{
  "index": "1988-A-4",
  "type": "COMB",
  "tag": [
    "COMB",
    "GEO"
  ],
  "difficulty": "",
  "question": "\\begin{enumerate}\n\\item[(a)] If every point of the plane is painted one of three colors,\ndo there necessarily exist two points of the same color exactly one\ninch apart?\n\\item[(b)] What if ``three'' is replaced by ``nine''?\n\\end{enumerate}",
  "solution": "Solution.\n\nFIGURE 6.\n(a) Suppose, for the sake of obtaining a contradiction, that we have a coloring of the plane with three colors such that any two points at distance 1 have different colors. If \\( A \\) and \\( B \\) are two points at distance \\( \\sqrt{3} \\), then the circles of radius 1 centered at \\( A \\) and \\( B \\) meet in two points \\( P, Q \\) forming equilateral triangles \\( A P Q \\) and \\( B P Q \\). (See Figure 6.) The three points of each equilateral triangle have different colors, and this forces \\( A \\) and \\( B \\) to have equal colors. Now consider a triangle \\( C D E \\) with \\( C D=C E=\\sqrt{3} \\) and \\( D E=1 \\). (See Figure 7.) We know that \\( C, D \\) have the same color and \\( C, E \\) have the same color, so \\( D, E \\) have the same color, contradicting our hypothesis about points at distance 1 .\n\nFIGURE 7.\n(b) For \\( P=(x, y) \\in \\mathbb{R}^{2} \\), define\n\\[\nf(P)=(\\lfloor(3 / 2) x\\rfloor \\bmod 3,\\lfloor(3 / 2) y\\rfloor \\bmod 3) \\in\\{0,1,2\\}^{2} .\n\\]\n\nColor the nine elements of \\( \\{0,1,2\\}^{2} \\) with the nine different colors, and give \\( P \\) the color of \\( f(P) \\). (Thus the plane is covered by blocks of \\( 3 \\times 3 \\) squares, where the little squares have side length \\( 2 / 3 \\). See Figure 8.) If \\( P \\) and \\( Q \\) are points with the same color, either they belong to the same little square, in which case \\( P Q \\leq(2 / 3) \\sqrt{2}<1 \\), or to different little squares. In the latter case, either their \\( x \\)-coordinates differ by at least \\( 4 / 3 \\), or their \\( y \\)-coordinates differ by at least \\( 4 / 3 \\), so \\( P Q \\geq 4 / 3>1 \\).\n\\begin{tabular}{l|l|l|l|l|l} \n& & & & & \\\\\n\\hline & \\( (0,0) \\) & \\( (1,0) \\) & \\( (2,0) \\) & \\( (0,0) \\) & \\\\\n\\hline & \\( (0,2) \\) & \\( (1,2) \\) & \\( (2,2) \\) & \\( (0,2) \\) & \\\\\n\\hline & \\( (0,1) \\) & \\( (1,1) \\) & \\( (2,1) \\) & \\( (0,1) \\) & \\\\\n\\hline & \\( (0,0) \\) & \\( (1,0) \\) & \\( (2,0) \\) & \\( (0,0) \\) & \\\\\n\\hline & & & & &\n\\end{tabular}\n\\[\n\\text { I } 3 / 2\n\\]\n\nFIGURE 8.\n\nRemark. Part (a) appears in [New, p. 7], as does the following variation: Assume that the points of the plane are each colored red or blue. Prove that one of these colors contains pairs of points at every distance.\n\nRemark. Figure 9 shows a well-known coloring of the plane with seven colors such that points at distance 1 always have different colors. Such a coloring can be constructed as follows. Let \\( \\omega=\\frac{-1+\\sqrt{-3}}{2} \\), and view the ring of Eisenstein integers \\( \\mathbb{Z}[\\omega]=\\{a+b \\omega: a, b \\in \\mathbb{Z}\\} \\) as a lattice in the complex plane. Then \\( \\mathfrak{p}=(2-\\omega) \\mathbb{Z}[\\omega] \\) is a sublattice of index \\( |2-\\omega|^{2}=7 \\). (In fact, \\( \\mathfrak{p} \\) is one of the two prime ideals of \\( \\mathbb{Z}[\\omega] \\) that divide (7).) Give each coset of \\( \\mathfrak{p} \\) in \\( \\mathbb{Z}[\\omega] \\) its own color. Next color each point in \\( \\mathbb{C} \\) according to the color of a nearest point in \\( \\mathbb{Z}[\\omega] \\). This gives a coloring by hexagons. The diameter of each hexagon equals \\( \\sqrt{4 / 3} \\), and if two distinct hexagons have the same color, the smallest distance between points of their closures is \\( \\sqrt{7 / 3} \\). Hence if \\( \\sqrt{4 / 3}<d<\\sqrt{7 / 3} \\), no two points in the plane at distance \\( d \\) have the same color. Scaling the whole picture by \\( 1 / d \\), we find a coloring in which no two points in the plane at distance 1 have the same color.