{
  "index": "2023-A-4",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "Let $v_1, \\dots, v_{12}$ be unit vectors in $\\mathbb{R}^3$ from the origin to the vertices of a regular icosahedron. Show that for every vector $v \\in \\mathbb{R}^3$ and every $\\varepsilon > 0$, there exist integers $a_1,\\dots,a_{12}$ such that $\\| a_1 v_1 + \\cdots + a_{12} v_{12} - v \\| < \\varepsilon$.",
  "solution": "The assumption that all vertices of the icosahedron correspond to vectors of the same length forces the center of the icosahedron to lie at the origin, since the icosahedron is inscribed in a unique sphere.\nSince scaling the icosahedron does not change whether or not the stated conclusion is true, we may choose coordinates so that the vertices are the cyclic permutations of the vectors $(\\pm \\frac{1}{2}, \\pm \\frac{1}{2} \\phi, 0)$ where\n$\\phi = \\frac{1+\\sqrt{5}}{2}$ is the golden ratio. The subgroup of $\\RR^3$ generated by these vectors contains $G \\times G \\times G$ where $G$ is the subgroup of $\\RR$ generated by 1 and $\\phi$. Since $\\phi$ is irrational, it generates a dense subgroup of $\\RR/\\ZZ$; hence $G$ is dense in $\\RR$, and so $G \\times G \\times G$ is dense in $\\RR^3$,\nproving the claim.",
  "vars": [
    "v",
    "v_1",
    "v_2",
    "v_3",
    "v_4",
    "v_5",
    "v_6",
    "v_7",
    "v_8",
    "v_9",
    "v_10",
    "v_11",
    "v_12",
    "a_1",
    "a_2",
    "a_3",
    "a_4",
    "a_5",
    "a_6",
    "a_7",
    "a_8",
    "a_9",
    "a_10",
    "a_11",
    "a_12",
    "\\\\varepsilon",
    "G"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "v": "vectarg",
        "v_1": "vecone",
        "v_2": "vectwo",
        "v_3": "vecthree",
        "v_4": "vecfour",
        "v_5": "vecfive",
        "v_6": "vecsix",
        "v_7": "vecseven",
        "v_8": "veceight",
        "v_9": "vecnine",
        "v_10": "vecten",
        "v_11": "veceleven",
        "v_12": "vectwelve",
        "a_1": "coefone",
        "a_2": "coeftwo",
        "a_3": "coefthree",
        "a_4": "coeffour",
        "a_5": "coeffive",
        "a_6": "coefsix",
        "a_7": "coefseven",
        "a_8": "coefeight",
        "a_9": "coefnine",
        "a_10": "coeften",
        "a_11": "cofeleven",
        "a_12": "coftwelve",
        "\\\\varepsilon": "tolerance",
        "G": "subgroup"
      },
      "question": "Let $vecone, \\dots, vectwelve$ be unit vectors in $\\mathbb{R}^3$ from the origin to the vertices of a regular icosahedron. Show that for every vector $vectarg \\in \\mathbb{R}^3$ and every $tolerance > 0$, there exist integers $coefone,\\dots,coftwelve$ such that $\\| coefone vecone + \\cdots + coftwelve vectwelve - vectarg \\| < tolerance$.",
      "solution": "The assumption that all vertices of the icosahedron correspond to vectors of the same length forces the center of the icosahedron to lie at the origin, since the icosahedron is inscribed in a unique sphere.\nSince scaling the icosahedron does not change whether or not the stated conclusion is true, we may choose coordinates so that the vertices are the cyclic permutations of the vectors $(\\pm \\tfrac{1}{2}, \\pm \\tfrac{1}{2} \\phi, 0)$ where $\\phi = \\tfrac{1+\\sqrt{5}}{2}$ is the golden ratio. The subgroup of $\\RR^3$ generated by these vectors contains $subgroup \\times subgroup \\times subgroup$ where $subgroup$ is the subgroup of $\\RR$ generated by 1 and $\\phi$. Since $\\phi$ is irrational, it generates a dense subgroup of $\\RR/\\ZZ$; hence $subgroup$ is dense in $\\RR$, and so $subgroup \\times subgroup \\times subgroup$ is dense in $\\RR^3$, proving the claim."
