Dear Ji Chen. This problem is dedicated to you. This is also inspired from an old unsolved inequality dedicated to Manlio in this forum of mine.

$\bigstar$
Given positive real numbers $a,b,c,d,e$ with sum $5$, prove that
\[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}+\frac{8\sqrt{5}}{\sqrt{a^{2}+b^{2}+c^{2}+d^{2}+e^{2}}}\ge 13.\]

A generalization of $n$ variable is not so hard, but I like the case $n=5$ with the numbers $13$ (alternatively, an attempt to make it stronger is not very necessary, I think). Could you please post a complete solution to this problem, Ji Chen?

I know it`s not dedicated to me so I am going to hide the solution.

[hide]use: a^2 + b^2 + c^2 + d^2 + e^2 >= (a+b+c+d+e)^2/5  and  1/a + 1/b + 1/c + 1/d +1/e>= 25/ a+b+c+d+e. and know is very simple
    [/hide]

sorry, i did a little mistake :D

[quote="Zamfirmihai"]I know it`s not dedicated to me so I am going to hide the solution.

[hide]use: a^2 + b^2 + c^2 + d^2 + e^2 >= (a+b+c+d+e)^2/5  and  1/a + 1/b + 1/c + 1/d +1/e>= 25/ a+b+c+d+e. and know is very simple
    [/hide][/quote]
i think you are wrong :D
[hide]with $a^{2}+b^{2}+c^{2}+d^{2}+e^{2}\geq$ .... you only obtain $\frac{8\sqrt{5}}{\sqrt{a^{2}+b^{2}+c^{2}+d^{2}+e^{2}}}\leq$ ....[/hide]

So we are waiting Ji Chen shows his idea  :blush:

... Do you have a less complicated proof? When I post this problem, I know your approach $(x,x+s,x+t)$ is strong enough to use, but I still hope a nice proof from you.

Thank you.

I think maybe Mr.Chen can solve any 3-variables inequality with his computer and his software in 1 minute.
I really like this method(though I don't know how the computer works),becuase It makes all things more simple.

[quote="hungkhtn"]Do you have a less complicated proof? [/quote]The maximum is $\frac{121808+8309\sqrt[3]{28}-5936\sqrt[3]{98}}{9375}$ for the condition $x+y+z=3$, and then it's impossibility to find a less complicated proof.

There is elegance to inequalities that makes them very attractive. It isn't aesthetic such as $13,\frac{166}{13},\frac{2987}{234},\cdots$ instead of the best constant.

The following inequality with $5+8=13$ dedicated to you :

If  $x,y,z$ be nonnegative real numbers such that $x+y+z=\frac{3}{2}$, then
$(x+y^{2})(y+z^{2})(z+x^{2})\leq\frac{5}{8}$,
with equality if $x=0,y=1,z=\frac{1}{2}$.

[b]Proof.[/b] Let $\frac{F(x,y,z)}{9}$

$\equiv \frac{5}{8}\left[\frac{2(x+y+z)}{3}\right]^{6}-\left[\frac{2x(x+y+z)}{3}+y^{2}\right]\left[\frac{2y(x+y+z)}{3}+z^{2}\right]\left[\frac{2z(x+y+z)}{3}+x^{2}\right]$.

Without losing generality, we may assuming $x=\min\{x,y,z\}$.

$F(x,y,z)=F(x,x+s,x+t)$

$=117x^{6}+234(s+t)x^{5}+3(59s^{2}+136st+59t^{2})x^{4}$

$+2(37s^{3}+108s^{2}t+135st^{2}+37t^{3})x^{3}+(22s^{4}+40s^{3}t+99s^{2}t^{2}+94st^{3}+22t^{4})x^{2}$

$+\frac{2(22s^{5}-10s^{4}t+49s^{3}t^{2}+76s^{2}t^{3}+98st^{4}+22t^{5})}{9}x$

$+\frac{2(s+t)(20s^{3}+18s^{2}t+30st^{2}+5t^{3})}{81}(s-2t)^{2}\geq0$,

which is clearly true for $22s^{5}-10s^{4}t+49s^{3}t^{2}=5s^{3}(s-t)^{2}+17s^{5}+44s^{3}t^{2}\geq0$.

[quote="Ji Chen"][quote="hungkhtn"]Do you have a less complicated proof? [/quote]The maximum is $\frac{121808+8309\sqrt[3]{28}-5936\sqrt[3]{98}}{9375}$ for the condition $x+y+z=3$, and then it's impossibility to find a less complicated proof.

There is elegance to inequalities that makes them very attractive. It isn't aesthetic such as $13,\frac{166}{13},\frac{2987}{234},\cdots$ instead of the best constant.

The following inequality with $5+8=13$ dedicated to you :

If  $x,y,z$ be nonnegative real numbers such that $x+y+z=\frac{3}{2}$, then
$(x+y^{2})(y+z^{2})(z+x^{2})\leq\frac{5}{8}$,
with equality if $x=0,y=1,z=\frac{1}{2}$.[/quote]

A nice proof exists, Ji Chen,

[quote="Ji Chen"][quote="hungkhtn"]Do you have a less complicated proof? [/quote]The maximum is $\frac{121808+8309\sqrt[3]{28}-5936\sqrt[3]{98}}{9375}$ for the condition $x+y+z=3$, and then it's impossibility to find a less complicated proof. [/quote]

although you play with the number $\frac{121808+8309\sqrt[3]{28}-5936\sqrt[3]{98}}{9375}$, we can find this value if we can change the inequality to the case $abc=0$. It is my solution, but I like to use the number $13$, and in this case, a very nice solution exists.

The same thing happens to your problem dedicating me. I solve it by the same way as I did to solve my problem.  :lol:


If solutions are submitted far before the deadline, then will they be graded anyways?
I'm pretty sure the answer to that one is yes, but it leads to the next question:

If my solutions are graded, then will graders' comments and scores be available to me before the round is over?

USAMTS tradition is that you must always wait an excruciatingly long time for your scores.

You can submit solutions as early as you like, but nothing will get graded early.

The main reason that the papers take so long to grade is that NSA mathematicians do a large chunk of the grading, and the papers have to be physically shipped from our offices in California to NSA's offices near Washington, DC, and back again.  The other reason is that we're not just assigning scores, we're writing comments, and this is time-consuming.


trigonometry
geometry
function
calculus
inequalities proposed
inequalities

find $A$ such that  $x\cos x\leq A$ where    $x\in [0,\frac{\pi}{2}]$

$\sin x \geq x \cos x$
Draw a unit circle and compare the areas of the segment and triangle

[quote="silouan"]The thing you wrte is true but I don't want something like this . I want a number as a function of $\pi$. But this I want I have not written it above so sorry.
As a conclusion
 I want a number $A$ as a function of $\pi$[/quote]
Why have to write it so bulky? In short, you are just looking for the maximum.

For A=0,65 see
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=52830
for A=0.5610963382 use Maple.
So when doing a synthetic proof, don't try values like 0,55 :).
I would like to see a synthetic proof for A better then 0,6...

Hi

  I think it is an interesting puzzle.  I still don't have the solution for it.
But I think A is unpredictable. Because as I use caculus , I have a very tough equation

Here is mine
$F(x) = xcos(x)$
$F'(x) = cos(x)- xsin(x)$
We will concentrate on $F'(x)$ only
$G(x)=F'(x)= cos(x)- xsin(x)$
$G'(x) = -2 sin(x) - xcos(x)$
I think $sin(x)>xcos(x)$ when it is in $ [0,\frac{\Pi}{2}] $so $G'(x)<0 $
It means that $G(x)$ is an decrease function  when $x=[0..\frac{\Pi}{2}]$ so G(x)=0 has only 1 root
  I call that root    d
so $0< d<\frac{\Pi}{2}    $
I put some value of x ( x between $[0..d]$ and $[d..\frac{\Pi}{2}$) and fingure out that $F(x)$ is  concave down. So $max(F(x)) =F(d)$

So $A= F(d)$ with d is a root of equation $cos(x)-xsin(x)$ (x is in $[0..\frac{\Pi}{2}$)

The key point is to solve that equation $L(x)=cos(x)-xsin(x)$ (x is in $ [0..\frac{\Pi}{2}$)

I don't know how to solve L(x) ( I have an approximate value of  d but I think it won't help you ) .

  It is interesting . I hope s.o could help me .
Bye bye

[b]This problem must appear in the section "Calculus computational" and not here !.[/b]

[b]Solution.[/b] $\left( \forall \right) x\in \left[ 0,\frac {\pi}{2} \right],\ 0\le x\cos x\le \frac {a^2}{\sqrt {1+a^2}}$, [b]where the number[/b] $a$ [b]is the unique solution of the equation[/b] $\tan x=\frac 1x,\ x\in \left(0,\frac {\pi}{2}\right).$


floor function
search
combinatorics unsolved
combinatorics

Let $ a>1$ be a positive integer. Prove that every non-zero positive integer N has a multiple in the sequence $ (a_{n})_{n\ge1}$,$ a_{n}\equal{}\left\lfloor\frac{a^{n}}{n}\right\rfloor$.


As we have for the past two years, AoPS will host Math Jams for each of the AMC exams and each of the AIME exams.  The Math Jam schedule can be found [url=http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php]here[/url].

The AMC and AIME Math Jams have been some of our largest in the past and we do our best to prepare guided solutions that help students see how they can find the solutions themselves next time for the ones that were just out of reach.  Students might find it helpful to study for the 2006 exams by reading through the [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php]AMC/AIME Math Jams from the past two years[/url].

Mr.Crawford, are you going to shut down AMC forum while AMC is tested this year?


Prove or disprove that $n$ $0$'s surrounded by a $1$ on either end is composite for $n > 1$.

Please do not post topics with very long titles.  Most people have to scroll to the right to see all the properties of a thread until this thread goes to page 2.  Please edit the title as soon as possible.

I've edited the title.  Like boxedexe says, don't do this, it's extremely annoying.

[quote="Kouichi Nakagawa"]Let $k \in \mathbb{N}, a \not = 1$.
if $a=2k+1$ then $11|10^{a}+1$
if $a=4k+2=2(2k+1)$ then $101|10^{a}+1$  ($\because 10^{a}=10^{2(2k+1)}=100^{2k+1}$)
if $a=8k+4=4(2k+1)$ then $10001|10^{a}+1$  ($\because 10^{a}=10^{4(2k+1)}=10000^{2k+1}$)
It is similar as follows.
($\because n\in\text{odd}\Rightarrow a^{n}+b^{n}= (a+b)(a^{n-1}-a^{n-2}b+\cdots+n^{n-1})$)

Therefore, A number of a meaning of a question is a composite number.[/quote]

What about something like $10^{64}+1$ ?

Hmm.  For numbers of the form $10^{2^{n}}+1$ the issue does not seem easy to resolve.  I see no good reason why a counterexample of this form would not exist.

FWIW, of course, $10^{64}+1$ is not prime because it is divisible by $1265011073$:)

And it may be a little hard for this thread, but can some one show that $10^{{2^{15}}}+1 =10^{32768}+1$ is [b]not[/b] prime.
(Hint; It is not that hard, if one  is able to  think about  the right trick!)

[quote]And it may be a little hard for this thread, but can some one show that $10^{{2^{15}}}+1 =10^{32768}+1$ is [b]not[/b] prime.
(Hint; It is not that hard, if one  is able to  think about  the right trick!)[/quote]

Just curious .. No takers to do the  problem posted in the above post..? or no interest?

(Hint:
[hide]These are  not primes either .. $10^{8}+1$ or $10^{128}+1$[/hide]

Okay here are a few hints:

(BTW, I was not sure how difficult or easy the problem is/was  as I thought of  the idea of this problem only after seeing these posts  )

You may like  to peek at  Hints 1 to 3 before the Big Hint which is given first...

Big Hint: (Peek , if you  do want to , or later  after looking at Hints 1 to 4)

[hide]Fermat’s  primes   (${2^{2^{n}}}+1$ , and for $n>1$  That is   (17, 257, 65537 )   all  are  $17 \bmod 40$
(We do know these above three are primes  - and only know  fermat’s primes, I might add)[/hide]

Hint 1 ( Not  worth hiding)
Let $p$  be a prime, greater than $5$ , obviously odd.
$10^{p-1}\equiv 1 \bmod p$

Hint 2  (Will not hide this too)
$10^{\frac{p-1}{2}}\equiv \pm 1 \bmod p$

Hint 3  (Peek with minimal risk to spoil the fun - Simple facts about quadratic residue)

[hide]Now further assume $p$ is prime (>5)  and  $p \equiv 1 \bmod 8$ AND $p \equiv 2 \bmod 5$  (Or in other words  $p$ is prime and  $17 \bmod 40$

Then:
$10^{\frac{p-1}{2}}\equiv-1 \bmod p$
because ...
[hide]$2^{\frac{p-1}{2}}\equiv+1 \bmod p$  (Because p is  1 mod 8 and thus 2 is a quadratic residue p)
$5^{\frac{p-1}{2}}\equiv-1 \bmod p$ (Because 5 is not a quadratic residue p) [/hide]
(For more explainations see topics under quardratic residue etc)    [/hide]

Hint 4; (Gives away the remaining steps of  the solution)
[hide]Now let us consider the prime $257$ and note that $(257-1)/2 = 128$
[b]And this  prime  is indeed 17 mod 40[/b]

(Easy to see that $2^{2^{n}}$ with $n>1$ is 0 mod 8 and 1  mod 5)
$10^{128}\equiv-1 \bmod 257 \implies 257 | (10^{128}+1)$
Similarly
$17| (10^{8}+1)$ and $65537 | (10^{32768}+1)$ [/hide]


The midpoints of sides AC and AB of triangle ABC are E and F respectively. A straight line is drawn through vertex A cutting CF, BE, and BC at P, Q, and R respectively. Prove that: $ \dfrac{1}{AP} \plus{} \dfrac{1}{AQ} \equal{} \dfrac{3}{AR}$.

