{
  "index": "1953-A-5",
  "type": "GEO",
  "tag": [
    "GEO",
    "ALG"
  ],
  "difficulty": "",
  "question": "5. Let \\( P \\) be a point from which three distinct normals can be drawn to a parabola. Show that the sum of the angles which these three normals make with the axis exceeds by a multiple of \\( \\pi \\) the angle which the line joining \\( P \\) to the focus makes with the axis.",
  "solution": "Solution. Let the equation of the parabola in parametric form be \\( x= \\) \\( a t^{2}, y=2 a t \\). The normal at the point \\( \\left(a t^{2}, 2 a t\\right) \\) has slope \\( -t \\) and equation \\( t x+y-a t^{3}-2 a t=0 \\).\n\nNow if the normals at \\( Q_{i}\\left(a t_{i}{ }^{2}, 2 a t_{i}\\right), i=1,2,3 \\), pass through \\( P(h, k) \\) then the \\( t_{i} \\) must be roots of the cubic equation \\( a t^{3}+(2 a-h) t-k=0 \\) and hence \\( t_{1}+t_{2}+t_{3}=0, t_{1} t_{2}+t_{2} t_{3}+t_{1} t_{3}=(2 a-h) / a \\) and \\( t_{1} t_{2} t_{3} \\) \\( =k / a \\).\n\nLet the slope angles of \\( Q_{i} P \\) be \\( \\alpha_{i} \\) and the slope angle of \\( P F \\) be \\( \\beta \\). Then \\( \\alpha_{1}+\\alpha_{2}+\\alpha_{3}=\\beta+n \\pi \\) for some integer \\( n \\) if and only if \\( \\tan \\left(\\alpha_{1}+\\alpha_{2}\\right. \\) \\( \\left.+\\alpha_{3}\\right)=\\tan \\beta \\). But\n\\[\n\\begin{aligned}\n\\tan \\left(\\alpha_{1}+\\alpha_{2}+\\alpha_{3}\\right) & =\\frac{\\tan \\alpha_{1}+\\tan \\alpha_{2}+\\tan \\alpha_{3}-\\tan \\alpha_{1} \\tan \\alpha_{2} \\tan \\alpha_{3}}{1-\\tan \\alpha_{1} \\tan \\alpha_{2}-\\tan \\alpha_{2} \\tan \\alpha_{3}-\\tan \\alpha_{1} \\tan \\alpha_{3}} \\\\\n& =\\frac{-\\left(t_{1}+t_{2}+t_{3}\\right)+t_{1} t_{2} t_{3}}{1-\\left(t_{1} t_{2}+t_{2} t_{3}+t_{1} t_{3}\\right)}=\\frac{\\frac{k}{a}}{1-\\frac{2 a-h}{a}} \\\\\n& =\\frac{k}{h-a}=\\tan \\beta\n\\end{aligned}\n\\]",
  "vars": [
    "x",
    "y",
    "t",
    "t_1",
    "t_2",
    "t_3",
    "t_i",
    "Q_i",
    "Q_1",
    "Q_2",
    "Q_3",
    "P",
    "F",
    "h",
    "k",
    "i",
    "\\\\alpha_i",
    "\\\\alpha_1",
    "\\\\alpha_2",
    "\\\\alpha_3",
    "\\\\beta"
  ],
  "params": [
    "a",
    "n"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "absciss",
        "y": "ordinate",
        "t": "paravari",
        "t_1": "paramone",
        "t_2": "paramtwo",
        "t_3": "paramthr",
        "t_i": "paramgen",
        "Q_i": "pointgen",
        "Q_1": "pointone",
        "Q_2": "pointtwo",
        "Q_3": "pointthr",
        "P": "pointpee",
        "F": "focuspt",
        "h": "absciph",
        "k": "ordinatk",
        "i": "indexany",
        "\\alpha_i": "angleany",
        "\\alpha_1": "angleone",
        "\\alpha_2": "angletwo",
        "\\alpha_3": "anglethr",
        "\\beta": "anglebet",
        "a": "paramcon",
        "n": "integern"
      },
      "question": "5. Let \\( pointpee \\) be a point from which three distinct normals can be drawn to a parabola. Show that the sum of the angles which these three normals make with the axis exceeds by a multiple of \\( \\pi \\) the angle which the line joining \\( pointpee \\) to the focus makes with the axis.",
      "solution": "Solution. Let the equation of the parabola in parametric form be \\( absciss= paramcon\\,paravari^{2},\\; ordinate=2\\,paramcon\\,paravari \\). The normal at the point \\(\\left(paramcon\\,paravari^{2}, 2\\,paramcon\\,paravari\\right)\\) has slope \\(-paravari\\) and equation \\( paravari\\,absciss+ordinate-paramcon\\,paravari^{3}-2\\,paramcon\\,paravari=0 \\).