Let $a_0, a_1, a_2, \ldots$ be an arbitrary infinite sequence of positive numbers.  Show that the inequality $1 + a_n > a_{n-1} \sqrt[n]{2}$ holds for infinitely many positive integers $n$.

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

We assume that the opposite inequality holds for all $n\ge n_0$.

Let $\ell=\liminf_{n\to\infty}a_n$. We have $\ell\ge 0$. We can find a subsequence $a_{k_n}\to \ell$ as $n\to\infty$, and we have $a_{k_n+1}\le a_{k_n}\sqrt[k_n+1]2-1$, but the RHS tends to $\ell-1$ as $n\to\infty$, so $a_{k_n+1}$ will eventually become $<\ell-\frac 12$, which is a contradiction.

[Edit: this only works for $\ell<\infty$]

:cool:  I'm not convinced !   What about  ${ \ell= +\infty} $  :cool:
 :D

Assume, as before, that for $n\ge n_0$ what we want to show doesn't hold anymore. We have $a_{n+k}\le S_k=a_n2^{H_{n+k}-H_n}-2^{H_{n+k}-H_{n+1}}-\ldots-2^{H_{n+k}-H_{n+k-1}}-1$, where $H_n=1+\frac 12+\ldots+\frac 1n$.

Let's fix $n$ and make $k\to\infty$. It would be enough to show that $S_k\to-\infty$ to obtain a contradiction. Clearly, $2^{H_{n+k}-H_n}\to\infty$ as $k\to\infty$, so it's enough to show that $\frac{2^{H_{n+k}-H_{n+1}}+\ldots+2^{H_{n+k}-H_{n+k-1}}+1}{2^{H_{n+k}-H_n}}\to\infty$ as $k\to\infty$ (remember, $n$ is fixed). We can multiply that fraction by $2^{H_n}$, which is constant, to get a nicer expression. We can also add the constant term $2^{-H_1}+\ldots+2^{-H_n}$, to get, in the end, $\sum_{t\ge 1} 2^{-H_t}=\infty\ (*)$ (this is what we want to prove). Since $\sum_{t\ge 1}2^{-\ln t}=\sum_{t\ge 1}\frac 1{t^{\ln 2}}=\infty$, we can use Cesaro-Stolz to prove $(*)$.

It looks rather nasty. Does it work?

Basic calculus shows that $ 2^{\frac{1}{n}}\le 1\plus{}\frac{1}{n}\equal{}\frac{n\plus{}1}{n}$ for all $ n\ge 1$.
Thus, we need to show there exists no sequence such that for all $ n\ge k$, $ 1\plus{}a_n\le a_{n\minus{}1}(2^\frac{1}{n})\le a_{n\minus{}1}\frac{n\plus{}1}{n}$.
Recursively, we can find that $ a_{n\minus{}1\plus{}m}\le a_{n\minus{}1}\frac{n\plus{}m}{n}\minus{}\frac{n\plus{}m}{n\plus{}1}\minus{}\frac{n\plus{}m}{n\plus{}2}\minus{}\cdots \minus{}\frac{n\plus{}m}{n\plus{}m}$.
It suffices, to get a contradiction, to show that the RHS is eventually negative, i.e. $ a_{n\minus{}1}\frac{n\plus{}m}{n}\le \frac{n\plus{}m}{n\plus{}1}\plus{}\cdots \plus{}\frac{n\plus{}m}{n\plus{}m}$ for some large finite $ m$.
Dividing out by $ n\plus{}m$ we get that $ \frac{a_{n\minus{}1}}{n}\le \frac{1}{n\plus{}1}\plus{}\cdots \frac{1}{n\plus{}m}$, which is true for some finite $ m$ by the divergence of the harmonic series.

[quote="GoldenFrog1618"]Basic calculus shows that $ 2^{\frac{1}{n}}\le 1\plus{}\frac{1}{n}\equal{}\frac{n\plus{}1}{n}$ for all $ n\ge 1$.[/quote]
Alternatively, $\left(1 + \frac{1}{n}\right)^n = \sum_{k=0}^n \frac{\binom{n}{k}}{n^k} = 2 + \sum_{k=2}^n \frac{\binom{n}{k}}{n^k} \geq 2$ (where $n < 2$ implies an empty sum.) :P

On the contrary we assume its true for finitely many $n$, hence after some finite terms (say $N$ ) it no longer holds.
We have $a_{N}\leq 2^{\frac{1}{N}}a_{N-1}-1\leq 2^{\frac{1}{2}}a_{N-1}-1$.
Hence we may safely assume a new sequence $b_i := a_{i+N}$.

We have $b_n\leq 2^{\frac{1}{n}}b_{n-1}-1$.
Let $S_n=2^{1+\cdots+\frac{1}{n-1}+\frac{1}{n}}+2^{\frac12+\cdots+\frac{1}{n-1}+\frac{1}{n}}+\cdots+2^{\frac{1}{n}}+1$.

Note that we have \[b_n\leq 2^{H_n}b_0-S_n\] I shall be proving that \[\frac{S_n}{2^{H_n}}>\frac{S_{n-1}}{2^{H_{n-1}}}\]\[\iff S_n>2^{\frac{1}{n}}S_{n-1}\] which is obviously true.

Thus $b_0-\frac{S_n}{2^{H_n}}<0$ after some time and we get a contradiction.

I solved this problem a long time before,probably this was my solution:

First of all,by Bernoulli inequality we have $(1+\frac{1}{n})^n \ge 2 \Rightarrow 1+\frac{1}{n} \ge 2^{\frac{1}{n}}$
Thus it is enough to show that $1+a_n > (1+\frac{1}{n})a_{n-1}$ for infinitely many $n$.

We assume just the opposite for the sake of contradiction,i.e,there exists an $M$ such that $\forall n \ge M$ we have $1+a_n \le (1+\frac{1}{n})a_{n-1}$ or $\frac{1}{n+1}+\frac{a_n}{n+1} \le \frac{a_{n-1}}{n}$.Summing this for $M,M+1,\cdots,m$ we obtain $\frac{a_{m-1}}{m} \le \frac{a_{M-1}}{M}-(\frac{1}{m}+\frac{1}{m+1}+\cdots+\frac{1}{M+1})$.But it is a well known consequence of AM-HM inequality that the harmonic sum on the RHS increases without bound,so letting $m \rightarrow \infty$ we get $a_{m-1}<0$.This is a contradiction since all the numbers in the sequence are positive.

As grobber pointed out, the problem is solved if we can show that $\sum_{k=1}^{\infty}(\frac{1}{2})^{H_k}$ diverges, where $H_k$ is the $k$th harmonic number. This can be done with some stupid bounds. We first give a upper bound for $H_k$. If $2^a \le i <2^{a+1}$ then $\frac{1}{i}\le \frac{1}{2^a}$ so $\sum_{i=2^a}^{2^{a+1}-1}\frac{1}{i} \le 1$. It follows that $H_{2^k-1} \le k$ for $k \ge 1$. Then if $2^{k-1} \le i < 2^k$ then $H_i \le H_{2^k-1} \le k$ so $(\frac{1}{2})^{H_i} \ge (\frac{1}{2})^{H_{2^k-1}} \ge (\frac{1}{2})^k$. So the sum is at least $\sum_{k=1}^{\infty}2^{k-1}(\frac{1}{2})^k=\sum_{k=1}^{\infty}\frac{1}{2}$ which obviously diverges.

The problem seems a bit easy so probably my solution is wrong ...
Do check and point out :)

[b]Solution :[/b]
Let there be finitely many such $n$.
Let $m-1$ be the largest integer such that $1+a_{m-1} > a_{m-2} \sqrt[m-1]{2}$
\[\implies  1+a_n \le a_{n-1}\sqrt[n]{2} \]
$\forall n \ge m$

\[1+a_m \le a_{m-1}\sqrt[m]{2} \]

\[1+a_{m+1} \le a_{m}\sqrt[m+1]{2} \]
\[ \cdots\]
\[1+a_{m+k} \le a_{m+k-1}\sqrt[m+k]{2} \]

Adding all of them,

\[k+\sum _{i=m}^{m+k}a_i \le \sum _{i=m}^{m+k}a_{i-1}\sqrt[i]{2}\]

But this is clearly not possible for a sufficiently large choice of $k$ and since $\sqrt[j]{2}<\sqrt{2}$ (use AM-GM)

Thus a contradiction.

Hence there must exist infinitely many such $n$

[quote="utkarshgupta"]
But this is clearly not possible since $\sqrt[j]{2}<1$ and $a_{m-1}<k$
[/quote]

Nope. First,  there doesn't have to exist such  $m$ (if the inequality holds for all $n$), but more importantly: $\sqrt[j]{2}>1$!

I assumed there are finitely many such $n$ so there must exist a largest ...
And as for the second mistake, i have corrected it so please relook :)
Thanks :)

Note that by Bernoulli's inequality,
\[
(1+\frac{1}{n})^{n}\ge 2\Rightarrow \sqrt[n]{2}\le 1+\frac{1}{n}
\]
for each $n\ge 1$. Now suppose that there is some $m$ with $a_{n}\le a_{n-1}\sqrt[n]{2}-1$ for all $n>m$. By our previous estimate, this implies
\[
a_{n}\le a_{n-1}\frac{n+1}{n}-1
\]
for each $n>m$. So
\begin{align*}
a_{m+1}&\le a_{m}\frac{m+2}{m+1}-1\\
\Rightarrow a_{m+2}&\le a_{m+1}\frac{m+3}{m+2}-1\le a_m\frac{m+3}{m+1}-\frac{m+3}{m+2}-1\\
\Rightarrow a_{m+3}&\le a_{m+2}\frac{m+4}{m+3}-1\le a_m\frac{m+4}{m+1}-\frac{m+4}{m+2}-\frac{m+4}{m+3}-1
\end{align*}
and in general, for each $k\ge 1$,
\[
a_{m+k}\le (m+k+1)\left(\frac{a_m}{m+1}-\frac{1}{m+2}-\frac{1}{m+3}-\cdots - \frac{1}{m+k+1}\right).
\]
But as $k\to \infty$, the second term tends to $O(1)-O(\log k)$, which is a contradiction if $a_k>0$ always.

Suppose that there existed an integer $N$ such that for all $n>N$ the inequality was false. Let $b_n=a_n+\sum_{i=1}^{n}2^{H_n-H_i}$, where $H_k=\sum_{i=1}^{k}\frac{1}{i}$. Note that the condition now becomes $b_n=2^\frac{1}{n}b_{n-1}$, so we have that $b_m=2^{H_m-H_N}b_N=c2^{H_m}$ for some constant $c$, so $a_m=c2^{H_m}-\sum_{i=1}^{m}2^{H_m-H_i}$. We claim that $\lim_{m\rightarrow\infty}\frac{\sum_{i=1}^{m}2^{H_m-H_i}}{2^{H_m}}=\infty$, which will imply that for some $m$ $\frac{\sum_{i=1}^{m}2^{H_m-H_i}}{2^{H_m}}>c$ so $a_m<0$. But this is equivalent to $\sum2^{-H_i}$ diverging. Since $H_i<\log i$ by integration, $\sum2^{-H_i}>\sum2^{-\log i}=\sum e^{-\log 2\log i}=\sum i^{-\log 2}>\sum i^{-1}$ which diverges as desired.

Routine Approximations:

Suppose for sake of contradiction that for all $m\ge n$, we have $1+a_m\le a_{m-1}\sqrt[m]{2}$.

[u]Lemma 1:[/u] We have $a_{n+N}\le a_n2^{\frac{1}{n+1}+\cdots+\frac{1}{n+N}}-2^{\frac{1}{n+2}+\cdots+\frac{1}{n+N}}-\cdots-2^{\frac{1}{n+N}}-1$.

[u]Proof of Lemma 1:[/u] Iterate the inequality $a_m>a_{m-1}2^{1/m}-1$ $N$ times. For small values of $N$, we have
\begin{align*}
a_{n+1}&\le a_n 2^{\frac{1}{n+1}}-1 \\
a_{n+2}&\le\left(a_n 2^{\frac{1}{n+1}}-1\right)2^{\frac{1}{n+2}}-1 \\
a_{n+3}&\le\left[\left(a_n 2^{\frac{1}{n+1}}-1\right)2^{\frac{1}{n+2}}-1\right]2^{\frac{1}{n+3}}-1 \\
&\,\,\vdots
\end{align*}

[u]Proof of Lemma 2:[/u] $\log(b/a)-1<\frac{1}{a+1}+\cdots+\frac{1}{b}<\log(b/a)+1$ for positive integers $a,b$.

[u]Proof of Lemma 2:[/u] Follows from bounding $\int_a^b\frac{1}{x}\,dx$ by left and right Riemann sums.

By applying lemma 1 and lemma 2 many times, we see that
\begin{align*}
a_{n+N} &\le a_n 2^{\log\left(\frac{n+N}{n}\right)+1}-\left[2^{\log\left(\frac{n+N}{n+1}\right)-1}+\cdots+2^{\log\left(\frac{n+N}{n+N}\right)-1}\right] \\
&= (n+N)^{\log 2}\left[\frac{2a_n}{n^{\log 2}}-\frac{1}{2}\left[(n+1)^{-\log 2}+\cdots+(n+N)^{-\log 2}\right]\right].
\end{align*}
But we know that $a_{n+N}>0$, so we must have that
\[(n+1)^{-\log 2}+\cdots+(n+N)^{-\log 2}<\frac{4a_n}{n^{\log 2}}\]
for all $N$. But Since $\log 2<1$, we then have that the left side diverges to $+\infty$ as $N\to\infty$ by series comparison with $(n+1)^{-1}+(n+2)^{-1}+\cdots$. Therefore, it cannot be bounded above, so we have the desired contradiction.

Really good problem.

