{
  "index": "2008-B-3",
  "type": "GEO",
  "tag": [
    "GEO",
    "ANA"
  ],
  "difficulty": "",
  "question": "What is the largest possible radius of a circle contained in a 4-dimensional\nhypercube of side length 1?",
  "solution": "The largest possible radius is $\\frac{\\sqrt{2}}{2}$.\nIt will be convenient to solve\nthe problem for a hypercube of side length 2 instead, in which case\nwe are trying to show that the largest radius is $\\sqrt{2}$.\n\nChoose coordinates so that the interior of the hypercube\nis the set $H = [-1,1]^4$ in $\\RR^4$. Let $C$ be a circle\ncentered at the point $P$. Then $C$ is contained both in $H$\nand its reflection across $P$; these intersect in a rectangular\nparalellepiped each of whose pairs of opposite faces are at most\n2 unit apart. Consequently, if we translate $C$ so that its center\nmoves to the point $O = (0,0,0,0)$ at the center of $H$,\nthen it remains entirely inside $H$.\n\nThis means that the answer we seek equals the largest possible radius\nof a circle $C$ contained in $H$ \\emph{and centered at $O$}.\nLet $v_1 = (v_{11}, \\dots, v_{14})$ and $v_2 = (v_{21},\\dots,v_{24})$\nbe two points on $C$ lying on radii perpendicular to each other.\nThen the points of the circle can be expressed as\n$v_1 \\cos \\theta + v_2 \\sin \\theta$ for $0 \\leq \\theta < 2\\pi$.\nThen $C$ lies in $H$ if and only if for each $i$, we have\n\\[\n|v_{1i} \\cos \\theta + v_{2i} \\sin \\theta|\n\\leq 1 \\qquad (0 \\leq \\theta < 2\\pi).\n\\]\nIn geometric terms, the vector $(v_{1i}, v_{2i})$ in $\\RR^2$\nhas dot product at most 1 with every unit vector. Since this holds\nfor the unit vector in the same direction as\n$(v_{1i}, v_{2i})$, we must have\n\\[\nv_{1i}^2 + v_{2i}^2 \\leq 1 \\qquad (i=1,\\dots,4).\n\\]\nConversely, if this holds, then the Cauchy-Schwarz inequality\nand the above analysis imply that $C$ lies in $H$.\n\nIf $r$ is the radius of $C$, then\n\\begin{align*}\n2 r^2 &= \\sum_{i=1}^4 v_{1i}^2 + \\sum_{i=1}^4 v_{2i}^2 \\\\\n&= \\sum_{i=1}^4 (v_{1i}^2 + v_{2i}^2) \\\\\n&\\leq 4,\n\\end{align*}\nso $r \\leq \\sqrt{2}$.\nSince this is achieved by the circle\nthrough $(1,1,0,0)$ and $(0,0,1,1)$,\nit is the desired maximum.\n\n\\textbf{Remark:}\nOne may similarly ask for the radius of the largest $k$-dimensional\nball inside an $n$-dimensional unit hypercube; the given problem is\nthe case $(n,k) = (4,2)$.\nDaniel Kane gives the following argument to show that the maximum radius\nin this case is $\\frac{1}{2} \\sqrt{\\frac{n}{k}}$.\n(Thanks for Noam Elkies for passing this along.)\n\nWe again scale up by a factor of 2, so that we are trying to show that\nthe maximum radius $r$ of a $k$-dimensional ball contained in the hypercube\n$[-1,1]^n$ is $\\sqrt{\\frac{n}{k}}$. Again, there is no loss of generality\nin centering the ball at the origin. Let $T: \\RR^k \\to \\RR^n$ be a\nsimilitude carrying the unit ball to this embedded $k$-ball.\nThen there exists a vector $v_i \\in \\RR^k$ such that\nfor $e_1,\\dots,e_n$ the standard basis of $\\RR^n$,\n$x \\cdot v_i = T(x) \\cdot e_i$ for all $x \\in \\RR^k$.\nThe condition of the problem is equivalent to requiring\n$|v_i| \\leq 1$ for all $i$, while the radius $r$ of the embedded ball\nis determined by the fact that for all $x \\in \\RR^k$,\n\\[\nr^2 (x \\cdot x) = T(x) \\cdot T(x) = \\sum_{i=1}^n x \\cdot v_i.\n\\]\nLet $M$ be the matrix with columns $v_1,\\dots,v_k$; then $MM^T = r^2 I_k$,\nfor $I_k$ the $k \\times k$ identity matrix. We then have\n\\begin{align*}\nkr^2 &= \\Trace(r^2 I_k) = \\Trace(MM^T)\\\\\n&= \\Trace(M^TM) = \\sum_{i=1}^n |v_i|^2 \\\\\n&\\leq n,\n\\end{align*}\nyielding the upper bound $r \\leq \\sqrt{\\frac{n}{k}}$.\n\nTo show that this bound is optimal, it is enough to show that one can\nfind an orthogonal projection of $\\RR^n$ onto $\\RR^k$ so that the\nprojections of the $e_i$ all have the same norm (one can then rescale\nto get the desired configuration of  $v_1,\\dots,v_n$). We construct\nsuch a configuration by a ``smoothing'' argument. Startw with any\nprojection.\nLet $w_1,\\dots,w_n$ be the projections of $e_1,\\dots,e_n$.\nIf the desired condition is not\nachieved, we can choose $i,j$ such that\n\\[\n|w_i|^2 < \\frac{1}{n} (|w_1|^2 + \\cdots + |w_n|^2) < |w_j|^2.\n\\]\nBy precomposing\nwith a suitable rotation that fixes $e_h$ for $h \\neq i,j$,\nwe can vary $|w_i|, |w_j|$ without varying $|w_i|^2 + |w_j|^2$\nor $|w_h|$ for $h \\neq i,j$. We can thus choose such a rotation to\nforce one of $|w_i|^2, |w_j|^2$ to become equal to\n$\\frac{1}{n} (|w_1|^2 + \\cdots + |w_n|^2)$.\nRepeating at most $n-1$ times gives the desired configuration.",
  "vars": [
    "r",
    "C",
    "H",
    "P",
    "O",
    "v_1",
    "v_2",
    "v_1i",
    "v_2i",
    "v_11",
    "v_12",
    "v_13",
    "v_14",
    "v_21",
    "v_22",
    "v_23",
    "v_24",
    "x",
    "T",
    "w_i",
    "w_j",
    "w_h",
    "M",
    "I_k",
    "e_i",
    "i",
    "j",
    "k",
    "n"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "r": "radius",
        "C": "circle",
        "H": "hypercube",
        "P": "centroid",
        "O": "originpoint",
        "v_1": "radiusvecone",
        "v_2": "radiusvectwo",
        "v_1i": "veconecomp",
        "v_2i": "vectwocomp",
        "v_11": "veconeone",
        "v_12": "veconetwo",
        "v_13": "veconethree",
        "v_14": "veconefour",
        "v_21": "vectwoone",
        "v_22": "vectwotwo",
        "v_23": "vectwothree",
        "v_24": "vectwofour",
        "x": "pointx",
        "T": "transform",
        "w_i": "projvectori",
        "w_j": "projvectorj",
        "w_h": "projvectorh",
        "M": "matrixm",
        "I_k": "identityk",
        "e_i": "basisvectori",
        "i": "indexi",
        "j": "indexj",
        "k": "indexk",
        "n": "dimensionn"
      },
      "question": "What is the largest possible radius of a circle contained in a 4-dimensional\nhypercube of side length 1?",
