1. That is pretty true, referencing this article from NBC: [url]http://www.nbcolympics.com/destinationbeijing/blogs/postid=221688.html#were+no+u+s+china+chase+spot[/url]
"The U.S., virtually alone among nations, counts total medals won, while China, like the rest of the world focuses on golds. So for now, both can declare victory."
It basically admits that the U.S.  method (the method they use to rank countries) isn't accepted by the rest of the world, but that they use it so the U.S.  can be on top.

2. In this day and age, everything is so international, team's are all shuffled.  In many instances, people leave a country b/c their program is too good, and go to others where they can still make the team, for eg, in the case of ping-pong, chinese players play for many countries:
[url]http://www.nbcolympics.com/tabletennis/news/newsid=223558.html#chinas+hand+downs[/url]

1. USA isn't playing fair. USA must be first, and they use any methods to achieve it.

2. 20,000,000 Chinese are playing table tennis profesionally so there is very big competition for the members of the represenation, so Chinese go to other countries and play for them.

1. the international olympic committee does not keep official records of global country ranking.  thus, any tally is unofficial and subject to an arbitrary weighting procedures.  moreover, canada uses the U.S. convention, too (check the CBC's site).  thus, this is more of a north american thing than some kind of ulterior plot to manipulate the games.  finally, i think that your claim that golds are worth more than silvers and bronzes is contentious.  for example, was michael phelps' gold medal in the 100m butterfly worth infinitely more than the silver medal of the serbian who finished 0.01 seconds behind him?

2. stop being racist.  believe it or not, some people who aren't white consider themselves american, not members of their ancestors' nation(s).  also, quickster's explanation is false; many athletes from the united states compete under different countries because it's easier to qualify for the olympic team that way.

[quote="Squarerootofisoscelestria"]2. Some of the U.S. olympic teams have people from OTHER COUNTRIES in them.  For example, the men's gymnastics team has a Chinese person in it and their women's synchronized swimming team has a japanese girl.  WHAT IS UP WITH THAT??!! :huh:[/quote]

congratulations, this is the dumbest thing i have read in the last week.

by the way, it's not like either of these things have anything to do with fairness in the olympics, so great thread OP!

[quote="Squarerootofisoscelestria"]Some of the U.S. olympic teams have people from OTHER COUNTRIES in them.  For example, the men's gymnastics team has a Chinese person in it and their women's synchronized swimming team has a japanese girl.  WHAT IS UP WITH THAT??!! :huh:[/quote]

This is true of countries other than the US...for example Chris Kaman, an American basketball player, is on the German olympic team because his great-grandfather was born in Germany (there's a YahooSports article about this). This is certainly more of a stretch than the Korean guy (I believe he is Korean) on the US gymnastics team, who is a naturalized citizen or came to the US at a very young age (he speaks perfect English iirc). The same is true of the Indian alternate (first name Raj) on the US team.

Saying that this isn't allowed is like saying Alex Zhai shouldn't be allowed on the USA IMO team because he's Chinese.

[quote="Squarerootofisoscelestria"]In my view, the United States are not playing the Olympics fair for two reasons:

1. NBC ranks countries based on HOW MANY MEDALS THEY GET only so the United States could be ranked #1, even though the OFFICIAL OLYMPIC WEBSITE ranks countries on HOW MANY GOLDS THEY GET.  The way the United States ranks countries is not fair.  They are basically treating every medal with the same value, even though it's pretty obvious that gold is far better than silver or bronze.   :mad:

2. Some of the U.S. olympic teams have people from OTHER COUNTRIES in them.  For example, the men's gymnastics team has a Chinese person in it and their women's synchronized swimming team has a japanese girl.  WHAT IS UP WITH THAT??!! :huh:[/quote]

1. I'm aware of this, but because they're ranking things differently doesn't mean they aren't playing unfairly. You can see it as the US being optimistic.

2. Why would that make it unfair?
Btw, irrelevant, but the Chinese guy you were referring to (Kevin Tan) graduated from my school (as did kristi yamaguchi and don hertzfeldt)

also, using gold medals as the only metric of a country's olympic performance is really silly, as is a system counting all medals equally.  i'd prefer an even more complex system which awarded (varying numbers of) points to everyone making the top 8, although this probably would be too incomprehensible to most followers of the olympics.

A universal method would to point base the medals (gold=3, silver=2, bronze=1) it seems a little more fair that way.

[quote="Squarerootofisoscelestria"]In my view, the United States are not playing the Olympics fair for two reasons:

1. NBC ranks countries based on HOW MANY MEDALS THEY GET only so the United States could be ranked #1, even though the OFFICIAL OLYMPIC WEBSITE ranks countries on HOW MANY GOLDS THEY GET.  The way the United States ranks countries is not fair.  They are basically treating every medal with the same value, even though it's pretty obvious that gold is far better than silver or bronze.   :mad:

2. Some of the U.S. olympic teams have people from OTHER COUNTRIES in them.  For example, the men's gymnastics team has a Chinese person in it and their women's synchronized swimming team has a japanese girl.  WHAT IS UP WITH THAT??!! :huh:[/quote]

It's just globalization. Also, it's about the athletes, not the countries is what olympics is about nowadays since many wars are gone.

Hmm...... I just read the news that all US track participants had food poisoning from chinese food (after US rejected their food, china complained and US took that back......)

[quote="(^_^)"]In my view, China is not playing the Olympics fair for two reasons:

1. CCTV ranks countries based on HOW MANY GOLD MEDALS THEY GET only so China could be ranked #1, even though A LOT OF WEBSITES ranks countries on HOW MANY TOTAL MEDALS THEY GET. The way China ranks countries is not fair. They are basically treating silver and bronze with no value, even though it's pretty obvious that gold is not infinitely times more valuable than silver and bronze.

2. Some of the Chinese olympic organizers have people from OTHER COUNTRIES in them. For example, the chinese asked Steven Spielberg to be their artistic advisers and they had bunch of western architectures design their olympic stadiums. WHAT IS UP WITH THAT??!![/quote]

I am hoping that that is sarcasm...

What we can say is that there are always certain rankings that rank this country first, while another ranking would rank that country first, it is almost impossible to rank things in a way such that some people won't think is "unfair". Let's stop ranking these countries and just give out the medal counts.

[quote="iostream.h"]also, quickster's explanation is false; many athletes from the united states compete under different countries because it's easier to qualify for the olympic team that way.[/quote]
[quote="quickster94"]In many instances, people leave a country b/c their program is too good, and go to others where they can still make the team[/quote]

these statements seem pretty similar to me?

They should do 5 pts for gold 3 for silver and 1 for bronze, see who's most pts.

go to my blog and I've ranked the countries on all those weighting systems


injective function
counting
combinatorics
Marcin Kuczma
combinatorial inequality
IMO
IMO 1989

A permutation $ \{x_1, x_2, \ldots, x_{2n}\}$ of the set $ \{1,2, \ldots, 2n\}$ where $ n$ is a positive integer, is said to have property $ T$ if $ |x_i \minus{} x_{i \plus{} 1}| \equal{} n$ for at least one $ i$ in $ \{1,2, \ldots, 2n \minus{} 1\}.$ Show that, for each $ n$, there are more permutations with property $ T$ than without.

Let $a(n)$ be the number of permutations $\{ x_{i}| 1 \leq i \leq 2n \}$ of $\{ i | 1 \leq i \leq 2n \}$
such that $\left| x_{i}-x_{i+1}\right| = n$ for some $1\leq i<2n$

Consider the $(2n+2)$-element permutations

If the elements $\{n+1,2n+2\}$ are adjacent,
There are exactly $2(2n+1)!$ possible permutations

If the elements $\{n+1,2n+2\}$ are not adjacent,
There exist adjacent elements $(p,q)$ such that $p-q=n+1$ for $n+2 \leq p \leq 2n+1$, $1 \leq q \leq n$
Consider the ordered $(2n)$-element subset $\{ y_{i}| 1 \leq i \leq 2n \}$ without the elements $\{n+1,2n+2\}$
Construct the permutation $\{ z_{i}| 1 \leq i \leq 2n \}$ of $\{ i | 1 \leq i \leq 2n \}$ where $z_{i}=\left\{{\begin{tabular}{lll}y_{i}& if & y_{i}<n+1 \\ y_{i}-1 & if & y_{i}>n+1 \\ \end{tabular}}$
This permutation also satisfies the property because the elements $\{p,q\}$ will become $\{p-1,q\}$ and will remain adjacent in the new permutation
Note that the number of permutations $\{x_{i}\}$ that map to the same permutation $\{z_{i}\}$ is exactly $(2n)(2n-1)$ because there are exactly $(2n)(2n-1)$ ways to insert the elements $\{n+1,2n+2\}$ between adjacent elements in the permutation $\{z_{i}\}$ except between elements $\{p-1,q\}$

Therefore $a(n+1) = a(n) \times (2n)(2n-1)+2(2n+1)!$
$a(1) = 2$

$\frac{ a(n+1) }{ (2n)! }= \frac{ a(n) }{ (2n-2)! }+2(2n+1)$

Let $b(n) = \frac{ a(n+1) }{ (2n)! }$
$b(0) = 2$

$b(n) = b(n-1)+(4n+2)$
$b(n) = b(0)+\sum_{i=1}^{n}{4i+2}= 2+2n(n+2)$
$b(n-1) = 2 n^{2}$

$a(n) = b(n-1) (2n-2)! = 2 n^{2}(2n-2)!$

It is now obvious that $a(n) > \frac{1}{2}(2n)(2n-1) (2n-2)! = \frac{1}{2}(2n)!$

The number of ways $P$ such that the pairs $\{ 1, n+1\},\ldots ,\{n,2n\}$ do not appear as adjacent entries, i.e. the number of ways without property $T$, is by the inclusion-exclusion principle equal to \[P=\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}(2n-k)!2^{k}\leq \sum_{k=0}^{2}(-1)^{k}\binom{n}{k}(2n-k)!2^{k}<\frac{(2n)!}{2}\]
And the conclusion follows. $\blacksquare$

[quote="boxedexe"]$\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}(2n-k)!2^{k}\leq \sum_{k=0}^{2}(-1)^{k}\binom{n}{k}(2n-k)!2^{k}$[/quote]

This inequality is far from obvious.

I read about Benferroni's Inequality of the Exclusion and Inclusion principle.  If in an inclusion-exclusion, the sum ends after you add, the sum is less than or equal to the finite sum, and if it ends after you subtract, the infinite sum is greater than or equal to the finite sum.

Is this solution correct? Seems rather simple (found in about 20 minutes just now):

Let S be the set of permutations without property T, and T the set with.

I define the function $ f: S \rightarrow T$ by $ f((x_1,x_2,...,x_{2n})) \equal{} (x_2,x_3,....,x_k,x_1,x_{k \plus{} 1},...,x_{2n})$ where k is the unique positive integer such that $ |x_1 \minus{} x_k| \equal{} n$.

We first note that the image of f is a set of permutations with only one k such that $ |x_k \minus{} x_{k \plus{} 1}| \equal{} n$, because its domain was S and we have only created one such pair in applying the function. It is therefore a proper subset of T (not containing, for example (1,n+1,2,n+2,....,2n).

We next note that it is injective, because we can define an inverse function $ g((y_1,y_2,...,y_{2n})) \equal{} (y_{k \plus{} 1},y_1,y_2,....,y_{2n})$ where k is the unique integer such that $ |x_k \minus{} x_{k \plus{} 1}| \equal{} n$ defined on the image of f.

Thus there exists an injection from S into a proper subset of T, so $ |S| < |T|$ as required.

I found another solution but also am not sure if it's correct:

We count those permutations without property $ T$ and show that it is less than $ n(2n\minus{}1)!$.  At first we take out the numbers from $ n\plus{}1$ to $ 2n$. There are then $ n!$ ways to permute the numbers $ 1$ through $ n$.  Let a permutation of these numbers be $ \{a_1, a_2, ..., a_n\}$.  Now we begin inserting the rest of the numbers.  There are currently $ n\plus{}1$ "slots" existing before $ a_1$, between $ a_1$ and $ a_2$, between $ a_2$ and $ a_3$, ..., and after $ a_n$.  The number $ a_1\plus{}n$ must be inserted in one of the $ n\minus{}1$ slots not bordering $ a_1$.  Its insertion then makes a total of $ n\plus{}2$ slots.  Similarly, $ a_2\plus{}n$ must be inserted in one of the $ n$ slots not bordering $ a_2$, and so on, so the number of ways to insert the numbers $ n\plus{}1$ to $ 2n$ is $ (n\minus{}1)(n)(n\plus{}1)\cdots (2n\minus{}2)$, so the total number of permutations without property $ T$ is $ n(n\minus{}1)(2n\minus{}2)!<n(2n\minus{}1)(2n\minus{}2)!$.

[quote="ZzZzZzZzZzZz"]There are currently $ n \plus{} 1$ "slots" existing before $ a_1$, between $ a_1$ and $ a_2$, between $ a_2$ and $ a_3$, ..., and after $ a_n$.  The number $ a_1 \plus{} n$ must be inserted in one of the $ n \minus{} 1$ slots not bordering $ a_1$.  Its insertion then makes a total of $ n \plus{} 2$ slots.  Similarly, $ a_2 \plus{} n$ must be inserted in one of the $ n$ slots not bordering $ a_2$, and so on, so the number of ways to insert the numbers $ n \plus{} 1$ to $ 2n$ is $ (n \minus{} 1)(n)(n \plus{} 1)\cdots (2n \minus{} 2)$[/quote]

Sorry for the silly revive, but I just feel like posting right now. Unfortunately, this is incorrect, because if we consider $ n\equal{}2$ and the permutation $ (a_1,a_2)\equal{}(1,2)$, then your algorithm cannot yield $ (1,4,3,2)$ since in doing so it would require the first insertion to make it $ (1,3,2)$.

