Given $a\in R,a\neq 0 , n\in N , n\ge 2$
The equation $ax^{n}+x+1=0$ have $n$ real roots $x_{1},x_{2},..,x_{n}$
Prove that there exits $i\in N$ such as $|x_{i}|\le 2$

Warning : if $n\geq 4$ then the equation $ax^{n}+x+1=0$ can't have $n$ real roots !!!
 :cool:

Maybe he meaned the polynomial $P(x)=ax^{n}+x+1$ have $n$ real roots
If so $P(x)=a(x-x_{1})(x-x_{2})...(x-x_{n})$ and $x_{1}x_{2}...x_{n}=1$ hence $\exists i\in N$ such that $|x_{i}|<1$

actually the roots of the polynomial are the same with the roots of the equation, Stun
and even so check that $x_{1}x_{2}\cdots x_{n}=(-1)^{n}\frac{1}{a}$ :lol:


Let $ ABC$ be a triangle. Circle $ \omega$ passes through $ A$ and $ B$ and meets sides $ AC$ and $ BC$ at $ D$ and $ E$, respectively. Let $ F$ be the midpoint of segment $ AD$.
Suppose that there is a point $ G$ on side $ AB$ such that $ FG\perp AC.$ Prove that $ \angle{EGF}\equal{}\angle{ABC}$ if and only if $ \frac{AF}{FC}\equal{}\frac{BG}{GA}.$

($ \Longrightarrow$) If $ \angle EGF \equal{} \angle ABC$. From $ ABDE$-cyclic $ \Rightarrow$ $ \angle EDC \equal{} \angle ABC$. Hence $ FGED$ is a cyclic. If $ H$ is the second intersection of circumcircle $ FDEG$ with $ BC$ we obtain that $ \angle HGB \equal{} \alpha$ and $ FHG \equal{} \alpha$. Hence $ FH\parallel{}AB$. Analogically we get that $ GH\parallel{}AC$. Hence $ \frac {AF}{FC} \equal{} \frac {BH}{HC} \equal{} \frac {BG}{GA}$
($ \Longleftarrow$) If $ \frac {AF}{FC} \equal{} \frac {BG}{GA}$. Construct point $ H$ on side $ BC$ such that $ GH\parallel{}AC$. Hence $ \frac {BG}{GA} \equal{} \frac {BE}{CE}$ but $ \frac {AF}{FC} \equal{} \frac {BG}{GA}$ hence $ FH\parallel{}AB$ $ \Rightarrow \angle FHG \equal{} \alpha$. But $ \angle GDF \equal{} \alpha$ $ \Rightarrow$ $ GHDF$-cyclic $ \Rightarrow$ $ \angle DGH \equal{} \angle DFH \equal{} \alpha$ but $ \angle DEC \equal{} \alpha$ $ \Rightarrow$ $ DGHE$-cyclic $ \Rightarrow$ $ FGHE$-cyclic $ \Rightarrow$ $ \angle EGF \equal{} \angle FHE \equal{} \beta$


Let $ a,b,c>0$, s.t. $ a\plus{}b\plus{}c\equal{}3$.Prove that:
$ abc\plus{}\frac{12}{ab\plus{}bc\plus{}ca}\geq 5$

Follows Schur's Inequality $ (a\plus{}b\plus{}c)^3\plus{}9abc\geq4(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)$ we get $ 3abc\geq 4(ab\plus{}bc\plus{}ca)\minus{}9$.

 Thus , it suffices to show that  $ 4(ab\plus{}bc\plus{}ca)\minus{}9 \plus{}\frac{36}{ab\plus{}bc\plus{}ca}\geq15$  or $ (ab\plus{}bc\plus{}ca\minus{}3)^2\geq0$  which is true.

[quote="Inequalities Master"]Let $ a,b,c > 0$, s.t. $ a + b + c = 3$.Prove that:
$ abc + \frac {12}{ab + bc + ca}\geq 5$[/quote]
setting $ p=a+b+c=3, q=ab+bc+ca, r=abc$

$ \Leftrightarrow r+\frac{12}{q}\ge 5$

by Schur'ineq $ r\ge \frac{p(4q-p^2)}{9}=\frac{4q-9}{3}$

$ \Leftrightarrow \frac{4q-9}{3}+\frac{12}{q}\ge 5$

$ \Leftrightarrow (q-3)^2 \ge 0$

done

[quote=Inequalities Master]Let $ a,b,c>0$, s.t. $ a\plus{}b\plus{}c\equal{}3$.Prove that:
$$ abc\plus{}\frac{12}{ab\plus{}bc\plus{}ca}\geq 5$$[/quote]
Let $ a,b,c>0,a+b+c=3.$ [url=https://artofproblemsolving.com/community/c6h1136009p13963619]Prove that[/url]$$ abc+\frac{3(\lambda -1)^2}{4(ab+bc+ca)+3\lambda -15}\ge \lambda . $$Where $\lambda \geq 5.$
Let $a , b , c ,d>0$ and $a + b + c+d =4.$ [url=https://artofproblemsolving.com/community/c6h1086843p12625737]Prove that[/url]
$$ abcd + \frac{12}{abc + bcd+cda+dab} \geq 4.$$
Let $a,b,c > 0$ and $a+b+c=3$. [url=https://artofproblemsolving.com/community/c6h311130p2020300]Prove that[/url]
$$\sqrt a+\sqrt b +\sqrt c + \frac{6}{ab+bc+ca} \ge 5 $$


[quote="Fraxia"]What is the largest integer which must evenly divide all integers of the form $n^{5}-n$?[/quote]

Is there a restriction to $n$ because if $n=1$ then it wouldn't really make sense.

Yeah, it would since any number is a factor of 0.

[quote="Fraxia"]What is the largest integer which must evenly divide all integers of the form $n^{5}-n$?[/quote]

[hide]If $n$ is odd or even, then $n^{5}-n=n(n-1)(n+1)(n^{2}+1)$ is divisible by 2. It is also divisible by 3.  If $n=2$ then it is not divisible by 4 or 9. The expression is divisible by 5. Since when $n=2$ then $n^{5}-n=30$ and $2\times 3\times 5=30$ the largest integer is $\boxed{30}$.[/hide]

well actually, it does make sense for $n=1.$
We'd get $1-1=0$ and all numbers divide 0.
[hide]Factoring, $n^{5}-n=(n-1)n(n+1)(n^{2}+1)$
Checking in $\mod 2,3,5$, we find that $2,3,5$ always divide $n^{5}-n.$
We know that there are no more factors that divide all $n^{5}-n,$ because $2^{5}-2=30=2\cdot3\cdot5.$
So the answer is $30$[/hide]
edit: wtf 3 posts in 2 min? o_0

[hide="A simple approach"] $n = 0 \implies n^{5}-n = 0$
$n = 1 \implies n^{5}-n = 0$
$n = 2 \implies n^{5}-n = 30$
$n = 3 \implies n^{5}-n = 240$
$n = 4 \implies n^{5}-n = 1020$

Since $gcd(30, 240, 1020) = \boxed{30}$ we suspect that this is our answer.  In fact, since it is true for these five numbers, we know that $n^{5}-n \equiv 0 \bmod 5$ (since every other number is congruent to one of these $\bmod 5$), and similarly we know that $n^{5}-n \equiv 0 \bmod 2, 3$. [/hide] [hide="A more theoretical approach"] We will consider separately each prime for which this is true.  It is true for a prime $p$ if and only if $p-1 | 5-1$ (remember that this is [b]all[/b] integers of that form; for any prime that this is not true we can find primitive roots $\bmod p$).

This is only true for $p = 2, 3, 5$. [/hide]


Prove that for all integers $n$ greater than 2 and all primes $p$ the number
$n^{p^p}+p^p$ is composite


If $a,b,c,d \in R$ then:
$ 2(a^2-ab+b^2)(c^2-cd+d^2) \geq a^2c^2+b^2d^2$ . When do we have equality?

cheers! :D :D

One can easily prove that (a^2-ab+b^2)^2>= (a^4+b^4)/2. Thus, it suffices to show that (a^4+b^4)(c^4+d^4)>=a^2c^2+b^2d^2)^2, which is Cauchy

This reminds me of Arthur Engel, "Problem-Solving Strategies", problem 56 in chapter 7, where one has to prove the following inequality:

  a2b(a-b) + b2c(b-c) + c2a(c-a) >= 0,

where a, b, c are the sides of a triangle. A student - Bernhard Leeb - set a >= b and a >= c and reduced the inequality to the form

  a(b-c)2(b+c-a) + b(a-b)(a-c)(a+b-c) >= 0.

This is obvious for a triangle. But how on the earth did he find this form?

  Darij

[quote="Annika"]One can also kill the inequality by multiplying it  :D

2(a^2-ab+b^2)(c^2-cd+d^2)-a^2c^2-b^2d^2=
=a^2·c^2 - 2·a^2·c·d + 2·a^2·d^2 - 2·a·b·c^2 + 2·a·b·c·d - 2·a·b·d^2 + 2·b^2·c^2 - 2·b^2·c·d + b^2·d^2=
=(a - b)^2·(c - d)^2+(a·d - b·c)^2>=0

(But this would be a torture in any competition, I'm glad that I have my computer)[/quote]
$2(a^2-ab+b^2)(c^2-cd+d^2)-a^2c^2-b^2d^2$

$=a^2c^2 - 2a^2cd + 2a^2d^2 - 2abc^2 + 2abcd - 2abd^2 + 2b^2c^2 - 2b^2cd + b^2d^2$

$=(a - b)^2(c - d)^2+(ad - bc)^2\ge 0.$


Does anyone have an algebra 1 MAO Competition they could give me? I only need one...

And to make this post legit...

Let A be the x intercept of the equation
Let B be the vertex of this equation
Let C be the sum of the y intercepts of the equation
Let D be the product of the y intercepts if the equation

x=y^2+3y+1

Find A+B+C+D.

There is a [b]TON[/b] of stuff here. I'm not sure exactly how they managed to come up with that many problems.  :)

plugging in 0 for y gives us A=1

to find the y coordinate of the vertex, simply do $\frac{-b}{2a}$

$\frac{-6}{1} = -6 = B$

C and D, use the formula

sum of two roots = -b/a, product of two roots = c/a

C=$\frac{-3}{1} = -3$, D = $\frac{1}{1} = 1$

$A+B+C+D = 1 + -6 + -3 + 1 = -7$


Hi!
I want to be a mathematician.
I really love number theory even though I do not know much about it....^^
I would like to solve Riemann Hypothesis in near future.....^^;;
So, could you advise me on some topics for me to study
so that I could possibly(?) crack it??

Thanks

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=2049498914&t=90900

I have a feeling that might not be the first time this was asked.


I found a problem today and one friend told me that it can be solved by Lagrange interpolation formula. But i dont know this theorem and my friend didnt have the time to explain it to me. So, does anyone know what does this theorem say?

It's easy to prove that if given points $ (x_0,y_0), (x_1, y_1) \ldots (x_k,y_k)$ where no two $ x_i$ are equal, there is a unique polynomial of degree $ k$ passing through those points. Lagrange interpolation is nothing more than the way of calculating that polynomial.

Metaphorically it's done by beating the crap out of the polynomial with a sledge hammer until it does what you want. Basically you let the polynomial be the following expression
$ c_0 f_0(x) \plus{} c_1 f_1(x) \plus{} \ldots \plus{} c_k f_k(x)$
with $ c_i$ being constants and $ f_i$ being $ k$th degree polynomials, where $ f_i(x_j)$ is $ 0$ if $ i \neq j$ (1) and $ c_i f_i(x_i) \equal{} y_i$ for all $ i$ (2). These two conditions together ensure that this sum evaluates to $ y_i$ when $ x_i$ is plugged in because most of the terms vanish.

To get this, just let $ f_i(x)$ be the polynomial with all of $ x_0, x_1, \ldots x_k$ as roots (except $ x_i$). For example, let $ f_0(x) \equal{} (x\minus{}x_1)(x\minus{}x_2)\cdots (x\minus{}x_k)$. That takes care of (1), now just set $ c_i \equal{} y_i / f(x_i)$ to get (2). After simplifying all of this, you'll get the formula for Lagrange interpolation that is usually presented.

[quote="t0rajir0u"][url=http://mathworld.wolfram.com/LagrangeInterpolatingPolynomial.html]Mathematica[/url] has a decent statement.  See also the discussion [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=237669]here[/url].  What's the problem?[/quote]

The problem is:

@@Edit: Problem deleted...

P.S Please this problem is from an admission test, so it would be wise not to post your solution here. Thanks in advance :)

Oh.  If this problem's from an admission test, you probably shouldn't have posted it at all.


quadratics
advanced fields
advanced fields unsolved

Which prime numbers can be represented as $ x^{2}+xy+2y^{2}$ where $ x$ and $ y$ are integers?

$ p = x^{2}+xy+2y^{2}\implies$
$ 4p = 4x^{2}+4xy+8y^{2}= (2x+y)^{2}+7y^{2}\implies$
$ \left( \frac{2x+y}{y}\right)^{2}\equiv-7 \bmod (4p) \implies$

We know that $ -7 \equiv 1$ is a QR $ \bmod 4$, so it remains to find $ p$ such that $ -7$ is a QR $ \bmod p$.

Well,

$ \left( \frac{-7}{p}\right) = \left( \frac{-1}{p}\right) \left( \frac{7}{p}\right) = (-1)^{ \frac{p-1}{2}}\left( \frac{p}{7}\right) (-1)^{ \frac{3(p-1)}{2}}= \left( \frac{p}{7}\right)$

Since $ 1, 2, 4$ are the quadratic residues $ \bmod 7$, it's necessary that

$ p \equiv 1, 2, 4, 8, 9, 11 \bmod 14$

Sufficiency probably requires work in $ \mathbb{Z}[ \sqrt{-7}]$...

