{
  "index": "1942-B-1",
  "type": "GEO",
  "tag": [
    "GEO",
    "ANA"
  ],
  "difficulty": "",
  "question": "7. A square of side \\( 2 a \\), lying always in the first quadrant of the \\( X Y \\) plane, moves so that two consecutive vertices are always on the \\( X \\) - and \\( Y \\)-axes respectively. Prove that a point within or on the boundary of the square will in general describe a (portion of a) conic. For what points of the square does this locus degenerate?",
  "solution": "Solution. Let \\( A \\) and \\( B \\) be two consecutive vertices of the square lying on the \\( X \\) - and \\( Y \\)-axes, respectively, and let \\( P(x, y) \\) be a specified point in the square. The coordinates of \\( P \\) are given by\n\\[\n\\begin{array}{c}\nx=n \\sin \\theta+l \\cos \\theta \\\\\ny=m \\sin \\theta+n \\cos \\theta\n\\end{array}\n\\]\nwhere \\( n, l, m \\), and \\( \\theta \\) are defined implicitly in the diagram.\nWe solve these equations for \\( \\sin \\theta \\) and \\( \\cos \\theta \\) :\n\\[\n\\begin{aligned}\n\\left(l m-n^{2}\\right) \\cos \\theta & =m x-n y \\\\\n\\left(l m-n^{2}\\right) \\sin \\theta & =-n x+l y\n\\end{aligned}\n\\]\n\nHence\n\\[\n\\left(l m-n^{2}\\right)^{2}=(m x-n y)^{2}+(-n x+l y)^{2}\n\\]\nis an equation satisfied by the coordinates of \\( P \\) for any value of \\( \\theta \\). Now (2) reduces to\n\\[\n\\left(m^{2}+n^{2}\\right) x^{2}-2 n(l+m) x y+\\left(l^{2}+n^{2}\\right) y^{2}=\\left(l m-n^{2}\\right)^{2},\n\\]\nwhich in general represents a central conic. Its discriminant is\n\\[\n\\Delta=4 n^{2}(l+m)^{2}-4\\left(m^{2}+n^{2}\\right)\\left(l^{2}+n^{2}\\right)=-4\\left(l m-n^{2}\\right)^{2}\n\\]\n\nEvidently, \\( \\Delta \\) cannot be positive. If \\( \\Delta<0 \\), then (3) represents an ellipse or circle. It will be a circle if and only if \\( m^{2}=l^{2} \\) and \\( n(l+m)=0 \\). Since \\( l+m=A B \\) cannot be zero, a circle occurs if and only if \\( n=0 \\) and \\( l=m \\); i.e., \\( P \\) is the midpoint of \\( A B \\).\n\nIf \\( \\Delta=0 \\), the right member of (3) is also zero and (3) reduces to\n\\[\n\\frac{l+m}{m}(m x-n y)^{2}=0\n\\]\nwhich is equivalent to \\( y=(m / n) x \\), so \\( P \\) moves along a straight line.\nThe geometrical meaning of \\( \\Delta=0 \\), or \\( \\operatorname{lm}=n^{2} \\) is that \\( P \\) is on a semicircle of which \\( A B \\) is a diameter. [Note that \\( n \\) is the mean proportional between \\( l \\) and \\( m \\).] Notice that \\( l=m=n \\) makes \\( P \\) the center of the given square and the locus is a portion of the line \\( y=x \\), as already found in Problem 1 of this competition.\n\nIf \\( \\Delta \\neq 0 \\), i.e., if \\( \\operatorname{lm} \\neq n^{2} \\), the parametrization (1) is non-singular (that is, \\( d x / d \\theta \\) and \\( d y / d \\theta \\) do not vanish together) so the point \\( P \\) traverses its elliptical path smoothly as \\( \\theta \\) goes from 0 to \\( \\pi / 2 \\).\n\nIf \\( \\Delta=0 \\), then as noted above, \\( P \\) lies on a semicircle with diameter \\( A B \\). The other half of the circle will pass through the origin \\( O \\). Hence as \\( \\theta \\) varies from 0 to \\( \\pi / 2 \\), a point \\( P \\) on the semicircle will move away from the origin until \\( P O \\) is a diameter of this moving circle and then move back. To see this analytically, note that reversal can occur only when \\( x \\) has a critical value as a function of \\( \\theta \\). But \\( d x / d \\theta=0 \\) requires \\( \\tan \\theta=n / l \\), which means that \\( P B \\) is perpendicular to the \\( y \\)-axis and therefore \\( P O \\) is a diameter of the moving circle.\n\nRemarks. The problem could be interpreted as meaning that after \\( A \\) gets to the origin, it continues up the \\( y \\)-axis while \\( D \\) moves toward the origin along the \\( x \\)-axis, etc. With this interpretation, a point will, in general, describe portions of four different conics. This interpretation makes no difference if \\( P \\) is the center of the square, since all four partial conics degenerate to the same segment.\n\nThis problem, the first problem of the afternoon session, is a nice generalization of the first problem of the morning session.",
