The whole numbers are listed in order and placed in this arrangement.  Each row has two more entries in it than the row above it.  What is the first number in the row containing the integer 500?

\begin{tabular}{ccccccccccccccc}
 \[ &\] & \[ & $ 1$ & $ 2$ &\] & \[ &\] \\
 \[ &\] & $ 3$ & $ 4$ & $ 5$ & $ 6$ & \[ &\] \\
 \[ & $ 7$ & $ 8$ & $ 9$ & $ 10$ & $ 11$ & $ 12$ &\] \\
 $ 13$ & $ 14$ & $ 15$ & $ 16$ & $ 17$ & $ 18$ & $ 19$ & $ 20$ \\
  & & & $ \vdots$  & & &
\end{tabular}

It goes:

1 2
3 4 5 6
7 8 9 10 11 12
...

Sum of first $ n$ even numbers is $ n^2\plus{}n$, so $ n^2\plus{}n \ge 500$ or the $ 22$nd row.

[quote="GameBot"]The whole numbers are listed in order and placed in this arrangement.  Each row has two more entries in it than the row above it.  What is the first number in the row containing the integer 500?

\begin{tabular}{ccccccccccccccc}
 & & & 1 & 2 & & & \\
 & & 3 & 4 & 5 & 6 & & \\
 & 7 & 8 & 9 & 10 & 11 & 12 & \\
13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 \\
  & & & \vdots  & & &
\end{tabular}[/quote]

To figure out where the sum of the first $ n$ even numbers comes from, look at the last number in each row.  It also follows from how the arrangement is defined.

Nobody has yet provided a full solution to the problem. May I kindly ask for elaboration?

The last number of each row is $ 2, 6, 12, 20$.
$ 2\equal{}2$
$ 2\plus{}4\equal{}6$
$ 2\plus{}4\plus{}6\equal{}12$
$ 2\plus{}4\plus{}6\plus{}8\equal{}20$
So the last number of the nth row is the sum of the first n even numbers, which is $ n^2\plus{}n$
So, we want the smallest number that is greater than 500 that is the sum of the first $ n$ numbers.
We then need to solve the inequality $ n^{2}\plus{}n\ge 500$.
This yields $ n\equal{}22$. This means that 500 is on the 22nd row.
Then, the last number on the 21st row is $ 21^2\plus{}21\equal{}462$ so the first number on the 22nd row is $ 463.$


geometry
perimeter
trigonometry
percent
ratio
geometry unsolved

Show how to construct triangle $ABC$ if we know:
(i)  measure of angle at vertex $A$
(ii) length of median from vertex $A$
(iii) perimeter of triangle

Yes, easy indeed.
 We set the perimeter $p$, so $a+b+c=p$ and $m_{a}$ the length of the median.

 We have by cosine theorem: $a^{2}= b^{2}+c^{2}-2bc$ $cosA$ and we know that $m_{a}^{2}=\frac{2(b^{2}+c^{2})-a^{2}}{4}$.
 $b+c=p-a$ so $b^{2}+c^{2}=(p-a)^{2}-2bc \rightarrow a^{2}=(p-a)^{2}-2bc(cosA+1)$
 $\rightarrow bc=\frac{(p-a)^{2}-a^{2}}{2(cosA+1)}(1)$
 On the other hand, $a^{2}=2 (p-a)^{2}-4m_{a}^{2}+4bc$ but we already expressed $bc$ in $(1)$ so in the end we get a equation from which we get $a$. Sorry I didn't make all the computations...
Anyway, $a=...$ and then $bc=...$, we know $b+c=p-a$,so $b=..., c=...$
And so we construct the triangle as we know $a, b, c$ and angle $A$.

$a, b, c$ represent the lenghts of the sides of the triangle.

[quote]And so we construct the triangle as we know $b$ $c$  and angle $A$ .[/quote]

Sorry Can you just explain this part Is this something like theorem or...
(For example: How can we construct it if $b$ and $c$ are irrational numbers).

on a line determine $A,C$ s.t $AC=b$ then determine line $b$ s.t $\angle CAb=\angle A$ (its easy using compass)
then on line $b$ determine point $B$ s.t $BA=c$ then the points $A,B,C$ are the desired vertices...

to [b]BaBaK Ghalebi[/b]

Are you sure that, after you get length of $b$ you can construct these points
Example:
Construct these points if $b$=19,14547515 blabla...

first of all note that in mathematical constructions,ruler isnt any regular ruler,a ruler in geometric constructions is a tool which can draw a length $x \in \mathbb{R^{+}}$...  :)

[quote]first of all note that in mathematical constructions,ruler isnt any regular ruler,a ruler in geometric constructions is a tool which can draw a length ... [/quote]

OK[b]BaBaK Ghalebi[/b] I am not good at constructions so maybe that's the reason why I don't understynd you.

Can you just explain me how to construct number $pi$=[b]3.14....... [/b]by using your mathematical ruler.

[i][u]Solution should be 100 percent correct.[/u][/i]

as said before ruler can draw any segment with length $x\in \mathbb{R^{+}}$ but for example if you want todraw $\sqrt{5}$
then you can construct a right triangle with sides equal to $2,1$ and then the third side is equal to $\sqrt{5}$
and also for drawing a segment with length $\frac{p}{q}$ you van do as follow:
assume that we want to divide the segment $AB$ with ratio $\frac{p}{q}$ internal and external...
draw parallel lines $\ell,\ell'$ passing thruogh $A,B$ respectively,then take $F$ on $\ell$ s.t $AF=p$ and take $H,G$ on $\ell'$ s.t $H,F$ be on the same side of $AB$ and $G$ on the other side,and $BH=BG=q$ let $FG,FH$ intersect $AB$ at $C,D$ then the points $C,D$ are the desired points...
now according to the statement above we can draw any line with length $\frac{a}{b}$...
but you must notice that $ruler$ in mathematical constructions is a tool to draw a line between points $A,B$ in the plane...
so if you just take points $A,B$ with distance $\pi$ in the plane then the segment between them would have the length $\pi$ which can be drawn by $ruler$...
in mathematical constructions the length of a segment isnt anything important,they simply assume that they can draw a segment with length $x\in \mathbb{R^{+}}$...
a more interesting question [problem] about $ruler$ in mathematical constructions is this"
[b]in the constructions the $ruler$ we use has length as much as we want for example if $AB=100000000000000...0$ then we can draw the line with the ruler we have,somehow the length of our ruler is infinity,now the question is:what if the ruler we use had the length $\alpha$ then how would we draw a line between two points $A,B$ with distance $\alpha+k$ where $k\in \mathbb{R^{+}}$...?[/b]
this is an interesting question which you can think on and try to find an answer to it...  :)

[quote]Can you just explain me how to construct number pi=3.14....... by using your mathematical ruler. [/quote]

This was my question and I think that I didn't get answer...
Also it is well known that we can construct any segment of length $\frac{p}{q}$ by using your mathematical ruler but thanks for this construction.

as I told you (see my edited post above) the ruler can draw any length $x \in \mathbb{R^{+}}$ but for your question a logical answer might be:
[b]draw a circle with radius equal to $\frac{1}{2}$ now if we denote the perimeter of this circle by $p$ then we know that $p=2r\pi=\pi$ so if we take a string and simply construct this circle then the length of this string is equal to $\pi$...[/b]

OK Thank you But I think that we can't agree on this topic.
Also for last question I asked you to construct number 3.14... by using just your mathematical  ruler becouse you said that it can construct any length.

Selam to Iran. :)

your right about the topic...
but as said before in construcxtions you can assumne that our ruler would draw any segment with length $x \in \mathbb{R^{+}}$ its like a PRINCIPLE...  :wink:
as said we use the ruler in constructions to connect two points in the plane...
the distance between them isnt important...


an ideal monoatomic gas is taken through a process where the magnitude of $P$ and $V$ always remain same.Find the molar heat capacity of the gas

i didn't get that can u be a bit more clear on what symbols u use

[quote="Armenia"]latex{a_{n}^{5}} :blush:[/quote]
also use   "\$" in ur poat instead of latex before latex symbols(din't use the "" just the dollar sign )


geometry
geometric transformation
rotation

Given points $ C(5,0,c)$ and $ D(5,0,d)$, a rotation of $ \overline{CD}$ about the x-axis produces a figure with an area of $ 32 \pi$ units squared; a rotation of $ \overline{CD}$ about the z-axis produces a figure with a volume of $ 100 \pi$ units cubed. Solve for c and d.

[quote="Temperal"]Sorry, but is $ x$ supposed to be $ c$?[/quote]

Whoops, editted. Sorry  :blush: .

[hide]
$ d^2\pi\minus{}c^2\pi\equal{}32\pi$
$ \Rightarrow d^2\minus{}c^2\equal{}32$

$ 25\pi(d\minus{}c)\equal{}100\pi$
$ \Rightarrow d\minus{}c\equal{}4$

$ \Rightarrow d\plus{}c\equal{}8$
$ \Rightarrow d\equal{}6,c\equal{}2$[/hide]


Let $ S_2(\mathbb{C})$ $ \equal{}\{A \in M_2(\mathbb{C})|A\equal{}A^t\}$ and $ A_2(\mathbb{C})$ $ \equal{}\{A \in M_2(\mathbb{C})|A\equal{}\minus{}A^t\}$

$ a)$ if $ A,B \in S_2(\mathbb{C})$ then $ A\plus{}B \in S_2(\mathbb{C})$ and if $ A,B \in A_2(\mathbb{C})$ then $ A\plus{}B \in A_2(\mathbb{C})$

$ b)$ if $ A \in S_2(\mathbb{C}) \cap A_2(\mathbb{C})$ then $ A\equal{}O_2$

$ c)$ $ \forall M \in M_2(\mathbb{C})$ there exist the unique matrices $ S \in S_2(\mathbb{C})$ and $ A \in A_2(\mathbb{C})$ such that $ M\equal{}S\plus{}A$

Parts 1 and 2 are quite obvious. Part 3 follows from this identity: $ M \equal{} \frac {(M \plus{} M^{t})}{2} \plus{} \frac {(M \minus{} M^{t})}{2}.$ Uniqueness of this representation is a corollary of parts 1 and 2.