For $n=1$  this is obvious. For $n \ge 2$ let a "nice pair" in a permutation be a pair of consecutive integers with difference $n$. We construct an bijection from the set of permutations that don't have the property to the set of permutations that have exactly one nice pair that is not at the end. Since there are permutations with the property with more than one nice pair ($112233...nn$, here is where we need $n \ge 2$), we will be done. We will call the permutations without the property ugly permutations, and permutations with exactly one nice pair that is not at the end, nice permutations.

The bijection is: for each ugly permutation, take the last number $k$, and move it so that it is exactly after the number $k \pm n$. For example, for $n=4$, $17542386$ goes to $17542638$, where we moved the $6$. This produces a nice permutation, since we created exactly one nice pair (in the example this is $26$). Notice how removing the final number cannot create nice pairs since there are no numbers after the final number. Also, the nice pair in the resulting permutation is obviously not at the end, since if it were, there would be a nice pair at the end of the original ugly permutation.

This is a bijection, since, for each nice permutation, we can do the opposite: take the latter number of the nice pair and move it to the end, creating an ugly permutation. This is because the removal of that number $k$ cannot create more nice pairs because it was already paired up, and putting it at the end also does not create nice pairs, since $k \pm n$ isn't the next to last number (since the nice pair in the nice permutation isn't at the end).

So we're done.

P.S. 100th post :D

@JuanOrtiz
Can you explain to me how your bijection solves the problem?

Wait, correct me if I am wrong, but because we have this bijection which proves that the number of ways to permute the set such that there is property T and the number of ways to permute without any T is equal. So adding back in the number of ways to permute with property T at the end shows that the number of ways in total with property T is greater then the number of ways without T.

We first observe that the expected number of pairs $\{i, i+1\}$ for which $ | x_i - x_{i+1} | = n$ is $E = 1$. To see this note if $j$, $ 1 \leq j \leq n$, appears in position $1$ or $2n$ it's adjacent to one number, otherwise two. Thus the probability it's adjacent to its partner $j+n$ in a random permutation is
\begin{align*}
e_j &= \frac{2}{2n}\cdot \frac{1}{2n-1} + \frac{2n-2}{2n}\cdot \frac{2}{2n-1} \\
&= \frac{2(2n-1)}{2n(2n-1)} \\
&= \frac{1}{n}.
\end{align*}
By linearity of expectation we overall have the expected number of $j$ adjacent to its partner $j+n$ is $\sum_{j=1}^{n} e_j = n\cdot\frac{1}{n} = 1$.

More is true. By the same argument, if we remove any partner pair $\{j,j+n\}$, the expected number of partner pairs in a random permutation of the remaining integers is still 1. This is the critical observation.

Conditional on the partner pair $\{j,j+n\}$ appearing in a random permutation, what is the expected number of partner pairs $e$? Observe that if $n>1$ it must be less than 2, since as before the expected number of partner pairs ignoring $\{j,j+n\}$ is 1, and the probability the $\{j,j+n\}$ pair where they appear has separated another partner pair is greater than 0.

Putting this together, if $n=1$ the property $P$ obviously holds. For $n>1$, note the expected number of partner pairs $E = p\cdot e$, where $p$ is the probability that a random permutation has property $P$ and $e$ is as before. But we already know $E=1$, and by the previous argument if $n>1$ we have $e<2$, hence $ 1 = p\cdot e < 2p $ and we conclude $ p > \frac{1}{2}$.

Here's a different solution (Hopefully correct): For $n=1$, the statement is trivially true. So from now on assume that $n \geq 2$. Call a permutation $\mathbb{P}$ of $\{1,2, \dots ,2n\}$ $k$-[i]good[/i] if exactly $k$ values of $i$ satisfy the relation $|x_i-x_{i+1}|=n$, and [i]bad[/i] if it doesn't have property $T$. Let $f(T_k)$ be the number of $k$-[i]good[/i] sequences. Note that, as $n \geq 2$ by our assumption, so $f(T_2)>0$. Consider a $1$-[i]good[/i] sequence with $i \in \{1,2, \dots ,2n-1\}$ satisfying $|x_i-x_{i+1}|=n$. Interchange $x_{i+1}$ with $x_{i+2}$ to get a new sequence which is [i]bad[/i] (Here we take $x_{2n+1}=x_1$). For example, for $n=3$, the $1$-[i]good[/i] sequence $\{1,6,2,5,4,3\}$ becomes $\{1,6,2,4,5,3\}$, which is evidently [i]bad[/i]. It is easy to see that for two different $1$-[i]good[/i] sequences, it is possible to get the same [i]bad[/i] sequence after this process, but the converse is not true, i.e. we cannot get the same $1$-[i]good[/i] sequence from two distinct [i]bad[/i] sequences on applying this same process in the reverse order. This means that $f(T_1)$ cannot be less than the number of [i]bad[/i] sequences. But the total number of permutations which show property $T$ is $f(T_1)+f(T_2)+ \dots +f(T_n)$, which is definitely greater than $f(T_1)$. Thus, the total number of permutations with property $T$ is more than the number of permutations without property $T$. Hence, done. $\blacksquare$

Here's a solution which involves constructing a bipartite graph.

For $n \le 2$, it's easy to check the problem by simple computations. Let's assume now that $n \ge 3.$

First of all, we only care about the pairs of positions that $\{1, n+1\}, \{2, n+2\},$ etc. occupy. For instance, for the permutation $126435$, our pairs are $\{1, 4\}, \{2, 6\}, \{3, 5\}$. There are $n !!$ different ways to select pair up the positions. Call a pair [b] tasty [/b] if it is of the form $\{k, k+1\}$ for some $1 \le k < 2n.$ We just need to show that at least half of that $n!!$ pairing methods have at least one tasty pair in them. Let $S_x$ be the set of pairing methods with exactly $x$ tasty pairs, for each $x \in \mathbb{Z}_{\ge 0}.$

Let's construct a bipartite graph with $S_0$ on the left side and $S_1, S_2$ on the right side as follows. Consider some pairing $\{a_1, a_2\}, \{a_3, a_4\}, \cdots, \{a_{2n-1}, a_{2n}\} \in S_0$. We will connect it to $2n-1$ things in $S_1 \cup S_2$. For each $1 \le k \le 2n-1$, we know that $k, k+1$ are not in the same pair. Suppose that $\{a_{2j-1}, a_{2j}\} = \{k, x\}$ and $\{a_{2i-1}, a_{2i}\} = \{k+1, y\}.$ Then, connect our pairing with the one where we use $\{k, k+1\}$ and $\{x, y\}$ instead of $\{k, x\}, \{k+1, y\}.$ It's easy to see that this new pairing must be in $S_1 \cup S_2,$ and that each pairing in the left side has degree exactly $2n-1$ (note that if $y = x+1$ or something, then we allow a double edge).

Now, observe that each pairing $p \in S_1$ has degree at most $2(n-1)$, since if $\{a, a+1\}$ is the one tasty pair of $p$ then the only pairs which could map to $p$ are those where we replace $\{a, a+1\}, \{b, c\}$ with $\{a, b\}$ and $\{a+1, c\}$ or with $\{a, c\}$ and $\{a+1, b\}$, where $\{b, c\}$ is some other pair of $p$. This gives us at most $2(n-1)$ pairs since there's $n-1$ other pairs of $p$.

Finally, each pairing $p \in S_2$ has degree at most $4$. Suppose that $\{x, x+1\}, \{y, y+1\}$ are the two tasty pairs in $p$. Then, the only pairings which can map to $p$ are those which replace $\{x, x+1\}, \{y, y+1\}$ with $\{x, y\}$ and $\{x+1, y+1\}$ or with $\{x, y+1\}$ and $\{y, x+1\}$. This would imply that $p$ has degree at most $2$, however remember that in these cases a double edge would have been constructed. Therefore, we actually have that $p$ has degree at most $4$.

With the previous results, we have constructed a bipartite graph with $S_0$ on the left side, and $S_1 \cup S_2$ on the right side. This graph satisfies the property that every vertex on the left side has degree strictly larger than any vertex on the right side (since $2n-1 > 2n-2, 4$). Therefore, we have that $|S_0| < |S_1| + |S_2|.$ This implies that $|S_0| < |S_1| + |S_2| + \cdots + |S_n|$, which implies the result.

$\square$

The problem can be generalized for all $0\le r\le n$

Let $A$ be the set of permutations that do not satisfy $P(r)$, and let $B$ be the set of permutation that do satisfy $P(r)$. Define a function $f:A\longrightarrow B$ as
$$f(x_1x_2\ldots x_{2n})=x_2x_3\ldots x_1x_s\ldots x_{2n}$$
where $s=\min{\{x_s\,|\, |x_s-x_1|=r\}}$ which is possible because $r\le n$. We claim that $f$ is injective. Suppose that $f(a)=f(b)$ meaning
$$a_2a_3\ldots a_1a_s\ldots x_{2n}=b_2b_3\ldots b_1b_r\ldots x_{2n}$$
Upon noticing that $(a_1,a_s),(b_1,b_r)$ are the only pairs that satisfy $|x_s-x_1|=r$ we have $s=r$ and thus $a_i=b_i$ for all $i$, implying that $a=b$ and that $f$ is injective. However there are now permutations that have more than one pair of numbers that satisfy $P(r)$ which are obviously not attainable by $f$. Hence, $f$ is injective but not surjective, implying $|A|<|B|$.

Cute problem.

[hide=solution]Let $A_i$ denote the set of permutations such that the elements $i$ and $n+i$ are next to each other for all integers $i$ between $1$ and $n,$ inclusive. Note that the number of permutations with property $T$ is $|A_1\cup A_2\cup A_3\cup \cdots \cup A_n|.$ By Bonferroni's Inequalities, $$|A_1\cup A_2\cup A_3\cup \cdots \cup A_n|\ge \sum_{k=1} ^{n} |A_k|-\sum_{1\le i<j\le n} |A_i\cap A_j|.$$ We will show that the expression on the right hand side is greater than $\frac{(2n)!}{2}.$

First we compute $\sum_{k=1} ^{n} |A_k|.$ For any $k,$ $|A_k|$ denotes the number of permutations such that the elements $k$ and $n+k$ are next to each other. There are $2n-1$ to choose which two consecutive spots in the permutaion are occupied by $k$ and $n+k.$ Additionally, there are $2$ ways to choose which order $k$ and $n+k$ are in and $(2n-2)!$ ways to arrange the remaining numbers. Hence, $$|A_k|=(2n-1)\cdot 2\cdot (2n-2)!=2\cdot (2n-1)!$$ for all integers $k$ between $1$ and $n,$ inclusive. Therefore, $$\sum_{k=1} ^{n} |A_k|=n\cdot 2\cdot (2n-1)!=(2n)!.$$

Now, we compute $\sum_{1\le i<j\le n} |A_i\cap A_j|.$ By symmetry, all of the terms $|A_i\cap A_j|$ are equal, so $$\sum_{1\le i<j\le n} |A_i\cap A_j|=\binom{n}{2}\cdot |A_1\cap A_2|.$$ To compute $|A_1\cap A_2|,$ we will use casework on the position of $1$ and $n+1.$

Case 1: $1$ and $n+1$ occupy the $1^{\text{st}}$ and $2^{\text{nd}}$ positions of the permutation when read from left to right. There are then $2n-3$ different sets of two consecutive positions $2$ and $n+2$ can occupy. Then, there are $2\cdot 2=4$ ways to determine which orders the pairs $1,n+1$ and $2,n+2$ are in. Finally, there are $(2n-4)!$ ways to arrange the remaining numbers. Therefore, there are a total of $(2n-3)\cdot 4\cdot (2n-4)!=4\cdot (2n-3)!$ permutations in this case.

Case 2: $1$ and $n+1$ occupy the $(2n-1)^{\text{th}}$ and $(2n)^{\text{th}}$ positions of the permutation when read from left to right. This case is symmetrical to Case 1. Therefore, there are $(2n-3)\cdot 4\cdot (2n-4)!=4\cdot (2n-3)!$ permutations in this case.

Case 3: $1$ and $n+1$ occupy the $k^{\text{th}}$ and $(k+1)^{\text{th}}$ positions in the permutation when read from left to right, where $k$ is an integer between $2$ and $2n-2,$ inclusive. Now, note that $2$ and $n+2$ can occupy the $i^{\text{th}}$ and the $(i+1)^{\text{th}}$ positions in the permutation when read from left to right for all integers $i$ between $1$ and $2n-1,$ inclusive except for $i=k-1,k,k+1.$ Therefore, there are $2n-4$ different sets of two consecutive positions $2$ and $n+2$ can occupy. Then, there are $2\cdot 2=4$ ways to determine which orders the pairs $1,n+1$ and $2,n+2$ are in. Finally, there are $(2n-4)!$ ways to arrange the remaining numbers. Therefore, there are a total of $(2n-4)\cdot 4\cdot (2n-4)!$ permutations for each $k.$ There are $2n-3$ possible values of $k$ in this case, so there are a grand total of $(2n-4)\cdot 4\cdot (2n-4)!\cdot (2n-3)=(8n-16)\cdot (2n-3)!$ permutations in this case.

Altogether, $$|A_1\cap A_2|=4\cdot (2n-3)!+4\cdot (2n-3)!+(8n-16)\cdot (2n-3)!=(8n-8)\cdot (2n-3)!=4\cdot (2n-2)!$$ Multiplying this number by $\binom{n}{2}$ yields that $$\sum_{1\le i<j\le n} |A_i\cap A_j|=4\cdot (2n-2)!\cdot \frac{n(n-1)}{2}=2n\cdot (2n-2)!\cdot (n-1)=\frac{n-1}{2n-1}\cdot (2n)!.$$

Now, $$|A_1\cup A_2\cup A_3\cup \cdots \cup A_n|\ge \sum_{k=1} ^{n} |A_k|-\sum_{1\le i<j\le n} |A_i\cap A_j|=(2n)!-\frac{n-1}{2n-1}\cdot (2n)!=\frac{n}{2n-1}\cdot (2n)!>\frac{(2n)!}{2},$$ as desired.[/hide]

I won't bother proofreading this, so as a heads-up, expect a lot of typos.

