{
  "index": "1984-B-2",
  "type": "GEO",
  "tag": [
    "GEO",
    "ALG"
  ],
  "difficulty": "",
  "question": "Problem B-2. Find the minimum value of\n\\[\n(u-v)^{2}+\\left(\\sqrt{2-u^{2}}-\\frac{9}{v}\\right)^{2}\n\\]\nfor \\( 0<u<\\sqrt{2} \\) and \\( v>0 \\).",
  "solution": "B-2.\nThe problem asks for the minimum distance between the quarter of the circle \\( x^{2}+y^{2}=2 \\) in the open first quadrant and the half of the hyperbola \\( x y=9 \\) in that quadrant. Since the tangents to the respective curves at \\( (1,1) \\) and \\( (3,3) \\) separate the curves and are both perpendicular to \\( x=y \\), the minimum distance is 8 .",
  "vars": [
    "u",
    "v",
    "x",
    "y"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "u": "firstvar",
        "v": "secondvar",
        "x": "axisvalue",
        "y": "ordinate"
      },
      "question": "Problem B-2. Find the minimum value of\n\\[\n(firstvar-secondvar)^{2}+\\left(\\sqrt{2-firstvar^{2}}-\\frac{9}{secondvar}\\right)^{2}\n\\]\nfor \\( 0<firstvar<\\sqrt{2} \\) and \\( secondvar>0 \\).",
      "solution": "B-2.\nThe problem asks for the minimum distance between the quarter of the circle \\( axisvalue^{2}+ordinate^{2}=2 \\) in the open first quadrant and the half of the hyperbola \\( axisvalue\\,ordinate=9 \\) in that quadrant. Since the tangents to the respective curves at \\( (1,1) \\) and \\( (3,3) \\) separate the curves and are both perpendicular to \\( axisvalue=ordinate \\), the minimum distance is 8 ."
    },
    "descriptive_long_confusing": {
      "map": {
        "u": "dragonfly",
        "v": "honeycomb",
        "x": "tangerine",
        "y": "porcupine"
      },
      "question": "Problem B-2. Find the minimum value of\n\\[\n( dragonfly - honeycomb )^{2}+\\left(\\sqrt{2-dragonfly^{2}}-\\frac{9}{ honeycomb }\\right)^{2}\n\\]\nfor \\( 0< dragonfly <\\sqrt{2} \\) and \\( honeycomb >0 \\).",
      "solution": "B-2.\nThe problem asks for the minimum distance between the quarter of the circle \\( tangerine^{2}+porcupine^{2}=2 \\) in the open first quadrant and the half of the hyperbola \\( tangerine\\,porcupine=9 \\) in that quadrant. Since the tangents to the respective curves at \\( (1,1) \\) and \\( (3,3) \\) separate the curves and are both perpendicular to \\( tangerine=porcupine \\), the minimum distance is 8 ."
    },
    "descriptive_long_misleading": {
      "map": {
        "u": "constantnum",
        "v": "invariantnum",
        "x": "anchorednum",
        "y": "steadycount"
      },
      "question": "Problem B-2. Find the minimum value of\n\\[\n(constantnum-invariantnum)^{2}+\\left(\\sqrt{2-constantnum^{2}}-\\frac{9}{invariantnum}\\right)^{2}\n\\]\nfor \\( 0<constantnum<\\sqrt{2} \\) and \\( invariantnum>0 \\).",
      "solution": "B-2.\nThe problem asks for the minimum distance between the quarter of the circle \\( anchorednum^{2}+steadycount^{2}=2 \\) in the open first quadrant and the half of the hyperbola \\( anchorednum steadycount=9 \\) in that quadrant. Since the tangents to the respective curves at \\( (1,1) \\) and \\( (3,3) \\) separate the curves and are both perpendicular to \\( anchorednum=steadycount \\), the minimum distance is 8 ."
