{
  "index": "1946-B-2",
  "type": "GEO",
  "tag": [
    "GEO",
    "ALG"
  ],
  "difficulty": "",
  "question": "2. Let \\( A, B \\) be variable points on a parabola \\( P \\), such that the tangents at \\( A \\) and \\( B \\) are perpendicular to each other. Show that the locus of the centroid of the triangle formed by \\( A, B \\) and the vertex of \\( P \\) is a parabola \\( P_{1} \\). Apply the same process to \\( P_{1} \\), obtaining a parabola \\( P_{2} \\), and repeat the process, obtaining altogether the sequence of parabolas \\( P, P_{1}, P_{2}, \\ldots, P_{n} \\). If the equation of \\( P \\) is \\( y^{2}=m x \\), find the equation of \\( P_{n} \\).",
  "solution": "Solution. Since \\( P \\) is a parabola with equation \\( y^{2}=m x \\), any point of \\( P \\) has coordinates of the form ( \\( m t^{2}, m t \\) ) for some real \\( t \\), and conversely, every such point is on \\( P \\). The slope of the line tangent to \\( P \\) at \\( \\left(m t^{2}, m t\\right) \\) is \\( 1 / 2 t \\).\n\nLet \\( A \\) and \\( B \\) be the points \\( \\left(m s^{2}, m s\\right) \\) and ( \\( m t^{2}, m t \\) ), respectively. The tangents to \\( P \\) at \\( A \\) and \\( B \\) are perpendicular if and only if \\( (1 / 2 s)(1 / 2 t) \\) \\( =-1 \\), i.e., if and only if\n\\[\ns t=-\\frac{1}{4}\n\\]\n\nThe centroid of the points \\( A, B \\), and the vertex \\( (0,0) \\) of \\( P \\) is\n\\[\n\\left(\\frac{1}{3} m\\left(s^{2}+t^{2}\\right), \\quad \\frac{1}{3} m(s+t)\\right)\n\\]\nand this centroid lies on a new parabola\n\\[\ny^{2}=\\frac{1}{3} m\\left(x-\\frac{m}{6}\\right)\n\\]\nif the perpendicularity condition (1) is satisfied.\nConversely, any point ( \\( x, y \\) ) on the parabola (3) has the form (2), since the equations\n\\[\n\\begin{aligned}\n\\frac{1}{3} m(s+t) & =y \\\\\ns t & =-\\frac{1}{4}\n\\end{aligned}\n\\]\ncan always be solved to give real \\( s \\) and \\( t \\). Indeed, \\( s \\) and \\( t \\) are zeros of \\( S^{2}-(3 y / m) S-\\frac{1}{4} \\), which has positive discriminant. Hence the locus in question is the entire parabola \\( P_{1} \\) given by (3).\n\nNow \\( P_{1} \\) is obtained from \\( P \\) by changing the constant from \\( m \\) to \\( m / 3 \\), and displacing the vertex to the right by \\( m / 6 \\). Consequently, \\( P_{2} \\) can be obtained from \\( P_{1} \\) by changing the constant from \\( m / 3 \\) to \\( (m / 3) / 3 \\) and displacing the vertex \\( (m / 3) / 6 \\) further to the right. The equation of \\( P_{2} \\) is therefore\n\\[\ny^{2}=\\frac{1}{9} m\\left(x-\\frac{1}{6} m-\\frac{1}{18} m\\right)\n\\]\n\nContinuing this reasoning, we see that \\( P_{\\boldsymbol{n}} \\) has the equation\n\\[\n\\begin{aligned}\ny^{2} & =\\frac{1}{3^{n}} m\\left(x-\\frac{1}{6} m-\\frac{1}{6 \\cdot 3} m-\\frac{1}{6 \\cdot 3^{2}} m-\\cdots-\\frac{1}{6 \\cdot 3^{n-1}} m\\right) \\\\\n& =\\frac{1}{3^{n}} m\\left(x-\\frac{m}{4}\\left(1-\\frac{1}{3^{n}}\\right)\\right) .\n\\end{aligned}\n\\]",
  "vars": [
    "A",
    "B",
    "S",
    "s",
    "t",
    "x",
    "y"
  ],
  "params": [
    "P",
    "P_1",
    "P_2",
    "P_n",
    "m",
    "n"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "A": "pointalpha",
        "B": "pointbeta",
        "S": "auxscalar",
        "s": "parameta",
        "t": "parazeta",
        "x": "abscissa",
        "y": "ordinate",
        "P": "origparabola",
        "P_1": "firstparabola",
        "P_2": "secondparabola",
        "P_n": "nthparabola",
        "m": "paracoeff",
        "n": "indexcount"
      },
      "question": "2. Let \\( pointalpha, pointbeta \\) be variable points on a parabola \\( origparabola \\), such that the tangents at \\( pointalpha \\) and \\( pointbeta \\) are perpendicular to each other. Show that the locus of the centroid of the triangle formed by \\( pointalpha, pointbeta \\) and the vertex of \\( origparabola \\) is a parabola \\( firstparabola \\). Apply the same process to \\( firstparabola \\), obtaining a parabola \\( secondparabola \\), and repeat the process, obtaining altogether the sequence of parabolas \\( origparabola, firstparabola, secondparabola, \\ldots, nthparabola \\). If the equation of \\( origparabola \\) is \\( ordinate^{2}=paracoeff abscissa \\), find the equation of \\( nthparabola \\).",
