inequalities
trigonometry
Euler
geometry proposed
geometry

Prove that:
$ \frac {1}{l_a} \plus{} \frac {1}{l_b} \plus{} \frac {1}{l_c} \ge \frac {\sqrt {(a \plus{} b \plus{} c)(a \plus{} b)(b \plus{} c)(c \plus{} a)}}{\sqrt {2}abc}$
Where $ l_a,l_b,l_c$ be the lengths of the interior angle-bisectors from the three vertices $ A,B,C$ of a triangle $ ABC$ and $ a,b,c$ be the sides of a triangle.
It stronger than the well know:
$ l_al_b \plus{} l_bl_c \plus{} l_cl_a \ge 3\sqrt {3}S$

$\color{white}{...........................................................................................................................................................................................}$
[quote="quykhtn-qa1"]\[\triangle\, ABC\ \implies\ \boxed{\ \frac 1{l_a}+\frac 1{l_b}+\frac 1{l_c}\, \ge\, \frac {\sqrt{(a+b+c)(a+b)(b+c)(c+a)}}{abc\sqrt 2}\ }\][/quote]

[color=darkblue][u][b]Proof.[/b][/u] Firstly, let us note that: $\boxed{\sum\, \frac 1{l_a}}\stackrel{(1)}{=}\frac 1{2rs}\sum\, \frac {ah_a}{l_a}\stackrel{(2)}{=}\frac {\sum\, a\cos\frac {B-C}2}{2rs}\stackrel{(3)}{=}$ $\frac {\sum\, [2R\sin A(\cos\frac B2\cos\frac C2+\sin\frac B2\sin\frac C2)]}{2rs}\stackrel{(4)}{=}$

$\stackrel{(4)}{=}\frac {\sum\, [2R\sin A(\frac s{4R}\cdot\frac 1{\cos\frac A2}+\frac r{4R}\cdot\frac 1{\sin\frac A2})]}{2rs}$ $\stackrel{(5)}{=}\frac {\sum\, [2\sin\frac A2\cos\frac A2(\frac s2\cdot\frac 1{\cos\frac A2}+\frac r2\cdot\frac 1{\sin\frac A2})]}{2rs}=\boxed{\frac {s\sum\, \sin\frac A2+r\sum\, \cos\frac A2}{2rs}}$

where at steps $(1)\, ,\, (2)\, ,\, (3)\, ,\, (4)\, ,\, (5)$ I used the well known identities: $\left\|\begin{array}{cccc}
1\blacktriangleright & 2S=ah_a=2rs\ \ ;\ \ 2\blacktriangleright\ \frac {h_a}{l_a}=\cos\frac {B-C}2 \\ \\
3\blacktriangleright & \{\begin{array}{cccc}
a=2R\sin A \\ \\
\cos\frac {B-C}2=\cos\frac B2\cos\frac C2+\sin\frac B2\sin\frac C2\end{array} \\ \\
4\blacktriangleright & \begin{array}{cccc}
\prod\, \cos\frac A2=\frac s{4R}\ ;\ \prod\, \sin\frac A2=\frac r{4R}\end{array} \\ \\
5\blacktriangleright &\, \ \sin A=2\sin\frac A2\cos\frac A2\end{array}\right\|$

In addition, since $\boxed{\prod\, (b+c)=2s(s^2+r^2+2Rr)}$ and $\boxed{\prod\, \tan\frac A2=\frac rs}$ the proposed inequality is now equivalent to :

$\frac {s\sum\, \sin\frac A2+r\sum\, \cos\frac A2}{2rs}\, \ge\, \frac {\sqrt{2s^2(s^2+r^2+2Rr)}}{4Rrs}\iff s$ $\sum\, \sin\frac A2+r\sum\, \cos\frac A2\, \ge\, s\cdot\sqrt{\frac {s^2+r^2+2Rr}{2R^2}}$

$\iff \sum\, \sin\frac A2+\prod\, \tan\frac A2\sum\, \cos\frac A2\, \ge\, \sqrt{\frac {s^2+r^2+2Rr}{2R^2}}$ .

But since in any $\triangle\, ABC$ the following strong inequality holds : $\boxed{\ \sum\, \cos\frac {B-C}2\, \ge\, 1+\sqrt{\frac {s^2+r^2+2Rr}{2R^2}}\ }$ (see[/color] [url=http://www.artofproblemsolving.com/blog/53230][color=red][b]here[/b][/color][/url][color=darkblue])

it suffices to prove that : $\boxed{\ 1+\sum\, \sin\frac A2+\prod\, \tan\frac A2\sum\, \cos\frac A2\, \ge\, \sum\, \cos\frac {B-C}2\ }\ \ (\ast)$ . By making the following

substitutions : $\{\begin{array}{llll}
A & = & 180^{\circ}-2X \\
B & = & 180^{\circ}-2Y \\
C & = & 180^{\circ}-2Z\end{array}$ where $\|\begin{array}{cccc}
X,Y,Z\in(0,90^{\circ}) \\ \\
X+Y+Z=180^{\circ}\end{array}\|$ we will transform the inequality $(\ast)$ (in [b]any[/b] triangle)

into one restricted to an [b]acute-angled[/b] triangle : $\boxed{\ 1+\sum\, \cos X+\prod\, \cot X\sum\, \sin X\, \ge\, \sum\, \cos(Y-Z)\ }\ \ (\ast\ast)$ .

For convenience, we will use the notations $R$ , $r$ , $s$ for denoting the classical elements of $\triangle\, XYZ$ (which are now different from

those appearing above). Recall the identities : $\|\begin{array}{cccc}
\sum\, \cos X=1+\frac rR\ \ ;\ \ \prod\, \cot X=\frac {s^2-(2R+r)^2}{2rs} \\ \\
\sum\, \sin X=\frac sR\ ;\ \sum\, \cos(Y-Z)=\frac {s^2+r^2+2Rr-2R^2}{2R^2} \end{array}\|$ in order to see

that the inequality $(\ast\ast)$ is equivalent to: $\frac {2R+r}R+\frac {s^2-(2R+r)^2}{2rs}\cdot\frac sR\ge\frac {s^2+r^2+2Rr-2R^2}{2R^2}$ which further becomes:

$2Rr(2R+r)+R[s^2-(2R+r)^2]\ge r(s^2+r^2+2Rr-2R^2)$ $\iff$ $\boxed{\ s^2\ge\frac {4R^3-2R^2r+Rr^2+r^3}{R-r}\ }\ (G)$

Finally, we will show that the last "boxed" inequality is true in any non-obtuse triangle. To do this, we'll make use of two well known

results, namely : $\boxed{\ s^2\ge 2R^2+10Rr-r^2-\sqrt{4R(R-2r)^3}\ }$ ([u][b]Blundon[/b][/u]) and $\boxed{\ s\ge 2R+r\ }$ , with the latter holding

in [b]non-obtuse[/b] triangles. To be more specific, we will prove inequality $(G)$ in [u]two[/u] stages: in the [u]first[/u] one we show that the inequality $(G)$ holds for $R\le(1+\sqrt 2)r$ when compared to [u][b]Blundon[/b][/u], while in the [u]second[/u] stage we prove that $(G)$ is fulfilled for $R\ge(1+\sqrt 2)r$ when compared to $s\ge 2R+r$ . In the following lines we will use the substitution: $t=\frac Rr$ where $t\ge 2$ ([u][b]Euler[/b][/u]) in order to make the computations easier. Thus:
\[\boxed{\ \begin{array}{cccc}
(G_1) & 2R^2+10Rr-r^2-\sqrt{4R(R-2r)^3}\, \ge\, \frac {4R^3-2R^2r+Rr^2+r^3}{R-r} \\ \\
\iff & (t-1)(2t^2+10t-1-\sqrt{4t(t-2)^3}\, )\, \ge\, 4t^3-2t^2+t+1 \\ \\
\iff & (t-1)(2t^2+10t-1)-(4t^3-2t^2+t+1)\, \ge\, (t-1)\sqrt{4t(t-2)^3} \\ \\
\iff & t(t-2)(3-t)\, \ge\, (t-1)\sqrt{t(t-2)^3} \\ \\
\stackrel{\text{For}\  t\le 3}{\iff} & (t-2)^2(-t^2+2t+1)\, \ge\, 0 \\ \\
\implies & (G_1)\text{ holds for }\boxed{2\le t\le 1+\sqrt 2} \\ \\
\hline\hline \\\
(G_2) & (2R+r)^2\, \ge\, \frac {4R^3-2R^2r+Rr^2+r^3}{R-r} \\ \\
\iff & (t-1)(2t+1)^2\, \ge\, 4t^3-2t^2+t+1 \\ \\
\iff & t^2-2t-1\, \ge\, 0 \\ \\
\implies & (G_2)\text{ holds for }\boxed{t\ge 1+\sqrt 2}\end{array}\ }\]
We conclude that the previous inequality $(G)$ holds true in any non-obtuse triangle, which means that $(\ast\ast)$ is also true in such triangles
so our proof is now complete. Equality occurs in the original inequality if and if $\triangle\, ABC$ is equilateral.[/color]

[color=darkred][u][b]Remarks.[/b][/u] As said by [b]quykhtn-qa1[/b], the proposed inequality represents a refinement of an old inequality (which also appeared in

the Canadian magazine[/color] [url=http://www.math.ca/crux/][color=red][u][b]CRUX MATHEMATICORUM[/b][/u][/color][/url][color=darkred] long time ago - look for problem [b]2029*[/b]) : $\boxed{\ l_al_b+l_bl_c+l_cl_a\, \ge\, 3S\sqrt 3\ }$

I will now present a sketch of the aforementioned fact. Multiplying the initial inequality by $\boxed{\ \prod\, l_a=\tfrac {16Rr^2s^2}{s^2+r^2+2Rr}\ }$ one can easily

obtain the following equivalent statement i.e. $\boxed{\ l_al_b+l_bl_c+l_cl_a\, \ge\, \tfrac {4rs^2\sqrt 2}{\sqrt{s^2+r^2+2Rr}}\ }$ thus remaining to prove that :

$4s\sqrt 2\, \ge\, 3\sqrt{3(s^2+r^2+2Rr)}$ which simplifies to the following easy to prove inequality : $5s^2\, \ge\, 54Rr+27r^2$ . [/color]


inequalities algebra polynomial trigonometry

In my opinion, this is not a pre-olympiad problem.  In fact, I don't think I could solve this problem if I had never seen anything like it before.  Basically, this problem strikes at the core of Chebyshev polynomials.  There is a very elegant theorem called Chebyshev's Theorem which states that if $ p$ is a degree $ n$, monic polynomial, then $ max \{|p(x)|: x\in [-1,1]\}\ge \frac{1}{2^{n-1}}$.  The proof for this is in "The Book" for anyone who has it, but the more interesting thing is only one polynomial for each $ n$ achieves this minimum, and these are the Chebyshev polynomials.

It is not a hard corollary to the theorem (if I did computations correctly) to see that $ max \{|p(x)|:-a\le x\le a\}\ge 2\cdot \left (\frac{a}{2}\right )^{n}$ where $ p(x)=\prod_{i=1}^{n}(x-x_{i})$ for $ x_{i}\in [-a,a]$.  Equality occurs if and only if $ p(x)=\prod_{k=0}^{n-1}(x-a\cos (\frac{(2k+1)\pi}{2n}))$.  This means that the polynomial the question is looking for is

$ q(x) = \prod_{k=0}^{5}(x-5\cos (\frac{(2k+1)\pi}{12}))$

which satisfies $ max \{|q(x)|:-5\le x\le 5\}=2\cdot\left (\frac{5}{2}\right )^{6}$.


abstract algebra
group theory
algebra
polynomial
induction
advanced fields
advanced fields unsolved

Let $ R$ be a ring with a grading $ R \equal{} R_0\oplus R_1 \oplus ....$. i.e. $ R$ is a direct sum of abelian subgroups of $ R$ such that for each $ i,j$ $ R_iR_j\subseteq R_{i \plus{} j}$.

Let $ R_{ \plus{} }: \equal{} \bigoplus_{i > 0}R_i$. This is easily shown to be an ideal of $ R$.

Show that if $ R$ is commutative then:

If $ R_0$ is right noetherian and $ R_ \plus{}$ is finitely generated $ \Rightarrow$ $ R$ is right noetherian.

I am also looking for a counterexample to where this fails when R is not commutative...

Pick a set of homogeneous generators $ x_1,..,.x_n$ of $ R_\plus{}$ (assume degree of $ x_i$ is $ a_i$ and show that it generates $ R$ as a $ R_0$ algebra. To show the generating relation, it suffices to show it for the homogeneous elements. We can proceed by induction. So let's say for homogeneous elements of degree < k they can be generated by polynomials of $ x_1,...,x_n$ over $ R_0$. Consider a homogeneous element $ x$ in $ R_k$, $ k > 0$. Then $ x \equal{} \sum y_ix_i$, where $ y_i$ is of degree $ k \minus{} a_i$. By induction hypothesis, $ y_i$ can be generated by polynomials of $ x_1,...,x_n$ over $ R_0$. Put these polynomials together in the expression of $ x$, we can write $ x$ as a polynomial of $ x_1,...,x_n$. This completes the induction.


AMC
AIME
geometry
ratio
angle bisector
Pythagorean Theorem

ABC is a triangle with angle B = 90o. AD is an angle bisector. E lies on the side AB with AE = 3, EB = 9, and F lies on the side AC with AF = 10, FC = 27. EF meets AD at G. Find the nearest integer to area GDCF.

side BC works out nicely to be 35. I can then use the angle bisector theorem to find BD and DC (i'm assuming D is on BC). Now, I can find the area of ABD, ABC, and AEF (by multiplying the area of ABC by 3/12 and 10/37) I can also find the area of AFG as a multiple of AEG. (using the area formula absin:theta:/2, we see that since angle EGA =angle GAF, the area of triangle AFG is just 10/3 that of triangle AEG. These are enough facts to solve the problem; I do not wish to do needless computation.

Yea, i can do this problem as well, but rather messily, and the algebra and arithmetic gets boring

anyone know the clever approach?

