Split (1)

train · 617k rows

text stringlengths 2 132k	source dict
A f (w, x, y, z ) dw dx dy dz · · · > A f (x1, . . . , x k) (8.2) 8.3. Multiline subscripts and superscripts The \substack command can be used to produce a multiline subscript or superscript: for example \sum_{\substack{ 0\le i\le m\\ 0 0≤i≤m > 0 n ∗ ∗ > ∗ Short Math Guide for L ATEX, version 1.09 (2002-03-22) 16 ...	{ "page_id": null, "source": 6821, "title": "from dpo" }
(1 − qk) \frac{{\displaystyle\sum_{n > 0} z^n}} {{\displaystyle\prod_{1\leq k\leq n} (1-q^k)}} And if you want full-size symbols but with limits on the side, use the nolimits command also: > n> 0 zn > 1≤k≤n (1 − qk) \frac{{\displaystyle\sum\nolimits_{n> 0} z^n}} {{\displaystyle\prod\nolimits_{1\leq k\leq n} (1-q^k)...	{ "page_id": null, "source": 6821, "title": "from dpo" }
2001. [Lamport] Lamport, Leslie: LATEX: a document preparation system , 2nd edition, Addison- Wesley, 1994. [LC] Goossens, Michel; Mittelbach, Frank; Samarin, Alexander: The L ATEX Com- panion , Addison-Wesley, 1994. [LFG] LATEX3 Project Team: LATEX 2 ε font selection , fntguide.tex , 1994. [LGC] Goossens, Michel; Raht...	{ "page_id": null, "source": 6821, "title": "from dpo" }
Title: URL Source: Markdown Content: Remarks on a Bernstein-type operator of Aldaz, Kounchev and Render: Remarks on a Bernstein-type operator of Aldaz, Kounchev and Render =============== Remarks on a Bernstein-type operator of Aldaz, Kounchev and Render ================================================================...	{ "page_id": null, "source": 6821, "title": "from dpo" }
the convention $e_j(x)=x^j,\, \, j=0,1,\dots $ King modified the classical Bernstein operators as follows: \begin{equation} \label{e1} f\to \left(B_nf\right)\circ r_n,\, \, f\in C[0,1], \end{equation} 1 where \[ r_n(x)=\left\{ \begin{array}{ll} x^2,& n=1,\\ \tfrac {1}{2(n-1)}\left(-1+\sqrt{1+4n(n-1)x^2}\right), & n=2...	{ "page_id": null, "source": 6821, "title": "from dpo" }
] introduced a polynomial Bernstein operator $B_{n,0,j}: C[0,1]\to \Pi _n$ that fixes $e_0$ and $e_j$, and converges in the strong operator topology to the identity as $n\to \infty $. The operator is a linear combination of the classical Bernstein basis $\{ p_{n,k}\} _{k=0,\dots ,n}$, thus produces polyno...	{ "page_id": null, "source": 6821, "title": "from dpo" }
$\& $ Mond) was given by Cardenas-Morales _et al._ in [8-f(x)\|\leq \omega _1(f,\delta )\left(1+\tfrac {1}{\delta }\sqrt{2x(1-x)\tfrac {1-(1-x)^{n-1}}{n-1}}\right), \] $ f\in C[0,1],\, x\in [0,1]\textrm{ and } \delta {\gt}0. $ Moreover, in 2014 Finta showed (see [11 \[ \|B_{n,0,j}(f;x)-f(x)\|\leq 2\omega _1\left(f; ...	{ "page_id": null, "source": 6821, "title": "from dpo" }
[0,1]\), $1 \leq j \leq n$. Then * $\|B_{n,0,2}(f;x)\! -\! f(x)\|\! \leq \! \tfrac {1}{\sqrt{n-1}} d_n(x)\omega _1\left(f;\tfrac {1}{\sqrt{n\! -\! 1}}\right)\! +\! \left(1\! +\! xd_n(x)\right)\omega _2\left(f;\tfrac {1}{\sqrt{n\! -\! 1}}\right)$, $n\geq 5$, where $d_n(x):=(1-x)\left[1-(1-x)^{n-1}\right]$; * \(\|B_...	{ "page_id": null, "source": 6821, "title": "from dpo" }
â–¼ Remark 2 a) The right hand side of inequality (8 If one compares the estimate from proposition 2 and the estimate from proposition 1\) is always the worse (_i.e._, dominant) term. In case $f\in C^2[0,1]$ both inequalities give ${\mathcal O}\big(\tfrac {1}{n}\big)$. This cannot be reached by any of the previous ...	{ "page_id": null, "source": 6821, "title": "from dpo" }
obvious. In [14 \leq c \cdot \omega _2(f;\sqrt\delta ). \] 3 Iterates of $ B_{{\lowercase {n}},0,{\lowercase {j}}}$ ========================================================== Using the approach of Agratini and Rus (see [2 the following result for the sequence of iterates $(B_{n,0,j}^m)_{m\geq 1}$ is obtained. The...	{ "page_id": null, "source": 6821, "title": "from dpo" }
\lim _{n\to \infty } n(B_{n,0,j}(f;x)-f(x))=\tfrac {x(1-x)}{2}f^{\prime \prime }(x)-\tfrac {(j-1)(1-x)}{2}f^{\prime }(x). \end{equation} 9 A quantitative pre-Voronovskaya theorem for the Bernstein type operator $B_{n,0,j}$ was proved by Finta [12=\displaystyle \sup _{0{\lt}h\leq \delta }\sup _{x\pm \frac{1}{2}h\...	{ "page_id": null, "source": 6821, "title": "from dpo" }
[ ![Image 6: \includegraphics[scale=0.1]{ext-link.png}]( 7 [_D. Cárdenas-Morales, P. Garrancho, I. Raşa_, _Bernstein-type operators which preserve polynomials_, Comput. Math. Appl., 62 (2011), 158–163, [ ![Image 7: \includegraphics[scale=0.1]{ext-link.png}]( 8 [_D. Cárdenas-Morales, P. Garrancho, I. Raşa_, _Asympto...	{ "page_id": null, "source": 6821, "title": "from dpo" }
