{
  "index": "1981-A-4",
  "type": "GEO",
  "tag": [
    "GEO",
    "COMB"
  ],
  "difficulty": "",
  "question": "Problem A-4\nA point \\( P \\) moves inside a unit square in a straight line at unit speed. When it meets a corner it escapes. When it meets an edge its line of motion is reflected so that the angle of incidence equals the angle of reflection.\n\nLet \\( N(T) \\) be the number of starting directions from a fixed interior point \\( P_{0} \\) for which \\( P \\) escapes within \\( T \\) units of time. Find the least constant \\( a \\) for which constants \\( b \\) and \\( c \\) exist such that\n\\[\nN(T) \\leq a T^{2}+b T+c\n\\]\nfor all \\( T>0 \\) and all initial points \\( P_{0} \\).",
  "solution": "A-4.\nSet up coordinates so that a vertex of the given unit square is \\( (0,0) \\) and two sides of the square are on the axes. Using the reflection properties, one can see that \\( P \\) escapes within \\( T \\) units of time if and only if the (infinite) ray from \\( P_{0} \\), with the direction of the first segment of the path, goes through a lattice point (point with integer coordinates) within \\( T \\) units of distance from \\( P_{0} \\). Thus \\( N(T) \\) is at most the number \\( L(T) \\) of lattice points in the circle with center at \\( P_{0} \\) and radius \\( T \\). Tiling the plane with unit squares having centers at the lattice points and considering areas, one sees that\n\\[\nN(T) \\leq L(T) \\leq \\pi[T+(\\sqrt{2} / 2)]^{2}\n\\]\n\nHence there is an upper bound for \\( N(T) \\) of the form \\( \\pi T^{2}+b T+c \\), with \\( b \\) and \\( c \\) fixed. When just one coordinate of \\( P_{0} \\) is irrational,\n\\[\nN(T)=L(T) \\geq \\pi[T-(\\sqrt{2} / 2)]^{2}\n\\]\n\nThis lower bound for \\( N(T) \\) exceeds \\( a T^{2}+b T+c \\) for sufficiently large \\( T \\) if \\( a<\\pi \\); hence \\( \\pi \\) is the desired \\( a \\).",
  "vars": [
    "P",
    "P_0",
    "N",
    "T",
    "L"
  ],
  "params": [
    "a",
    "b",
    "c"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "P": "movingpoint",
        "P_0": "startpoint",
        "N": "directioncount",
        "T": "elapsedtime",
        "L": "latticecount",
        "a": "quadraticcoef",
        "b": "linearcoef",
        "c": "constantcoef"
      },
      "question": "Problem A-4\nA point \\( movingpoint \\) moves inside a unit square in a straight line at unit speed. When it meets a corner it escapes. When it meets an edge its line of motion is reflected so that the angle of incidence equals the angle of reflection.\n\nLet \\( directioncount(elapsedtime) \\) be the number of starting directions from a fixed interior point \\( startpoint \\) for which \\( movingpoint \\) escapes within \\( elapsedtime \\) units of time. Find the least constant \\( quadraticcoef \\) for which constants \\( linearcoef \\) and \\( constantcoef \\) exist such that\n\\[\ndirectioncount(elapsedtime) \\leq quadraticcoef\\, elapsedtime^{2}+linearcoef\\, elapsedtime+constantcoef\n\\]\nfor all \\( elapsedtime>0 \\) and all initial points \\( startpoint \\).",
      "solution": "A-4.\nSet up coordinates so that a vertex of the given unit square is \\( (0,0) \\) and two sides of the square are on the axes. Using the reflection properties, one can see that \\( movingpoint \\) escapes within \\( elapsedtime \\) units of time if and only if the (infinite) ray from \\( startpoint \\), with the direction of the first segment of the path, goes through a lattice point (point with integer coordinates) within \\( elapsedtime \\) units of distance from \\( startpoint \\). Thus \\( directioncount(elapsedtime) \\) is at most the number \\( latticecount(elapsedtime) \\) of lattice points in the circle with center at \\( startpoint \\) and radius \\( elapsedtime \\). Tiling the plane with unit squares having centers at the lattice points and considering areas, one sees that\n\\[\ndirectioncount(elapsedtime) \\leq latticecount(elapsedtime) \\leq \\pi[elapsedtime+(\\sqrt{2} / 2)]^{2}\n\\]\n\nHence there is an upper bound for \\( directioncount(elapsedtime) \\) of the form \\( \\pi elapsedtime^{2}+linearcoef\\, elapsedtime+constantcoef \\), with \\( linearcoef \\) and \\( constantcoef \\) fixed. When just one coordinate of \\( startpoint \\) is irrational,\n\\[\ndirectioncount(elapsedtime)=latticecount(elapsedtime) \\geq \\pi[elapsedtime-(\\sqrt{2} / 2)]^{2}\n\\]\n\nThis lower bound for \\( directioncount(elapsedtime) \\) exceeds \\( quadraticcoef\\, elapsedtime^{2}+linearcoef\\, elapsedtime+constantcoef \\) for sufficiently large \\( elapsedtime \\) if \\( quadraticcoef<\\pi \\); hence \\( \\pi \\) is the desired \\( quadraticcoef \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "P": "velocity",
        "P_0": "identity",
        "N": "curvature",
        "T": "magnitude",