\n\nRemark. Let \\( \\chi(n) \\) denote the minimum number of colors needed to color the points in \\( \\mathbb{R}^{n} \\) so that each pair of points separated by distance 1 have different colors. Parts (a) and (b) show \\( \\chi(2)>3 \\) and \\( \\chi(2) \\leq 9 \\), respectively. Because of the hexagonal coloring in the previous remark, we know \\( 4 \\leq \\chi(2) \\leq 7 \\). Ronald L. Graham is offering \\( \\$ 1000 \\) for an improvement of either bound.\n\nThe study of \\( \\chi(n) \\) goes back at least to [Had] in 1944. As \\( n \\rightarrow \\infty \\), one has\n\\[\n(1+o(1))(1.2)^{n} \\leq \\chi(n) \\leq(3+o(1))^{n},\n\\]\nthe lower and upper bounds being due to \\( [\\mathrm{FW}] \\) and \\( [\\mathrm{LR}] \\), respectively. (See the remark after the solution to 1988A3 for the meaning of the little-o notation.) For more on such questions, consult [Gr2] and the references listed there, and browse issues of the journal Geombinatorics.",
  "vars": [
    "A",
    "B",
    "C",
    "D",
    "E",
    "P",
    "Q",
    "x",
    "y",
    "f",
    "d",
    "n"
  ],
  "params": [
    "\\\\omega",
    "\\\\chi",
    "R",
    "Z"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "A": "pointalpha",
        "B": "pointbeta",
        "C": "pointgamma",
        "D": "pointdelta",
        "E": "pointepsilon",
        "P": "pointsigma",
        "Q": "pointtau",
        "x": "coordinatex",
        "y": "coordinatey",
        "f": "mapperfunct",
        "d": "distvalue",
        "n": "dimcount",
        "\\omega": "constomega",
        "\\chi": "constchi",
        "R": "constreal",
        "Z": "constinteger"
      },
      "question": "\\begin{enumerate}\n\\item[(a)] If every point of the plane is painted one of three colors,\ndo there necessarily exist two points of the same color exactly one\ninch apart?\n\\item[(b)] What if ``three'' is replaced by ``nine''?\n\\end{enumerate}",
      "solution": "Solution.\n\nFIGURE 6.\n(a) Suppose, for the sake of obtaining a contradiction, that we have a coloring of the plane with three colors such that any two points at distance 1 have different colors. If \\( pointalpha \\) and \\( pointbeta \\) are two points at distance \\( \\sqrt{3} \\), then the circles of radius 1 centered at \\( pointalpha \\) and \\( pointbeta \\) meet in two points \\( pointsigma, pointtau \\) forming equilateral triangles \\( pointalpha\\, pointsigma\\, pointtau \\) and \\( pointbeta\\, pointsigma\\, pointtau \\). (See Figure 6.) The three points of each equilateral triangle have different colors, and this forces \\( pointalpha \\) and \\( pointbeta \\) to have equal colors. Now consider a triangle \\( pointgamma\\, pointdelta\\, pointepsilon \\) with \\( pointgamma pointdelta=pointgamma pointepsilon=\\sqrt{3} \\) and \\( pointdelta pointepsilon=1 \\). (See Figure 7.) We know that \\( pointgamma, pointdelta \\) have the same color and \\( pointgamma, pointepsilon \\) have the same color, so \\( pointdelta, pointepsilon \\) have the same color, contradicting our hypothesis about points at distance 1 .\n\nFIGURE 7.\n(b) For \\( pointsigma=(coordinatex, coordinatey) \\in \\mathbb{constreal}^{2} \\), define\n\\[\nmapperfunct(pointsigma)=(\\lfloor(3 / 2) coordinatex\\rfloor \\bmod 3,\\lfloor(3 / 2) coordinatey\\rfloor \\bmod 3) \\in\\{0,1,2\\}^{2} .