Found problems: 85335
Determine the maximum possible number of points you can place in a rectangle with lengths $14$ and $28$ such that any two of those points are more than $10$ apart from each other.
Luna and Sam have access to a windowsill with three plants. On the morning of January $1$, $2011$, the plants were sitting in the order of cactus, dieffenbachia, and orchid, from left to right. Every afternoon, when Luna waters the plants, she swaps the two plants sitting on the left and in the center. Every evening, when Sam waters the plants, he swaps the two plants sitting on the right and in the center. What was the order of the plants on the morning of January $1$, $2012$, $365$ days later, from left to right?
$\text{(A) cactus, orchid, dieffenbachia}\qquad\text{(B) dieffenbachia, cactus, orchid}$
$\text{(C) dieffenbachia, orchid, cactus}\qquad\text{(D) orchid, dieffenbachia, cactus}$
$\text{(E) orchid, cactus, dieffenbachia}$
Given an acute triangle $ABC$. The incircle with center $I$ touches $BC,CA,AB$ at $D,E,F$ respectively. Let $M,N$ be the midpoint of the minor arc of $AB$ and $AC$ respectively. Prove that $M,F,E,N$ are collinear if and only if $\angle BAC =90$$^{\circ}$
Let $m > n$ be positive integers such that $3(3mn - 2)^2 - 2(3m -3n)^2 = 2019$. Find $3m + n$.
Let $a, b$ and $c$ be positive reals such that $a + b + c \ge abc$. Prove that $a^2 + b^2 + c^2 \ge \sqrt3 abc$.
Olya wrote fractions of the form $1 / n$ on cards, where $n$ is all possible divisors the numbers $6^{100}$ (including the unit and the number itself). These cards she laid out in some order. After that, she wrote down the number on the first card, then the sum of the numbers on the first and second cards, then the sum of the numbers on the first three cards, etc., finally, the sum of the numbers on all the cards. Every amount Olya recorded on the board in the form of irreducible fraction. What is the least different denominators could be on the numbers on the board?
Find the sum of the real roots of the polynomial \[ \prod_{k=1}^{100} \left( x^2-11x+k \right) = \left( x^2-11x+1 \right)\left( x^2-11x+2 \right)\dots\left(x^2-11x+100\right). \][i]Proposed by Evan Chen[/i]
A stage course is attended by $n \ge 4$ students. The day before the final exam, each group of three students conspire against another student to throw him/her out of the exam. Prove that there is a student against whom there are at least $\sqrt[3]{(n-1)(n- 2)} $conspirators.
Determine all three primes $(a, b, c)$ that satisfied the equality $a^2+ab+b^2=c^2+3$.
$16$ progamers are playing in another single elimination tournament. Each round, each of the remaining progamers plays against another and the loser is eliminated. Additionally, each time a progamer wins, he will have a ceremony to celebrate. A player's rst ceremony is ten seconds long, and afterward each ceremony is ten seconds longer than the last. What is the total length in seconds of all the ceremonies over the entire tournament?
Prove that for any prime number $p$ the equation $2^p+3^p=a^n$ has no solution $(a,n)$ in integers greater than $1$.
$\vartriangle ABC$ is a triangle and $I_b. I_c$ its excenters opposite to $B,C$. Prove that $\vartriangle ABC$ is right at $A$ if and only if its area is equal to $\frac12 AI_b \cdot AI_c$.
Each of the three large squares shown below is the same size. Segments that intersect the sides of the squares intersect at the midpoints of the sides. How do the shaded areas of these squares compare?
[asy]
unitsize(36);
fill((0,0)--(1,0)--(1,1)--cycle,gray);
fill((1,1)--(1,2)--(2,2)--cycle,gray);
draw((0,0)--(2,0)--(2,2)--(0,2)--cycle);
draw((1,0)--(1,2));
draw((0,0)--(2,2));
fill((3,1)--(4,1)--(4,2)--(3,2)--cycle,gray);
draw((3,0)--(5,0)--(5,2)--(3,2)--cycle);
draw((4,0)--(4,2));
draw((3,1)--(5,1));
fill((6,1)--(6.5,0.5)--(7,1)--(7.5,0.5)--(8,1)--(7.5,1.5)--(7,1)--(6.5,1.5)--cycle,gray);
draw((6,0)--(8,0)--(8,2)--(6,2)--cycle);
draw((6,0)--(8,2));
draw((6,2)--(8,0));
draw((7,0)--(6,1)--(7,2)--(8,1)--cycle);
label("$I$",(1,2),N);
label("$II$",(4,2),N);
label("$III$",(7,2),N); [/asy]
$\text{(A)}\ \text{The shaded areas in all three are equal.}$
$\text{(B)}\ \text{Only the shaded areas of }I\text{ and }II\text{ are equal.}$
$\text{(C)}\ \text{Only the shaded areas of }I\text{ and }III\text{ are equal.}$
$\text{(D)}\ \text{Only the shaded areas of }II\text{ and }III\text{ are equal.}$
$\text{(E)}\ \text{The shaded areas of }I, II\text{ and }III\text{ are all different.}$
Given a digits {$0,1,2,...,9$} . Find the number of numbers of 6 digits which cantain $7$ or $7$'s digit and they is permulated(For example 137456 and 314756 is one numbers).
