calculus
derivative
function
real analysis
real analysis unsolved

Consider the two function F and G whose derivatives F' and G' exist. The function have the properties listed below for all values of x. Prove G'(x)= -F(x).

I. (F(x))^2 + (G(x))^2 =1
II. F'(x)=G(x)
III. G(x) > 0
IV. F(0) = 0

F(0)^2 + (G(x))^2 = 1
So G(x) = 1

Since G(x) = 1 , F'(x) = 1

Thats as far as I could get.

$ (F(x))^2 \plus{} (G(x))^2 \equal{} 1\Rightarrow (G(x))^2 \equal{} 1 \minus{} (F(x))^2$
Differentiate both sides:
$ 2G(x)(G'(x)) \equal{} \minus{} 2F(x)(F'(x))$   (*)
Because $ F'(x) \equal{} G(x) > 0$ we can divide the left side of (*) by $ G(x)$ and the right side of (*) by $ F'(x)$ and get the following equality:
$ 2G'(x) \equal{} 2F(x)$
So, $ G'(x) \equal{} \minus{} F(x)$ QED.

[quote="xethissy"]What does (*) mean?[/quote]

simply [b]rem[/b] has diferentiated with Chain rule.

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


Find, with proof, all whole number solutions to x^3 = y^2 + 2.

[hide="partial solution"]
There are four cases:
1) $ x$ is odd and $ y$ is even.
2) $ y$ is odd and $ x$ is even.
3) Both $ x$ and $ y$ are even.
4) Both $ x$ and $ y$ are odd.

If we take case 1, then we have $ (2a \plus{} 1)^3 \equal{} (2b)^2 \plus{} 2$. Since the LHS is odd and the RHS is even, we can rule out case 1.
Similar reasoning rules out case 2.
For case 4, let $ x \equal{} 2a$ and $ y \equal{} 2b$. Then $ 8a^3 \equal{} 4b^2 \plus{} 2$ which simplifies to $ 4a^3 \equal{} 2b^2 \plus{} 1$, but then we run into the same problem as cases 1 and 2, and therefore we can rule out case 3.

I'm still working on case 4. I'll get back to you with that probably tomorrow (it's midnight by me, so... yeah). Hopefully. >_<
[/hide]

[quote="aznlord1337"]Find, with proof, all whole number solutions to $ x^3 \equal{} y^2 \plus{} 2.$[/quote]
It's a beautiful Fermat's equation.
 [hide="The answer"] The answer is $ x \equal{} 3$ and $ y \equal{} \pm5$ only.[/hide]

Fermat is credited with first considering this equation.  It is the [url=http://mathworld.wolfram.com/MordellCurve.html]Mordell curve[/url] with $ n \equal{} \minus{}2$.

[b]Anaxerzia:[/b]  The blunt modular approach only works when there aren't any solutions.  A Diophantine equation with integer solutions has integer solutions $ \bmod p$ for any $ p$, so that (by itself) doesn't do much.

A complete solution is available here:

http://planetmath.org/encyclopedia/Y2X32.html

There's obviously a [b]huge[/b] amount of literature on elliptic curves and the special case of Mordell curves.


AMC
AIME
trigonometry
MATHCOUNTS
complex numbers
\/closed

There are math jams, not really "classes" but still good stuff
[url]http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php[/url]
You have to pay for all the real classes, but they're relatively very cheap
[url]http://www.artofproblemsolving.com/Classes/AoPS_C_About.php[/url]
Also, if you make MOP, you get to take W00T class for free.
[url]http://www.artofproblemsolving.com/Classes/AoPS_C_WOOT.php[/url]

[quote="Karth"][quote="moogra"]Apparently, it costs $150... I thought ccy said it was$800 dollars for 10 sheets of paper...[/quote]

 :huuh:

Are you talking about WOOT? It's more like $\$$600 and it's not only for "10 sheets of paper..." It's more like 300 (including the transcripts, handouts, etc.). Also, WOOT is worth it... my AIME score is improving by the minute  :D[/quote]
I'm pretty sure he's not remembering things correctly, or ccy was in a very bad mood when he said that(very unlikely, as ccy knows it's a lot cheaper).

And I totally agree with what Karth has to say about w00t. My AIME score went from 5 last year to like 8 on the practice test, and now I feel like I can get 10 no problem if I can learn to add/multiply

And about 150\$....that's extremely cheap considering what you get out of it. You get 12 lectures that teach problem solving skills, about 10-15 problems each week(or more depending on the class), you get 2 problem sets to work on and get feedback on solutions(midterms and finals), and you get to keep the lecture stuff as transcripts.
My fear of trig and complex numbers completely went away after taking the intermediate trig/complex numbers class, and now it's like my favorite subject.

For WOOT, they let you take it to October 5 for free and you get a full refund back if you don't like it... :starwars:

[quote="yes_trig"]get a scholarship.

If you win mathcounts team, I heard you get 25 bucks?

You can get a scholarship for 150 if you live in NC[/quote]

Yeah, I got 150 for the state subject tests.


[hide]The numbers that don't have 9 are $8*9*9$ which is 648. Subtract from $9*10*10$, 900=252 numbers.[/hide]

[quote="redsmartie"][hide]The range of numbers is 111 to 999. There has to be at least one 9 in the number. If it starts with 109 and we superimpose the ones have to be a 9, there are 109,119,129.......889. Then there are 9 times 9 giving us 81 examples. If the tens were 9: 190,191......898 with 9 times 9 examples. We continue this giving us a total of 81+81+81+18+8+1=270   [/hide][/quote]

isnt the range 100-999? but that didnt affect your answer. i get

[hide]heres my approach.
for tens that arent 9s, you have 1.  in each hundred, you have 9 tens that arent 9s.  and you have 8 hundreds.  so 9*8=72.

now we have the 90s.  theres 10 in each.  so 10*8=80.

and of course 900-999 is another hundred. so 100

100+80+72=[b]252[/b][/hide]

[quote="redsmartie"][hide]The range of numbers is 111 to 999. There has to be at least one 9 in the number. If it starts with 109 and we superimpose the ones have to be a 9, there are 109,119,129.......889. Then there are 9 times 9 giving us 81 examples. If the tens were 9: 190,191......898 with 9 times 9 examples. We continue this giving us a total of 81+81+81+18+8+1=270   [/hide][/quote]

Your method counts the following numbers twice:
199, 299, 399, 499, ...............

and then 990,991,992.........................

so you need to subtract 18 from 270.

[quote="bpms"]To all you people out there the best way to do this is with complementary counting.[/quote]

In other words, count the three-digit numbers that don't satisfy the given conditions.

[quote="espark52"]how many three digit numbers contain at least one 9?[/quote]

[hide]

The amount of 3-digit numbers that do [i]not[/i] contain a 9 is $8\times 9\times 9$. So, you subtract that from 900, the number of 3 digit numbers to get the answer of $\mbox{252}$. [/hide]

[hide]i got 252 too  :lol:[/hide]

[size=75][color=darkred]Added hide tags.
-nebula42[/color][/size]


trigonometry
rational numbers
IMO Shortlist

Let $ \alpha$ be a rational number with $ 0 < \alpha < 1$ and $ \cos (3 \pi \alpha) \plus{} 2\cos(2 \pi \alpha) \equal{} 0$. Prove that $ \alpha \equal{} \frac {2}{3}$.

Letting $ x=\cos(\pi \alpha)$ and the equation turns into $ (2x+1)(2x^{2}+x-2)=0$. The only plausible solutions are $ x=\frac{-1}{2}, \frac{-1+\sqrt{17}}{4}$. $ x=-\frac{1}{2}\Rightarrow \alpha=\frac{2}{3}$ but how would you prove that $ \frac{\arccos(\frac{-1+\sqrt{17}}{4})}{\pi}$ is irrational?

Solution by [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=25787]olorin[/url]:

Ok, I have a solution using [url=http://en.wikipedia.org/wiki/Cyclotomic_fields]cyclotomic fields[/url].

$ t: = \cos\pi x$, then $ \cos 2\pi x = 2t^{2} - 1$ and $ \cos 3\pi x = 4t^{3} - 3t$.
So $ 0 = 4t^{3} + 4t^{2} - 3t - 2 = (2t + 1)(2t^{2} + t - 2)$ and $ t = - {1\over 2}$ or $ 2t^{2} + t - 2 = 0$.
$ x\mapsto \cos\pi x$ is a bijection $ [0,1]\to [ - 1,1]$ with $ \cos{2\pi\over 3} = - {1\over 2}$.
So $ \cos\pi x = t = - {1\over 2}$ and $ x\in[0,1]$ forces $ x = {2\over 3}$.
Assume therefore $ 2t^{2} + t - 2 = 0$ and $ t = { - 1\pm\sqrt {17}\over 4}$ and $ [\mathbb{Q}(t): \mathbb{Q}] = 2$.
Write $ {x\over 2} = {m\over n}$ with $ m\in\mathbb{Z},n\in\mathbb{N},\gcd (m,n) = 1$.
Then $ \zeta: = e^{2m\pi i/n}$ is a primitive $ n$-th root of unity with $ \zeta + \zeta^{ - 1} = 2\cos{2m\pi \over n} = 2t$.
So $ \zeta$ is root of $ (X - \zeta)(X - \zeta^{ - 1}) = X^{2} - 2tX + 1\in\mathbb{Q}(t)[X]$ and $ [\mathbb{Q}(\zeta): \mathbb{Q}(t)]\leq 2$,
and so $ \varphi (n) = [\mathbb{Q}(\zeta): \mathbb{Q}] = [\mathbb{Q}(\zeta): \mathbb{Q}(t)][\mathbb{Q}(t): \mathbb{Q}]\leq 4$.
If $ p|n$ for some prime $ p\geq 7$, then $ 4 < p - 1|\varphi (n)$. Contradiction!
If $ p^{2}|n$ for some prime $ p\geq 3$, then $ 4 < (p - 1)p|\varphi (n)$. Contradiction!
If $ pq|n$ for some primes $ p > q\geq 3$, then $ 4 < (p - 1)(q - 1)|\varphi (n)$. Contradiction!
So $ n = 2^\alpha$ or $ n = 2^\alpha\cdot 3$ or $ n = 2^\alpha\cdot 5$ for some $ \alpha\in\mathbb{N}_{0}$.
This gives $ n\in\{1,2,2^{2},2^{3},3,2\cdot 3,2^{2}\cdot 3,5,2\cdot 5\}$
and then $ t = \cos{2m\pi \over n}\in\{0,\pm 1,\pm{1\over 2},\pm{\sqrt {2}\over 2},\pm{\sqrt {3}\over 2},{\pm 1\pm\sqrt {5}\over 4}\}$,
which contradicts $ t = { - 1\pm\sqrt {17}\over 4}$.

Let $y=2x$, and suppose $2\cos\pi\alpha$ is a root of $f(y)=y^2+y-4=0$. If $\alpha=p/q$ and $\omega=e^{\frac{2\pi i}{2q}}$, however, then because $f$ is irreducible,
\[f(y)|g(y)=\prod_{1\le k\le2q,\;(k,2q)=1}(y-\omega^k-\omega^{-k})\in\mathbb{Z}[y].\]But all roots of $g$ have magnitude at most $2$, while $f$ has the root $\frac{-1-\sqrt{17}}{2}$ of magnitude larger than $2$, a contradiction.


There are 4 towns A, B, C and D such that ABCD is a square of side 1km. The local municipal councils want a system of roads that will connect all the 4 towns. Presently, there is no motorable road between these towns and it costs USD 1 million  per km to build such a road.

What is the least cost in which one may build a road system wherein all 4 towns will be connected?

If you make roads along the diagonals AC and BD, the cost will be = 2*sqrt(2)... but we must find the minimum value.

Anyone?

isn't the answer just $2\sqrt{2}km$? so thats like $2.82842712$km or $2828427.12$ dollars USD.

unless theres some trick im overlooking :P

Actually I am not aware of the answer, but I have read somewhere that this is not the correct answer. There is some trick involving taking a point inside the imaginary square boundary and then extending to one of the side and doing something...

unfortunately I dont remeber it now... so asked.

may be you are right, may be not.

