There are some markets in a city. Some of them are joined by one-way streets, such that for any market there are exactly two streets to leave it. Prove that the city may be partitioned into $1014$ districts such that streets join only markets from different districts, and by the same one-way for any two districts (either only from first to second, or vice-versa).

This is a very hard problem. I've seen somewhere on ML that no one solved it.
this is the official solution:
Firstly we prove that if from any square leave at most 2 streets then we can colour the squares in 13 colors s.t. from any square we cannot reach a square with same color, passing less then 3 streets.To do this we construct an oriented graph in which markets are connected iff there is a path from less than 3 streets joining them.
From any market leave at most 6 edges.
Now we color inductively the vertices.  there exists a vertex in which enter at most 6 roads. So there are at most 12 vertices joined to it. We can use induction.
the claim is proved.
Now for each color we divide the markets of this color into 78 types, according to the colours of streets where lead the streets from one market.
We get 13*78=1014 districts.

In fact we can divide the city into 625 such districts. First let us prove a lemma.

Lemma: If each square has exactly one street leaving it, we may divide the city into 5 districts with the same condition.
Proof: We will call our colors (districts) $0$, $1$, $2$, $\infty$, and $-1$. We know that a connected directed graph where each vertex has out degree one consists of a single cycle with trees pointing in. From the conditions of the problem, we know that the cycle in each of our connected components has size at least 3. Begin coloring it from an arbitrary vertex in the cycle so that the colors follow the pattern $0, 1, 2, 0, 1, 2, \ldots$. Stop when there are two vertices remaining, and color the last two vertices $\infty$ and $-1$ in that order. Now we color the trees pointing into the cycle starting from the vertices that point directly to the cycle and working outward. We do this with the following rule: If the vertex points to $0$, color it $2$. If the vertex points to $1$, color it $0$. If the vertex points to $2$, color it $1$. If the vertex points to $\infty$, color it $0$ (although in this case it does not matter which of $0, 1, 2$ it is). Finally, if the vertex points to $-1$, color it $\infty$.

It is clear that in this case no edge is between squares in the same district. Furthermore, we notice that the only possible edges are $0 \to 1$, $1 \to 2$, $2 \to 0$, $0 \to \infty$, $1 \to \infty$, $2 \to \infty$, $\infty \to -1$, and $-1 \to 0$. Therefore, no two districts have edges pointing opposite directions between them, and the problem statement holds.

Now in the original city we will color each street either red or blue such that every square has exactly one red street and one blue street leaving it. We then consider the following three cities:

1. The cities formed by the original city squares and all the red streets
2. The cities formed by the original city squares and all the blue streets
3. The cities formed by splitting each city square into two half-squares: one red and one blue. A street connects a red half-square to a blue half-square if a red street connected the corresponding squares in the original graph, and a street connects a blue half-square to a red half-square if a blue street connected the corresponding squares in the original graph.

We will divide each of these cities into five districts satisfying the problem statement using our lemma. Then we will divide the original city into $5^4 = 625$ districts by coloring each vertex with the ordered quadruple of its color in the red city, its color in the blue city, the color of its red half-square in the third city, and the color of its blue half-square in the third city. It is clear that no edge connects two squares of the same district.

Now suppose that we have two districts with edges going both ways between them. If they are both red, then we notice that the two edges go both ways between the two corresponding districts in the red graph. This contradicts the fact that we divided the red city to satisfy the problem statement. If they are both blue, we get a similar contradiction in the blue city. Finally, if one is red and one is blue, then we have that the two red half-squares are colored the same in the third city and that the two blue half-squares are colored the same in the third city. But these edges go opposite directions between those two districts, so we get a contradiction in the third city. As a result, no two districts can have edges going both ways between them.

I have a 'solution' which gives me a partition into at most 91 districts, so it must be wrong. Can someone point out my mistake?

Call two districts harmonious with each other if all edges between are in the same direction. (that is A,B harmonious means any edge between them is from A to B or B to A, but not both)

Say we partition into 13 districts as in first part of iura's solution, i.e. we have 13 districts $D_1, D_2,...,D_{13}$ such that from any vertex $v$ we cannot reach any other vertex in the same district with 1 or 2 edges.

This means that, for any two districts, D and D', and $v \in D$, then we have 3 cases:
1. $v$ is not connected to any vertex in D' at all.,
2. $v$ can reach some vertex in D' with one edge, but not reachable by any vertex in D'.
3. $v$ is reachable by some vertex in D', but cannot reach any.

Take district $D_1$. Look at its relation with $D_2$. Take the set $X_2$ which is the subset of vertices of $D_1$ that are reachable from $D_2$ (in one edge). Separate $D_1$ into $X_2$ and $D_1-X_2$.

Now repeat this process. Take $X_3$ to be the subset of vertices of $D_1$ that are reachable by $D_3$. Separate $D_1$ (which at this point is the union of $X_2$ and $D_1-X_2$) into $X_2, X_1\X_2, D_1-X_2-X_3$. Repeat with D_4 etc. By the end we have a partition of $D_1$ into (at most) 13 smaller districts, each of which is harmonious with all 12 other districts.

Repeat the process with $D_2$ (so look at ${D_2,D_3}$, then ${D_2,D_4}$,...), finally we end up with at most 13+12+...+1=91 districts.

I must be being stupid but I can't see my mistake. I was hoping that by typing it out I would see it but I don't.

Your construction actually divides $D_1$ into $\binom{13}{2}$ districts, giving the same bound of $1014$. Consider $D_1$ having vertices $v_{b_2b_3\ldots b_{13}}$ where if $b_i = 0$ then there is an incoming edge from $D_i$ and if $b_i = 1$ there is an outgoing edge to $D_i$. Assume there are $\binom{13}{2}$ such vertices indexed by the binary strings with one or two 1s (there can't be 0 because then both edges go to $D_1$). Then without further justification we might need to separate all of these vertices into different districts in order to make $D_1$ harmonious with the other districts. However, your argument that once $D_1$ is harmonious with $D_2$ we don't need to make $D_2$ harmonious with $D_1$ I think does show the tighter bound of $\binom{13}{2} + (\binom{12}{2} + 1) + (\binom{11}{2} + 1) + \cdots + (\binom{1}{2} + 1)$ (Note that after the first district there can be two 0s), which gives $\binom{14}{3} + 12 = 376$. I should think about if there is a similar optimization to the 625 solution, but this nicely captures the extra work that was done by the original solution. It's also possible that this is tight -- it looks about like what I estimated the real answer to be.

EDIT: I think my numbers are slightly wrong, but I do think the argument gives a bound somewhere in the 300-400 range. I'll update again if I get numbers that I'm more confident about.

This problem recently came up; let me attempt to give a solution for 225 districts.

[b]Lemma 1:[/b] If there exists exactly one street leaving every square, then the city can be partitioned into five districts.
This lemma was proved above by Hamster1800, but let me give another proof.
[i]Proof:[/i] Let's assume that ignoring the orientation of the streets, the city is connected - otherwise work with every connected component.
There must exist a cycle - start with any square, and keep moving via the unique street leaving every square. By finiteness, eventually we return to a previously visited square which yields a cycle.
Let street $XY$ be part of a cycle (so it leads from $X$ to $Y$). Remove it. The city (without the orientation) still stays connected because $X$ and $Y$ are still joined via the rest of the cycle. If there are $n$ markets, then there is one street leaving each market but $X$ which yields $n-1$ edges in the undirected corresponding graph. This means it's a tree, in particular there are no more cycles.
From every square $Z$, there is a unique path starting from $Z$ because there is a unique street leaving every square. This path is finite because there are no cycles, i.e. it will eventually hit $X$ and stop.
Consider the following three districts: district 0 consisting of $Z$ for which the above path has length divisible by 3, district 1 consisting of $Z$ for which the path has length 1 modulo 3, and district 2 consisting of $Z$ for which the length of the path is 2 modulo 3.
Clearly, the edges run from district 2 to 1, 1 to 0 and 0 to 2. It follows that this partition satisfies our condition for the city with street $XY$ removed. Now add $XY$ back in. It may screw up somehow, so let's create a fourth district and place $X$ there (originally $X$ was placed in district 0). This is not enough yet: if $SY$ is a street (so $S$ is in district 1) and $Y$ is also in district 1, then the streets between districts 1 and 4 run both ways. Correct it by placing $Y$ in a newly created fifth district.
The lemma is proved.

[b]Lemma 2:[/b] If there exists one street leaving every square, then one can partition the city into (at most) three districts such that no two squares from the same district are connected.
[i]Proof:[/i] Do the same thing as in the previous lemma: assume the undirected graph is connected and then remove an edge $XY$ as above. This time, partition the city into two districts by parity of the length the corresponding path. Adding $XY$ back may create a street between two cities in district 0; if that's the case move $X$ into a new third district.

Now let's come back to the proof.
For every square, color the two streets leaving it red and blue.
We thus create two cities: the red city, and the blue city, with the same squares but half the streets. Both satisfy the conditions of lemma 1, thus we can divide the squares into (at most) five blue districts and five red districts that satisfy the requirements with respect to blue and red streets.
Consider also a third orange city. An orange "street" $A\to B$ consists of a blue street $A\to C$ and a red street $C\to B$; there is exactly one orange street leaving every square. By lemma 2, we can partition the squares into three orange districts such that no two squares from the same district are connected with an orange "street".
[b]Remark:[/b] in the statement of the problem it is implicitly assumed that no two squares can be connected with two streets of opposite directions - otherwise the problem is clearly false. More specifically, such a situation would fail lemma 1, because we might have $S=Y$. In the orange city, however, there might be opposite streets but that's fine because lemma 2 still holds. We do, however, use the fact that the orange city has no loops otherwise lemma 2 would not hold either.
Classify the squares according to the following criteria: a square $X$ is described by
$\bullet$ The blue district to which $X$ belongs.
$\bullet$ The red district to which $X$ belongs.
$\bullet$ The orange district to which $X$ belongs.
$\bullet$ There is a unique red street leaving $X$ and entering another square $Y$. Consider the orange district to which $Y$ belongs.
This description partitions the squares into (at most) $5\cdot 5\cdot 3\cdot 3=225$ districts. We prove they satisfy the requirements.
No two cities from the same district can be connected by a street, because a blue street changes the blue district and a red street changes the red district. Similarly, the red streets between two districts all point the same way, and the blue streets between two districts all point the same way. The last case to exclude is when a red street and a blue street join two districts, in opposite directions.
Suppose we have a blue street from $A$ to $B$, and a red street from $C$ to $D$ where $A$ and $D$ (not necessarily distinct) belong to the same district and $B$ and $C$ also belong to the same district.
Let $BE$ be the red street leaving $B$. Because $B$ and $C$ belong to the same district, it follows that $D$ and $E$ belong to the same orange district (fourth criterion). But $D$ and $A$ belong to the same orange district (third criterion), so $A$ and $E$ belong to the same orange district. Further, $A\to B\to E$ is an orange street connecting $A$ and $E$. But this contradicts lemma 2.

It has very easy soloution. You can partition the city into $3$ group that any 2 city in one group dont join each other. And any 2 group math with same direction road.
If there is a graph with city are vertices. There are some cycle. You number the edge of cycle:
$i)$If even cycle: get first vertice number$1$ and after any one put$2$(implye that any two same number aren't Consecutive. Means dont math with edge in graph.
$ii)$if odd cycle: same to first case number vertices. But the last vertice cant be 1 or 2.so put 3 on it.
So you can make 3 groups(firs one the vertices with number 1.second one with vertices number 2 and third one woth vertices number 3)
You can see that theese three group have Condition of question . And you can Division any group to some another group. You can do division until reach 1014 group.
I think there is correct solution.and easy question


integration
function
real analysis
real analysis unsolved

What's the point in stacking one irrelevancy upon another? :? Cleaned of all crazy notation, the problem reads

Consider all functions $u(=x^{2}x')$ such that $\int_{0}^{T}u(t)\,dt=A(=\frac{1}{3}\sqrt[4]{T^{12}-\dots})$. Find $\min \int_{0}^{T}t^{-5}u(t)^{4}\,dt$.

That is classical, indeed: the name is Holder inequality.

:lol: . I can't clean notation because in sulution you also need to  use this notation (last notation to be more exact).
say answer and if it will be true than I will say that your solution is correct. :wink:
Idea is good  Holder Really, but how to use: this is more interesting

Well, if you insist, here goes: we have
\[\Bigl(\int_{0}^{T}t^{-5}u^{4}\Bigr)^{1/4}\Bigl(\int_{0}^{T}t^{5/3}\Bigr)^{3/4}\ge \Bigl|\int_{0}^{T}u\Bigr|=A\]
whence, since $\int_{0}^{T}t^{5/3}=\frac{3}{8}T^{8/3}$,
\[\min=\left(\frac{8}{3}\right)^{3}T^{-8}A^{4}=\frac{8^{3}}{3^{7}}(T^{4}-T^{2}+T+T^{-1}-T^{-2}+T^{-4})\]
To attain the equality, $u$ must be proportional to $t^{5/3}$, so $x^{3}$ must be proportional to $t^{8/3}$ and $x$ to $t^{8/9}$. If $x=ct^{8/9}$, we get $c^{12}T^{32/3}=T^{12}-\dots$, so $c=\pm T^{-8/9}\sqrt[12]{T^{12}-T^{10}+T^{9}+T^{7}-T^{6}+T^{4}}$ (two solutions: one positive and one negative). I still fail to see why one needs to carry around that crazy sum of powers of $T$ and why not to state the problem in terms of $x^{3}$ instead of $x$ :?

