{"year": "2015", "tier": "T3", "problem_label": "1", "problem_type": null, "exam": "Benelux_MO", "problem": "## Determine the smallest positive integer $q$ with the following property:\n\nfor every integer $m$ with $1 \\leqslant m \\leqslant 1006$, there exists an integer $n$ such that\n\n$$\n\\frac{m}{1007} q<n<\\frac{m+1}{1008} q .\n$$", "solution": ". For $m=1006$, we have\n\n$$\nq-q / 1007<n<q-q / 1008\n$$\n\nfor some integer $n$. If $q \\leqslant 1007$, then $q-q / 1007$ and $q-q / 1008$ are both numbers that are at least $q-1$ and smaller than $q$, so there can be no integer $n$ in between. Hence $q>1007$ and $q-q / 1008 \\leqslant q-1$, implying that $n<q-1$, and that $q-q / 1007<q-2$. By rearranging terms, we find $q>2014$, and hence $q \\geqslant 2015$.\n\nLet us prove that $q=2015$ works. Indeed, $m q / 1007=2 m+m / 1007<2 m+1$ and $(m+1) q / 1008=2(m+1)-(m+1) / 1008>2 m+1$ for $1 \\leqslant m \\leqslant 1006$. Hence each pair of inequalities can be satisfied by taking $n=2 m+1$. This completes the proof, and shows that the smallest possible of $q$ is indeed 2015.", "metadata": {"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2015-zz.jsonl", "problem_match": "# Problem 1.", "solution_match": "\nSolution 1"}}
{"year": "2015", "tier": "T3", "problem_label": "1", "problem_type": null, "exam": "Benelux_MO", "problem": "## Determine the smallest positive integer $q$ with the following property:\n\nfor every integer $m$ with $1 \\leqslant m \\leqslant 1006$, there exists an integer $n$ such that\n\n$$\n\\frac{m}{1007} q<n<\\frac{m+1}{1008} q .\n$$", "solution": ". For $m=1, \\ldots, 1006$, there must exist an integer $N$ divisible by $1007 \\cdot 1008$ satisfying the double inequality\n\n$$\n1008 m q<N<1007(m+1) q .\n$$\n\nSince $N$ is divisible by 1007 and 1008 , we may write\n\n$$\nN=1008 m q+1008 k=1007(m+1) q-1007 \\ell\n$$\n\nwhere $k, \\ell>0$ are integers. Hence\n\n$$\n(1007-m) q=1008 k+1007 \\ell \\geqslant 2015\n$$\n\nsince $k, \\ell>0$. Choosing $m=1006$ in this last inequality, it follows that $q \\geqslant 2015$. Conversely, for $q=2015=1008+1007,(*)$ can be satisfied by taking $k=\\ell=1007-m$. (Indeed, 1007 and 1008 are coprime, and so integer divisible by 1007 and 1008 is also divisible by their product.) Thus $q=2015$ has the desired property, and we are done.\n\n#", "metadata": {"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2015-zz.jsonl", "problem_match": "# Problem 1.", "solution_match": "\nSolution 2"}}
{"year": "2015", "tier": "T3", "problem_label": "2", "problem_type": null, "exam": "Benelux_MO", "problem": "Let $A B C$ be an acute triangle with circumcentre $O$. Let $\\Gamma_{B}$ be the circle through $A$ and $B$ that is tangent to $A C$, and let $\\Gamma_{C}$ be the circle through $A$ and $C$ that is tangent to $A B$. An arbitrary line through $A$ intersects $\\Gamma_{B}$ again in $X$ and $\\Gamma_{C}$ again in $Y$. Prove that $|O X|=|O Y|$.\n\nThe following solutions are valid for the configurations appearing in the diagrams.", "solution": ". Let $O_{B}$ and $O_{C}$ denote the respective centres of $\\Gamma_{B}$ and $\\Gamma_{C}$. We shall show that $X O_{B} O$ and $O O_{C} Y$ are congruent. Now $A O_{B} \\perp A C$ since $\\Gamma_{B}$ is tangent to $A C$ and $O O_{C} \\perp A C$ since $O O_{C}$ is the perpendicular bisector of $[A C]$. Hence $A O_{B} \\| O O_{C}$, and similarly, $A O_{C} \\| O O_{B}$. It follows that $A O_{B} O O_{C}$ is a parallelogram. In particular, $\\left|O_{B} X\\right|=\\left|A O_{B}\\right|=\\left|O O_{C}\\right|$ and $\\left|O_{C} Y\\right|=\\left|O_{C} A\\right|=\\left|O O_{B}\\right|$.