{
  "index": "1939-A-2",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "2. A point \\( P \\) is taken on the curve \\( y=x^{3} \\). The tangent at \\( P \\) meets the curve again at \\( Q \\). Prove that the slope of the curve at \\( Q \\) is four times the slope at \\( P \\).",
  "solution": "Solution. Let \\( P \\) have coordinates \\( \\left(x_{0}, y_{0}\\right) \\); then the slope at \\( P \\) is \\( 3 x_{0}{ }^{2} \\). The equation of the tangent at \\( P \\) is \\( y=3 x_{0}{ }^{2}\\left(x-x_{0}\\right)+x_{0}{ }^{3} \\). The points of intersection of the tangent and the original curve are determined by the relation\n\\[\nx^{3}=3 x_{0}^{2}\\left(x-x_{0}\\right)+x_{0}^{3},\n\\]\nwhich is equivalent to\n\\[\n\\left(x-x_{0}\\right)^{2}\\left(x+2 x_{0}\\right)=0\n\\]\n\nHence the second point of intersection is \\( \\left(-2 x_{0},-8 x_{0}{ }^{3}\\right) \\). The slope at this point is \\( 12 x_{0}{ }^{2} \\), which is four times the slope at \\( P \\), as was to be proved.\n\nIf \\( x_{0}=0 \\), the tangent does not really meet the curve again. However, since the tangent in this case has a triple point of intersection with the curve, instead of the usual double point of intersection, it is reasonable to say that it meets the curve \"again\" at \\( (0,0) \\).",
  "vars": [
    "x",
    "y"
  ],
  "params": [
    "P",
    "Q",
    "x_0",
    "y_0"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "abscissa",
        "y": "ordinate",
        "P": "startpoint",
        "Q": "endpoint",
        "x_0": "origxcoord",
        "y_0": "origycoord"
      },
      "question": "2. A point \\( startpoint \\) is taken on the curve \\( ordinate = abscissa^{3} \\). The tangent at \\( startpoint \\) meets the curve again at \\( endpoint \\). Prove that the slope of the curve at \\( endpoint \\) is four times the slope at \\( startpoint \\).",
      "solution": "Solution. Let \\( startpoint \\) have coordinates \\( \\left(origxcoord, origycoord\\right) \\); then the slope at \\( startpoint \\) is \\( 3\\,origxcoord^{2} \\). The equation of the tangent at \\( startpoint \\) is \\( ordinate = 3\\,origxcoord^{2}\\left(abscissa - origxcoord\\right)+origxcoord^{3} \\). The points of intersection of the tangent and the original curve are determined by the relation\n\\[\nabscissa^{3}=3\\,origxcoord^{2}\\left(abscissa - origxcoord\\right)+origxcoord^{3},\n\\]\nwhich is equivalent to\n\\[\n\\left(abscissa - origxcoord\\right)^{2}\\left(abscissa + 2\\,origxcoord\\right)=0\n\\]\n\nHence the second point of intersection is \\( \\left(-2\\,origxcoord,-8\\,origxcoord^{3}\\right) \\). The slope at this point is \\( 12\\,origxcoord^{2} \\), which is four times the slope at \\( startpoint \\), as was to be proved.\n\nIf \\( origxcoord = 0 \\), the tangent does not really meet the curve again. However, since the tangent in this case has a triple point of intersection with the curve, instead of the usual double point of intersection, it is reasonable to say that it meets the curve \"again\" at \\( (0,0) \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "harborage",
        "y": "sandstone",
        "P": "lighthouse",
        "Q": "windstorm",
        "x_0": "amberglow",
        "y_0": "cobblestone"
      },
      "question": "2. A point \\( lighthouse \\) is taken on the curve \\( sandstone=harborage^{3} \\). The tangent at \\( lighthouse \\) meets the curve again at \\( windstorm \\). Prove that the slope of the curve at \\( windstorm \\) is four times the slope at \\( lighthouse \\).",
