{
  "index": "2023-B-4",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "For a nonnegative integer $n$ and a strictly increasing sequence of real numbers $t_0,t_1,\\dots,t_n$, let $f(t)$ be the corresponding real-valued function defined for $t \\geq t_0$ by the following properties:\n\\begin{enumerate}\n\\item[(a)] $f(t)$ is continuous for $t \\geq t_0$, and is twice differentiable for all $t>t_0$ other than $t_1,\\dots,t_n$;\n\\item[(b)] $f(t_0) = 1/2$;\n\\item[(c)] $\\lim_{t \\to t_k^+} f'(t) = 0$ for $0 \\leq k \\leq n$;\n\\item[(d)] For $0 \\leq k \\leq n-1$, we have $f''(t) = k+1$ when $t_k < t< t_{k+1}$, and $f''(t) = n+1$ when $t>t_n$.\n\\end{enumerate}\nConsidering all choices of $n$ and $t_0,t_1,\\dots,t_n$ such that $t_k \\geq t_{k-1}+1$ for $1 \\leq k \\leq n$, what is the least possible value of $T$ for which $f(t_0+T) = 2023$?",
  "solution": "The minimum value of $T$ is 29.\n\nWrite $t_{n+1} = t_0+T$ and define $s_k = t_k-t_{k-1}$ for $1\\leq k\\leq n+1$. On $[t_{k-1},t_k]$, we have $f'(t) = k(t-t_{k-1})$ and so $f(t_k)-f(t_{k-1}) = \\frac{k}{2} s_k^2$. Thus if we define\n\\[\ng(s_1,\\ldots,s_{n+1}) = \\sum_{k=1}^{n+1} ks_k^2,\n\\]\nthen we want to minimize $\\sum_{k=1}^{n+1} s_k = T$ (for all possible values of $n$) subject to the constraints that $g(s_1,\\ldots,s_{n+1}) = 4045$ and $s_k \\geq 1$ for $k \\leq n$.\n\nWe first note that a minimum value for $T$ is indeed achieved. To see this, note that the constraints $g(s_1,\\ldots,s_{n+1}) = 4045$ and $s_k \\geq 1$ place an upper bound on $n$. For fixed $n$, the constraint $g(s_1,\\ldots,s_{n+1}) = 4045$ places an upper bound on each $s_k$, whence the set of $(s_1,\\ldots,s_{n+1})$ on which we want to minimize $\\sum s_k$ is a compact subset of $\\mathbb{R}^{n+1}$.\n\nNow say that $T_0$ is the minimum value of $\\sum_{k=1}^{n+1} s_k$ (over all $n$ and $s_1,\\ldots,s_{n+1}$), achieved by $(s_1,\\ldots,s_{n+1}) = (s_1^0,\\ldots,s_{n+1}^0)$. Observe that there cannot be another $(s_1,\\ldots,s_{n'+1})$ with the same sum, $\\sum_{k=1}^{n'+1} s_k = T_0$, satisfying $g(s_1,\\ldots,s_{n'+1}) > 4045$; otherwise, the function $f$ for $(s_1,\\ldots,s_{n'+1})$ would satisfy $f(t_0+T_0) > 4045$ and there would be some $T<T_0$ such that $f(t_0+T) = 4045$ by the intermediate value theorem.\n\nWe claim that $s_{n+1}^0 \\geq 1$ and $s_k^0 = 1$ for $1\\leq k\\leq n$. If $s_{n+1}^0<1$ then\n\\begin{align*}\n& g(s_1^0,\\ldots,s_{n-1}^0,s_n^0+s_{n+1}^0)-g(s_1^0,\\ldots,s_{n-1}^0,s_n^0,s_{n+1}^0) \\\\\n&\\quad = s_{n+1}^0(2ns_n^0-s_{n+1}^0) > 0,\n\\end{align*}\ncontradicting our observation from the previous paragraph. Thus $s_{n+1}^0 \\geq 1$. If $s_k^0>1$ for some $1\\leq k\\leq n$ then replacing $(s_k^0,s_{n+1}^0)$ by $(1,s_{n+1}^0+s_k^0-1)$ increases $g$:\n\\begin{align*}\n&g(s_1^0,\\ldots,1,\\ldots,s_{n+1}^0+s_k^0-1)-g(s_1^0,\\ldots,s_k^0,\\ldots,s_{n+1}^0) \\\\\n&\\quad= (s_k^0-1)((n+1-k)(s_k^0+1)+2(n+1)(s_{n+1}^0-1)) > 0,\n\\end{align*}\nagain contradicting the observation. This establishes the claim.\n\nGiven that $s_k^0 = 1$ for $1 \\leq k \\leq n$, we have\n$T = s_{n+1}^0 + n$ and\n\\[\ng(s_1^0,\\dots,s_{n+1}^0) = \\frac{n(n+1)}{2} + (n+1)(T-n)^2.\n\\]\nSetting this equal to 4045 and solving for $T$ yields\n\\[\nT = n+\\sqrt{\\frac{4045}{n+1} - \\frac{n}{2}}.\n\\]\nFor $n=9$ this yields $T = 29$; it thus suffices to show that for all $n$, \n\\[\nn+\\sqrt{\\frac{4045}{n+1} - \\frac{n}{2}} \\geq 29.\n\\]\nThis is evident for $n \\geq 30$. For $n \\leq 29$, rewrite the claim as\n\\[\n\\sqrt{\\frac{4045}{n+1} - \\frac{n}{2}} \\geq 29-n;\n\\]\nwe then obtain an equivalent inequality by squaring both sides:\n\\[\n\\frac{4045}{n+1} - \\frac{n}{2} \\geq n^2-58n+841.\n\\]\nClearing denominators, gathering all terms to one side, and factoring puts this in the form\n\\[\n(9-n)(n^2 - \\frac{95}{2} n + 356) \\geq 0.\n\\]\nThe quadratic factor $Q(n)$ has a minimum at $\\frac{95}{4} = 23.75$\nand satisfies $Q(8) = 40, Q(10) = -19$; it is thus positive for $n \\leq 8$ and negative for $10 \\leq n \\leq 29$.",
  "vars": [
    "n",
    "t_0",
    "t_1",
    "t_n",
    "t",
    "f",
    "k",
    "t_k",
    "g",
    "s_k",
    "s_1",
    "s_n+1",
    "T",
    "t_n+1",
    "T_0",
    "t_k-1",
    "t_k+1",
    "s_n",
    "s_n-1",
    "Q",
    "R"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "n": "segcount",
        "t_0": "starttime",
        "t_1": "firsttime",
        "t_n": "endtime",
        "t": "elapsedtime",
        "f": "funcvalue",
        "k": "stageindex",
        "t_k": "intermtime",
        "g": "quadraticsum",
        "s_k": "spandiff",
        "s_1": "firstspan",
        "s_n+1": "lastspan",
        "T": "totalspan",
        "t_n+1": "postendtime",
        "T_0": "minimumspan",
        "t_k-1": "previoustime",
        "t_k+1": "nexttime",
        "s_n": "endspan",
        "s_n-1": "penultspan",
        "Q": "auxquadratic",
        "R": "auxiliaryr"
      },
