function
number theory
number theory proposed

How many solutions has the equation

$a+2b+5c+10d+20e+50f+100g+200h=500$  where  $a,b,c,d,e,f,g,h$  are positive integers

Silouan, do you know the answer?

Using dynamic programming I found that there are $6,295,434$ solutions

If it is correct, I'll show my way

yes this is the correct answer and this is the only I know(only the result)

Could you post a full solution ?Thank you

Silouan, μπορείς να δεις τη λύση και   [b][url=http://www.mathlinks.ro/Forum/viewtopic.php?t=74799]εδώ[/url] [/b], γραμμένη στα ελληνικά

I rewrite the equation like this: $\sum_{i=1}^{8}{a_i \cdot x_i}=500$,

where
$a_i$ are the numbers $1,2,5,10,20,50,100,200$ and
$(x_1,x_2, \cdots, x_8)$ is any possible solution with $x_i\geq 0$

The variables $x_i$ are not [u]positive[/u] integers, but they can have the value $0$
(or else the problem reduces to $A+B+4C+9D+19E+49F+99G+199H=112$)

I suppose that this problem arose from this:
[color=blue]In how many ways can we obtain 500 euros if we use coins of 1,2,5,10,20,50,100, and 200€[/color]

Maybe there exist other solutions using combinatorics or number theory, I don't know.

As I said, I solved it with Dynamic Programming, which is one of my favorites!!! :)

Of course, I couldn't do the calculations by hand (because I had enough things to calculate), so I used Microsoft Excel.

I will try to explain my solution, as well as I can.

[color=blue][b]Solution[/b][/color]

I define the function: $f(n,k) = \{ \mbox{The number of solutions of the equation} \sum_{i=1}^{n}{a_ix_i}=k\}$

So, $f(n,k)$ means in how many ways we can obtain the sum $k$ using only the coins $a_1, \cdots, a_n$

Since there is only one solution to the equation $a_1x_1=k$, we have $f(1,k)=1,\ \forall k\geq 0$

Let's find $f(n,k)$ for $n>1$.

$\blacksquare$ If $k<a_n$, then we have $f(n,k)=f(n-1,k)$. Indeed:

If $x_n>0$, then $\sum_{i=1}^{n}{a_ix_i}\geq a_n>k$.

So, it is necessary that $x_n=0$.

But for $x_n=0$, the equation $\sum_{i=1}^{n}{a_ix_i}=k$ becomes $\sum_{i=1}^{n-1}{a_ix_i}=k$

And we defined that this equation has $f(n-1,k)$ solutions

$\blacksquare$ If $k>a_n$, then $f(n,k)=f(n-1,k)+f(n,k-a_n)$. Indeed:

In $\sum_{i=1}^{n}{a_ix_i}=k$ we could have either $x_n=0$ or $x_n>0$.

For $x_n=0$ we have $f(n-1,k)$ solutions, because the equation reduces to $\sum_{i=1}^{n-1}{a_i x_i}=k$

If $x_n>0$, we can make a reccursive secuence.

$\sum_{i=1}^{n}{a_i x_i}=k\Rightarrow$

$\sum_{i=1}^{n-1}{a_i x_i}+a_n(x_n-1)=k-a_n \ \ (1)$

The equation $(1)$ is the same as $\sum_{i=1}^{n}{a_i y_i}=k-a_n$ (with a proper substitution), which has $f(n,k-a_n)$ solutions.

So we have:
$f(n,k) = \{\begin{array}{ll} f(n-1,k) & \mbox{if }k<a_n \\ f(n-1,k)+f(n,k-a_n) & \mbox{if }k\geq a_n \end{array}$

[quote]
[b]For example:[/b] [color=red]In how many ways we can get a total of 11€, using only the coins of 1,2 and 5€ ?[/color]

$f(3,k) = \{\begin{array}{ll} f(2,k) & \mbox{if }k<5 \\ f(2,k)+f(3,k-5) & \mbox{if }k\geq 5 \end{array} \Rightarrow f(3,11) = f(2,11)+f(3,6)$

That is, we can obtain 11€ in two ways:

[list] use noone coin of 5€ and then we have to obtain 11€ by using only the coins of 1 and 2€ (so there are $f(2,11)$ ways to do that)[/list]

[list] use one coin of 5€ and we can obtain the rest 6€ (using the three types of coins) with $f(3,6)$ ways [/list]

We can use again and again the reccursive function so that we can finally express $f(3,11)$ in terms of $f(2,something)$.

Finally, all are expressed in terms of $f(1,x)$, which is $1$[/quote]

In my attachment, I created an array with 501 rows and 8 columns

 - Each row is for a different amount from 0 to 500€.
 - Each column is representing the coin

The array has the value $f(n,k)$ in the place $(k,n)$

The column of [b]1€[/b] is filled by [b]1[/b]s, because $f(1,k)=1$.

Take some other column
The column of [b]5€[/b], for example, consist of two parts:

The coloured part is for $k<5$ so we must have $f(3,k) = f(2,k)$.
In the cell E7 the code is "=D7"

The non-colored part is the case $k\geq 5$. So we have to write in these cells the formula $f(3,k) = f(2,k)+f(k-5)$

If you see on the code of the cell E8, it is "=D8+E3".

The answer to the problem is the value $f(8,200)$, which is on the right-bottom cell of the array

[color=red]P.S.

I attached an image so that you can preview the contents of the Excel document[/color]


hey, try this game. i dunno if it's ever turned up anywhere befr,
 but i sure liked it.
 the idea is for person to list three words, preferably nouns,
 and the next person tries to make a funny short  sentence out of those wods.
 then he suggests three more words an' this goes on.
 try to maintain some kind of continuity. u r allowed to make it a short para.

 I'll start with the words:
 how abt
 $chicken$, $Malaysia$ and $sky$??

'are chickens raining frm the sky over malaysia?'

football , brett lee , drain.

(don't make 'brett lee kicked the football into the drain'....be creative!!!??
 :lol: )

Brett Lee drained the bathtub and went out to play foot ball.

rubik's cube, jellybeans, online

I ate jellybeans while looking at rubik's cube times online.

Ok, that one was lame...
someone else go

GO to your room, or ELSE, SOMEONE gonna get hurt real bad...

banana, hammock, emu

that crazy emu is eating your banana, so go catch it with the hammock

yesterday box tug-of-war

[b]yesterday[/b], i had a fight with a bum over a cardboard [b]box[/b] on the street and we ended up having a [b]tug of war [/b]with the box. In the end, i lost, o well, gotta find a new home now.

Vanilla, Handball, Porno

It's fun to consume vanilla extract while watching porno and squeezing a handball...(for anyone who doesn't get that, it's better if you don't...but Chocoboom, you asked for it by putting up porno as a word)

Stone, stab, sleep

I will stab that person and then sleep on a stone.

paper, tree, baked potatoes

unlike baked potatoes, paper comes from trees (potatoes grow on potato bushes, obviously).

fire, 7, monkey

At building #7, a monkey made a fire.

writing, tree, elf

(don't say "The elf saw writing on the tree." be creative.)

the elf [b]didn't[/b] see writing on the tree (just kidding  :rotfl: )

as the elf was writing his will, he couldn't remember if he had friends/relatives, so he left all his stuff to a tree.

strange, mammoth, octopus

A strange mammoth saw an octopus in the forest.

owl, rain, piece of glass.

the owl was surfing in the rain on a piece of glass. (now there's an interesting image  :lol: )

computer, share, pillow

"Share the computer with your pillow!"

cold blanket, personal space, origami sheet

So i was feelin a little over zeolous so i says to erkle i says eat the camel dung and hes all like nuuuu

math, ghetto, banana hammock

I played a math game with Ghetto on a banana hammock.

itchy, knee, sun

if your knee is itchy, fly into the sun in a spaceship.

games, fun, factory

The games and fun factory is cool!!!!

one, two, three

u said " one two three"!!!!!

milkshake, milk, iceage :P

rule 12358: Do not double post.

In the iceage, milkshakes weren't as popular as milk.

rule, mile, court

since the court was a mile long, the jury couldn't hear some of the testimony and fell asleep when they should have been ruling.

light, house, ninja  :ninja:

sorry for reviving AND double posting, but this topic is too funny to waste!

[quote="nutz_for2.718281828"]A ninga took over the light house.

silicosis, the moon, Atlantis[/quote]
Aliens from Atlantis, the moon, have silicosis (a main component of sand).

smiley, bread, morning

[quote="i_like_pie"]Until people start questioning the fact that it's funny, it's funny.[/quote]
so it's not funny anymore right?

A smiley guy has bread in the morning.

complexion, sublime, meaning


algebra
polynomial
limit
complex analysis
function
complex analysis unsolved

Here a nice one that I found:
Let $ c_k$ non negative such that for all $ r > 1$, $ \lim_{k\rightarrow \infty} c_kr^k \equal{} 0$
Prove that $ f(z) \equal{} \sum_{k \equal{} 0}^\infty c_kz^k$ is a polynomial.

Note: for $ r > 1$ since $ c_k(2r)^k$ is bounded, say by $ C$, then $ c_kr^k\leq \frac {C}{2^k}$, this means that the series converges in the hole plane, then it defines an holomorphic function, so makes sense to ask when it's a polynomial.

consider $ c_{k}\equal{}\frac{1}{k!}$ then for all $ r>1$ we have $ \lim_{k}c_{k}r^{k}\equal{}\lim \frac{r^{k}}{k!} \to 0$
now consider $ f(r)\equal{}\sum c_{k}r^{k}$ it's wellknown that $ f(r)\equal{}e^{r}$ thus it's not polynomial. :(

I think that it's never a polynomial unless there exists $ k$ such that for every $ i>k$ we have $ c_i \equal{} 0$. Or am I wrong?


The [url=http://www.imo-2008.es/]International Mathematical Olympiad 2008 in Madrid, Spain[/url] takes place from 10th July to 22nd July 2008. There are 97 participating countries and 535 students. The maximum possible score is 42, i.e. 7 points for each of the 6 problems. 47 gold medals were awarded to student 31 or more points, 100 silver medals to students with 22 or more points, 120 bronze medals to students with 15 and more points. 103 students received a honourable mention (who have solved at least one problem completely). The [url=http://www.imo-2008.es/paises/USA.html]American delegation (students, leaders, observers)[/url] was led by Zuming Feng and Razvan Gelca. The [url=http://www.imo-official.org/team_r.aspx?code=USA&year=2008]USA achieved team rank 3.[/url] The team rank is an inofficial cumulative score of all contestants' score of the respective country. [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=2426]Alex Zhai[/url] got a perfect score of 42 points, i.e. solving all six problems completely.

[b]Alex Zhai[/b],	42 points, gold medal
[b]Colin Sandon[/b],	35 points, gold medal
[b]Krishanu Roy Sankar[/b],	 32 points, gold medal
[b]Shaunak Kishore[/b], 32 points, gold medal
[b]Evan O'Dorney[/b],	 26 points, silver medal
[b]Paul Christiano[/b],	 23 points, silver medal

oh, congratulations to the USA team!!  everyone medaled!!

Whoa perfect! He's a beast!

I'm surprised Evan didn't get a gold though

"I don't like setting specific numbers as goals, but I'd like to try to get a perfect score on the IMO."

Apparently a success!

Other noteworthy mentions.

- The US team got a perfect 42 on problem 5. I, a lot of MOPpers here at Mathcamp, and probably our IMO team too, thought this problem was very easy and misplaced (when you can get a problem in your head...). But a lot of other teams seem to have screwed up on it. China had someone 0 it. I can't figure out where its challenge came from.
- Only thirteen people received more than 2 points for problem 6 in the whole IMO - there was one 6 and twelve 7s. Zhai and Krishanu were two of the 7s. This also pushed our score up a lot.
- There were three perfect scores. Two were from China. Alex is the third.

Big congratulations to the team.

Good job Alex! I was expecting a perfect score :wink:

also, Krishaunak lived up to their name by tying each other :D

Good job to USA!

Congrats to Evan O'Dorney, being as young as he is!

I KNOW HIM FROM MY MATH CIRCLE!!!!!!

Evan also won BAMO last year, bay area math olympiad.

So Reid Barton's record of four golds survives again... but congratulations to the whole team! Excellent finish; there was no way to reach China (217?!?), and while Russia was within reach, you all did very well. Of course Krishaunak tied, what else could happen with them? And congratulations to Colin on his perfect first day! Oh yeah, forgot about Alex...top 3 isn't too bad.

I still haven't gotten a chance to do the problems and see if I would have helped at all, but it looks like that's going to be tough. We haven't gotten four golds since 2005...good job again!

[quote="SamE"]So Reid Barton's record of four golds survives again[/quote]

I don't know you always seems to offer us some semi-knowledge. The case with Reid Barton is true but actually some people think that Christian Reiher is the leader who also got four gold medals AND a bronze one:

- http://www.mathematik-olympiaden.de/IMOs/topimo.html

The Hall of Fame is here: [url]http://www.imo-official.org/hall.aspx[/url]
I think SamE was talking about the IMO record held by a USA contestant.

P6 was really tough for my country, Korea :(
congrats to those who participated!

[quote="MellowMelon"]- The US team got a perfect 42 on problem 5. I, a lot of MOPpers here at Mathcamp, and probably our IMO team too, thought this problem was very easy and misplaced (when you can get a problem in your head...). But a lot of other teams seem to have screwed up on it. China had someone 0 it. I can't figure out where its challenge came from.[/quote]
I thought #2 was a little more guilty of this than #5, although it seems to have tripped up some people.  #6, on the other hand, is exactly where it should be :|

[color=blue][size=200][b]Welcome everybody to the next IMO 2009 in Germany!!![/b][/size][/color]

[quote="Valentin Vornicu"]You can notice that from the first 7 hall of fame IMO people, 4 are German :)[/quote]

And what do you think is the German trainer of [url=http://www.imo-official.org/participant_r.aspx?id=15545]Lisa Sauermann[/url] and [url=http://www.imo-official.org/participant_r.aspx?id=15547]Georg Schröter[/url] these days?  :D It is our own [url=http://www.imo-official.org/participant_r.aspx?id=10024]Wolfgang Burmeister![/url]  :)

The host of the next [url=http://www.imo2009.de/index.html]50th International Mathematical Olympiad 2009[/url] will be in [b]Germany.[/b] It will be hosted at the rising [url=http://math.jacobs-university.de/]Jacobs University Bremen.[/url] It will also be the 20th anniversary after the last IMO in Braunschweig, Germany in 1989.