  "vars": [
    "x",
    "y",
    "\\\\theta"
  ],
  "params": [
    "a",
    "n",
    "l",
    "m",
    "\\\\Delta"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "abscissa",
        "y": "ordinate",
        "\\theta": "\\rotation",
        "a": "sidehalf",
        "n": "meanprop",
        "l": "basexdist",
        "m": "baseydist",
        "\\Delta": "\\discrimin"
      },
      "question": "7. A square of side \\( 2\\, sidehalf \\), lying always in the first quadrant of the \\( X Y \\) plane, moves so that two consecutive vertices are always on the \\( X \\) - and \\( Y \\)-axes respectively. Prove that a point within or on the boundary of the square will in general describe a (portion of a) conic. For what points of the square does this locus degenerate?",
      "solution": "Solution. Let \\( A \\) and \\( B \\) be two consecutive vertices of the square lying on the \\( X \\) - and \\( Y \\)-axes, respectively, and let \\( P(abscissa, ordinate) \\) be a specified point in the square. The coordinates of \\( P \\) are given by\n\\[\n\\begin{array}{c}\nabscissa = meanprop \\sin \\rotation + basexdist \\cos \\rotation \\\\\nordinate = baseydist \\sin \\rotation + meanprop \\cos \\rotation\n\\end{array}\n\\]\nwhere \\( meanprop, basexdist, baseydist \\), and \\( \\rotation \\) are defined implicitly in the diagram.\nWe solve these equations for \\( \\sin \\rotation \\) and \\( \\cos \\rotation \\) :\n\\[\n\\begin{aligned}\n\\left(basexdist\\, baseydist - meanprop^{2}\\right) \\cos \\rotation & = baseydist\\, abscissa - meanprop\\, ordinate \\\\\n\\left(basexdist\\, baseydist - meanprop^{2}\\right) \\sin \\rotation & = -\\, meanprop\\, abscissa + basexdist\\, ordinate\n\\end{aligned}\n\\]\n\nHence\n\\[\n\\left(basexdist\\, baseydist - meanprop^{2}\\right)^{2} = (baseydist\\, abscissa - meanprop\\, ordinate)^{2} + (-\\, meanprop\\, abscissa + basexdist\\, ordinate)^{2}\n\\]\nis an equation satisfied by the coordinates of \\( P \\) for any value of \\( \\rotation \\). Now (2) reduces to\n\\[\n\\left(baseydist^{2} + meanprop^{2}\\right)\\, abscissa^{2} - 2\\, meanprop(basexdist + baseydist)\\, abscissa\\, ordinate + \\left(basexdist^{2} + meanprop^{2}\\right)\\, ordinate^{2} = \\left(basexdist\\, baseydist - meanprop^{2}\\right)^{2},\n\\]\nwhich in general represents a central conic. Its discriminant is\n\\[\n\\discrimin = 4\\, meanprop^{2} (basexdist + baseydist)^{2} - 4\\left(baseydist^{2} + meanprop^{2}\\right)\\left(basexdist^{2} + meanprop^{2}\\right) = -\\, 4\\left(basexdist\\, baseydist - meanprop^{2}\\right)^{2}\n\\]\n\nEvidently, \\( \\discrimin \\) cannot be positive. If \\( \\discrimin < 0 \\), then (3) represents an ellipse or circle. It will be a circle if and only if \\( baseydist^{2} = basexdist^{2} \\) and \\( meanprop(basexdist + baseydist) = 0 \\). Since \\( basexdist + baseydist = A B \\) cannot be zero, a circle occurs if and only if \\( meanprop = 0 \\) and \\( basexdist = baseydist \\); i.e., \\( P \\) is the midpoint of \\( A B \\).\n\nIf \\( \\discrimin = 0 \\), the right member of (3) is also zero and (3) reduces to\n\\[\n\\frac{basexdist + baseydist}{baseydist} (baseydist\\, abscissa - meanprop\\, ordinate)^{2} = 0\n\\]\nwhich is equivalent to \\( ordinate = (baseydist / meanprop)\\, abscissa \\), so \\( P \\) moves along a straight line.\nThe geometrical meaning of \\( \\discrimin = 0 \\), or \\( basexdist\\, baseydist = meanprop^{2} \\) is that \\( P \\) is on a semicircle of which \\( A B \\) is a diameter. [Note that \\( meanprop \\) is the mean proportional between \\( basexdist \\) and \\( baseydist \\).] Notice that \\( basexdist = baseydist = meanprop \\) makes \\( P \\) the center of the given square and the locus is a portion of the line \\( ordinate = abscissa \\), as already found in Problem 1 of this competition.