{
  "index": "1996-A-2",
  "type": "GEO",
  "tag": [
    "GEO"
  ],
  "difficulty": "",
  "question": "Let $C_1$ and $C_2$ be circles whose centers are 10 units apart, and\nwhose radii are 1 and 3. Find, with proof, the locus of all points $M$\nfor which there exists points $X$ on $C_1$ and $Y$ on $C_2$ such that\n$M$ is the midpoint of the line segment $XY$.",
  "solution": "Let $O_1$ and $O_2$ be the centers of $C_1$ and $C_2$, respectively.\n(We are assuming $C_1$ has radius 1 and $C_2$ has radius 3.)\nThen the\ndesired locus is an annulus centered at the midpoint of $O_1O_2$, with\ninner radius 1 and outer radius 2.\n\nFor a fixed point $Q$ on $C_2$, the locus of the midpoints of the\nsegments $PQ$ for $P$ lying on $C_1$ is the image of $C_1$ under a\nhomothety centered at $Q$ of radius $1/2$, which is a circle of radius\n$1/2$. As $Q$ varies, the center of this smaller circle traces out a\ncircle $C_3$ of radius $3/2$ (again by homothety). By considering the two\npositions of $Q$ on the line of centers of the circles, one sees that\n$C_3$ is centered at the midpoint of $O_1O_2$, and the locus is now\nclearly the specified annulus.",
  "vars": [
    "C_1",
    "C_2",
    "M",
    "X",
    "Y",
    "O_1",
    "O_2",
    "Q",
    "P",
    "C_3"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "C_1": "circleone",
        "C_2": "circletwo",
        "M": "midpoint",
        "X": "firstpoint",
        "Y": "secondpoint",
        "O_1": "centerone",
        "O_2": "centertwo",
        "Q": "varypoint",
        "P": "pointonone",
        "C_3": "circlethree"
      },
      "question": "Let $circleone$ and $circletwo$ be circles whose centers are 10 units apart, and\nwhose radii are 1 and 3. Find, with proof, the locus of all points $midpoint$\nfor which there exists points $firstpoint$ on $circleone$ and $secondpoint$ on $circletwo$ such that\n$midpoint$ is the midpoint of the line segment $firstpoint secondpoint$.",
      "solution": "Let $centerone$ and $centertwo$ be the centers of $circleone$ and $circletwo$, respectively.\n(We are assuming $circleone$ has radius 1 and $circletwo$ has radius 3.)\nThen the\ndesired locus is an annulus centered at the midpoint of $centeronecentertwo$, with\ninner radius 1 and outer radius 2.\n\nFor a fixed point $varypoint$ on $circletwo$, the locus of the midpoints of the\nsegments $pointonone varypoint$ for $pointonone$ lying on $circleone$ is the image of $circleone$ under a\nhomothety centered at $varypoint$ of radius $1/2$, which is a circle of radius\n$1/2$. As $varypoint$ varies, the center of this smaller circle traces out a\ncircle $circlethree$ of radius $3/2$ (again by homothety). By considering the two\npositions of $varypoint$ on the line of centers of the circles, one sees that\n$circlethree$ is centered at the midpoint of $centeronecentertwo$, and the locus is now\nclearly the specified annulus."
    },
    "descriptive_long_confusing": {
      "map": {
        "C_1": "pineapple",
        "C_2": "salamander",
        "M": "telescope",
        "X": "raincloud",
        "Y": "honeycomb",
        "O_1": "carousel",
        "O_2": "paperclip",
        "Q": "butterfly",
        "P": "notebook",
        "C_3": "flashlight"
      },
      "question": "Let $pineapple$ and $salamander$ be circles whose centers are 10 units apart, and\nwhose radii are 1 and 3. Find, with proof, the locus of all points $telescope$\nfor which there exists points $raincloud$ on $pineapple$ and $honeycomb$ on $salamander$ such that\n$telescope$ is the midpoint of the line segment $raincloudhoneycomb$.",
      "solution": "Let $carousel$ and $paperclip$ be the centers of $pineapple$ and $salamander$, respectively.\n(We are assuming $pineapple$ has radius 1 and $salamander$ has radius 3.)\nThen the\ndesired locus is an annulus centered at the midpoint of $carouselpaperclip$, with\ninner radius 1 and outer radius 2.\n\nFor a fixed point $butterfly$ on $salamander$, the locus of the midpoints of the\nsegments $notebookbutterfly$ for $notebook$ lying on $pineapple$ is the image of $pineapple$ under a\nhomothety centered at $butterfly$ of radius $1/2$, which is a circle of radius\n$1/2$. As $butterfly$ varies, the center of this smaller circle traces out a\ncircle $flashlight$ of radius $3/2$ (again by homothety). By considering the two\npositions of $butterfly$ on the line of centers of the circles, one sees that\n$flashlight$ is centered at the midpoint of $carouselpaperclip$, and the locus is now\nclearly the specified annulus."
