id int64 41 3.22M | problem stringlengths 38 4.44k | solution stringlengths 8 38.6k | 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 |
|---|---|---|---|---|---|
619,484 | $n$ is a natural number. We shall call a permutation $a_1,\dots,a_n$ of $1,\dots,n$ a quadratic(cubic) permutation if $\forall 1\leq i \leq n-1$ we have $a_ia_{i+1}+1$ is a perfect square(cube).
$(a)$ Prove that for infinitely many natural numbers $n$ there exists a quadratic permutation.
$(b)$ Prove that for no natur... | For part $(a)$, the identity $(x-1)(x+1) + 1 = x^2$ is important. This gives us the following construction (for any positive integer $k$)
\[ 2, 4, 6, \ldots, 4k^2+4k, 1, 3, 5, \dots, 4k^2+4k-1 \]
For part (b) take the largest power of $2$ less than or equal to $n$. Say that it's $2^k$. Note that $2^k > n/2$. If $2^k ... | [
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619,487 | Consider $n$ segments in the plane which no two intersect and between their $2n$ endpoints no three are collinear. Is the following statement true?
Statement: There exists a simple $2n$-gon such that it's vertices are the $2n$ endpoints of the segments and each segment is either completely inside the polygon or an edge... | Answer is no (unless there's a typo or smth, if you allow ALL SEGMENTS TO BE OUTSIDE the polygon the answer may be different, i believe that may be the typo?)
Counterexample:
[img]https://media.discordapp.net/attachments/864501962753703958/1092855532395708557/IMG_20230404_185723.jpg?width=383&height=681[/img] | [
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619,489 | Given a set $X=\{x_1,\ldots,x_n\}$ of natural numbers in which for all $1< i \leq n$ we have $1\leq x_i-x_{i-1}\leq 2$, call a real number $a$ [b]good[/b] if there exists $1\leq j \leq n$ such that $2|x_j-a|\leq 1$. Also a subset of $X$ is called [b]compact[/b] if the average of its elements is a good number.
Prove th... | Here is a different proof.
Fix some even $2 \le k \le n.$ We will use the term $k-$subsets to denote subsets of $X$ of size $k.$ We will show that the number of compact $k-$subsets is at least $\binom{n-2}{k-1}$. This would imply the desired bound since $\sum_{\textit{k odd}} \binom{n-2}{k} = 2^{n-3}.$
So let's sho... | [
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588,967 | Suppose a class contains $100$ students. Let, for $1\le i\le 100$, the $i^{\text{th}}$ student have $a_i$ many friends. For $0\le j\le 99$ let us define $c_j$ to be the number of students who have strictly more than $j$ friends. Show that \begin{align*} & \sum_{i=1}^{100}a_i=\sum_{j=0}^{99}c_j \end{align*} | A simple solution ,
The person having the $a_i$ friends is counted in $c_0,c_1, ....c_{a_i - 1}$ .i.e $a_i$ times . so in the RHS too it is counted $a_i$ times . so these are two different ways of counting the sum $ \sum _{i=1}^{100} a_i $ and hence it is proved | [
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588,972 | Consider $f(x)=x^4+ax^3+bx^2+cx+d\; (a,b,c,d\in\mathbb{R})$. It is known that $f$ intersects X-axis in at least $3$ (distinct) points. Show either $f$ has $4$ $\mathbf{distinct}$ real roots or it has $3$ $\mathbf{distinct}$ real roots and one of them is a point of local maxima or minima. | $f$ has at most 4 roots of which 3 are real. Hence the fourth is also real but not necessarily distinct from the other three. If it is distinct then $f$ has four distinct roots else if it is same as some other root (say $\alpha$) then $f(\alpha )=f'(\alpha )=0$ but $f''(\alpha)\neq 0$. Done | [
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588,973 | Let us consider a triangle $\Delta{PQR}$ in the co-ordinate plane. Show for every function $f: \mathbb{R}^2\to \mathbb{R}\;,f(X)=ax+by+c$ where $X\equiv (x,y) \text{ and } a,b,c\in\mathbb{R}$ and every point $A$ on $\Delta PQR$ or inside the triangle we have the inequality:
\begin{align*} & f(A)\le \text{max}\{f(P),f(... | Note that we can write each point $A$ on or inside the triangle $PQR$ as
\[
A=w_1P+w_2Q+w_3R
\]
Where $w_1,w_2,w_3$ are nonnegative weights adding to $1$ and $P,Q,R,A$ are viewed as vectors. This implies
\[
f(A)=w_1f(P)+w_2f(Q)+w_3f(R)
\]
If $f(A)>f(P),f(Q),f(R)$, then
\[
f(A)=w_1f(P)+w_2f(Q)+w_3f(R)<w_1f(A)+w_2f(A)+w_... | [
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588,980 | Let $f,g$ are defined in $(a,b)$ such that $f(x),g(x)\in\mathcal{C}^2$ and non-decreasing in an interval $(a,b)$ . Also suppose $f^{\prime \prime}(x)=g(x),g^{\prime \prime}(x)=f(x)$. Also it is given that $f(x)g(x)$ is linear in $(a,b)$. Show that $f\equiv 0 \text{ and } g\equiv 0$ in $(a,b)$. | $f(x)g(x)= ax+b, $, Again, we have, $f^{\prime}(x) g(x)+ g^{\prime}(x)f(x)=a$ ,
Differentiating again we get, $ f^{\prime \prime}(x) g(x)+g^{\prime \prime}(x)f(x)+2f^{\prime}(x)g^{\prime}(x)=0$
we have,$ f^{\prime \prime}(x)=g(x),g^{\prime \prime}(x)=f(x),$, so,
${f(x)}^2+{g(x)}^2+2f^{\prime}(x)g{\prime}(x)=0$,
We ... | [
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588,984 | Prove that sum of $12$ consecutive integers cannot be a square. Give an example of $11$ consecutive integers whose sum is a perfect square. | Clearly the sum of $12$ consecutive numbers cannot be odd.. Let the numbers be
$k+1,k+2, \cdots, k+12$.
$\therefore$ The sum is $12k+78$.
Now we know that any even square number is either $0$ or $4$ modulo $12$.
But, $12k+78 \equiv 6\pmod{12}$
$\implies$ The sum of $12$ consecutive numbers can never be a perfec... | [
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588,986 | Define $\mathcal{A}=\{(x,y)|x=u+v,y=v, u^2+v^2\le 1\}$. Find the length of the longest segment that is contained in $\mathcal{A}$. | Consider the linear transformation $T(u,v) = (u+v,v).$ Let $D_t$ be the diameter from $(-\cos t, -\sin t)$ to $(\cos t, \sin t).$ The length of $T(D_t)$ equals $2\sqrt {(\cos t +\sin t)^2 + \sin^2t}.$ Find the maximum value of the last expression over $[0,2\pi].$ Then you need an argument that this is the answer to the... | [
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588,989 | Let $f: [0,\infty)\to \mathbb{R}$ a non-decreasing function. Then show this inequality holds for all $x,y,z$ such that $0\le x<y<z$.
\begin{align*} & (z-x)\int_{y}^{z}f(u)\,\mathrm{du}\ge (z-y)\int_{x}^{z}f(u)\,\mathrm{du} \end{align*} | Rewriting, we want \[(1)\,\,\,\,\frac{1}{z-y}\int_{y}^{z}f(u)\,du \ge \frac{1}{z-x}\int_{x}^{z}f(u)\,du.\] This is the comparison of two averages. The second average is over a larger range of $f$-values. And each additional $f$-value is $\le$ every $f$-value in the first average. Intuitively the second average is less ... | [
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588,991 | $n(>1)$ lotus leaves are arranged in a circle. A frog jumps from a particular leaf from another under the following rule:
[list]
[*]It always moves clockwise.
[*]From starting it skips one leaf and then jumps to the next. After that it skips two leaves and jumps to the following. And the process con... | [hide="Solution"]Suppose $n$ is odd. Label the petals $1, 2, \cdots , n.$ After the $k-1$th move, the frog is on the petal $1 + 2 + \cdots + k = \dfrac{k(k+1)}{2},$ where the indices are taken modulo $n.$ After $n-1$ moves, the frog is on petal $\dfrac{n(n+1)}{2},$ which is equivalent to $n \pmod{n}$ since $\dfrac{n+1}... | [
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1,891,376 | Consider the number $\left(101^2-100^2\right)\cdot\left(102^2-101^2\right)\cdot\left(103^2-102^2\right)\cdot...\cdot\left(200^2-199^2\right)$.
