greedy
algorithm
IMO
IMO 2003
independent sets
combinatorics
IMO Shortlist

Let $A$ be a $101$-element subset of the set $S=\{1,2,\ldots,1000000\}$. Prove that there exist numbers $t_1$, $t_2, \ldots, t_{100}$ in $S$ such that the sets  \[ A_j=\{x+t_j\mid x\in A\},\qquad j=1,2,\ldots,100  \]  are pairwise disjoint.

Notice that if instead of taking an arbitrary t_{k+1}, you always take the smallest t_{k+1} that is not forbidden, you will only have to make sure that t_{k+1} is distinct from all t_i+a_j-a_l with a_j>a_l.  This is true because the values with a_j<a_l are all <=t_k, and so have been forbidden at some earlier step.

This gives you ceiling(1000000/5051) = 198 t_i's.

However, Fedor's proof also shows that you can choose 100 t_i's such that the sets are parwise disjoint even when taken mod 10^6.

Let A = {a1 < a2 < ... < a101}.

Draw an undirected graph and put a vertex between i and j iff the sets (A+i) and (A+j) are disjoint . The graph will have 10^6 vertices . For an arbitrary vertex x to be joined with y , x-y must not be one of the numbers ai - aj (i<>j) , which are 101*100 numbers . So the vertex x has degree at least 10^6 - 101*100 .

But this means that the graph has at least 10^6(10^6 - 101*100)/2 edges. The problem asks to prove that there is an 100-clique in the graph. But by Turan's theorem , there is a k-clique in a graph with n vertices iff the number of edges is strictly greater than :

M(n,k) = (k-2)/(k-1) * (n^2 - r^2)/2 + r*(r-1)/2

where we have taken r to be the remainder of n when divided by k-1.

In our case n=10^6 , k=100 and r=1. A simple calculation shows that the number of vertices is greater than M(10^6,100)+1 and thus we are done.

P.S. : in an IMO paper , should one prove Turan's theorem or not ? I guess so ..

[quote="Fedor Petrov"]Assume that we have already found t_1, t_2, ..., t_k (k<=99) and are searching for t_{k+1}. We can not take t_{k+1} only of form
t_i+a_j-a_l, where 1<=i<=k and a_j and a_l are elements of A.
So, we have k*101*100 forbidden values of t_{k+1}, which correspond to a_j<>a_l, and k forbidden values which correspond to a_j=a_l. So, at most 99*101*100+99=1000000-1 forbidden values, and at least 1 admissible.[/quote] so it is right to assume that for every K there exists a number such that \lim  \lambda

[quote="Fedor Petrov"]Assume that we have already found $\{t_k\}_{1=k}^{99}$ and are searching for $t_{k+1}$. We can not take $t_{k+1}$ only of form
$t_i+a_j-a_l \ | \ 1\leq i\leq k \ \ a_j, a_l \in A$.
So, we have $k*101*100$ forbidden values of $t_{k+1}$, which correspond to $a_j<>a_l$, and $k$ forbidden values which correspond to $a_j=a_l$. So, at most $99*101*100+99=1000000-1$ forbidden values, and at least $1$ admissible.[/quote]
[quote="Alison"]Notice that if instead of taking an arbitrary $t_{k+1}$, you always take the smallest $t_{k+1}$ that is not forbidden, you will only have to make sure that $t_{k+1}$ is distinct from all $t_i+a_j-a_l \ | \ a_j>a_l$.  This is true because the values with $a_j<a_l$ are all $\leq t_k$, and so have been forbidden at some earlier step.

This gives you $\lceil 1000000/5051 \rceil = 198 t_i's$.

However, Fedor's proof also shows that you can choose $100 t_i's$ such that the sets are parwise disjoint even when taken $mod 10^6$.[/quote]
[quote="Guest"]Let $A = {a_1 < a_2 < ... < a_{101}}$.

Draw an undirected graph and put a vertex between $i, j$ iff the sets $(A+i)$ and $(A+j)$ are disjoint . The graph will have $10^6$ vertices . For an arbitrary vertex $x$ to be joined with $y$ , $x-y$ must not be one of the numbers $ai - aj (i<>j)$ , which are $101*100$ numbers . So the vertex $x$ has degree at least $10^6 - 101*100$ .

But this means that the graph has at least $10^6(10^6 - 101*100)/2$ edges. The problem asks to prove that there is an $100-clique$ in the graph. But by Turan's theorem , there is a $k-clique$ in a graph with $n$ vertices iff the number of edges is strictly greater than :

$M(n,k) = (k-2)/(k-1) * (n^2 - r^2)/2 + r*(r-1)/2$

where we have taken $r$ to be the remainder of $n$ when divided by $k-1$.

In our case $n=10^6 , k=100 , r=1$. A simple calculation shows that the number of vertices is greater than $M(10^6,100)+1$ and thus we are done.

P.S. : in an IMO paper , should one prove Turan's theorem or not ? I guess so ..[/quote]

This problem has been proposed by Carlos Gustavo Tamm de Araujo Moreira, from Brazil.

http://www.artofproblemsolving.com/Wiki/index.php/IMO_Problems_and_Solutions

It already says so in the Wiki. If you know an author which isn't listed there, please contribute!

May be my solution is wrong but it seems to me that we just need $|S|\geq 99\binom{101}{2}+100$. Please help me if you find any bug in my solution
SOLUTION
[hide]

Let $x+t_i\in A_i$ and $y+t_j\in A_j$
Then $x+t_i=y+t_j\Leftrightarrow|x-y|=|t_i-t_j|$
So, there are at most $\binom{101}{2}$ values of $|x-y|$ with distinct $x,y\in A$
Now choose,$t_1=1$
and Make a set $S_1\subset S$ such that $S_1$ doesn't contain 1 and any element $K$ satisfying $K-1=|x-y|$ for any   distinct $x,y\in A$ But $S_1$ contain all other elements in $S$. As there is at most $\binom{101}{2}$ such K, $|S_1|\geq |S|-\binom{101}{2}-1$

Now for any $i\geq1$ Choose the smallest element from $S_i$ for  $t_{i+1}$
and again
Make a set $S_{i+1}\subset S_i$ such that $S_{i+1}$ doesn't contain $t_{i+1}$ and any element $K$ satisfying $K-t_{i+1}=|x-y|$ for any   distinct $x,y\in A$ But $S_{i+1}$ contain all other elements in $S_{i}$. As there is at most $\binom{101}{2}$ such K, $|S_{i+1}|\geq |S_i|-\binom{101}{2}-1$

This algorithm ensures that all $t_i$ for all $1\leq i\leq 100$ satisfy the condition and the interesting fact is for choosing all $t_i$ we need $|S|\geq 99\binom{101}{2}+100$ far less then $10^6$
[/hide]

[quote="Mahi"][quote="Dragonboy"]
Make a set $S_{i+1}\subset S_i$ such that $S_{i+1}$ doesn't contain $t_{i+1}$ and any element $K$ satisfying $K-t_{i+1}=|x-y|$ for any   distinct $x,y\in A$
[/quote]

In here, you are just striking out the elements such that $K-t_{i+1}=|x-y|$, but the case remains where $K-t_{i+1}=-|x-y|$, which can also satisfy $y+t_i=x+t_j$ instead of $y+t_j=x+t_i$. So the limit is (about) doubled and it reaches near $10^6$, which can be achieved with a little more strict bounding.

Although, nice approach with algorithmic way :)[/quote]
I think i have mentioned that $t_{i+1}$ is the smallest element in $S_i$ and any other element $K$ in $S_i$ is greater than $t_{i+1}$.So, there is no $K$ satisfying $K-t_{i+1}=-|x-y|$

[quote="Mahi"]Yes, if you choose $t_i$'s in increasing sequence, then it reduces to $t_i-t_j=y-x$ where $i>j$ which implies $y>x$ and thus the strategy is optimized by two :)[/quote]
I'm not understanding what you're saying. Let me make it more clear for you (As much as i can)
After choosing all $t_i$ by the algorithm, for the sake of contradiction , let's assume there exists $t_i>t_j$ and $x>y$ such that $t_i-t_j=x-y$.
But It's not possible since we've banished such $t_i$ when we choose $t_{j+1}$ (According to algorithm).
Is it clear now?

In my last post, I just shared my opinion about the algorithm. I understood it earlier. It was clear to me after I noticed the part "greatest in the set $S_i$". Thanks for your concern.
By the way, something similar was also told by Alison in a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=342#p342]previous post[/url].

Let each element of $ S $ be the vertex of a graph where two vertices $ u, v $ are connected by an edge if and only if the sets $ A + u $ and $ A + v $ are disjoint.  Consider an arbitrary vertex $ v $.  Since the $ |A + v| = 101 $ the maximum number of vertices $ w $ such that sets $ A + v $  and $ A + w $ are not disjoint is $ 100 * 101 $.  Therefore every vertex of the graph has degree at least $ 10^6 - 100*101 - 1. $  Therefore the graph has at least $ \frac{10^6(10^6 - 100*101 - 1)}{2} $ edges.  It suffices to show that this graph contains $ K_{100} $ as a subgraph.

Now, by Turan's Theorem, the maximum number of edges a graph with $ 10^6 $ vertices that does not contain $ K_{100} $ may contain is obtained when the graph is a complete $ 99 $-partite graph with $ 98 $ independent sets of size $ 10101 $ and $ 1 $ independent set of size $ 10102 $.  It is easy to compute that this graph has $ \binom{98}{2}10101^2 + 98 \cdot 10101 \cdot 10102 = 494949494949 $ edges.  But since $ \frac{10^6(10^6 - 100*101 - 1)}{2} = 494949500000 > 494949494949 $ we have the desired result.

Pretty similar to 2020 N1 (you select values and then eliminate possibilities over and over).

Suppose that we have selected $t_1,t_2,\dots,t_n$ such that they satisfy the condition, where $1\le n\le 99$. We show that we can select $t_{n+1}$ to satisfy the condition. This will imply the conclusion, since it's easy to show that we can select only $t_1$ and satisfy the condition.

Suppose that the elements in $A$ are $a_1,a_2,\dots,a_{101}$. Then since $A_i$ and $A_{n+1}$ are disjoint for some $1\le i\le n$ we have that
\[t_{n+1}\neq t_i+a_m-a_n\]
for all $1\le m,n\le 101$. Then each set $A_i$ will remove at most $101^2-100$ possibilities for $t_{n+1}$ (we subtract $100$ because the $m=n$ case has overlap). But now it suffices to show that $n(101^2-100)<1000000$ which is trivial since $99(101^2-100)=999999<1000000$. We are done. $\blacksquare$

If $a_i+t_j=a_k+t_l$, then $a_i-a_k=t_i-t_j$. This can be rephrased as the pairwise differences of the $t$ sequence cannot overlap with the pairwise differences of the $a$ sequence.

Note that the $t$ sequence must also consist of pairwise distinct elements. We will also have the $t$ sequence never overlap with the $a$ sequence. Therefore, for $t_1$, we have $999899$ choices. For $t_2$, our restrictions are that $t_2$ is not in the $a$ sequence, $t_2\neq t_1$, and that $t_2-t_1$ is not one of the pairwise differences of the $a$ sequences, which there are 5050 of. Therefore, there are at least $1000000-101-1-5050$ choices for $t_2$. For $t_3$, it is the same except $t_3-t_2$ and $t_3-t_1$ both cannot be a pairwise difference of $a$, which removes at most $5050\cdot2$ choices, so $t_3$ has at least $1000000-101-2-5050\cdot2$ choices. It follows that $t_n$ has at least $1000000-101-(n-1)-5050(n-1)$ choices, and this is positive for all $n\leq100$, so we are done.

Suppose $A = \{x_1, x_2, \ldots, x_{101} \}$ where $x_1 < x_2 < \ldots < x_{101},$ and define $$D = \{x_n - x_m \mid 1 \le m < n \le 101\}.$$ In addition, we assume $t_1 < t_2 < \ldots < t_{100}$ and let $$T = \{t_q - t_p \mid 1 \le p < q \le 100\}.$$ Clearly, it suffices to ensure $D$ and $T$ are disjoint sets.

[b][color=#f00]Claim:[/color][/b] WLOG, set $t_1 = 1$, and assume we always pick the least possible value for $t_2, t_3, \ldots, t_{100}$. Then, $t_j \le 5051j - 5050$.

[i]Proof.[/i] Define the [i]kill set[/i] of $t_j$ as $$K_j = \{t_j + r \mid r \in D \}.$$ Because $t_1 < t_2 < \ldots < t_{100}$, we know it's necessary and sufficient to ensure $$t_j \not\in U_{j} = \{K_1, K_2, \ldots, K_{j-1}, \{t_1, t_2, \ldots, t_{j-1} \} \}.$$ Now, observe $$\vert U_j \vert \le \vert D \vert \cdot (j-1) + (j - 1)$$ $$\le \binom{101}{2} (j - 1) + (j - 1) = 5051j - 5051$$ so $$t_j = \vert U_j \vert + 1 \le 5051j - 5050$$ as required. $\square$

It follows that $$t_{100} \le 5051 \cdot 100 - 5050 = 500,050 < 10^6$$ so we're done. $\blacksquare$

We take the maximal $k$, such that the differences in the set $\{t_1,t_2, \cdots, t_k\}$ are disjoint from those in $A$. Thus, for any other $x$, ($10^6-k$ possibilities), there is some $t_i$ ($k$ possibilities), such that $x=t_i+d$, where $d$ is a difference of two elements of $A$ ($101 \cdot 100$ possibilities), so we have that $10^6-k \leq 101 \cdot 100 \cdot k$, which rearranges to $k \geq 100$ and we are done.