Thank you.  :D  :D

[quote="Nawahd Werdna"] [color=darkred]The midpoints of sides $ [AC]$ , $ [AB]$ of triangle $ ABC$ are $ E$ and $ F$ respectively. A straight line is drawn

through vertex $ A$  cutting $ (CF)$ , $ (BE)$  and $ (BC)$ at $ P$ , $ R$ and $ D$ respectively. Prove that $ \frac {1}{AP} \plus{} \frac {1}{AR} \equal{} \frac {3}{AD}$ .[/color] [/quote]
[quote="Virgil Nicula"] [color=darkred][b][u]An easy extension.[/u][/b] $ \left\|\begin{array}{c} \boxed {\ \triangle\ ABC\ } \\
\ \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \\
\ D\in BC\ \ ,\ \ E\in CA\ \ ,\ \ F\in AB \\
 \\
P\in AD\cap CF\ \ ,\ \ R\in AD\cap BE\end{array}\right\|\ \implies\ \frac {\overline {AE}}{\overline {EC}}\cdot\frac {1}{\overline {AR}} \plus{} \frac {\overline {AF}}{\overline {FB}}\cdot\frac {1}{\overline {AP}} \equal{}$ $ \left(1 \plus{} \frac {\overline {AE}}{\overline {EC}} \plus{} \frac {\overline {AF}}{\overline {FB}}\right)\cdot\frac {1}{\overline {AD}}$ .[/color] [/quote]

[color=darkblue][b][u]Proof.[/u][/b] Apply the [b]Menelaus' theorem[/b] to the mentioned transversals and the adequate triangles :

$ \left\|\begin{array}{ccccc} \overline {BRE}/ \triangle ACD & : & \frac {\overline {BD}}{\overline {BC}}\cdot\frac {\overline {EC}}{\overline {EA}}\cdot\frac {\overline {RA}}{\overline {RD}} \equal{} 1 & \implies & \frac {\overline {EA}}{\overline {EC}}\cdot\frac {\overline {RD}}{\overline {RA}} \equal{} \frac {\overline {BD}}{\overline {BC}} \\
 \\
\overline {CPF}/\triangle ABD & : & \frac {\overline {CD}}{\overline {CB}}\cdot\frac {\overline {FB}}{FA}\cdot\frac {\overline {PA}}{\overline {PD}} \equal{} 1 & \implies & \frac {\overline {FA}}{\overline {FB}}\cdot\frac {\overline {PD}}{\overline {PA}} \equal{} \frac {\overline {DC}}{\overline {BC}} \end{array}\right\|\ \implies$ $ \boxed {\ \frac {\overline {EA}}{\overline {EC}}\cdot\frac {\overline {RD}}{\overline {RA}} \plus{} \frac {\overline {FA}}{\overline {FB}}\cdot\frac {\overline {PD}}{\overline {PA}} \equal{} 1\ }\ (*)$ $ \implies$

$ \frac {\overline {EA}}{\overline {EC}}\cdot\left(1 \plus{} \frac {\overline {AD}}{\overline {RA}}\right) \plus{} \frac {\overline {FA}}{\overline {FB}}\cdot\left(1 \plus{} \frac {\overline {AD}}{\overline {PA}}\right) \equal{} 1$ $ \implies$ $ \frac {\overline {EA}}{\overline {EC}}\cdot\frac {1}{\overline {AR}} \plus{} \frac {\overline {FA}}{\overline {FB}}\cdot\frac {1}{\overline {AP}} \equal{} \left(\frac {\overline {EA}}{\overline {EC}} \plus{} \frac {\overline {FA}}{\overline {FB}} \minus{} 1\right)\cdot\frac {1}{\overline {AD}}$ $ \implies$

$ \boxed {\ \frac {\overline {AE}}{\overline {EC}}\cdot\frac {1}{\overline {AR}} \plus{} \frac {\overline {AF}}{\overline {FB}}\cdot\frac {1}{\overline {AP}} \equal{} \left(1 \plus{} \frac {\overline {AE}}{\overline {EC}} \plus{} \frac {\overline {AF}}{\overline {FB}}\right)\cdot\frac {1}{\overline {AD}}\ }$ .

[b][u]Particular cases.[/u][/b]

$ 1\ \blacktriangleright\ \frac {\overline {EA}}{\overline {EC}} \equal{}\frac {\overline {FA}}{\overline {FB}} \equal{}m\ \implies\ \frac {1}{\overline {AP}} \plus{} \frac {1}{\overline {AR}} \equal{}\frac {2m\minus{}1}{m}\cdot  \frac {1}{\overline {AD}}$ .

$ 2\ \blacktriangleright\ \left\|\begin{array}{c} \frac {\overline {EA}}{\overline {EC}} \equal{} \minus{} 1 \\
 \\
\frac {\overline {FA}}{\overline {FB}} \equal{} \minus{} 1\end{array}\right\|\ \implies\ \frac {1}{AP} \plus{} \frac {1}{AR} \equal{} \frac {3}{AD}$ (the [b]proposed problem[/b] - the previous case for $ m\equal{}\minus{}1$ ).

$ 3\ \blacktriangleright\ P\equiv R\ \stackrel {(*)}{\implies}\ \frac {\overline {PA}}{\overline {PD}} \equal{} \frac {\overline {EA}}{\overline {EC}} \plus{} \frac {\overline {FA}}{\overline {FB}}$ (the remarkable [b]Van Aubel's relation[/b]).[/color]


Why is the bottom different [url=http://www.artofproblemsolving.com/Forum/album.php?mode=pic&id=5744]here[/url] than [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1002064#1002064]here[/url]?

The first is an album; the second a topic.

Or do you mean they [b]should[/b] look the same?

Wait, what's so different about it? Only the buttons and permissions are different. tHe buttons obviously should be, and the missing permissions in the album one don't apply to albums.


The table below displays some of the results of last summer's Frostbite Falls Fishing Festival, showing how many contestants caught $n$ fish for various values of $n$.

\[ \begin{array}{|c|c|c|c|c|c|c|c|c|} \hline n & 0 & 1 & 2 & 3 & \dots & 13 & 14 & 15 \\ \hline \text{number of contestants who caught} \ n \ \text{fish} & 9 & 5 & 7 & 23 & \dots & 5 & 2 & 1 \\ \hline \end{array}  \]

In the newspaper story covering the event, it was reported that
(a) the winner caught 15 fish;
(b) those who caught 3 or more fish averaged 6 fish each;
(c) those who caught 12 or fewer fish averaged 5 fish each.

What was the total number of fish caught during the festival?

[hide]Let $f$ be the number of fish caught and $n$ be the number of contestants. We get two linear equations in terms of $f$ and $n$.

From the 6 fish average, we see that $6(n-21)=f-19$, where $n-21$ is the number of contestants who caught $3$ or more fish. Similarly, we get $5(n-8)=f-108$ from the 5 fish average. Solving, $f=\boxed{943}$. That's a lot of fish. :P[/hide]

[quote=OlympusHero]A wild [b]nsato[/b] appears!

[hide=Solution]Let $x$ be the amount of caught fish and let $y$ be the amount of contestants. From the first condition, we have $6(y-9-5-7)=x-0 \cdot 9 - 1 \cdot 5 - 2 \cdot 7 = 6(y-21)=x-19$. From the second condition, we have $5(y-1-2-5)=x-15 \cdot 1 - 14 \cdot 2 - 13 \cdot 5=5(y-8)=x-108$. Solving these two equations gives $x=\boxed{943}, y=175$.[/hide][/quote]

I keep having to remind myself that the old mods were active in the olden days.


All the logs you posted can be computed by $ \log ab \equal{} \log a \plus{} \log b$,  $ \log a^b\equal{}b\log a$ and $ \frac{\log_a b}{\log_a c} \equal{} \log_c b$. I'll just give you the solutuions for the first two and maybe you'll get the hang of it ;)

[hide="Solution 1"]
Let's set $ x\equal{}\log_9 27$. By the definition of the log this means $ 9^x\equal{}27\Rightarrow \left(3^2\right)^x\equal{}3^3\Rightarrow 3^{2x}\equal{}3^3$ (recall that $ \left(a^b\right)^c\equal{}a^{bc}$). Comparing the exponents we get $ 2x \equal{} 3\Rightarrow x \equal{} \frac 32 \Rightarrow \log_9 27 \equal{} \frac 32.$
[/hide]

[hide="Solution 2"]
By our third law, $ \log_{27} 3 \equal{} \frac{\log_3 3}{\log_3 27} (a\equal{}3, b\equal{}3, c\equal{}3)$. But $ \frac{\log_3 3}{\log_3 27} \equal{} \frac 13\Rightarrow \log_{27} 3 \equal{} \frac 13.$
[/hide]

thank you very much for your help i think i get it now. Do you now how to figure out what N is? Because im kinda stuck on those two

Use this: $ \log_a b \equal{} c\Leftrightarrow a^c\equal{}b$.

thank you for your help guys you saved me my teacher was gonna kill me if i didnt do the work
thanks
Magic man

[quote="lingomaniac88"]Use this: $ \log_a b \equal{} c\Leftrightarrow a^c \equal{} b$.[/quote]

Yes, in the end you can get all the log rules just from its definition... so if you have something like $ \log_a x \equal{} b$ just apply the function $ f(x) \equal{} a^x$ to both sides which would lead to the equation $ x\equal{}a^b$, and that's quite a trivial equation ;)
Unfortunately this trick doesn't work for an equation like $ \ln x \equal{} ax$ but I don't think there's any way to solve such an equation  :(

[quote="Magic Man"]I'm a little bit lost on how to find the answer to these logs:

log9 27 = ?
log27 3 = ?
log10 0.1 = ?
log27 9 = ?
log25 5 = ?

What is N?

log1/2N = -2
log8 N = 1/3

If someone could explain them to me thank you very much for any help
Magic Man[/quote]
Here are the answers in order: 3/2, 1/3, -1, 2/3, 1/2, N=4, N=2


Using the letters $ A$, $ M$, $ O$, $ S$, and $ U$, we can form $ 120$ five-letter "words". If these "words" are arranged in alphabetical order, then the "word" $ USAMO$ occupies position

$ \textbf{(A)}\ 112 \qquad
\textbf{(B)}\ 113 \qquad
\textbf{(C)}\ 114 \qquad
\textbf{(D)}\ 115 \qquad
\textbf{(E)}\ 116$

Going backwards from 120, we have
USOMA -120
USOAM -119
USMOA -118
USMAO -117
USAOM -116
USAMO -115
$ \boxed D$

Another way to do it:

[hide]Since any permutation beginning with U will come after any permutation not beginning with U, first we count the ones not beginning with U, which is equivalent to $ 4 * 4 * 3 * 2 * 1 \equal{} 96$ . Next, we count the permutations which begin with U and come before a permutation starting with U S. The number of these permutations is equivalent to $ 3 * 3 * 2 * 1 \equal{} 18$ . Adding this to 96, there are 114 permutations which come before any permutation starting with U S. Because the remaining letters, A M O, are already in alphabetical order, then U S A M O will be the very next one, or $ 114 \plus{} 1 \equal{} 115 \Rightarrow D$[/hide]


analytic geometry
geometry
circumcircle
geometric transformation
inequalities
number theory
prime numbers

Hey Galois, here are some french2003 problems

(1)
A lattice point in the coordinate plane is a point with integer coordinates.
A lattice point is called invisible if the segment between the point and the
origin contains at least one other lattice point.Prove that for any positive integer
n, there exists a square of side lengh n so that all its interior lattice points are
invisible.(it's the same as romania2003 5th round)

(2)
Let ABC be a triangle and M a point lying in ABC.
D is the point where the line (AM) cuts the circumcircle of ABC.
Find point(s) M so that MB*MC/MD is minimal.
(try this problem,no one among the 20 french guys did it)

(3)
let p1,p2,..,pn be distinct prime  numbers>=5.
Show that 2^(p1*p2*..*pn)+1  has at least 4^n divisors.
(shortlist)

(4)
Ten cities are connected by one way airlines(you can travel in only one
way) such that you can go from any city to any other. Let's denote n the
minimal number of travel you have to do in order to visit every city and come back
to the first one(a cycle). The number n depends of how the airlines are constructed.
-> find the maximum n (and we had to draw the airlines corresponding to this
value of n during the exam).

there were also 2 other problem but there were much too easy to be write here.
(Why are countries like Russia,Romania,Bulgaria,etc.. so strong at maths olympiad?
What is their maths program at school?? )

the first problem was also used in the Romanian Selection Test this year. It was taken from the France selection Tests.

Alekk, did you participate in that exam ?

yes ... but i didn't qualify  :(  :( (it was the final IMO selection and there remained only 20 guys)

don't worry you will qualify this year ! :D

You know, I met some of the French guys in Japan. I remember Olivier Benoist (he's very strong, I think he got 25 points), Raphal Beuzart Plessis (he has a good chance of making the team next year, isn't it ?) (he was very drunk during the last night), Vladimir ... (I forgot his second name), Pierre ... (?), I guess you know them all ? Are you a candidate for the team next year ? I mean, are you still in secondary education ? If so, we could meet at the IMo in Greece the following year - if we're both selected of course, for me that should be the case since I was selected this year but you never know what happens ...  :D

I like the problem about the invisible lattice points !
It's a bit surprising an it has an elegant solution ...
I love the Chinese Remainder theorem  :)

Hey Alekk, I found the other two exercises on the internet (the ones you said were easy).
The first one caught my interest.

Find the minimal value of the expression
abc + def + ghi
if you know that the sets {1, 2, 3, 4, 5, 6, 7, 8, 9} and {a, b, c, d, e, f, g, h, i} are equal.

Now my own solution is very, very clumsy.
So do you have a short an elegant solution ?

The other one is also nice I think :
Let ABC be a triangle and let C1 and C2 be two circles.
C1 touches AB in B and C2 touches AC in C.
Suppose that C1 and C2 touch each other externally in D.
Then ABCD is a convex quadrilateral.
Prove that the circumcenter of BCD lies on the circumcircle of ABC.

The geometry question above is indeed quite trivial.
The main difficulty seems to be making an accurate drawing  :D

Arne for your first question, we can use the average inegality, it seems to be like that : p_1 + p_2 +...+  p_n / n > (p_1*p_2*...*p_n)^(1/n)
i dont know if my translation are good but i think that is the inegality of the geometrical and arithmetical(?) average...

I also thought of that but my solution was quite long.
By the way, moyenne = mean (=/= average) and ingalit = inequality (=/= inegality)  :D

Moubi & thn, did you participate in this test ?

Moubi is a teacher  :D
----------
I did the test too but didn't qualify (I got 14 and I'm in 2nde so I have 2 more years). Olivier Besnoist, Denis Boisnault and Raphael Beuzart Plessis got around 28.
Olivier got 26 at the IMO.

yeah - so one might say he was involved with the test, but didn't participate :D


factorial
limit
calculus
calculus computations

what's up cool dudes ?  :D

find $ \boxed{\left(\minus{}\frac{1}{4}\right)!}$

$ \left(\minus{}\dfrac{1}{4}\right)! \equal{} \Gamma\left(\dfrac{3}{4}\right) \equal{} \dfrac{\sqrt{2}\pi}{\Gamma\left(\dfrac{1}{4}\right)}$

$ \Gamma\left(\frac {1}{4}\right) \equal{} \lim_{m\to\infty}{\frac {m!\sqrt [4]{m}}{\prod_{k \equal{} 0}^{m}{(\frac {1}{4} \plus{} k)}}}\approx 3.625609909$

As I know there are no formula for $ \Gamma (z)$ for any $ z\in \left(0,\frac{1}{2}\right)$  :(


vector
algebra
function
domain
Support
abstract algebra
linear algebra

[color=darkred]Any help is highly appreciated  :) [/color]

1. Let $L_0\subset L, M$ be vector spaces. Prove that the correspondence $(l+L_0)\otimes m\rightarrow l\otimes m+L_0\otimes M$
extends to a well defined isomorphism $(L/L_0)\otimes M\rightarrow (L\otimes M)/(L_0\otimes M)$

2.Prove starting from the definition of a tensor product that $L\otimes 0=0$ where $0$ is the vector space with one vector.