[list=a]
[*] Determine its units digit.
[*] Determine its tens digit.
[/list] | [b]Part a:[/b]
Every factor in the product is an odd integer.
The third factor $103^2-102^2$ is a multiple of $5$.
Hence the product is an odd integer that is divisible by $5$.
Hence the unit digit is $5$.
[b]Part b:[/b]
We pair up the $k$-th factor from the left $(100+k)^2-(99+k)^2=199+2k$
with the $k$-th factor fro... | [
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1,891,380 | Let $ABCDEF$ be a convex hexagon. In the hexagon there is a point $K$, such that $ABCK,DEFK$ are both parallelograms. Prove that the three lines connecting $A,B,C$ to the midpoints of segments $CE,DF,EA$ meet at one point. | By vector, $K=A+C-B$ and $K=D+F-E$ thus $A+C-B = D+F-B$ or $A+C+E = B+D+F$. This means $\triangle ACE$ and $\triangle BDF$ have the same centroid so this is the concurrency point. | [
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1,891,382 | We are given a row of $n\geq7$ tiles. In the leftmost 3 tiles, there is a white piece each, and in the rightmost 3 tiles, there is a black piece each. The white and black players play in turns (the white starts). In each move, a player may take a piece of their color, and move it to an adjacent tile, so long as it's n... | Let the player moving the white pieces be called "Bob" and the other player be called "Rob."
We claim that Bob wins when $n$ is even and Rob wins when $n$ is odd.
Let's label the tiles $1, 2, \cdots, n$ from left to right. Let $b$ be a variable corresponding to the sum of the labels of the tiles containing Bob's piec... | [
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1,891,385 | Let $p$ be a polynomial with integer coefficients satisfying $p(16)=36,p(14)=16,p(5)=25$. Determine all possible values of $p(10)$. | To show that all multiples of $120$ are possible, take $p(x)=(x-10)^2 + c(x-5)(x-14)(x-16), c \in \mathbb{Z}$ for which $p(10) = 120c.$
On the other hand, supposing $p$ satisfies the problem conditions, define $q(x) = p(x) - (x-10)^2$ so that $5, 14, 16$ are roots of $q.$ Consequently, $q(x) = (x-5)(x-14)(x-16)r(x)$ ... | [
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1,891,387 | Let $n$ be a positive integer. Find the maximal real number $k$, such that the following holds:
For any $n$ real numbers $x_1,x_2,...,x_n$, we have $\sqrt{x_1^2+x_2^2+\dots+x_n^2}\geq k\cdot\min(|x_1-x_2|,|x_2-x_3|,...,|x_{n-1}-x_n|,|x_n-x_1|)$ | We will consider only $n>1$, since the case when $n=1$ is absurd.
We claim that the answer is $\frac{\sqrt{n}}{2}$ if $n$ is even and $\frac{\sqrt{n+5}}{2}$ if $n$ is odd.
If $n$ is even, this is clearly achieved when $(x_1, x_2, \cdots, x_n) = (1, -1, 1, -1, \cdots, 1, -1)$. If $n$ is odd, then this is achieved wit... | [
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614,655 | For any positive integer $n$, let $D_n$ denote the greatest common divisor of all numbers of the form $a^n + (a + 1)^n + (a + 2)^n$ where $a$ varies among all positive integers.
(a) Prove that for each $n$, $D_n$ is of the form $3^k$ for some integer $k \ge 0$.
(b) Prove that, for all $k\ge 0$, there exists an integer... | 1. So $D_n \mid a^n + (a + 1)^n + (a + 2)^n$ and $D_n \mid (a + 1)^n + (a + 2)^n + (a + 3)^n$, thus $D_n \mid (a + 3)^n - a^n$.
Let $p$ be a prime dividing $D_n$. Then $(a + 3)^n \equiv a^n \pmod{p}$ for all $a$, in particular for $a=p$, but then $3^n \equiv 0\pmod{p}$, forcing $p=3$.
2. For $k=0$ any even $n$ is goo... | [
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-0.001... | 2025-08-11 |
615,935 | Prove that there exists a positive integer that can be written, in at least two ways, as a sum of $2014$-th powers of $2015$ distinct positive integers $x_1 <x_2 <\cdots <x_{2015}$. | Suppose, on the contrary, this is not true, and every integer can be represented at most once as a sum of $2014$-th powers of $2015$ distinct positive integers.
Fix some positive integer $N$. There are $\binom{N}{2015}$ ways of choosing different $x_i \,,\, 1\le x_i \le N, i=1,2,\ldots,2015 $. Then, $\sum_{i=1}^{2015... | [
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571,040 | Find $ \sum_{a+b+c=5,\ a,\ b,\ c \geq 0} \binom {17} {a} \cdot \binom {17} {b} \cdot \binom {17} {c}$ | Yes, the answer is $_{51}C_5=\binom {51} {5}=2349060$. | [
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571,046 | 6 points $A,\ B,\ C,\ D,\ E,\ F$ are on a circle in this order and the segments $AD,\ BE,\ CF$ intersect at one point.
If $AB=1,\ BC=2,\ CD=3,\ DE=4,\ EF=5$, then find the length of the segment $FA$. | The relation $AB\cdot CD\cdot EF=BC\cdot DE\cdot FA$ is valid for any convex hexagon with $\hat A+\hat C+\hat E=\hat B+\hat D+\hat F=360^\circ$ and main diagonals concurrent (if İ am not wrong).
Best regards,
sunken rock | [
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571,087 | Given two blackboards $A,\ B$ on which distinct integers that are more than or equal to 2 and less than or equal to 20, are written, respectively.
You are to take out each number written on $A,\ B$ such that these two numbers are coprime.
Find the possible maximum value of the product of the number of integers written... | The answer is [hide]$65$.[/hide] | [
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571,099 | A school has a committee with 4 members. The committee has 4 duties and distinct duties are assigned to each member.
If each member has 2 duties that they desire and there are exactly 2 possibilities of ways by which all members has their own desirable duty,
then How many possible combinations for desire of 4 members a... | The answer is [hide]$936$.[/hide] | [
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-... | 2025-08-11 |
571,117 | Let $m$ be a positive integer with 1000 digits that each digit has no zero, and let $n$ be a positive integer less than or equal to $m$.
Find the possible maximum value of the number of the digits with zero in $\left[\frac{m}{n}\right].$
Note : For real number $r$, $[r]$ denotes the greatest integer not exceeding $r$... | The answer is [hide]$939$.[/hide] | [
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571,121 | For a grid of $55\times 55$, consider operation as follows:
Operation : Select one region of rectangles formed by some grids, then colour the region in one color, white or black.
Find the minimum number of times for operation needed to the state as below from the state that all grids are coloured in white.
$\bullet$... | The answer is [hide]784.[/hide] | [
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571,132 | Given a grid of $6\times 6$, write down integer which is more than or equal to 1 and less than or equal to 6 in each grid. For integers $1\leq i,\ j\leq 6$, denote
denote by $i \diamondsuit j$ the integer written in the entry in the $i-$ th row and $j-$ th colum of the grid.
How many ways are there to write down integ... | I tackled this problem for half day and finally I solved!!
Let $S=\{1,2,3,4,5,6\}$ , left coset $aS$ be $\{a\diamondsuit s | s\in S\}$ and right coset $Sa$ be $\{s\diamondsuit a | s\in S\}$.
We show these lemmas.
(1) For any two left cosets, they are same or disjoint.
Also, for any two right cosets, they are same... | [
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571,144 | Find the sum $\sum_{d} \frac{1}{d+\sqrt{10!}}$ for all positive divisors $d$ of $10!$. | Let
\[S=\sum_{d|10!} \frac{1}{d+\sqrt{10!}}=\sum_{d|10!}\frac{1}{\frac{10!}{d}+\sqrt{10!}}=\sum_{d|10!}\frac{d}{\sqrt{10!}}\cdot \frac{1}{d+\sqrt{10!}}.\]
Then $2S=\sum_{d|10!}\frac{1}{d+\sqrt{10!}}\left (1+\frac{d}{\sqrt{10!}}\right )=\sum_{d|10!}\frac{1}{\sqrt{10!}}=\frac{\tau(10!)}{\sqrt{10!}}$, but
$\tau(10!)=\tau... | [
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576,476 | Find all ordered triplets of positive integers $(a,\ b,\ c)$ such that $2^a+3^b+1=6^c$. | Notice that if we examine $\pmod 3$ we obtain $2^a+3^b+1\equiv(-1)^a+1\pmod 3$ which forces $a$ to be odd.
