inequalities
trigonometry
geometry
absolute value

im having trouble with these absolute value inequalities, i cant remember what the problems were exactly but if i could get the general idea of how to solve problems like this that would be great

1.  |x+2| :le: |x-5|

2. |n| < |n-3| < 9

3. |x-3| + |2x+5| > 6

I'm not sure whether this will help but I was looking at my trig book and found this:

If |a|<  x then it's -a < x < a
If |a| > x then it's a > x or a < -x

[quote="TripleM"]All you need to do is split each one up into cases. For each part in absolute value, think "what values of x would make this positive, which values would make it negative". Then work on each case separately.[/quote]

how would you do this? could you give me an example?

Let's solve:
|5 - x| < 9.

The absolute value only changes sign when x = 5, so we need consider two cases,

Case 1: x < 5

|5 - x| = 5 - x
|5 - x| < 9 [b]-->[/b] 5 - x < 9 [b]-->[/b] x > -4

Case 2: x :ge: 5
|5 - x| = x - 5
|5 - x| < 9 [b]-->[/b] x - 5 < 9 [b]-->[/b] x < 14.

Thus, we get from the first that -4 < x < 5 works and from the second that 5 :le: x < 14 works.  Thus, all possibilities are -4 < x < 14.

Make sense?

That makes sense to me (after i read a few times). I thank anyone who posts some good algebra because I have forgotten a lot of the more complex Algebra I learned in Algebra I, and am about to take Algebra II, so I really need to remember the forgotten elements. (I just finished geometry this year, which was 8th grade.)

One way to solve inequality problems like your first two is to square both sides. This works because if |x| :le: |y|, then x2 :le: y2. (But be careful!! This only works when both sides are positive).

So for the first one you would have:

1. |x+2| :le: |x-5|
(x+2)2 :le: (x-5)2
x2 + 4x + 4 :le: x2 - 10x + 25
14x :le: 21
x :le: 3/2

The upside to this method is that you don't have to consider cases. The downside is that - well, you have to square things and do some algebra.

For the second one, this is method little more complicated because you have two inequalities to deal with simultaneasly - but it still works.

2. |n| < | n-3| < 9
n2 < (n-3)2 < 92
n2 < n2 - 6n + 9 < 81
So 6n < 9 and n2 - 6n -72 < 0
Which means n < 3/2 and -6 < n < 12.

Together these imply -6 < n < 3/2.

For the third one, I think you do have to consider the cases when x :le: -5/2, -5/2 :le: x :le: 3, and x :ge: 3, and see what the resulting sum of those absolute values will be. Then solve in each case.

3. For x :le: -5/2,
-(2x+5) - (x-3) > 6
-3x - 2 > 6
x < -8/3
So, x < -8/3

For -5/2 :le: x :le: 3,
(2x+5) - (x-3) > 6
x + 8 > 6
x > -2
So, -2 < x :le: 3

For x :ge: 3,
(2x+5) + (x-3) > 6
3x + 2 > 6
x > 4/3
So, x :ge: 3

Therefore putting these all together, we get x < -8/3 or x > -2.

Another good way to solve inequality problems is to graph them. Looking at the graphs you can usually see where the solutions should be. This can at the least makes the case work easier. This works well on all three of these problems. Try it out.

Here's how I do these type of problems.

(1) Split it up into cases.  When you have |x|=5, x can be -5 or 5, logically, right?  So when you split it up, you just take x=5 or -x=5, and solve those individually.
(2) graph the solutions on a number line.  If you get x>5 you shade the part where x is greater than 5 on the number line, and sometimes the two solutions will tell you it is between two values.

I have never seen Gauss's method used before but it looks really good!


Bleh.

1. Five workers paint four houses in six days. Working at the same rate as these workers, how many workers are needed to paint 12 houses in three days?

2. Bob can dig a 3 x 4 x 5 hole in one day. His partner, John, can dig a 6 x 4 x 2 hole in 2 days. How many days will it take them, if they work together on one hole, to dig a hole that is 10 x 10 x 10.

(#2 was a made-up problem. #1 was from Math Counts 2009 Chapter Count Down.)

it's just painting three times as many houses in 1/2 the time so its 6 times as many workers or 30.

For #2, bob can dig a total of 60 units in 1 day, and john is 48/2, which is 24. so that's 84 total. 1000/84 is about 12 days?

mathboy is correct, we see that Bob can dig 60 units in one day, and John can dig 24 units in one day. Thus working together, they can dig 84 units in one day. $ 1000/84\approx12$

Please post a solution...not just answers  :wink:  :huh:

I believe the previous posts qualified as solutions, but I'll just post them in one post :)

[hide="1)"]
They must build three times as many houses in half the time, giving $ 5\cdot3\cdot2=\boxed{30}$ needed workers.

I agree that I have strange logic, but that's just how I approach them.[/hide]

[hide="2)"]
Bob can dig a volume of $ 3\times4\times5=60$ while John can dig a volume of $ \frac{6\times4\times2}{2}=24$ in one day. Working together, they can dig a volume of $ 24+60=84$ units. Using \[ (\text{work force})\times(\text{time})=\text{amount done},\] we get \begin{align*}1000\text{ units}&=\frac{84\text{ units}}{\text{day}}\times t\\t=\frac{1000\text{ units}}{\frac{84\text{ units}}{\text{day}}}&=\boxed{\frac{250}{21}\text{ days}}.\end{align*}[/hide]


geometry
perpendicular bisector
geometry unsolved

find the center of a fixed circle,with a fixed circle that nyou have (you can move this fixed circle),if you can.and if you can't prove that you can't. :D

What do you mean the second circle is fixed? Since we can move it, it's not fixed!
I guess that it has a constant radius.

By the way, doesn't matter if the second circle has a constant radius. The only thing we need is the two circles intersect at two different points $A,B$.

While the second circle is moving, the perpendicular bisector of $AB$ always pass through the centers of the circles, so we find the center of the unknown circle

If I understand well the problem......

i mean you have a circle with fixed radius in your hand and you can put it in the plane and use it,like ruler and compas.thanks.

Ok.

So, is this a solution for the problem?


How many subsets of the set $ \{1,2,3,4,...,30\}$ have the property that the sum of the elements of the subset is greater than 232?

(It looks like it shouldn't be all that hard, but I can't really think of a good solution besides semi-brute force.)

[hide="Hint"]
The lowest sum attainable is $ 0$, the highest is $ \frac {30 \cdot 31}{2} \equal{} 465$. Their average is $ 232.5$.

Given any subset with sum less than $ 232.5$, its complement has a sum greater than $ 232.5$. Now use the symmetry.
[/hide]

[hide="Answer"]
$ 2^{29}$
[/hide]

[hide="Solution"]The lowest possible value is 0. The highest possible is $ 1\plus{}2\plus{}\dots \plus{} 30\equal{}465$. The average is $ 232.5$. By symmetry, we have $ 2^{30}/2$ for the answer which is $ \boxed{2^{29}}$.[/hide]

*facepalm* Duh...
Sorry about the dumb question.  :roll:


Find all functions $ f: \mathbb{R}\to\mathbb{R}$ such that
a)$ f(x)\geq e^{2004x}$ for all $x\in\mathbb{R}$, and
b)$ f(x+y)\geq f(x)f(y)$ for all $x,y\in\mathbb{R}$.

[quote="N.T.TUAN"]Find all functions $ f: \mathbb{R}\to\mathbb{R}$ such that
a)$ f(x)\geq e^{2004x}\forall x\in\mathbb{R}$ and
b)$ f(x+y)\geq f(x)+f(y)\forall x,y\in\mathbb{R}$.[/quote]

From a), we get $ f(0)\geq 1$
From b) we get $ f(x+0)\geq f(x)+f(0)$ and so $ f(0)\leq 0$

So we have a contradiction and no such function exists.

I am wrong! I edited it , please continuou!  :wink:

[quote="N.T.TUAN"]Find all functions $ f: \mathbb{R}\to\mathbb{R}$ such that
a)$ f(x)\geq e^{2004x}\forall x\in\mathbb{R}$ and .[/quote]

I think $ f(x)\geq 1+2004x$. :maybe:


Prove that:

sh(ax)-(ax)ch(a)<=ach(a)-sh(a); a>0 and x from  [-1,1].
When does the equality hold?

cheers!

let f(x)=sh(ax)-(ax)ch(a)

f'(x)=a*ch(ax)-a*ch(a)=a*( ch(ax)-ch(a) )

since a>0 and x belongs to [-1,1], f'(x) < 0.

Then f is a decreasing function and f(-1)=a*ch(a)-sh(a)
that's all ;-)
(hope it's coreect)


Let $ n$ is positive integer such that $ n \ge 2$, prove that:

$ \left( {\frac{{n^2  \minus{} 1}}{{n^2 }}} \right)^{2n} .\left( {\frac{{n \plus{} 1}}{{n \minus{} 1}}} \right)^2  < 1$

[quote="Russian_Man"]If $ n \equal{} 2$ we have
$ (3/4)^{2*2}*(3)^2 < 1$
$ 27/16 < 1$ It's not true...... :huh:[/quote]

Sr, but I cannot edit this problem! :(

The true question is:

Prove that: $ \left( {\frac{{n^2 \minus{} 1}}{{n^2 }}} \right)^{2n} .\left( {\frac{{n \plus{} 1}}{{n \minus{} 1}}} \right) < 1$

OK! :(


Do you have a solution? If so, could you post it? It is a nice problem! ;)

[quote="Ezbakhe Yassin"][quote="arqady"]I found a very nice solution. I want you have fun to find it. :)[/quote]

I do not think I can have more fun with it since I have tried to solve it for more than two hours without success. :(  :(

Do you have a hint?[/quote]

I think you should have perseverance in solving problems. Two-hour trial is obviously very short.

[quote="Soarer"]I think you should have perseverance in solving problems. Two-hour trial is obviously very short.[/quote]

I think you are right. I will continue with it. :)

[quote="arqady"]$f(x)=1+asinx+bcosx+csin2x+dcos2x$
Known that $\forall x f(x)>0.$ Prove that $\forall x f(x)<3.$[/quote]
I find $b+d=-1$
is it useful?

Maybe I am wrong.
but I think this problem is impossible. :(
let $x=\pi$ in $xf(x) >0$
we get  $1-b+d>0$
and
let $x=-\pi$ in $xf(x) >0$
we  get $1-b+d<0$
it is impossible :)

or do you mean for $\forall x$  then $f(x)>0$?

[quote="zhaobin"]
or do you mean for $\forall x$  then $f(x)>0$?[/quote]
Obvious!
$f(x)=1+a\cdot sinx+b\cdot cosx+c\cdot sin2x+d\cdot cos2x.$
Prove that if for all $x\in\mathbb R$ exist $f(x)>0,$
 then for all $x\in\mathbb R$ exist $f(x)<3.$

arqady,Can you post the solution now?
Thanks,I can't solved it  :(

[quote="arqady"]$f(x-\frac{2\pi}{3})+f(x)+f(x+\frac{2\pi}{3})=3.$   :lol:[/quote]
Nice :lol:


[quote="ashwinkrishnan"]Find the smallest 5 digit number of the form AABCD such that it leaves a remainder of 67 when divided by 8, a remainder of 8 when divided by 9 , a remainder of 9 when divided by 10 and a remainder of 10 when divided by 11.[/quote]

If you wanted to say "remainder of $7$ (instead of $67$) when divided by $8$", then here is the solution:

$AABCD+1$ is divisible by $8$, $9$, $10$ and $11$ so it is of the form $2^{3}.3^{2}.5.11n=3960n$. After checking some $n$ we get that then $n=3$, $AABCD+1=11880$. So $AABCD=11879$


email
USAMTS
geometry
geometric transformation
reflection

I faxed my solution but I'm not sure if it as arrived yet. What would happen if for some reasons, solutions do not reach the USAMTAS headquarter, either through mailing, emailing or faxing? Would the student get a chance to clarify this?

Within a few days of sending your solutions, you should receive an email verifying that it has been received.  By the end of the week, the My USAMTS Scores page will also reflect submissions received.


hi to all
sory because i cannot write english in good way
===
i am a mechanical engeneering student and i  have a project
i want some from u if  u dont minde
my project is measuring sphere volume and the effect of raduis
==
if eny body can help
i need a good introduction
if u know eny website that can help me
plz give it to me
sooory about evrey thing
seee u later

You know, I am not sure I understand your problem. Any way, if your english is not quite good, you can either see if there is a regional comunity inside the forum or if post here the language(s) that you do speak (or at least write and read) fluently, and I am sure there should be someone here who can help you in your own language. English is also an option, but it could be a little slower.

Best,

thanxxxx my brother djimenez
i am studing in the universtiy and my major is mechanical engenereeing
i have a project and my project  is
measurment of sphere volume and th effect of raduis.
i need any help
the project consest from introduction,theory,objective,planning etc....

since $V = \frac{4}{3}\pi r^3$, volume varies cubically in respect to the radius

[quote="Treething"]Nope, it's something called chatspeak.

Chatspeakers are basically the shyt of the Internet, and they breed four times as fast as normal people. (A cookie to you if you know where I got this quote from.)[/quote]

This seems quite off topic to me...[img]http://img115.exs.cx/img115/5941/offtopic8ys.gif[/img]

I think he was saying that you shouldn't answer his questions becuase he writes in chatspeak.

[quote="djimenez"]C'mon, the same guy says that his english is not good enough, let's try to have patience with him. Remember that less than 5% of the world population has english as native language (well, at least if we exclude India, but I don't know how is exactly the language distribution there)

Just don't hesitate. And Basim, what is your mother language? If it is spahish, I can help directly on it, otherwise, someone else would!

