| {"year": "2013", "tier": "T2", "problem_label": "1", "problem_type": null, "exam": "Nordic_MO", "problem": "Let $\\left(a_{n}\\right)_{n \\geq 1}$ be a sequence with $a_{1}=1$ and\n\n$$\na_{n+1}=\\left\\lfloor a_{n}+\\sqrt{a_{n}}+\\frac{1}{2}\\right\\rfloor\n$$\n\nfor all $n \\geq 1$, where $\\lfloor x\\rfloor$ denotes the greatest integer less than or equal to $x$. Find all $n \\leq 2013$ such that $a_{n}$ is a perfect square.", "solution": "We will show by induction that $a_{n}=1+\\left\\lfloor\\frac{n}{2}\\right\\rfloor\\left\\lfloor\\frac{n+1}{2}\\right\\rfloor$, which is equivalent to $a_{2 m}=1+m^{2}$ and $a_{2 m+1}=1+m(m+1)$. Clearly this is true for $a_{1}$. If $a_{2 m+1}=1+m(m+1)$ then\n\n$$\na_{2 m+2}=\\left\\lfloor m^{2}+m+1+\\sqrt{m^{2}+m+1}+\\frac{1}{2}\\right\\rfloor\n$$\n\nand since $m+\\frac{1}{2}<\\sqrt{m^{2}+m+1}<m+1$ (easily seen by squaring), we get $a_{2 m+2}=\\left(m^{2}+m+1\\right)+(m+1)=1+(m+1)^{2}$.\n\nAnd if $a_{2 m}=1+m^{2}$ then\n\n$$\na_{2 m+1}=\\left\\lfloor m^{2}+1+\\sqrt{m^{2}+1}+\\frac{1}{2}\\right\\rfloor\n$$\n\nand here $m<\\sqrt{m^{2}+1}<m+\\frac{1}{2}$, so $a_{2 m+1}=\\left(m^{2}+1\\right)+m=1+m(m+1)$.\n\nIf $m \\geq 1$ then $m^{2}<1+m^{2}<(m+1)^{2}$ and $m^{2}<m^{2}+m+1<(m+1)^{2}$, so $a_{n}$ cannot be a perfect square if $n>1$. Therefore $a_{1}=1$ is the only perfect square in the sequence.", "metadata": {"resource_path": "Nordic_MO/segmented/en-2013-sol.jsonl", "problem_match": "\n\nPRoblem 1.", "solution_match": "\n\nSolution."}} |
| {"year": "2013", "tier": "T2", "problem_label": "2", "problem_type": null, "exam": "Nordic_MO", "problem": "In a football tournament there are $n$ teams, with $n \\geq 4$, and each pair of teams meets exactly once. Suppose that, at the end of the tournament, the final scores form an arithmetic sequence where each team scores 1 more point than the following team on the scoreboard. Determine the maximum possible score of the lowest scoring team, assuming usual scoring for football games (where the winner of a game gets 3 points, the loser 0 points, and if there is a tie both teams get 1 point).", "solution": "Note that the total number of games equals the number of different pairings, that is, $n(n-1) / 2$. Suppose the lowest scoring team ends with $k$ points. Then the total score for all teams is\n\n$$\nk+(k+1)+\\cdots+(k+n-1)=n k+\\frac{(n-1) n}{2}\n$$\n\nSome games must end in a tie, for otherwise, all team scores would be a multiple of 3 and cannot be 1 point apart. Since the total score of a tie is only 2 points compared to 3 points if one of the teams wins, we therefore know that\n\n$$\nn k+\\frac{(n-1) n}{2}<3 \\cdot \\frac{n(n-1)}{2}\n$$\n\nso $n k<n(n-1)$, and hence $k<n-1$. It follows that the lowest scoring team can score no more than $n-2$ points.\n\nWe now show by induction that it is indeed possible for the lowest scoring team to score $n-2$ points.\n\nThe following scoreboard shows this is possible for $n=4$ :\n\n| - | 3 | 1 | 1 | 5 |\n| :---: | :---: | :---: | :---: | :---: |\n| 0 | - | 1 | 3 | 4 |\n| 1 | 1 | - | 1 | 3 |\n| 1 | 0 | 1 | - | 2 |\n\nNow suppose we have a scoreboard for $n$ teams labelled $T_{n-2}, \\ldots, T_{2 n-3}$, where team $T_{i}$ scores $i$ points. Keep the results among these teams unchanged while adding one more team.