[b][color=#f00]Case 1:[/color][/b] $a=1$
Plugging $a=1$ we obtain
$3+3^b=6^c\Longrightarrow1+3^{b-1}=3^{c-1}\cdot2^c$ however notice that $1+3^{b-1}\equiv1\pmod 3$ which forces $3^{c-1}\cdot2^c\equiv1\pmod 3$ wh... | [
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589,861 | In a school, there are $n$ students and some of them are friends each other. (Friendship is mutual.) Define $ a, b $ the minimum value which satisfies the following conditions:
(1) We can divide students into $ a $ teams such that two students in the same team are always friends.
(2) We can divide students into $ b $ t... | We can prove by induction on $n$ that $N\leq n+1$. This is trivial for $n=1$. Consider a graph $G$ with $|V(G)|=n+1$, and a vertex $v\in V(G)$ with $\deg_G v = d$. Also consider the graph $G'= G - v$, with $|V(G')|=n$.
Say $a(G')>n-d$; then there cannot exist a vertex $v_i$ in each of the $a(G')$ cliques so that $vv_... | [
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589,863 | Let $ \Gamma $ be the circumcircle of triangle $ABC$, and let $l$ be the tangent line of $\Gamma $ passing $A$. Let $ D, E $ be the points each on side $AB, AC$ such that $ BD : DA= AE : EC $. Line $ DE $ meets $\Gamma $ at points $ F, G $. The line parallel to $AC$ passing $ D $ meets $l$ at $H$, the line parallel to ... | not really adding anything new, but here it is.
Let $X$ be the point on $\overline{BC}$ satisfying $\tfrac{BD}{DA} = \tfrac{AE}{EC} = \tfrac{BX}{XC}$. Then $\overline{XD} \parallel \overline{AC}$ so $X$ lies on line $HD$; likewise $X$ lies on line $IE$.
Now from
\[\measuredangle HXB
= \measuredangle ACB
= \measuredan... | [
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591,696 | In a triangle $ABC$, the external bisector of $\angle BAC$ intersects the ray $BC$ at $D$. The feet of the perpendiculars from $B$ and $C$ to line $AD$
are $E$ and $F$, respectively and the foot of the perpendicular from $D$ to $AC$ is $G$. Show that $\angle DGE + \angle DGF = 180^{\circ}$. | Purely Projective!
Let $X$ be the intersection of $A-$angle bisector with $BC$, then,
$$-1=(D,X;B,C) \overset{\infty_{AX}}{=} (D,A;E,F) \implies \angle EGA=\angle FGA \implies \angle DGF+\angle DGE=180^{\circ}$$ | [
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594,595 | Find all pairs $(m, n)$ of positive integers satsifying $m^6+5n^2=m+n^3$. | Another simple solution: from the condition, number m^6-m is of the form n^3-5n^2. For m>=4 we have that (m^2+1)^3-5(m^2+1)<m^6-m<(m^2+2)^3-5(m^2+2), so no solutions because n^3-5n^2 is strictly increasing in naturals which are >=5. After checking other cases we get that (m,n)={(1,5),(3,11)}. | [
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1,883,527 | 1. What is the probability that a randomly chosen word of this sentence has exactly four letters? | [hide=Solution]
There are $16$ words in the sentence and $5$ of them have four letters, so our desired probability is $$\boxed{\frac{5}{16}}.$$ $\square$
[/hide] | [
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1,883,536 | A palindrome is a word that reads the same backwards as forwards, such as “eye”, “race car”, and “qwertyytrewq”.
How many letters are in the smallest palindrome containing the letters b, o, g, t, r, and o, not necessarily in that order
and not necessarily adjacent? | Does it have to be a real word?
If not then [hide=the answer is]$9$.[/hide] | [
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1,883,537 | Alex the Kat has written $61$ problems for a math contest, and there are a total of $187$ problems submitted. How many more problems does he need to write (and submit) before he has written half of the total problems? | Suppose that Alex needs to submit $x$ more problems. Then we know that $2(61 + x) = 187 + x$, so that $122 + 2x = 187 + x$. By subtracting $122 + x$ from both sides, we get $x = 65$, so Alex needs to submit $65$ more problems. | [
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1,883,547 | Let $n!=n\cdot (n-1)\cdot (n-2)\cdot \ldots \cdot 2\cdot 1$.For example, $5! = 5\cdot 4\cdot 3 \cdot 2\cdot 1 = 120.$ Compute $\frac{(6!)^2}{5!\cdot 7!}$.
| We have $\frac{6! \cdot 6!}{5! \cdot 7!} = \frac{6 \cdot 5! \cdot 6!}{5! \cdot 7 \cdot 6!} = \frac{6 \cdot \cancel{5!} \cdot \cancel{6!}}{\cancel{5!} \cdot 7 \cdot \cancel{6!}} = \boxed{\frac{6}{7}}.$
edit: simplified solution | [
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1,883,551 | Find the sum of the greatest common factor and the least common multiple of $12$ and $18$.
| [hide = solution] $GCD(12, 18) = 6$ and $LCM(12, 18) = 36$ adding gives $42$ | [
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1,883,552 | What number is exactly halfway between $\frac 1 6$ and $\frac 1 4$? | [hide=Best Solution] The problem asks for the number perfectly between $\frac{1}{4}$ and $\frac{1}{6}$. This can be found by taking the average which adds the numbers up and divides by how many numbers there are. So the average must be $$\frac{\frac{1}{4}+\frac{1}{6}}{2} = \frac{5}{24}.$$ $\square$ | [
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1,883,560 | The sum of two integers is $8$. The sum of the squares of those two integers is $34$. What is the product of the two
integers? | [hide = solution] We can say that $x+y = 8$ and $x^2 + y^2 = 34$ squaring the first equation gives $x^2+2xy+y^2 = 64$ subtracting and dividing by $2$ gives $xy = 15$ | [
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1,883,583 | If a triangle has three altitudes of lengths $6, 6, \text{and} 6,$ what is its perimeter? | [hide=solution]Since the altitudes are equal, it is an equilateral triangle. Let $x$ represent the side length. Using the pythagorean theorem, we have \begin{align*}6^2+(\frac{1}{2}x)^2&=x^2 \\
x^2-(\frac{1}{2}x)^2=36\end{align*}
$(\frac{1}{2}x)^2$ is 1/4 of $x^2$, so we have $$\frac{3}{4}(x^2)=36.$$
Isolating $x^2$, w... | [
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594,818 | Find all triples of primes $(p,q,r)$ satisfying $3p^{4}-5q^{4}-4r^{2}=26$. | For storage. Classic modulo application.
[b][color=#f00]Claim.[/color][/b] $p=5$
Modulo $5$ takes us to congruence $3p^{4}+r^{2}\equiv 1\pmod 5$.
Suppose $p\neq 5$, then by Fermat's little theorem, $3p^{4}\equiv 3\pmod 5$, which means that $r^{2}\equiv 3\pmod 5$, but that's a contradiction, since $r^{2}\equiv -1,0,1\p... | [
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594,819 | For positive real numbers $a,b,c$ with $abc=1$ prove that $\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2}\geq 3(a+b+c+1)$ | By applying $x^2+y^2+z^2 \ge xy+yz+zx$ for all $x,y,z$ we have \[\sum \left( a+ \frac{1}{b} \right)^2 \ge \sum \left( a+ \frac 1b \right) \left(b+ \frac 1c \right)=\sum ab+ \sum \frac ac+ \sum \frac{1}{ab}+3.\]
Since $abc=1$ so $\sum \frac{1}{ab}=\sum a$. Hence, we need to prove $\sum ab+ \sum \frac ac \ge 2 \sum a. \q... | [
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594,827 | For a positive integer $n$, two payers $A$ and $B$ play the following game: Given a pile of $s$ stones, the players take turn alternatively with $A$ going first. On each turn the player is allowed to take either one stone, or a prime number of stones, or a positive multiple of $n$ stones. The winner is the one who take... | Answer $n-1$.
Proof: If $A$ has a winning strategy then $n$ is good otherwise $n$ is bad. Clearly from each good number you can get to at least one bad number with a move and from each bad number you ca only move to good numbers. Clearly for each class mod $n$ there is at most $1$ bad number since otherwise you could m... | [
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594,828 | Consider an acute triangle $ABC$ of area $S$. Let $CD \perp AB$ ($D \in AB$), $DM \perp AC$ ($M \in AC$) and $DN \perp BC$ ($N \in BC$). Denote by $H_1$ and $H_2$ the orthocentres of the triangles $MNC$, respectively $MND$. Find the area of the quadrilateral $AH_1BH_2$ in terms of $S$. | [b]Solution:[/b]
[b]Claim: [/b]$H_1MDN$ and $H_2MCN$ are parallelograms.
