id int64 8 3.28M | problem stringlengths 27 6.88k | solution stringlengths 2 18.5k |
|---|---|---|
2,938,164 | $A$ and $B$ are any two subsets of $\{1, 2,...,n - 1\}$ such that $|A| +|B|> n - 1$. Prove that one can find $a$ in $A$ and $b$ in $B$ such that $a + b = n$. | solve by pigeonhole theorem |
2,938,166 | $n$ and $d$ are positive integers such that $d$ divides $2n^2$. Prove that $n^2 + d$ cannot be a square. | Solved [url=https://artofproblemsolving.com/community/c6h3217220p29434468]here[/url] |
2,938,175 | Prove that if the two angles on the base of a trapezoid are different, then the diagonal starting from the smaller angle is longer than the other diagonal.
[img]https://cdn.artofproblemsolving.com/attachments/7/1/77cf4958931df1c852c347158ff1e2bbcf45fd.png[/img] | Let $AD = x, AA' = DD' = h$ such that $AA', DD' \perp BC$. If $\angle{BCD} = \alpha, \angle{CBA} = \beta$ such that $0< \alpha < \beta < \frac{\pi}{2}$, then the diagonals $AC, BD$ have lengths:
$AC = \sqrt{(AA')^2 + (A'C)^2} = \sqrt{h^2 + (x + h \cdot cot(\alpha))^2}$ (i);
$BD = \sqrt{(DD')^2 + (D'B)^2} = \sqrt{h^2... |
2,938,197 | Show that if $m$ and $n$ are integers such that $m^2 + mn + n^2$ is divisible by $9$, then they must both be divisible by $3$. | See [url]https://artofproblemsolving.com/community/c6t29718f6h3195541_divisibility_property[/url] |
2,939,024 | What integers $a, b, c$ satisfy $a^2 + b^2 + c^2 + 3 < ab + 3b + 2c$ ? | First we deal with the five trivial cases $a=-2,-1,0,1,2$.
If $a=-2$ the given inequality rearranges to:
$$\frac{23}4+\left(b-\frac12\right)^2+(c-1)^2<0$$
which is false.
If $a=-1$ the given inequality rearranges to:
$$2+(b-1)^2+(c-1)^2<0$$
which is false.
If $a=0$ the given inequality rearranges to:
$$(b-1)(b-2)+(c-1)... |
2,939,043 | Show that if $2 + 2\sqrt{28n^2 + 1}$ is an integer, then it is a square (for $n$ an integer). | solved [url=https://artofproblemsolving.com/community/c6h114176p648653]here[/url] |
2,941,134 | A straight line cuts the side $AB$ of the triangle $ABC$ at $C_1$, the side $AC$ at $B_1$ and the line $BC$ at $A_1$. $C_2$ is the reflection of $C_1$ in the midpoint of $AB$, and $B_2$ is the reflection of $B_1$ in the midpoint of $AC$. The lines $B_2C_2$ and $BC$ intersect at $A_2$. Prove that $$\frac{sen \, \, B_1A_... | posted for the image links |
2,941,117 | For what positive integers $n, k$ (with $k < n$) are the binomial coefficients $${n \choose k- 1} \,\,\, , \,\,\, {n \choose k} \,\,\, , \,\,\, {n \choose k + 1}$$ three successive terms of an arithmetic progression? | Let us begin by expanding each of the above combinatorial terms according to:
$\frac{n!}{(k - 1)!(n - k + 1)(n - k)(n - k - 1)!}, \frac{n!}{k(k - 1)!(n - k)(n - k - 1)!}, \frac{n!}{(k + 1)k(k - 1)!(n - k - 1)!}$ (i).
Each term in (i) has the common factor $\frac{n!}{(k - 1)! \cdot (n - k - 1)!}$, which reduces our ... |
2,941,142 | A triangle has inradius $r$ and circumradius $R$. Its longest altitude has length $H$. Show that if the triangle does not have an obtuse angle, then $H \ge r+R$. When does equality hold? | solved [url=https://artofproblemsolving.com/community/c6h294413p1593465]here[/url] |
2,941,144 | $f$ is a real-valued function defined on the reals such that $f(x) \le x$ and $f(x + y) \le f(x) + f(y)$ for all $x, y$. Prove that $f(x) = x$ for all $x$. | solved [url=https://artofproblemsolving.com/community/c6h1109982p5052117]here[/url] and [url=https://artofproblemsolving.com/community/c6h257315p1403063]here[/url] |
2,937,791 | The points of space are coloured with five colours, with all colours being used. Prove that some plane contains four points of different colours. | solved [url=https://artofproblemsolving.com/community/c6h42710p269993]here[/url] and [url=https://artofproblemsolving.com/community/c6h181160p995809]here[/url] |
2,937,995 | Prove that $$AB + PQ + QR + RP \le AP + AQ + AR + BP + BQ + BR$$ where $A, B, P, Q$ and $R $ are any five points in a plane. | solved [url=https://artofproblemsolving.com/community/c6h411690p2310245]here[/url]
similar but not the same
[quote]Given five points $A,B,C,D,E$ in a plane, no three of which are collinear, prove the inequality
\[AB+BC+CA+DE\le AD+AE+BD+BE+CD+CE \]
[/quote]
[url=https://artofproblemsolving.com/community/c6h390986p2172... |
2,937,996 | Let $n > 2$ be an even number. The squares of an $n\times n$ chessboard are coloured with $\frac12 n^2$ colours in such a way that every colour is used for colouring exactly two of the squares. Prove that one can place $n$ rooks on squares of $n$ different colours such that no two of the rooks can take each other. | solved [url=https://artofproblemsolving.com/community/c6h6164p21027]here[/url], [url=https://artofproblemsolving.com/community/c6h181157p995801]here [/url] and [url=https://artofproblemsolving.com/community/c6h2087281p15058748]here[/url] |
2,938,017 | Given are $n + 1$ points $P_1, P_2,..., P_n$ and $Q$ in the plane, no three collinear. For any two different points $P_i$ and $P_j$ , there is a point $P_k$ such that the point $Q$ lies inside the triangle $P_iP_jP_k$. Prove that $n$ is an odd number. | solved [url=https://artofproblemsolving.com/community/c6h34783p215998]here[/url] |
2,938,020 | Writing down the first $4$ rows of the Pascal triangle in the usual way and then adding up the numbers in vertical columns, we obtain $7$ numbers as shown above. If we repeat this procedure with the first $1024$ rows of the Pascal triangle, how many of the $2047$ numbers thus obtained will be odd?
[img]https://cdn.arto... | posted for the image links |
599,789 | We reflected each vertex of a triangle on the opposite side. Prove that the area of the triangle formed by these three reflection points is smaller than the area of the initial triangle multiplied by five. | Label $\triangle ABC$ the given triangle with circumcircle $(O,R)$ and $A',B',C'$ denote the reflections of $A,B,C$ on $BC,CA,AB.$ $\triangle A'B'C'$ is homothetic to the pedal triangle $\triangle XYZ$ of its 9-point center $N$ through the homothety with center $G,$ the centroid of $\triangle ABC,$ and coefficient 4 (t... |
599,790 | We have triangulated a convex $(n+1)$-gon $P_0P_1\dots P_n$ (i.e., divided it into $n-1$ triangles with $n-2$ non-intersecting diagonals). Prove that the resulting triangles can be labelled with the numbers $1,2,\dots,n-1$ such that for any $i\in\{1,2,\dots,n-1\}$, $P_i$ is a vertex of the triangle with label $i$. | I have a hunch I've seen this problem somewhere before, but anyway:
[hide=Solution]
Easy induction. $n=2$ is trivial.