Take set $B$ of all positive pairwise differences between elements in $A$. We have $$|B|\le\binom{101}{2}=5050.$$ Now, the problem is equivalent to choosing distinct $t_1,t_2,\dots,t_{100}$ such that no two have a positive difference in $B$. We claim that the following procedure works:

Choose $t_1=1$. Then, choose $t_2$ to be the lowest number that works(doesn't collide with $t_1$ with the differences), then choose $t_3$ to be the smallest that doesn't collide with $t_1,t_2$, etc.

First, we can see that the placement of $t_1$ brought at most $5051$ "bad numbers" that no other $t$ can use($5050$ from differences in $B$ because all other $t$ must be larger than $t_1$, and $1$ from distinct from $t_1$). Then, $t_2$ also brings at most $5051$ "bad numbers" because nothing lower than it can be chosen anymore($t_1$ is still valid because $t_2$ was chosen to not be a bad number). Repeating this $99$ times, we need at least $$99\cdot5051+1$$ numbers for this algorithm to work, and that number is clearly less than $10^6$, QED.

The condition that the sequences are disjoint is equivalent to the pairwise differences of the elements of $A$ being disjoint from the pairwise differences of the elements of $T$.\\

There are 5050 pairwise differences in the set $A$. We start with $10^6$ different options for the next element of $T$. Each time we pick one element to put in $T$, we will of course lose that element as a future choice, as well as losing at most 5050 other choices on each side of the chosen element (corresponding to the chosen element $\pm$ the 5050 pairwise differences of $A$), so we lose at most 10101 choices per iteration. Thus, after 99 iterations, we still have at least $$10^6-10101\cdot99=1$$ remaining choices for the final element, so we are done.

The condition is equivalent to $t_i + a_j \neq t_k + a_l$ for distinct $i,k$. We can pick $t$ constructively. After picking a value of $t$, we lose at most $101\cdot 100 + 1$ choices for future $t$, since we can pick $j\neq l$ in $101\cdot 100$ ways and if $j=l$, then one value of $t$ is lost. But after picking $99$ values, we still have at least $10^6 - 99(101\cdot 100 + 1) = 1$ value left to choose from, so we are done.

Solved on [b]OTIS Excerpts.[/b]
Very standard and straightforward counting problem. There will only be no way to add another $t_j$, from where there are $10^6$ choices, into the set T that contains all $t_k$s if the value is already in the set, or if there would then be some overlap between two sets $A_i$ and $A_j$; that is, $x+t_i=y+t_j$, for some distinct x and y in A. There are k of the first type; of the second type, |A|(|A|-1) ways for x and y, and |T| ways for there to exist some t_i such that there would be some overlap. Hence we have $$k+|A|(|A|-1)|T|=k+101*100k\ge 10^6.$$ It is now easily checked (by, for example, estimating the LHS=10^4k+k) that k=99 is too small, hence $\boxed{k\ge 100}$, as desired. $\blacksquare$

Suppose that we have chosen a maximal number $k$ of $t_i$. This means that for all $10^6-k$ numbers $m$ in $S$ not among the $t_i$, there exists $t_i$ and $x, y \in A$ such that $t_i + x = m + y$. On the other hand, the number of triplets $(t_i, x, y)$ is bounded above by $10100k$, so it follows that $10100k \geq 10^6 - k$, enough to imply $k \geq 100$ (rounding up).

We can see that for $a_x, a_y \in A$. We want no pairs such that $a_x - a_y = t_i - t_j$.

We choose our $t_i$ greedily and by induction. Firstly we choose $t_1 =1$.

For our induction step we let $t_{n+1}$ be the first value such that $t_{n+1} - t_j \neq a_x - a_y$ for all $j \in [1, n]$. We can see that there are $\binom{101}{2}$ values of $a_x - a_y$, and there are $n$ values of $j$. Thus the number of values that cannot be obtained by $t_{100}$ is bounded above by $\binom{101}{2} \cdot 100 \leq 10^6$. $\blacksquare$

The "disjoint" condition is satisfied iff $a_w + t_x \neq a_y + t_z$ for all valid $(w,x,y,z)$; in other words, $|a_w - a_y| \neq |t_x - t_z|$ for all valid $w, x, y$ and $z$. There are up to $\binom{101}{2} = 5050$ distinct absolute differences between two elements of $A$.

Consider just arithmetic sequences $t_1, t_2, \dots, t_{100}$ with common difference $d$, where $1 \leq d \leq 10101$. The possible values of $|t_x - t_z|$ are $\{ d, 2d, \dots, 99d \}$; as $d$ varies from $1$ to $10101$, there must be some (in fact, at least $5051$) values of $d$ for which none of $\{ d, 2d, \dots, 99d \}$ are the absolute difference between any two elements of $A$, so we just pick $d$ to be any of those values.

Let $k$ be the largest size of such a set $t$. Let $t_1, t_2, \cdots, t_k$ be the numbers, and let $A_1, A_2, \cdots, A_{101}$ be the numbers. Let $n$ be the number of numbers of the form $x - y + t_j$, where $x, y  \in A$. Then, we must have the inequality $10^6 - k \le n,$ since if $10^6 - k > n,$ we have at least one more valid choice to add to $t$. But, in turn, $n \le 101 \cdot 100 \cdot k$, which gives the desired inequality.

EDIT: This is incorrect. I'll edit in the correct solution sometime soon.
EDIT 2: Done.

Suppose that we pick $n$ numbers $t_1, t_2, \ldots, t_n$ satisfying the problem conditions for maximal $n$. Then for $x \in S$ and $x \neq t_k$ for some $k$, we must have $x - t_{\ell} = a_i - a_j$ or $x = t_{\ell} + a_i - a_j$ for some integers $1 \le \ell \le n$ and $1 \le i \neq j \le 100$. For a fixed $t_\ell$, there are $101 \cdot 100 = 10100$ ways to choose $(a_i, a_j)$, which is enough to imply that $10100n \ge 10^6 - n$, or $n \ge 100$.

We can prove this by contradiction. Suppose there is a $101$-element subset $A$ of $S = \{1,2,\ldots,1000000\}$ such that, for any choice of $100$ numbers $t_1, t_2, \ldots, t_{100}$ from $S$, there exist indices $i$ and $j$ ($1 \leq i < j \leq 100$) such that $A_i$ and $A_j$ have at least one common element.

Let $B_1, B_2, \ldots, B_{100}$ be the sets $A_1, A_2, \ldots, A_{100}$ respectively. For each $j$, the set $B_j$ can be expressed as $B_j = A + t_j = \{x + t_j \mid x \in A\}$.

Now, consider the union of all $B_j$'s for $j = 1, 2, \ldots, 100$:

\[ B = B_1 \cup B_2 \cup \ldots \cup B_{100} \]

Since $B_j = A + t_j$, the union $B$ is essentially the set of all elements in $S$ that can be obtained by adding an element from $A$ to any of the $100$ chosen $t_j$'s.

Now, by our assumption, there must exist two indices $i$ and $j$ ($1 \leq i < j \leq 100$) such that $B_i \cap B_j \neq \emptyset$. This implies that there exists an element $x$ in $A$ such that $x + t_i = x + t_j$, which means $t_i = t_j$.

However, this contradicts our assumption that any choice of $100$ numbers $t_1, t_2, \ldots, t_{100}$ must yield distinct sets $B_j$. Therefore, our assumption must be false, and there must be a $101$-element subset $A$ with the desired property.

Keep on choosing $t_i$ until we get stuck(we can't pick any more $t$) for some $i = k$. This implies that for all $t \in S$, it has already been picked or that for some $a$, $b$, $t_i \in S$ we have $t = b + t_i - a$.  So then there are at most $k \cdot 10100$($101$ choices for $b$ and $100$ for $a$) values for $t$, so $k \geq 100$ as desired.


Please try  solvingthe following problem without inversion:
Let ABC be a triangle. Suppose that  the circle through C tangent to AB at A and the circle through B tangent to AC at A have different radii, and let D be their second intersection. Let E be the point on the ray AB such that AB=BE. Let F be the second intersection of the ray CA with the circle through A, D E. Prove that AF=AC.

Sorry for my latex, i forgot how to use it... lol  ;)

Well, let $F$ be the point on the ray $CA$ such that $FA=AC$, then we want to prove that $DEFA$ is cyclic. It is true  iff  $<FDE=<FAD$ but $<FAD=<ADB$, so its is cyclic  iff  triangle $FDA$ is similar to triangle $BED$ iff $BE/AF=BD/AD$ but $BE/AF=AB/AC$

And $BD/AD=AB/AC$ is true due the fact that triangle $ADB$ is similiar to triangle $ADC$

Note: This can be done simply by angle chasing. $<BAD=<ACB$ and $<ABD=DAC$


graph theory
combinatorics unsolved
combinatorics

A [i]king[/i] in a tournament is a player $ a$ such that for every player $ b$ either $ a$ beats $ b$ or there exists $ c$ such that $ a$ beats $ c$ and $ c$ beats $ b$.  Describe the set of integers $ n$ such that there exists a tournament with $ n$ players in which every player is a king.

Other recent problems involving kings in tournaments:
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=143571]Existence[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=159625]If 2 then at least 3[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=162421]If only 1, then undefeated[/url]

[hide]
Let us translate this into graph theory language.  Any tournament can be represented uniquely as a complete, directed graph.  We want to know for what $ n$ is it possible to have a complete, directed graph such that from any vertex, there is either a direct path or a two-edge path to every other vertex.  The answer: all $ n \ge 3$.

First we show that if it is possible for some $ n$, then it is possible for $ n+2$.  If we have such a graph $ \mathcal{G}$ with $ n$ vertices with the desired property, then consider adding two more vertices $ A$ and $ B$ with the following properties (we say one vertex "beats" another if the edge connecting those vertices leads from the one who beat to the one who was beaten):

$ A$ beats $ B$
$ B$ beats everyone in $ \mathcal{G}$
Everyone in $ \mathcal{G}$ beats $ A$

We then have that for any $ V \in \mathcal{G}$, we have $ A, B, V$ are in a cycle of length three.  Therefore, we can get from any of $ A, B, V$ to any other in two moves or fewer, producing $ \mathcal{G}'$ with $ n+2$ vertices with the desired properties.

Now it is a simple matter to generate working graphs for $ n=3, 4$.  For $ 3$, just consider a cycle: A beats B, B beats C, C beats A.  For $ n=4$, consider a graph with $ A, B, C, D$ as follows:

$ A$ beats $ B, C$
$ B$ beats $ C$
$ C$ beats $ D$
$ D$ beats $ A, B$

This work.  Therefore, each $ n \ge 3$ has the desired property.[/hide]

[quote="JBL"]A nice start.  Unfortunately, the tournament given by
[quote="K81o7"]$ A$ beats $ B, C$
$ B$ beats $ C$
$ C$ beats $ D$
$ D$ beats $ A, B$[/quote]
has only 3 kings: there is no path of length 2 or less from $ B$ to $ A$.  (In fact, there is no tournament with 4 kings on 4 vertices.)[/quote]

Oh...in that case, I believe I have a working one for $ n=6$.  I believe I checked every ordered pair of people, and I believe this works (barely, but it does), but let me know if there's something wrong with it.

People $ A, B, C, D, E, F$

$ A$ beats $ B, E$
$ B$ beats $ C, E, F$
$ C$ beats $ A, E, F$
$ D$ beats $ A, B, C$
$ E$ beats $ D, F$
$ F$ beats $ A, D$

If this tournament does in fact work, then the final answer is $ n \ge 5$ or $ n=3$.

A proof that $ n=4$ does not work to come...though some second thought I suppose you could just check all $ 64$ possiblities if you wanted to.

EDIT: Okay, a simple proof that $ n=4$ does not work.
Clearly, there is someone who beat at least two other people.  And clearly, nobody can beat everyone else, or we have a problem.  So consider a four-person tournament where $ A$ beats $ B, C$ and loses to $ D$.  WLOG, $ B$ beats $ C$.  Then it is impossible to have a path of length 2 or fewer from both $ C$ to $ B$ and from $ B$ to $ D$.

Nice.

Also, there are only four tournaments (up to graph isomorphism) on four vertices, and three of them have either someone who wins all games or someone who loses all games, so there is only actually one tournament to check (rather than 64).


Anyone had an easier way of doing it without programming than the official solutions?

I jumped straight to directed areas, which removed some steps. I used a different simplification of the summation, instead relying on trig identities which were less algebraically complex to implement. Basically, start with $ E \equal{} \frac{\displaystyle\sum^{2006}_{a\equal{}1}\displaystyle\sum^{a}_{b\equal{}1} (\sin{\frac{a\pi}{1004}} \plus{} \sin{\frac{b\pi}{1004}} \plus{} \sin{\frac{(2008\minus{}a\minus{}b)\pi}{1004}})}{2007\cdot2006}$. Then $ E \equal{} \frac{1004}{669\cdot2006}\left(1\plus{} \displaystyle\sum^{501}_{a\equal{}1} 2\sin{\frac{a\pi}{1004}}\right)$, which is easily recognized as $ E \equal{} \frac{1004}{669\cdot2006}\left(1\plus{} \displaystyle\sum^{501}_{a\equal{}1} 2\sin{\frac{a\pi}{1004}}\right)$. Applying that $ 1\plus{}2\displaystyle\sum_{k\equal{}1}^n\cos(kx)\equal{}\frac{\sin\left(\left(n \plus{}\frac{1}{2}\right)x\right)}{\sin(x/2)}$, we get to $ E \equal{} \frac{502}{671007}\cdot\frac{\sin{\frac{1003\pi}{2008}}}{\sin{\frac{\pi}{2008}}}$, and from there, $ E \equal{} \frac{502}{671007}\cot{\frac{\pi}{2008}}$. I left out a good deal of the algebra and the proof that directed areas and $ 1\plus{}2\displaystyle\sum_{k\equal{}1}^n\cos(kx)\equal{}\frac{\sin\left(\left(n \plus{}\frac{1}{2}\right)x\right)}{\sin(x/2)}$ work for this post. If you want more detail on how those work, feel free to ask.