[b]Proof: [/b]Clearly $CMDN$ is cyclic. So, $\angle MNC = \angle MDC = 90-\angle DCM=A$. $\angle H_1MN=90-A=\angle MCD=\angle MND$. So, $MH_1||ND$. Same way we can get $H_1N||DM$, which gives us $H_1MDN$ is a parallelogram. Same way we can get ... | [
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581,900 | Given a scalene triangle $ABC$. Incircle of $\triangle{ABC{}}$ touches the sides $AB$ and $BC$ at points $C_1$ and $A_1$ respectively, and excircle of $\triangle{ABC}$ (on side $AC$) touches $AB$ and $BC$ at points $ C_2$ and $A_2$ respectively. $BN$ is bisector of $\angle{ABC}$ ($N$ lies on $BC$). Lines $A_1C_1$ and ... | let $M$ is midpoint of smaller arc $AC$ of $ (ABC) $. Then $M,P_1,K_1$ and $M,P_2,K_2$ are collinear. Since $K_1A=K_2C$ we get that $MK_1=MK_2$ and $ \angle MK_1A=\angle MK_2C $. Hence $ \angle P_1BN=\angle P_2BN $ and $ \angle ABP_1=\angle CBP_2 $ $ \Rightarrow $ $ AP_1=CP_2 $. :wink: | [
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582,514 | The triangle $ABC$ is inscribed in a circle $w_1$. Inscribed in a triangle circle touchs the sides $BC$ in a point $N$. $w_2$ — the circle inscribed in a segment $BAC$ circle of $w_1$, and passing through a point $N$. Let points $O$ and $J$ — the centers of circles $w_2$ and an extra inscribed circle (touching side $BC... | Amazing problem. Let $S$ be the center of $\omega_1$, $E$ the point where $\omega_2$ touches $\omega_1$. Obviously $E$ is the center of homothety between $\omega_2$ and $\omega_1$ so $E,D,N$ are collinear where $D$ is midpoint of an arc $BC$ of $\omega_1$ which doesn't contain $A$.
We will prove that points $O,I,N$ a... | [
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581,895 | Let $ABC$ be a isosceles triangle with $ AC = BC > AB$. Let $ E, F $ be the midpoints of segments $ AC, AB$, and let $l$ be the perpendicular bisector of $AC$. Let $ l $ meets $ AB$ at $K$, the line through $B$ parallel to $KC$ meets $AC$ at point $L$, and line $FL$ meets $ l$ at $W$. Let $ P $ be a point on segment $B... | So I solved this problem again today so I might as well make the above proof rigorous.
Start with the obvious observation of $\triangle ABC \sim \triangle ACK \sim \triangle ALB \sim \triangle AFE$.
Now we have $\frac{AL}{AB} = \frac{AF}{AE}$, so $\frac{AL}{AF} = \frac{AB}{AE}$, giving $\triangle ALF \sim \triangle AB... | [
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581,898 | There are $n$ students sitting on a round table. You collect all of $ n $ name tags and give them back arbitrarily. Each student gets one of $n$ name tags. Now $n$ students repeat following operation:
The students who have their own name tags exit the table. The other students give their name tags to the student who is... | I found a similar problem asking about the number of ways of giving cards to students so that exactly $k$ operations are needed for everyone to get their own cards. If we call it $f(n, k)$, the answer to the original problem will be $n!-\sum_{i=0}^{k} f(n, i)$.
That similar problem was appeared in the [url=https://mat... | [
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581,999 | Let $p>5$ be a prime. Suppose that there exist integer $k$ such that $ k^2 + 5 $ is divisible by $p$. Prove that there exist two positive integers $m,n$ satisfying $ p^2 = m^2 + 5n^2 $. | Since $p>5$ we have $gcd(k,p)=1$ .
By Thue's lemma since $gcd(k,p)=1$ there exist positive integers $a,b<\sqrt p$ such that $p|k^2a^2-b^2$ .
Condition says that $p|k^2+5$ so we have $p|k^2a^2+5a^2-(k^2a^2-b^2)=5a^2+b^2$ , so $5a^2+b^2=pt<5p+p=6p$ or $t<6$ . Checking $t=1,2,3,4,5$ we already have $p=x^2+5y^2$ for some... | [
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582,002 | In an island there are $n$ castles, and each castle is in country $A$ or $B$. There is one commander per castle, and each commander belongs to the same country as the castle he's initially in. There are some (two-way) roads between castles (there may be roads between castles of different countries), and call two castle... | For a set $S$ of castles, define $N_A(S)$ to be the set of $A$'s castles either in $S$ or adjacent to a castle in $S$. Similarly define $N_B(S)$.
(1) implies (2): For any subset $X$ of $A$'s castles, suppose $B$ makes an attack against a subset of those castles. $B$ can use up to $|N_B(X)|$ castles total in the attack... | [
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612,983 | $AB$ is a chord of $O$ and $AB$ is not a diameter of $O$. The tangent lines to $O$ at $A$ and $B$ meet at $C$. Let $M$ and $N$ be the midpoint of the segments $AC$ and $BC$, respectively. A circle passing through $C$ and tangent to $O$ meets line $MN$ at $P$ and $Q$. Prove that $\angle PCQ = \angle CAB$. | Let the circle passing through $C$ and tangent to $(O)$ be $\omega$, let $\omega \cap (O)=X, AX\cap \omega=Q', BX\cap \omega=P'$. Considering the homothety centered at $X$ carrying $O$ to $\omega$, we have $AB//P'Q'$.
$\angle Q'AC=\angle XBA=\angle XP'Q'=\angle XCQ'$, hence $Q'C^2=Q'X\times Q'A$, so $Q$ lies on the r... | [
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612,991 | There is a city with $n$ metro stations, each located at a vertex of a regular n-polygon. Metro Line 1 is a line which only connects two non-neighboring stations $A$ and $B$. Metro Line 2 is a cyclic line which passes through all the stations in a shape of regular n-polygon. For each line metro can run in any direction... | For the sake of contradiction, assume the contrary and say $(n,k)$ does not satisfy the problem statement.
Let $gcd(n,k) = g, n=ga, k=gb,$ where $gcd(a,b)=1, b<a$, and $g,a,b$ are odd numbers.
Number the stations, $1$ to $n$, in a counterclockwise order.
WLOG, Metro Line $1$ connects station $1$ and $l+1$.
WLOG, $l \le... | [
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621,837 | For $x, y$ positive integers, $x^2-4y+1$ is a multiple of $(x-2y)(1-2y)$. Prove that $|x-2y|$ is a square number. | Let $a=x-2y \ne 0$ (otherwise all is trivial) and $b=2y-1$. Then $x^2-4y+1=(a+b+1)^2-2b-1=a^2+b^2+2ab+2a$. Hence, the number \[t=\frac {a^2+b^2+2a}{ab}\] must be integer. Particularly, $a \mid b^2$. Let $a=c^2f$, where $c$ and $f$ are nonzero integers and $f$ is square-free. Then $b=cfd$ for some integer $d$. Now rewri... | [
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621,840 | Prove that there exists a function $f : \mathbb{N} \rightarrow \mathbb{N}$ that satisfies the following
(1) $\{f(n) : n\in\mathbb{N}\}$ is a finite set; and
(2) For nonzero integers $x_1, x_2, \ldots, x_{1000}$ that satisfy $f(\left|x_1\right|)=f(\left|x_2\right|)=\cdots=f(\left|x_{1000}\right|)$, then $x_1+2x_2+2^2x... | Every nonzero integer $n$ can be uniquely represented as \[n=2^{1000k+r}d\,,\] where $k$ is a nonnegative integer, $r=0,1,\dots,999$ and $d$ is an odd integer. With this in mind, put \[f(n)=r+1.\]Clearly, $\{f(n) \mid n \in \mathbb N\}=\{1,2,\dots,1000\}$. Now assume that we have taken 1000 numbers (including negative(... | [
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621,845 | Let $x, y, z$ be the real numbers that satisfies the following.
$(x-y)^2+(y-z)^2+(z-x)^2=8, x^3+y^3+z^3=1$
Find the minimum value of $x^4+y^4+z^4$. | Let $u=x+y+z$, $v=xy+yz+zx$, $w=xyz$.