[color=cyan]Actually doing the computation is important when you're talking about compeition problems, or even mathematics in general -- because knowing you can do something is nice, but no one else can use it unless you actually do.  Plus, it's just good practice.  Anyhow, I think this is a slightly easier way to do this:

Given 2 equal angles GHI and JKL, the ratio of the areas of the triangles is GH*HI to JK*KL.  So, using [XYZ] to represent the area of triangle XYZ,
[ADC]:[ABD] = 37:12
[ADC]:[ABC] = 37:49

[AGF]:[AEG] = 10:3
[AGF]:[AEF] = 10:13
[AEF]:[ABC] = 3*10:12*37 = 5:74
[AGF]:[ABC] = 25:481

[GFCD]:[ABC] = 37/49 - 25/481

BC = 35 by Pythagorean Theorem, so [ABC] = 6*35
[GFCD] = 6*37*5/7 - 25*6*35/(13*37) = 497160/3367
which is closest to 148.  I think there's a clever way to do the arithmetic, too, but I couldn't figure it out.
[/color]

[color=cyan]I think it was pretty similar, but you did some stuff computing lengths like BD and CD.  It's not substantially different, I guess.  I like doing it all with the ratios, though.  If anyone could figure out a good enough estimate on the fractions that would be quick to come up with, that would be cool.[/color]


Hello, I am new to this place.
I am quite interested in participating in a contest held here, but I don't know the basics about them.
What I have made out till now is that valentino Vornicu (sorry if i misspelt it) organises all the contests and seeing that he visits this thread, I am posting my elementary doubts here.
1. How do you make out whether a contest is currently open?
2. Where do you post your solutions?
3. Is there a contest currently open where I can send my slutions?

Hi, I've splitted your posts, since it was not relevant to the 7th round of the contest that just ended.

For question 1) read the regulations. Each .pdf file with problems has dates on it showing until when the problems can be submitted to be taken into account (you can submit solutions if you want to previous problems, but of course that won't affect the rankings of previous competitions).

At this moment there is no contest (the 4th edition has just ended). Another one will start in a couple of months so that it will end (the spring competition) before the IMO takes place.

I've already answered questions 2 and 3 above. Please read carefully the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=3683]REGULATIONS[/url] of the MathLinks Contest :)


Points $A$ and $B$ are $5$ units apart. How many lnes in a given plane containing $A$ and $B$ are $2$ units from $A$ and $3$ units from $B$?

$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ \text{more than 3}$

Yeah, I got the same answer.

[hide="hint"]
Draw circles whose centers are $A$ and $B$.
[/hide]

[hide]Tangents. One internal + two external = $3 \implies \boxed{E}$
Edit upon realizing that 5 = 3 + 2.[/hide]

Sorry, but I don't exactly understand this problem.  Can someone please show me a picture?  :?  :(

What does it mean by 2 units from A and 3 units from B?

The distance from a point to a line is the length of the perpendicular line dropped from the point to the line.

So we need to find the number of lines that satisfy the two conditions (2 units apart from A and 3 units apart from B) simultaniously.

Hope this helps...

Right, the plane contains A and B, not the lines  :oops:  :D


Is therre going to be a JUne or JUly FTW tourney?

I hope so! Because I want to participate.

Well, [i]someone[/i] has to organize such a tournament.... who?
And I'm in. :D

I guess I'll organize it. Give me a bit to get set up  :lol: .

MustItAlwaysBeMe and I are probably going to run a July one as we have done this before and already have a website and all to communicate everything.

Oh so should I not waste my time? Or can I run the June one? If so, I'll begin set up when I get back from tennis

You can do whatever you want with June; I was just letting you know we would probably doing something for July.
If you want me to post anything on the website, just let me know and I'll do so.   :)

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=209166[/url]

just 2 link em


For $ a,b,c>0$ and $ abc\equal{}1$ prove that holds

$ a^{\sqrt[3]3}\plus{}b^{\sqrt[3]3}\plus{}c^{\sqrt[3]3}>a^{\sqrt2}\plus{}b^{\sqrt2}\plus{}c^{\sqrt2}$

after normalization its clear by Muirhead's theorem :wink:

its stronger
 if abc=1 then (a^x+b^x+c^x>=a^y+b^y+c^y) <==>(x>=y ):)


calculus
number theory
number theory unsolved

This is the wrong forum for this, since it's not a calculus-analysis problem. How about Advanced Section -> Number Theory as the right place?

I would rather put it to the Intermediate or Pre-Olympiad topics!

OK, I'm moving it to Olimpiad/Number Theory/Unsolved Problems (hope, it'll be solved there soon :) )

[quote="bodhi"]let $b_n$ be the last digit of the number $1^1+2^2+3^3+...+n^n$
prove that the sequence $b_1,b_2,b_3,...$ is periodic with period $100$.[/quote]

First note that $n^{n}$ has the same units digit as $(n+20)^{n+20}$, because $n+20$ and $n$ obviously have the same units digit, and that $\displaystyle n^{4a+1}=n\mod 10$ for any integers a and n because $\phi(10)=4$
That equality then implies that the sum of any 20 consecutive terms of this sequence -- i.e., $n^{n}+(n+1)^{n+1}+...+(n+19)^{n+19}$ for all integers $n$ is invariant.  We can calculate the units digit of the sum of the first 20 terms with a little work by reducing the exponents mod 4 when necessary -- the sum is $1+4+7+6+5+6+3+6+9+0+1+6+3+6+5+7+4+9=4$) .  Since the sum of any 20 consecutive terms is 4 mod 10, then the sum of any 100 consecutive terms is 0 mod 10.  This implies the desired result.

With a little more thought, this argument can be extended to show that the last $k$ digits of the sum, with $k$ an arbitrary integer, is also periodic.


let $(\omega)$ be circumcircle of $\triangle ABC$ and $AM,BE,CF$ medians of $\triangle ABC$ extend $AM,BE,CF$ to meet  $(\omega)$ again at $A',B',C'$ repectively if we have three points $A',B',C'$ draw $\triangle ABC$

i don't quite understand the problem. clearify better.

The problem is to construct a triangle ABC, given the points where its medians intersect its circumcircle (apart from the points A, B, C, of course).

Paul Yiu has kindly forwarded the question to the Hyacinthos newsgroup, where it received very interesting answers. See [url=http://groups.yahoo.com/group/Hyacinthos/messages/11214?xm=1&viscount=4&m=e]Hyacinthos messages #11214-#11217[/url]. The problem has also been discussed on Hyacinthos before, e. g. in [url=http://groups.yahoo.com/group/Hyacinthos/messages/10906?xm=1&viscount=2&m=e]Hyacinthos messages #10906, #10907[/url] and in [url=http://groups.yahoo.com/group/Hyacinthos/messages/3890?xm=1&viscount=53&m=e]Hyacinthos messages #3890, #3892, #3925, #3928, #3936, #3943[/url].

The answer to the problem is:

If the medians of a triangle ABC from the vertices A, B, C, respectively, intersect the circumcircle of triangle ABC at the points X, Y, Z, then, in order to construct triangle ABC back from the points X, Y, Z, one has to locate the foci of [url=http://mathworld.wolfram.com/SteinerInellipse.html]the Steiner inellipse[/url] of the triangle XYZ, to denote one of them by G, and to intersect the lines XG, YG, ZG with the circumcircle of triangle XYZ. The new points of intersection are the required points A, B, C. (Of course, the point G is therefore the centroid of triangle ABC.)

The construction problem therefore has two solutions, since the Steiner inellipse of a non-equilateral triangle has two focii. These focii can be constructed by compass and ruler; a construction was given in

  [url=http://forumgeom.fau.edu/FG2003volume3/FG200307index.html]Paul Yiu, [i]On the Fermat Lines[/i], Forum Geometricorum 3 (2003), pp. 73–81[/url],

§6, (14).

Thanks to Paul Yiu, Jean-Pierre Ehrmann and others for their help in closing this problem.

  Darij


integration
calculus
calculus computations

determine whether the series

$ \sum_{1}^{\infty}\int_{0}^{1/n}\frac{\sqrt{x}}{1+x^{2}}\, dx$

converges or diverges

It converges by limit comparison to $ \sum_{n=1}^{\infty}\frac1{n^{3/2}}.$

Loosely and informally: since we're integrating near zero, that denominator of $ 1+x^{2}$ might as well be $ 1.$ So we're integrating $ \sqrt{x},$ and that gives us the $ \frac32$ power, hence the claim above.

If you don't like that reasoning, write your own formalization.


algorithm AMC USA(J)MO USAMO search \/closed

  Could we not develop a good "Mathematical[b] Algorithm[/b]"whose purpose is to:1)To sort out all 'present-day'posted message.2)To make sure that any other "ALREADY"posted message in the past(With its ideas

  [/b][b],suggestions[/b],problems[b],and questions[/b]ask,etc.)will not be posted again.(But give the aops 'members[b]'hint that such ideas,concept,suggestions,question raised aws already posted before[/b].)
                        These are just my ideas related to my suggestion that 'aops[b] founders[/b],directors[b]'will make effort to develop(If it is convenient,intellectually,financially.)a kind of 'special[/b] algorithm[b]' only for the purpose of sorting.
Probably,i will send my my ideas,or 'post' here in our site.

                                                            Respectfully,
                                                           A.Eddington

                  [/b]

Well I can tell you right now that any problem at an elementary level (say before USAMO) [b]has been[/b] posted and/or discussed before on the Forum :) Where ... ah there's the problem. You have to use keen search words, Darij has a topic about this somewhere ...


geometry
rectangle
perpendicular bisector

A rectangular sheet of paper measures 12 inches by 9 inches. One corner is folded onto the diagonally opposite corner and the paper is creased. What is the length in inches of the crease? Express your answer as a mixed number.

[hide] Is $\frac{75}{8}$ the correct answer? Oops. I mean $9\frac{3}{8}$.[/hide]

[quote="Iron Fist"][hide="odd"]
That's weird, I got 45/4, or [b]11 and 1/4...[/b]
[/hide][/quote]
I also got $11 \frac{1}{4}$.

[hide="solution"]After you fold, you can see a right triangle with a leg and hypotnuse that add up to 12 and a leg that is 9 so:

$81+x^2=(12-x)^$
$81+x^2=144-24x+x^2$
$81-144-24x$
$24x=63$
$x=21/8$

Then to find the length of the crease you do

$12-21/4=6.75$
Since you have a calculator:

$\sqrt[2]{6.75^2+81}=11.25$

$\displaystyle{\boxed{11\frac{1}{4}}}$[/hide]

Edit: I agree with Iron fist

Err, there's a ten times easier way, it involves easier similar triangles...

[hide="Solution"]
Draw the diagonal to the large rectangle - the length is 15. Now draw the perpendicular bisector, and draw a triangle from the crease (perpendicular bisector) so that one leg is parallel to the side with length 9 and the other leg is on the side with length 12.

This newly formed triangle is similar to the 9-12-15 triangle from the beginning. Setting up proportions gets us:

hypotenuse / long leg

15 / 12 = x / 9
12x = 135
4x = 45
x = 45/4 or 11 and 1/4.

Note that the hypotenuse of the smaller triangle is the crease itself, and also note that the longer leg of the smaller triangle is the same as one of the length-9 sides of the rectangle.

[i]This method looks long, as it has a lot of explanations, but in practice it'll take under a minute, and it doesn't even require a calculator.[/i]
[/hide]

[hide]The answer is 45/4, I mean 11.25, I mean 11 and 1/4.

Sorry[/hide]

Another one I got wrong.  :roll:  :cursing:

[hide]The diagonal of the rectangle is 15 from 3-4-5 triangle. The crease=x. um...two similar rectangles
$\implies \frac{15}{12}=\frac{x}{9}\implies x=\frac{45}{4}=11\frac{1}{4}$.[/hide]


Give $ a,b,c$ be real numbers. Prove that

$ 4(a^2\plus{}b^2\plus{}c^2)^2 \ge 9(a\plus{}b\plus{}c)(b\plus{}c\minus{}a)(c\plus{}a\minus{}b)(a\plus{}b\minus{}c)$

[quote="arqady"][quote="dduclam"]Give $ a,b,c$ be real numbers. Prove that

$ 4(a^2 \plus{} b^2 \plus{} c^2)^2 \ge 9(a \plus{} b \plus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)(a \plus{} b \minus{} c)$[/quote]
Why not $ (a^2 \plus{} b^2 \plus{} c^2)^2 \ge 3(a \plus{} b \plus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)(a \plus{} b \minus{} c)$ $ ?$ :wink:[/quote]

Yes, you are right, arqady. I think so  :).

[quote="dduclam"]Give $ a,b,c$ be real numbers. Prove that

$ 4(a^2 \plus{} b^2 \plus{} c^2)^2 \ge 9(a \plus{} b \plus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)(a \plus{} b \minus{} c)$[/quote]

If $ a,b,c \ge 0$ then we just have to prove it in the case $ a,b,c$ are length sides of triangle. But it's easy, we have

$ a^2 \plus{} b^2 \plus{} c^2 \ge \frac {1}{3}\left( {a \plus{} b \plus{} c} \right)^2$

$ \left( { \minus{} a \plus{} b \plus{} c} \right)\left( {a \minus{} b \plus{} c} \right)\left( {a \plus{} b \minus{} c} \right) \le abc$

Then we just need to show $ 4\left( {a \plus{} b \plus{} c} \right)\left( {a^2 \plus{} b^2 \plus{} c^2 } \right) \ge 27abc$

By AM-GM we have $ 4\left( {a \plus{} b \plus{} c} \right)\left( {a^2 \plus{} b^2 \plus{} c^2 } \right) \ge 36abc \ge 27abc$

So the inequality is proved. Now we check other cases.

