{
  "index": "1965-A-1",
  "type": "GEO",
  "tag": [
    "GEO"
  ],
  "difficulty": "",
  "question": "A-1. Let \\( A B C \\) be a triangle with angle \\( A< \\) angle \\( C<90^{\\circ}< \\) angle \\( B \\). Consider the bisectors of the external angles at \\( A \\) and \\( B \\), each measured from the vertex to the opposite side (extended). Suppose both of these line-segments are equal to \\( A B \\). Compute the angle \\( A \\).",
  "solution": "A-1. Suppose the bisector of the exterior angle at \\( A \\) intersects line \\( B C \\) at \\( X \\) and the bisector of the exterior angle at \\( B \\) meets the line \\( A C \\) at \\( Y \\). The assumption that \\( C \\) is between \\( B \\) and \\( X \\) contradicts the fact that \\( \\angle B\\rangle \\angle C \\) so we may assume that \\( B \\) is between \\( X \\) and \\( C \\). Similarly, we conclude that \\( C \\) is between \\( A \\) and \\( Y \\) because \\( \\angle A<\\angle C \\).\n\nIf \\( Z \\) is a point on line \\( A B \\) with \\( B \\) between \\( A \\) and \\( Z \\), we have from triangle \\( A B Y \\) that \\( \\angle Z B Y=2 A \\). Hence, \\( \\angle B X A=\\angle A B X=\\angle Z B C=2 \\angle Z B Y=4 A \\), and the angle sum of triangle \\( A B X \\) is \\( 90^{\\circ}-\\frac{1}{2} A+8 A \\). Thus, \\( A=12^{\\circ} \\).",
  "vars": [
    "X",
    "Y",
    "Z"
  ],
  "params": [
    "A",
    "B",
    "C"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "X": "pointx",
        "Y": "pointy",
        "Z": "pointz",
        "A": "vertexa",
        "B": "vertexb",
        "C": "vertexc"
      },
      "question": "A-1. Let \\( vertexa vertexb vertexc \\) be a triangle with angle \\( vertexa< \\) angle \\( vertexc<90^{\\circ}< \\) angle \\( vertexb \\). Consider the bisectors of the external angles at \\( vertexa \\) and \\( vertexb \\), each measured from the vertex to the opposite side (extended). Suppose both of these line-segments are equal to \\( vertexa vertexb \\). Compute the angle \\( vertexa \\).",
      "solution": "A-1. Suppose the bisector of the exterior angle at \\( vertexa \\) intersects line \\( vertexb vertexc \\) at \\( pointx \\) and the bisector of the exterior angle at \\( vertexb \\) meets the line \\( vertexa vertexc \\) at \\( pointy \\). The assumption that \\( vertexc \\) is between \\( vertexb \\) and \\( pointx \\) contradicts the fact that \\( \\angle vertexb\\rangle \\angle vertexc \\) so we may assume that \\( vertexb \\) is between \\( pointx \\) and \\( vertexc \\). Similarly, we conclude that \\( vertexc \\) is between \\( vertexa \\) and \\( pointy \\) because \\( \\angle vertexa<\\angle vertexc \\).\n\nIf \\( pointz \\) is a point on line \\( vertexa vertexb \\) with \\( vertexb \\) between \\( vertexa \\) and \\( pointz \\), we have from triangle \\( vertexa vertexb pointy \\) that \\( \\angle pointz vertexb pointy=2 vertexa \\). Hence, \\( \\angle vertexb pointx vertexa=\\angle vertexa vertexb pointx=\\angle pointz vertexb vertexc=2 \\angle pointz vertexb pointy=4 vertexa \\), and the angle sum of triangle \\( vertexa vertexb pointx \\) is \\( 90^{\\circ}-\\frac{1}{2} vertexa+8 vertexa \\). Thus, \\( vertexa=12^{\\circ} \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "A": "gemstone",
        "B": "avalanche",
        "C": "hurricane",
        "X": "sunflower",
        "Y": "rectangle",
        "Z": "butterfly"
      },
