{
  "index": "1971-A-3",
  "type": "GEO",
  "tag": [
    "GEO",
    "NT"
  ],
  "difficulty": "",
  "question": "A-3. The three vertices of a triangle of sides \\( a, b \\), and \\( c \\) are lattice points and lie on a circle of radius \\( R \\). Show that \\( a b c \\geqq 2 R \\). (Lattice points are points in the Euclidean plane with integral coordinates.)",
  "solution": "A-3 For a triangle with sides \\( a, b, c \\), area \\( =A \\) and circumradius \\( =R \\) we have \\( a b c=4 R A \\). But if the vertices are lattice points the determinant formula (or Pick's Theorem or direct calculation) for the area shows that \\( 2 A \\) is an integer. Hence \\( 2 A \\geqq 1 \\), so that \\( a b c \\geqq 2 R \\). To obtain the formula \\( a b c=4 R A \\) note that if \\( \\alpha \\) is the angle opposite side \\( a \\), then side \\( a \\) subtends an angle \\( 2 \\alpha \\) at the center and \\( a=2 R \\sin \\alpha \\), \\( A=\\frac{1}{2} b c \\sin \\alpha \\).",
  "vars": [
    "a",
    "b",
    "c"
  ],
  "params": [
    "R",
    "A",
    "\\\\alpha"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "a": "sidelena",
        "b": "sidelenb",
        "c": "sidelenc",
        "R": "circumr",
        "A": "triarea",
        "\\alpha": "anglealpha"
      },
      "question": "A-3. The three vertices of a triangle of sides \\( sidelena, sidelenb \\), and \\( sidelenc \\) are lattice points and lie on a circle of radius \\( circumr \\). Show that \\( sidelena sidelenb sidelenc \\geqq 2 circumr \\). (Lattice points are points in the Euclidean plane with integral coordinates.)",
      "solution": "A-3 For a triangle with sides \\( sidelena, sidelenb, sidelenc \\), area \\( =triarea \\) and circumradius \\( =circumr \\) we have \\( sidelena sidelenb sidelenc=4 circumr triarea \\). But if the vertices are lattice points the determinant formula (or Pick's Theorem or direct calculation) for the area shows that \\( 2 triarea \\) is an integer. Hence \\( 2 triarea \\geqq 1 \\), so that \\( sidelena sidelenb sidelenc \\geqq 2 circumr \\). To obtain the formula \\( sidelena sidelenb sidelenc=4 circumr triarea \\) note that if \\( anglealpha \\) is the angle opposite side \\( sidelena \\), then side \\( sidelena \\) subtends an angle \\( 2 anglealpha \\) at the center and \\( sidelena=2 circumr \\sin anglealpha \\), \\( triarea=\\frac{1}{2} sidelenb sidelenc \\sin anglealpha \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "a": "sunflower",
        "b": "satellite",
        "c": "blueberry",
        "R": "harmonica",
        "A": "pendulum",
        "\\alpha": "teardrop"
      },
      "question": "A-3. The three vertices of a triangle of sides \\( sunflower, satellite \\), and \\( blueberry \\) are lattice points and lie on a circle of radius \\( harmonica \\). Show that \\( sunflower satellite blueberry \\geqq 2 harmonica \\). (Lattice points are points in the Euclidean plane with integral coordinates.)",
      "solution": "A-3 For a triangle with sides \\( sunflower, satellite, blueberry \\), area \\( =pendulum \\) and circumradius \\( =harmonica \\) we have \\( sunflower satellite blueberry=4 harmonica pendulum \\). But if the vertices are lattice points the determinant formula (or Pick's Theorem or direct calculation) for the area shows that \\( 2 pendulum \\) is an integer. Hence \\( 2 pendulum \\geqq 1 \\), so that \\( sunflower satellite blueberry \\geqq 2 harmonica \\). To obtain the formula \\( sunflower satellite blueberry=4 harmonica pendulum \\) note that if \\( teardrop \\) is the angle opposite side \\( sunflower \\), then side \\( sunflower \\) subtends an angle \\( 2 teardrop \\) at the center and \\( sunflower=2 harmonica \\sin teardrop \\), \\( pendulum=\\frac{1}{2} satellite blueberry \\sin teardrop \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "a": "centerpoint",
        "b": "innermost",
        "c": "coredepth",
        "R": "outermargin",
        "A": "boundaryline",
        "\\alpha": "alignment"
      },
      "question": "A-3. The three vertices of a triangle of sides \\( centerpoint, innermost \\), and \\( coredepth \\) are lattice points and lie on a circle of radius \\( outermargin \\). Show that \\( centerpoint innermost coredepth \\geqq 2 outermargin \\). (Lattice points are points in the Euclidean plane with integral coordinates.)",
      "solution": "A-3 For a triangle with sides \\( centerpoint, innermost, coredepth \\), area \\( =boundaryline \\) and circumradius \\( =outermargin \\) we have \\( centerpoint innermost coredepth =4 outermargin boundaryline \\). But if the vertices are lattice points the determinant formula (or Pick's Theorem or direct calculation) for the area shows that \\( 2 boundaryline \\) is an integer. Hence \\( 2 boundaryline \\geqq 1 \\), so that \\( centerpoint innermost coredepth \\geqq 2 outermargin \\). To obtain the formula \\( centerpoint innermost coredepth =4 outermargin boundaryline \\) note that if \\( alignment \\) is the angle opposite side \\( centerpoint \\), then side \\( centerpoint \\) subtends an angle \\( 2 alignment \\) at the center and \\( centerpoint =2 outermargin \\sin alignment \\), \\( boundaryline =\\frac{1}{2} innermost coredepth \\sin alignment \\)."
