{
  "index": "2011-A-1",
  "type": "COMB",
  "tag": [
    "COMB",
    "GEO",
    "NT"
  ],
  "difficulty": "",
  "question": "of points with integer coordinates $P_0 = (0,0), P_1, \\dots, P_n$ such\nthat $n \\geq 2$ and:\n\\begin{itemize}\n\\item the directed line segments $P_0 P_1, P_1 P_2, \\dots, P_{n-1} P_n$\nare in the successive coordinate directions east (for $P_0 P_1$), north,\nwest, south, east, etc.;\n\\item the lengths of these line segments are positive and strictly\nincreasing.\n\\end{itemize} [Picture omitted.]  How many of the points $(x,y)$ with\ninteger coordinates $0\\leq x\\leq 2011, 0\\leq y\\leq 2011$ \\emph{cannot}\nbe the last point, $P_n$ of any growing spiral?",
  "solution": "y\\leq 2011$ that cannot be the last point of a growing spiral are as follows:\n$(0,y)$ for $0\\leq y\\leq 2011$; $(x,0)$ and $(x,1)$ for $1\\leq x\\leq 2011$;\n$(x,2)$ for $2\\leq x\\leq 2011$; and $(x,3)$ for $3\\leq x\\leq 2011$. This\ngives a total of\n\\[\n2012 + 2011 + 2011 + 2010 + 2009 = 10053\n\\]\nexcluded points.\n\nThe complement of this set is the set of $(x,y)$ with $0<x<y$, along with\n$(x,y)$ with $x \\geq y \\geq 4$. Clearly the former set is achievable as\n$P_2$ in a growing spiral, while a point $(x,y)$ in the latter set is $P_6$\nin a growing spiral with successive lengths $1$, $2$, $3$, $x+1$, $x+2$,\nand $x+y-1$.\n\nWe now need to rule out the other cases. Write $x_1<y_1<x_2<y_2<\\ldots$\nfor the lengths of the line segments in the spiral in order, so that $P_1 =\n(x_1,0)$, $P_2 = (x_1,y_1)$, $P_3 = (x_1-x_2,y_1)$, and so forth. Any point\nbeyond $P_0$ has $x$-coordinate of the form $x_1-x_2+\\cdots+(-1)^{n-1}\nx_n$ for $n\\geq 1$; if $n$ is odd, we can write this as $x_1 + (-x_2+x_3)\n+ \\cdots + (-x_{n-1}+x_n) > 0$, while if $n$ is even, we can write this as\n$(x_1-x_2) + \\cdots + (x_{n-1}-x_n) < 0$. Thus no point beyond $P_0$ can have\n$x$-coordinate $0$, and we have ruled out $(0,y)$ for $0\\leq y\\leq 2011$.\n\nNext we claim that any point beyond $P_3$ must have $y$-coordinate either\nnegative or $\\geq 4$. Indeed, each such point has $y$-coordinate of the\nform $y_1-y_2+\\cdots+(-1)^{n-1} y_n$ for $n \\geq 2$, which we can write\nas $(y_1-y_2) + \\cdots + (y_{n-1}-y_n) < 0$ if $n$ is even, and\n\\[\ny_1 +(-y_2+y_3) + \\cdots + (-y_{n-1}+y_n) \\geq y_1 + 2 \\geq 4\n\\]\nif $n \\geq 3$\nis odd. Thus to rule out the rest of the forbidden points, it suffices to\ncheck that they cannot be $P_2$ or $P_3$ for any growing spiral. But none\nof them can be $P_3 = (x_1-x_2,y_1)$ since $x_1-x_2<0$, and none of them\ncan be $P_2 = (x_1,y_1)$ since they all have $y$-coordinate at most equal\nto their $x$-coordinate.",
  "vars": [
    "x",
    "y",
    "n",
    "P_0",
    "P_1",
    "P_2",
    "P_3",
    "P_6",
    "P_n",
    "P_n-1",
    "x_1",
    "y_1",
    "x_2",
    "y_2",
    "x_n",
    "y_n",
    "x_n-1",
    "y_n-1"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "xcoord",
        "y": "ycoord",
        "n": "segcount",
        "P_0": "startpt",
        "P_1": "pointone",
        "P_2": "pointtwo",