Best,[/quote]

I agree. Lots of people would be willing to help with language and that kind of stuff on the forums. It is an international forum, and there are people from all across the world. One place you could get some help with english or just to talk to others who speak your first language is the international forums.

Everybody speaks about the international forums, however I see no-one gives a link (since before this is an international forum, it is first a HUGE forum :) :P). Take a look at my signature and choose your language after the flag ;)

Sorry about that  :blush:  but I was just interpreting what Treething said.

[quote="Valentin Vornicu"] Take a look at my signature and choose your language after the flag ;)[/quote]

The Spanish Communities are not there!! :P :D . I will try to add it to my signature.

Best,

<_<
>_>
<_<
>_>
<_<
>_>
<_<
>_>
shifty eyes

People who aren't good at English have deformed sentence structure, not chatspeak.

Oh well.

[quote="Treething"]People who aren't good at English have deformed sentence structure, not chatspeak.Oh well.[/quote]

You are not taking in consideration the posibility that many people never received a formal instruction in English, and therefore their main practice of english is ... do you guess??? Yes, CHATS, and therefore, it is not surprising that you find a lot of people who thinks that chatspeak is cool and a usual (and widely acceptable) informal way of written English, even when some people (and I include myself) find it annoying and disgusting.

Best,

[quote="mcalderbank"]what exactly constitutes chatspeak?[/quote]
Shortening words, eliminating vowels, making the text very hard to read for most people that don't use chatspeak. Something we deeply despise in Romania, and it is banned on all(most) Romania forums. :)

[quote="djimenez"][quote="mcalderbank"]what exactly constitutes chatspeak?[/quote]

I suppose that they refer to the substitutions of [b]you[/b] by [b]u[/b], [b]i[/b] instead of the grammatically correct [b]I[/b] for the personal pronoun, [b]plz[/b] instad of [b]please[/b] and that kind of stuff.

Best,[/quote]
lol u purdy mch hit teh na1l 0/\/ teh hedd!!!!!!!!!!!!!!!!
^^^like that.
It gets really annoying when I have to read through paragraphs of that.


algebra
polynomial
search
integration
algebra unsolved

Determine the smallest closed subinterval [a, b] of [0,1] such that if p is a polynomial of degree at most 6 that increase on [a, b] then necessary p(1)>=p(0).

[i]AMM 6449 by Kent D.Boklan[/i]

I need some clarification.

We search for particular interval $[a,b]$ or for the smallest $d$ s.t. if $P(x)$ increses on $[a,a+d]$ then $P(1)>P(0)$?

$[a,b]=[0,1]$.
Let $p'(x)=y^{k}z^{m}, y=\frac{4(b-x)(x-a)}{(b-a)^{2}}, z=\frac{2x-a-b}{b-a},$ were k is big odd number, m is big even number. Then \[p(1)-p(0)=\int_{0}^{1}y^{k}z^{m}dx<0\] for big k and m if $[a,b]\not [0,1].$

[quote="Myth"]Firstly, $\deg p\leq 6$ by the condition.

Secondly, your $p'(x)$ is not positive on the whole $[a,b]$.[/quote]
I say, that m is even. But I forget about degree.


There are $ 2005$ points in a plane s.t. no $ 3$ of them are collinear.Prove that there exists a circle in the plane which contains
inside $ 1002$ points and leaves outside $ 1002$ points.

See, for example, here: [url]http://www.cut-the-knot.org/Generalization/scircles.shtml[/url].


Find all functions $ f : R$→ $ R$ which satisfy
$ f($−$ x) \equal{} f(x)$ and $ f(x \plus{} 1) \equal{} f(x) \plus{} 1$ for all $ x$, and
$ f(1/x) \equal{} f(x)/x^{2}$ for all non zero $ x$

My source states the first equation as
$ f(x)\equal{}f(\minus{}x)$
but it is probably wrong as $ f(x)\equal{}\minus{}f(\minus{}x)$ seems to make the problem a little challenging.

sorry about the mistake

[hide="do not click this"] :harhar: ha ha ha I told you not to click this  :rotfl: [/hide]


trigonometry
inequalities
geometry proposed
geometry

(a) If $\displaystyle ABC$ is a triangle and $\displaystyle M$ is a point from its plane, then prove that
\[ \displaystyle AM \sin A \leq BM \sin B + CM \sin C . \]

(b) Let $\displaystyle A_1,B_1,C_1$ be points on the sides $\displaystyle (BC),(CA),(AB)$ of the triangle $\displaystyle ABC$, such that the angles of $\triangle A_1 B_1 C_1$ are $\widehat{A_1} = \alpha, \widehat{B_1} = \beta, \widehat{C_1} = \gamma$. Prove that
\[ \displaystyle \sum A A_1 \sin \alpha \leq \sum BC \sin \alpha . \]

[i]Dan Ştefan Marinescu, Viorel Cornea[/i]

I think Ptolemy works for a and that we can get b from a :) Don't have time for the details ;)


geometry
circumcircle
Gauss
number theory
greatest common divisor
projective geometry
radical axis

Let $ ABCD$ be a concyclic quadrilateral inscribed in $ (O)$. Let $ E$ be the intersection of $ AB$ and $ CD$. Let $ M$ be the second intersection of circumcircles of $ \triangle EAD$ and $ \triangle EBC$. Let $ P_b$, $ P_c$ respectively be the poles of $ CD$ and $ AB$ wrt $ (O)$. Prove that $ \angle OMP_c \equal{} \angle OMP_b$.

It is well-known that EM,AD,BC,PcPb are concurrent at one point N
Therefore we have to show NPcEPb are harmonic division
It is also well-known that E is the polar of PcPb  and  AC,BD,PcPb,XY are concurrent at one point E
thus if we can prove PcD,XE,PbA are concurrent the problem is done

yet, notice in ⊿AEPc,⊿CEPb we use sin law  , we have PcA/CPb=PcE/PbE
in ⊿XPcPb  we use ceva theory we obtain PcD,XE,PbA are concurrent
thus ,according to analysis cited above we complete the proof

[hide]dear [b]mathVNpro[/b],thanks for your nice problem[/hide]

$ (P_cP_bEN)\equal{}\minus{}1\Leftrightarrow C(P_cP_bEN)\equal{}\minus{}1\Leftrightarrow C(P_cP_bAB)\equal{}\minus{}1$
It's true from harmonic quadrilateral.

Dear [b]plane geometry[/b], thank you very much. Here is my solution:

Let $ F\equiv AD\cap BC$ and $ G\equiv AC\cap BD$. We have $ P_cP_b$ is the polar of $ E$ wrt $ (O)$ and $ FG$ is the polar of $ E$ wrt $ (O)$ $ \Longrightarrow$ $ F,P_c,G,P_b$ are collinear. It is well- known that $ G$ is the orthocenter of $ \triangle OEF$. Then we let $ P$ be the midpoint of $ FG$, it is followed then $ PM$ is a tangent from $ P$ to the circle with diameter $ OE$. Now, consider the inversion through pole $ O$, power $ k \equal{} R^2$, we have $ \mathcal {I}(O,k): P_c\mapsto M$, $ P_b\mapsto N$. Then $ M,N$ respetively are the midpoints of $ AB$ and $ CD$. By $ \text {Gauss theorem}$ for the complete quadrilateral $ (DA,DB,CA,CB)$, it is followed that $ P,M,N$ are collinear. Thus $ PM^2 \equal{} \overline {PM}.\overline {PN}$. But $ P_c,M,N,P_b$ are concyclic $ \Longrightarrow$ $ \overline {PP_c}.\overline {PP_b}$ $ \equal{} \overline {PM}.\overline {PN}$. Hence $ \overline {PP_b}.\overline {PP_c} \equal{} PM^2$. But it is also well- known that $ PM \equal{} PF \equal{} PG$. Hence $ \overline {PP_b}.\overline {PP_c} \equal{} PF^2 \equal{} PG^2$, which implies $ (FGP_cP_b) \equal{} \minus{} 1$ $ \Longrightarrow$ $ M(FGP_cP_b) \equal{} \minus{} 1$. But $ OG\equiv GM$ $ \perp$ $ MF$, which implies $ MG$ is the internal bisector of $ \angle P_cMP_b$. The result is lead as followed. $ \square$

$ \bullet$ [b]REMARK[/b]

$ 1)$ From the proof above, we find out that $ PM$ is also a tangent from $ P$ to $ (MP_cP_b)$. Therefore, $ (MP_cP_b)$ and $ (OMN)$ internally tangent to each other at $ M$. Hence, under the inversion through pole $ O$, power $ k \equal{} R^2$, we have $ (MP_cP_b)\mapsto (GMN)$, $ (OMN)\mapsto P_cP_b\equiv FG$; then $ \boxed {\text {FG is a tangents from F to (GMN)}}$.

$ 2)$ Now, consider the inversion through pole $ G$, power $ r \equal{} \overline {GA}.\overline {GC} \equal{} \overline {GD}.\overline {GB}$, we have $ \mathcal {I}(G,r): GF\mapsto GF$, $ AB\mapsto (GCD)$, $ CD\mapsto (GAB)$, hence if $ N\mapsto G_b$ and $ M\mapsto G_b$. It is followed then $ G_c\in (GAB)$ and $ G_b\in (GCD)$; further, $ (GMN)\mapsto G_bG_c$. But from [b]remark $ 1$[/b], we have found out that $ FG$ tangents to $ (GMN)$. Then under the inversion $ \mathcal {I}(G,r)$, it is followed immediately that $ \boxed {G_bG_c\parallel{} GF}$.

$ 3)$ The [b]Remark $ 2$[/b] can be proved without using all the following facts above. Try one then!

[hide]p/s: This is my $ 333^{th}$ post on mathlinks (such a pretty number), which also my last post on http://www.mathlinks.ro site!  :
So, I wish a good luck for every mathlinkers. Goodbye![/hide]

The result $ MO,AC,BD$ are concurrent can be proved by using the concept of radical center of $ (MAOC), (MDOB)$ and $ (O)$.
Here is another result about this beauty picture:
Let $ (MBD)$ cut $ EC$ at $ X, (MAC)$ cut $ EC$ at $ Y, AC\cap BD \equal{} \{L\}, S$ is the midpoint of $ DC$. Prove that $ AX, BY, SL$ are concurrent.
[hide]@[b]MathvnPro[/b]: Good luck to you, my friend!  :) [/hide]


geometry
trigonometry
circumcircle
parallelogram
projective geometry
geometry proposed

A triangle $ ABC$  is inscribed in a circle and $ BD,CE$ are two altitudes. The tangents to the circle at $ B,C$ intersect each other at $ M$. Prove that $ AM$ bisects $ DE$.

 [i][color=blue]I have made a short proof to this problem , but not .. so elementary(...antiparellel and symmedian). Do you have any simpler solution?[/color][/i]

 Babis

Very easy:(I guess that it is your solution too)
Denote $ F\equal{}\omega\cap AM$,where $ \omega$ is circumscribed of $ \triangle ABC$.Then $ ABFC$ is harmonic quadrilateral,so it is well-known that $ AF$ is symmedian of $ \triangle ABC$,then it is a median in $ ADE$.
[hide="in russian"]При поворте при котором $ B\rightarrow C$,$ AF$ станет медианой,но при этом $ B'C'$ станет параллельна $ DE$[/hide]

I'm not sure,but i think that you mean that $ DE$ and $ BC$ antiparallel,i ask about it,because i don't know what does antiparalle means(but i guess that it is opposite sides of cyclic quadrilateral).Thank you.And here is an image for your nice problem:
[img]http://img221.imageshack.us/img221/5465/stergiuzf8.png[/img]

[quote="Virgil Nicula"][color=darkblue]Let $ ABC$ be a triangle. Let $ M\in AB$, $ N\in AC$ ($ AB$, $ AC$ are sidelines !) be two points so that $ \widehat {AMN}\equiv\widehat {ACB}$, i.e. $ \triangle\ AMN\sim \triangle\ ACB\ .$ In this case we say that [b]the line $ MN$ is an antiparallel line to the sideline $ BC\ .$[/b] A same line has many and nice properties which are used often.[/color][/quote]
Thank you very much :) !

[quote="stergiu"][color=darkred]Let $ ABC$ be a triangle and denote $ \{\begin{array}{c} D\in AC\ ,\ BD\perp AC \\
\ E\in AB\ ,\ CE\perp AB\end{array}$. The tangents $ BB$ and $ CC$ to the circumcircle $ w$ of $ ABC$ meet in the point $ M$. Prove that $ AM$ bisects $ DE\ .$[/color][/quote]
[color=darkblue]This problem is evidently for who knows the properties of the antiparallels and symmedians.

It is a good problem for the local stage of M.O. (juniors). Here is an elementary proof.

Denote $ \|\begin{array}{c} R\in AM\cap DE \\
\ \{A,F\} = AM\cap w\end{array}\|$. Observe that $ MB = MC$. Therefore,

$ \|\begin{array}{ccc} \frac {RE}{RD} = \frac {AE}{AD}\cdot\frac {\sin\widehat {RAE}}{\sin\widehat {RAD}} & \implies & \frac {RE}{RD} = \frac {AE}{AD}\cdot \frac {BF}{CF} \\
 \\
\triangle\ ADE\sim\triangle\ ABC & \implies & \frac {AE}{AD} = \frac {AC}{AB} \\
 \\
\triangle\ MBF\sim\triangle\ MAB & \implies & \frac {BF}{AB} = \frac {MF}{MB} \\
 \\
\triangle\ MCF\sim\triangle\ MAC & \implies & \frac {AC}{CF} = \frac {MC}{MF}\end{array}\|$ $ \bigodot$ $ \implies$ $ RE = RD$.[/color]

[quote="Erken"]Very easy:(I guess that it is your solution too)
Denote $ F \equal{} \omega\cap AM$,where $ \omega$ is circumscribed of $ \triangle ABC$.Then $ ABFC$ is harmonic quadrilateral,so it is well-known that $ AF$ is symmedian of $ \triangle ABC$,then it is a median in $ ADE$.
[hide="in russian"]При поворте при котором $ B\rightarrow C$,$ AF$ станет медианой,но при этом $ B'C'$ станет параллельна $ DE$[/hide][/quote]

 Exactly !

