id int64 2 3.28M | problem stringlengths 30 5.17k | solution stringlengths 8 20.3k | problem_vector listlengths 1.02k 1.02k | solution_vector listlengths 1.02k 1.02k | last_modified stringdate 2025-08-11 00:00:00 2025-08-11 00:00:00 |
|---|---|---|---|---|---|
57,271 | Let $k$ be a positive integer and $M_k$ the set of all the integers that are between $2 \cdot k^2 + k$ and $2 \cdot k^2 + 3 \cdot k,$ both included. Is it possible to partition $M_k$ into 2 subsets $A$ and $B$ such that
\[ \sum_{x \in A} x^2 = \sum_{x \in B} x^2. \] | The answer is positive. Take \[ A = \{2k^2 + k,\ 2k^2 + k + 1,\ \dots,\ 2k^2 + 2k\} \] and \[ B = \{2k^2 + 2k + 1,\ 2k^2 + 2k + 2,\ \dots,\ 2k^2 + 3k\}. \]
Then \[ \sum_{x \in A}x^2 = \sum_{x \in B}x^2 = \frac{k}{6}\left(24k^4 + 60k^3 + 50k^2 + 15k + 1\right). \] | [
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57,272 | An $ n \times n, n \geq 2$ chessboard is numbered by the numbers $ 1, 2, \ldots, n^2$ (and every number occurs). Prove that there exist two neighbouring (with common edge) squares such that their numbers differ by at least $ n.$ | Here's an alternate way to use the idea in love_sc1's solution, where it suffices to partition the board into two consecutive intervals of numbers with at least $n$ borders. Let $S_k$ be the set of squares containing a number at most $k$. We consider the maximal $k$ such that $S_k$ contains neither a full row or a full column.
WLOG $S_{k+1}$ contains a full row, and $k+1$ is in row $r$, column $c$. The missing square in row $r$ gives us one vertical border between $S_k$ and its complement. We also know that $S_k$ has no full columns, so the other $n-1$ columns besides column $c$ are missing at least one square but do have a square in row $r$. We get at least one horizontal border in each of those $n-1$ columns. That's at least $n$ borders total. | [
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57,273 | Let $ n$ be an even positive integer. Let $ A_1, A_2, \ldots, A_{n \plus{} 1}$ be sets having $ n$ elements each such that any two of them have exactly one element in common while every element of their union belongs to at least two of the given sets. For which $ n$ can one assign to every element of the union one of the numbers 0 and 1 in such a manner that each of the sets has exactly $ \frac {n}{2}$ zeros? | First of all, the incidence structure is determined uniquely by the given conditions: every element of $A=\bigcup A_i$ belongs to [i]precisely[/i] two $A_i$'s, and $|A|=\frac{n(n+1)}2$. We can see this as follows:
Suppose we can find $x\in A_1\cap A_2\cap A_3$. Then $A_1\cup A_2\cup A_3$ has, except for $x$, at least $3(n-1)$ elements. Every $A_i,\ i>3$ covers at most $3$ of these (one from each of $A_1,A_2,A_3$), so in order to cover these elements one more time we would need at least $n-1$ more $A_i$'s, but we don't have that many. From this, the fact that $A$ has $\binom{n+1}2$ elements follows by a simple counting argument.
Now let's observe that $n$ being a multiple of $4$ is a necessary condition for the existence of the $0-1$ labeling. This is beause if we count the $0$-labeled elements every time they appear in a set we get $\binom{n+1}2$ (since there are $n+1$ sets, each one containing $\frac n2$ of them), and an even number, since every $0$-labeled element appears in precisely two sets, so it's counted twice. This shows that $\binom{n+1}2$ must be even, which, in turn, means that $4|n$.
Conversely, there is such a $0-1$ labeling when $4|n$, and we can prove it inductively. First of all, let's construct an $(n+1)\times n$ table $T_n$ containing numbers between $1$ and $\binom{n+1}2$ as follows:
The rows of the table will be the sets $A_i$ in order, and the numbers in a row will be the elements in the respective set. The first row is just $1,2,\ldots,n$. In the $i$'th step (with $i\in\overline{2,n+1}$) we construct the $i$'th row by setting $a_{i,j}=a_{j,i-1}$ when $j\in\overline{1,i-1}$, and making $a_{i,i},a_{i,i+1},\ldots,a_{i,n}$ the smallest consecutive numbers larger than all the numbers which have appeared so far.
The subtable formed by the last $n+1-4$ rows and $n-4$ columns has the same structure as $T_{n-4}$ (i.e. we can construct an isomorphism from $T_{n-4}$ to it induced by a bijection between the numbers $1,2,\ldots,\binom{n-4+1}2$ and those which appear in the mentioned subtable), so, by the induction hypothesis, we can find the desired $0-1$ labeling of this subtable. We divide the rest in three other subtables: $X$, the subtable formed by the first $4$ lines and $3$ columns, which we color $\begin{pmatrix}1&1&0\\1&0&1\\1&0&0\\0&1&0\end{pmatrix}$, $Y$, the one formed by the first $4$ lines and $n-3$ columns, whose columns we color alternately $\begin{pmatrix}0\\0\\1\\1\end{pmatrix}$ and $\begin{pmatrix}1\\1\\0\\0\end{pmatrix}$ (starting with the first one), and the rest of the initial table, which lies in the lower-left corner, and is just the transpose of $Y$ (so its coloration is determined).
