Could someone please help me with this question?  Thanks!

1) Let C be the unknown gas. Convert 1.39 g of C into chemical quantity: $ n_C \equal{} 0.0224\, mol$.

2) Convert 1.21 g of water into chemical quantity: $ n_{H_2O} \equal{} 0.0672\, mol$.

3) Now we can see that $ n_{H_2O} \equal{} 3 n_C$, which means that each molecule of our gas has 3 hydrogens (or a multiple of 3).

4) Now we run into problem. The method would be to substract 3 hydrogens (or a multiple) from the molar mass and see if the resultant is the molar mass (or a multiple) of a known element. However, I made a search and there is no binary compound with 62.09 as molar mass. Is there some typo?

On the other hand, after reading your post [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=227550]here[/url], it turns out that $ \ce{B5H9}$ has a molar mass close to 62.09 (63.127, to be precise). Is this the expected answer?

$ N_4 H_6$ could be an answer ... not precisely 62.09 .. but i dont think its $ B_5 H_9$

Interesting. And what kind of compound is exactly $ \ce{N4H6}$?

No idea . from what I looked up its called tetrazane.

[url]http://article.pubs.nrc-cnrc.gc.ca/ppv/RPViewDoc?issn=1480-3291&volume=41&issue=4&startPage=1045[/url]


solve the system of equation:
$ (i): x^2\plus{}y^2\equal{}\minus{}y(x\plus{}z)$
$ (ii): x^2\plus{}x\plus{}y\equal{}\minus{}2yz$
$ (iii): 3x^2\plus{}8y^2\plus{}8xy\plus{}8yz\equal{}2x\plus{}4z\plus{}2$

1.$ x\equal{}0$ give $ z\equal{}\minus{}\frac 12$ $ y\equal{}0$ or $ y\equal{}\frac 12$.
2.$ x\not \equal{}0$ give $ z\equal{}\minus{}\frac{5x^2\plus{}2x\plus{}2}{4},y\equal{}frac{2x\plus{}2}{5x\plus{}2}$ and
$ 2x^2(5x\plus{}2)^2\plus{}8(x\plus{}1)^2\plus{}(x\plus{}1)(5x\plus{}2)(5x^2\plus{}6x\plus{}2)\equal{}0$.


Let $P\in R[X]$ a polynomial of degree greater than $1$ such that $P$ has all its roots real and distinct. Prove that we can find a rational nonzero number $a$ such that the polynomial $P(x+a)-P(x)$ has only real roots.
The result is still true if the roots of P are not distinct?

Isn't it kind of obvious (when roots are distinct) or am I doing something wrong.  :maybe:

Take any $a$ sufficiently small. Let $x_{1},...,x_{r}$ be points in which $P(x)$ has local extremas. If we draw a tangent at $x_{i}$ and start to lower it (or higher, depending wheter it's minimum or  maximum) then by continuity we get a line which intersects $P(x)$ in two points $y_{0}, y_{1}$ which satisfies $y_{0}+a=y_{1}$. Also we get $P(y_{0}+a)=P(y_{1})$ (as tangent at $x_{i}$ is parallel to $x$-axis and we moved it parallely). Now as all roots distinct $r=deg \ P(x)-1= deg \ (P(x+a)-P(x))$, hence this $a$ work

Yes, it's obvious. Alternatively, one can show that $P(x+a)-P(x)$ changes its sign $n-1$ times (where $n=\deg P$) if $a$ is small enough.
If the roots of $P$ are not distinct, the result is not true, as one can check considering $P(x)=x^{3}$.


So how did the other NJers do on the USAMO?
I am Max Chang from Whippany Park, student 29, and got a 14 in my first attempt. Very happy with my results.

My student number is exactly 400 greater than yours  :) . And I got the same score as you did (and happy with it also).

This is a completely random and useless revival .. but I just realized that I've actually heard of your name before, through a friend who attends your school.


Dear everyone.

a easy but not bad problem

prove that the following.

$ a,b,c \in{R^{ \plus{} }}$

$ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a}\leq\frac {(a \plus{} b \plus{} c)^{2}}{6abc}$

Thanks for reading. :)

[hide="Not full solution"]Rearrange the inequality as

$ 2(a\plus{}b\plus{}c)(\frac{1}{a\plus{}b}\plus{}\frac{1}{b\plus{}c}\plus{}\frac{1}{c\plus{}a}) \leq \frac{(a\plus{}b\plus{}c)^3}{3abc}$

By Cauchy,

$ 2(a\plus{}b\plus{}c)(\frac{1}{a\plus{}b}\plus{}\frac{1}{b\plus{}c}\plus{}\frac{1}{c\plus{}a}) \geq 9$

By AM-GM,

$ \frac{(a\plus{}b\plus{}c)^3}{3abc} \geq 9$

So I guess I'm really close but I have to go now. Maybe I can finish it later.[/hide]

[quote="modularmarc101"][hide="Not full solution"]Rearrange the inequality as

$ 2(a \plus{} b \plus{} c)(\frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a}) \leq \frac {(a \plus{} b \plus{} c)^3}{3abc}$

By Cauchy,

$ 2(a \plus{} b \plus{} c)(\frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a}) \geq 9$

By AM-GM,

$ \frac {(a \plus{} b \plus{} c)^3}{3abc} \geq 9$

So I guess I'm really close but I have to go now. Maybe I can finish it later.[/hide][/quote]

What you did doesn't help with nothing .

As for the proposed problem it is very easy by SOS :

\[ RHS\minus{}LHS\equal{}\sum_{cyc}(a\minus{}b)^2\cdot \left(\frac{a\plus{}b\plus{}3c}{3abc(a\plus{}b)}\right)\]

or you can prove it as :

\[ LHS\le \frac{1}{2}\left(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\right)\equal{}\frac{ab\plus{}bc\plus{}ca}{2abc}\le RHS\]

Using what modularmarc101 wrote:

$ \frac {(a \plus{} b \plus{} c)^{3}}{3abc}\geq 9$

Is same as $ (a \plus{} b \plus{} c)^{3} \geq 27abc$

$ a \plus{} b \plus{} c \geq 3(abc)^{1/3}$

Which is true by AM-GM

EDIT: Sorry Pain rinnegan, I didnt read the entire solution by modularmarc101, since I assumed that he didnt make a mistake.

[quote="Tomekk"]Using what modularmarc101 wrote:

$ \frac {(a \plus{} b \plus{} c)^{3}}{3abc}\geq 9$

Is same as $ (a \plus{} b \plus{} c)^{3} \geq 27abc$

$ a \plus{} b \plus{} c \geq 3(abc)^{1/3}$

Which is true by AM-GM[/quote]

I know it's true , but how does this help ?


Let $n$ be a positive integer. Is it possible for $6n$ distinct straight lines in the plane to be situated so as as to have at least $6n^2-3n$ points where exactly three of these lines interesect and at least $6n+1$ points where exactly two of these lines intersect?

I think not. Double-counting the quantity $x$ of points where two lines intersect, we must have ${6n\choose 2} \geq x \geq 3(6n^2-3n)+6n+1$, since every intersection of three lines gets counted three times. So ${6n\choose 2}=18n^2-3n \geq 18n^2-3n+1$, which I find unlikely to be true.


calculus
integration
logarithms
function
algebra
polynomial
calculus computations

How can I do this?

[list]
$\int{\frac{logx}{e^{-2x}}dx}$
[/list]

I mean there is no way to write the antiderivative as a combination of logarithms, exponential functions, trigonometric functions, polynomials, etc.  It is possible to express it in terms of an [url=http://mathworld.wolfram.com/ExponentialIntegral.html]exponential integral[/url], but you're not going to get anything "nicer" than that.


LaTeX
algebra
polynomial
limit
algebra unsolved

let p(x)=Anx^n+...+A1x+A0 be a poly. with real coeff. s.t.
 An,A0>0 and summation Ai^2 , i going from 1 to n-1 <=4min(A0^2,An^2)/n-1.
 show p(x)>=0 for all real x.
   :welcome:

[quote="lordWings"]It's a little hard to understand, so I'll $\LaTeX$ it:
Let $p(x)=A_{n}x^{n}+...+A_{1}x+A_{0}$ be a polynomial with real coefficients such that
$A_{n}>0$, $A_{0}>0$ and $\sum_{i=1}^{n-1}{{A_{i}}^{2}}\leq\frac{4}{n-1}\min\{{A_{0}}^{2},{A_{n}}^{2}\}$.
Show $p(x)\geq 0$ for all real $x$.[/quote]
It is true for $x\ge0$ or if n is even.

rust your condition isnt neccessary?
any idea? :maybe:

If n is odd then exist largge x, suth that P(-x)<0, because $\lim_{x\to-\infty }P(x) =-\infty .$


induction
combinatorics proposed
combinatorics

[i]Biribol[/i] is a game played between two teams of 4 people each (teams are not fixed). Find all the possible values of $ n$ for which it is possible to arrange a tournament with $ n$ players in such a way that every couple of people plays a match in opposite teams exactly once.

Consider the graph where each vertex is a person and a edge means the two respective players have already played in oposite teams. At each game 16 edges are added so 16|n(n-1)/2. But at each game each vertex does not change the parity of it's numbers of edges (it either does not change or 4 is added). So 2|(n-1). So it is only possible if n=32k+1.
If n=33 the tournament consist of 33 games. Number the players as 1,2,...,33 and consider the 33 games:
(i,i+1,i+2,i+3) X (i+4,i+8,i+12,i+16), where the index are modulo 33. It is a possible tournament.
Now induction: If it is possible for 32k+1 it is for 32(k+1)+1:
Consider 32(k+1)+1 people and now choose one to be the leader. Choose 32k more people to play a complete tournament (with 32k+1 people) with the leader. Now the 32 people left and the leader play a complete tournament (with 33 people).
Now divide the 32k first people into 8k teams of 4 people, let these be the kind A teams. Now the 32 last people are divided into 8 teams of 4 people, let these be the kind B teams. Now all kind A teams play with all kind B teams and the tournament is complete.
So the answer is 32k+1 for k>0.

What is the motivation for the construction with $33$ players?


Given $ a, b, c \geq\ 0$. Prove that:
$ \frac{a^3}{2a^2\plus{}b^2} \plus{} \frac{b^3}{2b^2\plus{}c^2} \plus{} \frac{c^3}{2c^2\plus{}a^2}  \geq\ \frac{a\plus{}b\plus{}c}{3}$

And the general problem is so interesting:
Find the best constant $ k$ satisfy:
$ \frac{a^3}{ka^2\plus{}b^2} \plus{} \frac{b^3}{kb^2\plus{}c^2} \plus{} \frac{c^3}{kc^2\plus{}a^2} \geq\ \frac{a\plus{}b\plus{}c}{k\plus{}1}$
 :)

[quote="nguoivn"]Given $ a, b, c \geq\ 0$. Prove that:
$ \frac {a^3}{2a^2 \plus{} b^2} \plus{} \frac {b^3}{2b^2 \plus{} c^2} \plus{} \frac {c^3}{2c^2 \plus{} a^2} \geq\ \frac {a \plus{} b \plus{} c}{3}$

[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22937

Sorry, I don't know it has been posted before  :oops:
My another result:
Given $ a, b, c \geq\ 0$. Prove that:
$ \frac {3a^3 \plus{} abc}{b^3 \plus{} c^3} \plus{} \frac {3b^3 \plus{} abc}{c^3 \plus{} a^3} \plus{} \frac {3c^3 \plus{} abc}{a^3 \plus{} b^3} \geq\ 6$
 :)

[quote="nguoivn"]Given $ a, b, c \geq\ 0$. Prove that:
$ \frac {3a^3 \plus{} abc}{b^3 \plus{} c^3} \plus{} \frac {3b^3 \plus{} abc}{c^3 \plus{} a^3} \plus{} \frac {3c^3 \plus{} abc}{a^3 \plus{} b^3} \geq\ 6$
 :)[/quote]

It's a old problem of mine, nguoivn  :wink:

Besides, we have a similar result

$ \frac {2a^2 \plus{} bc}{b^2 \plus{} c^2} \plus{} \frac {2b^2 \plus{} ca}{c^2 \plus{} a^2} \plus{} \frac {2c^2 \plus{} ab}{a^2 \plus{} b^2} \ge\frac{9}2$

General problem for all $ k\ge0$

$ \sum_{cyc}\frac{a^k(3a^2\plus{}bc)}{b^{k\plus{}2}\plus{}c^{k\plus{}2}}\ge6$

[quote="dduclam"][quote="nguoivn"]Given $ a, b, c \geq\ 0$. Prove that:
$ \frac {3a^3 \plus{} abc}{b^3 \plus{} c^3} \plus{} \frac {3b^3 \plus{} abc}{c^3 \plus{} a^3} \plus{} \frac {3c^3 \plus{} abc}{a^3 \plus{} b^3} \geq\ 6$
 :)[/quote]

It's a old problem of mine, nguoivn  :wink:

[/quote]
Dear my brother, with some problems, accidentally, both of us found them.
This ineq has posted in my topic for a long time. You can see it here (problem 68)
http://www.maths.vn/forums/showthread.php?t=9244
I also have used Cauchy-Schwarts to solve it  :)

[quote="dduclam"]Besides, we have a similar result

$ \frac {2a^2 \plus{} bc}{b^2 \plus{} c^2} \plus{} \frac {2b^2 \plus{} ca}{c^2 \plus{} a^2} \plus{} \frac {2c^2 \plus{} ab}{a^2 \plus{} b^2} \ge\frac {9}2$

[/quote]
It's an old result of hungkhtn's vol 1.
The proof in his book used SOS. But indeed, it's weaker than Iran98  :)


geometry
perimeter
rectangle
combinatorics proposed
combinatorics

1.Prove that every convex polygon of area S and perimeter P contains a circle of radius greater than S/P

2.A convex polygon of area S and perimeter P contains a convex polygon of area T and perimeter Q,

Prove that 2S/P > T/Q.

(Use part 1 to prove part 2)

It's easier to deduce the second one from the first one:

Let $\ell_i,\ i\in\overline{1,n}$ be the sides of the polygon. If $O$ is a point inside, then the radius $r$ of a circle centered at $O$ and contained inside the polygon is $\le\min_i(d(O,\ell_i))$. This means that $r\le \frac{\sum_i d(O,\ell_i)\cdot \ell_i}{\sum_i \ell_i}=\frac {2S}P$. Now, since the second, smaller polygon contains, according to the first problem, a circle of radius $r>\frac TQ$, we get $\frac {2S}P\ge r>\frac TQ$, and we are done.

Now for the first problem:

Using arguments similar to those above, we can see that all we need to do is show that there is a point $O$ inside the polygon s.t. $d(O,\ell_i)>\frac SP,\ \forall i\in\overline{1,n}$. First draw lines $\ell_i'\|\ell_i$ s.t. the distance between $\ell_i$  and $\ell_i'$ is $d$, and $\ell_i'$ lies on the same side of $\ell_i$ as the interior of the polygon, for small $d$. It's easy to see that the area enclosed by $\ell_i,\ell_i'$ is smaller than $P\cdot d\ (*)$. At first, there's still some space inside the polygon formed by $\ell_i'$. We want to show that there's still space there when $d=\frac SP$. From $(*)$ we see that we can increase $d$ until it becomes $\frac SP$, because when it's $\frac SP$, the area enclosed by $\ell_i,\ell_i'$ is $<P\cdot \frac SP=S$, so we haven't yet covered the initial polygon.

