if the username you chose had already been taken what would u have been? i would have been chessblitz

Dragon96?
If you say thats taken, probably my pass (I'm not telling you) and my pass would be Dragon96

(The difference is the d is capitalized)

hmm..id say crazy<fill in with noun>maniac.

I'd probably be quickster9464, 64 is my fav number, or like quickster194, or quickster95

[quote="FantasyLover"]my username won't change.

id pick like Fantasy_Lover or FantasyLover(number)[/quote]

mine would be FantasyHater :P ;]

yangosplat

cuz thats where mine came from anyway

Mine would probably have be Eddydojokim.

hm... either FullMetalAlchemist or anime4evr or TheShamanQueen or CharaTransform or something else having to do with anime... or possibly darkmagic22 or coldfire22 or something else that just plain sounds cool...

morris142857 as it is my more frequently used username.

i actually want to change mine to this.

Well, I would have chosen a different variation on my name- when I couldn't have jmerry as an email handle, I went with merryj.

i'd hunt down the person with my chosen username eat them and take the username over  :rotfl:

I would have been "squarerootofirregulartetrahedron" or "pneumonoultramicroscopicsilicovolcanoconiosis"  :D


i tried to find a simple postable,readable solution but i don't think i was able to do better than this;

suppose the numbers in the first row are $a_1,a_2,\ldots a_n$ in descending order. also assume that $a_1,a_2,\ldots,a_m\geq 1/2;a_{m+1},\ldots,a_n<1/2$. wlog $m\leq n/2$ else we flip the two rows.

suppose $k$ is an integer such that $|k-\sum a_i|\leq 1/2$. if we choose $k$ terms from the second row and $n-k$ from the first and let the row sums of the chosen elements be $r_1,r_2$ then we have $|r_1-r_2|\leq 1/2$. if further $r_1+r_2\leq n/2$ then we are through since if say $r_1\geq r_2$ then we have $r_1-r_2\leq 1/2,r_1+r_2\leq n/2\Rightarrow 2r_1\leq (n+1)/2$.

now firstly note that $k-1/2\leq \sum a_i = \sum_{i=1}^m a_i+\sum_{i=m+1}^n a_i < m+(n-m)/2=(n+m)/2$ so that $2k\leq m+n$.
we claim that the following choice for the two rows is good:
 row 1 : $a_{k+1},a_{k+2},\ldots,a_n$
 row 2 : $1-a_1,1-a_2,\ldots,1-a_k$.
now firstly if $k\geq m$ then note that we have the following situation:
    Row 1:
    $a_1\geq\cdots\geq a_{2k-n}\geq \overbrace{a_{2k-n+1}\geq\cdots a_m\geq \cdots\geq a_k}\geq \overbrace{a_{k+1}\geq \cdots\geq a_n}$;     ( the number of terms is $n-k$ in both)
    Row 2:
                $(1-a_1)\leq (1-a_2)\leq\cdots\leq (1-a_{2k-n})\leq \underbrace {(1-a_{2k-n+1})\leq\cdots\leq (1-a_m)\leq\cdots\leq (1-a_k)}\leq\underbrace{(1-a_{k+1})\leq\cdots\leq (1-a_n)}$
now it is easy to see that the sum of the terms included in $r_1+r_2$ is $\leq$ the terms that are not included in this sum(the terms $a_{2k-n+1},\ldots a_k$ dominate the terms $a_{k+1},\ldots,a_n$;a likewise thing for the terms of row 2; finally $\forall i\in \{1,2,\ldots,2k-n\}, 1-a_i\leq a_i$ so that the claim is verified) and since sum of all the terms is $n$, it follows that $r_1+r_2\leq n/2$ as required.
if $m>k$ then note that the terms $a_i\geq 1-a_i ,i \in\{1,2,\ldots,2k-m\}$; also $a_i,i\in \{2k-m+1,\ldots,k\}$ dominate over the terms $a_i,i \in \{k+1,\ldots,m\}$; finally for $i>m, a_i<1-a_i$ and so the above inequality,$r_1+r_2\leq n/2$ is satisfied again.

[quote="grobber"]In a $2\times n$ array we have positive reals s.t. the sum of the numbers in each of the $n$ columns is $1$. Show that we can select a number in each column s.t. the sum of the selected numbers in each row is at most $\frac{n+1}4$.[/quote]

Here is my solution:

We denote the numbers from the first row by $a_1$, $a_2$, ..., $a_n$ in increasing order: $a_1\leq a_2\leq ...\leq a_n$. Then, the corresponding numbers from the second row are obviously $1-a_1$, $1-a_2$, ..., $1-a_n$.

Now, let $k$ be the largest index satisfying $a_1+a_2+...+a_k\leq\frac{n+1}{4}$. We WLOG assume that $k \neq n$ (since otherwise, we can simply select all numbers from the first row and no numbers from the second row, and we are done). Thus, $k+1 \leq n$, so that $a_{k+1}$ is well-defined. Moreover, $a_1+a_2+...+a_{k+1}>\frac{n+1}{4}$ (since $k$ was chosen to be the [b]largest[/b] index satisfying $a_1+a_2+...+a_k\leq\frac{n+1}{4}$). Now, we are going to prove that $\left(1-a_{k+1}\right)+\left(1-a_{k+2}\right)+...+\left(1-a_n\right)\leq\frac{n+1}{4}$.

In fact, recall that $a_1\leq a_2\leq ...\leq a_n$. Hence, the arithmetic mean of the numbers $a_{k+1}$, $a_{k+2}$, ..., $a_n$ is greater or equal than the number $a_{k+1}$ (the smallest of the numbers $a_{k+1}$, $a_{k+2}$, ..., $a_n$). In other words,

$\frac{a_{k+1}+a_{k+2}+...+a_n}{n-k}\geq a_{k+1}$.

On the other hand, recall again that $a_1\leq a_2\leq ...\leq a_n$. Hence, the arithmetic mean of the numbers $a_1$, $a_2$, ..., $a_{k+1}$ is smaller or equal than the number $a_{k+1}$ (the greatest of the numbers $a_1$, $a_2$, ..., $a_{k+1}$). In other words,

$\frac{a_1+a_2+...+a_{k+1}}{k+1}\leq a_{k+1}$.

Thus,

$\frac{a_{k+1}+a_{k+2}+...+a_n}{n-k}\geq a_{k+1} \geq \frac{a_1+a_2+...+a_{k+1}}{k+1}$,

and thus

$a_{k+1}+a_{k+2}+...+a_n\geq\left(n-k\right)\cdot\frac{a_1+a_2+...+a_{k+1}}{k+1} \geq \left(n-k\right)\cdot\frac{\left(\frac{n+1}{4}\right)}{k+1}$

(since $n-k\geq 0$ and $a_1+a_2+...+a_{k+1}>\frac{n+1}{4}$). In other words,

$a_{k+1}+a_{k+2}+...+a_n\geq \left(n-k\right)\cdot\frac{\left(\frac{n+1}{4}\right)}{k+1} = \frac{\left(n+1\right)\left(n-k\right)}{4\left(k+1\right)}$.

Hence,

$\left(1-a_{k+1}\right)+\left(1-a_{k+2}\right)+...+\left(1-a_n\right)$
$=\left(n-k\right)-\left(a_{k+1}+a_{k+2}+...+a_n\right)\leq\left(n-k\right)-\frac{\left(n+1\right)\left(n-k\right)}{4\left(k+1\right)}$.

Thus, in order to show that $\left(1-a_{k+1}\right)+\left(1-a_{k+2}\right)+...+\left(1-a_n\right)\leq\frac{n+1}{4}$, it will be enough to prove that $\left(n-k\right)-\frac{\left(n+1\right)\left(n-k\right)}{4\left(k+1\right)}\leq\frac{n+1}{4}$. This, however, is straightforward:

$\left(n-k\right)-\frac{\left(n+1\right)\left(n-k\right)}{4\left(k+1\right)}\leq\frac{n+1}{4}\ \ \ \ \ \Longleftrightarrow\ \ \ \ \ n-k\leq\frac{n+1}{4}+\frac{\left(n+1\right)\left(n-k\right)}{4\left(k+1\right)}$
$\Longleftrightarrow\ \ \ \ \ n-k\leq\frac{n+1}{4}\left(1+\frac{n-k}{k+1}\right)\ \ \ \ \ \Longleftrightarrow\ \ \ \ \ n-k\leq\frac{n+1}{4}\cdot\frac{n+1}{k+1}$
$\Longleftrightarrow\ \ \ \ \ n-k\leq\left(\frac{n+1}{2}\right)^2\cdot\frac{1}{k+1}\ \ \ \ \ \Longleftrightarrow\ \ \ \ \ \left(n-k\right)\left(k+1\right)\leq\left(\frac{n+1}{2}\right)^2$.

But this is clear from AM-GM: $\left(n-k\right)\left(k+1\right)\leq\left(\frac{\left(n-k\right)+\left(k+1\right)}{2}\right)^2 = \left(\frac{n+1}{2}\right)^2$.

So we have proved the inequality $\left(1-a_{k+1}\right)+\left(1-a_{k+2}\right)+...+\left(1-a_n\right)\leq\frac{n+1}{4}$. Together with $a_1+a_2+...+a_k\leq\frac{n+1}{4}$, this shows that if we choose the numbers $a_1$, $a_2$, ..., $a_k$ from the first row and the numbers $1-a_{k+1}$, $1-a_{k+2}$, ..., $1-a_n$ from the second row, then the sum of the chosen numbers in each row is $\leq\frac{n+1}{4}$. And the problem is solved.

-------

Unrelated to my solution above, here is a question on seshadri's solution: In the case $k\geq m$, how do we make sure that $k\geq n/2$ (we need this to speak of $a_{2k-n+1}$)? Maybe $m\leq n/2$ should be replaced by $m\geq n/2$ ? In this case, seshadri's solution seems to work (one minor mistake, though: the proof of $2k\leq n+m$ does not work in the special case when $n=m$, because we get a $\leq$ sign instead of the $<$ sign in $k-1/2\leq \sum a_i = \sum_{i=1}^m a_i+\sum_{i=m+1}^n a_i < m+(n-m)/2=(n+m)/2$; but fortunately there is an easy reason why $2k\leq n+m$ holds in the case when $n=m$).

  Darij

I've got this solution just few minutes ago.....

Assume the numbers in first row are $a_i's$ with non decreasing order , similarly define $b_i's$ for second row.
Let $k$ be the least such that $\sum_{r=1}^{k} a_r >\frac {n+1}{4}$
Now $a_k\geq \frac {\sum_{r=1}^{k} a_r }{k}>\frac {n+1}{4k}$
Now $\sum_{r=k}^{n}b_r\leq (n+1-k)(b_k)<(n+1-k)(1-\frac {n+1}{4k})=x$
Now note $x=\frac {5}{4}(n+1)-\frac {(n+1)^2+4k^2}{4k}\leq \frac {5}{4}(n+1)-\frac {4k(n+1)}{4k}=\frac {n+1}{4}$
Also by our assumption we had $\sum_{r=1}^{k-1}a_i\leq \frac {n+1}{4}$. So done.

it can be done beautifully by greedy algorithm
just take the row with the less sum and add it's min to it :lol:  :D  :starwars:

[hide=" Solution"]
Let the numbers in the first row be $a_1, a_2, \cdots , a_n,$ where WLOG $a_1 \leq a_2 \cdots \leq a_n.$ Let the numbers written in the second row be $b_1, b_2, \cdots, b_n$ where $a_i+b_i= 1 \quad \forall \ i.$ WLOG assume $\displaystyle \sum_{i=1}^{n} a_i \leq \displaystyle \sum_{i=1}^{n} b_i.$ Since $\displaystyle \sum_{i=1}^{n} a_i+ \displaystyle \sum_{i=1}^{n} b_i = n,$ $\displaystyle \sum_{i=1}^{n} a_i$ must be at least $\dfrac{n}{2}.$

Let $k$ be the largest number satisfying $\displaystyle \sum_{i=1}^{k} a_i \leq \dfrac{n+1}{4}$ (since the total sum of the $a_i$'s is greater than $\dfrac{n+1}{4}$ and each $a_i$ is at most 1, such a $k$ must exist).

Consider the numbers $b_{k+1}, b_{k+2}, \cdots , b_n.$ I claim that $\displaystyle \sum_{i=k+1}^{n} b_i \leq \dfrac{n+1}{4}.$ We have that $\displaystyle \sum_{i=k+1}^{n} b_i = \displaystyle \sum_{i=k+1}^{n} 1-a_i = (n-k) - \displaystyle \sum_{i=k+1}^{n} a_i.$ Hence, it suffices to show that $\displaystyle \sum_{i=k+1}^{n} a_i \geq n-\dfrac{n+1}{4}-k.$ Since the sequence $a_i$ is increasing, we have $\displaystyle \sum_{i=k+1}^{n} a_i \geq (n-k)a_{k+1}$ and $\dfrac{n+1}{4} \leq \displaystyle \sum_{i=1}^{k+1} a_{k+1} \leq (k+1) a_{k+1}.$ Combining them, it follows that
\[\displaystyle \sum_{i=k+1}^{n} a_i \geq \dfrac{(n+1)(n-k)}{4(k+1)}.\]
It is easy to show that the RHS is always at least $\dfrac{3n-1}{4}-k$, so we are done. $\blacksquare$ [/hide]

Sort the array by the top entry. Let the top entries form the non-decreasing sequence $(a_i)$. Let the corresponding bottom entries form the sequence $(b_i)$. Consider $k$ such that
$$\sum_{i=1}^{k} a_i \leq \frac{n+1}{4} \text{ and } \sum_{i=1}^{k+1} a_i > \frac{n+1}{4}$$
It follows that $a_{k+1}> \frac{n+1}{4(k+1)}$ implying that $b_{k+1}<\bigg(1-\frac{n+1}{4(k+1)}\bigg)$. Since for any $m\geq k+1$ we have that $b_m\leq b_{k+1}$, it suffices to show that
$$(n-k)\bigg(1-\frac{n+1}{4(k+1)}\bigg)\leq \frac{n+1}{4}$$
Expanding and simplifying, we have see that it suffices to show that
$$(n-2k)(n-2k-2)+1\geq 0$$
Considering the cases $n-2k=1$, $n-2k=0$, $n-2k<0$ and $n-2k>0$ gives the desired result. $\square$

[hide=Making Titu Proud]
Sort the columns in increasing order by top number in each column. Choose $k, m$ maximal such that the first $k$ entries in the first row and the last $m$ entries in the second row sum to at most $\frac{n+1}{4}$. Now suppose $k + m < n$. Let $a_1, a_2, \dots, a_n$ be the entries in the first row starting from the left, and $b_1, b_2, \dots, b_n$ be the entries in the second row starting from the right. We have $(k+1) + (m+1) \le n+1$, so by increasing condition, $a_{k+1} + b_{m+1} \le 1$. Furthermore, we have $a_1 + a_2 + \dots + a_{k+1} > \frac{n+1}{4}$ by maximality, so $a_{k+1} > \frac{n+1}{4(k+1)}$. Similarly, $b_{m+1} > \frac{n+1}{4(m+1)}$. Therefore, by Titu's Lemma, we have
$$1 > \frac{n+1}{4(k+1)} + \frac{n+1}{4(m+1)} \ge \frac{4(n+1)}{4(k+m+2)} \ge 1,$$
contradiction. Thus, $k + m \ge n$, and so the choice $a_1, a_2, \dots, a_k, b_{n-k}, b_{n-k-1}, \dots, b_1$ is valid.
[/hide]

Let $x_1,\ldots,x_n$ be the numbers in the top row. Our goal is to show that there is some $I\subset[n]$ such that \[\sum_{i\in I}x_i\le\frac{n+1}{4}\] and \[\sum_{i\not\in I}(1-x_i)\le\frac{n+1}{4}.\] WLOG suppose $x_1\le x_2\le\cdots\le x_n$. Then, pick $I=[m]$ where $m$ is the index such that \[x_1+\cdots+x_m\le\frac{n+1}{4}\quad\text{ and }\quad x_1+\cdots+x_{m+1}>\frac{n+1}{4}.\] If this index does not exist, then $I=[n]$ works just fine.