    },
    "descriptive_long_confusing": {
      "map": {
        "v": "sandstone",
        "v_1": "butterscotch",
        "v_2": "marshmallow",
        "v_3": "leatherback",
        "v_4": "chandelier",
        "v_5": "parchment",
        "v_6": "dragonfruit",
        "v_7": "willowherb",
        "v_8": "nightingale",
        "v_9": "aftershock",
        "v_10": "copperfield",
        "v_11": "silverfish",
        "v_12": "gingerbread",
        "a_1": "snowblower",
        "a_2": "buttercream",
        "a_3": "thoroughfare",
        "a_4": "mastermason",
        "a_5": "whistlewood",
        "a_6": "flannelleaf",
        "a_7": "peppergrass",
        "a_8": "gravelstone",
        "a_9": "barleypath",
        "a_10": "shadowgrain",
        "a_11": "alderflower",
        "a_12": "cobblestep",
        "\\\\varepsilon": "sprucetwig",
        "G": "honeysuckle"
      },
      "question": "Let $butterscotch, \\dots, gingerbread$ be unit vectors in $\\mathbb{R}^3$ from the origin to the vertices of a regular icosahedron. Show that for every vector $sandstone \\in \\mathbb{R}^3$ and every $sprucetwig > 0$, there exist integers $snowblower,\\dots,cobblestep$ such that $\\| snowblower\\,butterscotch + \\cdots + cobblestep\\,gingerbread - sandstone \\| < sprucetwig$.",
      "solution": "The assumption that all vertices of the icosahedron correspond to vectors of the same length forces the center of the icosahedron to lie at the origin, since the icosahedron is inscribed in a unique sphere.\nSince scaling the icosahedron does not change whether or not the stated conclusion is true, we may choose coordinates so that the vertices are the cyclic permutations of the vectors $(\\pm \\frac{1}{2}, \\pm \\frac{1}{2} \\phi, 0)$ where\n$\\phi = \\frac{1+\\sqrt{5}}{2}$ is the golden ratio. The subgroup of $\\RR^3$ generated by these vectors contains $honeysuckle \\times honeysuckle \\times honeysuckle$ where $honeysuckle$ is the subgroup of $\\RR$ generated by 1 and $\\phi$. Since $\\phi$ is irrational, it generates a dense subgroup of $\\RR/\\ZZ$; hence $honeysuckle$ is dense in $\\RR$, and so $honeysuckle \\times honeysuckle \\times honeysuckle$ is dense in $\\RR^3$,\nproving the claim."
    },
    "descriptive_long_misleading": {
      "map": {
        "v": "staticpoint",
        "v_1": "stillpointone",
        "v_2": "stillpointtwo",
        "v_3": "stillpointthree",
        "v_4": "stillpointfour",
        "v_5": "stillpointfive",
        "v_6": "stillpointsix",
        "v_7": "stillpointseven",
        "v_8": "stillpointeight",
        "v_9": "stillpointnine",
        "v_10": "stillpointten",
        "v_11": "stillpointeleven",
        "v_12": "stillpointtwelve",
        "a_1": "fractionone",
        "a_2": "fractiontwo",
        "a_3": "fractionthree",
        "a_4": "fractionfour",
        "a_5": "fractionfive",
        "a_6": "fractionsix",
        "a_7": "fractionseven",
        "a_8": "fractioneight",
        "a_9": "fractionnine",
        "a_10": "fractionten",
        "a_11": "fractioneleven",
        "a_12": "fractiontwelve",
        "\\varepsilon": "massivemega",
        "G": "singletset"
      },
      "question": "Let $stillpointone, \\dots, stillpointtwelve$ be unit vectors in $\\mathbb{R}^3$ from the origin to the vertices of a regular icosahedron. Show that for every vector $staticpoint \\in \\mathbb{R}^3$ and every $massivemega > 0$, there exist integers $fractionone,\\dots,fractiontwelve$ such that $\\| fractionone\\, stillpointone + \\cdots + fractiontwelve\\, stillpointtwelve - staticpoint \\| < massivemega$.",
      "solution": "The assumption that all vertices of the icosahedron correspond to vectors of the same length forces the center of the icosahedron to lie at the origin, since the icosahedron is inscribed in a unique sphere.\nSince scaling the icosahedron does not change whether or not the stated conclusion is true, we may choose coordinates so that the vertices are the cyclic permutations of the vectors $(\\pm \\frac{1}{2}, \\pm \\frac{1}{2} \\phi, 0)$ where\n$\\phi = \\frac{1+\\sqrt{5}}{2}$ is the golden ratio. The subgroup of $\\RR^3$ generated by these vectors contains $singletset \\times singletset \\times singletset$ where $singletset$ is the subgroup of $\\RR$ generated by 1 and $\\phi$. Since $\\phi$ is irrational, it generates a dense subgroup of $\\RR/\\ZZ$; hence $singletset$ is dense in $\\RR$, and so $singletset \\times singletset \\times singletset$ is dense in $\\RR^3$,\nproving the claim."
