{
  "index": "1974-B-1",
  "type": "GEO",
  "tag": [
    "GEO",
    "COMB"
  ],
  "difficulty": "",
  "question": "B-1. Which configurations of five (not necessarily distinct) points \\( p_{1}, \\cdots, p_{\\text {s }} \\) on the circle \\( x^{2}+y^{2}=1 \\) maximize the sum of the ten distances\n\\[\n\\sum_{i<1} d\\left(p_{i}, p_{i}\\right) ?\n\\]\n[Here \\( d(p, q) \\) denotes the straight line distance between \\( p \\) and \\( q \\).]",
  "solution": "B-1.\nSince the \\( p_{i} \\) need not be distinct, the sum is a continuous function on the compact set \\( C \\times C \\times C \\times C \\times C \\), where \\( C \\) is the circle. Hence maxima exist. One proves that the maximum occurs when the \\( p_{i} \\) are the vertices of a regular pentagon by showing that this configuration simultaneously maximizes both of the sums:\n\\[\n\\begin{array}{l}\nS=d\\left(p_{1}, p_{2}\\right)+d\\left(p_{2}, p_{3}\\right)+d\\left(p_{3}, p_{4}\\right)+d\\left(p_{4}, p_{5}\\right)+d\\left(p_{5}, p_{1}\\right), \\\\\nT=d\\left(p_{1}, p_{3}\\right)+d\\left(p_{2}, p_{4}\\right)+d\\left(p_{3}, p_{5}\\right)+d\\left(p_{4}, p_{1}\\right)+d\\left(p_{5}, p_{2}\\right) .\n\\end{array}\n\\]\n\nFor \\( S \\) or \\( T \\), one can fix four of the points; then the varying part of the sum is of the form.\n\\[\nD=d(p, a)+d(p, b), \\text { with } a \\text { and } b \\text { fixed. }\n\\]\n\nUsing the Law of Sines, one shows that \\( D \\) is a constant times \\( \\sin \\alpha+\\sin \\beta \\) where \\( \\alpha=\\Varangle p a b \\), \\( \\beta=\\Varangle p b a \\), and \\( \\alpha+\\beta \\) is constant. Then it is easy to show that \\( D \\) is not a maximum unless \\( p \\) is symmetrically situated with respect to \\( a \\) and \\( b \\).",
  "vars": [
    "x",
    "y",
    "p",
    "p_i",
    "p_1",
    "p_2",
    "p_3",
    "p_4",
    "p_5",
    "q",
    "i",
    "j",
    "s",
    "S",
    "T",
    "D",
    "\\\\alpha",
    "\\\\beta"
  ],
  "params": [
    "a",
    "b"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "xcoord",
        "y": "ycoord",
        "p": "pointgen",
        "p_i": "pointi",
        "p_1": "pointone",
        "p_2": "pointtwo",
        "p_3": "pointthr",
        "p_4": "pointfor",
        "p_5": "pointfiv",
        "q": "pointque",
        "i": "indexi",
        "j": "indexj",
        "s": "indexs",
        "S": "sumsucc",
        "T": "sumother",
        "D": "distvar",
        "\\alpha": "angleone",
        "\\beta": "angletwo",
        "a": "parpointa",
        "b": "parpointb"
      },
      "question": "B-1. Which configurations of five (not necessarily distinct) points \\( pointone, \\cdots, pointgen_{\\text{indexs}} \\) on the circle \\( xcoord^{2}+ycoord^{2}=1 \\) maximize the sum of the ten distances\n\\[\n\\sum_{indexi<1} d\\left(pointi, pointi\\right) ?\n\\]\n[Here \\( d(pointgen, pointque) \\) denotes the straight line distance between \\( pointgen \\) and \\( pointque \\).]",