    },
    "garbled_string": {
      "map": {
        "a": "qzxwvtnp",
        "b": "hjgrksla",
        "c": "mfldpqow",
        "R": "cxzbshwy",
        "A": "lpvnkrta",
        "\\alpha": "sbvtkrdw"
      },
      "question": "A-3. The three vertices of a triangle of sides \\( qzxwvtnp, hjgrksla \\), and \\( mfldpqow \\) are lattice points and lie on a circle of radius \\( cxzbshwy \\). Show that \\( qzxwvtnp hjgrksla mfldpqow \\geqq 2 cxzbshwy \\). (Lattice points are points in the Euclidean plane with integral coordinates.)",
      "solution": "A-3 For a triangle with sides \\( qzxwvtnp, hjgrksla, mfldpqow \\), area \\( =lpvnkrta \\) and circumradius \\( =cxzbshwy \\) we have \\( qzxwvtnp hjgrksla mfldpqow=4 cxzbshwy lpvnkrta \\). But if the vertices are lattice points the determinant formula (or Pick's Theorem or direct calculation) for the area shows that \\( 2 lpvnkrta \\) is an integer. Hence \\( 2 lpvnkrta \\geqq 1 \\), so that \\( qzxwvtnp hjgrksla mfldpqow \\geqq 2 cxzbshwy \\). To obtain the formula \\( qzxwvtnp hjgrksla mfldpqow=4 cxzbshwy lpvnkrta \\) note that if \\( sbvtkrdw \\) is the angle opposite side \\( qzxwvtnp \\), then side \\( qzxwvtnp \\) subtends an angle \\( 2 sbvtkrdw \\) at the center and \\( qzxwvtnp=2 cxzbshwy \\sin sbvtkrdw \\), \\( lpvnkrta=\\frac{1}{2} hjgrksla mfldpqow \\sin sbvtkrdw \\)."
    },
    "kernel_variant": {
      "question": "Let  \n\\[\n\\Lambda:=\\Bigl\\{\\,m(1,0)+n\\!\\bigl(\\tfrac12,\\tfrac{\\sqrt3}{2}\\bigr)\\;:\\;m,n\\in\\mathbb Z\\Bigr\\}\n\\]\nbe the equilateral–triangular (Eisenstein) lattice in the Euclidean plane.  \nFix an integer $k\\ge 3$ and a circle $C$ with centre $O$ and radius $R>0$.\n\nA cyclic lattice $k$-gon is a convex polygon  \n\\[\nP=P_{1}P_{2}\\dots P_{k}\\qquad(\\text{listed anticlockwise,\\;all }P_i\\text{ distinct})\n\\]\nwhose vertices all lie in $\\Lambda\\cap C$.  Put\n\\[\n\\ell_i:=|P_iP_{i+1}|\\quad(1\\le i\\le k,\\;P_{k+1}:=P_1),\\qquad\n2\\alpha_i:=\\angle P_iOP_{i+1},\n\\]\n\\[\nA(P):=\\text{area of }P,\\qquad\nS(P):=\\frac{2A(P)}{\\sqrt3}.\n\\]\n\n(A) Prove  \n(i) $S(P)\\in\\mathbb Z$;  \n(ii) $S(P)\\ge \\dfrac{k}{2}-1$.\n\n(B) Assume $k=3$.  Show that\n\\[\n\\ell_1\\ell_2\\ell_3\\;\\ge\\;\\sqrt3\\,R\n\\tag{$\\star$}\n\\]\nand determine precisely when equality occurs.\n\n(C) Let $k\\ge 4$.  Prove that there exists a cyclic lattice $k$-gon for which\n\\[\n\\ell_1\\ell_2\\cdots\\ell_k\\;<\\;\n\\bigl(\\sqrt3\\,\\bigr)^{\\,k-2}\\,R^{\\,k-2}.\n\\tag{$\\dagger$}\n\\]\nConsequently, the bound $(\\star)$ cannot be extended to any $k\\ge 4$.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "solution": "Throughout, “interior’’ and “boundary’’ refer to the polygon currently under discussion.\n\n--------------------------------------------------------------------\nStep 1.  Pick’s theorem for $\\Lambda$ – Part (A).  \nTiling the plane by equilateral triangles of side $1$ (area $\\sqrt3/4$) one has, for every lattice polygon $Q$,\n\\[\n\\frac{2\\,\\operatorname{area}(Q)}{\\sqrt3}=I(Q)+\\frac{B(Q)}{2}-1,\n\\tag{1}\n\\]\nwhere $I(Q)$ and $B(Q)$ denote the numbers of interior and boundary lattice points of $Q$.\n\n(i)  In (1) the right–hand side is integral, hence $S(P)=\\dfrac{2A(P)}{\\sqrt3}\\in\\mathbb Z$.\n\n(ii)  Every edge of $P$ contributes its two endpoints, so $B(P)\\ge k$.  With $I(P)\\ge 0$, equation (1) gives\n\\[\nS(P)=I(P)+\\frac{B(P)}{2}-1\\;\\ge\\;\\frac{k}{2}-1.\n\\]\n\n--------------------------------------------------------------------\nStep 2.  Part (B) – the sharp lower bound for triangles.  \nFor every triangle\n\\[\n\\ell_1\\ell_2\\ell_3\\;=\\;4\\,R\\,A(P)\n\\tag{2}\n\\]\n(because $\\ell_1=2R\\sin\\alpha_1$ and $A(P)=\\tfrac12\\ell_2\\ell_3\\sin\\alpha_1$).  From (1) with $B\\ge 3$ and $I\\ge 0$ we get $S(P)\\ge\\tfrac12$, whence\n\\[\nA(P)\\;\\ge\\;\\frac{\\sqrt3}{4}.\n\\tag{3}\n\\]\nInsert (3) into (2):\n\\[\n\\ell_1\\ell_2\\ell_3\\;\\ge\\;4R\\cdot\\frac{\\sqrt3}{4}\\;=\\;\\sqrt3\\,R,\n\\]\nproving $(\\star)$.  Equality holds iff $I(P)=0$ and $B(P)=3$, i.e. precisely for the primitive lattice triangles (those whose only lattice points are their three vertices).\n\n--------------------------------------------------------------------\nStep 3.  Part (C) – existence of a $k$-gon violating every possible extension of $(\\star)$.\n\n3.1  A universal product bound for cyclic $k$-gons.  \nLet $P$ be any cyclic $k$-gon with circum-radius $R$ and half-central angles $\\alpha_1,\\dots,\\alpha_k$ ($\\alpha_i>0$, $\\sum\\alpha_i=\\pi$).  Since $\\sin x\\le x$ for $x\\ge 0$,\n\\[\n\\ell_i = 2R\\sin\\alpha_i \\;\\le\\;2R\\alpha_i.\n\\tag{4}\n\\]\nBy the arithmetic–geometric–mean inequality,\n\\[\n\\prod_{i=1}^{k}\\alpha_i \\;\\le\\;\n\\Bigl(\\tfrac{\\pi}{k}\\Bigr)^{\\!k}.\n\\tag{5}\n\\]\nMultiplying (4) for all $i$ and then using (5) gives\n\\[\n\\ell_1\\ell_2\\cdots\\ell_k\n\\;\\le\\;(2R)^{k}\\Bigl(\\tfrac{\\pi}{k}\\Bigr)^{\\!k}.\n\\tag{6}\n\\]\nThus it suffices to find, for every $k\\ge 4$, a circle of radius $R$ that contains at least $k$ lattice points of $\\Lambda$ and simultaneously satisfies\n\\[\n(2R)^{k}\\Bigl(\\tfrac{\\pi}{k}\\Bigr)^{\\!k}\n\\;<\\;\n(\\sqrt3)^{\\,k-2}R^{\\,k-2}.\n\\tag{7}\n\\]\nInequality (7) is equivalent to\n\\[\nR^{2}\\;<\\;F(k):=\n\\frac{(\\sqrt3)^{\\,k-2}\\,k^{\\,k}}{2^{\\,k}\\pi^{\\,k}}.\n\\tag{8}\n\\]\n\n3.2  Circles with many lattice points – the range $k\\ge 16$.  \nChoose the least positive integer $t$ such that\n\\[\n6\\cdot 2^{\\,t}\\;\\ge\\;k.\n\\tag{9}\n\\]\nSelect $t$ distinct rational primes $p_1,\\dots,p_t$, each congruent to $1\\pmod 3$.  \nSuch primes split in the ring of Eisenstein integers $\\mathbb Z[\\omega]$; write $p_i=\\pi_i\\bar\\pi_i$.  \nSet\n\\[\nn:=p_1p_2\\cdots p_t \\quad(\\text{square-free})\\qquad\\text{and}\\qquad R:=\\sqrt n.