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[/i] | 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
$$\sqrt{a^2+Cac+c^2}+\sqrt{b^2+Cbd+d^2}<\sqrt{a^2+Cad+d^2}+\sqrt{b^2+Cbc+c^2}$$
Since $C\in(-2,2)$ we can write it as $C=-2\cos\theta$ for some $\theta\in(0,\pi)$. Therefore we can interpret $f(x,y)$ as the distance $XY$ of the points $X$ and $Y$ on two halflines with an angle $\theta$ between them, with the same origin $O$, so that $OX=x$ and $OY=y$.
So if $O,A,B$ are on the half line $Or$ in this order and $O,C,D$ in this order on the half line $Os$, it suffices to show $AC+BD<AD+BC$. To do this simply pick $X=AD\cap BC$, and by triangle inequality we have the strict inequality
$$AC+BD<AX+CX+BX+DX=AD+BC$$
as wanted. |
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 $x\geq y$.
$\textbf{Case 1.}$ $y=1$.
$x^2+x+z+2=xz+1 \Rightarrow (x-1)z=x^2+x+1>0 \Rightarrow x>1$.
$z=g(x,1)=\frac{x^2+x+1}{x-1}=x+2+\frac{3}{x-1} \in \mathbb{Z}$
$\Rightarrow x=2, 4$. $(x,y,z)=(2,1,7), (4,1,7)$.
$\textbf{Case 2.}$ $x=y\geq 2$.
$g(x,x)=\frac{2x^2+2x-1}{x^2-1}=2+\frac{2x+1}{x^2-1} \Rightarrow 2x+1 \geq x^2-1 \Rightarrow x=2$.
But when $x=y=2$, $g(x,x)=2+\frac{5}{3} \notin \mathbb{N*}$, so no triplets satisfy (*) in this case.
Thus we have $x>y$.
$\textbf{Case 3.}$ $y=2$.
$g(x,2)=\frac{x^2+x+5}{2x-1}$.
$4g(x,2)=\frac{4x^2+4x+20}{2x-1}=2x+3+\frac{23}{2x-1}\in \mathbb{N*}$
$\Rightarrow (2x-1)|23 \Rightarrow x=12\Rightarrow (x,y,z)=(12,2,7)$.
$\textbf{Case 4.}$ $y\geq 3 \wedge x=y+1$.
$g(y+1,y)=\frac{2y^2+4y+1}{y^2+y-1}=2+\frac{2y+3}{y^2+y-1}\in \mathbb{N*}$
$\Rightarrow 2y+3\geq y^2+y-1$ which contradicts with $y\geq 3$.
$\textbf{Case 5.}$ $y\geq 3 \wedge x\geq y+2$.
Suppose $(x,y,z)$ satisfies $x^2+y^2+x+y+z=xyz+1 (*)$.
$z=g(x,y)=\frac{x^2+y^2+x+y-1}{xy-1}\geq \frac{2xy+x+y-1}{xy-1} = 2+\frac{x+y+1}{xy-1} > 2 \Rightarrow z\geq 3$.
Fix $y,z$, and then $a^2-(yz-1)a+y^2+y+z-1=0$ is a quadratic, with one root $x$ and so the other is $\frac{y^2+y+z-1}{x}=yz-x-1$. Because $y^2+y+z-1 > 0$, $yz-x-1 \in \mathbb{N*}$.
\begin{align*}
y'<y\quad \Leftrightarrow\quad &yz-1-x<y \\
\Leftrightarrow\quad &y\frac{x^2+y^2+x+y-1}{xy-1}<x+y+1 \\
\Leftrightarrow\quad &x^2y+y^3+xy+y^2-y<x^2y+xy^2+xy-x-y-1 \\
\Leftrightarrow\quad &y^3+y^2-y<xy^2-x-y-1 \\
\Leftarrow\quad &y^3+y^2-y<(y+2)(y^2-1)-y-1 \\
\Leftrightarrow\quad &y^2-y-3\geq 0\\
\Leftarrow\quad &y\geq 3.
\end{align*}
\begin{align*}
x'-y'<x-y\quad\Leftrightarrow\quad &y-(yz-x-1)<x-y \\
\Leftrightarrow\quad &yz>2y+1 \\
\Leftarrow\quad &z\geq 3 \wedge y\geq 2
\end{align*}
Define $f(x,y,z):=(y,yz-x-1,z)$.
Then if $(x,y,z)$ satisfies (*), so does $f(x,y,z)$.
Suppose $(x,y,z)$ satisfies (*), and then we can replace it with $f(x,y,z)$ finite times until it does not satisfy the condition $x-2\geq y\geq 3$, because each time $y$ and $x-y$ strictly decreases. At last we will certainly have $(x,y,z)=(2,1,7),(4,1,7)$ or $(12,2,7)$. Surprisingly $f(12,2,7)=(2,1,7)$. Thus for all $(x,y,z)$ satisfying (*), we can replace it with $f(x,y,z)$ finite times to become (2,1,7) or (4,1,7). (Thus $z=7$.)
If $(x,y,z)\in \mathbb{N*}^3$, then $f^{-1}(x,y,z)=(xz-y-1,x,z)\in \mathbb{N*}^3$. Let $(x_0,y_0)=(2,1)$ or $(4,1)$.
Let $(x_{n+1},y_{n+1})=(7x_n-y_n-1,x_n)$. Then the two sequences of triplets $(x_n,y_n,7)$ (with different $(x_0,y_0)$) contains "all" the triplets satisfying (*). Be careful! $(y_n,x_n,7)$ satisfies (*) as well, because WLOG at the beginning.
We have $y_{n+1}=x_n$ and so $y_{n+2}=7y_{n+1}-y_n-1$. The only thing left to do is to solve two sequences.
[/hide] |
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 $S_2$.
[i]Proposed by Viktor Simjanoski[/i] | 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,
$B \leq \dfrac{n(3n+1)}{2} <\dfrac{3n(n+1)}{2} \leq 3A$
Since $A \mid B$, this implies that $B \in \{A,2A \}$. We distinguish two cases.
[b]Case 1:[/b] $B=A$. Then, $A=\dfrac{n(2n+1)}{2},$ and so $n$ must be even. For all even $n$, we may take $S_1=\{1,2n \} \cup \{2,2n-1 \} \cup \ldots \cup (\dfrac{n}{2},(2n+1)-\dfrac{n}{2})$. It is straightforward to check that $|S_1|=n$ and $A=n(2n+1)$.
[b]Case 2:[/b] $B=2A$. Then, $A=\dfrac{n(2n+1)}{3}$, and so $3 \mid n(2n+1)$, i.e. $n \not\equiv 2 \pmod 3$. Consider the collection $\mathcal{F}$ of all sets $X \subseteq \{1,2,\ldots, 2n \}$ such that $|X|=n$. Note that the minimum sum of the elements of a set belonging in $\mathcal{F}$ is $m=1+2+\ldots+n=\dfrac{n(2n+1)}{2}$, and the maximum is $M=2n+(2n-1)+\ldots+(n+1)=\dfrac{n(3n+1)}{2}$. Note that $m<A<M$.
We claim that all intermediate sums in $[m,M]$ can be achieved by a set in collection $\mathcal{F}$. Indeed, assume all sums in $[m,t]$ have be achieved for some $t \geq m$. If $t=M$, we are done. If not, we want to find a set that achieves $t+1$. Let $T=\{x_1,\ldots,x_n \}$ be a set such that its elements sum to $t$.
If there are two elements of $T$ that are not consecutive, we may increment the smallest one of them by one and finish. Moreover, if $x_n \neq 2n$, we may increment $x_n$ by one and finish. If neither of these happens, set $T$ must necessarily be $\{n+1,n+2,\ldots,2n \}$, which is a contradiction as we assumed $t \neq M$.
To sum up, the working $n$ are the evens and the $n \not\equiv 2 \pmod 3$, that is all $n \not\equiv 5 \pmod 6$. |
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 the form $m\sqrt n$, where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$.
[asy]
defaultpen(linewidth(0.6)+fontsize(11));
size(8cm);
pair A,B,C,D,P,Q;
A=(0,0);
label("$A$", A, SW);
B=(6,15);
label("$B$", B, NW);
C=(30,15);
label("$C$", C, NE);
D=(24,0);
label("$D$", D, SE);
P=(5.2,2.6);
label("$P$", (5.8,2.6), N);
Q=(18.3,9.1);
label("$Q$", (18.1,9.7), W);
draw(A--B--C--D--cycle);
draw(C--A);
draw(Circle((10.95,7.45), 7.45));
dot(A^^B^^C^^D^^P^^Q);
[/asy] | 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 = (0,0), B = (2,5), X = (B.x,0), C = (12,5), D = A+C-B;
real r = inradius(A,B,X);
path P = scale(B.y/(2*r))*incircle(A,B,X);
pair I = B.y/(2r) * incenter(A,B,X);
pair X = foot(I,A,D), Y = foot(I,B,C), Z = foot(I,A,B), R = foot(C,A,D);
draw(X--Y^^C--R--D^^rightanglemark(C,R,D),gray(0.5));
draw(A--B--C--D--A);
draw(P);
pair[] inter = intersectionpoints(P,A--C);
draw(A--C);
label("$A$",A,SW);
label("$B$",B,NW);
label("$C$",C,NE);
label("$D$",D,SE);
label("$P$",inter[1],dir(250));
label("$Q$",inter[0],dir(330));
label("$X$",X,S,gray(0.5));
label("$Y$",Y,N,gray(0.5));
label("$Z$",Z,NW,gray(0.5));
label("$R$",R,S,gray(0.5));
[/asy]
By Power of a Point, $AX = \sqrt{AP\cdot AQ} = 6$ and $CY = \sqrt{CQ\cdot CP} = 20$. This means $AR = AX + YC = 26$, so
\[
XY = CR = \sqrt{AC^2 - AR^2} = \sqrt{28^2 - 26^2} = 6\sqrt 3.
\]
Now let $BY = BZ = t$; then $AB = 6 + t$, and so another application of Pythagorean Theorem yields
\[
6\sqrt 3 = \sqrt{(6+t)^2 - (6-t)^2} = 2\sqrt{6t}.
\]
Thus $t = \tfrac 92$, so $BC = \tfrac{49}2$ and the area of $ABCD$ is $\tfrac{49}2\cdot 6\sqrt 3 = 147\sqrt 3$. The requested sum is $\boxed{150}$.
[/hide] |
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 primes. As it is impossible for all of $p_1$, $p_2$, and $p_3$ to be odd, as an odd number plus an odd number is an even number, either $p_1=2$ and $p_2,p_3$ are twin primes or $p_2=2$ and $p_1,p_3$ are twin primes.
Write $(p_1,p_2,p_3)$ as $(2,p,p+2)$ or $(p,2,p+2)$. It can be seen that $p$ can take on any of $3$, $5$, $11$, or $17$. Given a fixed $p$, there are $18-p$ values for the largest label $c$, as it ranges from $p+3$ to $20$. Then, $b$ can either be $c-2$ or $c-p$, and $a$ is fixed accordingly, so we have
$$2[(18-3)+(18-5)+(18-11)+(18-17)]=2[15+13+7+1]=\boxed{72}$$
total triangles, as requested. |
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 neighbor? | 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 working $10$-meter configuration, and from any $10$-meter configuration we can remove a meter between every pair of birds to get a configuration of $10$ birds on the first meter.
Thus the probability we are looking for is the same as the probability that each of the birds land on a specific $\frac{1}{10}$ of the strip. Since we have $10$ birds, that probability is $\frac{1}{10^{10}}$. |
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 $a_{2020} = a_1$ for all starting values $a_1$ different from $-1$ and $1$. | [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 $\frac43 \le 1+\frac{1}{a_n + 1} \leq \frac{3}{2} \Longleftrightarrow 1 \leq a_n\leq 2$, which is true. Thus, $\boxed{a_1=[1,2]}$.
[b](b)[/b] $a_{n+1} =\frac{a_n -3}{a_n + 1}$
We want to show that $a_4=a_1$, then $a_4=\frac{\frac{a_2 -3}{a_2 + 1}-3}{\frac{a_2 -3}{a_2 + 1} + 1}=\frac{\frac{a_2 -3-3(a_2+1)}{a_2 + 1}}{\frac{a_2 -3+(a_2+1)}{a_2 + 1}}=\frac{a_2 -3-3(a_2+1)}{a_2 -3+(a_2+1)}=\frac{-2a_2-6}{2a_2 -2}=-\frac{a_2+3}{a_2 -1}=-\frac{\frac{a_1 -3}{a_1 + 1}+3}{\frac{a_1 -3}{a_1 + 1}-1}=-\frac{\frac{a_1 -3+3(a_1+1)}{a_1 + 1}}{\frac{a_1 -3-1(a_1+1)}{a_1 + 1}}=-\frac{a_1 -3+3(a_1+1)}{a_1 -3-1(a_1+1)}=-\frac{4a_1}{-4}=a_1$.
Notice that we cannot have $a_2=-1\Longleftrightarrow a_1=1$ or $a_1=-1$.
Hence, we have showed that $a_4=a_1$, hence $a_{3k+1}=a_1$ and we are done. |
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)
$16a+4b=2020$
$4a+b = 505$
$3a+b=6$
$a+b+c=-990$
[rule]
case $f(0)=20$
$f(4) = 16a+4b+20 \equiv 0 \mod 4$
forcing f(4) = 20 contradiction
case $f(0) = 202$
$f(3) = 9a+3b+202 \equiv 1 \mod 3$
forcing $f(3) = 202$ contradiction
case $f(0) = 2022$
$f(3) = 9a+3b+2022 \equiv 0 \mod 3 $
forcing $f(3) = 2022$ contradiction |
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 $NC=NM$, so $ACXM$ is a parallelogram, so
\[\angle CAM=\angle CAX+\angle XAM=\angle CBX+\angle CXA=\angle CBX+\angle CBA=\angle ABX=\angle AMN=\angle AMC\Longrightarrow CA=CM\]
$\textit{Proof 2:}$ Let $S$ be the midpoint of $AC$. Now $AMONS$ is cyclic, but $SN$ is a midsegment in $\triangle AMC$, so $SN\parallel AM$ and $AMNS$ is cyclic, therefore $AMNS$ is an isosceles trapezoid. Now $AS=NM\Longrightarrow CA=CM$.
Now, if $a=BC$, $b=AC=CM$, $c=AB$ we can use the median formula ($m_{c}=\frac{1}{2}\sqrt{2a^2+2b^2-c^2}$) to get that:
\[b^2=CM^2=\frac{1}{4}(2a^2+2b^2-c^2)\Longrightarrow c^2=2(a-b)(a+b)\]
We want $a,b,c$ to be positive integers with minimal sum (the perimeter of $\triangle ABC$). If $\gcd(a,b)=d>1$, then $\left(\frac{a}{d},\frac{b}{d},\frac{c}{d}\right)$ are the sidelengths of a triangle which satisfies the conditions in the problem with smaller perimeter. Thus we can assume that $a,b,c$ are two by two coprime. Now we have two cases:
$\textbf{Case 1:}$ $a-b=2s^2$, $a+b=4t^2$ for positive integers $s,t$.
Then $a=s^2+2t^2$, $b=2t^2-s^2$, $c=4st$, so the perimeter is $4t^2+4st$. If $s=1,t=1$ we get $(a,b,c)=(3,1,4)$, impossible. If $s=1, t=2$, then $(a,b,c)=(9,7,8)$, so the least possible perimeter in this case is $24$.
$\textbf{Case 2:}$ $a-b=4s^2$, $a+b=2t^2$ for positive integers $s,t$.
Then $a=2s^2+t^2$, $b=t^2-2s^2$, $c=4st$, so the perimeter is $2t^2+4st$. If $s=1,t=2$ we get $(a,b,c)=(6,2,8)$, impossible. If $s=1, t=3$, then $(a,b,c)=(11,7,12)$, so the least possible perimeter in this case is $30$.
In both cases, it should be briefly explained why those are the minimal perimeter cases though it's pretty obvious. Finally, the answer is $\boxed{24}$ achieved when $(a,b,c)=(9,7,8)$.[/hide]
[hide=Note]Although it may be intuitive to think that $\triangle ABC$ is acute, this is not necessarily the case. We can have $AC<BC$, $\angle C>90^{\circ}$ and $(AMO)$ bisecting $CM$. This, however, isn't a problem for the solution above.[/hide] |
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 $p$, then $x+y \geq 2p > p+y$, so $x>p$ and $p+49>2p^2$, case check. So $x+y=p, y-x=\frac {p-1}{2}$, calculate $x, y$ in terms of $p$ and plug in the system, done. |
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 every two students sitting next to one another have different numbers.