\n\nIf \\( \\discrimin \\neq 0 \\), i.e., if \\( basexdist\\, baseydist \\neq meanprop^{2} \\), the parametrization (1) is non-singular (that is, \\( d\\,abscissa / d\\,\\rotation \\) and \\( d\\,ordinate / d\\,\\rotation \\) do not vanish together) so the point \\( P \\) traverses its elliptical path smoothly as \\( \\rotation \\) goes from 0 to \\( \\pi / 2 \\).\n\nIf \\( \\discrimin = 0 \\), then as noted above, \\( P \\) lies on a semicircle with diameter \\( A B \\). The other half of the circle will pass through the origin \\( O \\). Hence as \\( \\rotation \\) varies from 0 to \\( \\pi / 2 \\), a point \\( P \\) on the semicircle will move away from the origin until \\( P O \\) is a diameter of this moving circle and then move back. To see this analytically, note that reversal can occur only when \\( abscissa \\) has a critical value as a function of \\( \\rotation \\). But \\( d\\,abscissa / d\\,\\rotation = 0 \\) requires \\( \\tan \\rotation = meanprop / basexdist \\), which means that \\( P B \\) is perpendicular to the \\( Y \\)-axis and therefore \\( P O \\) is a diameter of the moving circle.\n\nRemarks. The problem could be interpreted as meaning that after \\( A \\) gets to the origin, it continues up the \\( Y \\)-axis while \\( D \\) moves toward the origin along the \\( X \\)-axis, etc. With this interpretation, a point will, in general, describe portions of four different conics. This interpretation makes no difference if \\( P \\) is the center of the square, since all four partial conics degenerate to the same segment.\n\nThis problem, the first problem of the afternoon session, is a nice generalization of the first problem of the morning session."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "radiusnine",
        "y": "velocityup",
        "\\theta": "momentumz",
        "a": "densityval",
        "n": "pressureio",
        "l": "masspoint",
        "m": "entropyair",
        "\\Delta": "gradientv"
      },
      "question": "7. A square of side \\( 2 densityval \\), lying always in the first quadrant of the \\( X Y \\) plane, moves so that two consecutive vertices are always on the \\( X \\) - and \\( Y \\)-axes respectively. Prove that a point within or on the boundary of the square will in general describe a (portion of a) conic. For what points of the square does this locus degenerate?",
      "solution": "Solution. Let \\( A \\) and \\( B \\) be two consecutive vertices of the square lying on the \\( X \\) - and \\( Y \\)-axes, respectively, and let \\( P(radiusnine, velocityup) \\) be a specified point in the square. The coordinates of \\( P \\) are given by\n\\[\n\\begin{array}{c}\nradiusnine=pressureio \\sin momentumz+masspoint \\cos momentumz \\\\\nvelocityup=entropyair \\sin momentumz+pressureio \\cos momentumz\n\\end{array}\n\\]\nwhere \\( pressureio, masspoint, entropyair \\), and \\( momentumz \\) are defined implicitly in the diagram.\nWe solve these equations for \\( \\sin momentumz \\) and \\( \\cos momentumz \\) :\n\\[\n\\begin{aligned}\n\\left(masspoint entropyair-pressureio^{2}\\right) \\cos momentumz & =entropyair radiusnine-pressureio velocityup \\\\\n\\left(masspoint entropyair-pressureio^{2}\\right) \\sin momentumz & =-pressureio radiusnine+masspoint velocityup\n\\end{aligned}\n\\]\n\nHence\n\\[\n\\left(masspoint entropyair-pressureio^{2}\\right)^{2}=(entropyair radiusnine-pressureio velocityup)^{2}+(-pressureio radiusnine+masspoint velocityup)^{2}\n\\]\nis an equation satisfied by the coordinates of \\( P \\) for any value of \\( momentumz \\). Now (2) reduces to\n\\[\n\\left(entropyair^{2}+pressureio^{2}\\right) radiusnine^{2}-2 pressureio(masspoint+entropyair) radiusnine velocityup+\\left(masspoint^{2}+pressureio^{2}\\right) velocityup^{2}=\\left(masspoint entropyair-pressureio^{2}\\right)^{2},\n\\]\nwhich in general represents a central conic. Its discriminant is\n\\[\ngradientv=4 pressureio^{2}(masspoint+entropyair)^{2}-4\\left(entropyair^{2}+pressureio^{2}\\right)\\left(masspoint^{2}+pressureio^{2}\\right)=-4\\left(masspoint entropyair-pressureio^{2}\\right)^{2}\n\\]\n\nEvidently, \\( gradientv \\) cannot be positive. If \\( gradientv<0 \\), then (3) represents an ellipse or circle. It will be a circle if and only if \\( entropyair^{2}=masspoint^{2} \\) and \\( pressureio(masspoint+entropyair)=0 \\). Since \\( masspoint+entropyair=A B \\) cannot be zero, a circle occurs if and only if \\( pressureio=0 \\) and \\( masspoint=entropyair \\); i.e., \\( P \\) is the midpoint of \\( A B \\).