      "solution": "The largest possible radius is $\\frac{\\sqrt{2}}{2}$. It will be convenient to solve\n the problem for a hypercube of side length 2 instead, in which case\n we are trying to show that the largest radius is $\\sqrt{2}$.\n\nChoose coordinates so that the interior of the hypercube\nis the set $hypercube = [-1,1]^4$ in $\\RR^4$. Let $circle$ be a circle\ncentered at the point $centroid$. Then $circle$ is contained both in $hypercube$\nand its reflection across $centroid$; these intersect in a rectangular\nparalellepiped each of whose pairs of opposite faces are at most\n2 unit apart. Consequently, if we translate $circle$ so that its center\nmoves to the point $originpoint = (0,0,0,0)$ at the center of $hypercube$,\nthen it remains entirely inside $hypercube$.\n\nThis means that the answer we seek equals the largest possible radius\nof a circle $circle$ contained in $hypercube$ \\emph{and centered at $originpoint$}.\nLet $radiusvecone = (veconeone, \\dots, veconefour)$ and $radiusvectwo = (vectwoone,\\dots,vectwofour)$\nbe two points on $circle$ lying on radii perpendicular to each other.\nThen the points of the circle can be expressed as\n$radiusvecone \\cos \\theta + radiusvectwo \\sin \\theta$ for $0 \\leq \\theta < 2\\pi$.\nThen $circle$ lies in $hypercube$ if and only if for each $indexi$, we have\n\\[\n|veconecomp \\cos \\theta + vectwocomp \\sin \\theta|\n\\leq 1 \\qquad (0 \\leq \\theta < 2\\pi).\n\\]\nIn geometric terms, the vector $(veconecomp, vectwocomp)$ in $\\RR^2$\nhas dot product at most 1 with every unit vector. Since this holds\nfor the unit vector in the same direction as\n$(veconecomp, vectwocomp)$, we must have\n\\[\nveconecomp^2 + vectwocomp^2 \\leq 1 \\qquad (indexi=1,\\dots,4).\n\\]\nConversely, if this holds, then the Cauchy-Schwarz inequality\nand the above analysis imply that $circle$ lies in $hypercube$.\n\nIf $radius$ is the radius of $circle$, then\n\\begin{align*}\n2 radius^2 &= \\sum_{indexi=1}^4 veconecomp^2 + \\sum_{indexi=1}^4 vectwocomp^2 \\\\\n&= \\sum_{indexi=1}^4 (veconecomp^2 + vectwocomp^2) \\\\\n&\\leq 4,\n\\end{align*}\nso $radius \\leq \\sqrt{2}$.\nSince this is achieved by the circle\nthrough $(1,1,0,0)$ and $(0,0,1,1)$,\nit is the desired maximum.\n\n\\textbf{Remark:}\nOne may similarly ask for the radius of the largest indexk-dimensional\nball inside an dimensionn-dimensional unit hypercube; the given problem is\nthe case $(dimensionn,indexk) = (4,2)$.\nDaniel Kane gives the following argument to show that the maximum radius\nin this case is $\\frac{1}{2} \\sqrt{\\frac{dimensionn}{indexk}}$.\n(Thanks for Noam Elkies for passing this along.)\n\nWe again scale up by a factor of 2, so that we are trying to show that\nthe maximum radius $radius$ of a indexk-dimensional ball contained in the hypercube\n$[-1,1]^{dimensionn}$ is $\\sqrt{\\frac{dimensionn}{indexk}}$. Again, there is no loss of generality\nin centering the ball at the origin. Let $transform: \\RR^{indexk} \\to \\RR^{dimensionn}$ be a\nsimilitude carrying the unit ball to this embedded indexk-ball.\nThen there exists a vector $v_i \\in \\RR^{indexk}$ such that\nfor $e_1,\\dots,e_{dimensionn}$ the standard basis of $\\RR^{dimensionn}$,\n$pointx \\cdot v_i = transform(pointx) \\cdot basisvectori$ for all $pointx \\in \\RR^{indexk}$.\nThe condition of the problem is equivalent to requiring\n$|v_i| \\leq 1$ for all $indexi$, while the radius $radius$ of the embedded ball\nis determined by the fact that for all $pointx \\in \\RR^{indexk}$,\n\\[\nradius^2 (pointx \\cdot pointx) = transform(pointx) \\cdot transform(pointx) = \\sum_{indexi=1}^{dimensionn} pointx \\cdot v_i.\n\\]\nLet $matrixm$ be the matrix with columns $radiusvecone,\\dots,v_{indexk}$; then $matrixm matrixm^T = radius^2 identityk$,\nfor $identityk$ the indexk \\times indexk identity matrix. We then have\n\\begin{align*}\nindexk radius^2 &= \\Trace(radius^2 identityk) = \\Trace(matrixm matrixm^T)\\\\\n&= \\Trace(matrixm^T matrixm) = \\sum_{indexi=1}^{dimensionn} |v_i|^2 \\\\\n&\\leq dimensionn,\n\\end{align*}\nyielding the upper bound $radius \\leq \\sqrt{\\frac{dimensionn}{indexk}}$.\n\nTo show that this bound is optimal, it is enough to show that one can\nfind an orthogonal projection of $\\RR^{dimensionn}$ onto $\\RR^{indexk}$ so that the\nprojections of the basisvectori all have the same norm (one can then rescale\nto get the desired configuration of  $radiusvecone,\\dots,v_{dimensionn}$). We construct\nsuch a configuration by a ``smoothing'' argument. Startw with any\nprojection.\nLet $w_1,\\dots,w_{dimensionn}$ be the projections of $e_1,\\dots,e_{dimensionn}$.\nIf the desired condition is not\nachieved, we can choose indexi,indexj such that\n\\[\n|projvectori|^2 < \\frac{1}{dimensionn} (|w_1|^2 + \\cdots + |w_{dimensionn}|^2) < |projvectorj|^2.\n\\]\nBy precomposing\nwith a suitable rotation that fixes $e_h$ for $h \\neq indexi,indexj$,\nwe can vary $|projvectori|, |projvectorj|$ without varying $|projvectori|^2 + |projvectorj|^2$\nor $|projvectorh|$ for $h \\neq indexi,indexj$. We can thus choose such a rotation to\nforce one of $|projvectori|^2, |projvectorj|^2$ to become equal to\n$\\frac{1}{dimensionn} (|w_1|^2 + \\cdots + |w_{dimensionn}|^2)$.\nRepeating at most dimensionn-1 times gives the desired configuration."
    },
    "descriptive_long_confusing": {
      "map": {
        "r": "marmalade",
        "C": "windstorm",
        "H": "neckbeard",
        "P": "toadflax",
        "O": "lighthouse",
        "v_1": "barrelful",
        "v_2": "blackbird",
        "v_1i": "shoelaces",
        "v_2i": "thumbtack",
        "v_11": "dragonfly",
        "v_12": "buttercup",
        "v_13": "afterglow",
        "v_14": "goldcrest",
        "v_21": "brickwork",
        "v_22": "whitecaps",
        "v_23": "clipboard",
        "v_24": "smalltalk",
        "x": "cornflakes",
        "T": "snowdrift",
        "w_i": "hatchling",
        "w_j": "floodgate",
        "w_h": "springbok",
        "M": "houseplant",
        "I_k": "parchment",
        "e_i": "dreamland",
        "i": "conductor",
        "j": "rainstorm",
        "k": "grassland",
        "n": "arrowhead"
      },
      "question": "What is the largest possible radius of a circle contained in a 4-dimensional\nhypercube of side length 1?",
      "solution": "The largest possible radius is $\\frac{\\sqrt{2}}{2}$.  \nIt will be convenient to solve the problem for a hypercube of side length 2 instead, in which case we are trying to show that the largest radius is $\\sqrt{2}$.  \n\nChoose coordinates so that the interior of the hypercube is the set $neckbeard = [-1,1]^4$ in $\\RR^4$.  Let $windstorm$ be a circle centered at the point $toadflax$.  Then $windstorm$ is contained both in $neckbeard$ and its reflection across $toadflax$; these intersect in a rectangular paralellepiped each of whose pairs of opposite faces are at most 2 unit apart.  Consequently, if we translate $windstorm$ so that its center moves to the point $lighthouse = (0,0,0,0)$ at the center of $neckbeard$, then it remains entirely inside $neckbeard$.  \n\nThis means that the answer we seek equals the largest possible radius of a circle $windstorm$ contained in $neckbeard$ \\emph{and centered at $lighthouse$}.  Let $barrelful = (dragonfly, \\dots, goldcrest)$ and $blackbird = (brickwork,\\dots,smalltalk)$ be two points on $windstorm$ lying on radii perpendicular to each other.  