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);
$BD = \sqrt{(DD')^2 + (D'B)^2} = \sqrt{h^2 + (x + h \cdot cot(\beta))^2}$ (ii).
If $AC > BD$, then:
$ \sqrt{h^2 + (x + h \cdot cot(\alpha))^2} > \sqrt{h^2 + (x + h \cdot cot(\beta))^2}$;
or $h^2 + x^2 + 2hx \cdot cot(\alpha) + h^2 \cdot cot^{2}(\alpha) > h^2 + x^2 + 2hx \cdot cot(\beta) + h^2 \cdot cot^{2}(\beta)$;
or $2hx \cdot cot(\alpha) + h^2 \cdot cot^{2}(\alpha) > 2hx \cdot cot(\beta) + h^2 \cdot cot^{2}(\beta)$;
or $2hx[cot(\alpha) -cot(\beta)] + h^2[cot^{2}(\alpha)- cot^{2}(\beta)] > 0$;
or $h[cot(\alpha) - cot(\beta)][h(cot(\alpha) +cot(\beta)) +2x] > 0$ (iii).
Since $h, x > 0$ and $cot(t) \in (0, \infty)$ & is strictly monotonically decreasing over $t \in \left(0, \frac{\pi}{2}\right) \Rightarrow cot(\alpha) - cot(\beta) > 0$ holds true for $0< \alpha < \beta < \frac{\pi}{2} \Rightarrow AC > BD$ holds true.
QED |
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:
$$(b-1)(b-2)+(c-1)^2<0$$
which is false
If $a=1$ the given inequality rearranges to:
$$(b-2)^2+(c-1)^2<1$$
which is only true for $(a,b,c)=\boxed{(1,2,1)}$.
If $a=2$ the given inequality rearranges to:
$$(2b-5)^2+4(c-1)^2<1$$
which is only true for $b=2.5$, no integer solutions.
Otherwise, the given inequality rearranges to:
$$\frac12(a-b)^2+\frac12(b-3)^2+(c-1)^2+\frac12(a^2-5)<0$$
which is false. |
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 $$\frac{sen \, \, B_1A_1C}{sen\, \, C_2A_2B} = \frac{B_2C_2}{B_1C_1}$$
[img]https://cdn.artofproblemsolving.com/attachments/3/8/774da81495df0a0f7f2f660ae9f516cf70df06.png[/img] | 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 triplet down to:
$\frac{1}{(n - k + 1)(n - k)}, \frac{1}{k(n - k)}, \frac{1}{k(k + 1)}$ (ii).
If the terms in (ii) are in arithmetic progression, then we obtain the following system of equations:
$\frac{1}{k(n - k)} = \frac{1}{(n - k + 1)(n - k)} + d$ (iii);
$\frac{1}{k(k + 1)} = \frac{1}{(n - k + 1)(n - k)} + 2d$ (iv)
and substituting (iii) into (iv) yields:
$\frac{1}{k(k + 1)} = \frac{1}{(n - k + 1)(n - k)} + 2[\frac{1}{k(n - k)} - \frac{1}{(n - k + 1)(n - k)}]$;
or $\frac{1}{k(k + 1)} = \frac{2}{k(n - k)} - \frac{1}{(n - k + 1)(n - k)}$;
or $\frac{n - k}{k(k + 1)} = \frac{2(n - k + 1) - k}{k(n - k + 1)}$;
or $(n - k)(n - k + 1) = (k + 1)(2n - 3k + 2)$;
or $n^2 - 2kn + k^2 + n - k = 2kn - 3k^2 + 2k + 2n - 3k + 2$;
or $n^2 - 4kn + 4k^2 = n + 2$;
or ($n - 2k)^2 = n + 2$;
or $n - 2k = \pm \sqrt{n + 2}$;
or $\frac{n \pm \sqrt{n + 2}}{2} = k$ (NOTE: the positive sign is eliminated since we require $k < n$.)
or $k = \frac{n - \sqrt{n + 2}}{2}$ (v).
Solving for $k < n$ yields:
$\frac{n - \sqrt{n + 2}}{2} < n$;
or $-\sqrt{n + 2} < n$;
or $0 < n^2 - n - 2$;
or $0 < (n - 2)(n + 1)$;
or $n \in (-\infty, -1) \cup (2, \infty)$;
or $n > 2$ (since we require $n \in \mathbb{N}$). If we substitute $n = t^2 - 2$ (for $t = \{3, 4, 5, ...\}$) into (v), then we obtain the solution set:
$n = t^2 - 2; k = \frac{t^2 - t - 2}{2}$ for $t = \{3, 4, 5, .....\}$. |
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/c6h390986p2172824]here[/url] |
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.artofproblemsolving.com/attachments/8/a/4dc4a815d8b002c9f36a6da7ad6e1c11a848e9.png[/img] | 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 (this has been posted before) $\Longrightarrow$ $[A'B'C']=16[XYZ].$ Hence, it suffices to prove that $16[XYZ] <5[ABC].$ By Euler's theorem for $\triangle XYZ,$ we get
$\frac{[XYZ]}{[ABC]}= \frac{|p(N,(O))|}{4R^2} \Longrightarrow 4\cdot |p(N,(O))|< 5R^2.$
Assume that $N$ lies outside $(O),$ otherwise the inequality is trivial. Then $4(ON^2-R^2)<5R^2$ $\Longrightarrow$ $ON^2 < \frac{_9}{^4}R^2.$ Since $ON=\frac{_3}{^2}OG,$ it follows that $\frac{_9}{^4}OG^2< \frac{_9}{^4}R^2$ $\Longrightarrow$ $OG <R,$ which is certainly true as $G$ is always inside $(O).$ |
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 shared with the polygon meet at $P_i$. If $1\le i\le n-1$, then label $\triangle$ with $i$ and get rid of it, and apply the induction hypothesis to polygon $P_0P_1\dots P_{i-1}P_{i+1}\dots P_{n-1}P_n$. Else if $i=0$ or $i=n$, then... whatever, we cannot have our only two special triangles at $P_0$ and $P_n$, because then both would have side $P_0P_n$, so it's possible to choose $\triangle$ such that this doesn't hold. Done.
[/hide]
[hide=Overkill]
We essentially want to arrange a pairing between the triangles and the labels $1,2,\dots,(n-1)$, which can be done via [url=https://en.wikipedia.org/wiki/Hall's_marriage_theorem]Hall's theorem[/url]. Set up a bipartite graph with one vertex set the triangles and the other vertex set the labels, where a triangle $\triangle$ is connected to label $i$ with an edge iff $P_i$ is a vertex of $\triangle$.
Considering $t$ triangles, they have a total of at least $t+2$ different vertices, and thus will contain at least $t$ vertices from $P_1,\dots,P_{n-1}$. Indeed, let's start from a triangle in the triangulation and paint it red, and from now in each step keep on picking another triangle that has a common edge with one of the red triangles, and paint that red as well. In each step, the cardinality of the vertex set of the red triangles increases by one. In fact, in each step when we paint one of our $t$ triangles red, we find another vertex of our $t$ triangles. Hence, the $t$ triangles have at least $3+(t-1)=t+2$ vertices, and this shows they satisfy Hall's criterion.
Btw this is a generalization I think - we did not use that the two labels left out were assigned to consecutive vertices.
[/hide] |
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 infinitely many positive integers $n$. | 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 sum minus the number of $k$-element subsets with odd sum. The values of $k$ we want are those where the $y^k$ coefficient is 0.
We quickly find that $f(-1,y) = (1-y^2)^{50}$, and the binomial theorem immediately tells us that the coefficient of $y^k$ is zero if and only if $k$ is odd.
Another method for showing $k$ odd works is to pair every subset $A$ with $\{101 - x \mid x \in A \}$. When $A$ has an odd number of elements, this partitions all of the subsets into pairs of subsets whose sums have opposite parity. But I wasn't able to extend this idea to show that $k$ even failed.
[b]EDIT[/b]: Here's how to use the second idea to finish $k$ even. For a subset $A$, call $x \in A$ unmatched if $101 - x \not\in A$. If there are $m$ unmatched elements, then we get a family of $2^m$ subsets (one of them $A$) by independently choosing to replace each unmatched element $x \in A$ with $x$ or $101 - x$. When $m > 0$, the parities of subset sums in the family are split in half.
The only time $m = 0$ is when $x \in A$ if and only if $101 - x \in A$. That is, we have $k/2$ pairs of elements summing to 101. There are a positive number of these subsets, and their sum is always odd if $k \equiv 2 \pmod{4}$ and always even if $k \equiv 0 \pmod{4}$. So there is not an equal number of subsets with even/odd sum. (This method can also obtain the exact difference between the two counts as easily as the generating functions argument did.) |
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 $d=1 \Rightarrow c=5 \Rightarrow (a,b,c,d) = (3,2,5,1)$
$\rightarrow$ if $d=2 \Rightarrow c=2 \Rightarrow (a,b,c,d) = (2,2,2,2)$
3)$b=3$ , $d=1 \Rightarrow c+d+9 = 3cd$ no solutions in this case (note that $a \ge b$).
4)$b=4$ , $d=1 \Rightarrow c+d+16 = 4cd$ no solutions in this case.
