \\). So we assume that \\( -1 / 2<m<1 / 2 \\) and assume the analogous facts for support lines containing \\( (1,0) \\), \\( (0,-1) \\), and \\( (-1,0) \\). At least one of the angles of the quadrilateral formed by these four support lines is not acute; we may take this angle \\( \\alpha \\) to be at a vertex ( \\( h, k \\) ) in the first quadrant. Then \\( \\alpha \\geqslant \\pi / 2 \\) implies that \\( h+k<2 \\), and this in turn implies that the area of \\( C \\) in the first quadrant does not exceed 1 . Hence \\( A(C) \\leqslant 4 \\).",
  "vars": [
    "C",
    "A",
    "m",
    "h",
    "k"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "C": "convexset",
        "A": "areafunc",
        "m": "slopeval",
        "h": "firstcoord",
        "k": "secondcoord"
      },
      "question": "In the plane, let \\( convexset \\) be a closed convex set that contains \\( (0,0) \\) but no other point with integer coordinates. Suppose that \\( areafunc(convexset) \\), the area of \\( convexset \\), is equally distributed among the four quadrants. Prove that \\( areafunc(convexset)<4 \\).",
      "solution": "B-5.\nA support line for \\( convexset \\) is a straight line touching \\( convexset \\) such that one side of the line has no points of \\( convexset \\). There is a support line containing \\( (0,1) \\); let its slope be \\( slopeval \\). If \\( slopeval>1 / 2 \\), the part of the area of \\( convexset \\) in the fourth quadrant is no more than 1 and we are done. Similarly, if \\( slopeval<-1 / 2 \\). So we assume that \\( -1 / 2<slopeval<1 / 2 \\) and assume the analogous facts for support lines containing \\( (1,0) \\), \\( (0,-1) \\), and \\( (-1,0) \\). At least one of the angles of the quadrilateral formed by these four support lines is not acute; we may take this angle \\( \\alpha \\) to be at a vertex ( \\( firstcoord, secondcoord \\) ) in the first quadrant. Then \\( \\alpha \\geqslant \\pi / 2 \\) implies that \\( firstcoord+secondcoord<2 \\), and this in turn implies that the area of \\( convexset \\) in the first quadrant does not exceed 1. Hence \\( areafunc(convexset) \\leqslant 4 \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "C": "perimeter",
        "A": "diameter",
        "m": "latitude",
        "h": "cushions",
        "k": "lanterns"
      },
      "question": "Problem B-5\nIn the plane, let \\( perimeter \\) be a closed convex set that contains \\( (0,0) \\) but no other point with integer coordinates. Suppose that \\( diameter(perimeter) \\), the area of \\( perimeter \\), is equally distributed among the four quadrants. Prove that \\( diameter(perimeter)<4 \\).",
      "solution": "B-5.\nA support line for \\( perimeter \\) is a straight line touching \\( perimeter \\) such that one side of the line has no points of \\( perimeter \\). There is a support line containing \\( (0,1) \\); let its slope be \\( latitude \\). If \\( latitude>1 / 2 \\), the part of the area of \\( perimeter \\) in the fourth quadrant is no more than 1 and we are done. Similarly, if \\( latitude<-1 / 2 \\). So we assume that \\( -1 / 2<latitude<1 / 2 \\) and assume the analogous facts for support lines containing \\( (1,0) \\), \\( (0,-1) \\), and \\( (-1,0) \\). At least one of the angles of the quadrilateral formed by these four support lines is not acute; we may take this angle \\( \\alpha \\) to be at a vertex ( \\( cushions, lanterns \\) ) in the first quadrant. Then \\( \\alpha \\geqslant \\pi / 2 \\) implies that \\( cushions+lanterns<2 \\), and this in turn implies that the area of \\( perimeter \\) in the first quadrant does not exceed 1 . Hence \\( diameter(perimeter) \\leqslant 4 \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "C": "openscatter",
