Suppose $p$ is a prime greater than $3.$ Find all pairs of integers $(a,b)$ satisfying the equation

           $a^{2}+3ab+2p(a+b)+p^{2}=0.$

[quote="t0rajir0u"][hide="Hint"] Consider as a quadratic in $p$. [/hide][/quote]
by using quadratice i get:
$p=-(a+b)+\sqrt{b^{2}-ab}$....

This problem was posted at
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=344670#p344670[/url]
Actually, the inmo prbelms from past 10 years have been posted on the forum; you can look in contest resources to find them

Thanks, [b]Rem[/b], I appreciate it. I had no idea. Should have put more effort in locating the problems.

That's ok. Thee have been tonnes and tonnes of problems posted on Aops, and it is very easy to post something already posted. Just look into the olympiad section. About every 10th problem there has been already posted. :lol:
Actually posting a problem again is not a bad idea, I think. Many people have not seen it but mght benefit from working on it. :)


Hi Could anyone help me prove that " an open interval A is connected". I'm taking my first course in topology and I've been doing well so far until the last three theorems we saw and this is one of them.
I tried proving it by contradiction and had 3 cases for A and taking two open sets U=(a,b) and V=(c,d)
case 1: b<c and showed it was impossible for A to be in U union V
case 2: b=c similarly
case 3: b>c similarly

But my professor did not accept it.

Thank you

It doesn't suffice to take $ U, V$ open intervals; they can be [b]any[/b] open set.  What does the complement of an open set look like?

I understand that U and V need to be any interval and that's how they are in my proof. The complement of an open set is a closed set but how does that help me?

[quote="romy"]I understand that U and V need to be any interval[/quote]
There are open sets other than intervals.

[quote="romy"]The complement of an open set is a closed set but how does that help me?[/quote]
Suppose $ A \equal{} U \cup V$.  If $ U, V$ are disjoint and non-empty then the complement of $ U$ is closed and not all of $ \mathbb{R}$, hence not open.  Show that it contains a nontrivial boundary in $ A$ which can't be in $ V$.

This probably isn't the easiest way to do it.

It's probably more productive to directly invoke completeness of $ \mathbb{R}$; you need it anyway. Suppose $ a\in U$. What can you say about the least upper bound of $ V\cap (\minus{}\infty,a)$?

We have not seen completeness and we're only working on the linearly ordered set X not R. All I have is that X is connected.

Wouldn't this work?

[hide]Path connectedness implies connectedness.

Take some open interval $ A \subset \mathbb{R}$. Then $ A \equal{} (a,b)$ for some $ a,b$. Take arbitrary points $ x,y \in A$. Then the function $ f : [0,1] \to A$ given by $ f(z) \equal{} x \plus{} z(y \minus{} x)$ is continuous, connecting $ x$ and $ y$ (i.e. $ f(0) \equal{} x$, $ f(1) \equal{} y$). Thus it's path connected.[/hide]

Ah yes, circular logic.

Well, you can always resort to things about a linear continuum.

So, if I understand correctly, the problem is the following, right?

Let $ X$ be a linearly ordered set, that is connected as a topological space. Let $ A$ be any interval of $ X$. Then prove that $ A$ is connected as well.

Got a solution, but it might be horribly inefficient and awful.

[hide="Hint"]
The case when $ A$ is a closed interval is easier in my eyes. Do that one first, and then think about how to apply that special case in the general case.
[/hide]

[hide="Solution"]
First off, suppose that $ A$ is a closed interval, say $ [a,b]$. Suppose that $ A$ is not connected, then there are two non-empty disjoint open sets $ U, V$ in $ A$ such that $ U \cup V \equal{} A$. We want to consider $ U, V$ as subsets of $ X$, but the problem is, as such, they are not open. To solve that problem, we consider the open intervals $ O_a \equal{} \{x \in X: x < a\}$, $ O_b \equal{} \{x \in X: x > b\}$. If $ a \in U$, then we add $ O_a$ to $ U$. (And if $ a \in V$, then we add $ O_a$ to $ V$ instead, and we do the same for $ b$.) Now you can check that you have made $ U$ and $ V$ two disjoint non-empty open sets that cover $ X$. By connectedness of $ X$, this gives a contradiction. Hence $ A$ is connected.

So now we know that every closed interval is connected. We turn to the general case. Let $ A$ be any interval. Then we can write $ A$ as a union of a chain of closed intervals $ (C_i)_{i \in I}$. Now suppose that $ A$ is not connected. Then we have two disjoint non-empty open subsets $ U,V$ such that $ U \cup V \equal{} A$. Note that there is a $ C_i$ that contains points of both $ U$ and $ V$. Intersecting with this $ C_i$, we see that $ U,V$ remain non-empty, disjoint, open and that they still cover $ A$. But since $ A$ is closed, and hence connected, we arrive at a contradiction. Hence $ A$ must be connected.
[/hide]

Yea, I need more efficient suggestions =p

EDIT: Heh, I just saw the problem was only about open intervals. Oh well.

A more efficient route:
- Choose two points $ a,b$ in the open interval $ I$.
- Show that $ [a,b]$ is connected.
- Conclude that if $ I$ is written as the disjoint union of open sets $ U$ and $ V$, $ a$ and $ b$ are in the same set.
- Repeat for all pairs; we conclude that one of $ U$ or $ V$ contains all of $ I$.

Well thanks but this does not help because I don't have such continuous functions and cannot just make it up.

Any other suggestion, I'm really stuck on this one.

Thanks

It also would have helped to have a complete and correct statement from the start. In a general ordered set with the order topology, intervals aren't connected. What exactly is the extra hypothesis?

[quote="t0rajir0u"][quote="romy"]Well thanks but this does not help because I don't have such continuous functions and cannot just make it up.[/quote]
What is this even in reference to?  There have been several viable solutions given in this thread and you've ignored all of them.[/quote]

Sorry I meant to respond to Tyl's thread and it ended up in the wrong place. Yes most of the suggestions are helping me a lot!

Thanks all!

[quote="Jinbi Jin"]So, if I understand correctly, the problem is the following, right?

Let $ X$ be a linearly ordered set, that is connected as a topological space. Let $ A$ be any interval of $ X$. Then prove that $ A$ is connected as well.

Got a solution, but it might be horribly inefficient and awful.

[hide="Hint"]
The case when $ A$ is a closed interval is easier in my eyes. Do that one first, and then think about how to apply that special case in the general case.
[/hide]

[hide=Solution]
First off, suppose that $ A$ is a closed interval, say $ [a,b]$. Suppose that $ A$ is not connected, then there are two non-empty disjoint open sets $ U, V$ in $ A$ such that $ U \cup V \equal{} A$. We want to consider $ U, V$ as subsets of $ X$, but the problem is, as such, they are not open. To solve that problem, we consider the open intervals $ O_a \equal{} \{x \in X: x < a\}$, $ O_b \equal{} \{x \in X: x > b\}$. If $ a \in U$, then we add $ O_a$ to $ U$. (And if $ a \in V$, then we add $ O_a$ to $ V$ instead, and we do the same for $ b$.) Now you can check that you have made $ U$ and $ V$ two disjoint non-empty open sets that cover $ X$. By connectedness of $ X$, this gives a contradiction. Hence $ A$ is connected.[/quote]

This seems to work at first but then when you look closely, it could be that O_a intersects with V and O_b with U because we don't know much about U and V. I mean we do not have much control over them. In this case, the two open sets that make up would not be disjoint. Right?

May be this helps you http://www.mathlinks.ro/viewtopic.php?t=271056, but it is a little messy.

:)


Let $ a,b,c$ be nonnegative real numbers. Prove that if $ x \ge 1,$ then
$ \left( a^{\frac {1}{x}} \plus{} b^{\frac {1}{x}} \plus{} c^{\frac {1}{x}}\right)^{2x} \ge a^2 \plus{} b^2 \plus{} c^2 \plus{} (4^x \minus{} 2)(ab \plus{} bc \plus{} ca) \plus{} 3(3^{2x \minus{} 1} \minus{} 4^x \plus{} 1)\sqrt [3]{a^2b^2c^2}.$
The equality holds iff $ x \equal{} 1$ or $ (a,b,c) \sim (1,1,1)$ or $ (a,b,c) \sim (1,1,0)$ or $ (a,b,c) \sim (1,0,0).$
:)

For more information about this inequality. Actually, I created it when trying to find a nice solution for the following inequality:
$ \sqrt[n]{\frac{a}{b\plus{}c}}\plus{}\sqrt[n]{\frac{b}{c\plus{}a}}\plus{}\sqrt[n]{\frac{c}{a\plus{}b}} \ge \sqrt[n]{2^n\plus{}\frac{4(3^n\minus{}2^{n\plus{}1})abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}}$ (*)
where $ a,b,c \ge 0$ and $ n>0$.

And the interesting thing is, with the original inequality, we get can a very nice solution for (*) in the case $ n \ge 2.$ :) For $ 2>n>0,$ my solution is very complicated. :(

can_hang2007, i want to know your solution for (*). Can you post it?


Solve the following simultaneous equations:

  $ a\sqrt{a}\plus{}b\sqrt{b}\equal{}183$
  $ a\sqrt{b}\plus{}b\sqrt{a}\equal{}182$

we have $ a,b>0$ and $ (\sqrt{a}\plus{}\sqrt{b})^3\equal{}a\sqrt{a}\plus{}b\sqrt{b}\plus{}3(b\sqrt{a}\plus{}a\sqrt{b})\equal{}183\plus{}3.182$
then $ \sqrt{a}\plus{}\sqrt{b}\equal{}\sqrt[3]{183\plus{}3.182}$
$ (...)$

[hide]Adding those two up gives $ (\sqrt {a} \plus{} \sqrt {b})(a \plus{} b) \equal{} 365$

notice that $ (\sqrt {a} \plus{} \sqrt {b})^3 \equal{} a\sqrt {a} \plus{} b\sqrt {b} \plus{} 3(a\sqrt {b} \plus{} b\sqrt {a}) \equal{} 182 \plus{} 3\times182 \equal{} 729$

$ \sqrt {a} \plus{} \sqrt {b} \equal{} 9.$
And it follows that $ a \plus{} b \equal{} \frac {365}{9}$

$ a \equal{} 81 \plus{} b \minus{} 18\sqrt {b}$
$ 2b \plus{} \frac {364}{9} \equal{} 18\sqrt {b}$
$ b \plus{} \frac {182}{9} \equal{} 9\sqrt {b}$
$ b^2 \plus{} \frac {364b}{9} \plus{} \frac{33124}{81} \equal{} 81b$
$ 81b^2 \minus{} 3285b \plus{} 33124 \equal{} 0$
$ b \equal{} \frac {196}{9}$ or $ \frac {169}{9}$

$ a \equal{} \frac {169}{9}$ or $ \frac {196}{9}$

so the solution set (a,b) is $ (\frac {169}{9},\frac {196}{9})$, $ (\frac {196}{9},\frac {169}{9})$[/hide]


For each real number $x$, let $f(x)$ be the minimum of the numbers $4x+1$, $x+2$, and $-2x+4$. Then the maximum value of $f(x)$ is
$\text{(A)} \ \frac 13 \qquad \text{(B)} \ \frac 12 \qquad \text{(C)} \ \frac 23 \qquad \text{(D)} \ \frac 52 \qquad \text{(E)} \ \frac 83$

Below is a hint. Try solving the problem before resorting to it, though.

[hide="big hint"]graph[/hide]

[hide="Solution"]
The three lines intersect at $x=\frac{1}{3},\frac{1}{2},\frac{2}{3}$.

In the interval $-\infty <x\le \frac{1}{3}$, $4x+1$ is minimum. Thus the maximum in this interval is $4\left(\frac{1}{3}\right)+1=\frac{7}{3}$.

In the same manner...
In the interval $\frac{1}{3}\le x\le \frac{2}{3}$, $x+2$ is minimum and its maximum value is $\frac{2}{3}+2=\frac{8}{3}$.
In the interval $\frac{2}{3}\le x< \infty$, $-2x+4$ is minimum and its maximum is also $\frac{8}{3}$ at $x=\frac{2}{3}$.

Therefore, the answer is $\frac{8}{3}$, or $\boxed{\text{(E)}}$.
[/hide]

[quote="frt"][quote="4everwise"][quote="catcurio"][hide]Graph the lines $y=4x+1$, $y=x+2$, and $y=-2x+4$. They intersect at $\left(\frac{1}{3},\frac{7}{3}\right)$, $\left(\frac{1}{2}, 3\right)$, and $\left(\frac{2}{3},\frac{8}{3}\right)$. The smallest x-value is $\frac{1}{3}\Rightarrow\boxed{A}$.[/hide][/quote]It asks for the maximum. :)[/quote]
You should change fix your title ("post subject") so that people won't be confused ;)[/quote]Um... Oops.  :rotfl:


probability geometry trapezoid inequalities

Alex needs to catch a train. The train arrives randomly some time between 1:00 and 2:00, waits for 10 minutes, and then leaves. If Alex also arrives randomly between 1:00 and 2:00, what is the probability that the train will be there when Alex arrives?

Basically, you try to illustrate the probability with geometry.

First, divide the $ x$ and $ y$ axes into $ 6$ congruent parts.  They will represent $ 10$ minute sectors.

Then, we need to draw $ 2$ lines.

The first boundary occurs when Alex arrives at the same time as the train.  This is simply the line $ y \equal{} x$.

The second boundary occurs when Alex arrives $ 10$ minutes after the train.  This is simply the line $ y \equal{} x \plus{} 10$.

[asy]draw((0,0)--(6,0)--(6,6)--(0,6)--cycle);
fill((0,0)--(6,6)--(5,6)--(0,1)--cycle,red);
label("$1:00$",(0,0),SW);
label("$2:00$",(6,0),SE);
label("$2:00$",(0,6),NW);
label("$1:30$",(3,0),S);
label("$1:30$",(0,3),W);[/asy]

Now, we must find the area of the trapezoid.

Okay, so the longer base is simply $ \sqrt {60^2 \plus{} 60^2}$ (do you see the right triangle?), so that base is $ 60\sqrt2$.  The shorter base is $ \sqrt {50^2 \plus{} 50^2}$ (do you see that one?), or $ 50\sqrt2$.  Thus, the average of the bases is $ 55\sqrt2$.