\n\\]\n\nColor the nine elements of \\( \\{0,1,2\\}^{2} \\) with the nine different colors, and give \\( pointsigma \\) the color of \\( mapperfunct(pointsigma) \\). (Thus the plane is covered by blocks of \\( 3 \\times 3 \\) squares, where the little squares have side length \\( 2 / 3 \\). See Figure 8.) If \\( pointsigma \\) and \\( pointtau \\) are points with the same color, either they belong to the same little square, in which case \\( pointsigma pointtau \\leq(2 / 3) \\sqrt{2}<1 \\), or to different little squares. In the latter case, either their \\( coordinatex \\)-coordinates differ by at least \\( 4 / 3 \\), or their \\( coordinatey \\)-coordinates differ by at least \\( 4 / 3 \\), so \\( pointsigma pointtau \\geq 4 / 3>1 \\).\n\\begin{tabular}{l|l|l|l|l|l} \n& & & & & \\\\\n\\hline & \\( (0,0) \\) & \\( (1,0) \\) & \\( (2,0) \\) & \\( (0,0) \\) & \\\\\n\\hline & \\( (0,2) \\) & \\( (1,2) \\) & \\( (2,2) \\) & \\( (0,2) \\) & \\\\\n\\hline & \\( (0,1) \\) & \\( (1,1) \\) & \\( (2,1) \\) & \\( (0,1) \\) & \\\\\n\\hline & \\( (0,0) \\) & \\( (1,0) \\) & \\( (2,0) \\) & \\( (0,0) \\) & \\\\\n\\hline & & & & &\n\\end{tabular}\n\\[\n\\text { I } 3 / 2\n\\]\n\nFIGURE 8.\n\nRemark. Part (a) appears in [New, p. 7], as does the following variation: Assume that the points of the plane are each colored red or blue. Prove that one of these colors contains pairs of points at every distance.\n\nRemark. Figure 9 shows a well-known coloring of the plane with seven colors such that points at distance 1 always have different colors. Such a coloring can be constructed as follows. Let \\( constomega=\\frac{-1+\\sqrt{-3}}{2} \\), and view the ring of Eisenstein integers \\( \\mathbb{constinteger}[constomega]=\\{a+b constomega: a, b \\in \\mathbb{constinteger}\\} \\) as a lattice in the complex plane. Then \\( \\mathfrak{p}=(2-constomega) \\mathbb{constinteger}[constomega] \\) is a sublattice of index \\( |2-constomega|^{2}=7 \\). (In fact, \\( \\mathfrak{p} \\) is one of the two prime ideals of \\( \\mathbb{constinteger}[constomega] \\) that divide (7).) Give each coset of \\( \\mathfrak{p} \\) in \\( \\mathbb{constinteger}[constomega] \\) its own color. Next color each point in \\( \\mathbb{C} \\) according to the color of a nearest point in \\( \\mathbb{constinteger}[constomega] \\). This gives a coloring by hexagons. The diameter of each hexagon equals \\( \\sqrt{4 / 3} \\), and if two distinct hexagons have the same color, the smallest distance between points of their closures is \\( \\sqrt{7 / 3} \\). Hence if \\( \\sqrt{4 / 3}<distvalue<\\sqrt{7 / 3} \\), no two points in the plane at distance \\( distvalue \\) have the same color. Scaling the whole picture by \\( 1 / distvalue \\), we find a coloring in which no two points in the plane at distance 1 have the same color.\n\nRemark. Let \\( constchi(dimcount) \\) denote the minimum number of colors needed to color the points in \\( \\mathbb{constreal}^{dimcount} \\) so that each pair of points separated by distance 1 have different colors. Parts (a) and (b) show \\( constchi(2)>3 \\) and \\( constchi(2) \\leq 9 \\), respectively. Because of the hexagonal coloring in the previous remark, we know \\( 4 \\leq constchi(2) \\leq 7 \\). Ronald L. Graham is offering \\$ 1000 for an improvement of either bound.\n\nThe study of \\( constchi(dimcount) \\) goes back at least to [Had] in 1944. As \\( dimcount \\rightarrow \\infty \\), one has\n\\[\n(1+o(1))(1.2)^{dimcount} \\leq constchi(dimcount) \\leq(3+o(1))^{dimcount},\n\\]\nthe lower and upper bounds being due to \\( [\\mathrm{FW}] \\) and \\( [\\mathrm{LR}] \\), respectively. (See the remark after the solution to 1988A3 for the meaning of the little-o notation.) For more on such questions, consult [Gr2] and the references listed there, and browse issues of the journal Geombinatorics."
