{
  "index": "1958-B-1",
  "type": "GEO",
  "tag": [
    "GEO"
  ],
  "difficulty": "",
  "question": "1. (i) Given line segments \\( A, B, C, D \\), with \\( A \\) the longest, construct a quadrilateral with these sides and with \\( A \\) and \\( B \\) parallel, when possible.\n(ii) Given any acute-angled triangle \\( A B C \\) and one altitude \\( A H \\), select any point \\( D \\) on \\( A H \\), then draw \\( B D \\) and extend until it intersects \\( A C \\) in \\( E \\), and draw \\( C D \\) and extend until it intersects \\( A B \\) in \\( F \\). Prove angle \\( A H E= \\) angle \\( A H F \\).",
  "solution": "Solution. Construct a triangle \\( P Q R \\) with \\( |P Q|=A-B,|P R|=C \\), and \\( |Q R|=D \\). This is possible if and only if these lengths satisfy the strict triangle inequality, i.e., the sum of the two shorter exceeds the longest. Extend \\( P Q \\) to \\( S \\) so that \\( |Q S|=B \\), and construct a parallelogram SQRT. Then PSTR is a quadrilateral satisfying the conditions of the problem.\n\nA second solution can be obtained by extending \\( P Q \\) in the opposite direction, but it is congruent to the first.\n\nIf \\( A=B \\) (it is not clear if this is supposed to be ruled out by the hypothesis \" \\( A \\) the longest\"), the triangle constructed above will be degenerate\nand the triangle inequality becomes \\( C=D \\). In this case there are infinitely many non-congruent solutions: any parallelogram with adjacent sides of lengths \\( A \\) and \\( C \\) will do.\n\nFirst Solution. Draw \\( l \\) through \\( A \\) parallel to \\( B C \\). Since angles \\( A B C \\) and \\( A C B \\) are acute, the foot \\( H \\) of the altitude falls between \\( B \\) and \\( C \\). Assuming \\( D \\neq A, H \\), which are trivial cases, complete the diagram as shown.\n\nConsidering several pairs of similar triangles, we see that\n\\[\n\\begin{array}{l} \n\\frac{|A X|}{|B H|}=\\frac{|A F|}{|B F|}=\\frac{|A W|}{|B C|}, \\quad \\frac{|A Y|}{|C H|}=\\frac{|A E|}{|C E|}=\\frac{|A Z|}{|C B|} \\\\\n\\frac{|A W|}{|H C|}=\\frac{|A D|}{|H D|}=\\frac{|A Z|}{|H B|}\n\\end{array}\n\\]\n\nTherefore, \\( |A X| \\cdot|B C|=|A W| \\cdot|B H|=|A Z| \\cdot|H C|=|A Y| \\cdot|B C| \\), whence \\( |A X|=|A Y| \\). So right triangles \\( A H X \\) and \\( A H Y \\) are congruent and \\( \\angle A H X=\\angle A H Y \\), as required.\n\nSecond Solution. Choose \\( B C \\) and \\( A H \\) as cartesian axes. Let \\( A=(0, a) \\), \\( B=(b, 0), C=(c, 0), D=(0, d) \\). Then the equation of \\( B D \\) is \\( d x+b y=b d \\), and the equation of \\( A C \\) is \\( a x+c y=a c \\). These lines meet at\n\\[\nE=\\left(\\frac{b c(a-d)}{a b-c d}, \\frac{a d(b-c)}{a b-c d}\\right)\n\\]\n\nHence the slope of \\( H E \\) is\n\\[\n\\frac{a d(b-c)}{b c(a-d)}\n\\]\n\nInterchanging \\( b \\) and \\( c \\), we find\n\\[\n\\text { slope } H F=\\frac{a d(c-b)}{b c(a-d)}=- \\text { slope } H E\n\\]\n\nTherefore, \\( \\angle A H E=\\angle A H F \\), as required.\nThe acute angle hypothesis shows that \\( b \\) and \\( c \\) have opposite signs, while \\( a \\) and \\( d \\) have the same sign (or \\( d=0 \\) ). Hence \\( a b-c d \\neq 0 \\), so \\( E \\) exists. Also \\( a c-b d \\neq 0 \\), so \\( F \\) exists. If \\( a=d \\), the lines \\( H E \\) and \\( H F \\) both coincide with \\( A H \\), so \\( \\angle A H E=\\angle A H F \\) anyway. The proof shows that we can take \\( D \\) anywhere on the line \\( A H \\) as long as neither \\( a b-c d \\) nor \\( a c-b d \\) is zero.\n\nThird Solution. The diagonal \\( B D \\) of the complete quadrilateral \\( B F, F D \\), \\( D H, H B \\) is divided harmonically by the other diagonals \\( H F \\) and \\( A C \\) at \\( G \\) and \\( E \\). Therefore, \\( H F, H E ; H B, H D \\) is a harmonic pencil. But \\( H B \\) and \\( H \\underset{\\sim}{\\boldsymbol{D}} \\) are perpendicular and hence they bisect the angles formed by \\( H F \\) and HE. See N, A. Court, College Geometry, 2nd ed., Barnes and Noble, New York, 1952.\n\nRemark. It is interesting to compare the special cases that arise in the different proofs. The case \\( D=H \\) is exceptional in the synthetic and projective arguments, but not in the analytic argument, while the case \\( B D \\| A C \\) is exceptional for the synthetic and analytic arguments, but not the projective one. In the synthetic proof, the figure changes its appearance markedly if \\( D \\) is outside \\( A H \\).",
  "vars": [
    "A",
    "B",
    "C",
    "D",
    "E",
    "F",
    "H",
    "P",
    "Q",
    "R",
    "S",
    "T",
    "X",
    "Y",
    "W",
    "G",
    "l"
  ],
  "params": [
    "a",
    "b",
    "c",
    "d"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "A": "pointa",
        "B": "pointb",
        "C": "pointc",
        "D": "pointd",
        "E": "pointe",
        "F": "pointf",
        "H": "pointh",
        "P": "pointp",
        "Q": "pointq",
        "R": "pointr",
        "S": "points",
        "T": "pointt",
        "X": "pointx",
        "Y": "pointy",
        "W": "pointw",
        "G": "pointg",
        "l": "linell",
        "a": "consta",
        "b": "constb",
        "c": "constc",
        "d": "constd"
      },