Then the points of the circle can be expressed as $barrelful \\cos \\theta + blackbird \\sin \\theta$ for $0 \\leq \\theta < 2\\pi$.  Then $windstorm$ lies in $neckbeard$ if and only if for each $conductor$, we have\n\\[\n|shoelaces \\cos \\theta + thumbtack \\sin \\theta| \\leq 1 \\qquad (0 \\leq \\theta < 2\\pi).\n\\]\nIn geometric terms, the vector $(shoelaces, thumbtack)$ in $\\RR^2$ has dot product at most 1 with every unit vector.  Since this holds for the unit vector in the same direction as $(shoelaces, thumbtack)$, we must have\n\\[\nshoelaces^2 + thumbtack^2 \\leq 1 \\qquad (conductor=1,\\dots,4).\n\\]\nConversely, if this holds, then the Cauchy–Schwarz inequality and the above analysis imply that $windstorm$ lies in $neckbeard$.  \n\nIf $marmalade$ is the radius of $windstorm$, then\n\\begin{align*}\n2\\,marmalade^2 &= \\sum_{conductor=1}^{4} shoelaces^2 + \\sum_{conductor=1}^{4} thumbtack^2\\\\\n&= \\sum_{conductor=1}^{4} (shoelaces^2 + thumbtack^2) \\\\\n&\\le 4,\n\\end{align*}\nso $marmalade \\le \\sqrt{2}$.  Since this is achieved by the circle through $(1,1,0,0)$ and $(0,0,1,1)$, it is the desired maximum.  \n\n\\textbf{Remark:}  One may similarly ask for the radius of the largest $grassland$-dimensional ball inside an $arrowhead$-dimensional unit hypercube; the given problem is the case $(arrowhead,grassland) = (4,2)$. Daniel Kane gives the following argument to show that the maximum radius in this case is $\\frac12\\sqrt{\\frac{arrowhead}{grassland}}$. (Thanks to Noam Elkies for passing this along.)  \n\nWe again scale up by a factor of 2, so that we are trying to show that the maximum radius $marmalade$ of a $grassland$-dimensional ball contained in the hypercube $[-1,1]^{arrowhead}$ is $\\sqrt{\\frac{arrowhead}{grassland}}$. Again, there is no loss of generality in centering the ball at the origin.  Let $snowdrift:\\RR^{grassland}\\to\\RR^{arrowhead}$ be a similitude carrying the unit ball to this embedded $grassland$-ball.  Then there exists a vector $v_{\\text{conductor}}\\in\\RR^{grassland}$ such that for $dreamland_1,\\dots,dreamland_{arrowhead}$ the standard basis of $\\RR^{arrowhead}$,\n$cornflakes\\cdot v_{\\text{conductor}} = snowdrift(cornflakes)\\cdot dreamland$ for all $cornflakes\\in\\RR^{grassland}$.  The condition of the problem is equivalent to requiring $|v_{\\text{conductor}}| \\le 1$ for all $conductor$, while the radius $marmalade$ of the embedded ball is determined by the fact that for all $cornflakes \\in \\RR^{grassland}$,\n\\[\nmarmalade^2 (cornflakes \\cdot cornflakes) = snowdrift(cornflakes) \\cdot snowdrift(cornflakes) = \\sum_{conductor=1}^{arrowhead} cornflakes \\cdot v_{\\text{conductor}}.\n\\]\nLet $houseplant$ be the matrix with columns $barrelful,\\dots,v_{grassland}$; then $houseplant houseplant^T = marmalade^2\\,parchment$, for $parchment$ the $grassland \\times grassland$ identity matrix.  We then have\n\\begin{align*}\ngrassland\\,marmalade^2 &= \\Trace(marmalade^2\\,parchment) = \\Trace(houseplant houseplant^T)\\\\\n&= \\Trace(houseplant^T houseplant) = \\sum_{conductor=1}^{arrowhead} |v_{\\text{conductor}}|^2 \\\\\n&\\le arrowhead,\n\\end{align*}\nyielding the upper bound $marmalade \\le \\sqrt{\\frac{arrowhead}{grassland}}$.  \n\nTo show that this bound is optimal, it is enough to show that one can find an orthogonal projection of $\\RR^{arrowhead}$ onto $\\RR^{grassland}$ so that the projections of the $dreamland_{\\text{conductor}}$ all have the same norm (one can then rescale to get the desired configuration of $barrelful,\\dots,v_{arrowhead}$). We construct such a configuration by a “smoothing” argument.  Start with any projection.  Let $w_1,\\dots,w_{arrowhead}$ be the projections of $e_1,\\dots,e_{arrowhead}$.  If the desired condition is not achieved, we can choose $conductor,rainstorm$ such that\n\\[\n|hatchling|^2 < \\frac1{arrowhead} (|w_1|^2 + \\cdots + |w_{arrowhead}|^2) < |floodgate|^2.\n\\]\nBy precomposing with a suitable rotation that fixes $e_h$ for $h \\neq conductor,rainstorm$, we can vary $|hatchling|, |floodgate|$ without varying $|hatchling|^2 + |floodgate|^2$ or $|springbok|$ for $h \\neq conductor,rainstorm$. We can thus choose such a rotation to force one of $|hatchling|^2, |floodgate|^2$ to become equal to $\\frac1{arrowhead} (|w_1|^2 + \\cdots + |w_{arrowhead}|^2)$. Repeating at most $arrowhead-1$ times gives the desired configuration."
    },
    "descriptive_long_misleading": {
      "map": {
        "r": "cornerlength",
        "C": "squareloop",
        "H": "hypersphere",
        "P": "vastplane",
        "O": "infinitypoint",
        "v_1": "scalarnumber",
        "v_2": "constantvalue",
        "v_1i": "overallfactor",
        "v_2i": "singularterm",
        "v_11": "collectiveq",
        "v_12": "aggregatez",
        "v_13": "totalityaa",
        "v_14": "wholenessb",
        "v_21": "fragmented",
        "v_22": "dispersedc",
        "v_23": "particulars",
        "v_24": "componentz",
        "x": "permanent",
        "T": "stagnation",
        "w_i": "expansionone",
        "w_j": "expansiontwo",
        "w_h": "expansionthr",
        "M": "singularity",
        "I_k": "zeroarray",
        "e_i": "randomvector",
        "i": "totalindex",
        "j": "combinedex",
        "k": "voidness",
        "n": "mindimnn"
      },
      "question": "What is the largest possible radius of a circle contained in a 4-dimensional hypercube of side length 1?",
      "solution": "The largest possible radius is $\\frac{\\sqrt{2}}{2}$.\\nIt will be convenient to solve\\nthe problem for a hypercube of side length 2 instead, in which case\\nwe are trying to show that the largest radius is $\\sqrt{2}$.\\n\\nChoose coordinates so that the interior of the hypercube\\nis the set $hypersphere = [-1,1]^4$ in $\\RR^4$. Let $squareloop$ be a circle\\ncentered at the point $vastplane$. Then $squareloop$ is contained both in $hypersphere$\\nand its reflection across $vastplane$; these intersect in a rectangular\\nparalellepiped each of whose pairs of opposite faces are at most\\n2 unit apart. Consequently, if we translate $squareloop$ so that its center\\nmoves to the point $infinitypoint = (0,0,0,0)$ at the center of $hypersphere$,\\nthen it remains entirely inside $hypersphere$.\\n\\nThis means that the answer we seek equals the largest possible radius\\nof a circle $squareloop$ contained in $hypersphere$ \\emph{and centered at $infinitypoint$}.\\nLet $scalarnumber = (collectiveq, \\dots, wholenessb)$ and $constantvalue = (fragmented,\\dots,componentz)$\\nbe two points on $squareloop$ lying on radii perpendicular to each other.