\n\nIf \\( gradientv=0 \\), the right member of (3) is also zero and (3) reduces to\n\\[\n\\frac{masspoint+entropyair}{entropyair}(entropyair radiusnine-pressureio velocityup)^{2}=0\n\\]\nwhich is equivalent to \\( velocityup=(entropyair / pressureio) radiusnine \\), so \\( P \\) moves along a straight line.\nThe geometrical meaning of \\( gradientv=0 \\), or \\( \\operatorname{masspoint entropyair}=pressureio^{2} \\) is that \\( P \\) is on a semicircle of which \\( A B \\) is a diameter. [Note that \\( pressureio \\) is the mean proportional between \\( masspoint \\) and \\( entropyair \\).] Notice that \\( masspoint=entropyair=pressureio \\) makes \\( P \\) the center of the given square and the locus is a portion of the line \\( velocityup=radiusnine \\), as already found in Problem 1 of this competition.\n\nIf \\( gradientv \\neq 0 \\), i.e., if \\( \\operatorname{masspoint entropyair} \\neq pressureio^{2} \\), the parametrization (1) is non-singular (that is, \\( d radiusnine / d momentumz \\) and \\( d velocityup / d momentumz \\) do not vanish together) so the point \\( P \\) traverses its elliptical path smoothly as \\( momentumz \\) goes from 0 to \\( \\pi / 2 \\).\n\nIf \\( gradientv=0 \\), then as noted above, \\( P \\) lies on a semicircle with diameter \\( A B \\). The other half of the circle will pass through the origin \\( O \\). Hence as \\( momentumz \\) varies from 0 to \\( \\pi / 2 \\), a point \\( P \\) on the semicircle will move away from the origin until \\( P O \\) is a diameter of this moving circle and then move back. To see this analytically, note that reversal can occur only when \\( radiusnine \\) has a critical value as a function of \\( momentumz \\). But \\( d radiusnine / d momentumz=0 \\) requires \\( \\tan momentumz=pressureio / masspoint \\), which means that \\( P B \\) is perpendicular to the \\( velocityup \\)-axis and therefore \\( P O \\) is a diameter of the moving circle.\n\nRemarks. The problem could be interpreted as meaning that after \\( A \\) gets to the origin, it continues up the \\( velocityup \\)-axis while \\( D \\) moves toward the origin along the \\( radiusnine \\)-axis, etc. With this interpretation, a point will, in general, describe portions of four different conics. This interpretation makes no difference if \\( P \\) is the center of the square, since all four partial conics degenerate to the same segment.\n\nThis problem, the first problem of the afternoon session, is a nice generalization of the first problem of the morning session."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalaxis",
        "y": "horizontalaxis",
        "\\\\theta": "lineardist",
        "a": "radiusvalue",
        "n": "variedness",
        "l": "heightless",
        "m": "breadthless",
        "\\\\Delta": "sameness"
      },
      "question": "7. A square of side \\( 2 radiusvalue \\), lying always in the first quadrant of the \\( X Y \\) plane, moves so that two consecutive vertices are always on the \\( X \\) - and \\( Y \\)-axes respectively. Prove that a point within or on the boundary of the square will in general describe a (portion of a) conic. For what points of the square does this locus degenerate?",