        "A": "perimeter",
        "m": "xintercept",
        "h": "depthvertical",
        "k": "widthhorizontal"
      },
      "question": "<<<\nProblem B-5\nIn the plane, let \\( openscatter \\) be a closed convex set that contains \\( (0,0) \\) but no other point with integer coordinates. Suppose that \\( perimeter(openscatter) \\), the area of \\( openscatter \\), is equally distributed among the four quadrants. Prove that \\( perimeter(openscatter)<4 \\).\n>>>",
      "solution": "<<<\nB-5.\nA support line for \\( openscatter \\) is a straight line touching \\( openscatter \\) such that one side of the line has no points of \\( openscatter \\). There is a support line containing \\( (0,1) \\); let its slope be \\( xintercept \\). If \\( xintercept>1 / 2 \\), the part of the area of \\( openscatter \\) in the fourth quadrant is no more than 1 and we are done. Similarly, if \\( xintercept<-1 / 2 \\). So we assume that \\( -1 / 2<xintercept<1 / 2 \\) and assume the analogous facts for support lines containing \\( (1,0) \\), \\( (0,-1) \\), and \\( (-1,0) \\). At least one of the angles of the quadrilateral formed by these four support lines is not acute; we may take this angle \\( \\alpha \\) to be at a vertex ( \\( depthvertical, widthhorizontal \\) ) in the first quadrant. Then \\( \\alpha \\geqslant \\pi / 2 \\) implies that \\( depthvertical+widthhorizontal<2 \\), and this in turn implies that the area of \\( openscatter \\) in the first quadrant does not exceed 1 . Hence \\( perimeter(openscatter) \\leqslant 4 \\).\n>>>"
    },
    "garbled_string": {
      "map": {
        "C": "vqmzdbef",
        "A": "xyjtrnwa",
        "m": "ptlksdqe",
        "h": "bnvaduwq",
        "k": "risplzco"
      },
      "question": "Problem B-5\nIn the plane, let \\( vqmzdbef \\) be a closed convex set that contains \\( (0,0) \\) but no other point with integer coordinates. Suppose that \\( xyjtrnwa(vqmzdbef) \\), the area of \\( vqmzdbef \\), is equally distributed among the four quadrants. Prove that \\( xyjtrnwa(vqmzdbef)<4 \\).",
      "solution": "B-5.\nA support line for \\( vqmzdbef \\) is a straight line touching \\( vqmzdbef \\) such that one side of the line has no points of \\( vqmzdbef \\). There is a support line containing \\( (0,1) \\); let its slope be \\( ptlksdqe \\). If \\( ptlksdqe>1 / 2 \\), the part of the area of \\( vqmzdbef \\) in the fourth quadrant is no more than 1 and we are done. Similarly, if \\( ptlksdqe<-1 / 2 \\). So we assume that \\( -1 / 2<ptlksdqe<1 / 2 \\) and assume the analogous facts for support lines containing \\( (1,0) \\), \\( (0,-1) \\), and \\( (-1,0) \\). At least one of the angles of the quadrilateral formed by these four support lines is not acute; we may take this angle \\( \\alpha \\) to be at a vertex ( \\( bnvaduwq, risplzco \\) ) in the first quadrant. Then \\( \\alpha \\geqslant \\pi / 2 \\) implies that \\( bnvaduwq+risplzco<2 \\), and this in turn implies that the area of \\( vqmzdbef \\) in the first quadrant does not exceed 1. Hence \\( xyjtrnwa(vqmzdbef) \\leqslant 4 \\)."
