logarithms
ratio
limit
calculus
integration
calculus computations

Determine wether the following series are [b]absolutely convergent[/b]  or [b]conditionally convergent [/b]or [b]divergent[/b]:

$1.\sum^{\infty}_{n=2}(\ln n)^{p}$

$2.\sum^{\infty}_{n=1}n^{3}e^{-n}$

$3.\sum^{\infty}_{n=2}n^{(-1-\frac{1}{n})}$

$4.\sum^{\infty}_{n=2}\frac{1}{(\ln n)^{\ln n}}$

$5.\sum^{\infty}_{n=3}\frac{1}{n\ln n\left(\ln\ln n\right)^{p}}$

$6.\sum^{\infty}_{n=2}n^{p}\left(\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\right)$

$7.\sum^{\infty}_{n=1}\left(\sqrt[n]{n}-1\right)^{n}$

$8.\sum^{\infty}_{n=1}n^{p}\left(\sqrt{n+1}-2\sqrt{n}+\sqrt{n-1}\right)$

I would assume that for numbers 1, 5, 6, and 8, the question should actually be phrased as "for which real values of $p$ is this series convergent and for which values of $p$ is it divergent?"

Oh, and let's take "conditionally convergent" off the table as a possible response for any of these problems. Each of these is a series whose terms are eventually positive (except for #8 whose terms are all negative); such series are subject to the "monotone sequence alternative": they either converge or go to infinity. We have either absolute convergence or divergene in all of these cases.

Oh, and I assume that in #5 the sum starts with $n=3.$

[hide="Answers - sort of."] In three cases, converges. In two cases, diverges. In one case, converges iff $p>1.$ In two cases, converges iff $p<\frac12.$ Now all you have to do is figure out which is which, and what the reasons are.[/hide]

[hide="Some justifications."]
1. $p \geq 0$ - obviously diverges;
$p < 0$ - diverges by the Cauchy condensation test.

2. converges by the ratio test.

3. diverges by the limit comparison test: $\lim_{n \to \infty}\frac{n^{-1}}{n^{-1-\frac1{n}}}= \lim_{n \to \infty}\sqrt[n] n = 1$.

4. converges by the condensation + root tests (I hope so).

5. should converge iff $p > 1$ (I've excluded all the other options).

6. we have that $\lim_{n \to \infty}\frac{n^{p}\left( \frac1{\sqrt{n-1}}-\frac1{\sqrt n}\right)}{n^{p-\frac32}}\neq 0,$ so it converges iff $p-\frac32 <-1 \, \, \Longleftrightarrow \, \, p < \frac12$.

7. converges by the root test.

8. we have that $\lim_{n \to \infty}\frac{n^{p}\left( \sqrt{n+1}-2 \sqrt n+\sqrt{n-1}\right)}{n^{p-\frac32}}\neq 0,$ so it converges iff $p-\frac32 <-1 \, \, \Longleftrightarrow \, \, p < \frac12$.

PS: For 6 & 8 you can use Kent's favourite tool.
[/hide]
PPS: Post edited according to Kent's observation.

[hide="And a new justification."]
5. condensation test twice.
[/hide]

Yes, that's all correct, including #4 which you seemed to be wondering about. (Well, in #8 that limit should be negative. But it's the right idea.)

Here's a different justification for #4:

$(\ln n)^{\ln n}=e^{(\ln n)(\ln\ln n)}=n^{\ln\ln n}$

For all sufficiently large $n,$ $\ln\ln n\ge 2.$ Hence

$\sum_{n}\frac1{(\ln n)^{\ln n}}$ converges by comparison to $\sum_{n}\frac1{n^{2}}.$

thank you very much guys :) , i learned a lot :) ,beautiful solutions,
about number (1.) and (5.) perfect-radio, you said we use Cauchy condensation test,let me remind myself what it was:
If $\mu$ is a positive number bigger than $1$ then the two series $\sum a_{n}$ and $\sum\mu^{n}a_{\mu^{n}}$ both converge or both diverge.am I right?

for u=2 this theorem is proofed in Apostel Book vol(1).
i think This is true for all (u) Because u=2 has not any specifiec property in his proof.
See also:
Y.Tabesh and[....] Topics and Problems in Mathematical olympiads.Sharif university of technology.
or:
problems in Analyses &Calculus .by:P.P.Demidvich.Mir Publisher.1976.
                                                                          Be Succesful.

Of course, the first application of the condensation test we know of is Oresme's proof (in the 14'th century) that the harmonic series diverges. It's a classic argument, well worth remembering.

[quote="Perfect_radio"]5. condensation test twice.[/quote]

For #5, it could perhaps be more straighforward to use the Cauchy Integral Test to conclude that the series converges only for p > 1.


This is double posting! Should be deleted.

taking the log of both sides we get

$\log{\frac{x}{20}}= \log_{x}{50}\cdot\log{\frac{5}{2}= \frac{\log{50}}{\log{x}}\cdot\log{\frac{5}{2}}}$
$\log{x}\cdot(\log{x}-\log{20}) = \log{50}(\log{5}-\log{2})$

EDIT: Ya I had more stuff but it was all wrong so no point of it showing...because i kept mixing up $\log{x}^{2}$ with $(\log{x})^{2}$

I just took a look at your simplification.  It is incorrect. Also, $\log x\cdot\log x\neq 2\log x$.

o right... $\log{x}\cdot\log{x}= (\log{x})^{2}$..i mistakenly said that it was $\log{x}^{2}$...

$x=\frac{2}{5}$ is the other solution.
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=126085]Full Solution[/url]

[quote="lotrgreengrapes7926"]$x=\frac{2}{5}$ is the other solution.[/quote]

Oh yeah... I discounted the 'negative' too quickly. $\log{x}=-\log{\frac{5}{2}}= \log{\frac{2}{5}}$


Give an approximate but reasoned answer to the following question.

How much cold water could be brought to the boil by the energy dissipated in stopping a car travelling at 80km/hr

$K=Q$

$\frac{1}{2}M_{car}v^{2}=cm_{watter}(100 ^{\circ}C-T_{0})+m_{wv}c_{v}$

$m_{w}$-->mass of water
$c_{v}$-->heat of vaporization
$T_{0}$-->initial temperature of watwer
$m_{wv}$-->mass that was vaporized

:bye:


algebra
function
domain
system of equations

[hide="Answer, but no proof"]$(3,3,3)$ but I'm not sure how to prove that there is no other answer (are there any others?). Domain constraint: $x,y,z<6$ [/hide]

First note that $x , y , z < 6$.
Rewrite the equations as
$y=6-3^\frac{x}{\sqrt{x^2-2x+6}}$,
$z=6-3^\frac{y}{\sqrt{y^2-2y+6}}$,
$z=6-3^\frac{z}{\sqrt{z^2-2z+6}}$.
Let $g(t)=6-3^\frac{t}{\sqrt{t^2-2t+6}}$.
The system of equations can be written as
$y=g(x), z = g(y),x = g(z)$
Let $f(t)=\frac{t}{(t^2-2t+6)^{\frac12}}$
Then $f'(t)=\frac{6-t}{(t^2-2t+6)^{\frac32}}>0$ for all t < 6.
Therefore $g(t) = 6 - 3^{f(t)}$ is strictly drecreasing for t < 6.
We claim that $x = y = z$.
For otherwise, say, if $y < x$, then
 ${y<x}\Longrightarrow g(y)>g(x)\Longrightarrow z > y \Longrightarrow g(z) < g(y) \Longrightarrow x < z \\\Longrightarrow g(x) > g(z) \Longrightarrow y > x$
which is  a contradiction!
This proves the uniqueness of solution for the given system of equations.


linear algebra
matrix
superior algebra
superior algebra solved

What is an efficient way to tackle the following problem?

Find the order of the group $GL_3(Z/4Z)$.

I've tried to approach this using a counting technique similar to that for $GL_n(Z/pZ)$, but the fact that $Z/4Z$ is not a field makes it a little harder. I have a possible answer, but I'd be quite sure there is some mistake in my counting.

Thanks!

A matrix with entries $\mod p^n $ is invertible if and only if it is invertible modulo $p$.  So I think the answer is to count invertible matrices modulo $p$ and then lift to a matrix modulo $p^k$ in all possible ways ($p^{(k-1)n^2}$ possible liftings).

Thanks for the suggestion, [i]fleeting_guest[/i]. I'm not sure I understand the lifting idea. But considering $GL_3(Z/2Z)$, the order is $(2^3-1)(2^3-2)(2^3-2^2) = 168$, which I believe is a well-known result. Could you help me proceed further from here?

Thanks!

Lifting means finding a matrix modulo $p^k$ whose reduction mod $p$ is a given matrix.
This can be done in $p^{k-1}$ ways for each of the $n^2$ entries, so the result is

$|GL_n (Z/{p^k})| = p^{(k-1)n^2} | GL_n (Z/p)|$.

Thanks again! That really helps --- I did, in fact, know lifting but was thinking in a wrong way till I looked at it again. So let's deem this post closed, with the final answer as $2^9 \cdot 168$.

Appreciate your help once again!


I hope you guys had an enjoyable summer.  This year, it looks like I will be coaching Dalton's math team,  and I was thinking of signing them up for the New York City Interscholastic Math League (NYCIML).  Anyway, I was going through last year's problems from the Soph/Frosh and Junior divisions, and I had questions on the wording of some of them.  Feel free to respond to my comments and/or post your solutions to the problems.

[color=blue]1.  In a game of chess, two rooks are considered to be attacking each other if they were placed in the same column or the same row, but not both because each rook must occupy its own square.  Compute the probability that if two rooks were randomly placed on an 8 x 8 checkerboard, they would be attacking each other.[/color] [From Spring 2005, Soph/Frosh Division, Problem 2.]

When the problem says that the "two rooks were randomly placed", I wasn't clear whether the rooks could be at the same square.  It's true that the previous sentence said that "each rook must occupy its own square", but I thought that was merely defining the notion of "attacking"; that is, the sentence was merely saying that two rooks at the same square are, by definition, not attacking each other.

[color=blue]2.  Luis traveled from point New City to Oldtown.  Luis got a ride in a car for half of the trip.  The care went ten times as fast as Luis could have walked.  For the second half of the trip, Luis had to take the bus, and traveled only twice as fast as he could have walked.  Compute how many times as fast he completed the trip, than if he had walked the whole way.[/color]  [From Spring 2005, Soph/Frosh Division, Problem 13.]

When the second sentence says "half of the trip", I wasn't sure whether that meant "half of the distance" or "half of the time".  The solution given seems to assume "half of the distance".  Would either interpretation be considered correct?

[color=blue]3.  Compute the number of positive integers $\le 101$ that are multiples of perfect squares, but are not perfect squares themselves. [/color] [From Fall 2005, Junior Division, Problem 16.]

Because 1 is a perfect square, the question becomes quite easy.  The solution given seems to assume that 1 is not a perfect square.  Which interpretation is correct?  If a student assumed that 1 is a perfect square, would he or she be marked wrong?

[color=blue]4. Compute $x$: $
2^x =
  \frac{1}{4} \cdot \frac{2}{6} \cdot \frac{3}{8} \cdot \frac{4}{10}
  \cdot \ldots \cdot \frac{126}{154} \cdot \frac{127}{256}$.[/color]  [From Spring 2005, Junior Division, Problem 4.]

First, I think the "154" in the denominator should be "254".  Second, the answer given is [hide="Answer"]${-135}$[/hide] but I don't think that is correct.  Which answer was the accepted one?

[color=blue]5.  Compute the real number represented by $\sqrt[3]{20 + 14\sqrt{2}} \cdot \sqrt[3]{20 - 14\sqrt{2}}$.[/color]  [From Spring 2005, Junior Division, Problem 6.]

That's the way the problem is written in my copy from Mr. Linker, but the solution seems to assume that the multiplication sign $\cdot$ is actually an addition sign $+$.  Which way was the problem actually presented to students?

[color=blue]6.  Compute: $
  1 + \frac{1}{2} + \frac{2}{2} + \frac{1}{3} + \frac{2}{3} + \frac{3}{3}
    + \frac{1}{4} + \frac{2}{4} + \frac{4}{4} + \frac{4}{4}
    + \ldots \frac{20}{20}$. [/color] [From Spring 2005, Junior Division, Problem 8.]

The first appearance of $\frac{4}{4}$ should actually be $\frac{3}{4}$, right?  Which way was the problem presented to students?

[color=blue]7.  Kevin is twice as old as Jack was when he was as old as Jack is now.  If Kevin is now 24 years old, compute Jack's age. [/color] [From Spring 2005, Junior Division, Problem 9.]

In the first sentence, I find the word "he" to be ambiguous.  I think "he" was supposed to stand for "Kevin", but I don't think it is clear from the sentence structure.

i dont know about the frosh/soph division, but as for the junior problems:
3: not sure
4: it should be $254$, else the pattern doesnt exist...i don't know the official answer offhand
5. it is definitely $+$ instead of $\cdot$.
6. it should be $\frac 34$... if there's a pattern, there's a pattern
7. i hated the wording of this one...but it is as written

sorry this is so late, but thought it might still be of some help. (i haven't taken the soph/frosh division last year)
1. i believe that rooks can't have the same spot, each has its own square and attacks as real rooks do!

2. Half=half distance, you probably could ask mr liker for details on appeals i guess :)

3. well, 1 is a perfect square itself, and the question wants numbers that are non perfect square multiples of perfect squares ($cx^2$ where $c$ is square free... c can't be 1... or am i misunderstanding your question about this one?)

1. I think it is assumed that two rooks cannot be placed on the same square--thus, it should be OK.  Any person that plays chess even casually knows what is meant by the problem.

2. In general, "half" as meant in that problem should mean half the distance; otherwise, it would have been explicitly stated that the problem referred to half of the total travel time.

