inequalities
trigonometry
geometry
inradius
circumcircle
function
complex numbers

Assume that ABC is not acuted-triangle. Find the maximum value of:
$a) \ sin^{2}A+sin^{2}B+sin^{2}C$
$b) \ sinA+sinB+sinC$
$c) \ cosA+cosB+cosC$

You can show (for example, via complex numbers) that $OH^{2}=9R^{2}-(a^{2}+b^{2}+c^{2})$ from there it follows: $\sin^{2}A+\sin^{2}B+\sin^{2}C \leq \frac{9}{4}$.
By AQ inequallity, we have $\sin A+\sin B+\sin C \leq 3 \sqrt{\frac{\sin^{2}A+\sin^{2}B+\sin^{2}C}3}= \frac{3\sqrt{3}}2$.
For sum of $\cos$ you can show that $\cos A+\cos B+\cos C = 1+\frac{r}{R}$ where $r$ and $R$ are inradius and circumradius, in that order. From there it stands that $\sum \cos A \leq \frac{3}{2}$ bacause $IO^{2}=R^{2}-2Rr$, in other word $R\geq 2r$.

Bye

I don't think so.
Ur inequality holds when ABC is equaliteral triangle, which means that $A=B=C=60^{o}$, but the problem assumes that ABC is not an acuted-triangle :D
Retry :)

[quote="gemath"]WLOG $A\ge 90\ge B\ge C$
1) $\sum\sin^{2}A=2+2\prod\cos A$ but $A\ge 90\ge B\ge C\Rightarrow \prod\cos A\le 0\Rightarrow \sum\sin^{2}A\le 2$ equality holds iff $A=90,B=C=45$.
2)$\sum\sin A\le 1+\sin B+\sin C=1+2\cos\frac{A}{2}\cos\frac{B-C}{2}\le 1+\sqrt{2}$ because $A\ge 90\Rightarrow \frac{A}{2}\ge 45 \Rightarrow\cos\frac{A}{2}\le \frac1{\sqrt{2}}$ equality holds iff $A=90,B=C=45$.
3)$\sum\cos A=1-2\sin^{2}\frac{A}{2}+2\sin\frac{A}{2}\cos\frac{B-C}{2}\le 1-2\sin^{2}\frac{A}{2}+2\sin\frac{A}{2}$ Let $t=\sin\frac{A}{2}\Rightarrow \frac1{\sqrt{2}}\le t\le 1$ Consider function $f(t)=-2t^{2}+2t+1$ on $[\frac1{\sqrt{2}},1]$ easily seen $\max_{t\in [\frac1{\sqrt{2}},1]}f(t)=\sqrt{2}$ equality holds iff $A=90,B=C=45$.[/quote]
That's true :D.

And these are real well-known inequality: :D
a) $\sum cos^{2}\frac{A}{2}+\sum tan^{2}\frac{A}{2}\ge \frac{13}{4}$
b) $1+cosAcosBcosC \ge \sqrt{3}sinAsinBsinC$
c) $cos\frac{A-B}{2}+cos\frac{B-C}{2}+cos\frac{C-A}{2}\le \sum cos[\frac{1}{3}.(A-60^{o}]$

[quote="Linkin_Park"]Oh, I have a question that I think it's true:
$cos^{2}\frac{A}{2}+cos^{2}\frac{B}{2}+cos^{2}\frac{C}{2}\le \frac{9}{4}$ for all arbitrary triangle ABC?
Is it true? If that, the pro.1 will be solved obviously[/quote]
Your problem is true for all triangle (By apply $\sum\sin^{2}A\le \frac{9}{4}$ for triangle with three angles $\frac{\pi-A}{2},\frac{\pi-B}{2},\frac{\pi-C}{2}$) but what do you mean pro.1?
 :D

We can express $tanA,tanB,tanC$ by $cosA,cosB,cosC$ and then apply AM-GM ineq :D
$cos\frac{A-B}{2}+cos\frac{B-C}{2}+cos\frac{C-A}{2}\le \sum cos[\frac{1}{3}.(A-60^{o}]$
This ineq is hard. Are there any ideas?


Is there a definite method to find whether a given number is prime?

isn't it something like for any number the biggest prime you have to check is the one less $\sqrt{n}$...
so if $n$ is not divisibly by any primes, from 2 to $\sqrt{n}$, then the number is prime

(If it's even its not a prime greater than $2$  :P )
Well, theres the really tedious way of finding out whether or not its prime: Just divide it by all numbers less than its square root. If nothing divides evenly then you've got yourself a prime number.
Here's some other prime tests:
[url=http://en.wikipedia.org/wiki/Prime_number#Primality_tests]Wikipedia's Prime Tests[/url]

Yeah I remember that method but isn't that still huge for big numbers?

Yes AKS Test ( paper written by two undergraduates,  and their professor from my Alma mater  IIT Kanpur  :)  ) , BTW , is not too  hard to understand, though it is less  practical then, other  straight forward (simple)  tests which are close to but not 100% correct.

A good reference is  [url=http://primes.utm.edu/prove/index.html]Proving Primality[/url]  from http://primes.utm.edu

The IIT Kanpur paper can be found from

[url=http://www.cse.iitk.ac.in/users/manindra/primality_v6.pdf]AKS  test from IIT site[/url]


modular arithmetic
number theory
prime numbers
number theory proposed

[quote="dblues"]Let $k=p^2, a=(p-1)$, then we have $p\mid k$ and $a^k-a = (p-1)^{p^2} - (p-1) \equiv 2 \pmod{p}$, implying $\gcd (a^k-a,k) = 1$ for all odd primes $p$. As for $p=2$, we just let $k=4, a=3$.

Thus there are no such primes.[/quote]
I think you have maked a mistake!
You have written that
$(p-1)^{p^2} - (p-1) \equiv 2 \pmod{p}$
But I think $(p-1)^{p^2} - (p-1) \equiv 0 \pmod{p}$,isn't it? ;)

[quote="dblues"]Oops! $k$ should be $2p$ instead of $p^2$

Since $\gcd (p-1,p) = 1$, by Fermat's Little Theorem, $(p-1)^p \equiv p-1 \pmod{p}$, thus $(p-1)^{2p} \equiv 1 \pmod{p}$. Consequently, $(p-1)^{2p} - (p-1) \equiv 2 \pmod{p}$.[/quote]
Now again you havn't solve my problem :!:  :P
Becaus $((p-1)^{2p} - (p-1));(2p))=2>1$ ;)

[quote="Tiks"]Find all prime numbers $p$,such that there aren't existe any positive integer number $k$ that
1)$p|k$
2)$(a^k-a;k)=1$ for some $a\in{N}$.[/quote]
Guys don't you want to solve this problem? :? ,I think it is really intersting!;)

[quote="dblues"]Ok, since $p$ is prime, $\exists$ a primitive root $g$ of $p$, so let $q$ be the smallest positive integer satisfying the congruence $g^q \equiv g+1 \pmod{p}$. We then note that $g^{pq} \equiv g^q \equiv g+1 \pmod{p}$, thus $g^{pq} - g \equiv 1 \pmod{p}$, and $\gcd (g^{pq}-g, pq)=1$.[/quote]
Can you explain,why $\gcd (g^{pq}-g, q)=1$? :?:

Ok, I must apologize for all the nonsense I've typed.. After rereading through my posts, I'm kinda embarrassed that I was actually wrong three times!  :blush:  :blush:  :blush:

Ok, let me write a full solution, which should be correct now.

First we assume $p>3$. We will show that for each $p>3$, we can always find $a,k$ such that $p\mid k$ and $\gcd (a^k-a,k) = 1$.

Since $p$ is prime, $\exists$ a primitive root $g$ of $p$. Let $q$ be any positive integer coprime to $p-1$ satisfying $1<q<p-1$. We note that such a $q$ exists since $\phi(x) \geq 2 \forall x >2$. Since $\gcd (q,p-1) = 1$, we know from Dirichlet's Theorem that the infinite sequence $q, q+(p-1), q+2(p-1), q+3(p-1), \ldots$ contains infinitely many primes. Let $q_0 = q+r_0(p-1)$ be any arbitrary prime in this infinite sequence that is coprime to $g$ and that satisfies $q_0>g^{p-1}-1$. Note that $q<p$ and $q+(p-1)>p$,  so we also have $q_0 \neq p$.

We now consider $a=g, k=pq_0$.

By Fermat's Little Theorem, $g^{pq_0} \equiv (g^p)^{q_0} \equiv g^{q_0} \equiv g^q \pmod{p}$. Since $g$ is a primitive root and $1<q<p-1$, we know that $g^q \not\equiv g \pmod{p}$, so we have $\gcd (g^{pq_0} - g, p) = 1$.

Next, we notice that since $q_0$ is prime, we have $g^{pq_0} \equiv (g^{q_0})^p \equiv g^p \pmod{q_0}$. This implies $g^{pq_0} - g \equiv g^p-g \equiv g(g^{p-1}-1) \pmod{q_0}$. We note that $\gcd (g, q_0) = 1$ by definition.

Finally, we want to prove that $q_0 \nmid g^{p-1} - 1$. Since $1 < g^{p-1}-1 < q_0$ and $q_0$ is prime, we have $\gcd (g^{p-1}-1, q_0) = 1$.

With that, the case for $p>3$ is proven to yield no solutions.

For the case $p=2$, we note that letting $k=2k_0$ for some integer $k_0$, we have $ak-a=a(a^{2k_0-1}-1)$, which is even since $a$ and $a^{2k_0-1}-1$ have different parities, so $\gcd (a^k-a, k) \geq 2$.

As for the case $p=3$, we note that if for some $a,k$, if $3\mid a^k-a$, then we are done, so we assume $3 \nmid a^k-a$. Let $k=3k_0$ for some integer $k_0$. We have $a^k \equiv a^{3k_0} \equiv a^{k_0} \pmod{3}$. Since $a^{k_0}-a \equiv 0 \pmod{3}$ for all odd $k_0$, we must have $k_0$ is even. Yet this means $a^{3k_0}-a = a(a^{3k_0-1}-1)$ which is even as $a$ and $a^{3k_0-1}-1$ have different parities, so we have $\gcd (a^k-a, a) \geq 2$.

Therefore, we only have $p=2$ or $p=3$ as solutions.

[i]Tiks[/i], could you share with us your solution?

Now your solution is right,:lol:
Here is hint to find my solution.
From Bertran's theorem for any $p>3$ we can find a prime number $q$ so that $p\leq q\leq 2p-3$.
Now it isn't hard to show that $(a^{pq}-a;pq)=1$ for all $a\in{N}$ ;)

Your condition should be $p < q \leq 2p-3$ instead of $p \leq q \leq 2p-3$.

Could you show us your solution in full instead? After all, we both have the idea that $k$ is a product of two primes.

I think my solution has the added advantage that the idea still work for stricter conditions, as it can be easily extended to show the following:

For each positive integer $n>1$, there exists a sufficiently large prime $p_0$ such that $\forall$ primes $p>p_0$, there exists some positive integers $a,k$, with $k$ having exactly $n$ distinct prime divisors, satisfying the condition that $\gcd (a^k-a, k) = 1$.


trigonometry function AMC AIME LaTeX quadratics USA(J)MO

What are some of the trig functions I should memorize to save time on the AIME, besides the 30-45-60 ones.

Also, could you post what they are equal to also (such as $\tan{15}= 2-\sqrt{3}$)

Thanks :)

I don't think you should bother memorizing anything beyond the 30-45-60 ones. It's not a good use of your memory space, as they can be easily derived.

I know 15-30-45-60-75, but that's from doing a lot of problems involving 15 and 75 degrees in them. More important (especially with a longer test such as the AIME) is that you know how to derive them if you need to, for example, $\sin(15^\circ)=\sin(45^\circ-30^\circ)=\sin(45^\circ)\cos(30^\circ)-\sin(30^\circ)\cos(45^\circ)$

Definitely know the angle sum and difference formulas, and you can derive specific things like $\sin 15^{\circ}$ from there.
$\sin(A+B)=\sin A\cos B+\cos A\sin B$
$\cos(A+B)=\cos A\cos B-\sin A\sin B$
$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$

law of sines help a lot. especially if you have easy angles with 15n degrees (n being an integer) and know two angles and one side or two sides and one angle, you can warp around and find all of the angles and sides to the whole triangle.

never seen a law of cosines used before in a solution though, but might be useful.

[quote="t0rajir0u"]Although I always thought that $\cos 36^{\circ}= \frac{\phi}{2}$ was cool.[/quote]
That, along with sin 18n, can be easily derived.

Just memorize sin 15 and cos 15, and I think on a long test like the AIME you will have time to use the addition formulas to get what you need. The tangents can be found by sin/cos, of course. I personally just remember a 30-60-90 triangle, and a 45-45-90 triangle, and figure stuff out from there.

Next year's AMC though, I'd probably recommend knowing all the multiples of 15 for every trig function, but I don't think that knowledge will be required.

I don't know how helpful these would be in AIME, but these are things I'm learning right now...

i'm using @ instead of theta

sin (-@) = -sin(@)
cos (-@) = cos (@)
tan (-@) = -tan(@)
csc (-@) = -csc(@)
sec (-@) = sec (@)
cot (-@) = -cot (@)

sin (90 - @) = cos (@)
cos (90 - @) = sin (@)
tan (90 - @) = cot (@)
csc (90 - @) = sec (@)
sec (90 - @) = csc (@)
cot (90 - @) = tan (@)

argh..here goes some latex

$\frac{sin\theta}{cos\theta}= tan\theta$
and when you flip the fraction over you get what cot@ is

Pythagorean Identities
$sin^{2}\theta+cos^{2}\theta = 1$
$1+cot^{2}\theta = csc^{2}\theta$
$tan^{2}\theta+1 = sec^{2}\theta$

Know the Law of Sines and Law of Cosines

$cos(\alpha-\beta) = cos\alpha cos\beta+sin\alpha sin\beta$
$cos(\alpha+\beta) = cos\alpha cos\beta-sin\alpha sin\beta$

$sin(\alpha+\beta) = sin\alpha cos\beta+cos\alpha sin\beta$
$sin(\alpha-\beta) = sin\alpha cos\beta-cos\alpha sin\beta$

$tan(\alpha+\beta) = \frac{tan\alpha+\beta}{1-tan\alpha tan\beta}$
$tan(\alpha-\beta) = \frac{tan\alpha-\beta}{1+tan\alpha tan\beta}$

Double-Angle Formulas
$sin(2A) = 2sinAcosA$
$cos(2A) = cos^{2}A-sin^{2}A$
$cos(2A) = 1-2sin^{2}A$
$cos (2A) = 2cos^{2}A-1$
$tan(2A) = \frac{2tanA}{1-tan^{2}A}$

$sin(\frac{\theta}{2}) = \pm\sqrt{\frac{1-cos\theta}{2}}$
$cos(\frac{\theta}{2}) = \pm\sqrt{\frac{1+cos\theta}{2}}$
$tan(\frac{\theta}{2}) = \pm\sqrt{\frac{1-cos\theta}{1+cos\theta}}$
$tan(\frac{\theta}{2}) = \frac{sin\theta}{1+cos\theta}$
$tan(\frac{\theta}{2}) = \frac{1-cos\theta}{sin\theta}$

sin(x+y) + sin(x-y) = 2sinxcosy
sin(x+y) - sin(x-y) = 2cosxsiny
cos(x+y) + cos(x-y) = 2cosxcosy
cos(x+y) - cos(x-y) = -2sinxcosy

$sinX+sinY = 2sin\frac{X+Y}{2}cos\frac{X-Y}{2}$
$sinX-sinY = 2cos\frac{X+Y}{2}sin\frac{X-Y}{2}$
$cosX+cos Y = 2cos\frac{X+Y}{2}cos\frac{X-Y}{2}$
$cosX-cos Y =-2sin\frac{X+Y}{2}sin\frac{X-Y}{2}$

EDIT: sorry, latexing problem on my part

EDIT:  also, I do know how to derive these things, but that would have been way too long to post.