        "L": "perimeter",
        "a": "particle",
        "b": "fragment",
        "c": "material"
      },
      "question": "Problem A-4\nA point \\( velocity \\) moves inside a unit square in a straight line at unit speed. When it meets a corner it escapes. When it meets an edge its line of motion is reflected so that the angle of incidence equals the angle of reflection.\n\nLet \\( curvature(magnitude) \\) be the number of starting directions from a fixed interior point \\( identity \\) for which \\( velocity \\) escapes within \\( magnitude \\) units of time. Find the least constant \\( particle \\) for which constants \\( fragment \\) and \\( material \\) exist such that\n\\[\ncurvature(magnitude) \\leq particle magnitude^{2}+fragment magnitude+material\n\\]\nfor all \\( magnitude>0 \\) and all initial points \\( identity \\).",
      "solution": "A-4.\nSet up coordinates so that a vertex of the given unit square is \\( (0,0) \\) and two sides of the square are on the axes. Using the reflection properties, one can see that \\( velocity \\) escapes within \\( magnitude \\) units of time if and only if the (infinite) ray from \\( identity \\), with the direction of the first segment of the path, goes through a lattice point (point with integer coordinates) within \\( magnitude \\) units of distance from \\( identity \\). Thus \\( curvature(magnitude) \\) is at most the number \\( perimeter(magnitude) \\) of lattice points in the circle with center at \\( identity \\) and radius \\( magnitude \\). Tiling the plane with unit squares having centers at the lattice points and considering areas, one sees that\n\\[\ncurvature(magnitude) \\leq perimeter(magnitude) \\leq \\pi[magnitude+(\\sqrt{2} / 2)]^{2}\n\\]\n\nHence there is an upper bound for \\( curvature(magnitude) \\) of the form \\( \\pi magnitude^{2}+fragment magnitude+material \\), with \\( fragment \\) and \\( material \\) fixed. When just one coordinate of \\( identity \\) is irrational,\n\\[\ncurvature(magnitude)=perimeter(magnitude) \\geq \\pi[magnitude-(\\sqrt{2} / 2)]^{2}\n\\]\n\nThis lower bound for \\( curvature(magnitude) \\) exceeds \\( particle magnitude^{2}+fragment magnitude+material \\) for sufficiently large \\( magnitude \\) if \\( particle<\\pi \\); hence \\( \\pi \\) is the desired \\( particle \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "P": "stillpoint",
        "P_0": "finalpoint",
        "N": "voidcount",
        "T": "timeless",
        "L": "continuum",
        "a": "maxspread",
        "b": "shiftvary",
        "c": "dynamics"
      },
      "question": "Problem A-4\nA point \\( stillpoint \\) moves inside a unit square in a straight line at unit speed. When it meets a corner it escapes. When it meets an edge its line of motion is reflected so that the angle of incidence equals the angle of reflection.\n\nLet \\( voidcount(timeless) \\) be the number of starting directions from a fixed interior point \\( finalpoint \\) for which \\( stillpoint \\) escapes within \\( timeless \\) units of time. Find the least constant \\( maxspread \\) for which constants \\( shiftvary \\) and \\( dynamics \\) exist such that\n\\[\nvoidcount(timeless) \\leq maxspread\\, timeless^{2}+shiftvary\\, timeless+dynamics\n\\]\nfor all \\( timeless>0 \\) and all initial points \\( finalpoint \\).",
      "solution": "A-4.\nSet up coordinates so that a vertex of the given unit square is \\( (0,0) \\) and two sides of the square are on the axes. Using the reflection properties, one can see that \\( stillpoint \\) escapes within \\( timeless \\) units of time if and only if the (infinite) ray from \\( finalpoint \\), with the direction of the first segment of the path, goes through a lattice point (point with integer coordinates) within \\( timeless \\) units of distance from \\( finalpoint \\). Thus \\( voidcount(timeless) \\) is at most the number \\( continuum(timeless) \\) of lattice points in the circle with center at \\( finalpoint \\) and radius \\( timeless \\). Tiling the plane with unit squares having centers at the lattice points and considering areas, one sees that\n\\[\nvoidcount(timeless) \\leq continuum(timeless) \\leq \\pi[timeless+(\\sqrt{2} / 2)]^{2}\n\\]\n\nHence there is an upper bound for \\( voidcount(timeless) \\) of the form \\( \\pi\\, timeless^{2}+shiftvary\\, timeless+dynamics \\), with \\( shiftvary \\) and \\( dynamics \\) fixed. When just one coordinate of \\( finalpoint \\) is irrational,\n\\[\nvoidcount(timeless)=continuum(timeless) \\geq \\pi[timeless-(\\sqrt{2} / 2)]^{2}\n\\]\n\nThis lower bound for \\( voidcount(timeless) \\) exceeds \\( maxspread\\, timeless^{2}+shiftvary\\, timeless+dynamics \\) for sufficiently large \\( timeless \\) if \\( maxspread<\\pi \\); hence \\( \\pi \\) is the desired \\( maxspread \\)."