        "P_3": "pointthr",
        "P_6": "pointsix",
        "P_n": "lastpt",
        "P_n-1": "prevlast",
        "P_{n-1}": "prevlast",
        "x_1": "firstlenx",
        "y_1": "firstleny",
        "x_2": "secondlx",
        "y_2": "secondly",
        "x_n": "lastlenx",
        "y_n": "lastleny",
        "x_n-1": "prevlenx",
        "x_{n-1}": "prevlenx",
        "y_n-1": "prevleny",
        "y_{n-1}": "prevleny"
      },
      "question": "of points with integer coordinates $startpt = (0,0), pointone, \\dots, lastpt$ such\nthat $segcount \\geq 2$ and:\n\\begin{itemize}\n\\item the directed line segments $startpt\\,pointone,\\ pointone\\,pointtwo, \\dots, prevlast\\,lastpt$\nare in the successive coordinate directions east (for $startpt\\,pointone$), north,\nwest, south, east, etc.;\n\\item the lengths of these line segments are positive and strictly\nincreasing.\n\\end{itemize} [Picture omitted.]  How many of the points $(xcoord,ycoord)$ with\ninteger coordinates $0\\leq xcoord\\leq 2011, 0\\leq ycoord\\leq 2011$ \\emph{cannot}\nbe the last point, $lastpt$ of any growing spiral?",
      "solution": "ycoord\\leq 2011$ that cannot be the last point of a growing spiral are as follows:\n$(0,ycoord)$ for $0\\leq ycoord\\leq 2011$; $(xcoord,0)$ and $(xcoord,1)$ for $1\\leq xcoord\\leq 2011$;\n$(xcoord,2)$ for $2\\leq xcoord\\leq 2011$; and $(xcoord,3)$ for $3\\leq xcoord\\leq 2011$. This\ngives a total of\n\\[\n2012 + 2011 + 2011 + 2010 + 2009 = 10053\n\\]\nexcluded points.\n\nThe complement of this set is the set of $(xcoord,ycoord)$ with $0< xcoord < ycoord$, along with\n$(xcoord,ycoord)$ with $xcoord \\geq ycoord \\geq 4$. Clearly the former set is achievable as\n$pointtwo$ in a growing spiral, while a point $(xcoord,ycoord)$ in the latter set is $pointsix$\nin a growing spiral with successive lengths $1$, $2$, $3$, $xcoord+1$, $xcoord+2$,\nand $xcoord+ycoord-1$.\n\nWe now need to rule out the other cases. Write $firstlenx<firstleny<secondlx<secondly<\\ldots$\nfor the lengths of the line segments in the spiral in order, so that $pointone =\n(firstlenx,0)$, $pointtwo = (firstlenx,firstleny)$, $pointthr = (firstlenx-secondlx,firstleny)$, and so forth. Any point\nbeyond $startpt$ has $x$-coordinate of the form $firstlenx-secondlx+\\cdots+(-1)^{segcount-1}\nlastlenx$ for $segcount\\geq 1$; if $segcount$ is odd, we can write this as $firstlenx + (-secondlx+x_3)\n+ \\cdots + (-prevlenx+lastlenx) > 0$, while if $segcount$ is even, we can write this as\n$(firstlenx-secondlx) + \\cdots + (prevlenx-lastlenx) < 0$. Thus no point beyond $startpt$ can have\n$x$-coordinate $0$, and we have ruled out $(0,ycoord)$ for $0\\leq ycoord\\leq 2011$.