{
  "index": "1971-B-3",
  "type": "COMB",
  "tag": [
    "COMB",
    "ALG"
  ],
  "difficulty": "",
  "question": "B-3. Two cars travel around a track at equal and constant speeds, each completing a lap every hour. From a common starting point, the first starts at time \\( t=0 \\) and the second at an arbitrary later time \\( t=T>0 \\). Prove that there is a total period of exactly one hour during the motion in which the first has completed twice as many laps as the second.",
  "solution": "B-3 At time \\( t \\), car 1 has conpleted [ \\( t \\) ] laps and car 2 has conpleted [ \\( t-T] \\) laps. The problem is to find values of \\( t \\geqq T \\) for which \\( [t]=2[t-T] \\).\n\nLet \\( T=k+\\delta \\), where \\( 0 \\leqq \\delta<1, k \\) an integer. Consider any integral interval [ \\( m, m+1 \\) ] and let \\( m \\leqq t<m+1 \\). Then \\( t=m+\\varepsilon \\), where \\( 0 \\leqq \\varepsilon<1 \\). Then the equation to be solved becomes\n\\[\n[t]=m=2[t-T]=2[m+\\varepsilon-(k+\\delta)]=2[m-k+\\varepsilon-\\delta] .\n\\]\n\nThus \\( m=2(m-k) \\), if \\( \\varepsilon \\geqq \\delta \\) and \\( m=2(m-k-1) \\), if \\( \\varepsilon<\\delta \\). If \\( 1>\\varepsilon \\geqq \\delta \\), then \\( m=2 k \\) and the equation is satisfied during [ \\( 2 k+\\delta, 2 k+1] \\), which has length \\( 1-\\delta \\).\n\nIf \\( 0 \\leqq \\varepsilon<\\delta \\), then \\( m=2 k+2 \\) and the equation is satisfied during [ \\( 2 k+2 \\), \\( 2 k+2+\\delta] \\) which has length \\( \\delta \\). Therefore the total length is \\( 1-\\delta+\\delta=1 \\).\n\nComment: The problem should have been more explicit by stating \"after the start of the second car\" instead of \"during the mction'. The solution is given for this interpretation, whereas, if \\( t<T,[t-T] \\) is negative but the second car would have completed zero laps.",
  "vars": [
    "t",
    "m",
    "\\\\varepsilon"
  ],
  "params": [
    "T",
    "k",
    "\\\\delta"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "t": "timevar",
        "m": "lapindex",
        "\\varepsilon": "smalloffset",
        "T": "delaytime",
        "k": "intshift",
        "\\delta": "smalldelta"
      },
      "question": "B-3. Two cars travel around a track at equal and constant speeds, each completing a lap every hour. From a common starting point, the first starts at time \\( timevar=0 \\) and the second at an arbitrary later time \\( timevar=delaytime>0 \\). Prove that there is a total period of exactly one hour during the motion in which the first has completed twice as many laps as the second.",
      "solution": "B-3 At time \\( timevar \\), car 1 has conpleted [ \\( timevar \\) ] laps and car 2 has conpleted [ \\( timevar-delaytime] \\) laps. The problem is to find values of \\( timevar \\geqq delaytime \\) for which \\( [timevar]=2[timevar-delaytime] \\).\n\nLet \\( delaytime=intshift+smalldelta \\), where \\( 0 \\leqq smalldelta<1, intshift \\) an integer. Consider any integral interval [ \\( lapindex, lapindex+1 \\) ] and let \\( lapindex \\leqq timevar<lapindex+1 \\). Then \\( timevar=lapindex+smalloffset \\), where \\( 0 \\leqq smalloffset<1 \\). Then the equation to be solved becomes\n\\[\n[timevar]=lapindex=2[timevar-delaytime]=2[lapindex+smalloffset-(intshift+smalldelta)]=2[lapindex-intshift+smalloffset-smalldelta] .\n\\]\n\nThus \\( lapindex=2(lapindex-intshift) \\), if \\( smalloffset \\geqq smalldelta \\) and \\( lapindex=2(lapindex-intshift-1) \\), if \\( smalloffset<smalldelta \\). If \\( 1>smalloffset \\geqq smalldelta \\), then \\( lapindex=2 intshift \\) and the equation is satisfied during [ \\( 2 intshift+smalldelta, 2 intshift+1] \\), which has length \\( 1-smalldelta \\).\n\nIf \\( 0 \\leqq smalloffset<smalldelta \\), then \\( lapindex=2 intshift+2 \\) and the equation is satisfied during [ \\( 2 intshift+2 \\), \\( 2 intshift+2+smalldelta] \\) which has length \\( smalldelta \\). Therefore the total length is \\( 1-smalldelta+smalldelta=1 \\).\n\nComment: The problem should have been more explicit by stating \\\"after the start of the second car\\\" instead of \\\"during the mction'. The solution is given for this interpretation, whereas, if \\( timevar<delaytime,[timevar-delaytime] \\) is negative but the second car would have completed zero laps."