There are $n-1$ triangles and $n+1$ edges of the polygon, so at least two triangles have two edges of the polygon as sides - call them special. Take one of these, call it $\triangle$, let its two sides ... |
599,791 | For every $n\in\mathbb{N}$, define the [i]power sum[/i] of $n$ as follows. For every prime divisor $p$ of $n$, consider the largest positive integer $k$ for which $p^k\le n$, and sum up all the $p^k$'s. (For instance, the power sum of $100$ is $2^6+5^2=89$.) Prove that the [i]power sum[/i] of $n$ is larger than $n$ for... | wow, this belongs in high-school intermediate. just consider $2^2+2,2^4+2,2^6+2$ and numbers like that. $2^{2k}+2$. The power sum is at least $2^{2k}
+3$ so we are done. |
599,785 | Any two members of a club with $3n+1$ people plays ping-pong, tennis or chess with each other. Everyone has exactly $n$ partners who plays ping-pong, $n$ who play tennis and $n$ who play chess.
Prove that we can choose three members of the club who play three different games amongst each other. | One possible graph for $n=2$ (attached). |
599,786 | A and B plays the following game: they choose randomly $k$ integers from $\{1,2,\dots,100\}$; if their sum is even, A wins, else B wins. For what values of $k$ does A and B have the same chance of winning? | The answer is $k$ an odd number with $1 \leq k \leq 99$. Consider the function $f(x,y) = (1+xy)(1+x^2y)(1+x^3y) \cdots (1+x^{100}y)$. The coefficient of $x^m y^k$ in $f$ is the number of $k$-element subsets with sum $m$. If we set $x = -1$, then the coefficient of $y^k$ is the number of $k$-element subsets with even su... |
599,787 | Let $n>2$ be a positive integer. Find the largest value $h$ and the smallest value $H$ for which
\[h<{a_1\over a_1+a_2}+{a_2\over a_2+a_3}+\cdots+{a_n\over a_n+a_1}<H\]
holds for any positive reals $a_1,\dots,a_n$. | $h_{max}=1$ and $H_{min}=n-1$. :maybe: |
599,782 | Find all quadruples of positive integers $(a,b,c,d)$ such that $a+b=cd$ and $c+d=ab$. | WLOG suppose that $a \ge b$ and $c \ge d$.
$\rightarrow 2a \ge cd$ , $2c \ge ab$ $\Rightarrow 4 \ge bd$ $\Rightarrow b,d \in {1,2,3,4}$
1)$b=1$ , $d={1,2,3,4} \Rightarrow c+d+1 = cd \Rightarrow c=3 \Rightarrow (a,b,c,d) = (5,1,3,2)$
2)$b=2$ , $d={1,2} \Rightarrow c+d+4 = 2cd$
$\rightarrow$ if $d=1 \Rightarrow... |
599,783 | Is there a set of points in space whose intersection with any plane is a finite but nonempty set of points? | I have seen this somewhere, but I don't remember. I think $S=\{(t^5,t^3,t)\lvert t\in \mathbb{R}\}$ work. To see why take any plane $ax+by+cz+d=0$. It will have an intersection with $S$ if $at^5+bt^3+ct+d=0$, which always has a real solution, whatever be $a,b,c,d$, hence non-empty, but at most $5$. Hence finite. |
598,811 | Prove that if there exists a point $P$ inside the convex quadrilateral $ABCD$ such that the triangles $PAB$, $PBC$, $PCD$, $PDA$ have the same area, then one of the diagonals of $ABCD$ bisects the area of the quadrilateral. | Suppose $P$ doesn't lie on $AC$ (if it did, we'd be done). Let $BP \cap AC = X$ and $DP \cap AC = Y$. Clearly, $X, Y$ are both midpoints of $AC$, so $P$ lies on $BD$, and we are done. |
598,812 | Set $T\subset\{1,2,\dots,n\}^3$ has the property that for any two triplets $(a,b,c)$ and $(x,y,z)$ in $T$, we have $a<b<c$, and also, we know that at most one of the equalities $a=x$, $b=y$, $c=z$ holds. Maximize $|T|$. | Of course, since you can pick $(a,b,c) = (x',y',z')$ and $(x,y,z) = (a',b',c')$ to get $x' < y' < z'$. |
598,808 | In the plane, two intersecting lines $a$ and $b$ are given, along with a circle $\omega$ that has no common points with these lines. For any line $\ell||b$, define $A=\ell\cap a$, and $\{B,C\}=\ell\cap \omega$ such that $B$ is on segment $AC$. Construct the line $\ell$ such that the ratio $\frac{|BC|}{|AB|}$ is maximal... | EDIT: ver 2(miss a:y=kx and correct mistakes)
let$D$ is the center of $w$, line $x||b$ through $D$, $O=x \cap a$,$y \perp x$ through $O, OD=a,E=w \cap x$,$DE=r \implies w: (x-a)^2+y^2=r^2 ,a:y=kx \implies \dfrac{BC}{AB}=\dfrac{2\sqrt{r^2-y^2}}{a-\dfrac{y}{k}-\sqrt{r^2-y^2}}=P$.
we want $P$ is max,$\implies \dfrac{1}{... |
598,809 | For any positive integer $n$ denote $S(n)$ the digital sum of $n$ when represented in the decimal system. Find every positive integer $M$ for which $S(Mk)=S(M)$ holds for all integers $1\le k\le M$. | Let $M=\sum_{i=0}^{n} a_i\cdot 10^{i},0\le a_1\le 9,\forall i\in\{0,1,2,...,n\},a_k\neq 0$.From the equation $S(M)=S(M(10^n+1))$ we easily see that $M$ must be of the form $M=10^l-1,l\in\mathbb{N}$.Now it's easy to prove that all the numbers of the form $10^l-1$ satisfy the given condition(just calculate the sum of the... |
598,810 | We play the following game in a Cartesian coordinate system in the plane. Given the input $(x,y)$, in one step, we may move to the point $(x,y\pm 2x)$ or to the point $(x\pm 2y,y)$. There is also an additional rule: it is not allowed to make two steps that lead back to the same point (i.e, to step backwards).
Prove th... | Let $A= \begin{pmatrix} 1& 2 \\
0 & 1 \end{pmatrix}$ and $B=\begin{pmatrix} 1& 0 \\
2 & 1 \end{pmatrix}.$ Since $\{1, \sqrt{2} \}$ are linearly independent over $\mathbb{Q}$, the problem reduces to proving that every reduced nonempty word in $A$ and $B$ is not the identity matrix. But this is just [url=https://artofpr... |
598,800 | Let $p>2$ be a prime number and $n$ a positive integer. Prove that $pn^2$ has at most one positive divisor $d$ for which $n^2+d$ is a square number. | [hide]
Prove by contradiction and suppose we have $n^2+d=x^2$, $n^2+e=y^2$ and $d \neq e$. Set $ad=be=pn^2$, and let $\gcd{(x,n)} = g$, $\gcd{(y,n)}=h$. We have
\[a(x+n)(x-n)=pn^2 \Longrightarrow a\left(\frac{x}{g}+\frac{n}{g}\right)\left(\frac{x}{g}-\frac{n}{g}\right)=p\left(\frac{n^2}{g^2}\right)\] It follows tha... |
598,806 | We would like to give a present to one of $100$ children. We do this by throwing a biased coin $k$ times, after predetermining who wins in each possible outcome of this lottery.
Prove that we can choose the probability $p$ of throwing heads, and the value of $k$ such that, by distributing the $2^k$ different outcomes ... | This problem is a gem, definitely worth trying to solve.
