{
  "index": "1984-A-4",
  "type": "GEO",
  "tag": [
    "GEO",
    "ANA"
  ],
  "difficulty": "",
  "question": "Problem A-4\nA convex pentagon \\( P=A B C D E \\), with vertices labeled consecutively, is inscribed in a circle of radius 1 . Find the maximum area of \\( P \\) subject to the condition that the chords \\( A C \\) and \\( B D \\) be perpendicular.",
  "solution": "A-4.\nLet \\( \\theta=\\operatorname{Arc} A B, \\quad \\alpha=\\operatorname{Arc} D E \\), and \\( \\beta=\\operatorname{Arc} E A \\). Then \\( \\operatorname{Arc} C D=\\pi-\\theta \\) and \\( \\operatorname{Arc} B C= \\) \\( \\pi-\\alpha-\\beta \\).\n\nThe area of \\( P \\), in terms of the five triangles from the center of the circle is\n\\[\n\\frac{1}{2} \\sin \\theta+\\frac{1}{2} \\sin (\\pi-\\theta)+\\frac{1}{2} \\sin \\alpha+\\frac{1}{2} \\sin \\beta+\\frac{1}{2} \\sin (\\pi-\\alpha-\\beta)\n\\]\n\nThis is maximized when \\( \\theta=\\pi / 2 \\) and \\( \\alpha=\\beta=\\pi / 3 \\). Thus, the maximum area is\n\\[\n\\frac{1}{2} \\cdot 1+\\frac{1}{2} \\cdot 1+\\frac{1}{2} \\frac{\\sqrt{3}}{2}+\\frac{1}{2} \\frac{\\sqrt{3}}{2}+\\frac{1}{2} \\frac{\\sqrt{3}}{2}=1+\\frac{3}{4} \\sqrt{3}\n\\]",
  "vars": [
    "P",
    "A",
    "B",
    "C",
    "D",
    "E",
    "\\\\theta",
    "\\\\alpha",
    "\\\\beta"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "P": "pentagon",
        "A": "vertexa",
        "B": "vertexb",
        "C": "vertexc",
        "D": "vertexd",
        "E": "vertexe",
        "\\theta": "angleth",
        "\\alpha": "angleal",
        "\\beta": "anglebe"
      },
      "question": "Problem A-4\nA convex pentagon \\( pentagon = vertexa vertexb vertexc vertexd vertexe \\), with vertices labeled consecutively, is inscribed in a circle of radius 1 . Find the maximum area of \\( pentagon \\) subject to the condition that the chords \\( vertexa vertexc \\) and \\( vertexb vertexd \\) be perpendicular.",
      "solution": "A-4.\nLet \\( angleth=\\operatorname{Arc} vertexa vertexb, \\quad angleal=\\operatorname{Arc} vertexd vertexe \\), and \\( anglebe=\\operatorname{Arc} vertexe vertexa \\). Then \\( \\operatorname{Arc} vertexc vertexd=\\pi-angleth \\) and \\( \\operatorname{Arc} vertexb vertexc=\\pi-angleal-anglebe \\).\n\nThe area of \\( pentagon \\), in terms of the five triangles from the center of the circle is\n\\[\n\\frac{1}{2} \\sin angleth+\\frac{1}{2} \\sin (\\pi-angleth)+\\frac{1}{2} \\sin angleal+\\frac{1}{2} \\sin anglebe+\\frac{1}{2} \\sin (\\pi-angleal-anglebe)\n\\]\n\nThis is maximized when \\( angleth=\\pi / 2 \\) and \\( angleal=anglebe=\\pi / 3 \\). Thus, the maximum area is\n\\[\n\\frac{1}{2} \\cdot 1+\\frac{1}{2} \\cdot 1+\\frac{1}{2} \\frac{\\sqrt{3}}{2}+\\frac{1}{2} \\frac{\\sqrt{3}}{2}+\\frac{1}{2} \\frac{\\sqrt{3}}{2}=1+\\frac{3}{4} \\sqrt{3}\n\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "P": "stonework",
        "A": "lavender",
        "B": "driftwood",
        "C": "moonlight",
        "D": "fernridge",
        "E": "copperton",
        "\\theta": "gatekeeper",
        "\\alpha": "floodplain",
        "\\beta": "springtime"
      },