The number of real solutions to the equation \[ \frac{x}{100} = \sin x \] is
$\text{(A)} \ 61 \qquad \text{(B)} \ 62 \qquad \text{(C)} \ 63 \qquad \text{(D)} \ 64 \qquad \text{(E)} \ 65$
Write $2013$ as a sum of $m$ prime numbers. The smallest value of $m$ is:
(A): $2$, (B): $3$, (C): $4$, (D): $1$, (E): None of the above.
Let $ABC$ be a triangle. Let $D, E, F$ be points respectively on the segments $BC, CA, AB$ such that $AD, BE, CF$ concur at the point $K$. Suppose $\frac{BD}{DC} = \frac {BF}{FA}$ and $\angle ADB = \angle AFC$. Prove that $\angle ABE = \angle CAD$.
The circle of center $I$ is inscribed in the convex quadrilateral $ABCD$. Let $M$ and $N$ be points on the segments $AI$ and $CI$, respectively, such that $\angle MBN = \frac 12 \angle ABC$. Prove that $\angle MDN = \frac 12 \angle ADC$.
Prove that if $n$ is a positive integer such that the equation \[ x^3-3xy^2+y^3=n \] has a solution in integers $x,y$, then it has at least three such solutions. Show that the equation has no solutions in integers for $n=2891$.
Let $n$ be a positive integer. Prove that $n$ is prime if and only if \[{{n-1}\choose k}\equiv (-1)^{k}\pmod{n}\] for all $k \in \{ 0, 1, \cdots, n-1 \}$.
Let $n>1$ be an integer and let $a_0,a_1,\ldots,a_n$ be non-negative real numbers. Definite $S_k=\sum_{i\equal{}0}^k \binom{k}{i}a_i$ for $k=0,1,\ldots,n$. Prove that\[\frac{1}{n} \sum_{k\equal{}0}^{n-1} S_k^2-\frac{1}{n^2}\left(\sum_{k\equal{}0}^{n} S_k\right)^2\le \frac{4}{45} (S_n-S_0)^2.\]
A positive integer \( n \) satisfies the following conditions:
[list]
[*] The number \( n \) has exactly \( 60 \) divisors: \( 1 = a_1 < a_2 < \cdots < a_{60} = n \);
[*] The number \( n+1 \) also has exactly \( 60 \) divisors: \( 1 = b_1 < b_2 < \cdots < b_{60} = n+1 \).
[/list]
Let \( k \) be the number of indices \( i \) such that \( a_i < b_i \). Find all possible values of \( k \).
[i]Note: Such numbers exist, for example, the numbers \( 4388175 \) and \( 4388176 \) both have \( 60 \) divisors.[/i]
[i]Proposed by Anton Trygub[/i]
is there a function $f:\mathbb{N}\rightarrow \mathbb{N}$ such that
$i) \exists n\in \mathbb{N}:f(n)\neq n$
$ii)$ the number of divisors of $m$ is $f(n)$ if and only if the number of divisors of $f(m)$ is $n$
Let $s(n)$ denote the smallest prime divisor and $d(n)$ denote the number of positive divisors of a positive integer $n>1$. Is it possible to choose $2023$ positive integers $a_{1},a_{2},...,a_{2023}$ with $a_{1}<a_{2}-1<...<a_{2023}-2022$ such that for all $k=1,...,2022$ we have $d(a_{k+1}-a_{k}-1)>2023^{k}$ and $s(a_{k+1}-a_{k}) > 2023^{k}$?
[i]Authored by Nikola Velov[/i]
In the adjoining diagram, $BO$ bisects $\angle CBA$, $CO$ bisects $\angle ACB$, and $MN$ is parallel to $BC$. If $AB=12$, $BC=24$, and $AC=18$, then the perimeter of $\triangle AMN$ is
[asy]
size(200);
defaultpen(linewidth(0.7)+fontsize(10));
pair B=origin, C=(24,0), A=intersectionpoints(Circle(B,12), Circle(C,18))[0], O=incenter(A,B,C), M=intersectionpoint(A--B, O--O+40*dir(180)), N=intersectionpoint(A--C, O--O+40*dir(0));
draw(B--M--O--B--C--O--N--C^^N--A--M);
label("$A$", A, dir(90));
label("$B$", B, dir(O--B));
label("$C$", C, dir(O--C));
label("$M$", M, dir(90)*dir(B--A));
label("$N$", N, dir(90)*dir(A--C));
label("$O$", O, dir(90));[/asy]
$\textbf {(A) } 30 \qquad \textbf {(B) } 33 \qquad \textbf {(C) } 36 \qquad \textbf {(D) } 39 \qquad \textbf {(E) } 42$