I have an ugly demonstration for $ c)$ with some calculation which also proves that those matrices are unique, without using $ a)$ and $ b)$


limit
real analysis
real analysis unsolved

Find the limit:

$ \lim\limits_{n\rightarrow \infty}\frac{\sqrt[n]{1^{n}(n\minus{}1)^{n}\plus{}2^{n}(n\minus{}2)^{n}\plus{}\cdots\plus{}(n\minus{}2)^{n}2^{n}\plus{}(n\minus{}1)^{n}1^{n}}}{n^{2}}$.

Yes, that's easy  :D
$ [\frac n2]^n(n\minus{}[\frac n2])^n  \leq \sum _{k\equal{}1}^n k^n(n\minus{}k)^n \leq n[\frac n2]^n(n\minus{}[\frac n2])^n$ so $ \frac{[\frac n2](n\minus{}[\frac n2])}{n^2} \leq \frac {\sqrt [n]{\displaystyle \sum _{k\equal{}1}^n k^n(n\minus{}k)^n}}{n^2} \leq \sqrt [n]n\frac{[\frac n2](n\minus{}[\frac n2])}{n^2}$
Finally the required limit is $ \frac 14$
 :cool:


Will there ever be posted solutions for the 1st edition of the contest?

I saw that there is only for the first round..

cheers! :D

there should have been ... but unfortunately I could write them up at that time, and I've got a lot of things to do ... most of all I've lost all the files from the people who send in solutions and I have to re-solve them :D

It will take some time but eventually there will be up! :D


Have numbers m  and n  ($ m \not \equal{} n$) the same sign if

$ a) \displaystyle \left |\frac {m}{n}\right | \equal{} \frac {m}{n}$

$ b) \displaystyle \left |\frac {m}{n}\right | \equal{} \minus{} \frac {m}{n}$

?

[quote="AndrewTom"]Try some numbers:

a): Yes; b): No.[/quote]
Some numbers is not equal all numbers .
How prove ?

Consider all possible combinations of signs for $ m$ and $ n$.

[quote="RoFlLoLcOpT"]Or we could recursively define absolute value.
\[ |x| \equal{} \begin{cases}x & \text{if }x\ge0 \\
\minus{} x & \text{if }x < 0\end{cases}\forall x\in\mathbb{R}
\]
If $ m$ and $ n$ have the same sign, then $ \frac {m}{n}$ must be positive.
Therefore,

A)Yes it's possible.
B)No it's not possible.

[hide="RESTRICTIONS"]You must also specify $ n\neq0$ otherwise the fraction would be undefined. You didn't have to specify $ m\neq n.$[/hide][/quote]
What is $ \forall x\in\mathbb{R}$

a) If m=n, $ \frac{m}{n}$ is always positive, so the absolute value doesn't matter.

b) An absolute value can never come out negative, unless there is another negative to cancel it out, which is impossible.

$ \forall x\in\mathbb{R}$ literally means "for all $ x$ in the set of real numbers."

@PowerOfPi: Try to make that a little more clear. The absolute value of a negative number is the opposite of it. In other words, if $ x<0,$ then the absolute value is $ \minus{}x.$

n=any number
|n| = n
|-n|= n
if a number is affected by absolute value, then it is positive.

a) |m/n| = m/n THIS IS POSSIBLE IF the result of m/n is a positive number
b) |m/n| = -m/n NOT POSSIBLE

Hope I helped! :starwars:

[quote="JokoMoko"]n=any number
|n| = n
|-n|= n
[/quote]

Unless n is negative...

[quote="JokoMoko"]
if a number is affected by absolute value, then it is positive.[/quote]
Also the value $ \minus{}\frac{m}{n}$ isn't necessarily negative.

Isn't the second one possible if $ m \equal{} 0$ and $ n \ne 0$?

However, the numbers are not of the same "sign" - 0 is neither positive nor negative.

a yes b no
you should figure out why

hide and go seek!


Ugh, I hate these types of questions...

During a basketball game the Lancers scored 16 points the points in the first quarter, 2/7 of their total score in the 2nd quarter, and 1/4 of their total score the third quarter. And the remaining 10 points in the final quarter. What was the total number of points Lancer scored?

A. 48
B. 53
C 56
D 60
E 65

Also, In a certain room the ratio of males to females is 4 to 5. After 8 males enter the room, the ratio of males to females goes up to 6:5. What is the total number of people in the room BEFORE the additional males enter?

Let the total number of points be $ x$.
Then...
First quarter-$ 16$ points
Second quarter-$ \frac27x$
Third quarter-$ \frac14x$
Fourth quarter-$ 10$ points.

Thus, we have $ 16 \plus{} \frac27x \plus{} \frac14x \plus{} 10 \equal{} x$, so $ 26 \plus{} \frac {15}{28}x \equal{} x\implies\frac {13}{28} \equal{} 26\implies x \equal{} \boxed{56} \equal{} \boxed{C}$.

Faster way/Cheat: Look at our choices. Notice that "...2/7 of their total score..." part means that the answer choice must be divisible by $ 7$. Our only possible choice is $ \boxed{C}$. :D

For the second one, let the original number of people in the room be $ x$. Thus, since the male/female ratio is $ 4: 5$, the number of males is $ \frac4{4\plus{}5}x\equal{}\frac49x$, and the number of females is $ \frac5{4\plus{}5}x\equal{}\frac59x$. Then, after the $ 8$ males enter, there are $ \frac49x\plus{}8$ males. Then, the ratio is $ \frac{\frac49x\plus{}8}{\frac59x}\equal{}\frac65$. Thus, $ \frac49x\plus{}8\equal{}\frac69x\implies8\equal{}\frac29x\implies x\equal{}\boxed{36}$.

It helps to know HOW to get the answer, b/c it might not always be multiple choice...

I like the second method better here, though.

[quote="math154"]Let the total number of points be $ x$.
Then...
First quarter-$ 16$ points
Second quarter-$ \frac27x$
Third quarter-$ \frac14x$
Fourth quarter-$ 10$ points.

Thus, we have $ 16 \plus{} \frac27x \plus{} \frac14x \plus{} 10 \equal{} x$, so $ 26 \plus{} \frac {15}{28}x \equal{} x\implies\frac {13}{28} \equal{} 26\implies x \equal{} \boxed{56} \equal{} \boxed{C}$.

Faster way/Cheat: Look at our choices. Notice that "...2/7 of their total score..." part means that the answer choice must be divisible by $ 7$. Our only possible choice is $ \boxed{C}$. :D[/quote]

HAH! Thats the bootleg explanation the book gave. It didn't even mention the "implicit" way of doing it. >.>.

[quote="AIME15"]I assume you mean the cheating way?

What book is this, btw?[/quote]

It's an SAT book I do occasional math out of. The name is "Acing the new SAT 1 Math." -GreenHall Publishing.

[quote="AIME15"]If you're studying for the SAT, these little "cheats" are useful to know.[/quote]

Oh no, I'm 2 years too early to take it. I'm a freshmen in high school!  :D

hehhehe...then the PSAT.

On multiple choice STANDARDIZED tests, these will be useful.

I don't think they will be on AMC though.

Faster Way to do Number 2:

Before the males enter, it was $ \frac45$, after 8 of them entered, it was $ \frac65$.  So it increased 2 over 8 males, then that means for every one is equal to 8.  $ 4(4)\plus{}5(4)\equal{}16\plus{}20\equal{}\boxed{36}$

A lot of math questions on tests such as the SAT involve using little tricks like this. For example, the question might ask,
$ x^2\plus{}y^2\equal{}1$
$ xy\equal{}1$

What is $ (x\plus{}y)^2$?

Rather than solving for $ x$ and $ y$, multiply the second equation by 2 on both sides, then add the equations to get $ x^2\plus{}2xy\plus{}y^2\equal{}3$

Factor.

$ (x\plus{}y)^2\equal{}3$

[quote="fishythefish"]A lot of math questions on tests such as the SAT involve using little tricks like this. For example, the question might ask,
$ x^2 \plus{} y^2 \equal{} 1$
$ xy \equal{} 1$

What is $ (x \plus{} y)^2$?

Rather than solving for $ x$ and $ y$, multiply the second equation by 2 on both sides, then add the equations to get $ x^2 \plus{} 2xy \plus{} y^2 \equal{} 3$

Factor.

$ (x \plus{} y)^2 \equal{} 3$[/quote]

The Department of Education of New York is so slick!  :mad:


x^3-6xy+y^3=2000, solve for x,y E Z

I hv a very troublesome way to do this, any gd approach?

Here is something that may help you, a simmilar problem.

http://www.kalva.demon.co.uk/imo/isoln/isoln824.html

mod 9 we have x 3 +3xy+y 3 =2(mod 9).
so x 3 +y 3 =2(mod 3)
If one of x,y is divisible by 3 then we have a 3 =2(mod 9), which is impossible.
Thus we have x,y=1(mod 3),but that'll give x 3 =1,y 3 =1 and 3xy=3(mod 9).
So we have 2=1+1+3=5(mod 9),contradiction.

o thx...
being stupid..
thought there's a solution... ><


calculus
integration
limit
geometry
3D geometry
sphere
Euler

An Interesting relation with integrals.

Suppose $ \int_0^{\infty } f(t) \, dt\equal{}L$ converges

Proove that:

$ \lim_{h\to \infty } \, \int_0^h f(t) \, dt\equal{}\lim_{h\to \infty } \, \frac{1}{h} \sum _{n\equal{}1}^h
   \int_0^n f(t) \, dt$

It's not really interesting. It's just a version of the fact that the Cesaro means of a convergent sequence converge.