\n\nWrite $n=3 q+r$ with $r \\in\\{1,-1,0\\}$, and let the new team tie with just one of the original teams, lose against $q$ teams, and win against the rest of them. The new team thus wins $n-1-q$ games, and gets $1+3(n-1-q)=3 n-2-3 q=2 n-2+r$ points.\n\nMoreover, we arrange for the $q$ teams which win against the new team to form an arithmetic sequence $T_{j}, T_{j+3}, \\ldots, T_{j+3(q-1)}=T_{j+n-r-3}$, so that each of them, itself having gained three points, fills the slot vacated by the next one.\n\n(i) If $r=1$, then let the new team tie with team $T_{n-2}$ and lose to each of the teams $T_{n-1}, T_{n+2}, \\ldots, T_{n-1+n-r-3}=T_{2 n-5}$.\n\nTeam $T_{n-2}$ now has $n-1$ points and takes the place vacated by $T_{n-1}$. At the other end, $T_{2 n-5}$ now has $2 n-2$ points, just one more than the previous top team $T_{2 n-3}$. And the new team has $2 n-2+r=2 n-1$ points, becoming the new top team. The teams now have all scores from $n-1$ up to $2 n-1$.\n\n(ii) If $r=-1$, then let the new team tie with team $T_{2 n-3}$ and lose to each of the teams $T_{n-2}, T_{n+1}, \\ldots, T_{n-2+n-r-3}=T_{2 n-4}$.\n\nThe old top team $T_{2 n-3}$ now has $2 n-2$ points, and its former place is filled by the new team, which gets $2 n-2+r=2 n-3$ points. $T_{2 n-4}$ now has $2 n-1$ points and is the new top team. So again we have all scores ranging from $n-1$ up to $2 n-1$.\n\n(iii) If $r=0$, then let the new team tie with team $T_{n-2}$ and lose to teams $T_{n-1}, T_{n+2}, \\ldots, T_{n-1+n-r-3}=T_{2 n-4}$.\n\nTeam $T_{n-2}$ now has $n-1$ points and fills the slot vacated by $T_{n-1}$. At the top end, $T_{2 n-4}$ now has $2 n-1$ points, while the new team has $2 n-2+r=2 n-2$ points, and yet again we have all scores from $n-1$ to $2 n-1$.\n\nThis concludes the proof.", "metadata": {"resource_path": "Nordic_MO/segmented/en-2013-sol.jsonl", "problem_match": "\n\nProblem 2.", "solution_match": "\n\nSolution."}} |
| {"year": "2013", "tier": "T2", "problem_label": "3", "problem_type": null, "exam": "Nordic_MO", "problem": "Define a sequence $\\left(n_{k}\\right)_{k \\geq 0}$ by $n_{0}=n_{1}=1$, and $n_{2 k}=n_{k}+n_{k-1}$ and $n_{2 k+1}=n_{k}$ for $k \\geq 1$. Let further $q_{k}=n_{k} / n_{k-1}$ for each $k \\geq 1$. Show that every positive rational number is present exactly once in the sequence $\\left(q_{k}\\right)_{k \\geq 1}$.", "solution": "Clearly, all the numbers $n_{k}$ are positive integers. Moreover,\n\n$$\nq_{2 k}=\\frac{n_{2 k}}{n_{2 k-1}}=\\frac{n_{k}+n_{k-1}}{n_{k-1}}=q_{k}+1\n$$\n\nand similarly,\n\n$$\n\\frac{1}{q_{2 k+1}}=\\frac{n_{2 k}}{n_{2 k+1}}=\\frac{n_{k}+n_{k-1}}{n_{k}}=\\frac{1}{q_{k}}+1\n$$\n\nIn particular, $q_{k}>1$ when $k$ is even, and $q_{k}<1$ when $k \\geq 3$ is odd.\n\nWe will show the following by induction on $t=2,3,4, \\ldots$ :\n\nClaim: Every rational number $r / s$ where $r$, s are positive integers with $\\operatorname{gcd}(r, s)=$ 1 and $r+s \\leq t$ occurs precisely once among the numbers $q_{k}$.\n\nThe claim is clearly true for $t=2$, since then $r / s=1 / 1=1$ is the only possibility, and $q_{1}$ is the only occurrence of 1 in the sequence.\n\nNow, assume that $u \\geq 3$ and that the claim holds for $t=u-1$. Let $r$ and $s$ be positive integers with $\\operatorname{gcd}(r, s)=1$ and $r+s=u$.\n\nFirst, assume that $r>s$. We know that $r / s=q_{m}$ is only possible if $m$ is even. But\n\n$$\n\\frac{r}{s}=q_{2 k} \\Leftrightarrow \\frac{r-s}{s}=q_{k}\n$$\n\nby (1), and moreover, the latter equality holds for precisely one $k$ according to the induction hypothesis, since $\\operatorname{gcd}(r-s, s)=1$ and $(r-s)+s=r \\leq t$.