$ AM$ is the symmedian from $ A$ and it bisects all the antiparallel segments  - as $ DE$ - to $ BC$.

[quote="stephen38"]I also have a trigonometric solution,it is a loong solution. :wink:
Can someone tell me where i can learn more about things like symmedian,antiparallel,etc?
 :)[/quote]

 Of course in this forum and at '' Collenge Geometry , N.A. Court  , ed Dover 2007'' or '' Johnson , Modenr geometry, Dover 2007'' . Both nice and cheap books, form the best at this level.You can buy them by  http://www.amazon.com

 Of course you can find many thinks about these terms in internet too.

Babis

[color=darkred][b]At least for Stephen38 ...[/b][/color] [hide="An another short proof."] [color=darkblue][b]Lemma[/b] ([u]easily[/u]). Let $ ABC$ be a triangle. Then $ T\in AA\cap BC$ $ \implies$ $ \frac {TB}{TC}\equal{}\left(\frac {AB}{AC}\right)^2\ .$

Indeed, $ \frac {TB}{TC}\equal{}$ $ \frac {AB}{AC}\cdot \frac {\sin\widehat {TAB}}{\sin\widehat {TAC}}\equal{}$ $ \frac {AB}{AC}\cdot\frac {\sin C}{\sin B}\equal{}$ $ \left(\frac {AB}{AC}\right)^2\ .$

[b][u]An another short proof of the our problem.[/u][/b] Denote the second intersection of $ AM$ with the

circumcircle of $ \triangle\ ABC$. I"ll use an well-known property in the triangles $ ABM$ and $ ACM$ :

$ \left\{\begin{array}{c}
\triangle\ ABM\ : \ \frac {MF}{MA}\equal{}\left(\frac {BF}{BA}\right)^2\\\\
\triangle\ ACM\ : \ \frac {MF}{MA}\equal{}\left(\frac {CF}{CA}\right)^2\end{array}\right\|$ $ \implies$ $ \frac {BF}{CF}\equal{}\frac {AB}{AC}\ (*)$ ($ ABFC$ is [b]harmonically[/b]).

Denote $ S\in AM\cap BC$. Thus, $ \frac {SB}{SC}\equal{}\frac {BA\cdot BF}{CA\cdot CF}$. Using $ (*)$ obtain $ \boxed {\frac {SB}{SC}\equal{}\left(\frac {AB}{AC}\right)^2}$,

i.e. the ray $ [AS$ is the $ A$[b]- symmedian[/b] in the triangle $ ABC$ a.s.o.[/color][/hide]

Through the points $ D,$ $ E,$ we draw two lines parallel to $ AB,$ $ AC$ respectively and we denote as $ A',$ their intersection point.

Because of the line segment $ AA',$ bisects the segment $ DE,$ it is enough to prove that the point $ A',$ lies on the line segment $ AM,$ as the $ A \minus{}$symmedian of $ \bigtriangleup ABC,$ as well known.

Let $ D',$ $ E'$ be, the orthogonal projections of $ A',$ on the sidelines $ AC,$ $ AB,$ respectively.

From the similar right triangles $ \bigtriangleup D'A'D,$ $ \bigtriangleup E'A'E$ and parallelogram $ ADA'E,$ we have that $ \frac {A'D'}{A'E'} \equal{} \frac {A'D}{A'E} \equal{} \frac {AE}{AD}$ $ ,(1)$

From the similar right triangles $ \bigtriangleup DAB,$ $ \bigtriangleup EAC,$ we have that $ \frac {AE}{AD} \equal{} \frac {AC}{AB}$ $ ,(2)$

From $ (1),$ $ (2)$ $ \Longrightarrow$ $ \frac {A'D'}{A'E'} \equal{} \frac {AC}{AB}$ $ ,(3)$ and the proof is completed.

Kostas Vittas.

PS. Without any reference to the [b][size=100]Symmedians theory[/size][/b], we can easy to prove that $ \frac {MC'}{MB'} \equal{} \frac {AC}{AB}$ $ ,(4)$

$ ($ because of the similar right triangles $ \bigtriangleup B'BM,$ $ \bigtriangleup DCB$ and $ \bigtriangleup C'CM,$ $ \bigtriangleup EBC,$ where $ B',$ $ C',$ are the orthogonal

projections of $ M,$ on $ AB,$ $ AC$ respectively, we have $ \frac {MC'}{CE} \equal{} \frac {MC}{BC} \equal{} \frac {MB}{BC} \equal{} \frac {MB'}{BD}$ $ \Longrightarrow$ $ \frac {MC'}{MB'} \equal{} \frac {CE}{BD} \equal{} \frac {AC}{AB}$ $ ).$

So, from $ (3),$ $ (4),$ $ \Longrightarrow$ $ \frac {MC'}{MB'} \equal{} \frac {AC}{AB} \equal{} \frac {A'D'}{A'E'}$ and hence, we conclude the collinearity points $ A,$ $ A',$ $ M.$

Let $ S$ be the intersection of $ BC$ with the tangent at $ A$ of the circumcircle $ (O)$ of triangle $ ABC$, and $ T$ be the intersection of $ AM$ with $ BC$. We know that $ AM$ is the $ A$ - symmedian of triangle $ ABC$, so $ \frac {TB}{TC} \equal{} \frac {AB^2}{AC^2}$ $ (1)$

On the other hand, $ \triangle SAB\sim\triangle SCA$, therefore $ \frac {SA}{SC} \equal{} \frac {SB}{SA} \equal{} \frac {AB}{CA}$, which implies that $ \frac {SB}{SC} \equal{} \frac {AB^2}{AC^2}$ $ (2)$

From $ (1)$ and $ (2)$ we conclude that $ \frac {TB}{TC} \equal{} \frac {SB}{SC}$, i.e. the division $ (SBTC)$ forms a harmonic division, so the pencil $ A(SBTC)$ is harmonic, too. $ (*)$ Otherwise, $ OA\perp EF$, it's equivalent to $ EF\parallel AS$ $ (**)$

Combining $ (*)$ and $ (**)$, we get that $ AM$ passes through the middle point of the segment $ EF$ as desired.


$x,$ $y,$ $z$ are positive real numbers such, that $x+y+z=3.$ Prove that:
$\frac1{x+yz}+\frac1{y+zx}+\frac1{z+xy}\leq\frac{9}{2(xy+xz+yz)}.$
I think, this inequality can be strengthened.

Nice inequality!

It is equivalent to
$\frac{3}{2(xy+yz+zx)} \ge \sum_{cycl}\frac{1}{x(x+y+z)+3yz}$
We have
$\frac{1}{2(xy+yz+zx)}-\frac{1}{x(x+y+z)+3yz}=\frac{(x-y)(x-z)}{2(xy+yz+zx)(x(x+y+z)+3yz)}$
It suffices to prove
$\sum_{cycl}\frac{(x-y)(x-z)}{x(x+y+z)+3yz} \ge 0$
Assume $x\ge y\ge z$. Then
$\frac{(z-x)(z-y)}{z(x+y+z)+3xy} \ge 0$
Further,
$\frac{(x-y)(x-z)}{x(x+y+z)+3yz}+\frac{(y-z)(y-x)}{y(x+y+z)+3zx}=\frac{z(2x+2y-z)(x-y)^2}{(x(x+y+z)+3yz)(y(x+y+z)+3zx)} \ge 0$
and the proof is completed.

Yes!  Vornicu-Schur  does not help! :lol:

[quote="pvthuan"]Thanks Arqady. It is also included in my book.

Prove that if $x+y+z=1$, then
\[ \frac1{x+yz}+\frac1{y+zx}+\frac1{z+xy}\leq\frac9{4(xy+yz+zx)}.  \]
All we need is the identity $x+yz=(x+y)(x+z)$, and
\[ (x+y)(y+z)(z+x)\geq (8/9)(x+y+z)(xy+yz+zx)  \]
which is quite well kwown in this forum.[/quote]
No,I think they are different...

[quote="pvthuan"]Humm, they look almost the same. I will check it.[/quote]

but the condition is different. Then they are totally different.


AMC
USA(J)MO
USAMO
ARML
geometry
probability
expected value

I was wondering what scores on the USAMO usually are. Or, to be more specific,

1) What scores (on previous years' exams) put you in the top 100? Top 50? Top 12?

2) Just how strict are the grading policies when grading proofs? What is the expected value of getting each of the possible scores on a proof?

I tried looking this up online, but I couldn't find any scores from previous years.

If you go here:
[url]http://www.unl.edu/amc/e-exams/e8-usamo/archiveusamo.shtml[/url]
And click on the individual scoring, you can find all the scores for each year, 2007 and previous seem to have percentile data and the such.

Thanks guys, exactly what I needed.

@Hamster1800: One more thing. If I can roughly describe the answer to the question, and can outline the proof right now, can I expect to get a 6 or a 7 when I write it up? Assume I can write a proof well enough to get mostly full credit on the ARML power or Mandelbrot team.

@Quickster94: This is great. Thanks a lot.

It really depends on what you mean when you say you can "outline the proof."  Since USAMO is arguably the hardest high school math competition in the US, you can't compare it with others.  Furthermore, I think Hamster1800 means that you will get a 6 or 7 [b]if[/b] and only if your solution is rigorous and correct. Any good proof writer will undoubtedly know what a rigorous solution looks like, so judge your proof's rigor to determine whether you get a 6 or 7.

[quote="Hamster1800"]The score for top 12 is usually 28-30. I don't know about top 50 and top 100, but top 30 is usually around 20-22 and top 30 freshmen (red mop) is usually around 9.

As for how strict the grading is, it is stricter than most people realize. You should not expect to get a 3 or 4 unless the problem is split into two parts of approximately equal difficulty and you solve one of them. If you make a small mistake but fixing it is trivial and doesn't affect the flow of the proof, then you'll get a 6 (or [i]maybe[/i] a 7 if it's a very tiny mistake like a typo in an isolated position which is fixed later). Anything past that you're likely to not get more than 1 point. Getting a 2 or a 5 is possible, but unlikely.[/quote]

What if your answer is correct, but about a quarter of the proof is misguided due to one error that is fixed later on? (If that's even possible...) Would your score still be only 1 or 2, or would you merit a 3?

mop red only needs a 9?
hmmm....i have a chance now (thought it needed a score around 15-20 :oops: )

yrudoy: It depends on if your solution is correct. An incorrect solution which you can write up now will get a 0 or a 1, but a correct solution will get a 6 or a 7. The writeup should not cost any points.

sdkudrgn88: If your proof can be trivially modified to fix the error, you will get a 6 (so if a quarter of the proof is simply unnecessary but it has an error, then you will probably get a 6).

[quote="sdkudrgn88"]What if your answer is correct, but about a quarter of the proof is misguided due to one error that is fixed later on? (If that's even possible...) Would your score still be only 1 or 2, or would you merit a 3?[/quote]Unless the error is easily fixable, you get a 1. Last year, I had the wrong answer for #4 even though the vast majority of my proof was correct; I got a 1. On #1, I was unaware that a certain theorem was well-known and so tried to prove it myself. I didn't succeed, so they gave me a 6 since the theorem was well-known anyway.

Hmm...I know this kid that solved 4 problems on the USAMO last year and got..

a 2.

What's the deal with this
???????

Like, what kind of error is considered "easily correctable?"

I'm guessing that kid's definition of "solved" was slightly different than the graders'...

Sounds like he might have made a few observations of nontrivial value, and proved these observations no more than 2 out of 4 times...

[quote="blackbelt14253"]Hmm...I know this kid that solved 4 problems on the USAMO last year and got a 2.
What's the deal with this
???????
Like, what kind of error is considered "easily correctable?"[/quote]

The deal is that kid obviously didn't know what an easily correctable error was and didn't know how to write rigorous solutions.  "Solved" takes on different meanings depending on whether you're in geometry class at school or taking the USAMO.

[quote="e^i_penguin"]What if your solution was all correct except for 1 part that is not as rigorous as it should be (if the part was small compared to the whole problem)? How many points would that cost you?[/quote]
If it's actually small it should not cost many points, but the problem is many people think their errors are "small" when they are not (ouch that was painful saying that).

[quote="e^i_penguin"]What if your solution was all correct except for 1 part that is not as rigorous as it should be (if the part was small compared to the whole problem)? How many points would that cost you?[/quote]

What do you mean by a 'small part'? In general every part of the solution is essential, there's no way to judge what's a 'big part' and what's a 'small part' (basically the only reasonable way to define what a 'big part' of the solution is would be to say that the hardest parts are the big parts. So pretty much by definition if you're stuck on something then it's probably a big part of the problem). If you're missing part of the solution, then your solution is incomplete, it doesn't matter how big or small you think the gap is.

(Also, on this sort of competition 'not as rigorous as it should be' almost always means the same thing as 'wrong'.)

I would say a solution like that would get [b]at most[/b] at 2 (and much more likely a 0 or a 1).

