Consider the following operation on positive real numbers written on a blackboard:
Choose a number $ r$ written on the blackboard, erase that number, and then write a pair of positive real numbers $ a$ and $ b$ satisfying the condition $ 2 r^2 \equal{} ab$ on the board.

Assume that you start out with just one positive real number $ r$ on the blackboard, and apply this operation $ k^2 \minus{} 1$ times to end up with $ k^2$ positive real numbers, not necessarily distinct. Show that there exists a number on the board which does not exceed kr.

If we have $ r$ and write a pair of positive real numbers $ a$ and $ b$ satisfying the condition  on the board, we prove that
$ \frac{1}{a^2}\plus{}\frac{1}{b^2} \ge \frac{1}{r^2}$
$ \frac{1}{a^2}\plus{}\frac{1}{b^2} \equal{} \frac{a^2\plus{}b^2}{a^2b^2} \ge \frac{2ab}{a^2b^2}\equal{} \frac{2}{ab} \equal{} \frac {1}{r^2}$
qed
If $ a_i \equal{} min a_k$ for all $ k$ then
$ n\frac{1}{a_i^2}\ge \frac{1}{a_1^2}\plus{}...\plus{}\frac{1}{a_n^2}\ge \frac {1}{r^2}$
$ n\frac{1}{a_i^2}\ge \frac {1}{r^2}$
$ nr^2 \ge a_i^2$
$ \sqrt{n} r\ge a_i$
We have $ n\equal{}k^2$
$ rk\ge a_i$
qed

Surprisingly, induction works. Furthermore, we can strengthen it to if you apply the operation $ n$ times, there exists a number that does not exceed $ \sqrt{n+1}r$.

To do this, we apply induction to the above statement. For $ n=0$ it is trivial, $ n=1$ follows from the fact that the geometric mean of $ a,b$, which is $ \sqrt{ab}$ is larger than the smaller of the two, so the smaller of the two cannot exceed $ \sqrt{ab}=\sqrt{2}r$.

Now, assume the result holds true for $ n=k$. We will show the result to be true for $ n=k+1$. If we split the original number on the board $ r$ into $ a$ and $ b$, then we consider each of the numbers formed from $ a$, and each of the numbers formed from $ b$. If $ p$ of the $ k$ remaining operations were done to $ a$, and the remaining $ q$ operations were done to $ b$, then the induction hypothesis guarantees a number that does not exceed $ min(a\sqrt{p+1},b\sqrt{q+1})$ with $ p+q=k$, and $ ab=2r^2$.

Assume the contrary, so that $ a\sqrt{p+1} > (\sqrt{k+2})r$ and $ b\sqrt{q+1} > (\sqrt{k+2})r$

$ p+1 > (k+2)\frac{r^2}{a^2}$, $ q+1>(k+2)\frac{r^2}{b^2}$

Summing, we get $ 1> r^2(\frac{1}{a^2}+\frac{1}{b^2})$

$ 2> ab(\frac{1}{a^2}+\frac{1}{b^2})$

$ 2>\frac{a}{b}+\frac{b}{a}$, contradicting AM-GM on the L.H.S. which gives $ \frac{a}{b}+\frac{b}{a} \ge 2$.

Thus, we must have had $ min(a\sqrt{p+1},b\sqrt{q+1}) \le \(sqrt{k+2})r$, and the induction is complete.

Of course, this may just look like the way of Russian_Man, but if you were not aware of that invariant, then this method would be the next most logical.

Cheers,

Rofler

Monovariant on $\sum_{\text{current numbers on the board}} \frac{1}{k^2}$.
It is easy to see that $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{a^2} + \frac{a^2}{4r^4} \ge \frac{1}{r^2}$
If all numbers are larger than $kr$, we would have the sum of $\frac{1}{i^2}$ less than $\frac{1}{r^2}$, which is a contradiction.

It is clear that at each step we should increase the minimum :\\
When applying the process we have : $\min(a,b)\le \sqrt{2}r$
note that $\min(i+1)=\min(i)$ or $\min(i+1)\le \sqrt{2}\min(i)$ .
Now analyzing the pattern we see that the best algorithm to increase the minimum is :\\
$(r)\rightarrow (\sqrt{2}r,\sqrt{2}r)\rightarrow (2r,2r,\sqrt{2}r)\rightarrow (2r,2r,2r,2r).....$
which means  we have : $\min(2^t-1)\le \sqrt{2}^t$ and $\min(i)\le \sqrt{2}^{t-1}$ where $i<2^t-1$\\
So : let $t\in\mathbb{Z}$ such that : $2^t-1\le k^2-1<2^{t+1}-1$\\
$\implies \sqrt{2}^t\le k<\sqrt{2}^{t+1}\\
\implies \min(k^2-1)\le \sqrt{2}^t \le k$

We define, for any positive real $x$ written on the board, a "level" $l$ defined as follows. The starting number $r$ has level $0$. Also, if a number $x$ with level $l_x$ is split into $a$ and $b$, then $a$ and $b$ both have level $l_a=l_b=l_x+1$. It's easy to see that $$\sum_{x \text{ on blackboard}} \frac{1}{2^{l_x}} = 1$$

For a number $x$ with level $l$, define
$$f(x,l)= \frac{\log_2 x}{2^l} - \frac{l}{2^{l+1}}$$

A simple computation yields $f(x,l_x)=f(a,l_a)+(b,l_b)$, where $x$ with level $l_x$ is split into $a$ and $b$ with level $l_a=l_b=l_x+1$.

Therefore, if the $n$ numbers in the end are $a_1,a_2,\cdots a_{n}$ with levels $l_1, l_2, \cdots l_n$, then $\sum_{i=1}^n \frac{1}{2^{l_i}}=1$, and
$$\sum_{i=1}^n f(a_i,l_i) = f(r,0)=\log_2 r$$

Suppose $a_i > \sqrt{n}r$ for all $i$. Then
$$\log_2 r > \sum_{i=1}^n \frac{\log_2 r\sqrt{n}}{2^{l_i}} - \sum_{i=1}^n \frac{l_i}{2^{l_i+1}}$$
which simplifies to
$$\sum_{i=1}^n \frac{l_i}{2^{l_i}} > \log_2 n$$

However, the function $g(x)=-x\ln x$ is concave on positive reals, so

$$\sum_{i=1}^n g\left ( \frac{1}{2^{l_i}} \right ) \le n \cdot g\left ( \frac{1}{n} \cdot \sum_{i=1}^n \frac{1}{2^{l_i}} \right )$$
which gives
$$\sum_{i=1}^n \frac{l_i}{2^{l_i}} \le \log_2 n,$$
a contradiction.

Solution from [i]Twitch Solves ISL[/i]:

The problem follows from one observation:

[b][color=red]Claim:[/color][/b]  	[Main claim] 	The sum of the squares of reciprocals does not change 	under this operation.
[i]Proof.[/i]  	Clear. $\blacksquare$

Hence, if $x_1$, \dots, $x_n$ are the numbers on the board, when $n = k^2$, they satisfy \[ \frac{1}{r^2} = \sum \frac{1}{x_n^2} \] which implies $\min(x_1, \dots, x_n) \le {\sqrt n}r = kr$.

I proved that the sum of the reciprocals in the $i-$th operation is always stricly bigger than $\sqrt(i)/r$. This clearly implies the result, as if all the numbers were bigger than $kr$, then the sum of the reciprocals would be less than $k^2/rk=k/r$.

We prove this by induction on the operation we're at:

In the first operation this is true as $1/r \ge 1/r$. Now, if $S$ is this sum in the $i-$th operation, then $S \ge \sqrt(i)/r$ and we want to prove that $S'$, which is the sum in the next operation, is bigger than $\sqrt(i+1)/r$. Notice that $S'=S-1/t+1/a+1/b$, where $t$ is the erased number. But notice:
$$ 1/a+1/b=a+b/ab=a+b/2t^2\ge \sqrt(ab)/t^2=t.\sqrt(2)/t^2=\sqrt(2)/t$$.

Thus $S' \ge S-1/t+\sqrt(2)/t \ge \frac{\sqrt(i)+\sqrt(2)-1}{t}$. Now, we prove the LHS is bigger than $\sqrt(i+1)/t$.

This occurs iff $\sqrt(i)+\sqrt(2)-1 > \sqrt(i+1)$, which occurs iff (by manipulating) $\sqrt(i^2+i) > \sqrt(2)-1$, which is true since $\sqrt(i^2+i) \ge \sqrt(2)$.

WLOG $r=1$ since we can just scale. Assign each number on the blackboard a weight such that the initial 1 starts with weight 1, and when a number with weight $w$ gets split into two numbers, each of the new numbers has weight $w/2$. Define the value of the configuration as the product of $$(number)^{weight}$$ over all numbers on the board. Note that the sum of weights is always 1. It suffices to show that after $k^2-1$ moves, the value of the configuration is at most $k$.\\

When a number $m$ with weight $w$ is split into $a$ and $b$, the value of the configuration is multiplied by $$\frac{(ab)^{w/2}}{m^w}=\frac{(2m^2)^{w/2}}{m^w}=2^{w/2}.$$ In particular, this does not depend on $m$, $a$, or $b.$\\

Define $$a_2=1/2,a_3=a_4=1/4,a_5=a_6=a_7=a_8=1/8\cdots$$ Now, define $$S_k=\sum_{i=2}^k a_k.$$

Claim 1: $$S_{k^2}\leq \log_2(k).$$ This is equivalent to $$S_{k}\leq \frac{1}{2}\log_2(k).$$ Note that when $k$ is a power of 2, equality holds. However, consider $k$ going up from one power of 2 to the next. The LHS will increase linearly by the same amount each time during this interval, while the RHS will increase concavely, and at the end of the interval, they become equal again. Therefore, the left side is at most the right side.\\

Since splitting a number with weight $w$ multiplies the value of the configuration by $2^{w/2}$, after $m$ moves, the total value is at most $$2^{S_{m+1}}.$$ Therefore, it suffices to show that $$2^{S_{k^2}}\leq k,$$ which is equivalent to Claim 1, so we are done.

[b][color=#f00]Claim :[/color] [/b] $\sum {\frac{1}{a_i^{2}}}$ is monovariant
$\textbf{proof:}$ $$\frac{1}{a^2}+\frac{1}{b^2} = \frac{a^2+b^2}{a^2b^2} \ge \frac{2ab}{a^2b^2}= \frac{2}{ab} = \frac {1}{r^2}$$

Hence $$k^2\frac{1}{(min a_i)^2}\ge \sum_{i=1}^{k^2}  \frac{1}{a_i^2} \ge \frac {1}{r^2} \implies kr \ge min a_i \blacksquare$$


Find all complex solutions of the system of equations
   $x_{1}^{k}+x_{2}^{k}+\ldots+x_{n}^{k}=0,\ k=1,2,\ldots,n.$

thank you soarer

Is there another proof ?


algebra
polynomial
inequalities
inequalities proposed

[quote="Vasc"]If $ a,b,c$ are nonnegative real numbers, then

$ \sum a^2(a \minus{} b)(a \minus{} c)(3a \minus{} 5b)(3a \minus{} 5c) \ge 0$.[/quote]
Let $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2,$ $ abc \equal{} w^3$ and $ u^2 \equal{} tv^2.$
Hence, $ \sum_{cyc} a^2(a \minus{} b)(a \minus{} c)(3a \minus{} 5b)(3a \minus{} 5c) \ge 0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(4.5a^6 \minus{} 24a^5b \plus{} 15a^4b^2 \plus{} 32a^4bc \minus{} 40a^3b^2c \plus{} 12.5a^2b^2c^2)\geq0\Leftrightarrow$
$ \Leftrightarrow49w^6 \plus{} 112(3u^2 \minus{} 4v^2)uw^3 \plus{} 9(9u^2 \minus{} 2v^2)(3u^2 \minus{} 4v^2)^2\geq0.$
$ \frac {\Delta}{14} \equal{} (3u^2 \minus{} 4v^2)^2(18v^2 \minus{} 17u^2).$ Thus, for $ t\geq\frac {18}{17}$ the inequality is true.
While for $ 1\leq t\leq\frac {18}{17}$ it's enough to prove that
$ w^3\leq\frac {8u(4v^2 \minus{} 3u^2) \minus{} (4v^2 \minus{} 3u^2)\sqrt {18v^2 \minus{} 17u^2}}{7}.$
$ (a \minus{} b)^2(a \minus{} c)^2(b \minus{} c)^2\geq0$ gives $ w^3\leq3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}.$
Hence, it remains to prove that
$ 3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}\leq\frac {8u(4v^2 \minus{} 3u^2) \minus{} (4v^2 \minus{} 3u^2)\sqrt {18v^2 \minus{} 17u^2}}{7},$
which is equivalent to $ (4v^2 \minus{} 3u^2)\sqrt {18v^2 \minus{} 17u^2}\leq11uv^2 \minus{} 10u^3 \minus{} 14\sqrt {(u^2 \minus{} v^2)^3}.$
$ 11uv^2 \minus{} 10u^3\geq14\sqrt {(u^2 \minus{} v^2)^3}$ is true for $ 1\leq t\leq\frac {18}{17}.$
Hence, $ (4v^2 \minus{} 3u^2)\sqrt {18v^2 \minus{} 17u^2}\leq11uv^2 \minus{} 10u^3 \minus{} 14\sqrt {(u^2 \minus{} v^2)^3}\Leftrightarrow$
$ \Leftrightarrow\left((4v^2 \minus{} 3u^2)\sqrt {18v^2 \minus{} 17u^2}\right)^2\leq\left(11uv^2 \minus{} 10u^3 \minus{} 14\sqrt {(u^2 \minus{} v^2)^3}\right)^2\Leftrightarrow$
$ \Leftrightarrow449t^3 \minus{} 1378t^2 \plus{} 1413t \minus{} 484\geq28(11 \minus{} 10t)\sqrt {t(t \minus{} 1)^3}\Leftrightarrow$
$ \Leftrightarrow449t^2 \minus{} 929t \plus{} 484\geq28(11 \minus{} 10t)\sqrt {t(t \minus{} 1)}\Leftrightarrow$
$ \Leftrightarrow(3t \minus{} 4)^2(117t \minus{} 121)^2\geq0,$ which is true. Done!

Very nice your proof, Arqady.  :lol:
This is a very strong inequality. Actually, the inequality holds

$ \sum a^2(a \minus{} b)(a \minus{} c)(a \minus{} kb)(a \minus{} kc) \ge 0$

for all $ 0\le k\le \frac 5{3}$, with equality for $ (a,b,c)\equal{}(1,1,1)$, or $ (a,b,c)\equal{}(0,1,1)$, or $ (a,b,c)\equal{}(k,1,1)$.