Maybe it's not that rigorous, but I think something like that definitely works.

For the first one (which is Grobber's way, but maybe a bit clearer :P )

On each side of the given polgon, we draw internally a rectangle with height $\frac S p.$ Since the polygon is convex some of these rectangles have some common interior point, and the sum of the areas of these rectangles is $S$. It follows that at least one point, say $M$, of the interior point of the polygon is not covered by the rectangles, which means that $M$ is at distance at least $\frac S p$ from each sides so that the disk with center $M$ and radius $\frac S p$ gives a solution.

Pierre.

Moreover, Pierre's solution is, actually, the official one.


vector
algebra
linear equation
geometry unsolved
geometry

hi all,
I have 3 points A = (a1, a2, ..., an), B = (b1, b2, ..., bn), C = (c1,c2, ..., cn)
What is the simpliest way to obtain altitude from point A?
I need just direction of the AH vector, not length.

Thank you in advance, D.

Vector cross product is still fine for $ n$-dimensional space.
So it suffices to find a value of $ t$, so that the point $ K \equal{} ( t b_{1} \plus{} (1 \minus{} t) c_{1}, t b_{2} \plus{} (1 \minus{} t) c_{2}, ... , t b_{n} \plus{} (1 \minus{} t) c_{n} )$ is the foot of the altitude: This equates to $ AK$ and $ BC$ being perpendicular, or that the vectors which they represent are orthogonal. This gives you an equation that you can solve to find $ T$, and then find the $ AH$ vector. The equation for $ t$ is:
$ \sum_{i \equal{} 1}^{n} ( a_{i} \minus{} (t b_{i} \plus{} (1 \minus{} t) c_{i}))(b_{i} \minus{} c_{i}) \equal{} 0$.
It is a linear equation after expansion, so you can solve it easily.


I have a power set problem about a fixed point

$ we\ have\ a\ function\ g: \textsl{P(Q)}\rightarrow\textsl{P(Q)} \\
With\ \forall A,B \in \textsl{P(Q)}, A\subset B\ implies\ g\left(A\right) \subset g\left(B\right) \\
prove\ that\ \exists \ an\ A\in {P(Q)}\ such\ that\ g\left(A\right)\equal{}A \\
Now\ show\ also\ that\ \overline{Q} \equal{} \cap A \equal{}\subset A$

So P(Q) is the power set of Q and Q overline is the fixed point

plz help  :blush:

Let $ A \equal{} \bigcup\,\{X\in\mathcal{P}(Q): X\subseteq\mathrm{g}(X)\}$.

For all $ X$ such that $ X\subseteq\mathrm{g}(X)$, $ X\subseteq A$ and so $ X\subseteq\mathrm{g}(X)\subseteq\mathrm{g}(A)$. Hence $ A \equal{} \bigcup\,X\subseteq\mathrm{g}(A)$.

So $ \mathrm{g}(A)\subseteq\mathrm{g}(\mathrm{g}(A))$; hence $ \mathrm{g}(A)\in\{X\in\mathcal{P}(Q): X\subseteq\mathrm{g}(X)\}$; therefore $ \mathrm{g}(A)\subseteq A$.

Since $ A\subseteq\mathrm{g}(A)$ and $ \mathrm{g}(A)\subseteq A$, we have $ A \equal{} \mathrm{g}(A)$.

In general, if $ L$ is a complete lattice and $ \mathrm{f}: L\to L$ is an order-preserving function, then there always exists $ A\in L$ such that $ \mathrm{f}(A) \equal{} A$ – by the [url=http://www.cas.mcmaster.ca/~forressa/academic/701-talk.pdf]Knaster–Tarski fixed-point theorem for complete lattices[/url]. :)


Let circle $ (O)$ be inscribeb the $ ABC$ triangle. Let $ M, N, P$ are points of contact with $ BC, CA, AB$. Let $ MM'$ is a diameter of circle $ (O)$ and $ MM'$ meet $ PN$ at $ Q$. $ AQ$ meet $ BC$ at $ E$. Prove that $ E$ is point center of $ BC$.
---------------------------------------
Thanks every body. $ 1 \equal{} \minus{}1?$

[quote="thanhnam2902"]Let circle $ (O)$ be inscribeb the $ ABC$ triangle. Let $ M, N, P$ are points of contact with $ BC, CA, AB$. Let $ MM'$ is a diameter of circle $ (O)$ and $ MM'$ meet $ PN$ at $ Q$. $ AQ$ meet $ BC$ at $ E$. Prove that $ E$ is point center of $ BC$.
---------------------------------------
Thanks every body. $ 1 \equal{} \minus{} 1?$[/quote]

Please , who has a little time to search? We have the solution in an older post. I'll try too , but later.

 Babis

thank you very much for your reply. I wish you good luck.


inequalities
trigonometry
geometry
circumcircle
geometry proposed

\[ \prod \frac{a}{b+c}= \prod \frac{\sin{A}}{\sin{B}+\sin{C}} = \frac{\sin{\frac{A}{2}}\cos{\frac{A}{2}}}{\frac{\sin{B}+\sin{C}}{2}} \]
$\sin{x}$ is concave on $[0, \pi]$, thus:
\[ \frac{\sin{B}+\sin{C}}{2} \leq \sin{\frac{B+C}{2}}=\cos{\frac{A}{2}} \]
Pluge it in we get the desired ineqaulity.

Equality holds if and only if $\angle{A}=\angle{B}=\angle{C}$.

[b]Lemma.[/b] $\triangle ABC\Longrightarrow \boxed {\ \sin \frac A2\le \frac{a}{b+c}\ }\ \ (*)\ .$

[u]The first method.[/u] Define: the second intersection $A'$ between the circumcircle $C(O,R)$ and the bisector $[AI$ of the angle $\widehat {BAC}$; the intersection $D\in BC\cap AI$ $(l_a=AD)$. From the well-known relations $l_a=\frac{2bc}{b+c}\cdot \cos \frac A2$ and $AA'\cdot AD=bc$ $(ABA'\sim ADC)$ results:
$AA'\le 2R$ $\Longrightarrow$ $bc\le 2Rl_a$ $\Longleftrightarrow$ $bc\le 2R\cdot \frac{2bc}{b+c}\cdot\cos\frac A2$ $\Longleftrightarrow$ $b+c\le 4R\cos \frac A2$  $\Longleftrightarrow$ $\frac{b+c}{a}\le 2\cdot \frac{\cos \frac A2 }{\sin A}$ $\Longleftrightarrow$ $\frac{b+c}{a}\le \frac{1}{\sin \frac A2}$  $\Longleftrightarrow$ $\sin \frac A2\le \frac{a}{b+c}\ .$ Therefore, the inequality $(*)$ is equivalently with $AA'\le 2R\ !$

[u]The second method.[/u] Using the identity $bc=p(p-a)+(p-b)(p-c)$ results:
$(*)\Longleftrightarrow \sqrt{\frac{(p-b)(p-c)}{bc}}\le \frac{a}{b+c}$ $\Longleftrightarrow$ $(p-b)(p-c)(b+c)^2\le a^2bc$ $\Longleftrightarrow$ $(p-b)(p-c)(b+c)^2\le a^2 [p(p-a)+(p-b)(p-c)]$ $\Longleftrightarrow$ $(p-b)(p-c)\left[(b+c)^2-a^2\right]\le a^2p(p-a)$ $\Longleftrightarrow$ $4p(p-a)(p-b)(p-c)\le a^2p(p-a)\Longleftrightarrow 4(p-b)(p-c)\le a^2\Longleftrightarrow$ $a^2-(b-c)^2\le a^2\Longleftrightarrow (b-c)^2\ge 0$, what is truly.

[b]Remark.[/b] Using this lemma, the proposed inequality is a immediate consequence.
The equality holds if and only if $a=b=c$. Moreover prove easily that
$\boxed {\frac 12\left(\frac rR\right)^2\le \boxed {\prod \sin \frac A2\le \prod \frac{a}{b+c}}\le \frac 18}\ .$

[quote="Cezar Lupu"]Well, I want to point some remarks here. This inequality is equivalent with

                          \[ \frac{R}{r}\geq\frac{(a+b)(b+c)(c+a)}{4abc}. \]
  If I'm not mistaken this was proposed by Dorin Andrica in 1984 in Romanian Olympiad.
  Anyway, related to this one here is a stronger one, but all acute-angled triangle:

                                    \[ \frac{R}{r}\geq\sqrt{\frac{(a^{2}+b^{2})(b^{2}+c^{2})(c^{2}+a^{2})}{2a^{2}b^{2}c^{2}}}. \]

  This inequality has been proposed for the first time by Tudorel Lupu and Dumitru Anton in local olympiad in Constantza in 1988. In 1997, the same problem appears in American Mathematical Monthly due to Mihaly Bencze and
  Florin Popovici. :)[/quote]
\[\frac{R}{r} \ge \frac{1}{2}\left(\frac{64a^2b^2c^2}{(4a^2-(b-c)^2)(4b^2-(c-a)^2)(4c^2-(a-b)^2)}\right)^2.\]
(India tst 2006 p7)
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=485987

[quote="Amir.S"]Show that for any arbitrary triangle $ABC$, we have
\[\sin\left(\frac{A}{2}\right) \cdot \sin\left(\frac{B}{2}\right) \cdot \sin\left(\frac{C}{2}\right) \leq \frac{abc}{(a+b)(b+c)(c+a)}.\][/quote]
Show that for any arbitrary triangle $ABC$, we have
\[\cos^2A+\cos^2B+\cos^2C\geq \frac{1}{2}(\frac{a^2}{b^2+c^2}+\frac{b^2}{c^2+a^2}.+\frac{c^2}{a^2+b^2}).\]

[quote="vslmat"]Sqing, a nice problem! Here is the solution:[/quote]
Very Good.
But ,only for acute triangle was founded.

$a^2=(c\cdot\cos B+b\cdot\cos C)^2\stackrel{\mathrm{(C.B.S)}}{\ \le\ }\left(c^2+b^2\right)\left(\cos^2B+\cos^2C\right)\implies$

$\sum\left(\cos^2B+\cos^2C\right)\ge \sum \frac {a^2}{b^2+c^2}\implies$ $\sum\cos^2A\ge\frac 12\cdot \sum \frac {a^2}{b^2+c^2}$ .

[quote="vslmat"]Oh, Sqing, sorry for being so careless! The second part of the solution is here[/quote]
Dear  [b]vslmat [/b].
Right triangle?

$\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}=\frac{\cosh+\cosh+\cosh-1}{4}$
Thus it suffices to prove that
$\sum{\frac{a^2+b^2-c^2}{8ab}}-\frac{1}{4} \le \frac{abc}{(a+b)(b+c)(c+a)}$
$\Leftrightarrow \sum_{%Error. "cyclic" is a bad command.
}{a^2b}+\sum_{%Error. "cyclic" is a bad command.
}{ab^2}-\sum{a^3}-2abc \le \frac{8a^2b^2c^2}{(a+b)(b+c)(c+a)}$

Expanding both the sides and using Muirhead's theorem does the job!!

[quote=sqing]
Show that for any arbitrary triangle $ABC$, we have
\[\cos^2A+\cos^2B+\cos^2C\geq \frac{1}{2}(\frac{a^2}{b^2+c^2}+\frac{b^2}{c^2+a^2}.+\frac{c^2}{a^2+b^2}).\][/quote]
[url=https://artofproblemsolving.com/community/c6h1573841p9678280][b]International Zhautykov Olympiad 2018/1[/b]
[/url]

[quote=sqing]
Show that for any arbitrary triangle $ABC$, we have
\[\cos^2A+\cos^2B+\cos^2C\geq \frac{1}{2}(\frac{a^2}{b^2+c^2}+\frac{b^2}{c^2+a^2}.+\frac{c^2}{a^2+b^2}).\][/quote]
Proof of [b]zhumazhenis[/b]:
$$\cos^2A+\cos^2B=\frac{\left(c^2+b^2-a^2\right)^2}{4b^2c^2}+\frac{\left(c^2+a^2-b^2\right)^2}{4a^2c^2}\geq\frac{4c^4}{4c^2(a^2+b^2)}=\frac{c^2}{a^2+b^2}$$
Very nice.

[quote=mudok][quote=sqing]
Show that for any arbitrary triangle $ABC$, we have
\[\cos^2A+\cos^2B+\cos^2C\geq \frac{1}{2}(\frac{a^2}{b^2+c^2}+\frac{b^2}{c^2+a^2}.+\frac{c^2}{a^2+b^2}).\][/quote]

it was here

http://artofproblemsolving.com/community/c6h423450p2394999[/quote]
[url=https://artofproblemsolving.com/community/c6h423450p10114231]Yeah.[/url]

[quote=Evenprime123]\(\prod \frac{a}{b+c}= \prod \frac{\sin{A}}{\sin{B}+\sin{C}}\)
Could you explain how you can conclude this?[/quote]

It is just the sine rule


invariant
algorithm
number theory
Euclidean algorithm

Starting with any 35 integers, you may select 23 of them and add 1 to each. By repeating this step, one can make all 35 integers equal. Prove this. Now replace 35 and 23 by $m$ and $n$, respectively. What condition must $m$ and $n$ satisfy to make the equalization still possible?

[hide]
Yes, it's from Engel, so an alternative solution (that I could understand  :blush: ) would be great.

[/hide]

[hide="Well..."] $(m, n) = 1$.

In order that the $m$ numbers are equal, their sum must be divisible by $m$; since their initial sum is arbitrary, we need to be able to cover every residue class $\bmod m$ by repeated addition of $n$, which we can do if and only if $(m, n) = 1$.

The proof that this is sufficient as well as necessary is annoying, and I leave it to someone else to fill in the meat  :P [/hide]

[quote="Phelpedo"]If gcd(m,n)=1, then by the Euclidean Algorithm we can find integers such that $nx=my+1$. So pick the least number, and choose the first n, then shift over one spot and do the same thing, etc. until you've done it x times, at which point you've increased that first number by 1 more than everything else.

I'll have to think a little more about the necessary component...[/quote]

For the necessary part, suppose that $\gcd (m,n)>1$, and suppose one of our initial $m$ integers is one less than the rest.  Assume for the sake of contradiction that there is a sequence of moves that increases the smaller integer by $k+1$ and the rest of the integers by $k$.  Then all the integers increase by a total of $k+1+k(m-1)=km+1$.  Since each move increases the sum by $n$, we have $km+1=jn$, and $\gcd (m,n)=1$, a contradiction.


Quite nice one . :)

By AM-GM we have
$LHS\geq 9\sqrt[3]{(xyz)^{2}(x+y)^{4}(y+z)^{4}(z+x)^{4}}$

Now use
$(x+y)(y+z)(z+x)\geq 8xyz$  and
$9(x+y)(y+z)(z+x) \geq 8(x+y+z)(xy+yz+zx)$

 :)

Thank your for your nice solution! Hope it is nice as you wrote!  :D


inequalities
trigonometry
inequalities proposed

If $ A$, $ B$ and $ C$ are measures of angles of triangle prove the inequality:

$ [\sqrt {\sin (\frac {A}{2}) \sin (\frac {B}{2}) \sin (\frac {C}{2})}][\frac {1}{\sin \frac {A}{2} \plus{} \sin \frac {B}{2} \plus{} \sin \frac {C}{2}}] \leq \frac {1}{3\sqrt {2}}$

Oh I am sorry for mistake that I did :( .It should be:

$ [\sqrt {\sin (\frac {A}{2}) \sin (\frac {B}{2}) \sin (\frac {C}{2})}]$ instead of

$ [\sqrt {\sin (\frac {A}{2})\plus{} \sin (\frac {B}{2}) \plus{} \sin (\frac {C}{2})}]$ that I posted before (I have edited my post).