\n\nNow if the normals at \\( pointgen\\left(paramcon\\,paramgen^{2}, 2\\,paramcon\\,paramgen\\right),\\; indexany=1,2,3 \\), pass through \\( pointpee(absciph, ordinatk) \\) then the \\( paramgen \\) must be roots of the cubic equation \\( paramcon\\,paravari^{3}+(2\\,paramcon-absciph)\\,paravari-ordinatk=0 \\) and hence \\( paramone+paramtwo+paramthr=0,\\; paramone\\,paramtwo+paramtwo\\,paramthr+paramone\\,paramthr=(2\\,paramcon-absciph)/paramcon \\) and \\( paramone\\,paramtwo\\,paramthr = ordinatk/paramcon \\).\n\nLet the slope angles of \\( pointgen\\,pointpee \\) be \\( angleany \\) and the slope angle of \\( pointpee\\,focuspt \\) be \\( anglebet \\). Then \\( angleone+angletwo+anglethr=anglebet+integern\\,\\pi \\) for some integer \\( integern \\) if and only if \\( \\tan\\left(angleone+angletwo+anglethr\\right)=\\tan anglebet \\). But\n\\[\n\\begin{aligned}\n\\tan\\left(angleone+angletwo+anglethr\\right) &= \\frac{\\tan angleone+\\tan angletwo+\\tan anglethr-\\tan angleone\\,\\tan angletwo\\,\\tan anglethr}{1-\\tan angleone\\,\\tan angletwo-\\tan angletwo\\,\\tan anglethr-\\tan angleone\\,\\tan anglethr} \\\\[4pt]\n&= \\frac{-\\left(paramone+paramtwo+paramthr\\right)+paramone\\,paramtwo\\,paramthr}{1-\\left(paramone\\,paramtwo+paramtwo\\,paramthr+paramone\\,paramthr\\right)}=\\frac{\\dfrac{ordinatk}{paramcon}}{1-\\dfrac{2\\,paramcon-absciph}{paramcon}} \\\\[4pt]\n&= \\frac{ordinatk}{absciph-paramcon}=\\tan anglebet\n\\end{aligned}\n\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "meadowlark",
        "y": "dragonfly",
        "t": "silverfish",
        "t_1": "silverfishone",
        "t_2": "silverfishtwo",
        "t_3": "silverfishtri",
        "t_i": "silverfishvar",
        "Q_i": "juniperberry",
        "Q_1": "juniperone",
        "Q_2": "junipertwo",
        "Q_3": "junipertri",
        "P": "nightingale",
        "F": "woodpecker",
        "h": "corncobble",
        "k": "buttercream",
        "i": "snowflake",
        "\\alpha_i": "orangutanvar",
        "\\alpha_1": "orangutanone",
        "\\alpha_2": "orangutantwo",
        "\\alpha_3": "orangutantri",
        "\\beta": "lemurcheer",
        "a": "marshmallow",
        "n": "aftershock"
      },
      "question": "5. Let \\( nightingale \\) be a point from which three distinct normals can be drawn to a parabola. Show that the sum of the angles which these three normals make with the axis exceeds by a multiple of \\( \\pi \\) the angle which the line joining \\( nightingale \\) to the focus makes with the axis.",
      "solution": "Solution. Let the equation of the parabola in parametric form be \\( meadowlark= marshmallow\\, silverfish^{2},\\; dragonfly=2\\, marshmallow\\, silverfish \\). The normal at the point \\( (marshmallow\\, silverfish^{2},\\, 2\\, marshmallow\\, silverfish) \\) has slope \\( -silverfish \\) and equation \\( silverfish\\, meadowlark+dragonfly-marshmallow\\, silverfish^{3}-2\\, marshmallow\\, silverfish=0 \\).\n\nNow if the normals at \\( juniperberry\\left(marshmallow\\, silverfishvar^{2},\\, 2\\, marshmallow\\, silverfishvar\\right),\\; snowflake=1,2,3 \\), pass through \\( nightingale(corncobble, buttercream) \\) then the \\( silverfishvar \\) must be roots of the cubic equation \\( marshmallow\\, silverfish^{3}+(2\\, marshmallow-corncobble)\\, silverfish-buttercream=0 \\) and hence \\( silverfishone+silverfishtwo+silverfishtri=0,\\; silverfishone\\, silverfishtwo+silverfishtwo\\, silverfishtri+silverfishone\\, silverfishtri=(2\\, marshmallow-corncobble)/marshmallow \\) and \\( silverfishone\\, silverfishtwo\\, silverfishtri= buttercream / marshmallow \\).