Oh, you meant that the minimum should be taken with respect to $T$ as well?  :lol: Then you are right, one more step is due.

[quote="fedja"]Oh, you meant that the minimum should be taken with respect to $T$ as well?  :lol:[/quote]
yes,  :yup:


Let $ n$ be a positive integer, and $ S_n,$ be the set of all positive integer divisors of $ n$ (including 1 and itself). Prove that at most half of the elements in $ S_n$ have their last digits equal to 3.

[quote="ZetaX"]It was posted recently.[/quote]
Where?

[quote=Andy Loo]
Where?[/quote]
See [url=https://artofproblemsolving.com/community/c6h1110603]here[/url].


A 1-permutation takes the last digit of a number and puts it in front of the number. For example, the 1-permutation of 3124 is 4312. Find a number which, when multiplied by 4, equals its 1-permutation.

Let $ N$ be the original number and $ N'$ be the new number. Then, it is clear that the last two digits of $ N'$ must be divisible by 4, so the the number formed by the third-last and second-last digits of $ N$ must be divisible by 4 as well.

The last digit must be at least 4. So the numbers are:

102564 -> 410256
128205 -> 512820
153846 -> 615384
179487 -> 717948
205128 -> 820512
230769 -> 923076

And the other solutions are $ \sum_{i \equal{} 0}^k N\cdot 10^{6k}$ for any $ k \geq 1$.

An interesting to note is that these numbers are all formed by fractions with period 6:

102564 -> $ \frac 1{975}$
128205 -> $ \frac 1{78}$
153846 -> $ \frac 1{65}$
179487 -> $ \frac 7{39}$
205128 -> $ \frac 8{39}$
230769 -> $ \frac 3{13}$

It appears 13 is a common factor of the denominators of these fractions... Hmm... try to find out why :idea:

thanks hmmmm I'm not sure why $ 13$ is always in the denominator, but I'll read up on continued fractions and have a go


Each of the 30 students in a physical education class plays the same number of minutes of full-court basketball in a class period. Ten students play at a time, and the class period lasts 51 minutes. What is the number of minutes each student plays?

There are 30 students, and 10 that play at one time, so there are 3 "groups" playing. These 3 groups take a total of 51 minutes, so each group takes $ \frac{51}{3}\equal{}\boxed{17\text{ min}}$


calculus
derivative
real analysis
real analysis theorems

What is Newton's method on sequence Convergence.
Can anyone help me to find a good material that can teach me about that?
Thanks in advance.

It's the iteration $ x_{n\plus{}1}\equal{}x_n\minus{}\frac{f(x_n)}{f'(x_n)}$, for finding roots of $ f(x)\equal{}0$. If $ g(x)\equal{}x\minus{}\frac{f(x)}{f'(x)}$ and $ f(x_0)\equal{}0, f'(x_0)\neq 0$, $ g(x_0)\equal{}x_0$ and $ g'(x_0)\equal{}0$. By the general theory of fixed point iterations, that derivative condition means that the sequence $ x_n$ converges very rapidly to $ x_0$ if it starts close enough.


At most one of $ a+b-c$, $ b+c-a$, $ c+a-b$ is $ \leq 0$, if one is then we are done. Now suppose that $ a+b-c$, $ b+c-a$ and $ c+a-b$ are $ >0$. Now we put: $ a=x+y$, $ b=y+z$ and $ c=z+x$ for some $ x,y,z \in \mathbb{R}^{+}$ and let $ S=x+y+z$. Now we have to prove:
\[ xyz(x+S)(y+S)(z+S)\leq (x+y)^{2}(y+z)^{2}(z+x)^{2}= (S-x)^{2}(S-y)^{2}(S-z)^{2}\ (*)\]
W.L.O.G. we may assume that $ S=1$.
[hide="Why?"] $ x=S\cdot x_{1}$, $ y=S\cdot y_{1}$ and $ z=S\cdot z_{1}$, we have $ x_{1}+y_{1}+z_{1}=1$, putting that in $ (*)$ we have:
\[ S^{3}\cdot x_{1}y_{1}z_{1}\cdot S^{3}(x_{1}+1)(y_{1}+1)(z_{1}+1)\leq S^{6}\cdot (1-x_{1})^{2}(1-y_{1})^{2}(1-z_{1})^{2}\]  [/hide]
Thus we have to prove:
\[ xyz(x+1)(y+1)(z+1)\leq (1-x)^{2}(1-y)^{2}(1-z)^{2}\]
with condition $ x+y+z=S=1$.
\[ xyz(xyz+xy+yz+zx+2)\leq (xy+yz+zx-xyz)^{2}\]
\[ (xyz)^{2}+xyz(xy+yz+zx)+2xyz\leq (xy+yz+zx)^{2}-2xyz(xy+yz+zx)+(xyz)^{2}\]
\[ 3xyz(xy+yz+zx)+2xyz\leq (xy)^{2}+(yz)^{2}+(zx)^{2}+2xyz(y+z+x)\]
\[ (xy)^{2}+(yz)^{2}+(zx)^{2}\geq 3xyz(xy+yz+zx)\]
We put $ xy=k$, $ yz=l$, $ zx=m$; $ k,l,m\in \mathbb{R}^{+}$. We have: $ kl+lm+mk=xyz(y+z+x)=xyz$, thus we must prove:
\[ k^{2}+l^{2}+m^{2}\geq 3(kl+lm+mk)(k+l+m)\]
We know that $ k^{2}+l^{2}+m^{2}\geq kl+lm+mk$, because of that it is enough to prove:
\[ 3(k+l+m)\leq 1 \Leftrightarrow k+l+m\leq \frac{1}{3}\Leftrightarrow xy+yz+zx\leq \frac{1}{3}\]
Suppose $ xy+yz+zx> \frac{1}{3}$ $ \Rightarrow x^{2}+y^{2}+z^{2}> \frac{1}{3}$ $ \Rightarrow 1=(x+y+z)^{2}=(x^{2}+y^{2}+z^{2})+2(xy+yz+zx)> \frac{1}{3}+\frac{2}{3}=1$ $ \Rightarrow 1>1$ $ \longrightarrow$ contradiction.
\[ Q.E.D.\]

[quote="10maths_tp0609"][quote="nguoigo"]Let a,b,c>0,prove that:
$ (a+b-c)(b+c-a)(c+a-b)(a+b)b+c)(c+a)\leq 8a^{2}b^{2}c^{2}$ :)[/quote]

Assume that $ c=min(a,b,c)$
We have:
$ RHS-LHS=M(a-b)^{2}+N(a-c)(b-c)$
In which:
$ M=2c(a+b-c)\left((a-b)^{2}+4ab-c^{2}\right) \ge 0$
$ N=(a+b+c)\left((a+b-2c)(a-b)^{2}+(a+b)c^{2}\right) \ge 0$
Done :D[/quote]

Very nice! But how did you came to such factorization?  :blush:


analytic geometry
geometry
parallelogram

The coordinates of the vertices of a parallelogram are (10, 1),
(7, -2), (4, 1) and ($ x, y$). What is the sum of the distinct
possible values for $ x$ ?

You just add all of the give x-cooordinates $ 10\plus{}7\plus{}4\equal{}21$


Find all positive integers with the following property: the number can be expressed as the sum of the squares of its four smallest divisors (including 1).

We say that n is even .why? because if n is odd  then $a^2,b^2,...,d^2 $are odd so their sum is even and it contradition.so n is even and $a=1 , b=2$
now $c$ if $4$ or it is a prime number, because if $c=2^k(2m+1)$ then $2m+1$ is divisor of n.so $2m+1=1$ that cause $c=2^a$ that means $c=4$, or $k=0$,that in this case, and because $c$ is the least divisor of $n$ after $2$, $c$ is prime.
First :if $c = 4$ :
$n = 1 + 4 + 16 + d^2$, so, because $d|4$ and $d|d^2$ we have: $d|21$ , so $d=7 or d=21$
that we put them it give us $n=70$ and $n=462$ that none of them are correct.
Second: if $c = p$:
in this case $n = 5 + p^2 + d$ cause that $d$ is even.so $d=4$ or $d=2p$.if $d=4$ then $c=3$
so $n = 30$ and it is not correct. So $d=2p$,so $n = 1 + 4 + p^2 + 4p^2 = 5(p^2+1)$ and because $p|n$ so $p|5$, so $p=5$ and $n=130$ and it is the only answer of problem.
[b]Just n=130 is the answer[/b]

I want to say one more thing, the problem is from Iran olympiad :lol:


geometry
3D geometry
probability
function
combinatorics unsolved
combinatorics

In relation with a probabilistic problem for concrete numbers I came upon the generalization of an idea and there's a problem:

Let's suppose we are throwing by $n$ special cubes with $m$ sides numbered serially by 1,2,...,m and the probability of toss of each number is the same according to the classical definition of probability). Let $k$ denote the sum of all numbers on the sides of all cubes shortly after the last toss. Which sum $k$ has the highest probability.

Let $X_1,X_2,...,X_n$ be random variables that represents the value of the toss on cubes $1,2,...,n$ in this order and $X = X_1 + X_2 +...+ X_n$. Then $E X$ should be the value of the most probable sum on the sides of cubes (or in case that $E X$ isn't integer let's take the closest two integers which E X is situated between). At last it's clear that $E X_1 = E X_2 = ... = E X_n = (m + 1)/2$, then:
$E X = E X_1 + E X_2 + ... + E X_n = n(m + 1)/2$

But I would like to solve it more classically and exactly, so first I need to solve this:

Let $k, m, n$ be some natural numbers such that $n <= k <= mn$, then the esencial question is how many (orderly) $n-$somes $(x_1,x_2,...,x_n)$ does there exist such that each x_i is a natural number from 1 to $m$ maximally and:
$x_1 + x_2 + ... + x_n = k$.
Could anybody give me any analytic formula (depending on $k,m,n$) for number of all convenient $n-$some (or any hint).

Thank you....

Well you could use the following generating function and then look at the coefficient of $ x^k $ in $ f(x) $:

$ f(x) = ( x + x^2 + \cdots + x^m)^n $,

which can of course be simplified but the reasoning for it is clearer in this form.

The problem with such an approach is that it might not shed any light on a direct formula, if a nice one exists.


I never thought too much math was possible, but I just accidentally put "Cosine" at the top of my paper instead of "Carlos". :lol:

Wow. I'm jealous of people who end up doing that though, it means they're actually getting in a reasonable amount of study time, while I, for whatever reason, always end up not finishing my homework till around 10 or 11 and then of course I'd rather go to bed than do work in a book that's never due. Still, even I have gotten to the point where I use LaTeX to typeset school math homework, just because I find using LaTeX fun...

[quote="LynnelleYe"]Still, even I have gotten to the point where I use LaTeX to typeset school math homework, just because I find using LaTeX fun...[/quote]

Hey, thats a good idea... if only it didnt take so much more time because my LaTeX isnt very good and i usually start/finish hw in class anyway.

At one point or another, I'm pretty sure I've written a persuasive english essay using if then arguments.

on all papers i write (excluding typed assignments) i use a + sign instead of the word "and."


linear algebra
matrix
invariant
linear algebra unsolved

Let $ V$ be the space of all $ n\times n$ matrices over a field $ F$ and let $ P$ be a fixed $ n\times n$ matrix. Let $ T$ be the linear operator on $ V$ defined by $ T(A)\equal{}PA$ and let $ [T]_{B}$ be the matrix of $ T$ in the ordered base $ B$.           Prove that $ tr([T]_{B})\equal{}n \cdot tr(P)$

Since the trace is an invariant, we can choose the standard basis $ \{E(r,s)\}$ of $ V$, i.e. $ E(r,s)_{ij}\equal{}\delta_{ir}\delta_{sj}$. From $ (T(E(r,s)))_{ij}\equal{}P_{ir}\delta_{sj}$ we conclude $ T(E(r,s))\equal{}\sum_{i\equal{}1}^nP_{ir}E(i,s)$ and $ tr(T)\equal{}\sum_{1\leq r,s\leq n}P_{rr}\equal{}n\, tr(P)$


geometry
perimeter
probability
ratio
analytic geometry
geometric sequence
arithmetic sequence

$ \rightarrow$What is the greatest three-digit number "$ abc$" such that $ 4, a, b$ forms a geometric sequence and $ b, c, 5$ forms an arithmetic sequence?

$ \rightarrow$Answer: [hide]$ \boxed{697}$[/hide]

$ \rightarrow$The square with verticies $ (\minus{}a, \minus{}a), (a, \minus{}a),(\minus{}a, a)$ and $ (a, a)$ is cut by the  line $ y\equal{}\frac{x}{2}$ into congruent quadrilaterals.  What is the number of units in the perimeter of each quadrilateral?  Express your answer in simplified radical form in terms of $ a$.

$ \rightarrow$Answer: [hide]$ \boxed{4a \plus{} a\sqrt{5} \ \text{units}}$[/hide]

$ \rightarrow$A fair, standard six-faced die is tossed eight times.  The sequence of eight results is recorded to form an eight-digit number.  What is the probability that the number formed is a multiple of eight?  Express your answer as a common fraction.