Good luck!

Nice solution.

Can anybody prove stronger result:

$ [\sqrt {\sin (\frac {A}{2}) \sin (\frac {B}{2}) \sin (\frac {C}{2})}][[\frac {1}{\sin \frac {A}{2}} \plus{}\frac{1} {\sin \frac {B}{2}} \plus{}\frac{1} {\sin \frac {C}{2}}]\leq \frac{3 \sqrt{2}}{2}$

useless comment by me  :(
EDIT:Due to delegat  :)

Oh I think that you haven't seen difference between first and second inequality [b]pardesi[/b]

In  second inequality it is:

$ [\frac {1}{\sin \frac {A}{2}} \plus{} \frac {1} {\sin \frac {B}{2}} \plus{} \frac {1} {\sin \frac {C}{2}}]$

In the first one it is:

$ [\frac {1}{\sin \frac {A}{2} \plus{} \sin \frac {B}{2} \plus{} \sin \frac {C}{2}}]$

Good luck!


1)You can buy $3$ chocolates for 1 unit of currency.If you give back $3$ chocolate wrappers back,you get $1$ chocolate.If you are given with $10$ units of currency,what is the maximum number of chocolates you can buy?(I dunno the answer,so give your methods)

2)You have a $12*12$ room and a $16*9$ carpet.How can you cut the carpet into $2$ parts so that it covers the $12*12$ room floor?

 :)

Wrt to the second problem  some interesting item:
There is a remarkable  theorem by Hilbert: If  two polygons are equivalent (same area) then it is possible then it is possible to transfer either polygon to the other by cutting it into finite number of cuts and rearranging the pieces.

Any rectangle where length is  not greater than  4 times the width, can be cut and made into a square with just two straight  cuts.
(I understand that it may produce 3 parts,  solution for two parts only requires a cut which is not straight - I think the question has been asked here before )

[quote="mbabbitt2003"]The first one is easy.

[hide]Lets consider 1 unit of currency 1 cent. You have 10 cents, so you can buy $30$ Chocs.
You now have 30 wrappers. You give back all of them to get $10$ more chocs.
You now have 10 wrappers. You give 9 back to get $3$ chocs.
You now have 4 wrappers (you didn't use one last time) You give 3 back to get $1$ more choc.
You have two wrappers and you throw them away since you can't buy a fraction of a choc.
So you have $30+10+3+1$ chocs, which is $44$[/hide]

I couldn't do the 2nd one though .[/quote]

I think that's wrong(at least my friend told me that when I told him this answer)

[quote="Gyan"][quote="mbabbitt2003"]The first one is easy.

[hide]Lets consider 1 unit of currency 1 cent. You have 10 cents, so you can buy $30$ Chocs.
You now have 30 wrappers. You give back all of them to get $10$ more chocs.
You now have 10 wrappers. You give 9 back to get $3$ chocs.
You now have 4 wrappers (you didn't use one last time) You give 3 back to get $1$ more choc.
You have two wrappers and you throw them away since you can't buy a fraction of a choc.
So you have $30+10+3+1$ chocs, which is $44$[/hide]

I couldn't do the 2nd one though .[/quote]
FWIW: One way to thnk about this, is
[hide]from 1 coin you can buy $3+(1/3)+(1/3^2)+\dots = 9/2$
 (mathematician don't have to follow silly rules like chock. have to be integer :) ) So theoretically upper limit of the answer is ($10 \cdot (9/2)= 45$) Just find a creative way to think if that answer is possible!... Hint.. Hint ... [/hide][/quote]

[hide] Actually, the answer is 44. There is no creative way to make 45.  But you can figure out why. :D  [hide="hint"] think of the last candy/wrapper [/hide][/hide] :lol:

[quote] Actually, the answer is 44. There is no creative way to make 45. But you can figure out why :D [/quote]
I don't have to!  the problem is fairly old, and  many, even those who hear the problem for the first time, have been known to find the answer without too much difficuly. It's really not that  hard to figure out 'how' , specially if you know the answer.

[quote="lingomaniac88"]Here's how you make 45:
[hide]When someone else buys one, ask for his/her wrapper.  Give in that wrapper plus the two you have for a 45th chocolate.  Give him back the wrapper when you're done with it.[/hide][/quote]
Yes, nice one.  [hide=" and for that matter,"]  you can walk in, pick  a chocolate, unwrap it, and on  your way out, pay   with the three wrappers you now have. (Assuming you started with two wrappers when you walked in)  [/hide]

The second one has been answered  above.

[quote="shadysaysurspammed"][quote="Adunakhor"][hide="2"]
Start $4$ units to the left of the top left corner of the rectangle (the horizontal lengths are $16$ units).

Make a cut $3$ units down, then make a cut $4$ units to the left, and then repeat this zig-zag pattern $2$ more times.

[/hide][/quote]

I don't understand that :? .....[/quote]

A quick diagram:
[img]http://reisque.com/rjt/images/cutsquare.png[/img]

Btw, excellent answer Adunakhor :)


analytic geometry
graphing lines
slope
complex numbers

Let $ w,x,y,z$ be the complex numbers that correspond to points $ W,X,Y,Z$, respectively. Prove that $ WX\parallel{}YZ$ iff $ \frac{x\minus{}w}{z\minus{}y}$ is real.

Low tech proof... for one direction. I'm not sure how to do the other way quite yet.

[hide]
Let $ WX$ and $ YZ$ be parallel
Let $ k$ be the slope of $ WX$ and $ YZ$
Let $ a,b,c,d,e,f,g,h\in\Re$
if $ x \equal{} a \plus{} bi, w \equal{} c \plus{} di, z \equal{} e \plus{} fi, y \equal{} g \plus{} hi$

the slope of $ WX \equal{} \frac {b \minus{} d}{a \minus{} c} \equal{} k$
so $ b \minus{} d \equal{} (a \minus{} c)k$
the slope of $ YZ \equal{} \frac {f \minus{} h}{e \minus{} g} \equal{} k$
so $ f \minus{} h \equal{} (e \minus{} g)k$

$ \frac {x \minus{} w}{z \minus{} y} \equal{} \frac {(a \plus{} bi) \minus{} (c \plus{} di)}{(e \plus{} fi) \minus{} (g \plus{} hi)} \equal{} \frac {(a \minus{} c) \plus{} (b \minus{} d)i}{(e \minus{} g) \plus{} (f \minus{} h)i}$
$ \equal{} \frac {(a \minus{} c) \plus{} (a \minus{} c)ki}{(e \minus{} g) \plus{} (e \minus{} g)ki} \equal{} \frac {(a \minus{} c)(1 \plus{} ki)}{(e \minus{} g)(1 \plus{} ki)} \equal{} \frac {a \minus{} c}{e \minus{} g}$
$ \frac {a \minus{} c}{e \minus{} g}$ is real so $ \frac {x \minus{} w}{z \minus{} y}$ is real[/hide]

Conversely, suppose that $ \frac{x\minus{}w}{z\minus{}y}\equal{}r\in\mathbb{R}$. Then, if we let $ x\equal{}x_r\plus{}x_ii$, $ w\equal{}w_r\plus{}w_ii$, $ z\equal{}z_r\plus{}z_ii$, $ y\equal{}y_r\plus{}y_ii$,

$ x\minus{}w\equal{}r(z\minus{}y)\Rightarrow (x_r\minus{}w_r)\plus{}(x_i\minus{}w_i)i\equal{}r(z_r\minus{}y_r)\plus{}r(z_i\minus{}y_i)i$. Therefore,

$ x_r\minus{}w_r\equal{}r(z_r\minus{}y_r)$ and $ x_i\minus{}w_i\equal{}r(z_i\minus{}y_i)$. Thus,

$ r\equal{}\frac{x_r\minus{}w_r}{z_r\minus{}y_r}\equal{}\frac{x_i\minus{}w_i}{z_i\minus{}y_i}\Rightarrow \frac{z_i\minus{}y_i}{z_r\minus{}y_r}\equal{}\frac{x_i\minus{}w_i}{x_r\minus{}w_r}$

Since the slopes of $ WX$ and $ YZ$ are equal, $ WX\parallel YZ$.


[quote="hungkhtn"]Prove that for all $a,b,c,d \ge 0$ and $a+b+c+d=4$ then
\[ (1+3a)(1+3b)(1+3c)(1+3d) \le 125+131abcd  \][/quote]
I think this is true. But I found a very ugly proof only. :(

Of course It's true, and the solution is very nice, too/

seems to be quite similar to IMO Short-listed, 1993...

$abc+bcd+cda+dab \leq \frac{1}{27}+\frac{176}{27}abcd$

where  $a+b+c+d=1$ and $a,b,c,d>0$

OK, and same as Tukervici to

\[ a^4+b^4+c^4+d^4+2abcd \ge a^2b^2+b^2c^2+c^2d^2+d^2a^2+a^2c^2+b^2d^2 \]

But... as far as I can say, It's more difficult.

[quote="hendrata01"]did anyone notice that the equality also holds when $a=0,b=c=d=\frac{4}{3}$?
It's really peculiar how to get that form....[/quote]Well, that can explain of course the 125 on the RHS - and probably be a hint that Vasc's n-1EQ principle might apply here :)

Nice!  :ninja:
Do you have a proof for this one? :)

[quote="hungkhtn"]Yes, very nice solution. It's also a solution in my book of a friend. But we have even nicer solution :D.[/quote]

Dear Hung, you are right !  :)
After a little minutes :D , I have another solution and I wonder if it is your solution.

When c,d is fixed then a+b is also fixed and:
$f(a,b,c,d)=(1+3a)(1+3b)(1+3c)(1+3d)-125-131abcd$
$=(1+3(a+b)+9t)(1+3c)(1+3d)-125-131cdt=g(t)$ with $t=ab$

$g(t)$ is linear with $t$ and $0\le t\le (\frac{a+b}{2})^2$, so
$g(t) \le max(g(0),g((\frac{a+b}{2})^2)) \le max(0,f(\frac{a+b}{2},\frac{a+b}{2},c,d))$
(because $g(0) \le 0$ (AM-GM) and $g((\frac{a+b}{2})^2)= f(\frac{a+b}{2},\frac{a+b}{2},c,d)$)

The rest is complete similar with my upper solution.

----

With this method, it is easy to generalize as Vasc's problem  :D :
[quote="Vasc"]We can generalize the inequality as follows:
$[1+(n-1)a_1][1+(n-1)a_2]...[1+(n-1)a_n] \leq (n^n-p)a_1a_2...a_n +p$,
where $p=(n+1)^{n-1}$ and $a_1+...+a_n=n$.[/quote]

[quote="hendrata01"]hmm maybe I'm missing something, but why does $t$ has to be $t \leq \frac{a+b}{2}$?
because if it's AM-GM then $t \leq (\frac{a+b}{2})^2$[/quote]

Yes, this is typo mistake  :D
I corrected my post, thank you!

[quote="Vasc"]We can generalize the inequality as follows:
$[1+(n-1)a_1][1+(n-1)a_2]...[1+(n-1)a_n] \leq (n^n-p)a_1a_2...a_n +p$,
where $p=(n+1)^{n-1}$ and $a_1+...+a_n=n$.[/quote]

Oh, It's one of my generation in my book, too. Even that we can replace $n-1$ by $m$. The solution for those and the first problem I posted are the same.


I dont know but plugging in 3 for n in the first equation produces a contradiction


Two numbers are chosen at random, with replacement, from the set $ \{1, 2, 3, 4\}$. The two numbers are used as the numerator and denominator of a fraction. What is the probability that the fraction represents a whole number? Express your answer as a common fraction.

we count the possibilities :)
denom.=1: 1,2,3,4
2: 2,4
3: 3
4: 4
obviously this is it, so we add $ 4\plus{}2\plus{}1\plus{}1$ different fractions.
that's 8 fractions.
there are $ 4\cdot 4$ fractions total able to be chosen, so the answer is $ \frac{8}{16}$ or
1/2


A circle is divided into $13$ segments, numbered consecutively from $1$ to $13$. Five fleas called $A,B,C,D$ and $E$ are sitting in the segments $1,2,3,4$ and $5$. A flea is allowed to jump to an empty segment five positions away in either direction around the circle. Only one flea jumps at the same time, and two fleas cannot be in the same segment. After some jumps, the fleas are back in the segments $1,2,3,4,5$, but possibly in some other order than they started. Which orders are possible ?

I remember having solved this before the crash.

Represent the motion on a circle numbered differently, so that each movement on the initial circle is translated to the second circle as one step (instead of $5$ steps at a time). We number the $13$ segmnents of the new circle in the following way, in a counterclockwise direction:

$1,6,11,3,8,13,5,10,2,7,12,4,9$. We can now see that because of the constraint that no two fleas can share a square, the fleas cannot jump over one another, so the only possible permutations are the cyclic permutations, and only those.

It was one of the easiest problems on Baltic Way 2004. Even my little students solved it.  ;)  :D


Determine all positive integers $n$ such that $n$ divides $3^n - 2^n$.

Let $p$ be the smallest prime divisor of $n$. Then $p\mid 3^n-2^n$ and $p \mid 3^{p-1}-2^{p-1}$, so $p \mid 3^{\gcd (n,p-1)}-2^{\gcd (n,p-1)}$ because $a_n = 3^n-2^n$ is a Mersenne sequence (well-known) $= 3-2 = 1$, so the only $n$ is 1.

So, for any a, b, un = an - bn is a Mersenne sequence?

Strange, I didn't know that, I only knew it is the case for an - 1.

n=1 is a solution.
If n>1 3 n-2 n is odd, so n is odd.
Let p be the smallest prime divisor of n.
We have
p|n|3 n-2 n
p|3 p-1-2 p-1 .
Because 3 n-2 n is a Mersene sequence, we get
p|3 gcd(n,p-1)-2 gcd(n,p-1).
Because p was the smallest prime divisor of n, gcd(n,p-1)=1, so
p|3-2=1, so p=1. This is a contradiction.
So the only solution is n=1.

I'm not sure right now, but I think a and b must be coprime.

Seems like grobber posted the solution while I was typing mine.

a and b need not be coprime. Just distinct.

So, for any a, b, gcd(an - bn, am - bm) = ad - bd where d = gcd(m, n).

Is this only known in Romania or so? :) I only saw this theorem on this forum.
Is it difficult to prove? Also, can you assume it as 'known' at an IMO?

Yes, I think you're right, sorry about the mistake (I'm only human! :D )

I appologise for my mistake, but I change my statement to a and b coprime.

Suppose n>m.
gcd(a^n-b^n,a^m-b^m)=gcd(a^n-b^n-a^n-b^m*a^n-m, a^m-b^m)=gcd(b^m*(b^(n-m)-a^(n-m)),a^m-b^m).
Continue using this by following Euclid's algorithm for (n,m) and you should come to
gcd(a m- bm,a n- bn)=a gcd(m,n)- bgcd(m,n).

I think it's Ok to use this in contests without proof. From what I've seen, it's pretty well-known (whether it's with a and b coprime or not  :D , because I'm still not sure how it should be).

grobber, we keep posting at the same time.