[b]Remark:[/b]  This is basically greedy algorithm. In fact, it is easy to show that this is forced, but we omit it for brevity.

We claim that $I=[m]$ works. Indeed, $\sum_{i\in I}x_i\le\frac{n+1}{4}$ by construction, so it suffices to show that $\sum_{i\not\in I}(1-x_i)\le\frac{n+1}{4}$.

We see that \[(m+1)x_{m+1}\ge x_1+\cdots+x_{m+1}>\frac{n+1}{4},\] so $x_{m+1}>\frac{n+1}{4(m+1)}$. Note that $x_i\ge x_{m+1}>\frac{n+1}{4(m+1)}$ for $i\ge m+1$, so \[S:=(1-x_{m+1})+\cdots+(1-x_n)\le(n-m)(1-x_{m+1})<(n-m)\left(1-\frac{n+1}{4(m+1)}\right).\] Let $k=\frac{m+1}{n+1}$. Then, we have \[S<(n+1)(1-k)\left(1-\frac{1}{4k}\right)=(n+1)\left(\frac{5}{4}-k-\frac{1}{4k}\right)\le\frac{n+1}{4},\] where the last step is by AM-GM. Thus, we have $S\le\frac{n+1}{4}$, so the proof is complete.

Let the numbers in the top row be $a_1 \le \cdots \le a_n$. We will choose $a_1,\ldots,a_k$ in the first row, such that
\[ a_1+\cdots+a_k \le \frac{n+1}{4}, \qquad a_1+\cdots+a_{k+1} >\frac{n+1}{4}. \]
We want to show $(1-a_{k+1})+\cdots+(1-a_n) \le \tfrac{n+1}{4}$. Note
\[ (1-a_{k+1})+\cdots+(1-a_n) \le (n-k)(1-a_{k+1}),\]
so we want to show
\[ a_{k+1} \ge 1-\frac{n+1}{4(n-k)}. \]
We have
\[ (k+1)a_{k+1} \ge a_1+\cdots+a_{k+1} > \frac{n+1}{4} \implies a_{k+1} > \frac{n+1}{4(k+1)}. \]
Now it suffices to show
\[ \frac{n+1}{4(k+1)} \ge 1-\frac{n+1}{4(n-k)}, \text{ i.e. } (n+1)^2 \ge 4(k+1)(n-k),\]
which is true by AM-GM.

[hide=Remarks]
Greedy is the obvious strategy and checking it works is not hard.
[/hide]

Let the numbers in the top row be \(a_1\le a_2\le\cdots\le a_n\) and let the numbers in the bottom row be \(1-a_1\), \(1-a_2\), \(\ldots\), \(1-a_n\). Consider the index \(k\) with the property that \[a_1+\cdots+a_k\le\tfrac{n+1}4<a_1+\cdots+a_{k+1}.\] Then it follows that \begin{align*}     (1-a_{k+1})+\cdots+(1-a_n)     &\le(n-k)(1-a_{k+1})\\     &\le(n-k)\left(1-\frac{n+1}{4(k+1)}\right)\le\frac{n+1}4, \end{align*} where the last inequality holds since \[\frac{n+1}{4(n-k)}+\frac{n+1}{4(k+1)}\ge1.\]

We sort the top row in ascending order, and denote them $a_1<a_2<a_3<\cdots < a_n$. Assume $A$ satisfies
\[\sum_{i=1}^A a_i > \frac{n+1}{4}\]
This implies $A\cdot a_A > \frac{n+1}{4}$ as well, and by extension $b_A < 1 - \frac{n+1}{4A}$. Then, we note that
\begin{align*}
    \sum_{i=A}^n b_i &< (n-A+1) b_A\\
    &<((n+1)-A) \left(1- \frac{n+1}{4A}\right)\\
    % &= \frac{1}{4A} ((n+1)-A)(4A - (n+1))
    &= (n+1) - A - \frac{(n+1)^2}{4A} + \frac{n+1}{4}\\
    &= -\frac{((n+1) + 2A)^2}{4A} + \frac{n+1}{4}\\
    &< \frac{n+1}{4}
\end{align*}
Thus, if we select the largest $X$ such that $\sum_{i=1}^X a_i< \frac{n+1}{4}$, then $\sum_{j=1}^{X+1} b_i< \frac{n+1}{4}$, so we have select the $a$ in the first $X$ columns, and $b$ in the last $n-X$ columns, and the sum in each row will always be $< \frac{n+1}{4}$.

Suppose in the top row our numbers are $a_1\le a_2\le\dots\le a_n$. Let $m$ be the index such that \[a_1+\dots+a_m\le\frac{n+1}{4}<a_1+\dots+a_{m+1}\] It suffices to show \[(1-a_{m+1})+(1-a_{m+2})+\dots+(1-a_n)\le\frac{n+1}{4}\] Notice that \[a_{m+1}\ge\frac{a_1+\dots+a_{m+1}}{m+1}>\frac{n+1}{4(m+1)}\] so \[(1-a_{m+1})+\dots+(1-a_n)\le (n-m)-\frac{(n-m)(n+1)}{4(m+1)}\le\frac{n+1}{4}\] by expanding and using AM-GM. $\square$

$\color{blue} \boxed{\textbf{SOLUTION}}$

Let $a_1 \le a_2 ... \le a_n, b_1\ge b_2... \ge b_n$ be the numbers in column $1$ to $n$
There exist some $k$ for which $$\sum_{i=1}^k a_i \le \frac{n+1}{4}$$
But
$$\sum_{i=1}^{k+1} a_i >\frac{n+1}{4}$$

$$a_{k+1} \ge \frac{\sum_{i=1}^{k+1} a_i}{k+1} > \frac{n+1}{4(k+1)}$$

$$\sum_{i=k+1}^n b_i < b_{k+1}(n-k) < (1-a_{k+1})(n-k)$$

We need to show,
$$\left(n-k\right)-\frac{\left(n+1\right)\left(n-k\right)}{4\left(k+1\right)}\leq\frac{n+1}{4}$$

Which is true because -
$$ n-k\leq\frac{n+1}{4}\left(1+\frac{n-k}{k+1}\right)\ \ \ \ \ \Longleftrightarrow\ \ \ \ \ n-k\leq\frac{n+1}{4}\cdot\frac{n+1}{k+1} \Longleftrightarrow\ \ \ \ \ n-k\leq\frac{n+1}{4}\cdot\frac{n+1}{k+1} \Longleftrightarrow\ \ \ \ \ n-k\leq\left(\frac{n+1}{2}\right)^2\cdot\frac{1}{k+1}\ \ \ \ \ \Longleftrightarrow\ \ \ \ \ \left(n-k\right)\left(k+1\right)\leq\left(\frac{n+1}{2}\right)^2$$

This is true by $\textbf{AM-GM Inequality}$
$$\left(n-k\right)\left(k+1\right)\leq\left(\frac{\left(n-k\right)+\left(k+1\right)}{2}\right)^2 = \left(\frac{n+1}{2}\right)^2 \blacksquare$$

Here's a pretty insane greedy solution.

We do the following: in the beginning, no columns are crossed out. We select the smallest number not crossed out on the array, providing that this does not violate the condition. Then we cross out the column that number is in.

At some point, this process must terminate, i.e. selecting the smallest remaining number will cause the corresponding row to have a sum greater than $\frac{n+1}4$. It follows that before that number was selected, the sum of that row $r$ is greater than $\frac{n-1}4$, as the selected number is at most $\frac 12$. Suppose that, at this point, $k$ numbers were selected in $r$.

Then, consider the remaining $n-k$ columns (note that some of the columns may be crossed out already, but that doesn't matter). Suppose for the sake of contradiction that the numbers in those $n-k$ columns $c_1, c_2, \dots, c_{n-k}$ not in $r$ sum to more than $\frac{n+1}4$. Then, the numbers in the $c_i$ in row $r$ must sum to less than $n-k-\frac{n+1}4$.

However, note that by our greedy choice, in row $r$, every element that is in a column $c_i$ must be greater than any element that is not in any of the $c_i$; otherwise, we would have chosen the element in $c_i$ over that other element. Let $C$ be the set of $c_i$. It follows that we have the inequality $$\frac{n-1}{4k} < \operatorname{Avgnum}(\overline C) < \operatorname{Avgnum}(C) < 1-\frac{n+1}{4k}.$$ This simplifies to $$k(n-k)(2k-n)(2k-n+1) > 0,$$ i.e. $\frac {n-1}2 < k < \frac n2$. There is no integer in that open interval, hence contradiction.

Order the top row so that the numbers are in increasing order and are $a_1 , a_2 ... , a_n$.
The algorithm is that we choose $a_1,a_2 ... a_r$ till the sum is $\le \frac{n+1}{4}$ and $a_1+ a_2 ... + a_r + a_{r+1} > \frac{n+1}{4}$. We claim that this algorithm works.
So we need to show
$$n-r - a_{r+1} - a_{r+2} ... - a_{m} \le \frac{n+1}{4} \iff a_{r+1} + a_{r+2} ... + a_{m} > \frac{3n-4r-1}{4}$$
For the sake of contradiction assume otherwise. We have
$$ a_r(n-r) \le a_{r+1} + a_{r+2} ... + a_{m} < \frac{3n-4r-1}{4}$$ and
$$a_{r+1} > \frac{n+1}{4} - a_1 ... - a_r \ge \frac{n+1}{4} - a_r r \Rightarrow \frac{3n-4r-1}{4} > (n-r)(\frac{n+1}{4} - a_r r) $$
Isolating $a_r$ on both inequalities gives
$$\frac{n^2 - 2n +3r-rn+1}{4r(n-r)} < a_r \le \frac{3n-4r-1}{4(n-r)} \Rightarrow n^2 -2n +4r -4rn + 4r^2 + 1 = (n - 2r - 1)^2 < 0$$ which is a contradiction.


I did it mod 20, any nicer solution?
I added question down...

Just use brute force. Here is brute force for 3:

[hide="look in here"]If a pentagonal number ends in 3, then twice that pentagonal number ends in 6. Therefore, n(3n-1) ends in 6. We can plug in the last digits of n and find out that it doesn't work:

0,0
1,2
2,0
3,5
4,4
5,0
6,2
7,0
8,4
9,4[/hide]

P.S. How did you do it in mod 20, or is it too long to type?

Mod 5, could be a little easier,
$(3n^{2}-n)/2 \equiv (-2n^{2}+4n)/2 \equiv 2n-n^{2}$
You can try  (0,1,2,3,4) here or ...
May be even $\equiv 1-(n-1)^{2}\equiv 1-(0,1,-1) \equiv (1,0,2)$

(Square of a number mod 5 could only be 0, +1 or -1 )
You don't see 3 or 4 mod 5 ... So you don't get (3,4,8,9 as unit digits) .

Working modulo 10 works fine by just factoring into a square and considering quadratic residues modulo 10.:)

$\frac{n(3n-1)}{2}=10x+k,n(3n-1)=20x+2k$
concerning the square problem, I found the answer easily by seeing the pattern, how to find the answer without it?


number theory
greatest common divisor
Euler
quadratics
modular arithmetic
number theory unsolved

given $ \phi(5^{m}-1)=5^{n}-1$ show $ (m,n)\not =1$.

$ m,n\in\mathbb{N}$ and $ \phi$ is the euler phi function.

I HAVE ONLY SOME IDEAS & A PARTIAL PROOF:

Observe that $ 5^{t}-1=4(5^{t-1}+5^{t-2}+\cdots+1)$ and $ (5^{n}-1,5^{m}-1)=5^{(m,n)}-1$.

CASE 1: $ n$ is even.
Suppose $ 5^{n}-1=\phi(5^{m}-1)$. So, $ 4\mid 5^{n}-1$ implies that $ 8\mid 5^{m}-1$.  Hence from our first observation, $ m$ is even.  Thus, $ (m,n)\ge 2$.

CASE 2: $ n$ is odd.
Again, $ 4 \mid 5^{n}-1$ but moreover, $ 8 \nmid 5^{n}-1$ (i.e. $ 4\parallel 5^{n}-1$) since $ n$ is odd.  Now, it must be that $ 8 \parallel 5^{m}-1$ which forces $ m$ to be even.  So, $ 8 \parallel 5^{m}-1$ and $ m=2k$ yields
$ 8\parallel 5^{m}-1=4(5^{k}+1)(5^{k-1}+5^{k-2}+\cdots+1)$,
i.e. $ 2\parallel (5^{k}+1)(5^{k-1}+5^{k-2}+\cdots+1)$ which forces $ k$ to be odd.  Now, maybe this will help...

hi baz,
could u explain why 4|5^n-1 implies that 8|5^m-1 in the first case.

sunset,  CORRECTION...Sorry how about this:

Observe that $ 5^{t}-1=4(5^{t-1}+5^{t-2}+\cdots+1)$.

CASE 1': Suppose a solution exists for $ m\ge 2$ is even.  Assume $ (m,n)=1$.  Then, $ n$ is odd.  So, $ 8\mid 5^{m}-1$ and $ 4\mid\mid 5^{n}-1$ via the observation.  The first fact implies that $ 2^{3}\mid \phi(5^{m}-1)$ since $ 2^{4}\mid 5^{m}-1$...Why?  Because, for $ m\ge 2$, $ 5^{m}-1$ has an odd prime divisor $ \ell$ which contributes another factor of 2 to $ \phi(5^{m}-1)$.

Again, I don't have a complete solution.  These are just ideas to help.  I posted these since you were anxious to solve this.  Hope this helps. :)

I'll try to give some further help.

Let's suppose $ (m,n)=1$ and $ \phi (5^{m}-1)=5^{n}-1$

If m is even then $ 8=\phi (24)|\phi (5^{m}-1)$ since $ 24|5^{m}-1$, and so also n is even, absurd...

Let's say $ \frac{5^{m}-1}{4}=\prod_{i=1}^{a}p_{i}^{k_{i}}$ where all $ p_{i}$ are odd primes. Modulo $ p_{i}$ 5 is a quadratic residue, for $ ord_{p_{i}}(5)|m$ so it's odd. Then $ p_{i}\equiv \pm 1\pmod{5}$. If $ p_{i}\equiv 1\pmod 5$ then $ 5|\phi(5^{m}-1)$, absurd. So we have:

$ \prod_{i=1}^{a}p_{i}^{k_{i}}=\frac{5^{m}-1}{4}\Rightarrow$ $ \prod_{i=1}^{a}(-1)^{k_{i}}\equiv 1\pmod 5$

So $ \sum_{i=1}^{a}k_{i}$ is even.