$b)$ $7$ of the students are boys, the rest are girls, and there isn't a boy and a girl with the same number. | 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] $k=4$. I follow in general the author's solution. We omit the counterexample that it is not possible for $k=3$ and let's prove that it's always possible when $k=4$. Consider a bipartite graph $G(S, N)$ where $S=\{1,2,\dots,14\}$ are the students and $N=\{n_1,n_2,\dots,n_k\}$ are all the numbers the students like. A vertex (student) $s\in S$ is connected with $n_i\in N$ if $n_i$ is a favorite number of $s$. The given condition says that every $s$ is connected with at least $4$ vertices in $N$, i.e. $d(s)\ge 4, \forall s\in S$. Let us partition the vertices $S$ into two groups $S_0$ (boys) and $S_1$ (girls). The problem actually asks to prove the following claim.
$\textbf{Claim.}$ The vertices in $N$ can be partitioned into two disjoint sets $N_0$ and $N_1$, such that each vertex in $S_0$ is connected with a vertex in $N_0$ and each vertex in $S_1$ is connected with a vertex in $N_1$.
$\textit{Proof.}$ Each partition of $N$ into two sets can be interpret as assigning to each $n_i\in N$ either a value $0$ if $n_i\in N_0$ or $1$ in case $n_i\in N_1$. Clearly the family $\mathcal{A}$ of all assignments (partitions) consists of $2^k$ elements.
For each $s\in S$ let us denote by $B_s\subset \mathcal{A}$ the set of "bad" assignments for the student $s$. That is, in case $s\in S_0$, $B_s$ consists of all assignments for which to all neighbors of $s$ is assigned $1$, and in case $s\in S_1,$ $B_s$ are those assignments for which $s$ is connected only with $0$'s. It means
$$|B_s|= \frac{2^k}{2^d}$$
where $d=d(s)$. Since $d(s)\ge 4$ we get $|B_s|\le 2^{k-4}$. Let $B:=\bigcup_{s\in S}B_s$ be all bad assignments. It yields
$$|B|=\left|\bigcup_{s\in S}B_s\right|\le \sum_{s\in S}|B_s|\le 14\cdot 2^{k-4}<2^k.$$
Therefore $|B|<|\mathcal{A}|$ which means there exists an assignment
$A\in \mathcal{A}\setminus B$. Obviously, $A$ comply with all the requirements of the Claim.
$\textbf{Comment}.$ Actually, it's a disguised probabilistic approach (or vise versa :) )Note that in the above proof it's not essential the students are $7$ boys and $7$ girls. The size of $S_0$ and $S_1$ can be whatever we want, providing $|S_0\cup S_1|=14$. Moreover, the same holds even if the number of students is $16$. But, in this case a further argument is needed. Namely, for any $i,j\in S$ that are of the same sex, that is either $i,j\in S_0$ or $i,j\in S_1$, it holds $B_i\cap B_j\neq \emptyset$. |
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 $AIN = \angle ACI$.
Let $I'$ be reflection of $I$ across $N$ and $I_c$ be reflection of $C$ across $I$. Note that $CI=2IL$ so $IL = LI_c$. Now we have $L$ is center of $AIBI_c$.
Claim : $I'$ lies on $AIBI_c$.
Proof : we had $ANI$ and $AIC$ are similar and $I'N = NI , I_cI = IC$ so $AI'I$ and $AI_cC$ are similar so $\angle AI'I = \angle AI_cI$ so $I'$ lies on $AIBI_c$.
Claim : $AN = NH$.
Proof : we had $\angle HIN = \angle HBI$ so $NH.NB = NI^2 = NI.NI' = NA.NB \implies NH = NA$.
with same approach we have $MH = MB$ so $MN = \frac{AB}{2}$.
we're Done. |
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 $2x^2 - 4x + 3 \equiv 0 \pmod 5$, impossible by direct check. For $p=7$ we get $2x^2 - 2x + 6 \equiv 0 \pmod 7$, again impossible by direct check. For $p=11$ we get $4x^2 + 2 \equiv 0 \pmod {11}$, with $x\equiv 4 \pmod {11}$ as a solution. To ensure that $k$ and $\ell$ are coprime just pick $x=59$ (hmm even $x=15$ works actually). |
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 official solution.\\
$\textbf{First observation}$: If $ M'$ is a "sum-free" set of integers modulo $ p$ the same holds for $ s\cdot M':=\{s\cdot m : m\in M\}$, where $ s$ is a natural number, and vise versa. \\
$\textbf{Second observation}$: The set $ I:= \{i : i\in [0..p-1], \frac{p}{2n}\le i\le \frac{p}{n}\}$ is sum-free modulo $ p$. Indeed, summing up two or more (but at most $ n$) elements of $ I$ will result in a sum that is in the interval $ \left(\frac{p}{n},p-1\right]$.\\
$\textbf{The key idea is}$ to find an integer $ s\in [1.. p-1]$ such that $ M_s:=s\cdot M$ has large number of elements in $ I$.
[b]Solution[/b]. We can assume all numbers in $ M$ are positive integers, otherwise they can be made such by multiplication by an appropriate integer. Let us take a prime number $ p$, greater than all the elements in $ M$. Further we consider $ M$ as a set of residues modulo $ p$. Denote by $ I$ the interval $ \displaystyle \left[\frac{p}{2n-1}, \frac{2p}{2n-1}\right].$ Note that the set $ M'$ of all integers inside $ I$ have the property that any sum of no more than $ n$ of them is not an element of $ M'$ modulo $ p$. For each $ a\in M$ let $ 1_{a, I}$ be the random variable that takes value $ 1$ if for a randomly taken integer $ s\in[1,p-1]$ we have $ s\cdot a\in I \pmod{p}$, and value $ 0$ otherwise. It holds
$$ \displaystyle \mathbb{E}[1_{a,I}]> \frac{ \frac{2p}{2n-1}- \frac{p}{2n-1}-1 }{p-1}$$
which yields
$$ \displaystyle \mathbb{E}[1_{a,I}]>\frac{1}{2n-1}-\frac{1}{p}$$
Let $ X$ be a random variable which equals $ \#\{a\in M : s\cdot a\in I \pmod{p}\}$ where $ s$ is a randomly taken integer in $ [1,p-1]$. Clearly, $X=\sum_{a\in M} 1_{a,I}.$ Using the linearity of expectation, we get
$$ \displaystyle \mathbb{E}[X]=\sum_{a\in M}\mathbb{E}[1_{a,I}] > m\cdot \left(\frac{1}{2n-1}-\frac{1}{p}\right)$$
Suppose we have already checked that
$$ \displaystyle m\cdot \left(\frac{1}{2n-1}-\frac{1}{p}\right)> n-1 \qquad(1)$$
It means $ \displaystyle \mathbb{E}[X]>n-1,$ therefore $X$ takes $ n$ as a value. Thus, there exists $ s\in [1..p-1]$ such that the set $ M'_s:=\{a\in M : s\cdot a\in I\}$ satisfies $ |M'_s|\ge n$. Hence, the elements of $ M'_s$ satisfy the requirement of the problem.
It remains to check that $ (1)$ holds. Since the prime $ p$ can be chosen as large as we want, it is enough to check
$$ \displaystyle m\cdot \left(\frac{1}{2n-1}\right)> n-1$$
which is a matter of simple calculation.
[b]Comment[/b]. The restriction of numbers being positive is redundant. The same proof can be used, just omitting the word "positive".
The official solution is like in $\#$2 (without the glitches and typos). Btw, no one managed to solve it at the competition - all the contestants scored only 1 point. The claim also holds for any real numbers. More details and comments can be found [url=https://dgrozev.wordpress.com/2022/04/01/bulgarian-2022-spring-math-competition-part-1/]in my blog.[/url] |
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 $VQ'=VQ$ and $VN'=VN$. Then $Q'N'$ and $MP$ also intersect at $H.$ We have now simplified a 3D problem into a 2D problem and can be graphed like so: [asy]defaultpen(fontsize(10pt));
pair V,Q,M,A,H,P,N,C;
V=(0,0);Q=(0,12/5);M=(0,4);A=(0,6);H=(4/3,4/3);P=(2,0);N=(3,0);C=(6,0);dot("$V(0,0)$",V,SW);dot("$A(0,6)$",A,W);dot("$M(4,0)$",M,W);dot("$Q'$",Q,W);draw(A--V);
dot("$P(2,0)$",P,S);dot("$N'(3,0)$",N,S);dot("$C(6,0)$",C,S);draw(C--V--H);draw(Q--N);draw(M--P);dot("$O$",H,NE);[/asy]
Now we can do some light coordinate bashing to find that $O=\left(\tfrac43,\tfrac43\right),$ and then from there $Q'=\left(0,\tfrac{12}5\right).$
So $VQ:QD=VQ':Q'A=\frac{12}5:6-\frac{12}5=\fbox{2:3}.$[/hide]
I haven't done math in years; hope this solution is ok |
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 (adjacent meaning sharing an edge) have the same $A_i$ color or the same $B_i$ color, and there is a region colored $A_i, B_j$ for any combination of $A_i, B_j$.
A number $n$ is [i]colourable[/i] if there is a coloring for any set of $n$ good lines. Find all colourable $n$. | 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 numbers get written in regions of both colors.
[*] Two adjacent regions don't have the same number written in them.
[/list]
It isn't hard to verify all $n \le 4$ don't work. This is true even when lines are not allowed to be parallel! For $n \le 3$, this is easy to see. For $n=4$, we do the work below:
[img]https://i.imgur.com/MyjygkE.png[/img]
Assume contrary. Note any configuration of four lines will look something like above. We color the regions red and blue. WLOG assume that the inner red quadrilateral region has $1$ written in it. Now $2,3$ should also be written in red regions. The only four red regions are around $1$. Now $2,3$ must be written diagonally opposite regions, otherwise a blue region sandwiched between three of those red regions can have no number written in it. So we have two choices. Both of them are shown below.
[img]https://i.imgur.com/hU9Brtx.png[/img]
And the other case is:
[img]https://i.imgur.com/1if2xjD.png[/img]
In both cases, numbers of rest of regions get automatically determined and $1$ gets written in none of the of the blue regions, which is a contradiction. $\blacksquare$
[rule]
Now we prove that for each $n \ge 5$, a construction exists.
[b][color=#f00]Claim: [/color][/b] If we could find a red region $R_0$ and a blue region $B_0$ such that their distance (wrt $G$) is $>3$, then we are done (i.e. such a construction exists).
[i]Proof: [/i] Intially write a $1$ in all red regions and a $2$ in all blue regions. Denote by $N(v)$ the set of neighbours of $v$. We do some rewriting:
[list]
[*] Rewrite a $3$ in all regions of $N(R_0) \cup N(B_0)$.
[*] Rewirte a $2$ in $R_0$.
[*] Rewrite a $1$ in $B_0$.
[/list]
Note that all numbers get written in regions of both colors. So we just have to note no two adjacent regions have same number written in them. By construction, we just have to verify that no two regions in $N(R_0)$ and $N(B_0)$ are adjacent, and that just follows by the fact that distance between $R_0$ and $B_0$ is $>3$. $\square$
So lastly, to solve the problem it suffices to show that we can always find two regions that have distance $\ge 5$ among them. To prove this, we use a similar idea as in the [url=https://cms.math.ca/wp-content/uploads/2022/03/2022CMO-exam-en.pdf]official solution[/url]. Due to rotation WLOG assume that none of the lines is horizontal. Further by translation we may assume all regions lie above the $x$-axes (for infinite regions, we only look at their relevant part). Let $R_1$ denote the region to the almost right and $R_2$ denote the region to the almost left. Then $R_1,R_2$ have a distance of $\ge n$ among them, because of the following: For any line $\ell$ among those $n$ lines, $R_1,R_2$ lie on opposite sides of $\ell$. So to go from $R_1$ to $R_2$, we have to cross all the $n$ lines. In a step we can jump over at most one line, so at least $n$ edges are required. $\blacksquare$
This completes the proof!
|
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:
$$LHS = 2+\sum_{i=1}^{n-1}\frac{1}{{n \choose i}} \leq 2+(n-2)\frac{2}{n(n-1)} < 2.022$$
which implies that we must show that, there exists an $n$ such that:
$$\frac{n-2}{n(n-1)} < 0.011$$
but notice that we have that:
$$\lim_{n \rightarrow \infty} \frac{n-2}{n(n-1)} = 0$$
Which means that as $n$ gets bigger the fraction gets smaller and smaller, thus there exists an $n$ so that the sum is less than $2.022$ |
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 $j ~(1 \le j \le 240)$ such that $z_j \in \Gamma$.
(2) For any open arc $\gamma$ of length $\pi/3$ on $C$, there are at most $120$ of $j ~(1 \le j \le 240)$ such that $z_j \in \gamma$.
Find the maximum of $|z_1+z_2+\ldots+z_{240}|$. | 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+$$z_{k+40}+z_{k+80}+z_{k+120}+z_{k+160}+z_{k+200},(1\leqslant k \leqslant 40).$For the complex number $\omega_k=a_{0 k}+a_{1 k}+a_{2 k}+a_{3 k}+a_{4 k}+a_{5 k},(1\leqslant k \leqslant 40),a_{i j}=z_{40 i+j}$,it should satisfy the following $2$ conditions:
(1) For any open arc $\Gamma$ of length $\pi$ on the unit circle, at most $5$ of $a_i (1 \leqslant i \leqslant 6)$ are on $\Gamma$.
(2) For any open arc $\gamma$ of length $ \frac \pi 3$ on the unit circle, at most $3$ of $a_i (1 \leqslant i \leqslant 6)$ are on $\gamma$.
So we have$|\omega|=|a_0+a_1+a_2+a_3+a_4+a_5| \leqslant |(a_0+a_5)+(a_1+a_4)+a_2+a_3|$
$\leqslant |2\cos(\frac \pi 2+\alpha)+2\cos(\frac \pi 6+\beta)+1+1|\leqslant 2+\sqrt3(\alpha,\beta\in[0,\frac\pi2])$
so$\displaystyle|\sum_{i=1}^{240}z_i|=|\sum_{i=1}^{40}\omega_i|\le40(2+\sqrt3).$
And when $z_1=z_2=\cdots=z_{40}=i,z_{41}=z_{42}=\cdots=z_{80}=-i,z_{81}=z_{82}=\cdots=$$z_{120}=\frac{\sqrt3}2+\frac 12i,$
$z_{121}=z_{122}=\cdots=z_{160}=\frac{\sqrt3}2-\frac 12i,z_{161}=z_{162}=\cdots=z_{240}=1$,we have$\displaystyle|\sum_{i=1}^{240}z_i|=80+40\sqrt3$
. |
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 of $A_1,A_2,\ldots,A_m$ contains both black and white elements. | [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 black and white such that each vertex in $S$ has at least one black neighbor and at least one white neighbor. The edges will be denoted by ordered pairs $(s,t)$ where $s\in S$ and $t\in T$.
By Hall's marriage theorem, there exists an injection $M\colon S\to T$ such that $M(s)\in N_G(s)$ for all $s\in S$. Let $E(M)$ denote the set of pairs $\{(s,M(s))\mid s\in S\}$.
The main claim is as follows:
[b]Claim.[/b] Let $k$ be an integer with $0\le k\le m$. Then there exists a subgraph $G_k(S,T,E_k)\subset G(S,T,E)$ and a subset $S_k\subset S$ such that the following conditions hold:
[list=1][*] $|S_k|=k$,
[*] $E(M)\subset E_k$,
[*] for a vertex $s\in S$ we have $\deg_{G_k}(s)=\begin{cases}2 & s\in S_k \\ 1 & s\notin S_k\end{cases}$,
[*] $N_{G_k}(S_k)\cap M(S\setminus S_k)=\emptyset$, and
[*] there exists a coloring of $N_{G_k}(S_k)$ in black and white such that for each vertex $s\in S_k$, the two neighbors of $s$ in $G_k$ are of distinct colors.
