inequalities
algebra
polynomial
inequalities unsolved

suppose that x,y,z>0 and we have $ x\plus{}y\plus{}z\equal{}3$
prove that

$ \sum \frac {x^3}{y^3 \plus{} 8} \geq \frac {1}{9} \plus{} \frac {2}{27} \times(xy \plus{} yz \plus{} zx)$

[quote="sumita"]suppose that x,y,z>0 and we have $ x \plus{} y \plus{} z \equal{} 3$
prove that

$ \sum \frac {x^3}{y^3 \plus{} 8} \geq \frac {1}{9} \plus{} \frac {2}{27} \times(xy \plus{} yz \plus{} zx)$[/quote]
By Cauchy - Schwartz $ \sum_{cyc} \frac {x^3}{y^3 \plus{} 8} \equal{} \sum_{cyc}\frac {x^4}{xy^3 \plus{} 8x}\geq\frac {(x^2 \plus{} y^2 \plus{} z^2)^2}{x^3z \plus{} y^3x \plus{} z^3y \plus{} 24}.$
By Vasc's inequality $ x^3z \plus{} y^3x \plus{} z^3y\leq\frac {1}{3}(x^2 \plus{} y^2 \plus{} z^2)^2.$
Hence, it remains to prove that
$ (x^2 \plus{} y^2 \plus{} z^2)^2\geq\left(\frac {1}{9} \plus{} \frac {2}{27}(xy \plus{} xz \plus{} yz)\right)\left(\frac {1}{3}(x^2 \plus{} y^2 \plus{} z^2)^2 \plus{} 24\right).$
But $ (x^2 \plus{} y^2 \plus{} z^2)^2\geq\left(\frac {1}{9} \plus{} \frac {2}{27}(xy \plus{} xz \plus{} yz)\right)\left(\frac {1}{3}(x^2 \plus{} y^2 \plus{} z^2)^2 \plus{} 24\right)\Leftrightarrow$
$ \Leftrightarrow81(x^2 \plus{} y^2 \plus{} z^2)^2\geq(3 \plus{} 2(xy \plus{} xz \plus{} yz))((x^2 \plus{} y^2 \plus{} z^2)^2 \plus{} 72)\Leftrightarrow$
$ \Leftrightarrow39(x^2 \plus{} y^2 \plus{} z^2)^2\geq(x^2 \plus{} y^2 \plus{} z^2)^2(xy \plus{} xz \plus{} yz) \plus{} 72(xy \plus{} xz \plus{} yz) \plus{} 108,$
which is true because $ (x^2 \plus{} y^2 \plus{} z^2)^2(xy \plus{} xz \plus{} yz)\leq3(x^2 \plus{} y^2 \plus{} z^2)^2$
and $ 72(xy \plus{} xz \plus{} yz) \plus{} 108\leq324\leq36(x^2 \plus{} y^2 \plus{} z^2)^2.$

furthermore we have:

$ \sum \frac {x^3}{y^3 \plus{} 8}\geq \frac {1}{3} \geq \frac {1}{9} \plus{} \frac {2}{27} \times (xy \plus{} yz \plus{} zx)$

and is easy to prov both sides

by AM-GM we have :
$ \frac{x^{3}}{y^{3}\plus{}8}\plus{}\frac{y\plus{}2}{27}\plus{}\frac{y^{2}\minus{}2y\plus{}4}{27} \geq \frac{x}{3}$
now adding similiar inequalities and u get result .

$ \sum \frac {x^3}{y^3 \plus{} 8} \geq \frac {{(x^3 \plus{} y^3 \plus{} z^3)}^2}{(x^3y^3 \plus{} y^3z^3 \plus{} z^3x^3) \plus{} 8(x^3 \plus{} y^3 \plus{} z^3)}$$ \geq\frac {{(x^3 \plus{} y^3 \plus{} z^3)}^2}{\frac {1}{3}{(x^3 \plus{} y^3 \plus{} z^3)}^2 \plus{} 8(x^3 \plus{} y^3 \plus{} z^3)}$.
$ \frac {{(x^3 \plus{} y^3 \plus{} z^3)}^2}{\frac {1}{3}{(x^3 \plus{} y^3 \plus{} z^3)}^2 \plus{} 8(x^3 \plus{} y^3 \plus{} z^3)}\geq \frac {1}{3}\longleftrightarrow$
$ (x^3 \plus{} y^3 \plus{} z^3)\geq3$

you can use couchy and also generalized jensen for convexity of $ \frac {1}{x}$ to prove first inequality


hi folks,

i've just arrived here. the campus sucks, the hosting conditions are pretty bad. however the city looks nice, and there are a lot of nice opportunities to have fun :)

after a 10-hour drive from Bucharest, in which I succesfully passed through 4 police radars without being caughts over-speeding :D:D ... I am feeling pretty tired, so that's today's report :)

other people from IMC wanting to say something, I welcome you here :)

to all good luck at the contest (for which I have not yet prepared :D;) ) ...

[quote="Moubinool"]Post the problems day 1 and 2 as soon it is possible.
thx[/quote] I have just done that. Have fun.

PS I have solved 1,2,3 and 4. Spend a lot of time on 5, but unfortunately I did not get it :(

[quote="Valentin Vornicu"]
PS I have solved 1,2,3 and 4. Spend a lot of time on 5, but unfortunately I did not get it :([/quote]

Don't worry about problem no5. - you've done well anyway. I wish I've done so good    :(


If $ n$ is a positive integer, then
$ \frac{1}{\sqrt{3n\plus{}1}}(\frac{2n\plus{}1}{2n\plus{}2})\leq\frac{1}{\sqrt{3n\plus{}4}}$

multiply both sides by $ \sqrt {3n \plus{} 1}$ (eheh) to get $ \frac {2n \plus{} 1}{2n \plus{} 2} \le \frac {\sqrt {3n \plus{} 1}}{\sqrt {3n \plus{} 4}}$
square both side, which is valid since both sides are positive
$ \frac {4n^4 \plus{} 4n \plus{} 1}{4n^2 \plus{} 8n \plus{} 4} \le \frac {3n \plus{} 1}{3n \plus{} 4}$
multiply both sides by $ (4n^2 \plus{} 8n \plus{} 4)(3n \plus{} 4)$, which is positive
ugh
*grind*
$ 12n^3 \plus{} 28n^2 \plus{} 19n \plus{} 4 \le 12n^3 \plus{} 28n^2 \plus{} 20n \plus{} 4$
$ 0 \le n$
which is true
reverse all the steps to get the original inequality

Don't you mean multiply both sides by $ \sqrt{3n \plus{} 1}$ instead of $ \frac{1}{\sqrt{3n \plus{} 1}}$?


there are no rules except for the following.
only say the word/phrase,
say the correct word/phrase to get a point, and
only i make the unknowns.
here is the first unknown:

this one is easy

i reveal points/answers each day.

lingomaniac88:1 point

unknown 2:

a theif :ninja: stole the keys to a house. when he tried to open the door, it didnt work. he yelled "the key isn't working!" "thats because its a fake!" he turns around to face the police.

morale: things aren't always as they seem.

maybe this would do better in puzzles and brainteasers?


calculus
integration
derivative
calculus computations

hi can anyone help me solving this..
the question is to find the lenght of the arc of f(x)=x^(2/3)-1 between 8 and 27. after using the arc lenght formula i found that the length is the definite integral from 8 to 27 [1+(4/9x^(2/3))]^(1/2) dxbut i am unable to integrate that.
thanks

[quote="Umi"] the length is the definite integral from 8 to 27 [1+(4/9x^(2/3))]^(1/2) dx but i am unable to integrate that.
[/quote]
Actually it is $\int_8^{27}\left[1+\frac 49 x^{-\frac 23}\right]^{\frac 12}\,dx=\int_8^{27}\left[x^{\frac 23}+\frac 49 \right]^{\frac12}x^{-\frac13}\,dx$. Now notice that the derivative of the expression in the brackets in the last integral is $\frac23 x^{-\frac  13}$, which is almost the same as what stands outside the brackets. You should be able to continue from here.

oh yes true i was looking for the du but couldnt see it. thanks a lot for your help.


finding zeros of functions

many forms of finding solutions to equations of form f(x)=0. one mwthod requires the equation to be expressed in the form x=g(x) so that asolution can be interpreted as the value of x where the line x=y intersect the curve y=g(x). this is described as the fixed point of the function g(x).
let g(x)=cosx. show graphically that g(x) has one fixed point between x=0 and x=2.

falula11,
You should make sure that your post belongs in the forum you are posting to, and this is completely the wrong forum.

You may wish to try this one instead: [url=http://www.mathlinks.ro/Forum/index.php?f=296]www.mathlinks.ro/Forum/index.php?f=296[/url]


Let $ a,b,c\geq0$ s.t. $ ab\plus{}bc\plus{}ca\equal{}3$ and let $ r\geq1$.Prove that:
$ \dfrac{1}{r\plus{}a^2\plus{}b^2}\plus{}\dfrac{1}{r\plus{}b^2\plus{}c^2}\plus{}\dfrac{1}{r\plus{}c^2\plus{}a^2}\leq \dfrac{3}{r\plus{}2}$

Sorry, but no one can solve?:?:

Where are you, arqady?


Find the number of ordered pairs of positive integers $(a,b)$ such that $a+b=1000$ and neither $a$ nor $b$ has a zero digit.

That only covers the three-digit numbers.  You missed cases like $(1,999)$.  Also, a little more explanation would be useful.

[hide]We'll start with the case where both a and b are 3 digit numbers.  b=(1000-a) will have no zero digits as long as the hundreds digit of a is 1-8, the tens digit of a is 1-8, and the units digit of a is 1-9.  This gives 8*8*9=576 possibilities.  If a is two digits, a can have a tens digit 1-8 and a units digit 1-9.  This gives 8*9=72 possibilities.  If a is one digit, a can be 1-9, for 9 possibilities.  So the total number of ordered pairs (a,b) is 576+72+9=657.[/hide]

I'll take a stab at this

[hide]
if we're looking for 2 numbers that add up to 1000, we can just count the number in the range 511-999 that satisfy the conditions: no zero, no zero in pair

for 511-899, we can use $4$ possible hundreds digits, $8$ possible tens digits, and $9$ possible units digits, for $4 \cdot 8 \cdot 9 = 288$

for 911-999, we can use $9$ possible tens and $9$ possible units, for $81$

$288 + 81 = \boxed{369}$

[i]edit (lol)[/i], at first I didn't realize how an answer could exceed 500- the a and b values can just be switched, so the final answer is $2 \cdot 369 = \boxed{738}$.
[/hide]

[hide]@E^(pi*i)=-1: You forgot when b is the number with the smaller number of digits. Just add 72 and 9 an extra time, to get 738.[/hide]

[hide]We're going to count the number of pairs  where there is a zero digit in them, then subtract them out from , because I'm only considering , and thus  total pairs.

By PIE, we know the total number of ordered pairs with a zero in them is

For the  variable, for each  digit number, there are  possible numbers such that there is NO zero in it. This is because we can choose any number one through nine for each digit. So for a, there are  numbers in the interval without a zero, and conversly there are  numbers with a zero in them.

For , the process is identical, so we end up with another  pairs.

Now, to find out how many pairs have both variables containing a zero, we need to consider how we can place them. If there is a zero in the units digit of , there MUST be a zero in 's unit digit, otherwise there won't be a zero in the final sum's unit digit (we aren't carrying anything over). If we have a zero in the ten's digit of , we either have a zero or a nine in the ten's digit of . If we have a , then we had to have carried from the unit's column, and thus they both have nonzero digits there, and clearly we can't have a zero digit in the hundred's column, so only one of the numbers will have a zero in the ten's digit. If we have a zero in both variables' ten's digits, then we aren't carrying, and thus we had zero in both of the unit's as well.

Thus, we basically consider all the possible pairs where there is a zero in both of their unit's digit, and it's clear there are  posibilities here.

So the total number of ordered pairs with a zero in them is , and the number of ordered pairs WITHOUT a zero in them is .

[/hide]

heres another approach

[hide]
Start with 999 total ordered pairs and subtract from it:
there are 9 numbers of the form a0
81 of the form ab0 (with no other nonzero digits)
9 of the form a00
In each of these types, the numbers pair also has a zero, so we subtract them only once.

In numbers of the form
a0b (81 of these)
the other number will have digits

[9-a]9[10-b], none of which can be zero, (except possible the first which is ok)

So, we must subtract these ones twice, since this type of ordered pair occurs with a number like this as both the first and second term.
999-9-81-9-2*81=738

[/hide]
Edit:  I guess this is basically the same as the above post but with less formal explanation.

[quote="Pocket Sand"]for 511-899, we can use $ 4$ possible hundreds digits, $ 8$ possible tens digits, and $ 9$ possible units digits, for $ 4 \cdot 8 \cdot 9 \equal{} 288$

for 911-999, we can use $ 9$ possible tens and $ 9$ possible units, for $ 81$

$ 288 \plus{} 81 \equal{} \boxed{369}$[/quote]

Why did you make two intervals, 511 - 899 and 911-999, instead of just 511-999?  Thanks!

Start with 999 possibilites for (a,b) such that a+b=1000. First find all of those with 0s in them. These will only be of the form xy0 or x0y.

Start with xy0. This gives 10r=a and 10s=b. SInce both would have to be multiples of 10 to obtain 1000 as a sum the following is derived.
r+s=100
This gives 99 possibilities.

Now find those of the form x0y. Since these will all follow x00<x0y<x10, there are 9 values for each x. SInce there are 9 values for x, there are 9*9=81 possibilities. Symmetry between a and b yields 162 possibities.

Therefore the number of (a,b) is 999 - 99 - 162 = 738.

[hide=Solution]
[b]Case 1:[/b] One number contains a $0.$

Notice this only happens if and only if the tens digit of the three digit number is a $0.$ The hundred’s digit and the one’s digit can be anywhere from $1$ to $9.$ So, $9 \cdot 1 \cdot 9=81$ possibilities. We have to multiply this by $2$ since this can be either $a$ or $b.$

[b]Case 2:[/b] Both numbers contains a $0.$

This happens if and only if the one’s digit is $0.$ The tens and hundreds digit can be $0$ to $9.$ However the number can’t be $0.$ So, $10 \cdot 10-1=99.$

The total number of ways is
$$\binom{n-1}{r-1}=\binom{1000-1}{2-1}=\binom{999}{1}=999$$
Thus, $999-99-81\cdot 2=\boxed{738}.$

We use casework with complementary counting.