\\nThen the points of the circle can be expressed as\\n$scalarnumber \\cos \\theta + constantvalue \\sin \\theta$ for $0 \\leq \\theta < 2\\pi$.\\nThen $squareloop$ lies in $hypersphere$ if and only if for each $totalindex$, we have\\n\\[\\n|overallfactor \\cos \\theta + singularterm \\sin \\theta|\\n\\leq 1 \\qquad (0 \\leq \\theta < 2\\pi).\\n\\]\\nIn geometric terms, the vector $(overallfactor, singularterm)$ in $\\RR^2$\\nhas dot product at most 1 with every unit vector. Since this holds\\nfor the unit vector in the same direction as\\n$(overallfactor, singularterm)$, we must have\\n\\[\\noverallfactor^2 + singularterm^2 \\leq 1 \\qquad (totalindex=1,\\dots,4).\\n\\]\\nConversely, if this holds, then the Cauchy–Schwarz inequality\\nand the above analysis imply that $squareloop$ lies in $hypersphere$.\\n\\nIf $cornerlength$ is the radius of $squareloop$, then\\n\\begin{align*}\\n2 cornerlength^2 &= \\sum_{totalindex=1}^4 overallfactor^2 + \\sum_{totalindex=1}^4 singularterm^2 \\\\n&= \\sum_{totalindex=1}^4 (overallfactor^2 + singularterm^2) \\\\n&\\leq 4,\\n\\end{align*}\\nso $cornerlength \\leq \\sqrt{2}$.\\nSince this is achieved by the circle\\nthrough $(1,1,0,0)$ and $(0,0,1,1)$,\\nit is the desired maximum.\\n\\n\\textbf{Remark:}\\nOne may similarly ask for the radius of the largest $voidness$-dimensional\\nball inside an $mindimnn$-dimensional unit hypercube; the given problem is\\nthe case $(mindimnn,voidness) = (4,2)$.\\nDaniel Kane gives the following argument to show that the maximum radius\\nin this case is $\\frac{1}{2} \\sqrt{\\frac{mindimnn}{voidness}}$.\\n(Thanks for Noam Elkies for passing this along.)\\n\\nWe again scale up by a factor of 2, so that we are trying to show that\\nthe maximum radius $cornerlength$ of a $voidness$-dimensional ball contained in the hypercube\\n$[-1,1]^{mindimnn}$ is $\\sqrt{\\frac{mindimnn}{voidness}}$. Again, there is no loss of generality\\nin centering the ball at the origin. Let $stagnation: \\RR^{voidness} \\to \\RR^{mindimnn}$ be a\\nsimilitude carrying the unit ball to this embedded $voidness$-ball.\\nThen there exists a vector $overallfactor_{totalindex} \\in \\RR^{voidness}$ such that\\nfor $randomvector_1,\\dots,randomvector_{mindimnn}$ the standard basis of $\\RR^{mindimnn}$,\\n$permanent \\cdot overallfactor_{totalindex} = stagnation(permanent) \\cdot randomvector_{totalindex}$ for all $permanent \\in \\RR^{voidness}$.\\nThe condition of the problem is equivalent to requiring\\n$|overallfactor_{totalindex}| \\leq 1$ for all $totalindex$, while the radius $cornerlength$ of the embedded ball\\nis determined by the fact that for all $permanent \\in \\RR^{voidness}$,\\n\\[\\ncornerlength^2 (permanent \\cdot permanent) = stagnation(permanent) \\cdot stagnation(permanent) = \\sum_{totalindex=1}^{mindimnn} permanent \\cdot overallfactor_{totalindex}.\\n\\]\\nLet $singularity$ be the matrix with columns $overallfactor_1,\\dots,overallfactor_{voidness}$; then $singularity singularity^{stagnation} = cornerlength^2 zeroarray$,\\nfor $zeroarray$ the $voidness \\times voidness$ identity matrix. We then have\\n\\begin{align*}\\nvoidness\\,cornerlength^2 &= \\Trace(cornerlength^2 zeroarray) = \\Trace(singularity singularity^{stagnation})\\\\n&= \\Trace(singularity^{stagnation} singularity) = \\sum_{totalindex=1}^{mindimnn} |overallfactor_{totalindex}|^2 \\\\n&\\leq mindimnn,\\n\\end{align*}\\nyielding the upper bound $cornerlength \\leq \\sqrt{\\frac{mindimnn}{voidness}}$.\\n\\nTo show that this bound is optimal, it is enough to show that one can\\nfind an orthogonal projection of $\\RR^{mindimnn}$ onto $\\RR^{voidness}$ so that the\\nprojections of the $randomvector_{totalindex}$ all have the same norm (one can then rescale\\nto get the desired configuration of  $overallfactor_1,\\dots,overallfactor_{mindimnn}$). We construct\\nsuch a configuration by a ``smoothing'' argument. Startw with any\\nprojection.\\nLet $w_1,\\dots,w_{mindimnn}$ be the projections of $randomvector_1,\\dots,randomvector_{mindimnn}$.\\nIf the desired condition is not\\nachieved, we can choose $totalindex,combinedex$ such that\\n\\[\\n|expansionone|^2 < \\frac{1}{mindimnn} (|w_1|^2 + \\cdots + |w_{mindimnn}|^2) < |expansiontwo|^2.\\n\\]\\nBy precomposing\\nwith a suitable rotation that fixes $randomvector_h$ for $h \\neq totalindex,combinedex$,\\nwe can vary $|expansionone|, |expansiontwo|$ without varying $|expansionone|^2 + |expansiontwo|^2$\\nor $|expansionthr|$ for $h \\neq totalindex,combinedex$. We can thus choose such a rotation to\\nforce one of $|expansionone|^2, |expansiontwo|^2$ to become equal to\\n$\\frac{1}{mindimnn} (|w_1|^2 + \\cdots + |w_{mindimnn}|^2)$.\\nRepeating at most $mindimnn-1$ times gives the desired configuration."
    },
    "garbled_string": {
      "map": {
        "r": "lonqjpfk",
        "C": "xazmteuv",
        "H": "qbvridyl",
        "P": "gwouxnse",
        "O": "kpryldav",
        "v_1": "bhqwnsri",
        "v_2": "nzxofuel",
        "v_1i": "bhqwnsrii",
        "v_2i": "nzxofueli",
        "v_11": "bhqwnsriaa",
        "v_12": "bhqwnsriab",
        "v_13": "bhqwnsriac",
        "v_14": "bhqwnsriad",
        "v_21": "nzxofuelaa",
        "v_22": "nzxofuelab",
        "v_23": "nzxofuelac",
        "v_24": "nzxofuelad",
        "x": "tldwqzsa",
        "T": "ugprhcei",
        "w_i": "dfyqmsvo",
        "w_j": "qzldnbpr",
        "w_h": "klmroest",
        "M": "jyshncvk",
        "I_k": "ftrbeimq",
        "e_i": "chzparwl",
        "i": "prxgmahu",
        "j": "vkyntewo",
        "k": "slcfzqbn",
        "n": "yvkatmeg"
      },
      "question": "What is the largest possible radius of a circle contained in a 4-dimensional hypercube of side length 1?",