The condition gives us $p^2-3q=4, 4p+3r=1$, so $q=\frac{p^2-4}{3}, r=\frac{1-4p}{3}$.
$f(t)=t^3-ut^2+vt-w$ needs to have $3$ real solutions.
Therefore, if $a,b$ are two real solutions to $f'(t)=3t^2-2ut+v$, $a<b$, $f(a) \ge 0$, $f(b) \le 0$.
We get $a,b=\frac{u \pm \sqrt{u^2-3v}}... | [
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621,847 | Determine all the functions $f : \mathbb{R}\rightarrow\mathbb{R}$ that satisfies the following.
$f(xf(x)+f(x)f(y)+y-1)=f(xf(x)+xy)+y-1$ | Let $P(x,y)$ be the assertion $f(xf(x)+f(x)f(y)+y-1)=f(xf(x)+xy)+y-1$
$P(0,y+1)$ $\implies$ $f(f(0)f(y+1)+y)=f(0)+y$ $\implies f$ is [b]surjective[/b].
That's why we only need a number $a$ such that $f(a)=0$.
So, $P(a,1)$ $\implies$ $f(0)=0$
$P(0,y+1)$ $\implies f(y)=y$, that is, $\boxed{f(x)=x}$, which is indeed a so... | [
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622,219 | How many one-to-one functions $f : \{1, 2, \cdots, 9\} \rightarrow \{1, 2, \cdots, 9\}$ satisfy (i) and (ii)?
(i) $f(1)>f(2)$, $f(9)<9$.
(ii) For each $i=3, 4, \cdots, 8$, if $f(1), \cdots, f(i-1)$ are smaller than $f(i)$, then $f(i+1)$ is also smaller than $f(i)$. | This is pretty much the same solution as JBL's.
Let $a_n$ be the answer for this question for $n$ instead of $9$.
We begin with a key observation.
If $f(k)=n$, $3\le k \le n-1$, note that the second condition for $i = k+1, k+2 \cdots , n$ are pretty much useless, since $f(k)>f(k+(i-k))$.
Also, we take note that for $f... | [
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585,181 | In a plane, 2014 lines are distributed in 3 groups. in every group all the lines are parallel between themselves. What is the maximum number of triangles that can be formed, such that every side of such triangle lie on one of the lines? | Obviously you can only pick one line from each group, so we want to find max of
abc, if a+b+c=2014. Note abc must be a whole number. "a" means number of lines in group 1, etc.
AMGM gives us 2014 >= 3sqrt^3(abc)
abc<=(2014/3)^3
Since it is AMGM, max must be a,b,c almost the same. (671, 671, 672) is the best which resul... | [
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585,182 | Solve the following equation in $\mathbb{Z}$:
\[3^{2a + 1}b^2 + 1 = 2^c\] | This is old, but I will post a full solution because I couldn't find one. Notice that if $b=0$, then $c=0$, and $a$ is an arbitrary integer. Now, assume that $b$ is distinct from zero. We will divide the problem into a few cases:
$1)$ $c\leq0$. Then ${2}^{c}\leq1$, so ${3}^{2a+1}{b}^{2}\leq0$, impossible.
$2)$ $c=1$. N... | [
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585,184 | Let $a,b,c$ be real numbers such that $a+b+c = 4$ and $a,b,c > 1$. Prove that:
\[\frac 1{a-1} + \frac 1{b-1} + \frac 1{c-1} \ge \frac 8{a+b} + \frac 8{b+c} + \frac 8{c+a}\] | [hide="Tangent Line method"]Let $(x,y,z) = ( a-1, b-1, c-1)$, then $x+y+z = 1$ and $0<x,y,z<1$. The inequality can be rewritten as
\[
S = \sum_{cyc} (\frac{1}{x} - \frac{8}{3-x} ) \geq 0
\]
Let $f(t) = \frac{1}{t} - \frac{8}{3-t}$, then we find that $f(t) \geq - \frac{81}{8} t + \frac{27}{8}$ ($*$proved below) with eq... | [
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613,831 | Say that a convex quadrilateral is [i]tasty[/i] if its two diagonals divide the quadrilateral into four nonoverlapping similar triangles. Find all tasty convex quadrilaterals. Justify your answer. | We claim that all such quadrilaterals are all rhombi and cyclic kites. It is not difficult to check that both types are tasty.
Call the quadrilateral $ABCD$ and the two diagonals intersect at $O$. If $\angle{AOB}\not=90^{\circ}$, then $\angle{BOC}\not=\angle{AOB}$, so either $\angle{BCO}$ or $\angle{CBO}$ is equal to ... | [
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613,838 | Let $n$ be a positive integer. A 4-by-$n$ rectangle is divided into $4n$ unit squares in the usual way. Each unit square is colored black or white. Suppose that every white unit square shares an edge with at least one black unit square. Prove that there are at least $n$ black unit squares. | For $1 \leq i \leq n$, let there be $a_i$ black squares in column $i$. We will only use the following "structure" of the original problem: if some $1 \leq i \leq n$ has $b_i=0$, then we require $b_{i-1}+b_{i+1}\geq 4$, where we define $b_0=b_{n+1}=0$ (but do not consider these as normal terms); we would like to show th... | [
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607,976 | Four different positive integers less than 10 are chosen randomly. What is the probability that their sum is odd? | [hide="Casework"]
We have two cases:
Case 1: 1 odd, 3 even.
Choose 1 odd in $\binom 51 = 5$ ways, 3 even in $\binom 43 = 4$ ways. This case gives a total of $5 \cdot 4 = 20$ ways.
Case 2: 3 odd, 1 even.
Choose 3 odd in $\binom 53 = 10$ ways, 1 even in $\binom 41 = 4$ ways. This case gives a total of $10\cdot 4 = 40... | [
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608,000 | A triangle has sides of length $\sqrt{13}$, $\sqrt{17}$, and $2 \sqrt{5}$. Compute the area of the triangle. | Heron's formula can be written in the form \[
\frac14 \sqrt{2\left( a^2b^2+b^2c^2+c^2a^2 \right)-\left( a^4+b^4+c^4 \right)}
= \frac14 \sqrt{\left( a^2+b^2+c^2 \right)^2 - 2\left( a^4+b^4+c^4 \right)}.
\] For this particular problem the answer is just \[ \frac14 \sqrt{50^2-2(13^2+17^2+20^2)} = \frac14 \sqrt{784} = 7.... | [
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608,003 | An ant is on one face of a cube. At every step, the ant walks to one of its four neighboring faces with equal probability. What is the expected (average) number of steps for it to reach the face opposite its starting face? | [hide]
Suppose that the ant starts on face $A$. Then let $B_1, B_2, B_3,$ and $B_4$ be the faces of the cube that are adjacent to $A$, and let $C$ be the face opposite $A$.We are looking for $E(A)$, the expected number of steps that the ant would take to get to C from A.
$E(C) = $expected number of steps to reach C fro... | [
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608,004 | Let $R$ be the set of points $(x, y)$ such that $\lfloor x^2 \rfloor = \lfloor y \rfloor$ and $\lfloor y^2 \rfloor = \lfloor x \rfloor$. Compute the area of region $R$. Recall that $\lfloor z \rfloor$ is the greatest integer that is less than or equal to $z$. | No solution to this one yet?
[hide="Solution"]
We first do a bit of bounding. Let $\lfloor x\rfloor = m$ for some integer $m$. Note that $m$ must be positive; otherwise, the conditions cannot hold.