If $ a \ge b \ge 0,c \le 0$ then setting $ a_1 \equal{} a,b_1 \equal{} b,c_1 \equal{} \minus{} c$ we have

$ \begin{array}{l} a \plus{} b \plus{} c \equal{} a_1 \plus{} b_1 \minus{} c_1 \\
\minus{} a \plus{} b \plus{} c \equal{} \minus{} a_1 \plus{} b_1 \minus{} c_1 \\
a \minus{} b \plus{} c \equal{} a_1 \minus{} b_1 \minus{} c_1 \\
a \plus{} b \minus{} c \equal{} a_1 \plus{} b_1 \plus{} c_1 \\
\end{array}$

So

$ \left( {a \plus{} b \plus{} c} \right)\left( { \minus{} a \plus{} b \plus{} c} \right)\left( {a \minus{} b \plus{} c} \right)\left( {a \plus{} b \minus{} c} \right)$
$ \equal{} \left( {a_1 \plus{} b_1 \plus{} c_1 } \right)\left( { \minus{} a_1 \plus{} b_1 \plus{} c_1 } \right)\left( {a_1 \minus{} b_1 \plus{} c_1 } \right)\left( {a_1 \plus{} b_1 \minus{} c_1 } \right)$

Hence we can do the same way as the case $ a,b,c \ge 0$.  :)

Similar for $ a \ge 0,b,c \le 0$ and $ a,b,c \le 0$

[quote="arqady"][quote="dduclam"]Give $ a,b,c$ be real numbers. Prove that

$ 4(a^2 \plus{} b^2 \plus{} c^2)^2 \ge 9(a \plus{} b \plus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)(a \plus{} b \minus{} c)$[/quote]
Why not $ (a^2 \plus{} b^2 \plus{} c^2)^2 \ge 3(a \plus{} b \plus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)(a \plus{} b \minus{} c)$ $ ?$ :wink:[/quote]

Thank you arqady,that is inequality which I want post. The first,equality don't occur  :wink:

Thank you Honey_S for solution. This inequality is easy enough. But i think the following stronger inequality also true:

$ (a\plus{}b\plus{}c)^5\ge 81 (a^2\plus{}b^2\plus{}c^2)(a\plus{}b\minus{}c)(b\plus{}c\minus{}a)(c\plus{}a\minus{}b)$

[quote="dduclam"][quote="arqady"][quote="dduclam"]Give $ a,b,c$ be real numbers. Prove that

$ 4(a^2 \plus{} b^2 \plus{} c^2)^2 \ge 9(a \plus{} b \plus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)(a \plus{} b \minus{} c)$[/quote]
Why not $ (a^2 \plus{} b^2 \plus{} c^2)^2 \ge 3(a \plus{} b \plus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)(a \plus{} b \minus{} c)$ $ ?$ :wink:[/quote]

Thank you arqady,that is inequality which I want post. The first,equality don't occur  :wink:

Thank you Honey_S for solution. This inequality is easy enough. But i think the following stronger inequality also true:

$ (a \plus{} b \plus{} c)^5\ge 81 (a^2 \plus{} b^2 \plus{} c^2)(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)$[/quote]

And I think that you need a condition for $ a,b,c$ (they are the side lenghts of triangle), because if not your inequality will be wrong :).

For example, $ a\equal{}2,b\equal{}\minus{}2,c\equal{}\minus{}5$.

[quote="dduclam"]
Thank you Honey_S for solution. This inequality is easy enough. But i think the following stronger inequality also true:

$ (a \plus{} b \plus{} c)^5\ge 81 (a^2 \plus{} b^2 \plus{} c^2)(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)$[/quote]
The Schur's Inequality for third degree gives us
\[ abc \ge (a\plus{}b\minus{}c)(b\plus{}c\minus{}a)(c\plus{}a\minus{}b)\]
It suffices to prove that
\[ (a\plus{}b\plus{}c)^5 \ge 81abc(a^2\plus{}b^2\plus{}c^2)\]
Using the known (Newton's Inequality), we have
\[ 3abc \le \frac{(ab\plus{}bc\plus{}ca)^2}{a\plus{}b\plus{}c}\]
It suffices to prove that
\[ (a\plus{}b\plus{}c)^6 \ge 27(a^2\plus{}b^2\plus{}c^2)(ab\plus{}bc\plus{}ca)^2\]
By the AM-GM, we get
\[ 27(a^2\plus{}b^2\plus{}c^2)\cdot (ab\plus{}bc\plus{}ca) \cdot (ab\plus{}bc\plus{}ca) \le (a^2\plus{}b^2\plus{}c^2\plus{}2(ab\plus{}bc\plus{}ca))^3\equal{}(a\plus{}b\plus{}c)^6\]
we have done. :)

[quote="Honey_S"][/quote][quote="dduclam"][quote="arqady"][quote="dduclam"]Give $ a,b,c$ be real numbers. Prove that

$ 4(a^2 + b^2 + c^2)^2 \ge 9(a + b + c)(b + c - a)(c + a - b)(a + b - c)$[/quote]
Why not $ (a^2 + b^2 + c^2)^2 \ge 3(a + b + c)(b + c - a)(c + a - b)(a + b - c)$ $ ?$ :wink:[/quote]

Thank you arqady,that is inequality which I want post. The first,equality don't occur  :wink:

Thank you Honey_S for solution. This inequality is easy enough. But i think the following stronger inequality also true:

$ (a + b + c)^5\ge 81 (a^2 + b^2 + c^2)(a + b - c)(b + c - a)(c + a - b)$[/quote][quote="Honey_S"]

And I think that you need a condition for $ a,b,c$ (they are the side lenghts of triangle), because if not your inequality will be wrong :).

For example, $ a = 2,b = - 2,c = - 5$.[/quote]

Oh,$ a,b,c\ge0$,of course! But don't need $ a,b,c$ are the side lenghts of triangle.
If they aren't the side lenghts of triangle,inequality's also true  :wink:

[quote="Altheman"]Lemma 1: If $ a,b,c,(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)$ are positive, then $ (a \plus{} b \minus{} c)$, $ (b \plus{} c \minus{} a)$, and $ (c \plus{} a \minus{} b)$ are positive.
Proof: The conditions are symmetric, so WLOG $ a\ge b\ge c > 0$. Then $ a \plus{} b \minus{} c\ge b > 0$, and $ c \plus{} a \minus{} b\ge c > 0$. So the product of those three items are positive, so the third one must be positive also.

Problem: Let $ a,b,c > 0$, prove that
\[ (a \plus{} b \plus{} c)^5\ge 81 (a^2 \plus{} b^2 \plus{} c^2)(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)
\]
Proof:

Case 1: $ (a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b) < 0$. But this is obvious since the LHS>= 0 >= RHS.

Case 2: $ (a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b) > 0$. By Lemma 1, $ (a \plus{} b \minus{} c)$, $ (b \plus{} c \minus{} a)$, and $ (c \plus{} a \minus{} b)$ are positive. Let $ 2x \equal{} b \plus{} c \minus{} a > 0$, etc. So $ y \plus{} z \equal{} a$, and the inequality becomes:
\[ (2x \plus{} 2y \plus{} 2z)^5\ge 81((x \plus{} y)^2 \plus{} (y \plus{} z)^2 \plus{} (z \plus{} x)^2)(2x)(2y)(2z)
\]

\[ (x \plus{} y \plus{} z)^5\ge \frac {81}{2}(x^2 \plus{} y^2 \plus{} z^2 \plus{} xy \plus{} yz \plus{} zx)(xyz)
\]
Denote the LHS minus the RHS by $ t(x,y,z)$. It is easy to verify that
\[ t(x,2y,2z) \minus{} t(x,y \plus{} z,y \plus{} z) \equal{} \frac {81}{2}\cdot (y \minus{} z)^2(x)(x^2 \plus{} 2xy \plus{} 2xz \plus{} 3y^2 \plus{} 2yz \plus{} 3z^2)\ge 0
\]
Since $ x \plus{} y \plus{} z > 0$, so we can normalize to $ x \plus{} y \plus{} z \equal{} 2$, so it suffices to prove that $ t(2x,1 \minus{} x,1 \minus{} x)\ge 0$. But this is equivalent to
\[ (1 \minus{} 3x)^2(32 \minus{} 51x \plus{} 54x^2 \minus{} 27x^3) \equal{} (1 \minus{} 3x)^2(32(2 \minus{} x) \plus{} 13x \plus{} 27x^2(2 \minus{} x))\ge 0
\]
[/quote]
Thanks for your solution,Altheman. I'll check it late.

[quote="can_hang2007"][quote="dduclam"]
Thank you Honey_S for solution. This inequality is easy enough. But i think the following stronger inequality also true:

$ (a \plus{} b \plus{} c)^5\ge 81 (a^2 \plus{} b^2 \plus{} c^2)(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)$[/quote]
The Schur's Inequality for third degree gives us
\[ abc \ge (a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)
\]
It suffices to prove that
\[ (a \plus{} b \plus{} c)^5 \ge 81abc(a^2 \plus{} b^2 \plus{} c^2)
\]
Using the known (Newton's Inequality), we have
\[ 3abc \le \frac {(ab \plus{} bc \plus{} ca)^2}{a \plus{} b \plus{} c}
\]
It suffices to prove that
\[ (a \plus{} b \plus{} c)^6 \ge 27(a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca)^2
\]
By the AM-GM, we get
\[ 27(a^2 \plus{} b^2 \plus{} c^2)\cdot (ab \plus{} bc \plus{} ca) \cdot (ab \plus{} bc \plus{} ca) \le (a^2 \plus{} b^2 \plus{} c^2 \plus{} 2(ab \plus{} bc \plus{} ca))^3 \equal{} (a \plus{} b \plus{} c)^6
\]
we have done. :)[/quote]

Nice,Can  :)


Let ABCD be a quadrilateral inscribed in a circle (O) and let a, b be the perpendicular bisectors of the line segments OC, OD respectively. M is a point on the circle (O). The line MA meets b at E, the line MB meets a at F. Prove that the line EF is always tangent to a fixed circle when M moves on the circle (O)

Are you sure? It looked like an ellipse when I did it.


Among $ n>2$ points on the plane (not all on a line), every point is assigned a number from the set $ \{\minus{}1,0,1\}$ so that the numbers assigned to the points lying on a straight line sums to $ 0$. Prove that if there are more than $ 2$ points on a line, then all the points on the straight line were assigned $ 0$.

i didn't understand your problem,can u clarify?thx!(sorry for my poor english :blush: )
i've heard a problem which is like this:
we have some points on the plane (not all on a line) and we give to every point a number in such a way that each line passing through at least two of them will have the sum of the numbers on it equal to $ 0$
prove that we have given $ 0$ to all points.(aro 1977)
is that what u mean?


Three people— A, B and C —sit in a circle and successively toss a fair coin in the order A, B, C, A, B, C,etc.
The winner is the first person to throw a “head.”
What is the probability that person A wins this game?

The probability that A wins on the first flip is $\frac{1}{2}$.  Then the probability that he will win on the second flip is $\left(\frac{1}{2}\right)^3\times \frac{1}{2}$, because A, B, and C have to get tails on their first flips and A has to get heads on his second.  Continuing in this way, we find that the probability is $\sum_{k=0}^\infty \frac{1}{2^{3k+1}}=\frac{\frac{1}{2}}{1-\frac{1}{8}}=\frac{4}{7}$.  Is that right?

[quote="samath"] Is that right?[/quote]
yes
  :10:


For some $ p$, the equation $ x^3 \plus{} px^2 \plus{} 3x \minus{} 10 \equal{} 0$ has three real solutions $ a,b,c$ such that $ c \minus{} b \equal{} b \minus{} a > 0$. Determine $ a,b,c,$ and $ p$.

c-b=b-a  means   a+c=2b  which  means  a,b,c  are in AP.

       now  we  have a+b+c=p  which  means  b=-p/3

ab+bc+ca=3     and abc=10  which  means  ac=10/b =-30/p

   b(a+c) -30/p =3   $ \rightarrow$ 3b-30/p=3   $ \rightarrow$ -p-30/p=3

                 now  u have  a quadratic  in  p , solve it  to get p  and hence others


Fix $\alpha >0$ and choose $x_{1}>\sqrt{\alpha}$.  Define $\left\{x_{n}\right\}$ by the recursion formula

\[x_{n+1}=\frac{1}{2}\left(x_{n}+\frac{\alpha}{x_{n}}\right).\]
Show that $\lim x_{n}=\sqrt{\alpha}$.

I have shown that $x_{n}>\sqrt{\alpha}$ and $x_{n+1}<x_{n}$ for each $n\in\mathbb{N}$, but am unsure of how to procede from here.

Consider $y_{n}= \frac{x_{n}}{ \sqrt{\alpha}}$.

well you have shown that it is bounded and monotonic so we know that it has a limit.
one way to find the limit is to not that as $n\to\infty$ the terms $x_{n}$ and $x_{n-1}$ will be "the same" (really they just closer and closer together but lets just say that they are the same)
then we have

$L=\frac{1}{2}\left(L+\frac{\alpha}{L}\right)$
where $L$ is the limit of the sequence.

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=121408]this other topic[/url]. There's some redundancy between these two topics.

[quote="amcavoy"][color=darkred]Fix $a >0$ and choose $x_{1}>0$.  Define the sequence $\{x_{n}\}$
by the recursion formula $x_{n+1}=\frac{1}{2}(x_{n}+\frac{a}{x_{n}})\ .$ Show that $\lim x_{n}=\sqrt a$.[/color][/quote]

[color=darkblue][b][u]Proof.[/u][/b] Observe that for any $n\in N^{*}\ ,\ x_{n}>0\ .$ Therefore,

$\{\begin{array}{cccc}1.\blacktriangleright & x_{n+1}-\sqrt a=\frac{x^{2}_{n}-2x_{n}\sqrt a+a}{2x_{n}}=\frac{(x_{n}-\sqrt a)^{2}}{2x_{n}}\ge 0& \Longrightarrow & (\forall) n\ge 2\ ,\ x_{n}\ge \sqrt a\\\\ 2.\blacktriangleright & x_{n+1}-x_{n}=\frac{a-x^{2}_{n}}{2x_{n}}=\frac{(\sqrt a+x_{n})(\sqrt a-x_{n})}{2x_{n}}\le 0 & \Longrightarrow & (\forall) n\ge 2\ ,\ x_{n+1}\le x_{n}\end{array}$

$\Longrightarrow$ $\sqrt a\le \ldots\le x_{n+1}\le x_{n}\le \ldots\le x_{3}\le x_{2}\ ,\ x_{1}$ $\Longrightarrow$

([u]Weierstrass[/u]) the our sequence is convergent, i.e. exists $l\in [\sqrt a,x_{2})$ so that $x_{n}\searrow l\ .$

For $n\to\infty$ in the recursion definition of this sequence obtain

$l=\frac{1}{2}\cdot(l+\frac{a}{l})$ $\Longrightarrow$ $\{\begin{array}{c}l^{2}=a\\\\ l\ge \sqrt a\end{array}\|$ $\Longrightarrow$ $\lim_{n\to\infty}x_{n}=\sqrt a\ .$[/color]


Find all positive integer solutions to $ y^{2}(x\plus{}1)\equal{}1576\plus{}x^{2}$

idea:
Maybe it would be easier to see if we replace $ x\plus{}1$ with $ w$ then we have
$ y^{2}w \equal{} 1576\plus{}(w\minus{}1)^{2}\equal{} w^{2}\minus{}2w\plus{}1577$
and note that $ 1577 \equal{} 19\cdot 83$.
This choices for $ w$ are limited.
Is this enough help?


linear algebra
matrix
parameterization
LaTeX
vector
algebra
system of equations

Hi,

I just started taking Linear Algebra in college and am very confused by the conditions upon homogeneous and inhomogeneous matrices are solvable. I am more of the type of student to see an specific example done and then understand the concept, but our teacher refuses to give us a concrete example.