I am hoping that some other members can also analyze it...

is that ok?

i've seen something like this and i think the answer involves 120 degree angles.

if you put points E and F inside the square such that $\angle AED$ and $\angle BFC$ are 120 degrees each,then the lengths DE, AE, BF, and CF all measure $\frac{1}{\sqrt{3}}$ and the length EF is $1-\frac{1}{\sqrt{3}}$.  so the total length is
$\frac{4}{\sqrt{3}}+1-\frac{1}{\sqrt{3}}=1+\frac{3}{\sqrt{3}}=1+\sqrt{3}$ which is approximately 2.7320508

i don't know why the answer is to use 120 degree angles, anyone have a proof that this always minimizes the length needed?

[quote="perfectnumber628"]i've seen something like this and i think the answer involves 120 degree angles.

if you put points D and E inside the square such that $\angle AED$ and $\angle BFC$ are 120 degrees each,then the lengths DE, AE, BF, and CF all measure $\frac{1}{\sqrt{3}}$ and the length EF is $1-\frac{1}{\sqrt{3}}$.  so the total length is
$\frac{4}{\sqrt{3}}+1-\frac{1}{\sqrt{3}}=1+\frac{3}{\sqrt{3}}=1+\sqrt{3}$ which is approximately 2.7320508

i don't know why the answer is to use 120 degree angles, anyone have a proof that this always minimizes the length needed?[/quote]

can you mention what are the square end points? (A, B , C F) or something else?

[quote="perfectnumber628"]the corners of the square are A, B, C, and D.  and i may have had a typo in that earlier post when i said D and E are inside the square, because it should be E and F.  i'll edit that.[/quote]

can anyone tell why use 120 degree angle?

any reason  behind that?


Sale prices at the Ajax Outlet Store are $ 50 \%$ below original prices.  On Saturdays an additional discount of $ 20 \%$ off the sale price is given.  What is the Saturday price of a coat whose original price is $ \$180$?

\[ \textbf{(A)}\ \$54 \qquad
\textbf{(B)}\ \$72 \qquad
\textbf{(C)}\ \$90 \qquad
\textbf{(D)}\ \$108 \qquad
\textbf{(E)}\ \$110
\]

[hide="I have no idea where you got those numbers..."]
The sale price of the coat is $ 50 \%$ of $ \$180$ or $ \$90$.  The additional discount is $ 20 \%$ of $ \$90$ or $ \$18$, so the Saturday price is $ \$90\minus{}\$18\equal{}\$72$, or answer choice $ \boxed{\textbf{(B)}}$.
[/hide]

[hide="Alternate Solution"]
The Saturday price is $ 80 \%$ of the sale price and the sale price is $ 50 \%$ of the original price, so the Saturday price is $ .8(.5(\$180))\equal{}.8(\$90)\equal{}\$72$, or $ \boxed{\textbf{(B)}}$.
[/hide]

Half off 180 is 90. 20% off of 90 is 80% of it so it's 0.8*90=72.

Hello PlaneGod. You have recently "bumped", or pushed some posts to the top of the forum list. Although your solutions are correct and new, try posting on newer problems. This keeps the forum's feed filled with new topics, and makes the forum easier to navigate. Thanks!


conics
hyperbola
geometry
geometric transformation
reflection
homothety
geometry unsolved

Denote $F'$ the another focus. Since $PF'- PF=a,$ then construct circle $\mathcal{C}$ with center $P$ and radius $PF+a$ $\Longrightarrow$ $F'$ lies on $\mathcal{C}.$ Construct orthogonal projection $Q$ of $F$ on the given tangent. The center $O$ of the hyperbola must lie on the circle $\mathcal{K}$ with center $Q$ and radius $\tfrac {1}{2}a,$ but the reflection $F'$ of $F$ about $O$ lies on $\mathcal{C}.$ Hence, construct the image $\mathcal{K}'$ of $\mathcal{K}$ under the homothety with center $F$ and coefficient 2. $ \mathcal{K}'$ and $ \mathcal{C}$ meet at the focus $F'$ (consider the one according to the sketch). The object hyperbola is now defined.


For $ x \in \mathbb{N}, 1 \le x \le n$ this is the binomial expansion of $ (1 \minus{} 1)^x \equal{} 0$.  Since the equation is a polynomial with at most $ n$ roots, we are done.


inequalities algebra polynomial calculus derivative

Find the range of $ m$ such that the line $ y\equal{}mx\minus{}1$ doesn't have intersection points with both curves $ y\equal{}\sqrt{x\minus{}1}$ and $ y\equal{}x^2\plus{}1$.

[hide].. which means both two following equations have no real root :
$ mx\minus{}1\equal{}x^2\plus{}1$ and $ mx\minus{}1\equal{}\sqrt{x\minus{}1}$
the first one gives $ \minus{}2\sqrt{2} \leq m \leq 2\sqrt{2}$
for the second one , denote $ \sqrt{x\minus{}1}\equal{}a$ then $ x\equal{}a^2\plus{}1$
$ m(a^2\plus{}1)\minus{}1\equal{}a$ we need this equations has no positove real root , rewrite the equation as :
$ ma^2\minus{}a\plus{}m\minus{}1\equal{}0$, this equation has no positive real root , equivalent to $ \delta <0 ; S <0 ; P > 0$ gives $ \frac{\minus{}1\minus{}\sqrt{2}}{2} <m<0$
compare two obtained condition of m gives the final result $ \frac{\minus{}1\minus{}\sqrt{2}}{2} <m<0$[/hide]

Sorry, I can't understand your solution.
Your answer is incorrect.

[quote="kunny"]Find the range of $ m$ such that the line $ y \equal{} mx \minus{} 1$ doesn't have intersection points with both curves $ y \equal{} \sqrt {x \minus{} 1}$ and $ y \equal{} x^2 \plus{} 1$.[/quote]

[hide="If you ignore the hand-waving, I think this is right..."]
We first consider the graphs $ y\equal{}mx\minus{}1$ and $ y\equal{}\sqrt{x\minus{}1}$. If they do not intersect, then the equation $ mx\minus{}1\equal{}\sqrt{x\minus{}1}$ has no real roots, and:
$ m^2x^2\minus{}2mx\plus{}1\equal{}x\minus{}1$
$ \implies (2m\plus{}1)^2\minus{}8m^2\equal{}\minus{}4m^2\plus{}4m\plus{}1<0$
$ \implies m\in \left( \minus{}\infty,\frac{1\minus{}\sqrt{2}}{2} \right) \cup \left( \frac{1\plus{}\sqrt{2}}{2},\infty \right)$

However, this set is too restrictive, since squaring the original inequality caused some potential solutions to be removed from the actual solution set. Geometric arguments (I don't know how to show this algebraically  :( ) will show that the actual set is actually:

$ m\in \left( \minus{}\infty,0 \right) \cup \left( \frac{1\plus{}\sqrt{2}}{2},\infty \right)$

For the second equation, we seek $ m$ such that $ mx\minus{}1\equal{}x^2\plus{}1$ has no real solutions:
$ mx\minus{}1\equal{}x^2\plus{}1$
$ \implies m^2<8$
$ \implies m\in \left(\minus{}2\sqrt{2},2\sqrt{2}\right)$

Therefore:

$ m\in \left(\minus{}2\sqrt{2},2\sqrt{2}\right) \cap \left( \left( \minus{}\infty,0 \right) \cup \left( \frac{1\plus{}\sqrt{2}}{2},\infty \right) \right)$
$ \boxed{m\in \left( \minus{}2\sqrt{2},0 \right) \cup \left( \frac{1\plus{}\sqrt{2}}{2},2\sqrt{2} \right)}$
[/hide]

You get 17 points of full marks 20.
It seems that we can't get full marks easily.

We can use tangent lines: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=194721&start=7]How to take a polynomial derivative without calculus[/url]


Gcd(a,b)=1. Find all integers that they satisfy the equation $\frac{a+b}{a^{2}+ab+b^{2}}=\frac{3}{13}$

Let $a+b=m$ and $a^{2}+ab+b^{2}=n$
From $\gcd(a,b)=1$ neither $a|m$ or $a|n$. (similarly with $b$).
Suppose $\frac{m}{n}$ is not in its lowest terms: that is $gcd(m,n)>1$. Specifically, we suppose that $p|m$ and $p|n$. Then $a \equiv-b \mod p$. So $n \equiv a^{2}-a^{2}+a^{2}\mod p$. But since $p|n$ this is $0$ and so $a^{2}\equiv a \equiv 0 \mod p$. But then $b \equiv 0 \mod p$ contradicting $\gcd(a,b)=1$. So $m=3$, $n=13$.
$b=3-a$
$a^{2}+a(3-a)+(3-a)^{2}=13$
$a^{2}-3a+9=13$
$a=4$ or $-1$.

$a=4,b=-1$


calculus
integration
function
calculus computations

show that/////////
the integral of an even function from -a to a is twice from 0 to a

$ \int_{\minus{}a}^a f(x)dx \equal{} \int_{\minus{}a}^0 f(x)dx \plus{} \int_0^a f(x)dx\equal{}$

$ \int_a^0 f(\minus{}x)d(\minus{}x)\plus{} \int_0^a f(x)dx \equal{} 2\int_0^a f(x)dx$


trigonometry
inequalities
function
logarithms
inequalities unsolved

[quote="Tiks"]Problem 3.
Prove that if $n\geq 28$ and $0<x<\frac{\pi}{2n}$
than $\frac{sin2x}{sinx}+\frac{sin3x}{sin2x}+...+\frac{sin(n+1)x}{sinnx}<2ctgx$(we know that $\pi<3.15$)[/quote]
Hay I want to know any solution :oops:  :roll:

Oops, I was wrong.
On Ukrainian TST 1999 there was this problem
$\frac{\sin{2x}}{\sin{x}}+...+\frac{\sin{(n+1)x}}{\sin{nx}}<2\frac{\cos{x}}{\sin^2{x}}$
for $0<x<\frac{\pi}{2n}$. They look very similiar :) .

[quote="Flatter"]Try such inequality:
$\frac{\sin(n+1)x}{\sin{nx}}<\sin(n+1)x\sin{nx}$
[/quote]
It is obvious wrong..
I can't prove this inequality.
I think it is hard to use the fact $n \ge 28$ and $\pi <3.15$ :(

$\frac{\sin2x}{sinx}+...+\frac{\sin{(n+1)x}}{sin{nx}}=n\cos{x}+\sin{x}(ctg{x}+ctg{2x}+...+ctg{nx})<2n\cos{x}$
So we have to prove:
$2n\cos{x}<2ctgx$ which is equal to
$\sin{x}<\frac{1}{n}$
$\sin{\frac{\pi}{2n}}<\frac{1}{n}$
So we have to prove $\sin{\frac{\pi}{2n}}<\frac{1}{n}$, for $n\geq 28$.

unfortunately for flatter, from jordan's inequality we have that for $0<\alpha<\frac{\pi}{2}$

$\sin \alpha>\frac{2\alpha}{\pi}$, which implies $\sin \frac{\pi}{2n}>\frac{1}{n}$... :(

Another mistake! When I was checking inequality for n=28, i miscalculated, and thought that inequality is true. Now I think that problem is more difficult than from the first view.

[quote="Tiks"]Problem 3.
Prove that if $n\geq 28$ and $0<x<\frac{\pi}{2n}$
than $\frac{sin2x}{sinx}+\frac{sin3x}{sin2x}+...+\frac{sin(n+1)x}{sinnx}<2ctgx$(we know that $\pi<3.15$)[/quote]
In AMO I writed $\frac{sin2x}{sinx}+\frac{sin3x}{sin2x}+...+\frac{sin(n+1)x}{sinnx}= ncosx+sinx(ctgx+ctg2x+...+ctgnx)$then I used  Jensen's ineq. for sum$(ctgx+ctg2x+...+ctgnx)$,but I don't know can we continius this solution :( (this problem was be unsloved in AMO :(  :D )

[quote="Tiks"]Problem 3.
Prove that if $n\geq 28$ and $0<x<\frac{\pi}{2n}$
than $\frac{sin2x}{sinx}+\frac{sin3x}{sin2x}+...+\frac{sin(n+1)x}{sinnx}<2ctgx$(we know that $\pi<3.15$)[/quote]
Hwo can  solve this problem?I realy want to see a solution :ninja:

[quote="bomb"]What is ctgx?