    },
    "descriptive_long_confusing": {
      "map": {
        "A": "turquoise",
        "B": "backyard",
        "C": "sunlight",
        "D": "waterfall",
        "E": "moonstone",
        "P": "pineapple",
        "Q": "quartzite",
        "x": "xylophone",
        "y": "yesteryear",
        "f": "fireplace",
        "d": "drainpipe",
        "n": "necklaces",
        "\\omega": "whirlpool",
        "\\chi": "chocolate",
        "R": "rainstorm",
        "Z": "zephyrus"
      },
      "question": "\\begin{enumerate}\n\\item[(a)] If every point of the plane is painted one of three colors,\ndo there necessarily exist two points of the same color exactly one\ninch apart?\n\\item[(b)] What if ``three'' is replaced by ``nine''?\n\\end{enumerate}",
      "solution": "Solution.\n\nFIGURE 6.\n(a) Suppose, for the sake of obtaining a contradiction, that we have a coloring of the plane with three colors such that any two points at distance 1 have different colors. If \\( turquoise \\) and \\( backyard \\) are two points at distance \\( \\sqrt{3} \\), then the circles of radius 1 centered at \\( turquoise \\) and \\( backyard \\) meet in two points \\( pineapple, quartzite \\) forming equilateral triangles \\( turquoise\\, pineapple\\, quartzite \\) and \\( backyard\\, pineapple\\, quartzite \\). (See Figure 6.) The three points of each equilateral triangle have different colors, and this forces \\( turquoise \\) and \\( backyard \\) to have equal colors. Now consider a triangle \\( sunlight\\, waterfall\\, moonstone \\) with \\( sunlight\\, waterfall = sunlight\\, moonstone = \\sqrt{3} \\) and \\( waterfall\\, moonstone = 1 \\). (See Figure 7.) We know that \\( sunlight, waterfall \\) have the same color and \\( sunlight, moonstone \\) have the same color, so \\( waterfall, moonstone \\) have the same color, contradicting our hypothesis about points at distance 1.\n\nFIGURE 7.\n(b) For \\( pineapple=(xylophone, yesteryear) \\in \\mathbb{rainstorm}^{2} \\), define\n\\[\nfireplace(pineapple)=\\left(\\lfloor(3 / 2) xylophone\\rfloor \\bmod 3,\\lfloor(3 / 2) yesteryear\\rfloor \\bmod 3\\right) \\in\\{0,1,2\\}^{2} .\n\\]\n\nColor the nine elements of \\( \\{0,1,2\\}^{2} \\) with the nine different colors, and give \\( pineapple \\) the color of \\( fireplace(pineapple) \\). (Thus the plane is covered by blocks of \\( 3 \\times 3 \\) squares, where the little squares have side length \\( 2 / 3 \\). See Figure 8.) If \\( pineapple \\) and \\( quartzite \\) are points with the same color, either they belong to the same little square, in which case \\( pineapple\\, quartzite \\leq(2 / 3) \\sqrt{2}<1 \\), or to different little squares. In the latter case, either their \\( xylophone \\)-coordinates differ by at least \\( 4 / 3 \\), or their \\( yesteryear \\)-coordinates differ by at least \\( 4 / 3 \\), so \\( pineapple\\, quartzite \\geq 4 / 3>1 \\).\n\\begin{tabular}{l|l|l|l|l|l}\n& & & & & \\\\\n\\hline & \\((0,0)\\) & \\((1,0)\\) & \\((2,0)\\) & \\((0,0)\\) & \\\\\n\\hline & \\((0,2)\\) & \\((1,2)\\) & \\((2,2)\\) & \\((0,2)\\) & \\\\\n\\hline & \\((0,1)\\) & \\((1,1)\\) & \\((2,1)\\) & \\((0,1)\\) & \\\\\n\\hline & \\((0,0)\\) & \\((1,0)\\) & \\((2,0)\\) & \\((0,0)\\) & \\\\\n\\hline & & & & &\n\\end{tabular}\n\\[\n\\text { I } 3 / 2\n\\]\n\nFIGURE 8.\n\nRemark. Part (a) appears in [New, p. 7], as does the following variation: Assume that the points of the plane are each colored red or blue. Prove that one of these colors contains pairs of points at every distance.\n\nRemark. Figure 9 shows a well-known coloring of the plane with seven colors such that points at distance 1 always have different colors. Such a coloring can be constructed as follows. Let \\( whirlpool=\\frac{-1+\\sqrt{-3}}{2} \\), and view the ring of Eisenstein integers \\( \\mathbb{zephyrus}[whirlpool]=\\{a+b\\, whirlpool: a, b \\in \\mathbb{zephyrus}\\} \\) as a lattice in the complex plane. Then \\( \\mathfrak{p}=(2-whirlpool) \\mathbb{zephyrus}[whirlpool] \\) is a sublattice of index \\( |2-whirlpool|^{2}=7 \\). (In fact, \\( \\mathfrak{p} \\) is one of the two prime ideals of \\( \\mathbb{zephyrus}[whirlpool] \\) that divide (7).) Give each coset of \\( \\mathfrak{p} \\) in \\( \\mathbb{zephyrus}[whirlpool] \\) its own color. Next color each point in \\( \\mathbb{C} \\) according to the color of a nearest point in \\( \\mathbb{zephyrus}[whirlpool] \\). This gives a coloring by hexagons. The diameter of each hexagon equals \\( \\sqrt{4 / 3} \\), and if two distinct hexagons have the same color, the smallest distance between points of their closures is \\( \\sqrt{7 / 3} \\). Hence if \\( \\sqrt{4 / 3}<drainpipe<\\sqrt{7 / 3} \\), no two points in the plane at distance \\( drainpipe \\) have the same color. Scaling the whole picture by \\( 1 / drainpipe \\), we find a coloring in which no two points in the plane at distance 1 have the same color.\n\nRemark. Let \\( chocolate(necklaces) \\) denote the minimum number of colors needed to color the points in \\( \\mathbb{rainstorm}^{necklaces} \\) so that each pair of points separated by distance 1 have different colors. Parts (a) and (b) show \\( chocolate(2)>3 \\) and \\( chocolate(2) \\leq 9 \\), respectively. Because of the hexagonal coloring in the previous remark, we know \\( 4 \\leq chocolate(2) \\leq 7 \\). Ronald L. Graham is offering \\$ 1000 for an improvement of either bound.\n\nThe study of \\( chocolate(necklaces) \\) goes back at least to [Had] in 1944. As \\( necklaces \\rightarrow \\infty \\), one has\n\\[\n(1+o(1))(1.2)^{necklaces} \\leq chocolate(necklaces) \\leq(3+o(1))^{necklaces},\n\\]\nthe lower and upper bounds being due to \\([\\mathrm{FW}]\\) and \\([\\mathrm{LR}]\\), respectively. (See the remark after the solution to 1988A3 for the meaning of the little-o notation.) For more on such questions, consult [Gr2] and the references listed there, and browse issues of the journal Geombinatorics."