Case 1: $a$ has one digit and $b$ has three digits.

We can easily see that there are $0$ ways here.

Case 2: $a$ has two digits and $b$ has three digits.

This can work in two ways; Either the last digit of both is $0$ or the middle digit of $b$ is $0$. The first option can happen in $9$ ways ($a=10,20,\ldots,90$) and the second option can happen in $9$ ways $(b=901, 902,\ldots,909)$, for a total of $18$ ways.

Case 3: $a$ has three digits and $b$ has three digits.

Note that this will only occur if $a \in [100,900]$. We can have both $a,b$ ending in $0$, for $90-10+1=81$ ways. Or, we can have one of $a,b$ have middle digit $0$. For right now say it's $a$ - we'll have to multiply by $2$ at the end, because we're looking for ordered pairs and it's possible for $b$ to have middle digit $0$ too. Our possibilities for $a$ that weren't counted in our previous subcase (the one with $81$ ways) are $101, 102, \ldots 109, 201, 202, \ldots 209 \ldots 809$, for a total of $9 \cdot 8 = 72$ ways. Doubling and adding to $81$ gives $225$ ways here.

Case 4: $a$ has three digits and $b$ has two digits.

This is just case $2$ flipped, so $18$ ways.

Case 5: $a$ has three digits and $b$ has one digit.

This is just case $1$ flipped, so $0$ ways.

Summing gives $18+225+18=261$. Our answer is $999-261 = \boxed{738}$.

OMG I GOT IT RIGHT YUHHHHH I WAS SO SURE I'D GET IT WRONG :D

Clearly, there's exactly $99$ ordered pairs where at least one of $a, b$ has a units digit of zero. (In fact, the only possibility is neither or both of $a, b$ having units digit $0$.)

Now, we consider all ordered pairs where at least one of $a, b$ has a tens digit of zero, but neither integer has a units digit of $0$. Taking the given equation modulo $100$ implies that it's impossible for both integers to have their tens digit be zero, so symmetry implies the answer is $$(1000 - 1) - 99 - 2(9^2) = 738.$$

[b]Remark:[/b] This solution is cleaner and "more accurate" than PIE, although the mod $100$ observation is somewhat hard to come up with.


[i][b]~hihihihihihi :D everyone!~

i need help on another problem -_____-"[/b][/i]
[b]Problem:
Study the numbers below and continue the pattern by listing the next five terms in the sequence:

1,1,2,2,8,10,3,27,30,4,64,68,5,....[/b]

[i][u]thanks! :)[/u][/i]

[b]NEVERMIND!!
hahahahah:]]
i figured it out by myself
but THANKS for the ones who replyed 2 da post :]]]]]][/b]

[quote="andersonw"][hide="pattern"]
The pattern is take 1, cube it, and then add 1. Then take 2, cube it, and then add 2. Then take 3, cube it, and then add 3 etc.[/hide]

[hide="solution"]
125, 130, 6, 216, 222[/hide][/quote]

....O.O...
ahhh i still dont get it :(
i have never been good at sequences ^_^

[quote="stevenmeow"]it looks like u just looked up "math help" and entered this problem... it doesn't even show evidence if u tried it!!!
besides your only posts are about questions...[/quote]

well thats cuz i dont get those questions
&& how am i supposed to show that i tried it? i cant put all my work on the forum x]]

Try starting with 1 and doing the pattern andersonw said. You get the same sequence. Method to get it: involves some guess and check. However, notice that after the terms increasing, it suddenly goes back. You can see that the "small" terms are 1, 2, 3... You should also notice that 8 and 27 are the perfect cubes of 2 and 3. 64 is also the cube of 4. Now we just need to find the relationship between the other numbers. Try subtracting the cubes from the leftover numbers. The differences are 1, 2, 3... So now you can find the pattern. Now just apply it and find the next terms.

If you still don't get it

1 -- random number
1 --1^3 , 1 is from the provious number
2 --1^3 +1 = 2 adding the original number to its cube
4 --2^3 , 2 is from the provious number
8 --2^3 +2 = 2 adding the original number to its cube
:
:
n
n^3
n^3 + n

where did u get this prob anyway?

wait is the $ n^{th}$ number in the sequence $ n^3 \plus{} n$ or is the previous number in the sequence n.

Actually, this is the function

$ \text{If }x\equiv 1\text{ (mod 3)},\text{ then }f(x)\equal{}\frac{x\plus{}2}{3}$.
$ \text{If }x\equiv 2\text{ (mod 3)},\text{ then }f(x)\equal{}\left(\frac{x\plus{}1}{3}\right)^3$
$ \text{If }x\equiv 0\text{ (mod 3)},\text{ then }f(x)\equal{}\left(\frac{x}{3}\right)^3\plus{}\frac{x}{3}$

let me know if this isn't the case for something

btw, this may be a bit too complicated for classroom math, but I brought it up because the $ n\plus{}2$nd term is not $ n^3\plus{}3$.  It fails for $ n\equal{}1$


Just some first remarks:

Take $x=y=t^{-1}, z=t^2$. Assume that $n>4$. Then you can show that the inequality gets wrong for $t\to \infty$.

But if $2m+n\leq 6$, we can prove the inequality:

Clearly, it is equivalent to

$\sum \frac{x^m-x^n}{(x^m+y+z)(x^n+y+z)}\geq 0$.

But similarly to iura's solution for IMO 2005/3, we have

$\frac{x^m-x^n}{(x^m+y+z)(x^n+y+z)} \geq \frac{x^m-x^n}{(x^m+x^{m-1}y+x^{m-1}z)(x^n+x^{n-1}y+x^{n-1}z)}$ $= \frac 1{(x+y+z)^2}(x^{2-n}-x^{2-m})$.

Therefore it suffices to show $\sum x^{2-n}\geq \sum x^{2-m}(xyz)^{\frac{m-n}{3}}\geq \sum x^{2-m}$. The first follows from muirhead, the second follows immediately from $xyz\geq 1$.

It is perhaps interesting to note that the case $m=\frac 52$, $n=1$ is just IMO 2005/3. And the solution is just iura's solution :).

Edit: Computer checks look like $n=1$ always gives a true inequality. Who can prove that? It would be quite a nice generalization...

Edit2: Some more numerical results let me conjecture that the inequality is true iff $n\leq 2$. I'm quite sure about counterexamples for $n>2$, and also quite sure that it works for $n=2$ which is the interesting case by what i have already shown.

Peter

[quote="Peter Scholze"]
...

Edit: Computer checks look like $n=1$ always gives a true inequality. Who can prove that? It would be quite a nice generalization...

Edit2: Some more numerical results let me conjecture that the inequality is true iff $n\leq 2$. I'm quite sure about counterexamples for $n>2$, and also quite sure that it works for $n=2$ which is the interesting case by what i have already shown.

...[/quote]

 :D


There are $n$ boys $B_1, B_2, ... , B_n$ and $n$ girls $G_1, G_2, ... , G_n$. Each boy ranks the girls in order of preference, and each girl ranks the boys in order of preference. Show that we can arrange the boys and girls into n pairs so that we cannot find a boy and a girl who prefer each other to their partners.

For example if $(B_1, G_3)$ and $(B_4, G_7)$ are two of the pairs, then it must not be the case that $B_4$ prefers $G_3$ to $G_7$ and $G_3$ prefers $B_4$ to $B_1$.

this one is already in the forum.search for it!


Around a month has elapsed since I posted the ELJMO problems on this forum, so I figure now is about the right time to post the set of solutions.

Thanks to all of you who sent in your work on the problems--it was an absolute if perverse pleasure to grade the submissions and bask in the knowledge that I had successfully wasted your time!  Congratulations to SorcererofDM who earned the high score of 21, beating out nearly all the MOPpers who took the official exam.

The attached solutions document is a compilation of labor on the part of the problem proposers, who each submitted official solutions, and the participants, the best of whose solutions were picked out and included after some simplification.  Credit is also due to Sherry Gong, who assisted me in typesetting the complete manuscript in LaTeX and who made numerous editorial suggestions.  I hope those of you who engaged the questions on the exam will study the manual thoroughly and gain a deeper understanding of the underlying mathematics by examining the convergence between the multiple methods of attack for each problem provided.

Enjoy!

--DBR

I am not sure if this has been posted anywhere yet, but it is possible to, in problem 4, replace $ n^2$ by $ 2n^2$ to obtain the following problem:

Let $ n$ be a positive integer. Given $ 2n^2$ points in a unit square, prove that there exists a broken line of length $ 2n\plus{}1$ that passes through all the points.

The solution to this harder version is a modification of the solution to the problem that appeared on the ELJMO. See if you can find it. Additionally, several of the coordinators were interested in the following extension:

These two problems (on the ELJMO and the harder version above), yield for $ n$ points a bound on the shortest path asymptotically equivalent to $ 2\sqrt{n}$ and $ \sqrt{2n}$. What is the smallest $ c$ for which there is always a path of length $ c\sqrt{n} \plus{} o(\sqrt{n})$? So far, we know that $ c \leq \sqrt{2}$ and by considering a grid, $ c \geq 1$, but we have no idea what the actual number is. A couple people feel like the smallest $ c$ should be $ 1$ as the grid pattern is intuitively the worst case, but I am not convinced. Any insight into this would be very interesting.

The constant c is known to be between 1.07 and 1.4, but is not known exactly.  For a discussion, see
[url]http://mathworld.wolfram.com/TravelingSalesmanConstants.html[/url].
It might make a good research problem to get better bounds.


Let $ a_{1}<a_{2}<\cdots<a_{n}$. For which permutation of $ a_{i}$'s like $ (b_{1},b_{2},\cdots,b_{n})$,

$ \prod^{n}_{i=1}\left(a_{i}+\frac{1}{b_{i}}\right)$ will have its [b]Max[/b]imum value?

[quote="hungkhtn"]I think this solution is not true because of case of equality.[/quote]
I had doubts about my decision, but I can not yet found another.  :(

[quote="N.T.TUAN"]$ a_{1},a_{2},...,a_{n}\in (0,\infty)$. Right?  :maybe:[/quote]

I also think so, but not mentioned in the Book! :maybe:


limit
function
topology
absolute value
algebra
floor function
calculus

Given $ \lim_{x \to 0} \frac {f(x)}{x^2} \equal{}2$ , then evaluate the following limits, giving explicit reasoning.

(i) $ \lim_{x \to 0} [f(x)]$
(ii) $ \lim_{x \to 0} [\frac{f(x)}{x}]$ where $ [x]$ denotes the greatest int function.

Also can anyone tell me what does length of an interval mean?

Let’s e.g. f(x)=2x^2 then f(x)/x=2x  .In the vicinity of 0 ,2x can be either a negative or positive real number with a small absolute value .In this case [f(x)/x] may be equal to 0 or -1 .Thus it has not a limit at 0 .

f(x)/x^2 = 2 + err(x)  lim err(x) = 0  ( when x---->0)
 f(x) =  ( 2 + err(x)) x^2 = 2 x^2 + x^2 err(x)
err  function can be regarded as a bounded function e.g. 0 <=|err|< 1 in the vicinity of 0 so for every x in this neighborhood :
0 < f(x) < 3 x^2   then    0 < =  [f(x)] < = [3 x^2   ]
    Since   lim [3 x^2]  ( when x---->0) = 0 , lim [f(x)]  ( when x---->0) = 0
Which  interval do you mean? Length of open ,semi open or infinite intervals is transformed to measure of such intervals in topology .

Clearly $ \lim_{x \to 0}f(x) \equal{} 0$.  Suppose it did not equal 0.  Then let $ \lim_{x \to 0}f(x) \equal{} L$.  Then, $ \lim_{x \to 0}\dfrac{f(x)}{x^2}$ tends to $ \dfrac{L}{0}$ which will go to infinity.

So, if $ \lim_{x \to 0}f(x) \equal{} 0$, there are only 2 options for the first limit (think positive or negative).  I think once you figure this out, the second limit will make more sense.

Graphing the floor function might help you think about it.

takiguy wrote :"Clearly . Suppose it did not equal 0. Then let limf(x)=L "
As a contradictory proof when we suppose a limit is not equal to a special value e.g. 0 it is possible that the limit doesn't exist .We should regard this case .


Avem $ x_{n}\ge n$ si prin inductie $ x_{n}<n+1$, rezulta $ \lim_{n\rightarrow\infty}\frac{x_{n}}{n}=1$, iar $ x_{n}-n=\cfrac{\cfrac{x_{n-1}}{n}}{1+\cfrac{x_{n}}{n}}$, deci $ \lim_{n\rightarrow\infty}x_{n}-n=\frac{1}{2}$.
Folosind acestea avem $ \lim_{n\rightarrow\infty}\left(\frac{n}{x_{n}}\right)^{x_{n}}=\lim_{n\rightarrow\infty}\left(1-\cfrac{x_{n}-n}{x_{n}}\right)^{\cfrac{x_{n}}{x_{n}-n}\cdot\cfrac{(x_{n}-n)\cdot x_{n}}{x_{n}}}=e^{-\frac{1}{2}}$


[b]Bueno, no falta nada para la IX OMCC, sólo quería darle la mejor de las suertes a los participantes :wink:
... nos vemos en Mérida :P
saludos :D [/b]

:lol: Aquí van los ganadores de la IX OMCC.