And:

$ \phi(5^{m}-1)=5^{n}-1\Rightarrow$ $ 2\cdot\prod_{i=1}^{a}(-2)(-1)^{k_{i}-1}\equiv 2^{a+1}\cdot\prod_{i=1}^{a}(-1)^{k_{i}}$ $ \equiv 2^{a+1}\equiv-1\pmod 5$

So $ a$ is odd.

This mean that there exists a $ k_{i}\equiv 0\pmod 2$ so $ p_{i}|5^{m}-1$ and $ p_{i}|5^{n}-1$ hence $ p_{i}|4$, absurd.

This problem is from Taiwan National Olympiad, year 2000.


[quote="Kastryas"][b]Let[/b] $ a; b; c > 0$ [b]and[/b] $ a \plus{} b \plus{} c \equal{} 1$
[b]Prove that:[/b]

$ \frac {\sqrt {a^2 \plus{} abc}}{ab \plus{} c} \plus{} \frac {\sqrt {b^2 \plus{} abc}}{bc \plus{} a} \plus{} \frac {\sqrt {c^2 \plus{} abc}}{ca \plus{} b}\le \frac {1}{2\sqrt {abc}}$[/quote]
Maybe it works:

$ \sum_{cyc}\frac {\sqrt {a^2 \plus{} abc}}{ab \plus{} c}\equal{}\sum_{cyc}\frac{\sqrt{a(a\plus{}b)(a\plus{}c)}}{(a\plus{}c)(b\plus{}c)}\equal{}\sum_{cyc}a\sqrt{\frac{a\plus{}b}{a(a\plus{}c)(b\plus{}c)^2}}\leq$
$ \leq\sqrt{\sum_{cyc}\frac{a\plus{}b}{(a\plus{}c)(b\plus{}c)^2}}.$ Thus, it remains to prove that:
$ (a\plus{}b)^2(a\plus{}c)^2(b\plus{}c)^2(a\plus{}b\plus{}c)\geq4abc\sum_{cyc}(a\plus{}b)^3(a\plus{}c).$ :maybe:

[quote="arqady"][quote="Kastryas"][b]Let[/b] $ a; b; c > 0$ [b]and[/b] $ a \plus{} b \plus{} c \equal{} 1$
[b]Prove that:[/b]

$ \frac {\sqrt {a^2 \plus{} abc}}{ab \plus{} c} \plus{} \frac {\sqrt {b^2 \plus{} abc}}{bc \plus{} a} \plus{} \frac {\sqrt {c^2 \plus{} abc}}{ca \plus{} b}\le \frac {1}{2\sqrt {abc}}$[/quote]
Maybe it works:

$ \sum_{cyc}\frac {\sqrt {a^2 \plus{} abc}}{ab \plus{} c} \equal{} \sum_{cyc}\frac {\sqrt {a(a \plus{} b)(a \plus{} c)}}{(a \plus{} c)(b \plus{} c)} \equal{} \sum_{cyc}a\sqrt {\frac {a \plus{} b}{a(a \plus{} c)(b \plus{} c)^2}}\leq$
$ \leq\sqrt {\sum_{cyc}\frac {a \plus{} b}{(a \plus{} c)(b \plus{} c)^2}}.$ Thus, it remains to prove that:
$ (a \plus{} b)^2(a \plus{} c)^2(b \plus{} c)^2(a \plus{} b \plus{} c)\geq4abc\sum_{cyc}(a \plus{} b)^3(a \plus{} c).$ :maybe:[/quote]

Can u give me a full and clear solution

[quote="Kastryas"][b]Let[/b] $ a; b; c > 0$ [b]and[/b] $ a \plus{} b \plus{} c \equal{} 1$
[b]Prove that:[/b]

$ \frac {\sqrt {a^2 \plus{} abc}}{b \plus{} ca} \plus{} \frac {\sqrt {b^2 \plus{} abc}}{c \plus{} ab} \plus{} \frac {\sqrt {c^2 \plus{} abc}}{a \plus{} bc}\le \frac {1}{2\sqrt {abc}}$[/quote]
This problem is unsolved in some forums. Let I post my solution for it
Denote $ q \equal{} ab \plus{} bc \plus{} ca,r \equal{} abc$, we have the following lemma
[b]Lemma.[/b]
\[ r \le \frac {q^2(1 \minus{} q)}{2(2 \minus{} 3q)}
\]
By The Cauchy Schwarz Inequality, we get
\[ \left[ \sum \frac {\sqrt {a^2 \plus{} abc}}{(b \plus{} c)(b \plus{} a)} \right]^2 \le \left[ \sum \frac {a}{(a \plus{} b)(b \plus{} c)}\right]\left(\sum \frac {a \plus{} c}{b \plus{} c}\right) \equal{} \frac {\sum a^2 \plus{} \sum ab}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}\left( \sum \frac {a \plus{} c}{b \plus{} c}\right)
\]
Furthermore,
\[ \sum \frac {a \plus{} c}{b \plus{} c} \equal{} \sum \frac {1}{b \plus{} c} \minus{} \sum \frac {b}{b \plus{} c} \le \sum \frac {1}{b \plus{} c} \minus{} \frac {(a \plus{} b \plus{} c)^2}{\sum a^2 \plus{} \sum ab}
\]
It suffices to prove that
\[ \frac {\sum a^2 \plus{} \sum ab}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}\left[ \frac {1}{b \plus{} c} \minus{} \frac {1}{\sum a^2 \plus{} \sum ab}\right] \le \frac {1}{4abc}
\]

\[ \Leftrightarrow \frac {1 \minus{} q}{q \minus{} r} \left( \frac {1 \plus{} q}{q \minus{} r} \minus{} \frac {1}{1 \minus{} q}\right) \le \frac {1}{4r}
\]

\[ \Leftrightarrow \frac {4(1 \minus{} q^2)}{q \minus{} r} \minus{} 4 \le \frac {q \minus{} r}{r}
\]

\[ \Leftrightarrow \frac {4(1 \minus{} q^2)}{q \minus{} r} \minus{} \frac {q}{r} \le 3
\]
Using our lemma, we get
\[ LHS \le \frac {4(1 \minus{} q^2)}{q \minus{} \frac {q^2(1 \minus{} q)}{2(2 \minus{} 3q)}} \minus{} \frac {q}{\frac {q^2(1 \minus{} q)}{2(2 \minus{} 3q)}} \equal{} 3 \minus{} \frac {q(1 \minus{} 3q)(5 \minus{} 7q)}{(1 \minus{} q)(4 \minus{} 7q \plus{} q^2)} \le 3
\]
We have done.

Problem for own study: Prove the above lemma. :)
One more question: how can we figure out the lemma? :)
Regards


Find all functions $f: \mathbb R \to \mathbb R$ such that \[ f((x\plus{}z)(y\plus{}z))\equal{}(f(x)\plus{}f(z))(f(y)\plus{}f(z)), \quad x,y,z \in \mathbb R.\]

[quote="Pirkuliyev Rovsen"]Find all functions $ f: R\rightarrow R$ such that $ f((x \plus{} z)(y \plus{} z)) \equal{} (f(x) \plus{} f(z))(f(y) \plus{} f(z))$,$ x,y,z \in R$
[/quote]
Let $ P(x,y,z)$ be the assertion $ f((x\plus{}z)(y\plus{}z))\equal{}(f(x)\plus{}f(z))(f(y)\plus{}f(z))$

1) Suppose $ \exists u$ such that $ f(u)\plus{}f(\minus{}u)\ne 0$. Then $ P(x,u,\minus{}u)$ $ \implies$ $ f(0)\equal{}(f(x)\plus{}f(\minus{}u))(f(u)\plus{}f(\minus{}u))$ and so $ f(x)\equal{}c$ with $ c\equal{}4c^2$ hence two solutions : $ f(x)\equal{}0$ and $ f(x)\equal{}\frac 14$

2) consider now $ f(x)\plus{}f(\minus{}x)\equal{}0$ $ \forall x$
This implies $ f(0)\equal{}0$
$ P(x,y,0)$ $ \implies$ $ f(xy)\equal{}f(x)f(y)$ and $ P(x,y,z)$ becomes $ f(x\plus{}z)f(y\plus{}z)\equal{}(f(x)\plus{}f(z))(f(y)\plus{}f(z))$

Using then $ x\equal{}y$ in this equality, we get $ f(x\plus{}z)^2\equal{}(f(x)\plus{}f(z))^2$ and so $ |f(x\plus{}z)|\equal{}|f(x)\plus{}f(z)|$

But $ f(xy)\equal{}f(x)f(y)$ implies $ f(x)\ge 0$ $ \forall x\ge 0$ and so $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ $ \forall x,y\ge 0$

This is a classical Cauchy's equation with bounds, and so $ f(x)\equal{}ax$ and so $ f(x)\equal{}x$ (using $ f(xy)\equal{}f(x)f(y)$) and this, indeed, is a solution.

Hence the three solutions :
$ f(x)\equal{}x$

$ f(x)\equal{}0$

$ f(x)\equal{}\frac 14$

[quote=pco]and so $ f(x)=c$ with $ c=4c^2$ [/quote]
sorry, I don't understand how you get this.

$f(0)=(f(x)+f(-u))(f(u)+f(-u))$ and $f(u)+f(-u)\ne 0$ implies

$f(x)=\frac{f(0)}{f(u)+f(-u)}-f(-u)=c$ constant $\forall x$

Plugging this back in original equation, we get $c=(c+c)(c+c)=4c^2$


Given a circle $\omega$ and chord $CD$ on it. and $M$ is the midpoint of $CD$. Chord $AB$ intersects $CD$ such that $CD$ bisects the $\angle AMB$. if $N$ is the midpoint of $AB$ prove that $AB$ bisects the $\angle CND$. :D

Let the parallel from $B$ to $CD$, intersect $\omega$ at point $X$.
Then $\triangle{CMX}=\triangle{DMB}$, so $\angle{CMX}=\angle{DMB}=\angle{AMD}$, hence points $A,M,X$ are collinear. So we have $S_\triangle{ACX}=S_\triangle{ADX}$, or:
$AC\cdot{CX}=AD\cdot{DX}$ (1) and since $CXBD$ is an isosceles trapezium $BC=DX$ and $CX=BD$ $\rightarrow$ (1) becomes $AC\cdot{BD}=AD\cdot{BC}$ (2).
But the condition $AB$ bisects $\angle{CND}$ is also equivalent to (2) (this can be proven in a similar way). Hence we are done.

That's a really strange way to formulate the problem from http://www.mathlinks.ro/Forum/viewtopic.php?t=6557 (by the way, this topic contains some facts about harmonic quadrilaterals, just without using the word harmonic quadrilateral)...

  darij

i did not formulate this problem, i saw this problem in my class and i even di not know it has formulated.anyway thank you so much. :D

Sailor your solution is very cool and you do it well.excellent!!.. ;)


[quote="easternlatincup"]Find all $f: N\rightarrow N$ such that
(i) $f(n+m) = f(n)f(m)$ for all $m,n\in N$
(ii) $f(f(k)) = f(k)^{2}$ for at least one $k\in N$[/quote]
From (i) we have $f(n)=a^{n}, a=f(1)$
If $a=1$, then $\forall k\in N \ 1=f(f(k))=(f(k))^{2}$.
If $a>1$ from (ii) we have exist k: $a^{k}=2k\Longrightarrow a=2, k=1 \ or \ 2.$


If $y=\sqrt{1+x^2}$, find the range of $\frac{y}{x+1}$.

wouldn't u jsut substitute the y in the problem for that which it equals?
i don't know how to get teh signs in and stuff but

(1+x)^.5
-----------
(x+1)

and range is all x values?

so
 -1<x ?

No. We are not finding the range of $x$, but the range of the whole expression. Think more.

Let $k=\frac{y}{x+1}=\frac{\sqrt{x^2+1}}{x+1}$. So $(k^2-1)x^2+2k^2x+(k^2-1)=0$
Leading coefficient cannot be zero, so $k\ne-1,1$. For real values of $x$ to exist, discriminant needs to be nonnegative, so $(2k^2)^2-4(k^2-1)^2=8k^2-4\ge0\Rightarrow k^2\ge\frac12$

So range is $(-\infty,-1)\cup(-1,-\frac1{\sqrt2}]\cup[\frac1{\sqrt2},1)\cup(1,+\infty)$

You still have some problems in your answer. Think of the title.


Let $ a,b,c$ be positive reals. Prove that

$ \prod(a^2\plus{}ab\plus{}b^2)\geq6(\sum a^2b)(\sum ab^2)\minus{}3(\sum a^2b^2)(\sum a^2)$

[quote="Ligouras"]Let $ a,b,c$ be positive reals. Prove that

$ \prod(a^2 \plus{} ab \plus{} b^2)\geq6(\sum a^2b)(\sum ab^2) \minus{} 3(\sum a^2b^2)(\sum a^2)$[/quote]

 The following inequality is equivalent to

$ 4(a\minus{}b)^2(\minus{}c\plus{}b)^2(c\minus{}a)^2\plus{}3\sum{b^2c^2(a\minus{}b)(a\minus{}c)}\plus{}3abc\sum{a(a\minus{}b)(a\minus{}c)}\geq 0$

$ \sum{b^2c^2(a\minus{}b)(a\minus{}c)}\geq 0$   and   $ \sum{a(a\minus{}b)(a\minus{}c)}\geq 0$

true due to schur


it's an ancient thread... I found it and thought it'd be cool.
I'm rebooting it for passing-the-time purposes
I'll give an example

[hide="I"][hide="love"][img]http://i124.photobucket.com/albums/p4/moxproti33/800px-apple_pie.jpg[/img][/hide][/hide]

I don't really know if the thread's going to live or not... but... let's give it a try  :P

[hide="chicken"]
[hide="is"]
[hide="better"]

rawr

[/hide]
[/hide]
[/hide]

I think the format is supposed to be "...<picture whose noun completes the sentence>".

[hide="I"]
[hide="just"]
[hide="ate"]
[img]http://www.thaitable.com/images/ingredients/pictures/5jasmine%20rice.jpg[/img]
[/hide]
[/hide]
[/hide]

[hide="I"]I like[/hide]
[hide="like"]veggies, and rice.... dunno how to put picture[/hide]

lol nice

[hide="I"]
[hide="is"]
[hide="coolio"]

[url=http://www.youtube.com/watch?v=mR0C55F8Eqg]cooliofied[/url]

[/hide]
[/hide]
[/hide]


[quote="mathmanman"]Please discuss it at http://www.mathlinks.ro/Forum/viewtopic.php?t=85724[/quote]
Thank you, but I don't think it has got the right solution


superior algebra
superior algebra solved

A is in M3(C) s.t. trA=trA^2=0. Show that det(A^2+I)=(detA)^2+1.

Let P(x) = det(A-xI_3) have the roots u,v,w.
Tr(A)=u+v+w=0
Tr(A^2)=u^2+v^2+w^2 => uv + vw + wu = 0
But P(x) = x^3 - (u+v+w) x^2 + (uv+vw+wu) x - (uvw) = x^3 - det(A)
Now compute P(i)*P(-i) and you're done.


