{
  "index": "1965-B-6",
  "type": "GEO",
  "tag": [
    "GEO"
  ],
  "difficulty": "",
  "question": "B-6. If \\( A, B, C, D \\) are four distinct points such that every circle through \\( A \\) and \\( B \\) intersects (or coincides with) every circle through \\( C \\) and \\( D \\), prove that the four points are either collinear (all of one line) or concyclic (all on one circle).",
  "solution": "B-6. Suppose \\( A, B, C, D \\) are neither concyclic nor collinear. Then \\( p \\), the perpendicular bisector of segment \\( A B \\), cannot coincide with \\( q \\), the perpendicular bisector of segment \\( C D \\). If the lines \\( p \\) and \\( q \\) intersect, their common point is the center of two concentric circles, one through \\( A \\) and \\( B \\), the other through \\( C \\) and \\( D \\). If instead \\( p \\) and \\( q \\) are parallel, so also are the lines \\( A B \\) and \\( C D \\). Consider points \\( P \\) and \\( Q \\), on \\( p \\) and \\( q \\) respectively, midway between the parallel lines \\( A B \\) and \\( C D \\). Clearly, the circles \\( A B P \\) and \\( C D Q \\) have no common point.",
  "vars": [
    "p",
    "q",
    "P",
    "Q"
  ],
  "params": [
    "A",
    "B",
    "C",
    "D"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "A": "pointalpha",
        "B": "pointbeta",
        "C": "pointgamma",
        "D": "pointdelta",
        "p": "bisectorone",
        "q": "bisectortwo",
        "P": "midpointone",
        "Q": "midpointtwo"
      },
      "question": "B-6. If \\( pointalpha, pointbeta, pointgamma, pointdelta \\) are four distinct points such that every circle through \\( pointalpha \\) and \\( pointbeta \\) intersects (or coincides with) every circle through \\( pointgamma \\) and \\( pointdelta \\), prove that the four points are either collinear (all of one line) or concyclic (all on one circle).",
      "solution": "B-6. Suppose \\( pointalpha, pointbeta, pointgamma, pointdelta \\) are neither concyclic nor collinear. Then \\( bisectorone \\), the perpendicular bisector of segment \\( pointalpha pointbeta \\), cannot coincide with \\( bisectortwo \\), the perpendicular bisector of segment \\( pointgamma pointdelta \\). If the lines \\( bisectorone \\) and \\( bisectortwo \\) intersect, their common point is the center of two concentric circles, one through \\( pointalpha \\) and \\( pointbeta \\), the other through \\( pointgamma \\) and \\( pointdelta \\). If instead \\( bisectorone \\) and \\( bisectortwo \\) are parallel, so also are the lines \\( pointalpha pointbeta \\) and \\( pointgamma pointdelta \\). Consider points \\( midpointone \\) and \\( midpointtwo \\), on \\( bisectorone \\) and \\( bisectortwo \\) respectively, midway between the parallel lines \\( pointalpha pointbeta \\) and \\( pointgamma pointdelta \\). Clearly, the circles \\( pointalpha pointbeta midpointone \\) and \\( pointgamma pointdelta midpointtwo \\) have no common point."
    },
    "descriptive_long_confusing": {
      "map": {
        "p": "dictionary",
        "q": "pendulum",
        "P": "alligator",
        "Q": "notebook",
        "A": "harboring",
        "B": "skeleton",
        "C": "pineapple",
        "D": "waterfall"
      },
      "question": "B-6. If \\( harboring, skeleton, pineapple, waterfall \\) are four distinct points such that every circle through \\( harboring \\) and \\( skeleton \\) intersects (or coincides with) every circle through \\( pineapple \\) and \\( waterfall \\), prove that the four points are either collinear (all of one line) or concyclic (all on one circle).",
      "solution": "B-6. Suppose \\( harboring, skeleton, pineapple, waterfall \\) are neither concyclic nor collinear. Then \\( dictionary \\), the perpendicular bisector of segment \\( harboring skeleton \\), cannot coincide with \\( pendulum \\), the perpendicular bisector of segment \\( pineapple waterfall \\). If the lines \\( dictionary \\) and \\( pendulum \\) intersect, their common point is the center of two concentric circles, one through \\( harboring \\) and \\( skeleton \\), the other through \\( pineapple \\) and \\( waterfall \\). If instead \\( dictionary \\) and \\( pendulum \\) are parallel, so also are the lines \\( harboring skeleton \\) and \\( pineapple waterfall \\). Consider points \\( alligator \\) and \\( notebook \\), on \\( dictionary \\) and \\( pendulum \\) respectively, midway between the parallel lines \\( harboring skeleton \\) and \\( pineapple waterfall \\). Clearly, the circles \\( harboring skeleton alligator \\) and \\( pineapple waterfall notebook \\) have no common point."