I have got the following similar result:
If $ a,b,c$ are the sidelengths of a triangle, then
$ \sum a^2(a\minus{}b)(a\minus{}c)(a\minus{}2b)(a\minus{}2c) \ge 3(a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2,$
with equality holds when $ a\equal{}b\equal{}c$ or $ a\equal{}b\plus{}c$ or $ b\equal{}c\plus{}a$ or $ c\equal{}a\plus{}b.$ :)

[quote="Vasc"]Arqady, works your method for the inequality

$ \sum a^2(a \minus{} b)(a \minus{} c)(2a \minus{} 3b)(2a \minus{} 3c) \ge 0$ ?[/quote]-
Not yet.  :(

This is a very strong inequality.

Let $ a,b,c$ be nonnegative real numbers. If $ k\le 4$, then

$ \sum a^2(a \minus{} b)(a \minus{} c)(a \minus{} kb)(a \minus{} kc) \ge (5 \minus{} 3k)(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2$.

The cases $ k\equal{}4$ and $ k\equal{}\frac 5{3}$ are special.

For $ k\equal{}5$, the inequality

$ \sum a^2(a \minus{} b)(a \minus{} c)(a \minus{} 5b)(a \minus{} 5c) \plus{}10(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2\ge 0$.

doesn't hold for any nonnegative real numbers $ a,b,c$, although it holds when one of  $ a,b,c$ is 0, and also when two of $ a,b,c$ are equal.

I think the best inequality of this type is the out of one's reach inequality

$ \sum a^2(a \minus{} b)(a \minus{} c)(a \minus{} 5b)(a \minus{} 5c) \plus{}(10\plus{}\frac 5{64})(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2\ge 0$.

[quote="Vasc"]For $ k \equal{} 5$, the inequality

$ \sum a^2(a \minus{} b)(a \minus{} c)(a \minus{} 5b)(a \minus{} 5c) \plus{} 10(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2\ge 0$.

doesn't hold for any nonnegative real numbers $ a,b,c$, although it holds when one of  $ a,b,c$ is 0, and also when two of $ a,b,c$ are equal.

I think the best inequality of this type is the out of one's reach inequality

$ \sum a^2(a \minus{} b)(a \minus{} c)(a \minus{} 5b)(a \minus{} 5c) \plus{} (10 \plus{} \frac 5{64})(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2\ge 0$.[/quote]
The following two both are not bad too, are they?

$ \sum a^2(a \minus{} b)(a \minus{} c)(a \minus{} 6b)(a \minus{} 6c) \plus{} (13\plus{}\frac{12}{25})(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2\ge 0$.

$ \sum a^2(a \minus{} b)(a \minus{} c)(a \minus{} 7b)(a \minus{} 7c) \plus{} (17\plus{}\frac{5}{16})(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2\ge 0$.

The constants $ 13\plus{}\frac{12}{25}$  and  $ 17\plus{}\frac{5}{16}$ both are best possible.

[quote="Guest"] The following two both are not bad too, are they?

$ \sum a^2(a \minus{} b)(a \minus{} c)(a \minus{} 6b)(a \minus{} 6c) \plus{} (13 \plus{} \frac {12}{25})(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2\ge 0$.

$ \sum a^2(a \minus{} b)(a \minus{} c)(a \minus{} 7b)(a \minus{} 7c) \plus{} (17 \plus{} \frac {5}{16})(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2\ge 0$.

The constants $ 13 \plus{} \frac {12}{25}$  and  $ 17 \plus{} \frac {5}{16}$ both are best possible.[/quote]
I think the same.
And, for $ k\ge 4$, we have

$ \sum a^2(a \minus{} b)(a \minus{} c)(a \minus{} kb)(a \minus{} kc) \plus{} \frac{k^4\plus{}4k^2\minus{}12k\minus{}20}{4(k\minus{}1)^2}(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2\ge 0$.

[quote="Vasc"][quote="Guest"] The following two both are not bad too, are they?

$ \sum a^2(a \minus{} b)(a \minus{} c)(a \minus{} 6b)(a \minus{} 6c) \plus{} (13 \plus{} \frac {12}{25})(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2\ge 0$.

$ \sum a^2(a \minus{} b)(a \minus{} c)(a \minus{} 7b)(a \minus{} 7c) \plus{} (17 \plus{} \frac {5}{16})(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2\ge 0$.

The constants $ 13 \plus{} \frac {12}{25}$  and  $ 17 \plus{} \frac {5}{16}$ both are best possible.[/quote]
I think the same.
And, for $ k\ge 4$, we have

$ \sum a^2(a \minus{} b)(a \minus{} c)(a \minus{} kb)(a \minus{} kc) \plus{} \frac {k^4 \plus{} 4k^2 \minus{} 12k \minus{} 20}{4(k \minus{} 1)^2}(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2\ge 0$.[/quote]
Wonderful! This is an extremely nice result. I just verified it by a package for inequality-proving in few seconds, but I cannot deduce it so far.

[quote="tuantam1lan"][quote="Vasc"]
I think the same.
And, for $ k\ge 4$, we have

$ \sum a^2(a \minus{} b)(a \minus{} c)(a \minus{} kb)(a \minus{} kc) \plus{} \frac {k^4 \plus{} 4k^2 \minus{} 12k \minus{} 20}{4(k \minus{} 1)^2}(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2\ge 0$.[/quote]
you can give me a good view of it?
I do not understand why you found a great result like!
I really want to know .[/quote]
You can find the necessary condition $ K\ge\frac {k^4 \plus{} 4k^2 \minus{} 12k \minus{} 20}{4(k \minus{} 1)^2}$ by replacing $ a \equal{} k$, $ b \equal{} 1 \plus{} \delta$ and $ c \equal{} 1 \minus{} \delta$ in
$ \sum a^2(a \minus{} b)(a \minus{} c)(a \minus{} kb)(a \minus{} kc) \plus{} K(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2\ge 0$. The inequality becomes $ K_2\delta^2 \plus{} K_3\delta^3 \plus{} ... \plus{} K_6\delta^6 \ge 0$, which yields $ K_2\ge 0$, where $ K_2 \equal{} 4K(k \minus{} 1)^2 \minus{} (k^4 \plus{} 4k^2 \minus{} 12k \minus{} 20)$.

Recently, I found a proof for this inequality, based on some new theoretical considerations.

[quote="can_hang2007"]I have got the following similar result:
If $ a,b,c$ are the sidelengths of a triangle, then
$ \sum a^2(a \minus{} b)(a \minus{} c)(a \minus{} 2b)(a \minus{} 2c) \ge 3(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2,$
with equality holds when $ a \equal{} b \equal{} c$ or $ a \equal{} b \plus{} c$ or $ b \equal{} c \plus{} a$ or $ c \equal{} a \plus{} b.$ :)[/quote]
Nice result, Can_hang207. :lol:
More general,

$ \sum a^2(a \minus{} b)(a \minus{} c)(a \minus{} kb)(a \minus{} kc) \ge (k^2\minus{}3k\plus{}5)(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2,$

for any real $ k$ and any triangle.

[quote="Vasc"]If $ a,b,c$ are nonnegative real numbers, then

$ \sum a^2(a \minus{} b)(a \minus{} c)(3a \minus{} 5b)(3a \minus{} 5c) \ge 0$.[/quote]

Your inequality is also  true for all $a,b,c$ reals

This is another problem
If $x,y,z$ are  reals, then
\[ \sum_{cyc} x^2(x+y)(x-z)(2x+y)(2x+z) \geq 0 \]

[quote="My_dad"][quote="Vasc"]If $ a,b,c$ are nonnegative real numbers, then

$ \sum a^2(a \minus{} b)(a \minus{} c)(3a \minus{} 5b)(3a \minus{} 5c) \ge 0$.[/quote]

Your inequality is also  true for all $a,b,c$ reals
[/quote]
What is your proof of this statement? Thank you.

[quote="arqady"] What is your proof of this statement? Thank you.[/quote]
My solution to the inequality $f(a,b,c)\ge 0$ for $k<4$, where \[f(a,b,c)=\sum a^2(a-b)(a-c)(a-kb)(a-kc)+(3k-5)(a-b)^2(b-c)^2(c-a)^2,\] is based on the highest coefficient cancellation method, and is given in
http://www.sciencedomain.org/abstract.php?iid=360&id=6&aid=2653
We have \[f(a,1,1)=a^2(a-1)^2(a-k)^2.\] Let $p=a+b+c$, $q=ab+bc+ca$, $r=abc$.  Since \[a(a-b)(a-c)=2a^3-pa^2+r, \ \ \ a(a-kb)(a-kc)=(k+1)a^3-kpa^2+k^2r,\] $\sum a^2(a-b)(a-c)(a-kb)(a-kc) $ has the same highest coefficient $A_1$ as $g(a,b,c)$, where

$\quad g=\sum (2a^3+r)[(k+1)a^3+k^2r]=2(k+1)\sum a^6+(2k^2+k+1)r\sum a^3+3k^2r^2;$
 that is, \[A_1=6(k+1)+3(2k^2+k+1)+3k^2=9(k^2+k+1).\] Therefore, $f$ has the highest coefficient \[A_2=9(k^2+k+1)-27(3k-5)=9(4-k)^2.\]
Case 1: $p^2\le 4q$. Since $f(1,1,1)=f(0,1,1)=f(k,1,1)$, define
\[P(a,b,c)=r+Bp^3+Cpq+Dq^2/p\] such that $P(1,1,1)=P(0,1,1)=P(k,1,1)$; that is
\[P(a,b,c)=r+\frac{2p^3-(10+k)pq+4(2+k)q^2/p}{3(4-k)},\] \[P(a,1,1)=\frac{2a(a-1)^2(a-k)}{3(4-k)(a+2)}.\] We will prove the sharper inequality $f_6(a,b,c)\ge 0$, where
\[f_6(a,b,c)=f(a,b,c)-9(4-k)^2P^2(a,b,c).\]
Since $f_6$ has the highest coefficient equal to zero, it suffices to show that $f_6(a,1,1)\ge 0$ for
$(a+2)^2\le 4(2a+1)$; that is for $0\le a\le 4$. Indeed,
\[f_6(a,1,1)=f(a,1,1)-9(4-k)^2P^2(a,1,1)=\frac{3a^2(a-1)^2(a-k)^2(4-a)}{(a+2)^2}\ge 0.\]
Case 2: $p^2> 4q$. We will prove the sharper inequality $g_6(a,b,c)\ge 0$, where \[g_6(a,b,c)=f(a,b,c)-9(4-k)^2a^2b^2c^2.\] Since $g_6$ has the highest coefficient equal to zero, it suffices to show that $g_6(a,1,1)\ge 0$ and $g_6(0,b,c)\ge 0$ for $a>4$, $b\ge 0$, $c\ge 0$. Indeed, we have  \[g_6(0,b,c)=(b-c)^4[(b-c)^2+(5-k)bc]\ge 0\] and \[g_6(a,1,1)=a^2[(a-1)^2(a-k)^2-9(4-k)^2]>0,\]
since $(a-1)^2>9$ and $(a-k)^2>(4-k)^2$.

[quote="arqady"]Thank you! I think for all reals $a$, $b$ and $c$ this way does not work. :maybe:[/quote]
Have you a symmetric and homogeneous inequality $f_6(a,b,c)\ge 0$ for which this method fails?

[quote="Vasc"]
Have you a symmetric and homogeneous inequality $f_6(a,b,c)\ge 0$ for which this method fails?[/quote]
Let $f(a,b,c)=\sum_{cyc}a^2(a-b)(a-c)(2a-3b)(2a-3c)$.
By your method we'll try to prove that
$\sum_{cyc}a^2(a-b)(a-c)(2a-3b)(2a-3c)\geq\frac{19\left(\sum\limits_{cyc}(4a^3-7a^2b-7a^2c+10abc)\right)^2}{441}$,
which is wrong. :(
For non-negatives $a$, $b$ and $c$ we know that $f\geq0$.
Why $f\geq0$ for all reals $a$, $b$ and $c$?

[quote="arqady"] By your method we'll try to prove that
$\sum_{cyc}a^2(a-b)(a-c)(2a-3b)(2a-3c)\geq\frac{19\left(\sum\limits_{cyc}(4a^3-7a^2b-7a^2c+10abc)\right)^2}{441}$,
which is wrong. :(
For non-negatives $a$, $b$ and $c$ we know that $f\geq0$.
Why $f\geq0$ for all reals $a$, $b$ and $c$?[/quote]
This inequality holds for $p^2\le 4q$. For $p^2\ge 4q$ we have
\[\sum_{cyc}a^2(a-b)(a-c)(2a-3b)(2a-3c)\geq 171\left(r+\frac 1{9}p^3-\frac 4{9}pq\right)^2.\] Therefore, the inequality holds for all real $a,b,c$.
Actualy, for all real $a,b,c$, the following sharper inequality holds
$\sum_{cyc}a^2(a-b)(a-c)(2a-3b)(2a-3c)\geq 2(a-b)^2(b-c)^2(c-a)^2.$
The proof is similar.

[quote="Vasc"][quote="arqady"] By your method we'll try to prove that
$\sum_{cyc}a^2(a-b)(a-c)(2a-3b)(2a-3c)\geq\frac{19\left(\sum\limits_{cyc}(4a^3-7a^2b-7a^2c+10abc)\right)^2}{441}$,
which is wrong. :(
For non-negatives $a$, $b$ and $c$ we know that $f\geq0$.
Why $f\geq0$ for all reals $a$, $b$ and $c$?[/quote]
This inequality holds for $p^2\le 4q$. For $p^2\ge 4q$ we have...[/quote]
For $b=c=1$ we have $(a-1)^2(2a-3)^2(365a^2+152a-76)\geq0$,
which is wrong for $(a+2)^2\leq4(2a+1)$ and for $(a+2)^2\geq4(2a+1)$. :wink:

[quote="Vasc"]
Actualy, for all real $a,b,c$, the following sharper inequality holds
$\sum_{cyc}a^2(a-b)(a-c)(2a-3b)(2a-3c)\geq 2(a-b)^2(b-c)^2(c-a)^2.$
The proof is similar.[/quote]
I have proven it for all non-negatives $a$, $b$ and $c$.

[quote="arqady"] For $b=c=1$ we have $(a-1)^2(2a-3)^2(365a^2+152a-76)\geq0$,
which is wrong for $(a+2)^2\leq4(2a+1)$, that is $0\le a\le 4$. :wink:[/quote]
Indeed, for $p^2\le 4q$, we need to use
$\sum_{cyc}a^2(a-b)(a-c)(2a-3b)(2a-3c)\geq\ 171P^2(a,b,c),$
where
$P(a,b,c)=r+\frac{4p^3-23pq+28q^2/p}{15}$
such that $P(1,1,1)=P(0,1,1)=P(3/2,1,1)$.
 By this method we can prove the general inequality for all real $a,b,c$:
\[\sum a^2(a-b)(a-c)(a-kb)(a-kc)\ge \alpha_k(a-b)^2(b-c)^2(c-a)^2,\]
where
\[
\alpha_k=\left\{
\begin{array}{ll}
\dfrac{20+12k-4k^2-k^4}{4(1-k)^2}, & k\in (-\infty,-2]\cup [4,\infty) \\[10pt]
1+k, & k\in [-2,1] \\[10pt]
5-3k, & k\in [1,4]
\end{array},
\right.
\]
with equality for $a=b=c$, for $a=0$ and $b=c$ $($or any cyclic permutation$)$, for $a/k=b=c$ $($or any cyclic permutation$)$ if $k\neq 0$, and for $a=0$ and $b+c=0$ $($or any cyclic permutation$)$ if $k\in [-2,1]$.

