{
  "index": "1952-B-1",
  "type": "GEO",
  "tag": [
    "GEO",
    "ALG"
  ],
  "difficulty": "",
  "question": "1. A mathematical moron is given two sides and the included angle of a triangle and attempts to use the Law of Cosines: \\( a^{2}=b^{2}+c^{2}-2 b c \\cos A \\), to find the third side \\( a \\). He uses logarithms as follows. He finds \\( \\log b \\) and doubles it; adds to that the double of \\( \\log c \\); subtracts the sum of the logarithms of \\( 2, b, c \\), and \\( \\cos A \\); divides the result by 2 ; and takes the antilogarithm. Although his method may be open to suspicion, his computation is accurate. What are the necessary and sufficient conditions on the triangle that this method should yield the correct result?",
  "solution": "Solution. The given sequence of operations will produce the correct answer for side \\( a \\) if and only if\n\\[\n\\frac{b^{2} c^{2}}{2 b c \\cos A}=b^{2}+c^{2}-2 b c \\cos A\n\\]\n\nThis yields\n\\[\n(b-2 c \\cos A)\\left(b-\\frac{c}{2 \\cos A}\\right)=0 .\n\\]\n\nHence either \\( b=2 c \\cos A \\) or \\( c=2 b \\cos A \\).\n\nThe condition \\( b=2 c \\cos A \\) implies that \\( D \\), the foot of the altitude from \\( B \\), is the midpoint of \\( \\overline{A C} \\), whence \\( \\angle A=\\angle C \\).\n\nConversely, if \\( \\angle A=\\angle C \\), then \\( b=2 c \\cos A \\).\nSimilarly, the condition \\( c=2 b \\cos A \\) is equivalent to \\( \\angle A=\\angle B \\).\nHence, in order for the moron's procedure to lead to the correct answer it is necessary and sufficient that the triangle \\( A B C \\) be isosceles with one of the base angles at \\( A \\).",
  "vars": [
    "a",
    "b",
    "c",
    "A",
    "B",
    "C",
    "D"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "a": "sidethird",
        "b": "sidefirst",
        "c": "sidesecond",
        "A": "angleone",
        "B": "angletwo",
        "C": "anglethree",
        "D": "footpoint"
      },
      "question": "1. A mathematical moron is given two sides and the included angle of a triangle and attempts to use the Law of Cosines: \\( sidethird^{2}=sidefirst^{2}+sidesecond^{2}-2 sidefirst sidesecond \\cos angleone \\), to find the third side \\( sidethird \\). He uses logarithms as follows. He finds \\( \\log sidefirst \\) and doubles it; adds to that the double of \\( \\log sidesecond \\); subtracts the sum of the logarithms of 2, sidefirst, sidesecond, and \\( \\cos angleone \\); divides the result by 2; and takes the antilogarithm. Although his method may be open to suspicion, his computation is accurate. What are the necessary and sufficient conditions on the triangle that this method should yield the correct result?",
      "solution": "Solution. The given sequence of operations will produce the correct answer for side \\( sidethird \\) if and only if\n\\[\n\\frac{sidefirst^{2} sidesecond^{2}}{2 sidefirst sidesecond \\cos angleone}=sidefirst^{2}+sidesecond^{2}-2 sidefirst sidesecond \\cos angleone\n\\]\n\nThis yields\n\\[\n(sidefirst-2 sidesecond \\cos angleone)\\left(sidefirst-\\frac{sidesecond}{2 \\cos angleone}\\right)=0 .\n\\]\n\nHence either \\( sidefirst=2 sidesecond \\cos angleone \\) or \\( sidesecond=2 sidefirst \\cos angleone \\).\n\nThe condition \\( sidefirst=2 sidesecond \\cos angleone \\) implies that \\( footpoint \\), the foot of the altitude from \\( angletwo \\), is the midpoint of \\( \\overline{angleone anglethree} \\), whence \\( \\angle angleone=\\angle anglethree \\).\n\nConversely, if \\( \\angle angleone=\\angle anglethree \\), then \\( sidefirst=2 sidesecond \\cos angleone \\).\nSimilarly, the condition \\( sidesecond=2 sidefirst \\cos angleone \\) is equivalent to \\( \\angle angleone=\\angle angletwo \\).\nHence, in order for the moron's procedure to lead to the correct answer it is necessary and sufficient that the triangle \\( angleone angletwo anglethree \\) be isosceles with one of the base angles at \\( angleone \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "a": "sunflower",
        "b": "waterfall",
        "c": "pineapple",
        "A": "parachute",
        "B": "notebook",
        "C": "rainstorm",
        "D": "quartzite"
      },