      "question": "For a nonnegative integer $\\segcount$ and a strictly increasing sequence of real numbers $\\starttime,\\firsttime,\\dots,\\endtime$, let $\\funcvalue(\\elapsedtime)$ be the corresponding real-valued function defined for $\\elapsedtime \\geq \\starttime$ by the following properties:\n\\begin{enumerate}\n\\item[(a)] $\\funcvalue(\\elapsedtime)$ is continuous for $\\elapsedtime \\geq \\starttime$, and is twice differentiable for all $\\elapsedtime>\\starttime$ other than $\\firsttime,\\dots,\\endtime$;\n\\item[(b)] $\\funcvalue(\\starttime) = 1/2$;\n\\item[(c)] $\\lim_{\\elapsedtime \\to \\intermtime^+} \\funcvalue'(\\elapsedtime) = 0$ for $0 \\leq \\stageindex \\leq \\segcount$;\n\\item[(d)] For $0 \\leq \\stageindex \\leq \\segcount-1$, we have $\\funcvalue''(\\elapsedtime) = \\stageindex+1$ when $\\intermtime < \\elapsedtime< \\nexttime$, and $\\funcvalue''(\\elapsedtime) = \\segcount+1$ when $\\elapsedtime>\\endtime$.\n\\end{enumerate}\nConsidering all choices of $\\segcount$ and $\\starttime,\\firsttime,\\dots,\\endtime$ such that $\\intermtime \\geq \\previoustime+1$ for $1 \\leq \\stageindex \\leq \\segcount$, what is the least possible value of $\\totalspan$ for which $\\funcvalue(\\starttime+\\totalspan) = 2023$?",
      "solution": "The minimum value of $\\totalspan$ is 29.\n\nWrite $\\postendtime = \\starttime+\\totalspan$ and define $\\spandiff = \\intermtime-\\previoustime$ for $1\\leq \\stageindex\\leq \\segcount+1$. On $[\\previoustime,\\intermtime]$, we have $\\funcvalue'(\\elapsedtime) = \\stageindex(\\elapsedtime-\\previoustime)$ and so $\\funcvalue(\\intermtime)-\\funcvalue(\\previoustime) = \\frac{\\stageindex}{2} \\spandiff^2$. Thus if we define\n\\[\n\\quadraticsum(\\firstspan,\\ldots,\\lastspan) = \\sum_{\\stageindex=1}^{\\segcount+1} \\stageindex\\spandiff^2,\n\\]\nthen we want to minimize $\\sum_{\\stageindex=1}^{\\segcount+1} \\spandiff = \\totalspan$ (for all possible values of $\\segcount$) subject to the constraints that $\\quadraticsum(\\firstspan,\\ldots,\\lastspan) = 4045$ and $\\spandiff \\geq 1$ for $\\stageindex \\leq \\segcount$.\n\nWe first note that a minimum value for $\\totalspan$ is indeed achieved. To see this, note that the constraints $\\quadraticsum(\\firstspan,\\ldots,\\lastspan) = 4045$ and $\\spandiff \\geq 1$ place an upper bound on $\\segcount$. For fixed $\\segcount$, the constraint $\\quadraticsum(\\firstspan,\\ldots,\\lastspan) = 4045$ places an upper bound on each $\\spandiff$, whence the set of $(\\firstspan,\\ldots,\\lastspan)$ on which we want to minimize $\\sum \\spandiff$ is a compact subset of $\\mathbb{\\auxiliaryr}^{\\segcount+1}$.\n\nNow say that $\\minimumspan$ is the minimum value of $\\sum_{\\stageindex=1}^{\\segcount+1} \\spandiff$ (over all $\\segcount$ and $\\firstspan,\\ldots,\\lastspan$), achieved by $(\\firstspan,\\ldots,\\lastspan) = (\\firstspan^0,\\ldots,\\lastspan^0)$. Observe that there cannot be another $(\\firstspan,\\ldots,\\lastspan)$ with the same sum, $\\sum_{\\stageindex=1}^{\\segcount'+1} \\spandiff = \\minimumspan$, satisfying $\\quadraticsum(\\firstspan,\\ldots,\\lastspan) > 4045$; otherwise, the function $\\funcvalue$ for $(\\firstspan,\\ldots,\\lastspan)$ would satisfy $\\funcvalue(\\starttime+\\minimumspan) > 4045$ and there would be some $\\totalspan<\\minimumspan$ such that $\\funcvalue(\\starttime+\\totalspan) = 4045$ by the intermediate value theorem.\n\nWe claim that $\\lastspan^0 \\geq 1$ and $\\spandiff^0 = 1$ for $1\\leq \\stageindex\\leq \\segcount$. If $\\lastspan^0<1$ then\n\\begin{align*}\n& \\quadraticsum(\\firstspan^0,\\ldots,\\penultspan^0,\\endspan^0+\\lastspan^0)-\\quadraticsum(\\firstspan^0,\\ldots,\\penultspan^0,\\endspan^0,\\lastspan^0) \\\\\n&\\quad = \\lastspan^0(2\\segcount\\endspan^0-\\lastspan^0) > 0,\n\\end{align*}\ncontradicting our observation from the previous paragraph. Thus $\\lastspan^0 \\geq 1$. If $\\spandiff^0>1$ for some $1\\leq \\stageindex\\leq \\segcount$ then replacing $(\\spandiff^0,\\lastspan^0)$ by $(1,\\lastspan^0+\\spandiff^0-1)$ increases $\\quadraticsum$:\n\\begin{align*}\n&\\quadraticsum(\\firstspan^0,\\ldots,1,\\ldots,\\lastspan^0+\\spandiff^0-1)-\\quadraticsum(\\firstspan^0,\\ldots,\\spandiff^0,\\ldots,\\lastspan^0) \\\\\n&\\quad= (\\spandiff^0-1)((\\segcount+1-\\stageindex)(\\spandiff^0+1)+2(\\segcount+1)(\\lastspan^0-1)) > 0,\n\\end{align*}\nagain contradicting the observation. This establishes the claim.\n\nGiven that $\\spandiff^0 = 1$ for $1 \\leq \\stageindex \\leq \\segcount$, we have\n$\\totalspan = \\lastspan^0 + \\segcount$ and\n\\[\n\\quadraticsum(\\firstspan^0,\\dots,\\lastspan^0) = \\frac{\\segcount(\\segcount+1)}{2} + (\\segcount+1)(\\totalspan-\\segcount)^2.\n\\]\nSetting this equal to 4045 and solving for $\\totalspan$ yields\n\\[\n\\totalspan = \\segcount+\\sqrt{\\frac{4045}{\\segcount+1} - \\frac{\\segcount}{2}}.\n\\]\nFor $\\segcount=9$ this yields $\\totalspan = 29$; it thus suffices to show that for all $\\segcount$,\n\\[\n\\segcount+\\sqrt{\\frac{4045}{\\segcount+1} - \\frac{\\segcount}{2}} \\geq 29.\n\\]\nThis is evident for $\\segcount \\geq 30$. For $\\segcount \\leq 29$, rewrite the claim as\n\\[\n\\sqrt{\\frac{4045}{\\segcount+1} - \\frac{\\segcount}{2}} \\geq 29-\\segcount;\n\\]\nwe then obtain an equivalent inequality by squaring both sides:\n\\[\n\\frac{4045}{\\segcount+1} - \\frac{\\segcount}{2} \\geq \\segcount^2-58\\segcount+841.\n\\]\nClearing denominators, gathering all terms to one side, and factoring puts this in the form\n\\[\n(9-\\segcount)(\\segcount^2 - \\frac{95}{2} \\segcount + 356) \\geq 0.\n\\]\nThe quadratic factor $\\auxquadratic(\\segcount)$ has a minimum at $\\frac{95}{4} = 23.75$ and satisfies $\\auxquadratic(8) = 40, \\auxquadratic(10) = -19$; it is thus positive for $\\segcount \\leq 8$ and negative for $10 \\leq \\segcount \\leq 29$."