Title: KbsdJames/Omni-MATH · Datasets at Hugging Face URL Source: Markdown Content: Subset (1) default·4.43k rows Split (1) \| domain sequence lengths 0 3 \| difficulty float64 1 9.5 \| problem string lengths 18 9.01k \| solution string lengths 2 11.1k \| answer string lengths 0 3.77k \| source string classes 67 values \| \| ...	{ "page_id": null, "source": 6821, "title": "from dpo" }
solution involves ensuring that the maximum number of indices $ k $ for which $ x_{ik} $ or $ x_{kj} $ is the maximum is minimized. By analyzing the constraints and constructing examples, it can be shown that the minimum $ m $ satisfying the conditions is: \[ m = 1 + \left\lceil \frac{n}{2} \right\rceil. \] Thu...	{ "page_id": null, "source": 6821, "title": "from dpo" }
Discrete Mathematics -> Graph Theory" ] \| 7 \| A tournament is a directed graph for which every (unordered) pair of vertices has a single directed edge from one vertex to the other. Let us define a proper directed-edge-coloring to be an assignment of a color to every (directed) edge, so that for every pair of directed e...	{ "page_id": null, "source": 6821, "title": "from dpo" }
the edge $u \to v$. Clearly, this works. We now prove the result by induction on $n$. It is trivial for $n=1$. Now say we want to prove the result for $n$, and assume without loss of generality that $n$ is even, say by deleting a vertex if needed. Fix a color, say red, and consider the set $S$ of all the ve...	{ "page_id": null, "source": 6821, "title": "from dpo" }
a_{\sigma(18)}x^{18} + \cdots + a_{\sigma(0)}\), for all permutations $\sigma$ of the numbers 0 to 19. We construct the coefficients $a_i$ in a specific manner. Let $a_i = 10000 + i\epsilon$ for $i = 0, 1, \ldots, 19$ and some small $\epsilon > 0$. This ensures that \(a_0 \|100^{20}\|, \|a_{18} \cdot 100^{18}\|,...	{ "page_id": null, "source": 6821, "title": "from dpo" }
while $P_\sigma(x)$ ($\sigma \neq \text{Id}$) takes both positive and negative values. Therefore, such positive reals $a_0, a_1, \ldots, a_{19}$ do exist. The answer is: \boxed{\text{Yes}}. \| \text{Yes} \| china_national_olympiad \| \| [ "Mathematics -> Algebra -> Abstract Algebra -> Group Theory" ] \| 7 \| Let $p$ be...	{ "page_id": null, "source": 6821, "title": "from dpo" }
$ A_{1\sigma(1)}, A_{2\sigma(2)}, \ldots, A_{p\sigma(p)} $ is constant for all $ \sigma \in S_p $. By comparing these equations, we find that the first row of the matrix is just a translation of the second row, i.e., $ A_{1i} - A_{2i} $ is constant for $ 1 \leq i \leq p $. This is true for any two other rows as...	{ "page_id": null, "source": 6821, "title": "from dpo" }
\). Then where is $ k+1 $? Yes, $ x_2 = k $ and $ k+1, \ldots, 2k $ lie below $ 1, 2, \ldots, k $. Playing around with possible positions, we arrive at the matrix: \[ M = \begin{bmatrix} 1 & 2 & \ldots & k & 2k+1 & 2k+2 & \ldots & 3k & 4k+1 & \ldots \\ k+1 & k+2 & \ldots & 2k & 3k+1 & 3k+2 & \ldots & 4k & 5k+1 ...	{ "page_id": null, "source": 6821, "title": "from dpo" }
$x_i = 1$ if it has an electron. Lemma: If there exists a permutation $\sigma \in S_n$ such that the physicist's knowledge is exactly \[ x_{\sigma(1)} \le x_{\sigma(2)} \le \cdots \le x_{\sigma(n)}, \] then firing a diode does not change this fact (though $\sigma$ may change). Proof of Lemma: If the physi...	{ "page_id": null, "source": 6821, "title": "from dpo" }
Bunbun the Bunny begins at $A_1$. She hops along $\gamma$ from $A_1$ to $A_2$, then from $A_2$ to $A_3$, until she reaches $A_{2022}$, after which she hops back to $A_1$. When hopping from $P$ to $Q$, she always hops along the shorter of the two arcs $\widehat{PQ}$ of $\gamma$; if $\overline{PQ}$ is a diameter of $\gam...	{ "page_id": null, "source": 6821, "title": "from dpo" }
$A_1A_3 \ldots A_{2021}$ forms a convex polygon. If the polygon includes point $P_{1012}$ or has points on both sides of the diameter $P_1P_{1012}$, the sum of arc lengths is $2022$. Otherwise, it is $P_1P_2P_3 \ldots P_{1011}$ or $P_1P_{2022}P_{2021} \ldots P_{1013}$, and the sum of arc lengths is $2020$...	{ "page_id": null, "source": 6821, "title": "from dpo" }
each other at most. 3. Players in the same pair do not play against each other when they pair with others respectively. Given these conditions, each vertex in $ \mathcal{G} $ can have a degree of at most 2. This implies that $ \mathcal{G} $ can be decomposed into disjoint cycles, paths, and isolated vertices. Let \...	{ "page_id": null, "source": 6821, "title": "from dpo" }
can ensure that the set of games corresponds to $ A $. Thus, the minimum number of players needed is: \[ \frac{1}{2} \max A + 3. \] The answer is: \boxed{\frac{1}{2} \max A + 3}. \| \frac{1}{2} \max A + 3 \| china_national_olympiad \| \| [ "Mathematics -> Geometry -> Plane Geometry -> Other", "Mathematics -> Discrete Mat...	{ "page_id": null, "source": 6821, "title": "from dpo" }
