id int64 2 3.28M | problem stringlengths 27 6.88k | solution stringlengths 2 38.6k |
|---|---|---|
2,759,383 | Given an integer $n\geq2$, let $x_1<x_2<\cdots<x_n$ and $y_1<y_2<\cdots<y_n$ be positive reals. Prove that for every value $C\in (-2,2)$ (by taking $y_{n+1}=y_1$) it holds that
$\hspace{122px}\sum_{i=1}^{n}\sqrt{x_i^2+Cx_iy_i+y_i^2}<\sum_{i=1}^{n}\sqrt{x_i^2+Cx_iy_{i+1}+y_{i+1}^2}$.
[i]Proposed by Mirko Petrusevski[/... | We can use a similar argument as in the proof of rearrangement inequality. Letting $f(x,y)=\sqrt{x^2+Cxy+y^2}$, it suffices to show the case $n=2$, which corresponds to a single transposition in the general case.
Al we have to show that if $a<b$ and $c<d$, then
$$\sqrt{a^2+Cac+c^2}+\sqrt{b^2+Cbd+d^2}<\sqrt{a^2+Cad+d^2... |
2,759,385 | Find all triplets of positive integers $(x, y, z)$ such that $x^2 + y^2 + x + y + z = xyz + 1$.
[i]Proposed by Viktor Simjanoski[/i] | [hide="Solution (using Vieta's Jumping Root Method)"]
$\wedge$ means 'and'.
$\mathbb{N*}$ means $\{n | n\in \mathbb{Z} \wedge n>0\}$.
$(*)$ stands for the equation $x^2+y^2+x+y+z=xyz+1$.
Define $g(x,y):=\frac{x^2+y^2+x+y-1}{xy-1}$.
WLOG assume $x\geq y$.
$\textbf{Case 1.}$ $y=1$.
$x^2+x+z+2=xz+1 \Rightarrow (x-1)z=x^2+... |
2,759,387 | Find all positive integers $n$ such that the set $S=\{1,2,3, \dots 2n\}$ can be divided into $2$ disjoint subsets $S_1$ and $S_2$, i.e. $S_1 \cap S_2 = \emptyset$ and $S_1 \cup S_2 = S$, such that each one of them has $n$ elements, and the sum of the elements of $S_1$ is divisible by the sum of the elements in $S_2$.
... | We claim the answer is all $n \not\equiv 5 \pmod 6$. Let $\sum_{i \in S_1} i=A$ and $\sum_{i \in S_2} i=B$. Then, $A+B=n(2n+1)$ and $A \mid B$. Note that
$A \geq 1+2+\ldots+n=\dfrac{n(n+1)}{2}$
and
$B \leq 2n+(2n-1)+\ldots+(n+1)=\dfrac{n(3n+1)}{2}.$
Therefore,
$B \leq \dfrac{n(3n+1)}{2} <\dfrac{3n(n+1)}{2} \leq 3... |
2,759,392 | We say that a positive integer $n$ is [i]memorable[/i] if it has a binary representation with strictly more $1$'s than $0$'s (for example $25$ is memorable because $25=(11001)_{2}$ has more $1$'s than $0$'s). Are there infinitely many memorable perfect squares?
[i]Proposed by Nikola Velov[/i] | $n^2=2^k \cdot a_k + ... + 2^1 \cdot a_1 + 2^0 a_0$
Next number$$\boxed {(2^{k+2} + 1)n}$$ |
2,777,211 | Quadratic polynomials $P(x)$ and $Q(x)$ have leading coefficients of $2$ and $-2$, respectively. The graphs of both polynomials pass through the two points $(16,54)$ and $(20,53)$. Find ${P(0) + Q(0)}$. | Mine.
[hide="Solution"]The polynomial $R(x) := P(x) + Q(x)$ is linear, and we may compute $R(16) = 108$ and $R(20) = 106$. Hence $R(0) = \boxed{116}$.[/hide] |
2,777,219 | Let $w = \frac{\sqrt{3}+i}{2}$ and $z=\frac{-1+i\sqrt{3}}{2}$, where $i=\sqrt{-1}$. Find the number of ordered pairs $(r, s)$ of positive integers not exceeding $100$ that satisfy the equation $i\cdot w^r=z^s$. | [hide=Sketch]Rewrite in exponential form using $e^{i\theta}=\cos{\theta}+i\sin{\theta}$. Take cases on $s$ being $0, 1, 2$ mod $3$. This gives $\boxed{834}$, which makes sense as its ~$\frac{10000}{12}$. |
2,777,232 | Let $ABCD$ be a parallelogram with $\angle BAD < 90^{\circ}$. A circle tangent to sides $\overline{DA}$, $\overline{AB}$, and $\overline{BC}$ intersects diagonal $\overline{AC}$ at points $P$ and $Q$ with $AP < AQ$, as shown. Suppose that $AP = 3$, $PQ = 9$, and $QC = 16$. Then the area of $ABCD$ can be expressed in th... | Mine.
[hide="Solution"]
Let $X$, $Y$, and $Z$ denote the tangency points of $\omega$ with $AD$, $BC$, and $AB$, respectively. Furthermore, let $R$ be the foot of the perpendicular from $C$ to $AD$.
[asy]
import olympiad;
size(250);
defaultpen(linewidth(0.7)+fontsize(10));
real h = 5, r = 3*sqrt(3), s = 26;
pair A =... |
2,782,948 | Twenty distinct points are marked on a circle and labeled $1$ through $20$ in clockwise order. A line segment is drawn between every pair of points whose labels differ by a prime number. Find the number of triangles formed whose vertices are among the original $20$ points. | Of the three labels $a<b<c$ at the vertices of such a triangle, let $p_1=b-a$ be the distance between the smaller two, $p_2=c-b$ that between the larger two, and $p_3=c-a$ that between the smallest and largest. Then, since $(b-a)+(c-b)=(c-a)$, we have that $p_1+p_2=p_3$, where we wish $p_1$, $p_2$, and $p_3$ to be prim... |
2,793,705 | Ten birds land on a $10$-meter-long wire, each at a random point chosen uniformly along the wire. (That is, if we pick out any $x$-meter portion of the wire, there is an $\tfrac{x}{10}$ probability that a given bird will land there.) What is the probability that every bird sits more than one meter away from its closest... | The probability of placing the birds on a $10$-meter wire such that they are all more than a meter away from their neighbor is in one to one correspondence to the probability that they all land on the first meter of the $10$ meter wire, since from there we can insert $1$ meter between every pair of birds to get a worki... |
2,352,936 | For a given value $t$, we consider number sequences $a_1, a_2, a_3,...$ such that $a_{n+1} =\frac{a_n + t}{a_n + 1}$ for all $n \ge 1$.
(a) Suppose that $t = 2$. Determine all starting values $a_1 > 0$ such that $\frac43 \le a_n \le \frac32$ holds for all $n \ge 2$.
(b) Suppose that $t = -3$. Investigate whether $a_{... | [b](a)[/b] $a_{n+1} =\frac{a_n +2}{a_n + 1}$
$\frac43 \le \frac{a_1 +2}{a_1 + 1} \le \frac32$
$4a_1+4 \le 3a_1+6 \implies a_1\le 2$ and $2a_1+4 \le 3a_1+3 \implies 1\le a_1$, thus $2\geq a_1\geq 1$.
Suppose $\frac43 \le a_{n}\le \frac32$, then we want to show that for all $n\geq 1$ we have $\frac43 \le 1+\frac{1}{a_n ... |
2,780,701 | A natural number is called [i]chaotigal [/i] if it and its successor both have the sum of their digits divisible by $2021$. How many digits are in the smallest chaotigal number? | solved also [url=https://artofproblemsolving.com/community/c6h2731190p23794410]here[/url] |
2,810,680 | Let $P=(x^4-40x^2+144)(x^3-16x)$.
$a)$ Factor $P$ as a product of irreducible polynomials.