    },
    "descriptive_long_confusing": {
      "map": {
        "n": "lanterns",
        "t_0": "meadowrise",
        "t_1": "harborsail",
        "t_n": "orchardgap",
        "t": "sundialer",
        "f": "maplewind",
        "k": "pebblestep",
        "t_k": "brooktrail",
        "g": "silkgrove",
        "s_k": "fernshadow",
        "s_1": "cedarbluff",
        "s_n+1": "pinehollow",
        "T": "glenstream",
        "t_n+1": "canyonridge",
        "T_0": "valleydawn",
        "t_k-1": "riverbend",
        "t_k+1": "hillcrest",
        "s_n": "thicketbay",
        "s_n-1": "grovefield",
        "Q": "woodlark",
        "R": "stonehaven"
      },
      "question": "For a nonnegative integer $lanterns$ and a strictly increasing sequence of real numbers $meadowrise,harborsail,\\dots,orchardgap$, let $maplewind(sundialer)$ be the corresponding real-valued function defined for $sundialer \\geq meadowrise$ by the following properties:\n\\begin{enumerate}\n\\item[(a)] $maplewind(sundialer)$ is continuous for $sundialer \\geq meadowrise$, and is twice differentiable for all $sundialer>meadowrise$ other than $harborsail,\\dots,orchardgap$;\n\\item[(b)] $maplewind(meadowrise) = 1/2$;\n\\item[(c)] $\\lim_{sundialer \\to brooktrail^+} maplewind'(sundialer) = 0$ for $0 \\leq pebblestep \\leq lanterns$;\n\\item[(d)] For $0 \\leq pebblestep \\leq lanterns-1$, we have $maplewind''(sundialer) = pebblestep+1$ when $brooktrail < sundialer< t_{k+1}$, and $maplewind''(sundialer) = lanterns+1$ when $sundialer>orchardgap$.\n\\end{enumerate}\nConsidering all choices of $lanterns$ and $meadowrise,harborsail,\\dots,orchardgap$ such that $brooktrail \\geq t_{k-1}+1$ for $1 \\leq pebblestep \\leq lanterns$, what is the least possible value of $glenstream$ for which $maplewind(meadowrise+glenstream) = 2023$?",
      "solution": "The minimum value of $glenstream$ is 29.\n\nWrite $canyonridge = meadowrise+glenstream$ and define $fernshadow = brooktrail-riverbend$ for $1\\leq pebblestep\\leq lanterns+1$. On $[riverbend,brooktrail]$, we have $maplewind'(sundialer) = pebblestep(sundialer-riverbend)$ and so $maplewind(brooktrail)-maplewind(riverbend) = \\frac{\\pebblestep}{2} fernshadow^{2}$. Thus if we define\n\\[\nsilkgrove(cedarbluff,\\ldots,pinehollow) = \\sum_{pebblestep=1}^{lanterns+1} pebblestep\\, fernshadow^{2},\n\\]\nthen we want to minimize $\\sum_{pebblestep=1}^{lanterns+1} fernshadow = glenstream$ (for all possible values of $lanterns$) subject to the constraints that $silkgrove(cedarbluff,\\ldots,pinehollow) = 4045$ and $fernshadow \\geq 1$ for $pebblestep \\leq lanterns$.\n\nWe first note that a minimum value for $glenstream$ is indeed achieved. To see this, note that the constraints $silkgrove(cedarbluff,\\ldots,pinehollow) = 4045$ and $fernshadow \\geq 1$ place an upper bound on $lanterns$. For fixed $lanterns$, the constraint $silkgrove(cedarbluff,\\ldots,pinehollow) = 4045$ places an upper bound on each $fernshadow$, whence the set of $(cedarbluff,\\ldots,pinehollow)$ on which we want to minimize $\\sum fernshadow$ is a compact subset of $\\mathbb{R}^{\\lanterns+1}$.\n\nNow say that $valleydawn$ is the minimum value of $\\sum_{pebblestep=1}^{lanterns+1} fernshadow$ (over all $lanterns$ and $cedarbluff,\\ldots,pinehollow$), achieved by $(cedarbluff,\\ldots,pinehollow) = (cedarbluff^{0},\\ldots,pinehollow^{0})$. Observe that there cannot be another $(cedarbluff,\\ldots,s_{n'+1})$ with the same sum, $\\sum_{pebblestep=1}^{n'+1} fernshadow = valleydawn$, satisfying $silkgrove(cedarbluff,\\ldots,s_{n'+1}) > 4045$; otherwise, the function $maplewind$ for $(cedarbluff,\\ldots,s_{n'+1})$ would satisfy $maplewind(meadowrise+valleydawn) > 4045$ and there would be some $glenstream<valleydawn$ such that $maplewind(meadowrise+glenstream) = 4045$ by the intermediate value theorem.\n\nWe claim that $pinehollow^{0} \\geq 1$ and $fernshadow^{0} = 1$ for $1\\leq pebblestep\\leq lanterns$. If $pinehollow^{0}<1$ then\n\\begin{align*}\n& silkgrove(cedarbluff^{0},\\ldots,grovefield^{0},thicketbay^{0}+pinehollow^{0})-silkgrove(cedarbluff^{0},\\ldots,grovefield^{0},thicketbay^{0},pinehollow^{0}) \\\n&\\quad = pinehollow^{0}(2\\lanterns\\, thicketbay^{0}-pinehollow^{0}) > 0,\n\\end{align*}\ncontradicting our observation from the previous paragraph. Thus $pinehollow^{0} \\geq 1$. If $fernshadow^{0}>1$ for some $1\\leq pebblestep\\leq lanterns$ then replacing $(fernshadow^{0},pinehollow^{0})$ by $(1,pinehollow^{0}+fernshadow^{0}-1)$ increases $silkgrove$:\n\\begin{align*}\n&silkgrove(cedarbluff^{0},\\ldots,1,\\ldots,pinehollow^{0}+fernshadow^{0}-1)-silkgrove(cedarbluff^{0},\\ldots,fernshadow^{0},\\ldots,pinehollow^{0}) \\\n&\\quad= (fernshadow^{0}-1)((\\lanterns+1-pebblestep)(fernshadow^{0}+1)+2(\\lanterns+1)(pinehollow^{0}-1)) > 0,\n\\end{align*}\nagain contradicting the observation. This establishes the claim.\n\nGiven that $fernshadow^{0} = 1$ for $1 \\leq pebblestep \\leq lanterns$, we have\n$glenstream = pinehollow^{0} + lanterns$ and\n\\[\nsilkgrove(cedarbluff^{0},\\dots,pinehollow^{0}) = \\frac{\\lanterns(\\lanterns+1)}{2} + (\\lanterns+1)(glenstream-\\lanterns)^2.\n\\]\nSetting this equal to 4045 and solving for $glenstream$ yields\n\\[\nglenstream = \\lanterns+\\sqrt{\\frac{4045}{\\lanterns+1} - \\frac{\\lanterns}{2}}.