Every year I see some people posting here who say something along the lines of: "I basically solved this problem. There was one [i]tiny[/i] part of my solution which wasn't [i]completely[/i] rigorous, but I'm still expecting a 6 or a 7." Pretty much without fail these people get 0s or 1s.


number theory
modular arithmetic
number theory unsolved

I am NOT a mathematician, so sorry if this sounds dumb to you. Could somebody tell me what I am dealing with here? I worked this out staring at a license plate in traffic the other day. Take any positive whole number and sum it again and again, reducing it to a single whole number between 1 and 9. For instance:

4318264 : 4+3+1+8+2+6+4 = 28
Then add that sum together so:
28 : 2+8 = 10
Then again:
10 : 1+0 = 1

That's as far as you can go, obviously - the sum is 1. Now try splitting the number differently, at random points, and adding it up again so:

e.g., 4318264 : 431+82 +64 = 577
577 : 57+7 = 64
64 : 6+4 = 10
10: 1+0 = 1

The ultimate sum for 4318264 will always be 1, no matter how you chop it up. Another iteration just to beat the thing to death:

431824 : 43+182+64 = 289
289: 2+89 = 91
91 : 9+1 = 10
10 : 1+0 = 1

Thus the sum for 23184 is always 9, the sum for 184 is always 4, the sum for 88926 is always 6, and so on. This also works no matter the order of the initial digits.

Also: take any two integers (a, b) and sum them as above (asum, bsum). Then multiply the two initial integers together (ab) and sum the result as above (absum) . The sums of the two initial integers multiplied together (asum.bsum) and then summed if needed to get a number between 1 and 9 will equal the sum of the multiplied integer (absum = asum.bsum).

What am I describing here? Where do I go to further look into it? Is  there a reference book I should read? Thanks in advance.

You are dealing with remainders upon division by $ 9$.

The first process you describe produces a number from $ 1$ to $ 9$ known as the [b]digital root[/b] of a number.

[b]Lemma:[/b]  The digital root $ d$ of a number $ n$ is the remainder when $ n$ is divided by $ 9$ if $ n$ is not divisible by $ 9$, and it is $ 9$ when $ n$ is divisible by $ 9$.

Discussion:  Write out a number $ n$ in terms of its digits:

$ n \equal{} d_0 \plus{} 10 d_1 \plus{} 10^2 d_2 \plus{} 10^3 d_3 \plus{} ... \plus{} 10^k d_k$

so that our number has $ k\plus{}1$ digits.  The sum of its digits is

$ S(n) \equal{} d_0 \plus{} d_1 \plus{} d_2 \plus{} d_3 \plus{} ... \plus{} d_k$.

Now, note that

$ 10 \minus{} 1 \equal{} 9$
$ 10^2 \minus{} 1 \equal{} 99$
$ 10^3 \minus{} 1 \equal{} 999$
$ 10^4 \minus{} 1 \equal{} 9999$

In other words, $ 9 | 10^n \minus{} 1$ for any value of $ n$.  We conclude that the number

$ n \minus{} S(n) \equal{} (10 \minus{} 1) d_1 \plus{} (10^2 \minus{} 1) d_2 \plus{} (10^3 \minus{} 1) d_3 \plus{} ... \plus{} (10^k \minus{} 1) d_k$

is divisible by $ 9$.  Now, this number also has the property that it will be much smaller than $ n$ (exactly how much smaller is not important).  This iterative process continues to decrease the number of digits in a number until we are down to one digit, and at each step the new sum of digits [b]still[/b] has the same remainder upon division by $ 9$ as each previous sum, therefore as the original sum.  :)

The second phenomenon you describe is a well-known property of [b]modular arithmetic[/b] (the arithmetic of remainders).  It states that if two numbers $ a, b$ leave remainders $ c, d$ respectively upon division by $ m$, then $ ab$ leaves remainder $ cd$ upon division by $ m$.  To see this, write

$ a \equal{} ma' \plus{} c$
$ b \equal{} mb' \plus{} d$

so that

$ ab \equal{} (ma' \plus{} c)(mb' \plus{} d) \equal{} m(m a'b' \plus{} a'd \plus{} b'c) \plus{} cd$

In modular notation, we write

$ a \equiv c \bmod m, b \equiv d \bmod m \implies ab \equiv cd \bmod m$

where $ a \equiv c \bmod m$ is to be read "$ a$ is congruent to $ c$ modulo $ m$," and means that $ a$ and $ c$ leave the same remainder upon division by $ m$ (in other words, $ m$ divides $ a \minus{} c$).

If you have further questions, feel free to ask.  If you want to consult an online resource, look up the term [i]modular arithmetic[/i].  :)

Absolutely fascinating - many thanks. It'll take a while for me to wade through your discussion, but the bulk of it makes sense.

It's amazing to me that these things are all just "out there", waiting to be seen on your nearest license plate!

Thanks again!


algebra
polynomial
function
geometry
geometric series
calculus
calculus computations

hi everyone, im having trouble doing these problems could anyone offer any help, thanks.

1.) the function f(x)= 3/(1+4x^2) is represented as a power series

sigma n=0 to infinity C[size=75]n[/size]x^n

find the first five coefficients in the power series and the radius of convergence R of the series.

2.) represent the function 2/(1-8x) as a power series

sigma n=0 to infinity C[size=75]n[/size]x^n

find the first five coefficients of the series and find the radius of convergence R.

any help would be great, thanks.

You have the Taylor formula (for expansions about $x=0$): $P_n(x)=\sum_{k=0}^n \frac{f^{(k)}(0)}{k!}x^k$; if $f$ is nice, these polynomials will converge to $f$ in some interval, and we then look at the infinite series. These are just calculations.

For the particular examples here, there is a shortcut; do you know the sum of a geometric series?

Do you know geometry series?
if $S=a+ar+ar^2+ar^3+ \cdots$ then $S= \frac{a}{1-r}$ with $|r|<1$ I thought you know it.
a) $f(x)= \frac{3}{1+4x^2} = \frac{3}{1- (4x^2)}$, here we use $a=3$ and $r=-4x^2$  and the radius of convergence is $4x^2 < 1 \Rightarrow - \frac{1}{2}<x< \frac{1}{2}$,as you can check in case $x = (+-) \frac{1}{2}$  the sum is 3 or 0, it is diverges. The series \begin{eqnarray*} f(x)= 3 - 12x^2 + 48x^4 -192x^6 + .... &=& \sum_{k=0}^{\infty} (-1)^k 3(2^{2k}) x^{2k} \\ &=& \sum_{k=0}^{\infty} (-1)^k (2^{2k+1}+2^{2k}) x^{2k} \end{eqnarray*}.
b) Use similar argument. ;)


From point $A$ outside circle we draw two tangents, which intersect circle in points $B$ and $C$. From a point $P$ on the line $BC$ we draw two tangents which intersect circle in points $Q$ an $R$. Prove that $A$, $Q$ and $R$ are collinear

Another simple fact on polars: $P$ lies on polar of $A$ so $A$ lies on polar of $P$ ie $A$,$Q$,$R$ are collinear

This is an old problem.
$AR$ intersects $(O)$ at $Q'$.$BC$ intersects $OA$ at $H$.We have $AR.AQ'=AC^2=AO.AH$.Thus the quadrilateral $OHQ'P$ is inscribed in a circle.But the quadrilateral $OHQR$ is inscribed in the circle with diameter $OP$.Therefore,$Q\equiv Q'$.(q.e.d)                                                                                [img]http://i43.photobucket.com/albums/e387/phamduyhiep/Mathlinks/TXuc.jpg[/img]


In chapter 17, there's this problem about dogs in the begining when they introduced this new method of counting. It's about a dog trainer wanting eight new dogs and can choose from three different types. I don't understand why the answer isn't 3^8.

The dogs aren't being placed into slots - i.e. we're just building a group of dogs.  In other words, 3 of breed A, 3 of breed B and 2 of breed C is one possibility, and we don't say AAABBBCC and BBBAACCA are two different possibilities.

isn't the answer 9C2?
There are 9 spots where 2 dividers could be filled
x = dog
o = space
o x o x o x o x o x o x o x o x o

9 spaces

No, because you configuration sets it so that at least one dog is from the second group of dogs (in between the two dividers)


sum of four squares of integers, without regard to order?

For example,  5^2 + 3^2 + (-1)^2 + 1^2  is counted as a

different way from  5^2 + 3^2 + 1^2 + 1^2.

$6^2+0^2+0^2+0^2$ gives us 2 ways.
$5^2+3^2+1^2+1^2$ gives us 12 ways.
$4^2+4^2+2^2+0^2$ gives us 6 ways.
$3^2+3^2+3^2+3^2$ gives us 5 ways.
Therefore, there are 25 ways(did I count everything?).

I expect +0 and -0 are not treated differently. Then 25 ways is the correct answer.  :D [/code]


number theory
greatest common divisor
least common multiple
floor function
logarithms
ceiling function
number theory proposed

The sequence of positive integers $ a_1,a_2,\ldots$ has the property
$ \min(a_i,a_j) \equal{} a_{\textrm{gcd}(i,j)},$  $ \max(a_i,a_j) \equal{} a_{\textrm{lcm}(i,j)}$  for every  $ i,j\ge1.$
Find the maximum possible number of distinct integers among $ a_1,a_2,\ldots,a_{2009}.$

[b]Problem.[/b] The sequence $ \{a_i\}_{i \in \mathbb{N}_0} \subseteq \mathbb{Z}$ has the property that $ a_{\text{gcd}(i,j)}=\min\{a_i,a_j\} \le \max\{a_i,a_j\}= a_{\text{lcm}(a_i,a_j)}$ for all $ (i,j) \in \mathbb{N}_0^2$. Find the maximum number $ f(n)$ of distinct integers between $ \{a_1,a_2,\ldots,a_n\}$ for some $ n \in \mathbb{N}_0$ fixed.
[i]Solution.[/i] Costruct the infinite sequence $ a_i=x^{\upsilon_2(i)}$ for all $ i \in \mathbb{N}_0$, where $ x \in \mathbb{N} \setminus \{0,1\}$ is fixed. It respects all assumption, so that $ f(n) \ge \left\lfloor \log_2(2x) \right\rfloor$. On the other hand, since ${ \text{gcd}(i,i+1})=1$ for all $ i \in \mathbb{N}_0$ then at least one between $ \{a_i,a_{i+1}\}$ is equal $ a_1$. It means that at least $ \left\lceil  2^{-1}n\right \rceil$ numbers are equal to $ a_1$. Then there exist integers $ 2\le i_1 < i_2 < \ldots < i_k \le n$ such that $ k \le \left \lfloor 2^{-1}n \right\rfloor$ and $ \prod_{1 \le j \le k}{(a_1-a_{i_j})^2} > 0$ and $ \text{gcd}(a_{i_1},a_{i_2},\ldots,a_{i_k}): =d >1$. Clearly $ \{a_{i_1},a_{i_2},\ldots,a_{i_k}\} \subseteq \{a_d,a_{2d},\ldots,a_{td}\}$ where $ t: =\left\lfloor nd^{-1} \right\rfloor$; since $ \text{gcd}(id,(i+1)d)=d$ we can say that at least $ \left\lceil 2^{-1}t \right\rceil$ numbers in the last sequence are equal to $ a_d$. Now obviusly $ d \ge 2$ and, so, there are at most $ \left\lfloor 2^{-1}t\right\rfloor$ numbers in the sequence $ \{a_{i_1},a_{i_2},\ldots,a_{i_k}\}$ such that $ \prod_{1 \le j \le k}{(a_1-a_{i_j})^2(a_d-a_{i_j})^2}>0$. And we can proceede with this algorithm. It is also clear now that we can maximize the number $ f(n)$ choosing $ d=2$, and it shows that $ f(n) \le \left\lfloor \log_2(2x) \right\rfloor$. It means that $ f(n)=\left\lfloor \log_2(2x) \right\rfloor$.[]


Find the Equation of the tangent plane to the paraholoid

                                Z=X^2+4y^2/10

at the point (2,-2,2).

/=means divided

The solution is that from  Z=F(X,Y)=(X^2+4Y^2)/10, we obtain

   Fx(X,Y)=X/5                   Fy(X,Y)=4Y/5

   Fx(2, -2)=2/5                      Fy(2,-2)=-8/5

Therefore, an equation of the tangent plane at (2,-2,2) is

                Fx(2,-2)(x-2)+Fy(2,-2)(y+2)-(z-2)=0
                2/5(x-2)-8/5(y+2)-(Z-2)=0
                                2x-8y-5z-10=0

Which is correct by my calculations.


prove that
$ \frac{\tan(x\minus{}y)}{\tan(x)}  \plus{}\frac{\sin^2(z)}{\sin^2(x)} \equal{}1    \Leftrightarrow  \tan(x)\tan(y)\equal{}\tan^2(z)$

[quote="greatestmaths"]prove that
$ \frac {\tan(x \minus{} y)}{\tan(x)} \plus{} \frac {\sin^2(z)}{\sin^2(x)} \equal{} 1 \Leftrightarrow \tan(x)\tan(y) \equal{} \tan^2(z)$[/quote]

$ \frac{1\plus{}cot^{2}x}{1\plus{}cot^{2}z}\equal{}\frac{\frac{1}{1\plus{}cot^{2}z}}{\frac{1}{1\plus{}cot^{2}x}}\equal{}\frac{sin^{2}z}{sin^{2}x}\equal{}1\minus{}\frac{tan(x\minus{}y)}{tanx}\equal{}\frac{tanx\minus{}tan(x\minus{}y)}{tanx}$

$ 1\plus{}cot^{2}z\equal{}\frac{(1\plus{}cot^{2}x)tanx}{tanx\minus{}tan(x\minus{}y)}$

$ cot^{2}z\equal{}\frac{tanx\plus{}cotx}{tanx\minus{}tan(x\minus{}y)}\minus{}1\equal{}\frac{cotx\plus{}tan(x\minus{}y)}{tanx\minus{}tan(x\minus{}y)}$

$ tan^{2}z\equal{}\frac{tanx\minus{}tan(x\minus{}y)}{cotx\plus{}tan(x\minus{}y)}\equal{}\frac{tanx\minus{}frac{tanx\minus{}tany}{1\plus{}tanxtany}}{\frac{1}{tanx}\plus{}\frac{tanx\minus{}tany}{1\plus{}tanxtany}}\equal{}\frac{(tanx\plus{}tan^{2}xtany\minus{}tanx\plus{}tany)tanx}{1\plus{}tanxtany\plus{}tan^{2}x\minus{}tanxtany}\equal{}\frac{(tan^{2}xtany\plus{}tany)tanx}{1\plus{}tan^{2}x}\equal{}tanxtany$ ---->

$ tan^{2}z\equal{}tanxtany$


why are lozenges (rhombus with 45 degree angles) important?

because they look :coool:

Really, I think so because they have interesting properties like they can be split into two right triangles >.<


trigonometry
inequalities open
inequalities

Find the minimum value of:
$ A\equal{}\sqrt{sin^4x\plus{}cos^4x}\plus{}cos^2x\plus{}cos^2y$

hello, the searched minimum is $ 1$.
Sonnhard.