$b)$ We write down the values of $P(10)$ and $P(91)$. What is the greatest common divisor of the two numbers? | a)P(x)=x(x+6)(x-6)(x+4)(x-4)(x+2)(x-2) b)(P(1),P(91))=3^2*5*7=245 |
2,810,686 | Let $f(x)$ be a quadratic function with integer coefficients. If we know that $f(0)$, $f(3)$ and $f(4)$ are all different and elements of the set $\{2, 20, 202, 2022\}$, determine all possible values of $f(1)$. | $f(x) = ax^2+bx+c$
$f(0) = c $
$f(3) = 9a+3b+c $
$f(4) = 16a+4b+c$
Case $f(0)=2$
$f(3) = 9a+3b+2 \equiv 20 \mod 3 $ ( 2 cannot be used since f(0)=2)
$9a+3b=18,3a+b=6$
$f(4) = 16a+4b+2 \equiv (202,2022) \mod 4$
[rule]
subCase (f(4)=202)
$16a+4b=200$
$4a+b = 50$
$a+b+c = 44+6-44*3+2 = 8-44*2 = -80$
subCase (f(4) 2022)
$... |
2,810,687 | Let $\triangle ABC$ have median $CM$ ($M\in AB$) and circumcenter $O$. The circumcircle of $\triangle AMO$ bisects $CM$. Determine the least possible perimeter of $\triangle ABC$ if it has integer side lengths. | [hide=Solution]$\underline{\textbf{Claim:}}$ $AC=CM$
$\textit{Proof 1:}$ Let the midpoint of $CM$ be point $N$. Since $\angle OMA=90^{\circ}$ and $N\in (AMO)$, this means that $\angle ONA=90^{\circ}$. If $\overline{AN}\cap (ABC)=X$, since $ON\perp AN$, this implies that $N$ is the midpoint of $AX$. Now $NA=NX$ and $NC=... |
2,810,688 | Find all primes $p$, such that there exist positive integers $x$, $y$ which satisfy
$$\begin{cases}
p + 49 = 2x^2\\
p^2 + 49 = 2y^2\\
\end{cases}$$ | Wow, this can be solved exactly like USAMO 2022/4, held two days ago. So, here is a [hide=sketch] Note that $p>y$, otherwise $y \leq 7$ and case check. Now, subtracting the two equations, we obtain $p(p-1)=2(y-x)(y+x)$. Obviously $p=2$ doesn't work, and $p$ can't divide $y-x$. So $p|y+x$ and if $x+y$ is not equal to $p... |
2,810,691 | 14 students attend the IMO training camp. Every student has at least $k$ favourite numbers. The organisers want to give each student a shirt with one of the student's favourite numbers on the back. Determine the least $k$, such that this is always possible if:
$a)$ The students can be arranged in a circle such that eve... | This nice problem was proposed by Miroslav Marinov. Part a) is just for warm up. Obviously, $k=1$ is not enough and it's possible for $k=2$. The worst case scenario is when all the students have the same favorite numbers, say $1$ and $2$. In this case we arrange the numbers as $1-2-1-\dots 2$.
[b]Part b).[/b] $k=4$. ... |
2,810,692 | If $x, y, z \in \mathbb{R}$ are solutions to the system of equations
$$\begin{cases}
x - y + z - 1 = 0\\
xy + 2z^2 - 6z + 1 = 0\\
\end{cases}$$
what is the greatest value of $(x - 1)^2 + (y + 1)^2$? | $y-x=z-1$
$xy=-2z^2+6z-1$
$(y-x)^2+4xy\geq 0\Rightarrow \frac{1}{7}\leq z\leq 3$
$(x-1)^2+(y+1)^2=(y-x)^2+2xy+2(y-x)+2=z^2-2z+1-4z^2+12z-2+2z-2+2=-3z^2+12z-1=-3(z-2)^2+11$
its maximum value is $11$ when $z=2$. |
2,810,693 | Let $\triangle ABC$ have incenter $I$. The line $CI$ intersects the circumcircle of $\triangle ABC$ for the second time at $L$, and $CI=2IL$. Points $M$ and $N$ lie on the segment $AB$, such that $\angle AIM =\angle BIN = 90^{\circ}$. Prove that $AB=2MN$. | Interesting Problem.
Let $IH$ be perpendicular to $AB$. First Lets have some angle chasing stuff. $\frac{\angle A}{2} = \angle HAI = \angle HIM,\frac{\angle B}{2} = \angle HBI = \angle HIN \implies \angle AIN = \frac{\angle C}{2} = \angle BIM$.
Claim : $ANI$ and $AIC$ are similar.
Proof : $\angle NAI = \angle IAC$ and ... |
2,810,695 | Find the smallest odd prime $p$, such that there exist coprime positive integers $k$ and $\ell$ which satisfy
\[4k-3\ell=12\quad \text{ and }\quad \ell^2+\ell k +k^2\equiv 3\text{ }(\text{mod }p)\] | No need to use overcomplicated ideas. (Hopefully I have not made a mistake in these rushed calculations.) Clearly $k=3x$ and then $\ell = 4x-4$, so the congruence becomes $37x^2 - 44x + 13 \equiv 0 \pmod p$. For $p=3$ we get $(x-1)^2 \equiv 0 \pmod 3$ and $k$ and $\ell$ are both divisible by $3$. For $p=5$ we get $2x^2... |
2,810,696 | Solve the equation
\[(x+1)\log^2_{3}x+4x\log_{3}x-16=0\] | $ x= \frac {1} {81} $ |
2,810,700 | Let $n \geq 2$ be a positive integer. The set $M$ consists of $2n^2-3n+2$ positive rational numbers. Prove that there exists a subset $A$ of $M$ with $n$ elements with the following property: $\forall$ $2 \leq k \leq n$ the sum of any $k$ (not necessarily distinct) numbers from $A$ is not in $A$. | The author of this problem is Aleksandar Ivanov. The main idea is to make the given numbers integers and work modulo $ p$ where $ p$ is a large (prime) number. If we manage to choose a large "sum-free" subset $ M'$ modulo $ p$ the same set $ M'$ will be sum-free. The plan follows as a whole the author's idea of the off... |
2,810,701 | $ABCD$ is circumscribed in a circle $k$, such that $[ACB]=s$, $[ACD]=t$, $s<t$. Determine the smallest value of $\frac{4s^2+t^2}{5st}$ and when this minimum is achieved. | Since $4s^2+t^2 \ge 4st$, the ratio is at least $4/5$. For an example, it suffices to consider $|AB|=|BC|=1$, $|CD|=|DA|=\sqrt{2}$, $\angle ADC = \angle ABC = \pi/2$. (Check that $|BD|=2\sqrt{2}/\sqrt{3}$ by Ptolemy and everything works out). |
2,810,702 | Let $ABCDV$ be a regular quadrangular pyramid with $V$ as the apex. The plane $\lambda$ intersects the $VA$, $VB$, $VC$ and $VD$ at $M$, $N$, $P$, $Q$ respectively. Find $VQ : QD$, if $VM : MA = 2 : 1$, $VN : NB = 1 : 1$ and $VP : PC = 1 : 2$. | [hide=Solution]Let $AB$ be of length $6$. The $VM=4,$ $VN=3,$ $VP=2.$
Let $O$ be the intersection of $MP$ and $QN.$ Then $O$ lies on the altitude of the pyramid, which also bisects the right angles $\angle MVP$ and $\angle QVN.$ Let points $Q'$ and $N'$ lie on segments $VA$ and $VC$, respectively, such that $VQ'=VQ$ an... |
2,799,975 | Call a set of $n$ lines [i]good[/i] if no $3$ lines are concurrent. These $n$ lines divide the Euclidean plane into regions (possible unbounded). A [i]coloring[/i] is an assignment of two colors to each region, one from the set $\{A_1, A_2\}$ and the other from $\{B_1, B_2, B_3\}$, such that no two adjacent regions (ad... | The answer is all $n \ge 5$.
Consider the corresponding graph $G$. As shown in above posts, $G$ is bipartite. We consider assigning :-
[list]
[*] $A_1,A_2$ as coloring the region with two colors red and blue.
[*] $B_1,B_2,B_3$ as writing $1,2,3$ in the regions.