\n\nNext we claim that any point beyond $pointthr$ must have $y$-coordinate either\nnegative or $\\geq 4$. Indeed, each such point has $y$-coordinate of the\nform $firstleny-secondly+\\cdots+(-1)^{segcount-1} lastleny$ for $segcount \\geq 2$, which we can write\nas $(firstleny-secondly) + \\cdots + (prevleny-lastleny) < 0$ if $segcount$ is even, and\n\\[\nfirstleny +(-secondly+y_3) + \\cdots + (-prevleny+lastleny) \\geq firstleny + 2 \\geq 4\n\\]\nif $segcount \\geq 3$\nis odd. Thus to rule out the rest of the forbidden points, it suffices to\ncheck that they cannot be $pointtwo$ or $pointthr$ for any growing spiral. But none\nof them can be $pointthr = (firstlenx-secondlx,firstleny)$ since $firstlenx-secondlx<0$, and none of them\ncan be $pointtwo = (firstlenx,firstleny)$ since they all have $y$-coordinate at most equal\nto their $x$-coordinate."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "marigold",
        "y": "pendulum",
        "n": "squirrel",
        "P_0": "sandstone",
        "P_1": "blueprint",
        "P_2": "labyrinth",
        "P_3": "nachosmile",
        "P_6": "grandpiano",
        "P_n": "bricklayer",
        "P_n-1": "buttercup",
        "x_1": "cucumber",
        "y_1": "windchime",
        "x_2": "toasteroven",
        "y_2": "gallivant",
        "x_n": "honeycomb",
        "y_n": "floodlight",
        "x_n-1": "raincloud",
        "y_n-1": "dandelion"
      },
      "question": "of points with integer coordinates $sandstone = (0,0), blueprint, \\dots, bricklayer$ such\nthat $squirrel \\geq 2$ and:\n\\begin{itemize}\n\\item the directed line segments $sandstone blueprint, blueprint labyrinth, \\dots, buttercup bricklayer$\nare in the successive coordinate directions east (for $sandstone blueprint$), north,\nwest, south, east, etc.;\n\\item the lengths of these line segments are positive and strictly\nincreasing.\n\\end{itemize} [Picture omitted.]  How many of the points $(marigold,pendulum)$ with\ninteger coordinates $0\\leq marigold\\leq 2011, 0\\leq pendulum\\leq 2011$ \\emph{cannot}\nbe the last point, $bricklayer$ of any growing spiral?",
      "solution": "pendulum\\leq 2011$ that cannot be the last point of a growing spiral are as follows:\n$(0,pendulum)$ for $0\\leq pendulum\\leq 2011$; $(marigold,0)$ and $(marigold,1)$ for $1\\leq marigold\\leq 2011$;\n$(marigold,2)$ for $2\\leq marigold\\leq 2011$; and $(marigold,3)$ for $3\\leq marigold\\leq 2011$. This\ngives a total of\n\\[\n2012 + 2011 + 2011 + 2010 + 2009 = 10053\n\\]\nexcluded points.\n\nThe complement of this set is the set of $(marigold,pendulum)$ with $0<marigold<pendulum$, along with\n$(marigold,pendulum)$ with $marigold \\geq pendulum \\geq 4$. Clearly the former set is achievable as\nlabyrinth in a growing spiral, while a point $(marigold,pendulum)$ in the latter set is grandpiano\nin a growing spiral with successive lengths $1$, $2$, $3$, $marigold+1$, $marigold+2$,\nand $marigold+pendulum-1$.\n\nWe now need to rule out the other cases. Write $cucumber<windchime<toasteroven<gallivant<\\ldots$\nfor the lengths of the line segments in the spiral in order, so that blueprint =\n$(cucumber,0)$, labyrinth = $(cucumber,windchime)$, nachosmile = $(cucumber-toasteroven,windchime)$, and so forth. Any point\nbeyond sandstone has $marigold$-coordinate of the form $cucumber-toasteroven+\\cdots+(-1)^{squirrel-1}\nhoneycomb$ for $squirrel\\geq 1$; if $squirrel$ is odd, we can write this as $cucumber + (-toasteroven+x_3)\n+ \\cdots + (-raincloud+honeycomb) > 0$, while if $squirrel$ is even, we can write this as\n$(cucumber-toasteroven) + \\cdots + (raincloud-honeycomb) < 0$. Thus no point beyond sandstone can have\n$marigold$-coordinate $0$, and we have ruled out $(0,pendulum)$ for $0\\leq pendulum\\leq 2011$.