      "solution": "Solution. Since \\( origparabola \\) is a parabola with equation \\( ordinate^{2}=paracoeff abscissa \\), any point of \\( origparabola \\) has coordinates of the form \\( ( paracoeff parazeta^{2}, paracoeff parazeta ) \\) for some real \\( parazeta \\), and conversely, every such point is on \\( origparabola \\). The slope of the line tangent to \\( origparabola \\) at \\( \\left(paracoeff parazeta^{2}, paracoeff parazeta\\right) \\) is \\( 1 / 2 parazeta \\).\n\nLet \\( pointalpha \\) and \\( pointbeta \\) be the points \\( \\left(paracoeff parameta^{2}, paracoeff parameta\\right) \\) and \\( \\left(paracoeff parazeta^{2}, paracoeff parazeta\\right) \\), respectively. The tangents to \\( origparabola \\) at \\( pointalpha \\) and \\( pointbeta \\) are perpendicular if and only if \\( (1 / 2 parameta)(1 / 2 parazeta) = -1 \\), i.e., if and only if\n\\[\nparameta\\,parazeta = -\\frac{1}{4}\n\\]\n\nThe centroid of the points \\( pointalpha, pointbeta \\), and the vertex \\( (0,0) \\) of \\( origparabola \\) is\n\\[\n\\left(\\frac{1}{3}\\,paracoeff\\left(parameta^{2}+parazeta^{2}\\right),\\;\\frac{1}{3}\\,paracoeff(parameta+parazeta)\\right)\n\\]\nand this centroid lies on a new parabola\n\\[\nordinate^{2}=\\frac{1}{3}\\,paracoeff\\left(abscissa-\\frac{paracoeff}{6}\\right)\n\\]\nif the perpendicularity condition (1) is satisfied.\n\nConversely, any point \\( ( abscissa, ordinate ) \\) on the parabola (3) has the form (2), since the equations\n\\[\n\\begin{aligned}\n\\frac{1}{3}\\,paracoeff(parameta+parazeta) & = ordinate \\\\\nparameta\\,parazeta & = -\\frac{1}{4}\n\\end{aligned}\n\\]\ncan always be solved to give real \\( parameta \\) and \\( parazeta \\). Indeed, \\( parameta \\) and \\( parazeta \\) are zeros of \\( auxscalar^{2}-(3\\,ordinate/paracoeff)\\,auxscalar-\\frac{1}{4} \\), which has positive discriminant. Hence the locus in question is the entire parabola \\( firstparabola \\) given by (3).\n\nNow \\( firstparabola \\) is obtained from \\( origparabola \\) by changing the constant from \\( paracoeff \\) to \\( paracoeff/3 \\), and displacing the vertex to the right by \\( paracoeff/6 \\). Consequently, \\( secondparabola \\) can be obtained from \\( firstparabola \\) by changing the constant from \\( paracoeff/3 \\) to \\( (paracoeff/3)/3 \\) and displacing the vertex \\( (paracoeff/3)/6 \\) further to the right. The equation of \\( secondparabola \\) is therefore\n\\[\nordinate^{2}=\\frac{1}{9}\\,paracoeff\\left(abscissa-\\frac{1}{6}\\,paracoeff-\\frac{1}{18}\\,paracoeff\\right)\n\\]\n\nContinuing this reasoning, we see that \\( nthparabola \\) has the equation\n\\[\n\\begin{aligned}\nordinate^{2} &= \\frac{1}{3^{indexcount}}\\,paracoeff\\left(abscissa-\\frac{1}{6}\\,paracoeff-\\frac{1}{6\\cdot3}\\,paracoeff-\\frac{1}{6\\cdot3^{2}}\\,paracoeff-\\cdots-\\frac{1}{6\\cdot3^{indexcount-1}}\\,paracoeff\\right) \\\\\n&= \\frac{1}{3^{indexcount}}\\,paracoeff\\left(abscissa-\\frac{paracoeff}{4}\\left(1-\\frac{1}{3^{indexcount}}\\right)\\right).\n\\end{aligned}\n\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "A": "lanterns",
        "B": "sailboat",
        "S": "grapevine",
        "s": "toffeejar",
        "t": "brickwall",
        "x": "catapult",
        "y": "lighthouse",
        "P": "chandelier",
        "P_1": "starfruit",
        "P_2": "raincloud",
        "P_n": "snowflake",
        "m": "horseshoe",
        "n": "dragonfly"
      },
      "question": "Let \\( lanterns, sailboat \\) be variable points on a parabola \\( chandelier \\), such that the tangents at \\( lanterns \\) and \\( sailboat \\) are perpendicular to each other. Show that the locus of the centroid of the triangle formed by \\( lanterns, sailboat \\) and the vertex of \\( chandelier \\) is a parabola \\( starfruit \\). Apply the same process to \\( starfruit \\), obtaining a parabola \\( raincloud \\), and repeat the process, obtaining altogether the sequence of parabolas \\( chandelier, starfruit, raincloud, \\ldots, snowflake \\). If the equation of \\( chandelier \\) is \\( lighthouse^{2}=horseshoe catapult \\), find the equation of \\( snowflake \\).",
