{
  "index": "2003-A-3",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG",
    "NT"
  ],
  "difficulty": "",
  "question": "Find the minimum value of\n\\[\n  | \\sin x + \\cos x + \\tan x + \\cot x + \\sec x + \\csc x |\n\\]\nfor real numbers  $x$.",
  "solution": "\\textbf{First solution:}\nWrite\n\\begin{align*}\nf(x) &= \\sin x + \\cos x + \\tan x + \\cot x + \\sec x + \\csc x \\\\\n&= \\sin x + \\cos x + \\frac{1}{\\sin x \\cos x} + \\frac{\\sin x + \\cos x}{\\sin x\n\\cos x}.\n\\end{align*}\nWe can write $\\sin x + \\cos x = \\sqrt{2} \\cos(\\pi/4 - x)$; this suggests\nmaking the substitution $y = \\pi/4 - x$. In this new coordinate,\n\\[\n\\sin x \\cos x = \\frac{1}{2} \\sin 2x = \\frac{1}{2} \\cos 2y,\n\\]\nand writing $c = \\sqrt{2} \\cos y$, we have\n\\begin{align*}\nf(y) &= (1 + c)\\left(1 + \\frac{2}{c^2 -1} \\right) - 1 \\\\\n&= c + \\frac{2}{c - 1}.\n\\end{align*}\nWe must analyze this function of $c$ in the range $[-\\sqrt{2}, \\sqrt{2}]$.\nIts value at $c=-\\sqrt{2}$ is $2 - 3\\sqrt{2} < -2.24$, and at\n$c = \\sqrt{2}$ is $2 + 3\\sqrt{2}>6.24$.\nIts derivative is $1 - 2/(c-1)^2$, which vanishes when\n$(c-1)^2 = 2$, i.e., where $c = 1 \\pm \\sqrt{2}$. Only the value\n$c = 1 - \\sqrt{2}$ is in bounds, at which\nthe value of $f$ is $1-2\\sqrt{2} > -1.83$. As for the pole at $c=1$,\nwe observe that $f$ decreases as $c$ approaches from below\n(so takes negative values for all $c<1$) and increases as $c$\napproaches from above (so takes positive values for all $c>1$); from\nthe data collected so far, we see that $f$ has no sign crossings, so\nthe minimum of $|f|$ is achieved at a critical point of $f$.\nWe conclude that the minimum of $|f|$ is $2 \\sqrt{2} - 1$.\n\nAlternate derivation (due to Zuming Feng): We can also minimize $|c + 2/(c-1)|$\nwithout calculus (or worrying about boundary conditions). For $c>1$, we have\n\\[\n1 + (c-1) + \\frac{2}{c-1} \\geq 1 + 2 \\sqrt{2}\n\\]\nby AM-GM on the last two terms, with equality for $c-1 = \\sqrt{2}$\n(which is out of range).\nFor $c<1$, we similarly have\n\\[\n-1 + 1-c + \\frac{2}{1-c} \\geq -1 + 2\\sqrt{2},\n\\]\nhere with equality for $1-c = \\sqrt{2}$.\n\n\\textbf{Second solution:}\nWrite\n\\[\nf(a,b) = a+b + \\frac{1}{ab} + \\frac{a+b}{ab}.\n\\]\nThen the problem is to minimize $|f(a,b)|$ subject to the constraint\n$a^2+b^2-1 = 0$. Since the constraint region has no boundary, it is\nenough to check the value at each critical point and each potential\ndiscontinuity (i.e., where $ab=0$) and select the smallest value\n(after checking that $f$ has no sign crossings).\n\nWe locate the critical points using the Lagrange multiplier condition:\nthe gradient of $f$ should be parallel to that of the constraint, which is\nto say, to the vector $(a,b)$. Since\n\\[\n\\frac{\\partial f}{\\partial a}\n= 1 - \\frac{1}{a^2 b} - \\frac{1}{a^2}\n\\]\nand similarly for $b$, the proportionality yields\n\\[\na^2 b^3 - a^3 b^2 + a^3 - b^3 + a^2 - b^2 = 0.\n\\]\nThe irreducible factors of the left side are $1+a$, $1+b$,\n$a-b$, and $ab-a-b$. So we must check what happens when any of\nthose factors, or $a$ or $b$, vanishes.\n\nIf $1+a = 0$, then $b=0$, and the singularity of $f$ becomes removable\nwhen restricted to the circle. Namely, we have\n\\[\nf = a + b + \\frac{1}{a} + \\frac{b+1}{ab}\n\\]\nand $a^2+b^2-1 = 0$ implies $(1+b)/a = a/(1-b)$. Thus we have\n$f = -2$; the same occurs when $1+b=0$.\n\nIf $a-b=0$, then $a=b=\\pm \\sqrt{2}/2$ and either\n$f = 2 + 3 \\sqrt{2} > 6.24$, or $f = 2 - 3 \\sqrt{2} < -2.24$.\n\nIf $a=0$, then either $b = -1$ as discussed above, or $b=1$. In the\nlatter case, $f$ blows up as one approaches this point, so there cannot\nbe a global minimum there.\n\nFinally, if $ab-a-b = 0$, then\n\\[\na^2b^2 = (a + b)^2 = 2ab + 1\n\\]\nand so $ab = 1 \\pm \\sqrt{2}$. The plus sign is impossible since\n$|ab| \\leq 1$, so $ab = 1 - \\sqrt{2}$ and\n\\begin{align*}\nf(a,b) &= ab + \\frac{1}{ab} + 1 \\\\\n&= 1 - 2 \\sqrt{2} > -1.83.\n\\end{align*}\nThis yields the smallest value of $|f|$ in the list (and indeed no sign\ncrossings are possible), so $2\\sqrt{2}-1$ is the desired minimum of $|f|$.\n\n\\textbf{Note:} Instead of using the geometry of the graph of $f$ to rule\nout sign crossings, one can verify explicitly that $f$ cannot\ntake the value 0. In the first solution, note that $c + 2/(c-1)=0$\nimplies $c^2 - c + 2 = 0$, which has no real roots.\nIn the second solution, we would have\n\\[\na^2 b + ab^2 + a + b =  -1.\n\\]\nSquaring both sides and simplifying yields\n\\[\n2a^3b^3 + 5a^2b^2 + 4ab = 0,\n\\]\nwhose only real root is $ab=0$. But the cases with $ab=0$ do not yield\n$f=0$, as verified above.",
  "vars": [
    "x",
    "y",
    "c",
    "a",
    "b"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "anglevar",
        "y": "shiftvar",
        "c": "coeffic",
        "a": "firstvar",
        "b": "secondv"
      },
      "question": "Find the minimum value of\n\\[\n  | \\sin anglevar + \\cos anglevar + \\tan anglevar + \\cot anglevar + \\sec anglevar + \\csc anglevar |\n\\]\nfor real numbers  $anglevar$.",