[hide=Solution (by Alex Gunning)]
Let $n=100$. Your probabilities are $p=\frac1{1+x}$ and $1-p=\frac{x}{1+x}$; then the possible outcomes include exactly $\binom kj$ possibilities with probability $\frac{x^j}{(1+x)^k}$ for $j=0,1,\dots,k$. We want to partition t... |
598,797 | Let $n$ be a positive integer, and $a,b\ge 1$, $c>0$ arbitrary real numbers. Prove that
\[\frac{(ab+c)^n-c}{(b+c)^n-c}\le a^n.\] | Alternative:
For $a=1$ we have equality and we can see the RHS grows faster than the LHS as function in a. |
598,798 | A convex polyhedron has two triangle and three quadrilateral faces. Connect every vertex of one of the triangle faces with the intersection point of the diagonals in the quadrilateral face opposite to it. Show that the resulting three lines are concurrent. | It's not hard to see that the polyhedron satisfying the restrictions of the problem belongs to the combinatorial type of triangle prism. Let it be $ABCA_1B_1C_1$ with face $ABC$ and edges $AA_1, BB_1,CC_1$. Denote by $C_0, A_0, B_0$ the diagonal intersection points of the quadrilaterals $AA_1B_1B, BB_1C_1C, CC_1A_1A$ r... |
598,792 | Define for $n$ given positive reals the [i]strange mean[/i] as the sum of the squares of these numbers divided by their sum. Decide which of the following statements hold for $n=2$:
a) The strange mean is never smaller than the third power mean.
b) The strange mean is never larger than the third power mean.
c) The s... | For n=2 the first statement is true.
It became equiavelnt with $[x-1]^4(x^2+x+1)\ge 0$ after homogenizing and working out. |
598,793 | For any positive integer $k$ define $f_1(k)$ as the square of the digital sum of $k$ in the decimal system, and $f_{n}(k)=f_1(f_{n-1}(k))$ $\forall n>1$. Compute $f_{1992}(2^{1991})$. | First we look modulo $9$ to see $f_{1992}(2^{1991}) \equiv 4 \pmod{9}$.
Because $f_{2k}(2^{1991}) \equiv 4 , f_{2k-1}(2^{1991}) \equiv 7 \pmod{9}$ for all $k \ge 1$.
Next we claim it will be $256$.
$f_1(2^{1991}) < (1000*9)^2<10^8$.
$f_2(2^{1991})<f_1(99999999)=72^2<9999$
$f_3(2^{1991})<36^2<1999$
$f_4(2^{1991})<28^2... |
598,789 | Let $a$ and $b$ be positive integers. Prove that the numbers $an^2+b$ and $a(n+1)^2+b$ are both perfect squares only for finitely many integers $n$. | [solution corrected]
Let $an^2+b=x^2$ and $a(n+1)^2+b=(x+k)^2$. Subtracting the latter from the former, we get $a(2n+1) = 2xk+k^2$ and thus $n=\frac{2xk+k^2-a}{2a}$. Plugging it into the first equation, we get $a\left( \frac{2xk+k^2-a}{2a} \right)^2 + b = x^2$, that is, $(2xk+k^2-a)^2 + 4ab = 4ax^2$ and thus
\[(\star)... |
598,791 | Let $n$ be a fixed positive integer. Compute over $\mathbb{R}$ the minimum of the following polynomial:
\[f(x)=\sum_{t=0}^{2n}(2n+1-t)x^t.\] | The identity
\begin{align*}
f(x)&=x^{2n}+2x^{2n-1}+3x^{2n-2}+\dots+(2n)x+(2n+1)\\
&= \left(x^n+x^{n-1}\right)^2+2\left(x^{n-1}+x^{n-2}\right)^2+\dots+n\left(x+1\right)^2+\left(n+1\right)
\end{align*}
implies that the minimum of $f$ is $\boxed{n+1}$, attained at $x=-1$. |
597,998 | Prove that in a trapezoid with perpendicular diagonals, the product of the legs is at least as much as the product of the bases. | Let $ ABCD $ be the trapezoid with $ AD \parallel BC $ and let $ AC \cap BD = P $. We want to prove that
\[ AB^2 \cdot CD^2 \ge AD^2 \cdot BC^2 \]
\[ \iff (PA^2 + PB^2)(PC^2 + PD^2) \ge (PA^2 + PD^2)(PB^2+PC^2) \]
\[ \iff (PA^2 - PC^2)(PD^2 - PB^2) \ge 0 \]
But since $ AD \parallel BC \implies \frac{PA}{PD} = \frac{PC... |
598,001 | Two countries ($A$ and $B$) organize a conference, and they send an equal number of participants. Some of them have known each other from a previous conference. Prove that one can choose a nonempty subset $C$ of the participants from $A$ such that one of the following holds:
[list][*]the participants from $B$ each know... | Let there be $n$ participants, so there are $2^n-1$ choices for $C$. Let the participants from $B$ be $b_1, b_2, \ldots, b_n$. Now, for every nonempty subset $C \subset A$, let $f(C) = (f_1, f_2, \ldots, f_n)$ where $f_i$ denotes the parity of the number of participants in $C$ that $b_i$ knows, $0$ for even and $1$ for... |
598,002 | Let $n$ and $k$ be arbitrary non-negative integers. Suppose we have drawn $2kn+1$ (different) diagonals of a convex $n$-gon. Show that there exists a broken line formed by $2k+1$ of these diagonals that passes through no point more than once. Prove also that this is not necessarily true when we draw only $kn$ diagonals... | I assume that the question means choosing $2k+1$ disjoint diagonals at the vertices, and not that diagonals are non intersecting.
Suppose otherwise, then we let the largest number of diagonals that can be chosen be $t$. Since $k<\frac{n-3}{4}$, we can always make $t=2k$ (by adjusting some diagonals, and also we are ass... |
597,995 | Let $p>2$ be a prime number and let $L=\{0,1,\dots,p-1\}^2$. Prove that we can find $p$ points in $L$ with no three of them collinear. | Just choose the points $ (x,y) $, $ 0 \le x \le p-1 $ and $ y \equiv x^2 ( \bmod p ) $. Then, for any three distinct points $ (x_1,y_1), (x_2,y_2), (x_3,y_3) $, the area of this triangle in $ \mathbb{F}_p $ is given by the determinant
\[ \begin{vmatrix}
1 & 1 & 1 \\
x_1 & x_2 & x_3 \\
x_1^2 & x_2^2 & x_3^2
\end{vmatri... |
597,996 | The center of the circumcircle of $\triangle ABC$ is $O$. The incenter of the triangle is $I$, and the intouch triangle is $A_1B_1C_1$. Let $H_1$ be the orthocenter of $\triangle A_1B_1C_1$. Prove that $O$, $I$, and $H_1$ are collinear. | Let $ AI \cap (O) =A_2 ,BI \cap (O) =B_2, CI \cap (O) =C_2 $ ,then the triangles $ A_1B_1C_1 $ and $ A_2B_2C_2 $ are similar homotetic with centered at $ I $ ( with parallel sides ).On the other hand $ I,O $ are orthocenter and circumcenter of the triangle $ A_2B_2C_2 $ , $ H_1 ,I $ are orthocenter and circumcenter of... |
597,997 | Prove that the vertices of any planar graph can be colored with $3$ colors such that there is no monochromatic cycle. | If the number of vertices, edges, and faces of a planar graph $\mathcal{G}$ is $V$, $E$, $F$, resp., then [url=https://en.wikipedia.org/wiki/Euler_characteristic#Planar_graphs]Euler's formula[/url] gives $V-E+F=2$. Adding up the number of edges for each face gives $2E\ge 3F$, so
\[
2+E-V=F\le \frac23 E\quad\implies \qu... |
597,991 | Is there an infinite sequence of positive integers where no two terms are relatively prime, no term divides any other term, and there is no integer larger than $1$ that divides every term of the sequence? | Take $6$, $10$, and $15p$ where $p$ runs through all the primes $\ge7$. |
597,992 | Prove that for every positive integer $n$, there exists a polynomial with integer coefficients whose values at points $1,2,\dots,n$ are pairwise different powers of $2$. | [u]Lemma[/u] [i]For every positive integer $n$, there exists a polynomial $P_n(x)$ with integer coefficients such that $P_n(x)=2^a$ when $x=-n,-n+1,\dots,-1,1,2,\dots,n$ and $P_n(0)=2^{a+b}$, where $a$ and $b$ are integers, $a \ge 0$ and $b>0$.[/i]
[i]Proof.[/i] Let $(n!)^2=2^a c$, where $a$ is nonnegative integer and ... |
597,993 | For which integers $N\ge 3$ can we find $N$ points on the plane such that no three are collinear, and for any triangle formed by three vertices of the points’ convex hull, there is exactly one point within that triangle? | Clearly, the convex hull most be an $n$-gon, and since a triangulation of this $n$-gon contains $n-2$ triangles, each with exactly one point in it, it follows that $N=2n-2$ for some $n$. We will now provide a construction for each even $N\ge 3$, i.e. for each $n\ge 3$, $N=2n-2$.