      "solution": "B-1.\nSince the \\( pointi \\) need not be distinct, the sum is a continuous function on the compact set \\( C \\times C \\times C \\times C \\times C \\), where \\( C \\) is the circle. Hence maxima exist. One proves that the maximum occurs when the \\( pointi \\) are the vertices of a regular pentagon by showing that this configuration simultaneously maximizes both of the sums:\n\\[\n\\begin{array}{l}\nsumsucc=d\\left(pointone, pointtwo\\right)+d\\left(pointtwo, pointthr\\right)+d\\left(pointthr, pointfor\\right)+d\\left(pointfor, pointfiv\\right)+d\\left(pointfiv, pointone\\right), \\\\\nsumother=d\\left(pointone, pointthr\\right)+d\\left(pointtwo, pointfor\\right)+d\\left(pointthr, pointfiv\\right)+d\\left(pointfor, pointone\\right)+d\\left(pointfiv, pointtwo\\right) .\n\\end{array}\n\\]\n\nFor \\( sumsucc \\) or \\( sumother \\), one can fix four of the points; then the varying part of the sum is of the form.\n\\[\ndistvar=d(pointgen, parpointa)+d(pointgen, parpointb), \\text { with } parpointa \\text { and } parpointb \\text { fixed. }\n\\]\n\nUsing the Law of Sines, one shows that \\( distvar \\) is a constant times \\( \\sin angleone+\\sin angletwo \\) where \\( angleone=\\Varangle pointgen\\ parpointa\\ parpointb \\), \\( angletwo=\\Varangle pointgen\\ parpointb\\ parpointa \\), and \\( angleone+angletwo \\) is constant. Then it is easy to show that \\( distvar \\) is not a maximum unless \\( pointgen \\) is symmetrically situated with respect to \\( parpointa \\) and \\( parpointb \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "tulipseed",
        "y": "cobaltash",
        "p": "glimmerfog",
        "p_i": "silentrune",
        "p_1": "whistleray",
        "p_2": "embermoss",
        "p_3": "lilacstone",
        "p_4": "harborquill",
        "p_5": "marblevine",
        "q": "zephyrnote",
        "i": "orchardmist",
        "j": "lanternbay",
        "s": "mulberrytide",
        "S": "frostedvale",
        "T": "crimsonhaze",
        "D": "meadowglen",
        "\\\\alpha": "sunlitcove",
        "\\\\beta": "rainwooden",
        "a": "thundersail",
        "b": "velvetpeak"
      },
      "question": "B-1. Which configurations of five (not necessarily distinct) points \\( whistleray, \\cdots, glimmerfog_{\\text {s }} \\) on the circle \\( tulipseed^{2}+cobaltash^{2}=1 \\) maximize the sum of the ten distances\n\\[\n\\sum_{orchardmist<1} d\\left(glimmerfog_{orchardmist}, glimmerfog_{orchardmist}\\right) ?\n\\]\n[Here \\( d(glimmerfog, zephyrnote) \\) denotes the straight line distance between \\( glimmerfog \\) and \\( zephyrnote \\).]",
      "solution": "B-1.\nSince the \\( glimmerfog_{orchardmist} \\) need not be distinct, the sum is a continuous function on the compact set \\( C \\times C \\times C \\times C \\times C \\), where \\( C \\) is the circle. Hence maxima exist. One proves that the maximum occurs when the \\( glimmerfog_{orchardmist} \\) are the vertices of a regular pentagon by showing that this configuration simultaneously maximizes both of the sums:\n\\[\n\\begin{array}{l}\nfrostedvale=d\\left(glimmerfog_{1}, glimmerfog_{2}\\right)+d\\left(glimmerfog_{2}, glimmerfog_{3}\\right)+d\\left(glimmerfog_{3}, glimmerfog_{4}\\right)+d\\left(glimmerfog_{4}, glimmerfog_{5}\\right)+d\\left(glimmerfog_{5}, glimmerfog_{1}\\right), \\\\\ncrimsonhaze=d\\left(glimmerfog_{1}, glimmerfog_{3}\\right)+d\\left(glimmerfog_{2}, glimmerfog_{4}\\right)+d\\left(glimmerfog_{3}, glimmerfog_{5}\\right)+d\\left(glimmerfog_{4}, glimmerfog_{1}\\right)+d\\left(glimmerfog_{5}, glimmerfog_{2}\\right) .\n\\end{array}\n\\]\n\nFor \\( frostedvale \\) or \\( crimsonhaze \\), one can fix four of the points; then the varying part of the sum is of the form.