      "solution": "Solution. Let \\( lighthouse \\) have coordinates \\( \\left(amberglow, cobblestone\\right) \\); then the slope at \\( lighthouse \\) is \\( 3 amberglow^{2} \\). The equation of the tangent at \\( lighthouse \\) is \\( sandstone=3 amberglow^{2}\\left(harborage-amberglow\\right)+amberglow^{3} \\). The points of intersection of the tangent and the original curve are determined by the relation\n\\[\nharborage^{3}=3 amberglow^{2}\\left(harborage-amberglow\\right)+amberglow^{3},\n\\]\nwhich is equivalent to\n\\[\n\\left(harborage-amberglow\\right)^{2}\\left(harborage+2 amberglow\\right)=0\n\\]\n\nHence the second point of intersection is \\( \\left(-2 amberglow,-8 amberglow^{3}\\right) \\). The slope at this point is \\( 12 amberglow^{2} \\), which is four times the slope at \\( lighthouse \\), as was to be proved.\n\nIf \\( amberglow=0 \\), the tangent does not really meet the curve again. However, since the tangent in this case has a triple point of intersection with the curve, instead of the usual double point of intersection, it is reasonable to say that it meets the curve \"again\" at \\( (0,0) \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "knownvertical",
        "y": "constanthoriz",
        "P": "straightline",
        "Q": "flatplane",
        "x_0": "specificvertical",
        "y_0": "specifichoriz"
      },
      "question": "2. A point \\( straightline \\) is taken on the curve \\( constanthoriz=knownvertical^{3} \\). The tangent at \\( straightline \\) meets the curve again at \\( flatplane \\). Prove that the slope of the curve at \\( flatplane \\) is four times the slope at \\( straightline \\).",
      "solution": "Solution. Let \\( straightline \\) have coordinates \\( \\left(specificvertical, specifichoriz\\right) \\); then the slope at \\( straightline \\) is \\( 3 specificvertical^{2} \\). The equation of the tangent at \\( straightline \\) is \\( constanthoriz=3 specificvertical^{2}\\left(knownvertical-specificvertical\\right)+specificvertical^{3} \\). The points of intersection of the tangent and the original curve are determined by the relation\n\\[\nknownvertical^{3}=3 specificvertical^{2}\\left(knownvertical-specificvertical\\right)+specificvertical^{3},\n\\]\nwhich is equivalent to\n\\[\n\\left(knownvertical-specificvertical\\right)^{2}\\left(knownvertical+2 specificvertical\\right)=0\n\\]\n\nHence the second point of intersection is \\( \\left(-2 specificvertical,-8 specificvertical^{3}\\right) \\). The slope at this point is \\( 12 specificvertical^{2} \\), which is four times the slope at \\( straightline \\), as was to be proved.\n\nIf \\( specificvertical=0 \\), the tangent does not really meet the curve again. However, since the tangent in this case has a triple point of intersection with the curve, instead of the usual double point of intersection, it is reasonable to say that it meets the curve \"again\" at \\( (0,0) \\)."
    },
    "garbled_string": {
      "map": {
        "x": "bnlgtiue",
        "y": "wzrduhke",
        "P": "qzxwvtnp",
        "Q": "hjgrksla",
        "x_0": "mprsjfdl",
        "y_0": "ckvdarmy"
      },
      "question": "2. A point \\( qzxwvtnp \\) is taken on the curve \\( wzrduhke=bnlgtiue^{3} \\). The tangent at \\( qzxwvtnp \\) meets the curve again at \\( hjgrksla \\). Prove that the slope of the curve at \\( hjgrksla \\) is four times the slope at \\( qzxwvtnp \\).",
      "solution": "Solution. Let \\( qzxwvtnp \\) have coordinates \\( \\left(mprsjfdl, ckvdarmy\\right) \\); then the slope at \\( qzxwvtnp \\) is \\( 3 mprsjfdl{ }^{2} \\). The equation of the tangent at \\( qzxwvtnp \\) is \\( wzrduhke=3 mprsjfdl{ }^{2}\\left(bnlgtiue-mprsjfdl\\right)+mprsjfdl{ }^{3} \\). The points of intersection of the tangent and the original curve are determined by the relation\n\\[\nbnlgtiue^{3}=3 mprsjfdl^{2}\\left(bnlgtiue-mprsjfdl\\right)+mprsjfdl^{3},\n\\]\nwhich is equivalent to\n\\[\n\\left(bnlgtiue-mprsjfdl\\right)^{2}\\left(bnlgtiue+2 mprsjfdl\\right)=0\n\\]\nHence the second point of intersection is \\( \\left(-2 mprsjfdl,-8 mprsjfdl{ }^{3}\\right) \\). The slope at this point is \\( 12 mprsjfdl{ }^{2} \\), which is four times the slope at \\( qzxwvtnp \\), as was to be proved.\n\nIf \\( mprsjfdl=0 \\), the tangent does not really meet the curve again. However, since the tangent in this case has a triple point of intersection with the curve, instead of the usual double point of intersection, it is reasonable to say that it meets the curve \"again\" at \\( (0,0) \\)."
