induction
recurrence relation
algebra
polynomial
Iteration
Sequence
IMO Shortlist

For an integer $x \geq 1$, let $p(x)$ be the least prime that does not divide $x$, and define $q(x)$ to be the product of all primes less than $p(x)$. In particular, $p(1) = 2.$ For $x$ having $p(x) = 2$, define $q(x) = 1$. Consider the sequence $x_0, x_1, x_2, \ldots$ defined by $x_0 = 1$ and \[ x_{n+1} = \frac{x_n p(x_n)}{q(x_n)} \] for $n \geq 0$. Find all $n$ such that $x_n = 1995$.

[quote=""ISL questions and solutions from 1995-2001" by A.Ajorloo"]
note that $ x_n$ is a natural and square free number,now denote the sequence of prime numbers as $ p_0 \equal{} 2,p_1 \equal{} 3,\ldots.$
now denote the sequence $ a_n \equal{} \left(s_m,s_{m \minus{} 1},\ldots ,s_0\right)$ in which each $ s_i$ is equal to $ 1$ iff $ p_i\mid x_n$,and $ 0$ otherwise.
now it can be easily proved,using induction,that $ a_n$ is the representation of $ n$ in base $ 2$,which yields that $ x_n \equal{} 1995$,$ a_n \equal{} (1,0,0,0,1,1,1,0)$ and $ n \equal{} 142$.

QED

[/quote]

[hide="Took Forever"]

Okay so note that $x_1=2$, $x_2=3$.  Then $x_3=\frac{3*2}{1}=6=2*3$.  We let this be $11$.  Then, $x_4=\frac{3*2*5}{3*2*1}=5$ let this be $100$.  I will prove in general the correspondence between binary numbers.

Assume via induction that this binary pattern holds.  The base case is proved above.  Now, assume it holds for the set $x_{2^k}$ to $x_{2^{k+1}}$ and I prove that it works for $x_{2^{k+1}}$ to $x_{2^{k+2}}$.

Note that the numbers without the factor of the $k+1$th prime (this is the leading factor) are going to create the group $2^{k+1}$ to $3*2^k$ by induction hypothesis.  Also, we get that for $3*2^k$ to $2^{k+2}$ , we get an added factor of the $k+1$th prime by all these numbers, therefore we are done by induction.

$1995=3*5*7*19$.  Therefore, putting this in this form of induction, we get $(19)(17)(13)(11)(7)(5)(3)(2)_2$ (where we only include digits which are in the prime factorization).  Therefore, we get $10001110_2$.  This is $2^7+2^3+2^2+2=\boxed{142}$. [/hide]

This problem happened to be reused in the 2014 AIME II (https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_15)

Easy problem, but unfortunately I didn't get the binary representation idea :oops:. Anyway, here's my solution: Let $p_k$ denote the $k^{\text{th}}$ prime, and define the primorial $p_k^*=p_1 \times p_2 \times \dots \times p_k$. Note that if at any moment $x_{n-1}=p_{k-1}^*$, then $x_n=p_k$. The main claim is as follows:-

[b]CLAIM:[/b] If $x_n=p_k$ for some $k,n \in \mathbb{N}$, then $x_{n+2^{k-1}-1}=p_k^*$.

[u][i]Proof of Claim[/u][/i] We proceed by strong induction on $k \geq 1$. The result easily follows for $k=1,2$. Suppose the result is true for $1,2, \dots ,k-1$. We will show that it is true for $k$ also. Notice that a prime greater than $p_k$ is not introduced into the sequence $(x_i)$ as long as we don't reach $p_k^*$. So till that instant (if we actually do reach $p_k^*$), we have that $p_k$ remains intact in the sequence. We claim that we eventually reach $p_ip_k$ for some $1 \leq i \leq k-1$. We again induct (nested induction :D) on $i \geq 1$. Note that $x_{n+1}=2p_k=p_1p_k$, so the result is true for $i=1$. Suppose we reach $p_ip_k$, i.e. $x_m=p_ip_k$ for some $m>n$. Then after $2^{i-1}-1$ steps (from the induction hypothesis of the outer induction, which is applicable here since $i<k$), we will get to $p_i^*p_k$, or equivalently $x_{m+2^{i-1}-1}=p_i^*p_k$. Then, using the defining recursion of the sequence $(x_i)$, we have $x_{m+2^{i-1}}=p_{i+1}p_k$, as desired. By putting $i=k-1$ in the above result, we get that we'll eventually encounter $p_{k-1}^*p_k=p_k^*$. Also the number of steps in reaching $p_k^*$ from $p_k$ can now be easily calculated, as the sum of the number of steps it takes to reach $p_{i+1}p_k$ from $p_ip_k$ for all $1 \leq i \leq k-1$. This evaluates to $2^0+2^1+ \dots +2^{k-2}=2^{k-1}-1$. Thus, $x_{n+2^{k-1}-1}=p_k^*$, as required. $\Box$
------

Return to the problem at hand. Note that $1995=3 \times 5 \times 7 \times 19$, which means we have $p_k=19$ with $k=8$. Also, if $x_n=1995$, then $x_{n+1}=p_4^* p_k$ (since $p_4=7$). Thus, we will reach $1995$ in the sequence when first we'll encounter $p_i^*$ for all $1 \leq i \leq 7$, then encounter $p_j^*p_k$ for all $1 \leq j \leq 4$, and finally remove the last step of reaching $p_4^*p_k$. Thus, the number of steps required is $$(2^0+2^1+ \dots +2^6)+1+(2^0+2^1+2^2+2^3)-1=142 \Rightarrow x_{142}=1995$$
Here, we add an intermediate $1$ to account for the step where we reach $p_k$ in the sequence.

For storage.

Computing the first few terms, we have
\begin{align*}
x_0 &= 1 \\
x_1 &= 2^1\\
x_2 &= 2^0 \times 3^1.\\
x_3 &= 2^1 \times 3^1\\
x_4 &=2^0 \times 3^0 \times 5^1\\
x_5 &= 2^1 \times 3^0 \times 5^1 .\\
\end{align*}

We see that the exponents in  the value fo $x_n$ form the binary representation of $n.$ We can easily prove this with induction.

So, we have $1995 = 3 \times 5 \times 7 \times 19,$ which means $n$ is $\overline{10001110}_2 = 142.$

I claim that if $n$ has a binary representation of the form $b_xb_{x-1}\ldots b_0$, and $p_i$ is the i-th prime (with $p_0=2$).
\[x_n = \prod_{i=0}^{x} p_i^{b_i}\]

We proceed with induction. The base case is trivial.

Inductive step. Note that adding $1$ to $n$ will remove however many 1s there are on the right side of $n$, and replace  the right-most 0 with a 1.

Additionally, $x_{n+1}=\frac{x_np(x_n)}{q(x_n)}$ clearly amounts to the same thing, $\frac{1}{q(x_n)}$ removes the series of 1s at the end, and $p(x_n)$ replaces the rightmost 0 with a 1. Thus, these two processes are the same so our induction holds. $\blacksquare$

Thus, $x_n=1995 = 2^03^15^17^119^1 \Longrightarrow n =\overline{10001110} = 128+8+4+2=142$

ISL Sequences is a thing?

Note that the sequence $x_i$ is unique.\\

We claim that the sequence is: for $x_n$, convert $n$ into binary, and then read off the last digit as $v_2(x_n),$ the second to last digit as $v_3(x_n)$, and so on. We will then only need to show that this sequence works since there is only one possible sequence.\\

We will use induction. This sequence clearly works at $n=0$ and $n=1$. Now, examine the equation $$x_{n+1}=x_n\cdot \frac{p(x_n)}{q(x_n)}.$$ This is essentially saying that goin from $x_n$ to $x_{n+1}$, we gain 1 factor of the smallest prime that $x_n$ doesn't have, but then we lose our factors of all primes before that smallest prime. This is isomorphic to adding 1 in binary: the rightmost 0 digit becomes a 1, but everything after that becomes 0. Therefore, we have shown our claim by induction.\\

Since $1995=3\cdot 5\cdot 7\cdot 19$, our answer is $$10001110_2=\boxed{142}.$$\\

The idea is to classify the sequence $x_n$. Suppose $n$ has $k$ digits in binary; then, for each $1 \leq i \leq k-1$, the $i$th digit from the right corresponds to the power of the prime $p_i$ in the sequence $\{x_n\}$. This is easy to show inductively.

Thus the answer is $ 10001110_2 = \boxed{142}$.

Let $P_i$ be the $i+1$th prime number. We claim that $\sum_{i=0}^\infty 2^iv_{P_i}(x_n)=n$. This is easy to prove by induction.

Since $1995=3\cdot5\cdot7\cdot19$, our answer is $2+4+8+128=142$.

We list out all the primes from greatest least. We claim that the binary representation of the exponents is the value of $n$ needed to obtain that number.

For example since $1995 = 19^1 \cdot 7^1 \cdot 5^1 \cdot 3^1$ and $19$, $7$, $5$, and $3$ are the eighth, fourth, third, and second primes respectively. We see the this would give $2^7 + 2^3 + 2^2  + 2^1 = \boxed{142}$.

We prove the result by induction. The base case is trivial since $x_1 = 2$.

The inductive is obviously just adding $1$ in binary since we replace all of the consecutive ones and carry over the last $1$. $\blacksquare$

I solved this in like $5$ min because I had already done AIME 2014 problem 15 :D

Just realized something called Sequences existed back then.

Let $p_1,p_2,\cdots$ be the sequence $2,3,\cdots$ of the primes in increasing order. Now, fist we have
$$x_0=1, x_1=2, x_2=3, x_3=6$$
Then, assume that for $i=2^{k_1}+2^{k_2}+\cdots + 2^{k_j}$
$$x_i = p_{k_1+1}p_{k_2+1}\cdots p_{k_j+1}$$
for all $i \leq 2^n-1$
Then, $$x_{2^n}=\frac{x_{2^n-1}p(x_{2^n-1})}{q(x_{2^n-1})} = p_{n+1}$$
Then, say for some $1\leq k < 2^n-1$ we have $$x_{2^n+k}=x_kp_{n+1}$$
for all $1, \cdots, k$
Thus, we obtain
$$ x_{2^n+k+1} = \frac{x_{2^n+k}p(x_{2^n+k})}{q(x_{2^n+k})} $$
Now, since $x_k<x_{2^n-1}=p_1p_2\cdots p_n$. There exists a prime $p \in \{p_1,\cdots,p_n\}$ such that $p \nmid x_k$. Thus, $p(x_{2^n+k})=p(x_kp_{n+1})=p(x_k)$.
Thus,
\begin{align*}
    x_{2^n+k+1} &= \frac{(x_kp_{n+1})p(x_kp_{n+1})}{q(x_kp_{n+1})}\\
    &= \frac{(x_kp_{n+1})p(x_k)}{q(x_k)}\\
    &= p_{n+1}x_{k+1}
\end{align*}
as needed. Thus by the Principle of Mathematical Induction we have that for all $1\leq k < 2^n$. Then, combining this with the previous fact we can again apply the Principle of Mathematical Induction to obtain that  for all $n \in \mathbb{N}$ if $n=2^{k_1}+\cdots + 2^{k_j}$ then $x_n = p_{k_1+1}\cdots p_{k_j+1}$. Thus, we obtain that since $1995 = 3\times 5 \times 7 \times 19$ which gives us that if $x_i = 1995$
$$i = 2^1+2^2+2^3+2^7 = \boxed{142}$$
which is the required answer.

For every base $10$ positive integer $n$, let $n_2$ be the binary representation of $n$, and $S_n$ be the set of all $i$ such that the $i$th digit of $n_2$ is $1$.

[b][color=#f00]Claim:[/color][/b] For $n \ge 1$ we have
$$x_n = \prod_{i \in S_n}p_i, $$
where $p_i$ is the $i$th prime.

[b][color=#f00]Proof:[/color][/b] We use induction on $n$. For the base case of $n = 1$, just note that $x_1 = 2$. For the inductive step, assume that the assertion holds for some positive integer $n$. Then to obtain $(n + 1)_2$, we need to:
[list]
[*]find the rightmost digit in $n_2$ which is a $0$ (if $n_2$ only contains $1$s, take the digit place to the left of its leftmost $1$)
[*]replace said digit with a $1$, and replace all of the digits to its right with a $0$
[/list]
However, by the given formula, this is also exactly what happens to obtain $x_{n + 1}$ with the product on the RHS of the assertion. This proves the claim.

Thus, since $1995 = 3 \cdot 5 \cdot 7 \cdot 19$, and $3 = p_2, 5 = p_3, 7 = p_4, 19 = p_8$, we can extract the final answer of $n = 2^1 + 2^2 + 2^3 + 2^7 = \boxed{142}$.

hmmm i didnt phrase my solution in terms of binary

The answer is $n=142$ only. We prove a series of claims.

[b]Claim:[/b] $x_{n+1}=2x_n$ when $n$ is even. Furthermore, all even-indexed terms are odd, and all odd-indexed terms are even. That is, they alternate parity.

[i]Proof.[/i] First, observe for any odd term $x_i,$ we have $p(x_i)=2,$ so $q(x_i)=1.$ Thus $\tfrac{x_ip(x_i)}{q(x_i)}=2x_i.$ If $x_i$ is an even term, then $p(x_i)$ is odd, and the prime factorization of $q(x_i)$ contains $2.$ Then the $2$'s in the prime factorization of $x_i$ and $q(x_i)$ cancel and the remaining number is odd.

[b]Claim:[/b] There can be no divisors of multiplicity $>1$ in any $x_i.$

[i]Proof.[/i] At each step, $x_i$ is multiplied by $p(x_i),$ which by definition is the smallest prime not in the prime factorization of $x_i.$ Dividing by $q(x_i)$ only removes existing divisors.

[b]Claim:[/b] Primes appear in the sequence in ascending order. In particular, the $i$th prime appears when $n=2^{i-1}.$

[i]Proof.[/i] Induct with base case $x_1=2.$ For inductive step, let $p_i$ be the $i$th prime and assume $x_{2^{i-1}}=p_i.$ This consequently implies all past prime factorizations involving divisors $1,2, \dots, p_{i-1}$ have appeared before this point. We prove $x_{2^{i}}=p_{i+1}.$ The key observation is to scale the procedure $x_0 \to x_1 \to \dots \to x_{2^{i-1}-1}$ up by $p_i.$ Then, after scaling, $x_0=x_{2^{i}},$ so we may shift this procedure so the starting point is at $x_{2^{i-1}}.$ Then, the endpoint is at $x_{2^{i}-1}=2 \cdot 3 \cdots p_i,$ so $x_{2^i}=p_{i+1}.$

[b]Claim:[/b] Every $x_i$ is unique.

