modular arithmetic
combinatorics unsolved
combinatorics

There are c red, p blue, and b white balls on a table. Two players A and B play a game by alternately making moves. In every move, a player takes two or three balls from the table. Player A begins. A player wins if after his/her move at least one of the three colors no longer exists among the balls remaining on the table. For which values of c, p, b does A have a winning strategy?
Please help me solve this problem. It is very hard for me.  :(

If one of $c, p, b$ is at most $3$ then obviously A wins.

If any of $c, p, b$ become $3$ or smaller at any point in the game, it's done. In other words, a player loses at once if he makes one of $c, p, b$ at most $3$. So we can assume $c, p, b \geq 4$ and take $C=c-4, P=p-4, B=b-4$. Now we can imagine that there are $C, P, B$ balls and the player who can't take his balls (i.e. is left with $0$ or $1$) loses. Note how the colors become irrelevant! Now suppose that $C+P+B \equiv 0, 1 \pmod{5}$. Then if $A$ takes $2$ balls, $B$ takes $3$ and viceversa, always handing $A$ a number of balls $\equiv 0, 1$. Eventually $A$ will get $0$ or $1$ balls and lose. On the other hand, if $C+P+B \equiv 2, 3, 4$ then $A$ can take 2, 2 or 3 balls on his first move and then play B's previous strategy. Since $C+P+B=c+p+b-12$, this translates to: $A$ wins if one of $c, p, b$ is $<4$, or if $c+p+b \equiv 0, 1, 4$, and $B$ wins if $c, p, b \geq 4$ and $c+p+b \equiv 2, 3$.

You're wrong, severius : if for example $c=p=1$ and $b=5$, A loses.

Umm... does he? Unless I misunderstood the condition for winning, A can take one of the $c$ balls and anything else, and win immediately.


logarithms geometry algebra function domain AMC

23. Square $ABCD$ has area 36, and $\overline{AB}$ is parallel to the x-axis. Vertices $A, B,$ and $C$ are on the graphs of $y = \log_{a}x, y = 2\log_{a}x,$ and $y = 3\log_{a}x$, respectively. what is $a$?

$(\text{A})\ \sqrt[6]3 \quad (\text{B})\ \sqrt3\quad (\text{C})\ \sqrt[3]6\quad (\text{D})\ \sqrt6 \quad (\text{E})\ 6$

[hide]
Clearly each side has length 6.

It's not hard to see that $A$ will be the lower right, $B$ will be the lower left, and $C$ will be the upper left. Let $B$ be $(x, y)$. We then have 3 equations for 3 variables.

$y = 2\log_{a}x$
$y = \log_{a}{(x+6)}$
$y+6 = 3\log_{a}x \implies y = 3\log_{a}x-6$

From the first and the second:
$2\log_{a}x = \log_{a}{(x+6)}$
$\log_{a}{\frac{x^{2}}{x+6}}= 0$
$\frac{x^{2}}{x+6}= 1$
$x^{2}= x+6$
$x^{2}-x-6 = 0 = (x-3)(x+2)$ Since $x>0$ because of the domain of a logarithm, $x = 3$.

Now, from the first and the third:
$2\log_{a}3 = 3\log_{a}3-6$
$6 = \log_{a}3$
$a^{6}= 3$
$a = \sqrt[6]3 = (\text{A})$
[/hide]

[hide="I liked this one."]
Say that $A=(x,y)$.  Then $B=(x+6,y)$ or $(x-6,y)$.  $\log_{a}x < 2\log_{a}x+6$, clearly, so $B=(x-6,y)$.  By the same reasoning, $C=(x-6,y+6)$. We thus have

$y = \log_{a}x$

$y = 2\log_{a}x-6$

$y+6 = 3\log_{a}x-6$

Subtracting the second equation from the third, we get

$6 = \log_{a}x-6\Rightarrow 2\log_{a}x-6 = y = 12$

With the first equation, this gives $x = a^{12}$.

The first two equations give $(x-6)^{2}=x$, for which the solutions are 9 and 4.  4 runs into problems with $\log_{a}x-6$, so $x=9$.

Thus $a^{12}=9=3^{2}$ and $a=\sqrt[6]{3}$.  A.

[/hide]

I liked it.

[hide]
Let A=(x,y).  Then B=(x-6,y) and C=(x-6, y+6).

From the problem we know that
$\log_{a}x = y \qquad\ \iff \qquad a^{y}=x$
$2\log_{a}{(x-6)}=y \qquad\iff\qquad a^{y}=(x-6)^{2}$
$3\log_{a}{(x-6)}=y+6\qquad\iff\qquad a^{y+6}=(x-6)^{3}$
From the first two we know that $x=4$ or $x=9$.
From the second and third:
$a^{y+6}=\left(a^{y}\right)\left(a^{6}\right)=(x-6)^{3}$
$(x-6)^{2}\cdot a^{6}=(x-6)^{3}$
$a^{6}=x-6=-2 \text{ or }3$
-2 doesn't work, so $a^{6}=3$ and $a=\sqrt[6]3$

[/hide]

[quote="1337h4x"]I didn't like it because being the idiot I am I placed vertex A to the left of Vertex B. That yielded an answer of -3.5 :roll:[/quote]

The way I solved it, that didn't matter.  Of course, I didn't finish solving it until 30 seconds after the test was over, but...


How many different ways can the five letters of the word STATE be scrambled if the two T's cannot be consecutive?

There are $ \frac{120}{2!}$ or $ 60$ ways to arrange the letters without restrictions.  From these, there are $ 4$ cases we must subtract:

Case 1:  TTxxx- $ 6$ ways

Case 2:  xTTxx- $ 6$ ways

Case 3:  xxTTx- $ 6$ ways

Case 4:  xxxTT- $ 6$ ways

Thus, the answer is $ 90 \minus{} 6(4)$, or $ \boxed{66}$.

You can just note that there are $ 4$ spots to put the double T, and $ 6$ ways to arrange the remaining letters, for $ 24$ invalid cases.  Also, it's $ 60\minus{}6(4)\equal{}\boxed{36}$.


calculus
integration
algebra
function
domain
Ring Theory
superior algebra

If $R=\mathbb{Z}[\sqrt{d}]$, then show that $F\cong \lbrace r+s\sqrt{d}|r,s\in\mathbb{Q}\rbrace$ where $R$ is an integral domain and $F$ is the field of quotients.

Is there an easier way to do this other than defining a map from $F$ to $\lbrace r+s\sqrt{d}|r,s\in\qq \rbrace$?  I foresee this becoming no more than a tedious task of basic algebraic manipulation...

why tedious? it's obvious :-)
and of course you can generalize it to adjunctions of integral elements.


What is the sum of all positive odd multiples of 3 that are less
than 100?

$ \frac {100} {3}$ $ \equal{}$ $ 33.333...\sim 33$ positive multiples of 3 less than 100. Since we start with an odd and end with an odd, we  add $ 33\plus{}1$ $ \equal{}$ $ 34$ and divide by $ 2$ in order to get the number of odd multiples of three. $ \frac {34} {2}$ $ \equal{}$ $ 17$. We add the first and last of such numbers. $ 3$ + $ 99$ $ \equal{}$ $ 102$. Now we do $ 17\minus{}1$ $ \equal{}$ $ 16$ and divide by two. $ \frac {16} {2}$ $ \equal{}$ $ 8$ and multiply it by $ 102$. $ 8$ times $ 102$ $ \equal{}$ $ 816$. But we still hhave the middle term of the sequence $ 3, 9, 15... 99$, so we divide $ 102$ in half (to average the first and last to get the middle). $ \frac {102} {2}$ $ \equal{}$ $ 51$, which we add to $ 816$ to get $ \boxed{\boxed{\boxed {867}}}$.

here's easier way:

the answer is 3*1 + 3*3 + ... + 3*33, which simplifies as 3*(1+3+...+33).

and there's total 17 odd numbers in the parentheses, which is $ 17^2$, 289.

soo the answer is 3*289=867.


Approximately how many people made it into the AIME in NY state this year? If you don't know, where can I find out?

You can find that by looking at the [url=http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2009-12a/09-1-1012-StateStats.shtml]state stats[/url].


[color=darkblue]Let $ S_1$ and $ S_2$ be circles meeting at the points $ A$ and $ B$. A line through $ A$ meets $ S_1$ at $ C$ and $ S_2$ at $ D$. Points $ M,N,K$ lie on the line segments $ CD, BC, BD$ respectively, with $ MN$ parallel to $ BD$ and $ MK$ parallel to $ BC$. Let $ E$ and $ F$ be points on those arcs $ BC$ of $ S_1$ and $ BD$ of $ S_2$ respectively that do not contain $ A$ . Given that $ EN$ is perpendicular to $ BC$ and $ KF$ is perpendicular to $ BD$ .Prove that $ \angle EMF \equal{} 90^0$.[/color]

[color=red][b]Pleases tell me another solution. Thank you very much.[/b][/color]

Thank you. But I read it some times, now I want to see another solutions.

There are 2 solutions there, and no need to open a new thread for the same question (and you knew for its existance!)


A college student drove his compact car 120 miles home for the weekend and
averaged 30 miles per gallon. On the
return trip the student drove his parents' SUV and averaged only 20 miles per gallon.
What was the average gas mileage, in miles per gallon, for the round trip?

We see that $ \frac{120}{30}\equal{}4$, and $ \frac{120}{20}\equal{}6$. Adding $ 4\plus{}6\equal{}10$, we get he used 10 gallons. Now dividing $ 240$ by $ 10$, we get $ 24$.


Find all triples of natural numbers $(x, y, n)$ such that $\frac{x!+y!}{n!}=3^{n}$

[hide]

Case 1: $x,y <n$

Since $(n+1)! > 2n!$ for all $n>1$, we need only check the case $n=1$.  This doesn't work though.

Case 2: $x<n, y \ge n$

This fails though since we need $n! | x!+y!$.

Case 3: $x,y \ge n$

We have that $\frac{x!}{n!}+\frac{y!}{n!}= 3^{n}$.  But each of those fractions is an integer in itself.  Notice that we must have the same number of powers of three that divide both of those integers.  Thus WLOG we can split into two subcases:

Subcase 1: $x=y$

This fails because then two is a factor of the LHS.

Subcase 2: $y= x+1$

We have that $(x+2)\frac{x!}{n!}= 3^{n}$ then.  Thus, since $x \ge n$ we have that $x+2$ is a power of 3.  This immediately implies that x is not divisible by three and that $1 = \frac{x!}{n!}$ or that $x=n$.  So we have that $n+2 = 3^{n}$ whose only solution is $n=1$.  Thus our only two solutions are $(1,2,1)$ and $(2,1,1)$.[/hide]

It is easy to check, that $x\not = y$. Let $x>y$ then
$[x(x-1)...(y+1)+1]y!=3^{n}n!$. Therefore $y=n$ or $y=n+1$.
If y=n we have $x(x-1)...(n+1)=3^{n}-1\Longrightarrow x\le n+2$.
In this case we have only one solution $x=2,y=n=1$.
Let $y=n+1$. It give $x(x-1)...(n+2)+1=\frac{3^{n}}{n+1}$. Therefore $n=3^{k}-1,x\le n+3$. It give $(3^{k}+1)(3^{k}+2)+1=3^{3^{k}-k-1}$
In this case there are not solutions.

[quote="pdiao"]I'm sorry, but I do not really understand why $y=n$ or $y=n+1$.  Could you explain?[/quote]
Because $[x(x-1)...(y+1)+1]y(y-1)...(n+1)=3^{n}$.
If $y>n+1$ then (n+1)(n+2) had prime divisor distinct 3.

Already posted, but I guess I'm a little bit too late  :wink:

http://www.mathlinks.ro/Forum/viewtopic.php?p=183463#183463


Okay, this is how my tournament works:

1) Peeps will sign up.
2) Those peeps will be sent the same problem, but with the numberings altered.
3) Those peeps will send me back a solution and an answer.

EDIT: Problems may not be the same, but I will try to have the difficulty the same.  Maybe Group A will be sent this problem, and Group B the other, like that. (using random.org of course.)

Grading:

If your answer is wrong, but I like your solution, I will award you 20 points.  If both answer and solution are god, you will get 50 points.  If the answer is good but not the solution, then you will get 10 points.

Rules:

the aid of a calculator is not permitted, neither is anything else.

So sign up by PMing me or posting here!!

NOTE: I will try to end reminder PM's.

Current sign-ups:

high_flyer10
AIME15
ghjk
BOGTRO
nikeballa96
AIME_is_hard

I've also added a new rule:

Inactivity will result in subtraction of points.  BTW: Each round will have like 3 questions and each one is worth 50 points.

[hide="because i really feel like making a hide thing"] NIKE JOINS! [/hide]

=]

Current sign-ups:

high_flyer10
AIME15
ghjk
BOGTRO
nikeballa96
AIME_is_hard
isabella2296
SuperMathBoy
Dojo
Twin Prime Conjecture

10 so far...need more!!

[quote="AIME_is_hard"][quote="007math"]Maybe Group A will be sent this problem, and Group B the other, like that. (using random.org of course.)
[/quote]

groups??[/quote]

Even if there are groups, the scores fo that one group will not effect on another.  You wouldn't even know what group you are in unless you figure out who has the same questions as you.  I need 8+ more people!!!

[quote="BOGTRO"]can someone give me the probs?[/quote]

I sent them to you, but I'll send them again.

Extending until Wednesday.

athunder turned in his work and got 100/100!!

Hurry up and submit!!!  You'll be eliminated if you don't!!

[quote="athunder"]How many rounds are you planning on having?[/quote]

UNtil I have a winner...

Round 2 ended today.  Scores will be up either tomorrow and Round 3 will be out Tomorrow also.