Yes, it's great!  :lol: I missed  $(1,1,0)$ equality case.
Thank you very much!

[quote="Vasc"]
Have you a symmetric and homogeneous inequality $f_6(a,b,c)\ge 0$ for which this method fails?[/quote]
We know that for $a+b+c=3$ the following inequality holds.
\[\sum_{cyc}\frac{1}{7+2a^2}\leq\frac{1}{3}\]
which after homogenization and full expanding gives $f(a,b,c)\geq0$, where $f(a,b,c)=$
$=\sum_{sym}(3.5a^6-21a^5b+177a^4b^2+205a^3b^3-178.5a^4bc-318a^3b^2c+132a^2b^2c^2)$.
The coefficient before $w^6$ is $1458$.
Let $F(a,b,c)=f(a,b,c)-\frac{1458\left(\sum\limits_{cyc}(a^3-3a^2b-3a^2c+5abc)\right)^2}{729}$.
Hence, $F(a,1,1)=(5a+22)(a+2 )(a-1)^2(a-4)^2$, which is true for $p^2\leq4q$.
What is your proof for $p^2\geq4q$? Thank you!


[size=150]The horizontal axis is marked in degrees. Find a+b+c where one period of the sinusoid
y=a+bsinc(x-45^0 )begins at (45^0  ,11) and ends at (135^0  ,11) with horizontal tangent lines at (67.5^0  ,16) and (112.5^0  , 6)

a) 21.5
b)21
c)20.5
D) 20
e)none of these[/size]

Thanks!

Is the eqn $y=a+b\sin (c(x-45^o))$ or $y=a+b(\sin c)(x-45^o)$?

The first one.  The second one is linear.


IF $a,b,c\in Z$, $(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})\in Z$ and
$(\frac{b}{a}+\frac{c}{b}+\frac{a}{c})\in Z$ prove that
$|a|=|b|=|c|$

[quote="Labiliau"]Let $v_p(a)=x$, $v_p(b)=y$,$v_p(c)=z$.

[/quote]
What symbolise these?

$v_p(a)$ is called the $p$-adic valuation ($p$ being a prime here) of the rational number $a$. It gives you the multiplicativity in that $p$ occurs in the factorisation of $a$. E.g. $v_2(24)=3$, since $24=2^3 \cdot 3$ and also $v_5(0.06)=-2$, since $0.06=2^{-1} \cdot 3 \cdot 5^{-2}$.
So you can say that $a = \prod_{p \in \mathbb{P} } p^{v_p(a)}$.


$ n \equal{} 3^k$.
now if $ n \neq 3^k$ assume that $ s$ is the greatest number such that $ 3s \le n$ then if the equality doesn't occur(i mean $ 3 \nmid n$) we choose $ i \equal{} 3s$ ,obviously $ 3 \nmid \binom{n}{i}$.so we must have $ 3 | n ,3s \equal{} n$.
let's suppose that $ 3^t \parallel{} s$.
we have just to choose $ i \equal{} 3^{t \plus{} 1}$ then $ 3 \nmid \binom{n}{i}$.we r done!
(note that we can replace $ 3$ by a prime number $ p$,the solution will be the same).


Prove that for any natural number $ n>1$,

$ \frac{1}{n\plus{}1} \plus{} \frac{1}{n\plus{}2} \plus{} \frac{1}{n\plus{}3} \plus{} ... \plus{} \frac{1}{2n} > \frac{13}{24}$.

hello, we have
$ \sum_{i\equal{}1}^{n}\frac{1}{n\plus{}i}>\frac{n}{n\plus{}1}>\frac{13}{24}$ because $ n>1$
Sonnhard.

Let $ u_n \equal{} \sum_{i\equal{}1}^n \frac{1}{n\plus{}i}$ Since $ u_2>\frac{13}{24}$ we only need to prove that this is an increasing sequence

We have $ u_{n\plus{}1} \minus{} u_n \equal{} \frac{1}{2n\plus{}1} \plus{} \frac{1}{2n\plus{}2}\minus{} \frac{1}{n\plus{}1} \equal{} \frac{1}{2n\plus{}1} \minus{} \frac{1}{2n\plus{}2}>0$

Hence proved

Thanks.

We could also prove it by mathematical induction, since (denoting the left-hand-side by $ S_{n}$), we have $ S_{2} \equal{} \frac{7}{12} \equal{} \frac{14}{24} > \frac{13}{24}$ and

$ S_{k\plus{}1} \minus{} S_{k} \equal{} \frac{1}{2k\plus{}1}  \frac{1}{2k\plus{}2} \minus{} \frac{1}{k\plus{}1} \equal{} \frac{1}{2(k\plus{}1)(2k\plus{}1)} > 0$ for every natural number $ k$.

So, $ S_{k\plus{}1} > S_{k} > \frac{13}{24}$ , by the inductive assumption.


Liyi wrote

"Chinese undergraduates camp starts from 1987. it is in summer. only about fourty students attend it. it is held by Math Institute, Chinese Academy of Science."

Thank you Liyi.

Can I have more informations about this camp ? It interests me  :) (price, where, french are they allowed to go to this camp, how etc....)

Thanks  :)

why is it that the chinese seems to know math alot

Chinese Underdraduate Contest has been canceled since a few years ago, in fact, the history of it is only 8 years long.

hey samebody past to problems china student olympiad problems


If the probability of winning the lead role in the school play is $ \frac{2}{9}$, what are the odds against getting the lead role? Express your answer in the form a:b.

It asked for the odds and not the probability so the answer is $7:2$.


calculus
integration
limit
logarithms
calculus computations

Find $ \lim_{x\to\infty} \int_{e^{\minus{}x}}^1 \left(\ln \frac{1}{t}\right)^ n\ dt\ (x\geq 0,\ n\equal{}1,\ 2,\ \cdots)$.

Setting $ t\equal{}e^{\minus{}u}$

$ \lim_{x\rightarrow \infty }\int_{e^{\minus{}x}}^{1}\left( \ln \frac{1}{t}\right) ^{n}dt\equal{}\allowbreak \int_{0}^{\infty }u^{n}e^{\minus{}u}\,du\equal{}n!$

by the time i refreshed he had already posted so couldn't see .Sorry .should i delete??


calculus
derivative
inequalities proposed
inequalities

Let $x,\ y,\ z$ be real numbers such that $x>0,\ xyz=1,\ xy+3yz+zx=1.$
Find the range of the value that $y+z$ can be valued.

As I see the thing....

rewrite as $x(y+z)=1-3yz=yz(x-3)=\frac{1}{x}(x-3)$

From this it follows that $y+z=\frac{1}{x^{2}}(x-3)$

So if we define $y+z=f(x)$, it sufficies to study the first derivative to find out that for $x=6$ we have the absolute maximum, so that $y+z<\frac{1}{12}$ (here the equality is not possible just by checking manually the case)
However, as $x\rightarrow 0$, $f(x)\rightarrow-\infty$, so that we can conclude that

$y+z\in \left(-\infty;\frac{1}{12}\right)$

Please correct me if i'm wrong :blush:

In your solution, what's the values of $y,\ z$ for $x=6?$

It's impossible to reach the value $\frac{1}{12}$ for $y+z$ because we would have a system for $y,z$ in which $y+z=\frac{1}{12}$ and $yz=\frac{1}{6}$. That's why i decided not to include that value in the range
 :)


Well, if it is not clear to the point that they mis-understand you, you definitely can't get full credit...

You want to make it as clear as possible since it also might give you another commended solutions.  (I have a couple solutions last year that got commended not because it's more elegant than others, but rather even aspect is fully explained.)

What is the advantage of a commended solution other than self-satisfaction?

Well, you get your name on the leader board  :wink:

So I guess if you're trying to compete with someone, and you both get a perfect score, you can have an advantage over them with commended solutions.

It has a chance of being posted on the web by the USAMTS people. Also, on the Leaders page, there is a section for most commended solutions.

[quote="Calculus"]Are our solutions graded solely based on accuracy or does clarity matter? Or as long as clarity doesn't affect the reader from understanding the solution?[/quote]

These are math questions, so we expect the answer to be absolutely accurate. It is not like chemistry or physics, where answers are rounded to the nearest thousandth.

Clarity matters a lot. The other important points are completeness and a strong line of reasoning. Elegance can earn a commendation, but it is not required for points.

Unclear solutions also have the risk that the grader won't be able to find all the details in the solution that he or she is looking for, and will mistakenly mark the solution as incomplete. There is a grade-dispute system that can correct that score later, but it is much easier on everyone if the solution is clear in the first place.

Erin Schram


I am almost certain that my calculations are wrong, could someone help?

Suppose there is a cart (mass .492 kg) on a table that is being tugged on by a hanging mass (25g).  {The cart is connected to a string, which goes along a pulley and is tied to the hanging mass, which is off the edge of the table .  Since gravity will pull the mass downward, the cart will move laterally along the table.}  From the pulley, we know that the distance that the mass falls (but not touch the ground) is .89 meters, which implies that the cart moves .89 meters laterally across the table.  Starting from rest (inital velocity = 0 m/s), it takes both the cart and the hanging mass 2.86 seconds to go .89 meters.

Using kinematics, $\Delta x = \frac 12(v_i + v_f)t$.  Which I plug in values to get final velocity of 0.62 m/s.

Then, given that (I solved for them) $W=\Delta KE + F_k\Delta x, \Delta KE = \frac 12mv^2, \Delta PE = mgh, \Delta E_{thermal} = F_k\Delta x$  I have to compute the force of friction between the cart and the table.

We know that (by conservation of energy) $\Delta E = 0$, so

$\Delta E = \Delta KE + \Delta PE + \Delta E_{thermal}$
$0=(\frac 12mv^2_f - \frac 12mv^2_0) + (mgh_f - mgh_0) + F_k\Delta x$
Solving for $F_k$, we get

$F_k = \frac{-m(\frac12(v^2_f - v^2_0) + g(h_f - h_0))}{\Delta x}$

Then, I plug in values to get:

$F_k = \frac{-0.492(\frac12(.062^2 - 0^2) + (-9.81)(0))}{0.89} \approx 0.11N$ (since we want positive magnitude).

Am I doing it right?

Your method is ok, but there are a couple of things/ommisions.

There are two masses, $m_1$ and $m_2$, but you have labelled them both as $m$.

This line,

$F_k = \frac{-m(\frac12(v^2_f - v^2_0) + g(h_f - h_0))}{\Delta x}$

should be,

$F_k = \frac{-m_1\frac12(v^2_f - v^2_0) - m_2g(h_f - h_0)}{\Delta x}$

And, in this line,

$F_k = \frac{-0.492(\frac12(.062^2 - 0^2) + (-9.81)(0))}{0.89} \approx 0.11N$

You have missed out putting in any value for $h_f$. You also put in 0.062 instead of 0.62 for the speed of $m_1$, but that's just a typo.

Your final expression shpould be,

$F_k = \frac{-0.492\frac12(.62^2 - 0^2) - 0.025(9.81)(0 - 0.89)}{0.89} \approx 0.139N$


Prove that for any integer polynomial $P(x)$ there exists $k$ such
that $Q(x)=2^{k}P(x)-1$ is irreducible over $Z$ (or $Q$).
Is it true that Q(x) must be irreducible for all sufficiently large $k$?

[quote="shasha"]Prove that for any integer polynomial $P(x)$ there exists $k$ such
that $Q(x)=2^{k}P(x)-1$ is irreducible over $Z$ (or $Q$). [/quote]

[hide]
Sorry if this spoiled the fun, but the basic technique of $2$-adic Newton polygons (the criterion of Eisenstein-Dumas) easily shows that $2^{k}P(x)-1$ is irreducible for [i]all[/i] sufficiently large $k$ coprime with $\deg{P}$ (for "sufficiently large" it suffices that the complex zeros of $Q$ lie in the circle $|z| < 1$, which certainly holds for large enough $k$). I will leave the proof to someone else.
[/hide]

[quote]Is it true that Q(x) must be irreducible for all sufficiently large $k$?[/quote]

As stated, clearly no, as for example $2^{k}x^{2}-1$ is reducible for all even $k$. However, if $P(x)$ is not of the shape $cR(z)^{m}$ with $R \in \mathbb{Z}[x]$ $m > 1$, then, yes $Q(z)$ is reducible for only finitely many $k$. The proof, though, is not so easy, as the Newton polygons approach breaks down in the case when $k$ and $\deg{P}$ have a common factor, so one must look for a different approach. The question is relevant to the so-called [i]problem of universal Hilbert sets[/i]. There is a general theorem of Zannier and Corvaja on that subject. It implies, for instance: let $F(x,y) \in \mathbb{Z}[x,y]$ be a polynomial such that $F(x,y^{m})$ is irreducible over $\mathbb{C}[x,y]$ for every $m > 0$. Then there are only finitely many $k$ for which $F(x,2^{k}) \in \mathbb{Z}[x]$ is reducible. ("the set $\{2^{k}\mid k \in \mathbb{N}\}$ is an universal Hilbert set"). Your question is the special case $F(x,y) : = yP(x)-1$.

Thanks, Vess, you didnt spoil the fun at all, moreover you motivated me to learn
that great Dumas criterion. Really, if k is large enough and (k+m,deg P)=1, where
m is exponent of 2 in oldest coefficient of P,
then Newton diagram of Q consists only of one segment with endpoints
(0,k+m) and (deg P,0) which doesnt contain any other integer point.
So, by Dumas criterion, Q is irreducible.
I used here that k+m and deg P are coprime, not k and deg P, as you said.
Am I wrong somewhere?

[quote="shasha"] Really, if k is large enough and (k+m,deg P)=1, where
m is exponent of 2 in oldest coefficient of P,
then Newton diagram of Q consists only of one segment with endpoints
(0,k+m) and (deg P,0) which doesnt contain any other integer point.
So, by Dumas criterion, Q is irreducible.
I used here that k+m and deg P are coprime, not k and deg P, as you said.
Am I wrong somewhere?[/quote]

Yes, this is absolutely the correct condition! For some reason I assumed that your polynomial $P(x)$ was monic...  :blush: You are absolutely right!


invariant
calculus
integration
algebra
function
domain
Ring Theory

Prove that if $ n\geq 1$ and $ G$ is finite, then $ \mathbb{C}[x_1,\cdots,x_n]^G\ne \mathbb{C}$.
Give a counterexample for the case $ G$ is infinite.

Thks in advanced/

There are a lot of ways to prove that. Here just one way: Define $ K \equal{} Quot(\mathbb C[X_1,\ldots,X_n])$. Then $ K^G\subset K$ is a (finite) algebraic field extension, by Galois. If $ \mathbb C[X_1,\ldots,X_n]^G \equal{} \mathbb C$, then $ \mathbb C \equal{} K^G\subset K$ is algebraic, hence $ K \equal{} \mathbb C$. Contradiction.

