{
  "index": "1958-A-7",
  "type": "GEO",
  "tag": [
    "GEO",
    "COMB"
  ],
  "difficulty": "",
  "question": "7. Show that ten equal-sized squares cannot be placed on a plane in such a way that no two have an interior point in common and the first touches each of the others.",
  "solution": "Solution. Suppose that such a set of unit squares exists. Let \\( S, S_{1}, S_{2} \\) be three of the non-overlapping squares with centers respectively at \\( C, C_{1}, C_{2} \\) and suppose \\( S \\) touches both \\( S_{1} \\) and \\( S_{2} \\).\n\nLet \\( \\theta=\\angle C_{1} C C_{2}, a=\\left|C_{1} C\\right|, b=\\left|C C_{2}\\right|, c=\\left|C_{2} C_{1}\\right| \\). Then \\( 1 \\leq a \\leq \\sqrt{2} \\), \\( 1 \\leq b \\leq \\sqrt{2} \\), and \\( c \\geq 1 \\). By the law of cosines\n\\[\n\\begin{aligned}\n\\cos \\theta & =\\frac{a^{2}+b^{2}-c^{2}}{2 a b} \\leq \\frac{a^{2}+b^{2}-1}{2 a b}=f(a, b) . \\\\\n\\frac{\\partial f}{\\partial a} & =\\frac{a^{2}-b^{2}+1}{2 a^{2} b} \\text { and } \\frac{\\partial f}{\\partial b}=\\frac{b^{2}-a^{2}+1}{2 a b^{2}} .\n\\end{aligned}\n\\]\n\nBoth of these derivatives are non-negative throughout the allowable \\( (a, b) \\) domain, so it follows that the maximum value of \\( f \\) on that domain is \\( f(\\sqrt{2}, \\sqrt{2})=3 / 4 \\). Thus \\( \\cos \\theta \\leq 3 / 4 \\) and \\( \\theta \\geq \\operatorname{arc} \\cos 3 / 4 \\). Let \\( \\alpha= \\) \\( \\arccos 3 / 4 \\). Then\n\\[\n\\cos 3 \\alpha=4 \\cos ^{3} \\alpha-3 \\cos \\alpha=-\\frac{9}{16}<-\\frac{1}{2}=\\cos \\frac{2 \\pi}{3} .\n\\]\n\nHence \\( 3 \\alpha>2 \\pi / 3 \\), and so \\( \\angle C_{1} C C_{2}=\\theta \\geq \\alpha>2 \\pi / 9 \\).\nNow suppose that \\( S, S_{1}, S_{2}, S_{3}, \\ldots, S_{9} \\) are the non-overlapping squares, with \\( S \\) touching each of the others. Let \\( C, C_{1}, C_{2}, C_{3}, \\ldots, C_{9} \\) be the respective centers. Choose polar coordinates with center at \\( C \\), and number the squares so that the angular coordinates of \\( C_{1}, C_{2}, \\ldots, C_{9} \\) increase monotonically between 0 and \\( 2 \\pi \\). Then\n\\[\n\\angle C_{1} C C_{2}+\\angle C_{2} C C_{3}+\\cdots+\\angle C_{9} C C_{1}=2 \\pi,\n\\]\nso at least one of these nine angles is no greater than \\( 2 \\pi / 9 \\), contrary to what we have proved above. Hence the configuration is impossible.\n\nRemark. Suppose \\( S^{*} \\) is a square of side 2 concentric with \\( S \\) as shown.\nBy a rather long analysis of cases, it can be shown that any unit square which touches \\( S \\) but contains no interior point of \\( S \\) must cut off from \\( S^{*} \\) an arc (for example \\( A B C \\) in the figure) of length at least 1 . Since the perimeter of \\( S^{*} \\) is 8 , at most 8 squares with disjoint interiors can touch \\( S \\). This proof has the advantage that it extends easily to prove that the only configuration of eight unit squares touching \\( S \\) is the familiar checkerboard arrangement.",
  "vars": [
    "S",
    "S_1",
    "S_2",
    "S_3",
    "S_4",
    "S_5",
    "S_6",
    "S_7",
    "S_8",
    "S_9",
    "C",
    "C_1",
    "C_2",
    "C_3",
    "C_4",
    "C_5",
    "C_6",
    "C_7",
    "C_8",
    "C_9",
    "a",
    "b",
    "c",
    "f",
    "\\\\theta",
    "\\\\alpha"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "S": "central",
        "S_1": "squareone",
        "S_2": "squaretwo",
        "S_3": "squarethree",
        "S_4": "squarefour",
        "S_5": "squarefive",
        "S_6": "squaresix",
        "S_7": "squareseven",
        "S_8": "squareeight",
        "S_9": "squarenine",
        "C": "center",
        "C_1": "centerone",
        "C_2": "centertwo",
        "C_3": "centerthree",
        "C_4": "centerfour",
        "C_5": "centerfive",