The last two groups, as well as the double angle formulas (which you can get the half-angle formulas from as well), can easily be derived from the angle addition formulas.
The first two groups can be derived from drawing an unit circle... :P
The second and third Pythagorean Identities can be obtained by dividing the first by $\sin^{2}\theta$ and $\cos^{2}\theta$.
So you pretty much only need to remember the Law of Sines and Cosines, angle addition formulas , the first Pythagorean identity, and the defintions of trig functions to get everything in your post.
EDIT: And is it just me, or are your angle addition formulas wrong? See [url=http://en.wikipedia.org/wiki/Trigonometric_identity#Angle_sum_and_difference_identities]here.[/url]

[quote="davidyko"]And is it just me, or are your angle addition formulas wrong? See [url=http://en.wikipedia.org/wiki/Trigonometric_identity#Angle_sum_and_difference_identities]here.[/url][/quote]

They are wrong.

man don't memorize stuff

just learn how to derive them

you pretty much need cos(a+b)=...
sin(a+b)=...

Actually, all you really need is $e^{i\theta}=\cos\theta+i\sin\theta$.

Sorry for this revive, but I just don't understand:

How can you derive stuff like sin(18) or stuff like cos(18n)?

[hide="Or"]Take a 36-72-72 isosceles triangle, and draw the angle bisector of one of the 72 degree angles. This should yield three isosceles triangles. You can now solve for sin(18) using similar triangles and the quadratic formula.[/hide]

[quote="flamefoxx99"][quote="thecmd999"][hide="Solution"]Let $x=18$; then we have $3x=90-2x$. Taking the sine of both sides and solving with the double and triple angle identities yields the result.[/hide][/quote]

Sorry for the revive, but what are the "triple angle identities" and how are they derived? Are they useful in competition?[/quote]
$\cos 3x = 4 \cos^3 x - 3 \cos x$ and something similar for sine. You can derive any of them by putting
\[ \cos nx + i \sin nx  = (\cos x + i \sin x)^n \]
and expanding the right-hand side with BT, and then using $\sin^2 x + \cos^2 x =1$.  (Of course the triple angle one is the special case $n=3$.)

Alternately, without the use of complex numbers, write $\cos 3x = \cos(2x+x)$ and use the addition formulas, the double-angle formulas, etc. repeatedly. (But the complex numbers way is much nicer!)

[quote="AMN300"]This might not answer your question, flamefoxx99, but this link is related.
https://brilliant.org/discussions/thread/deriving-trigonometric-identities-part-1/?ref_id=113694[/quote]
Ah shoot, I was going to post that. After all, I wrote it  :lol:

EDIT: completely unrelated, but I would like a few opinions and I thought this was too insignificant to post a new topic:

I recently got a 10 on the 2009 AIME II. What score do you think I will get on this year's AIME? Also, what score would I need to get into JMO, considering the low difficulty o the AMC 10? My AMC 10 score was 139.5.

I'm trying to make USAJMO or USAMO this year as well. I would think an 8 would most likely do it, but aim higher in case you fall low!


If $ \frac{1}{x} \plus{} \frac{1}{y} \equal{} 3$ and $ \frac{1}{x} \minus{} \frac{1}{y} \equal{} \minus{}7$ what is the value of $ x \plus{} y$. Express your answer as a common fraction.

hello, adding the two given equations we get
$ \frac{2}{x}\equal{}\minus{}4$, from here we get $ x\equal{}\minus{}\frac{1}{2}$, setting this in the first equation we get
$ \minus{}2\plus{}\frac{1}{y}\equal{}3$ hence $ y\equal{}\frac{1}{5}$ so our sum is $ \minus{}1/2\plus{}1/5\equal{}\minus{}3/10$.
Sonnhard.


If $ a,b,c$ are positive real, then prove that $ (1\plus{} \frac{a}{b})(1\plus{} \frac{b}{c})(1\plus{} \frac{c}{a}) \geq 2(1\plus{} \frac{a\plus{}b\plus{}c}{\sqrt[3]{abc}})$

[quote="hell_ever"]If $ a,b,c$ are positive real, then prove that $ (1 + \frac {a}{b})(1 + \frac {b}{c})(1 + \frac {c}{a}) \geq 2(1 + \frac {a + b + c}{\sqrt [3]{abc}})$[/quote]

Set $ x^3=a,y^3=b,z^3=c$ and after clearing denominators the inequality becomes
\begin{align*}
2x^3y^3z^3+\sum_{sym} x^6y^3 &\ge 2x^3y^3z^3 + \sum_{sym} x^5y^2z^2 \\
\sum_{sym}x^6y^3 &\ge \sum_{sym}x^5y^2z^2
\end{align*}
which holds by Muirhead. :D

we  have $ (1\plus{} \frac{a}{b})(1\plus{} \frac{b}{c})(1\plus{} \frac{c}{a}) \equal{}2\plus{}\frac{a\plus{}b}{c}\plus{}\frac{c\plus{}b}{a}\plus{}\frac{a\plus{}c}{b}$,so
$ (1\plus{} \frac{a}{b})(1\plus{} \frac{b}{c})(1\plus{} \frac{c}{a}) \geq 2(1\plus{} \frac{a\plus{}b\plus{}c}{\sqrt[3]{abc}})$
$ \Longleftrightarrow 2\plus{}\frac{a\plus{}b}{c}\plus{}\frac{c\plus{}b}{a}\plus{}\frac{a\plus{}c}{b} \ge  2(1\plus{} \frac{a\plus{}b\plus{}c}{\sqrt[3]{abc}})$.
WLOG let $ a\ge b \ge c$  then $ a\plus{}b \ge c\plus{}a \ge b\plus{}c$ and $ \frac{1}{c} \ge \frac{1}{b} \ge \frac{1}{a}$ and so by chebychev's theorem we have $ 2\plus{}\frac{a\plus{}b}{c}\plus{}\frac{c\plus{}b}{a}\plus{}\frac{a\plus{}c}{b} \ge 2\plus{}\frac{2}{3}(a\plus{}b\plus{}c)(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c})$
$ \ge 2(1\plus{} \frac{a\plus{}b\plus{}c}{\sqrt[3]{abc}})$ (by AM-GM)


calculus
derivative
function
limit
real analysis
real analysis unsolved

Let $ f: \mathbb R\to\mathbb R$ be a continuous function so that for all $ x\in\mathbb R$
\[ \lim_{h\to 0}\frac{f(x+2h)-f(x+h)}h=0\]
Prove that $ f$ is constant.
Remark: we do not know whether the function $ f$ is derivable or not!

This problem is a twin of [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=155036[/url].


[quote="GameBot"]Of a group of boys and girls at Central Middle School's after-school party, 15 girls left early to play in a volleyball game. The ratio of boys to girls then remaining was 2 to 1. Later, 45 boys left for a football game. The ratio of girls to boys was then 5 to 1. How many students attended the party?[/quote]

$ \frac{b}{g\minus{}15}\equal{}2$
$ \frac{g\minus{}15}{b\minus{}45}\equal{}5$
$ g\minus{}15\equal{}5b\minus{}225$
$ b\equal{}2g\minus{}30$
$ g\minus{}15\equal{}10g\minus{}375$
$ 9g\equal{}360$
$ g\equal{}40$
$ b\equal{}50$
$ 40\plus{}50\equal{}90$.


how would you find the index of refraction of red light if you are given the wavelengths and index of refraction of white light?

thanks

White light does not have an index of refraction since it is actually all the colors? Shorter wavelengths have a higher index of refraction than longer wavelengths if that is what you mean.


quadratics
trigonometry
algorithm
function
algebra
quadratic formula
geometry

I'm fairly good at writing in TI Basic. I recently made a program that does the quadratic formula, afterwards factoring or displaying the complex fraction or numbers perfectly. To do this fully, I developed a process that will simplify [i]any radical[/i] if one is contained.

To my knowledge, there is only one anomaly to this harmonious system of if-thens and Lbl's:

[hide="Trigonometry"]In The Law of Sines or The Law of Cosines, The denominator is unpredictable (so it seems) and my algorithm is failing to display correctly. For instance, $ 4\sqrt{3}$ is no problem, but $ \frac{\sqrt{3}}{2}$ looks like a total catastrophe. It actually displays [i]decimals[/i].

[b]Ideally, I would like to get the denominator value totally under my control.[/b]

Is there a Trig Identity, or perhaps an altered form of one that can yield the net numerator value to any trig function?  I can solve for the angles, but it's a guess and check process, and very time consuming.
[/hide]
 :help:

P.S.

-Let me know if you need my current approach to help.


trigonometry
inequalities
inequalities unsolved

Suppose that $a_1,a_2,...,a_n\in [0;\pi]$ and  $n+\sum_{i=1}^n\cos a_i$ is an odd integer.
Prove that $\sum_{i=1}^n\sin a_i\ge 1$

Lemma 1: $cosA+senA\geq 1$ becuase of triangle inequality.

So we have $\sum cos a_i+sen a_1 \geq n$

but we have $n-1\geq \sum cos a_i$, becuase we have $n+\sum cos a_i$ is odd integer less than $2n$, so it implies that $\sum sen a_1 \geq 1$.

[quote="Stun"]I'm sorry but I did a stupid mistake. I have just corrected it.
Now the problem is $a_1,a_2,...,a_n\in [0,\pi]$[/quote]

Let $x_i=cos^2 \frac{a_i}{2}$. Then $x_i\in[0,1]$

$n+\sum cos a_i=\sum (1+cos a_i)= \sum 2cos^2 \frac{a_i}{2}=2\sum x_i$
Hence, $\sum x_i =\frac{n+\sum cos a_i}{2}=\frac{2k+1}{2}=k+\frac{1}{2}$ for some non-negative integer k.

Note that $\sum sin a_i =\sum 2cos \frac{a_i}{2} sin \frac {a_i}{2}=2 \sum \sqrt{x_i (1-x_i)}$.

Let $f(x)= \sqrt{x (1-x)}$ on [0,1]. Then f(x) is a concave function. As $(1,1,1,...,1,\frac{1}{2},0,...,0)$ (k 1's, one $\frac{1}{2}$ and (n-k-1) 0's) majorises $(x_i,x_2,...,x_n)$, by majorisation inequality we have

$\sum sin a_i =2 \sum \sqrt{x_i (1-x_i)}=2\sum f(x_i) \geq 2 [kf(1) +f(\frac{1}{2})+(n-k-1)f(0)]=2[k(0)+\frac{1}{2}+(n-k-1)(0)]=1$


geometry
rectangle
ratio
similar triangles

There exists a rectangle ABCD, where AB = 8 and BC = 6

The diagonal BD is constructed. Square PQRS is inscribed in triangle ABD, such that P is touching AB, Q is touching AD, and R and S are along the diagonal BD.

Find the length of a side in square PQRS

I got: [hide]120/37
So... what I did was... well, I think I may have made it difficult for myself. But I used the similar triangles and their ratios.\[ \begin{array}{l} {{\rm We\; see\; that\; }\Delta AQP{\rm \; is\; similar\; to\; triangle\; }\Delta SPB{\rm \; and\; they\; are\; both\; similar\; to\; }\Delta ADB.} \\ {\overline{AB}=8} \\ {\overline{AD}=6} \\ {\overline{BD}=10} \\ {} \\ {{\rm Let: \; }\overline{AQ}=x{\rm ,\; }\overline{AP}=y{\rm ,\; }\overline{BS}=z,{\rm \; and\; length\; of\; the\; side\; of\; the\; square\; be\; }s.} \\ {{\rm Using\; the\; ratios\; of\; the\; similar\; triangles,\; }AQP{\rm \; to\; }ADB: \frac{x}{6} =\frac{y}{8} =\frac{s}{10} =a} \\ {{\rm Using\; the\; ratios\; of\; the\; similar\; triangles,\; }SPB\; {\rm to\; }ADB: \frac{s}{6} =\frac{z}{8} =\frac{8-y}{10} =b} \\ {{\rm Equation\; }1: y=8a=8-10b} \\ {{\rm Equation\; }2: s=10a=6b} \\ {5\times {\rm Equation\; 1: 4}0a=40-50b} \\ {4\times {\rm Equation\; 2: }40a=24b} \\ {74b=40} \\ {b=\frac{20}{37} } \\ {s=6b} \\ {s=6\left(\frac{20}{37} \right)} \\ {s=\frac{120}{37} } \end{array} \][/hide]

by the way, could someone tell me if i'm right or not?

and if i'm right, what's a better way to do it?