As counterexample let $ G \equal{} \mathbb C$ operate on $ \mathbb C[X]$ by $ \alpha\cdot f(X) \equal{} f(X \plus{} \alpha)$. Then $ \mathbb C[X]^G \equal{} \mathbb C$.

Thank you very much.

Could you please show me another way to solve this first question not using Galois theory. And could you recommend me some introductory books in Invariant Theory I should read. I am very new in this stuff.

Thanks a lot.

$ \prod_{\sigma \in G}\sigma(X_1) \not\in \mathbf{C}$.

There is Mumford, Geometric Invariant Theory.

Here another approach in a more general setting: Let $ R$ be an integral domain finitely generated over a field $ k$ together with a finite group $ G$ operating on $ R$ over $ k$. Then $ R^G\subset R$ is integral which implies that $ \dim (R^G)\equal{}\dim(R)$ - where the Krull dimension is meant. In your setting we get $ 0\equal{}\dim \mathbb C\equal{}\dim R^G\equal{}\dim R\equal{}n\ge 1$ - hence a contradiction.

Note that the dimension of $ R$ is equal to the transcendence degree of $ Quot(R)$ over $ k$. This allows you to work with the field extension $ Quot(R^G)\subset Quot(R)$ which is algebraic. Again with Galois theory you can get the result.

Me too, I had Mumford's text book immediately in mind, but unfortunately it is pretty much advanced - and it assumes a lot of basic knowledge not only on invariant rings.


Take a square $ABCD$.
$M$ and $N$ are midpoints of $AB$ and $BC$ respectfully.
$P$ is the intersection of $AN$ and $DM$.
$R$ is the intersection of $CM$ and $DN$.
$Q$ is the intersection of $AN$ and $CM$.

Now prove that $[AMP]$ + $[BMQN]$ + $[CNR]$ = $[DPQR]$, where $[\cdots]$ represents the area of the polygon.

wow wouldn't it be sexy if M and N can be taken arbitrarily on those sides and that result still holds...HINT: consider triangles MCD and NDA, the sum of their areas is equal to the whole, then what can you say about their overlap vs. the area they do not cover inside the square?

[quote="bthings_2000"]Not a very elegent method, but it works:

[hide]
Cord bash. That is, let D be the origin, A=(0,a), B=(a,a), C=(a,0). You can easily make a generalization argument for a=1 to simplify things considerably. Assign equations to the lines, find the coordinates of P, Q, and R, and use one-half base x height to calculate the area of the triangles, which is easy. Then use the determinant method for DPQR to find its area. Should work out.

I'd be interested to see a more elegent geometric solution, which can't be that far off, since any other solution would probably be more elegent.

[/hide][/quote]

dude did you read my post?

Yes...Altheman's method is very good.

Since DMC and AND are both half the area of the square (which would also be the case if M and N were not midpoints), their sum is the entire area.

[DMC]+[AND]=[ABCD].

But,

[AMP]+[BMQN]+[CNR]+[DMC]+[AND]-[DPQR]=[ABCD].

So I guess the conclusion follows.

Aha! Altheman wins; definately a good solution. But since you asked:

[quote]dude did you read my post?[/quote]

I barely ever read your posts because you tend to write solutions in a kindof lazy scatterbrained manner. You're virtually always right, and you're a better problem-solver than me for sure, but your solutions are quite a bother to read. In this case it was simple enough and I ought to have read it.

[quote]wow wouldn't it be sexy if M and N can be taken arbitrarily on those sides and that result still holds...[/quote]

Following your reasoning, that extension clearly holds because triangles DMC and AND each occupy half the area of the square no matter their location on those sides. If you're not convinced: draw out the figure, give each region a lable,  add the three regions that comprise each of the triangles DMC and AND (you should add DPQR twice), and set that sum equal to the sum of all the regions. You easily get the desired result.

[quote="bthings_2000"]Following your reasoning, that extension clearly holds because triangles DMC and AND each occupy half the area of the square no matter their location on those sides. If you're not convinced: draw out the figure, give each region a lable,  add the three regions that comprise each of the triangles DMC and AND (you should add DPQR twice), and set that sum equal to the sum of all the regions. You easily get the desired result.[/quote]
Ok...did you read MY post?  :ninja:

[quote="bthings_2000"]
If you're not convinced: draw out the figure, give each region a lable,  add the three regions that comprise each of the triangles DMC and AND (you should add DPQR twice), and set that sum equal to the sum of all the regions. You easily get the desired result.[/quote]

Apparently sarcasm does not work well on forums...Obviously I know that the result holds for and M and N...I basically gave the proof.

lol sorry...i ono...the thing is i am kind of procrastinating from my apush test tomorrow [i still have to review like 1900- on...ahhhhh]...also this method is from math olympiad treasures

edit: i might be becoming like arnav...this has to stop lol


The base of a triangle is $ 15$ inches. Two lines are drawn parallel to the base, terminating in the other two sides, and dividing the triangle into three equal areas. The length of the parallel closer to the base is:
$ \textbf{(A)}\ 5\sqrt{6}\text{ inches} \qquad\textbf{(B)}\ 10\text{ inches} \qquad\textbf{(C)}\ 4\sqrt{3}\text{ inches} \qquad\textbf{(D)}\ 7.5\text{ inches}\\
\textbf{(E)}\ \text{none of these}$

[quote="Carbon57"][hide]
The ratio of one side of the middle-sized triangle to one side of the large triangle is:

$ \sqrt {\frac {2}{3}} \equal{} \frac {\sqrt {6}}{3}$

$ 15\cdot\frac {\sqrt {6}}{3} \equal{} 5\sqrt {6}\implies\boxed{\text{A}}$[/hide][/quote]

iknow the ratio of areas is 2/3 but why is the ratio of sides sqrt(2/3)?

nevermined....got it!


I've been working on this one for awhile and am not really getting anywhere.  The 9/8 is a bit frustrating - it looks like Holder should work well, but I am finding it difficult to get xyz terms out of the RHS.

Edit - does this inequality even hold?  Some investigation seems to indicate that it may not.  Will try to find a counterexample.

I have been strugging with the trinomial principle , but due to huge computations I gave up! :(

I am sure there is a brilliant solution!

I am going to prove the inequality for two variables. If we let [tex]z=1[/tex], in
[tex]\displaystyle(x^{2}y^{2}z^{2}+xyz+1)^{2} \leq \frac{9}{8}(x^{4}+1)(y^{4}+1)(z^{4}+1)[/tex]
the inequality becames:

[tex]\displaystyle(x^{2}y^{2}+xy+1)^{2} \leq \frac{9}{4}(x^{4}+1)(y^{4}+1)[/tex] --------------(*)

Dividing [tex]x^2y^2[/tex] on both sides of (*),  (*) is equivalent to

[tex]\displaystyle (xy+\frac{1}{xy}+1)^{2} \leq \frac{9}{4}(x^{2}+\frac{1}{x^2})(y^{2}+\frac{1}{y^2})[/tex] --------------(**)

Let [tex]u = \frac{1}{2}(x^{2}+\frac{1}{x^2})\ge1[/tex] and [tex]v= \frac{1}{2}}(y^{2}+\frac{1}{y^2})\ge1[/tex],

Then RHS of (**) is [tex]9uv[/tex], the LHS is

[tex]\displaystyle (xy+\frac{1}{xy}+1)^{2} = (xy+\frac{1}{xy})^2+2 (xy+\frac{1}{xy})+1 \leq 4uv + 4\sqrt{uv} + 1[/tex] by Cauchy-Schwarz.

So (**) follows from [tex]4uv + 4\sqrt{uv} + 1 \leq 9uv[/tex] which is equivalent to [tex](\sqrt{uv} -1) (5\sqrt{uv} + 1) \ge0[/tex]. The last inequality is true since both [tex]u,v[/tex] are greater than or equal to 1. This proves that (*) is true.

I have been working on this problem for a couple of hours. Not going anywhere. There are three cases:

1. all x, y, z are  :ge: 1

2. all x, y, z are  :le: 1

3. one of x, y, z is  :le: 1 and another is  :ge: 1

Case 3 can be proved. Case 2 is equivalent to case 1 by considering the reciprocals.

So case 1 or 2 is the only thing remaining to be proved.

[quote="fuzzylogic"][quote="blahblahblah"]
If you let xyz=1 and expand, ...[/quote]

Why can you assume xyz=1?

I haven't got the proof yet. I just made up some story for two variables :-)[/quote]

Whoops.  I got so carried away in seeing your explanation that I magically made the inequality homogeneous in my head.

Wow! I finally found a solution!  :lol:

We first prove that

$\frac{9}{8}(x^4+1)^3 \geq (x^6+x^3+1)^2$

In fact,

$9(x^4+1)^3-8(x^6+x^3+1)^2$

$=(x-1)^4(x^8+4x^7+10x^6+4x^5-2x^4+4x^3+10x^2+4x+1)$

which is nonnegative since $x^8+1 \geq 2x^4$ by AM-GM.

Using the result above, with Holder's inequality, we have

$\frac{9}{8}(x^4+1)(y^4+1)(z^4+1) \geq [(x^6+x^3+1)^{1/3}(y^6+y^3+1)^{1/3}(z^6+z^3+1)^{1/3}]^2$

$\geq (x^2y^2z^2+xyz+1)^2$,

as desired. $\blacksquare$

This definitely does not belong in Intermediate...  ;)

Uh, wow this was almost two years ago?

Basically you expect that inequality to hold and $x = 1$ gives equality so you long divide by $(x-1)^{2}$ until you can't anymore...

Well since Iwasn't actually the one who revived it I don't feel bad at all...

my question was, you didn't know for sure that this inequality was seperable by variable, right?
So did you just apply this method because you were stuck and it worked?

Or is there some key observation that I'm missing?


modular arithmetic
function
arithmetic sequence
number theory proposed
number theory

Given a sequence $(a_n)$ of real numbers such that the set $\{a_n\}$ is finite.
If for every $k>1$ subsequence $(a_{kn})$ is periodic, is it true that the sequence $(a_n)$ must be periodic?

I know the following problem.
Given a sequence $\{a_n\}$ of natural numbers s.t. $\{a_{kn}\}$ is periodic for all $k>1$. Is it true that  $\{a_n\}$ is periodic?

In both problems, you can take the sequence $a_p$ with prime $p$ however you want. In the first problem the set must have more than one element, of course.

Ops. In my problem the essential condition is "$a_{kn}$ is pure periodic".
Sorry, I forgot to write it.  :blush:

purely periodic (?): $a_{kn}=a_{k(n+T)}$ for all $n$.

But what's the difference between this and periodic? What does the "purely" stand for? :? Sorry, I've never heard about this.

[quote="Myth"]periodic sequnce: $a_{kn}=a_{k(n+T)}$ for all $n>n_0$. It is like periodic fractions.[/quote]

Really? I've never used that terminology. I always thought periodic meant what you call purely periodic.

Anyway, I rushed into it. Let me think about it some more :).

What do you mean? :(
And what have you posted in reply #3?

I just wanted to see whether everything I wanted to do worked, that's all. When I posted post number $3$ I didn't make sure that everything was sound, so I was wondering whether I had made some mistakes.

First of all, we can see that no matter how we color $a_p$ with prime $p$ (assume we have two colors, black and white), we will be able to complete the sequence s.t. it satisfies the conditions.

Now, for each $T\ge 2$, choose two primes $p_T,q_T$ larger than any of $p_i,q_i$ with $i\in\overline{2,T-1}$, and s.t. $p_T\equiv q_T\equiv 1\pmod T$ and color them differently. This will ensure the fact that the sequence is not periodic with period $T$ for any $T$. After that, we color the rest of the primes however we want.

Does it work? :?

[quote="grobber"]First of all, we can see that no matter how we color $a_p$ with prime $p$ (assume we have two colors, black and white), we will be able to complete the sequence s.t. it satisfies the conditions.[/quote] I am too sleepy to see it.

Good night :)

The condition reduces to the fact that for all primes $p$, the sequence $(a_{pn})_n$ is periodic. We can achieve this one prime at a time. Assuming we have this for all primes $p_1,\ldots,p_{n-1}$, when we go to $p_n$, the problem is equivalent to the following:

Given the sequence $a_n$ s.t. for all $p\in\{p_1,\ldots,p_{n-1}\}$ the sequence $a_{pn}$ is periodic, show that we can complete it to a periodic sequence (assume that initially we are only given $a_m$ for $m$ being a multiple of at least one of the $p_i,\ i\in\overline{1,n-1}$). If $T_i$ is the period of $a_{p_in}$, then we can make $a_n$ periodic with period $p_1\ldots p_{n-1}\cdot T_1\ldots T_{n-1}$, I think.

[quote="sbos"]If for every $k>1$ subsequence $(a_{kn})$ is periodic, is it true that the sequence $(a_n)$ must be periodic?[/quote]

Consider all the relations $a_i = a_j$ imposed by periodicities.  If the period of $a_{pn}$ is a rapidly increasing function of $p$ (and sufficiently divisible), then the number of independent relations will be too small for number of equivalence classes to be bounded.

Can you explain better your solution grobber, give explicitely {a_n} if you can.

By the way, I think that  $a_n$ = ($f(n)$ mod 4)  where $f(n)$ is the smallest prime divisor of $n$, should be an example with 3 colors.  Clearly $a_{kn}$ is periodic.  Let us see that $4T$ cannot be a period of the sequence, for any $T$.  Let $q = 4m-1$ be a prime number that does not divide $T$.  By Dirichlet's theorem on primes in arithmetic progression, we can find a prime number $p$ so that:
$p + 4T = 0$ mod $q$,
$p = 1$ mod $4$,  and
$p + 4T = 1$ modulo all primes $2 < q' < q$.
Then $a_p = $ (1 mod 4)  but $a_{p+4T} = a_q = $ (-1 mod 4).


If $(AB)^n=O_n$ then $(BA)^n=O_n$ because $P_{AB}=P_{BA}$ so $AB$ and $BA$ have the same eigenvalues?... :blush:

I assume that $A$ and $B$ are both $n\times n$ matrices, where that's the same $n$ in the power $(AB)^n.$

In that case, yes, your argument is sound. Both $AB$ and $BA$ have the same characteristic polynomial, and that polynomial is $x^n.$

The series continue : suppose $A\in\mathcal l M_n(C)$ has the eigenvalues $\alpha_i,i=\overline{1,n}$ then the eigenvalues of $A^*$(the adjoint) are $\frac{\alpha_1\alpha_2...\alpha_n}{\alpha_i},i=\overline{1,n}$?

[quote="spix"]The series continue : suppose $A\in\mathcal M_n(C)$ has the eigenvalues $\alpha_i,i=\overline{1,n}$ then the eigenvalues of $A^*$(the adjoint) are $\frac{\alpha_1\alpha_2...\alpha_n}{\alpha_i},i=\overline{1,n}$?[/quote]

I don't know to prove it but  I think is true (it worked for some problems :D). Could somebody tell us why? Thanks!