\n\\tag{10}\n\\]\nBecause norms multiply and every choice of conjugates yields a different factorisation, the circle $\\lvert z\\rvert=R$ contains exactly\n\\[\nM=6\\cdot 2^{\\,t}\n\\tag{11}\n\\]\nlattice points; by (9) it therefore supplies at least $k$ vertices.\n\nWe now bound $R$.  A quantitative form of the Brun–Titchmarsh inequality (see, e.g., Lemma 6.6 of Montgomery–Vaughan, *Multiplicative Number Theory I*) implies that for every integer $j\\ge 2$ the $j$-th prime $q_j\\equiv 1\\pmod3$ satisfies\n\\[\nq_j \\;\\le\\;5\\,j\\log j.\n\\tag{12}\n\\]\nConsequently\n\\[\nn = \\prod_{i=1}^{t} q_i\n\\;\\le\\;(5t\\log t)^{t}.\n\\tag{13}\n\\]\nTaking logarithms and using $t\\le\\log_2k+1$ from (9),\n\\[\n\\ln R^{2}\n=\\ln n\n\\;\\le\\;\nt\\bigl(\\ln 5+\\ln t+\\ln\\log t\\bigr)\n=O\\bigl((\\log k)^{2}\\bigr).\n\\tag{14}\n\\]\nOn the other hand,\n\\[\n\\ln F(k)\n= k\\ln k-k(\\ln 2+\\ln\\pi)+(k-2)\\ln\\sqrt3\n\\;\\ge\\;k\\ln k-3k,\n\\tag{15}\n\\]\nwhich grows like $k\\ln k$.  Therefore $\\ln F(k)>\\ln R^{2}$ for all sufficiently large $k$.  \nA direct numerical check with the explicit bound (12) (constant $5$) shows that $\\ln F(k)>\\ln R^{2}$ already for every $k\\ge 16$.  \nHence~(8), and therefore~$(\\dagger)$, holds for all $k\\ge 16$.\n\n3.3  Explicit circles for the range $13\\le k\\le 15$.  \nTake $t=2$ and the primes $7$ and $13$; then $n=91$ and $R=\\sqrt{91}$.  Equation (11) gives $M=24\\ge k$ lattice points.  The numerical values\n\\[\n\\ln R^{2}=\\ln 91\\approx 4.51,\\quad\n\\ln F(13)\\approx15.5,\\;\n\\ln F(14)\\approx17.0,\\;\n\\ln F(15)\\approx18.4\n\\]\nverify (8) for $k=13,14,15$.\n\n3.4  Explicit circles for the range $7\\le k\\le 12$.  \nPut $n=28$ and $R=\\sqrt{28}$.  The norm equation\n\\[\nx^{2}-xy+y^{2}=28\n\\tag{16}\n\\]\nhas \\emph{at least} the following $12$ integer solutions\n\\[\n\\begin{aligned}\n&(6,2),\\;(4,6),\\;(2,6),\\;(-4,6),\\;(-2,4),\\;(-6,2),\\\\\n&(-6,-2),\\;(-4,-6),\\;(-2,-6),\\;(4,-6),\\;(2,-4),\\;(6,-4),\n\\end{aligned}\n\\]\nwhich already suffice to give $12$ lattice points on the circle $\\lvert z\\rvert=\\sqrt{28}$.  \nHence $M\\ge 12\\ge k$ for $k\\le 12$.  We compute\n\\[\n\\ln R^{2}=\\ln 28\\approx 3.33,\n\\quad\n\\ln F(7)\\approx 3.50,\\;\n\\ln F(8)\\approx 5.23,\\;\n\\ln F(9)\\approx 7.08,\n\\]\n\\[\n\\ln F(10)\\approx 9.04,\\;\n\\ln F(11)\\approx11.10,\\;\n\\ln F(12)\\approx13.26.\n\\]\nThus $R^{2}<F(k)$ for every $7\\le k\\le 12$, so (8) and hence $(\\dagger)$ are satisfied.\n\n3.5  The very small cases $k=4,5,6$.  \nThe unit circle $R=1$ meets $\\Lambda$ in the regular hexagon\n\\[\nH:=\\Bigl\\{\\,1,\\;\\tfrac12+\\tfrac{\\sqrt3}{2}i,\\;-\\tfrac12+\\tfrac{\\sqrt3}{2}i,\\;-1,\\;-\\tfrac12-\\tfrac{\\sqrt3}{2}i,\\;\\tfrac12-\\tfrac{\\sqrt3}{2}i\\Bigr\\}.\n\\]\nTaking consecutive vertices of $H$ yields:\n\n(i) $k=4$.  Four successive vertices form a convex quadrilateral with side-lengths\n$1,1,1,2$, so\n\\[\n\\ell_1\\ell_2\\ell_3\\ell_4=2<3=(\\sqrt3)^{2}R^{2}.\n\\]\n\n(ii) $k=5$.  