$ \rightarrow$Answer: [hide]$ \boxed{\frac{1}{8}}$[/hide]

[hide="1"]we know b has to be odd. and a has to be even. if a werent even, the common ratio in the geometric sequence would be something/4, which would make a odd, and b not being a digit. therefore, its something/2(something/1 wont work obviously). we want something to be as large as ppossible. the largest possible value of something is 3. so we have a=6 and b=9. c is the average of b and 5 so c=7 abc=697[/hide]

#3

[hide]Answer: $ \boxed{\frac{1}{8}}$

Solution:

the first thing that i notice is that to be divisible by 8 only the last three digits need to be divisibly by eight.  for example 1289756293874888 is divisible by eight because 888 the last 3 digits is divisible by eight.  And because there are only the digits 1-6 the only 3 digit numbers that we can make with the rolls are 111-166 211-266 311-366 411-466 511-566 611-666. we can divid this into the subsets.  I know that i may seem that i am including alot of information that really does not need to be said but i think that if someone doesnt understand (usually me) its nice to have someone who spells it all out.

111-166 there are 7 multiples of 8 that are in this subset

211-266 there are also 7 multiples of 8 in this subset

311-366 im kinda going fast here but you could find these multiples just by plugging the numbers into your calculator.  in this subset there are also 7 multiples of eight and this patter continues.  There are infact 7 multiples of 8 in all 6 subsets.

so we divide this by the total number of 3 digit numbers possible to get our probability. this is 56 the number of numbers in each subset times 6 the number of subsets. so we get $ \frac{7(6)}{56(6)} \equal{} \boxed{\frac{1}{8}}$[/hide]

[quote="abacadaea"] we know b has to be odd. and a has to be even. if a werent even, the common ratio in the geometric sequence would be something/4, which would make a odd, and b not being a digit. therefore, its something/2(something/1 wont work obviously).[/quote]

Could you please explain this a little more?  Thanks.

a geometric sequence is a sequence where each to consecutive terms share a common ratio. for example, the sequence 1,2 ,4 ,8... has a common ratio of 2. we want the greatest 3 digit number so we want "a" to be as large as possible, but b has to be less than 10. then something/2 i was talking about was the common ratio.

the common ratio cannot be something/4 because b will not be a digit. if something were even, then something/4 would be simplifiable. we need something to be odd. if that something we odd, then a would also be odd because a=4*something/4=something. then, to find b, we do it again. b=a/4=something/4. something is odd so something/4 is not an integer.
if the common ratio were just "something", it wouldnt work. i would be too big and b would be greater than 10.

we know that if "something" is even, something over 2 can be simplified like 6/2=3. the largest odd number we can find over 2 is 3. 5 is too big.so we have the common ratio is 3/2.

[hide="2"] Notice that the shape that is cut is simply a square with centerpoint at the origin and side length 2a. Here it is helpful to sketch a picture. Sketch the equation y=x/2. Now, it is helpful to find the points at which the square intersects the equation. They are (a, a/2) (a, -a) (-a, -a) and (-a, -a/2). The perimeter of the shape is $ 2a\minus{}a/2\plus{}2a \plus{} a/2 \plus{} \sqrt{(\minus{}a\minus{}a)^2\plus{}(\frac{3a}{2}\minus{}\frac{a}{2})^2}$
Simplifying yields $ 4a \plus{} a\sqrt{5}$
[/hide]

I understand up to the part where you find the points... But how did you get the things for the perimeter?

[hide="#1"]well common ratio has to be less than 2, and 2 is the only factor of 4, so try 3/2. so the number is 697[/hide]

the solution for #2 is pretty well written above me, but basically look for intersection points, and then you can find the lengths.

#3 can be done very quickly if you notice that the numbers 1-8 are 1,2,3,4,5,6,7,0 mod 8, so the probability is distributed equally. then its just 1/8

@georgiamathdude : If you drew a picture, you would easily see that there were two vertical lines, one slanted line and one horizontal. It is easy to see tha the horizontal line is 2a, and the 2 vertical lines are just 2a - a certain amount of distance, which is why we found the points. The oblique line can be found using the distance formula.

[quote="unimpossible"]#3 can be done very quickly if you notice that the numbers 1-8 are 1,2,3,4,5,6,7,0 mod 8, so the probability is distributed equally. then its just 1/8[/quote]

I don't get this part.  Is it a special property of mod that I should know?

Thanks.

To calculate the perimeter, I found the following lengths.  Are they correct?

[list]$ 2a$

$ 2a \plus{} \frac{a}{2}$

$ 2a \minus{} \frac{a}{2}$

$ \sqrt{(a\minus{}(\minus{}a))^{2} \plus{} (\frac{a}{2} \minus{} (\minus{}\frac{a}{2})^{2}}$[/list]

... I wouldn't know. Find the coordinates of the intersection points... and the vertices of the quadrilateral are: drumroll please.

$ (\minus{}a,a), (\minus{}a,\frac{\minus{}a}{2}),(a,a),(a,\frac{a}{2})$

Find lengths and sum them... you get...

$ 4a\plus{}\sqrt{a^2\plus{}4a^2}\equal{}4a\plus{}a\sqrt{5}$

[quote="mathcrazed"]... I wouldn't know. Find the coordinates of the intersection points... and the vertices of the quadrilateral are: drumroll please.

$ ( \minus{} a,a), ( \minus{} a,\frac { \minus{} a}{2}),(a,a),(a,\frac {a}{2})$

Find lengths and sum them... you get...

$ 4a \plus{} \sqrt {a^2 \plus{} 4a^2} \equal{} 4a \plus{} a\sqrt {5}$[/quote]

Do you know what I did wrong?  Thanks.

You got it! Your lengths are ok, except it's $ 2a \minus{} \frac {3a}{2} \equal{} \frac {a}{2}$ . Just don't make any silly addition errors, and you'll be fine.

Also note that when the lengths of the two vertical lines are added together, the sum must be 2a, as one is the difference of 2a and the length of the other. ; knowing that can save you an extra step.


ratio
symmetry
geometry
geometric transformation
homothety
3D geometry
sphere

This is a very hard question. Some help will be GREATLY appreciated.

[b]Given 3 congruent circles, such that they share 3 tangents, prove:
1) The triangle formed by joining the circle centres is equilateral
2) The triangle formed by the intersection of the tangent lines is equilateral
3) The diameter of the circle circumscribed about the equilateral triangle is in the ratio of 4:3 with the diameter of the original circles. [/b]

Ok, I have worked on a diagram for this and I am CERTAIN that I have made a correct picture. It is attached. To sum it up for you, the point is to prove that the blue triangle is equilateral, the green triangle is equilateral, and the diameter of the red circle is in the ratio of 4:3 with that of the black ones. Note, the black circles are all of equal radius.

This is not really a hard geometric problem.
HINT Circles' Radii

Someone correct me if this is not rigorous enough, but it should be:

[hide]
1. Symmetry, therefore equilateral. There is nothing unique about the position of any one circle so the triangle formed by joining their centers must be equilateral.
2. Symmetry, therefore equilateral. There is nothing unique about any of the three intersections of the tangents so each point must be equidistant from the other two.
3. The last one is trickier. Once you realize the green triangle has vertices on the midpoints of the blue triangle, and that the centroid of the triangles is the same, and that the distance from the midpoint of a side of a triangle to its centroid is 1/3 the distance from that midpoint to the opposite vertex, you realize the radius of the red circle is 1+1/3=4/3 the radius of each black circle.

The difficulty is in proving that the lines connecting the centers of the circles intersect the intersections of the tangent lines. This does not follow from the trivial symmetry argument because the center connections must all hit the tangent lines the same way, but may not necessarily hit their intersections. Hmm. Anyone?

[/hide]

That's just a well known property of circles. The internal centre of similitude (which is the intersection of the tangent lines) is collinear with the two centres. To see why, consider the homothety of negative factor sending one circle to another. It sends one centre to another, and it sends one point of tangency to another corresponding point of tangency.

"consider the homothety of negative factor sending one circle to another"

Can you expand on that? I've never heard of homothety.

You take a point O and transform the whole plane by sending P to the point P' on the line OP such that $OP'=kOP$ where $k$ is a constant, and distances are directed.

that symmetry stuff is bs...there is no reason why the diagram has to be symmetrical to have the given properties...

Suppose we have the inner triangle ABC, and O_a, etc is the circle opposite A,

the tangent from A to O_a has a length of s, the semiperimeter of ABC, likewise, all these tangents are equal

i am pretty sure that you can compute the sides of ABC in terms of the radii...it should follow that they are equal...i will get back to this

once you establish the green triangle is equilateral, the rest is trivial

[quote="bthings_2000"]As though O were a point on a sphere and the radius of the sphere were increased or decreased such that at O it appears everything is basically being uniformly stretched on the plane... I think I get homothety, but I don't follow the argument that homothety demonstrates that the intersection of tangent lines is collinear with the circles' radii. Can you expand on that?[/quote]

Ok. If I take a pair of nonintersecting circles, and let their common internal tangents (the ones that cross in the middle of the circles) be AB and CD, meeting at S. Can you see why some homothety centre S sends A to B and C to D? If you really need to, you can use a similar triangles argument to convince yourself. Furthermore, the homothety centre S sends the whole first circle to the whole second circle, and it therefore also sends the centre of one circle to the centre of another, implying that S is collinear with the centres.

The similar triangles argument makes perfect sense, and I'm now convinced of that property of circles. The homothety argument makes sense too, but I only see it working for circles of equal radii-- does it also hold for circles of nonequal radii? I don't see how it can, but this particular problem does specify equal radii.

Altheman, you really need to be better about giving more explicit and professional solutions, especially given how good you are at this. I mean, seriously:
[quote]that symmetry stuff is bs...[/quote]
You can do better.

Aside from the stylistic issues, you tend to be right, so I ask that you clarify. We have three congruent circles sharing three tangents. How can that figure possibly not be symmetrical? In particular, how would you differentiate one circle from the other two to show each needs individual treatment?

In any case, you made a symmetry argument yourself:
[quote]
Suppose we have the inner triangle ABC, and O_a, etc is the circle opposite A,

the tangent from A to O_a has a length of s, the semiperimeter of ABC, likewise, all these tangents are equal
[/quote]

See? "etc","likewise"=symmetry.

Yes. It does work for all circles. If they have equal radii, the external centre of similitude has the rather unfriendly property of being at infinity, however.

http://mathworld.wolfram.com/InternalSimilitudeCenter.html
http://mathworld.wolfram.com/HomotheticCenter.html

[quote="bthings_2000"]We have three congruent circles sharing three tangents. How can that figure possibly not be symmetrical?[/quote]

This essentially says, "Well, [i]of course[/i] it will be equilateral!"  To say, "the condition is symmetric in the three circles, therefore the triangle will be equilateral" is invalid; this type of reasoning sometimes turns out to be correct, but as a general rule, it fails -- the fact that the condition treats the three circles the same does not exclude the possibility of solutions in which the three circles are treated differently.

Altheman's use of symmetry is different: he says, "I can calculate the length of one side of the triangle.  This length turns out to be independent of the particular circle I started with -- the result is symmetric in the three circles.  Therefore all three sides have the same length."

Symmetry doesn't work that way. Here's an example that makes that obvious.

Choose three points randomly in the plane, and connect them to form a triangle. Since the points were chosen arbitrarily, and there is nothing unique about the way we chose a particular point (i.e., they were all chosen completely arbitrarily), each point must be equidistant from the other two. Therefore, the triangle is equilateral. Since the points were chosen randomly, all triangles are equilateral.

As you can see this argument does not work.

That's absolutely not the same thing. Three congruent circles sharing three tangents have exact relative locations, whereas three points in the plane are not bound by their connected tangents, etc.

I appreciate everyones comments but the fact of the matter is, this is a proof. We have not been taught to prove a problem using symmetry. We need to use a Euclidean style proof, namely either using vectors, properties, or coordinates (using the Cartesian plane). Anything along those lines will help me a LOT. Thnx in advance.

ah, but symmetry can be extremely useful! I accept that I might have used it incorrectly here, but no one has yet given a logical reason why. The one attempt given to show why symmetry was invalid was itself invalid.

Yes, I gave a reason why.  Your entire argument is, "The condition (on the three circles) is symmetric, therefore all realizations of this condition must be symmetric (in the three circles), therefore we have equilateral triangles."  This reasoning is [i]faulty[/i], because there are symmetric conditions with asymmetric realizations.  The reason you have difficulty seeing how bad this "proof" is is that in this particular circumstance, the conclusion (that the three circles must be arrayed symmetrically) happens to be true.  But its truth does not follow merely from the symmetry of the condition, as you claim.

There is at least a glimmering of this basic point which is already obvious to you, because of your rejection of TheAmazingOne's example.  His is example is total nonsense: the symmetry of the condition does not impose symmetry on the locations of the resulting three points.  You correctly rejected his example by noting that in this case, we have additional information (how the circles lie relative to certain lines, etc., etc.)  However, you [i]didn't actually use[/i] any of that information in your "proof" -- you claimed that the only necessary fact was the symmetry of the given condition.

Unfortunately, I don't have any nice geometric examples, and I don't know if you'll find the example I do have compelling.  Consider the following polynomial equation, symmetric in three real variables, for which $(a, a, a)$ is not a solution for any $a \neq 0$ (so, it has one symmetric realization, but only accidentally):

\[0 =\left(\left(y-2z\right)^{2}+\left(2z-3x\right)^{2}+\left(3x-y\right)^{2}\right) \cdot\left(\left(z-2x\right)^{2}+\left(2x-3y\right)^{2}+\left(3y-z\right)^{2}\right) \cdot\left(\left(x-2y\right)^{2}+\left(2y-3z\right)^{2}+\left(3z-x\right)^{2}\right) \cdot\left(\left(y-2x\right)^{2}+\left(2x-3z\right)^{2}+\left(3z-y\right)^{2}\right) \cdot\left(\left(z-2y\right)^{2}+\left(2y-3x\right)^{2}+\left(3x-z\right)^{2}\right) \cdot\left(\left(x-2z\right)^{2}+\left(2z-3y\right)^{2}+\left(3y-x\right)^{2}\right)\]

This is courtesy of Darij, in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=134436]this[/url] thread.  (The thread was started by a particularly obnoxious poster who claimed that every symmetric function $f(x, y, z)$ achieved its extrema along the line $x = y = z$.  In other words, he claimed that the symmetry of the condition $f(x, y, z) = m$ implied that we must have symmetry in the realization.  What it actually implies is that we must have symmetric points of realization, i.e. if $(a, b, c)$ is a solution then so must be $(c, a, b)$, $(b, a, c)$, etc.)