From the second equality we obtain $\lfloor y^2\rfloor = m\implies m \leq y^2 < m + 1$. To deal with the first equali... | [
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608,008 | Deepali has a bag containing 10 red marbles and 10 blue marbles (and nothing else). She removes a random marble from the bag. She keeps doing so until all of the marbles remaining in the bag have the same color. Compute the probability that Deepali ends with exactly 3 marbles remaining in the bag. | Note that the final four marbles should either be either RBBB or BRRR. These two cases are symmetric, so we will just consider RBBB. There are $\tbinom{16}{9}$ ways to pick the first 16 marbles as we must pick 9 red's and 7 blue's. Note that the probability this occurs is simply:
\[ \dfrac{10! \cdot (10!/3!)}{20!/3!} =... | [
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608,011 | A triangle has area 114 and sides of integer length. What is the perimeter of the triangle? | Yes, only that for $(x,y,z) = (38,36,2)$ the sides of the triangle will be $\left (\dfrac {x+y}{2}, \dfrac {y+z}{2}, \dfrac {z+x}{2} \right ) = (37, 19, 20)$, with $19+20>37$, ok. | [
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608,044 | If $\sin x + \sin y = \frac{96}{65}$ and $\cos x + \cos y = \frac{72}{65}$, then what is the value of $\tan x + \tan y$? | [hide="Cheap"]
Note that 16-63-65 and 33-56-65 are Pythagorean triples, so the following combination works: \[ \left( \sin x, \cos x, \sin y, \cos y \right) = \left( \frac {63}{65}, \frac {16}{65}, \frac {33}{65}, \frac {56}{65} \right). \]Hence, the answer is $ \frac {63}{16} + \frac {33}{56} = \boxed {\frac {507}{112... | [
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608,045 | Let $ABC$ be a triangle. Points $D$, $E$, and $F$ are respectively on the sides $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$ of $\triangle ABC$. Suppose that
\[
\frac{AE}{AC} = \frac{CD}{CB} = \frac{BF}{BA} = x
\]
for some $x$ with $\frac{1}{2} < x < 1$. Segments $\overline{AD}$, $\overline{BE}$, and $\ov... | lol ok what is this
[hide="Solution"]
We use barycentric coordinates. Let $t$ be the ratio $\frac{AE}{CE}$. Let $G = AD \cap BE$, $H = BE \cap CF$, and $I = CF \cap AD$. Then $D = (0 : t : 1)$, $E = (1: 0 : t)$, $F = (t : 1 : 0)$, $G = (1 : t^2 : t)$, $H = (t : 1 : t^2)$, and $I = (t^2 : t : 1)$. Let $S = [ABC]$. Now ... | [
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608,047 | Let $n$ be a positive integer. Let $(a, b, c)$ be a random ordered triple of nonnegative integers such that $a + b + c = n$, chosen uniformly at random from among all such triples. Let $M_n$ be the expected value (average value) of the largest of $a$, $b$, and $c$. As $n$ approaches infinity, what value does $\frac{... | [hide]The number of triples is $\binom{n+2}{2}=\frac12n^2+o(n^2)$. Thus the desired limit is equal to \[
\lim \frac2{n^3}\sum_{(a,b,c)}\max(a,b,c).\] We're going to use this little fudge factor approach for all that it's worth: the number of triples in which the maximum is achieved by two different entries simultaneou... | [
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608,048 | How many complex numbers $z$ such that $\left| z \right| < 30$ satisfy the equation
\[
e^z = \frac{z - 1}{z + 1} \, ?
\] | Let $z = a+bi$.
'[b][color=#f00]Claim.[/b][/color] $a=0$.
[i]Proof.[/i] The given equation rewrites to $$e^a(e^{bi}) = \frac{a-1+bi}{a+1+bi}.$$ Taking the magnitude of both sides, $$e^{2a} = \frac{a^2+b^2-2a+1}{a^2+b^2+2a+1} = 1 - \frac{4a}{a^2+b^2+2a+1}.$$ Notice that if $a > 0$, then the LHS is greater than 1, whi... | [
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592,838 | Let $a_1,\ldots,a_n$ and $b_1\ldots,b_n$ be $2n$ real numbers. Prove that there exists an integer $k$ with $1\le k\le n$ such that
$ \sum_{i=1}^n|a_i-a_k| ~~\le~~ \sum_{i=1}^n|b_i-a_k|.$
(Proposed by Gerhard Woeginger, Austria) | Yes.To be more precise if $a_s,a_l$ be the smallest and largest of the $a_i$'s and $b_s,b_l$ be the smallest and largest of the $b_i$'s then
$|a_l-a_i|+|a_i-a_s|=a_l-a_s \le |b_i-a_l|+|b_i-a_s|$.Summing over all $i$ we get $\sum_{i=1}^{n}{|a_i-a_l|}+\sum_{i=1}^{n}{|a_i-a_s|} \le \sum_{i=1}^{n}{|b_i-a_l|}+\sum_{i=1}^{n... | [
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592,840 | Consider increasing integer sequences with elements from $1,\ldots,10^6$. Such a sequence is [i]Adriatic[/i] if its first element equals 1 and if every element is at least twice the preceding element. A sequence is [i]Tyrrhenian[/i] if its final element equals $10^6$ and if every element is strictly greater than the su... | I will prove that the number of sequences of both types is equal for every natural number.
Let us fix a certain number $n$. Let $A$ be the set of all [i]Adriatic[/i] sequences with elements from $1,...,n$, and $T$ be the set of all [i]Tyrrhenian[/i] sequences with elements from $1,...,n$. We will make a bijection from... | [
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592,844 | In triangle $ABC$ let $A'$, $B'$, $C'$ respectively be the midpoints of the sides $BC$, $CA$, $AB$. Furthermore let $L$, $M$, $N$ be the projections of the orthocenter on the three sides $BC$, $CA$, $AB$, and let $k$ denote the nine-point circle. The lines $AA'$, $BB'$, $CC'$ intersect $k$ in the points $D$, $E$, $F$. ... | By Menelaus' Theorem, it is equivalent to prove that $\frac{LR}{MR} \cdot \frac{NQ}{LQ} \cdot \frac{MP}{NP} = 1.$ By Power of Point and Law of Sines,
\[\begin{aligned}
\left(\frac{RL}{RM} \cdot \frac{QN}{QL} \cdot \frac{PM}{PN}\right)^2 = 1 & \iff \frac{RL \cdot RM}{RM^2} \cdot \frac{QL \cdot QN}{QL^2} \cdot \frac{PM \... | [
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-... | 2025-08-11 |
613,419 | Each of the integers from 1 to 4027 has been colored either green or red. Changing the color of a number is making it red if it was green and making it green if it was red. Two positive integers $m$ and $n$ are said to be [i]cuates[/i] if either $\frac{m}{n}$ or $\frac{n}{m}$ is a prime number. A [i]step[/i] consists ... | If $2014$ is red, toggle $1007$ and $2014 = 1007 \cdot 2$. Now $2014$ is green.
If $1007$ is red, toggle $1007$ and $3021 = 1007 \cdot 3$. Now $1007$ is green.
Now, for each $k$ of $1,2,3,\ldots,2013$ in that order, if $k$ is red, then toggle $k$ and $2k$. Since $k \le 2013$, we have $2k \le 4026$ so it's still color... | [
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613,677 | Let $d(n)$ be the number of positive divisors of a positive integer $n$ (including $1$ and $n$). Find all values of $n$ such that $n + d(n) = d(n)^2$. | [b]Lemma 1:[/b] Let $m \in \mathbb{N}$. The number of divisors of $m$ that's between 1 and $\lfloor \sqrt{m} \rfloor$ is equal to $d(m)/2$ if $m$ isn't a perfect square, and equal to $(d(m)+1)/2$ if $m$ is a perfect square.
[b]Proof:[/b] $a$ is a divisor of $m$ if, and only if $m/a$ is a divisor of $m$ as well. Thus, ... | [
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614,026 | Let $a, b, c$ be positive reals such that $a + b + c = 3$. Prove:
\[ \frac{a^2}{a + \sqrt[3]{bc}} + \frac{b^2}{b + \sqrt[3]{ca}} + \frac{c^2}{c + \sqrt[3]{ab}} \geq \frac{3}{2} \]
And determine when equality holds. | [hide=Solution (to original problem)]
By Titu's Lemma, $$\frac{a^2}{a+\sqrt[3]{bc}}+\frac{b^2}{b+\sqrt[3]{ca}}+\frac{c^2}{c+\sqrt[3]{ab}}\geq\frac{\left(a+b+c\right)^2}{a+b+c+\sqrt[3]{ab}+\sqrt[3]{bc}+\sqrt[3]{ca}}=\frac{9}{3+\sqrt[3]{ab}+\sqrt[3]{bc}+\sqrt[3]{ca}}.$$ Now, by Hölder's Inequality, $$\left(1+1+1\right)\l... | [
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606,973 | For integers $n \ge k \ge 0$ we define the [i]bibinomial coefficient[/i] $\left( \binom{n}{k} \right)$ by
\[ \left( \binom{n}{k} \right) = \frac{n!!}{k!!(n-k)!!} .\]
Determine all pairs $(n,k)$ of integers with $n \ge k \ge 0$ such that the corresponding bibinomial coefficient is an integer.
[i]Remark: The double fa... | [hide="Solution"]Introduce the notation $f(n,k):=\left(\binom{n}{k}\right)$. Trivially, if $k=0$ or $k=n$, then $f(n,k)$ is an integer, so from now on we assume that $0<k<n$.