MEDALLA DE ORO

Alejandro Jiménez Martínez (MEX)
Antonio Ismael García Rodríguez (CUB)
Luis Ángel Isaías Castellanos (MEX)

MEDALLA DE PLATA

José Gabriel Pérez Clark (CUB)
Nahomy Jhopselyn Hernández Cruz (SAL)
Sara Rodríguez Martínez (PRC)
Daniel Otero Baguer (CUB)
Manuel Guillermo López Buenfil (MEX)
Paula Valeria Barreras Cortés (COL)
Hayden Liu Weng (COL)
Héctor Enmanuel Alberti Arroyo (SAL)

MEDALLA DE BRONCE

Iván Alejandro Solano Oviedo (COL)
Daniel Emilio Esquivel Barrantes (CRC)
Antonio Fan (PAN)
Heui Beom Lee (GUA)
Julio César Ayala Menjivar (SAL)
Ricardo Wong (PAN)
Aravind Arun (PRC)
David Urdaneta Sandoval (VEN)
Jonathan Valverde Lizano (CRC)

MENCIÓN HONORÍFICA

Alejandro José Vargas de León (GUA)
Sergio David Manzanárez Elvir (HON)
Elvira Loana Vega Castro (NIC)
Jesús Alberto Coello (HON)
Caleb De la Cruz Paulino (DOM)
Estefanía Ordaz Maurera (VEN)
Wendy Katiuska Mejia Mercedes (DOM)
Julio Andrés Vargas Ramírez (CRC)

Felicidades a todos.. saludos, se les quiere! :P

México, y de segundo Cuba, aunque no sé con cuántos puntos cada uno.

podrias postear el examen?  felicidades a mexico!!!

Ah si, que bien!!! Felicidades a México!!!!!!!!!

Hey, además, la Copa El Salvador otorgada al país con el mayor progreso se la llevó Honduras, al igual que la sede para la X OMCC en el 2008. Felicidades a ellos :wink:


geometry
circumcircle
parallelogram
rectangle
geometric transformation
homothety
symmetry

hi all,

umm, this problem is a bit of a generalization of another problem i posted in the geometry - > proposed problems section .. so here goes:

Define two circles, one inside the other, to be 'good', if we can find a quadrilateral ABCD s.t. it is bicentric, with one of the circles being the incircle of ABCD and the other being the circumcircle of ABCD. Now,
(i) Prove if we choose another point A', and B', D' on the bigger circle s.t. A'B' and A'D' are tangent to the smaller circle. Choose C' s.t. C'B' is tangent to the smaller circle and D' is on the larger circle. Now prove C'D' is tangent to the smaller circle. [ special case of Poncelet's Thm, I think. ]
(ii) Prove that the lines AC, BD, A'C', B'D' and IO meet at a point T.
Maybe also, find some relationship b/w the points IO, and the radii of the two circles, which might be helpful. Since the diagonals of the quadrilateral are fixed (that is what we want 2 prove in (i)), mayb we can find an algebraic expression for IT and OT, which involves only r and R. That would prove (ii).

Let $E, F, G, H$ be the tangency points of the sides $AB, BC, CD, DA$ of a bicentric quadrilateral $ABCD$ with the incircle $(I)$, forming a cyclic contact quadrilateral $EFGH$.

[color=blue][b](1)[/b][/color] The diagonals $EG \perp FH$ are perpendicular to each other: In an inversion with respect to the incircle $(I)$, the vertices $A, B, C, D$ are carried into the midpoints $A', B', C', D'$ of the chords $HE, EF, FG, GH$, forming a parallelogram $A'B'C'D'$ with opposite sides $A'B', C'D'$ and $B'C', D'A'$ parallel to the diagonals $FH, EG$, respectively. This parallelogram is inscribed in the inverted circumcircle $(O)$ of $ABCD$, hence, it is a rectangle and $EG \perp FH$, because $EG \parallel B'C', FH \parallel A'B', B'C' \perp A'B'$.

[color=blue][b](2)[/b] [/color]Let $P$ be the intersection of the rectangle diagonals $A'C', B'D'$, identical with its circumcenter. The circumcircles $(O), (P)$ of $ABCD$ and $A'B'C'D'$ are centrally similar with homothety center at the inversion center $I$. Thus the points $O, I, P$ are collinear. The point $P$ is the centroid of the cyclic quadrilateral $EFGH$ with circumcircle $(I)$. The centroid of a cyclic quadrilateral is the midpoint between its circumcenter and anticenter (the intersection of its maltitudes). By one of Brahmagupta's theorems, the diagonal intersection of a cyclic quadrilateral with perpendicular diagonals is identical with its anticenter. Thus the diagonal intersection $X \equiv EG \cap FH$ is collinear with the circumcenter $I$ and the centroid $P$ of $EFGH$. As a result, the points $O, I, X$ are collinear.

[color=blue][b](3)[/b][/color] Let 2 perpendicular lines meeting at the diagonal intersection $X$ intersect the circle $(I)$ at points $E_1, G_1$ and $F_1, H_1$, forming a cyclic quadrilateral $E_1F_1G_1H_1$ with perpendicular diagonals $E_1G_1 \perp F_1H_1$, the same circumcircle $(I)$ and the same diagonal intersection $X$ as $EFGH$. Let $A_1', B_1', C_1', D_1'$ be the midpoints of the chords $H_1E_1, E_1F_1, F_1G_1, G_1H_1$. Since the cyclic quadrilaterals  $EFGH$ and $E_1F_1G_1H_1$ have also the same centroid (the midpoint $P$ of $IX$), the segments $A_1'C_1'$ and $IX$ cut each other in half at the point $P$. Hence, the quadrilateral $IA_1'XC_1'$ is a parallelogram. Using the parallelogram rule,

$A_1'C_1'^2 + IX^2 = 2(IA_1'^2 + XA_1'^2)$

$PA_1'^2 + PX^2 = \frac{IA_1'^2 + XA_1'^2}{2}$

The segment $XA_1'$ is the median of the right angle triangle $\triangle H'A_1'E'$, hence, $H'A_1' = E'A_1' = XA_1'$. Since $A_1'$ is the midpoint of the chord $H'E'$ of the circle $(I)$, $IA_1' \perp E'F'$ and

$IA_1'^2 + XA_1'^2 = IA_1'^2 + E'A_1^2 = IE'^2 = r^2$

where $r$ is radius of the circle $(I)$. Consequently, the distance $PA_1'$ is given by

$PA_1'^2 = \frac{r^2}{2} - PX^2$

and similarly,

$PB_1'^2 = PC_1'^2 = PD_1'^2 = \frac{r^2}{2} - PX^2$

Thus the midpoints $A_1', B_1', C_1', D_1'$ all lie on the circle $(P)$ with center and radius independent on the direction of the perpendicular lines $E_1G_1, F_1H_1$ meeting at the diagonal intersection $X$. But the circle $(P)$ is an image of the same circle $(O)$ in an inversion in the circle $(I)$ and the midpoints $A_1', B_1', C_1', D_1'$ are images of vertices of a bicentric quadrilateral $A_1B_1C_1D_1$ having the same incircle $(I)$ and circuncircle $(O)$ as the bicentric quadrilateral $ABCD$.

[color=blue][b](4)[/b][/color] Since the points $O, I, X$ are collinear, we can choose one of the perpendicular lines passing through $X$ also passing through $O, I$, for example, the line $EG$ (skipping the indices 1). Because of symmetry of this figure, the resulting bicentric quadrilatreral $ABCD$ (skipping the indices 1) is an isosceles trapezoid with the bases $AB \parallel CD$ and  shoulders $BC = DA$. The tangency points $E, G$ are the midpoints of the bases $AB, CD$. Let $M, N$ be the midpoints of the shoulder sides $BC, DA$. Since $MN = \frac{AB + CD}{2}$ for a trapezoid and $AB + CD = BC + DA$ for a tangential quadrilateral, the trapezoid midline is equal to its shoulder sides, $MN = BC = DA$. The incenter $I$ is the midpoint of the midline $MN$ and consequently, the median $IM$ of the triangle $\triangle BCI$ is equal to half of its side $BC$, i.e., the angle $\measuredangle BIC = 90^o$ is right. Since the angles $\measuredangle BIE + \measuredangle CIG = 180^o - \measuredangle BIC = 90^o$ add up to a right angle, the right angle triangles $\triangle BIE \sim \triangle ICG$ are similar and $\frac{BE}{IE} = \frac{IG}{CG}$. Let $R, r$ be the circumradius and inradius of the bicentric trapezoid $ABCD$ and $d = IO$ the distance between the incenter and circumcenter. Then

$BE^2 = OB^2 - OE^2 = OB^2 - (IB - IO)^2 = R^2 - (r - d)^2$

$CG^2 = OC^2 - OG^2 = OC^2 - (IC + IO)^2 = R^2 - (r + d)^2$

$\left(\frac{BE}{IE}\right)^2 = \left(\frac{IG}{CG}\right)^2\ \ \Rightarrow\ \ \frac{R^2 - (r - d)^2}{r^2} = \frac{r^2}{R^2 - (r + d)^2}$

$(R^2 - r^2 - d^2)^2 - 4r^2d^2 = r^4$

$R^4 + d^4 - 2R^2r^2 - 2R^2d^2  - 2r^2d^2 = 0$

$(R^2 - d^2)^2 = r^2(2R^2 + 2d^2) = r^2\left[(R + d)^2 + (R - d)^2\right]$

$\frac{1}{r^2} = \frac{(R + d)^2 + (R - d)^2}{(R + d)^2 (R - d)^2} = \frac{1}{(R - d)^2} + \frac{1}{(R + d)^2}$

which is similar to the relation between the circumradius $R$, inradius $r$ and the incenter-circumcenter distance $d = IO$ for a triangle: $\frac 1 r = \frac{1}{R - d} + \frac{1}{R + d}$.

hey,

thanx 4 that, the proof to the first 3 were really really cool, and the last one was alrite, just a bit bashy.
One result that i think you didnt mention, you probably dismissed as being too obvious to bother writing, that sorta
ties everything together is this: AC,BD,EG and FH are concurrent. This is really easy to prove using Brianchons.
I think (1) can also be proved really easily with angle chasing but the inversion was pretty cool.
Back to the original problem, however, I'm not quite sure that you can get (ii) completely from your results. (1) and (2)
obviously prove that the points \displaystyle X \equiv AC \cap BD, I and O are collinear. But Im not sure that (3) proves the point where the diagonals intersect as ABCD sorta moves around between the two circles is fixed. But it would be evident if you proved that:
The tangents at E1 and H1, to the smaller circle, meet on the larger circle.
I don't see how this is immediately obvious .. mayb im just missing sumfin tho .. so yah cud sum1 post a proof of that?

so yah, thx again 4 all that, it was reali cool and interesting, and also easy to understand.

hey again,

actually i fink worked out a way to finish off (ii), its a bit dodgy tho.
u prove that given I,O, and the circles with centres I and O, you can construct the point X. We do this as follows:
first consider the circle that you obtain by inverting the circumcircle in the incircle. Let the centre of this circle be P.
then choose X s.t. IP=PX.
This works bcoz the circle through A1,B1,C1 and D1 has centre P and is the image of circle (O) inverted in circle (I). IP=PX,
coz X is the intersection of diagonals, and anticenter, P is the centroid and I is the circumcenter of the cyclic quad EFGH.
btw, could someone also prove poncelet's for n=4?
thx in advance ~!!

[quote="vinoth_90_2004"]... One result that i think you didnt mention, you probably dismissed as being too obvious to bother writing, that sorta ties everything together is this: AC,BD,EG and FH are concurrent. This is really easy to prove using Brianchons...[/quote]
Or by repeated use of Pascal's theorem for the diagonals, 2 chords and 2 tangents of $EFGH$, by projecting the line $KL$ to infinity, where $K \equiv EF \cap GH$, $L = FG \cap HE$, or by angle chasing.

[quote="vinoth_90_2004"]
... I think (1) can also be proved really easily with angle chasing but the inversion was pretty cool...[/quote]
See [url=http://www.mathlinks.ro/Forum/topic-33205.html]Please help to prove this[/url], part (2).

[quote="vinoth_90_2004"]...
Back to the original problem, however, I'm not quite sure that you can get (ii) completely from your results. (1) and (2) obviously prove that the points $X \equiv AC \cap BD$, I and O are collinear. But Im not sure that (3) proves the point where the diagonals intersect as ABCD sorta moves around between the two circles is fixed. But it would be evident if you proved that:
The tangents at E1 and H1, to the smaller circle, meet on the larger circle...[/quote]
The midpoints $A_1', B_1', C_1', D_1'$ of the chords $H_1E_1, E_1F_1, F_1G_1, G_1H_1$ are on the circle $(P)$, which is an image of the circle $(O)$ in the inversion in the circle $(I)$. The inverted points $A_1, B_1, C_1, D_1$ on $(O)$ are poles of the chords $H_1E_1, E_1F_1, F_1G_1, G_1H_1$ with respect to the inversion circle $(I)$. Consequently, tangents to $(I)$ from $A_1$ touch this circle at the points $H_1, E_1$, etc.

The bicentric quadrilateral $A_1B_1C_1D_1$ has the fixed circumcenter $O$ and incenter $I$. Thus $P$, the center of inverted circle $(O)$, identical with the centroid of the contact quadrilateral  $E_1F_1G_1H_1$, is fixed. The direction $OIP$ is also fixed. This implies that the anticenter of the contact quadrilateral is fixed as well. The diagonals of the contact quadrilateral are perpendicular to each other, i.e., they meet at the fixed anticenter $X$. The diagonals of a tangential quadrilateral and its contact quadrilateral meet at the same point. Hence, the diagonals of $A_1B_1C_1D_1$ meet at fixed point $X$ (just like in your 2nd reply).