Assume by way of contradiction that $N$ was the largest value for which this holds. Then for all $k> N$, we have $1+a_k\le a_{k-1}\sqrt[k]{2}$. First, we may show by induction that $$1+a_{N+m}\le a_N2^{1/(N+1)+\cdots +1/(N+m)}-\left(2^{1/(N+2)+\cdots+1/(N+m)}+2^{1/(N+3)+\cdots+1/(N+m)}+\cdots +2^{1/(N+m)}\right)$$
Therefore, the RHS is greater than $1$. This implies
\begin{align*}
a_N &>\frac{ 1+2^{1/(N+2)+\cdots+1/(N+m)}+2^{1/(N+3)+\cdots+1/(N+m)}+\cdots +2^{1/(N+m)} }{2^{1/(N+1)+\cdots +1/(N+m)}}\\
&=\frac{1}{2^{1/(N+1)}}+\frac{1}{2^{1/(N+1)+1/(N+2)}}+\cdots+\frac{ 1 }{2^{1/(N+1)+\cdots +1/(N+m)}}
\end{align*}
Standard calculus techniques show that $$\frac{1}{a}+\frac{1}{a+1}+\cdots +\frac{1}{b}\le 1+\ln\left(\frac{b}{a}\right)\le 1+\log_2\left(\frac{b}{a}\right).$$
Therefore,
\begin{align*}
a_N &>\frac{1}{2^{1/(N+1)}}+\frac{1}{2^{1/(N+1)+1/(N+2)}}+\cdots+\frac{ 1 }{2^{1/(N+1)+\cdots +1/(N+m)}}\\
&\ge \frac{N+1}{2(N+1)}+\frac{N+1}{2(N+2)}+\cdots+\frac{N+1}{2(N+m)}\\
&= \frac{N+1}{2}\left(\frac{1}{N+1}+\frac{1}{N+2}+\cdots+\frac{1}{N+m}\right)\end{align*}
But as $m$ approaches infinity, the RHS diverges, contradiction since $a_N<\infty$.

For contradiction, $\exists N>0$ such that $ 1+a_n>a_{n-1} \sqrt[n]{2} \forall n \ge N$
We have $$ 1+a_n \le 2^{\frac{1}{n}}a_{n-1} \forall n le N$$
Let $(x_n)$ such that $$\left\{ \begin{matrix}
   {{x}_{1}}=1  \\
   {{x}_{n}}={{2}^{\frac{-1}{2}-\frac{-1}{3}-.....-\frac{1}{n}}},\forall \,n\ge 2  \\
\end{matrix} \right.$$
Then, we have $$ x_n+x_na_n \le x_{n-1}a_{n-1}, \forall n \ge N$$
So $$ x_{N-1}a_{N-1} \ge x_{N+t}a_{N+t}+x_N+x_{N+1}+\ldots+x_{N+t} (1). $$
We have $x_N+x_{N+1}+\ldots+x_{N+t} \to +\infty$ when $t\to\infty$, contradiction

Suppose for the sake of contradiction there exists some $N$ such that for all $n\ge N$, we have $1+a_n \le 2^{1/n}\cdot a_{n-1}$. It is easy to show by induction that for all $k\ge 0$, we have
\begin{align*}
    a_{N+k} &\le \left( 2^{ \tfrac{1}{N} + \tfrac{1}{N+1} + \cdots + \tfrac{1}{N+k}}\right)a_{N-1} - \left( 1 + \sum_{i=1}^k 2^{ \tfrac{1}{N+i} + \cdots + \tfrac{1}{N+k}} \right)
\end{align*}
Let $C=a_{N-1}$ be a constant. We want to find some $k$ for which $a_{N+k}$ is negative, as this produces a contradiction. By the above, $a_{N+k}<0$ if
\[ 1 + \sum_{i=1}^k 2^{ \tfrac{1}{N+i} + \cdots + \tfrac{1}{N+k}} > C\cdot 2^{ \tfrac{1}{N} + \tfrac{1}{N+1} + \cdots + \tfrac{1}{N+k}}. \qquad (\heartsuit) \]
We will find a $k\le N$ satisfying the above.
For $0<\varepsilon<1$, one can show $2^\varepsilon >1+0.01\varepsilon$. Since $\tfrac{1}{N}+\cdots+\tfrac{1}{N+k}<1$ for $k\le N$, we can lower bound the LHS of $(\heartsuit)$:
\begin{align*}
 1 + \sum_{i=1}^k 2^{ \tfrac{1}{N+i} + \cdots + \tfrac{1}{N+k}} &> (k+1) + 0.01\sum_{i=1}^k \left( \frac{1}{N+i} + \cdots + \frac{1}{N+k} \right) \\
 &= (k+1) + 0.01\left[ \frac{1}{N+1}+\frac{2}{N+2}+\cdots+\frac{k}{N+k} \right] \\
 &= (1.01k+1) - 0.01N\left[\frac{1}{N+1} + \frac{1}{N+2}+\cdots + \frac{1}{N+k} \right].
\end{align*}
Similarly, for $0<\varepsilon<1$, one can show that $2^\varepsilon < 1+\varepsilon$, so we can upper bound the RHS of $(\heartsuit)$:
\[ C\cdot 2^{ \tfrac{1}{N} + \tfrac{1}{N+1} + \cdots + \tfrac{1}{N+k}} < C \left( 1 + \frac{1}{N} + \frac{1}{N+1} +\cdots+\frac{1}{N+k}\right). \]
Therefore, it suffices to find a $k\le N$ for which
\[ (1.01k+1) - 0.01N\left[\frac{1}{N+1} +\cdots + \frac{1}{N+k} \right] > C \left( 1 + \frac{1}{N} +\cdots+\frac{1}{N+k}\right),\]
i.e.
\begin{align*}
    1.01k+1-C-\frac{C}{N} &> (C+0.01N)\left( \frac{1}{N+1} +\cdots + \frac{1}{N+k} \right) \\
    &= (C+0.01N)(H_{N+k}-H_N).
\end{align*}
Take $k=N$. Then $H_{N+k}-H_N = H_{2N}-H_N < \ln(2)$, and since $1.01N-C-C/N > (C+0.01N)\ln(2)$ for large enough $N$, we are done.

[hide=Remarks]
The main idea is to write out what $a_{N+k}$ would be, repeatedly using the contrapositive bound. Then, logically reducing and making estimates along the way[i] after writing out the formula[/i] is not too hard. Note that $0.01$ could be replaced with any number smaller than $\ln(2)$.
[/hide]

Fakesolve or not? I can't find an error, but the other solutions seem longer and I can't find a similar solution to mine.

Suppose that the conclusion is not true. Then there exists a positive integer $N$ such that for all $n>N$, $1 + a_n \le a_{n-1} \sqrt[n]{2}$.

Define a new sequence $b_n = \frac{a_n}{n}$, and notice that this is always positive. By Bernoulli's inequality $\left(1+\frac{1}{n}\right)^n>2$ and thus $1+\frac{1}{n} > \sqrt[n]{2}$. Then notice that for all $n>N$, \begin{align*} b_n &= \frac{a_n}{n} \\ & \le \frac{a_{n-1}\sqrt[n]{2}-1}{n} \\ & = \frac{(n-1)b_{n-1}\sqrt[n]{2}-1}{n} \\ & <\frac{(n-1)b_{n-1}\left(1+\frac{1}{n}\right)-1}{n} \\ &= \frac{\frac{n^2-1}{n}b_{n-1}-1}{n} \\ &= b_{n-1} - \frac{b_{n-1}}{n^2}-\frac{1}{n} \\ &\le b_{n-1}-\frac{1}{n}.\end{align*} This implies that for $x>N$, $$b_x < b_N - \sum_{i=N+1}^x \frac{1}{i}.$$ As the harmonic series diverges, for sufficiently large $x$, $b_x$ will be negative, contradiction.

Just like #21, this seems a lot shorter (in idea, not actual post length) compared to the others on the thread. The problem is almost two separate problems put together but the conclusion is relatively cool nevertheless.
[hide=Solution]
Let $\mathbb{H}_k = \sum_{i=1}^{k} \frac{1}{i}$ for brevity. \\
Assume for the sake of contradiction that for all $n \geq N_0, n \in \mathbb{N}$, $1+a_n \leq a_{n-1} \cdot 2^{\frac{1}{n}}$. \\

$\textbf{Lemma 1:}$ \\
$$a_n > \sum_{i=n+1}^{n+k} 2^{-(\mathbb{H}_i-\mathbb{H}_n)}$$ for all $k \in \mathbb{N}$ and all $n \geq N_0-1$, $n \in \mathbb{N}$ \\
$\textbf{Proof)}$ \\
We will prove the lemma by inducting on $n$ and $n+1$ simultaneously, the base case is immediate. \\
Assume that the above assertion holds for some $k \in \mathbb{N}$ for all $n \geq N_0$, then notice that $$a_{n-1} \geq 2^{-\frac{1}{n}} \cdot (1+a_n) > 2^{-\frac{1}{n}} \cdot (1+ \sum_{i=n+1}^{n+k} 2^{\mathbb{H}_i-\mathbb{H}_n}) = \sum_{i=n}^{n+k} 2^{\mathbb{H}_i-\mathbb{H}_n}$$
which proves the assertion for $k+1$ as well as for all $n \geq N_0$ due to our initial assumption that for all $n \geq N_0$, $a_{n-1} \cdot 2^{\frac{1}{n}}$, completing the induction and concluding the lemma. $\blacksquare$ \\

The remaining part is routine analysis. We shall prove the following lemma which will allow us to show that for all $C \in \mathbb{R}^{+}$, $a_n > C$, a clear contradiction. \\

$\textbf{Remark:}$ We will be multiplying the inequality in $\textbf{Lemma 1}$ by $2^{-\mathbb{H}_n}$, the divergence of the sum will not change. \\

$\textbf{Lemma 2:}$ \\
For all $C \in \mathbb{R}^{+}$ there exists some $k \in \mathbb{N}$ such that
$$ \sum_{i=1}^{k} 2^{-\mathbb{H}_n} > C$$
$\textbf{Proof)}$ \\
Notice that for all $T \in \mathbb{N}$ $$T \geq \mathbb{H}_{2^{T-1}}, \cdots \mathbb{H}_{2^T-1}$$ by induction as $$\sum_{i=2{T-1}}^{2^T-1} \mathbb{H}_i < 2^{T-1}$$ and consequently $$2^{-\mathbb{H}_n}>2^{-(log_2(n)+1)} = 2^{-log_2(2n)} = \frac{1}{2^{log_2(2n)}} = \frac{1}{2n}$$

Now, take some $n \in \mathbb{N}$ such that $\mathbb{H}_n > 2C$ which is possible by the divergence of the harmonic series, then
$$\sum_{i=1}^{k} 2^{-\mathbb{H}_n} > \frac{1}{2} \cdot \mathbb{H}_n > C$$ which immediately allows us to complete the proof by $\textbf{Lemma 1}$. $\blacksquare$ \\
[/hide]

Assume that there exists a maximum $n$ for which the inequality holds. For all $n\ge k, 1 + a_n \le a_{n-1}\sqrt[n]{2}$ Note that by Bernoulli's inequality, $\sqrt[n]{2}\le 1+\frac{1}{n},$ thus, \[1+a_n\le \frac{n+1}{n}a_{n-1}\]for all $n\ge k.$

Then, \[a_{k+i}\le \frac{k+i+1}{k}a_{k-1}-\frac{k+i+1}{k+1}-\frac{k+i+1}{k+2}-\frac{k+i+1}{k+3}-\ldots-\frac{k+i+1}{k+i+1}\]
We claim that for sufficiently large $i$ this number is negative.

Note that this is equivalent to proving that $\frac{1}{k+1}+\frac{1}{k+2}+\dots+\frac{1}{k+i+1}$ diverges, which is true.

I think you may have posted this in the wrong thread.

Assume for the sake of contradiction that the inequality is only statisfied for finitely many $n$. That is, there exists an index $N$ such that $$a_n \leq a_{n-1}\sqrt[n]{2}-1$$ for all $n\geq N$. Note that $\sqrt[n]{2}\leq 1+\frac{1}{n}$, so we have $a_n \leq a_{n-1}(1+\frac{1}{n})-1$ for all $n\geq N$. We then have $a_{N+1}\leq 2a_N-1$, $a_{N+2}\leq 3a_N-\frac{5}{2}$, $a_{N+3}\leq 4a_N-\frac{26}{6}$, and so on. In general, we have $a_{N+k}\leq (k+1)a_N-\frac{S_k}{k!},$ where $S_k$ denotes the $k$th Sterling number. This can be shown by induction, as that term multiplies by $\frac{k+1}{k}$ and subtracts 1 each time. It is well known that $\frac{S_k}{k!}$ eventually dominates any linear function, so this eventually becomes negative, contradiction.

Assume not. Then there is a number $k$ such that $$a_n \le a_{n-1} \sqrt[n]{2} -1,$$ for all $n \ge k$. Using the binomial theorem, we get $\sqrt[n]{2} \le 1 + \frac{1}{n}$, so $a_n \le a_{n-1} \left( 1 + \frac{1}{n} \right) -1$ for all $n \ge k$. Let $b_n := \frac{a_n}{n+1}$. Using this notation we get $$b_n \le b_{n-1}-\frac{1}{n+1} \implies 0 \le b_n \le b_k -\frac{1}{k+2}-\frac{1}{k+3} - \cdots - \frac{1}{n+1} \implies \frac{1}{k+2}+\frac{1}{k+3} + \cdots +\frac{1}{n+1} \le b_k.$$ We know that the harmonic series is unbounded and so taking big enough $n$ produces a contradiction and we are done.

Suppose otherwise. Then there exists some constant $c$ such that $1+a_n\le a_{n-1}\sqrt[n]{2}$ for all $n\ge c$. Fix $a_c$ and consider the sequence $b_c,b_{c+1},\dots$ such that $b_c=a_c$ and $1+b_n=b_{n-1}\sqrt[n]{2}$. I claim that $b_n\ge a_n$. We show this by induction. Clearly the base case $n=c$ holds. Then, suppose that $b_n\ge a_n$. Then $a_{n+1}\le a_n\sqrt[n+1]{2}-1\le b_n\sqrt[n+1]{2}-1=b_{n+1}$.