Let la,lb,lc be the mesures of angle bissectors of a triangle ABC.
Prove that:
 la+lb+lc <=3/2sqrt(a^2+b^2+c^2)

That's absolutely easy: In http://www.mathlinks.ro/Forum/viewtopic.php?t=805 posts #6-#7 and in http://www.mathlinks.ro/Forum/viewtopic.php?t=14431 , it was proven that $l_a+l_b+l_c\leq p\sqrt3$, where $p=\frac{a+b+c}{2}$ is the semiperimeter of triangle ABC. But by the AM-QM inequality, $\frac{a+b+c}{3}\leq\sqrt{\frac{a^2+b^2+c^2}{3}}$, so that $a+b+c\leq 3\sqrt{\frac{a^2+b^2+c^2}{3}}=\sqrt{3\left(a^2+b^2+c^2\right)}$, and $p=\frac{a+b+c}{2}\leq\frac{\sqrt{3\left(a^2+b^2+c^2\right)}}{2}$, so that $l_a+l_b+l_c\leq p\sqrt3\leq \frac{\sqrt{3\left(a^2+b^2+c^2\right)}}{2}\sqrt3=\frac{3\sqrt{a^2+b^2+c^2}}{2}=\frac32\sqrt{a^2+b^2+c^2}$. And the problem is solved.

  Darij


Find a polynomial $ p(x)$ of degree 4 with $ p(1)\equal{} \minus{}1,p(2)\equal{}2,p(\minus{}3)\equal{}4,p(4)\equal{}5$ and $ p(5)\equal{}8$

By the Lagrange's interpolation we obtain:

$ P(x) \equal{} \sum_{i \equal{} 0}^{\deg{P(x)}} \left(f(i)\prod_{i,j \equal{} 0,i\neq j}^{\deg{P(x)}}\frac {x \minus{} x_j}{x_i \minus{} x_j}\right)$

Put $ i \equal{} 1,2, \minus{} 3,4,5$ we have desire.


[quote="xspxspxspxsp"]Let a,b,c be positive real numbers,
Find the maximum value of k
so that
look the a2.jpg[/quote]

It's VN TST 2009


calculus
integration
logarithms
absolute value
calculus computations

Calculate $ \int \frac {dx}{x^{2008}(1 \minus{} x)}$

OK. But You should add the sign of absolute value\ to anti-logarithm.

Since $ \frac{1}{x^m(1\minus{}x)}\equal{}\frac{1}{x}\plus{}\frac{1}{x^2}\plus{}\cdots \plus{}\frac{1}{x^m}\plus{}\frac{1}{1\minus{}x}\ (m\equal{}2,\ 3,\ ,\cdots),$

$ \int \frac{1}{x^m(1\minus{}x)}\ dx\equal{}\ln \left|\frac{x}{1\minus{}x}\right|\minus{}\frac{1}{x}\minus{}\frac{1}{2x^2}\minus{}\frac{1}{3x^3}\minus{}\cdots \minus{}\frac{1}{(m\minus{}1)x^{m\minus{}1}}\plus{}Const.$


Could someone please use inversive geometry to prove the following?  Thanks in advance.

[i]Triangle $DEF$ is acute. Circle $C_{1}$ is drawn with $DF$ as its
diameter and circle $C_{2}$ is drawn with $DE$ as its diameter.
Points $Y$ and $Z$ are on $DF$ and $DE$ respectively so that $EY$ and
$FZ$ are altitudes of triangle $DEF$. $EY$ intersects $C_{1}$ at $P$, and $FZ$
intersects $C_{2}$ at $Q$. $EY$ extended intersects $C_{1}$ at $R$, and $FZ$
extended intersects $C_{2}$ at $S$. Prove that $P$, $Q$, $R$, and $S$ are
concyclic points.[/i]

Masoud Zargar

$\triangle DPR$ is isosceles with DP = DR. Inversion in a circle (D) with this radius carries the lines DE, DF passing through the inversion center D into themselves. As the points $P, R \in (D)$ are carried into themselves, the circle $\mathcal C_{1}$ passing through the inversion center D is carried into the E-altitude line $PR \equiv EY$ and the E-altitude line $PR \equiv EY$ into the circle $\mathcal C_{1}.$ The circle $\mathcal C_{2}$ also passing through the inversion center D is carried into a line perpendicular to DE. Since $Y \in \mathcal C_{2}$ goes to F, this line is the F-altitude line FZ. It follows that the inversion in (D) carries the altitude line FZ in the circle $\mathcal C_{2},$ so that the intersections $Q, S \equiv FZ \cap \mathcal C_{2}$ are carried into themselves, which means that they also lie on the inversion circle (D). In conclusion, $P, Q, R, S \in (D).$


vector
linear algebra
matrix
analytic geometry

i read a proof that  there do no exist n+1 linearly independent vectors in a n-dimensional space but i dont quite get some bits
(1)Let  $ \bf{v}_{1}, {v}_{2},{v}_{3},...,{v}_{n \plus{} 1}$ be linearly independent vectors in n dimensional space
(2)then $ \bf{v}_{1}, {v}_{2},{v}_{3},...,{v}_{n}$ must be linearly independent
(3)Hence the matrix A whose columns are  $ \bf{v}_{1}, {v}_{2},{v}_{3},...,{v}_{n}$ is nonsingular ie the only solution to $ A\bf {c} \equal{} \bf0$ is $ \bf{c} \equal{} \bf0$ , where is a column vector  THIS I GET
(4)then the books says therefore the equation  $ A\bf c \equal{} \bf{v}_{n \plus{} 1}$ has unique soln for [b]c[/b] ie. $ {\bf{v}_{n \plus{} 1}} \equal{} {c}_{1}{\bf{v}_{1}} \plus{} {c}_{2}{\bf {v}_{2}} \plus{} {c}_{3}{\bf{v}_{3}} \plus{} ... \plus{} {c}_{n}{\bf {v}_{n}}$   CAN SOMEONE PLEASE EXPLAIN HOW THIS IS, as in how does  $ A\bf c \equal{} \bf{v}_{n \plus{} 1}$ have a soln??
(5)then it just follows that $ \bf{v}_{1}, {v}_{2},{v}_{3},...,{v}_{n \plus{} 1}$ would be linearly dependent and contary to assumption which i get...
so can someone please please please explain how line (4) came about

If $ \mathbf{Ac} \equal{} \mathbf{v}_n$ has two solutions $ \mathbf{c}_1, \mathbf{c}_2$ then $ \mathbf{A} (\mathbf{c}_1 \minus{} \mathbf{c}_2) \equal{} \mathbf{0} \implies \mathbf{c}_1 \equal{} \mathbf{c}_2$ by part (3).

i dont quite get it and why do you 1st assume there are 2 solns?? and sorry i mistyped in my original one should be Ac=v_n+1 not v_n

I am quite sure this is not an Open problem, and we have a forum for linear algebra; next times, please post it in an appropriate Forum (like the one it is now).

Sorry, that wasn't quite enough work.  There are a lot of properties equivalent to $ \mathbf{Ac} \equal{} \mathbf{0}$ having only the trivial solution:

- $ \mathbf{A}$ has nonzero determinant
- $ \mathbf{A}$ is invertible
- $ \mathbf{Ac} \equal{} \mathbf{b}$ has a unique solution for every $ \mathbf{b}$.

The third one is the one we want.

but that will mean that in a similar manner i can prove that there do not exist n linearly independent vectors in a n-dimensional space
Let$ \bf v_1, v_2,...v_n$  be linearly independent vectors in n dimensional space
then $ \bf v_1, v_2,...v_{n \minus{} 1}$must also be linearly independent
then again matrix whose columns are $ \bf v_1, v_2,...v_{n \minus{} 1}$ is nonsing. and have only have $ {\bf c} \equal{} {\bf 0}$ as soln to $ A{\bf c} \equal{} {\bf 0}$
then by your explanation $ A{\bf c} \equal{} {\bf {v}_{n}}$ will also have a unique soln and v_n can be expressed as a linear combination of the others and that means there cannot exist n linearly independent vectors in a n-dimensional space which does not makes sense...

[quote="calculus_jy"]then again matrix whose columns are $ \bf v_1, v_2,...v_{n \minus{} 1}$ is nonsing. and have only have $ {\bf c} \equal{} {\bf 0}$ as soln to $ A{\bf c} \equal{} {\bf 0}$[/quote]
My apologies; the properties I gave above are all only true of square matrices.  One cannot define the determinant of a nonsquare matrix, nor is there an obvious sense in which a nonsquare matrix is invertible.  (It is possible to show that if $ \mathbf{A}$ is not square, then it is impossible for $ \mathbf{Ac} \equal{} \mathbf{0}$ to have only the trivial solution.)

so now what??/ how to i prove the non existence of the n+1th vector??

[quote="t0rajir0u"](It is possible to show that if $ \mathbf{A}$ is not square, then it is impossible for $ \mathbf{Ac} \equal{} \mathbf{0}$ to have only the trivial solution.)[/quote]
Is that what you meant to say? Because that's not a correct statement. $ \mathbf{Ac} \equal{} \mathbf{0}$ has only the trivial solution precisely when the columns of $ \mathbf{A}$ are independent, which is something we can easily arrange for a matrix that is taller than it is wide.

does that mean my prove that the non exsistance of the nth lineraly independent vector in a n dimensional space is correct????/ i still dont get it???

My apologies again.  In either case, this discussion doesn't affect the validity of the original proof, and no, your proof is not correct; of course there are $ n$ linearly independent vectors in an $ n$-dimensional space. This is, in some sense, what it means to be $ n$-dimensional.

Do you understand the connection between $ \mathbf{Ac} \equal{} \mathbf{0}$ and $ \mathbf{Ac} \equal{} \mathbf{b}$ having unique solutions?  Relevant topics to explore include the rank-nullity theorem, if you want to look at it that way, or determinants and Cramer's rule if you want to look at it another way.

Here's how I would do it. Your $ n$-dimensional vector space might as well be $ \mathbb{R}^n$ (or whatever your field is). After all, saying it the space is $ n$ dimensional is saying it has a basis with $ n$ elements, and we can find for any vector the column vector of its coordinates in terms of that basis.

So assume WLOG that our $ n \plus{} 1$ vectors are given as $ n \plus{} 1$ columns, each $ n\times 1.$ Stack these columns together to form a single $ n\times (n \plus{} 1)$ matrix $ A$. Perform Gaussian elimination on $ A$, getting that it is row equivalent to a matrix in row echelon form. The matrix in row echelon form cannot have more pivot elements than it has rows, so there are at most $ n$ pivot elements. Since there are $ n \plus{} 1$ columns, then there is at least one column without a pivot element. The variable (the coefficient of a linear combination) associated with that column is then a free variable that can be set equal to any real number. Set it equal to an arbitrary nonzero real number, and use that to find a solution to $ Ax \equal{} 0$ with $ x\ne 0.$ But this is precisely a nontrivial trivial linear combination of the columns of $ A$ that equals the zero vector. If the columns of $ A$ were independent, then only the trivial linear combination of them could be zero. Hence, the columns of $ A$ are dependent.


calculus
integration
trigonometry
calculus computations

Set $x= 2+\sin \theta$ .Then the integral becomes

$\\ \int_{-\frac{ \pi }{6}}^{0}(\sin \theta+1) \, d \theta = [ \theta-\cos \theta]_{-\frac{ \pi }{6}}^{0}= \frac{ \sqrt{3}}{2}-1+\frac{ \pi }{6}$

[hide="Another hint"]$\sqrt{\frac{x-1}{3-x}}= \frac{x-1}{\sqrt{(3-x)(x-1)}}= \frac{(x-2)+1}{\sqrt{1-(x-2)^{2}}}= \frac12 \cdot \frac{2(x-2)}{\sqrt{1-(x-2)^{2}}}+\frac1{\sqrt{1-(x-2)^{2}}}.$

And, now use suitable substitutions to evaluate the original integral.

[b]Ramanujan's [/b]solution is, of course, a slick one![/hide]

Haha that problem lasted a long time  :blush: .  How about this one, maybe it will pose a challenge:

$\int^{q}_{p}{\left(\frac{x-a}{b-x}\right)^{\frac12}}\,dx$ $a<b$

where $p=\frac{3a+b}{4}$ and $q=\frac{a+b}{2}$

[quote="Ramanujan"]$\\ \frac{x-a}{b-x}= \frac{x-b+b-a }{b-x}= \frac{b-a}{b-x}-1$ ....

or you mean $\int^{q}_{p}{\left(\frac{x-a}{b-x}\right)^\frac{1}{2}}\,dx$     :)[/quote][hide="Hint"]Assuming the integral we need to evaluate is the one Ramanujan wrote, we note that the technique I posted (two posts above) works quite well again.

$\sqrt{\frac{x-a}{b-x}}= \frac{x-a}{(b-x)(x-a)}= \frac{(x-\frac{a+b}{2})+\frac{b-a}{2}}{\sqrt{(\frac{b-a}{2})^{2}-(x-\frac{a+b}{2})^{2}}}$[/hide]

Or we set  $x= \frac{b-a}{2}\cdot \sin \theta+\frac{b+a}{2}$

and the integral is $\frac{b-a}{2}\int_{-\frac{ \pi}{6}}^{0}(1+\sin \theta ) \, d \theta$

or something like that  ...     :D

Hehe sorry! i did mean to the power of $\frac12$. Doh! :lol: But yes, your method is what is needed.

Also suitable $x-a \, = \, (b-a) \sin u$

$\int \sqrt{\frac{x-a}{b-x}}\cdot dx \, = \, 2(b-a) \int \sin^{2}u \cdot du$


Hi, I derived the forumula below.

(-)=Theta

(-)r=sin^(-1)((n2/n1)(sin(-180+2a+sin^(-1)((n1/n2)sin((-)i))))

From a physics lab, the following variables were determined:
(-)i=57
(-)r=44
n1=1
n2=?

I was able to simplify the equation to:

sin^(-1)(0.8387/n2)-sin^(-1)(0.647/n2) = 60

Is it possible to solve for n2 or simplify the original equation?

i think it should be, you need to change from the $sin^{-1}$ on the LHS to $sin(60)$ on the RHS which should then give you something to solve for.

Hi,

Do you mean: if i take the sin^{-1} of everything, will it leave:

0.8387/n2 - 0.647/n2 = sin(60)
This doesn't seem to give a realistic answer...
n2 must be greater than one and less than 3.
Thanks,

Scott

Is there another way to solve this?  I'm confused...

Scott

That is true, except I forgot to mention that the angles come from a prism.  Therefore, the 57 is the 1st incident angle and the 44 is the second refracted angle.  That's why I had  to come up with a new equation...

The a value is the internal angle of the prism.  It looked like this:

[img]http://sgateman.awardspace.com/Raydiagram-prism2.jpg[/img]

The a is in the lower left hand corner.  This is a top down view.

i think the refractive index should be about 1.52.


Given reals 0 \leq x1,x2,x3 < \pi /2 and tan x1 tan x2 tan x3 =2 \sqrt 2.

Prove that  \sum cos xi <2.

do you have a neat solution to that? That's CMO Problem 3.

I and my classmates have invented a "Jensen inequality with one inflection point" to solve out this problem.

I'm really interested in this method, since I saw hard inequalities that could be solved like that. Could you please post it?
  Thank you.

I ask you again and I beg you to post that method. This is really extremely interesting. Thank you.

there must be something wrong with my proof it can't be so easy.
cotx1cotx2cotx3=1/2sqrt2
By chebyshev's inequality,
1/2sqrt2>=(cosx1+cosx2+cosx3)(1/sinx1+1/sinx2+1/sinx3)/3
cosx1+cosx2+cosx3<=3/2sqrt2(1/sinx1+1/sinx2+1/sinx3)<2
<=>1/sinx1+1/sinx2+1/sinx3>3/4sqrt2
While 1/sinx1+1/sinx2+1/sinx3>3
so it holds..
Very strange.