3. Given that the problem is...the fourth in a Junior division set, I think the writer just overlooked the fact that 1 is a perfect square.  Obviously, a shrewd contestant would know this and score an easy point, but given its position on the meet and the general difficulties of the questions in each position, I think the writer intended to neglect the fact that 1 is a square.

4. I don't see why you think the answer is incorrect.  I just did it myself in a couple of minutes and found that indeed, the answer you posted is correct.  I'm not good with the coding, so PM me and I can show you my quick solution...also, it's definitely a typo, because otherwise the terms wouldn't drop out properly.

5. I've seen problems similar to this from when I took these contests, and it should definitely be a +.

6. Definitely a typo.  The problem tests the patterns.

7. I thought that too, since "he" and Jack were used to refer to both people; thus, "he" must be Kevin.  It's probably better if the problem made that more clear, though.

For Question 4, I get an answer of ${-134}$.  I cancelled each numerator (after the first) with the preceding denominator.  That left me with 126 factors of 2 in the denominator, along with $256 = 2^8$.  So all together, I found the product to be $2^{-134}$.

Oh dear, you seem to be correct.  I redid the problem, and I seemed to assume the existence of an extra factor of 2 in the denominator.

The numerator turns out to be 127!.  The denominator is 128!, with a factor of two multiplied in to every term BUT the one.  When I initially did the problem, I mistakenly assumed that the one was included.  Thus, there are 127 factors of 2 included in the denominator.  Thus, the fraction turned out to be:

*forgive me, but I cannot figure out the LaTeX coding yet*

(127!)/(128! * 2^127)

The terms cancel out, and what is left is 2^-127 * 2^-7 = 2^-134.

Okay, it looks like we are in agreement.

By the way, you are quite close to using LaTeX.  All you have to do is put a dollar sign before each expression and a dollar sign after.  For example, you wrote:
[quote="Midnight"](127!)/(128! * 2^127)[/quote]
If you put dollar signs around that expression, you get $(127!)/(128! * 2^127)$.  The only thing wrong is the exponent 127.  We have to put it in braces {127} so that LaTeX knows the extent of the exponent.  We end up with $(127!)/(128! * 2^{127})$.  That already looks pretty good.

You can always use the Preview button to see if LaTeX rendered as you wanted.  Also, you can put your cursor over someone else's LaTeX to see the underlying code.


trigonometry function calculus integration

[quote="Geeimatree"]hey, does anyone know how to derive the general formula for finding the period of the function sin(ax/b)+cos(cx/d), where a, b, c, d are integers.[/quote]

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=160294[/url]

could you also explain to me why the period has to be an integral multiple of the periods of sinx and cosx

thanks


1. Let
\[ A_{m,n}=\bigg(\sum_{|k|\leq n}a^{k}\bigg)^{m}, \]
where $ a^{k}=a_{1}^{k_{1}}a_{2}^{k_{2}}a_{3}^{k_{3}}, a_{i}\in \mathbb{R}, k_{i}\in \mathbb{N}.$

$ A_{m,n}=?$

2. Let
\[ S_{m,n}=\sum_{m}\bigg(\sum_{|k|\leq n}a^{k}\bigg)^{j}, \]
where $ a^{k}=a_{1}^{k_{1}}a_{2}^{k_{2}}a_{3}^{k_{3}}, a_{i}\in \mathbb{R}, k_{i}\in \mathbb{N}.$

$ S_{m,n}=?$

When I do a problem, I see the second. However I think that we must do 1. first.

Please help me.

Thanks.

Dear Sir,

Any information about my difficulties?

[quote="Kuba"]$ |k|=k_{1}+k_{2}+k_{3}$ ?[/quote]

Ok. I'm looking forward to hearing from you and all. Thanks

[quote="livenju"][quote="NguyenThanh_545"]Rather hard.[/quote]

Let try[/quote]

How? Sir.


can you please post your questions in the right place? These questions should be placed in "Unsolved problems". I don't think they are open.

Also, this problem has already been discussed here ;)
(at least twice)

It's an easy one:)

We will prove that:

\[ a^b\geq \frac{a}{a+b} \]
\[ b^a\geq \frac{b}{a+b} \]

Using Bernoulli Inq we have:

\[ (1+\frac{b}{a})^{\frac{1}{b}}\geq 1+\frac{1}{a}>\frac{1}{a} \]
\[ 1+\frac{b}{a}>(\frac{1}{a})^b \]
\[ \frac{a}{a+b}\leq a^b \]

Doing the same we have $b^a\geq \frac{b}{a+b}$. Adding both inequalities we get

\[ a^b+b^a\geq 1 \]


algebra
polynomial
inequalities
algebra solved

Let $ b>0 $ and   $ n\; , (n\ge 5) ,$ be an  integer . Suppose that  $ f(x)=
\displaystyle x^n-10b x^{n-2}+\sum\limits_{k=3}^n a_kx^{n-k} .$  If $\; f(x_k)=0\;  ,\; 1\le k\le n,  $ where
\[ -\infty<x_n\le x_{n-1}\le ...\le x_1 < \infty \;  \; ,\; \; \;  x_5=-4 \sqrt{\frac{5b}{n(n-4)}}\; ,\]

[b]Question: [/b]   Find $\displaystyle a_3\; . $

Note that $x_5 < 0$, then for each $k \geq 5$ we have $x_k \leq x_5 < 0$ so that $x_k^2 \geq x_5^2$.
Let $\dsp S = \sum_{k=5}^n x_k$ and $\dsp T = \sum_{k=1}^n x_k^2$.

From above we have $S \leq (n-4)x_5 < 0$ and $S^2 \geq (n-4)^2x_5^2$.

Moreover, we have $\dsp \sum_{k=1}^n x_k = 0 $ and $\dsp \sum_{i<j} x_ix_j = -10b$.
It follows that $T = 20b$ (1) and
$\dsp \sum_{k=1}^4 x_k^2 = \sum_{k=1}^4 (x_k + \frac S 4 - \frac S 4)^2 = \frac {S^2} 4  + \sum_{k=1}^4 (x_k + \frac S 4)^2 - \frac S 2 \sum_{k=1}^4 (x_k + \frac S 4).$
But $\dsp \sum_{k=1}^4 (x_k + \frac S 4 ) = \sum_{k=1}^n x_k = 0$ then $\dsp \sum_{k=1}^4 x_k^2 = \frac {S^2} 4 + \sum_{k=1}^4 (x_k + \frac S 4)^2$.
We deduce that :
$\dsp T = \frac {S^2} 4 + \sum_{k=1}^4 (x_k + \frac S 4)^2 + \sum_{k=5}^n x_k^2 \geq \frac {n(n-4)} 4 x_5^2$, (2)
and equality occurs if and only if $x_5 = x_6 = \cdots = x_n$ and $\dsp x_1 = x_2 = x_3 = x_4 = - \frac S  .$

But, from (1) and (2), we are in the equality case.
Let $\dsp \alpha = -4 \sqrt { \frac {5b} {n(n-4)} }$ and $\dsp \beta = - \frac S 4 = - \frac {n-4} 4 \alpha$.
We then have :
$f(x) = (x - \alpha )^{n-4}(x- \beta)^4.$
Expanding, it leads to
\[ \dsp a_3 =- \frac  {20b(n-8)} 3 \sqrt { \frac {5b} {n(n-4)} }.\]

Pierre.


What are user groups for? Can I create a private one? (for that matter, a public one)

They are for easier administration and organisation. And no, you can't create one.


let $n\geq 2$ be an integer,and $x_i \in [0,1]$ for all $i=1,\dots ,n$ and $S=\sum_{i=1}^{n} x_i $ prove that :
$\frac 12(1+\frac {1}{S}) \leq \frac{\sum ix_i}{S^2}\leq  \frac 12(\frac {2n+1}{S}-1)$

it is hard for to solve it who has easier solution than I saw?

Either I am too idiot, or this problem is immediate by a convexity argument, meaning that it is easy to see that the extremal values of the expressions that appear there are attained when the variables are 0 or 1. But taking into account the magnificient level of difficulty of the problems you^post, samn, I am probably wrong.

maybe it is beacause of my weakness at inequality


Let $ a,\ b,\ m,\ n$ be postive numbers such that $ m\plus{}n\equal{}1$. Prove that $ \sqrt{\frac{m}{a}}\plus{}\sqrt{\frac{n}{b}}\geq\frac{m}{\sqrt{a}}\plus{}\frac{n}{\sqrt{b}}$.

[hide="Solution"]Write the inequality in the equivalent form $ (m\plus{}n)\left(\sqrt{\frac{m}{a}}\plus{}\sqrt{\frac{n}{b}}\right)^{2}\ge\left(\frac{m}{\sqrt{a}}\plus{}\frac{n}{\sqrt{b}}\right)^{2}$, which is $ 2(m\plus{}n)\sqrt{\frac{mn}{ab}}\plus{}\frac{mn}{a}\plus{}\frac{mn}{b}\ge\frac{2mn}{\sqrt{ab}}$, or equivalently $ 2(m\plus{}n)\sqrt{mnab}\plus{}mn(a\plus{}b)\ge 2mn\sqrt{ab}$. However, $ 2(m\plus{}n)\sqrt{mnab}\ge 4mn\sqrt{ab}$ and $ mn(a\plus{}b)\ge 2mn\sqrt{ab}$, so $ 2(m\plus{}n)\sqrt{mnab}\plus{}mn(a\plus{}b)\ge 4mn\sqrt{ab}\plus{}2mn\sqrt{ab}\equal{} 6mn\sqrt{ab}> 2mn\sqrt{ab}$, which was to be proved. There is no equality case.
[/hide]

[hide]
Since $ m,n$ positive with sum 1, $ m,n\in (0,1)$ so $ \sqrt{m}> m$ and $ \sqrt{n}> n$. Therefore
\[ \frac{\sqrt{m}}{\sqrt{a}}\plus{}\frac{\sqrt{n}}{\sqrt{b}}>\frac{m}{\sqrt{a}}\plus{}\frac{n}{\sqrt{b}}\]
[/hide]

This problem is not difficult for MathLinkers. Can someone solve this problem?

Sorry for that I have made a typo. The original problem was as follows.

Meiji University of Pharmacy entrnace exam 1986

[color=blue]Let $ a,\ b,\ m,\ n$ be postive numbers such that $ m\plus{}n \equal{} 1$. Prove that $ \sqrt{\frac{m}{a}\plus{}\frac{n}{b}}\geq\frac{m}{\sqrt{a}}\plus{}\frac{n}{\sqrt{b}}$.[/color]

We have to prove :
$ mb\plus{}an\geq m^{2}b\plus{}n^{2}a\plus{}2mn\sqrt{ab}$
Since $ 2\sqrt{ab}\leq a\plus{}b$,we have to prove
$ mb\plus{}an\geq m^{2}b\plus{}n^{2}a\plus{}mn(a\plus{}b)$
$ \Leftrightarrow (1\minus{}m\minus{}n)(mb\plus{}an)\geq 0$. :)

Now I have 3 solutions including your solution,stephen38 . :)

[quote="PhilAndrew"][quote="kunny"]Sorry for that I have made a typo. The original problem was as follows.
Meiji University of Pharmacy entrnace exam 1986
Let $ a,\ b,\ m,\ n$ be postive numbers such that $ m\plus{}n \equal{} 1$. Prove that $ \sqrt{\frac{m}{a}\plus{}\frac{n}{b}}\geq\frac{m}{\sqrt{a}}\plus{}\frac{n}{\sqrt{b}}$.[/quote]
Then, it's just a trivial application of Cauchy-Schwarz:
$ \left\{\begin{array}{c}(m\plus{}n)\left(\frac{m}{a}\plus{}\frac{n}{b}\right)\ge\left(\frac{m}{\sqrt{a}}\plus{}\frac{n}{\sqrt{b}}\right)^{2}\\ m\plus{}n \equal{} 1\end{array}\right|\Rightarrow\frac{m}{a}\plus{}\frac{n}{b}\ge\left(\frac{m}{\sqrt{a}}\plus{}\frac{n}{\sqrt{b}}\right)^{2}$.
Taking the square root of the both sides we obtain your claim. Equality holds iff $ a \equal{} b$.[/quote]

That's right. Here is more basic solution.

$ \left(\sqrt{\frac{m}{a}\plus{}\frac{n}{b}}\right)^{2}\minus{}\left(\frac{m}{\sqrt{a}}\plus{}\frac{n}{\sqrt{b}}\right)^{2}$

$ \equal{}\frac{m(1\minus{}m)}{a}\plus{}\frac{n(1\minus{}n)}{b}\minus{}\frac{2mn}{\sqrt{ab}}$

$ \equal{}mn\left(\frac{1}{\sqrt{a}}\minus{}\frac{1}{\sqrt{b}}\right)^{2}\geq 0$.


integration function calculus derivative algorithm Support logarithms

[quote="Kent Merryfield"]From one of my more recent posts:

[quote="Kent Merryfield"]Do you have some [i]a priori[/i] reason to believe that this antiderivative is an elementary function? Or did you just make it up as something you didn't immediately see a path for? Because if you just make up problems like that, then 99% of the time the answer is going to be "not an elementary function."[/quote][/quote]
Actually I do not know, but the main problem asks to to find $ \int_0^1 x^{e^x} \; dx$.
BTW, I have a question here, how do I know whether a function has antiderivative in an elementary function or not?

[quote="Moonmathpi496"]BTW, I have a question here, how do I know whether a function has antiderivative in an elementary function or not?[/quote]
See R. H. Risch (1969). "The Problem of Integration in Finite Terms". Transactions of the American Mathematical Society 139: 167–189 and R. H. Risch (1970). "The solution of the Problem of Integration in Finite Terms". Bulletin of the American Mathematical Society 76 (3): 605–608.

