{
  "index": "2007-A-6",
  "type": "COMB",
  "tag": [
    "COMB",
    "GEO"
  ],
  "difficulty": "",
  "question": "A \\emph{triangulation} $\\mathcal{T}$ of a polygon $P$ is a finite\ncollection of triangles whose union is $P$, and such that the\nintersection of any two triangles is either empty, or a shared vertex,\nor a shared side. Moreover, each side is a side of exactly one triangle\nin $\\mathcal{T}$. Say that $\\mathcal{T}$ is \\emph{admissible} if every\ninternal vertex is shared by 6 or more triangles. For example, [figure\nomitted.] Prove that there is an integer $M_n$, depending only on $n$,\nsuch that any admissible triangulation of a polygon $P$ with $n$ sides\nhas at most $M_n$ triangles.",
  "solution": "For an admissible triangulation $\\mathcal{T}$, number the\n  vertices of $P$ consecutively $v_1,\\dots,v_n$, and let $a_i$ be the\n  number of edges in $\\mathcal{T}$ emanating from $v_i$; note that\n  $a_i \\geq 2$ for all $i$.\n\n  We first claim that $a_1+\\dots+a_n \\leq 4n-6$. Let $V,E,F$ denote\n  the number of vertices, edges, and faces in $\\mathcal{T}$. By\n  Euler's Formula, $(F+1)-E+V=2$ (one must add 1 to the face count for the\n  region exterior to $P$). Each face has three edges, and each edge but the\n  $n$ outside edges belongs to two faces; hence $F = 2E-n$. On the\n  other hand, each edge has two endpoints, and each of the $V-n$\n  internal vertices is an endpoint of at least $6$ edges; hence\n  $a_1+\\dots+a_n+6(V-n) \\leq 2E$. Combining this inequality with the\n  previous two equations gives\n\\begin{align*}\na_1+\\dots+a_n &\\leq 2E+6n-6(1-F+E) \\\\\n&= 4n-6,\n\\end{align*}\nas claimed.\n\nNow set $A_3 = 1$ and $A_n = A_{n-1}+2n-3$ for $n \\geq 4$; we will\nprove by induction on $n$ that $\\mathcal{T}$ has at most $A_n$\ntriangles. For $n=3$, since $a_1+a_2+a_3=6$, $a_1=a_2=a_3=2$ and hence\n$\\mathcal{T}$ consists of just one triangle.\n\nNext assume that an admissible triangulation of an $(n-1)$-gon has at\nmost $A_{n-1}$ triangles, and let $\\mathcal{T}$ be an admissible\ntriangulation of an $n$-gon. If any $a_i = 2$, then we can remove the\ntriangle of $\\mathcal{T}$ containing vertex $v_i$ to obtain an\nadmissible triangulation of an $(n-1)$-gon; then the number of\ntriangles in $\\mathcal{T}$ is at most $A_{n-1}+1 < A_n$ by induction.\nOtherwise, all $a_i \\geq 3$. Now the average of $a_1,\\dots,a_n$ is\nless than $4$, and thus there are more $a_i = 3$ than $a_i \\geq 5$. It\nfollows that there is a sequence of $k$ consecutive vertices in $P$\nwhose degrees are $3,4,4,\\ldots,4,3$ in order, for some $k$ with\n$2\\leq k\\leq n-1$ (possibly $k=2$, in which case there are no degree\n$4$ vertices separating the degree $3$ vertices). If we remove from\n$\\mathcal{T}$ the $2k-1$ triangles which contain at least one of these\nvertices, then we are left with an admissible triangulation of an\n$(n-1)$-gon. It follows that there are at most $A_{n-1}+2k-1 \\leq\nA_{n-1}+2n-3 = A_n$ triangles in $\\mathcal{T}$. This completes the\ninduction step and the proof.\n\n\\textbf{Remark:}\nWe can refine the bound $A_n$ somewhat.\nSupposing that $a_i \\geq 3$ for all\n$i$, the fact that $a_1 + \\cdots + a_n \\leq 4n-6$ implies that there are at least\nsix more indices $i$ with $a_i = 3$ than with $a_i \\geq 5$. Thus there exist\nsix sequences with degrees $3,4,\\dots,4,3$, of total length at most\n$n+6$. We may thus choose a sequence of length $k \\leq \\lfloor \\frac{n}{6}\n\\rfloor + 1$, so we may improve the upper bound to $A_n = A_{n-1} +\n2 \\lfloor \\frac{n}{6} \\rfloor + 1$, or asymptotically\n$\\frac{1}{6} n^2$.\n\nHowever (as noted by Noam Elkies), a hexagonal swatch of a triangular\nlattice, with the boundary as close to regular as possible,\nachieves asymptotically $\\frac{1}{6} n^2$ triangles.",
  "vars": [
    "P",
    "T",
    "v_1",
    "v_n",
    "v_i",
    "a_i",
    "V",
    "E",
    "F",
    "k",
    "i"
  ],
  "params": [
    "n",
    "M_n",
    "A_3",
    "A_n",
    "A_n-1"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "P": "polygonset",
        "T": "triangset",
        "v_1": "vertexone",
        "v_n": "vertexn",
        "v_i": "vertexvar",
        "a_i": "emanating",
        "V": "vertcount",
        "E": "edgecount",
        "F": "facecount",
        "k": "seqlength",
        "i": "indexvar",
        "n": "sidescount",
        "M_n": "boundval",
        "A_3": "basetriang",
        "A_n": "recurtriang",
        "A_n-1": "prevtriang"
      },