Sort of. Let C be the center of the circle; split XYZ into XYC, YZC, XZC. Calculate their areas by $ A \equal{} \frac{1}{2}ab\sin{c}$, which reduces to $ A_{PQC} \equal{} \frac{\sin{\angle PCQ}}{2}$. Use the measure of the arc between two given points(not though the third one, this is important) to get that the area of the triangle is half the sum of the sines of the arcs determined by $ XY$, $ YZ$, and $ ZX$ that do not pass through any other labeled point. Some of these sines will be negative, and therefore the area is "directed."

Oh, I see now. I guess I did essentially that, though instead, I use a counting argument to find how many times each $ k\pi/2004$ angle subtending $ \triangle XYZ$ piece occurs in the sum, and used $ \sum_{k\equal{}1}^{n\minus{}1} \sin(k\pi/n) \equal{} \cot (\pi/(2n))$.

[quote="123456789"]Does "directed areas" just mean the areas of the triangles?[/quote]
See [url]http://softsurfer.com/Archive/algorithm_0101/algorithm_0101.htm#2D%20Polygons[/url] for details: In this case, we are taking P to be the center of our unit circle.


vector
abstract algebra
analytic geometry
superior algebra
superior algebra open

Here's what I was wondering (sorry if the answer is obvious or something like that :)):

Given two groups (the question holds for other structures as well) $G_1,G_2$, if there are two injective morphisms $f_{12}:G_1\to G_2$ and $f_{21}:G_2\to G_1$, does it folow that the groups are isomorphic?

A very partial answer in a very particular case: Suppose these are vector spaces, and the morphisms are linear mappings. Then each vector space must have a (Hamel) basis, and given the two isomorphisms, we can prove that the two bases must have the same cardinality. Thus (by the original Schrder-Bernstein or Cantor-Bernstein theorem), there's a one-to-one and onto mapping of the bases that can be extended to a one-to-one and onto linear mapping of the vector spaces.

But that's all very particular to vector spaces. I don't see how it help with groups.

No, it doesn't work. Here's an example, with two abelian groups:
Let $A$ be the set of all sequences $(a_1,a_2,\dots,a_n,\dots)$ with $a_n \in \mathbb{Z}/2^{n+1}\mathbb{Z}$
and let $B$ be the set of all sequences $(b_1,b_2,\dots,b_n,\dots)$ with $b_n \in \mathbb{Z}/2^n\mathbb{Z}$
In both cases, the group operation is pointwise addition of sequences.

Let $f:A\to B$ be defined by $f(a_1,a_2,a_3,\dots,a_n,\dots)=(0,a_1,a_2,\dots,a_{n-1},\dots)$
and $g:B\to A$ be defined by $g(b_1,b_2,\dots,b_n,\dots)=(2b_1,2b_2,\dots,2b_n,\dots)$.

Both $f$ and $g$ are injective homomorphisms, but the groups are not isomorphic: every element of order two in $A$ is twice some other element, but $(1,0,0,0,\dots)$ is an element of order two in $B$ which cannot be halved.

We could also restrict $A$ and $B$ to sequences which are eventually zero; this would make the groups countable without changing the argument.

a similar one:
let E and F be two compact metric spaces. Prove that if there is one isometry from E to F and one isometry from F to E then E and F are isometric. I mean that there is one bijection h from E to F that is an isometry too.

[quote="jmerry"]No, it doesn't work. Here's an example, with two abelian groups:
Let $A$ be the set of all sequences $(a_1,a_2,\dots,a_n,\dots)$ with $a_n \in \mathbb{Z}/2^{n+1}\mathbb{Z}$
and let $B$ be the set of all sequences $(b_1,b_2,\dots,b_n,\dots)$ with $b_n \in \mathbb{Z}/2^n\mathbb{Z}$
In both cases, the group operation is pointwise addition of sequences.

Let $f:A\to B$ be defined by $f(a_1,a_2,a_3,\dots,a_n,\dots)=(0,a_1,a_2,\dots,a_{n-1},\dots)$
and $g:B\to A$ be defined by $g(b_1,b_2,\dots,b_n,\dots)=(2b_1,2b_2,\dots,2b_n,\dots)$.

Both $f$ and $g$ are injective homomorphisms[/quote]
Is $f$ injective?
$f(1,1,1,\dots,) = (0,1,1,1,\dots)$
and
$f(1+2^n,1+2^n,1+2^n,\dots,) = (0,1,1,1,\dots)$

[quote="liyi"]$f(1+2^n,1+2^n,1+2^n,\dots,) = (0,1,1,1,\dots)$[/quote]
No, $f(1+2^n,1+2^n,1+2^n,\dots,)=(0,1+2^n,1+2^n,1+2^n,\dots,)$
If you remove the first coordinate of $B$, the quotient structure is naturally isomorphic to $A$, and $f$ is the natural isomorphism.

Didn't you mean $f(1+2,1+4,1+8,\dots,1+2^n,\dots)$ instead? It's still injective; there's room if we shift.

[quote="jmerry"]No, $f(1+2^n,1+2^n,1+2^n,\dots,)=(0,1+2^n,1+2^n,1+2^n,\dots,)$
If you remove the first coordinate of $B$, the quotient structure is naturally isomorphic to $A$, and $f$ is the natural isomorphism.[/quote]
But you defined $b_n\in \mathbb{Z}/2^n\mathbb{Z}$ and $1+2^n=1$ in $\mathbb{Z}/2^n\mathbb{Z}$

Look carefully at what $n$ means. It's an index, and it's being shifted.

Actually, writing $(1+2^n,1+2^n,\dots)$ is just confusing, since we want to use $n$ as an index and later shift it. If $A=(a_1,a_2,\dots)$ with $a_n=1+2^n$, then $B=f(A)=(b_1,b_2,\dots)$ has $b_1=0$ and $b_n=2^{n-1}+1$ for $n\ge2$.

I could have eliminated the doubling and given the images of both homomorphisms finite index by defining $A'=(a_1,a_2,a_3,\dots)$ $a_n\in\mathbb{Z}/4\mathbb{Z}$ for all $n$, $B'=(b_1,b_2,b+3,\dots)$ with $b_1\in \mathbb{Z}/2\mathbb{Z}$ and $b_n\in\mathbb{Z}/4\mathbb{Z}$ for $n\ge2$. The homomorphisms here are $f(a_1,a_2,a_3,\dots)=(0,a_1,a_2,\dots)$ and $g(b_1,b_2,b_3,\dots)=(2b_1,b_2,b_3,\dots)$.

Okay. I finally understand.
I misunderstood it  :blush: I thought $n$ a constant in $\mathbb{Z}/2^n\mathbb{Z}$

There are many examples of structures where the analogue of Cantor-Bernstein fails.

Groups have been asked for, here is another example which I think is easier to understand:
$\mathbb{Q}^\mathbb{N}$ and $\mathbb{Z} \times \mathbb{Q}^\mathbb{N}$ with componentwise addition
$(q_1, ...) \mapsto (0, q_1, ...)$
$(z, q_1, ...)\ {{<}{-}{|}}\ (z, q_1, ...)$
That the groups are nonisomorphic is established with the same argument jmerry used.

Rings (commutative, unitary):
same example, componentwise addition and multiplication

Even Fields:
$\mathbb{C}$ and $\mathbb{C}(X)$
$\mathbb{C}$ is isomorphic to the algebraic closure of $\mathbb{C}(X)$, let $\varphi: \textrm{cl}(\mathbb{C}(X)) \to \mathbb{C}$ be an isomorphism.
The embedding $\mathbb{C} \to \mathbb{C}(X)$ and the restriction of $\varphi$ to $\mathbb{C}(X)$ do the job.

But there are some structures where two monomorphisms induce the existence of an isomorphism.
Vectorspaces have been mentioned.
Well ordered sets are a rather trivial example.

I just ask myself wether the theorem holds for partially or for totally ordered sets.

LiMa

Another example of this kind. Let $Q_1$ and $Q_2$ be injective modules over some ring $R$. If there exist monomorphisms $\mu_1\colon Q_1\to Q_2$ and $\mu_2\colon Q_2\to Q_1$ then $Q_1\cong Q_2$.

@LiMa: $\mathbb{Z} \times \mathbb{Q}^{\mathbb{N}} \not\cong \mathbb{Q}^{\mathbb{N}}$ simply follows from the divisibility of the right group and the non-divisibility of the left group.
@lofar: Is this true or just a conjecture? :)

It is true. In particular, Cantor-Bernstein theorem for morphisms holds for divisible abelian groups.

I've just solved it for $R=\mathbb{Z}$ with the help of the structure theorem of divisible groups, but I have no idea concerning the general case ... could you give me a hint?

The proof is wery similar to the one of the Cantor-Bernsterin theorem for sets.

Suppose, $Q_1$ and $Q_2$ are injective $R$-modules. Let $\mu_1\colon Q_1\to Q_2$ and $\mu_2\colon Q_2\to Q_1$ be monomorphisms. We need to prove that $Q_1\cong Q_2$.

[hide="Step 1."]W.l.o.g we can assume that $Q_2$ is a submodule of $Q_1$ and that $\mu_2=\iota\colon Q_2\to Q_1$ is a natural imbedding. In this case $Q_1=Q_2\oplus A$. Consider monomorphism $\mu\colon Q_1\to Q_1$, $\mu=\iota\mu_1$.[/hide]
[hide="Step 2."]Let $B=A+\mu(A)+\mu^2(A)+\ldots$ Then $\mu(B)=\mu(A)+\mu^2(A)+\ldots$ and $\mu(B)=B\cap Q_2$. Thus $B=A\oplus(B\cap Q_2)$. Since $B\cong\mu(B)=B\cap Q_2$ we have $(B\cap Q_2)\oplus A\cong B\cap Q_2$.[/hide]
[hide="Step 3."]$B\cap Q_2$ is a submodule of the injective module $Q_2$, hence there exists an injective hull $C$ for $B\cap Q_2$ such that $B\cap Q_2\leqslant C\leqslant Q_2$. The module $A$ is injective, thus $C\oplus A$ is an injective hull for $(B\cap Q_2)\oplus A$. The isomorphism $(B\cap Q_2)\oplus A\cong B\cap Q_2$ yields an isomorphism for the injective hulls: $C\oplus A\cong C$. Let $Q_2=L\oplus C$ then $Q_1=L\oplus C\oplus A$. The result follows.[/hide]


I have been scouring the internet for the past 6 hours trying to solve this one, maybe someone here can help me and have some fun at the same time.  I'm trying to find a point on a triangle.  I drew a diagram to help explain what I know and what I'm trying to find.  I know the x,y for points A and B but I need to find x and y for point C.  Can anyone help me with the process required to solve this problem?

[img]http://webfluid.com/etc/Drawing.png[/img]

[hide="hint"]Distance Formula[/hide]

[hide="partial solution"]$ 38 \equal{} \sqrt {(x \minus{} 6)^{2} \plus{} (y \minus{} 12)^{2}} \rightarrow 1444 \equal{} (x \minus{} 6)^{2} \plus{} (y \minus{} 12)^{2}$
$ \sqrt {1573.96} \equal{} \sqrt {(x \minus{} 17)^{2} \plus{} (y \minus{} 15)^{2}} \rightarrow 1573.96 \equal{} (x \minus{} 17)^{2} \plus{} (y \minus{} 15)^{2}$

I may finish this later, but all you have to do from here is solve for $ x$ and $ y$.
Using a graphing program and estimating these circle intersections I got two possible points $ (\minus{}4, 48.66)$ and $ (16, \minus{}24.66)$
[/hide]

If I was able to create another right triangle and use 38 as the hypotenuse, could I solve for x of the new point (which would be x for point C) and then solve for y without doing circle intersections?

Yes you could, and no you don't need to graph the circles and find where they intersect its probably more straight forward to solve as a system of equations anyway. I was just lazy ;p.

There is an easier way to do this.

First find the hypotenuse: $ \sqrt{38^2\plus{}11.40^2}\equal{}\sqrt{\displaystyle\frac{39349}{25}}$

Now use the distance formula from point $ A$ to point $ C$ and set it equal to $ 38$: $ \sqrt{(6\minus{}x)^2\plus{}(12\minus{}y)^2}\equal{}38$

Do the same for point $ B$ to point $ C$: $ \sqrt{(17\minus{}x)^2\plus{}(15\minus{}y)^2}\equal{}\sqrt{\displaystyle\frac{39349}{25}}$

Now you have a system of equations and can solve for $ x$ and $ y$.

You can solve this system by using combination. If you need help just post.

[quote="JZambrano201"]
Now you have a system of equations and can solve for $ x$ and $ y$.

You can solve this system by using combination. If you need help just post.[/quote]

I think that's the same thing SnipedYou had in their first post.

Are there any assumptions I can make against a right triangle if it shared points A and C, it's third point (apposite the hypotenuse) would have a y value of 12?


integration
trigonometry
calculus
calculus computations

If $ f\left( x \right) \equal{} \int\limits_{ \minus{} 2}^2 {\frac{{{e^u}\cos \left( {xu} \right)}}{{{e^u} \plus{} 1}}du,x \in R}$
then find maximum and minimum values (if they exist)

$ f(x) \equal{} \left\{\begin{array}{ll} 2\ (x \equal{} 0) & \quad \\
\frac {\sin 2x}{x}\ (x\neq 0) & \quad \end{array} \right.$


calculus
integration
logarithms
trigonometry
calculus computations

Hi I hope you can help me because I got test in this Thursday  :(

What´s the procedure to calculate Integral of sen(Inx)?