Consider some convex $n$-gon $P$ and ma... |
597,986 | For any positive integer $m$, denote by $d_i(m)$ the number of positive divisors of $m$ that are congruent to $i$ modulo $2$. Prove that if $n$ is a positive integer, then
\[\left|\sum_{k=1}^n \left(d_0(k)-d_1(k)\right)\right|\le n.\] | We swap the order of summation ([url=http://www.mit.edu/~evanchen/handouts/Summation/Summation.pdf]see here[/url]):
\begin{align*}
S:&=-\sum_{k=1}^n (d_0(k)-d_1(k)) =-\sum_{d|k,\,1\le d,k\le n} (-1)^d = -\sum_{d=1}^n \sum_{1\le j\le \frac{n}{d}} (-1)^d \\
&= -\sum_{d=1}^n (-1)^d \left\lfloor \frac{n}{d}\right\rfloor = ... |
597,987 | Given a triangle on the plane, construct inside the triangle the point $P$ for which the centroid of the triangle formed by the three projections of $P$ onto the sides of the triangle happens to be $P$. | Overkill: In general, if $A',B',C'$ are the projections of $P$ on $BC,CA,AB,$ then the image of $P$ under the affine homography $\{A'B'C' \} \mapsto \{ABC \}$ is nothing but the isogonal conjugate of $P$ WRT $\triangle ABC.$ Thus, $P$ is centroid of $\triangle A'B'C'$ $\Longleftrightarrow$ $P$ is the isogonal conjugate... |
597,988 | We are given more than $2^k$ integers, where $k\in\mathbb{N}$. Prove that we can choose $k+2$ of them such that if some of our selected numbers satisfy
\[x_1+x_2+\dots+x_m=y_1+y_2+\dots+y_m\]
where $x_1<\dots<x_m$ and $y_1<\dots<y_m$, then $x_i=y_i$ for any $1\le i\le m$. | [hide=Solution]
We prove by induction on $k$ that given $2^k+1$ integers, we may choose a $(k+2)$-element subset of them satisfying the property $P$ that for any $m\le k+2$, no two sums of $m$ elements is equal. This is clear for $k=1$, as given $3$ numbers, $a<b<c$, we have $a+b<a+c<b+c$ as well. Assume the claim hold... |
597,720 | Paint the grid points of $L=\{0,1,\dots,n\}^2$ with red or green in such a way that every unit lattice square in $L$ has exactly two red vertices. How many such colorings are possible? | My answer is $2^{n+2}-2$
This problem is similar to USAMO 2019 P4 |
597,721 | Let $ABC$ be a non-equilateral triangle in the plane, and let $T$ be a point different from its vertices. Define $A_T$, $B_T$ and $C_T$ as the points where lines $AT$, $BT$, and $CT$ meet the circumcircle of $ABC$. Prove that there are exactly two points $P$ and $Q$ in the plane for which the triangles $A_PB_PC_P$ and ... | Circumcevian triangle is similar to pedal, so $P, Q$ are appolonius points, you can note they are isogonal conjugates of fermat points, and there are only $2$ ;) $\odot ABC$ is orthogonal to all the appolonian circle, and they are coaxial at $P, Q$. Hence, $P, O, Q$ are collinear too. |
597,722 | Let $k\ge 0$ be an integer and suppose the integers $a_1,a_2,\dots,a_n$ give at least $2k$ different residues upon division by $(n+k)$. Show that there are some $a_i$ whose sum is divisible by $n+k$. | The statement actually holds even if there are at least $k+1$ different residues. This stronger version was on the Komal competition. (December 2000, A.252.)
The solution only uses a pigeonhole argument. Let $a_1, \cdots, a_{k+1}$ be the numbers with different residues, and for the sake of convenience, let $S = a_1 + ... |
597,716 | Let $k\ge 3$ be an integer. Prove that if $n>\binom k3$, then for any $3n$ pairwise different real numbers $a_i,b_i,c_i$ ($1\le i\le n$), among the numbers $a_i+b_i$, $a_i+c_i$, $b_i+c_i$, one can find at least $k+1$ pairwise different numbers. Show that this is not always the case when $n=\binom k3$. | What a silly problem. To investigate the situation more effectively, focus on the $n$ triples $\{a_i,b_i,c_i\}$ and define
\[
S_i=\{a_i+b_i,a_i+c_i,b_i+c_i\}.
\]
Then each $S_i$ has three different elements, and $S_i$ uniquely determines $\{a_i,b_i,c_i\}$. Since the $a_i,b_i,c_i$'s are pairwise different, so are the $... |
597,717 | In a square lattice let us take a lattice triangle that has the smallest area among all the lattice triangles similar to it. Prove that the circumcenter of this triangle is not a lattice point. | Suppose that $ \triangle ABC $ is a lattice triangle and its circumcenter, $ O $ is also a lattice point. WLOG, we can assume $ A $ to be $ (0,0) $ and let $ B,C,O $ have coordinates $ (p,q) , (r,s) , (x,y) $ respectively. We have
\[ OA=OB \implies x^2+y^2 = (x-p)^2 + (y-q)^2 \]
so $ p^2 + q^2 $ is even $ \implies ... |
596,991 | We have an acute-angled triangle which is not isosceles. We denote the orthocenter, the circumcenter and the incenter of it by $H$, $O$, $I$ respectively. Prove that if a vertex of the triangle lies on the circle $HOI$, then there must be another vertex on this circle as well. | Well, $HOBC$ cyclic $\iff \angle BHC =\angle BOC\iff \pi-\angle BAC =2\angle BAC\iff \angle BAC=\tfrac{\pi}{3}$. Then drop perpendiculars from $I$ meeting $AB, AC, BC$ at $D,E,F$. Hence $\angle EID =\tfrac{2\pi}{3}$ and $\triangle ECI\cong \triangle FCI,\triangle DBI\cong \triangle FBI$, so $\angle BIC = \tfrac{1}{2}... |
596,994 | Prove that the edges of a complete graph with $3^n$ vertices can be partitioned into disjoint cycles of length $3$. | In other words: given the numbers $0,1,\ldots,3^n-1$, prove that we can pick several subsets of size $3$ such that every pair of numbers belong in exactly one subset together. Each subset translates into one cycle, whose elements are equal to the elements of the subset. As each edge belongs to one cycle, it means each ... |
596,986 | Draw a circle $k$ with diameter $\overline{EF}$, and let its tangent in $E$ be $e$. Consider all possible pairs $A,B\in e$ for which $E\in \overline{AB}$ and $AE\cdot EB$ is a fixed constant. Define $(A_1,B_1)=(AF\cap k,BF\cap k)$. Prove that the segments $\overline{A_1B_1}$ all concur in one point. | Let $A_1B_1 \cap EF =X$, and let $w$ be circle with center $F$, radius $EF$. $A_1,A$ are inverses with respect to $w$, as are $B_1,B$. So the inverse of $A_1B_1$ is the circumcircle of $\triangle ABF$, which means that if $Y$ is the inverse of $X$, $Y$ is the intersection of this circumcircle with $EF$. Then $AE * BE... |
596,987 | Prove that if a graph $\mathcal{G}$ on $n\ge 3$ vertices has a unique $3$-coloring, then $\mathcal{G}$ has at least $2n-3$ edges.