If gcd(a,b)=d and a=da1 and b=db1, we get t=gcd(a^n-b^n,a^m-b^m)=d n gcd(a1^n-b1^n, d m-n*(a1^m-b1^m)), for m>n.
We can have p|d and p|a1^n-b1^n and obtain a multiple of a^gcd(n,m)-b^gcd(n,m), larger that a^gcd(n,m)-b^gcd(n,m) (at least p times larger than it).

So gcd(a,b)=1 if we want a Mersene sequence.
I think Mersene sequences are IMO material.

[quote="grobber"] ... because $a_n = 3^n-2^n$ is a Mersenne sequence (well-known) ...[/quote] I think you meant: $a_n = 3^n-2^n$ is a [b]Lucas [/b]Sequence.
Look at P. Ribenboim's "The Little Book of Bigger Primes" (2004) page 44 : $U_n(P,Q)=\frac{\alpha^n-\beta^n}{\alpha-\beta}$ with $\alpha=3$ and $\beta=2$ and thus $P=5$ and $Q=6$.

$\'{E}$douard Lucas studied the work of Fibonacci and provided a primality test for Mersenne (and Fermat and other) numbers based on what we now call Lucas Sequences, because he studied them in depth, but he did not discover them.

Tony


Άλλη μια όμορφη ανισότητα :

Αν  $ a\ge b\ge c > 0$  και $ 0\le m \le 2$ δείξτε ότι

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

* Μετά από μια δεύτερη ματιά νομίζω ότι δεν υπάρχει πρόβλημα αν $ m\ge0$ . Eπίσης τα b,c μπορεί να έχουν οποιαδήποτε διάταξη. Το σωστότερο θα ήταν να είχαμε για υπόθεση $ a\ge b$, $ a\ge c$, με b,c > 0.

Let $ \frac{a}{b}\equal{}x,\ \frac{a}{c}\equal{}y\ (x\geq 1,\ y\geq 1)$, we can rewrite the inequality as $ x^3\plus{}y^3\plus{}(m^2\minus{}1)\left(\frac{1}{x\plus{}m}\plus{}\frac{1}{y\plus{}m}\right)\geq 2m$ for $ 0\leq m\leq 2.$

Θα δείξουμε ότι:
$ \frac {a^3}{b^3} \ge \frac {b \plus{} ma}{a \plus{} mb} \Leftrightarrow a^4 \plus{} ma^3b \ge b^4 \plus{} mab^3$
που ισχύει γιατί
$ a \ge b \ge 0 \Rightarrow a^4 \ge b^4$
και
$ ma^3b \ge mab^3 \Leftrightarrow mab(a^2 \minus{} b^2) \ge 0 \Leftrightarrow mab(a \plus{} b)(a \minus{} b) \ge 0$
που ισχύει γιατί $ a \ge b \ge 0$ και $ m \ge 0$

άρα $ \frac {a^3}{b^3} \ge \frac {b \plus{} ma}{a \plus{} mb}$
όμοια $ \frac {a^3}{c^3} \ge \frac {c \plus{} ma}{a \plus{} mc}$
προσθέτοντας κατά μέλη έχουμε το ζητούμενο


geometry
ratio
calculus
algebra
polynomial
calculus computations

[quote="Robert0201"]The line tangent to x^3 at point P also intersects x^3 at point Q.  Let A be the region bounded by the curve and the line segment PQ.  Let B be the area of the region defined in the same way with Q instead of P.  What is the relationship between A and B?[/quote]
I think  visual view of the problem would be very helpful.Can you use some tool, for example sketchpad to draw a graph?

It's clear enough without the picture. I'm also moving it to the "Computations and Tutorials" forum. Of course, point $P$ cannot be the origin, but everything else works.

[hide]If point $P$ occurs when $x=x_0$, then point $Q$ has $x=-2x_0.$ The area between the tangent line and the curve is then proportional to $x_0^4.$ To be exact, it is $\frac{27}4\,x_0^4$ but you don't need the coefficient to get the ration called for in the the problem.

The relationship called for is that $A:B\,\text{as}\,1:16.$

Nothing fancy for the techniques; just find the tangent line, find the other point, and compute the integral. It's a calculus exercise.[/hide]

how can you show that [tex]x=-2x_0[/tex]?

[quote="Robert0201"]how can you show that $x=-2x_0$?[/quote]
Suppose that the tangent line at $x_0$ has the form $y=ax+b.$ Then the places at which this line intersects the curve are solutions to the equation $x^3-ax-b=0.$ The roots of this equation are $x_0, x_0,$ and ?. We know there's a double root at $x_0$ because of the tangency. By Viete's relations for the roots of a polynomial, and since the $x^2$ term of this polynomial is zero, we know the sum of the three roots is zero. That means the other root must be $-2x_0.$

What's interesting about this problem is that I get the sense that it could have been solved before there was such a thing as calculus in the mathematical world. In particular, it seems well within the mathematical reach of someone like Descartes.

Of course, if you'd rather use calculus, you could explicitly find the equation of that tangent line and factor the resulting cubic.

can you prove this using more elementary methods?


geometry
3D geometry
number theory proposed
number theory

Hello guys,
                      Recently, I stumled upon this curious problem. At the outset let me define an Armstrong number. Numbers of the form "a1a2a3..an" (where ai's are the digits) are called Armstrong numbers if the sum a1^3+a2^3+..an^3 = a1a2a3..an.
(i.e the sum of the cube of digtis of the number equals the number itself)

 The smallest Armstrong number is 153 (1^3+5^3+3^3 = 153).

Now, can anyone help me prove the following:-

           When the cubes of the digits of any number, that is, a multiple of 3 are added, and then this process is repeated, the final result is 153, where the process ends because 153 = 13 + 53 + 33
For example: Take the number 108

13 + 03 + 83 = 513 and 53 + 13 + 33 = 153

So, the number 108 reaches 153 in two cycles and it can be represented as

108-->513-->153.

Any help received will be appreciated.

-Thank You

The number $a_1a_2a_3 \cdots a_n$

Goes to the number $10(a_1+a_3+ \cdots a_n)+3n$

Look at this mod 3.  $10(a_1+a_3+ \cdots a_n)+3n= a_1+a_2+a_3+ \cdots a_n= \text{The previous number}$

So the remainder on division by 3 is invarient. So the answer is no. For example, 1 will not end up at 153.

HI

Sorry, I am a bit lazy. But the numbers of our concern are only those which happen to be a multiple of 3 which is failed by the number  ---  1.

For all n (natural numbers of course) such that n = 3k the stated result holds. U can check for af ew right here to convince urself.

For eg:- Consider 3

3^3=27

2^3+7^3=351

3^3+5^3+1^3=153

Also,

Say we consider 36

3^3+6^3=243

2^3+4^3+3^3=99

9^3+9^3=1458

1^3+4^3+5^3+8^3=702

7^3+0^3+2^3=351

3^3+5^3+1^3=153

Well I experimented with many multiples of 3 and the found the claim to be surprisingly true. Please explain this mysterious result to me.
I am waitin' for someone to solve it....

There is a brutal way to show it:
The number $\sum_{k=0}^m a_k 10^k$ is mapped to $\sum_{k=0}^m a_k^3 \leq (m+1)9^3$.
So let $f$ be this mapping, and you get $f(n) \leq 9^3(1+log_{10}(n))$, so every number will for sure end up to fulfill $n \leq 9^3(1+log_{10}(n))$ (only true for $n \leq 3293$), which is just fulfileld for finitely many numbers, so you are left with checking them.


Alicia receives 2/5 of a \$120 gift, and Jason receives 1/3 of the gift. What is the ratio of Jason's share to Alicia's share? Express your answer as a common fraction.

Another way:

Alicia received 48 dollars, and Jason received 40 dollars. Thus, the answer is $ \frac {40}{48}$ or $ \boxed{\frac{5}{6}}$.

To simplify a complex fraction, as fishy has done, invert the denominator of the large fraction and multiply it by the numerator, or...

$ \frac{\frac{a}{b}}{\frac{c}{d}} \equal{} \frac{a}{b} \times \frac{d}{c}$

Which is basically [b]dividing[/b] fractions...

I think this is enough information for the asker, so let's lock this.  :)


Express the sum of one meter and one centimeter as a decimal number of meters to the nearest hundredth.

A meter is 100 centimeters, so we have $ 1.01$.


trigonometry
inequalities
geometry proposed
geometry

Indeed, it is an easy and [b]nice[/b] problem !

[b]Proof.[/b] Prove easily that $\{\begin{array}{c}ab\cdot\cos C-b^{2}=ac\cdot\cos B-c^{2}\\\\ r(ABC)=p-a\Longleftrightarrow A=90^{\circ}\end{array}$. Observer that

$\{\begin{array}{ccccc}PA^{2}=AB_{1}\cdot AC & , & AB_{1}=c\cdot\cos A & \Longrightarrow & \boxed{PA^{2}=bc\cdot\cos A}\\\\ PC^{2}=CB_{1}\cdot CA & , & CB_{1}=a\cdot\cos C & \Longrightarrow & PC^{2}=ab\cdot \cos C\\\\ QA^{2}=AC_{1}\cdot AB & ,& AC_{1}=b\cdot \cos A & \Longrightarrow & \boxed{QA^{2}=bc\cdot\cos A}\\\\ QB^{2}=BC_{1}\cdot BA & , & BC_{1}=a\cdot\cos B & \Longrightarrow &QB^{2}=ac\cdot\cos B\end{array}\|\Longrightarrow$

$\{\begin{array}{c}2r(APC)=PA+PC-AC=\sqrt{bc\cdot\cos A}+\sqrt{ab\cdot\cos C}-b\\\\ 2r(AQB)=QA+QB-AB=\sqrt{bc\cdot\cos A}+\sqrt{ac\cdot\cos B}-c\end{array}$. Therefore,

$r(APC)=r(AQB)$ $\Longleftrightarrow$ $\boxed{\sqrt{ab\cdot\cos C}-b=\sqrt{ac\cdot\cos B}-c}$ $\Longleftrightarrow$

$\frac{ab\cdot\cos C-b^{2}}{\sqrt{ab\cdot\cos C}+b}=\frac{ac\cdot\cos B-c^{2}}{\sqrt{ac\cdot\cos B}+c}$ $\Longleftrightarrow$

$\boxed{\sqrt{ab\cdot\cos C}+b=\sqrt{ac\cdot\cos B}+c}$ $\Longleftrightarrow$ $b=c$.

[quote="stergiu"][quote="Shishkin"]Triangle $ABC$ is acute-angled.$BB_{1}$ and $CC_{1}$ are it's altitudes.$P \in BB_{1}$, $Q \in CC_{1}$, $\measuredangle{APC}=\measuredangle{AQB}=\frac{\pi}{2}$, inradiuses of $\triangle{APC}$ and $\triangle{AQB}$ are equal.Prove that  $AB=AC$.[/quote]

[b]Hint[/b]

 I proved that $AP=AQ$(using metric relation for right triangles and power from $A$ for the inscribed quadrilateral $BC_{1}B_{1}C$ )  , so triangles $APC , AQB$ are equal , since they have a vertical(kathete) side equal and equal incircles.
  Is your proof on the same way?

Babis[/quote]
Your way is the best and the shortest.But I use some inequalitis to solve it.My way is too long.

[quote="dragonfire"]@Shishkin: Can you post all problems from Ukraine National Contest?[/quote]
OK.I'll post them.Also I have already posted the best problem on this contest
http://www.mathlinks.ro/Forum/viewtopic.php?t=141416
and this one
http://www.mathlinks.ro/Forum/viewtopic.php?p=800547#800547

[quote="stergiu"][quote="stergiu"][quote="Shishkin"]Triangle $ABC$ is acute-angled.$BB_{1}$ and $CC_{1}$ are it's altitudes.$P \in BB_{1}$, $Q \in CC_{1}$, $\measuredangle{APC}=\measuredangle{AQB}=\frac{\pi}{2}$, inradiuses of $\triangle{APC}$ and $\triangle{AQB}$ are equal.Prove that  $AB=AC$.[/quote]

[b]Hint[/b]

 I proved that $AP=AQ$(using metric relation for right triangles and power from $A$ for the inscribed quadrilateral $BC_{1}B_{1}C$ )  , so triangles $APC , AQB$ are equal , since they have a vertical(kathete) side equal and equal incircles.
  Is your proof on the same way?

Babis[/quote]

To be more precise , we have :

$AP^{2}= AB_{1}\cdot AC$ and $AQ^{2}= AC_{1}\cdot AB$
But
  $AB_{1}. AC = AC_{1}.AB$ , so $AP=AQ$ .

 With some elementary theory on equal right trinagles we find that triangles
    $APC , AQB$ are equal, so $AB=AC$
 How would  you prove that :

 [color=red]If two right triangles have a corespondent side equal and equal inradii , then they are equal.[/color].
 Problably Virgil can help us using some of the million relations that he knows about triangles :) .Otherwise , I will have only my own simple but not very-very short proof.

[u] Babis[/u][/quote]
It is equivalent to prove that there is only two tangents to circle from point. :wink:

[quote="Shishkin"][/quote][quote="stergiu"][quote="stergiu"][quote="Shishkin"]Triangle $ABC$ is acute-angled.$BB_{1}$ and $CC_{1}$ are it's altitudes.$P \in BB_{1}$, $Q \in CC_{1}$, $\measuredangle{APC}=\measuredangle{AQB}=\frac{\pi}{2}$, inradiuses of $\triangle{APC}$ and $\triangle{AQB}$ are equal.Prove that  $AB=AC$.[/quote]

[b]Hint[/b]

 I proved that $AP=AQ$(using metric relation for right triangles and power from $A$ for the inscribed quadrilateral $BC_{1}B_{1}C$ )  , so triangles $APC , AQB$ are equal , since they have a vertical(kathete) side equal and equal incircles.
  Is your proof on the same way?

Babis[/quote]

To be more precise , we have :

$AP^{2}= AB_{1}\cdot AC$ and $AQ^{2}= AC_{1}\cdot AB$
But
  $AB_{1}. AC = AC_{1}.AB$ , so $AP=AQ$ .

 With some elementary theory on equal right trinagles we find that triangles
    $APC , AQB$ are equal, so $AB=AC$
 How would  you prove that :

 [color=red]If two right triangles have a corespondent side equal and equal inradii , then they are equal.[/color].
 Problably Virgil can help us using some of the million relations that he knows about triangles :) .Otherwise , I will have only my own simple but not very-very short proof.

[u] Babis[/u][/quote][quote="Shishkin"]
It is equivalent to prove that there is only two tangents to circle from point. :wink:[/quote]

Any way ! I think you have done the same solution !

[u] Babis[/u]


Prove that for any x and y positive integers 4xy-y-x is not a square. For instance if y=1 Its easy to show 3x-1 is never a square.

Equivalent: $z^2=4xy-x-y = \left( 2x-\frac{1}{2} \right) \left( 2y-\frac{1}{2} \right) - \frac{1}{4}$.
Or: $(2z)^2+1 = (4x-1)(4y-1)$.
Now $4x-1$ has a prime divisor $p \equiv -1 \mod 4$. Then $(2z)^2 \equiv -1 \mod p$, but $t^2 \equiv -1 \mod p$ doesn't have a solution since that would give:
$1 \equiv t^{p-1} = \left( t^2 \right)^{\frac{p-1}{2}} \equiv (-1)^\text{odd number} = -1 \mod p$.


