calculus
inequalities proposed
inequalities

Let $x$, $y$, $z$ pozitive reals such that: $\frac{x^{2}}{y^{2}}+\frac{y^{2}}{z^{2}}+\frac{z^{2}}{x^{2}}=\alpha , \alpha\geq 3.$
Find the maximum value of: $\frac{z-y}{x}+\frac{x-z}{y}+\frac{y-x}{z}.$

[quote="neworder"]I'm not sure if it leads anywhere, but try substituting $\frac{x}{y}=a, \frac{y}{z}=b, \frac{z}{x}=c$, so you have: $abc=1, a^{2}+b^{2}+c^{2}=\alpha$, and you have to find the maximum value of $a+b+c-(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$, which is equivalent (with abc=1) to $a+b+c-ab-bc-ca$.[/quote]

Now if we put $t=ab+bc+ca$, by a well-known result, we have $t\leq \alpha$, and by AM-GM, $t\geq3$. Let $f(t)=\sqrt{\alpha+2t}-t$. We need to maximize $f(t)$ for $\alpha\geq t\geq 3$.

Basic calculus proves that f(t) is strictly decreasing, so it achieves its maximum for t=3, and then $f(t)=(\alpha+6)^{\frac{1}{2}}-3$.


The Incircle ⊙$ I$ of Δ$ ABC$ slice $ BC$ at $ D$，the nine points circle---⊙ω of Δ$ ABC$ meet $ BC$ at $ E,F$；$ DI$ insecet ⊙$ I$ at $ G$ .
$ Fe$ is the Feuerbach point，$ E$ is the midpoint，$ L$ is on $ BC$，and $ EL \equal{} DE$；$ GE,GF$ separately insecet ⊙$ I$ at $ M,N$ .
And $ DM,DN$ separately insecet $ IE,IF$ at $ P,Q$，$ IS$ is the diameter of ⊙$ IEF$，Prove that：

$ (1): Fe,D,S$ is collinear .
$ (2):$ Let $ DFe$ meet ⊙ω at $ T$，then $ IT$⊥$ PQ$ .
$ (3):$ Let $ DFe$ meet $ PQ$ at $ W$，$ EW,FW$ separately insecet $ IF,IE$ at $ V,U$，Then ⊙ω，⊙$ I$，⊙$ UVW$ are tangent in $ Fe$ .
$ (4): AI$ insecet $ BC$ at $ J$，let $ JFe,TL$ insecet at $ K$，$ AK$ insecet ⊙$ I$ at $ X$，then $ XFe$⊥$ KFe$ .

Propositions (1),(2) and (3) follow from this general configuration:

In a triangle $\triangle ABC,$ circle $(J)$ is internally tangent to its circumcircle $(O)$ at $A$ and tangent to $BC$ at $D$ (it's well-known that $AD$ is internal bisector of $\angle BAC$). $G$ is the antipode of $D$ on $(J)$ and $GB,GC$ cut $(J)$ again at $M,N.$ $DM,DN$ cut $JB,JC$ at $P,Q.$ Then we have:
[b]a)[/b] $A,D$ and the antipode $S$ of $J$ on $\odot(JBC)$ are collinear.
[b]b) [/b]If $T$ is the midpoint of the arc $BC$ of $(O),$ then $JT \perp PQ.$
[b]c) [/b]If $W \equiv AD \cap PQ,$ $U \equiv JB \cap CW$ and $V \equiv JC \cap BW,$ then $\odot(UVW)$ is tangent to $(O),(J)$ at $A.$

[b]a) [/b]Redefine $U$ and $V$ as the projections of $D$ on $JB,JC.$ From the problem [url=http://www.artofproblemsolving.com/community/c6h1137864]Small problem about Incenter[/url] (post #6), we get that $B,C,U,V$ lie on a same circle with center the midpoint $T$ of the arc $BC$ of $(O).$ Thus the perdendicular bisector of $CV$ is the midparallel of $DV \parallel SC$ and the perpendicular bisector of $BU$ is the midparallel of $DU \parallel SB$ $\Longrightarrow$ $T$ is the midpoint of $DS$ $\Longrightarrow$ $A,D,T,S$ are collinear.

[b]b)[/b] Since $BMUD$ is cyclic on account of the right angles at $M,U,$ then $PB \cdot PU=PD \cdot PM$ $\Longrightarrow$ $P$ has equal power WRT $(J)$ and $\odot(BUVC)$ and so does $Q$ similarly $\Longrightarrow$  $PQ$ is radical axis of $(J)$ and $\odot(BUVC)$ perpendicular to their center line $JT,$ i.e. $JT \perp PQ.$

[b]c)[/b] Redefine $W \equiv BV \cap CU.$ Since $WB \cdot WV=WC \cdot WU,$ then $W$ has equal power WRT $(J)$ and $\odot(BUVC),$ hence it lies on their radical axis $PQ.$ Now from [url=http://www.artofproblemsolving.com/community/c6h1137864]Small problem about Incenter[/url], we get that $W \in AD$ and $\odot(UVW)$ is tangent to $(O),(J)$ at $A.$

Solution to proposition (4):

Let $(N)$ and $(I_a)$ be the 9-point circle and A-excircle of $\triangle ABC$ externally tangent at the A-Feuerbach point $Fa.$ Obviously $(I_a)$ is tangent to $BC$ at $L.$ Since $(NT \parallel I_aL) \perp BC$ $\Longrightarrow$ $Fa \in TL$ and since $Fe$ is the exsimilicenter of $(I) \sim (N)$ and $J$ is the insimilicenter of $(I_a) \sim (I),$ then by Monge & d'Alembert theorem, $JFe$ passes through the insimilicenter $Fa$ of $(I_a) \sim (N).$ Consequently $K$ is identical with $Fa.$

Let $S$ be the antipode of $K$ on $(N)$ and label $\tau_K,\tau_S,\tau_X$ the tangents of $(N)$ at $K,S,X,$ respectively. Since $A$ is the exsimilicenter of $(I) \sim (I_a)$ and $K$ is the insimilicenter of $(N) \sim (I_a),$ it follows that $\overline{AXK}$ passes through the insimilicenter of $(I) \sim (N)$ $\Longrightarrow$ $\tau_X \parallel \tau_K.$ But obviously $\tau_S \parallel \tau_K$ $\Longrightarrow$ $\tau_X \parallel \tau_S$ $\Longrightarrow$ $X$ and $S$ are homologous points under the direct homothety that takes $(I)$ into $(N)$ $\Longrightarrow$ $Fe,X,S$ are collinear $\Longrightarrow$ $\angle KFeX \equiv \angle KFeS=90^{\circ},$ i.e. $XFe \perp KFe.$


In [u]concave[/u] hexagon ABCDEF, $\angle A = \angle B = \angle C = 90^\circ$, $\angle D = 100^\circ$, and $\angle F = 80^\circ$. Also, $CD=FA$, $AB=7$, $BC=10$, and $EF+DE=12$. Compute the area of the hexagon.

well at first i did: wlog $<F=<D=90$ [we still have the interior angles sum to 180 as before], $2AF+12=17$, $AF=\frac{5}{2}$, area is $70-(10-\frac{5}{2})(7-\frac{5}{2})=\frac{145}{4}$

i guess you can be like <E=90... $AF=x$, then the area is $70-\frac{1}{2}((10-x)(7-x)+DE*EF)$, note $DE^{2}+EF^{2}=(10-x)^{2}+(7-x)^{2}=144-2DE*EF$, substitute, $x$ cancels, $\frac{145}{4}$...

[quote="i_like_pie"][quote="Altheman"]well at first i did: wlog $<F=<D=90$[/quote]
[quote="Allan Z"]$\angle D = 100^\circ$, and $\angle F = 80^\circ$[/quote]
[/quote]

you did not read my post

the point is that all we need is the interior angles to sum to 180, the 80 degrees is a misnomer

the area is constant for any pair of angles that add to 180, this is evident from the second half of my post

so it is easiest to just take 90,90

stuff cancels either way...


We say that a group of $k$ boys is $n-acceptable$ if removing any boy from the group one can always find, in the other $k-1$ group,  a group of $n$ boys such that everyone knows each other. For each $n$, find the biggest $k$ such that in any group of $k$ boys that is $n-acceptable$ we must always have a group of $n+1$ boys such that everyone knows each other.

Do you mind posting the complete set of TST problems and your national olympiad ?  :)

$k=2n-1$. For $k=2n$, there is an obvious counterexample ( two complete graphs with $n$ vertices). For $k=2n-1$, it is a known problem. In Russia, it was proposed by Vladimir Leonidovich Dolnikov to 239 olympiad in 2000-2001 season. As I know, it is even called "a Dolnikov lemma" in graph theoretical community. I wonder to know an Argentinian source.

although it is "known", it is no well-known at all, and is interesting to think about it (in particular, i like my solution, which unfortunately is veeery long). Anyway, thanks for the information.

 To Orl: I posted only this problem because is the only one that is "interesting" to this forum, we can say that the other five are trivial for you, and I'm sure you will find another source of the problems... Our national olympiad takes place in October, remember we are in the South...

[quote="RaMlaF"]

 To Orl: I posted only this problem because is the only one that is "interesting" to this forum, we can say that the other five are trivial for you, and I'm sure you will find another source of the problems... Our national olympiad takes place in October, remember we are in the South...[/quote]

Who says you are forced to post it in the olympiad section?  ;)

[quote="pbornsztein"]Me....
Because they are olympiad problems. Even if they are not that hard...
This is the olympiad section, not the difficult section...

Pierre.[/quote]

OK. Though that should not prevent him from posting the other problems.  :)

So, I wonder, will somebody post a solution?
I found in a book something that is the "official" solution. It uses Hall's theorem, to show that for a minimal $n$ that doesn't satisfy the conditions, there exists a smaller $k$ that also doesn't satisfy the condition, so conclusion!

My solution is quite long, so i don't want to write it here... but I can tell you the basic idea: For each group $A$ of $2n-1$ boys consider $f(A)$ as the number of complete $n$ graphs among that group. Now, if there is a counterexample for some $n$ choose the one that has minimal $f(A)=k$, and using that for any $k-1$ complete $n$ graph from  there there has to be non-empty intersection, one can do some things to show that either the configuration is not a counterexample, or there is a counterexample with $f(A)=k-1$...

it's used for our training also
but i don't know the offical solution
can anyone post it

~bump ,
Can someone write a complete solution?

Can someone plz post a complete solution to this? :help:
Am weeek in combi

2 [url=https://www.omaforos.com.ar/viewtopic.php?p=3157#p3157]solutions [/url] from Argentina

[quote=spacemaven]i play minecraft lol[/quote]

nice same dude


limit
function
real analysis
real analysis unsolved

Does $ \mathop{\lim }\limits_{x\to\infty } \ f(x)\plus{}f'(x) \ < \infty$  imply   $ \mathop{\lim }\limits_{x\to\infty} \ f'(x)\equal{}0$ ? What if f' is continuous?

Assuming $ f$ to be continuous. $ \ f(x) \plus{} f'(x) < \infty$ is given i assume $ |f(x)| < M$ where M forms a bound for $ f(x)$. Thus for an arbitary $ \epsilon > 0$

$ |f(x) \plus{} f'(x)| < \epsilon, \forall x \ge C$

Now,

$ |f(x) \plus{} f'(x)| \le |M \plus{} f'(x)| < \epsilon$

or,

$ |f'(x)| \le \epsilon \minus{} M ,\forall x \ge C$

Since $ \epsilon > 0$ is arbitrary $ \mathop{\lim }\limits_{x\to\infty}\ f'(x) \equal{} 0$ follows. :blush:  :maybe:

 If $ f'(x)$ were continuous the function is monotonically decreasing in whole of its domain. :lol:


kasi mitoone in ro hesab kone? :)
$\int\frac{dx}{\sqrt[6]{x^2+5}}$

intor antegrala fekr nakonam ba dast hal beshan :(

chera hal mishan.hamin zemestoon so'ale payan terme ryiazi yeki az daneshgah ha boode :)


I am using the natbib package but it only show items in the bibliography if I use commands as \citet. How can I force $ \text{\LaTeX}$ to show all bibtex entries in my .bib-file without citing them in my document?

[quote="stevem"]Use \nocite{*}[/quote]

Excellent! Thanks!  :)


function
geometry
circumcircle
number theory
relatively prime
national olympiad

These are problems from 18th Korean Mathematics Olympiad - second round. It was written on September 7th, 2004(I know. This was a long time ago :) . But no one have posted this yet). I didn't work on the problems yet.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=224763#p224763]Problem 1[/url]
For arbitary real number x, f : R->R satisfies $f(f(x))-x^2+x+3=0$. Show that the function f does not exist(R is a set of all real numbers).

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=224776#p224776]Problem 2[/url]
x and y are positive and relatively prime and z is an integer. They satisfy $(5z-4x)(5z-4y)=25xy$. Show that at least one of $10z+x+y$ or quotient of this number divided by 3 is a square number.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=35903]Problem 3[/url]
Positive real numbers, $a_1, .. ,a_6$ satisfy $a_1^2+..+a_6^2=2$. Think six squares that has side length of $a_i$(i=1,2, .. ,6). Show that the squares can be packed inside a square of length 2, without overlapping.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=224817#p224817]Problem 4[/url]
Let k and N be positive real number which satisfy $k\leq N$. For $1\leq i \leq k$, there are subsets, $A_i$, of {1,2,3,...,N} that satisfy the following property.

Property: For arbitary subset of $\{ i_1, i_2, .. , i_s \} \subset \{ 1, 2, 3, ..., k \} $, $A_{i_1} \triangle A_{i_2}   \triangle ... \triangle A_{i_s}$is not an empty set.