Well, $ \mathbb{Z}\left[\frac{1+\sqrt{-7}}{2}\right]$, to be precise :D

There is, in fact, another way to prove sufficiency. Let me give a sketch. It is not hard to check that $ \left(\frac{1,7}{p}\right) = 1$ for the values of $ p$ you mentioned. I am using the Hilbert symbol here - see Serre, a course in arithmetic. Noticing that $ x^{2}-xy+2y^{2}= \frac14\left(\left(2x+y\right)^{2}+7y^{2}\right)$ and invoking the Davenport-Cassels theorem then gives the result.


geometry
angle bisector
geometry proposed

Let L be a point on the side BC,K be a point on the side AB of $ \bigtriangleup ABC$ such that AL bisector of $ \measuredangle  BAC$,$ \measuredangle  ACK\equal{} \measuredangle   ABC$,$ \measuredangle  CLK\equal{} \measuredangle   BKC$.Prove that $ AC\equal{}KB$.  :?:

_________________________________
$ \textbf{Azerbaijan Land of Fire }$

[hide]Note that \[ \angle LKC\equal{}180^{\circ}\minus{}\angle KCL\minus{}\angle CLK\equal{}180^{\circ}\minus{}\angle KCB\minus{}\angle BKC\equal{}\angle KBC\equal{}\angle KCA\] hence $ KL\parallel AC$. Thus $ \frac{BK}{BA}\equal{}\frac{BL}{BC}\equal{}\frac{c}{b\plus{}c}$, which gives $ BK\equal{}\frac{c^2}{b\plus{}c}$. Also note that $ \angle ACK\angle ABC$ and $ \angle KAC\equal{}\angle BAC$, hence $ \triangle BAC\sim\triangle CAK$, so $ \frac{AK}{AC}\equal{}\frac{AC}{BA}$, hence $ AK\equal{}\frac{b^2}{c}$. But we have $ AK\plus{}BK\equal{}AB$, therefore $ \frac{b^2}c\plus{}\frac{c^2}{b\plus{}c}\equal{}c$. This gives $ b^3\plus{}b^2c\plus{}c^3\equal{}c^2b\plus{}c^3$, or $ b^2\plus{}bc\equal{}c^2$, or $ c^2\minus{}b^2\equal{}bc$. Thus $ BK\equal{}c\minus{}AK\equal{}c\minus{}\frac{b^2}c\equal{}\frac{c^2\minus{}b^2}{c}\equal{}\frac{bc}c\equal{}b\equal{}AC$, as desired.

Q.E.D. :D [/hide]

From the second given relation, we get $ \triangle KBC \sim  \triangle LKC$, i.e. $ \angle KBC \equal{} \angle CKL$; this one, with the first given relation, gives $ AC \parallel KL$, from which $ \frac{BL}{CL}\equal{}\frac{BK}{AK}$, but from $ AL$ angle bisector, $ \frac{BL}{CL}\equal{}\frac{AB}{AC}$, so $ \frac{BK}{AK}\equal{}\frac{AB}{AC}$ ( 1 ). From the first given relation, $ AC$ is tangent to the circle $ (BCK)$, i.e. $ AC^2\equal{} AK\cdot AB$ ( 2 ), from $ (1)$ and $ (2)$ getting $ BK \equal{} AC$.

Best regards,
sunken rock


It's our way of resolving issues that we simply can't resolve otherwise.  I mean, this is why the UN was created and all, but you still can't prevent war copmletely with all the difference between ideologies and governments around the world.

War is peace.
Freedom is slavery.
Ignorance is strength.

and big brother is watching you too, ccy.

I like the show Big Brother.  Sometimes I watch it after dinner.

they are talking about the book 1984 by George Orwell

[quote="wobster109"]why is it that human society, at this advanced stage, must still resolve political issues by pouring money and resources into shooting down others, when children are hungry and people are poor?[/quote]
Well you need to rephrase that a bit.  For instance when USA and the UK go to war for oil or whatever (to serve the wallet of their own population and especially their ruling class) it's usually kids in Congo, Iraq,... who are starving or sick.
Selfishness (of individuals or populations) is the main reason.  It's the same with all those bloody conflicts in Africa.  Many of them are power-hungry and greedy leaders who couldn't care less about the people in their own group or others groups.

[quote="wobster109"]why is it that human society, at this advanced stage, must still resolve political issues by pouring money and resources into shooting down others, when children are hungry and people are poor?[/quote]

Because humans are stupid. I think we discussed this in the topic about how the world will end.

[quote="deimos"][quote="wobster109"]why is it that human society, at this advanced stage, must still resolve political issues by pouring money and resources into shooting down others, when children are hungry and people are poor?[/quote]

there's no logic in war. but, humans are stupid (as mustafa has noticed). we are stupid. all of us. better world without war? there's no such place as long as humans exist.[/quote]
I can only speak from a country that hasn't seen a war since the second World War II (only my grandparents have seen it) but I don't think it's because of our species being a dumb species.  Aggression because of selfishness happens all the time among animals (some tiger mothers destroy the babies of another mother just to eliminate competition for their own), it's because of the intelligence of humans that we can take our conflicts from biting each other and throwing stones at each other to destroying entire cities...
I think many wars are fueled by a desire for security, land to live, water, food, oil,... either by such a desire of an entire group or a few corrupt leaders :(

true frederik. but only human would start a war with others just because they don't think the way he does (religion, nationality, political views) or they don't have the same colour of skin, hair, or they wear glasses.
or revenge (i believe that monkeys do that too. aren't they? bad evil monkeys.)

I think you know by now what I think of this, so I leave it for you to guess. Good luck guessing!


Is the ultimate goal to add the Contests section to the Wiki so that they can be completed more quickly? Or were they meant to be kept separate?

From the announcement topic: "Such information will be easier to update (since I won't have to update it all) and should be better organized so that students can find what they are looking for."

and also

"The truth is that nobody knows quite what will become of the AoPS wiki, but the goal is to create an online organized resource for students to which students (and teachers and anyone interested) can contribute. "

So we don't yet know what will ultimately happen, but due to the advantages of the wiki it may eventually come into more prominent use or entirely replace the resources section.

Though don't take my word for it, I'm just interpreting the announcement.

Definitely not. The contests section (also known as the Resources section on MathLinks skin) is not to be Wikipedized, just becase I do not want all the people to be able to edit it.


The sum of the first $ n$ terms in the infinite geometric sequence $ \left\{\frac{1}{4},\frac{1}{8},\frac{1}{16},\dots \right\}$ is $ \frac{63}{128}$. Find $ n$.

The formula for a finite geometric sequence is $ a_1 (\frac{1\minus{}r^n}{1\minus{}r})$, where $ a_1$ is the first term, $ r$ is the common ratio, and $ n$ is the number of terms. Therefore,

$ \frac{1}{4} \plus{} \frac{1}{8} \plus{} \frac{1}{16} \plus{} ... \equal{} \frac{1}{4} (\frac{1\minus{}\frac{1}{2^n}}{1\minus{}\frac{1}{2}}) \equal{} \frac{1}{2}(1 \minus{} \frac{1}{2^n})$

So we have that

$ \frac{1}{2} ( 1\minus{} \frac{1}{2^n}) \equal{} \frac{63}{128}$

$ 1\minus{} \frac{1}{2^n} \equal{} \frac{63}{64}$

$ \frac{1}{2^n} \equal{} \frac{1}{64}$

$ \therefore \boxed{n\equal{}6}$


integration
logarithms
calculus
calculus computations

$\int ln(1+x^2)dx$and then use integration by parts???

[quote="sky_sc"]how about
1. ln(a+b) = ln(a) . ln(b)

2. ln(1) treated as constant[/quote]
1. ln(a+b) = ln(a) . ln(b) does it work ?

no it wont work

the correct law is
log(xy) = log(x) + log(y)

[quote="Kent Merryfield"]First step: $\ln\left(1+\frac1{x^2}\right)=\ln(1+x^2)-2\ln x.$[/quote]
$\ln\left(1+\frac1{x^2}\right)=\ln(1+x^2)-2\ln |x|.$

integration by parts give
$\int \ln(1+\frac{1} {x^2})dx = xln(1+x^2)-2x+2arctan(1+x)-xln(x^2)+2x +C$


Sarah has four decks of 15 cards each.  One deck is red, one is
blue, one is yellow and one is green.  Each deck’s cards are
numbered 1 through 15.  Sara combines all four decks to form
one large deck.  When selected at random and without
replacement, what is the probability that Sarah will first choose a
red card with a prime number and then choose a card with a
non-prime number on her second selection?  Express your
answer as a common fraction.

The prime numbers are 2,3,5,7,11,13 (6 of them). In the entire deck of 60 cards, there are 6 red prime numbers, and choosing one is a probability of 6/60 or 1/10. There are 4(15-6)=36 non-prime numbers. The chance of drawing htat is 36/59. Thus, the total probability is 18/295.

Note: In the last step dragon mutliplies the probabilities together.


$ a_1, a_2, \cdots, a_n$ is a sequence of real numbers, and $ b_1, b_2, \cdots, b_n$ are real numbers that satisfy the condition $ 1 \ge b_1 \ge b_2 \ge \cdots \ge b_n \ge 0$. Prove that there exists a natural number $ k \le n$ that satisifes $ |a_1b_1 \plus{} a_2b_2 \plus{} \cdots \plus{} a_nb_n| \le |a_1 \plus{} a_2 \plus{} \cdots \plus{} a_k|$

$\sum_{k=1}^{n}a_kb_k=\sum_{i=1}^{n}A_i\lambda_{i}$.(By Abel's Summation formula)
where $b_{n+1}=0$; $A_i=\sum_{k=1}^{i}a_k$ and $\lambda_i=b_i-b_{i+1}$.

Now $\sum_{i=1}^{n}\lambda_i=b_1\le 1$.
So, take $b_1'=1$ and $\lambda_1'=b_1'-b_2$.
Fixing the sum above, $\sum_{i=1'}^{n}\lambda_i=b_1'=1$.

Hence, \[\left|\sum_{k=1}^{n}a_kb_k\right|=\left|\sum_{i=1}^{n}A_i\lambda_{i}\right|\le\sum_{i=1}^{n}\left|A_i\right|\lambda_{i}\le\sum_{i=1'}^{n}\left|A_i\right|\lambda_{i}\]
So, by PHP on weighted mean we get there exists some $1\le k\le n$ such that
$ |a_1b_1+a_2b_2+\cdots+a_nb_n|\le |a_1+a_2+\cdots+a_k| $.


algebra
polynomial
calculus
integration
algebra unsolved

let $A ={a_{1},...,a_{n}}$ is set include n distinct integer number.($n\geq 3$). suppose max element of $A$ is $M$ and min element of $A$ is $m$ .suppose there polynomial $P(x)$with integral coefficients st
1)$m<P(a)<M$for a be long to $A$
2) $P(m) <P(a)$ for a belong to $A$ and a distinct $m,M$
prove that $n\leq  5$

what do you think of it? :lol:

http://www.mathlinks.ro/Forum/viewtopic.php?t=42413

noone, I think it is hard,hope some wil solution

noone I belives some one will post solution :(


articles
geometry
inequalities
SFFT
special factorizations

Consider the article on [url=http://www.artofproblemsolving.com/Wiki/index.php/Factoring]Factoring[/url], whose first sentence is, "Factoring is an essential part of problem solving," or articles like those titled "area" and "imaginary unit" in which a note about how frequently the notion appears in contests is set within the first sentence or two.  For contrast, the pages on Cauchy-Schwarz Inequality and Simon's Favorite Factoring Trick contain similar notes, but have them substantially lower down the page, in sections with names like "applications" or "contest problem solving."

My inclination is to simply strike those sentences in the Factoring, Area and Imaginary unit pages.  What do you think?

I would favor striking those sentences.  Of course, if they contain important information, that information should still be incorporated into the article somehow.

I agree with striking them.  The explanation of a concept should have nothing to do with contest aspects.  Contest problems are often great ways to learn about those concepts, and as such can be mentioned in the bodies of such articles without taking away from the focus of the concepts themselves.


calculus
integration
trigonometry
logarithms
calculus computations

show :D

$ \int_{0}^{\pi}\;\;\frac{x\cdot\sin x}{\frac{5}{4}\;\minus{}\;\cos x}\;\;\textbf dx\;\;\equal{}\;\;\boxed{2\pi\cdot\ln\left(\frac{3}{2}\right)}$

$ \int_{0}^{\pi}\ln(1\minus{}2\alpha\cos x\plus{}\alpha^{2})\, dx \equal{} 0\quad\text{for}\quad 0 <\alpha < 1$

In particular for $ \alpha \equal{}\frac{1}{2}\qquad\int_{0}^{\pi}\ln\left(\frac{5}{4}\minus{}\cos x\right) dx \equal{} 0$

Maybe that was meant by kunny.

Use integration by parts

$ \int_{0}^{\pi}x\cdot\frac{\sin x}{\frac{5}{4}\minus{}\cos x}dx \equal{}\underbrace{\left[x\cdot\ln\left(\frac{5}{4}\minus{}\cos x\right)\right]_{0}^{\pi}}_{ \equal{}\pi\ln\left(\frac{9}{4}\right)}\minus{}\underbrace{\int_{0}^{\pi}\ln\left(\frac{5}{4}\minus{}\cos x\right) dx}_{ \equal{} 0}$

How would you prove Dr. Does integral identity?

For $ 0 <\rho < 1\quad\sum_{k \equal{} 1}^{\infty}\frac{(\varrho e^{i\varphi})^{k}}{k}\equal{}\minus{}\ln(1\minus{}\varrho e^{i\varphi}) \equal{}\minus{}\ln(1\minus{}\varrho\cos\varphi\minus{}i\varrho\sin\varphi)$

Compare the real parts: $ \sum_{k \equal{} 1}^{\infty}\frac{\varrho^{k}\cos k\varphi}{k}\equal{}\minus{}\ln\sqrt{(1\minus{}\varrho\cos\varphi)^{2}\plus{}(\varrho\sin\varphi)^{2}}\equal{}\minus{}\frac{1}{2}\ln(1\minus{}2\varrho\cos\varphi\plus{}\varrho^{2})$

and integrate from $ 0$ to $ \pi$ with respect to $ \varphi$ : $ \sum_{k \equal{} 1}^{\infty}\frac{\varrho^{k}}{k}\underbrace{\int_{0}^{\pi}\cos k\varphi\, d\varphi}_{ \equal{} 0}\equal{}\minus{}\frac{1}{2} I\Rightarrow I \equal{} 0$


Suppose $ (2\plus{}\sqrt{3})^{2r\minus{}1} \equal{} 1\plus{}m\plus{}n\sqrt{3}$, where m,n,r are positive integers. Then prove that m has an odd number of divisors.