[/list]
Then we have to ensure that:
[list]
[*] All num... |
2,799,974 | Let $\triangle{ABC}$ has circumcircle $\Gamma$, drop the perpendicular line from $A$ to $BC$ and meet $\Gamma$ at point $D$, similarly, altitude from $B$ to $AC$ meets $\Gamma$ at $E$. Prove that if $AB=DE, \angle{ACB}=60^{\circ}$
(sorry it is from my memory I can't remember the exact problem, but it means the same) | Let $X$ be the foot from $A$ to $BC$ and $Y$ be the foot from $B$ to $AC$. By reflecting the orthocenter, $XY=\frac{1}{2}ED=\frac{1}{2}AB$. Since $AYXB$ is cyclic, we have $\triangle CYX\sim \triangle CBA$ so $CX=\frac{1}{2}AC$ and hence $\cos C=\frac{1}{2}\implies C=60^{\circ}$. |
2,800,863 | Positive integers $a$, $b$, $c$ are given. It is known that $\frac{c}{b}=\frac{b}{a}$, and the number $b^2-a-c+1$ is a prime. Prove that $a$ and $c$ are double of a squares of positive integers. | This is quite simple. Simply note that $b^2=ac$. Thus,
\[ac-a-c+1=(a-1)(c-1)\]
is a prime. This requires one of $a,c$ to be $2$. WLOG, let $a=2$. Note that then, $2\mid b$. Then,
\[c=\frac{b^2}{2} = 2\left(\frac{b}{2}\right)^2\]
Thus, $a,c$ are both twice a perfect square as claimed. |
2,800,872 | Do there exist 2021 points with integer coordinates on the plane such that the pairwise distances between them are pairwise distinct consecutive integers? | Compare with [url="https://artofproblemsolving.com/community/c6h2800893_pairwise_distances_are_consecutive_numbers"]this problem[/url] |
2,800,885 | Paul can write polynomial $(x+1)^n$, expand and simplify it, and after that change every coefficient by its reciprocal. For example if $n=3$ Paul gets $(x+1)^3=x^3+3x^2+3x+1$ and then $x^3+\frac13x^2+\frac13x+1$. Prove that Paul can choose $n$ for which the sum of Paul’s polynomial coefficients is less than $2.022$. | As stated in the title of the problem we need to show that we can choose $n$, such that:
$$\sum_{i=0}^{n}\frac{1}{{n \choose i}} < 2.022$$
Notice that for each $i$, $1 < i < n$, we have that:
$$\frac{1}{{n \choose 2}} \geq \frac{1}{{n \choose i}}$$
thus we have that:
$$LHS = 2+\sum_{i=1}^{n-1}\frac{1}{{n \choose i}} \l... |
2,800,893 | Do there exist 100 points on the plane such that the pairwise distances between them are pairwise distinct consecutive integer numbers larger than 2022? | Compare with [url="https://artofproblemsolving.com/community/c6h2800872_pairwise_distances_of_lattice_points_are_consecutive_numbers"]this problem[/url] |
2,808,538 | Let $C=\{ z \in \mathbb{C} : |z|=1 \}$ be the unit circle on the complex plane. Let $z_1, z_2, \ldots, z_{240} \in C$ (not necessarily different) be $240$ complex numbers, satisfying the following two conditions:
(1) For any open arc $\Gamma$ of length $\pi$ on $C$, there are at most $200$ of $j ~(1 \le j \le 240)$ suc... | Here is my [hide=solution]
Let the $240$ plurals be $e^{\theta_1 i},e^{\theta_2 i}, \cdots ,e^{\theta_{240}i},0 \leqslant \theta_1 \leqslant \theta_2 \leqslant \cdots \leqslant \theta_{240} \leqslant 2 \pi ,z_k=e^{\theta_k i}$. And let $\omega_k=z_k+$$z_{k+40}+z_{k+80}+z_{k+120}+z_{k+160}+z_{k+200},(1\leqslant k \leqsl... |
2,808,539 | Let $m$ be a positive integer, and $A_1, A_2, \ldots, A_m$ (not necessarily different) be $m$ subsets of a finite set $A$. It is known that for any nonempty subset $I$ of $\{1, 2 \ldots, m \}$,
\[ \Big| \bigcup_{i \in I} A_i \Big| \ge |I|+1. \]
Show that the elements of $A$ can be colored black and white, so that each ... | [hide=Solution]Consider a bipartite graph $G(S,T,E)$ with $S=\{A_1,\cdots,A_m\}$ representing the sets, $T=\bigcup_{j=1}^m A_j$ representing the elements and $E$ representing the edges. Connect an edge between $A_j$ and $b\in T$ if and only if $b\in A_j$. It is equivalent to prove that there exists a coloring of $T$ in... |
2,808,541 | Let $p$ be a prime, $A$ is an infinite set of integers. Prove that there is a subset $B$ of $A$ with $2p-2$ elements, such that the arithmetic mean of any pairwise distinct $p$ elements in $B$ does not belong to $A$. | Consider the following process:
I start with a copy $A_1$ of $A$. Then, on the $i$th iteration of the process:
If $A_i$ contains at least $p-1$ of any two residues $x,y$ mod $p$, then choose $p-1$ numbers from each residue to form a valid subset $B_i$ of $A_i$. This ends the process.
Else, only finitely many element... |
2,811,451 | Find all pairs of positive integers $(m, n)$, such that in a $m \times n$ table (with $m+1$ horizontal lines and $n+1$ vertical lines), a diagonal can be drawn in some unit squares (some unit squares may have no diagonals drawn, but two diagonals cannot be both drawn in a unit square), so that the obtained graph has an... | @above I don't think you can argue that two paths between four vertices of the same color do not intersect.
[hide = My solution]It is well known that a graph has an Eulerian cycle if and only if all vertices have even degree. Let $G$ be the graph consisting only diagonal edges, and call a vertex $\emph{strange}$ if it... |
2,811,452 | Given a non-right triangle $ABC$ with $BC>AC>AB$. Two points $P_1 \neq P_2$ on the plane satisfy that, for $i=1,2$, if $AP_i, BP_i$ and $CP_i$ intersect the circumcircle of the triangle $ABC$ at $D_i, E_i$, and $F_i$, respectively, then $D_iE_i \perp D_iF_i$ and $D_iE_i = D_iF_i \neq 0$. Let the line $P_1P_2$ intersec... | Great problem! My solution uses angle chasing to find interesting carectization of the points, and then the solution is quite natural!
[i][color=#00f]Key Lemma:[/color] Let $K$ be intersection of the tangents from $B$ and $C$ to $(ABC)$ and then let $\omega$ be the circle with center $K$ passing through $B.$ Further,... |
2,812,219 | Given a positive integer $n$, let $D$ is the set of positive divisors of $n$, and let $f: D \to \mathbb{Z}$ be a function. Prove that the following are equivalent:
(a) For any positive divisor $m$ of $n$,
\[ n ~\Big|~ \sum_{d|m} f(d) \binom{n/d}{m/d}. \]
(b) For any positive divisor $k$ of $n$,
\[ k ~\Big|~ \sum_{d|k}... | [hide=Solution]
For convenience, write $g(d)=\sum_{e \mid d} f(e)$ so that $f(d)=\sum_{e \mid d} \mu\left(\frac{d}{e}\right) g(e)$ by Möbius Inversion. Then we need to prove that $k \mid g(k)$ for all $k$ iff
\[n \mid \sum_{d \mid m} \binom{n/d}{m/d} \sum_{e \mid d} \mu\left(\frac{d}{e}\right) g(e)=\sum_{e \mid m} g(e)... |
2,835,360 | Given two circles $\omega_1$ and $\omega_2$ where $\omega_2$ is inside $\omega_1$. Show that there exists a point $P$ such that for any line $\ell$ not passing through $P$, if $\ell$ intersects circle $\omega_1$ at $A,B$ and $\ell$ intersects circle $\omega_2$ at $C,D$, where $A,C,D,B$ lie on $\ell$ in this order, then... | Let $P$ be a point such that after inversion wrt $P$ $\omega_1^*$ and $\omega_2^*$ are concentric, which is well known to exist (let $O$ be the new common center).
Now, the angle condition gets simply translated to $\angle A^*PC^*=\angle B^*PD^*$, where $(A^*B^*C^*D^*P)$ is cyclic. But this cyclicity + the two concentr... |
2,835,370 | Given a positive integer $n \ge 2$. Find all $n$-tuples of positive integers $(a_1,a_2,\ldots,a_n)$, such that $1<a_1 \le a_2 \le a_3 \le \cdots \le a_n$, $a_1$ is odd, and
(1) $M=\frac{1}{2^n}(a_1-1)a_2 a_3 \cdots a_n$ is a positive integer;
(2) One can pick $n$-tuples of integers $(k_{i,1},k_{i,2},\ldots,k_{i,n})$ fo... | [hide=Easy for P3, easiest problem on T3 for me]
Let $a$ be the number of odd terms in $a_i$. The answer is $2^a|a_1-1$ only.
Call points $(x_1, \cdots, x_n)$.
Claim: The maximum number of points that can be placed in $\times_{j=1}^n \mathbb{Z}_{a_j}$ is at most $\frac{1}{2^n} (a_1-1) \prod\limits_{j=2}^n a_j$
Proo... |
2,835,373 | Find all positive integer $k$ such that one can find a number of triangles in the Cartesian plane, the centroid of each triangle is a lattice point, the union of these triangles is a square of side length $k$ (the sides of the square are not necessarily parallel to the axis, the vertices of the square are not necessari... | Claim: the only such numbers are the multiples of $3$.
If $k=3t$ pick an axis and grid aligned $k\times k$ square divided in $t\times t$ square each of side $3\times 3$, and divide these in two triangles. In this way it is straight forward enough to see that the barycenters must have integer coordinates.
