{
  "index": "1946-B-1",
  "type": "GEO",
  "tag": [
    "GEO",
    "ANA"
  ],
  "difficulty": "",
  "question": "1. Let \\( K \\) denote the circumference of a circular disc of radius one, and let \\( k \\) denote a circular arc that joins two points \\( a, b \\) on \\( K \\) and lies otherwise in the given circular disc. Suppose that \\( k \\) divides the circular disc into two parts of equal area. Prove that the length of \\( k \\) exceeds 2 .",
  "solution": "Solution. If \\( a \\) and \\( b \\) were diametrically opposite on \\( K \\), there would exist no circular arc from \\( a \\) to \\( b \\) that bisects \\( K \\). Hence we may choose coordinates such that \\( K \\) is the unit circle \\( x^{2}+y^{2}=1 \\) and \\( a \\) and \\( b \\) have coordinates \\( (c, d) \\) and ( \\( c,-d \\) ), respectively, where \\( c<0 \\).\n\nNow the arc \\( k \\) divides the circular disc into two parts of equal area, and hence it must intersect the positive \\( x \\) axis at a point \\( e \\). If \\( O \\) is the origin, we get length \\( k>2 a e>2 a O=2 \\).\n\nRemark. The requirement that \\( k \\) be a circular arc is not really important: Any curve \\( k \\) that bisects the disc has length at least 2 with equality if and only if it is a diameter. If the endpoints \\( a \\) and \\( b \\) are diametrically opposite, the conclusion is immediate; otherwise, we can start as above. Then \\( k \\) must contain at least one point \\( p \\) of the open right half-plane, so its length is at least\n\\[\na p+p b>a O+O b=2 .\n\\]",
  "vars": [
    "a",
    "b",
    "c",
    "d",
    "e",
    "x",
    "y",
    "O",
    "p"
  ],
  "params": [
    "K",
    "k"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "a": "pointalpha",
        "b": "pointbeta",
        "c": "coordcharlie",
        "d": "coorddelta",
        "e": "pointecho",
        "x": "axisx",
        "y": "axisy",
        "O": "originpoint",
        "p": "pointphi",
        "K": "bigcircumference",
        "k": "smallarc"
      },
      "question": "1. Let \\( bigcircumference \\) denote the circumference of a circular disc of radius one, and let \\( smallarc \\) denote a circular arc that joins two points \\( pointalpha, pointbeta \\) on \\( bigcircumference \\) and lies otherwise in the given circular disc. Suppose that \\( smallarc \\) divides the circular disc into two parts of equal area. Prove that the length of \\( smallarc \\) exceeds 2 .",
      "solution": "Solution. If \\( pointalpha \\) and \\( pointbeta \\) were diametrically opposite on \\( bigcircumference \\), there would exist no circular arc from \\( pointalpha \\) to \\( pointbeta \\) that bisects \\( bigcircumference \\). Hence we may choose coordinates such that \\( bigcircumference \\) is the unit circle \\( axisx^{2}+axisy^{2}=1 \\) and \\( pointalpha \\) and \\( pointbeta \\) have coordinates \\( (coordcharlie, coorddelta) \\) and ( \\( coordcharlie,-coorddelta \\) ), respectively, where \\( coordcharlie<0 \\).\n\nNow the arc \\( smallarc \\) divides the circular disc into two parts of equal area, and hence it must intersect the positive \\( axisx \\) axis at a point \\( pointecho \\). If \\( originpoint \\) is the origin, we get length \\( smallarc>2 pointalpha pointecho>2 pointalpha originpoint=2 \\).\n\nRemark. The requirement that \\( smallarc \\) be a circular arc is not really important: Any curve \\( smallarc \\) that bisects the disc has length at least 2 with equality if and only if it is a diameter. If the endpoints \\( pointalpha \\) and \\( pointbeta \\) are diametrically opposite, the conclusion is immediate; otherwise, we can start as above. Then \\( smallarc \\) must contain at least one point \\( pointphi \\) of the open right half-plane, so its length is at least\n\\[\npointalpha pointphi+pointphi pointbeta>pointalpha originpoint+originpoint pointbeta=2 .\n\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "a": "sunflower",
        "b": "lighthouse",
        "c": "pendulum",
        "d": "sugarcane",