      "solution": "\\textbf{First solution:}\nWrite\n\\begin{align*}\nf(anglevar) &= \\sin anglevar + \\cos anglevar + \\tan anglevar + \\cot anglevar + \\sec anglevar + \\csc anglevar \\\\\n&= \\sin anglevar + \\cos anglevar + \\frac{1}{\\sin anglevar \\cos anglevar} + \\frac{\\sin anglevar + \\cos anglevar}{\\sin anglevar\n\\cos anglevar}.\n\\end{align*}\nWe can write $\\sin anglevar + \\cos anglevar = \\sqrt{2} \\cos(\\pi/4 - anglevar)$; this suggests\nmaking the substitution $shiftvar = \\pi/4 - anglevar$. In this new coordinate,\n\\[\n\\sin anglevar \\cos anglevar = \\frac{1}{2} \\sin 2 anglevar = \\frac{1}{2} \\cos 2 shiftvar,\n\\]\nand writing $coeffic = \\sqrt{2} \\cos shiftvar$, we have\n\\begin{align*}\nf(shiftvar) &= (1 + coeffic)\\left(1 + \\frac{2}{coeffic^2 -1} \\right) - 1 \\\\\n&= coeffic + \\frac{2}{coeffic - 1}.\n\\end{align*}\nWe must analyze this function of $coeffic$ in the range $[-\\sqrt{2}, \\sqrt{2}]$.\nIts value at $coeffic=-\\sqrt{2}$ is $2 - 3\\sqrt{2} < -2.24$, and at\n$coeffic = \\sqrt{2}$ is $2 + 3\\sqrt{2}>6.24$.\nIts derivative is $1 - \\frac{2}{(coeffic-1)^2}$, which vanishes when\n$(coeffic-1)^2 = 2$, i.e., where $coeffic = 1 \\pm \\sqrt{2}$. Only the value\n$coeffic = 1 - \\sqrt{2}$ is in bounds, at which\nthe value of $f$ is $1-2\\sqrt{2} > -1.83$. As for the pole at $coeffic=1$,\nwe observe that $f$ decreases as $coeffic$ approaches from below\n(so takes negative values for all $coeffic<1$) and increases as $coeffic$\napproaches from above (so takes positive values for all $coeffic>1$); from\nthe data collected so far, we see that $f$ has no sign crossings, so\nthe minimum of $|f|$ is achieved at a critical point of $f$.\nWe conclude that the minimum of $|f|$ is $2 \\sqrt{2} - 1$.\n\nAlternate derivation (due to Zuming Feng): We can also minimize $|coeffic + 2/(coeffic-1)|$\nwithout calculus (or worrying about boundary conditions). For $coeffic>1$, we have\n\\[\n1 + (coeffic-1) + \\frac{2}{coeffic-1} \\geq 1 + 2 \\sqrt{2}\n\\]\nby AM-GM on the last two terms, with equality for $coeffic-1 = \\sqrt{2}$\n(which is out of range).\nFor $coeffic<1$, we similarly have\n\\[\n-1 + 1-coeffic + \\frac{2}{1-coeffic} \\geq -1 + 2\\sqrt{2},\n\\]\nhere with equality for $1-coeffic = \\sqrt{2}$.\n\n\\textbf{Second solution:}\nWrite\n\\[\nf(firstvar,secondv) = firstvar+secondv + \\frac{1}{firstvar secondv} + \\frac{firstvar+secondv}{firstvar\nsecondv}.\n\\]\nThen the problem is to minimize $|f(firstvar,secondv)|$ subject to the constraint\n$firstvar^2+secondv^2-1 = 0$. Since the constraint region has no boundary, it is\nenough to check the value at each critical point and each potential\ndiscontinuity (i.e., where $firstvar secondv=0$) and select the smallest value\n(after checking that $f$ has no sign crossings).\n\nWe locate the critical points using the Lagrange multiplier condition:\nthe gradient of $f$ should be parallel to that of the constraint, which is\nto say, to the vector $(firstvar,secondv)$. Since\n\\[\n\\frac{\\partial f}{\\partial firstvar}\n= 1 - \\frac{1}{firstvar^2 secondv} - \\frac{1}{firstvar^2}\n\\]\nand similarly for $secondv$, the proportionality yields\n\\[\nfirstvar^2 secondv^3 - firstvar^3 secondv^2 + firstvar^3 - secondv^3 + firstvar^2 - secondv^2 = 0.\n\\]\nThe irreducible factors of the left side are $1+firstvar$, $1+secondv$,\n$firstvar-secondv$, and $firstvar secondv-firstvar-secondv$. So we must check what happens when any of\nthose factors, or $firstvar$ or $secondv$, vanishes.\n\nIf $1+firstvar = 0$, then $secondv=0$, and the singularity of $f$ becomes removable\nwhen restricted to the circle. Namely, we have\n\\[\nf = firstvar + secondv + \\frac{1}{firstvar} + \\frac{secondv+1}{firstvar secondv}\n\\]\nand $firstvar^2+secondv^2-1 = 0$ implies $(1+secondv)/firstvar = firstvar/(1-secondv)$. Thus we have\n$f = -2$; the same occurs when $1+secondv=0$.\n\nIf $firstvar-secondv=0$, then $firstvar=secondv=\\pm \\sqrt{2}/2$ and either\n$f = 2 + 3 \\sqrt{2} > 6.24$, or $f = 2 - 3 \\sqrt{2} < -2.24$.\n\nIf $firstvar=0$, then either $secondv = -1$ as discussed above, or $secondv=1$. In the\nlatter case, $f$ blows up as one approaches this point, so there cannot\nbe a global minimum there.\n\nFinally, if $firstvar secondv-firstvar-secondv = 0$, then\n\\[\nfirstvar^2 secondv^2 = (firstvar + secondv)^2 = 2 firstvar secondv + 1\n\\]\nand so $firstvar secondv = 1 \\pm \\sqrt{2}$. The plus sign is impossible since\n$|firstvar secondv| \\leq 1$, so $firstvar secondv = 1 - \\sqrt{2}$ and\n\\begin{align*}\nf(firstvar,secondv) &= firstvar secondv + \\frac{1}{firstvar secondv} + 1 \\\\\n&= 1 - 2 \\sqrt{2} > -1.83.\n\\end{align*}\nThis yields the smallest value of $|f|$ in the list (and indeed no sign\ncrossings are possible), so $2\\sqrt{2}-1$ is the desired minimum of $|f|$.\n\n\\textbf{Note:} Instead of using the geometry of the graph of $f$ to rule\nout sign crossings, one can verify explicitly that $f$ cannot\ntake the value 0. In the first solution, note that $coeffic + 2/(coeffic-1)=0$\nimplies $coeffic^2 - coeffic + 2 = 0$, which has no real roots.\nIn the second solution, we would have\n\\[\nfirstvar^2 secondv + firstvar secondv^2 + firstvar + secondv =  -1.\n\\]\nSquaring both sides and simplifying yields\n\\[\n2 firstvar^3 secondv^3 + 5 firstvar^2 secondv^2 + 4 firstvar secondv = 0,\n\\]\nwhose only real root is $firstvar secondv=0$. But the cases with $firstvar secondv=0$ do not yield\n$f=0$, as verified above."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "tangerine",
        "y": "blueberry",
        "c": "honeycomb",
        "a": "waterfall",
        "b": "starlight"
      },
      "question": "Find the minimum value of\n\\[\n  | \\sin tangerine + \\cos tangerine + \\tan tangerine + \\cot tangerine + \\sec tangerine + \\csc tangerine |\n\\]\nfor real numbers  $tangerine$.",