Scores for Round 2:

[hide="Round 1 Scores"]
BOGTRO: 150
athunder: 150
Discrete_Math : 150
high_flyer10: 150
Dojo: 150
AIME15: 150
ghjk : 140
someperson01 : 135
isabella2296: 110
Twin Prime Conjecture: 110
[/hide]

[hide="Round 2 Scores"]
Twin Prime Conjecture: 100
high_flyer10: 100
Discrete_Math: 100
athunder : 100
BOGTRO: 100
ghjk : 85

[/hide]

[hide="Official Current Scores"]
BOGTRO: 250
athunder: 250
Discrete_Math : 250
high_flyer10: 250
Twin Prime Conjecture: 210
ghjk : 225
Dojo: 150
AIME15: 150
someperson01 : 135
isabella2296: 110
[/hide]

The eliminated people are:

someperson01
isabella2296
Dojo (random.org got 4 heads for AIME and 1 for you....sorry...)

Now the scores are:
BOGTRO: 250
athunder: 250
Discrete_Math : 250
high_flyer10: 250
Twin Prime Conjecture: 210
ghjk : 225
AIME15: 150

Round 3 will be sent.

These are the Questions for Round 3...

1. I go to school and open locker 1.  Then I open locker two until I get to locker 50.  I go back to locker 1 and close every locker number which is a multiple of 2.  I go back to locker 1 and close a locker if it is open or open a locker if it is closed if the number is a multiple of 3.  I do this until the multiples of 6.  How many lockers are open?

2. My friend has 6 apples.  I have 9 banana.  My other friend has 10 oranges.  I decided I want to distribute these fruits among ourselves, how many ways are there to do this if each type of fruit is indistinguishable?

3. One day, Billy99 decided to inscribe his name on a door and install it in his bedroom.  The current door he has he can sell for 50 dollars.  Billy99 will have to buy a special door costing 200 dollars, and for each capital letter, it costs him the positive value of $ x$ in $ x^2 \minus{} 2x \minus{} 143$.  For every lowercase letter, it costs him $ xy \minus{} y \equal{} 108$ dollars where $ x$ is $ x$ from the previous equation and where $ y$ is the cost of a lowercase letter.  For each different type of character (ex. a number) it costs him $ xy^2 \plus{} x^2y$ dollars where $ x$ and $ y$ are the cost of uppercase letters and lowercase ones respectfully.  But if Billy99 takes two of the same letter or number, he gets a 10% discount four those 2 letters or numbers.  What is the amount of money it will cost him to get his name on a door if he sells his old one and buys the special one?

These are due on [b]January 9, 2009[/b]

[quote="AIME15"]For number 3, there are an infinite amount of possible values of x^2-2x-143.  Do you mean x^2-2x-143=0[/quote]

Yeah... :fool:

[quote="007math"]To end this quickly, the person with the highest score will win.  This time I will grade very aggressively, I'm going take off points for incorrect grammar too and more...[/quote]
What? Grammar is also counted? By the way, you already graded me harshly from the beginning though. Will you show your solution to every problems you give us at the end? :)

If you miss a comma, misspell a few words is okay.  But making it very hard to understand is what i mean (though many off you don't do that).  Because of (someones's...) request, I will post solutions.

Like see "Writing a Solution"...

Answers are due tomorrow!!  I will not be extending the deadline, so turn in, or th current scores will be used to determine winner!!!

Todays the last day..you have just a few hours until this end!!!!!!!!!

Well, no one turned in answers, so...THE WINNERS OF 007MATH'S TOURNEY!! (I used time to put places)

[hide="In 3rd Place came....."]

Discrete_Math  :winner_third: Good Job!!

[hide="IN 2nd place...."]

athunder :winner_second: Good Job

[hide="And in 1st place, the person who turned in their answers first (being one of the first) was..."]

BOGTRO!!!!  :first: GOOd Job!!

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

Good Job to everyone who participated.  ANd for those solutions...no one turned in answers for Round 3, so I guess there is no reason to post them.  They weren't that hard anyway.


ratio
trigonometry
trig identities
Law of Cosines

Source: Massachusetts Mathematics Leagues States 2004

Regular hexagon $ABCDEF$ is inscribed in a circle. $X$ is the midpoint of arc $\widehat{DC}$. Determine the exact value of $\frac{AX}{DX}$.

That is wrong.

[hide]If the radius of the circle is $r$, then by the Cosine Law we have

$AX^2=r^2+r^2-2r^2\cos 150^\circ=2r^2(1+\frac{\sqrt{3}}{2})=r^2(2+\sqrt{3})$

$DX^2=r^2+r^2-2r^2\cos 30^\circ=2r^2(1-\frac{\sqrt{3}}{2})=r^2(2-\sqrt{3})$

Therefore $\frac{AX^2}{DX^2}=\frac{2+\sqrt{3}}{2-\sqrt{3}}=\frac{(2+\sqrt{3})^2}{1}=(2+\sqrt{3})^2$, giving

$\frac{AX}{DX}=2+\sqrt{3}$[/hide]

[hide]$\widehat{DC}=60 \rightarrow \widehat{DX}=30$. Inscribed angles are half of their intercepted arcs, so $\angle{XAD}=15$, $\angle{XDA}=75 \rightarrow \angle{AXD}=90$. Right triangle $ADX$ is formed, $\frac{AX}{DX}=\tan{75}=\boxed{2+\sqrt3}$[/hide]
(edit: forgot hat)

[quote="Pocket Sand"]This may be a little off topic, but does anyone have a link that explains the properties of a 15/75 right triangle? I never learned them and it might be a good idea.

thanks[/quote]
I dont have a link but i can explain the basics....

First, lets get those trig ratios (which can be found this way or by using the half-angle identities and denesting the radicals:
$\sin{75}=\sin{45+30}=\sin{45}\cos{30} + \cos{45}\sin{30} = \boxed{\frac{\sqrt{6}+\sqrt{2}}{4}}$.
$cos{75}=\cos{45}\cos{30}-\sin{45}\sin{30}=\boxed{\frac{\sqrt{6}-\sqrt{2}}{4}}$.
$tan(75)=\frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}}=\boxed{2+\sqrt{3}}$. (this one is the most useful)

The sides are then $x(\frac{\sqrt{6}-\sqrt{2}}{4}),$ $x(\frac{\sqrt{6}+\sqrt{2}}{4}),$ $x$. I'm sure you could google a web page that explains more properties, but what you can do if you need the properties and dont remember the numbers is draw a 30-75-75 triangle, call a leg a simple number like 1 or 2 (or whatever value you need), and then use the Law of Cosines to solve for the other values, dividing the base in two when you draw the altitude to get the desired triangle:
$a^2=b^2+c^2-2bc\cos{A} \rightarrow \boxed{a^2=2b^2(1-\cos{30^\circ)})}$, where a is the base and b is the isosceles legs. remember then that the 15 degree side of the 15-75-90 triangle then is $\frac{a}{2}$, not $a$.

Hope this helps!  :D


What is the coefficient of $ x^2y^2$ in the expansion of $ (x\plus{}y)^4\plus{}(x\plus{}2y)^4$?

This would be kinda hard to do during a countdown, though he is right.  Instead, we can use the [url=http://www.artofproblemsolving.com/Wiki/index.php/Binomial_Theorem]Binomial Theorem[/url].  So we have $ \binom{4}{2}(x)^2(y)^2\plus{}\binom{4}{2}(x)^2(2y)^2\equal{}6x^2y^2\plus{}6x^2\cdot4y^2\equal{}\boxed{30}x^2y^2$.  Sorry if it seems kind of confusing for those who don't know the Binomial Theorem, it's kind of hard to explain  :oops:

Jasonion,

the Binomial Theorem states that

$(x+y)^n=\binom{n}{0}x^n+\binom{n}{1}x^{n-1}y+\binom{n}{2}x^{n-2}y^2+\cdots+\binom{n}{n}y^n$, where $\binom{n}{k}=\frac{n!}{k!(n-k)!}$

Edit: Thanks for correction, BOGTRO!

[quote="usamo42j"]Jasonion,

the Binomial Theorem states that

$(x+y)^n=\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\cdots+\binom{n}{n}$, where $\binom{n}{k}=\frac{n!}{k!(n-k)!}$[/quote]

The Binomial Theorem states that

$(x+y)^n=\binom{n}{0}x^ny^0+\binom{n}{1}x^{n-1}y^1+\hdots+\binom{n}{n}x^0y^n$.


[quote="Smartguy"][b]IMPORTANT[/b]: im looking for hints ONLY!!!
THat means no solutions and no answers!

im trying to avoid looking at the solution and i dont want to put up the problem again but here it is

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=455310#455310[/quote]Since the solutions in that topic are all hidden, why would you start another topic? You can ask for hints in that topic and ask for no solutions or hide solutions. This topic can be locked, it's pointless and considered SPAM
 :spam:

mods: please dont lock this!!
i meant this to be a post for hints, i didn't want to revive an old thread and have people be yelling at me for doing that...

can someone give some hints anyways?

Thanks!

[hide="A starting direction"]Let $ \theta \equal{} \text{m} \angle B$, and let $ a_1$ and $ a_2$ be the sidelengths of squares $ S_1$ and $ S_2$ (do not compute $ a_1,2$ yet). Use these values to find expressions for the areas of the two triangles. What happens when you equate them?[/hide]

People should be ashamed of for yelling at other individual for reviving an old topic because he wanted hints and [u]needed the necessary help to learn[/u]. And this should be addressed to [u]everyone[/u] on AoPS/ML and [i]MUST BE reminded that there is NO such rule for not reviving old topics out of necessity[/i].

[quote="Smartguy"]thank you for your insight TZF, i have solved it thanks to you and a friend  :D
 :D
 :D

Thank you for spamming in this post, 10000th user![/quote]Your welcome, creator of a spammable topic!  :D


For  positive  reals  a , b  show  that
[(a+1)/(b+1)]^(n+1) >=(a/b)^b

24  th  swedish  mo , 1984
thanks...

This can't be true.
Choose any a,b > 0 such that a < b, and choose n sufficientely large, you will obtain the reverse inequality.

Pierre.


[quote="silouan"]This problem is wrong...
The true problem is
$\frac{a}{b^{2}+1}+\frac{b}{c^{2}+1}+\frac{c}{a^{2}+1}\geq\frac{3(a\sqrt{a}+b\sqrt{b}+c\sqrt{c})^{2}}{4}$

and it is very easy just Cauchy in Engel form  :wink: .[/quote]

 I think you are right. I found the same result , by writing  $\frac{a}{b^{2}+1}= \frac{a^{3}}{a^{2}b^{2}+a^{2}}= \frac{(a \cdot \sqrt a)^{3}}{a^{2}b^{2}+a^{2}}$ and then applying  Caushy  in  Engel - Andreescu form. But I could not find the original  source , to ensure the problem.
 Thank you Silouan for your nice information !

[quote="stergiu"][quote="silouan"]This problem is wrong...
The true problem is
$\frac{a}{b^{2}+1}+\frac{b}{c^{2}+1}+\frac{c}{a^{2}+1}\geq\frac{3(a\sqrt{a}+b\sqrt{b}+c\sqrt{c})^{2}}{4}$

and it is very easy just Cauchy in Engel form  :wink: .[/quote]

 I think you are right. I found the same result , by writing
  $\frac{a}{b^{2}+1}= \frac{a^{3}}{a^{2}b^{2}+a^{2}}= \frac{(a \cdot \sqrt a)^{2}}{a^{2}b^{2}+a^{2}}$ and then applying  Caushy  in  Engel - Andreescu form. But I could not find the original  source , to ensure the problem.
 Thank you Silouan for your nice information ![/quote]

I have the original text in my house .
The correct problem is the above...... :)


function
inequalities unsolved
inequalities

Prove that if $x,y,z\geq0$, then
\[\frac1{\sqrt{4x^{2}+yz}}+\frac1{\sqrt{4y^{2}+zx}}+\frac1{\sqrt{4z^{2}+xy}}\geq\frac2{\sqrt{xy+yz+zx}}.\]

[quote="pvthuan"]Prove that if $ x,y,z\geq0$, then
\[ \frac1{\sqrt {4x^{2} \plus{} yz}} \plus{} \frac1{\sqrt {4y^{2} \plus{} zx}} \plus{} \frac1{\sqrt {4z^{2} \plus{} xy}}\geq\frac2{\sqrt {xy \plus{} yz \plus{} zx}}.\]
[/quote]
Let $ x\plus{}y\plus{}z\equal{}1.$ Since, $ f(t)\equal{}\frac{1}{\sqrt t}$ is convex function we obtain:
$ \sum_{cyc}\frac1{\sqrt {4x^{2} \plus{} yz}}\equal{}\sum_{cyc}\frac{\frac{y\plus{}z}{2}}{\sqrt {\frac{(y\plus{}z)^2(4x^{2} \plus{} yz)}{4}}}\geq\frac{1}{\sqrt{\sum\frac{(y\plus{}z)^3(4x^2\plus{}yz)}{8}}}.$
Id est, it remains to prove that
$ 2(x\plus{}y\plus{}z)^3(xy\plus{}xz\plus{}yz)\geq\sum_{cyc}(y\plus{}z)^3(4x^2\plus{}yz),$
which is equivalent to  $ \sum_{sym}(x^4y\minus{}x^3y^2\plus{}7x^3yz)\geq0,$ which is obvious.

See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=422538


$ a,b,c \ge 0$
$ a \plus{} b \plus{} c \equal{} 1,F(a,b,c) \equal{} \sqrt {\frac {1 \minus{} a}{1 \plus{} a}} \plus{} \sqrt {\frac {1 \minus{} c}{1 \plus{} c}} \plus{} \sqrt {\frac {1 \minus{} c}{1 \plus{} c}}$
Find the max and min vaule of F(a,b,c).

[hide]
the Min value is f(1,0,0),and Max is F(0.5,0.5,0)
[/hide]

For max try $ a$ close to $ \minus{}1$, $ b\equal{}1,c$ close to $ 1$

I am sorry for missing a condition that $ a,b,c>0$

Take $ f(x)\equal{}\sqrt{\frac{1\minus{}x}{1\plus{}x}}$ and this problem could be solved by Karamata Inequality (though not so obvious and direct).

Notice that

$ f''(x) \equal{} \frac {1 \minus{} 2x}{\sqrt{(1 \plus{} x)^{5}(1 \minus{} x)^{3}}}$.

WLOG we assume $ a\le b\le c$,

[b]Case 1[/b]: $ a \plus{} b\le \frac {1}{2}$.

