{
  "index": "2016-B-3",
  "type": "GEO",
  "tag": [
    "GEO",
    "COMB"
  ],
  "difficulty": "",
  "question": "Suppose that $S$ is a finite set of points in the plane such that the area of triangle\n$\\triangle ABC$ is at most 1 whenever $A$, $B$, and $C$ are in $S$. Show that there exists a triangle of area 4 that (together with its interior) covers the set $S$.",
  "solution": "Since $S$ is finite, we can choose three points $A,B,C$ in $S$ so as to maximize the area of the triangle $ABC$. Let $A', B', C'$ be the points in the plane such that $A,B,C$ are the midpoints of the segments $B'C', C'A', A'B'$; the triangle $A'B'C'$ is similar to $ABC$ with sides twice as long, so its area is 4 times that of $ABC$ and hence no greater than 4.\n\nWe claim that this triangle has the desired effect; that is, every point $P$ of $S$ is contained within the triangle $A'B'C'$. \n(To be precise, the problem statement requires a triangle of area exactly 4, which need not be the case for $A'B'C'$, but this is trivially resolved by scaling up by a homothety.)\nTo see this, note that since the area of the triangle $PBC$ is no more than that of $ABC$, $P$ must lie in the half-plane bounded by $B'C'$ containing $B$ and $C$. Similarly, $P$ must lie in the half-plane bounded by $C'A'$ containing\n$C$ and $A$, and the half-plane bounded by $A'B'$ containing $A$ and $B$. These three half-planes intersect precisely in the region bounded by the triangle $A'B'C'$, proving the claim.",
  "vars": [
    "A",
    "B",
    "C",
    "P"
  ],
  "params": [
    "S"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "A": "alphapoint",
        "B": "betapoint",
        "C": "gammapoint",
        "P": "generalp",
        "S": "pointsset"
      },
      "question": "Suppose that $pointsset$ is a finite set of points in the plane such that the area of triangle\n$\\triangle alphapoint betapoint gammapoint$ is at most 1 whenever $alphapoint$, $betapoint$, and $gammapoint$ are in $pointsset$. Show that there exists a triangle of area 4 that (together with its interior) covers the set $pointsset$.",
      "solution": "Since $pointsset$ is finite, we can choose three points $alphapoint,betapoint,gammapoint$ in $pointsset$ so as to maximize the area of the triangle $alphapoint betapoint gammapoint$. Let $alphapoint', betapoint', gammapoint'$ be the points in the plane such that $alphapoint,betapoint,gammapoint$ are the midpoints of the segments $betapoint'gammapoint', gammapoint'alphapoint', alphapoint'betapoint'$; the triangle $alphapoint'betapoint'gammapoint'$ is similar to $alphapoint betapoint gammapoint$ with sides twice as long, so its area is 4 times that of $alphapoint betapoint gammapoint$ and hence no greater than 4.\n\nWe claim that this triangle has the desired effect; that is, every point $generalp$ of $pointsset$ is contained within the triangle $alphapoint'betapoint'gammapoint'$. \n(To be precise, the problem statement requires a triangle of area exactly 4, which need not be the case for $alphapoint'betapoint'gammapoint'$, but this is trivially resolved by scaling up by a homothety.)\nTo see this, note that since the area of the triangle $generalp betapoint gammapoint$ is no more than that of $alphapoint betapoint gammapoint$, $generalp$ must lie in the half-plane bounded by $betapoint'gammapoint'$ containing $betapoint$ and $gammapoint$. Similarly, $generalp$ must lie in the half-plane bounded by $gammapoint'alphapoint'$ containing\n$gammapoint$ and $alphapoint$, and the half-plane bounded by $alphapoint'betapoint'$ containing $alphapoint$ and $betapoint$. These three half-planes intersect precisely in the region bounded by the triangle $alphapoint'betapoint'gammapoint'$, proving the claim."
    },
    "descriptive_long_confusing": {
      "map": {
        "A": "giraffeneck",
        "B": "pianochord",
        "C": "lanternpost",
        "P": "ribbontwist",
        "S": "marblecrate"
      },
      "question": "Suppose that $marblecrate$ is a finite set of points in the plane such that the area of triangle $\\triangle giraffeneck\\,pianochord\\,lanternpost$ is at most 1 whenever $giraffeneck$, $pianochord$, and $lanternpost$ are in $marblecrate$. Show that there exists a triangle of area 4 that (together with its interior) covers the set $marblecrate$.",
      "solution": "Since $marblecrate$ is finite, we can choose three points $giraffeneck, pianochord, lanternpost$ in $marblecrate$ so as to maximize the area of the triangle $giraffeneck\\,pianochord\\,lanternpost$. Let $giraffeneck', pianochord', lanternpost'$ be the points in the plane such that $giraffeneck, pianochord, lanternpost$ are the midpoints of the segments $pianochord'lanternpost',\\, lanternpost'giraffeneck',\\, giraffeneck'pianochord'$; the triangle $giraffeneck'pianochord'lanternpost'$ is similar to $giraffeneck\\,pianochord\\,lanternpost$ with sides twice as long, so its area is 4 times that of $giraffeneck\\,pianochord\\,lanternpost$ and hence no greater than 4.\n\nWe claim that this triangle has the desired effect; that is, every point $ribbontwist$ of $marblecrate$ is contained within the triangle $giraffeneck'pianochord'lanternpost'$. (To be precise, the problem statement requires a triangle of area exactly 4, which need not be the case for $giraffeneck'pianochord'lanternpost'$, but this is trivially resolved by scaling up by a homothety.) To see this, note that since the area of the triangle $ribbontwist\\,pianochord\\,lanternpost$ is no more than that of $giraffeneck\\,pianochord\\,lanternpost$, $ribbontwist$ must lie in the half-plane bounded by $pianochord'lanternpost'$ containing $pianochord$ and $lanternpost$. Similarly, $ribbontwist$ must lie in the half-plane bounded by $lanternpost'giraffeneck'$ containing $lanternpost$ and $giraffeneck$, and the half-plane bounded by $giraffeneck'pianochord'$ containing $giraffeneck$ and $pianochord$. These three half-planes intersect precisely in the region bounded by the triangle $giraffeneck'pianochord'lanternpost'$, proving the claim."
