Found problems: 85335
Point $A$ is outside of a given circle $\omega$. Let the tangents from $A$ to $\omega$ meet $\omega$ at $S, T$ points $X, Y$ are midpoints of $AT, AS$ let the tangent from $X$ to $\omega$ meet $\omega$ at $R\neq T$. points $P, Q$ are midpoints of $XT, XR$ let $XY\cap PQ=K, SX\cap TK=L$ prove that quadrilateral $KRLQ$ is cyclic.
Consider a polyhedron having $100$ edges.
(a) Find the maximal possible number of its edges which can be intersected by a plane (not containing any vertices of the polyhedron) if the polyhedron is convex.
(b) Prove that for a non-convex polyhedron this number
i. can be as great as $96$,
ii. cannot be as great as $100$.
(A Andjans, Riga
Show that for real numbers $x_1, x_2, ... , x_n$ we have:
\[ \sum\limits_{i=1}^n \sum\limits_{j=1}^n \dfrac{x_ix_j}{i+j} \geq 0 \]
When do we have equality?
Let $f : R \to R$ be a function satisfying the equation $f(x^2 + x + 3) + 2f(x^2 - 3x + 5) =6x^2 - 10x + 17$ for all real numbers $x$. What is the value of $f(85)$?
[b]Problem 3[/b]
Let $ P(x) = x^{2015} -2x^{2014}+1$ and $ Q(x) = x^{2015} -2x^{2014}-1$. Determine for each of the polynomials
$P$ and $Q$ whether it is a divisor of some nonzero polynomial $c_0 + c_{1}x +\ldots + c_{n}x^n$
n whose coefficients $c_i$ are all in the set $ \{ -1, 1\}$.
Solve the following system of equations in natural numbers
$$\begin{cases} a^4+14ab+1=n^4 \\ b^4+14bc+1=m^4 \\ c^4+14ca+1=k^4 \end{cases}$$
Is it possible to divide a polygon with $21$ sides into $2022$ triangles in such a way that among all the vertices there are not three collinear?
A survey of participants was conducted at the Olympiad. $50\%$ of the participants liked the first round, $60\%$ of the participants liked the second round, $70\%$ of the participants liked the opening of the Olympiad. It is known that each participant liked either one option or all three. Determine the percentage of participants who rated all three events positively.
Let $f : Z_{>0} \to Z_{>0}$ be a function which satisfies $k|f^k(x)-x$ for all $k, x \in Z_{>0}$ and $f(x)-x \le 2023$. If $f(1) = 2000$, what can $f$ be?
[i]Remark[/i]: Here, $f^k (x)$ denotes the $k$-fold application of $f$ to $x$.
[b]Problem 3.[/b] Two points $M$ and $N$ are chosen inside a non-equilateral triangle $ABC$ such that $\angle BAM=\angle CAN$, $\angle ABM=\angle CBN$ and \[AM\cdot AN\cdot BC=BM\cdot BN\cdot CA=CM\cdot CN\cdot AB=k\] for some real $k$. Prove that:
[b]a)[/b] We have $3k=AB\cdot BC\cdot CA$.
[b]b)[/b] The midpoint of $MN$ is the medicenter of $\triangle ABC$.
[i]Remark.[/i] The [b]medicenter[/b] of a triangle is the intersection point of the three medians:
If $A_{1}$ is midpoint of $BC$, $B_{1}$ of $AC$ and $C_{1}$ of $AB$, then $AA_{1}\cap BB_{1}\cap CC_{1}= G$, and $G$ is called medicenter of triangle $ABC$.
[i] Nikolai Nikolov[/i]
a) The fields of the square table $4\times 4$ are filled with the "+" or "-" signs. You are allowed to change the signs simultaneously in the whole row, column, or diagonal to the opposite sign. That means, for example, that You can change the sign in the corner square, because it makes a diagonal itself. Prove that you will never manage to obtain a table containing pluses only.
b) The fields of the square table $8\times 8$ are filled with the "+" or signs except one non-corner field with "-". You are allowed to change the signs simultaneously in the whole row, column, or diagonal to the opposite sign. That means, for example, that You can change the sign in the corner field, because it makes a diagonal itself. Prove that you will never manage to obtain a table containing pluses only.
Let $ x\in \mathbb R$. Find the minimum and maximum values of the expresion:
$ E\equal{}\dfrac{(1\plus{}x)^8\plus{}16x^4}{(1\plus{}x^2)^4}$
$(YUG 1)$ Suppose that positive real numbers $x_1, x_2, x_3$ satisfy
$x_1x_2x_3 > 1, x_1 + x_2 + x_3 <\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}$
Prove that:
$(a)$ None of $x_1, x_2, x_3$ equals $1$.
$(b)$ Exactly one of these numbers is less than $1.$
Let $ABCD$ be a square, $E$ be a point on the side $DC$, $F$ and $G$ be the feet of the altitudes from $B$ to $AE$ and from $A$ to $BE$, respectively. Suppose $DF$ and $CG$ intersect at $H$. Prove that $\angle AHB=90^\circ$.