      "solution": "Solution. Let \\( A \\) and \\( B \\) be two consecutive vertices of the square lying on the \\( X \\) - and \\( Y \\)-axes, respectively, and let \\( P(verticalaxis, horizontalaxis) \\) be a specified point in the square. The coordinates of \\( P \\) are given by\n\\[\n\\begin{array}{c}\nverticalaxis=variedness \\sin lineardist+heightless \\cos lineardist \\\\\nhorizontalaxis=breadthless \\sin lineardist+variedness \\cos lineardist\n\\end{array}\n\\]\nwhere \\( variedness, heightless, breadthless \\), and \\( lineardist \\) are defined implicitly in the diagram.\nWe solve these equations for \\( \\sin lineardist \\) and \\( \\cos lineardist \\) :\n\\[\n\\begin{aligned}\n\\left(heightless breadthless-variedness^{2}\\right) \\cos lineardist & =breadthless verticalaxis-variedness horizontalaxis \\\\\n\\left(heightless breadthless-variedness^{2}\\right) \\sin lineardist & =-variedness verticalaxis+heightless horizontalaxis\n\\end{aligned}\n\\]\n\nHence\n\\[\n\\left(heightless breadthless-variedness^{2}\\right)^{2}=(breadthless verticalaxis-variedness horizontalaxis)^{2}+(-variedness verticalaxis+heightless horizontalaxis)^{2}\n\\]\nis an equation satisfied by the coordinates of \\( P \\) for any value of \\( lineardist \\). Now (2) reduces to\n\\[\n\\left(breadthless^{2}+variedness^{2}\\right) verticalaxis^{2}-2 variedness(heightless+breadthless) verticalaxis horizontalaxis+\\left(heightless^{2}+variedness^{2}\\right) horizontalaxis^{2}=\\left(heightless breadthless-variedness^{2}\\right)^{2},\n\\]\nwhich in general represents a central conic. Its discriminant is\n\\[\nsameness=4 variedness^{2}(heightless+breadthless)^{2}-4\\left(breadthless^{2}+variedness^{2}\\right)\\left(heightless^{2}+variedness^{2}\\right)=-4\\left(heightless breadthless-variedness^{2}\\right)^{2}\n\\]\n\nEvidently, \\( sameness \\) cannot be positive. If \\( sameness<0 \\), then (3) represents an ellipse or circle. It will be a circle if and only if \\( breadthless^{2}=heightless^{2} \\) and \\( variedness(heightless+breadthless)=0 \\). Since \\( heightless+breadthless=A B \\) cannot be zero, a circle occurs if and only if \\( variedness=0 \\) and \\( heightless=breadthless \\); i.e., \\( P \\) is the midpoint of \\( A B \\).\n\nIf \\( sameness=0 \\), the right member of (3) is also zero and (3) reduces to\n\\[\n\\frac{heightless+breadthless}{breadthless}(breadthless verticalaxis-variedness horizontalaxis)^{2}=0\n\\]\nwhich is equivalent to \\( horizontalaxis=(breadthless / variedness) verticalaxis \\), so \\( P \\) moves along a straight line.\nThe geometrical meaning of \\( sameness=0 \\), or \\( \\operatorname{heightless~breadthless}=variedness^{2} \\) is that \\( P \\) is on a semicircle of which \\( A B \\) is a diameter. [Note that \\( variedness \\) is the mean proportional between \\( heightless \\) and \\( breadthless \\).] Notice that \\( heightless=breadthless=variedness \\) makes \\( P \\) the center of the given square and the locus is a portion of the line \\( horizontalaxis=verticalaxis \\), as already found in Problem 1 of this competition.\n\nIf \\( sameness \\neq 0 \\), i.e., if \\( \\operatorname{heightless~breadthless} \\neq variedness^{2} \\), the parametrization (1) is non-singular (that is, \\( d verticalaxis / d lineardist \\) and \\( d horizontalaxis / d lineardist \\) do not vanish together) so the point \\( P \\) traverses its elliptical path smoothly as \\( lineardist \\) goes from 0 to \\( \\pi / 2 \\).\n\nIf \\( sameness=0 \\), then as noted above, \\( P \\) lies on a semicircle with diameter \\( A B \\). The other half of the circle will pass through the origin \\( O \\). Hence as \\( lineardist \\) varies from 0 to \\( \\pi / 2 \\), a point \\( P \\) on the semicircle will move away from the origin until \\( P O \\) is a diameter of this moving circle and then move back. To see this analytically, note that reversal can occur only when \\( verticalaxis \\) has a critical value as a function of \\( lineardist \\). But \\( d verticalaxis / d lineardist=0 \\) requires \\( \\tan lineardist=variedness / heightless \\), which means that \\( P B \\) is perpendicular to the \\( y \\)-axis and therefore \\( P O \\) is a diameter of the moving circle.\n\nRemarks. The problem could be interpreted as meaning that after \\( A \\) gets to the origin, it continues up the \\( y \\)-axis while \\( D \\) moves toward the origin along the \\( x \\)-axis, etc. With this interpretation, a point will, in general, describe portions of four different conics. This interpretation makes no difference if \\( P \\) is the center of the square, since all four partial conics degenerate to the same segment.\n\nThis problem, the first problem of the afternoon session, is a nice generalization of the first problem of the morning session."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "\\\\theta": "mnpqrsuv",