      "question": "A \\emph{triangulation} $\\text{triangset}$ of a polygon $\\text{polygonset}$ is a finite collection of triangles whose union is $\\text{polygonset}$, and such that the intersection of any two triangles is either empty, or a shared vertex, or a shared side. Moreover, each side is a side of exactly one triangle in $\\text{triangset}$. Say that $\\text{triangset}$ is \\emph{admissible} if every internal vertex is shared by 6 or more triangles. For example, [figure omitted.] Prove that there is an integer $\\text{boundval}$, depending only on $\\text{sidescount}$, such that any admissible triangulation of a polygon $\\text{polygonset}$ with $\\text{sidescount}$ sides has at most $\\text{boundval}$ triangles.",
      "solution": "For an admissible triangulation $\\text{triangset}$, number the vertices of $\\text{polygonset}$ consecutively $\\text{vertexone},\\dots,\\text{vertexn}$, and let $\\text{emanating}$ be the number of edges in $\\text{triangset}$ emanating from $\\text{vertexvar}$; note that $\\text{emanating} \\ge 2$ for all $\\text{indexvar}$.\\n\\nWe first claim that $a_1+\\dots+a_{\\text{sidescount}} \\le 4\\text{sidescount}-6$. Let $\\text{vertcount},\\text{edgecount},\\text{facecount}$ denote the number of vertices, edges, and faces in $\\text{triangset}$. By Euler\\'s Formula, $(\\text{facecount}+1)-\\text{edgecount}+\\text{vertcount}=2$ (one must add 1 to the face count for the region exterior to $\\text{polygonset}$). Each face has three edges, and each edge but the $\\text{sidescount}$ outside edges belongs to two faces; hence $\\text{facecount}=2\\text{edgecount}-\\text{sidescount}$. On the other hand, each edge has two endpoints, and each of the $\\text{vertcount}-\\text{sidescount}$ internal vertices is an endpoint of at least $6$ edges; hence $a_1+\\dots+a_{\\text{sidescount}}+6(\\text{vertcount}-\\text{sidescount}) \\le 2\\text{edgecount}$. Combining this inequality with the previous two equations gives\\n\\begin{align*}\\na_1+\\dots+a_{\\text{sidescount}} &\\le 2\\text{edgecount}+6\\text{sidescount}-6(1-\\text{facecount}+\\text{edgecount}) \\\\ &= 4\\text{sidescount}-6,\\end{align*}\\nas claimed.\\n\\nNow set $\\text{basetriang}=1$ and $\\text{recurtriang}=\\text{prevtriang}+2\\text{sidescount}-3$ for $\\text{sidescount}\\ge4$; we will prove by induction on $\\text{sidescount}$ that $\\text{triangset}$ has at most $\\text{recurtriang}$ triangles. For $\\text{sidescount}=3$, since $a_1+a_2+a_3=6$, $a_1=a_2=a_3=2$ and hence $\\text{triangset}$ consists of just one triangle.\\n\\nNext assume that an admissible triangulation of an $(\\text{sidescount}-1)$-gon has at most $\\text{prevtriang}$ triangles, and let $\\text{triangset}$ be an admissible triangulation of an $\\text{sidescount}$-gon. If any $\\text{emanating}=2$, then we can remove the triangle of $\\text{triangset}$ containing vertex $\\text{vertexvar}$ to obtain an admissible triangulation of an $(\\text{sidescount}-1)$-gon; then the number of triangles in $\\text{triangset}$ is at most $\\text{prevtriang}+1<\\text{recurtriang}$ by induction. Otherwise, all $a_{\\text{indexvar}}\\ge3$. Now the average of $a_1,\\dots,a_{\\text{sidescount}}$ is less than $4$, and thus there are more $a_{\\text{indexvar}}=3$ than $a_{\\text{indexvar}}\\ge5$. It follows that there is a sequence of $\\text{seqlength}$ consecutive vertices in $\\text{polygonset}$ whose degrees are $3,4,4,\\ldots,4,3$ in order, for some $\\text{seqlength}$ with $2\\le\\text{seqlength}\\le\\text{sidescount}-1$ (possibly $\\text{seqlength}=2$, in which case there are no degree $4$ vertices separating the degree $3$ vertices). If we remove from $\\text{triangset}$ the $2\\text{seqlength}-1$ triangles which contain at least one of these vertices, then we are left with an admissible triangulation of an $(\\text{sidescount}-1)$-gon. It follows that there are at most $\\text{prevtriang}+2\\text{seqlength}-1 \\le \\text{prevtriang}+2\\text{sidescount}-3=\\text{recurtriang}$ triangles in $\\text{triangset}$. This completes the induction step and the proof.\\n\\n\\textbf{Remark:} We can refine the bound $\\text{recurtriang}$ somewhat. Supposing that $a_{\\text{indexvar}}\\ge3$ for all $\\text{indexvar}$, the fact that $a_1+\\cdots+a_{\\text{sidescount}}\\le4\\text{sidescount}-6$ implies that there are at least six more indices $\\text{indexvar}$ with $a_{\\text{indexvar}}=3$ than with $a_{\\text{indexvar}}\\ge5$. Thus there exist six sequences with degrees $3,4,\\dots,4,3$, of total length at most $\\text{sidescount}+6$. We may thus choose a sequence of length $\\text{seqlength}\\le\\lfloor\\frac{\\text{sidescount}}{6}\\rfloor+1$, so we may improve the upper bound to $\\text{recurtriang}=\\text{prevtriang}+2\\lfloor\\frac{\\text{sidescount}}{6}\\rfloor+1$, or asymptotically $\\frac{1}{6}\\text{sidescount}^2$.\\n\\nHowever (as noted by Noam Elkies), a hexagonal swatch of a triangular lattice, with the boundary as close to regular as possible, achieves asymptotically $\\frac{1}{6}\\text{sidescount}^2$ triangles."