    },
    "descriptive_long_misleading": {
      "map": {
        "p": "curvedpath",
        "q": "roundedtrail",
        "P": "broadarea",
        "Q": "largezone",
        "A": "vastregion",
        "B": "spaciousland",
        "C": "massivezone",
        "D": "grandexpanse"
      },
      "question": "B-6. If \\( vastregion, spaciousland, massivezone, grandexpanse \\) are four distinct points such that every circle through \\( vastregion \\) and \\( spaciousland \\) intersects (or coincides with) every circle through \\( massivezone \\) and \\( grandexpanse \\), prove that the four points are either collinear (all of one line) or concyclic (all on one circle).",
      "solution": "B-6. Suppose \\( vastregion, spaciousland, massivezone, grandexpanse \\) are neither concyclic nor collinear. Then \\( curvedpath \\), the perpendicular bisector of segment \\( vastregion spaciousland \\), cannot coincide with \\( roundedtrail \\), the perpendicular bisector of segment \\( massivezone grandexpanse \\). If the lines \\( curvedpath \\) and \\( roundedtrail \\) intersect, their common point is the center of two concentric circles, one through \\( vastregion \\) and \\( spaciousland \\), the other through \\( massivezone \\) and \\( grandexpanse \\). If instead \\( curvedpath \\) and \\( roundedtrail \\) are parallel, so also are the lines \\( vastregion spaciousland \\) and \\( massivezone grandexpanse \\). Consider points \\( broadarea \\) and \\( largezone \\), on \\( curvedpath \\) and \\( roundedtrail \\) respectively, midway between the parallel lines \\( vastregion spaciousland \\) and \\( massivezone grandexpanse \\). Clearly, the circles \\( vastregion spaciousland broadarea \\) and \\( massivezone grandexpanse largezone \\) have no common point."
    },
    "garbled_string": {
      "map": {
        "p": "ksladnqr",
        "q": "bmrtozxc",
        "P": "vnghklsa",
        "Q": "dpshzxcv",
        "A": "qzxwvtnp",
        "B": "plkmnhgf",
        "C": "jxcrdsvb",
        "D": "flsnqkrt"
      },
      "question": "B-6. If \\( qzxwvtnp, plkmnhgf, jxcrdsvb, flsnqkrt \\) are four distinct points such that every circle through \\( qzxwvtnp \\) and \\( plkmnhgf \\) intersects (or coincides with) every circle through \\( jxcrdsvb \\) and \\( flsnqkrt \\), prove that the four points are either collinear (all of one line) or concyclic (all on one circle).",
      "solution": "B-6. Suppose \\( qzxwvtnp, plkmnhgf, jxcrdsvb, flsnqkrt \\) are neither concyclic nor collinear. Then \\( ksladnqr \\), the perpendicular bisector of segment \\( qzxwvtnp plkmnhgf \\), cannot coincide with \\( bmrtozxc \\), the perpendicular bisector of segment \\( jxcrdsvb flsnqkrt \\). If the lines \\( ksladnqr \\) and \\( bmrtozxc \\) intersect, their common point is the center of two concentric circles, one through \\( qzxwvtnp \\) and \\( plkmnhgf \\), the other through \\( jxcrdsvb \\) and \\( flsnqkrt \\). If instead \\( ksladnqr \\) and \\( bmrtozxc \\) are parallel, so also are the lines \\( qzxwvtnp plkmnhgf \\) and \\( jxcrdsvb flsnqkrt \\). Consider points \\( vnghklsa \\) and \\( dpshzxcv \\), on \\( ksladnqr \\) and \\( bmrtozxc \\) respectively, midway between the parallel lines \\( qzxwvtnp plkmnhgf \\) and \\( jxcrdsvb flsnqkrt \\). Clearly, the circles \\( qzxwvtnp plkmnhgf vnghklsa \\) and \\( jxcrdsvb flsnqkrt dpshzxcv \\) have no common point."