      "question": "Problem A-4\nA convex pentagon \\( stonework = lavender driftwood moonlight fernridge copperton \\), with vertices labeled consecutively, is inscribed in a circle of radius 1 . Find the maximum area of \\( stonework \\) subject to the condition that the chords \\( lavender moonlight \\) and \\( driftwood fernridge \\) be perpendicular.",
      "solution": "A-4.\nLet \\( gatekeeper=\\operatorname{Arc} lavender driftwood, \\quad floodplain=\\operatorname{Arc} fernridge copperton \\), and \\( springtime=\\operatorname{Arc} copperton lavender \\). Then \\( \\operatorname{Arc} moonlight fernridge=\\pi-gatekeeper \\) and \\( \\operatorname{Arc} driftwood moonlight= \\)\n\\( \\pi-floodplain-springtime \\).\n\nThe area of \\( stonework \\), in terms of the five triangles from the center of the circle is\n\\[\n\\frac{1}{2} \\sin gatekeeper+\\frac{1}{2} \\sin (\\pi-gatekeeper)+\\frac{1}{2} \\sin floodplain+\\frac{1}{2} \\sin springtime+\\frac{1}{2} \\sin (\\pi-floodplain-springtime)\n\\]\n\nThis is maximized when \\( gatekeeper=\\pi / 2 \\) and \\( floodplain=springtime=\\pi / 3 \\). Thus, the maximum area is\n\\[\n\\frac{1}{2} \\cdot 1+\\frac{1}{2} \\cdot 1+\\frac{1}{2} \\frac{\\sqrt{3}}{2}+\\frac{1}{2} \\frac{\\sqrt{3}}{2}+\\frac{1}{2} \\frac{\\sqrt{3}}{2}=1+\\frac{3}{4} \\sqrt{3}\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "P": "lineshape",
        "A": "voidpoint",
        "B": "nullcorner",
        "C": "centerpoint",
        "D": "midpoint",
        "E": "planespread",
        "\\theta": "straightang",
        "\\alpha": "zeroangle",
        "\\beta": "flatangle"
      },
      "question": "Problem A-4\nA convex pentagon \\( lineshape=voidpoint nullcorner centerpoint midpoint planespread \\), with vertices labeled consecutively, is inscribed in a circle of radius 1 . Find the maximum area of \\( lineshape \\) subject to the condition that the chords \\( voidpoint centerpoint \\) and \\( nullcorner midpoint \\) be perpendicular.",
      "solution": "A-4.\nLet \\( straightang=\\operatorname{Arc} voidpoint nullcorner, \\quad zeroangle=\\operatorname{Arc} midpoint planespread \\), and \\( flatangle=\\operatorname{Arc} planespread voidpoint \\). Then \\( \\operatorname{Arc} centerpoint midpoint=\\pi-straightang \\) and \\( \\operatorname{Arc} nullcorner centerpoint= \\) \\( \\pi-zeroangle-flatangle \\).\n\nThe area of \\( lineshape \\), in terms of the five triangles from the center of the circle is\n\\[\n\\frac{1}{2} \\sin straightang+\\frac{1}{2} \\sin (\\pi-straightang)+\\frac{1}{2} \\sin zeroangle+\\frac{1}{2} \\sin flatangle+\\frac{1}{2} \\sin (\\pi-zeroangle-flatangle)\n\\]\n\nThis is maximized when \\( straightang=\\pi / 2 \\) and \\( zeroangle=flatangle=\\pi / 3 \\). Thus, the maximum area is\n\\[\n\\frac{1}{2} \\cdot 1+\\frac{1}{2} \\cdot 1+\\frac{1}{2} \\frac{\\sqrt{3}}{2}+\\frac{1}{2} \\frac{\\sqrt{3}}{2}+\\frac{1}{2} \\frac{\\sqrt{3}}{2}=1+\\frac{3}{4} \\sqrt{3}\n\\]\n"
    },
    "garbled_string": {
      "map": {
        "P": "qjxvlaet",
        "A": "zgkormufi",
        "B": "phqivorun",
        "C": "mctaygrel",
        "D": "fwnasojid",
        "E": "blifvexun",
        "\\theta": "vifplogam",
        "\\alpha": "sduxaepri",
        "\\beta": "lxqemohat"
