{
  "index": "1994-B-2",
  "type": "ALG",
  "tag": [
    "ALG",
    "GEO",
    "ANA"
  ],
  "difficulty": "",
  "question": "For which real numbers $c$ is there a straight line that intersects the curve\n\\[ x^4 + 9x^3 + cx^2 + 9x + 4\n\\]\nin four distinct points?",
  "solution": "Solution 1 (geometric). The constant and linear terms of\n\\[\nP(x)=x^{4}+9 x^{3}+c x^{2}+9 x+4\n\\]\nare irrelevant to the problem; \\( y=P(x) \\) meets the line \\( y=m x+b \\) in four points if and only if \\( y=P(x)+9 x+4 \\) meets the line \\( y=(m+9) x+(b+4) \\) in four points.\n\nAlso, \\( y=P(x) \\) meets the line \\( y=m x+b \\) in four points if and only if \\( y=P(x-\\alpha) \\) meets the line \\( y=m(x-\\alpha)+b \\) in four points, so we may replace the given quartic with \\( P(x-9 / 4)=x^{4}+(c-243 / 8) x^{2}+\\cdots \\) (where we ignore the linear and constant terms).\n\nThe problem is then to determine the values of \\( c \\) for which there is a straight line that intersects \\( y=x^{4}+(c-243 / 8) x^{2} \\) in four distinct points. The result is now apparent from the shapes of the curves \\( y=x^{4}+a x^{2} \\). For example, when \\( a<0 \\), this \"W-shaped\" curve has a relative maximum at \\( x=0 \\), so the horizontal lines \\( y=-\\epsilon \\) for small positive \\( \\epsilon \\) intersect the curve in four points, while for \\( a \\geq 0 \\), the curve is always concave upward, so no line can intersect it in more than two points; see Figure 28.\n\nSolution 2 (algebraic). We wish to know if we can choose \\( m \\) and \\( b \\) so that\n\\[\nq(x)=x^{4}+9 x^{3}+c x^{2}+9 x+4-(m x+b)\n\\]\nhas four distinct real solutions \\( \\alpha_{1}, \\alpha_{2}, \\alpha_{3}, \\alpha_{4} \\). If we can find four distinct real numbers such that\n\\[\n\\begin{array}{c}\n\\alpha_{1}+\\alpha_{2}+\\alpha_{3}+\\alpha_{4}=-9 \\\\\n\\alpha_{1} \\alpha_{2}+\\alpha_{1} \\alpha_{3}+\\alpha_{1} \\alpha_{4}+\\alpha_{2} \\alpha_{3}+\\alpha_{2} \\alpha_{4}+\\alpha_{3} \\alpha_{4}=c\n\\end{array}\n\\]\nthen we can choose \\( m \\) and \\( b \\) appropriately so that \\( q(x) \\) has these four zeros (by the expansion of \\( \\left.\\prod_{i=1}^{4}\\left(x-\\alpha_{i}\\right)\\right) \\).\n\nThen from\n\\[\n0<\\sum_{i<j}\\left(\\alpha_{i}-\\alpha_{j}\\right)^{2}=3\\left(\\alpha_{1}+\\alpha_{2}+\\alpha_{3}+\\alpha_{4}\\right)^{2}-8 c\n\\]\nwe get \\( c<243 / 8 \\). Conversely, we must show that if \\( c<243 / 8 \\), then we can find distinct real \\( \\alpha_{1}, \\alpha_{2}, \\alpha_{3}, \\alpha_{4} \\) satisfying (1) and (2). This is indeed possible; try \\( \\left(\\alpha_{1}, \\alpha_{2}, \\alpha_{3}, \\alpha_{4}\\right)=(9 / 4-\\mu, 9 / 4+\\mu, 9 / 4-\\nu, 9 / 4+\\nu) \\) where \\( \\mu \\) and \\( \\nu \\) are distinct positive numbers. Then (1) is automatically satisfied, and we can choose \\( \\mu \\) and \\( \\nu \\) so that the right side of \\( (3) \\), which is \\( 8\\left(\\mu^{2}+\\nu^{2}\\right) \\), is any desired positive number.",
  "vars": [
    "x",
    "y",
    "P",
    "\\\\alpha",
    "\\\\alpha_1",
    "\\\\alpha_2",
    "\\\\alpha_3",
    "\\\\alpha_4",
    "\\\\mu",
    "\\\\nu",
    "\\\\epsilon",
    "i",
    "j"
  ],
  "params": [
    "c",
    "a",
    "m",
    "b"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "realvarx",
        "y": "realvary",
        "P": "polyfuncp",
        "\\alpha": "rootalpha",
        "\\alpha_1": "alphaone",
        "\\alpha_2": "alphatwo",
        "\\alpha_3": "alphathr",
        "\\alpha_4": "alphafor",
        "\\mu": "parammu",
        "\\nu": "paramnu",
        "\\epsilon": "smalleps",
        "j": "indexjay",
        "c": "paramsee",
        "a": "paramayy",
        "m": "slopeemm",
        "b": "interbbee"
      },
      "question": "For which real numbers $paramsee$ is there a straight line that intersects the curve\n\\[ realvarx^{4} + 9realvarx^{3} + paramsee realvarx^{2} + 9realvarx + 4 \\]\nin four distinct points?",