[hide]
Let $ r\equal{}s\plus{}1$. Conjugating with respect to $ \sqrt{3}$, we get $ \left( 2\minus{}\sqrt{3}\right)^{2s\plus{}1}\equal{}1\plus{}m\minus{}n\sqrt{3}$ so that
\[ m\equal{}\frac{\left( 2\plus{}\sqrt{3}\right)^{2s\plus{}1}\plus{}\left( 2\minus{}\sqrt{3}\right)^{2s\plus{}1}\minus{}2}{2}\]
\[ \equal{}\frac{\left(2\plus{}\sqrt{3}\right)^{2s}\left( 1\plus{}\sqrt{3}\right)^2 \plus{}\left( 2\minus{}\sqrt{3}\right)^{2s}\left( 1\minus{}\sqrt{3}\right)^2\minus{}4}{4}\]
\[ \equal{}\left( \frac{\left( 1\plus{}\sqrt{3}\right) \left( 2\plus{}\sqrt{3}\right)^s\plus{}\left( 1\minus{}\sqrt{3}\right) \left( 2\minus{}\sqrt{3}\right)^s}{2}\right)^2\]
The element under the square is of the form $ \frac{\alpha \plus{}\overline{\alpha}}{2}$ where $ \alpha \equal{}a\plus{}b\sqrt{3}$ for integers $ a,b$. So $ m\equal{}a^2$. It then follows easily that $ m$ has an odd number of divisors, since each divisor less than $ a$ can be paired with $ \frac{m}{a}$.
[/hide]


logarithms
inequalities unsolved
inequalities

Let $ 0 < a < b$, for which interval will $ f(x) \equal{} \frac {b^{x} \minus{} a^{x}}{x((a \plus{} b)/2)^{x}}$, $ x\in ( \minus{} \infty, \plus{} \infty)$, increasing (or decreasing)?

It is easy to check that if $ x>1$, then $ f(x)$ is monotone increasing. What about other cases?

hell, you have to consider
$ f'(x)\equal{}\frac{b^x(x\ln(b)\minus{}1)\plus{}a^x(1\minus{}x\ln(a))\plus{}x\ln\left(\frac{a}{2}\plus{}\frac{b}{2}\right)(a^x\minus{}b^x)}{x^2\left(\frac{a}{2}\plus{}\frac{b}{2}\right)^x}$
Sonnhard.


There are  identical beads of color red,blue and green. there numbers respectively p, q and r.
(i)find number of circular arrangements taking all of them at a time.
(ii)find number of necklace which can be formed by taking all of them at a time.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=255829]Once[/url] was really sufficient ;)


superior algebra
superior algebra unsolved

Characterize those finite groups $G$ having the following property: whenever $a,b\in G$ have coprime orders, the order of $ab$ is the product of the orders of $a$ and $b$.

is there any other sol except nilpotent matrices?

nilpotent matrices doesn't matter, because they don't form a group!

But the finite nilpotent groups have the mentioned property.

Clearly, all commutative groups have the property.
Also, any group of order p^n trivially has this property.
=Uncool-


Six points D,E,F,G,H,K in that order, on a circle (O,R), such that $ DE\equal{}FG\equal{}HK\equal{}R$. KD meets EF at A; EF meets GH at B; GH meets KD at C. Prove that: $ OA.BC\equal{}OB.CA\equal{}OC.AB$

[quote="hotboy12"]Six points D,E,F,G,H,K in that order, on a circle (O,R), such that $ DE \equal{} FG \equal{} HK \equal{} R$. KD meets EF at A; EF meets GH at B; GH meets KD at C. Prove that: $ OA.BC \equal{} OB.CA \equal{} OC.AB$[/quote]
It is a nice problem!

I think you knew it is a M&Y problem. So, this topic should be stopped.
I will give you a suggest: just prove that three midpoints of EF,GH,KD form an equilateral triangle.
Good luck !
P/S: Next time, you should wait solutions of Math&Youth problems in this magazine.


[quote="harwi"]Could anybody please help ?
where can I find ebooks of Math in Chinese( Pdf or Djvu)?[/quote]
Why don't you google it? I think it may help.

Hey!
 try this :
http://www.docin.com/p-57563829.html
(All China MO problems 1986-2009)


What is the mean of the following list?

\[ 4x \plus{} 3, 7, 3x \minus{} 2, 9x\]

Well, the first part of finding the mean is adding the terms together.  We have:
\[ 4x \plus{} 3 \plus{} 7 \plus{} 3x \minus{} 2 \plus{} 9x \equal{} 16x \plus{} 8\]
Since there are $ 4$ terms, we simply divide by $ 4$.
\[ \frac {16x \plus{} 8}{4} \equal{} \boxed{4x \plus{} 2}\]


[quote="BornForMath"]Write the shortest sentence you can think of that contains the 26 alphabets.

 E.g. $\text{The quick brown fox jumps over the lazy dog. }$
$\text{ Jackdaws love my big sphinx of quartz. }$
[hide]
I will be very amazed if you know where the two sentences I wrote come from.  :D
[/hide][/quote]

The quick brown fox jumps over A lazy dog.  ;)

[quote="metsfan001"][quote="BornForMath"]Write the shortest sentence you can think of that contains the 26 alphabets.

 E.g. $\text{The quick brown fox jumps over the lazy dog. }$
$\text{ Jackdaws love my big sphinx of quartz. }$
[hide]
I will be very amazed if you know where the two sentences I wrote come from.  :D
[/hide][/quote]

Schott's Original Miscellany?[/quote]
LOL  :lol:

Another one:
Pack my box with five dozen liquor jugs.

[quote="stupidkid"][quote="BornForMath"]Write the shortest sentence you can think of that contains the 26 alphabets.

 E.g. $\text{The quick brown fox jumps over the lazy dog. }$
$\text{ Jackdaws love my big sphinx of quartz. }$
[hide]
I will be very amazed if you know where the two sentences I wrote come from.  :D
[/hide][/quote]
When you open font files, most of them display the first one, but I have also downloaded fonts that display the second one.[/quote]

YES!! Thats where i found the two sentences, stupidkid.  :D


I once had similar questions, so I'll include the solution to the following "problem":

Find the general solution of a linear second order differential equation \[a \cdot y^{\prime \prime}+b \cdot y^\prime+c \cdot y = 0 ,\] where $a,b,c$ are some constants with $a \neq 0$.

[hide="Solution."]
Let $r_{1,2}$ be the two roots of the equation $ax^{2}+bx+c = 0$ and take $y = f \left( x \right) \cdot \exp \left( r_{1}x \right)$.

We have \[y^\prime = \exp \left( r_{1}x \right) \left( f^\prime \left( x \right)+r_{1}f \left( x \right) \right) ,\] so \[y^{\prime \prime}= \exp \left( r_{1}x \right) \left( f^{\prime \prime}\left( x \right)+r_{1}f^\prime \left( x \right)+r_{1}f^\prime \left( x \right)+r_{1}^{2}f \left( x \right) \right) .\] Thus, \[a \cdot y^{\prime \prime}+b \cdot y^\prime+c \cdot y = \exp \left( r_{1}x \right) \left[ a f^{\prime \prime}\left( x \right)+\left( 2a r_{1}+b \right) f^\prime \left( x \right)+\left( a r_{1}^{2}+b r_{1}+c \right) f \left( x \right) \right] .\] Since $r_{1}$ is a root of $ax^{2}+bx+c = 0$, \[a \cdot y^{\prime \prime}+b \cdot y^\prime+c \cdot y = \exp \left( r_{1}x \right) \left[ a f^{\prime \prime}\left( x \right)+\left( 2a r_{1}+b \right) f^\prime \left( x \right) \right] .\] By Viete we have $r_{1}+r_{2}=-\frac{b}{a}$, so $2 r_{1}+\frac{b}{a}= 2 r_{1}-\left( r_{1}+r_{2}\right) = r_{1}-r_{2}$.

Consequently, \[f^{\prime \prime}\left( x \right)+\left( r_{1}-r_{2}\right) f^\prime \left( x \right) = 0 \, \, \, \Longleftrightarrow \, \, \, \left( f^\prime \right)^\prime \left( x \right)+\left( r_{1}-r_{2}\right) f^\prime \left( x \right) = 0 .\] If $r_{1}\neq r_{2}$, this is a first $ODE$ and has the solution \[f^\prime \left( x \right) = \mathcal C \exp \left( \left( r_{2}-r_{1}\right) x \right) ,\] so \[f \left( x \right) = \frac{\mathcal C}{r_{2}-r_{1}}\exp \left( \left( r_{2}-r_{1}\right) x \right)+\mathcal A .\] Thus, \[y = \frac{\mathcal C}{r_{2}-r_{1}}\exp \left( r_{2}x \right)+\mathcal A \exp \left( r_{1}x \right) ,\] which can be rewritten as \[y = \mathcal A \exp \left( r_{1}x \right)+\mathcal B \exp \left( r_{2}x \right) .\] If $r_{1}= r_{2}$, then $f^{\prime \prime}\left( x \right) = 0$ yields $f \left( x \right) = \mathcal A x+\mathcal B$. Hence, \[y = \left( \mathcal A x+\mathcal B \right) e^{r_{1}x}.\] [/hide]

[quote="darkmaster"]I have just begun studying differential equations. And I have problem with this one:
  $y^{,,}-t^{2}y=0$. I know the solution is $y= ae^{tx}+be^{-tx}$ but I can't prove it. Help me, please.[/quote]
$\frac{d^{2}y}{dx^{2}}-t^{2}y=(a{t}^{2}{e^{tx}}+b{t}^{2}{e^{-tx}})-t^{2}(a{e^{tx}}+b{e^{-tx}) =0}$

Merci, perfect_radio. This is perfect! But I still have a question: Why do you know $r_{1}et r_{2}$ are two roots of the quadradric equation, which is the main idea of this problem. Do you have some method to know this?


Find the least positive integer $ n$ satisfies the following conditions: For any sequence of $ n$ positive integer terms $ a_{1}$, $ a_{2}$, $ \ldots$, $ a_{n}$ with $ \sum_{i=1}^{2007}a_{i}= 2007$, there exist some consecutive terms in them with sum $ 30$.

I assume it's supposed to be $ \sum_{i =1}^{n}a_{n}$, right?

$ 1018$ would be the answer.

when $ n \equal{} 1017$, (1,1,1,......,1,31,1,1,1,.....,31,1,1,1.......,31,1,1,1,1,.....,1) would not satisfy the condition.
(29 1's and 1 31 appear in turns and in the last, 27 1's appear)

now let's see the case of $ n \equal{} 1018$
let's put $ S_k \equal{} \displaystyle\sum_{i \equal{} 1}^{k}a_i$
if we suppose there is no consecutive terms in them with sum 30, we can know that there is no $ i, j$ such that $ S_i \minus{} S_j \equal{} 30$
therefore, $ S_1, S_2, ....., S_(1018), S_1 \plus{} 30, S_2 \plus{} 30, ......, S_(1018) \plus{} 30$ are different each other.
let's put $ A \equal{} {S_i}, B \equal{} {S_i \plus{} 30}$. then $ A \cap B \equal{} \phi$
we know that $ S_i$ and $ S_i \plus{} 30$ are lower than 2007+30=2037...
it is easy to check 30 doesn't belong to either $ A$ or $ B$
it mentions that all the integer from 1 to 2037 except 30 belong to $ A$ or $ B$, not both.
and because $ S_i$ is lower than 2008, 2008~2037 belong to $ B$
for similar reason, 1~29 belong to $ A$
however, it is also easy to chsck if $ a$ belongs to $ A$($ B$), $ a \plus{} 60$ belongs to $ A$($ B$).
so $ 60k \plus{} 28$ must belong to $ A$. it means 2008 must belong to $ A$, which is contradiction.


inequalities
geometry
circumcircle
incenter
Euler
inequalities proposed

Let $O$ be the circumcenter and $G$ be the centroid of a triangle $ABC$. If $R$ and $r$ are the circumcenter and incenter of the triangle, respectively,
prove that \[ OG \leq \sqrt{ R ( R - 2r ) } . \]
[i]Greece[/i]

Euler`s relation in a triangle gives us $OI^2=R(R-2r)$ then all we have to do is prove that $OG \leq OI$ .And this seems like a well-known inequality(it looks too beautifull).

well we do use that
\[ OG^2=R^2- \frac{1}{9}(a^2+b^2+c^2) \]
But
 \[ R(R-2r)=R^2-2Rr=R^2-2\frac{2S}{p} \cdot \frac{abc}{4S}=R^2- \frac{abc}{a+b+c} \]
So we need to demonstrate that
       \[ \frac{1}{9}(a^2+b^2+c^2) \geq \frac{abc}{a+b+c} \]
...

...

[quote="fuzzylogic"]$OG^2 = R^2 - \frac{1}{9}(a^2+b^2+c^2)$

$R(R-2r) = R^2-2Rr = R^2 - \frac{2R}{s}\cdot \frac{abc}{4R} =R^2 - \frac{abc}{a+b+c}$ since $rs = \frac{abc}{4R}$ is the area of the triangle.

So it reduces to prove:

$(a^2+b^2+c^2)(a+b+c) \geq 9abc$ which is true by AM-GM.[/quote]

 :roll:

[quote="tim1234133"]Yes, it is a (fairly) well-known fact that $I$ lies in the circle diameter GH,
 and with a diagram showing the Euler line and this circle,
$OG \leq OI$ becomes immediate[/quote]
What is the meaning of "$I$ lies in the circle diameter GH"
Could you explain more clearly,thank you. :lol:

On the Euler line,

[b]O[/b] (the circumcentre) lies at one end.
[b]H[/b] (the orthocentre) lies at the other.
[b]G[/b] (the centroid) lies one third of the way along, nearer O.
[b]N[/b] (the nine-point centre) lies half-way between O and H.

Imagine the point halfway between G and H (and so two-thirds of the way from O to H). If a circle is drawn, with a centre at this point, through G and H, then GH is its diameter.

It is known that the incentre ([b]I[/b]) lies within this circle. (See [url=http://mathworld.wolfram.com/OrthocentroidalCircle.html]here[/url] for example).
Clearly, [b]I[/b] lies on the diameter GH if and only if the triangle is isoceles. As [b]I[/b] tends towards the nine-point centre, the triangle tends towards being equilateral. For more on the this circle, and the points that lie within it, see [url=http://forumgeom.fau.edu/FG2006volume6/FG200607index.html]here[/url]

Moving back to the actual question, it is clear, since OGH is a diamater of the circle, G is the point on the circle that is closest to O. (Draw a picture of this circle and the Euler line- it should look like a lollipop).
Therefore $OI \geq OG$

[color=darkblue][b]Remark.[/b] It is well-known the $\triangle ABC$-  inequality $p^{2}\ge 3\sum (p-b)(p-c)$ $\Longleftrightarrow$ $p^{2}\ge 3r(4R+r)$. The proposed inequality $OG\le OI$ is equivalently with the inequality $p^{2}\ge 3\sum (p-b)(p-c)+r(R-2r)$ $\Longleftrightarrow$ $p^{2}\ge 13Rr+r^{2}$ what is stronger than the first.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=53811[/color]

[quote=Feridimo]https://artofproblemsolving.com/community/c1049667h1991756_distance_between_circumcenter_and_center
$ OG\leq R\leq \sqrt{R(R-2r)} \implies 2r\leq R $[/quote]

are you sure?


limit
probability
geometry
geometric transformation
homothety

A finite sequence of three-digit integers has the property that the tens and units digits of each term are, respectively, the hundreds and tens digits of the next term, and the tens and units digits of the last term are, respectively, the hundreds and tens digits of the first term. For example, such a sequence might begin with the terms 	$ 247,$ $ 275,$ and $ 756$ and end with the term $ 824.$ Let $ \mathcal{S}$ be the sum of all the terms in the sequence. What is the largest prime factor that always divides $ \mathcal{S}?$

$ \textbf{(A)}\ 3 \qquad \textbf{(B)}\ 7 \qquad \textbf{(C)}\ 13 \qquad \textbf{(D)}\ 37 \qquad \textbf{(E)}\ 43$

[hide="Solution"]The sequence can be represented most simply with three terms.