      "solution": "\\textbf{First solution:}\nWrite\n\\begin{align*}\nf(tangerine) &= \\sin tangerine + \\cos tangerine + \\tan tangerine + \\cot tangerine + \\sec tangerine + \\csc tangerine \\\\\n&= \\sin tangerine + \\cos tangerine + \\frac{1}{\\sin tangerine \\cos tangerine} + \\frac{\\sin tangerine + \\cos tangerine}{\\sin tangerine\n\\cos tangerine}.\n\\end{align*}\nWe can write $\\sin tangerine + \\cos tangerine = \\sqrt{2} \\cos(\\pi/4 - tangerine)$; this suggests\nmaking the substitution $blueberry = \\pi/4 - tangerine$. In this new coordinate,\n\\[\n\\sin tangerine \\cos tangerine = \\frac{1}{2} \\sin 2 tangerine = \\frac{1}{2} \\cos 2 blueberry,\n\\]\nand writing $honeycomb = \\sqrt{2} \\cos blueberry$, we have\n\\begin{align*}\nf(blueberry) &= (1 + honeycomb)\\left(1 + \\frac{2}{honeycomb^2 -1} \\right) - 1 \\\\\n&= honeycomb + \\frac{2}{honeycomb - 1}.\n\\end{align*}\nWe must analyze this function of $honeycomb$ in the range $[-\\sqrt{2}, \\sqrt{2}]$.\nIts value at $honeycomb=-\\sqrt{2}$ is $2 - 3\\sqrt{2} < -2.24$, and at\n$honeycomb = \\sqrt{2}$ is $2 + 3\\sqrt{2}>6.24$.\nIts derivative is $1 - 2/(honeycomb-1)^2$, which vanishes when\n$(honeycomb-1)^2 = 2$, i.e., where $honeycomb = 1 \\pm \\sqrt{2}$. Only the value\n$honeycomb = 1 - \\sqrt{2}$ is in bounds, at which\nthe value of $f$ is $1-2\\sqrt{2} > -1.83$. As for the pole at $honeycomb=1$,\nwe observe that $f$ decreases as $honeycomb$ approaches from below\n(so takes negative values for all $honeycomb<1$) and increases as $honeycomb$\napproaches from above (so takes positive values for all $honeycomb>1$); from\nthe data collected so far, we see that $f$ has no sign crossings, so\nthe minimum of $|f|$ is achieved at a critical point of $f$.\nWe conclude that the minimum of $|f|$ is $2 \\sqrt{2} - 1$.\n\nAlternate derivation (due to Zuming Feng): We can also minimize $|honeycomb + 2/(honeycomb-1)|$\nwithout calculus (or worrying about boundary conditions). For $honeycomb>1$, we have\n\\[\n1 + (honeycomb-1) + \\frac{2}{honeycomb-1} \\geq 1 + 2 \\sqrt{2}\n\\]\nby AM-GM on the last two terms, with equality for $honeycomb-1 = \\sqrt{2}$\n(which is out of range).\nFor $honeycomb<1$, we similarly have\n\\[\n-1 + 1-honeycomb + \\frac{2}{1-honeycomb} \\geq -1 + 2\\sqrt{2},\n\\]\nhere with equality for $1-honeycomb = \\sqrt{2}$.\n\n\\textbf{Second solution:}\nWrite\n\\[\nf(waterfall,starlight) = waterfall+starlight + \\frac{1}{waterfall starlight} + \\frac{waterfall+starlight}{waterfall\nstarlight}.\n\\]\nThen the problem is to minimize $|f(waterfall,starlight)|$ subject to the constraint\n$waterfall^2+starlight^2-1 = 0$. Since the constraint region has no boundary, it is\nenough to check the value at each critical point and each potential\ndiscontinuity (i.e., where $waterfall starlight=0$) and select the smallest value\n(after checking that $f$ has no sign crossings).\n\nWe locate the critical points using the Lagrange multiplier condition:\nthe gradient of $f$ should be parallel to that of the constraint, which is\nto say, to the vector $(waterfall,starlight)$. Since\n\\[\n\\frac{\\partial f}{\\partial waterfall}\n= 1 - \\frac{1}{waterfall^2 starlight} - \\frac{1}{waterfall^2}\n\\]\nand similarly for $starlight$, the proportionality yields\n\\[\nwaterfall^2 starlight^3 - waterfall^3 starlight^2 + waterfall^3 - starlight^3 + waterfall^2 - starlight^2 = 0.\n\\]\nThe irreducible factors of the left side are $1+waterfall$, $1+starlight$,\n$waterfall-starlight$, and $waterfall starlight-waterfall-starlight$. So we must check what happens when any of\nthose factors, or $waterfall$ or $starlight$, vanishes.\n\nIf $1+waterfall = 0$, then $starlight=0$, and the singularity of $f$ becomes removable\nwhen restricted to the circle. Namely, we have\n\\[\nf = waterfall + starlight + \\frac{1}{waterfall} + \\frac{starlight+1}{waterfall starlight}\n\\]\nand $waterfall^2+starlight^2-1 = 0$ implies $(1+starlight)/waterfall = waterfall/(1-starlight)$. Thus we have\n$f = -2$; the same occurs when $1+starlight=0$.\n\nIf $waterfall-starlight=0$, then $waterfall=starlight=\\pm \\sqrt{2}/2$ and either\n$f = 2 + 3 \\sqrt{2} > 6.24$, or $f = 2 - 3 \\sqrt{2} < -2.24$.\n\nIf $waterfall=0$, then either $starlight = -1$ as discussed above, or $starlight=1$. In the\nlatter case, $f$ blows up as one approaches this point, so there cannot\nbe a global minimum there.\n\nFinally, if $waterfall starlight-waterfall-starlight = 0$, then\n\\[\nwaterfall^2starlight^2 = (waterfall + starlight)^2 = 2waterfall starlight + 1\n\\]\nand so $waterfall starlight = 1 \\pm \\sqrt{2}$. The plus sign is impossible since\n$|waterfall starlight| \\leq 1$, so $waterfall starlight = 1 - \\sqrt{2}$ and\n\\begin{align*}\nf(waterfall,starlight) &= waterfall starlight + \\frac{1}{waterfall starlight} + 1 \\\\\n&= 1 - 2 \\sqrt{2} > -1.83.\n\\end{align*}\nThis yields the smallest value of $|f|$ in the list (and indeed no sign\ncrossings are possible), so $2\\sqrt{2}-1$ is the desired minimum of $|f|$.\n\n\\textbf{Note:} Instead of using the geometry of the graph of $f$ to rule\nout sign crossings, one can verify explicitly that $f$ cannot\ntake the value 0. In the first solution, note that $honeycomb + 2/(honeycomb-1)=0$\nimplies $honeycomb^2 - honeycomb + 2 = 0$, which has no real roots.\nIn the second solution, we would have\n\\[\nwaterfall^2 starlight + waterfall starlight^2 + waterfall + starlight =  -1.\n\\]\nSquaring both sides and simplifying yields\n\\[\n2waterfall^3starlight^3 + 5waterfall^2starlight^2 + 4waterfall starlight = 0,\n\\]\nwhose only real root is $waterfall starlight=0$. But the cases with $waterfall starlight=0$ do not yield\n$f=0$, as verified above."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalaxis",
        "y": "horizontalaxis",
        "c": "invariable",
        "a": "rigidconst",
        "b": "fixedscalar"
      },
      "question": "Find the minimum value of\n\\[\n  | \\sin verticalaxis + \\cos verticalaxis + \\tan verticalaxis + \\cot verticalaxis + \\sec verticalaxis + \\csc verticalaxis |\n\\]\nfor real numbers  $verticalaxis$.",