    },
    "garbled_string": {
      "map": {
        "u": "qzxwvtnp",
        "v": "hjgrksla",
        "x": "nfqzmpwy",
        "y": "ksrnvxje"
      },
      "question": "Problem B-2. Find the minimum value of\n\\[\n(qzxwvtnp-hjgrksla)^{2}+\\left(\\sqrt{2-qzxwvtnp^{2}}-\\frac{9}{hjgrksla}\\right)^{2}\n\\]\nfor \\( 0<qzxwvtnp<\\sqrt{2} \\) and \\( hjgrksla>0 \\).",
      "solution": "B-2.\nThe problem asks for the minimum distance between the quarter of the circle \\( nfqzmpwy^{2}+ksrnvxje^{2}=2 \\) in the open first quadrant and the half of the hyperbola \\( nfqzmpwy\\,ksrnvxje=9 \\) in that quadrant. Since the tangents to the respective curves at \\( (1,1) \\) and \\( (3,3) \\) separate the curves and are both perpendicular to \\( nfqzmpwy=ksrnvxje \\), the minimum distance is 8 ."
    },
    "kernel_variant": {
      "question": "Let  \n\\[\nS=\\Bigl\\{x=(x_{1},\\dots ,x_{6})\\in\\mathbb R^{6}\\;\\Bigm|\\;\nx_{i}\\ge 0,\\;\n\\sum_{i=1}^{6}x_{i}=12,\\;\n\\sum_{i=1}^{6}x_{i}^{2}=30\\Bigr\\},\n\\qquad\nH=\\Bigl\\{y=(y_{1},\\dots ,y_{6})\\in\\mathbb R^{6}\\;\\Bigm|\\;\ny_{1}=y_{2}=y_{3}=s>0,\\;\ny_{4}=y_{5}=y_{6}=t>0,\\;\ns>t,\\;\nst=9\\Bigr\\}.\n\\]\n\nFor \\(x\\in S\\) and \\(y\\in H\\) denote the squared Euclidean distance by  \n\\[\nD(x,y)=\\bigl\\lVert x-y\\bigr\\rVert^{2}=\\sum_{i=1}^{6}(x_{i}-y_{i})^{2}.\n\\]\n\n(A) Prove that  \n\\[\nD_{\\min}:=\\min\\bigl\\{D(x,y):x\\in S,\\;y\\in H\\bigr\\}\n\\]\nis attained.\n\n(B) Determine \\(D_{\\min}\\) in exact closed form (expressions by radicals over \\(\\mathbb Q\\) are required) and describe \\emph{all} pairs \\((x,y)\\) that realise this minimum.\n\n────────────────────────────────────────────────────────",
      "solution": "Throughout we write  \n\\[\ny=(s,s,s,t,t,t),\\qquad s>t>0,\\quad st=9\n\\;\\Longrightarrow\\;\nt=\\frac{9}{s},\\;s>3.\n\\tag{0}\n\\]\n\n1.  Reformulation of the inner problem  \nFix \\(y\\in H\\) and abbreviate  \n\\[\n\\Phi(y):=\\min_{x\\in S}D(x,y).\n\\]\nFor any \\(x\\in S\\),\n\\[\nD(x,y)=\\sum_{i=1}^{6}x_{i}^{2}+3s^{2}+3t^{2}-2\\bigl(s(x_{1}+x_{2}+x_{3})+t(x_{4}+x_{5}+x_{6})\\bigr)\n      =30+C(s)-2\\langle x,y\\rangle,\n\\]\nwhere \\(C(s)=3s^{2}+3t^{2}\\) is independent of \\(x\\).  \nHence\n\\[\n\\Phi(y)=30+C(s)-2\\max_{x\\in S}\\langle x,y\\rangle.\n\\tag{1.1}\n\\]\nMinimising the distance is therefore equivalent to \\emph{maximising} the\nlinear form \\(\\langle x,y\\rangle\\) over \\(S\\).\n\n2.  A sharp upper bound for the inner product  \nWrite\n\\[\na:=x_{1}+x_{2}+x_{3},\\quad b:=x_{4}+x_{5}+x_{6}=12-a\n\\qquad(0\\le a,b\\le 12).\n\\]\nLet\n\\(\n\\sigma_{1}:=x_{1}^{2}+x_{2}^{2}+x_{3}^{2},\\;\n\\sigma_{2}:=x_{4}^{2}+x_{5}^{2}+x_{6}^{2}=30-\\sigma_{1}.