        "C_6": "centersix",
        "C_7": "centerseven",
        "C_8": "centereight",
        "C_9": "centernine",
        "a": "distancea",
        "b": "distanceb",
        "c": "distancec",
        "f": "functionf",
        "\\theta": "angletheta",
        "\\alpha": "anglealpha"
      },
      "question": "7. Show that ten equal-sized squares cannot be placed on a plane in such a way that no two have an interior point in common and the first touches each of the others.",
      "solution": "Solution. Suppose that such a set of unit squares exists. Let \\( central, squareone, squaretwo \\) be three of the non-overlapping squares with centers respectively at \\( center, centerone, centertwo \\) and suppose \\( central \\) touches both \\( squareone \\) and \\( squaretwo \\).\n\nLet \\( angletheta=\\angle centerone\\, center\\, centertwo, \\; distancea=\\left|centerone\\, center\\right|, \\; distanceb=\\left|center\\, centertwo\\right|, \\; distancec=\\left|centertwo\\, centerone\\right| \\). Then \\( 1 \\leq distancea \\leq \\sqrt{2} \\), \\( 1 \\leq distanceb \\leq \\sqrt{2} \\), and \\( distancec \\geq 1 \\). By the law of cosines\n\\[\n\\begin{aligned}\n\\cos angletheta & =\\frac{distancea^{2}+distanceb^{2}-distancec^{2}}{2\\,distancea\\,distanceb} \\leq \\frac{distancea^{2}+distanceb^{2}-1}{2\\,distancea\\,distanceb}\n=functionf(distancea,distanceb). \\\\\n\\frac{\\partial functionf}{\\partial distancea} & =\\frac{distancea^{2}-distanceb^{2}+1}{2\\,distancea^{2}\\,distanceb}, \\qquad\n\\frac{\\partial functionf}{\\partial distanceb}= \\frac{distanceb^{2}-distancea^{2}+1}{2\\,distancea\\,distanceb^{2}} .\n\\end{aligned}\n\\]\n\nBoth of these derivatives are non-negative throughout the allowable \\( (distancea,distanceb) \\) domain, so it follows that the maximum value of \\( functionf \\) on that domain is \\( functionf(\\sqrt{2},\\sqrt{2})=3/4 \\). Thus \\( \\cos angletheta \\leq 3/4 \\) and \\( angletheta \\geq \\arccos 3/4 \\). Let \\( anglealpha = \\arccos 3/4 \\). Then\n\\[\n\\cos 3anglealpha = 4\\cos^{3} anglealpha-3\\cos anglealpha=-\\frac{9}{16}< -\\frac{1}{2}=\\cos \\frac{2\\pi}{3} .\n\\]\n\nHence \\( 3anglealpha>2\\pi/3 \\), and so \\( \\angle centerone\\, center\\, centertwo = angletheta \\geq anglealpha > 2\\pi/9 \\).\n\nNow suppose that \\( central, squareone, squaretwo, squarethree, \\ldots, squarenine \\) are the non-overlapping squares, with \\( central \\) touching each of the others. Let \\( center, centerone, centertwo, centerthree, \\ldots, centernine \\) be the respective centers. Choose polar coordinates with origin at \\( center \\), and number the squares so that the angular coordinates of \\( centerone, centertwo, \\ldots, centernine \\) increase monotonically between 0 and \\( 2\\pi \\). Then\n\\[\n\\angle centerone\\, center\\, centertwo + \\angle centertwo\\, center\\, centerthree + \\cdots + \\angle centernine\\, center\\, centerone = 2\\pi,\n\\]\nso at least one of these nine angles is no greater than \\( 2\\pi/9 \\), contrary to what we have proved above. Hence the configuration is impossible.\n\nRemark. Suppose \\( central^{*} \\) is a square of side 2 concentric with \\( central \\) as shown. By a rather long analysis of cases, it can be shown that any unit square which touches \\( central \\) but contains no interior point of \\( central \\) must cut off from \\( central^{*} \\) an arc (for example \\( A B C \\) in the figure) of length at least 1. Since the perimeter of \\( central^{*} \\) is 8, at most 8 squares with disjoint interiors can touch \\( central \\). This proof has the advantage that it extends easily to prove that the only configuration of eight unit squares touching \\( central \\) is the familiar checkerboard arrangement."