      "solution": "\\textbf{First solution:}\nWrite\n\\begin{align*}\nf(verticalaxis) &= \\sin verticalaxis + \\cos verticalaxis + \\tan verticalaxis + \\cot verticalaxis + \\sec verticalaxis + \\csc verticalaxis \\\\\n&= \\sin verticalaxis + \\cos verticalaxis + \\frac{1}{\\sin verticalaxis \\cos verticalaxis} + \\frac{\\sin verticalaxis + \\cos verticalaxis}{\\sin verticalaxis\n\\cos verticalaxis}.\n\\end{align*}\nWe can write $\\sin verticalaxis + \\cos verticalaxis = \\sqrt{2} \\cos(\\pi/4 - verticalaxis)$; this suggests\nmaking the substitution $horizontalaxis = \\pi/4 - verticalaxis$. In this new coordinate,\n\\[\n\\sin verticalaxis \\cos verticalaxis = \\frac{1}{2} \\sin 2verticalaxis = \\frac{1}{2} \\cos 2horizontalaxis,\n\\]\nand writing $invariable = \\sqrt{2} \\cos horizontalaxis$, we have\n\\begin{align*}\nf(horizontalaxis) &= (1 + invariable)\\left(1 + \\frac{2}{invariable^2 -1} \\right) - 1 \\\\\n&= invariable + \\frac{2}{invariable - 1}.\n\\end{align*}\nWe must analyze this function of $invariable$ in the range $[-\\sqrt{2}, \\sqrt{2}]$.\nIts value at $invariable=-\\sqrt{2}$ is $2 - 3\\sqrt{2} < -2.24$, and at\n$invariable = \\sqrt{2}$ is $2 + 3\\sqrt{2}>6.24$.\nIts derivative is $1 - 2/(invariable-1)^2$, which vanishes when\n$(invariable-1)^2 = 2$, i.e., where $invariable = 1 \\pm \\sqrt{2}$. Only the value\n$invariable = 1 - \\sqrt{2}$ is in bounds, at which\nthe value of $f$ is $1-2\\sqrt{2} > -1.83$. As for the pole at $invariable=1$,\nwe observe that $f$ decreases as $invariable$ approaches from below\n(so takes negative values for all $invariable<1$) and increases as $invariable$\napproaches from above (so takes positive values for all $invariable>1$); from\nthe data collected so far, we see that $f$ has no sign crossings, so\nthe minimum of $|f|$ is achieved at a critical point of $f$.\nWe conclude that the minimum of $|f|$ is $2 \\sqrt{2} - 1$.\n\nAlternate derivation (due to Zuming Feng): We can also minimize $|invariable + 2/(invariable-1)|$\nwithout calculus (or worrying about boundary conditions). For $invariable>1$, we have\n\\[\n1 + (invariable-1) + \\frac{2}{invariable-1} \\geq 1 + 2 \\sqrt{2}\n\\]\nby AM-GM on the last two terms, with equality for $invariable-1 = \\sqrt{2}$\n(which is out of range).\nFor $invariable<1$, we similarly have\n\\[\n-1 + 1-invariable + \\frac{2}{1-invariable} \\geq -1 + 2\\sqrt{2},\n\\]\nhere with equality for $1-invariable = \\sqrt{2}$.\n\n\\textbf{Second solution:}\nWrite\n\\[\nf(rigidconst,fixedscalar) = rigidconst+fixedscalar + \\frac{1}{rigidconst fixedscalar} + \\frac{rigidconst+fixedscalar}{rigidconst\nfixedscalar}.\n\\]\nThen the problem is to minimize $|f(rigidconst,fixedscalar)|$ subject to the constraint\n$rigidconst^2+fixedscalar^2-1 = 0$. Since the constraint region has no boundary, it is\nenough to check the value at each critical point and each potential\ndiscontinuity (i.e., where $rigidconst fixedscalar=0$) and select the smallest value\n(after checking that $f$ has no sign crossings).\n\nWe locate the critical points using the Lagrange multiplier condition:\nthe gradient of $f$ should be parallel to that of the constraint, which is\nto say, to the vector $(rigidconst,fixedscalar)$. Since\n\\[\n\\frac{\\partial f}{\\partial rigidconst}\n= 1 - \\frac{1}{rigidconst^2 fixedscalar} - \\frac{1}{rigidconst^2}\n\\]\nand similarly for $fixedscalar$, the proportionality yields\n\\[\nrigidconst^2 fixedscalar^3 - rigidconst^3 fixedscalar^2 + rigidconst^3 - fixedscalar^3 + rigidconst^2 - fixedscalar^2 = 0.\n\\]\nThe irreducible factors of the left side are $1+rigidconst$, $1+fixedscalar$,\n$rigidconst-fixedscalar$, and $rigidconst fixedscalar-rigidconst-fixedscalar$. So we must check what happens when any of\nthose factors, or $rigidconst$ or $fixedscalar$, vanishes.\n\nIf $1+rigidconst = 0$, then $fixedscalar=0$, and the singularity of $f$ becomes removable\nwhen restricted to the circle. Namely, we have\n\\[\nf = rigidconst + fixedscalar + \\frac{1}{rigidconst} + \\frac{fixedscalar+1}{rigidconst fixedscalar}\n\\]\nand $rigidconst^2+fixedscalar^2-1 = 0$ implies $(1+fixedscalar)/rigidconst = rigidconst/(1-fixedscalar)$. Thus we have\n$f = -2$; the same occurs when $1+fixedscalar=0$.\n\nIf $rigidconst-fixedscalar=0$, then $rigidconst=fixedscalar=\\pm \\sqrt{2}/2$ and either\n$f = 2 + 3 \\sqrt{2} > 6.24$, or $f = 2 - 3 \\sqrt{2} < -2.24$.\n\nIf $rigidconst=0$, then either $fixedscalar = -1$ as discussed above, or $fixedscalar=1$. In the\nlatter case, $f$ blows up as one approaches this point, so there cannot\nbe a global minimum there.\n\nFinally, if $rigidconst fixedscalar-rigidconst-fixedscalar = 0$, then\n\\[\nrigidconst^2 fixedscalar^2 = (rigidconst + fixedscalar)^2 = 2rigidconst fixedscalar + 1\n\\]\nand so $rigidconst fixedscalar = 1 \\pm \\sqrt{2}$. The plus sign is impossible since\n$|rigidconst fixedscalar| \\leq 1$, so $rigidconst fixedscalar = 1 - \\sqrt{2}$ and\n\\begin{align*}\nf(rigidconst,fixedscalar) &= rigidconst fixedscalar + \\frac{1}{rigidconst fixedscalar} + 1 \\\\\n&= 1 - 2 \\sqrt{2} > -1.83.\n\\end{align*}\nThis yields the smallest value of $|f|$ in the list (and indeed no sign\ncrossings are possible), so $2\\sqrt{2}-1$ is the desired minimum of $|f|$.\n\n\\textbf{Note:} Instead of using the geometry of the graph of $f$ to rule\nout sign crossings, one can verify explicitly that $f$ cannot\ntake the value 0. In the first solution, note that $invariable + 2/(invariable-1)=0$\nimplies $invariable^2 - invariable + 2 = 0$, which has no real roots.\nIn the second solution, we would have\n\\[\nrigidconst^2 fixedscalar + rigidconst fixedscalar^2 + rigidconst + fixedscalar =  -1.\n\\]\nSquaring both sides and simplifying yields\n\\[\n2rigidconst^3 fixedscalar^3 + 5rigidconst^2 fixedscalar^2 + 4rigidconst fixedscalar = 0,\n\\]\nwhose only real root is $rigidconst fixedscalar=0$. But the cases with $rigidconst fixedscalar=0$ do not yield\n$f=0$, as verified above."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "c": "bfkmrlzad",
        "a": "pqsjdlmno",
        "b": "zktvrweya"
      },