If $n$ is odd, then $k$ and $n-k$ have different parity, so one of them must be even. That means that the denominator of $f(n,k)=\frac{n!!}{k!!(... | [
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607,042 | Determine the lowest possible value of the expression
\[ \frac{1}{a+x} + \frac{1}{a+y} + \frac{1}{b+x} + \frac{1}{b+y} \]
where $a,b,x,$ and $y$ are positive real numbers satisfying the inequalities
\[ \frac{1}{a+x} \ge \frac{1}{2} \] \[\frac{1}{a+y} \ge \frac{1}{2} \] \[ \frac{1}{b+x} \ge \frac{1}{2} \] \[ \frac{1}{b... | Of course, not $\dfrac {1}{2} + \dfrac {1}{2} + \dfrac {1}{2} + 1 = \dfrac {5}{2}$. Say $a+x = s \leq 2$, $a+y = t \leq 2$, $b+x = u \leq 2$, $b+y = v \leq 1$. Then $s+v = t+u \leq 3$, so $(a+y) + (b+x) \leq 3$.
Thus $\dfrac {1}{a+y} + \dfrac {1}{b+x} \geq \dfrac {4}{(a+y)+(b+x)} \geq \dfrac {4}{3}$, with equality for... | [
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607,043 | Determine all functions $f : \mathbb{R} \to \mathbb{R}$ such that
\[ xf(xy) + xyf(x) \ge f(x^2)f(y) + x^2y \]
holds for all $x,y \in \mathbb{R}$. | Taking $(x,y)=(0,0)$ and $(x,y)=(1,1)$, we derive $f(0)=0$ and $f(1)=1$. Taking $y=1$ gives $$2xf(x)\ge f(x^2)+x^2.$$
Note that this gives $f(-1)\le -1$. Letting $y=x$ yields $0\ge (f(x^2)-x^2)(f(x)-x)$. If $x\ge 0$ and $f(x^2)\ge x^2$ for all $x\ge 0$, we may multiply both sides by $2x$ to obtain $$0\ge (f(x^2)-x^2)(2... | [
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607,045 | In Happy City there are $2014$ citizens called $A_1, A_2, \dots , A_{2014}$. Each of them is either [i]happy[/i] or [i]unhappy[/i] at any moment in time. The mood of any citizen $A$ changes (from being unhappy to being happy or vice versa) if and only if some other happy citizen smiles at $A$. On Monday morning there w... | Suppose every unhappy citizen carries a $0$ and every happy citizen carries a $1$ initially. When $A$ smiles to $B$, the number that $B$ carries is added by the number that $A$ carries. Then we can see the invariant that every happy citizen carries an odd number and every unhappy citizen carries an even number. Thus re... | [
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607,047 | Let the incircle $k$ of the triangle $ABC$ touch its side $BC$ at $D$. Let the line $AD$ intersect $k$ at $L \neq D$ and denote the excentre of $ABC$ opposite to $A$ by $K$. Let $M$ and $N$ be the midpoints of $BC$ and $KM$ respectively.
Prove that the points $B, C, N,$ and $L$ are concyclic. | Let $\ell$ be the perpendicular bisector of $BC$. Define points $K'$ and $N'$ as the images of $K$ and $N$, respectively in the symmetry about $\ell$. Let $T$ be a point on line $BC$ such that $(B, C; D, T)=-1$ and note that $TL$ is tangent to $\omega$.
It is well-known that $B, I, C, K$ are concyclic, so $B, I, C, K'... | [
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607,048 | A finite set of positive integers $A$ is called [i]meanly[/i] if for each of its nonempy subsets the arithmetic mean of its elements is also a positive integer. In other words, $A$ is meanly if $\frac{1}{k}(a_1 + \dots + a_k)$ is an integer whenever $k \ge 1$ and $a_1, \dots, a_k \in A$ are distinct.
Given a positive ... | Let the elements of $A$ be $1\leq a_1<a_2<\cdots < a_n$, for $n\geq 3$. Let $2\leq k\leq n-1$. Consider two indices $1\leq i < j \leq n$, and a set of indices $S\subseteq \{1,2,\ldots,n\}\setminus \{i,j\}$ with $|S|=k-1$. We need $\displaystyle k\mid a_i + \sum_{\ell \in S} a_\ell$ and $\displaystyle k\mid a_j + \sum_{... | [
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607,049 | Determine all quadruples $(x,y,z,t)$ of positive integers such that
\[ 20^x + 14^{2y} = (x + 2y + z)^{zt}.\] | [hide="Solution"]
Let $A=20^x$, $B=14^{2y}$, $C=(x+2y+z)^{zt}$.
Since the LHS is even, we have $2|x+2y+z$ hence $x$ and $z$ have the same parity.
If $x$ and $z$ are both even, then $A$ and $B$ are square numbers not divisible by $3$, so $A+B\equiv 2\pmod 3$, whereas $C$ is also a perfect square, contradiction.
So $x$ a... | [
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614,143 | Let $f : \mathbb{R} \to \mathbb{R}$ be a continuous function and let $g : \mathbb{R} \to \mathbb{R}$ be arbitrary. Suppose that the Minkowski sum of the graph of $f$ and the graph of $g$ (i.e., the set $\{( x+y; f(x)+g(y) ) \mid x, y \in \mathbb{R}\}$) has Lebesgue measure zero. Does it follow then that the function $... | We will prove the answer is negative.
Particularly, for $f:=|x|$, it's possible to find a suitable $g$ complying the requirements.
Sum of the graphs of $f$ and $g$ is a subset of two families of parallel lines. One of them consists of lines passing through $(x-g(x);0)\,,x\in \mathbb{R}$ and parallel to the line $y=x$... | [
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617,384 | Let $n\ge 1$ be a fixed integer. Calculate the distance $\inf_{p,f}\, \max_{0\le x\le 1} |f(x)-p(x)|$ , where $p$ runs over polynomials of degree less than $n$ with real coefficients and $f$ runs over functions $f(x)= \sum_{k=n}^{\infty} c_k x^k$ defined on the closed interval $[0,1]$ , where $c_k \ge 0$ and $\... | First some notations. Let $E_n(f)_{[0,1]}=\inf_{p}\sup_{x\in[0,1]}|f(x)-p(x)|$, where $\inf $ is taken over all polynomials of degree at most $n$.
Let $T_n$ be the Chebyshev polynomial (of first kind) and $x_0=-1,x_1,\ldots,x_{n-1}, x_n=1$ be its oscillating points, where it takes alternatively values $\pm 1$. Let $t_j... | [
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618,377 | Let $ k\geq 1 $ and let $ I_{1},\dots, I_{k} $ be non-degenerate subintervals of the interval $ [0, 1] $. Prove that
\[ \sum \frac{1}{\left | I_{i}\cup I_{j} \right |} \geq k^{2} \]
where the summation is over all pairs $ (i, j) $ of indices such that $I_i\cap I_j\neq \emptyset$. | I think this problem is very beautiful! I haven't read drkim's proof carefully, but it seems different with this at first sight.
First, since $|I_i \cup I_j| + |I_i \cap I_j| = |I_i| + |I_j|$ and $|I_i \cap I_j| \le \min \{|I_i|, |I_j|\}$, we can prove the following inequality.
$$|I_i \cup I_j| \cdot |I_i \cap I_j| \... | [
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618,491 | We have $4n + 5$ points on the plane, no three of them are collinear. The points are colored with two colors. Prove that from the points we can form $n$ empty triangles (they have no colored points in their interiors) with pairwise disjoint interiors, such that all points occurring as vertices of the $n$ triangles have... | Is $ 4n+5 $ the best number? | [
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-0.... | 2025-08-11 |
618,493 | For a positive integer $n$, define $f(n)$ to be the number of sequences $(a_1,a_2,\dots,a_k)$ such that $a_1a_2\cdots a_k=n$ where $a_i\geq 2$ and $k\ge 0$ is arbitrary. Also we define $f(1)=1$. Now let $\alpha>1$ be the unique real number satisfying $\zeta(\alpha)=2$, i.e $ \sum_{n=1}^{\infty}\frac{1}{n^\alpha}=2 $
Pr... | Let $g(n)=\sum_{j=1}^n f(j)$. Clearly:
\[g(n) =\left| \{(a_1,a_2,\ldots,a_k)\mid a_1a_2\dots a_k\leq n\,,\, a_j\geq 2\} \right|\]
To avoid unnecessary notations let's set $g(x)=g(\lfloor x \rfloor)\,,\, x>0$.