Use the transformation$$(x_1, x_2, \cdots, x_{2n}) = (x_2, \cdots, x_{ki1}, x_1, x_i, \cdots, x_{2n}),$$where $x_i$ is the element such that $|x_i - x_1| = n$, which is injective but not surjective.


We have $ \sum\frac{x_i}{x_j}\equal{}n^2\minus{}n\plus{}1$. By QM-AM, we also have $ \sum\frac{x_i^2}{x_j^2}\ge\frac{(n^2\minus{}n\plus{}1)^2}{\binom{n}2}\equal{}\frac{2(n^2\minus{}n\plus{}1)^2}{n(n\minus{}1)}$. So it is enough to prove that $ 2(n^2\minus{}n\plus{}1)^2>n^3(n\minus{}1)\plus{}4n(n\minus{}1)\plus{}2$, which is equivalent to $ n^4\minus{}3n^3\plus{}2n^2\ge0$, which is true since $ n>2$.

[quote="Johan Gunardi"]We have $ \sum\frac {x_i}{x_j} \equal{} n^2 \minus{} n \plus{} 1$. By QM-AM, we also have $ \sum\frac {x_i^2}{x_j^2}\ge\frac {(n^2 \minus{} n \plus{} 1)^2}{\binom{n}2} \equal{} \frac {2(n^2 \minus{} n \plus{} 1)^2}{n(n \minus{} 1)}$. So it is enough to prove that $ 2(n^2 \minus{} n \plus{} 1)^2 > n^3(n \minus{} 1) \plus{} 4n(n \minus{} 1) \plus{} 2$, which is equivalent to $ n^4 \minus{} 3n^3 \plus{} 2n^2\ge0$, which is true since $ n > 2$.[/quote]
Sorry,but I didn't think it is correct :(
I think $ \sum\frac {x_i^2}{x_j^2}\ge\frac {(n^2 \minus{} n \plus{} 1)^2}{\binom{n}2}$ should be $ \sum\frac {x_i^2}{x_j^2}\ge\frac {(n^2 \minus{} n \plus{} 1)^2}{n(n\minus{}1)}$,am I right?


So my high school math team drives up to Harvard for the HMMT contest, in a blizzard. We come in 13th place (not bad for a public non-magnet school) and we're all happy with ourselves and we stop at this buffet for dinner wearing our team t-shirts. While waiting on line to pay, this old guy asks us this simple math question, and nobody can solve it. It's embarrassing. HERE IT IS:

You have 12 billiard balls, but one of them weighs differently (maybe more, maybe less). You also have a 2-sided balance, but after it is used three times, it disappears. How can you determine which billiard ball weighs differently and whether it weighs more or less.

P.S. or explain why that's impossible, like i suspect it is.

Try going [url=http://www.iwriteiam.nl/Ha12coins.html]here[/url] for a much better explanation than I could ever give.

I had this problem given to my fellow classmates and I at the Northwestern CTD camp Persuasion and Debate course for logic practice, and the solution I got was the weigh 4 against 4 thing they had on that page brut3forc3 linked. Basically for these types of problems you try to weigh 2 groups such that there are some leftovers and the leftovers can be determined from the first weighing, and then it gets easier from there.


trigonometry
geometry
trig identities
Law of Sines

Tow men measure the angle of elevation of a helicopter at the same time. One angle is measured as 35 degrees, the other as 50 degrees. If the two men are 100 feet apart and the helicopter lies over the line joining them, how high is the helicopter?

[hide]Calling the the man who measured the angle of elevation as $ 50^{o}$ A, the other B, and the helicopter H, we can draw a triangle using them. Using law of sines, we have
$ \frac{100}{\sin95}=\frac{HB}{\sin50}$
$ HB=\frac{100\sin50}{\sin95}$
$ [HAB]=\frac{1}{2}\cdot100\cdot\frac{100\sin50}{\sin95}\cdot\sin35=\frac{5000\sin50\sin35}{\sin95}=$
$ \frac{1}{2}\cdot100\cdot h$
$ h\approx44.1$ feet.[/hide]

I took a different approach than mountain.dew
[hide]We can connect the 3 objects(or whatever you want to call them) to form a triangle. Since we know 2 of the angles, the third is $ 95^{o}$. Naming the $ 35^{o}$ man $ A$, $ 50^{o}$, $ B$, helicopter, $ H$ I drew an altitude that goes down from $ H$ to the 100 feet side of the triangle. We can name it $ C$. The topmost angle on $ A's$ side is $ 55^{o}$. The one on $ B$ is $ 40^{o}$. Both share the side $ HC$.

$ AC+BC=100 feet$
$ tan 55 = \frac{AC}{HC}$

$ tan 40 = \frac{BC}{HC}$

From here, just solve the equations to find out $ HC$ which would be the height of the helicopter.[/hide]


function calculus integration derivative

I learned of a cool way to do this summation, and I was wondering how else one might find the answer. So, if you're posting an answer, please say how you did it. (Even if it's guess and check). Post an inductive proof if you want, but please also post the derivation of the formula.

Evaluate $ \sum_{k \equal{} 0}^{n}k2^{k}$ in terms of $ n$.

Edit: I noticed the intermediate forum had a summation just like this, so try $ \sum_{k \equal{} 0}^{n}k^{2}2^{k}$. It's slightly harder.

Edit: Typo in the second summation. Thanks.

[quote="Hamster1800"]I learned of a cool way to do this summation, and I was wondering how else one might find the answer. So, if you're posting an answer, please say how you did it. (Even if it's guess and check). Post an inductive proof if you want, but please also post the derivation of the formula.

Evaluate $ \sum_{k \equal{} 0}^{n}k2^{k}$ in terms of $ n$.

Edit: I noticed the intermediate forum had a summation just like this, so try $ \sum_{k \equal{} 0}^{n}k^{2}2^{n}$. It's slightly harder.[/quote]

[hide="Hint"]Try first to evaluate the integrable function $ f(x)\equal{}\sum_{k \equal{} 1}^{n}kx^{k\minus{}1}$[/hide]

Firstly, for the second summation, I think you mean $ \sum_{k=0}^{n}k^{2}2^{k}$.  Otherwise, you could just factor out the $ 2^{n}$

For the first one, we get:
[hide]$ \sum_{k=1}^{n}=2+2\cdot 2^{2}+\cdots n\cdot 2^{n}$

Notice how I changed the lower limit to 1 because the first term is just 0.  We can assume that $ n>0$ with this changed limit.  If $ n=0$, then the sum is 0.

Sum ${ =(2+2^{2}+2^{3}+\cdots+2^{n})+(2^{2}+2^{3}+\cdots+2^{n})+\cdots+(2^{n-1}+2^{n}})+(2^{n})$
$ =(2^{n+1}-2)+(2^{n+1}-2^{2})+\cdots+(2^{n+1}-2^{n})$
$ =n2^{n+1}-(2^{n+1}-2)$
$ =\boxed{(n-1) 2^{n+1}+2}$

This can be proven easily by induction.[/hide]

I'll try the second one later, although I imagine it is done using a very similar method.

The answer should be $ (n^{2}\minus{}2n\minus{}3)\cdot2^{n\plus{}1}\plus{}3\cdot2^{n\plus{}2}\minus{}6$, or $ (n^{2}\minus{}2n\plus{}3)\cdot2^{n\plus{}1}\minus{}6$. You were only off by one :).

Your error is that $ \sum_{k \equal{} 1}^{n}2^{k}\equal{} 2^{n\plus{}1}\minus{}2$, not $ 2^{n\plus{}1}\minus{}1$.

[hide="Here is the way I recently learned"]
So basically you do this with the following identity (explanation will come later):
$ \sum_{n \equal{} a}^{b\minus{}1}u(n)(v(n\plus{}1)\minus{}v(n)) \equal{} u(b)v(b)\minus{}u(a)v(a)\minus{}\sum_{n \equal{} a}^{b\minus{}1}v(n\plus{}1)(u(n\plus{}1)\minus{}u(n))$
This is called "Summation by Parts" because the form is almost identical to Integration by Parts.

So now, you just let $ u(n) \equal{} k^{2}$ and $ v(n) \equal{} 2^{k}$
$ \sum_{k \equal{} 0}^{n}k^{2}2^{k}\equal{}\sum_{k \equal{} 0}^{n}k^{2}(2^{k\plus{}1}\minus{}2^{k}) \equal{} (n\plus{}1)^{2}2^{n\plus{}1}\minus{}0\minus{}\sum_{k \equal{} 0}^{n}2^{k\plus{}1}(2k\plus{}1)$

Then we do it again with $ \sum_{k \equal{} 0}^{n}2^{k\plus{}1}(2k\plus{}1)$.
$ u(n) \equal{} 2k\plus{}1$, $ v(n) \equal{} 2^{k\plus{}1}$
$ \sum_{k \equal{} 0}^{n}(2k\plus{}1)2^{k\plus{}1}\equal{}\sum_{k \equal{} 0}^{n}(2k\plus{}1)(2^{k\plus{}2}\minus{}2^{k\plus{}1})$
$ \equal{} (2(n\plus{}1)\plus{}1)2^{(n\plus{}1)\plus{}1}\minus{}(2(0)\plus{}1)2^{0\plus{}1}\minus{}\sum_{k \equal{} 0}^{n}2\cdot2^{k\plus{}2}$
$ \equal{} (2n\plus{}3)2^{n\plus{}2}\minus{}2\minus{}(2^{n\plus{}4}\minus{}2^{3}) \equal{} (2n\plus{}3)2^{n\plus{}2}\minus{}2^{n\plus{}4}\plus{}6$

Thus,
$ \sum_{k \equal{} 0}^{n}k^{2}2^{k}\equal{} (n\plus{}1)^{2}2^{n\plus{}1}\minus{}((2n\plus{}3)2^{n\plus{}2}\minus{}2^{n\plus{}4}\plus{}6) \equal{} (n^{2}\plus{}2n\plus{}1)2^{n\plus{}1}\minus{}(4n\plus{}6)2^{n\plus{}1}\plus{}8\cdot2^{n\plus{}1}\minus{}6 \equal{}\boxed{(n^{2}\minus{}2n\plus{}3)2^{n\plus{}1}\minus{}6}$

[hide="reason behind the identity"]
So, first, let me define some notation: $ \Delta_{n}f(n) \equal{} f(n\plus{}1)\minus{}f(n)$. Also, $ \sum_{n: a\to b}f(n) \equal{}\sum_{n \equal{} a}^{b\minus{}1}f(n)$.

Now, notice that the two are inverses, just like differentiation and integration (these are so-called discrete versions). That is $ \sum_{n: a\to b}\Delta_{n}f(n) \equal{} f(b)\minus{}f(a)$ and $ \Delta_{n}\sum_{k: a\to n}f(k) \equal{} f(n)$. These two can be seen easily by writing out the sums.

Thus, we just need to find a function $ f(n)$ such that $ \Delta_{n}f(n) \equal{} n^{2}2^{n}$.

To do this, I will derive the Summation by Parts above.
First notice that both of these operations are linear.
Then
$ \Delta_{n}(f(n)g(n)) \equal{} f(n\plus{}1)g(n\plus{}1)\minus{}f(n)g(n) \equal{} f(n\plus{}1)g(n\plus{}1)\minus{}f(n)g(n\plus{}1)\plus{}f(n)g(n\plus{}1)\minus{}f(n)g(n) \equal{} g(n\plus{}1)\Delta_{n}f(n)\plus{}f(n)\Delta_{n}g(n)$. This means that
$ \Delta_{n}(f(n)g(n))\minus{}g(n\plus{}1)\Delta_{n}f(n) \equal{} f(n)\Delta_{n}g(n)$. If we sum both sides, we get
$ \sum_{n: a\to b}(\Delta_{n}(f(n)g(n))\minus{}g(n\plus{}1)\Delta_{n}f(n)) \equal{}\sum_{n: a\to b}f(n)\Delta_{n}g(n)$
or $ f(b)g(b)\minus{}f(a)g(a)\minus{}\sum_{n: a\to b}g(n\plus{}1)\Delta_{n}f(n) \equal{}\sum_{n: a\to b}f(n)\Delta_{n}g(n)$

I remember this as $ \sum u\Delta v \equal{} uv\minus{}\sum v_{\plus{}1}\Delta u$
[/hide]
[/hide]


If you have some ideas,just write down,thanks !


Find the smallest number k such that:
[tex]\frac{a+b}{c+d}+\frac{c+d}{a+b} \le k\frac{a+c}{b+d}+\frac{4-k}{2}[/tex]
with all a,b,c in [1,2].

I think, Inequalities Open Questions is not an appropriate section for such a kind of problems.
I will move it to Inequalities Unsolved Problems.

If $a=b=1$ and $c=d=2$ then we get $\displaystyle k \geq 1$.

Also, I think we can prove that \[ \frac{a+b}{c+d}+\frac{c+d}{a+b} \leq \frac{a+c}{b+d} + \frac{3}{2} \, , \, \, \forall a,b,c,d \in \left[ 1,2 \right] \, , \] but I don't know how.


A ball with a density of 0.940 g/cm^3 and a volume of 1.4 x 10^4 cm^3 is placed in a fluid with a density of 0.870 g/cm^3.  Does the ball sink or float?  If the ball floats, calculate the volume of the displaced fluid.  If the ball sinks, calculate the magnitude of the apparent weight of the ball.

I know that the ball sinks because its density is greater than the density of the fluid.  However, I have no clue how to calculate the magnitude of the apparent weight of the ball.  Please help me.