(A graph is $3$-colorable when there exists a coloring of its vertices with $3$ colors such that no two vertices of the same color are connected by an edge. The graph can be $3$-colored uni... | Let the number of vertices of each colour (red,blue,yellow) be $a,b,c$.
Consider the graph of only red and blue vertices, and edges connecting red and blue only. If between some two of these vertices there is no path, then we can change one of the vertices' colour without affecting the other, then the colouring is no l... |
596,988 | Prove that the following inequality holds with the exception of finitely many positive integers $n$:
\[\sum_{i=1}^n\sum_{j=1}^n gcd(i,j)>4n^2.\] | Let $g_p(x,y)$ for prime $p$ and $x,y\in \mathbb Z^+$ be $p$ if $p\mid \gcd(x,y)$ and $0$ otherwise. Since $\gcd(x,y)\geq \sum_{\text{prime }p\mid \gcd(x,y)}p$, it suffices to show that $\mathbb E\left[\sum_{\text{prime }p}g_p(x,y)\right]>4$. In fact, by LoE we have $\mathbb E\left[\sum_{\text{prime }p}g_p(x,y)\right]=... |
597,724 | Given is a triangle $ABC$, its circumcircle $\omega$, and a circle $k$ that touches $\omega$ from the outside, and also touches rays $AB$ and $AC$ in $P$ and $Q$, respectively. Prove that the $A$-excenter of $\triangle ABC$ is the midpoint of $\overline{PQ}$. | Its the ex-mixtillinear circle, intraversion is equivalent to the incircle, and this is well known. Either way, $R$ is tangency point of $k$ and $w$, $I_A$ excentre. $RP, RQ$ intersect at midpoints of arcs $BCA, ABC$, let they be $M_C, M_B$. Pascal on $M_BRM_CCAB \implies I_A, P, Q$ are collinear. But, $AI_A \perp PQ$. |
597,725 | Find the smallest positive integer $n\neq 2004$ for which there exists a polynomial $f\in\mathbb{Z}[x]$ such that the equation $f(x)=2004$ has at least one, and the equation $f(x)=n$ has at least $2004$ different integer solutions. | [hide=Answer]$n=(1002!)^2+2004$[/hide]
[hide=Solution]Let $a$ be an integer such that $f(a)=2004$, and let $f(x)-n=(x-r_1)(x-r_2)\ldots (x-r_{2004})g(x)$ where $r_i$ are pairwise distinct integers.
The condition implies that
\begin{align}
n-2004=-\prod_{i=1}^{2004} (a-r_i) \cdot g(a)
\end{align}
Note that $g(a)\neq 0$... |
597,728 | Let $N>1$ and let $a_1,a_2,\dots,a_N$ be nonnegative reals with sum at most $500$. Prove that there exist integers $k\ge 1$ and $1=n_0<n_1<\dots<n_k=N$ such that
\[\sum_{i=1}^k n_ia_{n_{i-1}}<2005.\] | See also [url=https://artofproblemsolving.com/community/c3735646h3490611]here[/url].
------
Let $k\in\mathbb N_+$ such that $2^{k-1}< N\le 2^k.$ Let $n_0=1,$ $n_k=N.$ For $1\le j\le k-1,$ let $n_j$ be the index of minimum value in $\{a_{2^{j-1}+1},\ldots ,a_{2^j}\}.$ Now
\[\sum_{i=1}^k n_ia_{n_{i-1}}\le 2a_1+\sum_{j=1}... |
597,729 | A and B play tennis. The player to first win at least four points and at least two more than the other player wins. We know that A gets a point each time with probability $p\le \frac12$, independent of the game so far. Prove that the probability that A wins is at most $2p^2$. | The probability player $A$ wins by either 4:0, 4:1 or 4:2 is
$$p^4+\binom{4}{1}p^4(1-p)+\binom{5}{2}p^4(1-p)^2\le \frac{11}{8}p^2\qquad (1)$$
where we use the fact that terms $p^2(1-p)$ and $p^2(1-p)^2$ attain their maximum value in $p\in [0,1/2]$ when $p=1/2$.
Let us now estimate the probabilities $A$ wins with eithe... |
597,730 | We build a tower of $2\times 1$ dominoes in the following way. First, we place $55$ dominoes on the table such that they cover a $10\times 11$ rectangle; this is the first story of the tower. We then build every new level with $55$ domioes above the exact same $10\times 11$ rectangle. The tower is called [i]stable[/i] ... | Cool problem! The answer is $\boxed{5}$ stories. It's pretty easy to construct an example attaining $5.$ We'll show that $4$ story stable towers cannot exist.
Define an [b] interior segment [/b] to be any segment of length one connecting two lattice points of the $10 \times 11$ rectangle, so that the segment is not co... |
596,962 | We deal $n-1$ cards in some way to $n$ people sitting around a table. From then on, in one move a person with at least $2$ cards gives one card to each of his/her neighbors. Prove that eventually a state will be reached where everyone has at most one card. | Let's label the people $1, 2, \cdots, n$ in clockwise order, and let $a_i$ be a variable representing how many coins person $i$ has at a given moment. Suppose, for contradiction, that the problem was false. Then, since there are only finitely many "configurations" possible, there must exist some sequence of moves on so... |
596,956 | We have placed $n>3$ cards around a circle, facing downwards. In one step we may perform the following operation with three consecutive cards. Calling the one on the center $B$, the two on the ends $A$ and $C$, we put card $C$ in the place of $A$, then move $A$ and $B$ to the places originally occupied by $B$ and $C$, ... | [hide="Lets try this"]
Consider two parameters for each card: the displacement, or number of cards away from its original position, and the orientation, or which side is faced up. For orientation, define 0 to be card facing downwards, 1 to be card facing upwards.
If $n$ is odd, consider the orientation. In each s... |
596,958 | Prove that any finite set $H$ of lattice points on the plane has a subset $K$ with the following properties:
[list]
[*]any vertical or horizontal line in the plane cuts $K$ in at most $2$ points,
[*]any point of $H\setminus K$ is contained by a segment with endpoints from $K$.[/list] | We will construct the desired set $K$ with an algorithm. First, define a [b] row [/b] to be a nonempty set consisting of all points in $H$ which lie on a horizontal line. Similarly define a [b] column. [/b] Start by coloring the leftmost point in each row red, and the rightmost point in each row green. If a row has onl... |
596,647 | Denote by $d(n)$ the number of positive divisors of a positive integer $n$. Find the smallest constant $c$ for which $d(n)\le c\sqrt n$ holds for all positive integers $n$. | Thanks, I edited it.
It's interesting no one has posted a solution yet. It is quite straightforward.
We want to maximize $\frac{d(n)}{\sqrt n}$. Let $n=p_1^{\alpha_1}\dots p_k^{\alpha_k}$. Then $d(n)=\prod_{i=1}^k (\alpha_i+1)$, meaning
\[\frac{d(n)}{\sqrt n}=\prod_{i=1}^k\frac{\alpha_i+1}{p_i^{\alpha_i/2}}.\]
From no... |
596,648 | Let $n\ge 1$ and $a_1<a_2<\dots<a_n$ be integers. Let $S$ be the set of pairs $1\le i<j\le n$ for which $a_j-a_i$ is a power of $2$, and $T$ be the set of pairs $1\le i<j\le n$ with $j-i$ a power of $2$. (Here, the powers of $2$ are $1,2,4,\dots$.) Prove that $|S|\le |T|$. | EDIT: The following is false. Thanks math4444 for spotting the bug. (It is not true that "A difference $a_j - a_i$ (for $i < j$) is a power of $2$ if and only if the vertices corresponding to $i$ and $j$ are connected by an edge in $Q_k$".)