[quote="vinoth_90_2004"]... btw, could someone also prove poncelet's for n=4?...[/quote]
Once you have the relation between the circimradius $R$, inradius $r$ and the inradius-circumradius distance $d = IO$, it is a simple contradiction. Since $ABCD$ is bicentric,

$\frac{1}{r^2} =  \frac{1}{(R - d)^2} + \frac{1}{(R + d)^2}$

Pick a point $A_1$ on the circumcircle $(O)$ and draw tangents to the incircle $(I)$, intersecting the circumcircle at points $D_1, B_1$ different from $A_1$. From the point $B_1$, draw the remaining tangent to $(I)$ intersecting the circumcircle at a point $C_1$. Assume that $C_1D_1$ is not a tangent of the circle $(I)$, for example, it intersects it. Continuously contract the circle $(I, r' < r)$, while keeping its center fixed. The points $D_1', B_1', C_1'$ continuously move on the circle $(O)$, until the chord $C_1'D_1'$ becomes a tangent to $(I, r' < r)$. The quadrilateral $A_1B_1'C_1'D_1'$ is then bicentric, with the same circumradius $R$ and the same distance $d = IO$ as the bicentric quadrilateral $ABCD$, but with a smaller inradius $r' < r$. Hence,

$\frac{1}{r^2} < \frac{1}{r'^2} = \frac{1}{(R - d)^2} + \frac{1}{(R + d)^2}$

which is a contradiction. The possibility that the chord $C_1D_1$ of $(O)$ and the circle $(I)$ do not have a common point is disproved similarly, by expanding the circle $(I, r' > r)$, while keeping its center fixed.


geometry
circumcircle
incenter
geometric transformation
reflection
rotation
ratio

Let $ AD $ the bisector of $\angle  A $  of a $\triangle   ABC $ and $  O  , K  ,  L $ the circumcenters of  triangles $ ABC  ,  ABD $ and  $ACD $ respectively . Prove that
                          $ OK = OL $

    [i]I spent half an hour for this problem but last minute I saw that the problem says << circumcenters>> and not << incenters>> as I wrongly read in the begin. Now I send the problem to all of you , (  probably  some of you do not have Prasolov's book) and enjoy it. I ' ll try again making in CABRI a good figure , but later !!! I 'll be waiting for your nice solutions.[/i]
   Babis

Let $E$ be the second intersection of $AD$ with the circumcircle $(ABC)$. A simple angle chase will show that the chords $AB,AC,AD$ "play the same role" in the circumcircles $(ABD),(ACD),(ABC)$, in the sense that they subtend the same angle with the same orientation, which means that $OKL$ is obtained from $EBC$ through a spiral similarity centered at $A$, namely the spiral similarity which takes $E$ onto $O$. Now, since $EB=ED$, we get the desired result: $OK=OL$.

Moreover, we find that $AKOL$ is cyclic, and $AO$ is the angle bisector of $\angle KAL$.

I have the Prasolov's and Kontogianis's books. Please, I wish a [b]new or original (own) problems [/b]!.

[quote="stergiu"]Let $ AD $ the bisector of $\angle  A $  of a $\triangle   ABC $ and $  O  , K  ,  L $ the circumcenters of  triangles $ ABC  ,  ABD $ and  $ACD $ respectively . Prove that
                          $ OK = OL $[/quote]

The problem is more or less trivial: The circumcenters O and K of triangles ABC and ABD must both lie on the perpendicular bisector of the common side AB of these triangles. Thus, the line OK is the perpendicular bisector of the segment AB. Hence, $OK\perp AB$, so that < (OK; AB) = 90° with directed angles modulo 180°. Similarly, for the circumcenters K and L of triangles ABD and ACD, we obtain < (KL; AD) = 90°. Hence,

  < OKL = < (OK; KL) = < (OK; AB) + < (AB; AD) - < (KL; AD) = 90° + < (AB; AD) - 90° = < (AB; AD) = < BAD.

Similarly, < OLK = < CAD. Now, since the line AD is the angle bisector of the angle CAB, we have < BAD = < DAC, so that < BAD = - < CAD. Thus, < OKL = - < OLK. In other words, < OKL = < KLO, so the triangle KOL is isosceles and OK = OL.

This solves the problem, but of course it doesn't show the additional properties pointed out by Grobber in post #2. These follow from a more general theorem:

[color=blue][b]Theorem 1.[/b] Let D be an arbitrary point on the side BC of a triangle ABC. Let O, K and L be the circumcenters of triangles ABC, ABD and ACD, respectively. Let the reflection of the line AD with respect to the angle bisector of the angle CAB meet the circumcircle of triangle ABC at a point E (apart from A). Then, the quadrilaterals ABEC and AKOL are directly similar.[/color]

The [i]proof of Theorem 1[/i] is an easy modification of Grobber's argument: Since the line AE is the reflection of the line AD with respect to the angle bisector of the angle CAB, we have < CAE = - < BAD; in other words, < CAE = < DAB. Also, since the point E lies on the circumcircle of triangle ABC, we have < CEA = < CBA, what rewrites as < CEA = < DBA. From < CAE = < DAB and < CEA = < DBA, it follows that the triangles CAE and DAB are directly similar. Corresponding points in directly similar triangles form directly similar triangles themselves; hence, since the points O and K are corresponding points in the directly similar triangles CAE and DAB (in fact, they are the circumcenters of these triangles), they form directly similar triangles themselves, i. e. the triangles AOE and AKB are directly similar. Hence, we have $\frac{AK}{AB}=\frac{AO}{AE}$ and < EAO = < BAK, where we, as an exception, use directed angles modulo 360° rather than modulo 180°. Hence, the spiral similitude with center A, rotation angle < EAO and stretch ratio $\frac{AO}{AE}$ maps not only the point E to the point O, but also the point B to the point K. Similarly, we can show that this similitude maps the point C to the point L. Finally, this similitude maps the point A to the point A itself (since A is the center of this similitude). Hence, altogether, we see that our spiral similitude maps the points A, B, E, C to the points A, K, O, L, respectively. Hence, the quadrilaterals ABEC and AKOL are directly similar, and Theorem 1 is proven.

Now, if, as in the initial problem, the line AD is the angle bisector of the angle CAB, then, since the reflection of this angle bisector in itself is itself :) , the point E is the point of intersection (apart from A) of the angle bisector of the angle CAB with the circumcircle of triangle ABC. Since the angle bisector of an angle of a triangle bisects the arc of the circumcircle intercepted by the opposite side, the point E must thus be the midpoint of the arc BC of the circumcircle of triangle ABC; thus, EB = EC. As the quadrilaterals ABEC and AKOL are similar after Theorem 1, this entails OK = OL, and the problem is solved once again.

  darij


geometry
geometric transformation
reflection
parallelogram
trapezoid
geometry proposed

Prove the following formula of the area of $\triangle ABC$ in terms of its medians

\[ [\triangle ABC]^2\equal{}\frac{_2}{^9}({m_a}^2{m_b}^2\plus{}{m_a}^2{m_c}^2\plus{}{m_b}^2{m_c}^2)\minus{}\frac{_1}{^9}({m_a}^4\plus{}{m_b}^4\plus{}{m_c}^4) \]

Use the identities $ m_a\equal{}\frac14(b^2\plus{}c^2\minus{}a^2)$, $ m_b\equal{}\frac14(a^2\minus{}b^2\plus{}c^2)$, $ m_c\equal{}\frac14(a^2\plus{}b^2\minus{}c^2)$ (which can be proved using Stewart's Theorem) and Heron's Formula.

We'll show that the area of the triangle formed by the lengths of the 3 medians of ABC equals 3/4 of the area [ABC], then the result follows by Heron formula. Let $ D,E,F$ be the midpoints of $ BC,$ $CA,$ $AB$ and $ G$ the centroid. Let $ G'$ the reflection of $ G'$ across $ D.$ It's easy to see that $ GCG'B$ is a parallelogram such that $ BG \equal{} \frac {_2}{^3}m_b,$ $BG' \equal{} \frac {_2}{^3}m_c$ and $ GG' \equal{} \frac {_2}{^3}m_a.$ Since $ CG' \parallel BG,$ then we have $ [\triangle BGG'] \equal{} [\triangle BGC] \equal{} \frac {_1}{^3}[\triangle ABC].$

[quote="arcsin1.01"]Use the identities $ m_a \equal{} \frac14(b^2 \plus{} c^2 \minus{} a^2)$, $ m_b \equal{} \frac14(a^2 \minus{} b^2 \plus{} c^2)$, $ m_c \equal{} \frac14(a^2 \plus{} b^2 \minus{} c^2)$ (which can be proved using Stewart's Theorem) and Heron's Formula.[/quote]

Your identity isn't right, try $ a\equal{}b\equal{}c\equal{}2$

Sorry, it should be $ m_a^2\equal{}\frac14(2b^2\plus{}2c^2\minus{}a^2)$, etc.

Another solution: Let $M,N$ be the midpoints of $AC,AB.$ By Betschneider theorem for the trapezoid $BCMN,$ we get:

\[16 \cdot [BCMN]^2=4 \cdot BM^2 \cdot CN^2-(BC^2-CM^2+MN^2-BN^2)^2 \]


solve this equation:
\[ \frac{\tan x\plus{}\cot x}{\sin x\plus{}\cos x}\equal{} \sqrt{2}\]

We have

$ \frac{\tan x\plus{}\cot x}{\sin x\plus{}\cos x}\equal{}\frac{\frac{\sin x}{\cos x}\plus{}\frac{\cos x}{\sin x}}{\sin x\plus{}\cos x}\equal{}\frac{\sin^2x\plus{}\cos^2x}{\sin x\cos x(\sin x\plus{}\cos x)}\equal{}\sqrt2$

$ \implies\sin x\cos x(\sin x\plus{}\cos x)\equal{}\frac{\sqrt2}2\implies2\sin x\cos x(\sin x\plus{}\cos x)\equal{}\sqrt2$

$ \implies\sin{2x}(\sin x\plus{}\cos x)\equal{}\sqrt2\implies\sin^2{2x}(\sin^2x\plus{}\cos^2x\plus{}2\sin x\cos x)\equal{}2$

$ \implies\sin^2{2x}(1\plus{}\sin{2x})\equal{}2\implies\sin^3{2x}\plus{}\sin^2{2x}\minus{}2\equal{}0$.

Letting $ y\equal{}\sin{2x}$, we see that

$ y^3\plus{}y^2\minus{}2\equal{}(y\minus{}1)(y^2\plus{}2y\plus{}2)\equal{}0$.

However, since $ y\equal{}\sin{2x}$ is real, there are no $ x$ such that $ y^2\plus{}2y\plus{}2\equal{}0$, and the only solutions occur when

$ \sin{2x}\equal{}1\implies2x\equal{}\frac{\pi\pm2k\pi}2\implies x\equal{}\boxed{\frac\pi4\pm\frac{k\pi}2}$.


geometry
area of a triangle
algebra unsolved
algebra

Solve the equation:
\[ \sqrt{abx(x\minus{}a\minus{}b)}\plus{}\sqrt{bcx(x\minus{}b\minus{}c)}\plus{}\sqrt{cax(x\minus{}c\minus{}a)}\equal{}\sqrt{abc(a\plus{}b\plus{}c)}\], $ x \in R$.

Perhaps this will help:

The RHS is the area of a triangle with side lengths $ a \plus{} b, b \plus{} c, c \plus{} a$.  The LHS is the sum of the areas of triangles with side lengths $ a \plus{} b, x \minus{} a, x \minus{} b$, etc.

Given $ \triangle ABC$ such that $ AB \equal{} a \plus{} b, BC \equal{} b \plus{} c, CA \equal{} c \plus{} a$, we therefore want a point $ P$ in the interior such that $ PA \equal{} x \minus{} a, PB \equal{} x \minus{} b, PC \equal{} x \minus{} c$.


[quote="Michael Niland"]A heavy thin flexible string AB whose mass per unit length, L, is constant is  initially at rest in a straight line on a smooth plane horizontal surface.

A hand then moves from rest at A with constant acceleration F towards B collecting the string with it as it goes.

Find the force exerted by the hand on the string after time t has elapsed,the hand then lying between A and B.[/quote]

constant acceleration F? a strange and confusing convention... F = mF!

say the acceleration is a, then
v(t) = at
x(t) = (at^2)/2

no constants of integration as the hand begins at rest and I define the origin at A.

therefore the amount of string that must be accelerated at a, as a function of time t, is half x(t) as the hand is twice as far from point A as the newest section of string picked up (being a bti ambiguous here with the dx and dt).. so call the function s(t) the length of string that is being dragged at the time instant t.

s(t) = (at^2)/4

but this is confusing because we are not given the mass per unit length of the string and so cannot calculate the force at time t...

Hm, I thought that $L$ was the mass per unit length. But still, I don't quite understand what you're saying and how did you get that 1/2, perhaps if you tried to rephrase it...


inequalities
geometry
circumcircle
inradius
Euler
geometry proposed

Let $ a$, $ b$, $ c$ be three real numbers from the interval $ \left]0;\ 1\right[$ satisfying $ a\plus{}b\plus{}c \equal{} 2$. Show that

$ \frac{a}{1\minus{}a}\cdot\frac{b}{1\minus{}b}\cdot\frac{c}{1\minus{}c}\geq 8$.

mmh ... I try

the given inequality can be transformed in the following way

abc >= 8(1-a)(1-b)(1-c) = (2-2a)(2-2b)(2-2c) = (-a+b+c)(a-b+c)(a+b-c).

if a, b and c are the sides of a triangle and S is its area, we have the following well known formulas:

abc = S/(4R); S = (a+b+c)r/2; Erone formula; where R and r are the circumradius and inradius respectively.

Is easy to see that the above inequality is equivalent to the Euler theorem

R > 2r.

As it is, it is not complete. since we don't know that $a,b,c$ are the sides of a triangle.

We can see that at most one of $x=-a+b+c,y=a-b+c,z=a+b-c$ can be negative, so if one of them is negative, the RHS is negative (the RHS in the inequality written by sprmnt), and we trivially get the conclusion. If they're all positive, then we can apply AM-GM to $x,y$, then $y,z$, then $z,x$ in order to get $(x+y)(y+z)(z+x)\ge 8xyz$, which is, in fact, what we want (this is the same as $abc\ge (-a+b+c)(a-b+c)(a+b-c)$).

But a, b, c are from the interval ]0; 1[ satisfying a+b+c = 2.  Does that imply x, y and z are all positive already?

the conditions imply that a,b,c are the sides of a triangle

This problem was my first ever post here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=76511&highlight=#p76511

and the first in the long list of problems that I couldn't solve that were subsequently posted here.

Multiplying it out, the problem is equivalent to Schurs.


can you explain your problem in more details?

[quote="Behsan"]you can find angle with "Z" Cos or Sin[/quote]

By Moivre we know

 :!:  $e{}^x{}^i=cosx+isinx$

and it's easy to get:

$e{}^-{}^x{}^i=cosx-isinx$

so.