    },
    "descriptive_long_confusing": {
      "map": {
        "S": "hummingbird",
        "S_1": "rainmaker",
        "S_2": "starlight",
        "S_3": "moonshade",
        "S_4": "riverbank",
        "S_5": "cloudburst",
        "S_6": "stonework",
        "S_7": "driftwood",
        "S_8": "candlewick",
        "S_9": "goldsmith",
        "C": "metalcraft",
        "C_1": "goldfinch",
        "C_2": "parchment",
        "C_3": "blacksmith",
        "C_4": "journeyman",
        "C_5": "waterwheel",
        "C_6": "bricklayer",
        "C_7": "carpentry",
        "C_8": "feathering",
        "C_9": "glassware",
        "a": "blueberry",
        "b": "raspberry",
        "c": "strawberry",
        "f": "trelliswork",
        "\\theta": "chrysalis",
        "\\alpha": "wanderlust"
      },
      "question": "7. Show that ten equal-sized squares cannot be placed on a plane in such a way that no two have an interior point in common and the first touches each of the others.",
      "solution": "Solution. Suppose that such a set of unit squares exists. Let \\( hummingbird, rainmaker, starlight \\) be three of the non-overlapping squares with centers respectively at \\( metalcraft, goldfinch, parchment \\) and suppose \\( hummingbird \\) touches both \\( rainmaker \\) and \\( starlight \\).\n\nLet \\( chrysalis = \\angle goldfinch \\, metalcraft \\, parchment, blueberry = \\left| goldfinch\\, metalcraft \\right|, raspberry = \\left| metalcraft\\, parchment \\right|, strawberry = \\left| parchment\\, goldfinch \\right| \\). Then \\( 1 \\leq blueberry \\leq \\sqrt{2} \\), \\( 1 \\leq raspberry \\leq \\sqrt{2} \\), and \\( strawberry \\geq 1 \\). By the law of cosines\n\\[\n\\begin{aligned}\n\\cos chrysalis & = \\frac{blueberry^{2}+raspberry^{2}-strawberry^{2}}{2\\, blueberry\\, raspberry} \\leq \\frac{blueberry^{2}+raspberry^{2}-1}{2\\, blueberry\\, raspberry}=trelliswork(blueberry, raspberry) . \\\\\n\\frac{\\partial trelliswork}{\\partial blueberry} & = \\frac{blueberry^{2}-raspberry^{2}+1}{2\\, blueberry^{2}\\, raspberry} \\text { and } \\frac{\\partial trelliswork}{\\partial raspberry} = \\frac{raspberry^{2}-blueberry^{2}+1}{2\\, blueberry\\, raspberry^{2}} .\n\\end{aligned}\n\\]\n\nBoth of these derivatives are non\u00160negative throughout the allowable \\( (blueberry, raspberry) \\) domain, so it follows that the maximum value of \\( trelliswork \\) on that domain is \\( trelliswork(\\sqrt{2}, \\sqrt{2}) = 3/4 \\). Thus \\( \\cos chrysalis \\leq 3/4 \\) and \\( chrysalis \\geq \\operatorname{arc} \\cos 3/4 \\). Let \\( wanderlust = \\arccos 3/4 \\). Then\n\\[\n\\cos 3\\, wanderlust = 4 \\cos^{3} wanderlust - 3 \\cos wanderlust = -\\frac{9}{16} < -\\frac{1}{2} = \\cos \\frac{2 \\pi}{3} .\n\\]\n\nHence \\( 3\\, wanderlust > 2 \\pi / 3 \\), and so \\( \\angle goldfinch \\, metalcraft \\, parchment = chrysalis \\geq wanderlust > 2 \\pi / 9 \\).\n\nNow suppose that \\( hummingbird, rainmaker, starlight, moonshade, \\ldots, goldsmith \\) are the non-overlapping squares, with \\( hummingbird \\) touching each of the others. Let \\( metalcraft, goldfinch, parchment, blacksmith, \\ldots, glassware \\) be the respective centers. Choose polar coordinates with center at \\( metalcraft \\), and number the squares so that the angular coordinates of \\( goldfinch, parchment, \\ldots, glassware \\) increase monotonically between 0 and \\( 2 \\pi \\). Then\n\\[\n\\angle goldfinch \\, metalcraft \\, parchment + \\angle parchment \\, metalcraft \\, blacksmith + \\cdots + \\angle glassware \\, metalcraft \\, goldfinch = 2 \\pi,\n\\]\nso at least one of these nine angles is no greater than \\( 2 \\pi / 9 \\), contrary to what we have proved above. Hence the configuration is impossible.\n\nRemark. Suppose \\( S^{*} \\) is a square of side 2 concentric with \\( hummingbird \\) as shown. By a rather long analysis of cases, it can be shown that any unit square which touches \\( hummingbird \\) but contains no interior point of \\( hummingbird \\) must cut off from \\( S^{*} \\) an arc (for example \\( A B C \\) in the figure) of length at least 1. Since the perimeter of \\( S^{*} \\) is 8, at most 8 squares with disjoint interiors can touch \\( hummingbird \\). This proof has the advantage that it extends easily to prove that the only configuration of eight unit squares touching \\( hummingbird \\) is the familiar checkerboard arrangement."