      "question": "Find the minimum value of\n\\[\n  | \\sin qzxwvtnp + \\cos qzxwvtnp + \\tan qzxwvtnp + \\cot qzxwvtnp + \\sec qzxwvtnp + \\csc qzxwvtnp |\n\\]\nfor real numbers  $qzxwvtnp$.",
      "solution": "\\textbf{First solution:}\nWrite\n\\begin{align*}\nf(qzxwvtnp) &= \\sin qzxwvtnp + \\cos qzxwvtnp + \\tan qzxwvtnp + \\cot qzxwvtnp + \\sec qzxwvtnp + \\csc qzxwvtnp \\\\\n&= \\sin qzxwvtnp + \\cos qzxwvtnp + \\frac{1}{\\sin qzxwvtnp \\cos qzxwvtnp} + \\frac{\\sin qzxwvtnp + \\cos qzxwvtnp}{\\sin qzxwvtnp\n\\cos qzxwvtnp}.\n\\end{align*}\nWe can write $\\sin qzxwvtnp + \\cos qzxwvtnp = \\sqrt{2} \\cos(\\pi/4 - qzxwvtnp)$; this suggests\nmaking the substitution $hjgrksla = \\pi/4 - qzxwvtnp$. In this new coordinate,\n\\[\n\\sin qzxwvtnp \\cos qzxwvtnp = \\frac{1}{2} \\sin 2 qzxwvtnp = \\frac{1}{2} \\cos 2 hjgrksla,\n\\]\nand writing $bfkmrlzad = \\sqrt{2} \\cos hjgrksla$, we have\n\\begin{align*}\nf(hjgrksla) &= (1 + bfkmrlzad)\\left(1 + \\frac{2}{bfkmrlzad^2 -1} \\right) - 1 \\\\\n&= bfkmrlzad + \\frac{2}{bfkmrlzad - 1}.\n\\end{align*}\nWe must analyze this function of $bfkmrlzad$ in the range $[-\\sqrt{2}, \\sqrt{2}]$.\nIts value at $bfkmrlzad=-\\sqrt{2}$ is $2 - 3\\sqrt{2} < -2.24$, and at\n$bfkmrlzad = \\sqrt{2}$ is $2 + 3\\sqrt{2}>6.24$.\nIts derivative is $1 - \\frac{2}{(bfkmrlzad-1)^2}$, which vanishes when\n$(bfkmrlzad-1)^2 = 2$, i.e., where $bfkmrlzad = 1 \\pm \\sqrt{2}$. Only the value\n$bfkmrlzad = 1 - \\sqrt{2}$ is in bounds, at which\nthe value of $f$ is $1-2\\sqrt{2} > -1.83$. As for the pole at $bfkmrlzad=1$,\nwe observe that $f$ decreases as $bfkmrlzad$ approaches from below\n(so takes negative values for all $bfkmrlzad<1$) and increases as $bfkmrlzad$\napproaches from above (so takes positive values for all $bfkmrlzad>1$); from\nthe data collected so far, we see that $f$ has no sign crossings, so\nthe minimum of $|f|$ is achieved at a critical point of $f$.\nWe conclude that the minimum of $|f|$ is $2 \\sqrt{2} - 1$.\n\nAlternate derivation (due to Zuming Feng): We can also minimize $|bfkmrlzad + 2/(bfkmrlzad-1)|$\nwithout calculus (or worrying about boundary conditions). For $bfkmrlzad>1$, we have\n\\[\n1 + (bfkmrlzad-1) + \\frac{2}{bfkmrlzad-1} \\geq 1 + 2 \\sqrt{2}\n\\]\nby AM-GM on the last two terms, with equality for $bfkmrlzad-1 = \\sqrt{2}$\n(which is out of range).\nFor $bfkmrlzad<1$, we similarly have\n\\[\n-1 + 1-bfkmrlzad + \\frac{2}{1-bfkmrlzad} \\geq -1 + 2\\sqrt{2},\n\\]\nhere with equality for $1-bfkmrlzad = \\sqrt{2}$.\n\n\\textbf{Second solution:}\nWrite\n\\[\nf(pqsjdlmno,zktvrweya) = pqsjdlmno+zktvrweya + \\frac{1}{pqsjdlmno zktvrweya} + \\frac{pqsjdlmno+zktvrweya}{pqsjdlmno zktvrweya}.\n\\]\nThen the problem is to minimize $|f(pqsjdlmno,zktvrweya)|$ subject to the constraint\n$pqsjdlmno^2+zktvrweya^2-1 = 0$. Since the constraint region has no boundary, it is\nenough to check the value at each critical point and each potential\ndiscontinuity (i.e., where $pqsjdlmno zktvrweya=0$) and select the smallest value\n(after checking that $f$ has no sign crossings).\n\nWe locate the critical points using the Lagrange multiplier condition:\nthe gradient of $f$ should be parallel to that of the constraint, which is\nto say, to the vector $(pqsjdlmno,zktvrweya)$. Since\n\\[\n\\frac{\\partial f}{\\partial pqsjdlmno}\n= 1 - \\frac{1}{pqsjdlmno^2 zktvrweya} - \\frac{1}{pqsjdlmno^2}\n\\]\nand similarly for $zktvrweya$, the proportionality yields\n\\[\npqsjdlmno^2 zktvrweya^3 - pqsjdlmno^3 zktvrweya^2 + pqsjdlmno^3 - zktvrweya^3 + pqsjdlmno^2 - zktvrweya^2 = 0.\n\\]\nThe irreducible factors of the left side are $1+pqsjdlmno$, $1+zktvrweya$,\n$pqsjdlmno-zktvrweya$, and $pqsjdlmno zktvrweya-pqsjdlmno-zktvrweya$. So we must check what happens when any of\nthose factors, or $pqsjdlmno$ or $zktvrweya$, vanishes.\n\nIf $1+pqsjdlmno = 0$, then $zktvrweya=0$, and the singularity of $f$ becomes removable\nwhen restricted to the circle. Namely, we have\n\\[\nf = pqsjdlmno + zktvrweya + \\frac{1}{pqsjdlmno} + \\frac{zktvrweya+1}{pqsjdlmno zktvrweya}\n\\]\nand $pqsjdlmno^2+zktvrweya^2-1 = 0$ implies $\\frac{1+zktvrweya}{pqsjdlmno} = \\frac{pqsjdlmno}{1-zktvrweya}$. Thus we have\n$f = -2$; the same occurs when $1+zktvrweya=0$.\n\nIf $pqsjdlmno-zktvrweya=0$, then $pqsjdlmno=zktvrweya=\\pm \\sqrt{2}/2$ and either\n$f = 2 + 3 \\sqrt{2} > 6.24$, or $f = 2 - 3 \\sqrt{2} < -2.24$.\n\nIf $pqsjdlmno=0$, then either $zktvrweya = -1$ as discussed above, or $zktvrweya=1$. In the\nlatter case, $f$ blows up as one approaches this point, so there cannot\nbe a global minimum there.\n\nFinally, if $pqsjdlmno zktvrweya-pqsjdlmno-zktvrweya = 0$, then\n\\[\npqsjdlmno^2 zktvrweya^2 = (pqsjdlmno + zktvrweya)^2 = 2 pqsjdlmno zktvrweya + 1\n\\]\nand so $pqsjdlmno zktvrweya = 1 \\pm \\sqrt{2}$. The plus sign is impossible since\n$|pqsjdlmno zktvrweya| \\leq 1$, so $pqsjdlmno zktvrweya = 1 - \\sqrt{2}$ and\n\\begin{align*}\nf(pqsjdlmno,zktvrweya) &= pqsjdlmno zktvrweya + \\frac{1}{pqsjdlmno zktvrweya} + 1 \\\\\n&= 1 - 2 \\sqrt{2} > -1.83.\n\\end{align*}\nThis yields the smallest value of $|f|$ in the list (and indeed no sign\ncrossings are possible), so $2\\sqrt{2}-1$ is the desired minimum of $|f|$.\n\n\\textbf{Note:} Instead of using the geometry of the graph of $f$ to rule\nout sign crossings, one can verify explicitly that $f$ cannot\ntake the value 0. In the first solution, note that $bfkmrlzad + 2/(bfkmrlzad-1)=0$\nimplies $bfkmrlzad^2 - bfkmrlzad + 2 = 0$, which has no real roots.\nIn the second solution, we would have\n\\[\npqsjdlmno^2 zktvrweya + pqsjdlmno zktvrweya^2 + pqsjdlmno + zktvrweya =  -1.\n\\]\nSquaring both sides and simplifying yields\n\\[\n2pqsjdlmno^3 zktvrweya^3 + 5pqsjdlmno^2 zktvrweya^2 + 4pqsjdlmno zktvrweya = 0,\n\\]\nwhose only real root is $pqsjdlmno zktvrweya=0$. But the cases with $pqsjdlmno zktvrweya=0$ do not yield\n$f=0$, as verified above."