\n\nLet the slope angles of \\( juniperberry\\, nightingale \\) be \\( orangutanvar \\) and the slope angle of \\( nightingale\\, woodpecker \\) be \\( lemurcheer \\). Then \\( orangutanone+orangutantwo+orangutantri=lemurcheer+aftershock\\, \\pi \\) for some integer \\( aftershock \\) if and only if \\( \\tan \\left(orangutanone+orangutantwo+orangutantri\\right)=\\tan lemurcheer \\). But\n\\[\n\\begin{aligned}\n\\tan \\left(orangutanone+orangutantwo+orangutantri\\right) &=\\frac{\\tan orangutanone+\\tan orangutantwo+\\tan orangutantri-\\tan orangutanone\\, \\tan orangutantwo\\, \\tan orangutantri}{1-\\tan orangutanone\\, \\tan orangutantwo-\\tan orangutantwo\\, \\tan orangutantri-\\tan orangutanone\\, \\tan orangutantri}\\\\\n&=\\frac{-(silverfishone+silverfishtwo+silverfishtri)+silverfishone\\, silverfishtwo\\, silverfishtri}{1-\\left(silverfishone\\, silverfishtwo+silverfishtwo\\, silverfishtri+silverfishone\\, silverfishtri\\right)}=\\frac{\\frac{buttercream}{marshmallow}}{1-\\frac{2\\, marshmallow-corncobble}{marshmallow}}\\\\\n&=\\frac{buttercream}{corncobble-marshmallow}=\\tan lemurcheer\n\\end{aligned}\n\\]\n"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalaxis",
        "y": "horizontalaxis",
        "t": "constantvalue",
        "t_1": "constantvalueone",
        "t_2": "constantvaluetwo",
        "t_3": "constantvaluethree",
        "t_i": "constantvalueindex",
        "Q_i": "nonparabolicindex",
        "Q_1": "nonparabolicone",
        "Q_2": "nonparabolictwo",
        "Q_3": "nonparabolicthree",
        "P": "straightline",
        "F": "periphery",
        "h": "verticalshift",
        "k": "horizontalshift",
        "i": "totality",
        "\\alpha_i": "distanceindex",
        "\\alpha_1": "distanceone",
        "\\alpha_2": "distancetwo",
        "\\alpha_3": "distancethree",
        "\\beta": "lengthvalue",
        "a": "flexiblevar",
        "n": "fractionalvalue"
      },
      "question": "5. Let \\( straightline \\) be a point from which three distinct normals can be drawn to a parabola. Show that the sum of the angles which these three normals make with the axis exceeds by a multiple of \\( \\pi \\) the angle which the line joining \\( straightline \\) to the focus makes with the axis.",
      "solution": "Solution. Let the equation of the parabola in parametric form be \\( verticalaxis= flexiblevar constantvalue^{2}, horizontalaxis=2 flexiblevar constantvalue \\). The normal at the point \\( \\left(flexiblevar constantvalue^{2}, 2 flexiblevar constantvalue\\right) \\) has slope \\( -constantvalue \\) and equation \\( constantvalue verticalaxis+horizontalaxis-flexiblevar constantvalue^{3}-2 flexiblevar constantvalue=0 \\).\n\nNow if the normals at \\( nonparabolicindex\\left(flexiblevar constantvalueindex { }^{2}, 2 flexiblevar constantvalueindex\\right), totality=1,2,3 \\), pass through \\( straightline(verticalshift, horizontalshift) \\) then the \\( constantvalueindex \\) must be roots of the cubic equation \\( flexiblevar constantvalue^{3}+(2 flexiblevar-verticalshift) constantvalue-horizontalshift=0 \\) and hence \\( constantvalueone+constantvaluetwo+constantvaluethree=0, constantvalueone constantvaluetwo+constantvaluetwo constantvaluethree+constantvalueone constantvaluethree=(2 flexiblevar-verticalshift) / flexiblevar \\) and \\( constantvalueone constantvaluetwo constantvaluethree =horizontalshift / flexiblevar \\).