[i]Proof.[/i] Induct with primes. For the base case $x_1=2$ we see every term before it is unique. For inductive step, assume every term before $x_{2^{i-1}}=p_i$ is unique. Then from the same scaling and shifting as before we see every term up until $p_{i+1}$ is unique. Extending this infinitely gives the result.

Prime factorizing, $1995=3 \cdot 5 \cdot 7 \cdot 19.$ From our claims, we know $19$ appears at $x_{128}.$ It is easy to check from here that $1995$ is achieved at $x_{142}.$


Support
topology
advanced fields
advanced fields unsolved

Prove that every continious linear functional on $ C^{\infty}(X)$ is of the form $ f \to  uf$, where $ u$ is a distribution with compact support in $ X$.

What is $ X$? And what is the topology defined on $ C^{\infty}(X)$?

Sorry, i should have said that. Let $ X$ be an open subset of $ \mathbb{R}^{n}$. The topology on $ C^{\infty}$ is a locally convex topology, defined by the seminorms $ p_{N}= sup \{ | D^{\alpha}f(x)|, |\alpha| \leq N , x \in \bar{K_{n}}\}$, where $ K_{n}$ are precompact subsets of $ \mathbb{R}^{n}$, such that $ K_{n}\subset K_{n+1}$, and the union of all $ K_{n}$ is $ X$.


calculus
integration
logarithms
trigonometry
calculus computations

$ P \equal{}\int_{\log (\pi )}^{\alpha }\sqrt{\frac{e^{x}\plus{}\pi }{e^{x}\minus{}\pi }}\, dx$

$ x \equal{}\log\left(\pi\plus{}\frac{2\pi }{y^{2}\minus{}1}\right)$ , $ \beta \equal{}\sqrt{\frac{e^{\alpha }\plus{}\pi }{e^{\alpha }\minus{}\pi }}$

$ P \equal{}\int_{\infty }^{\beta }\minus{}\frac{4 y^{2}}{y^{4}\minus{}1}\, dy \equal{}\int_{\infty }^{\beta }\left(\frac{1}{y\plus{}1}\minus{}\frac{2}{y^{2}\plus{}1}\minus{}\frac{1}{y\minus{}1}\right)\, dy$

$ P \equal{}\minus{}2\tan^{\minus{}1}(\beta )\minus{}\log (\beta\minus{}1)\plus{}\log (\beta\plus{}1)\plus{}\pi$


Well, here's some thoughts on it:

A = (7^m + p*2^n) / (7^m - p*2^n)
(7^m - p*2^n)*A = (7^m + p*2^n)
7^m*(A - 1) = p*(A + 1)*2^n
n :ge: 1  so  A = 2B - 1 for some B integer.
7^m*(2B - 2) = p*(2B)*2^n
7^m*(B-1) = p*B*2^n
So 2^n | (B - 1),  B = C*2^n + 1
7^m = p*(C*2^n + 1)

p = 7

7^(m - 1) = C*2^n + 1

Plugging back in to A, we get
A = ((C + 1)*2^n + 1)/((C - 1)*2^n + 1)

I'm not sure where to proceed, and I also feel as if I've done something wrong.

[quote="grobber"]I have some problems from various Romanian selection tests:

P1

m,n are positive integers and p is a prime.
Prove that if A=(7^m+p*2^n)/(7^m-p*2^n) is an integer then it's a prime.[/quote]

A=1+ p*2^(n+1)/(7^m-p*2^n). So we want p*2^(n+1)/(7^m-p*2^n) to be an integer. hrmph. that means 7^m-p*2^n=p*2^k or 2^k, where k:le:n. Alrighty, in the first case, we have 7^m=p(2^k+2^n). the LHS will always be odd, the RHS will be odd if k=0. p must be 7. that makes 7^(m-1)=1+2^n. However, this will not be true, since if you look at it mod 6, that makes 2^n a multiple of 6, which is not likely to happen. The other case gives 7^m=2^k+p*2^n. Since the LHS is odd, 2^k must be odd (if 2^k were even, so would p*2^n, which is a contradiction). Now, we have k=0. Either n=0 and p=2, or n:ge:1 and p is something else. In the first, case, we have 7^m=3. Okay. not likely. In the second case, we have 7^m-1=p*2^n. Again, we look mod 6. p must be 3. Alright, we have 7^m-1=6*2^(n-1). um...more to follow.

Yeah, I think this one isn't too tough, it's just basic case consideration... That's the main idea. I hope you enjoyed it :D


I posted this problem last year but only posted my own solution,actually no one replied :D , now lets discuss about it maybe we can find some more nice solutions friends... :)

Let $ ABC$ be an acute-angled triangle such that $ AC<BC$ and $ \angle C=60^{\circ}$. $ D$ is a point on the side $ AC$( near to $ A$) such that $ DC=BC$ and $ M$ is the midpoint of side $ BC$. $ C_{1}$ and $ C_{2}$ with centers $ O_{1}$ and $ O_{2}$ respectively are the circumcircles of the triangles $ ABC$ and $ ADM$ respectively. $ C_{1}$ and $ C_{2}$ intersect at a point $ P$(other than $ A$). Prove that:

$ a) S_{ABC}= \frac{\sqrt{3}}{2}BC^{2}-4S_{ADM}$

$ b) (AM-O_{1}M)(O_{1}P+O_{1}O_{2}) < AB^{2}+AC^{2}$

[quote="sinajackson"][color=darkred]Let $ ABC$ be an acute triangle such that $ b<a$ and $ C=60^{\circ}$. Let $ D\in (CA$ such that $ CD=a$ and let $ M$ be the midpoint of the side $ [BC]$.

Denote the circumcircles $ C=C(O,R)$ and $ C_{1}=C(O_{1},R_{1})$ of the triangles $ ABC$ and $ ADM$ respectively.

Prove that: $ \begin{array}{cc}1\blacktriangleright & \cdot [ABC]= \frac{a^{2}\sqrt 3}{4}-2\cdot [ADM]\\\\ 2\blacktriangleright & (AM-OM)(R+OO_{1}) < b^{2}+c^{2}\end{array}$.[/color][/quote]
[color=darkblue][b]Proof.[/b] $ C=60^{\circ}\implies h_{b}=\frac{a\sqrt 3}{2}$. Therefore, $ \boxed{2\cdot [ABC]+4\cdot [ADM]}=bh_{b}+4\cdot \frac{(a-b)h_{b}}{4}=ah_{b}=\boxed{\frac{a^{2}\sqrt 3}{2}}$.[/color]
I"ll come back here soon ! The first point was wrong. See above and compare with the initial enunciation.


The incircle $o$ of a triangle $ABC$ is tangent to the sides $AB$ and $BC$ at $P$ and $Q$ respectively. The angle bisector at $A$ meets $PQ$ at point $S$. Prove $\angle ASC = 90^{\cdot}$.

See the extended problem in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=187743#p187743]http://www.mathlinks.ro/Forum/viewtopic.php?t=23677 post #7[/url].

  darij


Putnam
function
linear algebra
linear algebra unsolved

Let $ \sigma_n$ denote the set of all permutations of numbers $ 1,2,3,..,n$ .For  $ \sigma \in \sigma_n$
Put $ v(\sigma)$ denote the number of $ \sigma$ if $ \sigma(i) \equal{} i$
$ p(\sigma)$ denote the number of $ \sigma$ if  $ \sigma(i) < i$
$ q(\sigma)$ denote the number of  $ \sigma$  if $ \sigma(i) > i$
$ I(\sigma)$ denote the number of  $ \sigma$ if $ i < j$ then $ \sigma(i) > \sigma(j)$
For all $ (i,j) \in \{1,2,3,..,n\}^2$
 Calculate :
1) $ \sum_{\sigma \in \sigma_n}{\frac {( \minus{} 1)^{I(\sigma)}}{(v(\sigma) \plus{} q(\sigma) \plus{} 1)(p(\sigma) \plus{} 1)}}$
[hide="Answer"]$ \frac {n \plus{} 1 \plus{} ( \minus{} 1)^{n \plus{} 1}}{n(n \plus{} 1)(n \plus{} 2)}$[/hide]
2) $ \sum_{\sigma \in \sigma_n}{\frac {( \minus{} 1)^{I(\sigma)}}{(v(\sigma) \plus{} 1)(q(\sigma) \plus{} 1)(p(\sigma) \plus{} 1)}}$
[hide="Answer"]  :play_ball: ... It is very nice [/hide]

For $ 2$) It is don't easy in caculate . But answer of it is nice ! I hope you are interested it .

[color=red]I will post my solution If you post  your solution [b](wait you)[/b]  Thanks you very much if  you are interested in to it .....[/color]
[color=red]Add short problem for nice  problem  of putnam |Sorry by I think many about problem it | :
  1)$ \sum_{\sigma \in \sigma_n}{( \minus{} 1)^{I(\sigma)}\frac {(n \minus{} v(\sigma))!}{(v(\sigma)^{n \minus{} v(\sigma)}}} \equal{} ?$
  2)$ \sum_{\sigma \in \sigma_n}{( \minus{} 1)^{I(\sigma)}v(\sigma)!(n \minus{} v(\sigma))!} \equal{} ?$[/color]
[hide="Answer"][color=red]I learn to take one point are too hard [/color]....... ..:play_ball:..............Thanks for help .... [/hide]

[quote="QuyBac"]
  1)$ \sum_{\sigma \in \sigma_n}{( - 1)^{I(\sigma)}\frac {(n - v(\sigma))!}{(v(\sigma)^{n - v(\sigma)}}} = ?$
 [hide="Answer"][/hide][/quote]
 I don't remember this stupid  :blush:
 [quote="QuyBac"]2)$ \sum_{\sigma \in \sigma_n}{( - 1)^{I(\sigma)}v(\sigma)!(n - v(\sigma))!} = ?$[/quote]
[hide="Answer"] Simple but nice use function $ B,\gamma$ , [/hide]


If $ x$ and $ y$ are numbers such that $ (x \plus{} 3)(y\minus{}5) \equal{} 0$, what is the smallest possible value of $ x^2\plus{} y^2$?

Let $ x$ equal $ \minus{}3$ and let $ y$ equal $ 0$.  Thus, we have $ 3^2 \plus{} 0^2$, or $ \boxed{9}$.

wow haha, that is so obvious, i can't believe I missed it. Thanks!  :lol:

What...you guys are supposed to post QUALITY solutions and HIDE your solutions.
[hide="Solution 9967"]In order for $(x+3)(y-5)=0$, either $x+3$ or $y-5$ has to be 0. (Can you see why?) This means that either $x=-3$ so that $(x+3)$ can equal 0, or $y=5$ so that $(y-5)$ can equal 0. Since we want $x^2+y^2$ to be as small as possible, the other variable that is not assigned should be 0. (Can you see why?) So, if we use $x=-3$ and $y=0$, $x^2+y^2=9$. If we use $y=5$ and $x=0$, then $x^2+y^2=25$. Clearly, 9 is smaller than 25, so the answer that we want is $9$.[/hide]


function
limit
real analysis
real analysis unsolved

Let $F$ be a family of equicontinuous functions and $F'$ be the same family with all pointwise limits added in i.e. $f \in F'$ if there is a sequence ${f_n}$ in $F$ such that $f = \lim f_n$. Then $F'$ is equicontinuous.

how does one attack this problem?

Google "Arzela Ascoli" finds:

http://at.yorku.ca/p/a/c/a/23.htm

which has proof(s) of what you are looking for.


let x;y;z be three positive real numbers . Prove that :
$\frac{x}{2x^{2}+yz}+\frac{y}{2y^{2}+xz}+\frac{z}{2z^{2}+xy}\geq \frac{9xyz}{(xy+yz+xz)^{2}}$

let S = LHS!

by Cauchy we have $\sum{yz(2x^{2}+yz)}. S \geq (3\sqrt{xyz})^{2}= 9xyz$

but $\sum{yz(2x^{2}+yz)}= (xy+yz+zx)^{2}$  :)


$ a,b,c,d > 0,$,prove that
$ 0\leq a\plus{}b\plus{}c\plus{}d\minus{}3\,{\frac {ab}{a\plus{}b\plus{}c}}\minus{}3\,{\frac {bc}{b\plus{}c\plus{}d}}\minus{}3\,{\frac
{cd}{c\plus{}d\plus{}a}}\minus{}3\,{\frac {da}{d\plus{}a\plus{}b}}$

$ a,b,c,d > 0,$,prove that
$ a \plus{} b \plus{} c \plus{} d \minus{} \frac{3}{4}\,{\frac {\left( a \plus{} b \right) ^{2}}{a \plus{} b \plus{} c}} \minus{} \frac{3}{4}\,{\frac { \left( b \plus{} c \right) ^{2}}{b \plus{} c \plus{} d}} \minus{} \frac{3}{4}\,{\frac {\left( c \plus{} d \right) ^{2 }}{c \plus{} d \plus{} a}} \minus{} \frac{3}{4}\,{\frac {\left( d \plus{} a \right) ^{2}}{d \plus{} a \plus{} b}}\leq 0.$

[quote="xzlbq"]$ a,b,c,d > 0,$,prove that
$ a \plus{} b \plus{} c \plus{} d \minus{} \frac {3}{4}\,{\frac {\left( a \plus{} b \right) ^{2}}{a \plus{} b \plus{} c}} \minus{} \frac {3}{4}\,{\frac {\left( b \plus{} c \right) ^{2}}{b \plus{} c \plus{} d}} \minus{} \frac {3}{4}\,{\frac {\left( c \plus{} d \right) ^{2 }}{c \plus{} d \plus{} a}} \minus{} \frac {3}{4}\,{\frac {\left( d \plus{} a \right) ^{2}}{d \plus{} a \plus{} b}}\leq 0.$[/quote]
Just C-S

$ \sum \frac {3}{4}\,{\frac {\left( a \plus{} b \right) ^{2}}{a \plus{} b \plus{} c}}\ge \frac {3}{4}\,{\frac {(2\sum a) ^{2}}{3\sum a}} \equal{} \sum a$ :)

[quote="xzlbq"]$ a,b,c,d > 0,$,prove that
$ 0\leq a \plus{} b \plus{} c \plus{} d \minus{} 3\,{\frac {ab}{a \plus{} b \plus{} c}} \minus{} 3\,{\frac {bc}{b \plus{} c \plus{} d}} \minus{} 3\,{\frac {cd}{c \plus{} d \plus{} a}} \minus{} 3\,{\frac {da}{d \plus{} a \plus{} b}}$[/quote]
Actually, the following more general inequality holds:

Let $ a,$ $ b,$ $ c,$ $ d$ be positive real numbers. If $ k \ge \frac{2}{7},$ then
$ \frac{ab}{a\plus{}kb\plus{}c}\plus{}\frac{bc}{b\plus{}kc\plus{}d}\plus{}\frac{cd}{c\plus{}kd\plus{}a}\plus{}\frac{da}{d\plus{}ka\plus{}b} \le \frac{a\plus{}b\plus{}c\plus{}d}{k\plus{}2}.$

There is a nice proof by Cauchy-Schwarz for this result.