    },
    "kernel_variant": {
      "question": "Let  \n\\[\nf(x)=\\sin x+\\cos x+\\tan x+\\cot x+\\sec x+\\csc x ,\n\\]\nand put  \n\\[\nG(x)=\\bigl(\\sin x+\\cos x\\bigr)\\,f(x), \\qquad\nF(x)=\\lvert G(x)\\rvert .\n\\]\n\nThroughout the problem the real variable $x$ is restricted to  \n\\[\nD=\\bigl\\{x\\in\\mathbb R : \\sin x\\neq 0,\\; \\cos x\\neq 0,\\; \\sin x+\\cos x\\neq 0\\bigr\\}.\n\\]\n\n1)  Put  \n\\[\nc=\\sin x+\\cos x .\n\\]\nShow that $G(x)$ depends on $x$ only via $c$ and that  \n\\[\nG(x)=\\varphi(c)=c^{2}+\\frac{2c(1+c)}{c^{2}-1},\n\\tag{$\\spadesuit$}\n\\]\nwhere  \n\\[\nI=(-\\sqrt 2,\\sqrt 2)\\setminus\\{-1,0,1\\}.\n\\]\n\n2)  Prove that  \n\\[\n\\varphi'(c)=2c-\\frac{2(c+1)^{2}}{(c^{2}-1)^{2}},\\qquad c\\in I,\n\\tag{$\\heartsuit$}\n\\]\nand that $\\varphi'(c)<0$ for every $c\\in I$.  \nDeduce that $\\varphi$ is strictly decreasing on each connected component of $I$, determine the sign of $\\varphi(c)$ on these components, and hence describe completely the graph of  \n\\[\nF(c)=\\lvert\\varphi(c)\\rvert ,\\qquad c\\in I .\n\\]\n\n3)  Compute  \n\\[\n\\inf_{x\\in D}F(x)\n\\]\nand show that this infimum is not attained.  \nFinally, determine (modulo $2\\pi$) the full set of limit points of sequences $\\,(x_{k})\\subset D$ for which $F(x_{k})\\to\\inf_{D}F$.",
      "solution": "{\\bf Step 1.  Reduction to a single real variable.}\n\nWrite  \n\\[\nc=\\sin x+\\cos x, \\qquad s=\\sin x\\cos x .\n\\tag{1}\n\\]\nBecause $\\sin^{2}x+\\cos^{2}x=1$ we have  \n\\[\nc^{2}=1+2s\\quad\\Longrightarrow\\quad s=\\frac{c^{2}-1}{2}.\n\\tag{2}\n\\]\nThe three exclusions $\\sin x\\neq 0$, $\\cos x\\neq 0$, $c\\neq 0$ translate into  \n\\[\ns\\neq 0 \\Longleftrightarrow c\\neq\\pm 1,\\qquad c\\neq 0.\n\\]\nTogether with the elementary bound $\\lvert c\\rvert\\le\\sqrt 2$ we obtain  \n\\[\nc\\in I=(-\\sqrt 2,\\sqrt 2)\\setminus\\{-1,0,1\\}.\n\\tag{3}\n\\]\n\n{\\bf Step 2.  Expressing $G$ in terms of $c$.}\n\nFrom (1)–(2)\n\\[\n\\tan x+\\cot x=\\frac{\\sin^{2}x+\\cos^{2}x}{s}=\\frac{1}{s},\\qquad\n\\sec x+\\csc x=\\frac{\\sin x+\\cos x}{s}=\\frac{c}{s}.\n\\]\nHence  \n\\[\nf(x)=c+\\frac{1+c}{s},\n\\tag{4}\n\\]\nand multiplying (4) by $c$ gives  \n\\[\nG(x)=c\\,f(x)=c^{2}+\\frac{2c(1+c)}{c^{2}-1}=:\\varphi(c),\n\\]\nestablishing $(\\spadesuit)$.\n\n{\\bf Step 3.  Computing $\\varphi'$.}\n\nWrite $\\varphi(c)=c^{2}+N(c)/D(c)$ with  \n\\[\nN(c)=2c(1+c),\\qquad D(c)=c^{2}-1.\n\\]\nThen  \n\\[\n\\varphi'(c)=2c+\\frac{N'(c)D(c)-N(c)D'(c)}{D(c)^{2}}\n          =2c-\\frac{2(c+1)^{2}}{(c^{2}-1)^{2}},\n\\]\nwhich is $(\\heartsuit)$.\n\n{\\bf Step 4.  Sign of $\\varphi'$.}\n\nBecause $(c^{2}-1)^{2}>0$ on $I$, $\\varphi'$ and  \n\\[\n\\Psi(c):=c(c^{2}-1)^{2}-(c+1)^{2}\n\\tag{5}\n\\]\nhave the same sign.  Factor  \n\\[\n(c^{2}-1)^{2}=(c-1)^{2}(c+1)^{2}\n\\]\nto obtain  \n\\[\n\\Psi(c)=(c+1)^{2}\\bigl[c(c-1)^{2}-1\\bigr]=:(c+1)^{2}R(c),\n\\]\nso the sign of $\\varphi'$ equals that of  \n\\[\nR(c)=c(c-1)^{2}-1.\n\\tag{6}\n\\]\n\nDifferentiate:\n\\[\nR'(c)=3c^{2}-4c+1=(c-1)(3c-1).\n\\tag{7}\n\\]\nConsequently $R'(c)>0$ on $(-\\infty,\\tfrac13)\\cup(1,\\infty)$ and $R'(c)<0$ on $(\\tfrac13,1)$.  \nSince $R(1)=-1<0$ and $R(\\tfrac13)=\\tfrac{4}{27}-1<0$, the unique local maximum of $R$ on $\\bigl(-\\infty,1\\bigr]$ is negative.  Because $\\lvert c\\rvert<\\sqrt 2<2$, there is no chance for $R$ to become positive on $I$.  Hence\n\\[\nR(c)<0 \\quad\\forall\\,c\\in I,\n\\]\nand therefore\n\\[\n\\boxed{\\;\\varphi'(c)<0\\text{ for every }c\\in I\\;} .\n\\]\n\nThus $\\varphi$ is strictly decreasing on every connected component of $I$:\n\\[\n(-\\sqrt 2,-1),\\;(-1,0),\\;(0,1),\\;(1,\\sqrt 2).\n\\tag{8}\n\\]\n\n{\\bf Step 5.  Sign of $\\varphi$ and qualitative picture of $F(c)=\\lvert\\varphi(c)\\rvert$.}\n\nOne checks the boundary values\n\\[\n\\begin{aligned}\n\\lim_{c\\to 0^{-}}\\varphi(c)&=0^{+},&\n\\lim_{c\\to 0^{+}}\\varphi(c)&=0^{-},\\\\[4pt]\n\\lim_{c\\to -1^{-}}\\varphi(c)&=2, &\n\\lim_{c\\to -1^{+}}\\varphi(c)&=2,\\\\[4pt]\n\\lim_{c\\to 1^{-}}\\varphi(c)&=-\\infty, &\n\\lim_{c\\to 1^{+}}\\varphi(c)&=+\\infty .\n\\end{aligned}\n\\tag{9}\n\\]\nUsing (8) and (9) one finds\n\\[\n\\begin{aligned}\n(-\\sqrt 2,-1)&:\\ \\varphi>0,\\\\\n(-1,0)&:\\ \\varphi>0,\\\\\n(0,1)&:\\ \\varphi<0,\\\\\n(1,\\sqrt 2)&:\\ \\varphi>0.\n\\end{aligned}\n\\tag{10}\n\\]\n\nHence\n\\[\nF(c)=\n\\begin{cases}\n\\varphi(c) &\\text{on }(-\\sqrt 2,-1)\\cup(-1,0)\\cup(1,\\sqrt 2),\\\\[4pt]\n-\\varphi(c)&\\text{on }(0,1).\n\\end{cases}\n\\]\nBecause $\\varphi$ is strictly decreasing:\n\n• $F$ is strictly {\\em decreasing} on $(-\\sqrt 2,-1)$ and on $(-1,0)$;  \n• $F$ is strictly {\\em increasing} on $(0,1)$ (reflection of a decreasing negative function);  \n• $F$ is strictly {\\em decreasing} on $(1,\\sqrt 2)$.\n\nIn particular,\n\\[\n\\lim_{c\\to 0}\\!F(c)=0,\\qquad\n\\lim_{c\\to 1^{-}}\\!F(c)=\\infty,\\qquad\n\\lim_{c\\to (-1)^{\\pm}}\\!F(c)=2.