One approach to this problem is to recast it in graph-theoretical terms. We WL... |
596,649 | In a far-away country, travel between cities is only possible by bus or by train. One can travel by train or by bus between only certain cities, and there are not necessarily rides in both directions. We know that for any two cities $A$ and $B$, one can reach $B$ from $A$, [i]or[/i] $A$ from $B$ using only bus, or only... | This is quite a beautiful problem. Let us make our statement simpler, so build a graph $G$ whose vertices are our cities and draw a red arrow from $A$ to $B$ if there is a way to go from $A$ to $B$ by only using the bus. Draw a blue arrow if we can do this by only using the train. Now in this setting our hypothesis tel... |
596,950 | Let $n,k$ be arbitrary positive integers. We fill the entries of an $n\times k$ array with integers such that all the $n$ rows contain the integers $1,2,\dots,k$ in some order. Add up the numbers in all $k$ columns – let $S$ be the largest of these sums. What is the minimal value of $S$? | The sum of all the entries in the array is $n(1+2+\dots+k)=n\frac{k(k+1)}2$, hence the average of the $k$ column sums is $n\cdot\frac{k+1}2$. Now, for $n=1$, clearly $\min S=k$, but for $n\ge 2$, we will show that $\min S=\left\lceil n\cdot\frac{k+1}2\right\rceil$. To do this, it is enough to show that for $n\ge 2$, we... |
596,951 | Find all positive integer pairs $(a,b)$ for which the set of positive integers can be partitioned into sets $H_1$ and $H_2$ such that neither $a$ nor $b$ can be represented as the difference of two numbers in $H_i$ for $i=1,2$. | Let P(n) denote the power of 2 in n's factorization. We will prove that P(a)=P(b) . Let's say 1 is in $H_1$. Then $H_1$ contains (2ak+1) and (2bn+1). Their difference is 2(ak-bn). Let's say P(a)$\geq$P(b). 2(ak-bn)=a $\longrightarrow$ a=$\frac{2bn}{2k-1}$ has no solutions if P(a)=P(b) and has an infinite number of solu... |
596,954 | Find all functions $f:\mathbb{Z}\to \mathbb{Q}$ with the following properties: if $f(x)<c<f(y)$ for some rational $c$, then $f$ takes on the value of $c$, and
\[f(x)+f(y)+f(z)=f(x)f(y)f(z)\]
whenever $x+y+z=0$. | It seems only $ f \equiv 0$ works.
$P(0,0,0):$ $f(0)=0$ (it can't be $\pm sqrt{3}$ )
$P(x,-x,0):$ $f(x)=-f(-x)$
$P(-2x,x,x):$ $f(-2x)[f(x)^2-1]=2f(x)$
Hence $|f|$ cant obtain values bigger or equal to one, as $f(x)=\pm 1$ would give $f(2x)*0=2$, contradiction.
$|f(2x)|=|f(-2x)|>|2f(x)|$ when $0<|f(x) |<1$.
Hence, whe... |
596,946 | We have $n$ keys, each of them belonging to exactly one of $n$ locked chests. Our goal is to decide which key opens which chest. In one try we may choose a key and a chest, and check whether the chest can be opened with the key. Find the minimal number $p(n)$ with the property that using $p(n)$ tries, we can surely dis... | This problem is pretty hard to solve rigorously. I think a lot of issues regarding the logic can be cleared up if we treat this as a game against the devil: we need to guarantee that $(1)$ we have a strategy which opens the chest in $p(n)$ tries, but $(2)$ that the devil has a strategy to answer our queries in a way th... |
596,947 | Consider a triangle $ABC$, with the points $A_1$, $A_2$ on side $BC$, $B_1,B_2\in\overline{AC}$, $C_1,C_2\in\overline{AB}$ such that $AC_1<AC_2$, $BA_1<BA_2$, $CB_1<CB_2$. Let the circles $AB_1C_1$ and $AB_2C_2$ meet at $A$ and $A^*$. Similarly, let the circles $BC_1A_1$ and $BC_2A_2$ intersect at $B^*\neq B$, let $CA_... | Note that $A^*$ is inside $\angle BAC$ due to the condition. This means that $\angle B_iA^*C_i=\pi-\alpha$. Moreover, $\angle B_iC_iA^*=\angle B_iAA^*$ and $\angle C_iB_iA^*=\angle C_iAA^*$, which gives us by the Law of Sines on $\triangle B_iC_iA^*$:
$$\frac{B_iA^*}{C_iA^*}=\frac{\sin\angle B_iC_iA^*}{\sin\angle C_iB_... |
596,948 | For what positive integers $n$ and $k$ do there exits integers $a_1,a_2,\dots,a_n$ and $b_1,b_2,\dots,b_k$ such that the products $a_ib_j$ ($1\le i\le n,1\le j\le k$) give pairwise different residues modulo $nk$? | Answer is $\gcd(n,k)= 1$. Construction is to take
\begin{align*}
a_1 = k +1 , a_2 = 2k+1 , \ldots, a_n = nk +1 \qquad \text{in general } ~ a_i = ki+1 \\
b_1 = n+1, b_2 = 2n+1, \ldots, b_k = kn + 1 \qquad \text{in general } ~ b_j = nj + 1
\end{align*}
Observe that
$$ a_ib_j = (ki + 1)(nj + 1) \equiv ki + nj + 1 \pmod... |
2,019,731 | Given $2n$ points and $3n$ lines on the plane. Prove that there is a point $P$ on the plane such that the sum of the distances of $P$ to the $3n$ lines is less than the sum of the distances of $P$ to the $2n$ points. | Here's a sketch of a solution.
Let $Q$ be an arbitrary point in the plane, and consider a circle $\Omega$ centered at $Q$ with radius $R$ for some arbitrarily large $R.$
As $R$ is sufficiently large, the average distance from a point of $\Omega$ to any of the lines is approximately $R \cdot \frac{2}{\pi}$ because $\i... |
596,659 | Consider $n$ events, each of which has probability $\frac12$. We also know that the probability of any two both happening is $\frac14$. Prove the following.
(a) The probability that none of these events happen is at most $\frac1{n+1}$.
(b) We can reach equality in (a) for infinitely many $n$. | (b):
If $n$ is odd, let $n = 2k-1$.
With $\frac{n}{n+1}$ probability, have a random set of $k$ of the $n$ events happen, with all subsets equally likely to be chosen.
With $\frac{1}{n+1}$ probability, have no event happen.
Then any event has a probability of $\frac{2k-1}{2k} \cdot \frac{k}{2k-1} = \frac{1}{2}$ of hap... |
596,652 | Let $a,b$ be positive real numbers satisfying $2ab=a-b$. Denote for any positive integer $k$ $x_k$ and $y_k$ to be the closest integer to $ak$ and $bk$, respectively (if there are two closest integers, choose the larger one). Prove that any positive integer $n$ appears in the sequence $(x_k)_{k\ge 1}$ if and only if it... | Solving for $b$ gives $b=\frac {a}{2a+1}$.