    },
    "kernel_variant": {
      "question": "Let $n\\ge 1$ and fix real numbers  \n\\[\na=(a_{1},\\dots ,a_{n})\\in\\mathbf R^{\\,n},\\qquad d\\in\\mathbf R,\n\\qquad c_{1},\\dots ,c_{n}\\in\\mathbf R\\setminus\\{0\\}.\n\\]  \nDefine the cubic polynomial  \n\\[\nf:\\mathbf R^{\\,n}\\longrightarrow\\mathbf R,\\qquad \nf(x_{1},\\dots ,x_{n})=\\sum_{i=1}^{n}c_{i}\\,(x_{i}-a_{i})^{3},\n\\]  \nand the real hypersurface  \n\\[\n\\mathcal S=\\Bigl\\{(x,x_{\\,n+1})\\in\\mathbf R^{\\,n}\\times\\mathbf R\\ \\Bigm|\\\n           x_{\\,n+1}=d+f(x)\\Bigr\\}.\n\\]\n\nFix a point  \n\\[\nA=\\bigl(x_{1}^{(0)},\\dots ,x_{n}^{(0)},\\,d+f(x^{(0)})\\bigr)\\in\\mathcal S,\n\\qquad s_{i}:=x_{i}^{(0)}-a_{i}\\ (\\neq 0\\text{ for every }i).\n\\]  \nPut $f(A):=f(x^{(0)})=\\sum_{i=1}^{n}c_{i}s_{i}^{3}$ and assume  \n\n\\[\nf(A)\\neq 0\\qquad(\\ast)\n\\]  \n(so $A$ is not a flat point of $\\mathcal S$).\n\nIntroduce the auxiliary point and line  \n\\[\nB=(a_{1}-2s_{1},\\dots ,a_{n}-2s_{n},\\,d-8f(A)),\\qquad\n\\ell=\\{\\,A+t(B-A)\\mid t\\in\\mathbf R\\,\\}.\n\\]\n\nProblems  \n\n1. Local intersection of $\\mathcal S$ with the tangent hyper-plane at $A$.  \n   (a) Prove that $B\\in\\mathcal S$ and that $B$ lies on the tangent hyper-plane  \n       $\\Pi_{A}$ to $\\mathcal S$ at $A$.  \n   (b) Show that $\\ell\\subset\\Pi_{A}$ and that, under $(\\ast)$,\n        \\[\n           \\Pi_{A}\\cap\\mathcal S\\cap\\ell=\\{A,B\\},\n        \\]\n        where the intersection multiplicity of $A$ equals $2$ and that of\n        $B$ equals $1$.  Describe what happens if $f(A)=0$.\n\n2. First-order scaling.  \n   Compute the gradient $\\nabla f$ and prove\n   \\[\n      \\nabla f(B)=4\\,\\nabla f(A).\n   \\]\n\n3. Second-order scaling.  \n   Let $H(P)=D^{2}f(P)$ be the Hessian of $f$ at $P$.  Show that  \n   \\[\n      H(B)=-2\\,H(A).\n   \\]\n\n4. A {\\em surface-centred} cubic affine self-map.  \n   Define  \n   \\[\n      \\tau:\\mathbf R^{\\,n+1}\\longrightarrow\\mathbf R^{\\,n+1},\\qquad\n      \\tau(x,x_{\\,n+1})=\\bigl(a-2(x-a),\\,d-8(x_{\\,n+1}-d)\\bigr).\n   \\]\n   (a) Prove that $\\tau$ sends $\\mathcal S$ onto itself and that\n       $\\tau(\\Pi_{P})=\\Pi_{\\tau(P)}$ for every $P\\in\\mathcal S$.  \n\n   (b) Compute $\\tau^{2}$ and, more generally, show that for any\n       $m\\in\\mathbf Z$  \n       \\[\n          \\tau^{m}(x,x_{\\,n+1})\n          =\\bigl(a+(-2)^{\\,m}(x-a),\\;d+(-8)^{\\,m}(x_{\\,n+1}-d)\\bigr).\n       \\]\n\n   (c) For $Y(t)=A+t(B-A)$ verify that  \n       \\[\n          \\tau\\bigl(Y(t)\\bigr)=\n          A+(1-2t)(B-A)_{\\mathrm h}+(1-8t)(B-A)_{\\mathrm v},\n       \\]\n       where ${\\mathrm h}$ (resp.\\ ${\\mathrm v}$) denotes the first $n$\n       (resp.\\ the last) component of $B-A$.  Deduce in particular that\n       $\\tau(A)=B$, that $\\ell$ is {\\em not} $\\tau$-invariant and that\n       $\\tau$ never sends $B$ back to $A$.\n\n   (d) Show that\n       \\[\n         D\\tau(A)=\\operatorname{diag}(-2,\\dots ,-2,-8),\\qquad\n         \\nabla\\!\\bigl(f\\circ\\tau\\bigr)(A)=-8\\,\\nabla f(A),\n       \\]\n       and recover the scaling relations of parts\\,2 and\\,3 from this\n       differential identity.\n\n5. (Optional, harder: global dynamics of $\\tau$ on $\\mathcal S$.)  \n   (a) Prove that for every $(x,d+f(x))\\in\\mathcal S$ and every\n       $m\\in\\mathbf Z$  \n       \\[\n          \\tau^{m}\\bigl(x,d+f(x)\\bigr)=\n          \\Bigl(a+(-2)^{\\,m}(x-a),\\;\n                 d+(-8)^{\\,m}f(x)\\Bigr)\\in\\mathcal S.\n       \\]\n\n   (b) Classify the periodic points of $\\tau$ on $\\mathcal S$.\n\n   (c) Show that if $x\\neq a$ then the orbit\n       $\\{\\tau^{m}(A)\\mid m\\in\\mathbf Z\\}$ is unbounded; more precisely,\n       if $f(A)\\ne 0$ its Euclidean norm tends to $\\infty$ as\n       $|m|\\to\\infty$, whereas if $f(A)=0$ but $x\\neq a$ the horizontal\n       component diverges while the vertical one stays equal to~$d$.\n\n   (d) Prove that $\\tau$ possesses no non-trivial finite invariant\n       algebraic subset of~$\\mathcal S$.\n\n\\bigskip",
      "solution": "Throughout write $s=(s_{1},\\dots ,s_{n})=x^{(0)}-a$ and\n$f_{0}:=f(A)\\ (\\neq 0)$.\n\n\\medskip\\noindent\n{\\bf Step 0. Two elementary identities.}\nFor any $u,v\\in\\mathbf R$\n\\[\nu^{3}-v^{3}-3v^{2}(u-v)=(u-v)^{2}(u+2v),\\qquad\n(a_{i}-2s_{i})-x_{i}^{(0)}=-3s_{i}.\n\\tag{I}\n\\]\n\n%--------------------------------------------------------------------\n\\medskip\\noindent\n{\\bf 1. Local intersection.}\n\n\\smallskip\n1(a) {\\it The point $B$ lies both on $\\mathcal S$ and on $\\Pi_{A}$.}\nSince $(x_{i}(B)-a_{i})=-2s_{i}$,\n\\[\nf(B)=\\sum_{i=1}^{n}c_{i}(-2s_{i})^{3}\n     =-8\\sum_{i=1}^{n}c_{i}s_{i}^{3}=-8f_{0},\n\\]\nhence $B=(\\dots ,d-8f_{0})\\in\\mathcal S$.\nThe tangent hyper-plane at $A$ is\n\\[\n\\Pi_{A}:\\;\nx_{\\,n+1}=d+f_{0}+\\sum_{i=1}^{n}3c_{i}s_{i}^{2}\\bigl(x_{i}-x_{i}^{(0)}\\bigr).\n\\]\nSubstituting $x=B$ and using (I) gives\n\\[\nd+f_{0}+\\sum_{i}3c_{i}s_{i}^{2}(-3s_{i})\n=d+f_{0}-9f_{0}=d-8f_{0}=x_{\\,n+1}(B),\n\\]\nso $B\\in\\Pi_{A}$.\n\n\\smallskip\n1(b) {\\it Intersection multiplicities along $\\ell$.}  \nAlong $\\ell$ we have\n\\[\nx_{i}(t)=a_{i}+(1-3t)s_{i},\\qquad\nx_{\\,n+1}(t)=d+f_{0}-9tf_{0},\n\\]\nhence\n\\[\nf\\bigl(x(t)\\bigr)=(1-3t)^{3}f_{0}.\n\\]\nPutting\n\\[\nG(t):=x_{\\,n+1}(t)-\\bigl(d+f(x(t))\\bigr)\n     =f_{0}\\bigl[1-9t-(1-3t)^{3}\\bigr]=27f_{0}\\,t^{2}(t-1),\n\\]\nwe see that $G$ vanishes only at $t=0$ (double root, $A$) and\n$t=1$ (simple root, $B$).  If $f_{0}=0$ then $G\\equiv 0$ and the whole\nline $\\ell$ is contained in $\\mathcal S$; in that case multiplicities\ncollapse and $A$ is a higher-order contact point.\n\n%--------------------------------------------------------------------\n\\medskip\\noindent\n{\\bf 2. Gradient scaling.}\n\nBecause\n\\[\n\\nabla f(x)=\\bigl(3c_{1}(x_{1}-a_{1})^{2},\\dots ,3c_{n}(x_{n}-a_{n})^{2}\\bigr),\n\\]\nwe obtain\n\\[\n\\nabla f(A)=\\bigl(3c_{1}s_{1}^{2},\\dots ,3c_{n}s_{n}^{2}\\bigr),\\qquad\n\\nabla f(B)=\\bigl(3c_{1}(-2s_{1})^{2},\\dots ,3c_{n}(-2s_{n})^{2}\\bigr)\n           =4\\,\\nabla f(A).\n\\]\n\n%--------------------------------------------------------------------\n\\medskip\\noindent\n{\\bf 3. Hessian scaling.}\n\nMixed second derivatives vanish, whereas\n$\\partial^{2}f/\\partial x_{i}^{2}=6c_{i}(x_{i}-a_{i})$.  Therefore\n\\[\nH(A)=\\operatorname{diag}(6c_{1}s_{1},\\dots ,6c_{n}s_{n}),\\qquad\nH(B)=\\operatorname{diag}(6c_{1}(-2s_{1}),\\dots ,6c_{n}(-2s_{n}))\n     =-2\\,H(A).