Dear mathlinker, i made a mistake. that 's

Find the minimum value of:
$ A \equal{} \sqrt {sin^4x \plus{} sin^4x} \plus{} cos^2x \plus{} cos^2y$

[quote="micf"]Are you sure?  :maybe:

Or do you think about

$ \sqrt {sin^4{x} \plus{} sin^4{y}} \plus{} cos^2x \plus{} cos^2y$

 :?:[/quote]
Yes, i sure. the proof is easy . use $ cosx,sinx \in [\minus{}1,1]$


How do you find a mentor(for the purpose of creating a research project for competition)?

A teacher has suggested that I go email professors at the University who do research in the topics I am interested in(math, computer science) but didn't suggest what I should say in these emails. I am rather confused as to what to say to someone to make them your mentor in a science project.

if you know anyone at the college, you can talk to that person about specific interests of the professors in those departments.

I think this really depends on whether you already have an idea for a project or not.  In either case, searching personal/professional websites to see what faculty members' interests are should help.

If you know what you want to do, then you want to find someone nearby who is an expert in that subject.  Then, the email should start by introducing yourself, then mention your project idea, say that you understand that this person is an expert, and that you're looking for a research supervisor for your project, and might this person be interested?

If you don't know what you want to do, but searching some websites identifies  a person (or some people) who is doing something interesting, then the email will be different.  It would start with an introduction again, but then say that you're interested in doing a research project, you saw on the person's website that they do Blah, and you think that looks interesting, and (a) do they have any appropriate-level project ideas they'd be willing to share and (b) if so, would they be willing to supervise the project?  Also, ask if they aren't able/willing to supervise you, whether they have any suggestions for people you might contact.

If you don't know what you want to do and web searching doesn't help, then I'm not sure what to suggest (but that's just stuff I don't know, rather than my saying it's hopeless).


geometry
3D geometry
calculus
function
integration
calculus computations

I just forgot how to do this if you could, just solve it for me, thanks!

differentiate the funcmake up the large cube?

Your first two questions should have been posted in "Calculus Computations and Tutorials," which is where I'm moving them.

The third question is off topic. I'd suggest the "High School Basics" forum of the high school section for it.

Differentiate the function:
1. $y = e^{tanx}$
2. $y = \sqrt{1+2e{^{3}x}}$

Evaluate the integral:
3. $\int \frac{1}{x^{2}+a^{2}}dx$

For 1 and 2, you need to use the chain rule. Click below if you forgot the form:
http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/chainruledirectory/ChainRule.html

For number 3, if $a = 0$, then it is trivial, but if $a \neq 0$, then rewrite the integral as
$\int \frac{1}{a^{2}((\frac{x}{a})^{2}+1)}dx$ and then let $\frac{x}{a}= tan\theta$.


[quote="tao_goldi"]In a.d. 1054, Chinese astronomers observed the light from a supernova in the night sky. This supernova was the incredibly powerful origin of what is now the Crab Nebula. Using the velocity of Light ( 186,000 mile/sec), devise and execute a strategy for finding out when the supernova actually occurred, relative to when the chinese astronomers saw the explosion. What are some possible strategies for solving this problem?[/quote]

current year - 1054 = t

 4660000 x t x speed of light


This problem is pretty difficult. who can solve it?

a) It's not hard to see that if $x$ is a root then $2-x$ is a root
Then let $x-1\geq 0$.
we can get $x\leq 2$ from the condition $\sqrt{2x-x^{2}}$.
Hence $x-1\leq 1$
Let $cos{y}=x-1,0\leq y\leq pi/2$
$2x^{2}-4x+1=cos{2y},2x-x^{2}=sin^{2}{y}$
then the problem is to find $y$ satisfy the equation
$\sqrt{1-sin{y}}+\sqrt{1+sin{y}}=2cos^{4}{y}cos{2y}$
$2+2\sqrt{1-sin^{2}{y}}=4cos^{8}{y}cos^{2}{2y}$
$1+cos{y}=2cos^{8}{y}cos^{2}{2y}$
$1+cos{y}=2cos^{2}{\frac{y}{2}}$
$cos^{2}{\frac{y}{2}}=cos^{8}{y}cos^{2}{2y}$
if $0\leq y\leq pi/4$
then $2y\leq pi/2,cos{2y}\geq 0$
$cos{\frac{y}{2}}=cos^{4}{y}cos{2y}$
if $y>0$
then $\frac{y}{2}<y<2y\leq pi/2,cos{\frac{y}{2}}>cos{2y}$
but $cos^{4}{y}<1$
then $cos{\frac{y}{2}}<cos{2y}$ contract
so $y=0,x=2$ is a root
if $pi/4\leq y\leq pi/2$
$cos{\frac{y}{2}}=-cos^{4}{y}cos{2y}$
$-cos{2y}\leq 1$
$\frac{y}{2}<y\leq pi/2$
then $cos{\frac{y}{2}}>cos{y}\geq cos^{4}{y}\geq-cos^{4}{y}cos{2y}$
contract
so we have $x-1\geq 0$
then $x=2$
if $x-1\leq 0$
then $x=0$
the equation have two zeros $x=0,x=2$

This is a beautiful solution. how to solve (b)?

Don't know if the user "Who" can solve it.

But stop this multiposting!

if $x\geq0$  then there is no solution .to be continued (may be :lol: )

since iff $x=0$ is not a solution we have the equation is equivalente to
$\frac{2+\sqrt{9x^{3}+3}}{1+\sqrt{1+x+x^{2}}}=-\frac{4x+2}{3x}$
so $RHS$ must be positive so the set of solution is $(\frac{-1}{2},0)$
but :blush: ..

According to arkhammedos the solutions are in ]-0.5,0[.
According to the function's graph there exists only one solution: $\alpha\approx-0.20385$.
With Maple we find that the $\alpha$'s minimal polynomial is: $6561x^{8}-2592x^{7}-4928x^{6}-16634x^{5}-11608x^{4}-7760x^{3}-3799x^{2}-680x-32$.
     This exercise is without interest.


How do you typeset in Latex such that I have something like this:

              5, if x = 1
f(x) = {
              4, otherwise

,only nicer and with a big fancy left brace?

[code]\[
f(x) =
\begin{cases}
5 & \text{if (dollar sign) x=1 (dollar sign)} \\
4 & \text{otherwise}
\end{cases}
\][/code]

This is my exact code:

[code]\[
x_{jt} =
\begin{cases}
1 & \text{if job $j$ starts at time $t$}\\
0 & \text{otherwise}
\end{cases}
\][/code]

It gives me a variety of errors though,  such as "Missing } inserted" here and there or "Misplaced alignment character &"

Whats wrong with it? I think the \text thing might not work...I'm using TeXnic center by the way.

Do you have the amsmath package loaded? It's required for the cases environment. The following compiles correctly using TeXnicCenter:
[code]\documentclass[12pt]{article}
\usepackage{amsmath}
\begin{document}
\[
x_{jt} =
\begin{cases}
1 & \text{if job  starts at time }\\
0 & \text{otherwise}
\end{cases}
\]
\end{document}[/code] and gives \[
x_{jt} =
\begin{cases}
1 & \text{if job  starts at time }\\
0 & \text{otherwise}
\end{cases}
\] You may find it useful to save as a template in TeXnicCenter the following:
[code]\documentclass[12pt]{article}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsthm}
\begin{document}

\end{document}[/code]which will allow you to compile many types of documents


[geogebra]cc675c120a4b8daf2f2252b8005b1d4bb60d1068[/geogebra]
Prove that $ \overline{ED}\equal{}\overline{EB}$.

Please state the problem, don't just refer to the applet - this is not a Java decompiling contest...

  darij

$ AB$ is the diameter of the big circle with its center $ O_1$. $ O_2$ is in $ AO_1$ and it is the center of the small circle which radius is $ O_2A$. $ C$ is on the the small circle such that $ BC$ is a tangent. The intersection of line $ AC$ and the big circle is $ E$. $ D$ is on the the small circle such that $ ED$ is a tangent. Prove that $ \overline{ED}\equal{}\overline{EB}$.

Hint: $ ED^2\equal{}EA\cdot EC$, so you only need to prove $ EA\cdot EC\equal{}EB^2$, what is equivalent to $ \measuredangle ECB\equal{}\measuredangle EBA$, what easily follows from $ CD\parallel BE$.

  darij

Sorry, I made a bad sketch with two points almost coinciding. What I meant is that $ CU\parallel BE$, where $ U$ is the second point of intersection of the circle $ O_2\left(O_2A\right)$ with $ AB$. This is because the two circles are tangent.

  darij

It is a well known property that E is the middle of the arc of the circle $ O_1$ made by the chord BC, which does not contain A, hence there is no need that AB be a diameter.

Best regards,
sunken rock


[quote="wolf"]2/3 - 15(-1/2)^2 + (-1/2))^-2

i got 27.5..[/quote]
I can't quite understand what you are saying. Can you please put it in LaTex?


[quote="watchcatno1"]Does anyone know when the German TST's are usually held? If Peter Scholze or Darij or someone else might post the questions it'll be nice.
   Andrew[/quote]

For the German TST structure check the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=21011]description by Darij.[/url] Past problems can be retrieved in the [url=http://www.mathlinks.ro/Forum/index.php?f=122]download section[/url] and in the [url=http://www.mathlinks.ro/Forum/resources.php?c=64&cid=54]resoures section.[/url] I guess questions from the current year did not yet get posted as the TST is still running, people have other stuff to do (or lazy) and quite often the test paper contains many ISL problems.  :D

And I am not sure whether Darij has access as he graduated from high-school last year. He is now at the German 'elite' university LMU Munich where most IMO participants seem to join since the new millennium. A few years ago the university of Bonn was known for it.

[quote="orl"]For the German TST structure check the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=21011]description by Darij.[/url] Past problems can be retrieved in the [url=http://www.mathlinks.ro/Forum/index.php?f=122]download section[/url] and in the [url=http://www.mathlinks.ro/Forum/resources.php?c=64&cid=54]resoures section.[/url] I guess questions from the current year did not yet get posted as the TST is still running, people have other stuff to do (or lazy) and quite often the test paper contains many ISL problems.  :D[/quote]

The latter reason, mostly.

[quote="orl"]And I am not sure whether Darij has access as he graduated from high-school last year. He is now at the German 'elite' university LMU Munich where most IMO participants seem to join since the new millennium. A few years ago the university of Bonn was known for it.[/quote]

I doubt Peter is going to join ;) .
And Bonn is still a place where IMOers gather.

  darij

[quote="darij grinberg"]

[quote="orl"]And I am not sure whether Darij has access as he graduated from high-school last year. He is now at the German 'elite' university LMU Munich where most IMO participants seem to join since the new millennium. A few years ago the university of Bonn was known for it.[/quote]

I doubt Peter is going to join ;) .

  darij[/quote]

That leaves the questions where Peter considers to go to!  :wink:

BTW since NoName graduated from high-school we have [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=27055]no names  and scores of possibly IMO scores anymore.[/url] BTW Christian Sattler also went to LMU Munich.


Determine strictly positive real numbers $ a_{1},a_{2},...,a_{n}$ if for any $ n\in N^*$ takes
place equality:
$ a_{1}^2\plus{}a_{2}^2\plus{}...\plus{}a_{n}^2\equal{}a_{1}\plus{}a_{2}\plus{}...\plus{}a_{n}\plus{}\frac{n(n^2\plus{}6n\plus{}11)}{3}$

[quote="Syler"]Determine strictly positive real numbers $ a_{1},a_{2},...,a_{n}$ if for any $ n\in N^*$ takes
place equality:
$ a_{1}^2 \plus{} a_{2}^2 \plus{} ... \plus{} a_{n}^2 \equal{} a_{1} \plus{} a_{2} \plus{} ... \plus{} a_{n} \plus{} \frac {n(n^2 \plus{} 6n \plus{} 11)}{3}$[/quote]

So $ \sum_{k\equal{}1}^n(a_k^2\minus{}a_k)\equal{}\frac {n(n^2 \plus{} 6n \plus{} 11)}{3}$ and so $ \sum_{k\equal{}1}^{n\minus{}1}(a_k^2\minus{}a_k)\equal{}\frac {(n\minus{}1)((n\minus{}1)^2 \plus{} 6(n\minus{}1) \plus{} 11)}{3}$

Subtracting these two equalities implies $ a_n^2\minus{}a_n\equal{}n^2\plus{}3n\plus{}2$ $ \iff$ $ (a_n\minus{}n\minus{}2)(a_n\plus{}n\plus{}1)\equal{}0$ and so $ a_n\equal{}n\plus{}2$ (since $ a_n>0$), and it's immediate to check back that this indeed is a solution.