      "question": "1. (i) Given line segments \\( pointa, pointb, pointc, pointd \\), with \\( pointa \\) the longest, construct a quadrilateral with these sides and with \\( pointa \\) and \\( pointb \\) parallel, when possible.\n(ii) Given any acute-angled triangle \\( pointa pointb pointc \\) and one altitude \\( pointa pointh \\), select any point \\( pointd \\) on \\( pointa pointh \\), then draw \\( pointb pointd \\) and extend until it intersects \\( pointa pointc \\) in \\( pointe \\), and draw \\( pointc pointd \\) and extend until it intersects \\( pointa pointb \\) in \\( pointf \\). Prove angle \\( pointa pointh pointe = \\) angle \\( pointa pointh pointf \\).",
      "solution": "Solution. Construct a triangle \\( pointp pointq pointr \\) with \\( |pointp pointq| = pointa - pointb, |pointp pointr| = pointc \\), and \\( |pointq pointr| = pointd \\). This is possible if and only if these lengths satisfy the strict triangle inequality, i.e., the sum of the two shorter exceeds the longest. Extend \\( pointp pointq \\) to \\( points \\) so that \\( |pointq points| = pointb \\), and construct a parallelogram points pointq pointr pointt. Then pointp points pointt pointr is a quadrilateral satisfying the conditions of the problem.\n\nA second solution can be obtained by extending \\( pointp pointq \\) in the opposite direction, but it is congruent to the first.\n\nIf \\( pointa = pointb \\) (it is not clear if this is supposed to be ruled out by the hypothesis \" \\( pointa \\) the longest\"), the triangle constructed above will be degenerate and the triangle inequality becomes \\( pointc = pointd \\). In this case there are infinitely many non-congruent solutions: any parallelogram with adjacent sides of lengths \\( pointa \\) and \\( pointc \\) will do.\n\nFirst Solution. Draw \\( linell \\) through \\( pointa \\) parallel to \\( pointb pointc \\). Since angles \\( pointa pointb pointc \\) and \\( pointa pointc pointb \\) are acute, the foot \\( pointh \\) of the altitude falls between \\( pointb \\) and \\( pointc \\). Assuming \\( pointd \\neq pointa, pointh \\), which are trivial cases, complete the diagram as shown.\n\nConsidering several pairs of similar triangles, we see that\n\\[\n\\begin{array}{l} \n\\dfrac{|pointa pointx|}{|pointb pointh|}= \\dfrac{|pointa pointf|}{|pointb pointf|}= \\dfrac{|pointa pointw|}{|pointb pointc|}, \\quad\n\\dfrac{|pointa pointy|}{|pointc pointh|}= \\dfrac{|pointa pointe|}{|pointc pointe|}= \\dfrac{|pointa Z|}{|pointc pointb|} \\\\\n\\dfrac{|pointa pointw|}{|pointh pointc|}= \\dfrac{|pointa pointd|}{|pointh pointd|}= \\dfrac{|pointa Z|}{|pointh pointb|}\n\\end{array}\n\\]\n\nTherefore, \\( |pointa pointx| \\cdot |pointb pointc| = |pointa pointw| \\cdot |pointb pointh| = |pointa Z| \\cdot |pointh pointc| = |pointa pointy| \\cdot |pointb pointc| \\), whence \\( |pointa pointx| = |pointa pointy| \\). So right triangles \\( pointa pointh pointx \\) and \\( pointa pointh pointy \\) are congruent and \\( \\angle pointa pointh pointx = \\angle pointa pointh pointy \\), as required.\n\nSecond Solution. Choose \\( pointb pointc \\) and \\( pointa pointh \\) as cartesian axes. Let \\( pointa = (0, consta) \\), \\( pointb = (constb, 0), pointc = (constc, 0), pointd = (0, constd) \\). Then the equation of \\( pointb pointd \\) is \\( constd x + constb y = constb constd \\), and the equation of \\( pointa pointc \\) is \\( consta x + constc y = consta constc \\). These lines meet at\n\\[\npointe = \\left( \\frac{constb\\,constc( consta - constd)}{consta\\,constb - constc\\,constd},\\; \\frac{consta\\,constd( constb - constc)}{consta\\,constb - constc\\,constd} \\right)\n\\]\nHence the slope of \\( pointh pointe \\) is\n\\[\n\\frac{consta\\,constd( constb - constc)}{constb\\,constc( consta - constd)}\n\\]\nInterchanging \\( constb \\) and \\( constc \\), we find\n\\[\n\\text{ slope }\\, pointh pointf = \\frac{consta\\,constd( constc - constb)}{constb\\,constc( consta - constd)} = -\\, \\text{ slope }\\, pointh pointe\n\\]\nTherefore, \\( \\angle pointa pointh pointe = \\angle pointa pointh pointf \\), as required.\nThe acute-angle hypothesis shows that \\( constb \\) and \\( constc \\) have opposite signs, while \\( consta \\) and \\( constd \\) have the same sign (or \\( constd = 0 \\) ). Hence \\( consta\\,constb - constc\\,constd \\neq 0 \\), so \\( pointe \\) exists. Also \\( consta\\,constc - constb\\,constd \\neq 0 \\), so \\( pointf \\) exists. If \\( consta = constd \\), the lines \\( pointh pointe \\) and \\( pointh pointf \\) both coincide with \\( pointa pointh \\), so \\( \\angle pointa pointh pointe = \\angle pointa pointh pointf \\) anyway. The proof shows that we can take \\( pointd \\) anywhere on the line \\( pointa pointh \\) as long as neither \\( consta\\,constb - constc\\,constd \\) nor \\( consta\\,constc - constb\\,constd \\) is zero.\n\nThird Solution. The diagonal \\( pointb pointd \\) of the complete quadrilateral \\( pointb pointf, pointf pointd, pointd pointh, pointh pointb \\) is divided harmonically by the other diagonals \\( pointh pointf \\) and \\( pointa pointc \\) at \\( pointg \\) and \\( pointe \\). Therefore, \\( pointh pointf, pointh pointe ; pointh pointb, pointh pointd \\) is a harmonic pencil. But \\( pointh pointb \\) and \\( pointh \\underset{\\sim}{\\boldsymbol{pointd}} \\) are perpendicular and hence they bisect the angles formed by \\( pointh pointf \\) and \\( pointh pointe \\). See N. A. Court, College Geometry, 2nd ed., Barnes and Noble, New York, 1952.\n\nRemark. It is interesting to compare the special cases that arise in the different proofs. The case \\( pointd = pointh \\) is exceptional in the synthetic and projective arguments, but not in the analytic argument, while the case \\( pointb pointd \\parallel pointa pointc \\) is exceptional for the synthetic and analytic arguments, but not the projective one. In the synthetic proof, the figure changes its appearance markedly if \\( pointd \\) is outside \\( pointa pointh \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "A": "kingfisher",