[quote="can_hang2007"][quote="xzlbq"]$ a,b,c,d > 0,$,prove that
$ 0\leq a \plus{} b \plus{} c \plus{} d \minus{} 3\,{\frac {ab}{a \plus{} b \plus{} c}} \minus{} 3\,{\frac {bc}{b \plus{} c \plus{} d}} \minus{} 3\,{\frac {cd}{c \plus{} d \plus{} a}} \minus{} 3\,{\frac {da}{d \plus{} a \plus{} b}}$[/quote]
Actually, the following more general inequality holds:

Let $ a,$ $ b,$ $ c,$ $ d$ be positive real numbers. If $ k \ge \frac {2}{7},$ then
$ \frac {ab}{a \plus{} kb \plus{} c} \plus{} \frac {bc}{b \plus{} kc \plus{} d} \plus{} \frac {cd}{c \plus{} kd \plus{} a} \plus{} \frac {da}{d \plus{} ka \plus{} b} \le \frac {a \plus{} b \plus{} c \plus{} d}{k \plus{} 2}.$

There is a nice proof by Cauchy-Schwarz for this result.[/quote]
See my proof here: http://can-hang2007.blogspot.com/2010/01/inequality-22-v-q-b-can.html


Let P be the smaller of m-th root of n and the n-th root of m

Find the maximum value of P

Sorry,I forgot to mention that the numbers m and n are positive integers

hmm well no proof here but i just checked manually....

m=n=3 ..... 1.4422


modular arithmetic
algebra
binomial theorem
number theory

1.) If $ p$ is prime, show that $ (a \plus{} b)^p \equiv a^p \plus{} b^p (mod p)$.

2.) An integer p is prime if and only if $ 2(p\minus{}3)!\plus{}1 \equiv 0 (mod p)$

[hide="2 Not Rigorous, Maybe an outline"]
For prime $ p$ and $ p > 2$ that equivalence is $ 2 \left( p \minus{} 3 \right) ! \equiv \minus{} 1 \bmod p$.
Then $ \frac {p \minus{} 1}{2} \in \mathbb{Z}$ and is included in the fractoral.
The conclusion immediately follows.
[/hide]

AstroPhys: As I understand it, your reasoning is flawed.

Number 1
[hide]
Expand with binomial theorem: since $ \binom{p}{k} \equal{} \frac {p!}{k!(p \minus{} k)!}$ is always an integer, and for $ 2\leq k\leq p \minus{} 1$, it always contain a prime factor $ p$ which won't be cancelled out, all these coefficients will be divisible by $ p$. The result thus follows.
[/hide]

Number 2
[hide]
If $ p$ is a prime greater than two, then by Wilson's Theorem, $ (p \minus{} 1)!\equiv \minus{} 1 \pmod p$, so
$ 2(p \minus{} 3)! \plus{} 1\equiv 0 \pmod p,$
$ \iff 2(p \minus{} 3)! \minus{} (p \minus{} 1)!\equiv 0 \pmod p,$
$ \iff p(p \minus{} 3)(p \minus{} 3)!\equiv 0\pmod p.$

If $ p$ is not a prime, then suppose $ a$ is a factor of it (neither itself nor one). Then $ a\leq p \minus{} 3$ for $ p > 2$. Hence $ a|2(p \minus{} 3)!$, which means $ 2(p \minus{} 3)! \plus{} 1$ is not divisible by $ a$. Since divisibility of $ a$ is a necessary condition for divisibility of $ p$, we've proved the converse of the proposition.
[/hide]


if
$xy+yz+zx =3$and $x,y,z\geq 0$
how to prove (show) that ;
$xy(x + y) + yz(y + z) + zx(z + x) = 9 - 3xyz$

$xy+yz+zx=3,\ (x+y+z)^2\ge 3(xy+yz+zx)\Longrightarrow x+y+z\ge 3.$

$\sum xy(x+y)\ge 9-3xyz\Longleftrightarrow \sum xy\cdot \sum x\ge 9\Longleftrightarrow x+y+z\ge 3.$

was the problem asking us to show $xy(x+y) + ... = 9 - 3xyz$, or was it to show that $xy(x+y) + ... \ge 9 - 3xyz$?

cuz if it was the former, then i think ur proof is incomplete Ph-An

of course, my misteake $\geq$ - should be there

So, after the cheater finally received "his" solution, I think it would be just unfair to delete this topic - at least the others should not be disadvantaged compared to him...

[b]PS.[/b] The competition ends on Dec 5. This time I checked twice...

  darij

ummm
what competition are you talking about?

Polish MO (he changed a bit a problem, but it's almost the same)

.... if he puts the British flag, it doesn't mean he comes from Britain...

Originally, he had the Polish flag though, then he changed it.. (check his previous posts if you want)

Here is a post by a member called "death" on the Polish MO inequality. It was deleted because it was posted before the deadline of the competition, but now can be reposted since the competition is over. Note that his solution shows how the inequality posted in this topic is related to the Polish MO inequality.

[color=blue][b]Problem 10 from 57th Polish MO 2005, 1st round, 3th series.[/b] Let $a$, $b$, $c$ be positive real numbers such that $ab+bc+ac=3$. Prove that

$a^3+b^3+c^3+6abc\geq 9$.[/color]

[quote="death"]From Schur's Inequality we have

$a^3+b^3+c^3+3abc \geq a^2b+a^2c+ab^2+b^2c+ac^2+bc^2$.

Add $3abc$ to both sides yield

$a^3+b^3+c^3+6abc \geq a^2b+a^2c+ab^2+b^2c+ac^2+bc^2+3abc$
$= \left(a+b+c\right)\left(ab+ac+bc\right) = 3\left(a+b+c\right)$.

From maclaurin's inequality we have

$\left(a+b+c\right)^2 \geq 3\left(ab+ac+bc\right) = 9$.

So

$a+b+c \geq 3$.

Therefore

$a^3+b^3+c^3+6abc \geq 9$.

And we're done![/quote]


$ a,b,c >0$ and $ a\plus{}b\plus{}c\equal{}3$ Prove that !
$ (ab\plus{}bc\plus{}ac)^6 \le 3^6abc$

[quote="bigbang195"]$ a,b,c > 0$ and $ a \plus{} b \plus{} c \equal{} 3$ Prove that !
$ (ab \plus{} bc \plus{} ac)^6 \le 3^6abc$[/quote]

$ a \to 0$  :wink:


I have a hard probelm :
Find all prime p  for which the sequence $\{\mathrm{C}^n_{2n} \mod p\}_{n=1}^{\infty}$ is periodic

The sequence $\{C^n_{2n}\}$ contains a $k$ consecutive terms divisible by $p$ for all $k$. Implies $p=2$

Sory..The initial relation,here:$p$ is positive integer

To truong:Could you tell more details???I still misunderstood....

First let $p$ be a odd prime.
Then $p$ divides $\binom{2n}{n}$ for $n<p^k<2n$ and doesn't divide it for $n=p^k$ (just calculate).
The same happens for $p=4$ (more exactly, in this case it's divisble by $4$ iff $n$ is a power of two).
So there are arbitrary long sequences of zero's followed by something nonzero in the sequence, thus will never get periodic.

Now if $p$ is some integer. When it would get periodic $\mod p$, it would also get periodic modulo any divisor of $p$, so it's not possible when $p$ has an odd divisor $\neq 1$ or is divisible by $4$. So $p=1$ or $=2$, and it's simply true for both.


How to make $\frac{\pi}{12}$ and $\frac{\pi}3$ in the following fraction bigger?
$\frac{\left(\cos\frac{\pi}{12}+i\sin\frac{\pi}{12}\right)^2}{\cos\frac{\pi}3-i\sin\frac{\pi}3}$

Use \displaystyle:

[code]\[\frac{\displaystyle \left(\cos\frac{\pi}{12}+i\sin\frac{\pi}{12}\right)^2}{\displaystyle \cos\frac{\pi}3-i\sin\frac{\pi}3}\][/code]

\[ \frac{\displaystyle \left(\cos\frac{\pi}{12}+i\sin\frac{\pi}{12}\right)^2}{\displaystyle \cos\frac{\pi}3-i\sin\frac{\pi}3} \]

also, with the amsmath package, you can use \dfrac which do the same as \displaystyle\frac... but more convenient.


logarithms
parameterization
probability and stats

Really cool problem I heard from my friend Ori G: Let $ X, Y, Z$ be independent random variables uniformly distributed on the unit interval $ [0,1]$. Prove that $ (XY)^Z$ is also uniformly distributed (yes, I didn't believe it either).

This can be solved in several ways, some of which may be more illuminating than others. I don't know a really transparent argument that makes this "obvious".

This problem is not hard.
The main idea is to  to take a logarithm and proove that $ Z (lnX\plus{}lnY)$ is Exponentiallu distributed.
We know that they are independent. So we can calculate the disdtribution $ lnX\plus{}lnY$ as  the sum of 2 independent Exp and that $ Z\cdot(...)$ using standard technic of transformation.

There is a very short and transparant argument without any calculation, provided you know
- A RV X is uniform on $ (0,1)$ iff $ \minus{} \log X$ is exponential with parameter $ 1$,
- certain properties of the Poisson point process.

The argument goes as follows:
It is enough to show that $ \minus{}Z(\log X \plus{} \log Y)$ is an exponential RV with parameter 1, and  we are given that $ \minus{}\log X$ and $ \minus{}\log Y$ are independent exponential RVs with parameter 1. (This is what Tarasenya pointed out.) In a Poisson point process  with parameter 1 the waiting times between events are independent exponential RVs with parameter 1, so $ \minus{}(\log X \plus{} \log Y)$ can be interpreted to be the waiting time until the second event. It is also well known that the first event is uniformly distributed on $ (0,t)$ if you are given that the second event is at time $ t$. (More generally given that the time of the $ n$-th event is $ t$, the preceding $ n \minus{} 1$ events are uniformly distributed on $ (0,t)$ conditioned on increasing order.) Since $ \minus{}Z(\log X \plus{} \log Y)$ is uniformly distributed on $ (0,t)$ given that $ \minus{}(\log X \plus{} \log Y) \equal{} t$, it follows that $ \minus{}Z(\log X \plus{} \log Y)$ has the same distribution as the first event in the Poisson point process, i.e. it is an exponential RV with parameter 1.


[quote="mestav"]$ [5,1,0] \plus{} [2,2,2]\geq 2[3,3,0]$[/quote]
It's obviously wrong.


Gauss
advanced fields
advanced fields theorems

Who know where I can download the book  <Disquisitiones Arithmeticae>(in English written by Gauss).

I have got a latin one, but I can't understand it, so i need an english one

could anyone give a link for me to download

Don't triple-post!
Let's continue in http://www.mathlinks.ro/Forum/viewtopic.php?t=113749 .


real analysis
limit
function
integration
calculus
topology
real analysis unsolved

Suppose $ E$ is a given set, and $ O_n$ is the open set:
 $ O_n$={$ x: d(x,E)<1/n$}
Show:
(1) If $ E$ is compact, then $ m(E)$=$ lim_{n \to \infty} m(O_{n})$.
(2) However, the conclusion in (1) may be false for $ E$ closed and unbounded; or $ E$ open and bounded.

closed and unbounded:
$ \cup_{n\equal{}1}^\infty \cup_{k\equal{}\minus{}\infty}^\infty k2^{\minus{}n}$

(1) Let $ E \equal{} \mathbb{Z}$. then $ m(E) \equal{} 0 \neq \infty \equal{} \lim_{n\to\infty} m(O_n)$.
(2) Let $ \mathbb{Q} \cap [0, 1] \equal{} \{r_1, r_2, \cdots \}$, $ \epsilon$ be a sufficiently small positive number and $ E \equal{} \bigcup_{n\equal{}1}^{\infty} (r_n \minus{} \tfrac{\epsilon}{2^{n}}, r_n \plus{} \tfrac{\epsilon}{2^{n}})$. Then $ E$ is an open bounded set with measure not exceeding $ \epsilon$, but $ [0, 1] \subset O_n$ shows that $ m(O_n) \geq 1$. Similar construction gives an open unbounded counterexample.

[quote="Bradley"]Suppose $ E$ is a given set, and $ O_n$ is the open set:
 $ O_n$={$ x: d(x,E) < 1/n$}
Show:
(1) If $ E$ is compact, then $ m(E)$=$ lim_{n \to \infty} m(O_{n})$.[/quote]
Use the Monotone Convergence Theorem. If $ \chi_n$ is the characteristic function of $ O_n$, and $ g$ is the characteristic function of $ E$, then $ m(O_n) \equal{} \int_{O_1}\chi_n\to\int_{O_1}g \equal{} m(E_n)$. Note that this argument requires $ E$ to be bounded (so that the integrals are finite) and closed (so that $ \chi_n$ converges pointwise to $ g$).


The activation energy of a certain reaction is 65.7 kJ/mol. How many times faster will the reaction occur at 50 degrees Celsius than at 0 degrees Celsius.

directly use the arrenius equation this is very simple question

${ \{K}{=}{A}{e^{-\frac{E_a}{RT}}}$

and substitue the values and get the ratios ok

i think you can simply just say 2^5 or 32x fasterat 50 degrees since every 10 degrees the rate doubles

That's a pretty rough approximation as far as I remember. Applying the Arrhenius Equation is what I had do.


Well, I have a problem solving problem.  Well here it is.  There are 50 cards and a person lays them face up.  Then he goes and starts from teh second card and flips every second card.  He then starts at the third card and flips every third card.  Then fifth, sixth, seventh,...50th.  How many cards will be face up at the end?

Masoud Zargar

Where would I see that? Could you please give me a link?

Masoud Zargar

EDIT: I found it on Kalva.