    },
    "descriptive_long_confusing": {
      "map": {
        "P": "dandelion",
        "T": "strawberry",
        "v_1": "narwhalone",
        "v_n": "narwhalnest",
        "v_i": "narwhaliris",
        "a_i": "pomegranate",
        "V": "watermelon",
        "E": "buttercup",
        "F": "canarybird",
        "k": "platypusk",
        "i": "iguanaeye",
        "n": "lemontree",
        "M_n": "raspberry",
        "A_3": "blackberry",
        "A_n": "blueberrie",
        "A_n-1": "cranberrie"
      },
      "question": "A \\emph{triangulation} $\\mathcal{strawberry}$ of a polygon $dandelion$ is a finite\ncollection of triangles whose union is $dandelion$, and such that the\nintersection of any two triangles is either empty, or a shared vertex,\nor a shared side. Moreover, each side is a side of exactly one triangle\nin $\\mathcal{strawberry}$. Say that $\\mathcal{strawberry}$ is \\emph{admissible} if every\ninternal vertex is shared by 6 or more triangles. For example, [figure\nomitted.] Prove that there is an integer $raspberry$, depending only on $lemontree$,\nsuch that any admissible triangulation of a polygon $dandelion$ with $lemontree$ sides\nhas at most $raspberry$ triangles.",
      "solution": "For an admissible triangulation $\\mathcal{strawberry}$, number the\n  vertices of $dandelion$ consecutively $narwhalone,\\dots,narwhalnest$, and let $pomegranate$ be the\n  number of edges in $\\mathcal{strawberry}$ emanating from $narwhaliris$; note that\n  $pomegranate \\geq 2$ for all $iguanaeye$.\n\n  We first claim that $a_1+\\dots+a_{lemontree} \\leq 4lemontree-6$. Let $watermelon,buttercup,canarybird$ denote\n  the number of vertices, edges, and faces in $\\mathcal{strawberry}$. By\n  Euler's Formula, $(canarybird+1)-buttercup+watermelon=2$ (one must add 1 to the face count for the\n  region exterior to $dandelion$). Each face has three edges, and each edge but the\n  $lemontree$ outside edges belongs to two faces; hence $canarybird = 2buttercup-lemontree$. On the\n  other hand, each edge has two endpoints, and each of the $watermelon-lemontree$\n  internal vertices is an endpoint of at least $6$ edges; hence\n  $a_1+\\dots+a_{lemontree}+6(watermelon-lemontree) \\leq 2buttercup$. Combining this inequality with the\n  previous two equations gives\n\\begin{align*}\na_1+\\dots+a_{lemontree} &\\leq 2buttercup+6lemontree-6(1-canarybird+buttercup) \\\\\n&= 4lemontree-6,\n\\end{align*}\nas claimed.\n\nNow set $blackberry = 1$ and $blueberrie = blueberrie-1+2lemontree-3$ for $lemontree \\geq 4$; we will\nprove by induction on $lemontree$ that $\\mathcal{strawberry}$ has at most $blueberrie$\ntriangles. For $lemontree=3$, since $a_1+a_2+a_3=6$, $a_1=a_2=a_3=2$ and hence\n$\\mathcal{strawberry}$ consists of just one triangle.\n\nNext assume that an admissible triangulation of an $(lemontree-1)$-gon has at\nmost $blueberrie-1$ triangles, and let $\\mathcal{strawberry}$ be an admissible\ntriangulation of an $lemontree$-gon. If any $pomegranate = 2$, then we can remove the\ntriangle of $\\mathcal{strawberry}$ containing vertex $narwhaliris$ to obtain an\nadmissible triangulation of an $(lemontree-1)$-gon; then the number of\ntriangles in $\\mathcal{strawberry}$ is at most $blueberrie-1+1 < blueberrie$ by induction.\nOtherwise, all $pomegranate \\geq 3$. Now the average of $a_1,\\dots,a_{lemontree}$ is\nless than $4$, and thus there are more $pomegranate = 3$ than $pomegranate \\geq 5$. It\nfollows that there is a sequence of $platypusk$ consecutive vertices in $dandelion$\nwhose degrees are $3,4,4,\\ldots,4,3$ in order, for some $platypusk$ with\n$2\\leq platypusk\\leq lemontree-1$ (possibly $platypusk=2$, in which case there are no degree\n$4$ vertices separating the degree $3$ vertices). If we remove from\n$\\mathcal{strawberry}$ the $2platypusk-1$ triangles which contain at least one of these\nvertices, then we are left with an admissible triangulation of an\n$(lemontree-1)$-gon. It follows that there are at most $blueberrie-1+2platypusk-1 \\leq\nblueberrie-1+2lemontree-3 = blueberrie$ triangles in $\\mathcal{strawberry}$. This completes the\ninduction step and the proof.