      "question": "1. A mathematical moron is given two sides and the included angle of a triangle and attempts to use the Law of Cosines: \\( sunflower^{2}=waterfall^{2}+pineapple^{2}-2 \\, waterfall \\, pineapple \\cos parachute \\), to find the third side \\( sunflower \\). He uses logarithms as follows. He finds \\( \\log waterfall \\) and doubles it; adds to that the double of \\( \\log pineapple \\); subtracts the sum of the logarithms of \\( 2, waterfall, pineapple \\), and \\( \\cos parachute \\); divides the result by 2 ; and takes the antilogarithm. Although his method may be open to suspicion, his computation is accurate. What are the necessary and sufficient conditions on the triangle that this method should yield the correct result?",
      "solution": "Solution. The given sequence of operations will produce the correct answer for side \\( sunflower \\) if and only if\n\\[\n\\frac{waterfall^{2} \\, pineapple^{2}}{2 \\, waterfall \\, pineapple \\cos parachute}=waterfall^{2}+pineapple^{2}-2 \\, waterfall \\, pineapple \\cos parachute\n\\]\n\nThis yields\n\\[\n(waterfall-2 \\, pineapple \\cos parachute)\\left(waterfall-\\frac{pineapple}{2 \\cos parachute}\\right)=0 .\n\\]\n\nHence either \\( waterfall=2 \\, pineapple \\cos parachute \\) or \\( pineapple=2 \\, waterfall \\cos parachute \\).\n\nThe condition \\( waterfall=2 \\, pineapple \\cos parachute \\) implies that \\( quartzite \\), the foot of the altitude from \\( notebook \\), is the midpoint of \\( \\overline{parachute rainstorm} \\), whence \\( \\angle parachute=\\angle rainstorm \\).\n\nConversely, if \\( \\angle parachute=\\angle rainstorm \\), then \\( waterfall=2 \\, pineapple \\cos parachute \\).\nSimilarly, the condition \\( pineapple=2 \\, waterfall \\cos parachute \\) is equivalent to \\( \\angle parachute=\\angle notebook \\).\nHence, in order for the moron's procedure to lead to the correct answer it is necessary and sufficient that the triangle \\( parachute \\, notebook \\, rainstorm \\) be isosceles with one of the base angles at \\( parachute \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "a": "nonlength",
        "b": "centerline",
        "c": "emptiness",
        "A": "nonangle",
        "B": "edgeless",
        "C": "boundless",
        "D": "headpoint"
      },
      "question": "1. A mathematical moron is given two sides and the included angle of a triangle and attempts to use the Law of Cosines: \\( nonlength^{2}=centerline^{2}+emptiness^{2}-2\\,centerline\\,emptiness \\cos nonangle \\), to find the third side \\( nonlength \\). He uses logarithms as follows. He finds \\( \\log centerline \\) and doubles it; adds to that the double of \\( \\log emptiness \\); subtracts the sum of the logarithms of 2, centerline, emptiness, and \\( \\cos nonangle \\); divides the result by 2; and takes the antilogarithm. Although his method may be open to suspicion, his computation is accurate. What are the necessary and sufficient conditions on the triangle that this method should yield the correct result?",
      "solution": "Solution. The given sequence of operations will produce the correct answer for side \\( nonlength \\) if and only if\n\\[\n\\frac{centerline^{2}\\emptiness^{2}}{2\\,centerline\\,\\emptiness \\cos nonangle}=centerline^{2}+\\emptiness^{2}-2\\,centerline\\,\\emptiness \\cos nonangle\n\\]\n\nThis yields\n\\[\n(centerline-2\\emptiness \\cos nonangle)\\left(centerline-\\frac{\\emptiness}{2 \\cos nonangle}\\right)=0 .\n\\]\n\nHence either \\( centerline=2 \\emptiness \\cos nonangle \\) or \\( \\emptiness=2 centerline \\cos nonangle \\).\n\nThe condition \\( centerline=2 \\emptiness \\cos nonangle \\) implies that \\( headpoint \\), the foot of the altitude from \\( edgeless \\), is the midpoint of \\( \\overline{nonangle \\boundless} \\), whence \\( \\angle nonangle=\\angle \\boundless \\).\n\nConversely, if \\( \\angle nonangle=\\angle \\boundless \\), then \\( centerline=2 \\emptiness \\cos nonangle \\).\nSimilarly, the condition \\( \\emptiness=2 centerline \\cos nonangle \\) is equivalent to \\( \\angle nonangle=\\angle edgeless \\).\nHence, in order for the moron's procedure to lead to the correct answer it is necessary and sufficient that the triangle \\( nonangle edgeless \\boundless \\) be isosceles with one of the base angles at \\( nonangle \\)."