[quote="Moonmathpi496"]
...but the main problem asks to to find $ \int_0^1 x^{e^x} \; dx$.
BTW, I have a question here, how do I know whether a function has antiderivative in an elementary function or not?[/quote]

Some definite integrals can be evaluated without having to find the anti-derivative of the integrand, or even when there is no elementary anti-derivative. As an extreme example, you know that $ \int_{-1}^{1} x \, \,dx = 0$ without having to anti-differentiate. Or even better, you can show that $ \int_{-\infty}^{\infty} e^{-x^2} \, \,dx = \sqrt{\pi}$, even though $ e^{-x^2}$ has no elementary anti-derivative!

As for how do you know whether a function has an elementary anti-derivative? Well in theory there's a (fairly recently discovered) [url=http://en.wikipedia.org/wiki/Risch_algorithm]algorithm[/url] that will determine if a function has an elementary anti-derivative, but it's not practical to implement without a computer.

In practice, it mostly boils down to experience and having seen enough examples of integrals that are and are-not elementary to make an educated guess.

As for the integral that you posted, I'm fairly certain that neither the indefinite nor the definite integral have a closed elementary form. [url=http://www.wolframalpha.com/]Wolframalpha[/url] also supports that claim (though, I think there are some elementary integrals Wolframalpha cannot simplify).

[quote="Moonmathpi496"]
Actually I do not know, but the main problem asks to to find $ \int_0^1 x^{e^x} \; dx$.
BTW, I have a question here, how do I know whether a function has antiderivative in an elementary function or not?[/quote]

There is usually a big difference between the being able to calculate a definite integral vs being able to calculate the antiderivative. In this case, I am not sure even the definite integral would be easy to find without resorting to numerical methods.

Tarik, please provide more context to the problem. For instance, did you/a friend make this up, just for fun? What is the source of this?

[quote="Aryabhatta"]... resorting to numerical methods.[/quote]
The integrand isn't smooth at the $ x \equal{} 0$ end, which you can see by rewriting it as $ e^{e^x\ln x}.$ That gets in the way of the rate of convergence of something like Romberg integration. Going with Romberg anyway (with 65 function evaluations) I get $ 0.37367928.$ I'd take the last digit (or two) of that as uncertain. (I am quite sure that that's not $ e^{ \minus{} 1}.$)

Tried my ancient TI-85, which would use some version of Gaussian quadrature: $ 0.3736793618.$ That's probably a little better than the Romberg table.

[quote="kunny"]$ \int_0^1 x^{e^x}\ dx\geq \int_0^1 x^{x \plus{} 1}\ dx$[/quote]

Wouldn't that be the other way round?

Kunny, I have no idea what you're trying to say, but $ 0 < x \plus{} 1 \leq e^x$ implies $ x^{x \plus{} 1} \geq x^{e^x}$ for all $ x \in [0, 1]$.  Aryabhatta is exactly right.

Kent, Mathematica thinks it's 0.3736793582700676... (agreeing with your TI to 7 digits), and I don't know anything about what sort of methods it uses for numerical integration.

The integral can't be solved by calculation unless nemerical method or Maclaurin expansion or by the method by Diogene etc.

I just made a estimate by inequality as one example. The problem is very vague.


MATHCOUNTS calculus integration inequalities ratio probability geometry

I was working on the Mathcounts State problems, and for the Team Round 6~10, I got the right answers but it took me way too much time. Would somebody please tell me how you got the answer and the answer?

Questions:

6. For how many positive integral values of [i]a [/i]is it true that [i]x=2[/i] is the only positive integer solution of the system of inequalities
[i]2x>3x-3
3x-a>-6[/i]?

7. One night two cylindrical wax candles of different heights and different diameters were lit. One of the candles was 20cm taller than the other. They were both lit at the same time and each burned at a steady rate. Five hours after they were lit they were both the same height. The taller one burned all of its wax six hours after it was lit, and the shorter one burned all of its wax 10 hours after it was lit. What was the ratio of the original height of the shorter candle to the original height of the taller candle?

8. Each of three, standard, six-sided dice is rolled once. What is the probability that the three numbers rolled are the lenghts of the sides of an obtuse triangle? Express your answer as a common fraction.

9. In the figure, [i]m<A=28 [/i]degrees, [i]m<B=74[/i]degrees, and [i]m<C=26 [/i]degrees. If [i]x[/i] and [i]y[/i] are the measures of the angles in which they are shown, what is the value of [i]x+y[/i]?
*Attatchment included to view figure

10. A circular tabletop is divided into four congruent sectors by two diameters that are perpendicular with one of the four colors. How many distinct ways can the table be painted? (A color may be used on more than one sector, but paintings that are the same after a rotation are not considered distinct.)

I got all the answers, but they all took a lot of time! Please help! :lol:

[hide="6"]$ x < 3$ from the first inequality.

The second implies that $ x > \frac {a \minus{} 6}{3}$
$ 1 < \frac {a \minus{} 6}{3}\le2$

Since if it is greater than 1, we only have one possible value, and it must also be two or less.

Thus $ 3 < a \minus{} 6\le6$
$ 9 < a\le12$

Thus there are 3 values, $ 1,2$ and 3.[/hide]
[hide="7"]
Let the shorter candle have height $ h_s$ and radius $ r_s$.

The taller candles has height $ h_t$ and radius $ r_t$

We are given that $ h_t\equal{}h_s\plus{}20$

If five hours have passed, the taller one has $ \frac{1}{6}$ of its height left, while the shorter one has $ \frac{1}{2}$.

Thus, $ \frac{1}{6}h_t\equal{}\frac{1}{2}h_s\Longrightarrow \frac{h_s}{h_t}\equal{}\frac{1}{3}$.[/hide]

[hide="8"]If the triangle is obtuse, $ a^2\plus{}b^c<c^2$.

If $ c\equal{}1$, no solution.
$ c\equal{}2$, no solution because if $ a,b\equal{}1$, $ 1\plus{}1\not>2$
$ c\equal{}3$, $ a\equal{}2,b\equal{}2\longrightarrow$ gives us $ 3$ solutions.
$ c\equal{}4$, $ a\equal{}2,b\equal{}3\longrightarrow$ gives us $ 6$ solutions.
$ c\equal{}5$, $ a\equal{}3,b\equal{}3$ or $ a\equal{}2,b\equal{}4$.  This gives a total of $ 9$ solutions.
$ c\equal{}6$, $ a\equal{}4,b\equal{}4$ or $ a\equal{}2,b\equal{}5$ or $ a\equal{}3,b\equal{}4$ or $ a\equal{}3,b\equal{}5$.  This gives a total of $ 21$ solutions.

Thus there are a total of $ 39$ out of 216, giving us a probability of $ \frac{13}{72}$.[/hide]

[hide="9"]If $ \angle C\equal{}26^{\circ}$, the other angle within the triangle must have degree measure of $ 64^{\circ}$.
The vertical angle opposite also has degree measure $ 64^{\circ}$.
Let the marked angle within that triangle be $ x$.  Then the last angle in that triangle must be $ 116\minus{}x$ degrees.
If we draw the perpendicular from the other marked angle to the horizontal line below, we can see that the last angle within the quadrilateral containing point $ A$ must be $ 180\minus{}A\equal{}152^{\circ}$.

Going back to the other marked angle, $ x$, we can see that the angle opposite $ (116\minus{}x)^{\circ}$ must also have that measure.  Since $ \angle B\equal{}74^{\circ}$, we can find the last angle in the quadrilateral containing point $ B$.  This is $ 80\plus{}x$ degrees.
Since $ y\plus{}x\plus{}80\plus{}152\equal{}360$, $ x\plus{}y\equal{}128^{\circ}$[/hide]

#10 involves alot of cases and is hard to count.  It doesn't matter how you get it, just get it =P.

Edit: Added 7,8,9.

Yes, only because Xiaoyu was on your team... :P

#10 resembles a problem from the 2006 National Sprint round (Problem 24?), if I'm interpreting it correctly. The casework isn't too bad.


Let $ x,y,z > 0$ Prove that
\[ \sum_{cyc}\frac{x^2\plus{}1}{(x\plus{}y)(x\plus{}z)} \ge \frac{(\sqrt{x}\plus{}\sqrt{y}\plus{}\sqrt{z})^2}{2(x^2\plus{}y^2\plus{}z^2)}\]

by AM-GM we have that
$ \sum_{cyc}\frac{x^2\plus{}1}{(x\plus{}y)(x\plus{}z)}\geq \sum_{cyc}\frac{2x}{(x\plus{}y)(x\plus{}z)}$
thus we need to prove that
$ 4(x^2\plus{}y^2\plus{}z^2)\sum_{cyc}\frac{x}{(x\plus{}y)(y\plus{}z)}\geq (\sqrt{x}\plus{}\sqrt{y}\plus{}\sqrt{z})^2$
notice that $ 4(x^2\plus{}y^2\plus{}z^2)\geq\sum_{cyc} (x\plus{}y)(x\plus{}z)$ and Cauchy finish the proof :)

[quote="apollo"]Let $ x,y,z > 0$ Prove that
\[ \sum_{cyc}\frac {x^2 \plus{} 1}{(x \plus{} y)(x \plus{} z)} \ge \frac {(\sqrt {x} \plus{} \sqrt {y} \plus{} \sqrt {z})^2}{2(x^2 \plus{} y^2 \plus{} z^2)}
\]
[/quote]
By AM-GM and Cauchy-Schwarz inequalities, we have:
$ \sum_{cyc}\frac {x^2 \plus{} 1}{(x \plus{} y)(x \plus{} z)} \ge \sum \frac{2x}{(x\plus{}y)(x\plus{}z)} \ge \sum \frac{2(\sqrt{x}\plus{}\sqrt{y}\plus{}\sqrt{z})^2}{\sum (x\plus{}y)(x\plus{}z)}$
So we need to prove that:
$ \frac{2}{\sum (x\plus{}y)(x\plus{}z)} \ge \frac{1}{2(x^2\plus{}y^2\plus{}z^2)}$
$ \Leftrightarrow (x\minus{}y)^2\plus{}(y\minus{}z)^2\plus{}(z\minus{}x)^2 \ge 0$
Which is clearly true!

isn't that exactly what I wrote?  :P  :)

[quote="Albanian Eagle"]isn't that exactly what I wrote?  :P  :)[/quote]
Sorry! We have the same solution, but I rap at computer more slow than you :( !


'Prof Pi, a mathematician in love with prime numbers named his six children after the first six prime numbers as two, three, five, seven, eleven and thirteen. He had a good collection of gold coins which he divided among his six children. Each one of them got less than $ 1000$ coins. Two got least, three got one coin more than two, five got one more than three, seven got one more than five, eleven got one more than seven. Thirteen was the luckiest. He got the most, one more than eleven. Interestingly, two's share was divisible by $ 2$, three's share by $ 3$, five's share by $ 5$, seven's share by $ 7$, eleven's share by $ 11$ and thirteen's share by $ 13$.

How many gold coins did Prof Pi have in total ?

Let Two be named A, Three be named B and so on till we have
A, B, C, D, E, F

$ B \equal{} A\plus{}1$
$ C \equal{} A\plus{}2$
$ D \equal{} A\plus{}3$
$ E \equal{} A\plus{}4$
$ F \equal{} A\plus{}5$

$ A\equiv 0\mod 2$
$ A\plus{}1\equiv 0\mod 3$
$ A\plus{}2\equiv 0\mod 5$
$ A\plus{}3\equiv 0\mod 7$
$ A\plus{}4\equiv 0\mod 11$
$ A\plus{}5\equiv 0\mod 13$

Smack with Chinese Remainder Theorem.
:-)

[hide="solution"]
Let two receive $ x$ coins. Consider $ x$ in mod 1001, working with the last three moduli.
Then $ x\equiv4\mod7$, $ x\equiv7\mod11$, $ x\equiv8\mod13$
Working with the first two of these moduli, we have $ 4\plus{}7(0)\equiv4\mod11$, $ 4\plus{}7(1)\equiv0\mod11$, $ 4\plus{}7(2)\equiv7\mod11$.

Thus $ x\equiv4\plus{}7(2)\equiv18\mod77$

We know $ 77\equiv\minus{}1\mod13$, so $ 18\plus{}77k\equiv18\minus{}k\mod13$
Since $ x\equiv8\mod13$, we have $ k\equal{}10$.

Thus $ x\equal{}18\plus{}77(10)\equal{}788$. We can see that $ x\equal{}788$ also satisfies the necessary congruences in mods 2, 3, and 5.

Now the total number of coins is $ 6x\plus{}1\plus{}2\plus{}3\plus{}4\plus{}5\equal{}6(788)\plus{}15\equal{}\fbox{4743}$.[/hide]


$ f(x) \equal{} \sum_{i \equal{} 1}^N|X_i \minus{} x|^\beta \cdot \sum_{i \equal{} 1}^N|X_i \minus{} x|^{2\beta}$ where $ N$ is finite, $ X_i$ are given and $ \beta\geq 1$.

It looks like $ f(x)$ is convex (at least all plots suggested that!!), is there a way to prove it.

Also, is
$ g(x) \equal{} \frac {\sum_{i \equal{} 1}^N|X_i \minus{} x|^\beta \cdot \sum_{i \equal{} 1}^N|X_i \minus{} x|^{2\beta}}{1 \minus{} B\sum_{i \equal{} 1}^N|X_i \minus{} x|^\beta}$ convex?. Here $ B$ is positive constant such that denominator is strictly positive in the region of interest $ [X_1,  X_N]$.

Please give some comments on this problem.


Given that $ \triangle\plus{}q\equal{}59$ and $ (\triangle\plus{}q)\plus{}q\equal{}106$, what is the value of $ \triangle$?