    },
    "descriptive_long_misleading": {
      "map": {
        "C_1": "straightline",
        "C_2": "flatplane",
        "M": "endpoint",
        "X": "offcircle",
        "Y": "outsidearc",
        "O_1": "peripheryone",
        "O_2": "peripherytwo",
        "Q": "variablepoint",
        "P": "fixedpoint",
        "C_3": "trianglearea"
      },
      "question": "Let $straightline$ and $flatplane$ be circles whose centers are 10 units apart, and\nwhose radii are 1 and 3. Find, with proof, the locus of all points $endpoint$\nfor which there exists points $offcircle$ on $straightline$ and $outsidearc$ on $flatplane$ such that\n$endpoint$ is the midpoint of the line segment $offcircle outsidearc$.",
      "solution": "Let $peripheryone$ and $peripherytwo$ be the centers of $straightline$ and $flatplane$, respectively.\n(We are assuming $straightline$ has radius 1 and $flatplane$ has radius 3.)\nThen the\ndesired locus is an annulus centered at the midpoint of $peripheryone peripherytwo$, with\ninner radius 1 and outer radius 2.\n\nFor a fixed point $variablepoint$ on $flatplane$, the locus of the midpoints of the\nsegments $fixedpoint variablepoint$ for $fixedpoint$ lying on $straightline$ is the image of $straightline$ under a\nhomothety centered at $variablepoint$ of radius $1/2$, which is a circle of radius $1/2$. As $variablepoint$ varies, the center of this smaller circle traces out a circle $trianglearea$ of radius $3/2$ (again by homothety). By considering the two positions of $variablepoint$ on the line of centers of the circles, one sees that $trianglearea$ is centered at the midpoint of $peripheryone peripherytwo$, and the locus is now clearly the specified annulus."
    },
    "garbled_string": {
      "map": {
        "C_1": "qzxwvtnp",
        "C_2": "hjgrksla",
        "M": "nfrtbgcs",
        "X": "pvkldmqa",
        "Y": "swjqhmze",
        "O_1": "ltrsckhp",
        "O_2": "gmnxdjvo",
        "Q": "wycfslbe",
        "P": "bzqtrnmu",
        "C_3": "dksmpvha"
      },
      "question": "Let $qzxwvtnp$ and $hjgrksla$ be circles whose centers are 10 units apart, and\nwhose radii are 1 and 3. Find, with proof, the locus of all points $nfrtbgcs$\nfor which there exists points $pvkldmqa$ on $qzxwvtnp$ and $swjqhmze$ on $hjgrksla$ such that\n$nfrtbgcs$ is the midpoint of the line segment $pvkldmqa swjqhmze$.",
      "solution": "Let $ltrsckhp$ and $gmnxdjvo$ be the centers of $qzxwvtnp$ and $hjgrksla$, respectively.\n(We are assuming $qzxwvtnp$ has radius 1 and $hjgrksla$ has radius 3.)\nThen the\ndesired locus is an annulus centered at the midpoint of $ltrsckhp gmnxdjvo$, with\ninner radius 1 and outer radius 2.\n\nFor a fixed point $wycfslbe$ on $hjgrksla$, the locus of the midpoints of the\nsegments $bzqtrnmu wycfslbe$ for $bzqtrnmu$ lying on $qzxwvtnp$ is the image of $qzxwvtnp$ under a\nhomothety centered at $wycfslbe$ of radius $1/2$, which is a circle of radius\n$1/2$. As $wycfslbe$ varies, the center of this smaller circle traces out a\ncircle $dksmpvha$ of radius $3/2$ (again by homothety). By considering the two\npositions of $wycfslbe$ on the line of centers of the circles, one sees that\n$dksmpvha$ is centered at the midpoint of $ltrsckhp gmnxdjvo$, and the locus is now\nclearly the specified annulus."
