This website contains problems from math contests. Problems and corresponding tags were obtained from the Art of Problem Solving website.

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Found problems: 85335

Let $L$ be the foot of the internal bisector of $\angle B$ in an acute-angled triangle $ABC.$ The points $D$ and $E$ are the midpoints of the smaller arcs $AB$ and $BC$ respectively in the circumcircle $\omega$ of $\triangle ABC.$ Points $P$ and $Q$ are marked on the extensions of the segments $BD$ and $BE$ beyond $D$ and $E$ respectively so that $\measuredangle APB=\measuredangle CQB=90^{\circ}.$ Prove that the midpoint of $BL$ lies on the line $PQ.$
Let circles $\omega$ and $\Gamma$, centered at $O_1$ and $O_2$ and having radii $42$ and $54$ respectively, intersect at points $X,Y$, such that $\angle O_1XO_2 = 105^{\circ}$. Points $A$, $B$ lie on $\omega$ and $\Gamma$ respectively such that $\angle O_1XA = \angle AXB = \angle BXO_2$ and $Y$ lies on both minor arcs $XA$ and $XB$. Define $P$ to be the intersection of $AO_2$ and $BO_1$. Suppose $XP$ intersects $AB$ at $C$. What is the value of $\frac{AC}{BC}$? [i]Proposed by Puhua Cheng[/i]
[b]a)[/b] prove that the function $\zeta(s)=\sum_{n=1}^\infty\frac{1}{n^s}$ that is defined on the area $Re(s)>1$, is an analytic function. [b]b)[/b] prove that the function $\zeta(s)-\frac{1}{s-1}$ can be spanned to an analytic function over $\mathbb C$. [b]c)[/b] using the span of part [b]b[/b] show that $\zeta(1-n)=-\frac{B_n}{n}$ that $B_n$ is the $n$th bernoli number that is defined by generating function $\frac{t}{e^t-1}=\sum_{n=0}^{\infty}B_n\frac{t^n}{n!}$.
Ivan claims that for all positive integers $n$, $$\left\lfloor\sqrt[2]{\frac{n}{1^3}}\right\rfloor + \left\lfloor\sqrt[2]{\frac{n}{2^3}}\right\rfloor + \left\lfloor\sqrt[2]{\frac{n}{3^3}}\right\rfloor + \cdots = \left\lfloor\sqrt[3]{\frac{n}{1^2}}\right\rfloor + \left\lfloor\sqrt[3]{\frac{n}{2^2}}\right\rfloor + \left\lfloor\sqrt[3]{\frac{n}{3^2}}\right\rfloor + \cdots$$ Why is he correct? (Note: $\lfloor x \rfloor$ denotes the floor function.)
Find all functions $f,g,h: \mathbb{R} \mapsto \mathbb{R}$ such that $f(x) - g(y) = (x-y) \cdot h(x+y)$ for $x,y \in \mathbb{R}.$
assume that we have a n*n table we fill it with 1,...,n such that each number exists exactly n times prove that there exist a row or column such that at least $\sqrt{n}$ diffrent number are contained.
How many positive integers less than $10,000$ have an even number of even digits and an odd number of odd digits ? (Assume no number starts with zero.)
Let $ ABC$ be a triangle and let $ \mathcal{M}_{a}$, $ \mathcal{M}_{b}$, $ \mathcal{M}_{c}$ be the circles having as diameters the medians $ m_{a}$, $ m_{b}$, $ m_{c}$ of triangle $ ABC$, respectively. If two of these three circles are tangent to the incircle of $ ABC$, prove that the third is tangent as well.
Given the monic polynomial \begin{align*} P(x) = x^N +a_{N-1}x^{N-1} + \ldots + a_1 x + a_0 \in \mathbb{R}[x] \end{align*} of even degree $N$ $=$ $2n$ and having all real positive roots $x_i$, for $1 \le i \le N$. Prove, for any $c$ $\in$ $[0, \underset{1 \le i \le N}{\min} \{x_i \} )$, the following inequality \begin{align*} c + \sqrt[N]{P(c)} \le \sqrt[N]{a_0} \end{align*}
In triangle $ABC$, $AB=13, BC=14, CA=15$. Let $\Omega$ and $\omega$ be the circumcircle and incircle of $ABC$ respectively. Among all circles that are tangent to both $\Omega$ and $\omega$, call those that contain $\omega$ [i]inclusive[/i] and those that do not contain $\omega$ [i]exclusive[/i]. Let $\mathcal{I}$ and $\mathcal{E}$ denote the set of centers of inclusive circles and exclusive circles respectively, and let $I$ and $E$ be the area of the regions enclosed by $\mathcal{I}$ and $\mathcal{E}$ respectively. The ratio $\frac{I}{E}$ can be expressed as $\sqrt{\frac{m}{n}}$, where $m$ and $n$ are relatively prime positive integers. Compute $100m+n$. [i]Proposed by Yannick Yao[/i]
Let $x,y$ and $a_0, a_1, a_2, \cdots $ be integers satisfying $a_0 = a_1 = 0$, and $$a_{n+2} = xa_{n+1}+ya_n+1$$for all integers $n \geq 0$. Let $p$ be any prime number. Show that $\gcd(a_p,a_{p+1})$ is either equal to $1$ or greater than $\sqrt{p}$.