Bomb[/quote]
$ctgx=\frac{cosx}{sinx}$ :lol:

[quote="Tiks"]Problem 3.
Prove that if $n\geq 28$ and $0<x<\frac{\pi}{2n}$
than $\frac{sin2x}{sinx}+\frac{sin3x}{sin2x}+...+\frac{sin(n+1)x}{sinnx}<2ctgx$(we know that $\pi<3.15$)[/quote]
Is there anyone hwo can help us to solve this problem? :|  :roll:

[quote="Tiks"]Problem 3.
Prove that if $n\geq 28$ and $0<x<\frac{\pi}{2n}$
than $\frac{sin2x}{sinx}+\frac{sin3x}{sin2x}+...+\frac{sin(n+1)x}{sinnx}<2ctgx$(we know that $\pi<3.15$)[/quote]
At laste,can anyone solve this problem? :huh:

hello

i think i almost got it

since there is no term dependent on n in rhs so i try  to prove

\[ \frac{sin2x}{sinx} + \frac{sin3x}{sin2x} ........ \frac{sin(n+1)x}{sin(nx)}<2cot(\pi/2n)<2cotx \] :!:

let \[ f(n)=\frac{sin2x}{sinx} + \frac{sin3x}{sin2x} ........ \frac{sin(n+1)x}{sin(nx)}-2cot(\pi/2n) \]

i want to prove \[ f(n+1)<f(n) \]   for all n >=28

the above relation reduces to \[ sin(\pi/n)sin(\frac{\pi}{2(n-1)})<4sin(\frac{\pi}{2n(n-1)}) \] for n>=29

i cudnt prove this relation but i think it must be true as i tried it for many vals of n using a calci

hence if it is true then max val of func will be at min val of n possible ie n=28

now consider \[ g(x,n)=\frac{sin(n+1)x}{sin(nx)} \]
it can be seen that this function is decreasing for positive x taking n as a const

so g wud take max val at min possible x  thus x->0

so \[ lim(x\to 0)g(x,n)=\frac{n+1}{n} \]

so \[ f_{max}=f(28)\leq (\sum\frac{k+1}{k}) - 2cot(\pi/56)<0 \] as i got by a calculator

i wud definitely like someone to prove or disprove the ineq i used in the middle  :roll:

[quote]Problem 3. Prove that if $n\geq 28$ and $0<x<\frac{\pi}{2n}$, then

$\frac{\sin{2x}}{\sin{x}}+\frac{\sin{3x}}{\sin{2x}}+\cdots+\frac{\sin{(n+1)x}}{\sin{nx}}<2\cot{x}$

(we know that $\pi<3.15$)
[/quote]

[b] Ugly inequality deserves a ugly proof![/b]

[hide=" It is so ugly, I have to hide it"]

$\frac{\sin{2x}}{\sin{x}}+\frac{\sin{3x}}{\sin{2x}}+\cdots+\frac{\sin{(n+1)x}}{\sin{nx}}= n\cos{x}+\sin{x}\sum_{k=1}^n\cot{kx}$

$n\cos{x} < \frac{\pi}{2x}\cos{x} < \frac{\pi}{2}\cot{x}$

It suffices to prove:

(1) $\sin{x}\sum_{k=1}^n\cot{kx} \leq (2-\frac{\pi}{2})\cot{x}$

Since $\cot{t} < \frac{1}{t}$ for $t\in\bigl(0,\frac{\pi}{2}\big)$, we have

$\sin{x}\sum_{k=1}^n\cot{kx} < \sin{x} \sum_{k=1}^n\frac{1}{kx} < \sum_{k=1}^n\frac{1}{k} < \log_2{n} + 1$

By Jordan's Inequality, we have

$\cot{x} > \frac{\cos{x}}{x} = \frac{1}{x}\sin(\frac{\pi}{2}-x) \ge \frac{1}{x}\cdot\frac{2}{\pi} (\frac{\pi}{2}-x) = \frac{1}{x} - \frac{2}{\pi} > \frac{2n}{\pi} - \frac{2}{\pi} = \frac{2}{\pi}(n-1)$

So $(2-\frac{\pi}{2})\cot{x} > (2-\frac{\pi}{2}) \frac{2}{\pi}(n-1) = (\frac{4}{\pi} -1)(n-1)$

Therefore, (1) is reduced to prove:

(2) $f(n) = (\frac{4}{\pi} -1)(n-1) - (\log_2{n} + 1) \geq 0$ for $n\geq28$.

It's easy to verify that $f(28) \geq 0$. Since $f(n)$ is an increasing function,  $f(n)\geq f(28)\geq0$ for all $n\geq 28$. Q.E.D

Hope it's correct!

[/hide]

[quote="fuzzylogic"][quote="Tiks"]Please explain the following part of your solution.
$\frac{1}{x}\sin(\frac{\pi}{2}-x) \ge \frac{1}{x}\cdot\frac{2}{\pi} (\frac{\pi}{2}-x)$[/quote]

Jordan's Inequality, see here

http://planetmath.org/encyclopedia/JordansInequality.html[/quote]
OK,thanks ;).
I think your solution is correct.


Let $x<y$ be positive integers and $P=\frac{x^3-y}{1+xy}$. Find all integer values that $P$ can take.

If $x=p^{3}$, $y=p^{5}-p$, then $\frac{x^{3}-y}{1+xy}=\frac{p^{9}-p^{5}+p}{1+p^{8}-p^{4}}=p$ and $0<x<y$ for $p\geq 2$? so $p\geq 2$ are good numbers.
$p=0$ is also good (for example, $x=2, y=8$).
If $p\leq-1$,  then $y-x^{3}\geq 1+xy$ :arrow: $x>xy$ - contradiction.
If $p=1$, then $x^{3}-1=(x+1)y$ :arrow: $x+1\mid x^{3}-1$ and $x+1\mid 2$, because $x+1\mid x^{3}+1$  :arrow: $x=1$, hence $y=0$ - contradiction with beeing $y$ positive.
So, answer is [b]$p\geq 2$[/b] and [b]$p=0$[/b].


(I) A wee wee man
      In a red red coat
      Staff in my hand
      And a stone in my throat.

Who or what is the “I” referred to in the puzzle?

(II) Red as blood, sweet as wine
     My heart is hard but my flesh is fine.

Who or what is the “I” referred to in the puzzle?

Sources: Puzzle Periodical

True. The answer to (II) is [hide]A cherry[/hide]


1.$x\neq0$.  if   $u_{1}$=0  If  $u_{n+1} = (1-x)u_{n} + nx$ for all natural numbers n,

then  prove   that  by     mathmatical    induction     that

$u_{n} = \frac{1}{x}{nx-1+(1-x)^{n}}$  for    all    natural  numbers   n.

2.In    a      $\triangle$ABC  , $\angle$ABC=72$\circ$.Explain  how    you    locate   all

the    points   p    in     the    plane    such  that   $\angle APC$=112$\circ$.

[quote="paladin8"][quote="bkunchanapalli"]1.$x\neq0$.       If       $u_{n+1} = (1-x)u_{n} + nx$  for   all    natural numbers   $n$,

then  prove   that  by     mathmatical    induction     that

$u_{n} = \frac{1}{x}[nx-1+(1-x)^{n}]$  for    all    natural  numbers   $n$[/quote]

It's not quite properly defined without an initial condition of some sort, but oh well.

[hide]
With a given condition, we have a base case.

Then if we know $u_n = \frac{1}{x}[nx-1+(1-x)^n]$, we can show that

$u_{n+1} = \frac{1}{x}[nx-1+(1-x)^n](1-x)+nx = \frac{1}{x}(x-1)^{n+1}+\left(n-\frac{1}{x}\right)(1-x)+nx$
$= \frac{1}{x}[(n+1)x-1+(1+x)^{n+1}]$

once see the question again .a  slight modification

after expanding and refactoring, completing the induction.[/hide][/quote]

[quote="bkunchanapalli"][quote="paladin8"][quote="bkunchanapalli"]1.$x\neq0$.       If       $u_{n+1} = (1-x)u_{n} + nx$  for   all    natural numbers   $n$,

then  prove   that  by     mathmatical    induction     that

$u_{n} = \frac{1}{x}[nx-1+(1-x)^{n}]$  for    all    natural  numbers   $n$[/quote]

It's not quite properly defined without an initial condition of some sort, but oh well.

[hide]
With a given condition, we have a base case.

Then if we know $u_n = \frac{1}{x}[nx-1+(1-x)^n]$, we can show that

$u_{n+1} = \frac{1}{x}[nx-1+(1-x)^n](1-x)+nx = \frac{1}{x}(x-1)^{n+1}+\left(n-\frac{1}{x}\right)(1-x)+nx$
$= \frac{1}{x}[(n+1)x-1+(1+x)^{n+1}]$

after expanding and refactoring, completing the induction.[/hide][/quote][/quote]

once see the question again .a  slight modification


trigonometry
geometry
LaTeX
area of a triangle
trig identities

Knowing that: $ \cos\alpha\equal{}\frac{1}{\sqrt{5}}$ and $ \sin\alpha>0$. Determine  the numerical value of: $ \sin(2\alpha)$.

How to do it ?  :maybe:

First, note that $ \sin(2a) \equal{} 2 \sin(a) \cos (a) \equal{} \frac{2 \sin(a)}{\sqrt{5}}$.

So all that is left is to calculate $ \sin(a)$. Since $ \sin^2 x \plus{} \cos^2 x \equal{} 1$ for all $ x$,

$ \sin^2 a \plus{} \cos^2 a \equal{} 1$

$ \sin^2 a \plus{} (\frac{1}{\sqrt{5}})^2 \equal{} 1$

$ \sin^2 a \equal{} \frac{4}{5}$

$ \sin a \equal{} \frac{2}{\sqrt{5}}$

$ \therefore \sin {2a} \equal{} 2 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}} \equal{} \boxed{\frac{4}{5}}$

Ok, very good. But how do you know that: $ \sin(2a) \equal{} 2\sin(a)\cos (a) \equal{}\frac{2\sin(a)}{\sqrt{5}}$
(I'm very new in problems like this)

Bom dia, luis. See here: http://www.mathsrevision.net/alevel/pages.php?page=11

I also saw a nice proof somewhere but I can't remember where it was.  It may have been on the left-hand side panel of the front page of the art of problem solving forum.

It had a right-angled triangle $ XYZ$ with $ XY$ horizontal and $ Z$ directly above $ Y$ and another right-angled triangle  $ XYW$ with $ W$ directly below $ Y$. Angle $ ZXY \equal{} A$ and angle $ YXW \equal{} B$.

$ XY$ has length $ 1$, so $ XY \equal{} \sec A$, $ ZY \equal{} \tan A$, $ YW \equal{} \tan B$ and $ XW \equal{} \sec B$.

Then using the cosine rule, sine rule and the formulas area for the area of a triangle, one can prove the double angle formulas and deduce, in your case, that $ \sin 2A \equal{} 2\sin A \cos A$.

Hope this helps as I don't know how to sent a picture.

Ah nice site Andrew. Thank you very much :D
If you find the prove send me please.

[quote="luiseduardo"]Ok, very good. But how do you know that: $ \sin(2a) \equal{} 2\sin(a)\cos (a) \equal{} \frac {2\sin(a)}{\sqrt {5}}$
(I'm very new in problems like this)[/quote]

Well that's just one of the many trig identities. This one is derived from the sine addition formula: $ \sin(\alpha \plus{} \beta) \equal{} \sin(\alpha) \cos (\beta) \plus{} \sin (\beta) \cos (\alpha)$. When you put $ \alpha \equal{} \beta$, you get $ \sin(2\alpha) \equal{} 2\sin (\alpha) \cos (\alpha)$.

We got the $ \frac {2 \sin(a)}{\sqrt {5}}$ after plugging in the value given for $ \cos(a) \equal{} \frac {1}{\sqrt {5}}$. No trig identity here; just substituting that value. Now do you see?

So, first I recommend you learn the most important trig identities, but most importantly, learn to derive them. That way, they are a lot easier to remember.

Boa tarde, Luis.

The right-angles are $ XYZ$ and $ XYW$.