      "solution": "Solution 1 (geometric). The constant and linear terms of\n\\[\npolyfuncp(realvarx)=realvarx^{4}+9 realvarx^{3}+paramsee realvarx^{2}+9 realvarx+4\n\\]\nare irrelevant to the problem; \\( realvary=polyfuncp(realvarx) \\) meets the line \\( realvary=slopeemm realvarx+interbbee \\) in four points if and only if \\( realvary=polyfuncp(realvarx)+9 realvarx+4 \\) meets the line \\( realvary=(slopeemm+9) realvarx+(interbbee+4) \\) in four points.\n\nAlso, \\( realvary=polyfuncp(realvarx-rootalpha) \\) meets the line \\( realvary=slopeemm(realvarx-rootalpha)+interbbee \\) in four points if and only if \\( realvary=polyfuncp(realvarx-rootalpha) \\) meets the line \\( realvary=slopeemm(realvarx-rootalpha)+interbbee \\) in four points, so we may replace the given quartic with \\( polyfuncp(realvarx-9 / 4)=realvarx^{4}+(paramsee-243 / 8) realvarx^{2}+\\cdots \\) (where we ignore the linear and constant terms).\n\nThe problem is then to determine the values of \\( paramsee \\) for which there is a straight line that intersects \\( realvary=realvarx^{4}+(paramsee-243 / 8) realvarx^{2} \\) in four distinct points. The result is now apparent from the shapes of the curves \\( realvary=realvarx^{4}+paramayy realvarx^{2} \\). For example, when \\( paramayy<0 \\), this \"W-shaped\" curve has a relative maximum at \\( realvarx=0 \\), so the horizontal lines \\( realvary=-smalleps \\) for small positive \\( smalleps \\) intersect the curve in four points, while for \\( paramayy \\geq 0 \\), the curve is always concave upward, so no line can intersect it in more than two points; see Figure 28.\n\nSolution 2 (algebraic). We wish to know if we can choose \\( slopeemm \\) and \\( interbbee \\) so that\n\\[\nq(realvarx)=realvarx^{4}+9 realvarx^{3}+paramsee realvarx^{2}+9 realvarx+4-(slopeemm realvarx+interbbee)\n\\]\nhas four distinct real solutions \\( alphaone, alphatwo, alphathr, alphafor \\). If we can find four distinct real numbers such that\n\\[\n\\begin{array}{c}\nalphaone+alphatwo+alphathr+alphafor=-9 \\\\\nalphaone\\,alphatwo+alphaone\\,alphathr+alphaone\\,alphafor+alphatwo\\,alphathr+alphatwo\\,alphafor+alphathr\\,alphafor=paramsee\n\\end{array}\n\\]\nthen we can choose \\( slopeemm \\) and \\( interbbee \\) appropriately so that \\( q(realvarx) \\) has these four zeros (by the expansion of \\( \\prod_{i=1}^{4}(realvarx-rootalpha_{i}) \\)).\n\nThen from\n\\[\n0<\\sum_{i<indexjay}(rootalpha_{i}-rootalpha_{indexjay})^{2}=3(alphaone+alphatwo+alphathr+alphafor)^{2}-8\\,paramsee\n\\]\nwe get \\( paramsee<243 / 8 \\). Conversely, we must show that if \\( paramsee<243 / 8 \\), then we can find distinct real \\( alphaone, alphatwo, alphathr, alphafor \\) satisfying (1) and (2). This is indeed possible; try \\( (alphaone, alphatwo, alphathr, alphafor)=(9 / 4-parammu, 9 / 4+parammu, 9 / 4-paramnu, 9 / 4+paramnu) \\) where \\( parammu \\) and \\( paramnu \\) are distinct positive numbers. Then (1) is automatically satisfied, and we can choose \\( parammu \\) and \\( paramnu \\) so that the right side of (3), which is \\( 8(parammu^{2}+paramnu^{2}) \\), is any desired positive number."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "butternut",
        "y": "chandelier",
        "P": "dragonfly",
        "\\\\alpha": "lavender",
        "\\\\alpha_1": "blackbird",
        "\\\\alpha_2": "sparrowly",
        "\\\\alpha_3": "kingfisher",
        "\\\\alpha_4": "nightingale",
        "\\\\mu": "cinnamon",
        "\\\\nu": "cardamom",
        "\\\\epsilon": "turmeric",
        "i": "soapstone",
        "j": "parchment",
        "c": "honeycomb",
        "a": "riverside",
        "m": "pineapple",
        "b": "lighthouse"
      },
      "question": "For which real numbers $honeycomb$ is there a straight line that intersects the curve\n\\[ butternut^4 + 9 butternut^3 + honeycomb butternut^2 + 9 butternut + 4\n\\]\nin four distinct points?",