Now, we need to find the height.  You can notice on the bottom left hand corner of the trapezoid, you have a rt. isosceles triangle with hypotenuse $ 10$.   So, the height then must be $ 5\sqrt2$.  So, multiplying these two numbers reveals the area of the trapezoid is $ 550$.  Thus, the answer is $ \frac {550}{60^2}$, or $ \boxed{\frac {11}{72}}$.

Dang, this thing took a long time to type up.

If we let $ x$ represents the time for Alex to arrive and $ y$ for the train, shouldn't we only have the inequality $ y>x\minus{}10$ bounded by $ 0<x,y<60$?  I don't think we need $ y\equal{}x$ since Alex can arrive before the train (so that he waits for the train to arrive...).

The question asks for the probability that Alex arrives [b]while[/b] the train is there :wink:

But how do we know how long Alex stayed.  It never mentions that. Are we supposed to assume that he also stays for 10 minutes?

It doesn't matter.  If Alex gets there first, then the train comes while he's waiting, then Alex was there when the train arrived.  We want the other way around :wink:


Let $ a_1,a_2,\ldots,a_n$ be positive real numbers, we denote $ A\equal{}\sum^n_{i\equal{}1}a_i\minus{}n, B\equal{}\sum^n_{i\equal{}1}\frac{1}{a_i}\minus{}n$ and $ C\equal{}a_1a_2\cdots a_n$. Prove that
$ (A\minus{}1)(B\minus{}1)\plus{}2\sqrt{\left(A\plus{}\frac{1}{C}\minus{}1\right)\left(B\plus{}C\minus{}1\right)} \ge 1.$

Can we replace the constant $ 2$ by a smaller constant?

[quote="can_hang2007"]Let $ a_1,a_2,\ldots,a_n$ be positive real numbers, we denote $ A \equal{} \sum^n_{i \equal{} 1}a_i \minus{} n, B \equal{} \sum^n_{i \equal{} 1}\frac {1}{a_i} \minus{} n$ and $ C \equal{} a_1a_2\cdots a_n$. Prove that
$ (A \minus{} 1)(B \minus{} 1) \plus{} 2\sqrt {\left(A \plus{} \frac {1}{C} \minus{} 1\right)\left(B \plus{} C \minus{} 1\right)} \ge 1.$

Can we replace the constant $ 2$ by a smaller constant?[/quote]
The constant $ 2$ is the best. To prove this, it suffices to set $ a_1\equal{}...\equal{}a_{n\minus{}1}\equal{}1$.

Apparently, this inequality (1) is stronger than my inequality
$ (2)\ \ \ \ AB\plus{}C\plus{}\frac 1{C}\ge 2$,
since from (1) and AM-GM inequality we can obtain (2).
But from (2) we also can obtain (1). For this, write (2) for $ n\plus{}1$ numbers $ a_1,..., a_n,a_{n\plus{}1}$ and replace
$ a_{n\plus{}1}\equal{}\sqrt\frac{A\plus{}\frac 1{C}\minus{}1}{B\plus{}C\minus{}1}$. :)


Ring Theory
function
algebra
polynomial
superior algebra
superior algebra solved

(A,+,.) is a ring which has the property that for all x<>0 there exist m,n in N* s.t. x=(x^m+1)^n. Show that all endomorphisms of A are automorphisms.

m and n depend on x, right? That's what I understand, but I wanted to make sure.

let f be an endomorphism of A.

Supose there exists x<>0 such that f(x)=0.
There are m,n such that x=(x m+1) n.
Then f(x)=f((x m+1) n)=(f(x) m+1) n. This leads to a contradiction.

So ker(f)=0, so f is injective, so f is bijective.

Who said anything about the ring being finite?  :D  :D
You should prove that f is surjective.

I am ashamed, but I must confess that I am not sure about this.

If f is an injective endomorphism of A then A is isomorphic to Imf. Can you give an example of a ring isomorphic to one of its subrings?

Good question. My intuition says that such rings must exist, but I do not have any example yet. :)

Maybe the following is an example:
   Take A the ring of rational functions in x and the subring formed by the rational functions in x^2.

Your example looks OK, harazi. I should move this to Solved Problems, as I have a solution now. How do I do it?

I think that only the moderators on this subforum can do that.

I have a solution to the initial problem :) But I am waiting one from you guys also :)

Then it should go into the "Proposed Problems" section, not the "Solved Problems" section.

amfulger has already shown that any morphism is injective, so all we need now is to show surjectivity.

It's clear that (A,+,.) is a field because for any x<>0 we have 1=x-(xm+1)n+1, and the left part can be written as x*(something)=(something)*x, so all x<>0 are invertible.

Let Sm,n be the set of all roots of the polynomial (xm+1)n-x which belong to A, for any m,n\in N*. Since A is a field, it means that all such sets are finite.

Let f\in End(A). It's clear that if x\in Sm,n, then f(x)\in Sm,n as well, and since f is injective and Sm,n, it means that f induces a bijection from Sm,n to Sm,n for all pairs of positive integers m,n. Any x\in A* belongs to some set of the form Sm,n and 0\in Im(f), so any x\in A belongs to Im(f), so f is surjective.

It's a really nice problem!

It's the same solution I know. Have you found it yourself? I thought the idea not natural at all. :D  but indeed very nice. :D

Oh, I did find it myself, and I was quite surprised that I thought of it, because it seems unnatural to me too. I guess I was sort of lucky :).


geometry
circumcircle
perpendicular bisector

[color=darkblue]Let $ABC$ be a right triangle $(AB\perp AC)$. The its incircle $w=C(I,r)$ touches the sides $[AB]\ ,\ [AC]$ in the points $E\ ,\ F\ .$

Prove that the intersections of the line $EF$ with the lines $BI\ ,\ CI$ belong to the circumcircle of the triangle $ABC\ .$[/color]

Let $BI$ meets $EF$ at point $D$. Since $\angle{A}=90^{o}$, so $AFIE$ is a square. Thus $EF$ is the perpendicular bisector of $AI$, hence $\triangle{DIA}$ is an isoseles triangle. Because

$\angle{BDE}=180^{o}-135^{o}-\frac{\angle{B}}{2}=45^{o}-\frac{\angle{B}}{2}$

so

$\angle{ADB}=2\angle{BDE}=90^{o}-\angle{B}= \angle{C}$

Therefore $D$ lies on the circumcircle of triangle $ABC$.


number theory
least common multiple
number theory unsolved

Is there a natural number  $ n > 10^{1000}$ which is not divisible by 10 and which satisfies: in its decimal representation one can exchange two distinct non-zero digits such that the set of prime divisors does not change.

I assume the two non-zero digits have to be distinct....otherwise the problem is trivial :D

Pierre.

Obviously you are right. I forgot the word.  :)

Anyone would like to give me a hint on this?

Thanks in advance. :)

Clearly this is the only way to prove it:
Consider the $2$ numbers as: $\prod{p_i ^{a_i+b_i}}$ and $\prod{p_i ^{a_i+c_i}}$ with all $a_i \geq 1$, $b_i, c_i \geq 0$, and for each $i$ one of $b_i, c_i$ is $0$ but not all of them are $0$:
$10^k$ $<$ $\prod{p_i ^{a_i+c_i}}$  $<$ $\prod{p_i ^{a_i+b_i}}$  $<$ $10^{k+1}$, with $k \geq 1000$.
The difference is:
$\prod{p_i ^{a_i}}(\prod{p_i ^{b_i}}-\prod{p_i ^{c_i}})=d(10^m-1)10^n$, with $d \in ${$1,2,...,8$}, $m \geq 1$, $n \geq 0$.
Also we have $\prod{p_i ^{c_i}}$ $<$ $\prod{p_i ^{b_i}}$ $<$ $9 \prod{p_i ^{c_i}}$, then get some contradiction.

You can also see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=28220]this[/url] thread for another solution.

[quote]You can also see this thread for another solution.[/quote]

according to http://www.mathlinks.ro/Forum/viewtopic.php?t=28220   there are some numbers, but here [b]heartwork[/b]  got some contradiction!
I confused!!!!

[quote="heartwork"]Clearly this is the only way to prove it:
Consider the $2$ numbers as: $\prod{p_{i}^{a_{i}+b_{i}}}$ and $\prod{p_{i}^{a_{i}+c_{i}}}$ with all $a_{i}\geq 1$, $b_{i}, c_{i}\geq 0$, and for each $i$ one of $b_{i}, c_{i}$ is $0$ but not all of them are $0$:
$10^{k}$ $<$ $\prod{p_{i}^{a_{i}+c_{i}}}$  $<$ $\prod{p_{i}^{a_{i}+b_{i}}}$  $<$ $10^{k+1}$, with $k \geq 1000$.
The difference is:
$\prod{p_{i}^{a_{i}}}(\prod{p_{i}^{b_{i}}}-\prod{p_{i}^{c_{i}}})=d(10^{m}-1)10^{n}$, with $d \in${$1,2,...,8$}, $m \geq 1$, $n \geq 0$.
Also we have $\prod{p_{i}^{c_{i}}}$ $<$ $\prod{p_{i}^{b_{i}}}$ $<$ $9 \prod{p_{i}^{c_{i}}}$, then get some contradiction.[/quote]

I don't understand where's the contradiction ??  :maybe:

One of the solution can be:
$ (17\times \frac{10^{560t}-1}{9},71\times\frac{10^{560t}-1}{9} )$
Where $lcm[\varphi (17),\varphi (71)]=560$ and change the first and last digits.
inspired from [b]vess[/b] solution. :)

(Solution with niyu)

Let $\alpha = \dfrac{10^{1080}-1}{9}$. Then, the numbers $13 \alpha$ and $31 \alpha$ work. They only differ in the first and last digit which can be swapped.


Let  x,y,z>=0  and  x+y+z=1. Show  that  x^2*y+y^2*z+z^2*x<=4/27
canadian  mo , 1999
Thanks!!
I allready  know  the  official  solution  so  im  looking  for another  one
Thanks  again !

Two solutions can be found [url=http://db.math.ust.hk/resource_sharing/selection/se_5th_hkcmo.pdf]here[/url].

leepakhin, it's a different problem.  I have posted solutions to both, but am too lazy to find them now.

The key to nickolas' inequality is to note that if

$f(x,y,z)=x^2y+y^2z+z^2x$, then

$f(x+z,y,0)\geq f(x,y,z)$

From there, it becomes a two variable inequality which is easily handled by AM-GM.


For the  function $f: \mathbb{R}\rightarrow\mathbb{R}$ we have $f(x^2+y^2)=f(x)+f(y)$ for all $x,y\in\mathbb{R}.$ Prove that  $f(x)=0$ for all $x\in\mathbb{R}.$

Putting $x=y=0$ we get $f(0)=2f(0)$ or $f(0)=0.$

Then putting $y=0$ we get $f(x)=f(x^2)$ which yields $f(-x)=f((-x)^2)=f(x^2)=f(x)$, hence $f$ is even.

On the other hand, $f(x)=f(x^2)$ permits us to write $f(x^2+y^2)=f(x^2)+f(y^2)\implies f(x^2)=f(x^2+y^2)-f(y^2).$ Putting $u^2=x^2+y^2, v^2=y^2$, we get $f(u^2-v^2)=f(u^2)-f(v^2)$, which implies $f(-v^2)=-f(v^2)$, hence $f$ is odd.

The only function that's both even and odd is $f(x)\equiv 0.$


[quote=tdl]Let $ P \in Z[X]$ be a monic irreducible (over $ Z[X]$) polynomial so that $ |P(0)|$ is not a perfect square. Prove that $ P(x^2)$ is irreducible over $ Z[X]$.[/quote]
It was [url=http://artofproblemsolving.com/community/c6h130843]here[/url].


Hello,

I am having trouble expressing the following trig equation with only one variable. Any help or hints shall be appreciated.

Thanks!

---

1. Solve $ 2 \tan x \plus{} \sec x \equal{} 1$, for $ 0 \leq x \leq 2\pi$.

$ 2 \tan x \plus{} \sec x \equal{} 1$

$ 2 \frac{\sin x}{\cos x} \plus{} \frac{1}{\cos x} \equal{} 1$

$ \frac{1}{\cos x}(2 \sin x \plus{} 1 \minus{} \cos x ) \equal{} 0$

Now I am not sure how to continue. Do I state that $ \frac{1}{\cos x}$ = 0 and solve for this part separately since this trig equation is equal to 0?

Well, you can always multiply both sides with $ \cos x$ since $ \cos x$ isn't zero (if it were 0, the first equation could never be fulfilled).

(You can take $ \frac{1}{\cos x}\equal{}0$ as one case, but it definitely has no solutions.

Thanks for your reply, hsiljak.

However, if I multiply both sides by $ \cos x$, I would get:

$ 2 \sin x \plus{} 1 \equal{} \cos x$

$ 2 \sin x \minus{} \cos x \plus{} 1 \equal{} 0$.

There are still two trig functions in the equation.

$ 2 \sin x \minus{} \cos x \plus{} 1 \equal{} 0$ can be transformed to a form: $ \sin(x \plus{} \alpha) \equal{} \beta$ where $ \alpha,\beta$ are real constants. Can you find $ \alpha\beta$ (using additive formula: $ \sin x \cos \alpha \plus{} \cos x \sin \alpha \equal{} \sin(x \plus{} \alpha)$?

Alternatively, you could transform it to the analogue form $ \cos(x \plus{} \alpha) \equal{} \beta$, and use additive formula for the cosine of the sum to find $ \alpha$ and $ \beta$.

[hide="Solution"] We'll do it with cosines. Notice that the ratio between the coefficients with sines and cosines is 2 (ignore the minus, it is already in the additive formula), hence the angle which we are chasing ($ \alpha$) is $ arc \tan 2$. Dividing whole equation with $ \sqrt {1^2 \plus{} 2^2}$ (standard procedure, it gets us to the fact that now our new coefficients with sines and cosines have the nature of sines and cosines-you can look at first coefficients as legs of a right triangle, so the division is being made with the hypotenuse). Now, we have:

$ \frac {1}{\sqrt 5}\cos x \minus{} \frac {2}{\sqrt 5}\sin x \equal{} \frac {1}{\sqrt 5}$.