The cosine rule gives $ (\tan A \plus{} \tan B)^{2} \equal{} \sec^{2} A \plus{} \sec^{2} B \minus{} 2 \sec A \sec B \cos(A \plus{} B)$

$ \tan^{2} A \plus{} \tan ^{2} B \plus{} 2\tan A \tan B \equal{} 1 \plus{} \tan^{2} A \plus{} 1 \plus{} \tan^{2} B \minus{} 2 \frac {\cos (A \plus{} B)}{\cos A \cos B}$

$ \cos A \cos B \minus{} \sin A \sin B \equal{} \cos (A \plus{} B)$.

Area of triangle $ WXZ \equal{} \frac {1}{2} \sec A \sec B \sin (A \plus{} B) \equal{} \frac {1}{2} (\tan A \plus{} \tan B) \times 1$

$ \sin (A \plus{} B) \equal{} \sin A \cos B \plus{} \cos B \sin A$.

Hope this makes sense.

[quote="AndrewTom"]

$ \sin (A + B) = \sin A \cos B + \cos B \sinh$.

[/quote]

You have a small latex error.

Thanks, I've edited it. I didn't leave a space between sin and A.


Let $a, b, c$ three non-negative real numbers such that $abc=1$. Prove that the following inequality holds:

                       \[ \sqrt [3] {a}+\sqrt [3] {b}+\sqrt [3] {c}\leq\sqrt [3] {3(3+a+b+c+ab+bc+ca)}.  \]

Let $a=x^3,$ $b=y^3,$ $c=z^3.$
Hence, $\sqrt [3] {a}+\sqrt [3] {b}+\sqrt [3] {c}\leq\sqrt [3]{3(3+a+b+c+ab+bc+ca)}\Leftrightarrow$
$\Leftrightarrow3\cdot\sum_{cyc}(x^3y^3-x^3y^2z-x^3z^2y+x^2y^2z^2)+2xyz(x^3+y^3+z^3-3xyz)\geq0.$
The last inequality is obviously true by Schur and AM-GM.


Let $ a$, $ b$, $ c$ are sides of a triangle. Find the least possible value $ k$ such that the following inequality always holds:
$ \left|\frac{a\minus{}b}{a\plus{}b}\plus{}\frac{b\minus{}c}{b\plus{}c}\plus{}\frac{c\minus{}a}{c\plus{}a}\right|<k$
[i](Vitaly Lishunov)[/i]

[quote="torbich"]Let $ a$, $ b$, $ c$ are sides of a triangle. Find the least possible value $ k$ such that the following inequality always holds:
$ \left|\frac {a \minus{} b}{a \plus{} b} \plus{} \frac {b \minus{} c}{b \plus{} c} \plus{} \frac {c \minus{} a}{c \plus{} a}\right| < k$
[i](Vitaly Lishunov)[/i][/quote]

Let $ a\equal{}x\plus{}y$ ,$ b\equal{}y\plus{}z$ , $ c\equal{}z\plus{}x$ and see here http://www.artofproblemsolving.com/Forum/viewtopic.php?t=204183

[quote="Sergic Primazon"]
Let $ a \equal{} x \plus{} y$ ,$ b \equal{} y \plus{} z$ , $ c \equal{} z \plus{} x$ and see here http://www.artofproblemsolving.com/Forum/viewtopic.php?t=204183[/quote]
You are right, but your best constant can be simplified ;)

The answer is $ k \equal{} \frac {8\sqrt {2} \minus{} 5\sqrt {5}}{3}$.


Let $ r_1,\ldots,r_n$ be positive reals. Give a necessary and sufficient condition for $ x_1^2\plus{}\cdots\plus{}x_n^2\ge\left(r_1x_1\plus{}\cdots\plus{}r_nx_n\right)^2$ to hold for all $ x_1,\ldots,x_n$ positive reals.

[quote="lingomaniac88"][hide]By Cauchy's, $ \left(\sum r_i^2\right)\left(\sum x_i^2\right)\geq \left(\sum r_ix_i\right)^2$.

Being as strict as possible, assume equality.

Substitution yields:

$ \sum x_i^2\geq \left(\sum r_i^2\right)\left(\sum x_i^2\right)$.

Divide both sides by $ \sum x_i^2$ to get $ \boxed{\sum r_i^2\leq 1}$.[/hide][/quote]

Yes, that's my solution too ;) Although I used the case $ r_i\equal{}x_i$ to prove the necessissity of that statement. (As this is the equality case of CS, my statement is principally the same as yours though)


Prove that among any 16 distinct positive integers not exceedin 100, there are four diferent ones a,b,c,d such that a+b=c+d

[hide]Consider $ |x\minus{}y|$ for any distinct $ x,y$ elements of the 16 member set.  This is an integer between $ 1$ and $ 98$ because $ 0<x,y<100$.  Thus out of the $ \binom{16}{2}\equal{}120$ possible pair of elements there are at least 20 sets of 2 distinct pairs of integers $ a,b$ and $ c,d$ such that $ |a\minus{}b|\equal{}|c\minus{}d|$.  If $ a,b,c,d$ are distinct, then we're done.  If not, then suppose without loss of generality that $ a\equal{}c$.  Then, since the pair must be distinct, we have $ a\minus{}b\equal{}d\minus{}a$ or $ b\plus{}d\equal{}2a$.  Notice that there are 16 possibilities for $ a$ and 20 different pairs whose difference is the same.  Thus, there exist $ h,g$ and $ b,d$ such that for some $ a$, $ h\plus{}g\equal{}b\plus{}d\equal{}2a$.  We prove that $ h,g,b,d$ are distinct to conclude the proof.  If $ h\equal{}b$ then $ g\equal{}d$ which contradicts the fact that the two pairs are distinct.     [/hide]


I read the instructions from the abstract package site, but still can't produce a nice centered abstract, which should look exactly like the normal abstract, except that the word "Abstract." is not above the abstract, but at left, at the beginning.

I have to quote a japanese mathematician named Nono, with a cap like ^ on the first o, and I don't know how. Could you please give me some hints? Thanks a lot.

Jean

\usepackage[runin]{abstract}
...
\begin{abstract}\par
Abstract goes here
\end{abstract}

$ \^{o}$?

yes, that the character I need. Where is it?

Jean

A model of the abstract for this journal may be found here. The package you suggested still keeps the word Abstract above the text. They ask as in this model: http://www.sigma2006.com/model.PDF

Yes that is the o with a cap, but how did you produce it?

Jean

Removing the \par after \begin{abstract} puts Abstract at the start of the line, but it will be indented. So add something like \setlength{\absparindent}{0.5em} before \begin{abstract} to remove the indent. Change the value of the length to get it right.
[code]\setlength{\absparindent}{0.5em}
\begin{abstract}
...
\end{abstract}[/code]It's all in the [url=http://tug.ctan.org/tex-archive/macros/latex/contrib/abstract/abstract.pdf]abstract package documentation[/url].
The tooltip when you hover over $ \^{o}$ tells you it is \^{o}. Pressing the quote button also reveals the latex code.


[b][size=150][color=orange][i]Find the number of triangles that can be constructed so that the length of its sides are three consecutive odd integers and its perimeter is less than 1000.[/i][/color][/size][/b]

[hide="This belongs to HSB"]
Let the sides be n-2,n,n+2 with n odd. So we have conditions $ n\minus{}2\plus{}n>n\plus{}2, n\minus{}2\plus{}n\plus{}n\plus{}2<1000$.
This gives $ n>4,n<334$, so n can be $ 5,7,9,...,333$. There are 165 such triangles.[/hide]


hello everybody,
from what i know, the contest ended recently.
so i would like to know what everybody thought about the test.

feel free to say whatever you like   :D

Hmm.

Problem 1 was really random
Problem 2 was annoying
Problem 3 was simple. Programming kills it
Problem 4 was sooo annoying
Problem 5 was again killed by programming
Problem 6 was good
Problem 7 was kinda weird
Problem 8 was haardd
Problem 9 was kinda random
Problem 10 was b3ast
Problem 11 was ok
Problme 12 wasn't too hard
Problem 13 good, but programmable
Problem 14 was good
Problem 15 was really really random.
BTW, was the answer to 15c m? lol we saw 215cm.

#8 was easy. I finished it in two hours.

#15c was in the second page of the FAQs. 10.943.


Five numbers are arranged in increasing order. As they get larger the difference between adjacent numbers doubles.
The average of the five numbers is 11 more than the middle number. The sum of the second and fourth numbers is equal to the largest number.
What is the largest number?

Let the five numbers be $ x, x\plus{}a, x\plus{}3a, x\plus{}7a,x\plus{}15a$
Then the conditions are $ (x\plus{}\frac{26a}{5})\minus{}(x\plus{}3a)\equal{}11 \Leftrightarrow a\equal{}5$
$ (x\plus{}5)\plus{}(x\plus{}7(5))\equal{}x\plus{}15(5) \Leftrightarrow x\equal{}35$,
so the largest number$ \equal{}35\plus{}15(5)\equal{}110$


Let $ (a_n)$ be a sequence satisfying two conditions below:
- $ a_0 \equal{} a_1 \equal{} 1$
- $ (n \plus{} 3)a_{n \plus{} 1} \equal{} (2n \plus{} 3)a_n \plus{} 3na_{n \minus{} 1}$ for all n
Show that $ a_n$ is an integer for all positive number n

Could anyone help me to solve this problem???  :)


LaTeX
number theory unsolved
number theory

suppose a,b,c,d are real numbers .prove that there exist $x_0,y_0$ (they are not 0 Synchronously)such that : $|a\cdot x_0+b\cdot y_0| +|c\cdot x_0+d\cdot y_0| \leq \sqrt{2\cdot |a\cdot d-b\cdot c|}$

There must be some condition about x[0], y[0] or else we take them  tend to 0 and get the result!

oh excuse me iura :blush:  $x_0,y_0 $ are integer

Anyway, if we put ad=bc and a/b=c/d=irrational, we can't get the result!

[quote="sam-n"]they are not 0 Synchronously[/quote]
'Synchronously' sounds awkward here. 'Simultaneously' serves better  ;)
Time to abuse some $\LaTeX$ usage:
Another way of looking at it:
\[ |(a,b)\cdot(x_0,y_0)|+|(c,d)\cdot(x_0,y_0)|=\sqrt{2\cdot\left|\left| \begin{array}{cc} a&b\\c&d\\  \end{array}\right|\right| } \]

Thx Valentin, I just learned some TeX'ing from that latexhelp.pdf document.


modular arithmetic
number theory
relatively prime
number theory unsolved

$ S$ is a set of unsigned integers,we have:

1) $ |S| \ge 3$,

2) $ a \in S,b|a \Rightarrow b \in S$,

3) $ 1 < a < b, a,b \in S \Rightarrow 1\plus{}ab \in S$.

Prove that all integers are in $ S$.

This problem is here and my solution is here too http://mathsmaroc.jeun.fr/groupe-etudiants-du-t-s-m-f28/preparation-aux-olympiades-t13456-15.htm#117389 and I'll traducate it later if you don't understand French language   :)  !!

i can't go to your site('don't know why);so if your solution is the same ,excuse-me!
we can easily prove that $ 3,2$ are in $ S$.
then owing to the fact that $ 1< 2 < 3$ and by using $ n$ times the $ 3\minus{}rd$ assumption for $ a\equal{}2,b\equal{}3$ u get that the number $ 3 \times 2^{n\plus{}1}\plus{}2^n\plus{}2^{n\minus{}1}\plus{}...\plus{}2\plus{}1$ is in $ S$ and so are all its divisors.but this number is equal to $ 2^{n\plus{}3}\minus{}1$.but it is well-known that for any odd integer $ m$ there exists an integer $ n$ s.t. $ m | 2^{n}\minus{}1$.so we can conclude that all odd numbers are in $ S$.and if a number $ k$ is even for proving that this number exists in $ S$ just take a number like $ t$ which is relatively prime to $ k$ and also congruent to $ \minus{}1$ modulo $ k$.(this number obviously exists in $ S$).
then there exists another integer $ l$ s.t. $ tl \equiv \minus{}1 \pmod{k}$.(take $ l$ s.t. it is congruent to $ 1$ modulo $ k$ and greater than $ 1$).
so all integers are in $ S$.