\n\\]\nFor $\\lanterns=9$ this yields $glenstream = 29$; it thus suffices to show that for all $\\lanterns$, \n\\[\n\\lanterns+\\sqrt{\\frac{4045}{\\lanterns+1} - \\frac{\\lanterns}{2}} \\geq 29.\n\\]\nThis is evident for $\\lanterns \\geq 30$. For $\\lanterns \\leq 29$, rewrite the claim as\n\\[\n\\sqrt{\\frac{4045}{\\lanterns+1} - \\frac{\\lanterns}{2}} \\geq 29-\\lanterns;\n\\]\nwe then obtain an equivalent inequality by squaring both sides:\n\\[\n\\frac{4045}{\\lanterns+1} - \\frac{\\lanterns}{2} \\geq \\lanterns^2-58\\lanterns+841.\n\\]\nClearing denominators, gathering all terms to one side, and factoring puts this in the form\n\\[\n(9-\\lanterns)(\\lanterns^2 - \\frac{95}{2} \\lanterns + 356) \\geq 0.\n\\]\nThe quadratic factor $woodlark(\\lanterns)$ has a minimum at $\\frac{95}{4} = 23.75$\nand satisfies $woodlark(8) = 40, woodlark(10) = -19$; it is thus positive for $\\lanterns \\leq 8$ and negative for $10 \\leq \\lanterns \\leq 29$. "
    },
    "descriptive_long_misleading": {
      "map": {
        "n": "infinitevalue",
        "t_0": "finalmoment",
        "t_1": "distanttime",
        "t_n": "earliesttime",
        "t": "spacedimension",
        "f": "constantvalue",
        "k": "aggregate",
        "t_k": "irrelevanttime",
        "g": "difference",
        "s_k": "overlapvalue",
        "s_1": "overlapfirst",
        "s_n+1": "overlaplast",
        "T": "infiniteperiod",
        "t_n+1": "preinitialmoment",
        "T_0": "maximalperiod",
        "t_k-1": "nexttimeindex",
        "t_k+1": "previousinstant",
        "s_n": "overlapmiddle",
        "s_n-1": "overlappenultimate",
        "Q": "linearpart",
        "R": "imaginarynumbers"
      },
      "question": "For a nonnegative integer $\\infinitevalue$ and a strictly increasing sequence of real numbers $\\finalmoment,\\distanttime,\\dots,\\earliesttime$, let $\\constantvalue(\\spacedimension)$ be the corresponding real-valued function defined for $\\spacedimension \\geq \\finalmoment$ by the following properties:\n\\begin{enumerate}\n\\item[(a)] $\\constantvalue(\\spacedimension)$ is continuous for $\\spacedimension \\geq \\finalmoment$, and is twice differentiable for all $\\spacedimension>\\finalmoment$ other than $\\distanttime,\\dots,\\earliesttime$;\n\\item[(b)] $\\constantvalue(\\finalmoment) = 1/2$;\n\\item[(c)] $\\lim_{\\spacedimension \\to \\irrelevanttime^+} \\constantvalue'(\\spacedimension) = 0$ for $0 \\leq \\aggregate \\leq \\infinitevalue$;\n\\item[(d)] For $0 \\leq \\aggregate \\leq \\infinitevalue-1$, we have $\\constantvalue''(\\spacedimension) = \\aggregate+1$ when $\\irrelevanttime < \\spacedimension< \\previousinstant$, and $\\constantvalue''(\\spacedimension) = \\infinitevalue+1$ when $\\spacedimension>\\earliesttime$.\n\\end{enumerate}\nConsidering all choices of $\\infinitevalue$ and $\\finalmoment,\\distanttime,\\dots,\\earliesttime$ such that $\\irrelevanttime \\geq \\nexttimeindex+1$ for $1 \\leq \\aggregate \\leq \\infinitevalue$, what is the least possible value of $\\infiniteperiod$ for which $\\constantvalue(\\finalmoment+\\infiniteperiod) = 2023$?",
      "solution": "The minimum value of $\\infiniteperiod$ is 29.\n\nWrite $\\preinitialmoment = \\finalmoment+\\infiniteperiod$ and define $\\overlapvalue = \\irrelevanttime-\\nexttimeindex$ for $1\\leq \\aggregate\\leq \\infinitevalue+1$. On $[\\nexttimeindex,\\irrelevanttime]$, we have $\\constantvalue'(\\spacedimension) = \\aggregate(\\spacedimension-\\nexttimeindex)$ and so $\\constantvalue(\\irrelevanttime)-\\constantvalue(\\nexttimeindex) = \\frac{\\aggregate}{2} \\overlapvalue^2$. Thus if we define\n\\[\n\\difference(\\overlapfirst,\\ldots,\\overlaplast) = \\sum_{\\aggregate=1}^{\\infinitevalue+1} \\aggregate\\overlapvalue^2,\n\\]\nthen we want to minimize $\\sum_{\\aggregate=1}^{\\infinitevalue+1} \\overlapvalue = \\infiniteperiod$ (for all possible values of $\\infinitevalue$) subject to the constraints that $\\difference(\\overlapfirst,\\ldots,\\overlaplast) = 4045$ and $\\overlapvalue \\geq 1$ for $\\aggregate \\leq \\infinitevalue$.\n\nWe first note that a minimum value for $\\infiniteperiod$ is indeed achieved. To see this, note that the constraints $\\difference(\\overlapfirst,\\ldots,\\overlaplast) = 4045$ and $\\overlapvalue \\geq 1$ place an upper bound on $\\infinitevalue$. For fixed $\\infinitevalue$, the constraint $\\difference(\\overlapfirst,\\ldots,\\overlaplast) = 4045$ places an upper bound on each $\\overlapvalue$, whence the set of $(\\overlapfirst,\\ldots,\\overlaplast)$ on which we want to minimize $\\sum \\overlapvalue$ is a compact subset of $\\mathbb{\\imaginarynumbers}^{\\infinitevalue+1}$.\n\nNow say that $\\maximalperiod$ is the minimum value of $\\sum_{\\aggregate=1}^{\\infinitevalue+1} \\overlapvalue$ (over all $\\infinitevalue$ and $\\overlapfirst,\\ldots,\\overlaplast$), achieved by $(\\overlapfirst,\\ldots,\\overlaplast) = (\\overlapfirst^0,\\ldots,\\overlaplast^0)$. Observe that there cannot be another $(\\overlapfirst,\\ldots,\\overlaplast)$ with the same sum, $\\sum_{\\aggregate=1}^{\\infinitevalue+1} \\overlapvalue = \\maximalperiod$, satisfying $\\difference(\\overlapfirst,\\ldots,\\overlaplast) > 4045$; otherwise, the function $\\constantvalue$ for $(\\overlapfirst,\\ldots,\\overlaplast)$ would satisfy $\\constantvalue(\\finalmoment+\\maximalperiod) > 4045$ and there would be some $\\infiniteperiod<\\maximalperiod$ such that $\\constantvalue(\\finalmoment+\\infiniteperiod) = 4045$ by the intermediate value theorem.