\n\nNext we claim that any point beyond nachosmile must have pendulum-coordinate either\nnegative or $\\geq 4$. Indeed, each such point has pendulum-coordinate of the\nform $windchime-gallivant+\\cdots+(-1)^{squirrel-1} floodlight$ for $squirrel \\geq 2$, which we can write\nas $(windchime-gallivant) + \\cdots + (dandelion-floodlight) < 0$ if $squirrel$ is even, and\n\\[\nwindchime +(-gallivant+y_3) + \\cdots + (-dandelion+floodlight) \\geq windchime + 2 \\geq 4\n\\]\nif $squirrel \\geq 3$\nis odd. Thus to rule out the rest of the forbidden points, it suffices to\ncheck that they cannot be labyrinth or nachosmile for any growing spiral. But none\nof them can be nachosmile = $(cucumber-toasteroven,windchime)$ since $cucumber-toasteroven<0$, and none of them\ncan be labyrinth = $(cucumber,windchime)$ since they all have pendulum-coordinate at most equal\nto their marigold-coordinate."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalaxis",
        "y": "horizontalaxis",
        "n": "terminalcount",
        "P_0": "finalpoint",
        "P_1": "backpoint",
        "P_2": "retreatpoint",
        "P_3": "reversepoint",
        "P_6": "contradictorypoint",
        "P_n": "unknownpoint",
        "P_{n-1}": "followingpoint",
        "x_1": "verticalone",
        "y_1": "horizontalone",
        "x_2": "verticaltwo",
        "y_2": "horizontaltwo",
        "x_n": "verticalfinal",
        "y_n": "horizontalfinal",
        "x_{n-1}": "verticalnext",
        "y_{n-1}": "horizontalnext"
      },
      "question": "of points with integer coordinates $finalpoint = (0,0), backpoint, \\dots, unknownpoint$ such\nthat $terminalcount \\geq 2$ and:\n\\begin{itemize}\n\\item the directed line segments $finalpoint backpoint, backpoint retreatpoint, \\dots, followingpoint unknownpoint$\nare in the successive coordinate directions east (for $finalpoint backpoint$), north,\nwest, south, east, etc.;\n\\item the lengths of these line segments are positive and strictly\nincreasing.\n\\end{itemize} [Picture omitted.]  How many of the points $(verticalaxis,horizontalaxis)$ with\ninteger coordinates $0\\leq verticalaxis\\leq 2011, 0\\leq horizontalaxis\\leq 2011$ \\emph{cannot}\nbe the last point, $unknownpoint$ of any growing spiral?",
      "solution": "horizontalaxis\\leq 2011$ that cannot be the last point of a growing spiral are as follows:\n$(0,horizontalaxis)$ for $0\\leq horizontalaxis\\leq 2011$; $(verticalaxis,0)$ and $(verticalaxis,1)$ for $1\\leq verticalaxis\\leq 2011$;\n$(verticalaxis,2)$ for $2\\leq verticalaxis\\leq 2011$; and $(verticalaxis,3)$ for $3\\leq verticalaxis\\leq 2011$. This\ngives a total of\n\\[\n2012 + 2011 + 2011 + 2010 + 2009 = 10053\n\\]\nexcluded points.