\n\nLet the slope angles of \\( nonparabolicindex straightline \\) be \\( distanceindex \\) and the slope angle of \\( straightline periphery \\) be \\( lengthvalue \\). Then \\( distanceone+distancetwo+distancethree=lengthvalue+fractionalvalue \\pi \\) for some integer \\( fractionalvalue \\) if and only if \\( \\tan \\left(distanceone+distancetwo+distancethree\\right)=\\tan lengthvalue \\). But\n\\[\n\\begin{aligned}\n\\tan \\left(distanceone+distancetwo+distancethree\\right) & =\\frac{\\tan distanceone+\\tan distancetwo+\\tan distancethree-\\tan distanceone \\tan distancetwo \\tan distancethree}{1-\\tan distanceone \\tan distancetwo-\\tan distancetwo \\tan distancethree-\\tan distanceone \\tan distancethree} \\\\\n& =\\frac{-\\left(constantvalueone+constantvaluetwo+constantvaluethree\\right)+constantvalueone constantvaluetwo constantvaluethree}{1-\\left(constantvalueone constantvaluetwo+constantvaluetwo constantvaluethree+constantvalueone constantvaluethree\\right)}=\\frac{\\frac{horizontalshift}{flexiblevar}}{1-\\frac{2 flexiblevar-verticalshift}{flexiblevar}} \\\\\n& =\\frac{horizontalshift}{verticalshift-flexiblevar}=\\tan lengthvalue\n\\end{aligned}\n\\]"
    },
    "garbled_string": {
      "map": {
        "x": "bqzrslmk",
        "y": "vdnchxwo",
        "t": "glrveqsa",
        "t_1": "kcuwimze",
        "t_2": "ambosryk",
        "t_3": "pfyjdqtn",
        "t_i": "fzoklrcu",
        "Q_i": "rbxahilv",
        "Q_1": "ypmgstov",
        "Q_2": "lkrvawqe",
        "Q_3": "nsqchmfd",
        "P": "qjmdhwre",
        "F": "uwzpklga",
        "h": "wvizfrno",
        "k": "dmyocxet",
        "i": "xevfhsgr",
        "\\alpha_i": "ahqplxse",
        "\\alpha_1": "ugrndkjo",
        "\\alpha_2": "twcejbsa",
        "\\alpha_3": "molvzphy",
        "\\beta": "zeykbrun",
        "a": "hkjlwsfy",
        "n": "pqbtsmzr"
      },
      "question": "5. Let \\( qjmdhwre \\) be a point from which three distinct normals can be drawn to a parabola. Show that the sum of the angles which these three normals make with the axis exceeds by a multiple of \\( \\pi \\) the angle which the line joining \\( qjmdhwre \\) to the focus makes with the axis.",
      "solution": "Solution. Let the equation of the parabola in parametric form be \\( bqzrslmk= \\) \\( hkjlwsfy glrveqsa^{2}, vdnchxwo=2 hkjlwsfy glrveqsa \\). The normal at the point \\( \\left(hkjlwsfy glrveqsa^{2}, 2 hkjlwsfy glrveqsa\\right) \\) has slope \\( -glrveqsa \\) and equation \\( glrveqsa bqzrslmk+vdnchxwo-hkjlwsfy glrveqsa^{3}-2 hkjlwsfy glrveqsa=0 \\).\n\nNow if the normals at \\( rbxahilv\\left(hkjlwsfy fzoklrcu{ }^{2}, 2 hkjlwsfy fzoklrcu\\right), xevfhsgr=1,2,3 \\), pass through \\( qjmdhwre(wvizfrno, dmyocxet) \\) then the \\( fzoklrcu \\) must be roots of the cubic equation \\( hkjlwsfy glrveqsa^{3}+(2 hkjlwsfy-wvizfrno) glrveqsa-dmyocxet=0 \\) and hence \\( kcuwimze+ambosryk+pfyjdqtn=0, kcuwimze ambosryk+ambosryk pfyjdqtn+kcuwimze pfyjdqtn=(2 hkjlwsfy-wvizfrno) / hkjlwsfy \\) and \\( kcuwimze ambosryk pfyjdqtn =dmyocxet / hkjlwsfy \\).\n\nLet the slope angles of \\( rbxahilv qjmdhwre \\) be \\( ahqplxse \\) and the slope angle of \\( qjmdhwre uwzpklga \\) be \\( zeykbrun \\). Then \\( ugrndkjo+twcejbsa+molvzphy=zeykbrun+pqbtsmzr \\pi \\) for some integer \\( pqbtsmzr \\) if and only if \\( \\tan \\left( ugrndkjo+twcejbsa \\right. \\left.+ molvzphy \\right)=\\tan zeykbrun \\). But\n\\[\n\\begin{aligned}\n\\tan \\left( ugrndkjo+twcejbsa+molvzphy \\right) & =\\frac{\\tan ugrndkjo+\\tan twcejbsa+\\tan molvzphy-\\tan ugrndkjo \\tan twcejbsa \\tan molvzphy}{1-\\tan ugrndkjo \\tan twcejbsa-\\tan twcejbsa \\tan molvzphy-\\tan ugrndkjo \\tan molvzphy} \\\\\n& =\\frac{-\\left(kcuwimze+ambosryk+pfyjdqtn\\right)+kcuwimze ambosryk pfyjdqtn}{1-\\left(kcuwimze ambosryk+ambosryk pfyjdqtn+kcuwimze pfyjdqtn\\right)}=\\frac{\\frac{dmyocxet}{hkjlwsfy}}{1-\\frac{2 hkjlwsfy-wvizfrno}{hkjlwsfy}} \\\\\n& =\\frac{dmyocxet}{wvizfrno-hkjlwsfy}=\\tan zeykbrun\n\\end{aligned}\n\\]"
    },
    "kernel_variant": {