Well...the way you invoked symmetry was false.

The way I used symmetry was considering the verticies in a different manner...like if you prove that <OAI=|B-C|/2, then you can immediately claim that <OBI=|C-A|/2 since O, I are symmetrical in properties with respect to the verticies of a triangle.

The way you used symmetry was to claim that the circles being equal implied the triangle was symmetrical. This ended up to work out correctly in this problem. To understand why this logic is faulty, an equivalent idea is to find the minimum of a function by saying symmetric => min is when variables are all equal. [This has been discussed]

So...my statement was correct...I guess it was unclear...but you asked...is this rigirous or not? and the answer is: no.

[quote="bthings_2000"]JBL said it right. Thanks! Though I'd be interested in seeing an example of symmetric conditions with asymmetric realizations, like you say.[/quote]
Here's one: if $a,b,c$ are side lengths of $\triangle ABC$ with circumradius $R$, then $a^{2}+b^{2}+c^{2}=8R^{2}$ if and only if $\triangle ABC$ is [i]right[/i]. The equation is symmetric, but the triangle isn't.

ok i remembered the formula r_A=A/(s-a)...so r_A=r_B implies a=b, etc


induction
combinatorics unsolved
combinatorics

Tanya chose a natural number $X\le 100$, and Sasha is trying to guess this number. He can select two natural numbers $M$ and $N$ less than $100$, and ask for $\gcd (X+M,N)$. Show that Sasha can determine Tanya's number with at most seven questions.

Heres a sketch (I wrote up a complete solution but an error deleted it):

Let $R_{k}= X$ mod $2^{k}$.  By using induction, you can prove that for $k<7$, you can find $R_{k}$ in at most $k$ questions.  The base case $k=1$ is trivial (let $M=N=2$).  To do the inductive step, let $M=2^{6}-R_{k}$ and let $N=2^{6}$.  Then the value $gcd(X+M,N)$ in the form $2^{a}$ where $a$ is the greatest integer less than $7$ such that $2^{a}|(X-R_{k})$.  If $a>k$, then $R_{k+1}=R_{k}$ and if $k=a$ then it is not hard to see that $R_{k+1}=2^{k}+R_{k}$.  This gives us the desired induction.

So after at most $6$ turns, we have the value $R_{6}$.  Now, if $R_{6}\in\{1,2,\dots,35,36\}$, then $X\in\{R_{6}, R_{6}+2^{6}\}$, otherwise $X=R_{6}$.  In the former case, suppose $R_{6}$ has an odd factor $Z>1$.  Then by finding $gcd(X+Z,Z)$, we can determine $X$.  If $R_{6}$ does not have an odd factor, then $R_{6}=2^{m}$ for some integer $m<6$.  Then by finding the value $gcd(X+(1+2^{6-m}),1+2^{6-m})$ we can determine the correct value for $X$.  Because we have used at most $7$ steps, we are done.


Is there any polynomial $p(x)$ such that  all the coeffients of $(p(x))^{n}$ is positive( for all $n\geq 2$),But $p(x)$ has a negetive coefient?

I didn't find a nice title of the topic :blush:

[quote="loup blanc"]If $p^{2}$ and $p^{3}$ have positive coefficients then $\forall{n},p^{n}$ has positive coefficients. The Rust's example doesn't work because the $x^{6}$ coefficient of $p^{3}$ is $-1$.
The following works: $20+20x-x^{2}+20x^{3}+20x^{4}$.[/quote]
Thank you very much.I have understood. :)


if $ x,y,z>0$ --->  . $ P\equal{}\frac{1}{\sqrt{x\plus{}y\plus{}z}}(\frac{x}{\sqrt{x\plus{}y}}\plus{}\frac{y}{\sqrt{y\plus{}z}}\plus{}\frac{z}{\sqrt{z\plus{}x}}).$   find max P.

[quote="zaya_yc"]if $ x,y,z > 0$ --->  . $ P \equal{} \frac {1}{\sqrt {x \plus{} y \plus{} z}}(\frac {x}{\sqrt {x \plus{} y}} \plus{} \frac {y}{\sqrt {y \plus{} z}} \plus{} \frac {z}{\sqrt {z \plus{} x}}).$   find max P.[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=1459
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=88454
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=90923


[code]In[1]:= JordanDecomposition[{{1, 0, 0}, {0, 0, -1}, {0, 1, 0}}]

Out[1]= {{{1, 0, 0}, {0, -\[ImaginaryI], \[ImaginaryI]}, {0, 1,
   1}}, {{1, 0, 0}, {0, -\[ImaginaryI], 0}, {0, 0, \[ImaginaryI]}}}[/code]

Why are there imaginary numbers in the Jordan Canonical Form? I thought by definition Jordan canonical form contained only real numbers?

Forgive me for replying without having a clue what a Jordan Canonical Form is, but [url=http://mathworld.wolfram.com/JordanCanonicalForm.html]MathWorld[/url] seems to say nothing about needing to be real, in fact gives an example with complex values..[/url]

The Jordan canonical form contains the eigenvalues, which can be real or complex.  You may be thinking of [url=http://mathworld.wolfram.com/RationalCanonicalForm.html]rational (Frobenius) canonical form[/url].

I see. I was confusing something called [b] Real [/b] Jordan Canonical Form with Jordan Canonical Form.

Annoyingly, the only link I could find that is not a pdf is: [url=http://books.google.com/books?id=5U6loPxlvQkC&pg=PA155&lpg=PA155&dq=real+jordan+canonical+form&source=web&ots=bbRmipwAt9&sig=g5dzohoE8EIgXZSWdAq1zVBoX7k&hl=en&sa=X&oi=book_result&resnum=8&ct=result#PPA159,M1]Real Jordan Canonical Form[/url]

Does anyone know if Mathematica can put a matrix in real jordan canonical form?

You can deduce the real Jordan form from the usual Jordan form by pairing up complex conjugate eigenvalues (under the assumption that the original matrix has real entries).  It's not an essentially different concept.


It is possible to separate all the points in the chart below, using only 5 straight lines. Show that it cannot be done with less than five lines.

[code]
                 *
               *   *
             *   *   *
           *   *   *   *
[/code]

[quote="JBL"]What do you mean?  Draw five lines that divide the plane into different regions, at most one point to a region?[/quote]Yep! Exactly that!

[quote="probability1.01"]Are you sure you can with just 5?  I can prove you need at least five, but...[/quote]Yes, I already did. It is not that hard  ;)

I have a solution and will try to describe it to the best of my ability.

[hide="solution"]
                  1
                2  3
              4  5  6
            7  8  9  0

first draw 2 lines that would cross right below one, sectioning off into groups containing (1), (2,4,7), (0,3,6), and (5,8,9)
next draw a line between rows 2 and 3.
Finally draw lines that cross between 8 and 9 and go through the gaps of 6 and 9 on one side and 4 and 8 on the other.

Sorry but it's kind of hard to describe.  When you draw it you end up with a pentagon around 5
[/hide]

why dont you name the points so that it gets easier to explain how we can pass the lies because I got a way but explaining  it seems a tedious job.

[quote="Shogia"]I have a solution and will try to describe it to the best of my ability.

[hide="solution"]
[code]                  1
                2  3
              4  5  6
            7  8  9  0
[/code]
first draw 2 lines that would cross right below one, sectioning off into groups containing (1), (2,4,7), (0,3,6), and (5,8,9)
next draw a line between rows 2 and 3.
Finally draw lines that cross between 8 and 9 and go through the gaps of 6 and 9 on one side and 4 and 8 on the other.

Sorry but it's kind of hard to describe.  When you draw it you end up with a pentagon around 5
[/hide][/quote]Yep, thisone works. You can do an image and paste it here  :D . It is easier to understand.

And note that the perimeter dots form 9 "gaps" between adjacent dots.  No line can cover more than two such gaps, so there are at least 10/2 = 5 lines.

[quote="Shogia"]or you could use the formula that n lines can divide a plane into at most n(n+1)\2 + 1 regions.[/quote]Are you sure? I mean, not the formula, but the fact that it can be useful here. Note that for $n=4$ the formula give you 11 regions, but you have just 10 points. You should argue a little forward! (but if argumented, it can be used!) probability1.1's solution is a little more direct! ;)


find all f that: f R:->R:
f(f(x)+y)=2x+f( f(f(y))-x)

$ x \equal{} y \equal{} 0$  $ \implies$  $ f(f(0)) \equal{} f(f(f(0)))$
$ f(0) \equal{} c$  $ \implies$  $ f(c) \equal{} f(f(c))$
$ y \equal{} 0$  $ \implies$  $ f(f(x)) \equal{} 2x \plus{} f(f(c) \minus{} x)$
$ x \equal{} f(c)$  $ \implies$  $ f(c) \equal{} 2f(c) \plus{} c$  $ \implies$  $ f(c) \equal{} \minus{} c \equal{} f( \minus{} c)$
$ y \equal{} c,x \equal{} \minus{} c$  $ \implies$  $ f(0) \equal{} \minus{} 2c \plus{} f(0)$  $ \implies$  $ c \equal{} 0$
$ \implies$  $ f(0) \equal{} 0$
$ y \equal{} \minus{} f(x)$  $ \implies$  $ \minus{} 2x \equal{} f(something)$  $ \implies$  $ f$ is surjective
$ x \equal{} 0$  $ \implies$  $ f(y) \equal{} f(f(f(y)))$ , $ f(y) \equal{} k$  $ \implies$
$ f(f(k)) \equal{} k$  $ \implies$  $ f$ is bijective
$ y \equal{} 0$  $ \implies$  $ f(f(x)) \equal{} 2x \plus{} f( \minus{} x)$  $ \implies$
$ f( \minus{} x) \equal{} \minus{} x$  $ \implies$  $ f(x) \equal{} x$  QED.

if f(x)=x+C, it is still right.  :lol: there is a problem in your post, could you find it  :P

$ f(f(x)\plus{}y)\equal{}f(f(f(y))\minus{}x)\plus{}2x$
$ f(x)\equal{}x\plus{}c$  $ \implies$
$ x\plus{}y\plus{}2c\equal{}2x\plus{}f(y\plus{}2c\minus{}x)\equal{}2x\plus{}y\plus{}3c\minus{}x$  $ \implies$
$ 2c\equal{}3c$  $ \implies$  $ c\equal{}0$   :D

[quote=cnyd] $ f(c) \equal{} \minus{} c \equal{} f( \minus{} c)$[/quote]
(4th line in post 2)
How does $f( \minus{} c)$ get involved in this equation?

because of line 1
$f(-c)=f(f(c))=f(c)=-c$

Sorry if this seems foolish but I can’t get my head around how $f(-c) = f(f(c))$ :(

[quote=Nathanisme]$ x \equal{} y \equal{} 0$  $ \implies$  $ f(f(0)) \equal{} f(f(f(0)))$
$ f(0) \equal{} c$  $ \implies$  $ f(c) \equal{} f(f(c))$
$ y \equal{} 0$  $ \implies$  $ f(f(x)) \equal{} 2x \plus{} f(f(c) \minus{} x)$
$ x \equal{} f(c)$  $ \implies$  $ f(c) \equal{} 2f(c) \plus{} c$  $ \implies$  $ f(c) \equal{} \minus{} c $
I suppose this part  is ok then
because $f(0)=c$ and by first line $f(f(0))=f(f(f(0))) \implies f(c)=f(f(c))$
now by last line$f(c)=-c$
So we have $f(c)=f(f(c))=_2f(-c)$ because $f(c)=-c$ in $=_2$[/quote]

Thanks :thumbup:


A real-valued function $ f$ on $ \mathbb{Q}$ satisfies the following conditions for arbitrary $ \alpha, \beta \in \mathbb{Q}:$

[b](i)[/b] $ f(0) \equal{} 0,$
[b](ii)[/b] $ f(\alpha) > 0 \text{ if } \alpha \neq 0,$
[b](iii)[/b] $ f(\alpha \cdot \beta) \equal{} f(\alpha)f(\beta),$
[b](iv)[/b] $ f(\alpha \plus{} \beta) \leq f(\alpha) \plus{} f(\beta),$
[b](v)[/b] $ f(m) \leq 1989$ $ \forall m \in  \mathbb{Z}.$

Prove that \[ f(\alpha \plus{} \beta) \equal{} \max\{f(\alpha), f(\beta)\} \text{ if } f(\alpha) \neq f(\beta).\]

[quote="benimath"]The function $f$ is a valuation on $\Bbb{Q}$. [url=http://en.wikipedia.org/wiki/Ostrowski's_theorem]Ostrowski's theorem[/url] says that the only nontrivial valuations on $\Bbb{Q}$ are $|\cdot|_p$ for some prime $p$ or $p=\infty$ (or are equivalent to these).
The condition we need to prove says in fact that the valuation is non-archimedian. A necessary and suficient condition for that is that for all integers we have $f(n) \leq 1$, but this is obvious from the condition $f(m) \leq 1989$, because if we would have $f(k)>1$ for some $k$ then $f(k^n)=(f(k))^n$ and can get arbitrarily large.[/quote]
maybe $f(m) \leq 1989$ same $f(n) \leq 1$ ,I think

@pco do you have any idea?
I mean without using the ostrowski theorem


function
topology
real analysis
real analysis theorems

[b]DEF[/b]: Suppose $ X$ is a topological space, $ x_0$ is a point in $ X$ and $ f : X \to R \cup$ {$ \minus{}\infty,\plus{}\infty$} is an extended real-valued function. We say that $ f$ is upper semi-continuous at $ x_0$ if for every $ \epsilon > 0$ there exists a neighborhood $ U$ of $ x_0$ such that $ f(x) \le f(x_0) \plus{} \epsilon$ for all $ x$ in $ U$.
 The function $ f$ is called upper semi-continuous if it is upper semi-continuous at every point of its domain.