\n\nThe complement of this set is the set of $(verticalaxis,horizontalaxis)$ with $0<verticalaxis<horizontalaxis$, along with\n$(verticalaxis,horizontalaxis)$ with $verticalaxis \\geq horizontalaxis \\geq 4$. Clearly the former set is achievable as\nretreatpoint in a growing spiral, while a point $(verticalaxis,horizontalaxis)$ in the latter set is contradictorypoint\nin a growing spiral with successive lengths $1$, $2$, $3$, $verticalaxis+1$, $verticalaxis+2$,\nand $verticalaxis+horizontalaxis-1$.\n\nWe now need to rule out the other cases. Write $verticalone<horizontalone<verticaltwo<horizontaltwo<\\ldots$\nfor the lengths of the line segments in the spiral in order, so that backpoint =\n$(verticalone,0)$, retreatpoint = $(verticalone,horizontalone)$, reversepoint = $(verticalone-verticaltwo,horizontalone)$, and so forth. Any point\nbeyond finalpoint has $verticalaxis$-coordinate of the form $verticalone-verticaltwo+\\cdots+(-1)^{terminalcount-1}\nverticalfinal$ for $terminalcount\\geq 1$; if $terminalcount$ is odd, we can write this as $verticalone + (-verticaltwo+x_3)\n+ \\cdots + (-verticalnext+verticalfinal) > 0$, while if $terminalcount$ is even, we can write this as\n$(verticalone-verticaltwo) + \\cdots + (verticalnext-verticalfinal) < 0$. Thus no point beyond finalpoint can have\n$verticalaxis$-coordinate $0$, and we have ruled out $(0,horizontalaxis)$ for $0\\leq horizontalaxis\\leq 2011$.\n\nNext we claim that any point beyond reversepoint must have $horizontalaxis$-coordinate either\nnegative or $\\geq 4$. Indeed, each such point has $horizontalaxis$-coordinate of the\nform $horizontalone-horizontaltwo+\\cdots+(-1)^{terminalcount-1} horizontalfinal$ for $terminalcount \\geq 2$, which we can write\nas $(horizontalone-horizontaltwo) + \\cdots + (horizontalnext-horizontalfinal) < 0$ if $terminalcount$ is even, and\n\\[\nhorizontalone +(-horizontaltwo+y_3) + \\cdots + (-horizontalnext+horizontalfinal) \\geq horizontalone + 2 \\geq 4\n\\]\nif $terminalcount \\geq 3$\nis odd. Thus to rule out the rest of the forbidden points, it suffices to\ncheck that they cannot be retreatpoint or reversepoint for any growing spiral. But none\nof them can be reversepoint = $(verticalone-verticaltwo,horizontalone)$ since $verticalone-verticaltwo<0$, and none of them\ncan be retreatpoint = $(verticalone,horizontalone)$ since they all have $horizontalaxis$-coordinate at most equal\nto their $verticalaxis$-coordinate."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "n": "bmvqxzie",
        "P_0": "ckthwrfj",
        "P_1": "nlvmsqar",
        "P_2": "pjezdhok",
        "P_3": "srqvplue",
        "P_6": "yxzevdab",
        "P_n": "vmrtqowc",
        "P_n-1": "kpsfldni",
        "x_1": "zrbkylam",
        "y_1": "dsvngxju",
        "x_2": "hvpwquier",
        "y_2": "mcfrzpla",
        "x_n": "ljqedsth",
        "y_n": "nwyhklga",
        "x_n-1": "gxtvsoer",
        "y_n-1": "qbwalcmi"
      },