3. Prove that $\mathcal{L}(L\otimes M,N)$ is isomorphic to $\mathcal{L}(L,\mathcal{L}(M,N))$.

4. Let $\dim L<\infty$.Prove that the map $\phi: L\otimes L^*\rightarrow End L$, $\phi(l\otimes\varphi)=(l'\rightarrow\varphi(l')l)$ extends to a well-defined isomorphism of algebras i.e. that $\phi$  is a vector space isomorphism such that $\phi(A\cdot B)=\phi(A)\cdot\phi(B)$, for $A,B\in L\otimes L^*$

5.Let $\sigma\in S_q$.Show that for any vector space L, the linear operator $f_\sigma: L\otimes ...\otimes L\rightarrow L\otimes ...\otimes L$ is well-defined were the tensor products are taken $q$ times in the domain and codomain.

6.Prove that $\dim S_0^q(L)=$ ${n+q-1}\choose{q}$, where $n=\dim L$.

7. Prove that $\dim\Lambda_0^q(L)=$ ${n}\choose{q}$, where $n=\dim L$.

8. Prove that  vectors $l_1,l_2,...,l_n$ are linearly independant if and only if $l_1\wedge l_2\wedge...\wedge l_n\neq 0$.

9.Prove that, for any vector space $L$ we have $T_0^2=S_0^2(L)\oplus \Lambda_0^2(L)$ (direct sum of subspaces).

10. What goes wrong in the $9^{th}$ problem if one replaces 2 by 3.

[color=darkred]That's it. Mesmerize me with your solutions please. The exam is in 10 hours and counting  :?[/color]

[quote="-oo-"]What about your own ideas? Most of these exercises are straight-forward ...[/quote]

It might be so, I don't doubt it, and me and a few others are working on them too but the ideeas and solutions of some persons who have worked with these kind of things before may be more concluding than the sketches we succed to do with our fresh understanding of the given topics.

I don't think so. Who is motivated to write down the proofs of these 0815 - exercises? :)
Well, [i]you[/i] should be, so that we can check your proofs.

The ideea is that I should get some sleep before the exam and get some help concerning these straight-forward problems as you sad at the same time, as extra support for my fragile knowledge. If that is a too egocentric thought and nobody feels like helping me I will try to post the solutions as soon as I can.
Is that explanatory enough?


Choose $ a$ and $ b$ positve integer, let be $ n \equal{} a \plus{} b$and $ x_{i}$ ($ i \equal{} 1,2,...,n$) positive numbers.

Prove that:

$ \sum^{n}_{j \equal{} 1}{\frac {\sum^{j \plus{} a \minus{} 1}_{k \equal{} j}{x_{k}}}{\sum^{j \plus{} a \plus{} b \minus{} 1}_{l \equal{} j \plus{} a}{x_{l}}}}\ge\frac {a}{b}n$

As always this may be ugly but if you analize the small cases you will understand it well.
(Example: $ a\equal{}1$ and $ b\equal{}2$ $ \equiv$  Nesbit's)

 :lol:

I assume the indices are taken modulo $n$. In this case assume
wlog that $\sum_{k=1}^n x_k=1$. Adding $n$ to both sides we get
to prove that

\[
\sum_{j=1}^n\frac{1}{\sum_{l=j+a}^{j+n-1}x_l}\geq \frac{n^2}{n-a}.
\]

This follows directly from Cauchy-Schwarz (or AM-HM).  :D


MATHCOUNTS AMC AIME calculus USA(J)MO USAMO search

Well first I'd like to say that I really admire you guys, and I wish I could be as smart as you all.

So here's my math life story. I was in Mathcounts, never had any success with that. I took the AMC the past two years, and I qualified for the AIME, mostly by luck. I embarrassed myself with a score of 0 last year, a decrease from last year's score of 2.

Please don't take the following as showing off. I took Calculus BC as a sophomore and easily made a 5. I probably got around 90 % right.

But when it comes to AIME-level math, I just have no idea where to start. Every single year  I crack open Art Of Problem Solving Vol. 1 and read the first few chapters, but to no avail (as you can see, my scores decreased). You may say that it's good to qualify for AIME, but I really would like to at least make USAMO, to have that sense of triumph when solving difficult problems.

I have the exact same problem with physics. I know I will easily make a 5 on the Physics B/C tests this year, but the Olympiad problems last year--I got only 1 of the free responses right.

I have only 1 more year to actually prove myself. Plus, my brother is entering high school next year, and I would hate for him to have as bad a start in terms of contest math as I did. If it's some spark that takes a long time to develop, I want at least him to get that spark.

Do any of you USAMOers and the like have any advice as to how I can remedy my situation? I would really appreciate any replies. I would especially like to know how you developed your problem-solving abilities to such an awesome level, and please be detailed ie hours studied, books read. Thanks so much in advance. I really want to tear down that wall that's preventing me from doing well.

[quote="cooljoe"]Every single year  I crack open Art Of Problem Solving Vol. 1 and read the first few chapters, but to no avail (as you can see, my scores decreased). [/quote]

Perhaps your issue is in your approach to the material.  AoPS Vol. 1 is not light reading.  It needs to be read intensively, and you need to make sure you understand what's being said in every paragraph before moving on to the next, as well as making sure you do every exercise and/or problem until you're sure you understand the chapter.

It might also not be a bad idea to finish the book.  Nothing comes easy.

As much as competition math is about training and experience, there is an aspect, once you get to AIME level and higher, of creativity, and this is not something that is easily taught.  I've never been able to give good advice about how to nurture this.

a friend of mine who made blue mop last year said that he only started competition math in 8th grade...the main reason he succeeded, he said, was because he finished so many problem books...vol. 1 not working? try russian math circles...or something of similar difficulty...and don't think you can just whiz by vol. 1. like torajiru said, it is INTENSIVE reading. it took me ~2 months to finish it in the summer. working 2-4 hours per day. as you see, you can't expect to learn it like a calculus book.
DO A LOT OF PROBLEMS! and go over teh solutions very carefully. im sure there are times when everybody doesn't get one problem or more...make sure you understand why you didn't get it.

and remember, it's not just about competition math, it's about developing a lifelong passion for such a wonderful subject. how many mathematicians that have solved great theorems, ectr. never participated in any math competition. they just developed a love for mathematics. that's what its about. LOVE and PASSION

hey, can someone tell me what the different colored mops are?

RED lowest level.
BLUE 2nd level
BLACK highest level (12 USAMO winners, PWNAGE)

MOP = mathematics olympiad program
MOSP = mathematics olympiad summer program
two abbreviations for the same thing. its a summer program where qualifing USAMO test takers are invited to. From there, you go on to the IMO team picking, etc.

yeah i know but what does mihail mean when he uses it. is it like score or something. how do you know which mops you are in?

The search function is your friend :wink:
Anyway,
RED=9th grade only, usually between 8 and 11 (10?) cutoff
BLUE=Anyone besides seniors, around 20
BLACK-Top 12, anybody, um...around 25-28?

indeed..abacadea..basically ur grouped into the levels of MOP based on your USAMO score..this year all 9th graders were in Red...and the 12 USAMO winners are always in black..the middle-scorers are in Blue.
you can check it out on the AMC website

Thanks for the response t0ijaru.

So would my best bet be just to do problems? I think the problem is that I am not consistent--I work on AoPS during the summer for a couple weeks, then give up because I'm not improving.

any more advice would be greatly appreciated.

consistency is perhaps one of the most important (and perhaps only) factor. ur not going to succeed if ur not consistent.


$ 1)$ Find the minimum value of the function $ f(x)$ if $ \minus{}6x\plus{}4f(x)\equal{}2x^2\minus{}7$

$ 2)$ If $ f(x)\equal{}\frac{(x\plus{}1)(x\plus{}3)(x\plus{}5)}{(x\minus{}2)(x\minus{}3)(x\minus{}4)}$, find $ f^{\minus{}1}(\minus{}8)$

$ 3)$ Let $ f(x)\equal{}\frac{x}{x\minus{}1}$. Find the composition $ f^{\minus{}1}(f^{\minus{}1}(f^{\minus{}1}(x)))$.

I know that making a function inverse, like with $ f^{\minus{}1}(x)$ of $ f(x)\equal{}3x$ is $ y\equal{}x/3$, but I just don't get it on these kinds of problems.

I don't know of your skill level but the first one is easiest by using calculus.

[hide]1) $ f(x)\equal{}\frac{2x^{2}\plus{}6x\minus{}7}{4}$.  Then $ f'(x)\equal{}x\plus{}\frac{3}{2}$.  Setting it equal to zero, $ x\equal{}\minus{}\frac{3}{2}$.  This is a critical point and testing it shows that it is indeed a minimum.  So the minimum occurs when $ x\equal{}\minus{}\frac{3}{2}$, and $ f(\minus{}\frac{3}{2})\equal{}\minus{}\frac{23}{8}$.[/hide]

For the first:

This rearranges to $ f(x) \equal{} \frac{1}{2} x^2 \plus{} \frac{3}{2} x \minus{} \frac{7}{4} \equal{} \frac{1}{2} \left( x^2 \plus{} 3x \minus{} \frac{7}{2} \right) \equal{} \frac{1}{2} \left( x^2 \plus{} 3x \plus{} \frac{9}{4} \minus{} \frac{23}{4} \right)$

$ \equal{} \frac{1}{2} \left( x^2 \plus{} 3x \plus{} \frac{9}{4} \right) \minus{} \frac{23}{8} \equal{} \frac{1}{2} \left( x \plus{} \frac{3}{2} \right) ^2 \minus{} \frac{23}{8} \geq \minus{} \frac{23}{8}$

It can be safely assumed that common high-school knowledge about $ f(x)\equal{}ax^2\plus{}bx\plus{}c$ includes the fact that it reaches its extreme at $ \left(\minus{}{b\over 2a},\minus{}{D\over 4a}\right)$

2) $ f^{\minus{}1}(\minus{}8)\equal{}x\Leftrightarrow f(x)\equal{}\minus{}8$
$ \displaystyle\frac{(x\plus{}1)(x\plus{}3)(x\plus{}5)}{(x\minus{}2)(x\minus{}3)(x\minus{}4)}\equal{}\minus{}8\Leftrightarrow$
$ \Leftrightarrow3x^3\minus{}21x^2\plus{}77x\minus{}59\equal{}0\Leftrightarrow (x\minus{}1)(3x^2\minus{}18x\plus{}59)\equal{}0\Rightarrow x\equal{}1$

3) $ \displaystyle f^{\minus{}1}(x)\equal{}\frac{x}{x\minus{}1}\equal{}f(x)$.

Then $ f^{\minus{}1}(f^{\minus{}1}(f^{\minus{}1}(x)))\equal{}f^{\minus{}1}(f^{\minus{}1}(f(x)))\equal{}f^{\minus{}1}(x)\equal{}f(x)$

[quote="Bicameral"]I know that making a function inverse, like with $ f^{ \minus{} 1}(x)$ of $ f(x) \equal{} 3x$ is $ y \equal{} x/3$, but I just don't get it on these kinds of problems.[/quote]

The general idea is that the inverse of $ y \equal{} f(x)$ can be found by writing it as $ x \equal{} f(y)$ and then solving for $ y$.  So, our original function being

$ y \equal{} \frac{x}{x \minus{} 1}$

We can rewrite the inverse function as

$ x \equal{} \frac{y}{y\minus{}1} \Leftrightarrow$
$ xy \minus{} x \equal{} y \Leftrightarrow$
$ y(x \minus{} 1) \equal{} x \Leftrightarrow$
$ y \equal{} \frac{x}{x \minus{} 1}$

So the inverse function turns out to be the same as the original function.


How many integers $ n$ are there such that $ 3 \leq n \leq 10$ and 121 base $ n$ (or $ 121_n$) is a perfect square?

We know that any number in the form $ 121_n$ is a perfect square
because it can be factored into $ (n \plus{} 1)^2$
So each value of $ n$ from $ 3$ to $ 10$ will work.

There are a total of $ \boxed{8}$ possible values.

@izzy: i think you did some calculations wrong...

$ 121_n\equal{}n^2\plus{}2n\plus{}1_{10}\equal{}(n\plus{}1)^2_{10}$
So Ca Math is right.
$ 121_6\equal{}49$
$ 121_7\equal{}64$
$ 121_8\equal{}81$
$ 121_9\equal{}100$
$ 121_10\equal{}121$
Izzy's mistake: $ \overline{abc}_n\equal{}(n^2\cdot a\plus{}n\cdot b\plus{}c)_{10}$, it's not $ \overline{abc}_n\equal{}(n^2\cdot a\plus{}n\cdot b\plus{}c\cdot n)_{10}$
Answer: $ \boxed{8}$

This may be considered spam, but...

[code]$121_{10}$[/code]

will come out as $ 121_{10}$.


[quote="bui_haiha2000"]Let a,b,c>0 satisfy that a+b+c=6.Find min:
   P=(a+1/a)(b+1/b)(c+1/c)[/quote]

i dont know how to use Latex.

actually, you can assume that c is a 'fixed value'. Therefore a+b=6-c..
Now we want to discuss the minimum value of P if c is a 'fixed value'.

We are going to find the minimum value of (a+1/a)(b+1/b)
WLOG. we can assume a+b=2
So, a=1+t, b=1-t (t is a positive real and t<1)
(a+1/a)(b+1/b)=((ab)^2+b^2+a^2+1)/ab=(t^4+4)/(1-t^2)
you can see when t trend to 1, (a+1/a)(b+1/b) trend to infinity..
therefore t should be 0, and we get a=b=1.
The results is:the minimum value of (a+1/a)(b+1/b) occurs when a=b

Hence, we can also get the minimum value of (a+1/a)(c+1/c) occurs when a=c , and vice versa..

Finally, we get P(min)=(5/2)^3, when a=b=c

Actually, you can assume that c is a 'fixed value'. Therefore a+b=6-c..
Now we want to discuss the minimum value of P if c is a 'fixed value'.

We are going to find the minimum value of (a+1/a)(b+1/b)
WLOG. we can assume a+b=2

=========
I dont think it is legal to assume that a+b=2.


analytic geometry
graphing lines
slope
LaTeX

This is a question I thought of that tests your basic knowledge of algebra.

Line $l$ has a equation of $y = kx-1$ for some real $k$ and passes through $(-1,2)$. Line $ll$, which is perpendicular to line $l$, passes through $(2,-1)$. If line $ll$ is in form of $y = \frac{m}{n} x - \frac{o}{p}$, compute $m+n+o+p$.

Enjoy!