Consider the function $ f(x)\equal{}\frac{(x\plus{}k)^2}{x^2\plus{}1},x\subset (\minus{}\infty,\infty)$, where $ k$ is a positive integer. Show that $ f(x)\le k^2\plus{}1$.

[hide="Short"]Consider sequences $ \{ x, 1 \}$ and $ \{ 1, k \}$ By Cauchy-Schwarz:
$ (x^2 \plus{} 1^2)(1^2 \plus{} k^2) \geq (x \cdot 1 \plus{} 1 \cdot k)^2 \iff (x^2 \plus{} 1)(k^2 \plus{} 1) \geq (x \plus{} k)^2$
$ \implies k^2 \plus{} 1 \geq \frac {(x \plus{} k)^2}{x^2 \plus{} 1} \equal{} f(x)$, as desired.

Equality occurs when $ \frac{x}{1} \equal{} \frac{1}{k} \iff xk \equal{} 1$
[/hide]


calculus
derivative
function
limit
calculus computations

Could someone please start me off on the right track?

Let f be a function that is defined and twice differentiable for values of x and has the following properties:

f(1) = 3
lim x-> 0+ f(x) = ∞
f'(x) < 0 when x > 0
f'(x) > 0 when x < 0
lim x-> ∞ f(x) = 2

g(x) = f(1/x)
find g(1) and lim x-> 0+ g(x)


linear algebra
matrix
combinatorics unsolved
combinatorics

[quote="spdf"]Let n>1 be a positive integer 1, and let n circles of same radius be given in the plane such that no two of them are tangent and the subset of the plane formed by the union of them is connected. Prove that the number of intersection points is at least n.[/quote]
denote our circles by $C_1,C_2,...,C_m$ and our points by $A_1,A_2,...,A_n$ we now have to show that $m\geq n$
now build a $n*m$ matrix such that if point $A_i$ is the intersection of $k$ circles then put $\frac{1}{k}$ in the intersection of the $i$th row and the $k$ columns that denote the $k$ circles
now assume that circle $C_1$ has $a$ points on it
hence the sum of the numbers of column $C_1$ is equal to $\sum_{i=1)^a \frac{1}}{k_i)}$
now we know that there doesnt exists a point $A_i$ on circle $C_i$ such that $A_i$ is the intersection of more than $a$ circles because $A_i$ is on circle $C_1$ therefor every circle that contains $A_i$ other than $C_1$ creates another point $A_j$ on $C_1$ so $C_1$ should have at least $a+1$ points contradiction
thus $a\geq k_i$ for $1\leq i\leq a$ so we have:
$\sum_{i=1}^a \frac{1}{k_i}> \underbrace{\left(\left(\frac{1}{a}\right)+\left(\frac{1}{a}\right)+...+\left(\frac{1}{a}\right)\right)}_{a\geq k_i}=1$
thus the sum of every column is at least $1$ and the sum of every row is equal to $1$ therefor $m\geq n$

sorry if I didnt explain it well
did you get it or should I explain it more clearly??

yes.it's also my solution.it also appear in VietnamTST 2000


Let ABC be a triangle $w_C=30 ,w_B=14\sqrt{6} and |AB|=35$ then find  lengths of  |AC|.w denotes length of bisector.

Have you got any solution?I can't manage that.I will be dying for solution.İs anybody has a solution?

use Stewart theorem .and Calculae lenght of bisector.as I can remmembr .
$w_a^2=t_a^2=bc[1-(\frac{a}{b+c})^2]$


geometry
incenter
geometric transformation
reflection
geometry proposed

Let I be the incentre of triangle ABC. Let D and E be points on AB and AC, respectively such that DE is tangent to the incircle and parallel to BC. Define similarly F and G, and H and J. Now suppose DE :cap: FG = Z, FG :cap: HJ = Y, HJ :cap: DE = X. Show that XYZ is identical to ABC. (too lazy to check if the vertices are in the right order, sorry... :lol: )

It's not hard to see, that the angles of XYZ are the same as in ABC; thus they're similar. Now use that the incircle of ABC is also incircle of XYZ and you've got congruency.

You don't even need to compute the angle!

Just note that, if two tangents to a circle are parallel, then they are reflections of each other in the center of the circle. Thus, as the lines DE and BC both are tangents to the incircle of triangle ABC and are parallel to each other, they are symmetric to each other with respect to the center I of the incircle. In other words, the line DE is the reflection of the line BC in the point I. Similarly, the lines FG and HJ are the reflections of the lines CA and AB in the point I. In other words, the sidelines of triangle XYZ are the reflections of the sidelines of triangle ABC in the point I. And therefore, the triangle XYZ is the reflection of the triangle ABC in the point I. What yields that the two triangles are congruent. $\blacksquare$

  Darij


search
number theory unsolved
number theory

let$ p$  prime number$ (p >2)$. prove that :
$ \frac{m}{n} \equal{} 1 \plus{}\frac{1}{2} \plus{}\frac{1}{3} \plus{}...\plus{}\frac{1}{p\minus{}1}$
where  $ \frac{m}{n}$ is irreducible fraction.
prove that :numerator of fraction $ m$ divided by $ p$

i think ur experienced enuf in this forum to be knowing where to post this??

It is not true for $ p\equal{}2$ and for $ p>3$ we get $ p^2|m$.

[quote="Rust"]It is not true for $ p \equal{} 2$ and for $ p > 3$ we get $ p^2|m$.[/quote]
 exactly  where  link?
 i don't understand that  how search  on this web ?
       let  send for me link of this  problem !
 PS:  to Rust  i have notice that : $ p > 2$

let $ Q(x)\equal{}\prod_{k\equal{}1}^{p\minus{}1}(x\minus{}k)\equal{}\sum_{j\equal{}0}^{p\minus{}1}a_jx^j$
so we have $ \frac{m}{n}\equal{}\sum_{k\equal{}1}^{p\minus{}1}\frac{1}{k}\equal{}\frac{(\minus{}1)^{p\minus{}2}a_1}{(p\minus{}1)!}\equal{}\minus{}\frac{a_1}{(p\minus{}1)!}$
then $ \minus{}a_1n\equal{}(p\minus{}1)!m$

$ (p\minus{}1)!\equal{}Q(p)\equal{}(p\minus{}1)!\minus{}a_1p\plus{}p^{2}\sum_{j\equal{}2}^{p\minus{}1}a_jp^{j\minus{}2}$
so $ p|a_1 \Rightarrow p|m$

More than :
$ \frac{m}{n}\equal{}\frac{1}{1}\plus{}...\plus{}\frac{1}{p}$
Prove that $ p^4|m\minus{}np$

[quote="TTsphn"]More than :
$ \frac {m}{n} \equal{} \frac {1}{1} \plus{} ... \plus{} \frac {1}{p}$
Prove that $ p^4|m \minus{} np$[/quote]
You are wrong, because $ p\not |m$.
Exactly $ p^4|mp \minus{} n \equal{} p^2m_0$, were $ \frac {m_0}{n_0} \equal{} \sum_{i \equal{} 1}^{p \minus{} 1}\frac 1i  \ p>3.$


geometry
circumcircle
trigonometry
trig identities
Law of Cosines

The triangle $ABC$ has $CA = CB$. $P$ is a point on the circumcircle between $A$ and $B$ (and on the opposite side of the line $AB$ to $C$). $D$ is the foot of the perpendicular from $C$ to $PB$. Show that $PA + PB = 2 \cdot PD$.

Extend $PB$ to $A'$ such that $BA'=PA$
Then compare triangles $CAP$ and $CBA'$

In the contest I solved it by aid of trig!  :)

[hide]$CA = CB = x$

Since $APBC$ is cyclic we have $AB \cdot CP = AP \cdot CB + PB \cdot CA$ $\Rightarrow$ $AB \cdot CP = (PA + PB)x$.

$PA + PB = \frac{2AH \cdot CP}{x}$ where $H$ is the midpoint of $AB$.

Also $\angle BPC = \angle BAC$ and for costruction $\angle AHC = \angle PDC$.

Hence $\Delta AHC \sim \Delta PDC$.

So $\frac{AH}{x} = \frac{PD}{CP}$ $\Rightarrow$ $\frac{2AH \cdot CP}{x} = 2PD = PA +PB$.[/hide]

I note the middlepoint $M$ of the side $AB$.

$\widehat {CAB}\equiv \widehat {CPB}\equiv \widehat {CPD}\Longrightarrow \triangle CAM\sim\triangle CPD\Longrightarrow$

$\frac{CA}{AM}=\frac{CP}{PD}\Longrightarrow PC=\frac{2a}{c}\cdot PD$.

From the Ptolemeu's theorem results the relation

$PA\cdot BC+PB\cdot AC=PC\cdot AB\Longrightarrow a\cdot (PA+PB)=c\cdot PC\Longrightarrow$

$PA+PB=\frac ca\cdot PC=\frac ca\cdot \frac{2a}{c}\cdot PD\Longrightarrow PA+PB=2PD$.

Well, first note that we can reformulate the problem using an arithmetic trick, writing, then:

$PA+PB = PA+PD+DB=2PD \Leftrightarrow PA+DB = PD$

This is nice, because it leads us to another interpretation (which I think is geometric meaningfull). Geomtrically, use the compass on the point P and make a circle with radius $\overline{PA}$. Ok, name the point of intersectiojn between the circle and the segment $\overline{PB}$, M. Draw the line $PC$. Observe that, because of the property of the triangle and other properties on angles and circles, we garantee that $\angle{APC}=\angle{CPB}$ (actually, because they "look" to the same arc). It's wasy to see that our nice $PC$ is notheing less then the mediatrice of the segment $AM$. Now, draw $CM$ and observe the following equality: $CA=CM=CB$. Bingo! $CD$ is mediatrice of the segment $MB$ which completes the proof =]]]

Nice!
Let X be the foot of the altitude from C to AP. Then $ \triangle CAX$ and $ \triangle CBD$ are congruent since $ \angle CAX\equal{}\angle CBD$ and $ CA\equal{}CB$, and $ \triangle CPX$ is congruent to $ \triangle CPD$ since $ CX\equal{}CD$. So $ PX\equal{}PD$, $ AX\equal{}BD$, and it is obvious since $ PA\plus{}PB\equal{}PA\minus{}AX\plus{}PB\plus{}BD\equal{}2PD$

[hide="Fairly simple"]Since $APBC$ is cyclic, Ptolemy gives us $CA\cdot PB+CB\cdot PA=PC\cdot AB$ and since $CA=CB$, $ CA(PA+PB)=PC\cdot AB$. Thus $PA+PB=\frac{PC\cdot AB}{CA}$.

Since $\angle{CAB}$ and $\angle{BPC}$ are inscribed in the same arc, $\angle{CAB}=\angle{BPC}$. Then Law of Cosines gives us $CA^2+AB^2-2\cdot CA\cdot AB=CB^2=CA^2$, so $AB^2=2\cdot CA \cdot AB$ and $\cos{BPC}=\frac{AB}{2\cdot AC}$.

However, we have $PD=PC\cdot\cos{BPC}=PC\cdot\frac{AB}{2\cdot AC}=\frac{PC\cdot AB}{2\cdot AC}=2(PA+PB)$, so we are done.[/hide]

[color=darkred][b][u]An equivalent enunciation[/u].[/b] Let $ABC$ be a triangle with the circumcircle $w$ . Denote the midpoint $M$ of the side $[BC]$ and the diameter

$[NS]$ of $w$ so that line $BC$ separates $A$ and $S$ . Denote $X\in AB$ so that $SX\perp AB$ . Prove that $MX\perp AS$ and $AX=\frac {b+c}{2}$ .[/color]

[b][u]Proof 1[/u].[/b] Apply [u][b]Ptolemy[/b]'s theorem[/u] to $ABSC\ :\ (b+c)\cdot SC=a\cdot SA$ . Prove easily $\triangle AXS\sim\triangle CMS$ ,

 i.e. $\frac {AX}{CM}=\frac {AS}{SC}$ . Thus, $\frac {b+c}{a}=\frac {SA}{SC}=\frac 2a\cdot AX$ $\implies$ $AX=\frac {b+c}{2}$ . Since $MXBS$ is cyclically,

obtain that $\widehat{MXS}\equiv$ $\widehat{MBS}\equiv$ $\widehat{SAX}$ , i.e.in the $X$-right triangle $AXS$ we have $MX\perp AS$ .

[b][u]Proof 2[/u].[/b]  Denote $D\in BC\cap BC$ . Well-known relations $\left\{\begin{array}{c}
AD\cdot AS=bc\\\\
AD=\frac {2bc\cos\frac A2}{b+c}\end{array}\right|\implies$ $AX=AS\cdot \cos\frac A2=$  $\frac {bc\cdot\cos\frac A2}{AD}=$ $\frac {b+c}{2}$ .

Let $\begin{cases}\angle ACP=\theta\\ \angle BCP=\phi\end{cases}$.

Then $\begin{cases}PA = \sin\theta \dfrac{PC}{\sin(A+\phi)} \\ PB=\sin\phi \dfrac{PC}{\sin(A+\theta)}\\ PD=2PC\cos A\end{cases}$

To prove $\dfrac{\sin\theta}{\sin(A+\phi)}+\dfrac{\sin\phi}{\sin(A+\theta)}=2\cos A$, which is easy to prove given that $\theta+\phi = \pi-2A$.

Dear Mathlinkers,
this problem is immediat if we think to the Archimede broken chord theorem in the opposite version.
Sincerely
Jean-Louis

On $PB$ take $K$ such that $PK=PA$ now $ \angle APC= \angle KPC$, $PA=PK$ and & $PC=PC$ so triangles $APC$ and $KPC$ are congruent.  So $AC=DC=BC$ or $KD=DB$ so $PD=PK+KD=PA+BD $ adding $PD$ on both sides we get the answer.

Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/Regard%202.pdf  p. 21-22.