Now, we get

$ \cos(x \plus{} arc \tan 2) \equal{} \frac { 1}{\sqrt 5}$

There's your one-function equation.[/hide]

$ 2\tan{x} = 1 - \sec{x}$
$ 4\tan^2{x} = 1 - 2\sec{x} + \sec^2{x}$
Then with $ y = \sec{x}$
$ 3y^2 + 2y - 5 = 0\Rightarrow y = 1, - \frac {5}{3}$

$ \cos{x} = 1\Leftrightarrow x = 2\pi n$ and $ \cos{x} = - \frac {3}{5}$.
We know that squaring yields and extraneous root so checking for $ x = 0$ in $ 2\tan{x} + \sec{x} = 1$, we observe that tangent is undefined for any $ x = 2\pi n$. Using triangles in the unit circle, $ \cos{x} = - \frac {3}{5}\Leftrightarrow \sin{x} = \pm \frac {4}{5}$. We can plug that into the original equation to get $ \pm \frac {8}{3} - \frac {5}{3} = 1\Leftrightarrow \sin{x} = \frac {4}{5}$. Finally, $ \arcsin{\bigg(\frac {4}{5}\bigg) = 126.8698976^{\circ}}$

edits: thought $ \sqrt {64} = 4$ hehe  :oops:

[quote="igiul"]
$ 4\tan^2{x} \equal{} 1 \minus{} 2\sec{x} \plus{} \sec^2{x}$
Then with $ y \equal{} \sec{x}$
$ 3y^2 \plus{} 2y \minus{} 5 \equal{} 0\Rightarrow y \equal{} 1, \minus{} \frac {5}{3}$
[/quote]

Thank you for your replies.

Igiul, could you please explain the following step?

$ 4\tan^2{x} \equal{} 1 \minus{} 2\sec{x} \plus{} \sec^2{x}$

$ 4\sin^2 x \sec^2 x \equal{} 1 \minus{} 2\sec x \plus{} \sec^2 x$
$ 4(1 \minus{} \cos^2 x)\sec^2 x \equal{} 1 \minus{} 2\sec x \plus{} \sec^2 x$
$ 4\sec^2 x \minus{} 4 \equal{} 1 \minus{} 2\sec x \plus{} \sec^2 x$

Then with $ y \equal{} \sec{x}$
$ 3y^2 \plus{} 2y \minus{} 5 \equal{} 0\Rightarrow y \equal{} 1, \minus{} \frac {5}{3}$

I took the liberty to explain it instead of [b]igiul[/b].  :lol:

I think igiul is assuming that you know the identity $ 1\plus{}\tan^2x\equal{}\sec^2x$, a result of the pythagorean theorem.

$ 2\tan{x}\plus{}\sec{x}\equal{}1$
$ 2\tan{x}\equal{}1\minus{}\sec{x}$

Multiply both sides by $ \minus{}(1\plus{}\sec{x})$

$ 2\tan{x}(\minus{}(1\plus{}\sec{x}))\equal{}(1\minus{}\sec{x})(\minus{}(1\plus{}\sec{x})$
$ (2\tan{x}(\minus{}1\minus{}sec{x})\equal{}tan^2{x}$
$ 2(\minus{}1\minus{}sec{x})\equal{}tan{x}$

Now you can procede from there

Taking it from there,
\[ \minus{} 4(1 \plus{} \sec{x}) \equal{} 2\tan{x} \equal{} 1 \minus{} \sec{x}
\]

\[ \Leftrightarrow \sec{x} \equal{} \minus{} \frac {5}{3}
\]
And then the above result follows.
edit: what inspired multiplying by -(1+sec{x})?


This problem's been bugging me for a while, maybe someone here could help:

Given $(n-1)k+1$ positive integers $a_{1}, a_{2}, \ldots, a_{(n-1)k+1}$ such that the product contains $\le k$ prime factors, show that we can select some of the integers such that the product $a_{i_{1}}a_{i_{2}}\cdots a_{i_{j}}$ is an $n$-th power.

The weaker case for n prime is N8 in shortlist for 2003 IMO Japan if that will help.

Oh, thanks, that means my problem is a bit harder than I thought it was... :P
I'll check out the solution from the shortlist and see if it helps.


AMC
AIME
complex numbers
geometry
cyclic quadrilateral

Can someone please explain a couple of problems on the 1983 AIME, specifically 4, 5, and part of 14.  These come directly from [url]http://www.kalva.demon.co.uk/aime/aime83.html[/url].

4. A and C lie on a circle center O with radius $\sqrt{50}$. The point B inside the circle is such that angle ABC = 90, AB = 6, BC = 2. Find OB.
(see website for picture)
Here was what I did, which does not get the right answer:
[hide]I drew in AC=$\sqrt{40}$ then used Ptolemy's theorem to find OB.  This gives you $2\sqrt{5}$, which isn't correct.[/hide]

5.  w and z are complex numbers such that $w^2 + z^2 = 7, w^3 + z^3 = 10$. What is the largest possible real value of w + z?

  I get 14 but part of the solution leads me to suspect there is an easy way to solve it.
  The question is: 14. The distance AB is 12. The circle center A radius 8 and the circle center B radius 6 meet at P (and another point). A line through P meets the circles again at Q and R (with Q on the larger circle), so that QP = PR. Find QP^2. (see website for picture)
  And the solution (and my question) are hidden.
[hide]    Put QP = 2a, take perp dist A, B from QP be b, c. Take line parallel to QP through B meeting perp bisector of QP at D. Considering triangle ADB gives 4a^2 + (b-c)^2 = 144. Also b^2 = 64 - a^2, c^2 = 36 - a^2. Hence 4a^2 = 130.
     MY QUESTION: Is there an easy way to get from the equations to 4a^2=130?[/hide]

Also, does the AIME get harder after the first couple years?  Because it seems like the 2004 one was much harder than the 1983 one.

Ya the first few AIMEs are easy. The 1985 year was super easy, I got a 9 on that.

[quote]I drew in AC= then used Ptolemy's theorem to find OB. This gives you 14, which isn't correct.
[/quote]

Ptolemy's requires a cyclic quadrilateral. What makes you think this quadrilateral is cyclic?

[quote="Wumbate"]

5.  w and z are complex numbers such that $w^2 + z^2 = 7, w^3 + z^3 = 10$. What is the largest possible real value of w + z?
.[/quote]
[hide]$w+z=x, wz=\frac{x^2-7}{2}$

$(w+z)(w^2+z^2)=(w^3+z^3)+wz(w+z)$

$7x=10+\frac{x(x^2-7)}{2}$

$x^3-21x+20=0$

Since this is an AIME problem, that equation must factor.
Rational Root Thm
so that's $(x-1)(x-4)(x+5)=0$, largest possible value=4[/hide]

[quote="Wumbate"][quote]Ptolemy's requires a cyclic quadrilateral. What makes you think this quadrilateral is cyclic?[/quote]
It looked like it was.  What are the criteria for a cyclic quadrilateral?[/quote]

Opposite angles add up to 180 degrees, which means it can be circumscribed by a circle.

Ok thanks, now I understand why I got 4 wrong. Can anyone explain 5 and 14 (I think there's a shortcut in the algebra to 14)?

[quote="Wumbate"][quote]Ptolemy's requires a cyclic quadrilateral. What makes you think this quadrilateral is cyclic?[/quote]
It looked like it was.  What are the criteria for a cyclic quadrilateral?

And beta, where did you come up with $wz=\frac{x+7}2$?[/quote]

It was $wz=\frac{x^2-7}{2}$

[hide]the reason is because $w^2+z^2+2wz=(w+z)^2=x^2$,
since $w^2+z^2=7$, so $7+2wz=x^2$, or $wz=\frac{x^2-7}{2}$[/hide]

[quote="DPopov"]here is my solution...there is also another one using an isosceles triangle...

[hide]Call the center of the circle $O$. Extend $AB$ and $BC$ to become chords  of the cirlce and call their points on the circumference $G$ and $D$ respectively. Now draw the diameter of circle $O$ perpendicular to chord $AB$ (let $P$ and $Q$ be the left and right points of this diameter on the circumference) and call their intersection $F$. Draw the diameter of circle $O$ (let $W$ and $Z$ be the top and bottom points of this diameter on the circumference) perpendicular to chord $CD$ and call their intersection $M$. Let $OM=FB=x$ and $BM=OF=y$. Using the properties of chords we can conclude the following

\[(WM)(MZ)=(CM)(MD)\]
\[(\sqrt{50}-x)(\sqrt{50}+x)=(y+2)^2\]
\[50-x^2=(y+2)^2\]
\[x^2+(y+2)^2=50\]

Also

\[(AB)(BG)=(PF)(FQ)\]
\[6(6-2x)=2(2+2y)\]
\[18-6x=2+2y\]
\[y=8-3x\]

Subsituting this result into our previous result

\[x^2+(8-3x+2)^2=50\]
\[x^2+9x^2-60x-50=0\]
\[x^2-6x-5=0\]
\[(x-1)(x-5)=0\]

Hence $x=5$ or $x=1$. We quickly see that $x=1$ since $y$ must be positive and if $x=5$ then $y$ is negative. Thus, $x=1$ and $y=8-3=5$. Returning back to our picture we see

\[OM^2+BM^2=OB^2\]

Since $x=OM$ and $y=BM$

\[x^2+y^2=OB^2\]
\[OB^2=1+25=26\]

Therefore our answer is $26$.

[/hide][/quote]

To bump an old thread...how on earth is $(PF)=2$? It is bigger then the radius which is $\sqrt{50}$

Not to mention that expansion is all wrong

The answer is right, but he just miswrote some things.  First it should have been $(AB)(BG)=(BC)(DB)$, and then it should have been +50 instead of -50.  Which would make it $x^2-6x+5=0$ instead of $x^2-6x-5=0$ but he factored the correct one to get the right answer.

I think that the aime has gotten harder because calculators weren't widely used in the 1980s


1) Define two sequences of rational numbers as follows: let $a_{0}=2$ and $b_{0}=3$, and recursively define $a_{n}=\frac{a_{n-1}^{2}}{b_{n-1}}$ and $b_{n}=\frac{b_{n-1}^{2}}{a_{n-1}}$. Find $b_{8}$, leaving your answer in the exponential form $m^{n}/p^{q}$. (Mandelbrot #2)

2) If the sequence ${a_{n}}$ is recursively defined by $a_{1}=2$, and $a_{n+1}=a_{n}+2n$ for $n\ge 1$ then find $a_{100}$. (AHSME 1984)

3) Let a sequence ${u_{n}}$ be defined by $u_{1}=5$ and the relation $u_{n+1}-u_{n}=3+4(n-1)$, $n=1,2,3,...$. If $u_{n}$ is expressed as a polynomial in $n$, what is the algebraic sum of its coefficients? (AHSME 1969)

[hide="Hint and answer for #1"]Observe $a_{n}b_{n}$ and ${a_{n}\over b_{n}}$. The answer is $b_{8}=\frac{3^{3281}}{2^{3280}}$[/hide]


MATHCOUNTS
number theory
least common multiple
LaTeX
absolute value

83-84 State Team MathCounts

#4 Find how many pairs of integers (x,y) satistfy the following conditions:

[img]http://img.photobucket.com/albums/v221/Farazy/83_84_ST04.gif[/img]

What is the most systematic way of solving this problem?  I got it, but it took to me a lot longer then I think it should have.

84-85 Chapter Team MathCounts
#2 What is the sum of the three greatest consecutive integers less than 200 for which theleast number has 4 as a factor the next number has five as a factor and the greatest has 6 as a factor?

What is the most systematic way of solving this problem too?

I dont know if this was the right way of doing this but...

[hide="Click Here to View My Method"]
So middle number has to have a factor of 5.  It must end in a 5 or a 0.  If it ends in a 0 the preceding terms units digit will end in 9, and no multiple of 4 ends in 9.  So it has to end in 5.  Since we need the greatest I started from 195, and tested numbers. 194 isn't divisible by four so I went down 10 to 185 and 184 was divisible by 4 and 186 was divisible by 6.  184 + 185 + 186 = 555

Is this the best way?

[/hide]

#6 All the digits used to number the pages of a magazine were cut out and tossed in a box. If the box contained 204 digits, how many pages did the magazine have?

I keep on doing something wrong... Not really sure what, and I always get the wrong answer.

[hide="I think this is the correct answer (I didn't come up with this.  Can be wrong.)"]104 Pages[/hide]

[quote="Farazy"][hide="Click Here to View My Method"]
So middle number has to have a factor of 5.  It must end in a 5 or a 0.  If it ends in a 0 the preceding terms units digit will end in 9, and no multiple of 4 ends in 9.  So it has to end in 5.  Since we need the greatest I started from 195, and tested numbers. 194 isn't divisible by four so I went down 10 to 185 and 184 was divisible by 4 and 186 was divisible by 6.  184 + 185 + 186 = 555

Is this the best way?