        "B": "plainsong",
        "C": "starlight",
        "D": "moonstone",
        "E": "riverbank",
        "F": "tumbleweed",
        "H": "goldcrest",
        "P": "cloudburst",
        "Q": "riverdance",
        "R": "crimsonia",
        "S": "windswept",
        "T": "silhouette",
        "X": "rainflower",
        "Y": "lighthouse",
        "W": "thornberry",
        "G": "crosswinds",
        "l": "marigolds",
        "a": "pineapple",
        "b": "dragonfly",
        "c": "honeycomb",
        "d": "silverleaf"
      },
      "question": "1. (i) Given line segments \\( kingfisher, plainsong, starlight, moonstone \\), with \\( kingfisher \\) the longest, construct a quadrilateral with these sides and with \\( kingfisher \\) and \\( plainsong \\) parallel, when possible.\n(ii) Given any acute-angled triangle \\( kingfisher\\ plainsong\\ starlight \\) and one altitude \\( kingfisher\\ goldcrest \\), select any point \\( moonstone \\) on \\( kingfisher\\ goldcrest \\), then draw \\( plainsong\\ moonstone \\) and extend until it intersects \\( kingfisher\\ starlight \\) in \\( riverbank \\), and draw \\( starlight\\ moonstone \\) and extend until it intersects \\( kingfisher\\ plainsong \\) in \\( tumbleweed \\). Prove angle \\( kingfisher\\ goldcrest\\ riverbank= \\) angle \\( kingfisher\\ goldcrest\\ tumbleweed \\).",
      "solution": "Solution. Construct a triangle \\( cloudburst\\ riverdance\\ crimsonia \\) with \\( |cloudburst\\ riverdance|=kingfisher-plainsong,|cloudburst\\ crimsonia|=starlight \\), and \\( |riverdance\\ crimsonia|=moonstone \\). This is possible if and only if these lengths satisfy the strict triangle inequality, i.e., the sum of the two shorter exceeds the longest. Extend \\( cloudburst\\ riverdance \\) to \\( windswept \\) so that \\( |riverdance\\ windswept|=plainsong \\), and construct a parallelogram windswept riverdance crimsonia silhouette. Then cloudburst windswept silhouette crimsonia is a quadrilateral satisfying the conditions of the problem.\n\nA second solution can be obtained by extending \\( cloudburst\\ riverdance \\) in the opposite direction, but it is congruent to the first.\n\nIf \\( kingfisher=plainsong \\) (it is not clear if this is supposed to be ruled out by the hypothesis “\\( kingfisher \\) the longest”), the triangle constructed above will be degenerate and the triangle inequality becomes \\( starlight=moonstone \\). In this case there are infinitely many non-congruent solutions: any parallelogram with adjacent sides of lengths \\( kingfisher \\) and \\( starlight \\) will do.\n\nFirst Solution. Draw \\( marigolds \\) through \\( kingfisher \\) parallel to \\( plainsong\\ starlight \\). Since angles \\( kingfisher\\ plainsong\\ starlight \\) and \\( kingfisher\\ starlight\\ plainsong \\) are acute, the foot \\( goldcrest \\) of the altitude falls between \\( plainsong \\) and \\( starlight \\). Assuming \\( moonstone \\neq kingfisher, goldcrest \\), which are trivial cases, complete the diagram as shown.\n\nConsidering several pairs of similar triangles, we see that\n\\[\n\\begin{array}{l} \n\\frac{|kingfisher\\ rainflower|}{|plainsong\\ goldcrest|}=\\frac{|kingfisher\\ tumbleweed|}{|plainsong\\ tumbleweed|}=\\frac{|kingfisher\\ thornberry|}{|plainsong\\ starlight|}, \\quad \\frac{|kingfisher\\ lighthouse|}{|starlight\\ goldcrest|}=\\frac{|kingfisher\\ riverbank|}{|starlight\\ riverbank|}=\\frac{|kingfisher\\ Z|}{|starlight\\ plainsong|} \\\\ \n\\frac{|kingfisher\\ thornberry|}{|goldcrest\\ starlight|}=\\frac{|kingfisher\\ moonstone|}{|goldcrest\\ moonstone|}=\\frac{|kingfisher\\ Z|}{|goldcrest\\ plainsong|}\n\\end{array}\n\\]\n\nTherefore, \\( |kingfisher\\ rainflower| \\cdot|plainsong\\ starlight|=|kingfisher\\ thornberry| \\cdot|plainsong\\ goldcrest|=|kingfisher\\ Z| \\cdot|goldcrest\\ starlight|=|kingfisher\\ lighthouse| \\cdot|plainsong\\ starlight| \\), whence \\( |kingfisher\\ rainflower|=|kingfisher\\ lighthouse| \\). So right triangles \\( kingfisher\\ goldcrest\\ rainflower \\) and \\( kingfisher\\ goldcrest\\ lighthouse \\) are congruent and \\( \\angle kingfisher\\ goldcrest\\ rainflower=\\angle kingfisher\\ goldcrest\\ lighthouse \\), as required.