$e{}^x{}^i+e{}^-{}^x{}^i=2cosx$ thus $cosx=\frac{e{}^x{}^i+e{}^-{}^x{}^i} {2}$

developing we get $e{}^2{}^x{}^i-2cosx.e{}^x{}^i+1=0$ and inmediatelly we get

$e{}^x{}^i=z \pm \sqrt{z^2-1}$ thus $x=-i.ln(z \pm \sqrt{z^2-1})$

If you were reffering to this.... :icecream:

[color=green][b]PD.[/b] You can get $e{}^x{}^i=cosx+isenx$ by [b]maclaurin[/b], finding the development for the $cosx$ and the development for the $i.sinx$ by derivates.[/color]


is there a formula for this?

\[ 3\minus{}\frac{2}{3\minus{}\frac{2}{3...}}\]

Well, if $ x\equal{}3\minus{}\frac{2}{3\minus{}\frac{2}{3\cdots}}$, we have:
$ 3\minus{}x\equal{}\frac{2}{3\minus{}\frac{2}{3\cdots}}$
$ (3\minus{}x)\left({3\minus{}\frac{2}{3\cdots}}\right)\equal{}2$
$ (3\minus{}x)x\equal{}2\implies x\equal{}1,2$.

I'm pretty sure $ x\equal{}2$ is the only possible value.

I don't understand what you mean by "formula."  But I can find a closed form (I see that grn_trtle can, too, but he did it incorrectly):

Let $ x\equal{}3\minus{}\dfrac{2}{3\minus{}\frac{2}{3\ldots}}$.  We can recognize the denominator as also being $ x$, so
$ x\equal{}3\minus{}\frac{2}{x}\Rightarrow x^2\minus{}3x\plus{}2\equal{}0\Rightarrow x\equal{}1,2$, but clearly $ x\equal{}\boxed{2}$ is the only solution (try evaluating the first few approximations of $ x$, i.e., evaluate $ 3,3\minus{}\frac{2}{3},3\minus{}\dfrac{2}{3\minus{}\frac{2}{3}},\ldots$, and you'll see that the value is converging to $ 2$ rather than $ 1$.

Note that I assumed convergence to begin with, but intuition and this type of assumption is all you need for contest problems like this (assuming that it is a contest problem, which I think it is).

by formula i mean like  for example the formula for an infinite radical series is the \[ \frac{\sqrt{4x\plus{}1}\plus{}1}{2}\]

is there something like that for an infinite fractional series?

I do not believe there is a formula, nor is it necessary. A general strategy is to set the entire series to a variable and look for repeats of the series. Then you solve the equation for your variable. Check your answers at the end by testing out some of the series to determine whether the answer is plausible. (Oftentimes, negative answers will not be possible for these questions...)

You don't need a formula for infinite square roots either, it can be derived the same way.
$ x\equal{}\sqrt{a\plus{}\sqrt{a\plus{}\cdots}}\implies x^2\minus{}a\equal{}\sqrt{a\plus{}\sqrt{a\plus{}\cdots}}\equal{}x$, and you go from there.

To prove convergence consider a sequence $ t_n$ where $ t_0 \equal{} 3 \minus{} \frac {2}{3}$ and $ t_{n \plus{} 1} \equal{} 3 \minus{} \frac {2}{t_n}$

Now we show by induction that $ t_n$ converges to $ 2$
$ t_0 \equal{} 3 \minus{} \frac {2}{3} > 2$

So assume it holds for $ t_n > 2$ then:
$ t_{n \plus{} 1} \equal{} 3 \minus{} \frac {2}{t_n} > 3 \minus{} \frac {2}{2} \equal{} 2$

So by the induction hypothesis every $ t_n > 2$

Now since the series converges we see that as $ n\to \infty, t_{n \plus{} 1} \equal{} t_n$
And the rest is the same as above

i mean, of course you dont need a formula, but it would never hurt to have one, right? :P


how do you do the combination symbols (like something take something)?

Do you mean
${8 \choose 3}$
$\binom{8}{3}$

[code]${8 \choose 3}$
$\binom{8}{3}$
[/code]


solve this equation and find tha answers in natural numbers:
$x^3-y^3=xy+61$

[hide]Look at this mod 3.  In mod 3 0^3=0 1^3=1 and 2^3=2.
The nine possibilities are 0-1 1-0 0-2 2-0 2-1 1-2 0-0 1-1 2-2.  61=1 mod 3.
0-1=2 and 0(1)+1=1 so (0,1) fails.
0-2=1 and 0(2)+1=1 so (0,2) works.
1-0=1 and 1(0)+1=1 so (1,0) works.
2-0=2 and 2(0)+1=1 so (2,0) fails.
1-2=2 and 1(2)+1=0 so (1,2) fails.
2-1=1 and 2(1)+1=0 so (2,1) fails.
0-0=0 and 0(0)+1=1 so (0,0) fails.
1-1=0 and 1(1)+1=2 so (1,1) fails.
2-2=0 and 2(2)+1=2 so (2,2) fails.
Notice that either x or y is a multiple of 3 so the sum is 1 mod 3.  Also, x>y.  Now, let's assume the case (0,2) works.  x doesn't equal zero so we try 3 but obviously, that's too small.  But (6,5) works since 216-125=30+61.[/hide]

ok at last you mean there is only one answer for this? :?:

I think it is.  A check tells us that for numbers 1 apart, this is the only solution since 9^3-8^3=72+61 is obviously false and 12^3-11^3=132+61 is even more false in terms of how far the numbers are separated.  In other words, this is diverging.  For the (1,0) mod 3 case, we have 4,3 being too small and 7,6 being too big.  Since the rest diverges other solutions 1 apart don't exist.  For numbers 4 apart, (0,2) definitely fails.  216-8=12+61 and things diverge even faster since the first number is bigger.  (1,0) won't work either since we can't pick 0, so we must pick 3 so 1 mod 3 is 7.  That's even bigger that the previous pair.  All others after that diverge even worse.


algorithm
number theory
greatest common divisor

I have not found enough fast algorithm for this problem. I have a given a random string of binary numbers, such as 1010001110010101. I have also given an operator $T(n)$ such that it changes bits in the indexes $n,2, n-1, n, n+1, n+2$ to their complement. Here indexes must be undestand as modulo string length.

Now, in what kind of algorithm I can transform an arbitrary string to "zero string" 00000 ... with a relatively few number of operations? String can be so long that it is not possible to try to test every single combination of transformations.

I thought the problem and found the following promising solution. First we transform the first non-zero bit $i_1$ to zero by $T(i_1+2)$, then we press find the next non-zero bit $i_2$ and do operation $T(i_2+2)$ and so on. The last four bits can probably be done by iterative deepening search.

Is there a fast algoritm to solve this related problem: Let $a_0,\cdots,a_{n-1}\in \mathbb{F}_2$. Assume $m\geq 0$ is a fixed integer. Find numbers $x_0,\cdots, x_{n-1}\in \mathbb{F}_2$ such that $x_k+x_{k+1}+\cdots +x_{k+m}=a_k$ for every $0\leq k \leq n-1$.Again indexes are modulo $n$.

I don't know what $\mathbb{F}_2$ is but notice that $x_{k+m+1}-x_k=a_{k+1}-a_k$. Therefore, if you choose some fixed value for $x_0$ you can easily calculate $x_{m+1}$, $x_{2m+2}$ and so on. If you choose fixed values for $x_0$, $x_1$, ... $x_{\text{gcd}(m+1,n)-1}$, you can calculate the rest of $x_i$s. Then simply modify some of the fixed values so that the sum of first $m$ numbers is $a_0$ and everything should be fine. Did I get anything wrong?

I have to try that. I denoted $\mathbb{F}_2$ by a field with two element, i.e. sums and products are taken modulo 2, distributive laws holds, $0=0+0=1+1$, $1=0+1=1+0$, and $1^{-1}=1$.


function
limit
complex analysis
complex analysis unsolved

On wikipedia, "The Jacobi theta function is the unique solution to the one-dimensional heat equation with periodic boundary conditions at time zero. This is most easily seen by taking z = x to be real, and taking τ = it with t real and positive. Then we can write"
$ \vartheta (x,it) \equal{} 1 \plus{} 2\sum_{n \equal{} 1}^\infty \exp( \minus{} \pi n^2 t) \cos(2\pi nx)$
and that:
$ \lim_{t\rightarrow 0} \vartheta(x,it) \equal{} \sum_{n \equal{} \minus{} \infty}^\infty \delta(x \minus{} n)$ with delta the dirac delta function.
For some reason (an extend brainfart?) I don't how the second conclusion is reached.  For integer x, it's obvious that $ \vartheta(z; \tau) \equal{} \sum_{n \equal{} \minus{} \infty}^\infty \exp (\pi i n^2 \tau \plus{} 2 \pi i n z)$ is infinity, since this is just an infinite sum of ones.  I don't see why for non-integers x the theta function necessarily takes on the value zero at t=0.

For some reason, the series $ \sum_{n \equal{} \minus{} \infty}^\infty \exp (2 \pi i n z)$ equals zero for non-integer real numbers, but I don't know why.

Intuitively, the points $ \exp(2 \pi i n z)$ are either a collection of roots of unity when $ z$ is rational or a dense subset of the unit circle otherwise, so they "average" to zero.  Strictly speaking, the series diverges, but the limit being taken is analogous to an [url=http://en.wikipedia.org/wiki/Abel_summation#Abel_summation]Abel mean[/url].


without calculator, can any body prove that
cos(2pi/5)+cos(4pi/5) = -1/2.

[hide="It is not that hard. Click here for a solution."]
Let $\omega = \exp \left( \frac{2i \pi}5 \right)$.
Then, $\sum_{k=1}^{5}\omega^{k}= \omega \cdot \frac{\omega^{5}-1}{\omega-1}= 0$, so $\textrm{Re}\left( \sum_{k=1}^{5}\omega^{k}\right) = 0$, i.e. $\sum_{k=1}^{5}\cos \left( \frac{2 k \pi}5 \right) = 0$.
That means \[\cos \left( \frac{2 \pi}5 \right)+\cos \left( \frac{4 \pi}5 \right)+\cos \left( \frac{6 \pi}5 \right)+\cos \left( \frac{8 \pi}5 \right)+\cos \left( \frac{10 \pi}5 \right) = 0 ,\] i.e. \[\cos \left( \frac{2 \pi}5 \right)+\cos \left( \frac{4 \pi}5 \right)+\cos \left( \frac{6 \pi}5 \right)+\cos \left( \frac{8 \pi}5 \right) =-1 .\] Since $\cos x = \cos \left( 2 \pi-x \right)$, that can be rewritten as \[2 \cos \left( \frac{2 \pi}5 \right)+2 \cos \left( \frac{4 \pi}5 \right) =-1 ,\] which yields \[\cos \left( \frac{2 \pi}5 \right)+\cos \left( \frac{4 \pi}5 \right) =-\frac12 .\] [/hide]


We simply add $ \binom{15}{6} \plus{} \binom{15}{7} \equal{} \binom{16}{7} \equal{} \boxed{11440}$.

One reason we add is because the Pascal's triangle is made of combinations. The first row is 0C1.
second row: 1C0 1C1
third row: 2C0 2C1 2C3

...and so on.

If you can picture the Pascal's triangle, 16C7 is below 15C6 and 15C7. Then, for the solving part, look at izzy's solution.

5005+6435=11440, by some weird property that:

$ {n \choose r}\plus{}{n \choose r\plus{}1}\equal{}{n\plus{}1 \choose r\plus{}1}$


inequalities
function
inequalities solved

Let p,q be positive real numbers and satisfy: p< q, n be natural number, n>=2. prove that if x1, x2, ...xn between inclusive p,q
 We have (x1+ x2+ ...+ xn)(1/x1 + 1/x2 + .. + 1/xn) <= n2 + Kn((p/q)1/2 - (q/p)1/2)2.
 Inthere: Kn= n2 when n even,Kn = (n2-1)/4 when n odd.
I've a solution, and you?
 :D Try to solv

Can you post your solution! I know this ineq and I kinda want to see a solution! :? it's discusting..(the ineq :D)

The first part(n even):
( \sum xi)*(  \sum pq/xi)<=( \sum (xi+pq/xi))^2/4. It's enough to prove that this one is smaller than that ugly thing :D.

(  \sum (xi+pq/xi))^2<=n^2*(p+q)^2 because (xi-p)*(xi-q)<=0 for any i.
But n^2*(p+q)^2<=4*n^2*(p^2+q^2-pq). This is really easy to prove by AM-GM between p^2 and q^2 so we got ( \sum xi)*(  \sum pq/xi)<=4*n^2*(p^2+q^2-pq) which is the inequality we want to prove, only multiplied by 4pq.

I hope it's right...

Here's an attempt to solve the second part(n odd):
Let's name E=( \sum xi)*( \sum 1/xi). We can assume that x1<=x2<=..<=xn. If we consider that x2,..,xn are fixed we can try to maximize E by changing x1 and xn. E=x1*( 1/x2+..+1/xn)+1+1/x1*(x2+..xn)+(x2+..+xn)*(1/x2+..+1/xn). 1 and the last term are fixed, so E is maximized when the sum between the first and the third term is maximum, but all the functions x/xi+xi/x are decreasing when x<=xi for i>=2, so E is maximized when x1 is the smallest, which is p.

In the same manner we show that E is maximized when xn=q. We replace x1 and xn by p and q, we right the new E, we fix the numbers x3,..,xn-2 and we see what happens when we change x2 and x(n-1). It's not hard to see that E is maximum when x2=p, x(n-1)=q, and we can repeat the process.

We will eventually reach x((n+1)/2), all the other terms being p or q, obtained by the method described above. We will easily see that E is maximized when x((n+1)/2) is either p or q. If we calculate E we will find the right part of the inequality, so it is proven.

The calculations are not that complex, but it takes a long time to write them on a computer key board. Besides, I'm a really slow typist :) I hope it's correct.

Did I lose my minds or the function in the LHS is convex in all its variables, so it attains its maximum value when all x_i are p or q?

Not yet Harazi :D
And that's the way this inequality is solved almost everywhere.

Pierre.


Rationalize the difference in boiling points between the members of the following pairs of substances?

a) HF (20 degrees C) and HCl (-85 degrees C)
b) CHCl3 (61 degrees C) and CHBr3 (150 degrees C)
c) Br2 (59 degrees C) and ICI (97 degrees C)
d) CH3CH2CH2CH3 (0 degrees C) and (CH3)3CH(-10 degrees C)

a) There are hydogen bonds between HF molecules.

b) Bromine atoms are more polarizable.

c) What is "ICI"?

d) Molecular weight is the same, but the molecular shape of the first is linear, while that of the second is egg-like. Linear shape means more surface area for contact, and so stronger London forces.