His post contains a solution; if $ a_n\to L$ then $ \frac {1}{n}\sum_{k\leq n} a_k \to L$ also (that is, if $ a_n$ converges then its cesaro means converge also and to the same limit), and the result follows upon taking $ a_n \equal{} \int^n_0 f(t)dt$.

what isn't obvious is why you're being a jerk about it.

edit: and why you are a horrible poster.

I'll add a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=167852]link[/url] to a proof of the theorem.

As for the personal attacks- I don't care about my post count. It's that high because I spend a lot of time here, and I've been a member for over three years.
I tend to approach this forum as a teacher, and that means I often post in the threads about classroom questions. That doesn't mean I'm particularly interested in the various integral questions that show up here. The harder questions I look at tend to come from other places.

[quote="Hetidemek"][quote="lordykelvin"]what is more waste of bit; my post that isnt interesting, or you post that doesnt have nothing to add?

dude, you dont gain money with more than 3 thousand posts. :)

you can try use Google system of add, with that you can gain money ;)

why don't you solve my thereom http://www.artofproblemsolving.com/Forum/viewtopic.php?t=168835?

just posting on easy topics eh? when have a difficult topic i don't see you :)

fake[/quote]
You should really watch your mouth man.

Do you have a kind of envy?

jmerry is a graduated student from the University of Washington, and he's an extremely huge mathematician, like Kent Merryfield, fedja and many other people.[/quote]

oh, jmerry, KM, fedja are extremely huge mathematicians, this is the first time I've known  :oops: this means that we are discussing with a lot of huge mathematicians..

[quote]he's an extremely huge mathematician[/quote]

In certain spheres of discourse that may be interpreted variably. ;)

if he was a huge mathematicians as you sayed he would understand my problem
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=168878

but he didn't

Euler, a great mathematician, infinitely better than anyone here, one question himself

$ 1\plus{} 1/2\plus{} 1/3 \ldots$  the sum of the first $ n$ numbers in this sequence.

n=1 the sum is 1, n=2 the sum is 3/2

but what would be if n = 1/2?!

the solution he gave was $ 2\minus{}\log (4)$

is the same thing as ask, what is the 1.5th fibonancci number?

he can't even thing as a old man about 2 century old.

i'm not in a univerty, i'm only a 17 years boy, and i didn't know about cesaro.

just because he know that, doesnt mean i knew it.

i develop this therem independently of cesaro.

and the first. if he didn't think this topic is interesting, why post here?

a lot of problems here in aops is trash, not interesting, etc.

but i dont go saying in each post "This problem is not interesting, is bad formulated etc..."

just rate it as Spam and turn around, Fake mathematician.

ugh you're such a bad poster that I'm tempted to not write anything here and just delete your last post but oh well.

[quote="lordykelvin"]i'm not in a univerty, i'm only a 17 years boy, and i didn't know about cesaro. [/quote]

believe me, it shows.

Continuing this topic would serve no useful purpose. Locked.


Given a tetrahedron, construct its centroid using only the ruler and the compasses.

Let me know when you crack this one.

14 year bump
I have like 8 now :D
Edit:7


Generalize 4/4/18 to an m by n array, in terms of m and n.

Here's a start (base case at least).

If $m = 2$, it is $\frac{(2+\sqrt{3})^{n}-(2-\sqrt{3})^{n}}{2 \sqrt{3}}$.

but that's simply the explicit formula for the recursion :wink:

it is right though

well I can find m=3, n=1/n=2:

3,1 => 1
3,2 is the same as 2,3 so 15


given $ dZ_t \equal{} X_tZ_tdB_t$, $ Z_0\equal{}1$ for $ X_t$ bounded adapted process, how can we show $ Z_t \ne 0$ a.s.

The solution is given explicitly as the Dolean exponent: $ Z_t\equal{}Z_0\exp[\int_0^tX_sdB_s\minus{}(1/2)\int_0^tX^2_sdx]$. Clearly, it is positive a.s.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Prove that the intersection $ K\cap H$ of subgroups of a group $ G$ is a subgroup of $ H$ and that if $ K$ is a normal subgroup of $ G$, then $ K\cap H$ is  a normal subgroup of $ H$.

Where are you stuck (this seems like a homework exercise, so show us you've tried something)? Try the subgroup test. Let $ x, y\in K\cap H$ and try to show that $ xy^{\minus{}1}\in K\cap H$.


If $ a,b,c > 0$ Show that
$ \sum\limits_{cyc} {\frac{{\frac{{4a}}
{{b \plus{} c}} \minus{} 1}}
{{2\left( {\frac{a}
{{b \plus{} c}}} \right)^2  \plus{} 1}}}  \leqslant 2$

$ \frac {a}{b \plus{} c} \equal{} x \: , \: \frac {b}{a \plus{} c} \equal{} y \: , \: \frac {c}{a \plus{} b} \equal{} z$

Inequality is equivalent to

$ \frac {4x \minus{} 1}{2x^2 \plus{} 1} \plus{} \frac {4y \minus{} 1}{2y^2 \plus{} 1} \plus{} \frac {4z \minus{} 1}{2z^2 \plus{} 1} \leq 2$ where $ xy \plus{} yz \plus{} zx \plus{} 2xyz \equal{} 1$

$ \Longleftrightarrow 8\lambda^2 \plus{} 24\mu^2 \plus{} 4\lambda\mu \minus{} 16\lambda \minus{} 44\mu \plus{} 21 \geq 0$ where $ \lambda \equal{} x \plus{} y \plus{} z \: , \: \mu \equal{} xy \plus{} yz \plus{} zx$

Since $ (xy \plus{} yz \plus{} zx)^2 \geq 3xyz(x \plus{} y \plus{} z) \: \Longleftrightarrow \: 2\mu^2 \geq 3\lambda(1 \minus{} \mu) \: \Longleftrightarrow \: \frac {8}{3}\mu^2 \plus{} 4\lambda\mu \geq 4\lambda$ and

Since $ 4\lambda^2 \geq 12\mu$ and $ 4\lambda^2 \minus{} 12\lambda \plus{} 9 \geq 0$

If we prove that $ \frac {64}{3}\mu^2 \minus{} 32\mu \plus{} 12 \geq 0$ then the proof is done.

The last one is equivalent to

$ \frac {4}{3}(4\mu \minus{} 3)^2 \geq 0$ which is obviously true.  :wink:


[hide="Solution"]First we determine the number of ways you can get all five games. We know that five of those boxes will always have the five games. So we consider the number of ways the toys could be distributed in the seven other boxes.

Case A: They are all the same game. They obviously have to be one of five, so $ 5$ cases.

Case B: They are two games.  We have 6 places to chose a spot for 1 divider. Then, we must select 2 out of the 5 games to be the 2 in this situation. $ 6{{5}\choose{2}}\equal{}60$

Case C: There are three games. We have 6 places to put 2 dividers, and we must select 3 out of the 5 games to be the 3 in this situation. $ {{6}\choose{2}}{{5}\choose{3}}\equal{}150$

Case D: There are four games. We have 6 places to put 3 dividers, and we must select 4 out of the 5 games to be the 4 in this situation. $ {{6}\choose{3}}{{5}\choose{4}}\equal{}100$

Case E: All five games are in the seven boxes, so we must put 5 dividers in 6 places. $ {{6}\choose{5}}\equal{}6$

We add all these together and get 321. We must put this number over the number of total possibilities for 12 cases. We get $ \frac {321}{244140625}$.

I have a bad feeling about that fraction.[/hide]

[quote="happysunnyshine"]5 different toys are offered as a promotion. One toy is randomly included with each box purchased.
Determine the probability of getting all 5 games if 12 are purchased.

Suppose you buy four boxes. Remember that each box has an equal probability of containing any one of the 7 toys.
What is the probability of getting two pairs of identical toys?

Please help me~[/quote]
By "games" do you mean toys? I don't really understand where the 7 toys come from in the second paragraph - I thought there are only 5...


geometry
3D geometry
sphere
calculus
integration
calculus computations

2 spheres $%Error. "mathdisplaystyle" is a bad command.
x^2+y^2+z^2=4$
$%Error. "mathdisplaystyle" is a bad command.
(x+2)^2+(y-1)^2+(z+2)^2=4$

find volume of solid created by intersection.  need a hint to do it without using double integrals:(
thanks,
Tr

You are to consider the section by the palne $z=t$ and to find the area of the section and integrate this.


la concursul ion barbu-dan barbilian de la calarasi s`a tinut o proba de baraj
postez aici problemele (din minte, deci s`ar putea sa sune enuntul putin modificat)

problema 1

fie $ABC$ un triunghi, $P$ un punct interior, $A_{1},B_{1},C_{1}$ proiectiile lui $P$ pe $BC,AC,AB$ si $A',B',C'$ mijloacele laturilor $[BC],[AC],[AB]$. fie $A_{2},B_{2},C_{2}$ proiectiile lui $P$ pe $B_{1}C_{1},C_{1}A_{1},A_{1}B_{1}$. sa se arate ca dreptele $A'A_{2}$, $B'B_{2}$ si $C'C_{2}$ sunt concurente

problema 2

sa se determine toate numerele naturale $n\geq 3$, care au urmatoarea prorpietate : poligonul regulat cu $n$ laturi poate fi triangulat astfel incat toate triunghiurile din triangulare sa fie isoscele. (triangulare inseamna o impartire a suprafetei poligonului in triunghiuri cu interioarele disjuncte, care au ca varfuri varfurile poligonului initial. sper ca nu m`am exprimat prea gresit ... )

problema 3

fie $p\geq 5$ un numar prim. sa se arate ca pentru orice intreg $x$, numarul $x^{p-1}+x^{p-2}+\ldots+x^{2}+x+2$ nu este patrat perfect.