Show that a subset $\{ j_1, j_2, .. ,j_t \} \subset \{ 1, 2, ... ,k \} $ exist that satisfies $n(A_{j_1} \triangle A_{j_2} \triangle ... \triangle A_{j_t})  \geq  k$. ($A \triangle B=A \cup B-A \cap B$)

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=224796#p224796]Problem 5[/url]
A, B, C, D are the four different points on the circle O in the order. Let the centre of the escribed circle of triangle ABC, which is tangent to BC, be $O_1$. Let the centre of the escribed circle of triangle ACD, which is tangent to CD, be $O_2$.
(1) Show that the circumcentre of triangle $ABO_1$ is on the circle O.
(2) Show that the circumcircle of triangle $CO_1O_2$ always pass through a fixed point on the circle O, when C is moving along arc BD.

I hope this is the correct way of posting.
-----------------------------------------------------------
I edited this so that the problems are showing. I am not a good translator(I will need some practice). Please ask me if you don't understand any of the problem. I'll try to clearify it. By the way, I am not sure how to add the .pdf file.

If possible in general also include the statements of all the problems in the post above and $\text{\TeX}$ a file that I provided with the problems and then attach the .pdf  (see http://www.mathlinks.ro/Forum/viewtopic.php?t=34216 for more info).


Hello,

I heard that one has to prove Muirhead on a competition, and you can always site AM-GM and other trivial inequalities. So could some one post here which inequalities can be just sited on a contest, and which have to be proved?

Thank you.

I guess you may want to email Prof. Barbeau. :)

yes...please let me know the results too thanks! :lol:

A good rule of thumb is, if a theorem/result appears somewhere in literature, you can probably cite it.  If the theorem/result has a name, cite it by name.  If not, state the theorem/result in full, so the grader knows what you are using.

However, if the problem is equivalent to the theorem/result, then you should probably prove it.

[quote="Prof. Edward Barbeau"]Generally speaking, if it is a named inequality - e.g. arithmetic-geometric means inequality, Cauchy-Schwarz inquality, power means inequality, rearrangement inequality, Minkowski inequality, Holder inequality, application of Jensen's inequality - it is enough to quote it "From the xxx inequality, we see that ..." Sometimes it is prudent to indicate how the situation at hand can be fed into the inequality "Taking xxx as yyy and applying the zz inequality, ... "
          If the inequality is more obscure, even if it has a name, it is wise to quote the inequality explicitly and then apply it. Whether or not you can prove it is a matter of judgment - if time is short, you may have other priorities;
the proof may not be so easy to construct or recall and you should not waste precious time on it. From the marker's point o view, the key thing is whether you are in command of the material. If it is clear that you have the essence of the situation and know what you are talking about, the marker probably will cut you quite a bit of slack.[/quote]

I once used Karamata inequality on the olymon. It was trivial problem if you knew Karamata.

I remember Dr. Barbeau replying back, saying something like 'don't ever do this again.' :rotfl:

So, how do you guys memorize all these formulas.

You either have a good memory of patterns, use them a lot and get used to them, or know why they are true.  The latter one is a much better strategy.

There's not that many to memorize... Just the most important ones -- there's only about 10 such formulas. And yes, memorization comes with a lot of practice.

Alex look at Valentin's (Valentin is an IMO coordinator)  post here (post 14,  )

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=tough&t=2619

which means heavy "Artillery" can be quite handy   :D


inequalities
linear algebra
matrix
combinatorics solved
combinatorics
Matrices

Let A1,A2,..,An be second order matrices with real entries. Prove that there is a choice  \varepsilon i in {1,-1} of the signs + and - such that :

det ( \varepsilon 1*A1+ \varepsilon 2*A2+..+ \varepsilon n*An)>=det(A1)+..+det(An) and there is another choice \varepsilon i' such that the converse inequality holds.

cheers! :D :D

I saw this in Gazeta Matematica, but I never gave it a try. However, I think it might have something to do with the formula (for A and B matrices of the second degree) det(A+B)+det(A-B)=2(detA+detB).

it follows easily using the formula grobber posted above and induction.

for n=2 it is obvious that from det (A+B ) +det (A-B) = 2(det A+ det B) then we either have det (A+B) \leq det A + det B or det (A-B) \leq det A + det B. Some for \geq sign.

suppose it's true for n. let us denote with A = \epsilon i Ai the matrix for \geq sign, and with A' the matrix for \leq sign, and also An+1 = B

Then det (A + B) + det (A - B)  = 2det A + 2det B \leq 2 \sum det A i , and for one of the choices +, - for B we get the one of required inequalities. Using A' we get the other one.


prove that:the medians of a triangle partitionits area into six equal parts.

The medians:
$|AE|, |BD|, |CF|$
Let $O$ be the point of intersections.
Let area of
AOD=a  AOC=b
COE=c  OEB=d
BOF=e  FOA=f
Area of ADB= Area of BDC
Area of AEC = Area ABE
Area of AFC = Area FBC, so

$b+c+d=e+f+a$
$f+a+b=e+d+c$
$c+b+a=e+d+f$

we get from the second and third equation $c=f$
OEC=OEB, $c=d$
OAF=BOF, $f=e$, so $d=e$, and $b=a$
Keep substituting into our three equations and you are done


In triangle $ ABC$, $ 3\sin A \plus{} 4\cos B \equal{} 6$ and $ 4\sin B \plus{} 3\cos A \equal{} 1$. Then $ \angle C$ in degrees is

$ \textbf{(A)}\ 30\qquad
\textbf{(B)}\ 60\qquad
\textbf{(C)}\ 90\qquad
\textbf{(D)}\ 120\qquad
\textbf{(E)}\ 150$

Square both equations, add, and simplify using $ \sin^2{\theta}\plus{}\cos^2{\theta}\equal{}1$ to get $ 25 \plus{}12(\sin{A}\cos{B} \plus{}\sin{B}\cos{A}) \equal{} 37$.

Using the sine addition identity, we simplify this further to $ \sin{(A\plus{}B)} \equal{} 1$, and find that the answer is $ \textbf{(C)}\ 90$

The 12 in your post should be a 24. But I am confused. Both 3sin A+4cos(30-A) and 3sinA+4cos(150-A) never reach 6. Is it a bad problem?

[hide="Click for solution"]
Square both sides of the equations and add the results:

$ 9(\sin^2A\plus{}\cos^2A)\plus{}16(\sin^2B\plus{}\cos^2B)\plus{}24(\sin A \cos B \plus{} \sin B \cos A)\equal{}37$.

Thus, $ 24 \sin (A\plus{}B)\equal{}12$.  Thus $ \sin C\equal{}\sin (180^\circ\minus{}A\minus{}B)\equal{}\sin(A\plus{}B)\equal{}\frac{1}{2}$, so $ \angle C\equal{}30^\circ$ or $ \angle C\equal{}150^\circ$.  The second one is impossible because it implies that $ A < 30^\circ$ and thus $ 3 \sin A\plus{}4 \cos B<3 \times \frac{1}{2}\plus{}4<6$, which is clearly incorrect.  Thus, $ \angle C\equal{}30^\circ$, or $ \boxed{\textbf{(A)}}$.
[/hide]


Find conditions for which the quadratic equation $ x^{2} \minus{}bx\plus{}c\equal{}0$ has real roots which are less than $ 1$ in magnitude.

This gives $ b^{2} \ge 4c$ and

$ \minus{}1 < \frac{b\minus{}\sqrt{b^{2}\minus{}4c}}{2}$ and $ \frac {b\plus{} \sqrt{b^{2}\minus{}4c}}{2} < 1$.


Let $ ABCD$ be a quadrilateral which has an incircle centered at $ I$.Prove that:
$ 2(IA.IC\plus{}IB.ID) \le AB.CD\plus{}AD.BC$

According to [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=331609[/url] we have
\[ IA\cdot IC+IB\cdot ID=\sqrt{AB\cdot BC\cdot CD\cdot DA} \]in a quadrilateral with an incircle. Now use the AM-GM inequality
\[ \sqrt{AB\cdot BC\cdot CD\cdot DA} \le \frac{AB\cdot CD+AD\cdot BC}{2} \]and your inequality follows.


(i) Show, with the aid of a sketch, that y > tanh (y/2) for y > 0 and deduce that
                      arcosh x > (x-1)/(sqrt(x^2 - 1)) for x > 1                                      (*)
(ii) By integrating (*), show that arcosh x > 2(x-1)/(sqrt(x^2 - 1)) for x > 1.
(iii) Show that arcosh x > 3(sqrt(x^2 - 1))/(x+2) for x > 1.
[Note:arcosh is another notation for cosh^-1(x).]

I drew the graphs of y=x and y=tanh x, showed that tanh(x/2) = 1 - 2/(exp(x) +1) which approaches 1 as y increases and that its derivative is <1 to show that y > tanh(y) for y > 0, that is x-1 > tanh((x-1)/2) for x >1, that is, x > 1+ tanh ((x-1)/2). This approaches 2 as x increases whereas cosh ((x-1)/(sqrt (x^2 -1)) approaches cosh 1 < 2 and both start at (1,1) and have positive derivatives.

Integrating by parts, (ii) is easy and (iii) easier. I'm bothered by my answer to (*): I can also do it by differentiating both sides and then starting by proving the resulting inequality, and integrating both sides of it to get (*). Perhaps I should have posted the question in College Playground as it involves calculus.


function
logarithms
real analysis
real analysis unsolved

Assume $ f(e^{x})\equal{}1\plus{}x$ , $ (0<x<\infty)$

if $ f[\varphi(x)]\equal{}1\plus{}x\plus{}\ln x$

find $ \varphi(x)$ :roll:

consider $ \varphi(x)\equal{}e^{x\plus{} \ln x}$

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

so $ f[\varphi(x)]\equal{}1\plus{}x\plus{} \ln x$

It seems right

but....Is there any other functions satisfy the condition?

or can we proof the function given above is the only solution? :idea:

Well, since $ f(e^x) \equal{} 1 \plus{} x$ in your interval we also have $ f(e^{\ln x}) \equal{} f(x) \equal{} 1 \plus{} \ln x$. Now you can substitute your second equation in directly and compare with your axiom that $ f(\varphi) \equal{} 1 \plus{} x \plus{} \ln x$ to get your answer.


hi! It should be consecutive positive EVEN integers.

[hide="My Solution"]
Consider the sequence of consecutive integers starting from $ k$ and ending with $ k \plus{} (j \minus{} 1)$. Clearly this sequence has $ j$ terms. Multiplying each term of this sequence by $ 2$, we have a sequence of $ j$ consecutive even integers. The sum of these terms is $ 2\cdot \frac{j(k \plus{} k \plus{} (j \minus{} 1))}{2} \equal{} j(2k \plus{} j \minus{} 1).$ We set this equal to $ 2010$ and attempt to maximize $ j$ by noting that $ 2010 \equal{} 2\cdot 3\cdot 5\cdot 67$. Clearly, $ j < 2010$ and $ j | 2010$ so we try $ j \equal{} 67$. It follows that $ k \equal{} 972$, which is valid. Thus, the answer is $ \boxed{67}$.
[/hide]

[quote="orangefronted"]If 2010 is expressed as a sum of consecutive positive even integers, what is the maximum possible number of terms?[/quote]
This should just be:
"..., what is the number of terms?" **

[quote="umesh123"]
first term
=38
n=number of terms=30
[/quote]
Supportive work for this user's conclusion:

Let $ 2(n)$ = the smallest term
Add $ 29$ to $ n$ to get the $ 30th$ term:
Then $ 2(n \plus{} 29)$  = the largest term
Let $ S$ = the sum of the terms

$ S \equal{} 2(n) \plus{} . . . \plus{} 2(n \plus{} 29)$
$ S \equal{} 2(n \plus{} 29) \plus{} . . . \plus{} 2(n)$
--------------------------------------
$ 2S \equal{} 2(2n \plus{} 29) \plus{} . . . \plus{} 2(2n \plus{} 29)$ ---> There were $ 30$ pairs added to get these $ 30$ sums.

$ 2S \equal{} [2(2n \plus{} 29)](30) \equal{} 2010(2)$

$ 60(2n \plus{} 29) \equal{} 4020$

$ 2n \plus{} 29 \equal{} 67$

$ 2n \equal{} 38$

The number of terms is $ 38$, because $ 2n$ is the first term.

**  There is only one $ n$ value that works for the given conditions.

[quote="TachyonPulse"][hide="My Solution"]
Consider the sequence of consecutive integers starting from $ k$ and ending with $ k \plus{} (j \minus{} 1)$. Clearly this sequence has $ j$ terms. Multiplying each term of this sequence by $ 2$, we have a sequence of $ j$ consecutive even integers. The sum of these terms is $ 2\cdot \frac {j(k \plus{} k \plus{} (j \minus{} 1))}{2} \equal{} j(2k \plus{} j \minus{} 1).$ We set this equal to $ 2010$ and attempt to maximize $ j$ by noting that $ 2010 \equal{} 2\cdot 3\cdot 5\cdot 67$. Clearly, $ j < 2010$ and $ j | 2010$ so we try $ j \equal{} 67$. It follows that $ k \equal{} 972$, which is valid. Thus, the answer is $ \boxed{67}$.
[/hide][/quote]Solution is not correct. Attempting to maximize without explanation may lead to mistakes, which makes this an example.

umesh123 wrote:

first term
=38
n=number of terms=30

What you did to notice that 38 was the first term?

Why is TaychonPulse answer incorrect?

Note that if there are an even number of terms, the mean is odd, and if there are an odd number of terms, the mean is even. Use the factors of 2010 to find the lowest, positive, possible starting value.
[quote="Chalao"]Why is TaychonPulse answer incorrect?[/quote]
His starting term is negative.

(Pre-note. This is not a hard explanation just because it is long, I just dumbed it down a lot for the people who learn slower, because some people might have a hard time understanding this)

Let's say that $ a$ is the number of terms and $ b$ is the mean of the terms. Then
$ ab\equal{}2010$ The object is to determine the maximum value of $ a$.

[u][b]Step 1: What can "a" not excede?[/b][/u]

Because there is only one factor of $ 2$ in $ 2010$, either $ a$ or $ b$ will be even, but not both.

First we figure out what $ a$ cannot excede. We know that $ a$ cannot be greater than $ a$. That is the first maximum value of $ a$. The second maximum value of $ a$ is in terms of $ b$...