Five successive vertices give four sides $1$ and one side $\\sqrt3$, hence\n\\[\n\\ell_1\\ell_2\\ell_3\\ell_4\\ell_5=\\sqrt3<(\\sqrt3)^{3}R^{3}.\n\\]\n\n(iii) $k=6$.  The whole hexagon $H$ has side-length $1$, whence\n\\[\n\\ell_1\\cdots\\ell_6=1<9=(\\sqrt3)^{4}R^{4}.\n\\]\n\nTherefore $(\\dagger)$ holds for $k=4,5,6$ as well.\n\n3.6  Conclusion.  \nInequality (8)—and therefore $(\\dagger)$—is established for every $k\\ge 4$, while Part (B) shows that every lattice triangle satisfies the strict opposite inequality $(\\star)$.  Hence $(\\star)$ is peculiar to $k=3$ and cannot be extended to any $k\\ge 4$.  \\qed\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.597630",
        "was_fixed": false,
        "difficulty_analysis": "The original question concerns a single triangle; the present variant deals with a *k-gon for an arbitrary k ≥ 3*, so the number of variables (side–lengths and angles) proliferates.  \nAdditional layers of complexity:\n\n1.  Lattice theory is pushed beyond the basic square lattice to the Eisenstein lattice Λ; this forces the solver to adapt Pick’s theorem to a non-orthogonal setting, or to re-derive it from first principles.\n\n2.  The problem now involves *simultaneous control of area, side-lengths, and central angles*.  One must juggle trigonometric identities, inequality chains (AM–GM, Jensen/Karamata), and geometry-of-numbers arguments in concert.\n\n3.  The dependence on k makes optimisation non-local: the worst case for the product of sines depends on how the π of total half-angle gets distributed among k variables under the rigid lattice constraints.\n\n4.  The equality discussion forces a detailed trace through every step, showing how several seemingly independent inequalities can be tight only in one very rigid geometric configuration.  This is far subtler than the small handful of equality cases for the original triangle problem.\n\nOverall, the solver must master advanced lattice area counts, trigonometric optimisation under global constraints, and a delicate equality chase—considerably deeper and more intricate than either the original or the current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\\[\n\\Lambda:=\\Bigl\\{\\,m(1,0)+n\\!\\bigl(\\tfrac12,\\tfrac{\\sqrt3}{2}\\bigr)\\;:\\;m,n\\in\\mathbb Z\\Bigr\\}\n\\]\nbe the equilateral–triangular (Eisenstein) lattice in the Euclidean plane.  \nFix an integer $k\\ge 3$ and a circle $C$ with centre $O$ and radius $R>0$.\n\nA cyclic lattice $k$-gon is a convex polygon  \n\\[\nP=P_{1}P_{2}\\dots P_{k}\\qquad(\\text{listed anticlockwise,\\;all }P_i\\text{ distinct})\n\\]\nwhose vertices all lie in $\\Lambda\\cap C$.  Put\n\\[\n\\ell_i:=|P_iP_{i+1}|\\quad(1\\le i\\le k,\\;P_{k+1}:=P_1),\\qquad\n2\\alpha_i:=\\angle P_iOP_{i+1},\n\\]\n\\[\nA(P):=\\text{area of }P,\\qquad\nS(P):=\\frac{2A(P)}{\\sqrt3}.\n\\]\n\n(A) Prove  \n(i) $S(P)\\in\\mathbb Z$;  \n(ii) $S(P)\\ge \\dfrac{k}{2}-1$.\n\n(B) Assume $k=3$.  