The bouyant force is equal to the weight of the fluid displaced.

So, find the volume of displaced fluid, convert that to a weight, subtract it from the real weight of the ball, and what you've got left is the apparent weight of the ball.

I found the mass of the ball to be 13.16 kg:
m=density*Volume = (.94)(1.4*10^4)=13,160g = 13.16 kg.

Then I found the weight of the ball by multiplying that number by gravity: (13.16)(9.81) = 129.1 N.

Now, how do I find the volume of the displaced fluid??  Please help.

Sorry about the other post. I didn't realise the apparent weight was negative.

On this question ...

The ball sinks, so the volume of the displaced fluid is the same as the volume of the ball.

m=density*Volume = (.87)(1.4*10^4)=12,180g = 12.18 kg.

So convert 12.18 kg to weight (119.49), then subtract that number from 129.1 N to get an apparent weight of 9.61 N?

Exactly. Or you could substract densities first and then multiply the result by volume and gravitational acceleration, it would be the same.


A college student drove his compact car $ 120$ miles home for the weekend and averaged $ 30$ miles per gallon. On the return trip the student drove his parents' SUV and averaged only $ 20$ miles per gallon. What was the average gas mileage, in miles per gallon, for the round trip?

$ \textbf{(A)}\ 22 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 26 \qquad \textbf{(E)}\ 28$

$there= 120 miles distance back= 120 miles distance$

It took $\frac{120}{30}$ gallons for the trip home, and $\frac{120}{20}$ gallons for the trip back.

so $4+6=10$

$\frac{240}{10}=24$

which is choice B

[quote="worthawholebean"]A college student drove his compact car $120$ miles home for the weekend and averaged $30$ miles per gallon. On the return trip the student drove his parents' SUV and averaged only $20$ miles per gallon. What was the average gas mileage, in miles per gallon, for the round trip?

$\text{(A)}\ 22 \qquad \text{(B)}\ 24 \qquad \text{(C)}\ 25 \qquad \text{(D)}\ 26 \qquad \text{(E)}\ 28$[/quote]
[hide]Harmonic mean.
$\frac{2(30)(20)}{20+30}=\frac{1200}{50}=24$.[/hide]


Putnam
calculus
integration
trigonometry
limit
function
real analysis

How would you go about solving the following problem.
Given that $ \displaystyle \int_0^\infty \frac{\sin(x)}{x}dx\equal{}A$ find $ \displaystyle \int_0^\infty\frac{\sin^2(x)}{x^2}dx$.

I've tried searching the forum but couldn't find this anywhere. if you do happen to find where else it has been posted a link would also be appreciated.

Suppose that $ I(a)=\int_0 ^{\infty} \frac{\sin^2(ax)}{x^2}dx$ for $ a\geq 0$. Then $ I^'(a)=\int_0 ^{\infty} \frac{\sin(2ax)}{x}dx$ by substituting $ y=2ax$ we have $ I^'(a)=\int_0 ^{\infty} \frac{\sin(y)}{y}dy=A \Longrightarrow I(a)=Aa+C$ and $ I(0)=0 \Longrightarrow I(a)=Aa \Longrightarrow I(1)=A$ :D

[quote="maokid7"][quote="Kent Merryfield"]1. Integrate by parts to get that $ \int_0^{\infty}\frac {\sin x}x\,dx \equal{} \int_0^{\infty}\frac {1 \minus{} \cos x}{x^2}\,dx$

2. Use the fact that $ \sin^2x \equal{} \frac12(1 \minus{} \cos 2x).$

3. Make the change of variables $ u \equal{} 2x.$[/quote]

how did you get (1)? i can't seem to get it when i integrate by parts.[/quote]

You let $ dv \equal{} \sin(x) dx$ such that $ v \equal{} 1 \minus{} \cos (x)$ and $ u \equal{} 1/x$ such that $ du \equal{} \minus{} dx/x^2$.

Then we have

$ \lim_{x \rightarrow \infty} (1 \minus{} \cos(x))/x \minus{} \lim_{x \rightarrow 0} (1 \minus{} \cos(x))/x \plus{} \int_0^{\infty} (1 \minus{} \cos(x))/x^2 dx.$

The rest follows.

how about finding
 :D

$ \displaystyle \int_0^\infty\frac{\sin^3(x)}{x^3}dx$

[hide="answer"] $ \frac{3\pi}{8}$ [/hide]

Using what I know about Fourier transforms, and the matched transform pair of the "sinc" function and the "box" function, and using the fact that multiplication on one side is convolution on the other, I arrive at the following formula:

For $ n\ge 1,$ $ \int_{\minus{}\infty}^{\infty}\left(\frac{\sin\pi x}{\pi x}\right)^ndx\equal{} \frac1{(n\minus{}1)!}\sum_{k\equal{}0}^{\lfloor n/2\rfloor}(\minus{}1)^k\binom{n}{k} \left(\frac n2\minus{}k\right)^{n\minus{}1}$

(With the $ 2\pi$ where I like to put it in the definition of the Fourier transform, I do define $ \text{sinc}(x)\equal{}\frac{\sin\pi x}{\pi x},$ extended to be continuous.)

Making the obvious simple change of variables, and dividing the integral of the even function in half, we get

$ \int_0^{\infty}\left(\frac{\sin x}{x}\right)^ndx\equal{}\frac{\pi}{2(n\minus{}1)!}\sum_{k\equal{}0}^{\lfloor n/2\rfloor}(\minus{}1)^k\binom{n}{k} \left(\frac n2\minus{}k\right)^{n\minus{}1}$

For $ n\equal{}1,$ this is $ \frac{\pi}2.$

For $ n\equal{}2,$ this is $ \frac{\pi}2.$

For $ n\equal{}3,$ this is $ \frac{\pi}{4}\left(\left(\frac32\right)^2\minus{} 3\left(\frac12\right)^2 \right)\equal{}\frac{3\pi}{8},$ as misan has indicated.

And so on.

I'll add that for large $ n,$ $ \int_0^{\infty}\left(\frac{\sin x}{x}\right)^ndx$ is asymptotic to $ \sqrt{\frac{3\pi}{2n}.}$

This comes from what is essentially the Central Limit Theorem.


trigonometry
rotation
geometry proposed
geometry

Let $ ABC$ be a triangle, $ \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}\in\mathbb{N}^*$ and Let $ M\in IntABC$, $ MB^2\plus{}MC^2\equal{}MA^2\plus{}MB\cdot MC$.

Compute $ \tan \angle BMC$.

Let $ a\equal{}z\plus{}y,\ b\equal{}x\plus{}z,\ c\equal{}y\plus{}x,\ x,y,z\geq 0$
After some computations we have
$ \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}\equal{}\frac{2x^2\plus{}2y^2\plus{}2z^2\plus{}2(xy\plus{}yz\plus{}zx)}{x^2\plus{}y^2\plus{}z^2\plus{}3(xy\plus{}yz\plus{}zx)}\equal{}1\plus{}\frac{x^2\plus{}y^2\plus{}z^2\minus{}xy\minus{}yz\minus{}zx}{x^2\plus{}y^2\plus{}z^2\plus{}3xy\plus{}3yz\plus{}3zx}\in \mathbb{N}$
But $ \frac{x^2\plus{}y^2\plus{}z^2\minus{}xy\minus{}yz\minus{}zx}{x^2\plus{}y^2\plus{}z^2\plus{}3xy\plus{}3yz\plus{}3zx}<1$, so we have $ x^2\plus{}y^2\plus{}z^2\minus{}xy\minus{}yz\minus{}zx\equal{}0$, meaning $ x\equal{}y\equal{}z$, so $ a\equal{}b\equal{}c$
So we obtain that triangle ABC is equilateral.
Let MA=a' MB=b' MC=c', let $ m(\widehat{BMC})\equal{}\varphi$
We rotate triangle ABC around point B by $ \frac{\pi}{3}$. A goes into A', C goes into A, and M into M'. Then $ m(\widehat{AM'B})\equal{}\varphi$
$ BM\equal{}BM'$ and $ m(\widehat{MBM'})\equal{}\frac{\pi}{3}$, then triangle BMM' is equilateral.
So we have AM'=c' MM'=b' AM=a'. Let $ m(\widehat{BMC})\equal{}\alpha$
Then we have $ \varphi\equal{}\alpha\plus{}\frac{\pi}{3}$
By cosine theorem in triangle AMM' we have
$ b'^2\plus{}c'^2\equal{}a'^2\plus{}2\cdot b'\cdot c'\cdot cos \alpha$
But the problem says $ b'^2\plus{}c'^2\equal{}a'^2\plus{}b'\cdot c'$. So we have $ cos \alpha\equal{}\frac{1}{2}$, so then $ \alpha\equal{}\frac{\pi}{3}$, and $ \varphi\equal{}\frac{2\pi}{3}$
Then $ tg(BMC)\equal{}\minus{}\sqrt{3}$


There are many numbers that divide 109 with a remainder of 4.  List all two-digit numbers that have that property.

[hide] if you divide into 109 with a remainder of 4, you must go into 105 perfectly. Thus we must simply list out the 2 digit facotrs of 105 which are: 15, 21 and 35 [/hide]

109 divided by x gives remainder 4
109= kx + 4
109-4= kx
105= kx

x divides 105
x is a two digit number
x = 15, 21, 35 :lol:


How do you add the degree symbol in $\LaTeX$ as in angles, temperature, etc...

Thanks!! I think I read the [code] \degree [/code] symbol in the comprehensiive list. What is it for???


Consider a pentagon regular ABCDE. An interior point of the pentagon is z so that angle AEZ =48 and angle BCZ =42 .Se asks for angle AZC.

( It is not allowed use trigonometry)

Circles (D), (E) with radii equal to the side of the regular pentagon meet at Z' inside the pentagon. The $\triangle DEZ'$ is equilateral, the isosceles $\triangle DCZ' \cong \triangle EAZ'$ are congruent. Using this, $\angle AEZ' = 48^\circ,$ $\angle BCZ' = \angle BAZ' = 42^\circ.$ It follows that $Z' \equiv Z$ are identical. As a result, $\angle AZC = 360^\circ-(\angle ABC+\angle BCZ+\angle BAZ) = 168^\circ.$

I like this solution, it is really very logical and also helped me for the solution of a similar question (With the help of another mathlinker)!!


The Fibonacci numbers are
$ F_1 \equal{} 1, F_2 \equal{} 1, F_3 \equal{} 2, F_4 \equal{} 3, F_5 \equal{} 5, F_6 \equal{} 8, F_7 \equal{} 13, . . .$
where the ﬁrst two are both equal to $ 1$, and from then on, each
one is the sum of the two preceding it. Of the ﬁrst $ 2004$ Fibonacci
numbers, how many have $ 2$ as their last digit?

The Fibonacci sequence mod 10 is periodic with period 60 ([url=http://www.mathlinks.ro/weblog_entry.php?t=156030]proof[/url]). Among the first 60 terms, 3 have 2 as their last digit, namely the 3rd, 36th and 54th terms. Since $ 2004\equal{}60\times 33\plus{}24$, the answer should be $ 33\times 3\plus{}1\equal{}34$.


When three spherical balls, each of radius 10cm, are placed in a hemi-spherical dish, it is noticed that the top of the balls are all exactly level with the top of the dish. Whwat is the radius, in centimetres, of the dish?


geometry
trigonometry
trig identities
Law of Sines

In triangle $ABC$, angles $A$ and $B$ measure 60 degrees and 45 degrees, respectively.  The bisector of angle $A$ intersects $\overline{BC}$ at $T$, and $AT=24.$  The area of triangle $ABC$ can be written in the form $a+b\sqrt{c},$ where $a$, $b$, and $c$ are positive integers, and $c$ is not divisible by the square of any prime.  Find $a+b+c.$

[hide="solution"]
draw the triangle first :)
in $\triangle BTA: \angle T=105$ and in  $\triangle CTA: \angle T=75\Rightarrow AC=AT=24$
by law of sines we have:

$\frac{\sin{30}}{BT}=\frac{\sin{45}}{AT}\Rightarrow BT=12\sqrt{2}$

$\sin{75}=\cos{75}=\sin{45+30}=\frac{\sqrt{2}+\sqrt{6}}{4}$
in $\triangle CTA$:
$\frac{\sin{75}}{AC}=\frac{\sin{30}}{TC}\Rightarrow TC=48(\sqrt{2}+\sqrt{6})$

let $AH$ be the altitude from A to BC:

then $\frac{AH}{24}=\cos{15}\Rightarrow AH=6(\sqrt{2}+\sqrt{6})$

area of triangle is: $AH\cdot BC=216+72\sqrt{3}\Rightarrow \boxed{291}$[/hide]

Sorryy to revive this but I have a solution without trig.

[hide="Special Right Triangles"] Let $D$ be the foot of perpendicular from $T$ to $AB$, since $\angle ATB = 105$, $TBD$ is a 45-45-90 right triangle and $ATD$  is a 30-60-90 right triangle. Since $AT = 24$ that means $TD = 12$ and $AD = 12\sqrt{3}$. So the area of $ATD$ is $72\sqrt{3}$.

Since $TD = 12$ then $BD$ must also be $12$ because $TBD$ is a 45-45-90. Therefore the area of $TBD$ is $72$.

Now if we let $E$ be the foot of the perpendicular from $T$ to $AC$ we have that $\angle ATE = 60$ and $ATE$ is a 30-60-90 right triangle,           $AT = 24$ therefore the length of $TE$ is $12$. Also $ATC$ is isosceles with $AT = AC=24$ because $\angle ATC = \angle ACT = 75$, therefore since $AC$ is $24$ and the perpendicular $TE$ is $12$ the area of $ATC$ is $144$.