      "solution": "The largest possible radius is $\\frac{\\sqrt{2}}{2}$.\\nIt will be convenient to solve\\nthe problem for a hypercube of side length 2 instead, in which case\\nwe are trying to show that the largest radius is $\\sqrt{2}$.\\n\\nChoose coordinates so that the interior of the hypercube\\nis the set $qbvridyl = [-1,1]^4$ in $\\RR^4$. Let $xazmteuv$ be a circle\\ncentered at the point $gwouxnse$. Then $xazmteuv$ is contained both in $qbvridyl$\\nand its reflection across $gwouxnse$; these intersect in a rectangular\\nparalellepiped each of whose pairs of opposite faces are at most\\n2 unit apart. Consequently, if we translate $xazmteuv$ so that its center\\nmoves to the point $kpryldav = (0,0,0,0)$ at the center of $qbvridyl$,\\nthen it remains entirely inside $qbvridyl$.\\n\\nThis means that the answer we seek equals the largest possible radius\\nof a circle $xazmteuv$ contained in $qbvridyl$ \\emph{and centered at $kpryldav$}.\\nLet $bhqwnsri = (bhqwnsriaa, \\dots, bhqwnsriad)$ and $nzxofuel = (nzxofuelaa,\\dots,nzxofuelad)$\\nbe two points on $xazmteuv$ lying on radii perpendicular to each other.\\nThen the points of the circle can be expressed as\\n$bhqwnsri \\cos \\theta + nzxofuel \\sin \\theta$ for $0 \\leq \\theta < 2\\pi$.\\nThen $xazmteuv$ lies in $qbvridyl$ if and only if for each $prxgmahu$, we have\\n\\[\\n|bhqwnsrii \\cos \\theta + nzxofueli \\sin \\theta|\\n\\leq 1 \\qquad (0 \\leq \\theta < 2\\pi).\\n\\]\\nIn geometric terms, the vector $(bhqwnsrii, nzxofueli)$ in $\\RR^2$\\nhas dot product at most 1 with every unit vector. Since this holds\\nfor the unit vector in the same direction as\\n$(bhqwnsrii, nzxofueli)$, we must have\\n\\[\\nbhqwnsrii^2 + nzxofueli^2 \\leq 1 \\qquad (prxgmahu=1,\\dots,4).\\n\\]\\nConversely, if this holds, then the Cauchy-Schwarz inequality\\nand the above analysis imply that $xazmteuv$ lies in $qbvridyl$.\\n\\nIf $lonqjpfk$ is the radius of $xazmteuv$, then\\n\\begin{align*}\\n2 lonqjpfk^2 &= \\sum_{prxgmahu=1}^4 bhqwnsrii^2 + \\sum_{prxgmahu=1}^4 nzxofueli^2 \\\\n&= \\sum_{prxgmahu=1}^4 (bhqwnsrii^2 + nzxofueli^2) \\\\n&\\leq 4,\\n\\end{align*}\\nso $lonqjpfk \\leq \\sqrt{2}$.\\nSince this is achieved by the circle\\nthrough $(1,1,0,0)$ and $(0,0,1,1)$,\\nit is the desired maximum.\\n\\n\\textbf{Remark:}\\nOne may similarly ask for the radius of the largest $slcfzqbn$-dimensional\\nball inside an $yvkatmeg$-dimensional unit hypercube; the given problem is\\nthe case $(yvkatmeg,slcfzqbn) = (4,2)$.\\nDaniel Kane gives the following argument to show that the maximum radius\\nin this case is $\\frac{1}{2} \\sqrt{\\frac{yvkatmeg}{slcfzqbn}}$.\\n(Thanks for Noam Elkies for passing this along.)\\n\\nWe again scale up by a factor of 2, so that we are trying to show that\\nthe maximum radius $lonqjpfk$ of a $slcfzqbn$-dimensional ball contained in the hypercube\\n$[-1,1]^{yvkatmeg}$ is $\\sqrt{\\frac{yvkatmeg}{slcfzqbn}}$. Again, there is no loss of generality\\nin centering the ball at the origin. Let $ugprhcei: \\RR^{slcfzqbn} \\to \\RR^{yvkatmeg}$ be a\\nsimilitude carrying the unit ball to this embedded $slcfzqbn$-ball.\\nThen there exists a vector $v_{prxgmahu} \\in \\RR^{slcfzqbn}$ such that\\nfor $chzparwl_1,\\dots,chzparwl_{yvkatmeg}$ the standard basis of $\\RR^{yvkatmeg}$,\\n$tldwqzsa \\cdot v_{prxgmahu} = ugprhcei(tldwqzsa) \\cdot chzparwl$ for all $tldwqzsa \\in \\RR^{slcfzqbn}$.\\nThe condition of the problem is equivalent to requiring\\n$|v_{prxgmahu}| \\leq 1$ for all $prxgmahu$, while the radius $lonqjpfk$ of the embedded ball\\nis determined by the fact that for all $tldwqzsa \\in \\RR^{slcfzqbn}$,\\n\\[\\nlonqjpfk^2 (tldwqzsa \\cdot tldwqzsa) = ugprhcei(tldwqzsa) \\cdot ugprhcei(tldwqzsa) = \\sum_{prxgmahu=1}^{yvkatmeg} tldwqzsa \\cdot v_{prxgmahu}.\\n\\]\\nLet $jyshncvk$ be the matrix with columns $bhqwnsri,\\dots,v_{slcfzqbn}$; then $jyshncvk jyshncvk^T = lonqjpfk^2 ftrbeimq$,\\nfor $ftrbeimq$ the $slcfzqbn \\times slcfzqbn$ identity matrix. We then have\\n\\begin{align*}\\nslcfzqbn lonqjpfk^2 &= \\Trace(lonqjpfk^2 ftrbeimq) = \\Trace(jyshncvk jyshncvk^T)\\\\\n&= \\Trace(jyshncvk^T jyshncvk) = \\sum_{prxgmahu=1}^{yvkatmeg} |v_{prxgmahu}|^2 \\\\n&\\leq yvkatmeg,\\n\\end{align*}\\nyielding the upper bound $lonqjpfk \\leq \\sqrt{\\frac{yvkatmeg}{slcfzqbn}}$.\\n\\nTo show that this bound is optimal, it is enough to show that one can\\nfind an orthogonal projection of $\\RR^{yvkatmeg}$ onto $\\RR^{slcfzqbn}$ so that the\\nprojections of the $chzparwl_{prxgmahu}$ all have the same norm (one can then rescale\\nto get the desired configuration of  $bhqwnsri,\\dots,v_{yvkatmeg}$). We construct\\nsuch a configuration by a ``smoothing'' argument. Startw with any\\nprojection.\\nLet $w_1,\\dots,w_{yvkatmeg}$ be the projections of $chzparwl_1,\\dots,chzparwl_{yvkatmeg}$.\\nIf the desired condition is not\\nachieved, we can choose $prxgmahu,vkyntewo$ such that\\n\\[\\n|dfyqmsvo|^2 < \\frac{1}{yvkatmeg} (|w_1|^2 + \\cdots + |w_{yvkatmeg}|^2) < |qzldnbpr|^2.\\n\\]\\nBy precomposing\\nwith a suitable rotation that fixes $e_h$ for $h \\neq prxgmahu,vkyntewo$,\\nwe can vary $|dfyqmsvo|, |qzldnbpr|$ without varying $|dfyqmsvo|^2 + |qzldnbpr|^2$\\nor $|klmroest|$ for $h \\neq prxgmahu,vkyntewo$. We can thus choose such a rotation to\\nforce one of $|dfyqmsvo|^2, |qzldnbpr|^2$ to become equal to\\n$\\frac{1}{yvkatmeg} (|w_1|^2 + \\cdots + |w_{yvkatmeg}|^2)$.\\nRepeating at most $yvkatmeg-1$ times gives the desired configuration."
    },
    "kernel_variant": {
      "question": "Let $n,k\\in\\mathbb N$ with $1\\le k<n$.  \nFor a matrix $A=[a_{ij}]\\in\\mathbb R^{\\,n\\times k}$ denote by  \n$a_i=(a_{i1},\\dots ,a_{ik})\\in\\mathbb R^{\\,k}$ its $i$-th row and put  \n\\[\nL_i(A)=\\lVert a_i\\rVert^{2}\\qquad(1\\le i\\le n).\n\\]\n\nThe real Stiefel manifold is  \n\\[\n\\operatorname{St}(n,k)=\\bigl\\{A\\in\\mathbb R^{\\,n\\times k}\\colon A^{\\mathsf T}A=I_k\\bigr\\}.\n\\]\n\nFor $A\\in\\operatorname{St}(n,k)$ let $\\Gamma(A)$ be the graph with vertex\nset $\\{1,\\dots ,n\\}$ in which $i\\neq j$ are adjacent iff\n$\\langle a_i,a_j\\rangle\\neq 0$.\nWrite $c(A)$ for the number of connected components of $\\Gamma(A)$.\n\nFor a positive weight vector $w=(w_1,\\dots ,w_n)$ satisfying  \n$\\sum_{i=1}^{n}w_i=k$ set  \n\\[\nF_w=\\Bigl\\{A\\in\\operatorname{St}(n,k)\\colon\n       L_1(A)=w_1,\\dots ,L_n(A)=w_n\\Bigr\\},\\qquad\nF_w^{\\operatorname{conn}}=\\bigl\\{A\\in F_w\\colon\n       \\Gamma(A)\\text{ connected}\\bigr\\}.\n\\]\n\nFurther, for a multiset $L=(L_1,\\dots ,L_n)$ of positive numbers define  \n\\[\nS_L(r)=\\sum_{i=1}^{n}\\frac{\\min\\{L_i,r^{2}\\}}{r^{2}}\n        \\quad(r>0),\\qquad\nr_{\\max}(L)=\\sup\\bigl\\{r>0\\colon S_L(r)\\ge n\\bigr\\}.\n\\]\n\nThe problem has three independent parts.\n\n1.  Basic properties of $S_L$.  \n    (a)  Prove that, writing $L_{\\min}:=\\min_{1\\le i\\le n}L_i$,  \n    \\[\n    S_L(r)=n\\ \\ (0<r\\le\\sqrt{L_{\\min}}),\\qquad\n    S_L(r_1)>S_L(r_2)\\ \\ \n              (\\sqrt{L_{\\min}}<r_1<r_2),\n    \\]\n    and deduce \n    \\[\n      \\lim_{r\\to\\infty}S_L(r)=0,\\qquad\n      r_{\\max}(L)=\\sqrt{L_{\\min}}.\n    \\]\n\n    (b)  For\n    \\[\n      \\bigl(L_i\\bigr)_{1\\le i\\le 12}=\n        (4,4,4,\\,2,2,2,2,\\,1,1,1,1,1)\n    \\]\n    compute $r_{\\max}(L)$ exactly.\n\n2.  Let $G\\colon\\operatorname{St}(n,k)\\longrightarrow\\mathbb R^{\\,n}$,\n    $G(A)=(L_1(A),\\dots ,L_n(A))$.  \n    Show that for every $A\\in\\operatorname{St}(n,k)$  \n    \\[\n        \\operatorname{rank}\\bigl(dG_A\\bigr)=n-c(A).\n    \\]\n\n3.  For the remainder fix $n=12$, $k=5$ and the uniform weight vector  \n    \\[\n      w_1=\\dots =w_{12}=\\frac{5}{12}.\n    \\]\n\n    (a)  Determine $r_{\\max}(w)$.\n\n    (b)  Prove that $F_w^{\\operatorname{conn}}\\neq\\varnothing$ and that the\n    quotient manifold $F_w^{\\operatorname{conn}}/\\mathrm O(k)$ has real\n    dimension $24$.\n\n\n\n--------------------------------------------------------------------",