[/list]
[b]Proof.[/b] We prove the claim by induction on $k$. For the base case $k=0$ choose $E_0=E(M)$, $S_0=\emptyset$ and the empty coloring. Now assume the claim is true for some $k<m$ and we will prove it for $k+1$. Pick a subgraph $G_k(S,T,E_k)\subset G(S,T,E)$, a subset $S_k\subset S$ and a coloring $c_k$ of $N_{G_k}(S_k)$ satisfying the induction hypothesis. Define $Y_k\doteqdot S\setminus S_k$. By the problem statement we have $N_G(Y_k)\ge |Y_k| + 1$. Hence there exist vertices $u_0\in Y_k$ and $v_0\in T\setminus M(Y_k)$ such that $e_0\doteqdot(u_0,v_0)\in E$. Moreover we have $M(u_0)\notin N_{G_k}(S_k)$ by condition (4) in the induction hypothesis. Now define a subgraph $G_{k+1}(S,T,E_{k+1})\subset G(S,T,E)$ by $E_{k+1}\doteqdot E_k\cup\{e_0\}$. Moreover define a subset $S_{k+1}\doteqdot S_k\cup\{u_0\}\subset S$. It is easily seen that conditions (1)-(4) hold. It remains to provide a coloring of $N_{G_{k+1}}(S_{k+1})$ satisfying the induction step. We divide into cases.
[i]Case 1.[/i] $v_0\in N_{G_k}(S_k)$. Then use the coloring $c_k$ and color $M(u_0)$ with the color not presented in $v_0$.
[i]Case 2.[/i] $v_0\notin N_{G_k}(S_k)$. Then use the coloring $c_k$, color $v_0$ white and $M(u_0)$ black.
In both cases the new coloring satisfies condition (5). Hence the induction step is proved. $\square$
To finish the proof of the problem, substitute $k=m$ in the claim. Then we have a graph $G_m(S,T,E_m)\subset G(S,T,E)$ such that $\deg_{G_m}(s)=2$ for every $s\in S$, and there exists a coloring of $N_{G_m}(S)$ such that for each vertex $s\in S$, the two neighbors of $s$ in $G_m$ are of distinct colors. Now color the rest of vertices in $T$ arbitrarily (e.g color all of them white). This coloring satisfies the problem statement.[/hide]
Here is an informal and more intuitive version of the solution for better understanding.
[hide=Informal solution]Let $X\doteqdot\bigcup_{i=1}^m A_i$. By Hall's marriage solution, there exists distinct elements $a_1,\ldots,a_m\in X$ such that $a_i\in A_i$ for every $i\in[m]$. By the conditions of the problem we have $|X|\ge m+1$. Hence there exists an element $b_1\in X$ such that $b_1\notin\{a_1,\ldots,a_m\}$. WLOG we have $b_1\in A_1$. By the conditions of the problem again we have $\left|\bigcup_{i=2}^m A_i\right|\ge m$. Hence there exists an element $b_2\in\bigcup_{i=2}^m A_i$ with $b_2\notin\{a_2,\ldots,a_m\}$. We can assume WLOG that $b_2\in A_2$. Continuing this way, after rearranging the $A_i$ we can assume that there exist $b_1,\ldots,b_n\in X$ such that $b_i\in A_i$ for every $i\in[m]$, and moreover $b_i\ne a_j$ for every $i,j\in[m]$ with $i\le j$. We will inductively color the pairs $\{a_1,b_1\},\{a_2,b_2\},\ldots,\{a_m,b_m\}$ such that at each step, if an element was colored previously then we will skip it. Moreover, we will make sure that for each $k\in[m]$, each pair among $\{a_1,b_1\},\{a_2,b_2\},\ldots,\{a_k,b_k\}$ is colored with both colors. At the first step, note that $a_1\ne b_1$ so we can color $a_1$ black and $b_1$ white. For the induction step, assume $2\le k\le m$ and $\{a_1,b_1\},\{a_2,b_2\},\ldots,\{a_{k-1},b_{k-1}\}$ are already colored. Since $a_k\notin\bigcup_{i=1}^{k-1}\{a_i,b_i\}$ we can color it as we wish. If $b_k$ is already colored, color $a_k$ with the opposite color. Otherwise color $a_k$ black and $b_k$ white. Now we have obtained a partial coloring of $X$ such that each set $A_i$ consists of elements of both colors. Then we can color the rest of elements of $X$ arbitrarily (e.g. color all of them white). This finishes the solution. |
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 elements belong to all but one residue mod $p$. shift the set so that infinitely many elements are multiples of $p$; then, divide every element by the GCD of everything if it isn't 1, then ignore all elements smaller than or equal to the greatest non-multiple of $p$, and divide remaining elements by $p$ to arrive at $A_{i+1}$.
If this process terminates, then we can reverse-engineer $B$ from our choice of $B_i$. Else, suppose that the process does not terminate after $p-1$ steps. Then, after appropriate shifting, for some positive integer $n\ge p-1$, only finitely many elements in $A$ are not divisible by $p^n$. Let the set of these elements be $C$; we know that $|C|\ge p-1$, and moreover $|\{\nu_p(c)|c\in C\}|\ge p-1$ by considering the nonzero residues "left behind" at each iteration of the process.
We are done now by picking $p-1$ arbitrary elements of $C$ with distinct $\nu_p$, and $p-1$ elements of $A\setminus C$ greater than $p$ times the supremum of $C$. |
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 Eulerian cycle. | @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 is on a side of the grid but is not a corner. Then all strange vertices have degree $1$ in $G$, and all other vertices have even degree.
[b]Claim:[/b] the set of strange vertices can be partitioned into pairs $(u_1,v_1)$, $(u_2,v_2), \dots, (u_k,v_k)$ such that there exists a path from $u_i$ to $v_i$ for all $1\le i\le k$, and these $k$ paths are disjoint.
[b]Proof:[/b] fix a strange vertex $u$, and walk along edges of $G$ until we are stuck (i.e. we are at a vertex $v$ such that all edges incident to $v$ have already been walked on). Then $v$ must be strange, so we can remove the path between $u$ and $v$ from $G$ and repeat this process $k-1$ times.
Define $P_i$ as the path between $u_i$ and $v_i$. For any $i$, $P_i$ divides the remaining strange vertices into two sets -- let $f(i)$ be the magnitude of the smaller of these two sets.
Now, color the vertices of the grid black and white in a checkerboard pattern. Then for all $1\le i\le k$, $u_i$ and $v_i$ are the same color. The condition on diagonals implies that no path between two black vertices can cross a path between two white vertices.
[b]Claim:[/b] for all $1\le i\le k$, $u_i$ and $v_i$ lie on different sides of the grid.
[b]Proof[/b]: Assume the contrary, and choose the $i$ with $f(i)$ minimal such that $u_i$ and $v_i$ lie on the same side of the grid. Let $S$ be the set of vertices between $u_i$ and $v_i$ on that side of the grid. WLOG let $u_i$ and $v_i$ be black. Then by the minimality of $f(i)$, any white vertex in $S$ is paired with a white vertex not in $S$ -- a contradiction and the claim is proved.
[b]Claim:[/b] either $m = n$ or for all $1\le i\le k$, $u_i$ and $v_i$ do not lie on opposite sides of the grid.
Proof: assume the contrary -- that $m\neq n$ and there exists some $i$ such that $u_i$ and $v_i$ lie on opposite sides of the grid. Let $S_1$ and $S_2$ be the sets of strange vertices on either side of $P_i$. Note that $|S_1|$ and $|S_2|$ are both odd. WLOG let $u_i$ and $v_i$ be black and lie on the top and bottom sides of the grid, respectively. We have two cases.
Case 1: there are an odd number of white vertices in $S_1$. Then at least one of these vertices must be paired with a white vertex in $S_2$, and we have a contradiction.
Case 2: there are an odd number of black vertices in $S_1$. Then at least one black vertex $u$ in $S_1$ must be paired with a black vertex $v$ in $S_2$. If either $u$ or $v$ lies on a vertical side of the grid, then $P_i$ and the path between $u$ and $v$ will "box in" at least one white strange vertex, forcing a contradiction to the previous claim. Hence, $u$ and $v$ lie on the left and right sides of the grid (in some order). Now consider the four quadrants into which $P(i)$ and the path between $u$ and $v$ divide the grid. In any one of these quadrants, the number of white strange vertices on the vertical side must equal the number of white strange vertices on the horizontal side. In particular, the total number of white strange vertices on the top and bottom sides of the grid equals the total number of white strange vertices on the left and right sides of the grid. From here, quick casework on the parities of $m$ and $n$ is enough to conclude that $m = n$, so we have a contradiction and the claim is proved.
Now, the claim implies that the total number of strange vertices on the top and bottom sides of the grid equals the total number of strange vertices on the left and right sides of the grid. This immediately implies $m = n$, as desired.
Finally, we have the following construction for $m = n = 6$ which generalizes readily for all $m = n$.[/hide] |
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 $P_1P_2$ intersects the circumcircle of $ABC$ at $Q_1$ and $Q_2$. The Simson lines of $Q_1$, $Q_2$ with respect to $ABC$ intersect at $W$.
Prove that $W$ lies on the nine-point circle of $ABC$. | 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, let $\gamma$ be the apolloniam circle of $A$ with respect to the segment $BC.$ Then, points $P_1$ and $P_2$ are the intersections of $\omega$ and $\gamma.$[/i]
[hide=proof]
This is just angle chasing. First let $X_1,Y_1,Z_1$ be the projections of $P_1$ with respect to $BC,AC,AB,$ respectively. It is well known that $\triangle X_1Y_1Z_1 \sim \triangle D_1E_1F_1$ (I know it as Lemma 2.5 from [i]Geometry of Conics, A. V. Akopyan, A. A. Zaslavsky;[/i] I hope this is well known; it is just angle chasing anyways ).
[asy]
import graph;
import geometry;
import olympiad;
size(200);
pair A,B,C,O,M,Ob,Oc,Oba,Oca,K,Oa,Oaa,L;
A=dir(126.791);
B=dir(200.87);
C=dir(-20.87);
O=(0,0);
M=B*0.5+C*0.5;
Ob=2*B-O;
Oc=2*C-O;
Oba=bisectorpoint(Ob,O);
Oca=bisectorpoint(Oc,O);
K=extension(B,Oba,C,Oca);
Oa=2*A-O;
Oaa=bisectorpoint(O,Oa);
L=extension(A,Oaa,B,C);
path u,p,q,r;
u=Circle(O,1,400);
p=Circle(K,distance(K,B),400);
q=Circle(L,distance(L,A),400);
r=circumcircle(B,O,C);
pair Px=intersectionpoints(p,q)[0];
pair Py=intersectionpoints(p,q)[1];
path l =Px--Py;
pair Qx=intersectionpoints(l,u)[0];
pair Qy=2*O-Qx;
pair Da,Ea,Fa;
Da=foot(Px,B,C);
Ea=foot(Px,A,C);
Fa=foot(Px,A,B);
fill(A--Fa--Px--Ea--cycle, 0.9*white+ 0.1*cyan);
fill(B--Da--Px--Fa--cycle, 0.9*white+0.1*green);
fill(C--Da--Px--Ea--cycle,0.9*white+0.1*magenta);
draw(A--B--C--cycle);
draw(Ea--Da--Fa--cycle);
draw(A--Px--C);
draw(Px--B);
draw(Ea--Px--Fa);
draw(Px--Da);
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$P_1$",Px,dir(140));
dot("$X_1$",Da,dir(Da));
dot("$Y_1$",Ea,dir(Ea));
dot("$Z_1$",Fa,dir(Fa));
[/asy]
By problem condition we must have that
$$ \measuredangle ABP_1 + \measuredangle P_1CA = \frac{\pi}{2}$$
and
$$ \measuredangle CBP_1 + \measuredangle CAP_1 = \measuredangle P_1CB + \measuredangle BAP_1$$
We can easily manipulate this to get $$\measuredangle BP_1C= \measuredangle A + \frac{\pi}{2} $$ so it follows easily that $K$ is indeed the circumcenter of $\triangle BP_1C.$ Finally, we have that by law of sines at $(BX_1P_1Z_1)$ and $(CX_1P_1Y_1),$
$$ \frac{P_1X_1}{X_1Y_1}=\frac{P_1X_1}{X_1Z_1} \iff \frac{\sin \angle P_1BC}{\sin \angle ABC} = \frac{\sin \angle P_1CB}{\sin \angle ACB} $$
so it follows that $P_1$ lies on $\gamma$. $\square.$
[/hide]
[asy]
import graph;
import geometry;
import olympiad;
size(300);
pair A,B,C,O,M,Ob,Oc,Oba,Oca,K,Oa,Oaa,L;
A=dir(126.791);
B=dir(200.87);
C=dir(-20.87);
O=(0,0);
M=B*0.5+C*0.5;
Ob=2*B-O;
Oc=2*C-O;
Oba=bisectorpoint(Ob,O);
Oca=bisectorpoint(Oc,O);
K=extension(B,Oba,C,Oca);
Oa=2*A-O;
Oaa=bisectorpoint(O,Oa);
L=extension(A,Oaa,B,C);
path u,p,q,r;
u=Circle(O,1);
p=Circle(K,distance(K,B));
q=Circle(L,distance(L,A));
r=circumcircle(B,O,C);
pair Px=intersectionpoints(p,q)[0];
pair Py=intersectionpoints(p,q)[1];
path l =Px--Py;
pair Qx=intersectionpoints(l,u)[0];
pair Qy=2*O-Qx;
pair Ota=2*B-Oba;
draw(u,red);
draw(p,green);
draw(q,blue);
draw(A--B--C--cycle);
draw(Px--Py--Qx--Qy--cycle,dotted+magenta);
draw(K--Ota,gray);
draw(K--Oca,gray);
dot("$A$",A,dir(90));
dot("$B$",B,dir(150));
dot("$C$",C,dir(30));
dot("$K$",K,dir(K));
dot("$O$",O,dir(90));
dot("$P_1$",Px,dir(150));
dot("$P_2$",Py,dir(Py));
dot("$Q_1$",Qx,dir(-90));
dot("$Q_2$",Qy,dir(Qy));
[/asy]
By the [i]Key Lemma,[/i] we must have that line $P_1P_2$ is the radical axes of $\omega$ and $\gamma.$ Hence, since $(ABC)$ is ortogonal to both $\omega$ and $\gamma,$ we must have that $O,$ the circumcenter of $\triangle ABC,$ must lie on $P_1P_2.$
For finish, let $H$ be the orthocenter of $\triangle ABC,$ $U$ be the intersection of the Simson lines of $Q_1$ and $Q_2,$ and $T$ be the reflection of $H$ across $U.$ By above, we must have that $Q_1,Q_2$ are diametrically oppositive and it follows easily from angle chasing that their Simson lines are perpendicular $(i)$. It is well known that the Simson line of $Q_1$(respectively $Q_2$) bissects segment $HQ_1$ (respectively $HQ_2$); so it follows that $TQ_1, TQ_2$ is parallel to the Simson line of $Q_1,Q_2,$ repsectively; then it follows from conclusion $(i)$ that $\angle Q_1TQ_2 = {\pi}{/2}.$ But since points $Q_1$ and $Q_2$ are diametrically oppositive, this implies that $T$ lies on $(ABC)$ and so $U$ must lie on the nine-point circle (beacause the homothety with ratio ${1}{/2}$ centered at $H$ sends $(ABC)$ to the nine-point circle). So done. |
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 $n$,
\[ k ~\Big|~ \sum_{d|k} f(d). \] | [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
\[n \mid \sum_{d \mid m} \binom{n/d}{m/d} \sum_{e \mid d} \mu\left(\frac{d}{e}\right) g(e)=\sum_{e \mid m} g(e)h(m/e,n/e)\]
where
\[h(m,n):=\sum_{d \mid m} \mu(d) \binom{n/d}{m/d}.\]
Now clearly if we can prove that $n \mid h(m,n)$ for all $m,n$, then we see that $e \mid g(e)$ for all $e$ implies that
\[n=e \cdot \frac{n}{e} \mid g(e)h(m/e,n/e)\]
for all $e$ and hence one direction follows. Conversely, if we assume the second condition and $n \mid h(m,n)$ for all $m,n$, then to prove $m \mid g(m)$ for all $m$, by induction we can assume that $n \mid g(m)h(1,n/m)=g(m) \cdot \frac{n}{m}$ and hence $m \mid g(m)$ as well.
So everything boils down to proving the claim that $n \mid h(m,n)$, in other words
\[n \mid \sum_{d \mid m} \mu(d)\binom{n/d}{m/d}.\]
But by Inclusion-Exclusion, the RHS counts the number of choices of a set of $m$ elements from $n$ objects in a circle such that the $n$ translates of this set are all distinct. But these sets of course fall into orbits of size $n$, so their number is clearly divisible by $n$, as desired. Done!