Hey. This is my first time posting on this board :P. I had done the problem out, and was just going to read up on how to use LaTeX to post the solution, but apparently I was too slow :(. That's the same solution I got, so I suppose it's right.

-BLt

Good job. That's the correct answer.  :)

Thebrokenlight, welcome to the AoPS Math Forums!  :welcomeani:


group theory
abstract algebra
superior algebra
superior algebra solved

$G$ is a group with $H \subseteq G$ and $K \subseteq G$. Show that if $K \subseteq H$ or $H \subseteq K$ the set $H \cup K \nsubseteq G$.

So what does one think? Lagrange's theorem maybe? If one donates the order of the different sets as $|G|=g, |H| = h$ and $|K| = k$ you know that $h | g$ and $k | g$ but $(h+k) \nmid g$. This implies that the number created by the addition is not one of the primefactors of $g$. One knows that $k$ and $h$ may very well have common factors. I belive that I should prove that

$\frac{g}{h+k}$ is not an integer when given the above, I just don't know where to start.

Suppose that it is an integer $m$? And show it with the knowledge that $m < g$ and come up to some kind of contradiction? Please help :)

what do you mean?
Do you want to prove that $(H\cup K)\leq G \Leftrightarrow H\subset K\mbox{ or }K\subset H$?

I want to prove that the group created when you take the union of $H$ and $K$ can't be a subgroup of $G$. Can't one express that like $(H \cup K) \nsubseteq G$?

This is the same thing as  saying that the number of elements $|H \cup K|$ is not a devisor of $|G|$. I still haven't figured it out. Pisses me off :)

M

Proof:

Let be $K \cup H$ a subgroup of $G$.
Assume that $K$ is not a subgroup/subset of $H$. Now you have for all $h \in H, k\in K \backslash H$ that $kh$ is not an element of $H$, so it's an element of $K$, so $h$ is an element of $K$, which gives imediatly that $H$ is a subgroup/subset of $K$.


Are there any math gatherings/contests/competitions/etc. that are offered for people at my age level (BTW, I'm 13-years-old)? I know of all the CEMC competitions and some IMO preparation stuff (Olymon, APMO, CMO, etc.). Is there anything else that can occupy my time?

[quote="TheDreamer"]It's honestly impossible to know about everything.  I know the feeling, I've always missed out on a lot of stuff in my past due to misinformation.  But it's gonna happen no matter what.   :(

Anyways, you have the main ones as far as I know; there's stuff like Math Battle and McTic, which are like team math competitions.

Best of luck in your mathematical endeavours.[/quote]

Hmm... I'll look those up.

[quote="d.lam.86"]
Hmm... I'll look those up.[/quote]

You won't find anything on McTic.  It's held in the spring... maybe reading some of the threads in this forum will help more than random intarweb searching ;)

Doesn't Dr. Rice hold some sort of Math circle in Toronto?  I swear he mentioned it at Seminar...but I zoned out since I'm not from there....

[quote="pengshi"]Well, if you live in Toronto. There is a "math circle" every monday from 4:30 pm to 6:30 pm at UTS (University of Toronto Schools). It starts in October I think. Last year, I remember that some 12-13 year old was there. The difficulty of the material ranges from Euclid to first 3 questions of CMO. I hope that this helps.[/quote]
That is good to know, but I think I'll have to wait a couple of years. I am definitely in no shape for Euclid or CMO.  :oops:  :lol:


Find the sum of the positive integer factors of 6552.

What's an easier way to do this besides adding them all together?

[quote="t0rajir0u"]Is this really classroom math?

[hide="Anyway..."] $6552 = 2^{3}\cdot 3^{2}\cdot 7 \cdot 13$

Its positive integer factors will look like $2^{a}3^{b}7^{c}13^{d}$ where $a, b, c, d$ range from $0$ to $3, 2, 1, 1$ respectively.  Their sum can be factored as

$(1+2+2^{2}+2^{3})(1+3+3^{2})(1+7)(1+13)$

(Can you see why this gives every possible factor?)  This can be evaluated in the general case with geometric series sums, but here it is just

$15 \cdot 13 \cdot 8 \cdot 14 = \boxed{21840}$ [/hide][/quote]

so.. say the number was 550... then the answer would be:
1116?


geometry
power of a point
geometry unsolved

Let $ (I)$ inscribed in $ \triangle ABC$. $ (I)$ touches $ BC,CA,AB$ at $ D,E,F$. $ AD \cap (I) \equal{} \{D,P\}$ such that $ \widehat{BPC}\equal{}90^o$. Prove that $ AE\plus{}AP\equal{}PD$.

A straightforward bash works. We have to show AE=PD.
Let x=AF=AE, y=BF=BD, z=CD=CE.
By Stewart's we can find AD. By Power of a point, we can find AP, and thus PD. By Stewart's we can find BP and PC, and then by Pythag on $ \triangle BPC$, we have the given relation is true. The algebra is atrocious, though. Is there a nicer solution?

We have to show AE + AP = PD, not AE = PD. Actually, 3 AE = 2 PD.
See [url]http://www.mathlinks.ro/viewtopic.php?t=147900[/url].


let $r,s \ge 1$ be intergers and $a_0,a_1,\cdots,a_{r-1},b_0,b_1,\cdots,b_{s-1}$ be real non-negative numbers such that:
$(a_0+a_1 x+\cdots+a_{r-1} x^{r-1}+x^r)(b_0+b_1 x+\cdots+b_{s-1} x^{s-1}+x^s)=1+x+x^2+\cdots+x^{r+s}$
Prove that each $a_i$ and each $b_j$ equals either 0 or 1.
thanks for help :)

goto http://www.ucl.ac.uk/~ucahjej/imc/imc2001/prob_sol2.pdf
 :cool:

sorry,I have the file.But I can't understand the proof in the last passage..
can you help me?

The coefficient of $X^i,\forall 1\le i \le s-1$ is $1=a_0b_i+a_1b_{i-1}+....\ge a_0b_i=b_i$.So, $b_i\le 1,\forall 1\le i \le s-1$. Do the same for $a_i$. This was problem?


This is IMO 1961 Problem 2. I hear that there are about 10 proof.

[quote=Kunihiko_Chikaya]This is [b]IMO 1961 Problem 2[/b]. I hear that there are about 10 proof.[/quote]
http://www.cut-the-knot.org/Curriculum/Geometry/Weitzenbock.shtml

[quote=Virgil Nicula]\[ \boxed {\ \triangle ABC\Longrightarrow a^2+b^2+c^2\ge 4S\sqrt 3\ }  \][hide="The my proof."]$4m^2_a=2\left(b^2+c^2\right)-a^2\Longrightarrow 2\left(a^2+b^2+c^2\right)=\left[2\left(b^2+c^2\right)-a^2\right]+3a^2\Longrightarrow$
$2\left(a^2+b^2+c^2\right)=4m^2_a+3a^2\ge 4am_a\sqrt 3\ge 8S\sqrt 3\Longrightarrow a^2+b^2+c^2\ge 4S\sqrt 3$,
with equality iff $2m_a=a\sqrt 3$ and $AM\perp BC$, where the point $M$ is the midpoint of the side $[BC]$ (i.e. $a=b=c$). [/hide][hide="A easy generalization."]$L\in [BC],\ \frac{LB}{LC}=m>0,\ AL=l\Longrightarrow al\ge S$ and from the
[u]Stewart's relation[/u] $(m+1)^2l^2+ma^2=(m+1)\left(mb^2+c^2\right)\Longrightarrow$
$(m+1)\left(a^2+mb^2+c^2\right)=(m+1)^2l^2+(2m+1)a^2\ge 2(m+1)al\sqrt{2m+1}\ge$
$4(m+1)S\sqrt{2m+1}\Longrightarrow \boxed {\ a^2+mb^2+c^2\ge 4S\sqrt{2m+1}\ }$, with equality iff $(m+1)\cdot AL=a\sqrt{2m+1}$ and $AL\perp BC\ .$[/hide][/quote]

[hide=Easy proof (method)]
The method :
Use Ravi Substitution
then simplification
then it can be easily proved using the AM-GM inequality.
[/hide]

[quote=adityaguharoy]Hadwiger-Finsler inequality is a stronger version of the Weitzenbock inequality
https://en.wikipedia.org/wiki/Hadwiger%E2%80%93Finsler_inequality[/quote]

The proof of this inequality is also easy and elementary.

[quote=Kunihiko_Chikaya]This is IMO 1961 Problem 2. I hear that there are about [b]10 proof[/b].[/quote]

Another proof is by some working with trigonometry.

https://artofproblemsolving.com/community/c6h31598
This link provides some other form of strong generalizations and discussion on the inequalities.

My stronger BF118:

\[a^2+\frac{4b^2c^2}{b^2+c^2}\geq 4\sqrt{3}\Delta\]

https://artofproblemsolving.com/community/c6h31598p6023172
this is another nice strong version.

There are other stronger versions by BQ sir I remember.

In a triangle,prove

\[2a^2+\frac{w_a^2}{m_a^2}(b+c)^2\geq 8\sqrt{3}\Delta\]
<=>

\[a^2+\frac{2bcs(s-a)}{m_a^2}\geq 4\sqrt{3}\Delta\]

\[\frac{a}{h_a}+\frac{2bc}{\sqrt{s(s-a)}a}\geq 2\sqrt{3}\]

[quote=achen29]I have discovered a truly remarkable proof of this theorem which this margin is too small to contain.[/quote]

You can put it in a file and share it if you want.

By Hadwiger-Finsler inequality ,have $$a^2+b^2+c^2\ge 4S\sqrt 3+2(a-b)^2$$
Amer Math Monthly 2017 , Problem 11990:
$$a^2+b^2+c^2\ge \sqrt 3\left(4S+(c-a)^2\right)$$


Let a trapez $ ABCD$ with $ AB // CD$ and the midpoint $ M$ of side $ CD$.A line through $ M$ meets diagonal $ AC$ at $ G$ and line $ AD$ (produced to $ D$) at $ F$. If lines $ BF,BG$ meet $ CD$ at points $ N,L$ , prove that $ DN\equal{}CL$.

[u] Babis[/u]

[quote="April"]Applying the [b]Menelaus theorem[/b] to triangle $ ACD$ with the transversal $ FMG$,
we get $ \frac {GC}{GA}\cdot\frac {FA}{FD} \equal{} 1$, i.e. $ \frac {GC}{GA} \equal{} \frac {FD}{FA}$ $ (1)$.

On the other hand, by [b]Thales theorem[/b], we have $ \frac {GC}{GA} \equal{} \frac {CL}{AB}$ and $ \frac {FD}{FA} \equal{} \frac {DN}{AB}$ $ (2)$.

From $ (1)$ and $ (2)$, we conclude that $ CL \equal{} DN$, as desired.[/quote]

 :) very nice !!!

 Babis


Two sides of an isosceles triangle are 15 cm and 10 cm. What is the greatest possible perimeter of this triangle, in centimeters?

The perimeter will be maximized when the triangle has side lengths of $ 10$, $ 15$, and $ 15$. Thus, the maximum perimeter is $ 10\plus{}15\plus{}15\equal{}\boxed{40}$.


Hi!

I have been trying to complete the series: 1, 8, 72, 46, 521, x ; where x is the number I need to get; but I can't seem to get it  :(

Can anyone help me on this one? Thanks!

[hide]Reverse the numbers to get 1,8,27,64,125...look familiar?[/hide]

[hide="Solution"]They are all cubes that are written backwards. The next term is $ 6^3\equal{}216$ written backwards, which is $ \boxed{612}$.[/hide]

How in the world do you get these? :huh:

[hide]they are all cube written backwards so the last one must be 6 ^ 3 which is 216 or 612 backwards[/hide]


geometry
circumcircle
rhombus
geometry proposed

Incircle of triangle $ ABC$ touch $ BC;CA;AB$ at $ D;E;F$
Suppose that $ T;T_1;T_2;T_3$ be circumcircle of triangle $ ABC;AEF;BDF;CDE$,respectively.
$ T$ cuts $ T_1$ at $ A$ and $ P$,$ T$ cuts $ T_2$ at $ B$ and $ Q$,$ T$ cuts $ T_3$ at $ C$ and $ R$
i)Prove that $ T_1;T_2;T_3$ meet together in one point.
ii)Prove that $ PD;QE;RF$ concurrent.

(a) is trivial, $ AEIF, BFID, CDIE$ are cyclic on account of right angles at $ D, E, F.$
(b): $ AI, BI, CI$ cut the circumcircle of $ \triangle ABC$ again at $ X, Y, Z.$
$ \angle FBP \equal{} \angle ABP \equal{} \angle ACP \equal{} \angle DCP$ and $ \angle AFP \equal{} \angle AEP$ $ \Longrightarrow$
$ \triangle FBP \sim \triangle ECP$ are similar, $ \frac{PB}{PC} \equal{} \frac{FB}{EC} \equal{} \frac{DB}{DC}$ $ \Longrightarrow$
$ PD$ bisects $ \angle CPB,$ goes through $ X.$ Similarly, $ QE, RF$ go through $ Y, Z.$
$ YI \equal{} YA$ and $ ZI \equal{} ZA,$ $ AYIZ$ is a rhombus, $ YZ \perp AI$ and similarly, $ ZX \perp BI, XY \perp CI$ $ \Longrightarrow$
$ \triangle XYZ \sim \triangle DEF$ are centrally similar, having parallel sides, lines $ PDX, QEY, RFZ$ concur at their similarity center.


[quote="metafor"]Am I right in assuming that, since $ x, y, z$ are integers, every square in the equation is an integer?[/quote]
Yes, since $ x\pm y\pm z$ will always evaluate to an integer if all three variables are integers themselves.

[quote="metafor"]If so, the only possible squares are $ 0$, $ 4$, and $ 36$, which means that $ (x \plus{} y \plus{} z) \equal{} 6$ (they are positive integers), and $ (x \minus{} y \plus{} z), (x \plus{} y \minus{} z)$ equal either $ 0$ or $ 2$ (but both cannot equal one of these).[/quote]
Also correct.

[quote="metafor"]From this, I can deduce that the solutions (expressed as triples) are all combinations of the set $ \{1, 2, 3\}$:

$ (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1)$.[/quote]
The permutations of the set $ \lbrace1,2,3\rbrace$ might not all be solutions. If you checked for this, then the entire solution is without flaws (I can't tell because you made no indication of doing this). If not, you should consider doing it from now on. :)

[quote="metafor"][b]Problem[/b]: Find all the positive integer triples $ (x, y, z)$ for the equation $ (x \minus{} y \plus{} z)^2 \plus{} (x \plus{} y \minus{} z)^2 \plus{} (x \plus{} y \plus{} z)^2 \equal{} 40$.