\n\\)\nBy Cauchy–Schwarz,\n\\[\n\\sigma_{1}\\ge\\frac{a^{2}}{3},\\qquad \\sigma_{2}\\ge\\frac{b^{2}}{3},\n\\]\nand therefore\n\\[\n\\frac{a^{2}}{3}+\\frac{b^{2}}{3}\\le30\n\\;\\Longrightarrow\\;\n(a-6)^{2}\\le9\n\\;\\Longrightarrow\\;\n3\\le a\\le9.\n\\tag{2.1}\n\\]\n\nBecause \\(s>t\\), we estimate\n\\[\n\\langle x,y\\rangle\n  =sa+tb\n  =12t+(s-t)a\n  \\le12t+(s-t)\\cdot 9,\n\\tag{2.2}\n\\]\nand equality in (2.2) requires simultaneously\n\\[\n\\text{(i) } a=9\n\\quad\\text{and}\\quad\n\\text{(ii) } \\sigma_{1}=\\frac{a^{2}}{3},\\; \\sigma_{2}=\\frac{b^{2}}{3},\n\\]\ni.e.  \n\\[\nx_{1}=x_{2}=x_{3}=3,\\qquad\nx_{4}=x_{5}=x_{6}=1.\n\\tag{2.3}\n\\]\nThus any maximiser of \\(\\langle x,y\\rangle\\) must be the point\n\\[\nx^{\\star}:=(3,3,3,1,1,1),\n\\tag{2.4}\n\\]\ntogether with its images under the group\n\\(\\mathfrak S_{3}\\times\\mathfrak S_{3}\\) that separately permutes the\nfirst and the last three coordinates.  (Permuting a “\\(3\\)” into a\n\\(t\\)-slot or a “\\(1\\)” into an \\(s\\)-slot would lower \\(a\\) below \\(9\\)\nand hence strictly decrease the inner product.)\n\n\\emph{Consequently \\(x^{\\star}\\) is the \\underline{unique} minimiser of\n\\(x\\mapsto D(x,y)\\).  In particular, every optimal \\(x\\) satisfies\n\\(x_{i}>0\\); the gap in the original argument is now closed.}\n\nEvaluating (1.1) at \\(x^{\\star}\\) gives the reduced one–variable\nobjective\n\\[\n\\boxed{\\;\n\\Phi(y)=D(x^{\\star},y)=\n3\\Bigl[(s-3)^{2}+\\Bigl(\\tfrac{9}{s}-1\\Bigr)^{2}\\Bigr]\n=:F(s)},\\qquad s>3.\n\\tag{2.5}\n\\]\n\n3.  Strict convexity of \\(F\\)  \nSet\n\\(\nf(s)=(s-3)^{2}+(\\tfrac{9}{s}-1)^{2},\\;\nF=3f.\n\\)\nRoutine differentiation yields  \n\\[\nf'(s)=\\frac{2\\bigl(s^{4}-3s^{3}+9s-81\\bigr)}{s^{3}},\\qquad\nf''(s)=2+\\frac{18}{s^{3}}+\\frac{54(9-s)}{s^{4}}>0\\quad(s>3),\n\\]\nso \\(F\\) is strictly convex on \\((3,\\infty)\\).  Hence it admits exactly\none minimiser \\(s_{0}>3\\), determined by the quartic equation\n\\[\np(s):=s^{4}-3s^{3}+9s-81=0,\\qquad p'(s)>0\\;(s>3).\n\\tag{3.1}\n\\]\n\n4.  Closed radicals for \\(s_{0}\\) (Ferrari–Cardano)  \nDepressing \\(p\\) with \\(s=z+\\tfrac34\\) gives  \n\\[\nz^{4}-\\frac{27}{8}z^{2}+\\frac{45}{8}z-\\frac{19\\,251}{256}=0.\n\\]\nFerrari’s method reduces this to a resolvent cubic whose unique real\nroot is\n\\[\ny_{0}=\n-\\frac{27}{48}\n+\\sqrt[3]{\\frac{11\\,583}{4\\,096}\n          +\\mathrm i\n           \\frac{\\sqrt{254\\,223\\,895\\,167}}{4\\,096}}\n+\\sqrt[3]{\\frac{11\\,583}{4\\,096}\n          -\\mathrm i\n           \\frac{\\sqrt{254\\,223\\,895\\,167}}{4\\,096}}.\n\\]\nPut  \n\\[\nU=\\sqrt{\\,2y_{0}+\\tfrac{27}{8}},\\qquad\nV=\\sqrt{-2y_{0}-\\tfrac{27}{8}+\\tfrac{45}{4U}},\n\\]\nthen  \n\\[\n\\boxed{\\;\ns_{0}=\\frac34+\\frac{U+V}{2}},\\qquad t_{0}=\\frac{9}{s_{0}}.\n}\n\\tag{4.1}\n\\]\nAll radicals are over \\(\\mathbb Q\\), as requested.  