\n\nIt will therefore suffice to show that the angles $\\angle O O_{B} X$ and $\\angle Y O_{C} O$ are equal. Noting that $\\angle A O_{B} O=\\angle A O_{C} O$ since $A O_{B} O O_{C}$ is a parallelogram, this follows by angle chasing:\n\n$$\n\\begin{aligned}\n\\angle X O_{B} O & =\\angle A O_{B} O-\\angle A O_{B} X=\\angle A O_{C} O-\\left(180^{\\circ}-2 \\angle O_{B} A X\\right) \\\\\n& =\\angle A O_{C} O-180^{\\circ}+2\\left(\\angle O_{B} A O_{C}-\\angle Y A O_{C}\\right) \\\\\n& =\\angle A O_{C} O-180^{\\circ}+2\\left(180^{\\circ}-\\angle A O_{C} O\\right)-\\left(180^{\\circ}-\\angle A O_{C} Y\\right) \\\\\n& =\\angle A O_{C} Y-\\angle A O_{C} O=\\angle O O_{C} Y .\n\\end{aligned}\n$$\n\n![](https://cdn.mathpix.com/cropped/2024_12_15_c84e14b61920ef432fecg-3.jpg?height=966&width=1438&top_left_y=1525&top_left_x=281)\n\nA Variant. As in the main solution, we note that $A O_{B} O O_{C}$ is a parallelogram. Notice that $O_{B}$ and $O_{C}$ lie on the respective perpendicular bisectors of $[A X]$ and $[A Y]$. The following lemma then implies that $O$ lies on the perpendicular bisector of $[X Y]$, completing the proof:\n\nLet $P_{1}, P_{2}, P_{3}$ be three points on a line. Let $O_{1}$ be a point on the perpendicular bisector of $\\left[P_{2} P_{3}\\right]$, let $O_{2}$ be a point on the perpendicular bisector of $\\left[P_{3} P_{1}\\right]$ and let $O_{3}$ be the point in the plane such that $O_{1} P_{3} O_{2} O_{3}$ is a parallelogram. Then $O_{3}$ lies on the perpendicular bisector of $\\left[P_{1} P_{2}\\right]$.\nProof. One can choose coordinates such that $P_{3}=(0,0)$ and $P_{1}=(2,0)$. Then $P_{2}=(2 a, 0), O_{1}=(a, b)$ and $O_{2}=(1, c)$ for some $a, b, c \\in \\mathbb{R}$. Hence $O_{3}=(a+1, b+c)$ lies on the perpendicular bisector of $\\left[P_{1} P_{2}\\right]$.", "metadata": {"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2015-zz.jsonl", "problem_match": "# Problem 2.", "solution_match": "\nSolution 1"}}
{"year": "2015", "tier": "T3", "problem_label": "2", "problem_type": null, "exam": "Benelux_MO", "problem": "Let $A B C$ be an acute triangle with circumcentre $O$. Let $\\Gamma_{B}$ be the circle through $A$ and $B$ that is tangent to $A C$, and let $\\Gamma_{C}$ be the circle through $A$ and $C$ that is tangent to $A B$. An arbitrary line through $A$ intersects $\\Gamma_{B}$ again in $X$ and $\\Gamma_{C}$ again in $Y$. Prove that $|O X|=|O Y|$.\n\nThe following solutions are valid for the configurations appearing in the diagrams.", "solution": ". Let $\\alpha=\\angle B A C$. Observe that $\\angle A X B=180^{\\circ}-\\alpha=\\angle C Y A$ by tangential angles. Let $Z$ be the intersection of $B X$ and $C Y$. Thus $Z X Y$ is isosceles with, in particular, $\\angle Z X Y=\\angle Z Y X=\\alpha$. It will thus suffice to show that $O Z$ bisects $\\angle X Z Y$.\n\nNow $\\angle B Z C=2 \\alpha=\\angle B O C$ since $O$ is the circumcentre of $A B C$, and hence $B Z O C$ is cyclic. In particular, since $|O B|=|O C|$, it follows that $\\angle O Z Y=\\angle O B C=90^{\\circ}-\\alpha$. But $\\angle X Z Y=180^{\\circ}-2 \\alpha$, and so $\\angle X Z Y=2 \\angle O Z Y$, which shows that $O Z$ bisects $\\angle X Z Y$.\n![](https://cdn.mathpix.com/cropped/2024_12_15_c84e14b61920ef432fecg-4.jpg?height=881&width=901&top_left_y=1408&top_left_x=566)", "metadata": {"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2015-zz.jsonl", "problem_match": "# Problem 2.", "solution_match": "\nSolution 2"}}