Hence the answer : $ \boxed{a_n\equal{}n\plus{}2}$

And I wonder how you could know that the problem is nice if you dont know the solution  :huh:

[quote="pco"][quote="Syler"]Determine strictly positive real numbers $ a_{1},a_{2},...,a_{n}$ if for any $ n\in N^*$ takes
place equality:
$ a_{1}^2 \plus{} a_{2}^2 \plus{} ... \plus{} a_{n}^2 \equal{} a_{1} \plus{} a_{2} \plus{} ... \plus{} a_{n} \plus{} \frac {n(n^2 \plus{} 6n \plus{} 11)}{3}$[/quote]

So $ \sum_{k \equal{} 1}^n(a_k^2 \minus{} a_k) \equal{} \frac {n(n^2 \plus{} 6n \plus{} 11)}{3}$ and so $ \sum_{k \equal{} 1}^{n \minus{} 1}(a_k^2 \minus{} a_k) \equal{} \frac {(n \minus{} 1)((n \minus{} 1)^2 \plus{} 6(n \minus{} 1) \plus{} 11)}{3}$

Subtracting these two equalities implies $ a_n^2 \minus{} a_n \equal{} n^2 \plus{} 3n \plus{} 2$ $ \iff$ $ (a_n \minus{} n \minus{} 2)(a_n \plus{} n \plus{} 1) \equal{} 0$ and so $ a_n \equal{} n \plus{} 2$ (since $ a_n > 0$), and it's immediate to check back that this indeed is a solution.

Hence the answer : $ \boxed{a_n \equal{} n \plus{} 2}$

And I wonder how you could know that the problem is nice if you dont know the solution  :huh:[/quote]
Thank you a lot.
I supposed that the problem is nice.


If [tex]x_1 = 2 [/tex] and [tex]\displaystyle x_{n+1} = \frac{x_n^4+1}{5x_n}[/tex] if [tex]n>0,[/tex] then prove that [tex]1/5 \leq x_n < 2[/tex] for all [tex]n>1[/tex].

I found Engel's solution to this one a bit sketchy at one point.

[quote="Ravi B"]Let f(x) = (x4 + 1) / (5x).  Show that if 1/5 :le: x :le: 2, then 1/5 :le: f(x) < 2.  Once you show that, you are done by induction.[/quote]

Yeah, that's what I tried to do. However, Engel showed it in two different steps, taking the cases when 1/5 <= x < 1 and 1 <= x < 2 separately. How come the induction works when we do the cases separately and not when we just try to do 1/5 <= x < 2 all at once? That seems so strange.

I haven't seen Engel's proof, so I can't comment on it.  But you can prove my assertion in several ways.  If Engel wants to prove it by breaking the interval into two cases, that's fine.  If you have a way to prove it without breaking into cases, that's fine too.

The way I would prove my assertion is by first showing that f is decreasing on (0, c] and increasing on [c, :inf:), where c = 1/31/4.

Well, to prove the f(x) < 2 part, Engel took two case:

1) 1 <= x_n <= 2
The conditions imply that f(x) = (1/5)(x_n^3 + 1/x_n) <= (8+1)/5 = 9/5 < 2

2) 1/5 <= x_n <= 1
The conditions imply that f(x) = (1/5)(x_n^3 + 1/x_n) <= (1+5)/5 = 6/5 < 2

However, if we try to apply the same method of reasoning to the overal case when 1/5 <= x_n <= 2, we find that the conditions imply that f(x) = (1/5)(x_n^3 + 1/x_n)  <= (8 + 5)/5 = 13/5 > 2. I just found this result somewhat weird/sketchy.

Yes, but all your last observation means is that simple methods to estimate the maximum of f on the interval [1/5, 2] do not lead to strong bounds.  Basically, f is the sum of two functions g(x) = x3/5 and h(x) = 1/(5x).  Your method relies on the fact that max f(x) = max (g(x) + h(x)) :le: max g(x) + max h(x).  That is true, but the resulting bound may not be very good, especially because g and h do not achieve their maxima at the same point.

That's why Engel broke up the interval into two subintervals.  He still uses the inequality max (g(x) + h(x)) :le: max g(x) + max h(x), but only on each subinterval, where the bound is a little sharper.  That's because on a subinterval, g and h achieve their maxima at points that are closer to each other.

Having said that, I don't think I would have found the max of f using Engel's method.  I would have just noticed that f is convex on (0, :inf:).  Thus the max of f on [1/5, 2] is achieved at the endpoints.  Because f(1/5) < f(2) = 1.7, the max of f on [1/5, 2] is 1.7.

[quote="Pork_Chop8"](x_i)^4 + 1 < 17 and 5x_i < 10, so ((x_i)^4 + 1)/(5x_i) < (17)/(10)[/quote]
No, that is not a valid step.  The inequality 5x_i < 10 (with x_i > 0) implies that 1/(5x_i) > 1/10, which is going in the wrong direction.


Find the product of $ 1\cdot3\cdot5\cdot7\cdot ... \cdot 9999$ in the simplest form.

You really have to specify what you mean by "simplest form."  One possible interpretation is "allowing factorials, using as few symbols as possible."

I ask the same question to myself but I find the question in this form?
So what is simplest form in this case? Can I just say $ 9999!!$?

$ 1\cdot3\cdot5...\cdot(2n\minus{}1)\equal{}\frac{(2n)!}{2\cdot4\cdot6...\cdot(2n)}\equal{}\frac{(2n)!}{2^n(n!)}$.

Not sure if that's what you were looking for.


AMC
AIME
LaTeX
modular arithmetic
function
Euler
search

6.  What is the remainder on dividing 6^83 + 8^83 by 49?

Is there a correct way to do this with mod's?
This is what i tried.
Sorry beforehand about not using the correct notation. I made --- equal to congruent.

So first I looked at the 6 and 8. I saw that 6---[-43(mod 49)] and 8---[-41(mod 49)] Next i looked at the exponents for each, which is obviously the same, 83. According to AOPS V1, a^c=b^c(mod m). So i used the converse to say  6^83---[-43^{83}(mod 49)]---6---[-43(mod 49)]. In other words, i neglected the exponents. Then, according to AOPS again, (a+b)(mod m)---[ a(mod m)+ b(mod m)]
It then follows that the expression 6^83 + 8^83 is equiv. to  -84 mod(49). Then i got stuck. Looking at the answer, it seems that if you make the 84 positive, and subtract the 49, you get the answer. Could someone please explain where I went wrong and/or what to do if possible.

[quote]Looking at the answer, it seems that if you make the 84 positive, and subtract the 49, you get the answer[/quote]

I think thats what you do (i'm kinda new to mods so you might want to get answers from someone with more experience).
You make it positive since remainders should be positive
And subtract 49 because $ a \equiv b \mod n$ then $ 0\leq b < n$

thanks for the help. I had no clue how to do it, didn't use  the binomial expansion since I didn't see where I could go with it. I had a feeling I could do it with mods, and I knew they were in V1, so I just went with it lol. Rambling aside, thanks also for using Latex; by looking at the code I will be more inclined to implement it next time.

[hide="big hint"] $ 6^{83}\equal{}(7\minus{}1)^{83}$ likewise $ 8^{83}\equal{}(7\plus{}1)^{83}$ now use binomial theorem and consider everything in mod 7. what terms are left?[/hide]

[quote="adrenalinefix36"]In other words, i neglected the exponents.[/quote] You really can't just ignore the exponents. If you ignore the exponents then you will get 14 every time. Note that $ \minus{} 84 \equiv \minus{} 84 \plus{} 2 \cdot 49 \equiv 14 \pmod{49}$, and this is the same as $ 6 \plus{} 8 \equal{} 14$ by ignoring the exponents. I suggest you go back to AoPS 1 and look at the particular conditions on a, b, c, and m in the equation you cited :wink:

In the meanwhile, here is a very straightforward solution that doesn't require any of that.

[hide="Solution by Binomial Theorem"]
$ 6^{83} \equal{} (7 \minus{} 1)^{83} \equiv \binom{83}{1} \cdot 7 \minus{} 1 \pmod{49}$
$ 8^{83} \equal{} (7 \plus{} 1)^{83} \equiv \binom{83}{1} \cdot 7 \plus{} 1 \pmod{49}$
[i]where the terms corresponding to $ 7^{83}$ down to $ 7^2$ became $ \equiv 0 \pmod{49}$[/i]

[i]Adding the two gives [/i]$ 2 \cdot 83 \cdot 7 \pmod{49}$.

[i]While the above is straightforward to do out by hand, it is even easier to note that [/i]$ 83 \equal{} 12 \cdot 7 \minus{} 1$ to get $ 2 \cdot 7 \cdot (12 \cdot 7 \minus{} 1) \equiv 2 \cdot 7 \cdot ( \minus{} 1) \equiv \minus{} 14 \equiv 35 \pmod{49}$
[/hide]

ok.  I did realize that i 14 (mod 49) is obtained, and I realized that everything seem awkward. I guess I was just hoping i was right in vain. On a side note, i think the lesson was pretty clear.

Another way I thought of was breaking it down:
(8^83)/49+(6^83)/49, then adding remainders.

[hide]Since $ \phi(49)\equal{}42$.

$ 6^83\equiv6^41\mod 49$
$ 8^83\equiv8^41\mod 49$

Since $ 6^42\equiv1\mod 49$, we try to divide the mod by 6.  note that $ 49 \equiv 1\mod 6$ so the first one that will be divisible by 6 is $ 49*5\plus{}1\equal{}246$.  This gives $ 41$ as a remainder.

Now for the second one, we need $ 49*7\plus{}1\equal{}344$, which gives $ 43$ as a remainder.  We thus have that the answer is $ \boxed{84\mod 49\equal{} 35}$[/hide]

mathcrazed, the latex code for your exponents is a little messed up.

mathcrazed, could you please explain a little clearer.

Thanks.

I solved this problem in a way very similar to what [b]i_like_pie[/b] did.

[b]Mathcrazed[/b], if you're gonna use the Totient function, you may as well go all the way with it :P

[hide="Solution with the Euler Totient Formula"]
Note that, by Euler's Totient formula, $ \varphi (49) = 42 \implies 6^{42} \equiv 1 \pmod{49}$, same with $ 8$.

Hence, $ 6 \cdot 6^{83} = 6^{84} \equiv 1 \pmod{49}$
Let $ a$ be the remainder upon division of $ 6^{83}$ by ${ 49}$; we have $ 6a \equiv 1 \pmod{49}$
We know $ 0 \leq a \leq 48 \implies 0 \leq 6a \leq 288$

At the same time, $ 6a \in \{ 50, 99, 148, 197, 246 \}$ because of the mod 49 requirement.

Checking, we must have $ 6a = 246 \implies a = 41$

Using a similar argument with $ 8^{83} \equiv b \pmod{49}$, we have:
$ 8b = 344 \implies b = 43$

Adding, $ 6^{83} + 8^{83} \equiv 41 + 43 \equiv 84 \equiv 35 \pmod{49}$
[/hide]

[quote="The Zuton Force"]$ \varphi (49) \equal{} 42 \implies 6^{42} \equiv 0 \pmod{49}$[/quote]Correction: $ \varphi (49) \equal{} 42 \implies 6^{42} \equiv 1 \pmod{49}$
@adrenalinefix36: please use the search function


geometry
3D geometry
sphere
analytic geometry
calculus
integration

Consider two solid spherical balls, one centered at $(0,0,\frac{21}{2})$ with radius $6$, and the other centered at $(0,0,1)$ with radius $\frac{9}{2}$. How many points $(x,y,z)$ with only integer coordinates (lattice points) are there in the intersection of the balls?

[hide]Is it 13?[/hide]

I'll post my solution if it's right...

[hide]
The first sphere gives us that the $z$ coordinate must be greater than or equal to $4.5$. The second sphere gives us that the $z$ coordinate must be less than or equal to $5.5$. Therefore the only integer $z$ coordinate in their intersection is $z=5$. Now, the intersection of the spheres is the intersection of the following:

$x^{2}+y^{2}+\left(z-\frac{21}{2}\right)^{2}\leq 36$
$x^{2}+y^{2}+(z-1)^{2}\leq \frac{81}{4}$

Substituting $z=5$, we obtain

$x^{2}+y^{2}\leq \frac{23}{4}$
$x^{2}+y^{2}\leq \frac{17}{4}$

We take the second equation, since it is more restrictive. Since $x^{2}+y^{2}$ is nonnegative and integral, we must have

$x^{2}+y^{2}= 0$ or $1$ or $2$ or $3$ or $4$

Taking each case, we find 6 solutions for $x$, $y$, and $z$ (which we already knew):

$(0,0,5), (0,1,5), (1,0,5), (1,1,5), (0,2,5), (2,0,5)$
[/hide]

Actually, check out the solution [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=419901#p419901]here[/url].  It's better than mine.

TheAmazingOne: you did it right, but you only took the solutions in the first quadrant or on the axes...