    },
    "descriptive_long_confusing": {
      "map": {
        "t": "lanternfish",
        "m": "toothbrush",
        "\\varepsilon": "raspberry",
        "T": "moonlight",
        "k": "limerick",
        "\\delta": "evergreen"
      },
      "question": "B-3. Two cars travel around a track at equal and constant speeds, each completing a lap every hour. From a common starting point, the first starts at time \\( lanternfish =0 \\) and the second at an arbitrary later time \\( lanternfish = moonlight >0 \\). Prove that there is a total period of exactly one hour during the motion in which the first has completed twice as many laps as the second.",
      "solution": "B-3 At time \\( lanternfish \\), car 1 has conpleted [ \\( lanternfish \\) ] laps and car 2 has conpleted [ \\( lanternfish - moonlight ] \\) laps. The problem is to find values of \\( lanternfish \\geqq moonlight \\) for which \\( [ lanternfish ] = 2[ lanternfish - moonlight ] \\).\n\nLet \\( moonlight = limerick + evergreen \\), where \\( 0 \\leqq evergreen < 1, limerick \\) an integer. Consider any integral interval [ \\( toothbrush , toothbrush +1 \\) ] and let \\( toothbrush \\leqq lanternfish < toothbrush +1 \\). Then \\( lanternfish = toothbrush + raspberry \\), where \\( 0 \\leqq raspberry < 1 \\). Then the equation to be solved becomes\n\\[\n[ lanternfish ] = toothbrush = 2[ lanternfish - moonlight ] = 2[ toothbrush + raspberry - ( limerick + evergreen ) ] = 2[ toothbrush - limerick + raspberry - evergreen ] .\n\\]\n\nThus \\( toothbrush = 2( toothbrush - limerick ) \\), if \\( raspberry \\geqq evergreen \\) and \\( toothbrush = 2( toothbrush - limerick - 1 ) \\), if \\( raspberry < evergreen \\). If \\( 1 > raspberry \\geqq evergreen \\), then \\( toothbrush = 2 limerick \\) and the equation is satisfied during [ \\( 2 limerick + evergreen, 2 limerick + 1 ] \\), which has length \\( 1 - evergreen \\).\n\nIf \\( 0 \\leqq raspberry < evergreen \\), then \\( toothbrush = 2 limerick + 2 \\) and the equation is satisfied during [ \\( 2 limerick + 2, 2 limerick + 2 + evergreen ] \\) which has length \\( evergreen \\). Therefore the total length is \\( 1 - evergreen + evergreen = 1 \\).\n\nComment: The problem should have been more explicit by stating \"after the start of the second car\" instead of \"during the mction\". The solution is given for this interpretation, whereas, if \\( lanternfish < moonlight ,[ lanternfish - moonlight ] \\) is negative but the second car would have completed zero laps."
    },
    "descriptive_long_misleading": {
      "map": {
        "t": "timeless",
        "m": "limitless",
        "\\varepsilon": "enormity",
        "T": "immediacy",
        "k": "fractional",
        "\\delta": "totality"
      },
      "question": "B-3. Two cars travel around a track at equal and constant speeds, each completing a lap every hour. From a common starting point, the first starts at time \\( timeless=0 \\) and the second at an arbitrary later time \\( timeless=immediacy>0 \\). Prove that there is a total period of exactly one hour during the motion in which the first has completed twice as many laps as the second.",
      "solution": "B-3 At time \\( timeless \\), car 1 has conpleted [ \\( timeless \\) ] laps and car 2 has conpleted [ \\( timeless-immediacy] \\) laps. The problem is to find values of \\( timeless \\geqq immediacy \\) for which \\( [timeless]=2[timeless-immediacy] \\).\n\nLet \\( immediacy=fractional+totality \\), where \\( 0 \\leqq totality<1, fractional \\) an integer. Consider any integral interval [ \\( limitless, limitless+1 \\) ] and let \\( limitless \\leqq timeless<limitless+1 \\). Then \\( timeless=limitless+enormity \\), where \\( 0 \\leqq enormity<1 \\). Then the equation to be solved becomes\n\\[\n[timeless]=limitless=2[timeless-immediacy]=2[limitless+enormity-(fractional+totality)]=2[limitless-fractional+enormity-totality] .\n\\]\n\nThus \\( limitless=2(limitless-fractional) \\), if \\( enormity \\geqq totality \\) and \\( limitless=2(limitless-fractional-1) \\), if \\( enormity<totality \\). If \\( 1>enormity \\geqq totality \\), then \\( limitless=2 fractional \\) and the equation is satisfied during [ \\( 2 fractional+totality, 2 fractional+1] \\), which has length \\( 1-totality \\).\n\nIf \\( 0 \\leqq enormity<totality \\), then \\( limitless=2 fractional+2 \\) and the equation is satisfied during [ \\( 2 fractional+2 \\), \\( 2 fractional+2+totality] \\) which has length \\( totality \\). Therefore the total length is \\( 1-totality+totality=1 \\).\n\nComment: The problem should have been more explicit by stating \"after the start of the second car\" instead of \"during the mction'. The solution is given for this interpretation, whereas, if \\( timeless<immediacy,[timeless-immediacy] \\) is negative but the second car would have completed zero laps."