      "solution": "Solution. Since \\( chandelier \\) is a parabola with equation \\( lighthouse^{2}=horseshoe catapult \\), any point of \\( chandelier \\) has coordinates of the form ( \\( horseshoe brickwall^{2}, horseshoe brickwall \\) ) for some real \\( brickwall \\), and conversely, every such point is on \\( chandelier \\). The slope of the line tangent to \\( chandelier \\) at \\( \\left(horseshoe brickwall^{2}, horseshoe brickwall\\right) \\) is \\( 1 / 2 brickwall \\).\n\nLet \\( lanterns \\) and \\( sailboat \\) be the points \\( \\left(horseshoe toffeejar^{2}, horseshoe toffeejar\\right) \\) and ( \\( horseshoe brickwall^{2}, horseshoe brickwall \\) ), respectively. The tangents to \\( chandelier \\) at \\( lanterns \\) and \\( sailboat \\) are perpendicular if and only if \\( (1 / 2 toffeejar)(1 / 2 brickwall)=-1 \\), i.e., if and only if\n\\[\ntoffeejar brickwall=-\\frac{1}{4}\n\\]\n\nThe centroid of the points \\( lanterns, sailboat \\), and the vertex \\( (0,0) \\) of \\( chandelier \\) is\n\\[\n\\left(\\frac{1}{3} horseshoe\\left(toffeejar^{2}+brickwall^{2}\\right), \\quad \\frac{1}{3} horseshoe(toffeejar+brickwall)\\right)\n\\]\nand this centroid lies on a new parabola\n\\[\nlighthouse^{2}=\\frac{1}{3} horseshoe\\left(catapult-\\frac{horseshoe}{6}\\right)\n\\]\nif the perpendicularity condition (1) is satisfied. Conversely, any point ( \\( catapult, lighthouse \\) ) on the parabola (3) has the form (2), since the equations\n\\[\n\\begin{aligned}\n\\frac{1}{3} horseshoe(toffeejar+brickwall) & =lighthouse \\\\\ntoffeejar brickwall & =-\\frac{1}{4}\n\\end{aligned}\n\\]\ncan always be solved to give real \\( toffeejar \\) and \\( brickwall \\). Indeed, \\( toffeejar \\) and \\( brickwall \\) are zeros of \\( grapevine^{2}-(3 lighthouse / horseshoe) grapevine-\\frac{1}{4} \\), which has positive discriminant. Hence the locus in question is the entire parabola \\( starfruit \\) given by (3).\n\nNow \\( starfruit \\) is obtained from \\( chandelier \\) by changing the constant from \\( horseshoe \\) to \\( horseshoe / 3 \\), and displacing the vertex to the right by \\( horseshoe / 6 \\). Consequently, \\( raincloud \\) can be obtained from \\( starfruit \\) by changing the constant from \\( horseshoe / 3 \\) to \\( (horseshoe / 3) / 3 \\) and displacing the vertex \\( (horseshoe / 3) / 6 \\) further to the right. The equation of \\( raincloud \\) is therefore\n\\[\nlighthouse^{2}=\\frac{1}{9} horseshoe\\left(catapult-\\frac{1}{6} horseshoe-\\frac{1}{18} horseshoe\\right)\n\\]\n\nContinuing this reasoning, we see that \\( snowflake \\) has the equation\n\\[\n\\begin{aligned}\nlighthouse^{2} & =\\frac{1}{3^{dragonfly}} horseshoe\\left(catapult-\\frac{1}{6} horseshoe-\\frac{1}{6 \\cdot 3} horseshoe-\\frac{1}{6 \\cdot 3^{2}} horseshoe-\\cdots-\\frac{1}{6 \\cdot 3^{dragonfly-1}} horseshoe\\right) \\\\\n& =\\frac{1}{3^{dragonfly}} horseshoe\\left(catapult-\\frac{horseshoe}{4}\\left(1-\\frac{1}{3^{dragonfly}}\\right)\\right) .\n\\end{aligned}\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "A": "voidpoint",
        "B": "nullplace",
        "S": "emptynode",
        "s": "hollowvar",
        "t": "vacantvar",
        "x": "antiabscis",
        "y": "antiordinate",
        "P": "antiparab",
        "P_1": "antiparabone",
        "P_2": "antiparabtwo",
        "P_n": "antiparabnth",
        "m": "immobile",
        "n": "uncounted"
      },