\n\\]\n\n%--------------------------------------------------------------------\n\\medskip\\noindent\n{\\bf 4. The surface-centred cubic map $\\tau$.}\n\n\\smallskip\n4(a)  Let $x' := a-2(x-a)$.  Then\n\\[\nf(x')=\\sum_{i}c_{i}\\bigl(-2(x_{i}-a_{i})\\bigr)^{3}\n      =-8f(x),\n\\]\nhence $\\tau(\\mathcal S)=\\mathcal S$.  For\n$P=(p,d+f(p))$ write $Q=\\tau(P)$.  The chain rule yields\n$D\\tau(P)^{T}\\nabla f(Q)=\\nabla\\!\\bigl(f\\circ\\tau\\bigr)(P)$, which\nimplies $\\tau(\\Pi_{P})=\\Pi_{Q}$.\n\n\\smallskip\n4(b)  A straightforward second iteration gives\n\\[\n\\tau^{2}(x,x_{\\,n+1})=\n\\bigl(4x-3a,\\;d+64(x_{\\,n+1}-d)\\bigr),\n\\]\nand by induction one finds for every $m\\in\\mathbf Z$\n\\[\n\\tau^{m}(x,x_{\\,n+1})\n=\\bigl(a+(-2)^{\\,m}(x-a),\\;d+(-8)^{\\,m}(x_{\\,n+1}-d)\\bigr).\n\\]\n\n\\smallskip\n4(c)  Since $B-A=(-3s,-9f_{0})$, substituting $Y(t)$ in $\\tau$ yields\n\\[\n\\tau\\bigl(Y(t)\\bigr)\n      =\\bigl(a-2s(1-3t),\\,d-8f_{0}(1-9t)\\bigr)\n      =A+(1-2t)(B-A)_{\\mathrm h}+(1-8t)(B-A)_{\\mathrm v},\n\\]\nso $\\tau(A)=B$, $\\ell$ is not $\\tau$-invariant and $\\tau$ never sends\n$B$ back to $A$.\n\n\\smallskip\n4(d)  The Jacobian at $A$ equals\n$D\\tau(A)=\\operatorname{diag}(-2,\\dots ,-2,-8)$, and\n$f\\circ\\tau=-8f$ implies\n\\[\n\\nabla(f\\circ\\tau)(A)=D\\tau(A)^{T}\\,\\nabla f(B)=-8\\,\\nabla f(A),\n\\]\nfrom which parts\\,2 and\\,3 are recovered exactly as before.\n\n%--------------------------------------------------------------------\n\\medskip\\noindent\n{\\bf 5. Global dynamics of $\\tau$ (optional).}\n\n\\smallskip\n5(a) {\\it Closed formula for $\\tau^{m}$ on $\\mathcal S$.}  \nFor $P=(x,d+f(x))\\in\\mathcal S$ one computes\n\\[\n\\tau^{m}(P)\n=(\\,a+(-2)^{\\,m}(x-a),\\;\n  d-8\\bigl(d+f(x)-d\\bigr)+\\cdots)\n =(a+(-2)^{\\,m}(x-a),\\;d+(-8)^{\\,m}f(x)),\n\\]\nand since $f\\!\\bigl(a+(-2)^{\\,m}(x-a)\\bigr)=(-8)^{\\,m}f(x)$, the image\nindeed lies on $\\mathcal S$.\n\n\\smallskip\n5(b) {\\it Periodic points.}  \nAssume $\\tau^{k}(P)=P$ for some $k\\ge 1$.  Writing\n$P=(x,d+f(x))$, the horizontal part yields\n$a+(-2)^{\\,k}(x-a)=x$, that is\n$\\bigl((-2)^{\\,k}-1\\bigr)(x-a)=0$.  Because\n$(-2)^{\\,k}\\neq 1$ for every $k\\ge 1$, we must have $x=a$.\nConsequently $f(x)=0$ and $P=(a,d)$.  Hence the unique periodic point\nof $\\tau$ on $\\mathcal S$ is its fixed point $(a,d)$.\n\n\\smallskip\n5(c) {\\it Growth of the orbit.}  \nLet $P=(x,d+f(x))\\in\\mathcal S$ with $x\\neq a$.\nThen $|x^{(m)}-a|=\\lvert(-2)^{\\,m}\\rvert\\,|x-a|$ tends to $\\infty$\nas $\\lvert m\\rvert\\to\\infty$.  If $f(P)\\neq 0$ then the vertical\ncomponent\n$\\lvert x_{\\,n+1}^{(m)}-d\\rvert = \\lvert(-8)^{\\,m}f(P)\\rvert$\nblows up even faster, so the whole orbit escapes every compact set.\nIf $f(P)=0$ but $x\\neq a$, the horizontal component still diverges\nwhile the vertical one is constantly $d$; the orbit is again unbounded.\n\n\\smallskip\n5(d) {\\it No finite invariant algebraic subset.}  \nAny $\\tau$-invariant algebraic subset $Z\\subset\\mathcal S$ would have\nto contain all iterates of each of its points.  By (b) and (c) that\nforces $Z$ either to be empty or to contain the whole infinite orbit of\nsome non-periodic point, hence to be Zariski dense in~$\\mathcal S$.