[quote="The Master"]could someone tell me if i'm right or not?[/quote]not

You wrote $\frac{s}{6} =\frac{z}{8} =\frac{8-y}{10} =b$

but then
$x=6a=8-10b$ and $s=10a=8b$

You sould write $y=8-10b$ and $s=6b$

ya i did the work on paper n then i think i switched the variables when i was writing it up on the computer........

let me see, now.......

it should be

[hide]\[ \begin{array}{l} {{\rm Using\; the\; ratios\; of\; the\; similar\; triangles,\; }AQP{\rm \; to\; }ADB: \frac{x}{6} =\frac{y}{8} =\frac{s}{10} =a} \\ {{\rm Using\; the\; ratios\; of\; the\; similar\; triangles,\; }SPB\; {\rm to\; }ADB: \frac{s}{6} =\frac{z}{8} =\frac{8-y}{10} =b} \\ {{\rm Equation\; }1: y=8a=8-10b} \\ {{\rm Equation\; }2: s=10a=6b} \\ {5\times {\rm Equation\; 1: 4}0a=40-50b} \\ {4\times {\rm Equation\; 2: }40a=24b} \\ {74b=40} \\ {b=\frac{20}{37} } \\ {s=6b} \\ {s=6\left(\frac{20}{37} \right)} \\ {s=\frac{120}{37} } \end{array} \][/hide]

well, same answer... i'm gonna edit my first post..


If $x,y,z$ are positive numbers prove that :
$\sum\frac{x}{x+\sqrt{(x+y)(x+z)}} \leq 1$

using lemma:

$\sqrt{(x+y)(x+z)}\geq\sqrt{xz}+\sqrt{xy}$

we get the proof :D

A refinement is

$ \sum{\frac {x}{x \plus{} \sqrt {(x \plus{} y)(x \plus{} z)}}}\leq\frac {2}{3}\sum{\frac {x}{\sqrt {(x \plus{} y)(x \plus{} z)}}}$

$ \leq\frac {8}{27}\sum{\frac {(x \plus{} y \plus{} z)^2}{(x \plus{} y \plus{} z)^2 \minus{} yz}} \leq 1.$

The last one is [b]Bài toán 4.18[/b] at page 130 of the following book:

Phạm Văn Thuận, Lê Vĩ, [url=http://www.mathlinks.ro/viewtopic.php?t=156796]Bất đẳng thức, Suy luận & Khám phá[/url], NHÀ XUẤT BẢN ĐẠI HỌC QUỐC GIA HÀ NỘI, 2007.

See also here : http://www.mathlinks.ro/viewtopic.php?t=124114


Hello everyone. Here is the question which I am trying to do. Thanks for your help.

Use the Biot-Savart law to prove the equation below, which expresses the magnetic field at the center of the Helmholtz coil apparatus as a function of the current, number of coils, and radius of the coil.

Thanks - topic answered
I really would appreciate anyone's help. The hint I was given was "The one piece of information you need is that the Helmholtz coil apparatus is set up such that there are two coils in parallel planes, and the perpendicular separation between the two coils' planes is equal to each coil's radius."

Thank you

Well, from Biot-Savart's law we can get that the magn. induction of a ring (through which passes the current $I$) in a point (call it $M$) on the central axis of the ring is \[ \Delta B=\frac{\mu_0}{2}I\frac{R^2}{(R^2+a^2)^{3/2}},  \] where $R$ is the radius of the ring, $\mu_0$ is the permeability of vacuum, and $a$ is the distance between $M$ and the center of the ring (do you require a proof of this formula?).

For a coil of $N$ turns we get that the induction is $B = N\Delta B = \frac{\mu_0}{2}NI\frac{R^2}{(R^2+a^2)^{3/2}}$. Thus, the total induction produced by two such coils (noting that point $M$ is $a=R/2$ away from each coil) is \[ 2B = \mu_0NI\left(\frac{4}{5}\right)^{3/2}.  \]


calculus
integration
conics
parabola
calculus computations

I am to switch the order of integration on this double integral

$ \int_{0}^{2\sqrt{3}}\int_{{y^{2}}/{6}}^{\sqrt{16-y^{2}}}dx dy$

I have sketched the region as it is and understand it is the intersection of a circle and a parabola in the first quadrant. I do not know how to switch the integral because when I try to, the regions that y varies from starts at $ 0$ and goes to $ \sqrt{6x}$ for the first half and then from $ 0$ to $ {\sqrt{16-x^{2}}}$ for the second part as x goes from $ 0$ to $ 4$

I do not know how to make that into one doulbe integral.

Any help is appreciated

So don't make it one double integral. Make it two; there's nothing wrong with splitting the region.


Wait... the polynomial is of the third degree, so it has 3 roots.  The problem says there are 2 real roots.  This means there is 1 imaginary root.  But don't imaginary roots have to come in pairs... complex conjugates?

Edit: oh, ok, I now see that it just says that they are real roots.  The third root is real as well.

[hide="Simplification"]Note that by Vieta's formula's, $ a\plus{}b\plus{}c \equal{} \minus{}1$ or $ a\plus{}b \equal{} \minus{}(c\plus{}1)$. Plugging into the desired expression, it simplifies to

$ \minus{}c^3 \minus{} c^2 \plus{} 2008c \plus{} 2008$

Note that since $ c$ is a root of $ x^3 \plus{} x^2 \minus{} 2009x \equal{} 232$, we can simplify it as

$ 1776 \minus{} c$[/hide]

It's not like we specified which of the roots a and b were (all 3 roots are real), so c could have 3 different values, and that would give 1776 - c three different values.


The first row of a table of size $ 2005\times5$ is filled with 1,2,3,4,5 so that every two neighbouring cells contain distinct numbers. Prove that it is possible to fill four other rows with 1,2,3,4,5 so that any neighbouring cells in them will contain distinct numbers as well as any cells of the same column will contain pairwise distinct numbers.

can you phrase the question a little more clearly?

I think it means something like...
The top 2005 cells of a 2005x5 table are filled with the integers 1 through 5, inclusive such that no two neighboring cells have the same number. Prove that it is possible to fill the other 4 rows similarly, which each column being a permutation of the integers from 1 through 5.

Am I missing something? seems rather simple. let the number in a cell be called x. write x+1 (mod 5) in the cell below (so beneath a 1 write 2, under 2 write 3.... under 5 write 1). do this for all cells. that way each colummn has five different numbers, and each row has no consecutive numbers that are the same.

BTW, I'm new here, how do you hide your solution?

[code][hide=Clicky me] or you won't see this[/hide][/code]

[hide="Clicky me"] or you won't see this[/hide]


calculus
integration
calculus computations

Integrate!

$ \int \sqrt{e^{2t}+e^{-2t}}\, dt$


Okay, so for some reason, if I don't get the solutions template to work for me, which it seems like now, can I just send the solutions individually in their own files, or is this unrecommended?

Also, since I suck at LaTex, if, for problem 1, I can't get the image to work out to give the example, can I just say "starting from the top corner in clockwise order" and then list the numbers or something or is that also unrecommended?

Finally, if I can't tex some up and just decide to finish it in Word, print it out, and mail it, can I submit some solutions by email along w/ a note saying that only some are by email and that you should grade the ones that come in the mail later on.  I want to do this just in case the mail version doesn't get there, so I at least get some points.

Oh yeah, one more thing.  What happens if we mail it on the right day, but it doesn't get postmarked until the next morning or something?  Do we just die or what?

If you take it to the post office before it closes, it will be postmarked that day.

[quote="vinny919"]Okay, so for some reason, if I don't get the solutions template to work for me, which it seems like now, can I just send the solutions individually in their own files, or is this unrecommended?[/quote]

It's not recommended to do it like that, but it's acceptable.

[quote="vinny919"]Also, since I suck at LaTex, if, for problem 1, I can't get the image to work out to give the example, can I just say "starting from the top corner in clockwise order" and then list the numbers or something or is that also unrecommended?[/quote]

That's fine.  Do note, the LaTeX source file for the Round 1 problems is on the Problems page, so you can use that to generate your image (you'll be able to edit it to stick the numbers you want on the pentagon).

[quote="vinny919"]Finally, if I can't tex some up and just decide to finish it in Word, print it out, and mail it, can I submit some solutions by email along w/ a note saying that only some are by email and that you should grade the ones that come in the mail later on.  I want to do this just in case the mail version doesn't get there, so I at least get some points.[/quote]

Ideally what you'll do here is send [b]everything[/b] by mail, and make clear in the email that your email is just a back-up.  I don't recommend this in general, but if it makes you feel better, ok.  You should [b]not[/b] send some problems by mail and others by email and expect us to grade both.

[quote="vinny919"]Oh yeah, one more thing.  What happens if we mail it on the right day, but it doesn't get postmarked until the next morning or something?  Do we just die or what?[/quote]

Mail it on Monday and you'll be fine.


Βρείτε όλους του μη αρνητικούς ακέραιους $ n$ τέτοιυς ώστε ο αριθμός το $ A_n \equal{} (2^{2n \plus{} 1})^2 \plus{} 1$ να διαρείται από 2 το πολύ δαιφορετικούς πρώτους.

[quote="maniopas"]$ A \equal{} 4(2^n)^4 \plus{} 1^4 \equal{} (2(2^n)^2 \plus{} 1)^2 \minus{} 4(2^n)^2 \equal{} (2(2^n)^2 \minus{} 4(2^n) \plus{} 1)(2(2^n)^2 \plus{} 4(2^n) \plus{} 1)$

Πρέπει τώρα $ p \equal{} 2(2^n)^2 \minus{} 4(2^n) \plus{} 1, q \equal{} 2(2^n)^2 \plus{} 4(2^n) \plus{} 1$ πρώτοι. Μετά;[/quote]

Δεν είναι ανάγκη να είναι πρώτοι. Θα μπορούσαν να είναι και δυνάμεις πρώτων!. Επίσης έχεις κάνει ένα μικρό λαθάκι στην παραγοντοποίηση. Το κανονικό είναι $ A_n\equal{}(2(2^n)^2 \minus{} 2(2^n) \plus{} 1)(2(2^n)^2 \plus{} 2(2^n) \plus{} 1)$


:ninja: need help with a calculus problem and the process to do so. Not neccesarily the answer however.

given that

f(1)=5
f(3)=21
f(a+b)-f(a)=kab+2b^2

find:
k when a=1 and b=2
f '(3)
f '(x), f "(x)

 :) thank you  :)

[quote="kunny"]You can use the definition of the derivative.

$\boxed{f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}}$[/quote]

got it....thank you.  :D


geometry
circumcircle
inequalities
search
national olympiad

1. Find all integer solutions of the equation \[ 12^x\plus{}y^4\equal{}2008^z.\]

2. Triangle $ ABC$ is given. Points $ D$ and $ E$ are chosen on the line $ AB$
such that $ D\minus{}A\minus{}B\minus{}E$, $ AD\equal{}AC$ and $ BE\equal{}BC$. Bisectors of inner angles at points
$ A$ and $ B$ meet opossite sides at points $ P$ and $ Q$, respectively, and circumcircle
of the triangle $ ABC$ at points $ M$ and $ N$, respectively. Lines that connect points
$ A$ and $ B$ with centers of circumcircles of triangles $ BME$ and $ AND$ meet at
point $ X$. Prove that $ CX\perp PQ$.

3. If  $ a$, $ b$ and $ c$ are positive numbers such that $ a\plus{}b\plus{}c\equal{}1$, prove the inequality
\[ \frac{1}{bc\plus{}a\plus{}\frac1a}\plus{}\frac{1}{ca\plus{}b\plus{}\frac1b}\plus{}\frac{1}{ab\plus{}c\plus{}\frac1c}\leq \frac{27}{31}.\]

Please post them in the appropriate subforums and after that write the corresponding links here.

1. Find all integer solutions of the equation
\[ 12^x \plus{} y^4 \equal{} 2008^z.
\]
http://www.mathlinks.ro/viewtopic.php?t=200035

2. Triangle $ ABC$ is given. Points $ D$ and $ E$ are chosen on the line $ AB$
such that $ D \minus{} A \minus{} B \minus{} E$, $ AD \equal{} AC$ and $ BE \equal{} BC$. Bisectors of inner angles at points
$ A$ and $ B$ meet opossite sides at points $ P$ and $ Q$, respectively, and circumcircle
of the triangle $ ABC$ at points $ M$ and $ N$, respectively. Lines that connect points
$ A$ and $ B$ with centers of circumcircles of triangles $ BME$ and $ AND$ meet at
point $ X$. Prove that $ CX\perp PQ$.
http://www.mathlinks.ro/viewtopic.php?t=199935

3. If  $ a$, $ b$ and $ c$ are positive numbers such that $ a \plus{} b \plus{} c \equal{} 1$, prove the inequality
\[ \frac {1}{bc \plus{} a \plus{} \frac1a} \plus{} \frac {1}{ca \plus{} b \plus{} \frac1b} \plus{} \frac {1}{ab \plus{} c \plus{} \frac1c}\leq \frac {27}{31}.
\]
http://www.mathlinks.ro/viewtopic.php?search_id=1640329173&t=199850

4. Each point of a plane is painted in one of three colors. Show that there exists a triangle such that:
$ (i)$ all three vertices of the triangle are of the same color;
$ (ii)$ the radius of the circumcircle of the triangle is $ 2008$;
$ (iii)$ one angle of the triangle is either two or three times greater than one of the other two angles.
http://www.mathlinks.ro/viewtopic.php?p=1107326#1107326

5. The sequence $ (a_n)_{n\ge 1}$ is defined by $ a_1 \equal{} 3$, $ a_2 \equal{} 11$ and $ a_n \equal{} 4a_{n\minus{}1}\minus{}a_{n\minus{}2}$, for $ n \ge 3$. Prove that each term of this sequence is of the form $ a^2 \plus{} 2b^2$ for some natural numbers $ a$ and $ b$.
http://www.mathlinks.ro/viewtopic.php?p=1107330#1107330

6. Let $ ABCDE$ be a convex pentagon in which $ AB \equal{} 1$, $ \angle BAE \equal{} \angle ABC \equal{} 120^{\circ}$, $ \angle CDE \equal{} 60^{\circ}$ and $ \angle ADB \equal{} 30^{\circ}$. Prove that the area of pentagon $ ABCDE$ is less than $ \sqrt 3$.
http://www.mathlinks.ro/viewtopic.php?p=1107334#1107334


integration
function
real analysis
real analysis unsolved

Let $ A: L^2(0,1)\rightarrow L^2(0,1)$ be defined $ Ax(t)\equal{}tx(t)$. Find the adjoint operator of $ A$? is there $ A\equal{}A*$

Yes, this is self-adjoint.

$ \langle Ax,y\rangle\equal{}\int_0^1tx(t)\overline{y(t)}\,dt\equal{}\int_0^1x(t)\overline{ty(t)}\,dt\equal{}\langle x,Ay\rangle.$

Note that pointwise multiplication by any real-valued function will be self-adjoint. (But bounded as an operator if and only if the multiplication is by a bounded ($ \L^{\infty}$ ) function.)