Now how would that help me solve this problem?....please help:)

Okay, my answer is $7$.  If this is correct, then I'll justify my answer. :)

Masoud Zargar

Hint: [hide] A  number has an even number of divisors, unless, of couse, it is a perfect square.  (Number of times  the  $n^{th}$ card card flips is equal to  the nunber of divisors of $n$ [/hide]

Yes, that's what I got.  Thank you.

Masoud Zargar


We are given:

A train travels $ 20$ feet in $ \frac{1}{5}$ second.

Assuming that the train is not accelerating, the $ \frac{\text{feet}}{\text{seconds}}$ will remain constant.  In this case, it is $ \frac{20}{\frac{1}{5}}\equal{}100$.

Because it travels for three seconds, we substitute:

$ \frac{\text{feet}}{3}\equal{}100 \implies \text{feet}\equal{}\boxed{300}$.

Using this formula, S=D/T

First of all we substitute our values into the formula S=D/T

Therfore, S=20/(1/5)
We can convert 1/5 to 0.2 if this is easier for you, (do this by dividing 1 by 5)

So therfore S=100

Now we have our S value

Substitute our newly found S value into the formula
100=x/3

Our x term is the unknown term, where 3 (given) is our time and 100 is our distance.
Therfore we times 3 by both sides, hence we understand our speed is 300 m/s.

isn't your solution the exact same as AIME15's... :(

OR

we could use ratios.
$ \frac{3\text{ sec}}{\frac{1}{5}\text{ sec}}\equal{}\frac{n\text{ meters}}{20\text{ meters}}\equal{}\implies15\equal{}\frac{n}{20}$. We finish up by multiplying both sides by $ 20$ to get $ n\equal{}\boxed{300}$, which is the same answer as the previous posters arrived at.

$ \frac{3}{\frac{1}{5}}\times20\equal{}\frac{3\times5}{1}\times20\equal{}15\times20\equal{}300$ feet

It is the simple solution :lol:

Seriously stop reviving old posts (2 months)
only do this if
1) answer is wrong
2) different solution
RoFlLoLcOpT doesn't seem like a person to get something wrong.

[quote="ballfot"]a train travels 20 feet in 1/5 second At this same speed, how many feet will it travel in
    three seconds? ___________ please explain how did you arrived and the answer.       thanks.[/quote]
20 x 15=300   300 ft per sec.     :rotfl:

Mods PLEASE lock.
Check the first date FEBRUARY.


geometry
perimeter
MATHCOUNTS
calculus
integration
arithmetic sequence
absolute value

[hide="Solution to #1 (not very rigorous)"]
There is DEFINITELY a better way to do this, but I think this is the fastest. Lets start with the formula:

Area of Triangle= (radius of inscribed circle) * semiperimeter of triangle

(A proof of this is given in Aops Vol 1. )

we know that the semiperimeter is 120/2 or 60 and the radius is 9.
Therefore the area of triangle is 60*9 or 540.
The triangle must be right so lets consider some base triples.

(off the top of my head)

3-4-5
8-15-17

since the perimeter of the first 3-4-5 is 12, we need to multiply each side by 10 for a perimeter of 120. That means that the legs are 30 and 40. But because the area is 30*40/2 or 600, we know this doesn't work. Using similar logic for 8-15-17, our sides would be 24-45-51. Note that 12*45= 540. Therefore the hypotenuse is 51.
[/hide]

Ummmm...

Is that formula one most Mathcounts people should know?

Is semiperimeter just the perimeter divided by two?

 :huh:

Yes, $ s\equal{}\frac{a\plus{}b\plus{}c}{2}$, where $ a$, $ b$, and $ c$ are the sides of the triangle.

[quote="TheCoffeeMonster"]I need help with a few questions...  :blush: 2 are (I think) geometry problems, and 2 are sequence problems.

1) A right triangle with integer length sides has a perimeter of 120 units and an inscribed circle with a radius of 9 units. What is the length of its hypotenuse?
[hide]Is this what it looks like?
 [geogebra]247041164d5f5679029216dc5e0e1254dfe23fce[/geogebra] [/hide]

2) Two members of an increasing arithmetic sequence are 1 and 27. The sequence has less than 40 terms and its median is 38. What is the value of the fourth term following 27?

3) Two circles with radii of 2 and 4 units are externally tangent to each other. The line containing their centers intersects a common external tangent line at point X, which is not on either of the circles. What is the shortest distance from X to the nearest circle?
[hide]Is this kind of what it looks like?
 [geogebra]a372ddb4228f166d74fab87d0b6a2cabadb2be20[/geogebra][/hide]

4) A sequence of positive integers begins 2009, 2008, 1, 2007, etc. Beginning with the third term, each term is the absolute value of the difference of the previous two terms. What is the 2009th term in the sequence?[/quote]

Um. For #2 do the numbers in the sequence have to be integers?

4.
2009,2008,1,2007,2006,1,2005,2004,1

Notice a pattern?

2009/3= 669 R 2. So $ 669\cdot 2\equal{}1338$ and we have $ 2009\minus{}1338\equal{}671$ then two remainder would get 1... I think. Someone please check my work.

[quote="aleph0"]In #4, term n is 1 if n is divisible by three and 2010-(n-int(n/3)) otherwise. Therefore term 2009 is 2010-(2009-int(2009/3))=2010-1340=670.[/quote]

Where did the 2010 come from?

[quote="TheCoffeeMonster"]I need help with a few questions...  :blush: 2 are (I think) geometry problems, and 2 are sequence problems.

1) A right triangle with integer length sides has a perimeter of 120 units and an inscribed circle with a radius of 9 units. What is the length of its hypotenuse?
[hide]Is this what it looks like?
 [geogebra]247041164d5f5679029216dc5e0e1254dfe23fce[/geogebra] [/hide]

2) Two members of an increasing arithmetic sequence are 1 and 27. The sequence has less than 40 terms and its median is 38. What is the value of the fourth term following 27?

3) Two circles with radii of 2 and 4 units are externally tangent to each other. The line containing their centers intersects a common external tangent line at point X, which is not on either of the circles. What is the shortest distance from X to the nearest circle?
[hide]Is this kind of what it looks like?
 [geogebra]a372ddb4228f166d74fab87d0b6a2cabadb2be20[/geogebra][/hide]

4) A sequence of positive integers begins 2009, 2008, 1, 2007, etc. Beginning with the third term, each term is the absolute value of the difference of the previous two terms. What is the 2009th term in the sequence?[/quote]

#3
[hide][img]http://i202.photobucket.com/albums/aa68/MBarracade/Project1.gif[/img]The two triangles are similar because they are right triangles and share another angle. So $ \frac{2}{4} \equal{} \frac{2\plus{}y}{8\plus{}y}$.
Now solve.[/hide]

For number 2, it's a similarity between triangles

x/2 = (x+6)/4 >>> x = 6

[quote="TheCoffeeMonster"]Oh, I see. So how did Dojo get 671?  :lol:[/quote]

Cuz im => :fool:

Basically I'm all brawn, no brain. :P

Okay, for number 2...

Is 1 the first term in the sequence?


we toss a coin, and the probability of Heads is p.
let Ak be the event that k consecutive heads appear between the toss 2^k and 2^(k+1) -1
show that P(limsup Ak) equals 0 or 1 if p<1/2 or >=1/2. and how does this result interesses you?

my approach is: cause (Ak)k are independent, P(limsup Ak) = 1 no matter what p is. Is that right?

The notation is a little odd, so I'll reinterpret. We have a sequence of independent events $ A_k$ with probability $ p_k$, and we want to calculate $ \lim_{n\to\infty} P\left(\bigcup_{k\equal{}n}^{\infty}A_k\right)\equal{}\lim_{n\to\infty} 1\minus{}\prod_{k\equal{}n}^{\infty}(1\minus{}p_k)$.
This limit is zero if $ \prod_k (1\minus{}p_k)$ converges to a nonzero value, and $ 1$ if that product diverges to zero.

Now, estimate $ p_k$.


I live in New Haven county, I know we have a large presence of people in Fairfield county on these boards. I want to see what the geographic distribution of AoPS CTers is like.

Hi Ignite pleased to meet you, i also live in the New Haven couty along with you.. :D  anway glad to be the first voter and replier.

Timecat8 :D

Does Connecticut only have 8 counties?  Wow.  SC is a small state, and we still have 40-something counties.  I wonder how may California has.

Hartford country.

MC chapter '05.
MC states '05.
MC nationals '05.

whee. Interestingly, I got like 7th at chapter and 4th at state.

i was at the mc states 05

yea r team was the one that sucked ;)

[quote="Timecat8"]it seems as though most of the people in connectiuct live in the New haven county... also becuase people didn't vote..  :wink:[/quote]
wow, this poll is looking really lopsided.......mostly because we only have 4 votes...

yeah... i know i mean i can't understand why people wont vote it's the easiest vote ever... you just click to show what county you live it...  :roll:

Chapters '05 (6th)
States '05 (8th)

Chapters '06 (2nd)
States '06 (3rd)
Nationals '06 (56th)

ummmmmmmmmmmmmmmmmmmmmm..............there's only 5 ct counties
-jorian

i thought we only had 5 districts
-jorian

We have 5 districts for representation for House of Representatives but we have 8 counties.

o, i thought u were talking about that
sry,
-jorian


calculus
derivative
geometry
similar triangles
calculus computations

[quote="cdon1490"]Water is being poured into a conical reservoir at the rate of 20 cubic feet per second. The reservoir has a radius of 10 feet across the top and a height of 24 feet. At what rate is the depth of the water increasing when the depth is 16 feet?[/quote]

[hide="Hint"]$ V \equal{} \frac \pi3r^2h$, use similar triangles to find $ r$ in terms of $ h$. Substitute, take the derivative, and then substitute the given values.[/hide]
We're not here to do your homework though ... at least post what part of the problem you're having trouble with. Also, I don't think this belongs to Intermediate Topics.


As I explore LaTeX, I always compare it to Word (or, OpenOffice Writer) and one particularly useful feature of it is the dropdown-list of the available fonts.

In LaTeX however, controling the font of my text is a great mystery to me, as I don't even know which fonts are available on my system, let alone how they are named. The only thing I know so far is that "Computer Modern Sans Serif" is called"cmss" (or so) for a \usefont command, and that pdftex spits out some "FontWarnings" when I try to use it. It just seems like one hell of a mess of tfm-files, mapping tables etc. to me.

Any advice what I can do to quickly find a specific font and its name? (as an example, I want to typeset the titlepage of my document in the font that is called "monotype corsiva" in Word)

There are some hints in the Latex Companion, but that's just a small selection of some standard fonts.
Is there a way of getting an overview of the currently installed fonts on my computer, some kind of table?

Any help would be appreciated.

Are you running MiKTeX?  If so, you can start up the MiKTeX Options program, then click on "Packages", then click on "Fonts" to see which fonts are installed.  And if you want a font that's not installed, I think you can install it from there too.

Uhm, no, I'm running TexLive, from the Latex Companion CD, which doesn't have this functionality. Only a simple package manager, as far as I have seen.

Is there some other way?


In the game Yahtzee, a player rolls five fair dice, and gets a Yahtzee if all five dice show the same number. After the initial roll, the player gets two chances to reroll some of the dice. What is the probability that, on the initial roll, at least two of the dice show the same number? Express your answer as a common fraction.

Conplementary Counting.

Together, there are 6^5, or 7776 total.

Find how many ways you do not get atleast a double.

You could get
12345
13456
12456
12346
23456
12356

6 ways

$ 7770/7776\equal{}1295/1296$

Or rather, we count the number of ways that all the numbers rolled are different.  There are $6$ choices for the first number, $5$ choices for the second number, $4$ choices for the third number, $3$ choices for the fourth,and $2$ choices for the fifth.

Then we calculate the number of possible outcomes.  This is just $6^6$ since each dice has $6$ possible outcomes.  Therefore, the probability is:

\begin{eqnarray*}1-\frac{\cancel{6}\times5\times\cancel4\times\cancel{3}\times\cancel{2}}{\cancel{6}\times\cancel{6}\times\cancelto36\times\cancelto36\times6}&=&1-\frac{5}{54}\\&=&\boxed{\frac{49}{54}}\\\end{eqnarray*}


Let $ P(x) = 2x^4 + 2x^3 + 3x^2 - 2x + 1 $
    Find $ n\in N $ such that $ \exists a\in N : P(a)\equiv 0 (mod n)  $

I misunderstood something probably, but solution is trivial if you have to find just one n e N; we have:

$P(a) = 2a^4 + 2a^3 + 3a^2 - 2a + 1 \equiv 0 (mod \ n)$

let's try [I begin from the end]
$a \equiv 1 (mod \ n) $
we have $ 2a^4 + 2a^3 + 3a^2 - 2a +1 \equiv 2 + 2 +3 - 2 +1 \equiv 6 (mod \ n)$

so we know that n may be 2,3 and 6. This way I even find 3 ns. It is incorrect, isn't it ?

No , it's incorrect  . May be you understood wrongly this problem .
   I meant we must find n such that : $ \exists a : P(a) \equiv 0 (mod n) $ .
 We don't know a

Yeah, I've begun it from the wrong side   :)


Find all functions $ f: \mathbb{R}\to \mathbb{R}$
so that $ f(f(x))\equal{}\minus{}x$

Posted before in the Olympiad Algebra forums, not to mention that it's far too difficult for this forum. All solutions are not continuous.

[quote="1.01math1.01"]But why are they non continuous? if a take a function composition with itself: $ f \circ f$ and this is continuous, then shouldnt the function itself be continuous?
if it has been discussed already please post a link :)  cause in the search function I cant input formulas I dont know why  :blush:[/quote]

$ 1)$ $ f(f(x))$ continuous does not imply $ f(x)$ continuous. Example : $ f(x) \equal{} 0$ $ \forall x\in\mathbb{Q}$ and $ f(x) \equal{} 1$ $ \forall x\notin\mathbb{Q}$ : $ f(x)$ is obviously everywhere non continuous and $ f(f(x)) \equal{} 0$ is everywhere continuous.

$ 2)$ Some links :

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=167038[/url]

[url]http://www.artofproblemsolving.com/viewtopic.php?search_id=1499275189&t=113408[/url].