\n\n\\textbf{Remark:}\nWe can refine the bound $blueberrie$ somewhat.\nSupposing that $pomegranate \\geq 3$ for all\n$iguanaeye$, the fact that $a_1 + \\cdots + a_{lemontree} \\leq 4lemontree-6$ implies that there are at least\nsix more indices $iguanaeye$ with $pomegranate = 3$ than with $pomegranate \\geq 5$. Thus there exist\nsix sequences with degrees $3,4,\\dots,4,3$, of total length at most\n$lemontree+6$. We may thus choose a sequence of length $platypusk \\leq \\lfloor \\frac{lemontree}{6}\n\\rfloor + 1$, so we may improve the upper bound to $blueberrie = blueberrie-1 +\n2 \\lfloor \\frac{lemontree}{6} \\rfloor + 1$, or asymptotically\n$\\frac{1}{6} lemontree^2$.\n\nHowever (as noted by Noam Elkies), a hexagonal swatch of a triangular\nlattice, with the boundary as close to regular as possible,\nachieves asymptotically $\\frac{1}{6} lemontree^2$ triangles."
    },
    "descriptive_long_misleading": {
      "map": {
        "P": "circleregion",
        "T": "chaosmesh",
        "v_1": "outsiderone",
        "v_n": "outsiderend",
        "v_i": "outsiderindex",
        "a_i": "gapcount",
        "V": "emptysize",
        "E": "voidlinks",
        "F": "hollowset",
        "k": "boundless",
        "i": "allindex",
        "n": "zerocount",
        "M_n": "leastbound",
        "A_3": "oppositeone",
        "A_n": "oppositeall",
        "A_n-1": "oppositeprev"
      },
      "question": "A \\emph{triangulation} $\\mathcal{chaosmesh}$ of a polygon $circleregion$ is a finite\ncollection of triangles whose union is $circleregion$, and such that the\nintersection of any two triangles is either empty, or a shared vertex,\nor a shared side. Moreover, each side is a side of exactly one triangle\nin $\\mathcal{chaosmesh}$. Say that $\\mathcal{chaosmesh}$ is \\emph{admissible} if every\ninternal vertex is shared by 6 or more triangles. For example, [figure\nomitted.] Prove that there is an integer $leastbound$, depending only on $zerocount$,\nsuch that any admissible triangulation of a polygon $circleregion$ with $zerocount$ sides\nhas at most $leastbound$ triangles.",
      "solution": "For an admissible triangulation $\\mathcal{chaosmesh}$, number the\nvertices of $circleregion$ consecutively $outsiderone,\\dots,outsiderend$, and let $gapcount$ be the\nnumber of edges in $\\mathcal{chaosmesh}$ emanating from $outsiderindex$; note that\n$gapcount \\geq 2$ for all $allindex$.\n\nWe first claim that $a_1+\\dots+a_{zerocount} \\leq 4zerocount-6$. Let $emptysize,voidlinks,hollowset$ denote\nthe number of vertices, edges, and faces in $\\mathcal{chaosmesh}$. By\nEuler's Formula, $(hollowset+1)-voidlinks+emptysize=2$ (one must add 1 to the face count for the\nregion exterior to $circleregion$). Each face has three edges, and each edge but the\n$zerocount$ outside edges belongs to two faces; hence $hollowset = 2voidlinks-zerocount$. On the\nother hand, each edge has two endpoints, and each of the $emptysize-zerocount$\ninternal vertices is an endpoint of at least $6$ edges; hence\n$a_1+\\dots+a_{zerocount}+6(emptysize-zerocount) \\leq 2voidlinks$. Combining this inequality with the\nprevious two equations gives\n\\begin{align*}\na_1+\\dots+a_{zerocount} &\\leq 2voidlinks+6zerocount-6(1-hollowset+voidlinks) \\\\\n&= 4zerocount-6,\n\\end{align*}\nas claimed.