    },
    "kernel_variant": {
      "question": "Let an integer $n\\ge 2$ be fixed and put  \n\\[\nr_{1}=4,\\qquad r_{2}=7,\\qquad d=18\\;(>r_{1}+r_{2}),\\qquad\n\\mathbf D=\\overrightarrow{O_{1}O_{2}},\\qquad |\\mathbf D|=d .\n\\]\nFor $i\\in\\{1,2\\}$ denote by  \n\\[\nS_{i}:=\\Bigl\\{P\\in\\mathbb R^{n}\\ \\Bigl|\\ |P-O_{i}|=r_{i}\\Bigr\\}\n\\]\nthe two $n$–spheres of radii $r_{1},r_{2}$.  \nFor every weight $\\lambda\\in(0,1)$ define the weighted–midpoint locus  \n\\[\nL_{\\lambda}:=\\Bigl\\{M\\in\\mathbb R^{n}\\ \\Bigl|\\ \\exists\\,X\\in S_{1},\\,Y\\in S_{2}\n        \\text{ such that } M=(1-\\lambda)X+\\lambda Y\\Bigr\\},\n\\]\nand put  \n\\[\nL:=\\bigcup_{0<\\lambda<1}L_{\\lambda}.\n\\]\n\n(a)  Show that, for every $\\lambda\\in(0,1)$, $L_{\\lambda}$ is a solid\n$n$–dimensional spherical shell (the closed region between two\nconcentric $(n-1)$–spheres).  Determine its centre $C_{\\lambda}$, inner\nradius $\\rho_{\\min}(\\lambda)$ and outer radius $\\rho_{\\max}(\\lambda)$\nexplicitly in terms of $\\lambda,r_{1},r_{2},d$.\n\n(b)  Prove that\n\\[\nL=\\Bigl\\{P\\in\\mathbb R^{n}\\ \\Bigl|\\ \\exists\\,s\\in[0,d]\\text{ with }\n      \\bigl|P-\\bigl(O_{1}+s\\,\\hat{\\mathbf D}\\bigr)\\bigr|\\le\n      4+\\dfrac{3s}{d}\\Bigr\\},\n\\]\nthat is, $L$ is the union of the closed balls whose centres run along\nthe segment $O_{1}O_{2}$ while their radii grow linearly from $4$ at\n$O_{1}$ to $7$ at $O_{2}$.  Deduce that $L$ itself is not a spherical\nshell.\n\n(c)  (i)  Show that $L_{\\lambda}=L_{\\mu}$ implies $\\lambda=\\mu$.  \n     (ii)  Prove that for $\\lambda\\neq\\mu$ neither of the two shells is\n     contained in the other; consequently the family\n     $\\{L_{\\lambda}\\}_{\\lambda\\in(0,1)}$ is not nested.\n\n(d)  Specialise to $n=3$ and compute the exact volume of the solid $L$\nfound in part (b).  Give your answer in the simplest\nrational–multiple–of–$\\pi$ form.\n\n\n\n--------------------------------------------------------------------",
      "solution": "Throughout boldface letters denote vectors, $|\\cdot|$ the Euclidean\nnorm and $\\hat{\\mathbf D}:=\\mathbf D/|\\mathbf D|$.\n\n(a)  Description of a single weighted–midpoint set $L_{\\lambda}$  \n\nWrite $X=O_{1}+r_{1}\\mathbf u$ and $Y=O_{2}+r_{2}\\mathbf v$ with\n$|\\mathbf u|=|\\mathbf v|=1$.  Fix $\\lambda\\in(0,1)$ and set  \n\\[\nC_{\\lambda}:=(1-\\lambda)O_{1}+\\lambda O_{2}=O_{1}+\\lambda\\mathbf D,\n\\qquad\nM:=(1-\\lambda)X+\\lambda Y=C_{\\lambda}+\\mathbf W,\n\\]\n\\[\n\\mathbf W:=(1-\\lambda)r_{1}\\mathbf u+\\lambda r_{2}\\mathbf v .\n\\]\nBecause $|\\mathbf u|=|\\mathbf v|=1$,\n\\[\n|\\mathbf W|^{2}=(1-\\lambda)^{2}r_{1}^{2}+\\lambda^{2}r_{2}^{2}\n                +2\\lambda(1-\\lambda)r_{1}r_{2}\\,(\\mathbf u\\cdot\\mathbf v).\n\\]\nPut $t:=\\mathbf u\\cdot\\mathbf v\\in[-1,1]$.  The right–hand side is\naffine in $t$, so its extrema occur at $t=\\pm1$:\n\\[\n\\bigl[(1-\\lambda)r_{1}-\\lambda r_{2}\\bigr]^{2}\\le\n|\\mathbf W|^{2}\\le\n\\bigl[(1-\\lambda)r_{1}+\\lambda r_{2}\\bigr]^{2}.\n\\]\nHence  \n\\[\n\\rho_{\\min}(\\lambda)=\\bigl|(1-\\lambda)r_{1}-\\lambda r_{2}\\bigr|,\n\\qquad\n\\rho_{\\max}(\\lambda)=(1-\\lambda)r_{1}+\\lambda r_{2},\n\\]\nand  \n\\[\nL_{\\lambda}=C_{\\lambda}+\n\\Bigl\\{\\mathbf w\\in\\mathbb R^{n}\\,\\Bigl|\\,\n      \\rho_{\\min}(\\lambda)\\le|\\mathbf w|\\le\\rho_{\\max}(\\lambda)\\Bigr\\},\n\\]\nso $L_{\\lambda}$ is indeed a closed $n$–dimensional spherical shell.