Am I right in assuming that, since $ x, y, z$ are integers, every square in the equation is an integer? If so, the only possible squares are $ 0$, $ 4$, and $ 36$, which means that $ (x \plus{} y \plus{} z) \equal{} 6$ (they are positive integers), and $ (x \minus{} y \plus{} z), (x \plus{} y \minus{} z)$ equal either $ 0$ or $ 2$ (but both cannot equal one of these). From this, I can deduce that the solutions (expressed as triples) are all combinations of the set $ \{1, 2, 3\}$:

$ (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1)$.

Are any of the previous assumptions true?[/quote]

Your first assumption seems correct.  x,y,z integers -> x+y-z integer -> (x+y-z)^2 integers
Anyway, with quadratic residues, all 3 squares must be either 0 or 1 mod 4.  They all have to be 0 because 40 is 0 mod 4.
Therefore your second assumption is also correct.

dang ilp beat me.

Yes, I meant [i]permutations[/i], and I found that only $ (1, 2, 3)$ and $ (1, 3, 2)$ work, because $ x$ is the only variable that is never negative. Sorry for not verifying before, I was sort of occupied when I made this thread.


trigonometry
rotation
AMC
AIME
trig identities
Law of Cosines

In a circle, parallel chords of lengths 2, 3, and 4 determine central angles of $\alpha$, $\beta$, and $\alpha + \beta$ radians, respectively, where $\alpha + \beta < \pi$.  If $\cos \alpha$, which is a positive rational number, is expressed as a fraction in lowest terms, what is the sum of its numerator and denominator?

[hide]If we let $\frac{\alpha}{2} = x$ and $\frac{\beta}{2} = y$ we have $\sin x = \frac 1r$, $\sin y = \frac 3{2r}$ and $\sin(x+y) = \frac 2r$.

$\sin x \cos y + \sin y \cos x = \frac 2r$

$\frac 1r \cdot \frac{\sqrt{r^2 - \frac 94}}{r} + \frac 3{2r} \cdot \frac{\sqrt{r^2 - 1}}{r} = \frac 2r$

$r=\frac 8{\sqrt{15}}$

$\cos \alpha = 2 \cos^2 x - 1$

$\cos x = \frac{\sqrt{r^2-1}}{r}$ so we have $\cos a = 1 - \frac 2{r^2} = 1- \frac{15}{32} = \frac{17}{32}$

The answer is $17+32=49$[/hide]

[hide="hint"]
Who says you can't move around the chords? :)
[/hide]

[hide]
We rearrange the chords such that an inscribed triangle is formed with side lengths $a = 2$, $b=3$, and $c=4$, and angles $A=\frac{\alpha}{2}$, $B=\frac{\beta}{2}$, and $C = 180^\circ - \frac{\alpha + \beta}{2}$.

Use the Law of Cosines,

$a^2 = b^2 + c^2 - 2bc\cos A$
$4 = 9 + 16 - 24\cos \frac{\alpha}{2}$
$\cos \frac{\alpha}{2} = \frac{21}{24} = \frac{7}{8}$

By a double-angle identity,

$\cos \alpha = 2\cos^2 \frac{\alpha}{2} - 1 = 2\left( \frac{49}{64} \right) - 1 = \frac{17}{32}$

Thus our answer is $17 + 32 = 49$.
[/hide]

[quote="TheAmazingOne"][hide]
We rearrange the chords such that an inscribed triangle is formed with side lengths $ a \equal{} 2$, $ b \equal{} 3$, and $ c \equal{} 4$, and angles $ A \equal{} \frac {\alpha}{2}$, $ B \equal{} \frac {\beta}{2}$, and $ C \equal{} 180^\circ \minus{} \frac {\alpha \plus{} \beta}{2}$.

Use the Law of Cosines,

$ a^2 \equal{} b^2 \plus{} c^2 \minus{} 2bc\cos A$
$ 4 \equal{} 9 \plus{} 16 \minus{} 24\cos \frac {\alpha}{2}$
$ \cos \frac {\alpha}{2} \equal{} \frac {21}{24} \equal{} \frac {7}{8}$

By a double-angle identity,

$ \cos \alpha \equal{} 2\cos^2 \frac {\alpha}{2} \minus{} 1 \equal{} 2\left( \frac {49}{64} \right) \minus{} 1 \equal{} \frac {17}{32}$

Thus our answer is $ 17 \plus{} 32 \equal{} 49$.
[/hide][/quote]

How do you know the angle A+B+C adds up to 180 dgrees?

[quote="math154"]They are the angles of a triangle...[/quote]

Since when did the three chords formed a triangle?  I don't get it.

I believe that the poster rotated the chords so that they formed a triangle.

See [[1985 AIME Problems/Problem 9]] for a diagram.

In the wiki solutions, how do they get the idea of making the segments into a triangle?


i have a stupid question:
$ f(x) \equal{} x \plus{} sin(x) \plus{} cos(x)$  when $ x \equal{} \pi$ then $ f(x) \equal{} ?$

i really want to undestand clearly!! thanks

[quote="J.Y.Choi"]$ \sin{\pi} \equal{} 0,\cos{\pi} \equal{} \minus{} 1$. So, $ f(\pi) \equal{} \pi \minus{} 1$. What do you want to understand 'clearly'? Do you want to know why $ \sin{\pi} \equal{} 0,\cos{\pi} \equal{} \minus{} 1$??[/quote]

i want to understand $ \pi\minus{}1\equal{}3,14116..\minus{}1$ or other?

Yes, $ \pi$ is $ 3.14159265358979\cdots$ in numerically.


inequalities calculus integration vector

Many of you that have some experience with elementary inequalities are aware of, and are able to apply, the Cauchy-Schwarz inequality, which states that for any $x_i,y_i\in\mathbb{R}$, we have that:

$\displaystyle\left(\sum^{n}_{i=0} x_i^2\right)\left(\sum^{n}_{i=0} y_i^2\right)\geq\left(\sum^{n}_{i=0} x_iy_i\right)^2$

with equality iff $\frac {a_i}{b_i}=\frac{a_j}{b_j}$ for all $i,j$ such that $0\leq\ i,j,\leq n$.

This inequality as stated above is due to Cauchy.  There is a version for integrals, due to Schwarz, which is:

$\displaystyle\left(\int^{b}_{a} f^2(x)dx\right)\left(\int^{b}_{a} g^2(x)dx\right)\geq\left(\int^{b}_{a} f(x)g(x)dx\right)^2$

with equality iff $f(x)=\lambda g(x)$ for some $\lambda\in\mathbb{R}$.

Now try and prove both parts!   The more proofs, the merrier.  In particular, there's a one line proof involving inner product spaces which takes care of both cases - it's tidy, but quite unenlightening, and I'd prefer NOT to see that proof.

Are you sure this ought to be put in Pre-Olympiad section ?  :maybe:

Why not? Cauchy-schwarz inequality has many many applications and is often found in pre as well as post (and present hehe) olympiad questions.

The proof that I am most familiar with is the boring vector one using dot prod.


(OBM/2008) The numbers a and b are the roots of the equation $ x{}^{2}\minus{}x\minus{}1\equal{}0$ . Calculate $ 13.a^{5}\plus{}5.b^{7}$.

$ a^2\equal{}a\plus{}1\Rightarrow a^3\equal{}a^2\plus{}a$

$ a^4\equal{}a^2\plus{}2a\plus{}1\Rightarrow a^5\equal{}a^3\plus{}2a^2\plus{}a\equal{}3a^2\plus{}2a\equal{}5a\plus{}3$

$ b^5\equal{}5b\plus{}3\Rightarrow b^7\equal{}5b^3\plus{}3b^2\equal{}5(b^2\plus{}b)\plus{}3(b\plus{}1)\equal{}13b\plus{}8$

Then, $ 13a^5\plus{}5b^7\equal{}13(5b\plus{}3)\plus{}5(13b\plus{}8)\equal{}65(a\plus{}b)\plus{}79\equal{}65\plus{}79\equal{}144$


In a triangle  $ABC$ with $AB = AC$ we take the midpoint $M$ of  $BC$  and let $MD \perp AC$ ($D$ is a point on $AC$). If $BD$  meets $AM$ at $E$ , which of the segments $AE$  and $AD$  is greater?

[u]Babis[/u]

[quote="ashegh"]i think they are equal.[/quote]

You are wrong.

why...

i think by telling the solution,u can answer me ;)

yes yes yes yes....

i found why.

thanks siluan ;)

Ashegh , now you can send your solution. Many ML  friends are waiting for it.

  [i]I must say that I needed 15 min to solve it( of course too much time for this problem !!!)[/i]

 [u]Babis[/u]

ok stergui...

i will post it as soon as i can...

suppose $ABC$ is a triangle.

$theorem$:if $\angle B>\angle C$,then we have:$b>c$.

look at the triangle$DMC$its clear thet $MC>MD$.but$MB=MC$

then as the theorem said,in triangle $DMB$ we have:$\angle BDM>\angle DBM$.

then:-$\angle BDM$$<$-$\angle DBM$

then:$90$-$\angle BDM$<$90$-$\angle DBM$

and then:$\angle ADE$<$\angle AED$

and according to the theorem we should have:$AD>AE$

ure right dear stergui itsnt so hard and doesnt need much time to prove ;)

did  u know that we call this theorem $(donkey theorem)$? :D

Thanks Ashegh!

 Now everybody can read the solution and the problem is CLOSED. I do not think that we can do a simpler solution.

[u]babis[/u]


Here is the sequence: $4, 8, 1, 2, 1, 6, 2, 0, 2, 4,...$

It is just multiples of $4$ but each digit is counted as a term,

Does anyone know how to find the formula for the $n\text {th}$ term of this sequence?

It has been bothering me for a while. Thanks for the help.

i dont have a cool surefire formula, but you could just take the 4nth term, and then count how many 2 digit numbers that are divis by 4 there are in front of the 4nth number, how many three digit numbers divis by 4, and subtract accordingly

There isn't a clean general formula because the answer depends on how many digits each multiple of $4$ has, and also on what those digits are.  Any "formula" you could give would have floors in it.

yes, this is true, you would need some floors, and probably some logs, but i stil cant think of anything good. probably the best place to start is to know which multiple of 4 you are on, any ideas?

[quote="Iversonfan2005"][quote="nat mc"]any more thought?   :?[/quote]

i knew you would eventually post this problem on aops to make sure your formula was right :D[/quote]
Yeah, I did lol. But, the problem is that I don't have a formula. I can see how to approach, but not how to acutally get it. I need some help


inequalities
calculus
derivative
inequalities unsolved

if $ a>0$,$ b>0$, prove that $ (a\plus{}b)^n<2^n(a^n\plus{}b^n)$

it 's just chebychev  :)

Assume $ a \geq b$ then we have:

$ a^n \plus{} b^n\geq\frac {1}{2}(a \plus{} b)(a^{n \minus{} 1} \plus{} b^{n \minus{} 1})\geq\frac {1}{2^2}(a \plus{} b)^2(a^{n \minus{} 2} \plus{} b^{n \minus{} 2})\geq....\frac {1}{2^{n\minus{}1}}(a \plus{} b)^n>\frac{1}{2^n}(a\plus{}b)^n$

after it's obvious  :lol:

It obviously follows from Holder:
\[ (a^n \plus{} b^n)^{\frac 1n}\underbrace{(1 \plus{} 1)^{\frac 1n}\cdots (1 \plus{} 1)^{\frac 1n}}\geq (a \plus{} b) \\
\therefore 2^{n \minus{} 1}(a^n \plus{} b^n) \ge (a \plus{} b)^n;\]
Also $ 2^n(a^n \plus{} b^n) > 2^{n \minus{} 1}(a^n \plus{} b^n) \ \left[\because a,b > 0\right];$ hence we are done.
[quote="kunny"]1.Jensen

2. Consider $ s,\ t$ such that $ s \plus{} t \equal{} 1,\ s > 0,\ t > 0$.

3. Let $ \frac {b}{a} \equal{} x$.[/quote]
The inequality is homogeneous so we can assume $ a \plus{} b \equal{} 1$ without loss of generality....

$ a\plus{}b\equal{}1$ is Good idea!

My solution requires differentiation, so I don't recommend the solution.

Anyway, Holder is strong.


Let $a,b,c$ be real of numbers such that
          $|a(b-c)| >|b^2-ac|+|c^2-ab|$
  and the equation  $ax^2+bx+c=0$ has root real.
    Prove that this equation has root  $0<x<\sqrt{3}-1$

[color=red]Edit by Megus: LaTeX edited[/color]

Any one can post the solution???Please help me

Can you post your solution more detail,Matematikator????
  It's clear that exist that x1 or x2 >0 but why  x1 or x2 <sqrt(3) -1. Help me


Solve in the set of reals numbers the following equation :

$ (x^2\plus{}2x\plus{}2)(3x^2\plus{}2x\plus{}4)\equal{}3(x^2\plus{}1)^2$

[hide]Bashing out the expansion, we get

$ 8x^3\plus{}8x^2\plus{}12x\plus{}5 \equal{} 0$

which we can take out a factor of $ (2x\plus{}1)$ from to get

$ (2x\plus{}1)(5 \plus{} 2 x \plus{} 4 x^2)$

Since the discriminant of the second quadratic is negative, the only real solution is $ \minus{}\frac{1}{2}$[/hide]

[hide="No bashing"]Substitute $ u \equal{} x^2 \plus{} 2x \plus{} 2, v \equal{} x^2 \plus{} 1$.

The equation becomes $ u(u \plus{} 2v) \equal{} 3v^2\iff u^2 \plus{} 2uv \minus{} 3v^2 \equal{} 0$

Solving it for $ u$, we get $ u \equal{} v\lor u \equal{} \minus{} 3v$

The first possibility yields $ 2x \plus{} 2 \equal{} 1\iff x \equal{} \minus{} {1\over 2}$

The second possibility yields $ x^2 \plus{} 2x \plus{} 2 \equal{} \minus{} 3x^2 \minus{} 3\iff 4x^2 \plus{} 2x \plus{} 5 \equal{} 0$, which has no real solutions.

Therefore the only solution is $ x \equal{} \minus{} {1\over 2}$.[/hide]

hello, rewriting your equation in the form $ (x^2\plus{}2x\plus{}2)(3x^2\plus{}2x\plus{}4)\minus{}3\cdot(x^2\plus{}1)^2\equal{}0$ and factorizing we get
$ (2x\plus{}1)(4x^2\plus{}2x\plus{}5)\equal{}0$, from here we have
$ x_1\equal{}\minus{}\frac{1}{2}$, $ x_2\equal{}\minus{}\frac{1}{4}\plus{}\frac{1}{4}\sqrt{19}i$, $ x_2\equal{}\minus{}\frac{1}{4}\minus{}\frac{1}{4}\sqrt{19}i$
as the solutions.
Sonnhard.

Let $ x^2\plus{}1\equal{}t$, the given equation can be written as $ (t\plus{}2x\plus{}1)(3t\plus{}2x\plus{}1)\equal{}3t^2$

$ \Longleftrightarrow (2x\plus{}1)(2x\plus{}1\plus{}4t)\equal{}0$, as Simon M said, yielding $ \boxed{x\equal{}\minus{}\frac{1}{2}}$.