Subtract the second equation from the first to get $ q \equal{} 106 \minus{} 59 \equal{} 47$. That means $ \triangle \equal{} 59 \minus{} 47 \equal{} \boxed{12}$.

[b]Another answer[/b]
$ \triangle \equal{} ( \triangle\plus{}q) \minus{} [\{(\triangle\plus{}q)\plus{}q\} \minus{} (\triangle\plus{}q)] \equal{} 59\minus{}(106\minus{}59) \equal{} 118\minus{}106\equal{}\boxed{12}$.


ARML AMC USA(J)MO USAMO MATHCOUNTS college email

This posting is for 2007 WA ARML Team Members.  I am helping Mr. Norris to put together some team practices.  By now you should have booked your flight tickets and sent your registration fee to Mr. Norris.

As mentioned in ARML Coach Manual: "Practices are a great way for the team members to get to know each other", ""...team practice leads to dramatic improvements on ARML exams, as three of the four ARML rounds involve teamwork.  Practicing will definitely pay off, in more ways than one".  With USAMO completed this week, hope team members are ready to meet each other and have some team practices.   Our first practice is set at Sunday May 6th 4pm at Redmond Library (check http://www.kcls.org if you need map / direction).  Please post here weekends that you have conflicts in rest of May (e.g. piggyinc has May 12 conflict with MATHCOUNTS National, many team members have May 19 BATH conflict, etc.), this will help to plan for rest of practices.  Sorry that the location may not work for remote team members like austinb.   We may have a guest lecture (by Chris J) but it is not confirmed yet.  As it is said in the coach manual "if you are eager to lecture, you're out of luck here.  Your students will be too busy working together on old ARML exams, forcing you to relax and watch them teach each other".   Both Chris and Tom mentioned that veteran ARML competitors (kamior and paladin8 especially) can organize the practices.  I will bring 2006 tests and some practice problems used by other ARML teams, it would be of great help if you have other ARML old tests to share (please send to me ahead of time so that I can make copies for everyone).  We also have parents who offered help (samath and piggyinc's parents) so we will bring pizza / food - please bring a few dollars.

Congratulations to Western Washington team members on winning 2006-2007 ARML Power Contest: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=140670.

See you on May 6th 4pm, and we will decide if we can meet again May 20th (Sunday) and during Memorial Day weekend.

If you absolutely can't make to 4pm on May 6th, we may consider a different time of the same day (at Microsoft).

May 6th is fine.  I have plans on the 12th and 19th.

im available 6th and 12th. But like you said on the 19th I will participate in BATH

I can probably go on the 6th and the 12th
Just curious, how long will it be?
Will we do stuff from both days of ARML?

We have the room booked from 4pm to 8pm, but if you need to leave early, feel free to do so after picking up the handouts / problems.  We have a complete set to work on (yes, Friday's Team Round and Power Question, as well as Saturday's Individual, Relay, and Tiebreaker).  With breaks and dinner (pizza most likely) we are likely to use all 4 hours.  I will bring 2006 set.  If you have not worked on it, please leave it for Sunday.  If you already finished this set, we will work on other similar problems, focusing on Team and Relay that can benefit most from team practices.

With SAT and AP tests going on, I understand this is a busy time.  There will be a sign up sheet for you to write down your next available practice time.   I just booked Bellevue Library room #3 for Saturday May 12 1-5pm as it looks like it is a pupular choice.   I hope May 20 remains open if you still have energy left after May 19 BHS BATH.

Bring your calculator, some scratch paper, a few dollars for pizza and/or snacks / treats that you would like to share.

We are in Conf Room #1.

I'm not going to be available on the 19th due to the BATH as well, but I can't be sure about some of the weekends later in May.  But I will be at the practice on Sunday.  Most of you probably don't know me because I'm new to the team this year.  In any case my name is Berkeley and I go to Bothell High.

One more update for Sunday May 6th:  I have received confirmation from the following (I am sure I missed a few from whom I did not have direct confirmation):

Adam, Andrew S., Berkeley, Joy, Lawrence, Patrick, Qiaochu, Sam, Tracey, Young.

Adam will coach the session.  Thanks Adam!

Also special thank goes to xxreddevilzxx for providing practice materials / previous test sets.

See you on Sunday!

If you have those handouts (lectures/problems) typed on the computer, maybe you could have Norris email them to everyone so that people who can't attend a practice can still see some of the work?  I would assume our teammates from East WA would benefit

That's a good idea.  I will do that (send files to Tom), probably tomorrow when I get my hands to it.

We had a great practice session today, with most of Western Washington team members participating.   We used a new set (no one worked on it before) so it is a good baseline, and the team did well (Individual, Team, Tie-Breaker, Relay, and won candy/chocolate in Super Relay).  We did not get time to complete Power Question from this set, which will be left to next week.

Our next practice is set at Saturday 5/12 1-5pm at Bellevue Library room #3.  With no dinner break we might be able to do more.  I will bring water and chocolate trouffle to practice.  From today's participants, I have Griffin, Berkeley (maybe), Adam, Young,  Meir, Qui, Andrew S. (maybe), Tracy, and Lawrence signed up.   If you plan to come and your name is not in the list, please post a confirmation here, or drop a quick note to my email (address is included in Mr. Norris' email to you with the handouts and updated roster) - this will help to  have enough copies of tests prepapred.  Some requested Sat evening, however for next week it does not work (Library closes on Saturday evenings, and parents who have access to Microsoft are not available next week).

See you next week!

At yesterday's practice, some students mentioned that instead of having next practice at May 20th Sunday, perhaps we can use the time after Bellevue BATH on May 19th Saturday.  If you have preference one way or another (Sat 5/19 5pm - 8:30pm or Sun 5/20 1-5pm), please respond (message to me, email to Tom or myself, or post here).  Thanks!

Yesterday we have 3 new students joining.  With new grouping, Relay and Super Relay rounds seemed to be struggle, although other sections' results were consistent with previous week.  We will try to do more relays in next two practices (5/19 - 5/20 weekend and the Memorial weekend).

I'd rather have the practice on Sunday... but I can see how it would be more convenient for people who don't live in the area to do both at the same time.

Saturday option is mostly for students traveling (e.g. from TJ, Garfield) as it will save a trip.  However, Jeffrey did mention that 5pm start sounds good except BHS students might be a little late (looks like he has tasked everyone to help after the BATH).  I also need to bring enough chocolate to keep the energy level up.  I am going to ask Tom to survey so that everyone can indicate preference if there is one.

We are set for Sunday 5/20 1pm - 5pm.  The location is Microsoft building 117:

http://www.microsoft.com/usability/directions117.mspx

See you there!

We are having next practice on Saturday 5/26 afternoon 2pm - 6pm at Microsoft building 117:

http://www.microsoft.com/usability/directions117.mspx

This is the last weekend and last practice before ARML competition.  Hope to see you all there.

Team T-shirt is finally ordered, many thanks to Richard's mom who pulled this together in a very tight schedule.  Expect another email from Mr. Norris this weekend about the cost and where to send your check.

While the T-Shirt order is in, the printing is going to be done next Wed.  We have up to next Tuesday to submit a design.    Our order is for one side printing (either front or back) and 1 color (outlined box OK), to save cost.  Sorry we did not give you chance to pick color, but you can still win design contest.  Please contribute your creative and artistic talent to send in proposed designs, logo, suggestions, and ideas.  Bring them to tomorrow's practice or email to us.  The proposed text is:

American Regions Math League
          ARML 2007
   TEAM WASHINGTON

Have a great weekend.  Goooooooo Washington!

We finished last team practice session yesterday with 15 students participating.  Thank you all for dedicating half day from your busy Memorial Day weekend schedule to prepare for the competition.  Each student picked up the additional practice problems yesterday as homework.  The remaing printed problem sets will be carried to LAS by Ying (Qiaochu's mom), in case anyone wants to take a look.  As to T-Shirts, the plan is that Richard's mom will pick them up from the vendor site on Thursday, I will deliver to Ying Thursday evening, and she will take them to LAS for Friday and Saturday competitions.  Thanks to all who turned in payments promptly.  I collected 7 payments yesterday and will pass to Rae on Thursday.

Thanks everyone for all you help.  Best of luck to Team Washington at ARML 2007.

How did it go? you guys better have won like first place in nation or something...  :)

I was not on-site, and learned that WA team had to bypass Award Ceremony due to delay to have lunch before going to airport.

What I gathered (not confirmed yet): Alabama won 1st place at LV site, WA team won 2nd place.  Nationally nothing posted but I heard that PEA won first.

I can confirm this much:

Alabama 149
Washington Gold 143
Southern California A 143
SFBA A 137

I don't yet know what that means in terms of national ranking - I'll probably find that out at about the same time you do.

We had a bad 2nd day. Relay score was 6 in both round, which was terrible compared to last year's 20. But we did fair okay on individuals. 4 of us scored 7 and went up for the tiebreak and many others scored in 6s and 5s.

Overall the contest was okay, except for some of us the flight was delayed for 3 hours so we had to take the friday night test in 30 minutes after we arrived  :mad:

You should be proud of the team achievements.  WA team was leading until Relay (would be nice if Relay result is better)

Congratulations to Adam, Sam, Young, and Meir for receiving top individual score in the nation (7 out from 8).

I agree with Mr. Norris that next couple of years look bright with so many young students in the team (9th and 8th grade) although we will miss Adam and Young.

The itinerary this year did make the schedule very tight and left very little room for unexpected situations.  Perhaps next year we should consider a more relaxed travel schedule.  It would be nice to settle Relay team grouping (subteams and the sequence) early next year for focused practices.

Congratulations to everyone for a job well done!

Good job guys :). Wish I could've made it.

Thanks to Dr. Merryfield (SoCal coach) and Mr. Norris who provided the details of the ARML West events and scores as they unfolded.

A look at the Division A race in the West as the students took the exams.

Analysis is done based on the order event occurred.  The top five teams turned out to be quite closely competitive: Alabama, Washington Gold, Southern California A, Colorado Granite, and SFBA (San Francisco Bay Area) A.

1. Team Round (Friday):

Wash: 28
SFBA: 28
SoCal: 24
Col: 24
Ala: 20

2.  Power Round (Friday) - a particular strength of both Washington and Colorado (see ARML Power Contest results). On that:

Wash: 32
Col: 32
SoCal: 30
Ala: 29
SFBA: 21

Standings at Friday night:

Wash: 60
Col: 56
SoCal: 54
Ala: 49
SFBA: 49

Note that at this point Alabama trailed Washington by 11 points. That would hold for some time.

3. Individual Round:

Individual problems 1 and 2 were a wash, with all five teams getting 27 points, ending score:

Wash: 87
Col: 83
SoCal: 81
Ala: 76
SFBA: 76

On individual problems 3 and 4, Washington gained ground and Alabama fell further behind:

Wash: 13 + 11 = 24
SoCal: 11 + 11 = 22
SFBA: 10 + 12 = 22
Col: 12 + 8 = 20
Ala: 11 + 8 = 19

That left the standings (half way through Individual Round):

Wash: 111
Col: 103
SoCal: 103
SFBA: 98
Ala: 95

Then came problems 5 and 6. SoCal shone on #5, and Alabama killed #6:

Ala: 5 + 13 = 18
SoCal: 9 + 8 = 17
Wash: 7 + 9 = 16
SFBA: 5 + 8 = 13
Col: 5 + 7 = 12

With that, Alabama overtook SFBA but was still 4th.  Standings after Individual 5&6:

Wash: 127
SoCal: 120
Col: 115
Ala: 113
SFBA: 111

Question #7 didn't make much difference, and #8 made no difference at all since no one got it. Ala, Col, and SFBA each got 12 on #7, with SoCal 11 and Wash 10.

Wash: 137
SoCal: 131
Col: 127
Ala: 125
SFBA: 123

4. Relay Round:
Relay #1 was a jaw-dropper (unfortunate for WA)

Ala: 18
SoCal: 10
Col: 8
SFBA: 8
Wash: 2

And the standings turned over (with WA dropped to #3):

Ala: 143
SoCal: 141
Wash: 139
Col: 135
SFBA: 141

Still completely up for grabs going into Relay #2:

Ala: 6
SFBA: 6
Wash: 4
SoCal: 2
Col: 2

Leaving the final standings:

Ala: 149
Wash: 143
SoCal: 143
Col: 137
SFBA: 137

National Results are posted.

In Division A, PEA's Red Team won 1st in the National level with a total score of 171.  Washington's Gold team ranked 15th in the Nation: http://arml.com/Results/2007a.pdf
(WA Gold had good scores in other rounds except Relay).

Division B result can be found at: http://arml.com/Results/2007b.pdf (Washington Red team is 28th).

Overall combined A and B division results: http://arml.com/Results/ARML2007Scores.xls


Show that for any positive integer $ k$ the equation $ a^2\plus{}b^2\equal{}c^k$ has positive integer solutions.