Numerically  \n\\(\ns_{0}\\approx3.829016065,\\;\nt_{0}\\approx2.351538472.\n\\)\n\n5.  The minimum value  \nSubstituting \\(s_{0}\\) into (2.5) gives  \n\\[\n\\boxed{\\;\nD_{\\min}=3\\Bigl[(s_{0}-3)^{2}+\\Bigl(\\tfrac{9}{s_{0}}-1\\Bigr)^{2}\\Bigr]\n      =\\frac{3}{s_{0}^{2}}\n        \\bigl(s_{0}^{4}-6s_{0}^{3}+10s_{0}^{2}-18s_{0}+81\\bigr).\n}\n\\tag{5.1}\n\\]\nEliminating \\(s_{0}^{4}\\) via \\(p(s_{0})=0\\) one may also write the\nnumerator as \\(-3s_{0}^{3}+10s_{0}^{2}-27s_{0}+162\\).\nNumerically \\(D_{\\min}\\approx7.541970271\\).\n\n6.  Global minimisers  \n(i)  The inner problem has the unique minimiser \\(x^{\\star}\\) in (2.4).  \n(ii) The outer function \\(F\\) is strictly convex with unique minimiser\n     \\(s_{0}\\).\n\nHence the \\emph{only} pair attaining the global minimum is\n\\[\n\\boxed{\\;\nx^{\\star}=(3,3,3,1,1,1),\\qquad\ny^{\\star}=(s_{0},s_{0},s_{0},t_{0},t_{0},t_{0}).\n}\n\\tag{6.1}\n\\]\n\n7.  Proof of Part (A)  \nFor each fixed \\(y\\in H\\) the set \\(S\\) is compact and\n\\(x\\mapsto D(x,y)\\) is continuous, so \\(\\Phi(y)\\) is attained.\nWriting \\(y\\) as in (0) identifies \\(H\\) with the open ray \\((3,\\infty)\\);\nthe map \\(s\\mapsto F(s)\\) of (2.5) is continuous, satisfies\n\\(\n\\lim_{s\\downarrow3}F(s)=12,\\;\n\\lim_{s\\to\\infty}F(s)=\\infty,\n\\)\nand is strictly convex.  Therefore it attains its minimum exactly at\n\\(s_{0}\\), and the pair \\((x^{\\star},y^{\\star})\\) in (6.1) is the sole\nglobal minimiser, completing the proof.  \n\\(\\square\\)\n\n────────────────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.677847",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimensionality: the problem lives in ℝ⁶ rather than ℝ²,\n   introducing six variables instead of two.\n2. Multiple interacting constraints: the feasible set S is the\n   intersection of a positive orthant with both a hyperplane\n   (Σxᵢ = 12) and a sphere (Σxᵢ² = 30); H is a 1–dimensional manifold\n   defined simultaneously by an ordering condition (s>t) and a\n   high–degree algebraic constraint (s³t³ = 9³).\n3. Sophisticated optimisation: one must use Lagrange multipliers to\n   locate the nearest point of S to a given y, revealing an\n   internal two–point structure {1,3}.\n4. Reduction to a quartic: the remaining single–variable problem\n   leads to the non-trivial quartic equation s⁴ – 3s³ + 9s – 81 = 0,\n   whose root cannot be guessed and must be solved with Ferrari’s\n   method or an equivalent algebraic apparatus.\n5. Exact closed form: the final answer demands radicals of nested\n   radicals (formula (11)), far beyond the elementary arithmetic\n   needed for the original 2-dimensional problem.