So I must beg of you, can you guys show me specific examples where homogeneous and inhomogeneous matrices have nontrivial solutions? And how to solve them step by step? Any help would be very very appreciated!

Linear algebra is a vocabulary-driven subject. The first thing you have to do is use the vocabulary correctly - and you're not doing that.

There's no such thing as a homogeneous matrix or a nonhomogenous matrix. The word "homogeneous" should not modify the word "matrix." Instead, the word "homogeneous" should modify "equation" or "system of equations."

That is, there are homogeneous systems of linear equations and there are nonhomogeneous systems of linear equations. A homogeneous system can be written in matrix form as $ Ax=0.$ A nonhomogeneous system can be written in matrix form as $ Ax=b,$ where $ b$ is not the zero vector.

Here's an example. Consider
\begin{align*}x_1-x_2+2x_3&=0\\2x_1+3x_2-x_3&=0\end{align*}
That's a homogeneous system. We can give it in the shorthand of an augmented matrix as
\[ \begin{bmatrix}1&-1&2&0\\2&3&-1&0\end{bmatrix}\]
or we may write it as a matrix equation as
\[ \begin{bmatrix}1&-1&2\\2&3&-1\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}\]

Now: try solving that. We know before we start that we have the trivial solution $ x_1=0,x_2=0,x_3=0.$ But by row elimination, can you find any solutions other than that?

Thanks for the response Kent.

You are totally correct, it should have been homogeneous system of equations, I should have been more careful with the language.

I have derived that matrix to be [10 10] in its row echelon form.
                                               [01-10]

As for solutions, it seems that x3 is the free parameter so another solution besides the trivial could be x3=x3, x2=x3, -x1=x3.
I did that by subbing in random numbers to x3. It seems like a homogeneous system of equations has a nontrivial solution as long as there is a column in the matrix without a leading element. Or in another works, rank of that matrix is less than columns.

However, besides randomly subbing in numbers, is there another way to find the solutions? My book goes into the "s" and "t" variables, which I don't quite get.

Also, more importantly, how about the solutions of non homogeneous equations?

Thank you so much again Kent.

I don't have time to read your answer right now since I have to go off to class - in fact, it's the same class you're taking, and there's a test next Monday. (One note: click "quote" to see the $ \text{\LaTeX}$ syntax for matrices; that would make your answer a whole lot easier to read.)

Here's an example, taken from a test a year or so ago, that I plan to use in today's lecture:

2. Let $ \mathbf{u} = \begin{bmatrix} - 2 \\
4 \\
9\end{bmatrix},\quad \mathbf{v} = \begin{bmatrix}2 \\
2 \\
3\end{bmatrix},\quad \mathbf{w} = \begin{bmatrix} - 10 \\
2 \\
9\end{bmatrix}.$

(a) Let $ \mathbf{b} = \begin{bmatrix}8 \\
2 \\
0\end{bmatrix}.$ Is $ \mathbf{b}\in\text{Span}\{\mathbf{u},\mathbf{v},\mathbf{w}\}?$

(b) Find [i]all[/i] possible ways that $ \mathbf{b}$ can be written as a linear combination of $ \mathbf{u},\mathbf{v},$ and $ \mathbf{w}.$

(c) Is the set $ \{\mathbf{u},\mathbf{v},\mathbf{w}\}$ independent or dependent?

(d) Is it possible to find a vector in $ \mathbb{R}^3$ which is not contained in $ \text{Span}\{\mathbf{u},\mathbf{v},\mathbf{w}\}?$ If it is possible to do so, write down one such example.


AMC
AIME
USA(J)MO
USAMO
USAMTS
Purple Comet

Are there any international contests that are free and are open to anyone?

I guess the best way to go to the international competition is this way (if you are in US) --> American Mathematics Contest (AMC), --> the American Invitational Mathematics Examination (AIME),-->  the United States of America Mathematical Olympiad (USAMO),-->  International Mathematical Olympiad (IMO)

I just hope I can make it into top 1% in this year's AMC...

Hi

I am going to study medicine in china .
i was given a top university in china.
Anybody has any idea on medical education in china.
I was given admission through this company http://www.studyoverseaschina.com
Please send me a sugesstion..

Regards
Thomas.A

Tournament of Towns is open to everyone and it is a very prestigious competition. Does somebody else know other prestigious competitions which are available for high-school students studying in Europe? :maybe:

Thank you,
Excellence

do you know a competition for international students which is similar to USATMS?
 :roll:

east europe, but i think this is not that important   :o   :huh:

no coz then i ll tell the competitons in ur region

as i said in my previous post, i'm from east europe (or i should be more specific? :) )

[url=http://en.wikipedia.org/wiki/List_of_mathematics_competitions]Here[/url] you can find an overview of math competitions. For Bulgaria I recommend:

- Bulgaria Mathematial Olympiad
- Balkan Mathematial Olympiad
- Tournament of Towns
- Mathlinks Contest

I know the iTest is a team-based test that includes a few IMO-level problems at the end (I think).. :oops:

[quote="rt_08"]hey is tournament of towns in india also?[/quote]

How about asking those people who are responsible for ToT?  :D

[quote="rt_08"]hey i dont know bout them but could u please help me in this problem?
'thanx[/quote]

This is the page for the [url=http://www.math.toronto.edu/oz/turgor/forms/start.php]ToT registration form from the University of Toronto.[/url]

You can find the problems for 1980-2007 on the Russian [url=http://www.turgor.ru/problems.php]Tugor site.[/url] For 2001 to 2008 you can find them with solutions in the English [url=http://www.math.toronto.edu/oz/turgor/archives.php]University Toronto archive.[/url] For the period 1980-2002 there are [url=http://www.mathpropress.com/books/Towns/index.html]five books for ToT.[/url]

The history of the "International Mathematics Tournament of Towns" can be found [url=http://www.amt.canberra.edu.au/imtot.html]here.[/url]

Besides ToT
Is there any more contests that are open for all to join?
Or for me to enroll in?

BTW, I'm from Hong Kong.

So how do I confirm if ToT is held in India or not?? And the history link doesn't work...

You can participate in the Purple Comet[color=indigo][size=150][i]![/i][/size][/color] Math Meet from anywhere. Form a team from one to six students and find an adult to supervise the team. Have them sign up at [url]http://purplecomet.org[/url]. It's free!


Let $ x;y;z$ be real numbers.Prove that:
$ \sum_{cyc} \frac{(x\minus{}y)^4}{(y\minus{}z)(z\minus{}x)} \leq \frac{15}{2}[(x\minus{}y)^2\plus{}(y\minus{}z)^2]$

nice, indeed!

note that $ \sum_{cyc}\dfrac{(x\minus{}y)^4}{(y\minus{}z)(z\minus{}x)}\equal{}5(a^2\plus{}ab\plus{}b^2)$, where $ a\equal{}x\minus{}y$ and $ b\equal{}y\minus{}z$, then, the inequality becomes $ 2(a^2\plus{}ab\plus{}b^2)\leq 3(a^2\plus{}b^2)$ which is easily proven.


ratio
probability
geometry
geometric transformation
rotation
induction
LaTeX

Here's how this works. One person posts a problem, a second posts a solution [b](not just an answer, a solution)[/b] and posts a second problem, a third posts a solution and posts a third problem, etc. These problems should be all COMBINATORICS. I shall start.

[b][color=red]Problem 1[/color][/b]
An unfair six-sided die with sides labeled 1 through 6 is weighted so that the ratio of the probability of rolling an $ x$ and the probability of rolling a $ y$ is equal to $ \frac{y}{x}$. Gurtej rolls the die two times. What is the probability that the sum of the results of the two rolls is even?

The probabilties of rolling a 1, 2, 3, 4, 5, or 6 are $ \frac{1}{21}$, $ \frac{2}{21}$, $ \frac{3}{21}$, $ \frac{4}{21}$, $ \frac{5}{21}$,  and $ \frac{6}{21}$ respectively. Thus, the probability of rolling a sum that is even is $ \frac{(1\plus{}3\plus{}5)^2\plus{}(2\plus{}4\plus{}6)^2}{21^2}\equal{}\frac{15^2}{21^2}\equal{}\frac{25}{49}$.

New problem: In how many ways can 8 people sit around a circular table if Abe and Bob refuse to sit next to each other but Chad and Dan insist on sitting next to each other?

1. I don't get it.
2. Hide your solution.
3. Why are you posting in a state that you don't even live in?

[hide]Let $ p$ be the probability of rolling a 1. Then the probabilities of rolling 2, 3, 4, 5, and 6 are, respectively, $ p/2$, $ p/3$, $ p/4$, $ p/5$, and $ p/6$. Then, we get $ p \plus{} p/2 \plus{} p/3 \plus{} p/4 \plus{} p/5 \plus{} p/6 \equal{} 1$, so $ p \equal{} 60/147$. The probability of rolling an odd number on one die is then $ 60/147 \plus{} 20/147 \plus{} 12/147 \equal{} 92/147$, and the probability of rolling an even number on one die is $ (147 \minus{} 92)/147 \equal{} 55/147$. The probability of getting the sum of two rolls even is then $ (\frac {92}{147})^2 \plus{} (\frac {55}{147})^2 \equal{} \frac {11489}{21609}$.[/hide]

[color=red][b]Problem 2[/b][/color]
When $ (a \plus{} 2b \plus{} \pi c)^{2009}$ is expanded and like terms combined, how many terms will there be?

[hide="solution"]Each term will be in the form $ k*a^{a_1}*b^{b_1}*c^{c_1}$, where $ k$ is some constant and $ a_1$, $ b_1$, and $ c_1$ are non-negative integers that sum to 2009. Therefore there is a one-to-one correspondence on the number of terms to the number of non-negative solutions to $ a_1\plus{}b_1\plus{}c_1\equal{}2009$. There are $ \binom{2009\plus{}3\minus{}1}{3\minus{}1}\equal{}\binom{2011}{2}\equal{}\frac{2011*2010}{2}\equal{}2021055$ solutions to that equation, and therefore there are that many terms in [/hide]

Problem three:
A hexagonal dartboard of side length six is colored three colors: red, blue, and green. The blue region is a hexagon centered at the center of the big hexagon and has a side length of two, the red region combined with the blue region is a hexagon with a side length of four, and the green region is everything outside the red and blue regions. Three darts are thrown at the board. What is the probability that two land on the blue region and one lands on the green region?
[/img]

[hide="Solution"] The probability that a dart lands in the blue region is $ (\frac{2}{6})^2 \equal{} \frac{1}{9}$. The probability that a dart lands in the green region is $ 1 \minus{} (\frac{4}{6})^2 \equal{} \frac{5}{9}$. Thus the desired probability is $ \binom{3}{1}(\frac{1}{9})^2(\frac{5}{9})^1 \equal{} \frac{5}{243}$. [/hide]

[color=red]Problem 4[/color]
How many integers can be expressed as a sum of two or more different members of the set {0, 1, 2, 4, 8, 16, 31}?

I don't live here, but I'm acting like AIME15 who always posts in our state forum: Since all the sums are going to be different, We have:
$ \binom{7}{2} \plus{} \binom{7}{3} \plus{} \binom{7}{4} \plus{} \binom{7}{5} \plus{} \binom{7}{6} \plus{} \binom{7}{7} \equal{} 120$

P.S.: I'm probably wrong.. Cause I come from Oklahoma.... :(

[quote]all the sums are going to be different[/quote]
$ 0 \plus{} 31 \equal{} 16 \plus{} 8 \plus{} 4 \plus{} 2 \plus{} 1$

[color=#666666]Hi.......[/color]

$ E$xcuse me I don't live in Upstate NY too but please allow me to go here :wink:
Answer $ P.4$ is $ 63$ because we can make from $ 0$ to $ 62$  easy.(very easy prove it) :lol:

[color=red]PROBLEM [size=200]5[/size][/color] (This is tinytim's problem in #2. It's very easy.I don't remember good problem :( ) [color=red]:[/color]
In how many ways can $ 8$ people sit around a circular table if Abe and Bob refuse to sit next to each other but Chad and Dan insist on sitting next to each other?

[color=#666666]Bye......[/color]

The answer to problem 4
[hide="is"]62. vandermond was correct up until he stated that 0-62 were attainable. 0 is in fact not attainable since we are talking about the sum of two or more numbers.[/hide]

[hide="Problem 5"]We can first assume that they stand in a line then divide by 8 to account for rotations of the table.

We can do complementary counting, subtract the number of ways that Abe and Bob sit together as well as Chad and Dan sitting together from the number of ways that Chad and Dan sit together.

the number of ways that Chad and Dan sit together: We can treat Chad and Dan as a block. There are two ways to arrange the people in the block and there are $ 7!$ ways to arrange the six other people plus the block. Therefore there are $ 7!*2\equal{}10080$ total ways to arrange them.

the number of ways that Abe and Bob sit together as well as Chad and Dan sitting together: Using the same method we have $ 6!*2*2\equal{}2880$.

$ 10080\minus{}2880\equal{}\boxed{7200}$[/hide]

[color=darkred]Problem 6[/color]

(a) If we remove two opposite corners from a standard 8 by 8 chessboard, is it possible to tile it with 2 by 1 dominoes?
(b) If we remove all four corners, is it possible?
(c) Jake removes two squares from a standard 8 by 8 chessboard randomly. What is the probability that he can tile the chessboard with dominoes [i]now[/i]?

[hide]
(a) No; every tile covers 1 black and 1 white square, but there are 32 of one color and 30 of another in the mutilated chessboard.
(b) yes:

First tile a 4x4 like this:

 11
2554
2664
 33

Now put four of these together and put 2 dominoes in the center.

(c)If two squares of the same color are chosen, then by (a) it cannot be tiled
Bleh, to prove that any chessboard with squares of different color squares removed can be tiled prove it for 2x2 and prove it for any $ 2^nx2^n$ board by induction on n.

 It's $ \frac{(64)(32)}{(64)(63)}\equal{}\frac{32}{63}$ [/hide]

[quote="1=2"]We can first assume that they stand in a line then divide by 8 to account for rotations of the table.