My solution is not the same as yours my friend  :lol:

if we suppose that not all elements of $ \mathbb{N}$ are not in $ S$ and let $ m$ The smallest one which is not in $ S$

so if $ m\equal{}2k\plus{}1$ and $ k \neq2$ we have $ k<m$ which means that $ k \in S$ and Since $ 2\in S$  then $ 2k\plus{}1 \in S$ which is absurd

if $ m\equal{}2k$  and $ k\neq 2$ then we have $ k<m \implies 2k\plus{}1 \in S$ and since $ 2k\minus{}1<m$ then $ 2k\minus{}1 \in S$ which implies that $ (2k\minus{}1)(2k\plus{}1)\plus{}1 \in S$ Hence $ 4k^2 \in S$ and so $ 2k \in S$ which is absurd

when $ k\equal{}2$ we can easly see that $ 2.2\plus{}1 \in S$ and $ 2.2 \in S$ after proving that $ 3 \in S$ in not trivial way

Hence the result  :lol:


Prove that 1/(sin x sin 2x) + 1/(sin 2x sin 3x) + 1/(sin 3x sin 4x) + ... n terms = cosec x(cot x - cot(n + 1)x).

$ \sum_{k \equal{} 1}^n\frac {\sin x}{\sin kx\sin (k \plus{} 1)x} \equal{} \sum_{k \equal{} 1}^n(\cot kx \minus{} \cot (k \plus{} 1)x) \equal{} \cot x \minus{} \cot (n\plus{}1)x$
The proof of the identity used is trivial.


function
quadratics
probability and stats

A multivariate Gaussian density function give an ellipsoidal distribution in $ n$ dimensions:
$ \frac{1}{(2\pi)^{n/2}|\Sigma|^{1/2}}\exp{\left(-\frac{1}{2}(x-\mu)^\top\Sigma^{-1}(x-\mu)\right)}$, where $ \Sigma$ is the covariance matrix.

For dimensions 2 and/or 3, what sort of density function gives a figure that is in the shape of a lune?

It seems that your variables are not independent. You should try nonlinear (quadratic?) [url=http://en.wikipedia.org/wiki/Nonlinear_regression]regression[/url]. Once you take out a suitable trend equation, the residuals may well be normally distributed.


inequalities
geometry
geometric transformation

Consider the number (3 + :sqrt:10)^2004. Determine

(a) the units digit ;
(b) the 1000th digit after the comma (= decimal point in USA)

in the decimal representation of this number. :)

To be continued ... I'll post some more problems later, but you'll have enough work with these and with the inequalities I think.

rofl, it took me a LONG while to figure out "after the comma".... then i realized you are in belgium and you reverse our system for decimals and comma's.... interesting problem.

Now I have not studied the notation from many other countries... but I believe that the number:

"one thousand five hundred fifty-five and two fifths"

is written in Belgium as: 1.555,4

in the US, we write it as: 1,555.4

We have them reversed.

So 1 million: 1,000,000 rather than 1.000.000

That's it ...
Yeah the 10 + :sqrt:3 was a typo ...

The 1003th digit (after the decimal point) of (3 + :sqrt:10)^2004 ? Dunno ... probably another 9 because of the estimate :)

[quote="Arne"]Should I post some other self - made problems or do you prefer the inequalities ? :)[/quote]

I prefer your self-made problems  :mrgreen:


search
combinatorics unsolved
combinatorics

we define A, B as following
A={$ (x,y)| x,y \in \mathbb Z^{ \plus{} } , 1 \le x \le 20, 1 \le y \le 20$}
B={$ (x,y)| x,y \in \mathbb Z^{ \plus{} } , 2 \le x \le 19, 2 \le y \le 19$}

the dots in A is colored with red or blue. There are 219 red dots and 180 red dots are in B.

Also, (1,1) (1,20) (20,1) (20,20) are blue.

Now we connect the neighboring dots in left-right or up-down, with red, blue, black lines.

Here we color red when both dots are red, color blue when both dots are blue, and color black when one is blue and one is red.

if there are 237 black lines, how many blue lines are there?

This looks like the same question, no?

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=2081917455&t=152961

What's the source?

(I found that thread by searching for "black lines" (without quotes).)


inequalities
calculus
derivative
inequalities unsolved

[quote="SAPOSTO"]Let $ a$,$ b$ and $ c$ be real numbers and $ a \plus{} b \plus{} c \equal{} 3$. Prove that
$ \frac {1}{5a^{2} \minus{} 4a \plus{} 11} \plus{} \frac {1}{5b^{2} \minus{} 4b \plus{} 11} \plus{} \frac {1}{5c^{2} \minus{} 4c \plus{} 11}\le \frac {1}{4}$.[/quote]
I think

$ f(x)\equal{}\frac{1}{5x^2\minus{}4x\plus{}11}$ is concave so the inequality is done by jensen.

You are wrong . The second derivative changes its sign at the internal [0,3] .
Here is my solution . Hope it holds ...
Let $ a$ be the max of the three numbers . Then $ b \plus{} c < 2$ so $ bc < 1$

We have that $ f(a,\frac {b \plus{} c}{2},\frac {b \plus{} c}{2}) \minus{} f(a,b,c) \equal{} \minus{} \frac {(b \minus{} c)^2(25b^2 \plus{} 25c^2 \plus{} 100bc \minus{} 60b \minus{} 60c \minus{} 78)}{A}$

where A is the common denominator and it is positive.

Now we have that

$ 25b^2 \plus{} 25c^2 \plus{} 100bc \minus{} 60b \minus{} 60c \minus{} 78\leq 25(b \plus{} c)^2 \plus{} 50 \minus{} 60b \minus{} 60c \minus{} 78 \equal{} 25a^2 \minus{} 90a \plus{} 17 \equal{} (5a \minus{} 1)(5a \minus{} 17) < 0$

So $ f(a,\frac {b \plus{} c}{2},\frac {b \plus{} c}{2})\geq f(a,b,c)$ So it remains to show that $ f(a,\frac {b \plus{} c}{2},\frac {b \plus{} c}{2})\leq\frac {1}{4}$

which is equivalent to $ \frac {(a \minus{} 1)^2(25a^2 \minus{} 80a \plus{} 103)}{B}\geq 0$ where B is the common denominator and it is positive.

The one holds and we are done


[quote="rofler"]f(x)=x^3 is concave on [0,infinity)
[/quote]

$ f(x)=x^{3}$ is convex on $ [0;\infty)$ so your proof is not correct.

Actually, whole inequality is incorrect.
Take $ a=1, b=c=3$ and inequality then is: $ \frac{19^{3}}{27}\geq\frac{783}{3}=\frac{7047}{27}$, but $ 7047>19^{3}=6859$

Thank you for your replies. Is this one true? If so, can anyone prove it? Thanks

Let $ a,b,c>0$. Prove that \[ (a^{2}+b^{2}+c^{2})(a+b+c)^{2}\ge 9(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})\]

[quote="nayel"]Thank you for your replies. Is this one true? If so, can anyone prove it? Thanks

Let $ a,b,c>0$. Prove that
\[ (a^{2}+b^{2}+c^{2})(a+b+c)^{2}\ge 9(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}) \]
[/quote]

Take $ a=1$ and $ b=c=7$. Then inequality is $ 22275\geq22491$ which is not true.


[quote="zaizai-hoang"]Let $ a,b,c,d\ge 0$ such that $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 4$. Determine the best constant k such that the following inequality holds:
\[ \frac {a^2bc}{k \minus{} bcd} \plus{} \frac {b^2cd}{k \minus{} cda} \plus{} \frac {c^2da}{k \minus{} dab} \plus{} \frac {d^2ab}{k \minus{} abc}\le \frac {4}{k \minus{} 1}
\]
In my opinion, it is really hard and nice. Let prove it.  :wink:[/quote]
Nice, Hoang. It holds for all $ k> \frac{8}{3\sqrt{3}}$. :)
And I like the following is the most
\[ \frac{a^2bc}{2\minus{}bcd}\plus{}\frac{b^2cd}{2\minus{}cda}\plus{}\frac{c^2da}{2\minus{}dab}\plus{}\frac{d^2ab}{2\minus{}abc} \le 4\]
:)

[quote="zaizai-hoang"][quote="can_hang2007"][quote="zaizai-hoang"]Let $ a,b,c,d\ge 0$ such that $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 4$. Determine the best constant k such that the following inequality holds:
\[ \frac {a^2bc}{k \minus{} bcd} \plus{} \frac {b^2cd}{k \minus{} cda} \plus{} \frac {c^2da}{k \minus{} dab} \plus{} \frac {d^2ab}{k \minus{} abc}\le \frac {4}{k \minus{} 1}
\]
In my opinion, it is really hard and nice. Let prove it.  :wink:[/quote]
Nice, Hoang. It holds for all $ k > \frac {8}{3\sqrt {3}}$. :)
And I like the following is the most
\[ \frac {a^2bc}{2 \minus{} bcd} \plus{} \frac {b^2cd}{2 \minus{} cda} \plus{} \frac {c^2da}{2 \minus{} dab} \plus{} \frac {d^2ab}{2 \minus{} abc} \le 4
\]
:)[/quote]
You have made mistake in typing  :P It should be:
\[ \frac {a^2bc}{2 \minus{} bcd} \plus{} \frac {b^2cd}{2 \minus{} cda} \plus{} \frac {c^2da}{2 \minus{} dab} \plus{} \frac {d^2ab}{2 \minus{} abc} \le 2
\]
Well, I know above inequality is not hard for you, it means you will solve it easily  :D  Can you post your solution.[/quote]
No, I have not made mistake in typing, Hoang.

[quote="zaizai-hoang"]oh, my stupid  :blush: I have mistake in seeing  :D Anyway I am waiting for your solution  :wink:[/quote]
My hint is the following Hoang :)
If $ a,b,c,d \ge 0,a^2\plus{}b^2\plus{}c^2\plus{}d^2\equal{}4$ then
\[ \frac{1}{2\minus{}abc}\plus{}\frac{1}{2\minus{}bcd}\plus{}\frac{1}{2\minus{}cda}\plus{}\frac{1}{2\minus{}dab} \le 1\plus{}\frac{3}{abcd}\]
:)


trigonometry
function
algebra unsolved
algebra

Let be given $\alpha,\beta\in [0,\frac{\pi}{3}]$ satisfying:
$\frac{\sin\alpha}{\sin\beta}=\frac{\sin (\frac{\pi}{3}-\beta)}{\sin (\frac{\pi}{3}+\beta)}.\frac{\sin (\frac{\pi}{3}+\alpha)}{\sin (\frac{\pi}{3}-\alpha)}$
Prove that:
$\alpha=\beta$ or $\alpha+\beta=\frac{2\pi}{3}$

Consider function:
$f( \alpha)=\frac{\sin\alpha \sin( \frac{\pi}{3}- \alpha)}{\sin( \frac{\pi}{3}+\alpha)}.$
Equation equavalent to:
$f(\alpha)=f(\beta).$
Trivial solution is $\alpha=\beta$
If $\alpha,\beta\in [0,\frac{\pi}{3}], \ \alpha+\beta=\frac{2 \pi}{3}$, then  $\alpha=\beta= \frac{\pi}{3}.$
$f(\alpha)$ is not monotonen in $\alpha\in (0,\frac{\pi}{3})$, therefore your condition is not correct.

Oh,sorry.The right problem is:
Let be given $\alpha,\beta\in [0,\frac{\pi}{3}]$ satisfying:
$\frac{\sin\alpha}{\sin\beta}=\frac{\sin (\frac{\pi}{3}-\beta)}{\sin (\frac{\pi}{3}+\beta)}.\frac{\sin (\frac{\pi}{3}+\alpha)}{\sin (\frac{\pi}{3}-\alpha)}$
Prove that:
$\alpha=\beta$ or $\alpha+\beta=\frac{\pi}{3}$


for  evry 3 nonnegative a,b,c
$a^{3}+b^{3}+c^{3}+6abc\geq \frac{1}{4}(a+b+c)^{3}$

[quote="ashrafmod"]for  evry 3 nonnegative a,b,c
$a^{3}+b^{3}+c^{3}\geq \frac{1}{4}(a+b+c)^{3}$[/quote]
It's wrong: $a=b=c=1.$


calculus
inequalities
derivative
Euler
integration
number theory
prime factorization

prove the divergence of the sums of the reciprocals of the primes (1/2+1/3+1/5+...) without calculus

I'll only use a bit of calculus...
[hide]
Let $p_1,p_2,\ldots,p_k$ be the primes in $[1,n]$.