Let $F_{n}(x)$ satisfies :$F_{n}(\cos x)=\cos nx$
there's only finite real number $x_{0}$with the property$|x_{0}|\leq 1$,
and for some $m,n$ $F_{n}(x_{0})=0,F_{m}(x_{0}^{2})=0$

Could you explain more? Thank you very much.
And I also have another supposition:
$F_{p}(x)$is irreducible. is this one right?

Take a look.

[url]http://mathworld.wolfram.com/ChebyshevPolynomialoftheFirstKind.html[/url]

Thank you ,kunny.But It is too hard to me! :(  :(


help!
i got the problems online all right but couldnt find the soulutions..
if you know a website that has the solutions please let me know
or if you can just give me the letter answers that'll be really appreciated!

[quote="british8985"]http://www.unl.edu/amc/e-exams/e6-amc12/archive12.html

it's always posted on the website^^[/quote]

I think he's looking for the solutions, not the answers ;) and yeah, where did you find them online :?

[quote="frt"][quote="h0425"]i got the problems online all right but couldnt find the soulutions..[/quote]
Just curious, where did you get the 2003/2004 AMC12 problems online?[/quote]

Before AMC-12, I'm going to try to post all AMC-12 problems from 2000. So, they'll be available in AoPS soon.

http://goodlymaths.0catch.com/contests/amc/amc_archive.htm

Problems from 2003 and 2004 are on there.  I know solutions for the later problems can usually be found in old transcripts.  Search the AoPS site for those.


function
calculus
derivative
algebra
functional equation
algebra unsolved

Find all functions $f: \mathbb R \to \mathbb R$ which satisfy \[ f(x\plus{}y)\equal{}f(x) \cdot e^{f(y)\minus{}1}\] for all $x,y \in \mathbb R$.

[quote="duythuc_lqd"]Find all function $ f: R\to R$ satisfy $ f(x \plus{} y) \equal{} f(x).e^{f(y) \minus{} 1}$[/quote]

Let $ P(x,y)$ be the assertion $ f(x\plus{}y)\equal{}f(x)\cdot e^{f(y)\minus{}1}$

If $ \exists a$ such that $ f(a)\equal{}0$, then $ P(a,x\minus{}a)$ $ \implies$ $ f(x)\equal{}0$ $ \forall x$

Consider now $ f(x)\neq 0$ $ \forall x$. Comparing $ P(x,y)$ and $ P(y,x)$, we get $ \frac{e^{f(x)\minus{}1}}{f(x)}\equal{}\frac{e^{f(y)\minus{}1}}{f(y)}$ and so $ e^{f(x)\minus{}1}\equal{}c\cdot f(x)$ $ \forall x$

But, for a given $ c$, the equation $ e^{x\minus{}1}\equal{}x$ has at most two solutions $ a_c$ and $ b_c$ and so  $ f(x)\in\{a_c,b_c\}$

Considering now two values $ x_1$ and $ x_2$ such that $ f(x_1)\equal{}f(x_2)$, $ P(x_1,x_2\minus{}x_1)$ implies $ f(x_2\minus{}x_1)\equal{}1$ and so $ c\equal{}1$ and $ f(x)\equal{}1$ $ \forall x$

Hence the two solutions :
$ f(x)\equal{}0$ $ \forall x$
$ f(x)\equal{}1$ $ \forall x$

[quote="pco"][quote="duythuc_lqd"]Find all function $ f: R\to R$ satisfy $ f(x \plus{} y) \equal{} f(x).e^{f(y) \minus{} 1}$[/quote]

Let $ P(x,y)$ be the assertion $ f(x \plus{} y) \equal{} f(x)\cdot e^{f(y) \minus{} 1}$

If $ \exists a$ such that $ f(a) \equal{} 0$, then $ P(a,x \minus{} a)$ $ \implies$ $ f(x) \equal{} 0$ $ \forall x$

Consider now $ f(x)\neq 0$ $ \forall x$. Comparing $ P(x,y)$ and $ P(y,x)$, we get $ \frac {e^{f(x) \minus{} 1}}{f(x)} \equal{} \frac {e^{f(y) \minus{} 1}}{f(y)}$ and so $ e^{f(x) \minus{} 1} \equal{} c\cdot f(x)$ $ \forall x$

But, for a given $ c$, the equation $ e^{x \minus{} 1} \equal{} x$ has at most two solutions $ a_c$ and $ b_c$ and so  $ f(x)\in\{a_c,b_c\}$

Considering now two values $ x_1$ and $ x_2$ such that $ f(x_1) \equal{} f(x_2)$, $ P(x_1,x_2 \minus{} x_1)$ implies $ f(x_2 \minus{} x_1) \equal{} 1$ and so $ c \equal{} 1$ and $ f(x) \equal{} 1$ $ \forall x$

Hence the two solutions :
$ f(x) \equal{} 0$ $ \forall x$
$ f(x) \equal{} 1$ $ \forall x$[/quote]

in fact there are no solutions because $ f: R\to R$

$ f : \mathbb{R} \to \mathbb{R}$ doesn't mean its surjective, just that it's a real-valued function on the reals. $ f(x) \equal{} 0$, and $ f(x) \equal{} 1$ are real-valued functions on the reals.

[quote="duythuc_lqd"]Find all function $ f: R\to R$ satisfy $ f(x \plus{} y) \equal{} f(x).e^{f(y) \minus{} 1} (1)$[/quote]
        [i]  Oh , easy problem

 :)   $ f \equiv 0$ is a root  . We will consider the case $ f \not \equiv 0$

      So , there is a real number $ x_0$ such that : $ f(x_0) \neq 0$

   Put $ x \equal{} x_0 \ ; y \equal{} 0$ into $ (1)$ : we have : $ f(x_0) \equal{} f(x_0)e^{f(0) \minus{} 1} \rightarrow e^{f(0) \minus{} 1} \equal{} 1$

  $ \boxed{\rightarrow f(0) \equal{} 1}$

  Put $ x \equal{} 0$ into $ (1)$ we have : $ f(y) \equal{} f(0)e^{f(y) \minus{} 1} \equal{} e^{f(y) \minus{} 1} \forall y \in \mathbb{R} (2)$

   Consider the function $ f(x) \equal{} e^{x \minus{} 1} \minus{} x$ , we can prove that :

   $ f(x) \ \geq \ 0 \ \ \forall x \in \mathbb{R}$ and  $ f(x) \equal{} 0 \iff x \equal{} 1$( Using derivative and variation board )

   So from $ (2)$ , we deduce that : $ f(y) \equiv 1 \rightarrow f(x) \equal{} 1 \forall x \in \mathbb{R}$

  By a direct check , we can see that $ f(x) \equal{} 1 \forall x \in \mathbb{R}$ is a root of our initial functional equation

    Done  :lol: [/i]


number theory
greatest common divisor
least common multiple
number theory unsolved

1.Show that $ \gcd(a,b)=\gcd(a+b\text{lcm}(a,b))$

2. Find two positive integers $ a,b$ such that a+b=186 and $ \text{lcm} (a,b)=1440.$

3. Let $ a,b$ be positive integers. Show that $ \gcd(a,b)$ and $ \text{lcm}(a,b)$ are perfect squares if and only if both $ a$ and $ b$ are perfect squares.

4. Find all positive integers $ a,b$ such that $ a^2+b^2=\text{lcm}(a,b)+7\gcd(a,b)$

5. Show that for all positive integers $ a,b,c$

$ \text{lcm}(a,b,c)*\gcd(a,b)*\gcd(b,c)*\gcd(c,a)=abc*\gcd(a,b,c,)$

[quote="kl2836"]1.Show that $ \gcd(a,b) = \gcd(a + b\text{lcm}(a,b))$

2. Find two positive integers $ a,b$ such that a+b=186 and $ \text{lcm} (a,b) = 1440.$
[/quote]

1. Let's denote $ a = dk$ and $ b = dl$,  $ k,l,d \in \mathbb{N}$ such that $ \gcd(k,l) = 1$. Than you have $ \gcd(a,b) = d$. As $ a + b = d(k + l)$ and $ \text{lcm}(a,b) = dkl$ we have $ \gcd(a + b, \text{lcm}(a,b)) = d$, because $ k + l$ and $ kl$ are coprime.

2. From the problem above we have $ \gcd(a,b) = \gcd(186, 1440) = 6$ now denote $ a = 6m$ and $ b = 6n$. As $ a + b = 186$ it's easy to see that $ m + n = 31$ and also as $ \text{lcm}(a,b) = dmn = 6mn = 1440$ we have $ mn = 240$. After solving this equations we see that our two numbers are $ 96$ and $ 90$.

[quote="kl2836"]
5. Show that for all positive integers $ a,b,c$,
$ \text{lcm}(a,b,c)*\gcd(a,b)*\gcd(b,c)*\gcd(c,a) = abc*\gcd(a,b,c,)$[/quote]
Let,
\[ a = \prod p_i^{\alpha_i} , \ \ \ b = \prod p_i^{\beta_i} ,\ \ \ c = \prod p_i^{\gamma_i}
\]
So the left hand side is equivalent to,
\[ \text{L.H.S} = \prod p_i^{\max{(\alpha_i,\beta_i,\gamma_i)}} \cdot \prod p_i^{\min{(\alpha_i,\beta_i)}} \cdot \prod p_i^{\min{(\beta_i,\gamma_i)}}\cdot \prod p_i^{\min{(\gamma_i,\alpha_i)}}
\]
And the right hand side is equivalent to,
\[ \text{R.H.S.} = \prod p_i^{(\alpha_i + \beta_i + \gamma_i)} \cdot \prod p_i^{\min{(\alpha_i,\beta_i,\gamma_i)}}
\]
It is enough to show that, for each non negative numbers, $ \alpha,\beta,\gamma$
\[ {\max{(\alpha,\beta,\gamma)} + {\min{(\alpha,\beta)} + {\min{(\beta,\gamma)} + {\min{(\gamma,\alpha)} = (\alpha + \beta + \gamma) + \min{(\alpha,\beta)}
}}}}\]
Wlog, we assume that, $ \alpha \le \beta \le \gamma$
So, $ \text{L.H.S.} = 2\alpha + \beta + \gamma$ and $ \text{R.H.S.} = 2\alpha + \beta + \gamma$

So, the equality is true.


AMC
AIME
algebra
difference of squares
special factorizations

This is just what the title said it would be a problem and method discussion!

I'll start things off: How many multiples of 7 are there in the first 1000 Fibonacci numbers?(easy)

You could start by finding a pattern. The first term of the Fibonacci sequence is mod 1, and so is the second, and if you write the fibonacci sequence like this;

1mod7
1mod7
2mod7
3mod7
5mod7
1mod7
6mod7
0mod7
6mod7
6mod7
5mod7
4mod7
2mod7
6mod7
1mod7
0mod7

So every eighth term is divisible by seven and it is shown here. The pattern keeps going on and on. So if every eighth term is divisible by seven, that means that there are 125 fibonacci numbers in the first 1000, that are divisible by 7.

that is correct game guy! Now you get to post a problem! :rotfl:  :rotfl:  :rotfl:

I like to laugh

What's with the sudden explosion of marathons on these forums?
Anyway, I guess I'll post a problem, since there isn't a current one.

[b]Problem.[/b] (AIME #3). For all $ x\in\mathbb{R}$, $ f(x)\plus{}f(x\minus{}1)\equal{}x^2$. Given that $ f(19)\equal{}94$, find the last three digits of $ f(94)$.

Hm. Apparently nobody's interested in this problem (or people forgot about this thread), so I'll solve it.

[hide="Solution"]$ f(x)\equal{}x^2\minus{}f(x\minus{}1)$. Plugging in $ x\equal{}20$ gives $ f(20)\equal{}20^2\minus{}f(19)$
Plugging in $ x\equal{}21$ gives $ f(21)\equal{}21^2\minus{}f(20)\equal{}21^2\minus{}20^2\plus{}f(19)$.
Plugging in $ x\equal{}22$ gives $ f(22)\equal{}22^2\minus{}21^2\plus{}20^2\minus{}f(19)$, and so on.

Then $ f(n)\equal{}n^2\minus{}(n\minus{}1)^2\plus{}(n\minus{}2)^2\minus{}\cdots\pm f(19)$.

In the case of $ n\equal{}94$, we have $ f(94)\equal{}94^2\minus{}93^2\plus{}92^2\minus{}91^2\plus{}\cdots\minus{}f(19)$.

From the difference of squares identity, $ f(94)\equal{}(94\plus{}93)(94\minus{}93)\plus{}(92\plus{}91)(92\minus{}91)\plus{}\cdots\equal{}94\plus{}93\plus{}\cdots\plus{}22\plus{}21\plus{}20^2\minus{}94$.

$ \frac{94\cdot 95}{2}\minus{}\frac{20\cdot 21}{2}\plus{}400\minus{}94\equal{}4561\equiv\boxed{561}\bmod 1000$.[/hide]

Someone else feel free to post a problem.

Actually I tried, but I didn't post. And yes this thread has been forgotten... Plz just start posting in one of my other threads.

[quote="grn_trtle"]
From the difference of squares identity, $ f(94) \equal{} (94 \plus{} 93)(94 \minus{} 93) \plus{} (92 \plus{} 91)(92 \minus{} 91) \plus{} \cdots \equal{} 94 \plus{} 93 \plus{} \cdots \plus{} 22 \plus{} 21 \plus{} 20^2 \minus{} 94$.

$ \frac {94\cdot 95}{2} \minus{} \frac {20\cdot 21}{2} \plus{} 400 \minus{} 94 \equal{} 4561\equiv\boxed{561}\bmod 1000$.[/hide]
[/quote]
How did you make this simplification? :maybe:

I used difference of squares to get $ (94\plus{}93)(94\minus{}93)\plus{}\cdots$. $ 94\minus{}93\equal{}1$, so those multipliers are meaningless, and we just end up with $ 94\plus{}93\plus{}\cdots\plus{}22\plus{}21$, plus some other things.