\n\nWe claim that $\\overlaplast^0 \\geq 1$ and $\\overlapvalue^0 = 1$ for $1\\leq \\aggregate\\leq \\infinitevalue$. If $\\overlaplast^0<1$ then\n\\begin{align*}\n& \\difference(\\overlapfirst^0,\\ldots,\\overlapmiddle^0+\\overlaplast^0)-\\difference(\\overlapfirst^0,\\ldots,\\overlapmiddle^0,\\overlaplast^0) \\\\\n&\\quad = \\overlaplast^0(2\\infinitevalue\\overlapmiddle^0-\\overlaplast^0) > 0,\n\\end{align*}\ncontradicting our observation from the previous paragraph. Thus $\\overlaplast^0 \\geq 1$. If $\\overlapvalue^0>1$ for some $1\\leq \\aggregate\\leq \\infinitevalue$ then replacing $(\\overlapvalue^0,\\overlaplast^0)$ by $(1,\\overlaplast^0+\\overlapvalue^0-1)$ increases $\\difference$:\n\\begin{align*}\n&\\difference(\\overlapfirst^0,\\ldots,1,\\ldots,\\overlaplast^0+\\overlapvalue^0-1)-\\difference(\\overlapfirst^0,\\ldots,\\overlapvalue^0,\\ldots,\\overlaplast^0) \\\\\n&\\quad= (\\overlapvalue^0-1)((\\infinitevalue+1-\\aggregate)(\\overlapvalue^0+1)+2(\\infinitevalue+1)(\\overlaplast^0-1)) > 0,\n\\end{align*}\nagain contradicting the observation. This establishes the claim.\n\nGiven that $\\overlapvalue^0 = 1$ for $1 \\leq \\aggregate \\leq \\infinitevalue$, we have\n$\\infiniteperiod = \\overlaplast^0 + \\infinitevalue$ and\n\\[\n\\difference(\\overlapfirst^0,\\dots,\\overlaplast^0) = \\frac{\\infinitevalue(\\infinitevalue+1)}{2} + (\\infinitevalue+1)(\\infiniteperiod-\\infinitevalue)^2.\n\\]\nSetting this equal to 4045 and solving for $\\infiniteperiod$ yields\n\\[\n\\infiniteperiod = \\infinitevalue+\\sqrt{\\frac{4045}{\\infinitevalue+1} - \\frac{\\infinitevalue}{2}}.\n\\]\nFor $\\infinitevalue=9$ this yields $\\infiniteperiod = 29$; it thus suffices to show that for all $\\infinitevalue$, \n\\[\n\\infinitevalue+\\sqrt{\\frac{4045}{\\infinitevalue+1} - \\frac{\\infinitevalue}{2}} \\geq 29.\n\\]\nThis is evident for $\\infinitevalue \\geq 30$. For $\\infinitevalue \\leq 29$, rewrite the claim as\n\\[\n\\sqrt{\\frac{4045}{\\infinitevalue+1} - \\frac{\\infinitevalue}{2}} \\geq 29-\\infinitevalue;\n\\]\nwe then obtain an equivalent inequality by squaring both sides:\n\\[\n\\frac{4045}{\\infinitevalue+1} - \\frac{\\infinitevalue}{2} \\geq \\infinitevalue^2-58\\infinitevalue+841.\n\\]\nClearing denominators, gathering all terms to one side, and factoring puts this in the form\n\\[\n(9-\\infinitevalue)(\\infinitevalue^2 - \\frac{95}{2} \\infinitevalue + 356) \\geq 0.\n\\]\nThe quadratic factor $\\linearpart(\\infinitevalue)$ has a minimum at $\\frac{95}{4} = 23.75$\nand satisfies $\\linearpart(8) = 40, \\linearpart(10) = -19$; it is thus positive for $\\infinitevalue \\leq 8$ and negative for $10 \\leq \\infinitevalue \\leq 29$.}",
      "confidence": 0.1
    },
    "garbled_string": {
      "map": {
        "n": "hqeslzpw",
        "t_0": "nlvqjfsa",
        "t_1": "rdzthmku",
        "t_n": "kjxmbtwe",
        "t": "gbsauvdy",
        "f": "zrtcqhyp",
        "k": "oayfnlze",
        "t_k": "wqghybru",
        "g": "vxsodmci",
        "s_k": "plcyrnuf",
        "s_1": "yzskdqme",
        "s_n+1": "ivbopgwl",
        "T": "keznprat",
        "t_n+1": "pxlrgkoi",
        "T_0": "syjpwqld",
        "t_k-1": "ujvrkcen",
        "t_k+1": "mgzfoswa",
        "s_n": "dwexfchu",
        "s_n-1": "qprbltaz",
        "Q": "xjzamvne",
        "R": "hadukiyr"
      },
      "question": "For a nonnegative integer $hqeslzpw$ and a strictly increasing sequence of real numbers $nlvqjfsa,rdzthmku,\\dots,kjxmbtwe$, let $zrtcqhyp(gbsauvdy)$ be the corresponding real-valued function defined for $gbsauvdy \\geq nlvqjfsa$ by the following properties:\n\\begin{enumerate}\n\\item[(a)] $zrtcqhyp(gbsauvdy)$ is continuous for $gbsauvdy \\geq nlvqjfsa$, and is twice differentiable for all $gbsauvdy>nlvqjfsa$ other than $rdzthmku,\\dots,kjxmbtwe$;\n\\item[(b)] $zrtcqhyp(nlvqjfsa) = 1/2$;\n\\item[(c)] $\\lim_{gbsauvdy \\to wqghybru^+} zrtcqhyp'(gbsauvdy) = 0$ for $0 \\leq oayfnlze \\leq hqeslzpw$;\n\\item[(d)] For $0 \\leq oayfnlze \\leq hqeslzpw-1$, we have $zrtcqhyp''(gbsauvdy) = oayfnlze+1$ when $wqghybru < gbsauvdy< mgzfoswa$, and $zrtcqhyp''(gbsauvdy) = hqeslzpw+1$ when $gbsauvdy>kjxmbtwe$.\n\\end{enumerate}\nConsidering all choices of $hqeslzpw$ and $nlvqjfsa,rdzthmku,\\dots,kjxmbtwe$ such that $wqghybru \\geq ujvrkcen+1$ for $1 \\leq oayfnlze \\leq hqeslzpw$, what is the least possible value of $keznprat$ for which $zrtcqhyp(nlvqjfsa+keznprat) = 2023$?",