      "question": "of points with integer coordinates $ckthwrfj = (0,0), nlvmsqar, \\dots, vmrtqowc$ such\nthat $bmvqxzie \\geq 2$ and:\n\\begin{itemize}\n\\item the directed line segments $ckthwrfj nlvmsqar, nlvmsqar pjezdhok, \\dots, P_{bmvqxzie-1} vmrtqowc$\nare in the successive coordinate directions east (for $ckthwrfj nlvmsqar$), north,\nwest, south, east, etc.;\n\\item the lengths of these line segments are positive and strictly\nincreasing.\n\\end{itemize} [Picture omitted.]  How many of the points $(qzxwvtnp,hjgrksla)$ with\ninteger coordinates $0\\leq qzxwvtnp\\leq 2011, 0\\leq hjgrksla\\leq 2011$ \\emph{cannot}\nbe the last point, $vmrtqowc$ of any growing spiral?",
      "solution": "hjgrksla\\leq 2011$ that cannot be the last point of a growing spiral are as follows:\n$(0,hjgrksla)$ for $0\\leq hjgrksla\\leq 2011$; $(qzxwvtnp,0)$ and $(qzxwvtnp,1)$ for $1\\leq qzxwvtnp\\leq 2011$;\n$(qzxwvtnp,2)$ for $2\\leq qzxwvtnp\\leq 2011$; and $(qzxwvtnp,3)$ for $3\\leq qzxwvtnp\\leq 2011$. This\ngives a total of\n\\[\n2012 + 2011 + 2011 + 2010 + 2009 = 10053\n\\]\nexcluded points.\n\nThe complement of this set is the set of $(qzxwvtnp,hjgrksla)$ with $0<qzxwvtnp<hjgrksla$, along with\n$(qzxwvtnp,hjgrksla)$ with $qzxwvtnp \\geq hjgrksla \\geq 4$. Clearly the former set is achievable as\n$pjezdhok$ in a growing spiral, while a point $(qzxwvtnp,hjgrksla)$ in the latter set is $yxzevdab$\nin a growing spiral with successive lengths $1$, $2$, $3$, $qzxwvtnp+1$, $qzxwvtnp+2$,\nand $qzxwvtnp+hjgrksla-1$.\n\nWe now need to rule out the other cases. Write $zrbkylam<dsvngxju<hvpwquier<mcfrzpla<\\ldots$\nfor the lengths of the line segments in the spiral in order, so that $nlvmsqar =\n(zrbkylam,0)$, $pjezdhok = (zrbkylam,dsvngxju)$, $srqvplue = (zrbkylam-hvpwquier,dsvngxju)$, and so forth. Any point\nbeyond $ckthwrfj$ has $qzxwvtnp$-coordinate of the form $zrbkylam-hvpwquier+\\cdots+(-1)^{bmvqxzie-1}\nljqedsth$ for $bmvqxzie\\geq 1$; if $bmvqxzie$ is odd, we can write this as $zrbkylam + (-hvpwquier+x_3)\n+ \\cdots + (-x_{bmvqxzie-1}+ljqedsth) > 0$, while if $bmvqxzie$ is even, we can write this as\n$(zrbkylam-hvpwquier) + \\cdots + (x_{bmvqxzie-1}-ljqedsth) < 0$. Thus no point beyond $ckthwrfj$ can have\n$qzxwvtnp$-coordinate $0$, and we have ruled out $(0,hjgrksla)$ for $0\\leq hjgrksla\\leq 2011$.\n\nNext we claim that any point beyond $srqvplue$ must have $hjgrksla$-coordinate either\nnegative or $\\geq 4$. Indeed, each such point has $hjgrksla$-coordinate of the\nform $dsvngxju-mcfrzpla+\\cdots+(-1)^{bmvqxzie-1} nwyhklga$ for $bmvqxzie \\geq 2$, which we can write\nas $(dsvngxju-mcfrzpla) + \\cdots + (y_{bmvqxzie-1}-nwyhklga) < 0$ if $bmvqxzie$ is even, and\n\\[\ndsvngxju +(-mcfrzpla+y_3) + \\cdots + (-y_{bmvqxzie-1}+nwyhklga) \\geq dsvngxju + 2 \\geq 4\n\\]\nif $bmvqxzie \\geq 3$\nis odd. Thus to rule out the rest of the forbidden points, it suffices to\ncheck that they cannot be $pjezdhok$ or $srqvplue$ for any growing spiral. But none\nof them can be $srqvplue = (zrbkylam-hvpwquier,dsvngxju)$ since $zrbkylam-hvpwquier<0$, and none of them\ncan be $pjezdhok = (zrbkylam,dsvngxju)$ since they all have $hjgrksla$-coordinate at most equal\nto their $qzxwvtnp$-coordinate."
