{
  "index": "1994-A-3",
  "type": "COMB",
  "tag": [
    "COMB",
    "GEO"
  ],
  "difficulty": "",
  "question": "Show that if the points of an isosceles right triangle of side length\n1 are each colored with one of four colors, then there must be two points\nof the same color whch are at least a distance $2 - \\sqrt{2}$ apart.",
  "solution": "Solution. Suppose that it is possible to color the points of the triangle with four colors so that any two points at least \\( 2-\\sqrt{2} \\) apart receive different colors. Suppose the vertices of the triangle are \\( A=(0,0), B=(1,0), C=(0,1) \\). Define also the points \\( D=(\\sqrt{2}-1,0) \\) and \\( E=(0, \\sqrt{2}-1) \\) on the two sides, and \\( F=(2-\\sqrt{2}, \\sqrt{2}-1) \\) and \\( G=(\\sqrt{2}-1,2-\\sqrt{2}) \\) on the diagonal; see Figure 27. Note that\n\\[\nB D=B F=C E=C G=D E=D G=E F=2-\\sqrt{2}\n\\]\n\nThe color of \\( B \\) must be different from the colors of the other named points. The same holds for \\( C \\). Thus the vertices of the pentagram \\( A G D E F \\) must each be painted one of the two remaining colors. Then two adjacent vertices of the pentagram must have the same color. But they are separated by at least \\( 2-\\sqrt{2} \\), so this is a contradiction.",
  "vars": [
    "A",
    "B",
    "C",
    "D",
    "E",
    "F",
    "G"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "A": "vertexa",
        "B": "vertexb",
        "C": "vertexc",
        "D": "vertexd",
        "E": "vertexe",
        "F": "vertexf",
        "G": "vertexg"
      },
      "question": "Show that if the points of an isosceles right triangle of side length\n1 are each colored with one of four colors, then there must be two points\nof the same color whch are at least a distance $2 - \\sqrt{2}$ apart.",
      "solution": "Solution. Suppose that it is possible to color the points of the triangle with four colors so that any two points at least \\( 2-\\sqrt{2} \\) apart receive different colors. Suppose the vertices of the triangle are \\( vertexa=(0,0), vertexb=(1,0), vertexc=(0,1) \\). Define also the points \\( vertexd=(\\sqrt{2}-1,0) \\) and \\( vertexe=(0, \\sqrt{2}-1) \\) on the two sides, and \\( vertexf=(2-\\sqrt{2}, \\sqrt{2}-1) \\) and \\( vertexg=(\\sqrt{2}-1,2-\\sqrt{2}) \\) on the diagonal; see Figure 27. Note that\n\\[\nvertexb vertexd=vertexb vertexf=vertexc vertexe=vertexc vertexg=vertexd vertexe=vertexd vertexg=vertexe vertexf=2-\\sqrt{2}\n\\]\n\nThe color of \\( vertexb \\) must be different from the colors of the other named points. The same holds for \\( vertexc \\). Thus the vertices of the pentagram \\( vertexa vertexg vertexd vertexe vertexf \\) must each be painted one of the two remaining colors. Then two adjacent vertices of the pentagram must have the same color. But they are separated by at least \\( 2-\\sqrt{2} \\), so this is a contradiction."
    },
    "descriptive_long_confusing": {
      "map": {
        "A": "blueberry",
        "B": "pinecone",
        "C": "rainstorm",
        "D": "horseshoe",
        "E": "goldfish",
        "F": "thumbtack",
        "G": "paintbrush"
      },
      "question": "Show that if the points of an isosceles right triangle of side length\n1 are each colored with one of four colors, then there must be two points\nof the same color whch are at least a distance $2 - \\sqrt{2}$ apart.",
      "solution": "Solution. Suppose that it is possible to color the points of the triangle with four colors so that any two points at least \\( 2-\\sqrt{2} \\) apart receive different colors. Suppose the vertices of the triangle are \\( blueberry=(0,0), pinecone=(1,0), rainstorm=(0,1) \\). Define also the points \\( horseshoe=(\\sqrt{2}-1,0) \\) and \\( goldfish=(0, \\sqrt{2}-1) \\) on the two sides, and \\( thumbtack=(2-\\sqrt{2}, \\sqrt{2}-1) \\) and \\( paintbrush=(\\sqrt{2}-1,2-\\sqrt{2}) \\) on the diagonal; see Figure 27. Note that\n\\[\npinecone\\ horseshoe = pinecone\\ thumbtack = rainstorm\\ goldfish = rainstorm\\ paintbrush = horseshoe\\ goldfish = horseshoe\\ paintbrush = goldfish\\ thumbtack = 2-\\sqrt{2}\n\\]\n\nThe color of \\( pinecone \\) must be different from the colors of the other named points. The same holds for \\( rainstorm \\). Thus the vertices of the pentagram \\( blueberry\\ paintbrush\\ horseshoe\\ goldfish\\ thumbtack \\) must each be painted one of the two remaining colors. Then two adjacent vertices of the pentagram must have the same color. But they are separated by at least \\( 2-\\sqrt{2} \\), so this is a contradiction."