      "solution": "The minimum value of $keznprat$ is 29.\n\nWrite $pxlrgkoi = nlvqjfsa+keznprat$ and define $plcyrnuf = wqghybru-ujvrkcen$ for $1\\leq oayfnlze\\leq hqeslzpw+1$. On $[ujvrkcen,wqghybru]$, we have $zrtcqhyp'(gbsauvdy) = oayfnlze(gbsauvdy-ujvrkcen)$ and so $zrtcqhyp(wqghybru)-zrtcqhyp(ujvrkcen) = \\frac{oayfnlze}{2} plcyrnuf^2$. Thus if we define\n\\[\nvxsodmci(yzskdqme,\\ldots,ivbopgwl) = \\sum_{oayfnlze=1}^{hqeslzpw+1} oayfnlze\\, plcyrnuf^2,\n\\]\nthen we want to minimize $\\sum_{oayfnlze=1}^{hqeslzpw+1} plcyrnuf = keznprat$ (for all possible values of $hqeslzpw$) subject to the constraints that $vxsodmci(yzskdqme,\\ldots,ivbopgwl) = 4045$ and $plcyrnuf \\geq 1$ for $oayfnlze \\leq hqeslzpw$.\n\nWe first note that a minimum value for $keznprat$ is indeed achieved. To see this, note that the constraints $vxsodmci(yzskdqme,\\ldots,ivbopgwl) = 4045$ and $plcyrnuf \\geq 1$ place an upper bound on $hqeslzpw$. For fixed $hqeslzpw$, the constraint $vxsodmci(yzskdqme,\\ldots,ivbopgwl) = 4045$ places an upper bound on each $plcyrnuf$, whence the set of $(yzskdqme,\\ldots,ivbopgwl)$ on which we want to minimize $\\sum plcyrnuf$ is a compact subset of $\\mathbb{R}^{hqeslzpw+1}$.\n\nNow say that $syjpwqld$ is the minimum value of $\\sum_{oayfnlze=1}^{hqeslzpw+1} plcyrnuf$ (over all $hqeslzpw$ and $yzskdqme,\\ldots,ivbopgwl$), achieved by $(yzskdqme,\\ldots,ivbopgwl) = (yzskdqme^{0},\\ldots,ivbopgwl^{0})$. Observe that there cannot be another $(yzskdqme,\\ldots,ivbopgwl)$ with the same sum, $\\sum_{oayfnlze=1}^{hqeslzpw'+1} plcyrnuf = syjpwqld$, satisfying $vxsodmci(yzskdqme,\\ldots,ivbopgwl) > 4045$; otherwise, the function $zrtcqhyp$ for $(yzskdqme,\\ldots,ivbopgwl)$ would satisfy $zrtcqhyp(nlvqjfsa+syjpwqld) > 4045$ and there would be some $keznprat<syjpwqld$ such that $zrtcqhyp(nlvqjfsa+keznprat) = 4045$ by the intermediate value theorem.\n\nWe claim that $ivbopgwl^{0} \\geq 1$ and $plcyrnuf^{0} = 1$ for $1\\leq oayfnlze \\leq hqeslzpw$. If $ivbopgwl^{0}<1$ then\n\\begin{align*}\n& vxsodmci(yzskdqme^{0},\\ldots,qprbltaz^{0},dwexfchu^{0}+ivbopgwl^{0})-vxsodmci(yzskdqme^{0},\\ldots,qprbltaz^{0},dwexfchu^{0},ivbopgwl^{0}) \\\n&\\quad = ivbopgwl^{0}(2 hqeslzpw \\, dwexfchu^{0}-ivbopgwl^{0}) > 0,\n\\end{align*}\ncontradicting our observation from the previous paragraph. Thus $ivbopgwl^{0} \\geq 1$. If $plcyrnuf^{0}>1$ for some $1\\leq oayfnlze \\leq hqeslzpw$ then replacing $(plcyrnuf^{0},ivbopgwl^{0})$ by $(1,ivbopgwl^{0}+plcyrnuf^{0}-1)$ increases $vxsodmci$:\n\\begin{align*}\n&vxsodmci(yzskdqme^{0},\\ldots,1,\\ldots,ivbopgwl^{0}+plcyrnuf^{0}-1)-vxsodmci(yzskdqme^{0},\\ldots,plcyrnuf^{0},\\ldots,ivbopgwl^{0}) \\\n&\\quad= (plcyrnuf^{0}-1)\\big((hqeslzpw+1-oayfnlze)(plcyrnuf^{0}+1)+2(hqeslzpw+1)(ivbopgwl^{0}-1)\\big) > 0,\n\\end{align*}\nagain contradicting the observation. This establishes the claim.\n\nGiven that $plcyrnuf^{0} = 1$ for $1 \\leq oayfnlze \\leq hqeslzpw$, we have\n$keznprat = ivbopgwl^{0} + hqeslzpw$ and\n\\[\nvxsodmci(yzskdqme^{0},\\dots,ivbopgwl^{0}) = \\frac{hqeslzpw(hqeslzpw+1)}{2} + (hqeslzpw+1)(keznprat-hqeslzpw)^2.\n\\]\nSetting this equal to 4045 and solving for $keznprat$ yields\n\\[\nkeznprat = hqeslzpw+\\sqrt{\\frac{4045}{hqeslzpw+1} - \\frac{hqeslzpw}{2}}.\n\\]\nFor $hqeslzpw=9$ this yields $keznprat = 29$; it thus suffices to show that for all $hqeslzpw$, \n\\[\nhqeslzpw+\\sqrt{\\frac{4045}{hqeslzpw+1} - \\frac{hqeslzpw}{2}} \\geq 29.\n\\]\nThis is evident for $hqeslzpw \\geq 30$. For $hqeslzpw \\leq 29$, rewrite the claim as\n\\[\n\\sqrt{\\frac{4045}{hqeslzpw+1} - \\frac{hqeslzpw}{2}} \\geq 29-hqeslzpw;\n\\]\nwe then obtain an equivalent inequality by squaring both sides:\n\\[\n\\frac{4045}{hqeslzpw+1} - \\frac{hqeslzpw}{2} \\geq hqeslzpw^2-58 hqeslzpw+841.\n\\]\nClearing denominators, gathering all terms to one side, and factoring puts this in the form\n\\[\n(9-hqeslzpw)(hqeslzpw^2 - \\frac{95}{2} hqeslzpw + 356) \\geq 0.\n\\]\nThe quadratic factor $xjzamvne(hqeslzpw)$ has a minimum at $\\frac{95}{4} = 23.75$ and satisfies $xjzamvne(8) = 40, xjzamvne(10) = -19$; it is thus positive for $hqeslzpw \\leq 8$ and negative for $10 \\leq hqeslzpw \\leq 29$.",
      "solution_format": "latex"
    },
    "kernel_variant": {
      "question": "Let n be a non–negative integer and let  \n\n  t₀<t₁<⋯<tₙ                    (★)\n\nbe real numbers that satisfy the spacing condition  \n\n  t_k−t_{k-1} ≥ 3  (1 ≤ k ≤ n).          (1)\n\nFor t ≥ t₀ define a real–valued function f by  \n\n(a)  f is continuous on [t₀,∞) and three times differentiable on (t₀,∞) except (possibly) at the mesh points t₁,…,tₙ;  \n\n(b)  f(t₀)=0 and, for every k=0,1,…,n,\n   lim_{t→t_k^{+}} f′(t)=0 and lim_{t→t_k^{+}} f″(t)=0;  \n\n(c)  for 1 ≤ k ≤ n the third derivative is constant on (t_{k-1},t_k) and equals  \n\n     f‴(t)=6k,  \n\n  whereas for t>tₙ it equals  \n\n     f‴(t)=6(n+1).                 (2)\n\n(The values of f‴ at the mesh–points t₀,t₁,…,tₙ are irrelevant.)\n\nAmong all admissible choices of n and of the sequence (t₀,…,tₙ) fulfilling (★)–(2)\n\n(i) determine the least real number T for which one can have  \n  f(t₀+T)=998 655;\n\n(ii) for that minimal value T list precisely the integers n for which such a construction is possible.",