    },
    "descriptive_long_misleading": {
      "map": {
        "S": "circlezone",
        "S_1": "circleone",
        "S_2": "circletwo",
        "S_3": "circlethree",
        "S_4": "circlefour",
        "S_5": "circlefive",
        "S_6": "circlesix",
        "S_7": "circleseven",
        "S_8": "circleeight",
        "S_9": "circlenine",
        "C": "borderpoint",
        "C_1": "borderpointone",
        "C_2": "borderpointtwo",
        "C_3": "borderpointthree",
        "C_4": "borderpointfour",
        "C_5": "borderpointfive",
        "C_6": "borderpointsix",
        "C_7": "borderpointseven",
        "C_8": "borderpointeight",
        "C_9": "borderpointnine",
        "a": "closenessa",
        "b": "closenessb",
        "c": "closenessc",
        "f": "constantvar",
        "\\theta": "straightangle",
        "\\alpha": "flatangle"
      },
      "question": "Problem:\n<<<\n7. Show that ten equal-sized squares cannot be placed on a plane in such a way that no two have an interior point in common and the first touches each of the others.\n>>>\n",
      "solution": "Solution. Suppose that such a set of unit squares exists. Let \\( circlezone, circleone, circletwo \\) be three of the non-overlapping squares with centers respectively at \\( borderpoint, borderpointone, borderpointtwo \\) and suppose \\( circlezone \\) touches both \\( circleone \\) and \\( circletwo \\).\n\nLet \\( straightangle=\\angle borderpointone \\, borderpoint \\, borderpointtwo, closenessa=\\left|borderpointone \\, borderpoint\\right|, closenessb=\\left|borderpoint \\, borderpointtwo\\right|, closenessc=\\left|borderpointtwo \\, borderpointone\\right| \\). Then \\( 1 \\leq closenessa \\leq \\sqrt{2} \\), \\( 1 \\leq closenessb \\leq \\sqrt{2} \\), and \\( closenessc \\geq 1 \\). By the law of cosines\n\\[\n\\begin{aligned}\n\\cos straightangle & =\\frac{closenessa^{2}+closenessb^{2}-closenessc^{2}}{2 closenessa closenessb} \\leq \\frac{closenessa^{2}+closenessb^{2}-1}{2 closenessa closenessb}=constantvar(closenessa, closenessb) . \\\n\\frac{\\partial constantvar}{\\partial closenessa} & =\\frac{closenessa^{2}-closenessb^{2}+1}{2 closenessa^{2} closenessb} \\text { and } \\frac{\\partial constantvar}{\\partial closenessb}=\\frac{closenessb^{2}-closenessa^{2}+1}{2 closenessa closenessb^{2}} .\n\\end{aligned}\n\\]\n\nBoth of these derivatives are non-negative throughout the allowable \\( (closenessa, closenessb) \\) domain, so it follows that the maximum value of \\( constantvar \\) on that domain is \\( constantvar(\\sqrt{2}, \\sqrt{2})=3 / 4 \\). Thus \\( \\cos straightangle \\leq 3 / 4 \\) and \\( straightangle \\geq \\operatorname{arc} \\cos 3 / 4 \\). Let \\( flatangle= \\) \\( \\arccos 3 / 4 \\). Then\n\\[\n\\cos 3 flatangle=4 \\cos ^{3} flatangle-3 \\cos flatangle=-\\frac{9}{16}<-\\frac{1}{2}=\\cos \\frac{2 \\pi}{3} .\n\\]\n\nHence \\( 3 flatangle>2 \\pi / 3 \\), and so \\( \\angle borderpointone \\, borderpoint \\, borderpointtwo=straightangle \\geq flatangle>2 \\pi / 9 \\).