To evaluate $ 94\plus{}\cdots\plus{}21$, we treat it as an arithmetic series. The sum is $ \sum_{k\equal{}21}^{94}k\equal{}\sum_{k\equal{}0}^{94}k\minus{}\sum_{k\equal{}0}^{20}k$ (Why?). Using the well-known closed form $ \sum_{k\equal{}0}^{n}k\equal{}\frac{n(n\plus{}1)}{2}$, this becomes $ \frac{94\cdot 95}{2}\minus{}\frac{20\cdot 21}{2}$.

Then I just added $ 20^2\minus{}94$.

Oh, I was missing the fact that $ 94\minus{}93\equal{}1$ and so on, sorry. :blush:


vector
linear algebra
matrix
linear algebra unsolved

let's consider the vector space $ \mathbb{R}^4$, with the bilinear (not degenerate) application $ g: \mathbb{R}^4 \times \mathbb{R}^4\minus{}>\mathbb{R}$ with matrix $ \begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&\minus{}c^2\end{bmatrix}$ in the standard base, with c the speed of light.
How can we prove that $ g(v,v)>0$, $ g(w,w)>0$ iff $ g(v,w)>0$
?

I don't think it's true.

To make things easy, let's adopt a system of measurement in which $ c\equal{}1.$

Let $ v\equal{}(2,0,0,1)^T$ and $ w\equal{}(0,2,0,1)^T.$

Then $ g(v,v)\equal{}3>0$ and $ g(w,w)\equal{}3>0,$ but $ g(v,w)\equal{}\minus{}1<0.$


algebra
polynomial
search
number theory
Diophantine equation
number theory proposed

Find all polynomials $f\in \mathbb{Z}[X]$ such that if $p$ is prime then $f(p)$ is also prime.

If $P\not\equiv x$ then there is exist $p$ is prime such that $P(p)=q\not =p$ is prime. Because $kq|P(p+kq)-P(p)\; \forall k\in\mathbb{N}$ and $q|kq\;\forall k\in\mathbb{N}$, we have $q|P(p+kq)\;\forall k\in\mathbb{N}$, but set $\{p+kq|k\in\mathbb{N}\}$ has infinite elements are prime, from this :$f(x)\equiv q$.

Final, we have answer $P(x)\equiv x \;\text{or}\; q$, where $q$ is prime.

My solution is exactly the same but is it allowed to use Dirichlet's theorem in competitions because I used it in a Bulgarian competition and was marked with 3 pts instead of 7 :(

If you live in Vietnam, you will have 0/7 if you use Dirichlet's theorem :P

[quote="dragonfire"]So is there a solution without it?  :play_ball:[/quote]
I think is yes, ctlhp said that maybe solve : Find all $P\in\mathbb{Z}[x]$ such that for every prime $p$  there are prime $q$ and positive integer $t$ satisfying $P(p)=q^{t}$.

[quote="TomciO"]It follows from the mentioned Dirichlet's theorem...[/quote]
You don't use it

[quote="tmbtw"][quote="N.T.TUAN"]set $\{p+kq|k\in\mathbb{N}\}$ has infinite elements are prime
[/quote]
You can post a proof for this[/quote]
You need a proof of Drichlet's theorem, Ok! Use Clytomic polynomials. Do you know them?

How do you want to proof it with cyclotomy only¿
Polynomial proof are only possible to show that there are $\infty$ many primes $\equiv a \mod n$ for those $a$ with $a^{2}\equiv 1 \mod n$.
Especially for $n$ prime only if $a \equiv \pm 1 \mod n$.

But proofs for the full theorem of Dirichlet are available online and were already discussed here.

[quote="ZetaX"]How do you want to proof it with cyclotomy only¿
Polynomial proof are only possible to show that there are $\infty$ many primes $\equiv a \mod n$ for those $a$ with $a^{2}\equiv 1 \mod n$.
Especially for $n$ prime only if $a \equiv \pm 1 \mod n$.[/quote]
Please see in : Lawrence C. Washington, Introduction to Cyclotomic Fields, Graduate text in Math.  :wink:

[quote]But proofs for the full theorem of Dirichlet are available online and were already discussed here.[/quote]
Where is it ?

I've read most of that book and looked at it again now. The proof there is far away from a purely cyclotomic proof, it uses analytic stuff et cetera.
And knowledge about cyclotomic polynomials is far away from sufficient knowledge to read that book.

Where: search for "Dirichlet" here on the forum, I think the title of at least one of those topics is very good.

[quote="ZetaX"]
And knowledge about cyclotomic polynomials is far away from sufficient knowledge to read that book.
[/quote] I know, and I reading it. :) So, I think tmbtw can't understand proof in that book.

[quote] Where: search for "Dirichlet" here on the forum, I think the title of at least one of those topics is very good.[/quote]
Ok, search for title 'Dirichlet' and Author ''ZetaX''?  :D

Without any author, it's theird from the top for me (by t0rajir0u).

When I Search for Keywords ''Dirichlet'' and
Author empty then I have http://www.mathlinks.ro/Forum/viewtopic.php?highlight=dirichlet&t=135341

Thanks  :wink:

P.S: Topic http://www.mathlinks.ro/Forum/viewtopic.php?highlight=dirichlet&t=100567 is nice too.

[quote="Erken"][quote="TomciO"]That's stupid since it is really famous, anyway it is always safe to give a source.[/quote]
Sorry for reviving this old thread.
I let myself to disagree with you.
Reasoning like you,we can say that Fermat's Great Theorem is really famous and instead of solving some particular case,we are allowed to refer to some source where Fermat's Great Theorem is proved.It is doubtful that someone would be able to prove this theorem during the contest,so having the problem,which can be solved by the use of Dirichlet's theorem often presumes existing another solution,probably much harder,but without any allusion to this theorem,or this is nothing more than just a particular case of Dirichlet's theorem,which can be solved by simple using of cyclotomic polynomials.
I had this problem at pre-last round of National olympiad and I've managed to solve all the problems,except this one.I still think that I should have had only 3 points or maybe even less but not 7 points...I've been really perplexed when I saw my result,happy,but ...
In any case,I think that this problem is too "bad" to be chosen,though,probably,somewhere using this theorem is allowed(like my country) and nobody has any doubt if this theorem can be used...[/quote]

Actually it's a good thing to discuss.
From the strictly mathematical (or rather problem-solving) point of view you're undoubtly right. I've always considered funny the situations where, for example, some diophantine equation is posted on the forums and then people start giving references to recently proved Catalan conjecture, even if the problem follows from mod $ 3$ -> mod $ 7$ -> mod $ 13$ or something similar. Destroying the problem automatically with famous (and hard) theorem just gives no benefits to the solver.
But on the contest, where we play for points, the situation is quite different. If the rules don't say clearly which theorems are excluded to refer (and it's rather impossible to make such list) I am convinced that every theorem should be accepted (assuming that the contestant gives the reference to the proof of theorem).  Obviously, the olympic problems shouldn't be the consequences of famous theorems but this is the task of the people who choose questions to avoid such situations - and that's it. It's very bad when the contestant isn't sure if he can refer to some well-known but hard results and he knows that it will help him with solving the problem.
It still happens from time to time that such problems appear and usually everything relies on the common sense of the contestant and the grader. Unfortunately they are not always the same. For example consider the Dirichlet's and Fermat's Theorem. In my opinion it should be allowed to use freely the first one and not the second. Why? Fermat's Theorem is just really not interesting in terms of possible consequences, while there are just plenty of very nice problems which couldn't be touched without the Dirichlet's Theorem. But someone can think other way. So, the most important is just choosing the right problems.

[quote="TomciO"]
Actually it's a good thing to discuss.
[/quote]
yup,but I am afraid that this discussion can not lead us to something really undoubted.I mean there are enough who thinks like "we must not use these theorems on olympiads" and like "what is the point of not using this theorem,if it has been proved,we definitely can use that".
[quote="TomciO"]
I've always considered funny the situations where, for example, some diophantine equation is posted on the forums and then people start giving references to recently proved Catalan conjecture, even if the problem follows from mod $ 3$ -> mod $ 7$ -> mod $ 13$ or something similar. Destroying the problem automatically with famous (and hard) theorem just gives no benefits to the solver.
[/quote]
Me either. :)
[quote="TomciO"]
But on the contest, where we play for points, the situation is quite different. If the rules don't say clearly which theorems are excluded to refer (and it's rather impossible to make such list) I am convinced that every theorem should be accepted (assuming that the contestant gives the reference to the proof of theorem).  Obviously, the olympic problems shouldn't be the consequences of famous theorems but this is the task of the people who choose questions to avoid such situations - and that's it. It's very bad when the contestant isn't sure if he can refer to some well-known but hard results and he knows that it will help him with solving the problem.
It still happens from time to time that such problems appear and usually everything relies on the common sense of the contestant and the grader. Unfortunately they are not always the same. For example consider the Dirichlet's and Fermat's Theorem. In my opinion it should be allowed to use freely the first one and not the second. Why? Fermat's Theorem is just really not interesting in terms of possible consequences, while there are just plenty of very nice problems which couldn't be touched without the Dirichlet's Theorem. But someone can think other way. So, the most important is just choosing the right problems.[/quote]
That's the point why we can not definitely say whether this theorem is allowed to use it during the olympiad or not.
Participating at "big" olympiad(like IMO) presumes having comprehensive experience with knowing,maybe it is better to say predicting or supposing,whether your solution will be accepted or not.I guess,nobody can be truly convinced that his solution will be accepted if he/she would use,for instance, Dirichlet theorem.
The dispute about this theme seems to me senseless.But if someone proposed a such problem,and you had to use some "probably forbidden" approach,you would worry about it during the contest and would spend time on thinking like "Maybe I should find another way,which avoid using of this theorem".And it anyhow can make your result much worse.
On the other hand proposer might not know that this problem could be solved using this theorem.

[quote="Erken"]On the other hand proposer might not know that this problem could be solved using this theorem.[/quote]
For this one here, it rather naturally occurs (similar for the $ 13|p^3\plus{}1$ problem where it is even more obvious). So such a problem may be interesting, but is not adequate for some contest. A problem should not get trivial by some theorem that can be easily proven in contest (yes, Dirichlet's theorem can), thus given those an advantage that just know some theorem.

[quote="ZetaX"][quote="Erken"]On the other hand proposer might not know that this problem could be solved using this theorem.[/quote]
For this one here, it rather naturally occurs (similar for the $ 13|p^3 \plus{} 1$ problem where it is even more obvious). So such a problem may be interesting, but is not adequate for some contest. A problem should not get trivial by some theorem that can be easily proven in contest (yes, Dirichlet's theorem can), thus given those an advantage that just know some theorem.[/quote]
May I know how can Dirichlet's theorem be easily proved(or at least just proved) by using only elementary math,avoiding using college-knowledges.I do really doubt it.
Nevertheless this is Olympiad Section,not College Playground or anything else.I mean,being older than someone must not give so much superiority over that someone,who hasn't studied non-elementary math yet.
I totally agree with you that a problem shouldn't get trivial if use any theorem,but how can we be sure that it is indeed so?

The non-elementary proof is in fact not too complicated to be written down in exam (maybe takes you 45 minutes, assuming you are prepared to do so).
You cannot be sure for a problem not be a easy corollary to something "known" (already because "known" means something different to eachperson), but at least in some attempts to solve the problem, one shouldn't encounter something that should really be well known.


geometry
incenter
LaTeX
angle bisector
geometry proposed

In the triangle $ ABC$, $ BC$ is the longest side and has the length equal with $ a$, and $ AB,AC$ have the length $ b,c$,respectively. Denote by $ D$ the intersection between $ AC$ and the angle bisector of $ \angle B$, and by $ E$ the intersection between $ AB$ and the angle bisector of $ \angle C$. $ M$ is the foot of the altitude from $ A$ to $ BD$, and $ N$ is the foot of the altitude from $ A$ to $ CE$. Prove that: $ MN\equal{}\frac{b\plus{}c\minus{}a}{2}$.

[size=150]considering O=BD intersection CE. since ang.AMO and ang. ANO are right angles. therefore AMNO is a cyclic quadrilateral.
THus MN= MO.AN+AM.NO/(AO) Ptolemy's theorem
            = c .sinACE. MO + b.sinABD.NO/(AO)
            let ang. ABD = o. angle ACE= a
             again MO= AO.sina and NO= AO sino because NAO= o and MAO= a
            AO cancels from both nr. and dr.
            = c.sin^2a + b.sin^o
            = c/2 . 1-cos2a + b/2. 1-cos2o
            as 2o and 2a are angle B and angle C resp. from ques as O is incentre
            = b+c-(b.cos B+ c.cosC)/(2)
            as  bcosB + c.cos C is a or BC. easily
            = we thus get b+c-a/(2) [/size]
   How can i write mathematical terms while posting?

Use [b]Latex[/b]. For example, if you want to write $ x^2$ then write $ \$$ x^2 $ \$$.

[url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideCommands.php][b][color=orange]Here[/color] [/b][/url] are some commands.

If AM, AN cut BC at P and Q, respectively, CAQ and ABP are isosceles triangles, while M and N are the midpoints of AP, AQ respectively, hence MN is half of PQ. From isosceles triangles BP = AB = c, CQ = AC = b, thence PQ = b + c - a and MN = (b + c - a)/2.

Best regards,
sunken rock