\n\n--------------------------------------------------------------------\n(b)  The global locus $L$\n\nFor $s\\in[0,d]$ define  \n\\[\nC(s):=O_{1}+s\\hat{\\mathbf D},\\qquad\nR(s):=4+\\frac{3s}{d}=4+\\frac{s}{6}.\n\\]\nWe claim  \n\\[\nL=\\bigcup_{s\\in[0,d]}B\\!\\bigl(C(s),R(s)\\bigr).\\tag{$\\ast$}\n\\]\n\nInclusion “$\\subset$”.  \nTake $M\\in L_{\\lambda}$ for some $\\lambda\\in(0,1)$.  Part (a) yields\n$|M-C_{\\lambda}|\\le\\rho_{\\max}(\\lambda)$.  Since\n$C_{\\lambda}=C(d\\lambda)$ and\n$\\rho_{\\max}(\\lambda)=4+\\dfrac{3(d\\lambda)}{d}=R(d\\lambda)$, we have\n$M\\in B\\!\\bigl(C(d\\lambda),R(d\\lambda)\\bigr)$; hence  \n$M\\in\\bigcup_{s}B\\!\\bigl(C(s),R(s)\\bigr)$.\n\nInclusion “$\\supset$”.  \nFix $P\\in B\\!\\bigl(C(s_{0}),R(s_{0})\\bigr)$ for some $s_{0}\\in[0,d]$\nand set  \n\\[\n\\lambda_{0}:=\\frac{s_{0}}{d},\\qquad\nh(\\lambda):=|P-C_{\\lambda}|-\\rho_{\\max}(\\lambda)\\quad(0\\le\\lambda\\le1).\n\\]\nBoth summands are continuous, hence $h$ is continuous.  \n\n(1)  At $\\lambda=\\lambda_{0}$ we have\n$|P-C_{\\lambda_{0}}|\\le R(s_{0})=\\rho_{\\max}(\\lambda_{0})$, so\n$h(\\lambda_{0})\\le0$.\n\n(2)  The balls $B(O_{1},4)$ and $B(O_{2},7)$ are disjoint\n($d=18>4+7$), hence $P$ cannot lie in both; therefore at least one of\n$h(0),h(1)$ is strictly positive.\n\n(3)  If $h(0)>0$ then $h(\\lambda_{0})\\le0<h(0)$, whereas if $h(1)>0$\nthen $h(\\lambda_{0})\\le0<h(1)$.  By continuity there is a\n$\\lambda_{*}\\in(0,1)$ with $h(\\lambda_{*})=0$, that is\n$|P-C_{\\lambda_{*}}|=\\rho_{\\max}(\\lambda_{*})$.  Because\n$\\rho_{\\min}\\le\\rho_{\\max}$ we also have\n$|P-C_{\\lambda_{*}}|\\ge\\rho_{\\min}(\\lambda_{*})$, whence\n\\[\nP\\in L_{\\lambda_{*}}\\subset L.\n\\]\nThus $(\\ast)$ holds.  Geometrically, $L$ is a “cigar’’ whose transverse\nradius grows linearly from $4$ at $O_{1}$ to $7$ at $O_{2}$; in\nparticular $L$ is not itself a spherical shell.\n\n--------------------------------------------------------------------\n(c)  Properties of the family $\\{L_{\\lambda}\\}$  \n\n(i)  Assume $L_{\\lambda}=L_{\\mu}$ with $\\lambda<\\mu$.  Put  \n\\[\nP:=C_{\\lambda}-\\rho_{\\max}(\\lambda)\\,\\hat{\\mathbf D}.\n\\]\nThen $P\\in L_{\\lambda}$, so $P\\in L_{\\mu}$ would imply\n$|P-C_{\\mu}|\\le\\rho_{\\max}(\\mu)$, but\n\\[\n|P-C_{\\mu}|=(\\mu-\\lambda)d+\\rho_{\\max}(\\lambda),\\qquad\n\\rho_{\\max}(\\mu)=\\rho_{\\max}(\\lambda)+(r_{2}-r_{1})(\\mu-\\lambda),\n\\]\n\\[\n|P-C_{\\mu}|-\\rho_{\\max}(\\mu)\n      =(\\mu-\\lambda)\\bigl[d-(r_{2}-r_{1})\\bigr]\n      =(\\mu-\\lambda)\\,(18-3)>0,\n\\]\na contradiction.  Hence $\\lambda=\\mu$.\n\n(ii)  For $\\lambda<\\mu$ the same point $P$ lies in\n$L_{\\lambda}\\setminus L_{\\mu}$, so neither shell is contained in the\nother.  Consequently the family $\\{L_{\\lambda}\\}_{\\lambda\\in(0,1)}$ is\nnot nested.\n\n--------------------------------------------------------------------\n(d)  Volume of $L$ for $n=3$  \n\nPlace $O_{1}$ at the origin and the $z$–axis along $\\hat{\\mathbf D}$ so\nthat $O_{2}=(0,0,18)$.  From part (b)\n\\[\nL=\\bigcup_{s=0}^{18}B\\Bigl((0,0,s),\\,R(s)\\Bigr),\n\\qquad\nR(s)=4+\\frac{s}{6}.\n\\]\n\n1.  