The corners of an $ 8^{\prime\prime} \times 12^{\prime\prime}$ rectangle are rounded with quarter circles having radius $ 1^{\prime\prime}$ as shown. How many square inches are in the area of the rounded rectangle? Express your answer in terms of $ \pi$.
[asy]import olympiad; import geometry_dev; import graph; size(90); defaultpen(linewidth(0.9));
draw((0,1)--(0,9)--(1,9)--(1,10)--(11,10)--(11,9)--(12,9)--(12,1)--(11,1)--(11,0)--(1,0)--(1,1)--cycle);
draw(Arc((1,1),1,180,270)); draw(Arc((1,9),1,90,180)); draw(Arc((11,9),1,0,90)); draw(Arc((11,1),1,270,360));[/asy]

The whole area is $ 96$.

We must find the area of the cut-off parts and then subtract from the total.
[asy]draw((0,0)--(0,1)--(1,1)--(1,0)--cycle); draw((0,0){up}..{right}(1,1)); draw((0,0)--(1,1));[/asy]

The area of the triangle is $ \frac {1\cdot1}{2}$.
The area of the quarter circle is $ \frac {1^2}{4}\pi$. The area of the small "half-leaf" region is $ \frac {2\pi \minus{} 1}{8}$.
Once again, the area of the triangle is $ 1/2$, so we subtract to get the area cut off:
\[ 1/2 \minus{} \frac {2\pi \minus{} 1}{8} \equal{} \frac {3 \plus{} 2\pi}{8}
\]
Since there are $ 4$ of these areas, the total is $ \frac {3 \plus{} 2\pi}{2}$. We subtract from $ 96$ to get
\[ \boxed{\frac {189 \minus{} 2\pi}{2}}
\]

The above answer is unfortunately incorrect (based on FTW).

Here's my solution and approach:

The corners of the rounded rectangle are quarter circles, and thus must have an area a quarter of that of a full circle. ($\frac{1}{4}A$, where A is the area of the full circle)
We have 4 quarter circles, which has the equivalent area of a full circle. ($\frac{1}{4}A \cdot 4 = \frac{4}{4}A = 1A$)
The area of a circle is $\pi{}r^{2}$, where $r$ is the length of the radius. The problem states that the radius of these circles are $1$ inch. Thus, the four quarter circles have an area of $1^{2}\pi{} = \pi{}$.
The rectangle's area, without the quarter circles, is $92$. We get this by multiplying the original rectangle dimensions ($8*12 = 96$) and subtracting $4(1)$ to make up for the replaced square corners. The quarter circles' areas are then added to the rectangle's area (without the quarter circles) to get the full area of the rounded rectangle, which is $\boxed{92+\pi{}} \textbf{ Ans.}$

Just find the area of the entire rectangle $ 12 \cdot 18 = 96 $
The find the area for one corner of the rectangle which is missing from the corner.
Area of the Square - Area of the quarter circle = $ 1 \cdot 1 - \frac { \pi }{4}$
Therefore you get: $ \frac {4- \pi }{4}$
There are four corners so multiply this by four, and subtract this from 96.
$96-(4- \pi )$
Answer is:
$ \boxed {92+ \pi{}}$


I don't know if this was mentioned before but I can't see any pictures in the Resources page.

It will be fixed. For now I am at the IMO in Slovenia (with Richard and Naoki) so we can't do much about it :)

They'll be back soon, when we bring the album back online.

Did you bring some pictures back? I can see a diagram in the 2006 AIME A page but the pictures of the B test are still missing.


inequalities
number theory unsolved
number theory

Several different positive integers lie strictly between two successive squares.Prove that their pairwise products are different.

Just want to confirm...
Those several positive integers between two consecutive perfect squares say $ n^{2}$ and $ (n \plus{} 1)^{2}$ will be $ ( n^{2} \plus{} 1 ,n^{2} \plus{} 2,........,n^{2} \plus{} 2n$ . Do you want to show that for some $ i,j,k,l$ with $ i \neq j$ and $ k \neq l$ and  $ 1 \le i,j,k,l \le  2n$ the following holds good -
 $ (n^{2} \plus{} i)(n^{2} \plus{} j) \neq (n^{2} \plus{} k)(n^{2} \plus{} l)$ where both $ i$ and $ j$ are not simultaneously equal to $ k$ and $ l$ . Isnt that obvious ? :blush:

@shoki:edited

[quote="srinath.r"]Just want to confirm...
Those several positive integers between two consecutive perfect squares say $ n^{2}$ and $ (n \plus{} 1)^{2}$ will be $ ( n^{2} \plus{} 1 ,n^{2} \plus{} 2,........,n^{2} \plus{} 2n$ . Do you want to show that for some $ i,j,k,l$ with $ i \neq j$ and $ k \neq l$ and  $ n^{2} \plus{} 1 \le i,j,k,l \le n^{2} \plus{} 2n$ the following holds good -
 $ (n^{2} \plus{} i)(n^{2} \plus{} j) \neq (n^{2} \plus{} k)(n^{2} \plus{} l)$ where both $ i$ and $ j$ are not simultaneously equal to $ k$ and $ l$ . Isnt that obvious ? :blush:[/quote]
no, i think he meant to prove $ (n^2\plus{}i)(n^2\plus{}j)\equal{}(n^2\plus{}k)(n^2\plus{}l)$ never occurs for $ 1 \le i,j,k,l \le 2n$ and $ \{i,j \} \neq \{k,l \}$
which can easily be proved by inequality.

oops ,sorry shoki that is what I meant ...
$ 1 \le i,j,k,l \le 2n$ ...typo :oops:
Yes..it can be proved say by method of contradiction also...

shoki: how exactly do you prove it can't hold by inequality?

Let $ (x,y,z,w) \in \left(\mathbb{Z} \cap (n^2,n^2 \plus{} 2n \plus{} 1) \right)^4$ fixed for some $ n \in \mathbb{N}_0$ s.t. $ x < y < z < w$ and $ xw \equal{} yz$. Since $ (y \minus{} x)(z \minus{} x) > 0$ then $ x \plus{} w > y \plus{} z$. Then we have $ 4n^2 < (x \plus{} y \plus{} z \plus{} w)(x \minus{} y \minus{} z \plus{} w) \equal{} (x \plus{} w)^2 \minus{} (y \plus{} z)^2$ $ < (x \plus{} w)^2 \minus{} 4yz \equal{} (x \plus{} w)^2 \minus{} 4xw \equal{} (x \minus{} w)^2 < (2n)^2$, that is a contradiction. []


Let $\frac{35x-29}{x^2-3x+2}=\frac{N_1}{x-1}+\frac{N_2}{x-2}$ be an identity in $x$. The numerical value of $N_1N_2$ is:

$\text{(A)} \ -246 \qquad \text{(B)} \ -210 \qquad \text{(C)} \ -29 \qquad \text{(D)} \ 210 \qquad \text{(E)} \ 246$

[hide] $\rightarrow \frac{ 35x - 29 } { (x-2)(x-1) } = \frac{N1(x-2)}{(x-2)(x-1)} + \frac{N2(x-1)}{(x-2)(x-1)}$
$\rightarrow 35x-29 = N1x + N2x - 2N1 - N2$
since we know the terms with the x must equal to the 35x etc. we can use linear combinations to figure it out... I decided to use substitution
$\rightarrow 35 = N1 + N2$
$\rightarrow -29 = -2N1 - 1N2$
$41 = N2$
$N1 = -6$
so... $-246$ or $\boxed{A}$
my arithmetic is not very good so you should probably check...
 :roll:
[/hide]


$a, b$ are the roots of equation $x^2-6x+1=0$
I've proved, that for any $n\in\mathbb{N}$  $a^n+b^n$ is a integer number.
But now I have to prove, that $a^n+b^n$ isn't divisible by $5$ for any natural $n$
Please help


induction
number theory proposed
number theory

Find all natural number n :
$ n|2^{n\minus{}1}\plus{}1$

Only n=1.
Obviosly n is odd, therefore from $ p_{i}|n$ we get $ p_{i}\equal{}1\mod 4$.
By induction for any k, $ 2^{k}|n\minus{}1$.

Can you post a full solution .I solved it very simple.
A problem has the same method(MOSP 2001)
 Let m,n is natural numbers ,p is a prime  satisfy
$ n|m^{(n\minus{}1)p}\minus{}1$
Prove that $ gcd(m^{n\minus{}1}\minus{}1,n)>1$

I give full solution. Obviosly $ 2|n\minus{}1$.
Let $ 2^{k}|n\minus{}1$ and $ p|n$, then for period T(p) we get $ 2^{k\plus{}1}|T(p)|p\minus{}1$, therefore $ 2^{k\plus{}1}|\prod_{p|n}p^{l(p)}\minus{}1\equal{}n\minus{}1.$

I will post my proof but I want you post a solution for second problem.

Conside n>1
Let p is prime divisor of n and $ e_{2}(p\minus{}1)$ min
Easy to check that $ e_{2}(n)\geq e_{2}(p\minus{}1)$
$ p|2^{p\minus{}1}\minus{}1$
$ p|2^{2(n\minus{}1)}\minus{}1$
Then $ p|gcd(2^{p\minus{}1}\minus{}1,2^{2(n\minus{}1)}\minus{}1)|2^{n\minus{}1}\minus{}1$
Then p|2 this tradition .
So n=1 is solution of problems
The second has the same method.


probability
geometry
3D geometry
algebra
function
domain
inequalities

Three number are chosen between 0 and 1, what is the proobalbyility that the differnece between the largest and smallest number is less than 1/2?

Uhh, a geometric interpretation would be the easiest way to solve if that helps you at all... I'm not all that great at probability but I'll give the problem a try later   :blush: .

[hide="Hint"]
A geometric interpretation would work well. How can we interpret the second restriction as a set of planes? :)[/hide]

Ok so I ujust did x<y<z
then 0<x<1
0<y<1
0<z<1

I graphed that first

Then I added the fact that z-x<1/2

Im horrible with 3D graphing but I got the area of the second one to be 1/12 and the area of the first one to be 1/6 so when I divied I get 1/2..
That doesnt sound right.  I think I have the right idea but I cant graph it properly.

So I made an attempt at simplifying the problem so as not to be using a cube, this could very well be wrong though so bear with me:

Since we are looking at the interval [0,1], if the difference between the smallest and the largest numbers picked is less than 1/2, then the difference between the smallest number and the middle number and the difference between the middle number and the largest number must also be less than 1/2. The probability that the difference between any two numbers is less than 1/2 is 3/4 (is it?). So the probability of this occurring all three times is:
3/4 * 3/4 * 3/4 = 27/64.

EDIT: We only need 3/4*3/4 because the third time is implied from the first two... So 9/16.
Wait, no, not it's not. I've confused myself now  :P.

... Nobody click on bob's link. Well... nobody click on the link with the intention of finding math. (It's a link to playboy, so.. if you have a credit card you could go nuts.)

I just drew a cube when I did this one, but I got it wrong, probably from messing up the volume calculation.

[quote="bob yates"]posted [url=http://www.playboy.com]here[/url], except its 1/3 but same idea[/quote]
Wtf?

Don't follow that link.

My solution isn't right in any way. Please disregard. Why is probability so deceiving  :mad:.

I'm not really sure about this
[hide]Of the three numbers chosen (assuming they are chosen randomly) the probability that we choose the max and the min is $\frac{1}{3}\cdot\frac{1}{2}$.  Let the numbers be $x_{1},x_{2},x_{3}$ and  let $x_{1}$ be the highest and $x_{3}$ be the lowest.  Then we want $P(x_{1}-x_{3}<\frac{1}{2}) = P(x_{1}<\frac{1}{2}+x_{3})$.  Then the domain of $x_{3}$ becomes $[0,\frac{1}{2}]$.  Then the area of the region bounded by $x_{1}= x_{3}+\frac{1}{2}$, $x_{3}= \frac{1}{2}, x_{1}= 0,x_{3}= 0$ is $.375 =$ the probability that $x_{1}-x_{3}<\frac{1}{2}$ GIVEN that $x_{1}$ and $x_{3}$ were in fact the the max and min.  So then $\frac{1}{3}\cdot\frac{1}{2}\cdot.375 = \frac{1}{16}$. [/hide]

[hide="further hint into geometric interpretation"] Consider what happens along one of the planes, say the $xy$-plane. What does $|x-y| < \frac{1}{2}$ look like? What happens when we combine the inequalities $|x-y| < \frac{1}{2}$, $|y-z| < \frac{1}{2}$, and $|x-z| < \frac{1}{2}$? I hope this works out... [/hide]

And for the record, the answer has to be more than $\frac{1}{4}$ since the probability that they are all $< \frac{1}{2}$ or all $> \frac{1}{2}$ is $\frac{1}{4}$.

EDIT: Not sure if this is right, but working through it the geometric way and the calculus way, I get

[hide] $\frac{1}{2}$ [/hide]

I got 1/2..but I dont have an answer key, mine wasnt a cube, it was 1/6 of a cube because i had set restricitons on x<y<z...and then I found out where the line z<x+1/2 wnet trhogh...but I dont know if i really did it right.

Solving this probability geometrically, heres what I got:

[hide]
First I let $x$ = the first number and $y$ = the second number.

I then graph an xy-plane such that the value of the variable x is on one axis and the value of variable y is on the other. Then I plot every point where (x, y) satisfies the condition, and got this:
[img]http://i32.photobucket.com/albums/d11/p4fn2w/graph.gif[/img]

If this graph is correct, then the area of the shaded region over the total area should be the answer.

$area of graph = 3/4$
$area of total area = 1$
$probability = 3/4$

The only thing that bothers me, though, is that the area would be 3/4 [i]if[/i] |x - y| is less than [i]or equal to[/i] 0.5, but the problem specifically states that states that |x - y| is less than 0.5, so the probability should be a bit smaller than 3/4. I'm not sure how to account for this restriction.
[/hide]

edit: Oops... I read the problem wrong and thought that there were only two variables. Sorry!

You have to account for all three inequalities, though. Your answer works for the two-variable case.

[quote="p4fn2w"]
The only thing that bothers me, though, is that the area would be 3/4 [i]if[/i] |x - y| is less than [i]or equal to[/i] 0.5, but the problem specifically states that states that |x - y| is less than 0.5, so the probability should be a bit smaller than 3/4. I'm not sure how to account for this restriction.
[/quote]

It doesn't matter.  In a continuous probability situation $P(X=x) = 0$. That is, the probability that the random variable $X$ will take on the value $x$ is $0$ since there are infinitely many points in the sample space.

You forgot to take into account that three number were chosen.  The question is equivalent to: find the probability that the range of the numbers chosen is less than $\frac{1}{2}$. So the greater the amount of numbers chosen the less the probability that the range is less then $\frac{1}{2}$.