\nTherefore no proper finite algebraic subset of $\\mathcal S$ can be\n$\\tau$-invariant.\n\n\\hfill$\\square$\n\n\\bigskip",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.348614",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension:  the original one-variable cubic curve is replaced by an arbitrary-dimension cubic hypersurface in ℝⁿ⁺¹.  \n• Additional structures:  gradients, tangent hyper-planes, intersection multiplicities, and second fundamental form all enter.  \n• Extra constraints:  a unique non-trivial real point of intersection must be located and proved to be the only one.  \n• Deeper theory:  the solution demands multivariable calculus, elementary algebraic geometry (multiplicity, Bezout-type counting), and differential-geometric notions (normal curvature, second fundamental form).  \n• More steps:  determining Π, analysing the polynomial system, proving uniqueness, scaling of gradients, and finally scaling of curvatures—each building on the previous—makes the argument substantially longer and conceptually richer than in the one-dimensional prototype."
      }
    },
    "original_kernel_variant": {
      "question": "Let n ≥ 1 and fix real numbers a₁,…,aₙ , d together with non-zero cubic\ncoefficients c₁,…,cₙ.  \nPut  \n\n  𝓢 = { (x , x_{n+1}) ∈ ℝⁿ×ℝ :\n      x_{n+1}=d+∑_{i=1}^{n} c_i (x_i-a_i)³ }                    (1)\n\nand abbreviate  \n\n  f(x₁,…,xₙ)=∑_{i=1}^{n} c_i (x_i-a_i)³.                         (2)\n\nChoose a point  \n\n  A=(x₁^{(0)},…,x_n^{(0)}, d+f(x^{(0)}))∈𝓢 , x_i^{(0)}≠a_i,\n\nand write s_i:=x_i^{(0)}-a_i, f(A)=∑ c_i s_i³.  \nDenote by Π_A the tangent hyper–plane to the graph of g=d+f at A.\nThroughout we assume  \n\n  f(A)≠0                                                         (∗)\n\n(hence A is not an inflexion point of 𝓢).\n\nIntroduce  \n\n  B:=(a₁-2s₁,…,a_n-2s_n , d-8f(A)),                              (3)\n\nand the affine line through A and B  \n\n  ℓ={ A+t(B−A) : t∈ℝ }.                                          (4)\n\nProblems  \n\n1. Local intersection (one–dimensional slice).  \n   (a) Prove that B lies in 𝓢 and in Π_A.  \n   (b) Show that ℓ⊂Π_A and that, under the hypothesis (∗),\n\n             Π_A∩𝓢∩ℓ={A,B},                                       (5)\n\n     with A counted with multiplicity 2 and B with multiplicity 1.  \n     (Describe what changes when f(A)=0.)\n\n2. First–order scaling.  \n   Show that the gradient of f satisfies  \n\n             ∇f(B)=4∇f(A).                                        (6)\n\n3. Second–order scaling.  \n   With H(P)=D²f(P) the Hessian matrix of f at P, prove  \n\n             H(B)=−2H(A).                                        (7)\n\n4. Cubic affine transformation centred at A.  \n   Define  \n\n        τ_A : ℝ^{n+1}→ℝ^{n+1},                                    (8)\n        τ_A(x,x_{n+1})=( a−2(x−a) , d−8(x_{n+1}−d) ).\n\n   (a) Show that τ_A sends 𝓢 onto itself and carries each tangent\n       hyper-plane Π_P to Π_{τ_A(P)}.  \n\n   (b) Compute τ_A² and prove the formula  \n\n             τ_A²(x,x_{n+1})=(4x−3a , d+64(x_{n+1}−d)).          (9)\n\n       Hence τ_A is not an involution; τ_A² is a homothety of\n       ratio 4 in the first n coordinates and of ratio 64 in the\n       vertical coordinate, followed by suitable translations.