\n\nSecond Solution. Choose \\( plainsong\\ starlight \\) and \\( kingfisher\\ goldcrest \\) as cartesian axes. Let \\( kingfisher=(0, pineapple) \\), \\( plainsong=(dragonfly,0), starlight=(honeycomb,0), moonstone=(0, silverleaf) \\). Then the equation of \\( plainsong\\ moonstone \\) is \\( silverleaf x+dragonfly y=dragonfly silverleaf \\), and the equation of \\( kingfisher\\ starlight \\) is \\( pineapple x+honeycomb y=pineapple honeycomb \\). These lines meet at\n\\[\nriverbank=\\left(\\frac{dragonfly honeycomb(pineapple-silverleaf)}{pineapple dragonfly-honeycomb silverleaf}, \\frac{pineapple silverleaf(dragonfly-honeycomb)}{pineapple dragonfly-honeycomb silverleaf}\\right)\n\\]\n\nHence the slope of \\( goldcrest\\ riverbank \\) is\n\\[\n\\frac{pineapple silverleaf(dragonfly-honeycomb)}{dragonfly honeycomb(pineapple-silverleaf)}\n\\]\n\nInterchanging \\( dragonfly \\) and \\( honeycomb \\), we find\n\\[\n\\text { slope } goldcrest\\ tumbleweed=\\frac{pineapple silverleaf(honeycomb-dragonfly)}{dragonfly honeycomb(pineapple-silverleaf)}=- \\text { slope } goldcrest\\ riverbank\n\\]\n\nTherefore, \\( \\angle kingfisher\\ goldcrest\\ riverbank=\\angle kingfisher\\ goldcrest\\ tumbleweed \\), as required.\nThe acute angle hypothesis shows that \\( dragonfly \\) and \\( honeycomb \\) have opposite signs, while \\( pineapple \\) and \\( silverleaf \\) have the same sign (or \\( silverleaf=0 \\) ). Hence \\( pineapple dragonfly-honeycomb silverleaf \\neq 0 \\), so \\( riverbank \\) exists. Also \\( pineapple honeycomb-dragonfly silverleaf \\neq 0 \\), so \\( tumbleweed \\) exists. If \\( pineapple=silverleaf \\), the lines \\( goldcrest\\ riverbank \\) and \\( goldcrest\\ tumbleweed \\) both coincide with \\( kingfisher\\ goldcrest \\), so \\( \\angle kingfisher\\ goldcrest\\ riverbank=\\angle kingfisher\\ goldcrest\\ tumbleweed \\) anyway. The proof shows that we can take \\( moonstone \\) anywhere on the line \\( kingfisher\\ goldcrest \\) as long as neither \\( pineapple dragonfly-honeycomb silverleaf \\) nor \\( pineapple honeycomb-dragonfly silverleaf \\) is zero.\n\nThird Solution. The diagonal \\( plainsong\\ moonstone \\) of the complete quadrilateral \\( plainsong\\ tumbleweed, tumbleweed\\ moonstone \\), \\( moonstone\\ goldcrest, goldcrest\\ plainsong \\) is divided harmonically by the other diagonals \\( goldcrest\\ tumbleweed \\) and \\( kingfisher\\ starlight \\) at \\( crosswinds \\) and \\( riverbank \\). Therefore, \\( goldcrest\\ tumbleweed, goldcrest\\ riverbank ; goldcrest\\ plainsong, goldcrest\\ moonstone \\) is a harmonic pencil. But \\( goldcrest\\ plainsong \\) and \\( goldcrest \\underset{\\sim}{\\boldsymbol{moonstone}} \\) are perpendicular and hence they bisect the angles formed by \\( goldcrest\\ tumbleweed \\) and \\( goldcrest\\ riverbank \\). See N, A. Court, College Geometry, 2nd ed., Barnes and Noble, New York, 1952.\n\nRemark. It is interesting to compare the special cases that arise in the different proofs. The case \\( moonstone=goldcrest \\) is exceptional in the synthetic and projective arguments, but not in the analytic argument, while the case \\( plainsong\\ moonstone \\| kingfisher\\ starlight \\) is exceptional for the synthetic and analytic arguments, but not the projective one. In the synthetic proof, the figure changes its appearance markedly if \\( moonstone \\) is outside \\( kingfisher\\ goldcrest \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "A": "minutiae",
        "B": "briefness",
        "C": "conciseness",
        "D": "diminutive",
        "E": "ephemeral",
        "F": "fleetingly",
        "H": "depthless",
        "P": "expansive",
        "Q": "placidity",
        "R": "restraint",
        "S": "stillness",
        "T": "tranquil",
        "X": "certainty",
        "Y": "familiar",
        "W": "wholeness",
        "G": "infinite",
        "l": "pointlike",
        "a": "ampleless",
        "b": "boundless",
        "c": "compactly",
        "d": "deficient"
      },
      "question": "1. (i) Given line segments \\( minutiae, briefness, conciseness, diminutive \\), with \\( minutiae \\) the longest, construct a quadrilateral with these sides and with \\( minutiae \\) and \\( briefness \\) parallel, when possible.\n(ii) Given any acute-angled triangle \\( minutiae\\, briefness\\, conciseness \\) and one altitude \\( minutiae depthless \\), select any point \\( diminutive \\) on \\( minutiae depthless \\), then draw \\( briefness diminutive \\) and extend until it intersects \\( minutiae conciseness \\) in \\( ephemeral \\), and draw \\( conciseness diminutive \\) and extend until it intersects \\( minutiae briefness \\) in \\( fleetingly \\). Prove angle \\( minutiae depthless ephemeral= \\) angle \\( minutiae depthless fleetingly \\).",
      "solution": "Solution. Construct a triangle \\( expansive\\, placidity\\, restraint \\) with \\( |expansive\\, placidity|=minutiae-briefness,|expansive\\, restraint|=conciseness \\), and \\( |placidity\\, restraint|=diminutive \\). This is possible if and only if these lengths satisfy the strict triangle inequality, i.e., the sum of the two shorter exceeds the longest. Extend \\( expansive\\, placidity \\) to \\( stillness \\) so that \\( |placidity\\, stillness|=briefness \\), and construct a parallelogram stillness placidity restraint tranquil. Then expansivestillnesstranquilrestraint is a quadrilateral satisfying the conditions of the problem.\n\nA second solution can be obtained by extending \\( expansive\\, placidity \\) in the opposite direction, but it is congruent to the first.