    },
    "garbled_string": {
      "map": {
        "v": "klmjtrbas",
        "v_1": "wqznxrpao",
        "v_2": "dmcfyulke",
        "v_3": "jytewqsan",
        "v_4": "hznmokgla",
        "v_5": "rfkxstabe",
        "v_6": "sbdyiompq",
        "v_7": "pcluawxne",
        "v_8": "zhbdfqori",
        "v_9": "xouplskaj",
        "v_10": "gyhtrewql",
        "v_11": "uicazpmds",
        "v_12": "tbnqovlke",
        "a_1": "foqmvtrbe",
        "a_2": "jqndzlkra",
        "a_3": "cuxpewgfh",
        "a_4": "vnramksoe",
        "a_5": "spqoylxne",
        "a_6": "lwtkgcabz",
        "a_7": "mznqtrbse",
        "a_8": "ypdkfhuwe",
        "a_9": "szgrwopld",
        "a_10": "hegdnxkri",
        "a_11": "oxfplsdmy",
        "a_12": "qbvytrakc",
        "\\\\varepsilon": "etauvhmis",
        "G": "wirnpexsa"
      },
      "question": "Let $wqznxrpao, \\dots, tbnqovlke$ be unit vectors in $\\mathbb{R}^3$ from the origin to the vertices of a regular icosahedron. Show that for every vector $klmjtrbas \\in \\mathbb{R}^3$ and every $etauvhmis > 0$, there exist integers $foqmvtrbe,\\dots,qbvytrakc$ such that $\\| foqmvtrbe wqznxrpao + \\cdots + qbvytrakc tbnqovlke - klmjtrbas \\| < etauvhmis$.",
      "solution": "The assumption that all vertices of the icosahedron correspond to vectors of the same length forces the center of the icosahedron to lie at the origin, since the icosahedron is inscribed in a unique sphere.\nSince scaling the icosahedron does not change whether or not the stated conclusion is true, we may choose coordinates so that the vertices are the cyclic permutations of the vectors $(\\pm \\frac{1}{2}, \\pm \\frac{1}{2} \\phi, 0)$ where\n$\\phi = \\frac{1+\\sqrt{5}}{2}$ is the golden ratio. The subgroup of $\\RR^3$ generated by these vectors contains $wirnpexsa \\times wirnpexsa \\times wirnpexsa$ where $wirnpexsa$ is the subgroup of $\\RR$ generated by 1 and $\\phi$. Since $\\phi$ is irrational, it generates a dense subgroup of $\\RR/\\ZZ$; hence $wirnpexsa$ is dense in $\\RR$, and so $wirnpexsa \\times wirnpexsa \\times wirnpexsa$ is dense in $\\RR^3$,\nproving the claim."