 Why is
 '$ f$ is upper semi-continuous if and only if {$ x \in X : f(x) \ge a$} is an closed set for every $ a \in \mathbb{R}$.'?

$ \{x \in X : f(x) \ge a\}$ is closed if and only if $ S : \equal{} \{x \in X : f(x) < a\}$ is open.

Fix $ x_0 \in S$. By upper semi - continuity, there exists a neighborhood$ U_0$ of $ x_0$ such that for all$ x\in U_0$, $ f(x) < f(x_0) \plus{} a \minus{} f(x_0) \equal{} a$ (that is, we chose $ a \minus{} f(x_0)$ as our $ \epsilon$). This shows that an arbitrary element of $ S$ has a neighborhood contained in $ S$, hence $ S$ is open.


search
inequalities
induction
inequalities unsolved

Let $ n\ge1$ Is it true that for any 2n+1 positive real numbers $ x_{1}, x_{2},...x_{2n\plus{}1}$ that:
$ \frac{x_{1}x_{2}}{x_{3}}\plus{}\frac{x_{2}x_{3}}{x_{4}}\plus{}...\plus{}\frac{x_{2n\plus{}1}x_{1}}{x_{2}}\ge x_{1}\plus{}x_{2}\plus{}...\plus{}x_{2n\plus{}1}$ With equality iff all the $ x_{i}$ are equal????

I am sure it was already solved by Valentin Vornicu many years ago.
Please, search in sections : Solved inequalities and Proposed inequalities.
Sorry, but I don't remember the title of the topic.

Perhaps an induction on n might work? For example... it is trivial to prove for n=1, as it is true by Cauchy.... :maybe:  :maybe:

well, I believe I have discovered a good solution.
multiply by x_1...x_(2n+1) and arrive at a cyclic inequality
sigma(x_1^2*x_2^2*x_4*...*x_(2n+1))>=sigma(x_1^2*x_2*...*x_2n+1)
we find appropiate weights to use Am-Gm
let us put x_1^2*x_2^2*x_4*...*x_(2n+1)=y_1
x_2^2*x_3^2*x_5*...*x_1=y_2, etc.
we try to find m_1,..,m_2n+1 such that
by applying the weighted Am-GM inequality m_1y_1+...+m_ny_n>=(m_1+...+m_n)x_1^2*x_2*...*x_2n+1
but such weights are easy to find
(ex)make m_i=f_(n+1-i) for i=1 to 2n, and m_(2n+1)=f_(n+1)-1 where f_n denotes the fibonnacci sequence)
and it is solved.

[color=red]Edit: Mr. Vasc says the inequality does not hold for n>=2 therefore this solution must be wrong
can anyone show me a counter example?[/color]

The inequality holds only for $ n\equal{}1$. :P

could anyone show me counterexample for n>=2?
I must have made a mistake somewhere but I can't see it :( :(

[color=red]Edit:I think Mr. Manlio was referring to http://www.mathlinks.ro/viewtopic.php?t=2827 which is a different inequality.
It took me a lot of time to search through every post of Mr. Valentin Vornicu's in Inequalities Solved and Proposed and Own. :oops: [/color]

[quote="ywl"]could anyone show me counterexample for n>=2?
I must have made a mistake somewhere but I can't see it :( :( [/quote]
For $ x_{1}\rightarrow\infty$, we must have

$ \frac{x_{2}}{x_{3}}\plus{}\frac{x_{2n\plus{}1}}{x_{2}}\ge 1$,
which is true only for $ n\equal{}1$. :D

I'm not sure I understand that... why does this disprove it?


inequalities
trigonometry
inequalities proposed

Let $ 0<x<\pi/2$. Show that:

$ x\minus{}\sin{x}<\equal{} \frac{1}{6}x^3$

[quote="Nemion"]Let $ 0 < x < \pi/2$. Show that:

$ x \minus{} \sin{x} < \equal{} \frac {1}{6}x^3$[/quote]

Verry old and easy inequality  :wink:

Put $ f(x) \equal{} x \minus{} \sin{x} \minus{} \frac {1}{6}x^3$

$ f''(x) \equal{} \sin x \minus{} x < \equal{} 0$ => $ f'(x)<\equal{}f'(0)\equal{}0$ => $ f(x) < \equal{} f(0) \equal{} 0$ ,q.e.d!


I just started learning about logarithms and I ran into a few problems I had trouble doing. Please post your solution and not just the answer. Any help is appreciated.

1. $ \log_{2}6 \cdot \log_{3}6 \minus{} (\log_{2}3\plus{}\log_{3}2)$

2. $ \log_{2}14 \cdot \log_{7}14 \minus{} (\log_{2}7\plus{}\log_{7}2)$

3. $ \log_{3}75 \cdot \log_{5}45 \minus{}(\log_{3}25 \plus{}\log_{5}9)$

The basic idea here was to use the base conversion formula and when you evaluate the log, it should come out to something clean like an integer.

Thanks in Advance

[quote="mathemagician1729"]

1. $ \log_{2}6 \cdot \log_{3}6 \minus{} (\log_{2}3 \plus{} \log_{3}2)$

2. $ \log_{2}14 \cdot \log_{7}14 \minus{} (\log_{2}7 \plus{} \log_{7}2)$

3. $ \log_{3}75 \cdot \log_{5}45 \minus{} (\log_{3}25 \plus{} \log_{5}9)$
[/quote]

What you have to do first is to put the different bases into one.

1. Put the base $ 3$ into $ 2$,and by formula we have

$ \log_{2}6 \cdot \log_{3}6 \minus{} (\log_{2}3 \plus{} \log_{3}2)$

$ \equal{}(\log2\plus{}\log3)\frac{\log6}{\log3}\minus{}(\log3\plus{}\frac{1}{\log3})$

$ \equal{}(1\plus{}\log3)\frac{1\plus{}\log3}{\log3}\minus{}\log3\minus{}\frac{1}{\log3}\equal{}2$

No2,and No3 can be solved in the same way.


geometry
geometric transformation
reflection
analytic geometry
graphing lines
slope
Pythagorean Theorem

One old target and one old sprint

1. [img]http://img134.imageshack.us/img134/8383/pointpqb5.png[/img]

2.
[img]http://img134.imageshack.us/img134/8271/1872zb0.png[/img]

1.

[hide]
The shortest chord that goes through P is 24 units long.
Use the Pythagorean Theorem to find this.
Next the diameter is the longest chord to go through P.
this is 30 units long.
There are two chords of lengths 25 through 29 each.  This is because of reflections across the 30 unit long diameter.
So there are 2*1+2*5 = 12 chords in total.
I think that that is the answer.
[/hide]
2.

[hide]
Drawing these lines shows that from the origin to the vertical line, there are 12 units.
Make this the base of the triangle.
That means that the height has to be 1872*2/12 = 312
The line goes 312 units vertically and 12 units horizontally.
This means that the slope of the line is 312/12 or 26.
[/hide]


function
calculus
derivative
inequalities
real analysis
real analysis solved

There's something I don't get:

f is continuous in x for all x in I, so f(x+y)-f(x)->0 as y->0, and f(x-y)-f(x)->0 as y->0, adn by adding these we get that the relation in the hypothesis is true for all x and for all continuous functions f, so we don't need it. This means that we must prove that for all continuous functions defined on I the set of pts of derivability is dense in I, but this isn't true (see, for example, Weierstrass' example, f:R->R which is everywhere continuou and nowhere derivable).

Maybe there's some mistake somewhere.

[quote="xxxxtt"]Let I an open interval and f:I->R a continuous function such that for every x in I we have limy->0 from f(x+y)+f(x-y)-2f(x) is 0.
Prove that the set of points in wich f is derivable is dense in I[/quote]
The set of points in which f is differentiable...Assume f is differentiable...prove this set of points Im(x)={f(x)|f is differentiable+conditions above} is dense in I. Weierstrass example does not qualify in this case. You are assuming f is differentiable in the first place. That's my interpretation of his statement.

I don't quite understand what you mean. Are you saying that the text of the problem is Ok? Doesn't differentiable mean that the derivative exists (not necessarily finite)? In Weierstrass' function the derivative doesn't exist for any x, meaning that [f(x+h)-f(x)]/h has no limit when h->0 for any x. Weierstrass' function is continuous in every point but nowhere differentiable, so it's still a counterexample.

What made me think there's something wrong in the first place isn't Weierstrass' example, but the fact that we are given a condition we don't need.

So how about it xxxxtt? Is the problem stated correctly?

Sorry, it was something missing in the text of the problem.The correct text is:
Let I an open interval and f:I->R a continuous function such that for every x in I we have limy->0 from[ f(x+y)+f(x-y)-2f(x)]/y is 0.
Prove that the set of points in wich f is derivable is dense in I

Here's a start:

For simplicity, we may assume that f isn't derivable in any point in I. Then it's clear that there are no local maximas, so the function must be strictly monotonic, so we may assume WLOG that f is strictly increasing on I.

This might work:

We have 2 cases:

[b]Case I[/b]: There exist 3 collinear points on the graph of the function, say (x,f(x)), (y,f(y)) and (z,f(z)) in this order, from left to right. Then take t a point in [y,z] s.t. (f(t)-f(x))/(t-x) is the maximum or minimum (whichever isn't reached only in y or z) (when t describes the segm [y,z]). Assume it's the maximum. This clearly exists because fx(t)=(f(t)-f(x))/(t-x) is continuous on [y,z]. It's then easy to show that f is derivble in t, and the derivative is (f(t)-f(x))/(t-x), because for h really close to 0 we have (f(t+h)-f(t))/h<=(f(t)-f(x))/(t-x), because otherwise we would have (f(t+h)-f(x))/(t+h-x)>(f(t)-f(x))/(t-x), which is impossible, because (f(t)-f(x))/(t-x) was the maximum (we assume h is small enough s.t. t+h, t-h are both in [y,z]). In the same way, (f(t)-f(t-h))/h>=(f(t)-f(x))/(t-x), and since lim{h->0}(f(t+h)-f(t))/h-(f(t)-f(t-h))/h=0, it means that when h->0, both of those tend to (f(t)-f(x))/(t-x).

[b]Case II[/b]: Ther are no 3 collinear pts on the graph of f, in which case it's easy to show that the function id either concave or convex, and it's easy to find a point of derivability from the monotony of the functions of the type fx described in Case I.

The inequalities in Case I may not all be obvious, but look at the graph of f (use angles and chords and stuff like that).


I applied for PROMYS this year and I mailed my application in late May...

But so far no response from PROMYS at all! I wonder how long it takes from them to grade an application?

The deadline is set at June 1. Which is quite late compared to Mathcamp. Hence I'm very afraid PROMYS may adopt "first come, first serve" principle, which means admission is automatically closed after all available places are filled. Rather than processing all applicaions after the dealine passed. (quite a number of you indicated that you have gotten into it)

Maybe my application already got rejected because of relatively late arrival date. I would be very disappointed if this were to come true. PROMYS seems my first choice and I've given up several other opportunities. I sacrificed a lot for it and tens of deadly hours were spent on the problems...

Hopefully I will be accepted.

PROMYS does have rolling admissions, so many people are probably accepted already. Late May wasn't really all that long ago, but if you really want to know I think you can try contacting them and asking about it.

Yeah man I'm basically in the same position as you. I sent in my app by email on June 1st (sad isnt it?) right before I left for ARML. I've emailed them and phoned them since then, but I haven't gotten any response (I left them a phone number). I'll post back here if I they contact me. My guess is that they're still debating, they probably get like a big pile of apps near the end from procrastinators like us. Good luck to you on getting in!

Yeah, I just called the promys office (617) 353-2563, The director was there. She said they were really really bogged down with apps, and that they hadn't made a decision on them yet.

I drove mine in on June 1st, and they still accepted me. You should receive a email if you do.
Hope you guys make it!


[quote="fireemblem13"]Recently, I've noticed that when I type in Microsoft Word, a buch of paragraph symbols, dots, and arrows appear that are really annoying. Does anyone know what I'm talking about? anyone have any advice? thx.[/quote]

Maybe you pressed a button which symbol is like a P looking at the other side, with a T after it. Look in a bar (up part, right side)


Expand $ (\text{cos} 13^{\cir})(\text{cos} 28^{\cir})$.

[hide="Solution"]Note that $ \cos \alpha\cos \beta\equal{}\frac{\cos(\alpha\plus{}\beta)\plus{}\cos(\alpha\minus{}\beta)}{2}$.

Hence, $ \cos 28\cos 13\equal{}\frac{\cos41\plus{}\cos15}{2}$.