      "solution": "Throughout we use only the symbols $\\times,\\cdot,\\in,\\forall,\\infty$ and\nwrite every formula in correct \\LaTeX.\n\n------------------------------------------------------------------\n1.  Properties of $S_L$.\n\n(a)  Constancy for small $r$.  \nIf $0<r\\le\\sqrt{L_{\\min}}$ then $L_i\\ge r^{2}$ for\nall $i$, hence $\\min\\{L_i,r^{2}\\}=r^{2}$ and\n\\[\nS_L(r)=\\sum_{i=1}^{n}\\frac{r^{2}}{r^{2}}=n.\n\\]\n\nStrict decrease for large $r$.  \nFix $\\sqrt{L_{\\min}}<r_1<r_2$.  For every index $i$ we have  \n\\[\n\\frac{\\min\\{L_i,r_1^{2}\\}}{r_1^{2}}\n      -\\frac{\\min\\{L_i,r_2^{2}\\}}{r_2^{2}}\n=\\begin{cases}\n      \\dfrac{L_i}{r_1^{2}}-\\dfrac{L_i}{r_2^{2}}>0,\n          &L_i<r_1^{2},\\\\[6pt]\n      1-\\dfrac{L_i}{r_2^{2}}\\ge 0,\n          &L_i\\ge r_1^{2}.\n  \\end{cases}\n\\]\nBecause at least one $L_i$ satisfies $L_i<r_1^{2}$,\nthe sum of the $n$ non–negative differences is strictly positive:\n$S_L(r_1)-S_L(r_2)>0$.\n\nLimit at infinity.  \nChoose $R>\\sqrt{\\max_i L_i}$.  \nFor $r\\ge R$ we have $\\min\\{L_i,r^{2}\\}=L_i$, whence  \n\\[\n0\\le S_L(r)=\\frac{\\sum_{i=1}^{n}L_i}{r^{2}}\n          \\xrightarrow{\\,r\\to\\infty\\,}0.\n\\]\nConsequently $S_L$ equals $n$ until $r=\\sqrt{L_{\\min}}$ and is already\nbelow $n$ for every larger $r$; therefore  \n$r_{\\max}(L)=\\sqrt{L_{\\min}}$.\n\n\\smallskip\n(b)  $\\;L_{\\min}=1$, hence $r_{\\max}(L)=1$.\n\n------------------------------------------------------------------\n2.  Rank of $dG_A$.\n\nWrite  \n\\[\nT_A\\operatorname{St}(n,k)=\n\\bigl\\{H\\in\\mathbb R^{\\,n\\times k}\\colon A^{\\mathsf T}H+H^{\\mathsf T}A=0\\bigr\\}.\n\\]\nFor $H\\in T_A\\operatorname{St}(n,k)$\n\\[\ndG_A(H)=2\\bigl(\\langle a_1,h_1\\rangle,\\dots ,\n               \\langle a_n,h_n\\rangle\\bigr)\\in\\mathbb R^{\\,n}.\n\\]\nPut $s_i(H)=\\langle a_i,h_i\\rangle$.\n\n------------------------------------------------------------------\n2.1  A \\emph{lower} bound for $\\dim\\ker dG_A$.\n\nFix a non–zero skew–symmetric matrix $S\\in\\mathbb R^{\\,k\\times k}$\n(so $S^{\\mathsf T}=-S$).  \nFor every connected component $C$ of $\\Gamma(A)$ define a matrix\n$H^{(C)}=[h^{(C)}_{ij}]$ by\n\\[\nh^{(C)}_i=\n\\begin{cases}\na_i\\,S, & i\\in C,\\\\[4pt]\n0, & i\\notin C.\n\\end{cases}\n\\]\nBecause $x^{\\mathsf T}Sx=0$ for all $x\\in\\mathbb R^{\\,k}$, we have\n$\\langle a_i,h^{(C)}_i\\rangle=\\langle a_i,a_iS\\rangle=0$; hence\n$dG_A\\bigl(H^{(C)}\\bigr)=0$.  On the other hand\n\\[\nA^{\\mathsf T}H^{(C)}=S\\sum_{i\\in C}L_i(A),\\qquad\nH^{(C)\\mathsf T}A=-S\\sum_{i\\in C}L_i(A),\n\\]\nso $A^{\\mathsf T}H^{(C)}+H^{(C)\\mathsf T}A=0$, i.e.\\\n$H^{(C)}\\in T_A\\operatorname{St}(n,k)$.\nDifferent components have disjoint supports, hence the\n$H^{(C)}$ are linearly independent.  Therefore\n\\[\n\\dim\\ker dG_A\\ge c(A)\\quad\\Longrightarrow\\quad\n\\operatorname{rank}(dG_A)\\le n-c(A).\n\\]\n\n------------------------------------------------------------------\n2.2  A \\emph{lower} bound for $\\operatorname{rank}(dG_A)$.\n\nFor every edge $e=(p,q)$ of $\\Gamma(A)$ define  \n\\[\nH^{(e)}\\in\\mathbb R^{\\,n\\times k}\n\\quad\\text{by}\\quad\nh^{(e)}_p=a_q,\\;h^{(e)}_q=-a_p,\\;\nh^{(e)}_t=0\\;(t\\neq p,q).\n\\]\nExactly as before $H^{(e)}\\in T_A\\operatorname{St}(n,k)$ and\n\\[\ndG_A\\bigl(H^{(e)}\\bigr)\n   =2\\langle a_p,a_q\\rangle(e_p-e_q)\\neq 0.\n\\]\n\nFix a component $C$.\nThe vectors $dG_A\\bigl(H^{(e)}\\bigr)$, $e$ edge of $C$,\nspan the codimension-$1$ subspace  \n$\\{x\\in\\mathbb R^{\\,C}\\colon\\sum_{i\\in C}x_i=0\\}$.\nSumming over all components gives  \n$\\operatorname{rank}(dG_A)\\ge n-c(A)$.\n\n------------------------------------------------------------------\n2.3  Equality.\n\nThe inequalities of 2.1 and 2.2 yield  \n\\[\nn-c(A)\\le\\operatorname{rank}(dG_A)\\le n-c(A),\n\\]\nhence\n\\[\n\\boxed{\\operatorname{rank}(dG_A)=n-c(A).}\n\\]\n\n------------------------------------------------------------------\n3.  The case $n=12$, $k=5$ and $w_i=5/12$.\n\n------------------------------------------------------------------\n3(a)  The maximal radius.\n\nAll $w_i$ coincide, so by Part 1  \n\\[\n\\boxed{r_{\\max}(w)=\\sqrt{\\dfrac{5}{12}}.}\n\\]\n\n------------------------------------------------------------------\n3(b)  Existence of a connected configuration and the dimension count.\n\n------------------------------------------------------------------\nStep 1.  Construction of a matrix in $F_w^{\\operatorname{conn}}$.\n\nBecause $12\\equiv0\\pmod4$, there exists a real $12\\times12$\nHadamard matrix $H$ whose entries are $\\pm1$ and whose rows are pairwise\northogonal.  Set  \n\\[\nO=\\frac1{\\sqrt{12}}\\,H,\\qquad\nA=\\text{first $5$ columns of }O.\n\\]\nThen $O$ is orthogonal, so its first five columns are orthonormal;\nhence $A\\in\\operatorname{St}(12,5)$.\n\nEvery entry of $A$ has magnitude $1/\\sqrt{12}$; therefore each row of\n$A$ has squared norm $5\\cdot(1/12)=5/12$, i.e.\\ $A\\in F_w$.\n\nIn $H$ any two different rows agree in $6$ positions and disagree in\n$6$; restricting to any $5$ columns produces an odd difference, hence  \n$\\langle a_r,a_s\\rangle\\neq0$ for $r\\neq s$.  Thus\n$\\Gamma(A)$ is the complete graph and $A\\in F_w^{\\operatorname{conn}}$.\n\n------------------------------------------------------------------\nStep 2.  Smooth structure and dimension of $F_w^{\\operatorname{conn}}$.\n\nFor every $A\\in F_w^{\\operatorname{conn}}$ one has $c(A)=1$, so\nby Part 2\n\\[\n\\operatorname{rank}(dG_A)=n-c(A)=12-1=11.\n\\]\nThe constant-rank theorem implies that\n$F_w^{\\operatorname{conn}}$ is a smooth submanifold of\n$\\operatorname{St}(12,5)$ of codimension $11$.\n\nThe dimension of the Stiefel manifold is\n\\[\n\\dim\\operatorname{St}(12,5)=12\\cdot5-\\frac12\\cdot5\\cdot6\n                           =60-15=45,\n\\]\nhence\n\\[\n\\dim F_w^{\\operatorname{conn}}=45-11=34.\n\\]\n\n------------------------------------------------------------------\nStep 3.  Quotient by the orthogonal group.\n\nThe free right action $A\\mapsto AQ$ for $Q\\in\\mathrm O(5)$ preserves\n$F_w^{\\operatorname{conn}}$.  Because $\\dim\\mathrm O(5)=\\tfrac12\\cdot5\\cdot4=10$,\nthe quotient is a smooth manifold of dimension\n\\[\n\\boxed{\\dim\\bigl(F_w^{\\operatorname{conn}}/\\mathrm O(5)\\bigr)=34-10=24.}\n\\]\n\n\\bigskip\nSummary for Part 3:\n\\[\nr_{\\max}(w)=\\sqrt{\\dfrac{5}{12}},\\qquad\n\\dim\\bigl(F_w^{\\operatorname{conn}}/\\mathrm O(5)\\bigr)=24.\n\\]\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.810930",