So the answers are : $(a,b,c,d) = (5,1,3,2),(1,5,3,2),(5,1,2,3),(1,5,2,3),(3,2,5,1),(2,3,5,1) , (3,2,1,5), (2,3,1,5),(2,2,2,2)$ |
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,$\implies \dfrac{1}{P}$ is min,$\implies \dfrac{ka-y}{\sqrt{r^2-y^2}}=\sqrt{\dfrac{(ka-y)^2}{r^2-y^2}}$ is min,
let
$t=\dfrac{(ka-y)^2}{r^2-y^2},(1+t)y^2-2kay+(ka)^2-tr^2=0,\Delta \ge0 \implies t \ge \dfrac{(ka)^2-r^2}{r^2},$
$\implies y=\dfrac{r^2}{(ka)}$ get $P_{max}$.
then construction is very easy:
$DF \perp OD, F=w \cap DF, GF \perp OF ,G=GF \cap OD, DH \perp a,H=FD \cap DH$,make $l$//$OD$ through $H$. |
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 digits of the number $t(10^l-1)=t\cdot 10^l-t,t\le 10^l-1$;it will be pretty clear that the sum of the digits of the number $t(10^l-1)$ is $9l$). |
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 that starting from the point $\left(1;\sqrt 2\right)$, we cannot return to it in finitely many steps. | 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://artofproblemsolving.com/community/c281674h1286696p6787256]CIIM 2015 Problem 3[/url]. |
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 that $a \cdot \left(\frac{x}{g}\right)^2 \equiv 0 \pmod{\frac{n^2}{g^2}}$. Since $\frac{x}{g}$ and $\frac{n}{g}$ are relatively prime, $a \equiv 0 \pmod{\frac{n^2}{g^2}}$. But then
\[\frac{a}{n^2/g^2} \cdot \left(\frac{x}{g}+\frac{n}{g}\right)\left(\frac{x}{g}-\frac{n}{g}\right) = p\] We must have $a = \frac{n^2}{g^2}$, $\frac{x}{g}+\frac{n}{g}=p$, $\frac{x}{g}-\frac{n}{g}=1$, so $\frac{n}{g} = \frac{p-1}{2}$, $\frac{x}{g}=\frac{p+1}{2}$, $d=pg^2$. $g$ is fixed by $n$ and $p$, so it follows that $g=h$. Then $e=pg^2=d$, contradiction.
[/hide] |
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 between the children in the right way, we can guarantee that each child has the same probability of winning. | 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 the possible outcomes into $n$ classes, with the probabilities in each class adding up to $\frac 1n$.
It suffices to take a big prime $q\equiv -1\pmod n$ (which exists by Dirichlet) and partition the $2^k$ probability values into $q+1$ equal classes for some $k$. Scale up by $(1+x)^k$, to have $\binom kj$ of $x^j$.
Let $k=q$, then $q$ divides each of
\[\binom k1,\binom k2,\dots,\binom k{k-1}.\]
Hence, we may form $q$ classes which are the same, and have the remaining $x^k+1$ put into one extra class. All the classes are the same if
\[
x^k+1 =\frac1q\left(\binom k1 x^1+\binom k2 x^2+\dots+\binom k{k-1}x^{k-1}\right).
\]
By IVT, this equation has a solution $x\in (0;1)$, because for $x=0$, $LHS=1>RHS=0$, but for $x=1$, $LHS=2<RHS=\frac1q(2^q-2)$ for $q$ large enough. This $x$ yields some probability $p$ which does the trick with $k=q$. $\blacksquare$
[/hide] |
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 $AA_1B_1B, BB_1C_1C, CC_1A_1A$ respectively. We need to show that $A_1A_0, B_1B_0, C_1C_0$ intersect at a point.
Because $B_1B_0$ and $C_1C_0$ lie in the plane $AB_1C_1$ they intersect. Analogously intersect also the pairs $A_1A_0, B_1B_0$ and $A_1A_0, C_1C_0$ thus the lines $A_1A_0, B_1B_0, C_1C_0$ intersect at a point (they share the point of intersection of the planes $AB_1C_1, BC_1A_1, CA_1B_1$) because these three lines don't lie in a plane. |
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 strange mean, depending on the given numbers, can be larger or smaller than the third power mean.
Which statement is valid for $n=3$? | 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 $256$.
$f_1(2^{1991}) < (1000*9)^2<10^8$.
$f_2(2^{1991})<f_1(99999999)=72^2<9999$
$f_3(2^{1991})<36^2<1999$
$f_4(2^{1991})<28^2<900$
$f_5( ) \le 900$
$f_6(2^{1991}) = (7+9t)^2$ where $7+9t$ is the sum of the digits of a square.
Hence it is $49,256$ or $625$
$f_7(2^{1991})=13^2=169$
$f_8=256$
and now it keeps repeating. |
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
\[(\star)\qquad 4(k^2-a)x^2 - 4(a-k^2)kx + (a-k^2)^2 + 4ab = 0. \]
It further implies that $4(k^2-a)x^2 - 4(a-k^2)kx < 0$.
Now, assume that there exists infinitely many suitable integers $n$. Then they can be arbitrarily large and so are the corresponding values of $x$. Since the last inequality holds for such $x$, we have $k^2-a \leq 0$, i.e., $k\leq\sqrt{a}$, which leaves only a finitely many possible values for $k$. For each such fixed value of $k$, the equation $(\star)$ represents a quadratic equation w.r.t. $x$ (and thus w.r.t. $n$) and has at most two solutions, unless $k=a=1$ (which further implies $b=0$ and thus is not possible). Therefore, the total number of suitable $n$ is finite, a contradiction to our assumption. |
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}{PB} = t $, the above inequality reduces to
\[ t^2(PD^2 - PB^2)^2 \ge 0 \]
which holds. |
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 an even number of people in $C$,
[*]the participants from $B$ each know an odd number of participants in $C$.[/list] | 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 odd. Then there are $2^n$ possible values of $f(C)$. Either all of these are distinct, and so at least one of them is either $(0,0,\ldots,0)$ or $(1,1,\ldots,1)$ and we're done, or else there are two distinct subsets $D,E$ such that $f(D) = f(E)$. But then take the symmetric difference $C$ of $D$ and $E$ (that is, people that belong in exactly one of these sets). This set is nonempty (because some element in $D,E$ doesn't belong in the other, otherwise the two sets are equal), and $f(C) = (0,0,\ldots,0)$. To prove this, let's take any $f_i$. Suppose $b_i$ knows $d_i$ people in $D$, $e_i$ people in $E$, and $c_i$ people in $D %Error. "intersect" is a bad command.
E$. Thus $f_i = d_i + e_i - 2c_i$. Taking modulo 2, we have $f_i \equiv d_i + e_i \pmod 2$, and since $d_i \equiv e_i \pmod 2$ ($f(D) = f(E)$), we have $f_i \equiv 0 \pmod 2$. Thus this $C$ is the set we seek. |
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 assuming the largest counter example). Let these diagonals be $A_iB_i$, $i=1$ to $2k$.
Now between the remaining $n-4k$ vertices there cannot be any diagonal joining two of them, and also if $C,D$ are two distinct vertices not in $A_i,B_i$, and $A_iC,B_iD$ are two drawn diagonals, then we pick them instead of $A_iB_i$, contradiction, hence between $A_i,B_i$ they only connect to $n+1-4k$ more diagonals among the remaining $n-4k$ vertices. Then among the $4k$ vertices we assume they are all connected (ie $K_{4k}$).
Hence the maximum number of diagonals is $C^{4k}_{2}+2k(n+1-4k)=8k^2-2k+2k(n+1)-8k^2=2kn<2kn+1$, contradiction.
To construct for $kn$, consider the bipartite graph $K_{2k,n-2k}$. Clearly only $2k$ diagonals can be picked. The number of edges is $2k(n-2k)=kn+k(n-4k)>kn$, so we are done. |
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
\[ \begin{vmatrix}
1 & 1 & 1 \\
x_1 & x_2 & x_3 \\
x_1^2 & x_2^2 & x_3^2
\end{vmatrix}
= (x_1-x_2)(x_2-x_3)(x_3-x_1) \neq 0 \]
So, no three points are collinear. |
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 the triangle $ A_1B_1C_1 $ , so $ O,I,H_1 $ are collinear. |
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
\[
2+E-V=F\le \frac23 E\quad\implies \quad E\le 3V-6,
\]
and the average degree of the vertices of our graph is $\frac{2E}{V}<6$. This means that $\mathcal{G}$ has a vertex $v$ with degree $\le 5$.
Remove $v$ from $\mathcal{G}$ to get $\mathcal{G}'$, and suppose we can $3$-color the vertices of $\mathcal{G}'$ in the described fashion. Then by Pigeonhole, some color $C$ appears among the neighbors of $v$ at most once; paint $v$ $C$. This way, no $C$-colored cycle can arise, as $v$ would need $\ge 2$ $C$ neighbors for that.