      "question": "2. Let \\( voidpoint, nullplace \\) be variable points on a parabola \\( antiparab \\), such that the tangents at \\( voidpoint \\) and \\( nullplace \\) are perpendicular to each other. Show that the locus of the centroid of the triangle formed by \\( voidpoint, nullplace \\) and the vertex of \\( antiparab \\) is a parabola \\( antiparabone \\). Apply the same process to \\( antiparabone \\), obtaining a parabola \\( antiparabtwo \\), and repeat the process, obtaining altogether the sequence of parabolas \\( antiparab, antiparabone, antiparabtwo, \\ldots, antiparabnth \\). If the equation of \\( antiparab \\) is \\( antiordinate^{2}=immobile antiabscis \\), find the equation of \\( antiparabnth \\).",
      "solution": "Solution. Since \\( antiparab \\) is a parabola with equation \\( antiordinate^{2}=immobile antiabscis \\), any point of \\( antiparab \\) has coordinates of the form ( \\( immobile vacantvar^{2}, immobile vacantvar \\) ) for some real \\( vacantvar \\), and conversely, every such point is on \\( antiparab \\). The slope of the line tangent to \\( antiparab \\) at \\( \\left(immobile vacantvar^{2}, immobile vacantvar\\right) \\) is \\( 1 / 2 vacantvar \\).\n\nLet \\( voidpoint \\) and \\( nullplace \\) be the points \\( \\left(immobile hollowvar^{2}, immobile hollowvar\\right) \\) and \\( \\left(immobile vacantvar^{2}, immobile vacantvar\\right) \\), respectively. The tangents to \\( antiparab \\) at \\( voidpoint \\) and \\( nullplace \\) are perpendicular if and only if \\( (1 / 2 hollowvar)(1 / 2 vacantvar)=-1 \\), i.e., if and only if\n\\[\nhollowvar vacantvar=-\\frac{1}{4}\n\\]\n\nThe centroid of the points \\( voidpoint, nullplace \\), and the vertex \\( (0,0) \\) of \\( antiparab \\) is\n\\[\n\\left(\\frac{1}{3} immobile\\left(hollowvar^{2}+vacantvar^{2}\\right), \\quad \\frac{1}{3} immobile(hollowvar+vacantvar)\\right)\n\\]\nand this centroid lies on a new parabola\n\\[\nantiordinate^{2}=\\frac{1}{3} immobile\\left(antiabscis-\\frac{immobile}{6}\\right)\n\\]\nif the perpendicularity condition (1) is satisfied.\nConversely, any point \\( (antiabscis, antiordinate) \\) on the parabola (3) has the form (2), since the equations\n\\[\n\\begin{aligned}\n\\frac{1}{3} immobile(hollowvar+vacantvar) & =antiordinate \\\\\nhollowvar vacantvar & =-\\frac{1}{4}\n\\end{aligned}\n\\]\ncan always be solved to give real \\( hollowvar \\) and \\( vacantvar \\). Indeed, \\( hollowvar \\) and \\( vacantvar \\) are zeros of \\( emptynode^{2}-(3 antiordinate / immobile) emptynode-\\frac{1}{4} \\), which has positive discriminant. Hence the locus in question is the entire parabola \\( antiparabone \\) given by (3).\n\nNow \\( antiparabone \\) is obtained from \\( antiparab \\) by changing the constant from \\( immobile \\) to \\( immobile / 3 \\), and displacing the vertex to the right by \\( immobile / 6 \\). Consequently, \\( antiparabtwo \\) can be obtained from \\( antiparabone \\) by changing the constant from \\( immobile / 3 \\) to \\( (immobile / 3) / 3 \\) and displacing the vertex \\( (immobile / 3) / 6 \\) further to the right. The equation of \\( antiparabtwo \\) is therefore\n\\[\nantiordinate^{2}=\\frac{1}{9} immobile\\left(antiabscis-\\frac{1}{6} immobile-\\frac{1}{18} immobile\\right)\n\\]\n\nContinuing this reasoning, we see that \\( antiparab_{\\boldsymbol{uncounted}} \\) has the equation\n\\[\n\\begin{aligned}\nantiordinate^{2} & =\\frac{1}{3^{uncounted}} immobile\\left(antiabscis-\\frac{1}{6} immobile-\\frac{1}{6 \\cdot 3} immobile-\\frac{1}{6 \\cdot 3^{2}} immobile-\\cdots-\\frac{1}{6 \\cdot 3^{uncounted-1}} immobile\\right) \\\\\n& =\\frac{1}{3^{uncounted}} immobile\\left(antiabscis-\\frac{immobile}{4}\\left(1-\\frac{1}{3^{uncounted}}\\right)\\right) .\n\\end{aligned}\n\\]"
    },
    "garbled_string": {
      "map": {
        "A": "zbxpfqle",
        "B": "ovdrnwhs",
        "S": "yjsdkmqt",
        "s": "qhcfrgza",
        "t": "mxjvluqi",
        "x": "hvlcyfpt",
        "y": "snrdmaok",
        "P": "kvgxtrda",
        "P_1": "jfwyhplq",
        "P_2": "xmtrcgwy",
        "P_n": "akptbrso",
        "m": "fepqsjln",
        "n": "rgtdkwvc"
      },