        "was_fixed": false,
        "difficulty_analysis": "•  Higher dimension and rectangular geometry.  \n The original problem dealt with a 2-ball inside a 4-cube; the present\n variant treats a k-ball inside an n-box with unequal side–lengths\n (and asks for both the general formula and a concrete 12×5 instance).\n\n•  Non-linear optimisation.  \n Instead of a single quadratic inequality, one must solve the\n piece-wise quadratic non-linear equation (†) and understand its unique\n root; an explicit computation is required for the given data.\n\n•  Frame theory and majorisation.  \n Showing sufficiency of (★) forces the solver to invoke deep facts about\n Parseval frames (or the Kadison–Singer/Carpenter theorem) rather than a\n simple Cauchy–Schwarz bound.\n\n•  Classification of extremals.  \n The problem asks not only for the maximal radius but also for the\n complete description of all embeddings that attain it, a step that is\n absent in the original setting and requires an understanding of tight\n frames up to orthogonal equivalence.\n\n•  Translation invariance.  \n Part (c) demands an argument that the optimum is unaffected by moving\n the centre, adding an extra conceptual layer.\n\n•  Concrete high-dimensional construction.  \n Producing an explicit 12×5 extremal matrix forces the solver to juggle\n norms of three different sizes and to weave them into a tight frame, a\n task much subtler than writing down two perpendicular diagonals in a\n cube.\n\nIn sum, the problem blends linear algebra, convex geometry, optimisation\nand frame theory, far surpassing the technical demands of both the\noriginal problem and the prior kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let $n,k\\in\\mathbb N$ with $1\\le k<n$.  \nFor a matrix $A=[a_{ij}]\\in\\mathbb R^{\\,n\\times k}$ write $a_i\\in\\mathbb R^{\\,k}$ for its $i$–th row and set  \n\\[\nL_i(A)=\\lVert a_i\\rVert^{2},\\qquad 1\\le i\\le n .\n\\]\n\nWrite  \n\\[\n\\operatorname{St}(n,k)=\\bigl\\{A\\in\\mathbb R^{\\,n\\times k}\\colon A^{\\mathsf T}A=I_k\\bigr\\}\n\\]\nfor the real Stiefel manifold, and for $A\\in\\operatorname{St}(n,k)$ let\n$\\Gamma (A)$ be the graph whose vertex set is $\\{1,\\dots ,n\\}$ and in which\n$i\\neq j$ are adjacent iff $\\langle a_i,a_j\\rangle\\neq 0$.\nGiven a positive weight vector $w=(w_1,\\dots ,w_n)$ with\n$\\sum_{i=1}^{n}w_i=k$, define  \n\\[\nF_w=\\Bigl\\{A\\in\\operatorname{St}(n,k)\\colon L_1(A)=w_1,\\dots ,L_n(A)=w_n\\Bigr\\},\n\\qquad\nF_w^{\\operatorname{conn}}=\\bigl\\{A\\in F_w\\colon\\Gamma(A)\\text{ connected}\\bigr\\}.\n\\]\n\nFor a multiset $L=(L_1,\\dots ,L_n)$ of positive numbers put  \n\\[\nS_L(r)=\\sum_{i=1}^{n}\\frac{\\min\\{L_i,r^{2}\\}}{r^{2}},\\qquad r>0 ,\n\\quad\\text{and}\\quad\nr_{\\max}(L)=\\sup\\bigl\\{r>0 \\colon S_L(r)\\ge n\\bigr\\}.\n\\]\n\nThe problem is divided into three independent parts.\n\n1.  Basic properties of $S_L$.  \n    (a)  Prove that\n    \\[\n       S_L(r)=n\\quad\\text{for }0<r\\le\\sqrt{L_{\\min}},\\qquad \n       L_{\\min}:=\\min_{1\\le i\\le n}L_i ,\n    \\]\n    and that $S_L$ is \\emph{strictly decreasing} on\n    $(\\sqrt{L_{\\min}},\\infty)$ — that is, whenever \n    $\\sqrt{L_{\\min}}<r_1<r_2$ one has $S_L(r_1)>S_L(r_2)$.  \n    Deduce $\\displaystyle\\lim_{r\\to\\infty}S_L(r)=0$ and\n    \\[\n        r_{\\max}(L)=\\sqrt{L_{\\min}} .\n    \\]\n\n    (b)  For\n    $\\bigl(L_i\\bigr)_{1\\le i\\le 12}=(4,4,4,\\,2,2,2,2,\\,1,1,1,1,1)$\n    compute $r_{\\max}(L)$ exactly.\n\n2.  Let $G\\colon\\operatorname{St}(n,k)\\longrightarrow\\mathbb R^{\\,n}$ be the\n    smooth map $G(A)=(L_1(A),\\dots ,L_n(A))$.  \n    Denote by $c(A)$ the number of connected components of $\\Gamma(A)$.\n    Prove that for every $A\\in\\operatorname{St}(n,k)$  \n    \\[\n       \\operatorname{rank}\\bigl(dG_A\\bigr)=n-c(A).\n    \\]\n\n3.  Fix $n=12$, $k=5$ and the weight vector  \n\\[\nw_1=w_2=w_3=1,\\qquad\nw_4=w_5=w_6=w_7=\\tfrac{8}{21},\\qquad\nw_8=w_9=w_{10}=w_{11}=w_{12}=\\tfrac{2}{21}.\n\\]\n\n    (a)  Determine $r_{\\max}(w)$.\n\n    (b)  Prove that $F_w^{\\operatorname{conn}}\\neq\\varnothing$ and that the\n    quotient space $F_w^{\\operatorname{conn}}/\\mathrm O(k)$ is a smooth\n    manifold of real dimension $24$.",
      "solution": "Throughout we only use the symbols $\\times,\\cdot,\\in,\\forall,\\infty$ and every\nformula is written in proper \\LaTeX.\n\n\\bigskip\n\\noindent\\textbf{1(a) Behaviour of $S_L$.}\n\nLet $L_{\\min}=\\min_i L_i$.\n\n\\emph{Plateau for small $r$.}\nIf $0<r\\le\\sqrt{L_{\\min}}$ then $\\min\\{L_i,r^{2}\\}=r^{2}$ for every\n$i$, hence $S_L(r)=n$.\n\n\\emph{Strict decrease for $r>\\sqrt{L_{\\min}}$.}\nFix $r>\\sqrt{L_{\\min}}$ and set\n$I(r)=\\bigl\\{i\\colon L_i<r^{2}\\bigr\\}$.  \nOn an interval that contains $r$ but avoids the finitely many points $\\sqrt{L_i}$ the set $I(r)$ is constant, so\n\\[\n   S_L(r)=\\sum_{i\\in I(r)}\\frac{L_i}{r^{2}}+\n          \\sum_{i\\notin I(r)}1 .\n\\]\nDifferentiating inside that interval gives\n\\[\n   S_L'(r)=\n      -\\frac{2}{r^{3}}\\sum_{i\\in I(r)}L_i<0,\n\\]\nbecause $I(r)\\neq\\varnothing$.\nHence $S_L$ is strictly decreasing on each such open interval.\nAt a breakpoint $r_0=\\sqrt{L_j}$ the function is continuous,\nbut $S_L'(r_0^{+})<0$, so the overall monotonicity is still strict:\nfor any $\\sqrt{L_{\\min}}<r_1<r_2$ one finds $S_L(r_1)>S_L(r_2)$.\n\n\\emph{Limit at infinity.}\nSince $\\min\\{L_i,r^{2}\\}\\le r^{2}$,\n\\[\n   0<S_L(r)\\le\\frac{n}{r^{2}}\\longrightarrow 0\n   \\quad(r\\to\\infty).\n\\]\n\n\\emph{Maximal admissible radius.}\nBecause $S_L(r)=n$ on $(0,\\sqrt{L_{\\min}}]$ and $S_L(r)<n$ for every\n$r>\\sqrt{L_{\\min}}$, the set $\\{r>0\\colon S_L(r)\\ge n\\}$ equals\n$(0,\\sqrt{L_{\\min}}]$.  Thus\n\\[\n   r_{\\max}(L)=\\sqrt{L_{\\min}} .\n\\]\n\n\\medskip\n\\noindent\\textbf{1(b) The concrete multiset.}\n\nHere $L_{\\min}=1$, hence  \n\\[\n   r_{\\max}(L)=1.\n\\]\n\n\\bigskip\n\\noindent\\textbf{2 Rank of $dG_A$.}\n\nFix $A\\in\\operatorname{St}(n,k)$ and\nlet $H\\in T_A\\operatorname{St}(n,k)$, so\n$A^{\\mathsf T}H+H^{\\mathsf T}A=0$.\nSet\n\\[\n   x:=dG_A(H)\\in\\mathbb R^{\\,n},\n   \\qquad x_i=2\\langle a_i,h_i\\rangle .\n\\]\n\n\\emph{Step 1: ${\\operatorname{Im}(dG_A)\\subseteq W}$, where}\n\\[\n   W=\\Bigl\\{y\\in\\mathbb R^{\\,n}\\colon\n          \\sum_{i\\in C}y_i=0\n          \\ \\text{for every component }C\\text{ of }\\Gamma(A)\\Bigr\\}.