The result is: (x/2){[cos (ln x)] - [sin (ln x)]} + C

But I never get there…

Thanks so much!

[quote]sen(Inx)[/quote]

[professor's peeve] That's not an "I"! That's an "L"! L as in [b]L[/b]ogarithm! [/peeve]

Try letting $w=\ln x.$ This will have to be a "forced" or "backward" substitution. Write it as $x=e^w$ from which you get $dx=e^wdw.$

Now, $\int\sin(\ln x)\,dx = \int e^w\sin(w)\,dw.$

Notice that this didn't actually solve your problem - it just reshaped it, into a shape we hope you've seen before. The standard Calculus II used gimmick to attack this problem is to integrate by parts twice and solve for the unknown integral.

And yes, you could have done the same double integration by parts without first doing that substitution, if you could see what you were doing. The main purpose of that substitution was to let you see more clearly.

If kenny is studing either in the UCR or the ITCR, then it should be a Calculus I trick for him :D . Nevertheless, kenny, if you are in Calc II in some of those places, may be you may want to do a "fancier" solution, and consider

$\displaystyle \int\sin(\ln x)\,dx = \int e^w\sin(w)\,dw = \Im\left( \int e^{w(1+i)}\,dw\right)$

I mean, usually all the complex stuff is in that course, and probably is what the professor could be expecting, even when the suggestion of Prof. Merryfield is totally right! By the way, it is nice to see another Costa Rican in the forum... Pura Vida Mae!


Is it correct to say

I will show by induction with respect to $ n$ that $ \sum_{i\equal{}1}^n i\equal{}\frac{n(n\plus{}1)}{2}$ or should we say by induction on $ n$?

"We can prove $ 1\plus{}2\plus{}\cdots\plus{}n\equal{}\frac{n(n\plus{}1)}{2}$ by using induction on $ n$.

That's just one way to say it.  You could probably say it however you want.  I've never heard "induction with respect to $ n$".  I get what it means, I've just never heard it.


Solve for $ x$: $ \frac{2}{5} \minus{} \frac{4}{3} \equal{} \frac{1}{x}$. Express your answer as a common fraction.

$ \frac{2}{5}\minus{}\frac{4}{3}\equal{}\frac{1}{x}$

$ \implies \minus{}\frac{14}{15}\equal{}\frac{1}{x} \implies x\equal{}\boxed{\minus{}\frac{15}{14}}$.


So, the solutions to part 1 of the University of Maryland Math Competition are now out at http://www.math.umd.edu/highschool/mathcomp/2007.shtml

Does anyone else from Maryland post on AoPS anymore (not that I've been very active lately)?

In any case, I got a 94 (#25 wrong), which hopefully will be top 5 at least.

wow this thread is old....
well i took it this school year =/
i only answered 14 questions cuz i was afraid of losing points
turns out i got all 14 of those right =)
meh i didnt do so well, i wonder if ill take it in 9th grade.


Let $ a,b,c,d$ be positive real numbers such that
\[ \frac1a+\frac1b+\frac1c+\frac1d=3.\]
Prove that
\[ \sqrt{a+b+c+d}\geq\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}+\sqrt{d-1}.\]

[quote="pvthuan"]Let $ a,b,c,d$ be positive real numbers such that
\[ \frac1a+\frac1b+\frac1c+\frac1d=3. \]
Prove that
\[ \sqrt{a+b+c+d}\geq\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}+\sqrt{d-1}. \]
[/quote]$ \frac1a+\frac1b+\frac1c+\frac1d=3\implies \frac{a-1}{a}+\frac{b-1}{b}+\frac{c-1}{c}+\frac{d-1}{d}=1$
$ \left(\frac{a-1}{a}+\frac{b-1}{b}+\frac{c-1}{c}+\frac{d-1}{d}\right)\left(a+b+c+d\right)\geq\left(\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}+\sqrt{d-1}\right)^{2}$
$ \sqrt{a+b+c+d}\geq\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}+\sqrt{d-1}$

We also have this problem:
Given $ \sum_{n}^{i=1}\frac{1}{a_{i}}=n-1$
Prove that $ \sqrt{\sum_{n}^{i=1}a_{i}}\geq \sum_{n}^{i=1}\sqrt{a_{i}-1}$
Proof similar to mimsoens's.

[quote="pvthuan"]Let $ a,b,c,d$ be positive real numbers such that
\[ \frac1a+\frac1b+\frac1c+\frac1d=3. \]
Prove that
\[ \sqrt{a+b+c+d}\geq\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}+\sqrt{d-1}. \]
[/quote]
I think, it's wrong! :D
Try $ a=b=c=3$ and $ d=\frac{1}{2}.$ :wink:

for $ d = \frac{1}{2}$ or $ 0.5$

$ \sqrt{d-1}= \sqrt{0.5-1}= \sqrt{-0.5}$ which is obviously a complex number. Perhaps the inequality only holds when no complex numbers are involved.

So I think we should have $ a_{i}\geq 1$

Can the new problem be answered with Cauchy-Schwarz inequality
CS ineq : sigma(xn.yn)<=sqrt(sigma(xn^2).sigma(yn^2))
Then we can put sqrt(ai-1)=xn.yn, then...


I have a very long and not very ingenious solution for this. I must mention I'm a big 0 at polynomials, so this might be easy.
Let $f_{1},f_{2},...,f_{n} \in {\rm Z}\left[X\right]$. Prove that there is a reducible polynomial $g$ (over ${\rm Z}\left[X\right]$) s.t. $f_{i}+g$ is ireducible for all $i=\overline{1,n}$.

I'm sure this was posted before, a long time ago.

Of course it is Iran 2003.It was our exam last year.Except einstein criterion there was a real easy solution using Roushe theorem (I don't know if my spell is not correct) given by Adel javanmard(One of people in Iran team for IMO 2004)

I'm incredibely stupid. I also used Eisenstein's criterion, but I just can't see any easy way to make sure that $g$ is reducible.
And thanks for the solution with Roushe Theorem (although I don't know it  :D ).

let $ f_i(x)=a_{im}x^m+\dots +a_{i0}$ for $1\leq i\leq n$take an integer k which is not  the root of $f_is$.for each i let $p_i$ be prime nuber such that $p_i$ doesnt divide $f_i(k)$,and also $p_i$ be all distinct.now by chinese reminder theorem the following systemof eqquations has a solution $b_j$ for every $1\leq j\leq m-1 $$b_j\equiv^{(p_1^2)} -a_{1j},\dots b_j\equiv^{(p_n^2)}-a_{nj}$ also following system for $b_m,
b_m\equiv ^{(p_1^2)} p_1-a_{1m}\dots b_m \equiv ^{(p_m^2)}p_n-a_{nm}$ now define $b_0=-kb_1-k^2b_2-\dots-k^mb_m$ and $g(x)=\sum_{i=0}^{m} b_ix^i$.we have $f_i(x)+g(x)=(b_0+a_{i0})+\dots +(b_m+a_{im})x^m$ now u can check witgh eisenstein criterion that $f_i(x)+g(x)$ is irriducible in $\mathbb{Z}[x]$ on the other hand $g(k)=0$ so it is reducible.

Roushe theorem:
If $f,g \in \mathbb{C}[x]$ and suppose $C$ is a circle on complex plane.
If $|f(x)|>|g(x)|$ for every $x$ on the circle $C$ the number of roots of $f$ and $f+g$ is equal in circle $C$

Thanks. And I promise I won't ask such idiot questions again.

Well think on the problem with Rousheh theorem.It's interesting.


Euler
LaTeX
real analysis
real analysis theorems

i saw it done one time in a pdf, but i dont know if it specific to the function.

what i was working off of was multiplication was merely repeated addition, however with each term of the eulor product the factors change.

(off topic where could i find the syntax to write the equations in the posts?)

It might help if you mentioned the definitions in question.

There is a $\LaTeX$ forum on this site with links to various guides and resources. Also, $\LaTeX$ math mode is enabled on this forum- put the appropriate code between dollar signs to use it. You can get a feel for how to write things by hovering over or copying from other people's posts- that's how I learned most of the commands.


calculus
integration
calculus computations

i may have posted this already, but not sure, anyone who can check my work here, i think i got it! :)

sorry, the pic didn't post, but the work is rather lengthy to type, so i made it in jpeg form. here it is: http://img462.imageshack.us/img462/8057/p17uy.jpg

thanks.


find the numbers set a(i) ,a(i) odd >=0 such that
a(1) +a(2)+a(3)+a(4)=98

Since $a(i)$ are odd positive integers, $a(i) = 2b(i)+1$ where $b(i)$ is a non-negative integer. Hence

$98 \;=\; \sum_{i=1}^{4}a(i) \;=\; \sum_{i=1}^{4}2b(i) \:+\: 1 \;=\; 4 \:+\: 2 \sum_{i=1}^{4}b(i)$,

i.e.

$(1) \;\; \sum_{i=1}^{4}b(i) \;=\; 47.$

From combinatorics theory it is known that the number of integer solutions of the equation

$(1) \;\; x_{1}+x_{2}+...+x_{n}= r, \;\; x_{i}\geq 0$

is ${\textstyle \binom{n+r-1}{r}.}$ In this case $n=4$ and $r=47$, so the number of integer solutions of equation (1) is

$\binom{4+47-1}{47}\;=\; \binom{50}{47}\;=\; 19600.$


Two positive numbers $x$ and $y$ are in the ratio $a: b$ where $0 < a < b$. If $x+y = c$, then the smaller of $x$ and $y$ is

$\textbf{(A)} \ \frac{ac}{b} \qquad \textbf{(B)} \ \frac{bc-ac}{b} \qquad \textbf{(C)} \ \frac{ac}{a+b} \qquad \textbf{(D)} \ \frac{bc}{a+b} \qquad \textbf{(E)} \ \frac{ac}{b-a}$

JesusFreak197, you didn't show that x was the smaller of x and y

[hide]From the ratio:
$x=y(\frac{a}{b})$
so $y(\frac{a}{b})+y=c$
$y=\frac{bc}{a+b}$

so $x=\frac{ac}{a+b}$

Since $a<b$, x is the smaller of the two so $\boxed{C}$[/hide]


inequalities
induction
inequalities unsolved

Let the sequences $\{a_{n}\}$ denoted by $a_{0}=0$,$a_{n+1}\geq a_{n}+1\forall n\in N$.
Prove that:$\sum\limits_{i=1}^{n}a_{i}^3\geq(\sum\limits_{i=1}^{n}a_{i})^2$


Phileas Fogg begins another journey around the world along the equator, finishing the voyage in eighty days. His companion, Passe-partout, makes a similar journey aroudn the world along the 30 degree latitude (30 degrees above the equator) and finishes his trip in eighty days as well. If the world has a radius of 20,000 miles, what is the sum of both the men's average travel speeds per day?

I don't see what the latitude fact can be applied to this problem.

Passe-partout makes a shorter journey.  (Find a globe and trace out his path!  It's not a great circle.)


Δίνονται 5 σημεία στο εσωτερικό τετραγώνου πλευράς 1. Αποδείξτε ότι δύο τουλάχιστον από αυτά απέχουν μεταξύ τους απόσταση μικρότερη του ρίζα δύο διά δύο.

[hide]Χωρίζουμε το τετράγωνο σε 4 μικρότερα ίσα τετραγωνάκια φέρνοντας τις παράλληλες στις πλευρές του τετραγώνου που διέρχονται απ' το κέντρο του.
Τότε 2 απ' τα σημεία βρίσκονται στο ίδιο τετράγωνο κι άρα απέχουν λίγοτερο από το ζητούμενο αφού  ρίζα2/2 είναι η διαγώνιος του κάθε μικρού τετραγώνου.[/hide]

Ναι, αυτη τη λυση ηξερα και εγω. Καλα ποτε προλαβες :!:  :?:


We have integer $ n>\equal{}2$. Let $ r_{1},r_{2},....,r_{n\minus{}1}$ be rests from division numbers $ 1, 1\plus{}2, 1\plus{}2\plus{}3, ...., 1\plus{}2\plus{}...\plus{}(n\minus{}1)$ by $ n$. Find all n that progression $ (r_{1},r_{2}, ...,r_{n\minus{}1})$ is permutation of progression $ (1,2,3,4,...,n\minus{}1)$

[b]Please, block this topic and do not reply to it. The problem is from the first round of Polish MO, which is currently running.[/b]


analytic geometry
geometry
function
algebra
polynomial
vector
AMC

Are you talking about coordinate geometry, functions, or complex number problems?
(actually they're quite tied together)

[quote="Calculus"]I have lots of difficulty handling problems dealing with graphs. Can anyone give me advice on how to tackle those and advice on improving?[/quote]

Your question is very vague. It's really hard to see what type of graph problems you mean because there are whole bunch (area, intersection, shape, type of the graphs of the equations that are combined, complex number, polynomial graph and relationship of it to roots, coordinate geometry, vectors in plane, etc..). Graph is very wide subject but if you mean AMC/AIME graph problems, remember that you're not required to use calculator to solve them so there must be "easy" way to do them. Think on what kind of situation you're in. If you're looking for area, for instance, the best way is to see the shape first. This can be done by rewriting the given equations in common form such as $(x-h)^2 + (y-k)^2 = r^2$. The best way is to just keep practicing the hard graphing problems. That's really only way advice that I can give you. Also, I believe Intermediate Algebra class will deal certain amount of those types of problems so that's something you can look into.