      "question": "2. Let \\( zbxpfqle, ovdrnwhs \\) be variable points on a parabola \\( kvgxtrda \\), such that the tangents at \\( zbxpfqle \\) and \\( ovdrnwhs \\) are perpendicular to each other. Show that the locus of the centroid of the triangle formed by \\( zbxpfqle, ovdrnwhs \\) and the vertex of \\( kvgxtrda \\) is a parabola \\( jfwyhplq \\). Apply the same process to \\( jfwyhplq \\), obtaining a parabola \\( xmtrcgwy \\), and repeat the process, obtaining altogether the sequence of parabolas \\( kvgxtrda, jfwyhplq, xmtrcgwy, \\ldots, akptbrso \\). If the equation of \\( kvgxtrda \\) is \\( snrdmaok^{2}=fepqsjln hvlcyfpt \\), find the equation of \\( akptbrso \\).",
      "solution": "Solution. Since \\( kvgxtrda \\) is a parabola with equation \\( snrdmaok^{2}=fepqsjln hvlcyfpt \\), any point of \\( kvgxtrda \\) has coordinates of the form ( \\( fepqsjln mxjvluqi^{2}, fepqsjln mxjvluqi \\) ) for some real \\( mxjvluqi \\), and conversely, every such point is on \\( kvgxtrda \\). The slope of the line tangent to \\( kvgxtrda \\) at \\( \\left(fepqsjln mxjvluqi^{2}, fepqsjln mxjvluqi\\right) \\) is \\( 1 / 2 mxjvluqi \\).\n\nLet \\( zbxpfqle \\) and \\( ovdrnwhs \\) be the points \\( \\left(fepqsjln qhcfrgza^{2}, fepqsjln qhcfrgza\\right) \\) and ( \\( fepqsjln mxjvluqi^{2}, fepqsjln mxjvluqi \\) ), respectively. The tangents to \\( kvgxtrda \\) at \\( zbxpfqle \\) and \\( ovdrnwhs \\) are perpendicular if and only if \\( (1 / 2 qhcfrgza)(1 / 2 mxjvluqi) \\) \\( =-1 \\), i.e., if and only if\n\\[\nqhcfrgza mxjvluqi=-\\frac{1}{4}\n\\]\n\nThe centroid of the points \\( zbxpfqle, ovdrnwhs \\), and the vertex \\( (0,0) \\) of \\( kvgxtrda \\) is\n\\[\n\\left(\\frac{1}{3} fepqsjln\\left(qhcfrgza^{2}+mxjvluqi^{2}\\right), \\quad \\frac{1}{3} fepqsjln(qhcfrgza+mxjvluqi)\\right)\n\\]\nand this centroid lies on a new parabola\n\\[\nsnrdmaok^{2}=\\frac{1}{3} fepqsjln\\left(hvlcyfpt-\\frac{fepqsjln}{6}\\right)\n\\]\nif the perpendicularity condition (1) is satisfied.\nConversely, any point ( \\( hvlcyfpt, snrdmaok \\) ) on the parabola (3) has the form (2), since the equations\n\\[\n\\begin{aligned}\n\\frac{1}{3} fepqsjln(qhcfrgza+mxjvluqi) & =snrdmaok \\\\\nqhcfrgza mxjvluqi & =-\\frac{1}{4}\n\\end{aligned}\n\\]\ncan always be solved to give real \\( qhcfrgza \\) and \\( mxjvluqi \\). Indeed, \\( qhcfrgza \\) and \\( mxjvluqi \\) are zeros of \\( yjsdkmqt^{2}-(3 snrdmaok / fepqsjln) yjsdkmqt-\\frac{1}{4} \\), which has positive discriminant. Hence the locus in question is the entire parabola \\( jfwyhplq \\) given by (3).\n\nNow \\( jfwyhplq \\) is obtained from \\( kvgxtrda \\) by changing the constant from \\( fepqsjln \\) to \\( fepqsjln / 3 \\), and displacing the vertex to the right by \\( fepqsjln / 6 \\). Consequently, \\( xmtrcgwy \\) can be obtained from \\( jfwyhplq \\) by changing the constant from \\( fepqsjln / 3 \\) to \\( (fepqsjln / 3) / 3 \\) and displacing the vertex \\( (fepqsjln / 3) / 6 \\) further to the right. The equation of \\( xmtrcgwy \\) is therefore\n\\[\nsnrdmaok^{2}=\\frac{1}{9} fepqsjln\\left(hvlcyfpt-\\frac{1}{6} fepqsjln-\\frac{1}{18} fepqsjln\\right)\n\\]\n\nContinuing this reasoning, we see that \\( akptbrso \\) has the equation\n\\[\n\\begin{aligned}\nsnrdmaok^{2} & =\\frac{1}{3^{rgtdkwvc}} fepqsjln\\left(hvlcyfpt-\\frac{1}{6} fepqsjln-\\frac{1}{6 \\cdot 3} fepqsjln-\\frac{1}{6 \\cdot 3^{2}} fepqsjln-\\cdots-\\frac{1}{6 \\cdot 3^{rgtdkwvc-1}} fepqsjln\\right) \\\\\n& =\\frac{1}{3^{rgtdkwvc}} fepqsjln\\left(hvlcyfpt-\\frac{fepqsjln}{4}\\left(1-\\frac{1}{3^{rgtdkwvc}}\\right)\\right) .\n\\end{aligned}\n\\]"
    },
    "kernel_variant": {
      "question": "Fix a positive real constant p and consider in ℝ³ the (upward–opening) paraboloid of revolution  \n\n    𝓟₀ :  z = (x² + y²)/(4p).\n\nFor two ordered points A,B ∈ 𝓟₀ denote by τ_A , τ_B the tangent planes to 𝓟₀ at A and B and by n_A , n_B any choice of normal vectors to these planes (their orientation is irrelevant for what follows).\n\n(A)  Impose the orthogonality condition  \n\n          n_A · n_B = 0        (†)\n\n(i)  Let V₀ = (0,0,0) be the vertex and F₀ = (0,0,p) the focus of 𝓟₀.  \n  Put G:=centroid(A,B,V₀,F₀).  \n  Show that the locus Γ₁ of G is again a paraboloid, call it 𝓟₁, and determine its Cartesian equation.\n\n(B)  Replace 𝓟₀ by 𝓟₁ and repeat the above construction (orthogonality (†) for the new tangent planes, centroid together with the new vertex V₁ and focus F₁).  \n  This produces a sequence of paraboloids  \n\n    𝓟₀, 𝓟₁, 𝓟₂, …, 𝓟_n      (n ∈ ℕ).\n\n(C)  Obtain a closed-form Cartesian equation of 𝓟_n in the original (x,y,z) coordinate system for arbitrary n ∈ ℕ.",