{"year": "2015", "tier": "T3", "problem_label": "2", "problem_type": null, "exam": "Benelux_MO", "problem": "Let $A B C$ be an acute triangle with circumcentre $O$. Let $\\Gamma_{B}$ be the circle through $A$ and $B$ that is tangent to $A C$, and let $\\Gamma_{C}$ be the circle through $A$ and $C$ that is tangent to $A B$. An arbitrary line through $A$ intersects $\\Gamma_{B}$ again in $X$ and $\\Gamma_{C}$ again in $Y$. Prove that $|O X|=|O Y|$.\n\nThe following solutions are valid for the configurations appearing in the diagrams.", "solution": ". Consider inversion $\\mathscr{I}$ in a circle centred at $A$. Under $\\mathscr{I}$,\n\n$$\nB \\mapsto B^{\\prime}, \\quad C \\mapsto C^{\\prime}, \\quad O \\mapsto O^{\\prime}, \\quad X \\mapsto X^{\\prime}, \\quad Y \\mapsto Y^{\\prime},\n$$\n\n$\\Gamma_{B} \\mapsto \\gamma_{B}$, a line through $B^{\\prime}$ parallel to $A C^{\\prime}$,\n$\\Gamma_{C} \\mapsto \\gamma_{C}$, a line through $C^{\\prime}$ parallel to $A B^{\\prime}$,\nNotice that $\\mathscr{I}$ sends the circumcircle of $A B C$ to the line $B^{\\prime} C^{\\prime}$, and hence maps $A O$ to a line perpendicular to $B C$. Further, if $[A D]$ is a diameter of this circumcircle and $D \\mapsto D^{\\prime}$ under $\\mathscr{I}$, then $D^{\\prime}$ lies on $B C$. Also, $|A D|=2|A O|$ implies $\\left|A O^{\\prime}\\right|=2\\left|A D^{\\prime}\\right|$, and hence $O^{\\prime}$ is the image of $A$ under reflection in $B C$. Finally, $\\angle O X A=\\angle X^{\\prime} O^{\\prime} A$ and $\\angle O Y A=\\angle Y^{\\prime} O^{\\prime} A$, and hence $|O X|=|O Y|$ if and only if $O^{\\prime} A$ bissects $\\angle X^{\\prime} O^{\\prime} Y^{\\prime}$ externally. We have thus reduced the problem to the following statement:\n\nIn triangle $A B C$, let $O$ be the reflection of $A$ over $B C, \\gamma_{B}$ be a line parallel to $A C$ through $B$, and $\\gamma_{C}$ be a line parallel to $A B$ through $C$. For an arbitrary line $\\ell$ passing through $A$, let $X$ and $Y$ be the intersections of $\\ell$ with $\\gamma_{B}$ and $\\gamma_{C}$, respectively. Prove that $O A$ bisects $\\angle X O Y$ externally.\n![](https://cdn.mathpix.com/cropped/2024_12_15_c84e14b61920ef432fecg-5.jpg?height=882&width=643&top_left_y=1196&top_left_x=692)\n\nLet $P=\\gamma_{A} \\cap \\gamma_{B}$. By construction, $P O B C$ is an isosceles trapezoid (and therefore cyclic), $\\angle O B X=\\angle O B P=\\angle O C P=\\angle O C Y$. Further, since $X B \\| A C$ and $Y C \\| A B$, we have $\\triangle A B X \\sim \\triangle Y C A$. Therefore, as $|O B|=|A B|$ and $|A C|=|O C|$ by construction,\n\n$$\n\\frac{|O B|}{|B X|}=\\frac{|A B|}{|B X|}=\\frac{|Y C|}{|C A|}=\\frac{|Y C|}{|C O|}\n$$\n\nIt follows that $\\triangle O B X \\sim \\triangle Y C O$. Hence\n\n$$\n\\frac{|O X|}{|O Y|}=\\frac{|X B|}{|O C|}=\\frac{|X B|}{|A C|}=\\frac{|X A|}{|A Y|}\n$$\n\nand so the result follows from the angle bisector theorem.\n\n#", "metadata": {"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2015-zz.jsonl", "problem_match": "# Problem 2.", "solution_match": "\nSolution 3"}}
{"year": "2015", "tier": "T3", "problem_label": "3", "problem_type": null, "exam": "Benelux_MO", "problem": "Does there exist a prime number whose decimal representation is of the form $3811 \\cdots 11$ (that is, consisting of the digits 3 and 8 in that order followed by one or more digits 1 )?", "solution": "Write\n\n$$\na(n)=38 \\underbrace{11 \\cdots 11}_{n \\text { digits } 1} .\n$$\n\nThere are three cases to consider, depending on the remainder of $n$ upon division by three.\n\n- If $n=3 k+1 \\equiv 1(\\bmod 3)$, then the sum of the digits of $a(n)$ is equal to $3(k+4)$, i.e. divisible by 3 , and hence so is $a(n)$.