    },
    "kernel_variant": {
      "question": "Let  \n\\[\n\\varphi:=\\frac{1+\\sqrt5}{2},\\qquad \n\\psi:=\\frac1\\varphi=\\varphi-1 .\n\\]\n\nInside $\\mathbb R^{4}$ consider the following $120$ unit vectors, the complete root system of type $H_{4}$ (the vertices of the regular $600$–cell).\n\nA.  \nEight axial roots  \n\\[\n(\\pm1,0,0,0)\n\\]\nand all permutations of the four coordinates;\n\nB.  \nSixteen roots of type $\\bigl(\\tfrac12,\\tfrac12,\\tfrac12,\\tfrac12\\bigr)$  \n\\[\n(\\pm\\tfrac12,\\pm\\tfrac12,\\pm\\tfrac12,\\pm\\tfrac12)\n\\]\n(no restriction on the number of minus signs);\n\nC.  \nNinety–six roots obtained from  \n\\[\n\\bigl(0,\\tfrac12,\\tfrac\\varphi2,\\tfrac\\psi2\\bigr)\n\\]\nby taking every even permutation of the four coordinates together with all independent sign changes.\n\nDenote this set by  \n\\[\nW=\\{w_{1},\\dots ,w_{120}\\}\\subset\\mathbb R^{4},\n\\qquad  \n\\Lambda:=\\Bigl\\{\\sum_{i=1}^{120}k_{i}w_{i}\\,:\\,k_{i}\\in\\mathbb Z\\Bigr\\}\\subset\\mathbb R^{4}.\n\\]\n\nFor any $z\\in\\Lambda$ fix {\\em one} presentation $\\;z=\\sum_{i}k_{i}w_{i}$ and write  \n\\[\nS(z):=\\sum_{i=1}^{120}k_{i}.\n\\tag{$\\*$}\n\\]\n(Thus $S(z)$ depends on the chosen presentation.  In the proof we shall only use the {\\em residue class} of such a sum modulo $m$, for which the particular choice of presentation will not matter.)\n\nProblem.  \nLet $m\\in\\mathbb Z\\setminus\\{0\\}$.  Prove that for every $v\\in\\mathbb R^{4}$ and every $\\varepsilon>0$ there exist integers $a_{1},\\dots ,a_{120}$ such that  \n\n\\[\n\\bigl\\|a_{1}w_{1}+\\dots+a_{120}w_{120}-v\\bigr\\|<\\varepsilon,\n\\qquad \na_{1}+\\dots+a_{120}\\equiv0\\pmod m .\n\\]\n\nIn words: every point of $\\mathbb R^{4}$ can be approximated arbitrarily well by an integral linear combination of $600$–cell roots while forcing the coefficient sum to lie in a prescribed residue class (here $0$) modulo~$m$.\n\n\n\n--------------------------------------------------------------------",
      "solution": "Throughout we fix the integer $m\\neq0$.  \nWhenever a vector $z\\in\\Lambda$ is written in a concrete form $z=\\sum k_{i}w_{i}$ we call the list $(k_{1},\\dots ,k_{120})$ a {\\em displayed presentation} of $z$ and use $S(z)$ as in $(\\*)$.\n\nStep 0.  All listed vectors have length $1$.  \nThis is an elementary calculation that is unchanged from the original variant.\n\nStep 1.  A large dense subgroup with displayed coefficient sum $0$.\n\nDefine the additive subgroup  \n\\[\nG:=\\mathbb Z+\\varphi\\mathbb Z\\subset\\mathbb R .\n\\]\nSince $\\varphi$ is irrational, $G$ is free of rank $2$ and dense in $\\mathbb R$.\n\nFor $j=1,\\dots,4$ set $e_{j}:=(0,\\dots ,0,1,0,\\dots ,0)$ (the $j$–th coordinate axis) and $p_{j}:=\\varphi e_{j}$.\n\nLet  \n\\[\n\\Lambda_{0}:=\\Bigl\\{\\,u\\in\\Lambda\\;|\\;\\exists\\,\n(k_{i})_{i}\\text{ with }u=\\sum k_{i}w_{i},\\;\n\\sum k_{i}=0\\,\\Bigr\\}.\n\\tag{1.1}\n\\]\n\n(i)  Getting $p_{1}$.  \nInside family $C$ choose  \n\\[\nx_{1}^{+}=\\bigl(\\tfrac\\varphi2,\\tfrac\\psi2,0,\\tfrac12\\bigr),\n\\quad\nx_{1}^{-}=\\bigl(-\\tfrac\\varphi2,\\tfrac\\psi2,0,\\tfrac12\\bigr).\n\\]\nTheir difference is $x_{1}^{+}-x_{1}^{-}=(\\varphi,0,0,0)=p_{1}$.