vertices are connected if the corresponding points are harmonic. We need to show that $G$ has no $K_5$ (complete graph on 5 vertices). Claim: $G$ has no $K_5$. Proof: Consider the following two facts: 1. If a coloring of the edges of $K_5$ with two colors does not produce a monochromatic triangle, the...	{ "page_id": null, "source": 6821, "title": "from dpo" }
\times 2} = 3750\). The answer is $\boxed{3750}$. \| 3750 \| usa_team_selection_test \| \| [ "Mathematics -> Geometry -> Plane Geometry -> Polygons", "Mathematics -> Number Theory -> Other" ] \| 7 \| Draw a $2004 \times 2004$ array of points. What is the largest integer $n$ for which it is possible to draw a convex $n$-gon...	{ "page_id": null, "source": 6821, "title": "from dpo" }
are all not friends. \| Given 30 students such that each student has at most 5 friends and for every 5 students there is a pair of students that are not friends, we need to determine the maximum $ k $ such that for all such possible configurations, there exists $ k $ students who are all not friends. In graph theory...	{ "page_id": null, "source": 6821, "title": "from dpo" }
To show that any graph satisfying the conditions has an independent set of size 6, we use Turán's Theorem. The complement graph $ \overline{G} $ has 30 vertices and at least 360 edges. If $ \overline{G} $ does not have a $ K_6 $, then by Turán's Theorem, $ G $ can have at most 360 edges, leading to a contradict...	{ "page_id": null, "source": 6821, "title": "from dpo" }
P(0), P(p^k + 1) - P(1), P(p^k + 2) - P(2), \ldots) \). We now claim that $ p^{k+1} \nmid P(p^k + 1) - P(1) $. Let $ Q(x) = \sum_{i=0}^m c_i x^i $, then \[ \begin{align*} P(p^k + 1) - P(1) &= (p^k + 1)^k Q(p^k + 1) - Q(1) \\ &= \left( \sum_{i=0}^k \binom{k}{i} (p^k)^i \right) Q(p^k + 1) - Q(1) \\ &\equiv (k p^k + 1...	{ "page_id": null, "source": 6821, "title": "from dpo" }
$ q \mid P(i + 1) - P(i) $ for all $ i $. But $ q \mid P(0) = 0 $, so $ q \mid P(i) $ for all nonnegative $ i $, which contradicts the hypothesis. Therefore, there are infinitely many $ n $ such that \[ \gcd(P(n) - P(0), P(n + 1) - P(1), P(n + 2) - P(2), \ldots) = n. \] The answer is: \boxed{\text{infinitel...	{ "page_id": null, "source": 6821, "title": "from dpo" }
trivially true for an equilateral triangle, but it is false for a regular pentagon (consider $ ABCDE $ and $ A'D'B'E'C' $). Consider now a regular $ 2n+1 $-gon for $ n \ge 3 $. Clearly, there are no right triangles. The number of obtuse triangles with a particular diagonal as the longest side is equal to the nu...	{ "page_id": null, "source": 6821, "title": "from dpo" }
smallest such $ a $, consider the following construction and proof: 1. Proof of Optimality: - Consider a triangle $ ABC $ inscribed in the unit circle with angles $ \angle A = 20^\circ $ and $ \angle B = \angle C = 80^\circ $. - The smallest equilateral triangle $ PQR $ containing $ \triangle ABC $ must...	{ "page_id": null, "source": 6821, "title": "from dpo" }
Given positive integers $ n $ and $ k $ such that $ n > k^2 > 4 $, we aim to determine the maximal possible $ N $ such that one can choose $ N $ unit squares in an $ n \times n $ grid and color them, with the condition that in any $ k $-group from the colored $ N $ unit squares, there are two squares wi...	{ "page_id": null, "source": 6821, "title": "from dpo" }
= c \), where $ O $ and $ \rho $ denote the circumcenter and circumradius of $\triangle XCD$, respectively. To solve this, we proceed as follows: 1. Define points $ B' $ and $ A' $ on $\Gamma$ such that $ BB' \parallel U_2V_2 $ and $ AA' \parallel U_1V_1 $. 2. Let $ K $ be the intersection point of li...	{ "page_id": null, "source": 6821, "title": "from dpo" }
\text{ and } c \text{ is a constant} \| usa_team_selection_test_for_imo \| \| [ "Mathematics -> Geometry -> Plane Geometry -> Other", "Mathematics -> Applied Mathematics -> Probability -> Other" ] \| 8 \| Find a real number $t$ such that for any set of 120 points $P_1, \ldots P_{120}$ on the boundary of a unit square, there...	{ "page_id": null, "source": 6821, "title": "from dpo" }
\). The key is that now \[ g_{\mathcal{U}}(Q) < g_{\mathcal{U}'}(Q') \quad \text{for all } Q \in \mathcal{U} \text{ and } Q' \in \mathcal{U}' \quad (\spadesuit). \] Let $ C_1, C_2, C_3, C_4 $ be the corners of the unit square $\mathcal{U}$ and $ M_1', M_2', M_3', M_4' $ the midpoints of the four sides of the unit...	{ "page_id": null, "source": 6821, "title": "from dpo" }