    },
    "descriptive_long_misleading": {
      "map": {
        "A": "emptiness",
        "B": "voidpoint",
        "C": "nullspot",
        "D": "blanknode",
        "E": "hollowplace",
        "F": "scarcity",
        "G": "nonentity"
      },
      "question": "Show that if the points of an isosceles right triangle of side length\n1 are each colored with one of four colors, then there must be two points\nof the same color whch are at least a distance $2 - \\sqrt{2}$ apart.",
      "solution": "Solution. Suppose that it is possible to color the points of the triangle with four colors so that any two points at least \\( 2-\\sqrt{2} \\) apart receive different colors. Suppose the vertices of the triangle are \\( emptiness=(0,0), voidpoint=(1,0), nullspot=(0,1) \\). Define also the points \\( blanknode=(\\sqrt{2}-1,0) \\) and \\( hollowplace=(0, \\sqrt{2}-1) \\) on the two sides, and \\( scarcity=(2-\\sqrt{2}, \\sqrt{2}-1) \\) and \\( nonentity=(\\sqrt{2}-1,2-\\sqrt{2}) \\) on the diagonal; see Figure 27. Note that\n\\[\nvoidpoint blanknode = voidpoint scarcity = nullspot hollowplace = nullspot nonentity = blanknode hollowplace = blanknode nonentity = hollowplace scarcity = 2-\\sqrt{2}\n\\]\n\nThe color of \\( voidpoint \\) must be different from the colors of the other named points. The same holds for \\( nullspot \\). Thus the vertices of the pentagram \\( emptiness nonentity blanknode hollowplace scarcity \\) must each be painted one of the two remaining colors. Then two adjacent vertices of the pentagram must have the same color. But they are separated by at least \\( 2-\\sqrt{2} \\), so this is a contradiction."
    },
    "garbled_string": {
      "map": {
        "A": "xqvmsldp",
        "B": "hregtcoa",
        "C": "pqzwkfnb",
        "D": "myjvnxsr",
        "E": "kuladqwe",
        "F": "zasbqmnr",
        "G": "vlochwer"
      },
      "question": "Show that if the points of an isosceles right triangle of side length\n1 are each colored with one of four colors, then there must be two points\nof the same color whch are at least a distance $2 - \\sqrt{2}$ apart.",
      "solution": "Solution. Suppose that it is possible to color the points of the triangle with four colors so that any two points at least \\( 2-\\sqrt{2} \\) apart receive different colors. Suppose the vertices of the triangle are \\( xqvmsldp=(0,0), hregtcoa=(1,0), pqzwkfnb=(0,1) \\). Define also the points \\( myjvnxsr=(\\sqrt{2}-1,0) \\) and \\( kuladqwe=(0, \\sqrt{2}-1) \\) on the two sides, and \\( zasbqmnr=(2-\\sqrt{2}, \\sqrt{2}-1) \\) and \\( vlochwer=(\\sqrt{2}-1,2-\\sqrt{2}) \\) on the diagonal; see Figure 27. Note that\n\\[\nhregtcoa\\ myjvnxsr = hregtcoa\\ zasbqmnr = pqzwkfnb\\ kuladqwe = pqzwkfnb\\ vlochwer = myjvnxsr\\ kuladqwe = myjvnxsr\\ vlochwer = kuladqwe\\ zasbqmnr = 2-\\sqrt{2}\n\\]\n\nThe color of \\( hregtcoa \\) must be different from the colors of the other named points. The same holds for \\( pqzwkfnb \\). Thus the vertices of the pentagram \\( xqvmsldp\\ vlochwer\\ myjvnxsr\\ kuladqwe\\ zasbqmnr \\) must each be painted one of the two remaining colors. Then two adjacent vertices of the pentagram must have the same color. But they are separated by at least \\( 2-\\sqrt{2} \\), so this is a contradiction."
    },
    "kernel_variant": {
      "question": "Let $n\\ge 2$ be an integer and put \n\\[\nS_{n}:=\\Bigl\\{(x_{1},\\dots ,x_{n})\\in\\mathbf R^{\\,n}:\\;\n           x_{i}\\ge 0\\;(1\\le i\\le n),\\;\n           x_{1}+\\dots +x_{n}\\le 1\\Bigr\\},\n\\]\nthe right–isosceles $n$–simplex whose $n$ edges issuing from the origin\nhave Euclidean length $1$.  \nFix an integer $k\\ge 1$ and colour every point of $S_{n}$ with one of\n(at most) $k$ colours.  For a colouring~$\\chi$ define\n\\[\nd_{\\max }(\\chi):=\\max_{\\substack{x,y\\in S_{n}\\\\\n                         \\chi(x)=\\chi(y)}}\\lVert x-y\\rVert ,\n\\qquad\n\\Delta_{n,k}:=\\inf_{\\chi\\text{ $k$–colouring of }S_{n}}d_{\\max }(\\chi).