Since $a_1$ can take values $2,3,\dots, \lfloor n/2 \rfloor$ we obtain:
\[g(n)= \sum_{j=2}^{\lfloor n/2 \rfloo... | [
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0.00... | 2025-08-11 |
618,534 | Let $\rho:G\to GL(V)$ be a representation of a finite $p$-group $G$ over a field of characteristic $p$. Prove that if the restriction of the linear map $\sum_{g\in G} \rho(g)$ to a finite dimensional subspace $W$ of $V$ is injective, then the subspace spanned by the subspaces $\rho(g)W$ $(g\in G)$ is the direct sum of ... | Solution:
Lemma 1
Let given field $\mathbb{F}$, $char\mathbb{F} = p$, and representation $\rho : G\to GL(V) $ if finite $p$ group $G$, $\dim V = n $. Prove that $\forall g_1,..., g_n \in G$, $\prod_{i = 1}^n (\rho(g_i) - E) = 0$.
Take $\bar\mathbb{F}$ and prove that in dome basis $\rho(G)$ are upper triangular matri... | [
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-0.... | 2025-08-11 |
618,537 | Let $\rho:\mathbb{R}^n\to \mathbb{R}$, $\rho(\mathbf{x})=e^{-||\mathbf{x}||^2}$, and let $K\subset \mathbb{R}^n$ be a convex body, i.e., a compact convex set with nonempty interior. Define the barycenter $\mathbf{s}_K$ of the body $K$ with respect to the weight function $\rho$ by the usual formula
\[\mathbf{s}_K=\frac{... | Here's a solution.
Define \[ g(t) = -\log\left( \int_K \rho(x+t) dx \right). \] One can note that the barycenter of $K + t$ ($K$ translated by $t$) is actually just $\nabla g$. It is known that the gradient of any strictly convex function is injective, so we now show that $g$ is convex. This reduces to showing that fo... | [
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-0.00... | 2025-08-11 |
579,002 | Let $a,b\in\mathbb{R}_+$ such that $a+b=1$. Find the minimum value of the following expression:
\[E(a,b)=3\sqrt{1+2a^2}+2\sqrt{40+9b^2}.\] | $3\sqrt{1+2a^2}\ge \frac{\sqrt{11}}{11}(6a+9)\iff (3a-1)^2\ge 0,$
$2\sqrt{40+9b^2}\ge \frac{\sqrt{11}}{11}(6b+40)\iff (3b-2)^2\ge 0,$
$3\sqrt{1+2a^2}+2\sqrt{40+9b^2}\ge \frac{\sqrt{11}}{11}(6a+9+6b+40)=5\sqrt{11},$
when $a=\dfrac{1}{3},\,b=\dfrac{2}{3},E(a,b)_{min}=5\sqrt{11}.$ | [
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0.0017427... | 2025-08-11 |
579,007 | Let $\triangle ABC$ be an acute triangle and $AD$ the bisector of the angle $\angle BAC$ with $D\in(BC)$. Let $E$ and $F$ denote feet of perpendiculars from $D$ to $AB$ and $AC$ respectively. If $BF\cap CE=K$ and $\odot AKE\cap BF=L$ prove that $DL\perp BF$. | Circle $\odot(AEDF)$ with diameter $\overline{AD}$ cuts $BC$ again at the projection $P$ of $A$ on it. Since $D$ is clearly midpoint of the arc $EPF,$ then $PD,PA$ bisect $\widehat{EPF}.$ Hence if $Q \equiv EF \cap BC,$ $R \equiv EF \cap AP,$ the pencil $P(E,F,R,Q)=-1$ is harmonic $\Longrightarrow$ $(B,C,P,Q)=-1.$ Thus... | [
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-0.000... | 2025-08-11 |
579,008 | Find all pairs of non-negative integers $(x,y)$ such that
\[\sqrt{x+y}-\sqrt{x}-\sqrt{y}+2=0.\] | Notice$\sqrt{x}+\sqrt{y} \ge 2.$
$\sqrt{x+y}-\sqrt{x}-\sqrt{y}+2=0 \iff \sqrt{x}+\sqrt{y}-2=\sqrt{x+y} \Longrightarrow$
${2(\sqrt{x}+\sqrt{y})=2+\sqrt{xy}\iff ( \sqrt{x}-2)(\sqrt{y}-2})=2 \Longrightarrow$
$\sqrt{x}-2=1,\sqrt{y}-2=2 $ or $\sqrt{x}-2=2 ,\sqrt{y}-2=1\Longrightarrow$
$ (x,y)=(9,16)$ or $(x,y)=(16,9).$ | [
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0.0019... | 2025-08-11 |
579,012 | Define $p(n)$ to be th product of all non-zero digits of $n$. For instance $p(5)=5$, $p(27)=14$, $p(101)=1$ and so on. Find the greatest prime divisor of the following expression:
\[p(1)+p(2)+p(3)+...+p(999).\] | This actually isn't all that hard to compute by hand. First, make the following observation: concatenation multiplies $p$, i.e., if $a$ and $b$ are natural numbers and $c$ is what you get by concatenating the two, we have $p(c) = p(a)p(b)$. For instance, $p(107) = p(10)p(7)$.
Now, for convenience's sake, we can define... | [
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... | 2025-08-11 |
583,202 | Consider $n \geq 2 $ positive numbers $0<x_1 \leq x_2 \leq ... \leq x_n$, such that $x_1 + x_2 + ... + x_n = 1$. Prove that if $x_n \leq \dfrac{2}{3}$, then there exists a positive integer $1 \leq k \leq n$ such that $\dfrac{1}{3} \leq x_1+x_2+...+x_k < \dfrac{2}{3}$. | If $x_n>\frac 13$, then $k=n-1$ done. $\frac 13 \le x_1+...+x_{n-1}=1-x_n<\frac 23$.
Else $x_n\le \frac 13.$ And exist k, suth that $\frac 13 <x_{k+1}+...+x_n\le \frac 23$ (because $x_i\le \frac 13$ for all i$).
It give suth k. | [
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-0... | 2025-08-11 |
583,206 | Consider $n \geq 2$ distinct points in the plane $A_1,A_2,...,A_n$ . Color the midpoints of the segments determined by each pair of points in red. What is the minimum number of distinct red points? | [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=42&t=369935[/url]
The problem is anyways a classic. The minimum number is $\boxed{2n-3}$, realized for example when all $n$ points are collinear, equally distanced. To prove this is a minimum, say the points are $A_1,A_2,\ldots,A_n$, with $A_1A_n \geq A_iA_... | [
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583,294 | Prove that there do not exist $4$ points in the plane such that the distances between any pair of them is an odd integer. | Oh, my ... such an old chestnut. If not mistaken, was asked in some Putnam. Simplest approach is to write the determinant giving the zero volume of the (degenerated) tetrahedron, and obtain a contradiction modulo $8$. | [
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-0... | 2025-08-11 |
583,295 | Find all functions $f:R \rightarrow R$, which satisfy the equality for any $x,y \in R$:
$f(xf(y)+y)+f(xy+x)=f(x+y)+2xy$, | The only solutions are $f(x)=x$ and $f(x)=-2x$.
$P(0,x) \rightarrow f(0)=0$
$P(1,y) \rightarrow f(f(y)+y)=2y \rightarrow f$ surjective
Thus, $\exists n$ such that $f(n)=1$.
$P(x, n) \rightarrow f(x)=cx$.
Plugging back in, we have $c = 1, -2 \rightarrow f(x)=x, f(x)=-2x$, as needed. $\blacksquare$ | [
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0.... | 2025-08-11 |
583,306 | On a circle $n \geq 1$ real numbers are written, their sum is $n-1$. Prove that one can denote these numbers as $x_1, x_2, ..., x_n$ consecutively, starting from a number and moving clockwise, such that for any $k$ ($1\leq k \leq n$) $ x_1 + x_2+...+x_k \geq k-1$. | By subtracting $1$ from each number the problem becomes equivalent to:
On a circle $n\geq1$ real numbers $a_{1},a_{2},...,a_{n}$ are written clockwise round the circle where $a_{1}+a_{2}+...+a_{n}=-1$. Prove that one can choose $j\in\{1,2,...,n\}$ such that $a_{j},a_{j}+a_{j+1},...,a_{j}+a_{j+1}+...+a_{j+n-1}$ are all... | [
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-0.00... | 2025-08-11 |
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