        "a": "xkjdploe",
        "n": "vchyteqr",
        "l": "zbrinxsw",
        "m": "pkdvoeun",
        "\\\\Delta": "rgfmslqa"
      },
      "question": "7. A square of side \\( 2 xkjdploe \\), lying always in the first quadrant of the \\( X Y \\) plane, moves so that two consecutive vertices are always on the \\( X \\)- and \\( Y \\)-axes respectively. Prove that a point within or on the boundary of the square will in general describe a (portion of a) conic. For what points of the square does this locus degenerate?",
      "solution": "Solution. Let \\( A \\) and \\( B \\) be two consecutive vertices of the square lying on the \\( X \\)- and \\( Y \\)-axes, respectively, and let \\( P(qzxwvtnp, hjgrksla) \\) be a specified point in the square. The coordinates of \\( P \\) are given by\n\\[\n\\begin{array}{c}\nqzxwvtnp=vchyteqr \\sin mnpqrsuv+ zbrinxsw \\cos mnpqrsuv \\\\\nhjgrksla=pkdvoeun \\sin mnpqrsuv+ vchyteqr \\cos mnpqrsuv\n\\end{array}\n\\]\nwhere \\( vchyteqr, zbrinxsw, pkdvoeun \\), and \\( mnpqrsuv \\) are defined implicitly in the diagram.\nWe solve these equations for \\( \\sin mnpqrsuv \\) and \\( \\cos mnpqrsuv \\) :\n\\[\n\\begin{aligned}\n\\left(zbrinxsw \\, pkdvoeun-vchyteqr^{2}\\right) \\cos mnpqrsuv & =pkdvoeun \\, qzxwvtnp-vchyteqr \\, hjgrksla \\\\\n\\left(zbrinxsw \\, pkdvoeun-vchyteqr^{2}\\right) \\sin mnpqrsuv & =-vchyteqr \\, qzxwvtnp+zbrinxsw \\, hjgrksla\n\\end{aligned}\n\\]\n\nHence\n\\[\n\\left(zbrinxsw \\, pkdvoeun-vchyteqr^{2}\\right)^{2}=(pkdvoeun \\, qzxwvtnp-vchyteqr \\, hjgrksla)^{2}+(-vchyteqr \\, qzxwvtnp+zbrinxsw \\, hjgrksla)^{2}\n\\]\nis an equation satisfied by the coordinates of \\( P \\) for any value of \\( mnpqrsuv \\). Now (2) reduces to\n\\[\n(pkdvoeun^{2}+vchyteqr^{2}) \\, qzxwvtnp^{2}-2 vchyteqr(zbrinxsw+pkdvoeun) \\, qzxwvtnp \\, hjgrksla+(zbrinxsw^{2}+vchyteqr^{2}) \\, hjgrksla^{2}=(zbrinxsw \\, pkdvoeun-vchyteqr^{2})^{2},\n\\]\nwhich in general represents a central conic. Its discriminant is\n\\[\nrgfmslqa=4 vchyteqr^{2}(zbrinxsw+pkdvoeun)^{2}-4(pkdvoeun^{2}+vchyteqr^{2})(zbrinxsw^{2}+vchyteqr^{2})=-4(zbrinxsw \\, pkdvoeun-vchyteqr^{2})^{2}\n\\]\n\nEvidently, \\( rgfmslqa \\) cannot be positive. If \\( rgfmslqa<0 \\), then (3) represents an ellipse or circle. It will be a circle if and only if \\( pkdvoeun^{2}=zbrinxsw^{2} \\) and \\( vchyteqr(zbrinxsw+pkdvoeun)=0 \\). Since \\( zbrinxsw+pkdvoeun=A B \\) cannot be zero, a circle occurs if and only if \\( vchyteqr=0 \\) and \\( zbrinxsw=pkdvoeun \\); i.e., \\( P \\) is the midpoint of \\( A B \\).\n\nIf \\( rgfmslqa=0 \\), the right member of (3) is also zero and (3) reduces to\n\\[\n\\frac{zbrinxsw+pkdvoeun}{pkdvoeun}(pkdvoeun \\, qzxwvtnp-vchyteqr \\, hjgrksla)^{2}=0\n\\]\nwhich is equivalent to \\( hjgrksla=(pkdvoeun / vchyteqr) \\, qzxwvtnp \\), so \\( P \\) moves along a straight line.\nThe geometrical meaning of \\( rgfmslqa=0 \\), or \\( zbrinxsw \\, pkdvoeun=vchyteqr^{2} \\) is that \\( P \\) is on a semicircle of which \\( A B \\) is a diameter. [Note that \\( vchyteqr \\) is the mean proportional between \\( zbrinxsw \\) and \\( pkdvoeun \\).] Notice that \\( zbrinxsw=pkdvoeun=vchyteqr \\) makes \\( P \\) the center of the given square and the locus is a portion of the line \\( hjgrksla=qzxwvtnp \\), as already found in Problem 1 of this competition.