\n\nNow set $oppositeone = 1$ and $oppositeall = oppositeprev+2zerocount-3$ for $zerocount \\geq 4$; we will\nprove by induction on $zerocount$ that $\\mathcal{chaosmesh}$ has at most $oppositeall$\ntriangles. For $zerocount=3$, since $a_1+a_2+a_3=6$, $a_1=a_2=a_3=2$ and hence\n$\\mathcal{chaosmesh}$ consists of just one triangle.\n\nNext assume that an admissible triangulation of an $(zerocount-1)$-gon has at\nmost $oppositeprev$ triangles, and let $\\mathcal{chaosmesh}$ be an admissible\ntriangulation of an $zerocount$-gon. If any $gapcount = 2$, then we can remove the\ntriangle of $\\mathcal{chaosmesh}$ containing vertex $outsiderindex$ to obtain an\nadmissible triangulation of an $(zerocount-1)$-gon; then the number of\ntriangles in $\\mathcal{chaosmesh}$ is at most $oppositeprev+1 < oppositeall$ by induction.\nOtherwise, all $gapcount \\geq 3$. Now the average of $a_1,\\dots,a_{zerocount}$ is\nless than $4$, and thus there are more $gapcount = 3$ than $gapcount \\geq 5$. It\nfollows that there is a sequence of $boundless$ consecutive vertices in $circleregion$\nwhose degrees are $3,4,4,\\ldots,4,3$ in order, for some $boundless$ with\n$2\\leq boundless\\leq zerocount-1$ (possibly $boundless=2$, in which case there are no degree\n$4$ vertices separating the degree $3$ vertices). If we remove from\n$\\mathcal{chaosmesh}$ the $2boundless-1$ triangles which contain at least one of these\nvertices, then we are left with an admissible triangulation of an\n$(zerocount-1)$-gon. It follows that there are at most $oppositeprev+2boundless-1 \\leq\noppositeprev+2zerocount-3 = oppositeall$ triangles in $\\mathcal{chaosmesh}$. This completes the\ninduction step and the proof.\n\n\\textbf{Remark:}\nWe can refine the bound $oppositeall$ somewhat.\nSupposing that $gapcount \\geq 3$ for all\n$allindex$, the fact that $a_1 + \\cdots + a_{zerocount} \\leq 4zerocount-6$ implies that there are at least\nsix more indices $allindex$ with $gapcount = 3$ than with $gapcount \\geq 5$. Thus there exist\nsix sequences with degrees $3,4,\\dots,4,3$, of total length at most\n$zerocount+6$. We may thus choose a sequence of length $boundless \\leq \\lfloor \\frac{zerocount}{6}\\rfloor + 1$, so we may improve the upper bound to $oppositeall = oppositeprev +\n2 \\lfloor \\frac{zerocount}{6} \\rfloor + 1$, or asymptotically\n$\\frac{1}{6} zerocount^2$.\n\nHowever (as noted by Noam Elkies), a hexagonal swatch of a triangular\nlattice, with the boundary as close to regular as possible,\nachieves asymptotically $\\frac{1}{6} zerocount^2$ triangles."
    },
    "garbled_string": {
      "map": {
        "P": "xbqvnepla",
        "T": "lkjrtmosa",
        "v_1": "sdfweqtra",
        "v_n": "plmzbxcva",
        "v_i": "vhqntysda",
        "a_i": "jzsorucna",
        "V": "oubxrmefa",
        "E": "fkyapudzs",
        "F": "wliqznmta",
        "k": "gphsraecl",
        "i": "bercuvhqa",
        "n": "yjkdapmen",
        "M_n": "qorxsuvna",
        "A_3": "kjdhlmawe",
        "A_n": "wprcxtneb",
        "A_n-1": "gnopvlsae"
      },
      "question": "A \\emph{triangulation} $\\mathcal{lkjrtmosa}$ of a polygon $xbqvnepla$ is a finite\ncollection of triangles whose union is $xbqvnepla$, and such that the\nintersection of any two triangles is either empty, or a shared vertex,\nor a shared side. Moreover, each side is a side of exactly one triangle\nin $\\mathcal{lkjrtmosa}$. Say that $\\mathcal{lkjrtmosa}$ is \\emph{admissible} if every\ninternal vertex is shared by 6 or more triangles. For example, [figure\nomitted.] Prove that there is an integer $qorxsuvna$, depending only on $yjkdapmen$,\nsuch that any admissible triangulation of a polygon $xbqvnepla$ with $yjkdapmen$ sides\nhas at most $qorxsuvna$ triangles.",