Sincerely
Jean-Louis

ptolemy and a pair of similar triangles solve it

[hide=Solution 1]
Extend $PB$ to $Q$ such that $BQ=PA$. Then $PQ=PB+BQ=PB+PA$, so we just have to prove that $D$ is the midpoint of $PQ$, or since $CD \perp PQ$ (and using Pythagoras’ Theorem), that $CP=CQ$.
[hide=Proof]
Consider $\triangle CAP, \triangle CBQ$. We will show that they are congruent.
$$CA=CB,AP=BQ$$ by definition and construction respectively.
Also $\angle CAP=\angle CBQ$ because $CAPB$ is cyclic ($BC$ separates $C,P$ avoids configuration issues here).
Thus the triangles are congruent by SAS and their corresponding sides $CP,CQ$ are equal.
[/hide]
$\blacksquare$
[/hide]
[hide=Solution 2]
Let us reformulate the problem to start with $P$ and construct $C$. It becomes:
In a $\triangle PAB$, let the midpoint of arc $AB$ of $\odot PAB$ not containing $P$ be $C$ and let $CD \perp PB,D \in PB$. Prove that $$PD=\frac{1}{2} \cdot (PA+PB)$$
[hide=Proof]
Assume $PA,PB$ are unequal. Drop $CE \perp PA,E \in PA$.
Since arcs $CA,CB$ are equal, $\angle APC=\angle CPB$. Thus $C$ is on the internal angle bisector of $\angle APB$ —(1),
so $CD=CE$—(2).
We will show that $PD=PE$ and $BD=AE$. Then [b]since[/b] $PA$ is not equal to $PB$, this implies $\{PA,PB\}=\{PD+BD,PD-BD\}$, so that $PD=\frac{1}{2} \cdot (PA+PB)$ as required.
Since $$\angle CDP=\angle CEP= \frac{\pi}{2}$$$$CP=CP$$$$\angle CPD=\angle CPE      (1)$$ $\triangle CPD,\triangle CPE$ are congruent, so $PD=PE$.
Since $$\angle CBD=\angle CAE$$(because $CAPB$ is cyclic, $AE=AP,BD=BP$ and both angles are almost $\frac{\pi}{2}$)$$\angle CDB=CEA=\frac{\pi}{2}$$$$CD=CE(2)$$$\triangle CBD,\triangle CAE$ are congruent, so $BD=AE$. Thus we are done.
[hide=If PAB is isosceles:]
Since arcs $CA=CB$ and $PA=PB$ of $\odot PAB$, so $C,P$ are diametrically opposite on the circle, so $D=B$. Thus $PD=PB=PA=\frac{1}{2} \cdot (PA+PB)$.
[/hide]
$\blacksquare$
[hide=Note]
This very construction actually features in a “spot-the-mistake” proof that every triangle is equilateral by glossing over the possibility that one of $PA,PB$ is the sum of $PD,DB$ and the other their difference.
[/hide]
[/hide]
[/hide]

Might be similar to others, but posting for storage :)

[hide=Solution][quote=JBMO 2002]$ABC$ is an isosceles triangle $(\overline{CA} = \overline{CB})$. Let $P$ be a point on the arc $\overarc{AB}$ on $(ABC)$ that doesn’t contain $C$. Let $D$ be the foot of the perpendicular from $C$ to $PB$. Show that $\overline{PA} + \overline{PB} = 2 \cdot \overline{PD}$.[/quote]

Apply Ptolemy Theorem to cyclic quadrilateral $ACBP$, we get :

$AB.PC=AC.BP+AP.BC$

$AB.PC=AC(PA+PB)$

$PA+PB=\frac{AB.PC}{AC}$

So, it suffices to show that :.

$AB.PC=2PD.AC$

Claim : $\cos\angle BAC=\frac{AB}{2AC}$

Drop a perpendicular $CD$ onto $AB$ hence :
$\cos\angle BAC=\frac{AD}{AC}\\ \implies \cos\angle BAC=\frac{AB}{2AC} \qquad \square$

Now, since $\angle CPD = \angle BAC$
Hence, $\cos\angle CPD=\frac{AB}{2AC}$.

However, $\cos\angle CPD = \frac{PD}{PC}$.
Thus, $\frac{AB}{2AC}= \frac{PD}{PC}\implies AB.PC=2PD.AC \qquad \blacksquare$


invariant
linear algebra
matrix
algebra
polynomial

Find $ m\in\mathbb{C}\ni M\equal{}[m_{ij}], m_{ij}\equal{}m\plus{}i\plus{}j \forall i,j \Rightarrow$ eigen value of M is an integer.

Do you want that
all the $ M$-eigenvalues are integers
 or
there exists at least one $ M$-eigenvalue that is an integer ?

Let $ M(m)\in M_n(\mathbb R)$ for $ n > 1$ with $ m\in\mathbb R$. Then $ M(m)$ is symmetric of rank 2: It is easy to see that $ x \equal{} (x_1,\ldots,x_n)^t\in \ker(M(m))\subset \mathbb R^n$ iff $ \sum_{k \equal{} 1}^nx_k \equal{} \sum_{k \equal{} 1}^nkx_k \equal{} 0$. In other words $ \ker(M(m)) \equal{} U^\perp$ where $ U$ is spanned by $ v \equal{} (1,\ldots,1)^t$ and $ w \equal{} (1,\ldots,n)^t$. Thus $ \dim\ker(M(m)) \equal{} n \minus{} 2\ge 0$ and $ rank(M(m)) \equal{} 2$. Since $ M(m)$ is symmetric, it leaves $ U$ as ortho complement of an eigen space of $ M(m)$ invariant ($ M(m)(U)\subset U$). The corresponding matrix of the induced linear mapping $ A: U\to U$ with respect to the basis $ (v,w)$ has the form
\[ \left(\begin{matrix}nm \plus{} \varphi_1(n) & n \\
m\varphi_1(n) \plus{} \varphi_2(n) & \varphi_1(n)\end{matrix}\right)\]
where $ \varphi_1(n) \equal{} \sum_{k \equal{} 1}^nk \equal{} \frac {1}{2}n(n \plus{} 1)$ and $ \varphi_2(n) \equal{} \sum_{k \equal{} 1}^nk^2 \equal{} \frac {1}{6}n(n \plus{} 1)(2n \plus{} 1)$.

$ A$ has the characteristic polynomial
\[ \chi_{A}(X) \equal{} X^2 \minus{} (nm \plus{} 2\varphi_1(n))X \plus{} \varphi_1(n)^2 \minus{} n\varphi_2(n) \equal{} (X \minus{} \alpha)^2 \minus{} \Delta\]
with $ \alpha \equal{} \frac {nm}{2} \plus{} \varphi_1(n)$ and $ \Delta \equal{} \left(\frac {nm}{2}\right)^2 \plus{} nm\varphi_1(n) \minus{} n\varphi_2(n)$.

Thus the eigen values of $ M(m)$ are 0 (with multiplicity $ n \minus{} 2$) and $ \alpha\pm\sqrt {\Delta}$.

The condition $ \alpha\pm\sqrt {\Delta}\in\mathbb Z$ is equivalent to $ \alpha \equal{} \frac {K}{2}$ (1) and $ 4\Delta \equal{} L^2$ (2) with $ K,L\in\mathbb N$. Then (1) gives
\[ nm \equal{} K \minus{} 2\varphi_1(n)\]
which plugged into (2) yields after some clearing
\[ K^2 \minus{} L^2 \equal{} K^2 \minus{} 4\Delta \equal{} 4\varphi_1(n)^2 \plus{} 4n\varphi_2(n) \equal{} \frac {1}{3}n^2(n \plus{} 1)(11n \plus{} 7)\in\mathbb N\,(*)\]
For every $ n > 1$ it can be checked that the RHS of (*) can be represented as a product of two even integers or two odd integers which is a criterion to be the difference of two integer squares. Altogether it is shown that there exist $ K,L\in\mathbb N$ such that (*) holds.

Note that $ m \equal{} \frac {K \minus{} 2\varphi_1(n)}{n}$ can be chosen rational such that the both non-zero eigen values of $ M(m)$ are integers.


function
combinatorics unsolved
combinatorics

Let $A=\{1,2,\cdots,n\}$.Find the number of functions $f: A\to A$ satysfying:
if $k,l\in A$ and $k\leq l$ , then $f(f(k))=f(k)\leq f(l)$ and each function achieve exactly $m$ values .(n,m are positive integers).

hello! (excuse me if my English is bad, because i'm French)I think that in the first line, k means n. The first condition shows that f=Id for exactly m values. The only condition on the n-m other values is the second condition. We should now examine how many increasing applications from n-m to m exist. I remember that it corresponds to n-m among n-1. Let multiply with how many m integers you can choose among n, and i think it's finished.
I usually make lots of mistakes, so, please, correct me, about my English too. :D
EDIT: I supposed that f is an application, because of the fact that we can consider every integer of A like an antecedent, am I wrong?

Hallo.You are right, in the first line schould be n instead of k(I have corrected it).I'm begginer,my English is not good.Could you explain me what do you mean here:"f=Id for exactly m values?"And what is application?

[quote]Hallo.[/quote] Bist du Deutsch? :D
"f=Id for exactly m values" means f(k)=k, and k can takes exactly m values. In fact, let consider a= f(k). Then, f(a)=a, and as f takes m values, this is true for m integers of A. An application is a function for which each antecedent gets a value. f(x)=1/x for x a real number isn't an application, because 0 gets no value. [/quote]


A book called The Penguin Dictionary of Curious and Interesting Numbers by David Wells (1986) contains fascinating numbers and their properties.  One of which is the only 3-digit numbers that are prime and all rearrangements of their digits are also prime are 113, 199, and 337.
Find all 2-digit numbers that have this property.
PLEASE HELP ME! :(

look in the back of your book.

theres no really good way of doing it except brute force..of course theres elimination methods like dont use any evens (so just use the 1s with 1,3,5,7,9).  Most likely if ab is prime, so is ba because of the rules..like a+b cannot be 0 mod 3..and so forth.


any one know who to prove m(a+b)=ma+mb, when a and b are vectors and m is equal or smaller than zero?


vector
invariant
abstract algebra
linear algebra

Let $ V$ be the vector space of sequences ($ a_n$) of Fibonacci numbers complex number .The shift operator $ S: V \to V$ is defined by  : $ S((a_1,a_2,a_3,...))\equal{}(a_2,a_3,a_4,...)$
1)Find a basis of $ V$ and find eigenvalues and eigenvectors of $ S$
2)Show that the subspace $ W$ consisting of sequences ($ x_n$) with $ x_{n\plus{}2}\equal{}x_{n\plus{}1}\plus{}x_n$ is two-dimensional , $ S$ invariant subspace of $ V$ and exhibit an explicit basis for $ W$

pls can you provide an accurate definition of the vector space $ V$, i.e. what are [quote]sequences ($ a_n$) of Fibonacci numbers complex number [/quote]

I think in this problem $ a_1 \equal{} 1,a_2 \equal{} 2,a_{n \plus{} 1} \equal{} a_n \plus{} a_{n \minus{} 1} ,\forall n\geq 2$

Looks to me like $ V$ should just be $ \{(a_1,a_2,\dots) : a_i \in \mathbb{C}\}$.

Then if $ S(a) \equal{} k a$, it better well be true that $ k \equal{} a_{n\plus{}1}/a_n$ for all $ n$, and so
$ a \equal{} (a_1, ka_1, k^2a_1, \dots)$.

$ W$ has the property that, if $ a \in W$, then $ (S^2 \minus{} S \minus{} 1) a \equal{} 0$; so $ W \equal{} \ker(S^2\minus{}S\minus{}1) \equal{} \ker\left((S\minus{}\phi)(S\plus{}1/\phi)\right)$,
which is the direct sum of the eigenspace belonging to $ \phi$ and that belonging to $ \minus{}1/\phi$, given above.

It's important to note that the operators $ S \minus{} \phi$ and $ S \minus{} \varphi$ commute; that's what makes computing the kernel particularly easy.


[hide]$ n \le k(k\minus{}1)$
implies
$ k\left(\frac{1}{k\minus{}1}\minus{}\frac{1}{n}\right )\le k\left(\frac{1}{k\minus{}1}\minus{}\frac{1}{k(k\minus{}1)}\right ) \equal{} 1$[/hide]


Most people say (or at least around here we say) "Red Sox and Yankees are having the best 4 out of 8" Why so? Are they being realistic?

really? well then, aren't they just messed up? 'cause you can get even split 4-4, and i think they mean whoever gets 4 first which would be best 4 out of 7...i'm confused

I know I know, but there's a mathematical reason behind it that I was impressed with so I was wondering if you guys would get it.

what? like the denominator is 2n-1. where n is how many games you have to win.

...never mind, here's the new question: What is the probability of having at least one tie in seven games of baseball if both teams were equally as strong?
What is the probability of having at least one tie in seven games of baseball if both teams have 50 % chance of winning?
Now do: What is the probability of having 8 games if the games end when one of the teams wins 4 times?

try your own question before you post it. i think you will find the first one very difficult.

... :-x

Hint: [hide]It's a trick question[/hide]

The answer to the first question: [hide]There is a high chance of having bad weather this season, the exact number you should look up.[/hide]

Sorry but I just help posting new questions under your baseball threads :P

You aim is to get the highest total. Would you rather choose: the sum of all your favourite teams number of runs each game this season, or the product?


Prove that $\forall a,b,c>0$we have
$\sum\frac{a^2-bc}{\sqrt{b^2+c^2+2a^2}}\ge 0$ $(1)$ :)
[u][b]NOTE[/b][/u]:I posted a similar problem:
$\sum\frac{a^2-bc}{b^2+c^2+2a^2}\ge 0\forall a,b,c>0$ :)  $(2)$
$argady$solved it.Now I introduce another solution:
$(2)\leftrightarrow\sum(a-b)^2.(c^2-ab)^2+(a^2+b^2+c^2).\sum(a-b)^4\ge 0$ :)
Finally,the following ineqs is true or false:
$\sum\frac{a^2-bc}{b^2+c^2+3a^2}\ge 0\newline\sum\frac{a^2-bc}{\sqrt{b^2+c^2+7a^2}}\ge 0$

[quote="nthd"]Prove that $\forall a,b,c>0$we have
$\sum\frac{a^2-bc}{\sqrt{b^2+c^2+2a^2}}\ge 0$ $(1)$ :)
[/quote]
Observe that
$\sum_{cycl}\frac{a^2-bc}{\sqrt{2a^2+b^2+c^2}}=\frac{1}{2}\sum_{cycl}\left(\sqrt{2a^2+b^2+c^2}-\frac{(b+c)^2}{\sqrt{2a^2+b^2+c^2}}\right)$
By Cauchy, $\sqrt{2(x^2+y^2)}\ge x+y$. Using this,
$\sum_{cycl}\sqrt{2a^2+b^2+c^2} \ge \frac{1}{\sqrt{2}}\sum_{cycl}(\sqrt{a^2+b^2}+\sqrt{a^2+c^2}) = \sqrt{2}\sum_{cycl}\sqrt{a^2+b^2}$
On the other hand, also by Cauchy
$\sum_{cycl}\frac{(b+c)^2}{\sqrt{2a^2+b^2+c^2}} \le \sqrt{2}\sum_{cycl}\frac{(b+c)^2}{\sqrt{a^2+b^2}+\sqrt{a^2+c^2}} \le \sqrt{2}\sum_{cycl}\left(\frac{b^2}{\sqrt{a^2+b^2}}+\frac{c^2}{\sqrt{a^2+c^2}}\right)=\sqrt{2}\sum_{cycl}\frac{a^2+b^2}{\sqrt{a^2+b^2}}=\sqrt{2}\sum_{cycl}\sqrt{a^2+b^2}$
Hence
$\sum_{cycl}\sqrt{2a^2+b^2+c^2} \ge \sqrt{2}\sum_{cycl}\sqrt{a^2+b^2} \ge \sum_{cycl}\frac{(b+c)^2}{\sqrt{2a^2+b^2+c^2}}$
from which the result follows.


geometry
ratio
rhombus
parallelogram
inradius
area of a triangle

I would gladly appreciate anyone answering the following questions: (Simple Quiz i suppose)

1. What is the formula to determine the area of a triangle?

2. Term for a parrallelogram with 4 equal sides?

3. How does one arrive at the value for pi?

4. What number must be added to 100 and to 20 so that the sums are in the ration of 3:1

5. What number does the binary number 11 represent?

------------------
Thank you, Kazoku

[quote="Kazoku"]I would gladly appreciate anyone answering the following questions: (Simple Quiz i suppose)

1. What is the formula to determine the area of a triangle?

2. Term for a parrallelogram with 4 equal sides?

3. How does one arrive at the value for pi?

4. What number must be added to 100 and to 20 so that the sums are in the ration of 3:1?

------------------
Thank you, Kazoku[/quote]
um.. ok then
[hide="1"]the area of the triangle is base times height over 2.[/hide]

[hide="2"]There is a square and a rhombus[/hide]

[hide="3"]There are many ways to arrive to the value of pi but im doing the way I learned the easiest. $\frac{22}{7}$[/hide]

[hide="4"]So its going to be $\frac{100+x}{20+x}=\frac{3}{1}$. Solving for x which is 20.[/hide]

Thank you! I added an extra one, if u dont mind answerin it =)

[hide="5"]ok, since binary represents base 2, we'll work with bases. I'm not sure if bases are allowed in this forum. Divide 11 by 2 and show the remainder in order starting from the first to the last. So we have 11/2=5 with r=1. Then 5/2=2, with r=1. 2/2=1 with no remainder. Finally 1/2=0 with remainer 1. Then list them from the beginning to the last to get the number. 1011 is the answer I think. [/hide]
I really think a mod should move this because bases aren't allowed in classroom math. Can a mod move this to hsb?