\n\\]\nLet $C$ be a component and let $V_C=\\operatorname{diag}(v)$ with\n$v_i=1$ for $i\\in C$ and $v_i=0$ otherwise.\nDefine\n\\[\n   M_C:=A^{\\mathsf T}V_CH-H^{\\mathsf T}V_CA .\n\\]\nBecause $V_C$ is diagonal, $M_C^{\\mathsf T}=-M_C$; hence\n$\\operatorname{tr}(M_C)=0$.  On the other hand\n\\[\n   \\operatorname{tr}(M_C)\n   =2\\sum_{i\\in C}\\langle a_i,h_i\\rangle\n   =\\sum_{i\\in C}x_i ,\n\\]\nso $x\\in W$.\n\n\\emph{Step 2: Surjectivity onto $W$.}\nChoose, for each connected component $C$ of $\\Gamma(A)$, a spanning\ntree $T_C$.  Denote by $\\mathcal E_C$ the edge set of $T_C$ and set\n$\\mathcal E=\\bigcup_C \\mathcal E_C$.  For each $\\{i,j\\}\\in\\mathcal E$\norient the edge from $j$ to $i$ if $y_i>0>y_j$ and from $i$ to $j$\notherwise; write the chosen orientation as $(p(e),q(e))$.\n\nFor every oriented edge $e=(p,q)\\in\\mathcal E$ define the matrix\n\\[\n   H^{(e)}:=e_{p}a_{q}^{\\mathsf T}-e_{q}a_{p}^{\\mathsf T}\\in\n            \\mathbb R^{\\,n\\times k},\n\\]\nwhere $e_{p}$ is the $p$–th standard basis vector of $\\mathbb R^{\\,n}$.\nOne checks directly that\n$A^{\\mathsf T}H^{(e)}\n     +{H^{(e)}}^{\\mathsf T}A=0$,\nso $H^{(e)}\\in T_A\\operatorname{St}(n,k)$, and\n\\[\n   dG_A\\bigl(H^{(e)}\\bigr)\n        =2\\langle a_{p},a_{q}\\rangle\n          \\bigl(e_{p}-e_{q}\\bigr).\n\\]\nBecause $\\langle a_{p},a_{q}\\rangle\\neq 0$ on every edge,\nthe vectors $dG_A(H^{(e)})$\nwith $e\\in\\mathcal E$ are linearly independent and there are\n\\[\n\\sum_{C}\\bigl(\\lvert C\\rvert-1\\bigr)=n-c(A)\n\\]\nof them.\nThey span $W$ (standard “flow–decomposition’’ along the trees).\nConsequently $dG_A$ is surjective onto $W$.\n\n\\emph{Step 3: Rank computation.}\nSince ${\\operatorname{Im}(dG_A)=W}$ and\n$\\dim W=n-c(A)$, we obtain\n\\[\n   \\operatorname{rank}\\bigl(dG_A\\bigr)=n-c(A).\n\\]\n\n\\bigskip\n\\noindent\\textbf{3(a) The maximal radius for the chosen weights.}\n\nWith $\\displaystyle\\min_{1\\le i\\le 12}w_i=\\tfrac{2}{21}$, Part~1 gives  \n\\[\n   r_{\\max}(w)=\\sqrt{\\tfrac{2}{21}} .\n\\]\n\n\\medskip\n\\noindent\\textbf{3(b) Non–emptiness of $F_w^{\\operatorname{conn}}$ and\nsmoothness of the quotient.}\n\n\\emph{Step 1: $F_w$ has positive dimension.}\nThe finite–dimensional Schur–Horn theorem ensures the existence of a rank–$5$ orthogonal projection $P$ on $\\mathbb R^{\\,12}$ with diagonal $(w_1,\\dots ,w_{12})$.  \nPick an ordered orthonormal basis $\\bigl(u_1,\\dots ,u_5\\bigr)$ for $\\operatorname{Im}P$ and set  \n\\[\nA:=[\\,u_1\\ \\dots\\ u_5\\,]\\in\\operatorname{St}(12,5).\n\\]\nRight–multiplying $A$ by any $Q\\in\\mathrm O(5)$ keeps $AA^{\\mathsf T}=P$, hence keeps the diagonal $(w_i)$.  Therefore\n\\[\n\\Psi\\colon\\mathrm O(5)\\longrightarrow F_w,\\qquad Q\\mapsto AQ ,\n\\]\nis an injective smooth map.  Since $\\dim\\mathrm O(5)=10$, the image $\\Psi(\\mathrm O(5))$ is a $10$–dimensional submanifold of $F_w$.  Thus every connected component of $F_w$ is positive–dimensional.\n\n\\emph{Step 2: Existence of an $A\\in F_w$ with $\\Gamma(A)$ connected.}\nFor $1\\le i<j\\le 12$ the map\n\\[\n    \\Phi_{ij}\\colon F_w\\longrightarrow\\mathbb R,\\qquad\n    \\Phi_{ij}(B)=\\langle b_i,b_j\\rangle ,\n\\]\nis real–analytic.\nIts zero set $\\Phi_{ij}^{-1}(0)$ is a proper real–analytic subset of\n$F_w$, hence of codimension $1$.\nA finite union of such codimension–$1$ subsets cannot cover a positive–dimensional real–analytic manifold, so there exists $A_0\\in F_w$ outside $\\bigcup_{i<j}\\Phi_{ij}^{-1}(0)$.  For this $A_0$ we have $\\langle a_{0i},a_{0j}\\rangle\\neq 0$ for all $i\\neq j$; hence $\\Gamma(A_0)$ is complete and therefore connected.  Thus $F_w^{\\operatorname{conn}}\\neq\\varnothing$.\n\n\\emph{Step 3: Smooth–manifold structure of $F_w^{\\operatorname{conn}}$.}\nFor every $A\\in F_w^{\\operatorname{conn}}$ one has $c(A)=1$, so\nPart~2 gives $\\operatorname{rank}(dG_A)=12-1=11$.\nHence the rank of $dG$ is constant and equal to $11$ along $F_w^{\\operatorname{conn}}$.  \nBy the constant–rank theorem $F_w^{\\operatorname{conn}}$\nis a smooth submanifold of $\\operatorname{St}(12,5)$ of codimension\n$11$.  Using \n\\[\n   \\dim\\operatorname{St}(n,k)=nk-\\frac12 k(k+1)\n\\]\nwe obtain\n\\[\n   \\dim F_w^{\\operatorname{conn}}\n      =(12\\cdot 5)-\\frac12\\cdot5\\cdot6-11\n      =60-15-11\n      =34 .\n\\]\n\n\\emph{Step 4: Passing to the quotient by $\\mathrm O(5)$.}\nRight multiplication by $\\mathrm O(5)$ leaves every $L_i$ unchanged,\nacts freely and properly on $F_w^{\\operatorname{conn}}$, and the\norbit space is the quotient\n\\[\n   F_w^{\\operatorname{conn}}/\\mathrm O(5).\n\\]\nBecause $\\dim\\mathrm O(5)=\\tfrac12\\cdot5\\cdot4=10$, it follows that\n\\[\n   \\dim\\bigl(F_w^{\\operatorname{conn}}/\\mathrm O(5)\\bigr)\n   =34-10=24 .\n\\]\nTherefore $F_w^{\\operatorname{conn}}/\\mathrm O(5)$ is a smooth\n$24$–dimensional manifold.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.619863",
        "was_fixed": false,
        "difficulty_analysis": "•  Higher dimension and rectangular geometry.  \n The original problem dealt with a 2-ball inside a 4-cube; the present\n variant treats a k-ball inside an n-box with unequal side–lengths\n (and asks for both the general formula and a concrete 12×5 instance).\n\n•  Non-linear optimisation.  \n Instead of a single quadratic inequality, one must solve the\n piece-wise quadratic non-linear equation (†) and understand its unique\n root; an explicit computation is required for the given data.\n\n•  Frame theory and majorisation.  \n Showing sufficiency of (★) forces the solver to invoke deep facts about\n Parseval frames (or the Kadison–Singer/Carpenter theorem) rather than a\n simple Cauchy–Schwarz bound.\n\n•  Classification of extremals.  \n The problem asks not only for the maximal radius but also for the\n complete description of all embeddings that attain it, a step that is\n absent in the original setting and requires an understanding of tight\n frames up to orthogonal equivalence.\n\n•  Translation invariance.  \n Part (c) demands an argument that the optimum is unaffected by moving\n the centre, adding an extra conceptual layer.\n\n•  Concrete high-dimensional construction.  \n Producing an explicit 12×5 extremal matrix forces the solver to juggle\n norms of three different sizes and to weave them into a tight frame, a\n task much subtler than writing down two perpendicular diagonals in a\n cube.\n\nIn sum, the problem blends linear algebra, convex geometry, optimisation\nand frame theory, far surpassing the technical demands of both the\noriginal problem and the prior kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation",
  "iteratively_fixed": true
}