    },
    "kernel_variant": {
      "question": "Growing spirals\n\nA finite sequence of lattice points\n        P₀ = (0,0), P₁ , … , Pₙ    (n ≥ 2)\nis called a growing spiral if it satisfies both conditions below.\n1.  (Direction pattern)\n    The directed segments make successive quarter–turns clockwise :\n        P₀P₁ points north,   P₁P₂ east,   P₂P₃ south,   P₃P₄ west,   P₄P₅ north, …\n2.  (Lengths)  The lengths of the segments\n        |P₀P₁| , |P₁P₂| , … , |P_{n-1}P_n|\n    are positive integers that are strictly increasing.\n\nFor how many ordered integer pairs (x , y) with\n        0 ≤ x ≤ 2023   and   0 ≤ y ≤ 2023\nis it impossible to occur as the last point Pₙ of a growing spiral?",
      "solution": "Let the (strictly increasing) lengths of the steps be \n        ℓ₁, ℓ₂, … , ℓₙ   with   ℓ₁<ℓ₂<⋯<ℓₙ .\nBecause the directions follow the cycle N,E,S,W, … we obtain\n\n(1) Xₙ = Σ_{j=1}^{⌊n/2⌋} (–1)^{j+1} ℓ_{2j},  Yₙ = Σ_{j=0}^{⌊(n–1)/2⌋} (–1)^{j} ℓ_{2j+1}. \n\nHere Xₙ and Yₙ denote the coordinates of the end–point Pₙ.\n\n────────────────────────────────────────────────────────\n1.  The spiral never meets either coordinate axis (apart from P₀).\n\nLemma.  For positive integers a₁<⋯<a_m set\n        S_m = a₁ – a₂ + a₃ – a₄ + ⋯ + (–1)^{m+1} a_m .\nThen S_m > 0 for m odd and S_m < 0 for m even; in particular S_m ≠ 0.\n\nProof.  By induction on m.\n•  m = 1:  S₁ = a₁ > 0.\n•  Induction step.  Suppose S_{m–1} has the required sign.  Because\n   a_m > a_{m–1}, we have\n      S_m = S_{m–1} + (–1)^{m+1} a_m\n   which necessarily flips the sign of S_{m–1}.  □\n\nIn (1) the numbers appearing are strictly increasing.  Hence each of Xₙ\nand Yₙ is a non–zero alternating sum and therefore\n        Xₙ ≠ 0, Yₙ ≠ 0 (n ≥ 1).\nConsequently after the origin the spiral never meets either axis.  This\nrules out the points\n  •  (0 , y) (0 ≤ y ≤ 2023)  2024 points,\n  •  (x , 0) (1 ≤ x ≤ 2023)  2023 points.\n\n────────────────────────────────────────────────────────\n2.  Small positive x–coordinates.\n\nPut m = ⌊n/2⌋ (the number of even–indexed steps).\n\nThe first time the x–coordinate can be positive is after two steps:\nP₂ = (ℓ₂ , ℓ₁) with ℓ₂ ≥ 2.\n\n•  n = 2 (m = 1).  The value X₂ = ℓ₂ can be any integer ≥ 2.\n•  n ≥ 4.  Then m ≥ 2.  From (1):\n        Xₙ = ℓ₂ – ℓ₄ + ℓ₆ – ⋯ + (–1)^{m+1} ℓ_{2m}.\n   If m is even, the lemma gives Xₙ < 0.\n   If m ≥ 3 is odd, use the smallest admissible even lengths\n        ℓ₂ = 2, ℓ₄ = 4, ℓ₆ = 6, …\n   to obtain the sharp lower bound\n        Xₙ ≥ 2 – 4 + 6 = 4,\n   and every additional pair (–ℓ_{2j}+ℓ_{2j+2}) adds at least 2, so\n        Xₙ ≥ 4.\n\n(a)  x = 1.  Impossible for n = 2 (since ℓ₂ ≥ 2) and for n ≥ 4 by the\n     preceding discussion. → 2023 points.\n\n(b)  x = 2.  Achievable only with n = 2 and (ℓ₁,ℓ₂) = (1,2), giving\n     P₂ = (2,1).  Thus (2 , y) with y ≥ 2 are impossible. → 2022 points.\n\n(c)  x = 3.  For n = 2 we can take (ℓ₁,ℓ₂) = (1,3) or (2,3) and obtain\n     (3,1) or (3,2).  