Therefore the total area of $ABC$ is $72+144+72\sqrt{3} = 216+72\sqrt{3}$ and $a+b+c = \boxed{291}$

 [/hide]

[quote=anandiyer12]I think this was a AIME problem from some year, I vaguely remember it.[/quote]

Yes its AIME I 2001 P4 I think

One more method is bit on a theory based as you would remember the length of the bisector is (2bc/a+c)cosa/2  putting those values in we would ab as 12√3+12 by applying sine rule for area of traingle you would get it as 1/2 24 12√3+12 simplfying give us 216+72√3
Which gives us 291


When you sing, the vocal folds in your throat act as a resonating cavity (one open end, one closed end) which allows only certain frequencies. Suppose you try to sing a pure tone which is detected by a microphone and measured to have a frequency 500.0 Hz. This is the resonance frequency of the vocal folds in your throat, which only depends on the tension and mass of the flesh in your vocal folds. If you now replace the air of density 1.000 kg/m3 with helium of density 0.1640 kg/m3, what frequency will the microphone measure?
You may assume that the bulk modulus remains constant in the two cases, and that only the density changes.

The speed of sound is $ c^2 \equal{} \left(\frac {\partial p}{\partial \varrho}\right)_S$ At $ S \equal{} \text{const}$ (adiabatic process), $ pV^\kappa \equal{} p \left(\frac {M}{\varrho}\right)^\kappa \equal{} \text{const},$ where $ M$ is the (kilo)molar mass and $ \kappa \equal{} \frac {C_P}{C_V}.$ Then $ c^2 \equal{} \left(\frac {\partial p}{\partial \varrho}\right)_S \equal{} \kappa\ \frac {p}{\varrho} \equal{} \kappa\ \frac {pV}{M},$ so that $ \frac {c^2_{He}}{c^2_{air}} \equal{} \frac {\kappa_{He}}{\kappa_{air}} \cdot \frac {M_{air}}{M_{He}}.$

$ M_{He} \equal{} 4.003\ \text{kg / kmol}$ and assuming air composition 71% $ N_2,$ 28% $ O_2,$ 1% $ Ar,$ $ M_{air} \equal{} 29.24\ \text{kg / kmol}.$

$ \kappa_{He} \equal{} \frac {5}{3}$ (monoatomic gas) and $ \kappa_{air} \doteq \frac {7}{5}$ (diatomic gas, neglecting 1% of Ar).

The frequency ratio is $ \frac {\nu_{He}}{\nu_{air}} \equal{} \frac {c_{He}}{c_{air}} \doteq 2.95,$ everyone sounds like Mickey Mouse.


function
integration
real analysis
real analysis unsolved

[quote="eugene"]Let $f$ be real function, $f\in {C^1([a,b])}$, such that there is not point $x$ that $f(x)=f'(x)=0$.Prove that there exists function $g\in {C^1([a,b])}$ that $fg'-f'g>0$ on $[a,b]$[/quote]

Take  g = exp, you have:

fg'-f'g = f*exp-f'*exp = (f-f')*exp. But f-f' cannot be 0 by continuity (otherwise there is a real t such that f(t)=f'(t)=0). Whence either exp or -exp answers the question.

Julien_Santini,what should we do if there exists some $x$ that $f'(x)=f(x) \neq 0$  ?

Suppose that $f>0$
$fg'-f'g=h$ or $g'-\frac{f'}{f}g=\frac{h}{f}$ is differential equation of order 1

$g(x)=Cf(x)+f(x)\int_{a}^{x}\frac{h(t)}{f(t)}dt$


The forum doesn't allow me to display symbols. So I have to put a link here.
[url]http://www.mp95.com/ChinaNationalMiddleSchoolMathTournament2009_session1.doc[/url]

[quote="Poincare"]Hmm...

I expected more of the chinese, the problems aren't very hard or am I overlooking stuff? What was the time limit?[/quote]

I will post it later

this is kinda spammy but
coudn't u have posted it when u syd that u would post it l8r?

Since $ |1 \minus{} 2a \plus{} b| \plus{} 2a \plus{} 1 \equal{} b^2 \minus{} a^2$, we have
$ b^2 \minus{} (a \plus{} 1)^2 \equal{} |1 \minus{} 2a \plus{} b|$ > or = 0
So |b| > or = |a+1|.
Since $ a^2 \plus{} b^2 \equal{} 1$ so $ a \equal{} 0$ and $ b \equal{} 1$ or $ b \equal{} \minus{} 1$.
Since $ a \equal{} 0$ we have $ |1 \plus{} b| \plus{} 1 \equal{} b^2$. So b must be -1.
Therefore the answer is: $ a \plus{} b \equal{} \minus{} 1$.

[quote="gauss202"]What are the rules of this contest and how are the students chosen for it?[/quote]

The candidates must be qualified in their own schools.  After that, they have to compete
in the county or in the city where they live. If they win, they will go to the province to take the test. If they win again, they will represent their province to join the national
competition. about 40 persons will be selected to train for international math olympics.

If a student wins medals in a province or the nation, usually, he or she doesn't have to
take the national college entrance exam (like SAT here, but tougher)  to get into a top college.

It's not that much harder than the  national Mathcounts test, and a lot of it is multiple choice, so I don't think it's that bad(though I still can't really solve any of them easily )

this MUST be competitions for junior 2 and junior 3 student; I am in senior 1  but I haven't heard of this competition at all when was in junior and I think the problems are  easier than CHINA JUNIOR MATH tournament （I don't know it's name exactly ）. I will post the problems soon . :blush:

First ,here are some geometry problems about the competition(I just post some of the problems :D ):
1.let polygon $ {ABCDEF}$ be inscribed in a circle such that $ {AB\equal{}CD\equal{}EF}$ and $ {AD}$,$ {BE}$,$ {CF}$  meet at one point called $ {Q}$  and $ {AD}$ and $ {CE}$ meet at $ {P}$ .prove that (1)$ {\frac {QD}{ED}\equal{}\frac{AC}{EC}}$;(2)$ {\frac{CP}{PE}\equal{}\frac{AC^2}{CE^2}}$.
2.$ {PA ,PB}$are tagent to $ {circle O}$. $ {PC\cap BA\equal{}D}$;$ {PC}$meet $ {circle O}$ at point$ {E}$. If :$ {PE\equal{}2,CD\equal{}1}$ ;find DE.
3.In $ {\triangle  ABC}$, $ {D\in BC}$;$ {E\in AB}$; and $ {\angle1\equal{}\angle2\equal{}\angle3}$ ;If the perimeter of $ {triangleABC; triangleEBD; triangle ADC}$ are $ {m,m1,m2 }$;prove that :$ {\frac{m1\plus{}m2}{m}<\equal{}\frac{5}{4}}$.
what do you think of these problems ? is it easy for junior or is it too difficult for junior students? :blush:
here is an image for problem1:

It is not so hard like the national round so I think it is suitable to test the ability of every individual contestant but the biggest challenge is timing .

I know it is easy for seniors ,but it is not too easy for junior students , can you solve the problems , dear mathlinkers? :)


limit
inequalities
topology
real analysis
real analysis unsolved

Let $ f \in C^1$ and  $ {\lim_{x \to \infty} {f'^2}(x) + {f^3}(x) = 0}$. Prove that $ {\lim_{x \to \infty} f(x) = 0}$.
Can anyone help me with this problem?

For any $ \epsilon>0$ there is $ x_0$ such that $ |f'^2(x) \plus{} {f^3}(x)|<\epsilon$ for $ x>x_0$. Consider the set $ U\equal{}\{x>x_0\colon |f(x)|>2\epsilon^{1/3}\}$. This is an open set on which $ f'(x)$ does not vanish, and therefore $ f$ is monotone on each of the open intervals that comprise $ U$. Try to find a contradiction here.

I do not understand what do you mean. Could you explain more?

This problem reminded me of the following one:
If $ \lim_{x->\inf} f'(x)+f(x)=a$, where $ a \in \mathbb{R}$, then $ \lim_{x->\inf} f(x)=a$.
To solve this one can reason as following:
${ \lim_{x->\inf} f(x)=\lim_{x->\inf} {{{e^x}f(x)} \over e^x}=\lim_{x->\inf} {{{e^x}(f'(x)+f(x))} \over e^x}=\lim_{x->\inf} f'(x)+f(x)}=a$
Where L'Hospitals rule is used in step two.

I thought something similar could be done to solve this problem, however I was not able to do it.
I also thought something about mean value theorems, however I was not able to use them either.

Let $ h(x) \equal{} \{ f'(x) \}^2 \plus{} \{ f(x) \}^3$. For any $ \epsilon > 0$, there is $ x_0$ such that for $ x>x_0$, $ |h(x)| < \epsilon$. Now I'll follow mlok's idea.

Consider the set $ U \equal{} \{ x>x_0 \, : \, |f(x)| > \sqrt[3]{2\epsilon} \}$. If $ U$ is empty, then the result immediately follows. So assume $ U$ is nonempty and choose any $ a \in U$. If $ f(a) > 0$, then $ \{ f(a) \}^3 > 2\epsilon$, so

$ \{ f'(a) \}^2 \plus{} 2\epsilon < h(a) < \epsilon,$

which is a contradiction. Thus $ f(a) < 0$ and $ \{ f(a) \}^3 < \minus{}2\epsilon$. Then

$ \minus{}\epsilon < \{ f'(a) \}^2 \plus{} \{ f(a) \}^3 < \{ f'(a) \}^2 \minus{} 2\epsilon,$

or $ \epsilon < \{ f'(a) \}^2$. Since $ f'$ is continuous, this inequality implies $ f'(a)$ does not vanish at any point in $ U$.

Now assume there is a point $ b \in U$ and $ x_0 < a < b$ such that $ f(b) < f(a)$, Applying intermediate value theorem, we may assume $ a$ is in the neighborhood of $ b$ in $ U$. By mean value theorem, there is $ c \in U$ such that $ f'(c) < 0$. Since $ f'$ cannot change its sign on $ U$, $ f$ is strictly decreasing on $ U$. Especially, $ I \equal{} [a, \infty) \subset U$.

Now let $ f(x) \equal{} \minus{} u^{\minus{}2}$, where $ u \equal{} u(x) > 0$ on $ I$. Then

$ h(x) \equal{} \{ f'(x) \}^2 \plus{} \{ f(x) \}^3 \equal{} 4 u'^2 u^{\minus{}6} \minus{} u^{\minus{}6}$,

or

$ 2u' \equal{} \minus{} \sqrt{1 \plus{} u^6 h(x)}$.

Since $ \epsilon < \{ f'(x) \}^2$ and $ f'(x) < 0$, we have $ \minus{}\sqrt{\epsilon} > f'(x) \equal{} 2u'u^{\minus{}3}$ and $ u$ is strictly decreasing. So $ u$ must converge and we have

$ \lim_{x\to\infty} 2u' \equal{} \minus{} \lim_{x\to\infty} \sqrt{1 \plus{} u^6 h(x)} \equal{} \minus{}1,$

which contradicts that $ u$ converges.

So, for all $ b \in U$ and $ x_0 < a < b$, $ f(b) \geq f(a)$. This means $ f$ is monotone increasing on $ U$. If $ f(x) < \minus{}\sqrt[3]{2\epsilon}$ for all $ x > x_0$, then $ f(x)$ must converge to some $ \minus{}L$ where $ L \geq \sqrt[3]{2\epsilon}$. But this implies $ \lim_{x\to\infty} f'(x) \equal{} \sqrt{L} > 0$, leading to a contradiction. Hence $ |f(x_1)| \leq \sqrt[3]{2\epsilon}$ for some sufficient large $ x_1$, and by our preceeding argument, $ |f(x)| \leq \sqrt[3]{2\epsilon}$ for all $ x \geq x_1$. So $ f(x) \to 0$ as $ x \to \infty$.

Suppose $ \lim_{x\to\infty}f(x)\ne 0$ then $ \exists \epsilon>0$ such that $ \forall x_0$ $ \exists x>x_0$ such that $ |f(x)|>\epsilon$  .....(1)

Since $ \lim_{x\to\infty}f'^2(x)\plus{}f^3(x)\equal{}0$ we have that $ \exists x_0$ such that $ \forall x>x_0$, $ |f'^2(x)\plus{}f^3(x)|<\frac{\epsilon^3}{2}$  ....(2)

Now, for such $ x_0$ as in (2), choose $ x>x_0$ as in (1) so that
$ \frac{\epsilon^3}{2}>|f'^2(x)\plus{}f^3(x)|\ge |f(x)|^3 > \epsilon^3$ , a contradiction.

[quote="packham"]$ |f'^2(x) \plus{} f^3(x)|\ge |f(x)|^3$[/quote]
That's not necessarily true if $ f(x)$ is negative.

Thank you all.
Can this problem be solved in more general therms? For example:
For what $ n,m \in \mathbb{N}$ does $ \lim_{x \minus{}> inf}f(x) \equal{} 0$ when $ \lim_{x \minus{}> inf} {{f'}^n}(x) \plus{} {f^m}(x) \equal{} 0$ if $ f \in C^1$?

The first thing to look at is the behaviour of solutions of the equation $ {{f'}^n}(x) \plus{} {f^m}(x) \equal{} 0$ at infinity. They are something like $ f(x)\equal{}(C\pm x)^{\frac{n}{n\minus{}m}}$. So the answer should be yes as long as $ n<m$. Try to see if the proof given by [b]sos440 [/b]works in this generality.


A small ball is thrown from a height of $ 1$ m, and $ 2$m away from a vertical wall. The ball bounces on the ground and on the vertical wall totally elastically and then it flies back to its initial position of throwing. Find the initial speed of the ball if it was thrown at an angle of $ 45$ degrees with the horizontal.

This looks easy, doesn't it, only two-dimensional motions...


AMC
AIME
geometry
3D geometry
USA(J)MO
USAMO
AIME I

[quote="ArcherPig"]I always have trouble visualizing 3-D figures in my head, unfortunately, there are a lot of cube/tetrahedral/pyramid etc. related questions on the AIME. Am I allowed to tear up scratch paper and build models for those questions? Since I'm the only one at my school taking it, it wouldn't bother anyone.[/quote]

i don't see why not.