Let $ a,b,c\geq0$.Prove that:
$ a^3\plus{}b^3\plus{}c^3\minus{}3abc\geq2(\dfrac{b\plus{}c}{2}\minus{}a)^3$

We can reduce all of the variables by min(a,b,c) and get a stronger inequality...so we have two cases: a=0, or b=0, both of which should be very easy.

[quote="Inequalities Master"]Let $ a,b,c\geq0$.Prove that:
$ a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc\geq2(\dfrac{b \plus{} c}{2} \minus{} a)^3$[/quote]
Setting $ t\equal{}\frac{b\plus{}c}{2}$, by Holder and AM-GM Inequality, we get
\[ b^3\plus{}c^3 \ge \frac{(b\plus{}c)^3}{4}\equal{}2t^3\]
\[ \minus{}bc \ge \minus{}\frac{(b\plus{}c)^2}{4} \equal{}\minus{}t^2\]
It suffices to prove that
\[ a^3\minus{}3at^2 \plus{}2t^3 \ge 2(t\minus{}a)^3\]
\[ \Leftrightarrow (a\minus{}t)^2(a\plus{}2t) \ge 2(t\minus{}a)^3\]
\[ \Leftrightarrow 3a(a\minus{}t)^2 \ge 0\]
which is true. :)
Equality holds if and only if $ a\equal{}b\equal{}c$ or $ a\equal{}0,b\equal{}c$ :)


inequalities
geometry
3D geometry
logarithms
inequalities unsolved

Prove that if x,y, and z are positive real numbers, such that $ xyz\equal{}1$, then

$ \sqrt{1\plus{}8x}\plus{}\sqrt{1\plus{}8y}\plus{}\sqrt{1\plus{}8z} \ge 9$.

Applying AM-GMinequality with 3 variables above we have:
LHS>=3 CUBE ROOT[square root(1+8x)(1+8y)(1+8z)](1)
Moreover I have:
(1+8x)(1+8y)(1+8z)=1+8(x+y+z)+64(xy+yz+xz)+8^3xyz>=1+24+192+8^3=729
Plug in this into (1) it's obviously true! :)

[quote="mathgeniuse^ln(x)"]Prove that if x,y, and z are positive real numbers, such that $ xyz \equal{} 1$, then

$ \sqrt {1 \plus{} 8x} \plus{} \sqrt {1 \plus{} 8y} \plus{} \sqrt {1 \plus{} 8z} \ge 9$.[/quote]
The following inequality is true too:
Let $ x,$ $ y$ and $ z$ are non-negative numbers such that $ \sqrt x\plus{}\sqrt y\plus{}\sqrt z\equal{}3.$ Prove that
\[ \sqrt {1 \plus{} 8x} \plus{} \sqrt {1 \plus{} 8y} \plus{} \sqrt {1 \plus{} 8z} \ge 9\]

[quote="arqady"][quote="mathgeniuse^ln(x)"]Prove that if x,y, and z are positive real numbers, such that $ xyz \equal{} 1$, then

$ \sqrt {1 \plus{} 8x} \plus{} \sqrt {1 \plus{} 8y} \plus{} \sqrt {1 \plus{} 8z} \ge 9$.[/quote]
The following inequality is true too:
Let $ x,$ $ y$ and $ z$ are non-negative numbers such that $ \sqrt x \plus{} \sqrt y \plus{} \sqrt z \equal{} 3.$ Prove that
\[ \sqrt {1 \plus{} 8x} \plus{} \sqrt {1 \plus{} 8y} \plus{} \sqrt {1 \plus{} 8z} \ge 9
\]
[/quote]

Would the best way of approaching this problem, be AM-GM, after expanding?

[quote="mathgeniuse^ln(x)"]

Would the best way of approaching this problem, be AM-GM, after expanding?[/quote]
Which expanding? I don't understand, what do you mean. :wink:

[quote="mathgeniuse^ln(x)"]Prove that if x,y, and z are positive real numbers, such that $ xyz \equal{} 1$, then

$ \sqrt {1 \plus{} 8x} \plus{} \sqrt {1 \plus{} 8y} \plus{} \sqrt {1 \plus{} 8z} \ge 9$.[/quote]

Let $ f(x) \equal{} \sqrt {1 \plus{} 8x} \minus{} \frac {4}{3}\ln x\ (x > 0)\Longrightarrow f'(x) \equal{} \frac {4(x \minus{} 1)}{3x(3x \plus{} \sqrt {1 \plus{} 8x})\sqrt {1 \plus{} 8x}}$, yielding $ f(x)\geq f(1) \equal{} 3.$
$ \therefore \sqrt {1 \plus{} 8x} \plus{} \sqrt {1 \plus{} 8y} \plus{} \sqrt {1 \plus{} 8z}\geq \frac {4}{3}\ln xyz \plus{} 3\cdot 3 \equal{} 9$. Q.E.D.

[quote="mathgeniuse^ln(x)"]Wow, nice.

Was there a reason that 4/3 came to mind, and not another number?

like, how did you know that 4/3 would work, and not anything else.[/quote]

put 4/3=a and choose a so it works.


A circle of unit radius is to be covered by three circles of equal radius.
Find the minimum radius required.

The minimum is m =  (\sqrt 3)/2.

Let D be a disk with unit radius.
It is not difficult to give an example of such a covering (but a little boring in html)

Now suppose that a covering does exist with three disks, say D_1, D_2 and D_3, each with radius r < ( \sqrt 3)/2.
Let D be the unit disk.
Suppose that D_1 covers an arc AB of the boundary of D whose length is at least 1/3 of the boundary of D, that is 2 \pi /3 which leads to AB >  \sqrt 3 > 2r = diameter of D, a contradiction.
Thus, D_1 covers an arc of the boundary of D which is less than 1/3 of the boundary of D. We have analoguous results for D_2 and D_3, and then these three disks cannot cover D. A contradiction.

Pierre.


let x, y be two positive odd integers. Can you show that it is impossible that the value of x^2+y^2 to be a perfect integer.

[hide]because $x$ and $y$ are odd, so we can say that $x^{2}=8k+1$ and $y^{2}=8q+1$. so we have $x^{2}+y^{2}=8(k+q)+2=2(4(k+q)+1)$ well its obvious that a number in this form can't be a perfect square....[/hide]

When you say 'a perfect integer', do you mean a perfect number such as 6, or a perfect square?

[hide="If the first"]
It is know that all odd square are congruent to $1 \mod 4$. Therefore $x^{2}+y^{2}\equiv 2 \mod 4$. Therefore, it is even. It is know that any even perfect number is of the form $2^{n-1}\cdot(2^{n}-1)$. Now since $x^{2}+y^{2}\equiv 2 \mod 4$, it only has one factor of 2. Therefore if it was of the form ${2^{n-1}(2^{n}-1)}$ we would have $n=2$, which gives 6. This is not expressible as the sum of odd squares.[/hide]

[hide="If the second"]
Again, $x^{2}+y^{2}\equiv 2 \mod 4$. But all squares are ccongruent to 0 or 1 mod 4, so $x^{2}+y^{2}$ is not asquare.[/hide]

[quote="sinajackson"]Is my solution wrong?
[hide]if $x$ is odd then $x^{2}$ can only be congurent to $1$ $mod$ $8$[/hide][/quote]

Why would it be wrong?

if $x$ is odd then: $x\equiv 1,3,5,7$  $(mod\ 8)$ and so $x^{2}\equiv 1,9,25,49\equiv 1,1,1,1$ $(mod\ 8)$. So my solution is correct too.... :)


There are exactly 3 integral values for x for which 4^x + 4^1993 + 4^1996 is a perfect square. Find these values of x.

[hide]If $ x<1993$, then $ 4^x(1\plus{}4^{1993\minus{}x}\plus{}4^{1996\minus{}x})$ is a perfect square. Since $ 4^x$ is a perfect square, then $ (1\plus{}4^{1993\minus{}x}\plus{}4^{1996\minus{}x})$ is a perfect square. We use the fact that $ 4y^2\plus{}4y\plus{}1\equal{}(2y\plus{}1)^2$. Hence, $ y\equal{}4^{1992\minus{}x}$ and $ y^2\equal{}4^{1995\minus{}x}$, so then $ 1995\minus{}x\equal{}2(1992\minus{}x)$ and $ \boxed{x\equal{}1989}$.

If $ 1993<x<1996$, then $ 4^{1993}(1\plus{}4^{x\minus{}1993}\plus{}4^3)$ is a perfect square. Using the same identity as above, $ (3\minus{}1)\equal{}2(x\minus{}1993\minus{}1)$ so $ \boxed{x\equal{}1995}$.

If $ x>1996$, then $ 4^{1993}(1\plus{}4^3\plus{}4^{x\minus{}1993})$ is a perfect square. Using the same identity as above, $ (x\minus{}1993\minus{}1)\equal{}2(3\minus{}1)$, so $ \boxed{x\equal{}1998}$.[/hide]


If $ \frac{x}{y}\equal{}\frac{3}{5}$ then compute $ \frac{3x\plus{}7y}{4x\plus{}2y}$.

hello, plugging $ x\equal{}\frac{3}{5}y$ in your term we get
$ \frac{\frac{9}{5}y\plus{}7y}{\frac{12}{5}y\plus{}2y}\equal{}\frac{9y\plus{}35y}{12y\plus{}10y}\equal{}\frac{44y}{22y}\equal{}2$.
Sonnhard.

$ x\equal{}\frac{3}{5}y$. Substituting it into the fraction gives

$ \frac{\frac{9}{5}y\plus{}7y}{\frac{12}{5}y\plus{}2y}\equal{}\frac{9\plus{}35}{12\plus{}10}\equal{}\frac{44}{22}\equal{}\boxed{2}$.

EDIT: beaten...blah

Assume $ x\equal{}3$ and $ y\equal{}5$, plug in, $ \frac{9\plus{}35}{12\plus{}10}\equal{}\frac{44}{22}\equal{}\boxed{2}$ :D


Can someone show me what to do here?

I'm given two circles (C1 and C2) and a line L. I need to find a line, which is parallel to L, such that the distance between the points where this line intersects C1 and C2 is equal to the length of a given segment (call it YZ).

Thanks for your help in advance.

I would think you make your ruler a marked ruler with YZ and try verging. Does this work?


Evaluate $ \int \frac{\sqrt{(x\minus{}1)^{3}}\plus{}\sqrt{(x\plus{}1)^{3}}}{\sqrt{x\left( x^{2}\minus{}1\right) }\left( \sqrt{x\plus{}1}\plus{}\sqrt{x\minus{}1}\right) }dx$      :ninja:

Using $ a^{3} \plus{} b^{3} \equal{} (a\plus{}b)(a^{2} \minus{} ab \plus{} b^{2})$, I get

$ I \equal{} \int\,\frac {(\sqrt {x \plus{} 1} \plus{} \sqrt {x \minus{} 1})(x \minus{} 1 \minus{} \sqrt {(x^{2} \minus{} 1)} \plus{} x \plus{} 1)}{\sqrt {x(x^{2} \minus{} 1)}(\sqrt {x \plus{} 1} \plus{} \sqrt {x \minus{} 1})}\, dx$

$ \equal{} \int\,\frac {2x \minus{} \sqrt {x^{2} \minus{} 1}}{\sqrt {x(x^{2} \minus{} 1)}}\, dx$

$ \equal{}\int\, \frac{2x}{\sqrt{x(x^{2}\minus{}1)}}\,dx \minus{}\int\frac{1}{\sqrt{x}}\,dx$

$ \equal{}\int\,\frac{2x}{\sqrt{x(x^{2}\minus{}1)}}\,dx \minus{} 2\sqrt{x}$

and have come to a stand-still.

Interesting approach, now you can work on the remaining integral.


My name is ugo,the math level in our nation is nothing to write home about.like we just concluded uor state qualifiers and i came third with a 65/100.i need help on how best to understand math. thanks.By the way check this out even books to read is a problem (no good books)is engeneering math good.?what of pure math for advance level?  :?:

If you're a beginner, get AoPS Volume 1 with the solutions book and once you finish that, go to AoPS Volume 2.  Or, consider getting the individual books.


topology
real analysis
real analysis unsolved

Let $(X,d)$ be a metric space. If $f: X \rightarrow X$ statisfies the condition \[d(f(x),f(y))=d(x,y)\]
for all $x,y \in X$, then $f$ is called an isometry of $X$. Show that if $f$ is an isometry  and $X$ is compact, then $f$ is bijective and hence homeomorphism.

A simple example to show that compactness is needed. Consider $X=[0,\infty)$ with the usual topology and let $f(x)=x+1.$ This is an isometry but not surjective.

This problem was discussed here: http://www.mathlinks.ro/Forum/viewtopic.php?t=117359 . (More precisely, it was proven there that $f$ is surjective. The fact that $f$ is injective is obvious.)


parameterization
vector
probability and stats

Suppose that X(i), i=1,…,n are independent Poisson random variables with common parameter T, and suppose that T is an exponential random variable with parameter L. Find:
1)	The distribution of the random vector (X(1), …., X(n)). (Your answer should depend on L but not T)
2)	The conditional distribution of T, given X(i)=k(i) for some fixed set of values k(i).

I am not sure how to tackle a vector distrbution of r.vs. I wrote out a vector of distributions, but that is all I have.

For the second part, I have not exactly attempted to fully write it out yet, but it will involve the joint density theorem.


Let a_i, i=1,n be the lengths of some segments.
Determine the maximum area of a poygon that can be obtained using these segments.

1 year...

It is time to give a hint.
[hide]Prove that it should be an inscribed polygon.[/hide]
Though I have some doubts we can find closed formula for the area.

[quote="DusT"]I don't want to seem  :moose: but is it at least possible for any $a_i$'s that respect the necessary inequalities for the polygon to exist, there also exists an inscribed polygon?[/quote]
As long as $\sum a_i>2a_i$(the condition for the polygon to exist) for any i, there exists a cyclic polygon with sides $a_i$.

Here's something which might be of interest:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=isoperimetric&t=22481[/url]


On average, are the AHSME tests from 1990-1999 harder, easier, or about the same as the AMC 12 tests.

Thanks, and Merry Christmas.

I would say probably easier, since the newer tests tend to have problems that test your creative problem abilities more, if you know what I mean.

[quote="MathTwo"]Thanks. Do you think it would be a good idea to do AHSME tests as preparation for the AMC 12 though.[/quote]

Yes.