\nNow suppose that \\( circlezone, circleone, circletwo, circlethree, circlefour, circlefive, circlesix, circleseven, circleeight, circlenine \\) are the non-overlapping squares, with \\( circlezone \\) touching each of the others. Let \\( borderpoint, borderpointone, borderpointtwo, borderpointthree, borderpointfour, borderpointfive, borderpointsix, borderpointseven, borderpointeight, borderpointnine \\) be the respective centers. Choose polar coordinates with center at \\( borderpoint \\), and number the squares so that the angular coordinates of \\( borderpointone, borderpointtwo, \\ldots, borderpointnine \\) increase monotonically between 0 and \\( 2 \\pi \\). Then\n\\[\n\\angle borderpointone \\, borderpoint \\, borderpointtwo+\\angle borderpointtwo \\, borderpoint \\, borderpointthree+\\cdots+\\angle borderpointnine \\, borderpoint \\, borderpointone=2 \\pi,\n\\]\nso at least one of these nine angles is no greater than \\( 2 \\pi / 9 \\), contrary to what we have proved above. Hence the configuration is impossible.\n\nRemark. Suppose \\( circlezone^{*} \\) is a square of side 2 concentric with \\( circlezone \\) as shown.\nBy a rather long analysis of cases, it can be shown that any unit square which touches \\( circlezone \\) but contains no interior point of \\( circlezone \\) must cut off from \\( circlezone^{*} \\) an arc (for example \\( A B C \\) in the figure) of length at least 1 . Since the perimeter of \\( circlezone^{*} \\) is 8 , at most 8 squares with disjoint interiors can touch \\( circlezone \\). This proof has the advantage that it extends easily to prove that the only configuration of eight unit squares touching \\( circlezone \\) is the familiar checkerboard arrangement."
    },
    "garbled_string": {
      "map": {
        "S": "qlxmfrta",
        "S_1": "zpkdwhsn",
        "S_2": "rgnvcieo",
        "S_3": "utjkbmza",
        "S_4": "yovslqpn",
        "S_5": "hxfmbtqe",
        "S_6": "iclrdpav",
        "S_7": "swqjzgey",
        "S_8": "damrhkib",
        "S_9": "bsyqvnet",
        "C": "lhnodfce",
        "C_1": "pkzfruay",
        "C_2": "wstghlqe",
        "C_3": "vzmdneko",
        "C_4": "jcfhrula",
        "C_5": "axqldpif",
        "C_6": "nfzvrboc",
        "C_7": "euwskhjm",
        "C_8": "gptalxri",
        "C_9": "omibhcyz",
        "a": "kjpqvhan",
        "b": "wqzmlrta",
        "c": "nvrxgase",
        "f": "tcswpljo",
        "\\theta": "zfoltrqk",
        "\\alpha": "hmajctup"
      },
      "question": "7. Show that ten equal-sized squares cannot be placed on a plane in such a way that no two have an interior point in common and the first touches each of the others.",