I remember going through the 3 variable case on one of my old aimes. It is quite difficult to envision the solid whose area you need to find. In my opinion, the calculus method is easier.

I did a very similar problem to this, except the parameters were $x+y < z, x+z < y, y+z < x$

(I was solving for the probability of three random numbers between 0 and 5 forming a triangle, BTW), and, after two hours of visualizing, sketching, and looking up formulas for volumes, myself and two friends got the (correct) answer to be $\frac{1}{2}$, which probably demonstrates our ineptitude for math (relatively speaking).  :blush:

My point? Don't try to find the volume geometrically, especially since your graph might be more complex than our own. Use calculus.

[quote="max_tm"]Hmmm would the calculus method be multivariable or would it be single-variable involving a solid of revolution?[/quote]

The multivariable way is simpler - I don't think a solid of revolution would even work. First assume $x < y < z$ and then consider the cases $x < \frac{1}{2}$ and $x > \frac{1}{2}$; we get

$\int_{0}^{\frac{1}{2}}\int_{x}^{x+\frac{1}{2}}\int_{y}^{x+\frac{1}{2}}dz dy dx$

and

$\int_{\frac{1}{2}}^{1}\int_{x}^{1}\int_{y}^{1}dz dy dx$,

which add up to $\frac{1}{12}$ if I calculated correctly. Since there are six permutations of the ordering of $x, y, z$ and they should all have the same probability by symmetry, the answer is $\frac{1}{2}$.

Was surprised by the simplicity of the answer. Then I've tried to take arbitrary $t$ instead of 1/2.

Using calculus (as paladin8) I've obtained P(R<t), where R is the distance between max and min.
The probability density function (via derivative) is very beautiful: $p(t)=6t(1-t)$  :!:

Maybe there is intuitive explanation? 6 is the number of permutations... $t(1-t)$ is?

[hide="quick and easy"]

[P(x>y>z)][P(x-z<.5)]=(1/6)(1/2)

Repeat x6 to cover all six linear orders of (x,y,z) and you get 6(1/6)(1/2)=1/2
[/hide]
not that bad

[quote="sapphyre571"]

[P(x>y>z)][P(x-z<.5)]=(1/6)(1/2)

Repeat x6 to cover all six linear orders of (x,y,z) and you get 6(1/6)(1/2)=1/2

not that bad[/quote]

Can you explain that a little more, especially with the [P(x>y>z)][P(x-z<.5)]=(1/6)(1/2) part.  Thanks

hold on - I major screwed up. I'll work on fixing that.


quadratics
Asymptote
induction
calculus
calculus computations

Hello.

I must find an easy way to prove that

$ 3e^{3x} \plus{} 8e^{2x \plus{} 1}(1 \minus{} x) \plus{} 6e^{x \plus{} 2} \plus{} 4e^3 > 0$

for all x.

And also

$ e^{3x}\plus{}4e^{2x\plus{}1}(x\minus{}2)\plus{}e^{x\plus{}2}(3\minus{}x)\plus{}e^3>0$

Thanks.

$ 3e^{3x} \plus{} 8e^{2x \plus{} 1}(1 \minus{} x) \plus{} 6e^{x \plus{} 2} \plus{} 4e^3 > 0, \forall x$

iff

$ 3e^{3 \minus{} 3x} \plus{} 8xe^{3 \minus{} 2x} \plus{} 6e^{3 \minus{} x} \plus{} 4e^{3 \minus{} 0x} > 0, \forall x$

iff

$ 3e^{ \minus{} 3x} \plus{} 8xe^{ \minus{} 2x} \plus{} 6e^{ \minus{} x} \plus{} 4 > 0, \forall x$

iff

$ 3 \plus{} 8xe^{x} \plus{} 6e^{2x} \plus{} 4e^{3x} > 0, \forall x$

let
$ f(x) \equal{} 3 \plus{} 8xe^{x} \plus{} 6e^{2x} \plus{} 4e^{3x}$

$ f'(x) \equal{} 8e^x \plus{} 8xe^x \plus{} 12e^{2x} \plus{} 12e^{3x} \equal{} 4e^x[2 \plus{} 2x \plus{} 3e^x \plus{} 3e^{2x}]$

$ f'(x) \equal{} 0$ when $ x \equal{} \frac {3}{2}e^x(1 \plus{} e^x) \minus{} 1$

so at its extrema,
$ f(x) \equal{} 3 \plus{} 8(\frac {3}{2}e^x(1 \plus{} e^x) \minus{} 1)e^x \plus{} 6e^{2x} \plus{} 4e^{3x} \equal{} 16e^{3x} \plus{} 18e^{2x} \minus{} 8e^x \plus{} 3 \equal{} g(x)$

$ g'(x) \equal{} 48e^{3x} \plus{} 36e^{2x} \minus{} 8e^x \equal{} 4e^x[12e^{2x} \plus{} 9e^x \minus{} 2]$ has one possible quadratic root at

$ e^x \equal{} \frac { \minus{} 9 \plus{} \sqrt {177}}{24}$ and maybe here you can number crunch it?

no promises this is the right way to approach it...

$ 3e^{3x} \plus{} 8e^{2x \plus{} 1}(1 \minus{} x) \plus{} 6e^{x \plus{} 2} \plus{} 4e^3 > 0$

Breaking it into three cases, $ x<0$, $ x\equal{}0$, $ x>0$

Case 1: $ x<0$ - This is obviously going to be greater than 0 for all negative x with an asymptote at $ y\equal{}4e^3$

Case 2: $ x\equal{}0$ - We get $ 3\plus{}8e\plus{}6e^2\plus{}4e^3 > 0$ so this is true.

Case 3: $ x>0$ - Can use induction to show it is increasing. Can also Use induction, but then subtract the normal equation and check it is always positive.


What is the exact value of sin10*sin50*sin70?

[hide="Hint"]$\sin 10^{\circ}\sin 50^{\circ}\sin 70^{\circ}=\cos 80^{\circ}\cos 40^{\circ}\cos 20^{\circ}$ [/hide]

[quote="GlenthemanN"]would this make use of the double angle identity?[/quote]
Yes. The 20, 40, 80 are pretty suggestive.

[quote="GlenthemanN"]would this make use of the double angle identity?[/quote]
Try multiplying the equation by $\frac{\sin 20}{\sin 20}$

[quote="GlenthemanN"]yeah I feel really dumb I still dont get it, and ive tried making sense of alltheman's response and cant follow it either :(

ive tried using double angle formulas and just end up with really ugly equivalences..[/quote]
[hide]
$\sin 10^{\circ}\sin 50^{\circ}\sin 70^{\circ}=\cos 80^{\circ}\cos 40^{\circ}\cos 20^{\circ}$

$\implies \sin 10^{\circ}\sin 50^{\circ}\sin 70^{\circ}\sin 20^{\circ}=\cos 80^{\circ}\cos 40^{\circ}\cos 20^{\circ}\sin 20^{\circ}$

$\implies 2\sin 10^{\circ}\sin 50^{\circ}\sin 70^{\circ}\sin 20^{\circ}=\cos 80^{\circ}\cos 40^{\circ}\sin 40^{\circ}$

$\implies 4\sin 10^{\circ}\sin 50^{\circ}\sin 70^{\circ}\sin 20^{\circ}=\cos 80^{\circ}\sin 80^{\circ}$

$\implies 8\sin 10^{\circ}\sin 50^{\circ}\sin 70^{\circ}\sin 20^{\circ}=\sin 160^{\circ}=\sin 20^{\circ}$

$\implies \sin 10^{\circ}\sin 50^{\circ}\sin 70^{\circ}=\frac{1}{8}$. [/hide]

well...my goal was to get you to do some of the thinking yourself...my post reduces to showing that $\sin 160=8\cos 80\cos 40\cos 20\sin 20$

yes I realize :P then I got fed up and asked :blush:


Please help with the following problem:

$ \text{If }x\text{ and }y\text{ are two non\minus{}zero real numbers which satisfy the equation}$\[ x^2\plus{}y^2\equal{}(x\plus{}y)^3,\]$ \text{then }x\plus{}y\text{ is at least: }$
$ \text{A: }\,\dfrac{1}{2}\qquad\text{B: }\,1\qquad\text{C: }\,2\qquad\text{D: }\,\sqrt{2}\qquad\text{E: }\,\text{none of the provided}$

Clearly $ x\plus{}y \neq 0$ or $ x\equal{}y\equal{}0$, contradiction.

So $ (x\minus{}y)^2 \ge 0 \Leftrightarrow x^2\plus{}y^2 \ge \frac{(x\plus{}y)^2}{2}$

$ (x\plus{}y)^3 \ge \frac{(x\plus{}y)^2}{2} \Rightarrow x\plus{}y \ge \frac{1}{2}$

Equality holds $ \Leftrightarrow x\equal{}y\equal{}\frac{1}{4}$

So the answer is $ A$


inequalities
function
inequalities unsolved

Let $x$, $y$, $z$ be three real numbers satisfying $x^{2}+y^{2}+z^{2}=9$.
Prove that $2\left(x+y+z\right) \leq xyz+10$.

[i]Alternative version.[/i] For any three reals $x$, $y$, $z$ satisfying $x^{2}+y^{2}+z^{2}=9$, show that $2\left(x+y+z\right)-xyz\leq 10$.

It's VMO 02 ! The most simple proof use Cauchy- Schwart's ineq .
See also
http://www.mathlinks.ro/Forum/viewtopic.php?t=52
http://www.mathlinks.ro/Forum/viewtopic.php?t=42995

Cauchy? Please enlighten (=

I was wondering, I read somewhere that in inequalities if you let $a = x+y+z, b=x^2+y^2+z^2, c=x^3+y^3+z^3$, but now I reach $39a - 2c - a^3 \leq 60$, and I don't know what to do...

have you done another problem:
if x,y,z satisfy $x^{2}+y^{2}+z^{2}=2$,prove $x+y+z \leq xyz+2$
http://www.mathlinks.ro/Forum/viewtopic.php?t=32399
I think they are similar. :)

If $x^2+y^2+z^2=9$ then prove that $2(x+y+z)\leq xyz+10$
 :) The inequality is equivalent to $2\ge xyz+(x-1)^2+(y-1)^2+(z-1)^2$
Suppose the contrary $2\leq xyz+(x-1)^2+(y-1)^2+(z-1)^2$
Then we have 3 cases:
(i)$x$,$y\leq 1$,$z\ge 1$,then $z^2\ge 7=>(z-1)^2\ge 2$ contradiction.
(ii)$x\ge y\ge z\ge 1$ then $x^2\ge 3=>xyz\ge \sqrt{3}$ and also $(x-1)^2\ge 2-\sqrt{3}$ contradiction.
(iii)$x\ge y\ge 1\ge z$,then $2x^2\ge x^2+y^2\ge 8=> x\ge 2=>xyz\ge 2yz$ and$ (x-1)^2\ge 1=>1\ge 2z+(y-1)^2+(z-1)^2=(y-1)^+z^2+1\ge 1$ contradiction.
Very strange that the equality doesn't occur.

[quote="hardsoul"]If $x^2+y^2+z^2=9$ then prove that $2(x+y+z)\leq xyz+10$
 :) The inequality is equivalent to $2\ge xyz+(x-1)^2+(y-1)^2+(z-1)^2$
Suppose the contrary $2\leq xyz+(x-1)^2+(y-1)^2+(z-1)^2$
Then we have 3 cases:
(i)$x$,$y\leq 1$,$z\ge 1$,then $z^2\ge 7=>(z-1)^2\ge 2$ [b]contradiction.[/b]
[/quote]
Hmm...i don't understand your solution.Why...contradition?

You wrote wrongly , Hardsoul .
 the ineq is equivalent to :
 $ 2 \leq xyz + \sum{(x-1)^2} $
  You should remember that : $ x, y ,z \in R $
 The equality holds $ x=-1 , y=z=2 $

I'm not sure, but maybe this solution works, it's ugly anyway:

1) we look through case when one or more of the variables is equal to zero
2) we show that if x,y,z < 0 or y,z < 0 and x > 0 then xyz + 10 >= 2(x+y+z)
3) we assume x >= y >= z > 0 and show that then f(x,y,z) = xyz + 10 - 2(x+y+z) > f(x,y,-z)
    proof: f(x,y,z) - f(x,y,-z) = 2xyz - 4z >=? 0 <=> yx >= 2. Now let's suppose xy < 2, then
    z =< x < 2/y -> 9 = x^2 + y^2 + z^2 <y^2 + 8/y^2  <=> 9y^2 < y^4 + 8 <=> (y^2 - 1)(y^2 - 8) > 0
    <=> y^2 > 8 or y^2 < 1. So unless y < 1 we get a contradiction. But if y,z < 1 then 2(x+y+z) < 2*5 < 10 +xyz,
    So we can assume y >= 1.
4) we show that if x >= y > 0 and z < 0, |z| <= |y| <= |x| then the inequality holds:
    proof: Lets take a = -z (then 0 < a <= y <= x), then we have to show that 10 + 2a >=? 2x + 2y + axy.
    now (x-2)^2 + (y-2)^2 + (a-1)^2 >= 0 <=> 4x + 4y + 2a <= x^2 + y^2 + a^2 + 9 = 18
    <=> 2x + 2y + a <= 9 so we have to prove that 9 - a + axy <=? 10 + 2a <=> 1 + 3a >=? axy
    <=> 2 + 6a >=? 2axy, now 2xy <= x^2 + y^2 = 9 - a^2, so we have to prove that 2 + 6a >=? 9a - a^3
    <=> a^3 + 2 - 3a >=? 0 but this holds for a = (0;sqrt(3)) so this case is proved. And from here we also get the
    equality case x = y = 2, z = -1
5) now we have to look through cases when y < 0 or x < 0 with assumption |x| >= |y| >= |z|
    if x < 0, then 2(x+y+z) <= y + z <= 2sqrt(3) <= 10 - 3sqrt(3) <= 10 + xyz
    if y < 0, then let b = -y. We have to prove that 10 + 2b >=? 2x + 2z + xbz, so it's enough to prove
    that 10 >=? 2x + xbz <=> 20 >=? x(4 + 2bz), but we know that 2bz <= 9 - x^2. So we have to prove
    20 >=? 13x - x^3 <=> x^3 - 13x + 20 >=? 0 and if i'm not mistaken, this one should hold for x = (0;3)

I would be very grateful if someone could verify if this proof is correct or show me my mistakes

The shortest solution for this problem :
 Suppose $ /x/ \leq /y/ \leq /z/  \to z^2 \geq 3 \to 2xy \leq x^2+y^2 \leq 6 $
 We have :
  $ [2(x+y+z) - xyz]^2 = [2(x+y) + z(2-xy)]^2 \leq [(x+y)^2+z^2][2^2 + (2-xy)^2] = 100 + (xy+2)^2(2xy-7) \leq 100 $

nttu, what inspired you to find this excellent solution?