      "solution": "Throughout write  \n\n  C:=998 655, t_{\\,n+1}:=t₀+T, s_k:=t_k−t_{k-1}\\;(1≤k≤n+1).  (3)\n\nHence  \n\n  T=\\sum_{k=1}^{n+1}s_k,  s_k≥3\\;(1≤k≤n),  s_{n+1}>0.  (4)\n\n\n\n1.  Cubic constraint.  \nOn (t_{k-1},t_k) we have f‴(t)=6k and f′(t_{k-1}^{+})=f″(t_{k-1}^{+})=0; three successive integrations give  \n\n  f(t_k)−f(t_{k-1})=k\\,s_k³.  (5)\n\nSummation over k=1,…,n+1 yields the single constraint  \n\n  g(s₁,…,s_{n+1}):=\\sum_{k=1}^{n+1}k\\,s_k³=C.  (6)\n\n\n\n2.  Existence of a minimiser.  \nBecause of (4)–(6) every s_k is bounded above and below; for fixed n the feasible set is compact.  Inequality (13) below bounds n, so a global minimiser exists.\n\n\n\n3.  Kuhn–Tucker equations.  \nIntroduce λ∈ℝ for (6) and μ_k≥0 for the side conditions s_k−3≥0 (1≤k≤n).  Stationarity gives  \n\n 1+3λk\\,s_k²+μ_k=0 (1≤k≤n),  (7a)  \n 1+3λ(n+1)\\,s_{n+1}²=0.    (7b)\n\nFrom (7b) λ<0.  Put  \n\n α:=-1/(3λ)>0, β:=α/(n+1).    (8)\n\n\n\n4.  Structure of an optimal vector.  \nIf s_k>3 then μ_k=0 and (7a) gives k s_k²=α,\nwhile for s_k=3 one has μ_k>0.  Thus the set  \n\n  F:={k\\mid s_k>3}  \n\nis an initial segment {1,…,m} (possibly empty) and  \n\n  s_k=√(α/k) (1≤k≤m),  s_k=3 (m<k≤n),  s_{n+1}=√β. (9)\n\n\n\n5.  Why m=0 (explicit computation).  \nAssume m≥1 and pick ℓ:=m, the largest index with s_ℓ>3.  \nSet  \n\n  δ:=s_ℓ−3>0,  a:=ℓ(s_ℓ³−27)/(n+1)>0.  \n\nKeep all spacings except the two concerned and define the competitor  \n\n  s′_ℓ:=3, s′_{n+1}:=(s_{n+1}³+a)^{1/3}, s′_k:=s_k (k≠ℓ,n+1). (10)\n\nBecause a exactly compensates the loss ℓ(s_ℓ³−27) in (6), the vector s′ is feasible.  \nIts travelling time satisfies  \n\n  ΔT:=T′−T=(s′_ℓ−s_ℓ)+(s′_{n+1}−s_{n+1})=−δ+[(s_{n+1}³+a)^{1/3}−s_{n+1}]. (11)\n\nThe map x↦x^{1/3} is concave, hence  \n\n  (s_{n+1}³+a)^{1/3}−s_{n+1} ≤ a/(3s_{n+1}²).  (12)\n\nUsing s_{n+1}²=β=α/(n+1) and s_ℓ²=α/ℓ we obtain  \n\n a/(3s_{n+1}²)=\\frac{ℓ(s_ℓ³-27)}{3(n+1)}\\frac{n+1}{α}\n      = \\frac{ℓ}{3α}(α^{3/2}ℓ^{-3/2}-27)\n      = \\frac{1}{3}\\bigl(√α/√ℓ-27ℓ/α\\bigr). (13)\n\nSince √α/√ℓ=s_ℓ>3, the right–hand side is strictly smaller than δ=s_ℓ−3, and therefore ΔT<0.  The competitor s′ has a shorter travelling time, contradicting optimality.  Thus  \n\n  m=0,\\quad s_k=3\\;(1≤k≤n),\\quad s_{n+1}>0.  (14)\n\n\n\n6.  Solving the reduced problem.  \nWith (14) the cubic constraint (6) becomes  \n\n  (n+1)s_{n+1}³+\\frac{27}{2}n(n+1)=C,  (15)\n\nwhence  \n\n  s_{n+1}³=Y_n:=\\frac{C-\\frac{27}{2}n(n+1)}{\\,n+1\\,}.  (16)\n\nPositivity of Y_n forces  \n\n  n(n+1)<\\frac{2C}{27}=73 974.44…,  0≤n≤271. (17)\n\nFor fixed n the travelling time is  \n\n  T(n)=3n+Y_n^{1/3}.       (18)\n\n\n\n7.  Monotonicity of T(n) for n≥5.  \nSet  \n\n  A_n:=Y_n-Y_{n+1}= \\frac{C}{(n+1)(n+2)}+\\frac{27}{2}>0. (19)\n\nBecause A_n>0, Y_n decreases with n.  By the mean–value theorem applied to t↦t^{1/3} we have  \n\n  Y_n^{1/3}-Y_{n+1}^{1/3}= \\frac{A_n}{Y_n^{2/3}+Y_n^{1/3}Y_{n+1}^{1/3}+Y_{n+1}^{2/3}}\n        ≤\\frac{A_n}{3Y_{n+1}^{2/3}}. (20)\n\nDefine  \n\n  ρ(n):=\\frac{A_n}{3Y_{n+1}^{2/3}}\\qquad(n≥5).  (21)\n\nTreating n as a real variable and differentiating ln ρ one finds  \n\n  ρ′(x)=\\frac{d}{dx}\\Bigl[\\ln A_x-\\frac{2}{3}\\ln Y_{x+1}\\Bigr]<0\\quad(x≥5), (22)\n\nbecause −A′/A dominates +(2/3)Y′/Y on that range.  Hence ρ(n) attains its maximum at n=5.  Direct computation gives  \n\n  ρ(5)=\\frac{23 791}{3·142 584^{\\,2/3}}≈2.879<3, so ρ(n)<3 (n≥5). (23)\n\nTherefore  \n\n  Y_n^{1/3}-Y_{n+1}^{1/3}<3  (n≥5)     (24)\n\nand consequently  \n\n  Δ_n:=T(n+1)-T(n)=3+Y_{n+1}^{1/3}-Y_n^{1/3}>0 (n≥5). (25)\n\nThus T(n) is strictly increasing for n≥5.\n\n\n\n8.  Small values of n.  \nUsing (16) one obtains  \n\n  T(0)=C^{1/3}≈99.96  \n  T(1)≈82.23 T(2)≈75.34 T(3)≈71.97  \n  T(4)≈70.45 T(5)=70.00 T(6)≈70.24.  (26)\n\nHence T(n) decreases strictly up to n=5 and increases afterwards.  The unique global minimum is  \n\n  T_{\\min}=70, attained at n=5.  (27)\n\n\n\n9.  Explicit extremal data.  \nFor n=5, equation (16) gives s₆³=166 375, whence s₆=55.\nWith s₁=⋯=s₅=3 we have  \n\n  t₀<t₁<⋯<t₆,\\; t_k−t_{k-1}=3\\;(1≤k≤5),\\;t₆−t₅=55, T=70. (28)\n\nThese values satisfy (★)–(2), and (5) confirms f(t₀+70)=998 655.\n\n\n\nAnswer.  \n(i) The least possible value of T is 70.  \n(ii) This minimum is attained only for n = 5.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.881410",
        "was_fixed": false,
        "difficulty_analysis": "• Higher-order derivatives:  the original problem deals with a piecewise constant acceleration (second derivative).  We raise the order to a piecewise constant jerk (third derivative), so every interval now involves cubic instead of quadratic motion, inflating algebraic complexity.\n\n• Stronger smoothness conditions:  the function must restart with BOTH zero velocity and zero acceleration after every break, introducing a second set of continuity constraints and forcing an additional optimisation layer.\n\n• Larger minimal spacing and a carefully chosen huge terminal value lead to non-trivial arithmetic (cubic roots of six-figure integers) instead of neat square roots.\n\n• Optimisation requires convex-analysis/majorisation reasoning or Lagrange multipliers in a non-quadratic setting, followed by a delicate one-dimensional minima search whose monotonicity is subtler than in the quadratic case.\n\n• Finally, the problem asks not only for the minimal time but also for a full classification of all indices n that realise it, compelling the solver to prove uniqueness and to exclude borderline competitors—steps absent from the original task.\n\nAll these ingredients interact, yielding a problem that is substantially more intricate and technical than the original."