      },
      "question": "Problem A-4\nA convex pentagon \\( qjxvlaet = zgkormufi phqivorun mctaygrel fwnasojid blifvexun \\), with vertices labeled consecutively, is inscribed in a circle of radius 1 . Find the maximum area of \\( qjxvlaet \\) subject to the condition that the chords \\( zgkormufi mctaygrel \\) and \\( phqivorun fwnasojid \\) be perpendicular.",
      "solution": "A-4.\nLet \\( vifplogam = \\operatorname{Arc} zgkormufi phqivorun, \\quad sduxaepri = \\operatorname{Arc} fwnasojid blifvexun \\), and \\( lxqemohat = \\operatorname{Arc} blifvexun zgkormufi \\). Then \\( \\operatorname{Arc} mctaygrel fwnasojid = \\pi - vifplogam \\) and \\( \\operatorname{Arc} phqivorun mctaygrel = \\pi - sduxaepri - lxqemohat \\).\n\nThe area of \\( qjxvlaet \\), in terms of the five triangles from the center of the circle is\n\\[\n\\frac{1}{2} \\sin vifplogam + \\frac{1}{2} \\sin (\\pi - vifplogam) + \\frac{1}{2} \\sin sduxaepri + \\frac{1}{2} \\sin lxqemohat + \\frac{1}{2} \\sin (\\pi - sduxaepri - lxqemohat)\n\\]\n\nThis is maximized when \\( vifplogam = \\pi / 2 \\) and \\( sduxaepri = lxqemohat = \\pi / 3 \\). Thus, the maximum area is\n\\[\n\\frac{1}{2} \\cdot 1 + \\frac{1}{2} \\cdot 1 + \\frac{1}{2} \\frac{\\sqrt{3}}{2} + \\frac{1}{2} \\frac{\\sqrt{3}}{2} + \\frac{1}{2} \\frac{\\sqrt{3}}{2} = 1 + \\frac{3}{4} \\sqrt{3}\n\\]"
    },
    "kernel_variant": {
      "question": "Let a convex hexagon \\(P=A_{1}A_{2}A_{3}A_{4}A_{5}A_{6}\\) (vertices listed counter-clockwise) be inscribed in a circle of radius \\(1\\).  \nDenote by  \n • \\(d_{1}=A_{1}A_{4},\\;d_{2}=A_{2}A_{5},\\;d_{3}=A_{3}A_{6}\\)  \nthe three “main” diagonals (each joins opposite vertices).\n\nAssume that  \n\n(i) the three diagonals are concurrent at an interior point \\(O\\);  \n\n(ii) the pairwise angles between the diagonals at \\(O\\) are all \\(60^{\\circ}\\).  \n\nDetermine the maximum possible area of the hexagon \\(P\\).\n\n——————————————————————————————————————",
      "solution": "Step 1.  Translating the angle conditions into relations between arcs  \nWrite the six consecutive central arcs as  \n\n\\[\n\\widehat{A_{1}A_{2}}=\\theta_{1},\\;\\widehat{A_{2}A_{3}}=\\theta_{2},\\;\\dots ,\\;\n\\widehat{A_{6}A_{1}}=\\theta_{6},\\qquad\n\\theta_{i}\\in(0,\\pi),\\quad\\sum_{i=1}^{6}\\theta_{i}=2\\pi .\n\\]\n\nFor two chords that intersect inside a circle the measure of the angle between\nthem equals one half of the sum of the measures of the two arcs subtended by\nthe opposite pairs of endpoints.  \nIn particular,\n\n• the angle between \\(d_{1}=A_{1}A_{4}\\) and \\(d_{2}=A_{2}A_{5}\\) equals  \n\\[\n\\frac12\\bigl(\\widehat{A_{1}A_{2}}+\\widehat{A_{4}A_{5}}\\bigr)=\\frac12\\,\n(\\theta_{1}+\\theta_{4}).\n\\]\n\n• the angle between \\(d_{2}\\) and \\(d_{3}=A_{3}A_{6}\\) equals  \n\\[\n\\frac12\\bigl(\\widehat{A_{2}A_{3}}+\\widehat{A_{5}A_{6}}\\bigr)\n=\\frac12\\,(\\theta_{2}+\\theta_{5}).\n\\]\n\n• the angle between \\(d_{3}\\) and \\(d_{1}\\) equals  \n\\[\n\\frac12\\bigl(\\widehat{A_{3}A_{4}}+\\widehat{A_{6}A_{1}}\\bigr)\n=\\frac12\\,(\\theta_{3}+\\theta_{6}).