    },
    "garbled_string": {
      "map": {
        "t": "blarmpqz",
        "m": "quostnex",
        "\\\\varepsilon": "vydricka",
        "T": "sweltruv",
        "k": "movtrens",
        "\\\\delta": "plinxode"
      },
      "question": "B-3. Two cars travel around a track at equal and constant speeds, each completing a lap every hour. From a common starting point, the first starts at time \\( blarmpqz=0 \\) and the second at an arbitrary later time \\( blarmpqz=sweltruv>0 \\). Prove that there is a total period of exactly one hour during the motion in which the first has completed twice as many laps as the second.",
      "solution": "B-3 At time \\( blarmpqz \\), car 1 has conpleted [ \\( blarmpqz \\) ] laps and car 2 has conpleted [ \\( blarmpqz-sweltruv] \\) laps. The problem is to find values of \\( blarmpqz \\geqq sweltruv \\) for which \\( [blarmpqz]=2[blarmpqz-sweltruv] \\).\n\nLet \\( sweltruv=movtrens+plinxode \\), where \\( 0 \\leqq plinxode<1, movtrens \\) an integer. Consider any integral interval [ \\( quostnex, quostnex+1 \\) ] and let \\( quostnex \\leqq blarmpqz<quostnex+1 \\). Then \\( blarmpqz=quostnex+vydricka \\), where \\( 0 \\leqq vydricka<1 \\). Then the equation to be solved becomes\n\\[\n[blarmpqz]=quostnex=2[blarmpqz-sweltruv]=2[quostnex+vydricka-(movtrens+plinxode)]=2[quostnex-movtrens+vydricka-plinxode] .\n\\]\n\nThus \\( quostnex=2(quostnex-movtrens) \\), if \\( vydricka \\geqq plinxode \\) and \\( quostnex=2(quostnex-movtrens-1) \\), if \\( vydricka<plinxode \\). If \\( 1>vydricka \\geqq plinxode \\), then \\( quostnex=2 movtrens \\) and the equation is satisfied during [ \\( 2 movtrens+plinxode, 2 movtrens+1] \\), which has length \\( 1-plinxode \\).\n\nIf \\( 0 \\leqq vydricka<plinxode \\), then \\( quostnex=2 movtrens+2 \\) and the equation is satisfied during [ \\( 2 movtrens+2 \\), \\( 2 movtrens+2+plinxode] \\) which has length \\( plinxode \\). Therefore the total length is \\( 1-plinxode+plinxode=1 \\).\n\nComment: The problem should have been more explicit by stating \"after the start of the second car\" instead of \"during the mction'. The solution is given for this interpretation, whereas, if \\( blarmpqz<sweltruv,[blarmpqz-sweltruv] \\) is negative but the second car would have completed zero laps."