[quote="Kazoku"]Sorry =( Im new here... but shouldn't it be 1101 instead of 1011? Since it's in order[/quote]
sorry, I meant the other way.

[quote="lifesapain"]So you have to convert 11 in base 2 to some number in base 10. Wouldn't that be 3?[hide][/hide][/quote]
I think so too. Others should have misread the question!

[quote="vamathletes"][quote="ckck"][quote="Kazoku"]I would gladly appreciate anyone answering the following questions: (Simple Quiz i suppose)

1. What is the formula to determine the area of a triangle?

2. Term for a parrallelogram with 4 equal sides?

3. How does one arrive at the value for pi?

4. What number must be added to 100 and to 20 so that the sums are in the ration of 3:1?

------------------
Thank you, Kazoku[/quote]

[hide="3"]There are many ways to arrive to the value of pi but im doing the way I learned the easiest. $\frac{22}{7}$[/hide]

[/quote]
Pi does not = $\frac{22}{7}$
First 4 digits of Pi: $3.1415$
First 4 digits of $\frac{22}{7}$: $3.1428$[/quote]
Its an approximation value of pi.  :wink:

[quote="8parks11"]I remember seeing this somewhere but

$\pi=4(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}.....$

and this could be proven by taking the tangent of $\frac{\pi}{4}$[/quote]

yes, i do too. i think it was on someone's signature...

You get pi by dividing the circumference of a circle by its diameter.  That shouldn't be hard.  And jka386, delete your post.

Pi=the ratio of a circle's circumference to it's diameter.

5: 3, because 1*2+1*1=3

[quote="Temperal"][quote="ckck"]
...
[hide="2"]There is a square and a rhombus[/hide]
...
[/quote]
A square has to have 1 right angle (and therefore 4).
Rhombus is the better answer.[/quote]
What are you talking about? It never spedified anything about right angles. A square has 4 sides. And so does a rhombus. All it said was a parallelogram with 4 equal sides.

[quote="frostmourne"]well, a square is a type of rhombus, so if there's only supposed to be one answer, then "rhombus" is the better answer[/quote]

Exactly. Squares are too specific. It's better to just give the right answer.

This is another formula: the area=inradius*semiperimeter

proof for that:
[hide][b]draw this out; don't be lazy because the extra 2 seconds can help you understand the concept. [/b]first, draw a triangle (preferrably scalene, but ok if equilateral or isosceles). label the triangle ABC, and the sides a, b, and c, where side a is opposite angle A, and b is opposite B, etc.

Now that you've got that done, draw the incircle of the triangle (this is the circle that is inscribed in triangle ABC). First, we will define the center of the incircle to be the intersection of the perpendicular bisectors of the sides of the triangle. Next, we proceed by drawing segments from the center O of the incircle to the corners of the triangle. We get three triangles. as our next step, we can draw perpendicular segments to the sides from the center because as we defined earlier, the center is the intersection of the perpendicular bisectors. these perpendicular segments, you will find, are the radii of the incircle. so, you know that the area of a triangle is base times height divided by 2. we now can find the areas of the little circles, and they are $\frac{ar}{2}$ , $\frac{br}{2}$ , and $\frac{cr}{2}$ .  If we add these together, we get $\frac{a+b+c}{2}*r$. that is $rs$, our final result.

[b]Again, if you don't get this proof, either go through the proof in AoPS I, or read it again and draw it out.[/b]
 :)  :wink:  :ninja:[/hide]


induction
counting
distinguishability
combinatorics unsolved
combinatorics

For natural number $ n$ compute the sum

$ \dbinom{n}{0} \cdot \dbinom{3n}{0}\plus{}\dbinom{n}{1} \cdot \dbinom{3n}{3}\plus{} \cdots \plus{}\dbinom{n}{n} \cdot \dbinom{3n}{3n}$

hello, prove by induction that
$ \sum_{i \equal{} 0}^n\binom{n}{i}\cdot\binom{3n}{i} \equal{} \frac {4(4n \minus{} 1)!}{3n!(3n \minus{} 1)!}$
It's not correct, i have calculated the wrong sum.
Sonnhard.

This is the coefficient of $ x^{3n}$ in that product $ (1 \plus{} x^3)^n \cdot (1 \plus{} x)^{3n}$, and it's easy to give a combinatorial interpretation (how many ways to make $ 3n$ dollars with $ n$ distinguishable 3 dollar coins and $ 3n$ distinguishable single dollar coins, for example), but it's not immediately obvious to me how to get a formula from that.  And, perhaps surprisingly, the [url=http://www.research.att.com/~njas/sequences]OEIS[/url] doesn't have anything to say about it.  Also, we can rule out a simple product-based formula, since for $ n \equal{} 4$ the value is $ 7306 \equal{} 2\cdot 13\cdot 281$, if I haven't made any errors.


Phương trình $ f(x)\equal{}3x^3\minus{}14x^2\plus{}x\plus{}62$ có nghiệm là $ a,b,c$. Tìm giá trị của
$ \frac{1}{a\plus{}3}\plus{}\frac{1}{b\plus{}3}\plus{}\frac{1}{c\plus{}3}$

[quote="doabli"]Phương trình $ f(x) \equal{} 3x^3 \minus{} 14x^2 \plus{} x \plus{} 62$ có nghiệm là $ a,b,c$. Tìm giá trị của
$ \frac {1}{a \plus{} 3} \plus{} \frac {1}{b \plus{} 3} \plus{} \frac {1}{c \plus{} 3}$[/quote]

Use Vietes Law, I think that! ;)

$ \frac {9x^2 \minus{} 28x \plus{} 1}{3x^3 \minus{} 14x^2 \plus{} x \plus{} 62} \equal{} \frac {1}{x \minus{} a} \plus{} \frac {1}{x \minus{} b} \plus{} \frac {1}{x \minus{} c}$.

Plugging $ x \equal{} \minus{} 3$ gives the answer $ \frac {83}{74}$.


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

$ \frac {13a\minus{}1}{ a^2\plus{}23} \plus{} \frac {13b\minus{}1}{ b^2\plus{}23} \plus{} \frac {13c\minus{}1}{ c^2\plus{}23} \le \frac 3{2}$ ,

with equality for $ a\equal{}b\equal{}c\equal{}1$, and again for $ a\equal{}\minus{}11$ and $ b\equal{}c\equal{}7$, or any cyclic permutation.

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

$ \frac {13a \minus{} 1}{ a^2 \plus{} 23} \plus{} \frac {13b \minus{} 1}{ b^2 \plus{} 23} \plus{} \frac {13c \minus{} 1}{ c^2 \plus{} 23} \le \frac 3{2}$ ,

with equality for $ a \equal{} b \equal{} c \equal{} 1$, and again for $ a \equal{} \minus{} 11$ and $ b \equal{} c \equal{} 7$, or any cyclic permutation.[/quote]
My pqr and CYH still works here. :)

My solution:
We have $ \frac{13a\minus{}1}{a^2\plus{}23}\le \frac{1}{2}a\Leftrightarrow (a\plus{}2)(a\minus{}1)^2\ge 0$ is true
$ \Rightarrow \frac{13a\minus{}1}{a^2\plus{}23}\plus{}\frac{13b\minus{}1}{b^2\plus{}23}\plus{}\frac{13c\minus{}1}{c^2\plus{}23}\le \frac{1}{2}(a\plus{}b\plus{}c)\equal{}\frac{3}{2}$
Have done!

[quote="vo thanh van"]My solution:
We have $ \frac {13a \minus{} 1}{a^2 \plus{} 23}\le \frac {1}{2}a\Leftrightarrow (a \plus{} 2)(a \minus{} 1)^2\ge 0$ is true
Have done![/quote]
It is wrong because $ a;b;c \in\ R$ :wink:


what is it that when u take away the whole u still have some left over?

[quote="randomdragoon"]Whole wheat: If you take away the whole you get some crappy, non-nutritional white wheat. :P[/quote]

good idea but no   :rotfl:


Express $ k$ as a common fraction in terms of $ n$. where $ n$ is an integer greater than 1.\\
$ \frac{k(n\minus{}2)!}{(n\plus{}1)!} \equal{} \frac{(n\minus{}1)!}{(n\plus{}2)!}$

First clear the denominators by multiplying by $ (n\plus{}2)!$. Then it becomes $ k(n\minus{}2)!(n\plus{}2)\equal{}(n\minus{}1)!$. Dividing by $ (n\minus{}2)!$, we are left with $ k(n\plus{}2)\equal{}(n\minus{}1)\implies k\equal{}\boxed{\frac{n\minus{}1}{n\plus{}2}}$


Vidim je na mnogo mesta, a ne znam koja je to nejednakost prevedena na jezike Balkana.
Da li neko zna?
http://en.wikipedia.org/wiki/Rearrangement_inequality
http://www.artofproblemsolving.com/Wiki/index.php/Rearrangement_Inequality

Čini se da još nismo još smislili neki inteligentan naziv:

http://hrcak.srce.hr/file/45104

u gornjem fileu je i profesor Brnetić, u članku na hrvatskom jeziku naziva Rearrangement inequality...

Na hrvatskom jeziku se ta nejednakost zove "monotono preuređenje vektora"  :)

U skripti profesora Brnetića se i pojavljuje taj naziv (na dnu 8. stranice): "Nejednakost (1) možemo dokazati [b]monotonim preuređenjem vektora[/b] (“rearrangement inequality”)."

Čisto iz razloga potpune preciznosti (imali smo danas Analizu, primi se za čovjeka to inzistiranje na preciznosti i ne možeš pobjeći ni da hoćeš  :D ), pozornim čitanjem jednog dijela navedenog članka imam dojam da ipak "monotono preuređenje vektora" nije u potpunosti ekvivalentno s "rearrangement inequality". Naime, na vrhu treće (tj. sedme, zbunjuju me te dvostruke oznake :D) stranice piše, citiram:"[i]Koristit ćemo i “rearrangement inequality” (monotono preuređenje vektora i pripadna nejednakost), čiji se dokaz može naći u [4] ili [6].[/i]".

Stoga, ja bih rekao da je monotono preuređenje vektora, kako da kažem, sam proces tog permutiranja, čega već, dok je nejednakost ("rearrangement inequality") usko povezana s tim, ali je to nejednakost, a ne proces. Možda griješim i činjenica jest da se "monotono preuređenje vektora" i "rearrangement inequality" manje-više upotrebljavaju kao sinonimi, i, eto, znam da me sad svi mrzite koliko sam sitničav... :D

Vjerojatno ovim postom i nisam previše pridonio raspravi o tvom pitanju, hm, Bugi? A dobro... slobodno zanemari moj odgovor.  :)

Sjajan post, sad ja ne moram da pišem istovjetnu stvar (a namjeravao sam  :lol:  )

Moj skromni prijedlog: Nazovimo ovu nejednakost Permutacijskom (permutacionom) nejednakošću, što više-manje odgovara nazivima

eng. Rearrangement inequality
rus. Перестановочное неравенство
njem. Umordnungs-Ungleichung

ili na kineskom, 排序不等式  :lol:

Na češkom je, koliko vidim na Wikipediji, "permutační nerovnost", što bi odgovaralo permutacijskoj nejednakosti na našem jeziku.

Na http://srb.imomath.com/dodatne/Nejednakosti2_ml.pdf (jest da je skripta za dodatnu nastavu, ali, eto, još jedan prijedlog) autor, ako se ne varam, zove to isto "lema o permutacijama".

Još ću pogledati malo kad se vratim s fakulteta... u Matematičkom rječniku I. Gusića ne nalazim ništa o tome. Također, siguran sam da postoje referentniji izvori, na um mi padaju prof. Pečarić i prof. Brnetić, koji se bave nejednakostima.

Hvala svima na odgovorima.
Sad cu je zvati ''lema o permutacijama'' :D

[quote="mornik"]na um mi padaju prof. Pečarić i prof. Brnetić, koji se bave nejednakostima.[/quote]

E da, naš akademik Pečarić je jedan od najvećih svjetskih stručnjaka za nejednakosti. (I pritom gorljivi desničar, smeta li ti to, morniče? :D)

Voliš provocirati, ha, djuro?  :lol:

Slušaj, čovjek je vodeći stručnjak na području nejednakosti i to treba cijeniti. Dok god njegova politička stajališta koja, naravno, ni najmanje ne dijelim i ne mogu si zamisliti što bi se trebalo dogoditi u mom životu da ih počnem dijeliti, ne ulaze u priču s nejednakostima (a ne vidim kako bi mogle ući), to me ne može smetati ništa više nego takvo političko stajalište drugih, "običnih" ljudi. Matematika je lijepa upravo jer je objektivna i ne podliježe oku pojedinca.

Činjenica je da čovjek ima veoma izražene, hm, stavove. Stavove onakvog tipa da bih se s tobom temeljito posvađao i vjerojatno bih rekao puno "lijepih" riječi na račun tebe kad bi mi ti "uletio" s takvim pričama. No, uzmimo za primjer René Descartesa... s jedne strane, čovjek je jedan od otaca analitičke geometrije, s druge strane čovjek je dao ontološki "dokaz" Boga. Ili Lav Tolstoj, izvanredan pisac koji je poslije postao svojevrsni religijski guru vlastitog kulta. No, to u svakom slučaju ne umanjuje njihov rad na područjima kojim su se bavili.

Uglavnom, dok je točno (koliko god to bilo teško vjerovati  :D) da ima ljudi koji ne razmišljaju kao ti, ja ili hsiljak  :lol: i mi zadržavamo pravo da kritiziramo njihova razmišljanja, ne treba diskreditirati ničiji profesionalni rad, ako je objektivno točan i kvalitetan.

Dok čekam ispit iz Električnih krugova (brrrr), moram se uključiti u ovu diskusiju.

Iskreno, divim se profesoru Pečariću kao naučnom radniku i intelektualcu. Dakako, poznata mi je politička orijentacija istog, no to ipak ne umanjuje njegovu veličinu.

1. J.P. je elektroinženjer  :lol: pa odmah ima plus kod mene
2. J.P. je vodeći stručnjak za nejednakosti
3. J.P. je iznimno društveno angažiran. Istina, ne slažem se s većinom njegovih istupa, povijesnih publikacija i razmišljanja, no ipak nalazim da je za svaku pohvalu činjenica da takav stručnjak u matematici gorljivo pokušava u dosta stranoj oblasti za matematičara nađe svoje mjesto.