If $ b$ is [u]even[/u], then [u]$ a$ cannot exceed $ \frac{b}{2}$[/u] because $ a$ cannot be greater than the amount of even numbers from $ 1$ to $ b$ .
If $ b$ is [u]odd[/u], then [u]$ a$ cannot exceed $ \frac{b\minus{}1}{2}$[/u] for the same reason.

So we have figured out that $ a$ cannot be greater than $ a$ or $ \frac{b}{2}$ if $ b$ is even, and that $ a$ cannot greater than $ a$ or $ \frac{b\minus{}1}{2}$ if $ b$ is odd.

[u][b]Step 2: Possible values for "a" and "b"[/b][/u]

The prime factors of $ 2010$ are $ 2*3*5*67$ . So, $ a$ and $ b$ are just going to different combinations of products of those prime factors (i.e. $ a$ might equal $ 2*3$ and $ b$ would equal $ 5*67$ , or $ a$ be $ 2*5*67$ and $ b$ be $ 3$, because in both, $ ab\equal{}2*3*5*67\equal{}2010$. Note that there are many more groupings not listed).

[u][b]Step 3: Plug and Chug[/b][/u]

So now we just choose values for $ a$ and $ b$ such that $ a$ is not greater than $ a$ nor $ \frac{b}{2}$ if $ b$ is even, and values for $ a$ and $ b$ such that a is not greater than $ a$ nor $ \frac{b\minus{}1}{2}$ if $ b$ is odd, keeping in mind that $ a$ and $ b$ are just different grouping of the numbers $ 2*3*5*67$.

We want to choose values of $ a$ and $ b$ such that both of the maximum values of $ a$ are as large as possible. Keep in mind that $ a$ must be less than or equal to both of the maximums. If one maximum is much larger than the other, that will cause the other maximum to be small, which will cause the maximum value of $ a$ to be small. So we want both of the maximums to be as close as possible.

First, we make $ b$ even, so $ b$ is going to have the factor of $ 2$. In order to make the maximums $ a$ and $ \frac{b}{2}$ as large as possible, we want to make them as close to equal to each other as possible. So we can make $ a\equal{}3*5$ and $ b\equal{}2*67$. After plugging the values in, we find that $ a$ cannot be greater than $ 15$ or $ 67$ when $ b$ is even, so the maximum value of $ a$, when $ b$ is even, is $ 15$.

Second, we make $ b$ odd, so $ a$ is going to have the factor of $ 2$. In order to make the maximums $ a$ and $ \frac{b\minus{}1}{2}$ as large as possible, we want to make them as close to equal to each other as possible. So we can make $ a\equal{}2*3*5$ and $ b\equal{}67$. After plugging the values in, we find that $ a$ cannot be greater than $ 30$ or $ 33$, so the maximum value of $ a$, when $ b$ is odd, is $ 30$.

Finally, of both situations, $ a$ has the highest maximum value when $ b$ is odd, which is $ \boxed{30}$.

[i]Post if this doesn't make sense..[/i]

[quote]If b is even, then a cannot exceed \frac{b}{2} because a cannot be greater than the amount of even numbers from 1 to b .
If b is odd, then a cannot exceed \frac{b-1}{2} for the same reason. [/quote]

Excuse me for being stupid, but I didn't understand that part.


If G of a chemical reactio is negative and the change in entropy is positive, what can one conclude about the reaction?
a) It is endergonic
b) It is exergonic
c) It requires energy
d) It is anabolic
e) It will not reach equilibrium

Answer: b
Can you please explain this to me?

A reaction is called exergonic if $ G$ is negative.  And you are given that $ G$ is negative.

energy is released so it wud b EXergonic--like RESPIRATION

Oh so G stands for free energy? And negative G represents a release of energy?

Okay, I got it. Thank you.

Instead of $ G$ we should have there $ \Delta G$.

G=G-product-G-reactants. if products have lower energy than reactants ,  G will be negative. this happens when reaction give energy outside,so that reaction is exergonic.


Given $ a, b ,c ,d >0$ and $ a^2\plus{}b^2\plus{}c^2\plus{}d^2 \ge \frac{a^3}{b^3}\plus{}\frac{b^3}{c^3}\plus{}\frac{c^3}{d^3} \plus{}\frac{d^3}{a^3}$
Prove that:
(i) $ a\plus{}b\plus{}c\plus{}d \ge \frac{1\plus{}a}{1\plus{}b}\plus{}\frac{1\plus{}b}{1\plus{}c}\plus{}\frac{1\plus{}c}{1\plus{}d}\plus{}\frac{1\plus{}d}{1\plus{}a}$
(ii) $ \frac{ab\plus{}1}{a\plus{}1}\plus{}\frac{bc\plus{}1}{b\plus{}1}\plus{}\frac{cd\plus{}1}{c\plus{}1}\plus{}\frac{da\plus{}1}{d\plus{}1} \ge  \frac{4(1\plus{}\sqrt{abcd})}{1\plus{}\sqrt[4]{abcd}}$
Bai de so voi anh nhung hi vong anh thich no  :)

[quote="nguoivn"][quote="quykhtn-qa1"]Given $ a, b ,c ,d > 0$ and $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \ge \frac {a^3}{b^3} \plus{} \frac {b^3}{c^3} \plus{} \frac {c^3}{d^3} \plus{} \frac {d^3}{a^3}$
Prove that:
(ii) $ \frac {ab \plus{} 1}{a \plus{} 1} \plus{} \frac {bc \plus{} 1}{b \plus{} 1} \plus{} \frac {cd \plus{} 1}{c \plus{} 1} \plus{} \frac {da \plus{} 1}{d \plus{} 1} \ge \frac {4(1 \plus{} \sqrt {abcd})}{1 \plus{} \sqrt [4]{abcd}}$
[/quote]
This ineq is obviously trues because $ abcd \geq\ 1$ and by Am-Gm : $ LHS \geq\ 4 \geq\ RHS$  :maybe:[/quote]
No ,it is false $ RHS \ge 4$  :wink:

[quote="nguoivn"][quote="quykhtn-qa1"]Given $ a, b ,c ,d > 0$ and $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \ge \frac {a^3}{b^3} \plus{} \frac {b^3}{c^3} \plus{} \frac {c^3}{d^3} \plus{} \frac {d^3}{a^3}$
Prove that:
(i) $ a \plus{} b \plus{} c \plus{} d \ge \frac {1 \plus{} a}{1 \plus{} b} \plus{} \frac {1 \plus{} b}{1 \plus{} c} \plus{} \frac {1 \plus{} c}{1 \plus{} d} \plus{} \frac {1 \plus{} d}{1 \plus{} a}$
[/quote]
From the hypothesis, we are easily to show that $ abcd \geq\ 1.$
So, (i) is obviously trues because:
$ \sum\ \frac {b(1 \plus{} a)}{1 \plus{} b} \geq\ 4$
[hide]Hiz, thuc. ra anh chi? thjk cac bdt dang. chuan?, dieu kien don gian, nhat la` dau bang` (2,1,0) va` (1,2,3) la` anh thjk nhat.
Bai` na`y gia thiet cua em mang tinh' "che giau" hoi nhieu`  :huh:  [/hide][/quote]
Nice proof
[hide] em co bai ma dau =(2;1;0) nhung de. http://www.mathlinks.ro/viewtopic.php?t=264218 :)  [/hide]

[quote="nguoivn"][quote="quykhtn-qa1"]

No ,it is false $ RHS \ge 4$  :wink:[/quote]
Sr, I'm so hurry. But I realize it's an old problem in "sang tao bdt". And I think it isn't easy.  :maybe:
[hide]Thuc su la` anh cung~ chua tim` ra loi` giai? nao` cho bai` toan nay`. Vi` anh dot' bdt 4 bien lam' em a.  :oops: [/hide][/quote]
yes, it is an old problem in "sang tao BDT"" :lol:
[hide] anh ko can khiem ton the dau :)  [/hide]


How can I prove this inequality using mathematical induction? What I find hard is that the RHS is a constant and not a term depending on n:

$ 1/1^2 \plus{} 1/2^2 \plus{} 1/3^2 \plus{} ...\plus{} 1/n^2 < 1.674$

[hide="hint"]

Prove a stronger Inequality

[hide="further hint"]

Prove

$ 1/1^{2}\plus{}1/2^{2}\plus{}1/3^{2}\plus{}...\plus{}1/n^{2}< 1.674 \minus{} \frac{1}{n\plus{}\frac{1}{2}}$

[/hide]

[/hide]

[hide="explanation sort-of"]We want to prove
$ \sum\frac {1}{n^2} < K$ where $ K \equal{} 1.674$.
In order to use induction, we must strengthen this inequality. A plausible attempt is to prove something like this:
$ \sum\frac {1}{n^2} < K \minus{} f(n)$
where $ f(n) \geq 0$ and $ \frac {1}{(n \plus{} 1)^2} \leq f(n) \minus{} f(n \plus{} 1)$. If such an $ f(n)$ could be found to satisfy the base case, then we are done. An educated guess is $ f(n) \equal{} \frac {1}{n \plus{} c}$ where $ c$ is some constant. Plugging this in, $ (n \plus{} c)(n \plus{} c \plus{} 1) < (n \plus{} 1)^2$ which simplifies to $ (2c \plus{} 1)n \plus{} (c^2 \plus{} c) \leq 2n \plus{} 1$, which gives $ c \equal{} \frac {1}{2}$ to be the best $ c$. Indeed, this holds for $ n \equal{} 1$, and we are done.
[/hide]


geometry
perimeter
pigeonhole principle
inequalities
congruent triangles
arithmetic series

How many different (i.e. non-congruent) triangles are there with integer side lengths and a perimeter of 2003?  (A mirror image is not distinct from the original.)

I got 83834 for my final answer using loss of generality and systematic listing, but is there any easier way to arrive at an answer.  I'm not even sure if I'm right, I didn't check yet.  I would have used combinations if it was more like x+y+z=50 and finding out how many solutions (x,y,z) there were, but in this case, loss of generality means I'm using x:le:y:le:z.  Is there an easier way to do these types of problems?

Actually, what I had up here before wasn't such a good idea -- instead, try looking for a pattern with the largest of the three numbers -- that is, set x :le: y :le: z, and then look for a pattern in the number of values when z varies.

Why isnt this correct then?

We have x + y + z = 2003.
x + y > z.
x  :le: y :le: z.
By pigeonhole, we get that z is at least 668.
And by the first inequality, we ahve 2003 - z > z, so 2003 > 2z. So z  :le: 1001. Now using x + y = 2003 - z, we have 1002  :le:  x + y  :le: 1335. Our number of solutions for x + y = 1002 is going to be (1001c1 + 1) / 2 = 501. Same number of solutions for x + y = 1003 (the middle term doesnt have to be counted again to divide by 2 this time, so theres just 1002/2 terms). Then when we get to 1334, the solution (by the same argument) is 667 and same for 1335. So our answer is 2((667(668))/2 - (500(501))/2), equals 667(668)-(500)(501) = 195056.


DPatrick said "It is strongly recommended that you not reveal your real name on the site."  So take that into consideration.

Some of us have our usernames fankly show our names, like my own (Alex from NY) and dingzhou's (Ding from TX), while others aren't as obvious.  if you wish to share your name, you may.

If you want to keep your privacy, that's fine too, but you will be left out.
DO NOT tell other people's usernames unless they give themselves away.
Oh, and please DO tell what state you're from.

To any mods: if this breaks some sort of code, then, by any means, remove this.

Disclaimer: This is optional.  Don't do it you don't want.  Neither AOPS nor I am responsible for your actions.

I was reading this, wondering, "how many people actually think my username was just a random combination of letters?"  :lol:

I'm Yongyi Chen, from MA.

I thought your username was random letters too.  :rotfl:

And I thought your username after the "alex" was random letters too :rotfl:  :rotfl:

My real name is Runpeng Liu. Not Tim!!!!! surprised? lol

Actually, I thought your first name was Tiny and your last name was Tim. :rotfl:

[quote="dynamo729"]My name is abcdefgh.  (Thaddeus) DONT TELL NO ONE

.... Did you believe me? lol No its ******. I'm not giving my name, sorry.[/quote]

I believe that there is some crazy mathematical formula that will map each character of your username to the corresponding letter of your real name.

my screen name and my real name share 1 letter. Problem: FIND MY REAL NAME!!!!

NOOOO! The MC site has my name....... oh welll.........I'm not from alabama (supposedly alabama is spelled wrong?) and I didnt get third. (sigh. I'm the only one without a totally cool last name.......)

I wonder if you can guess what MY name is...hahaha i won't tell anyone unless i make nats next year.... :wink:

It is [b]strongly recommended[/b] that you not reveal your real name on the site.

My username doesn't have my first name, just both of my parents' last name. Of course I have one of their last name, can you guess which one?

shen?

EDIT! the message is too small :P

[quote="alexhhmun"]Let me take this opportunity to apologize to any who think name-sharing online is bad, or did not make Mathcounts nationals.

I would just like to point out that my AOPS friends are the only new friends I've made in years.

I want the winning guy's autograph.   :P[/quote]

lol during ARML, we saw kevin chen and someone on our team asked him for his autograph.  :rotfl:

[quote="NightNinja94"]I didnt make nats can you people stop rubbing it in my face?[/quote]

Well don't read the nats-makers' topics if you don't want to be reminded you didn't make nats.

You won't get my name ever unless you take the nine out of my username and turn it backwards to get David Lu.

[quote="NightNinja94"]I didnt make nats can you people stop rubbing it in my face?[/quote]

What gives you the idea that we were even rubbing it in your face? If you don't want to see national topics, get out of the mathcounts forum.  If you're in 8th grade and didn't make nats, just get out.  It's harsh, but if you get mad at the people who actually made nats, and you didn't, there's no point to even visit this forum.