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=49191[/url]

$ 3)$ The fact that no continuous solution exist is quite simple to establish :
If $ f(f(x))\equal{}\minus{}x$, then obviously $ f(x)$ is bijective.
If $ f(x)$ is continuous, and since it is bijective, it must be a strictly monotonous function.
If $ f(x)$ is a strictly monotonous function, then $ f(f(x))$ is a strictly increasing function and so can't be equal to $ \minus{}x$

And so no continuous solution exist for the equation $ f(f(x))\equal{}\minus{}x$
But a lot of non continuous one exist (see links above).


The roots of equation $\displaystyle 2x^4 +2Ax^3+\left(A^2 -96\right)x^2 +Bx+C=0$ are  four integers.[b] Solve the equation.[/b]

[quote="arqady"]$\{0,_{\frac{+}{}}4,_{\frac{+}{}}4,_{\frac{+}{}}8\}$[/quote]
Indeed ! Please post the details as well as the rigurous justification  of the solution.

By DeMoivre's formulas we've got that

$x_1+x_2+x_3+x_4=-A \Rightarrow x_1^2+x_2^2+x_3^2+x_4^2+2\sigma_2 = (A^2-96)+96$

where $\sigma_2$ is product of the roots, taken two by two.

Again by DeMoivre's we've got

$2\sigma_2=A^2-96$

and so we conclude that

$\sum_{i=1}^4x_i^2=96$

It's easy to see that none of the roots might be equal or greater then 10 or even equal or less then -10.

Well, from this observation, I think it's not hard to guess [even testing] that the roots are $0,\pm 4,\pm 4, \pm 8$. Anyway, I'll try working on a more convincent argument.

[quote="aodeath"] ... we conclude that   $\sum_{i=1}^4x_i^2=96$ ....... [/quote]
[b]This equality is the key of the solution ! [/b]  Note that the only numbers which are sum of 4 squares in only one way
 are ( $96=3\cdot 2^{5}$)   \[ \{1,3,5,7,11,15,23, 2^{2k+1}, 3\cdot 2^{2k+1} , 7\cdot 2^{2k+1}\}\; .  \]

hmmmmmm

even simpler then I thought =]

I was trying to find a geometric argument, just for being visual...=]


geometry
trapezoid
rectangle
area of a triangle
Heron\u0027s formula

Let $ ABCD$ be a trapezium with sides $ AB$ and $ DC$ parallel. Suppose $ \mid AB \mid \equal{} 11, \mid BC \mid \equal{} 30, \mid CD \mid \equal{} 36$ and $ \mid DA \mid \equal{}25$. Find the area of the trapezium.

[hide=" One approach:"]

Drop perpendiculars from $ A$ and $ B$ to meet $ DC$ at $ E$ and $ F$ respectively. Let $ AE \equal{} BF \equal{} h$.

Take the rectangle $ ABFE$ out and slide the two triangles together to get a triangle with base $ 36\minus{}11\equal{} 25$ and altitude $ h$.

Heron's formula gives us the area of the triangle. Using half the base times the height gives the same area in terms of $ h$, so we can find $ h$ and thus the area of the rectangle. Adding this to the area of the (combined) triangle gives the area of the trapezium.
[/hide]

You can drop perpendiculars with length H from AB to DC. This separates it into 3 segments. The first segment will be S. The next segment is 11, because of the rectangle. The third segment is 25-S. Now you have two right triangles

$ S$ , $ H$ , $ 25$

$ 25\minus{}S$ , $ H$ , $ 30$

Then you have two equations, that you can solve for $ H$ . Then plug $ H$ back into the area of trapezoid formula. You will get $ [ABCD]$ = 564
Thank you, come again.

Thanks. I had the area of the triangle as $ \sqrt{40 \cdot 15 \cdot 15 \cdot 10} \equal{} 300 \equal{} \frac{1}{2} \cdot 25 \cdot h$ so that $ h\equal{}24$ and the area of the rectangle is $ 24 \cdot 11 \equal{} 264$, giving the area of the trapezium as $ 300 \plus{} 264 \equal{} 564$.

Using your method, I get $ x^{2} \plus{} h^{2} \equal{} 25^{2}$ and $ 25^{2} \minus{}50x \plus{} x^{2} \plus{} h^{2} \equal{} 30^{2}$.

So that $ x^{2} \plus{} h^{2} \minus{}50x\plus{}x^{2} \plus{} h^{2} \equal{} 30^{2}$

$ x^{2} \plus{} h^{2} \minus{} 25x \equal{} 30\cdot 15$

$ x(x\minus{}25) \plus{} 25^{2} \minus{} x^{2} \equal{} 30 \cdot 15$

$ \minus{}25x \plus{} 625 \plus{} 450$

$ x\equal{}7$

Therefore $ y\equal{}24$.

Therefore total area $ \equal{} \frac{1}{2} \cdot 7 \cdot 24 \plus{} \frac{1}{2} \cdot 18 \cdot 24 \plus{} 11 \cdot 24$

$ \equal{} 84 \plus{} 216 \plus{} 264$

$ \equal{} 564$.

Or just use trapezoid area formula, makes it quicker. $ \frac {47}{2} *24 \equal{} 564$


Is there exist function $f: \mathbb{N}\xrightarrow[1-1]{onto}\mathbb{N}$ such that,
i) $f(n+2006)=f(n)+2006$
ii) $f(f(n))=n+2$ for $n=1,2,\cdots ,2004$
iii) $f(2549)>2550$

Can you solve it? I'm not sure the problem is correct.

[quote="Rust"]for example $f(n)=n+3$ if n is odd and $f(n)=n-1$ if n is even.
It give
1) for any k $f(n+2k)=f(n)+2k$,
2) for any k $f(f(n))=n+2$,
3) for any odd k $f(k)=k+3>k+1$.[/quote]

But there is no $x$ such that $f(x)=2$


hi....again!

footnote this time...
well, as the subject says, I want to add a footnote in caption of a subfigure

when I do so, there's an error about too much } which is not true but I tried anyway to take some off but it didn't work either.

In my pdf, there's the $^{number}$ but there's no foot note

thanks!

ledurt

Oh, this is a bad thing to try to do.

\caption takes what is called a [i]moving argument[/i], because whatever is put into caption's argument is also put into the list of figures.  Putting cross-referencing commands like \footnote inside a moving argument will usually produce weird error messages and bizarre behaviors.

There are two things you might try:

(1) Try putting a \protect command before the footnote.  I don't think this will work, since I think this will give you the same footnot in your list of figures.

(2) Use the optional argument of \caption to explicitly tell what should be listed in the list of figures.  For example:
\caption[Map of California]{Map of California\footnote{not to scale}}
should print the caption with the footnote in the main text, but only list the caption without the footnote in the list of figures.  Not sure if this will work either.

ok, well, you were right...it didn't work...

but I haven't been clear in my question, and I don't think it'll help.  My footnote is in one of the caption of one of the subfigures

\begin{figure}
  \begin{center}
    \mbox{
      \subfigure[Facteur d'amplification dynamique [b]\protect{\footnote{Selon~\cite{meriam}}}[/b]]{      \includegraphics[width=.5\linewidth]{./dessins/Mfactor2.jpg} \label{fig:M}}
      %
      \subfigure[Transmissibilité [b]\protect{\footnote{Selon NOTES DYNAMIQUE}}]{      [/b]\includegraphics[width=.5\linewidth]{./dessins/TRbeau.jpg} \label{fig:TR}}}
    \caption{Comportement du système selon $\omega/\omega_n$}
    \label{fig:graph}
  \end{center}
\end{figure}

but maybe there's an other way...way different, and disapointing for a letex geek (i'm getting there too, but I just like it, not good enough to be called "geek")...
I want to add a footnote because the figures aren't from me and I want to give my sources so I'll just use \cite in the caption and it should do it... but not as nice as a footnote but it works...

thanks

ledurt

I've never heard of the \subfigure command, and it's not in my LaTeX book, so I can't help you there.

well, glad I can show you something!
I don't have a precise ref but here's an example :

\usepackage{subfigure}

\begin{figure}
  \begin{center}
    \mbox{
      \subfigure[Départ de la simulation.]{\includegraphics[width=.25\linewidth]{./img/init.eps} \label{fig-init}} \quad
      \subfigure[Défaillance du robot rond à t = 10s.]{\includegraphics[width=.25\linewidth]{./img/failure.eps} \label{fig-failure}}
      }
    \mbox{
      \subfigure[Le robot carré devient impatient car il n'a plus de nouvelle du robot rond par rapport à la tâche "pousser à droite" et finit par prendre le relais de ce dernier.]{\includegraphics[width=.25\linewidth]{./img/backup0.eps} \label{fig-backup0}} \quad
      \subfigure[Le robot carré exécutant la tâche "pousser à droite" à la place du robot rond. Ce dernier à été déplacé pour éviter des interférences non-représentatives à l'éxpérience.]{\includegraphics[width=.25\linewidth]{./img/backup1.eps} \label{fig-backup1}}
      }
    \mbox{
      \subfigure[Le robot carré devient impatient pour la tâche "pousser à gauche" et fini par reprendre sa tâche de départ. Il oscillera ainsi entre la tâche "pousser à gauche" et "pousser à droite".]{\includegraphics[width=.25\linewidth]{./img/backup2.eps} \label{fig-backup2}}
      }
    \caption{Captures d'écran lors d'une simulation Stage.}
    \label{fig:simulation}
  \end{center}
\end{figure}

*example by Philippe Cayouette*

so you have one number for the whole thing plus "subnumbers" for the subfigures, 1(a), 1(b)...

anyway, as I said, i have an other way

thank

ledurt


I have a problem, please help me :)  :lol:
Let $ a,b,c >0$, $ abc\equal{}1$
$ A \equal{} (a \plus{} b)(b \plus{} c)(c \plus{} a) \minus{} 4(a \plus{} b \plus{} c)$
$ minA \equal{} ?$

Use this one posted here ( [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=128529[/url]) am deduce that $ min \equal{} \minus{} 4$

It's a good solution :P but maybe it's complicated
so I want to see a simple solution

EDIT: Blah, I don't think I've ever posted a correct inequality solution.

[quote="MellowMelon"]Just AM-GM everything.

$ A \equal{} (a \plus{} b)(b \plus{} c)(c \plus{} a) \minus{} 4(a \plus{} b \plus{} c)$
$ \geq (2\sqrt {ab})(2\sqrt {bc})(2\sqrt {ca}) \minus{} 4(3\sqrt [3]{abc})$
$ \equal{} 8abc \minus{} 12\sqrt [3]{abc} \equal{} 8 \minus{} 12 \equal{} \minus{} 4$

with equality when $ a \equal{} b \equal{} c \equal{} 1$.[/quote]

I believe that you are wrong here since $ \minus{} (a \plus{} b \plus{} c)\leq \minus{} 3\sqrt [3]{abc}$ , so you cannot conclude that $ (a \plus{} b)(b \plus{} c)(c \plus{} a) \minus{} 4(a \plus{} b \plus{} c)$
$ \geq (2\sqrt {ab})(2\sqrt {bc})(2\sqrt {ca}) \minus{} 4(3\sqrt [3]{abc})$


Find all nonnegative real numbers $ x$ for which $ \sqrt[3]{13\plus{}\sqrt{x}}\plus{}\sqrt[3]{13\minus{}\sqrt{x}}$ is an integer.

[hide]Let $ N \equal{} \sqrt[3]{13\plus{}\sqrt{x}}\plus{}\sqrt[3]{13\minus{}\sqrt{x}}$.

Note that $ N > \sqrt[3]{\sqrt{x}}\plus{}\sqrt[3]{\minus{}\sqrt{x}} \equal{} 0$.

Cubing both sides yields:
$ N^3 \equal{} (13\plus{}\sqrt{x})\plus{}$ $ 3\left(\sqrt[3]{13\plus{}\sqrt{x}}\right)\left(\sqrt[3]{13\minus{}\sqrt{x}}\right)\left(\sqrt[3]{13\plus{}\sqrt{x}}\plus{}\sqrt[3]{13\minus{}\sqrt{x}}\right)$ $ \plus{}(13\minus{}\sqrt{x})$

$ N^3 \equal{} 26\plus{}3\left(\sqrt[3]{169\minus{}x}\right)N$

$ \sqrt[3]{169\minus{}x} \equal{} \dfrac{N^3\minus{}26}{3N}$

$ x \equal{} 169 \minus{} \left(\dfrac{N^3\minus{}26}{3N}\right)^3 > 0$

It is easy to see that if $ N \ge 5$, then $ x < 0$.

Therefore, $ N \in \{ 1,2,3,4 \}$.

$ N\equal{}1$ yields $ x \equal{} \dfrac{20188}{27} \equal{} 747 \frac{19}{27}$

$ N\equal{}2$ yields $ x \equal{} 196$

$ N\equal{}3$ yields $ x \equal{} \dfrac{123200}{729} \equal{} 168 \frac{168}{169}$

$ N\equal{}4$ yields $ x \equal{} \dfrac{29645}{216} \equal{} 137 \frac{53}{216}$[/hide]


What do you think the most deserted post on AOPS?

Post the state and give a reason if you don't pick Arizona.

Embarrasingly, it has only had 9 topics posted on it's forum.  It was last posted by johnnycade1592

on January 4.

Yeah math154, but colorado also has a private forum!!!!

So take that into account and.... :P

Wow, I thought this post would be interesting.

Not even any spam on this post?

Come on guys....  :rotfl:

[quote="justinahmann"]What else is there to do besides spam???   :clap2:

Technically, this isn't that much spam. It relates to the public forums.[/quote]

That doesn't make it not spam.

[quote][u][b]1.[/b] Does my post contain informative value?[/u]
This needs not be mathematical. There are informative things to be said about World Cup 2006 too. On the other hand, few informative things can be said about problem 5 of the IMO 2006 before the IMO has started.
[u][b]2.[/b] Could my post interest or be useful to anyone?[/u]
This is closely related to the question before. If a post doesn't interest anyone, why should it be posted?
A few tautologies: A solution of a problem surely interests people, at least the person who posted the problem. A problem, in most cases, also interests people. On the other hand, a problem from an ongoing competition may interest somebody a bit, but it is by no means useful; instead, it damages the competition.
By the way, you can well assume that a good joke will interest people. On the other hand, a statement of the kind "I think so too" is less likely to do that.
[u][b]3.[/b] Will anyone understand this post?[/u]
Yes, this is something many people just don't think about. If you know very well that your English sucks, try to write your problem in two different versions, or just explain it with more details, or use formula language instead of words where you can... And ffs check your questions for typos before posting them! Don't expect us to both correctly interpret and solve your problems!

If statement [b]3.[/b] is not fulfilled, or none of [b]1.[/b] and [b]2.[/b] is fulfilled, then the post is spam.[/quote]

Try to think about that.

