Found problems: 85335
PEN O Problems, 25
Let $A$ be a non-empty set of positive integers. Suppose that there are positive integers $b_{1}$, $\cdots$, $b_{n}$ and $c_{1}$, $\cdots$, $c_{n}$ such that [list] [*] for each $i$ the set $b_{i}A+c_{i}=\{b_{i}a+c_{i}\vert a \in A \}$ is a subset of $A$, [*] the sets $b_{i}A+c_{i}$ and $b_{j}A+c_{j}$ are disjoint whenever $i \neq j$.[/list] Prove that \[\frac{1}{b_{1}}+\cdots+\frac{1}{b_{n}}\le 1.\]
2005 All-Russian Olympiad Regional Round, 9.6
9.6, 10.6 Construct for each vertex of the trapezium a symmetric point wrt to the diagonal, which doesn't contain this vertex. Prove that if four new points form a quadrilateral then it is a trapezium.
([i]L. Emel'yanov[/i])
1986 Bulgaria National Olympiad, Problem 6
Let $0<k<1$ be a given real number and let $(a_n)_{n\ge1}$ be an infinite sequence of real numbers which satisfies $a_{n+1}\le\left(1+\frac kn\right)a_n-1$. Prove that there is an index $t$ such that $a_t<0$.
2025 Kosovo National Mathematical Olympiad`, P3
A subset $S$ of the natural numbers is called [i]dense [/i] for every $7$ consecutive natural numbers, at least $5$ of them are in $S$. Show that there exists a dense subset for which the equation $a^2+b^2=c^2$ has no solution for $a,b,c \in S$.
1997 Greece Junior Math Olympiad, 3
Establish if we can rewrite the numbers $1,2,3,4,5,6,7,8,9,10$ in a row in such a way that:
(a) The sum of any three consecutive numbers (in the new order) does not exceed $16$.
(b) The sum of any three consecutive numbers (in the new order) does not exceed $15$.
2012 Kyoto University Entry Examination, 1
Answer the following questions:
(1) Let $a$ be positive real number. Find $\lim_{n\to\infty} (1+a^{n})^{\frac{1}{n}}.$
(2) Evaluate $\int_1^{\sqrt{3}} \frac{1}{x^2}\ln \sqrt{1+x^2}dx.$
35 points
1989 USAMO, 2
The 20 members of a local tennis club have scheduled exactly 14 two-person games among themselves, with each member playing in at least one game. Prove that within this schedule there must be a set of 6 games with 12 distinct players.
2014 Dutch Mathematical Olympiad, 1
Determine all triples $(a,b,c)$, where $a, b$, and $c$ are positive integers that satisfy
$a \le b \le c$ and $abc = 2(a + b + c)$.
2010 ISI B.Math Entrance Exam, 5
Let $a_1>a_2>.....>a_r$ be positive real numbers .
Compute $\lim_{n\to \infty} (a_1^n+a_2^n+.....+a_r^n)^{\frac{1}{n}}$
2003 VJIMC, Problem 2
Let $A=(a_{ij})$ be an $m\times n$ real matrix with at least one non-zero element. For each $i\in\{1,\ldots,m\}$, let $R_i=\sum_{j=1}^na_{ij}$ be the sum of the $i$-th row of the matrix $A$, and for each $j\in\{1,\ldots,n\}$, let $C_j =\sum_{i=1}^ma_{ij}$ be the sum of the $j$-th column of the matrix $A$. Prove that there exist indices $k\in\{1,\ldots,m\}$ and $l\in\{1,\ldots,n\}$ such that
$$a_{kl}>0,\qquad R_k\ge0,\qquad C_l\ge0,$$or
$$a_{kl}<0,\qquad R_k\le0,\qquad C_l\le0.$$
LMT Speed Rounds, 2011.5
The unit of a screw is listed as $0.2$ cents. When a group of screws is sold to a customer, the total cost of the screws is computed with the listed price and then rounded to the nearest cent. If Al has $50$ cents and wishes to only make one purchase, what is the maximum possible number of screws he can buy?