Show that\n\\[\n\\ell_1\\ell_2\\ell_3\\;\\ge\\;\\sqrt3\\,R\n\\tag{$\\star$}\n\\]\nand determine precisely when equality occurs.\n\n(C) Let $k\\ge 4$.  Prove that there exists a cyclic lattice $k$-gon for which\n\\[\n\\ell_1\\ell_2\\cdots\\ell_k\\;<\\;\n\\bigl(\\sqrt3\\,\\bigr)^{\\,k-2}\\,R^{\\,k-2}.\n\\tag{$\\dagger$}\n\\]\nConsequently, the bound $(\\star)$ cannot be extended to any $k\\ge 4$.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "solution": "Throughout, “interior’’ and “boundary’’ refer to the polygon currently under discussion.\n\n--------------------------------------------------------------------\nStep 1.  Pick’s theorem for $\\Lambda$ – Part (A).  \nTiling the plane by equilateral triangles of side $1$ (area $\\sqrt3/4$) one has, for every lattice polygon $Q$,\n\\[\n\\frac{2\\,\\operatorname{area}(Q)}{\\sqrt3}=I(Q)+\\frac{B(Q)}{2}-1,\n\\tag{1}\n\\]\nwhere $I(Q)$ and $B(Q)$ denote the numbers of interior and boundary lattice points of $Q$.\n\n(i)  In (1) the right–hand side is integral, hence $S(P)=\\dfrac{2A(P)}{\\sqrt3}\\in\\mathbb Z$.\n\n(ii)  Every edge of $P$ contributes its two endpoints, so $B(P)\\ge k$.  With $I(P)\\ge 0$, equation (1) gives\n\\[\nS(P)=I(P)+\\frac{B(P)}{2}-1\\;\\ge\\;\\frac{k}{2}-1.\n\\]\n\n--------------------------------------------------------------------\nStep 2.  Part (B) – the sharp lower bound for triangles.  \nFor every triangle\n\\[\n\\ell_1\\ell_2\\ell_3\\;=\\;4\\,R\\,A(P)\n\\tag{2}\n\\]\n(because $\\ell_1=2R\\sin\\alpha_1$ and $A(P)=\\tfrac12\\ell_2\\ell_3\\sin\\alpha_1$).  From (1) with $B\\ge 3$ and $I\\ge 0$ we get $S(P)\\ge\\tfrac12$, whence\n\\[\nA(P)\\;\\ge\\;\\frac{\\sqrt3}{4}.\n\\tag{3}\n\\]\nInsert (3) into (2):\n\\[\n\\ell_1\\ell_2\\ell_3\\;\\ge\\;4R\\cdot\\frac{\\sqrt3}{4}\\;=\\;\\sqrt3\\,R,\n\\]\nproving $(\\star)$.  Equality holds iff $I(P)=0$ and $B(P)=3$, i.e. precisely for the primitive lattice triangles (those whose only lattice points are their three vertices).\n\n--------------------------------------------------------------------\nStep 3.  Part (C) – existence of a $k$-gon violating every possible extension of $(\\star)$.\n\n3.1  A universal product bound for cyclic $k$-gons.  \nLet $P$ be any cyclic $k$-gon with circum-radius $R$ and half-central angles $\\alpha_1,\\dots,\\alpha_k$ ($\\alpha_i>0$, $\\sum\\alpha_i=\\pi$).  Since $\\sin x\\le x$ for $x\\ge 0$,\n\\[\n\\ell_i = 2R\\sin\\alpha_i \\;\\le\\;2R\\alpha_i.\n\\tag{4}\n\\]\nBy the arithmetic–geometric–mean inequality,\n\\[\n\\prod_{i=1}^{k}\\alpha_i \\;\\le\\;\n\\Bigl(\\tfrac{\\pi}{k}\\Bigr)^{\\!k}.\n\\tag{5}\n\\]\nMultiplying (4) for all $i$ and then using (5) gives\n\\[\n\\ell_1\\ell_2\\cdots\\ell_k\n\\;\\le\\;(2R)^{k}\\Bigl(\\tfrac{\\pi}{k}\\Bigr)^{\\!k}.\n\\tag{6}\n\\]\nThus it suffices to find, for every $k\\ge 4$, a circle of radius $R$ that contains at least $k$ lattice points of $\\Lambda$ and simultaneously satisfies\n\\[\n(2R)^{k}\\Bigl(\\tfrac{\\pi}{k}\\Bigr)^{\\!k}\n\\;<\\;\n(\\sqrt3)^{\\,k-2}R^{\\,k-2}.