\n\nIf \\( minutiae=briefness \\) (it is not clear if this is supposed to be ruled out by the hypothesis \" \\( minutiae \\) the longest\"), the triangle constructed above will be degenerate and the triangle inequality becomes \\( conciseness=diminutive \\). In this case there are infinitely many non-congruent solutions: any parallelogram with adjacent sides of lengths \\( minutiae \\) and \\( conciseness \\) will do.\n\nFirst Solution. Draw \\( pointlike \\) through \\( minutiae \\) parallel to \\( briefness conciseness \\). Since angles \\( minutiae briefness conciseness \\) and \\( minutiae conciseness briefness \\) are acute, the foot \\( depthless \\) of the altitude falls between \\( briefness \\) and \\( conciseness \\). Assuming \\( diminutive \\neq minutiae, depthless \\), which are trivial cases, complete the diagram as shown.\n\nConsidering several pairs of similar triangles, we see that\n\\[\n\\begin{array}{l} \n\\frac{|minutiae\\, certainty|}{|briefness\\, depthless|}=\\frac{|minutiae\\, fleetingly|}{|briefness\\, fleetingly|}=\\frac{|minutiae\\, wholeness|}{|briefness\\, conciseness|}, \\quad \\frac{|minutiae\\, familiar|}{|conciseness\\, depthless|}=\\frac{|minutiae\\, ephemeral|}{|conciseness\\, ephemeral|}=\\frac{|minutiae\\, Z|}{|conciseness\\, briefness|} \\\\\n\\frac{|minutiae\\, wholeness|}{|depthless\\, conciseness|}=\\frac{|minutiae\\, diminutive|}{|depthless\\, diminutive|}=\\frac{|minutiae\\, Z|}{|depthless\\, briefness|}\n\\end{array}\n\\]\n\nTherefore, \\( |minutiae\\, certainty| \\cdot|briefness\\, conciseness|=|minutiae\\, wholeness| \\cdot|briefness\\, depthless|=|minutiae\\, Z| \\cdot|depthless\\, conciseness|=|minutiae\\, familiar| \\cdot|briefness\\, conciseness| \\), whence \\( |minutiae\\, certainty|=|minutiae\\, familiar| \\). So right triangles \\( minutiae depthless certainty \\) and \\( minutiae depthless familiar \\) are congruent and \\( \\angle minutiae depthless certainty=\\angle minutiae depthless familiar \\), as required.\n\nSecond Solution. Choose \\( briefness conciseness \\) and \\( minutiae depthless \\) as cartesian axes. Let \\( minutiae=(0, ampleless) \\), \\( briefness=(boundless, 0), conciseness=(compactly, 0), diminutive=(0, deficient) \\). Then the equation of \\( briefness diminutive \\) is \\( deficient x+boundless y=boundless deficient \\), and the equation of \\( minutiae conciseness \\) is \\( ampleless x+compactly y=ampleless compactly \\). These lines meet at\n\\[\nephemeral=\\left(\\frac{boundless\\, compactly(ampleless-deficient)}{ampleless\\, boundless-compactly\\, deficient}, \\frac{ampleless\\, deficient(boundless-compactly)}{ampleless\\, boundless-compactly\\, deficient}\\right)\n\\]\n\nHence the slope of \\( depthless\\, ephemeral \\) is\n\\[\n\\frac{ampleless\\, deficient(boundless-compactly)}{boundless\\, compactly(ampleless-deficient)}\n\\]\n\nInterchanging \\( boundless \\) and \\( compactly \\), we find\n\\[\n\\text{ slope } depthless\\, fleetingly=\\frac{ampleless\\, deficient(compactly-boundless)}{boundless\\, compactly(ampleless-deficient)}=-\\text{ slope } depthless\\, ephemeral\n\\]\n\nTherefore, \\( \\angle minutiae depthless ephemeral=\\angle minutiae depthless fleetingly \\), as required. The acute-angle hypothesis shows that \\( boundless \\) and \\( compactly \\) have opposite signs, while \\( ampleless \\) and \\( deficient \\) have the same sign (or \\( deficient=0 \\)). Hence \\( ampleless\\, boundless-compactly\\, deficient \\neq 0 \\), so \\( ephemeral \\) exists. Also \\( ampleless\\, compactly-boundless\\, deficient \\neq 0 \\), so \\( fleetingly \\) exists. If \\( ampleless=deficient \\), the lines \\( depthless\\, ephemeral \\) and \\( depthless\\, fleetingly \\) both coincide with \\( minutiae depthless \\), so \\( \\angle minutiae depthless ephemeral=\\angle minutiae depthless fleetingly \\) anyway. The proof shows that we can take \\( diminutive \\) anywhere on the line \\( minutiae depthless \\) as long as neither \\( ampleless\\, boundless-compactly\\, deficient \\) nor \\( ampleless\\, compactly-boundless\\, deficient \\) is zero.\n\nThird Solution. The diagonal \\( briefness\\, diminutive \\) of the complete quadrilateral \\( briefness\\, fleetingly, fleetingly\\, diminutive \\), \\( diminutive depthless, depthless briefness \\) is divided harmonically by the other diagonals \\( depthless\\, fleetingly \\) and \\( minutiae conciseness \\) at \\( infinite \\) and \\( ephemeral \\). Therefore, \\( depthless\\, fleetingly, depthless\\, ephemeral ; depthless\\, briefness, depthless\\, diminutive \\) is a harmonic pencil. But \\( depthless\\, briefness \\) and \\( depthless \\underset{\\sim}{\\boldsymbol{diminutive}} \\) are perpendicular and hence they bisect the angles formed by \\( depthless\\, fleetingly \\) and depthless ephemeral. See N, A. Court, College Geometry, 2nd ed., Barnes and Noble, New York, 1952.\n\nRemark. It is interesting to compare the special cases that arise in the different proofs. The case \\( diminutive=depthless \\) is exceptional in the synthetic and projective arguments, but not in the analytic argument, while the case \\( briefness\\, diminutive \\| minutiae conciseness \\) is exceptional for the synthetic and analytic arguments, but not the projective one. In the synthetic proof, the figure changes its appearance markedly if \\( diminutive \\) is outside \\( minutiae depthless \\)."