$A_{2}$ este proiectia lui $A'$ pe $B_{1}C_{1}$ si analoagele.Concluzia ramane aceeasi.
Si un hint:Aplicati Lema lui Carnot de 2 ori:o data directa si o data reciproca.

ups  :oops:
am scris cam aiurea acolo
ciupan zice bine

[quote="ciprian"]Mai interesante ar fi problemele 2 si 3 (mai ales 3 :P ). Vine cineva cu o solutie la ele? Ar fi interesant sa vad o solutie diferita la oricare dintre ele fata de cea pe cara o stiu. :D[/quote]

la 2, e cam prea lunga o rezolvare completa zic eu, dar in fine dadea $n=2^{a}(2^{b}+1)$, se arata ca alea $2^{k}$ sunt triangulabile , ca unesti din 2, in 2 si se obtine din aproape in aproape un poligon cu de 2 ori mai putine laturi si se ajunge la un patrat , iar apoi ca cele cu $2^{k}+1$ sunt triangulabile, pornind de la faptul ca ala cu $9$ e triangulabil, un rationament analog.

Pai atunci a ramas sa rezolvi 3-ul pohoatza :P

la 2
[hide="solutie prob 2"]raspuns : $n=2^{k}(2^{m}+1)$ , si $n\geq 3$.
demonstratie :
voi numi un nr $n$ natural bun, daca satisface proprietatea din ipoteza. numerele $1$ si $2$ le consider bune, insa nu sunt solutie.
apoi voi arata cateva afirmatii, care impreuna ma duc la raspuns
i) $n$ este bun daca si numai daca $2n$ este bun
evident, daca $n$ e bun atunci si $2n$ e bun, pentru ca se construiesc triunghiuri isoscele in exteriorul poligonului regulat, triangulat, cu $n$ laturi si se obtine unul cu $2n$ laturi deja triangulat.
pentru partea inversa, observ erva ca daca aleg o latura, in triunghiul isoscel ce o contine ea nu poate fi baza, deoarece pe mediatoarea sa nu se afla niciun varf al poligonului.
ii) $n$ impar este bun daca si numai daca este de forma $2^{k}+1$.
fie $n=2p+1$. atunci exista o latura a poligonului care se afla intr`un triunghi din triangulare format din doua diagonale si o latura, deoarece numarul de laturi este impar. (cazul $n=3$ se exclude ca $3$ e clar bun si e si $2^{1}+1)$. atunci, daca varful triunghiului este $A_{1}$, el este $A_{1}A_{p+1}A_{p+2}$. apoi, ma uit la poligonul $A_{1}A_{2}\ldots A_{p+1}$. si el trebuie triangulat, cu aceeasi proprietate. insa latura $A_{1}A_{p+1}$ este mai mare decat orice alta latura sau diagonala, deci este baza unui triunghi isoscel. de aici $p$ este par, si daca $p=2p_{1}$, atunci repet rationamentul pentru poligonul $A_{1}A_{2}\ldots A_{p_{1}+1}$. ma opresc cand ajung la un $p_{k}=1$, si deci $p=2^{s}$. deci daca $n$ este impar si bun, are forma dorita. apoi, din rationamentul de mai sus se deduce si triangularea pt un numar de forma asta
din i) si ii) rezulta concluzia[/hide]

la 3
[hide="solutie prob 3"]presupun ca numarul ala e un $y^{2}$. atunci fie $N=x^{p-1}+x^{p-2}+\ldots+x+1$. deci $N=(y-1)(y+1)$.
afirmatie : toti divizorii lui $N$ sunt congruenti cu $0$ sau $1$ modulo $p$.
demonstratie : fie $q$ un divizor prim al lui $N$.  voi lucra in $\mathbb{Z}_{q}$ :
$x=1$ implica $N=p$ , dar $N=0$, deci $p=q$.
$x\neq 0$ (pt ca $1\neq 0$)
fie $t$ ordinul lui $x$. atunci, cum $x^{p}=1$, rezulta $t|p$, si cum $t>1$ rezulta ca $t=p$. apoi $t|q-1$, deci $p|q-1$, deci $q$ e congruent cu $1$ modulo $p$. de aici rezulta afirmatia.
odata ce am ajuns aici, in concurs am zis "pas" rezolvarii :( (eu vedeam putin diferit lucrurile, nu stiu de ce). nici macar nu am scris asta pe foaie :(
$y-1$ si $y+1$ sunt divizori ai lui $N$, deci sunt congruenti cu $0$ sau $1$ modulo $p$, dar diferenta lor este $2$, si cum $p \geq 5$, acest lucru este imposibil. tare grea era concluzia asta :)[/hide]


If the roots of $x^3+ax^2+bx+c=0$ form an arithmetic progression, find $c$ in terms of $a$ and $b$.

[hide]The roots $x_1,\ x_2,\ x_3$ are in a arithmetic progression then:

$x_1+x_3=2x_2$, by Viete's formula we get:

$x_1+x_2+x_3=3x_2=-a \implies x_2=-\frac{a}{3}$,

$x_2(x_1+x_3)+x_1x_3=2x_2^2+x_1x_3=b \implies x_1x_3=b-\frac{2a^2}{9}$

$x_1x_2x_3=-c \implies c=\frac{a}{3}(b-\frac{2a^2}{9})$[/hide]


I tried to work this out but this doesn't seem to get me any far.

Can anyone give me HINTS? I want to solve this problem by myself, and hints will be greatly appreciated.

I haven't worked it out yet, but one thing that popped out at me: what would happen if you divided everything in the first equation by $2$?

[b]EDIT:[/b] Actually, you don't really need that... just try writing out the values, then substituting in the previous equation. For example, $12a_4=6a_3-2a_2$, and $6a_3=2a_2-a_1$, so $12a_4=-a_1$. You can probably find a pattern from there.

[quote="nsato"]Try working out $a_n$ for the first few values.[/quote]

Oops. I didn't do that.

Just wondering, for this type of problem, which one do you recommend more:

A. Try to place with the given equation and see if you can find any telescoping
B. Plug in numbers right away

I think sequence problems are quite popular and it was mentioned in first and second AoPS book as well. So, which one seemed to be better?

It really depends on the problem. Some, it's more useful to use telescoping, because that's the intended use, but other times it isn't, and there it may be just as well to work out the first few numbers and look for a pattern.

It depends on your starting point.  If you don't know the terms of the sequence or the sum of the series, you work out the first few terms until you see the pattern, and feel comfortable that you have the right formula.

Then, and only then, do you think about techniques that can help you prove your formula.


Perhaps he/she came to the conclusion that they only provide one answer sheet per problem in the USAMO Manual...

My teacher did that last year :[
so my handwriting was [i]sort of[/i] small

I don't think we ever imagined some one would interpret the fact that we put only one master solution page per problem to copy to mean that the solution could only go on one page.  Therefore, we don't  have explicit instructions to that regard.  However, there are plenty of implicit instructions in the USAMO Teachers' Manual.

Look at the italic instructions at the top of the T-III Certification Forms (pages 9-10) used as a cover page for the faxed solutions:  "                                                                                Please fill out one form for each student you have participating in the USAMO and indicate the number  of pages faxed for each problem:"  That surely indicates more than one page is possible.

Or, take a look at Section V, page 5:  "If requested, you may let the participant have unmarked scratch paper. If, for any reason, the student submits scratch paper instead of, or in addition to, an answer form, the Problem number and the Student Number, but NOT the student’s name or school, must be on each page submit-
ted. All papers must be anonymous at the time of the grading."

Also, it is very helpful to follow the directions on page 6:  "At the end of each session instruct your students to double-check that the student number on their answer forms matches the number assigned in the official invitation letter.  .... Please fax at least one official USAMO answer sheet
with student number for each problem, even if the answer sheet is
blank because the student did not attempt a solution. This assures
that we have received all problems from all students."

That should be clear enough, more than one page per problem is permitted.

As long as I have your attention, Please also " tell students to stay within the borders when
   writing their solutions and to write DARKLY and LEGIBLY so  the faxed copy can be clearly read."
This is really important!  This is the real reason we provide the outlined paper rather than telling people ot use blank paper.  And write darkly.  A black roller tip, black ballpoint pen, or medium felt tip faxes the best.  If we can't read it, we won't grade, it's that simple, we have too much to do to stop and agonize over something illegible, we move on.

Steve Dunbar
MAA Director, American Mathematics Competitions

Okay, thanks for that... It would be terrible if I was trying to squeeze a long solution into one page on the usamo...

If the master answer sheet contains "page:___ of ___" just underneath the problem #, it will leave less guess work for the proctors and students.

You're definitely allowed as many pages as you want.

I discovered my senior year that the bigger I wrote, the more legible it was, so I decided to take advantage of that.  I think one of my solutions ended up running seven pages.

[quote="tjhance"]My teacher says that on the USAMO, we're only allowed a one-page solution to this problem. This doesn't seem right, so I was wondering if anybody could clarify this?

I didn't see anything in the rules about it.[/quote]

O_O

WHAT????

no.. THAT IS WRONG


Not trivial but straightforward.

The [b]Cauchy-Schwarz inequality in the Engel form[/b] (this is, at least, the informal name this inequality has among the German IMO team) states that if $x_1$, $x_2$, ..., $x_n$ are arbitrary real numbers, and $y_1$, $y_2$, ..., $y_n$ are arbitrary positive real numbers, then

[tex]\frac{x_{1}^{2}}{y_{1}}+\frac{x_{2}^{2}}{y_{2}}+...+\frac{x_{n}^{2}}{y_{n}}\geq \frac{\left( x_{1}+x_{2}+...+x_{n}\right) ^{2}}{y_{1}+y_{2}+...+y_{n}}[/tex].

You can easily see that this is an equivalent form of the Cauchy-Schwarz inequality, but it is often more directly applicable to inequalities.

Now, we have

[tex]\frac{a}{b^{2}+bc+c^{2}}+\frac{b}{c^{2}+ca+a^{2}}+\frac{c}{a^{2}+ab+b^{2}}[/tex]
[tex]=\frac{a^{2}}{a\left( b^{2}+bc+c^{2}\right) }+\frac{b^{2}}{b\left( c^{2}+ca+a^{2}\right) }+\frac{c^{2}}{c\left( a^{2}+ab+b^{2}\right) }[/tex].