We can do complementary counting, subtract the number of ways that Abe and Bob sit together as well as Chad and Dan sitting together from the number of ways that Chad and Dan sit together.[/quote]

First of all, you neglected to count the possibilities where you split the circle into a line by dividing immediately between Abe and Bob.  Thus, you should also have subtracted the $ 2 \cdot 5! \cdot 2$ arrangements that we can get like that.

Second of all, you never divided by 8.  (This should have been obvious to you, since your answer is larger than the $ 7!$ distinct ways to seat eight people around a round table.)

Third of all, the use of either "*" or "x" for multiplication in LaTeX is really ugly: use \cdot or \times instead.

[color=#666666]Hi.........[/color]

[color=red]You[/color] don't write answer of Problem $ 5$. it's very easy and I don't think that you can't solve it or write false solution. Please in the first answer $ P.5$

[color=#666666]Bye........[/color]

So... no one decided to actually put up a new problem! ^^
An easy one, lovingly ripped off from an AAMC night and then changed:
[color=blue]Problem 7:[/color]
Solve for n<1500:

$ \binom{1663}{3000} \plus{} \binom{1664}{3000} \equal{} \binom{n}{3001}$

You appear to have your binomial coefficients upside down  :wink:  :P

[quote="BlueMouse"]So... no one decided to actually put up a new problem! ^^
An easy one, lovingly ripped off from an AAMC night and then changed:
[color=blue]Problem 7:[/color]
Solve for n<1500:

$ \binom{3000}{1663} \plus{} \binom{3000}{1664} \equal{} \binom{3001}{n}$[/quote]

Since $ \binom{y}{x}\plus{}\binom{y}{x\plus{}1}\equal{}\binom{y\plus{}1}{x\plus{}1}$, we have $ n\equal{}1664$ here.

Answer: 1664

Actually, n<1500, and $ \binom{3001}{1664}\equal{}\binom{3001}{1337}$, so 1337.

Yes that was definitely a coincidence.

[quote="gauss1181"][quote="BlueMouse"]So... no one decided to actually put up a new problem! ^^
An easy one, lovingly ripped off from an AAMC night and then changed:
[color=blue]Problem 7:[/color]
Solve for [b]n<1500[/b]:

$ \binom{3000}{1663} \plus{} \binom{3000}{1664} \equal{} \binom{3001}{n}$[/quote]

Since $ \binom{y}{x} \plus{} \binom{y}{x \plus{} 1} \equal{} \binom{y \plus{} 1}{x \plus{} 1}$, we have $ n \equal{} 1664$ here.

Answer: [b]1664[/b][/quote]

NOT READING INSTRUCTIONS PHAIL
Honestly, what is it we Upstate NY people are thinkiing half the time? We can't count, we don't read instructions...  :rotfl:
So, who is going to put up the next problem?

Um, I have no idea what's going on here.

Anyway, here's a problem:

So Bob is in the Cartesian plane, at the point $(0,0)$. Bob, who can only travel (in a "step") to any of the four points horizontally or vertically adjacent to his previous location, wants to visit every lattice point in (and on the boundaries of) the square with vertices at $(0,0)$, $(0,2012)$, $(2012,0)$, and $(2012,2012)$. Being a picky fictitious character, he also wants to do this in exactly $2013^{2}-1$ steps.

If Bob randomly chooses a path which satisfies all of the above conditions, what is the probability that he visits the point $(42,1337)$ before he visits the point $(2011,2011)$?

:D


Fie $G$ un grup in care nici un element nu are ordinul $2.$ Daca $(xy)^{2}=(yx)^{2}$ pentru orice $x, y\in G$,
 atunci grupul $G$ este abelian. ;)

Este suficient sa aratam ca functia $f: G\longrightarrow G$, $f(t)=t^2$ este injectiva. Fie $u,v\in G$ cu $u^2=v^2$.
Atunci $uv^{-1}=u^{-1}v$, astfel ca $\left(uv^{-1}\right)^2=\left(u^{-1}v\right)^2=\left(vu^{-1}\right)^2=\left(uv^{-1}\right)^{-2}$.
Rezulta ca $\left(\left(uv^{-1}\right)^2\right)^2=1$, deci $\left(uv^{-1}\right)^2=1$, deci $uv^{-1}=1$, adica $u=v$.
Functia $f$ este deci injectiva.

That's all, folks!

Da, este injectiva. Dar de unde reiese ca-i abelian? E ceva care-mi scapa? :blush:

it  is trivial to see that from $f(t)=t^{2}$ injective we and from the hypothesys $(xy)^{2}=(yx)^{2}$ it follows that $xy=yx$ and the conclusion...... :P

Nice Cezar, banuiesc ca asta inseamna sa fii cool. Stii ce ar fi frumos: sa postezi problema, apoi Razec sa o rezolve si apoi Upul sa aplaude reusita lui Razec. Ar fi cool, nu-i asa?  :?

Cred ca da. Tu ce zici? :D Good joke, Gabi. :rotfl:


Are there any details on when the problems are coming out? (Aside from "late January", as said on the website.)

I have nothing to add to "late January".  It will be announced here when the problems are available.


calculus
integration
function
logarithms
calculus computations

Define the function $ f(x)$ as : $ \displaystyle f(x)\equal{}\left\{
\begin{array}{ll}
\frac{\ln (1\plus{}x)}{1\plus{}x}\ (x\geq 0) &\quad  \\
\frac{\ln (1\minus{}x)}{1\minus{}x}\ (\minus{}1\leq x<0) &\quad
\end{array}
\right.$

1. Examine the variation of $ f(x)$ and find the maximum and minimum value of $ f(x)$.

2. Find the value of $ a$ for which $ \int_{\minus{}1}^{e\minus{}1} |f(x)\minus{}a|\ dx$ is minimized for $ 0\leq a\leq \frac{1}{e}$.

umm.. what do you mean by variation in $ f(x)$ ?? :maybe:

I meant that "variation" is the increse or decrease, what do you say in appropriate word? I used from French word : Tableau de variations.


symmetry calculus integration trigonometry

[quote="shreyas_patankar"]Use Biot-Savaart Law to calculate field at $ P$ due to one segment, take its normal component and multiply by 4.[/quote]

This is a correct approach. The reason why it works is because the segments are all the same length, and as the current travels around the loop, the symmetry is such that the components of field in the normal direction, $ \hat{x}$ according to the diagram, add (constructively interfere); while the components of parallel to the plane, in the $ \hat{y}$ and $ \hat{z}$ directions, cancel one another (destructively interfere). Which I believe is the point of the problem.

I know the approach, but I can't get the answer.Can someone post the full solution?

:dry: I was trying to avoid the slogging...
Anyway, you must be having trouble withthe integral (as there is nothing else that can possibly go wrong).
For the moment, let $ L \equal{} 2a$
Choose the midpoint of a side as the reference, and consider a small length of the wire $ dz$ at $ z$ from the reference.
By Biot-Savaart law, field due to the length is $ \frac {\mu_0}{4\pi}\frac {\sqrt {a^2 \plus{} x^2}idz}{\left( a^2 \plus{} x^2 \plus{} z^2 \right) ^{\frac {3}{2}}}$
using substitution $ z \equal{} \sqrt {a^2 \plus{} x^2}\tan\theta$
we get the integral as $ \vec{B_1} \equal{} \frac {\mu_0 i}{2\pi\sqrt {a^2 \plus{} x^2}}\frac {a}{\sqrt{x^2 \plus{} 2a^2}}$
Take the normal component of this and multiply by 4.


There is a very nice article of Valery Senderov and Alexander Spivak in [i]Quantum (year 2000)[/i]. One nice problem from this article:

For all pisitive integer s there exist number n such that S(n.s) = s, where
S(m) denote sum of digits of m.

Namdung

the answer is for all positive integers s. Let s= (10^t)*r
r is not divisible by 10. For s>=20 we can choose
n= (10^((s-S(r))/9) -1)*10^k + 1, where k is the number of the digits of r. By the fact that S((10^l-1)*x)= 9*l if 10^l -1 >= x and x is not divisible by 10, we see that S(n*s)=S(n*r)=9*(s-S(r))/9  + S(r) = s.
( for s>=20 we always have 10^((s-S(r))/9)-1 >= r )
It is also easy to see that for all 0< s <20 there exist n such that S(n*s)=s. ( by straight calculations).


The circles $C_{1},C_{2},C_{3}$ with radii $5$ and with centers $O_{1},O_{2},O_{3}$ respectively are drawn in plane, such that: $O_{1}O_{2}=6$, $O_{1}O_{3}=8$ and $O_{1}O_{3}\bot O_{1}O_{2}$. Find the area of that part of $C_{1}$ which is not common with $C_{2}$ and $C_{3}$.

There must be some mistake here - the circles don't intersect ($C_{1}$ and $C_{2}$ are tangent, $C_{1}$ and $C_{3}$ are disjunct), hence $A=9\pi$.

Oops :oops:  sorry, The radii of the circles are $5$....
sorryyyyyyyyyyy!

See the attached diagram.

Since $O_{2}O_{3}=\sqrt{6^{2}+8^{2}}=10$, circles $C_{2}$ and $C_{3}$ are tangent.

$BM=O_{1}N={O_{1}O_{2}\over 2}=3$, $BN=O_{1}M={O_{1}O_{3}\over 2}=4$

One of the two circular segments defined by the chord $AB$ has the area

$A_{1}={5^{2}\over 2}\cdot 2\arctan{3\over 4}-\frac{6\cdot 4}{2}=25\arctan{3\over 4}-12$

One of the two circular segments defined by the chord $BC$ has the area

$A_{2}={5^{2}\over 2}\cdot 2\arctan{4\over 3}-\frac{8\cdot 3}{2}=25\arctan{4\over 3}-12$

The total area of those segments is

$A'=2(A_{1}+A_{2})=50(\arctan{3\over 4}+\arctan{4\over 3})-48=50\cdot{\pi\over 2}-48=25\pi-48$

Hence the desired area is $A=5^{2}\pi-A'=48$


On an inﬁnite chessboard (whose squares are labeled by $ (x, y)$, where $ x$ and $ y$ range over all integers), a king is placed at $ (0, 0)$. On each turn, it has probability of $ 0.1$ of moving to each of the four edge-neighboring squares, and a probability of $ 0.05$ of moving to each of the four diagonally-neighboring squares, and a probability of $ 0.4$ of not moving. After $ 2008$ turns, determine the probability that the king is on a square with both coordinates even. An exact answer is required.

[hide="a start"]
...maybe

ignore the diagonal moves and the non moving, because they leave the parity of the sum alone
then theres a .4 chance it'll change the parity, because it starts at (0,0)
it must do this an even number of times
so how can we arrange this
it's $ (.4)^0(.5)^{2008}\binom{2008}{0} \plus{} (.4)^2(.5)^{2006}\binom{2008}{2} \plus{} \cdots \plus{} (.4)^{2008}(.5)^0\binom{2008}{2008}$

too sucky at math to follow up

[/hide]

[quote="Teki-Teki"]More precisely, it has a 0.4 probability of changing parity. (0.1 for [i]each[/i] of the edge squares)[/hide][/quote]

i see, i misread that
i still have no chance of calculating the exact probability though
i'll leave that to someone with a very powerful calculator or beastly number theory.

[hide="hint to finish"]Consider $ \frac{(1 \plus{} x)^n \plus{} (1 \minus{} x)^n}{2}$ for the appropriate values of $ n$ and $ x$.[/hide]

Teki-Teki, your answer is wrong.  That's the probability that the king ends up on a square with coordinates of the same parity, but we want the probability that it ends up on a square with both coordinates even.  Unfortunately I don't think you can get the answer directly from what you have.

On the test I assumed you could consider the $ x$ and $ y$ coordinates independently, which is wrong because the probability of the $ x$ coordinate changing parity is not independent of the probability of the $ y$ coordinate chaning parity.  :(

It looks like a recursive solution is also possible, although rather computational.

[hide="Hint for simplification"]
A vertical move changes the coordinates' parity the same way whether it's up or down. Similarly, we can treat horizontal moves equivalently and diagonal moves equivalently.

We simplify the probabilities to:
1/5 that we change the parity of the x coordinate
1/5 that we change the parity of the y coordinate
1/5 that we change the parity of both coordinates
2/5 that we maintain the parity of both coordinates

And now you can be like, number of vertical movements is V, number of horizontal is H, number diagonal is D, number with no movement is N, and V+H+D+N=2008, V+D is even, H+D is even.
[/hide]

[hide="hint"]
It is necessary and sufficient that the numbers of the four types of moves (change neither parity, change just x parity, change just y parity, and change both parities) are the same in parity.  I found this on the test, but didn't have time to calculate it out.

From here, you simply make a summation with $ \frac{1}{5^{2008}}$ in front, where the thing being summed is $ 2^i$ times $ \binom{2008}{i}$ times the number of ways to have $ 2008\minus{}i$ ordered objects of $ 3$ types so that the number of objects of each type has the same parity as $ i$.
[/hide]


Maybe some of you just haven't seen [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=317834]this[/url]. Namely prove that for every real number $ x,y,z$ we have
\[ \frac{|x\minus{}y|}{\sqrt{1\plus{}x^2}\sqrt{1\plus{}y^2}}\plus{}\frac{|y\minus{}z|}{\sqrt{1\plus{}y^2}\sqrt{1\plus{}z^2}}\geq \frac{|x\minus{}z|}{\sqrt{1\plus{}x^2}\sqrt{1\plus{}z^2}}.\]

[quote="MeKnowsNothing"]Maybe some of you just haven't seen [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=317834]this[/url]. Namely prove that for every real number $ x,y,z$ we have
\[ \frac {|x \minus{} y|}{\sqrt {1 \plus{} x^2}\sqrt {1 \plus{} y^2}} \plus{} \frac {|y \minus{} z|}{\sqrt {1 \plus{} y^2}\sqrt {1 \plus{} z^2}}\geq \frac {|x \minus{} z|}{\sqrt {1 \plus{} x^2}\sqrt {1 \plus{} z^2}}.\]
[/quote]
$ \sqrt {x^2 \plus{} 1} \equal{} |x \minus{} i|, \sqrt{y^2\plus{}1} \equal{} |y\minus{}i|, \sqrt{z^2\plus{}1} \equal{} |z\minus{}i|$, so it is equivalent to:
$ |(x \minus{} y)(z \minus{} i)| \plus{} |(y \minus{} z)(x \minus{} i)| \ge |(x \minus{} z)(y \minus{} i)|$
But $ (x \minus{} y)(z \minus{} i) \plus{} (y \minus{} z)(x \minus{} i) \equal{} (x \minus{} z)(y \minus{} i)$..


geometry
3D geometry
quadratics
induction
modular arithmetic
algebra
binomial theorem

Regional Mathematical Olympiad, 2004 was held on 5th December (this is the first level in the selection procedure for the Indian team to the IMO). Here are the problems (3 hours)

1. Consider in the plane a circle T with center O and a line l not intersecting T. Prove that there is a point Q on the perpendicular drawn from O to the line l, such that for any point P on the line l, PQ represents the length of the tangent from P to the circle T.