\n\nIf \\( rgfmslqa \\neq 0 \\), i.e., if \\( zbrinxsw \\, pkdvoeun \\neq vchyteqr^{2} \\), the parametrization (1) is non-singular (that is, \\( d qzxwvtnp / d mnpqrsuv \\) and \\( d hjgrksla / d mnpqrsuv \\) do not vanish together) so the point \\( P \\) traverses its elliptical path smoothly as \\( mnpqrsuv \\) goes from 0 to \\( \\pi / 2 \\).\n\nIf \\( rgfmslqa=0 \\), then as noted above, \\( P \\) lies on a semicircle with diameter \\( A B \\). The other half of the circle will pass through the origin \\( O \\). Hence as \\( mnpqrsuv \\) varies from 0 to \\( \\pi / 2 \\), a point \\( P \\) on the semicircle will move away from the origin until \\( P O \\) is a diameter of this moving circle and then move back. To see this analytically, note that reversal can occur only when \\( qzxwvtnp \\) has a critical value as a function of \\( mnpqrsuv \\). But \\( d qzxwvtnp / d mnpqrsuv=0 \\) requires \\( \\tan mnpqrsuv=vchyteqr / zbrinxsw \\), which means that \\( P B \\) is perpendicular to the \\( y \\)-axis and therefore \\( P O \\) is a diameter of the moving circle.\n\nRemarks. The problem could be interpreted as meaning that after \\( A \\) gets to the origin, it continues up the \\( y \\)-axis while \\( D \\) moves toward the origin along the \\( x \\)-axis, etc. With this interpretation, a point will, in general, describe portions of four different conics. This interpretation makes no difference if \\( P \\) is the center of the square, since all four partial conics degenerate to the same segment.\n\nThis problem, the first problem of the afternoon session, is a nice generalization of the first problem of the morning session."
    },
    "kernel_variant": {
      "question": "Let ℓ , c , d be fixed positive real numbers with\n    c  > ℓ and d  > ℓ.\nA square of side ℓ is allowed to move in the plane subject to the following three constraints.\n• One of its vertices, denoted A, is always situated on the vertical line x = –c.\n• The next vertex in counter-clockwise order, denoted B, is always situated on the horizontal line y =  d.\n• During the whole motion the entire square stays in the open second quadrant (x < 0 < y).\n\nFix once and for all a point P that lies in the interior of the square (the boundary is permitted).  When the square passes through every admissible position, P traces a curve in the plane.\n\na)  Prove that, for every interior point P except for a small exceptional set, its locus is an arc of one and the same central conic.\n\nb)  Determine precisely for which positions of P this conic\n  (i) is a circle, and\n  (ii) degenerates into one (repeated) straight line.\nFor each exceptional situation identify the corresponding locus of P.",
      "solution": "Throughout we write I := ( –c , d ) and translate the coordinate system so that I is the origin.  Capital letters (X , Y) will denote coordinates referred to this translated system,\n    X = x + c ,  Y = y – d.  (1)\n\n1.  Kinematics of the moving square\n-----------------------------------\nLet θ ∈ (0 , π/2) be the angle from the positive X–axis to the directed side AB.  Put\n  u = ( cosθ ,  sinθ ),  v = ( –sinθ , cosθ ).  (2)\nThus u is the unit vector from A to B and v the inward pointing unit normal to AB.  The four vertices of the square are\n  A , B = A + ℓu , D = A + ℓv , C = A + ℓu + ℓv.\nBecause B lies on the line Y = 0 we obtain\n  0 = Y_B = Y_A + ℓ sinθ ⇒ Y_A = –ℓ sinθ.  (3)\nSince X_A = 0 by construction,\n  A = ( 0 , –ℓ sinθ ).  (4)\n\nCondition c > ℓ guarantees that for every θ the x–coordinates of all four vertices are strictly negative; similarly d > ℓ guarantees that all y–coordinates are strictly positive, so the whole square indeed remains inside the open second quadrant.\n\n2.  Coordinates of a fixed interior point P\n------------------------------------------\nInside the square introduce the usual barycentric coordinates\n  AP = l u + n v with 0 ≤ l , n ≤ ℓ , and put m := ℓ – l.  (5)\nCombining (4) and (5) and resolving X , Y we obtain\n  X =  l cosθ – n sinθ ,\n  Y =  n cosθ – m sinθ.  (6)\nThe real constants l , m , n depend only on the chosen point P and remain fixed during the motion.\n\n3.  