    },
    "descriptive_long_misleading": {
      "map": {
        "A": "voidpoint",
        "B": "nullvertex",
        "C": "emptycorner",
        "P": "outsiderpt",
        "S": "unboundedset"
      },
      "question": "Suppose that $unboundedset$ is a finite set of points in the plane such that the area of triangle\n$\\triangle voidpoint nullvertex emptycorner$ is at most 1 whenever $voidpoint$, $nullvertex$, and $emptycorner$ are in $unboundedset$. Show that there exists a triangle of area 4 that (together with its interior) covers the set $unboundedset$.",
      "solution": "Since $unboundedset$ is finite, we can choose three points $voidpoint,nullvertex,emptycorner$ in $unboundedset$ so as to maximize the area of the triangle $voidpoint nullvertex emptycorner$. Let $voidpoint', nullvertex', emptycorner'$ be the points in the plane such that $voidpoint,nullvertex,emptycorner$ are the midpoints of the segments $nullvertex' emptycorner', emptycorner' voidpoint', voidpoint' nullvertex'$; the triangle $voidpoint' nullvertex' emptycorner'$ is similar to $voidpoint nullvertex emptycorner$ with sides twice as long, so its area is 4 times that of $voidpoint nullvertex emptycorner$ and hence no greater than 4.\n\nWe claim that this triangle has the desired effect; that is, every point $outsiderpt$ of $unboundedset$ is contained within the triangle $voidpoint' nullvertex' emptycorner'$. \n(To be precise, the problem statement requires a triangle of area exactly 4, which need not be the case for $voidpoint' nullvertex' emptycorner'$, but this is trivially resolved by scaling up by a homothety.)\nTo see this, note that since the area of the triangle $outsiderpt nullvertex emptycorner$ is no more than that of $voidpoint nullvertex emptycorner$, $outsiderpt$ must lie in the half-plane bounded by $nullvertex' emptycorner'$ containing $nullvertex$ and $emptycorner$. Similarly, $outsiderpt$ must lie in the half-plane bounded by $emptycorner' voidpoint'$ containing\n$emptycorner$ and $voidpoint$, and the half-plane bounded by $voidpoint' nullvertex'$ containing $voidpoint$ and $nullvertex$. These three half-planes intersect precisely in the region bounded by the triangle $voidpoint' nullvertex' emptycorner'$, proving the claim."
    },
    "garbled_string": {
      "map": {
        "A": "qzxwvtnp",
        "B": "hjgrksla",
        "C": "mfdqpezi",
        "P": "vyaltgbo",
        "S": "rcuqsdnm"
      },
      "question": "Suppose that $rcuqsdnm$ is a finite set of points in the plane such that the area of triangle\n$\\triangle qzxwvtnphjgrkslamfdqpezi$ is at most 1 whenever $qzxwvtnp$, $hjgrksla$, and $mfdqpezi$ are in $rcuqsdnm$. Show that there exists a triangle of area 4 that (together with its interior) covers the set $rcuqsdnm$.",
      "solution": "Since $rcuqsdnm$ is finite, we can choose three points $qzxwvtnp,hjgrksla,mfdqpezi$ in $rcuqsdnm$ so as to maximize the area of the triangle $qzxwvtnphjgrkslamfdqpezi$. Let $qzxwvtnp', hjgrksla', mfdqpezi'$ be the points in the plane such that $qzxwvtnp,hjgrksla,mfdqpezi$ are the midpoints of the segments $hjgrksla'mfdqpezi', mfdqpezi'qzxwvtnp', qzxwvtnp'hjgrksla'$; the triangle $qzxwvtnp'hjgrksla'mfdqpezi'$ is similar to $qzxwvtnphjgrkslamfdqpezi$ with sides twice as long, so its area is 4 times that of $qzxwvtnphjgrkslamfdqpezi$ and hence no greater than 4.\n\nWe claim that this triangle has the desired effect; that is, every point $vyaltgbo$ of $rcuqsdnm$ is contained within the triangle $qzxwvtnp'hjgrksla'mfdqpezi'$. \n(To be precise, the problem statement requires a triangle of area exactly 4, which need not be the case for $qzxwvtnp'hjgrksla'mfdqpezi'$, but this is trivially resolved by scaling up by a homothety.)\nTo see this, note that since the area of the triangle $vyaltgbohjgrkslamfdqpezi$ is no more than that of $qzxwvtnphjgrkslamfdqpezi$, $vyaltgbo$ must lie in the half-plane bounded by $hjgrksla'mfdqpezi'$ containing $hjgrksla$ and $mfdqpezi$. Similarly, $vyaltgbo$ must lie in the half-plane bounded by $mfdqpezi'qzxwvtnp'$ containing\n$mfdqpezi$ and $qzxwvtnp$, and the half-plane bounded by $qzxwvtnp'hjgrksla'$ containing $qzxwvtnp$ and $hjgrksla$. These three half-planes intersect precisely in the region bounded by the triangle $qzxwvtnp'hjgrksla'mfdqpezi'$, proving the claim."