By reducing coloring $\mathcal{G}$ to coloring $\mathcal{G}'$, we have proven the problem statement by induction. |
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 $c$ is odd positive integer. Let $b=\varphi(c)$, where $\varphi(c)$ is Euler's totient function. Then, by Euler's theorem, $2^b-1=kc$, where $k$ is an integer. So we can take \[P_n(x)=2^a+k\,(1^2-x^2)(2^2-x^2)\dots(n^2-x^2) \,.\] Indeed, $P_n(x)=2^a$ when $x=-n,\dots,-1,1,\dots,n$ as the addend in RHS becomes $0$, and $P_n(0)=2^a+k(n!)^2=2^a+2^a kc=2^a+2^a(2^b-1)=2^{a+b}$. $\Box$
For the purpose of the problem, we can take \[Q(x)=P(x-1) \, P^2(x-2) \, P^3(x-3) \, \dots \, P^n(x-n) \,,\] where $P(x)$ is $P_{n-1}(x)$ from the lemma, assuming that $n>1$ (the case $n=1$ is obvious). Indeed, then $Q(i)=2^{a+2a+\ldots+(i-1)a+(ia+ib)+(i+1)a+\ldots+na}=2^{s+ib}$ for any $i=1,2,\dots,n$, where $s=a \tfrac {n(n+1)} 2$ ($a$ and $b$ are the numbers from the lemma). $\blacksquare$
By the way, there is also a polynomial of exactly $n-1$-th degree having the same property, but this was not required in the statement. |
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 make a zigzag (i.e., a broken line) $v_1v_2\dots v_n$ as follows: let $v_1$ be any vertex of $P$, $v_2$ be its neighbor, and from then on, make $v_{i+1}$ be the vertex neighboring $v_{i-1}$ which hasn't been labelled yet. The zigzag partitions $P$ into the triangles $T_i=v_{i-1}v_iv_{i+1}$ ($2\le i\le n-1$). Now draw all the diagonals of $P$ - they split $P$ into a number of small polygons -, and place a point $p_i$ in the part inside $T_i$ closest to $v_i$. (It's easy to guarantee that no three points are collinear.)
This way, any triangle $\triangle=v_av_bv_c$ formed by three vertices of $P$ ($a<b<c$) contains exactly one of the $p_i$'s, namely $p_b$. This is because we chose each $p_i$ so close to $v_i$ that $p_i$ is inside some triangle only if $v_i$ is a vertex of that triangle. However, of $p_a,p_b,p_c$, only $p_b$ is inside $\triangle$, because the interior of $\triangle$ is disjoint from $T_a$ and $T_c$, but a sufficiently small part of $\angle v_{b-1}v_bv_{b+1}$ near $v_b$ is within $\triangle$, as this angle is within $\angle v_av_bv_c$.
We have thus provided a construction for each $N=2n-2$ and excluded the rest; therefore, exactly the even $N$ satisfy the conditions. $\blacksquare$ |
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 = \left\lfloor \frac{n}{1}\right\rfloor-\left\lfloor \frac{n}{2}\right\rfloor+\left\lfloor \frac{n}{3}\right\rfloor-\left\lfloor \frac{n}{4}\right\rfloor\pm\dots
\end{align*}
Since the function $x\mapsto \lfloor x\rfloor$ is increasing on $\mathbb{R}$, we may apply the following estimates to prove the given inequality:
\begin{align*}
S&=\underbrace{\left(\left\lfloor \frac{n}{1}\right\rfloor-\left\lfloor \frac{n}{2}\right\rfloor\right)}_{\ge 0}+\underbrace{\left(\left\lfloor \frac{n}{3}\right\rfloor-\left\lfloor \frac{n}{4}\right\rfloor\right)}_{\ge 0}+\dots &\ge 0, \\
S&=\left\lfloor \frac{n}{1}\right\rfloor-\underbrace{\left(\left\lfloor \frac{n}{2}\right\rfloor-\left\lfloor \frac{n}{3}\right\rfloor\right)}_{\ge 0}-\dots &\le n. \quad \blacksquare
\end{align*} |
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 of the centroid of $\triangle ABC,$ i.e. its symmedian point.
P.S. This is a rather old problem, for other proofs see
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=209057[/url]
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=258869[/url] |
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 holds for $2^k+1$ numbers, consider an arbitrary set $N$ of $2^{k+1}+1$.
Two trivial observations lead to the solution:
(1) if $P$ holds for some finite set $S$ of integers, then it remains true if we apply a linear transformation upon $S$, i.e. $P$ holds for the set $S'=\{as+b|s\in S\}$ for arbitrary constants $a\neq 0$, $b$.
(2) if $P$ holds for a set $S$ of even integers, then it also holds for $S\cup\{2\ell+1\}$ for $\ell\in\mathbb{Z}$.
If $N$ contains only even or odd numbers, then possibly add $1$ to each element to produce a set with only even numbers, and then divide each element by $2$. This step reduces the diameter of $N$ by half, so we may perform it only finitely many times, resulting in a set $N'$ with both even and odd numbers. It suffices to prove the claim for $N'$, by (1).
Since $|N'|=2^{k+1}+1$, by the box principle, there are $2^k+1$ elements of $N'$ with the same parity; taking $N'$ or $N'+1$, we may suppose there are $2^k+1$ even numbers. By the induction hypothesis, we may select a set of $k+1$ of them satisfying $P$, and because of (2), if we add another odd integer to this set, we arrive at $k+2$ integers satisfying $P$. Because of (1), we are done with the inductive step and the proof is complete.
[/hide] |
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 $A_QB_QC_Q$ are equilateral. Prove furthermore that line $PQ$ contains the circumcenter of $\triangle ABC$. | 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 $S = a_1 + \cdots + a_{k+1}$. We assume that there does not exist some $a_i$ whose sum is divisible by $n+k$.
Consider the numbers
X: $a_1, a_2, \cdots, a_{k+1}$
Y: $S-a_1, S-a_2, \cdots, S-a_{k+1}$
Z: $S, S+a_{k+2}, S+a_{k+2}+a_{k+3}, \cdots, a_1 + \cdots + a_n$.
First, for any $z_1, z_2 \in Z$, the difference is the sum of some $a_i$ and thus is not divisible by $n+k$. Hence every two elements of $Z$ have different residues. Likewise, for any $x \in X, z \in Z$, the numbers $x$ and $z$ also have different residues, and for any $y \in Y, z \in Z$, $y$ and $z$ also have different residues. For any $x_1, x_2 \in X$, they have different residues by the definition of $a_1, \cdots, a_{k+1}$, and also $y_1$ and $y_2$ have different residues for any $y_1, y_2 \in Y$. For $x \in X$ and $y \in Y$, they can have the same residue only if $x = a_i$ and $y = S-a_i$. But this occurs only when $2 a_i \equiv S \pmod{n+k}$, and thus there are at most $2$ such $i$.
From all the case-working above, we can deduce that the numbers of X, Y, Z all have different residues with the exception of at most two pairs, and thus, there are $(k+1)+(k+1)+(n-k)-2 = n+k$ all residues. Hence $0$ is one of them. |
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 $S_i$'s. It follows that if $\left|\bigcup_{i=1}^n S_i\right|\le k$, then the $S_i$'s can take on at most $\binom k3$ possible values, so $n>\binom k3$ implies $\left|\bigcup_{i=1}^n S_i\right|\ge k+1$.
We can reach $n=\binom{k}{3}$ by choosing a set $L$ with $k$ elements of $\mathbb{R}$ which are linearly independent over $\mathbb{Q}$. (Such a set $L$ may be chosen by adjoining newer and newer elements to it; we can always add another element because $\mathbb{R}$ is uncountable.) Then make the $S_i$'s the $3$-element subsets of $L$, and define the triples $\{a_i,b_i,c_i\}$ accordingly. Since $S_i=\{\alpha,\beta,\gamma\}$ defines
\[
\{a_i,b_i,c_i\}=\left\{\frac{\alpha+\beta-\gamma}2,\frac{\beta+\gamma-\alpha}2,\frac{\gamma+\alpha-\beta}2\right\},
\]
the linear independence implies that the $3n$ numbers $a_i,b_i,c_i$ are pairwise distinct. |
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 p+q , p-q $ are also even, i.e the point $ K \left( \frac{p+q}{2} , \frac{p-q}{2} \right ) $ is a lattice point. Similarly, $ L \left( \frac{r+s}{2} , \frac{r-s}{2} \right ) $ is also a lattice point. Then,
\[ AK^2 = \frac{p^2 + q^2}{2} = \frac{AB^2}{2} \]
\[ AL^2 = \frac{r^2 + s^2}{2} = \frac{AC^2}{2} \]
\[ KL^2 = \frac{ (p-r)^2 + (q-s)^2 }{2} = \frac{BC^2}{2} \]
so $ \triangle AKL \sim \triangle ABC $ and $ [\triangle AKL] < [\triangle ABC] $, a contradiction. |
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 $\angle BIC = \tfrac{1}{2}(2\pi - \angle EID )=\tfrac{2\pi}{3}=\angle BHC =\angle BOC$, so $HOIBC$ cyclic $\iff \angle BAC=\tfrac{\pi}{3}$. |
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 pair of numbers belong to one subset, hence we've proven that those problems are equivalent.