\n\\]\n\n1. (Exact values for the “few colours’’ regime.)  \n   Prove that, for every $n\\ge 2$,\n   \\[\n        \\boxed{\\;\n        \\Delta_{n,k}=\n        \\begin{cases}\n        \\sqrt{2}, & 1\\le k\\le n-1,\\\\[6pt]\n        1,        & k=n.\n        \\end{cases}}\n   \\]\n   Hence, a monochromatic pair of points at distance at least~$1$\n   is unavoidable as long as fewer than $n+1$ colours are available.\n\n2. (Uniformly positive gap for every finite number of colours.)  \n   Put\n   \\[\n      \\omega_{n}:=\\dfrac{\\pi^{\\,n/2}}{\\Gamma\\!\\bigl(\\tfrac n2+1\\bigr)},\\qquad\n      J_{n}:=\\sqrt{\\dfrac{n}{2(n+1)}}\\quad(n\\ge 2).\n   \\]\n   Show that for every integer $k\\ge n+1$\n   \\[\n          \\frac{1}{J_{n}\\,\\omega_n^{1/n}}\n          \\Bigl(\\tfrac{1}{n!\\,k}\\Bigr)^{1/n}\n          \\;\\le\\;\n          \\Delta_{n,k}\n          \\;\\le\\;\n          \\sqrt n\\,\n          \\frac{(n+1)^{1/n}}{k^{1/n}},\n          \\tag{$\\ast$}\n   \\]\n   and deduce that\n   \\[\n      0<\\liminf_{k\\to\\infty}k^{1/n}\\,\\Delta_{n,k}\\le\n      \\limsup_{k\\to\\infty}k^{1/n}\\,\\Delta_{n,k}<\\infty .\n   \\]\n   Consequently, $\\Delta_{n,k}>0$ for every finite $k$ and\n   $\\Delta_{n,k}= \\Theta(k^{-1/n})$ as $k\\to\\infty$.\n\n3. (Near–optimal colourings for $k=n$.)  \n   In order to obtain a genuine \\emph{partition} of $S_{n}$ we fix, once\n   and for all, the following tie–breaking convention:  \n   for $x\\in S_{n}$ let $\\max (x):=\\max\\{x_{1},\\dots ,x_{n}\\}$ and put\n   \\[\n       \\operatorname{dom}(x):=\n       \\min\\bigl\\{j:\\,x_{j}=\\max (x)\\bigr\\}.\n   \\]\n   Define the $n$ dominant–coordinate regions\n   \\[\n      \\mathcal R_{j}:=\\bigl\\{x\\in S_{n}:\\;\\operatorname{dom}(x)=j\\bigr\\},\n      \\qquad 1\\le j\\le n .\n   \\]\n\n   (a)  Prove that\n        \\[\n              \\lVert x-y\\rVert\\le 1\n              \\qquad(x,y\\in\\mathcal R_{j},\\;1\\le j\\le n),\n        \\]\n        and conclude that the colouring\n        $S_{n}=\\bigsqcup_{j=1}^{n}\\mathcal R_{j}$ satisfies\n        $d_{\\max }(\\chi)=1$, hence attains $\\Delta_{n,n}$.\n\n   (b)  For every $\\varepsilon>0$ construct an $n$–colouring\n        $\\chi_{\\varepsilon}$ of $S_{n}$ such that\n        $d_{\\max }(\\chi_{\\varepsilon})\\le 1+\\varepsilon$.\n        (Thus the bound $\\Delta_{n,n}=1$ is “stable’’.)\n\nRemark.  No claim is made that the diameter of every region\n$\\mathcal R_{j}$ equals $1$; in fact this is false whenever\n$j>1$.  What matters is the uniform bound of part~(a).",
      "solution": "Throughout write\n\\[\n\\mathbf 0:=(0,\\dots ,0),\\qquad\ne_{j}:=(0,\\dots ,0,\\underset{j}{1},0,\\dots ,0)\\;(1\\le j\\le n),\\qquad\n\\mathcal V:=\\{\\mathbf 0,e_{1},\\dots ,e_{n}\\}.\n\\]\nBecause $S_{n}=\\operatorname{conv}(\\mathcal V)$, its diameter equals\n$\\sqrt{2}$.\n\n\\bigskip\n\\textbf{A counting lemma needed for part 2.}  \nFor an integer $m\\ge 1$ subdivide the cube $[0,1]^{n}$ into\n$m^{\\,n}$ axis–parallel cubes of side–length $1/m$.  \nLet $N(m)$ be the number of these small cubes that meet $S_{n}$.  \nA cube is uniquely determined by its index vector\n$I=(i_{1},\\dots ,i_{n})$ with $0\\le i_{t}\\le m-1$, namely\n\\[\nI\\longmapsto\\prod_{t=1}^{n}\n      \\Bigl[\\tfrac{i_{t}}{m},\\,\\tfrac{i_{t}+1}{m}\\Bigr].\n\\]\nSuch a cube intersects $S_{n}$ iff\n\\[\n\\frac{i_{1}+\\dots +i_{n}}{m}\\le 1\n\\quad\\Longleftrightarrow\\quad\ni_{1}+\\dots +i_{n}\\le m-1 .\n\\]\nHence\n\\[\nN(m)=\\#\\Bigl\\{(i_{1},\\dots ,i_{n})\\in\\mathbf Z_{\\ge 0}^{\\,n}:\n              i_{1}+\\dots +i_{n}\\le m-1\\Bigr\\}\n      =\\binom{m-1+n}{n}.\n\\]\n\n\\emph{Two explicit upper bounds:}\n\\begin{align}\nN(m)&=\\binom{m-1+n}{n}\\le(n+1)\\,m^{\\,n}\\qquad(m\\ge 1),\\tag{1}\\\\\nN(m)&=\\binom{m-1+n}{n}\\le(n+1)\\,(m-1)^{\\,n}\\qquad(m\\ge 2).