We obtain

$ f(a) \plus{} f(b)\le f(0) \plus{} f(a \plus{} b)$.

[b]Case 2[/b]: $ a \plus{} b > \frac {1}{2}$.

We obtain

$ f(a) \plus{} f(b)\le f\left( a \plus{} b \minus{} \frac {1}{2}\right) \plus{} f\left( \frac {1}{2} \right)$.

So we may assume that one of $ a,b$ is $ 0$ or $ \frac {1}{2}$.

[b]Case 1[/b]: $ a \equal{} 0$.

$ f(a) \plus{} f(b) \plus{} f(c)\le 0 \plus{} 1 \plus{} 1 < 1 \plus{} \frac {2}{\sqrt {3}}$.

[b]Case 2[/b]: $ a \equal{} \frac {1}{2}$.

$ a \plus{} b \plus{} c\ge \frac {3}{2} > 1$ which contradicts.

[b]Case 3[/b]: $ b \equal{} 0$.

$ f(a) \plus{} f(b) \plus{} f(c) \equal{} 1 \plus{} 1 \plus{} 0 < 1 \plus{} \frac {2}{\sqrt {3}}$.

[b]Case 4[/b]: $ b \equal{} \frac {1}{2}$.

$ f(a) \plus{} f(b) \plus{} f(c)\le f(0) \plus{} f(b) \plus{} f(a \plus{} c) \equal{} 1 \plus{} \frac {2}{\sqrt {3}}$.


I left my TI-83+ at school and im on spring break. I have an 89 forwhatever reason. Will they let me use it?
Thanks.

[quote="abacadaea"]My friend has an nspire is he allowed to use that also?[/quote]
Oh, yeah. That's why they gave those out in Nats last year...... :D


calculus integration ratio geometry perimeter quadratics LaTeX

Show that the number of integral-sided right triangles whose ratio of area to semi-perimeter is $p^{m}$, where $p$ is a prime and $m$ is an integer, is $m+1$ if $p=2$ and $2m+1$ if $p \neq 2$.

The problem is equivalent to finding the number of solutions in positive integers to the following system of equations:
\[ \begin{cases} a^2 \plus{} b^2 \equal{} c^2 \\
ab \equal{} (a \plus{} b \plus{} c)p^m\end{cases}
\]
Denoting $ a \plus{} b \equal{} t$, we have $ t^2 \equal{} c^2 \plus{} 2(t \plus{} c)p^m$ implying that $ (t \plus{} c)(t \minus{} c) \equal{} 2(t \plus{} c)p^m$ and
\[ c \equal{} t \minus{} 2p^m.
\]
Therefore, the original system is equivalent to
\[ \begin{cases} a \plus{} b \equal{} t \\
ab \equal{} 2(t \minus{} p^m)p^m\end{cases}
\]
implying that $ a, b$ are the roots of the following quadratic equation:
\[ x^2 \minus{} t x \plus{} 2(t \minus{} p^m)p^m \equal{} 0
\]
which is solvable in integers only if its discriminant $ t^2 \minus{} 8 (t \minus{} p^m)p^m \equal{} (t \minus{} 4p^m)^2 \minus{} 8p^{2m}$ is a square.

If $ (t \minus{} 4p^m)^2 \minus{} 8p^{2m} \equal{} q^2$ then $ 8p^{2m} \equal{} (t \minus{} 4p^m \minus{} q)((t \minus{} 4p^m \plus{} q)$. All possible non-negative such $ q$ are obtained as $ q \equal{} \frac {d_2 \minus{} d_1}{2}$ where $ d_1 d_2 \equal{} 8p^{2m}$ and $ d_2 \geq d_1$ are of the same oddness.

If $ p \equal{} 2$ then $ 8p^{2m} \equal{} 2^{2m \plus{} 3}$ and $ q \equal{} \frac {2^k \minus{} 2^{2m \plus{} 3 \minus{} k}}{2} \equal{} 2^{k \minus{} 1} \minus{} 2^{2m \plus{} 2 \minus{} k}$ where $ k \equal{} m \plus{} 2\dots,2m \plus{} 2$. These $ m \plus{} 1$ distinct values of $ k$ give the following solutions:
\[ \begin{cases} a \equal{} 2^{k \minus{} 1} \plus{} 2^{m \plus{} 1} \\
b \equal{} 2^{2m \plus{} 2 \minus{} k} \plus{} 2^{m \plus{} 1} \\
c \equal{} 2^{k \minus{} 1} \plus{} 2^{2m \plus{} 2 \minus{} k} \plus{} 2^{m \plus{} 1}\end{cases}
\]
If $ p > 2$ then $ q \equal{} \frac {4p^k \minus{} 2 p^{2m \minus{} k}}{2} \equal{} 2 p^k \minus{} p^{2m \minus{} k}$ (for $ k \equal{} m,\dots,2m$) or $ q \equal{} \frac {2p^k \minus{} 4 p^{2m \minus{} k}}{2} \equal{} p^k \minus{} 2 p^{2m \minus{} k}$ (for $ k \equal{} m \plus{} 1,\dots,2m$). Therefore, the number of solutions here is $ 2m \plus{} 1$. Explicit solutions can be derived as above.

There is a MUCH simpler proof knowing that all Pythagorean triples can be parametrized as (k(m^2-n^2), 2mnk, k(m^2+n^2) where m, n, and k are positive integers such that m - n is odd and gcd(m, n) = 1. I won't write it here because I don't know how to use Latex and it will look horrible, but that idea just about solves the problem instantly.


vector
linear algebra
linear algebra unsolved

Let T be a linear transformation defined by a 4x5 matrix.  Can T be one to one? can T be onto?  why?

$T$ is a linear transformation $K^5 \to K^4$. If $T$ is one-to-one, then it's image would be a $5$-dimensional subspace of the $4$-dimensional vector space $K^4$. Dou you think this is possible? ;-)

For the case that $T$ is onto: There are many examples, e.g. $(x_1 , x_2 , x_3 , x_4 , x_5) \mapsto (x_1 , x_2 , x_3 , x_4)$.


floor function
combinatorics unsolved
combinatorics

[quote="kunny"]$ \sum_{n=1}^{\infty}\left\lfloor \frac{1000}{5^{n}}\right\rfloor= \left\lfloor \frac{1000}{5}\right\rfloor+\left\lfloor \frac{1000}{5^{2}}\right\rfloor+\left\lfloor \frac{1000}{5^{3}}\right\rfloor+\left\lfloor \frac{1000}{5^{4}}\right\rfloor+0+0+\cdots$
$ =200+40+8+1=\boxed{249}$[/quote]
Can you explain please?  :blush:

[quote="kunny"]From $ 10=2*5$, we are to consider how many $ 5, 5^{2}, 5^{3}, 5^{4}, \cdots$ can be included in $ 1000.$[/quote]
thx a lot kunny... :lol:

[quote="santosguzella"]How many zeros has $ 1000!$?[/quote]
Do you mean zeros at the end of $ 1000!$? That's what Kunny has found. How would you find the total number of zeros of $ 1000!$?


Where do you go to change your shoutbox colors? I've been looking all over blog options and I can't find it! Anyone know? Thanks a million bajillion times pi!

Odd numbered shouts (from the top):
Profile --> Blog Options --> Blog customization --> Block background color

Even numbered shouts (from the top):
Profile --> Blog Options --> Blog Style --> Entry background color

I would recommend using CSS rather than using the color options provided as this will give you much greater control over the look of your blog. There are numerous tutorials for CSS on the web.

In general you may have something like:

[code]#shoutsbox .row1 {
    background-color: red;
    color: black;
    font-weight: bold;
}

#shoutsbox .row2 {
    background-color: #2937ab;
    color: #283def;
    font-weight: normal;
}

#shoutsbox .shoutfooter {
    color: #cf4;
    font-family: serif;
    text-decoration: underline;
    font-size: 16px;
}[/code]

Thank you so very much!
I would've never figured that out!


geometry inradius circumcircle LaTeX inequalities

1.Let AD, BE, CF be angle-bisectors of triangle ABC. I is a point in triangle ABC such that ID is perpendicular to BC and IE is perpendicular to AC. Prove that IF is perpendicular to AC and (IA+IB+IC) is smaller or equal to 3R/2 with R is radius of (ABC).

What type of radius are you claiming that R IS of triangle ABC?

Are YOU claiming it to be an inradius or a circumradius?

Please go back and EDIT that into your post, if you are

allowed to.

by default a capital 'R' is used to denote the circumradius; this is a common convention (a lowercase 'r' is used for the inradius).

[color=darkblue]Oh! This problem $ R$ must be circumradius of $ \triangle ABC$. Hi [i][b]Nguyen Ngoc Linh[/b][/i]! Your problem is nice. Now I'm busy so I can't try to solve it. I think you should post it at forum Geometry Usolved Problem, you will get help more quickly.

And you should use latex in your post. It's very easy if you want.
You can see how to use Latex at this link: http://www.mathlinks.ro/viewtopic.php?t=165835

This is your problem use latex:

Let $ AD, BE, CF$ be angle-bisectors of triangle $ ABC$. Let $ I$ is a point in $ \triangle ABC$ such that $ ID \perp BC$ and $ IE \perp AC$. Prove that $ IF \perp AC$ and $ IA\plus{}IB\plus{}IC \le \frac{3R}{2}$ with $ R$ is radius of circumcircle of $ \triangle ABC$.[/color]

[quote="thanhnam2902"][color=darkblue]

Let $ AD, BE, CF$ be angle-bisectors of triangle $ ABC$. Let $ I$ be a point in $ \triangle ABC$ such that $ ID \perp BC$ and $ IE \perp AC$. Prove that $ IF \perp AC$ and $ IA \plus{} IB \plus{} IC \le \frac {3R}{2}$ with $ R$, the circumradius of the circumcircle of $ \triangle ABC$.[/color][/quote]
I made a few changes for grammar, and I allowed for redundancy
in describing the "R" variable to emphasize what it is.

IA, IB, and IC are the line segments, but if you want to be
correct in your inequality, you must have bars above each
of these three to indicate lengths of the line segments.

Perhaps you know the code(s) in Latex to place them there.


is there a proof for the shoelace algorithm?

I just googled it, so [url=http://www.mathreference.com/la-det,shoe.html]here[/url].


geometry
rotation
3D geometry
tetrahedron

Compute the area of tetragon if its sides are : $ a \equal{} 21.4 ,b \equal{} 13.8 .c \equal{} 22.25 , d \equal{} 14.5$
Is there any general formula to compute the area of iregular polygons ?

Shoelace would work, but it's annoying to set up when you have decimals...

Here's the link to the theorem:
http://www.artofproblemsolving.com/Wiki/index.php/Shoelace_Theorem

Actually for a tetragon/quadrilateral, I don't think there is a definite answer to every single set of side lengths. Take, for example, one with side lengths of $ 6$, $ 6$, $ 4$, and $ 4$. This is obviously a kite, but is it the convex version (the normal kite) or the concave version (with the cave-in, forming an arrowhead)? I think that in order to find the area of a quadrilateral, you need more than four sides...either information about the concavity of the quadrilateral or the length of the variable diagonal.

Actually, the shape is determined by the sides, since three sides determine a triangle (i.e. each face), and every orientation of the sides can be rotated to give the same figure.

You can find the volume of a general tetrahedron with the [url=http://en.wikipedia.org/wiki/Scalar_triple_product]triple scalar product[/url].

Um...math154...enndb0x said tetraGON, meaning quadrilateral, not tetrahedron

Hmm...  I think  Brahmagupta's Formula is applicable here since the length of the sides were given.[hide="That is..."]
$ \sqrt{(s\minus{}a)(s\minus{}b)(s\minus{}c)(s\minus{}d)}$
Where: $ s\equal{} \frac{a\plus{}b\plus{}c\plus{}d}{2}$

So, By Solving I've got $ s\equal{}35.975$
and Area of the tetragon is $ 304.64\approx305$

[/hide]

Brahmagupta's formula only works for cyclic quadrilaterals.
So you would need to prove that it is a cyclic quadrilateral.
If it is not, you need to find angle measures for the formula to work.
The formula is

$ \sqrt {(s\minus{}a)(s\minus{}b)(s\minus{}c)(s\minus{}d)\minus{}abcd cos^2 }\theta$.
Where $ \theta$ is half the sum of opposite angles.

sorry but I don't have the angles ...anyone any concret idea of solving this problem , or in general ?  :maybe:

[quote="r15s11z55y89w21"]Actually for a tetragon/quadrilateral, I don't think there is a definite answer to every single set of side lengths. ...
I think that in order to find the area of a quadrilateral, you need more than four sides...either information about the concavity of the quadrilateral or the length of the variable diagonal.[/quote]  Quoted for truth.  (Almost any other piece of information would suffice: the length of a diagonal or the measure of a single angle, for example.  It should be obvious that the edge-lengths don't determine the quadrilateral or its area: just think about the family of rhobuses, say.)


function
inequalities
integration
calculus
real analysis
real analysis unsolved

Let $ f$ and $ g$ be increasing functions on $ [a,b]$. Prove that
\[ \int_a^b f(x)g(a\plus{}b\minus{}x) dx \le \frac 1{b\minus{}a} \left( \int_a^b f(x) dx \right) \left( \int_a^b g(x) dx \right) \le \int_a^b f(x)g(x) dx.\]

This is an application of Chebyshev inequality. The last $ \leqq$ is directly followed. The first $ \leqq$ is obtained by a substitution.

I had not previously seen a proof of chebyshev's inequalities for integrals, and while you can probably get it from the discrete case using riemann sums, I labored for awhile to produce the following argument:

It suffices to assume that $ f(x)\geq 0$, since if the inequality holds for $ f(x) \plus{} c$, then it holds for $ f(x)$ as well.

Put
\[ \widetilde{g}(x) \equal{} \frac {1}{x \minus{} a}\int^x_a g(x)dx
\]
and define $ \widetilde{f}$ similarly.