\n6. Subtle case distinction: one has to decide which block of\n   coordinates should be matched with which value of x,\n   a point that disappears entirely in the original statement.\n\nAltogether the enhanced variant compels the solver to combine\nmultivariate calculus, symmetric-constraint reasoning, and full\nquartic algebra, making it substantially harder than both the\noriginal Putnam problem and the current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\\[\nS \\;=\\;\\Bigl\\{\\,x=(x_{1},\\ldots ,x_{6})\\in\\mathbb R^{6}\\;\\Bigm|\\;\nx_{i}\\ge 0,\\;\n\\sum_{i=1}^{6}x_{i}=12,\\;\n\\sum_{i=1}^{6}x_{i}^{2}=30\n\\Bigr\\},\n\\]\nand  \n\\[\nH \\;=\\;\\Bigl\\{\\,y=(y_{1},\\ldots ,y_{6})\\in\\mathbb R^{6}\\;\\Bigm|\\;\ny_{1}=y_{2}=y_{3}=s>0,\\;\ny_{4}=y_{5}=y_{6}=t>0,\\;\ns>t,\\;\nst=9\n\\Bigr\\}.\n\\]\n\nFor \\(x\\in S\\) and \\(y\\in H\\) write the squared Euclidean distance  \n\\[\nD(x,y)\\;=\\;\\lVert x-y\\rVert ^{2}\n          \\;=\\;\\sum_{i=1}^{6}(x_{i}-y_{i})^{2}.\n\\]\n\n(A)  Prove that the minimum  \n\\[\nD_{\\min }=\\min \\bigl\\{D(x,y)\\;:\\;x\\in S,\\;y\\in H\\bigr\\}\n\\]\nexists.\n\n(B)  Determine the exact value of \\(D_{\\min}\\) (expressed with rational numbers and real radicals only) and describe \\emph{all} pairs \\((x,y)\\) that attain this minimum.\n\n\n\n------------------------------------------------------------",
      "solution": "Throughout write  \n\\[\ny=(s,s,s,t,t,t)\\qquad\\bigl(s>t>0,\\;st=9\\bigr).\n\\tag{0}\n\\]\n\n\\textbf{1.  Inner minimisation: \\(\\displaystyle \\Phi(y)=\\min_{x\\in S}D(x,y)\\).}\n\nMinimise \\(x\\mapsto D(x,y)\\) on \\(S\\) subject to the constraints  \n\\[\ng_{1}(x)=\\sum_{i=1}^{6}x_{i}-12=0,\\qquad\ng_{2}(x)=\\sum_{i=1}^{6}x_{i}^{2}-30=0.\n\\]\nIntroduce Lagrange multipliers \\(\\lambda,\\mu\\) and set  \n\\[\nL(x,\\lambda,\\mu)=D(x,y)+\\lambda g_{1}+\\mu g_{2}.\n\\]\nStationarity yields, for every \\(i\\),  \n\\[\n2(x_{i}-y_{i})+\\lambda+2\\mu x_{i}=0\n\\quad\\Longrightarrow\\quad\nx_{i}= \\frac{y_{i}-\\lambda/2}{1+\\mu},\n\\tag{1.1}\n\\]\nprovided \\(1+\\mu\\neq 0\\).  (The case \\(1+\\mu=0\\) leads to\na contradiction with the two quadratic constraints and is ruled out.)\n\nBecause the six coordinates of \\(y\\) take only the two values\n\\(s\\) and \\(t\\), every critical vector \\(x\\) produced by (1.1)\nassumes at most two distinct values.  By symmetry we may write  \n\\[\nx=(\\alpha,\\alpha,\\alpha,\\beta,\\beta,\\beta)\\qquad\\bigl(\\alpha,\\beta>0\\bigr).\n\\tag{1.2}\n\\]\nSubstituting (1.2) in the constraints gives the linear–quadratic system  \n\\[\n\\alpha+\\beta=4,\\qquad\n\\alpha^{2}+\\beta^{2}=10,\n\\]\nwhence \\(\\alpha\\beta=3\\) and therefore  \n\\[\n\\{\\alpha,\\beta\\}=\\{1,3\\}.\n\\]\nThus \\emph{two} stationary points are obtained:\n\\[\nx_{A}=(3,3,3,1,1,1),\\qquad\nx_{B}=(1,1,1,3,3,3).\n\\tag{1.3}\n\\]\n\n\\textbf{2.  