      "solution": "Solution. Suppose that such a set of unit squares exists. Let \\( qlxmfrta, zpkdwhsn, rgnvcieo \\) be three of the non-overlapping squares with centers respectively at \\( lhnodfce, pkzfruay, wstghlqe \\) and suppose \\( qlxmfrta \\) touches both \\( zpkdwhsn \\) and \\( rgnvcieo \\).\n\nLet \\( zfoltrqk=\\angle pkzfruay\\, lhnodfce\\, wstghlqe, kjpqvhan=\\left|pkzfruay\\, lhnodfce\\right|, wqzmlrta=\\left|lhnodfce\\, wstghlqe\\right|, nvrxgase=\\left|wstghlqe\\, pkzfruay\\right| \\). Then \\( 1 \\leq kjpqvhan \\leq \\sqrt{2} \\), \\( 1 \\leq wqzmlrta \\leq \\sqrt{2} \\), and \\( nvrxgase \\geq 1 \\). By the law of cosines\n\\[\n\\begin{aligned}\n\\cos zfoltrqk & =\\frac{kjpqvhan^{2}+wqzmlrta^{2}-nvrxgase^{2}}{2 kjpqvhan\\, wqzmlrta} \\leq \\frac{kjpqvhan^{2}+wqzmlrta^{2}-1}{2 kjpqvhan\\, wqzmlrta}=tcswpljo(kjpqvhan, wqzmlrta) . \\\\\n\\frac{\\partial tcswpljo}{\\partial kjpqvhan} & =\\frac{kjpqvhan^{2}-wqzmlrta^{2}+1}{2 kjpqvhan^{2} wqzmlrta} \\text { and } \\frac{\\partial tcswpljo}{\\partial wqzmlrta}=\\frac{wqzmlrta^{2}-kjpqvhan^{2}+1}{2 kjpqvhan wqzmlrta^{2}} .\n\\end{aligned}\n\\]\n\nBoth of these derivatives are non-negative throughout the allowable \\( (kjpqvhan, wqzmlrta) \\) domain, so it follows that the maximum value of \\( tcswpljo \\) on that domain is \\( tcswpljo(\\sqrt{2}, \\sqrt{2})=3 / 4 \\). Thus \\( \\cos zfoltrqk \\leq 3 / 4 \\) and \\( zfoltrqk \\geq \\operatorname{arc} \\cos 3 / 4 \\). Let \\( hmajctup= \\) \\( \\arccos 3 / 4 \\). Then\n\\[\n\\cos 3 hmajctup=4 \\cos ^{3} hmajctup-3 \\cos hmajctup=-\\frac{9}{16}<-\\frac{1}{2}=\\cos \\frac{2 \\pi}{3} .\n\\]\n\nHence \\( 3 hmajctup>2 \\pi / 3 \\), and so \\( \\angle pkzfruay\\, lhnodfce\\, wstghlqe=zfoltrqk \\geq hmajctup>2 \\pi / 9 \\).\nNow suppose that \\( qlxmfrta, zpkdwhsn, rgnvcieo, utjkbmza, \\ldots, bsyqvnet \\) are the non-overlapping squares, with \\( qlxmfrta \\) touching each of the others. Let \\( lhnodfce, pkzfruay, wstghlqe, vzmdneko, \\ldots, omibhcyz \\) be the respective centers. Choose polar coordinates with center at \\( lhnodfce \\), and number the squares so that the angular coordinates of \\( pkzfruay, wstghlqe, \\ldots, omibhcyz \\) increase monotonically between 0 and \\( 2 \\pi \\). Then\n\\[\n\\angle pkzfruay\\, lhnodfce\\, wstghlqe+\\angle wstghlqe\\, lhnodfce\\, vzmdneko+\\cdots+\\angle omibhcyz\\, lhnodfce\\, pkzfruay=2 \\pi,\n\\]\nso at least one of these nine angles is no greater than \\( 2 \\pi / 9 \\), contrary to what we have proved above. Hence the configuration is impossible.\n\nRemark. Suppose \\( qlxmfrta^{*} \\) is a square of side 2 concentric with \\( qlxmfrta \\) as shown.\nBy a rather long analysis of cases, it can be shown that any unit square which touches \\( qlxmfrta \\) but contains no interior point of \\( qlxmfrta \\) must cut off from \\( qlxmfrta^{*} \\) an arc (for example \\( A B lhnodfce \\) in the figure) of length at least 1 . Since the perimeter of \\( qlxmfrta^{*} \\) is 8 , at most 8 squares with disjoint interiors can touch \\( qlxmfrta \\). This proof has the advantage that it extends easily to prove that the only configuration of eight unit squares touching \\( qlxmfrta \\) is the familiar checkerboard arrangement."