This topic is rather... pointless.

Haha, we have transformed this spammy post into greater spam.  :rotfl:

[quote="justinahmann"]Yeah, but you already know I contradicted those 4 states  :P

Nevada: [b]Gambling[/b]

Kentucky: Too much [b]fried chicken[/b] which makes people fat

Utah: This is very [b]"unoriginal"[/b] : Salt Lake

Minnesota: Too close to [b]Canada[/b], which may make it partly French who eat snails (

[i]escargot[/i], which is a French stereotype

But still,  :P   :P   :P   :P   :P[/quote]

This isn't that interesting. I sense some immaturity here.

Well moogra, other people may think the topic is interesting. Obviously, it is immature...

[quote="pythag011"][quote="justinahmann"]Uh, actually, that post has all those three characteristics.

Unless, you want to go and contradict atleast half of the posts on the missouri forum  :D .[/quote]

Well, Missouri forum has... a lot of spam...[/quote]

Obviously no one cares about spam anymore though. I bet I could easily spam 100 posts without anyone noticing.

[quote="math154"][quote="moogra"]I bet I could easily spam 100 posts without anyone noticing.[/quote]

Already done, except one week after a certain amount of spam, and, unfortunately, one math problem, piled into the Missouri forum, a mod or admin deleted it. We sort of went down by 5 pages and lost 2000+ posts and 100+ threads.[/quote]
My pentagon math problem? :)

Back to the thread, AZ is not boring or abandoned; it's one of the fastest growing states  :wink:

[quote="dnkywin"]My pentagon math problem? :)[/quote]

Haha, maybe; it might have been a couple of problems, actually. :P


advanced fields
advanced fields unsolved

Below there is an exercice which I'm trying to solve, but unfortunately I can't succeed. I'll be grateful for any help.

Suppose that $ C_1$ and $ C_2$ are convex sets in $ R^n$ such that $ C_1$ has interior points and $ C_2$ doesn't contain any interior points of $ C_1$. Prove that there is a hyperplane $ H$ in $ R^n$ such that $ C_1$ and $ C_2$ lie in the opposite closed half-spaces determined by $ H$, that is, there exist an $ a \neq 0$ in $ R^n$ and $ \alpha \in R^n$ such that $ x \cdot a \leq \alpha \leq y \cdot a$ for all $ x \in C_1$ and all $ y \in C_2$.

Cheers,

Nobody1111

An idea of a proof:
1) $ \bar{C_1},\bar{C_2}$ are convex sets which satisfy the same properties than $ C_1,C_2$. Then we may assume that $ C_1,C_2$ are closed.
2) If $ C_1\cap{C_2}\not\equal{}\oslash$ then $ C_1\cap{C_2}$ is a convex without interior point then it's included in a hyperplan $ H$ which is a correct solution
3)If $ C_1\cap{C_2}\equal{}\oslash$:
Case 1: there exists $ a\in{C_1},b\in{C_2}$ s.t. $ \parallel{}a\minus{}b\parallel{}$ is the distance  $ \delta\equal{}d(C_1,C_2)$. Take for $ H$ the mediator hyperplan of $ ab$.
Case 2: there don't exist such $ a,b$. Then there exist $ (a_n)_n\in{C_1}$ and $ (b_n)_n\in{C_2}$ s.t. $ lim(\parallel{}a_n\minus{}b_n\parallel{})\equal{}\delta\geq0$. $ \parallel{}a_n\parallel{}$ and $ \parallel{}b_n\parallel{}$ goes to $ \infty$; moreover if $ x\in{C_1}$ is an interior point then $ angle(a_n\minus{}x,a_n\minus{}b_n)>\pi{/2}$ (for a sufficiently big $ n$) and goes to $ \pi{/2}$ and if $ y\in{C_2}$ then $ angle(b_n\minus{}y,a_n\minus{}b_n)$ goes to $ \pi{/2}$. Let $ H_n$ be the mediator hyperplan of $ a_nb_n$ and $ H\equal{}lim(H_n)$ (if there exists). I hope that this $ H$ will be convenient.


Hi I was wondering if anyone here has any tips on preparing for the AMC 8, and how to do pretty well (>20).

Thanks,
lyra

Practice and study old AMC 8's if possible. AoPS Wiki has  a few AMC 10's. The first 15 or so problems on the AMC 10 are AMC 8 difficulty. Also, during the test, try not to make any careless mistakes.

BTW: Can AMC 8's be uploaded to the Wiki?

[quote="MysticTerminator"][quote="Elemennop"][quote="lyra"]and how to do pretty well (>20)[/quote]

Don't answer any questions?[/quote]

dude what I'm thinking 0 for not answering any questions

THAT IS NOT A GOOD STRATEGY FOR GETTING > 20[/quote]

no no i think elemmnop meant that they changed the scoring system so that you get a higher score if u get less questions right duh

[quote="lyra"]on the AMC 8, I thought the website said that you got 1 point for correct, 0 for incorect?

lyra[/quote]

...

oh

that's a shame

[quote="Xantos C. Guin"]Yes.

Elemennop might think that the AMC 8 is scored the same way as the AMC 10/12. If it was, by leaving all problems blank, you would recieve $25\cdot1.5=37.5>20$.

Unfortunately, that is not the case.[/quote]

that will be... stupid

In AMC 10/12 isn't the passing score 120 or something...

[quote="mathgeniuse^ln(x)"]I now have to get a perfect score on AmC 10 in order to do better than my score from this year.[/quote]
You can't even tie your old score.

[quote="mathgeniuse^ln(x)"]I now have to get a perfect score on AmC 10 in order to do better than my score from this year.[/quote]

Well, then you should take the AMC 12.  :wink:

The KSU AMC website has many tests and worksheets for each of the AMC 8,10 and 12.
[url=http://www.math.ksu.edu/main/events/hscomp/amc.htm]http://www.math.ksu.edu/main/events/hscomp/amc.htm[/url]


Given a point $D: (0,3)$ and an ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$. $M$ and $N$ are on the ellipse such that $\overrightarrow{DM}= \lambda \overrightarrow{DN}$. Find the range of $\lambda$.

Graphically, the range should be [$\frac15$,5].


inequalities
trigonometry
inequalities unsolved

Prove that $ \sin^{3}A\plus{}\sin^{3}B\plus{}\sin^{3}C<2$ in the $ ABC$ triangle.

[quote="Mateescu Constantin"]Prove that $ \sin^{3}A \plus{} \sin^{3}B \plus{} \sin^{3}C < 2$ in the $ ABC$ triangle.[/quote]

$ \sin^{3}A \plus{} \sin^{3}B \plus{} \sin^{3}C \equal{} \frac {s(s^2 \minus{} 6Rr \minus{} 3r^2)}{4R^{3}} \le \frac {s(2R \minus{} r)}{2R^2} < 2$ , $ (s^2 \le 4R^2 \plus{} 4Rr \plus{} 3r^2)$

Equality holdes for $ A \rightarrow \frac {\pi}{2}, B \rightarrow \frac {\pi}{2}, C \rightarrow 0$

[quote="Sergic Primazon"][quote="Mateescu Constantin"]Prove that $ \sin^{3}A \plus{} \sin^{3}B \plus{} \sin^{3}C < 2$ in the $ ABC$ triangle.[/quote]

$ \sin^{3}A \plus{} \sin^{3}B \plus{} \sin^{3}C \equal{} \frac {s(s^2 \minus{} 6Rr \minus{} 3r^2)}{4R^2} \le \frac {s(2R \minus{} r)}{2R^2} < 2$ , $ (s^2 \le 4R^2 \plus{} 4Rr \plus{} 3r^2)$

Equality holdes for $ A \equal{} B \equal{} \frac {\pi}{2}$[/quote]

I don't understand why $ \sin^{3}A \plus{} \sin^{3}B \plus{} \sin^{3}C \equal{} \frac {s(s^2 \minus{} 6Rr \minus{} 3r^2)}{4R^2}$. You should write at the denomiter $ 4R^{3}$, because $ \sin^{3} A\equal{}\frac{a^{3}}{8R^{3}}.$

The equality case is not correct; there exists no triangle ABC with $ \angle A\equal{}\angle B\equal{}\frac{\pi}{2}$
 :huh:


how do you do the problem:

when four coins are flipped, what is the probability that three of them are heads?

please include explanation

Ummmm..
[hide]there are $2^4$ different sequences of flips
$\binom{4}{3}$ ways to get $3$ heads.
$\binom{4}{3}=4$, $\frac{4}{16}=\frac{1}{4}$[/hide]

While mathematically, I agree with ch1n353ch3s54a1l. But couldn't you also argue that 4 heads is also counted towards the probability? It didn't say exactly 3 heads, so if you hit the $\frac{1} {16}$ chance that you get 4 heads, you've still got 3 heads.

So you can argue $\frac{5} {16}$ since it doesn't say exactly.

I think the problem meant "exactly".

[hide="so..."]$\frac{1}{2^4}(4)=\frac{1}{4}$ (since there are 4 ways to arrange the three heads in the sequences of flips). [/hide]

As the other posters have shown, this problem is made much easier by knowledge of the binomial probabilty distribution  :idea:


if we have r indistinguishable (identical)
balls and n different boxes, how many ways can we do this?

o..o|o...o|...|o..o

o - ball

There are r balls and n-1 borders.

We can arrange them in $ C^{r\plus{}n\minus{}1}_{n\minus{}1}$ ways.


In 1984, July 4 fell on a Wednesday. On what day of the week did July 4, 1990, fall?

6 years = 2190 days. $ \frac{2190}{7}$ has a remainder of 6, making the answer 6 days after Wednesday, or $ \boxed{Tuesday}$

Ebob, you are forgetting that 1988 is a leap year, so the day is actually Wednesday, not Tuesday.


[b] In a class 54% of the students are girls and 46 % are boys .I f the number of girls is 27 find the total number of students?[/b]

Please don't type the problem in boldface.

[hide="Solution"]Let $ s$ be the number of students in the classroom. Then, \[ \frac{54}{100}s\equal{}27\]Multiplying by $ 100$, we get \[ 54s\equal{}2700\] Next, dividing by $ 54$ yeilds \[ s\equal{}\boxed{50}\][/hide]


Hi fleeting_guest,

you asked once for the indiviudal scores and country ranking of Western Germany in the beginning of the 80s. I also have those of Eastern Germany. What do you want to know ?  :)

hello orl,  thanks for returning to this subject, which
was deleted when the web site shut down.

My question was about something we heard about
the old German IMO teams led by Artur Engel:
that he would select some 13-15 year olds
and take the same group to the IMO for several years.
This way, every few years a highly-trained team would
have a chance to place first with several gold medals.

I heard this from a non-German team leader, but maybe
he or I misunderstood.   Also was Engel leading the West
or East German teams?

[quote="fleeting_guest"]
My question was about something we heard about
the old German IMO teams led by Artur Engel:
that he would select some 13-15 year olds
and take the same group to the IMO for several years.
This way, every few years a highly-trained team would
have a chance to place first with several gold medals.
[/quote]

Well, next year I will come back with more details. There is a booklet called "The German Teams at the International Mathematical Olympiads 1959-1998" by Wolfgang Engel, Hans-Dietrich Gronau, Hanns-Heinrich Langmann and Host Sewerin. Next year I will provide an extract from the booklet. As images in a .zip-file or as .doc document with those images included. Then you can study the results more in detail. For now I just append some introducing words.  :)

If you take into consideration that FRG was in the TOP3 among the years 1979-1983. From 1986-1989 they were ranked in the TOP4.
I don't think this supports your statement that they got trained over years.

At the moment Dieter Gronau, the present German IMO team leader (ITL), is  involved in the IMO preparation and works together with FRG's former team leader  Arthur Engel (77-81,83-84,88, 89(chair of jury),90) and maybe also with Horst Sewerin (77-82,84-87,89-91,94-95), but I am not sure. Annually Engel guides the final IMO preparation week for the respective IMO team members.

Gronau took his doctor's degree under the former GDR IMO team coach Gustav Burosch (university Rostock). In the past Gustav Burosch accompanied the team among the years 1970 - 1990 (save 1976, for political reasons the GDR did not participate in IMO 1978/1981). Gronau's and Burosch's research in discrete mathematics place them at the world's highest level. In 1990 Burosch was removed from the IMO preparation and his university chair. Now he works as a teacher for Computer science/math in the private industry. By the way Gustav took his doctor's degree under the former GDR IMO team coach Wolfgang Engel (university Rostock) who accompanied the team among the years 1963 - 1964, 1965 (chair of the jury).

From 1966-1975 (1974 place 4) the GDR ranked in the TOP3. In 1972 Pawel Kröger from Leipzig got a perfect score (40 points) and a special prize. He was in grade 7 back then. He is one of the youngest students ever to manage that. In 1989 they ranked fourth and got three perfect scores by Frank Göring from Illmenau, Andreas Siebert Berlin and Gerard Zenkner from Chemnitz.

Beginning in 1991 a united German team participated in the IMO. In 1991 they ranked fourth and in 1993 second in the world. Ever since the performance got worse. And I always thought that uniting such strong countries would make Germany invincible. But today competitions are not only used as appreciation for superior students but also to select strong students. This was actually done in math circles some decades ago. But this got too expensive. In the GDR most students participarted in the first round of the German Mathematical Olympiad. Today most students don't even know it. Thus many talents get lost. Just look at Chinese Taipeh or Saint Petersburg what is possible if you put a lot of effort in it.

For any other questions just ask me or have a look on the [url=http://www.mathematik-olympiaden.de/IMOs/imo.htm]German IMO page.[/url]  :)

thanks orl.
There seem to be several people named Engel involved
which makes me less sure that the story pertained to
A. Engel and the FRG team.  After posting my original
query I saw the FRG results for some years and indeed
they fluctuated less than one would expect under the type
of system I heard about in the story.

Let me ask a broader question.  Was the team selection
in the old days always a matter of performance in a
pre-defined selection process (e.g. top 6-8 on national
olympiad + additional selection tests), or was it more
a personal selection by the team leader (informed but
not pre-determined by the test results)?

By "olden days" I mean, say, not later than the mid-1980's
(or maybe earlier if relevant).

[quote="fleeting_guest"]thanks orl.
There seem to be several people named Engel involved
which makes me less sure that the story pertained to
A. Engel and the FRG team.  After posting my original
query I saw the FRG results for some years and indeed
they fluctuated less than one would expect under the type
of system I heard about in the story.