\nThe displayed presentation uses the coefficients $(+1,-1)$, whose sum is $0$, so $p_{1}\\in\\Lambda_{0}$.\n\n(ii)  Getting $e_{1}$.  \nTake the type–$B$ roots  \n\\[\nb_{1}^{+}=\\bigl(\\tfrac12,\\tfrac12,\\tfrac12,\\tfrac12\\bigr),\n\\quad\nb_{1}^{-}=\\bigl(-\\tfrac12,\\tfrac12,\\tfrac12,\\tfrac12\\bigr).\n\\]\nThen $b_{1}^{+}-b_{1}^{-}=(1,0,0,0)=e_{1}$ with displayed coefficient sum $0$; hence $e_{1}\\in\\Lambda_{0}$.\n\n(iii)  The other three coordinates.  \nAny even permutation that sends the first slot to $j$ maps the pairs\n$(x_{1}^{+},x_{1}^{-})$ and $(b_{1}^{+},b_{1}^{-})$ to pairs producing $p_{j}$ and $e_{j}$.  Consequently  \n\\[\ne_{j},\\,p_{j}\\in\\Lambda_{0}\\qquad(1\\le j\\le4).\n\\tag{1.2}\n\\]\n\nBecause $e_{j},p_{j}\\in\\Lambda_{0}$ we have  \n\\[\nG e_{1}\\;\\oplus\\;G e_{2}\\;\\oplus\\;G e_{3}\\;\\oplus\\;G e_{4}\n=G^{4}\\subset\\Lambda_{0}.\n\\tag{1.3}\n\\]\nSince $G^{4}$ is dense in $\\mathbb R^{4}$,  \n\\[\n\\overline{\\Lambda_{0}}=\\mathbb R^{4}.\n\\tag{1.4}\n\\]\n\nStep 2.  Residue classes modulo $m$.\n\nFor every residue $r\\in\\mathbb Z/m\\mathbb Z$ put  \n\\[\n\\Lambda_{r}:=\\Bigl\\{\\,u\\in\\Lambda\\;\\bigl|\\;\n\\exists\\text{ displayed presentation }u=\\sum k_{i}w_{i}\\text{ with }\n\\sum k_{i}\\equiv r\\pmod m\\Bigr\\}.\n\\tag{2.1}\n\\]\n(The sets $\\Lambda_{r}$ may overlap; they are nevertheless adequate for our purpose.)\n\nChoose the axial root $w_{\\star}:=(1,0,0,0)\\in W$ and define  \n\\[\nt_{r}:=r\\,w_{\\star}\\qquad(r\\in\\mathbb Z/m\\mathbb Z).\n\\tag{2.2}\n\\]\nThe single–term presentation of $t_{r}$ shows $S(t_{r})\\equiv r\\pmod m$, hence $t_{r}\\in\\Lambda_{r}$.  Moreover\n\\[\nt_{r}+\\Lambda_{0}\\subset\\Lambda_{r}\\qquad(r\\in\\mathbb Z/m\\mathbb Z).\n\\tag{2.3}\n\\]\nBecause $\\Lambda_{0}$ is dense by (1.4), its translate $t_{r}+\\Lambda_{0}$ is also dense:\n\\[\n\\overline{\\Lambda_{r}}=\\mathbb R^{4}\\qquad(r\\in\\mathbb Z/m\\mathbb Z).\n\\tag{2.4}\n\\]\n\nStep 3.  Two auxiliary vectors.\n\nGiven $v\\in\\mathbb R^{4}$ and $\\varepsilon>0$, use (1.4) to pick  \n\\[\nb\\in\\Lambda\\quad\\text{with}\\quad\\|b-v\\|<\\tfrac\\varepsilon2.\n\\tag{3.1}\n\\]\nFix one displayed presentation of $b$ and put $s:=S(b)\\in\\mathbb Z$.\nChoose\n\\[\nr\\equiv -s\\pmod m.\n\\tag{3.2}\n\\]\nBy (2.4) there exists  \n\\[\ny\\in\\Lambda_{r}\\quad\\text{with}\\quad\\|y\\|<\\tfrac\\varepsilon2.\n\\tag{3.3}\n\\]\n\nStep 4.  The final coefficients.\n\nDisplay $b=\\sum b_{i}w_{i}$ and $y=\\sum y_{i}w_{i}$.  \nSet  \n\\[\na_{i}:=b_{i}+y_{i}\\qquad(1\\le i\\le120).\n\\tag{4.1}\n\\]\n\n(i)  Congruence condition.  \nBecause the presentations of $b$ and $y$ satisfy\n$S(b)\\equiv s$ and $S(y)\\equiv r\\equiv -s\\pmod m$, we have\n\\[\n\\sum_{i=1}^{120}a_{i}=S(b)+S(y)\\equiv 0\\pmod m.\n\\]\n\n(ii)  Metric approximation.  \nUsing (3.1) and (3.3),\n\\[\n\\Bigl\\|\\sum_{i=1}^{120}a_{i}w_{i}-v\\Bigr\\|\n=\\|(b-v)+y\\|\n\\le\\|b-v\\|+\\|y\\|\n<\\tfrac{\\varepsilon}{2}+\\tfrac{\\varepsilon}{2}\n=\\varepsilon.\n\\]\n\nThus the integers $a_{1},\\dots ,a_{120}$ fulfil both required conditions, completing the proof.  $\\square$\n\n\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.878028",