[quote="harazi"]It would have been to nice for this problem to be so easy. Unfortunately, when you apply Cebasev you get (sum cosx_i)*(sum 1/sinx_i)<=3sum ctgx_i, which is not 3/2sqrt(2). You only have that prod ctgx_i=1/2sqrt(2), not sum ctgx_i=1/2sqrt(2).[/quote]
o!!
o yes, i am stupid..

[quote="harazi"]Clearly false! This is not the answer to the problem![/quote]

Sorry may i know the ans to the problem??
After kalva removed cmo papers i dunno where to find the ans of cmo..

The answer is n-1 for all n greater than or equal to 3.

Could someone post the original, complete problem?

It seems very interesting.

Thanks!


Dear friends,
I am a newbie in Latex. I'm using TeXnicCenter to write & compile latex files. And I got an error:

Couldn't find input index file D:\LatexSamples\helloworld nor D:\LatexSamples\helloworld.idx
Usage: C:\Program files\MiKTeX 2.7\miktex\bin\makeindex.exe [-ilqrcgLT] [-s sty] [-o ind] [-t log] [-p num] [idx0 idx1 ..]

Can you explain and help me to fix this error ?
Thank you,

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]Minimal example please![/url]  - the log file would also be helpful.

Can you post your code (yes the whole document) so we can see how you got the error?

Hi idischan,

in TeXnicCenter open the menu "Output" and there "Define output profiles". In your profile(s) mark the checkbox "Don't use Makeindex in this profile".

Stefan

--
[url=http://texblog.net]TeXblog.net[/url]


MATHCOUNTS
probability
AMC
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[quote="mgao"]I was able to connect to our classroom of MATHCOUNT PROBLEM SERIES in the U.S., but now I am on vacation in Japan using a Dell laptop we bought in the U.S. before we left.  I got kicked out of the classroom each time when I clicked on classroom.  We tried to lower our security level and a few other things, but it did not work.  Could you help me with this problem?  Thanks.[/quote]

If you are using a wireless connection, that is likely the problem - they are pretty tricky to keep connected.  Get a land connection or be as close to the wireless connection point as possible.  If this isn't the problem, a firewall might be.  There is a link beneath the big box on the Classroom page (when the Classroom is open) that you should click to connect to the classroom from behind a firewall.

You can test it out during the classroom open times - there should be a post in the MATHCOUNTS Problem Series forum outlining at what times the Classroom is open for student discussion.  (We have times when the students can get together outside class time - this is a good time to test your connection.)


For a triangle ABC, let its circumcircle be (O) and a point P be on the small arc AB.
A line passing through P and perpendicular to OA meets AB, CA at D, E respectively such that PD=DE.
Draw a line passing through P and perpendicular to OB and let it meet AB, BC at F, G respectively.
Prove that PF=FG.

[quote="77ant"][color=darkred]For a triangle $ ABC$ let its circumcircle be $ (O)$ and a point $ P$ be on the small arc $ AB$ .

A line passing through $ P$ and perpendicular to $ OA$ meets $ AB$ , $ CA$ at $ D$ , $ E$ respectively .

A line passing through $ P$ and perpendicular to $ OB$ meets $ AB$ , $ BC$ at $ F$ , $ G$ respectively.

Prove that $ DP \equal{} DE\Longleftrightarrow FP \equal{} FG$ .[/color] [/quote]
[hide="An idea. I will come back soon."]
[color=darkblue]You can use a simple property of the median in a triangle. Consider a point $ M\in [BC]$ and denote $ \delta_d(X)$ - the [u]distance[/u] from $ X$ to the line $ d$ .[/color]

[quote] [color=darkred]$ MB \equal{} MC$ $ \Longleftrightarrow$  $ \delta_{AM}(B) \equal{} \delta_{AM}(C)$ $ \Longleftrightarrow$ $ AB\cdot\sin \widehat {MAB} \equal{} AC\cdot\sin\widehat {MAC}$ $ \Longleftrightarrow \sin C\cdot\sin\widehat {MAB} \equal{} \sin B\cdot\sin\widehat {MAC}$ .[/color] [/quote] [/hide]

Thanks for reading my posts and interests. :)

[quote][color=darkred]For a triangle $ ABC$ let its circumcircle be $ (O)$ and a point $ P$ be on the small arc $ AB$ .

A line passing through $ P$ and perpendicular to $ OA$ meets $ AB$ , $ CA$ at $ D$ , $ E$ respectively .

A line passing through $ P$ and perpendicular to $ OB$ meets $ AB$ , $ BC$ at $ F$ , $ G$ respectively.

Prove that $ DP \equal{} DE\Longleftrightarrow FP \equal{} FG$ $ \Longleftrightarrow$ the line $ AP$ is the $ C$-symmedian in $ \triangle ABC$ .[/color] [/quote]
[quote][color=darkblue][b][u]Lemma[/u].[/b]

In $ \triangle\ ABC$ consider a point $ M\in [BC]$ and denote $ \delta_d(X)$ - the [u]distance[/u] from $ X$ to the line $ d$ . Then $ MB \equal{} MC$ $ \Longleftrightarrow$

$ \delta_{AM}(B) \equal{} \delta_{AM}(C)$ $ \Longleftrightarrow$ $ AB\cdot\sin \widehat {MAB} \equal{} AC\cdot\sin\widehat {MAC}$ $ \Longleftrightarrow \sin C\cdot\sin\widehat {MAB} \equal{} \sin B\cdot\sin\widehat {MAC}$ .[/color] [/quote]

[color=darkblue][b][u]Proof of the proposed problem[/u].[/b]

Denote the midpoint $ M$ of the side $ [AB]$ , the intersections $ \left\|\begin{array}{c} X\in PE\cap OA \\
\ Y\in PG\cap OB \\
\ S\in CP\cap AB\end{array}\right\|$ and $ \left\|\begin{array}{c} m(\angle PAB) \equal{} x \\
\ m(\angle PBA) \equal{} y\end{array}\right\|$ .Observe that $ x \plus{} y \equal{} C$

and the quadrilaterals $ OXDM$ , $ OYFM$ are cyclically, i.e. $ m(\angle ADE) \equal{} m(\angle PDF) \equal{} m(\angle PFD) \equal{} m(\angle BFG) \equal{} C$ . Therefore,

$ PD \equal{} PF$ , $ \left\|\begin{array}{c} m(\angle APE) \equal{} C \minus{} x \equal{} y \\
\ m(\angle BPG) \equal{} C \minus{} y \equal{} x\end{array}\right\|$ and $ \left\|\begin{array}{c} m(\angle AEP) \equal{} B \\
\ m(\angle BGP) \equal{} A\end{array}\right\|$ (lines $ DE$ , $ FG$ are [u]antiparallels[/u] to $ BC$ , $ AC$ in $ \triangle\ ABC$ ).

Apply the upper lemma in the triangles $ PAE$ and $ PBG$ to the cevians $ AD$ , $ BF$ respectively :

$ \left\|\begin{array}{ccccc} DE \equal{} DP & \Longleftrightarrow & \sin\widehat {APE}\cdot\sin \widehat {DAE} \equal{} \sin\widehat {AEP}\cdot\sin\widehat {DAP} & \Longleftrightarrow & \sin y\cdot\sin A \equal{} \sin B\cdot\sin x \\
 \\
FG \equal{} FP & \Longleftrightarrow & \sin\widehat {BPG}\cdot\sin \widehat {FBG} \equal{} \sin\widehat {BGP}\cdot\sin\widehat {FBP} & \Longleftrightarrow & \sin x\cdot\sin B \equal{} \sin A\cdot\sin y\end{array}\right\|$ .

In conclusion, $ \boxed {\ DE \equal{} DP\ \Longleftrightarrow\ b\cdot\sin x \equal{} a\cdot \sin y\ \Longleftrightarrow\ FG \equal{} FP\ }$ . Observe that [u]in this case[/u]

$ \frac {SA}{SB} \equal{} \frac {CA}{CB}\cdot\frac {\sin \widehat {SCA}}{\sin\widehat {SCB}} \equal{} \frac ba\cdot\frac {\sin y}{\sin x} \equal{} \frac {b^2}{a^2}$ , i.e. [u]in this case[/u] the point $ S$ is the foot of the $ C$-symmedian in the triangle $ ABC$ .[/color]

[color=darkblue]Thank you[/color] [color=darkred][b]77Ant[/b][/color][color=darkblue], for your nice problem. I like it very much. You can apply this property to the problem from[/color] [url=http://www.mathlinks.ro/viewtopic.php?t=276014][color=red][b]here[/b][/color][/url].

As shown above, we first must prove that $APBC$ is harmonic.
We shall use the inversion of pole $A$ and power $AP^2$, which sends the circumcircle to the line $PE$, hence $\frac{DE}{BC}=\frac{AD}{AC}\ (\ 1\ )$, or $\frac{DE}{BC}=\frac{AD}{AC}\cdot\frac{AB}{AB}\ (\ 2\ )$, hence $DE=BC\cdot\frac{AD\cdot AB}{AC\cdot AB}\ (\ 3\ )$. Similarly, $DP=BP\cdot\frac{AD\cdot AB}{AP\cdot AB}\ (\ 4\ )$. With $DE=DP$, from $(3)$ and $(4)$ we get $\frac{BC}{AC}=\frac{BP}{AP}$, or $APBC$ is, indeed, harmonic.

Consequently $CP$ passes through $T$, intersection of the tangents to circumcircle at $A,B$ and $PE\parallel AT, PG\parallel BT$, thus $\frac{AE}{CE}=\frac{PT}{CP}=\frac{BG}{CG}\implies GE\parallel AB$, i.e. $DE=DP\iff FG=PF$.

Best regards,
sunken rock


calculus
integration
calculus computations

[b][u]without[/u][/b] explicitly solving either of the 2 intgerals on the LHS, show  :lol:

$ \int_0^1\dfrac{\textbf dx}{\sqrt{\;1\minus{}x^4 \;}}\;\cdot\;\int_0^1\dfrac{x^2\;\textbf dx}{\sqrt{\;1\minus{}x^4 \;}}\;\equal{}\;\boxed{\dfrac{\pi}{4}}$

[hide="Hint"]$ \int_0^1 \frac{x^{m\minus{}1}}{\sqrt{1\minus{}x^n}} dx \equal{} \frac{\sqrt{\pi}}{n}\frac{\Gamma(\frac{m}{n})}{\Gamma(\frac{m}{n}\plus{}\frac{1}{2})}$[/hide]

[hide="Answer"]Make the substitution $ x^4 \equal{} u$, $ dx \equal{} \frac{1}{4} u^{\minus{}3/4}du$: $ \int_0^1\frac{dx}{\sqrt{1\minus{}x^4}} \cdot \int_0^1\frac{x^2\,dx}{\sqrt{1\minus{}x^4}} \equal{}$

$ \frac{1}{4} \int_0^1 u^{\minus{}3/4} (1\minus{}u)^{\minus{}1/2}\,du \cdot \frac{1}{4} \int_0^1 u^{1/2} u^{\minus{}3/4} (1\minus{}u)^{\minus{}1/2}\,du$

$ \equal{} \frac{1}{4} \int_0^1 u^{\minus{}3/4} (1\minus{}u)^{\minus{}1/2}\,du \cdot \frac{1}{4} \int_0^1 u^{\minus{}1/4} (1\minus{}u)^{\minus{}1/2}\,du$

$ \equal{} \frac{1}{16} B\left(\frac{1}{4},\frac{1}{2}\right) \cdot B\left(\frac{3}{4},\frac{1}{2}\right) \equal{} \frac{1}{16} \frac{\Gamma \left(\frac{1}{4}\right) \Gamma \left(\frac{1}{2}\right)}{\Gamma \left(\frac{3}{4}\right)} \cdot \frac{\Gamma \left(\frac{3}{4}\right)\Gamma \left(\frac{1}{2}\right)}{\Gamma \left(1 \plus{} \frac{1}{4}\right)}$

$ \equal{} \frac{1}{16} \cdot \frac{\pi \Gamma \left(\frac{1}{4}\right)}{\frac{1}{4}\Gamma \left(\frac{1}{4}\right)} \equal{} \frac{\pi}{4}$[/hide]

Wouldn't the above two solutions actually be explicit solutions?

It's an interesting problem. There's probably a clever solution to tackle it, but I can't see it.

The problem can be rewritten as follows:

$ \int_0^1 {\int_0^1 {\frac{{x^2 }}
{{\sqrt {1 \minus{} x^4 } \sqrt {1 \minus{} y^4 } }}\,dx} \,dy}.$

This last double integral is doable, and its value is in efect, $ \frac\pi4.$


inequalities
calculus
derivative
search
inequalities solved

If $ a$, $ b$, $ c$, $ d$ are reals such that $ 0 < a\leq\frac12$, $ 0 < b\leq\frac12$, $ 0 < c\leq\frac12$, and $ 0 < d\leq\frac12$, then prove that

$ abcd\left(\left(1 \minus{} a\right)^4 \plus{} \left(1 \minus{} b\right)^4 \plus{} \left(1 \minus{} c\right)^4 \plus{} \left(1 \minus{} d\right)^4\right)\leq\left(1 \minus{} a\right)\left(1 \minus{} b\right)\left(1 \minus{} c\right)\left(1 \minus{} d\right)\left(a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4\right)$.

[i]Alternative formulation:[/i] If $ a$, $ b$, $ c$, $ d$ are reals from the interval $ \left]0,\frac12\right]$, then prove the inequality

$ \frac {a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4}{abcd}\geq\frac {\left(1 \minus{} a\right)^4 \plus{} \left(1 \minus{} b\right)^4 \plus{} \left(1 \minus{} c\right)^4 \plus{} \left(1 \minus{} d\right)^4}{\left(1 \minus{} a\right)\left(1 \minus{} b\right)\left(1 \minus{} c\right)\left(1 \minus{} d\right)}$.

in fact the problem is very simple, but long.  let a>=b>=c>=d
and notice that the derivative of x^4+c)/x is (3x^4-c)/x^2
let us search for the optimal configuration. let a,b,c be fixed.
Now let f(d) =d^4+c^4+b^4+a^4/dbca - (same thing for 1-a,1-b,1-c,...
hence f(d)=1/bca[(d^4+s1)/d  - ((1-d)^4+s2)t/(1-d)]
where t= bcd/(1-b)(1-c)(1-d) <=1
hence f'(d) = 1/bca( 3d^4 -s1/d^2  +  (3(1-d)^4 -s2)t/(1-d)^2)
where t=bcd/(1-b)(1-c)(1-d)
(notice that 3d^4<s1 and that 3(1-d)^4>s2) and t<=1
hence f'(d) < 1/bca( 3d^4-s1/d^2 +   3(1-d)^4-s2/(1-d)^2) <0
which shows that f(d) >= f(c) (that is we have to choose d=c)
 hence we have to prove that :
(a^4+b^4+2c^4)/abc^2 > (1-a)^4+.... / 1-a)(1-b)(1-c)^2 (*)
. Now we will try to optimize a,b,c. clearly for a fixed value of a,b.
2c^4<=a^4+b^4 and 2(1-c)^4>(1-a)^4+(1-b)^4
consider 2c^4+s/c^2 = 2c^2+s/c^2
and so  the difference of terms in (*) (left from right) equals
f(c)=1/ba(2c^4+s1/c^2 - (2(1-c)^4+s2/(1-c)^2 (t) where t<=1.
hence by exactly same argument as above f'(c)<0 and so
f(c) > f(b) that is in the optimal case we change c to be b.
hence we have shown that in the optimal configuration b=c=d.
and so we need to show that
(a^4+3b^4)/ab > (1-a)^4+3(1-a)^4/(1-a)(1-b)
but again b^4<a^4 and (1-b)^4>(1-a)^4 and using the same method as above you can prove that b=a .
so the difference of the two expressions given in your inequality is minimzed when a=b=c=d
but then difference =0 so the result follows immidiately.

compare the problem to the well-known key-fan inequality:

if xi are positive reals with xi<=1/2 then
 \pi xi )*(sum (1-xi) )^n<=  \pi (1-xi) * (sum xi)^n

to prove this one you just need to take f(x) = ln((1/x) -1) and observe that f'(x)<0 meaning that f(x) is decreasing on ]0, 1/2]. and so the inequality follows.