Now integrate $ (b \minus{} a)[f(x)g(x) \minus{} f(x)\widetilde{g}(x)]$ by parts over $ [a,b]$ to get
\[ (b \minus{} a)\int^b_a f(x)g(x) \minus{} \int^b_a f(x)dx\int^b_a g(x)dx \equal{} (*)
\]
where
\[ (*) \equal{} \minus{} (b \minus{} a)\int^b_a \widetilde{f}(x)(g(x) \minus{} \widetilde{g}(x))dx
\]
Since the integrand of $ (*)$ is clearly positive ($ \widetilde{f}$ is positive by our assumptions on $ f$ and the other term is positive since $ g$ is increasing), the result follows instantly.  A straightforward modification of the above shows that the opposite holds if $ g$ is decreasing, whence the inequality on the LHS follows by taking the decreasing function $ h(x) \equal{} g(a \plus{} b \minus{} x)$.

Is there a quicker proof?


Let d and z be integers such that d2 divides  z2. Prove that d divides z.

Look at the prime factors of d^2 and z^2. What do you know about them If $d^2|z^2$, then what do you know about d and z?

[quote="Misha123"]Write $kd^2=z^2$ and prove that k is a square.

Misha[/quote]

kd^2 = z^2
d:sqrt:k = z ?
ahhh I get it..

 :sqrt:k = z/d

so :sqrt: k is rational and so k must be a square!

[quote="Misha123"]Exactly!!! :idea:

Misha[/quote]

This can be extended to prove that if d^n divides z^n, d divides z, where n is any natural


MATHCOUNTS
percent
geometry
summer program
MathPath
AMC
USA(J)MO

I was wondering who made nationals after the state comp now that March is flying by.  Only take this poll if you've already done your state comp or didn't make state.  Thanks!

I know, but a lot of people, including me, didn't wait until after state until they voted, hence the title

5th place -- read my rant in the other thread.

I know, I heard.  I thoroughly feel your pain.  The only thing worse than getting last is getting 5th when it comes to mathcounts.

don't get down on yourself. You should be proud to get 5th place.

Well he's really smart so he deserves to go to nationals.

This is true. But there are also MANY other smart people (possibly smarter) that get weeded  out in other states, that may have had bad luck, not been feeling well, or even had an off day.

Besides, who are you to decide who "deserves" to go?

[quote="sferics"][quote="biffanddoc"]yeah. and the ranting was MUCH worse. there were at least 50 nat mc polls like "what place would i have come in you state" and "would i have won in your state" and "do you think its unfair that i didn't make nats" and "what was the top score in your state" and on and on and on and on.[/quote]

Easy for you to say. You live in one of the easiest states, from what I hear  :roll: .[/quote]
easy for me to talk about nat mc's ranting???
and yeah de is really easy

"How are you to say"

Well it's pretty obvious -- I mean, just take a look at stats. Nat_mc got perfect AMC 10 last year and beat a lot of other people.

And I'm this year's Indiana #5. Just wow, just wow. I had state and then missed #7 and #8. I CHANGED MY ANSWER FROM RIGHT TO WRONG ON 8 IN THE LAST SECOND. Literally, a SECOND. On the sprint, the only person I lost to was Nathan.

I'm pretty sure that's the worst.

In terms of a scripted loss -- I don't know the details of his. He told me he missed #7 and #8.

But coming within A SECOND of getting 4th or 3rd. ONE SECOND. and then changing your answer. WOW JUST WOW.

I'm sure that's worse if you take grade out of the equation.

Well, you did a good job. I'm over it now being in 9th grade (i just had to say it one more time), but you can still have a lot of fun if you go and watch it.  Also, you were easily the top 7th grader, so you're almost a number 1 lock in the state, and maybe a top 12 finish in nats next year. Don't get down on yourself like I did, it doesn't help.

I will. I just got back like ten minutes ago.

[quote="janieluvsmusic"]also, I think this year was the 1st year Kelsey ever did Mathcounts. It was really surprising when cowpi beat her in countdown round, just because she seemed so good at chapter countdown and she was up 2-0 before cowpi got 3 in a row to beat her.[/quote]

Really? Well that's too bad, because I think she can only compete one more time in Mathcounts. She's incredibly good though.

[quote="camathkid"]hahaha :rotfl:[/quote] What's so funny? :?

[quote] Really? Well that's too bad, because I think she can only compete one more time in Mathcounts. She's incredibly good though.[/quote]

yeah, she's usually first in math competitions. but I don't think she was in the top 10 at blaine this year and I know she was there. Last year she was 1st too. The only reason Kelsey knows about mathcounts is because she met Al Lippert at this math camp, and he formed a homeschool team this year.[/quote]

[quote="janieluvsmusic"]she met Al Lippert at this math camp[/quote]

What--MathPath?

http://www.mathpath.org

[quote="ehehheehee"]#1 I am in 8th grade
#2 I was 6th in chapter because I misread 3 questions
#3 I misread one question in target in state and missed another because I am bad at entering things into my calculator.  Four points could of given me so much more

Mathcounts is not the only competition in the world[/quote]

I did the same as you, I xxxx in chapter, xxx in state.

[Mod edit: Declare what year you are talking about, or it is assumed you are talking about this year. Which is illegal right now. ]

[quote="janieluvsmusic"]also, I think this year was the 1st year Kelsey ever did Mathcounts. It was really surprising when cowpi beat her in countdown round, just because she seemed so good at chapter countdown and she was up 2-0 before cowpi got 3 in a row to beat her.[/quote]
It is the first  year Kelsey was with Seattle Chapter, but she had two years before this (2004 and 2005) in Lake Washington Chapter (e.g. In 2004 she was 30th place in Lake Washington on Official team)

[quote="janieluvsmusic"]
yeah, she's usually first in math competitions. but I don't think she was in the top 10 at blaine this year and I know she was there. Last year she was 1st too. The only reason Kelsey knows about mathcounts is because she met Al Lippert at this math camp, and he formed a homeschool team this year.[/quote]
See my other posting - Kelsey started MATHCOUNTS in 2004 with Lake Washington Chapter.
The big surprise this year at Washington is that Kerry Xing did not make to National.  Kerry is a very strong mathlete both in written and in countdown (I watched Kerry edging out Samath in Countdown - both are incredible mathletes and made to this year's USAMO, the only two 8th graders made to that list from WA, I believe).  It was a shock to everyone since Kerry was in the National last year.

yeah, it was also really weird that kerry was 10th at blaine. It was weird because it meant I beat him.

Oh well, I made 19th chapters and placed 25th states.

funny quote: [quote]I"m sorry. Crap is not the right answer.[/quote]

[quote="Kamior"]Indeed, Kelsey is certainly not a ninth grader. Fortunately for a few previous posters, she'll be in Nevada next year, thus not competing against you unless you make nationals. It's a bit surprising that she lost in the State Countdown, but Washington did have its best team ever.[/quote]

How do you know she'll be in Nevada?

probably knows her personally or asked through AIM/MSN?


geometry
number theory
prime factorization
area of a triangle
Heron\u0027s formula
Pythagorean Theorem

(1) Find the number of inches in the length of the shortest altitude for a triangle whose side measure 5 inches, 7 inches and 8 inches.  express your answer in simplest radical form.

(2) The product of positive integers X, Y, and Z equals 2004.  What is the minimum possible value for the sum of X, Y, and Z?

(3) Using an array of 3x3 dots, how many different triangles can you make by picking and three of the nine dots at random and creating the vertices of a triangle?

(4) How many liters of water must be added to 8 liters of a 30% acid solution to make a 12% acid solution.

Thankx!![/hide]

[hide="4"]

The mixture is originally $ \frac{2.4}{8}$ acid.
We want a 12% acid solution, so:
$ \frac{2.4}{8\plus{}x}\equal{}\frac{3}{25}$
Cross multiplying, we have:
$ 60\equal{}24\plus{}3x \Rightarrow 36\equal{}3x \Rightarrow 12\equal{}x$
So $ \boxed{12 liters}$

[/hide]

[hide="1"]
The semiperimeter of this triangle is $ \frac {5 \plus{} 7 \plus{} 8}2 \equal{} 10$.

By Heron's, the area is $ \sqrt {10(10 \minus{} 5)(10 \minus{} 7)(10 \minus{} 8)} \equal{} \sqrt {2\cdot5\cdot5\cdot3\cdot2} \equal{} 10\sqrt3$.

Since $ A \equal{} \frac12bh$, $ h \equal{} \frac {2A}b \equal{} \frac {20\sqrt3}b$, and we quickly see that the shortest altitude occurs when $ b \equal{} 8$, and

$ h \equal{} \frac {20\sqrt3}8 \equal{} \boxed{\frac {5\sqrt3}2}$.
[/hide]
[hide="2"]
$ XYZ \equal{} 2004 \equal{} 2^2\cdot3\cdot167$, and since the minimum value of $ X \plus{} Y \plus{} Z$ occurs when $ X,Y,Z$ are as close as possible and $ \sqrt [3]{2004}\approx12.6$, we see that

$ 3\cdot4\cdot167$ gives the best result, $ 3 \plus{} 4 \plus{} 167 \equal{} \boxed{174}$.
[/hide]
[hide="3"]
First, we note that any three dots will form a triangle, so our preliminary count is $ \dbinom{3\cdot3}3 \equal{} 84$ triangles.

However, we are counting degenerate triangles, which can only be formed from all lines with three points in the array.

There are three rows, three columns, and two diagonals that form these degenerate triangles, so our final answer is

$ 84 \minus{} 3 \minus{} 3 \minus{} 2 \equal{} \boxed{76}$.
[/hide]
[hide="4"]
Currently, the solution is $ 0.3\cdot8 \equal{} 2.4$ liters acid.

Since we are only adding water, the amount of acid will stay the same, and we want

$ \frac {2.4}{8 \plus{} x} \equal{} 12\% \equal{} \frac3{25}\implies60 \equal{} 24 \plus{} 3x\implies x \equal{} \boxed{12}$ liters.
[/hide]

[hide="2"]

Taking the prime factorization of 2004, we have:

$ 2004\equal{}2 \times 2 \times 3 \times 167$

We want to minimize the sum of three factors, so take $ 167, 2 \times 2 \equal{}4,$ and $ 3$.

$ 167\plus{}3\plus{}4\equal{}\boxed{174}$

[/hide]

I don't understand #1 still, could you explain it by either (1) easier way or (2) explain each step more carefully?

First we find the area using Heron's Formula.  Heron's Formula states that the area of $ \triangle ABC\equal{}\sqrt{s(s\minus{}a)(s\minus{}b)(s\minus{}c)}$, where $ s\equal{}\frac{a\plus{}b\plus{}c}{2}$ and where $ a$, $ b$, and $ c$ are the side lengths of the triangle.  Then we have $ [ABC]\equal{}\sqrt{10(10\minus{}5)(10\minus{}7)(10\minus{}8)}\equal{}\sqrt{10 \times 5 \times 3 \times 2}\equal{}\sqrt{300}\equal{}10\sqrt{3}$.

Remember our formula $ A\equal{}\frac{1}{2}bh$.  Then we have $ 10\sqrt{3}\equal{}\frac{1}{2}bh$.  Because we want $ h$ to be minimized, we maximize $ b$.  The longest side is $ 8$, so we let that be $ b$.  Then we have $ 10\sqrt{3}\equal{}4h \implies \boxed{\frac{5\sqrt{3}}{2}\equal{}h}$.

math154, you forgot to multiply $ A$ by $ 2$.

[quote="math154"][url=http://www.artofproblemsolving.com/Wiki/index.php/Heron%27s_formula]Heron's Formula.[/url] (proof is not MC level, but the formula, at least, is good to know.)

As for the second part, i.e. minimizing $ \frac {10\sqrt3}b$, notice that the minimum of this occurs when $ b$ is at its maximum (which in this case, would be $ b \equal{} 8$).[/quote]

Ah...I got the Herons formula since I know it, but I had no idea where the 8 came from.  It came from the original side lengths, right?

[quote="AIME15"]math154, you forgot to multiply $ A$ by $ 2$.[/quote]

What are you talking about? I even wrote $ \frac{2A}b$. :P

[quote="007math"]Ah...I got the Herons formula since I know it, but I had no idea where the 8 came from.  It came from the original side lengths, right?[/quote]

Yes. :)

[hide="1 without herons"]
Label the triangle ABC, such that AB=7, BC=8, and CA=5.
Clearly, the smallest altitude is the one from A to BC, because BC is the longest side. Let the point that the altitude meets BC be called D.
Now, let the length CD be x, so that the length BD is 8-x.
From the pythagorean theorem on ACD, we know that $ x^2\plus{}AD^2\equal{}25$, or $ AD^2\equal{}25\minus{}x^2$
From the pythagorean theorem on ABD, we know that $ (8\minus{}x)^2\plus{}AD^2\equal{}49$, or $ AD^2\equal{}49\minus{}(8\minus{}x)^2$
Setting these two equal, we have $ 25\minus{}x^2\equal{}49\minus{}(8\minus{}x)^2 \iff 25\minus{}x^2\equal{}49\minus{}(64\minus{}16x\plus{}x^2) \iff 25\minus{}x^2\equal{}\minus{}15\plus{}16x\minus{}x^2 \iff 40\equal{}16x \iff x\equal{}\frac{5}{2}$.
Therefore, $ AD^2\equal{}25\minus{}(\frac{5}{2})^2\equal{}25\minus{}\frac{25}{4}\equal{}\frac{75}{4}$, so $ AD\equal{}\sqrt{\frac{75}{4}}\equal{}\frac{\sqrt{75}}{2}\equal{}\boxed{\frac{5\sqrt{3}}{2}}$
[/hide]


Factorize:

$ x^4 \plus{} 7x \plus{} 4y^2 \minus{} 8y \plus{} 4 \equal{} 0$

Second problem:

Find the domain and range of the following function:

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

For the second problem, the domain is $ \mathbb{R}$ since the function works for any $ x\in\mathbb{R}$.  The minimum value of the function occurs at $ x\equal{}\minus{}1$, where $ f(\minus{}1)\equal{}0$.  So the range is $ [0,\infty)$.


linear algebra
matrix
combinatorics proposed
combinatorics

[quote="edriv"][b]Problem 4[/b]
Today is Barbara's birthday, and Alberto wants to give her a gift playing the following game. The numbers 0,1,2,...,1024 are written on a blackboard. First Barbara erases $2^{9}$ numbers, then Alberto erases $2^{8}$ numbers, then Barbara $2^{7}$ and so on, until there are only two numbers a,b left. Now Barbara earns $|a-b|$ euro.
Find the maximum number of euro that Barbara can always win, independently of Alberto's strategy.[/quote]

I assume numbers are $1,...,1024$ and not $0,...,1024$ as stated.