    },
    "kernel_variant": {
      "question": "Corrected Problem\n\nLet D be a closed, bounded, convex subset of the Euclidean plane ℝ² whose (Lebesgue) area is denoted by A(D).  Assume that\n\n1.  the origin O = (0,0) is the only point of the integer lattice ℤ² that lies in D, and\n2.  the area of D is equally distributed among the four open coordinate quadrants\n      Q₁ = {x>0 , y>0}, Q₂ = {x<0 , y>0}, Q₃ = {x<0 , y<0}, Q₄ = {x>0 , y<0};\n   that is,\n      area ( D ∩ Qᵢ ) = ¼ · A(D) for i = 1,2,3,4.\n\nProve that A(D) < 8.\n\n(The original 2005 Putnam problem asks for the sharper bound A(D) < 4.  In this variant it suffices to establish the weaker bound 8.)",
      "solution": "Notation.  Throughout,  area(·) denotes Lebesgue planar measure.  We write\n        N = (0, 1), E = (1, 0), S = (0, −1), W = (−1, 0).\nAs usual, a straight line ℓ is called a supporting line of the convex set D if ℓ meets D and D is contained in one of the two closed half–planes bounded by ℓ.  The half–plane that contains D will be called the inward half–plane of ℓ.\n\n────────────────────────────────────────\n1.  Four special supporting lines\n────────────────────────────────────────\nLemma 1.  For every P ∈ {N, E, S, W} there exists a supporting line ℓ_P that\n(a) passes through P and\n(b) has its inward half–plane containing the origin O.\n\nProof.  Fix one of the four lattice points, call it P.  Because O is the only lattice point that lies in D, P ∉ D.  Let S¹ be the unit circle, and for any unit vector u ∈ S¹ put          \n            h(u) = sup_{x∈D} x·u                             (support function of D),\n            g(u) = h(u) − P·u.                              (1)\nFor each u the line ℓ(u) := {x : (x−P)·u = 0} passes through P and its inward half–plane is H(u) := {x : (x−P)·u ≤ 0}.  Observe that\n      D ⊂ H(u)   ⇔   g(u) ≤ 0,                        (2)\nand that u ↦ g(u) is continuous on the compact set S¹.\n\nBecause P ∉ D and D is compact and convex, there are directions u₀, u₁ ∈ S¹ with g(u₀) < 0 < g(u₁).  (Indeed, translate a very large circle centred at P toward D: at some orientation the whole of D lies inside the circle, giving g(u) < 0; half a turn later the circle lies on the other side of D, giving g > 0.)  Since S¹ is connected and g is continuous, there exists u* with g(u*) = 0.  For this u* we have D ⊂ H(u*) and P ∈ ℓ(u*), hence ℓ(u*) is the desired supporting line ℓ_P. ∎\n\nDenote the four lines obtained in this way by\n        ℓ_N , ℓ_E , ℓ_S , ℓ_W ,\nand write their slopes as\n        m_N , m_E , m_S , m_W .\n\n────────────────────────────────────────\n2.  The “steep–slope” case |m_P| ≥ ½\n────────────────────────────────────────\nAssume that for at least one of the four lines, say ℓ_N, we have |m_N| ≥ ½.  Then ℓ_N has equation y = m_N x + 1 and bounds a half–plane that contains both O and D.  The right triangle T with vertices O, N and X = (−1/m_N, 0) lies in the same half–plane and hence in one quadrant Q_j.  Consequently\n      area(D ∩ Q_j) ≤ area(T) = 1 / (2|m_N|) ≤ 1,\nso\n      A(D) = 4 · area(D ∩ Q_j) ≤ 4 < 8.\nThus the desired estimate is proved whenever one of the slopes satisfies |m_P| ≥ ½.\n\n────────────────────────────────────────\n3.  The “mild–slope” case |m_P| < ½ for every P\n────────────────────────────────────────\nFrom now on we assume\n      |m_N| , |m_E| , |m_S| , |m_W| < ½.        (3)\n\nUnder (3) we show that the four supporting lines occur around the origin in the cyclic order ℓ_N, ℓ_E, ℓ_S, ℓ_W and that no two neighbouring lines are parallel.\n\nLemma 2 (non–parallelism and order).  Under assumption (3)\n(a) no two adjacent lines among {ℓ_N, ℓ_E, ℓ_S, ℓ_W} are parallel, and\n(b) travelling clockwise around O one meets the lines in the order ℓ_N, ℓ_E, ℓ_S, ℓ_W.\n\nProof.  For each supporting line let n_P be the unit normal vector that is perpendicular to ℓ_P and points toward the origin O (so n_P is an inward unit normal).\n\nExplicit forms.  A quick computation gives\n      ℓ_N :  y = m_N x + 1          ⇒   n_N  ∝  (m_N , −1);\n      ℓ_E :  y = m_E (x−1)          ⇒   n_E  ∝  ( 1 ,  m_E);\n      ℓ_S :  y = m_S x − 1          ⇒   n_S  ∝  (−m_S,  1);\n      ℓ_W :  y = m_W (x+1)          ⇒   n_W  ∝  (−1 , −m_W).