For n ≥ 4 we have either Xₙ < 0 (m even) or\n     Xₙ ≥ 4 (m odd); hence no further point with x = 3 occurs.\n     So (3 , y) with y ≥ 3 are impossible. → 2021 points.\n\n────────────────────────────────────────────────────────\n3.  Realising all remaining lattice points.\n\n(a)  Points with 0 < y < x ≤ 2023.  Choose n = 2 and the lengths\n     (ℓ₁,ℓ₂) = (y , x); then P₂ = (x , y).\n\n(b)  Points with x ≥ 4 and y ≥ 3 (in either order).  Six steps work:\n          ℓ₁ = 1, ℓ₂ = 2, ℓ₃ = 3,\n          ℓ₄ = y + 1, ℓ₅ = y + 2, ℓ₆ = x + y – 1.\n     These are strictly increasing because x ≥ 4 and y ≥ 3.\n     Directly from (1):\n          X₆ = 2 – (y+1) + (x+y–1) = x,\n          Y₆ = 1 – 3 + (y+2)           = y,\n     so every such point is attainable.\n\n(c)  The three small first–quadrant points (2,1), (3,1), (3,2) were\n     realised in 2(b) and 2(c).\n\nThus every lattice point inside the 2023×2023 square that has not been\nruled out in Sections 1–2 \nreally is the end–point of a growing spiral.\n\n────────────────────────────────────────────────────────\n4.  Counting the impossible points.\n        (0 , y)          : 2024 points (0 ≤ y ≤ 2023)\n        (x , 0)          : 2023 points (1 ≤ x ≤ 2023)\n        (1 , y)          : 2023 points (1 ≤ y ≤ 2023)\n        (2 , y) , y ≥ 2  : 2022 points\n        (3 , y) , y ≥ 3  : 2021 points\nThese sets are disjoint, so the total number of impossible pairs is\n        2024 + 2023 + 2023 + 2022 + 2021 = 10113.\n\nAnswer: 10113.",
      "_meta": {
        "core_steps": [
          "Write x- and y-coordinates of P_n as alternating sums of successive edge-lengths.",
          "Parity of n forces x(P_n) ≠ 0, eliminating the entire y-axis (except the origin).",
          "For n ≥ 4 the same alternation shows y(P_n) < 0 or y(P_n) ≥ 4, so lattice points with 0 ≤ y ≤ 3 are reachable only (if at all) at n = 2 or 3.",
          "Directly check n = 2,3 to rule out the remaining low-y candidates; construct explicit length sequences (e.g. 1,2,3,x+1,x+2,x+y−1) to realize every other lattice point.",
          "Count the points ruled out by the two inequalities inside the chosen square to obtain the final answer."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Side length of the square in which P_n is to lie.",
            "original": "2011"
          },
          "slot2": {
            "description": "Orientation of the very first step (currently east); a uniform rotation of all four step directions leaves the argument intact up to the same rotation of the forbidden stripes.",
            "original": "east"
          },
          "slot3": {
            "description": "Lower bound from which y-coordinates become attainable again (comes from minimal possible y_1 + 2).",
            "original": "4"
          },
          "slot4": {
            "description": "Numerical totals arising only from the chosen square size (e.g., 2012, 2011, 10053).",
            "original": "10053"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "calculation",
  "iteratively_fixed": true
}