Now note that it suffices to prove the case where $c=0$; for all other cases, we can simply "work backwards" until we reach the $c=0$ case by constructing the terms $b_{c-1},b_{c-2},\dots$. Associate with each term $b_n$ a pair of numbers $c_{n,1}$ and $c_{n,2}$ such that $b_n=c_{n,1}b_0-c_{n,2}$. Then note that it suffices to show that $\frac{c_{n,2}}{c_{n,1}}$ is unbounded to achieve a contradiction. Note that
\[c_{n,1}=2^{\tfrac{1}{1}+\tfrac{1}{2}+\dots+\tfrac{1}{n}}\]
\[c_{n,2}=2^{\tfrac{1}{2}+\dots+\tfrac{1}{n}}+2^{\tfrac{1}{3}+\dots+\tfrac{1}{n}}+\dots+2^{\tfrac{1}{n}}+2^0\]
so it is enough to show that
\[\frac{1}{2^{\tfrac{1}{1}}}+\frac{1}{2^{\tfrac{1}{1}+\tfrac{1}{2}}}+\dots+\frac{1}{2^{\tfrac{1}{1}+\tfrac{1}{2}+\dots+\tfrac{1}{n}}}\]
is unbounded. Then it suffices to show that
\[\sqrt[n]{2}\le 1+\frac{1}{n}\]
since this will imply by induction that
\[2^{\tfrac{1}{1}+\tfrac{1}{2}+\dots+\tfrac{1}{n}}\le n+1.\]
But now by Bernoulli's Inequality we are done. $\blacksquare$

Suppose the problem is not true, then there exists $N\in \mathbb{N}$ such that for all $n>N$ we have $1+a_n\leq a_{n-1}\sqrt[n]{2}$
By an induction we can show that $a_{N+k}\leq a_N2^{\frac{1}{N+k}+...+\frac{1}{N+1}}-\sum_{i=0}^{k}2^{\frac{1}{N+k}+...+\frac{1}{N+k+1-i}}$
Let $H_n=\sum_{i=1}^{n}\frac{1}{i}$ be the $n$-th harmonic number ($H_0=0$)
Then $a_{N+k}\leq a_N2^{H_{N+k}-H_N}-\sum_{i=0}^{k}2^{H_{N+k}-H_{N+i}}=f(k)$
Using inequality $H_n<\ln n+1$ we get
$f(k)=2^{H_{N+k}}(\frac{a_N}{2^{H_N}}-\sum_{i=0}^{k}\frac{1}{2^{H_{N+i}}})<2^{H_{N+k}}(\frac{a_N}{2^{H_N}}-\sum_{i=0}^{k}\frac{1}{2^{\ln(N+i)+1}})=2^{H_{N+k}}(\frac{a_N}{2^{H_N}}-\frac{e}{2}\sum_{i=0}^{k}\frac{1}{N+i})=2^{H_{N+k}}(\frac{a_N}{2^{H_N}}-\frac{e}{2}(H_{N+k}-H_{N-1}))$
We want $f(k)<0$ so we want $\frac{a_N}{2^{H_N}}-\frac{e}{2}(H_{N+k}-H_{N-1})<0$
This is equivalent to $H_{N+k}>\frac{2a_N}{2^{H_N}e}+H_{N-1}$ but this holds true for sufficiently large $k$
Hence for sufficiently large $k$ we have $a_{N+k}<0$, contradiction

It's obvious that $(1+\frac{1}{n})^{n}\ge 2\implies\sqrt[n]{2}\le 1+\frac{1}{n}$. Then, take the largest number $a_l$ s.t. the inequality still holds (FTSOC). It follows that $$a_{l+1}\le\sqrt[l]{2}a_l-1\le a_l\frac{l+2}{l+1}-1\implies a_{l+2}\le a_{l+1}\frac{l+3}{l+2}-1\le a_l\frac{l+3}{l+1}-\frac{l+3}{l+2}-1,$$ which, by a simple induction, implies $$a_{l+k}\le(l+k+1)\left(\frac{a_l}{l+1}-\frac{1}{l+2}-\frac{1}{l+3}-\dots-\frac{1}{l+k+1}\right);$$ in particular, for sufficiently large k note that 1/1+1/2+...+1/n is unbounded, so even without the 1/1+1/2+...+1/{l+1} this sequence tends to infinity, which implies $\left(\frac{a_l}{l+1}-\frac{1}{l+2}-\frac{1}{l+3}-\dots-\frac{1}{l+k+1}\right)$ becomes negative, but this is a contradiction because all terms in the sequence are positive. $\blacksquare$

this should work but idk those types of wonky stuff im notrelaly good at

also the induction is provided for detail: Base case is done, inductive step: $$a_{l+k+1}\le \frac{l+k+2}{l+k+1}a_{l+k}-1\le(l+k+2)\left(\frac{a_l}{l+1}-\frac{1}{l+2}-\frac{1}{l+3}-\dots-\frac{1}{l+k+1}-\frac1{l+k+2}\right)$$


$A,P,Q,R,S$ are distinct points on a circle such that angle PAQ = angle RAS.Then show that $AR(AP+AR ) = AQ(AQ+AS)$

This is not true. $<PAQ=<QAR=<RAS$ is missing.
In that case, $PQ=QR=RS=a$, $PR=QS=b$ and, because Ptolomeo's theorem, we have:
$AR(AP+AR)a=AR.AQ.b=AQ(AQ+AS)a$


These are bonus problems from my math class which I found especially intruiguing (we recently started on combinatorics and set stuff).  I want to see some other solutions, though.  Just to reassure people, I've already solved them, written up solutions, and am going to turn them in tomorrow (Mods, if any sort of proof is needed, I can post up my solutions first).  I would post up the third, but I have to solve it first ;)  :D

1. Prove that \[ \sum\limits_{x=0}^{n}(\binom{n}{x}^2)=\binom{2n}{n}  \]

[hide="hint"]Prove #2 first[/hide]

2. Prove that \[ \binom{a}{b}=\sum\limits_{x=0}^{n}{\binom{a-n}{b-n+x}\binom{n}{x}}  \] for $a \ge b+n \ge b \ge n$

you sure about the first one?

\[ \sum\limits_{x=0}^n\binom{n}{x} = \binom{n}{0} + \binom{n}{1} + \cdots \binom{n}{n} = 2^n \]

which can be proven through binomial expansion of $(1 + 1)^{n}$

He probably means the square of each of those terms.

I agree with agolsme, I think you forgot something, especially with your hint to #2 which has products of 2 combinations.

On an unrelated note, the other day I just played around with $\binom{2n}{n}$ and found out $\binom{2n}{n} = \prod_{k=1}^n \frac{4k - 2}{k}$, which I think is pretty interesting.  I wonder if you can play with the RHS to get a combinatorics thing, which might be the proper LHS of your equation.

Yeah i forgot to post that...I thought I did put in the $^2$...i guess it got edited out or something.

For the first one, we can incorporate the binomial theorem:

[hide]

Find the coefficient n in the expansion of $(x+1)^{2n}$ :

$(x+1)^n(x+1)^n$

Pairing up the coefficients from both factors (e.g. x^1 * x^{n-1}), we get:

$= \binom{n}{0}\binom{n}{n} + \binom{n}{1}\binom{n}{n-1} + ... + \binom{n}{n}\binom{n}{0}$

$= \binom{n}{0}\binom{n}{0} + \binom{n}{1}\binom{n}{1} + ... + \binom{n}{n}\binom{n}{n}$

$= \sum_{k=0}^{n}[\binom{n}{k}]^2$

[/hide]


To determine whether an integer N is prime, it is sufficient to check whether  it has any prime factors. Not all the primes less than N need to be checked. What is the largest prime which must be checked out determine if 1009 is a prime.

I have to go to school but it would be 31. because you need to find the closest sqare (1024) which sqare root is 32. The closest prime lower than 32 is 31. So you must check primes up to 31

You only have to check up to the square root of a number because after the square root, it will just be numbers times a number already tested.  The square root of 1009 is closest to 32 (1024).  The greatest prime less than 32 is 31.


i asked myself a coupple of questions i don't have a clue how to answer. all help is very apriciated. i however don't know if what i conjecture is true or even if it is prooven.

if we look at the decimal representation of $ \sqrt {2}$ for example we see that every digit from $ 0$ to $ 9$ occurs. same is with square root of $ 3,5,6,7,8,10...$.

prove that in a decimal representation of $ \sqrt {n}$ (where $ n$ is a natural number and not a perfect square) every digit occurs infinitly many times.

(furthermore if upper is easy and it's just me who doesn't see the answer.. is it true for every algebraic number and for every number system, not just decimal? is it true not just for digits but for longer sequences of digits? can anyone come out with an inteligent real counterexmple (better than mine $ 0,12112111211112111112...$))

tnx for ur help

See [url]http://mathworld.wolfram.com/NormalNumber.html[/url]. I guess Issac's paper on the base 2 normality of $ \sqrt{n}$ is the most relevant piece of information.


Lines l and m are parallel lines 8 centimeters apart, and point P is
on line m. What is the total number of points that are equidistant from lines l and m and 5 centimeters from P?

all the points that are equidistant from the 2 lines is just a line 4 from each of the other lines, so 5 away would make a circle with r=5, so there would be
2 points

Is there a nice, easy website that explains equidistant questions?  I need to review this stuff.

one thing that helps is to draw out the situation


I recently deleted a blog and I want to create a new one, but all I can do is set the blog options. After I press submit, it says something like "must choose blog to edit", but I have no blog.

Re added your old blog. Dont delete it again.


There is a chessboard (8x8). If cut out two black corners of it. Can we cover it with  dominoes (2 celled dominos).And why?

[hide="Classic"]Every domino covers one white square and one black square.
From the beginning the number of black squares is not equal to the number of white squares, so it will be impossible to cover it.[/hide]  :P

You might want to hide your answers in the feature :wink:

[hide]We know that each domino can cover two squares, and no two adjacent squares are the same color. This means that each domino covers 1 black square and 1 white square. Because two black corners were cut out, there are 32 white squares and 30 black squares, and so it will be impossible to cover the entire board. [/hide]

shouldn't htis be in the brainteaser forum :P


Find all functions $ f: \mathbb{Z}\rightarrow \mathbb{Z}$ such that $ f(x\plus{}y)\plus{}f(xy)\equal{}f(x)f(y)\plus{}1$.
i think that $ \mathbb{Z}$ could be replaced with $ \mathbb{R}$ in my solution (in the hide), help me please

[hide]We have $ x\equal{}y\equal{}0\Rightarrow f(0)\equal{}1$. If $ y\equal{}1$ then $ f(x\plus{}1)\plus{}f(x)\equal{}f(x)f(1)\plus{}1$. If $ y\equal{}\minus{}x$ then $ f(\minus{}x^2)\equal{}f(x)f(\minus{}x)$ replacing $ x\equal{}1$ we have $ f(1)\equal{}1$. Then $ f(x\plus{}1)\plus{}f(x)\equal{}f(x)\cdot 1\plus{}1\Rightarrow f(x)\equal{}1$.[/hide]

[quote="pelao_malo"]If $ y \equal{} \minus{} x$ then $ f( \minus{} x^2) \equal{} f(x)f( \minus{} x)$ replacing $ x \equal{} 1$ we have $ f(1) \equal{} 1$.[/quote]
Or $ f( \minus{} 1) \equal{} 0$.  ;)  This problem has at least one non-constant solution, [hide="which is given by..."] $ f(x) \equal{} x \plus{} 1$. [/hide]


LaTeX
number theory
greatest common divisor
number theory unsolved

Yes.. its ok... and on the same test...
this is good:
"Find all solutions with m and n naturals: $n\cdot 2^{n-1} +1 = m^2$ "

[size=84][Admin Edit: Sometimes LaTeX makes equations look nicer ;)][/size]

I hope Pierre won't be angry with me if I answer in the same topic  ;)

n=0 is a solution, n=1 or 2 are not. Let's suppose n > 2.

n*2^(n-1) = (m-1)*(m+1) gives m = 2^(n-2)(2M+1) - 1 or m = 2^(n-2)(2M+1) + 1 since 4 does not divide gcd (m-1,m+1).

* If m = 2^(n-2)(2M+1) - 1 equation becomes n = (2M+1)(2^(n-3)(2M+1)-1)

But (2M+1)(2^(n-3)(2M+1)-1) >= 2^(n-3) - 1 so n < 6

* Case m = 2^(n-2)(2M+1) + 1 is similar : n = (2M+1)(2^(n-3)(2M+1)+1)

Checking "by hand" the case n=3..5 gives only 5 as solution.

So only 0,1 and 5,9 are solutions.

He he  :D  I discovered that moderators have the power to split topics....

Pierre.


Someone installs a new keyboard driver on Johnny's computer. His keyboard now works that each letter key on his keyboard inputs a different letter than what is labelled. So, for example, if he tries to type "hello world" normally he might end up typing "plzzx qxdzi" instead.

Rather than figure out what's wrong with his keyboard, he decides to type his homework normally, printing out the gibberish that results, and then typing in the gibberish, printing that out, etc., until the proper text comes out.

(1) Prove that eventually the proper text will appear,
(2) What is the maximum number of times he needs to type his homework?

Note: The English language has 26 letters in it.

I assume you mean for every letter to be used once -- otherwise, all the letters except P could map to Q and Q could map to P.