Lemma: given ... |
2,835,377 | Show that there exist constants $c$ and $\alpha > \frac{1}{2}$, such that for any positive integer $n$, there is a subset $A$ of $\{1,2,\ldots,n\}$ with cardinality $|A| \ge c \cdot n^\alpha$, and for any $x,y \in A$ with $x \neq y$, the difference $x-y$ is not a perfect square. | Choose the set $S=\{1,3,8\}.$ This set has the property that for any distinct $a,b\in S,$ $a-b$ is not a quadratic residue modulo $13$. Based on this observation, we will construct the following infinite set:\[N=\bigg\{\sum_{i=0}^\infty \lambda_i\cdot 13^i:\lambda_{2j}\in S, \ \lambda_{2j+1}\in\{0,1,\ldots,12\} \ \fora... |
2,835,389 | Initially, each unit square of an $n \times n$ grid is colored red, yellow or blue. In each round, perform the following operation for every unit square simultaneously:
[list]
[*] For a red square, if there is a yellow square that has a common edge with it, then color it yellow.
[*] For a yellow square, if there is a b... | I'm not quite convinced by the above proofs, as the logics don't seem to be clear enough. Instead I have the following argument.
We assume without loss of generality that all squares turn red in the end. For each square $X$, we define $s(X)$ as the smallest integer $k$ such that $X$ remains red from round $k$ until th... |
2,835,390 | Let $ABCD$ be a convex quadrilateral, the incenters of $\triangle ABC$ and $\triangle ADC$ are $I,J$, respectively. It is known that $AC,BD,IJ$ concurrent at a point $P$. The line perpendicular to $BD$ through $P$ intersects with the outer angle bisector of $\angle BAD$ and the outer angle bisector $\angle BCD$ at $E,F... | For completeness, here is the full solution:
Let $\omega_1$ and $\omega_2$ be the incircles of $\triangle ABC$ and $\triangle ADC$.
[b]Claim:[/b] $AB+AD=CB+CD$.
[b]Proof:[/b] Observe that $P$ is the center of negative homothety sending $\omega _1$ to $\omega_2$. Due to symmetry reasons, we may assume that $BC$ cuts $... |
2,835,392 | Find all functions $f: \mathbb R \to \mathbb R$ such that for any $x,y \in \mathbb R$, the multiset $\{(f(xf(y)+1),f(yf(x)-1)\}$ is identical to the multiset $\{xf(f(y))+1,yf(f(x))-1\}$.
[i]Note:[/i] The multiset $\{a,b\}$ is identical to the multiset $\{c,d\}$ if and only if $a=c,b=d$ or $a=d,b=c$. | Case $f(1)=1$ is almost trivial but does the above deal with the case $f(1)=-1$ ?\\
Let $M(x,y)$ denote the assertion that $\{f(xf(y)+1),f(yf(x)-1)\}=\{xf(f(y))+1,yf(f(x))-1\}$ ,that the multisets are equal but sometimes I will use just the fact that the sums are equal so let $S(x,y)$ denote $ f(xf(y)+1)+f(yf(x)-1)=xf... |
2,835,394 | Find all positive integers $a,b,c$ and prime $p$ satisfying that
\[ 2^a p^b=(p+2)^c+1.\] | My 300th post!!!
[b]Solution[/b]
Notice that $2\mid 2^ap^b,$ we have $p\equiv 1\pmod 2.$ Let $c=2^{\theta}r$ where $2\nmid r,$ then $(p+2)^{2^{\theta}}+1\mid 2^ap^b.$
Lemma $:$ $(p+2)^{2^{\theta}}+1$ is divisible by $p.$
If it is not true$,$ $(p+2)^{2^{\theta}}+1=2^{\beta},$ then $\beta >2.$ Therefore $(p+2)^{2^{\theta... |
2,835,400 | Let $n$ be a positive integer, $x_1,x_2,\ldots,x_{2n}$ be non-negative real numbers with sum $4$. Prove that there exist integer $p$ and $q$, with $0 \le q \le n-1$, such that
\[ \sum_{i=1}^q x_{p+2i-1} \le 1 \mbox{ and } \sum_{i=q+1}^{n-1} x_{p+2i} \le 1, \]
where the indices are take modulo $2n$.
[i]Note:[/i] If $q... | Here is another version of my proof above, which is simpler in my opinion. Moreover the gas station lemma is proved for completeness.
[hide=Solution]Let us write $[n]\doteqdot\{1,2,\ldots,n\}$ for brevity.
[i]Lemma.[/i] (gas station lemma) Let $n$ be a positive integer and let $a_1,\ldots,a_n$ be real numbers such th... |
2,835,403 | Given a positive integer $n$, let $D$ be the set of all positive divisors of $n$. The subsets $A,B$ of $D$ satisfies that for any $a \in A$ and $b \in B$, it holds that $a \nmid b$ and $b \nmid a$. Show that
\[ \sqrt{|A|}+\sqrt{|B|} \le \sqrt{|D|}. \] | Solved with Luke Robitaille, Justin Lee, and Espen Slettnes.
Let \[V=\bigcup_{a\in A}\{\text{divisors of }a\} \quad\text{and}\quad W=\bigcup_{a\in A}\{\text{multiples of }a\}\cap D\]
[color=red][b]Claim:[/b][/color] \(|V|\cdot|W|\ge|A|\cdot|D|\).
[i]Proof.[/i] We induct on the number of prime divisors... |
2,785,510 | Find all integer values of $x$ for which the value of the expression
\[x^2+6x+33\]
is a perfect square. | $(x+3)^2+24 =m^2$
$(m-x-3)(m+x+3) = 24$
$x = (-8,-4,-2,2)$ |
2,785,511 | Let $ABCD$ be a square. Let $E, Z$ be points on the sides $AB, CD$ of the square respectively, such that $DE\parallel BZ$. Assume that the triangles $\triangle EAD, \triangle ZCB$ and the parallelogram $BEDZ$ have the same area.
If the distance between the parallel lines $DE$ and $BZ$ is equal to $1$, determine the a... | Area of paralelogram $BEDZ$ is equal to $DE*1=DE$. On the other hand same area is equal to $EB*AB$. Area of whole square is $AB^2$ and Is also equal to sum of area of triangles $EAD, ZCB$ and parallelogram $BEDZ$ so we have $AB^2=3AB*EB$. Since $DE=EB*AB$ we have $DE=\frac{AB^2}{3}$. From triangle $EAD$ we have that $D... |
2,785,526 | Determine all pairs of prime numbers $(p, q)$ which satisfy the equation
\[
p^3+q^3+1=p^2q^2
\]
| A faster way (essentially optimized version of 2nd post).
Let $p=q$. Inspecting both sides modulo $p$, we find $p\mid 1$, which is absurd. Thus $p\ne q$. Let $p>q$ wlog. Inspecting both sides modulo $p^2$, we find that
\[
p^2\mid (q+1)(q^2-q+1).
\]
Now, let $p\mid q+1$. Then, as $p\ge q+1$, we must have $p=q+1$, whic... |
2,785,535 | Let $A$ be a subset of $\{1, 2, 3, \ldots, 50\}$ with the property: for every $x,y\in A$ with $x\neq y$, it holds that
\[\left| \frac{1}{x}- \frac{1}{y}\right|>\frac{1}{1000}.\]
Determine the largest possible number of elements that the set $A$ can have. | We begin with the following Claim:
[b]Claim:[/b] If $x,x+1 \in A$, then $x \leq 31$. Furthermore, if $x,x+2 \in A$, then $x \leq 43$.