[url=http://www.unl.edu/amc/e-exams/e9-imo/e9-1-imoarchive/2008-ia/2008imo-teamresults.shtml]Here[/url] you can find the 2008 IMO Team Results on the official USA math olympiad web page.


calculus
integration
real analysis
real analysis unsolved

Assume $f_{i}\in L^{p}$ and $f_{i}\to f$ in $L^{p}$. Show that if $f_{i}\to g$ a.e. then $f=g$ a.e.

Also if $f_{i}\in L^{p}\cap L^{q}$, $\parallel f_{i}-f\parallel _{p}\to 0$, and $\parallel f_{i}-g\parallel \to 0$ then $f=g$ a.e.

The problem is a simple application of the following $L^{p}$ integration theorem.

If $1\leq p\leq \infty$ and if $f_{n}$ is a Cauchy sequence in $L^{p}(\mu)$ with limit $f$, then $f_{n}$ has a subsequence which converges pointwise almost everywhere to $f(x)$.

Apply this theorem and your problem is solved.

And its proof can be found for example in Walter Rudin's book Real and complex analysis on pages 67-68.


geometry
rhombus
perimeter
trigonometry
LaTeX
geometry proposed

Let $ ABCD$ be a rhombus with $ \angle A\equal{}60^{\circ}$ and $ P$ the intersection of the two diagonals. Let $ Q,R,S$ be three points on the perimeter of the rhombus. If $ PQRS$ forms a rhombus, prove that exactly one of $ Q,R,S$ coincide with a vertex of $ ABCD$.

[size=150]first case
IF Q, R belong to AB, BC respectively.
A/Q.  considering Q, R,S not equal to any of A or B.
since QRS is the ext. angle. we get angle.QRS>B=120. Eventually the angles add up to more than 360.

second case
considering Q, R, S belongs to AD, AB, BS resp.
we use sine formula. DP/sinDQP = QP/sin60 and PS/sin60 = PB/sinPSB. thus we get ANG. DQP=angle PSB
if we look at polygon AQPSB and add up the angles we eventually get QPS as 180. Thus proved[/size]

 where can i download latex program?


AMC
USA(J)MO
USAMO
inequalities unsolved
inequalities

[quote="2006"]Let $\ x,y,z,a,b$ be real numbers interval$\ [p;q]$ and $\ p,q\leq0$
show that $\ (x+y+z+a+b)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{a}+\frac{1}{b})\leq 25+6(\sqrt{\frac{p}{q}}-\sqrt{\frac{p}{q}})^{2}$[/quote]
Are you sure it should be $\sqrt{\frac{p}{q}}-\sqrt{\frac{p}{q}}$? :maybe:

Well, $\sqrt{\frac{p}{q}}-\sqrt{\frac{p}{q}}=0$. Surely this can't be right... (since $p$ and $q$ cancel each other out and $(x,y,z,a,b)$ can take any values)   :rotfl:

If you replace $\sqrt{\frac{p}{q}}-\sqrt{\frac{p}{q}}$ with $\sqrt{\frac{p}{q}}-\sqrt{\frac{q}{p}}$ then you get a USAMO 1977 question which is also in Engel.
You can find the solution on [url=http://www.kalva.demon.co.uk/usa/usa77.html]Kalva.[/url]


calculus
integration
trigonometry
real analysis
real analysis unsolved

hello, sorry, it was the wrong post.
Sonnhard.

Please stop spouting meaningless answers like that, Sonnhard. A complicated expression spit out by a computer algebra system just isn't helpful. [The post this was referring to has since had its content removed]

Now, can we actually get somewhere?

Since there's no elementary antiderivative, the special limits are our only hope. First, note that $\cos(2x+2\sin 3x)$ is even, so $I=\int_0^{\pi}\cos(2x+2\sin 3x)\,dx = \frac12\int_{-\pi}^{\pi}\cos(2x+2\sin 3x)\,dx$.

That's an integral over a full period. Now we have some tricks. Trick #1: Shift it over. Substitute $t=x+\pi$ to get
$I=\frac12\int_0^{2\pi}\cos(2t-2\pi+2\sin(3t-3\pi))\,dt=\frac12\int_0^{2\pi}\cos(2t-2\sin 3t)\,dt$
$I=\frac12\int_{-\pi}^{\pi}\cos(2t-2\sin 3t)\,dt$
Combining the two forms and using the angle addition/subtraction identities for $\cos$, we get
$2I=\int_{-\pi}^{\pi}\cos(2x)\cos(2\sin 3x)\,dx$

Now, we shift again; let $u=x+\frac{2\pi}{3}$ and $v=x+\frac{4\pi}{3}$. We get
$2I=\int_{-\pi/3}^{5\pi/3}\cos\left(2u-\frac{4\pi}{3}\right)\cos(2\sin 3u - 2\pi)\,du$
$ = \int_{-\pi/3}^{5\pi/3}\cos\left(2u-\frac{4\pi}{3}\right)\cos(2\sin 3u)\,du$
$2I=\int_{\pi/3}^{7\pi/3}\cos\left(2v-\frac{8\pi}{3}\right)\cos(2\sin 3v - 4\pi)\,dx$
$ = \int_{-\pi/3}^{5\pi/3}\cos\left(2v-\frac{2\pi}{3}\right)\cos(2\sin 3v)\,dv$

Add them up, and
$6I=\int_0^{2\pi}\left(\cos 2x+\cos\left(2x-\frac{2\pi}{3}\right)+\cos\left(2x-\frac{4\pi}{3}\right)\right)\cos(2\sin 3x)\,dx$
$6I = \int_0^{2\pi} 0\,dx = 0$

Alternately, I could have expanded $\cos(2\sin 3x)$ in a series and appealed to Fourier theory. To generalize slightly,
$\int_0^{2\pi}\cos(ax+b\sin(cx))\,dx = 0$ whenever $a$ and $c$ are integers such that $a$ is not a multiple of $c$.


logarithms
inequalities
geometry
3D geometry

The most stupid solution : $x \cdot \ln x$ is convex + Jensen.

More inspired solutions may use Chebyshev or rearangement I think

I read this solution (not mine) one day, I don't remember where but I found it very smart  :o :

Wlog $a \geq b \geq c$ and taking cube of ineq :

$\frac{a^{3a}b^{3b}c^{3c}}{ (abc)^{(a+b+c)} } = (a/b)^{a-b} (b/c)^{b-c} (a/c)^{a-c} \geq 1$

Isn't it cool ?


induction
modular arithmetic
number theory
number theory proposed

Prove that for all odd $ n > 1$, we have $ 8n \plus{} 4|C^{4n}_{2n}$.

[quote="N.T.TUAN"]We have $ \dfrac{(2n \plus{} 2)\cdots (4n)}{(2n)!}$[/quote]
I think you must prove this number is an integer . An other approach :
$ \frac{{{4n} \choose {2n}}}{2n\plus{}1}\equal{}{{4n}\choose {2n}}\minus{}{{4n}\choose{2n\plus{}1}}$  ,thus it must be integer .
$ v_2((4n)!)\equal{}{4n\minus{}S(4n)}\equal{}4n\minus{}S(n)$
$ v_2((2n!)^2)\equal{}2(2n\minus{}S(2n))\equal{}2(2n\minus{}S(n))$
where $ S(n)$ is some digit of $ n$ when written in base 2. Because $ n$ is odd and $ n$ is great than $ 1$ so $ S(n)\geq 2$ . Thus $ 4|(\frac{4n}{2n})$ .
Note : In this problem I have used an old result in number theory .
$ v_p(n!)\equal{}\frac{n\minus{}S_p(n)}{p\minus{}1}$ where $ v_p(n)$ is the exponent of $ p$ and $ S_p(n)$ is sum digits of $ n$ in base p .

[quote="TTsphn"]
I think you must prove this number is an integer . An other approach  [/quote]
But it is easy. :D

[quote="N.T.TUAN"]We have $ C_{4n}^{2n} \equal{} \frac {(4n)!}{(2n)!(2n)!} \equal{} (2n \plus{} 1)\cdot \dfrac{(2n \plus{} 2)\cdots (4n)}{(2n)!}$ is divisble by $ (2n \plus{} 1)$. But it is easy to see that $ 4|C_{4n}^{2n}$, therefore we are done.[/quote]

Nice, but i see a typo $ C^{2n}_{4n}\not\equal{} C^{4n}_{2n}$ :P

To mayhem: Try induction to prove that $ \dfrac{(2n\plus{}2)\cdot...\cdot (4n)}{(2n)!}$ is an integer

[quote="TTsphn"][quote="N.T.TUAN"]We have $ \dfrac{(2n \plus{} 2)\cdots (4n)}{(2n)!}$[/quote]
I think you must prove this number is an integer . [/quote]
But this is just a Catalan number!

Vikernes:  was that a joke?  It's common notation in some countries.

[quote="t0rajir0u"]Vikernes:  was that a joke?  It's common notation in some countries.[/quote]

He means that some people denote $ \binom{a}{b}$ by $ C_a^b$, while others call it $ C_b^a$ (though all books I know use $ C_a^b$ only, but I have seen both notations on MathLinks - maybe the second one is a mistake every time it is posted). Anyway, there is no reason to use the $ C$ notation on the internet - we have enough space here.

  darij

[quote="t0rajir0u"]
Vikernes:  was that a joke?  It's common notation in some countries.[/quote]

Mmm i know that $ C^{n}_{k}\equal{}\binom{n}{k}$, but i thought that $ C^{2n}_{4n}\equal{}\binom{2n}{4n}$ also.

I mean that $ C_n^k\equal{}\frac{n!}{k!(n\minus{}k)!}$.

WOW!!
beautiful fact that  $ v_p(n!) \equal{} \frac {n \minus{} S_p(n)}{p \minus{} 1}$
I didn't know that.
SO I try this by lucas' theorem

by lucas' theorem
$ {{2m} \choose {2k}} \equiv {{m} \choose {k}}\pmod {2}$
$ {{2m} \choose {2k \plus{} 1}} \equiv 0 \pmod {2}$

$ {{4n} \choose {2n}} \equal{} 2^{4n} \minus{} 2({{4n} \choose {0}} \plus{} {{4n} \choose {1}} \plus{} \cdots \plus{} {{4n} \choose {2n \minus{} 1}})$

now we show $ {{4n} \choose {0}} \plus{} {{4n} \choose {1}} \plus{} \cdots {{4n} \choose {2n \minus{} 1}} \equiv 0 \pmod {2}$
$ {{4n} \choose {0}} \plus{} {{4n} \choose {1}} \plus{} \cdots {{4n} \choose {2n \minus{} 1}}$
$ \equiv {{2n} \choose {0}} \plus{} {{2n} \choose {1}} \plus{} \cdots \plus{} {{2n} \choose {n \minus{} 1}}$
$ \equiv {{n} \choose {0}} \plus{} {{n} \choose {1}} \plus{} \cdots \plus{} {{n} \choose {\frac {n \minus{} 1}{2}}}$
$ \equiv 2^{n \minus{} 1}$
$ \equiv 0\pmod {2}$

let $ \frac {(2n \plus{} 2)\cdot...\cdot (4n)}{(2n)!} \equal{} k$

then $ (2n \plus{} 1)k \equal{} C^{4n}_{2n} \in Z$

also $ (4n \plus{} 1)k \equal{} C^{4n \plus{} 1}_{2n} \in Z$

$ \Longrightarrow 2(2n \plus{} 1)k \minus{} (4n \plus{} 1)k \equal{} k \in Z$

Catalan number $ Ca_{2n}\equal{}\frac{1}{2n\plus{}1}C_{4n}^{2n}\in N$

$ C_{4n}^{2n}=\frac{4n!}{2n!(2n)!} $ it divides to $2n+1$ it's enough to show that $C_{4n}^{2n}$ divides by 4

[quote="mayhem"]Dear N T TUAN
How can we prove easy $ \dfrac{(2n \plus{} 2)\cdots (4n)}{(2n)!}$ is an integer?[/quote]
hi, $C_n^k$ means that in how many ways we can chose $k$ pencils from $n$ pencils, that's why it is integer and also positive, but by number theory sorry, I don't know........


Let $ a,$ $ b$ and $ c$ are real numbers such that $ a\plus{}b\plus{}c\equal{}\minus{}5.$ Prove that
\[ \sqrt{a^2\minus{}a\plus{}1}\plus{}\sqrt{b^2\minus{}b\plus{}1}\plus{}\sqrt{c^2\minus{}c\plus{}1}\geq7\]

[hide="Hint"]Let $ x\equal{}a\minus{}1/2$, etc. We have $ x\plus{}y\plus{}z\equal{} \minus{} 13/2$ and $ \sqrt{x^2\plus{} \frac{3}{4}} \plus{} \sqrt{y^2\plus{} \frac{3}{4}} \plus{} \sqrt{z^2\plus{} \frac{3}{4}}$.

And because the function uses only even powers, we can equivalently consider $ x\plus{}y\plus{}z\equal{}13/2$ instead.[/hide]

[quote="The Zuton Force"][hide="Hint"]Let $ x \equal{} a \minus{} 1/2$, etc. We have $ x \plus{} y \plus{} z \equal{} \minus{} 13/2$ and $ \sqrt {x^2 \plus{} \frac {3}{4}} \plus{} \sqrt {y^2 \plus{} \frac {3}{4}} \plus{} \sqrt {z^2 \plus{} \frac {3}{4}}$.

And because the function uses only even powers, we can equivalently consider $ x \plus{} y \plus{} z \equal{} 13/2$ instead.[/hide][/quote]
How it helps? Explain please. Thank you!