Sorry about that. The substance is ICl  :blush:

Iodine and chlorine have different electronegativities, therefore...

Does this mean the dipole-dipole forces are stronger, making the boiling point higher for ICl?

Because for molecules of approximately equal mass and size, dipole-dipole forces increase with increasing polarity right? This would then require more energy to overcome the stronger intermolecular forces when boiling the substance and forming a gas?

Is this correct?


If I am not making a mistake, this is the first inequality that I have created.

Prove that if $ a,b,c\in\mathbb{R}^ \plus{}$, $ \sum_{\text{cyc}}\sqrt [3]{\frac {1}{a} \plus{} nb} \equal{} 0$, and $ a \plus{} b \plus{} c \equal{} 3$ then
\[ \prod_{\text{cyc}}(1 \plus{} nab)\geq abc(1 \plus{} n)^3.
\]

I don't think that this problems makes a whole lot of sense. Since we have a squareroot of n in the inequality, then n>=0, but then a,b,c are positive reals so that sum (the given condition) cannot possibly be equal to zero.

My apologies for such ridiculous posts these days; my brain is malfunctioning.  I will fix it.


integration
trigonometry
calculus
calculus computations

Evaluate $ \int_{0}^{\frac{\pi}{2}} \frac{\sin^{3}x}{sin x \plus{} cos x}dx$.

hello, setting
$ \sin(x)\equal{}\frac{2z}{1\plus{}z^2}$
$ \cos(x)\equal{}\frac{1\minus{}z^2}{1\plus{}z^2}$
$ dx\equal{}\frac{2}{1\plus{}z^2}\,dz$
we get the indefinite integral
$ \minus{}16\int\frac{z^3}{(1\plus{}z^2)^3(z^2\minus{}2z\minus{}1)}\,dz$
Sonnhard.

Hi Sonnhard. Thanks for your reply. My solution is as follows:

[hide=" Click to reveal hidden solution"]

It seemed to me that the integral might be the same as $ \int_{0}^{\frac{\pi}{2}} \frac{\cos^{3}x}{\sin x \plus{} \cos x}dx$ and substituting $ y \equal{} \frac{\pi}{2} \minus{} x$, it is.

Therefore $ 2I \equal{} \int_{0}^{\frac{\pi}{2}} \frac{\sin^{3}x\plus{}\cos^{3}x}{\sin x \plus{} \cos x}dx$.

This looks like $ \frac{x^{3} \plus{} y^{3}}{x\plus{}y}$ which is $ x^{2} \minus{} xy \plus{} y^{2}$,

 so I got $ 2I \equal{} \int_{0}^{\frac{\pi}{2}}(\sin ^{2} x \plus{} \cos^{2} x \minus{} \sin x \cos x )dx$

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

$ \equal{} x \plus{}\frac{1}{4} \cos 2x$.

Putting in the limits, I got $ (\frac{\pi}{2} \minus{} \frac{1}{4}) \minus{} (0 \minus{} \frac{1}{4})$

$ \equal{}\frac{\pi}{2} \minus{} \frac{1}{2}$

and so I have $ I\equal{} \frac{\pi \minus{} 2}{4}$.

[/hide]

probably this can be done in a simpler manner..... :maybe:

Thanks buddhu. That's exactly what I did, as you can see by clicking on the blue text in my post.


conics
parabola
analytic geometry
graphing lines
slope
calculus
parameterization

Q:Prove that the locus of the mid points of the normal chords of the parabola y2=4ax is (y2/2a)+(4a3/y2)=x-2a.

Hi indialover, since no-one has tried this, I'll give it a go.

$ y^{2} \equal{} 4ax$ has parametric equation $ x \equal{} at^{2}, y \equal{} 2at$;

Differentiating, the slope of the tangent is $ \frac {1}{t}$ so that the slope of the normal is $ \minus{} t$.

Since it passes through $ (at^{2}, 2at)$, the equation of the normal is

$ y \equal{} \minus{} tx \plus{} 2at \plus{} at^{3}$.

This meets the parabola again at the point $ (as^{2}, 2as)$.

Substituting, $ 2as \equal{} \minus{} ats^{2} \plus{} 2at \plus{} at^{3} \equal{} 0$, with roots $ t$ and $ \frac {t^{2} \minus{} 2}{t}$, so

$ s \equal{} \frac {t^{2} \minus{} 2}{t}$.

Then the mid-point of the chord is $ (at^{2} \minus{} 2a \plus{} \frac {2a}{t^{2}}, 2at \minus{} \frac {2a}{t})$.

From this I get $ y^{2} \equal{} 4a^{2}(t^{2} \minus{} 2 \plus{} \frac {1}{t^{2}})$

and so

$ y^{2} \equal{} 4a(x \minus{} \frac {a}{t^{2}})$,

that is, since $ x \equal{} at^{2}$,

$ y^{2} \equal{} 4a(x \minus{} \frac {a{^2}}{x})$

I think this is different to your answer. I may well be wrong; intuitively it seems to me that the locus is another parabola.

I've done it twice and got the same answer each time but it doesn't seem to fit with yours which, unless I've misread it is

$ \frac{y^{2}}{2a} \plus{} \frac{4a^{3}}{y^{2}} \equal{} x\minus{}2a$

what do u mean by 'diffrentiating....'
pls explain that part.

Oh, 'diffrentiating' is a Calculus term. . . . an advance tool in solving such problems.

Hi AndrewTom
The solution you gave to the problem was very comprehensive and it helped me a lot. I was stuck on this question for a real long time.
But I guess you calculated the roots wrong, the other root besides t would be (-t2-2)/t. Using this root gives the equation Indialover mentioned in his question.


trigonometry
limit
trig identities
calculus
calculus computations

$ \lim_{h\to\.0} \frac{tan3(x+h) - tan(3x)}{h} $

I know the answer is $3sec^2(3x) $.  But I'm having a very difficult time finding why, could anyone help me?  What trig identities do you need?

Thanks in advanced!

Can't you understand $\frac{1}{\cos x}=\sec x$?

kunny

[quote="ecoli"][quote="ecoli"][quote="kunny"]$\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}=f'(x)$ so set $f(x)=\tan (3x)$[/quote]

I don't understand how that would help?[/quote]

Never mind, I got it!![/quote]

O.K. No problem. :)


Suppose that $ a,b,c$ is the three roots of cubic equation :$ x^3\minus{}8x^2\plus{}8x\minus{}1\equal{}0$

Find the unit digit of $ a^6\plus{}b^6\plus{}c^6$.

[quote="PKMathew"]Suppose that $ a,b,c$ is the three roots of cubic equation :$ x^3 \minus{} 8x^2 \plus{} 8x \minus{} 1 \equal{} 0$

Find the unit digit of $ a^6 \plus{} b^6 \plus{} c^6$.[/quote]

$ x^2\minus{}8x^2\plus{}8x\minus{}1\equal{}(x\minus{}1)(x^2\minus{}7x\plus{}1)\equal{}0$

Now WLOG let $ a\equal{}1$ and $ b,c$ be roots of $ x^2\minus{}7x\plus{}1\equal{}0$, then

$ a^6\plus{}b^6\plus{}c^6$
$ \equal{}1\plus{}(7b\minus{}1)^3\plus{}(7c\minus{}1)^3$
$ \equal{}1\plus{}(329b\minus{}48)(7b\minus{}1)\plus{}(329c\minus{}48)(7c\minus{}1)$
$ \equal{}1\plus{}(2\minus{}b)(7b\minus{}1)\plus{}(2\minus{}c)(7c\minus{}1)$
$ \equal{}1\plus{}(5\minus{}34b)\plus{}(5\minus{}34c)$
$ \equal{}1\plus{}6(b\plus{}c)$
$ \equal{}3(mod 10)$

So unit digit is $ 3$ (as $ b,c$ are positive).

[hide]
Suppose that $ a,b,c$ is the three roots of cubic equation :$ x^3 \minus{} 8x^2 \plus{} 8x \minus{} 1 \equal{} 0$

Find the unit digit of $ a^6 \plus{} b^6 \plus{} c^6$.

$ x^2 \minus{} 8x^2 \plus{} 8x \minus{} 1 \equal{} (x \minus{} 1)(x^2 \minus{} 7x \plus{} 1) \equal{} 0$

Now WLOG let $ a \equal{} 1$ and $ b,c$ be roots of $ x^2 \minus{} 7x \plus{} 1 \equal{} 0$, then

$ a^6\equal{}1$
$ b\plus{}c\equal{}7, bc\equal{}1$
$ b^6 \plus{} c^6\equal{}(b^3\plus{}c^3)^2\minus{}2b^3c^3\equal{}(b^3\plus{}c^3)^2\minus{}2(1)^3\equal{}[(b\plus{}c)^3\minus{}3bc(b\plus{}c)]^2\minus{}2$
$ \equal{}[7^3\minus{}21]^2\minus{}2\equiv 2\pmod {10}$

$ \therefore a^6 \plus{} b^6 \plus{} c^6$
$ \equiv 1\plus{}2\pmod {10}$

$ \equiv 3\pmod {10}$[/hide]

Here is my method:
We have
$ a \plus{} b \plus{} c \equal{} 8$
$ ab \plus{} bc \plus{} ca \equal{} 8$
$ abc \equal{} 1$
$ a^2 \plus{} b^2 \plus{} c^2 \equal{} (a \plus{} b \plus{} c)^2 \minus{} 2(ab \plus{} bc \plus{} ca) \equal{} 48$
Now note that we have
$ a^3 \minus{} 8a^2 \plus{} 8a \minus{} 1 \equal{} 0$
$ b^3 \minus{} 8b^2 \plus{} 8b \minus{} 1 \equal{} 0$
$ c^3 \minus{} 8c^2 \plus{} 8c \minus{} 1 \equal{} 0$
Adding them,
$ (a^3 \plus{} b^3 \plus{} c^3) \minus{} 8(a^2 \plus{} b^2 \plus{} c^2) \plus{} 8(a \plus{} b \plus{} c) \minus{} 3 \equal{} 0$, so $ a^3 \plus{} b^3 \plus{} c^3 \equal{} 323$
Also, $ a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \equal{} (ab \plus{} bc \plus{} ca)^2 \minus{} 2abc(a \plus{} b \plus{} c) \equal{} 48$
Now note that
$ a^3b^3 \plus{} b^3c^3 \plus{} c^3a^3 \minus{} 3a^2b^2c^2 \plus{} 3a^2b^2c^2 \equal{} (ab \plus{} bc \plus{} ca)(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \minus{} abc(a \plus{} b \plus{} c)) \plus{} 3 \equal{} 323$
Finally,
$ a^6 \plus{} b^6 \plus{} c^6 \equal{} (a^3 \plus{} b^3 \plus{} c^3)^2 \minus{} 2(a^3b^3 \plus{} b^3c^3 \plus{} c^3a^3) \equal{} 323^2 \minus{} 2 \cdot 323 \equiv 3 \pmod {10}$.

Of course now I see that I could have just found a recurrence by multplying the three equations by a, b, and c and adding, but oh well, I like my proof anyways  :D .


The expression $ \frac{7n+18}{2n+3}$ takes integer values for certain integer values of n. Find the sum of all such values of the expression.

[hide="Hint"]
$ =3+\frac{n+9}{2n+3}$
And now we can for example:
Let $ f(x)=\frac{x+9}{2x+3}$. We check $ x\in\{-9;-8;...;6\}$, and for $ x=\left\{-9;-4;-3;-2;-1;0;-1;6\right\}$ we've got what we want. For $ x>6$ we have $ f(x)\in\left(1;\frac{1}{2}\right)$ and for $ x<-9$ we have $ f(x)\in\left(0;\frac{1}{2}\right)$, so it's impossible to $ f(x)$ be an integer.[/hide]

well it simplifies to

$ \frac{7}{2}+\frac{25}{4n+6}$ but inorder for that whole thing to be an integer that second part must end in a $ .5$ so $ \frac{25}{2n+3}$ must an integer, thus when multiplying a $ \frac{1}{2}$ to it, it will become something with $ .5$ at the end.  Well, $ 15=3*5$ which means there are $ 4$ numbers that divide, it, times $ 2$ because they can be negative so it should be $ 8$ total numbers.

The solution set should be $ (-9,-4,-3,-2,-1,0,1,6)$ (this is extra not needed)

Now we want to find the sum of the values for the [i]expression[/i], not for $ n$.

Thus it since we have a negative and postive divisor of it...its $ 0+8(\frac{7}{2})$ because there are 8 divisors of it, so its $ 28$

thanks to 'ordinary boy'.

Nice solution`~

Im confused..I get some other answer.

I get $ 0$ because there are always postive and negative ones that divide it

Well, yes, but "positive" and "negative" are centered around $ \frac{7}{2}$, which is the average value.


Given a simple graph of $n$ vertices, a degree of each vetrex is at most 8. Is it true that we can color all vertices into 3 colors in such a way that the number of edges with monochromatic vertices will be at most $n$?

Very nice problem! Very simple too. To see a solution, click [hide="here"] Take the coloring with the minimal number of monochromatic edges. If some vertex belonged to at least 3 monochromatic edges, we could recolor it to make monochromatic edges fewer. So each vertex belongs to at most two monochromatic edges, from where it is easy to conclude that the total number of monochromatic edges does not exceed $n$. [/hide]

Yes, my solution was the same. It is a classical idea.
I came to it from the old USSR problem about parliament and fractions:
Each senator has 3 enemies in the parliament. Prove that it is possible to divide the parliament into two fractions in such a way that each senator will have at most 1 enemy in his fraction.