[/hide][/quote]

[hide="A Way Which Is Harder But Eliminates the Trial and Error"]The lowest number minus four equals the middle number minus five equals the highest number minus six. Let's call that number x. x must be divisible by four, five, and six, because the numbers four,five,and six above it are divisible by four,five,and six respectively. So x is the hightest number divisible by four,five,and six less than 200. That number is a multiple of 60, the LCM of four,five,and six. The highest multiple of 60 less than 200 is 180.[/hide]

[quote]#2. What is the sum of the three greatest consecutive integers less than 200
for which theleast number has 4 as a factor the next number has 5 as a factor
and the greatest has 6 as a factor?[/quote][hide]
$\text{From smallest to largest: let }a = 4p,\;b = 5q,\;c = 6r$

$\text{Since }c = b+1\text{ and }b = a+1:$
$.\quad\text{we have: }6r = 5q+1\;\;\Rightarrow\;\;q= \frac{6r-1}{5}\;\;[1]$
$.\qquad\;\;\text{and: }5q=4p+1\;\;\Rightarrow\;\;p = \frac{5q-1}{4}\;\;[2]$

$\text{Since }c < 200\;\;\Rightarrow\;\;6r < 200\;\;\Rightarrow\;\;r \leq 33,$

$\text{The largest value of }r\text{ that satisfies [1] is: }r = 31$

$.\qquad\text{Hence: }q = 37,\;p = 46$

$\text{Therefore: }a = 184,\;b = 185,\;c = 186.\;\;\text{Their sum is }555.$[/hide]

[hide]For your problem on the 3 integers, the answer is 555=184+185+186.  The easy way is to find numbers that are 4,5,6 more than the LCM of 4,5,6 which equals 60.  The highest CM under 200 is 180 so 184+185+186=555.[/hide]
[hide]  For the absolute value problem, it's not hard.  Remember that absolute value is always positive.  Since that's getting multiplied by -2, -2|x-5| will always have a negative number.  Since we don't want y to be negative, we need to find the numbers that do not make y negative.  The smallest x that satisfies the equation is x=2 and the highest is x=8.  So there's seven pairs, (2,0) (3,2) (4,4) (5,6) (6,4) (7,2) (8,0)[/hide]
   [hide]Page 1-9 has 9 digits.  Page 10-19 has 20 digits therefore, from 10-99 has 180 digits.  Since the magazine only has 204 digits, we're almost there.  Precisely, we have 15 digits left and each of the hundreds have 3 digits each so we have 5 numbers counting 100.  The magazine has 104 pages.[/hide]

[quote="Farazy"]I have a question...This is LaTeX related.  How do you guys convert the divs to gifs? I end up having to use the TeXer...Which is somewhat annoying to have to load to everytime.

Any alternatives?[/quote]

You can type the code you type into the TeXer directly into your post. (But with the dollar signs.)

$\alpha$

You can tell whether the LaTeX was rendered directly by the board or by another source like the TeXer is you hover your mouse over it. You can see the code if the board rendered it. You can click on long piece of code to get a window with the code which you can put on your clipboard.

$\alpha\beta\delta\gamma\zeta$

I mean generally.  If I want to get LaTex as a PNG or GIF image what do i do?

for the one about the number and thier sum:[hide]
the lowest number - 4 = middle number - 5 = highest number - 6 = x..... this means that 4,5,6|x... which means that 60|x.
we know that highest multiple of 60 less than 200 is 180, so our numbers are: 184, 185, and 186. sum is [b]555[/b][/hide]
for the one about pages:
[hide]I'll make a little chart so it is easier to under stand. chart in form: number1 - number 2 : digits

1-9: 9 digits
10-99: 180 digits
so far we have 189 digits... since box contained 204 digits we only have 15 more to go! which means that we have five more number left. these five numbers are: 100, 101, 102, 103, and [b]104[/b][/hide]
for the absolute value problem:
[hide]we know that all absolute values are positive, and we want y to be non-negative... so:
we can substitute here:
0 <= -2|x-5| +6
3 >= |x-5|
now we can easily solve this:
3 >= x-5             or -3 <= x-5
our solution set is:
2 <= x <= 8
there are 7 integer x's here, so there are [b]7[/b] solutions sets to the original question.[/hide]

Nobody has got the absolute value problem right yet.

good point... I only found the total number of x's:
the problem is easy enough now: plug the values of x in and see how many y's we get


limit
real analysis
real analysis unsolved

we know that
$\-1\leq cos(x) \leq 1$ for each x
then
$\lim_{ k \to\infty}(cos(\pi xn!))^{2k}=0$ or $\infty$ if k--->-00

[quote="mlok"]$1$ is $x$ is rational, $0$ if $x$ is irrational.[/quote]

I agree with this. :D


How many four digit numbers have exactly three distinct digits?

[quote="besttate"][hide]Case 1: The second occurrence of the non-disticnt digit is in the hundreds place.
9*1*8*7=504

Case 2: The second occurrence of the non-disticnt digit is in the tens place.
9*8*2*7= 1008

Case 3: The second occurrence of the non-disticnt digit is in the ones place.
9*8*7*3=1512

504+1008+1512=3024[/hide][/quote]

You didn't count the cases with $0$. That's the hardest part.

BTW, your stuff can be simplfied into: we choose three numbers, pick one to be the repeated one, and arrange them. This is

$\binom{9}{3}\cdot 3 \cdot \frac{4!}{2!}= 3024$.

Well including 0 shouldn't be that difficult. All we need to do is multiply by 9/7 and get 3888. Below is a more thorough solution however.

9*1*9*8 + 9*9*2*8 + 9*9*8*3= 648+1296+1944=$\boxed{3888}$


I suppose that means different. More general problem: find all integers $n$ such that $\frac{1-ni}{1+ni}$ is a root of unity. And this one in not the most general.  ;)

the original problem seems false...
but take a look here: http://www.mathlinks.ro/Forum/viewtopic.php?t=42893


LaTeX
induction
algebra proposed
algebra

I have a problem and I haven't got any idea how to solve it , so here it is :

Sequence ${x_n}$ is defined :

$x_0 \in [0,1]$ , $a > 2$ , $x_{n+1}=ax_n(1-x_n)$ .

Question : Depending on $a$ , for what values of $x_0$ sequence
$x_n$ is convergent ?

Hello.

Taking the limits in the definition of the sequence shows that a limit z must satisfy
z = az(1 - z). Thus 0 and 1 - 1/a are the only possible limits.
(They occur effectively. Take e.g. the initial value equal to one of these two values; then the sequence is constant.)

I think it's possible to prove that if the sequence converges towards one of the two values 0, 1 - 1/a, then the sequence takes this value (and thus becomes constant).

Let's suppose the sequence has limit 0.
We can choose c such that 0 < c < 1 - 1/a (because a is assumed > 2 and a fortiori > 1).
Since we assume the sequence has limit 0, there exists i such that for each j satisfying j >= i, we have X(j) <= c, thus 1 - X(j) >= 1 - c.
For such a number j, the definition X(j + 1) = a(1 - X(j) ) X(j) shows that
|X(j + 1)| >= a(1 - c)|X(j)|
(In fact, we could prove that the values of the sequence are all in [0, 1], so the || are nunnecessary.)
By recurrence on k, it follows that for each k >= 0, |X(i + k)| is at least equal to |X(i)| mutiplied by the k-th power of a(1 - c).
But we have chosen c such that a(1 - c) > 1. Since the sequence has limit 0, it follows that X(i) = 0.

We have thus proved : if the sequence has limit 0, it takes the value 0.

Assume now that the sequence has limit 1 - 1/a. Put Y(n) = X(n) - (1 - 1/a). Thus
Y(n) has limit 0.
The definition of the sequence X(n) gives :
Y(n + 1) = Y(n) (2 - a - aY(n) ).
A similar reasoning as above shows that Y(n) takes the value 0, in other words X(n) takes the value 1 - 1/a. (I will write the reasoning explicitly if somebody asks for it.)
(Edition : the proof is in a later post. In fact, it is valid only if a > 3.)

Thus the sequence X(n) has a limit if and only if X(0) can be found after successive solutions of :
ax(1 - x) = one of the already found solutions,
starting with "already found solution = 0" or "already found solution = 1 - 1/a".
(Edition : this must be corrected. If 2 < a <= 3 and X(n) -> 1 - 1/a, there can be other convenient values for X(0).)
The values 0, 1, 1 - 1/a and 1/a are convenient for X(0), but many others too, I think.

What are exactly the successive solutions ? The problem becomes tedious, don't you find ?

Panurge.

Thanks for reply , unfortunatelly - this problem is part of my homework...well it's additional work ... however I'm not sure whether the author of this problem thought through all possibilities :) .
Of course for $a \in [0,2]$ the limit is $\frac {a-1}{a}$or $0$-depends on the case  , what's easy to prove . Problems start when $a>2$ .
For instance let's put $a=4$ , $x_0=sin^2\alpha \in [0,1] $ . Then
$x_n=sin^22^n\alpha$.
For $x_n$ of the form $\frac{m\pi}{2^k}$ the limit is $0$ obviously ... but is there
another class of $x_0$ for which the limit is $0$ ? My intuition says that otherwise it
doesn't have a zero limit ...
The biggest problem is that we consider all $a>2$ . For $a>4$ we can have $x_n<0$ and then ${x_n}$ heads to $-\infty$.
By the way , for $a=3$ , if $x_0$ is very close to $\frac{2}{3}$ ( the point $g$ where $f(g)=g$ ) , it's limit is $\frac{2}{3}$ . I tried it out empirically :) on a picture :) .
...I think this problem hasn't got a precise solution ...

Hello Rafal.

You say that for a in [0, 2], the limit is 0 or 1 - 1/a.
Why do you restrict this to the case where a is in [0, 2] ?
Taking the limit in X(n + 1) = a X(n) (1 - X(n)) shows that the limit x satisfies :
x = ax(1 - x), thus the limit can only be 0 or 1 - 1/a.
Now, both limits occur really, e. g. the limit 0 for X(0) = 0 and the limit 1 - 1/a for X(0) = 1 - 1/a.
This doesn't depend on any hypothesis on a, I think (except that a is nonnull).
I think I have proved that if the sequence has a limit (thus converges towards 0 or 1 - 1/a), then the sequence takes this limit as one of its values and thus becomes constant after some index.
(Edition : see a later post. In fact, if the sequence has limit 1 - 1/a, the fact that it takes its limit as an effective value will be proved only if a > 3.)
It leads to a "solution" for X(0) in terms of successive solutions of " ax(1 - x) = one of the previous solutions ", starting with solution 0 or solution 1 - 1/a.

Now, there exists perhaps a more explicit and more elegant solution...

Panurge.

Excuse me but I think we are talking about two different things.
The problem is not whether the limit is 1-1/a or 0 . The problem is to find
all (!) x_0 for which {x_n} has a limit - depending on a (!). So actually we shouldn't even talk about the value of possible limits ( in this matter I agree with you in 100% ) but about their existance !
First we have to find all pairs (a,x_0) for which x_n is convergent - then we can
start thinking about the value of their limit  - this is the correct direction of solving the problem.
It's obvious that possible limit g must undo f(g)=g , cause this is a reccurency.
However there are of course values of x_0 for which x_n doesn't have a limit.
And that's the problem .
I say that for a>=4 and x_0=/=1-1/a or 0 or 1 or , {x_n} doesn't have a limit.
Proving it seems tedious of course , but I'll try .

This problem from my homework began
a) what happens when a=2 ( trivial one )
b) what happens when a=/=2 ? I've already checked a \in [0,2] , so now I wanted to check a>2 , and here is a problem ... that's why I've written it here.

ps. sorry I haven't used Latex but it's not necessary here .

[quote="panurge"]Hello Rafal.
This doesn't depend on any hypothesis on a, I think (except that a is nonnull).
I think I have proved that if the sequence has a limit (thus converges towards 0 or 1 - 1/a), then the sequence takes this limit as one of its values and thus becomes constant after some index.
Panurge.[/quote]

I wouldn't say so - for example a=2 and x_0=1/4 ...for instance , then
{x_n} is an increasing sequence and heads to 1-1/a=1/2 but never reaches
this value after some index - it's obvious .

Hello Rafal.

I have given the necessary and sufficient condition for X(0) : to be one of the successive solutions of " ax(1 - x) = one of the previous solutions ", starting with solution 0 or solution 1 - 1/a.

Do you find it's a bad definition ?

Panurge.

I wan't to know about all x_0 \in [0,1] ...

well ... there is no precise answer for this problem , hmmm...

Hello Rafal.

You give a counterexample with a = 2 but in the case where the limit is 1 - 1/a, I assume a > 2 to prove that the sequence has its limit as an effective value.
(Edition : and in fact the proof is valid only if a > 3.)
I have not posted the proof for this case, I will read it again to be sure.
But do you agree with the proof for the case where the limit is 0 ?

Panurge.

I don't think it's an easy (or even reasonable) task. Unless I'm mistaken, this is strongly related to chaos theory. Take a look [url=http://www.imho.com/grae/chaos/chaos.html]here[/url]. I didn't follow the discussion carefully so I don't know whether what you want is connected to what you'll find there, but it certainly bears a certain similarity.

P.S. Scroll down and take a look at that "next year's population = r * this year's population * (1 - this year's population)" thing.

yes , I agree with the 0 one - it's just the beginng:), but the one with 1-1/a can't be proved analogically , cause if a is a bit bigger than 2 ,"x_0" is a bit bigger 1/2 and smaller than 1-1/a then
lim x_n = 1-1/a but x_n never takes this value , using the Banach theorem
|f(x)-f(y)|<k|x-y| , where 1>k>0 , and for every x,y \in (1/2,2(1-1/a)-1/2) .
gosh...

if lim x_n = 0 then of course x_(n+1)>x_n for big n , so contradiction , your right.
I'm sure that this homework is mistaken ...

for a>4 another problem is to say somethin about x_0 for which there exists
x_n<0 ... then lim x_n = -\infty

Hello Rafal.

I'm glad you agree with the proof for the case where the limit is 0.
This gives a first set of values for X(0) : the two solutions of ax(1 - x) = 0, i. e. 0 and 1, and then the two solutions of ax(1 - x) = 1, and so on.

Now, you object to my argument in the case where the limit is 1 - 1/a.
Your objection is written in a concise form and I must confess that I don't understand it.

What do you think about the following proof ?

We thus assume that the sequence has limit 1 - 1/a. Put Y(n) = X(n) - (1 - 1/a).
Thus the sequence of the Y(n) has limit 0.
We have X(n) = Y(n) + 1 - 1/a, thus the hypothesis X(n + 1) = a X(n) (1 - X(n)) gives
Y(n + 1) = Y(n) (2 - a - aY(n)).  (relation 1)
(Edition. In the first version of this post, the proof was unaccurate. I assumed unconsciously a > 4. I give now a proof in the hypothesis a > 3.
Since a is assumed > 3, we have 2 - a < - 1.
Since we saw that Y(n) converges towards 0, 2 - a - aY(n) converges towards 2 - a < - 1, thus there exists i such that
|2 - a - aY(n)| >= 1 for each n >= i.
By our relation (1), we have thus, for each n >= i,
|Y(n + 1)| >= |Y(n).
Since the sequence Y(n) has limit 0, this implies Y(i) = 0, i. e. X(i) = 1 - 1/a. )

Do you object ? Thanks,
Panurge.
(Edition : I recall that the first version of the statement and of the proof was unaccurate.)

nice :) , however your idea is not the same with the 0 limit when for instance a \in (2,3) , then unfortunatelly b \in (0,1) , so lim b^k -> 0 , so that information doesn't give us much , of course I agree with the b>1 situation . So
there is still a small hole we have to fill :) . But I like it anyway ,  :cool:  :) .