\n\n   (c) Let Y(t)=A+t(B−A) be the parametrisation of ℓ.  Show that  \n\n             τ_A(Y(t)) = A+(1−2t)(B−A)_h + (1−8t)(B−A)_v,        (10)\n\n       where the subscripts “h’’ and “v’’ denote, respectively,\n       the first n and the (n+1)-st component of the vector B−A.\n       Deduce in particular that τ_A(A)=B, that ℓ is **not**\n       τ_A-invariant once n+1 dimensions are considered, and that\n       τ_A never sends B back to A.\n\n   (d) Compute the differential Dτ_A(A)=diag(−2,…,−2,−8) and\n       re-derive (6) and (7) from it.\n\n5. (Optional, harder)  Let A₁,…,A_k be points of 𝓢 whose first-coordinate\n    vectors a^{(j)}=(a₁^{(j)},…,a_n^{(j)}) form an affine basis of ℝⁿ.\n    Show that the subgroup of the affine group generated by the cubic\n    transformations τ_{A_j} acts transitively on 𝓢 inside each connected\n    component.  \n    (Hint: τ_P∘τ_Q is a homothety of ratio 4 followed by a translation,\n    and the translations generated under the stated hypothesis span\n    ℝⁿ.)\n\n──────────────────────────────────────────────────────────────────────",
      "solution": "Throughout we use (2), write s_i=x_i^{(0)}−a_i and set f₀:=f(A) (so\nf₀≠0 by (∗)).\n\nStep 0. Two algebraic identities.  \nFor u,v∈ℝ  \n\n u³−v³−3v²(u−v)=(u−v)²(u+2v),                               (I)  \n (a_i−2s_i)−x_i^{(0)}=−3s_i.                                (II)\n\n────────────────────────────────────────────────────────────  \n1. Local intersection.\n\n1(a) Membership of B in 𝓢 and Π_A.  \nSince x_i(B)−a_i=−2s_i,\n\n  f(B)=∑c_i(−2s_i)³=−8∑c_i s_i³=−8f₀,\n\nso x_{n+1}(B)=d−8f₀, exactly its (n+1)-st coordinate.  Thus B∈𝓢.\nThe tangent hyper-plane at A is\n\n  Π_A : x_{n+1}=d+f₀+∑_i 3c_i s_i²(x_i−x_i^{(0)}).          (11)\n\nWith (II) and ∑_i c_i s_i³=f₀ one obtains\n\n  RHS at x=B = d+f₀+∑_i 3c_i s_i²(−3s_i)\n             = d+f₀−9f₀=d−8f₀=x_{n+1}(B),\n\nso B∈Π_A.\n\n1(b) Intersection multiplicities on ℓ.  \nAlong ℓ put\n\n  x_i(t)=a_i+(1−3t)s_i, x_{n+1}(t)=d+f₀−9tf₀                (12)\n\nand compute\n\n  f(x(t))=(1−3t)³ f₀.                                       (13)\n\nDefine  \n\n  G(t):=x_{n+1}(t)−(d+f(x(t)))\n       =f₀[1−9t−(1−3t)³].                                   (14)\n\nExpanding (1−3t)³ gives 1−9t+27t²−27t³, hence\n\n  1−9t−(1−3t)³=27t²(t−1).                                  (15)\n\nBecause f₀≠0,  \n\n  G(t)=27f₀ t²(t−1).                                       (16)\n\nSo G(t)=0 ⇔ t∈{0,1}.  Therefore Π_A∩𝓢∩ℓ={A,B}.  The factor t² shows\nthat t=0 (point A) is a double zero, whereas t=1 (point B) is simple,\nestablishing (5).\n\n(Degenerate case f₀=0. Then G≡0 and the whole line ℓ is contained in 𝓢;\nA is an inflexion point and the “double–simple’’ description breaks\ndown.)\n\n────────────────────────────────────────────────────────────  \n2. Gradient scaling (6).\n\n∇f(x)=(3c_1(x₁−a₁)²,…,3c_n(x_n−a_n)²).\n\nAt A: ∇f(A)=(3c_1 s₁²,…,3c_n s_n²).  \nAt B: ∇f(B)=(3c_1(−2s₁)²,…,3c_n(−2s_n)²)=4∇f(A).\n\n────────────────────────────────────────────────────────────  \n3. Hessian scaling (7).\n\nMixed derivatives vanish and ∂²f/∂x_i²=6c_i(x_i−a_i), so\n\n  H(A)=diag(6c_i s_i), H(B)=diag(6c_i(−2s_i))=−2H(A).\n\n────────────────────────────────────────────────────────────  \n4. The cubic affine transformation τ_A.