\n\\]\n\nBecause each of these angles is \\(60^{\\circ}=\\pi/3\\), we obtain the linear\nconstraints  \n\n\\[\n\\boxed{\\;\n\\begin{aligned}\n\\theta_{1}+\\theta_{4}&=\\tfrac{2\\pi}{3},\\\\\n\\theta_{2}+\\theta_{5}&=\\tfrac{2\\pi}{3},\\\\\n\\theta_{3}+\\theta_{6}&=\\tfrac{2\\pi}{3}.%\n\\end{aligned}}\\tag{1}\n\\]\n\nStep 2.  Expressing the area  \nThe area of a polygon inscribed in a unit circle equals one half of the sum\nof the sines of its central arcs.  Hence  \n\n\\[\n\\operatorname{Area}(P)=\n\\frac12\\sum_{i=1}^{6}\\sin\\theta_{i}. \\tag{2}\n\\]\n\nSubject to (1) and \\(0<\\theta_{i}<\\pi\\), we must maximise (2).\n\nStep 3.  Pairwise optimisation  \nBecause the six variables occur only in the three pairs\n\\((\\theta_{1},\\theta_{4}),(\\theta_{2},\\theta_{5}),(\\theta_{3},\\theta_{6})\\),\nproblem (2) decouples into three identical two-variable problems.\nFix one pair, say \\(\\theta_{1}+\\theta_{4}=2\\pi/3\\), and let\n\n\\[\nf(x)=\\sin x+\\sin\\!\\left(\\tfrac{2\\pi}{3}-x\\right),\\qquad 0<x<\\tfrac{2\\pi}{3}.\n\\]\n\nBecause \\(f''(x)=-\\sin x-\\sin\\!\\bigl(\\tfrac{2\\pi}{3}-x\\bigr)<0\\),\n\\(f\\) is concave on the interval; hence it achieves its maximum at the unique\ncritical point where \\(f'(x)=0\\):\n\n\\[\nf'(x)=\\cos x-\\cos\\!\\left(\\tfrac{2\\pi}{3}-x\\right)=0\n\\iff x=\\tfrac{2\\pi}{3}-x\n\\iff x=\\tfrac{\\pi}{3}.\n\\]\n\nThus the maximum of \\(f\\) occurs at  \n\\[\n\\theta_{1}=\\theta_{4}= \\frac{\\pi}{3},\\quad\nf_{\\max}=2\\sin\\!\\left(\\tfrac{\\pi}{3}\\right)=\\sqrt3.\n\\]\n\nExactly the same reasoning yields  \n\n\\[\n\\theta_{2}=\\theta_{5}=\\frac{\\pi}{3},\\qquad\n\\theta_{3}=\\theta_{6}=\\frac{\\pi}{3}. \\tag{3}\n\\]\n\nStep 4.  Verification of feasibility  \nThe angles in (3) clearly satisfy (1) and the positivity constraints, so this\ncandidate is admissible.  Because each separate two-variable optimisation\nreached the global maximum for its pair, and the overall objective is the\nsum of the three pairwise sums, (3) realises the global maximum of (2).\n\nStep 5.  Computing the maximum area  \nWith (3) we have  \n\n\\[\n\\operatorname{Area}_{\\max}\n=\\frac12\\sum_{i=1}^{6}\\sin\\!\\left(\\tfrac{\\pi}{3}\\right)\n=\\frac12\\cdot6\\cdot\\frac{\\sqrt3}{2}\n=\\frac{3\\sqrt3}{2}.\n\\]\n\nEquivalently, \\(P\\) must be the regular hexagon of circumradius \\(1\\);\nindeed, its main diagonals concur at the centre and are separated by\n\\(60^{\\circ}\\), so all constraints are met.\n\nAnswer: \\(\\displaystyle\\operatorname{Area}_{\\max}=\\frac{3\\sqrt3}{2}.\\)\n\n——————————————————————————————————————",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.674780",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension / more variables: the pentagon of the original problem is\nreplaced by a hexagon, introducing six central angles instead of four or\nfive and requiring simultaneous optimisation over a three-dimensional\nfeasible region.\n\n• Additional constraints: instead of a single angle condition on two chords,\nthe problem imposes a concurrency requirement and forces all three pairwise\nangles between the main diagonals to be \\(60^{\\circ}\\).  This yields three\nindependent linear relations among the six variables.