[quote="spix"]The series continue : suppose $A\in\mathcal l M_{n}(C)$ has the eigenvalues $\alpha_{i},i=\overline{1,n}$ then the eigenvalues of $A^{*}$(the adjoint) are $\frac{\alpha_{1}\alpha_{2}...\alpha_{n}}{\alpha_{i}},i=\overline{1,n}$?[/quote]

Hello? Anyone?  :ninja: [b][color=red]BUZZ!!![/color][/b]

Let's suppose that the $\alpha_{i}$s are $\neq 0$. Then, $A$ is invertible. Since $A^\ast = \left( \det A \right) \cdot A^{-1}$ and $\det A = \prod_{k=1}^{n}\alpha_{k}$, it suffices to prove that the eigenvalues of $A^{-1}$ are $\frac1{\alpha_{i}}$, $i = \overline{1,n}$.

We have that
\begin{eqnarray*}\det \left( A-x I_{n}\right) &=& \det \, A \cdot \det \left( A^{-1}\right) \cdot \det \left( A-x I_{n}\right) \\ \ &=& \det \, A \cdot \det \left( I_{n}-x A^{-1}\right) \\ \ &=& \det \, A \cdot x^{n}\cdot \det \left( \frac1{x}I_{n}-A^{-1}\right),\end{eqnarray*}
which proves the desired claim.

The case when one of the $\alpha_{i}$s is $0$ is still under investigation...


Prove that $\forall a,b,c>0$ that satisfy $1+a>b$, $1+b>c$, $1+c>a$:

\[ \sqrt[3]{3}\leq \sqrt[3]{1+a-b}+\sqrt[3]{1+b-c}+\sqrt[3]{1+c-a}\leq 3  \]

[i]Could anyone improve the inequality?[/i]

[b]EDIT:[/b] Added the missing condition

LHS:
$\sqrt[3]{3} \leq x+y+z$ (where x,y,z are the square roots) is equivalent to $3 \leq (x+y+z)^{3} = (x+y+z)(x^{2}+y^{2}+z^{2}+2xy+2xz+2yz) = x^{3}+y^{3}+z^{3} + R$, where R is nonnegative. But $x^{3}+y^{3}+z^{3}=3$, QED.
RHS can be smashed using the Power Mean inequality:
$\frac{x+y+z}{3} \leq \sqrt[3]{\frac{x^{3}+y^{3}+z^{3}}{3}}$, x,y,z as above.

[quote="Ezbakhe Yassin"]Prove that $\forall a,b,c>0$:
\[ \sqrt[3]{3}\leq \sqrt[3]{1+a-b}+\sqrt[3]{1+b-c}+\sqrt[3]{1+c-a}\leq 3  \]
[/quote]
Try $a=1,b=c=8.$
$\sqrt[3]3\leq\sqrt[3]{1+1-8}+\sqrt[3]{1+8-8}+\sqrt[3]{1+8-1}\Leftrightarrow$
$\Leftrightarrow\sqrt[3]3\leq-\sqrt[3]6+3.$ This is not true. ;)

[quote="neworder"]LHS inequality is true only if 1+a>b, 1+b>c, 1+c>a.[/quote]

Yeah, I forgot that. However, it is evident since we need positive real numbers  ;)

[quote="neworder"]LHS inequality is true only if 1+a>b, 1+b>c, 1+c>a.[/quote]
Yes, you are right. But this is not given.
Moreover, $\forall k<0$ possibly to find $a>0,b>0,c>0,$ for them $\sum\sqrt[3]{1+a-b}<k.$  :)


Euler
geometry
parallelogram
search
geometry theorems

I got a PM request to prove Gossard theorem. (Never heard of it, but it can be googled.)

[color=red]Let $ \color{red}k$ be Euler line of a $ \color{red}\triangle ABC$ with sidelines $ \color{red}a \equiv BC, b \equiv CA, c \equiv AB.$ Euler lines $ \color{red}k_a, k_b, k_c$ of triangles $ \color{red}\triangle (b c k), \triangle (c a k), \triangle (a b k)$ form a triangle $ \color{red}\triangle (k_a k_b k_c)$ centrally congruent to the triangle $ \color{red}\triangle (a b c) \equiv \triangle ABC,$ with common Euler line $ \color{red}k,$ similarity center $ \color{red}S \in k$ and similarity coefficient $ \color{red} \minus{} 1.$[/color]

[b][color=blue]Lemma 1:[/color][/b] Let arbitrary line $ k$ cut $ CA, AB$ at $ E, F,$ let parallels to $ EF$ through $ B, C$ cut $ CA, AB$ at $ E', F',$ respectively, and let inversions with centers $ C, B$ and powers $ CE^2, BF^2$ take $ E', F'$ to $ E_a, F_a.$ Midpoint $ M_a$ of $ E_aF_a$ is then on $ EF.$

[i][color=blue]Proof:[/color][/i] $ \overline{FF_a} \equal{} \overline{FB} \plus{} \frac {FB^2}{\overline{BF'}} \equal{} \frac {\overline{FB} \cdot (\overline{FB} \plus{} \overline{BF'})} {\overline{BF'}} \equal{} \frac {\overline{FB} \cdot \overline{FF'}} {\overline{BF'}}$ and similarly, $ \overline{EE_a} \equal{} \frac {\overline{EC} \cdot \overline{EE'}} {\overline{CE'}}\ \Longrightarrow$

$ \frac {\overline{FF_a}}{\overline{FA}} \equal{} \frac {\overline{FB} \cdot \overline{FF'}} {\overline{FA} \cdot \overline{BF'}} \equal{} \frac {\overline{EC} \cdot \overline{EE'}} {\overline{EA} \cdot \overline{CE'}} \equal{} \frac {\overline{EE_a}}{\overline{EA}}.$

By Menelaus theorem for triangle $ \triangle AE_aF_a$ cut by the line $ EF,$ this line cuts its remaining side $ E_aF_a$ at its midpoint $ M_a.$ $ \square$

[b][color=blue]Lemma 2:[/color][/b] Let $ G$ be centroid and $ H$ orthocenter of the $ \triangle ABC$ and $ k \equiv GH$ its Euler line. Let $ k$ cut $ AB$ at $ F$ and let parallel $ b'$ to $ k$ through $ C$ cut $ AB$ at $ A'.$ Parallel $ k'$ to $ CA$ through $ F$ is then Euler line of $ \triangle A'BC.$

[i][color=blue]Proof:[/color][/i] Let $ k' \parallel CA$ cut $ AH$ at $ H'.$ Since $ H'H \perp FB, BH \perp H'F$ $ \Longrightarrow$ $ H$ is also orthocenter of the triangle $ \triangle FBH'$ and conversely, $ H'$ is orthocenter of the triangle $ \triangle FBH$ $ \Longrightarrow$ $ BH' \perp CA'$ and $ H'$ is also orthocenter of the triangle $ \triangle A'BC.$ Let parallel $ c'$ to $ c \equiv AB$ through $ G$ cut $ k' \equiv FH'$ at $ G'.$ Since triangles $ \triangle CA'A \sim \triangle FGG'$ with parallel sides are centrally similar with similarity coefficient $ \frac {\overline {FG}}{\overline{AA'}} \equal{} \minus{} 3,$ it follows that $ \overline{GG'} \equal{} \minus{} \overline{G'G} \equal{} \frac {_1}{^3} \overline {AA'}$ $ \Longrightarrow$ $ G'$ is centroid of $ \triangle A'BC$ and $ k' \equiv G'H'$ its Euler line. $ \square$

[i][color=red]Proof of theorem:[/color][/i] Let Euler line $ k$ of $ \triangle ABC$ cut $ BC, CA, AB$ at $ D, E, F.$ Since $ \triangle FBD \sim \triangle A'BC$ are centrally similar with center $ B$ $ \Longrightarrow$ Euler line $ k_b \parallel k'$ of $ \triangle FBD$ is also parallel to $ CA.$ Similarly, Euler lines $ k_c, k_a$ of $ \triangle DCE, \triangle EAF$ are parallel to $ AB, BC$ $ \Longrightarrow$ triangles $ \triangle(k_ak_bk_c) \equiv \triangle XYZ \sim \triangle ABC$ (where $ X \equiv K_b \cap K_c, y \equiv k_c \cap k_a, Z \equiv k_a \cap k_b$) are centrally similar and their Euler lines parallel.
If $ k_b$ cuts $ FB$ at $ F_a,$ then from central similarity of triangles $ \triangle FBD \sim \triangle A'BC$ $ \Longrightarrow$  $ \overline {BF_a} \equal{} \frac {BF^2}{\overline{BA'}}.$ In exactly the same way, if parallel $ a'$ to $ k$ through $ B$ cuts $ CA$ at $ B'$ and if Euler line $ k_a \parallel BC$ of $ \triangle FAD$ cuts $ CA$ at $ E_a,$ then $ \overline {CE_a} \equal{} \frac {CE^2}{\overline{CB'}}.$ Let $ k$ cut $ YZ, ZX, XY$ at $ U, V, W.$ By lemma 1, $ k$ goes through the diagonal intersection $ M_a$ of parallelogram $ AF_aXE_a$ $ \Longrightarrow$ $ \overline{XW} \equal{} \minus{} \overline {AF}.$ Similarly, $ \overline{YV} \equal{} \minus{} \overline {BE}, \overline{ZU} \equal{} \minus{} \overline {CD}$ $ \Longrightarrow$ triangles $ \triangle XYZ \cong \triangle ABC$ are centrally congruent with similarity coefficient $ \minus{} 1$ and with common Euler line $ k$ $ \Longrightarrow$ their similarity center $ S$ is on $ k.$ $ \square$

References:
[1][url]http://www.mathlinks.ro/viewtopic.php?t=284866[/url]
[2][url]http://www.mathlinks.ro/viewtopic.php?t=191643[/url]

Dear Vladimir,
very nice proof.
I want just to make a link
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=298479080&t=201805
Sincerely
Jean-Louis

I've got a simple good geometric way to prove [b]Gossard's theorem[/b]，and [b]The Gossard Perspector[/b] is on [b]Newton line [/b].
When $ k'//k$, wo can also get △$ a'b'c' \equal{}$△$ ABC$, the similarity center $ S'$ is on [b]Newton line[/b] which is formed by △$ ABC$ and $ k$ .


i downloaded latex, the file name was Miktex though, is it the same thing?
anyway, when i tried to install it, it asked for a Path to local package repository.  What do i put in there?  I'm allowed to browse but i dont know where to go.  Does anyone know?

[quote="Valentin Vornicu"]MikTex is a good instalation package. It is what you need. You downloaded a local package repository, so when you are installing it, you must point him back to itself :)[/quote]
It wont let me point it back at itself, it will only let me select desktop (but then it says error, doesnt represent a valiad package directory), and folders on the desktop, my computer, my documents, and my network places.  I tried creating a new folder in my computer->C drive-> program files, but it said it wasnt a valiad package directory


linear algebra
matrix
induction
linear algebra unsolved

Assume matrix $ M$ is unitary, upper triangular wih positive entries on the main diogonal. show that $ M$ is the identity matrix

Thanks

This is true for $ n\times n$ matrices for all $ n\in \mathbb{N},$ as we will prove by induction on $ n.$

Since the only $ 1\times 1$ matrix that is unitary and has a positive entry is the matrix with entry $ 1,$ which is the $ 1\times1$ identity, the theorem is true for

Suppose the theorem is true for $ n\times n$ matrices. Suppose $ M$ is an $ (n \plus{} 1)\times (n \plus{} 1)$ matrix that satisfies the hypotheses: unitary, upper triangular, positive diagonal entries. We can block structure it as
\[ M \equal{} \begin{pmatrix}a & u^* \\
0 & N\end{pmatrix}\]
where $ N$ is upper triangular and has positive diagonal entries. (We don't yet know that $ N$ is unitary.) We do know that $ M$ is unitary: that is, $ M^*M \equal{} I.$ Writing this out in block-structured form:
\[ M^*M \equal{} \begin{pmatrix}a & 0 \\
u & N^*\end{pmatrix}\begin{pmatrix}a & u^* \\
0 & N\end{pmatrix} \equal{} \begin{pmatrix}a^2 & au^* \\
au & uu^* \plus{} N^*N\end{pmatrix} \equal{} \begin{pmatrix}1 & 0 \\
0 & I\end{pmatrix}\]
Hence $ a^2 \equal{} 1,$ and since $ a > 0,$ then $ a \equal{} 1.$

Then $ au \equal{} u \equal{} 0.$ That means that $ uu^* \plus{} N^*N \equal{} N^*N \equal{} I,$ so $ N$ is unitary. The inductive hypotheses now apply to $ N$ and the theorem yields that $ N \equal{} I.$ That means we have found
\[ M \equal{} \begin{pmatrix}1 & 0 \\
0 & I\end{pmatrix}\]
which is the $ (n \plus{} 1)\times(n \plus{} 1)$ identity matrix and we are done.


this is my solution for number 4...

anyway... look at the end, where I have the two functions written down.

i have f(x) and f(y) both written as two completely different functions (you can see what i mean)

the answer is correct, but I should have written something like f(x) and g(y)....

How many points do you think they would take off?

Could you tell me what you think of my solution too?

Thanks! :D [/list]

I think you'll get full credit.  Assuming the logic leading up to your answer is correct (I didn't look at this...I only looked at the last part of your solution) I think they'll understand what you meant.  You have the right answer as well.  They usually only take off for a lack of justification, minor flaws that lead to a wrong answer, glaring flaws, or something along those lines.

What about for me?
I did make an error, but a minor one...so would they still give me a 3, or a 4?


For a natural number $n$, let $P(x)$ be the polynomial of $2n$−th degree such that:
$P(0) = 1$ and $P(k) = 2^{k-1}$ for $k = 1, 2, . . . , 2n$.
Prove that $2P(2n + 1) - P(2n + 2) = 1$.

P.S. I tried to prove it by firstly expressing this polynomial using Lagrange interpolation but get bored of computations - it seems like it can be done this way, but I'd like to see more 'clever' solution.  :)

We see that $Q(x)=P(x)-2x+1$ has the roots $1,2,\ldots ,2n$, therefore, since $degQ=2n$ we have
\[
Q(x)=a(x-1)(x-2)\ldots (x-2n).
\]
The condition $P(0)=1$ gives $a=\frac {2}{(2n)!}$, thus
\[P(x)=\frac {2}{(2n)!} (x-1)\ldots (x-2n)+2x-1.
\]
We find then $P(2n+1)=4n+3$ and $P(2n+2)=8n+5$. It is now easy to check the requested equality.

I'm so sorry [b] enescu [/b] :blush:  - I haven't checked after posting exactly what come out and here are the effects - you solved almost completely different problem - the most amusing of it all is that the hypothesis remained true even after I made such a terrible mistake  :rotfl: :rotfl:  :rotfl:  . I hope you're not very angry of writing your solution in vain - sorry again  :blush:  .