[hide="In other words..."] The [url=http://en.wikipedia.org/wiki/Permutation_group]permutation group[/url] on $ 26$ elements consists of elements of finite order.  [/hide]

Note: if you have a broken keyboard like that, it is much time-efficient to repair it than go through the (long) process of iteratively typing gibberish. And what if you make a mistake? :D

[hide="Part I outline"]Each letter will map to another letter. Since there are 26 letters in the alphabet, and no letter maps to itself, every letter will have to map to another letter. Also, no two letters may map to the same letter, as described in the problem. Therefore, by typing in a letter, and then typing in the letter that was outputted, one would have mapped the function twice. Likewise, if one kept doing so, they would eventually get mapped to the letter that the key represents.[/hide]

You haven't really proved anything.  The approach you suggest is [hide="essentially..."] showing that a permutation is a product of [url=http://mathworld.wolfram.com/PermutationCycle.html]cycles[/url]. [/hide]
This is actually more effort than necessary.  The fundamental point of the general proof [hide="is that..."] permutations are invertible.  This allows you to make the following observation:

Pick a letter in a particular position in his homework; call it $ x$.  Let $ P(x)$ be the letter to which the $ x$ key has been mapped.  Among $ \{ x, P(x), ... P^{26}(x) \}$ at least two letters must be the same.  What does this imply? [/hide]

If there are 3, 5, 7, and 11-cycles of letters being permuted, (Ex: A->B->C->A would be a 3-cycle), would that be 1155 iterations until he finally gets it right, and possibly the maximum?

(Edit:)  I was confused.   :blush:  The order of the permutation group is $ n!$.

Oh, well I just meant separate cycles like A-B-C (3), D-E-F-G-H (5), I-J-K-L-M-N-O (7), and P-Q-R-S-T-U-V-W-X-Y-Z (11). This would include all 26 letters. And since all of them are prime, wouldn't it take longer for the completely correct text to appear? (Though some of the letters may be right at a certain point in time.)


geometry
circumcircle
power of a point
radical axis
geometry proposed

ABC is a triangle. E is a point on the median from C. A circle through E touches AB at A and meets AC again at M. Another circle through E touches AB at B and meets BC again at N. Show that the circumcircle of CMN touches the two circles.

Let $M$ be the midpoint of $AB$. Since both $E$ and $M$ belong to the two circles, it means that the $C$-median of $ABC$ is the radical axis of the two. In particular, $C$ belongs to the radical axis, so $CM\cdot CA=CN\cdot CB$. This means that the inversion of pole $C$ and power $CM\cdot CA$ turns the circle $(CMN)$ into $AB$ and invariates the two circles. Since $AB$ is tangent to both the circles, $(CMN)$ must also be.

Let D  be the midpoint of AB,  let Ra be the point where DC cuts c(AEM)  and Rb be the point where DC cuts c(BEN). As DE*DRa = DA^2 = DB^2 = DE*DRb, then Ra==Rb==R.
Since NC*BC = EC*RC = MC*AC, ABMN is cyclic. Then BAC ~ MNC. Let P = NM^c(BEN), since <BPN = <BRN = <ABN, then PNB~MNC, so c(MNC) is tangent to c(PNB).
Similarly for the other hand.

That's too funny...

http://www.mathlinks.ro/Forum/viewtopic.php?t=17030

  darij


Show that 2n!/(n!^2) divides LCM(1,2,..,2n) for all positive integer n.

Just to elaborate on what Rust said:

It's enough to prove $\sum_{i=1}^{r}[\frac{2n}{p^{i}}]-2[\frac{n}{p^{i}}] \leq r$ (order of the $\binom{2n}{n}$ on the left and order of $lcm(1,...,2n)$ on the right), and it's obvious because:

$[\frac{2n}{p^{i}}]-2[\frac{n}{p^{i}}] \leq 1$

Anyway, it was posted before.


Indeed not a very hard one.  We shall prove that  \sum (x 3+1)/z 3  \geq 6, from which the conclusion follows immediately. By applying Chebysev's inequality, we have that \sum (x 3+1)/z 3 \geq  1/3(\sum x 3+3)  \sum 1/x 3 . It's easy to see that if we prove that  \sum 1/x 3 \geq 3, our inequality is proved. But, we have that  \sum 1/x 3 \geq 1/9(\sum 1/x) 3 \geq 1/9(9/(x+y+z)) 3=27/9=3, where we used the generalised means inequality.

As brought to my attention by Nicolas, the argument used in this solution is faulty.

Because $\{x^3+1, y^3+1, z^3+1\}$ and $\{1/z^3, 1/y^3, 1/x^3\}$ are not always similarly sorted (e.g. when $y<x<z$), the Chebyshev argument cannot be applied.

The problem is also discussed [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=20088]here[/url].


calculus
integration
abstract algebra
trigonometry
calculus computations

Familiarity with the Dirichlet kernel (from Fourier series) helps. Note that

$ \frac{\sin 2009x}{\sin x}\equal{}\sum_{n\equal{}\minus{}1004}^{1004}e^{2inx}\equal{}1\plus{}2\sum_{n\equal{}1}^{1004}\cos 2nx$

Note that $ \int_0^{\pi}\cos 2nx\,dx \equal{} 0$ for all integers $ n\ge 1.$ So if we interchange the sum and integral, the only term that contributes anything to the sum is the constant term. The answer is
\[ \int_0^{\pi}1\,dx\equal{}\pi.\]

Consider $ \boxed{I_n\equal{}\int_0^{\pi} \frac{\sin nx}{\sin x}\ dx\ (n\equal{}1,\ 2,\ \cdots)}$.

My answer covers that for odd $ n$ but not for even $ n.$

[quote="jagdish"][b](1) Integration of sin (2009*x) / (sin x)
here Limit  0 to pi[/b][/quote]
Let $ \ I_{n} \equal{} \int_{0}^{\pi}\frac {\sin nx}{\sin x}\ dx$

$ \implies I_{n} \minus{} I_{n \minus{} 2} \equal{} \int_{0}^{\pi}\frac {\sin nx \minus{} \sin (n \minus{} 2)x}{\sin x}\ dx$

$ \implies I_{n} \minus{} I_{n \minus{} 2} \equal{} \int_{0}^{\pi}\frac {2\cos (n \minus{} 1)x\sin x}{\sin x}\ dx$

$ \implies I_{n} \minus{} I_{n \minus{} 2} \equal{} 2\int_{0}^{\pi}\cos (n \minus{} 1)x\ dx$

$ \implies I_{n} \minus{} I_{n \minus{} 2} \equal{} 2\int_{0}^{\pi}\cos (n \minus{} 1)x\ dx$

$ \implies I_{n} \minus{} I_{n \minus{} 2} \equal{} 2\left(\frac {\sin (n \minus{} 1)x}{n \minus{} 1}\right)_{0}^{\pi}$

$ \implies I_{n} \minus{} I_{n \minus{} 2} \equal{} 0$

Now $ I_1 \equal{} \int_{0}^{\pi}\frac {\sin x}{\sin x}\ dx \equal{} \pi$ and $ \ I_2 \equal{} \int_{0}^{\pi}\frac {\sin 2x}{\sin x}\ dx \equal{} 2\int_{0}^{\pi}{\cos x}\ dx \equal{} 0$

$ \implies I_n \equal{} \begin{cases}\ \pi, & n\text{ is odd} \\
0, & n\text{ is even}\end{cases}$


geometry
3D geometry
rectangle
combinatorics unsolved
combinatorics

here is the generalization of the problem cube coloring which is posted to solved probelm.
we have 10*10*10 cube and we color 500 of 1*1*1 cube red and the remain blue prove that there are at least 100 rectangle cube 1*1*2 such that on of its cube is red and the other is blue and also these rectangle cube are pair wise disjointed.

and what is exactly the generalisation? I don't see any new elements? :?

oh ! maybe I am idiot  :maybe:  but 1*1*2 rectangle cube are disjointed (i think this condition makes diffrence)am I wrong?


Let $ \left\lfloor x\right\rfloor$ be floor function of $ x$, then how many possible solutions to $ 4x^2 \minus{} 40\left\lfloor x\right\rfloor \plus{} 51 \equal{} 0$ ?

 :lol:

[quote="vinskman"]Let $ \left\lfloor x\right\rfloor$ be floor function of $ x$, then how many possible solutions to $ 4x^2 \minus{} 40\left\lfloor x\right\rfloor \plus{} 51 \equal{} 0$ ?
[/quote]
$ 4x^2 \plus{} 51 \equal{} 40 \lfloor x \rfloor$
Hence, one necessary condition on $ x$ is that $ x \in \mathbb {R^\plus{}}$.
Now,
$ \frac{x^2}{10} \plus{} \frac{51}{40} \equal{} \lfloor x \rfloor$

$ 0.1x^2 \plus{} 1.275 \equal{} \lfloor x \rfloor$ which is essentially an integer.

See the graph below. That reveals the exact number of solutions, $ 4$.
(The graph at a broader level is a parabola.)


I have got ideas about how to prove that the sqare root of two is irrational.  :maybe:
How does a similar proof apply to pi being irrational?  :huh:
I have been curious about this for ages and was wondering if anyone could explain it to me please.  :) [/code]

The proof that the square root of two is irrational is number-theoretic; it essentially shows that $ \sqrt{2}$ is irrational but an [url=http://en.wikipedia.org/wiki/Algebraic_number]algebraic number[/url].  $ \pi$, on the other hand, is not algebraic but [url=http://en.wikipedia.org/wiki/Transcendental_number]transcendental[/url] (which immediately implies that it is irrational).

The proof that $ \pi$ is transcendental is difficult.  For a proof merely that $ \pi$ is irrational, see [url=http://www.lrz-muenchen.de/~hr/numb/pi-irr.html]here[/url] or [url=http://www.mcs.csuhayward.edu/~malek/Mathlinks/Pi.html]here[/url].


Pat squared two [i]different[/i] positive integers and added the results. Although Ali did the same thing and got the same sum, $S$, Ali's integers were different from Pat's. What is the least possible value of $S$?

[hide]
65.  Given $a^2+b^2=c^2+d^2$, $(a-c)(a+c)=(d-b)(d+b)$ Each factor must be of the same parity because there sum must be even (i.e $(a-c)+(a+c)=2a$ The smallest number which has two pairs of factors of same parity is 15.  $3+5=2a$, and $a=4$ likewise c=1, d=8, and b=7. Our desired sum is $4^2+7^2=1^2+8^2=65$ [/hide]
:D

hey diophantus, you think you could hide ur answers so others could try? thanks

[hide]Also discussed here many times - you want to find the product of two $(4k+1)$ primes, smallest ( distinct ) such primes are   $5$ and  $13$  so you multiply them and get 65.   Other answer  would be $5\cdot17$ etc ... of course you can also multiply any answer by 2 and get another answer ...

BTW if   zero was allowed  as one of the number, you cnould have used $5$ and $5$ to get $25=3^2+4^2=5^2+0^2$$
[/hide]

diophantus: you still have your answer posted, in the upper left.


MATHCOUNTS
AMC
AIME
geometry
ARML
trigonometry
number theory

This is the official invitation to the Columbus High School Invitational Mathematics Tournament. Attached documents include the invitation letter and the registration form. If you wish to attend, please print the registration form and send it to Mrs. Franklin at the school address indicated on the invitation letter.
Please remember that individuals are welcome to come!
Thanks for the time, and let me know if you have any problems downloading the files.
Mark

Download Details: Click on the link to download the registration form. In order to get a larger version of the invitation letter, right-click on the image and select "save target as." After selecting a location, save the file, and then open the file with Windows Picture and Fax Viewer. From here you should be able to print a full-size version of the letter. Ask if you need any assistance!

i'll see if i can convince our coach to go

ok. what school do you go to? you might already be on the invitation list.

i'm at lakeside this year. yeah would you check that for me?

yeah your school is on the list. (it's on briarecliff road right?)

but what you're school will get through the mail will be identicle to what's listed above, so you can print that stuff out if you need to.

You got Vestavia hooked just for countdown

What is the intended format of the countdown round?

oooooh countdown round? gosh that sounds exciting. is it like mathcounts countdown round or something?

Aaaah... countdown. I'm terrible at speed, but good with taking time. (Which is partly why I like the AIME more than the AMC though I tend to score higher on the AMC...)

hey! me too!! you know....139.5 on the AMC but only 12 on the AIME.....darn!

The countdown round will feature the top 10 scorers from the written round. You'll be in front of a large audience.

[quote="Elemennop"][quote="drunner2007"][quote="Cicatriz"]What are the problems going to be like?[/quote]

Well I know that you will probably be competing, so I can't get too specific  :) . Questions will not be extremely hard, but the time limit should make it fairly difficult, on the written test.

Countdown round should be tough just because of competition around you.[/quote]

What's the time limit and number of questions on written test? If they're not harder than UGA this year than time shouldn't be a problem...

Is Countdown like ciphering? If it is, hopefully it's harder than UGA ciphering questions this year...[/quote]

time limit for written test is 105 minutes for 35 questions (3 minutes per question).

most of the countdown questions should be harder than the uga ciphering that i saw. but just think about the competition. there will some good competition at the top, so you'll have someone buzzing in really fast. i would say that some questions can go down in 15 seconds. others will naturally take longer, and we'll put a final time limit of 2 minutes (which I doubt we'll reach).

[quote="Elemennop"]alright the written test seems fair timing

yea i know the countdown is cut throat, but I just hate when they have easy questions in that type of setting cause then it's just computation speed or carefulness as opposed to skill...best case scenario is something like ARML tiebreakers, those are about right difficulty to get it in a quick time[/quote]

I'm in the process of making some of the countdown questions more difficult. However, most of them are still not too challenging (you can't with a short time limit of course).

Some of them look a bit daunting at first glance, but they're fairly easy.

Vestavia had the top eight individuals. My school (Woodward) got 3rd place, but Vestavia B more than doubled our score, and Vestavia A more than tripled it. :rotfl: I tied for ninth individually (which was perfect for me because I really didn't want to do the countdown with the top eight) but lost the tie breaker. That was most likely the best written test I've ever taken because it wasn't ridiculously hard, but it was hard enough so that each time I got a problem right I felt good.

Can someone post the test/certain test questions?