What a hard question ! :D  Actually , I don't know what .....

Suppose $|x| \leq |y| \leq |z|$ . So $z^2 \geq 3$ and $2xy \leq x^2+y^2 =6$
 Apply Cauchy Schwarz 's ineq , we have :
 $[2(x+y+z)-xyz]^2 = [2(x+y)+z(2-xy)]^2 \leq [(x+y)^2+z^2][4+(2-xy)^2] = 100+(xy+2)^2(2xy-7) \leq 100$
 The equality holds for $x=-1 , y=2 , z=2$

$2(x+y+z)-xyz\leq 10$
$\Longleftrightarrow 2\cdot9(x+y+z)-9xyz\leq 90$
$\Longleftrightarrow 2(x+y+z)(x^{2}+y^{2}+z^{2})-9xyz\leq 90.$

[quote="Kapi"][i]Alternative version.[/i] For any three reals $x$, $y$, $z$ satisfying $x^{2}+y^{2}+z^{2}=9$, show that $2\left(x+y+z\right)-xyz\leq 10$.[/quote]

Let $x\leq y\leq z.$ If $x,$ $y$ and $z$ are non-negative or $x\leq y\leq0\leq z$ it's triviality.
Let $x\leq0$ but $y$ and $z$ are non-negative numbers. Then
$2(x+y+z)-xyz\leq2\left(x+\sqrt{2(y^{2}+z^{2})}\right)-x\cdot\frac{y^{2}+z^{2}}{2}=$
$=2\left(x+\sqrt{2(9-x^{2})}\right)-\frac{x(9-x^{2})}{2}.$
But $2\left(x+\sqrt{2(9-x^{2})}\right)-\frac{x(9-x^{2})}{2}\leq10\Leftrightarrow$
$\Leftrightarrow(x+1)^{2}(x^{4}-2x^{3}-7x^{2}-24x+112)\geq0,$ which obviously true.

Here my solution :
$\large [2(x+y+z)-xyz]^{2}= [x(2-yz)+2(y+z)]^{2}[x^{2}+(y+z)^{2}]. [(2-yz)^{2}+2^{2}]$, so $[2(x+y+z)-xyz]^{2}\leq 2t^{3}+t^{2}-20t+72$, where $\|t\|= \|yz\|$.
Without loss of generality, assume that $\|x\|\geq\|y\|\geq\|z\|$, since $x^{2}+y^{2}+z^{2}= 9$, we have $\|t\| \leq 3$.
Consider the function $f(t) = 2t^{3}+t^{2}-20t+72$, t belongs to $[-3,3]$, we easily find that $maxf(t) = f(-2) = 100$, so we have that result.

I'm sorry about my poor English!!


On her trip to Paris, Helene spent $ \$150$ on gifts. If the French franc was worth $ \$0.2148$, how many francs did she spend? Round your answer to the nearest franc.

Simply put Helene paid 150/.2148 francs, this is about 698 francs.

This is quite hard without a calculator in my opinion, but since this is a Target Round they expect you to use calcs.


function
real analysis
real analysis unsolved

[b]Try to prove (or disprove) :[/b]
[color=green][b] for every function $f$ $: {\mathbb R} \to {\mathbb R}$ there is a dense set $D\subset {\mathbb R}$ such that the restriction $f \!\left|_{D}: D \to {\mathbb R}\right.$ is continuous. [/b] [/color]

[size=75][b]see: Henry Blumberg, Trans.Amer.Math.Soc., 24(1922) 113-128. [/b] [/size]

This problem has already been posted by sam-n (see [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=30964[/url]).

but still no solution.. can you give us a hint, profesor Lupas?

[quote="xirti"]but still no solution...[/quote]

Here is the proof I found. I wonder how it relates to the original proof.

We'll start with a simple

Lemma: Suppose that a set $E\subset\mathbb R$ is such that, for every open interval $I$, there exists an open subinterval $J\subset I$ such that $J\cap E$ is of the first category. Then $E$ is of the first category.

Proof: consider the set $\mathcal J$ of all open intervals $J$ with rational endpoints such that $J\cap E$ is of first category. Let $E'=E\setminus \Bigl(\bigcup_{J\in\mathcal J}J\Bigr)$. The condition of the lemma implies that $E'$ is nowhere dense. But $\bigcup_{J\in\mathcal J}(J\cap E)$ is a countable union of sets of the first category and, thereby a set of the first category. The rest should be clear.

Let us call a set $D\subset\mathbb R$ [i]very dense[/i] if the intersection of $D$ with every open interval is of the second category. Our next step is to prove the following

Proposition: Let $D$ be a very dense set and $f : D\to\mathbb R$ be an arbitrary function. Then there exists a very dense set $D'\subset D$ and a point $p\in D'$ such that $f$ restricted to $D'$ is continuous at $p$. Moreover, we can arrange for the point $p$ to be contained in any given open interval.

Proof: Let $\varepsilon>0$. We shall call a point $x\in\mathbb D$ $\varepsilon$-regular if there exists $\delta>0$ such that, for every open subinterval $J$ of the interval $(x-\delta, x+\delta)$, the set $\{y\in J\cap D\, : \,|f(y)-f(x)|<\varepsilon\}$ is of the second category and $\varepsilon$-exceptional otherwise. Note that, to prove the proposition, it would suffice to find a point $p\in D$ that is $\varepsilon$-regular for every $\varepsilon>0$. Indeed, then we can take a sequence $\varepsilon_k$ decreasing to $0$, take the corresponding $\delta_k$, which, without loss of generality, we may consider decreasing to $0$ as well, and define $D'$ by the condition that $y\in D'$ if either $y=p$ or $y\in D$; $|y-p|\ge \delta_1$, or $y\in D$; $\delta_{k+1}\le |x-y|<\delta_k$; and $|f(y)-f(p)|<\varepsilon_k$. By construction, $D'$ is very dense and $f$ restricted to $D'$ is continuous at $p$. Now, to find a desired point $p$, it suffices to show that, for every $\varepsilon>0$, the set of $\varepsilon$-exceptional points is of the first category. To this end, let $D_k=\{x\in D : k\varepsilon\le f(x)<(k+1)\varepsilon\}$ and let $E_k=\{x\in D_k : x\text{ is }\varepsilon\text{-exceptional}\}$. We want to check that $E_k$ satisfy the assumptions of the lemma. Let $I$ be any open interval. If $I\cap E_k=\varnothing$, we can take $J=I$. Otherwise, there is some $x\in I\cap E_k$. Since $x$ is $\varepsilon$-exceptional and $I$ is an open neighborhood of $x$, there is an open subinterval $J\subset I$ such that $J\cap D_k$ is of the first category (note that all points $y\in D_k$ satisfy the condition $|f(y)-f(x)|<\varepsilon$. But then $J\cap E_k$ is also of the first category and we are done.

Now, let $I_k$ be some enumeration of open interval with rational endpoints. Using the proposition again and again, we can construct a sequence of very dense sets $\mathbb R=D_1\supset D_2\supset D_3\supset\dots$ and points $p_j\in I_j\cap D_j$ such that $f$ restricted to $D_j$ is continuous at $p_j$ for all $j$. Clearly, the set $P=\{p_j\}_{j\ge 1}$ is dense and $f$ restricted to $P$ is continuous.

I hope I haven't made a stupid mistake somewhere. :)  Note also that we constructed just a [i]countable[/i] dense set with the desited property. Whether we can construct a set of larger cardinality, remains unclear to me. :?


geometry
3D geometry
pyramid
linear algebra
matrix
calculus
integration

Consider the following equation pyramid:

$
\begin{matrix*}

3^2 + 4^2 = 5^2 \\
10^2 + 11^2 + 12^2 = 13^2 + 14^2 \\
21^2 + 22^2 + 23^2 + 24^2 = 25^2 + 26^2 + 27^2

\end{matrix*}
$

Is it possible to continue this idea that in the $n-$th equation ($n \geq 4$) we have $2 \cdot n + 1$ consecutive squares that the first $n+1$ squares are arranged on the LHS and the other $n$ squares occur on the RHS ? If it is possible please provide the $n-$th equation dependently on $n.$

[quote="tetrahedr0n"]Very nice problem.

[hide]Let's write this equation:

$(a-n)^2+(a-n+1)^2+...+a^2=(a+1)^2+...+(a+n)^2$.

We try to solve for $a$ in terms of $n$. Thus,

$(a-n)^2 = ((a+n)^2-a^2)+...+((a+1)^2-(a-n+1)^2)$ or
$(a-n)^2=n(2a+n+...+2a-n+1)$
$(a-n)^2=n(n(2a)+n)$
$(a-n)^2=n^2(2a+1)$
$a^2-2an+n^2=n^2=2an^2$
$a^2=2an^2+2an$
$a=2n^2+2n$
$a=2n(n+1)$

Thus it is sufficient to plug this clearly positive integral $a$ into our first equation, giving us an affirmative answer to the question.[/hide][/quote]

That is right.  :)


[quote="stephenyamqc"][hide]Is it 19603+19605+...+19997+19999?[/hide]

 :roll:  :roll:[/quote]
thats whatI got. Here:s how you do it.  After the $n^{\text{th}}$ term, there have been $n^2$ odd numbers. So after $99$ terms, the last odd number is $2\cdot99^2-1=19601$. So the starting odd for the $100^{\text{th}}$ term is $19603$. In the $n^{\text{th}}$ term, there are $2n-1$ odd numbers, so for $100$ there are $199$ odd numbers, the last one being $2\cdot199-1+19602=19999$

i got the same thing as ifai, except i did it this way:

the sum of whatever term is just the number of terms in there, times the middle term.

The nth term is going to have $2n-1$ terms. The nth term's middle term is going to be $1+4\frac{(n-1)n}{2}$

Plugging in 100:

$199(1+200*199)$

which is equivalent to ifai's expression

oh, i'm sorry.  :blush:
i misunderstood that [b]stephenyamqc[/b] and [b]nat mc[/b]'s answer is $\sum_{n=19603}^{19999}n$, not 19603+19605+19607+...+19999.

[quote="JesusFreak197"]19603, the first number in the sequence that will sum to the 100th term, is $2\cdot 99^2+1$, and 19999 is $2\cdot 100^2-1$. I don't think that's a coincidence. They should yield the same answer.[/quote]
and thanks for your reminder! :)


geometry
circumcircle
inequalities
cyclic quadrilateral
geometry proposed

$ ABC$ is a triangle such that $ AB \equal{} AC.$ Let $ D$ be a point on $ AB$ and let $ E$ be a point on $ AC$ produced such that $ DE \equal{} AC.$ Suppose $ DE$ meet the circumcircle of triangle $ ABC$ at $ T.$ Let $ P$ be a point on $ AT$ produced. Prove that $ PD\plus{}PE\equal{}AT$ if and only if $ P$ is on the circumcircle of triangle $ ADE.$

now as T is not equal to any of A, B or C. and T lies on the circumfernce of circumcircle ABC.  we surely have a cyclic quadrilateral i.e ACBT. we can see in triangle PDE and have the inequality PD+PE >DE. let DE=b=AB=AC. now, as from question we have that PD+PE=AT. thus we get AT>b. if we look in triangle ATB. and we alrady have AT>b=AB. thus angle ABT>angle ATB. again as ATB is 180-angle AEB. thus we get angle ABT+AEB> 180. which is impossible. thus now there is a confusion is position of P such that PDE does not form a triangle. if P does lie on the circuference then itss not possible. some confusion but maybe other mathlinkers can help. thanks !

typo and a slight mistake. its all right until we obttain from previous solution that AT>AC=AB. Thus now we can prove the impossibilty. as ACBT is cyclic. and 180-ATB>TBA. but also we have TBA>ATB. we get ATB acute. and thus we get ACB obtuse which impossible.


pigeonhole principle
ceiling function
graph theory

Prove that any graph with $ 11$ vertices, colored in three colors, contains a quadrilateral of same color.

[hide="Fast"]
By Pigeonhole, there exists at least one color for which there are at least $ \lceil 11/3 \rceil \equal{} 4$ points. These points can form a quadrilateral.
[/hide]

[quote="JBL"]Most likely, the intent was to color the edges rather than the vertices.[/quote]

Yes, you're right  :D

Is it a complete graph?  Otherwise I'm not sure what we're supposed to do, because we don't know anything about the number of edges.

Sorry for being unclear... it is a complete graph.


algebra polynomial quadratics Vieta symmetry

Find all solutions $(x,y)$ to
\[ \\x^5+y^5=33\\x+y=3 \]
Masoud Zargar

i believe one way to do it is to take $\frac{x}{y}=k$, and eliminate x and y, then you have that it $k$ is a solution, then $\frac{1}{k}$ is a solution, so it could be factored as $(k^2+ak+1)(k^2+bk+1)(k\pm1)$

$(1,2)$ is clearly a solution. Also \[ 11 = x^4-x^3y+x^2y^2+xy^3+y^4, \] but I don't think it helps much. $x,y\in\mathbb{Z}$?

[hide="hint"]
Symmetric Polynomials
[/hide]
Masoud Zargar

it actually is easier than i thought...

$x^5+(3-x)^5=33$

expand that, and factor out $(x-1)(x-2)$

then the rest is a quadratic that is easy to solve

[quote="luimichael"]
[b]Case 1[/b]
$xy=2$ and $x+y=3$
Solving $T^2-3T-2=0$ :? , we have $T=1 , 2$.
[/quote]

here is another solution

$x^5+(3-x)=33$

suppose that $r$ is a root, then $3-r$ is a root also

pairing those 2 factors, we have that it can be factored into (using vieta)

$15(x^2-3x+a)(x^2-3x+b)$

then expand and compare coefficients

The right answer is  $(x,y)=(1,2),\ (2,1),\ (\frac{3+\sqrt{19}i}{2},\ \frac{3-\sqrt{19}i}{2}),(\frac{3-\sqrt{19}i}{2},\ \frac{3+\sqrt{19}i}{2}).$

Remark the symmetry of the given expression for $x,y$ we set $xy=a,$ we have
$x^5+y^5=(x^3+y^3)(x^2+y^2)-x^2y^2(x+y)$
$=\{(x+y)^3-3xy(x+y)\}\{(x+y)^2-2xy\}-(xy)^2(x+y).$
Plugg $x+y=3,\ xy=a$ into this equation, we obtain $x^5+y^5=15a^2-135a+243=33\Longleftrightarrow a^2-9a+14=0,$ yielding $a=2,7.$ Then remark that $x,y$ are roots of the quadratic equation with respect to $t\ : t^2-3t+a=0.$ which follows [b]luimichael[/b]'s solution.


It's very easy because it's weaker than my old result:
$ \sum\ \frac{a}{a^2\plus{}2bc} \leq\ \frac{a\plus{}b\plus{}c}{ab\plus{}bc\plus{}ca}$
See the proofs here:
http://www.mathlinks.ro/Forum/viewtopic.php?t=224981