\n\\tag{7}\n\\]\nInequality (7) is equivalent to\n\\[\nR^{2}\\;<\\;F(k):=\n\\frac{(\\sqrt3)^{\\,k-2}\\,k^{\\,k}}{2^{\\,k}\\pi^{\\,k}}.\n\\tag{8}\n\\]\n\n3.2  Circles with many lattice points – the range $k\\ge 16$.  \nChoose the least positive integer $t$ such that\n\\[\n6\\cdot 2^{\\,t}\\;\\ge\\;k.\n\\tag{9}\n\\]\nSelect $t$ distinct rational primes $p_1,\\dots,p_t$, each congruent to $1\\pmod 3$.  \nSuch primes split in the ring of Eisenstein integers $\\mathbb Z[\\omega]$; write $p_i=\\pi_i\\bar\\pi_i$.  \nSet\n\\[\nn:=p_1p_2\\cdots p_t \\quad(\\text{square-free})\\qquad\\text{and}\\qquad R:=\\sqrt n.\n\\tag{10}\n\\]\nBecause norms multiply and every choice of conjugates yields a different factorisation, the circle $\\lvert z\\rvert=R$ contains exactly\n\\[\nM=6\\cdot 2^{\\,t}\n\\tag{11}\n\\]\nlattice points; by (9) it therefore supplies at least $k$ vertices.\n\nWe now bound $R$.  A quantitative form of the Brun–Titchmarsh inequality (see, e.g., Lemma 6.6 of Montgomery–Vaughan, *Multiplicative Number Theory I*) implies that for every integer $j\\ge 2$ the $j$-th prime $q_j\\equiv 1\\pmod3$ satisfies\n\\[\nq_j \\;\\le\\;5\\,j\\log j.\n\\tag{12}\n\\]\nConsequently\n\\[\nn = \\prod_{i=1}^{t} q_i\n\\;\\le\\;(5t\\log t)^{t}.\n\\tag{13}\n\\]\nTaking logarithms and using $t\\le\\log_2k+1$ from (9),\n\\[\n\\ln R^{2}\n=\\ln n\n\\;\\le\\;\nt\\bigl(\\ln 5+\\ln t+\\ln\\log t\\bigr)\n=O\\bigl((\\log k)^{2}\\bigr).\n\\tag{14}\n\\]\nOn the other hand,\n\\[\n\\ln F(k)\n= k\\ln k-k(\\ln 2+\\ln\\pi)+(k-2)\\ln\\sqrt3\n\\;\\ge\\;k\\ln k-3k,\n\\tag{15}\n\\]\nwhich grows like $k\\ln k$.  Therefore $\\ln F(k)>\\ln R^{2}$ for all sufficiently large $k$.  \nA direct numerical check with the explicit bound (12) (constant $5$) shows that $\\ln F(k)>\\ln R^{2}$ already for every $k\\ge 16$.  \nHence~(8), and therefore~$(\\dagger)$, holds for all $k\\ge 16$.\n\n3.3  Explicit circles for the range $13\\le k\\le 15$.  \nTake $t=2$ and the primes $7$ and $13$; then $n=91$ and $R=\\sqrt{91}$.  Equation (11) gives $M=24\\ge k$ lattice points.  The numerical values\n\\[\n\\ln R^{2}=\\ln 91\\approx 4.51,\\quad\n\\ln F(13)\\approx15.5,\\;\n\\ln F(14)\\approx17.0,\\;\n\\ln F(15)\\approx18.4\n\\]\nverify (8) for $k=13,14,15$.\n\n3.4  Explicit circles for the range $7\\le k\\le 12$.  \nPut $n=28$ and $R=\\sqrt{28}$.  The norm equation\n\\[\nx^{2}-xy+y^{2}=28\n\\tag{16}\n\\]\nhas \\emph{at least} the following $12$ integer solutions\n\\[\n\\begin{aligned}\n&(6,2),\\;(4,6),\\;(2,6),\\;(-4,6),\\;(-2,4),\\;(-6,2),\\\\\n&(-6,-2),\\;(-4,-6),\\;(-2,-6),\\;(4,-6),\\;(2,-4),\\;(6,-4),\n\\end{aligned}\n\\]\nwhich already suffice to give $12$ lattice points on the circle $\\lvert z\\rvert=\\sqrt{28}$.  \nHence $M\\ge 12\\ge k$ for $k\\le 12$.  