Last year on the official pamphlet of solutions, for number 15, it said "... or take a paper out and fold it yourself!"

Or maybe I'm remembering the math jam... I'm not sure, but you are absolutely allowed to do that.

[quote]
My friend had paper cubes all over his desk after the AMC 10A and the proctor didn't complain, so i would think that you can fold paper models on the AIME.[/quote]

omgsh...me too!  I spent the last 15-20 minutes of the contest building paper cubes out of my scratch paper =)

Does anyone know if you're allowed any (or all) of the following?

* Graph Paper
* Ruler (with numbers on it)
* Straightedge
* Protractor
* Compass

____________________
AMC 10A = 144/150
AMC 10B = 150/150
AIME I Goal = 7/15 (taken as practice)
AIME II Goal = 8/15 (real)

USAMO Goal = a multiple of 7

Sometimes if the problem is geometry and detailed enough, you can "cheat", by constructing a diagram with the given info and just measuring..., but that works better on the AMCs since its multiple choice so you can then choose whatever answer is closest to what you get.


Find all polynomials $P(x)$ such that

$P(x^2+1)=P(x)^2+1$


Putnam geometry conics ellipse integration calculus vector

Here is a problem in related to the original problem.

[quote="kunny"]Find the area of the region expressed by the system of inequality $ \frac {x^2}{4} \plus{} y^2\leq 1,\ \left(1 \plus{} \frac {\sqrt {3}}{2}\right)x \plus{} y\leq 1.$

1982 Waseda University entrance exam/Science and Technology[/quote]

[quote="Kent Merryfield"]It's a linear homothecy: compress the plane in the $ x$ direction by a factor of $ 2$. That changes the area of all figures in the plane by that factor of $ 2.$ You can generalize this for more general linear transformations, but the particular case was all we needed here. The motivation for doing that here was to convert that ellipse into a circle, because I knew that with a circle, I would have a good variety of geometric and trigonometric arguments to draw on. Note that I never even calculated an integral.[/quote]

Hey,

[quote="Kent Merryfield"]
Don't drive yourself crazy trying to do those particular integrals. There's a much better way - a way in which you might not write down any integrals. The key [hide="hint."]Try an affine transformation. [hide="Further elaboration."]Turn the ellipse into a circle.[/hide][/hide]
[/quote]
Thanks for the help Kent Merryfield. Are there any texts that cover the affine transformation from a problem solving point of view? Or at least show how to apply the affine transformation to exercises such as changing: circles to ellipses, ellipses to circles,...etc.?

From what I could find on the internet (ie Wikipedia, MathWorld, and searching AoPS) I found abstract treatments of the affine transformation and not so much (problem solving) practical applications. Although, I did find some applications of the affine transformation with respect to: vectors and matrices. However, nothing really useful with showing how to apply to siolving problems. That is, nothing showing how (step by step) to apply the affine transformation to say for example changing an ellipse in to a circle, are there any texts that cover this?

In the [i]Art and Craft of Problem Solving[/i], 2nd Edition by Paul Zeitz affine transformations aren't directly covered but are referred to...so something more in depth in to affine transformations and specifically their (geometric) applications would be nice, any suggestions?

Thanks,

-APSStudent

Sorry, can you understand that the area of an ellipse : $ \frac {x^2}{a^2} \plus{} \frac {y^2}{b^2} \equal{} 1\ (a > 0,\ b > 0)$ is $ \pi ab$?

,or prove it by $ \pi a^2\cdot \frac {b}{a}$ or $ \pi b^2\cdot \frac {a}{b}$?

If you wouldn't understand, then you don't need to use Affin Transformation for the original problem.

Anyway, without using Affin transformation, we can obtain the [hide="answer"] $ m \equal{} \frac {1}{6}$ [/hide] of the original problem within 10 minutes.

I must correct to say, since Japanese high school students arn't admitted to use the integral of arcsinx, so it would takes for them to solve about 15-20 minutes without using Affin Transformation.

Hey,

[quote="kunny"]
Anyway, without using Affin transformation, we can obtain the [hide="answer"] $ m \equal{} \frac {2}{9}$ [/hide] of the original problem within 5 minutes.
[/quote]
Thanks for the response kunny.

Exactly, how would you solve this problem without resorting to integration or using an affine transformation to transform the ellipse in to a circle?

Thanks,

-APSStudent

Before answering to your question, please take a look my post in #5.

Hey,

[quote="kunny"]
Sorry, can you understand that the area of an ellipse : $ \frac {x^2}{a^2} \plus{} \frac {y^2}{b^2} \equal{} 1\ (a > 0,\ b > 0)$ is $ \pi ab$?

,or prove it by $ \pi a^2\cdot \frac {b}{a}$ or $ \pi b^2\cdot \frac {a}{b}$?
[/quote]
I understand the trivial fact that the area of an ellipse is, $ {{\pi}ab}$. However, you seem to refer to a solution to this problem not involving an affine transformation nor integration centered around the property of an ellipse's area (ie. $ {{A}_{E}} \equal{} {{\pi}ab}$). To which I question what your solution actually is that avoids integration and affine transformation - what is your solution?

Thanks,

-APSStudent

[quote="APSStudent"]Hey,

[quote="kunny"]
Sorry, can you understand that the area of an ellipse : $ \frac {x^2}{a^2} \plus{} \frac {y^2}{b^2} \equal{} 1\ (a > 0,\ b > 0)$ is $ \pi ab$?

,or prove it by $ \pi a^2\cdot \frac {b}{a}$ or $ \pi b^2\cdot \frac {a}{b}$?
[/quote]
I understand the trivial fact that the area of an ellipse is, $ {{\pi}ab}$. [/quote]

Can you explain that the area can be expressed $ \pi a^2\cdot \frac {b}{a}$ or $ \pi b^2\cdot \frac {a}{b}$?

Sorry, I wouldn't like to teach the method using Affin transformation to those who wouldn't try to undesrtand the reason including my students.

I think what Kunny is trying to tell you is to think of an ellipse as a circle with radius $ a$ dilated in one direction by a factor of $ b/a$. When you apply that transformation, area is also scaled by $ b/a$. That gives the desired expression of the area of the ellipse, for one. And circles are easier to deal with than ellipses.

[quote="APSStudent"]Hey,
[i]The Art and Craft of Problem Solving[/i], 2nd Edition
By: Paul Zeitz

[b]1. The problem statement, all variables and given/known data[/b]
1.3.15
Find the positive value of $ {m}$ such that the area in the first quadrant enclosed by the ellipse $ {{{\frac {x^{2}}{9}} \plus{} {y^{2}}} \equal{} {1}}$, the $ {x}$-axis, and the line $ {{y} \equal{} {\frac {2x}{3}}}$ is equal to the area in the first quadrant enclosed by the ellipse $ {{{\frac {x^{2}}{9}} \plus{} {y^{2}}} \equal{} {1}}$, the $ {y}$-axis, and the line $ {{y} \equal{} {mx}}$.

Thanks,
-APSStudent[/quote]

Strange, I get $ m \equal{} \frac16$....

I followed Kent's suggestion and let $ u \equal{} \frac {x}3$. Now in the u-y axis we have the equations
$ u^{2} \plus{} y^{2} \equal{} 1$ and $ y \equal{} 2u$. So the area of sector of the circle bounded by u-axis is $ \frac12\tan^{ \minus{} 1} 2$ and the area of the circle bounded by the y-axis(the second bit) $ \frac12 \left(\frac {\pi}2 \minus{} \tan^{ \minus{} 1}3m\right)$. So:

$ \frac12 \left(\frac {\pi}2 \minus{} \tan^{ \minus{} 1}3m\right) \equal{} \frac12\tan^{ \minus{} 1} 2$

$ \frac {\pi}2 \equal{} \tan^{ \minus{} 1} 2 \plus{} \tan^{ \minus{} 1}3m \Rightarrow m \equal{} \frac16$

$ m\equal{}\frac16$ is correct. For a slightly simpler explanation:

Get to the point of having the line $ y\equal{}2u.$ Now, all we really need to do is reflect that line around the $ 45^{\circ}$ line $ y\equal{}u.$ That gives us the reciprocal of the slope, namely $ y\equal{}\frac12u$. Now undo the transformation, setting $ u\equal{}\frac{x}{3}$ to get $ u\equal{}\frac16x.$

It was perhaps surprising that this came in below the original line rather than above it - but such is the answer.

I probably made a pedagogical error in using the word "affine" because it made you say "aieee, I don't know that word." Perhaps I should have just said "stretch it or squash it." But don't be afraid of mere words - the underlying ideas are sometimes pretty simple.

For another problem in which the notion of affine transformations may be useful:

Consider $ \triangle ABC$ with vertices $ A(4,8),B(\minus{}1,2),$ and $ C(0,\minus{}3).$ Find the point $ D$ such that $ \triangle ABD,\triangle ACD$ and $ \triangle BCD$ all have the same area.

[quote="Kent Merryfield"]$ m \equal{} \frac16$ is correct. For a slightly simpler explanation:

Get to the point of having the line $ y \equal{} 2u.$ Now, all we really need to do is reflect that line around the $ 45^{\circ}$ line $ y \equal{} u.$ That gives us the reciprocal of the slope, namely $ y \equal{} \frac12u$. Now undo the transformation, setting $ u \equal{} \frac {x}{3}$ to get $ u \equal{} \frac16x.$[/quote]

:coolspeak: Beautiful solution  :lol:
You never cease to amaze me, Mr. Merryfield. What a slick solution!

[quote="Kent Merryfield"]
For another problem in which the notion of affine transformations may be useful:

Consider $ \triangle ABC$ with vertices $ A(4,8),B( \minus{} 1,2),$ and $ C(0, \minus{} 3).$ Find the point $ D$ such that $ \triangle ABD,\triangle ACD$ and $ \triangle BCD$ all have the same area.[/quote]

I am trying this problem from a long time. But I am not sure how to compress the axis. I thought of the idea that if I make the triangle equilateral by compressing it by some factor, then D is the circumcentre in that compressed plane. But I dont know how to make it equilateral. I will think a little more on this.

I am sure I can do this the brute force way, i.e finding the equations of the line AB,BC and CA. Then using perpendicular distance formula from D to the lines, thus getting the height.And then equating the areas and solving the system of equations to get D.

Thanks for these wonderful problems, Mr.Merryfield :D

[quote="Kent Merryfield"]

Consider $ \triangle ABC$ with vertices $ A(4,8),B( \minus{} 1,2),$ and $ C(0, \minus{} 3).$ Find the point $ D$ such that $ \triangle ABD,\triangle ACD$ and $ \triangle BCD$ all have the same area.[/quote]

Let $ A'(0,\ 0),\ B( \minus{} 5, \minus{} 6),\ C'( \minus{} 4, \minus{} 11),\ D'(a,\ b)$,

$ [A'B'D'] \equal{} [A'C'D'] \equal{} [B'C'D']$

$ \Longleftrightarrow \frac {1}{2}| \minus{} 5a \plus{} b| \equal{} \frac {1}{2}| \minus{} 4b \plus{} 11a| \equal{} \frac {1}{2}|(b \plus{} 6) \plus{} 5(a \plus{} 5)|$.

Solving this, the answer is $ D(a \plus{} 4,\ b \plus{} 8)$.

[quote="APSStudent"]Hey,
[i]The Art and Craft of Problem Solving[/i], 2nd Edition
By: Paul Zeitz

[b]1. The problem statement, all variables and given/known data[/b]
1.3.15
Find the positive value of $ {m}$ such that the area in the first quadrant enclosed by the ellipse $ {{{\frac {x^{2}}{9}} \plus{} {y^{2}}} \equal{} {1}}$, the $ {x}$-axis, and the line $ {{y} \equal{} {\frac {2x}{3}}}$ is equal to the area in the first quadrant enclosed by the ellipse $ {{{\frac {x^{2}}{9}} \plus{} {y^{2}}} \equal{} {1}}$, the $ {y}$-axis, and the line $ {{y} \equal{} {mx}}$.

-APSStudent[/quote]

This isn't the original problem of Putnam.

[url]http://www.unl.edu/amc/a-activities/a7-problems/putnam/-pdf/1994.pdf[/url]

[quote="Kent Merryfield"]Principles:

For an equilateral triangle all centers coincide. If the triangle were equilateral (which it isn't), the point $ D$ would be that center.

Given an arbitrary triangle, there is some affine transformation which would transform an equilateral triangle into this arbitrary triangle. (I do [i]not[/i] recommend trying to find that transformation in this specific case - just use that it exists.)

An affine transformation, which is just an arbitrary (invertible) linear transformation plus some constant point, transforms all areas by the same proportion. Hence, the point $ D$ is the point that the center of the equilateral triangle gets transformed to by an affine transformation.

Now: of all the centers that a triangle has, which one gets preserved by affine transformations? You might ask it this way: which center's coordinates can be obtained by finding a linear combination of the vertices?

(Another phrase to look up some time: [i]barycentric coordinates.[/i])[/quote]

wow. I knew the answer, of course, but this is a very very nice way to look at it.

[quote="Kent Merryfield"]
An affine transformation, which is just an arbitrary (invertible) linear transformation plus some constant point, transforms all areas by the same proportion. Hence, the point $ D$ is the point that the center of the equilateral triangle gets transformed to by an affine transformation.

Now: of all the centers that a triangle has, which one gets preserved by affine transformations? You might ask it this way: which center's coordinates can be obtained by finding a linear combination of the vertices?
[/quote]

Well...A centroid is a linear combination of its vertices... :huh:

But I am stumped, how will that help me get the point D, without actually finding the linear transformation?