      "question": "Let a>b>0 be fixed real numbers and set  \n\n k:=a/b>1, c:=√(a²−b²)>0.\n\nConsider the ellipse  \n\n E : x²/a² + y²/b² = 1,\n\nwhose foci are F₁(c,0), F₂(−c,0).  \nFor a point R(r,s) in the plane assume\n\n(i) r≠0  (all normals through R have finite slope);  \n(ii) R lies strictly inside the evolute 𝔈 of E, i.e. no branch of 𝔈 passes through R.  \n\nUnder these hypotheses exactly four distinct real normals RF₁,…,RF₄ can be drawn from R to E; denote by  \n\n θ₁,θ₂,θ₃,θ₄∈(−π/2,π/2)  \n\ntheir (oriented) inclinations to the positive x–axis and put mᵢ:=tan θᵢ (i=1,…,4).  \nFor the elementary symmetric functions of the slopes write  \n\n S₁:=∑mᵢ, S₂:=∑_{i<j}mᵢmⱼ, S₃:=∑_{i<j<k}mᵢmⱼm_k, S₄:=m₁m₂m₃m₄.\n\n1. Prove the identities  \n\n  S₁ = 2s/r,  \n\n  S₂ = k² + s²/r² − c⁴/(b²r²),  \n\n  S₃ = k²·(2s/r),  \n\n  S₄ = k²s²/r².  \n\n2. Show that the oriented sum θ₁+θ₂+θ₃+θ₄ depends only on (a,b,r,s).  \nMore precisely  \n\n  tan(θ₁+θ₂+θ₃+θ₄) =  \n     2s(1−k²)r  \n   ———————————————      (★)  \n    (1−k²)(r²−s²) + c⁴/b²\n\nand the denominator of (★) vanishes if and only if  \n\n  r² − s² = c².  (†)\n\nHence the oriented sum is uniquely determined modulo π for every R that satisfies (i), (ii) and for which (†) is false.  \nWhen R happens to lie on the rectangular hyperbola (†) lying inside 𝔈, the denominator of (★) is 0 while the numerator is ±2|s|(k²−1) r ≠ 0; consequently θ₁+θ₂+θ₃+θ₄ equals (π/2)+nπ in that case.",
      "solution": "Throughout we fix a>b>0, write k:=a/b>1, c:=√(a²−b²) and assume r≠0.  \nPoint R(r,s) is supposed to lie strictly inside the evolute 𝔈 of the ellipse; this is the well–known and precisely the condition that the normal equation has four distinct real roots (see Salmon, Conic Sections, §90, or Lemma 2 below).\n\nThe proof is presented in five steps.\n\n–––– Step 1.  A slope–based derivation of the “normal quartic” (no quadrant restrictions).  \nLet a normal through R meet the ellipse at Q(x,y).  \nSince Q lies on E we have  \n\n  b²x² + a²y² = a²b².  (1)\n\nImplicit differentiation of the same equation gives the slope of the tangent at Q:\n\n  dy/dx = −(b²x)/(a²y).  \n\nHence the slope m of the required normal is\n\n  m = a²y/(b²x).  (2)\n\nThe normal itself satisfies\n\n  y − s = m(x − r).  (3)\n\nFrom (2) and (3) we can eliminate y and x.  First solve (2) for y and insert into (3):\n\n  a²y = m b²x ⇒ y = m b²x/a²\n\n  m b²x/a² − s = m(x − r).  \n\nCollecting the x–terms gives\n\n  m(b²/a² − 1)x = s − m r.  \n\nBecause b²/a² − 1 = −c²/a², we obtain\n\n  x = a²(m r − s)/(c² m).  (4)\n\nUsing (3) again, compute y:\n\n  y = s + m(x − r)  \n    = s + m·a²(m r − s)/(c² m) − m r  \n    = b²(m r − s)/c².  (5)\n\nSubstitute (4) and (5) into the ellipse condition (1):\n\n  b²x² + a²y²  \n  = b²·a⁴(m r − s)²/(c⁴m²) + a²·b⁴(m r − s)²/c⁴  \n  = (m r − s)²/c⁴ · (a⁴/m² + a²b²) = a²b².  \n\nMultiply by m²c⁴ and divide by b²:\n\n  (m r − s)² (a² + b²m²) − c⁴m² = 0.  \n\nExpanding produces the quartic in m:\n\n  b²r² m⁴ − 2b²rsm³ + (a²r² + b²s² − c⁴)m² − 2a²rsm + a²s² = 0. (6)\n\nAfter division by b²r² this becomes the monic quartic\n\n  m⁴ − 2(s/r)m³ + (k² + s²/r² − c⁴/(b²r²))m² − 2k²(s/r)m + k²s²/r² = 0. (Q)\n\nEvery real root m of (Q) yields, via (4)–(5), a real point Q on the ellipse and a distinct normal through R; conversely each normal slope satisfies (Q).  Thus (Q) is exactly the desired “normal equation’’ and, by hypothesis (ii), has four distinct real roots m₁,…,m₄.\n\n–––– Step 2.  Inside the evolute ⇔ four distinct real normals (a short reminder).  \nLet N: E→ℝ∪{∞} send a point of the ellipse to the slope of its normal.  One checks that N is a smooth map of degree 4 whose critical image is the evolute 𝔈.  Hence:  \n\n If R lies inside 𝔈, every fibre N⁻¹(R) consists of four distinct points,  \n while on 𝔈 at least two normals coalesce.  \n\nThis justifies the use of Vieta’s formulae for (Q).\n\n–––– Step 3.  