[Edit] In other words - you showed that for a>=4 {x_n} is congruent
only when after a finite n x_n=1-1/a or x_n=0 , I agree . Otherwise it
can be ->-\infty or going crazy in (0,a/4) :) .
The only case is 4>a>2 , then , as you write b<1 ,so it has to be corrected.
What nationality are you btw. ?

ps. For a<0 the problem seems the same .

Hello !

You are right : I have unconsciously assumed a > 4.
I have edited the proof. I think it's accurate if a > 3.
Thus, the following question remains : if 2 < a <= 3, for which values of X(0) has the sequence X(n) a limit equal to 1 - 1/a ?

Panurge.

thanks anyway , I'm almost sure that for 2<a<3 there are certain x_0 for
which lim x_n=1-1/a but for all x_n : x_n=/=1-1/a - the answer depends on a , not mentioning of course
these x_0 for which after a finite number of recurrencies x_n=1-1/a .
For a<0 the proof is analogical to yours for a>4 , so the only problem
is 2<a=<3 , however it's (I think so) impossible to define such x_0 ...
I think that there is nothing to add ...

Thanks again , nice deduction :) .

Hello Rafal.

I think I have proved that if 2 < a <3, there exists an interval (with size > 0) in which all numbers are suitable for X(0). (So the set of suitable X(0) is not denumerable, so it isn't of the type we already know.) But I don't know what happens outside this interval.
I will read the proof again in the evening and if I find no errors, I will post it, unless somebody has posted a better one. It's no complete solution, but it can give you an idea.
I will try to be a little more critic regarding my deductions...

Panurge.

Hello Rafal !

Assume 2 < a < 3 and let us consider our favourite Y(n), defined recursively by
Y(n + 1) = Y(n) (2 - a - aY(n)).
I think that if |Y(0)| < (3 - a)/a, then Y(n) -> 0.
Put c = a|Y(0)| + a - 2.
Thus 0 <= c < 1.
Our aim is to prove that |Y(n + 1)| <= |Y(n)| c for each c.

We will first prove that, for all n,
|Y(n)| <= |Y(0)|.
It's clear for n = 0.
Assume it is true for some n :
|Y(n)| <= |Y(0)|            (hyp. rec. 1)
and let us prove
|Y(n + 1)| <= |Y(0)|.     (2 ?)
The recurrence hypothesis (1) gives
a|Y(n)| <= a|Y(0|,
which can be written
a|Y(n)| <= c + 2 - a.      (3)
This gives first
aY(n) <= c + 2 - a, otherly written
a - 2 + aY(n) <= c.         (4)
Relation (3) gives too :
- aY(n) <= c + 2 - a, otherly written
a - 2 - aY(n) <= c.
Since a > 2, this remains true if we subtract 2(a - 2) from the left member, thus
2 - a - aY(n) <= c.
Joined with (4), this gives
|2 - a - aY(n)| <= c.
Our recursive definition of Y(n) gives thus
|Y(n + 1) <= |Y(n)|c.     (5)
(This is our principal aim, but it depends still on our recurrence hypothesis |Y(n)| <= |Y(0)|, which is not yet proved for all n.)
Since c < 1, (5) gives
|Y(n + 1)| <= |Y(n)|, thus the recurrence hypothesis (1) gives
|Y(n + 1)| <= |Y(0)|.
So, we have proved by induction on n that
|Y(n)| <= |Y(0)| for all n.
The preceding reasoning shows that this implies (5), thus (5) is true for each n, thus, by recurrence on n,
|Y(n)| <= |Y(0)| multiplied by the n-th power of c.
Since c < 1, it follows that Y(n) -> 0.

We have thus proved that if 2 < a < 3 and Y(0) is in the interval
] - (3 - a)/a, (3 - a)/a [,
 then Y(n) -> 0.
In other words : if X(0) is in the interval
] (2a - 4)/a, 2/a[,
 then X(n) -> 1 - 1/a.

But what if X(0) is not in this interval ? And what if a = 3 ? I don't know.

Panurge.

Thanks for reply , I don't have much time at the moment but I'm sure
that the set of x_0 for which the limit of the sequence exists without taking it's value has a sup and inf - that's sure ... the only thing we could do is to
approximate that sup and inf - just what you have done . So my hipothesis
was indeed true :) . Today I finish writing the solution to this problem
and I've written all possibilities of convergence without the a \in (2,3] situation for limit 0 - adding my hipothesis :) , and the a \in [-1,0) situation for limit = 1-1/a . That should satisfy the person checking it , I hope so . As you said - the problem is tedious , thanks again :) .

This problem is  related to Chaos Theory as grobber said.
I give a hint.Let's solve the recursive formula as follows.
Completeing squaring RHS,

As $a>2,x_{n+1}=ax_n(1-x_n)\Longleftrightarrow\frac{1}{2}-{x_{n+1}}<a\left(\frac{1}{2}-x_n\right)^2$ Set $y_n=\frac{1}{2}-x_{n}$,we have .....

kunny

yes , however we can say much about the answer (not all unfortunately) :

for a<-1 :
{x_n} has limit 0 only if there exists such k , that x_k=0
{x_n} has limit 1-1/a only if there exists such k , that x_k=1-1/a
else
{x_n} hasn't got a limit ( total chaos... )

for -1=<a<0
{x_n} has limit 0 for a certain interval of x_0 - impossible to define it precisely
{x_n} has limit 1-1/a only if there exists such k , that x_k=1-1/a
else
{x_n} hasn't got a limit

for a=0
{x_n} has limit 0

for 0<a=<1
{x_n} has limit 0

for 1<a=<2
{x_n} has limit 1-1/a

for 2<a=<3
{x_n} has limit 0 only if there exists such k , that x_k=0
{x_n} has limit 1-1/a for a certain interval of x_0 - impossible to define it precisely
else
{x_n} hasn't got a limit

for 3<a
{x_n} has limit 0 only if there exists such k , that x_k=0
{x_n} has limit 1-1/a only if there exists such k , that x_k=1-1/a
else
{x_n} hasn't got a limit

I spoke with my teacher who occured to be that unfortunate author of this problem
and he said that he couldn't imagine we could say so much about the convergence.


I need some hard and complicated problems with answers about probability.
Can you help?Please ! ;)
                                      yours truley,
                                        Alissar

Misunderstood to have meant - I need help. Sorry about that.

[quote="white_horse_king88"]What are the problems?[/quote]

he says he needs some. :)

Perhaps you should give us some examples of what you consider hard and complicated.  It's possible that you should be asking this question in a more advanced forum.


[quote="Vasc"]If $ a,b,c$ are positive real numbers such that $ a \plus{} b \plus{} c \equal{} 3$, then

$ \frac {a \plus{} 1}{b} \plus{} \frac {b \plus{} 1}{c} \plus{} \frac {c \plus{} 1}{a} \ge 2(a^2 \plus{} b^2 \plus{} c^2)$.[/quote].
We have:
$ \frac {a \plus{} 1}{b} \plus{} \frac {b \plus{} 1}{c} \plus{} \frac {c \plus{} 1}{a} \ge 2(a^2 \plus{} b^2 \plus{} c^2) \Leftrightarrow \frac {a }{b} \plus{} \frac {b }{c} \plus{} \frac {c }{a} \plus{} \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \ge 2(a^2 \plus{} b^2 \plus{} c^2)$
$ \Leftrightarrow (\frac {a }{b} \plus{} \frac {b }{c} \plus{} \frac {c }{a} \minus{} 3) \plus{} (\frac {(ab \plus{} bc \plus{} ca)(a \plus{} b \plus{} c)}{3abc} \minus{} 3) \ge 2(a^2 \plus{} b^2 \plus{} c^2 \minus{} \frac {(a \plus{} b \plus{} c)^2}{3})$
$ \Leftrightarrow \frac {(a \minus{} b)^2}{ab} \plus{} \frac {(a \minus{} c)(b \minus{} c)}{ac} \plus{} \frac {2c(a \minus{} b)^2 \plus{} (a\plus{}b)(a \minus{} c)(b \minus{} c)}{3abc} \ge \frac {4}{3}((a \minus{} b)^2 \plus{} (a \minus{} c)(b \minus{} c)) \ge 0$
$ \Leftrightarrow (\frac {5}{3ab} \minus{} \frac {4}{3})(a \minus{} b)^2 \plus{} (\frac {a \plus{} b \plus{} 3c}{3abc} \minus{} \frac {4}{3})(a \minus{} c)(b \minus{} c)$
Assume c=max{a,b,c}, we have:
$ \frac {5}{3ab} \minus{} \frac {4}{3} \ge 0$
$ \frac {a \plus{} b \plus{} 3c}{3abc} \minus{} \frac {4}{3} \ge 0$
So we are done.

[quote="arqady"][quote="NguyenDungTN"]
$ (\frac {a }{b} \plus{} \frac {b }{c} \plus{} \frac {c }{a} \minus{} 3) \plus{} (\frac {(ab \plus{} bc \plus{} ca)(a \plus{} b \plus{} c)}{3abc} \minus{} 6) \ge 2(a^2 \plus{} b^2 \plus{} c^2 \minus{} \frac {(a \plus{} b \plus{} c)^2}{3})$
$ \Leftrightarrow \frac {(a \minus{} b)^2}{ab} \plus{} \frac {(a \minus{} c)(b \minus{} c)}{ac} \plus{} \frac {(a \plus{} b \minus{} c)(a \minus{} b)^2 \plus{} c(a \minus{} c)(b \minus{} c)}{3abc} \ge \frac {4}{3}((a \minus{} b)^2 \plus{} (a \minus{} c)(b \minus{} c)) \ge 0$
[/quote]
I think, here you are wrong. :wink:[/quote]
I'm sorry, I've editted my post!

[quote="NguyenDungTN"]
$ \frac {(a \minus{} b)^2}{ab} \plus{} \frac {(a \minus{} c)(b \minus{} c)}{ac} \plus{} \frac {2c(a \minus{} b)^2 \plus{} (a \plus{} b)(a \minus{} c)(b \minus{} c)}{3abc} \ge \frac {4}{3}((a \minus{} b)^2 \plus{} (a \minus{} c)(b \minus{} c)) \ge 0$
$ \Leftrightarrow (\frac {5}{3ab} \minus{} \frac {4}{3})(a \minus{} b)^2 \plus{} (\frac {a \plus{} b \plus{} 3c}{3abc} \minus{} \frac {4}{3})(a \minus{} c)(b \minus{} c)$
[/quote]
I think, it's wrong too. :wink:

Here is my proof:
$ \frac {a \plus{} 1}{b} \plus{} \frac {b \plus{} 1}{c} \plus{} \frac {c \plus{} 1}{a} \ge 2(a^2 \plus{} b^2 \plus{} c^2)\Leftrightarrow\sum_{cyc}(a^2c \plus{} ab \minus{} 2a^3bc)\geq0\Leftrightarrow$
$ \Leftrightarrow3(a^2c \plus{} b^2a \plus{} c^2b)(a \plus{} b \plus{} c)^2 \plus{} (ab \plus{} ac \plus{} bc)(a \plus{} b \plus{} c)^3\geq54\sum_{cyc}a^3bc\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^4b \plus{} 4a^4c \plus{} 6a^3b^2 \plus{} 9a^3c^2 \minus{} 41a^3bc \plus{} 21a^2b^2c)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}a(2b^2 \minus{} c^2 \minus{} 4ab \plus{} 5ac \minus{} 2bc)^2 \plus{} 3\sum_{cyc}(a^3bc \minus{} a^2b^2c)\geq0.$ :)
What is your proof, Vasc? Thank you.

Arqady, I proved this inequality by adding the inequalities
$ 6(\frac {a }{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a})\plus{}3 \ge 7(a^2 \plus{} b^2 \plus{} c^2)$
and
$ 6(\frac {1 }{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}) \ge 5(a^2 \plus{} b^2 \plus{} c^2)\plus{}3$.
To prove the first I used SOS method.

I think the inequality below is interesting for Arquady:

If $ a,b,c$ are positive real numbers such that $ a \plus{} b \plus{} c \equal{} 3$, then

$ 6(\frac {a \plus{} 1}{b} \plus{} \frac {b \plus{} 1}{c} \plus{} \frac {c \plus{} 1}{a})\plus{}3 \ge 13(a^2 \plus{} b^2 \plus{} c^2)$.