[color=red]This solution is wrong.[/color]
Let $f(a,b,c,d)=\frac{a^4+b^4+c^4+d^4}{abcd}$.
Keeping other variables constant, $f'(a)=-\frac{b^4+c^4+d^4-3a^4}{a^2bcd}$, so if a is the smallest among a,b,c,d, we have $f(a,b,c,d) \ge f(1-a,b,c,d)$, since f is decreasing.
Now, WLOG assume $a \le b \le c \le d$, then
$f(a,b,c,d) \ge f(1-a,b,c,d) \ge f(1-a,1-b,c,d) \ge f(1-a,1-b,1-c,d) \ge f(1-a,1-b,1-c,1-d)$.

[quote="Vasc"][quote="siuhochung"]Let $f(a,b,c,d)=\frac{a^4+b^4+c^4+d^4}{abcd}$.
Keeping other variables constant, $f'(a)=-\frac{b^4+c^4+d^4-3a^4}{a^2bcd}$, so if a is the smallest among a,b,c,d, we have $f(a,b,c,d) \ge f(1-a,b,c,d)$, since f is decreasing.
Now, WLOG assume $a \le b \le c \le d$.[/quote]
I think $f(x)$ is decreasing for $0<x \le b$. We have $f(a) \ge f(b)$, but we can't compare $f(a)$ with $f(1-a)$.[/quote]
but $a \le \frac{1}{2}$, so $1-a \ge a$

But let a=b=c=d=0.0001,clearly f(a,b,c,d)<f(1-a,b,c,d).

it have been post before.
but there is not a solutio,too.
I have an idea.
use K FAN inequality,we have(if i rememeber exactly):
$\frac{(1-a)(1-b)(1-c)(1-d)}{abcd} \geq \frac{1-a+1-b+1-c+1-d}{a+b+c+d}$
but can we prove:
$\frac{1-a+1-b+1-c+1-d}{a+b+c+d} \geq \frac{(1-a)^4+(1-b)^4+(1-c)^4+(1-d)^4}{a^4+b^4+c^4+d^4}$??

Prove it for a and b, then use the idea for  a,b,c,d by an algebraic relation between a,b,c,d.

Dear tnk,could you post your full solution? :)


Let $ x,y,z$ be positive real numbers such that
\[ \max\{x,y\} < z \le 1, \quad 2\sqrt {3}xz\leq\sqrt {3}x \plus{} z, \quad \sqrt {2}y \plus{} z\leq 2
\]
Find max:
\[ P \equal{} 3x^2 \plus{} 2y^2 \plus{} 5z^2
\]
Proof 1.
We will show that
\[ x^2 \plus{} z^2 \le \frac {4}{3}\qquad (1)
\]
and
\[ y^2 \plus{} z^2 \le \frac {3}{2}\qquad (2)
\]
Indeed, if $ z \le \frac {\sqrt {3} \plus{} 1}{2\sqrt {3}} < \frac {4}{5}$, then we have
\[ x^2 \plus{} z^2 \le 2z^2 < \frac {32}{25} < \frac {4}{3},\quad y^2 \plus{} z^2 \le 2z^2 < \frac {3}{2}
\]
If $ 1 \ge z \ge \frac {\sqrt {3} \plus{} 1}{2\sqrt {3}}$, then from the given condition, we have
\[ x \le \frac {z}{\sqrt {3}(2z \minus{} 1)}, \quad y \le \frac {2 \minus{} z}{\sqrt {2}}
\]
Thus
\[ x^2 \plus{} z^2 \le z^2 \plus{} \frac {z^2}{3(2z \minus{} 1)^2} \equal{} f(z)
\]

\[ f''(z) \equal{} 2 \plus{} \frac {2 \plus{} 8z}{3(2z \minus{} 1)^4} > 0
\]
Hence $ f(z)$ is convex, thus
\[ f(z) \le \max\left\{f(1),f\left(\frac {\sqrt {3} \plus{} 1}{2\sqrt {3}}\right)\right\} \equal{} \max\left\{\frac {4}{3},\frac {2 \plus{} \sqrt {3}}{3}\right\} \equal{} \frac {4}{3}
\]
And
\[ y^2 \plus{} z^2 \le \frac {(2 \minus{} z)^2}{2} \plus{} z^2 \equal{} \frac {3z^2 \minus{} 4z \plus{} 4}{2}
\]
Since $ z \ge \frac {\sqrt {3} \plus{} 1}{2\sqrt {3}} > \frac {2}{3}$, we have
\[ (z \minus{} 1)(3z \minus{} 2) \le 0\Rightarrow 3z^2 \le 5z \minus{} 2
\]
Thus
\[ y^2 \plus{} z^2 \le \frac {(5z \minus{} 2) \minus{} 4z \plus{} 4}{2} \equal{} \frac {z \plus{} 2}{2} \le \frac {3}{2}
\]
Now, we see that (1) and (2) is proved. It follows that
\[ 3x^2 \plus{} 2y^2 \plus{} 5z^2 \equal{} 3(x^2 \plus{} z^2) \plus{} 2(y^2 \plus{} z^2) \le 7
\]
Equality holds for $ x \equal{} 1,y \equal{} \frac {1}{\sqrt {2}},z \equal{} \frac {1}{\sqrt {3}}$. :)
Proof 2. We will use Abel's forumula. I will add it later. :)
Comment. The proof 1 is long but I like it since it a very natural way to prove this type of inequality and we can apply it to prove the others (for example the Vietnamese Mathematical Olympiad 2001). :)

[quote="can_hang2007"]Let $ x,y,z$ be positive real numbers such that
\[ \max\{x,y\} < z \le 1, \quad 2\sqrt {3}xz\leq\sqrt {3}x \plus{} z, \quad \sqrt {2}y \plus{} z\leq 2
\]
Find max:
\[ P \equal{} 3x^2 \plus{} 2y^2 \plus{} 5z^2
\]
[/quote]

Yes, a similar problem has appeared in Ho Chi Minh Team Selection (Round 1). :)
Let $ x,y,z$ be real numbers such that \[ \left\{ \begin{array}{l} x \ge y \ge z \ge 1 \\  2y \plus{} 3z \ge 6 \\  11x \plus{} 27z \ge 54 \\  \end{array} \right.\]
Then find max
\[ P \equal{} \frac{1}{{x^2 }} \plus{} \frac{{2007}}{{y^2 }} \plus{} \frac{{2008}}{{z^2 }}
\]
The proof's below. :)

[quote="Honey_S"][quote="can_hang2007"]Let $ x,y,z$ be positive real numbers such that
\[ \max\{x,y\} < z \le 1, \quad 2\sqrt {3}xz\leq\sqrt {3}x \plus{} z, \quad \sqrt {2}y \plus{} z\leq 2
\]
Find max:
\[ P \equal{} 3x^2 \plus{} 2y^2 \plus{} 5z^2
\]
[/quote]

Yes, a similar problem has appeared in Ho Chi Minh Team Selection (Round 1). :)
Let $ x,y,z$ be real numbers such that
\[ \left\{\begin{array}{l} x \ge y \ge z \ge 1 \\
2y \plus{} 3z \ge 6 \\
11x \plus{} 27z \ge 54 \\
\end{array} \right.
\]
Then find max
\[ P \equal{} \frac {1}{{x^2 }} \plus{} \frac {{2007}}{{y^2 }} \plus{} \frac {{2008}}{{z^2 }}
\]
The proof's below. :)[/quote]
Yes, Loc. The second proof of mine is similar to the proof in your attachment, the layout of these proofs is based on Abel's formula. :)


conics
parabola
algebra
polynomial
search
parameterization
system of equations

What is the minimum distance between two parabolas:

$ax^{2}+bx+c=y$ and $nx^{2}+px+q=y$?

In an earlier post this was stated by [b]pardesi[/b]:
[quote]...stated the minimum distanec between the parabolas is the ditance between the endpoints along the line joining their focus...[/quote]

Is it true? Does anyone have a proof?
Obviously in some cases it is not, like for example when they intersect,
but is there any substance in this? Perhaps it is true
when they don't intersect (of course  :P ) and $a=-n$?

[quote="color"]What is the minimum distance between two parabolas:

$ax^{2}+bx+c=0$ and $nx^{2}+px+q=0$?

[/quote]
It must is $ax^{2}+bx+c=y$ and $nx^{2}+px+q=y$

Since a line going through the focus of a parabola is almost never perpendicular to the parabola, the above claim is almost never true. I believe that there is no reasonable general formula for the distance. You can set up a system of equations, of course and reduce the problem to finding roots of some 9-th degree polynomial but that's about it. There seems to be nothing really special about the polynomial in question that would facilitate this root search (though I admit that it has only 3 really free parameters because the first parabola can always be assumed to be just $y=x^{2}$). At any rate, an attempt to find the distance between $y=x^{2}$ and $y=2x^{2}+1$ can already be discouraging enough to make one prefer to leave the problem alone. :(

it is the  distance between the endpoints of the common normal on each parabola :maybe:

[quote="Fedja"]You can set up a system of equations, of course and reduce the problem to finding roots of some 9-th degree polynomial but that's about it. There seems to be nothing really special about the polynomial in question that would facilitate this root search...[/quote]

That's why the purpose of the problem was only to derive the system of equations whose solutions would give the answer to the problem, and not to solve the equations.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Look at Higman's paper `A finitely generated infinite simple group', J. London Math. Soc. 26, (1951). 61--64.


could someone just post the basics of torques. i also want info abt the result of torques meaning what does torque cause.  :D

Torque is essentially the rotationtal analogue of force.  Suppose you have a lever arm (essentially a rod).  If you apply a force $ F$ to the rod some distance away from the pivot point, say $ \ell$, then $ \tau\equal{}\textbf{r}\times\textbf{F}$.  Don't let the notation get to you: if you rotate the rod some angle $ \theta$, then the magnitude of the torque is $ F\ell\sin{\theta}$.

Thanks JRav
It is essentially the cross product of displacement from the pivot point and the force applied right

Not essentially-it is the cross product of displacement and force.

[quote="amangupta2"]Torque , ALso called the Moment Of Force is the Rotational Counterpart of Force , which comes into play , while dealing With Rigid bodies , Torque (denoted by tau) is essentially the Cross Product of The Force applied about a Certain Point P and The Distance of F from the Point P.
 [u]Torque =rxF[/u][/quote]it wud b better if we say "Shortest Distance of F from the Point P" :wink:

[quote="amangupta2"]Torque (denoted by tau) is essentially the Cross Product of The Force applied about a Certain Point P and The Distance of F from the Point P.[/quote]
As I said before, it is not "essentially" the cross product.  It IS the cross product.  We define $ \tau\equal{}\mathbf{r}\times\mathbf{F}$.

[/quote]it wud b better if we say "Shortest Distance of F from the Point P" :wink:[/quote]

T=rxF = F(rsinA)
here r sinA denotes nothing but the shortest Distance of F from P.So There is nothing wrong in saying that Torque if the Cross Product Of r and F

The distance is implied to be the shortest straightline distance.  You don't need to specify that.  Also, note that $ \left|\tau\right|\equal{}rF\sin{\theta}$.


You're close nonie, your formula for the even numbers sum is incorrect.

You can also group the numbers so that it's $(1-2)+(3-4)+(5-6)+....+(99-100)$

there's 2 ways of doing this:
[hide]
you can find the sum of all the oven numbers and all the odd numbers

 the formula for the sum of the first n odd number is $n^2$, with $n$ being the $nth$ odd number, so it would be $50^2 = 2500$

the formula for the sum of the first nth even number is $n(n+1) = 50(50+1) = 2550$, and it would be $-2550$ because it was from -2 to -100

so $2500+(-2550) = -50$
[/hide]
 -------------------------------------------------------------or-------------------------------------------------------------

[hide]you can also do what 236factorial did.

As you can probably tell, if you regroup them, you'll get  $(1-2)+(3-4)+(5-6)+....+(99-100)$, each parenthesis turns out to be $-1$, and since there's $50$ groups of them, the answer is $-1*50 = -50$[/hide]

but i guess the second way is easier

[quote="nonie"]1to99= =2500

-2to-100 = = - 2501

so i think the answer is -1 [/quote][quote="Drunken_Math"]Evaluate:

$1+3+5+7+\cdots+97+99-2-4-6-8-\cdots-98-100$[/quote]

[hide=" updated answer"]Oops..... :blush:

its n(n+1) not n^2 +1 ...

Ok so then
50^2- 51^2= -50 :)[/hide]

[hide]using the formula nonie used, I get -50.[/hide]

[hide]if you group the numbers,  we get 50 groups of -1, so the answer is -50 :idea: [/hide]


In a rectangle ABCD with sides 28 and 15, and diagonal is drawn from A to C. The diagonal is cut into 7 equal pieces. Find the area enclosed by vertices B and D, and the points which the 3rd piece of the diagonal cuts the diagonal. Sorry for the language :(

I got [hide]60[/hide] but I need a simpler method.

[quote="Gib Z"] the points which the 3rd piece of the diagonal cuts the diagonal.
[/quote]
I don't understand this but I think we can call the points $ I$, $ O$

and $ \frac{S_{DIBO}}{S_{ABCD}}\equal{}\frac{OI}{AC}$


[quote="Toan_VN_LC"]let a,b,c be non-negative real numbers
prove that:
$ a\sqrt {16{a^4} \plus{} 65{b^3}c} \plus{} b\sqrt {16{b^4} \plus{} 65{c^3}a} \plus{} c\sqrt {16{c^4} \plus{} 65{a^3}b} \ge {(a \plus{} b \plus{} c)^3}$[/quote]

yes, it really seems a hard problem... i normalized it to $ a\plus{}b\plus{}c\equal{}1$ and mixed it up with convexities, but didn´t really get anything substantial...but i´ll try it more :mad:


[b]Probem 5.[/b] Given is $\triangle ABC$, and $\angle BAC=30^\circ$ and $\angle ABC=45^\circ$. Consider all couples of points $X$ and $Y$, such that $X$ and $Y$ are respectively on the rays $AC^{\rightarrow}$ and $BC^{\rightarrow}$, and $OX=BY$, where $O$ is the circumcenter of $\triangle ABC$. Proove that the bisectors of the segments $XY$ pass through fixed point.

[i]Emil Kolev[/i]

what is the bisector of $BC$ ??

the perpendicular on the middle ?

a bisector of line segment is line which passes through its middlepoint and is perpendicular to it.

See here for the solution
http://www.mathlinks.ro/Forum/viewtopic.php?t=88584


geometry
circumcircle
incenter
geometry proposed

Let $ABCD$ be a square. Let$M$ be an inner point on side $BC$ and $N$ be an inner point
on side $CD$ with $\angle MAN = 45\deg$ . Prove that the circumcenter of$AMN$ lies on $AC$.