2024 AMC 10, 15
A list of 9 real numbers consists of $1$, $2.2 $, $3.2 $, $5.2 $, $6.2 $, $7$, as well as $x, y,z$ with $x\leq y\leq z$. The range of the list is $7$, and the mean and median are both positive integers. How many ordered triples $(x,y,z)$ are possible?
$
\textbf{(A) }1 \qquad
\textbf{(B) }2 \qquad
\textbf{(C) }3 \qquad
\textbf{(D) }4 \qquad
\textbf{(E) infinitely many}\qquad
$
2022 MIG, 18
Two equilateral triangles are glued, and their opposite vertices are connected. If the larger equilateral triangle has an area of $225$ and the smaller equilateral triangle has an area of $100$, what is the area of the shaded region?
[asy]
size(4cm);
draw((0,0)--(3,0)--(3/2,3sqrt(3)/2)--(0,0));
draw((0,0)--(2,0)--(1,-sqrt(3))--(0,0));
draw((1,-sqrt(3))--(3/2,3sqrt(3)/2));
filldraw((0,0)--(6/5,0)--(3/2,3sqrt(3)/2)--cycle, gray);
[/asy]
$\textbf{(A) }60\qquad\textbf{(B) }90\qquad\textbf{(C) }96\qquad\textbf{(D) }108\qquad\textbf{(E) }120$
2014 China Team Selection Test, 5
Find the smallest positive constant $c$ satisfying: For any simple graph $G=G(V,E)$, if $|E|\geq c|V|$, then $G$ contains $2$ cycles with no common vertex, and one of them contains a chord.
Note: The cycle of graph $G(V,E)$ is a set of distinct vertices ${v_1,v_2...,v_n}\subseteq V$, $v_iv_{i+1}\in E$ for all $1\leq i\leq n$ $(n\geq 3, v_{n+1}=v_1)$; a cycle containing a chord is the cycle ${v_1,v_2...,v_n}$, such that there exist $i,j, 1< i-j< n-1$, satisfying $v_iv_j\in E$.
2010 Today's Calculation Of Integral, 611
Let $g(t)$ be the minimum value of $f(x)=x2^{-x}$ in $t\leq x\leq t+1$.
Evaluate $\int_0^2 g(t)dt$.
[i]2010 Kumamoto University entrance exam/Science[/i]
2009 South East Mathematical Olympiad, 5
Let $X=(x_1,x_2,......,x_9)$ be a permutation of the set $\{1,2,\ldots,9\}$ and let $A$ be the set of all such $X$ .
For any $X \in A$, denote $f(X)=x_1+2x_2+\cdots+9x_9$ and $ M=\{f(X)|X \in A \}$. Find $|M|$. ($|S|$ denotes number of members of the set $S$.)
1989 AMC 12/AHSME, 20
Let $x$ be a real number selected uniformly at random between 100 and 200. If $\lfloor {\sqrt{x}} \rfloor = 12$, find the probability that $\lfloor {\sqrt{100x}} \rfloor = 120$. ($\lfloor {v} \rfloor$ means the greatest integer less than or equal to $v$.)
$\text{(A)} \ \frac{2}{25} \qquad \text{(B)} \ \frac{241}{2500} \qquad \text{(C)} \ \frac{1}{10} \qquad \text{(D)} \ \frac{96}{625} \qquad \text{(E)} \ 1$
2004 National Chemistry Olympiad, 59
What substance is formed when $\ce{CF2=CF2}$ is polymerized?
$ \textbf{(A) } \text{Polyethylene} \qquad\textbf{(B) } \text{Polyurethane}\qquad\textbf{(C) } \text{PVC}\qquad\textbf{(D) } \text{Teflon}\qquad$
2005 France Team Selection Test, 3
In an international meeting of $n \geq 3$ participants, 14 languages are spoken. We know that:
- Any 3 participants speak a common language.
- No language is spoken more that by the half of the participants.