\n\\tag{11}\n\\]\n\n{\\bf Step 6.  Behaviour at the geometric ends $c=\\pm\\sqrt 2$.}\n\nDirect substitution yields\n\\[\n\\varphi(\\sqrt 2)=6+2\\sqrt 2,\\qquad\n\\varphi(-\\sqrt 2)=6-2\\sqrt 2,\n\\]\nhence $F$ attains the finite values $6\\pm2\\sqrt 2$ there.\n\n{\\bf Step 7.  The infimum of $F$.}\n\nFormula (11) shows $F(c)\\to0$ as $c\\to0$, whereas $c=0$ is forbidden by (3).  Therefore\n\\[\n\\boxed{\\;\\inf_{x\\in D}F(x)=0,\\;}\n\\qquad\\text{and this infimum is {\\em not} attained.}\n\\]\n\n{\\bf Step 8.  Limit points of minimising sequences.}\n\nLet $(x_{k})\\subset D$ satisfy $F(x_{k})\\to0$.  Then $c_{k}=\\sin x_{k}+\\cos x_{k}\\to0$, so\n\\[\n\\sin x_{k}+\\cos x_{k}\\to0,\\qquad\\sin x_{k}\\cos x_{k}\\neq0.\n\\]\nThe equation $\\sin x+\\cos x=0$ is equivalent to $\\tan x=-1$, i.e.\n\\[\nx\\equiv\\frac{3\\pi}{4}\\quad\\text{or}\\quad x\\equiv\\frac{7\\pi}{4}\\pmod{2\\pi}.\n\\]\nBoth points lie on the boundary of $D$, yet every neighbourhood of either contains points of $D$ with $F$ arbitrarily small.  Consequently the complete set of cluster points (modulo $2\\pi$) of all sequences achieving $F\\to0$ is\n\\[\n\\boxed{\\Bigl\\{\\tfrac{3\\pi}{4}+2\\pi m,\\;\\tfrac{7\\pi}{4}+2\\pi m : m\\in\\mathbb Z\\Bigr\\}.}\n\\]\n\n\\bigskip",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.779056",
        "was_fixed": false,
        "difficulty_analysis": "•  Additional variable interaction:  the quantity to be minimized is a product, not a simple sum, so zeros and poles of the two factors interfere in a non-trivial way.  \n•  Algebraic complexity:  after reduction the problem leads to analysing a rational function whose derivative is a quintic, forcing systematic algebra rather than short calculus tricks.  \n•  Boundary/critical‐point interplay:  besides interior critical points one must control behaviour near the forbidden values  c=0,1,±√2, which demands careful sign analysis.  \n•  Precise characterization of minimizers:  converting the single-variable condition back to trigonometric x requires inverse–cosine arguments and modular arithmetic on the circle, adding an extra geometric layer absent in the original exercise.  \nAltogether the problem uses the same “convert to  c=sin x+cos x” idea as the kernel variant, but the multiplicative structure and necessary quintic analysis make it considerably harder to solve completely."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\\[\nf(x)=\\sin x+\\cos x+\\tan x+\\cot x+\\sec x+\\csc x ,\n\\]\nand put  \n\\[\nG(x)=\\bigl(\\sin x+\\cos x\\bigr)\\,f(x), \\qquad\nF(x)=\\lvert G(x)\\rvert .\n\\]\n\nThroughout the problem the real variable $x$ is restricted to  \n\\[\nD=\\bigl\\{x\\in\\mathbb R : \\sin x\\neq 0,\\; \\cos x\\neq 0,\\; \\sin x+\\cos x\\neq 0\\bigr\\}.\n\\]\n\n1)  Put  \n\\[\nc=\\sin x+\\cos x .\n\\]\nShow that $G(x)$ depends on $x$ only via $c$ and that  \n\\[\nG(x)=\\varphi(c)=c^{2}+\\frac{2c(1+c)}{c^{2}-1},\n\\tag{$\\spadesuit$}\n\\]\nwhere  \n\\[\nI=(-\\sqrt 2,\\sqrt 2)\\setminus\\{-1,0,1\\}.\n\\]\n\n2)  Prove that  \n\\[\n\\varphi'(c)=2c-\\frac{2(c+1)^{2}}{(c^{2}-1)^{2}},\\qquad c\\in I,\n\\tag{$\\heartsuit$}\n\\]\nand that $\\varphi'(c)<0$ for every $c\\in I$.  \nDeduce that $\\varphi$ is strictly decreasing on each connected component of $I$, determine the sign of $\\varphi(c)$ on these components, and hence describe completely the graph of  \n\\[\nF(c)=\\lvert\\varphi(c)\\rvert ,\\qquad c\\in I .\n\\]\n\n3)  Compute  \n\\[\n\\inf_{x\\in D}F(x)\n\\]\nand show that this infimum is not attained.  \nFinally, determine (modulo $2\\pi$) the full set of limit points of sequences $\\,(x_{k})\\subset D$ for which $F(x_{k})\\to\\inf_{D}F$.",
      "solution": "{\\bf Step 1.  Reduction to a single real variable.}\n\nWrite  \n\\[\nc=\\sin x+\\cos x, \\qquad s=\\sin x\\cos x .\n\\tag{1}\n\\]\nBecause $\\sin^{2}x+\\cos^{2}x=1$ we have  \n\\[\nc^{2}=1+2s\\quad\\Longrightarrow\\quad s=\\frac{c^{2}-1}{2}.\n\\tag{2}\n\\]\nThe three exclusions $\\sin x\\neq 0$, $\\cos x\\neq 0$, $c\\neq 0$ translate into  \n\\[\ns\\neq 0 \\Longleftrightarrow c\\neq\\pm 1,\\qquad c\\neq 0.\n\\]\nTogether with the elementary bound $\\lvert c\\rvert\\le\\sqrt 2$ we obtain  \n\\[\nc\\in I=(-\\sqrt 2,\\sqrt 2)\\setminus\\{-1,0,1\\}.\n\\tag{3}\n\\]\n\n{\\bf Step 2.  Expressing $G$ in terms of $c$.}\n\nFrom (1)–(2)\n\\[\n\\tan x+\\cot x=\\frac{\\sin^{2}x+\\cos^{2}x}{s}=\\frac{1}{s},\\qquad\n\\sec x+\\csc x=\\frac{\\sin x+\\cos x}{s}=\\frac{c}{s}.\n\\]\nHence  \n\\[\nf(x)=c+\\frac{1+c}{s},\n\\tag{4}\n\\]\nand multiplying (4) by $c$ gives  \n\\[\nG(x)=c\\,f(x)=c^{2}+\\frac{2c(1+c)}{c^{2}-1}=:\\varphi(c),\n\\]\nestablishing $(\\spadesuit)$.\n\n{\\bf Step 3.  Computing $\\varphi'$.}\n\nWrite $\\varphi(c)=c^{2}+N(c)/D(c)$ with  \n\\[\nN(c)=2c(1+c),\\qquad D(c)=c^{2}-1.