\n\nNext, assume that $r<s$. We know that $r / s=q_{m}$ is only possible if $m$ is odd. But\n\n$$\n\\frac{r}{s}=q_{2 k+1} \\Leftrightarrow \\frac{s}{r}=\\frac{1}{q_{2 k+1}} \\Leftrightarrow \\frac{s-r}{r}=\\frac{1}{q_{k}}\n$$\n\nby (2), and moreover, the latter equality holds for precisely one $k$ according to the induction hypothesis, since $\\operatorname{gcd}(s-r, r)=1$ and $(s-r)+r=s \\leq t$.", "metadata": {"resource_path": "Nordic_MO/segmented/en-2013-sol.jsonl", "problem_match": "\n\nProblem 3.", "solution_match": "\n\nSolution."}} |
| {"year": "2013", "tier": "T2", "problem_label": "4", "problem_type": null, "exam": "Nordic_MO", "problem": "Let $A B C$ be an acute angled triangle, and $H$ a point in its interior. Let the reflections of $H$ through the sides $A B$ and $A C$ be called $H_{c}$ and $H_{b}$, respectively, and let the reflections of $H$ through the midpoints of these same sides be called $H_{c}^{\\prime}$ and $H_{b}^{\\prime}$, respectively. Show that the four points $H_{b}, H_{b}^{\\prime}, H_{c}$, and $H_{c}^{\\prime}$ are concyclic if and only if at least two of them coincide or $H$ lies on the altitude from $A$ in triangle $A B C$.", "solution": "If at least two of the four points $H_{b}, H_{b}^{\\prime}, H_{c}$, and $H_{c}^{\\prime}$ coincide, all four are obviously concyclic. Therefore we may assume that these four points are distinct.\n\nLet $P_{b}$ denote the midpoint of segment $H H_{b}, P_{b}^{\\prime}$ the midpoint of segment $H H_{b}^{\\prime}, P_{c}$ the midpoint of segment $H H_{c}$, and $P_{c}^{\\prime}$ the midpoint of segment $H H_{c}^{\\prime}$.\n\nThe triangle $H H_{b} H_{b}^{\\prime}$ being right-angled in $H_{b}$, it follows that the perpendicular bisector $\\ell_{b}$ of the side $H_{b} H_{b}^{\\prime}$ goes through the point $P_{b}^{\\prime}$. Since the segments $P_{b} P_{b}^{\\prime}$ and $H_{b} H_{b}^{\\prime}$ are parallel and $P_{b}^{\\prime}$ is the midpoint of the side $A C$, we then conclude that $\\ell_{b}$ also goes through the circumcentre $O$ of triangle $A B C$.\n\nSimilarly the perpendicular bisector $\\ell_{c}$ of the segment $H_{c} H_{c}^{\\prime}$ also goes through $O$. Hence the four points $H_{b}, H_{b}^{\\prime}, H_{c}$, and $H_{c}^{\\prime}$ are concyclic if and only if also the perpendicular bisector $\\ell$ of the segment $H_{b}^{\\prime} H_{c}^{\\prime}$ goes through the point $O$. Since $H_{b}^{\\prime} H_{c}^{\\prime}\\left\\|P_{b}^{\\prime} P_{c}^{\\prime}\\right\\| B C$, this is the case if and only if $\\ell$ is the perpendicular bisector $m$ of the segment $B C$.\n\nLet $k$ denote the perpendicular bisector of the segment $P_{b}^{\\prime} P_{c}^{\\prime}$. Since the lines $\\ell$ and $m$ are obtained from $k$ by similarities of ratio 2 and centres $H$ and $A$, respectively, they coincide if and only if $H A$ is parallel to $m$. Thus $H_{b}, H_{b}^{\\prime}, H_{c}$, and $H_{c}^{\\prime}$ are concyclic if and only if $H$ lies on the altitude from $A$ in triangle $A B C$.\n\n\n\nClick here to experiment with the figure in GeoGebra.", "metadata": {"resource_path": "Nordic_MO/segmented/en-2013-sol.jsonl", "problem_match": "\n\nProblem 4.", "solution_match": "\n\nSolution."}} |
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