\n\n• Deeper theoretical tools: the solution demands the general interior-angle\nformula for intersecting chords, systematic use of concavity/Jensen ideas on\npairs of variables, and a careful examination of global versus pairwise\nmaximisation.\n\n• Multiple interacting concepts: circle geometry (angle–arc relations),\ntrigonometric optimisation, concavity arguments, and polygon area formulae\nwork together.  Identifying that the optimum occurs when the hexagon is\nregular is far less immediate than in the original setup.\n\nThese layers of complexity make the enhanced variant significantly harder,\nboth conceptually and computationally, than the original pentagon problem\nand the current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let a convex hexagon \\(P=A_{1}A_{2}A_{3}A_{4}A_{5}A_{6}\\) (vertices listed counter-clockwise) be inscribed in a circle of radius \\(1\\).  \nDenote by  \n • \\(d_{1}=A_{1}A_{4},\\;d_{2}=A_{2}A_{5},\\;d_{3}=A_{3}A_{6}\\)  \nthe three “main” diagonals (each joins opposite vertices).\n\nAssume that  \n\n(i) the three diagonals are concurrent at an interior point \\(O\\);  \n\n(ii) the pairwise angles between the diagonals at \\(O\\) are all \\(60^{\\circ}\\).  \n\nDetermine the maximum possible area of the hexagon \\(P\\).\n\n——————————————————————————————————————",
      "solution": "Step 1.  Translating the angle conditions into relations between arcs  \nWrite the six consecutive central arcs as  \n\n\\[\n\\widehat{A_{1}A_{2}}=\\theta_{1},\\;\\widehat{A_{2}A_{3}}=\\theta_{2},\\;\\dots ,\\;\n\\widehat{A_{6}A_{1}}=\\theta_{6},\\qquad\n\\theta_{i}\\in(0,\\pi),\\quad\\sum_{i=1}^{6}\\theta_{i}=2\\pi .\n\\]\n\nFor two chords that intersect inside a circle the measure of the angle between\nthem equals one half of the sum of the measures of the two arcs subtended by\nthe opposite pairs of endpoints.  \nIn particular,\n\n• the angle between \\(d_{1}=A_{1}A_{4}\\) and \\(d_{2}=A_{2}A_{5}\\) equals  \n\\[\n\\frac12\\bigl(\\widehat{A_{1}A_{2}}+\\widehat{A_{4}A_{5}}\\bigr)=\\frac12\\,\n(\\theta_{1}+\\theta_{4}).\n\\]\n\n• the angle between \\(d_{2}\\) and \\(d_{3}=A_{3}A_{6}\\) equals  \n\\[\n\\frac12\\bigl(\\widehat{A_{2}A_{3}}+\\widehat{A_{5}A_{6}}\\bigr)\n=\\frac12\\,(\\theta_{2}+\\theta_{5}).\n\\]\n\n• the angle between \\(d_{3}\\) and \\(d_{1}\\) equals  \n\\[\n\\frac12\\bigl(\\widehat{A_{3}A_{4}}+\\widehat{A_{6}A_{1}}\\bigr)\n=\\frac12\\,(\\theta_{3}+\\theta_{6}).\n\\]\n\nBecause each of these angles is \\(60^{\\circ}=\\pi/3\\), we obtain the linear\nconstraints  \n\n\\[\n\\boxed{\\;\n\\begin{aligned}\n\\theta_{1}+\\theta_{4}&=\\tfrac{2\\pi}{3},\\\\\n\\theta_{2}+\\theta_{5}&=\\tfrac{2\\pi}{3},\\\\\n\\theta_{3}+\\theta_{6}&=\\tfrac{2\\pi}{3}.%\n\\end{aligned}}\\tag{1}\n\\]\n\nStep 2.  Expressing the area  \nThe area of a polygon inscribed in a unit circle equals one half of the sum\nof the sines of its central arcs.  Hence  \n\n\\[\n\\operatorname{Area}(P)=\n\\frac12\\sum_{i=1}^{6}\\sin\\theta_{i}. \\tag{2}\n\\]\n\nSubject to (1) and \\(0<\\theta_{i}<\\pi\\), we must maximise (2).\n\nStep 3.  Pairwise optimisation  \nBecause the six variables occur only in the three pairs\n\\((\\theta_{1},\\theta_{4}),(\\theta_{2},\\theta_{5}),(\\theta_{3},\\theta_{6})\\),\nproblem (2) decouples into three identical two-variable problems.