Without loss of generality, look at it like this:

123, 231, 312, and it would loop back.

You can write this WLOG as
$100a+10b+c+100b+10c+a+100c+10a+b$
$111a+111b+111c$
$111(a+b+c)$
Because 111 is $37 \cdot 3$,
The largest prime that divides it is choice $\boxed{D}$
[/hide]

Without loss of generality.

It means that I can generalize my results to all such sequences, because they can all be represented in the same way. The proof is trivial.

[quote="ckck"][quote="DiscreetFourierTransform"][hide="Solution"]The sequence can be represented most simply with three terms.

Without loss of generality, look at it like this:

123, 231, 312, and it would loop back.

You can write this WLOG as
$100a+10b+c+100b+10c+a+100c+10a+b$
$111a+111b+111c$
$111(a+b+c)$
Because 111 is $37 \cdot 3$,
The largest prime that divides it WLOG is choice $\boxed{D}$
[/hide][/quote]
What's WLOG?[/quote]

Without loss of generality.  That means you are going to set things equal to things that are not given in the original problem, but you won't set any new conditions that will affect the result.

EDIT: Discrete beat me to it!

[quote="4everwise"]A finite sequence of three-digit integers has the property that the tens and units digits of each term are, respectively, the hundreds and tens digits of the next term, and the tens and units digits of the last term are, respectively, the hundreds and tens digits of the first term. For example, such a sequence might begin with the terms 	$247,$ $\boxed{\boxed{275}},$ and $756$ and end with the term $824.$ Let $\mathcal{S}$ be the sum of all the terms in the sequence. What is the largest prime factor that always divides $\mathcal{S}?$

$\text{(A)}\ 3 \qquad \text{(B)}\ 7 \qquad \text{(C)}\ 13 \qquad \text{(D)}\ 37 \qquad \text{(E)}\ 43$[/quote]
I think that's 475.  :P

[hide="The power of the trivial case"] The trivial case is the sequence consisting of the single term $111$.  The answer must therefore either be $3$ or $37$. [/hide]

I can't believe I wasted like 5 minutes on this.. trying to provide a counter example -.-

Why a counter example?...
There is a $\lim_{x\to 0^{+}}x$ probability of a question being wrong on an AMC...

I liked this problem because the teacher had just called two mins left on the 10A and I had this one (22) as well as 21,23 left and all three problems were quick "glance at problems"... homothecy quick one for 21 then boom, 111$\to$37 on this one, then a couple second on the next one after counting factors of 24 (basically counting 1,2,3,4 then realizing, Done!)

None of that is needed
We see the example given in the question- try that sequence and use the choices.
 Now we see that 3 and 37 satisfy.
If we take the any other sequence we see that all satisfy 37
Answer [b]37[/b]

$f(x) = \begin{array}{llll}x \mbox{dla}\-1 < x < 1 \end{array}$

and incredibly easy solution is to take a repeating number, such as 111, and test which of the given primes will divide.  In this case, it is 3 and 37.  Because 37 is larger, D is the answer

Don't revive old topics, especially not to repeat exactly ehat others have already said.

[quote="ckck"][quote="IntellectMage"][quote="4everwise"]A finite sequence of three-digit integers has the property that the tens and units digits of each term are, respectively, the hundreds and tens digits of the next term, and the tens and units digits of the last term are, respectively, the hundreds and tens digits of the first term. For example, such a sequence might begin with the terms 	[u][i][b]247, 275,[/b][/i][/u] and 756 and end with the term 824.

$ \text{(A)}\ 3 \qquad \text{(B)}\ 7 \qquad \text{(C)}\ 13 \qquad \text{(D)}\ 37 \qquad \text{(E)}\ 43$[/quote]

Don't these two numbers violate the sequence? In 247, the tens and units digit are 4 and 7 respectively; but in 275, the hundreds digit and tens digit are 2 and 7...[/quote]
It was a mistype. And please don't revive very old topics.[/quote]

Sorry about reviving the old topic, but what I am supposed to do if I have a question about this problem? Also, if that's a mistype is there any where I can get the correct printing of the problem?

[quote="IntellectMage"][/quote][quote="ckck"][quote="IntellectMage"][quote="4everwise"]A finite sequence of three-digit integers has the property that the tens and units digits of each term are, respectively, the hundreds and tens digits of the next term, and the tens and units digits of the last term are, respectively, the hundreds and tens digits of the first term. For example, such a sequence might begin with the terms 	[u][i][b]247, 275,[/b][/i][/u] and 756 and end with the term 824.

$ \text{(A)}\ 3 \qquad \text{(B)}\ 7 \qquad \text{(C)}\ 13 \qquad \text{(D)}\ 37 \qquad \text{(E)}\ 43$[/quote]

Don't these two numbers violate the sequence? In 247, the tens and units digit are 4 and 7 respectively; but in 275, the hundreds digit and tens digit are 2 and 7...[/quote]
It was a mistype. And please don't revive very old topics.[/quote][quote="IntellectMage"]

Sorry about reviving the old topic, but what I am supposed to do if I have a question about this problem? Also, if that's a mistype is there any where I can get the correct printing of the problem?[/quote]

http://www.artofproblemsolving.com/Wiki/index.php/2007_AMC_12A_Problems/Problem_11


algebra
polynomial
Vieta
quadratics
number theory
IMO Shortlist

Prove that for any positive integer $ m$ there exist an infinite number of pairs of integers $ (x, y)$ such that

[i](i)[/i] $ x$ and $ y$ are relatively prime;
[i](ii)[/i] $ y$ divides $ x^2 \plus{} m$;
[i](iii)[/i] $ x$ divides $ y^2 \plus{} m.$
[i](iv)[/i] $ x \plus{} y \leq m \plus{} 1\minus{}$ (optional condition)

The last condition confuses me  :huh: If we have infinite pairs $ (x,y)$, then we simply make $ x$ and $ y$ negative so that $ x \plus{} y\le m \plus{} 1$. (I assume $ x$ and $ y$ are allowed to be negative, as otherwise we couldn't possibly have infinite pairs with that condition?)

Anyway, just using conditions (i), (ii), and (iii),

[hide]
Suppose

$ x^2 \plus{} y^2 \plus{} m \equal{} (m \plus{} 2)xy$ (1).

with $ \gcd(x,y) \equal{} 1$. Then

$ x|x^2 \plus{} y^2 \plus{} m\implies x|y^2 \plus{} m$ and $ y|x^2 \plus{} y^2 \plus{} m\implies y|x^2 \plus{} m$.

Therefore it suffices to find infinitely many solutions to (1). Clearly we have one solution $ (1,1)$. Now we use the standard technique of Vieta Jumping. Viewing (1) as a quadratic in $ x$:

$ x^2 \minus{} ((m \plus{} 2)y)x \plus{} (y^2 \plus{} m) \equal{} 0$.

If $ (x,y)$ is a solution (and assume without loss of generality that $ x\le y$) then we know that $ (x',y)$ is also a solution where

$ x' \equal{} (m \plus{} 2)y \minus{} x \ge (m \plus{} 1)y > y$

So given a solution $ (x,y)$ with $ x \le y$, we can find a bigger solution $ (x',y)$ with $ x' > y$, thus there are an infinite number of solutions.

Also note that all pairs $ (x,y)$ generated in this manner do indeed have $ \gcd(x,y) \equal{} 1$. If we assume that $ \gcd(x,y) \equal{} 1$, then $ \gcd(x',y) \equal{} \gcd((m \plus{} 2)y \minus{} x,y) \equal{} \gcd(x,y) \equal{} 1$.
[/hide]

[quote=AnonymousBunny]I claim that there exist infinitely many integers such that $x \mid y^2+m$ and $y \mid x^2+m.$ Indeed, if $(x,y)$ works, so does $(x, kx-y),$ where $k=m+2.$ Starting from the initial pair $(1,1),$ this generates infinitely many such coprime integers because $\gcd (kx-y, x)=1$ whenever $\gcd (x, y) = 1.$[/quote]

For any future visitor—this is just flat out wrong.  A counterexample is $(1, 2)$ with $m=3$.  $(2,9)$ doesn't work for obvious reasons.  Counterexamples exist whenever $m + 1$ is not prime.  I have no idea why this post has three upvotes lol


algebra
polynomial
Gauss
combinatorial geometry
algebra proposed

Hope they won't give this in TST. THis would be the 5-th problem already discussed on the forum and given in the TST:
  Find the least k with the property: if $f$ is a polynomial such that $ |f(z)|\leq 1$  if $|z|\leq 1$ and if all roots of $ (1+f(z))^n=1+f^n(z)$ are of modulus at most 1, then n is at most k.

We will prove that for $n\geq 8$ the equation $(1+f(z))^n=1+f(z)^n$ has a root $> 1$ if $|f(z)|\leq 1$ for all $|z|\leq 1$. Indeed, it sufficies to show that $(1+z)^n=1+z^n$ has a root $z_0$ s.t. $|z_0|>1$, then $f(z)=z_0$ has a root $z_1$ and it is necessary $|z_1|>1$!

So, consider $P(z)=(1+z)^n-(1+z^n)$. Due to Gauss-Lucas theorem roots of $P'(z)$ lies inside of convex hull of roots of $P(z)$, but $P'(z)=n(1+z)^{n-1}-nz^{n-1}$ has a root $q=\frac{1}{e^{2\pi i/(n-1)}-1}$, and it is easy to see that $|q|>1$ for $n\geq 8$. It means that $P(z)=0$ has a root outside of $|z|\leq 1$.

For $n=1,2,...,7$ you can [i]easily[/i] manually check, that all roots of $P(z)=0$ lies in $|z|\leq 1$.


quadratics
number theory proposed
number theory

For arbitrary positive integer $ n$, prove that there exist prime number $ p$ such that $ n$ is quadratic residue modulo $ p$

I think it can be proved by CRT
( if $ n \equal{} 2^s p_{1}^{e_i}...p_{k}^{e_k}$, we can assume that $ p \equal{} 1(mod8) ,p \equal{} 1 (mod p_i)$
but how can we prove existence of such prime number p??) :blush:

For some arbitrary $ x$, just choose $ p$ such that $ p|n\minus{}x^2$, then $ n\equiv x^2\mod p$ so $ n$ is a quadratic residue.


When I am hungry, I can often relieve the hunger by playing music. If I play for a while (like 20 or 30 minutes), usually I don't get hungry until it would be time for my next regular meal.

This seems REALLY weird. :what?:

Does this ever happen to anyone else on here??

[quote="karthik_k"]Once in a while it happens to me as well. :)
But not always ..... occassionally , I get more hungry after I sing...
 :D[/quote]

ditto!!!

Yes, that arrived to me several times.  Maybe when I work my violin, I do not feel the hunger to surely arrive (because of the concentration)and too, when i am very hungry, i made of the violin, that occupies and one does not see spending time (but also me with a piece of 24  pages hum...).
It's weird, but it's simply ( i think) because we are "so impassioned" and concentrate that play instrument relieve the hunger.

apparently, $H\\"andel$ did not eat for days whilst composing his "Messiah"

[quote="rookworm"]apparently, $H"andel$ did not eat for days whilst composing his "Messiah"[/quote]

The oratorio? That's really interesting... He must have basically rushed the last part to go eat  :roll:

And the dissapearing hunger doesn't really work for me, as you use your abdomen muscles to play clarinet... though I do lose track of time while practicing...

thats normal... it happens to me with computer :)


I just entered my first A.P. class this Senior year of High School; A.P. Calculus AB.

Our first assignment is obviously just review; however I seem to be stuck on this one problem:

|(x-3)/2| >= 5... I don't remember the rule for this type of problem.

Is it x>=a or x<=-a?

I got x>=28 or x<=-22; which is not right (checked answer section).

Well, you want $\frac{x-3}{2} \ge 5$ and $\frac{x-3}{2} \le -5$. That should give you the solutions you want.

:rotfl: Oh; man I need some more sleep; I did everything right; I just re-wrote the problem down wrong. I put (x-3)/5... Thanks.  :blush:


I'm trying to write a goog program to find the prime factorization of a number? Does anyone have one?

Do you prefer speed or simplicity? Do you want to be able to understand the program easily, or do you want it to be as fast as possible?

I have a program PRIMEFAC that does that, but it requires a free list, and it's not easy to understand.

Well George Silvis has a program for the smallest prime factor of a number, you write a program using that repeatedly.

I only want speed-this is for MC-my coach asked me to try and write/find one. If you coulod post that one it would be great.

using progs during MC competition:

since ur using a TI-84+ then heres what id do:

creat a backup program (with differnet name)

i spent weeks developing forumales into a prog and i went into the competition and my RAM cleared when i was entering something.

so creat a backup prog and archive that...so u can save ur hard work!


function
induction
algebra proposed
algebra

Find all real-valued functions $ f(x)$ on the rationals such that:
(1) $ f(x \plus{} y) \minus{} y f(x) \minus{} x f(y) \equal{} f(x) f(y) \minus{} x \minus{} y \plus{} xy$, for all $ x, y$
(2) $ f(x) \equal{} 2 f(x\plus{}1) \plus{} 2 \plus{} x$, for all $ x$ and
(3) $ f(1) \plus{} 1 > 0$.

I think posted before.
Consider $ g(x)\equal{}f(x)\plus{}x$
So $ g(x\plus{}y)\equal{}g(x)g(y)$ and $ g(x)\equal{}2g(x\plus{}1)$
Induction we have $ g(x)\equal{}\frac{1}{2^x}$


P is any point on a hyperbola ,one focus of which is  S and the corresponding directrix  denoted by D.

Lines through P parallel to the asymptotes meet D in points M and N.