[quote="Elemennop"]Pretty nice test, it was kinda cheesy and not really creative in some parts, but it was ENTIRELY CORRECT and the questions weren't entirely rote computation...The last question was especially nice

Countdown was intense, though a bit too easy in some parts (come on, that cosine sum?). I'm surprised it turned out the way it did though.

Relay is the one thing that DEFINITELY needed harder questions; aside from careless errors it was very easy to make a 400.[/quote]

Yeah I went in understanding relay was going to be easy. You have to realize that relay questions test a teams weakest link (if you have just one person mess up you're screwed), so you have to lower the difficulty. That's why ARML relay is pretty easy.

By the way, Vestavia was the only team to get relay questions right. I would have made relay quite a bit harder if I didn't have to think about giving another school some chances to get it.

I think countdown was about the right difficulty. True, some questions were a little easy.

I'm guessing you went through the Written Test pretty fast. How much time did you have to go back and check answers?

What were your favorite question (maybe top 5)?

Do you remember which ones you missed (I didn't have  chance to look at individual scantrons).

I'm going to post the test in another thread now.

[quote="drunner2007"]
I'm guessing you went through the Written Test pretty fast. How much time did you have to go back and check answers?[/quote]

I did most of it in about 1 hour or 1 hour and 15 minutes; the rest of the time was spent on the leftover questions (mostly the ones where I had to read alot...)

[quote]
What were your favorite question (maybe top 5)?
[/quote]

I don't remember much of the countdown, but I thought on the written test, 35 was really nice, 34 and 33 were pretty good (kinda nasty numbers though), 31 is nice WITH rigorous proof and not just assuming a=b, 21 was pretty good, and 9 was pretty interesting (easy but cool idea).

[quote]Do you remember which ones you missed (I didn't have  chance to look at individual scantrons).[/quote]

I missed 13 (found OP' instead of OP), 30 I got B right before the end of the test but the proctor never called 30 seconds or whatever so I didn't get around to putting it on the scantron, and 33 I had 4014 solutions but wasn't sure where another two came from so I skipped that one also (though logically I should've put 4016 anyways).

ok. you know that one of your teammates had a higher written test score than you. it was either kevin hu or ryan muller.

what did you think of the solution manual?

yea ryan had four points above me

the solution manual was fine, i didn't see anything wrong with it

TOP FIVE PROBLEMS OF ALL TIME (for small n)

1. 35
2. 34
3. 25
4. 31
5. 29

It would've been nice to compete against more of the Georgia people... oh well.

yeah lassiter was coming, but they had to back out last second because of some really important academic competition that happened to be scheduled on the same day.

that would have brought in miles (mosp participant), philip (1st/2nd in state last year), and nathan (GA ARML).

[quote="Karth"]So Vestavia ate the tournament then?  :lol:[/quote]

yeah. have you taken it yet?


hello, i have one question about sendin solution.

i wrote all the solutions in official solution paper.

and i want to send these solution by email.

cuz i have scanner in my home.

can i just scan the solutions, and send it?

[quote="USAMTS"]We will only accept the following file types: DVI, PDF, RTF, TEX, TXT, TIF, GIF, BMP, JPG, EPS[/quote]

So, yeah, you can scan the document; just make sure it's an acceptable file type.

Yes, you can scan handwritten solutions and email them to solutions@usamts.org.

We prefer PDF, but we'll accept a different format if that's better for you.  Make sure your entire submission is in one file (do not scan a separate file for each problem).


[quote="jackdillon"]Did you here about how Kramer(from seinfeld)went on his racist rant? It was just horrible!![/quote]

If we are going to make Round Table a news-forum, where every event brings forth a topic (which I think is the first step towards spam and eventually complete destruction)...let us at least give some background (like looking up his REAL NAME)
His name is Michael Richards, he played in the sitcom Seinfeld, and this is a Wikipedia article section about the controversy :

[url]http://en.wikipedia.org/wiki/Michael_Richards#2006_controversy[/url]

And it's "hear", not "here".

[quote="Ignite168"]There is a video of his rant here

http://youtube.com/watch?v=QgmCBKPHnSY

It was one of the funniest things I have ever seen on YouTube other than:

http://youtube.com/watch?v=QeH1TKmYd4c[/quote]
[url=http://www.youtube.com/watch?v=uTT5-FDxe5M&mode=related&search=]this[/url] and
[url=http://www.youtube.com/watch?v=uTT5-FDxe5M&mode=related&search=]this[/url] are probably funnier and definitely more absurd

About the Iraqi clip, well that's great.  Turning a country into a nightmare and then amusing yourself by making little boys run after you for a stupid bottle of water.  He could have been run over (which would definitely be the case here) .  It just shows how respectful they are.

[quote="fredbel6"]About the Iraqi clip, well that's great.  Turning a country into a nightmare and then amusing yourself by making little boys run after you for a stupid bottle of water.  He could have been run over (which would definitely be the case here) .  It just shows how respectful they are.[/quote]

To be fair, I don't think a majority of soldiers in Iraq are that disrespectful.  At least I hope not.  It's both shameful and disgraceful.

Both of those videos (Kramer and Iraq) were painful for me to watch.  I can have a pretty crude humor at times but there's a line.  And anything where humans are treated as less than humans definitely crosses that line.

About the racist ranting, I do believe it is over the line. However, many comedians talk about people's races, and get very, very, very close to the line, such as Russell Peters. It's just part of being funny. Many people racial stereotypes quite funny (asians carry calculators in their pockets, jews never buy stuff, etc.), so people reference them in comedy shows and stuff. Kramer was just trying to do that, I don't think he actually realized he was over the line until it actually came out of his mouth.

However, I am glad that he apologized, and I believe it completely.

There's nothing wrong, in my opinion, with making jokes which make fun of stereotypes of races.  This has been done a lot and a lot less people are offended by it.  First off, there did not seem to be any attempt at being funny in what Richards said.  In fact, if you listen to his [url=http://www.youtube.com/watch?v=EI5dGM_l9l8]apology[/url] it's pretty clear he just got pissed off.

You should also get the [url=http://www.youtube.com/watch?v=DVmtEqC6Fmc]other side[/url] of the story.

Yes, his apology seems to be genuine.  But what he's apologizing for is what he said [i]because[/i] of the possible ramifications involved.  Certainly he still has racism in him.  You can't expect that he would just suddenly see that racism is wrong and change the way that he views races.

Anyway, coming back to an earlier point, yes jokes about race are often times funny as can be backed up by the success of comedians such as Dave Chappelle, Carlos Mencia, Richard Pryor, and tons more, I'm sure.


geometry
quadratics
algebra
quadratic formula

The problem goes like this: "A rectangular picture with dimensions 5in. by 8in. is surrounded by a frame of uniform width. If the area of the frame (not including the space for the picture) is 114insquared greater than the area of the picture, find the width of the frame."

Ok!

From what I understand, There is a picture with a size of 5in by 8in, and a frame of which the picture is placed in. The size of the frame is equal to the size of the picture PLUS the 114insquared.

X=frame's width
Y=frame's lenght

x(y)=5(8)+114
x(y)=154

so now I am left with just x(y) but nothing else so I kind of just guessed using whole numbers that 11(14)=154. Since 11>5 and 14>8, this is quite possible but I think i just guessed the solution.

[hide]Let $x$ equal the thickness of the frame. We know that the area of the picture is $5 \cdot 8 = 40$ sq. in., so the area of the picture and frame together is $40+114 = 154$ sq. in. Assume the picture's height is 5 in. and its width is 8 in. Then, the height of the frame/picture is the height of the picture plus two thicknesses of the frame (top and bottom), or $5+2x$ in., and similarly, the width of the frame/picture is $8+2x$ in.

We know Area = Height * Width, so:

\begin{eqnarray*}154 & = & (5+2x)(8+2x) \\ & = & 40+26x+4x^{2}\\ 0 & = & 4x^{2}+26x-114 \\ & = & 2x^{2}+13x-57 \end{eqnarray*}

We can solve for x using the quadratic formula:

\begin{eqnarray*}x & = & \frac{-13 \pm \sqrt{13^{2}-4(2)(-57)}}{2(2)}\\ & = & \frac{-13 \pm \sqrt{169+456}}{4}\\ & = & \frac{-13 \pm \sqrt{625}}{4}\\ & = & \frac{-13 \pm 25}{4}\\ & = &-9.5, 3 \end{eqnarray*}

Since x must be positive, we have determined that x = 3. If the question is asking for the width (or thickness) of the frame, the answer is [b]3 inches[/b].

Furthermore, the height is $5+2(3) = 11$ in., and the width is $8+2(3) = 14$ in., so you guessed correctly.[/hide]

that's how you're supposed to do it

or

the length of the whole thing with the frame = (5 + 2x)(8 + 2x) where x is the width of the frame. That equals (5)(8) plus 114, so"

(5 + 2x)(8 + 2x) = 40 + 114
40 + 26x + 4x^2 = 40 + 114
4x^2 + 26x - 114 = 0
2x^2 + 13x - 57 = 0
(2x + 19)(x - 3) = 0

so x = either 3 or -9.5, and we know it can't be -9.5 inches wide, so the answer is 3 inches.

edit: he beat me to it

Ok, so I am wrong.

I am confused at "Then, the height of the frame/picture is the height of the picture plus two thicknesses of the frame (top and bottom), or 5+2x in., and similarly, the width of the frame/picture is 8+2x in. "

Why is it that it is 2x?

From "(5 + 2x)(8 + 2x)", the (5+2x) is the width, and (8+2x) is the lenght, but how is it that (5+2x) = width?(asking what 2x is)

draw it out, it might take a couple minutes but if you label the width of the frame to be x then you should be able to see it.

Call me stupid but I still can't see it.

Also you guys say the width of the frame is 3, but why is it when i guessed it, it was 11? or is 3 just the variable in the equation?

I don't see why 2(x) if x is the width of the frame.

See, if we let x equal the thickness of the frame, then the width of the whole thing will be composed of the picture, the left frame, and the right frame, which is $8+x+x = 8+2x$. It works the same way for the height.

The reason you guessed 11 for the width of the frame is that you thought it was asking for the width of the whole frame and picture. Not so, it is actually asking for the thickness of the frame, which we defined as x. Therefore, the answer you are looking for is $x = 3$ inches.

I guess that is where I missed up. I understood the problem wrong or thought it wrong. By "find the width of the frame", I though they mean the whole frame, which i thougth should include the picture.

So 8+2x = width of whole frame, while 5+2x= lenght of the whole frame, and they wanted to find out the two smaller lenght of the whole frame that doesn't include the picture's frame lenght.

[hide]Yes, but the way you did it, since you realized that it is 6, the thickness is just 6/2=3.[/hide]


superior algebra
superior algebra unsolved

Let $ G$ be a finite group operating on a finite set $ S$. For each $ x\in G$ define $ \alpha(x)\equal{}$number of elements $ s\in S$ s.t. $ xs\equal{}s$. Prove that the number of orbits of $ G$ in $ S$ is equal to $ \frac{1}{\#(G)}\sum_{x\in G}\alpha(x)$

That is the (not) Burnside lemma:

http://en.wikipedia.org/wiki/Burnside's_lemma

the pairs $ (x,s)$ such that $ xs \equal{} s$ may be enumerated in two ways: we choose $ x$, fix it and then vary $ s$, or vice versa. thus
\[ \sum_{x \in G} \alpha(x) \equal{} \# \{(x,s) \in G \times S : xs \equal{} s\} \equal{} \sum_{s \in G} \#\{x \in G : xs \equal{} s\} \equal{} \sum_{s \in G} \# G_s
\]
I think you can complete the proof using http://www.mathlinks.ro/Forum/viewtopic.php?p=1045630.

I think it's the old bug with links that have an apostrophe in their name.

In the past, the link was only shown halfways... Now it seems that it has aggravated :maybe:

The little workaround suggested by someone was:
[quote="fredbel6"]That is the (not) Burnside lemma:

[url]http://en.wikipedia.org/wiki/Burnside%27s_lemma[/url][/quote]
[Yes, the gool ol' ASCII.]


how can chinese remainder theorem be used? sorry ive never used it before.

lets say

$x\equiv7 \mod{11}$

$x\equiv13 \mod{17}$

Solve for $x$.

[quote="mathnerd314"]What is the easiest (fastest) way to solve simultaneous congruences? I've tried the Chinese Remainder Theorem but it takes forever.[/quote]
I'm fairly under the conviction that its Chinese Remainder theorem, because otherwise, it wouldnt be taught on AoPS

Wait, for the last one shouldn't it just be

[hide]
$\boxed{183}$ because $x\equiv -4\pmod{11}$ and $x\equiv -4\pmod{17}$ together imply $x\equiv -4\pmod{187}$?
[/hide]

Well, this is how I learned to solve simultaneous congruences: I'll use one where the shortcut above wouldn't work: $x\equiv 5\pmod{7}$ and $x\equiv 3\pmod{11}$ for example.

We want to find $a$ and $b$ such that $11|a$, $a\equiv 1\pmod{7}$, $b\equiv 1\pmod{7}$, and $b\equiv 1\pmod{11}$. Then $x\equiv 2a+3b\pmod{77}$.

Since $11\equiv 4\pmod{7}$ and $4\cdot 2\equiv 1\pmod{7}$, $a\equiv 11\cdot 2\equiv 22\pmod{77}$.

Clearly $b\equiv 1\pmod{77}$ so $2a+3b\equiv 2(22)+3(1)\equiv 47\pmod{77}\Rightarrow \boxed{x\equiv 47\pmod{77}}$.

to clarify, the chinese remainder theorem just tells us that a solution exists

here is the easy solution to the problem:
$-4\mod 13$ and $-4\mod 17$, so it is $-4 \mod 187 \equiv 183$

but that way is not that easy to apply normally, here is the general way to solve the problem, you can transform the congruences into: ($\{m,n\}\subset \mathbb{Z}$)

$x=7+11m=13+17n$
$11m=6+11n+6n$
$11(m-n)=6(n+1)$
$m-n=6$, $n+1=11$
$n=10$, and $m=16$
$x\equiv 183 \mod 187$

EDIT: woops... did not read scorpius' post  :blush:

whoops sorry, i edited the above post, i forgot to add 7 since x = 11k + 7.