\n\\[\nmeadowglen=d(glimmerfog, thundersail)+d(glimmerfog, velvetpeak), \\text { with } thundersail \\text { and } velvetpeak \\text { fixed. }\n\\]\n\nUsing the Law of Sines, one shows that \\( meadowglen \\) is a constant times \\( \\sin sunlitcove+\\sin rainwooden \\) where \\( sunlitcove=\\Varangle glimmerfog\\, thundersail\\, velvetpeak \\), \\( rainwooden=\\Varangle glimmerfog\\, velvetpeak\\, thundersail \\), and \\( sunlitcove+rainwooden \\) is constant. Then it is easy to show that \\( meadowglen \\) is not a maximum unless \\( glimmerfog \\) is symmetrically situated with respect to \\( thundersail \\) and \\( velvetpeak \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalcoor",
        "y": "horizontalcoor",
        "p": "voidentity",
        "p_i": "nullcollection",
        "p_1": "nullpointone",
        "p_2": "nullpointtwo",
        "p_3": "nullpointthr",
        "p_4": "nullpointfou",
        "p_5": "nullpointfiv",
        "q": "constantval",
        "i": "continuousvar",
        "j": "staticindex",
        "s": "infinitesize",
        "S": "differenceval",
        "T": "productval",
        "D": "proximity",
        "\\alpha": "straightness",
        "\\beta": "flatness",
        "a": "endpoint",
        "b": "midpoint"
      },
      "question": "B-1. Which configurations of five (not necessarily distinct) points \\( nullpointone, \\cdots, voidentity_{\\text {infinitesize }} \\) on the circle \\( verticalcoor^{2}+horizontalcoor^{2}=1 \\) maximize the sum of the ten distances\n\\[\n\\sum_{continuousvar<1} d\\left(nullcollection, nullcollection\\right) ?\n\\]\n[Here \\( d(voidentity, constantval) \\) denotes the straight line distance between \\( voidentity \\) and \\( constantval \\).]",
      "solution": "B-1.\nSince the \\( voidentity_{continuousvar} \\) need not be distinct, the sum is a continuous function on the compact set \\( C \\times C \\times C \\times C \\times C \\), where \\( C \\) is the circle. Hence maxima exist. One proves that the maximum occurs when the \\( voidentity_{continuousvar} \\) are the vertices of a regular pentagon by showing that this configuration simultaneously maximizes both of the sums:\n\\[\n\\begin{array}{l}\ndifferenceval=d\\left(nullpointone, nullpointtwo\\right)+d\\left(nullpointtwo, nullpointthr\\right)+d\\left(nullpointthr, nullpointfou\\right)+d\\left(nullpointfou, nullpointfiv\\right)+d\\left(nullpointfiv, nullpointone\\right), \\\\\nproductval=d\\left(nullpointone, nullpointthr\\right)+d\\left(nullpointtwo, nullpointfou\\right)+d\\left(nullpointthr, nullpointfiv\\right)+d\\left(nullpointfou, nullpointone\\right)+d\\left(nullpointfiv, nullpointtwo\\right) .\n\\end{array}\n\\]\n\nFor \\( differenceval \\) or \\( productval \\), one can fix four of the points; then the varying part of the sum is of the form.\n\\[\nproximity=d(voidentity, endpoint)+d(voidentity, midpoint), \\text { with } endpoint \\text { and } midpoint \\text { fixed. }\n\\]\n\nUsing the Law of Sines, one shows that \\( proximity \\) is a constant times \\( \\sin straightness+\\sin flatness \\) where \\( straightness=\\Varangle voidentity endpoint midpoint \\), \\( flatness=\\Varangle voidentity midpoint endpoint \\), and \\( straightness+flatness \\) is constant. Then it is easy to show that \\( proximity \\) is not a maximum unless \\( voidentity \\) is symmetrically situated with respect to \\( endpoint \\) and \\( midpoint \\)."