I was thinking along the lines of, that the function $ f(x)\equal{} \sqrt{x^2\plus{}3/4}$ is a convex hyperbola, and you can easily use a convexity argument to show that it's minimized when $ x\equal{}y\equal{}z$. To me it seemed easier to see when it was written as that form but maybe it does not make a difference to some people

Or we can use Minkowski's Theorem (this subcase can be easily proven using the distance formula):

$ \sqrt {x^2 \plus{} \frac {3}{4}} \plus{} \sqrt {y^2 \plus{} \frac {3}{4}} \plus{} \sqrt {z^2 \plus{} \frac {3}{4}}\ge\sqrt{(x\plus{}y\plus{}z)^2\plus{}(\frac{3\sqrt{3}}{2})^2}\equal{}7$.

Oh wow, how did I miss the distance formula solution  :(

[b]PS:[/b] [i]Minkowski's Inequality and Minkowski's Theorem are different[/i]


The sum of the numerator and denominator of a fraction is 160.  The fraction is equivalent to $ \frac{3}{7}$.  What is the original fraction?

or u can just see that the numerator and denominator add up to 10, so just multiply them by 16

then subtract


geometry
rhombus
linear algebra
matrix
algebra
polynomial
inequalities

I messed this one up pretty badly :D

(1)
Find the locus of points M from the plane of a rhombus ABCD for which MA*MC+MB*MD=AB 2 .

(2)
Consider the real numbers 1<=a_1<a_2<a_3<a_4, x_1<x_2<x_3<x_4 and M=(a_i x_j )1<=i,j<=4 a matrix of order 4. Show that detM>0.

(3)
f,g:[0,00)-.R, f is continuous and g is increasing and unbounded. Assume that for every sequence x_n of rational numbers s.t. x_n->00 as n->00 we have f(x_n)g(x_n)->1 as n->00. Show that f(x)g(x)->1 as x->00

(4)
Let A be a matrix in M_3(R). Show that:

a)If f is a polynomial in R[x] which doesn't have any real roots then f(A)=/= 0_3.
b)There is a positive integer n s.t. (A+A*) 2n = A 2n + (A*) 2n iff det(A)=0.

don't worry about it grob. I think you will do much better this year :)
worry more about the TST :)

[quote="grobber"]
(4)
Let A be a matrix in M_3(R). Show that:

a)If f is a polynomial in R[x] which doesn't have any real roots then f(A)=/= 0_3.
b)There is a positive integer n s.t. (A+A*) 2n = A 2n + (A*) 2n iff det(A)=0.[/quote]

a) We can  see A as a matrix in M_3(C), A is similar to a triangular superior
A~
(a * *)
(0 b *)
(0 0 c)

Suppose that f(A)=0, f(A) is similar to
(f(a) * *)
(0  f(b)*)
(0 0  f(c))
f(a)=f(b)=f(c)=0

The degree of the polynomial caracteristic of A is 3 at least one eigenvalue among a,b,c is real.This imply f has a real root contradiction.

So f(A)=/=0

b) We will use the relation A.A*=detA.I_3 where
A*=transpose Comatrix of A

Suppose detA=0, then A.A*=A*.A=0 these matrices commute
(A+A*)^2 = A^2+A*^2

If there exist n \in N  and detA=/=0 (A is invertible)
(1)  (A+A*)^(2n)=A^(2n)+A*^(2n) we multiply (1)  by A^(2n)
(A^2+k)^(2n) = A^(4n)+k^(2n)I
A vanishes the polynomial P(x)= (x^2+k)^(2n)-x^(4n)-k^(2n)
we verify that p(x)=kx^2.w(x) where w(x) has no real roots
kA^(2n).w(A)=0 but A is invertible, then w(A)=0 we get a contradiction with a)

Nice.

Try problem (1). We were all tempted to use analytic geometry and I got nowhere. There's a nice official solution using analytic geometry and a smart application of Cauchy's inequality, but a schoolmate showed me a very nice synthetic solution. Since I've seen 2 solutions I like, I drew the conclusion that this problem is very nice. :D

If you have the patience and determination you can do it by brute-force computations. I solved it like that in contest time, but unforunately I lost about 90 minutes on it. So I didn't have too much time to spend on the other three. Hope I won't make the same mistake this year  :)

Here's the main idea of the synthetic proof:

We translate the triangle AMD by the vector AB. Let M' be the image of M by this translaion. It's obvious that the images of A and D are B and C, respectively. The condition is MB*M'C+MC*M'B=BC*MM', so, from Ptolemy's thm, the quadrilateral MBM'C must be cyclic.

It's not hard to continue from here. Nice, huh? :D

well.. in my opinion problem 2) is the uglyest.. it's nasty... it's ... YAK :(.
what do u think about it?


calculus
integration
function
trigonometry
real analysis
real analysis solved

Find the continuous function $f(x)$ which satisfies for all real numbers $x$ the following equation.

\[\int^x_0 \sin t \cdot\ f(x-t)dt=f(x)-\sin x\]

Convolution again! Again, transform methods beckon. Let $F(s)$ be the Laplace transform of $f$; the transform of $\sin x$ is $\frac1{s^2+1}.$ Again, the transform of the convolution is the product of the transforms.

$\frac1{s^2+1}F(s)=F(s)-\frac1{s^2+1}$, which simplifies to

$F(s)=\frac1{s^2}$

and the inverse transform of that is $f(x)=x.$

That's right! Laplace transform is great isn't it, Kent Meryfield?

Thank you!

kunny


give a^2+b^2+c^2=3.find the max of
      ab+bc+ca-abc
where a,b,c is nonnegative numbers

Let us choose $ a\leq 1$. Then it's trivial to reduce $ ab+bc+ca-abc\leq 2a\cdot\sqrt{\frac{b^2+c^2}{2}}+(1-a)\cdot\frac{b^2+c^2}{2} $ to
 $ a(\sqrt{2b^2+2c^2}-b-c)\geq (1-a)\frac{(b-c)^2}{2} $ obvious. Then it is enough to prove that $ 2a\cdot\sqrt{3-a^2}{2}-\frac{a(3-a^2)}{2}+\frac{3-a^2}{2}\leq 2$. This comes down to $ -a^3+a^2+3a+1\geq 4a\cdot\sqrt{\frac{3-a^2}{2}} $. Just use the fact that $2\cdot\sqrt{\frac{3-a^2}{2}}\leq\frac{a(5-a^2)}{2} $.


floor function
number theory proposed
number theory

let $ p\geq 5$ is a prime number.
if $ \frac{a}{b}\equal{} 1\minus{} \frac{1}{2} \plus{} \frac{1}{3} \minus{} \frac{1}{4} \plus{}...\plus{}(\minus{}) \frac{1}{ [\frac{2p}{3}]}$
prove that:$ p^2$|$ a$

[quote="stvs_f"]let $ p\geq 5$ is a prime number.
if $ \frac {a}{b} \equal{} 1 \minus{} \frac {1}{2} \plus{} \frac {1}{3} \minus{} \frac {1}{4} \plus{} ... \plus{} ( \minus{} ) \frac {1}{ [\frac {2p}{3}]}$
prove that:$ p^2$|$ a$[/quote]

Wrong : let $ p\equal{}5$ and so $ [\frac {2p}{3}]\equal{}3$ and then $ 1 \minus{} \frac {1}{2} \plus{} \frac {1}{3}\equal{}\frac 56$ and we dont have $ p^2|a$

i'm really sorry,
1) $ \frac{a}{b}\equal{}1\plus{} \frac{1}{2}\plus{}...\plus{}\frac{1}{p\minus{}1} \Longrightarrow p^2|a$
2)$ \frac{a}{b}\equal{} 1\minus{}\frac{1}{2}\plus{}\frac{1}{3}\minus{}\frac{1}{4}\plus{}...\plus{}(\minus{})\frac{1}{ [\frac{2p}{3}]}  \Longrightarrow p|a$

The first one is [url=http://en.wikipedia.org/wiki/Wolstenholme%27s_theorem]Wolstenholme's theorem[/url].
For the second one, I distinguished two cases: $ p \equal{} 6k \plus{} 1$ and $ p \equal{} 6k \plus{} 5$ (the cases $ p \equal{} 2$ and $ p \equal{} 3$ are easy to check). I've rewritten $ \sum_{i \equal{} 1}^{n} ( \minus{} 1)^{i \plus{} 1} \frac {1}{i}$ as $ \sum_{i \equal{} \lfloor \frac {n}{2}\rfloor}^{n} \frac {1}{i}$ and then grouped the terms like that: $ \frac {1}{2} \sum_{i \equal{} \lfloor \frac {n}{2}\rfloor}^{n} \frac {1}{i} \plus{} \frac {1}{n \plus{} 1 \minus{} i}$.

[quote="FelixD"]The first one is [url=http://en.wikipedia.org/wiki/Wolstenholme%27s_theorem]Wolstenholme's theorem[/url].
[/quote]

[hide="@FelixD"]Oh bah bist du faul[/hide]

You could also get a factor p with summing $ \frac{1}{i}\plus{}\frac{1}{p\minus{}i}$ and then looking at the rest with modulo p, then with the use of multiple inverses you're done quite fast

[hide="@Rafikafi"]Ich hab das ja nur geschrieben, damit überhaupt wer ne Lösung reinschreibt^^ Aja, weißt eh wer ich bin?[/hide]

[hide="@FelixD"]Jup ich weiss wer du bist - Österreich + FelixD = Felix aus Österreich  :wink:

Eigentlich wollte ich gestern noch eine Lösung posten war aber dann auch zu faul :D [/hide]


Are you sure it's not (x+3)*sqrt((x+4)/(x+1))?

Well, it is the way I posted, no mistake this time!!!

cheers! :D:D

Sorry for my question:
   We have sqrt((x+i)/(x+i+1))>sqrt(x/(x+1)). Thus it is enough to prove that (4x+6)sqrt(x/(x+1))>4*(x^2+3x)/(1+sqrt(1+3x+x^2)).  Since 1+sqrt(1+3x+x^2)>x+2, it suffices to prove that (x+2)(2x+3)>(2x^2+6x)*sqrt(x/(x+1)), or 1+(x+6)/(2x^2+6x)>sqrt(x/(1+x). But (1+(x+6)/(2x^2+6x))^2>1+(2x+12)/(2x^2+6x)>1+1/x.
   I'm convinced that someone can find a better solution.


[quote="bénabar"]Hello!
How to use the function "search" to find the posts whose source is an oral examination of the ENS?
Thank you for your assistance. :blush:  ;)  :?  :)[/quote]
Just put in ENS in your search field and search. You might want to include "oral examination" in your search parameters.
I found [url=http://www.mathlinks.ro/Forum/search.php?search_id=1834624372]these problems[/url] ;)

[quote="Valentin Vornicu"][quote="bénabar"]Hello!
How to use the function "search" to find the posts whose source is an oral examination of the ENS?
Thank you for your assistance. :blush:  ;)  :?  :)[/quote]
Just put in ENS in your search field and search. You might want to include "oral examination" in your search parameters.[/quote]
I thought that it was possible to "refine" (:huh:) my research. Thank you all the same.  ;)  :no:


[quote="s.m.a"]let k,n positive integers, k>1 and n>2k-1.
prove that C(n,k) is not divisible by [n,n-1,...,n-k+1]. ;)[/quote]
What do you mean by [n,n-1,...,n-k+1]?

I think that he means the least common multiple of all those numbers.

We(maybe me) know this:

$\frac{1}{n+1}([1,2,...,n])=[\left(_0^n \right),\left(_1^n \right),...,\left(_n^n \right)]$

and from this we are done.

[quote="Mathx"]We(maybe me) know this:

$\frac{1}{n+1}([1,2,...,n,n+1])=[\left(_0^n \right),\left(_1^n \right),...,\left(_n^n \right)]$

and from this we are done.[/quote]


trigonometry function induction calculus derivative geometry Asymptote

1 - Transform in a product :

          f(x)=sin x + 1

2 - Solve the equation :

         (sqr3) sin x - cos x = 0

3 - Knowing that sin (a+b)=sin a cos b + cos a sin b , write analog expressions for

    a) sin (a-b)
    b) cos (a+b)

4- Solve the equations :

   a) sin (3x) + sin (5x) = cos x
   b) cos x + cos (3x)=cos(4x)+cos(2x)

5- Solve :

 a) tan^-1(sqr3/3)+cos^-1(-1/2)
 b) 2tan^1(-sqr3)-sin^-1(-1/2)
 c) 1-3tan^-1(3x)=pi(1+1/pi)

6 - Characterize the inverse function of :

 y(x) = pi/2 - (2tan^-1(1-x))/3

7 - Solve :

 a) cos [sin^-1(1/2)-sin^-1(3/5)]
 b) sin [cos^-1(-1/2)-2sin^-1(-4/5)]

8 - Consider the function f(x) = sin^-1((x^2/4-x)+pi/2

 a) Justify that f(x) is a continuous function

 b)  Prove,using the bolzano's theorem,that f(x) has a root in the partition
 [2,2+sqr2]

Quote : "There isn't more beautiful regime than democracy"- Plato

 And that's when we need to answer the following :

 1 - Differentiate :

  a ) ln tan (pi/4 + x/2)
  b ) (tan^-1 x)^sin x
  c ) cos [ln(tan^-1 x)]
  d) tan ^-1 (x) * cotan ^-1 (2x/(x+3))
  e) sin^2(2x^2-3)+2cos(3x^3-2x^2-1)
  f) tag^-1 (7th root) (1-sec^2(1/sqr(sin 2x))

 2 - Consider h(x) = 3xe^(x+1) + sin (2x).Prove,by induction,that
      (3e^(x+1)) * (x+n) * 2^n * sin (n*pi/2 + 2x) is h(x) order n derivative.