    },
    "descriptive_long_misleading": {
      "map": {
        "A": "notorigin",
        "B": "notcenter",
        "C": "outsider",
        "D": "unstable",
        "E": "unjoined",
        "P": "stationary",
        "Q": "stillness",
        "x": "constant",
        "y": "steadyval",
        "f": "unmapped",
        "d": "proximity",
        "n": "infinite",
        "\\omega": "realnumber",
        "\\chi": "monochrome",
        "R": "imaginary",
        "Z": "fractions"
      },
      "question": "\\begin{enumerate}\n\\item[(a)] If every point of the plane is painted one of three colors,\ndo there necessarily exist two points of the same color exactly one\ninch apart?\n\\item[(b)] What if ``three'' is replaced by ``nine''?\n\\end{enumerate}",
      "solution": "Solution.\n\nFIGURE 6.\n(a) Suppose, for the sake of obtaining a contradiction, that we have a coloring of the plane with three colors such that any two points at distance 1 have different colors. If \\( notorigin \\) and \\( notcenter \\) are two points at distance \\( \\sqrt{3} \\), then the circles of radius 1 centered at \\( notorigin \\) and \\( notcenter \\) meet in two points \\( stationary, stillness \\) forming equilateral triangles \\( notorigin\\;stationary\\;stillness \\) and \\( notcenter\\;stationary\\;stillness \\). (See Figure 6.) The three points of each equilateral triangle have different colors, and this forces \\( notorigin \\) and \\( notcenter \\) to have equal colors. Now consider a triangle \\( outsider\\;unstable\\;unjoined \\) with \\( outsider\\,unstable=outsider\\,unjoined=\\sqrt{3} \\) and \\( unstable\\,unjoined=1 \\). (See Figure 7.) We know that \\( outsider,\\,unstable \\) have the same color and \\( outsider,\\,unjoined \\) have the same color, so \\( unstable,\\,unjoined \\) have the same color, contradicting our hypothesis about points at distance 1 .\n\nFIGURE 7.\n(b) For \\( stationary=(constant, steadyval) \\in \\mathbb{imaginary}^{2} \\), define\n\\[\nunmapped(stationary)=(\\lfloor(3 / 2) constant\\rfloor \\bmod 3,\\lfloor(3 / 2) steadyval\\rfloor \\bmod 3) \\in\\{0,1,2\\}^{2} .\n\\]\n\nColor the nine elements of \\( \\{0,1,2\\}^{2} \\) with the nine different colors, and give \\( stationary \\) the color of \\( unmapped(stationary) \\). (Thus the plane is covered by blocks of \\( 3 \\times 3 \\) squares, where the little squares have side length \\( 2 / 3 \\). See Figure 8.) If \\( stationary \\) and \\( stillness \\) are points with the same color, either they belong to the same little square, in which case \\( stationary\\,stillness \\leq(2 / 3) \\sqrt{2}<1 \\), or to different little squares. In the latter case, either their \\( constant \\)-coordinates differ by at least \\( 4 / 3 \\), or their \\( steadyval \\)-coordinates differ by at least \\( 4 / 3 \\), so \\( stationary\\,stillness \\geq 4 / 3>1 \\).\n\\begin{tabular}{l|l|l|l|l|l} \n& & & & & \\\\\n\\hline & \\( (0,0) \\) & \\( (1,0) \\) & \\( (2,0) \\) & \\( (0,0) \\) & \\\\\n\\hline & \\( (0,2) \\) & \\( (1,2) \\) & \\( (2,2) \\) & \\( (0,2) \\) & \\\\\n\\hline & \\( (0,1) \\) & \\( (1,1) \\) & \\( (2,1) \\) & \\( (0,1) \\) & \\\\\n\\hline & \\( (0,0) \\) & \\( (1,0) \\) & \\( (2,0) \\) & \\( (0,0) \\) & \\\\\n\\hline & & & & &\n\\end{tabular}\n\\[\n\\text { I } 3 / 2\n\\]\n\nFIGURE 8.\n\nRemark. Part (a) appears in [New, p. 7], as does the following variation: Assume that the points of the plane are each colored red or blue. Prove that one of these colors contains pairs of points at every distance.