Samo se pitam kako jedan matematičar se može upustiti u prebrojavanje skupova čiji kardinalni broj ni izbliza ne poznaje (da, da, aludiram na to čudno štivo nazvano SMoJ, dotičnoga autora).

Što bi moj Stari rekao... "ne faleć' mu zakona, dobar je". Uostalom, kroz historiju Pečarić nije usamljen primjer (vidi pod Hasse, Helmut i slične individue)...

Odoh na ispit.


Consider the fumction $f: (-1,0)\bigcup(1,+\propto)\rightarrow\mathbb{R^*}$  such that $f(x)=-\frac{1}{x}.$ Find a  point $M(x,y)$  on the graph  of $f$  such $\frac{x^2+y^2}{x+y}$ is minimun.

I think you can just differentiate and turn the problem into solving an easy cubic.  On the interval that you specified, there are two points, namely (sqrt(3)+1)sqrt(2)/2 and -(sqrt(3)-1)sqrt(2)/2.

By the way, nickolas, I think you used the "proportional to" symbol, not infinity.

The extreme points are $U_{min}(a,b)$, and $V_{max}(b,a)$, where $a=\frac{\sqrt 2+\sqrt 6}{2},\ b=\frac{\sqrt 2-\sqrt6}{2}\ .$

i don't understand the domain of the function, how could a positive argument give you a positive value for $f(x)=-\frac{1}{x}$??

What do you mean?

Which is the positive argument and which is the positive value?

here is what i understand

$f(x)=-\frac{1}{x}$, with the domain $(-1,0)\cup(1,\infty)$ and the range of positive reals...or wait, does $R^*$ mean reals without 0? oh wait, that might be my problem...

[quote="pontios"]Yes, $\mathbb{R}^* = \{ x \in \mathbb{R} \ | \ x \neq 0 \}$

so, it's ok now?[/quote]

good problem! that trig substitution is clever...

[quote="Altheman"]good problem! that trig substitution is clever...[/quote]
Yeah....

[quote="pontios"]$x=2sin(-15^o)$ or  $x=2sin(75^o)$[/quote]

Actually, I didn't do this substitution $x=2\sin \theta$ to solve the problem :D

I just found the angles whose sine is $a=\frac{\sqrt 2+\sqrt 6}{2}$ and $b=\frac{\sqrt 2-\sqrt6}{2}$

But it would be a clever way anyway :)


Let ABC is a triangle wich is circumscribed (I).M is a point on AB, N is a point on AC.Prove that:
 $ I,M,N$ is collinear equiv to $ \frac{{\sin A}}{r} \equal{} \frac{1}{{AM}} \plus{} \frac{1}{{AN}}$ :)  :gathering:

Nice Problem!  :)

By theorem 1 of this topic: http://www.mathlinks.ro/viewtopic.php?p=570413#570413

We have that: $ \frac {1}{AM} \plus{} \frac {1}{AN} \equal{} \frac {AB \plus{} AC \plus{} BC}{AB \cdot AD}$.

Since $ AB \cdot AC \cdot \sin{A} \equal{} 2S \equal{} r(AB \plus{} AC \plus{} BC)$, so $ \frac {AB \plus{} AC \plus{} BC}{AB \cdot AD} \equal{} \frac {\sin{A}}{r}$.

[quote="SaYaT"][quote="shobber"]Nice Problem!  :)

By theorem 1 of this topic: http://www.mathlinks.ro/viewtopic.php?p=570413#570413

We have that: $ \frac {1}{AM} \plus{} \frac {1}{AN} \equal{} \frac {AB \plus{} AC \plus{} BC}{AB \plus{} AD}$.

Since $ AB \cdot AC \cdot \sin{A} \equal{} 2S \equal{} r(AB \plus{} AC \plus{} BC)$, so $ \frac {AB \plus{} AC \plus{} BC}{AB \plus{} AD} \equal{} \frac {\sin{A}}{r}$.[/quote]

I think you mean:
$ \frac {1}{AM} \plus{} \frac {1}{AN} \equal{} \frac {AB \plus{} AC \plus{} BC}{AB\cdot AC}$.  :?:[/quote]
Yeah, you are right it should be $ \frac {1}{AM} \plus{} \frac {1}{AN} \equal{} \frac {AB \plus{} AC \plus{} BC}{AB\cdot AC}$.Anyway, thanks Shober.I've also found that lemma some months ago by myself. And I think it is interesting because I found it after I solved another problem.Moreover, it is also the generalization of Menelaos. :wink: Thanks!

Thanks for pointing out the typo......I'm making a lot of mistakes in the Latex recently.   :blush:


a,b,c are positive reals and $a^2+b^2+c^2=1$
Prove or disprove
$a^4+b^4+c^4+4abc>1$

Does this help? \[a^4 + b^4 + c^4 + 4abc + 1 = \left(a^2 - b^2\right)^2 + 2\left(ab + c\right)^2 + \left(c^2 - 1\right)^2.\] Maybe you can proceed from here. Don't have time to think about it more right now.

Well it's false so it isn't strange you couldn't do it. Take $a = b$ and $c \rightarrow 0$.

I didn't understand you.I make this and I find $>1$

Sigh.

If $a = b = \frac{\sqrt{2}}{2}$ and $c = 0$ then $a^2 + b^2 + c^2 = 1$ and $a^4 + b^4 + c^4 + 4abc = \frac{1}{2}$.

Well, I know that $c$ is positive so $c$ cannot equal 0, so you just take $a = b$ and $c \rightarrow 0$. Do you get it?


inequalities
analytic geometry
number theory unsolved
number theory

Suppose that $n(r)$ denotes the number of points with integer coordinates on a circle of radius $r >1$.Prove that
                             $n(r)< 6\sqrt[3]{\pi.r^{2}}$

It is not true.
Easy can be proved $n(r)<\pi(r+1)^{2}<4\pi r^{2}\ (r>1)$.

In your post it does't seem that problem is false :D

it was posted before [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=circle+iran&t=56]here[/url] :wink:


inequalities
function
inequalities unsolved

Prove for all real numbers a,b,c,d such $ a^2\plus{}b^2\plus{}c^2\plus{}d^2\equal{}1$ iequality:

$ 8 \ge \frac{1}{1\minus{}ab}\plus{}\frac{1}{1\minus{}ac}\plus{}\frac{1}{1\minus{}ad}\plus{}\frac{1}{1\minus{}bc}\plus{}\frac{1}{1\minus{}bd}\plus{}\frac{1}{1\minus{}cd}$

[quote="MathUniverse"]Prove for all real numbers a,b,c,d such $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 1$ iequality:

$ 8 \ge \frac {1}{1 \minus{} ab} \plus{} \frac {1}{1 \minus{} ac} \plus{} \frac {1}{1 \minus{} ad} \plus{} \frac {1}{1 \minus{} bc} \plus{} \frac {1}{1 \minus{} bd} \plus{} \frac {1}{1 \minus{} cd}$[/quote]

Using Jensen's inequality on the convex function $ f(x)\equal{}\frac{1}{1\minus{}x}$ and the Mac-Laurin-Inequality, which gives us
$ \sqrt{\frac{\sum_{sym}ab}{6}}\leq\sqrt{\frac{a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2}{4}} \Leftrightarrow \frac{\sum_{sym}ab}{6}\leq\frac{1}{4}$,
we get
$ \sum_{sym} \frac {1}{1 \minus{} ab} \leq 6\cdot\frac{1}{1\minus{}\frac{\sum_{sym}ab}{6}} \leq 6\cdot \frac{1}{1\minus{}\frac{1}{4}}\equal{}8$.

 :)

[quote="Martin N."][quote="MathUniverse"]Prove for all real numbers a,b,c,d such $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 1$ iequality:

$ 8 \ge \frac {1}{1 \minus{} ab} \plus{} \frac {1}{1 \minus{} ac} \plus{} \frac {1}{1 \minus{} ad} \plus{} \frac {1}{1 \minus{} bc} \plus{} \frac {1}{1 \minus{} bd} \plus{} \frac {1}{1 \minus{} cd}$[/quote]

Using Jensen's inequality on the convex function $ f(x) \equal{} \frac {1}{1 \minus{} x}$ and the Mac-Laurin-Inequality, which gives us
$ \sqrt {\frac {\sum_{sym}ab}{6}}\leq\sqrt {\frac {a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2}{4}} \Leftrightarrow \frac {\sum_{sym}ab}{6}\leq\frac {1}{4}$,
we get
$ \sum_{sym} \frac {1}{1 \minus{} ab} \leq 6\cdot\frac {1}{1 \minus{} \frac {\sum_{sym}ab}{6}} \leq 6\cdot \frac {1}{1 \minus{} \frac {1}{4}} \equal{} 8$.

 :)[/quote]

"Mac-Laurin-Inequality"?
can you tell me more generaly of this inequality? ...

[quote="spanferkel"]see here:
http://www.artofproblemsolving.com/Wiki/index.php/Maclaurin%27s_Inequality
http://en.wikipedia.org/wiki/Maclaurin%27s_inequality[/quote]

thank you very much...
 :)

[quote="Martin N."][quote="MathUniverse"]Prove for all real numbers a,b,c,d such $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 1$ iequality:

$ 8 \ge \frac {1}{1 \minus{} ab} \plus{} \frac {1}{1 \minus{} ac} \plus{} \frac {1}{1 \minus{} ad} \plus{} \frac {1}{1 \minus{} bc} \plus{} \frac {1}{1 \minus{} bd} \plus{} \frac {1}{1 \minus{} cd}$[/quote]

Using Jensen's inequality on the convex function $ f(x) \equal{} \frac {1}{1 \minus{} x}$ and the Mac-Laurin-Inequality, which gives us
$ \sqrt {\frac {\sum_{sym}ab}{6}}\leq\sqrt {\frac {a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2}{4}} \Leftrightarrow \frac {\sum_{sym}ab}{6}\leq\frac {1}{4}$,
we get
$ \sum_{sym} \frac {1}{1 \minus{} ab} \leq 6\cdot\frac {1}{1 \minus{} \frac {\sum_{sym}ab}{6}} \leq 6\cdot \frac {1}{1 \minus{} \frac {1}{4}} \equal{} 8$.

 :)[/quote]

Now I see the mistake in your solution... Actually, I wasn't looking for mistake at the time you wrote this answer... :blush:

If $ f(x)$ is convex, than holds inequality: $ \frac {\sum_{i \equal{} 1}^{n}{f(x)_i} }{n} \geq f(\frac {\sum_{i \equal{} 1}^{n}{x_i}}{n})$ So this part is wrong:

$ \sum_{sym} \frac {1}{1 \minus{} ab} \leq 6\cdot\frac {1}{1 \minus{} \frac {\sum_{sym}ab}{6}}$

I'm very sorry for writing this now.

P.S. [b]Inequality is still unsolved[/b]... :)

Come on people, who can solve it, is it so hard?  :)

[quote="Martin N."]Thanks for pointing out my stupid mistake, MathUniverse...I think I considered the function as concave, I just saw $ f''(x) \equal{} \minus{} \frac {2}{(x \minus{} 1)^3}$ and was happy about the negative sign :wallbash: .

But however, your inequality was solved before [url=http://www.mathlinks.ro/viewtopic.php?t=109029]here[/url], #12:

[quote="zhaobin"][quote="pvthuan"]The trick also fails for 4 variable equivalent I proposed long time ago.
Let $ a,b,c,d$ be nonegative numbers such that
\[ a^{2} \plus{} b^{2} \plus{} c^{2} \plus{} d^{2} \equal{} 1,\]
prove that
\[ \frac1{1 \minus{} ab} \plus{} \frac1{1 \minus{} bc} \plus{} \frac1{1 \minus{} cd} \plus{} \frac1{1 \minus{} da} \plus{} \frac1{1 \minus{} bd} \plus{} \frac1{1 \minus{} ac}\leq8.\]
[/quote]
But we have:

$ \frac {ab}{1 \minus{} ab}\le \frac {2ab}{a^{2} \plus{} b^{2} \plus{} 2c^{2} \plus{} 2d^{2}}\le \frac {1}{2}(\frac {a^{2}}{a^{2} \plus{} c^{2} \plus{} d^{2}} \plus{} \frac {b^{2}}{b^{2} \plus{} c^{2} \plus{} d^{2}})$[/quote][/quote]

Nice solution. :)

This very nice method also works for the generalisation I proposed some time ago (and for which I thought I had a solution :blush: ):

[b]Problem[/b]
If $ x_i$, $ 1\leqslant i\leqslant n$ are $ n$ real numbers satisfying $ \sum x_i^2 \equal{} 1$, then
\[ \sum_{1\leqslant i < j\leqslant n}\frac {1}{1 \minus{} x_ix_j}\leqslant\frac {n^2}{2}\mbox{.}\]
[b]Proof[/b]
We use zhaobin's method to rewrite the inequality as
\[ \sum_{1\leqslant i < j\leqslant n}\frac {x_ix_j}{1 \minus{} x_ix_j}\leqslant \frac {n}{2}\mbox{.}\]
With
\[ \frac {x_ix_j}{1 \minus{} x_ix_j}\leqslant\frac {2x_ix_j}{(1 \minus{} x_i^2) \plus{} (1 \minus{} x_j^2)}\leqslant\frac {1}{2}\left(\frac {x_i^2}{1 \minus{} x_j^2} \plus{} \frac {x_j^2}{1 \minus{} x_i^2}\right)\mbox{,}\]
we have to prove
\[ \sum_{1\leqslant i < j\leqslant n}\frac {1}{2}\left(\frac {x_i^2}{1 \minus{} x_j^2} \plus{} \frac {x_j^2}{1 \minus{} x_i^2}\right)\leqslant\frac {n}{2}\mbox{.}\]
But we have equality here:
\[ \sum_{1\leqslant i < j\leqslant n}\left(\frac {x_i^2}{1 \minus{} x_j^2} \plus{} \frac {x_j^2}{1 \minus{} x_i^2}\right) \equal{} \sum_{i \not \equal{} j}\frac {x_i^2}{1 \minus{} x_j^2} \equal{} n\]
This completes the proof :) .


calculus
integration
calculus computations

please  can anyone tell me  what are the constants and what are the variables  inside the integral,  i am not able to get it.


limit
complex numbers
complex analysis
complex analysis unsolved

Suppose that $ \{a_k\}$ is a bounded sequence of complex numbers whose limit does not exist, and $ 0 < s < \frac{1}{2}$. Let

$ S_1(N) \equal{} \sum_{n\equal{}1}^N \frac{a_n}{n^{1 \plus{} s}}$

$ S_2(N) \equal{} \sum_{n\equal{}1}^N \frac{\overline{a_n}}{n^{2 \minus{} s}}$

Is it true that $ S_1$ and $ S_2$ must have different asymptotics as $ N \to \infty$?

Is there a reason to have complex conjugation in the second sum? It does not change the asymptotic, does it?

What do you mean by asymptotics? Both infinite sums converge.

Sorry, how about the exponents are $ s$ and $ 1 \minus{} s$...

I guess the conjugation doesn't matter -- it happens that in my particular problem those are the two sums we care about.

The sums may still be convergent, for example, if $ a_n\equal{}(\minus{}1)^n$. I guess your question should be stated as follows: is it possible that neither of the series is convergent, but $ \lim_{N\to\infty}\frac{S_1(N)}{S_2(N)}$ exists and is nonzero?