 3 - Determine one anti-derivative for :

  a) (cos (ln x))/x
  b) (sin (tg^-1 x))/(1+x^2)
  c) (2x+1)/cos^2(x^2+x)
  d) 3x/ sqr(16-25x)
  e) 1/sqr(x^4-x^2)

  4 - Write the the anti-derivatives general expression for :

 a) (xch*sqr(x^2+1))/3*sqr(x^2+1))
 b) cosech(1/x^2)*cotanh(1/x^2)*(1/x^3)

       "integrity or integrate"??<<---- nice question

1 - A ship is in the harbour.The distance h,of one given point of the ship's hull into the bottom of the sea,varies with the tide.
     Admit that h is given,in function of the time x, by h(x) = 10-3 cos (2x).
     What's the distance of that point of the hull to the bottom of the sea,at the high-tide  moment?

2 - The electrical intensity is given by : I(t)=75sin(100pi t + 0.32) amperes.
      Determine :

 a ) The period of the intensity
 b ) The range,or the set of values in wich the intensity varies.
 c ) The electrical intensity for t=0.005 s
 d ) The instants when the electrical intensity at its maximum level
 e)  The instants when it's null

These posts seem to be on the wrong forum.

I've moved the thread, though I suggest to the original poster that she/he not ask so many questions in one thread (use different threads) and that he/she put the questions in the appropriate fora.

Also, this is not a place where homework problems will be solved for you by others.

"political incompatibility between a vicious and futile life" <- Plato
 Let's go to what it matters :

 1 - Prove using the Lagrange theorem that

          sin^-1 (x) >1

 2 - Calculate the following limit :

 lim     (cotan x)^sin x
 x->0

1 - Consider the real function of real real variable f :

  f(x) = cos^2 (pi/2 - x) - cos (pi-x)

1.a) Prove that the function f has at least  one root in the partition [2pi/3 , 5pi/6]

1.b) Determine one tangent equation for the point  x=pi/2

1.c) Calculate lim    (f(x)-1)/2x
                      x->0

2- The area of a triangule is given by the equation
A(alpha)=25 sin(2alpha).
Calculate the value of alpha belonging to the partition ]0,pi/2[,for wich the triangule's area is at its maximum.

3- The electrical intensity in an circuit is given by :

 I(t) = 30 sin (50pi t +0.8)

 Calculate the instants when the electrical intensity is at its maximum.

 4- Consider h(x)= x + (cos x)/e^x

4.a)-Calculate the values of x for h''(x)=0

4.b) Investigate the existence of asymptotes

1 - Solve in R :

 | cos(5x) + cos(3x) | = | cos x |

2 - Beiing i the imaginary unit.Being W= 1-tan^2(betha)+(2tan(betha))i,express,in function of 2betha,the principal argument and the module of W.

3 - Solve,in R,the equation

 4[ sin (x/2)-2sin(x/2)*(sin (x/4))^2]- 1/(sin x) =1

4 - Determine the values phi and psi that verify simultaneously the equations :

 phi = 3 sin (psi/2)

 phi = 3 cos (psi/2) , with phi (>=) 0 , 0(<=) psi (<=)4pi

 "Fame costs and here you pay with ..." <- Fame tv series

1- Show that it's true the implication :

x+y=z => cos^2 x+cos^2 y+cos^2 z - 2cos x*cos y*cos x = 1

2- Calculate cos (3x) in function of cos (x)

3 - Show that,being A,B,C internal angles of a triangule,it verifies that :

 sin A + sin b + sin c = 4 cos (A/2)*cos (B/2)*cos(C/2)

4 - Relatively to the aplication f,of C in C, thus defined

 z= x + yi -> f(Z) = (x+y)+(x-y)i

a) Show that f is an automorphism of (c,+),where + designates the usual sum of complex numbers

b) Verify that (5pi/12) it's an argument of f(3-(sqr(3))i)

 "The true remedy is the indepht dive to the root of the problem"

[quote="Amnesty differentiator"][quote="Arne"]Why do you keep posting these? Nobody will reply.[/quote]

 If nobody will reply,why don't you reply!?[/quote]

 Belgians are a pain in the ass!!!

[quote="Amnesty differentiator"][quote="Altheman"]you are posting problems that do not require integration of concepts, they are homework problems, and you posted like 50 problems[/quote]

 Indeed,in those terms,you can go to the anti-derivatives section...[/quote]

 This nothing more then a jackass..

Bring me #9.2....

 Homework says the jackass.....


algebra
system of equations
algebra unsolved

Solve system of equations:

$ \begin {cases} \sqrt {3x} \left ( 1 \plus{} \frac {1}{x \plus{} y} \right ) \equal{} 2 \\
\sqrt {7y} \left ( 1 \minus{} \frac {1}{x \plus{} y} \right ) \equal{} 4\sqrt {2} \end {cases}$

hello, rewriting your system as follows
$ 1\plus{}\frac{1}{x\plus{}y}\equal{}\frac{2}{\sqrt {3x}}$
$ 1\minus{}\frac{1}{x\plus{}y}\equal{}\frac{4\sqrt{2}}{\sqrt{7y}}$
we get
$ 2\equal{}\frac{2}{\sqrt{3x}}\plus{}\frac{4\sqrt{2}}{\sqrt{7y}}$
let  $ a\equal{}\sqrt{3x}$ $ b\equal{}\sqrt{7y}$ the we get
$ a\equal{}\frac{b}{b\minus{}2\sqrt{2}}$
inserting this in
$ 1\plus{}\frac{1}{\frac{a^2}{3}\plus{}\frac{b^2}{7}}\equal{}\frac{2}{a}$
simplifying and factoring we get
$ (\minus{}b^2\minus{}4b\sqrt{2}\plus{}6)(\minus{}3b^2\plus{}12b\sqrt{2}\minus{}28)\equal{}0$
from here you will get $ b$ then $ y$ and $ x$.
Good luck!
Sonnhard.


geometry
rectangle
3D geometry
algebra
polynomial
Harvard
college

Heh, its back.

I have alot fewer questions this year  :lol:

[size=150]Comprehensive 2000[/size]

6. $r_{1}$, $r_{2}$, $r_{3}$, $r_{4}$, are the zeros of $x^{4}-4x^{3}+x+7$
Find $\frac{r_{2}+r_{3}+r_{4}}{r_{1}}+\frac{r_{1}+r_{3}+r_{4}}{r_{2}}+\frac{r_{1}+r_{2}+r_{4}}{r_{3}}+\frac{r_{1}+r_{2}+r_{3}}{r_{4}}$

A. $-1$
B. $\frac{-32}{7}$
C. $1$
D. $4$
E. None of These Answers

13. Jimme Shoe wants to know how many rectangles are in a 3x4 rectangular grid. What is the correct answer?

A. $24$
B. $36$
C. $47$
D. $60$
E. None of These Answers

15. What is the sum of the cubes of the distinct roots of the following equation?
$3x^{4}+2x^{3}-14x^{2}+3x+4 = 0$

A. $\frac{-341}{27}$
B. $\frac{286}{27}$
C. $\frac{118}{27}$
D. $\frac{-118}{27}$
E. None of These Answers

16. $\frac{31}{195}= \frac{1}{A+\frac{1}{B+\frac{1}{C+\frac{1}{D}}}}$
Find $A^{2}+B^{2}+C^{2}+D^{2}$, if {A,B,C,D} is a subset of the set of positive integers.

A. $46$
B. $65$
C. $71$
D. $97$
E. None of These Answers

[size=150]Comprehensive 1997[/size]
8. Find the area of triangle $ACD$, where in triangle $BAC$
$AC = 5$
$BC = \sqrt{39}$
$AD = 4$
$BD= 4$
D is on AB. So essentially CD is a median.

A. $\frac{5\sqrt{39}}{2}$
B. $\frac{15}{2}$
C. $4\sqrt{3}$
D. $2$
E. None of These Answers

12. The polynomial $f(x)=x^{4}+6x^{3}+ax^{2}-54x+c$ has four real roots. If $r_{1}+r_{2}=0$ and $r_{4}-r_{3}=4$, find the value of $a+c$.

A. $49$
B. $-49$
C. $41$
D. $-41$
E. None of These Answers

[size=150]Comprehensive 1999[/size]

5. Assume it is known that of the children of Harvard alumni, 80% go to Harvard, and the rest go to Yale. Of the children of Yale alumni, 40% go to Yale and the remainder are divided equally between Harvard and Dartmouth. Of the children of Darthmouth alumni, 70% go to Dartmouth, 20% to Harvard, and 10% to Yale. Find the probability that the great grandson of a Harvard alumni goes to Harvard.

A. .512
B. .608
C. .632
D. .644
E. None of These Answers

14. The roots of $x^{4}-7x^{3}+5x^{2}+31x-30=0$ are $a$, $b$, $c$, and $d$ where $a\leb\lec\le\d$. Find $2a+3b+4c+5d$.

A. 13
B. 36
C. 30
D. 5
E. None of These Answers

Thanks, and I apologize for posting so many problems in one thread... but it'll die in a few days anyway.

#13

[hide]since theres 5 lines on the width, and 4 on height, its 5 choose 2 times 4 choose 2.  4 choose 2=6, and 5 choose 2=10.  10*6=60[/hide]

[hide]

6. , so substituting it in for the numerator of each fraction, we get  and so on, so the answer is 4 times the sum of reciprocal of roots minus 4, which is B.

13. .

15. Newton's sums (look it up in aops). The answer's A.

16. . That's how you get A -- repeat to get B, C, D. The answer's B.

8. Picture? D is not unique.

12. , so . System of equations gives , . Then let , and use this on the THIRD Viete's sum to get , so the roots are . Then it's trivial to find , so the answer is B.

5. Just look at the possible sequences (where first is H = the great grandfather). The only possible sequences are HHHH (.512), HHYH (.048), HYHH (.048), HYDH (.012), HYYH  (.024). Adding them up yields .644, so D.

14. Rational Root Theorem, factors as , so , , , . Plug in, answer is B. [/hide]

[quote="Pakman2012"]
13. Jimme Shoe wants to know how many rectangles are in a 3x4 rectangular grid. What is the correct answer?

A. $24$
B. $36$
C. $47$
D. $60$
E. None of These Answers
[/quote][hide]The number of $n\times m$ rectangles is $(4-m)(5-n)$. Hence, there are \[\sum_{m=1}^{3}\sum_{n=1}^{4}(4-m)(5-n)\]rectangles. Set $x=4-m$ and $y=5-n$. Then the sum is \[\[\sum_{x=1}^{3}\sum_{y=1}^{4}xy=\sum_{x=1}^{3}x\sum_{y=1}^{4}y=\left(\frac{3\cdot4}{2}\right)\left(\frac{4\cdot5}{2}\right).\] There are $60$ rectangles: $\boxed{D}$.[/hide]

[hide="Comprehensive 2000, #13"]
[color=red]I am assuming that only rectangles with integer side length count here: this is because other rectangles have diagonals which do not 'appear' in the grid.[/color]

Number of 1x1 rectangles: 12
Number of 1x2 rectangles: 17
Number of 1x3 rectangles: 10
Number of 1x4 rectangles: 3
Number of 2x2 rectangles: 6
Number of 2x3 rectangles: 7
Number of 2x4 rectangles: 2
Number of 3x3 rectangles: 2
Number of 3x4 rectangles: 1
Total: 60
[/hide]

(wrote this one before others posted, I'll leave it here anyway)

[hide="Comprehensive 1999, #5"]
P(Harvard-Harvard-Harvard)$=0.8^{3}=0.512$
P(Harvard-Yale-Harvard)$=0.8 \cdot 0.2 \cdot 0.2=0.032$
P(Yale-Harvard-Harvard)$=0.2 \cdot 0.2 \cdot 0.8=0.032$
P(Yale-Yale-Harvard)$=0.2 \cdot 0.4 \cdot 0.2=0.016$
P(Yale-Dartmouth-Harvard$=0.2 \cdot 0.2 \cdot 0.2=0.008$
which would give a total of $0.6$: none of the above.
[/hide]

[hide]

8. Since D is on AB, then use either Stewart's Theorem or the more specific median length formula to get CD = 4, so then you use Heron's on CAD to get . D?

[/hide]

woot!!!!!! Vestavia on Saturdsay? who's going?

I AM! GEOMETRY! lol i'm sooo ready

[hide="2000 #13"]In an $m\times n$ grid, there are $\frac{mn(m+1)(n+1)}4$ rectangles. In a $3\times4$ grid, there are $\frac{3\cdot4(3+1)(4+1)}4=60$ rectangles. $\boxed{\text{D}}$[/hide]
[hide="1997 #8"]$A=\sqrt{\frac{13+\sqrt{39}}{2}\left(\frac{13+\sqrt{39}}{2}-5\right)\left(\frac{13+\sqrt{39}}{2}-8\right)\left(\frac{13+\sqrt{39}}{2}-\sqrt{39}\right)}=\frac{5\sqrt{39}}2\Rightarrow\boxed{\text{D}}$[/hide]


Prove that if: $ \frac{\sin ^4 \alpha}{a}\plus{}\frac{\cos ^4 \alpha}{b}\equal{}\frac{1}{a\plus{}b},$ where $ a,b \in \mathbb{R}^*_{\plus{}}$, then: $ \frac{\sin ^{2n\plus{}2} \alpha}{a^n}\plus{}\frac{\cos ^{2n\plus{}2} \alpha}{b^n}\equal{}\frac{1}{(a\plus{}b)^n}, \forall n \in \mathbb{N}^*,$ $ n$ odd.