$\sum_{i=1}^n 1/i\leq \prod_{i=1}^k (1-1/p_i)^{-1}\leq \exp\left(\sum_{i=1}^k 2/p_i\right)$

because we have the inequality $(1-x)^{-1}\leq e^{2x}$ for $x\in [0,1/2]$, which may be proved by noting that $d(e^{2x}(1-x))=(1-2x)e^{2x}\geq 0$, so that $e^{2x}(1-x)\geq e^{0}(1-0)=1$, and the result follows.

As $n\to\infty$, the LHS goes to $\infty$, so that the RHS must go to $\infty$ also, which means that

$\sum_{i=1}^{\infty} 2/p_i=\infty$[/hide]

[quote="blahblahblah"]I'll only use a bit of calculus...
[hide]
Let $p_1,p_2,\ldots,p_k$ be the primes in $[1,n]$.

$\sum_{i=1}^n 1/x_i\leq \prod_{i=1}^k (1-1/p_i)^{-1}\leq \exp\left(\sum_{i=1}^k 2/p_i\right)$

because we have the inequality $(1-x)^{-1}\leq e^{2x}$ for $x\in [0,1/2]$, which may be proved by noting that $d(e^{2x}(1-x))=(1-2x)e^{2x}\geq 0$, so that $e^{2x}(1-x)\geq e^{0}(1-0)=1$, and the result follows.

As $n\to\infty$, the LHS goes to $\infty$, so that the RHS must go to $\infty$ also, which means that

$\sum_{i=1}^{\infty} 2/p_i=\infty$[/hide][/quote]Very interesting. To complete the proof though for me I would need to know what the $x_i$ are and where the LHS of the inequality came from. I'm thinking maybe the LHS is the $n$th harmonic number? Also, I think there is a typo in \[d(e^{2x}(1-x))=(1-2x)e^{2x}\geq 0\] since this has an infinitesimal equal to a real quantity.

I just used $d()$ to denote derivatives, for brevity.  Anyways, the $x_i$'s should be $i$'s.

As for the leftmost inequality, it follows immediately from the Euler product formula, but the reasoning is along these lines.

Every number $\leq n$ has a prime decomposition containing primes $\leq n$.  Hence

$\frac {1}{n}=\prod_i \frac {1}{p_i^{r_i}}$ for some primes $p_i\leq n$ and some integral exponents $r_i$.

Hence $\sum_i\frac {1}{i}\leq \prod_j \sum_k \frac {1}{p_j^k}$ for appropriate bounds on $i,j,k$, since we see that each summand on the LHS is contained in the RHS by the previous remarks.

But since we can write $\sum_k \frac {1}{p_j^k}=\left(1-\frac {1}{p_j}\right)^{-1}$, the result follows.

[quote="blahblahblah"]But since we can write $\sum_k \frac {1}{p_j^k}=\left(1-\frac {1}{p_j}\right)^{-1}$, ...[/quote]I don't understand this part. Wouldn't that be the value of the infinite sum as $k\to\infty$?

Just to make sure I'm understanding the rest, the bounds for the summations and products would be: \[\sum_{i=1}^n\frac {1}{i}\leq\prod_{j=1}^m\sum_{k=0}^s\frac {1}{p_j^k},\]where $m$ is the number of primes $\leq{n}$ and $2^s$ is the greatest power of $2$ that is $\leq{n}$?

If we let $s=\infty$, then all we do is increase the RHS - that is, every number of the form $1/i$ where $i$ has a prime factorization only containing primes $\leq n$ is contained on the RHS, whereas the LHS only contains numbers of the form $1/i\leq n$.


let f be the unitary polynominal .if  $x_{1}>x_{2}.......>x_{n} \in Z$ prove that  $\exists k$ satisfy
$f(x_{k})\geq \frac{n!}{2^{n}}$

Another correction: $deg f$ must be $n$. Right. Or at least less than $n$. Else we can $f$ with integer rots and choose $x_i$ among it's roots.
By Lagrange th we get that $f(x)=\sum_{i=1}^n f(x_i)\Pi_{i\neq j}\frac{x-x_j}{x_i-x_j}$. The dominant coefficient (coefficient of $X^n$) of $f$ is
$1=\sum_{i=1}^n f(x_i) \Pi_{i\neq j}\frac{1}{x_i-x_j}\le \max_i f(x_i) \sum_{i=1}^n\Pi_{i\neq j}\frac{1}{x_i-x_j}\le$
$\max_i f(x_i) \sum_{i=1}^n\Pi_{i\neq j}\frac{1}{i-j} =\max_i f(x_i) \sum_{i=1}^n\frac{n!}{\binom n i}= \max_i f(x_i) \frac{n!}{2^n}$, which enstablishes our claim.

I don't understand the part $1 = \sum \prod$. Could you explain where it comes from?


Does anyone know how I can prepare for MATHCOUNTS state and national competitions? Thanks.

Also, if anyone knows where to find lots of past MATHCOUNTS problems, please let me know as well.

Do past problems. Ask around and you can trade with other people on this website to get some. Also, Alcumus has some, FTW and the marathons have a lot of problems.

Please [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1315377612&t=258459]search[/url].

(I'm becoming like math154, lol  :) )

See the sticky at the top of this forum:
MATHCOUNTS Resources HERE!
Or try the 230+ page mathcounts marathon...

Maybe your coach has some, also.

?: With only two weeks left, time is flying. What are you all doing to prepare? How much time a day?


[quote="digger"]$x^{y}=y^{x^{2}}$[/quote]
I don't think that this is an pre-olympiad problem :P .
See here[url]http://www.mathlinks.ro/Forum/resources.php?c=1&cid=16&year=1997&p=3845[/url]

Yes, it  remains only to find negative ones but it easy now.

Sorry but I was inspired by $x^{y}=y^{x}$ which was familiar to my grandfather.  :lol:


I found a proof for this very enlightening.

Prove for all integers $n$, such that $n>0$, that:

$\binom{n}{0} - \binom{n}{1} + \binom{n}{2} + \ldots + (-1)^n\binom{n}{n} = 0$

Sorry about that.

You forget to put the =0 at the end.  So you want to prove that:

\[\displaystyle \binom{n}{0} - \binom{n}{1} + \binom{n}{2} + \ldots + (-1)^n\binom{n}{n}=0.\]

Here's my favorite method:

[hide]Use binomial theorem to expand $(1-1)^n$[/hide]

[hide]Well, we can use to the binomial theorem:
$(1-1)^n = 0 = \binom{n}{0} - \binom{n}{1} + \binom{n}{2}+ ... + (-1)^n\binom{n}{n}$

We can also use the idea of symmetry. Using the fact that $\binom{n}{k} = \binom{n}{n-k}$, everything falls exactly into place.[/hide]

Symmetry only works, from what I have observed, for n is odd.

Yes, it would seem so. Does anyone know how to manipulate this idea a bit to make it work for evens, too?


Let $ a, b, c$ non-negative real numbers. Prove or disprove that

$ \frac{ab\plus{}3c\minus{}b}{a\plus{}b\plus{}2c} \plus{} \frac{bc \plus{} 3a \minus{} c}{2a\plus{}b\plus{}c} \plus{} \frac{ca \plus{} 3b \minus{} a}{a\plus{}2b\plus{}c} \le \frac{a\plus{}b\plus{}c\plus{}6}{4}$

_________
 :(

[quote="arqady"][quote="Ligouras"]Let $ a, b, c$ non-negative real numbers. Prove or disprove that

$ \frac {ab \plus{} 3c \minus{} b}{a \plus{} b \plus{} 2c} \plus{} \frac {bc \plus{} 3a \minus{} c}{2a \plus{} b \plus{} c} \plus{} \frac {ca \plus{} 3b \minus{} a}{a \plus{} 2b \plus{} c} \le \frac {a \plus{} b \plus{} c \plus{} 6}{4}$

_________
 :([/quote]
It's wrong. Try $ a \equal{} b \equal{} c \equal{} 0.$
For non-negative $ a,$ $ b$ and $ c$ such that $ a^2 \plus{} b^2 \plus{} c^2\neq0$
it follows from two obvious inequalities:
$ \sum_{cyc}\frac {ab }{a \plus{} b \plus{} 2c}\leq\frac {a \plus{} b \plus{} c}{4}$ and $ \sum_{cyc}\frac {3c \minus{} b}{a \plus{} b \plus{} 2c}\leq\frac {3}{2}.$

By the way, for all non-negative $ a,$ $ b$ and $ c$ such that $ a^2 \plus{} b^2 \plus{} c^2\neq0$ and $ k\geq \minus{} 0.6633749...$ the following inequality holds:
\[ \frac {kc \minus{} b}{a \plus{} b \plus{} 2c} \plus{} \frac {ka \minus{} c}{2a \plus{} b \plus{} c} \plus{} \frac {kb \minus{} a}{a \plus{} 2b \plus{} c} \le \frac {3(k \minus{} 1)}{4}
\]
[/quote]

Thanks Ted, your inequality is interesting & obvoius.... thank you


We have a hexagon such that all its edges touch a circle. If five of the edges have lengths 1,2,3,4, and 5 as on the figure, how long is the last edge?

[img]http://i250.photobucket.com/albums/gg265/geometry101/HexagonImage.jpg[/img]

A. 1
B. 3
C. 15/8
D. $ \sqrt{15}$
E. Not uniquely determined, more than one possibility

[hide]COmmon tangents are congruent, so the answer is 3[/hide]

[hide="Full Solution"]

Recall that common tangents, or tangents from a point to two points on the circle, have equal length. There are 6 different "kinds" (since there are six points) in this case in such way that:

$ x \plus{} r \equal{} 3$
$ x \plus{} y \equal{} 2$
$ y \plus{} z \equal{} 1$
$ z \plus{} p \equal{} ?$
$ p \plus{} q \equal{} 5$
$ r \plus{} q \equal{} 4$

Adding all these and dividing by 2, we get:

$ x \plus{} y \plus{} z \plus{} p \plus{} q \plus{} r \equal{} \frac {15 \plus{} ?}{2}$

Note that we're solving for $ ?$ so we want to remove all other variables than $ z$ and $ p$. So, we subtract $ (x \plus{} y)$ and $ (q \plus{} r)$ to get:

$ ? \equal{} z \plus{} p \equal{} \frac {15 \plus{} ?}{2} \minus{} 6$

Solving for $ ?$ gives value of 3, which is the answer. [/hide]


For $x,y,z$ positive reals such that $x+y+z=1$ prove that

$\frac{x}{xy+1}+\frac{y}{yz+1}+\frac{z}{zx+1} \geq \frac{36xyz}{13xyz+1}$

What's the difference between question 5 and question 6? Are you still holding another cool question for us right there? :D

Sorry, I copied two times the same ineq. Now I will change GM_5  with the right question.


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

If $ M$ is $ 3\times 3$ matrix with integer entries and determinant equal to $ \minus{} 1$, and all real eigenvalues are rational, what can we say about minimal polynomial and Jordan form? It is obvious that the characteristic polynomial is of the form $ ax^3 \plus{} bx^2 \plus{} cx \minus{} a$, and it can have 1 or 3 real roots, also we know that the root is of the form $ \frac {d_1}{d_2}$ where $ d_1, d_2$ are divisors of $ a$. Is there anything else to say?

Actually, that characteristic polynomial is $ x^3\plus{}ax^2\plus{}bx\plus{}1$. Rational eigenvalues are unlikely, since they could only be $ \pm 1$.

The minimal polynomial is guaranteed to be equal to the characteristic polynomial unless there's a double root. The only possible double roots are at $ \pm 1$.

Of course, it should be monic..then everything becomes clear. Thank you [b]Jmerry[/b]  :)


We have added a new community feature to the site.  You'll see a Blogs button above.  There we will building a ring of blogs of AoPS users.  If you maintain a blog and would like to share it with other problem solvers, please submit it to be included.  Our blogs rings supports a limited ratings system and a favorites list.  Please check it out and comment here if you find anything broken.  As you'll see from my (new) blog, I just built this over the last week, so there's bound to be plenty of little things that don't work quite right.