[quote="epitomy01"]What is the SOS method?[/quote]
Here you can find complete information about SOS
http://www.mathlinks.ro/Forum/viewtopic.php?t=80127

[quote="Vasc"]Arqady, I proved this inequality by adding the inequalities
$ 6(\frac {a }{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}) \plus{} 3 \ge 7(a^2 \plus{} b^2 \plus{} c^2)$
and
$ 6(\frac {1 }{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}) \ge 5(a^2 \plus{} b^2 \plus{} c^2) \plus{} 3$.
To prove the first I used SOS method.
[/quote]
By the way,
$ 6(\frac {1 }{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}) \ge 5(a^2 \plus{} b^2 \plus{} c^2) \plus{} 3\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^4b \plus{} a^4c \plus{} 3a^3b^2 \plus{} 3a^3c^2 \minus{} 17a^3bc \plus{} 9a^2b^2c)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}a(b \minus{} c)^2(b \plus{} c \minus{} 3a)^2 \plus{} \frac {1}{2}\cdot\sum_{cyc}(a \minus{} b)^2abc\geq0.$
This is a Sum Of Square method too. :D
Also we obtain: $ 6(\frac {a }{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}) \plus{} 3 \ge 7(a^2 \plus{} b^2 \plus{} c^2)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^4c \plus{} a^3b^2 \plus{} 2a^3c^2 \minus{} 8a^3bc \plus{} 4a^2b^2c)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}a(a \minus{} b)^2(b \minus{} 2c)^2\geq0.$ :)
[quote="Vasc"]

I think the inequality below is interesting for Arquady:

If $ a,b,c$ are positive real numbers such that $ a \plus{} b \plus{} c \equal{} 3$, then

$ 6(\frac {a \plus{} 1}{b} \plus{} \frac {b \plus{} 1}{c} \plus{} \frac {c \plus{} 1}{a}) \plus{} 3 \ge 13(a^2 \plus{} b^2 \plus{} c^2)$.[/quote]
Wow!
$ 6(\frac {a \plus{} 1}{b} \plus{} \frac {b \plus{} 1}{c} \plus{} \frac {c \plus{} 1}{a}) \plus{} 3 \ge 13(a^2 \plus{} b^2 \plus{} c^2)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}a(2b^2 \minus{} c^2 \minus{} 4ab \plus{} 5ac \minus{} 2bc)^2\geq0.$ :lol:

[quote="mathpk"][quote="arqady"]
$ \Leftrightarrow\sum_{cyc}(a^4b \plus{} 4a^4c \plus{} 6a^3b^2 \plus{} 9a^3c^2 \minus{} 41a^3bc \plus{} 21a^2b^2c)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}a(2b^2 \minus{} c^2 \minus{} 4ab \plus{} 5ac \minus{} 2bc)^2 \plus{} 3\sum_{cyc}(a^3bc \minus{} a^2b^2c)\geq0.$[/quote]

Btw how do you get that? (I mean how to factor them in sum of squares of such complex expression like that?)[/quote]
$ \sum_{cyc}(a^4b \plus{} 4a^4c \plus{} 6a^3b^2 \plus{} 9a^3c^2 \minus{} 41a^3bc \plus{} 21a^2b^2c) \equal{}$
$ \equal{} \sum_{cyc}a(2b^2\pm c^2 \plus{} xbc \plus{} yac \plus{} zab)^2 \plus{} A$
Case1.$ \sum_{cyc}(a^4b \plus{} 4a^4c \plus{} 6a^3b^2 \plus{} 9a^3c^2 \minus{} 41a^3bc \plus{} 21a^2b^2c) \equal{}$
$ \equal{} \sum_{cyc}a(2b^2 \minus{} c^2 \plus{} xbc \plus{} yac \plus{} zab)^2 \plus{} A,$ where $ x \plus{} y \plus{} z \plus{} 1 \equal{} 0.$
But $ \sum_{cyc}a(2b^2 \minus{} c^2 \plus{} xbc \plus{} yac \plus{} zab)^2 \equal{} \sum_{cyc}(a^4b \plus{} 4a^4c \plus{} (z^2 \minus{} 2y)a^3b^2 \plus{}$
$ \plus{} (y^2 \plus{} 4z)a^3c^2 \plus{} 2(yz \minus{} y \minus{} z \minus{} 1)a^3bc \plus{} (4y \minus{} 2z \minus{} 3 \minus{} (y \plus{} z)^2)a^2b^2c).$
Let $ z^2 \minus{} 2y \equal{} 6$ and $ y^2 \plus{} 4z \equal{} 9.$ It gives $ y \equal{} 5,$ $ z \equal{} \minus{} 4$ and $ x \equal{} \minus{} 2.$
Id est, we obtain that
$ \sum_{cyc}(a^4b \plus{} 4a^4c \plus{} 6a^3b^2 \plus{} 9a^3c^2 \minus{} 41a^3bc \plus{} 21a^2b^2c) \equal{}$
$ \equal{} \sum_{cyc}a(2b^2 \minus{} c^2 \minus{} 4ab \plus{} 5ac \minus{} 2bc)^2 \plus{} 3\sum_{cyc}(a^3bc \minus{} a^2b^2c).$
Case2. $ \sum_{cyc}(a^4b \plus{} 4a^4c \plus{} 6a^3b^2 \plus{} 9a^3c^2 \minus{} 41a^3bc \plus{} 21a^2b^2c) \equal{}$
$ \equal{} \sum_{cyc}a(2b^2 \plus{} c^2 \plus{} xbc \plus{} yac \plus{} zab)^2 \plus{} A.$
We don't need it already.
I think, it's a very strong method. Sometimes much stronger, than usual SOS. :wink:
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=84160
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=112423
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=176659


plz post ur thoughts on a mathcounts contest run by me. it will be a lot lyk topper's contests.

I think it's a good idea, but it would be best if there were a set date and time for the contest, for example: Every Monday, Wednesday and Friday afternoons at 3:30pm cst. Or something like that. Have it be at the same time every day or every other day or something so people can count on it, and not just hope that there are some questions left whehevever they're on. Alos, let the time be known to everybody.

What level of MathCounts are you thinking? Chapter, State, Nats???

well, mi teacher gave me lyk 5,000,000,000,000,... (u get da idea) of chapter problems, but i cud probably find sum state and national problems...

um.....
That would be a big advantage for me.....
I am in your class notfunny,
so is theone.
Maybe you could post some of the questions i had.
I won't participate but i'm sure theone will.
He never misses anything.
I could definitely contribute to the problems.
maybe Darkshadow could get some questions too.

I don't think it matters if people are in notfunny's class.  We all have access to lots of past MATHCOUNTS problems.  The real point is to get us to sit down & do some problems.  Nobody should really care about who "wins" in the end.  There are so many variables and advantages at play.  It is just kinda fun to do it all at the same time, not knowing who is answering what question, etc.

I say, choose a time.  Post like 20 questions at once, and be ready to quickly let us know if questions are answered correctly or incorrectly.

Just do it! :D

ok if i remember correctly... i might be able to post some problems... maybe tomorrow at around... say 11 PST? how does that work out for you guys? or i might be able to post some soon... maybe around 10:00 PST


In a violin ensemble, the players are seated in order of skill level.  For example, in any interval of twenty minutes, first chair has a $ \dfrac{1}{80}$ chance of making a mistake.  The violin ensemble has 20 people, seated from 1st to 20th chair, and the probability of each of them making a mistake in any interval of twenty minutes is $ \dfrac{1}{80}$, $ \dfrac{1}{79}$, $ \cdots \dfrac{1}{60}$ respectively.  On a 60 minute concerto, what is the probability that none of them make a mistake?

(How would you solve it?  I don't have the solution.)

[hide="Solution"]What is the probability the first person doesn't make a mistake? The probability he makes a mistake is $ \frac {1}{80}$, so the probability he doesn't is $ \frac {79}{80}$.

Similarly, the $ n$th person has probability $ \frac {80 \minus{} n}{81 \minus{} n}$ of not making a mistake.

The probability of all these events occurring, because they are independent, is simply the product of all the relevant probabilities. $ \frac {59}{60} \times \frac {60}{61} \times ... \times \frac {79}{80} \equal{} \boxed{\frac {59}{80}}$.[/hide]

[quote="darkdieuguerre"][hide="Solution"]What is the probability the first person doesn't make a mistake? The probability he makes a mistake is $ \frac {1}{80}$, so the probability he doesn't is $ \frac {79}{80}$.

Similarly, the $ n$th person has probability $ \frac {80 - n}{81 - n}$ of not making a mistake.

The probability of all these events occurring, because they are independent, is simply the product of all the relevant probabilities. $ \frac {59}{60} \times\frac {60}{61} \times ... \times \frac {79}{80} = \boxed{\frac {59}{80}}$.[/hide][/quote]

Fixed LaTeX
[hide="solution"]$ \frac {59}{60} \times\frac {60}{61} \times ... \times \frac {79}{80} = \boxed{\frac {59}{80}}$[/hide]

Is that still the solution even though it says over a 60 minute interval and it says over any 20 minutes?  Is that irrelevant?


So I broke three toes two Saturdays ago playing soccer. At the hospital I received a podiatric shoe and (more importantly) a pair of crutches.

Crutches are the best things ever. I'm splitting my time between whoring out attention and learning how to walk with using two crutches and no feet.

Discuss.

How do you play soccer with crutches????? :huh:

How did you break THREE toes?

Everybody knows crutches are awesome.

The Electronical Wheelchairs are fun too.

Maybe you can put bullets on the bottom of the crutches so you can use it as a gun?

You should add ROCKETS or something to the crutches so that you can fly

Why, wouldn't Lazarus be burnt, I like the gun idea better.

okay maybe it wouldn't work then

it was impractical anyway

How about putting something anti-gravitational there? Making it seem that the crutches are magical.

[quote="Lazarus"]Well I broke my toes playing soccer. Then I received crutches.[/quote]

I play soccer myself..So I know how it feels.

About 8 months ago, I broke my collar bone, playing soccer.
Man.. That made me cry...(a little bit..)
I was crying out for help, too. but no one listened..they thought I was diving..(if you know what diving is..you'll understand)! :mad:
I was calling for medics.. :lol:
I had to wear slings and this other thing...that made rash between my armpits..  :|
Anyways, from my experience of wearing cast, when you take them off, IT WILL SMELL!!!!
MARK MY WORDS MY FRIEND!! :D

XD everybody was like "dude just walk it off it's nothing major"

in any case if I was any good with technology I'd upload some of my crutchiness to youtube or something

I got crutches once when I was about...11?  I had a good brace on my knee, so I could do flying stuff with them.  I haven't checked, but I think I still have those crutches.  However, they're designed for someone 5' to 5'6"...so yeah...

Ehhhh my crutches are designed for people 5'2" to 5'10", and I'm somewhere between 5'11" and 6 foot, so they're fine with me.


[quote="ayush_2008"]i we wound a wire along a cylinder .what is the length of wire .is it 2 pi R or  2 pi (R+2r).where R IS RADIUS OF cylinder and r is radius of wire.pl explain clearly[/quote]
i think it's the circumference of the base...C=2piR

Well, if you take the width of the wire into account, you need to specify where you are measuring.

Assuming the wire is elastic enough that the outside can be a little longer than the inside, and so fit smoothly around the cylinder, then it just matters where you measure.

The inside length is $ 2\piR$.

The length in the middle is $ 2\pi(R+r)$.

The length on the outside is $ 2\pi(R+2r)$.


inequalities
calculus
calculus computations

Hey everyone, this is my first post of what will probably amount to hundreds; see my forum ID for clarification  :wink: .

Joking aside, I am unable to solve problems that seem to be quite simple to others.
[b]
1. If g(x)= 3 + x + e^x, find g^-1(4). [/b]

It seems as if I cannot isolate "x" by itself.

[b]2. lim x / ( √(1+3x) - 1 )
x⇒0[/b]

I know the answer is 2/3 (due to graphing), but I cannot prove this answer utilizing the Limit Laws.

[b]3. A Tibetan monk leaves the monastery at 7:00 A.M. and takes his usual path to the top of the mountain, arriving at 7:00 P.M. The following morning, he starts at 7:00 A.M. at the top and takes the same path back, arriving at the monastery at 7:00 P.M. Use the Intermediate Value Theorem to show that there is a point on the path that the monk will cross at exactly the same time of day on both days.[/b]

I am utterly stumped on this problem.  :oops:

[b]4. Find lim x⇒ ∞ f(x) if, for all x > 1,

5√(x) / (√(x-1)) < f(x) < (10e^x - 21) / (2e^x)[/b]

Once again, I am missing some concept for I do not understand how to solve this problem either.

I would appreciate it if someone could help me out with these problems- not just by providing answers, but rather demonstrating how to solve the problem. Thanks!

1: You can't. 4 is a special value, and you should be able to find one point in the inverse image. Then, show that it's the only point.

2: Rationalize the denominator. It's an indeterminate form, and you need some algebra before you can factor out the troublesome part.

3: Put the two days on top of each other. You want to show that the graphs cross.

4: Those inequalities are pointing in the wrong direction. Fix that, and use the squeeze theorem.


Solve the system of quadratic equations algebraically.

x^2 +4y^2 -2x -8y +1 = 0
-x^2 +2x -4y -1 = 0

anyone able to help and explain (i can figure it out if you show some work)
the answers i got are hidden below

[hide]i got (1,0) but the 2nd number comes out as imaginary[/hide]

[hide="Hint"] Add the equations. [/hide]

[hide]i did that and then took the y values i got and stuck them back in, but i cant graph a (1,0) and an imaginary point on the same plane, so i wanted to see what others got[/hide]

[quote="mkottman23"]awesome thanks guys, so the imaginary points wont be shown in a graph of the 2 will they?[/quote]

Um, maybe complex plane?

then i would need an x, y, and i plane
theres no way for there to be an intersection of two 2d figures on the (x,i) and (x,y).. is there?

Actually, you would need one plane for $x$ and another plane for $y$.  But no, on a simple two-dimensional graph the imaginary point won't show up.


A man makes an annual deposit of 200 units in an account whichs pays 5% intrest compounded annually.
How much money should be in the account immediately after the 10th deposit?.

Hi 21 7 15,

Immediately after the second deposit, the acount holds $ 200 \plus{}200 \times (1.05) \equal{} 200(1 \plus{} 1.05)$.

Immediately after the third deposit, it holds $ 200 \plus{} (200(1\plus{} 0.05))1.05 \equal{} 200(1 \plus{} 1.05 \plus{} 1.05^{2})$

Immediately after the fourth deposit, it holds $ 200(1 \plus{} 1.05 \plus{} 1.05^{2} \plus{} 1.05^{3})$

...

Immediatley after the tenth deposit, it holds $ 200(1 \plus{} 1.05 \plus{} 1.05^{2} \plus{} 1.05^{3} \plus{} ... \plus{} 1.05^{9})$

$ \equal{} 200 \frac{1.05^{10} \minus{} 1}{1.05\minus{}1}$, since we have a geometric series with ten terms,

 $ \equal{} 4000(1.05^{1} \minus{} 1)$

which is about $ 2515.58$


Let x be the three-digit number with digits ABC and y be the three-digit number with digits CBA.
The digits A and C are not 0. If x–y= 495, how many possibilities are there for x?