    },
    "kernel_variant": {
      "question": "Let A, B, C, D be four distinct points in the Euclidean plane. Assume that  \n\n(*) every circle that passes through A and B has at least one common point with every circle that passes through C and D.  \n\nProve that the four points are necessarily either collinear or concyclic.",
      "solution": "Assume that the four points are neither collinear nor concyclic; we shall derive a contradiction with property (*).\n\nDenote by p the perpendicular bisector of segment AB and by q the perpendicular bisector of segment CD.  \nBecause A, B, C, D are not concyclic, the two perpendicular bisectors cannot coincide: p ≠ q.\n\n------------------------------------------------------------\nCase 1.  p and q meet.\n------------------------------------------------------------\nLet O = p ∩ q.  Then OA = OB and OC = OD, so two different circles\n\n   Γ₁ : centre O, radius OA (through A and B),\n   Γ₂ : centre O, radius OC (through C and D)\n\nare concentric.  Concentric circles are disjoint, hence Γ₁ and Γ₂ have no common point, contradicting (*).\n\n------------------------------------------------------------\nCase 2.  p ∥ q.\n------------------------------------------------------------\nThen AB ∥ CD as well.  Let ℓ be the line that is parallel to AB and CD and midway between them (the two parallel segments have the same distance d from ℓ).  Because p ∥ q, the three lines p, q, ℓ are pairwise distinct.\n\nChoose the points\n\n   P = p ∩ ℓ, Q = q ∩ ℓ.\n\n(The two points are distinct because p ≠ q.)  We now study the two circles\n\n   Γ₁ : circle through A, B, P,   \n   Γ₂ : circle through C, D, Q.\n\n------------------------------------------------------------\nA geometric property of Γ₁.\n------------------------------------------------------------\nSince P ∈ p, we have PA = PB; therefore p is the perpendicular bisector of AB as well as of the chord AB of Γ₁.  Consequently the centre O₁ of Γ₁ lies on p.\n\nBecause ℓ ∥ AB and p ⟂ AB, we have p ⟂ ℓ.  The radius O₁P is contained in p, hence O₁P ⟂ ℓ.  Thus ℓ is tangent to Γ₁ at the point P.\n\nA tangent meets a circle in at most one point; therefore Γ₁ is entirely contained in the closed half-plane bounded by ℓ that contains AB and does not cross ℓ.\n\n------------------------------------------------------------\nA symmetric property of Γ₂.\n------------------------------------------------------------\nExactly the same argument shows that the centre O₂ of Γ₂ lies on q, that O₂Q ⟂ ℓ, and that ℓ is tangent to Γ₂ at Q.  Hence Γ₂ is contained in the opposite closed half-plane bounded by ℓ, the one that contains CD.\n\n------------------------------------------------------------\nDisjointness of Γ₁ and Γ₂.\n------------------------------------------------------------\nThe two circles lie in the two opposite half-planes determined by the common tangent ℓ and touch ℓ at different points P ≠ Q.  Consequently Γ₁ and Γ₂ have no point in common, contradicting (*).\n\n------------------------------------------------------------\nConclusion.\n------------------------------------------------------------\nBoth cases lead to a contradiction, so our initial assumption (that the points are neither collinear nor concyclic) is impossible.  Therefore A, B, C, D are necessarily either collinear or concyclic.",
      "_meta": {
        "core_steps": [
          "Assume the four points are neither collinear nor concyclic.",
          "Consider the perpendicular bisectors p of AB and q of CD; they are distinct.",
          "Case p ∩ q ≠ ∅: the intersection is a common centre, giving two concentric circles through AB and CD that cannot meet—contradiction.",
          "Case p ∥ q: then AB ∥ CD; choose P on p and Q on q equidistant from the two lines; circles ABP and CDQ are disjoint—contradiction.",
          "Thus the assumption fails; the points must be collinear or concyclic."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Formulation of the required overlap condition for the two families of circles (any wording meaning ‘have at least one common point’ works).",
            "original": "intersects (or coincides) with"
          },
          "slot2": {
            "description": "Exact choice of points P and Q on the bisectors in the parallel-bisector case (any symmetric placement that keeps the two new circles disjoint suffices; ‘midway’ is not essential).",
            "original": "P and Q taken midway between the parallel lines AB and CD"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}