Still waiting for the solution

OK, but now you do the maths taking $Q(x)=P(x)-f(x-1)$, where
\[f(x)=1+x+\frac {x(x-1)}{2}+\frac{x(x-1)(x-2)}{6}+\ldots +\frac{x(x-1)\ldots (x-2n+1)}{(2n)!}.
\]


floor function
function
quadratics
algebra

1) Prove that $ \forall n \in \mathbb{N}^*$ , the number $ \lfloor (2\plus{}\sqrt{3})^n \rfloor$ is odd , where $ \lfloor \rfloor$ is the floor function .
2.) Find the greatest $ k \in \mathbb{N}$ such that $ \lfloor (1\plus{}\sqrt{3})^n  \rfloor$ is divisible by $ 2^k$ .

[hide]I don't really know how to type floor function so I'll replace it by | | :
1)denote that $ x_1 \equal{} 2 \plus{} \sqrt {3}$ and $ x_2 \equal{} 2 \minus{} \sqrt {3}$ so obviously $ x_1$ and $ x_2$ are two roots of the quadratic :
$ x^2 \minus{} 4x \plus{} 1 \equal{} 0$
note that if we let $ S_n \equal{} x_1^n \plus{} x_2^n$ then we have $ S_{n \plus{} 1} \equal{} (x_1 \plus{} x_2).S_n \minus{} x_1.x_2.S_{n \minus{} 1}$ or $ S_{n \plus{} 1} \equal{} 4.S_n \minus{} S_{n \minus{} 1}$
secondly, $ S_0 \equal{} 2 ; S_1 \equal{} 4$ since $ S_2 \equal{} 4S_1 \minus{} S_0$ clearly $ 2|S_2$ and since $ S_3 \equal{} 4S_2 \minus{} S_1$ again clearly $ 2|S_3$ and so on we have $ 2|S_k$ for every natural k .
consider $ S_n \equal{} x_1^n \plus{} x_2^n \equal{} (2 \plus{} \sqrt {3})^n \plus{} (2 \minus{} \sqrt {3})^n$ , note that $ 0 < (2 \minus{} \sqrt {3})^n < 1$ therefore $ |(2 \minus{} \sqrt {3})^n| \equal{} S_n \minus{} 1$ since $ S_n$ is even , we have |$ (2 \minus{} \sqrt {3})^n|$ is odd . QED

2) problem 2 we can use the same method ,I think .[/hide]

[hide="proof 1)."]With the Newton binom we have that $ (2 \plus{} \sqrt {3})^n \equal{} x \plus{} y\sqrt {3}$ where $ m$ is integer as well as $ n$.
$ (2 \minus{} \sqrt {3})^n \equal{} m \minus{} n\sqrt {3}$
\[ (2 \plus{} \sqrt {3})^n \plus{} (2 \minus{} \sqrt {3})^n \equal{} 2m
\]
$ 0 < 2 \minus{} \sqrt {3} < 1$ $ \implies$ $ 0 < (2 \minus{} \sqrt {3})^n) < 1$ $ \implies$ $ [2m \minus{} (2 \minus{} \sqrt {3})^n] \equal{} [(2 \plus{} \sqrt {3})^n]$ that obviously is an odd numeber[/hide]


geometry
area of a triangle
Heron\u0027s formula
geometry proposed

Let  the  length of ABC triangle's sides  be  prime  numbers.Show  that the area  of ABC  triangle  can not integer.

I think this problem belongs on a less advanced forum.

[hide]If the three sides are all odd primes, then the s (semiperimeter) in Heron's formula is of the form (odd integer)/2, and the same can be said of s-a, s-b, and s-c. Hence the area is the sqaure root of (odd integer)/16, and that can't be an integer.

The same objection applies to using 2, 2, p (p an odd prime).

If one of the sides is 2 and the other two sides are two different odd primes, then that's not a triangle.

If the sides are 2, p, and p, then the area is sqrt(p^2-1). But (for p > 1), 1 less than a perfect square can't be a perfect square. This includes the equilateral trinangle 2, 2, 2.[/hide]


The terms a_n of a sequence of positive integers satisfy a_(n+3) = a_(n+2)(a_(n+1) + a_n) for n = 1,2,3...
If a_6 = 144, what is a_7 ?

:D  I think that Moubi will provide an answer to that one, since it already gave one in Crux...

Pierre.

I hope Moubi (who is usually known to be a human being by the way  :)) won't mind if I answer  :D

a6 = a3*(a2+a1)*(a3+a2)*(a3*(a2+a1)+a3)

Which solves easily (by tries) to a1=2, a2=1, a3=2.

a(7) = 3456

Good.
---------------------------------------------
Je vois que l'on se fait des reponses entre frogs
salut les amis passiones olympiad math style:)


do you have to solve all five problems when handing it in?  Can you hand in one, and then the other, and so on like that?  WHat if you cannot answer one question?

umm when you submit answers, you turn in all the solutions/work you have all at one time. It's ok if you can't solve all the problems, you may still get partial credit if you have a good start I think

You don't HAVE to solve all five. Solve as many as you can. However, you must send your solutions to the problems you solved at ONCE in ONE file. Also, you must put ID #, your name, Round #, Problem # on every page of your solution.

Hope I'm accurate on this.

That's correct - you don't have to do all 5 problems, but you do have to turn in all your work at once.  If you send in your work in pieces, only the first thing we receive will be graded (except in cases of fax machine failure, etc).


Let A be a 3 × 3 matrix in which each element is either 0 or 1 but is otherwise arbitrary.
(a) Prove that det(A) cannot be 3 or −3.
(b) Find all possible values of det(A) and prove your result.

For some reason I fear people will complain about this solution, though the approach is perfectly valid:

The determinant can only be one of: $ -2,-1,0,1,2$.

The following (unoptimized) Matlab code will generate all the possible determinants:

[code]
x(1,1:3)=[0 0 0];
x(2,1:3)=[0 0 1];
x(3,1:3)=[0 1 0];
x(4,1:3)=[0 1 1];
x(5,1:3)=[1 0 0];
x(6,1:3)=[1 0 1];
x(7,1:3)=[1 1 0];
x(8,1:3)=[1 1 1];
for i=1:8
    for j=1:8
        for k=1:8
           A=[x(i,:); x(j,:); x(k,:)];
           sol(i,j,k)=det(A);
        end
    end
end
sol;
[/code]

It's actually very simple. If $ A$ has only $ 1$'s, then $ \det A = 0$. Otherwise, there a $ 0$ on the $ j$-th row. Expand the determinant using that row. You'll get $ \pm \begin{vmatrix}? & ? \\ ? & ? \end{vmatrix}\pm \begin{vmatrix}? & ? \\ ? & ? \end{vmatrix}$, where each $ ?$ is either $ 0$ or $ 1$. Since every such determinant is $ -1$, $ 0$ or $ 1$, we get (a). For (b) you just have to give examples.

I guess I was right about the perceived response.  Within minutes my rating dropped by 0.2 and my response was equated with SPAM.  Gee, thanks.

Should I go into a spiel about the acceptance of the "proof-by-exhaustion" method, the success in proving the 4-color theorem via computer assistance, and the general accepted need in the "real" world to use computers to solve physical problems (e.g., through finite elements and CFD)?  No, I won't because [i]that[/i] would be SPAM...  :roll:

The following matrices are examples of determinants of $ -2,-1, 0, 1, 2$ :

$ \left[ \begin{array}{ccc}1&0&1\\\noalign{\medskip}0&1&1\\\noalign{\medskip}1&1&0\end{array}\right]$

$ \left[ \begin{array}{ccc}1&0&1\\\noalign{\medskip}0&0&1\\\noalign{\medskip}1&1&1\end{array}\right]$

$ \left[ \begin{array}{ccc}1&1&1\\\noalign{\medskip}1&1&1\\\noalign{\medskip}1&1&1\end{array}\right]$

$ \left[ \begin{array}{ccc}1&1&0\\\noalign{\medskip}0&1&0\\\noalign{\medskip}1&1&1\end{array}\right]$

$ \left[ \begin{array}{ccc}0&1&1\\\noalign{\medskip}1&0&1\\\noalign{\medskip}1&1&1\end{array}\right]$

[quote="edoo"]I guess I was right about the perceived response.  Within minutes my rating dropped by 0.2 and my response was equated with SPAM.  Gee, thanks.[/quote]
Actually, I gave it three hexagons. Although trivially correct, I considered your remark not to be in the spirit of the problem.

[quote="edoo"]Should I go into a spiel about the acceptance of the "proof-by-exhaustion" method, the success in proving the 4-color theorem via computer assistance, and the general accepted need in the "real" world to use computers to solve physical problems (e.g., through finite elements and CFD)?  No, I won't because [i]that[/i] would be SPAM...  :roll:[/quote]
Computer-aided proofs give one the certainty that a result is true, but in the same time may lead to discovering patterns (or whatever) that can help one find a proof without enormous case bashing. Most people prefer the latter, since it gives some insight into the problem.

PS: You worry too much about that bloody rating.

[quote="perfect_radio"]Computer-aided proofs give one the certainty that a result is true...[/quote]

Only if you can prove that your computer correctly computed what you told it to compute; and that is a problem of physics, not of mathematics.  (I actually have no problem with computer-aided proofs.)

Since you already have the program: what is the distribution of determinants of such matrices?  (How many 0s, 1s, etc.?)

[quote]Since you already have the program: what is the distribution of determinants of such matrices? (How many 0s, 1s, etc.?)[/quote]

The following list gives the number of matrices with each determinant value:

[u]Determinant[/u]                  [u]Number of Matrices[/u]
           $ -2 \hspace{20 mm}3$
           $ -1 \hspace{20 mm}84$
           $ 0 \hspace{22 mm}338$
           $ 1 \hspace{22 mm}84$
           $ 2 \hspace{22 mm}3$

The mean is $ 0$, the variance is $ 0.3757$, and the standard deviation is $ 0.6130$ (both to 4 decimal places).  So the values are symmetric and normally distributed.


Did anyone notice 2007 State Sprint #26 is the same as 1999-2000 Handbook WU 18 #7?

26.  The area of the semicircle in Figure A is half the area of the circle in Figure B.  The area of a square
inscribed in the semicircle, as shown, is what fraction of the area of a square inscribed in the circle?
Express your answer as a common fraction.

7. A square is inscribed in a semicircle, and a second square is inscribed in the whole circle, as shown below. What is the ratio of the area of the smaller square to the area of the larger square?  Express your answer as a common fraction.

[quote=math154]Not too surprising. When there are thousands of problems limited to a certain difficulty, repeats, nearly identical problems, or both are bound to occur.[/quote]

*cough cough*mathcounts state 2015 sprint #30*cough cough*

a problem in Mathcounts State CDR was a problem in the handbook for 2012-2013.


Bad title :P . Write $120=2^{3}\cdot 3\cdot 5$. Now, you need only prove that $2^{3},3$ and $5|A$. But this is easy! :)


define
K' = {n | Phi(n,n) doesnt halt.} - K' is the set of all program numbers
 (coded as godel numbers) that doesnt halt when they recieve thier own number as input
FIN = {n | Phi(x,n) halts for a finite number of x's} - FIN is the set of all
 program numbers that halt only for a finite number of inputs.

A<=mB iff there is a computable function f such that  A = {x | f(x) is in B}

i've proved that K' <=m FIN.

now they ask me whether FIN <=m K'.

I don't even have the intuition whether i should prove or show it is wrong. - any help, hint or a solution will be gladly accepted

thanks!

What is Phi? I know a little bit about computability so I might be able to make some naieve suggestions, but I don't know enough to understand your notation  :maybe: .

in my notation Phi is the universal function, Phy(x,y) runs the program whose number is y on the input x.

It seems hard to believe that FIN could mapping reduce to K'.  This is just intuition though.  Seems you would need a work-around method to get past the fact that FIN only halts for a finite number of inputs, which means it can only accept for a finite number of inputs.  K' can halt for an infinite number of inputs.  With this said, let me think about it for a little longer.


[quote="ashwath.rabindranath"]This was my solution during the test:
 Let $ p$ be a prime dividing $ \frac {x^7 \minus{} 1}{x \minus{} 1}$. SO, $ p$ must divide $ x^7 \minus{} 1$ which implies that $ x^7 \equiv 1 \bmod p$. Now let $ h$ be the order of $ x$ modulo 7. By common sense $ h | 7$. The only element with order $ 1$ is $ 1$ itself, so if $ x$ is not $ 1 \bmod 7$, $ h \equal{} 7$. As $ h | p \minus{} 1$, we have $ p \equiv 1 \bmod 7$. If it is 1 mod 7, p must be 7 itself (see why?)
Case 1: $ x \equiv 1 \bmod 7$ implies that $ \frac {x^7 \minus{} 1 } {x \minus{} 1} \equiv 0 \bmod 7$ (figure out why yourself :)) which implies that $ y^5 \equiv 1 \bmod 7$. which further implies that $ y \equiv 1 \bmod 7$. But this implies that $ y^4 \plus{} y^3 \plus{} y^2 \plus{} y \plus{} 1 \equiv 5 \bmod 7$ which means there must be a prime factor of $ \frac {x^7 \minus{} 1 } {x \minus{} 1}$ that is not $ 0$ or $ 1$ modulo 7, a contradiction to our earlier statement. Considering the other case, you get $ y^4 \plus{} y^3 \plus{} y^2 \plus{} y \plus{} 1 \equiv 3 \bmod 7$, which too is a contradiction. So, the equation has no solutions.[/quote]
thanks :D   :| By common sense $ h | 7$common sense is not so common and hence i dont get it

@Gaurav: That's because of the definition of the order of $ a$ modulo $ p$. It is is the least positive integer $ h$ such that $ a^h \equiv 1 \bmod p$. Now, use the division algorithm to prove that if $ a^n \equiv 1 \bmod p$,  $ h | n$. It's pretty straightforward and very useful. By the way, not everyone has seen the problem before, so please post a slightly more intelligible problem statement.


function
linear algebra
matrix
inequalities
geometry
linear algebra unsolved

Let $ \text{vol}_n (S)$ be a function that, given a contiguous subset $ S$ in $ \mathbb{R} ^n$, gives the n-dimensional "volume" of $ S$.

Let $ A$ be an n by n matrix with $ |A| \neq 0$, and let $ a_{i,j} > 1$ be the entry in the [i]i[/i]th row and [i]j[/i]th column.

[u]Part One:[/u] Show that:

$ \text{vol}_n \left( \left \{\vec{x} \in \mathbb{R}^n \; : \; \sum_{i \equal{} 1}^n \left| \sum_{j \equal{} 1}^{n} a_{i,j} x_j \right| \leq 1 \right \} \right) \equal{} \text{vol}_n \left( \left \{\vec{x} \in \mathbb{R}^n \; : \; \sum_{j \equal{} 1}^n \left| \sum_{i \equal{} 1}^{n} a_{i,j} x_i \right| \leq 1 \right \} \right)$

[u]Part Two:[/u] Ascertain the value of the above, in terms of $ A$

Any ideas?