May I wonder who locked this topic and why? I think it was an admin, but I don't know who or why and I'd like to know. Thanks :).


geometry
incenter
circumcircle
trigonometry
inequalities
trig identities
Law of Sines

If $I$ is the incenter of a triangle $ABC$ and $R$ is the radius of its circumcircle then \[AI+BI+CI\leq 3R\]

We have assertions following:
1)$r=IA\cdot \sin{\frac{A}{2}}=IB\cdot \sin{\frac{B}{2}}=IC\cdot \sin{\frac{C}{2}}$.
2)$r=4R\cdot\sin{\frac{A}{2}}\cdot\sin{\frac{B}{2}}\cdot\sin{\frac{C}{2}}$.
3)$\sin{\frac{A}{2}}\cdot\sin{\frac{B}{2}}+\sin{\frac{B}{2}}\cdot\sin{\frac{C}{2}}+\sin{\frac{C}{2}}\cdot\sin{\frac{A}{2}}\leq\frac{3}{4}$.

Now, problem is trivial.

[quote="andy_tok"]Nice, but how did you get 2)?[/quote]Because $r=\frac{S}{p}=\frac{abc}{2R(a+b+c)}=\frac{8R^{3}\cdot\prod\sin{A}}{4R^{2}\cdot\sum\sin{A}}$.

[quote="N.T.TUAN"][quote="andy_tok"]Nice, but how did you get 2)?[/quote]Because $r=\frac{S}{p}=\frac{abc}{2R(a+b+c)}=\frac{8R^{3}\cdot\prod\sin{A}}{4R^{2}\cdot\sum\sin{A}$.[/quote]
Nice, thanks :)

From the Law of Sines we get
\[AI=\frac{r}{\sin \frac{\alpha}{2}}=\frac{r\sqrt{bc}}{\sqrt{(p-b)(p-c)}}=\frac{\sqrt{bc(p-a)}}{\sqrt{p}}.\]

Using the principle of duality, we write $a=y+z, b=x+z, c=x+y$,
where $a,b,c$ represent the triangle's sides. Therefore
$p-a=x,p-b=y,p-c=z, p=x+y+z,$ and
\[R=\frac{abc}{4S}=\frac{abc}{4\sqrt{p(p-a)(p-b)(p-c)}}=\frac{(x+y)(y+z)(x+z)}{4\sqrt{xyz(x+y+z)}}.\]
Thus the inequality is equivalent to
\[\frac{\sqrt{x(x+y)(x+z)}}{\sqrt{x+y+z}}+\frac{\sqrt{y(y+x)(y+z)}}{\sqrt{x+y+z}}+\frac{\sqrt{z(z+x)(z+y)}}{\sqrt{x+y+z}}\leq \frac{3(x+y)(y+z)(x+z)}{4\sqrt{xyz(x+y+z)}}.\] Simplifying we get
\[x\sqrt{yz(x+y)(x+z)}+y\sqrt{xz(y+x)(y+z)}+z\sqrt{xy(z+x)(z+y)}\leq\]
\[\leq \frac{3}{4}(x+y)(y+z)(z+x).\] Using the AM-GM inequality we have
\[\sqrt{yz(x+y)(x+z)}=\sqrt{(yx+yz)(zx+zy)}\leq \frac{xy+2yz+zx}{2}.\] Analogously, \[\sqrt{xz(y+x)(y+z)}\leq \frac{xy+yz+2zx}{2}\text{ , }\sqrt{xy(z+x)(z+y)}\leq \frac{2xy+yz+zx}{2}.\] Thus it is enough to prove that
\[x(xy+2yz+zx)+y(xy+yz+2zx)+z(2xy+yz+zx)\leq \frac{3}{2}(x+y)(y+z)(z+x),\] which is equivalent to \[2(\sum_{sym}(x^{2}y+y^{2}x)+6xyz)\leq 3(\sum_{sym}(x^{2}y+y^{2}x)+2xyz).\] But that is
\[6xyz \leq \sum_{sym}(x^{2}y+y^{2}x)\] and easily follows from the
AM-GM inequality.


[quote="13375P34K43V312"]A five-digit number is called a mountain number if the first three digits increase and the last three digits decrease. How many five-digit numbers between 60,000 and 70,000 are there?[/quote]
[hide]Is it $336$? If it is I'll post my solution.[/hide]

I got...

[hide="probably wrong cuz preet got something different"]
78
[/hide]

and I got which I know is wrong cuz I'm stupid and the geniuses got different answers [hide]100[/hide]

Well the answer is obviously
[hide]
336
[/hide]
because that's what bpms got.

so true

I think there was a problem like this in intro to counting

Okay, I'll show my solution
[hide]Clearly the first digit is 6. Now we just need to count the number of ways to make a four number string since for every four digit string there is one arrangement of the numbers that works:
3*2*7*6=252. But I'm clearly undercounting since the numbers don't have to be distinct and both 8976 and 6987 work.[/hide]
Dang I'm wrong. Looking at my solution I found about 3 errors plus a multiplying error. See if someone can make a suggestion, I think I'm going in the right direction (though maybe not).

[hide="This is most likely wrong..."]
I got 77, ranging from 67821 to 68987[/hide]
Edit: I attached the numbers in an excel spreadsheet
Another Edit: I now have 100 numbers because I forgot to use "0"

[quote="mathgeniuse^ln(x)"][quote="13375P34K43V312"]A five-digit number is called a mountain number if the first three digits increase and the last three digits decrease. How many five-digit numbers between 60,000 and 70,000 are there?[/quote]

[hide]Okay, we have 6 in the first place automatically.

Then we know that there are 3 ways to put digits in the next two places.

If we do that, then either 8 is the middle or 9 is the middle.

If 8 is the middle there are $\binom{8}{2}=28$ choices

If 9 is the middle there are $\binom{9}{2}=36$ choices.

So, the number with 8 in the middle is $1*\binom{8}{2}=28$

The number with 9 in the middle is $2*\binom{9}{2}=72$

So the answer is 100

$100$[/hide][/quote]
this is exactly how I did it

Yes, ragnarok23's solution is correct.
Nice job.

I just saw a problem similar to this but with valley numbers. Here it is: A valley # is a 7-digit # such that the second digit is greater than the first and third, the sixth greater than the fifth and last, and the fourth less than the third and fifth. How many valley #s are there between 5,000,000 and 6,000,000?


ok, this is the same as the old version, except we don't use random hide tags, we use logic. And defenitely no long chains of hide tags and no posts of just click to reveal hidden content. ok, I will start.

remember the old marching scheme? left, right, left, right. now start at the beginning and you will have your ninja

[hide="left"] :ninja: [/hide][hide="right"] :ninja; [/hide]

left[hide="."]this is pointless, [color=white][size=9][hide=","]:ninja:[/hide][/size][/color]  right?[/hide]

[hide="left"].[/hide][hide="right"].[/hide]

you are supposed to use logic. oh, I would in this post it is the 2nd in binary

_________[hide="_"]  :ninja: [/hide]__________

[color=red][size=84]Please wait until the original thread is dead before making a revival, even if it's altered, and especially if it has to do with something fairly pointless. [/size][/color]


geometry
geometric transformation
homothety
power of a point
radical axis
geometry unsolved

Let a circle G and four circle G1, G2, G3, G4 internally tangent to G. Call (1,2) the radical axis of G1 and G2 and so on.

$ \blacktriangleright 1)$ (1,2), (2,3), (3,4), (4,1) concur iff G1, G2, G3, G4 are tangent exernally to a circle G'(different from G).
$ \blacktriangleright 2)$ call P the point of cuncurrence of (1,2), (2,3), (3,4), (4,1), call C the centre of G and C' the centre of G'. So prove that P, C and C' are collinear.

[b][size=100][color=DarkBlue]LEMMA. - A circle $ (O)$ is given with diameter $ AC$ and let $ O'$ be, an arbitrary point, between $ A,\ O,$ where $ O$ is the center of $ (O).$ We draw an arbitrary circle $ (O'),$ withe center $ O,$ which intersects $ AC,$ at points so be it $ B,\ D$ ( $ B,$ between $ A,\ O'$ ) and let $ (K)$ be, an arbitrary circle tangents externally to $ (O')$ at point $ E$ and also tangents internally to $ (O),$ at point $ F.$ Prove that $ P\equiv AC\cap EF,$ is a fixed point.[/color][/size][/b]

[b][size=100]PROOF.[/size][/b] Let be the point $ Z\equiv (O')\cap PF$ the other than $ E$ and it is easy to show that $ O'Z\parallel KF$ $ ,(1)$ where $ K$ is the center of the circle $ (K),$ because of the point $ E$ is the internal homothety center of the circles $ (O')$ and $ (K),$ as well.

From $ (1),$ and because of $ KF\equiv OF,$ we conclude that the point $ P\equiv OO'\cap ZF$ is fixed, as the internal homothety center of the circles $ (O),\ (O')$ and the proof of the [b][size=100]Lemma[/size][/b] is completed.

[b][size=100]REMARK.[/size][/b] - It is useful for the solution of the proposed problem, to say that $ P$ is the point of intersection of $ AC,$ from the radical axis of the circles $ (O),\ (O').$

Also, it is easy to show that the points $ C,\ D,\ E,\ F,$ are concyclic ( from $ O'Z\parallel OF$ $ \Longrightarrow$ $ \angle BO'Z \equal{} \angle COF$ $ \Longrightarrow$ $ \angle ZBD \equal{} \angle FCD$ $ \Longrightarrow$ $ \angle ZED \equal{} \angle FCD$ ). Similarly, the points $ A,\ B,\ E,\ F,$ are also concyclic.

Hence, we conclude that $ (PE)\cdot (PF) \equal{} (PA)\cdot (PB) \equal{} (PC)\cdot (PD) \equal{} u^{2}$ $ ,(2)$ ( the circles $ (O),\ (O'),$ are inverse each other with pole $ P$ and power $ u^{2}$ ).

$ \bullet$ Return now in to the proposed problem, based on the above [b][size=100]Lemma[/size][/b], we have that the line segments $ E_{1}F_{1}, \ E_{2}F_{2},\ E_{3}F_{3},\ E_{4}F_{4},$ pass through the point $ P,$ where $ E_{1},\ F_{1},$ are the tangency points of the circle $ (K_{1})$ ( instead of $ (G_{1})$ ), with $ (O')$ externally and also with $ (O)$ internally, respectively and so on.

Because of now the remark above, we have also $ (PE_{1})\cdot (PF_{1}) \equal{} (PE_{2})\cdot (PF_{2}) \equal{} (PE_{3})\cdot (PF_{3}) \equal{} (PE_{4})\cdot (PF_{4}) \equal{} u^{2}$ and so, we conclude that $ P$ is the point of intersection of the radical axes of the circles $ (K_{1}),\ (K_{2}),\ (K_{3}),\ (K_{4}),$ taken per two of them and the proof is completed.

[b][size=100]NOTE :[/size][/b] - It is clear that when two circles $ (O),\ (O')$ as the problem states are given, we can easy to construct the circle $ (K),$ tangented to $ (O')$ exteranally and also tangented to $ (O)$ internally, by the definition of the point $ P,$ as the point of intersection of the line segment $ OO',$ from the radical axis of the given circles.

Kostas Vittas.

Assume that $ \mathcal{G}_1,\mathcal{G}_2,\mathcal{G}_3,\mathcal{G}_4$ are externally tangent to a single circle $ \mathcal{G}'.$ Let $ P$ be the radical center of $ \mathcal{G}_1,$ $\mathcal{G}_2$ and $ \mathcal{G}_3.$ Inversion with center $ P$ and power equal to the power of $ P$ with respect to these three circles,  transforms $ \mathcal{G}_1,$ $\mathcal{G}_2,$ $\mathcal{G}_3$ into themselves and due to  the conformity, the inverse of $ \mathcal{G}$ is $ \mathcal{G'}.$ Thus,  circle $ \mathcal{G}_4$ is double $ \Longrightarrow$ $ P$ is the radical center of  $ \mathcal{G}_3,$ $\mathcal{G}_4$ and $ \mathcal{G}_1$ as well $ \Longrightarrow$ radical axes $ \mathcal{L}_{12},\mathcal{L}_{23},\mathcal{L}_{34}$ concur.

Similarly, if we suppose that four radical axes $ \mathcal{L}_{12},\mathcal{L}_{23},\mathcal{L}_{34}$ and $ \mathcal{L}_{41}$ concur, then inversion with center $ P$ and power equal to the power of $ P$ with respect to them, takes  $ \mathcal{G}$ into a circle $ \mathcal{G}'$ externally tangent to $ \mathcal{G}_1,\mathcal{G}_2,\mathcal{G}_3,\mathcal{G}_4,$ due to the conformity. We know that  the centers of $ \mathcal{G}$ and its inverse $ \mathcal{G}'$ must be collinear with the center of inversion $ P.$

Sorry dear all my friends but, as said a friend of mine [b][size=100]Kostas Elatiotis[/size][/b], there are some mistakes ( that colored red ) in what I wrote.
[quote="vittasko"][b][size=100]REMARK.[/size][/b] - It is useful for the solution of the proposed problem, to say that $ P$ is the point of intersection of $ AC,$ [color=red]from the radical axis of the circles [/color] $ (O),\ (O').$[/quote]
The correct is that $ P,$ is also the point of intersection of $ AC,$ from the radical axis of the circles with diameters $ AB$ and $ CD,$ because $ A,\ B$ and $ C,\ D,$ are the  pairs of points inverse each other, in the mentioned inversion.
[quote="vittasko"][b][size=100]NOTE :[/size][/b] - It is clear that when two circles $ (O),\ (O')$ as the problem states are given, we can easy to construct the circle $ (K),$ tangented to $ (O')$ exteranally and also tangented to $ (O)$ internally, [color=red]by the definition of the point[/color] $ P,$ [color=red]as the point of intersection of the line segment[/color] $ OO',$ [color=red]from the radical axis of the given circles.[/color][/quote]
The correct is, by the definition of the point $ P,$ as the internal homothety center of the given circles ( the definition of $ P$ as the point of intersection of $ AC,$ from the radical axis of the circles with diameters $ AB$ and $ CD$ is also correct, but it is more difficult to construct ).

Sorry again, Kostas Vittas.