Express the numbers in ternary. Each subset we pick will have the property that for each place, the three digits are all equal or all different. For example, the numbers $012_3, 110_3, 211_3$ form a subset, as the first digits are all different, the second digits are all equal, and the third digits are all different. We prove that this has each pair of numbers exactly once each.
Given any pair of numbers, we can construct exactly one subset including it. After all, the third number will be defined: for each place, if the two digits are equal, let the corresponding digit of the third number be the same; if they are different, the digit of the third number takes the remaining unused digit. So in this case, given the numbers $012_3, 110_3$, we can construct the third number: the first digit should be all different, so it should be $2$; the second digit should be all equal, so it should be $1$; the third digit should be all different, so it should be $1$. Thus we obtain the only number that fits in a subset with those, namely $211_3$.
Thus we have proven that each pair is included in exactly one subset, proving the corollary that I've stated in the first paragraph. |
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 $AE * BE =EY *EF$, so $Y$ is fixed, which means $X$ is as well. |
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 uniquely if there do not exist vertices $u$ and $v$ of the graph that are painted different colors in one $3$-coloring, yet are colored the same in another.) | 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 longer unqiue. Hence the vertices are connected, thus there are at least $a+b-1$ red-blue edges.
A similar argument for other types of edges gives that total number of edges is at least $a+b-1+b+c-1+a+c-1=2n-3$. |
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 $\mathbb E\left[\sum_{\text{prime }p}g_p(x,y)\right]=\sum_{\text{prime }p}\mathbb E\left[g_p(x,y)\right]=\sum_{\text{prime }p}\frac{1}{p}>4\ \Box$
Remark: This is not all that different from [b]randomusername[/b]'s solution above. |
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 $g(a)\neq 0$ and $a-r_i \neq 0$ for $1\leq i \leq 2004$ because otherwise $n=2004$ which is not what we want.
Also, since LHS in $(1)$ has at least $2002$ divisors greater than $1$ we must have $n\geq 2004$ because all numbers that are less than $2004$ and greater than $0$ can't have $2002$ factors greater than $1$.
Now, $$n-2004=|\prod_{i=1}^{2004}(a-r_i)\cdot g(a)| \geq (1002!)^2$$
because $a-r_i$'s are distinct integers and this implies $n\geq (1002!)^2+2004.$
To see that this is reachable take $f(x)=-\prod_{i=1}^{1002}(x-i-a) \cdot \prod_{i=1}^{1002} (x+i-a) +(1002!)^2+2004.$
[/hide] |
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
\[\sum_{i=1}^k n_ia_{n_{i-1}}\le 2a_1+\sum_{j=1}^{k-1}2^{j+1}\cdot\frac{a_{2^{j-1}+1}+\cdots +a_{2^j}}{2^{j-1}}\le 4(a_1+\cdots +a_N)\le 2000<2005.\Box\]
|
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 either $5:3\,;\,6:4\,;\,7:5\,;\dots.$ First, the probability the players reach a result $3:3$ equals $q:= \binom{6}{3}p^3(1-p)^3=20p^3(1-p)^3.$
If $A$ (resp. $B$) does not win with $4:2$ (resp. $2:4$) it means the result at some point was $3:3$.. Now, the possibility $A$ wins with result exactly $n+2:n, n\ge 3$ is the following chain to happen
$$\text{result }3:3\to 4:4\to\dots \to n:n\to n+1:n\to n+2:n\qquad (2)$$
The probability the event $(2)$ happens is
$$P(\text{result } [n+2:n])=q 2^{n-3}p^{n-3}(1-p)^{n-3}p^2=20p^2 2^{n-3}p^n(1-p)^n$$
Using $p(1-p)\le \frac{1}{4}$ when $p\in [1,1/2]$ it yields
$$P(\text{result } [n+2:n])\le 20p^2 2^{n-3}2^{-2n}=\frac{5}{2}p^22^{-n}\,;\, n\ge 3$$
Thus,
$$P(\text{result } [n+2:n], n\in\{3,4,\dots\})\le \frac{5}{2}p^2\sum_{n=3}^{\infty} 2^{-n}=\frac{5}{8}p^2$$
Combining this estimate with $(1)$ it follows
$$P(A\text{ wins})\le \left(\frac{11}{8}+\frac{5}{8}\right)p^2=2p^2.$$ |
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] if for every non-lattice point of the $10\times 11$ rectangle, we can find a domino that has an inner point above it. How many stories is the lowest [i]stable[/i] tower? | 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 contained in the perimeter of the rectangle. Call an interior segment [b] iffy [/b] if one of its endpoints is on the perimeter of the rectangle. There are $9 \cdot 11 + 10 \cdot 10 = 199$ of these in total. Observe that each level of dominoes covers exactly $55$ interior segments, and that we must cover all of them. This immediately shows that there must be at least four levels involved.
Suppose, for contradiction, that there was some way to cover all interior segments with only four levels. Call these levels $1, 2, 3, 4$ from bottom to top, and for every interior segment of the $10 \times 11$, we will mark it with $i$ if it is covered by level $i.$ Note that any four interior segments which form the sides of a unit square must all be be covered only once and be covered by distinct levels. This immediately implies that all interior segments which are not iffy can only be covered once.
We claim that iffy segments can only be covered at most twice. Indeed, for any iffy segment $s$, take three interior segments which comprise three sides of a unit square and include $s.$ Then, the other two segments must be covered at most once, so by Pigeonhole it is apparent that $s$ can be covered at most two times.
The final observation is the following. For any two iffy segments which are parallel to each other and a distance of one away from each other, at most one of them can be covered twice. Indeed, if both of them were, then we could apply Pigeonhole on the interior segments of the unit square including the two iffy sides.
These observations combine to show that at most $2(5+5) = 20$ interior segments can be covered twice, showing that the interior segments are covered a total of at most $199 + 20 = 219$ times. This is a clear contradiction, because they are covered precisely $4 \cdot 55 = 220$ times.
$\square$ |
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 some starting configuration $C$ so that the end configuration is also $C.$ Let $b_i$ be the initial value of $a_i$, for each $1 \le i \le n.$
In other words, the $a_i$'s are unaffected by the moves. Let the move where $a_i$ gives two cards be called $m_i$. Let our sequence of moves be $m_{b_1}, m_{b_2}, \cdots$. Firstly, it's clear that every move $m_i$ appears the same number of times in the sequence, say each move appears $t$ times. Now, we will say that $i$ [b] precedes [/b] $j$ if the $t$th appearance of $m_i$ in the sequence is before the $t$th appearance of $m_j$. For each $1 \le i \le n$, let $0 \le x_i \le 2$ be the number of $i-1, i+1$ which $i$ precedes. It's easy to see that $\sum x_i = n.$
Observe that at the instant that the $t$th $m_i$ occurs, $a_i$ is precisely $b_i - 2(n-1) + \alpha + \beta$, where $\alpha, \beta$ are the number of $m_{i-1}$'s and $m_{i+1}$'s respectively. This number must be at least $2$, and so from the definition of the $x_i$'s we have $b_i \ge x_i.$
However, summing this over all $1 \le i \le n$ yields an immedaite contradiction.
$\square$ |
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 $C$, respectively. Meanwhile, we flip the cards $A$ and $B$.
Using a number of these steps, is it possible to move each card to its original place, but facing upwards? | [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 step, the sum of orientations changes by an even number, since 2 cards are flipped. Since the orientations sum to 0 at the beginning, they can't sum to $n$, an odd number, at the end.
If $n$ is even, consider the parity of the sum of the (sum of displacements) and (sum of orientations). In each step, the C card's displacement changes by 2 and orientation changes by 0, the A card's displacement changes by 1 and orientation changes by 1, and the B card's displacement changes by 1 and orientation changes by 1. For each card, the parity of the sum of displacement and orientation is even. But in the scenario we want at the end, displacement is even and orientation is odd.
So using a number of these steps it is impossible to move each card to its original place but flipped over.
[/hide] |
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 only one point, then this point will be both red and green. Call a point [b] colorful [/b] if it is either red or green (or both). At any point in time, say that the set of colorful points is [b] happy [/b] if any other point is on a segment connecting two colorful points. Initially, it's clear that the colorful points are happy.
We will repeat the following operation as long as there exists a column with three colorful points. Take such a column. Let $a$ and $b$ be the highest and lowest colorful points in this column, respectively. Let $c$ be an arbitrary other colorful point in the column. If $c$ is red and green, then render $c$ colorless (by removing its red and green color). If $c$ is red (resp. green), then render $c$ colorless and color red (green) the leftmost (rightmost) point in $c$'s row which is to the right (left) of $c$ (if it exists).
First let us show that this operation preserves the happiness of the colorful points. We'll start by making some observations. First of all, make the trivial observation that there is always at most one green and at most one red point in each row. Furthermore, the red point in each row will be have an $x-$coordinate at most that of the green point in the row. It's now obvious that there is always either both a red and a green point in a row, or there is no colorful point in the row. Observe that the red point always travels to the right and the green point always travels to the left. This means that every point, once rendered colorless, will forever be colorless. Now, simply notice that every point which is ever rendered colorless will always be on a segment connecting two colorful points, since there will always exist a colorful point above and below it in its column. Combining the previous results, we can easily see that the happiness of the colorful points is preserved.