We have $ p^1 \equal{} m^2 \plus{} n^2$ where $ p$ is a prime, $ p\equiv1(mod 4)$.
If $ p^{k \minus{} 1} \equal{} s^2 \plus{} t^2$, then $ p^k \equal{} (ms \plus{} nt)^2 \plus{} (mt \minus{} ns)^2 \equal{} (ms \minus{} nt)^2 \plus{} (mt \plus{} ns)^2$ and at least one of the numbers $ mt \minus{} ns, ms \minus{} nt$ is not equal to $ 0$, because $ (mt \minus{} ns) \plus{} (ms \minus{} nt) \equal{} (m \minus{} n)(t \plus{} s)$ and $ m \neq n$.
So for every positive integer $ k$ this equation has infinitely many solutions.


inequalities
logarithms
inequalities proposed

Let $ a,b,c,d,e,f,g$ are positive real numbers. Prove that:
\[ \sqrt{\frac{a}{4a\plus{}5b}}\plus{}\sqrt{\frac{b}{4b\plus{}5c}}\plus{}\sqrt{\frac{c}{4c\plus{}5d}}\plus{}\sqrt{\frac{d}{4d\plus{}5e}}\plus{}\sqrt{\frac{e}{4e\plus{}5f}}\plus{}\sqrt{\frac{f}{4f\plus{}5g}}\plus{}\sqrt{\frac{g}{4g\plus{}5a}}\le \frac{7}{3}\]

I don't think so  :D  you are wrong.  :wink:  My inequality is correct  :)

I misunderstand what you wrote  :(
Try
\[ a \equal{} g \equal{} 1,b \equal{} 10^{ \minus{} 10}...
\]
we have:
\[ RHS \minus{} LHS \equal{} 9.399991896E \minus{} 11 > 0
\]
I don't think my inequality is wrong...  :D You should try to solve it and I think you shouldn't disprove it  :rotfl:

Can you give a hint for your inequality , Zaizai  :(

I cannot understand why I cannot disprove with these values;
$ a = 1,b = 10^{ - 10},c = 10^{ - 20},d = 10^{ - 30},e = 10^{ - 40},f = 10^{ - 50},g = 1$
$ \sqrt{\frac{a}{4a+5b}}=\sqrt{\frac{1}{4+5\frac{b}{a}}}$, and in this case $ \frac{b}{a}=\frac{c}{b}=\frac{d}{c}=\frac{e}{d}=\frac{f}{e}=10^{-10}$.
If you calculate $ \sqrt{\frac{1}{4+5\cdot 10^{-10}}}$, it will exceed 0.499, and since seven terms of $ LHS$ are all positive, LHS will exceed $ 5 \cdot 0.499$ which is larger than $ \frac{7}{3}$.
If I am wrong, can you please explain me why am I wrong?  :roll:

I don't think you are correct  :D I will check later but I used to computer to check it... and my inequality is not wrong with there value  :wink:

Your inequality is wrong, Hoang. Check it carefully. It is only true if you replace $ \frac{7}{3}$ by $ 3$.

Can you explain why my inequality is wrong, hungkhtn?
If replace $ 7/3$ by $ 3$, the inequality will be correct with 9 variables. I tried solve it and I was success. I don't think that they are different from each other  :blush:

Check the inequality carefully. My counterexample is right, and consider that the possible maximum value of $ \sqrt{\frac{a}{4a\plus{}5b}}$ is $ \frac{1}{4}$.
Can you post your solution?

what do you think about the following one?
Let $ a,b,c$ be non-negative numbers, to two of which are zero. Then:
\[ \sqrt{\frac{a}{4a\plus{}5b}}\plus{}\sqrt{\frac{b}{4b\plus{}5c}}\plus{}\sqrt{\frac{c}{4c\plus{}5a}}\le 1\]

Is it wrong, msceok?  :D
Hungkhtn, is it also true with 9-variables?  :?:

OMG, I have made mistaken in checking. Sorry everybody  :( But I think that It is correct with $ 4,5,6$ variables (I will try again) :) For n=9 (as what hungkhtn said ) it is wrong  :)

Let $ x_{i} \equal{} \ln(\frac {a_{i \plus{} 1}}{a_{i}})$,  $ \sum_{i \equal{} 1}^{n}x_{i} \equal{} 0$.
(Here $ a_{n \plus{} 1} \equal{} a_{1}$,  $ x_{n \plus{} 1} \equal{} x_{1}$)
$ \sum_{i \equal{} 1}^{n}(\frac {a_{i}}{4a_{i} \plus{} 5a_{i \plus{} 1}})^{\frac {1}{2}} \equal{} \sum_{i \equal{} 1}^{n}(4 \plus{} 5e^{x_{i}})^{\frac { \minus{} 1}{2}}$
$ ((4 \plus{} 5e^{x})^{\frac { \minus{} 1}{2}})'' \equal{} \frac {5e^{x}}{4}(5e^{x} \minus{} 8)(5e^{x} \plus{} 4)^{\frac { \minus{} 5}{2}}$
Which is positive, when $ e^{x} > \frac {8}{5}$ and negative when $ 0 < e^{x} < \frac {8}{5}$.
By Karamatas inequality we can assume that $ x_{1} \equal{} x_{2} \equal{} ... \equal{} x_{n \minus{} 1} < x_{n}$
(First we use it to those $ x_{i}$ which are more than $ \ln(\frac {8}{5})$ if they exist to get at list $ (n \minus{} 1) x_{i}$ which are not more than $ \ln(\frac {8}{5})$. For them we use Karamatas inequality again to make them equal.
In other words $ F(x_{1}, x_{2}, ... ,x_{n}) \le F(y, y, ..., \minus{} (n \minus{} 1)y)$ for some y.)
$ ((n \minus{} 1)(4 \plus{} 5e^{y})^{\frac { \minus{} 1}{2}} \plus{} (4 \plus{} 5e^{ \minus{} (n \minus{} 1)y})^{\frac { \minus{} 1}{2}})' \equal{} \frac {5}{2}(n \minus{} 1)e^{y}(5e^{y} \plus{} 4)^{\frac { \minus{} 3}{2}}(e^{ \minus{} ny}(5e^{ \minus{} ny} \plus{} 4)^{\frac { \minus{} 3}{2}} \minus{} 1)$
so minimum holds when $ e^{2ny}(5e^{ \minus{} ny} \plus{} 4)^{3} \equal{} 1$
$ 2e^{ \minus{} ny} \equal{} (5 \plus{} \sqrt {41})$
$ \sum_{i \equal{} 1}^{n}(4 \plus{} 5e^{x_{i}})^{\frac { \minus{} 1}{2}} \le (n \minus{} 1)(5(\frac {\sqrt {41} \minus{} 5}{8}) ^{\frac {1}{n}} \plus{} 4)^{\frac{\minus{}1}{2}} \plus{} (5(\frac {\sqrt {41} \plus{} 5}{2})^{\frac {n \minus{} 1}{n}} \plus{} 4)^{\frac{\minus{}1}{2}}$

I have some questions for you ...
When $ n \equal{} 1$ then $ \sqrt {\frac {a}{4a \plus{} 5b}}\le \frac {1}{3}$. Easy to see that $ a \equal{} b$
When $ n \equal{} 3$ then
\[ \sqrt {\frac {a}{4a \plus{} 5b}} \plus{} \sqrt {\frac {b}{4b \plus{} 5c}} \plus{} \sqrt {\frac {c}{4c \plus{} 5a}}\le 0,990877145
\]
when does the equality hold? In Vasc's book, we have the following problems:
\[ \sqrt {\frac {a}{4a \plus{} 5b}} \plus{} \sqrt {\frac {b}{4b \plus{} 5c}} \plus{} \sqrt {\frac {c}{4c \plus{} 5a}}\le 1
\]
The equality hold on $ a \equal{} b \equal{} c$
When $ n \equal{} 7$ then:
\[ f(a,b,c,d,e,f,g)\le 2,330067508 \le \frac{7}{3}
\]

But it is not correct  :( Can you explain clearly?


Let $ F$ be the midpoint of side $ BC$ of triangle $ ABC$. Construct isosceles right triangles $ ABD$ and $ ACE$ externally on sides $ AB$ and $ AC$ with the right angles at $ D$ and $ E$, respectively. Show that $ DEF$ is an isosceles right triangle.

Dear Mathlinkers,
complete the figure with the two square built on AB and AC externally...
Sincerely
Jean-Louis

Let $ D$ and $ E$ be the centers of the squares $ ABXY$ and $ ACPQ.$ Then $ \triangle ABQ$ and $ \triangle ACY$ are congruent by SAS criterion, due to $ AB \equal{} AY,$ $ AC \equal{} AQ$ and $ \angle BAQ \equal{} \angle YAC.$ Therefore, $ BQ \equal{} YC$ and $ \angle ACY \equal{} \angle AQB.$ If $ M \equiv{} YC \cap BQ,$ then quadrilateral $ AMCQ$ is cyclic $\Longrightarrow$ $\angle QMC$ is right. Lines $ FD$ and $ FE$ are respectively parallel to $ CY$ and $ BQ,$ since $ D , E$ are midpoints of $ BY,CQ.$ Hence, $FD \perp FE$ $\Longrightarrow$ $\triangle DEF$ is  isosceles right at $F.$

I have a problem which use moldovan's problem to solve :D
Let $ ABCD$ be a quadrilateral. Construct 4 squares $ ABB_1A_1, BCC_1B_2, CC_2D_1D, DD_2A_2A$ externally $ ABCD$. Let $ X, Y, Z, T$ are the midpoints of $ A_1A_2, B_1B_2, C_1C_2, D_1D_2$. Prove that $ XZ\equal{}YT$ and $ XZ\perp YT$


$ Y$ have pdf $ P(Y \equal{} k) \equal{} (k \plus{} 1)p^2 p^k$, with $ 0 < p < 1$ and $ q \equal{} 1 \minus{} p$ and $ k \equal{} 0,1,2,...$

Is easy prove that $ M_{Y}(t) \equal{} p^2 (1 \minus{} qe^t)^2$. Let $ X \equal{} Y_1 \plus{} \cdots Y_{24}$, with $ Y_{i}$ with same distribution of $ Y$

Calculate $ M_{X}(t)$

Note: $ M_Y (t) \equal{} E(e^{tk})$


Let $ a,b,c\geq 0$.Prove that $ \sum\frac{c^2(c^2\minus{}ab)}{c^4\plus{}ab(a^2\plus{}b^2)}\geq 0$

[quote="livetolove212"]Let $ a,b,c\geq 0$.Prove that $ \sum\frac {c^2(c^2 \minus{} ab)}{c^4 \plus{} ab(a^2 \plus{} b^2)}\geq 0$[/quote]

Apply AM-GM inequality, we have :
$ \frac {c^2(c^2 \minus{} ab)}{c^4 \plus{} ab(a^2 \plus{} b^2)} \geq \frac {c^2(c^2 \minus{} \frac {a^2 \plus{} b^2}{2})}{c^4 \plus{} \frac {(a^2 \plus{} b^2)^2}{2}}$
So the inequality is equivalent to :
$ \sum \frac {c^2(2c^2 \minus{} a^2 \minus{} b^2)}{2c^4 \plus{} (a^2 \plus{} b^2)^2} \geq 0$
Let $ a^2 \equal{} x,b^2 \equal{} y,c^2 \equal{} z$.WLOG assume $ x \plus{} y \plus{} z \equal{} 1$
$ ineq \Leftrightarrow \sum \frac {x(3x \minus{} 1)}{2x^2 \plus{} (1 \minus{} x)^2} \geq 0$
We have
$ \frac {x(3x \minus{} 1)}{2x^2 \plus{} (1 \minus{} x)^2} \geq \frac {3}{2}(x \minus{} \frac {1}{3})$
$ \Leftrightarrow \frac {(1 \minus{} x)(3x \minus{} 1)^2}{2(2x^2 \plus{} (x \minus{} 1)^2)} \geq 0$
which is obvious true .
$ \Rightarrow \sum \frac {x(3x \minus{} 1)}{2x^2 \plus{} (1 \minus{} x)^2} \geq \frac {3}{2}(x \plus{} y \plus{} z \minus{} 3.\frac {1}{3}) \equal{} 0$
Our proof is completed . :roll:

[quote="forevercontinue"][quote="livetolove212"]Let $ a,b,c\geq 0$.Prove that $ \sum\frac {c^2(c^2 \minus{} ab)}{c^4 \plus{} ab(a^2 \plus{} b^2)}\geq 0$[/quote]

Apply AM-GM inequality, we have :
$ \frac {c^2(c^2 \minus{} ab)}{c^4 \plus{} ab(a^2 \plus{} b^2)} \geq \frac {c^2(c^2 \minus{} \frac {a^2 \plus{} b^2}{2})}{c^4 \plus{} \frac {(a^2 \plus{} b^2)^2}{2}}$
[/quote]
Are you sure?

This is an example of wrong signs in inequality as Arquady has quoted. We should be careful when using cauchy  and especially pay a little bit attention on the signs.

I don't mean to spam but I have just found the solution for this interesting problem
We rewrite the original problem in the following form
$ \sum_{cyc} \frac{ab}{c^4\plus{}ab\left(a^2\plus{}b^2\right)} \le \frac{1}{a^2\plus{}b^2\plus{}c^2}$
Using Cauchy - Schwartz, we obtain
$ \sum_{cyc} \frac{ab}{c^4\plus{}ab\left(a^2\plus{}b^2\right)} \le \sum_{cyc}\frac{ab \left(1\plus{}\frac{a}{b}\plus{}\frac{b}{a}\right)}{\left(a^2\plus{}b^2\plus{}c^2\right)^2}$
It suffices to prove that
$ a^2\plus{}b^2\plus{}c^2 \ge ab\plus{}bc\plus{}ca$
We finish the proof!


hi,
we consider the sequence $ (a_{n})_{n \geq 1}$ with
\[ a_{n\plus{}1} \equal{} 2a_{n} \plus{} \sqrt{3a_{n}^{2} \minus{} 2}, a_{1} \equal{} 1.\]
Prove $ a_{n} \in \mathbb{N}$, $ \forall n \in \mathbb{N}$.


prove for any $ a,b,c>0$ that :
$ \frac{a^3}{(a\plus{}b)^3}\plus{}\frac{b^3}{(b\plus{}c)^3}\plus{}\frac{c^3}{(c\plus{}a)^3}\geq\frac{3}{8}$

It's the vietnamien olympiad 2005 :wink: .