What is the least value of $n$?
2014 Saudi Arabia Pre-TST, 2.4
Let $ABC$ be an acute triangle with $\angle A < \angle B \le \angle C$, and $O$ its circumcenter. The perpendicular bisector of side $AB$ intersects side $AC$ at $D$. The perpendicular bisector of side $AC$ intersects side $AB$ at $E$. Express the angles of triangle $DEO$ in terms of the angles of triangle $ABC$.
2018 Chile National Olympiad, 1
Is it possible to choose five different positive integers so that the sum of any three of them is a prime number?
1913 Eotvos Mathematical Competition, 2
Let $O$ and $O'$ designate two dìagonally opposite vertices of a cube. Bisect those edges of the cube that contain neither of the points $O$ and $O'$. Prove that these midpoints of edges lie in a plane and form the vertices of a regular hexagon
2009 Today's Calculation Of Integral, 454
Let $ a$ be positive constant number. Evaluate $ \int_{ \minus{} a}^a \frac {x^2\cos x \plus{} e^{x}}{e^{x} \plus{} 1}\ dx.$
2013 Sharygin Geometry Olympiad, 2
Two circles with centers $O_1$ and $O_2$ meet at points $A$ and $B$. The bisector of angle $O_1AO_2$ meets the circles for the second time at points $C $and $D$. Prove that the distances from the circumcenter of triangle $CBD$ to $O_1$ and to $O_2$ are equal.
2023 LMT Fall, 5C
In equilateral triangle $ABC$, $AB=2$ and $M$ is the midpoint of $AB$. A laser is shot from $M$ in a certain direction, and whenever it collides with a side of $ABC$ it will reflect off the side such that the acute angle formed by the incident ray and the side is equal to the acute angle formed by the reflected ray and the side. Once the laser coincides with a vertex, it stops. Find the sum of the smallest three possible integer distances that the laser could have traveled.
[i]Proposed by Jerry Xu[/i]
[hide=Solution]
[i]Solution.[/i] $\boxed{21}$
Whenever the laser hits a side of the triangle, reflect the laser's path over that side so that the path of the laser forms a straight line. We want the path of the laser to coincide with a vertex of one of the reflected triangles. Thus, we can restate the problem as follows:
Tessellate the plane with equilateral triangles of side length $3$. Consider one of these equilateral triangles $ABC$ with $M$ being the midpoint of $AB=2$. Find the sum of the three minimum integer distances from $M$ to any vertex in the plane.
[asy]
import geometry;
size(8cm);
pair A = (0,sqrt(3));
pair B = (-1,0);
pair C = (1,0);
pair M = (0,0);
for (int i = -1; i <= 2; ++i) {
draw((i-3,i*sqrt(3))--(-i+3,i*sqrt(3)));
draw(((i-1)*2,-sqrt(3))--(i+1,(2-i)*sqrt(3)));
draw((-i-1,(2-i)*sqrt(3))--((1-i)*2,-sqrt(3)));
}
draw(A--B--C--A, red);
dot(M);
label("$A$",A+(0,0.25),N);
label("$B$",B-(0.25,0),SW);
label("$C$",C+(0.25,0),SE);
label("$M$",M,S);
[/asy]
It is trivial to see that the vertical distance between $M$ and a given vertex is $n\sqrt{3}$ for $n \in \mathbb{N}^{0}$. If $n$ is even, the horizontal distance between $O$ and a given vertex is $1+2m$ for $m \in \mathbb{N}^{0}$. If $n$ is odd, the horizontal distance is $2m$ for $m \in \mathbb{N}^{0}$. We consider two separate cases:
$1.$ $n$ is even. We thus want to find $l \in \mathbb{N}$ such that
$$\left(n\sqrt{3}\right)^2+(1+2m)^2=l^2.$$Make the substitution $1+2m=k$ to get that
$$3n^2+k^2=l^2.$$Notice that these equations form a family of generalized Pell equations $y^2-3x^2=N$ with $N=k^2$. We can find some set of roots to these equations using the multiplicative principle: we will use this idea to find three small $l$ values, and that gives us an upper bound on what the three $l$ values can be. From there, a simple bash of lower $l$ values to see if solutions to each generalized Pell equation not given by the multiplicative principle exist finishes this case.