      "solution": "Throughout we write every paraboloid 𝓟_k in the form  \n\n  𝓟_k : z = α_k(x² + y²) + d_k  (α_k > 0),    (1)\n\nand denote  \n\n  V_k = (0,0,d_k),  p_k := 1/(4α_k),  F_k = (0,0,d_k + p_k).  (2)\n\nStep 1 – A convenient parametrisation of 𝓟_k.  \nFor (u,v) ∈ ℝ² set  \n\n  Φ_k(u,v) := (2p_k u, 2p_k v, p_k(u²+v²)+d_k).  (3)\n\nΦ_k is a global diffeomorphism ℝ² → 𝓟_k, so every point of 𝓟_k is represented once.\n\nStep 2 – Normal vectors and the orthogonality condition.  \n ∂_uΦ_k = (2p_k,0,2p_k u), ∂_vΦ_k = (0,2p_k,2p_k v),\n\nhence (a convenient normal choice)\n\n n(u,v) := ∂_uΦ_k × ∂_vΦ_k = 4p_k²(−u,−v,1).  (4)\n\nFor A = Φ_k(u₁,v₁) and B = Φ_k(u₂,v₂)\n\n n_A·n_B = 16p_k⁴(u₁u₂ + v₁v₂ + 1).  (5)\n\nThus condition (†) is equivalent to  \n\n  u₁u₂ + v₁v₂ = −1.           (6)\n\nStep 3 – Coordinates of A, B, V_k, F_k.  \nPut  \n\n  S := u₁² + v₁² + u₂² + v₂².  \n\nThen\n\nA = (2p_k u₁, 2p_k v₁, p_k(u₁²+v₁²)+d_k),  \nB = (2p_k u₂, 2p_k v₂, p_k(u₂²+v₂²)+d_k),  \nV_k = (0,0,d_k), F_k = (0,0,d_k + p_k).  (7)\n\nStep 4 – The centroid G of {A,B,V_k,F_k}.  \nAveraging the four position vectors gives  \n\n x_G = (p_k/2)(u₁+u₂), y_G = (p_k/2)(v₁+v₂),  (8)  \n z_G = d_k + (p_k/4)(S+1).          (9)\n\n(The “+1’’ comes from the extra p_k that raises the focus.)\n\nStep 5 – Eliminating the parameters u_i,v_i.  \nCompute\n\nx_G² + y_G² = (p_k²/4)[(u₁+u₂)²+(v₁+v₂)²]  \n            = (p_k²/4)[S + 2(u₁u₂+v₁v₂)]  \n            = (p_k²/4)(S−2)         (from (6)).  (10)\n\nFrom (9) we have p_k(z_G−d_k)= (p_k²/4)(S+1). Eliminating S between (10) and (9) gives  \n\n z_G = (x_G² + y_G²)/p_k + d_k + (3/4)p_k.  (11)\n\nHence the locus Γ₁ is contained in the paraboloid  \n\n  𝓟_{k+1} : z = α_{k+1}(x² + y²) + d_{k+1}  with  \n  α_{k+1} = 1/p_k = 4α_k,  d_{k+1}=d_k+(3/4)p_k.  (12)\n\nFor k=0 (α₀ = 1/(4p), p₀ = p, d₀ = 0) we obtain  \n\n  α₁ = 1/p, d₁ = (3/4)p, so 𝓟₁ : z = (x²+y²)/p + (3/4)p. (13)\n\nStep 6 – Surjectivity: every point of 𝓟_{k+1} arises.  \nLet G = (x,y,z) be an arbitrary point on the right-hand side of (11).  \nDefine  \n\n s := (2x/p_k, 2y/p_k) ∈ ℝ²,  σ := |s|.  (14)\n\nWe shall choose vectors a,b ∈ ℝ² with  \n (a) a + b = s,  \n (b) a·b = −1.                               (15)\n\nTake the unit vector e := s/σ if σ ≠ 0 (any unit vector if σ = 0) and set a := λe with a scalar λ to be determined.  \nThen b = s − a = (σ − λ)e and  \n\n a·b = λ(σ−λ)=−1 ⇔ λ² − σλ − 1 = 0.  (16)\n\nThe discriminant is σ²+4>0, so (16) has two real solutions; pick either.  \nDefine  \n\n (u₁,v₁)=a, (u₂,v₂)=b.  (17)\n\nCondition (b) holds by construction, i.e. (6) is satisfied.  \nMoreover |a|²+|b|² = σ²+2, hence (10) gives exactly the required z-coordinate (11).  \nThus there indeed exist A,B on 𝓟_k whose centroid is the prescribed point G, proving Γ₁ = 𝓟_{k+1}.\n\nStep 7 – Closed forms for p_k and d_k.  \nFrom (12) we get α_{k+1}=4α_k ⇒ α_k=4^{k-1}/p (k≥1) and α₀=1/(4p).  \nTherefore  \n\n p_k = 1/(4α_k)=p/4^{\\,k}.          (18)\n\nThe second recurrence d_{k+1}=d_k+(3/4)p_k with d₀=0 gives  \n\n d_n = ∑_{j=0}^{n-1}(3/4)p_j  \n      = (3p/4) ∑_{j=0}^{n-1}4^{−j}  \n      = p(1−4^{−n}).            (19)\n\nStep 8 – Explicit equation of 𝓟_n.  \nInsert (18) and (19) into (1):\n\n  𝓟_n : z = (4^{\\,n-1}/p)(x² + y²) + p(1 − 4^{−n}),  n ≥ 0. (20)\n\nFor n=0 this reads z=(x²+y²)/(4p), i.e. the original 𝓟₀, so (20) is valid for every n.