\n- If $n=3 k+2 \\equiv 2(\\bmod 3)$, then note that $a(2)=3811=3700+111$ is divisible by 37 . By induction, as $a(3 k+2)=1000 a(3 k-1)+111$, it follows that $a(3 k+2)$ is divisible by 37 for each $k \\geqslant 0$.\n- If $n=3 k \\equiv 0(\\bmod 3)$, observe that\n\n$$\n9 a(3 k)=342 \\underbrace{99 \\ldots 99}_{n \\text { digits } 9}=\\left(7 \\cdot 10^{k}\\right)^{3}-1\n$$\n\nwhich is properly divisible by $7 \\cdot 10^{k}-1$, a number that is larger than 9 . Hence $a(3 k)$ admits a non-trivial factor and so is not prime.\n\n#", "metadata": {"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2015-zz.jsonl", "problem_match": "# Problem 3.", "solution_match": "\nSolution."}}
{"year": "2015", "tier": "T3", "problem_label": "4", "problem_type": null, "exam": "Benelux_MO", "problem": "An arithmetic progression is a set of the form $\\{a, a+d, \\ldots, a+k d\\}$, where $a, d, k$ are positive integers and $k \\geqslant 2$. Thus an arithmetic progression has at least three elements and the successive elements have difference $d$, called the common difference of the arithmetic progression.\n\nLet $n$ be a positive integer. For each partition of the set $\\{1,2, \\ldots, 3 n\\}$ into arithmetic progressions, we consider the sum $S$ of the respective common differences of these arithmetic progressions. What is the maximal value $S$ that can attain?\n(A partition of a set $A$ is a collection of disjoint subsets of $A$ whose union is $A$.)", "solution": "The maximum value is $n^{2}$, which is attained for the partition into $n$ arithmetic progressions $\\{1, n+1,2 n+1\\}, \\ldots,\\{n, 2 n, 3 n\\}$, each of difference $n$.\n\nSuppose indeed that the set has been partioned into $N$ progressions, of respective lengths $\\ell_{i}$, and differences $d_{i}$, for $1 \\leqslant i \\leqslant N$. Since $\\ell_{i} \\geqslant 3$,\n\n$$\n2 \\sum_{i=1}^{N} d_{i} \\leqslant \\sum_{i=1}^{N}\\left(\\ell_{i}-1\\right) d_{i}=\\sum_{i=1}^{N} a_{i}-\\sum_{i=1}^{N} b_{i}\n$$\n\nwhere $a_{i}$ and $b_{i}$ denote, respectively, the largest and smallest elements of progression $i$. Now\n\n$$\n\\begin{aligned}\n& \\sum_{i=1}^{N} b_{i} \\geqslant 1+2+\\cdots+N=N(N+1) / 2 \\\\\n& \\sum_{i=1}^{N} a_{i} \\leqslant(3 n-N+1)+\\cdots+3 n=N(6 n-N+1) / 2\n\\end{aligned}\n$$\n\nand thus\n\n$$\n2 \\sum_{i=1}^{N} d_{i} \\leqslant N(3 n-N) \\leqslant 2 n^{2}\n$$\n\nas $N(3 n-N)$ is an increasing of $N$ on the interval $[0,3 n / 2]$ and since $N \\leqslant n$. This completes the proof.\n\nRemark. The maximising partition is in fact unique: from the above, it is immediate that all maximal partitions must have $n_{i}=3$ and hence $n=N$. It follows that the minimal and maximal elements of the arithmetic progressions of a maximal partition are precisely $1,2, \\ldots, n$ and $2 n+1,2 n+2, \\ldots, 3 n$, respectively. Consider the progression $\\{n+1-d, n+1, n+1+d\\}$ of difference $d$. Then $n+1-d \\geqslant 1$ and $n+1+d \\geqslant 2 n+1$,\nwhich implies $d=n$, and hence $\\{1, n+1,2 n+1\\}$ is an element of any maximal partition. By induction, it follows similarly that $\\{k, n+k, 2 n+k\\}$ is an element of any maximal partition for $1 \\leqslant k \\leqslant n$, since integers allowing for smaller or larger differences have already been used up. This proves the partition $\\{1, n+1,2 n+1\\}, \\ldots,\\{n, 2 n, 3 n\\}$ is the unique maximising partition, as claimed.", "metadata": {"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2015-zz.jsonl", "problem_match": "# Problem 4.", "solution_match": "\nSolution."}}