Sorry I meant problems like #4 on the 1991 AIME where you have to graph the function to find the solutions.


logarithms
trigonometry
calculus
calculus computations

So nobody, not even the teacher, can figure out this problem from a practice test we're completing as a class. Any help would be greatly appreciated:

15. The average (mean) value of $\tan(x)$ on the interval from $x=0$ to $x=\frac{\pi}{3}$ is

The answer given is $\frac{3}{\pi}\ln(2)$, but I can't figure out why for the life of me.

$-\ln\cos x$ is an antiderivative for the tangent. It's easy from there.

Oh wow I didn't even know that existed. Thanks a lot.


Circle k(S; r)is touching point A of line AB. Circle l(T; s)is touching point B of line AB and intersects circle k in the edge points C, D of its diameter. Prove that the intersection M of lines CD and AB is the centre of line AB.

Was it necessary to take $CD$ as diameter? The radical axis of 2 circles passes through the midpoint of the common tangent, be it external, or internal!

Best regards,
sunken rock


Patty has $ 20$ coins consisting of nickels and dimes. If her nickels were dimes and her dimes were nickels, she would have $ 70$ cents more. How much are her coins worth?

$ \textbf{(A)}\ \$1.15\qquad
\textbf{(B)}\ \$1.20\qquad
\textbf{(C)}\ \$1.25\qquad
\textbf{(D)}\ \$1.30\qquad
\textbf{(E)}\ \$1.35$

$ n \plus{} d \equal{} 20$

$ 10n \plus{} 5d \equal{} 5n \plus{} 10d \plus{} 70$

Solve this system of equations.

$ n \plus{} d \equal{} 20$

$ 5n \minus{} 5d \equal{} 70$

Divide the second equation by $ 5$.

$ n \plus{} d \equal{} 20$

$ n \minus{} d \equal{} 14$

Thus, $ n \equal{} 17$, $ d \equal{} 3$, and the amount of money is $ \boxed{\$1.15}$.


How many different paths are possible in moving from A to B given that you must move down to the right or down to the left?
(Sorry for the bad drawing)

Hey Sanntosh, look at the picture again.  :?
[hide]You have to go R 3 times and L 3 times, so the answer is c(6,3), or 20.[/hide]

Billy

[quote="Xantos C. Guin"][hide]No matter which way you move you will move 2 spaces down left and 2 spaces down right.
If L represents moving left and R represents moving right
The answer will be the # of ways to arrange 2 L's and 2 R's

This = 4!/(2!*2!) or 6[/hide][/quote]

My computer messed up the diagram that answer is correct.

Another way to do this without having to use any formulas is by that you could just draw on the diagram (me without the diagram because I don't know the fancy stuff to make one, so you'll just have to picture it.)

                                                  S
                                             1         1
                                        1        2         1
                                    1      3          3       1
                                        4        6         4
                                            10       10
                                                  20

Basically, you just start at the beginning and see how many ways you can get to the first two points. Then, as long as you can't go back up, or any other way than just down and to the right or left, you add the sum of the paths (the previous points directly above it, not all of the points up until that point, because you've already added those.)  going to your next point until you reach the bottom.

Also notice that the numbers are a chunck out of Pascal's Triangle.  :)

[hide]it's 6c3 or 6!/3!*3!=20[/hide]

elvenchamp777

There is a formula for this:
[hide]Situation: Go from A to B, can only go in 2 directions, both heading to B, paths are in a grid, l=lengh, w=width of grid.
Its l+w choose l or w
3+3 choose 3
6 choose 3
6*5*4/3*2*1
5*4
>>20<<[/hide]


What is the length of one edge of a regular tetrahedron inscribed in a sphere of unit radius?

im not sure if it would be useful in this problem, but i know that in chemistry the "radii" of the tetrahedral shape for a molecule (i.e. the line segments from the center of the tetrahedron to its vertices) form 109.5 degree angles. is this true of a regular tetrahedron?

The angle is only [b]approximately[/b] $109.5$, and no, that isn't useful.

[hide="Instead, consider..."] The "center" of the tetrahedron.  You know the length of the segments from the center to the vertices (which is $1$).  How might you go about calculating the side length now?
[hide="You might want to try..."] Considering appropriate triangles within the tetrahedron - of course $2$ dimensions is easier to work with than $3$! [/hide] [/hide]

[hide="Hint"]3-D problems are always easier if you reduce them to 2-D ones.[/hide]


[quote="hurdler"]this was on the "do you need this" test for the intermediate number theory seminar

3. The sequence 1, 3, 5, 7, 9, 15, 17, . . . consists of the positive integers whose binary representation
reads the same left to right as right to left. (Such as 9, which is written as 10012 in binary.)
Find the number of 1’s in the binary representation of the 1717th term of the sequence.

I can't seem to get the right answer, so can someone please point out my error? thanks:

[hide="my work"]
The end 2 digits are 1's, and for each digit in the first half of the "middle" (all digits except the 2 end ones), we have 2 choices (the second half of the middle 's digits are determined by these), and we make the chart:

number of digits (in binary), number of numbers in the given sequence:

1:1
2:1
3:2
4:2
5:4
6:4
7:8
8:8
n:$ 2^{\lceil \frac {n}{2} - 1\rceil}$
19:512
20:183

I may have messed up the the arithmetic above even though I checked about 3 times.

converting 183 to binary, and adding that to 2^9=1 followed by 9 zeroes, and then reflecting this number on its decimal point, we get 1000000000+10110111=1010110111 so answer is 10101101111110110101, which has 14 one's in it. I feel like I am probably making an off by one mistake somewhere but I can't find where

edit: the answer given was 12[/hide][/quote]

The first number in the sequence with 20 digits in binary expression is $ 10000000000000000001$ not $ 10000000011000000001$. So 1717th number is $ 10101101100110110101$ which consists of $ 12$ "1"s

[quote="hurdler"]

3. The sequence 1, 3, 5, 7, 9, 15, 17, . . . consists of the positive integers whose binary representation
reads the same left to right as right to left. (Such as 9, which is written as
$ 1001_2$ in binary.)

[/quote]

I didn't make sense out of your number example until I retyped
it as the example above in binary.


Suppose the graph of the function $ f: \mathbb{R}\rightarrow \mathbb{R}$ is path connected. Is $ f$ necessary continuous ?

Yes. Rough sketch: Suppose $ f$ is not continuous from the left at $ x_0$. By the assumption there is a continuous curve that can reach $ (x_0,f(x_0))$ from the left. Consider the first time the curve reaches $ (x_0,f(x_0))$.

Aha, I got the idea. I will try to proof it. Thanks a lot :)

Suppose $ g: [0, 1]\rightarrow Graph(f)$ is continuous with $ g(0)\equal{}(x_0\minus{}1,f(x_0\minus{}1))$ and $ g(1)\equal{}(x_0\plus{}1,f(x_0\plus{}1))$.
But how can we be sure that $ \{t |g(t)\equal{}(x_0,f(x_0))\}$ is not empty ?


Joe bought a pumpkin that cost 10 cents more per pound than his sister's. Together, the two pumpkins weighed 20 pounds, but Joe's pumpkin was heavier. Joe paid $ \$3.52$, and his sister paid $ \$0.48$. How many pounds did Joe's pumpkin weigh?

Let Joe's pumpkin weigh $ p > 10$ pounds so that his sister's weighs $ 20 \minus{} p$ pounds. Also, let his sister's pumpkin cost $ c$ cents so his costs $ c \plus{} 10$ cents. Then $ p(c \plus{} 10) \equal{} pc \plus{} 10p \equal{} 352$ and $ (20 \minus{} p)c \equal{} 20c \minus{} pc \equal{} 48$, so adding gives $ 10p \plus{} 20c \equal{} 400\implies p \plus{} 2c \equal{} 40$. Since $ p|352$ and $ 352 \equal{} 32\cdot11$, and for $ c$ to be an integer $ p$ must be even, one finds that $ p \equal{} \boxed{16}$ is the only possibility. This corresponds with $ c \equal{} 12$, which works.


real analysis
topology
real analysis unsolved

A is an open or closed set in R^2, with measure zero,
how to show, projection of A onto R^1 is of measure zero?

Depends on the kind of measure that you use.
If $A$ is closed and has planar Lebesgue measure zero, then its projection need not have linear Lebesgue measure zero. Example: $A=\mathbb R\times\{0\}$.
If $A$ is open and has planar Lebesgue measure zero, then $A$ is empty and so is its projection.
If $A$ has $1$-dimensional Hausdorff measure zero, then for any $\epsilon>0$ there is a covering of $A$ by disks $D(a_{i},r_{i})$ such that $\sum r_{i}<\epsilon$. Projections of these disks onto $\mathbb R$ cover the projection of $A$. Thus the projection of $A$ has linear measure zero.


limit
logarithms
real analysis
real analysis solved

$\lim_{n\rightarrow 0}\frac{\log_{2}\sum_{k=1}^{2^n}\sqrt{k}}{n}$

and,

$\lim_{n\rightarrow 0}\frac{\log_{2}\sum_{k=1}^{2^n}\sqrt{2k-1}}{n}$

What steps should I take in order to evaluate these?

Thanks.

man, it's already been posted...

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=33834

(is this from a test y'all are taking?... and you're trying to cheat?  tsk tsk tsk)


linear algebra
matrix
calculus
linear algebra unsolved

Let $ A$ be an $ n\times n$ matrix over the field of complex numbers. Prove that $ A$ is similar to a symmetric matrix with its matrix of change of basis being symmetric.

[hide=" PS."] A tough but nice problem. Isn't it? Later I will add some sources to this post.

In my opinion, the point in the difficulty of the problems regarding similarity is that we don't have sophisticated tools to check that whether two given matrices are similar or not, even with the aid of a computer. (Or at least I don't know one! :) Correct me if I'm wrong.) I know that we have necessary and sufficient conditions in that concepts, but they are rather theoretical. For example are these matrices similar over $ \mathbb{Z}_3$ or not?$ \begin{pmatrix}1 & 2 \\
1 & 1\end{pmatrix}, \begin{pmatrix}2 & 2 \\
1 & 0\end{pmatrix}.$ Can you give a simple answer to this one? Maybe these are worth discussing under a new topic in the linear algebra tutorial section. [/hide]

Hi Captain.
For your principal question: the result is "well-known" but I have lost the proof.
A simple remark: over $ \mathbb{C}$, a symmetric matrix is a matrix without any particular property and then the fact that the matrix of change of basis is symmetric is only a curiosity.

For your PS: if $ K$ is an extension of $ \mathbb{Q}$ of finite degree or if $ K\equal{}F_p$ then, using MAPLE, it is easy to test the similarity of 2 matrices $ A,B\in\mathcal{M}_n(\mathbb{K})$: if the rational canonical forms of $ A,B$ are the same then $ A,B$ are similar with a matrix $ P$ in $ \mathcal{M}_n(\mathbb{K})$. The calculus of $ P$ is very quick (in $ O(n^3)$).
  Your example over $ \mathbb{Z}^3$ is bad because $ det(A)\equal{}\minus{}1$ and $ det(B)\equal{}1$. Never mind: $ RCF(A)\equal{}\begin{pmatrix}0&1\\1&2\end{pmatrix}$ and $ RCF(B)\equal{}\begin{pmatrix}0&2\\1&2\end{pmatrix}$

with(LinearAlgebra):
a:=Matrix([[1,2],[1,1]]):
 F := Frobenius(a,'P') mod 3:

Thanks for your answer [b]loup blanc[/b]!
I don't think that this is a "well-known" result, because I saw it as a main theorem of a recent article. (Maybe it gives a new proof to an old result, but I hadn't seen it before.)

Thanks for your information in Maple, and sorry for the typo in my ps.

A partial result:
1) There exist $ E$ symmetric,$ S$ invertible s.t. $ A\equal{}SES^{\minus{}1}$.
Proof (in Horn and Jhonson,matrix analysis,p. 209-210):
It's sufficient to prove the result for $ A\equal{}N\equal{}[n_{ij}]$ with: if $ j\equal{}i\plus{}1$ then $ n_{ij}\equal{}1$ else $ n_{ij}\equal{}0$ ($ N$ is a nilpotent Jordan matrix).
Let $ B\equal{}[b_{ij}]$; if $ i\plus{}j\equal{}n\plus{}1$ then $ b_{ij}\equal{}1$ else $ b_{ij}\equal{}0$.
Then $ S_1\equal{}\dfrac{1}{\sqrt{2}}(I\plus{}iB)$ is s.t. $ S_1^{\minus{}1}\equal{}\bar{S_1}$.
$ S_1NS_1^{\minus{}1}$ is a symmetric matrix.   QED.
PS1: Obviously this result is false for real matrices.
PS2: $ S_1$ is symmetric but it isn't sufficient to obtain a symmetric $ S$.
Corollary:
2) There exist symmetric matrices $ B,C$ s.t. $ B$ or $ C$ is invertible and $ A\equal{}BC$:
Proof: $ A\equal{}SES^{\minus{}1}\equal{}(SES^T)(S^T)^{\minus{}1}S^{\minus{}1}\equal{}(SES^T)(SS^T)^{\minus{}1}$ or $ A\equal{}(SS^T)((S^{\minus{}1})^TES^{\minus{}1})$.
PS3: this last result is valid in any field:cf
Taussky,The role of symmetric matrices in the study of general matrices, Linear algebra and its applications, 5 (1972) 147-154.

The result mentioned by The Captain is true in any field and you can find a proof (by Taussky and Zassenhaus in 1959) in free access here:
http://projecteuclid.org/DPubS?verb=Display&version=1.0&service=UI&handle=euclid.pjm/1103039127&page=record


calculus
integration
trigonometry
logarithms
calculus computations

Calculate the following indefinite integrals.