[quote="chess64"][quote="scorpius119"]We want to find $a$ and $b$ such that $11|a$, $a\equiv 1\pmod{7}$, $b\equiv 1\pmod{7}$, and $b\equiv 1\pmod{11}$. Then $x\equiv 2a+3b\pmod{77}$.[/quote] Wait I don't really understand this part... :?[/quote]

If $a\equiv 1\pmod{7}$ and $b\equiv 1\pmod{7}$, then $2a+3b\equiv 5\pmod{7}$, and if $a\equiv 0\pmod{11}$ and $b\equiv 1\pmod{11}$, then $2a+3b\equiv 3\pmod{11}$. So $2a+3b$ satisfies the conditions that $x$ needs.

Oh. But how did you know that those a,b worked?


For what value of x does
$64 \cdot 4^4 = 4^x +4^x +4^x +4^x$ ?


Now that you can call to find out your AP results, let's see what everyone got.

I'll start, I took only BC Calc, 5 overall and 5 on the AB subsection.

You can call the number in your AP student booklet and find out. However, you have to pay 8 bucks.

[quote="bleumoose"]which is silly, imo. you will get it soon anyways. i would bet within another few days.[/quote]

Well since I'm going to mathcamp I'm not going to be home when I get them.

[quote="failure"]I think scores are supposed to come in the mail sometime around July 15.[/quote]

I'm pretty sure that's when they are mailing them out, a couple more days to get it.undefined

Ugh.. I'd rather not know my scores since I already know that they will be anything but 5s.

I called in and found out that I got:

Calc BC: 5
Statistics: 5
English Lit: 5
American Gov't: 5
Microecon: 5

Needless to say, I was very happy!  Too bad I don't really get to get credit, just advanced placement.

[quote="jeffrey"]I just got mine in the mail. 5 on Calc BC[/quote]  Was it mailed from New Jersey?  I believe that Spokane is in Washington, so I'm surprised that the first person to report receiving results is there.

It says New Jersey on the envelope, but I don't really know...

[quote="scm007"]5 here as well.[/quote]
... on what test?

missouri got it today...

US History, Comp Sci A, Biology, Statistics: 5/5/5/5.

I finally got my scores in the mail today, but I already posted them.  It does seem pretty fishy that you can find out your scores on July 1, but you don't get them in the mail until the 20th.  Hmmm....

I got mine today.  It turns out the questionable character was actually the letter "O," not the digit "0."

My scores:

Econ-Micro 5
Econ-Macro 5
Calculus BC 5
AB subscore 5
English Lit 4 (I couldn't have gotten credit even with a 5, anyway)
Physics C-Mechanics 5
Physics C-Electricity and Magnetism 4 (my one day independent study was a bit lacking, apparently)
Spanish Language-5

[quote="failure"]My scores:

Econ-Micro 5
Econ-Macro 5
Calculus BC 5
AB subscore 5
English Lit 4 (I couldn't have gotten credit even with a 5, anyway)
Physics C-Mechanics 5
Physics C-Electricity and Magnetism 4 (my one day independent study was a bit lacking, apparently)
Spanish Language-5[/quote]

I suppose you have some deep personal reason for choosing "failure" as a screen name, but I sure wouldn't guess it has anything to do with your AP scores. Congratulations.

[quote="tokenadult"]I suppose you have some deep personal reason for choosing "failure" as a screen name, but I sure wouldn't guess it has anything to do with your AP scores. Congratulations.[/quote]Haha, thanks.  I sometimes like to choose self-deprecating screen names in order to create low expectations for myself.  If I were to choose, for example, "SuperAwesomeMathGod," and I said something mathematically incorrect, I would look rather pompous.  With a name like "failure," it doesn't stand out so much.  It was more a spur of the moment thing, not related to any "deep personal reason."

As a sophomore, I only got to take one AP course: Physics.

AP Physics B: 5.

Yay.

And congratulations to all of you on your AP success.

5s on Physics C E&M, MicroEcon, US govt,English Lit and Chem... Good thing I'm not getting credit for any of them but chem.

Oh I got mine back last Saturday also

Calculus BC 5 (AB subscore 5)
Physics B 5
Physics C:Mechanics 5
US Gov & Politics 4

I have no idea how I got a 5 on the first three, and I have no idea how I did not get a 5 on the last one. I was more prepared for the last one than any of the other three, considering that I self-studied for the BC exam and my Physics B teacher didn't cover half of the B material so I had to make up for it myself in the span of several weeks.

I gpt a 5 in Calculus AB, a 4 in Physics B and only 2 in English lit ( I didn't expect to get higher). I am really glad that I got a 5 in Calculus because I think I made some stupid errors on the test and wasn't sure what I would get.

5 in chem and stats. *yay*. only need 20 more credit hours to start as a sophomore at my current favorite school. *yay again*

w00!  5 in comp sci AB!  And stats and world history.  But CS!

[quote="kentliu"]confuted, ur like the only person in the world who gets a 5 on EM and 4 on mech.. haha.. and how much did all those tests cost?? man.. i might go take a crap load of tests blind next year if i use a fee waver..
but the dean of math/science got mad at me for taking physics EM, cuz our school doesnt teach tat class..[/quote]
I was prepared to pay 82 dollars for each test, but I ended up getting a fee reduction and paying a total of 56 dollars. For over $13,000 in credits, that's not bad at all.

I was wondering, is it possible just to self study all the AP's you sign up for, or is it much better to take a school course, because if my school is willing to give me credit, I might just sign up for APs and take them. I am only a freshy, and I am also wondering which APs colleges look at the most, or is there no difference?

[quote="white_horse_king88"]I was wondering, is it possible just to self study all the AP's you sign up for, or is it much better to take a school course?[/quote]

What is better for you depends on how good your school courses are, what the barriers are to getting into those courses, and how many resources you will apply with good self-discipline if you self-study. It is DEFINITELY possible to self-study (including using distance learning) ANY subject on the list of AP tests, but you may desire to attend a class if one is available.

My son (homeschooled) is in a chemistry class this year such that SOME of the students enrolled will take the AP Chemistry exam this school year. I'm not sure yet whether or not I'll encourage my son to do that, but I probably will encourage him to take the AP Statistics exam after self-study this year, which is what most people would call his "seventh grade" year. A year or two later, at the most, he should be ready for Calculus BC (in that case, because of a classroom course).

What works for you depends on what courses you can get into, or what you desire to study on your own. It is DEFINITELY a good idea to have some AP scores of 5 already in hand by the end of your junior year (eleventh grade year) of high school, because then those scores can be mentioned in your college application.

long time no come
I took only AP Calc BC
congratulation to all ppl, u r so great.


Suppose $ A$, $ B$, and $ C$ are three numbers for which $ 1001C\minus{}2002A\equal{}4004$ and $ 1001B\plus{}3003A\equal{}5005$.The average of the three numbers $ A$, $ B$, and $ C$ is

$ \text{(A)}\ 1 \qquad
\text{(B)}\ 3 \qquad
\text{(C)}\ 6 \qquad
\text{(D)}\ 9 \qquad
\text{(E)}\ \text{not uniquely determined}$

[hide]Add to get $ 1001(A\plus{}B\plus{}C)\equal{}9009$, so A+B+C=9, so B.[/hide]

[quote="worthawholebean"]Suppose $ A$, $ B$, and $ C$ are three numbers for which $ 1001C \minus{} 2002A \equal{} 4004$ and $ 1001B \plus{} 3003A \equal{} 5005$.The average of the three numbers $ A$, $ B$, and $ C$ is

$ \text{(A)}\ 1 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ \text{not uniquely determined}$[/quote]
[hide]Dividing by 1001 for the first equation results in $ C\minus{}2A\equal{}4$. Dividing by 1001 for the second equation results in $ B\plus{}3A\equal{}5$. Adding both equations gives us $ A\plus{}B\plus{}C\equal{}9$, so the average is $ 3$, or $ B$. [/hide]


If $a,b,c>0$ then prove the inequality:

            $(a+b+c)^3\geq\frac{9}{4}\cdot ((a+b)^2c+((b+c)^2a+(c+a)^2b)$

:huh: could u write correctly....some paranthezes must be closed/opened or put on the right place...

Rewrite the ineq :
   $  (a+b+c)^3 \geq \frac{9}{4}\cdot ((a+b)^2c+(b+c)^2a+(c+a)^2b) $
 Suppose $ a+b+c =1 $
  The ineq is equivalent to
 $ 3(a+b+c)^2 + (a+b+c)^3 \geq 9(ab+ac+bc) + 27abc $ : true by AM-GM


$\{a_i\}_{i=1}^n\ ,\ \{b_i\}_{i=1}^n$ be two sequence of nonegetive integer and $b_i\le a_i\ \ ,\ \ 1\le i\le n$
Prove:

$\prod_{i-1}^n\binom {a_i}{b_i}\le\binom {a_1+\cdots+a_n}{b_1+\cdots+b_n}$

Let $S_1 = {a_1+\cdots+a_n}$ and $S_2 = {b_1+\cdots+b_n}$

$S_1 \geq S_2$ so $x^{S_2}$ may be found in binomial expansion of $(1+x)^{S_1} = \prod (1+x)^{a_i}$
$a_i \geq b_i$ so $x^{b_i}$ may be found in $(1+x)^{a_i} = 1 + \cdots + \binom {a_i}{b_i} \cdot x^{b_i} + \cdots + x^{a_i}$

Conclusion follows


$ ABC$ is an isosceles triangle with $ AC\equal{}BC$. The midpoint of $ AB$ is $ M$. $ H\in BC$ so that $ MH$ is perpendicular to $ BC$. the midpoint of $ MH$ is $ P$. Prove that $ AH \perp CP$.

Let $ K$ be midpoint of $ HB$.It's easy to see $ P$ is the orthocenter of $ \triangle KMC$ (QED) :)

The problem can be generalized: if $ ABC$ is just a triangle, $ CM$ altitude, $ H\in BC |  MH \perp BC$ and $ P\in MH | \frac{PH}{MP}\equal{}\frac{AM}{BM}$, then $ CP\perp AH$.
Below there are two ways to prove:

1) similar to the previous one: let $ K\in BH | \frac{KH}{BK}\equal{}\frac{AM}{MB}$, then $ PK \parallel{} AB \perp CM$, hence $ P$ is the orthocenter of $ \triangle CMK$.

2) Let $ D\in BC | AD \perp BC$, then $ \triangle ADB \sim  \triangle CHM$ and they have all their homologue sides perpendicular: $ AD\perp CH$, $ BD \perp MH$, $ H$ and $ P$ divide homologue and perpendicular sides $ BD$ and $ MH$ at the same ratio, hence $ CP \perp AH$.

Best regards,
sunken rock


Hey,

I was wondering if anyone knew when the Math is Cool competition is going to be during the day and where it is (for High School Students on Nov. 3rd).

[quote="xxreddevilzxx"]well.... it already happened, but fyi it was at Highline High School in Burien.[/quote]
HOw'd you do? My school won 4th... so we're going to the masters!

[quote="xxreddevilzxx"]we were third, and i was also third.... losing in the tiebreakers.[/quote]
Yeah, sorry for hogging the stage while we were taking pictures  :P and congrats!


Not that hard if you know this :wink:

In a quadrangle $ABCD$, let $\angle ADB=60^\circ$, $\angle BDC=60^\circ$, $AD=a$, $BD=b$, $CD=c$.
Then, $AB=\sqrt{a^{2}-ab+b^{2}}$, $BC=\sqrt{b^{2}-bc+c^{2}}$, $AC=\sqrt{a^{2}+ac+c^{2}}$

Since $AB+BC \geq AC$, $\sqrt{a^{2}-ab+b^{2}}+\sqrt{b^{2}-bc+c^{2}}\geq \sqrt{a^{2}+ac+c^{2}}$


In the adjoining figure the five circles are tangent to one another consecutively and to the lines $L_1$ and $L_2$ ($L_1$ is the line that is above the circles and $L_2$ is the line that goes under the circles). If the radius of the largest circle is 18 and that of the smallest one is 8, then the radius of the middle circle is

[asy]
size(250);defaultpen(linewidth(0.7));
real alpha=5.797939254, x=71.191836;
int i;
for(i=0; i<5; i=i+1) {
real r=8*(sqrt(6)/2)^i;
draw(Circle((x+r)*dir(alpha), r));
x=x+2r;
}
real x=71.191836+40+20*sqrt(6), r=18;
pair A=tangent(origin, (x+r)*dir(alpha), r, 1), B=tangent(origin, (x+r)*dir(alpha), r, 2);
pair A1=300*dir(origin--A), B1=300*dir(origin--B);
draw(B1--origin--A1);
pair X=(69,-5), X1=reflect(origin, (x+r)*dir(alpha))*X,
Y=(200,-5), Y1=reflect(origin, (x+r)*dir(alpha))*Y,
Z=(130,0), Z1=reflect(origin, (x+r)*dir(alpha))*Z;
clip(X--Y--Y1--X1--cycle);
label("$L_2$", Z, S);
label("$L_1$", Z1, dir(2*alpha)*dir(90));[/asy]

$\text{(A)} \ 12 \qquad \text{(B)} \ 12.5 \qquad \text{(C)} \ 13 \qquad \text{(D)} \ 13.5 \qquad \text{(E)} \ 14$

[hide]i think...if $r$ is the radius of the middle circle, then $\frac{r}{8}=\frac{18}{r}$, so $\boxed{r=12}$.[/hide]

since AMC is a fast test (well, to me) and I would like to answer as fast as possible, I will guess that the middle circle's radius is some mean of 8 and 18.  since $8=4*2$ and $18=9*2$, it looks logical to pick geometric mean because the 2 2's makes perfect square when multiplying.  If I ran out of time, I'll be 90% sure to guess 12.