[i]Proof:[/i] Easy to see from the given hypothesis $\blacksquare$
Therefore, by partitioning the elements of the set $X=\{1, 2, 3, \ldots, 50\}$ in 10 groups as follows,
[b]Group 1:[/... |
2,785,630 | Let
\[M=\{1, 2, 3, \ldots, 2022\}\]
Determine the least positive integer $k$, such that for every $k$ subsets of $M$ with the cardinality of each subset equal to $3$, there are two of these subsets with exactly one common element. | Say that a set of triples, $V$, is "good" iff each pair of triples $u,v\in V$ is disjoint or has two elements in common. Construct a graph $\mathcal{G}$ on good $V$ by drawing an edge $u\sim v$ between each $u,v\in V$ with $|u\cap v|=2$. The relation $\sim$ is obviously reflexive and, in a good graph, it's transitive... |
2,787,231 | Find all pairs of real numbers $(x,y)$ for which
\[
\begin{aligned}
x^2+y^2+xy&=133 \\
x+y+\sqrt{xy}&=19
\end{aligned}
\] | $x^2+y^2+xy=(x+y+\sqrt{xy})(x+y-\sqrt{xy})\implies \frac{133}{19}=7=x+y-\sqrt{xy}$
$(x+y+\sqrt{xy})+(x+y-\sqrt{xy})=19+7\implies x+y=13\implies \sqrt{xy}=6\implies xy=36$
By Vieta’s formula, $x,y$ are roots of $$a^2-13a+36=0\implies (a-4)(a-9)=0\implies a=4,9\implies \boxed{(x,y)=(4,9),(9,4)}$$ |
2,810,677 | Let $f:\mathbb{N}^*\rightarrow \mathbb{N}^*$ be a function such that $\frac{x^3+3x^2f(y)}{x+f(y)}+\frac{y^3+3y^2f(x)}{y+f(x)}=\frac{(x+y)^3}{f(x+y)},~(\forall)x,y\in\mathbb{N}^*.$
$a)$ Prove that $f(1)=1.$
$b)$ Find function $f.$ | Giả sử tồn tại hàm số thỏa mãn $$\frac{x^3+3x^2f(y)}{x+f(y)}+\frac{y^3+3y^2f(x)}{y+f(x)}=\frac{(x+y)^3}{f(x+y)},\forall x,y\in \mathbb{Z^+}.(1)$$
Từ $(1)$ cho $x=y=1$ ta được $$\frac{2(1+3f(1))}{1+f(1)}=\frac{8}{f(2)}$$
Hay ta được $$f(2)=\frac{4(1+f(1))}{1+3f(1)}.$$Suy ra $(1+f(1))|(4+4f(1))$ hay $(1+3f(1))|8$ suy ra ... |
2,810,679 | $a)$ Prove that $2x^3-3x^2+1\geq 0,~(\forall)x\geq0.$
$b)$ Let $x,y,z\geq 0$ such that $\frac{2}{1+x^3}+\frac{2}{1+y^3}+\frac{2}{1+z^3}=3.$ Prove that $\frac{1-x}{1-x+x^2}+\frac{1-y}{1-y+y^2}+\frac{1-z}{1-z+z^2}\geq 0.$ | $a)$ The expression factorizes as $(x-1)^2(2x+1)$ which is clearly positive for $x\geq 0$. Moreover, equality occurs for $x=1$.\\
$b)$ Using the result from point $a)$, we get $1-x^3\leq \frac{3}{2}(1-x^2)$, thus $$3=\sum_{cyc}\frac{2}{1+x^3}\implies 0=\sum_{cyc}\frac{1-x^3}{1+x^3}\leq \frac{3}{2}\sum_{cyc}\frac{1-x^2}... |
2,810,681 | $a)$ Solve over the positive integers $3^x=x+2.$
$b)$ Find pairs $(x,y)\in\mathbb{N}\times\mathbb{N}$ such that $(x+3^y)$ and $(y+3^x)$ are consecutive. | Part a)
$x=1$ only solution. For $x>1, LHS>RHS$
Part b)
If $(x+3^y)$ and $(y+3^x)$ are consecutive $\Longrightarrow x+3^y+1= y+3^x$
Case $x=y$
$\Longrightarrow x+3^x+1= x+3^x \Longrightarrow 1= 0$ impossible
Case $y>x$
Let $y=x+n\Longrightarrow x+3^n3^x+1=x+n+3^x\Longrightarrow 3^x=\frac{n-1}{3^n-1}<1$ Contrad... |
2,810,713 | Let $(G,\cdot)$ be a group and $H\neq G$ be a subgroup so that $x^2=y^2$ for all $x,y\in G\setminus H.$ Show that $(H,\cdot)$ is an Abelian group. | Let $x \in G \setminus H.$ For any $a \in H,$ we have $ax \notin H$ and so $(ax)^2=x^2,$ which gives $axa=x$ hence
$$a=xa^{-1}x^{-1}, \ \ \ \ \ \forall a \in H.$$
So if $a,b \in H,$ then
$$xb^{-1}a^{-1}x^{-1}=x(ab)^{-1}x^{-1}=ab=(xa^{-1}x^{-1})(xb^{-1}x^{-1})=xa^{-1}b^{-1}x^{-1},$$
which gives $b^{-1}a^{-1}=a^{-1}b^{... |
2,810,718 | Let $A,B\in\mathcal{M}_3(\mathbb{R})$ de matrices such that $A^2+B^2=O_3.$ Prove that $\det(aA+bB)=0$ for any real numbers $a$ and $b.$ | Interesting problem
$A^2=-B^2$ take the det then $det(A)^2=-det(B)^2$ ==> det(A)=det(B)=0 (1)
If a=0 or b=0 with (1) ==> det(aA+bB)=0
If a=/=0 and b=/=0 we must prove det(A+xB)=0 for any x in R
$Q(x)=det(A+xB)=det(B)x^3 + px^2 + qx +det(A)= px^2 + qx$ where p,q reals numbers
$Q(i)Q(-i)=|Q(i)|^2 = (-p+iq)(-p-iq) ... |
2,810,731 | Let $A\in\mathcal{M}_n(\mathbb{C})$ where $n\geq 2.$ Prove that if $m=|\{\text{rank}(A^k)-\text{rank}(A^{k+1})":k\in\mathbb{N}^*\}|$ then $n+1\geq m(m+1)/2.$ | Let $m = \{ d_1 , d_2, \dots, d_m\}$ we have
$$\frac{m(m+1)}{2}=1+2+\cdots + m \leq d_1 +d_2 +\cdots d_m$$
$$\leq \text{rank}(A^{i_1})-\text{rank}(A^{i_1+1})+\text{rank}(A^{i_2})-\text{rank}(A^{i_2+1})+\cdots +\text{rank}(A^{i_m})-\text{rank}(A^{i_m+1}) $$
$$\leq \text{rank}(A^{i_1})-\text{rank}(A^{i_m+1})\leq n$$
Tha... |
2,811,496 | Determine all $x\in(0,3/4)$ which satisfy \[\log_x(1-x)+\log_2\frac{1-x}{x}=\frac{1}{(\log_2x)^2}.\] | I claim $x=1/2$ is the only possibility.
We have
\begin{align*}
&\frac{\log_2(1-x)}{\log_2 x} + \log_2(1-x) = \log_2 x + \frac{1}{(\log_2 x)^2}\\
&\iff \log_2(1-x)\left(\frac{1+\log_2 x}{\log_2 x}\right) = \frac{(\log_2 x+1)((\log_2 x)^2-\log_2 x+1)}{(\log_2 x)^2}.
\end{align*}
If $\log_2 x+1=0$, that is if $x=1/2$, ... |
2,790,593 | Suppose that $P(x)$ is a monic quadratic polynomial satisfying $aP(a) = 20P(20) = 22P(22)$ for some integer $a\neq 20, 22$. Find the minimum possible positive value of $P(0)$.
[i]Proposed by Andrew Wu[/i]
(Note: wording changed from original to specify that $a \neq 20, 22$.) | The conditions of the problem allow us to define
\[ XP(X) = (X-a)(X-20)(X-22) + C. \]
Since $X$ divides the RHS, then $C = 440a$. So,
\begin{align*}
P(X) = \dfrac{1}{X} \left[(X-a)(X-20)(X-22) + 400a\right] = X^2 - (42+a)X + (440+42a).
\end{align*}
Henceforth, we want to minimize the value of $P(0) = 440+42a$ which is... |
2,796,089 | [b]p10 [/b]Kathy has two positive real numbers, $a$ and $b$. She mistakenly writes
$$\log (a + b) = \log (a) + \log( b),$$
but miraculously, she finds that for her combination of $a$ and $b$, the equality holds. If $a = 2022b$, then $b = \frac{p}{q}$ , for positive integers $p, q$ where $gcd(p, q) = 1$. Find $p + q$.
... | [hide=answer p10]$\frac{p}{q} = \frac{2023}{2022}$. Since they are relatively prime, $4045$.[/hide] |
2,790,101 | Let $n>4$ be a positive integer, which is divisible by $4$. We denote by $A_n$ the sum of the odd positive divisors of $n$. We also denote $B_n$ the sum of the even positive divisors of $n$, excluding the number $n$ itself. Find the least possible value of the expression $$f(n)=B_n-2A_n,$$
for all possible values of $n... | I claim the answer is $4$, and is attained iff $n=4p$, where $p>2$ is a prime or when $n=8$. Check that indeed when $n=4p$, then $A_n=1+p$ whereas $B_n=2+4+2p$, yielding $B_n-2A_n =4$.
Let $n=2^k\cdot m$, where $m$ is odd and $k\ge 2$. Notice that $A_n=\textstyle\sum_{d\mid m}d$. On the other hand, $B_n = 2A_n+4A_n+\c... |
2,790,106 | The positive real numbers $a,b,c,d$ satisfy the equality
$$a+bc+cd+db+\frac{1}{ab^2c^2d^2}=18.$$
Find the maximum possible value of $a$. | Using AM-GM on the last four terms, we obtain
\[
18\ge a+ 4a^{-\frac14}\iff t^4 +\frac4t-18\le 0 \qquad\text{where}\qquad a=t^4.