[1] $\int (\sin x+\cos x)^4 dx$

[2] $\int  \frac{e^{2x}}{e^x+1}dx$

[3] $\int \sin ^ 4 xdx$

[4] $\int \sin 6x\cos 2xdx$

[5] $\int \frac{x^2}{\sqrt{(x+1)^3}}dx$

[quote="amirhtlusa"]2
$e^x+1-\ln{(e^x+1)}$[/quote]

 :)

[quote="kunny"][quote="amirhtlusa"]just glancing through, #3 seems very easy, just use a half angle trign stuff..
ans:
$\frac{3x}{8}-\frac{\sin{2x}}{4}+\frac{\sin{4x}}{32}$[/quote]

 :([/quote]

It is correct kunny! (Minus the arbitary constant)

FOR 4
let takeI= $\int\sin2x\cos6xdx$ and J the integral you want.
calculate I+J and I-J it's too easy..
than you will got J.

IT'S TOO EASY. ;)

[quote="amirhtlusa"]yes i did it again and got the same thing. ;)[/quote]

I'm very sorry, you were RIGHT!  :blush:

kunny

[quote="staicovitch"]FOR 4
let takeI= $\int\sin2x\cos6xdx$ and J the integral you want.
calculate I+J and I-J it's too easy..
than you will got J.

Yes. I also have had your solution.


What are $\pi$ bonds and big $\pi$ bonds?

$\pi$ bonds are bonds made up by the lateral overlapping of two p orbitals.
Big $\pi$ bonds are $\pi$ bonds that are delocalized over three or more nuclei.


1. Deepali is four times as old as Ben.  In 20 years, she will be 30 years older than Ben will be.  How old is Ben now?

2. Jooho is 3 times as old as Molly.  In 15 years, he will be twice as old as Molly will be.  How old is Molly now?

3. Shanai is one third as old as Jose.  When Shanai is double her current age, how many times older than her will Jose be?

4. Natalya is three times as old as Frankie is.  In 18 years, the ratio of their ages will have been cut in half.  How old will Natalya be then?

1. Deepali is four times as old as Ben. In 20 years, she will be 30 years older than Ben will be. How old is Ben now?
[hide]4times as old is equal to 30 years older, that means Ben is 10. [/hide]

2. Jooho is 3 times as old as Molly. In 15 years, he will be twice as old as Molly will be. How old is Molly now?

[hide]J = 3x and x =Molly now.     3x + 15 = 2x +30 and you solve for x and x is equal to: 15[/hide]

3. Shanai is one third as old as Jose. When Shanai is double her current age, how many times older than her will Jose be?
[hide]
I'm not real sure about this one, but I'm gonna guess 2/3...[/hide]

4. Natalya is three times as old as Frankie is. In 18 years, the ratio of their ages will have been cut in half. How old will Natalya be then?

[hide]I did guess and check and got lucky and figured that 18 and 6'll work for their current age, so Natalya willl be 36 then[/hide]

Very good on #1 and #2!  Anyone want to give #3 a try or do #4 without guess-and-check?

[quote]I'm not real sure about this one, but I'm gonna guess 2/3...
[/quote]

I think you're right

because

S = J/3

and since it's

S + S = J/3 + J/3

and it's 2J/3

and it's 2/3 * J

but since we are looking for how many times of J, you can forget about it..

So, I got 2/3 also..

I didn't try #4 yet

4. Natalya is three times as old as Frankie is. In 18 years, the ratio of their ages will have been cut in half. How old will Natalya be then?

And here is numero cuatro...

[hide="And the answer is"]Okay, so let us begin by finding out what the original ratio between the ages is.
Very simply: $\frac{3x}{x}\Rightarrow\frac{3}{1}\Rightarrow3$

In 18 years, the ratio of their ages will be cut in half...

Thus: $\frac{3x+18}{x+18}=\frac{3}{2}\Rightarrow\frac{x+6}{x+18}=\frac{1}{2}$

Cross multiply and solve:  $2x+12=x+18 \Rightarrow x=6$

Natalya will be $3x+18$ then... So she is $3(6)+18\Rightarrow36$
[/hide]

[quote="shinwoo"][quote]4. Natalya is three times as old as Frankie is. In 18 years, the ratio of their ages will have been cut in half. How old will Natalya be then?

Click to reveal hidden content
I did guess and check and got lucky and figured that 18 and 6'll work for their current age, so Natalya willl be 36 then [/quote]

you're wrong..

I got some help from Silverfalcon.......

I didn't get the question, so I asked him....

He solved it in really neat way....

setting the equation,, Here's his work...

jwec05: N = Natayla's age
jwec05: F = Frankie's
jwec05: N = 3F
jwec05: N+18 = 2(F+18)
jwec05: 3F+18 = 2F+36
jwec05: F = 18
jwec05: N = 54

( ^- he's really smart )

so the Natayla's present age would be, 72...( 54 + 18 )[/quote]

I'd look over that... Because I think he did something wrong...Just get him over here to tell me why he did what he did, because his fourth step looks kind of odd...

Here is what I did (yet again)...

[hide="Is that your final answer?"]Okay, so let us begin by finding out what the original ratio between the ages is.
Very simply: $\frac{3x}{x}\Rightarrow\frac{3}{1}\Rightarrow3$

In 18 years, the ratio of their ages will be cut in half...

Thus: $\frac{3x+18}{x+18}=\frac{3}{2}\Rightarrow\frac{x+6}{x+18}=\frac{1}{2}$

Cross multiply and solve:  $2x+12=x+18 \Rightarrow x=6$

Natalya will be $3x+18$ then... So she is $3(6)+18\Rightarrow36$
[/hide]

How can that be wrong?

because I checked my his answer ..

if F = 18 and N = 54

when you add 18 to the both sides, you would get 36 and 72..... which is the twice as many as N..

so, it can't be wrong...

I think when it said that the ratio will be cut in half, it actually meant that it's gonna be 1:1.5 NOT 1:2

Lucy is correct on number 4, and Farazy's method shows how you can find the answer.  No one has answered number 3 correctly yet.  Keep trying :)  Make sure you read the questions very carefully -- I think that is part of the problem on both 3 and 4.

I just misunderstood the problem. Wording was confusing. ;)

It sounds like it was switched to half, not to 3/2. Oh well. I was doing my biology hwk when I did that anyway. :D

[quote]It sounds like it was switched to half, not to 3/2. Oh well. I was doing my biology hwk when I did that anyway. [/quote]

I hate ( I mean, really hate ) biology, so I usually copy off from other people's homework....

well, doing two things at a time can be really confusing.. :D

Number 3:

[hide]
Call Jose's age $j$ and Shanai's age $s$. After $s$ more years, Shanai will be double her current age, so we need to find:

$\frac{j+s}{s+s}$

We know that $j=3s$, so we can just sub in:

$\frac{(3s)+s}{s+s}=\frac{4s}{2s}=\frac{2}{1}=2$

Sheesh, this problem must have tricked me five times. :) [/hide]

[quote]3. Shanai is one third as old as Jose.
When Shanai is double her current age, how many times older than her will Jose be?[/quote][hide]
Let $x$ = Shanai's present age.
Then $3x$ = Jose's present age.

When Shanai doubles her age, she will be $2x$ years old.
$\text{This happens exactly }x\text{ years from now!}$

Then Shanai will be $x + x = 2x$ years old.
Hence, Jose will be $3x + x = 4x$ years old.

He will be [b]twice[/b] as old as Shanai.[/hide]


trigonometry
inequalities
geometry
circumcircle
incenter
function
trig identities

Prove that if $ \angle A, \angle B, \angle C$ are the interior angles of the nondegenerate $ \triangle ABC$,

$ \cos A \plus{} \cos B \plus{} \cos C \le \dfrac{3}{2}$

[hide="solution"]
Use the Law of Cosines:

$ \cos A \equal{} \dfrac{b^2\plus{}c^2\minus{}a^2}{2bc}$

$ \cos B \equal{} \dfrac{c^2\plus{}a^2\minus{}b^2}{2ca}$

$ \cos C \equal{} \dfrac{a^2\plus{}b^2\minus{}c^2}{2ab}$

Then the left hand side becomes

$ \dfrac{b^2\plus{}c^2\minus{}a^2}{2bc} \plus{} \dfrac{c^2\plus{}a^2\minus{}b^2}{2ca} \plus{} \dfrac{a^2\plus{}b^2\minus{}c^2}{2ab}$

$ \equal{} \dfrac{ab^2 \plus{} c^2a \minus{} a^3 \plus{} bc^2 \plus{} a^2b \minus{} b^3 \plus{} ca^2 \plus{} b^2c \minus{} c^3}{2abc}$

$ \equal{} \dfrac{\minus{} \displaystyle\sum_{cyc} [a(a\minus{}b)(a\minus{}c)] \plus{} 3abc}{2abc}$

By Schur's Inequality,

$ \sum_{cyc} [a(a\minus{}b)(a\minus{}c)] \ge 0$

Therefore,

$ \dfrac{\minus{} \sum_{cyc} [a(a\minus{}b)(a\minus{}c)] \plus{} 3abc}{2abc} \le \dfrac{3abc}{2abc} \equal{} \dfrac{3}{2}$

Equality holds in Schur's Inequality iff $ a \equal{} b \equal{} c$ or if two of the variables are equal and the other is equal to zero. However, since $ \triangle ABC$ is defined to be nondegenerate, the latter equality case is invalid.

Therefore, $ \cos A \plus{} \cos B \plus{} \cos C \equal{} \dfrac{3}{2} \Leftrightarrow \angle A \equal{} \angle B \equal{} \angle C \equal{} 60^\circ$.

[/hide]

hello, you can also prove the following equation
$ \cos(A)\plus{}\cos(B)\plus{}\cos(C)\equal{}1\plus{}\frac{r}{R}$, here $ R$ and $ r$ denote the radii of the circumcircle and the incircle of the given triangle.
Sonnhard.

[hide="Question?"]
So we have by Jensen's that:
$ \cos{A} + \cos{B} + \cos{C} \leq 3 \cdot \cos{\frac{180}{3}} = \frac{3}{2}$
So that looks fine and dandy.

But doesn't $ \cos$ change concavity in the interval $ [0, \pi]$ at ${ \frac{\pi}{2}}$? ..And Jensen's kind of requires concavity of the function to work.
Why does it work in this case? Because $ \cos{\theta} \in [-1,1]$ ?
[/hide]

[quote="probability1.01"][hide="another idea (fill in the details)"]
Let $ I$ be the incentre of $ ABC$ and $ r$ the inradius. Prove equivalence to
\[ AI \plus{} BI \plus{} CI \ge 6r
\]
which reduces to
\[ \csc A/2 \plus{} \csc B/2 \plus{} \csc C/2 \ge 6
\]
.

OK actually I just realized that if you already know Jensen then you might as well just use it directly? oh well in any case Kevin should stop doing geometry and inequalities.
[/hide][/quote]
umm what's Jensen? :?:

He gave you that problem on CSC? what's csc

[quote="Aryth"][quote="stuck"][hide="Question?"]
So we have by Jensen's that:
$ \cos{A} + \cos{B} + \cos{C} \leq 3 \cdot \cos{\frac {180}{3}} = \frac {3}{2}$
So that looks fine and dandy.

But doesn't $ \cos$ change concavity in the interval $ [0, \pi]$ at ${ \frac {\pi}{2}}$? ..And Jensen's kind of requires concavity of the function to work.
Why does it work in this case? Because $ \cos{\theta} \in [ - 1,1]$ ?
[/hide][/quote]

Do not question Alex Zhai  :ninja:  .
I think he was referring to the csc problem he gave.[/quote]

Oh, no. No one in his right mind would do that. I was simply wondering why it worked in this case when the function does not keep the same concavity over the entire interval.

[quote="Aryth"]$ \csc$: cosecant.

$ \csc(x) \equal{} \dfrac{1}{\sin x}$[/quote]
oh never mind i thought it stood for the name of a math contest
me am so stupid :mad:


I know you have provided less-and-less support for older browsers, but I recently had occasion to visit your site from a Windows 2000 machine running Netscape 4.73.  The Forums would not load, at all.  Are you aware of this, and is this your intent?

Separately, your site's clock is off by about 5 minutes.  If you run Linux, I would suggest NTP.  With your classrooms / chatrooms opening and closing automatically at specified times, the incorrect clock is more visible to users.

That is a [b]very, very, very[/b] old version of Netscape (currently at 7.2 if I'm not mistake-ing). I suggest you use IExplore instead if you have a Windows 2000 machine.

The forum's setup has been tested with older browsers also (such as Opera 6, IE 5), but it's hard to make everything perfect, so indeed, all optimization is made for the newest browsers (~95% of the site's visitators use IE 6, Netscape 7 or more, Opera 7 or more or Firefox).

[quote="rcv"]Separately, your site's clock is off by about 5 minutes.  If you run Linux, I would suggest NTP.  With your classrooms / chatrooms opening and closing automatically at specified times, the incorrect clock is more visible to users.[/quote]Thanks for pointing this out  -- I have reset our server's clock.

Your system clock seems to be off by about 5 minutes, again.  I still recommend running NTP.  Thank you.


function
abstract algebra
algebra
polynomial
rational function
superior algebra
superior algebra unsolved

Let me quote my book (Matsumura):

Let $ R \equal{} \bigoplus_{n \geq 0} R_n$ be a Noetherian graded ring with $ R_0$ Artinian and let $ M$ be a finitely generated graded $ R$-module. Suppose $ R \equal{} R_0[x_1,...,x_r]$ with $ x_i$ of degree $ d_i$, and suppose that $ P(M,t) \equal{} \sum_{n \equal{} 0}^{\infty} l(M_n) t^n$ then $ P(M,t)$ is a rational function of $ t$ and can be written as $ P(M,t) \equal{} f(t)/\prod_{i \equal{} 1}^r (1 \minus{} t^{d_i})$.