In general, it's a good idea to study for the tier one level higher than what you are hoping to pass. For example, if you want to do well on the AMC 12, it would be good to study AIME problems, and if you want to score high enough on the AIME to make USAMO, you should study USAMO or other olympiad-level problems. This way, you get challenged more, and can improve more quickly. If you feel that AHSME problems provide enough challenge, there's nothing wrong with studying them, though.

Merry Christmas everyone!

A little history: at first, calculators were prohibited on the AHSME, but a decision was made to begin allowing calculators. I believe this decision came into effect in 1994. Possibly because of that transition (or possibly for other reasons), they got the calibration of difficulty wrong for that 1994 test. That one year saw what was probably the single easiest AHSME/AMC12 ever. So if you're trying out a 1994 test, bear that in mind.

But the effect wasn't permanent; they got the difficulty level back under control immediately, and the AHSMEs of the late 90's were pretty challenging.

One thing I've noticed from the lists of perfect scores: apart from the 1994 aberration, it was significantly harder to get a perfect score on the AHSME than on the AMC12. I think the primary reason is simply the number of questions. The AHSME (at least in the last few years) was 30 questions in 90 minutes, as opposed to the AMC12 25 questions in 75 minutes. And it's just harder to get 30 questions right than 25. The back pages of those AHSME exams, the last five or so questions, usually overlapped with the AIME in terms of the challenge posed by each individual question.

Now the AHSME's of my generation (I was in high school in 1969-70-71) were significantly different, with more questions, questions of varying point values, and subject matter that was a little closer to standard classroom math than has been true in more recent years.


algebra
binomial theorem
Pascal\u0027s Triangle

What is $ {n\choose 0}+{n\choose 1} + {n\choose 2} + {n\choose 3} + ... {n\choose n} $ In terms of $n$

are you sure it's not
[hide] (2^n)-1? [/hide]

[quote="sirorange"]Ahh, chess, it starts at n choose 1, not n choose 0, but otherwise, you would be correct.[/quote]

Oops, I didn't see that. But if I remember correctly, the problem actually included $n \text { choose } 0$.

I didn't know MC would expect students to know the binomial theorem, but I suppose there are other counting approaches as well.

With the 3 minutes given, you could figure it out... (assuming it's target)... just plug in numbers... you would get 0,1,3,7,15... not that hard... you don't even need to know the binomial theorem to do it, although it would save you about 3 minutes...

[quote="chess64"][quote="biffanddoc"][quote="sirorange"]And wasn't it a team round problem?[/quote]Target[/quote]

Team. (1985 State Team #5)

MC's solution: "The middle-school approach to this problem is to solve it via pattern recognition...... A more advanced solution reuires recognizing that the expression is equivalent to the $n$th row of Pascal's triangle, and the sum of the numbers in the $n$th  row of Pascal's triangle is equal to $2^n$."[/quote]O u might be right. I think it was on HAYMATH #3, but thats really easy for a team round problem.

No such problem was on Haymath #3...

My coach talked a lot about Pascal's triangle. Lots of stuff in it.

Some old target or other contest. I forget where it was from

you can think of it in terms of coin flipping

If you flip a coin n times the possibilities for
0 heads=nC0/2^n
1 heads=nC1/2^n
2 heads=nC2/2^n
3 heads=nC3/2^n...

There must be a total of 2^n possibilities.


function
logarithms
algebra unsolved
algebra

Hello:

Suppose we had something like $ x^{2}=e^{x}$.  What more advanced methods could be used to solve this algebraically, besides graphing it and then zeroing in on the solutions by Intermediate Value, Newton's method, etc?.

Would the LambertW come into play here?.

Thank you

The Lambert $ W$ function is not so much an advanced algebraic method as a cop-out.  Since the solutions of many exponential equations can't be expressed in terms of elementary functions, we defined another function $ W$ to express them.  In any case, taking the natural logarithm yields $ 2 \ln x = x$.  Letting $ u =-\ln x$ we have

$ -2u = e^{-u}\Leftrightarrow$
$ ue^{u}=-\frac{1}{2}\Leftrightarrow$
$ u = W \left(-\frac{1}{2}\right) \Leftrightarrow$
$ x = e^{-W \left(-\frac{1}{2}\right) }$


[quote="GameBot"]Charles Schultz drew 18,000 comic strips over a 50-year period.
What is the mean number of comic strips he drew each year?[/quote]

Our answer is simply $ 18000 \div 50 \equal{} \boxed{360}$.


Discuss, well, what the topic says.

TAYLOR SWIFT!!!!!!!!! lol

Love Story.

Kevin Rudolf- "Let it Rock"
my favorite for now. beast song

a while back my favorite was "superstar" by lupe fiasco

TAYLOR SWIFT ALL THE WAY!!!!!!!!!!!! and...For favorite song, I have absolutely no idea... :P

hmm... tough question.

I like "Our house -- Crosby, Stills, and Nash", "Imagine -- John Lennon", Anything by the Beatles, "Summer of '69 -- Byran Adams", Various others that are too long to post here

Sunset Glow/Heaven--Big Bang

im not sure which one is better though

as for american songs/bands i have to say

[u]Bands[/u]
Sum41
Linkin Park
Avril Lavigne
OAR
All-American Rejects

Favorite songs i have no clue

er. more of an r/b and rap guy.

Mad-Ne-Yo, anything by Kanye or Lupe (Heartless, Love Lockdown, Dumb it Down, Superstar, Hip-Hop Saved my Life), Right Now (Na Na Na) by Akon, uhh... Remember the Name is a pretty good song.

idk. that's some.

Shout at the Devil-song
Vince Neil-singer

(Motley Crue)

oooo tough.
i have wayyyy too many; but i think all time fave is probably
graduation - vitamin c.

but i have 'phases' where i listen to a song over and over and over for a period of time. currently: if today was your last day-nickelback.

taylor swift is pretty amazing too. boys like girls.

Adam Gontier - best singer.

Best song: Bleeding Mascara (by Atreyu)

Wanted Dead or alive - Bon Jovi
or Face me - Edsim & Mahlta

I have too many bands :(

Oh, goodness...so many...I'll have to go with Miley Cyrus "Hoedown Throwdown".

Oh !
What about others here ?

I like "Nothing Else Matters", "Sweet Amber" and "Fade To Black" - All are Metallica!

Um... You just revived an one year old thread.

I don't have a fav singer, but I have a fav song: Just Dance by Lady Gaga. I reccomend it to everyone!!!!!!!!! Don't like it just becuz its by Lady Gaga. I'm not a fan of her, one of my relatives showed me this song and I got addicted. My 2nd fave = If I Can't Have You by Kelly Clarkson. I like it almost as much as Just Dance.

Hip Hop guy.

Eminem ft. Rihanna-Love the Way You Lie

It's a new one.
And it's gonna be hot.

Rucka Rucka Ali

Down

[url]http://www.youtube.com/watch?v=A-54SW02y6g&feature=related[/url]

hillarious

Beatles: ALL OF THEM  :w00t:
Michael Jackson: Smile, Beat It, all the good ones
Frank Sinatra: Summer Wind

Yeah, I like old songs.

Lady Gaga: Paparazzi, Poker Face
Don McLean: American Pie

This should be posted in the music section.


trigonometry
inequalities proposed
inequalities

[quote="kunny"][url=http://www.mathlinks.ro/Forum/viewtopic.php?t=142583]see here[/url][/quote]
Thanks,kunny :)


Suppose the sequence $ a_{1},a_{2},...,a_{n}$ satisfy
$ a_{1}\equal{}0,\left|a_{k\plus{}1}\right|\equal{}\left|a_{k}\plus{}1\right|$,for $ k\equal{}1,2,...,n\minus{}1$.
Prove that $ a_{1}\plus{}a_{2}\plus{}...\plus{}a_{n}\geq\minus{}\frac{1}{2}n$.

I solved the problem but in the case where k takes on the values through n.If you square both sides of the given equality and then sum,it is a telescoping sums where all the squares are deleted and the rest is the equality:
                               0=<a(n+1)^2=2(a1+a2+...+an)+n
as desired.
Is this problem solvable by considering the values of k through n-1?


[quote="saracampbellsoup"]hello, i'm in the new hampshire group.  right now it seems like only the exeter math club is using it, so i wanted to make a stickie so everyone will know who has what username.  do i have the power to do this?  i can't find a button for it =b
also is there any way i can see a list of the members?[/quote]

Only moderators and admins can see the members list.

If you make a post there and request that it gets stickied, the moderator can make it sticky.  However, we discourage students from revealing their identities in the forums, even in the local communities.

[quote="bubala"]Umm... Mr. Crawford, I can see the usernames of everyone in every national and local community by clicking on usergroups and then selecting the community.  Is this not supposed to happen?[/quote]You can only see the ones that are public. It is normal.


Asymptote
rotation
geometry
analytic geometry

I were Trying to make a asymptote code to make the first Fermat point.
But there were somethings I thougth could be improved.

See Image:
[asy]size(220);
void DrawTriangle(pair A, pair B, pair C, string a, string b, string c){
 D( A--B--C--cycle);
 MP(a,D(A),dir(C--A,B--A));
 MP(b,D(B),dir(C--B,A--B));
 MP(c,D(C),dir(A--C,B--C));}

path Equi(pair A, pair B, pair C, string m){
 pair M1,M2,M;
 M1 = IP(rotate(+60,A)*(A--B),rotate(-60,B)*(B--A));
 M2 = IP(rotate(-60,A)*(A--B),rotate(+60,B)*(B--A));
 if (d(M1,C) > d(M2,C)) {M=M1;} else {M=M2;}
 D(A--MP(m,D(M),dir(A--M,B--M))--B,red+dashed);
 D(circumcircle(A,B,M), green);
 path K = D(M--C,blue+dashdotted);
 return K;}

pair FermatPoint(pair A, pair B, pair C, string f){
 path N1,N2,N3;
 N1=Equi(A,B,C,"N_1");
 N2=Equi(C,A,B,"N_2");
 N3=Equi(C,B,A,"N_3");
 pair F = MP(f,D(IP(N1,N2)));
 return F;
}

pair A, B, C;
A = (0,0);
B = (5,2);
C = (3,4);
DrawTriangle(A,B,C,"A","B","C");
FermatPoint(A,B,C,"F");[/asy]

In the definition of $ M$ if one change > to <, it were expected to find the second Fermat point, but it did not find, because the path did not intersept.
I could make the compand IP3(A,B,C) (Intersection of 3 path, circles) based on IP. but this would make the code bigger.

Someone have any idea of how to solve this without IP3 or without extending the path $ AN_3$ ... to a bigger line?

Hi,

I do not answer to your question directly but I propose an alternative with the [url=http://svnweb.tuxfamily.org/listing.php?repname=piprim/asymptote&path=%2Fbranches%2F&rev=0&sc=0]latest version de geometry_dev[/url]
[code]
import geometry_dev;
size(12cm);

point A=(0,0), B=(5,2), C=(3,4);

triangle t=triangle(A,B,C),
et1=triangle(A,B,rotate(60,A)*B),
et2=triangle(B,C,rotate(60,B)*C),
et3=triangle(C,A,rotate(60,C)*A);

draw(et1^^et2^^et3, 0.8*red+dashed);
draw(t); label(t, alignFactor=2.5);
draw(circle(et1)^^circle(et2)^^circle(et3), 0.8*green);

draw(line(C,et1.C)^^line(A,et2.C)^^line(B,et3.C),
     0.8*blue+dashdotted);
label("$N_1$",et1.VC);
label("$N_2$",et2.VC);
label("$N_3$",et3.VC);

point[] F=fermat(t);
dot("$F_1$",F[0], S, red);
dot("$F_2$",F[1], W, purple);
[/code]

actually I did not find the Fermat Point threougth the constructions made. But this is simplistic to be donne. Thanks for the code.

In your code I only didn't understand 2 things.

et2.C and et1.VC.

What do C and VC make do? Because they seems the same thing.

And the last noob question. What's the difference between pair and point?

Thanks.

[quote="EulerIntegral"]actually I did not find the Fermat Point threougth the constructions made. But this is simplistic to be donne. Thanks for the code.[/quote]
In geometry_dev the definition of the routine [b]fermat(triangle t)[/b] is:
[code]
point[] fermat(triangle t)
{
  point[] P;
  real A=t.alpha(), B=t.beta(), C=t.gamma();
  P.push(point(trilinear(t,1/Sin(A+60), 1/Sin(B+60), 1/Sin(C+60))));
  P.push(point(trilinear(t,1/Sin(A-60), 1/Sin(B-60), 1/Sin(C-60))));
  return P;
}[/code]
Are there errors ?

[quote="EulerIntegral"]
et2.C and et1.VC.
What do C and VC make do? Because they seems the same thing.
[/quote]
triangle.C is a point (you can assimilate it to a pair in my exemple);
triangle.VC is a vertex, it refers to the triangle to which it belongs.
So, bisector(triangle.C) is a nonsense while bisector(triangle.VC) makes sense.
Here an example (also note triangle.AB):
[asy]size(8cm,0);
import geometry_dev;

triangle t=triangleabc(3, 4, 5);
draw(t);
draw(bisector(t.VA));
draw(bisector(t.VC,90), red);

draw(bisector(t.AB), blue);[/asy]

[quote="EulerIntegral"]
And the last noob question. What's the difference between pair and point?
[/quote]
A [i]pair[/i] is defined relatively to the default coordinates system of Asymptote while a [i]point[/i] is defined relatively the current coordinate system of geometry_dev: [b]currentcoordsys[/b].
geometry_dev is entirely coded with [i]point[/i] to allow to work in other coordinates systems.
Here a simple example:
[asy]size(8cm,0);
import geometry_dev;
show(currentcoordsys);

currentcoordsys=cartesiansystem((0.5,0.5), i=(0.7,0.7), j=(-0.7,0.7));
show("$O'$", "$\vec{u'}$", "$\vec{v'}$", currentcoordsys, xpen=invisible);

pair a=(0,0.5), b=(1,0.5);
point A=a, B=b;

dot("$pair(0,0.5)$",a);
dot("$pair(1,0.5)$",b);

dot("$point(0,0.5)$",A);
dot("$point(1,0.5)$",B);[/asy]

PS: my english is poor; so excuse me if I misunderstand your remarks.