[quote="MithsApprentice"]Does our blog have to be math-related or can it just be a regular old web journal?
[/quote]No one imposes any restrictions of what you write in your blogs. There are your blogs. We just provided the means to access them from the site. In plus you can easily find links to your user friends :)

[quote="Valentin Vornicu"][quote="Rep123max"]Is there a certain blog that you got that allows you to Tex? Does it cost money?[/quote]This addon is actually not a single feature of the site, but just a shell for the LiveJournal site. Since they don't have $\LaTeX$, it's impossible to have one.

Maybe we'll create a blog of our own, but that'll be a little bit later, when we'll have the time for it ;)[/quote]

Actually, this is a wrapper for more than LJ - it holds any site.

If you have your own server, you can make a blog that does LaTeX.  The same LaTeXrender that powers this board can be embedded into blogs such as WordPress (that's how mine has LaTeX - thanks, stevem)

[quote="MithsApprentice"]Does our blog have to be math-related or can it just be a regular old web journal?

And also, is there going to be a place to post links for already established AoPS webrings? For example, we have one going on Xanga that has become quite popular...

And then...what're those counters next to the links...

And how are "Favorites" done?[/quote]

Valentin took care of the first question.

Post a link to the ring here for now.

I'll be putting up more instructions about what the images are - that one depicts highly visited blogs (ones highly visited through the site).

The Favorites is so that when we have 150 blogs on the site, you can have a view that only includes those that are your favorites rather than having to scroll through the list to find your friends.

Aiighty

[url=http://www.xanga.com/groups/group.aspx?id=143909]Xanga AoPS Blogring[/url]


trigonometry
quadratics
function
algebra
quadratic formula

Hi, I need help with these trig related problems.. The problem is, I don't know how to approach these questions.
I need to hand these questions in to my teacher, so it is greatly appreciate if you [size=150][b]only give advice on how to approach these questions, and no exact solutions[/size][/b].. Thank you.

*doesn't seem very clear, but the π is pi. Only variable is x.
Prove:
1. (1+sinx+cosx) / (1+sinx-cosx)  = cot (x/2)
I tried using reciprocals of the half-angle identities of tan, but I have no idea what to do next..

2. [ (sin( x - π) / cos( π + x) ] – [ cos ((π/2)-x) / sin(-π-x) ] = (sinx – cosx) \ cosx
I tried using Sum and Difference Identities for the left side, but it doesn't seem to be going anywhere.

Solve for x
3. Tan^2 (2x – (π/12)) = 3
I think I need to use the Double-Angle identity for tan, but what do I do with the π/12?

4. Develop a formula for tan (a+b-c) in terms of tan a, tan b, tan c

5. Solve 4+3sin2x = 3sinx+4cos^2
I tried changing the cos^2 using pythagorean identities, and I made it all equal to 0, but it's not possible to use quadratic formula or factor. I also changed the sin2x using double-angle identities.

Am I on the right track for any of these problems? [/b]

EDIT: I just realized I can use the unit circle for #2..Can't believe I didn't think of it.
I also realized I can also use the sum dif identity for tan in #3 instead of double angle..but for some reason, I don't have it in my notes on what to do after I substitute it into the identity.

[hide="Problem 1"]instead of manipulating the right hand side, substitute $ 2\theta\equal{}x$ and use the identities for double angle on the left hand side. try to cancel the $ 1$s with different identities for $ \cos{x}$.[/hide]

LHS: Multiply numerator and denominator by $ ((1\plus{}\sin x)\plus{}\cos x)$ show that $ \frac{1\plus{}\cos x}{\sin x} \equal{} \cot \frac{x}{2}$

[hide="Problem 3"]$ \tan^2 \left(2x \minus{} \frac {\pi}{1}\right) \equal{} 3$
$ \tan (A \pm B)\equal{}\frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}$

You can use the sum of angle identity for the tangent function to "split" the argument of the function into two, so to speak. After that, it's easy.[/hide]

For #5, your method should be working. Expressing $ \cos^2 x$ as $ 1 \minus{} \sin^2 x$ (there's a $ 4$ on the left and a $ 4\cos^2 x$ on the right) and expressing $ \sin 2x \equal{} 2\sin x\cos x$ allows you to factor out $ \sin x$.


ratio geometry circumcircle trigonometry

The triangle ABC is right triangled at A. Ratio of the radius of the circle circumscribed to the radius of the circle ascribed to the hypotenuse is √2 : (√3 + √2). Find the angles B & C and the ratio of the two perpendicular sides of the triangle.

I assume it is the escribed circle (excircle-Впи́санная окружность).

(Hmm... "escribed" and "excircle" aren't in FF dictionary, but ascribed is  :) )

Впи́санная окружность = incircle...

Lapsus mentis, while typing:

Bневписанная окружность.  :blush:

So what does the problem mean?The ratio between the circumcircle and in circle is sqrt2:sqrt2+sqrt3??

Not the incircle, excircle!

Yes, the ratio between their radia is $ \sqrt{2}: (\sqrt2\plus{}\sqrt3)$. And you should know what the circumradius in right triangle is... And the formula for the excircle's radius. Or you can simply derive it for this particular case.

[hide="hint"] write the sides as $ a$ ,$ a\cos\theta$ and $ a\sin\theta$ and use $ a/2$ as circum radius and $ area/(s \minus{} a)$ as the exradius . and solve for $ \theta$ we r done [/hide]


A light bulb operating on 110 volts, has 0.50 ampere of current passing through it. How much energy does the bulb convert in 5 minutes?

[hide]Q=It=150 C. The energy it coverts is qV=16 500J[/hide]


calculus
derivative
function
search
real analysis
real analysis unsolved

Give an example of a function f:R-R that is continuous in every x ε R but
a)is not differentiable in any x in R
b)has no partial derivative in any spot x in R

[quote="jkarbas"]Give an example of a function f:R-R that is continuous in every x ε R but
a)is not differentiable in any x in R
[/quote]

Search for the "Weierstrauss Function".

i appreciate ur help...
i am aware that there are several functions that are nowhere differentiable but continuous(weierstrass,van der waerden,blackmange function)but is there any with no partial derivatives at all...
 :?:

What do you mean by [b]partial [/b]derivative? The usual meaning of that word in English applies to functions of several variables, which doesn't seem to be what we're discussing?

For existence the method outlined in my second post here: http://www.mathlinks.ro/Forum/viewtopic.php?t=150273 works fine.

[quote="spix"]I think he means derivatives to the left or to the right of the point.
$f(x)=\sum_{n=0}^{\infty}\frac{\sin((n!)^{2}x)}{n!}$ can this work?
For point a) http://www.math.tamu.edu/~tom.vogel/gallery/node7.html[/quote]

Yes, this http://www.math.tamu.edu/~tom.vogel/gallery/node7.html is really good example. it was also if I remember right the first example of functions with this property was represented by Euler.


How many ways are there to express 1000000 as a product of exactly three integers greater than 1? (For the purpose of this problem, $abc$ is not considered different from $bac$, etc.)

No, that's wrong, the actual number is less than 208.

I'm not very good at combinatorics (or at math in general compared to you guys), but would you please take some time to point out what is wrong with my idea for a solution?

$10^6 = 5^6\cdot2^6$. This means that there are twelve factors that need to be arranged into three different numbers.

If we start by investigating the amount of ways that we can arrange twelve unique factors into three different numbers, here is my proposed argument: If we are to put twelve objects in three different "boxes" without any restrictions this can be done in $3^{12}$ ways. However, we have to subtract the $3$ combinations where there are two empty boxes, and the $3\cdot(2^{12} - 2)$ combinations where there is one empty box*. Now, for the remaining combinations of "boxes" we can for any given combination rearrange the boxes themselves in $3! = 6$ different ways, so that only the order between the boxes has been changed, not the numbers they contain. Since we count for example $abc$ and $bac$ as the same three numbers, this means that each combination has $5$ other "equivalent" combinations. And so, we get the total number of ways of doing this as:
\[ \frac{3^{12} - 3 - 3\cdot(2^{12} - 2)}{6} = 86526  \]

Now, since we have 6 5-factors for any given combination we can rearrange the 5s in $6!$ ways without changing the numbers in any way. Similarly, the 2-factors can also be arranged in $6!$ different ways. So, we should divide by $720^2$, which is obviously wrong.

*My reasoning for this is: First choose the one box to be empty. Then the 12 objects are to be put in two different boxes, but we must subtract the two cases where there is another empty box.

Let's talk only about puting the 2's into the boxes. You suppose, that giving 3 into the first box, 2 into the second and 1 into the last can be done in $6!=720$ ways, but it can be done only in
$\binom 6 3 \binom 3 2 \binom 1 1=\frac{6!}{3!2!1!}=120$ ways. Do you see it?
After PMing Arne I know that 64 is also not the right answer. I've been looking for a mistake in my computations and founf it. I hope now I'm finaly right (3rd attempt must go, 3 is a lucky number). So:

Let a,b,c be ordered by the greatest power of two, that divides them. We can split it into three cases: all the powers are different-there are 3 ways T={(0,1,5), (0,2,4), (1,2,3)}. Two of them are different-also 3 ways D={(0,0,6),(1,1,4), (0,3,3)} and all are the same M={(2,2,2)}. With powers of five is it similar.

If powers of 2 in a,b,c are in T, and powers of 5 are from T, there are 6 ways to join the powers of 2 and 5, if powers of 5 are in D there are only 3 ways and if they are from M there's one way. So if powers of 2 are in T, there are 6.3+3.3+1=28 ways af for each triplet of powers of  2.

If powers of 2 in a,b,c are in D, and powers of 5 are from T, there are 3 ways to join the powers of 2 and 5,  if powers of 5 are in D there are only 2 ways and if they are from M there's one way. So if powers of 2 are in M, there are 3.3+2.3+1=16 ways af for each triplet of powers of  2.

If powers of 2 in a,b,c are in M, and powers of 5 are in T, D or M, there is always 1 way to join the powers of 2 and 5.  So if powers of 2 are in M, there are 1.3+1.3+1=7 ways af for each triplet of powers of  2.

So there should be 3.28+3.16+1.7=84+48+7=139 ways to fulfil the task, but it's easz to show that in 24 of them is exactly 1 of the numbers equal to 1 and in 1 are 2 of them aqual to 1. So there are 139-24-1=114 ways.

Yes, I do see why it is wrong now... is everything leading up to the last part correct, i.e. is that the correct way to calculate the amount of ways to put 12 different factors into 3 numbers?

This problem is way over my skill level :(

Yep, Kondr, you got the right answer now - 114!

A confusing problem, isn't it? It's so easy to make a mistake...

I thought of something else...

First we distribute the 6 2s among the three numbers a,b,c ( I like seeing these as boxes). This can be done by putting 8 positions in a row where in each position there is either a 2 or a seperator (my professor showed us this trick for a much simpler problem), so the 2s can be distributed in $\binom{8}{2}$ ways. Using the same reasoning for the 5s, we see that the amount of ways to distribute the 2s and the 5s is $\binom{8}{2}^2 = 784$.

Now, I'd like to just be able to divide by $3! = 6$ here by my reasoning above, but I realize that it is wrong since we have duplicates among the numbers that cannot be arranged in 6 different ways. There is only one combination with $a = b = c = 2^2*5^2$. For two equal numbers there are 45 different combinations (I counted them. This is obviously a very bad solution), where there are 15 "unique" ones if you don't count $abc$ and $bac$ as different arrangements. And so, we have that the total number of combinations WITHOUT duplicates as $784 - 45 - 1 = 738$. Now we can divide by six to get $738/6 = 123$.

Adding the duplicates again we see that the amount of arrangements not counting $abc$ and $bac$ as different is $123 + 15 + 1 = 139$.