[/quote]

Well obviously both countries tried hard to excel each other especially science and knowledge was the only issue they could show their skills. But look today. Compare the length of the training camps of countries like Russia, China, USA etc. with Germany, France and Switzerland. Especially the Asians often loose creativity as they proceed very technically in the choice of their training problems. But real mathematicians are meant to be creative.

Both Engel were awarded by the World Federation of National Mathematics Competitions (WFNMC). Arthur Engel received the David Hilbert Award in 1991 and Wolfgang Engel the Paul Erdös Award in 1998. Most of the award recipients were or are closely related as IMO team leader, deputy, jury member or former participant.

[quote="fleeting_guest"]
Let me ask a broader question.  Was the team selection
in the old days always a matter of performance in a
pre-defined selection process (e.g. top 6-8 on national
olympiad + additional selection tests), or was it more
a personal selection by the team leader (informed but
not pre-determined by the test results)?

By "olden days" I mean, say, not later than the mid-1980's
(or maybe earlier if relevant).[/quote]

Well, I think the method you mentioned first was applied. For more information read the booklet extract I appended above.  :)


Hi Guys I really in trouble with a problem about Dilworth's theorem

The problem states:

Let $ a_1, a_2,....a_{n^2\plus{}1}$ be a permutation of integers $ 1,2,...n^2\plus{}1$. Show that Dilwor theorem implies that the sequence has a subsequence of length $ n \plus{} 1$ that is monotone.

Thanks a lot to all of you

All the best

joksy

this is  a special case of Erdos-Szekeres theorem ....  :wink: i didn't heard about that theorem ... so can you explain to me what do you mean by Dilwor theorem ?  :blush:

I'm pretty sure it is just a typo, http://en.wikipedia.org/wiki/Dilworth%27s_theorem

Sorry guy in fact it is a type the theorem is Dilworth s theorem, my mistake.

anyone can give some hints..

For the sake of reference, here is the Erdös-Szekeres Theorem:

[color=blue][b]Erdös-Szekeres Theorem.[/b] Let $ n$ and $ m$ be two positive integers, and let $ a_1$, $ a_2$, ..., $ a_{mn \plus{} 1}$ be $ mn \plus{} 1$ elements of an ordered set (for instance, $ mn \plus{} 1$ real numbers). Then, the sequence $ \left(a_1,a_2,...,a_{mn \plus{} 1}\right)$ either has a monotonically increasing subsequence, or has a strictly monotonically decreasing subsequence.[/color]

The [url=http://en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Szekeres_theorem]Wikipedia article[/url] tells you all you need, maybe even too much. (Though it gives a weaker version of the Erdös-Szekeres Theorem, for whatever reason... but the same proof applies to the stronger one.)

  darij


Suppose positive variables $ x,y$ and $ z$ satisfy : $ x^2\plus{}y^2\plus{}z^2\equal{}1$ Prove that :
     $ \sqrt{\frac{x^2\plus{}(y^2\minus{}z^2)^2}{4}}\plus{}y\plus{}z \leq \sqrt{3}$

[quote="QuyBac"]Suppose positive variables $ x,y$ and $ z$ satisfy : $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 1$ Prove that :
     $ \sqrt {\frac {x^2 \plus{} (y^2 \minus{} z^2)^2}{4}} \plus{} y \plus{} z \leq \sqrt {3}$[/quote]

we prove that the stronger:

$ \sqrt {x^2 \plus{} \frac {(y^2 \minus{} z^2)^2}{4}} \plus{} y \plus{} z \le \sqrt{3}.$

let $ p\equal{}y\plus{}z,q\equal{}yz$, we have $ q \le \frac{p^2}{4}$, and

$ \sqrt {x^2 \plus{} \frac {(y^2 \minus{} z^2)^2}{4}} \plus{} y \plus{} z$

$ \equal{} \sqrt {1 \minus{} y^2  \minus{} z^2  \plus{} \frac{1}{4}(y^2  \plus{} z^2 )^2  \minus{} y^2 z^2 }  \plus{} y \plus{} z$

$ \equal{} \sqrt {1 \minus{} (y \plus{} z)^2  \plus{} 2yz \plus{} \frac{1}{4}((y \plus{} z)^2  \minus{} 2yz)^2  \minus{} y^2 z^2 }  \plus{} y \plus{} z$

$ \equal{} \sqrt {1 \minus{} p^2  \plus{} 2q \plus{} \frac{1}{4}(p^2  \minus{} 2q)^2  \minus{} q^2 }  \plus{} p$

$ \equal{} \sqrt {1 \minus{} p^2  \plus{} \frac{1}{4}p^4  \plus{} (2 \minus{} p^2 )q}  \plus{} p,$

because $ p^2\equal{}(y\plus{}z)^2 \le 2(y^2\plus{}z^2) \le 2$, so we have $ 2\minus{}p^2 \ge 0$, and

$ \sqrt {1 \minus{} p^2  \plus{} \frac{1}{4}p^4  \plus{} (2 \minus{} p^2 )q}  \plus{} p$

$ \le \sqrt {1 \minus{} p^2  \plus{} \frac{1}{4}p^4  \plus{} (2 \minus{} p^2 )\frac{{p^2 }}{4}}  \plus{} p$

$ \equal{} \sqrt {1 \minus{} \frac{{p^2 }}{2}}  \plus{} p$

$ \equal{} \sqrt {1 \minus{} \frac{{p^2 }}{2}}  \plus{} \sqrt {\frac{{p^2 }}{4}}  \plus{} \sqrt {\frac{{p^2 }}{4}}$

$ \le 3\sqrt {\frac{{1 \minus{} \frac{{p^2 }}{2} \plus{} \frac{{p^2 }}{4} \plus{} \frac{{p^2 }}{4}}}{3}}$

$ \equal{} \sqrt{3},$

done.


This is a well known fact, but prove it.

$\displaystyle\sum_{i=0}^{n} \binom{n}{i} = 2^n$

The numbers from $n\choose0$ to $n\choose{n}$ form the pascal's triangle, and every row of the triangle add up to a power of 2. I don't really know why that is though  :(

[quote="xang"]if we wrote $2^n$ as $(1+1)^n$ and then do binomial expansion?

but this isn't prooving anything , am i going in circles? :huuh:[/quote]
I think your approach is correct :)
If we expand $(1+1)^n$ using binomial expansion, we get $ _n C_0+\ _nC_1+...+\ _nC_n=2^n$

Does anyone know other way(s) to prove this property?

combinatorial proof?
ways to make all possible subsets in a set of n numbers
to make sets with one element $\binom{n}{1}$
to make sets with two elements $\binom {n}{2}$
...
to make sets with n elements $\binom {n}{n}$

another way to get all possible subsets is $2^n$, an element is either in or out of that subset

Take the elements of a set (1.2 . . .n), now, we shall form a subset.

In our subset we can either include 1, or not include one. We can include 2, or not include 2, and apply this reasoning all the way up to n.

Therefore we have 2^n

It's important to know a algebraic approach and a good combinatorics argument, as given by sirorange, when dealing with proof of combinatorics identities.

Good job everyone.


real analysis
limit
logarithms
real analysis unsolved

Put $ s_N \equal{} 1 \plus{} (1/2) \plus{} \cdots (1/N)$. Prove that

$ \lim_{N \to \infty} \left(s_N \minus{} \log N \right)$ exists.

I know this is on the internet, I just want a hint though.

I noticed that the limit can also be expressed as

$ 1 \plus{} \sum_{n \equal{} 2}^{\infty} \frac {1}{n} \minus{} \log \frac {n}{n \minus{} 1}$

so I have been trying to prove that converges by the comparison test. Its not hard to show that each term in the series is negative. Other than that I am stuck.

[quote="Extremal"]for beginners , I think this is best hint :
$ 1 \plus{} \sum_{n \equal{} 2}^{\infty} \frac {1}{n} \minus{} \log \frac {n}{n \minus{} 1}$[/quote]

You didn't read the end of my post.


#1
More vineger

3/4=6/8

so 5/8<6/8

#7

d=vXt
t=15/2 h
v=50 miles per hour

So

d=50X15/2
d=375 miles

[hide="number 1"]
more vinegar because
3/4=6/8
and 6/8 is bigger than 5/8 :)
[/hide]

[hide="number 2"]
5/8+6/8=11/8
which is

$1 \frac{3}{8}$ cups[/hide]

[/hide]

[hide="10"]

We want to do the operation
$16\frac{3}{8}-2\frac{7}{8}$

$16\frac{3}{8}=15\frac{11}{8}$ (regroup)
$15\frac{11}{8}-2\frac{7}{8}=13\frac{4}{8}=13\frac{1}{2}$.

if you still have trouble, you can ask in the classroom math forum. [/hide]

I want to know what sort of exactly problem for number 10th except the substraction fraction, 236factorial. Besides, it looks like I have one left question (number 9th) that need to be answer, hopefully. By the way, thank you all (felipesa, math92, templar, 236factorial) very much. You have been excellently helpful to me (I don't want to sound like take advantage of you, but i have to learn how to become one of good mathematicians like you...you know). Wow, I am so glad that all of you are really awesome and you have good intelligences! I guess I will use this site for all the time or when I have problems. Oh, I will post my future problems in the Math Links Forum of Classroom Math next time. Thank for informed me, 236factorial.

[quote="Templar"]
4. Truck tires are 2/3 foot, car tires are 5/8 of 2/3 ie 5/8*2/3=10/24=5/12 foot wide
[/quote]

Are you certain the number 4th is correct?

[quote="236factorial"][quote="Anonymous52111"]I want to know what sort of exactly problem for number 10th except the substraction fraction, 236factorial. [/quote]

I don't fully understand the statement, but if you want to know when to use subtraction in stock prices, look for the word fell or dropped.[/quote]

Sorry, I was trying to articulate my meaning. However, I want to know what's called for number 10th as the specific type of problem.  I hope this would clarify for you.

[quote="Anonymous52111"][b]Please help me to answer each question and describe why. I hope I will get crystal clear by how to figure this out. Thanks. [/b]
[b]
10 Questions:[/b]
1. A recipe for salad dressing calls for 5/8 of oil and 3/4 cups of vinegar. Does the recipe use more oil or vinegar?

2. How much salad dressing does the above recipe make?

3. Dan has 2/3 cup of oil. His recipe calls for 7/8 cup of oil. Does he have enough oil?

4. The tires on John Sauble's car are 5/8 as wide as the tires on his pickup truck. The truck tires are 2/3 of a foot wide. How wide are the car tires?

5. The fuel tank of John's semitrailer truck holds 3 1/3 times a much as the fuel tank of his car. The car fuel tank holds 14 1/2 gallons. How much does the truck fuel tank hold?

6. John drove his pickup truck 170 miles in 3 3/4 hours. What was his average speed?

7. At 50 miles per hour, how far can John drive in 7 1/2 hours?

8. John's semitrailer truck travels 8 1/3 miles per gallon of gas. How much fuel will be used during a 3200 mile trip?

9. Melinda bought stock when the selling price was 8 /4 dollars per share. During the next quarter (three months), it rose 4 1/2. Then, during the following quarter, it rose 4 3/4. What was the price of the stock at the end of two quarters?

10. At the beginning of the fourth quarter, the price of a stock was 16 3/8. During the quarter the price fell 2 7/8. What was the price at the end of the quarter?[/quote]

1. vinegar

[/hide]

[quote="Anonymous52111"][b]Please help me to answer each question and describe why. I hope I will get crystal clear by how to figure this out. Thanks. [/b]
[b]
10 Questions:[/b]
1. A recipe for salad dressing calls for 5/8 of oil and 3/4 cups of vinegar. Does the recipe use more oil or vinegar?

2. How much salad dressing does the above recipe make?

3. Dan has 2/3 cup of oil. His recipe calls for 7/8 cup of oil. Does he have enough oil?

4. The tires on John Sauble's car are 5/8 as wide as the tires on his pickup truck. The truck tires are 2/3 of a foot wide. How wide are the car tires?

5. The fuel tank of John's semitrailer truck holds 3 1/3 times a much as the fuel tank of his car. The car fuel tank holds 14 1/2 gallons. How much does the truck fuel tank hold?

6. John drove his pickup truck 170 miles in 3 3/4 hours. What was his average speed?

7. At 50 miles per hour, how far can John drive in 7 1/2 hours?

8. John's semitrailer truck travels 8 1/3 miles per gallon of gas. How much fuel will be used during a 3200 mile trip?

9. Melinda bought stock when the selling price was 8 /4 dollars per share. During the next quarter (three months), it rose 4 1/2. Then, during the following quarter, it rose 4 3/4. What was the price of the stock at the end of two quarters?

10. At the beginning of the fourth quarter, the price of a stock was 16 3/8. During the quarter the price fell 2 7/8. What was the price at the end of the quarter?[/quote]

1) you want to make the denominators common.  oil: $ \frac{5}{8} $  vinegar: $ \frac{6}{8} $
so the answer is vinegar

2) you want to add the fractions, and you still need common denominators.  $ \frac{5}{8} + \frac{6}{8} = \frac{11}{8} $
so the answer is $ \frac{11}{8} $ cups

3)you want to figure out if $\frac{2}{3} $is equal to or greater than $ \frac{7}{8} $ . again, common denominators.
$\frac{16}{24} < \frac{21}{24} $
so the answer is no

4) multiply: $\frac{5}{8} * \frac{2}{3} = \frac{5}{12} $

5) multiply: $ 48\frac{1}{3} $


geometry
similar triangles
power of a point

The tangents from an external point P touch a circle with centre O at A and B. AD is perpendicular to diameter BC. Prove that CP bisects AD.

Peng Shi

Could you perhaps provide a picture or a more detailed description? Because with the description you have so far, CP does not bisect AD in my drawing. Furthermore, where is D? is it on the circle? Clarification would be nice. Thanks!

I've added a diagram. I hope that it is clear now.

Peng Shi

****removed spolier, added a diagram, nothing else changed****

Let E be the intersection of PC and AD.

 :ang: ACD =  :ang: PAB =  :ang: POB since PAOB is cyclic.

So  :tri: ACD  :sim:  :tri: POB ==> AD/CD = PB/OB = PB/r

:tri: ECD  :sim:  :tri: PCB ==> ED/CD = PB/CB = PB/(2r)

Therefore AD=2ED, i.e. CP bisects AD.

As in the diagram, E is the intersection of PA and BC, M is the midpoint of PB, and X is the intersection of AD and BC. Since BM/BP = 1/2 = BO/BC, triangles MBO and PBC are similar, so <MOB = <XCD. So, right triangles MBO and XDC are similar.