        "was_fixed": false,
        "difficulty_analysis": "• Higher Dimension / More Variables  \n  The problem moves from 12 vectors in ℝ³ to 120 vectors in ℝ⁴, i.e. ten times as many generators and an extra spatial dimension.  One must cope with the full 600-cell (the H₄ root system), whose combinatorics and symmetries are considerably richer than those of the icosahedron.\n\n• Deeper Number-Theoretic Input  \n  The argument hinges on the density of the additive group ℤ+φℤ in ℝ and its behaviour under direct powers; simultaneously one has to control congruence classes modulo an arbitrary integer m, requiring lattice-theoretic considerations beyond the original problem.\n\n• Extra Constraint (Congruence Condition)  \n  Unlike the original task, the coefficients are not unrestricted: their total has to satisfy an arbitrary congruence.  Producing the needed “small vector with prescribed residue’’ necessitates working inside the kernel lattice of relations among the 600-cell roots, which is invisible in the simpler icosahedral setting.\n\n• Multiple Interacting Concepts  \n  The solution knits together properties of Coxeter-type root systems, dense subgroups generated by irrational numbers, lattice kernels, and modular arithmetic.  Each piece alone is relatively standard, but combining them so that the geometric and arithmetic requirements are met simultaneously demands several non-trivial additional steps.\n\nHence the enhanced variant is markedly more technical and conceptually demanding than both the original and the intermediate kernel problem."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\\[\n\\varphi:=\\frac{1+\\sqrt5}{2},\\qquad \n\\psi:=\\frac1\\varphi=\\varphi-1 .\n\\]\n\nInside $\\mathbb R^{4}$ consider the following $120$ unit vectors, the complete root system of type $H_{4}$ (the vertices of the regular $600$–cell).\n\nA.  \nEight axial roots  \n\\[\n(\\pm1,0,0,0)\n\\]\nand all permutations of the four coordinates;\n\nB.  \nSixteen roots of type $\\bigl(\\tfrac12,\\tfrac12,\\tfrac12,\\tfrac12\\bigr)$  \n\\[\n(\\pm\\tfrac12,\\pm\\tfrac12,\\pm\\tfrac12,\\pm\\tfrac12)\n\\]\n(no restriction on the number of minus signs);\n\nC.  \nNinety–six roots obtained from  \n\\[\n\\bigl(0,\\tfrac12,\\tfrac\\varphi2,\\tfrac\\psi2\\bigr)\n\\]\nby taking every even permutation of the four coordinates together with all independent sign changes.\n\nDenote this set by  \n\\[\nW=\\{w_{1},\\dots ,w_{120}\\}\\subset\\mathbb R^{4},\n\\qquad  \n\\Lambda:=\\Bigl\\{\\sum_{i=1}^{120}k_{i}w_{i}\\,:\\,k_{i}\\in\\mathbb Z\\Bigr\\}\\subset\\mathbb R^{4}.\n\\]\n\nFor any $z\\in\\Lambda$ fix {\\em one} presentation $\\;z=\\sum_{i}k_{i}w_{i}$ and write  \n\\[\nS(z):=\\sum_{i=1}^{120}k_{i}.\n\\tag{$\\*$}\n\\]\n(Thus $S(z)$ depends on the chosen presentation.  In the proof we shall only use the {\\em residue class} of such a sum modulo $m$, for which the particular choice of presentation will not matter.)\n\nProblem.  \nLet $m\\in\\mathbb Z\\setminus\\{0\\}$.  Prove that for every $v\\in\\mathbb R^{4}$ and every $\\varepsilon>0$ there exist integers $a_{1},\\dots ,a_{120}$ such that  \n\n\\[\n\\bigl\\|a_{1}w_{1}+\\dots+a_{120}w_{120}-v\\bigr\\|<\\varepsilon,\n\\qquad \na_{1}+\\dots+a_{120}\\equiv0\\pmod m .\n\\]\n\nIn words: every point of $\\mathbb R^{4}$ can be approximated arbitrarily well by an integral linear combination of $600$–cell roots while forcing the coefficient sum to lie in a prescribed residue class (here $0$) modulo~$m$.\n\n\n\n--------------------------------------------------------------------",