Let $ C(I,r)$ be the incircle of the triangle $ ABC$. Denote $ \|\begin{array}{c} \{A_b,A_c\}\subset (BC) \\
\ \{B_c,B_a\}\subset (CA) \\
\ \{C_a,C_b\}\subset (AB)\end{array}$ so that

$ I\in B_cC_b\cap C_aA_c\cap A_bB_a$ and $ \|\begin{array}{c} AI\perp B_cC_b \\
\ BI\perp C_aA_c \\
\ CI\perp A_bB_a\end{array}$. Prove that $ 2r\sqrt 3\le A_bA_c + B_cB_a + C_aC_b\le \frac {a + b + c}{3}\ .$

[hide]$ A_bA_c = BA_c + CA_b - BC$
$ A_bA_c = \frac {IB}{cos\frac {B}{2}} + \frac {IB}{cos\frac {C}{2}} - BC$
$ A_bA_c = \frac {2r}{sinB} + \frac {2r}{sinC} - a$
$ A_bA_c = 4Rr(\frac {1}{b} + \frac {1}{c}) - a$
$ \Rightarrow A_bA_c + B_cB_a + C_aC_b = 8Rr(\frac {1}{a} + \frac {1}{b} + \frac {1}{c}) - (a + b + c)$
$ \Rightarrow A_bA_c + B_cB_a + C_aC_b = \frac {4(ab + bc + ca)}{a + b + c} - (a + b + c)$[/hide]
If ı am not mıscalculated $ A_bA_c + B_cB_a + C_aC_b = \frac {4(ab + bc + ca)}{a + b + c} - (a + b + c)$
So rıght hand leads to $ \frac {(ab + bc + ca)}{a + b + c}\le \frac {a + b + c}{3}$
Whıch ıs obvıously true.
At the left hand
$ 2r\sqrt 3\le \frac {4(ab + bc + ca)}{a + b + c} - (a + b + c)$
$ 4\sqrt 3sr = 4\sqrt 3S \le4(ab + bc + ca) - (a + b + c)^2$
Whıch ıs whell-known
[hide]Squarıng and elemınatıng we get
$ \sum_{cyclic}{(a^3b + b^3a)} \le a^4 + b^4 + c^4 + abc(a + b + c)$
$ 0 \le \sum_{cyclic}{a^2(a - b)(a - c)}$[/hide]


1. A= the sum of the digits of 4444^4444
    B= the sum of the digits of A
    C= the sum of the digits of B

    Find the value of C

2. Find 1000 natural numbers such that their product equals their sum.

I'm sure the 1. appeared recently

The 2. is fun !

[hide]Take 998 '1', one '2' and '1000'.

Perhaps it's not fair to use '1' ??

[/hide]

belenos wrote:[hide]Perhaps it's not fair to use '1' ??
[/hide]

[hide]It's easy to prove that for any number of numbers greater than 2, it isn't possible without the 1's.[/hide]


Let $ (x,y,z)\in{R^3_ \plus{} },m,n\in{R_ \plus{} },$ prove that

$ \sum_{cyc}{\frac {x^m}{y^n \plus{} z^n}}\ge{\sum_{cyc}{\frac {y^m}{y^n \plus{} z^n}}}.$

The triples $ (x^m,\ y^m,\ z^m)$ and $ (\frac{1}{y^n\plus{}z^n},\ \frac{1}{z^n\plus{}x^n},\ \frac{1}{x^n\plus{}y^n})$ are equally ordered and the result follows from rearrangement inequality.

[quote="Martin N."]The triples $ (x^m,\ y^m,\ z^m)$ and $ (\frac {1}{y^n \plus{} z^n},\ \frac {1}{z^n \plus{} x^n},\ \frac {1}{x^n \plus{} y^n})$ are equally ordered and the result follows from rearrangement inequality.[/quote]

 Please further prove that

Let $ (x,y,z)\in{R^{3}_{ \plus{} }},m,n\in{R_{ \plus{} }}, 0\le{k}\le{1},$ then

  $ \sum_{cyc}{\frac {x^{m}}{y^{n} \plus{} kz^{n}}}\ge{\sum_{cyc}{\frac {y^{m}}{y^{n} \plus{}k z^{n}}}}.$


inequalities
triangle inequality
geometry solved
geometry

Let $D$ be a point on the side $BC$ of a triangle $ABC$ such that $AD > BC$. Let $E$ be a point on the side $AC$ such that $\frac{AE}{EC} = \frac{BD}{AD-BC}$. Show that $AD > BE$.

We will work with non-directed segments. Let T be the point on the segment AD such that AT = BC (in fact, since AD > BC, such a point T must exist on the segment AD). Then, TD = AD - AT = AD - BC, and thus,

$\frac{AE}{EC}\cdot\frac{BC}{BD}\cdot\frac{TD}{AT}=\frac{BD}{AD-BC}\cdot\frac{BC}{BD}\cdot\frac{AD-BC}{BC}=1$.

Since the points E and T lie inside the segments AC and AD, respectively, while the point B lies on the extension of the segment CD, we can thus apply the Menelaos theorem to the triangle ADC, and see that the points E, B and T must be collinear. Now, applying the Menelaos theorem to the triangle ATE and the collinear points B, C, D on its sidelines TE, AE, AT, respectively, we obtain $\frac{BE}{BT}\cdot\frac{TD}{AD}\cdot\frac{AC}{EC}=1$, so that $\frac{BE}{BT}=\frac{EC}{AC}\cdot\frac{AD}{TD}$. Now, $\frac{AE}{EC}=\frac{BD}{AD-BC}$ yields

$\frac{AC}{EC}=\frac{AE+EC}{EC}=\frac{AE}{EC}+1=\frac{BD}{AD-BC}+1=\frac{BD+\left(AD-BC\right)}{AD-BC}$
$=\frac{AD-\left(BC-BD\right)}{AD-BC}=\frac{AD-CD}{AD-BC}$,

so that $\frac{EC}{AC}=\frac{AD-BC}{AD-CD}$, and thus

$\frac{BE}{BT}=\frac{EC}{AC}\cdot\frac{AD}{TD}=\frac{AD-BC}{AD-CD}\cdot\frac{AD}{AD-BC}=\frac{AD}{AD-CD}$.

Hence, $BE=BT\cdot\frac{AD}{AD-CD}$. Therefore, the inequality in question, AD > BE, is equivalent to $AD>BT\cdot\frac{AD}{AD-CD}$, what simplifies to AD - CD > BT. But this is almost trivial: By the triangle inequality for triangle BTD, we have BT < BD + TD = (BC - CD) + (AD - BC) = AD - CD, and the proof is complete.

  Darij


Letters $A,B,C,$ and $D$ represent four different digits from 0,1,2,3...9. If $(A+B)/(C+D)$ is an integer that is as large as possible, what is the value of $A+B$?

$\textbf{(A)}\ 13 \qquad
\textbf{(B)}\ 14 \qquad
\textbf{(C)}\ 15 \qquad
\textbf{(D)}\ 16 \qquad
\textbf{(E)}\ 17$

[hide]
Ok so, (8+9)/(0+1) would be the largest, so A+B is [b]17[/b][/hide]

[hide]

Yep, just maximize the numerator and minimize the denominator  .

[/hide]


A is a set of five distinct positive integers, none of which exceeds 9. Prove that the sums of the non-empty subsets of A cannot all be distinct.

[hide]Denote the numbers by $ a,b,c,d,e$ and WLOG assume $ a < b < c < d < e$. The smallest sum is $ a$, and the greatest is $ a \plus{} b \plus{} c \plus{} d \plus{} e$, making at most $ b \plus{} c \plus{} d \plus{} e \plus{} 1\leqslant 6 \plus{} 7 \plus{} 8 \plus{} 9 \plus{} 1 \equal{} 31$ different sums. However, there are $ 2^5 \minus{} 1 \equal{} 31$ different subsets. So, if we actually have less than $ 31$ different sums, there must exist two subsets which yield a same sum. And if we assume we have exactly $ 31$ different sums, then $ b \equal{} 6, c \equal{} 7, d \equal{} 8, e \equal{} 9$, hence $ b \plus{} e \equal{} c \plus{} d$, which means we can't have $ 31$ different sums. QED[/hide]


limit
real analysis
real analysis unsolved

If $A$ is compact, show that sup $A\in A$.

We have $A$ are closed and bound. Choose sequence $\{x_{n}\}\subset A$ such that $\lim_{n\to\infty}x_{n}=\sup{A}$, we're done because $A$ is closed.

Directly from the definition of compactness:

If $\sup A\notin A$, then $A\subset \bigcup_{n=1}^{\infty}(-\infty,\sup A-1/n)$. Choosing a finite subcover, get $A\subset (-\infty,\sup A-1/N)$ for some $N$, a contradiction.


hello friends

 I really need to have Russian math olympiad problems by their solutions for 2000 ,2001,...,2007
 because i am not good in combinatorics and there are a lot of good combinatorics problems in Russian   olympiad  can anyone help me :?:

 thanks

This is the wrong forum for this request. Please post this in the National Olympiads forum.

C'mon guys (russian speaking ML's), help us with this please.

The Russian Mathematics competition questions are on the resources section of mathLinks (its the icon in the right corner) and its under national competition. The solutions are not there but some MathLinkers have attempted to give solutions.

You can find the solutions here in Russian language.

http://olympiads.mccme.ru/vmo/
Всероссийские олимпиады школьников по математике 1993–2006.

There are also solutions for 2007 and 2009, I download before but forget the link, so you have to find yourself. The only missing solution is 2008. I downloaded 2008 but there is some error, the solution is not the final stage Russian Olympiad.


algebra
polynomial
inequalities
induction
analytic geometry
real analysis
real analysis unsolved

I believe (one of) Weyl's theorem(s) is that the sequence $\{\alpha n^{2}\}, n \in \mathbb{Z}$ (where $\{x\}$ denotes the fractional part of $x$) is equidistributed in $[0, 1]$ if and only if $\alpha$ is irrational.

It is also true that if $P(n)$ is any polynomial that takes integer values whenever $n$ is an integer, the sequence $\{\alpha P(n)\}, n\in \mathbb{Z}$ is equidistributed in $[0, 1]$.

I have seen a dynamical proof of the first theorem which can be generalized to prove the second claim as well. (I mean one can construct a dynamical system in which the sequence in question arises, and use the properties of dynamical systems to arrive at the desired result.)

However, I don't remember the classical proof for the theorem, and I wonder if that can be easily generalized to the $P(n)$ case as well. So can anyone prove Weyl's theorem without using ergodic theory? ;)

[quote="harazi"]There exists a proof based on Van der Corput's theorem, but it is far from being easy. Basically, all you need is Weyl's criterion and the cited inequality. It has as consequence that if $x_{n+h}-x_{n}$ is equi-distributed for all $h$, then so is $x_{n}$. This and an inductive proof should provide an alternative proof of this theorem.[/quote]
Wow, that looks like a really cool approach! Is [url=http://mathworld.wolfram.com/vanderCorputsInequality.html]this[/url] the inequality you mean? I will have to try to work out this line of approach over my break next week. I suspect that it somehow is the same idea, just carried out in a much different way.

[quote="harazi"]I don't know ergodic theory (yet...), so I don't know if it is just a variant of a result that you saw in ergodic theory, but I found this approach purely beautiful. And what's even extraordinary is that the monstruous inequality of Van der Corput is just the Cauchy-Schwarz inequality! Again...[/quote]
:D
Unfortunately I have several papers due tomorrow and I can't think of a concise way to present the proof without citing lots of results from ergodic theory that we did earlier in the class. So after my papers are done I can present a sketch of the proof that does not require prior knowledge of ergodic theory.

Really the reason I said it seemed similar was the induction is the same. In the proof my professor gave, we define a transformation on the $d$-dimensional torus that is like $Tx_{j}= x_{j}+c_{j}x_{j-1}+\alpha$. $x_{j}$ are coordinates and $c_{j}$ are constants. Thereby for the $j^{th}$ coordinate we will have a polynomial of degree $j$ in $n$ when we apply the transformation $n$ times. By looking at what the transformation does to the torus we can prove that the sequence $T^{n}0$ is equidistributed in the torus, and that means the projection to each dimension is equidistributed in $[0, 1]$, and each coordinate will be a polynomial of $n$ times $\alpha$. So by inducting on the dimension we induct on the degree of the polynomial for victory.

To be honest, without knowing the proof you mentioned and with my current level of understanding of ergodic theory, I really shouldn't make claims that the important parts of each proof convey the same idea... :P I was just speculating. ;)


isolate x from the rest of the gunk

y=2x-1
y+1=2x
1/2(y+1)=x
1/2y+1/2=x

so

X = 1/2y+1/2

[quote="hsiljak"]Correct, ernie. Funny thing is, that the Q doesn't say "2x-1" but "2x+1". But there's no doubt about the solution, just the question remains, is the typo made in the statement of the question or in the statement of the solution? First option is more probable, since all three answers "have pluses in it".[/quote]

wow... probably was a typo  :D

oh well... i figured it out anyways  :D

u know y?

cuz im special! :D

Um.... on FTW, it says the answer is $x=\dfrac12$...

If you guys are disagreeing with the answer, this problem should be reported.

Srry for double posting, but this time i will actually solve the problem.

[quote="GameBot"]If $ y\equal{}2x\plus{}1$, which of the following equations is true?\\

$ x\equal{}2y\plus{}1$\\

$ x\equal{}\frac{1}{2}y \plus{} 1$\\

$ x \equal{} \frac{1}{2}y \plus{} \frac{1}{2}$[/quote]

If we isolate $x$, we would get $2x = y-1$.

Then we get $x=\frac{y}{2}-\frac{1}{2}$.