    },
    "kernel_variant": {
      "question": "Let eleven congruent squares, each of side-length 3, be placed in the plane in such a way that \n(1) the interiors of any two squares are disjoint, \n(2) all squares are parallel translates of each other (that is, their corresponding sides are parallel), and \n(3) one distinguished square touches every one of the other ten (two squares \"touch\" whenever they have at least one point in common while their interiors are disjoint).  \nShow that such a configuration is impossible.",
      "solution": "We keep the side–length 3 throughout and denote the distinguished square by S.  \nThe ten remaining squares are S₁ ,…, S₁₀ ; their centres are C₁ ,…, C₁₀ , while C denotes the centre of S.\n\nStep 1.  Distances between the centre of S and the centres of its neighbours.\nBecause every square is parallel to S, a neighbour Sᵢ can meet S in only two different ways.\n• If they share a whole side, the two centres differ by exactly the side–length:\n            |CCᵢ| = 3.\n• If they meet at only one vertex, the vector that joins the centres equals (±3, ±3), so\n            |CCᵢ| = 3√2.\nConsequently\n            3 ≤ |CCᵢ| ≤ 3√2.                                   (1)\n\nStep 2.  A lower bound for the angle formed by two neighbours at C.\nFix two distinct neighbours, S₁ and S₂.  Let\n            a = |C₁C| ,   b = |CC₂| ,   c = |C₁C₂| ,   θ = ∠C₁ C C₂.\nRelation (1) yields 3 ≤ a, b ≤ 3√2.\n\nBecause S₁ and S₂ are also parallel squares with disjoint interiors, their centres cannot be closer than one side–length:\n            c ≥ 3.                                             (2)\nIndeed, if |x(C₁) – x(C₂)| < 3 and |y(C₁) – y(C₂)| < 3, the two unit squares would overlap in a rectangle of positive area, contradicting the hypothesis that their interiors are disjoint.\n\nApplying the law of cosines in ΔC₁CC₂ and using (2) we obtain\n            cos θ = (a² + b² – c²)/(2ab) ≤ (a² + b² – 3²)/(2ab) =: f(a, b).\nOn the rectangle 3 ≤ a, b ≤ 3√2 one has\n            ∂f/∂a = (a² – b² + 9)/(2a²b) ≥ 0,   ∂f/∂b = (b² – a² + 9)/(2ab²) ≥ 0,\nso f attains its maximum when a and b are as large as possible, namely at a = b = 3√2.  Hence\n            cos θ ≤ f(3√2, 3√2) = 3/4,\nwhence\n            θ ≥ arccos(3/4) =: α (≈ 41.41°).\nUsing the triple–angle identity,\n            cos 3α = 4 cos³ α – 3 cos α = 4(3/4)³ – 3(3/4) = –9/16 < –1/2 = cos(2π/3),\nwhich implies\n            3α > 2π/3   and therefore   θ = ∠C₁CC₂ ≥ α > 2π/9.           (3)\n\nStep 3.  Summing the ten angles around C.\nOrder the neighbours cyclically around C so that the directed angles\n            ∠C₁ C C₂ , ∠C₂ C C₃ , … , ∠C₁₀ C C₁\nsum to 2π.  There are ten such angles, each of them strictly larger than 2π/9 by (3).  Consequently\n            2π = Σ_{i=1}^{10}∠Cᵢ C C_{i+1} > 10·(2π/9) = 20π/9,\na contradiction.\n\nStep 4.  Conclusion.\nThe assumption that eleven pairwise parallel congruent squares of side 3 can be arranged so that one touches all the others leads to an impossibility; therefore no such configuration exists.\n\nRemark on necessity of the “parallel” hypothesis.\nWithout requiring the squares to be parallel, the lower bound |CCᵢ| ≥ 3 in (1) fails (e.g. a square rotated by 45° can touch S while its centre is closer than 3 to C), and the foregoing argument breaks down.  The extra hypothesis included in the corrected statement guarantees the distance bounds employed above and is exactly what was tacitly used in the original solution.",
      "_meta": {
        "core_steps": [
          "Model each square by its center; touching implies center–center distance in [s, s√2] for common side-length s.",
          "For a square S that touches two others S₁, S₂, apply the Law of Cosines in △C₁C C₂ to bound cos θ and obtain θ ≥ α where α = arccos (3/4) > 2π⁄9.",
          "Observe that every pair of consecutive neighbours of S forms such an angle; therefore each of the (k−1) angles around C exceeds 2π⁄9.",
          "Sum of the (k−1) angles is 2π, so if (k−1)·2π⁄9 ≥ 2π (strict since θ>2π⁄9) a contradiction arises.",
          "Hence no configuration exists once k−1 ≥ 9, i.e. for ten or more equal squares touching a common one."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Common side-length of the squares (uniform scaling of the whole figure).",
            "original": "1 (\"unit\" squares)"
          },
          "slot2": {
            "description": "Total number k of squares in the statement; any k ≥ 10 still leads to the same contradiction.",
            "original": "10"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}