$n$ appears in $(x_k)\iff $ there exists $k$ such that $n-\frac {1}{2}\le ak<n+\frac {1}{2}\Rightarrow 0\le k-\frac {n-1}{2a}<\frac {1}{a}$
$n$ appears at least 3 times in $(y_k)\iff $there exists natural number $m$ such that $n-\frac {1}{2}\le m\frac {a}{2a+1}<n+\frac {1}{2... |
596,654 | Is it true that for integer $n\ge 2$, and given any non-negative reals $\ell_{ij}$, $1\le i<j\le n$, we can find a sequence $0\le a_1,a_2,\ldots,a_n$ such that for all $1\le i<j\le n$ to have $|a_i-a_j|\ge \ell_{ij}$, yet still $\sum_{i=1}^n a_i\le \sum_{1\le i<j\le n}\ell_{ij}$? | An equality case is when only the $\ell_{1i}$'s are positive and the rest are $0$, but for general $\ell_{ij}$'s, the condition seems to allow for quite a lot of space, so the answer seems to be yes (say, with $0.8$ probability).
Essentially, you have a complete graph on $n$ vertices, and a number on each edge repres... |
609,502 | Consider a company of $n\ge 4$ people, where everyone knows at least one other person, but everyone knows at most $n-2$ of the others. Prove that we can sit four of these people at a round table such that all four of them know exactly one of their two neighbors. (Knowledge is mutual.) | For any vertex $x\in G$ denote by $N(x)$ the set of its neighbours. Let $a\in G$ be a vertex of maximal degree $\deg a$, then take $b\in G\setminus N(a)$ and $c\in N(b)$.
If $c\in N(a)$, it means there must exist $d\in N(a)$ so that $d\not \in N(c)$, otherwise $\deg c = \deg a + 1$, contradicting the maximality of $a$... |
1,524,427 | Let $ABC$ be a triangle. Choose points $A'$, $B'$ and $C'$ independently on side segments $BC$, $CA$ and $AB$ respectively with a uniform distribution. For a point $Z$ in the plane, let $p(Z)$ denote the probability that $Z$ is contained in the triangle enclosed by lines $AA'$, $BB'$ and $CC'$. For which interior point... | Let $AZ\cap BC=D$, similarly define $E,F$.
Suppose $\frac{AF}{AB}=a,\frac{BD}{BC}=b,\frac{CE}{CA}=c$, we have $abc=(1-a)(1-b)(1-c)$.
Not hard to see that $p(Z)=abc+(1-a)(1-b)(1-c)=2abc$.
Let $x=\frac{a}{1-a},y=\frac{b}{1-b},z=\frac{c}{1-c}$, we get $xyz=1$.
Also, we want to maximize $2abc=\frac{2xyz}{(1+x)(1+y)(1+z)}\l... |
1,524,428 | Do there exist polynomials $p(x)$ and $q(x)$ with real coefficients such that $p^3(x)-q^2(x)$ is linear but not constant? | My solution also works using derivatives.
[hide]
Suppose that $p^3-q^2=\ell$ with some non-constant linear $\ell$.
First observe that $p$ and $q$ must be co-prime: if $d=gcd(p,q)$ then $d^2 \,\big|\, p^3-q^2=\ell$ so $\deg(d^2) \le \deg\ell=1$ and therefore $d$ is constant.
Moreover, none of $p$ and $q$ can be constan... |
1,524,431 | An $n$ by $n$ table has an integer in each cell, such that no two cells within a row share the same number. Prove that it is possible to permute the elements within each row to obtain a table that has $n$ distinct numbers in each column. | I solved it in the exam, it took about 2 hours fully. No, I never heard of this problem before. I first noticed that basically it should be somehow traced back to Hall's Marriage Theorem (which I worked with for half a month after a math camp and managed to find a fully own proof, I randomly noticed that a statement li... |
1,719,235 | In a village (where only dwarfs live) there are $k$ streets, and there are $k(n-1)+1$ clubs each containing $n$ dwarfs. A dwarf can be in more than one clubs, and two dwarfs know each other if they live in the same street or they are in the same club (there is a club they are both in).
Prove that is it possible to c... | I just realize there's even easier solution!
We try to inductively select one new (i.e. not selected before) dwarf from each club. If this can be carry out for all $k(n-1)+1$ clubs, PHP gives us there're $n$ from the selected $k(n-1)+1$ dwarfs that live on the same street, and so know each other, done. On the other ha... |
2,521,265 | Find all functions $f\colon \mathbb{Q}\to \mathbb{R}_{\geq 0}$ such that for any two rational numbers $x$ and $y$ the following conditions hold
[list]
[*] $f(x+y)\leq f(x)+f(y)$,
[*]$f(xy)=f(x)f(y)$,
[*]$f(2)=1/2$.
[/list] | The answer is $f(r)=2^{-\nu_2(r)}$ for $r\neq 0$ and $f(0)=0$. It is easy to check that this satisfies the conditions.
The second and third are trivial; as for the first, we know that $$\nu_2(x+y)\geq \min\{\nu_2(x),\nu_2(y)\}$$
so $$f(x+y)=2^{-\nu_2(x+y)}\leq 2^{-\min\{\nu_2(x),\nu_2(y)\}}\leq 2^{-\nu_2(x)}+2^{-\nu_2(... |
2,689,145 | Let $P_0=(a_0,b_0),P_1=(a_1,b_1),P_2=(a_2,b_2)$ be points on the plane such that $P_0P_1P_2\Delta$ contains the origin $O$. Show that the areas of triangles $P_0OP_1,P_0OP_2,P_1OP_2$ form a geometric sequence in that order if and only if there exists a real number $x$, such that
$$
a_0x^2+a_1x+a_2=b_0x^2+b_1x+b_2=0
$$ | Let areas $S_0 = [P_1OP_2]$, $S_1 = [P_0OP_2]$, and $S_2 = [P_0OP_1]$, and let vectors $p_0 = (a_0, b_0)$, $p_1 = (a_1, b_1)$, and $p_2 = (a_2, b_2)$.
We claim that $S_0p_0 + S_1p_1 + S_2p_2 = \vec{0}$.
Since $O$ is in the interior of $\triangle P_0P_1P_2$, $P_1$ and $P_2$ must be on different sides of line $OP_0$. So... |
2,689,170 | In neverland, there are $n$ cities and $n$ airlines. Each airline serves an odd number of cities in a circular way, that is, if it serves cities $c_1,c_2,\dots,c_{2k+1}$, then they fly planes connecting $c_1c_2,c_2c_3,\dots,c_1c_{2k+1}$. Show that we can select an odd number of cities $d_1,d_2,\dots,d_{2m+1}$ such that... | [url=https://dgrozev.wordpress.com/2020/03/10/romanian-master-of-mathematics-2020-problem-3/]RMM 2020, p3 [/url] |
2,936,638 | A square has been divided into $2022$ rectangles with no two of them having a common interior point. What is the maximal number of distinct lines that can be determined by the sides of these rectangles? | [hide=Solution]
We will prove that for any $n$ the maximum number of lines $f(n)=n+3$ (for a rectangle not square).
We will use strong induction to prove $f(n) \le n+3$.
Obvious for $n=1,2$.
Assume that it is true for $1,2,3,...,n-1$. If there is only one rectangle which has an edge on the left edge of main rectangle... |
2,936,642 | Let $p$ and $q$ be prime numbers of the form $4k+3$. Suppose that there exist integers $x$ and $y$ such that $x^2-pqy^2=1$. Prove that there exist positive integers $a$ and $b$ such that $|pa^2-qb^2|=1$. | Kinda cute. As @TheMathBob said, we should assume that $x$ and $y$ are non-negative (but $p\neq q$ is not needed).
Let $(x,y)$ be the pair satisfying $x^2-pqy^2=1$ with minimal $|xy|>0$. Note that $pqy^2=(x-1)(x+1)$.