I don't think you would get a prettier result than that.[/hide]

Thank you, FanFan.

And it's probably the answer since that's what it said in the back of the book. :P  (I just wanted to know how to get there.)


geometry
incenter
invariant
ratio
geometry unsolved

Problem: Let I be the incenter of triangle ABC.  Let the incircle of ABC touch the sides BC, CA, and AB at K, L, and M, respectively.  The line through B parallel to MK meets the lines LM and LK at R and S, respectively.  Prove that angle RIS is acute.

My idea: Let the midpoint of KM be X.  It's pretty clear that we are done if we show triangle XRS is self-polar.  Since RS is the polar of X, AX is the polar of R, and CX is the polar of S, it suffices to show that A,X,S or C,X,R are collinear (equivalent conditions).

But I can't seem to find a nice, non-bashful way to do this.  Can anyone help out?

$ MXK \parallel RBS \Longrightarrow$ $ Z \equiv MXK \cap RBS$ is at infinity. Project the incircle $ (I)$ to a circle and the line $ RBS$ to infinity. This makes the projected lines $ A'B' \parallel C'B'$ and $ M'K'$ diameter of the projected circle. $ Z' \in R'B'S'$ remains at infinity and the invariant cross ratio $ (Z, X, M, K) \equal{} (Z', X', M', K') \equal{} \minus{} 1$ keeps $ X'$ as the midpoint of $ M'K',$ i.e., makes $ X'$ center of the projected circle. The projected $ \triangle K'L'M'$ becomes right at $ L'$ and $ A'X' \parallel L'K',$ $ C'X' \parallel L'M'$ become its midlines $ \Longrightarrow$ the intersections $ A'X' \cap L'K',$ $ C'X' \cap L'M'$ are at infinity, i.e., on $ R'B'S'.$

See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=17895][color=red][b]here[/b][/color][/url] more proofs.


let $ x,y,z \geq 0$ show that: x+y+z=2;
prove that:
                  $ xy\plus{}yz\plus{}zx\leq \frac{92}{81}\plus{} \frac{2}{3}xyz$

We have: $ a^2\plus{}b^2\plus{}c^2\plus{}2abc\plus{}1\geq2(ab\plus{}bc\plus{}ca)$, a,b,c are positive real numbers.
The equality holds iff: $ a\equal{}b\equal{}c\equal{}1$
Put $ x\equal{}\frac{2a}{3},y\equal{}\frac{2b}{3},z\equal{}\frac{2c}{3}$.
It follows that: $ \frac{9(x^2\plus{}y^2\plus{}z^2)}{4}\plus{}\frac{27xyz}{4}\plus{}1\geq\frac{9(xy\plus{}yz\plus{}xz)}{2}$
or $ \frac{9(x\plus{}y\plus{}z)^2}{4}\plus{}\frac{27xyz}{4}\plus{}1\geq9(xy\plus{}yz\plus{}xz)$
or $ xy\plus{}yz\plus{}zx\leq\frac{10}{9}\plus{}\frac{3}{4}xyz$.
Addition we have: $ xyz\leq\frac{8}{27}$.


HELP needed
$ ABCDE$ is a regular penagon and $ P$ is a point  on its circumcircle between $ A$ & $ E$. show that
$ PA \plus{}PC \plus{} PE \equal{}PB \plus{}PD$
ISHFAQ


function
real analysis
real analysis unsolved

Let $\alpha,\beta>0$, $f(x)=x^{\alpha}cos{\frac{1}{x^\beta}}$ for $0<x\leq1,f(0)=0$. Prove that
a).$f\in{V[0,1]}$ iff $\alpha\ge\beta$
b).$f\in {Lip_{1}([0,1])}$ only when $\alpha>\beta+1$
c).if $\alpha<\beta+1$ then $f\in {Lip_{\gamma}([0,1])}$ where $\gamma=\frac{\alpha}{\beta+1}$
(Notice: we denote by the $V([a,b])$ the set of all functions that's have bounded variation)

[quote="eugene"] a).$f\in{V[0,1]}$ [/quote]
What does it mean ?!

 :cool:

[quote="Diogene"][quote="eugene"] a).$f\in{V[0,1]}$ [/quote]
What does it mean ?!
[/quote]
It means $f$ is bounded variation on $[0,1]$.
I think most books use $\mbox{BV}([0,1])$.

Maybe, I just wrote notation, that used in my book


Complete the square of each quadratic expression.  Then graph each function using the technique of shifting.

HINT GIVEN IN MATH BOOK:

If f(x) = x^2 - 2x, then f(x) = (x^2 - 2x + 1) - 1 = (x - 1)^2 - 1
================================================

QUESTIONS:

(1) f(x) = x^2 + 4x + 2

(2) f(x) = x^2 - x + 1

To graph these equations:

Example with 1:

$(x+2)^{2}-2=0$

Take the basic graph of $f(x)=x^{2}$, then move it 2 units to the left (Add to X, go west), then down 2 units.

All the answers are simplistic and to the point.

Thanks!

[quote="lotrgreengrapes7926"]Interval wants the graph.[/quote]

Yea but I can't find a suitable program online where you input an equation and it graphs it for you.

Here's the best I could find for online graphing programs:

[url=http://www.fooplot.com/]an online graphing calculator[/url]

Thank you b-flat for the link.

By the way, are you a musician?

To call yourself b-flat, I think about flat notes and sharps.

I play piano...

I'm not really sure why I choose the name b-flat originally...

b-flat plays piano?

I play classical guitar.

How about that?

Two musicians here.


Find the minimum value of $ (x\plus{}y)(y\plus{}z)$, where x, y, z are positive real numbers such that $ xyz(z\plus{}y\plus{}z)\equal{}1$

Is it $ xyz(z\plus{}y\plus{}z)\equal{}1$ or maybe $ xyz(x\plus{}y\plus{}z)\equal{}1$

It is $ xyz(x\plus{}y\plus{}z)\equal{}1$.  That's what makes this problem slightly more interesting, because the condition given is symmetric, but the expression you are asked to minimise isn't.

[quote="Machodog"]Find the minimum value of $ (x \plus{} y)(y \plus{} z)$, where x, y, z are positive real numbers such that $ xyz(z \plus{} y \plus{} z) \equal{} 1$[/quote]
By AM-GM , we have:
$ 1\equal{}xyz(x \plus{} y \plus{} z)\equal{}xz(y^2 \plus{} yz \plus{} yx)\le (\frac {y^2 \plus{} xy \plus{} yz \plus{} zx}{2})^2\equal{}(\frac {(x \plus{} y)(y \plus{} z)}{2})^2$
Therefore $ (x \plus{} y)(y \plus{} z)\ge 2$


A brief introduction: I'm a pure math major but I recently took an option on my degree that requires me to take a (relatively) long string of differential equations courses out of the department of applied math. I'm starting these next semester (which is earlier than I thought!) and I'm worried that I really don't have the physics background (having never taken physics :( ). Here is the course description to the first course:

Introduction to Differential Equations
Physical systems which lead to differential equations (examples include mechanical vibrations, population dynamics, and mixing processes). Dimensional analysis and dimensionless variables. Solving linear differential equations: first- and second-order scalar equations and first -order vector equations. Laplace transform methods of solving differential equations.

All that being said, I'd be very appreciative if you guys could direct me to some resources to catch up on my "world problem" skills.

just read some standard books of physics--like  Resnick Halliday
the initai few chaps of mechanics--it shud help u... :lol:


inequalities
trigonometry
geometry theorems
geometry

[b]Weitzenbock Inequality[/b]

Given an arbitrary triangle of side lengths $a,b,c$ we have

$a^{2}+b^{2}+c^{2}\geq ab+bc+ca \geq 4 \sqrt{3}|\triangle ABC|$.

Can anyone prove this? I have a proof but it takes ages so I was wondering if anyone has an elegant proof.

First ineq is trivial, For second ineq : Use $2S=bc\sin{A}=...$ and $\sum\sin{A}\leq \frac{3\sqrt{3}}{2}$.

[quote="N.T.TUAN"]$\sum\sin{A}\leq \frac{3\sqrt{3}}{2}$.[/quote]

How did you get this? Is it some well known trig relationship?

[quote="BanishedTraitor"] Is it some well known trig relationship?[/quote]
Yes , It is well know.

Mind providing a proof? An outline would be fine.

$\sin$ is concave on $(0,\pi)$ so $\sum \sin(A) \leq 3 \sin(\frac{\sum A}{3})=3\sin(\frac{\pi}{3})=\frac{3\cdot\sqrt{3}}{2}$


If $ a,b,c>0$ and $ a\plus{}b\plus{}c\equal{}3$, then
$ \sum\frac{1}{\sqrt{4a^2\plus{}bc\plus{}7a\plus{}8}}\geq\frac{2}{3}$

[quote="filomen"]If $ a,b,c > 0$ and $ a \plus{} b \plus{} c \equal{} 3$, then
$ \sum\frac {1}{\sqrt {4a^2 \plus{} bc \plus{} 7a \plus{} 8}}\geq\frac {2}{3}$[/quote]

The inequality is  holds


conics
parabola
analytic geometry
graphing lines
slope
geometry
circumcircle

Four distinct points $A,\ B,\ C, D$ on the parabola $y=x^{2}$ are concyclic.
Let $\alpha,\ \beta$ be the abcissae of $A,\ B$ respectively.Express the slope of the line $CD$ in terms of $\alpha,\ \beta.$

Denote the given parabola $y = x^{2}$ as $\mathcal P.$ Slope of AB is $m = \frac{\alpha^{2}-\beta^{2}}{\alpha-\beta}= \alpha+\beta.$ Let antiparallel to AB (with slope -m) through $C \in \mathcal P$ meet the parabola again at D' and let $\gamma, \delta$ be abscissas of C, D'. Slope of CD' is

$-(\alpha+\beta) =-m = \frac{\gamma^{2}-\delta^{2}}{\gamma-\delta}= \gamma+\delta$

Then slope of BC is $n = \beta+\gamma$ and slope of D'A is $n' = \delta+\alpha =-(\beta+\gamma) =-n,$ so that BC, D'A are also antiparallel, therefore A, B, C, D' are concyclic. But circumcircle of $\triangle ABC$ meets $\mathcal P$ again at a unique point, hence $D' \equiv D$ are identical.


geometry
analytic geometry
geometric transformation
reflection
geometry unsolved

Give $ \bigtriangleup{ABC}$, find $ M$ in triangle so that $ \ MA^2 \equal{} MB^2 \plus{} MC^2$ and $ \angle{AMC} \equal{} 90^{\circ}$

one of the problems of the book : An Intro. To The Modern Geometry Of The Triangle And The Circle,Nathan Altshiller Court,
says that we can find the locus of the points $ M$ for which we have $ AM^2 \equal{} BM^2 \plus{} CM^2$.
if we solve this question then the rest is trivial...
then we find the intersection(s) of this locus with the circle with diameter $ CA$ and we r done!
the part in which this question appears it explains the medians property and prove this well-know property of the triangle in which for all points $ M$ in the plane of $ \Delta ABC$ :
$ MA^2 \plus{} MC^2 \plus{} MB^2 \equal{} AG^2 \plus{} BG^2 \plus{} CG^2 \plus{} 3MG^2$ where $ G$ is the centroid of $ \Delta ABC$.
that will be kind if someone proves this problem that i mentioned without $ (x,y)$ coordinates....
thx in advance :)
my solution:
we use $ (x,y)$ coordinates and define $ A: (0,a);B: (b,0);C: (c,0)$ where $ a,c \ge 0 \ge b$.
assuming $ M: (x,y)$ the above formula give us :
$ 2MA^2 \minus{} 3MG^2 \equal{} const \implies 2(x)^2 \plus{} 2(y \minus{} a)^2 \minus{} 3(x \minus{} \frac {b \plus{} c}{3})^2 \minus{} 3(y \minus{} \frac {a}{3})^2 \equal{} const$

$ \implies 2x^2 \plus{} 2y^2 \plus{} 2a^2 \minus{} 4ay \minus{} 3x^2 \minus{} 3(\frac {b \plus{} c}{3})^2 \plus{} 2x(b \plus{} c) \minus{} 3y^2 \minus{} 3(\frac {a}{3})^2 \plus{} 2ay \equal{} const$

$ \implies \minus{} x^2 \minus{} y^2 \plus{} \frac {5a^2}{3} \minus{} 2ay \plus{} 2x(b \plus{} c) \minus{} \frac {(b \plus{} c)^2}{3} \equal{} const$

$ \implies \minus{} (x^2 \plus{} (b \plus{} c)^2 \minus{} 2x(b \plus{} c)) \minus{} (y^2 \plus{} a^2 \plus{} 2ay) \equal{} const \minus{} \frac {2(b \plus{} c)^2}{3} \minus{} \frac {8a^2}{3}$

$ \implies \minus{} (x \minus{} (b \plus{} c))^2 \minus{} (y \plus{} a)^2 \equal{} const \minus{} \frac {2(b \plus{} c)^2}{3} \minus{} \frac {8a^2}{3}$

$ \implies (x \minus{} (b \plus{} c))^2 \plus{} (y \plus{} a)^2 \equal{} \minus{} const \plus{} \frac {2(b \plus{} c)^2}{3} \plus{} \frac {8a^2}{3}$

note that the left side is representing a circle with center coordinates $ ((b \plus{} c), \minus{} a)$ and radius $ \sqrt { \minus{} const \plus{} \frac {2(b \plus{} c)^2}{3} \plus{} \frac {8a^2}{3}}$ also it's obvious that the radius of this circle is constant since $ a,b,c$ are constant.
so we have proven that the locus of the point $ M$ is a circle centered at $ ((b \plus{} c), \minus{} a)$ with a specific radius which is
$ const \equal{} AG^2 \plus{} BG^2 \plus{} CG^2 \equal{} \frac {1}{3}(AB^2 \plus{} BC^2 \plus{} CA^2)$

$ \equal{} \frac {1}{3}((a^2 \plus{} b^2) \plus{} (b \minus{} c)^2 \plus{} (a^2 \plus{} c^2))$

$ \implies \minus{} const \plus{} \frac {2(b \plus{} c)^2}{3} \plus{} \frac {8a^2}{3} \equal{} \frac {b^2 \plus{} c^2}{3} \plus{} 2a^2 \plus{} 2bc$

so the radius of this circle is $ \sqrt {\frac {b^2 \plus{} c^2}{3} \plus{} 2a^2 \plus{} 2bc}$
and the center of this circle can be found as follows:
we can assume that $ CH > BH$ where $ H$ is the projection of $ A$ on $ BC$.then u take the point $ T$ on $ BC$ such that $ T$ belongs to the [u] segment [/u] $ BC$ and $ CT \equal{} BH$.then we draw a line parrallel to $ BC$ passing through $ A$ and a line perpendicular to $ BC$ at $ T$.if the intersections of this two line is $ L$ then the reflection of $ L$ wrt $ BC$ called $ L'$ will be the center of the circle on which $ M$ varies.( i mean $ L'B \equal{} LB,L'C \equal{} LC,L' \neq L$)
so finally the center of this circle is $ L'$ ,as defined,and its radius is $ \sqrt {\frac {b^2 \plus{} c^2}{3} \plus{} 2a^2 \plus{} 2bc}$.
note that in order to find all $ M$ asked in the topic we must consider both cases $ BH > CH$ and $ BH < CH$.
now we can easily find the intersection(s) of this circle with the circle with diameter $ CA$.the rest is a matter of computations.