\nBecause |m_P| < ½, each n_P makes an angle of at most arctan(½) ≈ 26.6° with the following coordinate directions:\n      n_N with the negative y–axis,\n      n_E with the positive x–axis,\n      n_S with the positive y–axis,\n      n_W with the negative x–axis.\n\nLet C_P be the open circular cone of half–angle 30° about the indicated axis.  The four cones are pairwise disjoint because the coordinate axes meet at 90° while 30° + 30° = 60° < 90°.  Hence the four normals n_P belong to four disjoint cones, so they are pairwise non-collinear; therefore no two supporting lines are parallel.\n\nMoreover, when one moves clockwise around the origin one encounters the cones in the order C_N, C_E, C_S, C_W and hence the corresponding normals—and therefore the supporting lines—in exactly the same cyclic order. ∎\n\nBecause adjacent lines intersect, their successive intersections form a convex quadrilateral\n        R = ℓ_N ∩ ℓ_E ∩ ℓ_S ∩ ℓ_W,    with   D ⊂ R.      (4)\n\n────────────────────────────────────────\n4.  A non-acute interior angle\n────────────────────────────────────────\nThe interior angles of the quadrilateral R sum to 2π, so at least one of them is ≥ π/2.  Without loss of generality this non-acute angle is at\n        V = ℓ_N ∩ ℓ_E ∈ Q₁,\nand we write V = (h, k) with h > 0, k > 0.\n\nLemma 3.  If the interior angle of R at V is at least 90°, then h + k ≤ 2.\n\nProof.  The vectors\n        u = N − V = (−h, 1 − k),  v = E − V = (1 − h, −k)\npoint along the sides of R that meet at V.  Non-acuteness means u·v ≤ 0, i.e.\n      (−h)(1 − h) + (1 − k)(−k) ≤ 0\n⇔    h² − h + k² − k ≤ 0\n⇔    h² + k² ≤ h + k.                           (5)\n\nBy Cauchy–Schwarz,\n      (h + k)² ≤ 2(h² + k²).\nCombine this with (5) and divide by h + k (which is positive) to obtain\n      h + k ≤ 2. ∎\n\n────────────────────────────────────────\n5.  Bounding the area in the first quadrant\n────────────────────────────────────────\nBecause D ⊂ R and V ∈ Q₁, the portion D ∩ Q₁ is contained in the rectangle\n        [0, h] × [0, k],\nwhose area equals hk.  Subject to the constraints h ≥ 0, k ≥ 0 and h + k ≤ 2, the product hk is maximised when h = k = (h + k)/2 ≤ 1, so hk ≤ 1.  Therefore\n      area(D ∩ Q₁) ≤ hk ≤ 1,\nand consequently\n      A(D) = 4 · area(D ∩ Q₁) ≤ 4 < 8.          (6)\n\n────────────────────────────────────────\n6.  Conclusion\n────────────────────────────────────────\nThe ‘steep-slope’ situation of Section 2 immediately gives A(D) ≤ 4.  When all slopes are mild, Sections 3–5 again yield the same bound.  Hence in every case\n        A(D) ≤ 4 < 8,\nwhich is the inequality required. ∎",
      "_meta": {
        "core_steps": [
          "Take support lines to C through the four nearest lattice points on the axes and study their slopes.",
          "If any such slope lies outside a symmetric critical window, the corresponding quadrant’s slice of C is a triangle of area ≤1.",
          "Otherwise the four support lines form a convex quadrilateral around the origin.",
          "A quadrilateral in this position must have at least one non-acute angle; that angle forces h+k to be below a fixed bound.",
          "That coordinate bound again limits the chosen quadrant’s area to ≤1, so (equal-area hypothesis) A(C) < 4."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Numerical cut-off for the slope outside which the triangular area is forced to be ≤1.",
            "original": "1/2"
          },
          "slot2": {
            "description": "Choice of the four lattice points used to anchor the support lines (currently the axial neighbours of the origin).",
            "original": "(0,1), (1,0), (0,-1), (-1,0)"
          },
          "slot3": {
            "description": "Upper bound on the sum of coordinates of the relevant vertex derived from a non-acute angle.",
            "original": "2  (i.e. h + k < 2)"
          },
          "slot4": {
            "description": "Per-quadrant area cap that guarantees the global bound once areas are equal.",
            "original": "1"
          },
          "slot5": {
            "description": "Resulting total-area bound obtained by multiplying the per-quadrant cap by 4 quadrants.",
            "original": "4"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}