    },
    "garbled_string": {
      "map": {
        "P": "kzmpvqnw",
        "P_0": "vsnqtkgh",
        "N": "hjrfkdls",
        "T": "qzxwvtnp",
        "L": "bnxcfmtr",
        "a": "wplxzqme",
        "b": "rknldsvo",
        "c": "ghtmwqra"
      },
      "question": "Problem A-4\nA point \\( kzmpvqnw \\) moves inside a unit square in a straight line at unit speed. When it meets a corner it escapes. When it meets an edge its line of motion is reflected so that the angle of incidence equals the angle of reflection.\n\nLet \\( hjrfkdls(qzxwvtnp) \\) be the number of starting directions from a fixed interior point \\( vsnqtkgh \\) for which \\( kzmpvqnw \\) escapes within \\( qzxwvtnp \\) units of time. Find the least constant \\( wplxzqme \\) for which constants \\( rknldsvo \\) and \\( ghtmwqra \\) exist such that\n\\[\nhjrfkdls(qzxwvtnp) \\leq wplxzqme qzxwvtnp^{2}+ rknldsvo qzxwvtnp + ghtmwqra\n\\]\nfor all \\( qzxwvtnp>0 \\) and all initial points \\( vsnqtkgh \\).",
      "solution": "A-4.\nSet up coordinates so that a vertex of the given unit square is \\( (0,0) \\) and two sides of the square are on the axes. Using the reflection properties, one can see that \\( kzmpvqnw \\) escapes within \\( qzxwvtnp \\) units of time if and only if the (infinite) ray from \\( vsnqtkgh \\), with the direction of the first segment of the path, goes through a lattice point (point with integer coordinates) within \\( qzxwvtnp \\) units of distance from \\( vsnqtkgh \\). Thus \\( hjrfkdls(qzxwvtnp) \\) is at most the number \\( bnxcfmtr(qzxwvtnp) \\) of lattice points in the circle with center at \\( vsnqtkgh \\) and radius \\( qzxwvtnp \\). Tiling the plane with unit squares having centers at the lattice points and considering areas, one sees that\n\\[\nhjrfkdls(qzxwvtnp) \\leq bnxcfmtr(qzxwvtnp) \\leq \\pi[qzxwvtnp+(\\sqrt{2} / 2)]^{2}\n\\]\n\nHence there is an upper bound for \\( hjrfkdls(qzxwvtnp) \\) of the form \\( \\pi qzxwvtnp^{2}+ rknldsvo qzxwvtnp + ghtmwqra \\), with \\( rknldsvo \\) and \\( ghtmwqra \\) fixed. When just one coordinate of \\( vsnqtkgh \\) is irrational,\n\\[\nhjrfkdls(qzxwvtnp)=bnxcfmtr(qzxwvtnp) \\geq \\pi[qzxwvtnp-(\\sqrt{2} / 2)]^{2}\n\\]\n\nThis lower bound for \\( hjrfkdls(qzxwvtnp) \\) exceeds \\( wplxzqme qzxwvtnp^{2}+ rknldsvo qzxwvtnp + ghtmwqra \\) for sufficiently large \\( qzxwvtnp \\) if \\( wplxzqme<\\pi \\); hence \\( \\pi \\) is the desired \\( wplxzqme \\)."