Show that P is the centre of the circle through S,M and N

Do this parametrically (here we are going to take focus as $+ae,0$... of course, the negative case can be done in the same manner):

Let P be the point $(asec\theta, btan\theta)$,
then,

$|PS|=|\sqrt{(asec\theta-ae)^2+b^2tan^2\theta}|=|\sqrt{a^2sec^2\theta-2a^2esec\theta+a^2e^2+b^2sec^2\theta-b^2}|$
$=|\sqrt{a^2sec^2\theta-2a^2esec\theta+a^2+b^2+b^2sec^2\theta-b^2}|$
now using $b^2=a^2(e^2-1)$,
$=|\sqrt{a^2e^2sec^2\theta + a^2 -2a^2esec\theta|}$
$=|a(esec\theta-1)| ... (1)$

Using Point Gradient formula (we are going to take the gradient as $\frac{b}{a}$ - the negative case is done in the same manner -  and remembering that the directrix is $x=\frac{a}{e}$), we can determine M as,

$(\frac{a}{e}, \frac{b}{e}-bsec\theta+btan\theta)$,
so, $|MP|=|\sqrt{(\frac{a}{e}-asec\theta)^2+(\frac{b}{e}-bsec\theta)^2}|$
$=|(\frac1e-sec\theta)\sqrt{a^2+b^2}|=|(\frac1e-sec\theta)ae|=|a(1-esec\theta)|$

Therefore, |MP|=|PS|... this can then be repeated for |NP|.

Of course, since |MP|=|PS|=|NP|, then P is the centre of that circle with M, S, N on its circumference.

(P.S. can this be done using the focus-directrix property?)


function
inequalities
inequalities unsolved

Let $x,y,z$ positive real numbers. Determine the maximum value of $k$ such that:

$\frac{x}{\sqrt{y+z}}+\frac{y}{\sqrt{x+z}}+\frac{z}{\sqrt{x+y}}\geq  k\sqrt{x+y+z}$

We will prove that the best value of $k$ is $\sqrt{\frac32}$. Assuming that $x+y+z=1$, we get
$\sum \frac{x}{\sqrt{1-x}}\geqslant \sqrt{\frac32}$.
Since the function $f(t)= \frac{t}{\sqrt{1-t}}$ is convex, by Jensen:
$\sum\frac{x}{\sqrt{1-x}}\geqslant 3\cdot \frac{\frac13}{\sqrt{\frac23}}= \sqrt{\frac{3}{2}}$.
The equality holds iff $x=y=z$.


Let $ x,y,z$ be positive reals such that $ x\plus{}y\plus{}z\equal{}1$. Prove that

$ \frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z} \geq 24(x^2\plus{}y^2\plus{}z^2)\plus{}1$

[quote="4865550150"]Let $ x,y,z$ be positive reals such that $ x \plus{} y \plus{} z \equal{} 1$. Prove that

$ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \geq 24(x^2 \plus{} y^2 \plus{} z^2) \plus{} 1$[/quote]
Let $ x \plus{} y \plus{} z \equal{} 3u,$ $ xy \plus{} xz \plus{} yz \equal{} 3v^2,$ $ xyz \equal{} w^3$ and $ u^2 \equal{} tv^2.$
Hence, $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \geq 24(x^2 \plus{} y^2 \plus{} z^2) \plus{} 1\Leftrightarrow9u^3v^2\geq(25u^2 \minus{} 16v^2)w^3.$
But $ (x\minus{}y)^2(x\minus{}z)^2(y\minus{}z)^2\geq0$ gives $ w^3\leq3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}.$
Thus, it remains to prove that
$ 9u^3v^2\geq(25u^2 \minus{} 16v^2)\left(3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}\right).$
But $ 9u^3v^2\geq(25u^2 \minus{} 16v^2)\left(3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}\right)\Leftrightarrow$
$ \Leftrightarrow9t\geq(25t \minus{} 16)\left(3 \minus{} 2t \plus{} 2\sqrt {\frac {(t \minus{} 1)^3}{t}}\right)\Leftrightarrow(t \minus{} 1)^2(15t \minus{} 16)^2\geq0.$ :)

[quote="4865550150"]Let $ x,y,z$ be positive reals such that $ x \plus{} y \plus{} z \equal{} 1$. Prove that

$ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \geq 24(x^2 \plus{} y^2 \plus{} z^2) \plus{} 1$[/quote]

This inequality is equivalent to

$ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \plus{} 48(xy \plus{} yz \plus{} zx) \geq 25$

which was Kvant's, problem 2002. Indeed, Kvant problem formulated as

                   $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \geq \frac {25}{1 \plus{} 48xyz}$

Besides solution provided by arqady, we can use two other methods to solve it  1) Mixing variables 2) SOS.

Not that 48 is the best constant for the inequality of the form $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \plus{} k(xy \plus{} yz \plus{} zx) \geq 9 \plus{} k/3$


geometry
3D geometry
tetrahedron
ratio
vector
similar triangles

(Czech and Slovak Olympiad, 1999, number 2):

Define the median of a tetrahedron to be the line connecting a vertex to the centroid of the opposite face.  In tetrahedron ABCD, denote by E and F the midpoints of the medians from vertices A and D, respectively.  Determine with proof the ratio of the volumes of tetrahedrons BCEF and ABCD.

Let the altitude from D to BC meet BC at M. Then

[ABCD] = (1/3)(BC)(DM)h
[BCEF] = (1/3)(BC)(1/3*DM)(1/3*h) = 1/9*[ABCD]

(Its 1/3*h due to similar triangles)

Wow, I don't follow that at all.  First of all, what's h?  Second of all, shouldn't there be a 1/2 in the first expression?  Third of all, how do you get the second expression?

Yeah, I forgot the 1/2..though it doesn't affect the answer :)

Let the altitude from D to BC meet BC at M, and let the altitude from A and E to triangle BCD meet triangle BCD at N and P, respectively. Then

[ABCD] = (1/6)(BC)(DM)(AN)
[BCEF] = (1/6)(BC)(EM)(FP)

EM = 1/3*DM since medians divide one another into ratios of 2:1. If we let Q be the foot of the altitude from A to BC, we can see that triangles QAN and QFP are similar by AA similarity.

[BCEF] = (1/6)(BC)(1/3*DM)(1/3*AN) = 1/9*[ABCD]

I don't see the significance of EM or FP -- they aren't special lines of any sort -- not perpendicular to anything as far as I can tell, not special in any other notable way.  Did you mix up letters?  (For the record, my answer doesn't agree with yours, although I don't have a solution, so it is possible I did something wrong.)

blah, sorry, my mistake, for some reason I had used the assumption that one point was directly above another in my proof. poo, i shoulda known that the problem wouldn't be in the advanced section if it was this easy :D

I got 1/9 too, but for a different reason. I used vectors. In the following proof, . is the dot product and x is the cross product. Let MAB represent the midpoint of segment AB.

First, I dug up something from Mathworld: tetrahedron XYZW's volume is given by 1/6 * XY . (XZ x XW).

Then, tetrahedron ABCD's volume is

VABCD
= 1/6 * BC . (BA x BD)
= 1/6 * BC . [(BC + CA) x (BC + CD)]
= 1/6 * BC . [BC x (BC + CD + CA) + CA x CD]

BC . [BC x anything] = 0 (BC x anything will be perpendicular to BC), so

VABCD
[b]= 1/6 * BC . (CA x CD)[/b]

VBCEF
= 1/6 * BC . (BE x BF)
= 1/6 * BC . [(2/3 * BMCD) x (2/3 * BMAC)]
= 2/27 * BC . [BMCD x BMAC]
= 2/27 * BC . [(BC + 1/2 * CD) x (BC + 1/2 * CA)]
= 2/27 * BC . [BC x (BC + 1/2 * CD + 1/2 * CA) + 1/4 * CD x CA]
= 2/27 * BC . (1/4 * CD x CA)
[b]= 1/54 * BC . (CD x CA)[/b]

Thus VABCD : VBCEF = 9 : 1.

Still working through it ... one thing: you could have gotten to [quote="Osiris"]VABCD
[b]= 1/6 * BC . (CA x CD)[/b][/quote]
simply by applying the formula to vertex C instead of vertex B.  I got a different answer, but I'm still working on yours.  This post will be updated later.

Later edit:  how do you get
VBCEF = 1/6 * BC . [(2/3 * BMCD) x (2/3 * BMAC)]
from
VBCEF = 1/6 * BC . (BE x BF) ?  I don't see it.

Something [b]is[/b] screwed up in my solution though. I forgot to notice that for vectors [b]a[/b] and [b]b[/b], a x b = -(b x a). (That's apparent when you use the right hand rule). So for my solution, the ratio turns out to be negative? ...

E is the centroid of triangle BCD, which means that BMCD passes through E.

***IMPORTANT***

SCRAP EVERYTHING I HAVE WRITTEN; IT IS COMPLETE RUBBISH. FOR SOME STRANGE REASON I THOUGHT E AND F WERE THE CENTROIDS.

I'll try to fix the solution (not abandoning the vector approach) but it might take a while. A good while.

Let's try this again:

Recall line division by vectors: if segment XY is divided into XZ and ZY such that XZ:ZY = m:n, then for any point in the space O, OZ = n / (m+n) * OX + m / (m+n) * OY.

VABCD = 1/6 * BC . (BA x BD)

VBCEF
= 1/6 * BC . (BE x BF)
= 1/6 * BC . [(1/2 * BA + 1/2 * BR) x (1/2 * BD + 1/2 * BS)]
= 1/24 * BC . [(BA + BR) x (BD + BS)]
= 1/24 * BC . [(BA + 1/3 * BC + 1/3 * BD) x (BD + 1/3 * BC + 1/3 * BA)]

Expanding it will yield a [b]lot[/b] of terms. However, since BC . (BC x anything) = 0 and BA x kBA = 0 where k is some constant anyway,

VBCEF
= 1/24 * BC . (BA x BD + 1/9 * BD x BA)
= 1/24 * BC . (8/9 * BA x BD)
= 1/27 * BC . (BA x BD)

So VABCD : VBCEF = 9 : 2.

Is this right? ~_~

It looks good to me.  Here's my method:

V(ABCD) = 2V(ABCF), since by similar triangles the height from F is exactly half the height from D, while the two bases are the same.  So we need to find V(BCEF) as a fraction of V(ABCF), so we need to find the ratio of d(A, BCF) to d(E, BCF) (since they will still have the same bases).  To do this, we'll use mass-point geometry.  (This part isn't all that different from the vector part of your proof.) Let:
A = (1, 0, 0, 0)
B = (0, 1, 0, 0)
C = (0, 0, 1, 0)
D = (0, 0, 0, 1)
Let R be the centroid of triangle BCD and S be the centroid of triangle ABC.  Then R = (0, 1/3, 1/3, 1/3) and S = (1/3, 1/3, 1/3, 0).  Then E = (1/2, 1/6, 1/6, 1/6) and F = (1/6, 1/6, 1/6, 1/2).
BC can be parametrized as (0, t, 1 - t, 0) for real t.  Thus BCF can be paramatrized by (s/6,  s/6 + t(1 - s), s/6 + (1 - t)(1 - s), s/2) while AE can be parametrized by ((1 - u) + u/2, u/6, u/6, u/6).  We wish to find the value of u at the intersection of the plane and line.

Setting components equal, we see that we must have (1 - u) + u/2 = s/6 and u/6 = s/2.  Substituting, we get 1 - u/2 = u/18, or u = 9/5.  Thus the ratio d(E, BCF)/d(A, BCF) = (9/5 - 1)/(9/5 - 0) = 4/9.  So V(BCEF)/V(ABCD) = 1/2*4/9 = 2/9.


function
modular arithmetic
induction
algebra proposed
algebra

[u][i]   Problem   [/i] [/u]   [i] Let $ f: Z \minus{} > Z$ be a function satisfying the conditions

            $ f(0) \equal{} 1$ , $ f(f(x)) \equal{} x \plus{} 4f(x)$ for all $ x \in Z$

              Find all natural numbers $ n$ ($ n \geq 1$) such that

  $ f_n(0)$ is divisible by $ 20^{11^{2005}}$, where $ f_1(x) \equal{} f(x) , f_n{x}\equal{} f(f_{n\minus{}1}(x))$ for all $ n \geq 2$

   Love An forever[/i]

Let $ a_n\equal{}f_n(0)$ so that
\[ a_0\equal{}0,a_1\equal{}1,a_{n\plus{}2}\equal{}4a_{n\plus{}1}\plus{}a_n\]
In this case, let $ x\equal{}2\plus{}\sqrt{5},y\equal{}2\minus{}\sqrt{5}$ so that
\[ a_{n\plus{}2}\minus{}xa_{n\plus{}1}\equal{}y(a_{n\plus{}1}\minus{}xa_n)\Rightarrow a_{n\plus{}1}\minus{}xa_n\equal{}y^n\]
\[ a_{n\plus{}2}\minus{}ya_{n\plus{}1}\equal{}x(a_{n\plus{}1}\minus{}ya_n)\Rightarrow a_{n\plus{}1}\minus{}ya_n\equal{}x^n\]
Solving for $ a_n$ gives $ a_n\equal{}\frac{x^n\minus{}y^n}{x\minus{}y}$.

I'm splitting the power of 2 and power of 5 case separately. Here $ b_n\equal{}\frac{x^n\plus{}y^n}{2}$ so that $ b_0\equal{}1,b_1\equal{}2,b_{n\plus{}2}\equal{}4b_{n\plus{}1}\plus{}b_n$.
[hide="2"]Consider $ n\equal{}m\cdot 2^k$ with $ m$ odd. Then
\[ a_n\equal{}a_m\cdot 2^k\prod_{j\equal{}0}^{k\minus{}1} b_{m\cdot 2^j}\]
The common recurrence relation for the $ a$'s and $ b$'s modulo 4 implies that for odd $ m$ and $ j\geq 1$, $ a_m$ odd, $ b_m\equiv 2\pmod{4}$, and $ b_{m\cdot 2^j}$ odd. So the highest power of 2 dividing $ a_n$ is 0 if $ n$ odd and $ 2^{k\plus{}1}$ if $ n$ even. So $ a_n$ is divisible by enough powers of 2 if and only if $ n$ is divisible by $ 2^{2\cdot 11^{2005}\minus{}1}$.[/hide]

[hide="5"]I'll redefine $ m,k$ here (hopefully that's not confusing) with $ n\equal{}m\cdot 5^k$ where $ m$ not divisible by 5. In this case,
\[ a_n\equal{}a_m\cdot 5^k\prod_{j\equal{}0}^{k\minus{}1}c_{m\cdot 5^j}\]
Here
\[ c_n\equal{}\frac{x^{4n}\plus{}x^{3n}y^n\plus{}x^{2n}y^{2n}\plus{}x^ny^{3n}\plus{}y^{4n}}{5}\equal{}\frac{2b_{4n}\plus{}2(\minus{}1)^nb_{2n}\plus{}1}{5}\]
We will show $ c_n\equiv 1\pmod{5}$ by induction (in particular, there are no more factors of 5). We can check $ c_0\equal{}1$ and $ c_1\equal{}61$. Now we can show $ b_{2n\plus{}4}\equal{}18b_{2n\plus{}2}\minus{}b_{2n}$ and $ b_{4n\plus{}8}\equal{}322b_{4n\plus{}4}\minus{}b_{4n}$, so that modulo 5 (reduce numerator modulo 25)
\[ c_{n\plus{}2}\equiv \frac{2\cdot 22b_{4n\plus{}4}\minus{}2b_{4n}\plus{}2\cdot 7(\minus{}1)^{n\plus{}1}b_{2n\plus{}2}\minus{}2(\minus{}1)^nb_n\plus{}1}{5}\]
\[ \equiv b_{4n\plus{}4}\plus{}2c_{n\plus{}1}\minus{}c_n\minus{}1\]
By the inductive hypothesis and the fact that $ b_{4n}$ is congruent to 1 modulo 5 (using $ b_{4n\plus{}8}\equal{}322b_{4n\plus{}4}\minus{}b_{4n}$), we get $ c_{n\plus{}2}\equiv 1\pmod{5}$, as desired.