[/hide] |
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 $\angle APC=\angle BPD$. | 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 concentric circles implies by symmetry that the arc $A^*C^*$ is equal to the arc $B^*D^*$, so we are done. |
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 $(k_{i,1},k_{i,2},\ldots,k_{i,n})$ for $i=1,2,\ldots,M$ such that for any $1 \le i_1 <i_2 \le M$, there exists $j \in \{1,2,\ldots,n\}$ such that $k_{i_1,j}-k_{i_2,j} \not\equiv 0, \pm 1 \pmod{a_j}$. | [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 $\frac{1}{2^n} (a_1-1) \prod\limits_{j=2}^n a_j$
Proof: we induct on $n$.
Base Case: $n=1$, clear
Inductive Step: For $i=0,\cdots, a_n-1$, let $X_i$ denote the union of planes $x_n=2i, x_n=2i+1$. By inductive hypothesis, we can put at most $2^{1-n} (a_1-1)a_2 \cdots a_{n-1}$ points in $X_i$ (since which row you put don't matter). Summing $X_i$ over $0\le i\le n-1$ counts everything twice, and dividing by 2 yields the desired result.
Therefore this problem is about characterising all equality cases.
Claim: it suffices to solve the problem when $a_i$ is odd for all $i$.
Proof: if $a_i$ is even then we can split into $X_0, X_1, \cdots, X_{a_i/2-1}$ and use induction to complete our construction. Alternatively, if $(a_1,\cdots,a_{i-1}, a_{i+1},\cdots, a_n)$ fails, then $(a_1,\cdots,a_n)$ fails by mimicking the inductive argument.
At this point M must be an integer since it is an upper bound, so $2^n|a_1-1$ is necessary.
Now I claim this is also sufficient. We first solve the problem for $a_1=\cdots=a_n=k2^n+1$. On the line $x_2=c_2, x_3=c_3, \cdots, x_n=c_n$ for fixed $c_2,\cdots,c_n$ place points with $x_n\in \{0,2,\cdots, 2(k-1)\} + 2k\sum\limits_{j=1}^{n-1} 2^{j-1}c_{j+1}$.
We will prove there is no conflict between $x_2=c_2, \cdots, x_{n}=c_{n}$ and $x_2=c_1+e_2, \cdots, x_{n-1}=c_{n-1}+e_{n}$ for $e_j\in \{-1,0,1\}$ for $2\le j\le n$ and $e_j$ are not all zero.
Note the set of points on $x_2=c_2+e_1, \cdots, x_{n}=c_{n}+e_{n-1}$ is precisely the set of points on $x_2=c_2, \cdots, x_{n}=c_{n}$ shifted by $2k \sum\limits_{j=1}^{n-1} 2^{j-1}e_{j+1}$. Note $0<|\sum\limits_{j=1}^{n-1} 2^{j-1}e_{j+1}|\le 2^{n-1}-1$. For simplicity let $Y=\sum\limits_{j=1}^{n-1} 2^{j-1}e_{j+1}$
Let $[X,X+2k-2]$ be the interval of points on the line $x_2=c_2, \cdots, x_{n}=c_{n}$, then $[X+2kY, X+2kY+2k-2]$ is the interval of points on the line $x_2=c_2, \cdots, x_n=c_n$. WLOG $Y$ is positive, and we can easily check there is no conflict when $1\le Y\le 2^{n-1}-1$ by only checking $Y=1, Y=2^{n-1}-1$.
Now we extend to $a_2,\cdots,a_n$ being odd integers at least $a_1$. To go from $(a_1,\cdots,a_{i-1}, a_i, a_{i+1}, \cdots,a_n)$ to $(a_1,\cdots,a_{i-1}, a_i+2, a_{i+1}, \cdots,a_n)$, we place the same set of points on $x_i=a_i-1$ on the plane $x_i=a_i+1$ and the same set of points on $x_i=a_i$ to $x_i=a_i+2$. This way, there is no conflict with a fixed $x_i$, no conflict between planes $x_i=a_i-1, x_i=a_i$ implies no conflict between planes $x_i=a_i, x_i=a_i+1$ and no conflicts between $x_i=a_i+1, x_i=a_i+2$, and no conflict between planes $x_i=1, x_i=a_i$ implies no conflict between planes $x_i=1, x_i=a_i+2$, completing the extension step.
At each step, for a fixed value of $x_2,\cdots,x_n$ there are exactly $k=\frac{a_1-1}{2^n}$ points, so this construction works.
[/hide] |
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 necessarily lattice points), and the intersection of any two triangles is an empty-set, a common point or a common edge. | 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 triangles with integer centroids (calls such triangles good) $ABC$ and $A'BC$, the vertices $A$ and $A'$ differ by three times an integer vector
Proof: $A'-A=3(\frac{A'+B+C}{3}-\frac{A+B+C}{3})$. Since the two centroids have integer coordinates, their differences is an integer vector, so we are done.
Now, using the lemma, divide the vertices of the triangulation in ($\leq$) $3$ equivalence classes where the equivalence relation is differing by three times an integer vector [from any vertex we can reach a vertex of a fixed triangle, by multiple steps of the form $A\to A'$ if there is a segment $BC$ such that $ABC,A'BC$ both belong to the triangulation].
Since the square has $4>3$ vertices, at least two will be in the same of the three equivalence classes. Let $\Delta x$ be the difference between the $x$ coordinates of these two vertices, and $\Delta y$ that of $y$ coordinates. If the two vertices are opposite wrt the square we have $\Delta x^2+\Delta y^2=2k^2$. Since the two vertices are equivalent, $3|\Delta x,\Delta y\implies 3|\Delta x^2+\Delta y^2=2k^2\implies 3|k$. Similarly if the two vertices are adjacent wrt the square, $3|\Delta x^2+\Delta y^2=k^2\implies 3|k$. In both cases, $k$ is divisible by $3$, so we are done. |
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:\[N=\bigg\{\sum_{i=0}^\infty \lambda_i\cdot 13^i:\lambda_{2j}\in S, \ \lambda_{2j+1}\in\{0,1,\ldots,12\} \ \forall j\geq 0\bigg\}.\]Let's firstly show that $N$ is indeed square-difference-free. Assume this weren't the case. Therefore, for two sequences $(\alpha_i)$ and $(\beta_i)$ satisfying $\alpha_{2j},\beta_{2j}\in S$ and $\alpha_{2j+1},\beta_{2j+1}\in \{0,1,\ldots, 12\}$ we have \[\sum_{i=0}^\infty (\alpha_i-\beta_i)\cdot 13^i=K^2.\]Let $t$ be the smallest index for which $\alpha_t\neq\beta_t.$ Such an index clearly exists, as $(\alpha_i)\neq(\beta_i).$ It then follows that \[k\cdot 13^{t+1}+(\alpha_t-\beta_t)\cdot 13^t=K^2.\]If $t$ is odd then as $\nu_{13}(K^2)=t$ we reach a contradiction. On the other hand, if $t$ is even, then we get $13k+(\alpha_t-\beta_t)=K'^2$ which is another contradiction, since $\alpha_t\neq\beta_t\in S$ so $\alpha_t-\beta_t$ is not a quadratic residue modulo $13$.
Therefore $N$ is square-difference-free. It only remains to estimate the number of elements of $N$ smaller than some $t$. Note that $\lambda_i\leq 12$ for all indices, so we can infer that $\lambda_0\cdot 13^0+\cdots+\lambda_k\cdot 13^k\leq 12(13^0+\cdots+13^k)=13^{k+1}-1.$ Thus \begin{align*}N_t=\Bigg\{\sum_{i=0}^{\lfloor\log_{13}t\rfloor} \lambda_i\cdot 13^i:\lambda_{2j}\in S, \ \lambda_{2j+1}\in\{0,1,\ldots,12\} \ \forall j\geq 0\Bigg\}\subseteq \{1,2,\ldots,t\}.\end{align*}Bearing in mind that there are $\lfloor k/2\rfloor$ even numbers $\leq k$ and $\lfloor (k-1)/2\rfloor$ odd numbers $\leq k$ we can conclude that\begin{align*}|N_t|&=3^{\lfloor\lfloor\log_{13}t\rfloor/2\rfloor}\cdot 13^{\lfloor(\lfloor\log_{13}t\rfloor-1)/2\rfloor}\geq \sqrt{3^{\log_{13}t}\cdot 13^{\log_{13}t-1}}=1/\sqrt{13}\cdot \sqrt{t^{1+\log_{13}3}}.\end{align*}Note that $N_t\subset N$ so $N_t$ is square-difference-free. Therefore, by choosing $c=1/\sqrt{13}$ and $\alpha=(1+\log_{13}3)/2$ we win! |
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 blue square that has a common edge with it, then color it blue.
[*] For a blue square, if there is a red square that has a common edge with it, then color it red.
[/list]
It is known that after several rounds, all unit squares of this $n \times n$ grid have the same color. Prove that the grid has became monochromatic no later than the end of the $(2n-2)$-th round. | 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 the end.
-----
Claim 1: for each square $X$, either $s(X) = 0$ or there exists a square $Y$ next to $X$ with $s(Y) \leq s(X) - 1$.
Proof: Suppose $s(X) = k > 0$. Then $X$ turns from blue to red at round $k$, which means that at round $k - 1$ there is a red square $Y$ next to $X$.
The square $Y$ cannot change color after round $k - 1$: if it changed to yellow at some round $m \geq k$, then $X$ would also change to yellow at round $m + 1$, contradicting the defintion of $s(X)$.
Therefore $Y$ remains red after round $k - 1$, i.e. $s(Y) \leq k - 1$.
Claim 2: if two squares $X, Y$ are next to each other, then $|s(X) - s(Y)| \leq 1$.
Proof: Write $k = s(X)$ and suppose $s(Y) \leq k - 2$. The square $X$ must be blue at round $k - 1$, and hence must be either blue or yellow at round $k - 2$.
By assumption, the square $Y$ is already red at round $k - 2$.
If $X$ is blue at round $k - 2$, then it would turn into red at round $k - 1$, contradiction.
If $X$ is yellow at round $k - 2$, then $Y$ would turn into yellow at round $k - 1$, contradicting the assumption $s(Y) \leq k - 2$.
Thus we have contradiction in either case, proving the claim.
-----
The original question is then a consequence of the follow simple result:
Fill each squre in an $n \times n$ grid with a nonnegative integer, such that
- if a square has $k > 0$ in it, then one of its neighbor has $k - 1$;
- if two squares are next to each other, then their numbers differ at most by $1$.
Then the largest number in the grid is at most $2n - 2$.
Proof: the first requirement tells us that there is at least one square with $0$ in it; the second requirement tells us that any two squares have numbers that differ at most by their Manhattan distance. |
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 $E,F$, respectively. Show that $PE=PF$. | 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 $AD$ on the ray $(DA$ beyond $A$, in the point $M$. Consider $\omega$ the excircle of $\triangle DCM$ opposite angle $D$. Then $D$ is the center of positive homothety sending $\omega_2$ to $\omega$. Monge's theorem implies that the center of negative homothety sending $\omega_1$ to $\omega$ lies on line $BD$. However, $BC$ is the internal tangent of $\omega$ and $\omega_1$, thus $B$ is the center we are looking for. This further implies that $AB$ is tangent to $\omega$, making $\omega$ tangent to rays $(DA$ beyond $A$, $(DC$ beyond $C$, $(AB$ beyond $B$ and $(CB$ beyond $B$. Calling these tangency points $U,V,W,T$ respectively, we would have that $$DU=DA+AU=DA+AW=DA+AB+BW$$ and similarly $$DV=DC+CV=DC+CT=DC+CB+BT.$$ Since $DU=DV$ and $BW=BT$, we get the desired conclusion.
Call $\Gamma$ the ellipse with foci in $B$ and $D$ passing through $A$ and $C$. It's well known that the external angle bisectors of angles $BAD$ and $BCD$ are the tangents in $A$ and $C$ to $\Gamma$. Let $d$ be the line through $P$ perpendicular to $BD$, and set $d\cap \Gamma = \{R,S\}$. Since $BD$ is an axis of symmetry in $\Gamma$, $PR=PS$. But now, applying Desargues Involution Theorem to the degenerate quadrilateral $AACC$, conic $\Gamma$ and line $d$, we get that $(P,P)$, $(R,S)$ and $(E,F)$ are involution pairs on $d$, the conclusion following. |
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(f(y))+y(f(f(x))$\\
$M(0,0)$ gives that $\{f(1),f(-1)\}=\{1,-1\}$ so $f(f(1))=1$\\ $M(x-1,1)$ gives $\{f(xf(1)+1),f(f(x)-1)\}=\{x,f(f(x))-1\}$ and this implies that $f$ is [b]surjective[/b].\\ \\
[b]Claim 1:[/b] $f(c)=0 \Rightarrow c=0$[hide]First note that there is such a $c$.\\
$S(c,c)$ gives $0=f(1)+f(-1)=2cf(0)$ which in any case gives $f(0)=0$ .Now suppose that $c\neq 0$ \\ \\
$M(x,c)$ gives $\{ f(1),f(cf(x)-1)\}=\{1,cf(f(x))-1\}$ ,since $f$ is surjective this forces that $f(1)=1$ and $f(cf(x)-1)=cf(f(x))-1$ which because $f(x)$ takes all real values gives $f(cx-1)=cf(x)-1$\\
$M(c,x)$ similarly gives $f(cx+1)=cf(x)+1$ combining these we get $f(x+1)-f(x-1)=2$\\ \\
Now $M(c-1,1)$ gives $\{0,f(f(c-1)-1)\}=\{c,f(f(c-1))-1\}$ and as $c\neq 0$ we get $f(f(c-1))=1$\\
Similarly $M(1,c+1)$ gives $\{f(f(c+1)+1),0\}=\{f(f(c+1))+1,c\}$ and we get $f(f(c+1))=-1$\\
But this is clearly a contradiction since $f(f(c+1))=f(f(c-1)+2)=f(f(c-1))+2$
[/hide]
[b]Case $f(1)=1$ [/b]
For $y\neq 0$ , $M(-\frac{2}{f(y)},y)$ gives $\{-1,f(yf(-\frac{2}{f(y)})-1)\}=\{-\frac{2f(f(y))}{f(y)}+1,yf(f(-\frac{2}{f(y)}))-1\}$ for which in view of the previous claim the only possible case is $-1=-\frac{2f(f(y))}{f(y)}+1\Rightarrow f(f(y))=f(y)$ and due to surjectivity of $f$ we get $f(x)=x$.\\
For the rest we assume $f(1)=-1$ which is the hard case.\\
To begin with \\
$M(x,\frac{1}{f(x)})$ : $\{f(xf(\frac{1}{f(x)})+1),0)=\{xf(f(\frac{1}{f(x)}))+1,\frac{f(f(x))}{f(x)}-1\}$\\
$M(x,\frac{1}{f(f(x))})$: $\{ f(xf(\frac{1}{f(f(x))})+1),f(\frac{f(x)}{f(f(x))}-1)\}=\{xf(f(\frac{1}{f(f(x))}))+1,0\}$\\
$M(-\frac{1}{f(x)},x)$: $\{ 0,f(xf(-\frac{1}{f(x)})-1)\}=\{-\frac{f(f(x))}{f(x)}+1,xf(f(-\frac{1}{f(x)}))-1\}$ \\
$M(-\frac{1}{f(f(x))},x)$:$\{f(-\frac{f(x)}{f(f(x))}+1),f(xf(-\frac{1}{f(f(x))})-1)\}=\{0,xf(f(-\frac{1}{f(f(x))}))-1\}$
These give that either (claim 1 is being used)\\
[b]Claim 2:[/b]$f(f(x))=f(x)$ or all of these hold: $xf(f(\frac{1}{f(x)}))+1=xf(\frac{1}{f(f(x))})+1=xf(f(-\frac{1}{f(x)}))-1=xf(-\frac{1}{f(f(x))})+1=0$\\
[b]Claim 3:[/b] $f$ is injective [hide]Suppose that $f(a)=f(b)\neq 0$. We should have that $f(f(a))=f(a)$ or else use claim 2 to get that $a=b$.\\
$M(x,a)$: $\{ f(xf(a)+1),f(af(x)-1)\}=\{xf(a)+1,af(f(x))-1\}$\\
$M(x,b)$: $\{ f(xf(a)+1),f(bf(x)-1)\}=\{xf(a)+1,bf(f(x))-1\}$ so either $xf(a)+1$ is a fixed point or $f(xf(a)+1)=af(f(x))-1=bf(f(x))-1\Rightarrow f(f(x))=0\Rightarrow x=0$. So in particular $-1$ is a fixed point which is a contradiction.[/hide]
Injectivity with claim 2 show that either $f(f(x))=f(x)$ or $f(\frac{1}{f(x)})=\frac{1}{f(f(x)}$ and due to surjectivity this can be restated as \\
$f(x)=x$ or $f(\frac{1}{x})=\frac{1}{f(x)}$ . If we suppose that $f(x)=x$ then the previous statement can be applied to $\frac{1}{x}$ to show that $f(\frac{1}{x})=\frac{1}{x}$ so $$f\left( \frac{1}{x}\right) =\frac{1}{f(x)}$$
for every $x\neq 0$.