[hide]Let \[x=100A+10B+C\] \[y=100C+10B+A\] Then \[x-y=99(A-C)\] Thus, \[99(A-C)=495\] \[A-C=5\]
The values for $A$ are $6,7,8,9$.
There are $4$ values for $A$.
So, there is $1$ value for $C$ (for each $A$).
$B$ can be anything, so there are $10$ values for it.
Thus, there are $4\times 1\times 10=40$ values for $x$ (and $y$). EDIT: I made everything less "cluttery". Thanks, Ubemaya. [/hide]

You overcounted by 4 times, because all 4 possible A's correspond to exactly 1 C, so there's no need to multiply by 4 for C.

What do mean by overcount, ubemaya? It looks all right to me...

He already edited his solution, and now it is correct.


calculus
integration
calculus computations


If the sides $a,b,c$ of a triangle are roots of a cubic \[x^3-mx^2 + nx -p = 0.\] Find $m,n,p$ in terms of $R,r,s.$

Hi, I think I got this one:

In the following work, K = area.

m = a+b+c = [b]2s[/b].

p = abc = 4RK = [b]4Rsr[/b].

So, n is a bit tougher.

n = ab + ac + bc

K = sr -->
K 2 = s 2 r 2  -->
s(s-a)(s-b)(s-c) = s 2 r 2 -->
sr 2 = (s-a)(s-b)(s-c) -->
sr 2 = s 3 - s 2 (a+b+c) + s(ab+ac+bc) - abc -->
sr 2 = s 3 - s 2 (2s) + s(ab+ac+bc) - 4Rsr -->
r 2 = s 2 - 2s 2 + (ab+ac+bc) - 4Rr -->
ab + ac + bc = [b]r 2 + s 2 + 4Rr[/b].

How is that?


Hi all.

Just a bit stuck, well im not stuck, i know how to divide and stuff but i'd like to learn some easy methods to divide huge numbers. By this i mean:

23923.53 / 5039272678

Thats just an example.

But without using a calc how would be the easiest way to do that sum?

Thanks guys. I can use a calculator, but i don't carry one everywhere and its nice to know how to do it quickly on paper.

This rather depends how satisfied you are with an 'approximate' answer. For example, I would be very satisfied to do the following for getting a rough estimate:

23923.53 is about $ 2.4\times10^4$ and
5039272678 is about $ 5.0\times10^9$ so that

23923.53 / 5039272678 is about $ 4.8\times10^{\minus{}6}$

I chose 5.0 as a close approximation because division by 5 is like times 2 divided by 10, ie $ \frac15\equal{}\frac2{10}$.

So there is no way to do the sum and get a precise answer? Even if it does take a little longer than the estimation method?

[quote="mountain.dew"]You could always do standard pencil/paper division..[/quote]

Thats what i was trying to do, but doing the division straight is hard. What i was doing before is rounding the numbers to get an estimate, and doing some subtraction to get the remainders of the original number (not rounded) and going through it like that.

Im not sure if what i said makes sense, and its long winded. Sometimes ill get distracted and forget where i was, lol.

If it's easy, I suppose you could factor out easy numbers (2s, 5s, etc.) until you get to something easier to work with.

[quote="mihail911"]im just curious, but why is knowing an easy way to divide big numbers really necessary? it's not like there's going to be a problem on a competition that asks: divide 4567665.6/23.445
anyone can divide, and as far as i know, no one knows ( or maybe has even tried) to develop a trick because of the sheer uselessness and triviality (hope that's a word)[/quote]

Isn't that Vedic mathematics which uses tricks to do that. (But I completely agree its a waste of time (there's more to maths than calculating))


modular arithmetic
function
calculus
number theory
relatively prime
number theory proposed

Prove that

a) every positive integer has at least as many divisors of the form $ 4k\plus{}1$ as divisors of the form $ 4k\plus{}3$

b) there exist infinitely many positive integers which have as many divisors of the form $ 4k\plus{}1$ as divisors of the form $ 4k\plus{}3$

c) there exist infinitely many positive integers which have more prime divisors of the form $ 4k\plus{}1$ than divisors of the form $ 4k\plus{}3$.

a) Strong induction. For a power of a prime, the assertion is easy to check. Now we need to show the assertion is true for numbers with at least two distinct prime factors. Assume that we've checked the assertion for all numbers $ < n$. If $ n$ has at least two distinct prime divisors, we may write $ n \equal{} ab$, where $ a$ and $ b$ are relatively prime and both smaller than $ n$. Let $ u, u'$ denote the number of factors of $ a$ that are $ 1, \minus{} 1\pmod{4}$, respectively, $ v, v'$ the number of factors of $ b$ that are $ 1, \minus{} 1 \pmod{4}$. then

i) the number of factors of $ n$ that are $ 1\pmod{4}$ is $ uv \plus{} u'v'$
ii) the number of factors of $ n$ that are $ \minus{} 1\pmod{4}$ is $ uv' \plus{} u'v$

so we need to show that $ uv \plus{} u'v'\geq uv' \plus{} u'v$. but this is equivalent to $ (u \minus{} u')(v \minus{} v')\geq 0$, which is true by the inductive hypothesis.

b) There are infinitely many primes that are $ \minus{} 1 \pmod{4}$ (e.g., by Dirichlet's Theorem). Every one of them has 1 divisor that's $ 1\pmod{4}$ (namely, 1) and 1 divisor that's $ \minus{} 1 \pmod{4}$ (namely, itself).

c) There are infinitely many primes that are $ 1\pmod{4}$. Every one of them has no divisors that are $ \minus{} 1\pmod{4}$.

[quote="Agr_94_Math"]Prove that

a) every positive integer has at least as many divisors of the form $ 4k + 1$ as divisors of the form $ 4k + 3$

b) there exist infinitely many positive integers which have as many divisors of the form $ 4k + 1$ as divisors of the form $ 4k + 3$

c) there exist infinitely many positive integers which have more prime divisors of the form $ 4k + 1$ than divisors of the form $ 4k + 3$.[/quote]
I remember that I solved this problem before. My way is to consider the function ${ f: N\to \{0,1,-1}$ and $ f(d)=0$ iff $ d$ is even, $ f(d)=1$ if $ d\equiv 1(\mod 4)$ and $ f(d)=-1$ if $ d\equiv -1 (\mod 4)$.
It is easy to check that $ f(d)$ is a multiple function, for each pair of relative numbers $ (x,y)$ we have $ f(x.y)=f(x).f(y)$.
Thus, it is not very hard to show that $ G(n)=\sum_{d|n} f(d)$ is also a multiple function.
Therefore we only need to calculus $ g(p^a)$ and (ii),(iii) are followed after that.


symmetry
calculus
Support
inequalities
inequalities proposed

If $a>b>c,a+b+c=1,a^2+b^2+c^2=1$, find the range of value of the following equality.

$(1)\ a+b$
$(2)\ a^2+b^2$

Click:
[hide="maximum"]
Part 1: b is non-negative, $(a+b)^2 \le 2(a^2+b^2) \le 2$, so $(a+b)\le \sqrt{2}$, which is attained when $a=b=\frac{1}{\sqrt{2}},c=0$

Part 2: $a^2+b^2\le 1$ so the maximum is 1, attained when $a=1, b=c=0$[/hide]

[hide="minimum"]
It's clear that $a>b>0>c$. Due to symmetry we must have $a=b$ in the minimum, for both, so put $b=a$. (if you want you can prove by placing a=b+k and finding k=0)

Part 1: We get $2a^2+c^2=1, 2a+c=1$, so $2a^2+4ac=0$ so $a+2c=0$ which means $a=b=2/3,c=-1/3$.

Part 2: Exactly the same.[/hide]

Peter VDD, your answer is incorrect,you missed $a>b>c$. :(

No kunny, I didn't miss that one. If you forget that condition the answers for the minimum would be totally different. [e.g. $(0,0,1)$ would give 0 for both]

If I write somewhere a=b, that means here of course "a should be as close to b as possible". :)

The problem is that if you let $a=1$, the conditions can't be satisfied.  I think you need to do some more work, Peter.  (and don't just say that it's a limit - you need to make sure that the conditions are still satisfied)

For example, say that $a=1-\epsilon$.

Then we have $b+c=\epsilon$ and $b^2+c^2=2\epsilon-\epsilon^2$, so that $b^2+c^2=2(b+c)-\epsilon^2$.  As $\epsilon\to 0$, this means that $b^2+c^2=2(b+c)$.  But we also know that $b,c\to 0$, so that $b^2+c^2<2(b+c)$ for sufficiently small $b,c$, a contradiction.

[quote="blahblahblah"]For example, say that $a=1-\epsilon$.

Then we have $b+c=\epsilon$ and $b^2+c^2=2\epsilon-\epsilon^2$, so that $b^2+c^2=2(b+c)-\epsilon^2$.  As $\epsilon\to 0$, this means that $b^2+c^2=2(b+c)$.  But we also know that $b,c\to 0$, so that $b^2+c^2<2(b+c)$ for sufficiently small $b,c$, a contradiction.[/quote]
Hmm no... I'm afraid you have a little calculus mistake there, blahblahblah. You scratch epsilon squared, and then say the other side is larger? Yes, but it would not be with the epsilon squared. The difference also goes to 0 right?

Look, I'll support my claim with a numerical argument if you don't understand me:
a = .9000000000, b = .3541381265, c = -.2541381265
a = .9900000000, b = .1046242942, c = -.09462429423
a = .9990000000, b = .03211091584, c = -.03111091584
a = .9999000000, b = .01004962499, c = -.009949624993
a = .9999900000, b = .003167265802, c = -.003157265802
...

I see really no problem in it... this [i]is[/i] just a limit case? :?

Can anyone solve this problem? :)

I saw Inequality's experts in this site, maybe this problem seems to be easy. :D

Both $a+b$ and $a^2+b^2$ don't have maximum or minimum value. :(

No, just replace $\ge$ with $>$, $\le$ with $<$, $maximum$ with $supremum$, $minimum$ with $infimum$, $=$ with $approaches\ to$, for the nonequality in the statement. :)

[quote="gauss202"]I believe the answer to your first question is $1 < a + b < 4/3$, and the answer to your second is $8/9 < a^2 + b^2 < 1$. I used the method of Lagrange Multipliers to solve.[/quote]

Your answer is correct, gauss202 :)

Japanese high school students don't study Lagrange multipliers.

We can solve this problem in basic way.

kunny

from the given condition it follows that
$ab+bc+ca=0 \Rightarrow c\leq 0$

also, from $ab+bc+ca=0$ it follows that $ab=c(c-1)$

then, from $(a+b)^2\geq 4ab$ we have that
$(1-c)^2\geq 4c(c-1) \Rightarrow (1-c)(3c+1)\geq 0 \Rightarrow c\geq \frac{-1}{3}$

then $\frac{-1}{3}\leq c \leq 0 \Rightarrow 1\leq a+b \leq \frac{4}{3}$ and $\frac{8}{9}\leq a^2+b^2\leq 1$

if $a>b>c$, then the inequalities become strict... but they can be "achieved" by letting $b\rightarrow a$ or $c\rightarrow b$ or things by the way

for example: $a+b=\frac{4}{3}$ iff $c=\frac{-1}{3} \Rightarrow a^2+b^2=\frac{8}{9}$ which yields $(a,b,c)=(\frac{2}{3},\frac{2}{3},\frac{-1}{3})$, then we can let $b\rightarrow a=\frac{2}{3}$ to obtain the "equality"


AMC
AIME
MATHCOUNTS
geometry
3D geometry
pyramid
geometric transformation

Are you sure #9 is 737? And could you enlighten me on why?

And nutz, I'm almost sure that 13 is 80.

[quote="1=2"]13 isn't right. Its[hide]20. Everyone, even the seniors, who are pretty smart, agreed with my 20. I even have some people who got the top 10 at Nationals of mathcounts sometime in their lives. :wink: [/hide][/quote]

Hmm, well I got these from Alex Zhai, who coordinated with a couple other people.

woo 10 questions solved - 7 careless mistakes (including misreading, forgetting to divide by 4, multiplying by the wrong number, being off by 2, screwing up some signs, multiplying wrong, and subtracting wrong) = [b]3[/b]

13 definitely isn't 20... the intersection of the plane with the square pyramid doesn't only include the triangle that is formed from the three points where the plane intersects.

Messed up on 10, 14, 15.  #10, I just did some stupid stuff and was stupid (solved a different problem).  #14, I said 225 (was a guesstimate...but i guess the value is actually really really close to 224).  #15, I got 451...that's the only radical i got.

[quote="1/(ln x)"]Are you sure #9 (the tangent circles) problem is correct? I got $\frac{606}{41}$ for the radius. Which means the sum is 647.[/quote]

Well I got 51/8...

Hahaha.
1. Got it
2. Misread
3. Got it
4. Arithmetic
5. Got it
6. Arithmetic
7. Got it
8. Arithmetic
9. Missed
10. Got it
11. Did to 1000 instead of 2007
12. Nope
13. Got it
14. I scaled down the sequence and forgot to scale back up. But I think the answer is 225.
15. I kept getting that the length was just 9... somehow.

Score: $\boxed{6}$
Hooray!

8 answered - 4 mistakes (misread #2, solved #4 for collinear on the same side of the star, miscounted #5 by 1, and missed something (not sure what) on 6) + 1 lucky guess (somehow guessed #8, and was also only 2 off on #10 with another guess...so close to 2 guesses) = [b]5[/b]

I got 4 right, and was off by 1 on both 5 and 7...

Forgot 2^0=1  :mad:

I'm wondering a little on #5, I got 538 after the contest, but this is what I got from Feiqi/Zhai/others.

I got 80 for 13.

ok, so I think I got a ... 10? WTF? around 4 stupid mistakes... wow, man, wow

Yay - min. [b]7[/b].