2. Positive integers are written on all the faces of a cube, one on each. At each corner (vertex) of the cube, the product of the numbers on the faces that meet at the corner is written. The sum of the numbers written at all the corners is 2004. If T denotes the sum of the numbers on all the faces, find all possiblem values of T.

3. Let $\alpha$ and $\beta$ be the roots of the quadratic equation $x^2+mx-1=0$, where m is an odd integer. Let $\lambda_{n}=\alpha^n + \beta^n$, for $n>=0$. Prove that for $n>=0$,
   (a)$\lambda_n$ is an integer; and
   (b)gcd($\lambda_n, \lambda_{n+1}$) = 1

4. Prove that the number of triples (A, B, C) where A, B, C are subsets of {1,2,...,n} such that $A \cap B \cap C = \phi$, $A \cap B \neq \phi$, $B \cap C \neq \phi$ is $7^n - 2.6^n + 5^n$

5. Let ABCD be a quadrilateral; X and Y be the midpoints of AC and BD respectively; and the lines through X and Y respectively parallel to BD, AC meet in O. Let P, Q, R, S be the midpoints of AB, BC, CD, DA respectively. Prove that
.....(a) quadrilaterals APOS and APXS have the same area;
.....(B) the areas of the quadrilaterals APOS, BQOP, CROQ, DSOR are all equal equal

6. Let $<p_1, p_2, p_3,...,p_n,...>$ be a sequence of primes difined by $p_1=2$ and for $n>=1$,$ p_{n+1}$is the largest prime factor of $p_1p_2...p_n + 1$. (Thus $p_2=3$, $p_3=7$). Prove that $p_n \neq 5$ for any n.

7. Let x and y be positive real numbers such that $y^3+y<=x-x^3$. Prove that
...(a) $y < x < 1$; and
...(b) $x^2+y^2 < 1$.

3) the sequence $( \lambda_n )$ staifies the relation :
$\lambda_{n+2} = -m \lambda_{n+1} + \lambda_n $,
with $\lambda_0 = 2$ and $\lambda_1 = -m$.

Therefore all terms are clearly integers.
And two consecutive terms are coprime by an easy induction (the fact that $m$ is odd is useful for the initialization of the induction)

Pierre.

4) We may construct a 'good' triple as follows :
- First we select $B$ with $|B| = k \geq 2$.
There are $C_n^k$ such choices.
- For each of them, we select two disjoint and non-empty subsets $I,J$ which will be $I = A \cap B$ and $J = B  \cap C$. This is the most difficult part of the problem :
Note that since there is an order among $A,B,C$ there is an order among $I$ and $J$.
There are $C_k^i$ ways to select $I$ with $|I| = i \geq 1$. Then to choose $J$ we may choose any non-empty set in $B \setminus I$ so that there are $2^{k-i} - 1$ choices for $J$. It follows that for a given set $B$ as above, the number of choices for $I,J$ are :
$N_k = \dsp \sum_{i=1}^{k-1} (2^{k-i}C_k^i - C_k^i) = \sum_{i=0}^k (2^{k-i}C_k^i - C_k^i)  - (2^k - 1).$
Using the binomial theorem, we deduce that $N_k = 3^k - 2^{k+1} + 1.$
Note that $N_0=N_1 = 0$.

- Now we complete the sets $A$ and $C$ by any respective subsets of $\{1, \cdots ,n \} \setminus B$, thus there are $2^{n-k}$ ways to complete $A$ and the same for $C$.
Thus the number of good triples is :
$\dsp \sum_{k=2}^n N_kC_n^k 2^{n-k}2^{n-k} =  \sum_{k=0}^n N_kC_n^k 2^{n-k}2^{n-k} = 4^n  \sum_{k=0}^n (\frac 3 4)^kC_n^k -2(\frac 1 2)^kC_n^k  + (\frac 1 4)^kC_n^k ) = 4^n( ( 1 + \frac 3 4 )^n - 2(1+ \frac 1 2)^n + (1 + \frac 1 4)^n) = 7^n - 2 \cdot 6^n + 5^n$, as desired.

Pierre.

Ah, Pierre! It is [b]Pre-Olympiad[/b]! ;)

I can't understand 6.

Problem 2 is the same as baltic Way 2004, pr.6

For 6) I think it means $p_n \neq 5$.

Pierre.

Then problem is trivial, just prove by induction that $p_1p_2...p_n+1=3\pmod{4}$ for $n\geq 2$.

I didn't say it was not easy  :D

Pierre.

5)For a) it is sufficient to prove that $S_{PXS}=S_{POS}$ which is obvious because XO is parallel to PS.
b)Now it is easy to prove that  $S_{APXS}=\frac{S_{ABCD}}{4}$.From this and a) easily follows b). :)

for q7-

a. rewite as y/x<=(1-x^2)/(1+y^2)
since  y,x are positive,  x has to be <1, RHS has to be <1 and y has to be <x

b.again rewrite as y/(1-x^2)<=x/(1+y^2);
since RHS is +ve and less than 1;y<=1-x^2, and since y<1, y^2<1-x^2 or  x^2+y^2<1

1) The circles with centers $P$ on $\ell$ and passing through the points where the tangents from $P$ touch $\mathcal T$ are orthogonal to $\mathcal T$, so two such circles intersect in two points collinear with $O$ (it's clear that any two such circles do indeed intersect). Let $X$ be the foot of the perpendicular from $O$ to $\ell$, and let $t$ be the length of the tangent from $X$ to $\mathcal T$. It's easy to see now that the two points mentioned above must be the points where $(X,t)$ cuts the perpendicular from $O$ to $\ell$.

Thanks to everyone for their solutions

[quote]For 6) I think it means $p_n \neq 5$. [/quote]
Yes, sorry for the error. I'll correct that.

My solution to the 4th one was somewhat simpler:
First the number of triples which have $A \cap B \cap C = \phi$ is $7^n$ (there are 7 possibilities for all i). The number of triples which have $A \cap B = \phi or B \cap C = \phi$ or both is $6^n + 6^n - 5^n$. Subtracting we get the result.


For all positive real numbers $x,y,z$, its true:

$x+y+z=4xyz(\frac{1}{(x+y)^{2}}+\frac{1}{(x+z)^{2}}+\frac{1}{(y+z)^{2}})$

It must is

For all positive real numbers $x,y,z$, its true:

$x+y+z\geq 4xyz(\frac{1}{(x+y)^{2}}+\frac{1}{(x+z)^{2}}+\frac{1}{(y+z)^{2}})$

So if, use $(x+y)^{2}\geq 4xy$, we obtain $\frac{1}{xy}\geq \frac{4}{(x+y)^{2}}$.


we have two circles and a line $l$ find point $A$ on $l$ that when we tangent two line on two circles the sum of tangent lenghts be minimum.

that was realy nice problem and here is my solution:
we took points $P$,$Q$ which that the line $p$ be the radial axis of $P,C_1$ and $Q,C_2$. therefore: $AB+AC=AP+AQ$ and from here it's easy

This solution is more or less equivalent to the one I gave at http://www.mathlinks.ro/Forum/viewtopic.php?t=20752 . Note that it works only in the case when the line $l$ doesn't intersect the two circles; but I guess the problem was intended for this case only.

  Darij


Let $ a;b;c;k >0$ and $ abc\equal{}1$ . prove that:

$ \frac{k(ab\plus{}bc\plus{}ca)}{a^2\plus{}b^2\plus{}c^2}\plus{}\frac{1\plus{}kbc^2}{1\plus{}kb^2c}\plus{}\frac{1\plus{}kca^2}{1\plus{}kc^2a}\plus{}\frac{1\plus{}kab^2}{1\plus{}ka^2b} \le\ k\plus{}3$

I think It can be killed by SS
You problem can be rewritten
$ \frac{k(ab\plus{}bc\plus{}ca)}{a^2\plus{}b^2\plus{}c^2}\plus{}\frac{a\plus{}kc}{a\plus{}kb}\plus{}\frac{b\plus{}ka}{b\plus{}kc}\plus{}\frac{c\plus{}kb}{c\plus{}ka} \le\ k\plus{}3$
And it can be written in SS form
Nguyen Viet Anh had a problem Which is quite similar to yours :lol:

[quote="Allnames"]I think It can be killed by SS
You problem can be rewritten
$ \frac {k(ab \plus{} bc \plus{} ca)}{a^2 \plus{} b^2 \plus{} c^2} \plus{} \frac {a \plus{} kc}{a \plus{} kb} \plus{} \frac {b \plus{} ka}{b \plus{} kc} \plus{} \frac {c \plus{} kb}{c \plus{} ka} \le\ k \plus{} 3$
And it can be written in SS form
Nguyen Viet Anh had a problem Which is quite similar to yours :lol:[/quote]
Thank for your ideal . Although U know use SS but I think it very compound  with Coefficient k.
Detail it please !!! ( My proof also use SS  :lol:  )
Thanks!

I have just found
$ \frac{a+kc}{a+kb}+\frac{b+ka}{b+kc}+\frac{c+kb}{c+ka}-3=\frac{k^2}{(ka+c)(kb+c)}(a-b)^2+\frac{k[(k-1)a+(k^2-k+1)b+kc]}{(a+kb)(b+kc)(c+kb)}(a-c)(b-c)$
And $ \frac {k(ab + bc + ca)}{a^2 + b^2 + c^2} -k=k\frac{(a-b)^2+(a-c)(b-c)}{a^2+b^2+c^2}$
We have SS form
By the way $ \frac{a+kc}{a+kb}+\frac{b+ka}{b+kc}+\frac{c+kb}{c+ka}-3\le \frac{(%Error. "suma" is a bad command.
)^2}{\sum ab}$
Is it useful

[quote="onlylove_math"][quote="Allnames"]I have just found
$ \frac {a \plus{} kc}{a \plus{} kb} \plus{} \frac {b \plus{} ka}{b \plus{} kc} \plus{} \frac {c \plus{} kb}{c \plus{} ka} \minus{} 3 \equal{} \frac {k^2}{(ka \plus{} c)(kb \plus{} c)}(a \minus{} b)^2 \plus{} \frac {k[(k \minus{} 1)a \plus{} (k^2 \minus{} k \plus{} 1)b \plus{} kc]}{(a \plus{} kb)(b \plus{} kc)(c \plus{} kb)}(a \minus{} c)(b \minus{} c)$
And $ \frac {k(ab \plus{} bc \plus{} ca)}{a^2 \plus{} b^2 \plus{} c^2} \minus{} k \equal{} k\frac {(a \minus{} b)^2 \plus{} (a \minus{} c)(b \minus{} c)}{a^2 \plus{} b^2 \plus{} c^2}$
We have SS form
By the way $ \frac {a \plus{} kc}{a \plus{} kb} \plus{} \frac {b \plus{} ka}{b \plus{} kc} \plus{} \frac {c \plus{} kb}{c \plus{} ka} \minus{} 3\le \frac {(\sum a)^2}{\sum ab}$
Is it useful[/quote]
[quote="Allnames"]And $ \frac {k(ab \plus{} bc \plus{} ca)}{a^2 \plus{} b^2 \plus{} c^2} \minus{} k \equal{} k\frac {(a \minus{} b)^2 \plus{} (a \minus{} c)(b \minus{} c)}{a^2 \plus{} b^2 \plus{} c^2}$[/quote]
Are U sure ? I think it have to
$ \frac {k(ab \plus{} bc \plus{} ca)}{a^2 \plus{} b^2 \plus{} c^2} \minus{} k \equal{} \minus{} k\frac {(a \minus{} b)^2 \plus{} (a \minus{} c)(b \minus{} c)}{a^2 \plus{} b^2 \plus{} c^2}$ :)
And Can U prove that $ S_a;S_b \ge\ 0$ ?
with $ S_a \equal{} \frac {k}{a^2 \plus{} b^2 \plus{} c^2} \minus{} \frac {k^2}{(ka \plus{} c)(kb \plus{} c)} \ge\ 0$
       $ S_a \equal{} \frac {k}{a^2 \plus{} b^2 \plus{} c^2} \minus{} \frac {k[(k \minus{} 1)a \plus{} (k^2 \minus{} k \plus{} 1)b \plus{} kc]}{(a \plus{} kb)(b \plus{} kc)(c \plus{} kb)} \ge\ 0$
It isn't easy  :wink:
Thanks! :)[/quote]
It is not true! I think it should be
$ S_a \equal{} \frac {k}{a^2 \plus{} b^2 \plus{} c^2} \minus{} \frac {k^2}{(ka \plus{} c)(kb \plus{} c)}\le0$
$ S_a \equal{} \frac {k}{a^2 \plus{} b^2 \plus{} c^2} \minus{} \frac {k[(k \minus{} 1)a \plus{} (k^2 \minus{} k \plus{} 1)b \plus{} kc]}{(a \plus{} kb)(b \plus{} kc)(c \plus{} kb)}\le0$

[quote="Math pro"]
It is not true! I think it should be
$ S_a \equal{} \frac {k}{a^2 \plus{} b^2 \plus{} c^2} \minus{} \frac {k^2}{(ka \plus{} c)(kb \plus{} c)}\le0$
$ S_a \equal{} \frac {k}{a^2 \plus{} b^2 \plus{} c^2} \minus{} \frac {k[(k \minus{} 1)a \plus{} (k^2 \minus{} k \plus{} 1)b \plus{} kc]}{(a \plus{} kb)(b \plus{} kc)(c \plus{} kb)}\le0$[/quote]
Really, my friend ?   :maybe:  I don't spam but U are mistake .
I want prove LHS \le\ RHS . U should check carefully before U post a reply  :wink:

sorry for my mistake
$ S_a \equal{} \frac {k}{a^2 \plus{} b^2 \plus{} c^2} \minus{} \frac {k^2}{(ka \plus{} c)(kb \plus{} c)}\geq0$
let c = max{a,b,c}
case1:$ k\le1$  ineq < - >$ k^2.ab\plus{}ak(c\minus{}a)\plus{}bk(c\minus{}b)\plus{}(1\minus{}k)c^2\le0$ ->it is true
case2:$ k\geq1$
ineq<->$ \frac {k^2 \plus{} kc^3 \minus{} 1}{c} \plus{} ak(c \minus{} a) \plus{} bk(c \minus{} b)\geq0$- > it is true,too
$ S_b$ is the same way! :blush:

[quote="Allnames"]I think It can be killed by SS
You problem can be rewritten
$ \frac {k(ab \plus{} bc \plus{} ca)}{a^2 \plus{} b^2 \plus{} c^2} \plus{} \frac {a \plus{} kc}{a \plus{} kb} \plus{} \frac {b \plus{} ka}{b \plus{} kc} \plus{} \frac {c \plus{} kb}{c \plus{} ka} \le\ k \plus{} 3$
And it can be written in SS form
Nguyen Viet Anh had a problem Which is quite similar to yours :lol:[/quote]
The inequality holds just for $ k \equal{}0$. If $ k>0$, then take $ a\equal{}b\equal{}1,c\equal{}\frac{(k\plus{}1)(k\plus{}4)}{k}$. ;)

[quote="onlylove_math"][quote="Math pro"]
case2:$ k\geq1$
ineq<->$ \frac {k^2 \plus{} kc^3 \minus{} 1}{c} \plus{} ak(c \minus{} a) \plus{} bk(c \minus{} b)\geq0$- > it is true,too
 :blush:[/quote]
Sorry but it 's wrong . U should check carefully , my friend ! :wink:
[quote="Math pro"]$ S_b$ is the same way![/quote]
Why similar?  :oops:
Detail , detail it ?Pls
Thanks![/quote]
why it 's wrong?$ k\geq1$ and $ c\geq1$ :maybe:


trigonometry
trig identities
Law of Cosines

A convex hexagon is inscribed in a circle. If its successive sides are 2,2,7,7,11,11, compute the diameter of the circumscribed circle.