This problem is obviously very glitchy.


quadratics
algebra
quadratic formula
special factorizations
difference of squares

I'm not trying to be mean or anything but high school math teachers are not fit to be math teachers (atleast in my school) because i asked my teacher a simple question and she couldnt do it. Heres the question:

Since we were learning abt quadratics i asked her to prove the quadratic formula but the sad thing is that she couldn't.  :huh: She didnt know what i meant by prove it. So i told her solve for x and she used the quadratic formula.  :|

If you need a proof, here it is:

\begin{eqnarray*}ax^2+bx+c &=& a\left(x^2+{b\over a}x+{c\over a}\right)\\ &\stackrel{\text{completing the square}}{=}& a\left(x^2+{b\over a}x+\frac{b^2}{4a^2}-\frac{b^2}{4a^2}+{c\over a}\right)\\ &=& a\left[\left(x+{b\over 2a}\right)^2-\frac{b^2-4ac}{4a^2}\right]\\ &=& a\left[\left(x+{b\over 2a}\right)^2-\left(\frac{\sqrt{b^2-4ac}}{2a}\right)^2\right]\\ &\stackrel{\text{difference of squares}}{=}& a\left(x+{b\over 2a}-\frac{\sqrt{b^2-4ac}}{2a}\right)\left(x+{b\over 2a}+\frac{\sqrt{b^2-4ac}}{2a}\right)\\ &=& a\left(x-\frac{-b+\sqrt{b^2-4ac}}{2a}\right)\left(x-\frac{-b-\sqrt{b^2-4ac}}{2a}\right)\\ &=& a(x-x_1)(x-x_2) \end{eqnarray*}

[quote="Farenhajt"]Sounds terrifying... A teacher who can't derive a quadratic formula? Where does the world go...[/quote]
To AlgebraI teachers that are even worse:

A.
1. Enters Y1=3/5X on a TI-82's graph menu
2. Sees the graph is not linear
3. Teacher's Conclusion: "The calculator can not handle fractions."

B.
1. Enters $\sqrt{7}$ and gets 2.645751311
2. Enters 2.645751311^2 manualy and gets the displayed result of 7.
3. Teacher's Conclusion: "The square root of 7 is rational and Penguin is wrong."

Penguin's conclusions:
I: The teacher can not handle fractions. Using parenthesis would have solved the problem in scenario A.
II: The teacher must be very gulible to believe a TI-82. By entering Ans-7 would have yeilded the result -3.418E-10, indicating that 2.645751311 is not the square root of 7 and solving the problem in Scenario B.

Don't feel bad, my teacher could not derive the quadratic formula, the vertex formula, or alot of the other formulas we learned.

Me 2.
Because I go to a magnet school, I get smart teachers, and several Math competitions to compete in.

Ok guys this should enough ranting on teachers in this [b]forum[/b]. Whoever wants to continue or even start a topic about their own math teachers, please go to Round Table, this forum is for serious problem solving, let's be more responsible, shall we?

A proof:

$ax^2+bx+c$... Then multiply both sides by $4a$:

$4a^2x^2+4abx+4ac=0$. Then, add and rest $b^2$, and we have now

$4a^2x^2+4abx+b^2-b^2+4ac=0$.

Now, look at the first three terms; we have a "trinomio cuadrado perfecto" (I don't know how is it called in english)... Now, we have:

$(2ax+b)^2-b^2+4ac=0$

Now, we got that $(2ax+b)^2=b^2-4ac$

$2ax+b=\sqrt{b^2-4ac}$

As a sqare root have a positive and a negative result, we finally got that

$x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}$ and $x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$

[quote="236factorial"]Just out of curiosity, Farenhajt, how did you write those words over the equal signs?[/quote]This is a question that belongs to another forum as well but I'll answer it for you

[code]\stackrel{236factorial}{=}[/code]

produces $\stackrel{236factorial}{=}$


function
inequalities unsolved
inequalities

$ x, y, z \in [0, 1]$

Find the minimum value of $ \sum_{cyc} (1 \minus{} x \plus{} xy)^2$.

I do not think this problem can be solved beautiful
Max:$ \frac{27}{16}$,If I'm not mistaken.

hello, how did you get that result?
Sonnhard.

I have forgotten this topic
I really want feilz post  solutions your self.
Ps Dr Sonnhard Graubner:I using function convex.
function with variable x is convex,similary y,z.
Thus we find the max.But here isn't nice solution.

[quote="math10"][quote="Sergic Primazon"]$ x \equal{} \frac {1}{2} \plus{} t_1$ , $ y \equal{} \frac {1}{2} \plus{} t_2$ ,$ z \equal{} \frac {1}{2} \plus{} t_3$;

$ LHS \equal{} \frac {27}{16} \plus{} \frac {1}{2}(t_1 \plus{} t_2 \plus{} t_3)^2 \plus{} (t_1t_2)^2 \plus{} (t_2t_3)^2 \plus{} (t_3t_1)^2 \plus{} (t_1 \minus{} t_2)(t_2 \minus{} t_3)(t_3 \minus{} t_1)$[/quote]
I did not think to this equal.you can explain reasons???[/quote]

You can simply prove it  :)


Toss the triangle on the argand plane such that the circumcentre of $ \Delta ABC$ coincides with the origin.

Let the affix of $ A$ be $ Z_1$ and that of G be $ G$.
Then the affix of $ A_1$ will be $ \minus{}Z_1$ and that of $ G_1$ will be $ G\minus{}\frac 23Z_1$
Let the affix of P be $ Z$

equation of $ OG$ : $ Z \bar G\equal{}G \bar Z$
equation of $ AG_1$ : $ Z(\bar {Z_1}\minus{}\bar {G_1})\minus{}\bar Z (Z_1\minus{}G_1)\equal{}\bar {Z_1}G_1\minus{}\bar {G_1}Z_1$

sEE THAT $ \bar {Z_1}G_1\minus{}\bar {G_1}Z_1\equal{}\bar {Z_1}G\minus{}\bar {G}Z_1$

So substituting in the equation of $ AG_1$ : $ Z(\bar {Z_1}\minus{}\bar {G_1})\minus{}Z\frac {\bar G}{G} (Z_1\minus{}G_1)\equal{}\bar {Z_1}G\minus{}\bar {G}Z_1$

simplify the lhs to get  $ 5Z(\frac {\bar {Z_1}G\minus{}\bar {G}Z_1}{3G})\equal{}\bar {Z_1}G\minus{}\bar {G}Z_1$

sO, $ Z\equal{}3\frac G5$

$ \frac {PG}{PO}\equal{}\frac {|Z\minus{}G|}{|Z|}\equal{}\frac {|3\frac G5\minus{}G|}{|3\frac G5|}\equal{}\frac 23$

[quote="turcas_c"]Let $ O$ be the circumcenter of $ \triangle{ABC}$ , and  the point $ A_1$ the diametrically opposite point of $ A$ , on the circle . Let $ G,G_1$ the centroids of $ \triangle{ABC}$ and $ \triangle{A_1BC}$ and $ P \equal{} AG_1 \cap OG$ . Prove that $ \frac {PG}{PO} \equal{} \frac {2}{3}$ .[/quote]
[hide="Solution"]
Let $ M$ be the midpoint of $ BC$.  It follows that $ AO \equal{} A'O$, $ \frac {A'G_1}{G_1M} \equal{} 2$, and $ \frac {AG}{MG} \equal{} 2$.  Now, $ \frac {MG_1}{G_1A'}\cdot \frac {A'O}{AO}\cdot \frac {AG}{MG} \equal{} \frac {1}{2}\cdot 1\cdot 2 \equal{} 1$, and thus by Ceva's Theorem, we see that $ AG_1$, $ MO$, and $ A'G$ concur at a point, call it $ Q$.  Now, we use Mass Points to find $ \frac {PG}{PO}$.  Give $ A'$ a mass of $ 1$, so $ M$ has a mass of $ 2$ and then $ G_1$ has a mass of $ 3$.  Furthermore, $ A$ has a mass of $ 1$ and $ O$ will have a mass of $ 2$.  We also get that $ G$ has a mass of $ 3$.  Since $ G$ has a mass of $ 3$ and $ O$ has a mass of $ 2$, we see that $ \frac {GP}{PO} \equal{} \frac {2}{3}$, as desired. [/hide]

[hide="geometric "]Aply Menelaus in the triangle $ MOA_1$ for the transversal $ \overline{G_1QA}$ where $ M$ is the middle of $ BC$ and $ Q \in AG_1 \cap MO$ $ \implies$ $ \frac{G_1M}{G_1A_1}\cdot\frac{QO}{QM}\cdot\frac{AA_1}{AO}\equal{}1$ $ \implies \frac{QO}{QM}\equal{}1$
Then aply Menelaus in the triangle $ GOM$ for $ \overline{QPA}$ $ \implies$ $ \frac{QO}{QM}\cdot\frac{PG}{PO}\cdot\frac{AM}{AG}\equal{}1$ $ \implies \frac{PG}{PO}\equal{}\frac{2}{3}$[/hide]

See that $GG_1\parallel AA'$ and $\frac{GG_1}{AA'}=\frac{1}{3}$; with $2OA=AA'$ we get $\frac{GG_1}{OA}=\frac{2}{3}$, but $\frac{GG_1}{OA}=\frac{PG}{PO}$, so done!

Best regards,
sunken rock

M is the midpoint of BC.we can prove $AG_1$bisects the line $OG$.
so we use of menelaus theorom:($AG_1$meets$OM$ at point N)
$(ON/PN)(AM/AG)(GP/OP)$=1 thus$(GP/OP)$=2/3
so done it.


[quote=chien than]Let $a;b$ be positive real numbers.Prove that:
'$\frac{a}{\sqrt{a^{2}+b^{2}}}+\frac{b}{\sqrt{9a^{2}+b^{2}}}+\frac{2ab}{\sqrt{a^{2}+b^{2}}\sqrt{9a^{2}+b^{2}}}\leq \frac{3}{2}$[/quote]
Let $a=\frac{\sqrt{x}b}{\sqrt3},$ where $x>0$.
Thus, after squaring twice we obtain:
$$(x-1)^2(75x^2+106x+75)\geq0.$$

[quote=chien than]Let $a;b$ be positive real numbers.Prove that:
'$\frac{a}{\sqrt{a^{2}+b^{2}}}+\frac{b}{\sqrt{9a^{2}+b^{2}}}+\frac{2ab}{\sqrt{a^{2}+b^{2}}\sqrt{9a^{2}+b^{2}}}\leq \frac{3}{2}$[/quote]
[url=https://artofproblemsolving.com/community/c6h299899p1955849]Crux mathematicorum, Proposer: Jack Garfunkel[/url]
[url=https://math.stackexchange.com/questions/2154453/prove-that-fraca-sqrta2b2-fracb-sqrt9a2b2-frac2ab-sqrt?rq=1]here[/url]


algorithm search calculus integration Support

for those of you more advanced in the world of mathematics, what kind of computer do you need for scientific calculation ? I'm starting to look into buying a laptop and see if it's worth it. I use Maple and Mathlab but only use the basics so far. So for those of you who do research in maths, how much power is needed ? and do you prefer mac or pc ?

Well, what do [i]you[/i] want to do with your laptop?

If you just want to use MATLAB and Maple (I prefer Mathematica), or run some rudimentary simulations and simple algorithms, a multicore processor clocked at 2.0 GHz+ and 3GB (or even 2 GB) of memory should be sufficient. If you want to perform a heavy finite-element simulation or other highly intensive numerical processes, it certainly helps to have more memory. If you want to go prime-number hunting, you'd need a supercomputer. Etc.

I sometimes do some number theoretical computations (class numbers etc.), and I can just say: it's never enough. In fact I sometimes exceeded 2GB of RAM, and estimated computations to take days.

For multicores: A lot of programs cannot use them properly, so effectively have just one core! So for some programs a fast clocked single/dual core may be better than a quad- or octocore; this may change in future, but aware of this fact!

But more standard problems like umerical things and equation solving, I never needed them, so have no idea.

While it might not help to get 8 processor cores, it may be very useful to get four. Two is an absolute must. A few comments on multicores:

1. Programs such as the newer versions of Mathematica are adapting to become more efficient at breaking tasks into parallel threads, so they will in fact be able to take advantage of multicore. i cannot speak for MATLAB, but I'm sure it's on their agenda.

2. Having more cores allows you to do other things while running a computation in the background.

Note that you are not necessarily restricted to your own machine for computations. In a university or similar setting, you may have access to a computer system specifically set up for more involved computations.

At this point, just get something you're comfortable with. If you go deeper and find you need more from your computer, upgrade then.

Some pretty cheap laptops out there with 4GB RAM are appearing.
Also, people, notice he said LAPTOP and that means DUAL CORE.  Go search for a 4-core for laptops.
Look in the Dell Outlet or something.

I know Acer and Lenovo definitely have quadcore laptops. Of course, they have a disgusting price tag. But that's today. A few months from now, that will probably start to change.

Regarding PC vs Mac, I like to use PC's.
I remember trying out some things in Mathematica using the "Timing[]" function.
- I did some hardcore 3D graphing, some disgusting differential equations, sums, and integrals, array operations, and some factoring/simplification stuff.
- I compared my laptop, a Dell dv9000 with a 2.4 GHz dual-core processor and 2GB ram, with my math teacher's iMac, with a 2.5 GHz dual-core processor and 2GB ram.
- The general results were that things under 2 seconds were usually (about 2/3 of the time) done a bit more quickly by the Mac, whereas things over 2 seconds were almost always (9/10) done more quickly by my laptop.
- Obviously this is not statistically significant or necessarily representative, but it supports my lean towards using PC's.
- That said, if you want to use Limewire or whatever boho and get yourself a virus on your PC, then your PC's going to suck.


OK wth? What would even make you think that?

I'd have to say no on this one, because mewto and cherry often go into a password protected game and talk for a long time (actaully i dont know that they're talking, but if htey were the same they owuldn't go in the same room for so long(.

ok what's up with these spammy polls?? ik this is ur first one jj but still mewmew is definitely NOT cherry cuz id like to see someone type fast enough so they can enter something at the same time. anyway... i think ftwos needs some cleaning up

ok
first of all, i agree with all the people that have said that all these polls are getting out of hand
plz stop making threads just for the heck of it
it pointless and it's spam
mewto and cherry are [i]not[/i] the same person from my observations

okay sry everyone :blush: . lets stop this topic.

of course mewto and cherrys the same person, why else would mewto and cherry lock themself in a private room?

Many reasons.

Sorry for spamming and bumping.  :blush:


How many different triangles are there if all the sides are integer lengths and the perimeter is 48.