Now, let us show that the operation actually terminates. Notice that there can be at most finitely many operations where we render a point which is red and green colorless, because the number of colorful points strictly decreases when this happens. Hence, at some point, we only render colorless points which are only red or only green. Consider the sum over all rows of the $x-$coordinate of the green point minus the $x-$coordinate of the red point. This is clearly strictly decreasing under the operation of rendering colorless a point which is only red or only green. Hence, there can also only be finitely many operations of the second kind, and so the operation must terminate.
After the operation terminates, every row has $0$ or $2$ colorful points, while every column also has at most $2$ colorful points. Hence, since these colorful points are happy, we are also happy because we're done.
$\square$ |
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 now on it's a bunch of estimates.
First of all, $\frac{\alpha}{p^{(\alpha-1)/2}}>\frac{\alpha+1}{p^\alpha}$ is equivalent to $\sqrt p>\frac{\alpha+1}{\alpha}$, which holds for all $p\ge 5$ and $\alpha\in\mathbb{N}$, moreover for $p=3$ and $\alpha\ge 2$, and for $p=2$ and $\alpha\ge 3$. So the maximal $\frac{d(n)}{\sqrt n}$ is attained when $n=2^a\cdot 3^b$, where $a\le 2$ and $b\le 1$.
It is easy to check that from the divisors of $12$, $n=12$ gives the maximum, meaning the $c$ in question is
\[\frac{d(12)}{\sqrt{12}}=\frac{3\cdot 2}{2\sqrt 3}=\sqrt 3.\] |
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 WLOG assume that all $a_i$ are nonnegative integers (otherwise, just add a big number to each of them), and replace them by their base-$2$ representations. These base-$2$ representations are strings of $0$'s and $1$'s, and thus can be viewed as vertices of a hypercube graph $Q_k$ for a sufficiently large $k$. 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 $Q_k$. So the question is now the following: Given $n$ distinct vertices of $Q_k$, prove that the number of edges between these vertices is at most as large as the number of edges between the $n$ lexicographically smallest vertices of $Q_k$ (that is, the vertices corresponding to the integers $0, 1, \ldots, n-1$). But the latter number is clearly $\sum\limits_{i=0}^{n-1} s\left(i\right)$, where $s\left(i\right)$ is the sum of digits in the base-$2$ representation of $i$ (indeed, $s\left(i\right)$ is the number of edges from the vertex corresponding to $i$ to smaller vertices). Hence, the problem is transformed into a claim that appears, e.g., in the first paragraph of [url=https://doi.org/10.1016/j.ejc.2012.09.004]Geir Agnarsson, [i]Induced subgraphs of hypercubes[/i], European Journal of Combinatorics, Volume 34, Issue 2, February 2013, Pages 155-168[/url]. |
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 train rides. Prove that there exists a city such that any other city can be reached using only one type of vehicle (but different cities may be reached with different vehicles). | 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 tells us that $G$ is a tournament in which every edge is either red or blue. If we show that there is a vertex $v$ in $G$ such that we can reach any other vertex of $G$ from $v$ by a monochromatic path, then we've solved the problem.
We'll say a vertex $v$ [i]reaches[/i] a set $S\subset V(G)$ if for any vertex $b$ of $S$ there is a monochromatic path from $a$ to $b$ (of course $a$ reaches itself).
We are going to use induction on $n=|V(G)|$ to prove this result. For $n=1,2$ everything is clear. Suppose our result holds for any tournament with $n$ vertices and let's prove it for the case when $|V(G)|=n+1$. Let's suppose we can't find a vertex that reaches the whole graph. Now for any subset $A\subset V(G)$ of $n$ vertices there is a vertex $a\in A$ reaching it. If two different such subsets $A1$ and $A2$ have a common vertex that reaches them, then as $A1\cup A2 = V(G)$ we get a contradiction. So the map associating to each $n$-elements set $A\subset V(G)$ its reaching point $a$ is injective, hence surjective as it is a map between sets of the same cardinal.
Therefore, as a first observation, this implies that any vertex $a$ of $G$ reaches a set of vertices of cardinal $n$ (but not the whole graph).
Pick now a vertex $v_1$ and let $A_1$ be the set of $n$ vertices it reaches (from now on let $A_i$ be the set that $v_i$ reaches). Let $v_2$ be the only vertex in $V(G) - A_1$. Then $\overrightarrow{v_1v_2}\notin E(G)$ as we'd get that $v_1$ reaches the whole graph. Suppose now that $\overrightarrow{v_2v_1}$ is red. More generally suppose we've found vertices $v1,v2,...,v_k$ such that $\{ v_{i+1} \}= V(G) - A_i$ for all $i=1,k-1$ and the edges $\overrightarrow{v_kv_{k-1}},...,\overrightarrow{v_3v_2},\overrightarrow{v_2v_1} $ are all red. Now we claim that for $v_{k+1}$, which is also defined as $\{ v_{i+1} \}= V(G) - A_i$, we have $\overrightarrow{v_{k+1}v_k}$ colored red. Indeed note $v_{k+1}$ is in $A_{k-1}$ as $v_{k+1}\neq v_k$. If the path from $v_{k-1}$ to $v_{k+1}$ is red then by adding $\overrightarrow{v_kv_{k-1}}$ we'd get a red path from $v_k$ to $v_{k+1}$, contradiction. So this path is blue. But in this case if $\overrightarrow{v_{k+1}v_k}$ is blue we'd get by adding it a path from $v_{k-1}$ to $v_k$, again a contradiction and the claim follows.
To end this solution we just have to notice that by doing this process at some point one of the $v_i$'s is going to repeat and thus we create a red circuit in our graph. Just look at a vertex $v_j$ in this circuit and then by following its arrows we see that we can reach $v_{j+1}$ from $v_j$, which contradicts once again our assumption. Therefore there is a vertex that reaches all $V(G)$, done! |
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 may always construct the array such that the largest difference between two column sums is at most $1$.
There is an obvious $2\times k$ array with all the column sums the same, namely
\begin{tabular}{|c|c|c|c|}
\hline
$1$ & $2$ & $\dots$ & $k$ \\
\hline
$k$ & $k-1$ & $\dots$ & $1$\\
\hline
\end{tabular}
The existence of this table implies that it is enough to construct the required table for $n=3$, as we have done so for $n=2$ and $n=3$, and from then on we may expand it by these $2$ rows many times.
For the $n=3$ case, we differentiate between even $k$ and odd $k$. For even $k=2m$, we may take the following table:
\begin{tabular}{|c|c|c|c||c|c|c|c|}
\hline
$1$ & $2$ & $\dots$ & $m$ & $m+1$ & $m+2$ & $\dots$ & $2m$\\
\hline
$m+1$ & $m+2$ & $\dots$ & $2m$ & $1$ & $2$ & $\dots$ & $m$ \\
\hline
$2m$ & $2m-2$ & $\dots$ & $2$ & $2m-1$ & $2m-3$ & $\dots$ & $1$ \\
\hline
\end{tabular}
Here, the first $m$ columns have sum $3m+2$, and the next $m$ columns have sum $3m+1$ (they have the same sums because when moving one column to the right, the sum changes by $+1+1-2=0$). Hence, it works.
For odd $k=2m-1$, we may do something similar, with all the column sums all $3m$:
\begin{tabular}{|c|c|c|c||c|c|c|c|}
\hline
$1$ & $2$ & $\dots$ & $m$ & $m+1$ & $m+2$ & $\dots$ & $2m-1$ \\
\hline
$m$ & $m+1$ & $\dots$ & $2m-1$ & $1$ & $2$ & $\dots$ & $m-1$ \\
\hline
$2m-1$ & $2m-3$ & $\dots$ & $1$ & $2m-2$ & $2m-4$ & $\dots$ & $2$ \\
\hline
\end{tabular}
We have thus constructed the required tables, proving the solution is $k$ for $n=1$ and $\left\lceil n\cdot\frac{k+1}2\right\rceil$ for $n\ge 2$. |
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 solutions if P(a)>P(b). |
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, when a function value $f(a)$ isn t equal to one, $|f(2^k a)|>1$ for some k, contradiction. |
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 discover which key belongs to which chest. | 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 that guarantees that we cannot assign all the right keys to all the right chests in less than $p(n)$ tries.
Solution with Kálmán Lajkó (mainly by him):
Just arbitrarily label the keys $1,2,\dots,n$ and the chests $1,2,\dots,n$. An $(i,j)$-try is when we try opening chest $j$ with key $i$. The answer for $n=0$ and $n=1$ is $0$; from now on, $n\ge 2$.
$(1)$ is simple: we keep trying key $1$ on chests until it opens one or until we have made $(n-1)$ tries. We thus know which chest key $1$ opens in at most $(n-1)$ tries. Next, we do the same for key $2$, etc., which proves that
\[
p(n)\le (n-1)+(n-2)+\dots+2+1=\binom{n}{2}.
\]
$(2)$ is the hard part. The devil's strategy is to always answer 'no' to all of our $(i,j)$-tries. (If at some point, the devil had to answer 'yes' to our $(i,j)$-try, that means that we know that key $i$ opens chest $j$ for sure, and so this question is a waste: this is why we may without loss of generality assume that the devil always answers 'no'.)