    },
    "kernel_variant": {
      "question": "Let $d\\ge 2$ be an integer.  \nInside the open $d$–dimensional cube  \n\\[\nQ=(0,1)^{d}\\subset\\mathbb{R}^{d}\n\\]\nfix an interior point  \n\\[\nP_{0}=(\\xi_{1},\\dots ,\\xi_{d}),\\qquad 0<\\xi_{j}<1\\;(1\\le j\\le d).\n\\]\n\nA material point $P$ is launched from $P_{0}$ with unit speed in some direction $v\\in S^{d-1}$.  \nWhile $P$ moves, the following rules are enforced.\n\n1. (Specular reflections)  \n Whenever $P$ meets a $(d-1)$–dimensional face of $Q$ it is instantaneously reflected so that the angle of incidence equals the angle of reflection.\n\n2. (Escape)  \n If $P$ ever reaches a vertex of $Q$, that is, a point of the set $\\{0,1\\}^{d}$, it immediately leaves the cube and never returns.\n\nFor $T>0$ let $N_{d}(T;P_{0})$ be the number of \\emph{pairwise different} initial directions $v$ for which the trajectory that starts at $P_{0}$ with velocity $v$ reaches \\emph{some} vertex of $Q$ in time $\\le T$.\n\nDetermine, as an explicit function of $d$, the smallest real constant $a_{d}$ for which there exist real numbers $b_{d-1},b_{d-2},\\dots ,b_{0}$ – depending only on $d$ – such that for every interior starting point $P_{0}$ and every $T>0$\n\\[\n\\boxed{\\;N_{d}(T;P_{0})\\le a_{d}\\,T^{d}+b_{d-1}T^{d-1}+\\dots +b_{1}T+b_{0}\\;}\\tag{$\\star$}\n\\]\nholds.  Moreover, prove that no smaller leading constant than $a_{d}$ can satisfy $(\\star)$ for \\emph{all} interior points $P_{0}$ and all $T>0$.",
      "solution": "Throughout we write  \n\\[\nB_{d}(r)=\\{x\\in\\mathbb{R}^{d}:\\lVert x\\rVert\\le r\\},\\qquad \nV_{d}=\\operatorname{vol}B_{d}(1)=\\dfrac{\\pi^{d/2}}{\\Gamma(d/2+1)}.\n\\]\n\n--------------------------------------------------------------------\nStep 1 – Unfolding the billiard.  \n--------------------------------------------------------------------\nReflecting $Q$ successively in its $2d$ faces tiles $\\mathbb{R}^{d}$ by unit\ncubes with vertices in the lattice $\\mathbb{Z}^{d}$.  After this “unfolding’’ a\nreflected billiard trajectory is turned into the straight half–ray\n\\[\n\\ell(v)=\\{P_{0}+t v:\\;t\\ge 0\\}\\subset\\mathbb{R}^{d}.\n\\]\nCondition 2 becomes  \n\n$\\bullet$ the launched particle \\emph{escapes} in time $\\le T$  \n$\\Longleftrightarrow$ $\\ell(v)$ meets a lattice point $w\\in\\mathbb{Z}^{d}$ with  \n\\[\n0<\\lVert w-P_{0}\\rVert\\le T.\\tag{1}\n\\]\n\nDenote  \n\\[\n\\Lambda(T;P_{0})=\\{w\\in\\mathbb{Z}^{d}:0<\\lVert w-P_{0}\\rVert\\le T\\}.\n\\]\n\nA lattice point $w\\in\\Lambda(T;P_{0})$ is called \\emph{visible} (from $P_{0}$) if\nthe open segment $(P_{0},w)$ contains no lattice point.  Let  \n$V(T;P_{0})$ be the set of visible points.  By construction\n\\[\nN_{d}(T;P_{0})=\\#V(T;P_{0}).\\tag{2}\n\\]\n\n--------------------------------------------------------------------\nStep 2 – A universal polynomial upper bound.  \n--------------------------------------------------------------------\nTranslate every $w\\in\\mathbb{Z}^{d}$ by the closed cube\n$C=[-1/2,1/2]^{d}$; the family $\\{w+C:w\\in\\mathbb{Z}^{d}\\}$ is disjoint and\neach such cube has volume $1$.  If $\\lVert w-P_{0}\\rVert\\le T$ then\n\\[\nw+C\\subset P_{0}+B_{d}\\!\\bigl(T+\\sqrt{d}/2\\bigr),\n\\]\nhence  \n\\[\n\\#\\{w\\in\\mathbb{Z}^{d}:\\lVert w-P_{0}\\rVert\\le T\\}\\le\nV_{d}\\bigl(T+\\sqrt{d}/2\\bigr)^{d}.\\tag{3}\n\\]\nSince $V(T;P_{0})\\subset\\Lambda(T;P_{0})$, (3) implies\n\\[\nN_{d}(T;P_{0})\\le V_{d}\\bigl(T+\\sqrt{d}/2\\bigr)^{d}.