But by the Cauchy-Schwarz inequality in the Engel form, we have

[tex]\frac{a^{2}}{a\left( b^{2}+bc+c^{2}\right) }+\frac{b^{2}}{b\left( c^{2}+ca+a^{2}\right) }+\frac{c^{2}}{c\left( a^{2}+ab+b^{2}\right) }\geq \frac{\left( a+b+c\right) ^{2}}{a\left( b^{2}+bc+c^{2}\right) +b\left( c^{2}+ca+a^{2}\right) +c\left( a^{2}+ab+b^{2}\right) }[/tex].

Thus,

[tex]\frac{a}{b^{2}+bc+c^{2}}+\frac{b}{c^{2}+ca+a^{2}}+\frac{c}{a^{2}+ab+b^{2}}\geq \frac{\left( a+b+c\right) ^{2}}{a\left( b^{2}+bc+c^{2}\right) +b\left( c^{2}+ca+a^{2}\right) +c\left( a^{2}+ab+b^{2}\right) }[/tex]
[tex]=\frac{\left( a+b+c\right) ^{2}}{\left( a+b+c\right) \left( bc+ca+ab\right) }=\frac{\left( a+b+c\right) ^{2}}{3\left( bc+ca+ab\right) } \geq 1[/tex].

In the last step, we have used the well-known inequality $\left( a+b+c\right) ^{2}\geq 3\left( bc+ca+ab\right) $. Proof complete.

  Darij

[quote="RK"]This problem is similar to the one fuzzylogic posted, and it has a very elegant solution...

Prove that for arbitrary positive real numbers a, b, c, the following inequality holds.
[tex]\frac{a^3}{b^2+bc+c^2} + \frac{b^3}{c^2+ca+a^2} + \frac{c^3}{a^2+ab+b^2}\geq \frac{a+b+c}{3}. [/tex][/quote]

This is different, the problem I posted before is as follows (notice the differences in the bottom):

Let $a,b,c> 0$ with $a+b+c=1$. Prove that
$\displaystyle \frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2} \geq \frac{1}{3}.$

By rearrangement, this one is stronger than the one in your quote.

[quote="fuzzylogic"]the problem I posted before is as follows (notice the differences in the bottom):

Let $a,b,c> 0$ with $a+b+c=1$. Prove that
$\displaystyle \frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2} \geq \frac{1}{3}.$[/quote]

I have just found a simple proof of this. In fact, we have a tighter inequality:

$\displaystyle \frac{a^{3}}{a^{2}+ab+b^{2}}+\frac{b^{3}}{b^{2}+bc+c^{2}}+\frac{c^{3}}{c^{2}+ca+a^{2}}\geq a^{2}+b^{2}+c^{2}$

(again for positive a, b, c satisfying a + b + c = 1).

Now, $a^{2}+b^{2}+c^{2}\geq \frac{1}{3}$ (in fact, by the power mean inequality, $\displaystyle \sqrt{\frac{a^{2}+b^{2}+c^{2}}{3}}\geq \frac{a+b+c}{3}=\frac{1}{3}$, so that $\displaystyle \frac{a^{2}+b^{2}+c^{2}}{3}\geq \left( \frac{1}{3}\right) ^{2}$, and thus $a^{2}+b^{2}+c^{2}\geq 3\cdot \left( \frac{1}{3}\right) ^{2}=\frac{1}{3}$). Therefore, we conclude that Fuzzylogic's inequality

$\displaystyle \frac{a^{3}}{a^{2}+ab+b^{2}}+\frac{b^{3}}{b^{2}+bc+c^{2}}+\frac{c^{3}}{c^{2}+ca+a^{2}}\geq \frac{1}{3}$

holds.

Now here is the proof of the tighter inequality

$\displaystyle \frac{a^{3}}{a^{2}+ab+b^{2}}+\frac{b^{3}}{b^{2}+bc+c^{2}}+\frac{c^{3}}{c^{2}+ca+a^{2}}\geq a^{2}+b^{2}+c^{2}$

(not from me, but from one of the gurus, i. e. Harazi, Gil, Vasc, ...): We have

$\displaystyle \frac{a^{3}}{a^{2}+ab+b^{2}}+\frac{b^{3}}{b^{2}+bc+c^{2}}+\frac{c^{3}}{c^{2}+ca+a^{2}}$
$\displaystyle =\frac{a^{4}}{a\left( a^{2}+ab+b^{2}\right) }+\frac{b^{4}}{b\left( b^{2}+bc+c^{2}\right) }+\frac{c^{4}}{c\left( c^{2}+ca+a^{2}\right) }$
$\displaystyle =\frac{\left( a^{2}\right) ^{2}}{a\left( a^{2}+ab+b^{2}\right) }+\frac{\left( b^{2}\right) ^{2}}{b\left( b^{2}+bc+c^{2}\right) }+\frac{\left( c^{2}\right) ^{2}}{c\left( c^{2}+ca+a^{2}\right) }$
$\displaystyle \geq \frac{\left( a^{2}+b^{2}+c^{2}\right) ^{2}}{a\left( a^{2}+ab+b^{2}\right) +b\left( b^{2}+bc+c^{2}\right) +c\left( c^{2}+ca+a^{2}\right) }$

after the Cauchy-Schwarz inequality in the Engel form. But

$a\left( a^{2}+ab+b^{2}\right) +b\left( b^{2}+bc+c^{2}\right) +c\left( c^{2}+ca+a^{2}\right) =\left( a+b+c\right) \left( a^{2}+b^{2}+c^{2}\right) =a^{2}+b^{2}+c^{2}$

(since a + b + c = 1). Thus,

$\displaystyle \frac{a^{3}}{a^{2}+ab+b^{2}}+\frac{b^{3}}{b^{2}+bc+c^{2}}+\frac{c^{3}}{c^{2}+ca+a^{2}}\geq \frac{\left( a^{2}+b^{2}+c^{2}\right) ^{2}}{a^{2}+b^{2}+c^{2}}=a^{2}+b^{2}+c^{2}$,

qed..

  Darij

Very nice!!!

There are two other solutions posted at: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=100814


Find all functions $f: \mathbb Z \to \mathbb Z$ which satisfy \[f(m+ f(n)) = f(m) + n\] for all integers $m$ and $n$.

for $ m \equal{} 0$  $ f(f(n)) \equal{} n \plus{} f(0)$ so $ f$ is bijective

$ n \equal{} 0$ $ \implies f(m \plus{} f(0)) \equal{} f(m) \implies f(0) \equal{} 0$

replace in the original equation $ n$ by $ f(n)$  , $ f(m \plus{} n) \equal{} f(m) \plus{} f(n)$
wich is cauchy equation so $ f(n) \equal{} an \forall n\in\mathbb{Z}$ with $ a\in { \minus{} 1,1}$

[quote="b555"]Find all functions f : Z --> Z which satisfy f(m+ f(n)) = f(m) + n for all m, n belong to Z.[/quote]

$ m\equal{}0$ $ \implies$ $ f(f(n))\equal{}n\plus{}f(0)$
$ n\equal{}f(p)$ $ \implies$ $ f(m\plus{}p\plus{}f(0))\equal{}f(m)\plus{}f(p)$
$ m\equal{}x\minus{}f(0)$ an $ p\equal{}y\minus{}f(0)$ $ \implies$ $ f(x\plus{}y\minus{}f(0))\equal{}f(x\minus{}f(0))\plus{}f(y\minus{}f(0))$

Let $ g(x)\equal{}f(x\minus{}f(0))$. this last equation becomes $ g(x\plus{}y)\equal{}g(x)\plus{}g(y)$ and so $ g(x)\equal{}ax$ and s $ f(x)\equal{}ax\plus{}b$

Plugging this in the original equation, we get $ a(m\plus{}an\plus{}b)\plus{}b\equal{}am\plus{}b\plus{}n$ and so $ a\equal{}\pm 1$ and $ b\equal{}0$

Hence the two solutions :
$ f(n)\equal{}n$
$ f(n)\equal{}\minus{}n$


abstract algebra
group theory
superior algebra
superior algebra solved

[b][i]Prove[/i][/b] that if G and H are cyclic groups of orders m and n with (m,n) = 1, then G X H is cyclic.

Then How many different Generators does it have.

____________________

For the second part I think there would be m*n generators but besides that i am unsure or hwo to accomplish my first part

My teacher gave me hints about prove orger of GxH is mn and also try to prove the order of (a,b) in GxH is m*n.

I am unsure of how to mostly do the proof any help would be great![/b]

Let $G=\langle g \rangle$ and $H=\langle h\rangle$ be cyclic groups of orders $m,n$, where $(m,n)=1$. Consider the homomorphism $\mathbb{Z}\to G \times H$ such that $k\mapsto (g^k,h^k)$. An element in its kernel must be divisible by $m$ and $n$ thus, because $(m,n)=1$, $mn$. Conversely, $mn\mathbb{Z}$ is contained in the kernel of $g\times h$ so the kernel of $G\times H$ is $mn\mathbb{Z}$. By Chinese remainder theorem, the homomorphism $\mathbb{Z}\to G\times H$ is surjective.

Now we have found a cyclic group of order $mn$ thus it has $\varphi (mn)=\varphi (m)\varphi (n)$ generators.


Let A';B';C' be three points respectively locate on sides BC;CA;AB of triangle ABC and A",B",C" respectively be symetrical points of A;B;C to A';B';C'. Prove that: areaA"B"C" = 3areaABC + 4areaA'B'C'


i just cant figure this one out.......:wallbash:  :wallbash_red:

hope there's isnt some very very easy solution to it   :oops:

$3^x+4^x+5^x=6^x$

x?