A lot of information on the values of $ l(M_n)$ can be obtained from the above. Especially when $ d_1 \equal{} ... \equal{} d_r \equal{} 1$, so that $ R$ is generated over $ R_0$ by elements of degree $ 1$. In this case $ P(M,t) \equal{} f(t)(1 \minus{} t)^{ \minus{} r}$; if $ f(t)$ has $ (1 \minus{} t)$ as a factor, we can cancel to get $ P$ in the form:

$ P(M,t) \equal{} f(t)(1 \minus{} t)^{ \minus{} d}$

Write $ (1 \minus{} t)^{ \minus{} d} \equal{} \sum_{n \equal{} 0}^{\infty} { d \plus{} n \minus{} 1 \choose d \minus{} 1} t^n$.

If $ f(t) \equal{} a_0 \plus{} a_1 t \plus{} ... \plus{} a_s t^s$ then

$ l(M_n) \equal{} a_0 { d \plus{} n \minus{} 1 \choose d \minus{} 1} \plus{} ... \plus{} a_s { d \plus{} n \minus{} s \minus{} 1 \choose d \minus{} 1}$

here we set $ { m \choose d \minus{} 1} \equal{} 0$ for $ m < d \minus{} 1$. We can rearrange this as a polynomial in $ n$ with rational coefficients

 :?:  :rotfl:  :?:

$ \phi(X) \equal{} \frac {f(1)}{(d \minus{} 1)!}X^{d \minus{} 1}$ + terms of lower degree

Someone can clear up this last step?

This is strange, because the last step is probably the most elementary one in the proof (without looking up, I cannot even say what $ l\left(M_n\right)$ is, but I hope it is the dimension of $ M_n$ if $ R_0$ is a field...).

We have

$ l\left(M_n\right) \equal{} a_0 { d \plus{} n \minus{} 1 \choose d \minus{} 1} \plus{} a_1 { d \plus{} n \minus{} 2 \choose d \minus{} 1 } \plus{} ... \plus{} a_s { d \plus{} n \minus{} s \minus{} 1 \choose d \minus{} 1}$.

In other words, $ l\left(M_n\right) \equal{} \phi\left(n\right)$, where $ \phi$ is a polynomial defined by

$ \phi\left(X\right) \equal{} a_0 { d \plus{} X \minus{} 1 \choose d \minus{} 1} \plus{} a_1 { d \plus{} X \minus{} 2 \choose d \minus{} 1 } \plus{} ... \plus{} a_s { d \plus{} X \minus{} s \minus{} 1 \choose d \minus{} 1}$.

If we try to write out $ \phi$ explicitely as a sum of powers of $ X$, we get

$ \phi\left(X\right) \equal{} a_0 { d \plus{} X \minus{} 1 \choose d \minus{} 1} \plus{} a_1 { d \plus{} X \minus{} 2 \choose d \minus{} 1 } \plus{} ... \plus{} a_s { d \plus{} X \minus{} s \minus{} 1 \choose d \minus{} 1}$
$ \equal{} a_0 { X \plus{} d \minus{} 1 \choose d \minus{} 1} \plus{} a_1 { X \plus{} d \minus{} 2 \choose d \minus{} 1 } \plus{} ... \plus{} a_s { X \plus{} d \minus{} s \minus{} 1 \choose d \minus{} 1}$
$ \equal{} a_0 \frac {\left(X \plus{} d \minus{} 1\right)\left(X \plus{} d \minus{} 2\right)...\left(X \plus{} 1\right)}{\left(d \minus{} 1\right)!} \plus{} a_1 \frac {\left(X \plus{} d \minus{} 2\right)\left(X \plus{} d \minus{} 3\right)...X}{\left(d \minus{} 1\right)!} \plus{} ... \plus{} a_s \frac {\left(X \plus{} d \minus{} s \minus{} 1\right)\left(X \plus{} d \minus{} s \minus{} 2\right)...\left(X \minus{} s \plus{} 1\right)}{\left(d \minus{} 1\right)!}$
$ \equal{} a_0 \frac {X^{d \minus{} 1} \plus{} \text{ terms of lower degree in }X}{\left(d \minus{} 1\right)!} \plus{} a_1 \frac {X^{d \minus{} 1} \plus{} \text{ terms of lower degree in }X}{\left(d \minus{} 1\right)!} \plus{} ... \plus{} a_s \frac {X^{d \minus{} 1} \plus{} \text{ terms of lower degree in }X}{\left(d \minus{} 1\right)!}$
$ \equal{} \frac {a_0 \plus{} a_1 \plus{} ... \plus{} a_s}{\left(d \minus{} 1\right)!}X^{d \minus{} 1} \plus{} \text{ terms of lower degree in }X$
$ \equal{} \frac {f\left(1\right)}{\left(d \minus{} 1\right)!}X^{d \minus{} 1} \plus{} \text{ terms of lower degree in }X$

(because $ a_0 \plus{} a_1 \plus{} ... \plus{} a_s \equal{} f\left(1\right)$).

  darij


abstract algebra
number theory unsolved
number theory

Find all positive integers $ n$ that can be uniquely expressed as a sum of at most five nonzero perfect squares. (The order of the summands is irrelevant.)

suppose that $ n \ge 25$.the case $ n < 25$ is easy.
due to lagrange's theorem every integer can be represented as a sum of four perfect squares.
note that in the following cases $ x,y,z,w>0$ and also $ l_i,m_i,k_i \ge 0$.
1-if $ n$ can be written in this form $ n\equal{}x^2\plus{}y^2$ then since $ \{2,1\} \neq  \{x,y\}$ so we choose one of the number $ 1$ or $ 2$ which isn't in the set $ \{x,y\}$,suppose that this number is $ t \in \{1,2\}$,since $ n\minus{}t^2$  can be written as a sum of four perfect squares then we have $ x^2\plus{}y^2\equal{}n\equal{}t^2\plus{}(n\minus{}t^2)\equal{}t^2\plus{}l_1^2\plus{}l_2^2\plus{}l_3^2\plus{}l_4^2$
so the number $ n$ can be written in two different forms and we are done for this case.
2-now suppose that $ n$ can't be written as the sum of two perfect squares but can be written as the sum of three perfect squares .then we can do the same as the previous case.suppose that $ n\equal{}x^2\plus{}y^2\plus{}z^2$.then at least one of the number $ 1,2,3,4$ doesn't appear in the set $ \{x,y,z\}$.suppose that this number is $ t \in \{1,2,3,4\}$.then the number $ n\minus{}t^2$ can be written as the sum of four perfect squares so we get $ x^2\plus{}y^2\plus{}z^2\equal{}n\equal{}t^2\plus{}(n\minus{}t^2)\equal{}t^2\plus{}m_1^2\plus{}m_2^2\plus{}m_3^2\plus{}m_4^2$.obviously these two representation are different.so we are done for this case.
3-finally suppose that $ n$ can't be written as the sum of two or three perfect squares.suppose that $ n\equal{}x^2\plus{}y^2\plus{}z^2\plus{}w^2$.
again,one of the number $ 1,2,3,4,5$ doesn't appear in the set $ \{x,y,z,w\}$.let this number be $ t \in \{1,2,3,4,5\}$.then the number $ n\minus{}t^2$ can be written as the sum of four perfect squares so we get $ x^2\plus{}y^2\plus{}z^2\plus{}w^2\equal{}n\equal{}t^2\plus{}(n\minus{}t^2)\equal{}t^2\plus{}k_1^2\plus{}k_2^2\plus{}k_3^2\plus{}k_4^2$ and so we have two different representation for the number $ n$ and we r done for this case!


you're definitely right. i remember a space to choose your roommate. of course, i didn't know anybody on my team, but you probably do. If so, you can always just tell your coach.

I have been to nationals with at least one of my kids the last 3 years(and will again this year) as a coach and there was always a spot for choosing a roommate, but then again there were also three forms to fill out.  They consolidated the forms into one, but I guess there wasn't enough room for the roommate thing.  I just emailed the National office to request that the 2 kids from my school could room together.

Good luck at Nats,

frost

weird, we didn't get any forms to fill out.

[quote="captcha000"]Do any Mathcounts veterans remember the registration forms from last year? Because unless I remember incorrectly, on last year's forms, there was a space for you to choose your roommate. This year its gone![/quote]

Just ask your coach to do what frost did. I think his solution was pretty reasonable.

(See, captcha's afraid I'm going to "flood the bathroom" again...)

ml2, i believe the coordinator said he didn't have the forms yet, and that he'd have to contact nats coordinators to get them or something like that

So rooms will be 2 to a room? or maybe this year you just get bigger rooms and you all share it.


if     x+y+z=360

prove

(cos x)^2 +(cos y)^2 +(cos z)^2 =2cos x cos y cos z  + 1

Note that

$\\ 2 \cos x \cos y = \cos (x+y)+\cos(x-y)$ and

$\\ \cos^{2}x+\cos^{2}y-1 =\cos (x+y) \cos(x-y)$ .

Now the equality is written as

$\\ (\cos z-\cos (x+y) )(\cos z-\cos (x-y) )=0$

which is true if $x+y+z =2 \pi$ .


Let be given $n\in{\mathbb Z}_+$. Assume that $f,g$ are polynomials with integers coefficients not constants so that $f(k)/g(k)$ is a $n-$ power of an integer number for all $k\in\mathbb Z$. Prove or disprove that there is a polynomial $h(x)\in Z[x]$ so that
$f(x)=\Big(h(x)\Big)^3g(x)$ for all $x$.

Enjoy~  :)

[quote="enescu"]It's easy to see that $g$ divides $f$, then look at [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=51600]this topic[/url], posts #8-10.[/quote]

Suprise! Thank u for the nice link ;)

But what is $n$ in the statement? Is it $n=3$?

Pierre.

I assumed our friend made a typo, writing $3$ instead of $n$.


WOO I'M BORED I'LL DO AN AHSME QUESTION

[hide="DID I HEAR YOU SAY CHICKEN"]note that $\log_y x = \frac{1}{\log_x y}$, so in that first equation, we have that something plus its reciprocal is $\frac{10}{3}$ - it's pretty obvious that one of the things is $3$ and the reciprocal is $\frac{1}{3}$. This implies that $y = x^3$ (or vice versa, but it clearly doesn't matter as everything is symmetric). Thus $x^4 = 144 \Rightarrow x = 2\sqrt{3} \Rightarrow y = 24\sqrt{3} \Rightarrow \frac{x+y}{2} = 13\sqrt{3}$.[/hide]

as indicated by scientific data presented in beta's poll, the answer is obviously B

[hide="Solution"]
Denote $u = \log_x y$.  We have $u + \frac{1}{u} = \frac{10}{3} \implies u = 3$ so $y = x^3$ and $xy = x^4 = 144 \implies x = 2 \sqrt{3}, y = 24 \sqrt{3}$.  This gives $\frac{x+y}{2} = 13 \sqrt{3} \implies \boxed{B}$.
[/hide]

[hide]Let $a=\log_xy$.  Then $a+\frac{1}{a}=\frac{10}{3}$.  Multiplying by $a$ and solving the quadratic gives $a=3,\frac{1}{3}$.  WLOG let $a=3$.  Then by the definition of log, $x^3=y$.  Since $xy=144$, then $y=\frac{144}{x}$.  Substituting this gives $x^3=\frac{144}{x}$, therefore $x=\sqrt{12}$.  From this $y=12^{\frac{3}{2}}$.  Therefore $\frac{x+y}{2}=\frac{\sqrt{12}+12^{\frac{3}{2}}}{2}=13\sqrt{3}$, so the answer is $\text{B}$.[/hide]


algebra
polynomial
geometry
3D geometry
sphere
topology
function

a polynomial $P: \mathbb{C}\to \mathbb{C}$ induces a map on the Rieman sphere, hence can be seen as a map $\mathbb{S}^{2}\to \mathbb{S}^{2}$. Show that the degree of this application is exactly the degree of this polynomial.

Just a brief sketch:

Collapsing the terms of the polynomial of less-than-maximal degree in a continuous manner gives a homotopy of the polynomial into a function of the form $z\mapsto az^{n}$ (all maps are regarded as self-maps of the sphere). After the orientation-preserving change of coordinates $z\mapsto bz,\ b^{n}=a$, we can assume that the map we're dealing with is $z\mapsto z^{n}$. This is the suspension of the map $z\mapsto z^{n}$ on the unit circle, which has degree $n$, and since suspension preserves degrees, we're done.


linear algebra
matrix
group theory
abstract algebra
invariant
superior algebra
superior algebra solved

Let $G= \{ A \in \mathcal M_2 \left( \mathbb C \right) \mid |\det A| = 1 \}$ and $H =\{A \in \mathcal M_2 \left( \mathbb C \right) \mid \det A = 1 \}$. Prove that $G$ and $H$ together with the operation of matrix multiplication are two non-isomorphical groups.

$\infty?$ And what does that double $\in$ sign mean?

Thanks for noticing the typo :) It's fixed now :P

[quote="grobber"]This can quite effortlessly be extended to larger values of $n$ :).

Consider a subgroup of $GL_n(\mathbb C)$ made up of symmetries (i.e. matrices $A$ satisfying $A^2=1$). All these matrices are diagonalizable and they commute pairwisely, so they are simultaneously diagonalizable. This means that any such group is a conjugate of a subgroup of the diagonal matrices with diagonal entries equal to $\pm 1$. In particular, every such group is finite.

Now given a subgroup $K\le GL_n(\mathbb C)$, let $u(K)$ be the size of the largest subgroup of $K$ made up of isometries (largest meaning with the most elements, not in the sense of inclusion). $u$ is invariant to isomorphism, of course, but it's clear from the paragraph above that $u(G)=2u(H)$ (where $G,H$ are defined just like in the initial problem, but with $n$ instead of $2$).[/quote]

This problem was proposed by Marius Cavachi and it has a solution more fit to the XII-th grade level. Show some respect  :P

[edit]: Ok, I admit that the "elementary" proof is not easy to generalize.