\\tag{4}\n\\]\n\nExpanding $(T+\\sqrt{d}/2)^{d}$ via the binomial theorem gives\n\\[\nN_{d}(T;P_{0})\\le\nV_{d}\\sum_{k=0}^{d}\\binom{d}{k}\\Bigl(\\tfrac{\\sqrt{d}}{2}\\Bigr)^{k}T^{d-k}.\n\\]\nThus $(\\star)$ holds with\n\\[\na_{d}=V_{d},\\qquad \nb_{d-k}=V_{d}\\binom{d}{k}\\Bigl(\\tfrac{\\sqrt{d}}{2}\\Bigr)^{k}\\quad(1\\le k\\le d).\\tag{5}\n\\]\n\n--------------------------------------------------------------------\nStep 3 – Existence of starting points with \\emph{complete} visibility.  \n--------------------------------------------------------------------\nFor distinct lattice points $z,w\\in\\mathbb{Z}^{d}$ denote  \n\\[\nS_{z,w}=\\{\\,z+t(w-z):\\ 0<t<1\\,\\}\\subset\\mathbb{R}^{d}.\n\\]\n$S_{z,w}$ is a one–dimensional segment and hence has Lebesgue measure~$0$.\nIf $P_{0}\\in S_{z,w}$ then $w$ is \\emph{not} visible from~$P_{0}$ because $z$\nlies between them.\n\nDefine\n\\[\n\\mathcal{B}=\\bigcup_{\\substack{z,w\\in\\mathbb{Z}^{d}\\\\ z\\neq w}}S_{z,w}.\n\\]\nThe family $\\{S_{z,w}\\}_{z\\neq w}$ is countable, so $\\mathcal{B}$ also has\nmeasure~$0$.\n\nChoose an interior point $P_{0}\\in(0,1)^{d}\\setminus\\mathcal{B}$.\nThen for \\emph{every} lattice point $w\\neq P_{0}$ the open segment\n$(P_{0},w)$ contains no further lattice point, i.e.  \n\\[\nV(T;P_{0})=\\Lambda(T;P_{0})\\qquad\\text{for all }T>0.\\tag{6}\n\\]\n(Almost every interior point enjoys this property.)\n\n--------------------------------------------------------------------\nStep 4 – Lattice–point asymptotics for the chosen $P_{0}$.  \n--------------------------------------------------------------------\nBecause of (6), to estimate $N_{d}(T;P_{0})$ we may count \\emph{all} lattice\npoints within radius~$T$.\n\nLower bound.  \nFor $T\\ge\\sqrt{d}/2$,\n\\[\nP_{0}+B_{d}\\!\\bigl(T-\\tfrac{\\sqrt{d}}{2}\\bigr)\\subset\n\\bigcup_{\\lVert w-P_{0}\\rVert\\le T}(w+C),\n\\]\nand comparing volumes yields\n\\[\nN_{d}(T;P_{0})\\ge V_{d}\\bigl(T-\\tfrac{\\sqrt{d}}{2}\\bigr)^{d}.\\tag{7}\n\\]\n\nUpper bound.  \nEquation (4) obviously remains valid for the present $P_{0}$:\n\\[\nN_{d}(T;P_{0})\\le V_{d}\\bigl(T+\\tfrac{\\sqrt{d}}{2}\\bigr)^{d}.\\tag{8}\n\\]\n\nTherefore\n\\[\nV_{d}\\bigl(T-\\tfrac{\\sqrt{d}}{2}\\bigr)^{d}\\le N_{d}(T;P_{0})\\le\nV_{d}\\bigl(T+\\tfrac{\\sqrt{d}}{2}\\bigr)^{d}\\qquad(T\\ge\\sqrt{d}/2).\\tag{9}\n\\]\nDividing by $T^{d}$ and letting $T\\to\\infty$ we obtain\n\\[\n\\lim_{T\\to\\infty}\\dfrac{N_{d}(T;P_{0})}{T^{d}}=V_{d}.\\tag{10}\n\\]\n\n--------------------------------------------------------------------\nStep 5 – Optimality of the leading constant.  \n--------------------------------------------------------------------\nAssume, to obtain a contradiction, that $(\\star)$ is valid with some\n$a_{d}<V_{d}$.  Fix the point $P_{0}$ from Step~3.  \nBecause of (7) we have\n\\[\nN_{d}(T;P_{0})\\ge V_{d}\\bigl(T-\\tfrac{\\sqrt{d}}{2}\\bigr)^{d}\n               =V_{d}T^{d}+O_{d}(T^{d-1})\\qquad(T\\to\\infty).\\tag{11}\n\\]\nInsert this in $(\\star)$:\n\\[\nV_{d}T^{d}+O_{d}(T^{d-1})\\le a_{d}T^{d}+b_{d-1}T^{d-1}+\\dots +b_{0}.\\tag{12}\n\\]\nRearranging gives\n\\[\n\\bigl(V_{d}-a_{d}\\bigr)T^{d}\\le O_{d}(T^{d-1}),\\qquad T\\to\\infty.\\tag{13}\n\\]\nThe left–hand side dominates the right–hand side for large~$T$ unless\n$V_{d}-a_{d}\\le 0$.  Hence\n\\[\na_{d}\\ge V_{d}.\\tag{14}\n\\]\n\n--------------------------------------------------------------------\nStep 6 – Conclusion.  \n--------------------------------------------------------------------\nThe universal upper bound (4) shows that $(\\star)$ holds with\n$a_{d}=V_{d}$, while Step~5 proves that no smaller leading coefficient\ncan work.  Therefore  \n\\[\n\\boxed{\\,a_{d}=V_{d}=\\dfrac{\\pi^{d/2}}{\\Gamma(d/2+1)}\\,},\n\\]\nand the explicit coefficients $b_{d-1},\\dots ,b_{0}$ are those displayed in\n(5).  This completes the proof.  \\hfill$\\square$",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.657831",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher Dimension:  \n   The original square (d=2) problem is lifted to arbitrary dimension d ≥ 2, introducing sophisticated geometry and a dependence on d in both statement and proof.\n\n2. Additional Mathematical Structures:  \n   • The unit sphere S^{d-1}, primitive lattice vectors in ℤᵈ, and Γ-function formulas for the volume V_d.  \n   • Circle- and sphere-method lattice-point estimates, which are non-trivial analytic-number-theoretic tools when d ≥ 3.\n\n3. Deeper Theoretical Requirements:  \n   • Understanding the unfolding of a d-dimensional billiard;  \n   • Handling injectivity issues via primitivity and Diophantine conditions on P₀;  \n   • Employing geometry-of-numbers coverings and precise O(T^{d-1}) error analysis.\n\n4. Multiple Interacting Concepts:  \n   The solution blends combinatorial geometry (cube tilings), analytic number theory (lattice-point counting in growing d-balls), and dynamical systems/billiard theory (specular reflections in higher dimensions).\n\n5. Substantially More Work:  \n   Compared with the planar case, one must generalise every step, derive new constants, control error terms of order T^{d-1}, and argue minimality for all dimensions d—effort well beyond a routine extension."
      }
    },
    "original_kernel_variant": {
      "question": "Let $d\\ge 2$ be an integer.  \nInside the open $d$–dimensional cube  \n\\[\nQ=(0,1)^{d}\\subset\\mathbb{R}^{d}\n\\]\nfix an interior point  \n\\[\nP_{0}=(\\xi_{1},\\dots ,\\xi_{d}),\\qquad 0<\\xi_{j}<1\\;(1\\le j\\le d).\n\\]\n\nA material point $P$ is launched from $P_{0}$ with unit speed in some direction $v\\in S^{d-1}$.  \nWhile $P$ moves, the following rules are enforced.\n\n1. (Specular reflections)  \n Whenever $P$ meets a $(d-1)$–dimensional face of $Q$ it is instantaneously reflected so that the angle of incidence equals the angle of reflection.\n\n2. (Escape)  \n If $P$ ever reaches a vertex of $Q$, that is, a point of the set $\\{0,1\\}^{d}$, it immediately leaves the cube and never returns.\n\nFor $T>0$ let $N_{d}(T;P_{0})$ be the number of \\emph{pairwise different} initial directions $v$ for which the trajectory that starts at $P_{0}$ with velocity $v$ reaches \\emph{some} vertex of $Q$ in time $\\le T$.\n\nDetermine, as an explicit function of $d$, the smallest real constant $a_{d}$ for which there exist real numbers $b_{d-1},b_{d-2},\\dots ,b_{0}$ – depending only on $d$ – such that for every interior starting point $P_{0}$ and every $T>0$\n\\[\n\\boxed{\\;N_{d}(T;P_{0})\\le a_{d}\\,T^{d}+b_{d-1}T^{d-1}+\\dots +b_{1}T+b_{0}\\;}\\tag{$\\star$}\n\\]\nholds.  Moreover, prove that no smaller leading constant than $a_{d}$ can satisfy $(\\star)$ for \\emph{all} interior points $P_{0}$ and all $T>0$.",