contradicts $(\clubsuit)$! Therefore, such a $ t $ exists. In particular, we can show $ t = 30(1 + \sqrt{5}) $ by proving that $ t 30(1 + \sqrt{5}) $ fails from the midpoints bound; now, since we have shown at least one valid $ t $ exists, it must be the claimed value. The answer is: \(\boxed{30(1 + \sqrt{5}...	{ "page_id": null, "source": 6821, "title": "from dpo" }
= 180^\circ - \angle ABP = 105^\circ \), we can use SSA similarity (since $ 105^\circ > 90^\circ $) to conclude that $ \triangle AQR \sim \triangle ADC $. Thus, it follows that \[ \frac{AC^2}{AR^2} = \frac{2}{3}. \] The answer is: $\boxed{\frac{2}{3}}$. \| \frac{2}{3} \| usa_team_selection_test \| \| [ "Mathematics -...	{ "page_id": null, "source": 6821, "title": "from dpo" }
\frac{64 \cdot 63}{2} + 1 = 63 \cdot 2016 + 1 = 127008 + 1 = 127009. \] Thus, the maximum total amount the university could have paid is: \[ \boxed{127009}. \] \| 127009 \| usa_team_selection_test \| \| [ "Mathematics -> Discrete Mathematics -> Combinatorics" ] \| 7 \| Let $f:X\rightarrow X$, where $X=\{1,2,\ldots ,100\}$, b...	{ "page_id": null, "source": 6821, "title": "from dpo" }
and their image cover $ C $. We have the following key claim: Claim: We have $ \alpha(C) \geq \beta(C) - 1 $. Proof: It suffices to show that given a subset $ D \subseteq C $ such that $ D $ and $ f(D) $ cover $ C $, we can find a subset $ D' \subseteq C $ such that $ \|D'\| \leq \|D\| $ and such th...	{ "page_id": null, "source": 6821, "title": "from dpo" }
the set of bad edges becomes a strict subset of what it was before, so the sum of their weights goes down. If $ f(f(a)) $ is not already present, then the size of $ D $ doesn't change, and we lose at least one bad edge with weight $ \omega^k $, and potentially gain many bad edges with weights $ \omega^{k-1} $ o...	{ "page_id": null, "source": 6821, "title": "from dpo" }
any elements of $ D $ one level above it in the tree rooted at the vertex, or in other words, a vertex is deficient if it will not be covered by $ D \cup f(D) $ if we remove all the cycle elements from $ D $. Note that all elements of $ D $ on the cycle are deficient since there are no bad edges not on the cycl...	{ "page_id": null, "source": 6821, "title": "from dpo" }
\in \mathcal{C}} \alpha(C) \geq \sum_{C \in \mathcal{C}} \beta(C) - \|\mathcal{C}\|. \] If $ \|\mathcal{C}\| \leq 30 $, then we see that \[ \sum_{C \in \mathcal{C}} \beta(C) \leq 69, \] so we can select a subset $ B \subseteq X $ such that $ \|B\| \leq 69 $ and $ B \cup f(B) = X $. If $ \|\mathcal{C}\| \geq 31 $, the...	{ "page_id": null, "source": 6821, "title": "from dpo" }
component. This shows that the answer to the problem is $ \boxed{69} $. \| 69 \| china_national_olympiad \| \| [ "Mathematics -> Algebra -> Intermediate Algebra -> Inequalities", "Mathematics -> Discrete Mathematics -> Algorithms" ] \| 8 \| Consider pairs $(f,g)$ of functions from the set of nonnegative integers to itself ...	{ "page_id": null, "source": 6821, "title": "from dpo" }
\] First, we note that $x \geq 1$ because if $x = 0$, the left-hand side would be a fraction, which cannot equal 1. ### Case 1: $w = 0$ The equation simplifies to: \[ 2^x \cdot 3^y = 1 + 5^z. \] - Subcase 1.1: $z = 0$ \[ 2^x \cdot 3^y = 1. \] This implies $x = 0$ and $y = 0$, but $x \geq 1$, so this i...	{ "page_id": null, "source": 6821, "title": "from dpo" }
to be: \[ \boxed{(1, 1, 1, 0), (2, 2, 1, 1), (1, 0, 0, 0), (3, 0, 0, 1)}. \] \| (1, 1, 1, 0), (2, 2, 1, 1), (1, 0, 0, 0), (3, 0, 0, 1) \| china_national_olympiad \| \| [ "Mathematics -> Algebra -> Intermediate Algebra -> Complex Numbers" ] \| 8.5 \| Find the largest real number $\lambda$ with the following property: for any ...	{ "page_id": null, "source": 6821, "title": "from dpo" }
given conditions is: \[ \boxed{\sqrt{3}}. \| \sqrt{3} \| china_national_olympiad \| \| [ "Mathematics -> Discrete Mathematics -> Algorithms" ] \| 6 \| Find all functions $f\colon \mathbb{Z}^2 \to [0, 1]$ such that for any integers $x$ and $y$, \[f(x, y) = \frac{f(x - 1, y) + f(x, y - 1)}{2}.\] \| Let \( f \colon \mathbb{Z}^2 ...	{ "page_id": null, "source": 6821, "title": "from dpo" }
2^n \left( f(x, y) - f(x - 1, y + 1) \right) \leq \binom{n}{\left\lfloor \frac{n}{2} \right\rfloor}. \] For large $ n $, this inequality becomes impossible because $ 2^n $ grows exponentially while $ \binom{n}{\left\lfloor \frac{n}{2} \right\rfloor} $ grows polynomially. Therefore, we must have \( f(x, y) = f(x -...	{ "page_id": null, "source": 6821, "title": "from dpo" }
given by: \[ r = \frac{a + b - c}{2}. \] The points where the incircle touches the sides of the triangle are such that $EC = CD = r$. Since $AD$ is the angle bisector of $\angle BAC$, we can use the Angle Bisector Theorem and properties of the incircle to find relationships between the segments. Given that \(\ang...	{ "page_id": null, "source": 6821, "title": "from dpo" }