Nowerdays the things are compiled by AoPS. (You have to substitue the line [b][i]import geometry_dev;[/i][/b] by [b][i]import geometry;;[/i][/b])

[asy]
import geometry;//_dev;
size(12cm);

point A=(0,0), B=(5,2), C=(3,4);

triangle t=triangle(A,B,C),
et1=triangle(A,B,rotate(60,A)*B),
et2=triangle(B,C,rotate(60,B)*C),
et3=triangle(C,A,rotate(60,C)*A);

draw(et1^^et2^^et3, 0.8*red+dashed);
draw(t); label(t, alignFactor=2.5);
draw(circle(et1)^^circle(et2)^^circle(et3), 0.8*green);

draw(line(C,et1.C)^^line(A,et2.C)^^line(B,et3.C),
     0.8*blue+dashdotted);
label("$N_1$",et1.VC);
label("$N_2$",et2.VC);
label("$N_3$",et3.VC);

point[] F=fermat(t);
dot("$F_1$",F[0], S, red);
dot("$F_2$",F[1], W, purple);
[/asy]

[asy]size(8cm,0);
import geometry;//_dev;
triangle t=triangleabc(3, 4, 5);
draw(t);
//label(t);
draw(bisector(t.VA));
draw(bisector(t.VC,90), red);
draw(bisector(t.AB), blue);
[/asy]

[asy]
size(8cm,0);
import geometry;//_dev;
show(currentcoordsys);
currentcoordsys=cartesiansystem((0.5,0.5), i=(0.7,0.7), j=(-0.7,0.7)); show("$O'$", "$\vec{u'}$", "$\vec{v'}$", currentcoordsys, xpen=invisible);
pair a=(0,0.5), b=(1,0.5);
point A=a, B=b;
dot("$pair(0,0.5)$",a);
dot("$pair(1,0.5)$",b);
dot("$point(0,0.5)$",A);
dot("$point(1,0.5)$",B);
[/asy]


geometry parallelogram trapezoid trigonometry

In parallelogram $ ABCD$, $ AB$ = 16 cm, $ DA$ = $ 3\sqrt{2}$ cm, and
sides $ AB$ and $ DA$ form a 45-degree interior angle. In isosceles
trapezoid $ WXYZ$ with $ WX$ $ \neq$ $ YZ$, segment $ WX$ is the longer
parallel side and has length 16 cm, and two interior angles each
have a measure of 45 degrees. Trapezoid $ WXYZ$ has the same
area as parallelogram $ ABCD$. What is the length in centimeters of segment $ YZ$?

$ S_{ABCD}\equal{}AB\cdot AD\cdot sin 45\equal{}48$
$ S_{WXYZ}\equal{}\frac{WX\plus{}YZ}{2}\cdot (\frac{WX\minus{}YZ}{2}\cdot tg 45)\equal{}48$
So $ YZ\equal{}8$

I used $ S_{parallelogram}\equal{}ab sin\alpha$ where $ \alpha$ is angle between $ a$ and $ b$,
and also $ S_{trapezoid}\equal{}\frac{a\plus{}b}{2}\cdot h$ where $ h$ is height.

Shouldn't there be a way w/o trig, because MC isn't supposed to need it?

[color=blue][size=75]No need to quote the entire post before you. It is assumed you are responding to the post directly before you. -Isabella2296[/size][/color]

You can use that in right-angled triangles with one angle 45 degrees, 2 sides are equal.
What I used is the same, just I wrote it shorter (with trig)


What's your typing speed? Mine is around 105 WPM, and the highest i've hit is 124. My friend does 137. :o What's yours?

I"m not like addicted to typing, however, I do lots of programming so I type about 75-80 wpm.

I can normally get about 80 to 90, but if you're counting mistake correction, then about 75.

My record is 115.
I type fast-ish.. but typo a lot.

about 45-60.

I'm not a fast typer, yet I can type accurately.

I typo a lot, so 66-75

But I still haven't really gotten [, ], and most of the numbers/other symbols.

Mine's about 60 avg.

but the max I've hit is 146 on 'we went to the store'.

100 something WPM.
:)
Example;
[hide] I I I I I I I I I I I I I I I I I I I etc.... Just kidding.
Real Example;
[hide] I am typing a 100+ WPM. While I do this I will explain my day.... so like my mom went to work but first she dropped me off at my aunt's house where she then dropped me off at school. I had an ASB meeting so I went there... the art teacher is our advisor and she made CINNAMONBUNZ! Yay... so anyways I ate my cinnamon bun. It tasted like strawberries. HA. So then we discussed proposals. It was all like the secretary read them and we took a vote... etc. and like we had 2/10 agreed on. Nice right? So then after that I went to class.... wait is this already a 100+ words? Okay. Done :D
[/hide][/hide]
Har har.

[quote="bluecarneal"]In the poll ^   :wink:[/quote]

Smart fellow, do I vote for 76-85 or 85-99?

[quote="mathelete"]words per minute is not a good way, cause some words are longer. now if you are talking about small words like "work" or "walk", i would say about 80[/quote]
Yeah, so uh, like...
When we say "words" per minute, we are counting a word as five letters.

For some reason, I'm basically stuck at the 85 wpm level. Is their any way I can get up to 130?

34 wpm max the last time I checked (which was 3 years ago)

I'm too lazy to measure right now, since I memmed part of the typing test

103. I'm usually around 110-120, guess I'm a bit slow today. Record is 130ish.

I usually get around 80 to 90.
If the sentences are really easy, I get in the hundreds.

I'm a slow typer, usually somewhere around the 56-65. Record 78.

I average like 135-140 ish and I think my max was 155. I don't remember exactly what it was. I think it comes from playing the piano and violin haha.

YAY!
Improved 120 :)
But my max is like 160


inequalities function calculus algebra polynomial

The theorem in KK's packet (#9) about how any inequality in symmetric functions in n variables implies the same inequality in n+1 variables seems useful, but I don't understand the proof he gives...

Also, the proof uses calculus, so will the graders get angry if I use something like it on an olympiad?  What about inequalities like Jensen, if I prove a function is convex using calculus, will they hate me and give me a "7 or 0"?

Thank you!

----OR--

You can use calculus, or any other form of "higher mathematics", on olympiads.

Yes, calculus is okay.

I know that proof looks rather magical: it took me a bit to figure out the way I saw it.  The point is that if you have n positive reals, you can find n-1 positive reals that have the same symmetric averages up to d_(n-1) (of course, d_n will be 0 for n-1 reals).  The way you do that is take the polynomial with your n positive reals as roots, differentiate it, and take the roots of the new polynomial (this is where the Rolle's theorem part comes in, to show that you still get positive reals).  This works because the coefficients of the polynomial are symmetric sums of the roots (with proper signs).


limit
real analysis
calculus
calculus computations

Please help!I have a exam next day and i dont know how to prove it during revision.

Suppose that fn(x) converge to f(x) on [a,b] and that |f'n(x)|<=M for all n and x in[a,b]
show that fn(x) converges to f(x) uniformly on [a,b]

$|f_n(x) - f_n(y)| = |f'(\xi)||x-y| \leq M|x-y|$ so $f_n$ is uniformly continuous. $f$ is also uniformly continuous since $|f(x)-f(y)| = |\lim_{n\to\infty} (f_n(x) - f_n(y))| = \lim_{n\to\infty} |f_n(x) - f_n(y)| \leq M|x-y|$

So, choose $\epsilon > 0$ and $\delta > 0$ so that $|f(x)-f(y)|<\epsilon$ and  $|f_n(x)-f_n(y)|<\epsilon$ when $|x-y| < \delta$. Partition $[a,b]$ into intervals $a^1, a^2,...,a^k$ so that $|a^i| < \delta$ for every $i$. Now choose the midpoints $\xi_1, ..., \xi_k$ for every interval $a_i$. There exists an N such that $|f(\xi_i) - f_n(\xi_i)| < \epsilon$ if $n > N$ for every $i$ due to the pointwise convergence.

Consider a point $x \in a^i$. We have that $|f(x) - f_n(x)| \leq |f(x)-f(\xi_i)| + |f(\xi_i) - f_n(\xi_i)| + |f_n(\xi_i) - f_n(x)| \leq 3\epsilon$ if $n > N$. Therefore, for any point $x \in [a,b]$ $|f(x) - f_n(x)| < 3\epsilon$ if n > N. This is equivalent to the uniform convergence we want.


[quote="Utkirstudios"]If a,b,c,d are real numbers, prove that:
\[6(a^{2}+b^{2}+c^{2}+d^{2})+(a+b+c+d)^{2}\ge 12(ab+bc+cd) \]
[/quote]
Yes, it's true. The equality holds when $a=d$ and $b=c=2d.$

It's problem 38 from Algebraic Inequality (p. 10 and 43), where there are 2 solutions.

Vasc
how we can buy your book
we have visa card please help me to buy your books
I'm inerested in your books, but I can't found it
let's contact
my email: Utkirstudios@gmail.com
thanks !!!

[quote="Utkirstudios"]Vasc
how we can buy your book
[/quote]
You can buy the book from Gil Publishing House.
e-mail: sales@gil.ro
See
http://www.mathlinks.ro/Forum/viewtopic.php?t=100399


Given that x, y and z are positive real numbers satisfying xyz=32, find the minimum value of
x 2  + 4xy + 4y 2  + 2z 2

Fine, but how does one get this result??


linear algebra
matrix
analytic geometry
inequalities
triangle inequality

[u]Leaderboard[/u]
Mysticterminator: 11 points
Cats: 1 point
Not that scores matter...

[u]Round 4 (somewhat expanded)[/u]
1970 2) Let c>a:ge:b be the lengths of the sides of an obtuse triangle. Prove that c^n = a^n + b^n for no positive integer n.

1970 3) Suppose that P1=P2 :^2: +P3:^2: + P4:^2:, where P1, P2, P3 and P4 are primes. Prove that at least one of  P2, P3, P4 is equal to 3.

1970 5) Find all solutions to the following system of simultaneous equations:
x+y+z=7
x:^2:+y:^2:+z:^2: = 31
x:^3:+y:^3:+z:^3: = 154

1994 1)
Al usually arrives at the train station on the commuter train at 6:00, where his wife Jane meets him and drives him home. Today Al caught the early train and arrived at 5:00. Rather than waiting for Jane, he decided to jog along the route he knew Jane would take and hail her when he saw her. As a result, Al and Jane arrived home 12 minutes earlier than usual. If Al was jogging at a constant speed of 5 mph and Jane always drives at the constant speed that would put her at the station at 6:00, what was her speed, in mph?

1994 3)
a) Show that there is a triangle whose angles have measure arctan 1, arctan 2, and arctan 3
b) Find all values of k for which there is a triangle whose angles have measure arctan(1/2), arctan(1/2 + k), and arctan(1/2 + 2k)

1993 1) A matrix is a rectangular array of numbers. For example,
( 1 2 )
( 3 4 )
and
( 1 3 )
( 2 4 )
(forgive the drawing. You should be able to figure it out)
are 2x2 matrices. A saddle point in a matrix is an entry which is simultaneously the smallest number in its row and the largest number in its column.
a) Write down a 2x2 matrix which has a saddle point, and indicate which entry is the saddle point.
b) Write down a 2x2 matrix which has no saddle point.
c) Prove that a matrix of any size, all of whose entries are distinct, can have at most one saddle point.

1993 2) a) Find four different pairs of positive integers satisfying the equation 7/m + 11/n =1
b) Prove that the solutions you have found in part a are all possible pairs of positive integers satisfying the equation 7/m + 11/n = 1

Easy: 1992 1)
The English alphabet consists of 21 consonants and 5 vowels (y being a consonant)
a) Suppose all the letters are listed in arbitrary order. Prove that there must be 4 consecutive consonants.
b) Give a list to show that there need not be 5 consecutive consonants.
c) Suppose that all the letters are arranged in a circle. Prove that there must be 5 consecutive consonants.

2 [hide]All squares mod 3 are either 0 or 1, for multiples of 3 and nonmultiples of 3 respectively. We see that if none of p_2, p_3, or p_4 are multiples of 3, p_1 will have to be a multiple of 3 larger than 3, which is not possible as it is prime. Thus one of p_2, p_3, or p_4 is a multiple of 3, and as they are prime, is thus equal to 3.[/hide]

4. [hide]Let x be the distance that Al got from the train station. The difference in time between normal and this altered schedule is twice the time that Jane takes to drive x miles, so if her rate is r, it is 2x/r. As we are given this is 1/5, x/r=1/10. Meanwhile, the time it took Al to get home is (x/5+n-x/r-1)=n-1/5, where n is the normal time it takes to get home. From this, we have x-5x/r=4 => x-1/2=4 => x=9/2, r=45[/hide]

6. [hide]

(3 2)

(6 1) the 2 is a saddlepoint.

(1 1)

(1 1) no saddlepoint

let us assume there are 2 saddlepoints, value x at coordinates (a,b) and value y at (c,d). Then x is the smallest number in the ath row and the largest in the bth column and y is the smallest in the cth row and largest in the dth column. Ok, consider the value p at (c,b). y<p, x>p. Great now consider q at (a,d). x<q. y>q. So we have y<p<x and y>q>x. Contradiction.[/hide]

Announcement: Round 5, another 8 questions, will be posted as soon as Round 4 is finished, provided that I've returned home. I have to drive to McDonalds, though, so if Round 4 gets finished in the next few minutes, Round 5 will be a bit delayed.

I'm back, and a bit disappointed that nobody has tried these other than Mysticterminator. He's not even online right now, guys. You have time. Work on the problems.

the tangent one

[hide]

take tan(arctan1 + arctan 2 +arctan3).  treat arctan2 + arctan3 as one entity and use the tan(a+b) identity.  So we have (1-tan(arctan2+arctan3)/(1-1*tan(arctan2+arctan3).  Then we use the that again to simplify tan(arctan2+arctan3), which is (2+3)/(1-2*3) = -1.  So we have (1-1)/stuff, which equals 0.  From this we know tan(arctan1+arctan2+arctan3)=0, so arctan1+arctan2+arctan3 = 0 or pi.  But 0<arctan1, arctan2, arctan3, so the sum is pi, so there must be some triangle.

b)  we do the same thing...  some algebra though, i think k=11/4 is only soln.

[/hide]

last one

.[hide]

a)we can use the vowels such that the worst case scenario is that there are 6 sections.  In each of these 6 sections, if each had at most 3, we'd only use 6*3=18 consonants.  So we must have 4 consonants in some sections.

b)  cccVccccVccccVccccVccccVcc

c) if we divide it into a circle, the 5 vowels can only make 5 sections as opposed to the 6 in part a).  So using the same logic, if we only use 4 consonants for each section, there are only 20 consonants used.  So there must be at least 1 section with 5.[/hide]

Mystic gets 2,3,4,6
Cats gets 5 and 8

1 and 7 are still unsolved

7. [hide]mn=7n+11m => (m-7)(n-11)=77. As 77 can only be factored finitely many ways in the positive integers, we check all cases to get the following ordered pairs: (m,n)=(8, 88) (14, 22) (18, 18) (84, 12)[/hide]

1:

[hide]

The only way that C^n=a^n+b^n would be if n=2 and the triangle is right. Since the triangle is obtuse, it follows that C<A+B (triangle inequality), so they can never be equal.