      "solution": "Throughout we write  \n\\[\nB_{d}(r)=\\{x\\in\\mathbb{R}^{d}:\\lVert x\\rVert\\le r\\},\\qquad \nV_{d}=\\operatorname{vol}B_{d}(1)=\\dfrac{\\pi^{d/2}}{\\Gamma(d/2+1)}.\n\\]\n\n--------------------------------------------------------------------\nStep 1 – Unfolding the billiard.  \n--------------------------------------------------------------------\nReflecting $Q$ successively in its $2d$ faces tiles $\\mathbb{R}^{d}$ by unit\ncubes with vertices in the lattice $\\mathbb{Z}^{d}$.  After this “unfolding’’ a\nreflected billiard trajectory is turned into the straight half–ray\n\\[\n\\ell(v)=\\{P_{0}+t v:\\;t\\ge 0\\}\\subset\\mathbb{R}^{d}.\n\\]\nCondition 2 becomes  \n\n$\\bullet$ the launched particle \\emph{escapes} in time $\\le T$  \n$\\Longleftrightarrow$ $\\ell(v)$ meets a lattice point $w\\in\\mathbb{Z}^{d}$ with  \n\\[\n0<\\lVert w-P_{0}\\rVert\\le T.\\tag{1}\n\\]\n\nDenote  \n\\[\n\\Lambda(T;P_{0})=\\{w\\in\\mathbb{Z}^{d}:0<\\lVert w-P_{0}\\rVert\\le T\\}.\n\\]\n\nA lattice point $w\\in\\Lambda(T;P_{0})$ is called \\emph{visible} (from $P_{0}$) if\nthe open segment $(P_{0},w)$ contains no lattice point.  Let  \n$V(T;P_{0})$ be the set of visible points.  By construction\n\\[\nN_{d}(T;P_{0})=\\#V(T;P_{0}).\\tag{2}\n\\]\n\n--------------------------------------------------------------------\nStep 2 – A universal polynomial upper bound.  \n--------------------------------------------------------------------\nTranslate every $w\\in\\mathbb{Z}^{d}$ by the closed cube\n$C=[-1/2,1/2]^{d}$; the family $\\{w+C:w\\in\\mathbb{Z}^{d}\\}$ is disjoint and\neach such cube has volume $1$.  If $\\lVert w-P_{0}\\rVert\\le T$ then\n\\[\nw+C\\subset P_{0}+B_{d}\\!\\bigl(T+\\sqrt{d}/2\\bigr),\n\\]\nhence  \n\\[\n\\#\\{w\\in\\mathbb{Z}^{d}:\\lVert w-P_{0}\\rVert\\le T\\}\\le\nV_{d}\\bigl(T+\\sqrt{d}/2\\bigr)^{d}.\\tag{3}\n\\]\nSince $V(T;P_{0})\\subset\\Lambda(T;P_{0})$, (3) implies\n\\[\nN_{d}(T;P_{0})\\le V_{d}\\bigl(T+\\sqrt{d}/2\\bigr)^{d}.\\tag{4}\n\\]\n\nExpanding $(T+\\sqrt{d}/2)^{d}$ via the binomial theorem gives\n\\[\nN_{d}(T;P_{0})\\le\nV_{d}\\sum_{k=0}^{d}\\binom{d}{k}\\Bigl(\\tfrac{\\sqrt{d}}{2}\\Bigr)^{k}T^{d-k}.\n\\]\nThus $(\\star)$ holds with\n\\[\na_{d}=V_{d},\\qquad \nb_{d-k}=V_{d}\\binom{d}{k}\\Bigl(\\tfrac{\\sqrt{d}}{2}\\Bigr)^{k}\\quad(1\\le k\\le d).\\tag{5}\n\\]\n\n--------------------------------------------------------------------\nStep 3 – Existence of starting points with \\emph{complete} visibility.  \n--------------------------------------------------------------------\nFor distinct lattice points $z,w\\in\\mathbb{Z}^{d}$ denote  \n\\[\nS_{z,w}=\\{\\,z+t(w-z):\\ 0<t<1\\,\\}\\subset\\mathbb{R}^{d}.\n\\]\n$S_{z,w}$ is a one–dimensional segment and hence has Lebesgue measure~$0$.\nIf $P_{0}\\in S_{z,w}$ then $w$ is \\emph{not} visible from~$P_{0}$ because $z$\nlies between them.\n\nDefine\n\\[\n\\mathcal{B}=\\bigcup_{\\substack{z,w\\in\\mathbb{Z}^{d}\\\\ z\\neq w}}S_{z,w}.\n\\]\nThe family $\\{S_{z,w}\\}_{z\\neq w}$ is countable, so $\\mathcal{B}$ also has\nmeasure~$0$.\n\nChoose an interior point $P_{0}\\in(0,1)^{d}\\setminus\\mathcal{B}$.\nThen for \\emph{every} lattice point $w\\neq P_{0}$ the open segment\n$(P_{0},w)$ contains no further lattice point, i.e.  \n\\[\nV(T;P_{0})=\\Lambda(T;P_{0})\\qquad\\text{for all }T>0.\\tag{6}\n\\]\n(Almost every interior point enjoys this property.)\n\n--------------------------------------------------------------------\nStep 4 – Lattice–point asymptotics for the chosen $P_{0}$.  \n--------------------------------------------------------------------\nBecause of (6), to estimate $N_{d}(T;P_{0})$ we may count \\emph{all} lattice\npoints within radius~$T$.\n\nLower bound.  \nFor $T\\ge\\sqrt{d}/2$,\n\\[\nP_{0}+B_{d}\\!