Eliminating the parameter θ\n------------------------------\nSolve (6) for cosθ and sinθ (Cramer’s rule).  Let\n  Δ := n² – l m.  (7)\nIf Δ ≠ 0 we get\n  cosθ = ( –m X + n Y ) ⁄ Δ ,  sinθ = ( –n X + l Y ) ⁄ Δ.  (8)\nImposing cos²θ + sin²θ = 1 gives the quadratic equation\n  ( m X – n Y )² + ( n X – l Y )² = Δ²\n⇔ ( m² + n² ) X² – 2 n ( l + m ) X Y + ( l² + n² ) Y² = ( n² – l m )². (★)\nBecause no linear terms in X , Y appear, (★) is the equation in the (X , Y)–plane of a central conic with centre I.\n\n4.  Type of the conic when Δ ≠ 0\n--------------------------------\nIts discriminant is\n D = [ –2 n ( l + m ) ]² – 4 ( m² + n² )( l² + n² )\n   = –4 ( l m – n² )² ≤ 0.\nSince the quadratic part is definite (coefficients m²+n² and l²+n² are positive), the conic is an ellipse whenever Δ ≠ 0.\n\n5.  When does the ellipse become a circle?\n-----------------------------------------\nFor a central conic of the form a X² + 2 b X Y + c Y² = constant to be a circle we need b = 0 and a = c.\nHere b = –n ( l + m ), so b = 0 ⇔ n = 0.  With n = 0 the coefficients reduce to a = m² and c = l²; equality a = c forces l = m.  Because m = ℓ – l, we conclude\n  n = 0 and l = m = ℓ⁄2.  (9)\nGeometrically, P is the midpoint of the side AB.  In this case (★) becomes\n  X² + Y² = ( ℓ⁄2 )²,  (10)\na circle of radius ℓ⁄2 centred at I.\n\n6.  Degenerate cases (Δ = 0)\n----------------------------\nΔ = 0 is equivalent to l m = n².  Then the right–hand side of (★) is 0 and (★) factors as\n  ( m X – n Y )² + ( n X – l Y )² = 0\n⇔ m X – n Y = 0  = n X – l Y.  (11)\nBecause the two factors are proportional, the conic collapses to a single straight line through I.  Three sub-cases arise.\n\n(a)  n ≠ 0 (the generic degenerate case).\n     Equation (11) yields the oblique line\n         Y = ( m / n ) X.  (12)\n     The condition l m = n² implies that the point P lies on the semicircle having AB as diameter.\n\n(b)  n = 0 , l = 0 (so m = ℓ).  Then P = A.  From (6) we have X ≡ 0; the locus is the vertical line X = 0, i.e. the original line x = –c.\n\n(c)  n = 0 , m = 0 (so l = ℓ).  Then P = B.  From (6) we have Y ≡ 0; the locus is the horizontal line Y = 0, i.e. the original line y =  d.\n\nIn each of the three situations the conic degenerates to the repeated linear factor found above.\n\n7.  Summary of the loci\n-----------------------\n• For every interior point P with l m ≠ n² the locus is an arc of the fixed ellipse (★) centred at I.\n• If P is the midpoint of AB (condition (9)), the ellipse becomes the circle (10).\n• If P lies on the semicircle with diameter AB (condition l m = n²) the locus degenerates to a straight line as described in (a)–(c) above; for A and B themselves these lines coincide with the given supporting lines x = –c and y =  d.",
      "_meta": {
        "core_steps": [
          "Introduce a rotation parameter θ and write the coordinates of the chosen point P in the square as linear combinations of sin θ and cos θ with fixed offsets (l, m, n).",
          "Solve the two linear equations for sin θ and cos θ, then square-add to eliminate θ, obtaining a second-degree polynomial F(x, y)=0.",
          "Recognize F(x, y)=0 as the equation of a central conic and compute its discriminant Δ.",
          "Use the sign of Δ to decide: Δ<0 → ellipse (circle under an extra coefficient equality), Δ=0 → degenerate straight line; list the length-relation cases that make these happen."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "numeric value chosen for the square’s side length (pure scale factor)",
            "original": "2a"
          },
          "slot2": {
            "description": "stipulation that the square always stays in the first quadrant (any other fixed quadrant or half-plane would work just as well)",
            "original": "first quadrant of the XY-plane"
          },
          "slot3": {
            "description": "choice of the two perpendicular lines on which the sliding vertices lie",
            "original": "X-axis and Y-axis"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}