[b]Claim 4:[/b]$f$ has no fixed points apart from $0$[hide]$S(1,-1)$ gives that $f(2)=-2$ .Suppose there is a fixed point $z$ then $\frac{1}{z}$ is also a fixed point and $S(z,\frac{1}{z})$ gives $f(2)=2$[/hide]
So $f(f(x))=f(x)$ cannot hold unless $x=0$ and claim 2 gives $xf(\frac{1}{f(f(x))})+1=0\Rightarrow $
$$f(f(f(x)))=-x$$ It also gives that $f(-\frac{1}{f(x)})=-\frac{1}{f(f(x))} \Rightarrow f(-f(x))=-f(f(x))\Rightarrow $ $$f(-x)=-f(x)$$
[b]Claim 5:[/b] $f(x+1)+f(x-1)=-2f(f(x))$[hide] $S(x,-1): f(x+1)+f(-f(x)-1)=-x-f(f(x))$\\
$S(f(x),-1): f(f(x)+1)+f(-f(f(x))-1)=-f(x)+x$\\
$S(f(f(x)),-1): f(f(f(x))+1)+f(x-1)=-f(f(x))+f(x)$ add them.[/hide]
[b]Claim 6:[/b]$ f\left( \frac{x}{f(x)}\right) =-\frac{f(x)}{f(f(x))}$[hide]$S(\frac{1}{f(x)},-f(f(x)))$ gives:$f(\frac{x}{f(x)}+1)+f(-2)=-\frac{f^4(x)}{f(x)}-\frac{f(f(x)}{f^3(x)}=\frac{f(f(x))}{x}+1$\\
$S(-f(f(x)),\frac{1}{f(x)})$ gives: $f(0)+f(\frac{x}{f(x)}-1)=-\frac{f(f(x))}{f^3(x)}-\frac{f^4(x)}{f(x)}=\frac{f(f(x))}{x}+1$ add these to get \\
$2\frac{f(f(x))}{x}=f(\frac{x}{f(x)}+1)+f(\frac{x}{f(x)}-1)=-2f(f(\frac{x}{f(x)}))\Rightarrow f(f(\frac{x}{f(x)}))=-\frac{f(f(x))}{x}$ apply this relation to $x=:f(f(x))$ to get $f(f(-\frac{f(f(x))}{x}))=-\frac{f^4(x)}{f(f(x))}=\frac{f(x)}{f(f(x))}$ thus $\frac{f(x)}{f(f(x))}=f^4(\frac{x}{f(x)})=-f(\frac{x}{f(x)})$[/hide]
[b]Claim 7 :[/b]$f(2x)=2f(x)$[hide] $M(-\frac{2}{f(x)},x)$ gives: $\{1,f(xf(-\frac{2}{f(x)})-1)=\{-\frac{2f(f(x))}{f(x)}+1,xf(f(-\frac{2}{f(x)}))-1\}$ thus $xf(f(-\frac{2}{f(x)}))=2$ put $\frac{1}{x}$ instead of $x$: $f(f(-2f(x)))=2x$ and $f$- both sides to get the result.[/hide]
Now consider $M(-1,x)$: $$\{f(-f(x)+1),f(x-1)\}=\{-f(f(x))+1,-x-1\}$$
and $M(f(f(x)),\frac{x}{f(x)})$: (see claim 6) $$\{ f(-f(x)+1),f(-\frac{x^2}{f(x)}-1)\}=\{-\frac{f(f(x))^2}{x}+1,-x-1\}$$
If $f(-f(x)+1)$ weren't $-x-1$ then $f(-f(x)+1)=-f(f(x))+1=-\frac{f(f(x))^2}{x}+1\Rightarrow f(f(x))=x$ using $f^3(x)=-x$ you get $f(x)=-x$ then $f(x+1)=f(-f(x)+1)=-x+1$ but $M(-1,x)$ also gives $f(x-1)=-x-1$ so $f(f(x+1))=x+1$ which then gives $f(x+1)=-x-1$ a contradiction.\\
Thus $f(-f(x)+1)=-x-1$ always holds and we are almost done. Put $f(f(x))$ in place of $x$ to get $f(x+1)=-f(f(x))-1$ then \\
$f(x+2)=-f(f(x+1))-1=-f(-f(f(x))-1)-1=f(f(f(x))+1)-1=-f^4(x)-1-1=f(x)-2$\\ using claim 7 we also get $f(x+1)=f(x)-1$ and then $-f(f(x))=f(x)\Rightarrow f(x)=-x$.
|
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 $(p+2)^{2^{\theta}}+1\equiv 0\pmod 4,\theta =0,c=r,$
then $p+3=2^{\beta},$ therefore $p\equiv 5\pmod 8,$ so that $\left(\frac{-2}p\right)=-1.$
However $2^{r+1}\equiv -2\pmod p,$ contradiction! Therefore $(p+2)^{2^{\theta}}+1$ is divisible by $p.$
If $r\geq 3,$ let $N=\frac{(p+2)^{2^{\theta}r}+1}{(p+2)^{2^{\theta}}+1},$ then $N=\sum\limits_{j=0}^{r-1}(p+2)^{j\cdot 2^{\theta}}(-1)^j\equiv r\equiv 1\pmod 2,$ therefore $2\nmid N.$
From $N\mid 2^ap^b,$ $N=p^k,$ where $k\in\mathbb N.$ Using LTE Lemma$,$ $v_p(N)=v_p(r),$ therefore $N\leq r.$
Let $x=(p+2)^{2^{\theta}},$ then $x\geq 5,$ therefore $N=\frac{x^r+1}{x+1}\geqslant x^{r-1}-x^{r-2}\geqslant\frac 12x^{r-1}>r,$ contradiction!
Now $r=1.$ If $\theta =0,c=1,$ then $p+3=2^ap^b\equiv 0\pmod p,$ so $p=3,a=b=1,(a,b,c,p)=(1,1,1,3).$
Consider $\theta\geq 1,$ then $2^ap^b=(p+2)^{2^{\theta}}+1\equiv 2\pmod 4,$ therefore $a=1,2p^b=(p+2)^{2^{\theta}}+1\cdots (*).$
If $\theta =1,$ $2^ap^b=p^2+4p+5,$ therefore $p=5,$ we can get $b=2,(a,b,c,p)=(1,2,2,5).$
Now let $\theta\geq 2,$ then $2^{2^{\theta}}\equiv -1\pmod p,2^{2^{\theta +1}}\equiv 1\pmod p,$ so $\delta _p(2)=2^{\theta +1},2^{\theta+1}\mid (p-1).$
Therefore $p\equiv 1\pmod {16}.$ From $(*)$ we have $2(-1)^b\equiv 2\pmod {p+1},$ so $2\mid b.$
We can write $(*)$ as $2(p^b-1)=(p+2)^{2^{\theta}}-1.$
Using LTE Lemma$,$ $v_2\left((p+2)^{2^{\theta}}-1\right)=v_2\left((p+2)^2-1\right) +\theta -1=\theta +2.$
However$,$ $\theta +1=v_2(p^b+1)\geqslant v_2\left(p^2-1\right)\geqslant v_2(p-1)+v_2(p+1)\geqslant\theta +2,$ contradiction!
Therefore $\boxed{(a,b,c,p)=(1,1,1,3),(1,2,2,5)}.\blacksquare$ |
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 $q=0$, then $\sum_{i=1}^q x_{p+2i-1}=0$; if $q=n-1$, then $\sum_{i=q+1}^{n-1} x_{p+2i}=0$. | 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 that $\sum_{i=1}^n a_i=0$. Then there exists an integer $p$ such that $\sum_{i=1}^k a_{p+i-1}\ge 0$ for all positive integers $k\in[n]$, where the indices are taken modulo $n$.
[hide=Proof]Pick an integer $c$ such that $0\le c\le n-1$ and $\sum_{i=1}^c a_i$ is minimal, where the empty sum is $0$. We claim that $p\doteqdot c+1$ works. To prove this, let $k\in[n]$ be a positive integer. We wish to prove that $\sum_{i=1}^k a_{p+i-1}\ge 0$. Equivalently:
\begin{align*}
\iff&\sum_{i=c+1}^{c+k} a_{i}\ge 0\\
\iff&\sum_{i=1}^{c+k}a_i\ge\sum_{i=1}^{c}a_i.
\end{align*}
We distinguish between two cases.
[list]
[*] [i]Case 1:[/i] $c+k<n$. By minimality we obtain $\sum_{i=1}^{c+k}a_i\ge\sum_{i=1}^{c}a_i$.
[*] [i]Case 2:[/i] $c+k\ge n$. Note that $c+k\le (n-1)+n\le 2n-1$. Hence $0\le c+k-n\le n-1$. Since $\sum_{i=1}^n a_i=0$ and by minimality we obtain
$$\sum_{i=1}^{c+k}a_i=\sum_{i=1}^{c+k-n}a_i\ge\sum_{i=1}^{c}a_i.$$
[/list]
Since we have exhausted all cases, the lemma is proved.[/hide]
Now define real numbers $a_1,b_1,a_2,b_2,\ldots,a_n,b_n$ by $a_i\doteqdot x_{2i-1}$ and $b_i\doteqdot x_{2i}$ for all $i\in[n]$. It suffices to prove that there exist integers $p$ and $q$ such that $0\le q\le n-1$ and
$$\sum_{i=q+2}^{n} a_{p+i}\le 1\mbox{ and }\sum_{i=1}^q b_{p+i}\le 1,$$
where indices are taken modulo $n$. The assignment $(p,q)\mapsto (2p+1,q)$ solves the original problem.
Define $A\doteqdot\sum_{i=1}^n a_i$ and $B\doteqdot\sum_{i=1}^n b_i$. If $A\le 1$ then we can choose $(p,q)=(0,0)$. Similarly, if $B\le 1$ then we can choose $(p,q)=(0,n-1)$. So assume $A>1$ and $B>1$. Define real numbers $c_1,\ldots,c_n$ by $c_i\doteqdot Ba_i-Ab_i$ for all $i\in[n]$. Note that
$$\sum_{i=1}^n c_i=B\sum_{i=1}^n a_i-A\sum_{i=1}^n b_i=BA-AB=0.$$
By the gas station lemma, there exists an integer $p$ such that $\sum_{i=1}^k c_{p+i}\ge 0$ for all integers $k\in[n]$, where indices are taken modulo $n$. Let $q$ be the maximal integer $q$ such that $0\le q\le n$ and $\sum_{i=1}^q b_{p+i}\le 1$. Since we assume $B>1$ then $q\le n-1$. Assume by contradiction that $\sum_{i=q+2}^n a_{p+i}>1$. By maximality we obtain $\sum_{i=1}^{q+1} b_{p+i}>1$. Hence
$$0\le\sum_{i=1}^{q+1}c_{p+i}=B\sum_{i=1}^{q+1} a_{p+i}-A\sum_{i=1}^{q+1} b_{p+i}<B\sum_{i=1}^{q+1} a_{p+i}-A.$$
Therefore $\sum_{i=1}^{q+1} a_{p+i}>\frac{A}{B}$. Now we have
$$\sum_{i=q+2}^n c_{p+i}=\sum_{i=1}^n c_{p+i}-\sum_{i=1}^{q+1} c_{p+i}=-\sum_{i=1}^{q+1} c_{p+i}\le 0.$$
Hence
$$0\ge\sum_{i=q+2}^n c_{p+i}=B\sum_{i=q+2}^n a_{p+i}-A\sum_{i=q+2}^n b_{p+i}>B-A\sum_{i=q+2}^n b_{p+i}.$$
Therefore $\sum_{i=q+2}^n b_{p+i}>\frac{B}{A}$. Now by AM-GM
\begin{align*}\sum_{i=1}^n(a_{p+i}+b_{p+i})&=\sum_{i=1}^{q+1} a_{p+i}+\sum_{i=1}^{q+1} b_{p+i}+\sum_{i=q+2}^n a_{p+i}+\sum_{i=q+2}^n b_{p+i}\\
&>\frac{A}{B}+1+1+\frac{B}{A}\\
&>4,
\end{align*}
contradiction.[/hide] |
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 of \(n\), with base case \(n=1\) trivial.
For the inductive step, take a prime \(p\mid n\). For any set \(S\), let \(S_i=\{s\in S\mid \nu_p(s)=i\}\). Then \(|V_0|\ge|V_1|\ge\cdots\ge|V_n|\) and \(|W_0|\le|W_1|\le\cdots\le|W_n|\). By Chebyshev's inequality, \begin{align*} |V|\cdot|W|=\left(\sum_{i=0}^n|V_i|\right)\left(\sum_{i=0}^n|W_i|\right) &\ge(n+1)\sum_{i=0}^n|V_i|\cdot|W_i|\\ &\ge(n+1)\sum_{i=0}^n|A_i|\cdot|D_i| =|D|\sum_{i=0}^n|A_i|=|A|\cdot|D|. \end{align*} \(\blacksquare\)
We may assume that for any \(d\in D\), if \(d\) has a divisor and a multiple in \(A\), then \(d\in A\). Then \(V\cap W=A\) and \(|V\cup W|=|D|-|B|\), so \[|D|-|B|+|A|=|V|+|W|\ge2\sqrt{V|\cdot|W|}\ge2\sqrt{|A|\cdot|D|},\] which rearranges to the desired.
|
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 area of the square. | 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 $DE=\frac{AB\sqrt{13}}{3}$, so from last two equalities we get $AB=\sqrt{13}$ and so the area of square is $13$. |
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 $p=q+1$, which holds iff $(p,q)=(3,2)$. Assume $p\nmid q+1$. We then have $p^2\mid q^2-q+1$, in particular $p^2\le q^2-q+1<q^2$. As $p>q$ this is a contradiction. |
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:[/b] $X_1=\{1,2,\ldots, 31 \}$,
[b]Group 2:[/b] $X_2=\{32,33 \}$,
[b]Group 3:[/b] $X_3=\{34,35 \}$,
$\ldots$
[b]Group7:[/b] $X_7=\{42,43 \}$,
[b]Group 8:[/b] $X_8=\{44,45,46 \}$,
[b]Group 9:[/b] $X_9=\{47,48,49 \}$ και
[b]Group 10:[/b] $X_{10}=\{50\}$,
we note that $A$ can contain at most one element from each $X_i$, with $2 \leq i \leq 7$, and at most one element from each $X_i$, with $8 \leq i \leq 9$. In total, we have
$|A| \leq 31+(1+1+\ldots+1)+(1+1)+1=31+6+2+1=40$.
An example showing that $|A|=40$ is feasible, is the set $\{1,2,\ldots,31,32,34,36,38,40,42,44,47,50 \}$. To check that this set works, note that for all $a,b,c \in A$ with $a<b<c$, we have
$\dfrac{1}{a}-\dfrac{1}{c}=(\dfrac{1}{a}-\dfrac{1}{b})+(\dfrac{1}{b}-\dfrac{1}{c})>\dfrac{1}{a}-\dfrac{1}{b},$
so it suffices to check the hypothesis for each two consecutive (when written in increasing order) elements. This is immediate from the Claim. |
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 as well.
What do components of this graph look like? NB: we'll illustrate paths in what follows, not all other edges in the component.
[list][*]Case $1$: isolated vertices. This uses up $3$ elements of $M$ for the reward of only a single vertex (triple). Say it has "efficiency" $\eta_3 = \tfrac{1}{3}$, i.e. #triples divided by #$M$-elements.
[*]Case $2$: isolated component $u\sim v$. This has efficiency $\tfrac{2}{4}=\tfrac{1}{2}$. We'll never use this case.