Here's an example of the formula at work:

Suppose we have the matrix $ \left( \begin{array}{cc} 1 & \minus{} 1 \\
3 & 2 \end{array} \right)$.
-Taking the rows, we have the inequality $ |x \minus{} y| \plus{} |3x \plus{} 2y| \leq 1$
-Taking the columns, we have the inequality $ |x \plus{} 3y| \plus{} | \minus{} x \plus{} 2y| \leq 1$
The formula tells that the areas of $ \mathbb{R}^2$ contained by these inequalities are equal.


topology
advanced fields
advanced fields theorems

Let $(M,d)$ be a metric space with metric $d$

consider the topology induced by the metric : prove that this topological space is normal

a topological space is normal if for every two disjoint closed sets A and B, open disjoints U and V can be found such that A is in U and B is in V

now my professor claims that by realising every regular space with a countable basis is normal, you should get there, but does a metric space have a countable basis??

All comments are welcome

Take your favorite countable dense subset of $(0,\infty)$ - the rationals will do.

Consider balls with radii in that set. How far does that get you?

Well that is exactly what my professor said, but still the centers of those balls are uncountable in number?  I am working in a very general metric space, not just subsets or anything of $\mathbb{R}^n$ or is this irrelevant?

Of course not every metric space has a countable basis. Consider, for instance, the space of bounded sequences, with the supremum norm. The elements having having some entries equal to $1$ and some equal to $0$ have mutual distance $1$, but there are uncountably many of them.

In order to see that a metric space is normal, let $A,B$ be two disjoint losed subsets. Cover every $x\in A$ with a ball $\mathcal A_x$ centered at $x$ of radius less than $\frac 12$ times the distance between $x$ and $B$ (which is positive, since $x$ is a point not belonging to the closed set $B$), and do the same for the points of $B$.

nice
but could you just explain how that uncountable set of balls in your counterexample gives us that no countable basis exists?

I thought basis meant : every open set of the topology (not necessarily countable) union of the sets in the basis?

Around each of those points, take a ball of radius $\frac 12$. You have an uncountable set of mutually disjoint open balls. Obviously, a space with a countable basis cannot have an uncountable collection of disjoint open sets (each open set in the collection must have a basis element inside, and you get uncountably many basis elements).

thank you that actually was not difficult :blush:

Thanks for all the answers! :)

a friend of mine actually explained me another, perhaps less exciting, but easier example of a metric space with no countable basis : consider the discrete metric on the real numbers (thus one if the points are different, zero when equal)

every basis must at least contain all singletons

Or is this no good example?


Let $ a,b,c$ be the sides of a triangle. Prove that,
\[ 3\left(\dfrac{1}{a^2}\plus{}\dfrac{1}{b^2}\plus{}\dfrac{1}{c^2}\right)>\left(\dfrac{1}{a\plus{}b}\plus{}\dfrac{1}{a\plus{}c}\plus{}\dfrac{1}{b\plus{}c}\right)^2\]

By Triangle Inequality, $ a \plus{} b > c$, so $ \frac {1}{a \plus{} b} < \frac {1}{c}$. Hence,

$ (\sum \frac {1}{a})^2 > (\sum \frac {1}{a \plus{} b})^2$

So it suffices to prove that

$ 3 \sum \frac {1}{a^2} > (\sum \frac {1}{a})^2$ .

which is true by Cauchy-Shwartz.

QED

Yes, that works (in fact that was also my solution)

Edit: As far as I know yes, but it's probably been made before.


inequalities
trigonometry
geometry
inradius
inequalities proposed

[quote="vo thanh van"]9.3 (R.Pirkuliev) Prove the inequality
\[ \frac1{\sqrt {2\sin A}} \plus{} \frac1{\sqrt {2\sin B}} \plus{} \frac1{\sqrt {2\sin C}}\leq\sqrt {\frac {p}{r}},
\]
where $ p$ and $ r$ are the semiperimeter and the inradius of
triangle $ ABC$.[/quote]
we have $ \frac{p}{r}\equal{}\frac{(a\plus{}b\plus{}c)^2R}{abc}$,$ a\equal{}2sinA R,b\equal{}2sinB R,c\equal{}2sinC R$(The Sin theorem)
THe ineq becomes $ \sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca}\le a\plus{}b\plus{}c$ and it is true


Source: 360 PMC
How many 7-digit numbers that do not start nor end with 1 are there?

My reasoning may be flawed... [hide]

The first number cannot be $1$ or $0$.
The last number cannot be $1$.

Thus,
There are $8$ possible digits for the first number;
There are $10$ possible digits for the second through sixth numbers;
There are $9$ possible digits for the last number.

$8\times10\times10\times10\times10\times10\times9$
Or, $72\times10^5$[/hide]


calculus
integration
induction
trigonometry
calculus computations

Using mathematical induction prove that
\[ \underbrace{\int\sin^n x\;dx}_{\mathcal{I}_n} \equal{} \minus{}\frac{\sin^{n\minus{}1} x\cos x}{n} \plus{} \frac{n\minus{}1}{n}\int\sin^{n\minus{}2} x\;dx\]
It's kind of trivial and I can prove it but unfortunately not using induction :(
I tried integrating $ \mathcal{I}_{n\plus{}1}$ by parts - [i]u=sin^n x, dv=sin x dx[/i] but I have no idea where to use the fact that $ \mathcal{I}_{n} \equal{} \text{something}$   :(
Any ideas, please?

This seems a little ill-stated. The formula is not itself something to be proved by induction. Instead, it is an inductive or recursive step. The true induction here is the representation of $ I_n$ in terms of either $ I_0$ or $ I_1$ (depending on whether $ n$ is even or odd). And yes, integration by parts (plus a Pythagorean trig identity) is the way to prove this.

This is a classic text book example of integration by parts. Doing the same, we get:

$ \int \sin^n x \,dx \equal{} \minus{} \cos x \sin^{n\minus{}1}x \plus{} (n\minus{}1) \int \sin^{n\minus{}2} \cos^2 x\,dx$

Now, remember the identity:

$ \sin ^2 \plus{} \cos ^2 \equal{} 1$

and use that on the RHS. You should get the same thing you have.

[quote="valeriummaximum"]what  mathematical induction????can u elucidate please????...arew u talking abt the reduction formula...[/quote]

http://en.wikipedia.org/wiki/Mathematical_induction.

[quote="@petko"]That's right, but... I said that I can compute this that way, but I have to use math induction - that is my problem :([/quote]
Well, if somebody insists on solving something it this particular way, you can always prove any statement $ P(n)$ by induction like this:

Check the base $ P(1)$
Say the maic words "assume $ P(n\minus{}1)$ holds". Prove $ P(n)$ [i]without[/i] using $ P(n\minus{}1)$. Say that since $ P(n)$ is true, the implication $ P(n\minus{}1)\implies P(n)$ is also true.
Use the principle of induction to conclude that $ P(n)$ is true for all $ n$.

It all sounds idiotic of course, but I wish to see how somebody will argue that it is not a proof by induction.


$ \boxed {\ \left(\forall\right)\ \alpha >0\ \left(\exists\right)\ n_1\in\mathbb N^*\ \mathrm{so\ that}\ \left(\forall\right)\ n > n_1\ ,\ \left(\sum_{k \equal{} 1}^n\frac 1k\right)^{\alpha}\ < \ n\ }$ .

I think it is true even when p is a positive real number

You are right ! I modified the enunciation how you say.

It is even true that for any $ \alpha > 0$, we have $ \lim_{n\to\infty} \frac{1}{n} \left( \sum_{k\equal{}1}^{n} \frac{1}{k} \right)^{\alpha} \equal{} 0$, since the partial sum of the harmonic series has growth of logarithmic order.


[hide]Since $ 10^k\equiv 1\pmod 3$ for all $ k$, we get $ 123456\dots 2009\equiv 1 \plus{} 2 \plus{} \dots \plus{} 2009\pmod 3\equiv 2009\cdot 1005\equiv 0\pmod 3$.[/hide]

[hide="No."]
It is divisible by 3. Write it as $ \sum_{n=1}^{2009} n \times 10^{a_n}$ where $ a_n$ are appropriate powers of 10 we don't want to bother calculating. Working modulo 3 we notice that $ 10^{\mbox{whatever}}$ is 1, so the number is congruent mod 3 to $ \sum_{n=1}^{2009} n$. But this sum is $ 2009 \times 2010 / 2$ which is 0 mod 3.
[/hide]


Let $ A,B\in\mathcal{M}_{n}(\mathbb R) , A\neq B$ , so that $ A^3\equal{}B^3$ and $ A^2B\equal{}AB^2$.
It is possible that $ det(A^2\plus{}B^2)\neq0$?

Direct calculation yields $ A^4\equal{}AB^3\equal{}AB^2B\equal{}A^2BB\equal{}AAB^2\equal{}AA^2B\equal{}B^4$. The assumption $ \det(A^2\plus{}B^2)\not\equal{}0$ would therefore imply $ A^2\minus{}B^2\equal{}(A^4\minus{}B^4)(A^2\plus{}B^2)^{\minus{}1}\equal{}0$. This gives $ 0\not\equal{}\det(A^2\plus{}B^2)\equal{}\det(2A^2)$, i.e. $ A$ is invertible and therefore $ B$, too. Since the property $ A^2B\equal{}AB^2$ is equivalent to $ A(A\minus{}B)B\equal{}0$, this would yield the contradiction $ A\minus{}B\equal{}0$.

Sorry but I don't understand something. $ A^2\minus{}B^2\equal{}(A^4\minus{}B^4)(A^2\plus{}B^2)^{\minus{}1}$ is the same with $ (A^2\minus{}B^2)(A^2\plus{}B^2)\equal{}A^4\minus{}B^4$ and i think that this is true only if $ A^2B^2\equal{}B^2A^2$. How can I demonstrate this?

I am sorry, I forgot to transfer this part from my worksheet into the editor (here the long version):

$ \begin{array}{lcl}A^2B^2\minus{}B^2A^2 &\equal{}& (A^2B^2\minus{}B^2A^2)(A^2\plus{}B^2)(A^2\plus{}B^2)^{\minus{}1}\\
 & \equal{} & (A^2B^2A^2\minus{}B^2A^4\plus{}A^2B^4\minus{}B^2A^2B^2)(A^2\plus{}B^2)^{\minus{}1}\\
 & \equal{} & (AB^3A^2\minus{}B^6\plus{}A^6\minus{}B^2AB^3)(A^2\plus{}B^2)^{\minus{}1}\\
 & \equal{} & (A^6\minus{}B^6\plus{}A^6\minus{}B^2A^4)(A^2\plus{}B^2)^{\minus{}1}\\
& \equal{} & 2(A^6\minus{}B^6)(A^2\plus{}B^2)^{\minus{}1}\\
& \equal{}&2((A^3)^2\minus{}(B^3)^2)(A^2\plus{}B^2)^{\minus{}1}\equal{}0\end{array}$

There's a simpler argument - don't go above the third power.

$ (A^2\plus{}B^2)(A\minus{}B)\equal{}A^3\plus{}B^2A\minus{}A^2B\minus{}B^3\equal{}0$ by hypothesis.

If $ A^2\plus{}B^2$ were invertible, then we would have $ A\minus{}B\equal{}0.$

I don't see anything in this argument that depends on the particular field.


limit
induction
real analysis
real analysis solved

Well, it is easy to show that it converges, but I haven't been able to find the value of convergence. Just in case, if you assume $0<|x|<1$, it is quite easy to prove that

$S_n=(1-x)\sum_{k=0}^n\frac{x^{2^k}}{1-x^{2^{k+1}+1}}$

May be that turns to be easier for you to compute (but I couldn't :oops: )

Best,

Very nice, djimenez, but frankly, I'm just the same as you, this is one of those series that I'm unable to compute. Can someone compute it or show at least that it is not computable?

[quote="zhaobin"]if $y_n=y_{n-1}^2-2$,and $y_0 \geq 2$
$S_n=\frac{1}{y_0}+\frac{1}{y_0 y_1}+\cdots+\frac{1}{y_0 y_1 \cdots y_n}$
find ${\lim }\limits_{n \to \infty } S_n$
I let $y_0=x+\frac{1}{x}$,then know $y_n$,but I can't complete it.
can you help me :) ?[/quote]
You have done all the work!!!
Just consider $x-S_n$ and you will see that $x-S_n=\frac{x^{2^{n+1}}}{y_0y_1...y_n}$. Thus $S_n\to x$.

[quote="Myth"]you will see that $x-S_n=\frac{x^{2^{n+1}}}{y_0y_1...y_n}$. Thus $S_n\to x$.[/quote]

Myth, May be I am not watching something that is just in front of my nose, but I don't get any of the two statement you do above.

Best,

Zhao Bin,

Just in case, the sign of $y_0$ is really irrelevant, therefore you can assume that it is possitive. When you do the transformation $y_0=x+\frac1x$ you are assuming $y_0>2$. In this case, it converges, but something funny is that it does not converge in general. I haven't found cases when you let $y_0<2$ where it doesn't. It could be a nice excercise find for which values of $y_0$ the values or $S_n$ are not well defined (ex: $y_0=\sqrt2$), or the sequence does not converges,m (ex, $y_0=1$).

Myth, if the first of your two statements is true, the second part is true if $x\leq1$. Could you please post how to deduce the first of your affirmations, the one that says that $x-S_n=\frac{x^{2^{n+1}}}{y_0y_1\cdots y_n}$, I have tried in a few ways to deduce it, and I can't, but it is true for $x=1$, therefore, I tend to think that probably you're right.

Best,

Sorry for the double post,

If $x-S_n=\frac{x^{2^{n+1}}}{y_0y_1\cdots y_n}$ is true, then we would have that $S_n=x-\frac{x^{2^{n+1}+2^n}(1-x)}{1-x^{2^{n+1}+1}}$. I would like to do something more, but it probably will have to wait until tomorrow afternoon.

By the way, very interesting problem :D

Best,

Thank you everyone who have posted above. :)
BTW:$djimenez$,can you tell me how did you know this one at first:
$S_n=(1-x)\sum_{k=0}^n\frac{x^{2^k}}{1-x^{2^{k+1}+1}} $

It is actually not quite difficult. Note that $y_0y_1\cdots y_k=x^{2^{k}}+x^{2^{k}-1}+\cdots+x+1+x^{-1}+cdots+x^{-2^{k}}$. Then, factor $x^{-2^{k}}$ and you'll have $y_0y_1\cdots y_k = x^{-2^{k}}(x^{2^{k+1}}+x^{2^{k+1}-1}+\cdots+x+1) = x^{-2^k}\frac{x^{2^{k+1}+1}-1}{x-1}$ Then, invert that, and sum over all the $k$ up to $n$.

Best,


A generic function for {a[n]} is the function:  A(x)=a[0]+a[1]*x^1+a[2]*x^2+...+a[n]*x^n+...

and in the problem is given that A(x) is the generic function of {a[n]}

and that b[n]=a[2*n].