\n\n4(a) τ_A preserves 𝓢 and tangent hyper-planes.  \nGiven x′:=a−2(x−a),\n\n  f(x′)=∑c_i(−2(x_i−a_i))³=−8f(x),                         (17)\n\nwhence d+f(x′)=d−8f(x).  Thus τ_A(𝓢)=𝓢.\n\nFor P=(p,d+f(p))∈𝓢 set Q=τ_A(P).  Writing out Π_P and using (6) one\nverifies directly that τ_A(Π_P)=Π_Q.\n\n4(b) The square of τ_A.  \nThe computation\n\n  τ_A²(x,x_{n+1})=(4x−3a , d+64(x_{n+1}−d))                (18)\n\nfollows exactly as in the draft, proving (9).\n\n4(c) Action of τ_A on the line ℓ.  \nLet Y(t)=A+t(B−A) with B−A=(−3s,−9f₀).  Then\n\n  τ_A(Y(t)) = ( a−2(x(t)−a) , d−8(x_{n+1}(t)−d) )\n            = ( a−2s(1−3t) , d−8f₀(1−9t) ).               (19)\n\nSubtracting A and using B−A one gets the **componentwise** identity\n\n  τ_A(Y(t)) = A+(1−2t)(B−A)_h+(1−8t)(B−A)_v,              (20)\n\nwhere (⋯)_h (resp. (⋯)_v) denotes the horizontal (resp. vertical)\npart of the vector.  Consequently\n\n  • τ_A(A)=B, while τ_A(B)=A+(B−A)_h−7(B−A)_v≠A;  \n  • ℓ is **not** invariant under τ_A because the two scale factors\n    1−2t and 1−8t differ unless t∈{0,½}.  \n  • The first n coordinates of points on ℓ are transformed by\n    t↦1−2t, whereas the (n+1)-st coordinate follows t↦1−8t.\n\n4(d) Differential at A.  \nDτ_A(A)=diag(−2,…,−2,−8).  \nActing on the gradient (a covector) multiplies it by −2 twice, giving\nthe factor 4 of (6); acting on second derivatives introduces one more\n−2, giving (7).\n\n────────────────────────────────────────────────────────────  \n5. (Sketch of the optional part.)  \nFor P∈𝓢 write τ_P(x)=−2x+3a_P for the first n coordinates.  A direct\ncomputation shows\n\n  τ_P∘τ_Q (x)=4x+3(a_P−2a_Q).                              (21)\n\nHence any element of the subgroup G generated by the τ_{A_j} acts on\nthe first n coordinates as\n\n  x↦4^m x+u, m∈ℤ, u∈V:=⟨3(a^{(i)}−2a^{(j)})⟩_ℤ.          (22)\n\nBecause the vectors a^{(j)} form an affine basis of ℝⁿ, the subgroup\nV equals ℝⁿ, and multiplication by 4 together with these translations\ngive the full affine group of ℝⁿ.  Adding the vertical coordinate\n(determined uniquely once the horizontal part is fixed) one sees that\nG acts transitively on every connected component of 𝓢. ∎\n\n──────────────────────────────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.304046",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension:  the original one-variable cubic curve is replaced by an arbitrary-dimension cubic hypersurface in ℝⁿ⁺¹.  \n• Additional structures:  gradients, tangent hyper-planes, intersection multiplicities, and second fundamental form all enter.  \n• Extra constraints:  a unique non-trivial real point of intersection must be located and proved to be the only one.  \n• Deeper theory:  the solution demands multivariable calculus, elementary algebraic geometry (multiplicity, Bezout-type counting), and differential-geometric notions (normal curvature, second fundamental form).  \n• More steps:  determining Π, analysing the polynomial system, proving uniqueness, scaling of gradients, and finally scaling of curvatures—each building on the previous—makes the argument substantially longer and conceptually richer than in the one-dimensional prototype."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}