      "solution": "For an admissible triangulation $\\mathcal{lkjrtmosa}$, number the\n  vertices of $xbqvnepla$ consecutively $sdfweqtra,\\dots,plmzbxcva$, and let $jzsorucna$ be the\n  number of edges in $\\mathcal{lkjrtmosa}$ emanating from $vhqntysda$; note that\n  $jzsorucna \\geq 2$ for all $bercuvhqa$.\n\n  We first claim that $a_1+\\dots+a_n \\leq 4yjkdapmen-6$. Let $oubxrmefa,fkyapudzs,wliqznmta$ denote\n  the number of vertices, edges, and faces in $\\mathcal{lkjrtmosa}$. By\n  Euler's Formula, $(wliqznmta+1)-fkyapudzs+oubxrmefa=2$ (one must add 1 to the face count for the\n  region exterior to $xbqvnepla$). Each face has three edges, and each edge but the\n  $yjkdapmen$ outside edges belongs to two faces; hence $wliqznmta = 2fkyapudzs-yjkdapmen$. On the\n  other hand, each edge has two endpoints, and each of the $oubxrmefa-yjkdapmen$\n  internal vertices is an endpoint of at least $6$ edges; hence\n  $a_1+\\dots+a_n+6(oubxrmefa-yjkdapmen) \\leq 2fkyapudzs$. Combining this inequality with the\n  previous two equations gives\n\\begin{align*}\na_1+\\dots+a_n &\\leq 2fkyapudzs+6yjkdapmen-6(1-wliqznmta+fkyapudzs) \\\\\n&= 4yjkdapmen-6,\n\\end{align*}\nas claimed.\n\nNow set $kjdhlmawe = 1$ and $wprcxtneb = gnopvlsae+2yjkdapmen-3$ for $yjkdapmen \\geq 4$; we will\nprove by induction on $yjkdapmen$ that $\\mathcal{lkjrtmosa}$ has at most $wprcxtneb$\ntriangles. For $yjkdapmen=3$, since $a_1+a_2+a_3=6$, $a_1=a_2=a_3=2$ and hence\n$\\mathcal{lkjrtmosa}$ consists of just one triangle.\n\nNext assume that an admissible triangulation of an $(yjkdapmen-1)$-gon has at\nmost $gnopvlsae$ triangles, and let $\\mathcal{lkjrtmosa}$ be an admissible\ntriangulation of an $yjkdapmen$-gon. If any $jzsorucna = 2$, then we can remove the\ntriangle of $\\mathcal{lkjrtmosa}$ containing vertex $vhqntysda$ to obtain an\nadmissible triangulation of an $(yjkdapmen-1)$-gon; then the number of\ntriangles in $\\mathcal{lkjrtmosa}$ is at most $gnopvlsae+1 < wprcxtneb$ by induction.\nOtherwise, all $jzsorucna \\geq 3$. Now the average of $a_1,\\dots,a_n$ is\nless than $4$, and thus there are more $jzsorucna = 3$ than $jzsorucna \\geq 5$. It\nfollows that there is a sequence of $gphsraecl$ consecutive vertices in $xbqvnepla$\nwhose degrees are $3,4,4,\\ldots,4,3$ in order, for some $gphsraecl$ with\n$2\\leq gphsraecl\\leq yjkdapmen-1$ (possibly $gphsraecl=2$, in which case there are no degree\n$4$ vertices separating the degree $3$ vertices). If we remove from\n$\\mathcal{lkjrtmosa}$ the $2gphsraecl-1$ triangles which contain at least one of these\nvertices, then we are left with an admissible triangulation of an\n$(yjkdapmen-1)$-gon. It follows that there are at most $gnopvlsae+2gphsraecl-1 \\leq\ngnopvlsae+2yjkdapmen-3 = wprcxtneb$ triangles in $\\mathcal{lkjrtmosa}$. This completes the\ninduction step and the proof.\n\n\\textbf{Remark:}\nWe can refine the bound $wprcxtneb$ somewhat.\nSupposing that $jzsorucna \\geq 3$ for all\n$bercuvhqa$, the fact that $a_1 + \\cdots + a_n \\leq 4yjkdapmen-6$ implies that there are at least\nsix more indices $bercuvhqa$ with $jzsorucna = 3$ than with $jzsorucna \\geq 5$. Thus there exist\nsix sequences with degrees $3,4,\\dots,4,3$, of total length at most\n$yjkdapmen+6$. We may thus choose a sequence of length $gphsraecl \\leq \\lfloor \\frac{yjkdapmen}{6}\n\\rfloor + 1$, so we may improve the upper bound to $wprcxtneb = gnopvlsae +\n2 \\lfloor \\frac{yjkdapmen}{6} \\rfloor + 1$, or asymptotically\n$\\frac{1}{6} yjkdapmen^2$.\n\nHowever (as noted by Noam Elkies), a hexagonal swatch of a triangular\nlattice, with the boundary as close to regular as possible,\nachieves asymptotically $\\frac{1}{6} yjkdapmen^2$ triangles."