And the task is to be found the generic function  for b[n].

I will give u one example for a problem that i decided:
"to be found the generic function for b[n] ,  if it is given the deneric function A(s) for a[n] and a[n]=3*b[n]."
 here is the salution of the last task:
a[n]=3*b[n]  ----it is for n=0,1,2,3,...

and we made  the following:

b[0]=3*a[0]
b[1]*s=3*a[1]*s
b[2]*s^2=3*a[2]*s^2
............

b[n]*s^n=3*a[n]*s^n
.......

and after that we make the sum

 b[0]+b[1]*s^1+....+b[n]*s^n+....=3*A(s)

;;;
And for the first problem have to be made something like this but i do not decide it

 help with some idea  pls...

This is a hint in the  book about this problem, but i still can not use it... Can you write  how to apply this?

[hide]The problem wants only the even terms, so use $A(x)$ and $A(-x)$ which have opposite signs on odd terms.

$A(x)=a_0+a_1x+a_2x^2+a_3x^3+\cdots$

$A(-x)=a_0-a_1x+a_2x^2-a_3x^3+\cdots$

Add the two to clear terms with odd powers of $x$. $A(x)+A(-x)=2a_0+2a_2x^2+2a_4x^4+\cdots$.

The function to find is $a_0+a_2x+a_4x^2+\cdots$, which by inspection is $\boxed{\frac{A(\sqrt{x})+A(-\sqrt{x})}{2}}$.[/hide]


Hello, rrusczyk!

[quote]$\text{Find }x\text{ if }8^x = 32$[/quote]
We have: $(2^3)^x = 2^5\;\;\Rightarrow\;\;2^{3x} = 2^5$

Therefore: $3x = 5\;\;\Rightarrow\;\;x = \frac{5}{3}$

Very slightly different method

[hide]

8^x = 32

x = log(8)32

x = 5/3

[/hide]

To make fractions in $\LaTeX$ you use \frac{a}{b} for a/b.  If you just want to make it so that it is one character over another you can just do \frac 35. Otherwise, if you are doing something larger you need the {'s and }'s.  Here are some examples:

\frac 35 gives:

$\frac 35$

\frac ab gives
$\frac ab$

\frac{5}{11} gives
$\frac{5}{11}$

Note, that sometimes it looks better to not use the \frac command and just leave it like 3/5....it depends on the situation.
Also, instead of using * for the multiplication you can use \cdot:

x\cdot y gives:
$x\cdot y$

Or, in some situations, x\times y, which is
$x\times y$

This is very good question, Mr. Rusczyk.

8^x = 32

Before we even solve this problem, one thing we can realize is that this answer will have range between 0 to 2. The reason is this:

If it's 2, it's 64.

If it's negative number, example -1, it will get 1/8.

For ones who might not know the negative exponent, look this:

8^-2 = (1/8)^2 = 1/64

See that the negative sign on exponent disappeared as we made the inside to its reciprocal?

Since all the methods I can think of are given, I won't talk about the methods.

But just to show that you can figure out the range of this question without actually solving it

 ;)

Silverfalcon

hey Gupti Aga

what' the log?

and I like the british guy's answer the most out of all of them because it's easy and easy and easy to understand :D

Check this topic for an answer:

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=125246&highlight=#125246]Logarithms[/url]


All the students in an algebra class took a $ 100$-point test. Five students scored $ 100$, each student scored at least $ 60$, and the mean score was $ 76$. What is the smallest possible number of students in the class?

$ \textbf{(A)}\ 10 \qquad
\textbf{(B)}\ 11 \qquad
\textbf{(C)}\ 12 \qquad
\textbf{(D)}\ 13 \qquad
\textbf{(E)}\ 14$

To lower the average at a maximum rate,  let us say that the rest of the students scored a 60. The equation would then be:

$ \frac{(5\cdot100)+(60x)}{5+x}\&=76$

$ 500+60x\&=380+79x$

$ x\approx6.315.$

By this we know that if cannot be that 6 student, it has to be 7 as a minimum.

$ 7+5=12$ where 5 were the original students

easier (maybe):

$ 76x\minus{}100>60x$

$ 16x>100$

$ \boxed{x>6.25}$

knowing that a number of students cant be a fraction, adding those 7 students to the original 5, we get $ \boxed{\boxed {\textbf {(C)}    12}}$

The correct answer is D, 13. Nandubetax: (5+x)*76 = 380 + 76 x != 380 + 79 x.
I think it's easier if you use (500 + 60(x-5))/x = 76, to get x = 12.5 ==> 13.

[hide="My approach"]The problem is to find the smallest positive integer $ x$ such that

$ \frac {100(5) \plus{} 60(x \minus{} 5 \minus{} 1) \plus{} n}{x} \equal{} 76$

$ 500 \plus{} 60(x \minus{} 6) \plus{} n \equal{} 76x \Longleftrightarrow n \equal{} 16x \minus{} 140 \geq 60 \Longrightarrow 16x \geq 200$
$ \Longleftrightarrow x \geq 12.5 \Longrightarrow \boxed{x \equal{} 13}$, which gives $ n \equal{} 68$.[/hide]

Or, if you prefer mental math, you think $ 100 \minus{} 76 \equal{} 24$, $ 24\times5 students \equal{} 120$. $ 76 \minus{} 60 \equal{} 16$, divide $ 120$ by $ 16$ and round to $ 8$. $ 5 \plus{} 8 \equal{} \fbox{13}$.


Given a circle of radius $ R$ centred at $ O$. Two other circles touch the given circle internally and intersect at points $ A$ and $ B$. FInd the sum of radii of these 2 circles if angle $ OAB \equal{} 90$ degrees.


1. http://www.mathlinks.ro/Forum/viewtopic.php?p=746405#746405
In triangle $ABC$ right angled at $C$, the median through $B$ bisects the angle between $BA$ and the bisector of angle $B$. Prove that
$5/2 < AB/BC < 3.$

2. http://www.mathlinks.ro/Forum/viewtopic.php?t=131843
Let $n$ be a natural number such that $n= a^{2}+b^{2}+c^{2}$, for some natural numbers $a,b,c$ . Prove that $9n = (p_{1}a+q_{1}b+r_{1}c)^{2}+(p_{2}a+q_{2}b+r_{2}c)^{2}+(p_{3}a+q_{3}b+r_{3}c)^{2}$, where $p_{i}, q_{i}, r_{i}$ are all non zero integers. Further,if $3$ does not divide at least one of $a,b,c,$ prove that $9n$ can be expressed in the form $x^{2}+y^{2}+z^{2}$, where $x,y,z$ are natural numbers none of which is divisble by $3$.

3. http://www.mathlinks.ro/Forum/viewtopic.php?t=131840
Let $m$ and $n$ be positive integers such that the equation $x^{2}-mx+n = 0$ has real roots $a$ and $b$. Prove that $a$ and $b$ are integers if and only if $[ma]+[mb]$ is the square of an integer. Here,$[x]$ is the greatest integer not exceeding $x$.

4. http://www.mathlinks.ro/Forum/viewtopic.php?p=746406#746406
Let $A = (a_{1},a_{2},...,a_{n})$ be a permutation of $(1,2,...,n)$. A pair $(a_{i},a_{j})$ is said to correspond to an inversion of $A$, if $i<j$ but $a_{i}> a_{j}$. How many permutations of $(1,2,...,n), n\geq 3$ have exactly two inversions?

5. http://www.mathlinks.ro/Forum/viewtopic.php?t=132002
 Let $ABC$ be a triangle in which $AB = AC$. Let $D$ be the midpoint of $BC$ and $P$ be a point on $AD$. Suppose $E$ is the foot of perpendicular from $P$ on $AC$. If $AP/PD = BP/PE = m$ and $BD/AD = n$ and $z =$ $n^{2}$($1+m$), prove that $z^{2}$ - ($m^{3}$ -$m^{2}$ $-2$)$z+1 = 0$.
Hence,show that $m\geq 2$ and $m = 2$ iff ABC is equilateral.

6. http://www.mathlinks.ro/Forum/viewtopic.php?t=131825
If $x, y,z$ are positive real numbers, prove that
 $(x+y+z)^{2}(yz+zx+xy)^{2}\leq 3(y^{2}+yz+z^{2})(z^{2}+zx+x^{2})(x^{2}+xy+y^{2})(*)$.

P.S: Thanks to Sathej :)


Princeton college Harvard Yale Stanford MIT

[quote="zptlimyw"]Looking through number of different websites, I have realized that some schools posted #applied and #admitted [???/????]. However, I realized that none of the websites have actually broken the numbers down into domestics and internationals.

Does anyone know #applied (internationals) and # admitted (internationals) for Princeton, Harvard, Yale, Stanford, CalTech, MIT, etc..?[/quote]
MIT does break it down: http://www.mitadmissions.org/topics/apply/admissions_statistics/index.shtml (scroll down a bit).  In general, acceptance rates for international students are quite low, because the big schools never have more than 5-10% of their class as international students, and since international includes the rest of the world, that's a lot of competition.


search
number theory proposed
number theory

prove that the equation
$k^4+(k+1)^4+(k+2)^4=n^2+(n+1)^2$
has no integer solutions.

tips:
I find it can be solved only by (mod k)
you can guess which k is proper.

I'm near to sure it was already posted. But the way the search works at the moment, it's as good as impossible to find it...
I will look when it gets better.

Define $k+1=t$, than the equation takes this form:
$2x^2+2x-(3t^4+6t^2+1)=0$, now it's obvios that $t$ should be odd, then we get $x^2+x=1(mod4)$, this is impossible. You can also solve this one by chechking out the discriminant of final equation.

I took the $(mod 9)$
it can also be solved in this way


function
limit
real analysis
real analysis unsolved

[i]Let $ f: [0,\infty[\to [0,\infty[$ be a continuous function, with the following properties:

(1) $ x<y\Rightarrow f(x)<f(y)$
(2) $ \exists x_0>0$ such that $ f(x_0)\equal{}x_0$
(3) $ f(x)>x$ for all $ 0\leq x<x_0$
(4) $ f(x)<x$ for all $ x>x_0$

Let $ 0\leq c<x_0$. Define the sequence $ (a_n)_{n\geq 1}$ by $ a_1\equal{}c$ and $ a_{n\plus{}1}\equal{}f(a_n)$.[/i]

I already proved that $ (a_n)$ increases strictly, and it is bounded above by $ x_0$. (If $ a_k>x_0$ for some $ k$, then $ a_{k\plus{}1}\equal{}f(a_k)<a_k$ by property 4, in contradiction with the fact that $ (a_n)$ strictly increases, therefore $ a_n\leq x_0$ for all $ n$.)

I want to prove that $ x_0$ is the [i]least [/i]upper bound, i.e. $ x_0\equal{}\sup\{a_n|n\in\mathbb{N}\}$. If I draw a picture, it is intuitively clear [property 3 makes sure that [i]eventually [/i]you will come arbitrarily close to $ x_0$]. I tried:

Assume by contradiction that there is some $ x_1$ such that $ a_n\leq x_1<x_0$ for all $ n$. If $ a_k\equal{}x_1$ for some $ k$, then $ a_{k\plus{}1}\equal{}f(a_k)>a_k$ by property 3, so $ a_{k\plus{}1}>x_1$, contradiction. So we are left with the assumption that $ a_n<x_1<x_0$ for all $ n$.
All I got from this is $ x_1<f(x_1)<x_0$ and $ a_n<f(a_n)\equal{}a_{n\plus{}1}<f(x_1)$, but I can't get a contradiction (something like $ a_k>x_1$ for some $ k$).

Who can help? Thanks in advance.

At times like these, the best thing to do is usually to ask yourself [i]What haven't I used yet?[/i]. Here the answer is the continuity of $ f$.

Let $ l$ be the least upper bound. Assume that $ l < x_0$. By the monotonicity of the sequence and the definition of least upper bound, $ a_n$ converges to $ l$. That wouldn't be a problem, except that, by continuity,

$ l \equal{} \lim_{n\to \infty}a_{n \plus{} 1} \equal{} \lim_{n\to \infty} f(a_n) \equal{} f(\lim_{n\to \infty} a_n) \equal{} f(l)$

and this violates property (3).

:clap2:  Thanks a lot, it's perfectly clear!


ARML Local is back in two weeks (March 21st).  If you and your school are interested in competing and haven't registered, there is still time.  More information about the contest is available at http://arml.com/arml_local/ , and I am available at arml.local@gmail.com if there are any further questions.

I realize we're intersecting with spring breaks and other contests left and right.  Plan on ARML Local 2010 in February next year.

Paul Dreyer
ARML Local Head Coordinator

Neato! I bet the problems are gonna ROCK  :lol:


geometry
AMC
AIME
USA(J)MO
USAMO
probability
number theory

Eh?

I'd say Calc cant remember which though...

Right now, I'm doing an independent study for math.  I did geometry last year, and next year I'm doing precalc.  I decided to take a break from the normal curriculum to take some AoPS classes and study some other topics in math that I find interesting on my own.

BC Calc.  But they shouldn't be arranged in any specific order like in the poll.

I don't know what you mean by "Math 6 or below". We are doing stuff related to Pre-Algebra, though...

Why the order? The class I'm taking is Geometry, but it's stupendously boring. Maybe I'll actually learn something near the end of the year when we discuss how to count (I've always had trouble with that), but I doubt it.

Meh, next year I'm taking AP Calc (BC, probably, though I'm not entirely sure) by a lucky break, so I can't complain.

Is there an option "other"?  I took Calc BC last year and will probably start taking college courses next year, but currently, I'm kind of doing an AIME-USAMO self-preparation thingy.  It's great... no tests!  (except if you count the AMC, AIME, USAMO, and local competitions)

But more importantly, I'm glad I waited out for college courses because I am learning so much by exploring problems by myself (and with a teacher).  It's amazing how, once you start stepping out of the classroom and doing (extremely large amounts of) math for the fun of it, how much you can learn.

Uh, since when is discrete mathematics a class? It covers such an enormous range of subjects (from combinatorics to number theory to probability to field theory) that teaching it in only one year would be silly.


In triangle ABC, D lies on BC so that AC = 3, AD = 3, BD = 8, and CD = 1. What is the length of AB?

[color=darkblue]By cosine theorem in $ \triangle{ACD}$ we have $ 9 \equal{} AD^2 \equal{} AC^2 \plus{} CD^2 \minus{} 2AC\cdot CD\cos{\angle{ACD}} \equal{} \Leftrightarrow \cos{\angle{ACD}} \equal{} \frac {1}{6}$.

Then by cosine theorem in $ \triangle{ABC}$ we have $ AB \equal{} \sqrt {AC^2 \plus{} BC^2 \minus{} 2AC\cdot BC\cos{\angle{ACD}}} \equal{} 9$.

So $ \boxed{AB \equal{} 9}$[/color]

It also follows easily from [url=http://mathworld.wolfram.com/StewartsTheorem.html]Stewart's theorem[/url].