For those of you who disagree, note that this is a homothetic transformation of the first circle with geometrically increasing distance, so the circle's dimension will form a geometric sequence (think of a projector), which tells us that the middle circle's radius is indeed the geometric mean of 8 and 18.

it's a geometric series..that should be enough to solve the problem. the answer is A

Here's my proof...
[hide]Let $r_1$, $r_2$ and $r_3$ be the radii of the consecutive circles, where $r_1<r_2<r_3$.
$O_1$, $O_2$ and $O_3$ are the centers of the circles whose radii are $r_1$, $r_2$ and $r_3$, respectively.
$T_1$, $T_2$ and $T_3$ are the points of tangency of $O_1$, $O_2$ and $O_3$, respectively. (All of them are on the same line.)

Now, we have $O_1T_1=r_1$, $O_2T_2=r_2$ and $O_3T_3=r_3$.
Let $A$ be the intersection of $O_2T_2$ and the perpendicular line from $O_1$ to $O_2T_2$. Similarly, let $B$ be the intersection of $O_3T_3$ and the perpendicular line from $O_2$ to $O_3T_3$.

Since $\triangle O_1AO_2 \sim \triangle O_2BO_3$, we have $\frac{O_1A}{O_1O_2}=\frac{O_2B}{O_2O_3}$ $\Longleftrightarrow \frac{r_2-r_1}{r_2+r_1}=\frac{r_3-r_2}{r_3+r_2}$

Simplifying,
$(r_2-r_1)(r_3+r_2)=(r_2+r_1)(r_3-r_2)$

$r_2r_3+r_2^2-r_1r_3-r_1r_2= r+2r_3-r_2^2+r_1r_3-r_1r_2$

$2r_2^2=2r_1r_3$

$\therefore \frac{r_2}{r_1}=\frac{r_3}{r_2}$

Apparently, for the same reason, we have $\frac{r_3}{r_2}=\frac{r_4}{r_3}=\frac{r_5}{r_4}=... =\frac{r_n}{r_{n-1}}$.
Hence $r_1,r_2,r_3...$ is a geometric series.
[/hide]

(This [i]looks[/i] long, but about a half of this proof is actually just explanations of the setup.)

[hide="Answer"]Drawing the line connecting each of the radii and then the tangents to either line, we see that the radii are in geometric progression. Therefore, we have $8,a,b,c,18$, so $8b=a^2$, $ac=b^2$, and $18b=c^2$. Multiplying the first and third together, we have $144b^2=a^2c^2$. By the second, $b^2=144\Rightarrow b=12\Rightarrow \boxed{A}$.[/hide]

[quote="WarpedKlown1335"]Prove that it's a geometric series.[/quote]

We can actually generalize that in all homethetic transformations in which the images under transformation are all touched that their dimension forms a gemetric sequence.

For any sequence of image, draw a line between each of them (in this example it'll be a line through the tangent point of each circle).  For convenience, call each line $L_1$, $L_2$, etc., and their intersection with one of the homothetic lines (the external tangent line) $P_1$, $P_2$, etc., and finally the center of transformation is $O$, as always.

Let's say that the second circle has dimensions $k$ times greater than the first circle's, and the $OP_1 = l$, then $P_1P_2 = kl-l$ because $OP_2 = kl$ by definition (it should be obvious tho...).  so we wish to prove that $OP_3 = k^2l$ or $P_2P_3 = k^2l - kl$ and so on.  Notice that since $P_1P_2$ is the side of the first image (we can take it as a whole), so $P_2P_3$ must be $k$ times that because it's the side of thet second image and  that's what we stated in the beginning.  So we see that $P_2P_3 = k^2l - kl$ and the rest will follow by simple induction.

Believe me or not, I found this from building a tower... (For Science Olympiad if anyone knows what it is).  It is strange how math can pop out of no where :rotfl:  :rotfl:


[quote="aufha"]Having gone through AMC 10s 2000-2006 and first halves of the AIMEs 00-08, what would be the next logical course of action if I wanted to go through more old contests?

After a few short months of studying for math competitions to try to qualify for the AIME this year, it appears to me that math competition experience is much more important than knowing formulas and facts.[/quote]
go through AMC12's :P...but it is more important to push for AIME though...

Kalva has archives of TONS of old AIMEs/USAMOs. You might wanna go there and check out the old contests.


For each non-empty subset of {1, 2, 3, 4, 5, 6, 7}, arrange the members in decreasing order with alternating signs beginning with a minus sign and take the sum. For example, for the subset {5} we get 5. For {6, 3, 1} we get 6 - 3 + 1 = 4. Find the sum of all the resulting numbers from all the subsets.

[hide]
Consider a subset not containing $ 7$ and the same subset with $ 7$ added in. The sum of the results will be $ 7$ (everything cancels out except for the $ 7$ at the beginning of the second). Since this is a bijection between subsets without $ 7$s and subsets with $ 7$s, we get that we obtain $ 7$ for every pair of subsets. There are $ 2^7$ subsets, so $ 2^6$ pairs, so our sum is $ 7\cdot 2^6 \equal{} 448$.
[/hide]

Nice solution, but i had trouble even starting to see that =P How do you "see" this sort of bijection?

[hide]First we consider the subsets of $ {1,2,3,4,5,6}$, denoted by $ x$. If we add in  a 7, each  of $ x \plus{} (x\cup{7})$ is 7, because the terms of $ x\cup{7}$ have opposite signs, and cancel off. There are a total of $ 2^6 \equal{} 64$ subsets, but one is empty. Also, we forgot about the one just with 7, so there are 64. $ 64\cdot{7} \equal{} 448$.[/hide]

Not as immpresive as hamster's but I did my best.

Oh ok i get it.. the alternating signs change.  :) You did as well  :P


linear algebra
matrix
algebra
polynomial
superior algebra
superior algebra unsolved

What is the maximal order of an element of the group $ \mbox{GL}_n\left(\mathbb{F}_q\right)$?

[quote="Gockel"]Every matrix of order k is diagonalizable[/quote]

Huh?

  darij

[quote="Gockel"]f...

you're right, this holds only if the polynomial x^k-1 splits completely.[/quote]

Ehm... this is not the problem ;)
We have a lot of matrices with nontrivial Jordan cells in $ \text{GL}_n\left(\mathbb{F}_q\right)$.

  darij

[quote="Gockel"]It is q-1. Every matrix of order k is diagonalizable and on the diagonal are k-th roots of unity. Therefore the highest order is the highest order of an element in $ \mathbb{F}_q^\times$ which is q-1 as the latter is cyclic.[/quote]

There is an element of order $ q^n\minus{}1$ in $ \mbox{GL}_n(\mathbb{F}_q)$


Sign-up. My rating is 1300. When you pm me for sign-up, give me your rating if you have one.

Maybe it would help if you told us why you need our ratings and what we're signing up for?


When Julian spends his weekly allowance at the bookstore, his father lets him buy a book whose regular price is double his allowance. However, he gets a $ 10\%$ discount on the price of the book. What fraction of the final purchase price is Julian's weekly allowance? Express your answer as a common fraction.

yeah, but for some reason, on For the Win!, the answer key says the answer is 5/9....
i mean, i know it's hard to catch all these bugs, but maybe AoPS shouldn't have released the Alpha with so many glitches...

it is 5/9.
the price of the book is 9/5 times his allowance
so his allowance is 5/9 of the price. it's not a glitch.


function
algebra
polynomial
number theory
modular arithmetic
superior algebra
superior algebra unsolved

If you're familiar with modular arithmetic, this is kind of a trick question.  The function in question is obviously well-defined modulo 10, so it's certainly well-defined modulo 5.

If you don't know modular arithmetic, try looking at $ (x^2\plus{}2x\minus{}1) \minus{} (y^2\plus{}2y\minus{}1)$.  Can you factor this?  If the difference between $ x$ and $ y$ is divisible by $ 10$, is this divisible by $ 5$?

So does this work for the proof:

Suppose a=b (mod10)
10|(a-b)
10|(a^2+2a-1)-(b^2+2b-1)=(a-b)(a+b+2)
5|10|(a-b)(a+b+2)
5|(a-b)(a+b+2)
 therefore it is well defined?

Correct!

Notice that this works with any polynomial, because $ x\minus{}y$ divides $ x^n\minus{}y^n$.

[quote="VolleyCandi04"]Prove that the function p is well defined. [b]Z[/b]10>[b]Z[/b]5 defined by p([x]10)=[x^2+2x-1]5[/quote]

$ \rm HINT \quad 10\;|\;x\minus{}y \Rightarrow 10\;|\;x\minus{}y\;|\;p(x)\minus{}p(y)$ for all $ \rm p(x) \in Z[x]$ by the [b]Factor Theorem[/b]

More generally consider the universal property of quotient rings.


Call the matrix M= (a b)
                         (c d)   the companion matrix for the mapping  t:z -> (az+b)/(cz+d)

(a).  Prove that if  M(i) is the companion matrix for T(i)  (i=1,2), then M1M2 is the companion matrix for the mapping T1T2.


What is the sign of $ (a)^{b}$ where $ a \in \mathbb{Z^{-}}$ and b is irrational?

Example: is $ (-3)^{\pi}$ positive or negative, or is it impossible to define? If the latter is true, why can $ 3^{\pi}$ be defined?

Well for a negative number $ A$, $ A^{n}$ is positive if n is 0 mod 2 and negative if n is 1 mod 2. But if n is irrational, we cant define $ A^{n}$ in a fractional form with exponents, and n is neither 1 or 0 mod 2. then we have a problem. which is why the sign of $ A^{n}$ in this case is probably undefined. or defined in a special way.

I believe it is undefined.
Look at this:

$ 3^{3.14159}=3^{3}\cdot 3^{.14159}= 27 \cdot \sqrt[14159]{3^{100000}}$
which is positive.

$ -3^{3.14}=-3^{3}\cdot-3^{.14}=-27 \cdot \sqrt[14]{-3^{100}}=-27 \cdot \sqrt[7]{-3^{50}}$ , at whic point we have an odd root of a negative number.  (which is bad.)


inequalities
algebra
polynomial
inequalities proposed

Let $ a,b,c$ be the sidelengths of a triangle. Prove that
\[ \frac{3a^3\minus{}16abc}{(b\minus{}c)^2}\plus{}\frac{3b^3\minus{}16abc}{(c\minus{}a)^2}\plus{}\frac{3c^3\minus{}16abc}{(a\minus{}b)^2} \le \minus{}21(a\plus{}b\plus{}c).\]
Equality holds if and only if $ (a,b,c) \sim (1,2,3).$ ;) :)

I wonder if VasC's new theorem can solve this inequality? ;)

Very nice and very strong inequality, Can_hang. :lol:  Have you a solution for it ?
My theorem below is also very strong and solve it.

[b]Theorem.[/b] Any symmetric homogeneous polynomial of degree seven can ve written in the form
$ f(a,b,c)\equal{}Apx^2\plus{}g_1(p,q)x\plus{}g_2(p,q)$,
where $ A$ is a real number, and $ p\equal{}a\plus{}b\plus{}c$, $ q\equal{}ab\plus{}bc\plus{}ca$, $ x\equal{}abc$.
Assume that $ f(0,b,c)\ge 0$ and $ f(a,1,1) \ge0$ for any nonnegative real numbers $ a,b,c$.
The inequality $ f(a,b,c)\ge 0$ holds for any nonnegative real numbers $ a,b,c$ if one of the following conditions is fulfilled:

(a) there exists a real number $ a_1\ge 1$ such that

$ f(a,1,1)\ge \frac {4A(a\plus{}2)(a\minus{}1)^6}{729} \ \  for \ \ 1\le a\le a_1$,
and
$ \ \ \ f(a,1,1) \ge \frac {A(a\plus{}2)(a\minus{}1)^4}{27} \ \  for \ \ a\ge a_1$;

(b) there exists a real number $ \delta \neq \minus{}2$ such that

$ \ \ \ f(a,1,1)\ge \frac {4A(a\plus{}2)(a\minus{}\delta)^2(a\minus{}1)^4}{81(\delta\plus{}2)^2} \ \  for  \ \ a\ge 1$.

Very nice, VasC. From now, I can see that your theorem is so strong, what an incredible! :)

P/s: My solution is different with yours! ;)

[quote=can_hang2007]Let $ a,b,c$ be the sidelengths of a triangle. Prove that
\[ \frac{3a^3\minus{}16abc}{(b\minus{}c)^2}\plus{}\frac{3b^3\minus{}16abc}{(c\minus{}a)^2}\plus{}\frac{3c^3\minus{}16abc}{(a\minus{}b)^2} \le \minus{}21(a\plus{}b\plus{}c).\]
Equality holds if and only if $ (a,b,c) \sim (1,2,3).$ ;) :)
[/quote]
After using the Ravi's substitution and the [url=http://www.artofproblemsolving.com/community/c6h278791]uvw[/url]'s substitutions we need to prove that
$$9uw^6+(44u^4-61u^2v^2-v^4)w^3+9uv^2(2u^2-3v^2)^2\geq0$$
Let $u^2=tv^2$.
Hence,  it's enough to prove our inequality for $1\leq t<\frac{61+3\sqrt{433}}{88}=1.40...$ and it remains to prove that
$324t((2t-3)^2-(44t^2-61t-1)^2\geq0$, which is $(t-1)^2(1936t^2-2792t+1)\leq0$, which is true for $t\in\left[1, \frac{61+3\sqrt{433}}{88}\right)$.
Done!


Eliminate 'A' from the following equations:
sinA - cosA = p, and
cosecA - sinA = q

Source: Plane Trigonometry I by S.L.Loney.

Thanks!