\n\nThus the iterative “perpendicular-tangent/centroid’’ procedure produces exactly the sequence of paraboloids (20).",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.406201",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension: the original problem works in ℝ²; the variant elevates everything to ℝ³, replacing tangents (1-dimensional lines) by tangent planes (2-dimensional), so normal-vector geometry and 3-vector algebra are indispensable.\n\n2. More variables and parameters: each point on the paraboloid needs two independent parameters (u,v) instead of one, leading to systems with four real parameters (u₁,v₁,u₂,v₂) subject to a quadratic constraint.\n\n3. Additional structures: the condition “tangent planes are perpendicular’’ translates to orthogonality of normals, requiring manipulation of cross products and dot products, concepts absent from the planar version.\n\n4. Deeper iteration: the recurrence now involves two coupled sequences (α_n, d_n).  Solving them entails linear-non-homogeneous recurrences, substitution techniques, and summation of a truncated geometric series—all at a higher algebraic complexity than the single geometric progression of the planar case.\n\n5. Coordinate-free insight: recognising that the direction of the normal vector is k-independent is crucial; without it, elimination of the parameters becomes extremely messy.\n\n6. Non-trivial focal translation: besides the ever-shrinking focal length p_n = p/4^{n}, each iteration produces a vertical translation d_n following a non-obvious closed form (17), significantly complicating the final explicit equation.\n\nAltogether these elements make the enhanced problem substantially harder than both the original Olympiad problem and the previous kernel variant, demanding multivariable calculus, 3-space analytic geometry, and recurrence solving techniques beyond simple pattern recognition."
      }
    },
    "original_kernel_variant": {
      "question": "Fix a positive real constant p and consider in ℝ³ the (upward–opening) paraboloid of revolution  \n\n    𝓟₀ :  z = (x² + y²)/(4p).\n\nFor two ordered points A,B ∈ 𝓟₀ denote by τ_A , τ_B the tangent planes to 𝓟₀ at A and B and by n_A , n_B any choice of normal vectors to these planes (their orientation is irrelevant for what follows).\n\n(A)  Impose the orthogonality condition  \n\n          n_A · n_B = 0        (†)\n\n(i)  Let V₀ = (0,0,0) be the vertex and F₀ = (0,0,p) the focus of 𝓟₀.  \n  Put G:=centroid(A,B,V₀,F₀).  \n  Show that the locus Γ₁ of G is again a paraboloid, call it 𝓟₁, and determine its Cartesian equation.\n\n(B)  Replace 𝓟₀ by 𝓟₁ and repeat the above construction (orthogonality (†) for the new tangent planes, centroid together with the new vertex V₁ and focus F₁).  \n  This produces a sequence of paraboloids  \n\n    𝓟₀, 𝓟₁, 𝓟₂, …, 𝓟_n      (n ∈ ℕ).\n\n(C)  Obtain a closed-form Cartesian equation of 𝓟_n in the original (x,y,z) coordinate system for arbitrary n ∈ ℕ.",
      "solution": "Throughout we write every paraboloid 𝓟_k in the form  \n\n  𝓟_k : z = α_k(x² + y²) + d_k  (α_k > 0),    (1)\n\nand denote  \n\n  V_k = (0,0,d_k),  p_k := 1/(4α_k),  F_k = (0,0,d_k + p_k).  (2)\n\nStep 1 – A convenient parametrisation of 𝓟_k.  \nFor (u,v) ∈ ℝ² set  \n\n  Φ_k(u,v) := (2p_k u, 2p_k v, p_k(u²+v²)+d_k).  (3)\n\nΦ_k is a global diffeomorphism ℝ² → 𝓟_k, so every point of 𝓟_k is represented once.\n\nStep 2 – Normal vectors and the orthogonality condition.  \n ∂_uΦ_k = (2p_k,0,2p_k u), ∂_vΦ_k = (0,2p_k,2p_k v),\n\nhence (a convenient normal choice)\n\n n(u,v) := ∂_uΦ_k × ∂_vΦ_k = 4p_k²(−u,−v,1).  (4)\n\nFor A = Φ_k(u₁,v₁) and B = Φ_k(u₂,v₂)\n\n n_A·n_B = 16p_k⁴(u₁u₂ + v₁v₂ + 1).  (5)\n\nThus condition (†) is equivalent to  \n\n  u₁u₂ + v₁v₂ = −1.           (6)\n\nStep 3 – Coordinates of A, B, V_k, F_k.  \nPut  \n\n  S := u₁² + v₁² + u₂² + v₂².  \n\nThen\n\nA = (2p_k u₁, 2p_k v₁, p_k(u₁²+v₁²)+d_k),  \nB = (2p_k u₂, 2p_k v₂, p_k(u₂²+v₂²)+d_k),  \nV_k = (0,0,d_k), F_k = (0,0,d_k + p_k).  (7)\n\nStep 4 – The centroid G of {A,B,V_k,F_k}.  \nAveraging the four position vectors gives  \n\n x_G = (p_k/2)(u₁+u₂), y_G = (p_k/2)(v₁+v₂),  (8)  \n z_G = d_k + (p_k/4)(S+1).          (9)\n\n(The “+1’’ comes from the extra p_k that raises the focus.)