\\tag{2}\n\\end{align}\n\nInequality (1) is classical.  \nFor (2) put $R(m):=N(m)/(m-1)^{n}$ for $m\\ge 2$ and observe\n\\[\n\\frac{R(m+1)}{R(m)}\n   =\\Bigl(1+\\frac{n}{m}\\Bigr)\\Bigl(1-\\frac{1}{m}\\Bigr)^{n}\\le 1,\n\\]\nso $R(m)$ is decreasing and $R(m)\\le R(2)=n+1$.\n\n\\bigskip\n\\textbf{1. Exact values for $k\\le n$.}\n\n(i) $1\\le k\\le n-1$.  \nAmong the $n$ vertices $e_{1},\\dots ,e_{n}$ two share a colour, hence\n$d_{\\max }(\\chi)\\ge\\sqrt{2}$.  \nThe constant colouring gives equality, so\n$\\Delta_{n,k}=\\sqrt{2}$.\n\n(ii) $k=n$.  \n\\emph{Lower bound.}\nThe $n+1$ points of $\\mathcal V$ receive $n$ colours, so two share a\ncolour and are at distance at least $1$.  \nThus $\\Delta_{n,n}\\ge 1$.\n\n\\emph{Upper bound.}\nPart~3 (a) below constructs an $n$–colouring with\n$d_{\\max }=1$, whence $\\Delta_{n,n}=1$.\n\n\\bigskip\n\\textbf{2. Proof of $(\\ast)$ for $k\\ge n+1$.}\n\n\\emph{Lower bound (volume $+$ Jung).}  \nLet $\\chi$ be any $k$–colouring and put $\\delta:=d_{\\max }(\\chi)$.  \nBy Jung’s theorem every subset of diameter $\\delta$ is contained in a\nball of radius $J_{n}\\delta$, so every colour class has volume\n$\\le\\omega_{n}(J_{n}\\delta)^{n}$.  \nBecause $\\operatorname{vol}(S_{n})=1/n!$,\n\\[\nk\\,\\omega_{n}(J_{n}\\delta)^{n}\\ge\\frac{1}{n!}\n\\;\\Longrightarrow\\;\n\\delta\\ge\n\\frac{1}{J_{n}\\,\\omega_{n}^{1/n}}\n\\Bigl(\\tfrac{1}{n!\\,k}\\Bigr)^{1/n}.\n\\]\n\n\\emph{Upper bound (refined grid argument).}  \nSet \n\\[\nm:=\\max\\Bigl\\{2,\\bigl\\lceil(k/(n+1))^{1/n}\\bigr\\rceil\\Bigr\\}.\n\\]\nBy definition $m\\ge 2$ and from (3) below we have\n\\[\n(n+1)(m-1)^{\\,n}<k\\le(n+1)m^{\\,n}.\\tag{3}\n\\]\nTogether with~(2) this implies $N(m)<k$.  Colour each point of $S_{n}$\nby the \\emph{lexicographically} smallest cube (of side $1/m$) that\ncontains it.  \nTwo points of the same colour lie in one cube, whose diameter is\n$\\sqrt n/m$.  \nSince $m\\le(k/(n+1))^{1/n}+1$, the monotonicity of $x\\mapsto1/x$\ngives\n\\[\n\\frac{1}{m}\\le\\frac{(n+1)^{1/n}}{k^{1/n}},\n\\]\nhence\n\\[\nd_{\\max }(\\chi)\\le\\sqrt n\\,\\frac{(n+1)^{1/n}}{k^{1/n}},\n\\]\nwhich is the right–hand side of~$(\\ast)$.\n\n\\bigskip\n\\textbf{3. Near–optimal colourings for $k=n$.}\n\n\\medskip\n\\textbf{(a) The dominant–coordinate colouring attains $\\Delta_{n,n}$.}\n\nRecall\n\\[\n\\mathcal R_{j}:=\\bigl\\{x\\in S_{n}:\\;\n          x_{j}>x_{i}\\;(i<j),\\;\n          x_{j}\\ge x_{i}\\;(i>j)\\bigr\\},\n\\qquad\n\\chi(x):=\\operatorname{dom}(x).\n\\]\nPut\n\\[\n\\overline{\\mathcal R_{j}}\n :=S_{n}\\cap\\bigl\\{x\\in\\mathbf R^{\\,n}:x_{j}\\ge x_{i}\\;\\forall i\\bigr\\}.\n\\]\n(The over-bar denotes the closure; the strict inequalities are removed.)\nObserve that $\\overline{\\mathcal R_{j}}$ is the convex polytope  \n\\[\nP_{j}:=\\Bigl\\{x\\in\\mathbf R^{\\,n}:\\;\n            x_{i}\\ge 0,\\;\n            \\sum_{i=1}^{n}x_{i}\\le 1,\\;\n            x_{j}\\ge x_{i}\\;\\forall i\\Bigr\\}.\n\\]\n\n\\emph{Step 1: description of the extreme points of $P_{j}$.}  \nFor $k=0,\\dots ,n$ set  \n\\[\nv^{(k)}:=\\frac1k\\bigl(e_{j}+e_{i_{1}}+\\dots +e_{i_{k-1}}\\bigr)\n\\quad(k\\ge 1),\\qquad\nv^{(0)}:=\\mathbf 0,\n\\]\nwhere $\\{i_{1},\\dots ,i_{k-1}\\}\\subset\\{1,\\dots ,n\\}\\setminus\\{j\\}$.\nEvery $v^{(k)}$ lies in $P_{j}$, and a routine linear–programming\nargument shows that\n\\[\n\\operatorname{Ext}(P_{j})=\\bigl\\{v^{(k)}:0\\le k\\le n\\bigr\\}.\n\\]\n\n\\emph{Step 2: pairwise distances between extreme points.}\n\\[\n\\lVert v^{(0)}-e_{j}\\rVert=1,\\quad\n\\lVert v^{(0)}-v^{(k)}\\rVert=\\sqrt{\\tfrac1k}<1\\;(k\\ge 2),\\quad\n\\lVert e_{j}-v^{(k)}\\rVert\n   =\\sqrt{1-\\tfrac1k}<1\\;(k\\ge 2),\n\\]\nand, by the triangle inequality, every other pair of extreme points is\nalso at distance $<1$.\n\n\\emph{Step 3: diameter of $P_{j}$ and of $\\mathcal R_{j}$.