Found problems: 85335
2019 Iran Team Selection Test, 3
Point $P$ lies inside of parallelogram $ABCD$. Perpendicular lines to $PA,PB,PC$ and $PD$ through $A,B,C$ and $D$ construct convex quadrilateral $XYZT$. Prove that $S_{XYZT}\geq 2S_{ABCD}$.
[i]Proposed by Siamak Ahmadpour[/i]
2012 IFYM, Sozopol, 5
Let $\sum_{i=1}^n a_i x_i =0$, $a_i,x_i\in \mathbb{Z}$. It is known that however we color $\mathbb{Z}$ with finite number of colors, then the given equation has a monochromatic (of one color) solution. Prove that there is some non-empty sum of its coefficients equal to 0.
1982 AMC 12/AHSME, 6
The sum of all but one of the interior angles of a convex polygon equals $2570^\circ$. The remaining angle is
$\textbf{(A)} \ 90^\circ \qquad \textbf{(B)} \ 105^\circ \qquad \textbf{(C)} \ 120^\circ \qquad \textbf{(D)} \ 130^\circ \qquad \textbf{(E)} \ 144^\circ$
2014 Harvard-MIT Mathematics Tournament, 9
There is a heads up coin on every integer of the number line. Lucky is initially standing on the zero point of the number line facing in the positive direction. Lucky performs the following procedure:
$\bullet$ Lucky looks at the coin (or lack thereof) underneath him.
$\bullet \, - \, $ If the coin is heads, Lucky flips it to tails up, turns around, and steps forward a distance of one unit.
$ \qquad a -$ If the coin is tails, Lucky picks up the coin and steps forward a distance of one unit facing the same direction.
$ \qquad a -$ If there is no coin, Lucky places a coin heads up underneath him and steps forward a distance of one unit facing the same direction.
He repeats this procedure until there are 20 coins anywhere that are tails up. How many times has Lucky performed the procedure when the process stops?
2017 Switzerland - Final Round, 10
Let $x, y, z$ be nonnegative real numbers with $xy + yz + zx = 1$. Show that:
$$\frac{4}{x + y + z} \le (x + y)(\sqrt3 z + 1).$$
1962 AMC 12/AHSME, 36
If both $ x$ and $ y$ are both integers, how many pairs of solutions are there of the equation $ (x\minus{}8)(x\minus{}10) \equal{} 2^y?$
$ \textbf{(A)}\ 0 \qquad
\textbf{(B)}\ 1 \qquad
\textbf{(C)}\ 2 \qquad
\textbf{(D)}\ 3 \qquad
\textbf{(E)}\ \text{more than 3}$
KoMaL A Problems 2020/2021, A. 791
A lightbulb is given that emits red, green or blue light and an infinite set $S$ of switches, each with three positions labeled red, green and blue. We know the following:
[list=1]
[*]For every combination of the switches the lighbulb emits a given color.
[*]If all switches are in a position with a given color, the lightbulb emits the same color.
[*]If there are two combinations of the switches where each switch is in a different position, the lightbulb emits a different color for the two combinations.
[/list]
We create the following set $U$ containing some of the subsets of $S$: for each combination of the switches let us observe the color of the lightbulb, and put the set of those switches in $U$ which are in the same position as the color of the lightbulb.
Prove that $U$ is an ultrafilter on $S$. In other words, prove that $U$ satisfies the following conditions:
[list=1]
[*]The empty set is not in $U.$
[*]If two sets are in $U,$ their intersection is also in $U.$
[*]If a set is in $U,$ every subset of $S$ containing it is also in $U.$
[*]Considering a set and its complement in $S,$ exactly one of these sets is contained in $U.$
[/list]
2020 Iran Team Selection Test, 6
$p$ is an odd prime number. Find all $\frac{p-1}2$-tuples $\left(x_1,x_2,\dots,x_{\frac{p-1}2}\right)\in \mathbb{Z}_p^{\frac{p-1}2}$ such that
$$\sum_{i = 1}^{\frac{p-1}{2}} x_{i} \equiv \sum_{i = 1}^{\frac{p-1}{2}} x_{i}^{2} \equiv \cdots \equiv \sum_{i = 1}^{\frac{p-1}{2}} x_{i}^{\frac{p - 1}{2}} \pmod p.$$
[i]Proposed by Ali Partofard[/i]
2020 CHMMC Winter (2020-21), 1
A unit circle is centered at $(0, 0)$ on the $(x, y)$ plane. A regular hexagon passing through $(1, 0)$ is inscribed in the circle. Two points are randomly selected from the interior of the circle and horizontal lines are drawn through them, dividing the hexagon into at most three pieces. The probability that each piece contains exactly two of the hexagon's original vertices can be written as
\[ \frac{2\left(\frac{m\pi}{n}+\frac{\sqrt{p}}{q}\right)^2}{\pi^2} \]
for positive integers $m$, $n$, $p$, and $q$ such that $m$ and $n$ are relatively prime and $p$ is squarefree. Find $m+n+p+q$.
2013 NZMOC Camp Selection Problems, 4
Let $C$ be a cube. By connecting the centres of the faces of $C$ with lines we form an octahedron $O$. By connecting the centers of each face of $O$ with lines we get a smaller cube $C'$. What is the ratio between the side length of $C$ and the side length of $C'$?
2008 Mongolia Team Selection Test, 2
Given positive integers$ m,n$ such that $ m < n$. Integers $ 1,2,...,n^2$ are arranged in $ n \times n$ board. In each row, $ m$ largest number colored red. In each column $ m$ largest number colored blue. Find the minimum number of cells such that colored both red and blue.