    },
    "kernel_variant": {
      "question": "Let S be a compact set of points in the Euclidean plane such that the area of every triangle whose three vertices belong to S is at most 3. Prove that there exists a triangle of area 27 whose interior (together with its boundary) contains the whole set S.",
      "solution": "Because S is compact, the set of the areas of all triangles with vertices in S attains its maximum. Denote this maximum by M and choose A, B, C ∈ S such that [ABC] = M.\n\n--------------------------------------------------------------------\nCase 1.  M > 0 (S is not contained in a line)\n--------------------------------------------------------------------\n1.  Anticomplementary (\"double\") triangle.\n   Let G be the centroid of ΔABC and apply the homothety with centre G and ratio –2.  The images of A,B,C will be called A′, B′, C′.  Well-known facts about this map give\n      A′ = B + C – A, B′ = C + A – B, C′ = A + B – C,\n   AB′∥AB, etc., and\n      [A′B′C′] = 4[ABC] = 4M ≤ 12.                 (1)\n\n2.  The set S lies inside ΔA′B′C′.\n   We verify the claim for the side B′C′; the other two sides are treated analogously.\n   Since A is the midpoint of B′C′, the distance between the parallel lines BC and B′C′ equals h_A = d(A,BC).  If some point P ∈ S were situated in the half-plane bounded by B′C′ that does not contain B and C, then d(P,BC) > h_A, and\n         [PBC] = ½ · |BC| · d(P,BC) > ½ · |BC| · h_A = [ABC] = M,\n   contradicting the maximality of M.  Therefore every point of S lies in the half-plane bounded by B′C′ that contains B and C.  Repeating the argument for the other two sides shows S ⊆ ΔA′B′C′.\n\n3.  Expanding to area 27.\n   If [A′B′C′] = 27 we are finished.  Otherwise (1) gives 0 < [A′B′C′] ≤ 12 < 27, and the homothety with centre G and ratio\n         r = √(27 / [A′B′C′]) > 1\n   sends ΔA′B′C′ to a triangle of area r²·[A′B′C′] = 27.  Because G lies inside ΔA′B′C′, this homothety enlarges the region while keeping S inside it.  Hence S is contained in some triangle of area 27.\n\n--------------------------------------------------------------------\nCase 2.  M = 0 (all triangles from S have area 0)\n--------------------------------------------------------------------\nThen every triple of points of S is collinear, so S itself is contained in a line ℓ.\n\n•  If S consists of a single point P (or of two points), choose any triangle of area 27 that contains P in its interior; this clearly satisfies the requirement.\n\n•  Otherwise S has at least two distinct points.  Let X, Y ∈ S be the extreme points of S along ℓ, so that the segment XY contains S and |XY| = L > 0.  Place the point Z above the line ℓ at distance\n      h = 54 / L, so that area(ΔXYZ) = ½·L·h = 27.\n   Because XY is the entire base of ΔXYZ and S ⊆ XY, every point of S lies inside ΔXYZ.  Thus a triangle of area 27 covering S exists in this degenerate situation as well.\n\n--------------------------------------------------------------------\nConclusion.\nIn both cases we have produced a triangle of area 27 whose interior contains the compact set S.  This completes the proof.",
      "_meta": {
        "core_steps": [
          "Select a triangle of maximal area among the points of S (extremal principle).",
          "Form the triangle whose sides are parallel to, and twice as long as, those of the maximal triangle (midpoint/scale-2 construction).",
          "Compare areas to show any other point of S lies in the half-plane determined by each side of the enlarged triangle.",
          "The intersection of the three half-planes is exactly the enlarged triangle, so it contains S."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Common upper bound on areas of all triangles with vertices in S",
            "original": 1
          },
          "slot2": {
            "description": "Target area of the covering triangle guaranteed to exist",
            "original": 4
          },
          "slot3": {
            "description": "Linear scaling factor used when enlarging the maximal triangle (area factor is its square)",
            "original": 2
          },
          "slot4": {
            "description": "Assumption ensuring a maximal-area triangle is attained",
            "original": "finite"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}