}  \nThe squared Euclidean distance\n\\[\nD(x,y):=\\lVert x-y\\rVert^{2}\n\\]\nis jointly convex in $(x,y)$.  Over a compact convex set a convex\nfunction attains its maximum at an extreme point, whence\n\\[\n\\max_{(x,y)\\in P_{j}\\times P_{j}}D(x,y)\n      =\\max_{u,v\\in\\operatorname{Ext}(P_{j})}D(u,v)=1,\n\\]\nthe maximum being attained (uniquely) by the pair $(\\mathbf 0,e_{j})$.\nBecause $\\mathcal R_{j}\\subset P_{j}$ we conclude that\n\\[\n\\lVert x-y\\rVert\\le 1\n\\qquad(x,y\\in\\mathcal R_{j}),\n\\]\nand equality is indeed possible (take $x=\\mathbf 0$, $y=e_{j}$).\nConsequently $d_{\\max }(\\chi)=1$, so the dominant–coordinate\ncolouring realises $\\Delta_{n,n}$.\n\n\\medskip\n\\textbf{(b) Approximate sharpness (stability).}\n\nLet $\\varepsilon>0$ be given and put\n\\[\n\\gamma:=\\frac{\\varepsilon}{2\\sqrt{2}},\\qquad\n\\mathcal R_{j}^{\\gamma}:=\\Bigl\\{x\\in\\mathcal R_{j}:\\;\nx_{j}\\ge x_{i}+\\gamma\\;\\forall i\\neq j\\Bigr\\}.\n\\]\nThe point $e_{j}$ satisfies the above inequalities, hence every\n$\\mathcal R_{j}^{\\gamma}$ is non–empty.\nMoreover $\\operatorname{diam}(\\mathcal R_{j}^{\\gamma})\\le 1$\nbecause $\\mathcal R_{j}^{\\gamma}\\subset\\mathcal R_{j}$.\n\nDefine the colouring\n\\[\n\\chi_{\\varepsilon}(x):=\n\\begin{cases}\nj, & x\\in\\mathcal R_{j}^{\\gamma},\\\\[4pt]\n\\operatorname{dom}(x), & \\text{otherwise.}\n\\end{cases}\n\\]\n\n\\emph{Bringing a point close to $\\mathcal R_{j}^{\\gamma}$.}\nFix $x\\notin\\bigcup_{i}\\mathcal R_{i}^{\\gamma}$ and set\n$j=\\operatorname{dom}(x)$.\nSince $x\\notin\\mathcal R_{j}^{\\gamma}$ we have\n$d_{\\min}:=\\min_{i\\neq j}(x_{j}-x_{i})<\\gamma$.  \nDefine \n\\[\nx':=(1-\\gamma)\\,x+\\gamma\\,e_{j}.\n\\]\nBecause $x,e_{j}\\in S_{n}$, their convex combination $x'$ also lies in\n$S_{n}$.\nFurthermore\n\\[\nx'_{j}-x'_{i}=(1-\\gamma)(x_{j}-x_{i})+\\gamma\n              \\ge (1-\\gamma)d_{\\min}+\\gamma\\ge\\gamma\n\\quad(i\\neq j),\n\\]\nso $x'\\in\\mathcal R_{j}^{\\gamma}$.\nThe displacement is small:\n\\[\n\\lVert x-x'\\rVert\n     =\\gamma\\,\\lVert x-e_{j}\\rVert\n     \\le\\gamma\\,\\sqrt{2}\n     =\\frac{\\varepsilon}{2}.\n\\]\n\n\\emph{Bounding the monochromatic diameter.}\nTake $x,y$ with $\\chi_{\\varepsilon}(x)=\\chi_{\\varepsilon}(y)=j$.\nIf both $x,y$ already lie in $\\mathcal R_{j}^{\\gamma}$, then\n$\\lVert x-y\\rVert\\le 1$.\nOtherwise replace the offending points by $x',y'$ obtained through the\nabove procedure; these satisfy $x',y'\\in\\mathcal R_{j}^{\\gamma}$ and\n\\[\n\\lVert x-y\\rVert\n  \\le\\lVert x-x'\\rVert+\\lVert x'-y'\\rVert+\\lVert y'-y\\rVert\n  \\le\\frac{\\varepsilon}{2}+1+\\frac{\\varepsilon}{2}\n  =1+\\varepsilon .\n\\]\nThus $d_{\\max }(\\chi_{\\varepsilon})\\le 1+\\varepsilon$, completing the\nproof of stability.\n\n\\hfill$\\square$",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.735424",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension:  the problem is now set in $\\mathbb R^{\\,n}$ for arbitrary $n\\ge 2$, introducing $n$ variables instead of 2.  \n2. More colours and tighter bound:  precisely $n$ colours are allowed and an explicit dimension–dependent bound $d_n$ must be proved and shown optimal.  \n3. Additional tasks:  besides proving existence of a distant monochromatic pair, the solver must also construct an \\emph{almost extremal colouring}, establishing sharpness.  \n4. Deeper techniques:  the solution demands (i) a cleverly chosen $(n+1)$-point configuration with controlled pairwise distances, (ii) careful distance estimates in $\\mathbb R^{\\,n}$, (iii) a geometric pigeon-hole argument, and (iv) a two-part constructive sharpness proof involving both “corner caps’’ and fine barycentric tilings.  \n\nThese layers of abstraction and computation go well beyond the planar, single-bound, existence-only argument required in both the original and the current kernel variants, making the enhanced problem significantly more challenging."