\n\nRemark. Figure 9 shows a well-known coloring of the plane with seven colors such that points at distance 1 always have different colors. Such a coloring can be constructed as follows. Let \\( realnumber=\\frac{-1+\\sqrt{-3}}{2} \\), and view the ring of Eisenstein integers \\( \\mathbb{fractions}[realnumber]=\\{a+b\\,realnumber: a, b \\in \\mathbb{fractions}\\} \\) as a lattice in the complex plane. Then \\( \\mathfrak{p}=(2-realnumber)\\mathbb{fractions}[realnumber] \\) is a sublattice of index \\( |2-realnumber|^{2}=7 \\). (In fact, \\( \\mathfrak{p} \\) is one of the two prime ideals of \\( \\mathbb{fractions}[realnumber] \\) that divide (7).) Give each coset of \\( \\mathfrak{p} \\) in \\( \\mathbb{fractions}[realnumber] \\) its own color. Next color each point in \\( \\mathbb{C} \\) according to the color of a nearest point in \\( \\mathbb{fractions}[realnumber] \\). This gives a coloring by hexagons. The diameter of each hexagon equals \\( \\sqrt{4 / 3} \\), and if two distinct hexagons have the same color, the smallest distance between points of their closures is \\( \\sqrt{7 / 3} \\). Hence if \\( \\sqrt{4 / 3}<proximity<\\sqrt{7 / 3} \\), no two points in the plane at distance \\( proximity \\) have the same color. Scaling the whole picture by \\( 1 / proximity \\), we find a coloring in which no two points in the plane at distance 1 have the same color.\n\nRemark. Let \\( monochrome(infinite) \\) denote the minimum number of colors needed to color the points in \\( \\mathbb{imaginary}^{infinite} \\) so that each pair of points separated by distance 1 have different colors. Parts (a) and (b) show \\( monochrome(2)>3 \\) and \\( monochrome(2) \\leq 9 \\), respectively. Because of the hexagonal coloring in the previous remark, we know \\( 4 \\leq monochrome(2) \\leq 7 \\). Ronald L. Graham is offering \\$ 1000 for an improvement of either bound.\n\nThe study of \\( monochrome(infinite) \\) goes back at least to [Had] in 1944. As \\( infinite \\rightarrow \\infty \\), one has\n\\[\n(1+o(1))(1.2)^{infinite} \\leq monochrome(infinite) \\leq(3+o(1))^{infinite},\n\\]\nthe lower and upper bounds being due to \\( [\\mathrm{FW}] \\) and \\( [\\mathrm{LR}] \\), respectively. (See the remark after the solution to 1988A3 for the meaning of the little-o notation.) For more on such questions, consult [Gr2] and the references listed there, and browse issues of the journal Geombinatorics."
    },
    "garbled_string": {
      "map": {
        "A": "klmopqrw",
        "B": "xcvbmnkl",
        "C": "zpoiuytr",
        "D": "qwertyui",
        "E": "asdfghjk",
        "P": "lkjhgffd",
        "Q": "mnbvcxza",
        "x": "plokijuh",
        "y": "edcwsxza",
        "f": "gydlqjkr",
        "d": "nvreuioy",
        "n": "iruaytps",
        "\\omega": "\\snthplkj",
        "\\chi": "\\kjdlmreo",
        "R": "killoqpwe",
        "Z": "mpqognvx"
      },
      "question": "\n\\begin{enumerate}\n\\item[(a)] If every point of the plane is painted one of three colors,\ndo there necessarily exist two points of the same color exactly one\ninch apart?\n\\item[(b)] What if ``three'' is replaced by ``nine''?\n\\end{enumerate}\n",
      "solution": "\nSolution.\n\nFIGURE 6.\n(a) Suppose, for the sake of obtaining a contradiction, that we have a coloring of the plane with three colors such that any two points at distance 1 have different colors. If \\( klmopqrw \\) and \\( xcvbmnkl \\) are two points at distance \\( \\sqrt{3} \\), then the circles of radius 1 centered at \\( klmopqrw \\) and \\( xcvbmnkl \\) meet in two points \\( lkjhgffd, mnbvcxza \\) forming equilateral triangles \\( klmopqrw \\, lkjhgffd \\, mnbvcxza \\) and \\( xcvbmnkl \\, lkjhgffd \\, mnbvcxza \\). (See Figure 6.) The three points of each equilateral triangle have different colors, and this forces \\( klmopqrw \\) and \\( xcvbmnkl \\) to have equal colors. Now consider a triangle \\( zpoiuytr \\, qwertyui \\, asdfghjk \\) with \\( zpoiuytr qwertyui = zpoiuytr asdfghjk = \\sqrt{3} \\) and \\( qwertyui asdfghjk = 1 \\). (See Figure 7.) We know that \\( zpoiuytr, qwertyui \\) have the same color and \\( zpoiuytr, asdfghjk \\) have the same color, so \\( qwertyui, asdfghjk \\) have the same color, contradicting our hypothesis about points at distance 1 .