    },
    "kernel_variant": {
      "question": "Let a fixed checkpoint $P$ lie on a circular course that two identical autonomous drones travel at the same constant speed, each completing one full lap every hour.\n\n•  Drone A passes $P$ at the clock time $t=-T$ (with $T>0$) and afterwards continues indefinitely.  \n•  Drone B passes $P$ at the clock time $t=0$ and afterwards continues indefinitely.\n\nFor every real $t\\ge 0$ set  \n\\[\nA(t):=\\lfloor t+T\\rfloor ,\\qquad \nB(t):=\\lfloor t\\rfloor ,\n\\]\nso that $A(t)$ (resp. $B(t)$) equals the number of completed laps of Drone A (resp. Drone B) by the instant $t$.\n\nThroughout write  \n\\[\nT=k+\\delta \\quad\\text{with}\\quad k\\in\\mathbb Z,\\;k\\ge 0,\\;0\\le\\delta<1.\\tag{$\\ast$}\n\\]\n\nPart (a)  (Prologue — existence of one contiguous hour)  \nProve that a contiguous time-interval of length exactly one hour on which  \n\\[\nA(t)=2\\,B(t)\\qquad\\text{for every }t\\text{ in the interval}\n\\]\nexists if and only if $T$ is an integer (that is, $\\delta=0$ in $(\\ast)$).\n\nPart (b)  (Complete description of all solutions for the ratio $2:1$)  \nFor the representation $(\\ast)$ determine the set  \n\\[\nS:=\\{\\,t\\ge 0 : A(t)=2\\,B(t)\\,\\}.\n\\]\nShow that  \n\n(i)  If $\\delta=0$ then\\; $S=[k,k+1)$ (a single interval of length $1$).  \n\n(ii) If $0<\\delta<1$ then \n\\[\nS=S_{1}\\cup S_{2},\\qquad \nS_{1}=[\\,k,\\;k+1-\\delta),\\;\nS_{2}=[\\,k+2-\\delta,\\;k+2),\n\\]\nso $|S_{1}|=1-\\delta$, $|S_{2}|=\\delta$ and $|S|=1$.\n\nPart (c)  (Total measure and asymptotic density for the ratio $2:1$)  \nProve that for every $R>0$  \n\\[\n|S\\cap[0,R]|\\le 1,\n\\qquad\n\\lim_{R\\to\\infty}\\frac{|S\\cap[0,R]|}{R}=0.\n\\]\n\nPart (d)  (General integral ratio — harder extension)  \nFix an integer $r\\ge 2$ and keep the notation $A(t)=\\lfloor t+T\\rfloor$, $B(t)=\\lfloor t\\rfloor$.  \nDefine  \n\\[\nS_{r}:=\\{\\,t\\ge 0 : A(t)=r\\,B(t)\\,\\}.\n\\]\n\nProve the following complete description for all $T>0$.\n\n•  Put $k$ and $\\delta$ as in $(\\ast)$.  Then $S_{r}\\neq\\varnothing$ if and only if  \n\\[\n\\bigl(k\\equiv 0\\pmod{r-1}\\bigr)\\quad\\text{or}\\quad\n\\bigl(k\\equiv -1\\pmod{r-1}\\ \\text{ and }\\ \\delta>0\\bigr).\n\\]\n\n•  If $k\\equiv 0\\pmod{r-1}$ and $T=k+\\delta$ (with any $0\\le\\delta<1$) then  \n\\[\nS_{r}=\\Bigl[\\,\\dfrac{k}{\\,r-1\\,},\\; \\dfrac{k}{\\,r-1\\,}+1-\\delta\\Bigr),\n\\qquad |S_{r}|=1-\\delta.\n\\]\n\n•  If $k\\equiv -1\\pmod{r-1}$ and $\\delta>0$ then  \n\\[\nS_{r}=\\Bigl[\\;\\dfrac{k+1}{\\,r-1\\,}+1-\\delta,\\; \\dfrac{k+1}{\\,r-1\\,}+1\\Bigr),\n\\qquad |S_{r}|=\\delta.\n\\]\n(When $\\delta=0$ this interval degenerates and $S_{r}$ is empty.)\n\n•  For $r=2$ one has $r-1=1$, so every $k$ satisfies both congruence conditions.  \n   Consequently  \n   \\[\n   S_{2}= \\begin{cases}\n           [\\,k,k+1)\\;, &\\delta=0,\\\\[4pt]\n           [\\,k,k+1-\\delta)\\cup[\\,k+2-\\delta,k+2), &0<\\delta<1,\n          \\end{cases}\n   \\qquad |S_{2}|=1.\n   \\]\n\nIn all cases $|S_{r}|\\le 1$, and therefore  \n\\[\n\\forall r\\ge 2:\\qquad \n\\lim_{R\\to\\infty}\\frac{|S_{r}\\cap[0,R]|}{R}=0.\n\\]\n\n(Here $|\\;\\cdot\\;|$ denotes Lebesgue measure.)\n\n\n------------------------------------------------------------------",