$\frac{1}{\sin A}-\sin A = q \implies \sin^{2}A+q \sin A-1 = 0 \implies \sin A = \frac{-q \pm \sqrt{q^{2}+4}}{2}$

$p=\sin A \pm \sqrt{1-\sin^{2}A}\implies \boxed{p=\frac{-q \pm^{\left(1 \right)}\sqrt{q^{2}+4}}{2}\pm^{\left(2 \right)}\sqrt{1-\left( \frac{-q \pm^{\left(1 \right)}\sqrt{q^{2}+4}}{2}\right)^{2}}}$

Where $\pm^{\left(N \right)}$ is internally consistent


function
combinatorics unsolved
combinatorics

Need a little help here.

1. f(1)+f(2)+.....+f(n)=n²f(n)
f(1)=1999
find f(1998)

2. a²+b²=c²
b+2=c
find a Pythogoras triple fulfilling the characteristic.

3. x+1/x=7
what is x³+1/x³?

1)
f(1)+f(2)+.....+f(n)=n²f(n),
f(1)+f(2)+.....+f(n-1)=(n-1)²f(n-1), pluging into the first eq
(n-1)²f(n-1)+f(n)=n²f(n) (note this is equivalent to the given eq). From there
(n-1)²f(n-1)=(n²-1)f(n), dividing by n-1
(n-1)f(n-1)=(n+1)f(n).
Function $f(n)=\frac c{n(n+1)}$. Because f(1)=1999, we have c=2.1999 and f(1998)=2.1999/(1998.1999)=1/999.

2)
a²+b²=b²+4b+4
(1) a²=4b+4, so 2|a, a=2k, from (1) b=k²-1, c=b+2=k²+1.
For k>1 we have a+b>c, so the triplet a,b,c may form a triangle.

c)
x=a, 1/x=b. We have
a+b=7,
ab=1,
looking for $a^3+b^3=(a+b)((a+b)^2-3ab)=7.46=322$.


Hey everyone,

I wanted to let you know that TAMS (Texas Academy of Mathematics and Science) is going to be hosting another competition this year on October 24, 2009, at the University of North Texas campus in Denton, TX.

If you would like some more information about the contest, you can email tamstournament@gmail.com

I'll also upload the math tests from last year (there are also science events; each student picks one math and one science).

-sameer

apparently, the number of attachments is capped, so the rest of the tests are here and in the next post.

Basically, if you are currently in say Algebra II, you would take the Algebra II test, etc. etc.
Because it is relatively early in the school year, the Algebra II test, for example, covers mainly stuff from Geometry and some stuff from Algebra II.


geometry
3D geometry
algebra
polynomial
Rational Root Theorem

[hide="Hint"]Write $ 1$ as $ 3^{3}\minus{}\sqrt [3]{26^{3}}$[/hide]

[hide="Answer"]The value of the expression is $ 3$[/hide]

[hide="Solution"]Let $ y\equal{}\sqrt[3]{26}$

So if we let our equation equal x we get:

$ \sqrt[3]{1\minus{}27y\plus{}9y^{2}}\equal{}x\minus{}y$

Now as Farenhajt hinted let $ 1\equal{}27\minus{}y^{3}$ so we get:

$ \sqrt[3]{27\minus{}y^{3}\minus{}27y\plus{}9y^{2}}\equal{}x\minus{}y$

taking the cube of both sides we get:

$ 27\minus{}27y\plus{}9y^{2}\minus{}y^{3}\equal{}(x\minus{}y)^{3}$

A quick check reveals that 3 is a root of the RHS so we get

$ (y\minus{}3)(\minus{}y^{2}\plus{}6y\minus{}9)\equal{}(x\minus{}y)^{3}$
$ (3\minus{}y)(y\minus{}3)^{2}\equal{}(x\minus{}y)^{3}$
$ \minus{}1\cdot(y\minus{}3)^{3}\equal{}(x\minus{}y)^{3}$

Taking the cube root of both sides gives us

$ \minus{}(y\minus{}3)\equal{}x\minus{}y$
$ \minus{}y\plus{}3\equal{}x\minus{}y$
$ \boxed{3\equal{}x}$[/hide]

[quote="Icy"][quote="sonny"][hide="Solution"]Let $ y \equal{}\sqrt [3]{26}$

So if we let our equation equal x we get:

$ \sqrt [3]{1\minus{}27y\plus{}9y^{2}}\equal{} x\minus{}y$

Now as Farenhajt hinted let $ 1 \equal{} 27\minus{}y^{3}$ so we get:

$ \sqrt [3]{27\minus{}y^{3}\minus{}27y\plus{}9y^{2}}\equal{} x\minus{}y$

taking the cube of both sides we get:

$ 27\minus{}27y\plus{}9y^{2}\minus{}y^{3}\equal{} (x\minus{}y)^{3}$

A quick check reveals that 3 is a root of the RHS so we get

$ (y\minus{}3)(\minus{}y^{2}\plus{}6y\minus{}9) \equal{} (x\minus{}y)^{3}$
$ (3\minus{}y)(y\minus{}3)^{2}\equal{} (x\minus{}y)^{3}$
$ \minus{}1\cdot(y\minus{}3)^{3}\equal{} (x\minus{}y)^{3}$

Taking the cube root of both sides gives us

$ \minus{}(y\minus{}3) \equal{} x\minus{}y$
$ \minus{}y\plus{}3 \equal{} x\minus{}y$
$ \boxed{3 \equal{} x}$[/hide][/quote]

What check did you use to reveal that 3 is a root of RHS? And how did $ (3\minus{}y)$ become $ \minus{}1$?[/quote]

A mix of rational root theorem and synthetic division reveals that it is a root.

And $ (3\minus{}y)(y\minus{}3)^{2}\equal{}\minus{}1(y\minus{}3)(y\minus{}3)^{2}\equal{}\minus{}1(y\minus{}3)^{3}$ Yeah I guess those steps might be a little hard to follow.


calculus
integration
function
limit
LaTeX
trigonometry
analytic geometry

This isn't a particularly difficult problem, but I thought it was nice.

The function $f$ is continuous on $\mathbb{R}$. Let $0 < a < b$ and define for  $\delta > 0$
 $D_\delta = \{(x,y) \in \mathbb{R}^{2}\, ; \, a^{2}\delta^{2}\le x^{2}+y^{2}\le b^{2}\delta^{2}\}$

Calculate
$\lim_{\delta\to 0^{+}}\int \!\!\! \int_{D_\delta}\frac{f\left(\sqrt{x^{2}+y^{2}}\right)}{x^{2}+y^{2}}\,dx\,dy$

edit: Can someone teach me how to make integrals look nice in LaTeX? For one I'd like the integral signs closer to each other. I'd also like the dxdy to look nicer.

how do you mean "look nicer"?
aren't they nice enough like this?

as regarding the double (triple, ...) integral, you can use \iint (\iiint, \iiiint, \idotsint...) to get $\iint$ ($\iiint$, $\iiiint$, $\idotsint$...).
they don't look much closer, though... strange...

Ah, I have already messed around with the LaTeX a little bit. For the integrals I removed space in between them with \! . Although \\iint looks handy. For the dx dy stuff I added some spaces and they look a little bit different, but I still don't like the way they look. I want them to look more like $\cos$ than $cos$.

you can use \text{} then... but, personally, i prefer variables to be $x$ more than $\text{x}$... and i think $dx$ is better than $\text{d}x$...

coming to the problem...
call $L$ your limit; $f(r) = f(0)+o(1)$, as $r\to 0$.
change variable in your double integral (radial coordinates) and get: $L = 2\pi\lim_{\delta\to 0}\int_{a\delta}^{b\delta}\frac{f(r)}{r}dr = 2\pi\lim_{\delta\to 0}\int_{a\delta}^{b\delta}\frac{f(0)+o(1)}{r}dr = 2\pi f(0)\log\frac{b}{a}$.

Why can you just ignore the $o(1)/r$ part?

doesn't the $o$ pass out of integral?

but, anyway, call $f(x) = f(0)+\phi(x)$.
fix $\epsilon > 0$: since $\phi(x) \to 0$ as $x\to 0$, there is $\delta$ small enough to have $|\phi(y)|<\epsilon$ for $|y|<b\delta$.
then you have $\left|\int_{a\delta}^{b\delta}\frac{\phi(r)}{r}dr\right| < \int_{a\delta}^{b\delta}\frac{\epsilon}{r}dr = \epsilon\log\left(\frac{b}{a}\right)$.

i guess you mean the solution to kalle's problem with integrals in latex, moubinool, right? :)
anyway, thanks :)
if i recall correctly, something like one year and an half ago you posted a problem which was quite similar to this one, just tougher, which i solved with much help from two friends of mine... am i wrong?

Well, the first one is the best. I think there should be a smaller space between dx and dy, but that's easy to fix. However, it is still ugly how the integrand is taller than the integral signs.


In the diagram, AC=20, and the altitude DB of triangle ABC is extended 5 units to E. What is the area of the shaded region?

[hide]The area can be found by adding up the areas of $ \triangle EBC$ and $ \triangle EBA$:
\begin{align*}
&\frac{EB \cdot AD}{2} + \frac{EB \cdot DC}{2} \\
&=\frac{EB\left(AD+DC\right)}{2} \\
&=\frac{EB \cdot AC}{2} \\
&=\frac{5 \cdot 20}{2} \\
&=\boxed{50}
\end{align*}
[/hide]


The following problem from Mathlinks resources is missing answer choices:

10.  Two points $ B$ and $ C$ are in a plane.  Let $ S$ be a set of all points $ A$ in the plane which $ \triangle ABC$ has area $ 1$.  Which of the following describes $ S$?

PM me when you find stuff like this. No need to post a new thread.

Edit: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=251560]It isn't missing answer choices...[/url]


find the maximum of $k$ that we have:
$(a+b)^2+(a+b+4c)^2 \geq \frac{kabc}{a+b+c}$

answer k=100. Take all the expression to the left.
define y=c/x and define x=a+b . apply ab<=(a+b)^2/4 =c^2/4 and then write every thing in terms of y. you get
L.H.S = 4(2+10y+24y^2+16y^3 )/y = 4(1/4+...+1/4+y+...+y+4y^2+...+4y^2+16y^3)/y >=4*25/y (1 y^25)^1/25 =100. equality iff y=1/4 and there fore a=b=2c.
-Ali


Find a natural n and some divisors of n such as 2007 times the sum of those divisors equal to n.

How about all divisors of n?

[quote="mszew"]Find a natural n and some divisors of n such as 2007 times the sum of those divisors equal to n.[/quote]

Take $n=32112=2007*(4+12)$ and $4$ and $12$ are two divisors of $32112$.

[quote="mszew"] How about all divisors of n?[/quote]

Sum of all divisors of $n$ is greater than $n$, so $n$ can't be equal to $2007$ times the sum of all divisors of $n$.

--
Patrick

There are infinitely many solutions to your question of the form 2007*N! where N is a natural number.
This is true because N! can be expressed as sum of its different divisors for all N.
The last fact is not hard and can be prooved by indcuction.

Thanks for your responses,
i made some mistakes about the problem, i have edited it.

Find all natural n and some divisors of n (more than 1) such as 2007 times the sum of those divisors equal to n.

How about all proper ( less than n) divisors of n?


[quote="sirkuku"]Here is the question:

You walk into a hut abandoned with 7 beds and 7 chairs.
In the middle you see a basket of fruit filled with five different kids of fruits.
They are Apple, Banana, Pear, Strawberry, and Orange.
You must pick one, and by that choice I can predict your personality.

The answers:

   Apple: Person that likes apple.
Banana: Person that likes banana.
Pear: Person that likes pears.
Strawberry: Person that likes strawberry.
Orange: Person that likes Oranges

How do I do it???[hide]I DON'T KNOW!!![/hide][/quote] I don't know either

the pear. Yeah? tell me what im like.. i can bet you are gonna be wrong.

what if I hat all the listed fruit but was forced to choose one?

[quote="ToBe"]not true. I could pick the pear as a weapon, as ii'm in a war, and want to save the other fruits for actual eating.[/quote]

uhhh who uses a pear as a weapon

[quote="ToBe"]not true. I could pick the pear as a weapon, as ii'm in a war, and want to save the other fruits for actual eating.[/quote]

lol , thats really a funny reason you have there . Anyway , that " test " doesnt predict any " personality " at all ....

yeah what if the strawberries had a bunch of bugs on them so i didnt want to eat them so i chose something else at random

I thought it was going to be something like, if you didn't pick a banana or orange you would be considered lazy, because you went for the fruit you didn't have to peel.

some smart person tried to hold up  a bank with a zuchinni, im sure you could do it with a pear(forget the part about being arrested...)

or you could be in a rush if you eat strawberries because you dont have to peel and you can eat it in one bite wheras pear and apples you need to bite around a core...and if you pick pear or apples you dont have sensitive teeth

[quote="sirkuku"]with five different [b]kids [/b]of fruits.[/quote]

Is it just me, or did he say five different KIDS of fruits. Is that a spelling error?

Because in that case, I don't want any [u]kids[/u] of fruits. :D

I should win this! :yup:

What if I picked the pear so I could give it to JLI because I hate pears?

You guys are just looking for something to complain/argue about.

None of that really happened, I bet.

so
people like being right and someone else being wrong it is our nature

[quote="sirkuku"]Here is the question:

You walk into a hut abandoned with 7 beds and 7 chairs.
In the middle you see a basket of fruit filled with five different kids of fruits.
They are Apple, Banana, Pear, Strawberry, and Orange.
You must pick one, and by that choice I can predict your personality.

[hide]The answers:

   Apple: Person that likes apple.
Banana: Person that likes banana.
Pear: Person that likes pears.
Strawberry: Person that likes strawberry.
Orange: Person that likes Oranges

How do I do it?[/hide][/quote]

My answer: I don't walk into the hut.
No jk, but seriously, I could just pick one at random.  And when you say "kid of fruit", do you actually mean a kid who is like...nvm...

humanity is so overrated like totally
it says you do walk into the hut

Why do I bring 7 beds and 7 chairs along?  Are those of any significance?
Or did the person who abandon it use 7 beds and 7 chairs in the process?
Well I think I pick the watermelon just to be a rebel.