I have a simple stategy for Barbara which gives her at least 48 at the end, but I'm not sure it is the best :
Barbara just have to pick one number each two numbers for each of her 4 first turns and she takes the two middle numbers at her last turn. With this method :
Difference between two consecutive numbers is exactly 2 after the first Barbara's turn (it remains 512 numbers)
Difference between two consecutive numbers is at least 4 after the second Barbara's turn (it remains 128 numbers)
Difference between two consecutive numbers is at least 8 after the third Barbara's turn (it remains 32 numbers)
Difference between two consecutive numbers is at least 16 after the fourth Barbara's turn (it remains 8 numbers)
Difference between two consecutive numbers is at least 16 before the last Barbara's turn (it remains 4 numbers) and so, by picking the two middle numbers, the remaining difference is at least 48.

It's possible on the contrary to show that it exist strategies for Alberto which prevents Barbara going over 64 (or even 56) : take at each turn the numbers which minimize the "max - min" of remaining numbers.

So I can't prove 48 is the best sure result.

--
Patrick

[quote="mathrix"]i think that 32 its the best results.[/quote]

I'm sorry but the best result is between 48 and 64.

In my previous post, I gave a strategy for Barbara to have at least 48 and I said it exist a strategy for Alberto to leave at most 64 for Barbara (remove numbers in such a way max(reminder)-min(reminder) be minimized).

--
Patrick

[quote="Sepp"]I haven't read your proof but mathrix is right, the best is 32.[/quote]

OK. I thought that there was a mistake i the problem and that there was 1024 numbres (and not 1025).
With 1025, there is a round more and the strategy I gave lead to 32 min.

The strategy o Alberto divides max - min by two, and since Alberto now have 5 rounds, il leads to 32 max.

And he result is 32.

Sorry mathrix, you were right, indeed.  :blush:

--
patrick


i don't see the need for multiple steps...this is just one partial fractions decomposition

$\frac{(n-1)(n+1)}{(2n-1)(2n+1)}=\frac{3}{8}*( \frac{1}{2n+1}-\frac{1}{2n-1})+\frac{1}{4}$


Find the remainder when $ 3^{5^{100}}$ is divided by 7

$ 5^{2} \equiv 1 \mod 6$, hence $ 5^{100} \equiv 1 \mod 6.$

On the other hand, Fermat's little theorem allows us to ascertain that

$ 3^{6} \equiv 1 \mod 7.$

Therefore, there exists $ k \in \mathbb{N}$ such that

$ 3^{5^{100}} \equal{} 3^{6k \plus{} 1} \equal{} (3^{6})^{k}(3) \equiv 3 \mod 7$

and we're done: the remainder is [b]3[/b].

[quote="pankajsinha"]Thanks.
I would want an answer without using congruence modulo concept as the problem is intended for students who haven't studied congruence modulo.[/quote]

You can use coquitao's ideas without the congruence notation:

1) Notice that $ 3^6\equal{}729\equal{}7\cdot 104\plus{}1$
2) Notice that $ (7a\plus{}1)(7b\plus{}1)\equal{}7(ab\plus{}a\plus{}b)\plus{}1\equal{}7j\plus{}1$ then $ (7a\plus{}1)^n\equal{}7k\plus{}1$ for all natural $ n$ (by induction).
2*) Moreover using that approach then $ (ma\plus{}1)^n\equal{}mk\plus{}1$
3) Then $ (3^6)^l\equal{}(7\cdot 104\plus{}1)^l\equal{}7k\plus{}1$

4) Similarly $ 5^2\equal{}25\equal{}6\cdot4\plus{}1$
5) Using 2*) then $ 5^{100}\equal{}(5^2)^{50}\equal{}(6\cdot4\plus{}1)^{50}\equal{}6l\plus{}1$
6) All together now... $ 3^{{5}^{100}}\equal{}3^{6l\plus{}1}\equal{}(3^6)^l \cdot 3\equal{}(7k\plus{}1) \cdot 3\equal{} 7 (3k)\plus{}3$


function
algebra
functional equation
real analysis
real analysis unsolved

Determine whether there exists a nonconstant power series $ y \equal{} \sum a_n \frac{x^n}{n!}$ with $ y(0) \equal{} 1$ that satisfies, formally, the functional equation

$ y \equal{} e^{y\minus{}1}$.

[quote="t0rajir0u"]Determine whether there exists a nonconstant power series $ y \equal{} \sum a_n \frac {x^n}{n!}$ with $ y(0) \equal{} 1$ that satisfies, formally, the functional equation

$ y \equal{} e^{y \minus{} 1}$.[/quote]
I assume you are interested in a formal power series. If such a power series existed with finite radius of convergence, then within this disk we'd have $ y' \equal{} y' e^{y\minus{}1}$ and hence, since $ y \ne const$, $ 1 \equal{} e^{y\minus{}1}$ a.e., implying that $ y$ is constant after all.

Suppose now a formal power series with $ y\equal{}e^{y\minus{}1}, \, y(0) \equal{} 1$ exists and $ a_1 \equal{} a_2 \equal{} \dots \equal{} a_{n\minus{}1} \equal{} 0 \ne a_n.$. The $ 2n$th term of $ y$ is
\[ a_{2n} \frac{x^{2n}}{(2n)!} \]
On the other hand, the term of order $ 2n$ of $ e^{y\minus{}1}$ is
\[ a_{2n} \frac{x^{2n}}{(2n)!} \plus{} \frac12 \left( a_n \frac{x^n}{n!} \right)^2 \, .
\]
This implies $ a_n \equal{} 0$, a contradiction. Hence no such formal power series exists.

Doesn't it follow from the fact that $ e^{x}\equal{}1\plus{}x \implies x\equal{}0$ that $ y \equiv 1$ is the only solution?

It is slightly harder than that, but not by much:  a formal power series can have radius of convergence $ 0$.

To emphasize that point, here's a functional equation that does have a (unique) formal power series solution: $ f\left(\frac{z}{1\minus{}z}\right)\minus{}f(z)\equal{}z$, $ f(0)\equal{}1$. The function defined this way is meromorphic on $ \mathbb{C}$ except at the origin, where it has an essential singularity and infinitely many nearby poles.


Find the minimum value of the function $ f(x)\equal{}3(x^2\minus{}1)\plus{}2$.

a graph of a function reaches maximum/minimum when $ x \equal{} \minus{} \frac {b}{2a}$

$ f(x) \equal{} 3x^2 \minus{} 1$

EDIT: lol, $ x\equal{}\minus{}\frac{b}{2a}$

$ x\equal{}0$

plug it in, and the minimum of $ f(x)\equal{}\minus{}1$

answer : -1

multiplying the whole thing out and simplifying, we get $ f(x) \equal{} 3x^2 \minus{} 1$. If you have a sense of maxima and minima, you can pretty easily understand that the minimum of this graph will be at its vertex. Using the fact that the x-value of the vertex will always be $ x \equal{} \frac { \minus{} b}{2a}$, we can determine that the x value of the vertex is 0. Plugging this back into our original function we see that our answer is  $ \boxed{ \minus{} 1}$

EDIT: BEATEN TO IT


Prove that the equation $a^2 +b^2 =c^2 +3$ has infinitely many integer solutions $(a, b, c)$.

If $ a\plus{}1\equal{}n^{2}$, then $ (2n)^{2}\plus{}(2a)^{2}\equal{}(2a\plus{}1)^{2}\plus{}3$, since there are infinitely many $ a$ for which $ a\plus{}1$ is a perfect square, this solves the problem.

[quote="Johan Gunardi"]$ a \equal{} 2k, b \equal{} 2k^2 \minus{} 2,c \equal{} 2k^2 \minus{} 1$[/quote]I think that's not correct: look at the coefficient of $ k$...

I'm sorry, must have missed the squares. Never mind.


search
combinatorics unsolved
combinatorics

Consider the set of all discs in the plane with centres at lattice points and radius = $\frac{1}{1000}$ each.

(a) Prove that there exists an equilateral triangle whose vertices lie in distinct discs considered above.
(b) Prove that the side lengths of any such equilateral triangle is greater than $100$

IMO 2003 Shortlist, C5 if I am not mistaken. Search button...


$a_0, a_1,a_2,...a_n $ real numbers such that $a_0>=a_1>=a_2>=...>=a_n>0 $
and r is a root of the polynomial
$ P(z)=a_0z^n+a_1z^(n-1)+..+a_n $
prove that  $ r<=1 $

[quote="Misha123"]$P(z)>0$ if $z>0$ so $r\leq0<1$. Or do you meant $r\in C$ and $|r|\leq 1$?

Misha[/quote]
$a_i, r $ and$ z $ are all real numbers


Let $A$ and $B$ be matrices such that $A^{n}=B^{n}=0$ and $AB=BA$. Prove that $(A+B)^{n}=0$

Notice that we´ll have to assume that $n$ is the rank of our matrices $A$ and $B$.

So, we´ll have that all eigenvalues of $A$ will be equals to 0. That´s because if we considerate $x_{a}$ an eigenvector of $A$, we´ll have that $A^{n}x_{a}=0 x_{a}$ -> $t^{n}x_{a}=0$ -> $t^{n}=0$ -> $t=0$. That´s because $x_{a}$ is far from $0$.

By an analogue demonstration we´ll have that all eigenvalues of $B$ we´ll be equals to $0$.

Now we considerate $x_{b}$ an eigenvector of $B$. Notice that $A^{n}x_{b}=0$ -> $A(A^{n-1}x_{b})=0$. But $det(A)=0$ and so we´ll have that $A^{n-1}x_{b}$ isn´t the null vector. And I´ll prove now that this vector is an eigenvector of $(A+B)$. To do this observe that $(A+B)A^{n-1}x_{b}=A^{n}x_{b}+BA^{n-1}x_{b}=A^{n}x_{b}+A^{n-1}Bx_{b}=0$.

here I did not considerate that $A$ and $B$ have an commun eigenvector. If they have this eigenvector will be an eigenvector of $(A+B)$.

And we can make this for all eigenvector of $B$ and conclude now that all eigeinvalue of $(A+B)$ is zero and so by Cayley-Hamilton theorem we´ll have that $(A+B)^{n}=0$.


[b]For COP[/b]

The last round will be graded and result after about two weeks. The reason is because I won't have computer access at all. So as soon as I get my computer back, I'll grade the last round of COP and finish this.

[b]For Getting Started[/b]

Other moderator white_horse_king is currently in vacation. He sent me message on this matter and till now, it was okay since I was able to moderate here. But since I won't be able for about two weeks (hopefully this is it probably), I suggest you guys to do NOT SPAM.

When I get back and see some unncessary posts made often by certain users, I'll notify the administrator.

Not meaning that you can't post at all. Let's just not spam.  :)

Have a good summer!

[quote="IntrepidMath"]jli is at MathPath. He does not have ample access to a computer. None of us do.[/quote]
what is mathpath?

So, there will be no mods for a while?
[hide]Hmmm.... [evil_music]:evil_laugh: I could take over the forum! :evil_laugh:[/evil_music] [hide]Nah.... That takes too much work.[/hide] [/hide]

those administrators will still be there
nice tags Xantos

Mathpath is a math camp for middle schoolers. For more info about it, please do a search in the forum with that keyword.

[quote="IntrepidMath"]jli is at MathPath. He does not have ample access to a computer. None of us do.[/quote]

do not fear...jli is here. :D

No, he's still at Mathpath, as am I. But we have 15+ mins of computer time every day, and I guess that's enough for him to moderate??

I got a tab bit of computer access today, but the internet spontaneously shuts down... sigh.

Consider yourself lucky. The people at the campus we're at gave us computers that have monitors with the wrong power cords. So the monitors shut down after about two seconds and we have to keep turning them off and on. It's really annoying, but at least better than having the internet shut down spontaneously.

Wrong.

If you really want to use the computers, just go to the library after lunch or during free time. You lose.

I lose what?

Not all of us can go to the library at arbitrary time x:yz, where x, y, and z are positive integers x<13, y<6, z<9. Some of us have other things we have to do, and some
of us have to go to the library with someone else.

Camp's almost over anyway.

PMS affecting you, mathnerd?

Ad hominem aside, yeah, camp is almost over, so jli will be back to his moderater-ness in three or four days.

Treething, easy there.

I'm currently in my local town library and eh.. let's just say it will take sometime for it to get fixed.

There was major attack on Windows itself and I can't do anything at all. No internet, no simple documenting, etc.. Turning on the computer gives me warnings on how it can't function..

Hopefully it will be okay so I can start the COP as I planned. (So far, it will but there is chance that it won't and if it doesn't, I'll notify in here).

Silverfalcon


geometry rectangle calculus ratio trigonometry function

Sorry if my problem has been poorly evaluated considering difficulty.  Here's a problem that I really enjoyed, and I hope you do too.  Here it is.

What is the maximum area of a rectangle circumscribed about a fixed rectangle of length 8 and width 4?

Masoud Zargar

Calculus... *cough* Optimization *cough* *cough*  :)

I have a question.  Would you define very very basic defferentiation inappropriate for this forum?

Masoud Zargar

[quote="Altheman"]how do you define circumscribing about a rectangle[/quote]

Within a rectangle at an angle.

[quote="boxedexe"]I have a question.  Would you define very very basic defferentiation inappropriate for this forum?[/quote]Oh, not at all. I was just implying that I knew the calculus solution (didn't fail my math class).  :)

[hide]
Call the original rectangle ABCD and the new one WXYZ such that A lies on WX, B on XY, C on YZ, and D on WZ.

Let $AW = x$, $WD = y$. Since $DWA$ and $AXB$ are both right and $\angle WAD = 90 - \angle BAX = \angle ABX$, we have $DWA$ similar to $AXB$. But the hypotenuses are in ratio $1: 2$ so the sides are as well. Then $AX = 2y$, $BX = 2x$.