      "question": "A-1. Let \\( gemstone avalanche hurricane \\) be a triangle with angle \\( gemstone< \\) angle \\( hurricane<90^{\\circ}< \\) angle \\( avalanche \\). Consider the bisectors of the external angles at \\( gemstone \\) and \\( avalanche \\), each measured from the vertex to the opposite side (extended). Suppose both of these line-segments are equal to \\( gemstone avalanche \\). Compute the angle \\( gemstone \\).",
      "solution": "A-1. Suppose the bisector of the exterior angle at \\( gemstone \\) intersects line \\( avalanche hurricane \\) at \\( sunflower \\) and the bisector of the exterior angle at \\( avalanche \\) meets the line \\( gemstone hurricane \\) at \\( rectangle \\). The assumption that \\( hurricane \\) is between \\( avalanche \\) and \\( sunflower \\) contradicts the fact that \\( \\angle avalanche\\rangle \\angle hurricane \\) so we may assume that \\( avalanche \\) is between \\( sunflower \\) and \\( hurricane \\). Similarly, we conclude that \\( hurricane \\) is between \\( gemstone \\) and \\( rectangle \\) because \\( \\angle gemstone<\\angle hurricane \\).\n\nIf \\( butterfly \\) is a point on line \\( gemstone avalanche \\) with \\( avalanche \\) between \\( gemstone \\) and \\( butterfly \\), we have from triangle \\( gemstone avalanche rectangle \\) that \\( \\angle butterfly avalanche rectangle=2 gemstone \\). Hence, \\( \\angle avalanche sunflower gemstone=\\angle gemstone avalanche sunflower=\\angle butterfly avalanche hurricane=2 \\angle butterfly avalanche rectangle=4 gemstone \\), and the angle sum of triangle \\( gemstone avalanche sunflower \\) is \\( 90^{\\circ}-\\frac{1}{2} gemstone+8 gemstone \\). Thus, \\( gemstone=12^{\\circ} \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "A": "nonvertex",
        "B": "offvertex",
        "C": "voidspace",
        "X": "disjoint",
        "Y": "separate",
        "Z": "absentpt"
      },
      "question": "A-1. Let \\( nonvertex offvertex voidspace \\) be a triangle with angle \\( nonvertex< \\) angle \\( voidspace<90^{\\circ}< \\) angle \\( offvertex \\). Consider the bisectors of the external angles at \\( nonvertex \\) and \\( offvertex \\), each measured from the vertex to the opposite side (extended). Suppose both of these line-segments are equal to \\( nonvertex offvertex \\). Compute the angle \\( nonvertex \\).",
      "solution": "A-1. Suppose the bisector of the exterior angle at \\( nonvertex \\) intersects line \\( offvertex voidspace \\) at \\( disjoint \\) and the bisector of the exterior angle at \\( offvertex \\) meets the line \\( nonvertex voidspace \\) at \\( separate \\). The assumption that \\( voidspace \\) is between \\( offvertex \\) and \\( disjoint \\) contradicts the fact that \\( \\angle offvertex\\rangle \\angle voidspace \\) so we may assume that \\( offvertex \\) is between \\( disjoint \\) and \\( voidspace \\). Similarly, we conclude that \\( voidspace \\) is between \\( nonvertex \\) and \\( separate \\) because \\( \\angle nonvertex<\\angle voidspace \\).\n\nIf \\( absentpt \\) is a point on line \\( nonvertex offvertex \\) with \\( offvertex \\) between \\( nonvertex \\) and \\( absentpt \\), we have from triangle \\( nonvertex offvertex separate \\) that \\( \\angle absentpt offvertex separate=2 nonvertex \\). Hence, \\( \\angle offvertex disjoint nonvertex=\\angle nonvertex offvertex disjoint=\\angle absentpt offvertex voidspace=2 \\angle absentpt offvertex separate=4 nonvertex \\), and the angle sum of triangle \\( nonvertex offvertex disjoint \\) is \\( 90^{\\circ}-\\frac{1}{2} nonvertex+8 nonvertex \\). Thus, \\( nonvertex=12^{\\circ} \\)."