\]
From here, $t\le 2$ and therefore $a\le 16$.
We now give an example attaining the equality. Note that if $b=c=d$ and $b^2 = (ab^6)^{-1}\iff b^8 = \frac1a$, we enjoy equality. Namely, $a=1... |
2,801,549 | Let $ABC$ be a triangle with centroid $G$, and let $E$ and $F$ be points on side $BC$ such that $BE = EF = F C$. Points $X$ and $Y$ lie on lines $AB$ and $AC$, respectively, so that $X$, $Y$ , and $G$ are not collinear. If the line through $E$ parallel to $XG$ and the line through $F$ parallel to $Y G$ intersect at $P\... | Trivial. Equivalent to $S_{GPX} = S_{GPY}$ and the parallel lines rewrite the latter into $S_{GEX} = S_{GFY}$. But $GE \parallel AB$ (consider $M = CG \cap AB$ and note $BE/EM = CG/GM$) and similarly $GF \parallel AC$, so we reduce to $S_{GEB} = S_{GFC}$, which is clear. |
2,801,551 | Let $P(x) = x^4 + ax^3 + bx^2 + x$ be a polynomial with four distinct roots that lie on a circle in the complex plane. Prove that $ab\ne 9$. | Suppose by way of contradiction that some $a,b$ work. Let $0,p,q,r$ be the roots, which lie on the circle in that order. From Vieta, we have $pqr = -1$ and
\[ab = -(p + q + r)(pq + qr + rp) = (p + q + r)\left(\frac1p + \frac1q + \frac1r\right) = 9\text{.}\]
Now we deal with the circle condition. By Ptolemy, $0,p,q,r$ a... |
2,804,653 | Let $S$ be a set of size $11$. A random $12$-tuple $(s_1, s_2, . . . , s_{12})$ of elements of $S$ is chosen uniformly at random. Moreover, let $\pi : S \to S$ be a permutation of $S$ chosen uniformly at random. The probability that $s_{i+1}\ne \pi (s_i)$ for all $1 \le i \le 12$ (where $s_{13} = s_1$) can be written a... | [hide=Answer]1000000000004[/hide]
[hide=Solution]
Let's count the number $N$ of ordered pairs $(T, \pi)$, where $T$ is the 12-tuple, satisfying the given condition.
We will use inclusion-exclusion on the 12 constraints of the form $s_{i+1} \neq \pi(s_i)$. We have \[N = \sum_{A \subseteq \{1, 2, \dots, 12\}}(-1)^{|A|}... |
2,795,412 | Let $D$ be an interior point on the side $BC$ of an acute-angled triangle $ABC$. Let the circumcircle of triangle $ADB$ intersect $AC$ again at $E(\ne A)$ and the circumcircle of triangle $ADC$ intersect $AB$ again at $F(\ne A)$. Let $AD$, $BE$, and $CF$ intersect the circumcircle of triangle $ABC$ again at $D_1(\ne A)... | Clearly $I_1 = BE_1 \cap BF_1$ because $BE_1$ is the angle bisector of $\angle F_1E_1D_1$ as $$\angle FE_1B = \angle F_1CB = \angle FCD = \angle FAD = \angle BAD_1 = \angle BE_1D_1$$ and similarly $CF_1$ is the angle bisector of $\angle D_1F_1E_1$.
$\textbf{Lemma 1}$ \\
In any $\triangle XYZ$ with incenter $J$, $\ang... |
2,735,507 | Given an acute triangle $ABC$. with $H$ as its orthocenter, lines $\ell_1$ and $\ell_2$ go through $H$ and are perpendicular to each other. Line $\ell_1$ cuts $BC$ and the extension of $AB$ on $D$ and $Z$ respectively. Whereas line $\ell_2$ cuts $BC$ and the extension of $AC$ on $E$ and $X$ respectively. If the line th... | Same problem as [url=https://artofproblemsolving.com/community/c6h1137975p5325937]Iranian Geometry Olympiad 2015 Advanced P3[/url] |
2,735,508 | Prove that there exists a set $X \subseteq \mathbb{N}$ which contains exactly 2022 elements such that for every distinct $a, b, c \in X$ the following equality:
\[ \gcd(a^n+b^n, c) = 1 \] is satisfied for every positive integer $n$. | [hide=Solution][hide=FLT On Steroids]Let $p_1, p_2, \cdots, p_{2022}$ be the smallest odd primes and let \[ N = \prod_{i=1}^{2022} (p_i - 1). \] Then, the set $X = \{ p_1^N, p_2^N, \ldots, p_{2022}^N\}$ works. Since $p_i - 1 \vert N$ for every such $i$, by FLT we have that any distinct $a,b,c$ satisfies the following c... |
2,742,895 | Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that
$$(a + b + c)(ab + bc + ca) + 3\ge 4(a + b + c).$$ | $a+b+c=3x^2\geq3 => x\geq1$
$(ab+bc+ac)^2\geq3abc(a+b+c)$
$(ab+bc+ac)\geq3x$
$(a+b+c)(ab+bc+ac)+3\geq9x^3+3$
We prove that $9x^3+3\geq12x^2$
=> $(3x^2-x-1)(x-1)\geq0$
This is true $x\geq1$ |
2,744,240 | For each natural number $n$, let $f(n)$ denote the number of ordered integer pairs $(x,y)$ satisfying the following equation:
\[ x^2 - xy + y^2 = n. \]
a) Determine $f(2022)$.
b) Determine the largest natural number $m$ such that $m$ divides $f(n)$ for every natural number $n$. | Gosh, this is just Turkey 2000 TST just change into 2022:
https://artofproblemsolving.com/community/q2h396239p2203313 |
2,744,245 | Determine all functions $f : \mathbb{R} \to \mathbb{R}$ satisfying
\[ f(a^2) - f(b^2) \leq (f(a)+b)(a-f(b)) \] for all $a,b \in \mathbb{R}$. | Let $P(a,b)$ be the given assertion.
$P(0,0)\Rightarrow f(0)^2\le0\Rightarrow f(0)=0$
$P(x,0)\Rightarrow f(x^2)\le xf(x)$
$P(0,x)\Rightarrow-f(x^2)\le-xf(x)\Rightarrow f(x^2)\ge xf(x)$
So $f(x^2)=xf(x)$, hence $f$ is odd. Now $P(x,y)$ gives the new assertion $Q(x,y):f(x)f(y)\le xy$.
$Q(x,-y)\Rightarrow-f(x)f(y)\le-xy\R... |
2,783,539 | In parallelogram $ABCD$ with acute angle $A$ a point $N$ is chosen on the segment $AD$, and a point $M$ on the segment $CN$ so that $AB = BM = CM$. Point $K$ is the reflection of $N$ in line $MD$. The line $MK$ meets the segment $AD$ at point $L$. Let $P$ be the common point of the circumcircles of $AMD$ and $CNK$ such... | Let $\overline{KM}\cap (CNK)=O\neq K$ and $(CNK)\cap (AMD)=V\neq P$ and . Quick angle and side chasing gives that $CNOD$ and $BMOA$ are rhombi, so $O$ is the center of $(AMD)$. Now note that $KNMA$ is cyclic because $$\angle{MAN}=\angle{MAD}=\frac{1}{2}\angle{MOD}=\angle{NKM}.$$ Now, combining this with the similarity ... |
2,783,541 | In triangle $ABC$, a point $M$ is the midpoint of $AB$, and a point $I$ is the incentre. Point $A_1$ is the reflection of $A$ in $BI$, and $B_1$ is the reflection of $B$ in $AI$. Let $N$ be the midpoint of $A_1B_1$. Prove that $IN > IM$. | Notice that both $IN$ and $IM$ are medians, then we need to show that:
$$\frac{1}{2}\sqrt{2IB_1^2+2IA_1^2-A_1B_1^2} > \frac{1}{2}\sqrt{2IB^2+2IA^2-AB^2}$$
which implies that $AB > A_1B_1$, thus if we show this we are done.
By angle chase, we have that $\angle A_1IB_1 \equiv \frac{3}{2}\gamma - 90 \pmod{\pi}$ which is ... |
2,783,542 | 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? | For each $i$, draw a square of side length $1/\sqrt{i}$ with sides parallel to the coordinate axes centered at $(a_i,b_i)$. Then no squares overlap, since for $m>n$ we have $\max\{|a_m-a_n|,|b_m-b_n|\}>1/\sqrt{n}>1/(2\sqrt{n})+1/(2\sqrt{m})$. On the other hand, the sum of the areas of the squares diverges, hence their ... |
2,796,889 | Consider the set $\mathcal{T}$ of all triangles whose sides are distinct prime numbers which are also in arithmetic progression. Let $\triangle \in \mathcal{T}$ be the triangle with least perimeter. If $a^{\circ}$ is the largest angle of $\triangle$ and $L$ is its perimeter, determine the value of $\frac{a}{L}$. | A lengthier approach.