Draw an $n\times n$ table, and put an X in the cell in row $i$ column $j$ if and only if we have the information that key $i$ doesn't open chest $j$. Call a set of $n$ cells a [i]transversal[/i] when they are in pairwise different columns and rows; a transversal with no X's in it represents a possible set $(1,c_1)$, $(2,c_2)$, $\dots$, $(n,c_n)$ with key $i$ opening chest $c_i$. Hence, to prove that $p(n)\ge \binom{n}{2}$, we need to show that if there are less than $\binom{n}{2}$ X's in the table, and at least one transversal is possible, then we can find another transversal.
We prove this statement for all $n\ge 2$ by induction, the base case $n=2$ being easy to check. For $n>2$, either there is a column with $n-1$ X's in it, in which case we can delete the row and apply the induction hypothesis, or each column has at least $2$ cells with no X in it. Relabel the columns so that one possible transversal is $(1,1)$, $(2,2)$, $\dots$, $(n,n)$, and suppose that for $i=1,2,\dots,n$, $a_i\neq i$ is such that cell $(i,a_i)$ does not contain an X. In this case, say that $i\to a_i$. Then we have $1\to b_1\to b_2\to\dots$, and due to finiteness this sequence eventually enters a loop $b_i\to b_{i+1}\to\dots\to b_j\to b_i$. Then another transversal will be the same as before with $(b_i,b_i)$, $\dots$, $(b_j,b_j)$ replaced by $(b_i,b_{i+1})$, $(b_{i+1},b_{i+2})$, $\dots$, $(b_j,b_i)$, and we are done again. The induction is now complete, and we have thus proven $p(n)\ge \binom{n}{2}$ as well. $\blacksquare$ |
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 $CA_1B_1$ and $CA_2B_2$ intersect at $C^*\neq C$. Prove that the lines $AA^*$, $BB^*$, $CC^*$ are concurrent. | 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 $\triangle B_iC_iA^*$:
$$\frac{B_iA^*}{C_iA^*}=\frac{\sin\angle B_iC_iA^*}{\sin\angle C_iB_iA^*}=\frac{\sin\angle B_iAA^*}{\sin\angle C_iAA^*}=:\lambda_A.$$
Hence, there is a spiral similarity from $A^*$ with angle $\pi-\alpha$ and ratio $\lambda_A$ taking $\overline{C_1C_2}$ to $\overline{B_1B_2}$, which implies $\lambda_A=\frac{B_1B_2}{C_1C_2}$, so again we're done by TrigCeva.
Edit: yeah, this is basically XmL's solution. |
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
$$ a_ib_j = (ki + 1)(nj + 1) \equiv ki + nj + 1 \pmod{kn} $$
Now since $\gcd(k,n) = 1$, so
$$ \{ki + nj : 1 \le i \le n, 1 \le j \le k\} $$
is a complete residue class modulo $kn$, which shows are construction indeed works.
Next we show we must have $\gcd(k,n) = 1$.
[b][color=#f00]Claim 1:[/color][/b] $\gcd(a_1,kn) \le n ~~ \forall ~ i$ and $\gcd(b_i,nk) \le k ~~ \forall ~ j$.
[i]Proof:[/i] Fix $i$, we will show $\gcd(a_i,nk) \le n$. Observe that
$$ a_ib_1,a_ib_2,\ldots,a_ib_k $$
are distinct modulo $nk$. If $d = \gcd(a_i,nk)$, then
$$ b_1,b_2,\ldots,b_k $$
must be distinct modulo $\frac{nk}{d}$. So by PHP, $\frac{nk}{d} \ge k \implies d \le n$. $\square$
Recall $nk$ divides some $a_ib_j$, WLOG
$$ nk \mid a_1b_1 \implies nk \mid \gcd(a_1,nk) \cdot \gcd(b_1,nk) $$
combining this with [b]Claim 1[/b] forces $\gcd(a_1,nk) = n, \gcd(b_1,nk) = k$. We obtain
\begin{align*}
\{a_1,a_2,\ldots,a_n\} ~ \text{ is a complete residue class modulo } n \\
\{b_1,b_2,\ldots,b_k\} ~ \text{ is a complete residue class modulo } k
\end{align*}
[b][color=#f00]Claim 2:[/color][/b] If some prime $p \mid n$, then $p \nmid b_1b_2 \cdots b_k$. Similarly, if $p \mid k$, then $p \nmid a_1a_2 \cdots a_n$.
[i]Proof:[/i] Suppose $p \mid n$. Suppose $s$ elements of $\{b_1,b_2,\ldots,b_k\}$ are not divisible by $p$. By counting the number of elements of set
$$ \{a_ib_j : 1 \le i \le n, 1 \le j \le k\} \equiv \{1,2,\ldots,nk \} \pmod{p}$$
we obtain
$$ \left(n - \frac{n}{p} \right)s = \left(kn - \frac{kn}{p} \right) \implies s=k $$
Hence all elements of $\{b_1,b_2,\ldots,b_k\}$ are not divisible by $p$, as desired. $\square$
Using $\{b_1,b_2,\ldots,b_k\}$ is a complete residues class modulo $k$, we conclude any prime dividing $n$ does not divide $k$. Hence $\gcd(k,n) = 1$, as desired. $\blacksquare$
[b][color=#9a00ff]Motivation: [/color][/b] The construction is just motivated by [b]Claim 2[/b]. Now [b]Claim 2[/b] in some sense was motivated why $k=n=2$ doesn't work (since $\{a_ib_j\}$ would have a lot of even numbers, given $\{a_1,a_2\},\{b_1,b_2\}$ are complete residues classes modulo $2$). The stuff before [b]Claim 2[/b] wasn't hard to observe. |
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 $\int_{x =0}^{2\pi} |\sin x| dx = 4.$ Also, the average distance from a point of $\Omega$ to any of the points is approximately $R$.
So for sufficiently large $R$, we're done.
$\square$ |
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 happening, and any two events have a probability of $\frac{2k-1}{2k} \cdot \frac{k}{2k-1} \cdot \frac{k-1}{2k-2} = \frac{1}{4}$ happening together, which checks out.
(a): (far more easily motivated after figuring out what the solutions to (b) look like)
Let $k = \frac{n + 1}{2}$. Suppose that $v$ events happen; consider the random variable $(v - k)^2$. Note that $v = v_1 + v_2 + \cdots + v_n$ where $v_i$ is the indicator variable (?) of event $i$, i.e. 1 if the event occurs, 0 if it doesn't.
Then, by linearity of expectation, and noting that the problem conditions imply that
\begin{align*} \mathbb{E}(v_i) &= 1/2 \\
\mathbb{E}(v_i^2) &= 1/2 \\
\mathbb{E}(v_iv_j) &= 1/4 \qquad (i \neq j), \end{align*} we can expand and evaluate \begin{align*} \mathbb{E}((v - k)^2) &= \mathbb{E}((v_1 + v_2 + \cdots + v_n - k)^2) \\
&= \mathbb{E}\left(\sum_{1 \leq i \leq n} v_i^2 + 2\sum_{1 \leq i < j \leq n} v_iv_j - 2k\sum_{1 \leq i \leq n} v_i + k^2\right) \\
&= \left(\frac{n}{2} + 2\binom{n}{2}\cdot \frac{1}{4} - 2kn\cdot\frac{1}{2} + k^2\right) \\
&= \left(\frac{n}{2} + \frac{n(n-1)}{4} - \frac{(n+1)n}{2} + \frac{(n+1)^2}{4}\right) \\
&= \left(\frac{2n + n(n-1) - 2(n+1)n + (n+1)^2}{4}\right) \\
&= \frac{2n + n^2 - n - 2n^2 - 2n + n^2 + 2n + 1}{4} \\
&= \frac{n + 1}{4} = \frac{k}{2} \end{align*}
If the probability of none of these events happening is $p$, then the expected value of $(v - k)^2$ would be at least $pk^2$, since $(v - k)^2$ would be $k^2$ when no event happened and nonnegative if not. Therefore
\[ pk^2 \leq \frac{k}{2} \Longrightarrow p \leq \frac{1}{n+1}, \] as desired. |
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 appears at least three times in the sequence $(y_k)_{k\ge 1}$. | 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 $n-\frac {1}{2}\le m\frac {a}{2a+1}<n+\frac {1}{2}-\frac {2a}{2a+1}$ $\Rightarrow 0\le m-(2a+1)\frac {n-1}{2a}<\frac {1}{a}$.
The two statements are clearly equivalent when noted that $(2a+1)\frac {n-1}{2a}=n-1+\frac {n-1}{2a}$ hence we can just take $m=k+(n-1)$ and be done. |
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 represents the minimal distance between the vertices. You wish to embed the graph in the real number line as $n$ points with a sufficiently small diameter. To do so, you definitely need to apply some sort of algorithm (or induct, but algorithms are more general), which means adding each new group of points according to some rule. Things which may complicate the algorithm: (a) you may need to move the already existing points around, and (b) you may need to add multiple points at a time. These seem useless, though, as (b) gives very weak improvement, and (a) could also be dealt with by choosing all your earlier points smartly, in order to have some information about them later. In any case, we wish to embed the points in a manner which results in as small numbers and distances as possible, which motivates the following.