By the multiplicative principle some set of solutions $(x_n,y_n)$ to the above equation with sufficiently small $x_n$ follow the formula$$x_n\sqrt{3}+y_n=\left(x_0\sqrt{3}+y_0\right)\left(u_n\sqrt{3}+v_n\right),$$where $\left(x_0,y_0\right)$ is a solution to the generalized Pell equation and $\left(u_n,v_n\right)$ are solutions to the Pell equation $y^2-3x^2=1$. Remember that the solutions to this last Pell equation satisfy$$u_n\sqrt{3}+v_n=\left(u_0\sqrt{3}+v_0\right)^k$$where the trivial positive integer solution
$$\left(u_0, v_0\right)=(1,2)$$(this can easily be found by inspection or by taking the convergents of the continued fraction expansion of $\sqrt{3}$).
We thus get that$$\left(u_1,v_1\right)=(4,7),\left(u_2,v_2\right)=(15,26),\left(u_2,v_2\right)=(56,97)\dots$$(also don't forget that $(u,v)=(0,1)$ is another solution).
From here, note that $k$ must be odd since $k=1+2m$ for $m \in \mathbb{N}^{0}$. For $k=1$, the smallest three solutions to the Pell equation with $n$ even are
\begin{align*}
(x,y)&=(0,1),(4,7),(56,97) \\
\longrightarrow (n,m,l)&=(0,0,1),(4,0,7),(56,0,97)
\end{align*}Our current smallest three values of $l$ are thus $1,7,97$. A quick check confirms that all of these solutions are not extraneous (extraneous solutions appear when the path taken by the laser prematurely hits a vertex).
For $k=3$, using the multiplicative principle we get two new smaller solutions
\begin{align*}
(x,y)&=(0,3),(12,21) \\
\longrightarrow (n,m,l)&=(0,1,3),(12,1,21)
\end{align*}However, note that $(n,m,l)=(0,1,3)$ is extraneous since is equivalent to the path that is traced out by the solution $(n,m,l)=(0,0,1)$ found previously and will thus hit a vertex prematurely. Thus, our new three smallest values of $l$ are $1,7,21$.
For $k \ge 5$, it is evident that there are no more smaller integral values of $l$ that can be found using the multiplicative principle: the solution set $(n,m,l)=\left(0,\dfrac{k-1}{2},k\right)$ is always extraneous for $k > 1$ since it is equivalent to the path traced out by $(0,0,1)$ as described above, and any other solutions will give larger values of $l$.
Thus, we now only need to consider solutions to each generalized Pell equation not found by the multiplicative principle. A quick bash shows that $l=3,5,9,11$ gives no solutions for any odd $k$ and even $n$, however $n=13$ gives $k=11$ and $n=4$, a non-extraneous solution smaller than one of the three we currently have. Thus, our new three smallest $l$ values are $1,7,13$.
$2$. $n$ is odd. We thus want to find $l \in \mathbb{N}$ such that
$$\left(n\sqrt{3}\right)^2+(2m)^2=l^2.$$Make the substitution $2m=k$ to get that
$$3n^2+k^2=l^2.$$This is once again a family of generalized Pell equations with $N=k^2$, however this time we must have $k$ even instead of $k$ odd. However, note that there are no solutions to this family of Pell equation with $n$ odd: $k^2 \equiv 0 \text{ (mod }4)$ since $k$ is even, and $3n^2 \equiv 3 \text{ (mod }4)$ since $n$ is odd, however $0+3 \equiv 3 \text{ (mod }4)$ is not a possible quadratic residue mod $4$. Thus, this case gives no solutions.
Our final answer is thus $1+7+13=\boxed{21}$.
[/hide]