Vertical extent.  \nThe extreme balls are $B\\bigl((0,0,0),4\\bigr)$ and\n$B\\bigl((0,0,18),7\\bigr)$, so\n\\[\n-4\\le z\\le 25\\qquad(\\text{because }18+7=25).\n\\]\n\n2.  Maximal cross–sectional radius.  \nFor a fixed height $z\\in[-4,25]$ set  \n\\[\nf_{z}(s):=\\sqrt{\\,R(s)^{2}-(z-s)^{2}}\\quad(0\\le s\\le18).\n\\]\nBecause  \n\\[\nf_{z}'(s)=\\frac{R'(s)R(s)+(z-s)}{f_{z}(s)}\n         =\\frac{\\dfrac{1}{6}\\bigl(4+\\dfrac{s}{6}\\bigr)+(z-s)}\n                {f_{z}(s)},\n\\]\nthe unique critical point is  \n\\[\ns^{\\ast}(z)=\\frac{36z+24}{35}.\n\\]\nIt satisfies $0\\le s^{\\ast}(z)\\le18$ exactly for \n\\[\n-\\frac{2}{3}\\le z\\le\\frac{101}{6}.\n\\]\nHence the squared maximal radius is\n\\[\nr_{\\max }(z)^{2}=\n\\begin{cases}\n16-z^{2}, & -4\\le z\\le-\\dfrac23,\\\\[6pt]\n\\dfrac{(z+24)^{2}}{35}, & -\\dfrac23< z<\\dfrac{101}{6},\\\\[10pt]\n49-(z-18)^{2}, & \\dfrac{101}{6}\\le z\\le25 .\n\\end{cases}\n\\]\n\n3.  Volume integral.  \nUsing cylindrical slices,  \n\\[\n\\operatorname{Vol}(L)=\n\\pi\\!\\int_{-4}^{25}r_{\\max}(z)^{2}\\,dz\n=\\;I_{A}+I_{B}+I_{C},\n\\]\nwhere\n\n\\[\n\\begin{aligned}\nI_{A}&=\\pi\\!\\int_{-4}^{-2/3}\\!\\bigl(16-z^{2}\\bigr)\\,dz\n    =\\pi\\Bigl[16z-\\tfrac13 z^{3}\\Bigr]_{-4}^{-2/3}\n    =\\pi\\cdot\\frac{2\\,600}{81},\\\\[8pt]\nI_{B}&=\\pi\\!\\int_{-2/3}^{101/6}\\!\\frac{(z+24)^{2}}{35}\\,dz\n    =\\pi\\Bigl[\\tfrac{(z+24)^{3}}{105}\\Bigr]_{-2/3}^{101/6}\n    =\\pi\\cdot\\frac{37\\,975}{72},\\\\[8pt]\nI_{C}&=\\pi\\!\\int_{101/6}^{25}\\!\\bigl[49-(z-18)^{2}\\bigr]\\,dz\n    =\\pi\\Bigl[49z-\\tfrac13(z-18)^{3}\\Bigr]_{101/6}^{25}\n    =\\pi\\cdot\\frac{184\\,877}{648}.\n\\end{aligned}\n\\]\n\n4.  Summation and simplification.  \nConverting to the common denominator $648$,\n\\[\nI_{A}+I_{B}+I_{C}\n      =\\pi\\cdot\\frac{20\\,800+341\\,775+184\\,877}{648}\n      =\\pi\\cdot\\frac{547\\,452}{648}\n      =\\pi\\cdot\\frac{5\\,069}{6}.\n\\]\nBecause $\\gcd(5\\,069,6)=1$, this is already the simplest\nrational–multiple–of–$\\pi$ form:\n\n\\[\n\\boxed{\\displaystyle\\operatorname{Vol}(L)=\\frac{5\\,069}{6}\\,\\pi }.\n\\]\n\n(The numerical value $\\frac{5\\,069}{6}\\pi\\approx2\\,654.1$\nserves as a quick consistency check.)\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.746669",
        "was_fixed": false,
        "difficulty_analysis": "Compared with the original problem (two circles, simple midpoint) this variant is considerably harder for several reasons.\n\n1.  Higher dimension: the setting is $\\Bbb R^{n}$, not the plane.  \n2.  Weighted division: $M$ is not necessarily a midpoint; the ratio $\\lambda$ is arbitrary.  \n3.  Family of loci: one must study an entire one-parameter family $L_{\\lambda}$ and then their union, including inclusion relations.  \n4.  Abstract tools: the solution uses Minkowski sums, vector decomposition, and optimisation of dot products, techniques well outside the elementary geometry required for the original annulus.  \n5.  Additional quantitative task: the volume computation in $\\Bbb R^{3}$ forces the solver to translate geometric information into explicit integrals.