        "e": "teardrop",
        "x": "cobblestone",
        "y": "moonlight",
        "O": "daybreak",
        "p": "rainstorm",
        "K": "blueberry",
        "k": "sandcastle"
      },
      "question": "1. Let \\( blueberry \\) denote the circumference of a circular disc of radius one, and let \\( sandcastle \\) denote a circular arc that joins two points \\( sunflower, lighthouse \\) on \\( blueberry \\) and lies otherwise in the given circular disc. Suppose that \\( sandcastle \\) divides the circular disc into two parts of equal area. Prove that the length of \\( sandcastle \\) exceeds 2 .",
      "solution": "Solution. If \\( sunflower \\) and \\( lighthouse \\) were diametrically opposite on \\( blueberry \\), there would exist no circular arc from \\( sunflower \\) to \\( lighthouse \\) that bisects \\( blueberry \\). Hence we may choose coordinates such that \\( blueberry \\) is the unit circle \\( cobblestone^{2}+moonlight^{2}=1 \\) and \\( sunflower \\) and \\( lighthouse \\) have coordinates \\( (pendulum, sugarcane) \\) and ( \\( pendulum,-sugarcane \\) ), respectively, where \\( pendulum<0 \\).\n\nNow the arc \\( sandcastle \\) divides the circular disc into two parts of equal area, and hence it must intersect the positive \\( cobblestone \\) axis at a point \\( teardrop \\). If \\( daybreak \\) is the origin, we get length \\( sandcastle>2 sunflower teardrop>2 sunflower daybreak=2 \\).\n\nRemark. The requirement that \\( sandcastle \\) be a circular arc is not really important: Any curve \\( sandcastle \\) that bisects the disc has length at least 2 with equality if and only if it is a diameter. If the endpoints \\( sunflower \\) and \\( lighthouse \\) are diametrically opposite, the conclusion is immediate; otherwise, we can start as above. Then \\( sandcastle \\) must contain at least one point \\( rainstorm \\) of the open right half-plane, so its length is at least\n\\[\nsunflower rainstorm+rainstorm lighthouse>sunflower daybreak+daybreak lighthouse=2 .\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "a": "centerpoint",
        "b": "corevertex",
        "c": "positivex",
        "d": "stationary",
        "e": "outerpoint",
        "x": "verticalax",
        "y": "horizontal",
        "O": "infinitypt",
        "p": "exterior",
        "K": "interior",
        "k": "straightline"
      },
      "question": "1. Let \\( interior \\) denote the circumference of a circular disc of radius one, and let \\( straightline \\) denote a circular arc that joins two points \\( centerpoint, corevertex \\) on \\( interior \\) and lies otherwise in the given circular disc. Suppose that \\( straightline \\) divides the circular disc into two parts of equal area. Prove that the length of \\( straightline \\) exceeds 2 .",
      "solution": "Solution. If \\( centerpoint \\) and \\( corevertex \\) were diametrically opposite on \\( interior \\), there would exist no circular arc from \\( centerpoint \\) to \\( corevertex \\) that bisects \\( interior \\). Hence we may choose coordinates such that \\( interior \\) is the unit circle \\( verticalax^{2}+horizontal^{2}=1 \\) and \\( centerpoint \\) and \\( corevertex \\) have coordinates \\( (positivex, stationary) \\) and \\( (positivex,-stationary) \\), respectively, where \\( positivex<0 \\).\n\nNow the arc \\( straightline \\) divides the circular disc into two parts of equal area, and hence it must intersect the positive \\( verticalax \\) axis at a point \\( outerpoint \\). If \\( infinitypt \\) is the origin, we get length \\( straightline>2 centerpoint outerpoint>2 centerpoint infinitypt=2 \\).\n\nRemark. The requirement that \\( straightline \\) be a circular arc is not really important: Any curve \\( straightline \\) that bisects the disc has length at least 2 with equality if and only if it is a diameter. If the endpoints \\( centerpoint \\) and \\( corevertex \\) are diametrically opposite, the conclusion is immediate; otherwise, we can start as above. Then \\( straightline \\) must contain at least one point \\( exterior \\) of the open right half-plane, so its length is at least\n\\[\ncenterpoint exterior+exterior corevertex>centerpoint infinitypt+infinitypt corevertex=2 .\n\\]"