itself is called to be satisfying property $P(k)$, if for any $x,y\in S$, there exist $f_1, \cdots, f_k \in F$ (not necessarily different), such that $f_k(f_{k-1}(\cdots (f_1(x))))=f_k(f_{k-1}(\cdots (f_1(y))))$. Find the least positive integer $m$, such that if $F$ satisfies property $P(2019)$, then it also satisfies ...	{ "page_id": null, "source": 6821, "title": "from dpo" }
is at most the number of unordered pairs of distinct elements, which is exactly $ \binom{35}{2} $. To construct such a sequence, let $ S = \{0, 1, \ldots, 34\} $ and define two mappings $ f(x) $ and $ g(x) $ as follows: \[ f(x) = (x + 1) \pmod{35}, \] \[ g(0) = 1, \quad g(x) = x \text{ for all } 1 \leq x \leq 3...	{ "page_id": null, "source": 6821, "title": "from dpo" }
a_n \) and $ \sum_{i=1}^n a_i = \sum_{i=1}^n b_i = 1 $. Let $ A = \sum_{i=1}^n a_i^2 $, $ B = \sum_{i=1}^n b_i^2 $, and $ X = \sum_{i=1}^n a_i b_i $. We define the function: \[ f(X) = \frac{A + X}{B + X}. \] The derivative of $ f(X) $ is: \[ f'(X) = \frac{B - A}{(B + X)^2}. \] Since $ f'(X) B $, we want to...	{ "page_id": null, "source": 6821, "title": "from dpo" }
b_1 = 0 \), $ b_2 = \cdots = b_n = \frac{1}{n - 1} $. Thus, the maximum value of the given expression is: \[ \frac{\sum_{i=1}^n a_i(a_i + b_i)}{\sum_{i=1}^n b_i(a_i + b_i)} = n - 1. \] The answer is: \boxed{n - 1}. \| n - 1 \| china_national_olympiad \| \| [ "Mathematics -> Number Theory -> Prime Numbers" ] \| 7 \| Find al...	{ "page_id": null, "source": 6821, "title": "from dpo" }
$1^n - 1 = 0$ for any $n$, which trivially satisfies the condition. Combining these results, the complete solution set is: \[ (a, n) = (2, 6), (2^k - 1, 2), (1, n) \text{ for any } n \ge 1. \] The answer is: \boxed{(2, 6), (2^k - 1, 2), (1, n) \text{ for any } n \ge 1}. \| (2, 6), (2^k - 1, 2), (1, n) \text{ for any...	{ "page_id": null, "source": 6821, "title": "from dpo" }
which is an intersection of $ i $ segments belongs to $ i $ basic segments. Also, every basic segment belongs to two vertices. Hence, \[ 2N = 2s_2 + 3s_3 + 4s_4 = 8 + 3s_3 + 4s_4 \quad (1). \] Each vertex which is an intersection of $ i $ segments belongs to $ 1, 2, 4 $ rectangles, where $ i = 2, 3, 4 $ respe...	{ "page_id": null, "source": 6821, "title": "from dpo" }
(5), and applying the AM-GM inequality, we get: \[ 2016 \leq (a+1)(b+1) \leq \left( \frac{a+b}{2} + 1 \right)^2 \leq \left( \frac{s_3}{4} + 1 \right)^2, \] which implies \[ s_3 + 4 \geq \lceil \sqrt{32256} \rceil = 180, \] thus, \[ s_3 \geq 176. \] Therefore, \[ N = 4034 + \frac{s_3}{2} \geq 4034 + \frac{176}{2} = 4122...	{ "page_id": null, "source": 6821, "title": "from dpo" }
minimality of $m + n$, we must have: \[ m + n \le m' + n \implies m \le m'. \] However, since $m > n > 1$: \[ n(m - n) \ge 2 \implies mn - 2 \ge n^2 \implies m(mn - 2) \ge mn^2 > n^3, \] \[ n(m^2 - n^2) > 2m \implies m^2n > 2m + n^3, \] \[ 2m^2n - 2m > m^2n + n^3 \implies 2m(mn - 1) > n(m^2 + n^2) \implies m > cn -...	{ "page_id": null, "source": 6821, "title": "from dpo" }
\), then there is a block of (at least two) consecutive terms in the sequence with their (arithmetic) mean being an integer. We start by examining small values of $ n $: - For $ n = 2 $, consider the sequence $(1, 2)$. The sum is $1 + 2 = 3 = 2 \cdot 2 - 1$. The arithmetic mean of the block $(1, 2)$ is \(\fra...	{ "page_id": null, "source": 6821, "title": "from dpo" }
Using strong induction and the properties of sums of sequences, we can show that there will always be a block of consecutive terms whose arithmetic mean is an integer. ### Case 3: $ n = 4k + 2 $ If $ n = 4k + 2 $, the sum $ S(1, n) $ is odd and $ 6k + 5 \le S(1, n) \le 8k + 3 $. Similar to Case 2, using strong ...	{ "page_id": null, "source": 6821, "title": "from dpo" }
$n$ for which there exists a bijective function $g: \mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}$, such that the 101 functions \[g(x), \quad g(x) + x, \quad g(x) + 2x, \quad \dots, \quad g(x) + 100x\] are all bijections on $\mathbb{Z}/n\mathbb{Z}$. We claim that the answer is all numbers relatively prime to ...	{ "page_id": null, "source": 6821, "title": "from dpo" }
\text{ relatively prime to } 101! \| usa_team_selection_test_for_imo \| \| [ "Mathematics -> Algebra -> Intermediate Algebra -> Complex Numbers", "Mathematics -> Algebra -> Algebra -> Equations and Inequalities" ] \| 9 \| Let $C=\{ z \in \mathbb{C} : \|z\|=1 \}$ be the unit circle on the complex plane. Let $z_1, z_2, \ldots, ...	{ "page_id": null, "source": 6821, "title": "from dpo" }