      }
    },
    "original_kernel_variant": {
      "question": "Let $n\\ge 2$ be an integer and put  \n\\[\nS_{n}:=\\Bigl\\{(x_{1},\\dots ,x_{n})\\in\\mathbf R^{\\,n}:\\;\n           x_{i}\\ge 0\\;(1\\le i\\le n),\\;\n           x_{1}+\\dots +x_{n}\\le 1\\Bigr\\},\n\\]\nthe right–isosceles $n$–simplex whose $n$ edges issuing from the origin\nhave Euclidean length $1$.  \nFix an integer $k\\ge 1$ and colour every point of $S_{n}$ with one of\n(at most) $k$ colours.  For a colouring $\\chi$ define\n\\[\nd_{\\max }(\\chi):=\\max_{\\substack{x,y\\in S_{n}\\\\\n                         \\chi(x)=\\chi(y)}}\\lVert x-y\\rVert ,\n\\qquad\n\\Delta_{n,k}:=\\inf_{\\chi\\text{ $k$–colouring of }S_{n}}d_{\\max }(\\chi).\n\\]\n\n1. (Exact values for the “few colours’’ regime.)  \n   Prove that, for every $n\\ge 2$,\n   \\[\n        \\boxed{\\;\n        \\Delta_{n,k}=\n        \\begin{cases}\n        \\sqrt{2}, & 1\\le k\\le n-1,\\\\[6pt]\n        1,        & k=n.\n        \\end{cases}}\n   \\]\n   Hence, a monochromatic pair of points at distance at least $1$\n   is unavoidable as long as fewer than $n+1$ colours are available.\n\n2. (Uniformly positive gap for every finite number of colours.)  \n   Put\n   \\[\n      \\omega_{n}:=\\dfrac{\\pi^{\\,n/2}}{\\Gamma\\!\\bigl(\\tfrac n2+1\\bigr)},\\qquad\n      J_{n}:=\\sqrt{\\dfrac{n}{2(n+1)}}\\quad(n\\ge 2).\n   \\]\n   Show that for every integer $k\\ge n+1$\n   \\[\n          \\frac{1}{J_{n}\\,\\omega_n^{1/n}}\n          \\Bigl(\\tfrac{1}{n!\\,k}\\Bigr)^{1/n}\n          \\;\\le\\;\n          \\Delta_{n,k}\n          \\;\\le\\;\n          \\sqrt n\\,\n          \\frac{(n+1)^{1/n}}{k^{1/n}},\n          \\tag{$\\ast$}\n   \\]\n   and deduce that\n   \\[\n      0<\\liminf_{k\\to\\infty}k^{1/n}\\,\\Delta_{n,k}\\le\n      \\limsup_{k\\to\\infty}k^{1/n}\\,\\Delta_{n,k}<\\infty .\n   \\]\n   Consequently, $\\Delta_{n,k}>0$ for every finite $k$ and\n   $\\Delta_{n,k}= \\Theta(k^{-1/n})$ as $k\\to\\infty$.\n\n3. (Near–optimal colourings for $k=n$.)  \n   In order to obtain a genuine \\emph{partition} of $S_{n}$ we fix, once\n   and for all, the following tie–breaking convention:  \n   for $x\\in S_{n}$ let $\\max (x):=\\max\\{x_{1},\\dots ,x_{n}\\}$ and put\n   \\[\n       \\operatorname{dom}(x):=\n       \\min\\bigl\\{j:\\,x_{j}=\\max (x)\\bigr\\}.\n   \\]\n   Define the $n$ dominant–coordinate regions\n   \\[\n      \\mathcal R_{j}:=\\bigl\\{x\\in S_{n}:\\;\\operatorname{dom}(x)=j\\bigr\\},\n      \\qquad 1\\le j\\le n .\n   \\]\n\n   (a)  Prove that\n        \\[\n              \\lVert x-y\\rVert\\le 1\n              \\qquad(x,y\\in\\mathcal R_{j},\\;1\\le j\\le n),\n        \\]\n        and conclude that the colouring\n        $S_{n}=\\bigsqcup_{j=1}^{n}\\mathcal R_{j}$ satisfies\n        $d_{\\max }(\\chi)=1$, hence attains $\\Delta_{n,n}$.\n\n   (b)  For every $\\varepsilon>0$ construct an $n$–colouring\n        $\\chi_{\\varepsilon}$ of $S_{n}$ such that\n        $d_{\\max }(\\chi_{\\varepsilon})\\le 1+\\varepsilon$.\n        (Thus the bound $\\Delta_{n,n}=1$ is “stable’’.)\n\nRemark.  No claim is made that the diameter of every region\n$\\mathcal R_{j}$ equals $1$; in fact this is false whenever\n$j>1$.  What matters is the uniform bound of part (a).",
      "solution": "Throughout write\n\\[\n\\mathbf 0:=(0,\\dots ,0),\\qquad\ne_{j}:=(0,\\dots ,0,\\underset{j}{1},0,\\dots ,0)\\;(1\\le j\\le n),\\qquad\n\\mathcal V:=\\{\\mathbf 0,e_{1},\\dots ,e_{n}\\}.\n\\]\nBecause $S_{n}=\\operatorname{conv}(\\mathcal V)$, its diameter equals\n$\\sqrt{2}$.\n\n\\bigskip\n\\textbf{A counting lemma needed for part 2.}  \\\\\nFor an integer $m\\ge 1$ subdivide the cube $[0,1]^{n}$ into\n$m^{\\,n}$ axis–parallel cubes of side–length $1/m$.  \nLet $N(m)$ be the number of these small cubes that meet $S_{n}$.  \nA cube is uniquely determined by its index vector\n$I=(i_{1},\\dots ,i_{n})$ with $0\\le i_{t}\\le m-1$, namely\n\\[\nI\\longmapsto\\prod_{t=1}^{n}\n      \\Bigl[\\tfrac{i_{t}}{m},\\,\\tfrac{i_{t}+1}{m}\\Bigr].\n\\]\nSuch a cube intersects $S_{n}$ iff\n\\[\n\\frac{i_{1}+\\dots +i_{n}}{m}\\le 1\n\\quad\\Longleftrightarrow\\quad\ni_{1}+\\dots +i_{n}\\le m-1 .\n\\]\nHence\n\\[\nN(m)=\\#\\Bigl\\{(i_{1},\\dots ,i_{n})\\in\\mathbf Z_{\\ge 0}^{\\,n}:\n              i_{1}+\\dots +i_{n}\\le m-1\\Bigr\\}\n      =\\binom{m-1+n}{n}.\n\\]\n\n\\emph{Two explicit upper bounds:}\n\\begin{align}\nN(m)&=\\binom{m-1+n}{n}\\le(n+1)\\,m^{\\,n}\\qquad(m\\ge 1),\\tag{1}\\\\\nN(m)&=\\binom{m-1+n}{n}\\le(n+1)\\,(m-1)^{\\,n}\\qquad(m\\ge 2).\\tag{2}\n\\end{align}\n\nInequality (1) is classical.  \nFor (2) put $R(m):=N(m)/(m-1)^{n}$ for $m\\ge 2$ and observe\n\\[\n\\frac{R(m+1)}{R(m)}\n   =\\Bigl(1+\\frac{n}{m}\\Bigr)\\Bigl(1-\\frac{1}{m}\\Bigr)^{n}\\le 1,\n\\]\nso $R(m)$ is decreasing and $R(m)\\le R(2)=n+1$.\n\n\\bigskip\n\\textbf{1. Exact values for $k\\le n$.}\n\n(i) $1\\le k\\le n-1$.  \nAmong the $n$ vertices $e_{1},\\dots ,e_{n}$ two share a colour, hence\n$d_{\\max }(\\chi)\\ge\\sqrt{2}$.  \nThe constant colouring gives equality, so\n$\\Delta_{n,k}=\\sqrt{2}$.\n\n(ii) $k=n$.  \n\\emph{Lower bound.}\nThe $n+1$ points of $\\mathcal V$ receive $n$ colours, so two share a\ncolour and are at distance at least $1$.  \nThus $\\Delta_{n,n}\\ge 1$.\n\n\\emph{Upper bound.}\nPart 3 (a) below constructs an $n$–colouring with\n$d_{\\max }=1$, whence $\\Delta_{n,n}=1$.\n\n\\bigskip\n\\textbf{2. Proof of $(\\ast)$ for $k\\ge n+1$.}\n\n\\emph{Lower bound (volume $+$ Jung).}  \nLet $\\chi$ be any $k$–colouring and put $\\delta:=d_{\\max }(\\chi)$.  \nBy Jung’s theorem every subset of diameter $\\delta$ is contained in a\nball of radius $J_{n}\\delta$, so every colour class has volume\n$\\le\\omega_{n}(J_{n}\\delta)^{n}$.  \nBecause $\\operatorname{vol}(S_{n})=1/n!$,\n\\[\nk\\,\\omega_{n}(J_{n}\\delta)^{n}\\ge\\frac{1}{n!}\n\\;\\Longrightarrow\\;\n\\delta\\ge\n\\frac{1}{J_{n}\\,\\omega_{n}^{1/n}}\n\\Bigl(\\tfrac{1}{n!\\,k}\\Bigr)^{1/n}.\n\\]\n\n\\emph{Upper bound (refined grid argument).}  \nSet \n\\[\nm:=\\max\\Bigl\\{2,\\bigl\\lceil(k/(n+1))^{1/n}\\bigr\\rceil\\Bigr\\}.\n\\]\nBy definition $m\\ge 2$ and from (3) below we have\n\\[\n(n+1)(m-1)^{\\,n}<k\\le(n+1)m^{\\,n}.\\tag{3}\n\\]\nTogether with (2) this implies $N(m)<k$.  Colour each point of $S_{n}$\nby the \\emph{lexicographically} smallest cube (of side $1/m$) that\ncontains it.  \nTwo points of the same colour lie in one cube, whose diameter is\n$\\sqrt n/m$.  \nSince $m\\le(k/(n+1))^{1/n}+1$, the monotonicity of $x\\mapsto1/x$\ngives\n\\[\n\\frac{1}{m}\\le\\frac{(n+1)^{1/n}}{k^{1/n}},\n\\]\nhence\n\\[\nd_{\\max }(\\chi)\\le\\sqrt n\\,\\frac{(n+1)^{1/n}}{k^{1/n}},\n\\]\nwhich is the right–hand side of $(\\ast)$.\n\n\\bigskip\n\\textbf{3. Near–optimal colourings for $k=n$.}\n\n\\medskip\n\\textbf{(a) The dominant–coordinate colouring attains $\\Delta_{n,n}$.}\n\nRecall\n\\[\n\\mathcal R_{j}:=\\bigl\\{x\\in S_{n}:\\;\n          x_{j}>x_{i}\\;(i<j),\\;\n          x_{j}\\ge x_{i}\\;(i>j)\\bigr\\},\n\\qquad\n\\chi(x):=\\operatorname{dom}(x).\n\\]\nPut\n\\[\n\\overline{\\mathcal R_{j}}\n :=S_{n}\\cap\\bigl\\{x\\in\\mathbf R^{\\,n}:x_{j}\\ge x_{i}\\;\\forall i\\bigr\\}.\n\\]\n(The over-bar denotes the closure; the strict inequalities are removed.)\nObserve that $\\overline{\\mathcal R_{j}}$ is the convex polytope  \n\\[\nP_{j}:=\\Bigl\\{x\\in\\mathbf R^{\\,n}:\\;\n            x_{i}\\ge 0,\\;\n            \\sum_{i=1}^{n}x_{i}\\le 1,\\;\n            x_{j}\\ge x_{i}\\;\\forall i\\Bigr\\}.\n\\]\n\n\\emph{Step 1: description of the extreme points of $P_{j}$.}  \nFor $k=0,\\dots ,n$ set  \n\\[\nv^{(k)}:=\\frac1k\\bigl(e_{j}+e_{i_{1}}+\\dots +e_{i_{k-1}}\\bigr)\n\\quad(k\\ge 1),\\qquad\nv^{(0)}:=\\mathbf 0,\n\\]\nwhere $\\{i_{1},\\dots ,i_{k-1}\\}\\subset\\{1,\\dots ,n\\}\\setminus\\{j\\}$.