\nFix one pair, say \\(\\theta_{1}+\\theta_{4}=2\\pi/3\\), and let\n\n\\[\nf(x)=\\sin x+\\sin\\!\\left(\\tfrac{2\\pi}{3}-x\\right),\\qquad 0<x<\\tfrac{2\\pi}{3}.\n\\]\n\nBecause \\(f''(x)=-\\sin x-\\sin\\!\\bigl(\\tfrac{2\\pi}{3}-x\\bigr)<0\\),\n\\(f\\) is concave on the interval; hence it achieves its maximum at the unique\ncritical point where \\(f'(x)=0\\):\n\n\\[\nf'(x)=\\cos x-\\cos\\!\\left(\\tfrac{2\\pi}{3}-x\\right)=0\n\\iff x=\\tfrac{2\\pi}{3}-x\n\\iff x=\\tfrac{\\pi}{3}.\n\\]\n\nThus the maximum of \\(f\\) occurs at  \n\\[\n\\theta_{1}=\\theta_{4}= \\frac{\\pi}{3},\\quad\nf_{\\max}=2\\sin\\!\\left(\\tfrac{\\pi}{3}\\right)=\\sqrt3.\n\\]\n\nExactly the same reasoning yields  \n\n\\[\n\\theta_{2}=\\theta_{5}=\\frac{\\pi}{3},\\qquad\n\\theta_{3}=\\theta_{6}=\\frac{\\pi}{3}. \\tag{3}\n\\]\n\nStep 4.  Verification of feasibility  \nThe angles in (3) clearly satisfy (1) and the positivity constraints, so this\ncandidate is admissible.  Because each separate two-variable optimisation\nreached the global maximum for its pair, and the overall objective is the\nsum of the three pairwise sums, (3) realises the global maximum of (2).\n\nStep 5.  Computing the maximum area  \nWith (3) we have  \n\n\\[\n\\operatorname{Area}_{\\max}\n=\\frac12\\sum_{i=1}^{6}\\sin\\!\\left(\\tfrac{\\pi}{3}\\right)\n=\\frac12\\cdot6\\cdot\\frac{\\sqrt3}{2}\n=\\frac{3\\sqrt3}{2}.\n\\]\n\nEquivalently, \\(P\\) must be the regular hexagon of circumradius \\(1\\);\nindeed, its main diagonals concur at the centre and are separated by\n\\(60^{\\circ}\\), so all constraints are met.\n\nAnswer: \\(\\displaystyle\\operatorname{Area}_{\\max}=\\frac{3\\sqrt3}{2}.\\)\n\n——————————————————————————————————————",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.529443",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension / more variables: the pentagon of the original problem is\nreplaced by a hexagon, introducing six central angles instead of four or\nfive and requiring simultaneous optimisation over a three-dimensional\nfeasible region.\n\n• Additional constraints: instead of a single angle condition on two chords,\nthe problem imposes a concurrency requirement and forces all three pairwise\nangles between the main diagonals to be \\(60^{\\circ}\\).  This yields three\nindependent linear relations among the six variables.\n\n• Deeper theoretical tools: the solution demands the general interior-angle\nformula for intersecting chords, systematic use of concavity/Jensen ideas on\npairs of variables, and a careful examination of global versus pairwise\nmaximisation.\n\n• Multiple interacting concepts: circle geometry (angle–arc relations),\ntrigonometric optimisation, concavity arguments, and polygon area formulae\nwork together.  Identifying that the optimum occurs when the hexagon is\nregular is far less immediate than in the original setup.\n\nThese layers of complexity make the enhanced variant significantly harder,\nboth conceptually and computationally, than the original pentagon problem\nand the current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}