Rewrite $3^x+4^x+5^x=6^x$ as $(3/6)^x+(4/6)^x+(5/6)^x=1$

Now $f(x) = (3/6)^x+(4/6)^x+(5/6)^x$ is a strictly decreasing function, as so $f(x) = 1$ has at most one root.

$x = 3$ works and it is the only root.

[quote="varun"]how do you get 3?[/quote]

I trusted dakyru ;)

So only the guess and check mehtod works here ?

[quote="eminem"]hey guess and check in my opinion isnt a nice method for solving questions, cant someone tell some proper solution ??? or is guess and check the only way to do it??

hey dakyru how did you get 3 , ( not trial and error i hope :D )[/quote]

I don't think there is a way to explicitly solve such an equation.  That would require isolation of $x$...  Anyway, using the function that tutu defined and the fact it is strictly decreasing allows for you to guess, check and make bounds if you didn't see $x=3$ off the bat.

ok then , i guess thats the only way to do it :D

I only got the fact that $x$ can't be an even number...that's all I got. :?


Hello,

Prove that $ (\forall n \in N\minus{}\{0, 1\}) : 1\plus{}\frac{1}{2}\plus{}\frac{1}{3}\plus{} \dots \plus{}\frac{1}{n} \notin N$

[hide="hint"]See [url=http://www.exampleproblems.com/wiki/index.php/Bertrands_postulate]Bertrands postulate[/url][/hide]
[hide="solution"]Let in contrary $ m\equal{}1\plus{}\dots\plus{}\frac{1}{n}$ be integer for some $ n>1$. Choose prime $ p$ between $ \frac{n}{2}$ and $ n$

Then $ p$ devides none of $ 2, 3, \dots p\minus{}1, p\plus{}1, \dots n$

Multiplying by $ \frac{n!}{p}$ we have $ \frac{mn!}{p}\equal{}\frac{n!}{p}\plus{}\frac{n!}{2p}\plus{}\dots \plus{}\frac{n!}{p^2}\plus{}\dots \plus{}\frac{n!}{np}$
LHS is integer, the all members in RHS are integers exept $ \frac{n!}{p^2}$ which is not integer.[/hide]

[quote="PKMathew"][quote="Johan Gunardi"][hide="A more elementary approach"]Let $ m$ be the greatest integer such that $ 2^m\le n$. Hence all other numbers are of the form $ 2^k\cdot l$, where $ k < m$ and $ l$ is odd. So if we take the lowest common denominator, the denominator is of the form $ 2^m\cdot j$; and all numerators, except the $ \frac1m$, are even. So if we sum up, the numerator is odd and the denominator is even, and hence the value is not an integer.[/hide][/quote]

Would you explain it clearly and precisely, may be my stupid understanding?[/quote]

Which part do you not understand? I will repeat it more clearly.

Consider the number $ m$, such that $ 2^m<n<2^{m\plus{}1}$. For example, if $ n\equal{}2000$, we have $ m\equal{}10$, because $ 2^{10}<2000<2^{11}$.

So we consider the sum: $ 1\plus{}\frac12\plus{}\frac13\plus{}\ldots\plus{}\frac1{2^m}\plus{}\ldots\plus{}\frac1n$. Take the common denominator $ 2^m\cdot k$ ($ k$ is odd). If we sum up, the numerator is $ 2^m\cdot k\plus{}\frac{2^m\cdot k}2\plus{}\frac{2^m\cdot k}3\plus{}\ldots\plus{}\ldots\plus{}k\plus{}\ldots\plus{}\frac{2^m\cdot k}n$. So, all the terms except $ k$ are odd, and hence the sum is also odd. Since the denominator is even, it cannot be an integer.

[quote="Johan Gunardi"][quote="PKMathew"][quote="Johan Gunardi"][hide="A more elementary approach"]Let $ m$ be the greatest integer such that $ 2^m\le n$. Hence all other numbers are of the form $ 2^k\cdot l$, where $ k < m$ and $ l$ is odd. So if we take the lowest common denominator, the denominator is of the form $ 2^m\cdot j$; and all numerators, except the $ \frac1m$, are even. So if we sum up, the numerator is odd and the denominator is even, and hence the value is not an integer.[/hide][/quote]

Would you explain it clearly and precisely, may be my stupid understanding?[/quote]

Which part do you not understand? I will repeat it more clearly.

Consider the number $ m$, such that $ 2^m < n < 2^{m \plus{} 1}$. For example, if $ n \equal{} 2000$, we have $ m \equal{} 10$, because $ 2^{10} < 2000 < 2^{11}$.

So we consider the sum: $ 1 \plus{} \frac12 \plus{} \frac13 \plus{} \ldots \plus{} \frac1{2^m} \plus{} \ldots \plus{} \frac1n$. Take the common denominator $ 2^m\cdot k$ ($ k$ is odd). If we sum up, the numerator is $ 2^m\cdot k \plus{} \frac {2^m\cdot k}2 \plus{} \frac {2^m\cdot k}3 \plus{} \ldots \plus{} \ldots \plus{} k \plus{} \ldots \plus{} \frac {2^m\cdot k}n$. So, all the terms except $ k$ are odd, and hence the sum is also odd. Since the denominator is even, it cannot be an integer.[/quote]

$ \boxed{2^m\cdot k \plus{} \frac {2^m\cdot k}2 \plus{} \frac {2^m\cdot k}3 \plus{} \ldots \plus{} \ldots \plus{} k \plus{} \ldots \plus{} \frac {2^m\cdot k}n}$.

How can we know sum up each term in the numerator ,other than $ k$, is an odd integer?

[quote="PKMathew"][/quote][quote="Johan Gunardi"][quote="PKMathew"][quote="Johan Gunardi"][hide="A more elementary approach"]Let $ m$ be the greatest integer such that $ 2^m\le n$. Hence all other numbers are of the form $ 2^k\cdot l$, where $ k < m$ and $ l$ is odd. So if we take the lowest common denominator, the denominator is of the form $ 2^m\cdot j$; and all numerators, except the $ \frac1m$, are even. So if we sum up, the numerator is odd and the denominator is even, and hence the value is not an integer.[/hide][/quote]

Would you explain it clearly and precisely, may be my stupid understanding?[/quote]

Which part do you not understand? I will repeat it more clearly.

Consider the number $ m$, such that $ 2^m < n < 2^{m + 1}$. For example, if $ n = 2000$, we have $ m = 10$, because $ 2^{10} < 2000 < 2^{11}$.

So we consider the sum: $ 1 + \frac12 + \frac13 + \ldots + \frac1{2^m} + \ldots + \frac1n$. Take the common denominator $ 2^m\cdot k$ ($ k$ is odd). If we sum up, the numerator is $ 2^m\cdot k + \frac {2^m\cdot k}2 + \frac {2^m\cdot k}3 + \ldots + \ldots + k + \ldots + \frac {2^m\cdot k}n$. So, all the terms except $ k$ are odd, and hence the sum is also odd. Since the denominator is even, it cannot be an integer.[/quote][quote="PKMathew"]

$ \boxed{2^m\cdot k + \frac {2^m\cdot k}2 + \frac {2^m\cdot k}3 + \ldots + \ldots + k + \ldots + \frac {2^m\cdot k}n}$.

How can we know sum up each term in the numerator ,other than $ k$, is an odd integer?[/quote]All the terms are even integers [u]except[/u] $ k$ because $ 2^mk$ is a common denominator. $ k$ is an odd integer. odd+even+even+...+even=odd. I'm pretty sure Johan meant 'even' for all those terms except k.

[quote="10000th User"]I'm pretty sure Johan meant 'even' for all those terms except k.[/quote]Yes I did.  :oops:

[quote="Johan Gunardi"][quote="PKMathew"][quote="Johan Gunardi"][hide="A more elementary approach"]Let $ m$ be the greatest integer such that $ 2^m\le n$. Hence all other numbers are of the form $ 2^k\cdot l$, where $ k < m$ and $ l$ is odd. So if we take the lowest common denominator, the denominator is of the form $ 2^m\cdot j$; and all numerators, except the $ \frac1m$, are even. So if we sum up, the numerator is odd and the denominator is even, and hence the value is not an integer.[/hide][/quote]

Would you explain it clearly and precisely, may be my stupid understanding?[/quote]

Which part do you not understand? I will repeat it more clearly.

Consider the number $ m$, such that $ 2^m < n < 2^{m \plus{} 1}$. For example, if $ n \equal{} 2000$, we have $ m \equal{} 10$, because $ 2^{10} < 2000 < 2^{11}$.

So we consider the sum: $ 1 \plus{} \frac12 \plus{} \frac13 \plus{} \ldots \plus{} \frac1{2^m} \plus{} \ldots \plus{} \frac1n$. Take the common denominator $ 2^m\cdot k$ ($ k$ is odd). If we sum up, the numerator is $ 2^m\cdot k \plus{} \frac {2^m\cdot k}2 \plus{} \frac {2^m\cdot k}3 \plus{} \ldots \plus{} \ldots \plus{} k \plus{} \ldots \plus{} \frac {2^m\cdot k}n$. So, all the terms except $ k$ are odd, and hence the sum is also odd. Since the denominator is even, it cannot be an integer.[/quote]
How do we know, say, k divides 3 for \frac {2^m\cdot k}3 to be an (even) integer?

$ k$ is the common denominator; it's defined in a way so that 3 does divide it.


The altitude to the hypotenuse of a right triangle divides the hypotenuse into two line segments, one of length 2003 and the other of length 25. What is the area of the triangle?

Think of similar triangles. :)

Look for ratios.
I see 3 similar triangles.
b:a:2028
c:25:a
2003:c:b
to find $b^2$ cross multiply $\frac{b}{2028}$ and $\frac{2003}{b}$then$b^2=2003*2028$
to find $a^2$ cross multiply $\frac{a}{2028}$ and $\frac{25}{a}
$then$a^2=5*2028$
Multiply $a^2$ and $b^2$
so you get 2028*2028*25*2003
take the squareroot $ab=5*2028\sqrt{2003}$
area is half $\frac{ab}{2}=5070\sqrt{2003}$


[quote="USAMO13"]Can anyone help me with this problem:
with the property that x^y is rational? Can x^y be irrational?[/quote]

are these two separate problems, cuz how could both of these be satisfied? am i misreading the problem?