[*]Case $3$: $\{a,b,c\}\sim\{a,b,d\}\sim\{a,c,d\}\sim\{b,c,d\}$. Note that after $\{a,c,d\}$, no further triple involving a new element together with any of $(a,b,c,d)$ is allowed, since any of them would clash (have intersection size 1) with an existing triple:
\begin{tabular}{c|c}
new &existing\\ \hline
a,b,x & a,c,d \\
a,c,x & a,b,d \\
a,d,x & a,b,c \\
b,c,x & a,b,d \\
b,d,x & a,b,c \\
c,d,x & a,b,c
\end{tabular}where $x$ is new. However, we can add $\{b,c,d\}$ without a clash, giving $\eta_4=1$.
[*]Case $4$: $\{a,b,c\}\sim\{a,b,d\}\sim\{a,b,e\}\sim\ldots$. All subsequent terms must be of the form $\{a,b,\star\}$ in order to avoid a return to case $3$ (which would immediately lead to a contradiction). Supposing this path consists of $k-2\ge 3$ triples, its efficiency is $\eta_k=\tfrac{k-2}{k}$.[/list]
The most efficient way to group triples into components is therefore case 3, with $\eta_4=1$, followed by case 4, with case 1 useless unless we find ourselves with three elements of $M$ not yet assigned.
To finish: if there are $q$ "case 3" components, then there are $2022-4q=2$ (unusable) or $6,10,14,\ldots$ elements of $M$ remaining to be packed into a "case 4" component, making for $4q+(2022-4q)\eta_{2022-4q}=\boxed{2020}$ triples. |
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 $f(1)=1$ nên suy ra $f(2)=2.$Ta sẽ chứng minh rằng $f(x)=x,\forall x\in \mathbb{Z^+}.$Thật vật với $x=1,2$ thì khẳng định hiển nhiên đúng.Giả sử khẳng định đúng đến $x=k,k\in \mathbb{Z^+}.$Thì ta chứng minh rằng $k+1$ cũng đúng.Từ $(1)$ thay $x=k,y=1$ ta được $$\frac{k^3+3x^2}{k+1}+\frac{1+3k}{k+1}=\frac{(k+1)^3}{f(k+1)},\forall k\in \mathbb{Z^+}.$$Hay ta được $$f(k+1)=k+1,\forall k\in \mathbb{Z^+}.$$Vậy theo quy nạp ta có tất cả hàm số cần tìm là $$f(x)=x,\forall x\in \mathbb{Z^+}.$$ |
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 $$3=\sum_{cyc}\frac{2}{1+x^3}\implies 0=\sum_{cyc}\frac{1-x^3}{1+x^3}\leq \frac{3}{2}\sum_{cyc}\frac{1-x^2}{1+x^3}=\frac{3}{2}\sum_{cyc}\frac{1-x}{1-x+x^2}$$ which is exactly what we wanted to prove. Equality occurs for $x=y=z=1$ (which does verify the constraint over the variables). |
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 $y=x+n\Longrightarrow x+3^n3^x+1=x+n+3^x\Longrightarrow 3^x=\frac{n-1}{3^n-1}<1$ Contradiction
Case $x>y$
Let $x=y+n\Longrightarrow y+n+3^y+1=y+3^n3^y\Longrightarrow3^y=\frac{n+1}{3^n-1}$
$n=1\Longrightarrow 3^y=\frac{2}{2}\Longrightarrow 3^y=1\Rightarrow x=1, y=0$ is the only solution if $0\in Z$.
If $n>1, 3^y=\frac{n+1}{3^n-1}<1$. Contradiction. |
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 $b^{-1}a^{-1}=a^{-1}b^{-1}$ and so $ab=ba,$ i.e. $H$ is abelian.
The condition given in the problem is strong, and so I think we can say more about $G.$ For example, an obvious one is that $x^4=1$ for all $x \notin H$ because if $x \notin H,$ then $x^{-1} \notin H$ and so $x^2=x^{-2},$ which gives $x^4=1.$ |
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
$Q(i)Q(-i)=|Q(i)|^2 = (-p+iq)(-p-iq) = p^2 +q^2$ (2)
Q(i)Q(-i)=det((A+iB)(A-iB))=$det(A^2+B^2+iBA-iAB)=i^3det(BA-AB)$=-idet(BA-AB) (3)
(2) and (3) give p=q=det(BA-AB)=0
This prove for any x in R, det(A+xB)=0
Remark AB-BA is not invertible |
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
$$\frac{m(m+1)}{2}=1+2+\cdots + m \leq d_1 +d_2 +\cdots d_m$$
$$\leq \text{rank}(A^{i_1})-\text{rank}(A^{i_1+1})+\text{rank}(A^{i_2})-\text{rank}(A^{i_2+1})+\cdots +\text{rank}(A^{i_m})-\text{rank}(A^{i_m+1}) $$
$$\leq \text{rank}(A^{i_1})-\text{rank}(A^{i_m+1})\leq n$$
That is assuming $i_1\leq i_2 \leq \cdots \leq i_n$ correspond to $d_1, d_2, \dots , d_m$ in some order and noting the fact that for each $j$, $\text{rank}(A^{i_j+1})\geq \text{rank}(A^{i_{j+1}})$.
It remains to show the equality case can not hold. But the equality of the last inequality holds when $\text{rank}(A^{i_1})=n$ and $\text{rank}(A^{i_m+1})=0$ that requires $A$ to be singular and full rank which is not possible. The conclusion follows. |
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 $x=1/2$, we enjoy equality. Assuming this is not the case, we perform cancellations and get
\[
\log_2(1-x) = \frac{(\log_2 x)^2 -\log_2 x+1}{\log_2 x}.
\]
Now, set $x=2^t$. We then get
\[
\log_2 (1-x) =\frac{t^2-t+1}{t} \implies 1-x = 2^{\frac{t^2-t+1}{t}} \iff f(t)\triangleq 2^t + 2^{\frac{t^2-t+1}{t}}=1.
\]
It is readily verified $t\mapsto (t^2-t+1)/t$ is increasing on $(-\infty,-1]$ and decreasing on $(-1,0)$. Moreover, $f(-1)<1$, it follows that such a $t$, if exists, is in $(-1,0)$. Notice, however, that as $x<3/4$, we must have $2^{(t^2-t+1)/t}>1/4\iff (t^2-t+1)/t > -2$. Notice, however, that setting $t=-s$ for $s>0$ and recalling $s+1/s\ge 2$, we have
\[
\frac{t^2-t+1}{t} = t+\frac1t -1 = -1 - (s+1/s)\le -3,
\]
impossible. Hence, $x=1/2$ is the only solution. |
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 $\boxed{20}$ and occurs at $a = -10$. |
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 $p + q$.
[b]p11[/b] Points $X$ and $Y$ lie on sides $AB$ and $BC$ of triangle $ABC$, respectively. Ray $\overrightarrow{XY}$ is extended to point $Z$ such that $A, C$, and $Z$ are collinear, in that order. If triangle$ ABZ$ is isosceles and triangle $CYZ$ is equilateral, then the possible values of $\angle ZXB$ lie in the interval $I = (a^o, b^o)$, such that $0 \le a, b \le 360$ and such that $a$ is as large as possible and $b$ is as small as possible. Find $a + b$.
[b]p12[/b] Let $S = \{(a, b) | 0 \le a, b \le 3, a$ and $b$ are integers $\}$. In other words, $S$ is the set of points in the coordinate plane with integer coordinates between $0$ and $3$, inclusive. Prair selects four distinct points in $S$, for each selected point, she draws lines with slope $1$ and slope $-1$ passing through that point. Given that each point in $S$ lies on at least one line Prair drew, how many ways could she have selected those four points? | [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 $n$, as well as for which positive integers $n$ this minimum value is attained. | 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, $B_n = 2A_n+4A_n+\cdots+2^k A_n - n$. Assume first that $k=2$. Then, $B_n-2A_n = 4A_n - n =4(A_n-m)$. Since $A_n\ge m+1$ (as $n>4$), it follows $B_n-2A_n\ge 4$. Equality holds iff $A_n = m+1$, that is when $m$ itself is an (odd) prime.
Next, assume $k=3$. Then, $B_n=2A_n+4A_n+8A_n-n$, yielding $B_n-2A_n = 4A_n + 8(A_n-m)$. As $A_n\ge 1$ and $A_n-m\ge 0$, we find $B_n-2A_n\ge 4$ with equality iff $k=3$ and $m=1$, that is $n=8$.
Finally, let $k\ge 4$. Then, $B_n = 2A_n + 4A_n + 8A_n +2^k A_n - n$, yielding $B_n-2A_n\ge 4A_n+8A_n\ge 12$, yielding a strictly worse value. |
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, $a=16$ is indeed the maximal value, concluding the case. |
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 $P\ne G$, prove that $GP$ passes through the midpoint of $XY$. | 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, $0,p,q,r$ are concyclic iff
\[|q(r - p)| = |r(q - p)| + |p(r - q)| \iff |1/p - 1/r| = |1/p - 1/q| + |1/q - 1/r|\text{.}\]
But this is simply the equality case of the triangle inequality, so this implies $\frac1p,\frac1q,\frac1r$ are collinear. We see
\[\frac{\frac1p + \frac1q + \frac1r}{3}\]
is also on the line. But since the map $z \to \frac1z$ is a Mobius transformation, it sends clines to clines, so
\[\frac{3}{\frac1p + \frac1q + \frac1r} = \frac{p + q + r}{3}\]
is on the same circle as $0,p,q,r$. But this is impossible as this point is the centroid of triangle $PQR$. |
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 as $\frac{a}{b}$ where $a$ and $b$ are relatively prime positive integers. Compute $a$. | [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 \[N = \sum_{A \subseteq \{1, 2, \dots, 12\}}(-1)^{|A|} \text{(number of ways to choose}\ (T, \pi)\ \text{such that for all}\ i \in A, s_{i+1} = \pi(s_i)\text{)}.\] For a moment, assume $A \neq \{1, 2, \dots, 12\}$. Then we claim that there are $11! \cdot 11^{12-|A|}$ ways to choose $(T, \pi)$ for this particular $A$, since once one of the $11!$ possible $\pi$'s is chosen, every $s_i$ such that $i - 1 \not \in A$ ($12 \not \in A$ in the case of $i = 0$) can be independently assigned to an arbitrary element of $S$, allowing all other $s_i$ to be uniquely determined afterwards. Therefore we can rewrite this sum as \[N = \text{(number of ways to choose}\ (T, \pi)\ \text{such that for all}\ i, s_{i+1} = \pi(s_i)\text{)}\ + \sum_{A \subsetneq \{1, 2, \dots, 12\}}(-1)^{|A|}11!\cdot 11^{12-|A|},\] where the first term comes from the case $A = \{1, 2, \dots, 12\}$; call it $M$. The sum can be rewritten by summing over $x = |A|$: \[N = M + 11! \sum_{x=0}^{11} \binom{12}{x}(-1)^x11^{12-x}.\] By the binomial theorem, we have \[N = M + 11!((11 - 1)^{12} - 1) = M + 11! \cdot 10^{12} - 11!.\] Now it remains to evaluate $M$. Suppose that the permutation $\pi$ has been fixed. Then, all the elements of $T$ are uniquely determined by $s_1$. Also, in order for it to be the case that $s_1 = \pi(s_{12})$, the size of the cycle of $\pi$ that $s_1$ is located in must be a factor of 12 (that is, either 1, 2, 3, 4, 6; note that 12 is not possible since $\pi$ has only 11 elements). Therefore, the number of valid pairs for the fixed $\pi$ is the sum of the sizes of all cycles of $\pi$ whose size satisfies this condition.
Now, we evaluate the sum of this value over all $\pi$ by looking at one element of $S$ at a time and counting the number of permutations that place this element in a cycle of appropriate size, and summing over all elements (in effect, multiplying by 11).
[b]Lemma:[/b] The number of permutations that place a fixed element $x$ of $S$ in a cycle of length $k$ is $10!$.
[b]Proof:[/b] We need to choose $k-1$ elements from the 10 elements of $S \setminus \{x\}$ to form the cycle with $S$; then there are $(k-1)!$ ways to arrange the cycle and $(11-k)!$ ways to arrange the permutation's structure outside the cycle. This gives a count of $\binom{10}{k-1}(k-1)!(11-k)! = 10!$, as desired.
By the lemma, we have \[M = 11 \cdot (10! \cdot 5) = 11! \cdot 5,\] since there are 5 possible choices of cycle size. This gives \[N = 11!(10^{12} + 4),\] so the desired probability is \[\frac{N}{11!\cdot11^{12}} = \frac{10^{12} + 4}{11^{12}}.\] Since $\gcd(10^{12} + 4, 11^{12}) = 1$, our answer is $10^{12} + 4$.
[/hide] |
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 $D_1(\ne A)$, $E_1(\ne B)$ and $F_1(\ne C)$, respectively. Let $I$ and $I_1$ be the incentres of triangles $DEF$ and $D_1E_1F_1$, respectively. Prove that $E,F, I, I_1$ are concyclic. | 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 $J$, $\angle YJZ = 90^{\circ}+\frac{\angle YXZ}{2}$. \\
$\textbf{Proof}$ \\
Omitted, just angle chase. $\blacksquare$ \\
Now, notice that $$\angle EDF = \angle EDA + \angle ADF = \angle FCA + \angle EBA = \angle F_1CA + \angle ABE_1 = \angle E_1D_1F_1$$ using the two given cyclites, then using that $I_1 = FF_1 \cap EE_1$ and $\textbf{Lemma 1}$, $$\angle EI_1F = \angle E_1I_1F_1 = 90^{\circ} + \frac{\angle E_1D_1F_1}{2} = 90^{\circ} + \frac{\angle EDF}{2} = \angle EIF$$ giving the desired cyclicity. $\blacksquare$
Edit (regarding configuration issues pointed out below):
You can observe that both $I_1, I$ lie inside $\triangle DEF$ as $D$ lies in the interior of $BC$, I did not include it previously because it's an AoPS post. |
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 through $D$ and parallel to $AC$ and the line through $E$ parallel to $AB$ intersects at $Y$, prove that $X,Y,Z$ are collinear. | 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 congruence:
\[ a^n + b^n \equiv 1^n + 1^n \text{ (mod $\sqrt[N]{c}$)} \equiv 2 \text{ (mod $\sqrt[N]{c}$)}. \] Since all such $c$ are powers of primes, it means $\gcd(a^n + b^n, \sqrt[N]{c}) = 1 \implies \gcd(a^n+b^n, c)=1$, as desired. [/hide][/hide] |
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 $Q(x,y):f(x)f(y)\le xy$.
$Q(x,-y)\Rightarrow-f(x)f(y)\le-xy\Rightarrow f(x)f(y)\ge xy$
So equality holds in $Q(x,y)$, so $f(x)=\frac1{f(1)}x$ after setting $y=1$. Then $x=1$ yields the solutions $\boxed{f(x)=x}$ and $\boxed{f(x)=-x}$, which both work. |
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 that $A$ and $P$ share the same side of the line $MK$. Prove that $\angle CPM = \angle DPL$. | 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 of triangles $MNL$ and $OLD$ gives $$\frac{LA}{LK}\overset{(KNMA)}{=} \frac{LM}{LN}\overset{\triangle{MNL}\sim \triangle{OLD}}{=}\frac{LO}{LD}\Rightarrow LA\cdot LD=LO\cdot LK.$$So $OADK$ is cyclic and by $PoP$, $L$ lies on the radical axis of $(CNK)$ and $(ADM)$, i.e. $L\in \overline{PV}$. Now let $R$ be the point diametrically opposite to $M$ with respect to $(ADM)$. Since $\angle{MDR}=90$ and since $\overline{DM}$ bisects the angle $\angle{NDK}$, we have $(K,L;M,R)=-1$. Since $\angle{RPM}=90$ too, our previous cross-ratio gives that $\overline{PM}$ bisects the angle $\angle{LPK}$. Finally, note that $\angle{KPC}=\angle{KOC}=\frac{1}{2}\angle{MOD}=\angle{MPD}$ and thus $$\angle{DPL}=\angle{MPL}-\angle{MPD}=\angle{MPK}-\angle{KPC}=\angle{CPM}$$and we are done. |
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 in both cases less that $\angle AIB = 90 + \frac{1}{2}\gamma$.