= z_k + z_{k+40} + z_{k+80} + z_{k+120} + z_{k+160} + z_{k+200} \) for $ 1 \leq k \leq 40 $. Each $ \omega_k $ sums six complex numbers spaced by $ \frac{2\pi}{6} = \frac{\pi}{3} $ radians apart. Given the conditions: 1. For any open arc $\Gamma$ of length $\pi$ on the unit circle, at most 5 of $ z_i $ (whe...	{ "page_id": null, "source": 6821, "title": "from dpo" }
\) is: \[ \boxed{80 + 40\sqrt{3}}. \] \| 80 + 40\sqrt{3} \| china_team_selection_test \| \| [ "Mathematics -> Algebra -> Abstract Algebra -> Ring Theory" ] \| 8 \| Find $f: \mathbb{Z}_+ \rightarrow \mathbb{Z}_+$, such that for any $x,y \in \mathbb{Z}_+$, $$f(f(x)+y)\mid x+f(y).$$ \| We are tasked with finding a function \( f:...	{ "page_id": null, "source": 6821, "title": "from dpo" }
means that for any $ n $, the set $ \{f(1), f(1 + t), \dots, f(1 + nt)\} $ contains at least $ n + 1 $ numbers in the interval $[1, n + f(1)]$. If $ t \geq 2 $, this clearly violates $ f $ being injective. $\square$ We now use strong induction to prove $ f(n) = n $ for all $ n $. The base case \( n = ...	{ "page_id": null, "source": 6821, "title": "from dpo" }
\). This implies $ f(y + 1) = 1 $ for all $ y \geq 1 $. Clearly, $ f(1) $ can take any value. - If $ S = \{1, 2\} $, then $ 2 \mid (a - b) $ for any $ f(a) = f(b) $, so $ f(n) $ alternates between $ 1 $ and $ 2 $ for large enough $ n $. Plugging in $ (x, y) = (n, y) $, we get \[ f(y + f(n)) \mid f...	{ "page_id": null, "source": 6821, "title": "from dpo" }
& \text{if } x > 1 \text{ is odd} \\ 2 & \text{if } x \text{ is even} \end{cases} \text{ for any } n \text{ odd}}. \| f(x) = x \text{ or } f(x) = \begin{cases} n & \text{if } x = 1 \\ 1 & \text{if } x > 1 \end{cases} \text{ or } f(x) = \begin{cases} n & \text{if } x = 1 \\ 1 & \text{if } x > 1 \text{ is odd} \\ 2 & \tex...	{ "page_id": null, "source": 6821, "title": "from dpo" }
1 \le m \le 2 \left( \frac{4^k - 1}{3} \right) \), consider the expressions $ 2^l \pm m $ where $ l = 0, 1, \dots, 2k+1 $. Since $ 2^{2k-1} < 2 \left( \frac{4^k - 1}{3} \right) < 2^{2k} $, by this method we achieve an expression with $ k+1 $ terms for each positive integer less than or equal to \[ 2^{2k+1} + 2 ...	{ "page_id": null, "source": 6821, "title": "from dpo" }
\) and dividing by $ 2 $ yields that $ \frac{4^{k+1} - 1}{3} $ requires $ k $ summands minimum. This contradicts the fact that $ \frac{4^{k+1} - 1}{3} $ is not $ k $-good. Similarly, if it is a $ -1 $, then adding $ 1 $ and dividing by $ 2 $ contradicts the fact that \( 2 \left( \frac{4^{k+1} - 1}{3} \r...	{ "page_id": null, "source": 6821, "title": "from dpo" }
\). For instance, if $ n = 2 $, we can choose $ x = 1 $, $ y = 1 $, and $ z = 1 $, giving $ P(1,1,1) = 1^2 + 1^2 + 1^2 + 2 \cdot 1 \cdot 1 \cdot 1 = 4 $, which is not a perfect square. 2. If $ n $ is a perfect square: Conversely, if $ n $ is a perfect square, then for any positive integers \( x, y, z ...	{ "page_id": null, "source": 6821, "title": "from dpo" }
Lemma (USAMO 1991 P3): The sequence $ \{b_k\}_{k \geq 1} $ eventually becomes constant modulo $ a $. Proof of Lemma: We use strong induction on $ a $. For $ a = 1, 2 $, the result is obvious. Suppose our claim is true for $ 1, 2, \ldots, a-1 $. Consider the case when $ \gcd(a, b) = d > 1 $. Choose s...	{ "page_id": null, "source": 6821, "title": "from dpo" }
snake can [i]move[/i] to occupy $(s, s_1, \dots, s_{k-1})$ instead. The snake has [i]turned around[/i] if it occupied $(s_1, s_2, \dots, s_k)$ at the beginning, but after a finite number of moves occupies $(s_k, s_{k-1}, \dots, s_1)$ instead. Determine whether there exists an integer $n > 1$ such that: one can place so...	{ "page_id": null, "source": 6821, "title": "from dpo" }
\| usa_team_selection_test_for_imo \| \| [ "Mathematics -> Geometry -> Plane Geometry -> Triangulations" ] \| 7.5 \| Let $ABC$ be an acute scalene triangle and let $P$ be a point in its interior. Let $A_1$, $B_1$, $C_1$ be projections of $P$ onto triangle sides $BC$, $CA$, $AB$, respectively. Find the locus of points $P$ su...	{ "page_id": null, "source": 6821, "title": "from dpo" }
\cos x_3 = \cos y_1 \cos y_2 \cos y_3. \] Since the sum of all six angles is $ \pi $, we get: \[ x_1 + x_2 + x_3 = y_1 + y_2 + y_3 = \frac{\pi}{2}. \] Conversely, if $ P $ satisfies these three conditions, then $ P $ is in the locus (since Ceva's and Trigonometric Ceva's Theorems are if-and-only-if statements). I...	{ "page_id": null, "source": 6821, "title": "from dpo" }
x_2 \sin x_3. \] Dividing by $ \cos x_1 \cos x_2 \cos x_3 $ and rearranging, we get: \[ \frac{1}{\cos x_1 \cos x_2 \cos x_3} + \tan x_1 \tan x_2 \tan x_3 = \tan x_1 + \tan x_2 + \tan x_3. \] However, the same identity holds for $ y_1, y_2, y_3 $, and the left-hand side doesn't change when we replace \( x_1, x_2, x_...	{ "page_id": null, "source": 6821, "title": "from dpo" }