[quote="Cezar Lupu"]Mihai how do you know that this problem has been proposed by Marius Cavachi?[/quote]Given that he was his teacher for some time, I would guess they talk from time to time :)

My solution in the contest: Look for $Z(G)$ and $Z(H)$. $Z(G)$ has an infinity of elements, and $Z(H)$ just two. If $G$ and $H$ are isomorphical, then $Z(G)$ and $Z(H)$ are isom., which is false. And it is quite easy to generalise.

Can anybody post another solution?  :oops:

The author's solution was not that easy. It relied on characterizing the squares in each group.


Suppose that $M_i\subset${1,2,3,...,9} and $|M_i|=3$   $\forall i=1,..,k$
And $|M_i\cap M_j|\le 1$   $\forall i\ne j$
a)	Prove that $k\le 12$
b)	Find 12 sets satisfy the above conditions
c)	How many groups of 12 sets can we find?
a) and b) are easy but the final one I still haven’t solved yet

in the case of equality($k=12$) it simply implies that the structure you are looking at is the affine plane of order $3$.but the affine plane of order $3$ is unique upto relabelling of its elements.


search
function
number theory theorems
number theory

How can we prove that for each prime p, ord (k) = p-1 for some k?

so what? I cannot find a [i]proof[/i]...   :(

Look at http://www.mathlinks.ro/Forum/viewtopic.php?t=55473
or see the second last post in http://www.mathlinks.ro/Forum/viewtopic.php?t=75806 for why every finite group with at most $ n$ elements $ x$ with $ x^n$ is cyclic.

Ok thanks, the search function gave too many results for "primitive root".  :)

You are implying that k is a primitive root for all primes which is not true. Different primes have different primitive roots.


Solve for x,y
$ \frac{x}{2}\plus{}\frac{3}{2x}\equal{}y\minus{}1$
$ \frac{y}{2}\plus{}\frac{7}{2y}\equal{}x\minus{}1$

let $ \alpha\equal{}x\plus{}1 \text{ and } \beta\equal{}y\plus{}1$ so the equations is equivalent to $ 2(\alpha\minus{}1)(\beta\minus{}1)\equal{}\alpha^2\plus{}2\equal{}\beta^2\plus{}6$. From the last two members we know that $ \alpha^2\minus{}\beta^2\equal{}4$ that is a iperbole with vertex $ (\pm 2,0)$ flattened to the $ \alpha$-axis, in the plane $ \alpha 0 \beta$.

But it is also true that $ 2(\alpha\minus{}1)(\beta\minus{}1)\equal{}\frac{(\alpha^2\plus{}2)\plus{}(\beta^2\plus{}6)}{2}$. if we define $ S\equal{}\alpha\plus{}\beta$ and $ P\equal{}\alpha\beta$ so we obtain $ 4P\equal{}(S\plus{}2)^2$, and since the variables are reals so $ \delta\equal{}S^2\minus{}4P \ge 0\implies S \le \minus{}1$.

So, if a solution exists then it is on the left side of the iperbola. We need to intersect that curve with another one to find all interesection points, so, giving the first initial equation we get $ \beta\equal{}\alpha\plus{}3\plus{}\frac{3}{\alpha\minus{}1}$. We easily see that $ (\alpha,\beta)\equal{}(\minus{}2,0)$ is common to both curves. We can show that it is the unique.

In fact for every $ \alpha < \minus{}2$ we have $ \beta\equal{}\alpha\plus{}3\plus{}\frac{3}{\alpha\minus{}1}\equal{}\pm \sqrt{\alpha^2\minus{}4}$, but we obtain:
$ \sqrt{\alpha^2\minus{}4}> \minus{}\sqrt{\alpha^2\minus{}4} > \alpha\plus{}3\plus{}\frac{3}{\alpha\minus{}1}$ iff $ (\alpha\plus{}2)(6(\frac{5}{12}\minus{}\alpha)^2\plus{}\frac{23}{24})>0$, and we are done  :lol:

[quote="I"] [...]if we define $ S \equal{} \alpha \plus{} \beta$ and $ P \equal{} \alpha\beta$ so we obtain $ 4P \equal{} (S \plus{} 2)^2$[...][/quote]

It is wrong  :(  it is only $ 6P\equal{}(S\plus{}2)^2$ and all changes..

however, if we sostitute the expression of $ y$ in the second equation we obtain: $ P(x)\equal{}(x\plus{}3)(3x^3\minus{}9x^2\minus{}7x\minus{}3)\equal{}0$ which has two real solutions:
(x,y)=(-3,-1) and (x,y)=(3,70303921509.. , 3,25659228.. )


What is "Old and New Inequalities"???
(stupid question)

That is name of a book on inequalities. :)

I don't know how to buy a book from GIL, could you show me how?

[quote="Mircea Lascu"]"Old and New Inequalities" is actually a collection of problems about inequalities. Here are few details about the book
Authors: T. Andreescu, V. Cartoaje, G. Dospinescu, M. Lascu
Title: Old and New Inequalities
ISBN: 973-9417-35-3
Publisher: GIL Publishing House

There are also other books published by GIL and you can find them [url]http://www.gil.ro/en[/url].[/quote]
I can't go to http://www.gil.ro/en

[quote="Mircea Lascu"]"Old and New Inequalities" is actually a collection of problems about inequalities. Here are few details about the book
Authors: T. Andreescu, V. Cartoaje, G. Dospinescu, M. Lascu
Title: Old and New Inequalities
ISBN: 973-9417-35-3
Publisher: GIL Publishing House

There are also other books published by GIL and you can find them [url]http://www.gil.ro/en[/url].[/quote]

Hey Mircea can I buy that book in India?

[quote="Poincare"][hide]you can get it in the small stores for cheap, not the big ones like crossword though [/hide][/quote]

I live in Assam and I have searched whole NE India and cd nt fine that!Plz help!


Suppose $ \{N_n, \ n \ge 0\}$ is a sequence of normal random variables (random variables with normal distributions). Show $ N_n$ converges to $ N_0$ in distribution if and only if $ E(N_n) \to E(N_0)$ and $ Var(N_n) \to Var(N_0)$.

If I'm not mistaken, this is a straight forward application of equality between convergence in distribution and convergence ( by points ) of corresponding characteristic functions.


[quote="Swift"]First thank you to anyone who reads this in order to help me solve my problem.
I am being reminded that I do not understand how opposite and adjacent sides relate to the hypoteneus.
I have one side of a right triangle and two angles, my angles are 90º and 15, also the length of my opposite side is .75
Could anyone tell me what formula to use and why?  Thank you very much[/code][/quote]
I'm assuming the 15º is opposite to the 0.75 and that we are supposed to find the other side. In this case, we know that
$\tan a=\frac{opposite}{adjacent}$, in terms of some angle $a$. We know $0.75$ is opposite to the 15º, so
$\tan 15=\frac{0.75}{adjacent}$. now you solve for adjacent and then you're done. If you have any questions, feel free to ask. :)
By the way, this should be in the high school basics forum.

Actually I am sorry, 15 is not the opposite of .75, it is adjacent.  The opposite angle of that side would be 75º.

Regardless, If I used .75/tan-1(15)=.498
Tan(15)= is resulting in a numbe too large for my triangle.  I am sorry to be asking such a newbie question, its been so long since I have had to use these formula's.  Thanks guys and gal's for all the support.

again here are the specs of my triangle,  the opposite side is .75, the angle adjacent to that is 90 and 15, the opposite angle is 75.  I believe you may be totally right though with all the sides and I am the one incorrect Inspired by Nature.

With your result's though, I hate to have to ask but how would I still solve for the adjacent if tan15=.75/adjacent.

Thanks again for all the help.  Math Guru's = GOD.

EDIT:  I did some more calculations and I believe my final answer is going to AdjacentSide=.75/tan(75)=.2009618943

That small number looks fair enough to be the proper side length of my triangle

Now that gives me one new side of a new triangle with the same angle's only the sides are way shorter.  If .2 is my Opposite side and the opposite angle of .2 is 75 I would do the same formula again wouldnt I?  In actuallity I believe to find that new hypotneus side I must do (.75/tan(75))/sin(75)=HypotneussTriangle2

You all are greatly appreciated.  This website is a god-send to people like myself who are out of classes, but still have to use trigonomic functions in the work-world.  Thanks for the support

uuhh.. what are you trying to find? Because if I know, I can help you. :)

I am trying to find the length of the hypoteneus on a second triangle.  My first triangle I know one side is .75, the angles touching that side are 90 and 15.  The opposite angle would be 75 I believe.  The adjacent sides of the two triangles are equal.  The second triangle I also want to say has to same angles as the first triangle But the

[url=http://img357.imageshack.us/my.php?image=trianglerw8.png][img]http://img357.imageshack.us/img357/9538/trianglerw8.th.png[/img][/url]

Sorry for this very crude drawing.  But I am trying to solve for x.  I know the values of my first triangle, one side is .75 and the opposite angle of that is 75 also, because we know that its a right triangle.

Thats all for the first(RIGHTSIDE) triangle.  So I need to find the side that they both share in order to find the hypotenues of my second(LEFTSIDE) triangle which is X

I know this is very basic math for some of you guys but It's got me stomped.  Thanks a lot for the help.  Much appreciated.

Again I am sorry for the crude image, let me define..... the top number above the triangle you see would be .75.  The x is the letter on the far left of the picture by the hypotenues.  Solve for X

The angles on the left sided triangle like I said before are 90, 15, and 75.  The 15º is the angle on the top right of the triangle.

If anyone can solve for X(the left sided hypoteneus) and show me how they did that I would really be thankful.

With the information given this is impossible. We know one side and one angle of the left triangle. Are you sure there is nothing else? And $\tan(15)$ should be small, not large.

If $k$ is the length of the vertical line, then $k/.75=\tan(15)$.

Aha! Were you taking the tan in radians? If so you would have gotten about 140 degrees, which would be a negative number. That can't be it...


Given triangle $ ABC$ with interenal anlge bisector is $ w_{a},w_{b},w_{c}$ and semiperimeter $ s$ prove that
\[ \frac{1}{w_{a}^{2}}+\frac{1}{w_{b}^{2}}+\frac{1}{w_{c}^{2}}\geq\frac{9}{s^{2}}\]

The inequality can be written as
$  \frac{(b+c)^{2}}{4bcs(s-a)}+\frac{(a+c)^{2}}{4acs(s-b)}+\frac{(a+b)^{2}}{4abs(s-c)}\geq\frac{9}{s^{2}}$
We have $  \frac{(b+c)^{2}}{4bc}\geq 1,\frac{(a+c)^{2}}{4ac}\geq 1,\frac{(a+b)^{2}}{4ab}\geq 1$.
So we have to prove the inequality $  \frac{1}{s-a}+\frac{1}{s-b}+\frac{1}{s-c}\geq\frac{9}{s}$
We have $  [(s-a)+(s-b)+(s-c)]\left(\frac{1}{s-a}+\frac{1}{s-b}+\frac{1}{s-c}\right)\geq 9\Leftrightarrow$
$  \Leftrightarrow s\left(\frac{1}{s-a}+\frac{1}{s-b}+\frac{1}{s-c}\right)\geq 9\Leftrightarrow\frac{1}{s-a}+\frac{1}{s-b}+\frac{1}{s-c}\geq\frac{9}{s}$

[color=purple]The following problem is due to Walther Janous  :) :
   $ \frac{1}{l_{a}^{2}}\plus{}\frac{1}{l_{b}^{2}}\plus{}\frac{1}{l_{c}^{2}}\le \frac{1}{3r^{2}}$ [/color]

[quote="evarist"][color=purple]The following problem is due to Walther Janous  :) :
   $ \frac {1}{l_{a}^{2}} \plus{} \frac {1}{l_{b}^{2}} \plus{} \frac {1}{l_{c}^{2}}\le \frac {1}{3r^{2}}$ [/color][/quote]
It's equivalent to $ \sum_{cyc}(a \minus{} b)^2ab(3a \plus{} 3b \minus{} c)\geq0,$ which true for all non-negative $ a,$ $ b$ and $ c.$


when i select hide from the toolbar above the textbook, it say "click to reveal hidden content" in blue. How do i choose what that says?

Use hide=label inside the [] to get something like [hide="label"]stuff[/hide]

This isn't $ \text{\LaTeX}$.

[hide="thanks"]i know its not $ LaTeX$[/hide]


vector
linear algebra
linear algebra unsolved

true or false?

1.
let O be an homogen linear eauation's system and M is not.
both have m equations number and n number of variables,and they have exacly the same "minimized"(without results) matrix.
if O has infinite number of solutions than m<n or m=n.
2.
same data as q 1,
if m<n  or m=n than M has infinited number of solutions.
3.
same as q 1,
if vectors x and y are solutions for M and dalta1*x+dalta2*y  is a solution of M than dalta1+dalta2=1
4.same as q 1,
if any vector that runs in 4 solves O than M has no solutions.
5.same as q 1,
if O has infinite number of solutions than M has infinite number of solutions.
6.last question that is same as q 1,
if M has no solution than O might have infinite or one solution.

7.let A,B,C,D be nxn matrix.
if a*a=0 than a=0.
8.same as q 7,
if a*a=a than a=0 or a=i.
9.same as q 7,
if ab=0 than ba=0.
10.same as...
if ab=0 and b is not 0 than a=0
11.same
 a*a-b*b=(a-b)(a+b) if and only if(two sides) (a+b)*(a+b)=a*a+2ab+b*b