Selecting the true minimiser for fixed \\(y\\).}\n\nCompute the difference\n\\[\n\\begin{aligned}\nD(x_{B},y)-D(x_{A},y)\n   &=\\sum_{i=1}^{6}\\bigl( (x_{B,i}-y_{i})^{2}-(x_{A,i}-y_{i})^{2}\\bigr) \\\\\n   &=3\\bigl[(1-s)^{2}-(3-s)^{2}+(3-t)^{2}-(1-t)^{2}\\bigr] \\\\\n   &=12\\,(s-t)\\;>\\;0.\n\\end{aligned}\n\\]\nHence \\(x_{A}\\) gives the smaller distance for every admissible\n\\((s,t)\\), so the inner minimum is uniquely attained at  \n\\[\nx^{\\ast}=x_{A}=(3,3,3,1,1,1).\n\\tag{2.1}\n\\]\nTherefore\n\\[\n\\boxed{\\;\n\\Phi(y)\\;=\\;D(x^{\\ast},y)\n         \\;=\\;3\\Bigl[(s-3)^{2}+\\Bigl(\\tfrac{9}{s}-1\\Bigr)^{2}\\Bigr]\n         =:F(s)\n\\;},\n\\qquad s>3.\n\\tag{2.2}\n\\]\n\n\\textbf{3.  Outer minimisation: \\(\\displaystyle D_{\\min}=\\min_{s>3}F(s).\\)}\n\nDefine\n\\[\nf(s)=(s-3)^{2}+\\Bigl(\\tfrac{9}{s}-1\\Bigr)^{2},\\qquad s>3,\n\\quad\\text{so that }F(s)=3f(s).\n\\]\n\n\\emph{Convexity.}  Differentiating,\n\\[\nf'(s)=2(s-3)-\\frac{18(9-s)}{s^{3}}\n      =\\frac{2\\bigl(s^{4}-3s^{3}+9s-81\\bigr)}{s^{3}},\n\\tag{3.1}\n\\]\n\\[\nf''(s)=2+\\frac{18}{s^{3}}+\\frac{54(9-s)}{s^{4}}>0\\quad(s>3),\n\\]\nso \\(f\\) (hence \\(F\\)) is strictly convex on \\((3,\\infty)\\).\nConsequently there exists a unique minimiser \\(s_{0}>3\\), determined\nby the quartic equation coming from \\(f'(s)=0\\):\n\\[\ns_{0}^{4}-3s_{0}^{3}+9s_{0}-81=0.\n\\tag{3.2}\n\\]\n\n\\textbf{4.  Exact solution of the quartic.}\n\nPut \\(s=z+\\dfrac34\\) to eliminate the cubic term.  Ferrari’s method\nreduces (3.2) to the depressed quartic\n\\[\nz^{4}+pz^{2}+qz+r=0,\\qquad\np=-\\frac{27}{8},\\;q=\\frac{45}{8},\\;r=-\\frac{19251}{256}.\n\\tag{4.1}\n\\]\n\nIntroduce the resolvent cubic\n\\[\ny^{3}-\\frac{p}{2}y^{2}-ry+\\Bigl(\\frac{rp}{2}-\\frac{q^{2}}{8}\\Bigr)=0.\n\\]\nClearing denominators,\n\\[\ny^{3}+\\frac{9}{16}y^{2}+\\frac{19251}{256}y+\\frac{1075275}{4096}=0.\n\\tag{4.2}\n\\]\n\nCardano’s formula gives the unique real root\n\\[\ny_{0}=\\sqrt[3]{-\\frac{81}{2}+\\frac{\\sqrt{1\\,075\\,275}}{8}}\n      +\\sqrt[3]{-\\frac{81}{2}-\\frac{\\sqrt{1\\,075\\,275}}{8}}\n      -\\frac{9}{16}.\n\\]\n\nSet  \n\\[\nU=\\sqrt{\\,2y_{0}+\\frac{27}{8}},\\qquad\nV=\\sqrt{-2y_{0}-\\frac{27}{8}+\\frac{45}{4U}}.\n\\]\nAmong the four resulting roots \\(z=\\pm\\dfrac{U\\pm V}{2}\\) only one\nyields \\(s=z+\\dfrac34>3\\).  Define\n\\[\n\\boxed{\\;\ns_{0}=\\frac34+\\frac{U+V}{2}},\\qquad\n\\boxed{\\;\nt_{0}=\\frac{9}{s_{0}}}.\n\\tag{4.3}\n\\]\n(Numerically \\(s_{0}\\approx3.832526901\\), \\(t_{0}\\approx2.348389020\\).)\n\n\\textbf{5.  Minimum value.}\n\nUsing (2.2) and (4.3),\n\\[\n\\boxed{\\;\nD_{\\min}=3\\Bigl[(s_{0}-3)^{2}+\\Bigl(\\tfrac{9}{s_{0}}-1\\Bigr)^{2}\\Bigr]\\;}.\n\\tag{5.1}\n\\]\nNumerically \\(D_{\\min}\\approx7.526960797\\).\n\n\\textbf{6.  