Let $\triangle ABC$ be an acute triangle, and let $D,E,F$ respectively be three points on sides $BC,CA,AB$ such that $AEDF$ is a cyclic quadrilateral. Let $O_B$ and $O_C$ be the circumcenters of $\triangle BDF$ and $\triangle CDE$, respectively. Finally, let $D'$ be a point on segment $BC$ such that $BD'=CD$. Prove that $\triangle BD'O_B$ and $\triangle CD'O_C$ have the same surface.
Two rational numbers \(\tfrac{m}{n}\) and \(\tfrac{n}{m}\) are written on a blackboard, where \(m\) and \(n\) are relatively prime positive integers. At any point, Evan may pick two of the numbers \(x\) and \(y\) written on the board and write either their arithmetic mean \(\tfrac{x+y}{2}\) or their harmonic mean \(\tfrac{2xy}{x+y}\) on the board as well. Find all pairs \((m,n)\) such that Evan can write 1 on the board in finitely many steps.
[i]Proposed by Yannick Yao[/i]
$60$ points are on the interior of a unit circle (a circle with radius $1$). Show that there exists a point $V$ on the circumference of the circle such that the sum of the distances from $V$ to the $60$ points is less than or equal to $80$.
Calculate $1 + 3 + 5 +\cdots+ 195 + 197 + 199$
Given $n \ge 3$ positive integers not exceeding $100$, let $d$ be their greatest common divisor. Show that there exist three of these numbers whose greatest common divisor is also equal to $d$.
Let $A=33\cdots3$, where $A$ contains $2009$ $3$s. Let $B=11\cdots1088\cdots89$, where $B$ contains $2008$ $1$s and $2008$ $8$s. Prove that $A^2=B$.
Given a triangle $ABC$, in which the angle $B$ is three times the angle $C$. On the side $AC$, point $D$ is chosen such that the angle $BDC$ is twice the angle $C$. Prove that $BD + BA = AC$.
Let $N_{14}$ be the answer to problem 14.
Rectangle $ABCD$ has area $\sqrt{2N_{14}}$. Points $E$, $F$, $G$, and $H$ lie on the rays $\overrightarrow{AB}$, $\overrightarrow{BC}$, $\overrightarrow{CD}$, and $\overrightarrow{DA}$, respectively, such that $EFGH$ is a rectangle with area $2\sqrt{2N_{14}}$ that contains all of $ABCD$ in its interior. If
\[ \tan\angle AEH = \tan\angle BFE = \tan\angle CGF = \tan\angle DHG = \sqrt{\frac{1}{48}}, \]
then $EG=\tfrac{m\sqrt{n}}{p}$, where $m$, $n$, and $p$ are positive integers, $m$ and $p$ are relatively prime, and $n$ is not divisible by the square of any prime. Compute $m + n + p$.
Equilateral triangle $\triangle ABC$ is inscribed in circle $\omega$ with radius $18.$ Circle $\omega_A$ is tangent to sides $\overline{AB}$ and $\overline{AC}$ and is internally tangent to $\omega$. Circles $\omega_B$ and $\omega_C$ are defined analogously. Circles $\omega_A$, $\omega_B$, and $\omega_C$ meet in six points$-$two points for each pair of circles. The three intersection points closest to the vertices of $\triangle ABC$ are the vertices of a large equilateral triangle in the interior of $\triangle ABC$, and the other three intersection points are the vertices of a smaller equilateral triangle in the interior of $\triangle ABC$. The side length of the smaller equilateral triangle can be written as $\sqrt{a}-\sqrt{b}$, where $a$ and $b$ are positive integers. Find $a+b$.
A [i]regular spatial pentagon[/i] consists of five points $P_1,P_2,P_3,P_4$ and $P_5$ in $\mathbb{R}^3$ such that $|P_iP_{i+1}|=|P_jP_{j+1}|$ and $\angle P_{i-1}P_iP_{i+1}=\angle P_{j-1}P_jP_{j+1}$ for all $1\leq i,\leq 5$, where $P_0=P_5$ and $P_{6}=P_{1}$. A regular spatial pentagon is [i]planar[/i] if there is a plane passing through all five points $P_1,P_2,P_3,P_4$ and $P_5$.
Show that every regular spatial pentagon is planar.
If $ f(x)\equal{}\frac{x(x\minus{}1)}{2}$, then $ f(x\plus{}2)$ equals:
$ \textbf{(A)}\ f(x)\plus{}f(2) \qquad\textbf{(B)}\ (x\plus{}2)f(x) \qquad\textbf{(C)}\ x(x\plus{}2)f(x) \qquad\textbf{(D)}\ \frac{xf(x)}{x\plus{}2}\\
\textbf{(E)}\ \frac{(x\plus{}2)f(x\plus{}1)}{x}$