\n\nThese layers of abstraction and calculation make the enhanced variant significantly more technical and conceptually deeper than both the original problem and the current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let an integer $n\\ge 2$ be fixed and put  \n\\[\nr_{1}=4,\\qquad r_{2}=7,\\qquad d=18\\;(>r_{1}+r_{2}),\\qquad\n\\mathbf D=\\overrightarrow{O_{1}O_{2}},\\qquad |\\mathbf D|=d .\n\\]\nFor $i\\in\\{1,2\\}$ denote by  \n\\[\nS_{i}:=\\Bigl\\{P\\in\\mathbb R^{n}\\ \\Bigl|\\ |P-O_{i}|=r_{i}\\Bigr\\}\n\\]\nthe two $n$–spheres of radii $r_{1},r_{2}$.  \nFor every weight $\\lambda\\in(0,1)$ define the weighted–midpoint locus  \n\\[\nL_{\\lambda}:=\\Bigl\\{M\\in\\mathbb R^{n}\\ \\Bigl|\\ \\exists\\,X\\in S_{1},\\,Y\\in S_{2}\n        \\text{ such that } M=(1-\\lambda)X+\\lambda Y\\Bigr\\},\n\\]\nand put  \n\\[\nL:=\\bigcup_{0<\\lambda<1}L_{\\lambda}.\n\\]\n\n(a)  Show that, for every $\\lambda\\in(0,1)$, $L_{\\lambda}$ is a solid\n$n$–dimensional spherical shell (the closed region between two\nconcentric $(n-1)$–spheres).  Determine its centre $C_{\\lambda}$, inner\nradius $\\rho_{\\min}(\\lambda)$ and outer radius $\\rho_{\\max}(\\lambda)$\nexplicitly in terms of $\\lambda,r_{1},r_{2},d$.\n\n(b)  Prove that\n\\[\nL=\\Bigl\\{P\\in\\mathbb R^{n}\\ \\Bigl|\\ \\exists\\,s\\in[0,d]\\text{ with }\n      \\bigl|P-\\bigl(O_{1}+s\\,\\hat{\\mathbf D}\\bigr)\\bigr|\\le\n      4+\\dfrac{3s}{d}\\Bigr\\},\n\\]\nthat is, $L$ is the union of the closed balls whose centres run along\nthe segment $O_{1}O_{2}$ while their radii grow linearly from $4$ at\n$O_{1}$ to $7$ at $O_{2}$.  Deduce that $L$ itself is not a spherical\nshell.\n\n(c)  (i)  Show that $L_{\\lambda}=L_{\\mu}$ implies $\\lambda=\\mu$.  \n     (ii)  Prove that for $\\lambda\\neq\\mu$ neither of the two shells is\n     contained in the other; consequently the family\n     $\\{L_{\\lambda}\\}_{\\lambda\\in(0,1)}$ is not nested.\n\n(d)  Specialise to $n=3$ and compute the exact volume of the solid $L$\nfound in part (b).  Give your answer in the simplest\nrational–multiple–of–$\\pi$ form.\n\n\n\n--------------------------------------------------------------------",
      "solution": "Throughout boldface letters denote vectors, $|\\cdot|$ the Euclidean\nnorm and $\\hat{\\mathbf D}:=\\mathbf D/|\\mathbf D|$.\n\n(a)  Description of a single weighted–midpoint set $L_{\\lambda}$  \n\nWrite $X=O_{1}+r_{1}\\mathbf u$ and $Y=O_{2}+r_{2}\\mathbf v$ with\n$|\\mathbf u|=|\\mathbf v|=1$.  Fix $\\lambda\\in(0,1)$ and set  \n\\[\nC_{\\lambda}:=(1-\\lambda)O_{1}+\\lambda O_{2}=O_{1}+\\lambda\\mathbf D,\n\\qquad\nM:=(1-\\lambda)X+\\lambda Y=C_{\\lambda}+\\mathbf W,\n\\]\n\\[\n\\mathbf W:=(1-\\lambda)r_{1}\\mathbf u+\\lambda r_{2}\\mathbf v .\n\\]\nBecause $|\\mathbf u|=|\\mathbf v|=1$,\n\\[\n|\\mathbf W|^{2}=(1-\\lambda)^{2}r_{1}^{2}+\\lambda^{2}r_{2}^{2}\n                +2\\lambda(1-\\lambda)r_{1}r_{2}\\,(\\mathbf u\\cdot\\mathbf v).\n\\]\nPut $t:=\\mathbf u\\cdot\\mathbf v\\in[-1,1]$.  The right–hand side is\naffine in $t$, so its extrema occur at $t=\\pm1$:\n\\[\n\\bigl[(1-\\lambda)r_{1}-\\lambda r_{2}\\bigr]^{2}\\le\n|\\mathbf W|^{2}\\le\n\\bigl[(1-\\lambda)r_{1}+\\lambda r_{2}\\bigr]^{2}.\n\\]\nHence  \n\\[\n\\rho_{\\min}(\\lambda)=\\bigl|(1-\\lambda)r_{1}-\\lambda r_{2}\\bigr|,\n\\qquad\n\\rho_{\\max}(\\lambda)=(1-\\lambda)r_{1}+\\lambda r_{2},\n\\]\nand  \n\\[\nL_{\\lambda}=C_{\\lambda}+\n\\Bigl\\{\\mathbf w\\in\\mathbb R^{n}\\,\\Bigl|\\,\n      \\rho_{\\min}(\\lambda)\\le|\\mathbf w|\\le\\rho_{\\max}(\\lambda)\\Bigr\\},\n\\]\nso $L_{\\lambda}$ is indeed a closed $n$–dimensional spherical shell.