So $[WXYZ] = (AW+AX)(WD+BX) = (x+2y)(y+2x) = 2(x^2+y^2)+5xy$. Since $x^2+y^2 = 16$ and $xy \le \frac{x^2+y^2}{2} = 8$ by AM-GM, we have $[WXYZ] \le 2(16) + 5(8) = 72$, which is satisfied when WXYZ is a square.[/hide]

Nice solution.  I really like it.  Here's my solution.
[hide="Solution"]
Let $ABCD$ and $JKLM$ be the circumscribing rectangle and the fixed rectangle, respectively.  $J$ lies on $AB$, $K$ lies on $BC$, $L$ lies on $CD$, and $M$ lies on $AD$.  $AJ$ subtends $\angle\theta$, and $BK$ subtends another angle which is also $\angle\theta$.  We know that $|ML|=|JK|=8$ and $|MJ|=|KL|=4$.  We use trigonometry and get, $|AJ|=4\sin\theta$, $|BJ|=8\cos\theta$, $|AM|=4\cos\theta$, and $|MD|=|BK|=8\sin\theta$.

It is now clear that the area of the circumscribing rectangle is $A=(|AM|+|MD|)(|AJ|+|BJ|)$.  Let us define $A$ as a function of $theta$.
\[ \\A(\theta)=(4\cos\theta+8\sin\theta)(4\sin\theta+8\cos\theta)\\=16\sin\theta\cos\theta+32\sin^2\theta+32\cos^2\theta+64\sin\theta\cos\theta\\=80\sin\theta\cos\theta+32=40\sin 2\theta+32\implies 0<\theta<\frac{\pi}{2}  \]
\[ \\A'(\theta)=80\cos 2\theta\\0=\cos 2\theta\\2\theta=\frac{\pi}{2}\\\theta=\frac{\pi}{4}  \]
$\\|AJ|=4\sin\frac{\pi}{4}=2\sqrt{2}\\|BJ|=8\cos\frac{\pi}{4}=4\sqrt{2}\\|AM|=4\cos\frac{\pi}{4}=2\sqrt{2}\\|MD|=8\sin\frac{\pi}{4}=4\sqrt{2}$

We now determine the area of the circumscribing rectangle. $A=(|AM|+|MD|)(|AJ|+|BJ|)=(2\sqrt{2}+4\sqrt{2})(2\sqrt{2}+4\sqrt{2})=(6\sqrt{2})^2=\boxed{72}$.
We see that our dimensions satisfy teh dimensions of a square, which has the largest area.
$\mathbb{QED}$
[/hide]

Masoud Zargar


The area and volume of a sphere are given by $A\pi$ and $V\pi$ respectively.If the value of A and V are both 3 digit integers.What s the smallest possible radius of the sphere ?
 :?

$A\pi=4\pi r^2\Longrightarrow A=4r^2$
$V\pi=\frac{4}{3}\pi r^3\Longrightarrow V=\frac{4}{3}r^3$
Since $V$ is an integer, $r^3$ must be divisible by $3$, so $r$ must be divisible by $\sqrt[3]{3}$. Also, $r^3$ must be an integer.
Since $V$ is an integer, $r^2$ must be an integer.
If $r^3$ is an integer and $r^2$ is an integer, $r$ is an integer.
Because $4r^2\ge 100$, $r\ge 5$.
Since $r$ must be divisible by $3$, the least possible value for $r$ is $\boxed{6}$


Let be $ a,b,c$ the side lenghts of a triangle . Then prove that :

$ \frac{a\plus{}b\minus{}2c}{c^2\plus{}ab}\plus{}\frac{b\plus{}c\minus{}2a}{a^2\plus{}bc}\plus{}\frac{c\plus{}a\minus{}2b}{b^2\plus{}ca}\ge 0$

hello, your inequality is equivalent to
$ ba^4\minus{}b^3c^2\plus{}b^4c\minus{}c^2a^3\plus{}ab^4\minus{}c^3b^2\plus{}c^4a\minus{}a^2c^3\minus{}a^3b^2\minus{}b^3a^2\plus{}c^4n\plus{}a^4c\geq0$ this can be simplifyied to $ ab(a\minus{}b)(a^2\minus{}b^2)\plus{}bc(b^2\minus{}c^2)(b\minus{}c)\plus{}ac(a\minus{}c)(a^2\minus{}c^2)\geq0$
this is true.
Sonnhard.


A helicopter took of from Leningrad and flew north.  After five hundredkilometres it turned and flew 500 kilometres east.  After that it turned south and covered another 500 kilometres.  Then it flew 500 kilometres west, and landed.  The question is, where did it land: west, east, north, or south of Leningrad?

 I'm not sure if this is too simple........

that depends. It can be anywhere but north of lenningrad.

Say it were 500 miles south of the north pole. It can land either east or west of lenningrad or right on it.

say it were on the south pole. Then it would land right on lenningrad

say it were on the north pole. It can't go north, goes around the earth infinitely, goes south, goes west and is still south of lenningrad.

but if you mean anywhere else, it'll be east.

[hide="answer"]Since Leningrad is in the Northern hemisphere, and less than 500 km from the pole, the helicopter would end up East of Leningrad. The reason for this is that the surface of the earth is a sphere, and the closer to the pole you are, the smaller the distance around the earth. Thus, the 500 km east that the helicopter flies, being at a distance closer to the North Pole than the 500 km west, actually take the helicopter further around the earth in angular measure.

An easier example to see this effect is as follows. Imagine you are on the equator and go north till you reach the north pole. There you spin around at random and go south til you reach the equator. Even though all you did was go north and then go south the same distance, you will not necessarily end up in the place yuo started from! In fact, if you don't spin and keep going straight then you will end up on the opposide side of the globe!

Hope this helps.[/hide]

oh....lenningrad is an acutal place? i didn't know that... :oops:

correct, i thought most people would fall for the "  :huuh:  dosen't it land $in$ leningrad?"

well that would be possible...

if lenningrad were exactly 250 km south of the equator

[quote="math92"]would it land back in Leningrad?[/quote]
did u say that just to say it lol? ...  :P right after GUNIt
said:
correct, i thought most people would fall for the "  dosen't it land [b][i]in[/i][/b] leningrad?"

you might want to read more of the topic before you post math92. but i agree with JanSiwanowicz it would land east of Leningrad since as you go north the distance around earth is less.

how old is this problem...

Leninsgrad doesn't even exist anymore....it is called St. Petersburg now....

right on it...
you guys take to many chances... :rotfl:


In IMO2005 VISA form which the contestents (I) have to solve it is said that: on that the contestents have to sing. but i can't find any place for signing!

in one place of second page it is written that:
FIRMA DEL PROMOVENTE

i can easily guess that: FIRMA = sign, DEL= of the but PROMOVENTE=? I am not sure! can anyone help?

Mahbub

From my side i am 99.99% sure ( :D ) that it is the reader of the VISA letter! :D

Let me see if my guessing power is correct! :P

[quote="Pascual2005"]I am sure David will help you, becuase I am not very sure about "Promovente", it could be the person requaesting the Visa, or it could also be the person who will read the request and accept it, I dont know...[/quote]

Sorry for not being answering very fast lately, but my wrist was left wrist was almost broken in an accident. It is a little better tonight, but I still am too slow writting on a computer.

I searched on several dictionaries (included the dictionary of the [b]Real Academia Española[/b]), and the word is not there, hence I suppose it may be a mistranslation. Now, by the roots it looks like it is either the "Sponsor" if it applies, or the person who would potentially take care of the debt in case you can't pay (I suppose that most of the IMO participants are too young for VISA to give them directly a credit account to them). If that does not make sence, then probably it is actually your firm the one that has to go there.

Bye


The largest square possible is inscribed in a regular octagon. The longest diagonal of the octagon is 20 cm. What is the number of square centimeters in the area of the square?

[hide="Hint"]The biggest square is the one that connects four of the vertices. (I'm not sure how to prove this myself so if anyone knows how please tell :D )[/hide]

[quote="ritchjp"]From the hint that you posted, the diagonal of the octagon would also be the diagonal of the square.
[hide]
One side of the square would be [tex]\frac {20}{\sqrt 2}[/tex]

So the area would be [tex] 200 \ cm^2[/tex]

[/hide][/quote]

Right. :)


counting
derangement
search
floor function

Suppose a school has $2007$ students, each with a corresponding piece of paper with his/her name on it. How many ways can the papers be distributed to the students so that every student receives exactly one piece of paper and no student receives the paper with his/her name on it?

It seems like this is a little calculation intensive. Using the recurrence relation would take a long time to get up to 2007, and if I remember correctly there's a summation method of calculating it, but surely you have to have a calculator?

A link for anyone who doesn't know what derangements are: [url]http://en.wikipedia.org/wiki/Derangement[/url]
You could also search it on the forum.

It's going to be approximately $\frac{2007!}{e}$. Probably if you were to calculate that and then round to the nearest integer that would be the correct answer.

I agree with the approximation of $\frac{2007!}{e}$, but is this what drunner wants?

I can't prove it but I've managed to convince myself that it's exactly $\left\lfloor\frac{2007!}{e}\right\rfloor$, which is about $1.5835747690667932749504701216585 \times 10^{5758}$.

This question isn't designed for a tournament. Its just a problem.
The approximation should be $2007!(1-e^{-1})$. There is an exact answer though. You don't need to think recursively to get it. Just apply PIE

The exact answer is $\left\lfloor\frac{2007!}{e}\right\rfloor$  :P

yeah, I believe there's a non-recursive formula.  Actually, what is the recursive formula??? (I only know the formula from PIE)
[hide]
we'll count the number of ways that at least 1 student got his/her name.  Using PIE we have $2007! - \binom{2007}{2}2005! + \binom{2007}{3}2004! - \dots \pm \binom{2007}{k}(2007-k)! \mp \dots + 1$, and 2007! subtract this is $\binom{2007}{2}2005! - \binom{2007}{3}2004! + \dots -1$.
[/hide]

Let me post my answer.  :)

First the exact number of ways. It will be (total number of ways you can distribute paper to students)-(number of ways at least one person gets their own paper). Let's find the second quantity.
There are $\binom{2007}{1}2006!$ ways to guarantee that at least one person receives their hat since there are $\binom{2007}{1}$ ways to choose the person and $2006!$ ways to arrange the rest of the hats. But we've overcounted the cases where two people have received their hat. So we subtract $\binom{2007}{2}2005!$. Then we add back in the cases where three people receive their hat: $\binom{2007}{3}2004!$, and so on, giving $\binom{2007}{1}2006!-\binom{2007}{2}2005!+\binom{2007}{3}2004!-\ldots+\binom{2007}{2007}0!$ as the number of ways that at least one person receives their hat.
The total number of ways we can distribute the hats is $2007!$. So the number of ways that we can distribute the hats without one person receiving their own hat is $2007!-(\binom{2007}{1}2006!-\binom{2007}{2}2005!+\binom{2007}{3}2004!-\ldots+\binom{2007}{2007}0!)=\binom{2007}{0}2007!-\binom{2007}{1}2006!+\binom{2007}{2}2005!-\binom{2007}{3}2004!+\ldots-\binom{2007}{2007}0!$

Now let's find the approximation. Factor out $2007!$ to get $2007!(\binom{2007}{0}\frac{1}{0!}-\binom{2007}{1}\frac{1}{2007}+\binom{2007}{2}\frac{1}{2007*2006}-\binom{2007}{3}\frac{1}{2007*2006*2005}+\ldots-\binom{2007}{2007}\frac{1}{2007!}=2007!(\frac{1}{0!}-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\ldots-\frac{1}{2007!})$ which is approximately equal to $2007!(\frac{1}{0!}-\frac{1}{1!}+\frac{1}{2!}-\ldots)=2007!(e^{-1})$

Ah man, you were right! I accidentally wrote the last part as $2007!(1-e^{-1})$ the other day. Ok, I agree with you guys  :lol:

Yeah, I type till mid-way then I'm like "How do you continue simplifying again...???" :rotfl: thx, drunner by posting the real formula.

[quote="pkerichang"]what is the recursive formula???[/quote]It's on the wikipedia page in my first post. $d_n=(n-1)(d_{n-1}+d_{n-2})$

[quote="Altheman"]you have to be careful with the derangement formula, it is

$\frac{n!}{e}$ rounded, not floor,

the proof has something to do with the identity:

$e=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+...$

and comparing it to the formula from the inclusion-exclusion formula[/quote]

Well, you can verify that for odd $n$, it's floor, and for even $n$, it's floor + 1. That amounts to the same thing as rounding, or the floor of $\frac{n!}{e}+\frac{1}{2}$, though.


limit
superior algebra
superior algebra solved

I update this problem
Compute \lim_{x->0} (x^2003 - (tgx)^2003)/x^2005
Paris-2003

Rewrite the limit like so:

lim(x-->0) [1-(tan(x)/x)^1999]/x^2=lim(x-->0) [x-tan(x)]/x^3[1+tan(x)/x+..+(tan(x)/x)^1998]=1999lim(x-->0) [x-tan(x)]/x^3=-1999/3

lim(x-->0) (x-tan(x))/x^3=-1/3 by L'hopital.

I think that the other problem has the answer:
L=-2003/3

cheers! :D :D


How do you type a percent sign or approximation sign on latex? Thanks everybody.

Percent sign: It doesn't show up!---> $ 30%
$ (I typed "30%" in between 2 dollar symbols)

Approximation...I can get ~ but not two of those.... :(

Did you try looking at http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Symbols ?   :wink:

Specifically, it would be

\approx - $ \approx$

30 \% - $ 30 \%$

$ 30% x=y
$

I was just curious about what would happen on the forum; what I actually typed between the dollar signs there was 30% x=y. You'll notice that not only did the % sign disappear, but so also the x=y disappeared; that is what I expected to happen.

In a $ \text{\LaTeX}$ document - and remember, such a document is essentially a computer program - the % sign serves a function familiar to programmers: it starts a comment. Specifically, everything on a line after % stays in the source code as an annotation but is not executed and produces nothing on a document.