[quote="mr.inequalities"]prove for any $ a,b,c > 0$ that :
$ \frac {a^3}{(a \plus{} b)^3} \plus{} \frac {b^3}{(b \plus{} c)^3} \plus{} \frac {c^3}{(c \plus{} a)^3}\geq\frac {3}{8}$[/quote]
$ \sum_{cyc}\frac{a^3}{(a\plus{}b)^3}\equal{}\sum_{cyc}\frac{a^4}{a(a\plus{}b)^3}\geq\frac{(a^2\plus{}b^2\plus{}c^2)^2}{\sum a(a\plus{}b)^3}.$
But $ \frac{(a^2\plus{}b^2\plus{}c^2)^2}{\sum a(a\plus{}b)^3}\geq\frac{3}{8}\Leftrightarrow\sum_{cyc}(5a^4\minus{}9a^3b\plus{}7a^2b^2\minus{}3ab^3)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left((a\minus{}b)(5a^3\minus{}4a^2b\plus{}3ab^2)\minus{}(a^4\minus{}b^4)\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a\minus{}b)^2(4a^2\minus{}ab\plus{}b^2)\geq0.$
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=101425
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=36567


trigonometry
geometry
angle bisector
geometry proposed

Dear everyone.

Please read the following.

for a right angled triangle $ ABC$ ($ \angle{C}$=$ 90^{\circ}$) $ CG\bot{AB}$, $ CE$=bisector of ($ \angle{ACG}$).
$ D$=midpoint ($ CA$), $ DE\cap{CG}$=$ F$. parallel (to $ AC$) through $ E\ \cap{FB}$ = $ H$.
Prove that $ BF$=$ BH$.

I think this is a little nice. Thank you. :)

in fact $ CE \parallel{} FB$
this can be proved by using menelaus:
at first we draw the parrallel through $ B$ to $ CE$ and let it intersect $ CG$ at $ F'$ (note that $ BF'$ is the external angle bisector obviously)
then we use menelaus's theorem for $ DEF$ and $ \Delta ACG$ and we obtain the fraction $ GF/CF$ which is
$ \frac{CG}{AC}\equal{} \sin \angle A$
and for the point $ F'$ we know that $ BF'\parallel{} XE$ which implies $ \frac{GF'}{CF'}\equal{}\frac{GB}{EB}$
but we know that $ GB\equal{}\sin \angle A . CB$ and $ BE\equal{}CB$ since $ CE$ is perpendicular to the angle bisector of $ B$
so we are done for this part of the proof
now, note that we have :
$ \Delta ACE \sim \Delta EHB \implies \frac{BH}{AC}\equal{}\frac{BE}{EA}\equal{}\frac{CB}{EA} \implies BH\equal{}AC.\frac{CB}{EA}$
also for $ BF$ we obtain

$ \Delta CGE \sim \Delta BGF \implies \frac{BF}{CE}\equal{}\frac{BG}{GE} \implies BF\equal{}CE . \frac{BG}{GE}$

so we have to prove the equality :

$ AC.\frac{CB}{EA}\equal{}CE . \frac{BG}{GE}$
which is really easy:
we r done!

1) It's not a big deal to see that the triangle $ BCE$ is isosceles, so $ BE \equal{} BC$ ( 1 ).
2) $ \triangle ABC \sim  \triangle ACG \sim  \triangle CGB$ so $ \frac {AC} {CG} \equal{} \frac {BC}{BG}$ ( 2 )
3) Apply angle bisector theorem in $ \triangle ACG$ and get $ \frac {AE}{EG} \equal{} \frac {AC}{CG}$  ( 3 ).
4) Apply Menelaus theorem in $ \triangle ACG$ with the transversal $ DEF$ and will get, with $ AD \equal{} CD$: $ \frac {FC}{FG} \equal{} \frac {GE}{AE}$ or, with (2) and (3), $ \frac {FC}{FG} \equal{} \frac {BC}{BG}$ ( 4 ), showing that $ BF$ is an angle bisector of $ \angle CBE$, that is, $ BF \parallel{} CE$ [ from (1) ].
5) With $ CF$ and $ EH$ altitudes of the isosceles $ \triangle CBE$, it's now clear that the quad $ CEFH$ is symmetrical w.r.t. angle bisector of $ \angle CBE$, hence $ BF \equal{} BH$.

Best regards,
sunken rock

Dear everyone.

Here is my shoddy proof (only one for me) :(

[hide]
$ Menelaos$ ($ \triangle{ACK}$,$ \overline{DEF}$)$ \Longrightarrow\frac {CF}{FK}$=$ \frac {BE}{BK}$ ($ \because\triangle{BCE}$=isosceles)
$ \therefore\frac {CK}{FK}$=$ \frac {EK}{BK}\Longrightarrow\triangle{CKE}\sim\triangle{FKB}\Longrightarrow\angle{ECK}$=$ \angle{BFK}$

$ \therefore CE\parallel{BF}$

$ CG\cap{EH}$=$ I\Longrightarrow\mathcal{H}(F,I/G,C) \equal{} \minus{} 1\Longrightarrow{BF \equal{} BH}$[/hide]


calculus
real analysis
real analysis solved

Try looking at the topic "Dense" posted recently in the Calculus section. It's probably the fifth time this problem is posted.

[quote="pcalin"]Is it true that [tex]\{na\},\;n\in\mathbb{N}[/tex] is dense in [tex][0,1)[/tex] if [tex]a\in\mathbb{R}\backslash\mathbb{Q}[/tex], and [tex]\{x\}[/tex] denotes the fractional part of [tex]x[/tex]?[/quote]
Really, it is a good tradition to send this problems each week.  :)  ;)  :lol:  :lol:

A full proof was given by Kent Merryfield in [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=27475[/url]


Find the maximum possible value of $ x+y+z$ in the following system of equations:

$ \left\{\begin{array}{c}{1}
x^2-(y-z)x-yz=0\
y^2-\left(\frac{8}{z^2}-x\right)y-\frac{8x}{z^2}=0\
z^2-(x-y)z-xy=0
\end{array}\right.$
:welcomeani:

[hide]
These equations can be factored as
$ (x \minus{} y)(x \plus{} z) \equal{} 0.$

$ (y \minus{} \frac {8}{z^2})(y \plus{} x) \equal{} 0.$

$ (z \minus{} x)(z \plus{} y) \equal{} 0.$

Unfortunately, I don't have a better solution then casework on which thing is 0....

Consider the case x = y. Then, z = x or -x. If y+x = 0, then y+x = 0 and z = 0, which is impossible because z appears in the denominator. If y+x is nonzero, then y = $ \frac {8}{z^2}.$ $ z^2 \equal{} y^2,$ so $ y \equal{} \frac {8}{y^2}$ and $ y^3 \equal{} 8$ so y = 2. Also, x = 2 and z can either = 2 or -2. The sum of the numbers can either be 6 or 2.

Consider the case x+z = 0 and x = -z. z cannot be 0 because it appears in the denominator. Therefore, x cannot be 0, and x-z is nonzero, which implies z+y = 0 and y = -z = x. Therefore, we get to the same case above.

The maximum possible value of x+y+z is 6.[/hide]


Proove that 2,3 and 8,9 are the only pairs of consecutive numbers of the form 2^n, 3^m

That is a = 2^n, b = 3^m and b = a + 1.

The general problem is solved here http://www.mathlinks.ro/Forum/viewtopic.php?t=28571

It may  have been posted  in the above link, any way it is not hard, here is one  solution -  (Sorry if it is  already posted)

[hide]if $m \ge  3 $ and
$3^n - 1 = 2^m$
taking modulo 8, we see  $3^2 \equiv 1 \bmod 8  \  and  \ RHS \  is  \  0 \bmod 8 $
so $n$  must be even  say $n=2k$
so $(3^k +1)(3^k -1) = 2^m$
So each factor must be a power of 2 , which is impossible (  except for k=1) (as the differencebetween the factors is only 2)
QED.[/hide]

Yep, that's it. The standard is pretty high on these forums, eh?

There's another question you might be able to help me with which I asked on another thread but got no answer.

What kind of modulus has the least number of [b]unique[/b] quadratic residues? A prime p has (p - 1)/2 but a composite modulus has less usually.

I'm using (p1.p2.p3........pn)^2 where the p's are primes. Is there a better definition?

thanks.


Let $ k$ and $ d$ be integers such that $ k>1$ and $ \leq 0 < 9$.
Prove that there exists a positive integer $ n$ such that the decimal representation of the number $ 2^n$ has the digit $ d$ in the position $ k$(from the right)

I have some idea:
When n is sufficient large, the last k digits will become periodic, and they will go through all the less than k digit numbers that can be divided by 2^k. Since 2^k<10^(k-1), so their will be a 2^n with the kth digit d.


function
Support
calculus
integration
real analysis
real analysis unsolved

Prove that the set of infinitely differentiable functions with compact support is non-trivial.

This is classical stuff (even though I haven't looked at the proof yet). However, what can we do with such functions ? I mean, very often we need a $C^{\infty}$ function which is say $1$ on a compact set $K$, and whose support is included in another compact set $L$ ($K \subseteq L$).

How do we know such functions exist ?

Let $K\subset U\subset\bar U\subset L$, find a continuous function $f$ with support in $L$ and values 1 on $U$, then $f\ast\mu$ is the required one where $\mu$ is a test function with small support and has integration 1.

Regularization by convolution, I knew I shouldn't have skipped that  :D


search
superior algebra
superior algebra unsolved

Classify the groups of order $ pq$, where $ p$ and $ q$ are primes, $ p < q$ and $ p | (q \minus{} 1)$.


For some teams, going to ARML is quite an expensive thing. Does anyone coach or is a member of a team that receives corporate donations? I'm interested in how you guys received it. Thanks

I don't think our school district covers it...and there's like 10 kids from our district.

But the thing is, (I'm sure you got the letter from TML today or will get it tomorrow max) it would cost us at least 400 per person to go. That's a ridiculous amount for a math competition.

[quote="Rep123max"] That is a lot of money, we should get TI to sponsor ARML, their headquarters is in Dallas after all.[/quote]

Good idea. Next yr...but I wanna know who I'm supposed to contact and stuff...

Minnesota students have to put up $250, which includes competition fee, bus, dorm for 1 night, and all meals for Friday and Saturday.  In my 3 years of going, my school has covered none of the cost, half the cost, and all of it.  I don't think we get corporate sponsorship, but I'm pretty sure the state math league pays (or gets donations) for things like coaches, printing costs, use of practice space, etc.

My team (upstate NY ARML) used to get sponsorship. It was $60 for departure on Wednesday, two days of practice, and competition, with food, rooming and stuff. I think the funding was withdrawn this year, so it's $200 now for Thursday morning-competition. My school covered the fee last year, though I'm not sure they're willing to do the same this year after it's more than tripled.

New York City's ARML team is free to the students.  It involves lots of begging and pleading with the Board of Ed. and donations from City College.  We also have received donations from D.E. Shaw, because someone there is a former math team person.  That's really the trick if you want corporate sponsors -- you have to find someone who can related and is interested in math teams.

we get donations and plus each person pays 100 dollars which doesnt cover meals...some people who don't have five people from our school going actually are able to get their school to provide meals and everything

[quote="MithsApprentice"](I'm sure you got the letter from TML today or will get it tomorrow max) it would cost us at least 400 per person to go. That's a ridiculous amount for a math competition.[/quote]

Elessar - did you not know the price until today? It's up from last year, so I may have told you something on the low side. But it should have been in your invitation.

Most of the members of the Texas ARML team have their way paid by their schools, which sharply decreases the urgency to get funding for the team. The cost to the Texas team is probably higher than almost any other team because of the distance traveled, and the extra night in the dorm. The team has to charter a bus for 30 hours driving time each way, plus the drivers (two) need 2 nights in a hotel and have to get paid for sitting around waiting for the team from Thursday afternoon until Sat. afternoon. Teams within a few hours of their competition site have much lower expenses.

You shouldn't think of it as $400 for the 6 hours or so of the actual math competition. That's really just the culmination. It isn't even all that important how the team does. You need to think in terms of the entire 4 day experience. Most of the previous participants look forward to it all year as an opportunity to spend time with the other top math kids from all over the state, renew friendships, compare progress, envigorate each other mathematically. Many of the people on the team have few or no people like them at their schools. The trip serves the same sort of social support function that conferences serve for professional mathemeticians.

[quote="MithsApprentice"]For people that have mentioned donations, do u know how you obtained this stuff? Who you contacted? How?[/quote]

mostly the state coaches take care of that...however they do give us letters to give to the principals of our schools that ask for financial aid. schools give money, most of our contributors from companies have been contributing for a number of years now...again, the coaches mainly take care of that so i'm not so sure how it works...

[quote="texas137"]Elessar - did you not know the price until today? It's up from last year, so I may have told you something on the low side. But it should have been in your invitation.

Most of the members of the Texas ARML team have their way paid by their schools, which sharply decreases the urgency to get funding for the team. The cost to the Texas team is probably higher than almost any other team because of the distance traveled, and the extra night in the dorm. The team has to charter a bus for 30 hours driving time each way, plus the drivers (two) need 2 nights in a hotel and have to get paid for sitting around waiting for the team from Thursday afternoon until Sat. afternoon. Teams within a few hours of their competition site have much lower expenses.

You shouldn't think of it as $400 for the 6 hours or so of the actual math competition. That's really just the culmination. It isn't even all that important how the team does. You need to think in terms of the entire 4 day experience. Most of the previous participants look forward to it all year as an opportunity to spend time with the other top math kids from all over the state, renew friendships, compare progress, envigorate each other mathematically. Many of the people on the team have few or no people like them at their schools. The trip serves the same sort of social support function that conferences serve for professional mathemeticians.[/quote]

I understand where the costs are going and everything, and for a 4-day trip it's really not a lot of money. However, it still is quite a bit for "just a math competition". Of course it is worth the cost, but I was wondering if it would be possible to reduce the amount by any chance.