Yeah, so I need help solving problems like these. Any suggestions?

connecting each vertex of the hexagon to the center of the circle, i used the law of cosines on the triangles that had sides r,r,2 , r,r,7 , r,r,11, and after some manipulation i got (letting the angle of the triangles that had 2,7,11 be a,b,c respectively)

$ 2r \equal{} \dfrac{2}{\sin\dfrac{a}{2}} \equal{} \dfrac{7}{\sin\dfrac{b}{2}} \equal{} \dfrac{11}{\sin\dfrac{c}{2}}$

But from there i'm not sure on how to find the angles $ a,b,c$. We want to find $ d \equal{} 2r$, so if any of $ a,b,c$ is known, we are basically done from the equation above...


[b]Consider the set of integers S={1,2,3,...3000}
In how many ways can a subset be chosen from S, such that no two elements of that subset add up to 9 or its multiple?[/b]

I'm not sure how to approach this, but maybe Pigeonhole Principle with the residues $ \mod{9}$? :maybe:


In the figure, $ABCD$ is an isosceles trapezoid with side lengths $AD = BC = 5, AB = 4,$ and $DC = 10$. The point $C$ is on $\overline{DF}$ and $B$ is the midpoint of hypotenuse $\overline{DE}$ in the right triangle $DEF$. Then $CF =$
[asy]
size(200);
defaultpen(fontsize(10));
pair D=origin, A=(3,4), B=(7,4), C=(10,0), E=(14,8), F=(14,0);
draw(B--C--F--E--B--A--D--B^^C--D, linewidth(0.7));
dot(A^^B^^C^^D^^E^^F);
pair point=(5,3);
label("$A$", A, N);
label("$B$", B, N);
label("$C$", C, S);
label("$D$", D, S);
label("$E$", E, dir(point--E));
label("$F$", F, dir(point--F));
markscalefactor=0.05;
draw(rightanglemark(E,F,D), linewidth(0.7));[/asy]
$\text{(A)} \ 3.25 \qquad \text{(B)} \ 3.5 \qquad \text{(C)} \ 3.75 \qquad \text{(D)} \ 4.0 \qquad \text{(E)} \ 4.25$

I don't see the hint, but....
well, maybe the title is the hint, yes, that's what it is.

[hide]Draw a line from B perpendicular to CD and meets CD at G. BG is parallel to EF (since they're both perpendicular to CD), and $\triangle{DBG}=\triangle{DEF}$. Since DE=2DB, DF=2DG. Since DG=$\frac{1}{2}(10-4)+4=7$, DF=14. Thus CF = 14 - DC = 4. [/hide]

Since $ABCD$ is isosceles trapezoid, $\angle BDC = \angle ACD$.
Since $DEF$ is a right triangle, $\angle BDF = \angle BFD$.
Since $\angle ACD = \angle BFD$ and $AC \parallel BF$, $ABFC$ is a parallelogram and $AB=CF=4$.

PS. The lengths of $BC$ and $CD$ are unneccesary to solve the problem.


I want help in understanding the binomial theorem

Here is a nice description.  Essentially, we are using an approximation.  We can actually refine our estimate using Taylor expansion that requires calculus.

http://en.wikipedia.org/wiki/Binomial_theorem

I know the basics but i need help on a more advanced form,a brain twisting one

Are you talking about the [url=http://en.wikipedia.org/wiki/Multinomial_theorem]multinomial theorem[/url] or the [url=http://en.wikipedia.org/wiki/Binomial_theorem#Newton.27s_generalized_binomial_theorem]general binomial theorem[/url]?

I think just the binomial theorem, although you can still use the multinomial theorem with binomials.


function
algebra
polynomial
induction
algebra unsolved

Find all functions $ f$ $ N\to N$ such that
$ f(n\plus{}1)\equal{}f(f(n))\plus{}1$

Since f N\toN
We can assume that  f(x)=ax^n+...

Left hand side is a degree n polynomial while right hand side is a degree n^2 polynomial
so n=0 or 1

If n=0
f(x)=c
LHS=c
RHS=c+1
So n is not equal to 0
i.e. n=1

Let f(x)=ax+b
an+a+b=(a^2)n+ab+b+1
a=a^2
a=1
b=0
so f(x)=x

BTW, can anyone tell me how to use the codes?
Just send me a message

That was wrong, wendytmchk. Remember we are looking for [b]functions[/b], not [b]polynomials[/b].

Thank you
The problem that I received is only "prove that exist a function satisfying",But I think it is too easy
So I hope ,some one can help me kill that hard problem

[hide]Take any $ a$ such that $ f(a) \equal{} b$ where $ b$ is the least element in the range of $ f$. If $ a > 1$, then $ n \equal{} a\minus{}1$ gives $ b \minus{} 1 \equal{} f(f(a\minus{}1))$, contradicting $ b$ being least. So $ f(1)$ must take the lowest value in the range of $ f$, and be the only one that does.

Suppose $ f(1) > 1$. If $ f(1) \plus{} 1$ were in the range of $ f$, then for some $ a$ $ f(a) \equal{} f(1) \plus{} 1 \equal{} f(f(a\minus{}1)) \plus{} 1$ (note $ a > 1$), so $ f(f(a\minus{}1)) \equal{} f(1)$ and since $ f(1)$ is only taken once in the range of $ f$, $ f(a\minus{}1) \equal{} 1$. This contradicts $ f(1)$ being the minimal element in the range. So $ f(1) \plus{} 1$ is not in the range. Take the smallest element of the range $ r$ besides $ f(1)$, so that there is an $ a$ with $ f(a) \equal{} r$ (again $ a \neq 1$). Since $ f(1) \plus{} 1$ is not in the range, it follows $ r\minus{}1$ is not in the range since it would have to be equal to $ f(1)$. But $ n \equal{} a\minus{}1$ gives $ f(f(a\minus{}1)) \equal{} r\minus{}1$. Oops. This forces $ f(1) \equal{} 1$, and now it is obvious $ f(n) \equal{} n$ for all $ n$.[/hide]

MellowMelon beat me to it, but I thought it'll be nice to offer a slightly different approach.

[hide="Solution"]
Let $ s: \mathbb{N} \to \mathbb{N}$ be the successor function, $ s(n) \equal{} n\plus{}1$. We'll be using the following notation: whenever $ f$ and $ g$ are functions, $ fg$ is the function defined by $ fg(x)\equal{}f(g(x))$. Similarly, $ f^2\equal{}ff$ is just $ f(f(x))$ etc.

What we know about $ f$ can be written as

$ fs\equal{}sf^2$

so now we can have fun calculating

$ fs^2 \equal{} fss \equal{} sf^2s \equal{} sffs \equal{} sfsf^2 \equal{} s^2f^2f^2 \equal{} s^2f^4$

and it's easy to proceed by induction and show that, for all $ k \geq 1$,

$ fs^k \equal{} s^kf^{2^k}$,

which evaluated at $ n$ yields

$ f(n\plus{}k) \equal{} f^{2^k}(n)\plus{}k$. This holds for $ n \geq 1$ and $ k \geq 0$.

In particular, setting $ n\equal{}1$,

$ f(k\plus{}1) \geq k\plus{}1$.

Let $ f(1)\equal{}x$ and $ f(2)\equal{}y$. Then $ y\equal{}f(x)\plus{}1$. From this it follows that $ x \neq 2$, but if $ x \geq 3$ then

$ f(x) \equal{} f(2\plus{}x\minus{}2) \equal{} f^{2^{x\minus{}2}}(2)\plus{}x\minus{}2$ = $ f^{2^{x\minus{}2}\minus{}1}(f(2))\plus{}x\minus{}2$

so $ f^{2^{x\minus{}2}\minus{}1}(y) \equal{} y\minus{}x\plus{}1$, which is impossible since $ f^m(y) \geq y$ for any $ y$ and any $ m$ (this follows directly from $ f(k) \geq k$). So $ x$ must be 1, and from $ f(1)\equal{}1$ it follows immediately by induction that $ f(n)\equal{}n$ for all $ n \in \mathbb{N}$.

[/hide]

Thank randomgraph,MellowMelon.I have found that we only need $ f(n\plus{}1)>f(f(n))$ :lol: (an IMO problem which I can remember when


Let $a$, $b$, $c$ be three reals having the sum of their squares $9$.

prove that $2\left(a+b+c\right)-abc \leq 10$.

i have a proof using lagrange multipliers,which is pretty straightforward(in case it is correct ne1 with basic knowledge abt multipliers can get it,so right now i'm not posting it)but this problem duznt take me anywhere using anything "low-tech".
thanx
(is this an olympiad question,if so a proof with multipliers will not be preferred)

yes it is sort of an olympiad question - so a non-multiplier solution would be preferred.

How about subsituting a=3(Root(x/(x+y+z)), b=3(Root(y/(x+y+z)), c=3(Root(z/(x+y+z))? Haven't tried but seem so

nice ... indeed! hey hxtung where are you from?

nice for you - but where are you from? :)

let's not make this thread an a/s/l thread please :) - i've made another one at the non-math related stuff :)

[url=http://www.artofproblemsolving.com/community/c6h340344p1821613][b]Vietnam TST 2002[/b][/url]
Let $ a,b,c\in\mathbb{R}$ , $ a^{2}\plus{}b^{2}\plus{}c^{2} \equal{} 9$.  Prove that:\[ 2(a\plus{}b\plus{}c) \minus{} abc\leq  10\]
[quote=hxtung]Hey, this is my soln
WLOG, we suppose |a|<=|b|<=|c|. we have [2(a+b+c)-abc]2 = [2(a+b) +c(2-ab)]2 <= [(a+b)2 +c2][4+ (2-ab)2]
=(9+2ab)[8-4ab+(ab)2] = 2(ab)3 +(ab)2 -20ab +72 = (ab+2)2(2ab-7) +100.(*)
On the other hand, c2 >=3, then 2ab<=a2 +b2= 9-c2 <=6
(*), we get: [2(a+b+c)-abc]2 <=100 then 2(a+b+c)-abc <=10.
The proof compelte. It's very good?? Valentino Vornicu :D[/quote]
[b]Nice.[/b]
WLOG, we suppose$ |a|\leq|b|\leq|c| $. we have $\left(2(a+b+c)-abc\right)^2 = \left(2(a+b) +c(2-ab)\right)^2 \leq(a+b)^2 +c^2)(4+ (2-ab)^2)$
$=(9+2ab)\left(8-4ab+(ab)^2\right) = 2(ab)^3 +(ab)^2 -20ab +72 = (ab+2)^2(2ab-7) +100$ .$(*)$
On the other hand, $ c^2\geq 3$, then $2ab \leq a^2+b^2= 9-c^2\leq6 $  $(*)$, we get:$ \left(2(a+b+c)-abc\right)^2\leq100 $ , then$ 2(a+b+c)-abc \leq10$.

[url=http://www.artofproblemsolving.com/community/c6h271530p1470900]Vietnam Team Selection Tests 2002[/url]

[url=http://www.artofproblemsolving.com/community/c4h610752p3621683][b]Vietnam Inequality[/b][/url]
Let $x,y,z$ are real numbers such that $x^2+y^2+z^2=2$ . Prove that $$x+y+z-xyz \le 2$$[url=http://www.artofproblemsolving.com/community/c6h244272p1343204]Poland 1991[/url]


limit
search
real analysis
real analysis unsolved

It's not exactly trivial (in the context of an entrance exam or otherwise) :D See:
1. http://www.mathlinks.ro/viewtopic.php?search_id=2111690549&t=114142
2. http://pefmath2.etf.bg.ac.yu/files/118/868.pdf

[quote="Slizzel"]Compute : $ \lim_{n\to\infty}{\left(\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}\right)}$.[/quote]

that's a very nice limit !   :thumbup:


The projectile pass tangenting the top of the mountain, but dont touch it,Find the horizontal distance between M and P in function of R,d,h.

I saw this problem a few days ago (with concrete numbers) and I solved it then, but I don't have time to post a solution now. Basically you just need a lot of mathematics and manipulation of equations. It's not that hard. I might post a solution later.


analytic geometry
graphing lines
slope
function
probability

Hi all

I have been given two points

pnt1 is a random point
pnt2 is has a x value i know (23) and a Y value of 0

I need to create a 3rd point pnt3 that would some how fall along the line that would be created by pnt1 and pnt2

if i know the slope could i some how use that to create a new point?