    },
    "garbled_string": {
      "map": {
        "A": "qzxwvtnp",
        "B": "hjgrksla",
        "C": "fncmezod",
        "D": "plkturay",
        "E": "gsivnola",
        "F": "kdwqepim",
        "H": "xvpalbru",
        "P": "sorbdeqm",
        "Q": "meflarxd",
        "R": "jubncovy",
        "S": "wzeskiht",
        "T": "nahuvlpg",
        "X": "tqferdsm",
        "Y": "oxkclvwe",
        "W": "lirpzqun",
        "G": "ycvmrdah",
        "l": "nawcvpoh",
        "a": "krptlena",
        "b": "zgxeirof",
        "c": "whubtdam",
        "d": "vysloqne"
      },
      "question": "1. (i) Given line segments \\( qzxwvtnp, hjgrksla, fncmezod, plkturay \\), with \\( qzxwvtnp \\) the longest, construct a quadrilateral with these sides and with \\( qzxwvtnp \\) and \\( hjgrksla \\) parallel, when possible.\n(ii) Given any acute-angled triangle \\( qzxwvtnp hjgrksla fncmezod \\) and one altitude \\( qzxwvtnp xvpalbru \\), select any point \\( plkturay \\) on \\( qzxwvtnp xvpalbru \\), then draw \\( hjgrksla plkturay \\) and extend until it intersects \\( qzxwvtnp fncmezod \\) in \\( gsivnola \\), and draw \\( fncmezod plkturay \\) and extend until it intersects \\( qzxwvtnp hjgrksla \\) in \\( kdwqepim \\). Prove angle \\( qzxwvtnp xvpalbru gsivnola= \\) angle \\( qzxwvtnp xvpalbru kdwqepim \\).",
      "solution": "Solution. Construct a triangle \\( sorbdeqm meflarxd jubncovy \\) with \\( |sorbdeqm meflarxd|=qzxwvtnp-hjgrksla,|sorbdeqm jubncovy|=fncmezod \\), and \\( |meflarxd jubncovy|=plkturay \\). This is possible if and only if these lengths satisfy the strict triangle inequality, i.e., the sum of the two shorter exceeds the longest. Extend \\( sorbdeqm meflarxd \\) to \\( wzeskiht \\) so that \\( |meflarxd wzeskiht|=hjgrksla \\), and construct a parallelogram wzeskihtmeflarxdjubncovynahuvlpg. Then sorbdeqmwzeskihtnahuvlpgjubncovy is a quadrilateral satisfying the conditions of the problem.\n\nA second solution can be obtained by extending \\( sorbdeqm meflarxd \\) in the opposite direction, but it is congruent to the first.\n\nIf \\( qzxwvtnp=hjgrksla \\) (it is not clear if this is supposed to be ruled out by the hypothesis \" \\( qzxwvtnp \\) the longest\"), the triangle constructed above will be degenerate\nand the triangle inequality becomes \\( fncmezod=plkturay \\). In this case there are infinitely many non-congruent solutions: any parallelogram with adjacent sides of lengths \\( qzxwvtnp \\) and \\( fncmezod \\) will do.\n\nFirst Solution. Draw \\( nawcvpoh \\) through \\( qzxwvtnp \\) parallel to \\( hjgrksla fncmezod \\). Since angles \\( qzxwvtnp hjgrksla fncmezod \\) and \\( qzxwvtnp fncmezod hjgrksla \\) are acute, the foot \\( xvpalbru \\) of the altitude falls between \\( hjgrksla \\) and \\( fncmezod \\). Assuming \\( plkturay \\neq qzxwvtnp, xvpalbru \\), which are trivial cases, complete the diagram as shown.\n\nConsidering several pairs of similar triangles, we see that\n\\[\n\\begin{array}{l} \n\\frac{|qzxwvtnp tqferdsm|}{|hjgrksla xvpalbru|}=\\frac{|qzxwvtnp kdwqepim|}{|hjgrksla kdwqepim|}=\\frac{|qzxwvtnp lirpzqun|}{|hjgrksla fncmezod|}, \\quad \\frac{|qzxwvtnp oxkclvwe|}{|fncmezod xvpalbru|}=\\frac{|qzxwvtnp gsivnola|}{|fncmezod gsivnola|}=\\frac{|qzxwvtnp Z|}{|fncmezod hjgrksla|} \\\\\n\\frac{|qzxwvtnp lirpzqun|}{|xvpalbru fncmezod|}=\\frac{|qzxwvtnp plkturay|}{|xvpalbru plkturay|}=\\frac{|qzxwvtnp Z|}{|xvpalbru hjgrksla|}\n\\end{array}\n\\]\n\nTherefore, \\( |qzxwvtnp tqferdsm| \\cdot|hjgrksla fncmezod|=|qzxwvtnp lirpzqun| \\cdot|hjgrksla xvpalbru|=|qzxwvtnp Z| \\cdot|xvpalbru fncmezod|=|qzxwvtnp oxkclvwe| \\cdot|hjgrksla fncmezod| \\), whence \\( |qzxwvtnp tqferdsm|=|qzxwvtnp oxkclvwe| \\). So right triangles \\( qzxwvtnp xvpalbru tqferdsm \\) and \\( qzxwvtnp xvpalbru oxkclvwe \\) are congruent and \\( \\angle qzxwvtnp xvpalbru tqferdsm=\\angle qzxwvtnp xvpalbru oxkclvwe \\), as required.