(i probably trivialized this, but is it correct?

[/hide]

Mystic: correct on number 7. It also could have been solved like this:

[hide]

7/m - 11/n = 1

11m + 7n=mn

7n=(n-11)m

If m=7k then 7n=(n-11)7k, so n=11k/(k-1) giving one sol'n for k=2: n=22, m=14.

Also, if n-1 = 11j then 11(11j+1) / 11j = 11+1/j giving j=1: n=12, m=84. n is not an integer for any other j.

If n-11=7k then 7n=7km, 11+7k=km, so m=7+11/k. For k=1: m=18,n=18. For k=11: m=8, n=88. m is not an integer for any other k.[/hide]

1.

[hide]Since c is the longest side in an obtuse triangle, we have c < a + b and c2 > a2 + b2. Now suppose that for some positive integer k we have c2k :le: a2k + b2k. This implies that a2k + b2k > (a2 + b2)k = a2k + b2k + (a bunch of square terms), which is obviously false. So we have c2k > a2k + b2k for all positive integers k.

Multiply both sides of c2k > a2k + b2k by c to get c2k + 1 > c (a2k +b2k). For c2k + 1 :le: a2k + 1 + b2k + 1 to be true, we then need a2k + 1 + b2k + 1 > c (a2k +b2k). Rearrange to get a2k(a - c) + b2k(b - c) > 0, obviously false since a2k, b2k > 0 and a, b < c.

So, in conclusion, ck > ak + bk for all positive integers k.[/hide]

MagnusRex: You'll have to expand on that and show some more work. I'm assuming that you have more work than that, at least.


help me deleted this post,please! Thanks!

It is the longest question I haven't seen in my life. Thanks!


modular arithmetic
number theory unsolved
number theory

Prove that \[\sum_{k=1}^{[\frac{2p}{3}]}\binom{p}{k}\equiv 0\pmod{p^{2}},\]
here $p>3$ is a prime number.


Let $ a, b, c \in[0;1]$. Prove that: $ \frac {a}{b\plus{}c\plus{}1} \plus{} \frac{b}{c\plus{}a\plus{}1}\plus{}\frac{c}{a\plus{}b\plus{}1}\plus{}(1\minus{}a)(1\minus{}b)(1\minus{}c) \le 1$.

[quote="thanhnam2902"]Let $ a, b, c \in[0;1]$. Prove that: $ \frac {a}{b \plus{} c \plus{} 1} \plus{} \frac {b}{c \plus{} a \plus{} 1} \plus{} \frac {c}{a \plus{} b \plus{} 1} \plus{} (1 \minus{} a)(1 \minus{} b)(1 \minus{} c) \le 1$.[/quote]
Let $ f(a,b,c)\equal{}\frac {a}{b \plus{} c \plus{} 1} \plus{} \frac {b}{c \plus{} a \plus{} 1} \plus{} \frac {c}{a \plus{} b \plus{} 1} \plus{} (1 \minus{} a)(1 \minus{} b)(1 \minus{} c).$
Hence, $ \frac{\partial^2 f}{\partial a^2},$ $ \frac{\partial^2 f}{\partial b^2}$ and $ \frac{\partial^2 f}{\partial c^2}$ are non-negative numbers.
Id est, $ \max_{\{a,b,c\}\subset[0,1]}f\equal{}\max\{f(0,0,0),f(0,0,1),f(0,1,1),f(1,1,1)\}\equal{}1.$

Thank. You very much. You can slove my problem by a way easy more than?


Find all pairs of positive integer solutions $(x,y,z,t)$ of the equation\[(x+y)(y+z)(x+z)=txyz\]so that $(x,y)=(y,x)=(x,z)=1$

First note that $x=y=z=1,t=8$ is a solution.
Wlog, we may assume that $x>y \geq z$. And, if $y=z$ then $y=z=1$ (they are supposed to be coprime) which leads to the solution $(2,1,1,9)$.

Now assume that $x>y>z$.
Then $x$ divides $yz(y+z)$ and is coprime with $yz$ so that $x$ divides $y+z <2x$. Thus $y+z = x$ and the equation reduces to $(2y+z)(2z+y) = tyz.$
It follows that $y$ divides $2z^2$ and is coprime with $z$ so that $y=2$ (we cannot have $y=1$ otherwise $y=z=1$).
It easily follows that $z=1$ and $t=10,x=3$, which leads to the solution $(3,2,1,10).$

Pierre.


In the convex quadrilateral ABCD <ABC=<ADC=135, and there exist two points M and N, on the rays AB and AD respectively, such that <MCD=<NCB=90. The circumcircles of AMN and ABD meet at a point K and A. (AK is their common chord).
Prove that AK is perpendicular to KC

[color=red][Moderator edit: This problem was also discussed on http://www.mathlinks.ro/Forum/viewtopic.php?t=16899 .][/color]

It is from Iran 2002 first round!!!!!!
For proof you must prove that centers of these 3 circles are colinear.
Circle with diameter $AC$
Circle $AMN$
Circle $ABD$


function
parameterization
real analysis
advanced fields
advanced fields unsolved

Hallo,

I'm trying to evaluate the following sum :
$ n\sum_{i=1}^n \frac{1}{(n-i+1)^2}\binom{n}{i-1}x^{i-1}(1-x)^{n-i+1}$, $0<x<1$.
My guess is that for large n this tends to $\frac{1}{(1-x)^2}$, as $ n \rightarrow +\infty$
I know that
$n\sum_{i=1}^n \frac{1}{n-i+1}\binom{n}{i-1}x^{i-1}(1-x)^{n-i+1}\rightarrow \frac{1}{1-x}$  (1).
as $ n \rightarrow +\infty$.
My first attempt was to solve the differential equation (differentiate with respect to $x$, unknown function is the sum, in the parenthesis on the LHS)
$ \left (n\sum_{i=1}^n \frac{1}{(n-i+1)^2}\binom{n}{i-1}x^{i-1}(1-x)^{n-i+1}\right )'$=
$n\sum_{i=1}^n \frac{i-1}{(n-i+1)^2}\binom{n}{i-1}x^{i-2}(1-x)^{n-i+1}$+$n\sum_{i=1}^n \frac{1}{(n-i+1)}\binom{n}{i-1}x^{i-1}(1-x)^{n-i}$
which after a simple trick (plus and minus n on the numerator of the first sum on the RHS and substitute from the expressions above) becomes
$ \left (n\sum_{i=1}^n \frac{1}{(n-i+1)^2}\binom{n}{i-1}x^{i-1}(1-x)^{n-i+1})\right )'$=
$n^2\sum_{i=1}^n \frac{1}{(n-i+1)^2}\binom{n}{i-1}x^{i-2}(1-x)^{n-i+1}$
- $n\frac{1}{x(1-x)}$ +$n\frac{1}{(1-x)^2}$ as $ n \rightarrow +\infty$.
Unfortunately the solution is rather complex and it won't simplify any further.
Also tried direct computation in Maple v9.5, gave as a result generalized hypergeometric function with parameters [1,1,-n];[2,2];-x which again won't simplify.
Does anybody have any better idea on how I could go about it?
Thanks in advance,
Dimitris

I think there is a better way to attack this sum.
Unfortunately i haven't got pen and paper, so i will give you only an heuristic.
I think it is an exponential convolution, so it reduces to a multiplication of exponential generating functions. I will try to find something allowing me to write and then i will post something more precise.. :P

This is just a reminder that multiple posting is discouraged. If you want to further contribute to this topic, please look at [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=248684[/url] and post your ideas there (just to keep everything related to one problem in one place)  :).


MATHCOUNTS
geometry
3D geometry
videos
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For you guys who don't read the AMC forum, I thought it would be useful to post a link to my post about the [i]Arithmetic Game[/i]: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=136580[/url]. It's a fun (well, sort of) way to practice mental arithmetic—great for the Sprint and Countdown rounds. (direct link: [url]http://zetamac.com/arithmetic/[/url])

How high a score can you get?

Yes, I did that all morning before states (aka for 5 minutes), and a lot the night before.

Thanks to the creator! He made it not a day too soon, haha.

Copy & pasting from other thread:

High: 56
Low: 41
Average: 51

Ohmigosh this is addicting...  I got 43 on second try (lets just say the first try was bombed because I was kinda testing the format).

edit: ^_^  55 on third try

First two tries: 49, 52.

I seriously need to work at mental math/reacting faster.

EDIT: 3rd try, 54. Getting better :) .

yay I got a 68. check out this picture of my work.

how did you people get so fast? i can only get around 25-30.

My average is a 25, but I just got a 34 on it :D

Wow I'm REALLY bad at this...
Are we aloud to use scratch paper or is the point doing it in your head?

Cause on countdown and sprint you can use scratch paper.

You can use anything I guess....

Mine was without scratch paper :P

[quote="apple pi"]...  I thought it would be useful to post a link to my post about the [i]Arithmetic Game[/i]: ... It's a fun (well, sort of) way to practice mental arithmetic—great for the Sprint and Countdown rounds. (direct link: [url]http://zetamac.com/arithmetic/[/url]) ... [/quote]
I have spoken to the drill's author.  Since this game seems to be so popular, I'm "sticky"ing this thread, for a while.  [Maybe until the end of the State Competitions?]

I agree this drill can improve Sprint and Countdown performance.  There's a large statistical variation in problem difficulty, so don't dwell on your very low or very high scores, especially if you only play a few games,  If you play many games, with the sincere goal of improving, I think most people will see their average score improve significantly.

This drill might also show you how your performance varies when you are tired or distracted.

No single tool is sufficient to hone your skills for MATHCOUNTS, but this is another useful (and fun) tool.  Good luck.  And nice job, apple pi!

1. 29
2. 30
3. 25 :oops:
4. 34

average:

29.5   So it takes me roughly 4 seconds per question.

i used the numbers on the right of the keyboard, but i'm not used to using numbers on the keyboard in a fast way too much, so I can only get like 42 :(  :mad:
-jorian

[quote="xpmath"]I improved to:46. Not by much. My average is in the 40s now though[/quote]
Yay, I got a 49 now :lol:

aww i only got 35...you're right this is addicting
[hide=" :wow: "][hide=" :starwars: "] :rotfl: [/hide][/hide]

I improved to 38. Not a big difference. :(

geez im improving fast. i went from a 25 the day before to a 33 yesterday and today i got a 37.

my all-time high: 50.
but now i can't practice because i have a laptop and that slows my typing-skills even more which is kinda hard to imagine.

[quote="abacadaea"]how does it slow your typing skills. i know this will be marked spam[/quote]
you dont need to mention "i know this will be marked spam" in hoping people will not mark it as spam in getting your poster rating up.  i cant even vote so it doesnt apply to me and frankly i think the rating thing is pretty stupid since its so subjective.

Anyhow to answer your question, well my laptop for expample does not have a key pad and only the number line at the top so its harder to type with that.  i have to look down at the numbers sometimes and that slows me down.

anyways my highest is 61 but i average loike 45-52.

thats what i use. maybe thats why im so slow

It doesn't really matter, you get used to it.

Well, I always use the numbers at the top.

(Don't think that because I do it, you should too. I only get in the 30s!)

Meh. I go a 40. That's a huge improvement.

wow. ive goten a lot better

old stats:
high-34
low-19
average-25

now:
high-40
low-25
average-30

Argh.... I got 68. Then I got 593 after tweaking the settings...


A long conductive straight wire is in the same plane as a circle wire which has radius [b]r[/b]. The distance between the straight wire and the middle of the circle is [b]b[/b] ([b]b>a[/b]). Find the mutual inductance [b]L[/b] and the interaction force [b]F[/b], if the current in the straight wire is [b]I1[/b], and in the circle wire is [b]I2[/b].

I have a hunch that this can be solved by some clever use of magnetostatic energy, but that's only a hunch. I'll do it by straightforward integration.

The flux of the line's magnetic field through the circle is
\[\Phi = \int B\text dS = \frac{\mu_{0}I_{1}}{4\pi}\int_{0}^{\pi}\frac{2R^{2}\sin^{2}\phi\text d\phi}{b-R\cos\phi}= \frac{\mu_{0}I_{1}}{2}(b-\sqrt{(b+R)(bR-b-R)}). \]
Since $L = \Phi I_{2}$, now it's trivial.

Before I proceed, please tell me if I've made a mistake...

Heh... Your approach is much more ellegant than mine :) Good job! But:

1) I don't think you've gotten the integral correctly. Let's see:
The flux at a distance $r$ from the center of the circle is
\[\text d\Phi(r) = \frac{\mu_{0}I_{1}}{4\pi}\int_{0}^{2\pi}\frac{2r\pi \text dr \text d\phi}{b-r\sin\phi}= \frac{\mu_{0}I_{1}}{2}\frac{r\text dr}{b^{2}-r^{2}}. \]
The total flux is
\[\Phi = \frac{\mu_{0}I_{1}}{2}\int_{0}^{a}\frac{r\text dr}{b^{2}-r^{2}}= \frac{\mu_{0}I_{1}}{2}(b-\sqrt{b^{2}-a^{2}}). \]
2) The mutual inductance should be $\Phi/ I_{2}$, right?

3) How did you get the formula for the force? The gradient of magnetic energy?

Unfortunately, electromagnetism is my weak spot, so I may give a lot of incorrect responses and reasonings, probably like the one in my previous post :( But I still don't understand why my answer is wrong, though  :blush: Do you?