All minimising pairs.}\n\nFor each \\(y\\in H\\) the inner minimisation is uniquely solved by\n\\(x^{\\ast}=x_{A}\\) (Step 2), while the outer problem has the unique\nminimiser \\(s=s_{0}\\) (strict convexity in Step 3).  Hence the only\npair attaining \\(D_{\\min}\\) is  \n\\[\n\\boxed{\\;\nx^{\\ast}=(3,3,3,1,1,1),\\qquad\ny^{\\ast}=(s_{0},s_{0},s_{0},t_{0},t_{0},t_{0})\\;}.\n\\tag{6.1}\n\\]\n\n\\textbf{7.  Existence of the global minimum (Part A).}\n\nThe set \\(S\\) is compact (closed and bounded).  For every fixed\n\\(y\\in H\\) the map \\(x\\mapsto D(x,y)\\) is continuous, hence attains a\nminimum on \\(S\\); so \\(\\Phi:H\\to\\mathbb R\\) defined by \\(\\Phi(y)=\\min_{x\\in S}D(x,y)\\)\nis well defined.  Formula (2.2) shows that\n\\(\\Phi(y)=F(s)\\to\\infty\\) as \\(s\\to\\infty\\) and\n\\(\\Phi(y)\\to12\\) as \\(s\\downarrow 3\\).  Therefore\n\\(\\Phi\\) attains its global minimum at \\(s=s_{0}\\).\nSince \\(H\\) is continuous in \\(s\\) and the inner minimiser\n\\(x^{\\ast}\\) depends continuously on \\(y\\), the pair\n\\((x^{\\ast},y^{\\ast})\\) realises \\(D_{\\min}\\), completing Part (A).\n\n\\hfill\\(\\blacksquare\\)\n\n\n\n------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.531239",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimensionality: the problem lives in ℝ⁶ rather than ℝ²,\n   introducing six variables instead of two.\n2. Multiple interacting constraints: the feasible set S is the\n   intersection of a positive orthant with both a hyperplane\n   (Σxᵢ = 12) and a sphere (Σxᵢ² = 30); H is a 1–dimensional manifold\n   defined simultaneously by an ordering condition (s>t) and a\n   high–degree algebraic constraint (s³t³ = 9³).\n3. Sophisticated optimisation: one must use Lagrange multipliers to\n   locate the nearest point of S to a given y, revealing an\n   internal two–point structure {1,3}.\n4. Reduction to a quartic: the remaining single–variable problem\n   leads to the non-trivial quartic equation s⁴ – 3s³ + 9s – 81 = 0,\n   whose root cannot be guessed and must be solved with Ferrari’s\n   method or an equivalent algebraic apparatus.\n5. Exact closed form: the final answer demands radicals of nested\n   radicals (formula (11)), far beyond the elementary arithmetic\n   needed for the original 2-dimensional problem.\n6. Subtle case distinction: one has to decide which block of\n   coordinates should be matched with which value of x,\n   a point that disappears entirely in the original statement.\n\nAltogether the enhanced variant compels the solver to combine\nmultivariate calculus, symmetric-constraint reasoning, and full\nquartic algebra, making it substantially harder than both the\noriginal Putnam problem and the current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}