I knew the price from quite a while back, but it just struck me now that some teams have corporate sponsors and I wondered if we might possibly try to obtain some too.


geometry
circumcircle
function
national olympiad

[b]The problems were provided and translated by [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=3749]pigfly.[/url]  :) [/b]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=36502]1.[/url] Let $(I),(O)$ be the incircle, and, respectiely, circumcircle of $ABC$. $(I)$ touches $BC,CA,AB$ in $D,E,F$ respectively. We are also given three circles $\omega_{a},\omega_{b},\omega_{c}$, tangent to $(I),(O)$ in $D,K$ (for $\omega_{a}$), $E,M$ (for $\omega_{b}$), and $F,N$ (for $\omega_{c}$).

[b]a)[/b] Show that $DK,EM,FN$ are concurrent in a point $P$;

[b]b)[/b] Show that the orthocetner of $DEF$ lies on $OP$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=36213]2.[/url] Given $n$ chairs around a circle which are marked with numbers from 1 to $n$ .There are $k$, $k \leq 4 \cdot n$  students sitting on those chairs .Two students are called neighbours if there is no student sitting between them.  Between two neighbours students ,there are at less 3 chairs. Find the number of choices of $k$ chairs so that $k$ students can sit on those and the condition is satisfied.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=14908]3.[/url] Find all functions $f: \mathbb{Z}\mapsto \mathbb{Z}$ satisfying the condition: $f(x^{3}+y^{3}+z^{3})=f(x)^{3}+f(y)^{3}+f(z)^{3}.$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=36567]4.[/url] Let be given $a,b,c>0$. Prove that: $\sum \frac{a^{3}}{(a+b)^{3}}\geq \frac{3}{8}$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=36566]5.[/url] Let $p\in \mathbb P,p>3$. Calculate:

a)$S=\sum_{k=1}^{\frac{p-1}{2}}\left[\frac{2k^{2}}{p}\right]-2 \cdot \left[\frac{k^{2}}{p}\right]$  if $p\equiv 1 \mod 4$

b) $T=\sum_{k=1}^{\frac{p-1}{2}}\left[\frac{k^{2}}{p}\right]$   if $p\equiv 1 \mod 8$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=36564]6.[/url] $n$ is called [i]diamond 2005[/i] if $n=\overline{...ab999...99999cd...}$, e.g. $2005 \times 9$. Let $\{a_{n}\}: a_{n}< C\cdot n,\{a_{n}\}$ is increasing. Prove that $\{a_{n}\}$ contain infinite [i]diamond 2005[/i].

I don't think they are Vietnam TST 2005 problems.Because they are too old.When clicked the link I saw 3 was posted by Pigfly 10 months ago,similarly the genaral of 6 was also posted by Pigfly(!) long time ago, 2 is a consequence of problem 6 Vietnam TST 1997(Namanhams said that), 4 is very familiar.Probably the problems are training team problems for TST,right?

Yes, you are right. These problems are very old (I mean 2, 3, 5, 6, and may be 1, 4).

But it is fact that these 6 problems are Vietnam TST 2005, I confirm.

Namdung


Hi, everyone,

I was wondering what you would recommend for someone studying for next year's AP Chemistry exam. My local homeschooling group is organizing a chemistry group class, and the class will be using a college-level textbook and going to a local lab to do labs. The kids in the class are bright but young, mostly ages eleven to fourteen. Some of the kids will want to take the AP Chemistry exam next May, or maybe a year later. What would be the best AP Chemistry cram book to use just for test familiarity, and filling in gaps in the course syllabus?

Thanks for any advice you have. Are the recommendations radically different for SAT II chemistry cram books, or can one brand of cram book cover both tests?

You're teaching AP Chem to 11-14 year olds? I hope that their math skills are strong and that they are uncommonly mature for children their age; many of the required AP chem labs use highly concentrated acids and bases, which are not something so "play" with.

I don't know which brand of study book is the best - I've used a couple brands for different tests (when I was teaching myself the material), but I have no real "favorite" - they're all about the same.

Basically I would suggest that you use a college freshman chem textbook and compare it with the syllabus for AP chem which you can get from collegeboard.com.  I don't think prep books are that helpful.

Yea, uh make sure to buy goggles.

[quote="confuted"]You're teaching AP Chem to 11-14 year olds?[/quote]

I'm not personally doing the teaching. I'm part of a group of parents who interviewed, resume-checked, and hired a teacher who was recommended by the U of MN Chemistry Department. We have very strict rules about safety procedures and the like, and there is a complete set of goggles for the class, as well as parent lab monitors to make sure the safety rules are followed.

[quote="confuted"]You're teaching AP Chem to 11-14 year olds?[/quote]
Hey, I was 13 when I took the AP Chem exam.  I was homeschooled at that time and I was studying high school chemistry.  In about two weeks, I crammed for the AP Test.  I just did the best I could on the test, making educated guesses when I had to.  I certainly did not know all the material, but I guess that I knew enough, since I got a 5.
[quote="tokenadult"]What would be the best AP Chemistry cram book to use just for test familiarity, and filling in gaps in the course syllabus? [/quote]
I used the Barron's book, but I'm pretty sure that almost any AP cram book would work.  The only really bad AP cram books I've seen were for the AP Computer Science in Java that came out for the APs this past May.  I guess it was because it was the first year AP CS used Java.
[quote="confuted"]many of the required AP chem labs use highly concentrated acids and bases, which are not something so "play" with.[/quote]
When I was studying for AP Chem, I didn't do the labs at all.  First, I didn't have anywhere to do the labs and second, I didn't have the time.  What you could do (I did this for my AP Bio test) is just read about the labs in an AP review book.  On the test, just use common sense and your knowledge of the subject to do the best you can!

Reading Zumdahl, doing some of the problems, and working on several old AP tests should do the trick.

I used Princeton Review. I liked it a lot better than Kaplan's because it had more practice test taking in it. Also, all of the free-response questions up to this year have been released, you can probably find them online, and those are pretty much the best practice you can get.

I used Zumdahl as a textbook...

I used barron's, arco, and princeton review for cramming. (i love the library)

Barron's is quite lengthy, almost like a textbook, bunch of errors, but all in all a thorough review. Arco is similar, but more aesthetically pleasing.

Princeton review is pretty quick, easiest to go through, but i didn't feel like i got a thorough review from it, especially after doing Barron's and Arco.

the thing that helped me the most, i think, is using the practice problems from the previous years that can be found on the AP website.

another point, i had a chem lab with other homeschoolers, but i really didn't learn anything about chemistry or chem labs in it, and I still got a 5. so the lab part really isn't a huge part of the test.

[quote="blackeyejohn"]Hi , I'm a Chinese freshman to the university. And I'm also interested in chemistry.
During my high school time, I had taken part in the Chinese National Chemistry Olympiad. And I got a quite good place. I'm quite interested in the chemistry education in the USA. Could someone make an introduction to me? I'll be very grateful.[/quote]
What sort of an introduction would you like?

[quote="applepuppy"]I used Zumdahl as a textbook...

the thing that helped me the most, i think, is using the practice problems from the previous years that can be found on the AP website. [/quote]

I agree... Zumdahl was good and had more than enough practice problems (often 100+ at the end of each chapter).  I'm not sure, though, that Zumdahl would be appropriate for 11-14 year olds unless they have a really high reading level and good attention span; Zumdahl takes stamina!

Old tests are definitely the best review.  Our chemistry teacher has lots of old tests, so about 2 months before the exam she made us take one as a regular test.  Most of the people in my class did terrible, but it showed them how much and in what areas they needed to study.  We took about one per week between then and the exam so everybody pretty much knew how they were going to do when they took it for real.

My school's AP Chemistry program is quite good (around 50 people a yearg get 5s on the AP exam.) We used the Zumdahl textbook, like the majority of people posting. I really like the textbook. My teacher simply assigned homework problems (usually all the even problem at the end of a chapter,) and we would have a quiz where he would select 4 or 5 of the problems and score us on them. For tests, he used old AP questions from the multiple choice test. he als made us do 4/5 complete old AP tests for practice. This seemed to work very well for almost everybody. I also got some extra education using Barron's book, which was very good and covered pretty much everything.

[quote="kentliu"]u ever heard of the chemteam?
its like the best website online about chem.. we used to use it for chem honors since my teacher made the site.. i would think its perfect for ur kids
http://dbhs.wvusd.k12.ca.us/webdocs/ChemTeamIndex.html[/quote]

The Chem Team guy is your teacher? That's interesting. He and I corresponded by email many years ago. That is a very useful site.


geometry trigonometry perimeter inradius

So the area of any regular polygon can be expressed as (S^2N)/(4tan(pi/N))

BUT the issue is that the tangent is expressed in radians. Is there a way to express the area of any regular polygon that involves degrees, not radians?

I'm not very good at trigonometry but isn't $ 360^{\circ}\equal{}2\pi \text{radians}$? Can't you just convert it? Once again, I scuk at trig.

Yes. So you would multiply the quantity inside the tangent by $ \frac{180^\circ}{\pi}$ which would yield

$ \frac {S^2N}{4\tan{\left(\frac {180^\circ}{N}\right)}}$.

What is a radian? Google, for some reason, isn't working on my computer, so don't send a "there's something call  Google" please.

[quote="bengal"]What is a radian?[/quote]
One radian is equal to the angle formed by an arc on a circle with an arc length equal to the radius of the circle. Since the circumference of a circle is $ 2 \pi r$, there are $ 2 \pi$ radians in a full circle, so $ 2 \pi$ radians = $ 360 ^ \circ$ and $ \pi$ radians = $ 180 ^ \circ$.

There's also this formula:

$ A \equal{} \frac{1}{2}Pa$

$ A \equal{}$ area of regular polygon
$ P \equal{}$ perimeter of the polygon
$ a \equal{}$ apothem

It's a segment from the center of the polygon perpendicular to the side. I think it might be the inradius.

Since the perpendicular from the center to a side is the shortest distance and forms a right angle, it is indeed the inradius.


ratio
geometry
angle bisector
perpendicular bisector
geometry proposed

$\underline {\underline {\triangle ABC,\ b^2=c(b+c): }}$
$M\in [BC],MB=MC;N\in [AM],\widehat {ABN}\equiv\widehat {CAM}$ $\Longrightarrow m(\widehat {CNM})=\frac 12\cdot m(\widehat {BAC}).$

[i]Remark.[/i] The my last message from http://www.mathlinks.ro/Forum/viewtopic.php?t=54610 is the generalization of this problem.

$m(\angle CNM)$
What does $m$ mean?
Sorry if the question is stupid.  :oops:

It is O.K. $m(\angle \blacksquare )$ is the [u]measure[/u] of the angle $\angle \blacksquare$.

[quote="Virgil Nicula"]$\underline{\underline{\triangle ABC,\ b^{2}=c(b+c): }}$
$M\in [BC],MB=MC;N\in [AM],\widehat{ABN}\equiv\widehat{CAM}$ $\Longrightarrow m(\widehat{CNM})=\frac{1}{2}\cdot m(\widehat{BAC})$[/quote]

Let $z=(A,b)$. Then $C\in z.$ Let $D,E\in z\cap AB$

$b^{2}=c(b+c) \Rightarrow b^{2}-cb-c^{2}=0 \Rightarrow \boxed{(b-c)b=c^{2}}$ and $\frac{b}{c}=\frac{1+\sqrt 5}{2}=\phi$

Let $O$ on the ray $BA$ such that $OB=b$. Then $OA \cdot OB = (b-c)b = c^{2}$
We draw the circle $w=(O,c)$

Let $q$ be the circle which touches $AC$ at $A$ and passes through $B$
Let $X\in w\cap q$ such that $X,C$ are in the same side of $AB$. Then $\angle ABX = \angle CAX$
We will see that $X\equiv N$

$OA \cdot OB = OX^{2}\Rightarrow OX$ is tangent to $q$
Suppose that $DX$ intersects the circle $z$ at $F$
The triangles $DOX,DAF$ are similar, so $OX \parallel AF$.
But $OX$ is a tangent, so $X$ is the midpoint of the arc $AH$, so $\angle XAF = \angle XAC$
Since $AX$ is angle bisector of the isosceles $\triangle ACF$, it is also median and perpendicular bisector of $CF$
So $X\equiv N$.

But then $NC=NF$
The isosceles triangles $NCF,ACE$ are similar because $\angle CFD = \angle CED$

Finally, $\angle FNC = \angle EAC\Rightarrow \angle CNM = \frac{\angle BAC}{2}$

$\blacktriangle$ As $C$ is moving on $z$, th point $N$ is moving on the circle $w$, which is the apollonius circle of $AB$ with ratio $\frac{BN}{AN}= \phi$

Why ${\angle CNM=\frac{1}{2}\angle CNF}$  :?:

Hi, nobody1  :)

[quote="pontios"]Since $AX$ is angle bisector of the isosceles $\triangle ACF$, it is also median and perpendicular bisector of $CF$
So $X\equiv N$.