      "solution": "Solution 1 (geometric). The constant and linear terms of\n\\[\ndragonfly(butternut)=butternut^{4}+9 butternut^{3}+honeycomb butternut^{2}+9 butternut+4\n\\]\nare irrelevant to the problem; \\( chandelier = dragonfly(butternut) \\) meets the line \\( chandelier = pineapple butternut + lighthouse \\) in four points if and only if \\( chandelier = dragonfly(butternut)+9 butternut+4 \\) meets the line \\( chandelier = (pineapple+9) butternut+(lighthouse+4) \\) in four points.\n\nAlso, \\( chandelier=dragonfly(butternut) \\) meets the line \\( chandelier=pineapple butternut+lighthouse \\) in four points if and only if \\( chandelier=dragonfly(butternut-lavender) \\) meets the line \\( chandelier=pineapple(butternut-lavender)+lighthouse \\) in four points, so we may replace the given quartic with \\( dragonfly(butternut-9 / 4)=butternut^{4}+(honeycomb-243 / 8) butternut^{2}+\\cdots \\) (where we ignore the linear and constant terms).\n\nThe problem is then to determine the values of \\( honeycomb \\) for which there is a straight line that intersects \\( chandelier=butternut^{4}+(honeycomb-243 / 8) butternut^{2} \\) in four distinct points. The result is now apparent from the shapes of the curves \\( chandelier=butternut^{4}+riverside butternut^{2} \\). For example, when \\( riverside<0 \\), this \"W-shaped\" curve has a relative maximum at \\( butternut=0 \\), so the horizontal lines \\( chandelier=-turmeric \\) for small positive \\( turmeric \\) intersect the curve in four points, while for \\( riverside \\geq 0 \\), the curve is always concave upward, so no line can intersect it in more than two points; see Figure 28.\n\nSolution 2 (algebraic). We wish to know if we can choose \\( pineapple \\) and \\( lighthouse \\) so that\n\\[\nq(butternut)=butternut^{4}+9 butternut^{3}+honeycomb butternut^{2}+9 butternut+4-(pineapple butternut+lighthouse)\n\\]\nhas four distinct real solutions \\( blackbird, sparrowly, kingfisher, nightingale \\). If we can find four distinct real numbers such that\n\\[\n\\begin{array}{c}\nblackbird+sparrowly+kingfisher+nightingale=-9 \\\\\nblackbird\\,sparrowly+blackbird\\,kingfisher+blackbird\\,nightingale+sparrowly\\,kingfisher+sparrowly\\,nightingale+kingfisher\\,nightingale=honeycomb\n\\end{array}\n\\]\nthen we can choose \\( pineapple \\) and \\( lighthouse \\) appropriately so that \\( q(butternut) \\) has these four zeros (by the expansion of \\( \\left.\\prod_{soapstone=1}^{4}\\left(butternut-lavender_{soapstone}\\right)\\right) \\)).\n\nThen from\n\\[\n0<\\sum_{soapstone<parchment}\\left(lavender_{soapstone}-lavender_{parchment}\\right)^{2}=3\\left(blackbird+sparrowly+kingfisher+nightingale\\right)^{2}-8 honeycomb\n\\]\nwe get \\( honeycomb<243 / 8 \\). Conversely, we must show that if \\( honeycomb<243 / 8 \\), then we can find distinct real \\( blackbird, sparrowly, kingfisher, nightingale \\) satisfying (1) and (2). This is indeed possible; try \\( \\left(blackbird, sparrowly, kingfisher, nightingale\\right)=(9 / 4-cinnamon, 9 / 4+cinnamon, 9 / 4-cardamom, 9 / 4+cardamom) \\) where \\( cinnamon \\) and \\( cardamom \\) are distinct positive numbers. Then (1) is automatically satisfied, and we can choose \\( cinnamon \\) and \\( cardamom \\) so that the right side of \\( (3) \\), which is \\( 8\\left(cinnamon^{2}+cardamom^{2}\\right) \\), is any desired positive number."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalaxis",
        "y": "horizontalaxis",
        "P": "constantvalue",
        "\\alpha": "finalletter",
        "\\alpha_1": "finalletterone",
        "\\alpha_2": "finallettertwo",
        "\\alpha_3": "finalletterthree",
        "\\alpha_4": "finalletterfour",
        "\\mu": "macrosize",
        "\\nu": "megavalue",
        "\\epsilon": "largedeviation",
