{"year": "2011", "tier": "T3", "problem_label": "A-1", "problem_type": "Algebra", "exam": "BalticWay", "problem": "The real numbers $x_{1}, \\ldots, x_{2011}$ satisfy\n\n$$\nx_{1}+x_{2}=2 x_{1}^{\\prime}, \\quad x_{2}+x_{3}=2 x_{2}^{\\prime}, \\quad \\ldots, \\quad x_{2011}+x_{1}=2 x_{2011}^{\\prime}\n$$\n\nwhere $x_{1}^{\\prime}, x_{2}^{\\prime}, \\ldots, x_{2011}^{\\prime}$ is a permutation of $x_{1}, x_{2}, \\ldots, x_{2011}$. Prove that $x_{1}=x_{2}=\\cdots=x_{2011}$.", "solution": "For convenience we call $x_{2011}$ also $x_{0}$. Let $k$ be the largest of the numbers $x_{1}, \\ldots, x_{2011}$, and consider an equation $x_{n-1}+x_{n}=2 k$, where $1 \\leq n \\leq 2011$. Hence we get $2 \\max \\left(x_{n-1}, x_{n}\\right) \\geq x_{n-1}+x_{n}=2 k$, so either $x_{n-1}$ or $x_{n}$, say $x_{n-1}$, satisfies $x_{n-1} \\geq k$. Since also $x_{n-1} \\leq k$, we then have $x_{n-1}=k$, and then also $x_{n}=2 k-x_{n-1}=2 k-k=k$. That is, in such an equation both variables on the left equal $k$. Now let $\\mathcal{E}$ be the set of such equations, and let $\\mathcal{S}$ be the set of subscripts on the left of these equations. From $x_{n}=k \\forall n \\in \\mathcal{S}$ we get $|\\mathcal{S}| \\leq|\\mathcal{E}|$. On the other hand, since the total number of appearances of these subscripts is $2|\\mathcal{E}|$ and each subscript appears on the left in no more than two equations, we have $2|\\mathcal{E}| \\leq 2|\\mathcal{S}|$. Thus $2|\\mathcal{E}|=2|\\mathcal{S}|$, so for each $n \\in \\mathcal{S}$ the set $\\mathcal{E}$ contains both equations with the subscript $n$ on the left. Now assume $1 \\in \\mathcal{S}$ without loss of generality. Then the equation $x_{1}+x_{2}=2 k$ belongs to $\\mathcal{E}$, so $2 \\in \\mathcal{S}$. Continuing in this way we find that all subscripts belong to $\\mathcal{S}$, so $x_{1}=x_{2}=\\cdots=x_{2011}=k$.", "metadata": {"resource_path": "BalticWay/segmented/en-bw11sol.jsonl", "problem_match": "\n## A-1 DEN\n", "solution_match": "\nSolution 1"}} {"year": "2011", "tier": "T3", "problem_label": "A-1", "problem_type": "Algebra", "exam": "BalticWay", "problem": "The real numbers $x_{1}, \\ldots, x_{2011}$ satisfy\n\n$$\nx_{1}+x_{2}=2 x_{1}^{\\prime}, \\quad x_{2}+x_{3}=2 x_{2}^{\\prime}, \\quad \\ldots, \\quad x_{2011}+x_{1}=2 x_{2011}^{\\prime}\n$$\n\nwhere $x_{1}^{\\prime}, x_{2}^{\\prime}, \\ldots, x_{2011}^{\\prime}$ is a permutation of $x_{1}, x_{2}, \\ldots, x_{2011}$. Prove that $x_{1}=x_{2}=\\cdots=x_{2011}$.", "solution": "Again we call $x_{2011}$ also $x_{0}$. Taking the square on both sides of all the equations and adding the results, we get\n\n$$\n\\sum_{n=1}^{2011}\\left(x_{n-1}+x_{n}\\right)^{2}=4 \\sum_{n=1}^{2011} x_{n}^{\\prime 2}=4 \\sum_{n=1}^{2011} x_{n}^{2}\n$$\n\nwhich can be transformed with some algebra into\n\n$$\n\\sum_{n=1}^{2011}\\left(x_{n-1}-x_{n}\\right)^{2}=0\n$$\n\nHence the assertion follows.", "metadata": {"resource_path": "BalticWay/segmented/en-bw11sol.jsonl", "problem_match": "\n## A-1 