    },
    "kernel_variant": {
      "question": "Fix two non-negative integers  \n\n  g  (orientable genus) and b ≥ 1 (number of boundary components).\n\nFor 1 ≤ j ≤ b let P_j be a simple polygon with n_j ≥ 3 sides, drawn pairwise disjointly in the oriented surface Σ_g (of genus g) in such a way that the closure  \n\n  Ω := Σ_g \\ ⋃_{j=1}^{\\,b} Int P_j  \n\nis a compact surface whose boundary  \n\n  ∂Ω = P_1 ∪⋯∪ P_b  \n\nconsists of the N := n_1+⋯+n_b edges of the polygons P_j.  \nA finite family 𝒯 of (topological) triangles is called a triangulation of Ω if  \n\n(1) ⋃_{Δ∈𝒯} Δ = Ω;  \n(2) any two distinct triangles intersect, if at all, only in a common vertex or a whole edge;  \n(3) every boundary edge of Ω belongs to exactly one triangle of 𝒯.\n\nWrite F = |𝒯|, E for the number of edges, V for the number of vertices, and V_b := N for the boundary vertices.  \nCall 𝒯 hyper-admissible if each interior vertex is incident with at least  \n\n  d := 9 triangles.\n\nProve that every hyper-admissible triangulation of Ω satisfies  \n\n  F ≤ M_{N,g,b} := ⌊ (7N)/3 + 6b + 12g – 12 ⌋.\n\n(Consequently, for each fixed pair (g,b) the size of a hyper-admissible triangulation of Ω is bounded above by a constant depending only on N.)",
      "solution": "Step 0.  Notation and basic relations  \nLet E_i and E_b (= N) be the numbers of interior and boundary edges respectively, and put V_i := V – N.\n\nEuler characteristic.  \nBecause Ω has genus g and b boundary components,  \n\n  χ(Ω) = 2 – 2g – b.                                     (1)\n\nCounting edges through faces.  \nEvery triangle contributes three edges; an interior edge is shared by two triangles whereas a boundary edge belongs to only one. Hence  \n\n  3F = 2E_i + N ⇒ E = E_i+N = (3F+N)/2.                (2)\n\nCounting vertices with Euler.  \nUsing V – E + F = χ(Ω) together with (2) we obtain  \n\n V = χ(Ω) – F + E = χ(Ω) – F + (3F+N)/2  \n  = χ(Ω) + ½(F+N).                                      (3)\n\nStep 1.  Relating boundary degrees to F  \nDenote by S := ∑_{v∈∂Ω}deg(v) the sum of the degrees of the boundary vertices.  \nBecause every interior vertex has degree at least d = 9,\n\n  2E = Σ_{v all}deg(v) ≥ S + dV_i.                       (4)\n\nSubstituting V_i = V – N from (3) and E from (2) yields\n\n S ≤ 2E – d(V – N)  \n = (3F+N) – d(χ(Ω)+½(F+N) – N)  \n = (3 – d/2)F + (1 + d/2)N – dχ(Ω).                      (5)\n\nFor d = 9 the coefficient 3 – d/2 equals –3/2, i.e. negative.\n\nStep 2.  A crude lower bound on S  \nEach boundary vertex belongs to its two boundary edges, so S ≥ 2N.  \nCombining this with (5) gives  \n\n 2N ≤ (3 – d/2)F + (1 + d/2)N – dχ(Ω).                  (6)\n\nBecause 3 – d/2 = –(d/2 – 3), rewrite (6) as\n\n (d/2 – 3)F ≤ (d/2 – 1)N – dχ(Ω).                        (7)\n\nStep 3.  Substituting χ(Ω) and simplifying  \nInsert d = 9 (so d/2 = 9/2) into (7):\n\n (3/2)F ≤ (7/2)N – 9χ(Ω).   Multiply by 2:               (8)\n\n 3F ≤ 7N – 18χ(Ω).\n\nNow use χ(Ω) = 2 – 2g – b from (1):\n\n 3F ≤ 7N – 18(2 – 2g – b)  \n  = 7N – 36 + 36g + 18b.                                (9)\n\nStep 4.  Final bound  \nDividing by 3 gives\n\n F ≤ (7N)/3 + 12g + 6b – 12.                            (10)\n\nBecause F is an integer, (10) implies\n\n F ≤ ⌊ (7N)/3 + 6b + 12g – 12 ⌋ = M_{N,g,b},\n\nas desired. ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.806598",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher topological complexity.  \n   The setting now involves an arbitrary orientable genus g and an arbitrary number b of boundary components.  Handling Euler characteristic χ(Ω)=2–2g–b instead of the disk’s χ=1 requires extra care and introduces new terms (the 12g and 6b contributions) in the final bound.\n\n2. More variables.  \n   The bound must depend simultaneously on N, g and b, so the entire count has to track three independent parameters instead of one.\n\n3. Stronger local constraint.  \n   The minimum interior degree is raised from 6 (original) or 8 (kernel variant) to 9, altering all incidence inequalities and making the coefficients in the counting identities sensitive to sign.\n\n4. Deeper combinatorial algebra.  \n   Balancing the negative coefficient of F obtained in (5) against a \\emph{lower} bound for S is subtler than the original argument, which used an \\emph{upper} bound on the boundary-degree sum.  The new direction of the inequality forces a different manipulation and more intricate algebra.\n\n5. General formula.  \n   The solution produces an explicit closed form for M_{N,g,b} valid for every combination of (g,b), not just the planar case; deriving and simplifying this universal expression demands careful bookkeeping throughout.\n\nFor these reasons the enhanced variant is substantially more technical and conceptually richer than both the original problem and the previous kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Fix two non-negative integers  \n\n  g  (orientable genus) and b ≥ 1 (number of boundary components).