      "solution": "Preliminaries.  \nKeep the notation $(\\ast)$ and write every $t\\ge 0$ uniquely as  \n\\[\nt=m+\\varepsilon ,\\qquad m\\in\\mathbb Z,\\ m\\ge 0,\\ 0\\le\\varepsilon<1.\\tag{1}\n\\]\nThen  \n\\[\nA(t)=\\lfloor t+T\\rfloor =\\lfloor m+\\varepsilon+k+\\delta\\rfloor\n      =m+k+\\lfloor\\varepsilon+\\delta\\rfloor,\\tag{2}\n\\qquad\nB(t)=\\lfloor t\\rfloor =m.\\tag{3}\n\\]\n\nBecause $0\\le\\varepsilon,\\delta<1$,\n\\[\n\\lfloor\\varepsilon+\\delta\\rfloor=\n\\begin{cases}\n0,&\\varepsilon<1-\\delta,\\\\[4pt]\n1,&\\varepsilon\\ge 1-\\delta \\ (\\text{which implies }\\delta>0).\n\\end{cases}\\tag{4}\n\\]\n\n------------------------------------------------------------------\nPart (a)\n------------------------------------------------------------------\n($\\Rightarrow$)  Suppose an interval $J$ of length $1$ with $A=2B$ exists.  \nIf $\\delta\\neq 0$ the jump points of $A$ are the non-integers $\\mathbb Z-(k+\\delta)$, while the jump points of $B$ are the integers $\\mathbb Z$.  \nBecause these sets are disjoint, $A$ and $B$ cannot jump simultaneously.  \nConsequently $A-2B$ changes at every jump point of either function, so $A=2B$ cannot hold on an interval of length $1$.  Thus $\\delta\\neq 0$ is impossible, hence $\\delta=0$.\n\n($\\Leftarrow$)  If $\\delta=0$ then (2)–(3) give $A(t)=m+k$ and $B(t)=m$.  \nThe equality $A=2B$ reduces to $m=k$, with $\\varepsilon$ arbitrary, so the equality holds exactly on $[k,k+1)$, an interval of length $1$. ∎\n\n\n\n------------------------------------------------------------------\nPart (b)\n------------------------------------------------------------------\nInsert (4) into (2)–(3) for the ratio $2:1$:\n\\[\nm+k+\\lfloor\\varepsilon+\\delta\\rfloor=2m\n\\ \\Longleftrightarrow\\\nm=k+\\lfloor\\varepsilon+\\delta\\rfloor.\\tag{5}\n\\]\n\nCase 1: $\\delta=0$.  \nThen $\\lfloor\\varepsilon+\\delta\\rfloor=0$ for all $\\varepsilon$, so (5) forces $m=k$ and $t=m+\\varepsilon\\in[k,k+1)$.  Thus $S=[k,k+1)$.\n\nCase 2: $0<\\delta<1$.  \nIf $\\varepsilon<1-\\delta$ then $\\lfloor\\varepsilon+\\delta\\rfloor=0$ and (5) gives $m=k$, hence  \n$t\\in[k,k+1-\\delta)=S_{1}$.  \nIf $\\varepsilon\\ge 1-\\delta$ then $\\lfloor\\varepsilon+\\delta\\rfloor=1$ and (5) gives $m=k+1$, hence  \n$t\\in[k+2-\\delta,k+2)=S_{2}$.  \nThe two intervals are disjoint and exhaust $S$, with total length $1$. ∎\n\n\n\n------------------------------------------------------------------\nPart (c)\n------------------------------------------------------------------\nIn every case $S$ is either one interval of length $1$ or two disjoint intervals whose total length is $1$.  Therefore $|S\\cap[0,R]|\\le 1$ for every $R>0$, whence  \n\\[\n\\lim_{R\\to\\infty}\\frac{|S\\cap[0,R]|}{R}=0. \\qquad\\quad\\Box\n\\]\n\n------------------------------------------------------------------\nPart (d) (Integral ratio $r\\ge 2$)\n------------------------------------------------------------------\nLet $r\\ge 2$ be fixed.  From (2)–(3) we require\n\\[\nm+k+\\lfloor\\varepsilon+\\delta\\rfloor=r\\,m\n\\quad\\Longleftrightarrow\\quad\n(r-1)m=k+\\lfloor\\varepsilon+\\delta\\rfloor.\\tag{6}\n\\]\n\nPut  \n\\[\ns:=\\lfloor\\varepsilon+\\delta\\rfloor\\in\\{0,1\\}.\\tag{7}\n\\]\nThen (6) becomes  \n\\[\n(r-1)m=k+s.\\tag{8}\n\\]\n\nBecause $m$ must be an integer, $k+s$ must be divisible by $r-1$; conversely, whenever $k+s\\equiv 0\\pmod{r-1}$, (8) yields a suitable $m$.  Two sub-cases arise.