[hide]
So
Pt1=(x,y)
Pt2=(23,0)

just use the midpoint formula

$[(\frac{23+x}{2}),(\frac{y}{2})]$[/hide]

Though that would work i can use it because its not random.

what i was hopeing is that i could take a random power of slope and times it by the point or something like that.

What do you mean? Like an equation with a variable m (for slope) and n (for randomness) that gives you the coordinates of a point on a line in which you know one point and the slope?

I am writing a program

that program

1) creates a random point (pnt1)
2) takes input and makes a (pnt2) ( that point will be Xinput y0)
3) calculate the slope between pnt1 and pnt2
4) find a new point on the line that is formed between the two points(pnt1 and pnt2)
4a) this point has to be randomly pulled from the line

the out put of the program will be pnt1 and pnt2

that can later be used to equate the input by solveing for X

[quote="CircleSquared"]I don't really get it either, I'll try my best:
Math92's works.  In fact, it's great.

Alternatively, to get an infinite number of points, write the equation of the line in point-slope form and just find values of x and y that work.
Point slope form is:
$y-y_{0}=m(x-x_{0})$, where $(x_{0},y_{0})$ is your point and m is your slope.[/quote]

i'm not sure that makes sense to me

i was thinking something more along the lines of

           X1 =random.Next(1,10000); /random number between 1 and 10000
            Y1 =random.Next(1,10000);
            X2=Convert.ToInt16(textBox7.Text);
            Y2=0;
            X3=1; //holder value
            Y3=1;// holder value

            int slope = (Y2 - Y1) / (X2 - X1);
            int t = random.Next(10000);
            X3 = t + X2;
            Y3 = t * slope + Y2;

[quote="math92"]
So
Pt1=(x,y)
Pt2=(23,0)

just use the midpoint formula

$[(\frac{23+x}{2}),(\frac{y}{2})]$[/quote]

couldnt i just use that equation but rather than devide it by 2

divide it by a random number as long as its the same for both x and y

T= random number
$[(\frac{23+x}{t}),(\frac{y}{t})]$

[quote="crash893"]

couldnt i just use that equation but rather than devide it by 2

divide it by a random number as long as its the same for both x and y

T= random number
$[(\frac{23+x}{t}),(\frac{y}{t})]$[/quote]

I dont thiink so.

Lets say you have the points $(2,1) and (6,1)$
If we make $t=4$ then we get
that the point = $(2,.5)$ which is not on the line.
Lets make $t=2$
we then get the point $(4,1)$ which is on the line.
It has to be 2 because thats the actual thereom.

[quote="Davidd"]It might all have been said, but the first 'unknown' point will be random.  The second must be on a line with that and the given point (23,0).  A random such point need have only a random value for "x", and the "y" value would be found from the given point and the determined other random point using slope as you suggested.[/quote]

im not sure what your asking or saying?

[quote="Hamster1800"]Actually, you can do this. For any $p$, $(p\cdot x_{1}+(1-p)\cdot x_{2},p\cdot y_{1}+(1-p)\cdot y_{2})$ is a point on the line. Taking $p=\frac{1}{2}$ gives the midpoint.[/quote]

so if i got this right

p = random number between (.1 and .9)

pnt3 = x3 and y3

x3 = x1+(1-p)*x2
y3=p*y1+(1-p)*2

so i dont even need slop at all?

You are using Java, yes? (Maybe this should be in the Computer Science and Informatics section)

I don't know anything about the random.Next() function, but couldn't you just generate sets using the Math.random() function, use the $\frac{y_{1}-y_{2}}{x_{1}-x_{2}}$ slope formula, than use an endless for loop which test random coordinate sets in the equation $y=mx$ and breaks the loop if it matches?

[quote="crash893"]
[quote="Hamster1800"]Actually, you can do this. For any $p$, $(p\cdot x_{1}+(1-p)\cdot x_{2},p\cdot y_{1}+(1-p)\cdot y_{2})$ is a point on the line. Taking $p=\frac{1}{2}$ gives the midpoint.[/quote]

so if i got this right

p = random number between (.1 and .9)

pnt3 = x3 and y3

x3 = [b]p*[/b]x1+(1-p)*x2
y3=p*y1+(1-p)*[b]y[/b]2

so i dont even need slop at all?[/quote]

I believe that will work (provided you make the corrections I put in bold).

@ the post above me: Sure you can, but it will never terminate (the probability of hitting a line on a plane with a point is 0)

Well, but he could control it if the plane was limited, e.g. 10 by 10 like on graphing calculators.


I signed up for AMC12, and this is my first time.
I followed you guys' advice and studied with AoPS volume 1 for about a month.
The problem is I'm studying ch.18 right now and don't think I can finish all the chapters until February 10th.
How should I study at this point?
Should I try to finish as many chapters as I can? By the way, my current goal is to make AIME.
Please advise.

First, finish the geometry chapters and make sure you completely understand them. Next, do the counting chapters. There are usually a few important counting problems on the AMC12. Then, if you still have time, DO PROBLEMS! Do practice AMC12's under timed conditions.  When you're done, make sure you completely understand the solutions as if you could teach them tomorrow. Doing problems is the most helpful thing.

Additionally the pdfs [url=http://www.hiddendiscoveries.com/mathcompetition.html]here[/url] help a ton.

Finish the geometry chapters in volume 1 and make sure you really understand them. Then just do the past  AMC 12's. I think this the most important. For me, this helped a lot more than the books.


AMC
USA(J)MO
USAMO
AIME
email
ARML
AMC 12

We have determined that the 20 students from the 5% Club are not residents of the US, and therefore do not qualify for the USAMO under the rules.  (As explained in another thread, we had insufficient information about those students.  It appeared that based on the AMC 10/12 registration they were residents of Southern California.)

That means in turn that the AMC staff and I went back to the database, re-tabulated the information, and slightly modified the USAMO invitation levels.

For students who took the 2007 AMC 12 A or 2007 AMC 12 B and then subsequently took the AIME, the required index level is (still) 197.5.   This gives a total of 327 students.  (Three 5% Club students were dropped,   but 5 other students who took the AMC 12 B at a location different from their AMC 12 A location but were not updated due to inconsistent information in the database were found and added.)

The floor value remains at a score of 6 on the AIME.
Among students in 10th grade and below with an AIME score of 6 or greater, an index of 181.0 or greater is used as an invitation level to the USAMO.  This is a modification (down) of the previous index level.  This gives a total of 166 students in 10th grade and below added.  (17 students from the 5% Club were dropped, and then 12 new students added.)

The 8 states without qualifiers and the corresponding 9 high index students in that state remained the same.

This makes a total of 502 students invited to the USAMO.  The revised list will be posted at our website later today.  The email and paper invitation will go out Monday, April 9, because some of our staff have taken some holiday time off.

Steve Dunbar
AMC Director

Thanks for the information. This means that we should disregard the number given in the first email, correct?

Hmm, some of my friends are saying that perhaps you're just reassigning the 20 since only 17 new people are getting in. If that's the case, are we even going to get new emails?

[quote]This is great![/quote]

For people who are now getting in.  Probably not so great for the 20 people who are now out of USAMO.  Well, they probably knew that they were inelgible.

yeah - I have to admit that 20 kids whose names all sounded Korean qualifying from an unknown school with no prior history was pretty strange... but it's not like other strange things don't happen in USAMO.

I guess I was wishing for a miracle and that sort of was one, but I still missed the cutoff by one point. Either way, when I  saw the USAMO list and saw how many of 5% club got in, I thought it was pretty weird how this new school seemed to have just popped out from no where.

Revised tally of California USAMO qualifiers by ARML team region:

Northern California: 1
SFBA: 47
Southern California: 14
San Diego: 4

The changes made here removed the 20 from the 5% Club from Southern California and added one individual to SFBA.

SFBA rules.  :P
Well, now I know that if I hadn't made the two stupid mistakes on the AIME, I would've qualified with $112.5+70=182.5>181.0$.

Well, at least I now know that I have the ability to qualify (granted, very borderline, though). Now to actually do it next year.  :P

I guess I'm pretty lucky:

111 amc12 + 6 AIME correctly worked + 1 complete guess on AIME = 181 index

Why were the 5% club people dropped?
I guess that works for me, b/c 7th grade is now the youngest grade qualifying in the U.S.A. =D
(not counting that genius 4th grader kid from Canada)

All the 5% club members that qualified for USAMO reside in Korea, and are therefore ineligible to take the USAMO.

Maybe I'll get an email this time around?  :wink:


Can anyone show me a derivation of the Nernst Equation and post some problems using this equation.

$ E = E^{\circ}-\frac{RT}{nF}\ln \frac{a_{red}}{a_{ox}}$.

i think [url=http://en.wikipedia.org/wiki/Nernst_equation]this[/url] should help you.
as for problems lets see...


Determine with proof all natural numbers which are not in the set

{li(n + n^0.5 + 1/2)|n \in N}.

li(x) denotes the greatest integer \leq x.

[i]German IMO selection test 1980 problem 3.[/i]

This is a very well known result. The answer is : all natural numbers except the positive perfect squares. For this, let f(n)=[n+sqrt(n)+1/2]. It can be immediately checked that f(n+1)-f(n) is 1 or 2 and it is 2 exactly when n is of the form t^2+t. From here it follows easily the above assertion.

More generally, for each integer k  \geq 2, it is not difficult to prove that, if [.] denotes the integer part :
The set E(k) = {  n + [ (n + n 1/k ) 1/k  ] / n nonnegative integer} is the set of nonnegative integers except those of the form n k  for n > 0.

Pierre.

Wasn't the generalization a concours gnral problem? 1998?


What is the least value of $ x$ for which $ \frac{x\plus{}2}{3} \equal{} \frac{4}{x\plus{}1}$?

Simply cross-multiply and solve, and find the smallest solution.

$ \frac {x + 2}{3} = \frac {4}{x + 1}$
$ (x + 2)(x + 1) = 4(3)$
$ x^2 + 3x + 2 = 12$
$ x^2 + 3x - 10 = 0$
$ (x + 5)(x - 2) = 0$
$ x = 2$ or $ x = - 5$

Since there are no extraneous solutions here, and $ - 5 \& lt; 2$, the answer is $ \fbox{-5}$.

The values of $ x+2 $ and $ x+1 $ differ by 1. The least factors of 12 that differ by one are -3 and -4.
$ x+1 < x+2 $ and $ -4<-3 $ so $ x+1 = -4 $
$ x= \fbox{-5} $


IMO Longlist
Divisibility
factorial
number theory proposed
number theory

Let $m$ be a positive integer and define $f(m) =$ the number of factors of 2 in $m!$ (that is, the greatest positive integer $k$ such that $2^k|m!$). Prove that there are infinite positive integers $m$ such that $m - f(m) = 1989$.

This one is trivial if you know the appropriate tool :
It is a classical result, due to Legendre, that $f(m) = m - s_2(m)$, where $s_2(m)$ is the sum of the digits of $m$ in its expansion in base $2$, that is the number of $1$.
Therefore $m-f(m) = s_2(m)$ so that $m$ satisfies the equation if and only if it has exactly $1989$ digits $1$ in its binary expansion.
In particular, we may find infinitely many such $m$.

Pierre.

By the prime-dividing-a-factorial formula, we have that $m-f(m)=1$ when $m=2^a$. Thus when $m=2^a-1$, $m-f(m)=a$.

Now consider $m=2^b+c$, where $c < 2^b$. $(2^b+c)!=(2^b)!(2^b+1)(2^b+2)...(2^b+c)$, which becomes $(2^b)!c! \mod 2^b$. Hence $f(m)=2^b-1+f(c)$. Putting $c=2^{1988}-1$ we get $m-f(m)=2^b-(2^b-1)+c-f(c)=1+1988=1989$. We use values of $b > 1988$ to get our infinte set.


$ 1\leq a\sqrt{bc}+b\sqrt{ac}+c\sqrt{ab}\leqq\frac{b+c}{2}+b\frac{a+c}{2}+c\frac{a+b}{2}=ab+ac+bc\leq\frac{1}{3}(a+b+c)^{2}$  -----> $ a+b+c\geq\sqrt{3}$

[quote="noogaht"]1. let $ a,b,c > 0$and $ a \sqrt {bc} \plus{} b \sqrt {ca} \plus{} c \sqrt {ab} \geq 1$ prove : $ a \plus{} b \plus{} c \geq \sqrt {3}$
2. $ a,b > 0$ prove: $ (a^{2} \plus{} b \plus{} \frac {3}{4})(b^{2} \plus{} a \plus{} \frac {3}{4}) \geq (2a \plus{} \frac {1}{2})(2b \plus{} \frac {1}{2})$
3. $ a,b \geq 0$ prove: $ \sqrt {a(a \plus{} b)^{3}} \plus{} b\sqrt {a^{2} \plus{} b^{2}} \leq 3\sqrt {a^{4} \plus{} b^{4}}$[/quote]
2. By AM-GM inequality, we hvae:
$ a^2 \plus{} \frac {1}{4} \ge a, b^2 \plus{} \frac {1}{4} \ge b$
It follow that:
$ (a^{2} \plus{} b \plus{} \frac {3}{4})(b^{2} \plus{} a \plus{} \frac {3}{4}) \ge (a \plus{} b \plus{} \frac {1}{2})^2$
and: $ (2a \plus{} \frac {1}{2})(2b \plus{} \frac {1}{2}) \le\frac {(2a \plus{} 2b \plus{} 1)^2}{4} \equal{} (a \plus{} b \plus{} \frac {1}{2})^2$
So we are done!

[quote="noogaht"]3. $ a,b \geq 0$ prove: $ \sqrt {a(a \plus{} b)^{3}} \plus{} b\sqrt {a^{2} \plus{} b^{2}} \leq 3\sqrt {a^{4} \plus{} b^{4}}$[/quote]
We have:

$ \sqrt {a(a \plus{} b)^{3}} \plus{} b\sqrt {a^{2} \plus{} b^{2}} \equal{}a\sqrt{a(a\plus{}b)}\plus{}b\sqrt{a(a\plus{}b)}\plus{}b\sqrt{a^2\plus{}b^2} \leq \sqrt{a^2\plus{}2b^2}.\sqrt{3a^2\plus{}2ab\plus{}b^2}$
$ \leq \sqrt{a^2\plus{}2b^2}.\sqrt{4a^2\plus{}2b^2}$
We will prove that $ (a^2\plus{}2b^2)(4a^2\plus{}2b^2) \leq 9(a^4\plus{}b^4)$
or $ 5(a^4\plus{}b^4) \geq 10a^2b^2$(true by AM-GM)
Done!