After a modulo $4$ analysis, it follows that $x-1=2u^2\delta$ and $x+1=2v^2(pq/\delta)$ for some div... |
229,102 | Let an infinite sequence of measurable sets be given on the interval $ (0,1)$ the measures of which are $ \geq \alpha>0$. Show that there exists a point of $ (0,1)$ which belongs to infinitely many terms of the sequence. | [hide=To clarify Yustas' solution]
We use
[hide=Rudin RCA, Theorem 1.19 (e)]
Given measurable sets $B_1 \supset B_2 \supset B_3 \supset \cdots$ such that $m(B_1)$ is finite, we have $m(B_i) \to m(B_1 \cap B_2 \cap B_3 \cap \cdots)$ as $i \to \infty$
[/hide]
Let $B_i = A_i \cup A_{i+1} \cup A_{i+2} \cup \cdots$. Since $... |
229,105 | Let $ p$ be an odd prime number and $ a_1,a_2,...,a_p$ and $ b_1,b_2,...,b_p$ two arbitrary permutations of the numbers $ 1,2,...,p$ . Show that the least positive residues modulo $ p$ of the numbers $ a_1b_1, a_2b_2,...,a_pb_p$ never form a permutation of the numbers $ 1,2,...,p$. | The two $ p$'s must be at the same place (otherwise we already get two $ p$'s in $ a_1b_1, ..., a_pb_p$). W.l.o.g. let $ a_p \equal{} 0 \equal{} b_p$.
If now $ a_1b_1, ..., a_{p \minus{} 1}b_{p \minus{} 1}$ would be a permutation of $ 1,2,...,p \minus{} 1$ $ \mod p$, then
\[ \minus{} 1 \equiv (p \minus{} 1)! \equiv \... |
229,109 | Let $ f(x)$ be a polynomial of second degree the roots of which are contained in the interval $ [\minus{}1,\plus{}1]$ and let there be a point $ x_0\in [\minus{}1.\plus{}1]$ such that $ |f(x_0)|\equal{}1$. Prove that for every $ \alpha \in [0,1]$, there exists a $ \zeta \in [\minus{}1,\plus{}1]$ such that $ |f'(\zeta)|... | If $f(x)=a(x-b)(x-c)$ (we can assume wlog that $a \ge 0$ because $|f|$ and $|f'|$ do not change when $f$ is replaced by $-f$), then introduce $m = \tfrac{b+c}{2}$ and consider two intervals $[-1,m]$ and $[m,1]$. The longer of these two is of length at most $2$ and has `elevation change' of at least $1$ (either $f(m) \... |
229,121 | Let $ n$ and $ k$ be positive integers, $ n\geq k$. Prove that the greatest common divisor of the numbers $ \binom{n}{k},\binom{n\plus{}1}{k},\ldots,\binom{n\plus{}k}{k}$ is $ 1$. | if $d$ is the greatest common divisor of these numbers, then $d$ must divide their consecutive differences, i.e $d|\dbinom{n}{k-1},...,\dbinom{n+k-1}{k-1}$, so it can be done by induction. |
229,123 | Find the complex numbers $ z$ for which the series
\[ 1 \plus{} \frac {z}{2!} \plus{} \frac {z(z \plus{} 1)}{3!} \plus{} \frac {z(z \plus{} 1)(z \plus{} 2)}{4!} \plus{} \cdots \plus{} \frac {z(z \plus{} 1)\cdots(z \plus{} n)}{(n \plus{} 2)!} \plus{} \cdots\]
converges and find its sum. | Here's the key idea:
[b][color=red]Lemma:[/color][/b] Let $n$ be a positive integer and $z$ be a complex number. Then, \[ \prod_{k=1}^n \frac{k+z}{k} = e^{\gamma z} n^z (1+O_z(1/n)) \] where $\gamma = 0.577\ldots$ denotes the Euler-Mascheroni constant.
[i]Proof.[/i] Recall that $\log(1+x) = x + O(x^2)$. Thus if the pro... |
229,142 | Let $ \{k_n\}_{n \equal{} 1}^{\infty}$ be a sequence of real numbers having the properties $ k_1 > 1$ and $ k_1 \plus{} k_2 \plus{} \cdots \plus{} k_n < 2k_n$ for $ n \equal{} 1,2,...$. Prove that there exists a number $ q > 1$ such that $ k_n > q^n$ for every positive integer $ n$. | It is clear that $ k_n > 2^{n\minus{}2} k_1$ for all $ n > 1$ and that the sequence $ (k_n)$ could be make to be sufficiently near to the sequence $ ( 2^{n\minus{}2} k_1 )$. Hence, if $ q$ has such property, it must be the case that $ q<k_1$ and $ q < \sqrt[n]{2^{n\minus{}2} k_1}$ for all $ n$ (or $ q < \inf \{ \sqrt[n... |
229,164 | Find the polynomials $ f(x)$ having the following properties:
(i) $ f(0) \equal{} 1$, $ f'(0) \equal{} f''(0) \equal{} \cdots \equal{} f^{(n)}(0) \equal{} 0$
(ii) $ f(1) \equal{} f'(1) \equal{} f''(1) \equal{} \cdots \equal{} f^{(m)}(1) \equal{} 0$ | See a very much related question [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=67&t=463940[/url], with much more insight in the discussions following. |
229,165 | Prove that for every positive integer $ k$ there exists a sequence of $ k$ consecutive positive integers none of which can be represented as the sum of two squares. | Let $ \{ q_i \}$ be an enumeration of the primes congruent to $ 3 \bmod 4$ (recall that there are infinitely many such primes). The system of congruences
$ x \equiv q_1 \bmod q_1^2$
$ x \plus{} 1 \equiv q_2 \bmod q_2^2$
...
$ x \plus{} (k\minus{}1) \equiv q_k \bmod q_k^2$
has a unique solution $ \bmod q_1^2 q... |
229,167 | Let $ x$ be an arbitrary real number in $ (0,1)$. For every positive integer $ k$, let $ f_k(x)$ be the number of points $ mx\in [k,k \plus{} 1)$ $ m \equal{} 1,2,...$
Show that the sequence $ \sqrt [n]{f_1(x)f_2(x)\cdots f_n(x)}$ is convergent and find its limit. | Observe that $ f_n(x)$ is near to $ \lfloor \frac {1}{x} \rfloor$ as $ n \rightarrow \infty$.
Verify that $ mx \in [k,k \plus{} 1) \Leftrightarrow k \leq mx < k \plus{} 1 \Leftrightarrow \frac {k}{x} \leq m < \frac {k \plus{} 1}{x} \Leftrightarrow \lceil \frac {k}{x} \rceil \leq m \leq \lfloor \frac {k \plus{} 1}{x}... |
229,169 | Let $ A \equal{} (a_{ik})$ be an $ n\times n$ matrix with nonnegative elements such that $ \sum_{k \equal{} 1}^n a_{ik} \equal{} 1$ for $ i \equal{} 1,...,n$.
Show that, for every eigenvalue $ \lambda$ of $ A$, either $ |\lambda| < 1$ or there exists a positive integer $ k$ such that $ \lambda^k \equal{} 1$ | research "stochastic matrix" in the forum and you will get all the answers, and even more ..
eg: [url]http://www.mathlinks.ro/viewtopic.php?search_id=547181886&t=105426[/url] |
229,171 | Find the sum of the series
$ x\plus{}\frac{x^3}{1\cdot 3}\plus{}\frac{x^5}{1\cdot 3\cdot 5}\plus{}\cdots\plus{}\frac{x^{2n\plus{}1}}{1\cdot 3\cdot 5\cdot \cdots \cdot (2n\plus{}1)}\plus{}\cdots$ | If $ f(x) \equal{} x \plus{} \frac {x^3}{1\cdot 3} \plus{} \frac {x^5}{1\cdot 3\cdot 5} \plus{} \cdots \plus{} \frac {x^{2n \plus{} 1}}{1\cdot 3\cdot 5\cdot \cdots \cdot (2n \plus{} 1)} \plus{} \cdots$
then
$ \frac{d}{dx}f(x)\minus{}1 \equal{} xf(x)$.
This is easy to solve by standard techniques
The solution of $... |
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