\\bigl(T-\\tfrac{\\sqrt{d}}{2}\\bigr)\\subset\n\\bigcup_{\\lVert w-P_{0}\\rVert\\le T}(w+C),\n\\]\nand comparing volumes yields\n\\[\nN_{d}(T;P_{0})\\ge V_{d}\\bigl(T-\\tfrac{\\sqrt{d}}{2}\\bigr)^{d}.\\tag{7}\n\\]\n\nUpper bound.  \nEquation (4) obviously remains valid for the present $P_{0}$:\n\\[\nN_{d}(T;P_{0})\\le V_{d}\\bigl(T+\\tfrac{\\sqrt{d}}{2}\\bigr)^{d}.\\tag{8}\n\\]\n\nTherefore\n\\[\nV_{d}\\bigl(T-\\tfrac{\\sqrt{d}}{2}\\bigr)^{d}\\le N_{d}(T;P_{0})\\le\nV_{d}\\bigl(T+\\tfrac{\\sqrt{d}}{2}\\bigr)^{d}\\qquad(T\\ge\\sqrt{d}/2).\\tag{9}\n\\]\nDividing by $T^{d}$ and letting $T\\to\\infty$ we obtain\n\\[\n\\lim_{T\\to\\infty}\\dfrac{N_{d}(T;P_{0})}{T^{d}}=V_{d}.\\tag{10}\n\\]\n\n--------------------------------------------------------------------\nStep 5 – Optimality of the leading constant.  \n--------------------------------------------------------------------\nAssume, to obtain a contradiction, that $(\\star)$ is valid with some\n$a_{d}<V_{d}$.  Fix the point $P_{0}$ from Step~3.  \nBecause of (7) we have\n\\[\nN_{d}(T;P_{0})\\ge V_{d}\\bigl(T-\\tfrac{\\sqrt{d}}{2}\\bigr)^{d}\n               =V_{d}T^{d}+O_{d}(T^{d-1})\\qquad(T\\to\\infty).\\tag{11}\n\\]\nInsert this in $(\\star)$:\n\\[\nV_{d}T^{d}+O_{d}(T^{d-1})\\le a_{d}T^{d}+b_{d-1}T^{d-1}+\\dots +b_{0}.\\tag{12}\n\\]\nRearranging gives\n\\[\n\\bigl(V_{d}-a_{d}\\bigr)T^{d}\\le O_{d}(T^{d-1}),\\qquad T\\to\\infty.\\tag{13}\n\\]\nThe left–hand side dominates the right–hand side for large~$T$ unless\n$V_{d}-a_{d}\\le 0$.  Hence\n\\[\na_{d}\\ge V_{d}.\\tag{14}\n\\]\n\n--------------------------------------------------------------------\nStep 6 – Conclusion.  \n--------------------------------------------------------------------\nThe universal upper bound (4) shows that $(\\star)$ holds with\n$a_{d}=V_{d}$, while Step~5 proves that no smaller leading coefficient\ncan work.  Therefore  \n\\[\n\\boxed{\\,a_{d}=V_{d}=\\dfrac{\\pi^{d/2}}{\\Gamma(d/2+1)}\\,},\n\\]\nand the explicit coefficients $b_{d-1},\\dots ,b_{0}$ are those displayed in\n(5).  This completes the proof.  \\hfill$\\square$",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.518677",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher Dimension:  \n   The original square (d=2) problem is lifted to arbitrary dimension d ≥ 2, introducing sophisticated geometry and a dependence on d in both statement and proof.\n\n2. Additional Mathematical Structures:  \n   • The unit sphere S^{d-1}, primitive lattice vectors in ℤᵈ, and Γ-function formulas for the volume V_d.  \n   • Circle- and sphere-method lattice-point estimates, which are non-trivial analytic-number-theoretic tools when d ≥ 3.\n\n3. Deeper Theoretical Requirements:  \n   • Understanding the unfolding of a d-dimensional billiard;  \n   • Handling injectivity issues via primitivity and Diophantine conditions on P₀;  \n   • Employing geometry-of-numbers coverings and precise O(T^{d-1}) error analysis.\n\n4. Multiple Interacting Concepts:  \n   The solution blends combinatorial geometry (cube tilings), analytic number theory (lattice-point counting in growing d-balls), and dynamical systems/billiard theory (specular reflections in higher dimensions).\n\n5. Substantially More Work:  \n   Compared with the planar case, one must generalise every step, derive new constants, control error terms of order T^{d-1}, and argue minimality for all dimensions d—effort well beyond a routine extension."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}