        "j": "staticindex",
        "c": "unchangeable",
        "a": "fixednumber",
        "m": "flatness",
        "b": "gradient"
      },
      "question": "For which real numbers $unchangeable$ is there a straight line that intersects the curve\n\\[ verticalaxis^{4} + 9verticalaxis^{3} + unchangeable verticalaxis^{2} + 9verticalaxis + 4 \\]\nin four distinct points?",
      "solution": "Solution 1 (geometric). The constant and linear terms of\n\\[\nconstantvalue(verticalaxis)=verticalaxis^{4}+9 verticalaxis^{3}+unchangeable verticalaxis^{2}+9 verticalaxis+4\n\\]\nare irrelevant to the problem; \\( horizontalaxis=constantvalue(verticalaxis) \\) meets the line \\( horizontalaxis=flatness\\,verticalaxis+gradient \\) in four points if and only if \\( horizontalaxis=constantvalue(verticalaxis)+9 verticalaxis+4 \\) meets the line \\( horizontalaxis=(flatness+9) verticalaxis+(gradient+4) \\) in four points.\n\nAlso, \\( horizontalaxis=constantvalue(verticalaxis) \\) meets the line \\( horizontalaxis=flatness\\,verticalaxis+gradient \\) in four points if and only if \\( horizontalaxis=constantvalue(verticalaxis-finalletter) \\) meets the line \\( horizontalaxis=flatness(verticalaxis-finalletter)+gradient \\) in four points, so we may replace the given quartic with \\( constantvalue(verticalaxis-9 / 4)=verticalaxis^{4}+(unchangeable-243 / 8) verticalaxis^{2}+\\cdots \\) (where we ignore the linear and constant terms).\n\nThe problem is then to determine the values of \\( unchangeable \\) for which there is a straight line that intersects \\( horizontalaxis=verticalaxis^{4}+(unchangeable-243 / 8) verticalaxis^{2} \\) in four distinct points. The result is now apparent from the shapes of the curves \\( horizontalaxis=verticalaxis^{4}+fixednumber verticalaxis^{2} \\). For example, when \\( fixednumber<0 \\), this \"W-shaped\" curve has a relative maximum at \\( verticalaxis=0 \\), so the horizontal lines \\( horizontalaxis=-largedeviation \\) for small positive \\( largedeviation \\) intersect the curve in four points, while for \\( fixednumber \\geq 0 \\), the curve is always concave upward, so no line can intersect it in more than two points; see Figure 28.\n\nSolution 2 (algebraic). We wish to know if we can choose \\( flatness \\) and \\( gradient \\) so that\n\\[\nq(verticalaxis)=verticalaxis^{4}+9 verticalaxis^{3}+unchangeable verticalaxis^{2}+9 verticalaxis+4-(flatness verticalaxis+gradient)\n\\]\nhas four distinct real solutions \\( finalletterone, finallettertwo, finalletterthree, finalletterfour \\). If we can find four distinct real numbers such that\n\\[\n\\begin{array}{c}\nfinalletterone+finallettertwo+finalletterthree+finalletterfour=-9 \\\\\nfinalletterone finallettertwo+finalletterone finalletterthree+finalletterone finalletterfour+finallettertwo finalletterthree+finallettertwo finalletterfour+finalletterthree finalletterfour=unchangeable\n\\end{array}\n\\]\nthen we can choose \\( flatness \\) and \\( gradient \\) appropriately so that \\( q(verticalaxis) \\) has these four zeros (by the expansion of \\( \\left.\\prod_{i=1}^{4}\\left(verticalaxis-finalletter_{i}\\right)\\right) \\).\n\nThen from\n\\[\n0<\\sum_{i<staticindex}\\left(finalletter_{i}-finalletter_{staticindex}\\right)^{2}=3\\left(finalletterone+finallettertwo+finalletterthree+finalletterfour\\right)^{2}-8\\;unchangeable\n\\]\nwe get \\( unchangeable<243 / 8 \\). Conversely, we must show that if \\( unchangeable<243 / 8 \\), then we can find distinct real \\( finalletterone, finallettertwo, finalletterthree, finalletterfour \\) satisfying (1) and (2). This is indeed possible; try \\( \\left(finalletterone, finallettertwo, finalletterthree, finalletterfour\\right)=(9 / 4-macrosize, 9 / 4+macrosize, 9 / 4-megavalue, 9 / 4+megavalue) \\) where \\( macrosize \\) and \\( megavalue \\) are distinct positive numbers. Then (1) is automatically satisfied, and we can choose \\( macrosize \\) and \\( megavalue \\) so that the right side of \\( (3) \\), which is \\( 8\\left(macrosize^{2}+megavalue^{2}\\right) \\), is any desired positive number."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "P": "kfhudmea",