Ofcourse the triangle is a $3,5,7$ triangle. Thus using heron's formula, the area of the traingle is $$ \frac{\sqrt{(15)(9)(1)(5)}}{4} = \frac{1}{2} \times 5 \times 3 \times \sin (\angle \text{opposite to 7}).$$ Ofcourse the angle opposite to $7$ is the largest. $$\Longrightarrow \sin \theta = \f... |
2,796,894 | For any real number $t$, let $\lfloor t \rfloor$ denote the largest integer $\le t$. Suppose that $N$ is the greatest integer such that $$\left \lfloor \sqrt{\left \lfloor \sqrt{\left \lfloor \sqrt{N} \right \rfloor}\right \rfloor}\right \rfloor = 4$$Find the sum of digits of $N$. | The first nest of the floor and square root implies $$16 \leq \left \lfloor \sqrt{\left \lfloor \sqrt{N} \right \rfloor} \right \rfloor < 25.$$ The next nest of floor and square root implies $$256 \leq \left \lfloor \sqrt{N} \right \rfloor < 625$$ Thus, $$256^2 \leq N < 625^2.$$
The maximum value is $625^2 - 1 = 39062... |
2,779,324 | Find all pair of primes $(p,q)$, such that $p^3+3q^3-32$ is also a prime. | Simply note that when $p,q$ are odd, $p^3+3q^3-32$ is clearly even and hence must be 2. But clearly,
\[p^3+3q^3-32 \geq 4 \cdot 3^3 - 32 >2\]
So, there exists no solutions in this case. Thus, we must have one of $p,q$ being even. We check these two cases seperately.\\
[b] Case 1 : [/b] $p$ is even ($p=2$)... |
2,779,328 | In an acute triangle $ABC$, $AB<AC$. The perpendicular bisector of the segment $BC$ intersects the lines $AB,AC$ at the points $D,E$ respectively. Denote the mid-point of $DE$ as $M$. Suppose the circumcircle of $\triangle ABC$ intersects the line $AM$ at points $P$ and $A$, and $M,A,P$ are arranged in order on the lin... | Let $N$ be the midpoint of $\overline{BC}$. Redefine $P=(BNE) \cap (CND)\neq N$. Then
$$\angle BPC=\angle BPN+\angle CPN=\angle BEN+\angle CDN=(90^\circ-\angle C)+(90^\circ-\angle B)=\angle BAC,$$
hence $P$ lies on $(ABC)$ as well. It suffices to show $P,A,M$ are collinear.
First observe that
$$\angle NPA=\angle BPA-\... |
2,754,649 | In the trapezoid $ABCD$ with an inscribed circle, we have $AD\parallel BC$, and we know that angle $B$ and angle $C$ are both acute. Given $AB=7$, $CD=8$. Find the area of trapezoid $ABCD$. | By the Pitot's Theorem, AD+CB=15, and the height must be 6
So (15*6)/2=45 |
2,779,215 | Find all functions $f:\mathbb{Z}_+\to\mathbb{Z}_+$, such that for any two positive integers $m,n$, we have
$$f^{f(n)}(m)+mn=f(m)f(n)$$
where $f^k(n)=\underbrace{f(f(\ldots f(}_{k}n)\ldots))$. | Let $P(m,n)$ the assertion of this F.E.
Assume that there exists $a,b$ such that $f(a)=f(b)$ then by $P(m,a)-P(m,b)$
$$ma=mb \implies a=b \implies f \; \text{injective}$$
Before continuing we prove that $f(n) \ge n+1$ using $P(n,n)$
$$f(n)^2=n^2+f^{f(n)}(n)>n^2 \implies f(n)>n \implies f(n) \ge n+1$$
Now by $P(m,n)-P(n... |
2,779,218 | Find all positive integer pairs $(x,y)$ satisfying the following equation:
$$3^x-8^y=2xy+1$$ | There is no solution when $x$ or $y$ is equal to $1$. So assume $(x,y) \geq (2,2)$:
[hide=Claim1][b]Claim:[/b] $2\mid y$
[i]Proof:[/i] The equation is equivalent to $3^{x}-1=2xy+8^{y} \cdots \blacklozenge $
take $v_2$ for both sides:
$v_2(3^{x}-1)\overset{LTE}{=} v_2(2)+v_2(4)+v_2(x)-1=2+v_2(x)$
$v_2(RHS)\geq \min\{v_... |
2,802,471 | For positive integers $a$ and $b$, if the expression $\frac{a^2+b^2}{(a-b)^2}$ is an integer, prove that the expression $\frac{a^3+b^3}{(a-b)^3}$ is an integer as well. | WLOG $(a,b)=1$. We have
$$(a-b)^2|a^2+b^2-(a-b)^2=2ab$$
Also, $gcd((a-b)^2, a)= gcd((a-b)^2, b)=1$. Hence, $(a-b)^2|2$. Therefore, $a-b=\pm 1$. The rest follows. |
2,802,473 | For a real number $a$, $[a]$ denotes the largest integer not exceeding $a$.
Find all positive real numbers $x$ satisfying the equation
$$x\cdot [x]+2022=[x^2]$$ | Let $[x]=a$ and $x=a+b$ where $0\leq b<1$. We have
$$(a+b)a+2022=[(a+b)^2]\Rightarrow a^2+ab+2022=a^2+[2ab+b^2]\Rightarrow ab+2022=[2ab+b^2]$$
Clearly, RHS is an integer. Hence, $ab$ should be an integer too. Then, $[2ab+b^2]=2ab+[b^2]=2ab$. So the original equation becomes
$$ab+2022=2ab\Rightarrow 2022=ab\Rightarrow ... |
2,802,500 | Let $c$ be a real number. If the inequality
$$f(c)\cdot f(-c)\ge f(a)$$
holds for all $f(x)=x^2-2ax+b$ where $a$ and $b$ are arbitrary real numbers, find all possible values of $c$. | I claim the answer is $c=\pm 1/2$. This is easily seen to work. Next, note that $f(c)f(-c)\ge f(a)$ iff $g(b)\ge 0$ where
\[
g(b) = b^2 + (2c^2-1)b +c^4 -4a^2c^2 + a^2.
\]
We want the quadratic $g(b)\ge 0$ for all choices of $a,b$. This holds iff its discriminant is non-positive: $(2c^2-1)^2 - 4(c^4+a^2-4a^2c^2) = (4a... |
2,764,281 | Does there exist a quadratic trinomial $ax^2 + bx + c$ such that $a, b, c$ are odd integers, and $\frac{1}{2022}$ is one of its roots? | $2022^2c+2022b+a=0 \implies a$ is evan $\implies$ contradiction |
2,769,215 | Find all triples $(a, b, c)$ of positive integers for which $a + [a, b] = b + [b, c] = c + [c, a]$.
Here $[a, b]$ denotes the least common multiple of integers $a, b$.
[i](Proposed by Mykhailo Shtandenko)[/i] | Note that $a\mid a+[a,b]=c+[c,a]\implies a\mid c$. Therefore $a \mid c \mid b \mid a$, or $a=b=c$, which works. |
2,769,223 | Find all triples $(a, b, c)$ of positive integers for which $a + (a, b) = b + (b, c) = c + (c, a)$.
Here $(a, b)$ denotes the greatest common divisor of integers $a, b$.
[i](Proposed by Mykhailo Shtandenko)[/i] | Solved it ORALLY..
WLOG $gcd(a, b, c) =1$ , now if a prime $p \mid gcd(a, b) $ , then $p \mid c$ , which is not possible, so $a, b, c$ are pairwise coprime, hence condition is $a+1 = b+1 =c+1$ , implying $a=b=c$ , so only solutions that work are $(k, k, k) $ for all natural $k$... |
2,769,226 | $2022$ points are arranged in a circle, one of which is colored in black, and others in white. In one operation, The Hedgehog can do one of the following actions:
1) Choose two adjacent points of the same color and flip the color of both of them (white becomes black, black becomes white)
2) Choose two points of oppos... | No. Label the positions of the points consecutively $1,-1,1,-1,\dots$ and let $\ell_i,i=1,\dots,2022$ be the values od these labels. Set $c_i:=1$ if the color of the $i$-th point is white, otherwise $c_i=-1$. Consider $S:=\sum_{i=1}^{2022}\ell_i c_i.$ This value is invariant under the allowed recolorings. But $S$ has ... |
2,769,231 | Let $AH_A, BH_B, CH_C$ be the altitudes of triangle $ABC$. Prove that if $\frac{H_BC}{AC} = \frac{H_CA}{AB}$, then the line symmetric to $BC$ with respect to line $H_BH_C$ is tangent to the circumscribed circle of triangle $H_BH_CA$.