[b]Solution.[/b] Relabel the variables so that for each $k=1,2,\dots,n$, $\sum_{i<k}\ell_{ij}$ is minimal over $j\in\{k,k+1,\dots,n\}$ at $j=k$. (We can do this by recursively assigning the indexes $1,2,\dots$.) Let
\[
a_k=\sum_{i<k}\ell_{ik}\qquad \text{for}\quad k=1,2,\dots,n.
\]
That way, $\sum_{i=1}^n a_i=\sum_{1\le i<j\le n}\ell_{ij}$, and whenever $j<k$, the definition of $j$ implies
\begin{align*}
a_k&=\sum_{i<k}\ell_{ik}=\sum_{i<j}\ell_{ik}+\ell_{jk}+\sum_{j<i<k}\ell_{ik} \\
&\ge \sum_{i<j}\ell_{ij}+\ell_{jk}+0=a_j+\ell_{jk}. \quad \blacksquare
\end{align*} |
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 $a$. Then $abcd$ is such a seating.
If $c\not \in N(a)$, again 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 $a$. Then $abcd$ is such a seating. |
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 $Z$ in triangle $ABC$ is $p(Z)$ maximised? | 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 $2abc=\frac{2xyz}{(1+x)(1+y)(1+z)}\leq \frac{1}{4}$, by easy A.M.-G.M.
Equality hold at $x=y=z=1$, which gives $a=b=c=\frac{1}{2}$.
So $p(Z)$ attain it’s maximum value when $Z$ is the centroid of the triangle. |
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 constant, so $\deg p=2k$ and $\deg q=3k$ with some positive integer $k$.
Take derivatives and eliminate $\ell$:
$$ p^3-q^2 = \ell $$
$$ 3p^2p'-2qq' =\ell' $$
$$ (p^3-q^2)\ell' = (3p^2p'-2qq')\ell $$
$$ p^2(p\ell'-3p'\ell) = q(q\ell'-2q'\ell). $$
Since $p^2$ is coprime to $q$,
$$ p^2 \,\big|\, q\ell'-2q'\ell . $$
The leading term of $q\ell'-2q'\ell$ does not cancel out, so
$$ \deg(p^2)=4k>\deg(q\ell'-2q'\ell)=3k,$$
contradiction.
[/hide] |
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 like this should be true if another exercise was true so I looked it up and found this theorem), the question was how to manage to pair up the two problems. The solution was roughly this: we try to go row by row and try to permute the numbers in a proper way, we'll show that for each row it is possible, no matter how we arranged the numbers above. Permute the first row's numbers in a random way. Going from up to down in the rows, assume the $k$-th row can't be properly filled, we have a look at it. Draw a a bipartite graph such that one part is the numbers in the $k$-th row and the other part is the cells in the $k$-th row. We connect the numbers with the cells for which they can be placed in (which depends on the $k-1$ rows above, if for example one of the rows contained a number in the $t$-th column above, then we can't place that in the cell in the $t$-th column and $k$-th row). Now, we know Hall's Marriage Theorem so we assume that there are some $m$ amount of numbers that are only connected to $l$ amount of cells and that $m>l$. We note and proof the following statements:
-a number is connected to at least $n-(k-1)$ cells
-a cell is connected to at least $n-(k-1)$ numbers
-if a cell is connected to more than $n-(k-1)$ numbers then it is connected to a number which didn't appear in all the rows above (it's possible that it appeared in some, but not all)
With these, to show that our statement that $m>l$ is false, we do the following moves:
-Search for cells that have more than $n-(k-1)$ edges. By the statement that it is connected to at least one number that didn't appear in all the rows above, we're basically saying that it is connected to a number that also has more than $n-(k-1)$ edges. Delete that connection.
If the cell still has more than $n-(k-1)$ edges then it's still connected to a number that has more than $n-(k-1)$ edges, delete that too.
Keep doing till we get that the cell has exactly $n-(k-1)$ connections.
One thing to notice, we know that each cell is connected to at least $n-(k-1)$ numbers so after this algorithm, we will have exactly $n-(k-1)$ edges for all cells.
Other thing to notice, when we take away an edge, we won't make any number have less than $n-(k-1)$ edges. That is because we always take away 1 edge from a number that has more than $n-(k-1)$ edges.
The last thing to mention is that because at the start we had at least $n-(k-1)$ edges at all numbers, we'll still have atleast $n-(k-1)$ edges at all numbers.
Now, counting all the edges, (denote it $S$), we first get that $S=l*(n-(k-1))$, and second off that $S\geq m*(n-(k-1))$, from which $l*(n-(k-1))\geq m*(n-(k-1))$, and from this we get that $l\geq m$, which is contradiction.
(If you look closely, we can leave the Reductio ab absurdum and just directly prove the same way that for any $m$ amount of numbers, there are at least $m$ amount of cells.)
But because of this, we can use Hall's Marriage Theorem to make sure we'll get a proper solution for the $k$-th row. Keep going till you get to the last row then you're done.
(Look up the theorem on YT, although my proof is just simple contradiction with an easy algorithm. I also wrote my proof on the theorem on he exam paper for like 30 minutes.)
I also had other ideas on proving it with tracing it back to a special Latin square, but it had quite a few holes in it (with unproven known theorems, that I knew were true but didn't know the proof to it, and I think there was a not totally true statement I wrote too). I don't think I'll get much on that, but that's a shame because I spent a good time on the second solution too instead of solving the 2nd problem. I also messed up in the first problem, I misunderstood the question, I still got that the centroid will be the solution but instead of $p(Z)=1/4$ I got $p(Z)=1$.. If I didn't mess that up then I'd prob. race for 1st or 2nd place, nevermind now |
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 choose $n$ different dwarfs from $n$ different clubs (one dwarf from each club), such that the $n$ dwarfs know each other! | 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 hand, if this process stop, i.e. certain club can't nominate new dwarfs, it means that all the $n$ dwarfs in that club have been selected. These $n$ dwarfs know each other and also come from $n$ different clubs, done. |
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 $$f(x+y)=2^{-\nu_2(x+y)}\leq 2^{-\min\{\nu_2(x),\nu_2(y)\}}\leq 2^{-\nu_2(x)}+2^{-\nu_2(y)}=f(x)+f(y)$$
On to the proof that this is the only function that satisfies the conditions.
First, from the second condition, by considering the pairs $(x,y)=(0,2),(1,2)$, we get $f(0)=0,f(1)=1$. We also get $f(2^k)=\frac{1}{2^k}$. In fact, $f$ is multiplicative under $\mathbb{Q}$. Furthermore, we may obtain $f(-1)=1$ from $(x,y)=(-1,-1)$, and $f(-r)=f(r)$ for all rational numbers $r$.
From the first condition, we get $f(y)\geq |f(x+y)-f(x)|$ for all pairs $(x,y)$.
Set $y=2^a$. We get $\frac{1}{2^a}\geq |f(x+2^a)-f(x)|$. Choose an arbitrary positive integer $t$ and write it in the form $2^{a_1}+2^{a_2}+\cdots+2^{a_n}$. Using the triangle inequality, we obtain
$$|f(x+t)-f(x)|\leq |f(x+t)-f(x+t-2^{a_n})|+|f(x+t)-f(x+t-2^{a_n}-2^{a_{n-1}})|+\cdots+|f(x+2^{a_1})-f(x)|\leq \frac{1}{2^{a_n}}+\cdots+\frac{1}{2^{a_1}}<\frac{2}{2^{a_1}}$$
We now claim that for any prime $p\geq 3$, we have $f(p)=1$, which automatically implies the desired conclusion.
Let $t=p^u-p^v$ and $x=p^v$. Then we have
$$\frac{2}{2^{\nu_2(p^u-p^v)}}\geq |f(p)^u-f(p)^v|$$
If $f(p)>1$, then we fix $u-v$ and let $u$ be large. This yields a contradiction.
If $f(p)<1$, let $x=1$ and $t=p^k-1$. We have
$$\frac{2}{2^{\nu_2(p^k-1)}}\geq |1-f(p)^k|$$
Setting $k$ to be a large power of 2, we get a contradiction.
Hence, $f(p)=1$ and we are done. |
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 the components of $p_1$ and $p_2$ which are perpendicular to $p_0$ must be in opposite directions. Furthermore, the ratio of their magnitudes is $\frac{OP_1 \cdot sin\angle P_0OP_1}{OP_2 \cdot sin\angle P_0OP_2} = \frac{1/2 \cdot OP_0 \cdot OP_1 \cdot sin\angle P_0OP_1}{1/2 \cdot OP_0 \cdot OP_2 \cdot sin\angle P_0OP_2} = \frac{S_2}{S_1}$. Thus, $S_1p_1 + S_2p_2$ is parallel to $p_0$, which implies $S_0p_0 + S_1p_1 + S_2p_2$ is parallel to $p_0$. Similarly, $S_0p_0 + S_1p_1 + S_2p_2$ is parallel to $p_1$ and $p_2$. Therefore, $S_0p_0 + S_1p_1 + S_2p_2$ must be the zero vector $\vec{0}$.