Yeah..2 seperate probs..2 seperate proofs..

[hide="1"]$ \sqrt{2}^{\sqrt{2}}\equal{}x$
If x is rational, we're done with first problem. If not, then consider this:
$ \sqrt{2}^x\equal{}\sqrt{2}^{(\sqrt{2}^{\sqrt{2}})}\equal{}\sqrt{2}^{\sqrt{2}\cdot\sqrt{2}}\equal{}\sqrt{2}^2\equal{}2$

thus the second equation satisfies the condition.[/hide]

[quote="james4l"][hide="1"]$ \sqrt {2}^{\sqrt {2}} \equal{} x$
If x is rational, we're done with first problem. If not, then consider this:
$ \sqrt {2}^x \equal{} \sqrt {2}^{(\sqrt {2}^{\sqrt {2}})} \equal{} \sqrt {2}^{\sqrt {2}\cdot\sqrt {2}} \equal{} \sqrt {2}^2 \equal{} 2$

thus the second equation satisfies the condition.[/hide][/quote]

Dude...thats kinda confusing.. $ \sqrt {2}^{\sqrt {2}}$ is rational?

Not necessarily - however if it isn't then we try the second case which is guaranteed to work.

What if we let $ x \equal{} \sqrt2$ and $ y \equal{} \pi$

[quote="USAMO13"]Can anyone help me with this problem:
Are there two irrational numbers x and y with the property that x^y is rational? Can x^y be irrational?
Please provide an explanation.
Thanks![/quote]

first of all, your avatar is obnoxious and copied. and i'm pretty sure i know who you are.

and yes, x^y can be irrational. james4l has given a great example, read over it. i'm pretty sure $ \sqrt2^{\sqrt2}$ is irrational.

Trouble is, $ \sqrt2^{\sqrt2}$ is proven to be irrational (transcendental in fact) by the Gelfond–Schneider Theorem (http://www.math.sc.edu/~filaseta/gradcourses/Math785/Math785Notes8.pdf) which is a [sarcasm]bit[/sarcasm] above MC level...

[quote="USAMO13"]Thanks...but can you explicate a tad more on your solutions..thanks...

and BTW nikeballa96....I got this avatar from someone else....He said I could have it...and he said you copied him.[/quote]

okay, had this avie since last november, look at dojo's album, and the name of this avatar. its "NIKEBALLA96". so please change your avie, it's not really fair, considering i had it, and now its dumb.

and as james4l said, i can't really "explicate" the concept any further, it's a "bit" above MC level.

[quote="pythag011"]Well, there is an elementary nonconstructive argument using countable and uncountable sets. Should I post it?

Blah, anyways, consider the set of numbers that can be expressed as sqrt(2)^x for some irrational x. Since the irrationals are uncountable, this set is uncountable. Therefore, this set can't be a subset of the rationals since the rationals are countable, so there exists an irrational x such that sqrt(2)^x is irrational.[/quote]

that's not THAT advanced, it's actually rather simple. you just have to understand / know that there are uncountable irrationals (well, duh.), but countable rationals.

edit: actually knew that. whoops.

James4l's proof is incorrect, because he assumed that a^(b^c) was equal to (a^b)^c, which it is not. As a result his tower of root 2's is not equal to 2.

[quote="james4l"]Flaw: multiple x^y can map to the same number? (uncountably many)

@nike: there are an infinite number of rationals[/quote]

sqrt(2)^n, not x^n.

and @Aime: why?

@Countdown: whoops. Put sqrt(2) in the exponent and x in the base for desired results.
@AIME: inconclusive evidence, as of now

New Question:

Is it possible to prove the previous statement without using any math above MC level? (the concept of multiple infinities is [b]not[/b] MC level)

[quote="AIME15"]@nikeballa: I think you mean "uncountable" irrationals and "countable" rationals...there are infinite of each :wink:

BTW, it's obvious that USAMO13 = Gears.[/quote]

no, there's an ACT36 who stole my avie too, and he stole my signature. it's really pissing me off.

hm. as james4l said, $ (\sqrt2^{\sqrt2})^{\sqrt2}\equal{}\sqrt2^{\sqrt2\cdot\sqrt2}\equal{}\sqrt2^2\equal{}\boxed{2}$. yay?

Thanks ppl I kinda get it.
And BTW....ME IS NOT A MULTI :D


Delete from the harmonic series
1/1+1/2+1/3+ · · · +1/n+ · · ·
all terms whose denominators contain digit 9. Prove that the obtained series converge

[quote="nateharman1234"]Consider all 1 digit numbers with no digit 9, there are 8 of them the smallest being 1.  Consider all 2 digit numbers with no nines in them, there are 8x8=64 of them, the least being 10.  In general there are 8^n n digit numbers without a 9 digit, the least of which is 10^(n-1).  So when we sum we know that our sum is less than 8x(1/1)+64(1/10)+8^3(1/100)+8^4(1/1000)... and leik ZOMG its a gemetrik sequence with ratio less than 1![/quote]

Good method, but wrong calculus :

There are not $ 8*8 \equal{} 64$ two-digits such numbers, but $ 8*9 \equal{} 72$ (except for he leading digit which can be neiher 0, neither 9, so 8 possibilities, the second may be $ 0,1,2,3,4,5,6,7$ or $ 8$, so nine possibiliies).
And so n-digits such numbers are $ 8*9^{n\minus{}1}$, and not $ 8^{n}$

And so the upper bound is $ 8(1\plus{}\frac{9}{10}\plus{}(\frac{9}{10})^{2}\plus{}\ldots)$ which converges indeed.

And the real limit is very difficult to compute with precision : it is $ 22,920676619\ldots$


First, we have two comment:
1) For all real numbers $a,b,c$: $ab+bc+ca\leq a^{2}+b^{2}+c^{2}\Leftrightarrow 3(ab+bc+ca)\leq (a+b+c)^{2}$ Combine with $a+b+c=2$ infering:
$ab+bc+ca\leq \frac{4}{3}\Leftrightarrow-(ab+bc+ca)\geq-\frac{4}{3}$
2) For all triagle ABC then:
$abc\geq (a+b-c)(a+c-b)(b+c-a)$ Combine with $a+b+c=2$ infering:
$abc\geq 8(1-a)(1-b)(1-c)=8-8(a+b+c)+8(ab+bc+ca)-8abc$
$\Leftrightarrow 9abc\geq 8(ab+bc+ca)-8$
Therefore:
$a^{2}+b^{2}+c^{2}+2abc\geq a^{2}+b^{2}+c^{2}+\frac{16}{9}(ab+bc+ca)-\frac{16}{9}= (a+b+c)^{2}-2(ab+bc+ca)+\frac{16}{9}(ab+bc+ca)-\frac{16}{9}\geq-\frac{2}{9}(ab+bc+ca)+\frac{20}{9}\geq\frac{20}{9}-\frac{8}{27}=\frac{54}{27}$
So we have proved left hand of inequality! The remain to solve the right hand

[quote="April"]If triangle $ABC$ has $a+b+c=2$, then:
\[a^{2}+b^{2}+c^{2}+2abc<2 \]
[/quote]

just $\sum{a^{2}}=2p^{2}-2r^{2}-8Rr$ and $abc=4pRr$, and $a+b+c=2\Longleftrightarrow p=1$ , therefore $\sum{a^{2}}+2abc<2 \Longleftrightarrow r>0$

remark: for the left part we can do the same thing.

[quote="April"]If triangle $ABC$ has $a+b+c=2$, then:
\[\frac{52}{27}\le a^{2}+b^{2}+c^{2}+2abc \]
[/quote]
We have: $(1-a)(1-b)(1-c)\le\left(\frac{3-(a+b+c)}{3}\right)^{3}$
$\Longleftrightarrow 1+ab+bc+ca-(a+b+c)-abc\le\frac{1}{27}$
$\Longleftrightarrow ab+bc+ca-abc\le\frac{28}{27}$
$\Longleftrightarrow (a+b+c)^{2}-(a^{2}+b^{2}+c^{2})-2abc\le\frac{56}{27}$
$\Longleftrightarrow a^{2}+b^{2}+c^{2}+2abc\ge\frac{52}{27}$


limit logarithms LaTeX calculus derivative

hi all
can you help me to sove this limit
$ \lim_{x\rightarrow\infty} \frac{x}{\ln(\ln(x))}-\ln(x)$
thanks

[color=red]edit:  added $ \text{\LaTeX}$ - b-flat[/color]

[quote="kira of morocco"]hi all
can you help me to sove this limit
$ \lim_{x\rightarrow\infty} \frac {x}{\ln(\ln(x))} - \ln(x)$
thanks

[color=red]edit:  added $ \text{\LaTeX}$ - b-flat[/color][/quote]

As x approaches infinity, both ln(x) and ln(ln(x)) approach infinity, so I think it's $ 1 - \infty$ which is $ - \infty$?

[quote="kira of morocco"]
can you help me to sove this limit

$ \lim_{x\rightarrow\infty} \frac {x}{\ln(\ln(x))} \minus{} \ln(x)$

[/quote]
(Unless someone can do this without using derivatives (I mean a
non-calculus way), instead of how I did it, this problem should
be placed into a higher level mathematics section.)

Let $ ln(x) \equal{} y$.
Then $ x \equal{} e^y$.
And $ ln(ln(x)) \equal{} ln(y)$.