    },
    "garbled_string": {
      "map": {
        "a": "qzxwvtnp",
        "b": "hjgrksla",
        "c": "plmoknji",
        "d": "uhbvtgfr",
        "e": "zsxplmqa",
        "x": "qwertyui",
        "y": "asdfghjk",
        "O": "lkjhgfds",
        "p": "mnbvcxzq",
        "K": "poiulkjh",
        "k": "lkjhgfdp"
      },
      "question": "1. Let \\( poiulkjh \\) denote the circumference of a circular disc of radius one, and let \\( lkjhgfdp \\) denote a circular arc that joins two points \\( qzxwvtnp, hjgrksla \\) on \\( poiulkjh \\) and lies otherwise in the given circular disc. Suppose that \\( lkjhgfdp \\) divides the circular disc into two parts of equal area. Prove that the length of \\( lkjhgfdp \\) exceeds 2 .",
      "solution": "Solution. If \\( qzxwvtnp \\) and \\( hjgrksla \\) were diametrically opposite on \\( poiulkjh \\), there would exist no circular arc from \\( qzxwvtnp \\) to \\( hjgrksla \\) that bisects \\( poiulkjh \\). Hence we may choose coordinates such that \\( poiulkjh \\) is the unit circle \\( qwertyui^{2}+asdfghjk^{2}=1 \\) and \\( qzxwvtnp \\) and \\( hjgrksla \\) have coordinates \\( (plmoknji, uhbvtgfr) \\) and ( \\( plmoknji,-uhbvtgfr \\) ), respectively, where \\( plmoknji<0 \\).\n\nNow the arc \\( lkjhgfdp \\) divides the circular disc into two parts of equal area, and hence it must intersect the positive \\( qwertyui \\) axis at a point \\( zsxplmqa \\). If \\( lkjhgfds \\) is the origin, we get length \\( lkjhgfdp>2 qzxwvtnp zsxplmqa>2 qzxwvtnp lkjhgfds=2 \\).\n\nRemark. The requirement that \\( lkjhgfdp \\) be a circular arc is not really important: Any curve \\( lkjhgfdp \\) that bisects the disc has length at least 2 with equality if and only if it is a diameter. If the endpoints \\( qzxwvtnp \\) and \\( hjgrksla \\) are diametrically opposite, the conclusion is immediate; otherwise, we can start as above. Then \\( lkjhgfdp \\) must contain at least one point \\( mnbvcxzq \\) of the open right half-plane, so its length is at least\n\\[\nqzxwvtnp mnbvcxzq+mnbvcxzq hjgrksla>qzxwvtnp lkjhgfds+lkjhgfds hjgrksla=2 .\n\\]"
    },
    "kernel_variant": {
      "question": "Let  D  be the closed circular disc of radius 4 centred at the origin and let  K = ∂D  be its circumference.  A proper circular arc is an arc of some circle of finite radius whose interior points lie in the open disc  D°  (so a straight diameter is not considered a proper arc).\n\nLet  k  be a proper circular arc that joins two distinct points  a , b  on  K  and whose interior lies in  D°.  Suppose that the union of  k  with one (and hence only one) of the two arcs of  K  between  a  and  b  encloses exactly one half of the area of  D.  Prove that the length of  k  is strictly greater than 8.",
      "solution": "Write\n      D = { (x,y) ∈ ℝ² : x² + y² ≤ 16 },    K = { (x,y) : x² + y² = 16 }.\nDenote the endpoints of k by a and b and set |ab| for their Euclidean distance.  Because the radius of D equals 4 we always have |ab| ≤ 8, with equality precisely when a and b are antipodal.\n\nWe distinguish two situations.\n\n------------------------------------------------------------\n1.  a and b are diametrically opposite ( |ab| = 8 ).\n------------------------------------------------------------\nLet ρ be the radius and φ (0 < φ < 2π) the central angle of the circle that contains k; then the chord–length formula gives\n          |ab| = 2ρ sin(φ/2).                           (1)\nThe length of k is ρφ, hence\n      length(k) = ρφ = |ab| · φ /(2 sin(φ/2)).          (2)\nBecause 0 < φ/2 < π we have sin(φ/2) < φ/2, so the fraction in (2) exceeds 1.  With |ab| = 8 we obtain\n          length(k) > 8.\n(Straight diameters have length 8 but are excluded by the word “proper”.)  In this case the area-bisecting hypothesis is irrelevant; any proper circular arc with antipodal endpoints is already long enough.