        "\\alpha": "vqpkldse",
        "\\alpha_1": "vqpkldsea",
        "\\alpha_2": "vqpkldsfb",
        "\\alpha_3": "vqpkldsgc",
        "\\alpha_4": "vqpkldsjd",
        "\\mu": "rsewqlnm",
        "\\nu": "cmlgdxor",
        "\\epsilon": "wdnqjvkp",
        "i": "pxzealti",
        "j": "oqlsdnma",
        "c": "axmyehtr",
        "a": "gkzrlmnd",
        "m": "pvrldkqe",
        "b": "nhelfqaw"
      },
      "question": "For which real numbers $axmyehtr$ is there a straight line that intersects the curve\n\\[ qzxwvtnp^4 + 9qzxwvtnp^3 + axmyehtr qzxwvtnp^2 + 9qzxwvtnp + 4 \\]\nin four distinct points?",
      "solution": "Solution 1 (geometric). The constant and linear terms of\n\\[\nkfhudmea(qzxwvtnp)=qzxwvtnp^{4}+9 qzxwvtnp^{3}+axmyehtr qzxwvtnp^{2}+9 qzxwvtnp+4\n\\]\nare irrelevant to the problem; $ hjgrksla =kfhudmea(qzxwvtnp)$ meets the line $ hjgrksla =pvrldkqe qzxwvtnp+nhelfqaw$ in four points if and only if $ hjgrksla =kfhudmea(qzxwvtnp)+9 qzxwvtnp+4$ meets the line $ hjgrksla =(pvrldkqe+9) qzxwvtnp+(nhelfqaw+4)$ in four points.\n\nAlso, $ hjgrksla =kfhudmea(qzxwvtnp)$ meets the line $ hjgrksla =pvrldkqe qzxwvtnp+nhelfqaw$ in four points if and only if $ hjgrksla =kfhudmea(qzxwvtnp-vqpkldse)$ meets the line $ hjgrksla =pvrldkqe(qzxwvtnp-vqpkldse)+nhelfqaw$ in four points, so we may replace the given quartic with $ kfhudmea(qzxwvtnp-9 / 4)=qzxwvtnp^{4}+(axmyehtr-243 / 8) qzxwvtnp^{2}+\\cdots$ (where we ignore the linear and constant terms).\n\nThe problem is then to determine the values of $ axmyehtr$ for which there is a straight line that intersects $ hjgrksla =qzxwvtnp^{4}+(axmyehtr-243 / 8) qzxwvtnp^{2}$ in four distinct points. The result is now apparent from the shapes of the curves $ hjgrksla =qzxwvtnp^{4}+gkzrlmnd qzxwvtnp^{2}$. For example, when $ gkzrlmnd<0$, this \"W-shaped\" curve has a relative maximum at $ qzxwvtnp=0$, so the horizontal lines $ hjgrksla=-wdnqjvkp$ for small positive $ wdnqjvkp$ intersect the curve in four points, while for $ gkzrlmnd \\ge 0$, the curve is always concave upward, so no line can intersect it in more than two points; see Figure 28.\n\nSolution 2 (algebraic). We wish to know if we can choose $ pvrldkqe$ and $ nhelfqaw$ so that\n\\[\nq(qzxwvtnp)=qzxwvtnp^{4}+9 qzxwvtnp^{3}+axmyehtr qzxwvtnp^{2}+9 qzxwvtnp+4-(pvrldkqe qzxwvtnp+nhelfqaw)\n\\]\nhas four distinct real solutions $ vqpkldsea, vqpkldsfb, vqpkldsgc, vqpkldsjd$. If we can find four distinct real numbers such that\n\\[\n\\begin{array}{c}\nvqpkldsea+vqpkldsfb+vqpkldsgc+vqpkldsjd=-9 \\\\\nvqpkldsea vqpkldsfb+vqpkldsea vqpkldsgc+vqpkldsea vqpkldsjd+vqpkldsfb vqpkldsgc+vqpkldsfb vqpkldsjd+vqpkldsgc vqpkldsjd=axmyehtr\n\\end{array}\n\\]\nthen we can choose $ pvrldkqe$ and $ nhelfqaw$ appropriately so that $ q(qzxwvtnp)$ has these four zeros (by the expansion of $ \\prod_{pxzealti=1}^{4}(qzxwvtnp-vqpkldse_{pxzealti})$).\n\nThen from\n\\[\n0<\\sum_{pxzealti<oqlsdnma}(vqpkldse_{pxzealti}-vqpkldse_{oqlsdnma})^{2}=3(vqpkldsea+vqpkldsfb+vqpkldsgc+vqpkldsjd)^{2}-8 axmyehtr\n\\]\nwe get $ axmyehtr<243 / 8$. Conversely, we must show that if $ axmyehtr<243 / 8$, then we can find distinct real $ vqpkldsea, vqpkldsfb, vqpkldsgc, vqpkldsjd$ satisfying (1) and (2). This is indeed possible; try $ (vqpkldsea, vqpkldsfb, vqpkldsgc, vqpkldsjd)=(9 / 4-rsewqlnm, 9 / 4+rsewqlnm, 9 / 4-cmlgdxor, 9 / 4+cmlgdxor)$ where $ rsewqlnm$ and $ cmlgdxor$ are distinct positive numbers. Then (1) is automatically satisfied, and we can choose $ rsewqlnm$ and $ cmlgdxor$ so that the right side of (3), which is $ 8(rsewqlnm^{2}+cmlgdxor^{2})$, is any desired positive number."