So there are exactly $ k$ powers of 5 dividing $ a_n$. We require $ n$ to be divisible by $ 5^{11^{2005}}$.[/hide]

The end result is that $ \boxed{20^{11^{2005}}|a_n\iff 2^{2\cdot 11^{2005}\minus{}1}\cdot 5^{11^{2005}}|n}$.

[color=green]He edited it 5 minutes later your comment.[/color]


Counting with Restrictions Part 1

We count the number of ways to stack 6 different books, including 3 Art of Problem Solving Introduction series books, given that an Introduction book must be on the top and on the bottom.  In the process, we learn how to deal with restrictions in counting problems, and we see 6 really good books!

:D  :lol: Very good video tutor. Really helped me understand it. Good books too!


Please help me to solve that equation.
Problem: Let $ p \in P,q \in P$.Solve the equation
$ p^2\minus{}p\plus{}1\equal{}q^3.$

I spend 2 hours to solve it but It's very hard.  :(

lemma:suppose $ x,y\in N$ if we have $ \frac {x^2 \plus{} y^2 \plus{} x \minus{} y \plus{} 1}{xy} \equal{} k\in N$ we have $ K \equal{} 3$
proof:root jumping works here.assume $ k\neq 3$ take $ (x,y)$ with $ max(x,y)$ is minimum.assume $ (x > y)$
$ P(x) \equal{} x^2 \plus{} y^2 \plus{} x \minus{} y \plus{} 1 \minus{} kxy$ we know that $ P(ky \minus{} 1 \minus{} x) \equal{} 0$ but we prove that $ 0 < ky \minus{} 1 \minus{} x < x$
we have $ y^2 \minus{} y \plus{} 1 > 0$ so $ kxy > x^2 \plus{} x$ means $ ky \minus{} 1 \minus{} x > 0$
$ x^2 > y^2 \minus{} y \plus{} 1$ so we have $ 2x^2 \plus{} x > kxy$ means $ ky \minus{} 1 \minus{} x < x$.now assume $ (y > x)$
$ P(y) \equal{} x^2 \plus{} y^2 \plus{} x \minus{} y \plus{} 1 \minus{} kxy$ we know that $ P(kx \plus{} 1 \minus{} y) \equal{} 0$but we prove that $ 0 < kx \plus{} 1 \minus{} y < y$
we have $ x^2 \plus{} x \plus{}1 >0$ so $ kxy > y^2 \minus{} y$ means $ kx \plus{} 1 > y$
we want to prove $ y > kx \plus{} 1 \minus{} y$
suppose the inverse:
case 1:$ kx\plus{}1\equal{} 2y$ but this means $ x^2 \plus{} x \plus{} 1 \equal{} y^2$ contradicts.
case 2:$ kx\plus{}1>2y$ but this means $ x^2 \plus{} x \plus{} 1 > y^2\plus{}y$but this means$ x\equal{}1,k\equal{}3$
now assume $ x\equal{}y$ we have $ x\equal{}1,k\equal{}3$
so lemma is proved.
we have $ p|(q \minus{} 1)(q^2 \plus{} q \plus{} 1)$ but (p>q) so $ p|q^2 \plus{} q \plus{} 1$and we have $ q|p^2 \minus{} p \plus{} 1$ so$ pq|p^2 \plus{} q^2 \plus{} p \minus{} q \plus{} 1$
now from lemma we have $ 3pq \equal{} p^2 \plus{} q^2 \plus{} p \minus{} q \plus{} 1$so we have $ 3p \equal{} q^2 \plus{} q \plus{} 1$
we have $ 3(q^3 \minus{} 1) \equal{} 3p(p \minus{} 1) \equal{} (q^2 \plus{} q \plus{} 1)(p \minus{} 1) > (q^2 \plus{} q \plus{} 1)(\frac {(q \minus{} 1)^2}{3})$ so easily we have q<10.


algebra
polynomial
calculus
derivative
binomial coefficients
binomial theorem
Polynomials

Prove that for a prime $p$, $x^{p-1}+x^{p-2}+ \cdots +x+1$ is irreducible in $\mathbb{Q}[x]$.

Proof that cyclotomic polynomials are irreducible:

Lemma: If $\Phi_n(x) = f(x)g(x)$ is a factorisation of the $n$-th cyclotomic polynomial (over $\mathbb{Q}$) and $p$ a prime that doesn't divide $n$, then $f(\zeta)=0$ implies $f( \zeta ^p)=0$.

Proof:
Assume that $\zeta ^p$ is not a root of $f$, so it's a root of $g$.
So consider the polynomials modulo $p$ (this is possible since all coefficients are integers) to get (by writing out the polynomial in term's of binomial coefficients using the binomial theorem) that $f(\zeta ^p) \equiv f(\zeta)^p \equiv 0 \mod p$ (think of $\zeta \mod p$ as some number 'added' to the 'normal' residue classes $\mod p$). But since also $g(\zeta^p)\equiv 0 \mod p$, there is some root occuring twice $\mod p$. But this is impossible, since the derivative of $x^n-1 \mod p$ is $nx^{n-1}$ and so different from zero for $z\not\equiv 0 \mod p$ (it's the same like in the real numbers: you only can have a double root if the polynomial and it's derivative are both zero, the proof for that isn't that hard). This contradiction implies that $f(\zeta ^p)=0$.

To prove the theorem now, simple take some representing integer $k$ of a prime residue class $\mod n$ and factor it into primes (not dividing $n$). Now use the lemma inductively on the factors of $k$ to prove that with $\zeta$ being a root of $f$ also $\zeta^k$ is necessary a root and by this [u]all[/u] primitive $n$-th roots of unity are roots of $f$. But this shows that $f=\Phi_n$, so the cyclotomic polynomial is irreducible.

[quote="Ilthigore"]Let $x=y+1$. Then we get \[\frac{(y+1)^{p}-1}{y}= y^{p-1}+\binom{p}{1}y^{p-2}+.....+\binom{p}{p-1}\]

All the binomial coefficients are divisible by $p$, $\binom{p}{p-1}=p$ and therefore cannot be divisible by $p^{2}$, and the leading coefficient is 1, so it follows from Eisenstein's Criterion (see Q5).[/quote]

Do you have a paper where Eisenstein's Criterion is well explained?


A rectangular prism $ 77$ X $ 81$ X $ 100$ is cut into cubes of edge 1 by planes parallel to the faces of the prism. A given internal diagonal pierces how many of these cubes?

COuld you please explain how you've got the answer?

I posted the question because i don't know how to solve it.

Sure Ill post the solution

But first you gotta tell me if my answers right because tehres no point in posting a bad solution.

I don't have the official answer for it.
But 256 is indeed one of the five choices given in the test.

SO, you might as well post your solution~

Thanks!

Theres no way 256 is right, its too small..

The five choices given are
255, 256, 257, 258, and 259.

So, 256 is not "too small"~

isn't this like AIME 1996 #14? (the given is 150x324x375)

Thanks, pda.
It is indeed the same prob as AIME 1996 #4.
Check this link.
 [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=315374#315374[/url]

So, the answer to this question looks like to be 256!!

[quote="vikas"]but isnt there like a formula tofind the diagnonal likea2_b2_c2?[/quote]

The [i]length[/i] of the diagonal (which is not what the problem is asking for) is given by $ \sqrt{77^{2}+81^{2}+100^{2}}\approx 150.0$.

What the problem is asking for is how many cubes the diagonal passes through.  The length of the diagonal is a lower bound.

[quote="JBL"][quote="t0rajir0u"]The length of the diagonal is a lower bound.[/quote]

Not in general.  Consider, for example, a $ 1 \times 1 \times n$ brick or an $ n \times n \times n$ brick for any $ n$.[/quote]

Point taken.  So what I really mean is that the length of the diagonal over $ \sqrt{3}$ is a lower bound.   :oops:


Let a, b, c be positive reals such that abc = 1. Prove that
$ 1/sqrt(b\plus{}(1/a)\plus{}(1/2))\plus{}1/sqrt(c\plus{}(1/b)\plus{}(1/2))\plus{}1/sqrt(a\plus{}(1/c)\plus{}(1/2))\geq  sqrt(2)$

It was already posted :)

Let $ a \equal{}\frac{x}{y}$ $ b \equal{}\frac{y}{z}$ $ c \equal{}\frac{z}{x}$

It will become easier :D

How so? can you give me the link to the other post please? or a more complete solution?

Do NOT post anything further on this problem. It is an indian contest problem and any solution is cheating. Mod lock this topic.

[b]mnrc_2[/b]
i hope
[color=red][size=200]THIS IS THE LAST TIME U R DOING SO [/size][/color]

So do I. It's sad that people want to indulge in such malpractice when if they just sit down and try the problems they will in fact get them by themselves.
Ashwath

This is completely unacceptable. You cannot post IMOTC postal problems before the deadline is over! Since you are - as you claim - not an IMOTC student, you have no stake. So will you PLEASE wait till the deadline is over before posting the problems?


algebra
function
domain
trigonometry
polynomial
real analysis
real analysis unsolved

Hi.
Well, I should start off with saying that I have no clue if this is an appropriate forum for my question.
I have bothered all my friends with the subject to no avail. I believe it is merely a high school level equation that I (or no one I know) can't simplify and solve.
Please, let me know if there is a more appropriate forum here or on a different domain for such questions!  :blush:

anyhow, If that is the right place and everything, here goes...

y=s*tan(x)
y=m*(d-(cos(x)*r))

s,m,d and r, are all known variables.
I am trying to simplify this to one equation where I could find the angle (x).
(The equation will later be used in a computer program where the angle will be calculated for different variables values. I am aware that it's not solvable with every value.)
Thank you, and hoping I did not cause too much trouble around here...   :maybe:

So, it is just $ s\tan x\equal{}m(d\minus{}r\cos x)$, right? Just put $ z\equal{}\cos x$, write $ \tan x\equal{}\pm\frac{\sqrt{1\minus{}z^2}}z$, multiply by $ z$, square both sides and get a $ 4$th degree algebraic equation for $ z$. Now, there is a multitude of quick methods for finding all solutions of polynomial equations of low degree with one variable on the interval $ [\minus{}1,1]$. Also, there is Ferrari's formula, though I wouldn't use it for computational purposes. :)

[quote="Isachar"]I am aware that it's not solvable with every value.[/quote]In fact, this equation [i]always[/i] has at least one real solution in $ \left( \minus{} \frac {\pi}{2},\frac {\pi}{2}\right)$ (and as many as three), and at least two real solutions in $ [ \minus{} \pi ,\pi )$ (and as many as four), regardless of the choice of parameters.

To be more explicit about what fedja already said:
Let $ \alpha \equal{} \frac {md}{s}$, $ \beta \equal{} \frac {mr}{s}$, and $ z \equal{} \cos x$, then the equation becomes
\[ \frac {1 \minus{} z^2}{z^2} \equal{} (\alpha \minus{} \beta z)^2 \\
 \\
\Rightarrow \boxed{\beta ^2 z^4 \minus{} 2\alpha \beta z^3 \plus{} (\alpha ^2 \plus{} 1)z^2 \minus{} 1 \equal{} 0}
\]
If the roots of the above polynomial are $ \{ z_1,z_2,z_3,z_4 \}$, then each of $ \{\arccos z_1,\arccos z_2, \arccos z_3, \arccos z_4\}$ are solutions of the original equation (though as many as two of these may not be real).


function
geometry
geometric transformation
reflection
complex analysis
complex analysis unsolved

Let $ f$ be holomorfhic in $ \mathbb{C}$. Prove, by considering a suitable exponential, that the condition  that $ f(z)$ be real, when $ |z|\equal{}1$ forces $ f$ to be constant. :oops:

There are several ways to solve this one:

1. Consider $ g(z) \equal{} e^{if(z)}$ and apply maximum modulus principle.

2. Apply Poisson summation formula to $ f(z)$. What does this tell you about $ f$?

3. Consider $ f(z)$ restricted to the unit disc. Apply reflection principle across the boundary. Analytic continuation tells you that this function is $ f$; now what can you say about the behavior of $ f(z)$ as $ z \to \infty$?

Anyone have any others?

seriously, why do people name problems like this?  95% of problems in this subforum have to do with 'entire functions'.

A much better title would be 'f real on |z|=1 implies f constant'

Well, guido [b]is[/b] the only one to use the exact phrase "entire funcion" on the forum. That particular misspelling seems to come from Spanish speakers, and isn't very common.


In the AMC or some other math books, there are problems that say things like
John's house is five blocks below and six blocks to the left of Michael's house.
How many paths can John take to get to Michael's house?

I've seen this problem many times, and ended up drawing all these paths all over my paper
help please
thank you

Ok then so that is a 6 by 5 block box. 6 blocks or squares across and 5 blocks or squares down. In an x by y box, the formula is (idk latex sry)

                (x+y)!/x!y!

We will make x be 6 and y be 5.   So thats 11 factorial divided by 5 factorial and 6 factorial. After cancelling and computation, you get your answer. i dont wanna work it out right now sry. At least how i think it works

Well, if John is six blocks below and five to the left, and he can only go up and right, then he must choose five right choices out of six total, and so the answer is $ \binom{6}{5}$.

Else, lifeisapain is right.

[hide="Discussion"] There is a very general way to count paths:  put a $ 1$ at the point where you start, and at every other point put the number of ways you can get to that point using the allowable moves (here, movements down or to the left).  The key is that you can calculate this number based on the numbers on the points above and to the right (in particular, you add them).  So the grid you get grows like this:

[code]
1

1 1
  1

1 1 1
  2 1
    1

1 1 1 1
  3 2 1
    3 1
      1[/code]

Essentially, this kind of path-counting will just give you Pascal's triangle (for this problem), which is why you get binomial coefficients.  However, the value of this method is that it works even when the shape of the grid you're using is really weird.  For example, you can block out points and still count paths to get an answer.  AoPS v2 has a good demonstration of this.