As $ {x\rightarrow\infty}, {y\rightarrow\infty}$

$ \lim_{x\rightarrow\infty} [\frac {x}{\ln(\ln(x))} \minus{} \ln(x)] \equal{}$

$ \lim_{y\rightarrow\infty} [\frac {e^y}{\ln(y)} \minus{} y] \equal{}$

Join the quantities into one fraction:

$ \lim_{y\rightarrow\infty} \frac {e^y \minus{} yln(y)}{\ln(y)} \equal{}$

Because of the form, I am going to use L'Hopital's Theorem, now,
and in some following steps:

$ \lim_{y\rightarrow\infty} \frac {e^y \minus{} 1\cdot ln(y) \minus{} y\cdot \frac1y}{\frac1y} \equal{}$

$ \lim_{y\rightarrow\infty} \frac {e^y \minus{} ln(y) \minus{} 1}{\frac1y} \equal{}$

 $ \lim_{y\rightarrow\infty} \frac {e^y \minus{} \frac1y}{\frac{\minus{}1}{y^2}} \equal{}$

$ \lim_{y\rightarrow\infty} \frac {e^y \plus{} \frac{1}{y^2}}{\frac{2}{y^3}} \equal{}$

Multiply the both the numerator and the denominator by $ y^3$:

$ \lim_{y\rightarrow\infty} \frac {y^3e^y \plus{} y}{2} \equal{}$

$ \plus{}\infty$


hi everyone..

i'm trying to understand congruences from David Burton...the following is an extract....can any on tell me ho did we get "60" at the end from the array of numbers separated by "."

$ 5^{64\plus{}32\plus{}8\plus{}4\plus{}2}
\equal{} 5^{64} . 5^{32} . 5^8 . 5^4 . 5^2
\equal{} 105.74. 114. 101 .25 \equal{} 60 (mod\; 131)$

please also suggest a better resourse to study congruences..
i'm facing great difficulty..

hello, what is the original problem?
Sonnhard.

sir i'm not able to post it online ...i'll do it at ur mail id....

plz check it

hello, i have a short solution
$ 5^{10}\equiv 99 \mod 131$, so $ (5^{10})^2 \equiv 107 \mod 131$ and $ 5^{100} \equiv 80 \mod 131$,
from here we get $ 5^{110} \equiv 99\cdot80\equiv60 \mod 131$.
Sonnhard.

that is ...

i multiply 99 and 80
divide it by 131 and get the remainder...??

is that what is done in the book with so many numbers....?

also plz suggest me some good book from where i can study congruences all by myself...infact numer theory...it'll be of great help rather than posting silly questions which are and will be many...and then waiting for answers that never turn up and even if they do they do not explain everything to me...

plz answer this as well   http://www.mathlinks.ro/viewtopic.php?t=235714

hello, see here, i hope this will help you:
http://mathworld.wolfram.com/Congruence.html
Sonnhard.


Two circles $ C_{1}$ and $ C_{2}$ touch externally at $ A$. $ BC$ is a common tangent, touching $ C_{1}$ at $ B$ and $ C_{2}$ at $ C$. Prove that the common tangent at $ A$ bisects $ BC$.

The common tangent at $ A$ is the radical axis of the two circles, hence the result.

Best regards,
sunken rock

Thanks, sunken rock.

I can see that the common tangent at $ A$ is the locus of points from which tangents to the two circles have the same length; I cannot see how this common tangent bisects $ BC$.

If X is the intersection of BC and the common tangent at A then the tangents from X to one of the circles must be the same and the same for the tangents from X to the other circle, thus XA=XC and XB=XC, so XA=XB.

Of course. I'm not having a good day. Thanks, both of you.


By generalized mean inequality $ {(\frac {sin^{14}{x} + cos^{14}{x}}{2})}^{\frac {1}{14}}\geq {(\frac {sin^{2}{x} + cos^{2}{x}}{2})}^{\frac {1}{2}}$ holds (with the equality iff $ sin^{14}{x} = cos^{14}{x}$). Note that we don't have to worry about $ sin^{14}{x}$ being positive - it's a square so it is nonnegative and $ sin^{14}{x} = 0$ is obviously not a solution.
Because $ sin^2{x} + cos^2{x} = 1$, we have $ {(\frac {sin^{14}{x} + cos^{14}{x}}{2})}^{\frac {1}{14}}\geq \frac {\sqrt {2}}{2}$, which is equivalent to ${ sin^{14}{x} + cos^{14}{x}}\geq \frac {1}{64}$. Since the left side must equal $ \frac {1}{64}$, we can see this is exactly what we're looking for: equality only stands for $ sin^{14}{x} = cos^{14}{x}$. Now, obviously, $ sin^{2}{x} = 1 - sin^{2}{x}$, hence we have two solutions: $ sin{x} = \frac {\sqrt {2}}{2}$ and $ sin{x} = - \frac {\sqrt {2}}{2}$. We unite these solutions into this: $ x = \frac {\pi + 2k\pi}{4}$ ($ k$ is an integer).
We check and the solution is correct. That's it.


is there some equivalent forms for the Riemann hypothesis, that are simpler ?


function
algebra
functional equation
geometry
3D geometry

[quote="Silverfalcon"]Let $f(x)$ be a function with the two properties:

(a) for any two real numbers $x$ and $y$, $f(x+y) = x + f(y)$, and
(b) $f(0) = 2$

What is the value of $f(1998)$?

A. 0
B. 2
C. 1996
D. 1998
E. 2000[/quote]
[hide]
E. 2000

$f(1998)=f(1998+0)$
$f(1998+0)=1998+f(0)$
$1998+f(0)=2000$[/hide]

[hide="Just a remark"] Yeah, I think this "functional equation" is rather well-known: $f(x+y)=x+f(y)$. Everyone that saw it before, knows that, first we set $y=0$ and we get $f(x)=x+f(0)$, which is a function, and the "problem" is solved. Nothing important, though, just a little remark. [/hide]

[hide="Answer"]$f(0)=-1998+f(1998)\Rightarrow f(1998)=2+1998=2000\Rightarrow \boxed{E}$[/hide]

That was extremely easy for a #17. :huh:

[b]EDIT:[/b] LOL, my way was actually more complicated than the others. :P

[quote=Silverfalcon]Let $ f(x)$ be a function with the two properties:

[list=a]
[*] for any two real numbers $ x$ and $ y$, $ f(x \plus{} y) \equal{} x \plus{} f(y)$, and
[*] $ f(0) \equal{} 2$
[/list]
What is the value of $ f(1998)$?

$ \textbf{(A)}\ 0\qquad
\textbf{(B)}\ 2\qquad
\textbf{(C)}\ 1996\qquad
\textbf{(D)}\ 1998\qquad
\textbf{(E)}\ 2000$[/quote]

[hide="My solution"]
let $P(x,y)$ be the assersion, $$f(x+y)=x+f(y).$$ Then, $P(x,0)$ yields, $f(x)=x+f(0)=x+2.$ for all real numbers $x.$
 Therefore, $f(1998)=1998+2=2000.$ Which corresponds to answer choice $\textbf{(E)}$.
[/hide]


Let $ a,b,c\in \mathbb{R}^\plus{}$ such that $ abc\geq 1$.

Prove that

$ \frac{1}{a\plus{}b^4\plus{}c^4}\plus{}\frac{1}{b\plus{}a^4\plus{}c^4}\plus{}\frac{1}{c\plus{}a^4\plus{}b^4}\leq 1$

[quote="outback"]Let $ a,b,c\in \mathbb{R}^ \plus{}$ such that $ abc\geq 1$.

Prove that

$ \frac {1}{a \plus{} b^4 \plus{} c^4} \plus{} \frac {1}{b \plus{} a^4 \plus{} c^4} \plus{} \frac {1}{c \plus{} a^4 \plus{} b^4}\leq 1$[/quote]
$ \sum_{cyc}\frac {1}{a \plus{} b^4 \plus{} c^4}\leq\sum_{cyc}\frac{a^3\plus{}2}{(a^2\plus{}b^2\plus{}c^2)^2}\leq1.$ :wink:

\[ \begin{array}{l}
 (1){\rm{ }}Let{\rm{  }}x,y,z > 0{\rm{, }}xyz \ge 1,{\rm{ }} \minus{} 2 \le p \le 1 \Rightarrow \frac{{x^p }}{{x^p  \plus{} y \plus{} z}} \plus{} \frac{{y^p }}{{x \plus{} y^p  \plus{} z}} \plus{} \frac{{z^p }}{{x \plus{} y \plus{} z^p }} \le 1 \\
  \\
 (2){\rm{ }}Let{\rm{  }}x,y,z > 0{\rm{, }}xyz \ge 1,{\rm{ }}p \le  \minus{} 2{\rm{  }}or{\rm{  }}p \ge 1 \Rightarrow \frac{{x^p }}{{x^p  \plus{} y \plus{} z}} \plus{} \frac{{y^p }}{{x \plus{} y^p  \plus{} z}} \plus{} \frac{{z^p }}{{x \plus{} y \plus{} z^p }} \ge 1 \\
 \end{array}
\]


[hide]add all the equations up, divide by 6, then subtract this from each equation to get the values of x_1 to x_6[/hide]

I'm not sure but I think albertfermat1 posted same thing twice.


[quote="thanhnam2902"]Prove that:   $ \frac {20}{60} < sin20^0 < \frac {21}{60}$[/quote]
[hide="hint"]Use $ sin60^0 \equal{} \minus{} 4sin^320^0 \plus{} 3sin20^0 \equal{} \frac{\sqrt3}{2}$,then let $ sin20^0\equal{}sinx$
The last I think you can,my idol :) [/hide][/hide]

[quote="Allnames"][quote="thanhnam2902"]Prove that:   $ \frac {20}{60} < sin20^0 < \frac {21}{60}$[/quote]
[hide="hint"]Use $ sin60^0 \equal{} \minus{} 4sin^320^0 \plus{} 3sin20^0 \equal{} {\sqrt3}{2}$,then let sin20^0=sinx
The last I think you can,my idol :) [/hide][/hide][/quote]
Yes I see. I can done it by use your way.