Since triangles $AIB$ and $A_1IB_1$ have two same sides and because $\angle A_1IB_1 < \angle AIB$, we have that $AB > A_1B_1$ |
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 centers cannot be contained inside a bounded box, which clearly implies the conclusion. $\blacksquare$
[b]Remark:[/b] Hm this did not feel very difficult. I guess the main motivation was the following: this condition that "at least one of the two following inequalities holds" is really hard to do anything with, but the condition does imply that $(a_m-a_n)^2+(b_m-b_n)^2>1/n$ for all $m>n$. This lends to an obvious geometric interpretation: if we draw a circle of radius $1/\sqrt{n}$ around each point $(a_n,b_n)$, then any $(a_m,b_m)$ for $m>n$ can't lie inside this circle. But once we get this idea to think about points in the 2d plane it's obvious that we can replace these circles with squares. And then we probably want to have the squares not overlap at all, instead of this weird "points that come after can't lie inside the square", but this turns out to be very easy to do (the idea to take a smaller shape shows up fairly often). |
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. $$\Longrightarrow \sin \theta = \frac{\sqrt{3}}{2}.$$ Thus the angle is either $60^{\circ}$ or $120^{\circ}$. Ofcourse it can't be $60^{\circ}$ as then the total interior angle of the $\bigtriangleup$ will not be $180^{\circ}$. Thus the angle is $120^{\circ}$. And the perimeter is $15$. So the required answer is $\frac{120}{15} = \boxed{8}$. |
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 $625^2 - 1 = 390624$ for a total of $\boxed{24}.$ |
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 ($p=2$). In this case we obtain,
\[8+3q^3-32 = 3(q^3-8)\]
is a prime, which is clearly false for all odd $q\geq 3$ since then $p^3-8>1$. Thus, no solutions exist in this case as well.\\
[b] Case 2 : [/b] $q$ is even ($q=2$). This gives us that
\[p^3+24-32 = p^3-8 = (p-2)(p^2+2p+4)\]
which is clearly not prime for all $p>3$ (both factors are then greater than 1). Thus, the only possibility is when $p=3$ which clearly works since then,
\[p^3+3q^3-32 = 27 + 24 - 32 = 19\]
which is most certainly prime. Thus, the only pair of solutions which work is $ \boxed{(3,2)}$ . |
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 line. Prove that $\angle BPE=90^{\circ}$. | 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
$$\angle NPA=\angle BPA-\angle BPN=(180^\circ-\angle C)-\angle BEN=90^\circ.$$
Thus let $\overline{AP}$ intersect $(BNE)$ and $(CND)$ again at $X$ and $Y$ respectively. Then $X$ and $Y$ are the antipodes of $N$ in their respective circles, so $BNEX$ and $CNDY$ are rectangles. Hence $\overline{XE} \parallel \overline{YD}$, and $XE=BN=CN=YD$, implying $DXEY$ is a parallelogram, so $\overline{XYPA}$ bisects $\overline{DE}$ as desired. $\blacksquare$
[b]Remark:[/b] $\angle NPA=90^\circ$ is enough to imply $P$ is the $A$-queue point in the redefined version, so if you use the JBMO problem I think this just finishes. |
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 $P(m,n)-P(n,m)$
$$f^{f(n)}(m)=f^{f(m)}(n)$$
Now on this one the trick is to use injectivity and that it has no cycles since $f(n)>n$. By plugging $m=f^{a}(n)$ for any posititve integer $a$ we get that
$$f^{f(n)+a}(n)=f^{f^{a+1}(n)}(n) \implies f(n)+a=f^{a+1}(n)$$
On our last equation we will plug $a=f(n)-1$ and $n=m$ to get the following equation.
$$f^{f(n)}(m)=f(m)+f(n)-1$$
Now plugging this on the original F.E. we get
$$f(m)+f(n)+mn-1=f(n)f(m)$$
And here if we plug $n=m$ we have that
$$(f(n)-1)^2=n^2 \implies f(n)=n+1$$
Thus we are done :blush: |
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:
$v_2(3^{x}-1)\overset{LTE}{=} v_2(2)+v_2(4)+v_2(x)-1=2+v_2(x)$
$v_2(RHS)\geq \min\{v_2(2xy),v_2(8^y)\}=\min \{1+v_2(x)+v_2(y),3y\}$
[color=#ff9a00]Case1:[/color] $1+v_2(x)+v_2(y)=3y$
FTSoC $y$ isn't even. Then we have:
$1+v_2(x)+v_2(y)=3y\geq 2y +v_2(y) \Leftrightarrow v_2(x)\geq 2y-1 \Leftrightarrow \boxed{x \geq 2^{2y-1}}$
take the main equation $3^x-8^y=2xy+1$:
$LHS \geq 3^{2^{2y-1}}-8^y \geq 2^{2^{2y-1}}-2^{3y} \overset{y\geq 3}{\geq} 2^{2^{2y-1}-1} \geq 2^{4y}> 2^{3y}+1\geq 2^{2y}y+1=RHS$
A contradiction.
[color=#ff9a00]Case2:[/color] $1+v_2(x)+v_2(y) \neq 3y$
Back to $\blacklozenge$ we get:
$ 2+v_2(x)=v_2(RHS)=v_2(LHS)\leq 1+v_2(x)+v_2(y) \Leftrightarrow \boxed{v_2(y)\geq 1}$[/hide]
[hide=Claim2][b]Claim: [/b] $2\mid x$
[i]Proof:[/i] Back to main equation:
$LHS \equiv 3^x \equiv \{1,3\} \equiv RHS \equiv \{1,5\} \pmod{8} \Rightarrow 3^x \equiv 1\pmod{8}$
Hence $x$ is even. [/hide]
[hide=The inequality] Put $x=2u,y=2v$. Then we have: $(3^u-8^v)(3^u+8^v)=8uv+1$. And so
$$3^u+8^v \leq8uv+1$$
$\bullet$ $\forall u\geq 3$ we have the following: $u\sqrt{3}\leq 3^{\frac{u}{2}} $
this is true actually since $3\sqrt{3}=3^{\frac{3}{2}}$ and $4\sqrt{3}<3^{2}$, and the derivative of $3^{\frac{u}{2}}$ is strictly increasing.
$\bullet$ And also $\forall v\geq 1$ the following holds: $v\sqrt8 \leq 8^{\frac{v}{2}}$
the proof is the same as above.
So we can write for $u\geq 3, v\geq 1$:
$8uv+1 \leq 9uv < 2\sqrt{3.8}uv \leq 2\sqrt{3^u.8^v} \overset{GM-AM}{\leq} 3^u+8^v \leq 8uv +1$
A contradiction. And by checking manually the left cases we get that $(u,v)=(2,1)$ is the only solution.
And so $\boxed{(x,y)=(4,2)}$.$\blacksquare$ [/hide] |
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
$$ab+2022=2ab\Rightarrow 2022=ab\Rightarrow b=\frac{2022}a$$
Since $1>b$, we find that $a\ge 2023$.
Clearly, all numbers in the form $k+\frac{2022}k$ works where $k\ge 2023, k\in\mathbb{Z}$. So, we are done. |
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: $(2c^2-1)^2 - 4(c^4+a^2-4a^2c^2) = (4a^2-1)(4c^2-1)\le 0$. It is now clear that unless $4c^2=1$, one can choose $a$ such that ${\rm sgn}(4a^2-1)={\rm sgn}(4c^2-1)$ and make the discriminant positive, which would yield a contradiction. Thus, $c=\pm 1/2$ as claimed. |
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 opposite colors with exactly one point in between them, and flip the color of both of them
Is it possible to achieve a configuration where each point has a color opposite to its initial color with these operations?
[i](Proposed by Oleksii Masalitin)[/i] | 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 different values for the initial and desired configurations. |
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. I claim that at most 1 $i \geq 1011$ can have $b_i$ be an integer larger than 1.
First note that $b_i \leq \frac{n + (n-1) + \ldots + (n-i+1)}{1 + 2 + \ldots + i} = \frac{n + n - i + 1}{1 + i} \leq \frac{3034}{1012} < 3$ for $i \geq 1011$, so the only possible value of $b_i$ is 2.
Now note that if $b_i \leq 2$ then $b_{i+1} < 2$. We can see this because $\frac{a_{i+1}}{i+1} \leq \frac{2022}{i+1} \leq \frac{2022}{1012} < 2$ and as $b_{i+1}$ is the mediant of $b_i$ and $\frac{a_{i+1}}{i+1}$ this implies $b_{i+1} < 2$. This means that if $b_i = 2$ for any $i \geq 1011$ then all further $b_j, j > i$ will be less than $2$ and cannot be integers.
Thus the maximum number of $b_i$ that are integers larger than $1$, is the first $1010$ and at most $1$ from the remaining terms. Thus $k \leq 1011$ as desired.
[/hide] |
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 chair has exactly one occupant.
[*] Whenever she claps her hands, each guest looks at the number on the sheet attached to their current chair, and moves to the chair labelled with that number.
[/list]
Show that if $1 < m \leq n$, where $m$ is not a prime power, it is always possible for the PMO Magician to choose which sheet to attach to each chair so that everyone returns to their original seats after exactly $m$ claps. | 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 nonnegative integers $x,y$ so that $n = sx + ty$ by the chicken mc nugget theorem. Take $x$ cycles of length $s$ and $y$ cycles of length $t$ so the lcm of cycle lengths is indeed $m$ and we use all $n$ vertices.
In fact, if $m$ was a prime power of a prime say $p^k$ with $p$ not dividing $n$, then clearly this is not possible since each cycle length must be divisible by $p$ but $n$ is not. But if we had $p | n$, then take a cycle of length $p^k$, then $\frac{n-p^k}{p}$ cycles of length $p$, which works.
thanks @below - fixed now |
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 lost. | [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 cycle $A_1 \rightarrow A_2 \rightarrow \ldots \rightarrow A_m \rightarrow A_1$, we cancel one of the matches $A_i \rightarrow A_{i+1}$, where $A_{n+1} = A_1$. Then we can ignore the cycle, and just delete it. Repeat this process. We can see that it terminates and also at the end there cannot be any directed edges. Suppose not, let $A_1\rightarrow A_2 \rightarrow \ldots \rightarrow A_k$ be the longest directed path, by the assumption we know that there exists $A_r$ such that $A_k \rightarrow A_r$, because $A_k$ lost to $A_{k-1}$. We know that $r\neq 1,2,\ldots,k-2$, because it would form a cycle, so $A_1\rightarrow A_2\rightarrow\ldots\rightarrow A_k \rightarrow A_r$ is a longer that the longest, contradiction. So we left with an undirected graph with even degree of every vertex. So for every component there exist a Euler cycle, so for every Euler cycle $B_1 - B_2 - \ldots - B_s - B_1$, cancel matches $B_i - B_{i+1}$, where $B_{s+1} = B_1$. In that way we canceled exactly have matches.
[/hide] |
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 a quadratic equation we have ${(a+d),(b+c)} = {3,3}$.
Now $(a+b)+(a+c)-(b+c) = {0,4,2,6} \implies a = {0,1,2,3}$.
Applying that with other equations gives us $4 $ answers :
$(a,b,c,d) = (0, 1, 2, 3), (1, 0, 3, 2), (2, 3, 0, 1), (3, 2, 1, 0)$
I will do one of the quadratic equation parts for more clarification.
$(a+b)(c+d) = 5, (a+b) + (c+d) = 6 \implies (a+b) = \frac{5}{(c+d)} \implies \frac{(c+d)^2 + 5}{(c+d)} = 6 \implies (c+d)^2 - 6(c+d) + 5 = 0 \implies ((c+d)-5)((c+d)-1) = 0 \implies (c+d) = {1,5} \implies {(a+b),(c+d)} = {1,5}$. |
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)$, $ A_1Q$ is antiparallel of $ AA_2 \parallel BC$ WRT angle $ \angle (AB, A_2C)$ $ \Longrightarrow$ $ A_1BCQ$ is cyclic, and $ A_2P$ is antiparallel of $ AA_1 \parallel DC$ WRT angle $ \angle (DA, CA_1)$ $ \Longrightarrow$ $ CDA_2P$ is also cyclic. From $ \odot (A_1BCQ), \odot(CDA_2P),$
$ \angle QBA \equal{} \angle QBA_1 \equal{} \angle QCA_1 \equal{} \angle QCP \equal{} \angle A_2CP \equal{} \angle A_2DP \equal{} \angle ADP.$ |
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 $MN=|\overrightarrow{MN}|=|\tfrac{1}{2}(\overrightarrow{AD}+\overrightarrow{BC})| \le \tfrac{1}{2}(AD+BC)$.
[*] In a triangle $XYZ$, the projection of the midpoint of $YZ$ on the angle bisector of $\angle YXZ$ lies outside the triangle (or on the side $YZ$). (Obvious.) In this configuration, $|\text{the projection of } MN \text{ on the line } KOL| \ge KL$.
[/list]
Therefore, $2(OP+OR) \le 2(OK+OL)=2KL \le 2 \cdot |\text{the projection of } MN \text{ on the line } KOL| \le 2MN \le AD+BC$. |
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 $\omega$
[b]Claim:[/b] $\text{Pow}_{(ABC)}(P)=\text{Pow}_{(ABC)}(Q)$
[i]Proof[/i]: Since, $\overline{AP}=s-a$, $\overline{CP}=s-c$, $\overline{AQ}=s-c$ and $\overline{CQ}=s-a$, we have
\begin{align*}
\overline{AP}\cdot \overline{CP}=\overline{AQ}\cdot \overline{CQ}
\end{align*}
Then by Power of Point Theorem, we can say that
\begin{align*}
\text{Pow}_{(ABC)}(P)=\text{Pow}_{(ABC)}(Q) \quad \blacksquare
\end{align*}
Therefore,
\begin{align*}
\overline{BP}\cdot \overline{PP'}=\overline{AP}\cdot \overline{CP}=\overline{AQ}\cdot \overline{CQ}=\overline{BQ}\cdot \overline{QQ'}\\
\end{align*}
Which implies that,
\begin{align*}
\boxed{\overline{BP}\cdot \overline{PP'}=\overline{BQ}\cdot \overline{QQ'}}
\end{align*}
[b]Claim:[/b] $\overline{BQ}>\overline{BP}$
[i]Proof[/i]: By applying Stewart Theorem in $\Delta ABC$ we have
\begin{align*}
\overline{BP}^2=\frac{(s-a)a^2+(s-c)c^2}{b}-(s-a)(s-c)\\
\overline{BQ}^2=\frac{(s-a)c^2+(s-c)a^2}{b}-(s-a)(s-c)\\
\overline{BQ}^2-\overline{BP}^2=\frac{(a-c)^2(a+c)}{b}>0
\end{align*}
So we have $\overline{BQ}>\overline{BP}$. $\blacksquare$
If $\overline{BQ}>\overline{BP}$, therefore from equation (1) we will have $\overline{PP'}>\overline{QQ'}$ $\blacksquare$ |
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 = (4.39, 3.47);
pair B1 = (2.29, 1.12);
pair F = (1.01, 1.53);
draw(circumcircle(A, B, C));
draw(circumcircle(X, Ee, D));
draw(circumcircle(D, B1, A));
draw(X--A--C--cycle);
draw(X--B);
draw(D--B1--A--cycle);
draw(B1--Ee);
draw(C--F);
draw(X--D);
draw(rightanglemark(X, D, B1), blue);
label("$A$", A, SW);
label("$X$", X, N);
label("$B$", B, S);
label("$C$", C, SE);
label("$B'$", B1, SE+S+E*(0.5));
label("$E$", Ee, NE);
label("$F$", F, S*(0.2)+W);
label("$D \equiv D'$", D, NW);
draw(rightanglemark(X, Ee, B1), blue);
draw(rightanglemark(X, F, B1), blue);
pair huh = (B1+B)/2;
draw(rightanglemark(B1, huh, C), blue);
markscalefactor = 0.05;
draw(anglemark(A, D, B1), red);
draw(anglemark(C, A, B1), red);
[/asy]
For now, we build our setup void of $D$, and then include it in later. Let $\omega$ be the circumcircle of $\triangle ABC$. Extend $BB'$ to hit $\omega$ again at $X \ne B$. Then we notice that $B'$ is infact the orthocentre of $\triangle XAC$, as it is the reflection of $B$ over $AC$. Now let $E = AB' \cap XC$ and $F = CB' \cap XA$. Consider the circle with diameter $XB'$, note that it goes through $F$ and $E$. Let this circle instersect $\omega$ at $D' \ne X$. We claim that $D \equiv D'$ now. Notice that we are given that the product of the opposite sides of $ABCD$ are equal. Since $ABCD$ is inscribed in $\omega$, this in fact means that $ABCD$ is a harmonic quadrilateral. Thus it suffices to show that $ABCD'$ is harmonic as well, since that would force $D \equiv D'$. Notice that $D'$ is the intersection of circumcircles of $XFE$ and $XAC$. Since $AF \cap CE = X$, this means that $D'$ is the Miquel point of $AFEC$. In particular $D'AF \sim^{+} D'CE$ which implies that $\dfrac{D'A}{D'C} = \dfrac{AF}{EC}$. But clearly even $B'FA \sim B'EC$ and thus \[\frac{D'A}{D'C} = \frac{AF}{EC} = \frac{B'A}{B'C} = \frac{BA}{BC}\] where the last equality is simply using the fact that $B, B'$ are reflections of each other about $AC$. Thus $\dfrac{D'A}{D'C} = \dfrac{BA}{BC}$ which is the same as $(D'B;AC) = -1$, or that $ABCD'$ is harmonic, as needed. Thus $D \equiv D'$. We now want to show that $CA$ is tangent to the circumcircle of $DB'A$, which translates to showing $\angle ADB = \angle CAB'$ Note that $\angle XCA + \angle XDA = 180^{\circ}$ and since $D$ lies on the circle with diameter $XB'$ we also have $\angle XDB' = 90^{\circ}$. These together imply that $\angle XCA + \angle ADB + 90^{\circ} = 180^{\circ}$ or that \[\angle ADB = 90^{\circ}-\angle XCA = 90^{\circ}-\angle ECA = \angle EAC = \angle CAB'\]
Thus we have shown that $\angle CAB' = \angle ADB'$ as required.