\nEvery $v^{(k)}$ lies in $P_{j}$, and a routine linear‐programming\nargument (or an enumeration of tight constraints) shows that\n\\[\n\\operatorname{Ext}(P_{j})=\\bigl\\{v^{(k)}:0\\le k\\le n\\bigr\\}.\n\\]\n\n\\emph{Step 2: pairwise distances between extreme points.}\n\\[\n\\lVert v^{(0)}-e_{j}\\rVert=1,\n\\qquad\n\\lVert v^{(0)}-v^{(k)}\\rVert=\\sqrt{\\frac1k}<1\\;(k\\ge 2),\n\\qquad\n\\lVert e_{j}-v^{(k)}\\rVert\n   =\\sqrt{1-\\frac1k}<1\\;(k\\ge 2),\n\\]\nand, by the triangle inequality, every other pair of extreme points is\nalso at distance $<1$.\n\n\\emph{Step 3: diameter of $P_{j}$ and of $\\mathcal R_{j}$.}  \nThe squared Euclidean distance\n\\[\nD(x,y):=\\lVert x-y\\rVert^{2}\n\\]\nis jointly convex in $(x,y)$.  Over a compact convex set a convex\nfunction attains its maximum at an extreme point, whence\n\\[\n\\max_{(x,y)\\in P_{j}\\times P_{j}}D(x,y)\n      =\\max_{u,v\\in\\operatorname{Ext}(P_{j})}D(u,v)=1,\n\\]\nthe maximum being attained (uniquely) by the pair $(\\mathbf 0,e_{j})$.\nBecause\n$\\mathcal R_{j}\\subset P_{j}$, we conclude that\n\\[\n\\lVert x-y\\rVert\\le 1\n\\qquad(x,y\\in\\mathcal R_{j}),\n\\]\nand equality is indeed possible (take $x=\\mathbf 0$, $y=e_{j}$).\nConsequently $d_{\\max }(\\chi)=1$, so the dominant–coordinate\ncolouring realises $\\Delta_{n,n}$.\n\n\\medskip\n\\textbf{(b) Approximate sharpness (stability).}\n\nLet $0<\\varepsilon<1$ and put\n\\[\n\\gamma:=\\frac{\\varepsilon}{2\\sqrt n},\\qquad\n\\mathcal R_{j}^{\\gamma}:=\\Bigl\\{x\\in\\mathcal R_{j}:\\;\nx_{j}\\ge x_{i}+\\gamma\\;\\forall i\\neq j\\Bigr\\}.\n\\]\nBecause the set $\\{x_{j}-x_{i}\\mid x\\in\\mathcal R_{j}\\}$ contains\n$(0,1]$ and $\\gamma<1$, each $\\mathcal R_{j}^{\\gamma}$ is non-empty and\n$\\operatorname{diam}(\\mathcal R_{j}^{\\gamma})\\le 1$\n(as it is a subset of $\\mathcal R_{j}$).\n\nDefine the colouring\n\\[\n\\chi_{\\varepsilon}(x):=\n\\begin{cases}\nj, & x\\in\\mathcal R_{j}^{\\gamma},\\\\[4pt]\n\\operatorname{dom}(x), & \\text{otherwise.}\n\\end{cases}\n\\]\nFor $x$ outside every $\\mathcal R_{j}^{\\gamma}$ let\n$x':=(x+\\gamma e_{j})/(1+\\gamma)$ with \n$j=\\operatorname{dom}(x)$; then $x'\\in\\mathcal R_{j}^{\\gamma}$ and\n$\\lVert x'-x\\rVert\\le\\gamma\\sqrt n=\\varepsilon/2$.  \n\nIf $x,y$ share colour $j$, either both lie in $\\mathcal R_{j}^{\\gamma}$\nand $\\lVert x-y\\rVert\\le 1$, or we replace the offending points by\n$x',y'$ and obtain\n\\[\n\\lVert x-y\\rVert\n     \\le\\lVert x-x'\\rVert+\\lVert x'-y'\\rVert+\\lVert y'-y\\rVert\n     \\le\\frac{\\varepsilon}{2}+1+\\frac{\\varepsilon}{2}=1+\\varepsilon .\n\\]\nThus $d_{\\max }(\\chi_{\\varepsilon})\\le 1+\\varepsilon$, proving stability.\n\n\\hfill$\\square$\n\n\\bigskip",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.569585",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension:  the problem is now set in $\\mathbb R^{\\,n}$ for arbitrary $n\\ge 2$, introducing $n$ variables instead of 2.  \n2. More colours and tighter bound:  precisely $n$ colours are allowed and an explicit dimension–dependent bound $d_n$ must be proved and shown optimal.  \n3. Additional tasks:  besides proving existence of a distant monochromatic pair, the solver must also construct an \\emph{almost extremal colouring}, establishing sharpness.  \n4. Deeper techniques:  the solution demands (i) a cleverly chosen $(n+1)$-point configuration with controlled pairwise distances, (ii) careful distance estimates in $\\mathbb R^{\\,n}$, (iii) a geometric pigeon-hole argument, and (iv) a two-part constructive sharpness proof involving both “corner caps’’ and fine barycentric tilings.  \n\nThese layers of abstraction and computation go well beyond the planar, single-bound, existence-only argument required in both the original and the current kernel variants, making the enhanced problem significantly more challenging."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}