Let $I$ be the incenter of $\triangle MNC$, let $x = \angle BAM$, and let $y = \angle DAN$.  We have $x+y = 45^\circ$ so $\angle MIN = 135^\circ$.  Thus, $AMIN$ is cyclic.  It follows that $\angle NMC = 2x$ and $\angle MNC = 2y$ so $\angle AMB + \angle IMC = 90^\circ$.  Finally, note that this implies $\angle AMI = 90^\circ$ and similarly, $\angle ANI = 90^\circ$ so the midpoint of $AI$ is the circumcenter of $\triangle AMN$ and the result clearly follows.


The real numbers $x_0, x_1, ... , x_{n+1}$ satisfy $0 = x_0 < x_1 <\cdots < x_n < x_{n+1} = 1$ and

$\frac{1}{x_i - x_0} + \frac{1}{x_i - x_1} +\cdots + \frac{1}{x_i - x_{i-1}} + \frac{1}{x_i - x_{i+1}} + \frac{1}{x_i - x_{i+2}} +\cdots + \frac{1}{x_i - x_{n+1}} = 0$ for $i = 1, 2, ... , n.$

Show that $x_i + x_{n+1-i} = 1$ for all $i$.

I can't solve this.

source : http://www.kalva.demon.co.uk/short/sh86.html


[quote="weatherbottle"]Source: Massachusetts Mathematics League March 2003

If $P(x)$ is a third degree polynomial where $P(-2) = 0, P(1) = 0, P(4) = 0,$ and $P(2) = 6;$ what is the remainder when $P(x)$ is divided by $x^2+2x-3$.[/quote]

you can solve k by plugging 2 in for x and making P(x)=6

[quote="Vihang"][hide="Solution"]
[quote="weatherbottle"]Source: Massachusetts Mathematics League March 2003

If $P(x)$ is a third degree polynomial where $P(-2) = 0, P(1) = 0, P(4) = 0,$ and $P(2) = 6;$ what is the remainder when $P(x)$ is divided by $x^2+2x-3$.[/quote]

Hence by Remainder theorem :
(x+2)|P(x) and (x-1)|P(x) and (x-4)|P(x)

But P(x) is cubic

Thus P(x)=k(x+2)(x-1)(x-4)
Thus \[ P(x) = k(x^3 - 3x^2 - 6x + 8)  \]

Hence the remainder is k(-x+5)
[/hide][/quote]

What's the Remainder Theorem?

(x-a) divides p(x) iff p(a)=0

It's very easy to prove.

Ok.  I knew the theorem, just for some reason not the name... ;)


modular arithmetic
number theory proposed
number theory

Hope someone can help me with this question.

Let k be the smallest positive integer such that $ 10^{k} \equiv 1 \pmod p \quad \quad$. show that k divides p-1.

Thanks

Pretty obvious. Just use the title you made.

$ 10^{p\minus{}1}\equiv 1\pmod p$, except for $ p\equal{}2$ or $ p\equal{}5$, for which none $ k$ exist.

As $ k$ is the order of $ 10$ mod $ p$, we get that $ k\mid p\minus{}1$

more general statement: let $ a,b$ be integers, and $ k$ the smallest positive integer* for which $ a^k\equiv 1\pmod{b}$. if $ a^n\equiv 1\pmod{b}$, then $ k|n$.

Proof: assume that $ k\not| n$. then we can write $ n \equal{} kq \plus{} r$, where $ 0 < r < n$. then $ 1\equiv a^n \equal{} a^{kq}a^r \equal{} (a^k)^qa^r\equiv 1^qa^r \equal{} a^r \pmod{b}$, but this contradicts the minimality of $ k$. thus our initial assumption is false, and we must have $ k|n$.

*there exists $ k$ such that $ a^k\equiv 1\pmod{b}$ if and only if $ \gcd(a,b)\equal{}1$.

Yeah, that's what I used in my last line (the order of a number mod any divides any number which also satisfies the same relation as the order)


From a point P ,two tangents PA and PB are drawn to a circle with centr O.if OP=diameter of the circle show that triangle APB is equilateral.

[hide="Proof"]From the definition of tangents, OA=OB=r, $ \angle OAP \equal{}\angle OBP \equal{}90$, and $ AP\equal{}BP$. Therefore, OAP and OBP are 60-90-30 triangles, and $ \angle APB \equal{}60^{\circ}$.

Since $ AP\equal{}BP$, $ \triangle ABP$ is isosceles with base AB. Since P is $ 60^{\circ}$, $ \angle PAB \equal{}\angle PBA \equal{}60^{\circ}$, and $ \triangle PAB$ is equilateral.[/hide]


function
linear algebra
matrix
vector
logarithms
search
probability

Let $a_1,...,a_K$ be any given postive value.  and $\mathbf{H}$ is a given $N\times K$ matrix, denote its i-th column vector by $\mathbf{h}_i$. Function $f: \mathbb{R}_+^K \mapsto \mathbb{R}$ is given by

$f(\mathbf{p})=(\sum_{i=1}^K a_i) \log \det (\mathbf{I}+\sum_{i=1}^K p_i \mathbf{h}_i \mathbf{h}_i^T)
-\sum_{j=1}^K a_j \log \det (\mathbf{I}+\sum_{i \neq j} p_i \mathbf{h}_i \mathbf{h}_i^T)-\sum_{i=1}^K p_i$

Note the function is defined on  $\mathbb{R}_+^K$, that is $p_i \geq 0$.
The function is non-concave, so it might have multiple maxima.
The claim if a maximum is obtained. The number of nonzero components of $\mathbf{p}$ is at most $rank(\mathbf{H})$.

I used gradient method to do lots of numerical simulations and found that is always the case. But since the function is non-concave. the numerical method only converges to one local maximum.

Can anyone prove it or provide a counter example?

The function here is same as the one in this post:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=201906#p201906

It is not hard to prove that the number of non-zero components is not greater than $\frac {\text{rank}(H)(\text{rank}(H)+1)}2$, but this estimate is much weaker than your conjecture if the rank is greater than $1$. I would suggest an extensive search of $2\times 3$ matrices to see whether there is an example where all 3 $p_j$ are strictly positive at some point of (local) maximum. Meanwhile I'll try to see whether my estimate can be improved.

Thanks. I also proved a similar statement where my upper bound is ${\text{rank}(H)^2}$, which is even worse than yours. I conjecture the case in which we have more than rank(H) non-zero $p_i$ may happen. But the probability is just so small.

BTW: How to do a extensive search for matrix? What I was doing is to use random matrix with i.i.d guassian entries. Maybe my method is hard to generate a wierd matrix that contradicts my conjecture. The choice of $a_i$ is also a issue. what choice do you think is more likely to give a counter example?

Some thoughts about this problem: since  the gap between $\text{rank}(H)$ and $\frac {\text{rank}(H)(\text{rank}(H)+1)}2$ becomes large as $\text{rank}(H)$ increases. If the statement is not true, intuitively, the chance to observe a counter example should be more likely as $\text{rank}(H)$ is large.  But even for a large $\text{rank}(H)$, I didn't observe any
exception with numerical methods. the number of non-zero component is never greater than $\text{rank}(H)$, which is much smaller than $\frac {\text{rank}(H)(\text{rank}(H)+1)}2$, when $\text{rank}(H)$ is large.

But I still doubt this conjecture. What I was hoping is trying to give an explanation why it is true with most of cases. Maybe the statment  is just asymototically true as some parameter goes to a limit. But it turns out to be a good estimation even the parameter doesn't go to a limit.

Meanwhile, can you show me the scatch of your proof of the bound $\frac {\text{rank}(H)(\text{rank}(H)+1)}2$. I didn't even get that, guessing it is related to the degree of freedom of a diagnoal matrix.

Try the case $a_1=a_2=a_3=1$, $H=t\left(\begin{matrix} 2 & -1 & -1 \\ 0 & \sqrt 3 & -\sqrt 3  \end{matrix}\right)$ with $t>1/2$. I have some reasons to believe that in this case you'll have a local maximum on the diagonal $p_1=p_2=p_3$, which for $t$ close to $1/2$ may be even a global maximum. But I only did a "back of envelope" computation for this case, so I am not 100% sure...  :)

Yes!!!!  you are right.

I tried your matrix. and it seems to be a very ill-conditioned matrix. The cost function value is very small.

when I set t=0.5. My numeric methods never stoped??? Seems to go into a dead loop???

when I set t=0.6. One method gives 2 nonzero maxima, the other one gives 3 nonzero maxima. and their cost function are very close.

Amazingly, the cost function seems converge to 0 as t converges to 0.5.
But the 3 nonzero maxima is always a little little bit smaller than the 2 nonzero maxima.

When I get too close to 0.5, my numerical methods became malfunction.

Update: Just notice all methods didn't converge to a solution where $p_1=p_2=p_3$, I only get a solution where $p_1>p_2=p_3$. So there is still hope for it to be global maximum. Let me try again....

Why don't you try to find the maximum of the cost function on the diagonal (which is one-dimensional problem) and then check whether it is a local maximum. Meanwhile I'll see whether I can make my "back of envelope" computation into something rigorous.

Yes, that is what I am doing now. I will let you know soon.

I try the one dimensional optimization. I let t close to 0.5, say 0.51,0.501,0.5001.... and the cost becomes close to zero.
But in all cases, it turns out its maximum is always a little bit smaller than the maximum with two non zero components.

When t=0.5, all numerical methods didn't work.

Hopefully, my program is right.

[quote="remgenesis"]But in all cases, it turns out its maximum is always a little bit smaller than the maximum with two non zero components.
Hopefully, my program is right.[/quote]
This is extremely weird because I believe that I have a proof that, for $t$ close to $1/2$, there is a unique global maximum with $p_1=p_2=p_3>0$ :? (for $t=1/2$, the maximum is attained at $p_1=p_2=p_3=0$, which, probably, is the reason why your methods fail). The proof is based on 3 claims.

Claim 0: for all $t$, $f$ is a symmetric function of $p_1, p_2, p_3$. This is how $H$ was chosen and easy to check directly.

Claim 1: For every $\delta>0$ there exists $t_0>1/2$ such that, for $t\in (0.5, t_0)$, we have $f(p_1,p_2,p_3)<0$ if $p_1+p_2+p_3>\delta$ (so the maximum of $f$ can be attained only at a point very close to $0$ if $t$ is sufficiently close to $1/2$).
I hope that agrees with what you saw numerically, so I'll skip the proof.

Claim 2: There exists $\delta>0$ such that for all $t$ sufficiently close to $1/2$ the function $f$ is strictly concave in the $\delta$-neighborhood of $0$.
This, together with symmetry, directly contradicts existence of any 2 (or 1) non-zero  component local maxima. To check this claim, it is enough to show that for $t=1/2$, the second derivative of $f$ at $0$ is strictly negative definite. Then, by continuity, it will be negative definite for all $p$ close to $0$ and $t$ close to $1/2$. It is a little bit unpleasant exercise, but what I ended up with was a quadratic form proportional to $-p_1^2-p_2^2-p_3^2-(p_1+p_2+p_3)^2$. You may want to check this computation either by hand or numerically. Perhaps, I made some idiotic mistake somehwere, but, if not, you'll have to check your results. One simple way to check them is to check the symmetry between $p_1,p_2$ and $p_3$ (just to make sure there was no error in the definition of $f$ in your program) and then to take the 2 non-zero component maximum you get, replace the point with the point where all 3 components are equal and their sum is the same, and compare the values.

I made a stupidest mistake. one element in matrix H is wrongly inputed.  :blush:

After correcting this. the results are purely symmetirc as expected. Even multidimensional optimization converges to the maximum with $p_1=p_2=p_3$.

And all numerical methods converge to the same point. So I cannot find any 2 nonzero components local maxima now.

Update: just a conjecture, your matrix is purly symmetric, which rarely happens when $H$ is randomly generated as in my old program. You just showed that 3 non zero components maxima did exist.  But i think the measure of such matrix is just zero. That would explain why I never got counter example when $H$ is randomly generated.

Do you have any explanation for this phenomenon?

[quote="remgenesis"] But i think the measure of such matrix is just zero. That would explain why I never got counter example when $H$ is randomly generated.
Do you have any explanation for this phenomenon?[/quote]

The measure is not exactly zero, because the phenomenon is stable and you have an open set of such matrices. On the other hand, I also believe that the set of such matrices has a small measure but cannot prove it. Also it is merely an example for $\text{rank}H=2$. I do not have an example for higher ranks showing that my upper bound is attainable.

[quote="remgenesis"]
Meanwhile, can you show me the scatch of your proof of the bound $\frac{\text{rank}H(\text{rank}H+1)}{2}$. I didn't even get that, guessing it is related to the degree of freedom of a diagnoal matrix.
[/quote]
The idea is to fix $I+\sum_{i=1}^K p_i h_ih_i^T$. If $\text{rank}H=r$, it is a system of not more than $\frac{r(r+1)}{2}$ independent linear equations. With this sum fixed, the function becomes convex and its maximum can be attained only on the boundary. I suspect that your proof of the estimate $\text{rank} H^2$ is pretty much the same.  :)

Thanks. Even all my conjectures turn out to be false. I am happy that the case is now "temporarily closed".
I've wasted so much time on this problem. You've helped me so much!  :)

[quote="remgenesis"] I am happy that the case is now "temporarily closed".
[/quote]
So am I  :). BTW, a friend of mine suggested an example for $\text{rank}H\geqslant 7$ showing that my upper bound is sharp. I also found an example for $\text{rank}H=3$. This leaves unsettled the cases of rank 4,5 and 6, but I'm too lazy to think of these 3 possible exceptions to the general rule  :P. In all these cases the matrix $H$ is very special with a lot of symmetries, so to hit it by random choice is next to impossible.

I know why I get $rank(H) ^2$ instead of $rand(H) (rank(H)+1)/2$. I assume in my problem $H$ is a complex matrix. So by your method, the independent linear equations increase to $rank(H) ^2$.  Is that correct? Do you think for this complex matrix case $rank(H) ^2$ would be a tight bound?

You said that by fixing $\sum p_ih_i h_i^T$, the cost function is convex. I am not sure on what space it is convex? a space of  $rand(H) (rank(H)+1)/2+1$ dimensions? since I think the boundary should be $rand(H) (rank(H)+1)/2$ dimensional. How do you represent this space?

[quote="remgenesis"] I assume in my problem $H$ is a complex matrix. So by your method, the independent linear equations increase to $rank(H) ^2$.  Is that correct? [/quote] Yes, then I also get $\text{rank}(H)^2$. I thought that your matrices were real because you always wrote $h^T$ which usually means just "transpose" rather than "transpose conjugate" :blush:

[quote="remgenesis"]
Do you think for this complex matrix case $rank(H) ^2$ would be a tight bound?
[/quote]
I don't know. To prove it, it would suffice to find $n^2$ unit vectors $h_i$ in $\mathbb C^n$ such that the quadratic form $\sum_i x_i^2+4\sum_{i<j} |\langle h_i, h_j\rangle|^2 x_i x_j$ is positive definite. You may start with an attempt to find $4$ such vectors in $\mathbb C^2$.