    },
    "garbled_string": {
      "map": {
        "A": "qzxwvtnp",
        "B": "hjgrksla",
        "C": "mvnlkqre",
        "X": "tbsfwrkm",
        "Y": "lzcqohif",
        "Z": "pjakuedm"
      },
      "question": "A-1. Let \\( qzxwvtnp hjgrksla mvnlkqre \\) be a triangle with angle \\( qzxwvtnp< \\) angle \\( mvnlkqre<90^{\\circ}< \\) angle \\( hjgrksla \\). Consider the bisectors of the external angles at \\( qzxwvtnp \\) and \\( hjgrksla \\), each measured from the vertex to the opposite side (extended). Suppose both of these line-segments are equal to \\( qzxwvtnp hjgrksla \\). Compute the angle \\( qzxwvtnp \\).",
      "solution": "A-1. Suppose the bisector of the exterior angle at \\( qzxwvtnp \\) intersects line \\( hjgrksla mvnlkqre \\) at \\( tbsfwrkm \\) and the bisector of the exterior angle at \\( hjgrksla \\) meets the line \\( qzxwvtnp mvnlkqre \\) at \\( lzcqohif \\). The assumption that \\( mvnlkqre \\) is between \\( hjgrksla \\) and \\( tbsfwrkm \\) contradicts the fact that \\( \\angle hjgrksla\\rangle \\angle mvnlkqre \\) so we may assume that \\( hjgrksla \\) is between \\( tbsfwrkm \\) and \\( mvnlkqre \\). Similarly, we conclude that \\( mvnlkqre \\) is between \\( qzxwvtnp \\) and \\( lzcqohif \\) because \\( \\angle qzxwvtnp<\\angle mvnlkqre \\).\n\nIf \\( pjakuedm \\) is a point on line \\( qzxwvtnp hjgrksla \\) with \\( hjgrksla \\) between \\( qzxwvtnp \\) and \\( pjakuedm \\), we have from triangle \\( qzxwvtnp hjgrksla lzcqohif \\) that \\( \\angle pjakuedm hjgrksla lzcqohif=2 qzxwvtnp \\). Hence, \\( \\angle hjgrksla tbsfwrkm qzxwvtnp=\\angle qzxwvtnp hjgrksla tbsfwrkm=\\angle pjakuedm hjgrksla mvnlkqre=2 \\angle pjakuedm hjgrksla lzcqohif=4 qzxwvtnp \\), and the angle sum of triangle \\( qzxwvtnp hjgrksla tbsfwrkm \\) is \\( 90^{\\circ}-\\frac{1}{2} qzxwvtnp+8 qzxwvtnp \\). Thus, \\( qzxwvtnp=12^{\\circ} \\)."
    },
    "kernel_variant": {
      "question": "Let \\(\\triangle ABC\\) be a triangle whose angles satisfy\n\n\\[\\angle A<\\angle C<90^{\\circ}<\\angle B.\\]\n\nDenote by \\(X\\) the point at which the bisector of the exterior angle at \\(A\\) meets the line \\(BC\\), and by \\(Y\\) the point at which the bisector of the exterior angle at \\(B\\) meets the line \\(AC\\).  Assume that each of those two bisector–segments has the same length as the side \\(AB\\); that is,\n\\[AX = AB = BY.\\]\nDetermine the measure of \\(\\angle A\\).",
      "solution": "Write\n\\[A=\\angle A,\\quad B=\\angle B,\\quad C=\\angle C,\\qquad A<C<90^{\\circ}<B,\\qquad A+B+C=180^{\\circ}.\\]\n\n--------------------------------------------------\n1.  Where are the points \\(X\\) and \\(Y\\)?\n--------------------------------------------------\nFor an exterior–angle bisector we will always take the bisector of the angle formed by **one side of the triangle and the extension of the other side**.\n\n•  Exterior bisector at \\(B\\).  It is the bisector of the angle formed by the rays \\(BA\\) (extended beyond \\(B\\)) and \\(BC\\).  By the Exterior–Angle–Bisector Theorem we have an *external* division\n\\[\\frac{AY}{YC}=\\frac{AB}{BC}.\\tag{1}\\]\nBecause \\(B>C>A\\) (so \\(A\\) is the smallest angle), the opposite–side lengths satisfy\n\\[BC<AB<AC.\\tag{2}\\]\nThus \\(AB>BC\\), and (1) implies \\(AY/YC>1\\); consequently \\(Y\\) lies on the *ray* \\(CA\\) beyond \\(C\\):\n\\[A\\;\\;C\\;\\;Y.\\]\n\n•  Exterior bisector at \\(A\\).  An analogous application of the theorem gives\n\\[\\frac{BX}{XC}=\\frac{AB}{AC}<1\\quad\\bigl(\\text{by (2)}\\bigr),\\]\nso \\(BX < XC\\) and \\(X\\) must lie on the ray \\(CB\\) beyond \\(B\\):\n\\[C\\;\\;B\\;\\;X.\\]\n\n--------------------------------------------------\n2.  Two isosceles triangles\n--------------------------------------------------\nThe hypotheses give\n\\[AB=BY\\quad\\text{and}\\quad AB=AX,\\]\nso\n\\[\\triangle ABY\\text{ is isosceles with }AB=BY,\\qquad \\triangle ABX\\text{ is isosceles with }AB=AX.