over $ BC $ lies on $ (ABC) $. This implies that $ P $ lies on $ (BHC) $, and similarly it lies on $ (AHB) $ and $ (CHA) $, so $ P = H $ (the orthocenter). We have exhausted all cases for $ x_1 $, so the locus of points $ P $ is the set of the incenter, circumcenter, and orthocenter of \( \triangle AB...	{ "page_id": null, "source": 6821, "title": "from dpo" }
exist $ x, y \in X $ satisfying $ ax + y \equiv k \pmod{37} $. ### Proof of Optimality Let $ \omega $ be any primitive $ 37 $th root of unity. The condition can be translated into the polynomial identity: \[ \left( \sum_{t \in X} \omega^{at} \right) \left( \sum_{t \in X} \omega^t \right) = -1. \] In particular,...	{ "page_id": null, "source": 6821, "title": "from dpo" }
X} \omega^t \right\|^2 \), which is a contradiction. If $ a^3 \equiv 1 \pmod{37} $, we can get $ X = \{c, a^2 c, a^4 c, d, a^2 d, a^4 d\} $, so $ aX = X $, leading to a contradiction. #### Case 3: $ d = 2 $ Then $ a $ can be $ 6 $ or $ 31 \pmod{37} $. Check $ X = \{16, 17, 18, 19, 20, 21\} $ works. #### ...	{ "page_id": null, "source": 6821, "title": "from dpo" }
40 \), $ \|a_i - a_{i+1}\| \leq 1 $. We aim to determine the greatest possible values of: 1. $ a_{10} + a_{20} + a_{30} + a_{40} $ 2. $ a_{10} \cdot a_{20} + a_{30} \cdot a_{40} $ ### Part 1 Let $ s_1 = \frac{1}{2} a_5 + a_6 + a_7 + \cdots + a_{14} + \frac{1}{2} a_{15} $. Define $ s_2, s_3, s_4 $ similarly. Obs...	{ "page_id": null, "source": 6821, "title": "from dpo" }
\leq 10 \), so $ x - y \leq 20 $. Claim: $ x \leq 12.5 $. Proof: Suppose $ a_{10} + a_{20} > 12.5 $. Let $ t = a_{10} $ and $ u = a_{20} $. Then: \[ \frac{1}{2} a_{15} + a_{14} + a_{13} + \cdots + a_1 + a_{40} + a_{39} + \cdots + a_{36} + \frac{1}{2} a_{35} \geq 20t - 125, \] and similarly: \[ \frac{1...	{ "page_id": null, "source": 6821, "title": "from dpo" }
any fixed positive integer $ k $ and positive reals $\alpha$ and $\beta$: i) $\frac{\omega(n+k)}{\omega(n)} > \alpha$ ii) $\frac{\Omega(n+k)}{\Omega(n)} \alpha$ by choosing appropriate $\alpha$ and $\beta$. Thus, both statements are proven to be true. The answer is: \boxed{\text{True}}. \| \text{True} \| c...	{ "page_id": null, "source": 6821, "title": "from dpo" }
positive integers $\{a_n\}$ good if for any distinct positive integers $m,n$, one has $$\gcd(m,n) \mid a_m^2 + a_n^2 \text{ and } \gcd(a_m,a_n) \mid m^2 + n^2.$$ Call a positive integer $a$ to be $k$-good if there exists a good sequence such that $a_k = a$. Does there exists a $k$ such that there are exactly $2019$ $k$...	{ "page_id": null, "source": 6821, "title": "from dpo" }
have: \[ \nu_p(n) = \nu_p(\gcd(m, n)) \le \nu_p(a_m^2 + a_n^2) = \nu_p(a_n^2) \] and \[ \nu_p(a_n) \le \nu_p(\gcd(a_m, a_n)) \le \nu_p(m^2 + n^2) = \nu_p(n^2). \] This confirms the constraint. Finally, we check if there exists a $ k $ such that there are exactly 2019 $ k $-good positive integers. For each prime \( ...	{ "page_id": null, "source": 6821, "title": "from dpo" }
of $ m $. Proof: Let $ u = \mathrm{ord}_p(n) $. Then $ u \mid 2A $ but $ u \nmid A $ (since $ m $ is odd). Therefore, $ \nu_2(u) = \nu_2(A) + 1 $. Since $ u \mid p-1 $, we have $ \nu_2(p-1) \geq \nu_2(A) + 1 $. Let $ t = \nu_2(A) $. The claim implies $ m \equiv 1 \pmod{2^{t+1}} $. Using the Lift...	{ "page_id": null, "source": 6821, "title": "from dpo" }
] \| 7 \| Two incongruent triangles $ABC$ and $XYZ$ are called a pair of [i]pals[/i] if they satisfy the following conditions: (a) the two triangles have the same area; (b) let $M$ and $W$ be the respective midpoints of sides $BC$ and $YZ$. The two sets of lengths $\{AB, AM, AC\}$ and $\{XY, XW, XZ\}$ are identical $3$-e...	{ "page_id": null, "source": 6821, "title": "from dpo" }
the formula: \[ \text{Area} = \frac{1}{2} \sqrt{a^2c^2 - \left( \frac{a^2 + c^2 - b^2}{2} \right)^2}. \] Applying this to our triangles, we get: \[ [ABC] = \frac{1}{2} \sqrt{x^2 y^2 - \left( \frac{x^2 + y^2 - (2x^2 + 2y^2 - 4z^2)}{2} \right)^2} = \frac{1}{4} \sqrt{4x^2 y^2 - (4z^2 - x^2 - y^2)^2}. \] By symmetry: \[ [X...	{ "page_id": null, "source": 6821, "title": "from dpo" }
\| usa_team_selection_test \| \| [ "Mathematics -> Algebra -> Linear Algebra -> Matrices", "Mathematics -> Discrete Mathematics -> Graph Theory", "Mathematics -> Algebra -> Other (Matrix-related optimization) -> Other" ] \| 8 \| Find the greatest constant $\lambda$ such that for any doubly stochastic matrix of order 100, we...	{ "page_id": null, "source": 6821, "title": "from dpo" }
of Necessity We need to show that we can find a matching of 50 in any 150 cells that we pick such that each row and each column has a sum of picked cells at least $\lambda$. If $r_j$ or $c_k$ has exactly one chosen cell, the unique chosen cell on $r_j$ or $c_k$ is at least $\lambda$. Let $S$ be the set of...	{ "page_id": null, "source": 6821, "title": "from dpo" }