\n\nFor 1 ≤ j ≤ b let P_j be a simple polygon with n_j ≥ 3 sides, drawn pairwise disjointly in the oriented surface Σ_g (of genus g) in such a way that the closure  \n\n  Ω := Σ_g \\ ⋃_{j=1}^{\\,b} Int P_j  \n\nis a compact surface whose boundary  \n\n  ∂Ω = P_1 ∪⋯∪ P_b  \n\nconsists of the N := n_1+⋯+n_b edges of the polygons P_j.  \nA finite family 𝒯 of (topological) triangles is called a triangulation of Ω if  \n\n(1) ⋃_{Δ∈𝒯} Δ = Ω;  \n(2) any two distinct triangles intersect, if at all, only in a common vertex or a whole edge;  \n(3) every boundary edge of Ω belongs to exactly one triangle of 𝒯.\n\nWrite F = |𝒯|, E for the number of edges, V for the number of vertices, and V_b := N for the boundary vertices.  \nCall 𝒯 hyper-admissible if each interior vertex is incident with at least  \n\n  d := 9 triangles.\n\nProve that every hyper-admissible triangulation of Ω satisfies  \n\n  F ≤ M_{N,g,b} := ⌊ (7N)/3 + 6b + 12g – 12 ⌋.\n\n(Consequently, for each fixed pair (g,b) the size of a hyper-admissible triangulation of Ω is bounded above by a constant depending only on N.)",
      "solution": "Step 0.  Notation and basic relations  \nLet E_i and E_b (= N) be the numbers of interior and boundary edges respectively, and put V_i := V – N.\n\nEuler characteristic.  \nBecause Ω has genus g and b boundary components,  \n\n  χ(Ω) = 2 – 2g – b.                                     (1)\n\nCounting edges through faces.  \nEvery triangle contributes three edges; an interior edge is shared by two triangles whereas a boundary edge belongs to only one. Hence  \n\n  3F = 2E_i + N ⇒ E = E_i+N = (3F+N)/2.                (2)\n\nCounting vertices with Euler.  \nUsing V – E + F = χ(Ω) together with (2) we obtain  \n\n V = χ(Ω) – F + E = χ(Ω) – F + (3F+N)/2  \n  = χ(Ω) + ½(F+N).                                      (3)\n\nStep 1.  Relating boundary degrees to F  \nDenote by S := ∑_{v∈∂Ω}deg(v) the sum of the degrees of the boundary vertices.  \nBecause every interior vertex has degree at least d = 9,\n\n  2E = Σ_{v all}deg(v) ≥ S + dV_i.                       (4)\n\nSubstituting V_i = V – N from (3) and E from (2) yields\n\n S ≤ 2E – d(V – N)  \n = (3F+N) – d(χ(Ω)+½(F+N) – N)  \n = (3 – d/2)F + (1 + d/2)N – dχ(Ω).                      (5)\n\nFor d = 9 the coefficient 3 – d/2 equals –3/2, i.e. negative.\n\nStep 2.  A crude lower bound on S  \nEach boundary vertex belongs to its two boundary edges, so S ≥ 2N.  \nCombining this with (5) gives  \n\n 2N ≤ (3 – d/2)F + (1 + d/2)N – dχ(Ω).                  (6)\n\nBecause 3 – d/2 = –(d/2 – 3), rewrite (6) as\n\n (d/2 – 3)F ≤ (d/2 – 1)N – dχ(Ω).                        (7)\n\nStep 3.  Substituting χ(Ω) and simplifying  \nInsert d = 9 (so d/2 = 9/2) into (7):\n\n (3/2)F ≤ (7/2)N – 9χ(Ω).   Multiply by 2:               (8)\n\n 3F ≤ 7N – 18χ(Ω).\n\nNow use χ(Ω) = 2 – 2g – b from (1):\n\n 3F ≤ 7N – 18(2 – 2g – b)  \n  = 7N – 36 + 36g + 18b.                                (9)\n\nStep 4.  Final bound  \nDividing by 3 gives\n\n F ≤ (7N)/3 + 12g + 6b – 12.                            (10)\n\nBecause F is an integer, (10) implies\n\n F ≤ ⌊ (7N)/3 + 6b + 12g – 12 ⌋ = M_{N,g,b},\n\nas desired. ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.616536",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher topological complexity.  \n   The setting now involves an arbitrary orientable genus g and an arbitrary number b of boundary components.  Handling Euler characteristic χ(Ω)=2–2g–b instead of the disk’s χ=1 requires extra care and introduces new terms (the 12g and 6b contributions) in the final bound.\n\n2. More variables.  \n   The bound must depend simultaneously on N, g and b, so the entire count has to track three independent parameters instead of one.\n\n3. Stronger local constraint.  \n   The minimum interior degree is raised from 6 (original) or 8 (kernel variant) to 9, altering all incidence inequalities and making the coefficients in the counting identities sensitive to sign.\n\n4. Deeper combinatorial algebra.  \n   Balancing the negative coefficient of F obtained in (5) against a \\emph{lower} bound for S is subtler than the original argument, which used an \\emph{upper} bound on the boundary-degree sum.  The new direction of the inequality forces a different manipulation and more intricate algebra.\n\n5. General formula.  \n   The solution produces an explicit closed form for M_{N,g,b} valid for every combination of (g,b), not just the planar case; deriving and simplifying this universal expression demands careful bookkeeping throughout.\n\nFor these reasons the enhanced variant is substantially more technical and conceptually richer than both the original problem and the previous kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}