From power-of-a-point, (EC)(EB)=(EA)^2, which equals (ED)(EO) from similar triangles ADE and OAE. Therefore,

ED/EB = EC/EO. (*)

Note that triangles PBE and ADE are similar. We then have

AD/PB
= ED/EB (similar triangles)
= EC/EO (from *)
= (ED-EC)/(EB-EO)
= DC/BO
= XD/MB (similar triangles).

Therefore, XD/AD = MB/PB = 1/2, meaning X is the midpoint of AD, as desired. [tex]\blacksquare[/tex]

My first attempt was very close to [b]zabelman[/b]'s approach. Then I realized that a shorter proof exists as posted above. I just removed the spoiler and added a diagram  in the above post for clarity.

Hm, I think the problem has been posted few times after this posting, with nice proof antiparallel - symmedian: $CP$ symmedian of $ABC$, while $AD$ is antiparallel to $AB$ in $\triangle ABC$, so $CP$ is median of $\triangle ACD$.

Best regards,
sunken rock


function calculus conics hyperbola algebra domain

Given that $x(t) = e^{t} - e^{-t}$, $y(t) = e^{t} + e^{-t}$, express this relationship as a cartesian function $y = f(x)$. (p.s. calculus is allowed.)

$y^2-x^2=e^{2t}+2+e^{-2t}-(e^{2t}-2+e^{-2t})=4$, hence it's a hyperbola with the foci on $y$-axis.


calculus
integration
algebra
polynomial
calculus computations

Let $ f(x)$ be a polynomial and $ C$ be a real number.
Find the $ f(x)$ and $ C$ such that $ \int_0^x f(y)dy\plus{}\int_0^1 (x\plus{}y)^2f(y)dy\equal{}x^2\plus{}C$.

We know that $ \exists n\in\mathbb{N},\deg f(x)\equal{}n$.

Furthermore, $ \deg\left[\int_0^x f(y)dy\right]\equal{}n\plus{}1$ and $ \deg\left[\int_0^1 (x\plus{}y)^2f(y)dy\right]\equal{}2$.

Since we have $ \int_0^x f(y)dy\plus{}\int_0^1 (x\plus{}y)^2f(y)dy\equal{}x^2\plus{}C$, it follows that $ \deg f\equal{}1$.

Assuming $ f(x)\equal{}ax\plus{}b$ where $ a,b\in\mathbb{R}$. Hence,

$ x^{2}\plus{}C\equal{}\int_0^x f(y)dy\plus{}\int_0^1 (x\plus{}y)^2f(y)dy\equal{}\frac{ax^{2}}{2}\plus{}bx\plus{}\int_{0}^{1}(x^{2}\plus{}2xy\plus{}y^{2})(ay\plus{}b)dy$

$ x^{2}\plus{}C\equal{}(a\plus{}b)x^{2}\plus{}\left(\frac{2a}{3}\plus{}2b\right)x\plus{}\left(\frac{a}{4}\plus{}\frac{b}{3}\right)$

Therefore $ f(x)\equal{}\frac{3x\minus{}1}{2},C\equal{}\frac{5}{24}$. :)


RC-7th has submitted his answer and sam 3.14 didn't.

And RC-7th got number two right although he didn't get number one right.

So, the winner is RC-7th.

I'll try to build up a COP final soon.

 :)


geometry
area of a triangle
Heron\u0027s formula

I am trying to prove this :
If c^2=a^2 + b^2 , prove : [(a+b+c)(a+b-c)] [(b+c-a)(c+a-b)] = (2ab)^2

My textbook shows it like this :

1)= [ (a+b)^2 - (c)^2 ] [(c+a-b) (c-a+b)]
2) = [ 2ab ] [ (c+(a-b)) (c-(a+b) ] - I smell a rat here
3) = [ 2ab ] [ (c+a-b) (c-a-b) ]
4) = [ 2ab ] [ (c-b)^2 - (a)^2 ]
5) = [ 2ab ] [ 2ab]
6) = (2ab)^2

[Difficulty]
I Smell a rat. There is something fishy about Step2 where the brackets
are grouped. Is it permitted to group brackets in that way?
The grouping seems to work for this sum (by chance?). But i dont think
it will work for others.

Can tell why its wrong (or right)? And please give an example to show
that its wrong if it is.

[quote="mathguy999"]I am trying to prove this :
If c^2=a^2 + b^2 , prove : [(a+b+c)(a+b-c)] [(b+c-a)(c+a-b)] = (2ab)^2

My textbook shows it like this :

1)= [ (a+b)^2 - (c)^2 ] [(c+a-b) (c-a+b)]
2) = [ 2ab ] [ (c+(a-b)) (c-(a+b) ] - I smell a rat here
3) = [ 2ab ] [ (c+a-b) (c-a-b) ]
4) = [ 2ab ] [ (c-b)^2 - (a)^2 ]
5) = [ 2ab ] [ 2ab]
6) = (2ab)^2

Can tell why its wrong (or right)? And please give an example to show
that its wrong if it is.[/quote]

Step 2 is wrong as $ c\minus{}(a\plus{}b)\equal{}c\minus{}a\minus{}b \neq c\minus{}a\plus{}b$

the solution should be:

$ \equal{}[(a\plus{}b)^2\minus{}c^2][c^2\minus{}(a\minus{}b)^2]$
$ \equal{}[(a^2\plus{}b^2\minus{}c^2)\plus{}2ab] [(c^2\minus{}a^2\minus{}b^2)\plus{}2ab]$
$ \equal{}(2ab)(2ab)$
$ \equal{}(2ab)^2$

I guess this question came from Heron's formula. :)

[b]@ mathguy999:[/b] He just expanded $ (a\plus{}b)^2$ and $ (a\minus{}b)^2$. To expand, say $ (a\plus{}b)^2$, you can:
1. Write it as $ (a\plus{}b)(a\plus{}b)$
2. Use the distributive property to break up the first binomial: $ a(a\plus{}b)\plus{}b(a\plus{}b)$
3. Use the distributive property to break up the two remaining binomials: $ aa\plus{}ab\plus{}ba\plus{}bb$
4. Simplify: $ a^2\plus{}2ab\plus{}b^2$

The formula for the square of a sum is worth remembering.

[b]@ stephencheng:[/b] Good observation  :) To clarify for anyone who didn't get that the first time around:

Suppose we have a right triangle with legs $ a,b$ and hypotenuse $ c$. That is, $ a^2\plus{}b^2\equal{}c^2$.

It's area is $ \frac{ab}{2}$. But Heron's formula says the area is also:

$ \frac{1}{4} \sqrt{(a\plus{}b\plus{}c)(a\plus{}b\minus{}c)(a\minus{}b\plus{}c)(\minus{}a\plus{}b\plus{}c)}$

Equate, multiply by 4, and square.


Express -
$ 1\cdot a^0 \plus{} 2\cdot a^1 \plus{} 3\cdot a^2 \plus{} ....... \plus{} n\cdot a^{n \minus{} 1}$ as a general formula in $ n$.

[hide=solution ... and it works!!!!]
Let ,
$ A \equal{} \sum_{i \equal{} 1}^n ia^{i \minus{} 1}$
write $ 2A$ as,
$ 2A \equal{} \left[ \sum_{i \equal{} 1}^{n \minus{} 1} \left(ia^{i \minus{} 1} \plus{} (i \plus{} 1)a^i \right) \right] \plus{} na^{n \minus{} 1} \plus{} 1$

$ 2A \equal{} \left[ \sum_{i \equal{} 1}^{n \minus{} 1} \left(ia^{i \minus{} 1}(1 \plus{} a) \plus{} a^i \right) \right] \plus{} na^{n \minus{} 1} \plus{} 1$

$ 2A \equal{} (1 \plus{} a)\sum_{i \equal{} 1}^{n \minus{} 1} \left(ia^{i \minus{} 1}\right) \plus{} \sum_{i \equal{} 1}^{n \minus{} 1} a^i \plus{} na^{n \minus{} 1} \plus{} 1$

$ 2A \equal{} (1 \plus{} a)(A \minus{} na^{n \minus{} 1}) \plus{} \frac {a^{n \minus{} 1} \minus{} 1}{a \minus{} 1} \plus{} na^{n \minus{} 1}$

$ 2A \equal{} A(a \plus{} 1) \minus{} na^{n \minus{} 1} \minus{} na^n \plus{} na^{n \minus{} 1} \plus{} \frac {a^{n \minus{} 1} \minus{} 1}{a \minus{} 1}$

$ 2A \equal{} A(a \plus{} 1) \minus{} na^n \plus{} \frac {a^{n \minus{} 1} \minus{} 1}{a \minus{} 1}$

$ A(a \minus{} 1) \equal{} na^n \minus{} \frac {a^{n \minus{} 1} \minus{} 1}{a \minus{} 1}$

$ A(a \minus{} 1) \equal{} \frac {na^{n \plus{} 1} \minus{} (n \plus{} 1)a^n \plus{} 1}{a \minus{} 1}$

or,
$ A \equal{} \frac {na^{n \plus{} 1} \minus{} (n \plus{} 1)a^n \plus{} 1}{(a \minus{} 1)^2}$
done!!

[quote="TZF"]Here's a calculus solution, if you're interested:

\begin{eqnarray*} f_n (x) & = & \sum_{k = 1}^n k x^{k - 1} \\
& = & \frac {d}{dx} \left( \sum_{k = 1}^n x^k \right) \\
& = & \frac {d}{dx} \left( \frac {x^{n + 1} - x}{x - 1} \right) \\
& = & \frac {nx^{n + 1} - (n + 1)x^{n} + 1}{(x - 1)^{2}} \end{eqnarray*}[/quote]
 :( as much as i wish i can understand that i cant.
hoping one day i can do all this calculus stuff..

This Calculus Part isn't that hard. I understood it without have taken any form of formal training.
TZF integrated each of the terms individually since
$ \frac d{dx} x^k\equal{}k x^{k\minus{}1}$
and then derviated it back togther after finding the easy geometric sum

[quote="RobRoobiks"]This Calculus Part isn't that hard. I understood it without have taken any form of formal training.
TZF integrated each of the terms individually since
$ \frac d{dx} x^k \equal{} k x^{k \minus{} 1}$
and then derviated it back togther after finding the easy geometric sum[/quote]
ofcourse! i see you are right.. i apologize for my ignorance that made me omit anything in calculus... :oops:


IMO Shortlist
combinatorics unsolved
combinatorics

For any positive integer $n$, let $f_{1}(n)$ be the square of sum of the digits of $n$. Denote that $f_{k+1}(n)=f_{1}(f_{k}(n))$. Calculate $f_{2007}(2^{2006})$

I think this is not Hong Kong TST 1997, instead, it is 3rd Hong Kong TST 2007(As known as CHKMO 2007), Problem 2, which the problem can be find here: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=122164[/url]

BTW, this is a problem generate from 31st IMO shortlist  :P


[quote="Peter VDD"]You can just copy them as far as I know.[/quote]

Absolutely not!  We refuse to accept de facto the particular copyleft (viral) documentation policy of Wikipedia.  I've been told by many of their moderators that if they started over, they would choose a different documentation policy.

[quote="mysmartmouth"]Many Wikipedia articles are about the same topics we'd like to add.  Some information needs to be rewritten, however, some portions of the Wikipedia articles are perfect for AoPSWiki.  However, we obviously cannot copy them directly, and it would be pointless to sit there and paraphrase every line.  Is there any way we can integrate portions of Wikipedia, copy and cite, or something like that?  I am not talking about entire articles, but rather small portions that would contribute well to the AoPSWiki.  Perhaps I should contact Wikipedia...[/quote]

Use Wikipedia as a resource tool for appropriate topics.  AoPSWiki articles should wind up looking very different because it's meant not as an encyclopedia in the same sense -- the AoPSWiki's utility should be based on what is good for students -- and that often requires several levels of explanation for many topics.

[quote="Peter VDD"][quote="MCrawford"]Absolutely not!  We refuse to accept de facto the particular copyleft (viral) documentation policy of Wikipedia.  I've been told by many of their moderators that if they started over, they would choose a different documentation policy.[/quote] What do you mean exactly by this? It's not because we copy their content that we have to copy their policies, is it? :?

There is no way they can verify wether the topic author or someone else has submitted the info to the AoPS wiki... right?[/quote]

A wiki keeps a log of all content production, so any material copied can be traced to the starting point.

And yes, the Wikipedia viral copyleft license requires that reproduction of the material is free only the documentation license itself comes with it.


My first account was actually JWHooper, and I think one of the guy who runs this website just changed the password of the account so that I couldn't get in. Can one of you please give me the password (and I won't be a social retard from now on)? Please. I will act better, so please give me a chance.

Thanks,
J.

Related to http://www.mathlinks.ro/viewtopic.php?t=234400,you didn't want to be here anymore  :wink:

Also, they deleted your account.  If you click "Memberlist" at the top, and search JWHooper, it says "None found".

plus, you're not allowed to have multiple accounts
it says so in the agreement you agreed to when you made your account

Well now this is just hair splitting...
Who cares¿ Rules are there as a protection, not to annoy. Don't follow them word by word but get the meaning.
And his only other possibility would be to mail the admins.


I can prove it must be greater than $\ \frac{\sqrt 2}{2}$.

Edit: I can prove the original one too actually.

To prove it is greater than $\ \frac{\sqrt 2}{2}$ use AM-GM.  To prove the original problem statement use Cauchy.

[quote="xxxyyyy"][quote="Lovasz"]Prove:
\[ \sin x \sin y \sin z + \cos x \cos y \cos z \le 1  \][/quote]

what is x, y and z?[/quote]
They're arbitrary reals.


How do you bracket equations in a piece-wise function? I searched the LaTeX bar on the left but I couldn't find it.

Use array code. Here's an example:

$f(x)=\left\{\begin{array}{ccc}x+2&\text{if}& x\le0\\1&\text{if}&0<x<1\\3&\text{if}& x\ge 1\end{array}\right.$

That's a column for the values, a column for the "if", and a column for the condition. The big curly brace is \left\{ ; the \right. displays nothing, but indicates what to enclose.

Or \begin{cases} ... \end{cases}:

$f(x)=\begin{cases}x+2&\text{if}\ x\le0\\1&\text{if}\ 0<x<1\\3&\text{if}\ x\ge 1\end{cases}$

That's interesting- I'll have to remember that. Left-justified information in each column has it's uses.