Can you please tell me how you got that many solutions?
If you need the answer here it is, I need an explanation on
how you got all the solutions.

[hide]The answer is 48 different triangles.[/hide]

This post exists in two other places (Classroom Basics & Puzzles and Brainteasers).  Don't double-post!


linear algebra
matrix
algebra
system of equations
algebra unsolved

Can the system of equations
$x_{1}+x_{2}+x_{3}=a_{1}$

$x_{1}+x_{2}+x_{4}=a_{2}$

...

$x_{3}+x_{4}+x_{5}=a_{10}$,

where the equations take all the possible combinations of $x_{1},x_{2},...,x_{5}$, and $x_{i}, a_{i}\in R$ ($a_{i}$'s are fixed values), have more than one set of solutions?

Evidently LHS matrix has rank 5, so this system or has no solutions or has the unique solution depending on the RHS.

Well, I'm not very familiar with matrices.  :blush:

Is there an easier way to look at the problem?

Yes, using  5 equations only we can find unique candidat for solution. But this candidat doesn't obliged to satisfy all 10 equations.

For instance consider the following 5 equations with special RHS's=3 just for simplesity:

$x_{1}+x_{2}+x_{3}=3$
$x_{2}+x_{3}+x_{4}=3$
$x_{3}+x_{4}+x_{5}=3$
$x_{4}+x_{5}+x_{1}=3$
$x_{5}+x_{1}+x_{2}=3$

Their sum gives
(*) $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=5$

Now 1st+4th-(*) gives $x_{1}=1$. Similary all $x_{i}=1$
It is evident now that only possibility for this candidat to satisfy the remaining 5 equations is to have 3 as RHS's in all the remaining equations.


Where it says who is online, why are people who are on FTW considered to be in the Forum Index?

I think it's because FTW isn't part of the forum, it's just another part of the site.  So if you have another window up in the forum, it'll say you're in the forum index.

A lot of times, somebody in FTW will be considered offline in the forum, because there's no other windows up.

It's where you were active in the last 5 minutes.


A city has the shape of a square of side $ n\minus{}1$ divided by roads into $ (n\minus{}1)^2$ unit squares. Some two-way bus routes are to be established in such a way that every route has at most one turn and it is possible to travel between any two crossroads with at most one transfer. How many bus routes at least are necessary to achieve this?


Είχα δει κάτι παρόμοιο παλαιότερα εδώ, οπότε βάζω το συγκεκριμένο πρόβλημα στις ολυμπιάδες.

Έχεις μια βρύση συνεχούς ροής νερού και 2 δοχεία χωρητικότητας ν και κ λίτρων, όπου ν και ακέραιοι και ν>κ. Να βρεθεί ποιες είναι όλες οι ποσότητες νερού που μπορούν να παραχθούν.

Επιτρέπεται να γεμίζεις από τη βρύση και να ρίχνεις από το ένα δοχείο στο άλλο. Δεν επιτρέπεται να πεις «γεμίζω το ένα δοχείο μέχρι τη μέση» γιατί δεν ξέρεις που ακριβώς είναι η μέση ή κάποιο κλάσμα του. Ξέρεις μόνο πότε γέμισε όλο (δηλαδή ν ή κ).

ΥΓ
Ίσως έχετε ακούσει το πρόβλημα με δοχεία πχ 7 και 4 λίτρων που ζητάμε να παραχθεί η ποσότητα 5. Εδώ ουσιαστικά γενικεύουμε το πρόβλημα για κάθε ν και κ.

Εννοείται ότι έχουμε και έναν τρίτο άδειο "κουβά" στον οποίο "δημιουργούμε" την ποσότητα που θέλουμε ή περιοριζόμαστε μόνο σε αυτούς τους 2 κουβάδες? Για να είμαι πιο συγκεκριμένος, οι ποσότητες που παράγουμε είναι μικρότερες του ν (αυτό θα ίσχυε αν θέλαμε να δημιουργήσουμε την εκάστοτε ποσότητα μέσα σε αυτούς τους 2 κουβάδες) ή μπορούμε να παράγουμε ας πούμε και το 2ν (αυτό θα χρειαζόταν τον τρίτο κουβά σοτν οποίο αναφέρομαι)?

Εύχομαι να είναι κατανοητή η ερώτησή μου..

Είναι απόλυτα κατανοητή η ερώτησή σου.
Όχι δεν υπάρχει τρίτο δοχείο. Έχουμε μόνο τα 2 και άρα η μέγιστη ποσότητα που μπορούμε να φτιάξουμε είναι ν. Φυσικά μπορείς να γεμίζεις το ένα και να αδειάζεις το περιεχόμενο στο άλλο.

Για να δώσω κάποια εικόνα ας δούμε πως παράγεται η ποσότητα 5 αν έχουμε 2 δοχεία 7 και 4 λίτρων.

Γεμίζεις το 4λ δοχείο και το αδειάζεις μέσα στο 7λ
Ξαναγεμίζεις το 4λ δοχείο.
Αδειάζεις 3 λίτρα μέσα στο μεγάλο μέχρι να γεμίσει (μένει 1 λίτρο στο μικρό)
Αδειάζεις το μεγάλο δοχείο και ρίχνεις το 1 λίτρο από το μικρό στο μεγάλο (το μεγάλο έχει 1 λίτρο μέσα)
Γεμίζεις το 4λ δοχείο και το αδειάζεις μέσα στο 7λ

Και τώρα το μεγάλο έχει μέσα 5 λίτρα.

Βασικά ζητάμε τη γενική λύση του παραπάνω προβλήματος για τυχαίο ν και κ.

Ουσιαστικά πρέπει να περιγραφεί μαθηματικά το παραπάνω πρόβλημα και μετά είναι άσκηση θεωρίας αριθμών.
Για την ώρα είναι [b]πρόβλημα[/b] θεωρίας αριθμών

ουσιαστικά έχει να κάνει με τη διαδικασία εύρεσης του αντίστροφου ενός αριθμού $ (mod patates )$

Μπορούμε πάντα να πάρουμε ποσότητες της μορφής $ a_1 n \plus{} a_2 k$, όπου $ a_1,a_2$ ακέραιοι. Το πρόσημο $ " \minus{} "$ σημαίνει ότι αδειάζω ένα δοχείο. Αφού $ n\neq k$, άρα ο ελάχιστος αριθμός που μπορείς να πάρεις από τέτοιους γραμμικούς συνδυασμούς είναι ο Μέγιστος Κοινός Διαιρέτης $ (n,k)$ των $ n,k$. Άρα μπορούμε να μετρήσουμε ποσότητες  $ t\cdot (n,k)$, όπου $ t$ θετικός ακέραιος.

Αλέξανδρος

Σωστά Αλέξανδρε  :clap2:

Ότι κι αν κάνεις πάντα η ποσότητα που είναι στο δοχείο έχει σχηματιστεί βάζοντας κάμποσες φορές n και βγάζοντας k ή ανάποδα ανεξάρτητα από τη σειρά που γίνεται αυτό. Τελικά η ποσότητα είναι της μορφής που γράφεις και άρα παράγονται όλες οι ποσότητες που είναι πολλάπλάσια του ΜΚΔ των n και k και μόνο αυτές. Χρησιμοποίησες έτοιμη τη λύση της $ an\plus{}bk\equal{}d$


A 0.3 kg apple falls from rest through a height of 40 cm onto a flat surface. Upon impact, the apple comes to rest in 0.1 s, and 4 cm2 of the apple comes into contact with the surface during the impact. What is the average pressure exerted on the apple during the impact? Ignore air resistance.

I thought the answer would be [hide]210 Pa[/hide], but its actually [hide]21000Pa[/hide]. I've checked my units, does anyone know how to do this?

If you are just getting it wrong by a magnitude of 100, it may be a mistake in units.
Keep in mind, that $ 4 cm^2$ is NOT equal to $ .04 m^2$

You're right. That's it. I can't believe I forgot that. Thanks.

How would you do this problem?
I'm getting...

$ v_f^2 \equal{} 2gd \equal{} 2(10)(0.4) \equal{} 8 \rightarrow v_f \equal{} 2\sqrt {2}$.

$ I \equal{} \Delta p \equal{} (0.3)(2\sqrt {2}) \equal{} Ft \equal{} F(0.1) \rightarrow F \equal{} 6\sqrt {2}$.

$ F \equal{} PA \equal{} P(.04) \equal{} 6\sqrt {2}\Rightarrow P \equal{} 150\sqrt {2}$ ??

edit: nvm, i saw post 2 and they use g=9.8


:? In a game of chance each player throws two unbiased dice and scores the difference
between the larger and smaller numbers which arise.Two players compete and one or the other wins if,and only if,he scores at least 4 more than his opponent.Find the probability that neither player wins. :P

The total number of possibilities are $\binom{6}{2}\binom{6}{2}=225$
Also for winning the difference of scores must be atleast 4.
Thus the score of the winner can be only and only 5.[6-1]and the score of the loser can be only 1.[2-1,3-2,4-3,5-4,6-5].
Since anyone can win.
Number of possibilities in which one wins= 2x5x1=10.

Thus the probability that one wins is, say p, $p=\frac{10}{225}$

Hope, I am correct. :?:

Let me try:

We have $2$ dice, each with $6$ sides.  $6X6=36$ so there are $36$ different combinations of two dice.  The only combinations that will work are $(1,5),(1,6),(2,6)$ That makes $3$ combinations, but we multiply by 2, because it doesn't matter who wins.  That makes 6 combinations out of $36$.  $6/36 = 1/6$ But we want the complement, so the answer is $5/6$.

EDIT: Never mind, I misread the question.

I want to know the solution.  The answer is 0.914 (Enya said this).  Your result is $\frac{5}{6}$, so it is incorrect unless Enya made a mistake. :(

This is not a very hard problem, but I'm kind of tired so I won't type out. Let me tell you what is wrong with varun's approach: He doesn't take into account the fact that you can score two dice of the same number and have a difference of 0.. So you have to add those combinations, and then there are more possibilities for the other person to win as well (he can score 4 or 5 points).

Oh... I see what you are saying.  I think I got it!

...scrambles to find a pencil...


The president, Barbicane, of the Gun Club want to shot a ball to the space.

So, Suppose the Earth is a sphere of radius $ R$.
There is no Air. The initial Velocity of the ball is $ v$ the angle of shot is tangent on the earth..

So, for security of the population of the earth, barbicane want to know

1 - For what velocity the ball will not reach the earth? Or in another words, for what velocity the ball will no shot the earth?

2 - is There  a velocity for what the ball will be in orbit within the earth?

3 - is there a velocity for what the ball will reach zero velocity?

[hide="My Answers"](1): $ \forall v\geq v_{c}\equal{}\sqrt\frac{GM}{R}$
(2): $ \forall v$ such that $ v_{c}\geq v\geq 0$
(3): $ v\equal{}v_{e}\equal{}\sqrt\frac{2GM}{R}$ and $ v\equal{}0$; though I am not sure these are the only values.[/hide]


linear algebra
matrix
superior algebra
superior algebra unsolved

let $K$ be a compact subset of $R^n$ that contains 0 in its interior.
Let $S$ be a symetric positive definite matrix.
Define $B_S=\{ X \in R^n s.t. X^tSX \leq 1 \}$Show that there exists a unique
symetric positive definite matrix S of minimal volume that contains  $K$

First of all, let's notice that the volume of an ellipsoid is inversely proportional to the determinant of the corresponding symmetric positive definite matrix.

Since $K$ contains a ball centered at the origin $O$, the distances from $O$ to the points on the boundary of $K$ are bounded away from $0$, so the same must hold for the values of the semiaxes of the ellipsoids which contain $K$. This, in turn, means that if we only look at the ellipsoids that contain $K$ and have volume bounded from above by some value, this set is compact, so there is an ellipsoid of minimal volume containing $K$. We must now argue that it’s unique.

Suppose we have two ellipsoids with the same volume, both containing $K$. We have to show that there is an ellipsoid with smaller volume which contains $K$. In this problem, we can replace $K$ with the intersection of the two initial ellipsoids, we can assume WLOG that one of the ellipsoids is the unit ball, and also that the axes of the other one are oriented along the standard axes of $\mathbb R^n$.

The two matrices have eigenvalues $(1,1,\ldots,1)$ and $\lambda_i$ with $\lambda_i>0$ and $\prod\lambda_i=1$ (and the matrices are diagonal). We need to find another matrix with eigenvalues $\alpha_i$, all positive and with product $>1$ s.t. $\sum x_i^2\le 1,\sum\lambda_ix_i^2\le 1\Rightarrow\sum\alpha_ix_i^2\le 1$. To this end, we consider the diagonal matrix with eigenvalues $\frac{1+\lambda_i}2$.

I hope it's Ok.


algebra
polynomial
ratio
Rational Root Theorem

Let $f(x) = px^5 + qx^4 + rx^3 + sx^2 + tx + 1$ be a polynomial such that $f(1) = 4$, $f(2) = 11$ and all the numbers $p, q, r, s$ and $t$ are integers. Prove that the equation

$px^5 + qx^4 + rx^3 + sx^2 + tx + 1 = 0$

has no integer roots

what does it by mean by not having any integer roots? what is the difference between intger roots and real roots?  :?:

yikes...i really need help with this problem... :( is there no one who can help me? or at least give a me a hint?

well... i'm not that good with polynomials... but here's something:

suppose the equation has integer root $a$, then we know that $a|1 \Rightarrow a=\pm 1$ but $f(1)=4$, so $a=-1$
consequently, $-p+q-r+s-t+1=0$

from $f(1)=4$ we get that $p+q+r+s+t+1=4$
from $f(2)=11$ we get that $32p+16q+8r+4s+2t+1=11$

rearranging stuff gives me in the easiest form $4p+2q=t$

i don't know if you can use this, but as i said before... i'm not good with polynomials  :?

I found an elegant solution.

By the Rational Root Theorem, the only zeroes are +/- 1/p.  In order for this to be an integer, p = 1 or -1, making the root have to be either 1 or -1.  We know it is not 1, because f(1) = 4.