Determine the number of three digit integers that are equal to $19$ times the sum of its digits.
Each point in the Euclidean space is colored with one of $n \ge 2$ colors, and each of the $n$ colors is used. Prove that one can find a triangle such that the color assigned to the orthocenter is different from all the colors assigned to the vertices of the triangle.
There are $30$ students in a class. In an examination, their results were all different from each other. It is given that everyone has the same number of friends. Find the maximum number of students such that each one of them has a better result than the majority of his friends. PS. Here majority means larger than half.
[b]2/1/32.[/b] Is it possible to fill in a $2020$ x $2020$ grid with the integers from $1$ to $4,080,400$ so that the sum of each row is $1$ greater than the previous row?
Do there exist two bounded sequences $a_1, a_2,\ldots$ and $b_1, b_2,\ldots$ such that for each positive integers $n$ and $m>n$ at least one of the two inequalities $|a_m-a_n|>1/\sqrt{n},$ and $|b_m-b_n|>1/\sqrt{n}$ holds?
Look at these fractions. At firs step we have $ \frac{0}{1}$ and $ \frac{1}{0}$, and at each step we write $ \frac{a\plus{}b}{c\plus{}d}$ between $ \frac{a}{b}$ and $ \frac{c}{d}$, and we do this forever \[ \begin{array}{ccccccccccccccccccccccccc}\frac{0}{1}&&&&&&&&\frac{1}{0}\\ \frac{0}{1}&&&&\frac{1}{1}&&&&\frac{1}{0}\\ \frac{0}{1}&&\frac{1}{2}&&\frac{1}{1}&&\frac{2}{1}&&\frac{1}{0}\\ \frac{0}{1}&\frac{1}{3}&\frac{1}{2}&\frac{2}{3}&\frac{1}{1}&\frac{3}{2}&\frac{2}{1}&\frac{3}{1}&\frac{1}{0}\\ &&&&\dots\end{array}\] a) Prove that each of these fractions is irreducible. b) In the plane we have put infinitely many circles of diameter 1, over each integer on the real line, one circle. The inductively we put circles that each circle is tangent to two adjacent circles and real line, and we do this forever. Prove that points of tangency of these circles are exactly all the numbers in part a(except $ \frac{1}{0}$). [img]http://i2.tinypic.com/4m8tmbq.png[/img] c) Prove that in these two parts all of positive rational numbers appear. If you don't understand the numbers, look at [url=http://upload.wikimedia.org/wikipedia/commons/2/21/Arabic_numerals-en.svg]here[/url].
On the lines $AB$ are located $2009$ different points that do not belong to the segment $[AB]$. Prove that the sum of the distances from point $A$ to these points is not equal to the sum of the distances from point $B$ to these points.
In quadrilateral $ABCD$ with diagonals $\overline{AC}$ and $\overline{BD}$ intersecting at $O$, $\overline{BO}=4$, $\overline{AO}=8$, $\overline{OC}=3$, and $\overline{AB}=6$. The length of $\overline{AD}$ is: $\textbf{(A)}\ 9\qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ 6\sqrt{3}\qquad \textbf{(D)}\ 8\sqrt{2}\qquad \textbf{(E)}\ \sqrt{166}$
Find the sum of all the positive integers which have at most three not necessarily distinct prime factors where the primes come from the set $\{ 2, 3, 5, 7 \}$.
Let $ a$ be a positive integer. Consider the sequence $ (a_n)$ defined as $ a_0\equal{}a$ and $ a_n\equal{}a_{n\minus{}1}\plus{}40^{n!}$ for $ n > 0$. Prove that the sequence $ (a_n)$ has infinitely many numbers divisible by $ 2009$.
Find integer $n$ such that both $n-86$ and $n + 86$ are perfect squares.
On a circle with diameter $AB$ we marked an arbitrary point $C$, which does not coincide with $A$ and $B$. The tangent to the circle at point $A$ intersects the line $BC$ at point $D$. Prove that the tangent to the circle at point $C$ bisects the segment $AD$.
suppose that $G<S_n$ is a subgroup of permutations of $\{1,...,n\}$ with this property that for every $e\neq g\in G$ there exist exactly one $k\in \{1,...,n\}$ such that $g.k=k$. prove that there exist one $k\in \{1,...,n\}$ such that for every $g\in G$ we have $g.k=k$.(20 points)
In a triangle ABC we have $\angle C = \angle A + 90^o$. The point $D$ on the continuation of $BC$ is given such that $AC = AD$. A point $E$ in the side of $BC$ in which $A$ doesn’t lie is chosen such that $\angle EBC = \angle A, \angle EDC = \frac{1}{2} \angle A$ . Prove that $\angle CED = \angle ABC$. by Morteza Saghafian