2017 Romania Team Selection Test, P2
Determine all intergers $n\geq 2$ such that $a+\sqrt{2}$ and $a^n+\sqrt{2}$ are both rational for some real number $a$ depending on $n$
2021 Durer Math Competition (First Round), 3
Let $k_1$ and $k_2$ be two circles that are externally tangent at point $C$. We have a point $A$ on $k_1$ and a point $B$ on $k_2$ such that $C$ is an interior point of segment $AB$. Let $k_3$ be a circle that passes through points $A$ and $B$ and intersects circles $k_1$ and $k_2$ another time at points $M$ and $N$ respectively. Let $k_4$ be the circumscribed circle of triangle $CMN$. Prove that the centres of circles $k_1, k_2, k_3$ and $k_4$ all lie on the same circle.
Durer Math Competition CD 1st Round - geometry, 2016.C1
Let $P$ be an arbitrary point of the side line $AB$ of the triangle $ABC$. Mark the perpendicular projection of $P$ on the side lines $AC$ and $BC$ as $A_1$ and $B_1$ respectively. Denote $C_1$ he foot of the alttiude starting from $C$. Prove that the points $A_1$, $B_1$, $C_1$, $C$ and $P$ lie on a circle.
2014 Contests, 3
Positive real numbers $a, b, c$ satisfy $\frac{1}{a} +\frac{1}{b} +\frac{1}{c} = 3.$ Prove the inequality \[\frac{1}{\sqrt{a^3+ b}}+\frac{1}{\sqrt{b^3 + c}}+\frac{1}{\sqrt{c^3 + a}}\leq \frac{3}{\sqrt{2}}.\]
2011 Today's Calculation Of Integral, 743
Evaluate $\int_0^{\frac{\pi}{2}} \ln (1+\sqrt[3]{\sin \theta})\cos \theta\ d\theta.$
1986 Bundeswettbewerb Mathematik, 4
The sequence $a_1, a_2, a_3,...$ is defined by $$a_1 = 1\,\,\,, \,\,\,a_{n+1} =\frac{1}{16}(1 + 4a_n +\sqrt{1 + 24a_n}) \,\,\,(n \in N^* ).$$ Determine and prove a formula with which for every natural number $n$ the term $a_n$ can be computed directly without having to determine preceding terms of the sequence.
2016 BMT Spring, 19
Regular tetrahedron $P_1P_2P_3P_4$ has side length $1$. Define $P_i$ for $i > 4$ to be the centroid of tetrahedron $P_{i-1}P_{i-2}P_{i-3}P_{i-4}$, and $P_{ \infty} = \lim_{n\to \infty} P_n$. What is the length of $P_5P_{ \infty}$?
1979 IMO Longlists, 80
Prove that the functional equations
\[f(x + y) = f(x) + f(y),\]
\[ \text{and} \qquad f(x + y + xy) = f(x) + f(y) + f(xy) \quad (x, y \in \mathbb R)\]
are equivalent.
LMT Theme Rounds, 2023F 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]
2007 Balkan MO Shortlist, G3
Let $ A_{1}A_{2}A_{3}A_{4}A_{5}$ be a convex pentagon, such that
\[ [A_{1}A_{2}A_{3}] \equal{} [A_{2}A_{3}A_{4}] \equal{} [A_{3}A_{4}A_{5}] \equal{} [A_{4}A_{5}A_{1}] \equal{} [A_{5}A_{1}A_{2}].\]
Prove that there exists a point $ M$ in the plane of the pentagon such that
\[ [A_{1}MA_{2}] \equal{} [A_{2}MA_{3}] \equal{} [A_{3}MA_{4}] \equal{} [A_{4}MA_{5}] \equal{} [A_{5}MA_{1}].\]
Here $ [XYZ]$ stands for the area of the triangle $ \Delta XYZ$.
1988 Mexico National Olympiad, 7
Two disjoint subsets of the set $\{1,2, ... ,m\}$ have the same sums of elements. Prove that each of the subsets $A,B$ has less than $m / \sqrt2$ elements.
2023 VN Math Olympiad For High School Students, Problem 6
Prove that these polynomials are irreducible in $\mathbb{Q}[x]:$
a) $\frac{{{x^p}}}{{p!}} + \frac{{{x^{p - 1}}}}{{(p - 1)!}} + ... + \frac{{{x^2}}}{2} + x + 1,$ with $p$ is a prime number.
b) $x^{2^n}+1,$ with $n$ is a positive integer.
2005 Moldova Team Selection Test, 4
$n$ is a positive integer, $K$ the set of polynoms of real variables $x_1,x_2,...,x_{n+1}$ and $y_1,y_2,...,y_{n+1}$, function $f:K\rightarrow K$ satisfies
\[f(p+q)=f(p)+f(q),\quad f(pq)=f(p)q+pf(q),\quad (\forall)p,q\in K.\]
If $f(x_i)=(n-1)x_i+y_i,\quad f(y_i)=2ny_i$ for all $i=1,2,...,n+1$ and
\[\prod_{i=1}^{n+1}(tx_i+y_i)=\sum_{i=0}^{n+1}p_it^{n+1-i}\]
for any real $t$, prove, that for all $k=1,...,n+1$
\[f(p_{k-1})=kp_k+(n+1)(n+k-2)p_{k-1}\]
2017 Bosnia Herzegovina Team Selection Test, 6
Given is an acute triangle $ABC$. $M$ is an arbitrary point at the side $AB$ and $N$ is midpoint of $AC$. The foots of the perpendiculars from $A$ to $MC$ and $MN$ are points $P$ and $Q$. Prove that center of the circumcircle of triangle $PQN$ lies on the fixed line for all points $M$ from the side $AB$.