[quote="moldovan"]Prove that if: $ \frac {\sin ^4 \alpha}{a} \plus{} \frac {\cos ^4 \alpha}{b} \equal{} \frac {1}{a \plus{} b},$ where $ a,b \in \mathbb{R}^*_{ \plus{} }$, then: $ \frac {\sin ^{2n \plus{} 2} \alpha}{a^n} \plus{} \frac {\cos ^{2n \plus{} 2} \alpha}{b^n} \equal{} \frac {1}{(a \plus{} b)^n}, \forall n \in \mathbb{N}^*,$ $ n$ odd.[/quote]

$ cos^2 \alpha \equal{} t$----> $ \frac {(1 \minus{} t)^{2}}{a} \plus{} \frac {t^{2}}{b} \equal{} \frac {1}{a \plus{} b}$---> $ ((a \plus{} b)t \minus{} b)^{2} \equal{} 0$----> $ t \equal{} \frac {b}{a \plus{} b}$--->

$ cos^2 \alpha \equal{} \frac {b}{a \plus{} b}$ --> $ sin^2 \alpha \equal{} \frac {a}{a \plus{} b}$ ----->

$ \frac {sin^{2n \plus{} 2}\alpha}{a^n} \plus{} \frac {cos^{2n \plus{} 2}\alpha}{b^n} \equal{} \frac {(\frac {a}{a \plus{} b})^{n \plus{} 1}}{a^n} \plus{} \frac {(\frac {b}{a \plus{} b})^{n \plus{} 1}}{b^n} \equal{}
\frac {a}{(a \plus{} b)^{n \plus{} 1}} \plus{} \frac {b}{(a \plus{} b)^{n \plus{} 1}} \equal{} \frac {1}{(a \plus{} b)^n}$--->

$ \frac {sin^{2n \plus{} 2}\alpha}{a^n} \plus{} \frac {cos^{2n \plus{} 2}\alpha}{b^n} \equal{} \frac {1}{(a \plus{} b)^{n}}$


[quote="SkimGuy"]1. A locksmity has ten types of blanks for keys. Each blank has five different cutting positions and three different cutting depths at each position, except the first position, which only has two depths. How many different keys are possible with these blanks?

2. How many 5-digit numbers are there that include the digit 5 and exclude the digit 8?

If possible, could you show me how you got the answer? Thanks  :)[/quote]

1) 10(2 * 3 * 3 * 3 *3)? [b]1620[/b]? I don't really get this problem, I'm just multiplying the number of blanks by the number of ways each blank can be cut?

2) (8 * 9 * 9 * 9 * 9) - (7 * 8 * 8 * 8 * 8)? [b]23816[/b]? This is basically the number of five-digit numbers that don't have 8 minus the number of five-digit numbers that don't hav 8 or 5.

First one is confusing, I didn't understand the question (Answer is 140) But the second one is right. :D Thanks

If the answer is 140, then it would come from $140=(2+3+3+3+3)$ but this would imply that he could only cut in one of the five positions... strange.

Anyway I don't know how to make keys, so maybe they do only cut in one place  :)


geometry
geometric transformation
rotation

[quote="Neod4001"]Hello all, I am a British Maths GCSE student working towards the final exam and one day this question cropped up in a mock Maths exam I have done recently. I could not work it out and think it might have cost a percentage of the mark, and was wondering whether it would crop up again.

 I try to represent it here and wonder whether any of you guys could help with offering a solution as I could not for revision purposes.

Any help concerning the answer would be greatly appreciated.
Here goes.

The Earth’s circumference around the equator is 24907.55 miles

(1 mile = 1.609344 km)

A plane flies 11km above the surface of the Earth and flies westward following the line of the equator. If the Plane is travelling at 510km per hour, how long will it take for the plane to reach the same place above the Earth’s surface from where it started from? (i.e. at the beginning of the question)

I think you have to consider the rotation of the Earth and km vs miles.
And I think the solution lies in finding the radius of the Earth and adding 11km to it, but I'm not sure.
When it says "the same place above the surface", I believe it is talking about the same place over the Earth's surface at the beginning and at the end as opposed to a complete circumnavigation of the Earth.
i.e. Over Australia at the beginning and at the end.

Any help with this would be greatly appreciated.[/quote]

[hide="Super Hint"]Consider the angular velocities of the plane and the earth.[/hide]

[hide="Super Hint"]Consider the angular velocities of the plane and the earth.[/hide][/quote]

That is way more than is needed to solve this problem. This is a speed distance time job. And if that is needed, GCSEs have got harder since last year

It's a think for fadvanced fasttrack students. I'm 14 and so not yet fully able to do this 100% correctly, and I think the answer is 78 hrs 44 mins, but I'm not sure.

[quote="mthiyaga"][hide]This is a straight forward speed problem. Speed of the airplane is given relative to the earth. So no relative velocity is involved.

The radius of the earth is = $ \frac {24907.55}{2\pi} \equal{} 3964.1597$ miles. Hence the distance the plane has to travel for a complete circle is $ 2\pi \times \left (3964.1597 \plus{} \frac {11}{1.609344} \right ) \equal{} 24950.4961$ miles.

So the time taken is $ \frac {24950.4961}{\left (\frac {510}{1.609344}\right )} \equal{} 78.733$ hours or 78 hrs and 44 minutes.[/hide][/quote]

Thank you very much, this is the most helpful anyone has been in any of the places I have visited.
I rated your post 6 for excelent.


Here is a mthod. Though its a little long, a nice idea
We first have the following claim:
Let $ b$ and $ c$ be the lengths of the segments into which a focal chord is divided in by the focus. Then the latus rectum is the hormonic mean between $ b$ and $ c$.

Try proving the above, its simple enough.

Now the problem gives that $ b \plus{} c \equal{} \frac{25}{4}$.
Moreover by the above claim we get another equation from which we can solve for $ b,c$. Next by usng the result that ends of a focal chord are $ t$ and $ \minus{}t^{\minus{}1}$ and in conjunction with that any point on the chord shud be of the form $ (1\plus{} \cos \theta , \sin \theta )$.
it will give us the result.  :lol:

The radical centre of the circles drawn on the focal chord of  as diameters is?

Are you trying this sum??

The answer is really simple.. Just use the t and 1/t idea.

[quote="Euclidean Geometer"]no no... i was trying the one before that. the one involving the trapezium. the lemma i had used was nice. :D[/quote]

Sorry I got confused looking at the question number. :( Anyway, yes thats the nice method to solve this.. There is another mechanical method which will kill you. :)


algebra
polynomial
inequalities
inequalities unsolved

If $x^5 - x ^3 + x = a,$  prove that $x^6 \geq 2a - 1$.

Rewrite it as:
$x^6 \geq 2x^5 - 2x^3 + 2x - 1$
which is equivalent to:
$(x-1)^2((x^2-1)^2 + x^2) \geq 0$
and this inequality is obviously true.

nice solution, but a little mistake
$a=x\frac{x^6+1}{x^2+1}\neq x\frac{x^6+1}{x+1}$, but the rest is nice and elegant as you said.

Consider the graph of the function given and the graph of the function $x^6$..... !!

[quote=ashwath.rabindranath][hide]$a=x\frac{x^6 +1} {x+1}$
which implies that $a(x + \frac{1} {x}) =x^6 + 1$
  as $x + \frac{1} {x} \geq 2$
 $2a-1\leq x^6$
  and we are done.[/hide]
 Ashwath :D
P.S. nice and elegant ain't it?
[/quote]

This is not true because $x+\frac{1}{x} \leq -2$ for $x<0$..

[quote=madhusudan kale]Multiply both sides by $x^{2}$ and we get
$x^{7}-x^{5}+x^{3}=ax^{2}$
now substitute $x^{5}-x^{3}$ as $x^{7}-ax^{2}$
so we get
$x^{7}-ax^{2}+x^{1}=a.$
rearranging terms ,
$x(x^{6}+1)=a(x^{2}+1)$
applying AM-GM on $(x^{2}+1)$ we get
$x(x^{6}+1)>=2ax$
thus we get the desired result.[/quote]

What if x<0

[quote=ashwath.rabindranath][hide]$a=x\frac{x^6 +1} {x+1}$
which implies that $a(x + \frac{1} {x}) =x^6 + 1$
  as $x + \frac{1} {x} \geq 2$
 $2a-1\leq x^6$
  and we are done.[/hide]
 Ashwath :D
P.S. nice and elegant ain't it?

Ganitham Namaha[/quote]

 A small doubt:

What if $x + \frac{1}{x} \le -2$, won't we have to check the sign of $a$?


I know that there were 6 perfect individual scores... thats well known. Does anyone know how many 7's there were at your location? There were 3 at Iowa. Thanks!

There were two sevens at SJSU: Darren Yin (SFBA A) and Adam Hesterberg (Washington A). Who were the sevens at other sites?

There were only five perfect scores:
Jongmin Baek
Eric Price
Ricky Briggs
Aaron Pixton
Anders Kaseorg.

If you check the other ARML thread, you'll see that there were actually six perfect scores. As for the 7's at Penn State, I think there were around 10 -- two just on the TJ A team.

slightly off topic: was Tiankai Liu with Exeter at ARML?

[quote="paradox"]No, Tiankai and some other Exeter seniors were not at ARML due to a conflict with graduation.[/quote]

Which is why Exeter didn't perform all that well.

Brian Lawrence also received a score of 8.

There were several 7's at ARML -- Alison, of Upstate New York had one, as did Dmitriy Karabash of NYC A team.


AMC
USA(J)MO
USAMO
trigonometry
analytic geometry
AIME
inequalities

Recent USAMOs: http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=27

Older USAMOs: http://www.kalva.demon.co.uk/usa.html

[quote="D.P.L"]Does anyone know where i can get past usamo and mop problems? i m looking at 1960s-2005

Thanx in advance :lol:[/quote]
mop isn't a contest...

http://web.mit.edu/rwbarton/Public/mop/

This might help, these are Reid Barton's MOP notes. Included are some problems.

students.imsa.edu/~tliu/Math/mop05hwmain.pdf

This is the 2005 MOP Homework Problems. (I think :lol: )

[quote="Karth"]I think he means the homework. The 2002 homework is on the UNL website.[/quote]

There are actually a lot of problems given out during MOP: there are 7 mop tests with 8-10 questions each, three black group tests with 3 problems each, and two IMO tests with 3 problems each. Some of the problems are taken from various sources, but I had never seen any of them before and I get the feeling many or most are original.

I don't think these problems are available in compiled form anywhere for the general public (although they are sent to all the attendees).

Reid Barton's MOP notes are pretty nice...

Meh. The're mostly just a collection of a few problems.

Hmm... I have a question about Problem Solving Strategies by Engel. Which chapters should we read first if we're getting started on Olympiad math? Is there a particular order that anybody recommends?

1,2,3,4.... might be a good way to go, especially because the later chapters are much harder than the earlier ones.

I skipped the first 5 chapters because I don't like them and I'm reading and doing problems in chapter 6. And I don't really want to do the first 5 chapters....
Is that bad?

Are the chapters standalone? In other words, can you understand chapter 10 without having read chapter 1?

Also, certain problems in the book are designed such as I'd have no idea as to how they thought of the answer (such as the trig problem in the first chapter) How do I remedy this?

Do you need to know about polar coordinates?

Particularly $e^{i\theta}= \cos \theta+i \sin \theta$

Questions involving the complex plane have appeared many times on the AIME, so it's very useful.

Like AIME 1999, #9. :D

A question involving polar coordinates and de Moivre appears like every year, except for last year for some reason. They also didn't have an inequality on the USAMO, which was also weird.


Hey...does anyone know if we're sending teams to HMMT?

I don't think that the NYC math team sends a team.  Maybe it should?

My daughter's school, Hunter College High School, sent an 8-person team last year.  I'm not sure about other schools.  Maybe you can contact someone at Hunter or another school to see if you can piggyback?


It seems that IMO'2009 will be huge...  Currently (April 16) there are 99 registered countries:
   http://www.imo-official.org/year_reg.aspx?column=code&order=asc
Last year there were 97 participating countries, and it seems that this record number will be
broken in 2009.

* 92 out of the 97 countries from last year have also registered for IMO'2009.
* Missing from last year are still:  Cambodia;  Kuwait;  Mongolia;  Vietnam;  Saudi Arabia.
* Furthermore after several years of absence Algeria, Brunei, Nigeria, and Pakistan will participate again.
* Benin and Mauritania will participate for the first time with student teams (they had sent observers to IMO'2008).
* Zimbabwe will send an observer, and may send a student team from 2010 onwards.

Since Mongolia and Vietnam are regular IMO participants, I guess that this will be the first IMO with
students participating from at least 100 different countries!

as far as i understood imo-official right zimbabwe will send a student also this year...

Naphthalin

[quote="Naphthalin"]as far as i understood imo-official right zimbabwe will send a student also this year...[/quote]

Yes, I also noticed this and don't understand what's going on.

The regulations in http://www.imo-official.org/documents/InvitationIMO.pdf are very clear about
this point: Since Zimbabwe has never before sent an observer to an IMO, there is no way that it
could officially participate with a students team at IMO'2009.

Perhaps Zimbabwe wants to participate unofficially, as Ukraine and Kazakhstan and six
other countries did at IMO'1992 in Russia.  The Russian organizers allowed these unofficial
participations (against the IMO regulations), and this caused quite a mess.
For instance the listing at
http://www.imo-official.org/year_country_r.aspx?year=1992&column=total&order=desc
lists two types of participating countries (the unofficial ones do not show up in the ranking).

An update (May 14):

Now there are 105 registered countries:  http://www.imo-official.org/year_reg.aspx?column=code&order=asc

* 96 out of the 97 countries from last year have also registered for IMO'2009. (Only Mongolia is missing)
* Furthermore after some time of absence the following five countries will participate again: Algeria, Brunei, Nigeria, Panama, and Pakistan
* Benin and Mauritania will participate for the first time with student teams (they had sent observers to IMO'2008).
* Senegal and Zimbabwe show up for the first time at an IMO (According to the regulations, they can only send an observer)

So IMO'2009 will be the first IMO with at least 100 different countries!

Peru will send an 11 years old participant: Raúl Arturo Chávez Sarmiento
http://www.imo-official.org/year_reg_team.aspx?code=PER

And I guess that South-Korea will send Soo Hong Lee again, who is now 15.
http://www.imo-official.org/participant_r.aspx?id=15628

[quote="keyree10"][quote="Bugi"]I am 99% sure that he wasn't in the first 6 [b]Peruans[/b]...[/quote]

Just to be accurate...  :D
It is "Peruvian", not "Peruan".[/quote]

My mistake  :)

But really an 11 year old is a way too young for IMO. I'm not thinking anything bad or wrong, just  :huh:

[quote="tipe"][quote="Bugi"][quote="test20"]Peru will send an 11 years old participant: Raúl Arturo Chávez Sarmiento
[/quote]

Can anybody tell me what's the point of that?