In triangle $ABC$, prove that:
\[(p-a)^{2}\sin A+(p-b)^{2}\sin B+(p-c)^{2}\sin C=4r(2R-r)\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}\]

$1.\ \ \left\{\begin{array}{c}abc=4Rpr\\\\ab+bc+ca=p^{2}+r^{2}+4Rr\\\\ \sum a^{3}=\left(\sum a\right)^{3}-3\cdot\sum a\cdot\sum bc+3abc\end{array}\right\|\Longrightarrow$

$\sum a(p-a)^{2}=2pr(2R-r)\ .$

$2.\ \ \left\{\begin{array}{c}(p-a)(p-b)(p-c)=pr^{2}\\\\ \cos\frac{A}{2}=\sqrt{\frac{p(p-a)}{bc}}\mathrm{\ \ \ a.s.o.}\end{array}\right\|$ $\Longrightarrow$ $\prod\cos\frac{A}{2}=\frac{p}{4R}\ .$

$3.\ \ \left\{\begin{array}{c}\sin A=\frac{a}{2R}\mathrm{\ \ \ a.s.o.}\\\\ \prod\cos\frac{A}{2}=\frac{p}{4R}\\\\ \sum a(p-a)^{2}=2pr(2R-r)\end{array}\right\|\Longrightarrow$

$\sum (p-a)^{2}\sin A=4r(2R-r)\prod\cos\frac{A}{2}\ .$


Σε ένα σύνολο $A$, για τη σχέση $\sim$ ισχύουν (για κάθε $x,y,z$):

Για κάθε $x\in A$,
$x\sim x$ (1)
Αν $x\sim y$, τότε $y \sim x$ (2)
Αν $x\sim y$ και $y\sim z$, τότε $x\sim z$ (3)

Κάποιος ισχυρίζεται ότι:

"To (1) είναι άχρηστο, αφού προκύπτει από τις (2), (3) ως εξής:

Αφού ισχύει το $x\sim y$, βγάζουμε το $y\sim x$. Από την (3), συνδυάζοντας τα προηγούμενα, για $z=x$, προκύπτει $x\sim x$"

Έχει δίκαιο ή αδικο;

[quote="stedes"]Έχει άδικο και ο λόγος είναι ότι δεν ξέρουμε αν υπάρχει $y$ τέτοιο ώστε $x \sim y$.  :wink:[/quote]

Σίγουρα υπάρχει $x \sim y$  :|

Αλλού παίζει το λάθος

Εννοείς ότι για κάθε $x \in A$ υπάρχει $y \in A$ τέτοιο ώστε $x \sim y$ ;;

Π.χ. ένα αντιπαράδειγμα είναι το εξής:
$A=\{1,2,3\}$  και η σχέση $\sim$ ορίζεται από το σύνολο $S=\{(1,1),(2,2),(1,2),(2,1)\}$.
Σ'αυτή τη περίπτωση ισχύουν οι συνθήκες (2),(3), αλλά όχι η (1) γιατί $(3,3) \not \in S$.
Ο λόγος που η απόδειξή του δεν ισχύει, είναι ότι δεν υπάρχει $y \in A$ τέτοιο ώστε $3 \sim y$.

:lol: ....profanos+exi adiko...
des +ena alo antiparadigma.....
esto$A=R$(to sinolo ton pragmatikon..)
kai mia sxesi  sto $R$ gia tin opia gia ka8e $x,y,z$ pragmatikus orizume
$x$~$y$<=> $x*y>0$

tote isxii oti
an $x$~$y$ tote $y$~$x$ (gt an $x*y>0$ tote kai $y*x>0$)
an $x$~$y$ kai $y$~$z$ tote $z$~$x$ (gt an $x*y>0$ kai $y*z>0$  tote $x*z>0$)
ala dn isxii
$x$~$x$ diladi $x^{2}>0$ gt gia $x=0$ dini $0>0$..

 :D

the noob is here!! :D
τι σημαινει αυτο ~?

Το $\sim$ είναι σύμβολο όχι πράξη .

Αρκεστείτε στα δεδομένα της υπόθεσης και μόνο για να ανάγετε κάποιο [u]λογικό[/u] συμπέρασμα .

ps: Σα να μιλάμε για abstract algebra


function
integration
combinatorics unsolved
combinatorics

How do I find a closed-form solution to $ a_n\equal{}2xa_{n\minus{}1}\plus{}2na_{n\minus{}2}$?

I don't know any methods like characteristic equations for non-constant coeffs. What should I use?

[quote="Sage!"]Use generating functions.[/quote]

Ok, so, using the power series generating function method, I got the differential equation
\[ \minus{} 2z^3A'(z) \equal{} (2xz \minus{} 1)A(z) \minus{} 2a_0 x z \plus{} a_1z \plus{} a_0
\]
which leads to
\[ \frac {{\int (a_0 \plus{} a_1z \minus{} 2a_0xz)e^\frac {1 \minus{} 4xz}{4z^2}dz} \plus{} C}{e^\frac {1 \minus{} 4xz}{4z^2}}.
\]
Where do I go from there?


MATHCOUNTS
probability
geometry
3D geometry
octahedron
HCSSiM

If I am correct, yesterday was the last day for State Competitions.  So does that mean we can discuss the problems?

The MATHCOUNTS website lists one competition as being today actually.  And if there were any competitions cancelled due to snow or ice yesterday they may get rescheduled for a week later.  Better wait until we get some type or official information.

MATHCOUNTS has now posted the State Competition and Solutions on their Web site:  [url]http://www.mathcounts.org/Problems/2004Competitions.html[/url]

how do you do the last problem on the team? that's the one we got wrong, im sure.

Carlos has chosen 12 different CDs he would like to buy. Four are rap, 5 are country, and 3 are heavy metal music. Carlos then selects 5 of the 12 CDs to purchase. What's the probability that his purchase includes at least 1 cd from each of the 3 categories?

the answer is [hide] 295/396 [/hide]

[quote="pinkbubblez90"]how do you do the last problem on the team? that's the one we got wrong, im sure.[/quote]

Here's yet another method.  There are 12C5=792 total ways of picking 5 of 12 CDs.

Let the first 3 come from separate categories.  4*5*3=60 choices.  The number of CDs remaining in each category is now 3,4,2.

Allowing the next 2 CDs to come from the same category multiply by 3*2+4*3+2*1=20.  Divide by 6 to allow any permutation of three CDs from one category.   60*20/6=200.

Allowing the next 2 CDs to come from different categories, multiply by 3*4+4*2+2*3=26.  Divide by 4 to allow permutations of two CDs from each of the two categories.  60*26/4=390.

Add and solve.  (200+390)/792 = 590/792 = 295/396.

hmm...nope, that and the octahedron one was the only problem besides for target #3 (the one about the liscence plates) that i actually didn't know...

fortunately for me, a 26 on sprint was good enough (combined with 14 on target) to put me at fifth in the countdown round (and i ended up fourth)  But the one that is asking for the sum of the cubes, you know that x has to start as a negative number, but can't be less than -11, and then you can just test -10 and -9, because -8 is too small, and you can't reach -1197, and if -10 or -9 gets really close but is greater, then you can increase n, it would take longer trying to write it out algebraicly, so i guess you just have to figure out possible answers (and there aren't that many) and guess among those

[quote="Rep123max"]Can anyone think of a solution algebraically and not just guess and check?[/quote]

For Sprint #24 (the sum of the cubes), I think "guess and check" along with some basic "number sense" is the fastest way to solve this one.

Is x even or odd?  Why?  Now, number sense tells me that consecutive numbers change by small steps.  Consecutive squares change with bigger steps.  Consecutive cubes change with even bigger steps.  Since we need a sum near -1197, we need a cube close to -1197.  What are the nearby perfect cubes?  (-11^3=-1331, -10^3=-1000, -9^3=-729, -8^3=-512).  Number sense has lead me very near the number with the biggest magnitude.  Add terms one-by-one until you get the expected sum or exceed the expected sum.

[Alternate]  If you had time on your hands, you might have noticed that the sum of any two consecutive terms is even.  The sum of any three consecutive terms is a multiple of 3.  The sum of any four consecutive terms is a multiple of 4.  Etc.  So let's factor.  1197=3*3*7*19.  There are 12 total factors.  We might expect the series that sums to 1197 to be 1, 3, 7, 9, or etc. terms in length.  But remember the problem requires there be at least 4 terms in the series.

were all the teams really really bad this year? my team won the state with only 6 (!!!!!!!!!!!!!!!!!) right on the team round, and a team score of 42.25.

My team won with a 9 on the team round, and an average individual score of 38.75, to calculate team score, it's just average plus the team round? cuz our score was 47.75, i thought they were supposed to double the team round score....

They are supposed to count the team round double.  So either they miscalculated your team score, your average was 9 points lower, or you got less team rounds right.  probably the first one.  great score.  In New Hampshire we won with an average of 32.75 and 7 on team round for a 46.75

frost

Our team got 7 right on team round, 36.50 on average... so 50.50.

wow. connecticut sure is pathetic this year

don't say that. you can prove Conn.'s good at nats.

By the way, how can i tell who yall are at nats?

i'll be the one on my team trading all my badges away really quickly. anyone have any ideas about a meeting place? at dinner or in the lobby?

hmm, then my team's score was a 56.75, but anyways, we could all wear white shirts and write AOPS or something on them, there are 2 AOPSrs on Team Washington, so it shouldn't be too hard to find us, anybody have ideas on how to recognize other AOPSers?

Wow. Texas team is tougher than I thought compared to other states.  We had a 24, 29, 33 (me, only because of a 5 target), and a 34, for a average score of 30.  We got 8 on the team round, so our team score was a 46.  8th in Texas.


inequalities
Cauchy Inequality
inequalities proposed

Let $x,y,z$ be positive real numbers such that
\[x^{2}+y^{2}+z^{2}=1,\]
provet that
\[\frac34\leq\frac{x^{2}}{1+yz}+\frac{y^{2}}{1+zx}+\frac{z^{2}}{1+xy}\leq1.\]

[quote="pvthuan"]Let $x,y,z$ be positive real numbers such that
\[x^{2}+y^{2}+z^{2}=1, \]
provet that
\[\frac34\leq\frac{x^{2}}{1+yz}+\frac{y^{2}}{1+zx}+\frac{z^{2}}{1+xy}\leq1. \]
[/quote]
The right part is trivial.
The left part:
using Cauchy Inequality, we have
\[\sum\frac{x^{2}}{1+yz}\ge \frac{(x^{2}+y^{2}+z^{2})^{2}}{x^{2}+y^{2}+z^{2}+xyz(x+y+z)}=\frac{1}{1+xyz(x+y+z)}\geq \frac{1}{1+\frac{(x^{2}+y^{2}+z^{2})^{2}}{3}}=\frac{3}{4}\]

We are done. :)

[quote="pvthuan"]Let $x,y,z$ be positive real numbers such that
\[x^{2}+y^{2}+z^{2}=1, \]
provet that
\[\frac34\leq\frac{x^{2}}{1+yz}+\frac{y^{2}}{1+zx}+\frac{z^{2}}{1+xy}\leq1. \]
[/quote]
Use Chebychev inequality:
$\frac{x^{2}}{1+yz}+\frac{y^{2}}{1+xz}+\frac{z^{2}}{1+xy}\geq\frac{1}{3}(x^{2}+y^{2}+z^{2})(\frac{1}{1+xy}+\frac{1}{1+yz}+\frac{1}{1+zx}\geq\frac{3}{3+xy+yz+zx}\geq\frac{3}{4}$
The right part is equavalent to:
$x^{3}+y^{3}+z^{3}\leq 2(x+y+z)+xyz$(It's true)
We are done. :)


Source: WMTS 1994 Round 4 #4

Solve the equation
x:^4: + 1 = 2x(x:^2:+1)

For some reason I'm having trouble with this one. I believe I'm missing something. This is the first approach I used...

[hide] Subtract 2 from both sides, the equation becomes:

x^4-1= 2x (x^2+1)-2

(x^2-1)(x^2+1) - 2x(x^2+1)= -2

(x^2+1) (x^2-2x-1) = -2

Add 2(x^2+1) to both sides:

(x^2+1) (x^2-2x+1) = -2+ 2x^2 + 2 = 2x^2

(x^2+1)(x-1)(x-1)= 2x^2

[/hide]

Second approach...

[hide] (x^2+1)^2= x^4+2x^2+1

(x^2+1)^2-2x^2-2x(x^2+1)= 0

add 3x^2 to both sides:

(x^2+1)^2-2(x^2+1)(x)+ x^2= 3x^2

the left side is in the form a^2-2ab+b^2, or (a-b)^2, where a= (x^2+1) and b= x.

(x^2-x+1)^2= 3x^2.

[/hide]

There must be some easy trick I'm not missing, or a calculation error somewhere in there.

Did you try solving those equations? Is there any reason they can't both be correct?

ComplexZeta wrote:Here's a path to the solution (in spoiler).

[hide]x^4+1=2x^3+2x
Divide everything by x^2.
x^2+1/x^2=2x+2/x
(x+1/x)^2-2=2(x+1/x).
Can you solve it from there?[/hide]

Yea, we could now.

So we just let x+1/x=y, and we get a quadratic in y. Solve for y, and then plug the two values into the equation y=x+1/x and we get the answer. Nice.

complex, very nice... but I have a question, and it may be stupid but o well....

I would never have thought to factor x^2 + 1/x^2 into (x+1/x)^2-2

How did you think of that, and I can't even find a process for it.

Yes, I am bad at this stuff :(

When you have something really symmetric, see how you can factor it. If you have things with a(x^n+x^(-n)), there is usually a quick solution involving (x+1/x)^n.

While we're at it, I saw an indentity recently (probably on this forum) that if x+1/x=2cos theta, then x^n+1/x^n would be 2 cos (n*theta), or something similar. How would you prove that? Complex numbers?

Let x=e^(i*theta) and use Euler's identity.

I think a derivation of Demoivre's would work, too

Yes. That's the same thing. If you use Euler's Identity, DeMoivre's Theorem is a trivial corollary. It just follows from the laws of exponents.


[quote="MCrawford"]If $2^n$ is a factor of $20!$, what is the largest possible value of $n$?[/quote]

[hide]I think its 18 but I could be wrong[/hide]

it's similar to finding the amount of zeros ending a factorial amount

[quote="hwenterprise"]tarquin, I don't get it.[/quote]
You do 20 over the powers of 2, then add it together.

[quote="tarquin"][hide]$\frac {20}{16} + \frac {20}{8} + \frac {20}{4} + \frac {20}{2} = 18$[/hide][/quote]

That equals 18.75

[hide]I did it like this..