    },
    "garbled_string": {
      "map": {
        "x": "lqmrznxh",
        "y": "vgtsckbl",
        "p": "kwzdphqa",
        "p_i": "qnjtmsvo",
        "p_1": "afyerhqd",
        "p_2": "ghzlptwa",
        "p_3": "jcvmorbe",
        "p_4": "mydkseuh",
        "p_5": "ozwcrnfa",
        "q": "tbkglsme",
        "i": "rupdxeon",
        "j": "nhcivpaw",
        "s": "fqeodkri",
        "S": "bermwqlj",
        "T": "szxavcno",
        "D": "uwrqgpyh",
        "\\alpha": "qzxwvtnp",
        "\\beta": "hjgrksla",
        "a": "mtylgzsd",
        "b": "xprfucen"
      },
      "question": "B-1. Which configurations of five (not necessarily distinct) points \\(afyerhqd, \\cdots, kwzdphqa_{\\text {fqeodkri }}\\) on the circle \\(lqmrznxh^{2}+vgtsckbl^{2}=1\\) maximize the sum of the ten distances\n\\[\n\\sum_{rupdxeon<1} d\\left(qnjtmsvo, qnjtmsvo\\right) ?\n\\]\n[Here \\( d(kwzdphqa, tbkglsme) \\) denotes the straight line distance between \\( kwzdphqa \\) and \\( tbkglsme \\).]",
      "solution": "Since the \\( qnjtmsvo \\) need not be distinct, the sum is a continuous function on the compact set \\( C \\times C \\times C \\times C \\times C \\), where \\( C \\) is the circle. Hence maxima exist. One proves that the maximum occurs when the \\( qnjtmsvo \\) are the vertices of a regular pentagon by showing that this configuration simultaneously maximizes both of the sums:\n\\[\n\\begin{array}{l}\nbermwqlj=d\\left(afyerhqd, ghzlptwa\\right)+d\\left(ghzlptwa, jcvmorbe\\right)+d\\left(jcvmorbe, mydkseuh\\right)+d\\left(mydkseuh, ozwcrnfa\\right)+d\\left(ozwcrnfa, afyerhqd\\right), \\\\\nszxavcno=d\\left(afyerhqd, jcvmorbe\\right)+d\\left(ghzlptwa, mydkseuh\\right)+d\\left(jcvmorbe, ozwcrnfa\\right)+d\\left(mydkseuh, afyerhqd\\right)+d\\left(ozwcrnfa, ghzlptwa\\right) .\n\\end{array}\n\\]\n\nFor \\( bermwqlj \\) or \\( szxavcno \\), one can fix four of the points; then the varying part of the sum is of the form.\n\\[\nuwrqgpyh=d(kwzdphqa, mtylgzsd)+d(kwzdphqa, xprfucen), \\text { with } mtylgzsd \\text { and } xprfucen \\text { fixed. }\n\\]\n\nUsing the Law of Sines, one shows that \\( uwrqgpyh \\) is a constant times \\( \\sin qzxwvtnp+\\sin hjgrksla \\) where \\( qzxwvtnp=\\Varangle kwzdphqa mtylgzsd xprfucen \\), \\( hjgrksla=\\Varangle kwzdphqa xprfucen mtylgzsd \\), and \\( qzxwvtnp+hjgrksla \\) is constant. Then it is easy to show that \\( uwrqgpyh \\) is not a maximum unless \\( kwzdphqa \\) is symmetrically situated with respect to \\( mtylgzsd \\) and \\( xprfucen \\)."
    },
    "kernel_variant": {
      "question": "Let 𝒞 be the circle x² + y² = 4.  For seven distinct points P₁, P₂, … , P₇ on 𝒞 put\n        Σ  =  ∑_{1 ≤ i < j ≤ 7} d(Pᵢ , Pⱼ),\nthe sum of their \\(\\binom{7}{2}=21\\) pairwise Euclidean distances.\nDetermine   \n(a) the maximal possible value of Σ, and  \n(b) all configurations of the seven points that realise this maximum.",
      "solution": "Throughout write R = 2 and identify the circle 𝒞 with the complex numbers of modulus R.\n\nStep 1 – An angular description.\nAfter a rigid rotation we may assume\n        0 = θ₁ < θ₂ < ⋯ < θ₇ < 2π.                    (1)\nPut Pₖ = R e^{iθₖ}.  Introduce the successive gaps\n        αₖ = θ_{k+1} − θₖ   (k = 1,…,7 , θ₈ := θ₁+2π)            (2)\nso that\n        α₁+⋯+α₇ = 2π , αₖ > 0 .                               (3)\n\nFor two points whose (smaller) central angle is φ ∈ (0,2π) the chord length equals\n        d = 2R sin(φ/2) = 4 sin(φ/2).                          (4)\n\nFor the seven points every unordered pair is characterised by its cyclic index distance k = 1,2,3.  (Distance 4 gives the same chord as 3 because ‘going the other way round’ yields the complementary angle 2π−φ, but (4) shows the chord is the same.)  Set\n        β_{i,k} = α_i + α_{i+1} + … + α_{i+k-1}   (indices mod 7).  (5)\nWith f(t) :=  sin(t/2) we get\n        Σ(α₁,…,α₇) = 4  ∑_{k=1}^{3}  ∑_{i=1}^{7}  f(β_{i,k}).        (6)\nThe right–hand side contains exactly the 21 distances.\n\nStep 2 – A Jensen–type upper bound.\nThe function f(t)=sin(t/2) is concave on (0,2π) because f''(t)=−¼ sin(t/2)<0.  For every fixed k∈{1,2,3} we apply Jensen’s inequality to the seven numbers β_{1,k},…,β_{7,k} :\n        (1/7)∑_{i=1}^{7} f(β_{i,k}) ≤ f\\bigl((1/7)∑_{i=1}^{7} β_{i,k}\\bigr).     (7)\nEquality occurs iff β_{1,k}=⋯=β_{7,k}.\n\nBecause each α_j occurs exactly k times in the sum ∑_{i} β_{i,k},\n        ∑_{i=1}^{7} β_{i,k} = k(α₁+⋯+α₇)=k·2π,                 (8)\nso the average in (7) equals (k·2π)/7.  Insert this into (7) and multiply by 7:\n        ∑_{i=1}^{7} f(β_{i,k}) ≤ 7 f(k·2π/7).                     (9)\nUsing (6) we obtain the global inequality\n        Σ(α₁,…,α₇) ≤ 4·7·[ f(2π/7)+f(4π/7)+f(6π/7) ]              (10)\n        = 4·7·[ sin(π/7)+sin(2π/7)+sin(3π/7) ].\n\nStep 3 – Characterisation of the equality case.\nFor k = 1, inequality (9) reads ∑ f(α_i) ≤ 7 f(2π/7).  Equality in (10) therefore forces equality in the k=1 instance of (9), whence\n        α₁ = α₂ = ⋯ = α₇ = 2π/7.                               (11)\nConversely, the equal–gap point (11) indeed realises equality in (9) for k=1,2,3, hence attains the upper bound (10).\n\nThus the only maximising 7–tuple (α₁,…,α₇) is (2π/7,…,2π/7).  Up to a cyclic shift – which merely renames the points – there is no other possibility.\n\nStep 4 – Evaluation of the maximum.\nInsert (11) into (6).  Now β_{i,1}=2π/7, β_{i,2}=4π/7 and β_{i,3}=6π/7, so\n  Σ_max = 4 · 7 · ( sin(π/7) + sin(2π/7) + sin(3π/7) )\n         ≈ 4 · 7 · 2.1906419…\n         ≈ 61.34 .                                            (12)\n(Using sin(π/7)≈0.433883, sin(2π/7)≈0.781831, sin(3π/7)≈0.974928.)\n\nStep 5 – Geometric description of the maximisers.\nBecause Σ is invariant under rigid motions of 𝒞, rotating the unique angular configuration (11) around the centre produces every maximising 7–tuple (P₁,…,P₇).  In other words, the points must be the seven vertices of some regular heptagon inscribed in x²+y² = 4, and every such heptagon indeed gives the maximal value (12).\n\nAnswer.\n(a) The maximal possible value of Σ is\n        4·7·[ sin(π/7)+sin(2π/7)+sin(3π/7) ] ≈ 61.34.\n(b) It is attained exactly when the seven points are the vertices of a regular heptagon on the circle x² + y² = 4 (all such heptagons being congruent by rotation).",
      "_meta": {
        "core_steps": [
          "Use compactness of C^5 to guarantee a maximizing configuration exists.",
          "Write the total pair-sum as S (distances of neighbors) + T (distances of next-neighbors).",
          "With four points fixed, the part that varies is D = d(p,a)+d(p,b).",
          "Apply the Law of Sines to get D = k( sinα + sinβ ) with α+β fixed; this is maximized when α = β, i.e. p is symmetric w.r.t. a and b.",
          "Iterating that symmetry for every vertex yields equally spaced vertices ⇒ regular pentagon."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Radius of the circle; any rescaling multiplies every distance by the same constant and does not affect which configuration maximizes the sum.",
            "original": "x² + y² = 1 (unit circle)"
          },
          "slot2": {
            "description": "Cardinality of the point set; the same symmetry argument works for n points and would lead to a regular n-gon.",
            "original": "5 points (10 pairwise distances)"
          },
          "slot3": {
            "description": "Permission for coincident points; forbidding or allowing repeats does not change the maximizer because the optimum has distinct points anyway.",
            "original": "‘not necessarily distinct’"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}