I am 99% sure that he wasn't in the first 6 Peruans...[/quote]

I took part in the peruvian IMO TST process, and i can say that this process was fair. [b]Raul[/b] took the 5th place, and we are very impressed by him.

More information (in spanish) about our TST in [url]http://selectivos-peru.blogspot.com/2009/05/resultados-imo-2009.html[/url] The TST consists of 6 problems in two days and each problem is worth 10 points.

The problems from the First Day: [url]http://www.mathlinks.ro/viewtopic.php?search_id=1550938156&t=274271[/url]

$ Tipe$[/quote]

Lol, then it's a really smart kid  :)

[quote="tipe"][quote="Bugi"][quote="test20"]Peru will send an 11 years old participant: Raúl Arturo Chávez Sarmiento
[/quote]

Can anybody tell me what's the point of that?

I am 99% sure that he wasn't in the first 6 Peruans...[/quote]

I took part in the peruvian IMO TST process, and i can say that this process was fair. [b]Raul[/b] took the 5th place, and we are very impressed by him.

More information (in spanish) about our TST in [url]http://selectivos-peru.blogspot.com/2009/05/resultados-imo-2009.html[/url] The TST consists of 6 problems in two days and each problem is worth 10 points.

The problems from the First Day: [url]http://www.mathlinks.ro/viewtopic.php?search_id=1550938156&t=274271[/url]

$ Tipe$[/quote]

Another terry tao! :o

11 years - that's nothing.  I have 0  :P.

[url]http://www.imo-official.org/year_reg_team.aspx?code=LVA&language=en[/url]

Well, you're older than

[quote]Cillian Power
(Contestant 6)

• -1 years old[/quote]

http://www.imo-official.org/year_reg_team.aspx?code=IRL

Hello, Peteris, nice to see you here!  :lol:  You are probably the youngest participant in the whole IMO history )).

Another update (May 29):

Now there are 108 registered countries: http://www.imo-official.org/year_reg.aspx?column=code&order=asc

* All 97 countries from last year have also registered for IMO'2009.
* Furthermore after some time of absence the following 5 countries will participate again: Algeria, Brunei, Nigeria, Panama, and Pakistan
* The following 4 countries had sent observers to previous IMOs, and will participate for the first time with student teams: Benin, Ivory Coast, Mauritania, and Syria.
* Senegal and Zimbabwe show up for the first time at an IMO.

Every year, two or three registered countries drop out in the very last moment.
It seems that even these drop-outs cannot prevent IMO'2009 from being the first IMO with student teams from over 100 different countries!

Some more very young participants:

11 years:
--------------------------------------------------------------------------------------
Raúl Arturo Chávez Sarmiento (Peru)
  http://www.imo-official.org/year_reg_team.aspx?code=PER

13 years:
--------------------------------------------------------------------------------------
Jeck Lim (Singapore)
  http://www.imo-official.org/year_reg_team.aspx?code=SGP

14 years:
--------------------------------------------------------------------------------------
Nerses Srapionyan (Armenia)
  http://www.imo-official.org/year_reg_team.aspx?code=ARM
Pranon Rahman Khan (Bangladesh)
  http://www.imo-official.org/year_reg_team.aspx?code=BGD
Tham Ying Hong (Malaysia)
  http://www.imo-official.org/year_reg_team.aspx?code=MAS
Byambadorj Otgonsuren (Mongolia)
  http://www.imo-official.org/year_reg_team.aspx?code=MNG
Jombo Eniweke Eunice (Nigeria)
  http://www.imo-official.org/year_reg_team.aspx?code=NGA
Omer Cerrahoglu (Romania)
  http://www.imo-official.org/year_reg_team.aspx?code=ROU
Nipun Pitimanaaree (Thailand)
  http://www.imo-official.org/year_reg_team.aspx?code=THA
Ufuk Kanat (Turkey)
  http://www.imo-official.org/year_reg_team.aspx?code=TUR

--------------------------------------------------------------------------------------
And there are tons of students at age 15...

[quote="test20"]And I guess that South-Korea will send Soo Hong Lee again, who is now 15.
http://www.imo-official.org/participant_r.aspx?id=15628[/quote]

Well, it turns out that Soo Hong Lee is not on the South-Korean team for IMO'2009:
http://www.imo-official.org/year_reg_team.aspx?code=KOR

Does anybody know what happened to him?

Does anybody know what's going on with Bolivia?
Have they selected a team at all?
Will they show up in Germany?

Bolivia is the only country that has pre-registered for IMO'2009,
but has neither registered a team-leader nor a deputy nor a single
contestant nor an observer.
http://www.imo-official.org/year_reg_team.aspx?code=BOL

[quote="hsiljak"]I recall reading a text about him as the youngest student to enter University of Seoul, but I'm not sure since I can't find that text now. If it's correct, then that's the reason why he isn't in the team.

Edit: I still can't find that article, but here's a trace:

http://math.missouristate.edu/assets/Math/College_Mathematic_Journal_January_09.pdf

[quote]Soo HONG Lee (student), Inst, of Sei. Ed. for the Gifted and Talented, Yonsei U„ Seoul, Korea[/quote]

so he is at the university, according to this (print date: Jan 2009).[/quote]

[quote]Soo HONG Lee (student), Inst, of Sei. Ed. for the Gifted and Talented, Yonsei U„ Seoul, Korea[/quote] is not refering
to a university, but rather a program that the Yonsei Univerisity runs that he took part in some time ago.

However, he has graduated high school and is the youngest student to enter Seoul National University(Which is the most prestigous university in Korea).

And does anybody know what's going on with Turkmenistan, Syria,
United Arab Emirates, Saudi Arabia, and Trinidad?
http://www.imo-official.org/year_reg.aspx?column=code&order=asc

* Turkmenistan has only registered a team leader, but neither deputy nor contestants.
Do they have a team at all?

* Syria, United Arab Emirates, Trinidad have only registered team leader and deputy.
Will they also send a team?

* Saudi Arabia only has registered four observers.
One of the observers is their team leader from last year.
They probably did not understand/use the registration software correctly.
Will they be able to register a team?

Israel has only registered three contestants so far:
http://www.imo-official.org/year_reg_team.aspx?code=ISR

But usually they send a full team with six contestants:
http://www.imo-official.org/country_team_r.aspx?code=ISR

Does anybody know more about this?

IMO'2009 will be the first IMO with participants from over 100 different countries.

IMO'2009 will also have the highest number of contestants in history.
The old record was 535 from IMO'2008.
This year there are already 559 registered contestants, with the teams of Austria, Netherlands, and USA still to come.

http://www.imo-official.org/year_reg.aspx?column=code&order=asc

IMO'2009 will be the first IMO with over 80 full teams with 6 contestants!

Currently there are 85 full teams registered.

[quote="test20"]Another update (May 29):

Now there are 108 registered countries: http://www.imo-official.org/year_reg.aspx?column=code&order=asc

* All 97 countries from last year have also registered for IMO'2009.
* Furthermore after some time of absence the following 5 countries will participate again: Algeria, Brunei, Nigeria, Panama, and Pakistan
* The following 4 countries had sent observers to previous IMOs, and will participate for the first time with student teams: Benin, [b]Ivory Coast[/b], Mauritania, and Syria.
* Senegal and Zimbabwe show up for the first time at an IMO.

Every year, two or three registered countries drop out in the very last moment.
It seems that even these drop-outs cannot prevent IMO'2009 from being the first IMO with student teams from over 100 different countries![/quote]

Are you sure that Ivory Coast will participate this year for the first time with students? Because so far they have not registred any students.
Also they did not send an observer (at least not last year, or the year before).

Do you happen to know when they've sent an observer for the first time?
Thank you in advance

[quote="test20"][quote="ermath"]Are you sure that Ivory Coast will participate this year for the first time with students? Because so far they have not registred any students.
Also they did not send an observer (at least not last year, or the year before).

Do you happen to know when they've sent an observer for the first time?
Thank you in advance[/quote]

You are definitely right: Ivory Coast shows up for the first time, and will only send some oberservers.

The correct situation seems to look like this:

* 96 out of the 97 countries from last year will also send student teams to IMO'2009.
* Saudi-Arabia has sent contestants last year, but will only send observers this year.
* Furthermore after some time of absence the following 5 countries will participate again: Algeria, Brunei, Nigeria, Panama, and Pakistan
* The following 3 countries had sent observers to previous IMOs, and will participate for the first time with student teams: Benin, Mauritania, and Syria.
* Zimbabwe shows up for the first time, but seems to have special permission to also send students.
* Senegal and Ivory Coast show up for the first time at an IMO.

All in all, this yields
official student teams from 96+5+3=104 countries,
one student team participating unofficially,
and three countries that only send oberservers.[/quote]

OK thank you for the correction. By the way it's a shame that Ivory Coast can not participate in this year's IMO. Did you know that last month Ivory Coast Won the Pan African Mathematical Olympiad (PAMO)?

http://www.samf.ac.za/pamo2009.html
Take a look at this nice site.
On the right you see results and there you can see that Ivory Coast took the overall championship and also the individual championship. Very impressive if you ask me.

I was really looking forward to see what the champion of Africa could do on the international stage. They even beat South Africa (which has been participating for years now at the IMO's and has much more facilities than Ivory Coast).

I guess we just have to wait until next year.

[quote="test20"]* Saudi Arabia only has registered four observers.
One of the observers is their team leader from last year.
They probably did not understand/use the registration software correctly.
Will they be able to register a team?[/quote]
They intend not to send anyone this year until they figure out their (or probably our  :( ) problem. Last year they though that Wael AlGhamdi who is also a Mathlinker will get a nice medal because he make full marks in the two Saudi Olympiad exams (links to Saudi MO [url=http://www.kfupm.edu.sa/mo/eng/WeeklyProblems_Exams/MO3_2008/MO3_2008_English_Stage1_Exam.pdf]Stage 1[/url] and Stage 2 [url=http://www.kfupm.edu.sa/mo/eng/WeeklyProblems_Exams/MO3_2008/MO3_2008_English_Stage2_Part1_Exam.pdf]Part 1[/url] and [url=http://www.kfupm.edu.sa/mo/eng/WeeklyProblems_Exams/MO3_2008/MO3_2008_English_Stage2_Part2_Exam.pdf]Part 2[/url] (can someone give me his opinion about them?)) and then he gets 6/42. In my view it's a very nice mark in a society that didn't help you at all in mathematics there is no available books for mathematics at bookstores and no one cares about young matematicians (in fact even old). So they are thinking of making a strong centers and programs in order to make our performance better.

[quote="hasan4444"]They [Saudi Arabia] intend not to send anyone this year until they figure out their (or probably our  :( ) problem. Last year they though that Wael AlGhamdi who is also a Mathlinker will get a nice medal because he make full marks in the two Saudi Olympiad exams and then he gets 6/42.[/quote]

Most of the students at the IMO perform excellently at their national competitions...
6/42 is not a bad score at an IMO.

I think you should first find out why Wael did not manage to solve problem 1. This was a standard geometry problem that allowed lots of different solutions; basically every approach lead to a solution.
It probably is something like drawing the right picture, or writing down all the information, or digesting the problem statement, or setting up the proof in several steps, etc.
And then set up your training around similar problems at a similar level, and try to repair this gap.

The problem is not with Wael for me at least he is a smart in my view but the major problem is with the whole situation. For example you are very interested and smart in school mathematics so you've decided to make more effort outside school books what you would do if you only know Arabic? No books available! No e-books available! so what might I do I will be tired and return to the school stuff. Remember not everyone will keep searching for things especially if the society will not help you they would say why are you turning in mathematics leave it for the smart countries. Most students hear this phrase so they get a huge shock and stop. Now they are seriously considering a new vision I hope they continue it  :)

[quote="hasan4444"]The problem is not with Wael for me at least he is a smart in my view [/quote]

Sorry, this was not my point.
Wael clearly is a smart guy.
He scored points on problem 4, but for some reason he got stuck with the much easier problem 1.
If you investigate into this, you might get some insight about the current shortcomings of your students.

Well maybe for some silly reason that's happen sometime like fear or misunderstanding  :wink:

There will be at least ten contestants in Bremen that are of age 14 or younger.

I find the following three contestants extremely interesting (and very promising for future IMOs!!!).
All three are from relatively strong countries, where it is fairly difficult to qualify for the team.
Hence I would guess that all three will end up with more than 25 points in Bremen:

Raúl Arturo Chávez Sarmiento (Peru), 11 years
  http://www.imo-official.org/year_reg_team.aspx?code=PER

Omer Cerrahoglu (Romania), 14 years
  http://www.imo-official.org/year_reg_team.aspx?code=ROU

Ufuk Kanat (Turkey), 14 years
  http://www.imo-official.org/year_reg_team.aspx?code=TUR

He test20

can you tell me what you think of the questions that were on the Pan African Mathematical Olympiad?

Do you consider them difficult or pretty easy?

http://www.samf.ac.za/pamo2009.html

On the right of that page you see question papers.

[quote="khashi70"][quote="test20"]Under the current political situation in Iran:
Does anybody know whether the Iranian team will participate in Germany as planned?[/quote]

Yes we will participate  :)[/quote]

Great!   :)

[quote="test20"][quote="ermath"]He test20

can you tell me what you think of the questions that were on the Pan African Mathematical Olympiad?
Do you consider them difficult or pretty easy?

http://www.samf.ac.za/pamo2009.html[/quote]

I find them "easy for a well-trained student".

I think the main issue is that most of them are mainstream or well-known, and can be settled by standard techniques:
* I have seen problem 1 in many variations at many competitions.
* Problem 2 is nice, but also known.
* Problem 3 is a well-known exercise; it is at the starting point of approximation theory.
* Problem 4 deals with so-called "total colorings" of the complete graph; I do not recall this problem from competitions, but it is a standard exercise in graph theory courses
* Problem 5 is an old functional equation that showed up at many competitions.
* Problem 6 is very nice (but I have definitely seen it before).