Here's the fun part.  You can easily show that, for all integers a,b, a - b | f(a) - f(b) since a-b | a^n-b^n.  Suppose f(-1) = 0.  Then 2-(-1) | f(2) - f(-1) => 3 | 11.  Contradiction.

That was fun :D

[quote="nr1337"]
a - b | f(a) - f(b)
[/quote]

thx, your solution's really elegant! but, what does the above quoted mean? what is this sign "|"?

'|' means 'divides' :)

in this case, it's the same as: $f(a)-f(b)\equiv 0 \mod (a-b)$

[quote="beaucoa"]woah...i seem to be the only one who still does not understand this... could you explain further? maybe in more details? would be eternally grateful if you could kindly do so!  :blush:[/quote]

"a - b | f(a)-f(b)" simply means that f(a)-f(b) is a multiple of a-b.

yikes, i really mean that i do not understand the whole thing! not just that part...sorry...  :blush:

nrleet, you should be careful -- you misapplied the rational root theorem (as you don't know what factors p has).
Instead, argue that if $|a| > 0$ then $f(a) \equiv 1 \mod a$ and so $f(a) \neq 0$ and then proceed.  (What psycho did, in other words.)

[quote="nr1337"]
By the Rational Root Theorem, the only zeroes are +/- 1/p.  In order for this to be an integer, p = 1 or -1, making the root have to be either 1 or -1.  We know it is not 1, because f(1) = 4.
[/quote]

This really should be written: By the rational root theorem, the only zeros are $\pm$ 1/[a factor of $p$].  In order for this to be an integer, a factor of $p$ must be 1 or -1.  Since these are factors of all integers, the argument still holds.  All other factors of $p$ will make the ratio 1/[a factor of $p$], be a non-integer.  Correct me if I'm wrong, I'm new to this stuff.

Oops.  But no matter what p is, as calculuslover800 said, the only factor q of p such that 1/q is an integer is +/- 1 anyway, so the proof is the same from there.


Given a,b,c satisfying -1≤a,b,c≤1 and ab+bc+ca=1.
Prove that $ \sum \frac{a^2\plus{}b^2}{(1\minus{}a^2)(1\minus{}b^2)}$≥$ \frac{9}{2}$

[quote="SUPERMAN2"]Given a,b,c satisfying -1≤a,b,c≤1 and ab+bc+ca=1.
Prove that $ \sum \frac {a^2 \plus{} b^2}{(1 \minus{} a^2)(1 \minus{} b^2)}$≥$ \frac {9}{2}$[/quote]
It's obvious after expanding. :wink:

Is there another way to solve this?I think that is not a nice solution


Determine if there is a way to tile $5x6$ unit square board by dominos such that one can not use a needle to peer through the tiliing ? Determine if there is a way to tile $5x6$ unit square board by dominos such that one can use a needle to through the tiling ? What if it is a $6x6$ board ?

It's clearly possible to have a needle peer through to the other side of some configuration (namely, if $5\times 6$ means 5 rows, 6 columns, then having all dominos horizontal works).

As for if it's possible to block the needle:
[hide="?"]
It's possible for $5\times 6$ board! Use the following tiling:
[asy]draw((0,0)--(120,0)); draw((20,20)--(100,20)); draw((0,40)--(60,40)); draw((100,40)--(120,40)); draw((0,60)--(40,60)); draw((60,60)--(100,60)); draw((40,80)--(120,80)); draw((0,100)--(120,100)); draw((0,0)--(0,100)); draw((20,0)--(20,40)); draw((20,60)--(20,100)); draw((40,40)--(40,100)); draw((60,0)--(60,80)); draw((80,20)--(80,60)); draw((80,80)--(80,100)); draw((100,0)--(100,80)); draw((120,0)--(120,100));[/asy]

Considering the general case of an $a\times b$ board, suppose it's possible to block the needle. Then in each of the $a-1$ spaces between rows, there can be at most $2(b-1)$ edges (each one counted twice), and there are $b$ edges on each horizontal border. Also, in each of the $b-1$ spaces between columns, there can be at most $2(a-1)$ edges, and there are $a$ edges on each vertical border. On the other hand, there are $\frac{ab}{2}$ dominos, each with 6 edges, so there are $3ab$ edges total. Therefore, we have
\[3ab\leq 4(a-1)(b-1)+2a+2b\Rightarrow (a-2)(b-2)\geq 0 \]
But this isn't nearly strong enough to arrive at a contradiction... maybe there's some way based on colorings to determine how what combinations of horizontals and verticals are possible, and use this to sharpen the bound. Or maybe it is possible to block the needle in the $6\times 6$ case, but I can't find a way yet.
[/hide]

Forgive the question, but what does "a needle peering through the tiling" mean?

If my memory serves me correctly, the $ 6\times 6$ case is discussed in the "coloring proofs" chapter of Engels.


It is a very difficult problem to show that $\binom{2n}{n}$ is not squarefree for all sufficiently large $n$. On the other hand, let $A$ be the set of those such $n$ (for which the coefficient is squarefree). It is clear that only powers of $2$ can be in $A$ (why?). Prove that the density of those $n$ for which $2^{n}$ is in $A$ is zero.

$ord_{p}((\frac{2n}{n}))=m$, were m is number of p-adic digits $\ge (p/2)$. Therefore if $n\=2^{k}$, then $4|(\frac{2n}{n}).$
Consider $n=2^{k}, \ p=3$ give density $\{k|ord_{3}((\frac{2n}{n}))\le 1, n=2^{k}\}$ is 0. I think can be prove, that exist only finite number of n, suth that $(\frac{2n}{n})$ square-free.

Let's see a proof. But please, write it in a clear way, otherwise it's useless to post it. I mean, by posting only cryptic formulas with no explanation will not help the other people understand a proof or understand if the proof is correct or not.

It is easy to show, that $ord_{p}(n!)=[\frac{n}{p}]+[\frac{n}{p^{2}}]+...$
It give $ord_{p}(C_{n}^{k})=m$, were m is number of transfers p-adic digits, when k added with n-k. For example, if $n_{(3)}=1112022, \ (n=1115)$ $(n+n)_{(3)}=10001121$ had 6 transfers 3-adic digits and $ord_{3}({2230\choose 1115})=6$.
It prove, that $4|{2n\choose n}$ if $n\not =2^{k}$.
Let $n=2^{k}$. Consider 3-adic digits of n.  $2^{k_{1}}=2^{k_{2}}(mod \ 3^{m})$ if and only if $k_{1}=k_{2}(mod \ 2*3^{m-1}).$ Let $N=2*{}^{3}{}^{m-1}$. We have $2^{m}-1$ residues, suth that all last m 3-adic digits of $2^{k}$ are 0 or 1 (last digit $\not =0$) and $m*2^{m-2}$ residues mod N, when $(n+n)_{(3)}$ had 1 transfer in last m digits. Therefore density of k, when $9\not |{2^{k+1}\choose 2^{k}}$ less than $\frac{(m+4)2^{m-2}}{2*3^{m-1}}\to 0$, when $m\to \infty$.


Please choose the time that it is now, GMT. This is to find out when most people are online and when least people are online

i think this poll is just gonna be loaded with spam......but, its 2330 GMT anyway.

Wouldn't the fact that AoPS members are scattered across the world be a problem if GMT is used?  I suppose not, if all you want is the GMT when the most people are on.  But I think it'd be more interesting to see what [i]local[/i] time is most common.  Oh well.

Ah yes, it's 1148 GMT

This topic is more appropriate for Games & Fun Factory forum and right now is 23:58 GMT

bubka are you in New Dehli or something?


logarithms
calculus
calculus computations

$ \sum_{n\equal{}2}^{\infty} \sum_{m\equal{}1}^{n\minus{}1}\frac{(\minus{}1)^n}{mn}$

I think the answer is a half of $ (\ln{2})^2$ ... but i'm not sure

How do you think? :)

$ S \equal{} \sum_{n \equal{} 2}^{\infty} \sum_{m \equal{} 1}^{n \minus{} 1}\frac {( \minus{} 1)^n}{mn}$

$ S \equal{} \sum_{1 \le m < n}\frac {( \minus{} 1)^n}{mn}$

$ S \equal{} \sum_{m \equal{} 1}^{\infty} \sum_{n \equal{} m\plus{}1}^{\infty} \frac {( \minus{} 1)^n}{mn}$

$ S \equal{} \sum_{m \equal{} 1}^{\infty} \sum_{k \equal{} 1}^{\infty} \frac {( \minus{} 1)^{m\plus{}k}}{m(m\plus{}k)}$

$ S \equal{} \sum_{m \equal{} 1}^{\infty} \sum_{k \equal{} 1}^{\infty} ( \minus{} 1)^{m\plus{}k}(\frac {1}{mk} \minus{} \frac {1}{(m\plus{}k)k})$

$ S \equal{} \sum_{m \equal{} 1}^{\infty} \sum_{k \equal{} 1}^{\infty} ( \minus{} 1)^{m\plus{}k}\frac {1}{mk} \minus{}  \sum_{m \equal{} 1}^{\infty} \sum_{k \equal{} 1}^{\infty} \frac{( \minus{} 1)^{m\plus{}k}}{(m\plus{}k)k}$

$ S \equal{} (\sum_{k \equal{} 1}^{\infty}( \minus{} 1)^{k}\frac {1}{k})^2 \minus{} S \equal{} (\minus{}ln2)^2 \minus{} S$

$ S \equal{}  \frac{1}{2}(ln2)^2$

What justifies your passage from your first step above to your third step? You switched the order of summation in a double sum. The two hypotheses most commonly used to justify that are either that the terms are nonnegative or that the entire sum is absolutely convergent. Neither one of those apply to this.

By the same token, at the end of your argument, you arrived at an algebraic expression with the unknown sum $ S$ on both sides, which you solved for $ S.$ How do you know that $ S$ exists? One could imagine scenarios in which a divergent $ S$ would destroy such an argument.

yes, you are right to ask for justification.

I wont go into laborious details but roughly...

The original expression converges because the inner sum has alternating signs and is $ O(\frac{\ln{n}}{n})$ which tends to zero.

The change of order can be justified by taking a partial sum up to a limit $ n < N$, then there are a finite number of terms and the order can be reversed. The limit can then be taken to recover the result for the infinite sums.

When splitting the sum in two it is also necessary to show that both parts converge individually, but that is not difficult.


geometry
circumcircle
inequalities proposed
inequalities

In any triangle

$r_a+r_b+r_c \geq 3(R+r)$

where R= radius of circumcircle
           r=  radius of incircle
           r_a,r_b,r_c= radii of excircles

Trivial !!!!!!

We have $r_a+r_b+r_c=r+4R$, so that it remains to show that $r+4R\geq 3\left(R+r\right)$. This is equivalent to $R\geq 2r$, what is more than well-known.

  Darij

Oh!!!!!  Sorry. :blush:

I had to complete it in this way. :lol:

$r_a+r_b+r_c \geq 3(R+r) \geq w_a+w_b+w_c \geq h_a+h_b+h_c$

where $w_a,w_b,w_c$ are the bisectors and $h_a,h_b,h_c$ the altitudes


In how many ways can we choose 3 distinct letters of the alphabet, without regard to order, if we must choose 1 vowel (A, E, I, O, or U) and 2 consonants?

I was being stupid and forgot about the word "distinct". :(

[quote=hilarious]Sorry for bumping, but can someone explain?
My solution is that we choose 2 consonants and 1 vowel in [hide=my INCORRECT solution]$\frac{21*20*5}{3!}$[/hide] ways, because there are $3!$ ways to order the letters, but the solution only accounts for order when we choose the consonants.
The solution says [hide=Sppoiler alert]We have 5 options for the choice of the vowel, and we must make 2 choices out of the remaining 21 letters, for a total of $\binom{21}{2} = 210$ choices for the consonants. This gives a total of $5 \times 210 = \boxed{1050}$.
But this is saying that if we have ABC, then ABC and ACB are the same, BAC and CAB are the same, and CBA and BCA are the same; but also saying that ABC and BAC and CBA are different. [/hide]
Sorry about this being messy, I wasn't sure how to format it.[/quote]

Do you still need help with this? If you do, I can help explain.

Can anybody help me with this I got 2100( 5*21*20)

[font=TIMES NEW ROMAN][hide=Easy Solution.]The question says "without regard to order," so we can use choose functions to solve this.

We start off with the vowel; there are $\tbinom{5}{1}$ choices for the vowel. Then there are the consonants. There are $26-5=21$ consonants in the alphabet, and we must choose 2. So we have $\tbinom{21}{2}$ choices.

In total, there are $\tbinom{5}{1}\tbinom{21}{2}=\boxed{1050}$ choices. $$\definecolor{color}{rgb}{0.9, 0.7, 0.0} \scriptsize{\colorbox{color}{\color{black}{A pifinity solution}}}$$

[quote=hhu98473]why isn't it 5*21*20[/quote]

because order of the 2 consonants don't matter, so you end up overcounting.

to correct overcounting you divide by 2


calculus
integration
logarithms
calculus computations

Evaluate the integral

$\int\sqrt{a^{2}+\sqrt{b^{2}+\frac{c}{x}}}\,dx$

Yes the integral is right. Try the case a =1, b = 1 and c = 1.


geometry
inequalities proposed
inequalities

[quote="manlio"]P is a point inside the triangle ABC; the intersection of AP,BP,CP and the opposite sides is D,E,F. The area of DEF is K. Prove that

  AD*BE*CF  \geq 4Ks

where s is the semiperimeter.

Source: NieuW  Arch. Wisk. 14(1966)[/quote]


Is it known if for any k, there exist two primes p and q such that p-q = 2k?

I imagine that it should be true, but I don't have a proof.

If you have a proof or know of one, please post.

Like Goldbach's, it is still a conjecture...

How do you know this is still a conjecture? (post please)

Every even number is the difference of two primes.
er....I think it is a conjecture as some one claimed it in the discussion of the Goldbach's conjecture
This may [url=http://primes.utm.edu/notes/conjectures/]Help[/url]
Something similar to this  there exists a russian olympiad problem -Prove that every natural number can be expressed as a difference of two other natural numbers which have the same number of prime factors .