\n\\]\nThen  \n\\[\n\\varphi'(c)=2c+\\frac{N'(c)D(c)-N(c)D'(c)}{D(c)^{2}}\n          =2c-\\frac{2(c+1)^{2}}{(c^{2}-1)^{2}},\n\\]\nwhich is $(\\heartsuit)$.\n\n{\\bf Step 4.  Sign of $\\varphi'$.}\n\nBecause $(c^{2}-1)^{2}>0$ on $I$, $\\varphi'$ and  \n\\[\n\\Psi(c):=c(c^{2}-1)^{2}-(c+1)^{2}\n\\tag{5}\n\\]\nhave the same sign.  Factor  \n\\[\n(c^{2}-1)^{2}=(c-1)^{2}(c+1)^{2}\n\\]\nto obtain  \n\\[\n\\Psi(c)=(c+1)^{2}\\bigl[c(c-1)^{2}-1\\bigr]=:(c+1)^{2}R(c),\n\\]\nso the sign of $\\varphi'$ equals that of  \n\\[\nR(c)=c(c-1)^{2}-1.\n\\tag{6}\n\\]\n\nDifferentiate:\n\\[\nR'(c)=3c^{2}-4c+1=(c-1)(3c-1).\n\\tag{7}\n\\]\nConsequently $R'(c)>0$ on $(-\\infty,\\tfrac13)\\cup(1,\\infty)$ and $R'(c)<0$ on $(\\tfrac13,1)$.  \nSince $R(1)=-1<0$ and $R(\\tfrac13)=\\tfrac{4}{27}-1<0$, the unique local maximum of $R$ on $\\bigl(-\\infty,1\\bigr]$ is negative.  Because $\\lvert c\\rvert<\\sqrt 2<2$, there is no chance for $R$ to become positive on $I$.  Hence\n\\[\nR(c)<0 \\quad\\forall\\,c\\in I,\n\\]\nand therefore\n\\[\n\\boxed{\\;\\varphi'(c)<0\\text{ for every }c\\in I\\;} .\n\\]\n\nThus $\\varphi$ is strictly decreasing on every connected component of $I$:\n\\[\n(-\\sqrt 2,-1),\\;(-1,0),\\;(0,1),\\;(1,\\sqrt 2).\n\\tag{8}\n\\]\n\n{\\bf Step 5.  Sign of $\\varphi$ and qualitative picture of $F(c)=\\lvert\\varphi(c)\\rvert$.}\n\nOne checks the boundary values\n\\[\n\\begin{aligned}\n\\lim_{c\\to 0^{-}}\\varphi(c)&=0^{+},&\n\\lim_{c\\to 0^{+}}\\varphi(c)&=0^{-},\\\\[4pt]\n\\lim_{c\\to -1^{-}}\\varphi(c)&=2, &\n\\lim_{c\\to -1^{+}}\\varphi(c)&=2,\\\\[4pt]\n\\lim_{c\\to 1^{-}}\\varphi(c)&=-\\infty, &\n\\lim_{c\\to 1^{+}}\\varphi(c)&=+\\infty .\n\\end{aligned}\n\\tag{9}\n\\]\nUsing (8) and (9) one finds\n\\[\n\\begin{aligned}\n(-\\sqrt 2,-1)&:\\ \\varphi>0,\\\\\n(-1,0)&:\\ \\varphi>0,\\\\\n(0,1)&:\\ \\varphi<0,\\\\\n(1,\\sqrt 2)&:\\ \\varphi>0.\n\\end{aligned}\n\\tag{10}\n\\]\n\nHence\n\\[\nF(c)=\n\\begin{cases}\n\\varphi(c) &\\text{on }(-\\sqrt 2,-1)\\cup(-1,0)\\cup(1,\\sqrt 2),\\\\[4pt]\n-\\varphi(c)&\\text{on }(0,1).\n\\end{cases}\n\\]\nBecause $\\varphi$ is strictly decreasing:\n\n• $F$ is strictly {\\em decreasing} on $(-\\sqrt 2,-1)$ and on $(-1,0)$;  \n• $F$ is strictly {\\em increasing} on $(0,1)$ (reflection of a decreasing negative function);  \n• $F$ is strictly {\\em decreasing} on $(1,\\sqrt 2)$.\n\nIn particular,\n\\[\n\\lim_{c\\to 0}\\!F(c)=0,\\qquad\n\\lim_{c\\to 1^{-}}\\!F(c)=\\infty,\\qquad\n\\lim_{c\\to (-1)^{\\pm}}\\!F(c)=2.\n\\tag{11}\n\\]\n\n{\\bf Step 6.  Behaviour at the geometric ends $c=\\pm\\sqrt 2$.}\n\nDirect substitution yields\n\\[\n\\varphi(\\sqrt 2)=6+2\\sqrt 2,\\qquad\n\\varphi(-\\sqrt 2)=6-2\\sqrt 2,\n\\]\nhence $F$ attains the finite values $6\\pm2\\sqrt 2$ there.\n\n{\\bf Step 7.  The infimum of $F$.}\n\nFormula (11) shows $F(c)\\to0$ as $c\\to0$, whereas $c=0$ is forbidden by (3).  Therefore\n\\[\n\\boxed{\\;\\inf_{x\\in D}F(x)=0,\\;}\n\\qquad\\text{and this infimum is {\\em not} attained.}\n\\]\n\n{\\bf Step 8.  Limit points of minimising sequences.}\n\nLet $(x_{k})\\subset D$ satisfy $F(x_{k})\\to0$.  Then $c_{k}=\\sin x_{k}+\\cos x_{k}\\to0$, so\n\\[\n\\sin x_{k}+\\cos x_{k}\\to0,\\qquad\\sin x_{k}\\cos x_{k}\\neq0.\n\\]\nThe equation $\\sin x+\\cos x=0$ is equivalent to $\\tan x=-1$, i.e.\n\\[\nx\\equiv\\frac{3\\pi}{4}\\quad\\text{or}\\quad x\\equiv\\frac{7\\pi}{4}\\pmod{2\\pi}.\n\\]\nBoth points lie on the boundary of $D$, yet every neighbourhood of either contains points of $D$ with $F$ arbitrarily small.  Consequently the complete set of cluster points (modulo $2\\pi$) of all sequences achieving $F\\to0$ is\n\\[\n\\boxed{\\Bigl\\{\\tfrac{3\\pi}{4}+2\\pi m,\\;\\tfrac{7\\pi}{4}+2\\pi m : m\\in\\mathbb Z\\Bigr\\}.}\n\\]\n\n\\bigskip",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.596672",
        "was_fixed": false,
        "difficulty_analysis": "•  Additional variable interaction:  the quantity to be minimized is a product, not a simple sum, so zeros and poles of the two factors interfere in a non-trivial way.  \n•  Algebraic complexity:  after reduction the problem leads to analysing a rational function whose derivative is a quintic, forcing systematic algebra rather than short calculus tricks.  \n•  Boundary/critical‐point interplay:  besides interior critical points one must control behaviour near the forbidden values  c=0,1,±√2, which demands careful sign analysis.  \n•  Precise characterization of minimizers:  converting the single-variable condition back to trigonometric x requires inverse–cosine arguments and modular arithmetic on the circle, adding an extra geometric layer absent in the original exercise.  \nAltogether the problem uses the same “convert to  c=sin x+cos x” idea as the kernel variant, but the multiplicative structure and necessary quintic analysis make it considerably harder to solve completely."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}