\n\n------------------------------------------------------------\n2.  a and b are not diametrically opposite ( |ab| < 8 ).\n------------------------------------------------------------\nBy a rigid motion of the plane we may suppose that a and b are symmetric with respect to the y-axis and lie in the lower half-plane.  Thus\n          a = ( d , −h ),    b = ( −d , −h ),\nwith d > 0,  h > 0 and d² + h² = 16.\n\nExistence of an intersection with the positive y-axis\n–––––––––––––––––––––––––––––––––––––––––––––––––––––––\nBecause the endpoints are symmetric, the perpendicular bisector of the chord ab is the y-axis x = 0.  Let C be the centre of the circle that carries k.  This centre lies on the perpendicular bisector, hence on the y-axis.  As a consequence the entire arc k is symmetric with respect to the y-axis and therefore must meet this axis.\n\nSuppose, for a contradiction, that every point of k on the y-axis satisfies y ≤ 0.  Then k together with the lower semicircle of K ( y ≤ 0 ) would bound a region contained strictly inside that lower semicircle.  The lower semicircle already has area 8π, i.e. exactly one half of the disc, so the region just described would have area strictly smaller than 8π.  This contradicts the hypothesis that k together with some sub-arc of K encloses one half of the disc.  Hence k must meet the y-axis at some point whose y-coordinate is positive.\n\nDenote by\n          e = (0, t),     t > 0,\nthe first such point encountered when we travel along k from a to b.  Split k into two sub-arcs\n          k₁ : a → e,          k₂ : e → b.\n\nReflection argument and a lower bound for |ae|\n–––––––––––––––––––––––––––––––––––––––––––––––\nReflect k₂ across the y-axis to obtain a new arc k₂′ from e back to a.  Reflection is an isometry, hence\n          length(k) = length(k₁) + length(k₂) = length(k₁) + length(k₂′).\nEach of k₁ and k₂′ is a proper circular arc, so each is strictly longer than the straight segment connecting its endpoints.  Therefore\n          length(k) > |ae| + |ea| = 2|ae|.            (3)\n\nWe next estimate |ae|.  From a = (d,−h) and e = (0,t)\n      |ae|² = d² + (h + t)²\n             = (d² + h²) + 2ht + t²\n             = 16 + 2ht + t²    (> 16),\nbecause h > 0 and t > 0.  Consequently |ae| > 4.\n\nCombining this with (3) yields\n          length(k) > 2 · 4 = 8.\n\n------------------------------------------------------------\nConclusion\n------------------------------------------------------------\nIn both cases we have established length(k) > 8.  Hence every proper circular arc that bisects the area of the disc of radius 4 has length strictly greater than 8. ∎",
      "_meta": {
        "core_steps": [
          "If endpoints were opposite, a diameter would already bisect the disc; otherwise place the disc so the endpoints are symmetric with respect to an axis (say the x–axis).",
          "Because the curve cuts the area exactly in half while both endpoints lie on the same side of the chosen diameter, the curve must cross that diameter at some interior point e in the opposite half–plane.",
          "Apply the triangle‐inequality (using the reflection of the sub-arc below the axis) to get length(k) > 2·AE > 2·AO, where AO is the radius; thus length(k) > 2·radius ( = 2 when the radius is 1)."
        ],
        "mutable_slots": {
          "radius": {
            "description": "Common scale of the disc; the whole argument is scale-invariant.",
            "original": "1"
          },
          "chosen_axis": {
            "description": "Diameter used as the symmetry/area-counting line crossed by k (x-axis in the write-up).",
            "original": "x-axis"
          },
          "endpoint_side": {
            "description": "Which half-plane (left, right, top, …) initially contains both endpoints so that the curve must enter the opposite half-plane.",
            "original": "c < 0  (left half-plane)"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}