If $S_2, S_1, S_0$ are in a geometric sequence, then $S_1 = S_2r$ and $S_0 = S_2r^2$ for some positive real $r$. So $S_0p_0 + S_1p_1 + S_2p_2 = \vec{0} \iff r^2p_0 + rp_1 + p_2 = \vec{0} \iff r^2a_0 + ra_1 + a_2 = r^2b_0 + rb_1 + b_2 = 0$, which means there exists a real number $x$ such that $a_0x^2 + a_1x + a_2 = b_0x^2 + b_1x + b_2 = 0$.
If there exists a real number $x$ such that $a_0x^2 + a_1x + a_2 = b_0x^2 + b_1x + b_2 = 0$, then $x^2p_0 + xp_1 + p_2 = \vec{0}$. Then $$\vec{0} = S_2\vec{0} - \vec{0} = S_2(x^2p_0 + xp_1 + p_2) - (S_0p_0 + S_1p_1 + S_2p_2) = (S_2x^2 - S_0)p_0 + (S_2x - S_1)p_1$$ but $p_0$ and $p_1$ cannot be of the same or opposite directions, so $S_2x^2 - S_0 = 0 \iff S_0 = S_2x^2$ and $S_2x - S_1 = 0 \iff S_1 = S_2x$, which implies $S_2, S_1, S_0$ are in a geometric sequence. |
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 we can fly $d_1\rightarrow d_2\rightarrow\dots\rightarrow d_{2m+1}\rightarrow d_1$ while using each airline at most once. | [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, apply induction to get $f(n) \le f(1)+f(n-1)-3 \le n+3$. If there are at least two choose one at the bottom and call it $A$. Draw the line which is determined by the top side of $A$. Assume that "this line" intersects $a$ rectangles and partitons main rectangle into $X, Y$ big rectangles (we now have $2a$ rectangles). Apply induction to big rectangles to get $f(n) \le f(X)+f(Y)-a-3 \le X+Y-a+3=n+3$ (there are $a$ left lines of these $a$ rectangles which is counted twice, "this line" and two sides of the main rectangle)
For construction divide main rectangle into rows.
[/hide] |
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 divisor $\delta$ of $pq$ and integers $u,v$ satisfying $2uv=y$. Therefore, \[v^2\cdot\frac{pq}{\delta}-u^2\cdot\delta=1.\]If $\delta\in\{p,q\}$ then we get the desired identity. If $\delta=pq$ then as $|uv|=|y|/2<|xy|$ we reach a contradiction. Finally, if $\delta=1$ then $pq\mid u^2+1$ which is impossible, as $p\equiv q\equiv 3\bmod{4}$, so the proof is finished. |
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 $m(A_i) < 1$, we know that $m(B_1) < 1 < \infty$, and we also know that $B_1 \supset B_2 \supset B_3 \supset \cdots$ so the theorem applies and we know that there exists $0<x<1$ such that $x \in B_i$ for all $i$. Now it is easy to construct a sequence $\{k_i\}^{\infty}_1$ such that $x\in A_{k_i}$ for all $i$, which finishes the proof.
[/hide] |
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
\[ \minus{} 1 \equiv (p \minus{} 1)! \equiv \prod_{i \equal{} 1}^{p \minus{} 1} a_i b_i \equal{} \prod_{i \equal{} 1}^{p \minus{} 1} a_i \cdot \prod_{i \equal{} 1}^{p \minus{} 1} b_i \equiv (p \minus{} 1)!^2 \equiv 1 \mod p,
\]
being a contradiction. |
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)|\equal{}\alpha$ and that this statement is not true if $ \alpha>1$. | 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 $f(m) \le -1$ and $f(\pm 1) \ge 0$ or $f(m) \le 0$ and $f(\pm 1) \ge 1$). Shifting vertically and horizontally appropriately (so that $(m, f(m))$ becomes the origin) we get that $g(x) = ax^2$ and $g(2) \ge 1$, implying $a \ge \tfrac{1}{4}$ and $g'(x) = 2ax$ taking values from $0$ to $4a \ge 1$. The example when $|f'|$ is bounded by $1$ is $f(x) = \tfrac{1}{4} (x+1)^2$. Indeed, $f(1) = 1$ and $f'(x) = \tfrac{x+1}{2} \in [0, 1]$. |
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 product is denoted $P$, we have \[ \log P = \sum_{k=1}^n \log\left(1 + \frac zk\right) = \sum_{k=1}^n \left(\frac zk + z^2O(k^{-2})\right) = z H_n + O(z^2/n). \] Recall that $H_n = \log n + \gamma + O(1/n)$. Thus, \[ \log P = z \log n + \gamma z + O_z(1/n) \implies P = e^{\gamma z} n^z (1 + O_z(1/n)), \] as desired. $\square$
-------------------
[color=blue][b]Convergence.[/b][/color] Note that when $z = 1$ we get the harmonic series, which diverges. Let $w = z-2$ and multiply by $w+1$, so that the series rewrites as \[ \sum_{n=1}^\infty \prod_{k=1}^n \frac{k+w}{k}. \] By the previous lemma, this sum is \[ e^{\gamma w} \sum_{n=1}^\infty n^w (1+O(1/n)). \] When $\Re(w) > -1$ the partial sums of the series are $\tfrac{n^{w+1}}{w+1} + O(n^w)$, which diverges. Otherwise if $\Re(w) \leq -1$ then the error term converges, so we may ignore it. The rest converges only when $\Re(w) < -1$.
In sum, the series converges exactly when $\Re(z) < 1$. $\blacksquare$
-------------------
[color=blue][b]Evaluating the sum.[/b][/color] Rewrite the sum from the previous section as \[ \sum_{n=1}^\infty \binom{n+w}{n}, \] where we define $\tbinom{\zeta}{k} = \zeta^{\underline{k}}/k!$ for any complex $\zeta$ by using falling factorials. One can check \[ \binom{n+w}{n} = \binom{n+w+1}{n} - \binom{n+w}{n-1}, \] with which the desired sum telescopes: \[ \sum_{n=1}^N \binom{n+w}{n} = \binom{N+1+w}{N}\, - \,\binom{1+w}{0} = \biggl(\prod_{k=1}^N \frac{k+(w+1)}{k}\biggr)\, -\, 1 \xrightarrow{N \to \infty} -1, \] since the previous lemma implies the above product is $\Theta(n^{w+1}) \to 0$ due to $\Re(w+1) < 0$.
Dividing by $z-1$, we find that the original series converges to $\tfrac{1}{1-z}$. $\blacksquare$ |
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]{2^{n\minus{}2} k_1} \}$).
We have $ \sqrt[n]{2^{n\minus{}2} k_1} \leq \sqrt[n\plus{}1]{2^{n\minus{}1} k_1}$ if and only if $ k_1 \leq 4$. So, for $ 1<k_1<4$, the sequence is increasing and we could choose to be any number less than $ \sqrt{k_1}$.
In case $ k_1 \geq 4$, we have $ \sqrt[n]{2^{n\minus{}2} k_1} \geq 2$ for all $ n$. Choose $ n$ to be $ \frac{3}{2}$ is sufficient. |
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_2^2 ... q_k^2$, and the integers $ x, x\plus{}1, ... x\plus{}(k\minus{}1)$ are all divisible by a square congruent to $ 3 \bmod 4$ with multiplicity $ 1$. It's well-known that no such number is the sum of two squares. |
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} \rfloor$ (from the hypothesis that $ x \in (0,1)$)
From the final relation, we deduce that $ f_k (x) \equal{} \lfloor \frac {k \plus{} 1}{x} \rfloor \minus{} \lceil \frac {k}{x} \rceil \plus{} 1$.
Now, consider two cases:
Note that from the equation, if $ x$ is irrational, $ f_n (x)$ is $ \lfloor \frac {k \plus{} 1}{x} \rfloor \minus{} \lfloor \frac {k}{x} \rfloor$ oscillates between $ \lfloor \frac {1}{x} \rfloor$ and $ \lceil \frac {1}{x} \rceil$. If $ x$ is rational or $ x\equal{}\frac{a}{b}$ then $ f_{ka}(x) \equal{} \lfloor \frac {ka \plus{} 1}{x} \rfloor \minus{} \lfloor \frac {ka}{x} \rfloor \equal{} \frac{b}{a} \equal{} \lfloor \frac{b}{a} \rfloor \plus{} 1$ still oscillates between $ \lfloor \frac {1}{x} \rfloor$ and $ \lceil \frac {1}{x} \rceil$. So I guess that the limit is $ \frac{1}{x}$. To prove this, just consider how many terms of $ \lfloor \frac {1}{x} \rfloor$ and $ \lceil \frac {1}{x} \rceil$. |
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 $ \frac{d}{dx}g(x) \equal{} x*g(x)$ is $ g(x) \equal{} Ce^{\frac12 x^2}$.
Substitution of $ f(x)\equal{}C(x)e^{\frac12 x^2}$ yields
$ f(x)\equal{}\sqrt{\frac{\pi}{2}}\mbox{ erf}\left(\frac{x}{\sqrt{2}}\right)e^{\frac12 x^2}$. |
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