    },
    "kernel_variant": {
      "question": "Let  \n\n  C_{p,q} :  y = x^{6} + p\\,x^{4} + q\\,x^{2},  p,q \\in \\mathbb R ,  \n\nbe the even sextic curve that depends on the real parameters p and q.  \n\nDetermine explicitly the set  \n\n  Ω = { (p , q) ∈ ℝ² : there exists a real number k for which the horizontal line y = k intersects C_{p,q} in six distinct real points }.  \n\n(Only horizontal lines are to be taken into account; vertical lines meet C_{p,q} in at most one point, and oblique non-vertical lines are irrelevant.)",
      "solution": "Fix a pair (p , q) ∈ ℝ² and look for a real k such that the equation  \n\n  x^{6} + p\\,x^{4} + q\\,x^{2} = k   (1)\n\npossesses six distinct real solutions.  \nBecause the left–hand side is even in x, equation (1) has 0, 2, 4, or 6 real roots; the last alternative occurs precisely when it has three distinct positive roots, whose negatives provide the other three.\n\nIntroduce the substitution  \n\n  t = x²   (t ≥ 0)\n\nand write  \n\n  φ_k(t) = t^{3} + p\\,t^{2} + q\\,t − k   (2)\n\nso that (1) ⇔ φ_k(t) = 0.  \nDefine  \n\n  F(t) := t^{3} + p\\,t^{2} + q\\,t   (3)\n\nso that φ_k(t) = F(t) − k.  \nEquation (1) has six distinct real solutions  ⇔ φ_k has three distinct positive zeros and φ′_k has no multiple root in common with φ_k.\n\nStep 1. Critical points of F  \nThe derivative  \n\n  F′(t) = 3t^{2} + 2p\\,t + q =: Q(t)   (4)\n\nis a quadratic, whose discriminant  \n\n  Δ = 4p² − 12q\n\ncontrols the number of positive critical points.  Denote the roots of Q by t₁ < t₂.\n\nQ has two distinct positive zeros  ⇔   \n\n  Δ > 0,  t₁ + t₂ = −2p/3 > 0,  t₁ t₂ = q/3 > 0.  (5)\n\nConsequently  \n\n  p < 0  and  0 < q < p²/3.   (N)\n\nHence (N) is necessary for F to have the required “rise–fall–rise’’ shape on (0,∞).\n\nStep 2. Values of F at its critical points  \nPut  \n\n  M := F(t₁),  m := F(t₂).   (6)\n\nBecause t₁ is a local maximum and t₂ a local minimum of F the integral\n\n  m − M = ∫_{t₁}^{t₂} Q(s) ds = 3∫_{t₁}^{t₂} (s-t₁)(s-t₂) ds < 0\n\nimplies  \n\n  M > m.   (7)\n\nWe now prove that the larger critical value is in fact positive.\n\nLemma.  Under condition (N) one has M = F(t₁) > 0.\n\nProof.  \nSince t₁ satisfies Q(t₁)=0,  \n\n  3t₁^{2} + 2p t₁ + q = 0 ⇒ q = −3t₁^{2} − 2p t₁.  (8)\n\nSubstituting (8) into (3) gives  \n\n  F(t₁) = t₁^{3} + p t₁^{2} + q t₁\n     = t₁^{3} + p t₁^{2} + t₁(−3t₁^{2} − 2p t₁)\n     = −2 t₁^{3} − p t₁^{2}\n     = −t₁^{2}(2t₁ + p).   (9)\n\nBecause t₁>0 and p<0, it suffices to show 2t₁ + p < 0.  \nUsing again Q(t₁)=0 and solving for 2t₁ + p we obtain  \n\n  2t₁ + p = (t₁^{2} - q)/(2t₁).   (10)\n\nWith q = 3t₁ t₂ (from (5)),\n\n  t₁^{2} - q = t₁(t₁ - 3t₂) < 0 (as t₂ > t₁ > 0).  \n\nHence 2t₁ + p < 0, and (9) yields M = F(t₁) > 0. ∎\n\nStep 3. Choosing the height k  \nLet (p , q) satisfy (N).  \nBecause M > m and M > 0, the open interval  \n\n  (m , M)  (11)\n\nis non-empty and contains positive numbers.  Pick any k with  \n\n  m < k < M and k > 0.  (12)\n\nObserve that\n\n  φ_k(0) = −k < 0,  φ_k(t₁) = M − k > 0,  φ_k(t₂) = m − k < 0,\n\nwhile  \n\n  lim_{t→∞} φ_k(t) = +∞.  (13)\n\nIntermediate–Value arguments now yield  \n• a root in (0 , t₁) (sign change − → +);  \n• a root in (t₁ , t₂) (+ → −);  \n• a root in (t₂ , ∞) (− → +∞).\n\nBecause k was chosen strictly between m and M and because Q(t)≠0 at those three zeros, all roots are simple and positive; reverting to x = ±√t gives six distinct real solutions of (1).  \nTherefore every pair (p , q) that fulfils (N) admits at least one horizontal line with six distinct intersections.\n\nStep 4. Necessity of (N)  \nConversely, suppose that some horizontal line y = k meets C_{p,q} in six distinct real points.  Then φ_k possesses three distinct positive zeros, whence φ′_k (= Q) has two distinct positive zeros by Rolle’s Theorem; conditions (5) follow and force (N).  Thus (N) is also necessary.\n\nStep 5. The parameter domain  \nCombining Steps 3 and 4 we reach the final description  \n\n  Ω = { (p , q) ∈ ℝ² : p < 0 and 0 < q < p²/3 }. □",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.737069",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher degree: The polynomial is sextic instead of quartic, so Bézout permits up to 6 intersections, doubling the combinatorial complexity.  \n2. Two independent parameters (p,q) must be classified, producing a two-dimensional answer set rather than a simple interval.  \n3. The solution requires calculus (critical-point analysis), algebra (discriminant/Vieta), and geometric reasoning about the global shape of even polynomials—several interacting techniques.  \n4. The necessary–sufficient proof demands both directions: analysing why six intersections force two positive critical points, and showing that these conditions indeed suffice by constructing an explicit line.  \n5. The final description is a curved region (a strict parabola-shaped inequality) in ℝ², markedly more intricate than the single inequality c < 243/8 in the original problem.\n\nThus the variant is substantially harder, involving higher degree, more variables, deeper analytic geometry, and a full classification in the parameter plane."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\n  C_{p,q} :  y = x^{6} + p\\,x^{4} + q\\,x^{2},  p,q \\in \\mathbb R ,  \n\nbe the even sextic curve that depends on the real parameters p and q.  \n\nDetermine explicitly the set  \n\n  Ω = { (p , q) ∈ ℝ² : there exists a real number k for which the horizontal line y = k intersects C_{p,q} in six distinct real points }.  \n\n(Only horizontal lines are to be taken into account; vertical lines meet C_{p,q} in at most one point, and oblique non-vertical lines are irrelevant.)",
      "solution": "Fix a pair (p , q) ∈ ℝ² and look for a real k such that the equation  \n\n  x^{6} + p\\,x^{4} + q\\,x^{2} = k   (1)\n\npossesses six distinct real solutions.  \nBecause the left–hand side is even in x, equation (1) has 0, 2, 4, or 6 real roots; the last alternative occurs precisely when it has three distinct positive roots, whose negatives provide the other three.\n\nIntroduce the substitution  \n\n  t = x²   (t ≥ 0)\n\nand write  \n\n  φ_k(t) = t^{3} + p\\,t^{2} + q\\,t − k   (2)\n\nso that (1) ⇔ φ_k(t) = 0.  \nDefine  \n\n  F(t) := t^{3} + p\\,t^{2} + q\\,t   (3)\n\nso that φ_k(t) = F(t) − k.  \nEquation (1) has six distinct real solutions  ⇔ φ_k has three distinct positive zeros and φ′_k has no multiple root in common with φ_k.\n\nStep 1. Critical points of F  \nThe derivative  \n\n  F′(t) = 3t^{2} + 2p\\,t + q =: Q(t)   (4)\n\nis a quadratic, whose discriminant  \n\n  Δ = 4p² − 12q\n\ncontrols the number of positive critical points.  Denote the roots of Q by t₁ < t₂.\n\nQ has two distinct positive zeros  ⇔   \n\n  Δ > 0,  t₁ + t₂ = −2p/3 > 0,  t₁ t₂ = q/3 > 0.  (5)\n\nConsequently  \n\n  p < 0  and  0 < q < p²/3.   (N)\n\nHence (N) is necessary for F to have the required “rise–fall–rise’’ shape on (0,∞).