Find all of the numbers containing factors of 2 in 20!
$2\cdot4\cdot6\cdot8\cdot10\cdot12\cdot14\cdot16\cdot18\cdot20$
Then find all of the factors of the two in these numbers (respectively..)
$2^1\cdot2^2\cdot2^1\cdot2^3\cdot2^1\cdot2^2\cdot2^1\cdot2^4\cdot2^1\cdot2^2$
Add the exponents and get
$2^{18}$
[/hide]

[quote="xxreddevilzxx"]something like this:

if $2^n \mid 20!$, then

$\displaystyle n=\lfloor{\frac{20}{2^1}}\rfloor + \lfloor{\frac{20}{2^2}}\rfloor + \lfloor{\frac{20}{2^3}}\rfloor + \lfloor{\frac{20}{2^4}}\rfloor$
$ = 10 + 5 + 2 + 1 = 18$.[/quote]
[hide]the problem is that 20/2^3=5/2, and 20/2^4=5/4[/hide]


logarithms
number theory proposed
number theory

Let $ (u_n)$ be a sequence given recursively as follow
\[ u_1 \equal{} \minus{} 2; u_2 \equal{} \minus{} 3
\]

\[ nu_{n \plus{} 2} \plus{} 2(n \plus{} 1)u_n \equal{} (3n \minus{} 1)u_{n \plus{} 1}
\]
Let
\[ S \equal{} \sum_{n \equal{} 1}^{2009} u_n \plus{} 2(2^{2009} \minus{} 1)
\]
Prove that S is divisible by $ 2009$
I have just edited, sorry!

[quote="Brown"]Let $ (u_n)$ be a sequence given recursively as follow
\[ u_1 \equal{} \minus{} 2; u_2 \equal{} \minus{} 3
\]

\[ nu_{n \plus{} 2} \equal{} 2(n \plus{} 1)u_n \equal{} (3n \minus{} 1)u_{n \plus{} 1}
\]
[...]
Prove that S is divisible by $ 2009$[/quote]
you want say $ nu_{n \plus{} 2} \equal{} 2(n \plus{} 1)u_n\plus{} (3n \minus{} 1)u_{n \plus{} 1}$ ???


arithmetic sequence
number theory
prime factorization

For each positive integer $k$, let $S_k$ denote the increasing arithmetic sequence of integers whose first term is $1$ and whose common difference is $k$. For example, $S_3$ is the sequence $1,4,7,10,...$. For how many values of $k$ does $S_k$ contain the term $2005$?

[hide="Comprehensive Solution"]Because each arithmetic sequence has a common difference of $k$, $1+nk=2005$ for some integer value of k.

This also means that $nk=2004$ for some integer value of $k$, and the $k$ must be a factor of 2004.

The prime factorization of $2004$ is $2^2*3*167$ and there are $3*2*2=12$ factors.

The answer is 12.[/hide]

[hide]It is easy to see that $k | 2004$. Since $2004 = 2^2 \cdot 3 \cdot 167$, it has $3 \cdot 2 \cdot 2 = \fbox {12}$ factors.
EDIT: Darn, my little brother was jumping on me and 4everwise beat me! :D
[/hide]

[hide="Answer"]The first term is $1$, so we are adding a total of $2004$ to reach $2005$. $2004=2^2\cdot 3\cdot 167$, so we have $3(2)(2)=\boxed{12}$.[/hide]


This is an example of such problem:

Hi Bob,

      well I think I have a better way to do this. Since the problem is caused by having 2 folders doing similar things (e.g.  downloads folder and shared video folder), I think I will just make a word file that explains the situation.

      The problem is where I should put such word file. Should it be just under the "My document" folder (this name need to be changed since inside the folder there are a lot of things that are not document.), "Doc," or "Cheerful Coffin's files," or else? As you can see, this is very confusing, especially trying to find the file afterwards.

Regards,
Cheerful Coffin

Basically I have way too many folders doing similar things. And sometimes I don't even know where I should put my files into. You guys may probably say "use your intuition." However, I think a better way to do it is to see if other people have some sort of filing system that I can just simply duplicate and use it.

After all, this isn't rocket science.

thank you for your help in advance :) (if there is any :huh:...)

"Use your intuition." ;)

But seriously, I just organize everything into semi-rigorous categories like "Pictures", etc. If something fits into two categories, I randomly choose one. If I can't find something in one folder, I look in another one that makes sense.

Well, the best solution would be to...
[size=75]
Get a mac[/size]

Spotlight clears that problem up on macs. But instead of using the system folders, make your own folders, right on your desktop. Much easier  ;) .

use shortcuts; if a file fits in 2 or more folder then place it in one and then create shortcuts in the other folders

there. see my attached files, a template of how I'm going to organize the files

I have a new organizational scheme on my computer! All of my files except for temporary ones go on a 50GB FAT32 partition that I can access from both Windows and Linux. Programs install into Program Files.

I just dump everything on my desktop. 3/5 of my desktop is filled with icons.


modular arithmetic
number theory proposed
number theory

Let $ x,y$ and $ z$ are natural numbers greater than 1 such that $ xy\minus{}1$ is divisible by $ z$, $ yz\minus{}1$ is divisible by $ x$ and $ zx\minus{}1$ is divisible by $ y$. Find $ x,y,z$.

let $ x\geq y \geq z$,$ xy \minus{} 1\equiv 0\pmod{z},xz \minus{} 1\equiv 0\pmod{y},yz \minus{} 1\equiv 0\pmod{x}$  $ \implies$
$ xy \plus{} yz \plus{} zx \minus{} 1\equiv 0\pmod{xyz}$  $ \implies$
$ xy \plus{} yz \plus{} zx \equal{} (xyz).k \plus{} 1$  $ \implies$
$ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \equal{} k \plus{} \frac {1}{xyz}$

$ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z}\leq \frac {3}{2}$  $ \implies$  $ k \equal{} 1$

$ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \equal{} 1 \plus{} \frac {1}{xyz}$

if $ z > 3$ $ \implies$

$ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} < 1$  contradiction!

$ \implies$  $ z \equal{} 3$ or $ z \equal{} 2$
$ z \equal{} 2$  $ \implies$ $ 2y \plus{} 2x \plus{} xy \equal{} 2xy \plus{} 1$
$ z \equal{} 3$   $ \implies$  $ 3y \plus{} 3x \plus{} xy \equal{} 3xy \plus{} 1$   :wink:

Let wlog $ y$ be the smallest number among $ x$, $ y$ and $ z$. We have $ xz \mid xy\plus{}yz\minus{}1<2xz$, hence $ xz\equal{}xy\plus{}yz\minus{}1$. This gives $ y \mid xz\minus{}1\equal{}y(x\plus{}z)\minus{}2$ and hence $ y\equal{}1$ or $ y\equal{}2$. If $ y\equal{}1$, it follows $ (x\minus{}1)(z\minus{}1)\equal{}0$ and hence $ x\equal{}1$ and $ z$ can be chosen arbitrarily (it's easy to proof that this is a solution indeed). If $ y\equal{}2$, we have $ x(z\minus{}2)\equal{}2z\minus{}1$ and hence $ z\minus{}2 \mid 2z\minus{}1$. So $ z\minus{}2 \mid 2z\minus{}1\minus{}2(z\minus{}2) \equal{} 3$. Thus, $ z\equal{}3$ or $ 5$, as $ y \le z$. $ z\equal{}3$ gives $ x\equal{}5$ and $ z\equal{}5$ gives $ x\equal{}3$. Hence the set of solutions: $ (x,y,z)\equal{}\{(1,1,t), (2,3,5) \}$ and their permutations.


Is there any more LaTex codes than what are under the LaTex PHP? :?:

There are dozens, if not hundreds, of LaTeX packages with many LaTeX commands.  If you are looking for something specific, ask in the LaTeX message board on this forum.


geometry
trigonometry
quadratics
Pythagorean Theorem
trig identities
Law of Cosines
Law of Sines

In $ \triangle ABC$, $ \angle ABC\equal{}120^\circ$, $ AB\equal{}3$ and $ BC\equal{}4$. If perpendiculars constructed to $ \bar{AB}$ at $ A$ and to $ \bar{BC}$ at $ C$ meet at $ D$, find $ CD$.

edit: how to make the long overbar for lines?

\overline{AB} produces $ \overline{AB}$.
[hide="Edit: Might as well post solution while here"]
[asy]defaultpen(fontsize(10));
pair E=(0,0), B=(6,0), C=(10,0), D=(20/(3^0.5))*expi(pi/6), A=(3*3^.5)*expi(pi/6);
draw(A--B--C--D--cycle);
draw(A--E--B,dashed);
label("\(A\)",A,NW);
label("\(B\)",B,S);
label("\(C\)",C,S);
label("\(D\)",D,NW);
label("\(E\)",E,S);
label("\(3\)",(A+B)/2,SW);
label("\(4\)",(B+C)/2,S);
draw(rightanglemark(D,A,B));
draw(rightanglemark(B,C,D));
draw(anglemark(C,B,A));
label("\(120^{\circ}\)",B+(0.1,0.1),NE);[/asy]
Extend $ \overline{AD}$ and $ \overline{BC}$ to meet at $ E$; then since $ \angle BAE \equal{} \angle DCE \equal{} 90^{\circ}$ it follows that $ \triangle BAE \sim \triangle DCE$, which are thus $ 30 \minus{} 60 \minus{} 90$ right triangles. So $ BE \equal{} 6$ and $ CD \equal{} \frac 1{\sqrt {3}} \cdot (4\plus{}6) \equal{} \boxed{\frac {10}{\sqrt {3}}}$.

[hide="Or.."]
Alternatively, note that $ ABCD$ is cyclic and use Law of Cosines to find $ AC$, then Extended Law of Sines to find radius (as $ BD$ is diameter), then Pythagorean Theorem on $ \triangle BCD$.
[/hide]

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Edit: simpler?  :wink:

Another solution(I think it's simpler than azips's solution):
Applying  Laws of cosine we have: AC=square root 37
Since angleBAD+angle(BCD)=180, we have: ADCB is cyclic.Therefore, angle(ADB)=60
Since BD is the hypotenuse of 2 square triangles, we have: CD^2+16=AD^2+9
Applying Laws of cosine again we have: 2CD.AD=CD^2+AD^2-37
Solving the system equation we got the result(it's a quadratic equation with CD is a variable and CD^2>=15)


Let $ a, b$ are real numbers. Find the rank of the matrix:

$ \left( \begin{array}{cccc} 1 & \minus{} 1 & 0 & 1 \\
2 & a \plus{} b \minus{} 2 & a \plus{} b & \minus{} a \minus{} b \minus{} 2 \\
1 & \minus{} a \minus{} b \minus{} 1 & \minus{} a \minus{} b \minus{} 1 & a \plus{} b \minus{} 1 \\
\minus{} 1 & \minus{} a \minus{} b \plus{} 1 & \minus{} a \minus{} b \minus{} 1 & 2a \minus{} 3 \end{array} \right)$

I did some row elimination - easy steps, no division - and got that this matrix is row equivalent to
\[ \begin{bmatrix}1&\minus{}1&0&1\\ 0&a\plus{}b&a\plus{}b&\minus{}a\minus{}b\minus{}4\\ 0&0&\minus{}1&\minus{}6\\ 0&0&0&a\minus{}b\end{bmatrix}\]
So if both $ a\plus{}b$ and $ a\minus{}b$ are nonzero - that is, $ a^2\ne b^2$ - then this matrix is invertible (rank 4).

If either $ a\equal{}b$ or $ a\equal{}\minus{}b,$ but $ a$ and $ b$ are nonzero, then the matrix has rank 3.

Finally if $ a\equal{}b\equal{}0,$ then the first three rows are still independent and it still has rank 3.

The determinant of the original matrix is $ b^2\minus{}a^2.$ You get that from the upper triangular form shown above and from the fact that I used only "type III" row operations (no interchanges or scalar multiplications) to get there.


analytic geometry
geometry proposed
geometry

Introduction: Let internal and external angle bisectors of $ \angle AXB$ intersect $ AB$ at $ M$ and $ N$. The circle $ (MXN)$ is called the Apollonius circle of $ \angle AXB$.

$ \triangle ABC$ and a point $ P$ is given. Draw Apollonius circles of $ \angle APB$,  $ \angle APC$, and $ \angle CPB$. Prove that these three circles pass through a common point other than $ P$.

$ \mbox{Let internal and external angle bisectors of } \angle{BPC} \mbox{ line BC at points }A^\prime \mbox{ and } A^{\prime\prime}. \\
\mbox{If denote } |PA| \equal{} a_1, |PB| \equal{} b_1, |PC| \equal{} c_1 \mbox{, then in barycentric coordinate we: } \\
A^\prime(0;c_1;b_1); A^{\prime\prime}(0; \minus{} c_1;b_1) \mbox{ and for midpoint of the }[A^\prime A^{\prime\prime}] \mbox{ we: }O_a(0; \minus{} c_1^2;b_1^2). \\
\mbox{Similarly define the points } B^\prime, B^{\prime\prime}, C^\prime, C^{\prime\prime} \mbox{ and points }O_b(c_1^2;0; \minus{} a_1^2); O_c( \minus{} b_1^2;a_1^2;0). \\
\mbox{Because } \left|\begin{array}{ccc} 0 & { \minus{} c_1^2} & {b_1^2} \\
{c_1^2} & 0 & { \minus{} a_1^2} \\
{ \minus{} b_1^2} & {a_1^2} & 0 \end{array}\right| \equal{} 0\Rightarrow\mbox{ the points } O_a, O_b, O_c \mbox{ coliniar}\Rightarrow \\
\Rightarrow \mbox{ these three circles pass through a other common point Q} \\
\mbox{ (this point is the symetric of point P be line: }O_a \minus{} O_b \minus{} O_c).$

Once we have $ \frac{BO_a}{CO_a}\equal{}\frac{BP^2}{CP^2}$, $ \frac{CO_b}{AO_b}\equal{}\frac{CP^2}{AP^2}$, $ \frac{AO_c}{BO_c}\equal{}\frac{AP^2}{BP^2}$.  Then apply inverse menelaus for points $ O_a,O_b,O_c$ on $ \triangle ABC$. So $ O_a, O_b, O_c$ lie on a common line $ l$.