\n\nFIGURE 7.\n(b) For \\( lkjhgffd = (plokijuh, edcwsxza) \\in \\mathbb{killoqpwe}^{2} \\), define\n\\[\ngydlqjkr(lkjhgffd) = (\\lfloor(3 / 2) \\, plokijuh \\rfloor \\bmod 3,\\lfloor(3 / 2) \\, edcwsxza \\rfloor \\bmod 3) \\in\\{0,1,2\\}^{2} .\n\\]\n\nColor the nine elements of \\( \\{0,1,2\\}^{2} \\) with the nine different colors, and give \\( lkjhgffd \\) the color of \\( gydlqjkr(lkjhgffd) \\). (Thus the plane is covered by blocks of \\( 3 \\times 3 \\) squares, where the little squares have side length \\( 2 / 3 \\). See Figure 8.) If \\( lkjhgffd \\) and \\( mnbvcxza \\) are points with the same color, either they belong to the same little square, in which case \\( lkjhgffd mnbvcxza \\leq(2 / 3) \\sqrt{2}<1 \\), or to different little squares. In the latter case, either their \\( plokijuh \\)-coordinates differ by at least \\( 4 / 3 \\), or their \\( edcwsxza \\)-coordinates differ by at least \\( 4 / 3 \\), so \\( lkjhgffd mnbvcxza \\geq 4 / 3>1 \\).\n\\begin{tabular}{l|l|l|l|l|l} \n& & & & & \\\\\n\\hline & \\( (0,0) \\) & \\( (1,0) \\) & \\( (2,0) \\) & \\( (0,0) \\) & \\\\\n\\hline & \\( (0,2) \\) & \\( (1,2) \\) & \\( (2,2) \\) & \\( (0,2) \\) & \\\\\n\\hline & \\( (0,1) \\) & \\( (1,1) \\) & \\( (2,1) \\) & \\( (0,1) \\) & \\\\\n\\hline & \\( (0,0) \\) & \\( (1,0) \\) & \\( (2,0) \\) & \\( (0,0) \\) & \\\\\n\\hline & & & & &\n\\end{tabular}\n\\[\n\\text { I } 3 / 2\n\\]\n\nFIGURE 8.\n\nRemark. Part (a) appears in [New, p. 7], as does the following variation: Assume that the points of the plane are each colored red or blue. Prove that one of these colors contains pairs of points at every distance.\n\nRemark. Figure 9 shows a well-known coloring of the plane with seven colors such that points at distance 1 always have different colors. Such a coloring can be constructed as follows. Let \\( \\snthplkj=\\frac{-1+\\sqrt{-3}}{2} \\), and view the ring of Eisenstein integers \\( \\mathbb{mpqognvx}[\\snthplkj]=\\{a+b \\snthplkj: a, b \\in \\mathbb{mpqognvx}\\} \\) as a lattice in the complex plane. Then \\( \\mathfrak{p}=(2-\\snthplkj) \\mathbb{mpqognvx}[\\snthplkj] \\) is a sublattice of index \\( |2-\\snthplkj|^{2}=7 \\). (In fact, \\( \\mathfrak{p} \\) is one of the two prime ideals of \\( \\mathbb{mpqognvx}[\\snthplkj] \\) that divide (7).) Give each coset of \\( \\mathfrak{p} \\) in \\( \\mathbb{mpqognvx}[\\snthplkj] \\) its own color. Next color each point in \\( \\mathbb{C} \\) according to the color of a nearest point in \\( \\mathbb{mpqognvx}[\\snthplkj] \\). This gives a coloring by hexagons. The diameter of each hexagon equals \\( \\sqrt{4 / 3} \\), and if two distinct hexagons have the same color, the smallest distance between points of their closures is \\( \\sqrt{7 / 3} \\). Hence if \\( \\sqrt{4 / 3}<nvreuioy<\\sqrt{7 / 3} \\), no two points in the plane at distance \\( nvreuioy \\) have the same color. Scaling the whole picture by \\( 1 / nvreuioy \\), we find a coloring in which no two points in the plane at distance 1 have the same color.\n\nRemark. Let \\( \\kjdlmreo(iruaytps) \\) denote the minimum number of colors needed to color the points in \\( \\mathbb{killoqpwe}^{iruaytps} \\) so that each pair of points separated by distance 1 have different colors. Parts (a) and (b) show \\( \\kjdlmreo(2)>3 \\) and \\( \\kjdlmreo(2) \\leq 9 \\), respectively. Because of the hexagonal coloring in the previous remark, we know \\( 4 \\leq \\kjdlmreo(2) \\leq 7 \\). Ronald L. Graham is offering \\( \\$ 1000 \\) for an improvement of either bound.\n\nThe study of \\( \\kjdlmreo(iruaytps) \\) goes back at least to [Had] in 1944. As \\( iruaytps \\rightarrow \\infty \\), one has\n\\[\n(1+o(1))(1.2)^{iruaytps} \\leq \\kjdlmreo(iruaytps) \\leq(3+o(1))^{iruaytps},\n\\]\nthe lower and upper bounds being due to \\( [\\mathrm{FW}] \\) and \\( [\\mathrm{LR}] \\), respectively. (See the remark after the solution to 1988A3 for the meaning of the little-o notation.) For more on such questions, consult [Gr2] and the references listed there, and browse issues of the journal Geombinatorics.\n"