[quote="remgenesis"]
You said that by fixing $\sum p_ih_i h_i^T$, the cost function is convex. I am not sure on what space it is convex? a space of  $rand(H) (rank(H)+1)/2+1$ dimensions? [/quote]
No, it'll be the $K-\frac{rank(H) (rank(H)+1)}2$-dimensional set of the non-negative solutions of the system of equations used to fix the "full" determinant, which is just a polyhedron. When saying that the maximum must be attained on the boundary, I was not precise. I should say "on the set of the vertices of the corresponding polyhedron". Hope, it's clear now.


Ley a,b,c,d are real numbers such that $ x \plus{} y\ge 0$, where x,y distinct and $ x,y\in$ {a,b,c,d}.
Find the min value of
$ A \equal{} \frac {a \plus{} b}{a \plus{} d \plus{} 2c} \plus{} \frac {b \plus{} c}{b \plus{} a \plus{} 2d} \plus{} \frac {c \plus{} d}{c \plus{} b \plus{} 2a} \plus{} \frac {d \plus{} a}{a \plus{} b \plus{} 2c}$

By "not nice," do you mean "not even close to cyclic"? (as there is no $ 2b$ term anywhere?)

Try putting $ a_1\equal{}a\plus{}b,a_2\equal{}b\plus{}c,a_3\equal{}c\plus{}d,a_4\equal{}d\plus{}a,a_5\equal{}a\plus{}c,a_6\equal{}b\plus{}d$.


trigonometry
analytic geometry
geometry solved
geometry

I found this question in some other place and it looked really interesting to me.

I'll write it here as well:

From a certain point on a circular track 2 runners, A and B, start at the same time, B faster than A (both have constant speeds). What's the locus of the midpt of AB as they move along the circle?

Some raindrop like stuff

I think the only way to paste values on it will be to say vB = x.vB and work with that

should work, but it far from beautiful

Without loss of generality, we may assume that the radius of the circle is $1$. Since $B$ is faster than $A$, at a certain time they will necessarily overlap: let's call this time $t=0$. Now, suppose that $u$ and $v$ are the angular speeds of $A$ and $B$, respectively. The parametric equations for $A$, $B$ and their midpoint $M$ are

$A=(\cos ut, \sin ut)$, $B=(\cos vt, \sin vt)$,

$M=\left(\frac{\cos ut+\cos vt}{2}, \frac{\sin ut+\sin vt}{2} \right)=\left(\cos\left(\frac{u+v}{2}t\right)\cos\left(\frac{u-v}{2}t\right), \sin\left(\frac{u+v}{2}t\right)\cos\left(\frac{u-v}{2}t\right)\right)$.

Finally, switching to polar coordinates, we get

$\rho_M=\cos\left(\frac{u-v}{u+v}\theta\right)$,

which is a curve called [i]rose[/i]. To view a few examples of roses, go to Mathworld: http://mathworld.wolfram.com/Rose.html .

[quote="grobber"]Thanks! I guess the calculations weren't as gruesome as I initially thought :).[/quote]
Sure they weren't!  :lol:
I've just omitted the calculations to switch from parametric to polar coordinates, but they are short:

$\rho^2 = x^2+y^2 = \cos^2\left(\frac{u-v}{2}t\right) \left( \cos^2\left(\frac{u+v}{2}t\right)+\sin^2\left(\frac{u+v}{2}t\right) \right)$,

so $\rho = \cos\left(\frac{u-v}{2}t\right)$.

$\theta = \arctan \frac y x = \arctan \frac{\sin\left(\frac{u+v}{2}t\right)}{\cos\left(\frac{u+v}{2}t\right)} = \frac{u+v}{2}t$.

Then $t=\frac{2}{u+v}\theta$, and finally $\rho = \cos\left(\frac{u-v}{u+v}\theta\right)$.


What is the thirteenth term in the arithmetic sequence -5, -2, 1, 4, $ \ldots$?

The common difference is $ 3$. Thus $ 12 \times 3 \plus{} (\minus{}5) \equal{} \boxed{31}$.


does anyone here know the fridrich method completely?

fredrich method is too hard to learn completely. I know all the important F2L, Cross OLL, and 17/21 PLL and the other 4 I can do with 2 algos. I'm officially not learning anymore algorithms

I've memorized all the PLL, but there are only 13 i think (excluding fliping or inverting) because i know them all. Should I learn the F2L ones? I usually just put in the corner, then put in the edge. I can do the cross OLL ones in 1 or 2 algorithms. Should I learn each cross OLL in one algorithm?

well since you can already do PLL, just go ahead and learn all the cross oll.


inequalities
calculus
logarithms
function
inequalities unsolved

Let a,b,c are real numbers in (0.1). Prove that |a-b|/(1-a*b)<=|a^c-b^c|/(1-a^c*b^c)

Does anybody can solve this inequality???  :(  It's hard problem using classical inequalities and I'm not sure that there is a solution with this.

[quote="bilarev"] It's hard problem using classical inequalities ...[/quote]
Certainly true. With calculus, it's of course very easy.
Wlog $a>b$. Put $p=\ln\sqrt{\frac ab}>0$ and $q=-\ln\sqrt{ab}>0$ and just observe \[ \frac{a^c-b^c}{1-a^cb^c}=\frac{\sqrt{\frac ab}^c -\sqrt{\frac ba}^c}{\frac1{\sqrt{ab}^c}-\sqrt{ab}^c}=\frac{\sinh(cp)}{\sinh(cq)} \] which is decreasing as a function of $c$ because of $p<q$. (For $c>1$, the inequality flips.)


What is the equation of the line parallel to $ 4x\plus{}2y\equal{}8$ and passing through the point $ (0,1)$? Write the equation in slope-intercept form.

We solve for the slope of the line:

\[ 4x\plus{}2y\equal{}8 \implies 2y\equal{}\minus{}4x\plus{}8 \implies y\equal{}\minus{}2x\plus{}4
\]

In slope-intercept form, the equation of the line is $ y\equal{}mx\plus{}b$.  Thus:

\[ y\equal{}mx\plus{}b \implies 1\equal{}\minus{}2(0)\plus{}b \implies 1\equal{}b
\]

Thus,

\[ \boxed{y\equal{}\minus{}2x\plus{}1}
\]


What is the positive value of m such that the triangle bounded
by the lines $ y \equal{} 0, x \equal{} 12$ and $ y \equal{} mx$ has an area of
1872 square units?

$ A\equal{}\frac{1}{2}bh\equal{}1872\equal{}\frac{1}{2}\cdot 12h\Rightarrow h\equal{}312$
The line $ y\equal{}mx$ must satisfy $ 312\equal{}m(12)\Rightarrow m\equal{}26$.


geometry
circumcircle
incenter
rhombus
trapezoid
geometry unsolved

Let $ ABCD$ be a bicentric quadrilateral inscribed in a circle with center $ O$ and circumcribed about a circle with center $ I$. A line through $ O$, parallel to a side of $ ABCD$, intersects its two opposite sides at $ M$ and $ N$. Prove that the length of MN does not depend on the choice of side to which the line is parallel.

ABCD is bicentric quadrilateral with circumcenter O and incenter I. The proposition holds for parallels to AB, BC, CD, DA through the incenter I, not for parallels through the circumcenter O.

Let parallel to AB through I cut BC, DA at M, N and let parallel to DA through I cut AB, CD at K, L. AKIN is a rhombus, IK = IN. IMCL is cyclic, having equal angles as the bicentric ABCD, and CI bisects the angle <MCL => IM = IL. Adding, KL = MN.

Let ABCD be bicentric trapezoid with BC || DA. Let M, N, K, L be defined as before. Let parallel to AB through O cut BC, DA at M', N' and let parallel to BC through O cut AB, CD at K', L'. Then K'L' > KL = MN = M'N' or K'L' < KL = MN = M'N', unless O, I are identical, ABCD is a square.


the number of permutations of letters of the word PARALLEL  taken four at a time are ......................

pl suggest a shorter way

1  .A music academy  has to send 10 members to 5 shows ,two in each show. two of the shows are in city and the 3 are outside .two of the five members prefer to work in city while  3 prefer to work  outside  city..the number of ways in which five shows can be held if preference are to be satisfied .
  options are 5200, 5400,5472,5250

Q2  There are two students  A and B having a and b numbers of books respectively.the number of ways in which they can exchange their their books so that after exchange they have same number of books with them but not he same books is .............

 options are a+b^(C)a ,(a+b^C a)-1,a+b^(P)a,( a+b^P a)-1

I didn't know there was an Australian Forum :) , anyway

[hide="permutation question"]
What do you mean when you say "taken in groups of four?"

$ \dfrac{8!}{3!2!}$...perhaps divide that by $ 4$
[/hide]

[hide="1"]
Well there are $ \binom{5}{2} \equal{} 10$ ways to split the musicians into groups of $ 4: 6$ and maintain their work preferences
Then there are $ \binom{4}{2}$ ways to split up the musicians in the city and $ \binom{6}{2} . \binom{4}{2}$ ways to split them up outdoors

$ \Rightarrow \binom{5}{2}.\binom{6}{2}^2 . \binom{4}{2} \equal{} 5400$
[/hide]

[hide="2"]
I think you need to clarify this question a little more. What do we define an exchange to be? swaping one book or infinite number of books? Also if two exchanges end up with the same books for A and B then do we consider these the same?[/hide]

EDIT: I just realised this was posted in october, haha


modular arithmetic
number theory unsolved
number theory

[color=darkblue]Let $ n$ is a integer, ($ n \in Z$). Let $ n \equal{} 7p \plus{}5, p \in Z$ and $ n \equal{} 13q\plus{}4, q \in Z$.

Let $ n\equal{}91r\plus{}s$, with $ r, s \in Z$. Find $ s$ ?[/color]

Obviosly $ s\equal{}\minus{}14\plus{}5\equal{}\minus{}9\equal{}82\mod 91\equal{}7*13$.

[quote="Rust"]Obviosly $ s \equal{} \minus{} 14 \plus{} 5 \equal{} \minus{} 9 \equal{} 82\mod 91 \equal{} 7*13$.[/quote]

Can you give me the full solution?

$ n\equiv 5 \pmod{7} \\ n\equiv 4\pmod{13}$

$ 13n \equiv 65 \pmod{91} \\ 7n\equiv 28\pmod{91}$

$ 13n \equiv 65 \pmod{91} \\ 14n \equiv 56 \pmod{91}$

$ n\equal{}14n\minus{}13n\equiv 56 \minus{} 65 \equiv 82 \pmod{91} \\ \boxed{s\equal{}82}$


geometry
circumcircle
combinatorics unsolved
combinatorics

Find all sets X of points in a plane, not all collinear, such that:
For any two distinct circumferences, each contains three points of X, its intersection points are points of X.

Just an idea

If we have convex cuadrilaterals, then we can choose one of them with the smallest area. It doesnt contains more interior points, let it be $ABCD$ and let $X$ be a external point then maybe, i dont know the circumcircles of $ACX$ and $BDX$ intersect in $Y$ in the interior of $ABCD$

or i dont know anyway a good set must have lots of collinear points then...


a,b,c>0 Show that
Σ[a 2 /(a 2 +2bc)] \geq 1 \geq Σ[bc/(a 2 +2bc)]

I think this one has been posted before. Anyway it's not very hard

Yes  you  are  right..not  difficult  at all!

But why do you post it as an unsolved, advanced problem? :?
It's better to [b]propose[/b] this in the beginner's corner.

By Cauchy inquality:
            b^2+c^2>=2bc.
=>Sum(a^2/(a^2+2bc))>=Sum(a^2/(a^2+b^2+c^2)=1.
Besides:Sum(a^2/(a^2+2bc))+Sum(2bc/(a^2+2bc))=3.
=>1>=sum(bc/(a^2+2bc))
The equality happen when a=b=c(dpcm).


geometry
Harvard
college
MIT
3D geometry
sphere
calculus

These are stolen from the general and geometry portions of this year's Harvard-MIT math tournament.

1. In a classroom, 34 students are seated in 5 rows of 7 chairs. The place at the center of the room is unoccupied. A teacher decides to reassign the seats such that each student will occupy a chair adjacent to his/her present one (i.e. move one desk forward, back, left or right). In how many ways can this reassignment be made?

2. Take a clay sphere of radius 13 and drill a circular hole of radius 5 through its center. Take the remaining bead and mold it into a new sphere. What is this sphere's radius?

3. In triangle ABC, angle ABC= 50 degrees and angle ACB= 70 degrees. Let D be the midpoint of side BC. A circle is tangent to BC at B and is also tangent to segment AD; this circle intersects AB again at P. Another circle is tangent to BC at C and is also tangent to segment AD; this circle intersects AC again at Q. Find angle APQ.

I really didn't like #2 because calculus would get you the answer easily, and calculus should not be used on the non-calculus portion of the HMMT. Just my opinion.

You can do number 2 without formal calculus, but it isn't easy.  Unless there's a emthod I don't know about for tackling it.

Still, even with another method, a cheap, standard calculus method would get you the answer in a minute.

For the 3'rd one:

X is the pt of tangency of the first circle and AD and Y is the same for the second circle. Since the tangents drwan from a pt to a circle are equal it means that DB=DX and DC=DY, but DC=DB, so X=Y. Let's look at the powers of A with respect to the 2 circles: AP*AQ=AX^2=AQ*AC (the 2 powers are equal), so AP*AB=AQ*AC, so AP/AC=AQ/AB and because A is a common angle of triangles APQ and ACB it means that these are similar (in this order of vertices), so angle APQ=angle ACB=70 deg. I think we only needed angle ACB (we didn't use angle ABC).

[quote="314159265358979323846"]http://web.mit.edu/hmmt/www/datafiles/solutions/2003/sgeom03.pdf

While there is a working noncalculus method, it takes quite a bit of ingenuity to find the solution (taking cross-sections of the sphere and comparing areas)...[/quote]

That's what I call ``calculus without doing calculus.''

You could use sphere sectors and it's easy from there, in fact...I think I'll just do it right now... *Btw...the volume of a sphere sector is 2:pi:r:^2:h/3.  I suggest you look on http://www.mathworld.wolfram.com for more info.  (Or I think it has it there...)

Volume of the sphere with radius of 13

13:^3:4:pi:/3 = 8788:pi:/3

Volume of hole of radius 5 drilled through the sphere

Volume of the cylinder of radius 5 and height of 24 = 600:pi:
*A 5-12-13 right triangle is formed by half of the height, the hole's radius, and the sphere's radius.*

Volume of the two spherical sectors = 2:pi:13:^2:/3 * 2 = 676:pi:/3
*The height of both spherical sectors was each 1 unit because it was (radius of the sphere) - (half the cylinder's height) = 13-12=1*

The volumes of the cones who's bases intersect the edges of the whole need to be subtracted though so the volumes of the two cones with radiuses of 5 and heights of 12 will be 2 * 1/3 * 25:pi: * 12 = 200:pi:  Notice I multiplied by two because there were two congruent cones.

 Now subtract the cones from the sphere sectors and you'll get the volumes of the spherical caps on both ends.  (676/3 - 200):pi: = 76:pi:/3.  Add that to the volume of the cylinder to get (76/3 + 600):pi: = 1876:pi:/3.  That's the volume of the drilled hole.

8788:pi:/3 - 1876:pi:/3 = 6912:pi:/3 = 2304:pi:
That's the volume of the remaining sphere.

4:pi:r:^3:/3 = 2304:pi:
r:^3: = 1728
r = 12

Voila!  That's the radius of the new sphere!  12!

Sorry, this problem just took me an eternity to explain using my method.