Let $ a,b,c$ positive reals. Prove that

$ (\sum\frac{a}{b})^2\plus{}(\sum\frac{a}{b})\geq\frac{5}{3}(\sum a)(\sum\frac{1}{a})\minus{}\sqrt{(\sum a^2)(\sum\frac{1}{a^2})}$

We have $ \sqrt{a^2\plus{}b^2\plus{}c^2}\ge\frac{a\plus{}b\plus{}c}{\sqrt3}$ and $ \sqrt{\frac1{a^2}\plus{}\frac1{b^2}\plus{}\frac1{c^2}}\ge\frac{\frac1a\plus{}\frac1b\plus{}\frac1c}{\sqrt3}$, so $ \sqrt{\left(a^2\plus{}b^2\plus{}c^2\right)\left(\frac1{a^2}\plus{}\frac1{b^2}\plus{}\frac1{c^2}\right)}\ge\frac{\left(a\plus{}b\plus{}c\right)\left(\frac1a\plus{}\frac1b\plus{}\frac1c\right)}3$. Thus it is enough to prove that \[ (\sum\frac{a}{b})^{2}\plus{}(\sum\frac{a}{b})\geq\frac{4}{3}(\sum a)(\sum\frac{1}{a})\] After expansion and simplification, this inequality becomes \[ 3\frac{a^2}{b^2}\plus{}3\frac{b^2}{c^2}\plus{}3\frac{c^2}{a^2}\plus{}2\frac{a}c\plus{}2\frac{b}a\plus{}2\frac{c}b\ge12\plus{}\frac{a}b\plus{}\frac{b}c\plus{}\frac{c}a\] This is obviously true because $ \sum\frac{\frac{a^2}{b^2}\plus{}\frac{a^2}{b^2}\plus{}\frac{b}a}3\ge\sum\frac{a}b$ and $ \sum\frac73\frac{a^2}{b^2}\plus{}\frac53\frac{a}c\ge12$.


function
real analysis
real analysis unsolved

A monotonic function $ f$ on $ [a,b]$ is called [b]singular[/b] if $ f' = 0$ a.e.

Let $ < f_n >$ be a sequence of nondecreasing singular functions on $ [a,b]$ such that the function
\[ f(x) = \sum f_n(x)
\]
is everywhere finite. Prove that $ f$ is singular.

------------------------------------------------------------

There is another proposition about singular function: ( condition (S) )

$ f$ be a nondecreasing function on $ [a,b]$ is singular iff given $ \epsilon > 0$ and $ \delta > 0$, there is a finite collection $ \{[y_k,x_k]\}$ of nonoverlapping intervals such that
\[ \sum{(x_k - y_k)} < \delta
\]
and
\[ \sum{(f(x_k) - f(y_k))} > f(b) - f(a) - \epsilon
\]
-------------------------------------------------------------------

One can easily find a wrong proof as follows:
for each $ f_n$, find the collection which length $ < \delta / 2^n$ and sum $ > f_n(b) - f_n(a) - \epsilon / 2^n$, union all the intervals and delete the tailling enumerable intervals with sum very small to get the finite collection.

But the collection is not of nonoverlapping intervals.

Given $ \epsilon > 0$, $ \delta > 0$, $ \exists N$, such that $ \sum_{n \equal{} N \plus{} 1}^\infty{f_n(b) \minus{} f_n(a)} < \epsilon/2$. Let $ h \equal{} \sum_{n \equal{} N \plus{} 1}^\infty{f_n(x)}$. Then $ h$ is nondecreasing and $ h(b) \minus{} h(a) < \epsilon/2$. Let $ g \equal{} f \minus{} h \equal{} \sum_{n \equal{} 1}^N{f_n(x)}$. It's obviously $ g' \equal{} \sum_{n \equal{} 1}^N{f'_n} \equal{} 0$ a.e., so that $ g$ is nondecreasing singular function. So that there exists a finite collection intervals $ \{[y_k, x_k]\}$ such that
\[ \sum{x_k \minus{} y_k} < \delta
\]
and
\[ \sum{g(x_k) \minus{} g(y_k)} > g(b) \minus{} g(a) \minus{} \epsilon/2,
\]
so that
\[ \sum{f(x_k) \minus{} f(y_k)} \ge \sum{g(x_k) \minus{} g(y_k)} > g(b) \minus{} g(a) \minus{} \epsilon/2 \equal{} f(b) \minus{} f(a) \minus{} (h(b) \minus{} h(a) \plus{} \epsilon/2) > f(b) \minus{} f(a) \minus{} \epsilon.
\]
Then $ f$ satisfy the condition (S), hence $ f$ is singular.


Billy Bob came home after school on a warm sunny day.
His mom said "clean your room before you go play."
so...he...showed...
re-spon-si-bili-bili-bility
(clap clap calp clap)
re-spon-si-bili-bili-bility
(clap clap calp clap)
re-spon-si-bili-bili-bility
(clap clap calp clap)
responsibility...

suppose Billy Bob decides not to be responsible and decides to sing this song instead. it takes him 20 seconds to sing the song. his mom has a 15% chance every minute of coming in to check his progress. if his mom catches him singing the song he gets in trouble. his mom will leave to go shopping in half an hour.

1) what is the chance that Billy Bob will get out of cleaning his room without getting in trouble?

2) Billy Bob has a wonderfully brilliant excuse to use if his mom catches him singing.  it will only work once, though. now what is the chance he will get out of cleaning the room without getting in trouble?

[quote="lotrgreengrapes7926"]Does he sing the song over and over for half an hour? Or only once for 20 seconds?[/quote]

he sings it over and over.

this was a real song we had to sing, among others, such as "increase the peace," "germs, my invisible dog," and "purple people eater."  it was kind of weird.

[quote="Captain Sugar"]suppose Billy Bob decides not to be responsible and decides to sing this song instead. it takes him 20 seconds to sing the song. his mom has a 15% chance every minute of coming in to check his progress. if his mom catches him singing the song he gets in trouble. his mom will leave to go shopping in half an hour.

1) what is the chance that Billy Bob will get out of cleaning his room without getting in trouble?

2) Billy Bob has a wonderfully brilliant excuse to use if his mom catches him singing.  it will only work once, though. now what is the chance he will get out of cleaning the room without getting in trouble?[/quote]
[hide="1"] The probability that his mom will not come up 30 minutes in a row is $.85^{30}\approx.763\%$[/hide]
[hide="2"] The probability that his mom will not come up 30 minutes in a row is $.85^{30}\approx.763\%$. The probability that his mom will come exactly once is $\binom{30}{1}(.15)(.85)^{29}\approx4.04\%$. $.763+4.04=\boxed{4.803\%}$. [/hide]

Nice Problem!!

[hide="1"]The probability that his mom does not come in each minute is $1-15\%=85\%$.  Therefore, the probability of his mom not coming in in 30 minutes is $(85\%)^{30}\approx0.763\%$[/hide]

[hide="2"]There are 2 ways he could not get caught:

Way 1:
Mom doesn't check, where the probability is $\85\%^{30}\approx0.763\%$

Way 2:
Mom checks exactly once.  There are 30 minutes in which she could check, and the probability of this happening is $(85\%)^{29}\cdot15\%$

$\therefore30\cdot(85\%)^{29}\cdot15\%+85\%^{30}\approx4.8\%$[/hide]

[size=75][color=darkred]Fixed your coding
-nebula42[/color][/size]

[quote="daermon"][hide="1"].85^30, so about 76%.[/hide]
does it even matter how long it takes to sing the song?
I'm still stuck on #2  :huh:[/quote]

#1 is 76%

#2, it's ok for his mom to come up once, not twice.


[quote="thanhtrung"]find n is a integer n \geq 1 so that
    x^2+y^2+z^2+t^2-nxyzt -n=0  has x,y,z,t is a prosive integer
 :)  :D  :D  :(  :)[/quote]

One solution is $x=y=z=t=1$ and $n=2$.

RX the Crazy Math

i am almost sure it already appeared in the forum

It's a problem from Vietnam TST 1994. I bet it appeared also in AMM 1989 or 1990, you can find the proof on AMM.

thanhtrung!
Please, read http://www.mathlinks.ro/viewtopic.php?p=140098 and name problems properly next time!
Regards,
Myth


Let [tex]a,b,c[/tex] be  rational numbers such that
   [tex]a + b\sqrt[3] 2+c\sqrt[3] 4= 0[/tex]Prove that [tex]a=b=c=0[/tex]

it's sufficient to proove for $a,b,c$ integers and assume that $c$ is nonzero (when $c=0$ it's trivial), so
$\sqrt[3]{2}b+\sqrt[3]{4}c$ must be an integer so $(\sqrt[3]{2}b+\sqrt[3]{4}c)^2$ must be an integer thus
$\sqrt[3]{4}b^2+2\sqrt[3]{2}c^2$ must be an integer so $(\sqrt[3]{4}b^2+2\sqrt[3]{2}c^2)^2$ must be an integer what implies that $2\sqrt[3]{2}b^4+4\sqrt[3]{4}c^4$ must be an integer so
$2b^3(\sqrt[3]{2}b+\sqrt[3]{4}c)-2\sqrt[3]{2}b^4-4\sqrt[3]{4}c^4$ must be an integer what implies that
$2c\sqrt[3]{4}(b^3-2c^3)$ must be an integer so $\sqrt[3]{4}(b^3-2c^3)$ must be a rational number
so $b^3-2c^3=0$ (cause if not then $4(b^3-2c^3)^3=t^3$, but the numbers on the left and on the right differ $mod$ $2$ with their exponents of $2$) so $b^3=2c^3$, so $b=c=0$ (the same argumentation), concluding
$a=b=c=0$, q.e.d.

Actually there is a more general subject  :D  : $(R,+,*) $ = the usual real numbers  set  ;   $(A,+,*)$ = sub-field  of $R\ $ ; $\ n = $ positive integer, $ n > 1$.

Let $U$ be a real number such that $\ U^n\in A ,\  U^k\notin A \ \forall k\in\{1,...,n-1\}$

Prove that $(a_0,...,a_{n-1}\in A)\& (\sum_{k=0}^{n-1}a_kU^k=0) \Longrightarrow a_0=a_1=...=a_{n-1}=0$

 :cool: need you a hint ??!! :cool:

[quote="Diogene"]  :cool: need you a hint ??!! :cool:[/quote]

[quote="Rafal"]who ? you need a hint ?;) [/quote]

Tell me, Rafal ,  what's the difference between  "need you a hint ?" and "I need a hint " :?:
Anyway, try to solve the general subject :D

:cool:

Let $U$ be a real number such that $U\notin Q ,U^2\notin Q,U^3\in Q\ ;\ a,b,c\in Q$

$ \ a + bU + cU^2 = 0 \Longrightarrow  b(a+bU+cU^2)-cU(a + bU + cU^2)=0 $ $\Longrightarrow ab-c^2U^3 + (b^2-ac)U = 0$

But $U\notin Q ,U^3\in Q$, so   $b^2-ac=0,ab-c^2U^3=0$, so $b^3= c^3U^3 $ so $b=cU$.

But $ U\notin Q$, so $b=c=0 $ and finally $a=0$ .

:cool:


abstract algebra
group theory
advanced fields
advanced fields unsolved

(I saw this theorem in ASCII-format, so I'm not sure about the formulation)

Prove that every finitely generated abelian group $G$ can be represented as $G=T \times \mathbb{Z}^{r}$ for some positive integer $r$, where $T$ is a torsion subgroup of $G$.

Please post this kind of stuff into the superior algebra forum.

And this theorem is correct, and we can decompose $T$ further into a product of finitely many cyclic finite groups.
But the proof I know is rather lengthy...


Student tickets for a concert cost $2 while adult tickets cost $3.50. If 400 tickets were sold and total receipts for the concert were $935, how many adult tickets were sold?
--> What I'd like to know is if it matters where I put the X in the questions.
1) Student tickets = $2
2) Adult tickets = $3.50
3) 400t sold
4) How many adult tickets were sold?

1: (x)2 + (400-x)3.50 = 935
2: 2x + 1400-3.50x = 935
3: -3.50x + 2x = - 465
4: -1.5x = -465
5: y(-1)
6: 1.5x = 465
7: 1.5x/1.5 = 465/1.5
8: x = 310

But that's student tickets sold, right? Because I put the X next to the student ticket cost?

$x$ would not matter at all in this equation.

You can either do $2(x)+3.5(400-x)=935$ or do $2(400-x)+3.5(x)=935$

The important thing is that you have to identify what they are asking for. Since in this case they are asking for adult tickets
it would be better to use the second equation and putting $x$ to the $3.5$dollars. If you put it for the $2$ dollars equation just remember to subtract $400$ from it since adult tickets were $400-x$ for that case.

In any 2 variable system of equations, it doesn't matter where the variables are. just alter the equations to fit what you want to find.


It's in your textbook. Do your own homework.

[quote="sharkman"]I need to find the surface area of a piece of wire with a length of 12mm and a diameter of 0.7mm.

I just want the formula needed to find the surface area.[/quote]

$ 2\pi r^{2}\plus{}2\pi r h$
now plug in the values


group theory
abstract algebra
linear algebra
matrix
induction
superior algebra
superior algebra solved

Let $ A,B$ two matrices nxn with integer entries such that $ \det A, \ detB$ is 1 or -1. Suppose that there exists a positive integer $ k>2$ such that all entries of $ A-B$ are multiples of k. Then prove that the subgroup generated by A and B in $ GL_n(\mathbb{C})$ is infinite.

I'm a little bit confused... if $A=B=I_n$ then the subgroup generated by $A,B$ is finite  :?
or if $A=B$ equal to a matrice obtained from $I_n$ by permutation of two lines(or columns) we get also that the subgroup is finite...

I just realized that the solution I gave for a proposed problem of alekk works very well for this one too. So, it is not anymore unsolved.

By working with $AB^{-1},1$ instead of $A,B$, it suffices to treat the case $B=1$ (I'll denote $I_n$ by $1$, because it's easier to type :)).

We know that $A=1+pD$ for some non-zero $D\in\mathcal M_n(\mathbb Z)$, and we want to prove that we cannot have $A^t=1$ for some positive integer $t$. Since every power of a matrix of the form $1+pD$ has the same form, it suffices to prove the assertion above for $p\not|t$ and for $t=p^k$. In turn, the latter case reduces, by induction on $k$, to the case $t=p$.

If $p\not|t$ then from $1+\binom t1pD+\ldots+\binom ttp^tD^t=1$ we find $p|D$ (after subtracting $1$ and dividing by $p$). After that we divide by $p$ again to get $p^2|D$, and so on. We get $p^k|D,\ \forall k\Rightarrow D=0$, contradicting the fact that $D$ was non-zero. The exact same reasoning works for $t=p$.

My solution: if $A$ is of finite order and $A=I+dB$ with $d>2$ and $B$ integer matrix, then clearly the spectral radius of $B$ is strictly smaller than $1$ (all eigenvalues of $A$ have modulus 1) and thus $B^k$ converges to 0 at infinity and thus $B$ is nilpotent. But it is also diagonalisable ($A$ is) and thus it is the zero matrix.

Why does it suffice working with $AB^{-1},I_n$? I can't figure out why the two generated subgroups are the same...