There is a very specific way to do this problem:  let $ D$ represent a down move and $ L$ represent a down move.  A path from John's house to Michael's house can be represented by a string of five $ D$s and six $ L$s:

$ DDLDLLDDLLL$

The total number of different paths is the total number of different ways to arrange this sequence of $ 11$ letters.  Since $ D$ is repeated five times and $ L$ is repeated six times, a well-known expression that counts the number of different ways to arrange a word is

$ \frac{11!}{5! 6!}$

A brief explanation:  pretend that the $ D$s are all different and that the $ L$s are all different (for example, you can number or color them).  Then the total number of arrangements is just $ 11!$.  But there are $ 5!$ ways to arrange the $ D$s in order (just looking at the $ D$s and ignoring the $ L$s), so you need to divide out because you overcounted.  Similarly, there are $ 6!$ ways to arrange the $ L$s in order, so we need to divide that out too. [/hide]


If it isn't too much trouble (and if you're allowed to tell us this early), what books are you expecting to be prizes this year so far? I'm just curious so that I know what to buy now and what I might get for free. Thanks!

We're still working on it!  Expect the selection to be very similar to last year.


If $ \frac {x^2 \minus{} bx}{ax \minus{} c} \equal{} \frac {m \minus{} 1}{m \plus{} 1}$ has roots which are numerical equal but of opposite signs, the value of $ m$ must be:

$ \textbf{(A)}\ \frac {a \minus{} b}{a \plus{} b} \qquad\textbf{(B)}\ \frac {a \plus{} b}{a \minus{} b} \qquad\textbf{(C)}\ c \qquad\textbf{(D)}\ \frac {1}{c} \qquad\textbf{(E)}\ 1$

[hide="Click for solution"]
This equation is equivalent to $ x^2\minus{}\left (b\plus{}\frac{m\minus{}1}{m\plus{}1}a \right )x\plus{}c \left (\frac{m\minus{}1}{m\plus{}1} \right )\equal{}0$.  Since the coefficient of $ x$ is equal to the sum of its roots, it must be zero.  We have $ b\plus{}\frac{m\minus{}1}{m\plus{}1}a\equal{}0 \implies bm\plus{}b\plus{}ma\minus{}a\equal{}0 \implies m\equal{}\frac{a\minus{}b}{a\plus{}b}$, or $ \boxed{\textbf{(A)}}$.
[/hide]


finding trouble solving this one:
Show that √x + √y + √z ≥ xy + yz + zx for positive reals x, y, z with sum 3

[quote="gautam.bt2006"]finding trouble solving this one:
Show that √x + √y + √z ≥ xy + yz + zx(1) for positive reals x, y, z with sum 3[/quote]
$\sqrt{x}=a$
------------
------------
I have : $a^{2}+b^{2}+c^{2}=3$
 (1) is : $a+b+c \geq (ab)^{2}+(bc)^{2}+(ca)^{2}+(ba)^{2}[/tex] [tex] \leftrightarrow a+b+c \ge \frac{9-a^{4}+b^{4}+c^{4}}{2}$

$\leftrightarrow 2( a+b+c)+a^{4}+b^{4}+c^{4}\geq 9$

but I have :
 $a^{4}+a+a \ge 3a^{2}$ .
And : $b,c \to$  ....


how many sequences can we make from the numbers 1 to 24.if we make only four term of the sequence ?

$ 1,2,3,4,5,6,7,8,9,10,.............................................24$

examples
1st sequence $ 1,2,3,4$
2nd sequence $ 1,4,9,16$
3rd sequence $ 16,9,4,1$

how many sequence can we make from those numbers if we only make 4 terms. is there any formula for that?

Um... unless you restrict the sequences somehow, the answer is $ 24^{4}$, since there are 24 ways to choose each term.  Is this what you want?

The number of strictly increasing sequences is $ {24 \choose 4}$; note that any $ 4$-number subset uniquely determines a strictly increasing sequence.  The number of sequences with distinct elements is $ 4! {24 \choose 4}$, since there are $ 4!$ ways to permute a $ 4$-number subset.  Equivalently, it's $ \frac{24!}{20!} \equal{} 24 \cdot 23 \cdot 22 \cdot 21$ because there are $ 24$ ways to choose the first element, $ 23$ ways to choose the second, and so forth.

The last interpretation I will leave as an exercise:  what is the number of sequences $ a, b, c, d$ such that $ a \le b \le c \le d$?

Case 1: All four numbers are distinct.
Like t0rajir0u said, the number of such sequences is $ \binom{24}{4}$.
Case 2: a,b,c,d has 3 distinct values.
There are $ \binom{24}{3}$ ways to choose the three values, and $ 3$ ways to choose two variables to be equal, so there are $ 3\cdot\binom{24}{3}$ sequences total.
Case 3: a,b,c,d has 2 distinct values.
There are $ \binom{24}{2}$ ways to choose the two values. To make a,b,c,d have 2 distinct values, either $ a\equal{}b\equal{}c$, $ a\equal{}b$ and $ c\equal{}d$, or $ b\equal{}c\equal{}d$. So there are $ 3\cdot\binom{24}{2}$ sequences total.
Case 4: a=b=c=d.
Clearly, there are $ 24$ such sequences.
Adding up all cases yields $ \binom{24}{4} \plus{} 3\cdot(\binom{24}{3} \plus{} \binom{24}{2}) \plus{} 24 \equal{} 10626 \plus{} 6900 \plus{} 24 \equal{} \boxed{17550}$.


email
inequalities
function
analytic geometry
geometry
3D geometry
octahedron

Hello,

February may seem like a long time away but in actuality it is very close (especially if you are preparing for the AMC s ; ) ).  Sometime in the middle of my busy week I found time to become bored and thought of making something which I call AMC-a-Day.  Its nothing really special but hopefully it will help everyone practice for the AMC s and thus raise their scores.

So heres how it will work:

I will post a problem in the morning every day.  The difficulty will be pretty much random and ranges from the easiest to hardest AMC problems.  Then everyone can work on it and send me their answer by PM, AIM, or email( jlaurendi@gmail.com ).  Since all of these are multiple choice no work needs to be shown just A,B,C,D or E.  Also, please do not discuss the problem until the next days problem has been posted.

I will keep score as follows:

1 point for a correct answer

0 points for no answer

-1 point for an incorrect answer

I made the 1 point to discourage ppl swamping my PM box with random guesses.

At the end of each week I will post the top scorers from the week.

Also, I will be using problems from old AHSME s/ AMC s (and possibly other sources if I run out).  So please try to do the problem on your own instead of going through all the old tests.  There are no prizes for the top scorers, unfortunately, so there is no reason to do that.

I would also like to mention that if you really want to practice for the AMC I strongly encourage you to consider taking either the AMC 10 Problem Series or the AMC 12 problem series that is offered by AoPS.  These classes are far superior to what I am doing here.  So if you really want good practice I would enroll in either of those classes.

I think thats just about all the things that I needed to cover.  So heres the first problem!

Day 1

A positive number x satisfies the inequality $\sqrt{x}<2x$ if and only if

(A) x> 1/4    (B) x>2   (C) x>4   (D) x< 1/4  (E) x<4

Yesterday the answer was A.   The first few problems will be on the easier spectrum of the AMC but then they will start to get more random in difficulty.

Day 2

The largest integer n for which $n^{200}<5^{300}$ is

(A) 8  (B)  9  (C)  10  (D)  11 (E)  12

Great idea  :P .  Would it be possible for someone to post their solution after each problem's answer has been announced?  That way we know how the problem was solved and might even learn a different approach to the problem.
I don't mean for this topic to get crowded with solutions.  I think one per question would work just fine.

The answer was D for yesterday.

Day 3

Let f be a function satisfying f(xy)=f(x)/y for all positive real numbers x and y. If f(500)=3, what is the value of f(600)?

(A) 1    (B) 2   (C) 5/2   (D)   3  (E)  18/5

Solution to day 2:
Raising both sides to the 1/200 power, you get $n<5^{300/200} => n<5^{3/2}$.  Since $5^3=125$ and $\sqrt{121}<\sqrt{125}<\sqrt{144}$, 11 is the smallest integer.

The answer yesterday was C 5/2.

Day 4

A circle with center O is tangent to the coordinate
axes and to the hypotenuse of the 30_-60_-90_ triangle
ABC as shown, where AB = 1. To the nearest
hundredth, what is the radius of the circle?
(A) 2.18 (B) 2.24 (C) 2.31
(D) 2.37 (E) 2.41

Yesterday the answer was D

Day 5

A regular octahedron is formed by joining the centers of adjoining faces of a cube.  The ratio of the volume of the octahedron to the volume of the cube is

(A)	$\frac{\sqrt{13}}{12}$  (B)  $\frac{\sqrt{6}}{16}$  (C)  1/6  (D) $\frac{\sqrt{2}}{8}$  (E)

Solution for Day 4:

Refer to the below diagram.  Let D, E, and F be the points of tangency to the y-axis, CB, and the x-axis, respectively.  Angles ADO, AFO, and DAF are right, so ADOF is a rectangle, and DO = OF, so ADOF is a square.  Thus we have AF = AD = r.  So BF = r - 1.

By the Power of a Point theorem, EF = BF and CE = CD.  So BE = r - 1, and CE = BC - BE = 2 - (r-1) = 3 - r.  So CD = 3 - r.  So AD = AC + CD = $\sqrt{3}$ + 3 - r.

But AD also equals r.  Thus,
$\sqrt{3}$ + 3 - r = r
2r = $\sqrt{3}$ + 3 $\approx$ 4.732
r $\approx$ 2.37.

Ummm, the answer you got wasn't one of the choices.  You might want to check your calculations; $3+\sqrt{3}\approx 4.732\neq 2.732$.

Anyway, this is what I did (refering to your diagram).  Seems kinda similar to what you did, except that I just got CB in terms of $r$, rather than work more.

It's easy to see that AFOD is a square, so BF=$r-1$ and CD=$r-\sqrt{3}$ (AC=$\sqrt{3}$ because of the 30-60-90 triangle).  Because two tangents from the same point to the same circle are equal, we know that CE=CD=$r-\sqrt{3}$ and BE=BF=$r-1$.  Thus, BC=$2r-1-\sqrt{3}$.  We also know that BC=2, again, because of the 30-60-90 triangle.  Thus, we have $2r-1-\sqrt{3}=2$, so $r=\frac{3+\sqrt{3}}{2}\approx 2.36$, so the answer is D.

Oops.  I think I added 1 instead of 3 by accident.  Fixed now.

Yesterday was C.

Day 6

A spider has one sock and one shoe for each of its eight legs. In how many
different orders can the spider put on its socks and shoes, assuming that, on
each leg, the sock must be put on before the shoe?

(A) $8!$ (B) $2^8\cdot 8!$  (C) $(8!)^2$  (D) $16!/(2^8)$  (E) $16!$

Solution to yesterday's problem:

WLOG assume the length of the cube is 2
A regular octahedron is basically two regular square pyramid sharing the same base. The base is a square with side sqrt(2), height of one of the pyramid is 1.  So volume of one of the pyramid=2/3,  volume of the octahedron=4/3, volume of the cube=8,

so the ratio=1/6

Wow, not very many ppl got yesterday's problem.  Only 4 ppl got it and 2 ppl got it wrong.  The answer was D and the problem was #16 from the 2001 AMC 12.  Btw, I m going to start trying to post all the sources for problems the day after.  There may not be some for the next couple because I didn't write down the source when I was looking for them.

Day 7

How many positive integers not exceeding 2001 are multiples of 3 or 4 but not 5?
(A) 768 (B) 801 (C) 934 (D) 1067 (E) 1167

I got e.  If f(-3) is 2, you have ax^4-bx^2+x+5, or ax^4-bx^2+2.  That means that ax^4-bx^2 = 0.  ax^4-bx^2 will be the same for 3, since even exponents always get a postive answer.  THat means that all you have left is x+5 = 3+5 = 8.

I think that I'm right becuase I asked my dad who is also a math professor and he got the same answer as I did.  Did I do something wrong?

I was referring to the Day 21.  The number theory one.  Sorry for the confusion.  Could someone edit hspotter's solution beta will still accept answers for that one today?

[quote="joml88"]Call a positive real number special if it has a decimal representation that consists entirely of digits 0 and 7. ... What is the smallest n such that 1 can be written as a sum of n special numbers?[/quote]
I didn't think too carefully, but maybe this works.  If 1 can be written as a sum of n special numbers, then 1/7 can be written as the sum of n numbers using only the digits 0 and 1.  The decimal expansion of 1/7 is .142857... The only way to pick up the digit 8 is to use at least 8 numbers with digits 0 and 1.  And it's easy to find a way to express .142857 using 8 such numbers.

Did that make sense?

well, for that problem...

you must view the one as 9/9 or .99999999999999999

so that in the tenth digit there is only going to be 1-7.

then the hundredth is going to make the 7 in the tenth digit a 9 by carryin so it is going to be 4-7.

then you make the 8 from the hundredth digit a 9 by using 2-7 in the thousandth digits.

continue and you will get 8-7 for the next, 5-7 for the next, and then 7-7 for the next.

then we have a 9 so we can continue this problem by repeating the digits.

NOCTICE two things [b]8-7[/b] have 8 sevens so the least is 8 numbers but ALSO see that

[b]1[/b]-7 [b]4[/b]-7 [b]2[/b]-7 [b]8[/b]-7 [b]5[/b]-7 [b]7[/b]-7
are just like wut someone pointed out being 1/7 or .142857 with 8 being the number of numbers it take to make 1.

It is not so tough, only took a few minutes, but most ppl r probably too lazy to do it.

Yesterday's answer was E.

Day 23

In a geometric sequence of real numbers, the first term is 2 and the sum of the reciprocals of its third and fourth terms is 2 more than its second term. Find the sum of all possible values for its seventh term.

A)$\frac{1}{8}$

B)$\frac{3}{4}$

C)2

D)$\frac{17}{8}$

E) 3

You can send me the answers by pm, email(feiqijiang@yahoo.com), or AIM (Feiqi J1989)

Yesterday's answer is D.

Day 24

If $a$ and $b$ are positive real numbers and each of the equations $x^2+ax+2b=0$ and $x^2+2bx+a=0$ has real roots, then the smallest possible value of $a+b$ is

A)2

B)3

C)4

D)5

E)6

Yesterday's answer is E, only one person turned in the answer :(

Day 25

Find the sum of all solutions to

$x^{\log x}=\frac{x^3}{100}$

A)100
B)1010
C)1100
D)110
E)10

Yesterday's answer is D

Day 26

If the line y=mx+1 intersects the ellipse x^2+4y^2=1 exactly once, then find m^2

A)$\frac{1}{4}$

B)$\frac{3}{4}$

C)$\frac{3}{8}$

D)$\frac{1}{2}$

E)1

Yesterday's answer is B

Day 27

If x, y, z are real numbers satisfying

x+1/y=4,  y+1/z=1, z+1/x=7/3

Find xyz

A)2/3

B)1

C)4/3

D)2

E)7/3

Some answers.