\n\nStep 5 – Eliminating the parameters u_i,v_i.  \nCompute\n\nx_G² + y_G² = (p_k²/4)[(u₁+u₂)²+(v₁+v₂)²]  \n            = (p_k²/4)[S + 2(u₁u₂+v₁v₂)]  \n            = (p_k²/4)(S−2)         (from (6)).  (10)\n\nFrom (9) we have p_k(z_G−d_k)= (p_k²/4)(S+1). Eliminating S between (10) and (9) gives  \n\n z_G = (x_G² + y_G²)/p_k + d_k + (3/4)p_k.  (11)\n\nHence the locus Γ₁ is contained in the paraboloid  \n\n  𝓟_{k+1} : z = α_{k+1}(x² + y²) + d_{k+1}  with  \n  α_{k+1} = 1/p_k = 4α_k,  d_{k+1}=d_k+(3/4)p_k.  (12)\n\nFor k=0 (α₀ = 1/(4p), p₀ = p, d₀ = 0) we obtain  \n\n  α₁ = 1/p, d₁ = (3/4)p, so 𝓟₁ : z = (x²+y²)/p + (3/4)p. (13)\n\nStep 6 – Surjectivity: every point of 𝓟_{k+1} arises.  \nLet G = (x,y,z) be an arbitrary point on the right-hand side of (11).  \nDefine  \n\n s := (2x/p_k, 2y/p_k) ∈ ℝ²,  σ := |s|.  (14)\n\nWe shall choose vectors a,b ∈ ℝ² with  \n (a) a + b = s,  \n (b) a·b = −1.                               (15)\n\nTake the unit vector e := s/σ if σ ≠ 0 (any unit vector if σ = 0) and set a := λe with a scalar λ to be determined.  \nThen b = s − a = (σ − λ)e and  \n\n a·b = λ(σ−λ)=−1 ⇔ λ² − σλ − 1 = 0.  (16)\n\nThe discriminant is σ²+4>0, so (16) has two real solutions; pick either.  \nDefine  \n\n (u₁,v₁)=a, (u₂,v₂)=b.  (17)\n\nCondition (b) holds by construction, i.e. (6) is satisfied.  \nMoreover |a|²+|b|² = σ²+2, hence (10) gives exactly the required z-coordinate (11).  \nThus there indeed exist A,B on 𝓟_k whose centroid is the prescribed point G, proving Γ₁ = 𝓟_{k+1}.\n\nStep 7 – Closed forms for p_k and d_k.  \nFrom (12) we get α_{k+1}=4α_k ⇒ α_k=4^{k-1}/p (k≥1) and α₀=1/(4p).  \nTherefore  \n\n p_k = 1/(4α_k)=p/4^{\\,k}.          (18)\n\nThe second recurrence d_{k+1}=d_k+(3/4)p_k with d₀=0 gives  \n\n d_n = ∑_{j=0}^{n-1}(3/4)p_j  \n      = (3p/4) ∑_{j=0}^{n-1}4^{−j}  \n      = p(1−4^{−n}).            (19)\n\nStep 8 – Explicit equation of 𝓟_n.  \nInsert (18) and (19) into (1):\n\n  𝓟_n : z = (4^{\\,n-1}/p)(x² + y²) + p(1 − 4^{−n}),  n ≥ 0. (20)\n\nFor n=0 this reads z=(x²+y²)/(4p), i.e. the original 𝓟₀, so (20) is valid for every n.\n\nThus the iterative “perpendicular-tangent/centroid’’ procedure produces exactly the sequence of paraboloids (20).",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.347481",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension: the original problem works in ℝ²; the variant elevates everything to ℝ³, replacing tangents (1-dimensional lines) by tangent planes (2-dimensional), so normal-vector geometry and 3-vector algebra are indispensable.\n\n2. More variables and parameters: each point on the paraboloid needs two independent parameters (u,v) instead of one, leading to systems with four real parameters (u₁,v₁,u₂,v₂) subject to a quadratic constraint.\n\n3. Additional structures: the condition “tangent planes are perpendicular’’ translates to orthogonality of normals, requiring manipulation of cross products and dot products, concepts absent from the planar version.\n\n4. Deeper iteration: the recurrence now involves two coupled sequences (α_n, d_n).  Solving them entails linear-non-homogeneous recurrences, substitution techniques, and summation of a truncated geometric series—all at a higher algebraic complexity than the single geometric progression of the planar case.\n\n5. Coordinate-free insight: recognising that the direction of the normal vector is k-independent is crucial; without it, elimination of the parameters becomes extremely messy.\n\n6. Non-trivial focal translation: besides the ever-shrinking focal length p_n = p/4^{n}, each iteration produces a vertical translation d_n following a non-obvious closed form (17), significantly complicating the final explicit equation.\n\nAltogether these elements make the enhanced problem substantially harder than both the original Olympiad problem and the previous kernel variant, demanding multivariable calculus, 3-space analytic geometry, and recurrence solving techniques beyond simple pattern recognition."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}