But then $NC=NF$
[/quote]

$NM$ is the perpendicular bisector of $CF$, so it is also an angle bisector of the isosceles $\triangle CNF$.
Then $\angle CNM=\frac{1}{2}\angle CNF$

[quote="nobody1"]Hi, pontios!
But why $NM\perp CF?$ I don't know why $M$ lies on the perpendicular bisector of $CF$
Thanks.[/quote]
You are right! I'm sorry

I suppose that I thought $M$ as the midpoint of $CF$ instead of $AC$
[quote="pontios"][color=blue]Since $AX$ is angle bisector of the isosceles $\triangle ACF$, it is also median and perpendicular bisector of $CF$ => correct[/color]
[color=red]So $X\equiv N$=>wrong[/color]
[/quote]

Well, $OX\parallel AF \Rightarrow \frac{DX}{XF}= \frac{DO}{OA}= \frac{c}{b-c}$
On the other hand
$\frac{DA}{AB}=\frac{b}{c}= \frac{c}{b-c}\Rightarrow$

$\frac{DA}{AB}=\frac{DX}{XF}\Rightarrow AX\parallel BF \Rightarrow BF\perp CF,$

so $\triangle BCF$ is right and consequently the midpoint $M$ of its hypotenuse belongs to the perpendicular bisector of $CF$,
so $A,X,M$ collinear, so $X\equiv N$

$\star$ The point $X$ does not appear in the diagram, it is the same with the point $N$

[size=150]Lemma[/size]
Let $\triangle ABC$. Let $N$ be a point such that $\widehat{CAN}= \widehat{ABN}$. Let $M\in AN\cap BC$. Then $\boxed{\frac{MC}{MB}= \frac{AC}{AB}\cdot \frac{NA}{NB}}$

[size=150][hide="Proof of the lemma"][/size]
The parallel line to $AC$ through $B$ intersects $AN$ at $F$. Then $\angle AFB=\angle CAN\Rightarrow$
$\angle AFB=\angle ABN$. Also $\angle FAB \equiv \angle BAN$

So $\triangle AFB \sim \triangle ABN\Rightarrow \boxed{\frac{AB}{FB}=\frac{NA}{NB}}\ \ (1)$

[img]9436[/img]

$\triangle AMC \sim \triangle FMB\Rightarrow \boxed{\frac{MC}{MB}= \frac{AC}{FB}}\ \ (2)$

$(1) \wedge (2)\Rightarrow \boxed{\frac{MC}{MB}= \frac{AC}{AB}\cdot \frac{NA}{NB}}$
[/hide]
$\hline$
Now, to our problem

[quote="Virgil Nicula"]$\underline{\underline{\triangle ABC,\ b^{2}=c(b+c): }}$
$M\in [BC],MB=MC;\ N\in [AM],\ \widehat{ABN}\equiv\widehat{CAM}$ $\Longrightarrow m(\widehat{CNM})=\frac{1}{2}\cdot m(\widehat{BAC}).$[/quote]
It is $\frac{AC}{AB}=\phi$

From the lemma we get ${1 =\phi\cdot \frac{NA}{NB}}\Rightarrow \frac{NB}{NA}=\phi$

Let $D$ be the point on the ray $BA$ such that $AD=AC$

[img]9440[/img]

$\boxed{\boxed{\triangle DNB, \triangle CNA: }}\ \frac{BD}{AC}=\frac{b+c}{b}=\phi =\frac{NB}{NA}$ and $\angle DBN = \angle CAN\Rightarrow \triangle DNB\sim \triangle CNA\Rightarrow$

$\angle ADN=\angle ACN\Rightarrow DANC$ cyclic $\Rightarrow \angle CNA = 180-\angle CDA = 180-\frac{\angle BAC}{2}$

So, if $N\in [AM]$ then $\angle CNM = \frac{\angle BAC}{2}$

[quote="Virgil Nicula"]$\underline{\underline{\triangle ABC,\ b^{2}=c(b+c): }}$
$M\in [BC]: \,MB=MC;\,N\in [AM]: \,\widehat{ABN}\equiv\widehat{CAM}$ $\Longrightarrow m(\widehat{CNM})=\frac{1}{2}\cdot m(\widehat{BAC}).$
[/quote]
[color=red][size=150]Very nice problem, Virgil Nicula![/size][/color]
[color=darkblue][b]Proof.[/b] We have: $\widehat{BNM}=\widehat{BAN}+\widehat{NBA}=\widehat{BAN}+\widehat{CAM}=\widehat{BAC}$
So we have to prove that: $\widehat{CNM}=\frac{1}{2}\cdot\widehat{BNM}$

Take the point $P$ lies on the segment $BN$ such that $PB=\frac{b}{b+c}\cdot AN$
We have:
\[\triangle ABP\sim\triangle CAN \]\[\implies\frac{[ABP]}{[CAN]}=\left(\frac{b}{b+c}\right)^{2}\]\[\implies\frac{[ABP]}{[ABN]}=\left(\frac{b}{b+c}\right)^{2}\]\[\implies\frac{BP}{BN}=\left(\frac{b}{b+c}\right)^{2}\]\[\implies PB=\frac{b}{b+c}\cdot PN \]
Hence $NA=NP\implies\widehat{BNM}=2\cdot\widehat{NPA}=2\cdot\widehat{CNM}$
$\mathbb{QED}$[/color]

[quote="At the topic http://www.mathlinks.ro/Forum/viewtopic.php?t=54610, Virgil Nicula"][size=134][color=darkred][b][u]Generalization.[/u][/b] $\left\{\begin{array}{c}\underline{\underline{\triangle ABC}}\\\\ M\in (BC)\ ,\ \frac{MB}{MC}=\frac{c(b+c)}{b^{2}}\\\\ N\in (AM)\ ,\ m(\widehat{ABN})=m(\widehat{CAM})\end{array}\right\|$ $\implies$ $m(\widehat{CNM})=\frac{A}{2}$[/color][/size][/quote]
[color=indigo][b]Proof.[/b] We have to pove that $\widehat{CNM}=\frac{1}{2}\cdot\widehat{BNM}$

Denote $P\in BN: \,BP=\frac{c}{b}\cdot AN$
We have: \[\triangle ABP\sim\triangle CAN\] \[\implies\frac{[ABP]}{[CAN]}=\left(\frac{c}{b}\right)^{2}\] \[\implies\frac{[ABP]}{\frac{b^{2}}{c\cdot (b+c)}\cdot [ABN]}=\left(\frac{c}{b}\right)^{2}\] \[\implies\frac{c\cdot (b+c)}{b^{2}}\cdot\frac{BP}{BN}=\frac{c^{2}}{b^{2}}\] \[\implies \frac{BP}{PN}=\frac{c}{b}\]
Hence $NA=NP\implies\widehat{BNM}=2\cdot\widehat{APN}=2\cdot\widehat{CNM}$[/color]


Slove system of equation:
$ \sqrt{x}\plus{}\sqrt[4]{32\minus{}x}\minus{}y^2\equal{}\minus{}3$
$ \sqrt[4]{x}\plus{}\sqrt{32\minus{}x}\plus{}6y\equal{}24$

Use the inequalities to prove
$ (x,y)\equal{}(16,3)$ is unique solution.

Could you explain the process in detail?

we have :
 $ \sqrt{x} \plus{} \sqrt{32\minus{}x} \leq 8$ (  by AM-GM )
 and$ \sqrt[4]{x} \plus{}\sqrt[4]{32\minus{}x} \leq 4.$
 but we also have : $ y^2 \minus{}6y \plus{}21 \equal{}(y\minus{}3)^2\plus{}12 \geq 12.$
 hence : to "=" when only when :
 x=16 and y =3

[quote="Hong Quy"]we have :
 $ \sqrt {x} \plus{} \sqrt {32 \minus{} x} \leq 8$ (  by AM-GM )
 [/quote]

Sorry but I don't see how you are getting the above inequality using AM-GM.

Thanks but is $ \sqrt[4]{x} \plus{}\sqrt[4]{32\minus{}x} \leq 4$ also obtained using AM-GM?

[quote="mathwizarddude"]Thanks but is $ \sqrt [4]{x} \plus{} \sqrt [4]{32 \minus{} x} \leq 4$ also obtained using AM-GM?[/quote]
Cauchy can do it
$ \sqrt [4]{x} \plus{} \sqrt [4]{32 \minus{} x} \le \sqrt{(\sqrt {x} \plus{} \sqrt {32 \minus{} x})(1\plus{}1)}\le \sqrt{8\cdot 2}\equal{}2$

Also the other one:
$ \sqrt {x} \plus{} \sqrt {32 \minus{} x}\le \sqrt{(x\plus{}(32\minus{}x))(1\plus{}1)}\equal{}8$

[quote="mathwizarddude"]What are the names of these inequalities?[/quote]
These are direct application of [url=http://en.wikipedia.org/wiki/Cauchy_Schwarz_Inequality#Notable_special_cases]Cauchy Schwartz inequality[/url]
look at the first statement in the "notable special cases" section.
I used $ i\equal{}2$ which means $ (x_1y_1\plus{}x_2y_2)^2\le (x_1^2\plus{}x_2^2)(y_1^2\plus{}y_2^2)\iff (x_1y_1\plus{}x_2y_2)\le \sqrt{(x_1^2\plus{}x_2^2)(y_1^2\plus{}y_2^2)}$
In the first inequality
$ x_1\equal{}\sqrt [4]{x},x_2\equal{}\sqrt [4]{32 \minus{} x},y_1\equal{}1,y_2\equal{}1$
and in the second one
$ x_1\equal{}\sqrt {x},x_2\equal{}\sqrt {32 \minus{} x},y_1\equal{}1,y_2\equal{}1$

Also if you want a pure AM>GM proof then
AM>GM $ \implies 2\sqrt [4]{x}\sqrt [4]{32 \minus{} x}\le \sqrt {x} \plus{} \sqrt {32 \minus{} x}$
$ \implies \sqrt {x} \plus{} \sqrt {32 \minus{} x}\plus{}2\sqrt [4]{x}\sqrt [4]{32 \minus{} x}\le 2(\sqrt {x} \plus{} \sqrt {32 \minus{} x})$
$ LHS\equal{}(\sqrt [4]{x} \plus{} \sqrt [4]{32 \minus{} x})^2$ and $ RHS\le 2\cdot 8\equal{}16$
$ \implies (\sqrt [4]{x} \plus{} \sqrt [4]{32 \minus{} x})^2\le 16\implies \sqrt [4]{x} \plus{} \sqrt [4]{32 \minus{} x}\le 4$


function
induction
limit
algebra unsolved
algebra

Find all functions $f: R-\{0\} \rightarrow R$ such that
(i)$f(1) = 1$
(ii)$f(\frac{1}{x+y}) = f(\frac{1}{x})+f(\frac{1}{y})$ for $x,y,x+y \neq 0$
(iii)$[(x+y)f(x+y)]^2 = xyf(x)f(y)$ for $x,y,x+y\neq 0$

Put $g(x)=xf(x)$ $g(1)=1$
From (iii) $g^2(x+y)=g(x)g(y)$
1) Take $n$ a positive integer , $g^2(n+1)=g(n)$
Sup $g(n)=1$ then $g(n+1)=1$ cos $g(n+1)=-1=g^2(n+2)$
Take $n\geq 2$ , $1=g^2(1)=g^2((1-n)+n)=g(1-n)g(n)=g(1-n)$
Then for all integers $n$ , $f(n)=\frac{1}{n}$

2)From (ii) $f(\frac{1}{nx+x})=f(\frac{1}{nx})+f(\frac{1}{x})$ so by induction $f(\frac{1}{nx})=nf(\frac{1}{x})$
Take $x=1$ then $f(\frac{1}{n})=n$
Take $x=\frac{1}{m}$ then $f(\frac{m}{n})=nf(m)=\frac{n}{m}$ so for all rationnals $r$ , $f(r)=\frac{1}{r}$

3)From (ii) we have $g(\frac{1}{x+y})=\frac{x}{x+y}g(\frac{1}{x})+\frac{y}{x+y}g(\frac{1}{y})$
Take $x$  real non rational &  $y$ rational & smal as possible for example $y=\frac{1}{n}$ then
$g(\frac{1}{x+y})=\frac{x}{x+y}g(\frac{1}{x})+\frac{y}{x+y}$ and then
$\lim_{n\to\infty} g(\frac{1}{x+\frac{1}{n}})=g(\frac{1}{x})$ so $g$ is continious
Now take $r_{n}=\frac{[10^nx]}{10^n}$ Obviously $r_{n}$ is a rationnal and tend to $x$ real
$\lim_{n\to\infty} g(r_{n})=g(\lim_{n\to\infty} r_{n})=g(x)=1$ and then $g(x)=1$

So for all real $x$ , $f(x)=\frac{1}{x}$

[quote="AYMANE"]
3)From (ii) we have $g(\frac{1}{x+y})=\frac{x}{x+y}g(\frac{1}{x})+\frac{y}{x+y}g(\frac{1}{y})$
Take $x$  real non rational &  $y$ rational & smal as possible for example $y=\frac{1}{n}$ then
$g(\frac{1}{x+y})=\frac{x}{x+y}g(\frac{1}{x})+\frac{y}{x+y}$ and then
$\lim_{n\to\infty} g(\frac{1}{x+\frac{1}{n}})=g(\frac{1}{x})$ so $g$ is continious
[/quote]

No, that's not sufficient to prove the continuity of $g$ .. :D
Try again ..
 :cool:

[quote="AYMANE"]Ok
$\lim_{n\to\infty}g({\frac{1}{x+\frac{1}{n}}})= g({\frac{1}{x})= \lim_{n\to\infty}g({\frac{1}{x-\frac{1}{n}}})}$ :)[/quote]

That's still insufficient! Be careful, the continuity by sequences is:   $\forall z_n \ convergent\ to \ 0 , \lim_{n\to\infty}g({\frac{1}{x+z_n})= g({\frac{1}{x})}}$

 :cool:

Yes , I agree with you
That's why i said for example.
Any way, you are right ;)

Let $ g(x)\equal{}f(\frac1x)$. We have
$ g(x\plus{}y)\equal{}g(x)\plus{}g(y)$ for $ x,y,x\plus{}y\ne 0$

Let $ x\equal{}1$ and $ y>0$ in iii), We see $ f(y)\ge 0$.

Hence $ g$ has cauchy equation and $ g$ is also increasing. Thr two conditions guarantee $ g$ to be linear. Then it is trivial to show $ f(x)\equal{}\frac 1x$.