All in all, these are excellent puzzles for contestants that HAVE NOT GOTTEN much training.
Their solution requires mathematical insight and talent.

However, IMO contestants from countries like China/Russia/Korea/Vietnam/Iran would probably solve these six problems altogether within 10-30 minutes.[/quote]

Man 10-30 minutes? Wow that's impressive. It shows you how advanced those countries are (in IMO terms).

I really wonder why for instance a country like Iran is doing so unbelievably good, yet neighboring Pakistan (which has double the population) is nowhere near Iran's performances.
Can you tell me why that is. And why for instance India is not as good as you would expect (given their strong pool of engineers and scientists).

[quote="ermath"]I really wonder why for instance a country like Iran is doing so unbelievably good, yet neighboring Pakistan (which has double the population) is nowhere near Iran's performances.
Can you tell me why that is. And why for instance India is not as good as you would expect (given their strong pool of engineers and scientists).[/quote]The key for doing good in the IMO as a country is your ability to [b]find[/b] and thereafter [b]train[/b] the best students in your country.

I imagine that to be quite difficult in India, due to the vast population and huge restrictions in infrastructure compared to other countries. China on the other hand has similar problems in selecting the 6 students, [b]however[/b] due to the extremely well though training program, any of the 30 or so students that get selected for the final stages can be part of the team. They are all about the same level, and that's why you see them getting 5-6 gold medals every year.

In smaller countries such as Bulgaria, Romania, Hungary, you will see good results, which are mainly due to a good training and selection system, but you will also see some variation (in 96 Romania is #1, but in 2001 it went down to #15 and so on) in their results because the pool of the students is rather small compared with countries with 10 times the population (or more) that come in top 5 every year: China, US, Russia, Korea.

[quote="Valentin Vornicu"][quote="ermath"]
In smaller countries such as Bulgaria, Romania, Hungary, you will see good results, which are mainly due to a good training and selection system, but you will also see some variation (in 96 Romania is #1, but in 2001 it went down to #15 and so on) in their results because the pool of the students is rather small compared with countries with 10 times the population (or more) that come in top 5 every year: China, US, Russia, Korea.[/quote][/quote]

Hmm I think Korea can be considered simillar to Romania. Although Korea's population is double that of Romania's, there hasn't been an IMO team member who wasn't from Seoul (or its suburbs) in a few years. All the math-Olympiad people are from the capital city. So the selection pool-for the IMO is limited.

With a population of over 10 million, it is one of the world's largest cities and the second largest mega city in the world (has 24.5 million inhabitants). Only Seoul by itself is bigger than Romania :D

Quote:
Soo HONG Lee (student), Inst, of Sei. Ed. for the Gifted and Talented, Yonsei U„ Seoul, Korea

PRO! I used to be in this team :)

[quote=test20][quote="ermath"]He test20

can you tell me what you think of the questions that were on the Pan African Mathematical Olympiad?
Do you consider them difficult or pretty easy?

http://www.samf.ac.za/pamo2009.html[/quote]

I find them "easy for a well-trained student".

I think the main issue is that most of them are mainstream or well-known, and can be settled by standard techniques:
* I have seen problem 1 in many variations at many competitions.
* Problem 2 is nice, but also known.
* Problem 3 is a well-known exercise; it is at the starting point of approximation theory.
* Problem 4 deals with so-called "total colorings" of the complete graph; I do not recall this problem from competitions, but it is a standard exercise in graph theory courses
* Problem 5 is an old functional equation that showed up at many competitions.
* Problem 6 is very nice (but I have definitely seen it before).

All in all, these are excellent puzzles for contestants that HAVE NOT GOTTEN much training.
Their solution requires mathematical insight and talent.

However, IMO contestants from countries like China/Russia/Korea/Vietnam/Iran would probably solve these six problems altogether within 10-30 minutes.[/quote]

I don't think that IMO problems in recently years is quite easy to be solved in only 30 minutes!


(A+B)^3=A^3+3A^2B+3AB^2+B^3 -> AB=BA

please disprove this or find a counterexample

[quote="Green29"]$ (A \plus{} B)^3 \equal{} A^3 \plus{} 3A^2B \plus{} 3AB^2 \plus{} B^3$ -> $ AB \equal{} BA$

please disprove this or find a counterexample[/quote]
 We have $ (A \plus{} B)^3 \equal{} A^3 \plus{} 3A^2B \plus{} 3AB^2 \plus{} B^3$
Take $ n\equal{}2$
 $ A \equal{}\left[\begin{array}{cc} 1 & 0 \\ 0 & \minus{}2\end{array}\right]$
 $ B \equal{}\left[\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right]$
Then $ 3A^2B \equal{} 3B ,B^2\equal{}0 ,B^3\equal{}0$ and
 $ (A\plus{}B)^3\equal{}\left[\begin{array}{cc} 1 & 3 \\
0 & \minus{}8 \end{array}\right]\equal{} \left[\begin{array}{cc} 1 & 0 \\
0 & \minus{}8 \end{array}\right]\plus{}3\left[\begin{array}{cc} 0 & 1 \\
0 & 0 \end{array}\right]\equal{}A^3\plus{}3A^2B\plus{}3AB^2\plus{}B^3$
 But $ AB\equal{}\left[\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right]$
       $ BA\equal{}\left[\begin{array}{cc} 0 & \minus{}2 \\ 0 & 0 \end{array}\right]$
and $ AB \neq BA$

You've got a mistake, $ ABA \plus{} BAB \plus{} BA^2 \plus{} B^2A \equal{}A$

:blush: [b].... You are so fast to I edit  ....[/b] :blush:


geometry
trapezoid
rectangle
projective geometry
geometry proposed

AB is diameter in circle (K) . X is a point in the circle (k) , the tangent in A at (K) intersects (XB) in Y and the tangent in B at (K) intersects (XA) in Z , $t_x$ is the tangent in X at (K) .
prove that (YZ) , $t_x$ and (AB) are parallel or concurrent.

This was posted in the high school section as [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=74956]Easy concurency with tangents[/url]. Let the tangents $t_a, t_b$ at A, B meet the tangent $t_x$ at X at points C, D, respectively. As CA = CX and $\angle AXY = 90^\circ$, C is the midpoint of AY. As DX = DB and $\angle BXZ = 90^\circ$, D is the midpoint of of BZ. The line $CD \equiv t_x$ connecting the midpoints of the bases AY, BZ of the trapezoid ABZY passes through the intersection of its shoulder side lines AB, YZ. Unless the trapezoid is a rectangle, in which case $AB \parallel CD \parallel YZ.$

The proposition is true even if AB is not a circle diameter. Considering the triangle $\triangle ABX$ as a degenerate cyclic hexagon $AA'BB'XX'$, by Pascal's theorem, the intersections $Y \equiv t_a \cap BX,\ Z \equiv t_b \cap AX,\ P \equiv t_x \cap AB$ are collinear. Unless P is at infinity, in which case $AB \parallel t_x \parallel YZ.$


geometry
conics
geometric transformation
rotation
calculus
trigonometry
function

Well, it's very hard to actually define what falls in the range of "precalc."  I guess what is important for competitions is the analytic geometry part, such as conics and their rotations and whatnot.

Good luck!  I skippied precalc and took calculus in 9th grade too :D.

Ummm... what about memorization?  I heard people saying you gotta memorize a lot for trig.  What do you have to memorize?  Sine/Cosine values listed in the table in AoPS volume 1?  What else?

I suppose you've taken Algebra II. Pre-calc is an extension of that. Additionally, you might learn more about complex numbers, polar graphs, and that good stuff.  :)

Well, I don't think you have to [i]memorize[/i] trig functions and identities. Once you've worked with them a bit and know their origin, you won't need to memorize them. It'll just come to you.

Excuse me for double posting. I was too lazy to click the edit button.  :P

As far as Mu Alpha Theta goes,

Precalculus is defined as:

[b]1) Trigonometry[/b]
i.e. - the unit circle, the pythagorean identities, and other identities (for example, sin2A = 2sinAcosA)
[b]2) Analytic Geometry[/b]
i.e. - rotation of axes, conics, finding the asymptote, etc.
[b]3) Polar Coordinates[/b]
i.e. - Polar Graphs (like the cardiod), conversions to Cartesian Coordinates, Cylindrical Coordinates, etc.
[b]4) Complex Numbers[/b]
i.e. - DeMoivre's Theorem, Euler's Theorem, taking the root of a complex, etc.
[b]5) Matrices & Vectors[/b]
i.e. - row echelon form, dot product, cross product, box product, projections, etc.
[b]6) Parametric Equations[/b]
i.e. - the Cycloid x = a(T - sinT) & y = a(1 - cosT)
[b]7) Limits & Derivatives[/b]
i.e. - derivative of sine, limit as x tends towards infinity of some equation, very basic stuff and not calculus
[b]8) More Algebra[/b]
i.e. - equations & inequalities, logs & exponents, etc.
[b]9) Higher Level Geometry[/b]
i.e. - pretty much geometry (not olympiad level)

The list goes on forever into less important topics. The point is, there's a little bit of everything, but it focuses on the first 7 listed above.

Precalculus also can be put to practical applications in physics, especially AP Physics B, for example: finding forces using vector addition.

For the AMC competitions, I've found it useful to ignore #6 and #7. The others would definitely help.

Don't be intimidated by the complex names. It's not as bad as it sounds. :P

[quote="mysmartmouth"]I am gifted, but not exceptionally gifted.  I am in 8th grade, and I want to take precalculus.  However, for weird reasons, I cannot take precal at school this early.  So I want to do self study.  While I do not have a tutor, I have AoPS, as well as an excellent math coach I can consult every so often.  I just went and bought a precalculus book.  So here's my problem:  I don't even know exactly what precalculus is.  I want to learn precal for math competitions, not to pass an exam.  With this in mind, can someone explain to me what precal is, and what I need to learn for a math competition.  I have already acquired a precal textbook, and I just need to know how to get started.

Thanks![/quote]
What are you taking now? When I was in 8th grade, I was in geometry....

work on trigonometry especially, its an integral part of precalc.

Precalc at my school included just about all of beginning calculus, including integrals (which I hadn't seen mentioned before). It's not too complicated, and you can learn almost all of it out of AoPS volume 2 except for any calculus ideas, in which case I'd get a calc book for that.


Ceea ce mi-am propus sa prezint ar fi fost mai bine plasat intr-o subsectiune [b]"Intermediate topics"[/b] pentru gimnaziu, insa asa ceva nu exista pe Site - ul national (sugerez [u]administratiei[/u] introducerea acestei subsectiuni).

[color=red]Dragii mei, voi stiti sa rezolvati un sistem de cel mult doua ecuatii liniare cu cel mult doua necunoscute (sau un sistem reductibil la acesta), chiar prin cel putin doua metode ( [u]metoda[/u] este un [u]artificiu[/u] aplicat de cel putin doua ori). Dar daca va intalniti intr-o modelare matematica (in fizica, chimie etc) cu un sistem de $m\ge 1$ ecuatii cu $n\ge 1$ necunoscute ? Stiti sa va descurcati ? Iata ce doresc sa ating in final prin aceasta [u]lectie-mataron.[/u] Astfel veti fi mai bine pregatiti pentru a intelege ce veti face in clasa a XI - a la algebra, sau cel putin pentru examenul de capacitate.[/color]

[color=blue]Vom nota prin $S$ solutia unui sistem de ecuatii liniare. Pentru inceput vom discuta [b]metoda substitutiei[/b]. Va ofer cateva exemple pe care le vom comenta mai tarziu. Unora s-ar putea sa li se para simple, ceea ce nu este rau, insa ii rog sa vada in ele cum "filogeneza repeta ontogeneza", adica intr-o evolutie finita se vede toata desvoltarea de la origini pana azi ( generalul poarta in sine aspectele esentiale ale particularului ).

$\underline {1.\ \ m=1\ ;\ n=2\ .}$

$\blacktriangleright\ 1.1.\ \ 2x-3y=6\ .$
$\blacktriangleright\ 1.2.\ \ (x-1)y+2=x(y+1)-y\ .$
$\blacktriangleright\ 1.3.\ \ x^2-y^2=(x+y)(x-1)\ .$
$\blacktriangleright\ 1.4.\ \ 4+x+y=3(x-y)-2x+4(y+1)\ .$
$\blacktriangleright\ 1.5.\ \ 2+x+y=2(x-y)-x+3y\ .$

$\underline {2.\ \ m=2\ ;\ n=1\ .}$

$\blacktriangleright\ 2.1.\ \ 3x=6\ ;\ 2x+1=x+3\ .$
$\blacktriangleright\ 2.2.\ \ 2x-4=x+1\ ;\ x-3=4x\ .$
$\blacktriangleright\ 2.3.\ \ x+3=2(x+1)\ ;\ 3x+4=x+6\ .$
[hide="Aici sunt raspunsurile."]
$S_{1.1.}=\{(3m,2m-2)\ |\ m\in R\}\ ;$
$S_{1.2.}=\{(2,m)\ |\ m\in R\}\ ;$
$S_{1.3.}=\{(m,-m)\ |\ m\in R\}\cup \{(n,1)\ |\ n\in R\}\ ;$
$S_{1.4.}=R^2\ ;$
$S_{1.5.}=\emptyset\ .$
$S_{2.1.}=\{2\}\ ;$
$S_{2.2.}=\emptyset\ ;$
$S_{2.3.}=R\ .$[/hide]
$\underline {3.\ \ m=n=2\ .}$