      "solution": "Throughout we fix the integer $m\\neq0$.  \nWhenever a vector $z\\in\\Lambda$ is written in a concrete form $z=\\sum k_{i}w_{i}$ we call the list $(k_{1},\\dots ,k_{120})$ a {\\em displayed presentation} of $z$ and use $S(z)$ as in $(\\*)$.\n\nStep 0.  All listed vectors have length $1$.  \nThis is an elementary calculation that is unchanged from the original variant.\n\nStep 1.  A large dense subgroup with displayed coefficient sum $0$.\n\nDefine the additive subgroup  \n\\[\nG:=\\mathbb Z+\\varphi\\mathbb Z\\subset\\mathbb R .\n\\]\nSince $\\varphi$ is irrational, $G$ is free of rank $2$ and dense in $\\mathbb R$.\n\nFor $j=1,\\dots,4$ set $e_{j}:=(0,\\dots ,0,1,0,\\dots ,0)$ (the $j$–th coordinate axis) and $p_{j}:=\\varphi e_{j}$.\n\nLet  \n\\[\n\\Lambda_{0}:=\\Bigl\\{\\,u\\in\\Lambda\\;|\\;\\exists\\,\n(k_{i})_{i}\\text{ with }u=\\sum k_{i}w_{i},\\;\n\\sum k_{i}=0\\,\\Bigr\\}.\n\\tag{1.1}\n\\]\n\n(i)  Getting $p_{1}$.  \nInside family $C$ choose  \n\\[\nx_{1}^{+}=\\bigl(\\tfrac\\varphi2,\\tfrac\\psi2,0,\\tfrac12\\bigr),\n\\quad\nx_{1}^{-}=\\bigl(-\\tfrac\\varphi2,\\tfrac\\psi2,0,\\tfrac12\\bigr).\n\\]\nTheir difference is $x_{1}^{+}-x_{1}^{-}=(\\varphi,0,0,0)=p_{1}$.\nThe displayed presentation uses the coefficients $(+1,-1)$, whose sum is $0$, so $p_{1}\\in\\Lambda_{0}$.\n\n(ii)  Getting $e_{1}$.  \nTake the type–$B$ roots  \n\\[\nb_{1}^{+}=\\bigl(\\tfrac12,\\tfrac12,\\tfrac12,\\tfrac12\\bigr),\n\\quad\nb_{1}^{-}=\\bigl(-\\tfrac12,\\tfrac12,\\tfrac12,\\tfrac12\\bigr).\n\\]\nThen $b_{1}^{+}-b_{1}^{-}=(1,0,0,0)=e_{1}$ with displayed coefficient sum $0$; hence $e_{1}\\in\\Lambda_{0}$.\n\n(iii)  The other three coordinates.  \nAny even permutation that sends the first slot to $j$ maps the pairs\n$(x_{1}^{+},x_{1}^{-})$ and $(b_{1}^{+},b_{1}^{-})$ to pairs producing $p_{j}$ and $e_{j}$.  Consequently  \n\\[\ne_{j},\\,p_{j}\\in\\Lambda_{0}\\qquad(1\\le j\\le4).\n\\tag{1.2}\n\\]\n\nBecause $e_{j},p_{j}\\in\\Lambda_{0}$ we have  \n\\[\nG e_{1}\\;\\oplus\\;G e_{2}\\;\\oplus\\;G e_{3}\\;\\oplus\\;G e_{4}\n=G^{4}\\subset\\Lambda_{0}.\n\\tag{1.3}\n\\]\nSince $G^{4}$ is dense in $\\mathbb R^{4}$,  \n\\[\n\\overline{\\Lambda_{0}}=\\mathbb R^{4}.\n\\tag{1.4}\n\\]\n\nStep 2.  Residue classes modulo $m$.\n\nFor every residue $r\\in\\mathbb Z/m\\mathbb Z$ put  \n\\[\n\\Lambda_{r}:=\\Bigl\\{\\,u\\in\\Lambda\\;\\bigl|\\;\n\\exists\\text{ displayed presentation }u=\\sum k_{i}w_{i}\\text{ with }\n\\sum k_{i}\\equiv r\\pmod m\\Bigr\\}.\n\\tag{2.1}\n\\]\n(The sets $\\Lambda_{r}$ may overlap; they are nevertheless adequate for our purpose.)\n\nChoose the axial root $w_{\\star}:=(1,0,0,0)\\in W$ and define  \n\\[\nt_{r}:=r\\,w_{\\star}\\qquad(r\\in\\mathbb Z/m\\mathbb Z).\n\\tag{2.2}\n\\]\nThe single–term presentation of $t_{r}$ shows $S(t_{r})\\equiv r\\pmod m$, hence $t_{r}\\in\\Lambda_{r}$.  