\n\n--------------------------------------------------------------------\n(b)  The global locus $L$\n\nFor $s\\in[0,d]$ define  \n\\[\nC(s):=O_{1}+s\\hat{\\mathbf D},\\qquad\nR(s):=4+\\frac{3s}{d}=4+\\frac{s}{6}.\n\\]\nWe claim  \n\\[\nL=\\bigcup_{s\\in[0,d]}B\\!\\bigl(C(s),R(s)\\bigr).\\tag{$\\ast$}\n\\]\n\nInclusion “$\\subset$”.  \nTake $M\\in L_{\\lambda}$ for some $\\lambda\\in(0,1)$.  Part (a) yields\n$|M-C_{\\lambda}|\\le\\rho_{\\max}(\\lambda)$.  Since\n$C_{\\lambda}=C(d\\lambda)$ and\n$\\rho_{\\max}(\\lambda)=4+\\dfrac{3(d\\lambda)}{d}=R(d\\lambda)$, we have\n$M\\in B\\!\\bigl(C(d\\lambda),R(d\\lambda)\\bigr)$; hence  \n$M\\in\\bigcup_{s}B\\!\\bigl(C(s),R(s)\\bigr)$.\n\nInclusion “$\\supset$”.  \nFix $P\\in B\\!\\bigl(C(s_{0}),R(s_{0})\\bigr)$ for some $s_{0}\\in[0,d]$\nand set  \n\\[\n\\lambda_{0}:=\\frac{s_{0}}{d},\\qquad\nh(\\lambda):=|P-C_{\\lambda}|-\\rho_{\\max}(\\lambda)\\quad(0\\le\\lambda\\le1).\n\\]\nBoth summands are continuous, hence $h$ is continuous.  \n\n(1)  At $\\lambda=\\lambda_{0}$ we have\n$|P-C_{\\lambda_{0}}|\\le R(s_{0})=\\rho_{\\max}(\\lambda_{0})$, so\n$h(\\lambda_{0})\\le0$.\n\n(2)  The balls $B(O_{1},4)$ and $B(O_{2},7)$ are disjoint\n($d=18>4+7$), hence $P$ cannot lie in both; therefore at least one of\n$h(0),h(1)$ is strictly positive.\n\n(3)  If $h(0)>0$ then $h(\\lambda_{0})\\le0<h(0)$, whereas if $h(1)>0$\nthen $h(\\lambda_{0})\\le0<h(1)$.  By continuity there is a\n$\\lambda_{*}\\in(0,1)$ with $h(\\lambda_{*})=0$, that is\n$|P-C_{\\lambda_{*}}|=\\rho_{\\max}(\\lambda_{*})$.  Because\n$\\rho_{\\min}\\le\\rho_{\\max}$ we also have\n$|P-C_{\\lambda_{*}}|\\ge\\rho_{\\min}(\\lambda_{*})$, whence\n\\[\nP\\in L_{\\lambda_{*}}\\subset L.\n\\]\nThus $(\\ast)$ holds.  Geometrically, $L$ is a “cigar’’ whose transverse\nradius grows linearly from $4$ at $O_{1}$ to $7$ at $O_{2}$; in\nparticular $L$ is not itself a spherical shell.\n\n--------------------------------------------------------------------\n(c)  Properties of the family $\\{L_{\\lambda}\\}$  \n\n(i)  Assume $L_{\\lambda}=L_{\\mu}$ with $\\lambda<\\mu$.  Put  \n\\[\nP:=C_{\\lambda}-\\rho_{\\max}(\\lambda)\\,\\hat{\\mathbf D}.\n\\]\nThen $P\\in L_{\\lambda}$, so $P\\in L_{\\mu}$ would imply\n$|P-C_{\\mu}|\\le\\rho_{\\max}(\\mu)$, but\n\\[\n|P-C_{\\mu}|=(\\mu-\\lambda)d+\\rho_{\\max}(\\lambda),\\qquad\n\\rho_{\\max}(\\mu)=\\rho_{\\max}(\\lambda)+(r_{2}-r_{1})(\\mu-\\lambda),\n\\]\n\\[\n|P-C_{\\mu}|-\\rho_{\\max}(\\mu)\n      =(\\mu-\\lambda)\\bigl[d-(r_{2}-r_{1})\\bigr]\n      =(\\mu-\\lambda)\\,(18-3)>0,\n\\]\na contradiction.  Hence $\\lambda=\\mu$.\n\n(ii)  For $\\lambda<\\mu$ the same point $P$ lies in\n$L_{\\lambda}\\setminus L_{\\mu}$, so neither shell is contained in the\nother.  Consequently the family $\\{L_{\\lambda}\\}_{\\lambda\\in(0,1)}$ is\nnot nested.\n\n--------------------------------------------------------------------\n(d)  Volume of $L$ for $n=3$  \n\nPlace $O_{1}$ at the origin and the $z$–axis along $\\hat{\\mathbf D}$ so\nthat $O_{2}=(0,0,18)$.  From part (b)\n\\[\nL=\\bigcup_{s=0}^{18}B\\Bigl((0,0,s),\\,R(s)\\Bigr),\n\\qquad\nR(s)=4+\\frac{s}{6}.\n\\]\n\n1.  Vertical extent.  \nThe extreme balls are $B\\bigl((0,0,0),4\\bigr)$ and\n$B\\bigl((0,0,18),7\\bigr)$, so\n\\[\n-4\\le z\\le 25\\qquad(\\text{because }18+7=25).\n\\]\n\n2.  Maximal cross–sectional radius.  \nFor a fixed height $z\\in[-4,25]$ set  \n\\[\nf_{z}(s):=\\sqrt{\\,R(s)^{2}-(z-s)^{2}}\\quad(0\\le s\\le18).\n\\]\nBecause  \n\\[\nf_{z}'(s)=\\frac{R'(s)R(s)+(z-s)}{f_{z}(s)}\n         =\\frac{\\dfrac{1}{6}\\bigl(4+\\dfrac{s}{6}\\bigr)+(z-s)}\n                {f_{z}(s)},\n\\]\nthe unique critical point is  \n\\[\ns^{\\ast}(z)=\\frac{36z+24}{35}.\n\\]\nIt satisfies $0\\le s^{\\ast}(z)\\le18$ exactly for \n\\[\n-\\frac{2}{3}\\le z\\le\\frac{101}{6}.\n\\]\nHence the squared maximal radius is\n\\[\nr_{\\max }(z)^{2}=\n\\begin{cases}\n16-z^{2}, & -4\\le z\\le-\\dfrac23,\\\\[6pt]\n\\dfrac{(z+24)^{2}}{35}, & -\\dfrac23< z<\\dfrac{101}{6},\\\\[10pt]\n49-(z-18)^{2}, & \\dfrac{101}{6}\\le z\\le25 .\n\\end{cases}\n\\]\n\n3.  Volume integral.  \nUsing cylindrical slices,  \n\\[\n\\operatorname{Vol}(L)=\n\\pi\\!\\int_{-4}^{25}r_{\\max}(z)^{2}\\,dz\n=\\;I_{A}+I_{B}+I_{C},\n\\]\nwhere\n\n\\[\n\\begin{aligned}\nI_{A}&=\\pi\\!\\int_{-4}^{-2/3}\\!\\bigl(16-z^{2}\\bigr)\\,dz\n    =\\pi\\Bigl[16z-\\tfrac13 z^{3}\\Bigr]_{-4}^{-2/3}\n    =\\pi\\cdot\\frac{2\\,600}{81},\\\\[8pt]\nI_{B}&=\\pi\\!\\int_{-2/3}^{101/6}\\!\\frac{(z+24)^{2}}{35}\\,dz\n    =\\pi\\Bigl[\\tfrac{(z+24)^{3}}{105}\\Bigr]_{-2/3}^{101/6}\n    =\\pi\\cdot\\frac{37\\,975}{72},\\\\[8pt]\nI_{C}&=\\pi\\!\\int_{101/6}^{25}\\!\\bigl[49-(z-18)^{2}\\bigr]\\,dz\n    =\\pi\\Bigl[49z-\\tfrac13(z-18)^{3}\\Bigr]_{101/6}^{25}\n    =\\pi\\cdot\\frac{184\\,877}{648}.\n\\end{aligned}\n\\]\n\n4.  Summation and simplification.  \nConverting to the common denominator $648$,\n\\[\nI_{A}+I_{B}+I_{C}\n      =\\pi\\cdot\\frac{20\\,800+341\\,775+184\\,877}{648}\n      =\\pi\\cdot\\frac{547\\,452}{648}\n      =\\pi\\cdot\\frac{5\\,069}{6}.\n\\]\nBecause $\\gcd(5\\,069,6)=1$, this is already the simplest\nrational–multiple–of–$\\pi$ form:\n\n\\[\n\\boxed{\\displaystyle\\operatorname{Vol}(L)=\\frac{5\\,069}{6}\\,\\pi }.\n\\]\n\n(The numerical value $\\frac{5\\,069}{6}\\pi\\approx2\\,654.1$\nserves as a quick consistency check.)\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.576717",
        "was_fixed": false,
        "difficulty_analysis": "Compared with the original problem (two circles, simple midpoint) this variant is considerably harder for several reasons.\n\n1.  Higher dimension: the setting is $\\Bbb R^{n}$, not the plane.  \n2.  Weighted division: $M$ is not necessarily a midpoint; the ratio $\\lambda$ is arbitrary.  \n3.  Family of loci: one must study an entire one-parameter family $L_{\\lambda}$ and then their union, including inclusion relations.  \n4.  Abstract tools: the solution uses Minkowski sums, vector decomposition, and optimisation of dot products, techniques well outside the elementary geometry required for the original annulus.  \n5.  Additional quantitative task: the volume computation in $\\Bbb R^{3}$ forces the solver to translate geometric information into explicit integrals.\n\nThese layers of abstraction and calculation make the enhanced variant significantly more technical and conceptually deeper than both the original problem and the current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}