Precisely because the % symbol has that specialized use, you need to use \% to actually use it as a symbol in a document.


A move consists of joining two clusters, including clusters of just one piece. What is the minimum number of moves required to complete a two-thousand piece jigsaw puzzle that is separated into individual pieces?

Every time you join two clusters you reduce the number of clusters by $ 1$.  We need to reduce it from $ 2000$ to $ 1$, so $ 2000\minus{}1\equal{}\boxed{1999}$ moves are needed.


calculus
integration
geometry
parameterization
trigonometry
analytic geometry
3D geometry

A hemisphere with radius $R$ lies in the $xy$-plane.
Let $\bar{A}$ be a vectorfield $\bar{A}=-y\hat{x}+x\hat{y}+f(z)\hat{z}$

Calculate $\oint_\gamma \bar{A}\cdot d\bar{r}$

and $\iint_\Sigma (\bar{\nabla}\times \bar{A}) \cdot dS$

Where $\gamma$ is the large circle (base) of the hemisphere lying in the xy-plane (+ orientation) and $\Sigma$ the surface of the hemisphere with $\gamma=\partial \Sigma$. $f(z)$ is an arbitrary nice function.

By Stokes' Theorem, the two integrals are equal, so we only need do one of them. Look at the line integral: since $\hat{z}\cdot\vec{dr}=0$ on that curve, we may safely ignore $f(z).$ I'd suggest just parameterizing the curve and computing it - it's straightforward.

But an alternative: A third integral that is equal to the other two is the integral of $\nabla\times A$ over the flat disk in the $xy$-plane bounded by $\gamma.$ That's also equal to the line integral by Stokes' Theorem. But since the normal to the surface is $\hat{z},$ we need only compute the $\hat{z}$ component of the curl (again, that ignores $f(z).$) This will turn into the integral of a constant over that area, once you compute that component of the curl.

But i get different values for each of them :( .

[b]The first one:[/b]
first I find a parametrization: $\bar{r}(t)=(R\cos{t},R\sin{t},0)$, $t\in[0,2\pi]$
\[\oint_\gamma \bar{A}\cdot d\bar{r}=\int_{0}^{2\pi}\bar{A}(\bar{r}(t))\cdot(\bar{r}'(t))=\ldots \]
And we have: $\bar{r}'(t)=R(-\sin{t}, \cos{t},0)$
and also: $\bar{A}(\bar{r}(t))=R(-\sin{t}, \cos{t}, f(0))$
so scalar prod. yields
\[\ldots=\int_{0}^{2\pi}R^{2}(-\sin{t}, \cos{t}, f(0))\cdot R(-\sin{t}, \cos{t},0)=2\pi R^{2}\]

[b]The second one:[/b]
Here we switch to spherical polar coordinates:
$\bar{r}(\theta, \varphi)=R[(\sin{\theta}\cos{\varphi}),(\sin{\theta}\sin{\varphi}),(\cos{\theta})]$
And since $\mbox{rot}\bar{A}=(0,0,2)$ our integral becomes
\[\int_{0}^{2\pi}\int_{0}^\pi (0,0,2)\cdot(|\bar{r}'_\theta\times\bar{r}'_\varphi|d\theta d\varphi) \]
Where $d\bar{S}=|\bar{r}'_\theta \times\bar{r}'_\varphi|d\theta d\varphi$
This can also be expressed as $\bar{n}dS$ where $\bar{n}$ is the normal to the sphere aand could simpler be calculated as $\frac{\bar{r}}{|\bar{r}|}=\frac{\bar{r}}{R}$

Thus the integral becomes
\[\int_{0}^{2\pi}\int_{0}^\pi (0,0,2)\cdot(\frac{\bar{r}}{R}d\theta d\varphi) =2\int_{0}^{2\pi}\int_{0}^\pi \cos{\theta}d\theta d\varphi=0 \]

[color=blue]So where is the problem?
I appreciate som help with the details.[/color]

[quote="Kent Merryfield"]The first one, the line integral, is correct.

Your problem in the second one is that by letting $\varphi$ go from $0$ to $\pi$ you integrated over the whole sphere. You should have only the upper hemisphere, in which $\varphi$ ranges from $0$ to $\frac{\pi}2.$[/quote]

Of course! yes! How stupid of me, I did not even consider that! :blush:  :blush:  Thanks a lot Mr. Merryfield !

[quote="Kent Merryfield"]The first one, the line integral, is correct.

Your problem in the second one is that by letting $\varphi$ go from $0$ to $\pi$ you integrated over the whole sphere. You should have only the upper hemisphere, in which $\varphi$ ranges from $0$ to $\frac{\pi}2.$[/quote]

is it correct to take abs signs in ($r\times r \ldots$)? And is the geometrical
method $\frac{\bar{r}}{R}$ equivalent? Does $r\times r \ldots$ have anything to do with the jacobian for the parametrization?


I am taking an Algebra 2 course and just was told that you can't take the log of negative numbers. Can't you do $ log_\minus{}3 \minus{}27$? The answer would be negative 3. PLease explain why this is not allowed.

The answer would be $ 3$.  However, note that your base is also negative.  In general, the base of a logarithm is positive, and once we add that condition, $ \log_3 (\minus{}27)$, for example, is not defined.

Why are negative bases discouraged?  The simple answer is that exponential functions with negative bases are very badly behaved.  Try to graph $ y \equal{} (\minus{}3)^x$ on your calculator.

The more complicated answer is that in analysis, we define every logarithm in terms of the natural logarithm $ \ln x \equal{} \log_e x$, and a negative base (or taking the logarithm of a negative number) requires you to define $ \ln \minus{}1$, which is not real.


Hey, you know what's frustrating sometimes is trying to split topics but not being able to. People sometimes address two issues in one post and then you can't move that post b/c it's relevant to the old topid but you can't not move it b/c it's part of the new one.

Anyways, what I'm saying is, is there a way we can make it so mods can split posts as well as threads? That would be most awesome.

Just a suggestion ;-)

Hehe, splitting posts is and probably will not be possible very soon. We'll see maybe later about merging topics, and then, you split out a certain post and merge it with the previously splitted topic.

Merging topics would be good too. *nods* Oodles of convenience for us.

Merging topics would be great as there are so many places where the same question gets asked.  It would be great if we could respond by tacking their post onto the previous answer with the response "see above."

The only problem is that our priority list is currently 13.2 miles long.

[quote="MCrawford"]The only problem is that our priority list is currently 13.2 miles long.[/quote]Actually it's $10 + \pi$ long, but 13.2 will do as an approximation :P


the \[ x_i\] are all distinct .
so the given expression can never be 0.

i think some sorta inequality will do the trick :wink:

We have the minimum of:

$ ( {x_{i_1 } - x_{i_2 } } )^2 + ( {x_{i_2 } - x_{i_3 } } )^2 + ( {x_{i_3 } - x_{i_4 } } )^2 + ......... + ( {x_{i_{n - 1} } - x_{i_n } } )^2 + ( {x_{i_n } - x_{i_1 } } )^2$

When $ A=x_{i_1 }.x_{i_2 }+x_{i_2 }.x_{i_3 }+...+x_{i_n }x_{i_1 } \hspace{1cm}$ is maximum.

We will prove by induction that we have the maximum of $ A$ when:
$ A=x_nx_{n-1}+x_{n-1}x_{n-3}+x_{n-3}x_{n-5}+...+x_1x_2+...+x_{n-4}x_{n-2}+x_{n-2}x_n$

For $ n=3$ is obvious.

Suppose that it is true for $ n=k$

Let see $ A$ like a string. We choose the position of $ x_1$ in any string with the numbers $ x_2,x_3,...,x_k$ such that $ A$ is maximum.

Suppose that the string after we choose the position of $ x_1$ is $ ....x_j,x_1,x_p....$

Then when we add $ x_1$ to the string the value of $ A$ increase by:
$ B=x_jx_1+x_1x_p-x_jx_p$
$ B=-(x_1-x_j)(x_1-x_p)+x_1^2$
We have the maximum of $ B$ when:
$ C=(x_1-x_j)(x_1-x_p)$ is minimum
Then we have the minimum of C when $ x_j=x_2$ and $ x_p=x_3$.

By the supposition, the maximum of $ A$ with the numbers $ x_2,x_3,...,x_k,x_{k+1}$ is:
$ x_{k+1}x_k+x_kx_{k-2}+x_{k-2}x_{k-4}+...x_2x_3+...+x_{k-3}x_{k-1}+x_{k-1}x_{k+1}$

And choosing the best position of $ x_1$:
$ A=x_{k+1}x_k+x_kx_{k-2}+x_{k-2}x_{k-4}+...x_2x_1+x_1x_3+...+x_{k-3}x_{k-1}+x_{k-1}x_{k+1}$
Then it is true for $ n=k+1$ and by the induction is true for all $ n$.

Then the rearrange is:

${ ( {x_{n } - x_{n-1 } } )^2 + ( {x_{n-1 } - x_{n-3 } } )^2 + ( {x_{n-3 } - x_{n-5 } } )^2 + ......... + ( {x_{n-4} } - x_{n-2 } } )^2 + ( {x_{n-2 } - x_{n } } )^2$


3 countries $A, B, C$ participate in a competition where each country has 9 representatives.  The rules are as follows: every round of competition is between 1 competitor each from 2 countries.  The winner plays in the next round, while the loser is knocked out.  The remaining country will then send a representative to take on the winner of the previous round. The competition begins with $A$ and $B$ sending a competitor each.  If all competitors from one country have been knocked out, the competition continues between the remaining 2 countries until another country is knocked out.  The remaining team is the champion.

[b]I.[/b] At least how many games does the champion team win?

[b]II.[/b] If the champion team won 11 matches, at least how many matches were played?

Suppose $A$ is the champion. $A$ plays $a$ times and loses $b$ times ($b\le8$).

$A$ does not play when: (1) the first game starts with $B$ and $C$; (2) $A$ loses. So the number of games $A$ does not play is $1+b$ or $b$. The total number of games is equal to the total number of loses, which is $b+9+9=b+18$. Thus we get $1+b+a=b+18$ or $b+a=b+18$, whence $a=18$ or $a=17$.

I. The number of wins for $A$ is $a-b\ge17-8=9$.

Construction: $B$ and $C$ starts the game. $B,C,A$ loses repeatedly in this order.

II. We have $a-b=11$, so $b=a-11=6$ or $7$. Thus the number of games is $b+18\ge24$.

Construction: $B$ and $C$ starts the game. $B$ beats 6 representatives of $A$ and 7 representatives of $C$ in the first 13 games. Then $A$ beats all 11 remaining representatives from $B$ and $C$ in the last 11 games.


geometry
parallelogram
calculus
integration
trapezoid
complex numbers
national olympiad

I have post CWMO 2001, 2003 and 2004 problems, I'll post CWMO 2002 problem here, which is also in Chinese.
Have a good day!

I'll translate it, but later perhaps, because I really should be asleep at this hour of the day :blush: .

There you go, Myth! :)

1) Find all positive integers $n$ such that $n^4 - 4n^3 + 22n^2 -36n + 18$ is a perfect square.

2) Let $O$ be the circumcentre of $\triangle ABC$ and $P$ be a point in the interior of $\triangle AOB$.  Let $D, E, F$ be the feet of the perpendiculars from $P$ onto $BC, CA, AB$ respectively.  Prove that the parallelogram with adjacent sides $FE$ and $FD$ is situated within $\triangle ABC$.

3) Consider a square in the complex plane.  The four complex numbers that its vertices represent happen to be the four roots of the equation $x^4 + px^3 + qx^2 + rx + s = 0$ which has integral coefficients.  Find the smallest possible area of such a square.

(I really hope that I translated this problem correctly...)

4) Let $n$ be a positive integer.  The sets $A_1, A_2, \ldots, A_{n+1}$ are $n + 1$ non-empty subsets of $\{1, 2, \ldots, n\}$.  Prove that there exist two disjoint subsets $\{i_1, i_2, \ldots, i_k\}$ and $\{j_1, j_2, \ldots, j_m\}$ of $\{1, 2, \ldots, n + 1\}$ such that $A_{i_1} \cap A_{i_2} \cap \cdots \cap A_{i_k} = A_{j_1} \cap A_{j_2} \cap \cdots \cap A_{j_m}$.

5) In a given trapezium $ABCD$, $AD$ is parallel to $BC$, $E$ is a variable point on $AB$, $O_1, O_2$ are the circumcentres of $\triangle AED, \triangle BEC$ respectively.  Prove that $O_1O_2$ is a constant.

6) Let $n$ ($n \geq 2$) be a given integer.  Find all sets of integers $\{a_1, a_2, \ldots, a_n\}$ that satisfy the following conditions:
(I) $a_1 + a_2 + \ldots + a_n \geq n^2$
(II) $a_1^2 + a_2^2 + \ldots + a_n^2 \geq n^3 + 1$.

7) Let $\alpha, \beta$ be the roots of the equation $x^2 - x - 1 = 0$, and let $a_n = \frac{\alpha^n - \beta^n}{\alpha - \beta}, n = 1, 2, \ldots$.
a) Prove that for any positive integer $n$, $a_{n+2} = a_{n+1} +a_n$
b) Find all positive integers $a, b$ $(a < b)$ such that for any positive integer $n$, $b$ divides $a_n - 2na^n$.

8) Let $S = (a_1, a_2, \ldots, a_n)$ be a longest possible sequence formed from the numbers $0, 1$ satisfying the following condition: No $2$ pairs of $5$ consecutive numbers in $S$ are the same, that is for any $1 \leq i < j \leq n - 4, a_i, a_{i+1}, a_{i+2}, a_{i+3}, a_{i+4}$ is not the same as $a_j, a_{j+1}, a_{j+2}, a_{j+3}, a_{j+4}$.  Prove that the first $4$ terms of $S$ are the same as the last $4$ terms of $S$.


Do you know any TST's or Olympiads which had problems from Balkan shortlist?(I know that UK has problems from balkan shortlist)


factorial AMC AIME blogs LaTeX FTW ratio

Guess who will post next.

Keep track of points in each post.

You get 1 point for guessing.

The person who is guessed get's 2 points.

You get 5 additional points for getting your guess right.