$ N$ is midpoint of $ BC$ and $ P$ is the reflection of $ A$ about $ N.$ $ MN$ is common M-median of $\triangle AMP$ and $\triangle MBC.$

$ MB^2 \plus{} MC^2 \equal{} 2MN^2 \minus{} \frac {_1}{^2}a^2 \ , \ MN^2 \equal{} \frac {_1}{^2}(MA^2 \plus{} MP^2) \minus{} AN^2.$

Combining both expresions yields

$ MB^2 \plus{} MC^2 \minus{} MA^2 \equal{} MP^2 \minus{} 2AN^2 \plus{} \frac {_1}{^2}a^2$

$ MB^2 \plus{} MC^2 \minus{} MA^2 \equal{} MP^2 \minus{} 2 \left [\frac {_1}{^2}(b^2 \plus{} c^2) \minus{} \frac {_1}{^4}a^2 \right] \plus{} \frac {_1}{^2}a^2$

$ MB^2 \plus{} MC^2 \minus{} MA^2\equal{} MP^2 \minus{} b^2 \minus{} c^2 \plus{} a^2.$

Hence, locus of $ M$ that satisfies $ MA^2 \equal{} MB^2 \plus{} MC^2$ is a circumference $ (P)$ centered at the reflection of $ A$ about the midpoint of $ BC$ with  radius $ \sqrt {b^2 \plus{} c^2 \minus{} a^2},$ as long as $ b^2 \plus{} c^2 \ge a^2.$ Now, draw this circumference and cut it with the circumference with diameter $ AC$ to get the wanted points $ M.$


Round 1 results are now available.  After logging in, visit My Scores to see your scores and comments.  As this is our first time using this software for parsing the scores, please let us know if you see anything amiss.

Also, check out the leaders and the statistics pages!  (Note: Not all participants are included in the statistics yet.)

Props to Benjamin Lee.

Wow, 5 commended solutions.  Nice job.

Heh, nice... I was kinda surprised that I got 4 commended solutions.  Well anyways, I'm happy. :)

Yes!  25 points!

The USAMTS feels a lot more like a math competition now that it has a leaderboard.

[quote]Heh, nice... I was kinda surprised that I got 4 commended solutions. Well anyways, I'm happy. Smile[/quote]

Congrats!

[quote]Congrats![/quote]

Indeed.  And congratulations to the AoPS team and all the graders for giving us our results so quickly.  Thanks!

I got 22 with one commendation. I guess that's alright, but I know I could have gotten a 25. I tried to learn LaTeX and type up my solutions in one day, which was really stupid.

25 with 1 commended solution!

I'm very happy;  I expected to get about 22, with no commendations.[/hide]

Wow, getting 5 points for the process actually isn't so difficult! Just have to make sure not to skil any steps! Maybe not commended but still good.

I got 25 with 3 commended. I'm thinking other rounds are going to be harder (I still haven't come up with a proof for one of the round 2 problems) so I shouldn't get too comfortable here. I thought some of the round 1 problems were a little too easy.

Haha in problem 4 when I wrote up it up on MS Word it auto formated the beginning of my equation to a capital N instead of leaving it as n standing for the number of sides(-1 point). Curse you MS Word!

25, 1 and 2 were commended.  Yippee!

Edit: The official solutions aren't up yet, are they?

Yes, 25!!  Yesterday I only had one commended solution(#4), but today I have two commended solutions(#3, #4)!

[quote="mathfanatic"]Indeed.  And congratulations to the AoPS team and all the graders for giving us our results so quickly.  Thanks![/quote]

Exactly.  I think both the AoPS staff and students have done an excellent job!  :)

Im very happy with my 25/2 commendations.  But, I looked at Beta's solutions to the problems on which I was commended (#1,2) and it was nearly exactly the same.  Conversely, I look at other solutions for 3 through 5 and find that I did not get commended while somebody else did.  I used Benjamin Lee's solution on #4...

I do admit it is hard to pick some solutions specially when they all seem the same, not to mention the separate graders.

Now, I know how to get commended.  Before, I thought the generalization thing just took up space.  But now, I realize I can play with it and write it in my solution!  That's the funnest thing on Earth.

Thank you very much AoPS and graders!

[quote="beta"]Yeah, getting "commended" does seem random...[/quote]

Not entirely random, but there is not a perfect algorithm for it.  Some graders are stingier than others with their commendations - they have much stricter criteria for the grading.  A lot of commendation comes down to style, as well.  Solutions which are really well written (i.e. sufficient explanation, but not overly wordy - a tough balance) potentially get commended.  Novel approaches that are effectively conveyed are as well.


find a expression of genaral term to these sequence
$ x_{1}\equal{}4 ,x_{n}\equal{}3x_{n\minus{}1}\plus{}\frac{n\plus{}2}{n\plus{}1}(n\ge1)$
$ y_{1}\equal{}2,y_{2}\equal{}3,y_{3}\equal{}4,y_{4}\equal{}5,y_{n\plus{}4}\equal{}2y_{n\plus{}3}\minus{}y_{n\plus{}1}\minus{}\frac{1}{4}y_{n}(n\ge1)$
and i appreciate to someone who give the general solution to 2 kinds of sequence

hello for 1) Mathematica gives the following term
$ \left\{\left\{x(n)\to \frac{1}{6} \left(\minus{}2 \Phi
   \left(\frac{1}{3},1,n\plus{}2\right)\plus{}2 3^n\plus{}2 3^{n\plus{}2} \log
   \left(\frac{3}{2}\right)\minus{}3\right)\right\}\right\}$
Sonnhard.

can you explain more detailed Dr Sonnhard Graubner please!
what does the sign ${ \Phi(\frac{1}{3},1,n+2})$ mean

[quote="hoangvn.fix"]find a expression of genaral term to these sequence
$ x_{1} \equal{} 4 ,x_{n} \equal{} 3x_{n \minus{} 1} \plus{} \frac {n \plus{} 2}{n \plus{} 1}(n\ge1)$
$ y_{1} \equal{} 2,y_{2} \equal{} 3,y_{3} \equal{} 4,y_{4} \equal{} 5,y_{n \plus{} 4} \equal{} 2y_{n \plus{} 3} \minus{} y_{n \plus{} 1} \minus{} \frac {1}{4}y_{n}(n\ge1)$
and i appreciate to someone who give the general solution to 2 kinds of sequence[/quote]

Before I tried to solve, please could you kindly confirm that is is a real problem (and not a problem transformed by yourself) and you are sure that it exists a solution.

Thanks in advance  for your answer.

dear mr pco actually the first one is a problem which i transform but i would be glad to see your solution for counting approximately
the second one is from my teacher  and i am sure this is the real problem thanh you for helping

$ x_{1} \equal{} 4$

 $ x_{n} \equal{} 3x_{n \minus{} 1} \plus{} \frac {n \plus{} 2}{n \plus{} 1}$

divide by $ 3^n$ , you have telescoping

$ \frac{x_{n}}{3^n}\minus{}\frac{x_{n\minus{}1} }{3^{n\minus{}1}}\equal{}\frac {n \plus{} 2}{(n \plus{} 1)3^n}\equal{}\frac{1}{3^n}\plus{}\frac{1}{3^n(n\plus{}1)}$

I let you continue..

$ y_{1} \equal{} 2,y_{2} \equal{} 3,y_{3} \equal{} 4,y_{4} \equal{} 5,y_{n \plus{} 4} \equal{} 2y_{n \plus{} 3} \minus{} y_{n \plus{} 1} \minus{} \frac {1}{4}y_{n}(n\ge1)$

Consider the characteric equation $ t^4\minus{}2t^3\plus{}t\plus{}1/4\equal{}0$ etc...

hello, the solution of the second equation is terrible, i let Mathematica do this stupid work, here is the solution
\left\{\left\{y(n)\to \frac{4 \left(2911 \sqrt{3}
   \left(\frac{1}{2}-\frac{\sqrt{3}}{2}\right)^n n-5042
   \left(\frac{1}{2}-\frac{\sqrt{3}}{2}\right)^n n+151316 \sqrt{3}
   \left(\frac{1}{2}+\frac{\sqrt{3}}{2}\right)^n n-262087
   \left(\frac{1}{2}+\frac{\sqrt{3}}{2}\right)^n n-43456 \sqrt{3}
   \left(\frac{1}{2}-\frac{\sqrt{3}}{2}\right)^n+75268
   \left(\frac{1}{2}-\frac{\sqrt{3}}{2}\right)^n-162180 \sqrt{3}
   \left(\frac{1}{2}+\frac{\sqrt{3}}{2}\right)^n+280904
   \left(\frac{1}{2}+\frac{\sqrt{3}}{2}\right)^n\right)}{\left(\sqrt{3}
   -1\right) \left(7 \sqrt{3}-12\right) \left(11 \sqrt{3}-19\right)
   \left(41 \sqrt{3}-71\right)}\right\}\right\}
this formula seems to be to big for the editor, you can copy and paste it for yourself.
Sonnhard.

[quote="hoangvn.fix"]find a expression of genaral term to these sequence
$ x_{1} \equal{} 4 ,x_{n} \equal{} 3x_{n \minus{} 1} \plus{} \frac {n \plus{} 2}{n \plus{} 1}(n\ge1)$[/quote]

And then

[quote="hoangvn.fix"]dear mr pco actually the first one is a problem which i transform [/quote]

So I will not look at it. Post it in "open questions" forum and surely some mathlinkers will study it.

[quote="hoangvn.fix"]find a expression of genaral term to these sequence
$ y_{1} \equal{} 2,y_{2} \equal{} 3,y_{3} \equal{} 4,y_{4} \equal{} 5,y_{n \plus{} 4} \equal{} 2y_{n \plus{} 3} \minus{} y_{n \plus{} 1} \minus{} \frac {1}{4}y_{n}(n\ge1)$
[/quote]

And then

[quote="hoangvn.fix"]the second one is from my teacher  and i am sure this is the real problem thanh you for helping[/quote]

The characteristic equation is $ x^4\minus{}2x^3\plus{}x\plus{}\frac 14\equal{}(x^2\minus{}x\minus{}\frac 12)^2$ so two double roots $ \frac{1\plus{}\sqrt 3}2$ and $ \frac{1\minus{}\sqrt 3}2$

So the general term : $ y_n \equal{} (an \plus{} b)\left(\frac {1 \plus{} \sqrt 3}2\right)^n$ $ \plus{} (cn \plus{} d)\left(\frac {1 \minus{} \sqrt 3}2\right)^n$

And you just have to find $ a,b,c,d$ solving the system :

$ y_1 \equal{} 2 \equal{} (a \plus{} b)\left(\frac {1 \plus{} \sqrt 3}2\right)$ $ \plus{} (c \plus{} d)\left(\frac {1 \minus{} \sqrt 3}2\right)$

$ y_2 \equal{} 3 \equal{} (2a \plus{} b)\left(\frac {1 \plus{} \sqrt 3}2\right)^2$ $ \plus{} (2c \plus{} d)\left(\frac {1 \minus{} \sqrt 3}2\right)^2$

$ y_3 \equal{} 4 \equal{} (3a \plus{} b)\left(\frac {1 \plus{} \sqrt 3}2\right)^3$ $ \plus{} (3c \plus{} d)\left(\frac {1 \minus{} \sqrt 3}2\right)^3$

$ y_4 \equal{} 5 \equal{} (4a \plus{} b)\left(\frac {1 \plus{} \sqrt 3}2\right)^4$ $ \plus{} (4c \plus{} d)\left(\frac {1 \minus{} \sqrt 3}2\right)^4$