Thanks for the hints,
Spai

The centroid before the transformation was $ \frac13(u\plus{}v\plus{}w)$ where $ u,v,$ and $ w$ are the vertices expressed as vectors. The transformation $ L$ is given by $ L(x)\equal{}Ax\plus{}b$ where $ A$ is some matrix and $ b$ is some vector. Then we follow what that does to the centroid:

$ L\left(\frac13(u\plus{}v\plus{}w)\right)\equal{}\frac13A(u\plus{}v\plus{}w)\plus{}b\equal{}$

$ \equal{}\frac13\left((Au\plus{}b)\plus{}(Av\plus{}b)\plus{}(Aw\plus{}b)\right)\equal{}\frac13(L(u)\plus{}L(v)\plus{}L(w)).$

Just average the three vertices you're given.


If I understand your question correctly,  after the shuffle, a card which is at $n$ position is now at $2n$ (Modulo 53 if there are 52 cards in the deck) right? so after $k$  shuffle, nth card is at $n 2^{k}$  th position. right?  so all you have to find a $k$ such that $2^{k}\equiv 1 \bmod 53$

you do know that, 53 is prime so,  $2^{52}\equiv 1 \bmod 53$ so $52$  is the answer (Or multiple of  the answer -  but quick check gives, neither $2^{4}$  nor $2^{26}$   is 1 mod 53 - so 52 is the answer - actually a little knowledge of NT is enough to deduce that for a prime like 8k+5, eg 53, 2 indeed is a primitive root)

Similar trick for $26$ cards, 27 is not prime, but $\phi(27) = 18$ and $18$ indeed is the answer, (Quick check for the factors of 18 )

Hope this helps.

"a card which is at $n$ position is now at $2n$ " I don;t think so, for exmaple, a pack of 8 cards, divided to 4 and 4, the card of the RHS will be in the following order-1,1+2,1+2+3

May be I am mistaken so can you make it clear:
In your opinion, if the original order is (1,2,3,4,5,6,7,8), how will it look after  one shuffle?

This , is nothing but the same (similar to ) as what I posted above, [b]except [/b]that the  top and  the bottom cards remains at the same place. (Think of 6 cards shuffled as my first post and then add top and bottom card - that would produce your 8 card shuffle)

So the above logic stands, except you change  52 cards to a 50 card deck and then use  the logic given above.
[hide](The answer would be to find a $k$  such that,  $2^{k}\equiv 1 \bmod 51$
Since 51=17*3, and $2^{4}\equiv-1 \bmod 17$ (and of course $2^{4}\equiv 1 \bmod 3$   so one can see,  that the smallest value of k is $8$ ...  $2^{k}\equiv 1 \bmod 51$

Similary for 26 cards problem reduce to doing math mod 25, and answer would be 20.

 [/hide]

Hope that helps.


I tested in Number Theory, and the counter increased, but didn't decrease after deleting (and yes, I first tried here, but then saw what you mentioned).

I see ...  :maybe: ... I'll take a look.

You were right, due to a small mistake a new bug was applied only when deleting posts. Now it's (finally I hope :) ) fixed.


number theory
Diophantine equation
relatively prime
greatest common divisor
number theory proposed

Solve the diophantine equation: $x^a-1=y^b$ where $x,y,a,b>1$ and $x\equiv1[y]$.
(related question:)[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=28543][u]here[/u][/url]

Hé hé...I still have your mail from May 1999 with your proof... :D  :P

Pierre.

I'am sure somebody here will find a more elegant proof ;)

I am a specialist in making stupid mistakes while solving diophantine equations, so please excuse me for the possibly long and useless post:
  Consider $p$ a prime factor of and put $ a'=a/p$. If we have solved the equation with a prime then we find that $ (x^{a'})^p-1=y^b$ and $ x^{a'}=1 (mod y)$ thus we obtain $ x^{a'}=3$ and $ y=2, b=3$. Thius shows that we can assume from the beggining that a is prime. Let $ x=1+ky$ and write the equation in the form $ k[1+(1+ky)+...+(1+ky)^{a-1}]=y^{b-1}$. Observe that the greatest common divisor of k and $ 1+(1+ky)+...+(1+ky)^{a-1}$ is 1 or a.
   First case: the greatest common divisor is 1. We deduce the existence of u,v such that $ k=u^{b-1}$ and $ 1+(1+ky)+...+(1+ky)^{a-1}=v^{b-1}$ and $ y=uv$. Then $ ky=u^b v$ and then the equation becomes $ 1+(1+u^b v)+...+(1+u^b v)^{a-1}= v^{b-1} $.  Taking this modulo v we get that v=1 or v=a. It is clear that $v>1$ thus $v=a$ and $ 1+(1+u^b a)+...+(1+u^b a)^{a-1}=a^{b-1}$ The case $b=2$ is not difficult, so let b$ b>2$. Taking the last relation mod $a^2$ we get a=2 (if not the left hand side is congruent a a mod $a^2$, while the right hand side is 0 mod $a^2$) and we have the equation $ 2^{b-2}-1=u^b$. We find that $u$ is at most 3 and then a simple check shows that $b=3$ and we find the well-known solution.
   Second case: the greatest common divisor is a. Write $ k=au$ and $ 1+(1+ky)+...+(1+ky)^{a-1}=av$ (*)with u,v relatively prime. Then $ a^2 uv=y^{b-1}$ and thus a divides y. Let $ y=a^{t}r$ with r relatively prime with a. Wed obtain $ uv= a^{t(b-1)-2} r^{b-1}$. The relation (*) shows that v cannot be a multiple of a if a is not 2 ( again, take the relation modulo $a^2$). I will neglect the case $a=2$. So, v is relatively prime with a and thus u is divisible by $ a^{t(b-1)-2}$, let $ u=a^{t(b-1)-2}s$. Thus, $ sv=r^{b-1}$ and since s,v are relatively prime we have $ s=A^{b-1}$ and $ v=B^{b-1}$ for A,B natural numbers. Of course, $r=AB$. Then $ ky= a^{tb-1} A^b B$ and the relation (*) becomes $ 1+(1+ a^{tb-1} A^b B)+...+(1+ a^{tb-1} A^b B)^{a-1}=aB^{b-1}$, which can be rewritten $ (1+ a^{tb-1} A^b B)^a=1+a^{tb} A^b B^b$. Develop the left hand side with Newton and simplify by $ a^{tb} A^b B$. We find $ 1+ a(a-1)/2 a^{tb-2} A^b B+...=B^{b-1}$ from where we see that $B=1$. Return in the last relation with this value and you will find that a sum of positive numbers is 0, false. Thus necessarily we have $a=2$. This case is not difficult and my you have already found five major mistakes till here, so I will stok before making the sixth.

Each time it is so hard to read such proofs! (this is not because of you, Harazi, just the problem assumes that solution should be boring).

Thank you, Harazi, for the great job!

I wouldn't say that before being sure the solution is not wrong.  ;)

wwell use the following: if $A^m-1$ and $A^n-1$ share prime factors $m>n$, $A>1$, then $m=2$, $n=1$ and $A$ is of the form $2^j-1$. It is easy to prove, using "Hensels" lemma in theorems section

Pascual, could you please detail your proof?

I will try to do it as soon as i got time.

Great! I understood what you want and I found a short solution. You don't need to use the result you said (which is anything but not easy, a last problem from ISL). Thanks a lot for the idea.
 :)

HArazi, see this topic: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=20031[/url]

I know to solve your problem. And probably for you it's easy, but for many it wasn't.  ;)  :P

Jaja, Harazi! It wasnt easy for me! What happens is that I love problems with $A^n-1$. you already told me you found a solution, but if there is anything in which i could help you, i will do it :)

It wasn't me who told it's nice and easy.  ;)  And to be sincere, I also adore such problems.  ;) Thank you, you already helped me with your nice idea to solve this beautiful problem. By the way, what does Jaja mean? If it is bad, I prefer not to know.  :D

By Zsigmondy's Theorem there must exist a prime factor $p|x^a-1=y^b$ not dividing $x-1$, but $p|y|x-1$. Thus we only need to check the three exceptions:

Case 1: $a=1, x-1=1$. Absurd.

Case 2: $a=6, x=2$, then $y^b=63$. Absurd.

Case 3: $a=2, x+1=2^k(k\geq 2)$. Then
$$(x+1)(x-1)=y^b$$
Since $4|x+1$, we have $x-1=2m^b$ where $m$ is odd.
$$2^{k+1}m^b=y^b$$
Let $y=2^dm$ then $y|x-1$ implies $2^dm|2m^b$ and thus $d=1$.
$$2^{k+1}m^b=2^bm^b$$
$$k=b-1$$
Now since $2^{b-1}-2m^b=2$, by simple bounding we get $m=1, b=3$.

Thus $(x,y,a,b)=(3,2,2,3)$.


function
algebra
polynomial
quadratics
inequalities
floor function
analytic geometry

Given a is a positive real number and the equation: $ a^5 \minus{} a^3 \plus{} a \equal{} 2$. Find $ [a^6]$ ($ [x]$ is the greatest integer function of x)(No calculator please!)
My point here is how you can estimate the upper bound of a in order to have: $ k < a^6 < k \plus{} 1$ with k is a positive integer. I can find the lower bound easily but can't find the most strict upper bound of a. If this is the proof question, then it's much easier!

[quote="Sage!"][hide="Solution"]
Think of the left side as a polynomial of variable $ a$.

$ f(a) \equal{} a^5 \minus{} a^3 \plus{} a \minus{} 2$
$ f'(a) \equal{} 5a^4 \minus{} 3a^2 \plus{} 1$

After substitution $ t \equal{} x^2$ it becomes a quadratic equation of variable $ t$.

$ 5t^2 \minus{} 3t \plus{} 1 \equal{} 0$
$ \Delta \equal{} 9 \minus{} 4\cdot5 < 0$

Since $ 5 > 0$ then $ 5t^2 \minus{} 3t \plus{} 1 > 0$ and indicates that $ f(a) \equal{} a^5 \minus{} a^3 \plus{} a \minus{} 2$ is strictly increasing.

But $ f(1) \equal{} \minus{} 1 < 0, f(2) \equal{} 24 > 0$ so all roots of polynomial $ f$ are in $ [1,2)$ interval.

If $ a \in [1,2)$ then $ [a^6] \in [1,63]$ so the most strict upper bound for $ [a^6]$ is $ 63$
[/hide][/quote]
The question asks about $ [a^6]$, not its upper bound. In fact, your upper bound is not strict at all.The upper bound of $ [a^6]$ is 4 in the interval (1,2](proving it by AM-GM's inequality). The lower bound, on the other hand, is 3(we can assume this value and prove it as a possible lower bound value).We can also point out that $ f(1.207)<0$, which immediately leads to $ 1.207^6>3$(I;m not able to find out this "strange" value by hand though).The point now is how can we find out the number 2>a>1 which is close enough to satisfy both inequalities:
$ f(a)<0$
$ a^6>3$

Am I missing something? Because $ a^5\minus{}a^3\plus{}a\minus{}2\equal{}(a^2\minus{}a\plus{}1)(a^3\plus{}a^2\minus{}a\minus{}2)\equal{}0$ means that both non-real solutions to $ a^3\equal{}1$ work, giving $ a^6\equal{}1$ (so I don't see how the lower bound is $ 3$).

[quote="math154"]Am I missing something? Because $ a^5 \minus{} a^3 \plus{} a \minus{} 2 \equal{} (a^2 \minus{} a \plus{} 1)(a^3 \plus{} a^2 \minus{} a \minus{} 2) \equal{} 0$ means that both non-real solutions to $ a^3 \equal{} 1$ work, giving $ a^6 \equal{} 1$ (so I don't see how the lower bound is $ 3$).[/quote]
You're completely right! However, I forgot to add the restriction a is a positive real number, so you still have one more steps to do though. Sorry very much! :oops:

[b]EDIT[/b]: Your factorization almost trivializes this question because of Cardano's formula for cubic equation. I now wonder whether we have a nicer way to do it or not at this step. Nice observation math 154!

[hide]Note that $ a \in ( 6/5, \; 5/4)$, and also that $ (6/5)^3 \approx 2.986$ and $ (5/4)^6 \approx 3.815$. It suffices to push our lower bound up by enough to make $ \lfloor (\text{lower bound})^6 \rfloor \equal{} 3$.

Letting $ f \equal{} x^3 \plus{} x^2 \minus{} x \minus{} 2$, we find that $ f' \equal{} 3x^2 \plus{} 2x \minus{} 1$. Because our cubic $ f$ is convex, we know that the maximum slope on $ (6/5, \; 5/4)$ is $ f'(5/4) \equal{} 99/16$, thus the line through $ (6/5, \; f(6/5))$ with slope $ f'(5/4) \equal{} 99/16$ crosses the $ x$ axis to the left of where $ f \equal{} 0$ crosses the $ x$ axis. We set this root as the new lower bound. We can use a slightly looser lower bound by using $ 8$ as the slope of this line instead of $ 99/16 \approx 6$.

$ f(6/5) \equal{} \minus{} 4/125$. So, this line must increase by $ 4/125$ over some distance of $ \Delta x$. Because the slope is $ 8$, $ \Delta x \equal{} \frac {1}{8} \cdot \frac {4}{125} \equal{} \frac {1}{250}$

$ x^6$ increases at a rate $ 6 x^5$, so the increase in $ x^6$ over a distance of $ \Delta x$ is at least $ 6 \cdot (6/5)^5 \cdot \Delta x \equal{} (6/5)^6 \cdot \frac {1}{50}$, an increase of $ 2 \%$.

It is easy to see that a 2% increase from $ (6/5)^6 \approx 2.986$ will tip $ a^6$ over $ 3$.[/hide]


[quote="osmantelli"]ABC is isosceles triangle(AB=AC) and D is a point on AC such that BAC=40 and DBC=x.   x=?[/quote]

Correct it!! :?