    },
    "kernel_variant": {
      "question": "Let the unit of length be the centimetre.\n\n(a)  Every point of the Euclidean plane is coloured with one of three colours.  Prove that there necessarily exist two points of the same colour whose distance is exactly $2\\,\\mathrm{cm}$.\n\n(b)  Show that the assertion in part (a) is no longer true if the word \"three\" is replaced by \"sixteen\"; that is, exhibit an explicit colouring of the plane with sixteen colours in which no two points of the same colour are $2\\,\\mathrm{cm}$ apart.",
      "solution": "Throughout, all distances are measured in centimetres.\n\n(a) Suppose, to the contrary, that the plane has been 3–coloured in such a way that no two points of the same colour are 2 cm apart.\n\nStep 1: A 2√3–lemma. Pick two points A, B with AB=2√3. Consider the circles of radius 2 centred at A and B. These circles intersect in two points P, Q, and ΔAPQ and ΔBPQ are equilateral of side 2 (since AP=AQ=BP=BQ=2 and the distance between P and Q is 2). Hence the three vertices of each equilateral triangle are pairwise 2 cm apart, so they must all have different colours. In particular, P and Q have the two colours different from that of A. Now in ΔBPQ the points P and Q already use those two colours, so B cannot use either of them; hence B must share the colour of A. We conclude:\n\n    (1) If AB=2√3, then A and B share a colour.\n\nStep 2: A 2√3–2 isosceles triangle. Take an isosceles triangle CDE with\n    CD=CE=2√3,\n    DE=2.\nBy (1) we know C and D have the same colour and C and E have the same colour, whence D and E must also be the same colour. But DE=2, contradicting the assumption that no monochromatic pair is 2 cm apart. This contradiction shows that no 3–colouring of the plane can avoid a monochromatic pair at distance 2.\n\n(b) A 16–colouring with no monochromatic 2 cm segment. Partition the plane into unit squares by the integer grid, and assign to the square [m,m+1)×[n,n+1) the colour given by the ordered pair ((m mod 4),(n mod 4)) in {0,1,2,3}². This uses exactly 16 colours, repeating every 4 squares in each direction.\n\nIf P and Q have the same colour then ⌊xₚ⌋≡⌊x_q⌋ (mod 4) and ⌊yₚ⌋≡⌊y_q⌋ (mod 4).\n  • If P,Q lie in the same unit square, then |PQ| ≤ √2 < 2.\n  • Otherwise, the floor‐index differences in x or y are nonzero multiples of 4, so the corresponding coordinate difference is at least 4–1 = 3, giving |PQ| ≥ 3 > 2.\n\nIn either case PQ ≠ 2, so no monochromatic pair is exactly 2 cm apart. Thus sixteen colours suffice to avoid distance 2.",
      "_meta": {
        "core_steps": [
          "Assume a 3-coloring with no monochromatic unit segment and study its consequences.",
          "Intersecting unit circles about two points force those points (√3 apart) to share a color.",
          "Use an isosceles triangle with sides (√3, √3, 1) to obtain a monochromatic unit segment—contradiction.",
          "For 9 colors, color each point by the pair (⌊(3/2)x⌋ mod 3 , ⌊(3/2)y⌋ mod 3), i.e. by its 3×3 square in a grid of side 2⁄3.",
          "Show that two points of the same color are either nearer than 1 or farther than 1, so no unit-distance pair is monochromatic."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Chosen unit length that the argument is scaled around (all distances scale with it).",
            "original": "1 (\"one inch\")"
          },
          "slot2": {
            "description": "Number of available colors in part (b); any larger set can replicate the 9-color grid coloring.",
            "original": "9"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}