No double posting.

No guessing yourself.

-----------------------------------------------------------

I'll guess Zephyredx.

iLord: 1
Zephyredx: 2

iLord: 1
Zephyredx: 2
manofdayear:1
1=2: 2

I guess 1=2

iLord: 1
Zephyredx: 2
manofdayear:1
1=2: 4
myyellowducky82:1

i guess 1=2

iLord: 1
Zephyredx: 2
manofdayear:2
1=2: 4
myyellowducky82:1
Perfect628: 2

perfect628

iLord: 1
Zephyredx: 2
manofdayear:2
1=2: 4
myyellowducky82:2
Perfect628: 2
crazychinesemanic: 2

i guess crazychinesemanic.

Ummm... manofdayear = crazychinesemanic ...

iLord: 1
Zephyredx: 2
1=2: 4
myyellowducky82: 4
Perfect628: 2
crazychinesemanic: 4
chenhsi: 1

i guess myyellowducky

I'll guess chenhsi!

iLord: 2
Zephyredx: 2
1=2: 4
myyellowducky82: 4
Perfect628: 2
crazychinesemanic: 4
chenhsi: 3

[quote="chenhsi"]Ummm... manofdayear = crazychinesemanic ...[/quote]

how do u know my runescape sn...i dont even know you
EDIT: O wait...my bad  :oops:

iLord: 4
Zephyredx: 2
1=2: 4
myyellowducky82: 4
Perfect628: 2
crazychinesemanic: 5
chenhsi: 3

iLord

I'll guess Chenhsi again.

iLord: 5
Zephyredx: 2
1=2: 4
myyellowducky82: 4
Perfect628: 2
crazychinesemanic: 5
chenhsi: 5

w00t i guessed right!!!!

iLord: 7
Zephyredx: 2
1=2: 4
myyellowducky82: 4
Perfect628: 2
crazychinesemanic: 10
chenhsi: 5

i guess iLord again

Yep! I'll guess Crazychinesemaniac!

iLord: 8
Zephyredx: 2
1=2: 4
myyellowducky82: 4
Perfect628: 2
crazychinesemanic: 17
chenhsi: 5

err...scoreboard is messed up. My fault, fixed score board is here.

iLord: 17
Zephyredx: 2
1=2: 4
myyellowducky82: 4
Perfect628: 2
crazychinesemanic: 19
chenhsi: 5

I guess...lol 3 times the charm iLord!

iLord: 18
Zephyredx: 2
1=2: 4
myyellowducky82: 4
Perfect628: 2
crazychinesemanic: 26
chenhsi: 5

Yeah, let's rack up points!

I'm waiting for CCM!

[quote="iLord"]iLord: 20
Zephyredx: 2
1=2: 4
myyellowducky82: 4
Perfect628: 2
crazychinesemanic: 26
chenhsi: 5
alexhhmun: 1

Yeah, let's rack up points!

I'm waiting for CCM![/quote]

I guess iLord.  Please?

P.S.  It's kinda sad that 1=2 has never posted here, yet is still beating me...

iLord: 20
Zephyredx: 2
1=2: 4
myyellowducky82: 4
Perfect628: 2
crazychinesemanic: 26
chenhsi: 5
alexhhmun: 1
Budi713: 3

No rule against double posting but that would be cheap so i will not do that but is will still
guess that budi713 is next giving me 3 points.

[sarcasm]Good job, guys. [/sarcasm]You have revived a thread that already has a revival thread. Not only that, there are no such AoPS users, at least for the moment being, named "r15" or "r15somethingsomethingsomething."

Good job King!.  You would've been in trouble if it was a girl but luckily for you, I'm male.
My guess: Powerofpi

Hi. 1/2 year bump. Don't post in dead threads.

This is a very old thread. It died for a reason. Let dead threads lie in peace. Can this be locked, please?

It the fun factory so they (mods) dont ussually do that.
Do you poeple just trawle though the old threads and randomly revive one?

Phire's usually pretty good at locking old threads.

New members usually don't know the revival rules (I know when I was new I revived a lot), and it looks like this guy is new, so we can cut him some slack.


Can have a triangle the lenghts of sides 3k+1,3l+1,3m+1 or 3k+2,3l+2,3m+2 where $ l,m,n \in Z$?

Yes. For example when $ k \equal{} l \equal{} m$ then the triangle is equilateral.


2001! For the first time, someone has attained this truly remarkable rating. After AIME's defeat when he had 2000, he slowly worked his way back to 2000, culminating his epic climb to the top with a series of 11 cds against hurdler! Congratulations AIME, a true hero!

Depends. I think hurdy logged out but other people might "help the cause".

Gee...this is weird!  Somehow, once I reached 2000 again, RG comes on!  And plays mewto a whole bunch of times to get to 2004!

Somebody is a hypocrite...

Huh? When did i play rg? I was just cding myself a bunch. RG is still asleep right now...

And mewto has 2008 now.
AIME15, something is weird (OK, we all know what is it).


Prove that the remainder when any prime number is divided by 30 is either 1 or a prime number.

[hide]
Suppose $ p\equal{}30k\plus{}n$, where $ 0\leq n<30$. Since $ p$ is prime, we have $ 30\not|p$, so $ n\neq 0$. We also have $ \gcd(30,n)\equal{}1$, so $ n$ has no factors of $ 2,3,$ or $ 5$. But then, any such composite $ n$ is at least $ 7\cdot 7\equal{}49$, which is larger than 30, so $ n$ must be prime or $ 1$.[/hide]

In fact, you can say more: $ 30$ is the largest modulus with this property.


superior algebra
superior algebra unsolved

Is it true that for any $ n\geqslant 3$, there exist $ x,y\in D_{n}$ such that $ \left|x\right|\equal{}\left|y\right|\equal{}2$ and $ \left|xy\right|\equal{}n$?  If so, how do we prove it?

Think of the group as the symmetries of a regular $ n$-gon, and look at two adjacent reflections.


calculus
integration
trigonometry
logarithms
calculus computations

I'm having trouble with this integral. I usually find calculus fairly easy (classroom calculus :D ) but I'm not getting anywhere with this problem.

Find: $\displaystyle \int \tan^2 x \sec x \ dx$

I tried many things such as U-Substitution, using $\tan^2 x = \sec^2 x - 1$ as substitution but I'm not getting anywhere. Can anyone help me? I appreciate this. Thanks.

Rewrite the integrand as $\sec^3x-\sec x$, and integrate $\sec^3x$ by parts.

$\begin{eqnarray*} \int \tan^2 x \, \sec x \, dx &=& \int \sec^3 x -\sec x \, dx \\ &=& \int \sec^3 x \, dx - \int \sec x \, dx \\ &=& \sec x \tan x - \int \tan^2 x \sec x\, dx - \ln |\sec x + \tan x| +C$

So $2 \int \tan^2 x \, \sec x \, dx = \sec x \tan x - \ln |\sec x + \tan x |$, and $\int \tan^2 x \, \sec x \, dx = \frac{\sec x \tan x - \ln |\sec x + \tan x |}{2}$

Yeah that's what I did.

Only thing was that we didn't learn that rule in the class and when we went over it, our substitute teacher just stopped and said, "We'll go over it after the final."

I don't know why the problem was assigned for Final Review when it was going to be discussed AFTER the final but yeah. Thanks for the help.  :)


[hide="easy"]1 user=1% there are 100-20=80 users who remain.=> 80%[/hide]

[hide="solution"]There are 20% banned, so 80% remain[/hide]

Hix very easy
[hide] $80\%$ [/hide]

[size=75][color=green]Use \% for percents
LaTeX edited[/color][/size]

[hide][quote="anirudh"]If there are 100 users of a website, and 20 are banned, what is the percentage of those who remain?[/quote]

4/5 self explanatory[/hide]


function
trigonometry
real analysis
real analysis unsolved

Consider the function $ f(x) \equal{} \sin{\frac {1}{x}}$ mapping the open interval $ (0,1)$ into $ \mathbb{R}$. Prove that it is not uniformly continuous.

Two possible approaches:

1. If it were uniformly continuous, it could be extended to a continuous function on the closure of that domain.

2. More explicitly: can you find a sequence $ x_n$ such that $ |x_n\minus{}x_{n\plus{}1}|\to 0$ but $ |f(x_n)\minus{}f(x_{n\plus{}1})|$ does not tend to zero?


geometry
3D geometry
tetrahedron
octahedron
icosahedron
graph theory
advanced fields

Hi, can anyone help me with the following question in graph theory?

We want to consider the planar embedding of the tetrahedron, octahedron, and icosahedron, the three platonic solids with triangular faces. We know that it is a planar graph because they are convex polyhedra.

A [i]Hamiltonian path[/i] is a path that visits each vertex exactly once. A [i]traceable graph[/i] is a graph that contains a Hamiltonian path.

The problem is to show that [b]if $ T$ is a tree contained in a planar embedding $ G$ of these planar graphs, then we can always find some [i]traceable subgraph[/i] $ H$ of of $ G$ with $ V(H) \equal{} V(T)$ and $ E(H) \subseteq E(G).$[/b]

[i]i.e., [/i] We want to find some traceable subgraph $ H$ such that it spans the vertices of $ T$ with edges contained in the edges of $ G$.

My strategy is to first construct the closure $ \text{cl}(T)$ of $ T$ by adding for all nonadjacent pairs of vertices $ u$ and $ v$ with $ (u,v) \in E(G)$ the new edge $ (u,v)$ to the set $ E(T).$ And then use some theorem to show that we can always find a Hamiltonian path.

Actually thinking about it, there is a simpler way to put it.

Show that any connected subgraph of a planar embedding of these planar graphs is also a traceable subgraph.

I'm a little confused here, so maybe I'm misunderstanding.

[quote="lenny"]
The problem is to show that [b]if $ T$ is a tree contained in a planar embedding $ G$ of these planar graphs [/quote]

What's the point of looking at planar embeddings? A tree being contained in a graph has nothing to do with if or how the graph is embedded in the plane.

[quote] then we can always find some [i]traceable subgraph[/i] $ H$ of of $ G$ with $ V(H) \equal{} V(T)$ and $ E(H) \subseteq E(G).$[/b] [/quote]

so let's say the tree $ T$ consists of a a vertex $ v$ of degree 3 along with its 3 neighbors $ u_1,u_2,u_3$ (which obviously exists in the cube, tetrahedron, icosahedron or any other cubic graph). You want a subgraph $ H$ whose vertex set is precisely $ \{v,u_1,u_2,u_3\}$, which implies that it's either disconnected or it's the whole tree $ T$. Obviously this $ T$ contains to spanning graph with a Hamiltonian path.

[quote]What's the point of looking at planar embeddings? A tree being contained in a graph has nothing to do with if or how the graph is embedded in the plane. [/quote]

You have the point, there isn't much advantage in doing this. I guess it just makes things looks easier by reducing the problem to planar graphs.

[quote]which implies that it's either disconnected or it's the whole tree .[/quote]

No, I can add edges besides the edges in $ T$ (i.e., besides $ vu_1, vu_2, vu_3$.) to be included in $ H,$ as long as those edges are in $ G.$

[quote]Obviously this $ T$ contains to spanning graph with a Hamiltonian path.[/quote]

This $ T$ does not contain a spanning graph with a Hamiltonian path. Do you mean the closure of $ T$ as I defined in the first entry? Even so, I don't see why it is obvious and will be happy to see your proof of it. =)

never mind. I just found an counter example.


is there a closed form for
$\sum_{k=1}^{n} k*y^{k}$ for $y \ne 1$
i.e. a function that gives the same result as the summation but as a function of n and y

$S_n  = \sum\limits_{k = 1}^n {k \cdot y^k  = y \cdot \sum\limits_{k = 1}^n {\left[ {\left( {k - 1} \right) + 1} \right]}  \cdot y^{k - 1}  = } $

$ = y \cdot \sum\limits_{k = 1}^n {\left( {k - 1} \right) \cdot y^{k - 1} }  + y \cdot \sum\limits_{k = 1}^n {y^{k - 1} }  = $

$ = y \cdot \sum\limits_{k = 1}^{n - 1} {k \cdot y^k  + y \cdot \sum\limits_{k = 1}^n {y^{k - 1} }  = } $

$ = y \cdot \sum\limits_{k = 1}^n {k \cdot y^k }  - ny^{n + 1}  + y \cdot \frac{{1 - y^n }}{{1 - y}} = $

$ = y \cdot S_n  + \frac{{n \cdot y^{n + 2}  - \left( {n + 1} \right) \cdot y^{n + 1}  + y}}{{1 - y}}$

Then:
$S_n  = \frac{{n \cdot y^{n + 2}  - \left( {n + 1} \right) \cdot y^{n + 1}  + y}}{{\left( {1 - y} \right)^2 }}$

$\sum_{k=1}^{n} k*y^{k}$
$y* \sum_{k=1}^{n} k*y^{k-1}$
think of $D_y(y^k) = k*y^{k-1}$, since derivitives are distributive over additon,
$y* D_y(\sum_{k=1}^{n} y^{k})$ since this is a geometric sequence,
$y* D_y(\frac{y-y^{n+1}}{1-y})$, using the derivitative rule for quotients...
$y* \frac{ (1-y)(1-(n+1)y^{n}) - (y-y^{n+1})(-1)}{(1-y)^{2}}$implifying gives
$\frac{y(n*y^{n+1} - (n+1)y^{n}+1)}{(y-1)^{2}}$
i don't think that this simplifies...somebody already got this, but i thought that this was a pretty clever solution


The radius of a circle is 2 inches. When the radius is doubled,
by how many square inches is the area increased?  Express your
answer in terms of $ \pi$.

The original circle's area is $ 2^2\pi \equal{} 4\pi$

The new radius is $ 2*2 \equal{} 4$
Its area is $ 4^2\pi \equal{} 16\pi$
So the area increased is $ (16\minus{}4)\pi \equal{} \boxed{12\pi}$

$\text{Radius doubled meaning the area increased by a factor of }4\text{ (because a circle is }2\text{D).}$

$\text{Original area is }2^2\pi=4\pi$

$4-1=3$

$3\times (4\pi)=\boxed{12\pi}$