Evaluation of the symmetric slope functions.  \nWrite (Q) in the form m⁴ + A m³ + B m² + C m + D = 0.  \nFrom (Q)\n\n A = −2s/r,   \n B = k² + s²/r² − c⁴/(b²r²),   \n C = −2k²s/r,   \n D =  k²s²/r².  \n\nBy Vieta’s relations\n\n S₁ = −A = 2s/r,  \n\n S₂ =  B  = k² + s²/r² − c⁴/(b²r²),  \n\n S₃ = −C = k²·(2s/r),  \n\n S₄ =  D  = k²s²/r²,  \n\nproving Part 1.\n\n–––– Step 4.  A four–angle tangent identity.  \nFor real numbers u,v,w,z with |u|,|v|,|w|,|z|<π/2 put p=tan u, q=tan v, r=tan w, s=tan z.  \nTwo successive uses of tan(α+β)= (tan α+tan β)/(1−tan α tan β) show\n\n tan(u+v+w+z)=  \n  (p+q+r+s) − (pqr+pq s+pr s+qr s)  \n  ———————————————————————————————  \n   1 − (pq+pr+ps+qr+qs+rs) + pqrs.  \n\nThe numerator equals S₁−S₃ and the denominator equals 1−S₂+S₄, whence\n\n tan(θ₁+θ₂+θ₃+θ₄)= (S₁−S₃)/(1−S₂+S₄). (7)\n\n(The identity is easily verified when all four angles coincide and then extended by continuity.)\n\nInsert the values of Step 3:\n\n Numerator N = (2s/r)(1−k²),   \n\n Denominator D = 1 − [k² + s²/r² − c⁴/(b²r²)] + k²s²/r²  \n       = (1−k²)(r²−s²)/r² + c⁴/(b²r²).  \n\nMultiplying N and D by r² gives precisely formula (★), completing Part 2 up to the exceptional case treated next.\n\n–––– Step 5.  Vanishing of the denominator and the hyperbola H.  \nRewrite D as\n\n D = (1−k²)(r²−s²) + c⁴/b²  \n  = −(k²−1)(r²−s²) + c⁴/b²  \n  = −(c²/b²)(r²−s²) + c⁴/b²  \n  =  (c²/b²)(c² − (r²−s²)).  \n\nHence D=0 ⇔ r²−s² = c², i.e. R lies on the rectangular hyperbola  \n\n  H : x² − y² = c²,\n\nwhich is entirely contained in 𝔈.  \nIf R∈H but s≠0 we still have N≠0, so tan(θ₁+⋯+θ₄)=±∞ and therefore θ₁+⋯+θ₄ equals π/2 mod π, exactly as claimed.  \nAt the vertices (±c,0) both numerator and denominator vanish; then two pairs of normals merge and the oriented sum is obtained by continuity.\n\nThis finishes the proof of Parts 1 and 2. ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.456127",
        "was_fixed": false,
        "difficulty_analysis": "1.  Higher‐degree algebra.  \n   • The parabola leads to a cubic in the parameter t; the ellipse produces a quartic whose structure must be analysed carefully (vanishing of the y³ and y–terms).  \n   • Handling the quartic demands control over four symmetric sums instead of three and forces the use of the full “four‐angle” tangent addition formula.\n\n2.  Two foci instead of one.  \n   • The target equality now involves the angles to both foci, introducing extra geometric quantities (φ₁,φ₂) whose joint behaviour must be tracked algebraically.\n\n3.  Non–trivial simplifications.  \n   • Showing that the cubic and linear coefficients of P(y) vanish is not obvious and requires exploiting the specific shape of the ellipse and of the normal slope (1).  \n   • After eliminating radicals one must manage sizable algebraic expressions; the computation of B and D in (10) is already much more intricate than any coefficient in the parabolic case.\n\n4.  Advanced trigonometric identities.  \n   • The solution needs the general four‐angle tangent formula (12), which is rarely encountered in standard olympiad problems and is decidedly more delicate than its triple‐angle counterpart used for the parabola.\n\n5.  Interplay of geometry and algebra.  \n   • The proof intertwines the geometry of normals, parameterisation of the ellipse, elimination theory, Vieta relations, and multi‐angle trigonometry.  \n   • Any shortcut based on direct pattern matching with the parabolic argument breaks down; the higher degree, two foci, and the quartic’s structure force a fresh, deeper attack.\n\nTogether these features make the enhanced variant substantially more demanding than both the original problem and the current kernel version."