\\]\n\n--------------------------------------------------\n3.  Angles in \\(\\triangle ABY\\)\n--------------------------------------------------\nBecause ray \\(AY\\) is the continuation of ray \\(AC\\), we have\n\\[\\angle BAY=\\angle BAC=A.\\]\nThe base angles of the isosceles triangle \\(ABY\\) are equal; hence\n\\[\\angle AYB=\\angle BAY=A.\\tag{3}\\]\n\nLet \\(Z\\) be the point on line \\(AB\\) such that \\(B\\) is between \\(A\\) and \\(Z\\).  In \\(\\triangle ABY\\) the exterior angle at \\(B\\) is\n\\[\\angle ZBY=\\angle BAY+\\angle AYB=A+A=2A.\\tag{4}\\]\n\n--------------------------------------------------\n4.  Angles in \\(\\triangle ABX\\)\n--------------------------------------------------\nThe exterior angle at \\(A\\) equals \\(180^{\\circ}-A\\); its bisector therefore makes\n\\[\\angle BAX=\\tfrac12(180^{\\circ}-A)=90^{\\circ}-\\tfrac12A.\\tag{5}\\]\n\nRay \\(BA\\) coincides with ray \\(BZ\\), and ray \\(BX\\) with ray \\(BC\\); hence\n\\[\\angle ABX=\\angle ZBC=2\\,\\angle ZBY=4A\\quad\\bigl(\\text{by (4)}\\bigr).\\tag{6}\\]\nBecause \\(AB=AX\\), triangle \\(ABX\\) is isosceles with base \\(BX\\), so\n\\[\\angle BXA=\\angle ABX=4A.\\tag{7}\\]\n\n--------------------------------------------------\n5.  The decisive equation\n--------------------------------------------------\nSum of the angles in \\(\\triangle ABX\\):\n\\[\\angle BAX+\\angle ABX+\\angle BXA=180^{\\circ}.\\]\nSubstituting (5), (6) and (7) we obtain\n\\[(90^{\\circ}-\\tfrac12A)+4A+4A=180^{\\circ}\\;\\Longrightarrow\\;90^{\\circ}+\\tfrac{15}{2}A=180^{\\circ}\\;\\Longrightarrow\\;\\frac{15}{2}A=90^{\\circ}\\;\\Longrightarrow\\;A=12^{\\circ}.\\]\n\n--------------------------------------------------\n6.  Consistency check (existence of such a triangle)\n--------------------------------------------------\nTake \\(A=12^{\\circ}\\).  Choose, for instance, \\(C=36^{\\circ}\\); then \\(B=132^{\\circ}>90^{\\circ}\\) and the angle ordering \\(A<C<B\\) is fulfilled.\n\nLet the side lengths be scaled so that \\(AB=1\\).  By the Law of Sines,\n\\[\\frac{BC}{\\sin A}=\\frac{AB}{\\sin C}=\\frac{1}{\\sin 36^{\\circ}}\\quad\\Longrightarrow\\quad BC=\\frac{\\sin 12^{\\circ}}{\\sin 36^{\\circ}}\\approx0.362.\\]\nSimilarly,\n\\[AC=\\frac{\\sin 132^{\\circ}}{\\sin 36^{\\circ}}\\approx1.532.\\]\nWith these numbers relation (1) indeed gives an external division point \\(Y\\) on ray \\(CA\\) with \\(BY=1\\); an analogous computation for the bisector at \\(A\\) yields \\(AX=1\\).  Hence a concrete triangle satisfying all requirements exists, so the obtained value of \\(A\\) is attainable.\n\n--------------------------------------------------\nAnswer\n--------------------------------------------------\n\\[\\boxed{\\angle A=12^{\\circ}}\\]",
      "_meta": {
        "core_steps": [
          "Equal-length exterior bisectors give AB = AX and AB = BY, so ΔABX and ΔABY are isosceles.",
          "Isosceles property ⇒ base angles equal; in ΔABY this makes the exterior angle at B (∠ZBY) equal to 2A.",
          "Exterior-angle-bisector fact at A gives ∠BAX = 90° – A⁄2.",
          "Angle chase with the two bisectors yields ∠ABX = ∠BXA = 4A.",
          "Triangle-angle-sum in ΔABX: (90° – A⁄2) + 8A = 180° ⇒ A = 12°."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Right-angle figure (90°) appearing only in the hypothesis C<90°<B; any fixed angle separating the acute C from the obtuse B would serve.",
            "original": "90°"
          },
          "slot2": {
            "description": "Stated ordering A < C; the proof needs merely C ≠ A for point placement, so this inequality could be reversed without changing the angle-chase logic.",
            "original": "A < C"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "calculation",
  "iteratively_fixed": true
}