DEN\n", "solution_match": "\nSolution 2"}} {"year": "2011", "tier": "T3", "problem_label": "G-1", "problem_type": "Geometry", "exam": "BalticWay", "problem": "Geometry\n\nLet $A B$ and $C D$ be two diameters of the circle $\\mathcal{C}$. For an arbitrary point $P$ on $\\mathcal{C}$, let $R$ and $S$ be the feet of the perpendiculars from $P$ to $A B$ and $C D$, respectively. Show that the length of $R S$ is independent of the choice of $P$.", "solution": "Let $O$ be the centre of $\\mathcal{C}$. Then $P, R, S$, and $O$ are points on a circle $\\mathcal{C}^{\\prime}$ with diameter $O P$, equal to the radius of $\\mathcal{C}$. The segment $R S$ is a chord in this circle subtending the angle $B O D$ or a supplementary angle. Since the angle as well as radius of $\\mathcal{C}^{\\prime}$ are independent of $P$, so is $R S$.\n\n![](https://cdn.mathpix.com/cropped/2024_04_17_ecac827bee541f9d910cg-11.jpg?height=514&width=577&top_left_y=945&top_left_x=745)", "metadata": {"resource_path": "BalticWay/segmented/en-bw11sol.jsonl", "problem_match": "\n## G-1 FIN\n", "solution_match": "\nSolution."}} {"year": "2011", "tier": "T3", "problem_label": "G-3", "problem_type": "Geometry", "exam": "BalticWay", "problem": "Let $E$ be an interior point of the convex quadrilateral $A B C D$. Construct triangles $\\triangle A B F$, $\\triangle B C G, \\triangle C D H$ and $\\triangle D A I$ on the outside of the quadrilateral such that the similarities $\\triangle A B F \\sim \\triangle D C E, \\triangle B C G \\sim \\triangle A D E, \\triangle C D H \\sim \\triangle B A E$ and $\\triangle D A I \\sim \\triangle C B E$ hold. Let $P$, $Q, R$ and $S$ be the projections of $E$ on the lines $A B, B C, C D$ and $D A$, respectively. Prove that if the quadrilateral $P Q R S$ is cyclic, then\n\n$$\nE F \\cdot C D=E G \\cdot D A=E H \\cdot A B=E I \\cdot B C .\n$$", "solution": "We consider oriented angles modulo $180^{\\circ}$. From the cyclic quadrilaterals $A P E S$, $B Q E P, P Q R S, C R E Q, D S E R$ and $\\triangle D C E \\sim \\triangle A B F$ we get\n\n$$\n\\begin{aligned}\n\\angle A E B & =\\angle E A B+\\angle A B E=\\angle E S P+\\angle P Q E \\\\\n& =\\angle E S R+\\angle R S P+\\angle P Q R+\\angle R Q E \\\\\n& =\\angle E S R+\\angle R Q E=\\angle E D C+\\angle D C E \\\\\n& =\\angle D E C=\\angle A F B,\n\\end{aligned}\n$$\n\nso the quadrilateral $A E B F$ is cyclic. By Ptolemy we then have\n\n$$\nE F \\cdot A B=A E \\cdot B F+B E \\cdot A F .\n$$\n\nThis transforms by $A B: B F: A F=D C: C E: D E$ into\n\n![](https://cdn.mathpix.com/cropped/2024_04_17_ecac827bee541f9d910cg-13.jpg?height=914&width=922&top_left_y=1662&top_left_x=567)\n\nSince the expression on the right of this equation is invariant under cyclic permutation of the vertices of the quadrilateral $A B C D$, the asserted equation follows immediately.", "metadata": {"resource_path": "BalticWay/segmented/en-bw11sol.jsonl", "problem_match": "\n## G-3 DEN\n", "solution_match": "\nSolution."