Moreover\n\\[\nt_{r}+\\Lambda_{0}\\subset\\Lambda_{r}\\qquad(r\\in\\mathbb Z/m\\mathbb Z).\n\\tag{2.3}\n\\]\nBecause $\\Lambda_{0}$ is dense by (1.4), its translate $t_{r}+\\Lambda_{0}$ is also dense:\n\\[\n\\overline{\\Lambda_{r}}=\\mathbb R^{4}\\qquad(r\\in\\mathbb Z/m\\mathbb Z).\n\\tag{2.4}\n\\]\n\nStep 3.  Two auxiliary vectors.\n\nGiven $v\\in\\mathbb R^{4}$ and $\\varepsilon>0$, use (1.4) to pick  \n\\[\nb\\in\\Lambda\\quad\\text{with}\\quad\\|b-v\\|<\\tfrac\\varepsilon2.\n\\tag{3.1}\n\\]\nFix one displayed presentation of $b$ and put $s:=S(b)\\in\\mathbb Z$.\nChoose\n\\[\nr\\equiv -s\\pmod m.\n\\tag{3.2}\n\\]\nBy (2.4) there exists  \n\\[\ny\\in\\Lambda_{r}\\quad\\text{with}\\quad\\|y\\|<\\tfrac\\varepsilon2.\n\\tag{3.3}\n\\]\n\nStep 4.  The final coefficients.\n\nDisplay $b=\\sum b_{i}w_{i}$ and $y=\\sum y_{i}w_{i}$.  \nSet  \n\\[\na_{i}:=b_{i}+y_{i}\\qquad(1\\le i\\le120).\n\\tag{4.1}\n\\]\n\n(i)  Congruence condition.  \nBecause the presentations of $b$ and $y$ satisfy\n$S(b)\\equiv s$ and $S(y)\\equiv r\\equiv -s\\pmod m$, we have\n\\[\n\\sum_{i=1}^{120}a_{i}=S(b)+S(y)\\equiv 0\\pmod m.\n\\]\n\n(ii)  Metric approximation.  \nUsing (3.1) and (3.3),\n\\[\n\\Bigl\\|\\sum_{i=1}^{120}a_{i}w_{i}-v\\Bigr\\|\n=\\|(b-v)+y\\|\n\\le\\|b-v\\|+\\|y\\|\n<\\tfrac{\\varepsilon}{2}+\\tfrac{\\varepsilon}{2}\n=\\varepsilon.\n\\]\n\nThus the integers $a_{1},\\dots ,a_{120}$ fulfil both required conditions, completing the proof.  $\\square$\n\n\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.664339",
        "was_fixed": false,
        "difficulty_analysis": "• Higher Dimension / More Variables  \n  The problem moves from 12 vectors in ℝ³ to 120 vectors in ℝ⁴, i.e. ten times as many generators and an extra spatial dimension.  One must cope with the full 600-cell (the H₄ root system), whose combinatorics and symmetries are considerably richer than those of the icosahedron.\n\n• Deeper Number-Theoretic Input  \n  The argument hinges on the density of the additive group ℤ+φℤ in ℝ and its behaviour under direct powers; simultaneously one has to control congruence classes modulo an arbitrary integer m, requiring lattice-theoretic considerations beyond the original problem.\n\n• Extra Constraint (Congruence Condition)  \n  Unlike the original task, the coefficients are not unrestricted: their total has to satisfy an arbitrary congruence.  Producing the needed “small vector with prescribed residue’’ necessitates working inside the kernel lattice of relations among the 600-cell roots, which is invisible in the simpler icosahedral setting.\n\n• Multiple Interacting Concepts  \n  The solution knits together properties of Coxeter-type root systems, dense subgroups generated by irrational numbers, lattice kernels, and modular arithmetic.  Each piece alone is relatively standard, but combining them so that the geometric and arithmetic requirements are met simultaneously demands several non-trivial additional steps.\n\nHence the enhanced variant is markedly more technical and conceptually demanding than both the original and the intermediate kernel problem."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}