The exercise problem in chapter 6 friction in h.c.verma problem 23 , the problem with the three blocks makes no sense. Could anyone plz post a detailed solution. Also the practise problem 9 in the same chapter doesn't take any psuedo force into account. :( One more thing , could anyone explain how gradients can be used to find whether a force is conservative or not. :blush:

http://www.physicsforums.com/showthread.php?t=247620

if u want me to give a detailed soln , do tell

btw , are u refferring to solved example 9 ?

Yes i want a detailed solution. Of course i refer to solved example number 9 :blush:

I think curl can be used to determine whether any force is conservative or not.

If curl of any force is $ 0$,force is conservative. :)

and curl is nothing but cross product of del operator and force

Ex 9
well they have solved in inertial frame
we consider pseudo force in non inertial frames :)

[hide]sorry for not being active for some time[/hide]

Thanks slogger. But I have one more doubt though. For the second block the force by the topmost block is 4N and the maximum frictional force is 15N. Now why can't the net force on the second block be 15-4 = 11N to the left and the net force on the lowermost block be 15N to the right and in this case the 2nd and 3rd blocks will have different accelerations. :blush:

this should help u ->
[hide]
friction is a self adjusting force [/hide]


Can any of u give me some good suggestions to study for AIME and USAMO?
How many hours u have to practice a day?
what type of problem?
What else u can do to qualify for USAMO?
also,can anyone give me a practice test from the past USAMO?

THANKY0U. :wink:

Doing old exams is a nice start.  As is reading the AoPS books.  You can find old exams for many contests if you look under the "Contests" tab on the forum page.  And it doesn't matter how long you study-quality over quantity.


Find a function $f:R->R$ that maps any open interval $(a,b)$ onto $R$ (independently of $a,b$).

Similar stuff has been discussed [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=31493]here[/url].


analytic geometry
calculus
calculus computations

Find the minimum value of $ k$ such that there exists real number satisfying $ k(x\plus{}3)\plus{}1\equal{}\sqrt{x}.$

hello kunny, i have got
$ x_1\equal{}\frac{\minus{}1\minus{}3k\plus{}\frac{1}{2}\frac{1\plus{}\sqrt{1\minus{}12k^2\minus{}4k}}{k}}{k}$
$ x_2\equal{}\frac{\minus{}1\minus{}3k\minus{}\frac{1}{2}\frac{\minus{}1\plus{}\sqrt{1\minus{}12k^2\minus{}4k}}{k}}{k}$
Sonnhard.

i think it is $ k\equal{}\frac{\minus{}1}{3}$

That's correct.  :) The real numbers satisfying the given equation is the $ x$-coordinate of the intersection point of the curve $ y \equal{} \sqrt {x}$ and the line $ y \equal{} k(x \plus{} 3) \plus{} 1$ which passes through the fixed point $ ( \minus{} 3,\ 1)$, yielding

$ k_{min} \equal{} \minus{} \frac 13$ when the line passes through the origin.

How about $ k_{max}?$


Hi all,

Please help me to write the symbol in the picture attached in Latex!!!!!

Many thanks,

Joel


trigonometry
quadratics
induction
function
strong induction
algebra
quadratic formula

If $ \theta$ is a constant such that $ 0<\theta <\pi$ and $ x\plus{}\frac{1}{x}\equal{}2\cos{\theta}$, then for each positive integer $ n$, $ x^n\plus{}\frac{1}{x^n}$ equals

(A) 2cos$ \theta$
(B) $ 2^n$cos$ \theta$
(C) $ 2\cos^n$ $ \theta$
(D) 2cos$ n\theta$
(E)$ 2^n$ $ \cos^n$ $ \theta$

Thanks in advance!

[hide="HINT"]Find the root of the quadratic[/hide]

You can get pretty far by process of elimination alone.  Ignore that first condition; this identity clearly has to hold everywhere.  When $ x \equal{} 1$ we have $ \theta \equal{} 0$ and $ x^n \plus{} \frac {1}{x^n} \equal{} 2$, which eliminates B) and E).  When $ x \equal{} \minus{} 1$ we have $ \theta \equal{} \pi$ and $ x^n \plus{} \frac {1}{x^n}$ equalling either $ 2$ or $ 0$ depending on the parity of $ n$, which eliminates A) and C).  Hey!

[hide="Explanation"] $ x \equal{} \cos \theta \plus{} i \sin \theta$ and $ \frac {1}{x} \equal{} \cos \theta \minus{} i \sin \theta$.  See [url=http://en.wikipedia.org/wiki/Euler%27s_formula]Euler's formula[/url]. [/hide]

[quote="elcric"]If $ \theta$ is a constant such that $ 0 < \theta < \pi$ and $ x + \frac {1}{x} = 2\cos{\theta}$, then for each positive integer $ n$, $ x^n + \frac {1}{x^n}$ equals

(A) 2cos$ \theta$
(B) $ 2^n$cos$ \theta$
(C) $ 2\cos^n$ $ \theta$
(D) 2cos$ n\theta$
(E)$ 2^n$ $ \cos^n$ $ \theta$

Thanks in advance![/quote]

[hide="Outline for proof."]
=====
The answer is $ \boxed{(D)}$. We would proceed by strong induction, noting that:
$ x^n + \frac {1}{x^n} = \left(x^{n - 1} + \frac {1}{x^{n - 1}}\right)\left(x + \frac {1}{x}\right) - \left(x^{n - 2} + \frac {1}{x^{n - 2}}\right)$,

and:

\begin{eqnarray*} 2\cos n\theta & = & (2\cos (n - 1)\theta)(2\cos \theta) - (2\cos (n - 2)\theta) \\
& = & 2(\cos n\theta + \cos (n - 2)\theta) - 2\cos (n - 2)\theta \\
& = & 2\cos n\theta \end{eqnarray*}
=====
[/hide]

EDIT: I think this proof was in Engel.

EDIT2:

[quote="t0rajir0u"]You can get pretty far by process of elimination alone.  Ignore that first condition; this identity clearly has to hold everywhere.  When $ x = 1$ we have $ \theta = 0$ and $ x^n + \frac {1}{x^n} = 2$, which eliminates B) and E).  When $ x = - 1$ we have $ \theta = \pi$ and $ x^n + \frac {1}{x^n}$ equalling either $ 2$ or $ 0$ depending on the parity of $ n$, which eliminates A) and C).  Hey!
[/quote]

That's clever! How do you convince yourself that the identity has to hold for all $ \theta$?

Continuity.  If any of those identities holds for $ \theta$ very close to $ 0$ or $ \theta$ very close to $ \pi$, they have to hold at $ 0$ and $ \pi$ since all of the functions involved are continuous.  (I suppose it's not completely trivial to convince yourself that $ x$ varies continuously as $ \theta$ varies, but the quadratic formula should settle this.)


$a,b,x,y$ satisfy that $a+b=3,ab=1,x+y=4,xy=2.$Find the value of $(ax+by)^3+(ay+bx)^3.$

If you factorise it as the difference of two cubes, the first bracket factorises perfectly to give an easy value. Not so sure bout the second one, im doing this in my head....

dondigo,your answer is correct. :)

Let $A=ax+by,B=ay+bx$, we have $(ax+by)^3+(ay+bx)^3=A^3+B^3$


If $ p>5$,prove that the equation  $ (p\minus{}1)!\plus{}1\equal{}p^n$   doesn't have solutions in set of $ N$.

I think this involves Wilson theorem.

It states that if $ p>1$ is in integer, then $ (p\minus{}1)!\plus{}1$ is a multiple of $ p$ if and only if $ p$ is a prime. So it doesn't have a solution since you didn't state that p was prime. If p was prime it would have solutions..

[quote="modularmarc101"]If p was prime it would have solutions..[/quote]
$ p$ dividing the LHS is a necessary but not sufficient condition for solutions to exist.  The point of the problem is that there are no solutions even if $ p$ is prime.


Let $ a,b$ and $ c$ be positive real numbers whose $ a\plus{}b\plus{}c\equal{}3$. Prove that

$ (ab\plus{}bc\plus{}ca)(a^2b\plus{}b^2c\plus{}c^2a)\le9$

The following inequality also true
$ (a\plus{}b\plus{}c)(a^2b\plus{}b^2c\plus{}c^2a)\ge 9$ where $ a,b,c\ge0$ such that $ ab\plus{}bc\plus{}ca\equal{}3$

[quote="tardi"]Let $ a,b$ and $ c$ be positive real numbers whose $ a \plus{} b \plus{} c \equal{} 3$. Prove that

$ (ab \plus{} bc \plus{} ca)(a^2b \plus{} b^2c \plus{} c^2a)\le9$[/quote]
[url=http://www.mathlinks.ro/viewtopic.php?t=193697]See here[/url] :)


[quote="quangtrung"]prove that there exists three different integer numbers $a,b$ and $c$ in the set
{$n^2$, $n^2+1$,...,$n^2+n+3\sqrt{n}$} so that $c$ divides $ab$[/quote]
Let c=(n-x)(n+x+2),a=(n-x)(n+x+1),b=(n-x-1)(n+x+2). If n-3sqrt(n)<=x(x+1)<=n a,b,c  satisfyie condition.  x= [sqrt(n)-0.5] satisfyes n-3sqrt(n)<=x(x+1)<=n


A tetrahedron $ ABCD$ has edges of total length $ 1$.  The angles at $ A$ are all $ 90^{\circ}$.  Find the maximum volume of this tetrahedron.

Suppose that lengths of the edges adjacent to $ A$ are $ x$, $ y$, and $ z$. The other sides are $ \sqrt {y^2 \plus{} z^2}$, etc.

Then we have $ 1 \equal{} \sum [x \plus{} \sqrt {y^2 \plus{} z^2}]\ge \sum [x \plus{} \sqrt {2yz}]\ge 3\sqrt [3]{xyz} \plus{} 3\sqrt [3]{\sqrt {8}\sqrt {yz}\sqrt {zx}\sqrt {xy}}$
$ \equal{} (3 \plus{} 3\sqrt {2})*\sqrt [3]{xyz}$
where all the steps are by the AM-GM.

It follows that $ V \equal{} \frac {xyz}{6}\le \frac {5\sqrt {2} \minus{} 7}{162}$ with equality when $ x \equal{} y \equal{} z \equal{} \frac {1}{3 \plus{} 3\sqrt {2}}$

Comment: If you consider what types of symmetric expressions are the smallest, products are the smallest symmetric sum, so it should be easy to put an upper bound on a product.


Anyone have good LaTex books to recommend?

I bought copies of LaTeX: A document preparation system, The LaTeX Companion, and Math into LaTeX this year to learn TeX...(haven't exactly read thru any of them though)...any more I need to add?

I have the source (Donald Knuth's [i]The [tex]\TeX[/tex]book[/i]) and [i]The [tex]\LaTeX[/tex] Companion[/i] By Goossens, Mittelbach, and Samarin, and I think they are both very good.

I should get a book and figure out how to use TeX.. tell me which you find the easiest to use/understand =)

Get Maple. Type it in Maple. And then use the Latex() function. And you get your Latex stuff. You will never need to learn Latex (but it will gradually come to you)!

No, that's silly. Typing in [tex]\LaTeX[/tex] is actually fun. Here's the way I learned [tex]\LaTeX[/tex], more or less:

Go to the AMC website and pick a random USAMO. Print out the pdf and tex files from the same year. See if you can figure out how they did what they did. Then pick another year. Print out only the pdf. See if you can figure out how to make what they did on your own. Then see how they did it.

[quote="MithsApprentice"]I bought copies of LaTeX: A document preparation system, The LaTeX Companion, and Math into LaTeX this year to learn TeX...(haven't exactly read thru any of them though)...any more I need to add?[/quote]
That's plenty of books.  (I have all three too.)  Among those, I would start by reading the numbered chapters of Lamport's "LaTeX: A Document Preparation System".  That should be plenty of information to get you started.  When you need further information on a particular topic, you can read particular chapters of the other books, or chapter C of Lamport.

[quote="rrusczyk"]And when you get stuck, nothing beats google.[/quote]

Except for asking Simon what the command for such-and-such is (translation: bugging Simon via IM)

I'm sure I'll avail myself of that frequently when he's here this summer :)

[quote="ComplexZeta"]Keep in mind that if you ask me for weird things, I just look on my wall right next to my computer, since I posted all the math [tex]\TeX[/tex] macros. I don't have them all memorized... yet. :)[/quote]

cheater :-P my parents wont let me stick anything on the walls next to the computers...lol

[quote="MithsApprentice"][quote="mathfanatic"]Except for asking Simon what the command for such-and-such is (translation: bugging Simon via IM)[/quote]

when he's not online (which is almost never except when he's sleeping or at math club or classes) there's always confuted u could bug...[/quote]

I do that, too.  Except confuted knows only marginally more (unless he's been studying recently) than I do.

Does anyone know how to make nice graphics in TeX?

I know how to make commutative diagrams. I'm trying to figure out how to get image imports to work properly. They keep coming out too small.

[quote="Fermatprime"][quote="ComplexZeta"]Go to the AMC website and pick a random USAMO. Print out the pdf and tex files from the same year. See if you can figure out how they did what they did. Then pick another year. Print out only the pdf. See if you can figure out how to make what they did on your own. Then see how they did it.[/quote]

The problem with that being that I'm neither crazy nor a complete masochist.

So, what are good books for learning [tex]LaTeX[/tex] (or better yet, websites that you can print out the commands from and stick them on the wall next to the computer)?[/tex][/quote]

whats wrong with that?

That's how I learned HTML, CSS, and JavaScript: by looking at finished products and trying to reproduce them on my own.

Its a very effective method, if you ask me.

There's nothing wrong, per se, with that method of learning. Really, it often works well. It's just not the way I learn best, personally, and I'm not alone there either. (Actually, I do use this method sometimes; that's how I taught myself some programming techniques. But those examples are rare for me.)