Deoarece acest caz va este familiar, nu-l voi ilustra cu exemple, mai ales ca voi cunoasteti si o frumoasa interpretare geometrica in plan (drepte concurente, paralele sau confundate). Intrebarea este de ce sa ne oprim aici ? Mai exista si spatiul, chiar dincolo de el (veti invata mai tarziu) exista si alte entitati geometrice ("hiperspatii"). Noi insa ne intoarcem la aspectul algebric al unui sistem de ecuatii liniare. Urmarind cu atentie solutiile sistemelor de tip $(m,n)\ ,$ unde $1\le m,n\le 2$ se observa ca in rezolvarea acestora  nu exista decat trei [b]"iesiri"[/b] : sistemul nu are solutii (imposibil, [u]incompatibil[/u] etc), adica $S=\emptyset$ ; sistemul are cel putin o solutie si aceasta este unica ([u]compatibil unic determinat[/u]), adica multimea $S$ are un singur element (multime atomica) ; sistemul are cel putin doua solutii si in acest caz va avea o (simpla sau dubla) infinitate de solutii ([u]compatibil nedeterminat[/u]) si in acest caz $S=R$ (dreapta sau o dreapta in plan) sau $S=R^2$ (planul).[/color]

[color=darkred]Sa consideram asadar un sistem concret de [b]patru[/b] ecuatii liniare cu [b]cinci[/b] necunoscute in a carui rezolvare vom proceda identic, prin metoda substitutiei, ca la sistemele de tip $(m,n)\ ,$ unde $1\le m,n\le 2\ :$
$\{\begin{array}{cccc} 2a-b+c-3d-e=-2\\\ a+b+2c-2d-2e=5\\\ 4a+b+5c-7d-5e=8\\ a+4b+5c-3d-5e=17\end{array}$
Alegem o ecuatie din care sa eliminam una dintre necunoscute. De exemplu, din prima ecuatie eliminam necunoscuta $b$ si o introducem in celelalte trei ecuatii ramase : $\boxed {\ b=2a+c-3d-e+2\ }\ .$ Vom obtine un sistem de trei ecuatii liniare cu patru necunoscute (cu o ecuatie mai putin si cu o necunoscuta mai putin decat sistemul initial : odata cu necunoscuta $b$ care dispare temporar din sistem si la care ne vom intoarce dupa ce vom determina necunoscutele $a,c,d,e$ dispare definitiv prima ecuatie din care am facut eliminarea necunoscutei $b$) :
$\{\begin{array}{ccc} 3a+3c-5d-3e=3\\ 6a+6c-10d-6e=6\\ 9a+9c-15d-9e=9\end{array}$
Reluam procesul cu acest ultim sistem obtinut. Din prima ecuatie (de exemplu) eliminam necunoscuta $a$ si o introducem in celelalte doua ecuatii ramase : $\boxed {\ 3a=-3c+5d+3e+3\ }\ .$ Vom obtine un sistem de doua ecuatii liniare cu trei necunoscute :
$\{\begin{array}{cc} 0\cdot c+0\cdot d+0\cdot e=0\\ 0\cdot c+0\cdot d+0\cdot e=0\end{array}$
Acest ultim sistem are solutia $R^3\ ,$ adica necunoscutele (variabilele) $c,d,e$ sunt libere (pot lua orice valori reale). Se observa ca in acest proces (algoritm) nu am intalnit nici o contradictie (de exemplu $0=1$), ceea ce inseamna ca sistemul este compatibil. Necunoscutele care se exprima in functie de altele (vezi relatiile inpachetate)  le vom numi [b]necunoscute dependente (principale)[/b] iar celelalte le vom numi [b]necunoscute independente (secundare)[/b]. In cazul nostru necunoscutele $a,b$ sunt principale iar necunoscutele $c,d,e$ sunt secundare, adica solutia sistemului are trei [u]grade de libertate[/u] (numarul variabilelor independente). [u]Numarul necunoscutelor secundare masoara[/u] [b]gradul de nedeterminare (gradele de libertate) al sistemului[/b]. In cazul nostru, [u]sistemul are atatea solutii cate puncte in spatiu exista ![/u] Vom nota $c=m$, $d=n$, $e=p$ unde marimile $m,n,p$ parcurg independent una de cealalta multimea numerelor reale $R$ (axa reala). Acestea le vom introduce in relatiile de dependenta ale necunoscutelor $a,b$ (de la ultima catre prima) si vom obtine :
$a=-m+\frac 53 n+p+1$ si $b=-m+\frac 13n+p+4\ .$
In concluzie, sistemul initial este [b]compatibil[/b] $3$ [b]- nedeterminat[/b] si solutia acestuia este :
$a=-m+\frac 53 n+p+1\ ;\ b=-m+\frac 13n+p+4\ ;\ c=m\ ;\ d=m\ ;\ e=p\ ,$ unde parametrii $m,n,p$ apartin multimii $R$. Cu alte cuvinte, solutia $S$ a sistemului initial este :\[ \boxed {\ \{(-m+\frac 53 n+p+1,-m+\frac 13n+p+4,m,n,p)\ |\ m,n,p\in R\}\ }\ .  \][b]Observatie.[/b] Noutatea acestei extinderi consta de fapt in introducerea unor notiuni noi (daca sistemul este compatibil) : necunoscuta (variabila) principala (dependenta) ; necunoscuta (variabila) secundara (independenta) ; grad de nedeterminare al sistemului (grade de libertate ale solutiei) ; hiperspatiul  de dimensiune $n\ \ ,\ R^n=\{\ (x_1\ ,x_2,\ \ldots \ ,\ x_n)\ |\ x_1,x_2,\ldots x_n\in R\}$ : in particular, planul este $R^2$ si spatiul este $R^3$ . Pentru moment nu avem un test de compatibilitate al sistemului; raspunsul se va obtine printr-un proces "orb" de cautare a solutiei. Am putea sa avem surpriza lui Columb, care a descoperit America printr-o intamplare fericita : de fapt el si-a propus, pornind spre Vest, sa ajunga in India (la Est) ! Aceasta problema, de a ne convinge mai intai ca sistemul este compatibil si apoi sa-i determinam solutia, va fi obiectul de studiu la Algebra (liniara) in clasa a XI - a. [b][u]Dovada existentei unui obiect matematic precede (logic) intotdeauna determinarea (constructia) acestuia ![/u][/b][/color]

Ai [b]rabdare[/b], Sorine, vine cat de curand si randul tau ! Poti sari peste etapele preliminare. Dar ai sesizat esentialul (ceea ce este comun, indiferent de dimensiunile sistemului) ? Poti sa-l enunti macar ?

Domnule profesor Nicula, nu prea am rabdare cu problemele astea.Sunt cam simple...dar, cum ati zis, sper sa vina si randul meu... :)

Dar nu este vorba sa le rezolvi, ci sa lecturezi raspunsurile. Vezi esentialul ? Asta-i problema pentru tine ! Sorry, nu am timp $\Longleftrightarrow$ nu stiu.

Aici nu inteleg.Cum sa [b]lecturez[/b] raspunsurile? :? La ce va referiti?

Intoarce-te la mesajul initial si vei capata raspunsul.

Domnule Nicula, din cate am observat pana acum, aveti un stil interesant de a explica notiunile legate de matematica, respectiv orice alt tip de notiune :) . Nu vreau sa va intrerup din aceasta lectie pe care ati inceput-o, insa vreau sa va intreb un lucru. Dupa cum ati punctat adineauri (intr-un raspuns pt. Sorin) [u]nu este indicat[/u] sa trecem repede cu vederea, peste ceea ce este mai ushor(in sens matematic, bineinteles). Deci daca la un moment dat invatzam niste notiuni noi, urmate de un set imens de aplicatii are rost sa fie parcurse toate? (aceasta cand se vede ca gradul de dificultate al acelor aplicatii nu prea creshte, adica avem de-a face cu cateva ore de monotonie pt a le rezolva). Sunt de acord ca: [color=red][b]repetarea e mama invatzaturii[/b][/color] sau alftel spus [b][color=red]practice makes perfect[/color][/b] dar as vrea sa aud parerea dvs. Va multumesc anticipat! :)

Stimabile domn profesor V.Nicula, nu va continuati lectia? :) Mie nu prea imi place cand cineva incepe un lucru si nu-l continua. :(

[quote="Virgil Nicula"][color=red]Ceea ce mi-am propus sa prezint ar fi fost mai bine plasat intr-o subsectiune [b]"Intermediate topics"[/b] pentru gimnaziu, insa asa ceva nu exista pe Site - ul national (sugerez [u]administratiei[/u] introducerea acestei subsectiuni).[/color][/quote] [color=darkblue][i]Dragii mei Stizzel si Sorin[/i], ma bucur ca sunteti nerabdatori sa continui pentru a intelege ce doresc cu aceasta lectie maraton.

$\blacktriangleright$ Nu ati citit insa cu atentie ceea ce am scris mai sus.
$\blacktriangleright$ Va rog sa-mi acordati circumstante atenuante pentru intarzierea aparitiei mele pe acest post din urmatoarele trei motive :
$1.$ Sunt bunic si am doi nepotei ;
$2.$ Am cateva ore la un liceu si se apropie sfarsitul anului (extemporale, teze de corectat, diverse situatii de incheiat, pregatire pentru Bac, meditatii, consultatii, cerc de matematica  etc.) ;
$3.$ Aceasta lectie este gratuita.

Va rog sa apreciati pe toti cei care prezinta aspecte teoretice pe acest Site (vezi subsectiunile finale) si solutii detaliate la problemele propuse (de exemplu Darij Grinberg, Yetti etc) - [u]gratuit[/u].

In concluzie, voi intrerupe aceasta lectie sine die.
Cand va apare "Intermediate Topics" pentru Gimnaziu voi reveni.[/color]

Am inteles. :) Succes la liceu si sa va traiasca nepoteii! :lol:

[quote="stancioiu sorin"][b][u]Stimabile[/u][/b] domn profesor V.Nicula,...... [b][u]mie nu prea imi place cand cineva incepe un lucru si nu-l continua[/u][/b]. [/quote][color=darkred]Dragul meu, iti multumesc de urari. Citeste ceea ce este ingrosat si subliniat in textul de sus care iti apartine. Numai daca este ceva in neregula, trimite-mi te rog, doar un scurt mesaj privat (cel mult cinci cuvinte si eventuale semne de punctuatie).[/color]

ce urat e atunci cand elevii nu au un minim bun simt/respect fata
de un profesor.


Find the number of 15-centimetre square tiles required to cover a rectangular floor 5.4m long and 4.05m wide.

plz do not take ppl's ans and make it ur own. thanxx :)


I'm a little messed up with mods after sitting and thinking about them for two hours straight...

If you have a^c is congruent to b^c mod d then is a^2c congruent to b^2c mod d^2?
And
if you were using a base higher than 10, would you have to consider letters as last digits?

1. No, $a^c \equiv b^c \pmod {d} \implies a^{2c} \equiv b^{2c} \pmod {d}$.

2. If you consider it in mod 16, for example, and you find that $3^{11} \equiv 11 \pmod {16}$, then the last digit of $3^{11}$ when written in base 16 is B.

or, more generally,

$a\equiv b\pmod d \implies a^n\equiv b^n \pmod d$


[quote="E^(pi*i)=-1"]Square ABCD has area 3.  E is on the extension of AD such that BE intersects CD at F and angle BFC is 60 degrees.  Find the length of DE.[/quote]

Using a bit of trig, I got $\sqrt{3}(\sqrt{3}-1)$

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=757[/img]
[url=http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=757]Link if image doesn't work[/url]
(It is a PNG Image)
[hide="Click to reveal solution"]
[ABCD]=3, thus AB=BC=CD=AD=$\sqrt{3}$.
The interior angles of a square are 90 degrees so we have a 30-60-90 Right Triangle in the corner. BC=$\sqrt{3}$, thus FC=$1$ and BF=$2$.
We now know that DF=$\sqrt{3}-1$.
Angle EDF=90 degrees and <DFE=60 degrees. We have another 30-60-90 Right Triangle. DF=$\sqrt{3}-1$ thus DE=$\sqrt{3}(\sqrt{3}-1)=3-\sqrt{3}$
[/hide]


Let $ A_1,A_2,...,A_n$ be the sets consisting of $ n$ segments on a given line. Prove that the intersection $ A_1\cap A_2\cap ...\cap A_n$ contains not more than $ n^2\minus{}n\plus{}1$ disjoint segments.

Anybody?

I think this is from Belarus, if I'm not wrong.  :lol:


if you take p/q where $p=q=1$, and follow it by the fraction (p+2q)/(p+q), why does this sequence converge to $\sqrt{2}$?

I'm not quite sure what you mean... What, specifically, is the sequence (or is it a series)?

this is equivalent to having x, then take it to: $\frac{x+2}{x+1}$

but if x is negative, then it goes to $-\sqrt{2}$

also it does not oscillate, since y^n(x)=x where y(x)=(x+2)/(x+1) has the same solutions [nth composition]

Oh, so he's asking why the sequence $a_{n}=f^{n}(1)$ where $f(n)=\frac{x+2}{x+1}$ converges to $\sqrt{2}$?

[quote="iamagenius"]why is it that this is equivalent to having x, then take it to: $\frac{x+2}{x+1}$?[/quote]

If the number is $x=\frac{p}{q}$, the function is $f(x)= \frac{p+2q}{p+q}= \frac{ \frac{p}{q}+2}{\frac{p}{q}+1}= \frac{x+2}{x+1}$.

As far as the continued fraction goes, this is exactly the computation that produces the convergents of the continued fraction of $\sqrt{2}$, as The Zuton Force noticed.

[b]Follow-up:[/b]  Describe the set of all algebraic integers with a continued fraction of period $1$.


Let Xn be sequence of random variables such that
P[ Xn=0] = 1- (1/n) and P[Xn=1] = 1/n   for n=1,2,3,....

Prove that Xn doesn't converge a.s.

A hint is enough.

Sorry...... the sequence is an independent sequence.

do you know the "second" Borel-Cantelli lemma: [url]http://en.wikipedia.org/wiki/Borel-Cantelli_lemma[/url] ? It should help ..

Thanks Mr. Alekk. In fact I got the solution well before your reply. I just applied the Borel Cantelli lemma and  lim sup P[An] < = P[An i.o.]


Can a good math teacher get his or her message across to students without using technical math jargon?

Your view?

I think a good teacher does their job by making students understand technical math jargon