}} {"year": "2011", "tier": "T3", "problem_label": "G-4", "problem_type": "Geometry", "exam": "BalticWay", "problem": "The incircle of a triangle $A B C$ touches the sides $B C, C A, A B$ at $D, E, F$, respectively. Let $G$ be a point on the incircle such that $F G$ is a diameter. The lines $E G$ and $F D$ intersect at $H$. Prove that $C H \\| A B$.", "solution": "We work in the opposite direction. Suppose that $H^{\\prime}$ is the point where $D F$ intersect the line through $C$ parallel to $A B$. We need to show that $H^{\\prime}=H$. For this purpose it suffices to prove that $E, G, H^{\\prime}$ are collinear, which reduces to showing that if $G^{\\prime} \\neq E$ is the common point of $E H^{\\prime}$ and the incircle, then $G^{\\prime}=G$.\n\n![](https://cdn.mathpix.com/cropped/2024_04_17_ecac827bee541f9d910cg-14.jpg?height=588&width=765&top_left_y=837&top_left_x=651)\n\nNote that $H^{\\prime}$ and $B$ lie on the same side of $A C$. Hence $C H^{\\prime} \\| A B$ gives $\\angle A C H^{\\prime}=180^{\\circ}-\\angle B A C$. Also, some homothety with center $D$ maps the segment $B F$ to the segment $C H^{\\prime}$. Thus the equality $B D=B F$ implies that $C H^{\\prime}=C D=C E$, i.e. the triangle $E C H^{\\prime}$ is isosceles and\n\n$$\n\\angle H^{\\prime} E C=\\frac{1}{2}\\left(180^{\\circ}-\\angle E C H^{\\prime}\\right)=\\frac{1}{2} \\angle B A C .\n$$\n\nBut $G^{\\prime}$ and $H^{\\prime}$ lie on the same side of $A C$, so $\\angle G^{\\prime} E C=\\angle H^{\\prime} E C$ and consequently\n\n$$\n\\angle G^{\\prime} F E=\\angle G^{\\prime} E C=\\angle H^{\\prime} E C=\\frac{1}{2} \\angle B A C\n$$\n\nso that\n\n$$\n\\angle G^{\\prime} F A=\\angle G^{\\prime} F E+\\angle E F A=\\frac{1}{2} \\angle B A C+\\frac{1}{2}\\left(180^{\\circ}-\\angle F A E\\right)=90^{\\circ} .\n$$\n\nHence $F G^{\\prime}$ is a diameter of the incircle and the desired equality $G^{\\prime}=G$ follows.\n\nRemark. A similar proof also works in the forward direction: one may compute $\\angle E H D=$ $\\frac{1}{2} \\angle A C B$. Hence $H$ lies on the circle centred at $C$ that passes through $D$ and $E$. Consequently the triangle $E H C$ is isosceles, wherefore\n\n$$\n\\angle E C H=180^{\\circ}-2 \\angle G E C=180^{\\circ}-\\angle B A C .\n$$\n\nThus the lines $A B$ and $C H$ are indeed parallel.", "metadata": {"resource_path": "BalticWay/segmented/en-bw11sol.jsonl", "problem_match": "\n## G-4 POL\n", "solution_match": "\nSolution."}} {"year": "2011", "tier": "T3", "problem_label": "G-5", "problem_type": "Geometry", "exam": "BalticWay", "problem": "Let $A B C D$ be a convex quadrilateral such that $\\angle A D B=\\angle B D C$. Suppose that a point $E$ on the side $A D$ satisfies the equality\n\n$$\nA E \\cdot E D+B E^{2}=C D \\cdot A E .\n$$\n\nShow that $\\angle E B A=\\angle D C B$.", "solution": "Let $F$ be the point symmetric to $E$ with respect to the line $D B$. Then the equality $\\angle A D B=\\angle B D C$ shows that $F$ lies on the line $D C$, on the same side of $D$ as $C$. Moreover, we have $A E \\cdot E D