23. [hide]Let the ratio of successive terms be r.  Then the second term is 2r, the third term is 2r^2, the fourth term is 2r^3, and the seventh term is 2r^6.  From the given requirement, we have 1/(2r^2) + 1/(2r^3) = 2 + 2r, or multiplying both sides by 2r^3, r + 1 = 2(r+1)*2r^3.  Thus, (4r^3 - 1)(r+1) = 0, so either r^3 = 1/4 or r = -1.  In the former case, 2r^6 = 2(r^3)^2 = 1/8; in the latter case, 2r^6 = 2*(-1)^6 = 2.  The sum of the solutions is then 1/8 + 2 = 17/8, D.[/hide]

24. [hide]In order for a quadratic to have real roots, the discriminant must be greater than or equal to 0, so a^2 - 8b  0 and 4b^2 - 4a  0.  Rearranging these equations yields a^2  8b and b^2  a, so that a^4  64b^2  64a, i.e. a  4.  Similarly, b^4  a^2  8b, so b  2.  The minimum sum is then 4 + 2 = 6, E.[/hide]

25. [hide]In order for log x to be defined, x > 0.  Thus, we can say that x = 10^y for some real number y.  Substituting this into the given equation yields (10^y)^y = (10^y)^3/100, or 10^(y^2) = 10^(3y - 2), or y^2 = 3y - 2, or y^2 - 3y + 2 = 0, or (y-2)(y-1) = 0.  Thus, y = 1 or 2.  If y = 1, then x = 10; if y = 2, then x = 100; the sum of these is 110, D.[/hide]

26. [hide]First we substitute y = mx + 1 into x^2 + 4y^2 = 1 to find the points of intersection for an arbitrary m.  We have x^2 + 4(mx+1)^2 = 1, or x^2 + 4m^2*x^2 + 8mx + 4 = 1, or (4m^2+1)x^2 + 8mx + 3 = 0, or x^2 + 8m/(4m^2+1)*x + 3/(4m^2+1) = 0.  Now, in order for there to be exactly one intersection point, our last equation (a quadratic) must have one solution, so it must be a perfect square.  Thus we must have 3/(4m^2+1) = [4m/(4m^2+1)]^2, or 3 = 16m^2/(4m^2+1), or 12m^2 + 3 = 16m^2, or 3 = 4m^2, or m^2 = 3/4, B.[/hide]

Yesterday's answer is B

Day 28

If $z(z+i)(z+3i)=2002i$ has a zero in the form of $a+bi$, where $a$ and $b$ are positive real numbers, find $a$.

A)$\sqrt{118}$
B)$\sqrt{210}$
C)$2\sqrt{210}$
D)$\sqrt{2002}$
E)$100\sqrt{2}$

27.  [hide]I brute-forced this...anyone have an easier way?  Of course, the ultimate easy way is to type solve(x+1/y=4 and y+1/z=1 and z+1/x=7/3,{x,y,z}) into your TI-89 and voila![/hide]

mathfanatic wrote:27.  [hide]I brute-forced this...anyone have an easier way?  Of course, the ultimate easy way is to type solve(x+1/y=4 and y+1/z=1 and z+1/x=4/3,{x,y,z}) into your TI-89 and voila![/hide]

multiply all equations together you get x+y+z+1/x+1/y+1/z+xyz+1/xyz=28/3

add all equations together you get

x+y+z+1/x+1/y+1/z=22/3

so

22/3+xyz+1/xyz=28/3, simple solving get xyz=1

Is this ended?

I think it was pretty cool but then it stopped...


AMC
AMC 10
AMC 10 A
AMC 12
USA(J)MO
USAMO
AMC 12 B

[url]http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2007-12a/07-1-1012-dir,pic.shtml[/url]

So, how did everyone do compared to the nation?

Someone in my grade beat me, so I'm 2nd in 8th grade for 12A, and tied with someone for 1st in 8th grade for 12B.

Edit: removed link from "Subject description" line, it was messing around with my screen width.

*shrugs* Top 40-something in the nation (9th, AMC 12A). Nothing special, might've done better had I actually studied for it.

Top 9 (9th, AMC 12A), Top 18 (9th, AMC 12A&B)

tied for first in the nation on the amc10a

meh

Tied in late 20's/Early 30's in Nation (10B)
1st in NY State!

[quote="MysticTerminator"]tied for highest male scorer in NC[/quote]
Yeah same here.

dude you're not in NC what

also tied for highest score on AMC 12A for 17-year old males who are shorter than 6 feet but weigh more than 120 pounds and who may or may not have significant facial hair on any given day

[quote="Ignite168"]If you notice, they let in more than 1% for both 10 contests (lowering the cut off to lett in 3% for the A!) and more than 5% for the 12A. The AMC committee has been more than generous enough, maybe we should stop complaining as they are obviously trying to do anything in their power to let as many people qualify as possible while still being fair.[/quote]

Uh, it's always been their policy to let the top 5% on the AMC12 take the AIME.

Which is why he's saying that they're giving [b]more[/b] than the top 5% on the AMC12 an opportunity to take the AIME...

[quote="Xantos C. Guin"]21st 10th grade and under on the 12A.

Why do they bother putting rows for scores that aren't multiples of 1.5?[/quote]
I was wondering that too, they probably did that so it's easier to read or something.

tied for first in U.S. on AMC 10A...yay...

[quote="perfectnumber628"]Wow, 3 people on the 12A and 1 on the 12B managed to get a 0.[/quote]

Dude, that takes skill.

So, I got 9th in Ohio for the 12A, don't feel like counting for in the country.  Pitiful, I know...

[quote="paladin8"]Tied for lowest score on the AMC 12B among all previous USAMO qualifiers at my school.[/quote]

Whoa, me too.   :o

18th or something for 9th grade in AMC12B, and 32nd or something counting A and B. disgraceful

How many perfect scores for the AMC 10A were there and what was the grade distribution for them?

Well, you could just look, but I'll be nice.  :wink:
AMC 10A: 39 total perfect scores.
8th grade: 9
9th grade: 12
10th grade: 12
?th grade: 6

Male: 34
Female: 4

Some information extracted from the tables of question difficulty, with 2006 for comparison:

--------Right--Blank--Wrong--Old---New
06 12A--9.88--10.57--4.55--85.71--75.13
06 12B--9.32--11.85--3.83--85.52--73.67
07 12A--8.30--11.31--5.38--78.11--66.80
07 12B--9.59--10.34--5.07--83.40--73.06

That's the average number of right, blank and wrong answers for all official takers, with the average scores under old and new systems. There does appear to have been somewhat more guessing this year- the number of wrong answers is up noticeably.

Also, the difference in scores between this year's B test and last year's tests is almost entirely due to the change in scoring system. The question difficulties were only noticeably higher on this year's A test.


The cubic $ x^3 \minus{} 2x^2 \minus{} 3x \minus{} 4 \equal{} 0$ has three roots, $ a$, $ b$, and $ c$. Compute $ a^4 \plus{} b^4 \plus{} c^4$.

Do you mean $ x^3\minus{}2x^2\minus{}3x\plus{}4\equal{}0$?

Problem has been corrected (I also mistyped the constant). My apologies.

[hide] From Vieta, we get that

$ a\plus{}b\plus{}c\equal{}2$

$ abc\equal{}4$

$ ab\plus{}bc\plus{}ac\equal{}3$

Now, $ (a\plus{}b\plus{}c)^2\equal{}a^2\plus{}b^2\plus{}c^2\plus{}2ab\plus{}2ac\plus{}2bc$

$ \implies 4\equal{}a^2\plus{}b^2\plus{}c^2\plus{}2(3) \implies a^2\plus{}b^2\plus{}c^2\equal{}\minus{}2$.

(A graphing calculator check confirmed that two of the roots are complex).

So, $ (a^2\plus{}b^2\plus{}c^2)^2\equal{}a^4\plus{}b^4\plus{}c^4\plus{}2(a^2b^2\plus{}c^2a^2\plus{}b^2c^2)$.

We also have,

$ (ab\plus{}ac\plus{}bc)^2\equal{}a^2b^2\plus{}b^2c^2\plus{}a^2c^2\plus{}2a^2bc\plus{}2b^2ac\plus{}2c^2ab$

$ \implies 9\equal{}a^2b^2\plus{}a^2c^2\plus{}b^2c^2\plus{}2abc(a\plus{}b\plus{}c)$

$ \implies 9\equal{}a^2b^2\plus{}b^2c^2\plus{}a^2c^2\plus{}2(4)(2) \implies a^2b^2\plus{}a^2c^2\plus{}b^2c^2\equal{}\minus{}7$.

Subbing this into the equation above,

$ 4\equal{}a^4\plus{}b^4\plus{}c^4\plus{}2(\minus{}7) \implies a^4\plus{}b^4\plus{}c^4\equal{}\boxed{18}$.[/hide]

[hide="Solution"]
Note that $ a^4$, $ b^4$, and $ c^4$ are the roots of
$ x^{3/4} \minus{} 2 \sqrt {x} \minus{} 3x^{1/4} \minus{} 4 \equal{} 0$
$ x^{1/4}(\sqrt {x} \minus{} 3) \equal{} 2(\sqrt {x} \plus{} 2)$
Squaring both sides,
$ \sqrt {x}(x \minus{} 6 \sqrt {x} \plus{} 9) \equal{} 4(x \plus{} 4 \sqrt {x} \plus{} 4)$
$ \sqrt {x}(x \minus{} 7) \equal{} 2(5x \plus{} 8)$
Squaring both sides,
$ x(x \minus{} 7)^2 \equal{} 4(5x \plus{} 8)^2$
$ x(x \minus{} 7)^2 \minus{} 4(5x \plus{} 8)^2 \equal{} 0$
Now this is a cubic. The sum of it's roots, which is $ a^4 \plus{} b^4 \plus{} c^4$, is the opposite of the $ x^2$ coefficient by Vieta's formulas, which is $ \minus{} 114$, so $ a^4 \plus{} b^4 \plus{} c^4 \equal{} 114$.[/hide]
For Gaussintraining's solution, it should be $ ab \plus{} bc \plus{} ca \equal{} \minus{} 3$. And if you fix it, the answer indeed comes out to the same as mine.

[hide="Quick solution"]Simply apply [url=http://www.artofproblemsolving.com/Wiki/index.php/Newton's_sums]Newton's sums[/url].

$ S_1\equal{}2$ by Vieta.

$ S_2\minus{}2S_1\minus{}6\equal{}0\implies S_2\equal{}10$

$ S_3\minus{}2S_2\minus{}3S_1\minus{}12\equal{}0\implies S_3\equal{}38$

$ S_4\minus{}2S_3\minus{}3S_2\minus{}4S_1\equal{}0\implies S_4\equal{}\boxed{114}$[/hide]

@dgreenb801 Yeah, you're right, I made a stupid mistake with Vieta.  :oops:


It is quite normal and easy but nice :
For $x,y,z \ge 0$prove that

$1 \ge \frac{x}{x+\sqrt{(x+y)(x+z)}}+\frac{y}{y+\sqrt{(y+z)(y+x)}}+\frac{z}{z+\sqrt{(z+y)(z+x)}}$

$\sqrt{(x+y)(x+z)} \ge \sqrt{xy} + \sqrt{xz}$ -> done. It's probably been discussed before.

Here is a part of my solution:

$\sqrt{(x+y)(x+z)} \ge x+\sqrt{yz}$ so

$1 \ge \frac{x}{2x+\sqrt{yz}}+\frac{y}{2y+\sqrt{xz}}+\frac{z}{2z+\sqrt{xy}}$

So...... the inequality reduces to: $\frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c}\leq 1$ plus the condition $abc=1$ , which has been discussed before. ;)

[quote="lomos_lupin"]Let $ x,y,z > 0$, prove that

$ \frac {x}{x \plus{} \sqrt {(x \plus{} y)(x \plus{} z)}} \plus{} \frac {y}{y \plus{} \sqrt {(y \plus{} z)(y \plus{} x)}} \plus{} \frac {z}{z \plus{} \sqrt {(z \plus{} x)(z \plus{} y)}}\leq 1$[/quote]This was proposed by Walther Janous, see:

Problem $ 1654^{*}$, [url=http://journals.cms.math.ca/CRUX/]CRUX Mathematicorum[/url], 1991, No.6, p. 171.

Actually, the following stronger inequality holds :

$ \sum{\frac {x}{x \plus{} \sqrt {(x \plus{} y)(x \plus{} z)}}}\leq \sum \frac {3x^2}{(2x \plus{} y)(2x \plus{} z)},$

with equality if $ x\equal{}1\plus{}\sqrt{3},y\equal{}z\equal{}1$.

See: http://www.mathlinks.ro/Forum/viewtopic.php?t=66660


The day a man runs out of soap, he immediately makes a mental note to buy a new bar the next day. However, there is a 1/2 chance that he forgets, and he always realizes that he forgot right when the store closes. The next day, there is a 1/2 chance again that he forgets, and so on. What is the expected number of days that will pass before he remembers to and buys a bar of soap?

[hide="Solution"]So you are basically asking us to compute the value of $S=\sum_{k=0}^{\infty}\frac{k}{2^{k}}=\sum_{k=1}^{\infty}\frac{k}{2^{k}}$

Note that $S/2 = \sum_{k=0}^{\infty}\frac{k}{2^{k+1}}=\sum_{k=1}^{\infty}\frac{k-1}{2^{k}}$

Thus, $S-S/2 = \sum_{k=1}^{\infty}\frac{k}{2^{k}}-\sum_{k=1}^{\infty}\frac{k-1}{2^{k}}=\sum_{k=1}^{\infty}\frac{1}{2^{k}}=1$
$\implies S=2$

So the expected time that will pass is 2 days.
[/hide]

[b]Edit:[/b] Ok, so Bill the Maniac beat me to it. Or at least, I'm assuming you tried to do the same thing that I did... something fishy with your coding.

Yup, exactly.

(in your solution, take the "inf" and put it on top of the summation where the P is, I think that's what you meant)

You may want to use "\infty" for infinity and "\frac{ numerator }{ denominator }" for fractions  :) that's just me, though.

[hide="Wait..."]
Doesn't this follow directly from one of the definitions of expected value?
\[E = \frac{1}{p}= \boxed{2}\]where $p=\text{probability that he does remember}$
[/hide]

Does it really? The only definition I knew was $E=\sum_{n}\text{Value}(n) P(n)$. Cool