\n\n------------------------------------------------------------------\nCase A: $s=0$  (this needs no restriction on $\\delta$)\n------------------------------------------------------------------\nThe divisibility condition is  \n\\[\nk\\equiv 0\\pmod{r-1}.\\tag{9A}\n\\]\nWhen (9A) holds, (8) gives $m=k/(r-1)$.  \nBecause $s=0$ corresponds to $\\varepsilon<1-\\delta$ (see (4)), we obtain\n\\[\nS_{r}^{(0)}=\n\\Bigl[\\,\\frac{k}{\\,r-1\\,},\\;\\frac{k}{\\,r-1\\,}+1-\\delta\\Bigr),\n\\qquad |S_{r}^{(0)}|=1-\\delta.\\tag{10A}\n\\]\n\n------------------------------------------------------------------\nCase B: $s=1$  (possible only if $\\delta>0$)\n------------------------------------------------------------------\nNow $k+s=k+1$ must satisfy  \n\\[\nk\\equiv -1\\pmod{r-1},\\tag{9B}\n\\]\nand $\\varepsilon\\ge 1-\\delta$ by (4), which forces $\\delta>0$.  \nWhen both conditions hold, (8) gives $m=(k+1)/(r-1)$, and therefore\n\\[\nS_{r}^{(1)}=\n\\Bigl[\\,\\frac{k+1}{\\,r-1\\,}+1-\\delta,\\;\\frac{k+1}{\\,r-1\\,}+1\\Bigr),\n\\qquad |S_{r}^{(1)}|=\\delta.\\tag{10B}\n\\]\nIf $\\delta=0$ the interval in (10B) collapses and $S_{r}^{(1)}$ is empty.\n\n------------------------------------------------------------------\nSummary\n------------------------------------------------------------------\nThus\n\\[\nS_{r}=\n\\begin{cases}\nS_{r}^{(0)}, & k\\equiv 0\\pmod{r-1},\\\\[6pt]\nS_{r}^{(1)}, & k\\equiv -1\\pmod{r-1}\\text{ and }\\delta>0,\\\\[6pt]\n\\varnothing, &\\text{otherwise}.\n\\end{cases}\\tag{11}\n\\]\nConsequently $|S_{r}|\\le 1$ in every case.\n\n------------------------------------------------------------------\nSpecialisation $r=2$\n------------------------------------------------------------------\nHere $r-1=1$, so $k\\equiv 0\\pmod{1}$ and $k\\equiv -1\\pmod{1}$ are both automatically true.  From (10A)–(10B):\n\n•  If $\\delta=0$ only Case A contributes, giving  \n   \\[\n   S_{2}=[\\,k,k+1),\\qquad |S_{2}|=1.\n   \\]\n\n•  If $0<\\delta<1$ both cases contribute and  \n   \\[\n   S_{2}=[\\,k,k+1-\\delta)\\cup[\\,k+2-\\delta,k+2),\\qquad |S_{2}|=1.\n   \\]\n\n------------------------------------------------------------------\nDensity estimate\n------------------------------------------------------------------\nBecause $S_{r}$ is either empty, one interval of length $\\le 1$, or (for $r=2$ with $\\delta>0$) two disjoint intervals whose total length is $1$, we have $|S_{r}\\cap[0,R]|\\le 1$ for every $R>0$.  Hence\n\\[\n\\forall r\\ge 2:\\qquad\n\\lim_{R\\to\\infty}\\frac{|S_{r}\\cap[0,R]|}{R}=0.\\qquad\\quad\\Box\n\\]",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.600025",
        "was_fixed": false,
        "difficulty_analysis": "[解析失败]"
      }
    },
    "original_kernel_variant": {
      "question": "Two identical autonomous drones fly around the same circular course at a constant speed, completing one full lap every hour.\n\n•  Drone A passes the fixed checkpoint P at the clock time t = –T  (with T > 0) and afterwards continues indefinitely.  \n•  Drone B passes P at the clock time t = 0 and afterwards continues indefinitely.\n\nFor every real t ≥ 0 define  \n  A(t) := ⌊t+T⌋ and B(t) := ⌊t⌋,  \nso that A(t)   (resp. B(t)) is the number of completed laps of Drone A (resp. B) by the instant t.\n\nThroughout write  \n  T = k + δ with k ∈ ℤ, k ≥ 0, 0 ≤ δ < 1.    (∗)\n\nPart (a) (Prologue — existence of one contiguous hour)  \nProve that a contiguous time–interval of length exactly one hour on which  \n   A(t) = 2 B(t)  for every instant t in the interval  \nexists if and only if T is an integer (i.e. δ = 0 in (∗)).\n\nPart (b) (Complete description of all solutions for the ratio 2 : 1)  \nFor the representation (∗) determine the set  \n  S := { t ≥ 0 : A(t) = 2 B(t) }.  \nShow that\n\n (i) If δ = 0 then S = [k, k+1).               (a single interval of length 1)\n\n (ii) If 0 < δ < 1 then S is the disjoint union of the two intervals  \n   S₁ = [k, k+1−δ) and S₂ = [k+2−δ, k+2),  \nwhose lengths are 1−δ and δ, respectively.  \n\nIn every case |S| = 1.\n\nPart (c) (Total measure and asymptotic density for the ratio 2 : 1)  \nProve that, for every R > 0,  \n  |S ∩ [0,R]| ≤ 1, and hence lim_{R→∞} |S ∩ [0,R]| / R = 0.\n\nPart (d) (General integral ratio — corrected harder extension)  \nFix an integer r ≥ 2 and keep the notation A(t) = ⌊t+T⌋, B(t) = ⌊t⌋.  \nDefine  \n  S_r := { t ≥ 0 : A(t) = r B(t) }.  \n\nProve the following complete description for all T > 0:\n\n• Put k and δ as in (∗).  The set S_r is non–empty if and only if  \n k ≡ 0 or k ≡ −1 ≡ r−2 (mod r−1).\n\n• If k ≡ 0 (mod r−1) and T = k+δ then  \n  S_r = [ k/(r−1), k/(r−1) + 1 − δ ),  |S_r| = 1 − δ.\n\n• If k ≡ −1 (mod r−1) and T = k+δ then  \n  S_r = [ (k+1)/(r−1) + 1 − δ, (k+1)/(r−1) + 1 ),  |S_r| = δ.\n\n• For r = 2 one has r−1 = 1, so both congruence conditions are always satisfied; hence  \n  S₂ = [k, k+1−δ) ∪ [k+2−δ, k+2),  |S₂| = 1.\n\nConsequently, for every integer r ≥ 2,  \n |S_r| ≤ 1 and lim_{R→∞} |S_r ∩ [0,R]| / R = 0.",
      "solution": "Preliminaries.  \nRetain the notation (∗) and write every t ≥ 0 uniquely as  \n  t = m + ε with m ∈ ℤ, m ≥ 0, 0 ≤ ε < 1.     (1)\n\nThen  \n A(t) = ⌊t+T⌋ = ⌊m+ε+k+δ⌋ = m + k + ⌊ε+δ⌋,             (2)  \n B(t) = ⌊t⌋ = m.                                         (3)\n\nBecause 0 ≤ ε,δ < 1,  \n  ⌊ε+δ⌋ = 0  iff ε < 1−δ,         (4a)  \n  ⌊ε+δ⌋ = 1  iff ε ≥ 1−δ.        (4b)\n\n\n\n------------------------------------------------------------------\nPart (a)\n------------------------------------------------------------------\n(⇒) Suppose a one–hour interval J exists on which A = 2 B.  \nIf δ ≠ 0 then the “jump points’’ of A are the set ℤ − T = ℤ − (k+δ), i.e. the non–integers …, −δ, 1−δ, 2−δ, ….  \nThe jump points of B are the integers ℤ.  \nBecause δ ∉ ℤ, the two jump sets are disjoint; hence A and B can never jump simultaneously.  \nWithin any open sub–interval of J that contains no jump point both A and B are constant, so equality persists.  \nWhen J meets a jump point of either function equality is destroyed, contradicting the assumed length |J| = 1.  \nTherefore δ ≠ 0 is impossible; hence δ = 0.\n\n(⇐)  If δ = 0 then (2)–(3) give A(t) = m+k and B(t) = m.  \nEquality A = 2 B becomes m+k = 2m, i.e. m = k, and ε is unrestricted.  \nThus A = 2 B exactly on [k,k+1), an interval of length 1. ∎\n\n\n\n------------------------------------------------------------------\nPart (b)\n------------------------------------------------------------------\nInsert (4) into (2)–(3) (with r = 2):\n\n m + k + ⌊ε+δ⌋ = 2m  ⇔  m = k + ⌊ε+δ⌋.             (5)\n\nCase 1: δ = 0.  \nThen ⌊ε+δ⌋ = 0 for all ε, so (5) forces m = k and hence  \n t = m+ε ∈ [k, k+1).  \nTherefore S = [k,k+1).\n\nCase 2: 0 < δ < 1.  \nUsing (4a)–(4b) in (5) one obtains\n\n • If ε < 1−δ  (so ⌊ε+δ⌋ = 0) then m = k, giving  \n  t ∈ [k, k+1−δ).  \n\n • If ε ≥ 1−δ (so ⌊ε+δ⌋ = 1) then m = k+1, giving  \n  t ∈ [k+2−δ, k+2).\n\nThe two intervals are disjoint and exhaust S; their lengths are 1−δ and δ, so |S| = 1. ∎\n\n\n\n------------------------------------------------------------------\nPart (c)\n------------------------------------------------------------------\nIn every case S is the union of at most two intervals whose total length is 1.  \nHence for every R > 0 we have |S ∩ [0,R]| ≤ 1 and therefore  \n  lim_{R→∞} |S ∩ [0,R]| / R = 0. ∎  \n\n\n\n------------------------------------------------------------------\nPart (d) (Integral ratio r ≥ 2)\n------------------------------------------------------------------\nLet r ≥ 2 be fixed.  With (1)–(4) we require\n\n m + k + ⌊ε+δ⌋ = r m  ⇔  (r−1)m = k + ⌊ε+δ⌋.     (6)\n\nPut  \n  s := ⌊ε+δ⌋ ∈ {0,1}.                             (7)\n\nThen (6) becomes  \n  (r−1)m = k + s.                                (8)\n\nFor m to be integral we must have  \n\n  k + s ≡ 0 (mod r−1).                           (9)\n\nBecause s = 0 or 1, exactly the following two congruence classes can occur:\n\n • s = 0 ⇒ k ≡ 0 (mod r−1),                     (10a)  \n • s = 1 ⇒ k ≡ −1 ≡ r−2 (mod r−1).             (10b)\n\nNo other values of k permit solutions, so S_r = ∅ unless (10a) or (10b) is satisfied.\n\n–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––\nCase A: k ≡ 0 (mod r−1) (s = 0).\n–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––\nThen (8) gives m = k/(r−1), an integer.  \nCondition (4a) (ε < 1−δ) is equivalent to (7) with s = 0, so\n\n  t = m + ε ∈ [ k/(r−1), k/(r−1) + 1 − δ ).     (11)\n\nSet  \n  S_r^{(0)} := [ k/(r−1), k/(r−1) + 1 − δ ), |S_r^{(0)}| = 1−δ.   (12)\n\n–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––\nCase B: k ≡ −1 (mod r−1) (s = 1).\n–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––\nNow (8) gives m = (k+1)/(r−1).  \nCondition (4b) (ε ≥ 1−δ) combines with (7) to give\n\n  t = m + ε ∈ [ m + 1 − δ, m + 1 )  \n    = [ (k+1)/(r−1) + 1 − δ, (k+1)/(r−1) + 1 ).     (13)\n\nSet  \n  S_r^{(1)} := [ (k+1)/(r−1) + 1 − δ, (k+1)/(r−1) + 1 ),  \n  |S_r^{(1)}| = δ.                                             (14)\n\n–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––\nCase C: k ≡ 2,…,r−3 (mod r−1) (possible only for r ≥ 4).\n–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––\nNeither value of s makes k+s divisible by r−1, so S_r = ∅.        (15)\n\nPutting the pieces together,\n\n S_r = \n  ⎧ S_r^{(0)}                           if (10a) holds and (10b) fails,  \n  ⎨ S_r^{(1)}                           if (10b) holds and (10a) fails,  \n  ⎩ S_r^{(0)} ∪ S_r^{(1)} (= two intervals) if r = 2.                (16)\n\nLengths.  Each of S_r^{(0)}, S_r^{(1)} has length ≤ 1, and their intersection is empty.  Therefore |S_r| ≤ 1 in all cases; equality holds only for r = 2 or when δ = 0 with k ≡ 0 (mod r−1).\n\nDensity.  Because S_r is either empty, a single interval of length ≤ 1, or (when r = 2) two disjoint intervals whose total length is 1, we have |S_r ∩ [0,R]| ≤ 1 for every R > 0.  Hence  \n  lim_{R→∞} |S_r ∩ [0,R]| / R = 0. ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.480332",
        "was_fixed": false,
        "difficulty_analysis": "[解析失败]"
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}