\n\nSecond Solution. Choose \\( hjgrksla fncmezod \\) and \\( qzxwvtnp xvpalbru \\) as cartesian axes. Let \\( qzxwvtnp=(0, krptlena) \\), \\( hjgrksla=(zgxeirof, 0), fncmezod=(whubtdam, 0), plkturay=(0, vysloqne) \\). Then the equation of \\( hjgrksla plkturay \\) is \\( vysloqne x+zgxeirof y=zgxeirof vysloqne \\), and the equation of \\( qzxwvtnp fncmezod \\) is \\( krptlena x+whubtdam y=krptlena whubtdam \\). These lines meet at\n\\[\ngsivnola=\\left(\\frac{zgxeirof whubtdam(krptlena-vysloqne)}{krptlena zgxeirof-whubtdam vysloqne}, \\frac{krptlena vysloqne(zgxeirof-whubtdam)}{krptlena zgxeirof-whubtdam vysloqne}\\right)\n\\]\n\nHence the slope of \\( xvpalbru gsivnola \\) is\n\\[\n\\frac{krptlena vysloqne(zgxeirof-whubtdam)}{zgxeirof whubtdam(krptlena-vysloqne)}\n\\]\n\nInterchanging \\( zgxeirof \\) and \\( whubtdam \\), we find\n\\[\n\\text { slope } xvpalbru kdwqepim=\\frac{krptlena vysloqne(whubtdam-zgxeirof)}{zgxeirof whubtdam(krptlena-vysloqne)}=- \\text { slope } xvpalbru gsivnola\n\\]\n\nTherefore, \\( \\angle qzxwvtnp xvpalbru gsivnola=\\angle qzxwvtnp xvpalbru kdwqepim \\), as required.\nThe acute angle hypothesis shows that \\( zgxeirof \\) and \\( whubtdam \\) have opposite signs, while \\( krptlena \\) and \\( vysloqne \\) have the same sign (or \\( vysloqne=0 \\) ). Hence \\( krptlena zgxeirof-whubtdam vysloqne \\neq 0 \\), so \\( gsivnola \\) exists. Also \\( krptlena whubtdam-zgxeirof vysloqne \\neq 0 \\), so \\( kdwqepim \\) exists. If \\( krptlena=vysloqne \\), the lines \\( xvpalbru gsivnola \\) and \\( xvpalbru kdwqepim \\) both coincide with \\( qzxwvtnp xvpalbru \\), so \\( \\angle qzxwvtnp xvpalbru gsivnola=\\angle qzxwvtnp xvpalbru kdwqepim \\) anyway. The proof shows that we can take \\( plkturay \\) anywhere on the line \\( qzxwvtnp xvpalbru \\) as long as neither \\( krptlena zgxeirof-whubtdam vysloqne \\) nor \\( krptlena whubtdam-zgxeirof vysloqne \\) is zero.\n\nThird Solution. The diagonal \\( hjgrksla plkturay \\) of the complete quadrilateral \\( hjgrksla kdwqepim, kdwqepim plkturay \\), \\( plkturay xvpalbru, xvpalbru hjgrksla \\) is divided harmonically by the other diagonals \\( xvpalbru kdwqepim \\) and \\( qzxwvtnp fncmezod \\) at \\( ycvmrdah \\) and \\( gsivnola \\). Therefore, \\( xvpalbru kdwqepim, xvpalbru gsivnola ; xvpalbru hjgrksla, xvpalbru plkturay \\) is a harmonic pencil. But \\( xvpalbru hjgrksla \\) and \\( xvpalbru \\underset{\\sim}{\\boldsymbol{plkturay}} \\) are perpendicular and hence they bisect the angles formed by \\( xvpalbru kdwqepim \\) and \\( xvpalbru gsivnola \\). See N, A. Court, College Geometry, 2nd ed., Barnes and Noble, New York, 1952.\n\nRemark. It is interesting to compare the special cases that arise in the different proofs. The case \\( plkturay=xvpalbru \\) is exceptional in the synthetic and projective arguments, but not in the analytic argument, while the case \\( hjgrksla plkturay \\| qzxwvtnp fncmezod \\) is exceptional for the synthetic and analytic arguments, but not the projective one. In the synthetic proof, the figure changes its appearance markedly if \\( plkturay \\) is outside \\( qzxwvtnp xvpalbru \\)."