Correct: 1, 3, 5, 6, 7, 8, 10, maybe 12 (I guessed something that ended in 75 and b/w 600 to 900)

Like half of everyone here, I misread # 2 and got 210 for #4

#11 forgot to multiply by 2 at the end (guess that makes my only careless mistake)

For #13, I got 28, as did my friend (wow…)

[quote="paladin8"][quote="junggi"][quote="paladin8"]

[hide="awesome"] Find the equations of the lines, calculate the intersection points, and use... Shoelace Formula! [/hide][/quote]
oh
my
god
why did i not see that? i spent like 30 minutes working through really yucky algebra and made a computation error. wait dude that puts me at 8 stupid mistakes[/quote]

Oh, well that was intended to be a statement rife with sarcasm, but I'm glad you liked my idea. Considering I had to multiply trinomial radicals, I'm guessing it's not the best method. In fact, I found a better method, but it gives a wrong answer of $925$ (I convinced myself that this was the right answer, but apparantly not).[/quote]

ya ya

the BEST way

i used 5 sheets of paper

and this wasn't normal paper

it was extra long for extra bash

i would have to agree, although i will probably do just as bad (maybe even worse) if i am allowed to take the second AIME   :rotfl:

I brought a stack of graph paper and worked through about 6-8 sheets front and back but leaving a lot of space, making diagrams large, etc. to try to minimize careless errors. I ended up with 6 though  :(

EDIT: I actually felt (during the test) that this was much much easier than last year, when I froze up and could only attempt 5 because my interpretation just didn't work on many of them. Just being able to confidently solve 11 without issues and educated guess another made me ecstatic (not really) after the test, and I was sure I did well...

#15 is 989.. My math team coach used a ti-89 to solve a bunch of equations to get 989. while I used similar triangles to get 989.. so I am pretty sure it's 989

It is 989. It's been posted on the AMC website.  :roll:

I posted my solution for #15 here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=138740

messy.. messy.. lol

[quote="paladin8"][quote="Elemennop"]ya ya

the BEST way

i used 5 sheets of paper

and this wasn't normal paper

it was extra long for extra bash[/quote]

Haha. Props to you. I think I only used one sheet, front and back :wink:.[/quote]

We're getting a little off-topic here, but I think I used a total of 3 sheets front and back for the whole AIME. That was a bad idea. By the end of the test I was finding any blank space to do my work... at least I didn't have to trace over existing work. The #12 bash took about one side of a page; I calculated everything in terms of the area of ABC, which helped a bit. After that it was just a matter of keeping track of equations and carrying out piecemeal simplification. Still kind of surprised I got this right though.

[quote="probability1.01"]We're getting a little off-topic here, but I think I used a total of 3 sheets front and back for the whole AIME. That was a bad idea. By the end of the test I was finding any blank space to do my work... at least I didn't have to trace over existing work. The #12 bash took about one side of a page; I calculated everything in terms of the area of ABC, which helped a bit. After that it was just a matter of keeping track of equations and carrying out piecemeal simplification. Still kind of surprised I got this right though.[/quote]

I think I used 4. It took about 2 pages to do all of them except 10, 12, 13, and 15. The I used 1 page each on 12 and 13, but I never finished 13. I definitely should have tried something better than Heron's on nested radicals... *was not thinking straight by the time I got through 12*.

I did a law of sines, half-angle formula thing on Number 9, and I keep getting 408/47 as the radius. Is there an easy way to do Number 9...?

There's quite a few "nicer" ways.  I used pythagorean theorem and solved a system of 2 equations.  It seemed like this problem was too much algebra than geometry though.

$AB = 34$.  Drop perpendiculars $P_{1}$ and $P_{2}$ from $O_{1}$ and $O_{2}$ to $AB$.  Then you get $\angle AO_{1}P_{1}= \angle BAC / 2$ and similarly for $P_{2}$.  Then you can use half-angle formulas for tangents to get $AP_{1}$, $P_{2}B$ in terms of $r$.  Since $P_{1}P_{2}= 2r$, you now have a linear equation for $r$ since you know the trigonometric functions of triangle $ABC$.


1)equilateral triangle eql is inscribed in a circle with radius 1. Find the area of the segment of the circle bounded by segment EQ and minor arc EQ

2) Three vertices of the kite KITE lie on a circle and its diagonals intersect at the center of the circle. The other vertex E lies outside of the circle. SEGMENT EI intersects the circle at point X. The area of the kite is 24, and EX = 8. When is the product (KI)(KE) is written in simplest radical form as m root n, find m + n

 :o

The area of this minor section is one third of the difference of area btw. the circle and the triangle.
Area of the circle: $\pi$
Area of the triangle: Slightly more difficult.
Use a bit of trig to determine the sides of the triangle, then use some nice geometry and find the area.

Trig is not really necessary. Just observe that the circle center and triangle centroid coincide and go from there.

[hide="1"]Since $\triangle EQL$ is equilateral, the circle and the triangle are concentric. Let's call the circle O. We know that EO=QO=LO=1. But since circle $O$ and $\triangle EQL$ are concentric, we can extend QO to M in LE and make an altitude with length 1.5. Therefore,

$\frac{s^{2}\sqrt3}{4}=\frac{3s}{4}$

$4s^{2}\sqrt{3}=12s$

$s\sqrt{3}=3$

$s=\sqrt{3}$

We want the area of the circle minus the area of the triangle  all divided by three. We can get that:

$\frac{\pi}{3}-\frac{\sqrt{3}}{4}$[/hide]

[hide="2"]Since it is a kite, we know the area to be $1/2(IE)(KT)$. $IE = IX+8$. If we call the center of the circle $O$, we then know that $(OK)=(OI)=(OT)=(OX)$. Let us denote this length of one of those 4 segments to be $L$. Thus, the area of our kite is $1/2(2L)(2L+8)=2L^{2}+8L$. We know this to be equal to $24$, so we set it equal to $24$ and solve for $L$.

$2L^{2}+8L=24$
$2L^{2}+8L-24=0$
$L^{2}+4L-12=0$
$(L+6)(L-2)=0)$
$L=2$ or $L=-6$

Since lengths cannot be negative, we know that $L=2$. We are now looking for $(KI)(KE)$.

$(KI)=L\sqrt{2}=2\sqrt{2}$
$(KE)=\sqrt{L^{2}+(L+8)^{2})}=\sqrt{104}=2\sqrt{26}$
$2\sqrt{2}\cdot 2\sqrt{26}=8\sqrt{13}$

Thus, our final answer is $8+13=\boxed{21}$[/hide]


find x,y,z are positive integer sastisfy :
x!+y!=z!

That's one of the most stupid titles I've ever seen.


calculus
integration
function
absolute value
calculus computations

How do you calculate definite integrals?
FOr example:
Find the integral lower bound:-3
uper bound: 2
of the function y=abs(x+1)

Thanks

The absolute value function is a bit tricky since it can take on different values.  I could solve it for you but instead, I think an easier example would be better:

Let's instead compute
$ \int_{\minus{}2}^{2}x^{2}\ dx$
By the fundamental theorem of calculus,
$ \int_{\minus{}2}^{2}x^{2}\ dx\equal{}f(2)\minus{}f(\minus{}2)\equal{}\left(\frac{x^{3}}{3}\right)_{\minus{}2}^{2}\equal{}\frac{8}{3}\minus{}\frac{\minus{}8}{3}\equal{}\frac{16}{3}$

This is the third calc question I've solved for you, and no offense, but these are very, very elementary questions so you might want to check your understanding.


If $ a,b,c \ge 1$, then:

\[ \boxed{\frac{1}{a\plus{}bc}\plus{}\frac{1}{b\plus{}ca}\plus{}\frac{1}{c\plus{}ab} \ge \frac{3}{1\plus{}abc}}\]

We have :

$ \sum_{cyc}\left(\frac{1\plus{}abc}{a\plus{}bc}\minus{}1\right)\equal{}\sum_{cyc}\frac{(a\minus{}1)(bc\minus{}1)}{a\plus{}bc} \ge 0$

Nice!
What about this one:

    $ \frac{1}{a\plus{}ab}\plus{}\frac{1}{b\plus{}bc}\plus{}\frac{1}{c\plus{}ca} \geq \frac{3}{1\plus{}abc}$ where $ a,b,c$ are positive numbers.

[quote="new_member"]Nice!
What about this one:

    $ \frac {1}{a \plus{} ab} \plus{} \frac {1}{b \plus{} bc} \plus{} \frac {1}{c \plus{} ca} \geq \frac {3}{1 \plus{} abc}$ where $ a,b,c$ are positive numbers.[/quote]
Cauchy-Schwartz: $ LHS \ge \frac {9}{\sum_{cyc} a \plus{} ab} \ge \frac {3}{1 \plus{} abc} \iff$ $ \sum_{cyc} abc \plus{} 1 \minus{} a \minus{} bc \equal{} \sum_{cyc} (a \minus{} 1)(bc \minus{} 1)  \ge 0$.

But what if $ a \equal{} \frac {1}{2}$ and $ b \equal{} c \equal{} 2$.Then it's negative.
You proved the inequality only for $ a,b,c > \equal{} 1$ but this inequality is true for every positive real $ a,b,c$
hence you also need to examine the remaining cases.

[quote="new_member"]Nice!
What about this one:

    $ \frac {1}{a \plus{} ab} \plus{} \frac {1}{b \plus{} bc} \plus{} \frac {1}{c \plus{} ca} \geq \frac {3}{1 \plus{} abc}$ where $ a,b,c$ are positive numbers.[/quote]

Use $ \sum_{cyc}\left(\frac{abc\plus{}1}{a\plus{}ab}\plus{}1\right)\equal{}\sum_{cyc}\left(\frac{ab(c\plus{}1)}{a(b\plus{}1)}\plus{}\frac{a\plus{}1}{a(b\plus{}1)}\right) \ge 6$

Acording The AM GM inequality

[quote="new_member"]Nice!
What about this one:

    $ \frac {1}{a \plus{} ab} \plus{} \frac {1}{b \plus{} bc} \plus{} \frac {1}{c \plus{} ca} \geq \frac {3}{1 \plus{} abc}$ where $ a,b,c$ are positive numbers.[/quote]
It's equivalent to $ \sum_{cyc}bc(1\plus{}c)(1\minus{}ab)^2\geq0.$

[quote="new_member"]Nice!
What about this one:

    $ \frac {1}{a \plus{} ab} \plus{} \frac {1}{b \plus{} bc} \plus{} \frac {1}{c \plus{} ca} \geq \frac {3}{1 \plus{} abc}$ where $ a,b,c$ are positive numbers.[/quote]
this is the famous AMS inequality, a similar one posted by Walther Janous -
$ (\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c})(\frac{1}{1\plus{}a} \plus{}\frac{1}{1\plus{}b}\plus{}\frac{1}{1\plus{}c}) \ge \frac{9}{1\plus{}abc}$ :)

[quote="new_member"]Arqady,could you please become more specific?How did you end up to these factorizations?
Can you please give the solution writing step by step your answer?Thanks a lot

ps The solution i had in mind was the same as abdek's so I am looking forward to seeing exactly
how you got these factorizations because i can't get them[/quote]
Look, after beautiful  abdek's proof my proof seems very ugly.
But it needs 10 minutes to find and to write the following:
$ \sum_{cyc}\frac {1}{a \plus{} ab} \geq \frac {3}{1 \plus{} abc}\Leftrightarrow$
$ \Leftrightarrow(1 \plus{} abc)\sum_{cyc}bc(1 \plus{} a)(1 \plus{} c)\geq3abc\prod_{cyc}(1 \plus{} a)\Leftrightarrow$
$ \Leftrightarrow(1 \plus{} abc)\sum_{cyc}(ab \plus{} abc \plus{} a^2c \plus{} a^2bc)\geq\sum_{cyc}(abc \plus{} a^2b^2c^2 \plus{} 3a^2bc \plus{} 3a^2b^2c)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(ab \plus{} abc \plus{} a^2c \plus{} a^2bc \plus{} a^2b^2c \plus{} a^2b^2c^2 \plus{} a^3c^2b \plus{} a^3b^2c^2)\geq$
$ \geq\sum_{cyc}(abc \plus{} a^2b^2c^2 \plus{} 3a^2bc \plus{} 3a^2b^2c)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(ab \plus{} a^2c \plus{} a^3c^2b \plus{} a^3b^2c^2 \minus{} 2a^2bc \minus{} 2a^2b^2c)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(ab \minus{} 2a^2bc \plus{} a^3c^2b \plus{} b^2a \minus{} 2a^2b^2c \plus{} a^3b^2c^2)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(ab \plus{} a^2b)(1 \minus{} ac)^2\geq0\Leftrightarrow\sum_{cyc}bc(1 \plus{} c)(1 \minus{} ab)^2\geq0.$

[quote="Dimitris X"][quote="arqady"][quote="new_member"]Arqady,could you please become more specific?How did you end up to these factorizations?
Can you please give the solution writing step by step your answer?Thanks a lot

ps The solution i had in mind was the same as abdek's so I am looking forward to seeing exactly
how you got these factorizations because i can't get them[/quote]
Look, after beautiful  abdek's proof my proof seems very ugly.
But it needs 10 minutes to find and to write the following:
[/quote]

Well it takes 10 minutes for you dear arqady not for us :wink: :D  :lol:  ......

SORRY FOR SPAM....[/quote]
It can take less than 10 minutes. Let me show how:

The inequality is equivalent to
$ \sum \left[ \frac{abc\plus{}1}{a(1\plus{}b)}\minus{}1\right] \ge 0.$
We have
$ \frac{abc\plus{}1}{a(1\plus{}b)} \minus{}1\equal{}\frac{abc\plus{}1\minus{}a\minus{}ab}{a(1\plus{}b)} \equal{}\frac{a(bc\minus{}1)\plus{}(1\minus{}ab)}{a(1\plus{}b)}.$
Therefore, the quantity on the left hand side of the above inequality can be written as
$ \sum \frac{bc\minus{}1}{1\plus{}b} \minus{}\sum \frac{ab\minus{}1}{a(1\plus{}b)}.$
Since the sum is cyclic, we can write it as
$ \sum \frac{bc\minus{}1}{1\plus{}b} \minus{}\sum \frac{bc\minus{}1}{b(1\plus{}c)} \equal{}\sum \frac{(bc\minus{}1)^2}{b(1\plus{}b)(1\plus{}c)} \ge 0.$
This shows that the desired inequality holds.

[quote="new_member"]Nice!
What about this one:

    $ \frac {1}{a \plus{} ab} \plus{} \frac {1}{b \plus{} bc} \plus{} \frac {1}{c \plus{} ca} \geq \frac {3}{1 \plus{} abc}$ where $ a,b,c$ are positive numbers.[/quote]

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