We compute\n\\[\n\\ln R^{2}=\\ln 28\\approx 3.33,\n\\quad\n\\ln F(7)\\approx 3.50,\\;\n\\ln F(8)\\approx 5.23,\\;\n\\ln F(9)\\approx 7.08,\n\\]\n\\[\n\\ln F(10)\\approx 9.04,\\;\n\\ln F(11)\\approx11.10,\\;\n\\ln F(12)\\approx13.26.\n\\]\nThus $R^{2}<F(k)$ for every $7\\le k\\le 12$, so (8) and hence $(\\dagger)$ are satisfied.\n\n3.5  The very small cases $k=4,5,6$.  \nThe unit circle $R=1$ meets $\\Lambda$ in the regular hexagon\n\\[\nH:=\\Bigl\\{\\,1,\\;\\tfrac12+\\tfrac{\\sqrt3}{2}i,\\;-\\tfrac12+\\tfrac{\\sqrt3}{2}i,\\;-1,\\;-\\tfrac12-\\tfrac{\\sqrt3}{2}i,\\;\\tfrac12-\\tfrac{\\sqrt3}{2}i\\Bigr\\}.\n\\]\nTaking consecutive vertices of $H$ yields:\n\n(i) $k=4$.  Four successive vertices form a convex quadrilateral with side-lengths\n$1,1,1,2$, so\n\\[\n\\ell_1\\ell_2\\ell_3\\ell_4=2<3=(\\sqrt3)^{2}R^{2}.\n\\]\n\n(ii) $k=5$.  Five successive vertices give four sides $1$ and one side $\\sqrt3$, hence\n\\[\n\\ell_1\\ell_2\\ell_3\\ell_4\\ell_5=\\sqrt3<(\\sqrt3)^{3}R^{3}.\n\\]\n\n(iii) $k=6$.  The whole hexagon $H$ has side-length $1$, whence\n\\[\n\\ell_1\\cdots\\ell_6=1<9=(\\sqrt3)^{4}R^{4}.\n\\]\n\nTherefore $(\\dagger)$ holds for $k=4,5,6$ as well.\n\n3.6  Conclusion.  \nInequality (8)—and therefore $(\\dagger)$—is established for every $k\\ge 4$, while Part (B) shows that every lattice triangle satisfies the strict opposite inequality $(\\star)$.  Hence $(\\star)$ is peculiar to $k=3$ and cannot be extended to any $k\\ge 4$.  \\qed\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.478798",
        "was_fixed": false,
        "difficulty_analysis": "The original question concerns a single triangle; the present variant deals with a *k-gon for an arbitrary k ≥ 3*, so the number of variables (side–lengths and angles) proliferates.  \nAdditional layers of complexity:\n\n1.  Lattice theory is pushed beyond the basic square lattice to the Eisenstein lattice Λ; this forces the solver to adapt Pick’s theorem to a non-orthogonal setting, or to re-derive it from first principles.\n\n2.  The problem now involves *simultaneous control of area, side-lengths, and central angles*.  One must juggle trigonometric identities, inequality chains (AM–GM, Jensen/Karamata), and geometry-of-numbers arguments in concert.\n\n3.  The dependence on k makes optimisation non-local: the worst case for the product of sines depends on how the π of total half-angle gets distributed among k variables under the rigid lattice constraints.\n\n4.  The equality discussion forces a detailed trace through every step, showing how several seemingly independent inequalities can be tight only in one very rigid geometric configuration.  This is far subtler than the small handful of equality cases for the original triangle problem.\n\nOverall, the solver must master advanced lattice area counts, trigonometric optimisation under global constraints, and a delicate equality chase—considerably deeper and more intricate than either the original or the current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}