[i](Proposed by Mykhailo Bondarenko)[/i] | Let $D$ lie on $BC$ s.t. $AC//DH_C$.
By the condition, we have $DH_B//AB$.
This implies $(AH_BH_C)$ is symmetric to $(DH_BH_C)$ wrt $H_BH_C$.
Since $AH\bot BC$, we have the tangent at $D$ wrt $(DH_BH_C)$ is parallel to $BC$, which implies $BC$ is tangent to $(DH_BH_C)$.
So we're done. |
2,769,234 | Find the largest $k$ for which there exists a permutation $(a_1, a_2, \ldots, a_{2022})$ of integers from $1$ to $2022$ such that for at least $k$ distinct $i$ with $1 \le i \le 2022$ the number $\frac{a_1 + a_2 + \ldots + a_i}{1 + 2 + \ldots + i}$ is an integer larger than $1$.
[i](Proposed by Oleksii Masalitin)[/i] | [hide=Answer]$k = 1011$[/hide]
[hide=Solution]
Let $b_i = \frac{a_1 + a_2 + \ldots + a_i}{1 + 2 + \ldots + i}$.
Now, note that $k = 1011$ is achievable by letting $a_i = 2i~~\forall i \leq 1011$ and completing the permutation in any way. This will have $b_i = 2$ for $i \leq 1011$.
It remains to show the upper bound. ... |
2,804,299 | The PMO Magician has a special party game. There are $n$ chairs, labelled $1$ to $n$. There are $n$ sheets of paper, labelled $1$ to $n$.
[list]
[*] On each chair, she attaches exactly one sheet whose number does not match the number on the chair.
[*] She then asks $n$ party guests to sit on the chairs so that each cha... | Really, the problem is asking for us to create a bunch of cycles (of length $\ge 2$) on $n$ vertices such that every vertex is part of exactly one cycle and such that the lcm of cycle lengths is $m$. If $m$ is not a prime power, say its $m = st$ where $s,t$ are coprime and $> 1$, since $m > st - s - t$, there exist non... |
2,779,165 | $n$ players took part in badminton tournament, where $n$ is positive and odd integer. Each two players played two matches with each other. There were no draws. Each player has won as many matches as he has lost. Prove that you can cancel half of the matches s.t. each player still has won as many matches as he has los... | [hide]
Define a graph, where there is a directed edge $A \rightarrow B$, when $A$ won 2 times with $B$, and undirected edge $A - B$, when $A$ won one time and $B$ also won one time. Note that for every pair of vertices there exist exactly one (un)directed edge. Now let's deleted directed cycles, i.e. for a directed cyc... |
2,779,436 | Find all real quadruples $(a,b,c,d)$ satisfying the system of equations
$$
\left\{ \begin{array}{ll}
ab+cd = 6 \\
ac + bd = 3 \\
ad + bc = 2 \\
a + b + c + d = 6.
\end{array} \right.
$$ | $(a+b)(c+d) = ac + ad + bc + bd = 5, (a+b) + (c+d) = 6$ so with making a quadratic equation we have ${(a+b),(c+d)} = {1,5}$.
$(a+c)(b+d) = ab + ad + bc + dc = 8, (a+c) + (b+d) = 6$ so with making a quadratic equation we have ${(a+c),(b+d)} = {2,4}$.
$(a+d)(b+c) = ab + ac + bd + cd = 9, (a+d) + (b+c) = 6$ so with making... |
2,779,448 | Positive integers $a,b,c$ satisfying the equation
$$a^3+4b+c = abc,$$
where $a \geq c$ and the number $p = a^2+2a+2$ is a prime. Prove that $p$ divides $a+2b+2$. | A bit more straightforward: The equation is equivalent to
\[(ab-1)(ac-4)=a^4+4=(a^2-2a+2)(a^2+2a+2)\]
(classical SFFT+Sophie-Germain). Hence $p \mid ab-1$ or $p \mid ac-4$.
However, $0<ac-4<p$ since $c \le a$ whence $p \mid ab-1$. Hence
\[p \mid (a^2+2a+2)+2(ab-1)=a(a+2b+2)\]
and hence $p \mid a+2b+2$. |
276,011 | Let be given a parallelogram $ ABCD$ and two points $ A_1$, $ C_1$ on its sides $ AB$, $ BC$, respectively. Lines $ AC_1$ and $ CA_1$ meet at $ P$. Assume that the circumcircles of triangles $ AA_1P$ and $ CC_1P$ intersect at the second point $ Q$ inside triangle $ ACD$. Prove that $ \angle PDA \equal{} \angle QBA$. | Let $ (X) \equiv \odot(AA_1P)$ cut $ DA$ again at $ A_2$ and let $ (Y) \equiv \odot (CC_1P)$ cut $ CD$ again at $ C_2.$
$ \angle CQP \equal{} \angle PC_1B \equal{} \angle PAA_2 \equal{} \pi \minus{} \angle A_2QP$
$ \Longrightarrow$ $ C, Q, A_2$ are collinear and similarly, $ A, Q, C_2$ are collinear. From $ (X)$,... |
2,781,225 | Let $ABCD$ be a convex quadrilateral and let $O$ be the intersection of its diagonals. Let $P,Q,R,$ and $S$ be the projections of $O$ on $AB,BC,CD,$ and $DA$ respectively. Prove that \[2(OP+OQ+OR+OS)\leq AB+BC+CD+DA.\] | By symmetry, we only need to prove $2(OP+OR) \le AD+BC$. It follows from these facts:
[list]
[*] Let the angle bisectors of $\angle AOB$ and $\angle COD$ be $OK$ and $OL$ ($K,L$ lie on $AB,CD$, respectively). Then $OP \le OK$, $OQ \le OL$.
[*] Let $M,N$ be the midpoint of $AB,CD$, then $MN=|\overrightarrow{MN}|=|\tfrac... |
2,781,232 | Let $p\geq 3$ be an odd positive integer. Show that $p$ is prime if and only if however we choose $(p+1)/2$ pairwise distinct positive integers, we can find two of them, $a$ and $b$, such that $(a+b)/\gcd(a,b)\geq p.$ | when I see the problem,I notice that it is same as 2007 CMO P2.
|
2,794,130 | The incircle and the excircle of triangle $ABC$ touch the side $AC$ at points $P$ and $Q$ respectively. The lines $BP$ and $BQ$ meet the circumcircle of triangle $ABC$ for the second time at points $P'$ and $Q'$ respectively.
Prove that
$$PP' > QQ'$$ | I actually used power of point theorem and stewarts theorem to do this problem([url=https://drive.google.com/file/d/1PWgLmq5UlpxswUbVHiBdnoVo4Kh7-Qq7/view?usp=sharing]Problem 6[/url])
Let, $\overline{AB}=c,\overline{BC}=a,\overline{CA}=b$ and $\text{Pow}_\omega(X)$ denote the power of $X$ with respect to a circle $\o... |
2,794,161 | The products of the opposite sidelengths of a cyclic quadrilateral $ABCD$ are
equal. Let $B'$ be the reflection of $B$ about $AC$. Prove that the circle passing through $A,B', D$ touches $AC$ | so this is mainly about completing well known configurations for instance consider Brazil 2011/5
[hide = solution]
[asy]
unitsize(1.2cm);
defaultpen(fontsize(10pt));
markscalefactor = 0.03;
pair X = (2.26, 5.36);
pair A = (0.06, -1.4);
pair B = (2.34, -3.88);
pair C = (9.78, -1.32);
pair D = (0.16, 3.13);
pair Ee = (4.... |
2,794,309 | Let $K$, $L$, $M$, $N$ be the midpoints of sides $BC$, $CD$, $DA$, $AB$ respectively of a convex quadrilateral $ABCD$. The common points of segments $AK$, $BL$, $CM$, $DN$ divide each of them into three parts. It is known that the ratio of the length of the medial part to the length of the whole segment is the same for... | @above a much simpler counterexample exists, but ok. Here is my full solution, including the motivation behind how we come up with a counterexample:
The answer is $\textbf{No}$.
$\textbf{Solution:}$ I wanted to present my thought process and how I came up with the counterexample. In what's presented below I explain ... |
End of preview. Expand in Data Studio
README.md exists but content is empty.
- Downloads last month
- 8