    },
    "garbled_string": {
      "map": {
        "a": "qzxwvtnp",
        "b": "hjgrksla",
        "c": "mvdlkqre",
        "A": "nxfjgual",
        "B": "wqrzovkm",
        "C": "tpubmxsa",
        "D": "cgrylpqe"
      },
      "question": "Problem:\n<<<\n1. A mathematical moron is given two sides and the included angle of a triangle and attempts to use the Law of Cosines: \\( qzxwvtnp^{2}=hjgrksla^{2}+mvdlkqre^{2}-2 hjgrksla mvdlkqre \\cos nxfjgual \\), to find the third side \\( qzxwvtnp \\). He uses logarithms as follows. He finds \\( \\log hjgrksla \\) and doubles it; adds to that the double of \\( \\log mvdlkqre \\); subtracts the sum of the logarithms of \\( 2, hjgrksla, mvdlkqre \\), and \\( \\cos nxfjgual \\); divides the result by 2 ; and takes the antilogarithm. Although his method may be open to suspicion, his computation is accurate. What are the necessary and sufficient conditions on the triangle that this method should yield the correct result?\n>>>\n",
      "solution": "Solution:\n<<<\nSolution. The given sequence of operations will produce the correct answer for side \\( qzxwvtnp \\) if and only if\n\\[\n\\frac{hjgrksla^{2} mvdlkqre^{2}}{2 hjgrksla mvdlkqre \\cos nxfjgual}=hjgrksla^{2}+mvdlkqre^{2}-2 hjgrksla mvdlkqre \\cos nxfjgual\n\\]\n\nThis yields\n\\[\n(hjgrksla-2 mvdlkqre \\cos nxfjgual)\\left(hjgrksla-\\frac{mvdlkqre}{2 \\cos nxfjgual}\\right)=0 .\n\\]\n\nHence either \\( hjgrksla=2 mvdlkqre \\cos nxfjgual \\) or \\( mvdlkqre=2 hjgrksla \\cos nxfjgual \\).\n\nThe condition \\( hjgrksla=2 mvdlkqre \\cos nxfjgual \\) implies that \\( cgrylpqe \\), the foot of the altitude from \\( wqrzovkm \\), is the midpoint of \\( \\overline{nxfjgual tpubmxsa} \\), whence \\( \\angle nxfjgual=\\angle tpubmxsa \\).\n\nConversely, if \\( \\angle nxfjgual=\\angle tpubmxsa \\), then \\( hjgrksla=2 mvdlkqre \\cos nxfjgual \\).\nSimilarly, the condition \\( mvdlkqre=2 hjgrksla \\cos nxfjgual \\) is equivalent to \\( \\angle nxfjgual=\\angle wqrzovkm \\).\nHence, in order for the moron's procedure to lead to the correct answer it is necessary and sufficient that the triangle \\( nxfjgual wqrzovkm tpubmxsa \\) be isosceles with one of the base angles at \\( nxfjgual \\).\n>>>\n"
    },
    "kernel_variant": {
      "question": "Let △PQR have side-lengths\n   p = QR, q = RP, r = PQ\nopposite the angles\n   α = ∠P, β = ∠Q, γ = ∠R,\nrespectively.  The two side-lengths p and r together with their included angle β (known at the vertex Q) are given, and 0 < β < 90° so that cos β > 0.\n\nA student wishes to determine the third side q and employs the following ‘natural–logarithm recipe’.\n1. Evaluate ln p and ln r.\n2. Double each logarithm and add the two results.\n3. Subtract ln 2 + ln p + ln r + ln(cos β).\n4. Divide the difference by 2.\n5. Take the exponential of the obtained number.\n\nDenote by q_calc the number produced after the five steps.  For which (acute-angled) triangles does one always have q_calc = q, i.e. when does the method miraculously give the correct length of the third side?",