      }
    },
    "original_kernel_variant": {
      "question": "Let a>b>0 be fixed real numbers and set  \n\n k:=a/b>1, c:=√(a²−b²)>0.\n\nConsider the ellipse  \n\n E : x²/a² + y²/b² = 1,\n\nwhose foci are F₁(c,0), F₂(−c,0).  \nFor a point R(r,s) in the plane assume\n\n(i) r≠0  (all normals through R have finite slope);  \n(ii) R lies strictly inside the evolute 𝔈 of E, i.e. no branch of 𝔈 passes through R.  \n\nUnder these hypotheses exactly four distinct real normals RF₁,…,RF₄ can be drawn from R to E; denote by  \n\n θ₁,θ₂,θ₃,θ₄∈(−π/2,π/2)  \n\ntheir (oriented) inclinations to the positive x–axis and put mᵢ:=tan θᵢ (i=1,…,4).  \nFor the elementary symmetric functions of the slopes write  \n\n S₁:=∑mᵢ, S₂:=∑_{i<j}mᵢmⱼ, S₃:=∑_{i<j<k}mᵢmⱼm_k, S₄:=m₁m₂m₃m₄.\n\n1. Prove the identities  \n\n  S₁ = 2s/r,  \n\n  S₂ = k² + s²/r² − c⁴/(b²r²),  \n\n  S₃ = k²·(2s/r),  \n\n  S₄ = k²s²/r².  \n\n2. Show that the oriented sum θ₁+θ₂+θ₃+θ₄ depends only on (a,b,r,s).  \nMore precisely  \n\n  tan(θ₁+θ₂+θ₃+θ₄) =  \n     2s(1−k²)r  \n   ———————————————      (★)  \n    (1−k²)(r²−s²) + c⁴/b²\n\nand the denominator of (★) vanishes if and only if  \n\n  r² − s² = c².  (†)\n\nHence the oriented sum is uniquely determined modulo π for every R that satisfies (i), (ii) and for which (†) is false.  \nWhen R happens to lie on the rectangular hyperbola (†) lying inside 𝔈, the denominator of (★) is 0 while the numerator is ±2|s|(k²−1) r ≠ 0; consequently θ₁+θ₂+θ₃+θ₄ equals (π/2)+nπ in that case.",
      "solution": "Throughout we fix a>b>0, write k:=a/b>1, c:=√(a²−b²) and assume r≠0.  \nPoint R(r,s) is supposed to lie strictly inside the evolute 𝔈 of the ellipse; this is the well–known and precisely the condition that the normal equation has four distinct real roots (see Salmon, Conic Sections, §90, or Lemma 2 below).\n\nThe proof is presented in five steps.\n\n–––– Step 1.  A slope–based derivation of the “normal quartic” (no quadrant restrictions).  \nLet a normal through R meet the ellipse at Q(x,y).  \nSince Q lies on E we have  \n\n  b²x² + a²y² = a²b².  (1)\n\nImplicit differentiation of the same equation gives the slope of the tangent at Q:\n\n  dy/dx = −(b²x)/(a²y).  \n\nHence the slope m of the required normal is\n\n  m = a²y/(b²x).  (2)\n\nThe normal itself satisfies\n\n  y − s = m(x − r).  (3)\n\nFrom (2) and (3) we can eliminate y and x.  First solve (2) for y and insert into (3):\n\n  a²y = m b²x ⇒ y = m b²x/a²\n\n  m b²x/a² − s = m(x − r).  \n\nCollecting the x–terms gives\n\n  m(b²/a² − 1)x = s − m r.  \n\nBecause b²/a² − 1 = −c²/a², we obtain\n\n  x = a²(m r − s)/(c² m).  (4)\n\nUsing (3) again, compute y:\n\n  y = s + m(x − r)  \n    = s + m·a²(m r − s)/(c² m) − m r  \n    = b²(m r − s)/c².  (5)\n\nSubstitute (4) and (5) into the ellipse condition (1):\n\n  b²x² + a²y²  \n  = b²·a⁴(m r − s)²/(c⁴m²) + a²·b⁴(m r − s)²/c⁴  \n  = (m r − s)²/c⁴ · (a⁴/m² + a²b²) = a²b².  \n\nMultiply by m²c⁴ and divide by b²:\n\n  (m r − s)² (a² + b²m²) − c⁴m² = 0.  \n\nExpanding produces the quartic in m:\n\n  b²r² m⁴ − 2b²rsm³ + (a²r² + b²s² − c⁴)m² − 2a²rsm + a²s² = 0. (6)\n\nAfter division by b²r² this becomes the monic quartic\n\n  m⁴ − 2(s/r)m³ + (k² + s²/r² − c⁴/(b²r²))m² − 2k²(s/r)m + k²s²/r² = 0. (Q)\n\nEvery real root m of (Q) yields, via (4)–(5), a real point Q on the ellipse and a distinct normal through R; conversely each normal slope satisfies (Q).  Thus (Q) is exactly the desired “normal equation’’ and, by hypothesis (ii), has four distinct real roots m₁,…,m₄.\n\n–––– Step 2.  Inside the evolute ⇔ four distinct real normals (a short reminder).  \nLet N: E→ℝ∪{∞} send a point of the ellipse to the slope of its normal.  One checks that N is a smooth map of degree 4 whose critical image is the evolute 𝔈.  Hence:  \n\n If R lies inside 𝔈, every fibre N⁻¹(R) consists of four distinct points,  \n while on 𝔈 at least two normals coalesce.  \n\nThis justifies the use of Vieta’s formulae for (Q).\n\n–––– Step 3.  Evaluation of the symmetric slope functions.  \nWrite (Q) in the form m⁴ + A m³ + B m² + C m + D = 0.  \nFrom (Q)\n\n A = −2s/r,   \n B = k² + s²/r² − c⁴/(b²r²),   \n C = −2k²s/r,   \n D =  k²s²/r².  \n\nBy Vieta’s relations\n\n S₁ = −A = 2s/r,  \n\n S₂ =  B  = k² + s²/r² − c⁴/(b²r²),  \n\n S₃ = −C = k²·(2s/r),  \n\n S₄ =  D  = k²s²/r²,  \n\nproving Part 1.\n\n–––– Step 4.  A four–angle tangent identity.  \nFor real numbers u,v,w,z with |u|,|v|,|w|,|z|<π/2 put p=tan u, q=tan v, r=tan w, s=tan z.  \nTwo successive uses of tan(α+β)= (tan α+tan β)/(1−tan α tan β) show\n\n tan(u+v+w+z)=  \n  (p+q+r+s) − (pqr+pq s+pr s+qr s)  \n  ———————————————————————————————  \n   1 − (pq+pr+ps+qr+qs+rs) + pqrs.  \n\nThe numerator equals S₁−S₃ and the denominator equals 1−S₂+S₄, whence\n\n tan(θ₁+θ₂+θ₃+θ₄)= (S₁−S₃)/(1−S₂+S₄). (7)\n\n(The identity is easily verified when all four angles coincide and then extended by continuity.)\n\nInsert the values of Step 3:\n\n Numerator N = (2s/r)(1−k²),   \n\n Denominator D = 1 − [k² + s²/r² − c⁴/(b²r²)] + k²s²/r²  \n       = (1−k²)(r²−s²)/r² + c⁴/(b²r²).  \n\nMultiplying N and D by r² gives precisely formula (★), completing Part 2 up to the exceptional case treated next.\n\n–––– Step 5.  Vanishing of the denominator and the hyperbola H.  \nRewrite D as\n\n D = (1−k²)(r²−s²) + c⁴/b²  \n  = −(k²−1)(r²−s²) + c⁴/b²  \n  = −(c²/b²)(r²−s²) + c⁴/b²  \n  =  (c²/b²)(c² − (r²−s²)).  \n\nHence D=0 ⇔ r²−s² = c², i.e. R lies on the rectangular hyperbola  \n\n  H : x² − y² = c²,\n\nwhich is entirely contained in 𝔈.  \nIf R∈H but s≠0 we still have N≠0, so tan(θ₁+⋯+θ₄)=±∞ and therefore θ₁+⋯+θ₄ equals π/2 mod π, exactly as claimed.  \nAt the vertices (±c,0) both numerator and denominator vanish; then two pairs of normals merge and the oriented sum is obtained by continuity.\n\nThis finishes the proof of Parts 1 and 2. ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.389419",
        "was_fixed": false,
        "difficulty_analysis": "1.  Higher‐degree algebra.  \n   • The parabola leads to a cubic in the parameter t; the ellipse produces a quartic whose structure must be analysed carefully (vanishing of the y³ and y–terms).  \n   • Handling the quartic demands control over four symmetric sums instead of three and forces the use of the full “four‐angle” tangent addition formula.\n\n2.  Two foci instead of one.  \n   • The target equality now involves the angles to both foci, introducing extra geometric quantities (φ₁,φ₂) whose joint behaviour must be tracked algebraically.\n\n3.  Non–trivial simplifications.  \n   • Showing that the cubic and linear coefficients of P(y) vanish is not obvious and requires exploiting the specific shape of the ellipse and of the normal slope (1).  \n   • After eliminating radicals one must manage sizable algebraic expressions; the computation of B and D in (10) is already much more intricate than any coefficient in the parabolic case.\n\n4.  Advanced trigonometric identities.  \n   • The solution needs the general four‐angle tangent formula (12), which is rarely encountered in standard olympiad problems and is decidedly more delicate than its triple‐angle counterpart used for the parabola.\n\n5.  Interplay of geometry and algebra.  \n   • The proof intertwines the geometry of normals, parameterisation of the ellipse, elimination theory, Vieta relations, and multi‐angle trigonometry.  \n   • Any shortcut based on direct pattern matching with the parabolic argument breaks down; the higher degree, two foci, and the quartic’s structure force a fresh, deeper attack.\n\nTogether these features make the enhanced variant substantially more demanding than both the original problem and the current kernel version."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}