[/hide] |
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 all segments. Does this yield that $ABCD$ is a parallelogram? | @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 this.
Let $X=AK\cap BL$, $Y=BL\cap CM$, $Z=CM\cap DN$, $W=DN\cap AK$. Let $\frac{XY}{BL}=\frac{YZ}{CM}=\frac{ZW}{DN}=\frac{WX}{AK}=\lambda$. We'll denote the area of $\triangle A_{1}A_{2}A_{3}$ by $S_{A_{1}A_{2}A_{3}}$ or $S_{\triangle A_{1}A_{2}A_{3}}$ and respectively for quadrilaterals.
$\textbf{Step 1:}$ Let $G$ be the centroid of $ABCD$, that is, the center of mass for the four points $A,B,C,D$ and also the intersection of $MK$ and $NL$. Then notice that $G$ lies strictly inside $XYZW$ (because $G=MK\cap NL$), so:
$$S_{XYZW}=S_{\triangle XGY}+S_{\triangle YGZ}+S_{\triangle ZGW}+S_{\triangle WGX}=$$
$$=\lambda(S_{\triangle BGL}+S_{\triangle CGM}+S_{\triangle DGN}+S_{\triangle AGK})=$$
$$=\frac{\lambda}{2}(S_{\triangle BLN}+S_{\triangle CMK}+S_{\triangle DLN}+S_{\triangle AKM})=$$
$$=\frac{\lambda}{4}(S_{\triangle ALB}+S_{\triangle BMC}+S_{\triangle CND}+S_{\triangle DKA})=$$
$$=\frac{\lambda}{8}\Big((S_{\triangle DAB}+S_{\triangle ABC})+(S_{\triangle ABC}+S_{\triangle BCD})+$$
$$+(S_{\triangle BCD}+S_{\triangle CDA})+(S_{\triangle CDA}+S_{\triangle DAB})\Big)=$$
$$=\frac{\lambda}{8}\times 4S_{ABCD}=\frac{\lambda}{2}S_{ABCD}$$
\[\Longrightarrow S_{XYZW}=\frac{\lambda}{2}S_{ABCD}\quad (1)\]
We can also show that:
$$S_{\triangle AZW}+S_{\triangle CXY}=\lambda(S_{\triangle AND}+S_{\triangle CLB})=\frac{\lambda}{2}(S_{\triangle ABD}+S_{\triangle BCD})=\frac{\lambda}{2}S_{ABCD}$$
\[\Longrightarrow S_{\triangle AZW}+S_{\triangle CXY}=\frac{\lambda}{2}S_{ABCD}\quad (2)\]
$$S_{\triangle BXW}+S_{\triangle DYZ}=\lambda(S_{\triangle ABK}+S_{\triangle CDM})=\frac{\lambda}{2}(S_{\triangle ABC}+S_{\triangle CDA})=\frac{\lambda}{2}S_{ABCD}$$
\[\Longrightarrow S_{\triangle BXW}+S_{\triangle DYZ}=\frac{\lambda}{2}S_{ABCD}\quad (3)\]
Also, because $S_{AKCM}=S_{\triangle AKC}+S_{\triangle CMA}=\frac{1}{2}(S_{\triangle ABC}+S_{\triangle CDA})=\frac{1}{2}S_{ABCD}$ and $S_{DNBL}=S_{\triangle DNB}+S_{\triangle BLD}=\frac{1}{2}(S_{\triangle DAB}+S_{\triangle BCD})=\frac{1}{2}S_{ABCD}$ we have that:
$$S_{\triangle BXC}+S_{\triangle AZD}=2(S_{\triangle CXK}+S_{\triangle AZM})=$$
$$=2(S_{AKCM}-S_{XYZW}-(S_{\triangle AZW}+S_{\triangle CXY}))\overset{(2)}{=}$$
$$\overset{(2)}{=}2\left(\frac{1}{2}S_{ABCD}-\frac{\lambda}{2}S_{ABCD}-\frac{\lambda}{2}S_{ABCD}\right)=(1-2\lambda)S_{ABCD}$$
\[\Longrightarrow S_{\triangle BXC}+S_{\triangle AZD}=(1-2\lambda)S_{ABCD}\quad (4)\]
$$S_{\triangle DYC}+S_{\triangle AWB}=2(S_{\triangle DLY}+S_{\triangle BNG})=2(S_{BLDN}-S_{XYZW}-(S_{\triangle BXW}+S_{\triangle DYZ})\overset{(3)}{=}$$
$$\overset{(3)}{=}2(\frac{1}{2}S_{ABCD}-\frac{\lambda}{2}S_{ABCD}-\frac{\lambda}{2}S_{ABCD})=(1-2\lambda)S_{ABCD}$$
\[\Longrightarrow S_{\triangle DYC}+S_{\triangle AWB}=(1-2\lambda)S_{ABCD}\quad (5)\]
Now, using $(1),(2),(3),(4),(5)$ we can evaluate $\lambda$:
$$S_{ABCD}=S_{XYZW}+(S_{\triangle BXW}+S_{\triangle DYZ})+(S_{\triangle AZW}+S_{\triangle CXY})+$$
$$+(S_{\triangle CXB}+S_{\triangle DZA})+(S_{\triangle CYD}+S_{\triangle AGB})=$$
$$=\frac{\lambda}{2}S_{ABCD}+\frac{\lambda}{2}S_{ABCD}+\frac{\lambda}{2}S_{ABCD}+(1-2\lambda)S_{ABCD}+(1-2\lambda)S_{ABCD}=$$
$$=\left(2-\frac{5}{2}\lambda\right)S_{ABCD}\Longrightarrow \boxed{\lambda=\frac{2}{5}}$$
From here on we'll use the results above with $\lambda=\frac{2}{5}$ as well as use the ratios, which $\lambda$ represents.
$\textbf{Step 2:}$ Now look at $AKCM$ and let $\frac{AW}{AK}=a<1-\lambda=\frac{3}{5}$ and $\frac{CY}{CM}=b<1-\lambda=\frac{3}{5}$. Then we have that:
$$\frac{2}{5}S_{AKCM}=\frac{1}{5}S_{ABCD}=S_{XYZW}=S_{\triangle XYW}+S_{\triangle YZW}=\frac{2}{5}S_{\triangle AYK}+\frac{2}{5}S_{\triangle CWM}=$$
$$=\frac{2}{5}\left(bS_{AMK}+(1-b)S_{ACK}+aS_{CKM}+(1-a)S_{AMC}\right)=$$
$$=\frac{2}{5}\left(bS_{AMCK}-bS_{CMK}+(1-b)S_{AKCM}-(1-b)S_{AMC}+aS_{CKM}+(1-a)S_{AMC}\right)=$$
$$=\frac{2}{5}S_{AKCM}+\frac{2}{5}(a-b)(S_{\triangle CKM}-S_{\triangle AMC})$$
$$\Longrightarrow \frac{2}{5}(a-b)(S_{\triangle CKM}-S_{\triangle AMC})=0$$
There are two cases: $a=b$ or $S_{\triangle CKM}=S_{\triangle AMC}$ and in the second case that'd mean that $CM\parallel AK$ as $\text{dist}(A,MC)\cdot MC=2S_{\triangle AMC}=2S_{\triangle CKM}=\text{dist}(K,MC)\cdot MS\Longrightarrow \text{dist}(A,MC)=\text{dist}(K,MC)\Longrightarrow AK\parallel CM$. An analogical argument can be said about the quadrilateral $BLDN$. Now we just have to examine the cases.
$\textbf{Step 3}$ If $AK\parallel CM$ and $BL\parallel DN$, then $XYZW$ is a prallelogram and so $AK=\frac{5}{2}XW=\frac{5}{2}YZ=CM$ and analogically $DN=BL$. But then $AK\parallel CM$ and $AK=CM$, so $AKCM$ is a parallelogram and similarly $BNDL$ is a parallelogram as well. But then $AM\parallel CK$ and $DL\parallel BN$, so $ABCD$ is a parallelogram. Therefore this case doesn't lead to any non-trivial solutions for $ABCD$.\newline\newline
If WLOG $\frac{AW}{AK}=\frac{CY}{CM}=a$, then from (2) in $\textbf{Step 1}$ we have that:
$$\frac{1}{5}S_{ABCD}=S_{\triangle AWZ}+S_{\triangle CXM}=a(S_{\triangle AZK}+S_{\triangle CXM})=\frac{2a}{5}(S_{\triangle ZXY}+S_{\triangle XWZ})=$$
$$=\frac{2a}{5}S_{XYZW}=\frac{1}{5}S_{ABCD}\Longrightarrow a=\frac{2}{5}$$
Thus $AW=\frac{2}{5}AK=WX$, but also $AN=NB$, so $DN\parallel BL$. It seems like we're so close to proving that $ABCD$ has to be a parallelogram, but we can make no further progress, due to the existence of counterexamples, which we presented in the next "chapter" of the solution. This was the thought process behind finding the counterexample.
$\textbf{Counterexample:}$ Consider the coordinate plane and let $A=(0,0)$, $B=(1,1)$, $C=(3,-1)$, $D=(1,-2)$. The quadrilateral $ABCD$ is convex. Also $K=(2,0)$, $L=(2,-\frac{3}{2})$, $M=(\frac{1}{2},-1)$, $N=(\frac{1}{2},\frac{1}{2})$. Let $X=AK\cap BL$, $Y=BL\cap CM$, $Z=CM\cap DN$, $W=DN\cap AK$ be the common points of the segments $AK,BL,CM,DN$. Then $X,W\in AK\Longrightarrow X=(x_{0},0)$ and $W=(w_{0},0)$ for some $0<x_{0},w_{0}<2$ and $Y,Z\in CM\Longrightarrow Y=(y_{0},-1)$ and $Z=(z_{0},-1)$ for some $\frac{1}{2}<z_{0},y_{0}<3$. Now we will calculate $x_{0},y_{0},z_{0},w_{0}$. We'll do this by calculating the slope of the lines $DN$ and $BL$. Let's denote $\ell_{1}$ to be the line $DN$ and $BL$ to be the line $\ell_{2}$ for shortness. Firstly, let's calculate the slope of $\ell_{1}$ and let's call it $s_{1}$ (we'll denote by $(x_{t},y_{t})$ the point $T$ even if we know the exact coordinates of said point just so we write out the slope of line fomrula clearly):
$$D\in\ell_{1}, N\in\ell_{1}\Longrightarrow s_{1}=\frac{y_{n}-y_{d}}{x_{n}-x_{d}}=\frac{\frac{1}{2}-(-2)}{\frac{1}{2}-1}=-5$$
$$Z\in \ell_{1},D\in \ell_{1}\Longrightarrow -5=s_{1}=\frac{y_{z}-y_{d}}{x_{z}-x_{d}}=\frac{-1-(-2)}{z_{0}-1}=\frac{1}{z_{0}-1}\Longrightarrow z_{0}=\frac{4}{5}$$
$$W\in \ell_{1},D\in \ell_{1}\Longrightarrow -5=s_{1}=\frac{y_{w}-y_{d}}{x_{w}-x_{d}}=\frac{0-(-2)}{w_{0}-1}=\frac{2}{w_{0}-1}\Longrightarrow w_{0}=\frac{3}{5}$$
$$\Longrightarrow W=\left(\frac{3}{5},0\right),\quad Z=\left(\frac{4}{5},-1\right)$$
$$B\in\ell_{2}, L\in\ell_{2}\Longrightarrow s_{2}=\frac{y_{b}-y_{l}}{x_{b}-x_{l}}=\frac{1-(-\frac{3}{2})}{1-2}=-\frac{5}{2}$$
$$X\in \ell_{2},B\in \ell_{2}\Longrightarrow -\frac{5}{2}=s_{2}=\frac{y_{x}-y_{b}}{x_{x}-x_{b}}=\frac{0-1}{x_{0}-1}=\frac{-1}{x_{0}-1}\Longrightarrow x_{0}=\frac{7}{5}$$
$$Y\in \ell_{2},B\in \ell_{2}\Longrightarrow -\frac{5}{2}=s_{2}=\frac{y_{y}-y_{b}}{x_{y}-x_{b}}=\frac{-1-1}{y_{0}-1}=\frac{-2}{y_{0}-1}\Longrightarrow y_{0}=\frac{9}{5}$$
$$\Longrightarrow X=\left(\frac{7}{5},0\right),\quad Y=\left(\frac{9}{5},-1\right)$$
Now we can calculate the ratios of the medial parts to the whole segments of $AK,BL,CM,DN$. For the segments $AK$ and $CM$ the calculations are easier, because all $y$ values of points on either of those lines are equal, but for $BL$ and $DN$ we'll need to use the Pythagorian theorem. Here are the claculations:
$$AK=2, \quad XW=x_{x}-x_{w}=\frac{7}{5}-\frac{3}{5}=\frac{4}{5}\Longrightarrow \frac{XW}{AK}=\frac{2}{5}$$
$$CM=x_{c}-x_{m}=3-\frac{1}{2}=\frac{5}{2},\quad YZ=x_{y}-x_{z}=\frac{9}{5}-\frac{4}{5}=1\Longrightarrow \frac{YZ}{CM}=\frac{2}{5}$$
$$BL=\sqrt{(x_{b}-x_{l})^2+(y_{b}-y_{l})^2}=\sqrt{1+\frac{25}{4}}=\frac{\sqrt{29}}{2}$$
$$XY=\sqrt{(x_{x}-x_{y})^2+(y_{x}-y_{y})^2}=\sqrt{\frac{4}{25}+1}=\frac{\sqrt{29}}{5}\Longrightarrow \frac{XY}{BL}=\frac{2}{5}$$
$$DN=\sqrt{(x_{d}-x_{n})^2+(y_{d}-y_{n})^2}=\sqrt{\frac{1}{4}+\frac{25}{4}}=\frac{\sqrt{26}}{2}$$
$$ZW=\sqrt{(x_{z}-x_{w})^2+(y_{z}-y_{w})^2}=\sqrt{\frac{1}{25}+1}=\frac{\sqrt{26}}{5}\Longrightarrow \frac{ZW}{DN}=\frac{2}{5}$$
Therefore we've shown that $\frac{XW}{AK}=\frac{YZ}{CM}=\frac{XY}{BL}=\frac{ZW}{DN}=\frac{2}{5}$ and $ABCD$ isn't a parallelogram (the simplest proof of this is the fact that $AB=\sqrt{(x_{a}-x_{b})^2+(y_{a}-y_{b})^2}=\sqrt{2}$ and $CD=\sqrt{(x_{c}-x_{d})^2+(y_{c}-y_{d})^2}=\sqrt{5}$, so $AB\neq CD\Longrightarrow ABCD$ isn't a parallelogram), so the answer to the problem is negative, the conditions don't yield that $ABCD$ is a parallelogram, because we showed a counterexample. |
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