      "solution": "Write q_calc for the output of the student’s procedure.\n\n1.  Reconstructing the logarithmic computation\n   ln q_calc = ½[ 2 ln p + 2 ln r − ( ln 2 + ln p + ln r + ln(cos β ) ) ]\n              = ½ ln( p²r² / (2pr cos β) )\n              = ln √( pr / (2 cos β) ).\n   Hence                                        (1)\n   q_calc = √( pr / (2 cos β) ).\n\n2.  True value of the third side\n   By the Law of Cosines\n        q = √( p² + r² − 2pr cos β ).             (2)\n\n3.  Equality condition\n   The recipe succeeds exactly when (1) = (2):\n        pr /(2 cos β) = p² + r² − 2pr cos β.\n   Multiply by 2 cos β (> 0):\n        0 = 2p² cos β + 2r² cos β − 4pr cos²β − pr.\n   Rearranging and factorising,\n        0 = 2 cos β ( p − 2r cos β ) ( p − r /(2 cos β) ).      (3)\n   Because cos β > 0, equality (3) forces\n        p = 2r cos β  or  p = r /(2 cos β)\n   (the second form is equivalently r = 2p cos β).\n\n4.  Geometric interpretation\n   Apply the Law of Sines p / sin α = q / sin β = r / sin γ = 2R (R the circum-radius).\n\n   • Case 1: p = 2r cos β → from (2) one finds q = r.  Sides q and r are opposite β and γ, respectively; therefore β = γ and ∠Q = ∠R.  The triangle is isosceles with vertex P.\n\n   • Case 2: r = 2p cos β → from (2) one finds q = p.  Sides q and p are opposite β and α; hence β = α and ∠Q = ∠P.  The triangle is isosceles with vertex R.\n\nConversely, if the triangle is isosceles with β = γ (∠Q = ∠R) or with β = α (∠Q = ∠P), the corresponding relation p = 2r cos β or r = 2p cos β holds, making (1) and (2) identical.  Therefore the student’s logarithmic procedure works precisely in these two cases.\n\nFinal statement\nWith 0 < β < 90°, the recipe gives the correct third side if and only if the triangle is isosceles and the known angle β equals one of the other two angles, i.e.\n                     β = γ or β = α.",
      "_meta": {
        "core_steps": [
          "Translate the ‘log-trick’ into a closed-form value:  a_moron = √[(b^2 c^2)/(2 b c cos A)].",
          "Impose correctness by equating a_moron with the Law-of-Cosines value √[b^2 + c^2 − 2 b c cos A].",
          "Square both sides and factor to get (b − 2 c cos A)(b − c/(2 cos A)) = 0.",
          "Interpret b = 2 c cos A ⇔ ∠A = ∠C and c = 2 b cos A ⇔ ∠A = ∠B (via Law of Cosines / basic isosceles criterion).",
          "Conclude: triangle is isosceles with vertex A (and conversely), hence the moron’s answer is correct ⇔ ∠A equals one of the other two angles."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Choice of which side is to be found and which angle is the included one; any cyclic relabeling of (a,b,c) and (A,B,C) preserves the argument.",
            "original": "Unknown side a opposite angle A; known sides b and c adjacent to A."
          },
          "slot2": {
            "description": "Base (or notation) of the logarithms; the cancellation laws work for any positive base ≠1.",
            "original": "Implicitly common (base-10) logarithms denoted ‘log’."
          },
          "slot3": {
            "description": "Letter names for sides/angles used throughout the algebra and geometry.",
            "original": "Sides labelled a,b,c and opposite angles A,B,C."
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}