The system is
$ (a \plus{} b)(1 \plus{} \sqrt 3)$ $ \plus{} (c \plus{} d)(1 \minus{} \sqrt 3) \equal{} 4$
$ (2a \plus{} b)(1 \plus{} \sqrt 3)^2$ $ \plus{} (2c \plus{} d)(1 \minus{} \sqrt 3)^2 \equal{} 12$
$ (3a \plus{} b)(1 \plus{} \sqrt 3)^3$ $ \plus{} (3c \minus{} d)(1 \plus{} \sqrt 3)^3 \equal{} 32$
$ (4a \plus{} b)(1 \plus{} \sqrt 3)^4$ $ \plus{} (4c \plus{} d)(1 \minus{} \sqrt 3)^4 \equal{} 80$

$ (a \plus{} b)(1 \plus{} \sqrt 3) \plus{} (c \plus{} d)(1 \minus{} \sqrt 3) \equal{} 4$
$ (2a \plus{} b)(2 \plus{} \sqrt 3) \plus{} (2c \plus{} d)(2 \minus{} \sqrt 3) \equal{} 6$
$ (3a \plus{} b)(5 \plus{} 3\sqrt 3) \plus{} (3c \plus{} d)(5 \minus{} 3\sqrt 3) \equal{} 16$
$ (4a \plus{} b)(7 \plus{} 4\sqrt 3) \plus{} (4c \plus{} d)(7 \minus{} 4\sqrt 3) \equal{} 20$

Setting $ u \equal{} a \plus{} c$, $ v \equal{} b \plus{} d$, $ w \equal{} (a \minus{} c)\sqrt 3$ and $ t \equal{} (b \minus{} d)\sqrt 3$, this becomes :

$ u \plus{} v \plus{} w \plus{} t \equal{} 4$
$ 4u \plus{} 2v \plus{} 2w \plus{} t \equal{} 6$
$ 15u \plus{} 5v \plus{} 9w \plus{} 3t \equal{} 16$
$ 28u \plus{} 7v \plus{} 16w \plus{} 4t \equal{} 20$

$ u \plus{} v \plus{} w \plus{} t \equal{} 4$
$ 3u \plus{} v \plus{} w \equal{} 2$
$ 12u \plus{} 2v \plus{} 6w \equal{} 4$
$ 24u \plus{} 3v \plus{} 12w \equal{} 4$

$ u \plus{} v \plus{} w \plus{} t \equal{} 4$
$ 3u \plus{} v \plus{} w \equal{} 2$
$ 6u \plus{} 4v \equal{} 8$
$ 12u \plus{} 9v \equal{} 20$

$ u \plus{} v \plus{} w \plus{} t \equal{} 4$
$ 3u \plus{} v \plus{} w \equal{} 2$
$ 6u \plus{} 4v \equal{} 8$
$ v \equal{} 4$

$ u \equal{} \minus{} \frac 43$
$ v \equal{} 4$
$ w \equal{} 2$
$ t \equal{} \minus{} \frac 23$

$ a \equal{} \frac {\sqrt 3 \minus{} 2}3$
$ b \equal{} \frac {18 \minus{} \sqrt 3}9$
$ c \equal{} \minus{} \frac {2 \plus{} \sqrt 3}3$
$ d \equal{} \frac {18 \plus{} \sqrt 3}9$

So the general term : $ y_n \equal{} \left(\frac {\sqrt 3 \minus{} 2}3n \plus{} \frac {18 \minus{} \sqrt 3}9\right)\left(\frac {1 \plus{} \sqrt 3}2\right)^n$ $ \minus{} \left(\frac {2 \plus{} \sqrt 3}3n \minus{} \frac {18 \plus{} \sqrt 3}9\right)\left(\frac {1 \minus{} \sqrt 3}2\right)^n$

And I think your teacher gives some rathr ugly problems  :huh:

$ x_{1} \equal{} 4$

 $ x_{n} \equal{} 3x_{n \minus{} 1} \plus{} \frac {n \plus{} 2}{n \plus{} 1}$

divide by $ 3^n$ , you have telescoping

$ \sum_{k\equal{}2}^{n}(\frac {x_{k}}{3^k} \minus{} \frac {x_{k \minus{} 1} }{3^{k \minus{} 1}}) \equal{} \sum_{k\equal{}2}^{n}\frac {k \plus{} 2}{(k \plus{} 1)3^k})$

$ \frac {x_{n}}{3^n} \minus{} \frac {x_{1} }{3}) \equal{}\sum_{k\equal{}2}^{n}\frac {k \plus{} 2}{(k \plus{} 1)3^k})$

this gives x(n) in term of n


Say you tried to solve a question, but you didn't arrive at the answer.  Should you submit your solution still, or just not submit anything?

I'd submit the work. Graders are known to be lenient with giving out partial credit. You'll get some partial credit if you show you understood what the problem was asking, etc.

Unless you want to mail it, the solution deadline for the online upload is passed.

But yes, I think you get credit for as much work as you do.

I'm pretty sure you can get 1 point at least with some nontrivial progress.


superior algebra
superior algebra solved

[color=green]Suppose that R and S are rings with identity. Let (r, s) in R × S. Show that
(r, s) is unit in R × S <=> r is a unit in R and s is a unit in S.

Also, if (r, s) is a unit in R×S, show that inverse of (r, s) = (inverse of r, inverse of s)
(Hint: It is efficient to prove the last statement and the implication “<=” in the first statement at the
same time.)[/color]

um....can anyone help!!! urgent!!! Thanks very much  :blush:

Because I want to move this exhilarating problem, I will post its incredible solution.
$(r,s)$ is a unit iff there exist $r'\in R$ and $s'\in S$ such that $(r,s)(r',s')=(r',s')(r,s)=(1,1)$. This is equivalent to $(rr',ss')=(r'r,s's)=(1,1)\Leftrightarrow rr'=r'r=1$ and $ss'=s's=1$ which in turn is equivalent to $r$ and $s$ being units.


algebra
polynomial
induction
algebra unsolved

I usually do problems in my head for fun (or when I don't have pen and paper) and I came across this problem in a book which did not have a solution. It took only about an hour to figure out all the details in my head, so it shouldn't be too hard for any of you :)

Let $ p(x)$ be a polynomial with coefficients in $ Z$ such that $ p(n)>n$ for all $ n$ where $ n$ is a positive integer, and for every $ N$ there exists a term in the sequence $ 1,p(x),p(p(x)),...$ which is divisible by N. Prove that $ p(x)\equal{}x\plus{}1$

There is a three line solution, believe me :P

[quote="rofler"]
There is a three line solution, believe me :P[/quote]

So, why did you post it in the unsolved problems section... :wink:

Pierre.

Because who checks the solved section? Lol  :roll:

Not rigourous, don't want to :P but is the idea to show that since p(x) = x+1, and there is a sequence 1,p(x),p(p(x), ....,
then well.. trying those first examples yeilds 1,x+1,x+2,(...), x+3,x+4,... etc.
Only question is is the sequence up to a finite iteration count? or indefinitely... because then this question is quite simple :P

Something quick:
Since p(n) > n for all n > 0 (element of) Intergers
P(n) has a constant. Otherwise p(1) cannot be greater than 1.
Therefore p(x) = x^n + c, for some n, c.
For some N to divide into it evenly, let n = 1, c = 1, and therefore ... yea straight forward.

just an idea:

for $ n\in\mathbb{N}\ arbaitrary$ if we take $ \sigma \in S_n$ such that $ P(i)\equiv \sigma(i)(mod\ n)$ then $ \exists (a_1,a_2,...,a_n)$ permutation of $ (1,2,....,n)$ such that $ \sigma(a_i)\equal{}a_{i\plus{}1}$ (with $ a_{n\plus{}1}\equal{}a_1$)
and $ P(i)\equal{}\sigma(i)\plus{}nQ_n(i)$ for some $ Q_n\in Z[X]$

Euclid's Turtle: it is an arbitrary polynomial, not necessarily x^n+c

[hide="solution"]
Let the first term be $ x_{0}\equal{}1$ then $ x_{1}\equal{}p(x)$..... Note that inductively, $ x_{k}\geq k\plus{}1$ for all positive k.

Let m be the first number s.t. $ x_{m} \geq m\plus{}2$ then consider the terms mod $ x_{m}\minus{}1$

By construction, no previous number is greater than or equal to $ x_{m}\minus{}1$ Therefore, none of the previous terms is congruent to 0 mod $ x_{m}\minus{}1$

$ x_{m}$ is congruent to 1 mod $ x_{m}\minus{}1$, which was what the first term was congruent to. Therefore, $ x_{k}$ cycles mod $ x_{m}\minus{}1$ going 1,.......,$ x_{m\minus{}1},x_{m}$=1,...... and will consequently never hit 0.

Therefore, $ k\plus{}1\le x_{k}<k\plus{}2$ so $ x_{k}\equal{}k\plus{}1$ and then the result readily follows by induction (or whatever you want to do, the problem is solved).[/hide] Any comments on the soln?

This problem with solution was posted earlier:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=107368[/url].

Rofler, what do you mean in regards to my sol'n it is an arbitrary polynomial ? You want a_1x^n + a_2x^n-1+...+a_0 ?? Skipped that part, made it easy with just some x^n +c  if so..
And yea with your solution, I don't get the firstline :( Where did you come up with x_k >= k+1 from? Induction isn't very good because you have to know what you're looking for exactly to use it... a little bias I guess.


n a positive integer
compare the following numbers..
$ [ \sqrt(4n+1)]; [ \sqrt(4n+2)]; [ \sqrt(4n+3)];  [ \sqrt(n)+ [ \sqrt(n+1)] $

[quote="prowler"]It easy to prove
√(4n+1) ≤ √n + √(n+1) ≤ √(4n+2) < √(4n+3)
You can easy to see that 4n+2 and 4n+3 can't be perfect squares.
So if you suppose
  4n+1 ≥ t2 then 4n+3 < (t+1)2
Now you can take integer part:
t ≤ [√(4n+1)] ≤ [√n + √(n+1)] ≤ [√(4n+2)] < [√(4n+3)] < (t+1).
it means all of them are equal.             q.e.d.[/quote]
i have just a remarque about the solution
can't we also use:
√(4n+1) ≤ √n + √(n+1) ≤ √(4n+2) < √(4n+3)
and because Floor(x) is increasing for every x in R ( so in R+ too)
we have
[√(4n+1)] ≤ [√n + √(n+1)] ≤ [√(4n+2)] < [√(4n+3)]    ?


calculus
derivative
trigonometry
logarithms
calculus computations

(1) Let $ y \equal{} \alpha \cos x \plus{} 2 \minus{} (\cos x)\ln \frac {1 \plus{} \cos x}{1 \minus{} \cos x}\ (0 < x < \pi)$, where $ \alpha$ is a constant.

Find $ \frac {\frac {d^2y}{dx^2} \plus{} \frac {\cos x}{\sin x}\cdot \frac {dy}{dx}}{y}$.

(2) Let $ x \equal{} e^t\cos t,\ y \equal{} e^t\sin t$, find $ \frac {d^2y}{dx^2}$.

hello, your first result is
$ \frac {\frac {d^2y}{dx^2} \plus{} \frac {\cos x}{\sin x}\cdot \frac {dy}{dx}}{y}\equal{}\minus{}2$.
Sonnhard.

hello, for your second problem i have found
$ \frac{d^2y}{dx^2}\equal{}\frac{2}{1\minus{}\sin(2t)}$.
Sonnhard.

The first is correct, but the second is incorrect. :(

hello, sorry, i have made a mistake, the result is
$ \frac{1}{2\sin^2(t)}$.
Sonnhard.

For the second one:

$ \frac{d^2y}{dx^2}\equal{}\frac{d}{dx}\left(\frac{dy}{dx}\right)\equal{}\frac{d}{dx}\left(\frac{dy}{dt}\cdot \frac{dt}{dx}\right)\equal{}\frac{d}{dt}\left(\frac{dy}{dt}\cdot \frac{dt}{dx}\right)\cdot \frac{dt}{dx}$

We have

$ \frac{dy}{dt}\equal{}e^t (\sin(t)\plus{}\cos(t))$

$ \frac{dt}{dx}\equal{}\frac{1}{e^t (\cos(t)\minus{}\sin(t))}$

so

$ \frac{d}{dt}\left(\frac{dy}{dt}\cdot \frac{dt}{dx}\right) \equal{} \frac{d}{dt}\left(\frac{\sin(t)\plus{}\cos(t)}{\cos(t)\minus{}\sin(t)}\right)\equal{}\frac{2}{(\cos(t)\minus{}\sin(t))^2}$

and finally

$ \frac{d^2y}{dx^2} \equal{} \frac{2}{(\cos(t)\minus{}\sin(t))^2} \cdot \frac{dt}{dx} \equal{} \frac{2}{e^t(\cos(t)\minus{}\sin(t))^3}$.

Hope it's OK.


Runners A and B practice in a round track going in opposite directions. Runner A takes 40 minutes to go once around the track. The two begin at the same time and meet every 15 minutes. How long does B take to go once around the track?..

[quote="AmericanPi"]after fifteen minutes of running, A meets B. A has run 15/40=3/8 of the track. thus B has run 5/8 of the track in 15 minutes, or 1/8 every three. 3 x 8 = 24 minutes for B to run one lap.[/quote]
Quite correct. I think this has been used in a mathcounts before, so it's not really a question for everyone, it's a question that's already been given to everyone. :huh:

i don't think so...maybe a problem similar to this appeared in mathcounts...but not exactly...