      }
    },
    "original_kernel_variant": {
      "question": "Let n be a non–negative integer and let  \n\n  t₀<t₁<⋯<tₙ                    (★)\n\nbe real numbers that satisfy the spacing condition  \n\n  t_k−t_{k-1} ≥ 3  (1 ≤ k ≤ n).          (1)\n\nFor t ≥ t₀ define a real–valued function f by  \n\n(a)  f is continuous on [t₀,∞) and three times differentiable on (t₀,∞) except (possibly) at the mesh points t₁,…,tₙ;  \n\n(b)  f(t₀)=0 and, for every k=0,1,…,n,\n   lim_{t→t_k^{+}} f′(t)=0 and lim_{t→t_k^{+}} f″(t)=0;  \n\n(c)  for 1 ≤ k ≤ n the third derivative is constant on (t_{k-1},t_k) and equals  \n\n     f‴(t)=6k,  \n\n  whereas for t>tₙ it equals  \n\n     f‴(t)=6(n+1).                 (2)\n\n(The values of f‴ at the mesh–points t₀,t₁,…,tₙ are irrelevant.)\n\nAmong all admissible choices of n and of the sequence (t₀,…,tₙ) fulfilling (★)–(2)\n\n(i) determine the least real number T for which one can have  \n  f(t₀+T)=998 655;\n\n(ii) for that minimal value T list precisely the integers n for which such a construction is possible.",
      "solution": "Throughout write  \n\n  C:=998 655, t_{\\,n+1}:=t₀+T, s_k:=t_k−t_{k-1}\\;(1≤k≤n+1).  (3)\n\nHence  \n\n  T=\\sum_{k=1}^{n+1}s_k,  s_k≥3\\;(1≤k≤n),  s_{n+1}>0.  (4)\n\n\n\n1.  Cubic constraint.  \nOn (t_{k-1},t_k) we have f‴(t)=6k and f′(t_{k-1}^{+})=f″(t_{k-1}^{+})=0; three successive integrations give  \n\n  f(t_k)−f(t_{k-1})=k\\,s_k³.  (5)\n\nSummation over k=1,…,n+1 yields the single constraint  \n\n  g(s₁,…,s_{n+1}):=\\sum_{k=1}^{n+1}k\\,s_k³=C.  (6)\n\n\n\n2.  Existence of a minimiser.  \nBecause of (4)–(6) every s_k is bounded above and below; for fixed n the feasible set is compact.  Inequality (13) below bounds n, so a global minimiser exists.\n\n\n\n3.  Kuhn–Tucker equations.  \nIntroduce λ∈ℝ for (6) and μ_k≥0 for the side conditions s_k−3≥0 (1≤k≤n).  Stationarity gives  \n\n 1+3λk\\,s_k²+μ_k=0 (1≤k≤n),  (7a)  \n 1+3λ(n+1)\\,s_{n+1}²=0.    (7b)\n\nFrom (7b) λ<0.  Put  \n\n α:=-1/(3λ)>0, β:=α/(n+1).    (8)\n\n\n\n4.  Structure of an optimal vector.  \nIf s_k>3 then μ_k=0 and (7a) gives k s_k²=α,\nwhile for s_k=3 one has μ_k>0.  Thus the set  \n\n  F:={k\\mid s_k>3}  \n\nis an initial segment {1,…,m} (possibly empty) and  \n\n  s_k=√(α/k) (1≤k≤m),  s_k=3 (m<k≤n),  s_{n+1}=√β. (9)\n\n\n\n5.  Why m=0 (explicit computation).  \nAssume m≥1 and pick ℓ:=m, the largest index with s_ℓ>3.  \nSet  \n\n  δ:=s_ℓ−3>0,  a:=ℓ(s_ℓ³−27)/(n+1)>0.  \n\nKeep all spacings except the two concerned and define the competitor  \n\n  s′_ℓ:=3, s′_{n+1}:=(s_{n+1}³+a)^{1/3}, s′_k:=s_k (k≠ℓ,n+1). (10)\n\nBecause a exactly compensates the loss ℓ(s_ℓ³−27) in (6), the vector s′ is feasible.  \nIts travelling time satisfies  \n\n  ΔT:=T′−T=(s′_ℓ−s_ℓ)+(s′_{n+1}−s_{n+1})=−δ+[(s_{n+1}³+a)^{1/3}−s_{n+1}]. (11)\n\nThe map x↦x^{1/3} is concave, hence  \n\n  (s_{n+1}³+a)^{1/3}−s_{n+1} ≤ a/(3s_{n+1}²).  (12)\n\nUsing s_{n+1}²=β=α/(n+1) and s_ℓ²=α/ℓ we obtain  \n\n a/(3s_{n+1}²)=\\frac{ℓ(s_ℓ³-27)}{3(n+1)}\\frac{n+1}{α}\n      = \\frac{ℓ}{3α}(α^{3/2}ℓ^{-3/2}-27)\n      = \\frac{1}{3}\\bigl(√α/√ℓ-27ℓ/α\\bigr). (13)\n\nSince √α/√ℓ=s_ℓ>3, the right–hand side is strictly smaller than δ=s_ℓ−3, and therefore ΔT<0.  The competitor s′ has a shorter travelling time, contradicting optimality.  Thus  \n\n  m=0,\\quad s_k=3\\;(1≤k≤n),\\quad s_{n+1}>0.  (14)\n\n\n\n6.  Solving the reduced problem.  \nWith (14) the cubic constraint (6) becomes  \n\n  (n+1)s_{n+1}³+\\frac{27}{2}n(n+1)=C,  (15)\n\nwhence  \n\n  s_{n+1}³=Y_n:=\\frac{C-\\frac{27}{2}n(n+1)}{\\,n+1\\,}.  (16)\n\nPositivity of Y_n forces  \n\n  n(n+1)<\\frac{2C}{27}=73 974.44…,  0≤n≤271. (17)\n\nFor fixed n the travelling time is  \n\n  T(n)=3n+Y_n^{1/3}.       (18)\n\n\n\n7.  Monotonicity of T(n) for n≥5.  \nSet  \n\n  A_n:=Y_n-Y_{n+1}= \\frac{C}{(n+1)(n+2)}+\\frac{27}{2}>0. (19)\n\nBecause A_n>0, Y_n decreases with n.  By the mean–value theorem applied to t↦t^{1/3} we have  \n\n  Y_n^{1/3}-Y_{n+1}^{1/3}= \\frac{A_n}{Y_n^{2/3}+Y_n^{1/3}Y_{n+1}^{1/3}+Y_{n+1}^{2/3}}\n        ≤\\frac{A_n}{3Y_{n+1}^{2/3}}. (20)\n\nDefine  \n\n  ρ(n):=\\frac{A_n}{3Y_{n+1}^{2/3}}\\qquad(n≥5).  (21)\n\nTreating n as a real variable and differentiating ln ρ one finds  \n\n  ρ′(x)=\\frac{d}{dx}\\Bigl[\\ln A_x-\\frac{2}{3}\\ln Y_{x+1}\\Bigr]<0\\quad(x≥5), (22)\n\nbecause −A′/A dominates +(2/3)Y′/Y on that range.  Hence ρ(n) attains its maximum at n=5.  Direct computation gives  \n\n  ρ(5)=\\frac{23 791}{3·142 584^{\\,2/3}}≈2.879<3, so ρ(n)<3 (n≥5). (23)\n\nTherefore  \n\n  Y_n^{1/3}-Y_{n+1}^{1/3}<3  (n≥5)     (24)\n\nand consequently  \n\n  Δ_n:=T(n+1)-T(n)=3+Y_{n+1}^{1/3}-Y_n^{1/3}>0 (n≥5). (25)\n\nThus T(n) is strictly increasing for n≥5.\n\n\n\n8.  Small values of n.  \nUsing (16) one obtains  \n\n  T(0)=C^{1/3}≈99.96  \n  T(1)≈82.23 T(2)≈75.34 T(3)≈71.97  \n  T(4)≈70.45 T(5)=70.00 T(6)≈70.24.  (26)\n\nHence T(n) decreases strictly up to n=5 and increases afterwards.  The unique global minimum is  \n\n  T_{\\min}=70, attained at n=5.  (27)\n\n\n\n9.  Explicit extremal data.  \nFor n=5, equation (16) gives s₆³=166 375, whence s₆=55.\nWith s₁=⋯=s₅=3 we have  \n\n  t₀<t₁<⋯<t₆,\\; t_k−t_{k-1}=3\\;(1≤k≤5),\\;t₆−t₅=55, T=70. (28)\n\nThese values satisfy (★)–(2), and (5) confirms f(t₀+70)=998 655.\n\n\n\nAnswer.  \n(i) The least possible value of T is 70.  \n(ii) This minimum is attained only for n = 5.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.666257",
        "was_fixed": false,
        "difficulty_analysis": "• Higher-order derivatives:  the original problem deals with a piecewise constant acceleration (second derivative).  We raise the order to a piecewise constant jerk (third derivative), so every interval now involves cubic instead of quadratic motion, inflating algebraic complexity.\n\n• Stronger smoothness conditions:  the function must restart with BOTH zero velocity and zero acceleration after every break, introducing a second set of continuity constraints and forcing an additional optimisation layer.\n\n• Larger minimal spacing and a carefully chosen huge terminal value lead to non-trivial arithmetic (cubic roots of six-figure integers) instead of neat square roots.\n\n• Optimisation requires convex-analysis/majorisation reasoning or Lagrange multipliers in a non-quadratic setting, followed by a delicate one-dimensional minima search whose monotonicity is subtler than in the quadratic case.\n\n• Finally, the problem asks not only for the minimal time but also for a full classification of all indices n that realise it, compelling the solver to prove uniqueness and to exclude borderline competitors—steps absent from the original task.\n\nAll these ingredients interact, yielding a problem that is substantially more intricate and technical than the original."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}