\frac{17}{1900}\) if $26 \leq j \leq 100$ and $25 \leq k \leq 100$. We can see that for any $\lambda > \frac{17}{1900}$, the construction fails to meet the conditions. ### Proof of Optimality Consider a bipartite graph with vertices $\{r_1, \ldots, r_{100}\}$ representing rows and $\{c_1, \ldots, c_{100}\}$ r...	{ "page_id": null, "source": 6821, "title": "from dpo" }
we can see an edge $AB$ of the regular octahedron from point $P$ outside if and only if the intersection of non degenerate triangle $PAB$ and the solid regular octahedron is exactly edge $AB$. \| To determine the maximum number of edges of a regular octahedron that can be seen from a point outside the octahedron, we sta...	{ "page_id": null, "source": 6821, "title": "from dpo" }
B_1C_1, C_1A_1 \) - $ A_1B_2, A_1C_2, B_1A_2, B_1C_2, C_1A_2, C_1B_2 $ Thus, we can see a total of 9 edges from such a point $ P $. Therefore, the maximum number of edges of a regular octahedron that can be seen from a point outside the octahedron is: \[ \boxed{9} \] \| 9 \| china_team_selection_test \| \| [ "Mathemati...	{ "page_id": null, "source": 6821, "title": "from dpo" }
sequence $ a_n := \left\lceil kn^2 \right\rceil + n $ for $ k \frac{b_{n+1}}{(n+1)^2} $ for $ n \frac{\left\lceil k(n+1)^2 \right\rceil + n + 1}{(n+1)^2}. \] Since \( \left\lceil kn^2 \right\rceil \geq kn^2 $, \[ \frac{\left\lceil kn^2 \right\rceil + n}{n^2} \geq k + \frac{1}{n}, \] and since \( \left\lceil k(n...	{ "page_id": null, "source": 6821, "title": "from dpo" }
$f_{2020}$. (2) Find the minimum of $f_{2020} \cdot f_{2021}$. \| Let $\{ z_n \}_{n \ge 1}$ be a sequence of complex numbers, whose odd terms are real, even terms are purely imaginary, and for every positive integer $k$, $\|z_k z_{k+1}\|=2^k$. Denote $f_n=\|z_1+z_2+\cdots+z_n\|,$ for $n=1,2,\cdots$. 1. To find the...	{ "page_id": null, "source": 6821, "title": "from dpo" }
$f \colon \mathbb{R} \to \mathbb{R}$ that satisfy the inequality \[ f(y) - \left(\frac{z-y}{z-x} f(x) + \frac{y-x}{z-x}f(z)\right) \leq f\left(\frac{x+z}{2}\right) - \frac{f(x)+f(z)}{2} \] for all real numbers $x Number Theory -> Congruences", "Mathematics -> Algebra -> Algebra -> Polynomial Operations" ] \| 8 \| Find a...	{ "page_id": null, "source": 6821, "title": "from dpo" }
numbers $P(0)$, $P^2(0)$, $\ldots$, $P^{m-1}(0)$ are not divisible by $n$; and [*]$P^m(0)$ is divisible by $n$. [/list] Here $P^k$ means $P$ applied $k$ times, so $P^1(0) = P(0)$, $P^2(0) = P(P(0))$, etc. [i]Carl Schildkraut[/i] \| We need to find all integers $ n \ge 2 $ for which there exists an integer $ m $ and ...	{ "page_id": null, "source": 6821, "title": "from dpo" }
m/n \in \{1, 2, \ldots, p\} \), which implies the claim. $\blacksquare$ Now, order all primes $ p_1 0 $. Since $ m > 1 $, there exists an index $ i $ such that the length $ \ell $ of the cycle of $ P $ modulo $ p_i^{a_i} $ that contains $ 0 $ is $ > 1 $. However, by the lemma, $ \ell $ must be \( p...	{ "page_id": null, "source": 6821, "title": "from dpo" }
$ p^t $. $\blacksquare$ Take $ Q $ from above and let $ Q(x) = \sum_{i=0}^{d} a_i x^i $. We construct $ b_i $ such that \[ \begin{align} b_i &\equiv 0 \pmod{\frac{n}{p^t}}, \\ b_i &\equiv a_i \pmod{p^t}, \end{align} \] and let $ P(x) = \sum_{i=0}^d b_i x^i $. It's clear that modulo $ \frac{n}{p^t} $, \(...	{ "page_id": null, "source": 6821, "title": "from dpo" }
\equiv \frac{a+1}{a} \pmod{p}\). Then: \[ z^n \equiv 1 \pmod{p}. \] By Fermat's Little Theorem, we know: \[ z^{p-1} \equiv 1 \pmod{p}. \] Since $z^n \equiv 1 \pmod{p}$, it follows that: \[ z^{\gcd(n, p-1)} \equiv 1 \pmod{p}. \] Given that $p$ is the smallest prime divisor of $n$, we have $\gcd(n, p-1) = 1$. The...	{ "page_id": null, "source": 6821, "title": "from dpo" }
1. \] The answer is: \boxed{2 \binom{100}{50} + 2 \binom{100}{49} + 1}. \| 2 \binom{100}{50} + 2 \binom{100}{49} + 1 \| china_team_selection_test \| \| [ "Mathematics -> Number Theory -> Congruences", "Mathematics -> Number Theory -> Greatest Common Divisors (GCD)" ] \| 7 \| An integer $n>1$ is given . Find the smallest posi...	{ "page_id": null, "source": 6821, "title": "from dpo" }
$ x + cy \equiv 0 \pmod{2n} $. Choose $ y = \left\lfloor \frac{2n}{c} \right\rfloor $ and $ x = 2n - c \left\lfloor \frac{2n}{c} \right\rfloor $. - Subcase 2.1: If \( 2 Algebra -> Intermediate Algebra -> Permutations and Combinations -> Other" ] \| 8 \| Let $a_1,a_2,\cdots,a_n$ be a permutation of $1,2,\cdots,...	{ "page_id": null, "source": 6821, "title": "from dpo" }

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