Actually drawing these tables should help clarify the whole thing. | [
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57,274 | In a given tedrahedron $ ABCD$ let $ K$ and $ L$ be the centres of edges $ AB$ and $ CD$ respectively. Prove that every plane that contains the line $ KL$ divides the tedrahedron into two parts of equal volume. | Let $p$ be the plane through $AB$ parallel to $CD$. Any plane $p'$ parallel to $p$ which intersects the segment $KL$ will intersect the tetrahedron in a parallelogram with center on the line $KL$. Any plane $q$ through $KL$ will pass through the center of the parallelogram and hence divide it into two equal parts. So suppose $q$ divides the tetrahedron into two parts $T$ and $T'$. We have established that the area of the intersection of $p'$ and $T$ is equal to the area of the intersection of $p'$ and $T'$. It follows from Cavalieri's principle that $T$ and $T'$ have equal volume.
(To see that $PQRS$ is a parallelogram, note that $PQ$ is parallel to $CD$, and $RS$ is parallel to $CD$, so $PQ$ and $RS$ are parallel. Similarly $QR$ and $PS$ are parallel to $AB$ and hence to each other.) | [
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57,276 | "Let $ a$ be the greatest positive root of the equation $ x^3 \\minus{} 3 \\cdot x^2 \\plus{} 1 \\eq(...TRUNCATED) | "Here's a sketch of an answer. Let $a > b > c$ be the three real roots of $f(x) = x^3 - 3x^2 + 1$ a(...TRUNCATED) | [-0.005721735768020153,-0.023692406713962555,-0.002080986276268959,0.026376424357295036,0.0082322834(...TRUNCATED) | [-0.009241919033229351,-0.021124469116330147,-0.0035529611632227898,0.02683972381055355,-0.007079737(...TRUNCATED) | 2025-08-11 |
57,277 | "If $a_0$ is a positive real number, consider the sequence $\\{a_n\\}$ defined by:\r\n\r\n\\[ a_{n+1(...TRUNCATED) | "I claim $a=2$. We need to show that conditions (i) and (ii) are satisfied. \n\n[b]Condition (i).[/b(...TRUNCATED) | [0.022382721304893494,-0.00932216551154852,0.0017708478262647986,0.0176790002733469,-0.0168904587626(...TRUNCATED) | [0.005641154479235411,0.014216535724699497,-0.0007001379271969199,0.043039772659540176,-0.0421609766(...TRUNCATED) | 2025-08-11 |
57,278 | "Let $ u_1, u_2, \\ldots, u_m$ be $ m$ vectors in the plane, each of length $ \\leq 1,$ with zero su(...TRUNCATED) | "I think post #5 is missing some detail.\n\n[hide =Solution]\n\nWLOG suppose that $u_1 + u_2 + \\cdo(...TRUNCATED) | [0.051676906645298004,-0.0011020437814295292,-0.006477417424321175,0.03320226073265076,0.01035994756(...TRUNCATED) | [0.02195059135556221,-0.011205174960196018,-0.0034953956492245197,0.025424722582101822,0.01784763485(...TRUNCATED) | 2025-08-11 |
57,279 | "Show that there do not exist more than $27$ half-lines (or rays) emanating from the origin in the $(...TRUNCATED) | "The area of the whole unit sphere is less than $26.28$ times the area of a spherical cap of (spheri(...TRUNCATED) | [-0.017196130007505417,-0.01568634994328022,-0.007116302847862244,0.058195408433675766,0.05397160723(...TRUNCATED) | [0.011304437182843685,-0.044200289994478226,-0.008504729717969894,0.06981875002384186,0.048961095511(...TRUNCATED) | 2025-08-11 |
57,284 | "Let $T$ be a triangle with inscribed circle $C.$ A square with sides of length $a$ is circumscribed(...TRUNCATED) | "This isn't true when the square shares a side with $T$, verify it with an equilateral triangle. Alt(...TRUNCATED) | [0.026185594499111176,0.02002521976828575,-0.001863322569988668,0.030465321615338326,0.0338117182254(...TRUNCATED) | [0.0413215272128582,0.008137323893606663,0.0027474169619381428,0.057471245527267456,0.01918337866663(...TRUNCATED) | 2025-08-11 |
57,285 | "Let $1 \\leq k \\leq n.$ Consider all finite sequences of positive integers with sum $n.$ Find $T(n(...TRUNCATED) | Bump bump ! | [-0.0008027150179259479,-0.04909360781311989,-0.004574501421302557,-0.012738458812236786,-0.03427037(...TRUNCATED) | [-0.004749286454170942,-0.0357050821185112,-0.014455660246312618,-0.000979508156888187,0.05963527411(...TRUNCATED) | 2025-08-11 |
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