    },
    "kernel_variant": {
      "question": "1.  (i)  Four positive real numbers a , b , c , d are given, c being the greatest of the four.  Decide for which values one can construct a CONVEX quadrilateral ABCD with consecutive sides\n|AB| = a ,  |BC| = b ,  |CD| = c ,  |DA| = d\nand with AB ∥ CD.\nWhenever the construction is possible, describe a straight-edge–and-compass construction of such a quadrilateral.\n\n(ii)  Let △ABC be acute-angled and let BH be the altitude from B to AC (so H ∈ AC and BH ⟂ AC).  Fix an arbitrary point D on BH (the limit positions D = B and D = H are allowed).  Put\nE = AD ∩ BC , F = CD ∩ AB .  Prove that ∠BHE = ∠BHF .",
      "solution": "Part (i)\n\nPut Δ = |c − a|  (Δ ≥ 0).  We first find a necessary condition, then show that it is also sufficient, keeping the figure non-degenerate.\n\n1.  Necessary condition\n\nPlace AB on the x-axis so that →AB = (a,0).  Let 𝐯 = →BC and 𝐰 = →DA.\nBecause →AB + 𝐯 + →CD + 𝐰 = 0 and AB ∥ CD, we have →CD = (−c,0), hence\n                  𝐯 + 𝐰 = (c − a, 0) = (±Δ,0).            (1)\nwith |𝐯| = b and |𝐰| = d.\n\nThus 𝐯 , 𝐰 and (Δ,0) are the consecutive sides of a triangle (possibly degenerate) written head-to-tail, so they satisfy the (non-strict) triangle inequalities\n                  |b − d| ≤ Δ ≤ b + d .                    (2)\n\nIf Δ = 0 the vector sum in (1) is the zero vector; then 𝐯 and 𝐰 are opposite, forcing b = d.  In that situation a parallelogram with opposite sides a and b exists, but it will be convex only when the two pairs of opposite sides are not collinear.\n\nFor a convex quadrilateral we must exclude degeneracy; i.e. 𝐯 , 𝐰 and (Δ,0) must form a non-degenerate triangle.  Consequently\n                  |b − d| < Δ < b + d   when   Δ > 0.      (3)\n\nSummarising, a necessary condition for a CONVEX quadrilateral is\n  (C*)   either  (i)  c ≠ a and  |b − d| < |c − a| < b + d,      or\n          (ii)  c = a and b = d   (parallelogram case).\n\n2.  Sufficiency and constructions\n\nA.  The trapezoid case  (c ≠ a, condition (3) holds).\n\nBecause of (3) the three positive numbers Δ , b , d are the side lengths of a non-degenerate triangle.  The following construction (almost identical to the one in the original solution) produces the desired convex trapezoid.\n\n1.  Draw a horizontal line ℓ.  Mark points D , C on ℓ with DC = c.\n2.  On ℓ mark a point C′ between D and C such that CC′ = a.  Then DC′ = c − a = Δ.\n3.  Draw the circle k₁ with centre D and radius d and the circle k₂ with centre C′ and radius b.  Condition (3) guarantees that the two circles intersect in TWO distinct points because Δ , b , d satisfy the strict triangle inequalities.  Choose the intersection A that lies above ℓ (the other one gives the mirror image).\n4.  Through A draw the line m parallel to ℓ.  On m lay off AB = a to the right, obtaining the point B.\n5.  Join B with C and A with D.  By construction\n       AB = a ,  BC = b ,  CD = c ,  DA = d   and   AB ∥ CD,\n   while the choice of A above ℓ keeps the four vertices in strictly convex position.\n\nHence a convex quadrilateral exists whenever (3) holds.\n\nB.  The parallelogram case  (c = a and b = d).\n\nTake any segment AB of length a.  At A construct a segment AD of length b that is NOT collinear with AB.  Through B draw a line parallel to AD and through D a line parallel to AB; they meet at C.  The figure ABCD is a parallelogram with\n       AB = CD = a ,   BC = DA = b ,   AB ∥ CD ,\nand is convex because AD was chosen non-collinear with AB.\n\n3.  Conclusion\n\nCondition (C*) is both necessary and sufficient for the existence of a convex quadrilateral with the prescribed side lengths and AB ∥ CD.  When c ≠ a the strict inequalities |b − d| < |c − a| < b + d are required; when c = a one must have b = d.\n\n\nPart (ii)\n\n(The proof in the original text is already correct; only a streamlined version is reproduced.)\n\nChoose an orthonormal coordinate system with origin H, x-axis along AC and y-axis along BH.  Write\n        A = (−a,0),   C = (c,0) (a, c > 0),\n        B = (0,b) (b > 0),     D = (0,d) (0 ≤ d ≤ b).\n\n1.  E = AD ∩ BC.\n   AD: y = d(x + a)/a,      BC: y = b − (b/c)x.\n   Solving, we get\n        x_E = ac(b−d)/(ab+cd),     y_E = db(a+c)/(ab+cd).\n\n2.  F = CD ∩ AB.\n   CD: y = d − (d/c)x,      AB: y = b(x + a)/a.\n   Solving, we get\n        x_F = ac(d−b)/(bc+ad),     y_F = bd(a+c)/(bc+ad).\n\n3.  Angles at H.\n   tan∠(HB,HE) = |x_E|/|y_E| = ac|b−d|/[bd(a+c)],\n   tan∠(HB,HF) = |x_F|/|y_F| = ac|d−b|/[bd(a+c)].\n   Hence tan∠BHE = tan∠BHF and, since both angles are acute, ∠BHE = ∠BHF.\n   The equality persists in the limit cases d = 0 (D = H) and d = b (D = B).\n\nTherefore part (ii) is proved.",
      "_meta": {
        "core_steps": [
          "Replace the desired quadrilateral by a triangle whose sides are |longer‒shorter| and the two non-parallel given lengths.",
          "Invoke the strict triangle-inequality to decide existence of that triangle.",
          "Attach a segment equal to the shorter parallel side and complete a parallelogram to recover the quadrilateral with parallel edges.",
          "In the triangle problem, draw a line through the altitude’s vertex parallel to the opposite side and compare right-angled sub-triangles.",
          "Use chains of similar triangles to prove two right triangles are congruent, yielding ∠AHE = ∠AHF."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Which of the four given side lengths is declared the “longest”; only its being longer than the side that will be made parallel to it matters.",
            "original": "Side A"
          },
          "slot2": {
            "description": "Which adjacent pair of sides is prescribed to be parallel in the quadrilateral construction.",
            "original": "Sides A and B are parallel"
          },
          "slot3": {
            "description": "Which altitude of the triangle is chosen as the working line for part (ii).",
            "original": "Altitude AH from vertex A"
          },
          "slot4": {
            "description": "The acute-angle assumption; any condition guaranteeing the foot of the chosen altitude lies on the opposite side (so directed segments behave) suffices.",
            "original": "Triangle ABC is acute"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}