\n\nStep 2. Values of F at its critical points  \nPut  \n\n  M := F(t₁),  m := F(t₂).   (6)\n\nBecause t₁ is a local maximum and t₂ a local minimum of F the integral\n\n  m − M = ∫_{t₁}^{t₂} Q(s) ds = 3∫_{t₁}^{t₂} (s-t₁)(s-t₂) ds < 0\n\nimplies  \n\n  M > m.   (7)\n\nWe now prove that the larger critical value is in fact positive.\n\nLemma.  Under condition (N) one has M = F(t₁) > 0.\n\nProof.  \nSince t₁ satisfies Q(t₁)=0,  \n\n  3t₁^{2} + 2p t₁ + q = 0 ⇒ q = −3t₁^{2} − 2p t₁.  (8)\n\nSubstituting (8) into (3) gives  \n\n  F(t₁) = t₁^{3} + p t₁^{2} + q t₁\n     = t₁^{3} + p t₁^{2} + t₁(−3t₁^{2} − 2p t₁)\n     = −2 t₁^{3} − p t₁^{2}\n     = −t₁^{2}(2t₁ + p).   (9)\n\nBecause t₁>0 and p<0, it suffices to show 2t₁ + p < 0.  \nUsing again Q(t₁)=0 and solving for 2t₁ + p we obtain  \n\n  2t₁ + p = (t₁^{2} - q)/(2t₁).   (10)\n\nWith q = 3t₁ t₂ (from (5)),\n\n  t₁^{2} - q = t₁(t₁ - 3t₂) < 0 (as t₂ > t₁ > 0).  \n\nHence 2t₁ + p < 0, and (9) yields M = F(t₁) > 0. ∎\n\nStep 3. Choosing the height k  \nLet (p , q) satisfy (N).  \nBecause M > m and M > 0, the open interval  \n\n  (m , M)  (11)\n\nis non-empty and contains positive numbers.  Pick any k with  \n\n  m < k < M and k > 0.  (12)\n\nObserve that\n\n  φ_k(0) = −k < 0,  φ_k(t₁) = M − k > 0,  φ_k(t₂) = m − k < 0,\n\nwhile  \n\n  lim_{t→∞} φ_k(t) = +∞.  (13)\n\nIntermediate–Value arguments now yield  \n• a root in (0 , t₁) (sign change − → +);  \n• a root in (t₁ , t₂) (+ → −);  \n• a root in (t₂ , ∞) (− → +∞).\n\nBecause k was chosen strictly between m and M and because Q(t)≠0 at those three zeros, all roots are simple and positive; reverting to x = ±√t gives six distinct real solutions of (1).  \nTherefore every pair (p , q) that fulfils (N) admits at least one horizontal line with six distinct intersections.\n\nStep 4. Necessity of (N)  \nConversely, suppose that some horizontal line y = k meets C_{p,q} in six distinct real points.  Then φ_k possesses three distinct positive zeros, whence φ′_k (= Q) has two distinct positive zeros by Rolle’s Theorem; conditions (5) follow and force (N).  Thus (N) is also necessary.\n\nStep 5. The parameter domain  \nCombining Steps 3 and 4 we reach the final description  \n\n  Ω = { (p , q) ∈ ℝ² : p < 0 and 0 < q < p²/3 }. □",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.570652",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher degree: The polynomial is sextic instead of quartic, so Bézout permits up to 6 intersections, doubling the combinatorial complexity.  \n2. Two independent parameters (p,q) must be classified, producing a two-dimensional answer set rather than a simple interval.  \n3. The solution requires calculus (critical-point analysis), algebra (discriminant/Vieta), and geometric reasoning about the global shape of even polynomials—several interacting techniques.  \n4. The necessary–sufficient proof demands both directions: analysing why six intersections force two positive critical points, and showing that these conditions indeed suffice by constructing an explicit line.  \n5. The final description is a curved region (a strict parabola-shaped inequality) in ℝ², markedly more intricate than the single inequality c < 243/8 in the original problem.\n\nThus the variant is substantially harder, involving higher degree, more variables, deeper analytic geometry, and a full classification in the parameter plane."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}