Now the only problem is that we allowed "boxes" to be empty in the beginning of the problem, i.e. we have allowed a,b and c to be 1, so we must count all combinations where one of the factors is 1. If we think about the boxes again and let one of them be empty (which one doesn't matter because we don't count $abc$ and $bac$ as different) we must now arrange the 2s and 5s into the other two boxes, which by pretty much the same arguments above all over again can be done in 24 ways if we count $ab$ and $ba$ as the same numbers (I also manually counted these at first :o). There is also one combination where two numbers are equal to $1$, so we get that the total number of ways of arranging the numbers without letting factors be 1 is $139 - 25 = 114$.

edit: Away with all the edits. I am pretty convinced it's right now. If it isn't I'll cry. This was quite the way to spend 2.5 hours of the evening. I would never have solved this problem during a time constrained test (plus I cheated and looked at Kondr's solution to see what I was doing wrong towards the end :blush:).

Well, the problem is solved, but both my and Kalles solutions use a lot of manual counting.
Is there a nice formula for the number of  triplets $(a,b,c)\in \mathbb N^3$,$a\leq b \leq c$ such that
$abc=\prod_{i=1}^{n} {p_i}^{e_i}$? Then the task will be easily finished by the number $X$ of pairs
$(x,y)\in \mathbb N^2$, $x\leq y$ for which $xy=\prod_{i=1}^{n} {p_i}^{e_i}$, because we know
$X=ceil(\frac 1 2\prod_{i=1}^{n} {\left(e_i+1\right)})$, where ceil(x) is a the least integer greater or equal to x.

The only solution I knew is with a lot of "manual" counting as well...

Is this considered a difficult or easy competition problem? I just feel so inferior to and jealous of people who can solve difficult problems like this in high school. I can barely solve any of the problems on these forums, and I'm finished with high school. :(

Do you have a third solution, Arne? I'd like to see it.

The official solution is very similar to yours.

Well, this is definitely not a very hard olympiad problem, but I wouldn't call it easy either. It's certainly not my type of question :D


If a,b,c real numbers

1)$ \sum (a \minus{} b)^6 \minus{} 3(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2 \equal{} 2(\sum a^2 \minus{} \sum ab )^3$

2)$ \sum (a \minus{} b)^7 \equal{} 7(a \minus{} b)(b \minus{} c)(c \minus{} a)(\sum a^2 \minus{} \sum ab)^2$

Try, if possible, to find a solution without expansion. :)

I found a proof using the Log function Expansion Method.


Find all $ n$ natural such that $ 2^{n}-1$ is divisible by $ 7$ and
Show that there's no natural $ n$ such that $ 2^{n}+1$ is divisible by $ 7$ ...

[hide]
$ n=3k$
and $ 2^{n}=1,2,4 mod7$
might be wrong im quite new to mods
[/hide]


This is the miquel point of a complete quadrilateral.

[quote="pianoforte"]Given concave quad ABCD with concave angle C, let E and F be the points AB and AD such that they are on the extensions of DC and BC. Prove that the circumcircles of AFB, ADE, BCE, DCF, are concurrent.[/quote]
[hide="Solution"]
Denote $ P$ to be the intersection between the circumcircles of $ \triangle BCE$ and $ \triangle DFC$.  Thus, $ \angle CPF \equal{} \angle CDA \equal{} \angle EDA$ and $ \angle CPB \equal{} \angle AED$.  It follows that $ \angle BPE \equal{} \angle AED \plus{} \angle ADE \equal{} 180 \minus{} \angle DAB$, so $ ABPF$ is cyclic.  Hence, $ P$ lies on the circumcircle of $ \triangle ABF$ and thus the circumcircles of $ \triangle ABF$, $ \triangle BCE$, and $ \triangle DCF$ are concurrent at $ P$.  Similar reasoning gives that the circumcircle of $ \triangle ADE$ also passes through $ P$ so the circumcircles of all four triangles are concurrent at $ P$, as desired. [/hide]

A very nice apllication for this problem is [b]Miquel's pentagram[/b]
[color=brown]It is given a pentagon and the circumscribe circles of the triangles formed by extending the sides of the pentagon are intersecting two by two in 5 concyclical points[/color]

Dear Mathlinkers,
for the last message, you can see
http://jl.ayme.pagesperso-orange.fr/   vol. 4, Les pentagrammes de Miquel et Morley, p. 1-3.
Sincerely
Jean-Louis


When I'm looking at the list of blogs, and click on rrusczyk's name as the author of the current board announcement, I am actually sent to my own user profile. This is not the case when I click on rrusczyk on the forum index.

Heh. When I do that, I get sent to my own profile as well. I think some bug makes it so that it links to the profile of the logged-in user.


Hello Everyone,

My 7th grade daughter is practicing for her FCAT, and she asked me for assistance with some of her math problems. I know how to do a few problems, but, I have trouble with the others  :(  . It's been awhile (almost 25 years)  since I've used these formulas. I was wondering if someone could help me with following problems. These problems deal with formulas:

1. A can of soup is cylindrical in shape. Find the total surface area of this right circular cylinder can of soup that is 3 inches high and has a 1.4 inch radius. Let Pi equal 3.14.

2. An architect has designed a circular hot tub. She wants to surround the hot tub with a decorative border. If the hot tub has a radius of 7 ft, what is the lenth of the decorative border? Let Pi equal 3.14.

3. A square room has an area of 144 square feet. What is the perimeter of the square room?

4. A circular patio has a circumference of 154 feet. What is the diameter of the patio? Use 22/7 for Pi.

5. The perimeter of a retangular piece of property is 1100 yards. If the width of the property is 250 yards, what is the area of the property?

I hope this is not asking too much. Any assistance on any (if not all) of the above problems is greatly appreciated. Thanks in advance for your help.  :)

Ann

[hide=""Cryptic hints only""]
$S=2\pi rh+2\pi r^2$

$C=2\pi r$

$P=4s$

$d=\frac{C}{\pi}$

$l=\frac12 P-w$, $A=lw$[/hide]

Thanks for the hints 3X.lich.

I think I may have the solved problems 1, 4, and 5.  However, I still have trouble with problems with 2 and 3.

Which formula would I use for number 2?

I think I would use P=4s for number 3, but what is the area of a square?

Need help  :(  ! Thanks again

Number 2 is there. $C=2\pi r$

Area of square $A=s^2$ so $s=\sqrt{A}$ and so $P=4s=4\sqrt{A}$

(maybe that's what she intended to do)
j/k

3. A square room has an area of 144 square feet. What is the perimeter of the square room?

[hide="liek omg t3h solution"]
Well, the formula for a square is s:^2: = a, where s = side length and a = area
The formula for the perimeter of a square is p = 4s, where p = perimeter and s = side length.

a = 144
s:^2: = 144
s = 12

p = 4s
s = 12
p = 4(12)
p = [b]48 feet[/b]
[/hide]

I think this person is new...

so we should welcome her..

I agree (I removed the earlier posts).  Note that some people are too shy to post initially and thus their parents will help them.

hey wait a minute FCAT and 7th grade.
lol i m in 7th grade and i take FcAT's
coz its a Florida test and i live in
Florida!
i would be glad to help her!
hey dont u think i sound a bit dumb
 :blush:

Wow...what a small world.  I am from Florida and I never would have imagined that a lot of people from Florida came to this site. What a small world eh?

If you daughter still needs FCAT practice, I would reccomend you go to http://fcatexplorer.com. Your daughter's teacher should be able to get her an account if she needs one.  That site has good problems and has helpful explanations. :D

(For those of you unfamiliar with the FCAT...It is Florida's standardized assesment test.)

Oh yeah! I almost forgot! Welcome! Hope you enjoy this site as much as I do! :lol:  :P  :D

What's the difficulty for FCAT???

if it's about the level of middle schoolers, I would recommend you to get problems from past MATHCOUNTS problems...
( Farazy has a great website on his own..Has all the past practices, more than MC websites )

and by the way,

practicing right before the test won't help at all...

you have to do math constantly

The FCAT is standardized...Of course, I dont have a copy of the test to give you guys (that would be illegal) but it basically encompasses what an average kid in a specified grade learns (not gifted or advanced).  They have different tests for different grade levels.  The test really isn't extremely hard if you have been taught well by your teachers (and pay attention in class), and the test has no major impact on you unless you miserably fail it (you can get put in remediation classes).  The only time it makes a BIG difference is when you take it for the last time in highschool in 10th grade.  That score decides whether you graduate from highschool or not...and I think it goes on your college resume (not entirely sure though).

In short...the test is definately easy for any AoPSer.  The majority of the other kids do well too...Unless they don't care to do well (and bubble in random answers *rollseyes*) or their teachers don't teach them well enough.

But the test isn't close to MatchCounts at all... It's a lot less difficult.  MathCounts is something that your daughter might like...she has one more year ya know ;) She should give it a try :D

But [url=http://fcatexplorer.com]FCAT Explorer[/url] is more close to the actual test.

:D  :P Oh yeah...If anyone has some MathCounts stuff they would like to contribute to my site...email me or pm me.  I am trying to create a really good site to give practice to kids without it avaliable to them (as was the case for many people on my team. our coach is [size=59]horrible[/size]).

BY THE WAY SOMETHING VERY IMPORTANT...

[size=200]THEY HAVE A FORMULA SHEET IN THE TEST FOR YOUR DAUGHTER TO USE INCASE SHE FORGETS ANYTHING! :D [/size]

Hey, is it what we call CAT???

California Assesment Test??

I failed that test last year :D

last year i ran out of time
coz i had this article
on white house and
abraham lincoln
and i just hate politics
and i had to start reading
it all over again coz i
always forgot wat
i just read
i hate those articeles
but still i did a gud job!
out of 5 i got 4 in math
and 4 in reading
coz i just came to USA
last year and i couldnt
believe i got a 4!

hey im not shy 2 post or anythin! prolly coz i hve a bigmouth and talk to much!  :D

that's kinda hard to read with all the slang stuff...

How can you fail a standardized test? They're waaaaaaaaay too easy.

Every year when I take the MSA (for MD) I always finish each section at least 30 minutes before we're supposed to, and then I use my scratch paper and compass to experiment with mathematical theorems about geometry. It's fun! I've made up lots of formulas during the time I have left after MSA  :D  :D


[quote="Hong Quy"].$A$ and $B$ are two subsets of $N$ such that $A\cap B=\phi$ and $A\cup B=N$.
Suppose both $A+A$ and $B+B$contain finite prime numbers. (here $A+A=\{a+b|a;b\in A\}$)
Prove that:
$A =\{2k|k\in N\}$or $A =\{2k-1|k\in N\}$.[/quote]

Yes man and i think you're surely know why and where u can find the only solution, other hand, it is on AMM 1984, proposed by PE


I know I'm going to qualify for the AIME because I have for the last 2 years, but I was wondering if I would qualify for the usamo regardless of my amc 10 and amc 12 scores if I hit the AIME floor score since I am in 10th grade.  Also, what is the floor score usually?  I know it was 6 last year but I couldn't find what it was other than that year.

1. No. See [url]http://www.unl.edu/amc/e-exams/e8-usamo/usamo.shtml[/url].
2. In 2006, it was 8... it has always been between 6 and 8, if I'm not mistaken. However, it will likely be closer to 6 now that 500 people qualify for USAMO...

I have a question too.
Let's say you qualify through the 12,not the 10, even if you're below 10th grade. Will the floor value still count if you didn't make it through the 10?

What do you mean? If you qualify through the 12... you're in...

Argh, letme try again.
If you qualify to the [b]AIME[/b] through the 12, but not through the 10, and you get exactly the floor score on AIME. If you're in less than 10th grade, will the floor still count for you (because you didn't make it through the 10)?

He means:

Lets say on the AMC 12 you get a 103.  That is enough for qualifying for the AIME through the 12.

Then you take the AMC 10 (B test) and you get a 114.  That 114 is not good enough to qualify you for the AIME.

However, at the end when they do the AMC + AIME floor regradless of wheter it is AMC 10 or AMC 12, can they still use that 114 AMC 10 score?

Even though that AMC 10 score was not good enough to qualify you for the AIME, can they still use it since out of your two AMC scores, that one is still the highest?

Cause if they do, that just shows the loophole in that AMC + AIME system regradless of the AMC type.

Another loophole whould be the fact that even if you get like 2 above the AIME floor, you might still not qualify because you didnt get high enough on the AMC (if you took the 12 since someone with 103 on 12 and 8 on AIME is considered to not get in but a person with 126 on AMC 10 and 6 on AIME will get in if the floor is 186 which is totally unfair since the first clearly is more impressive)

NOTE: even if you get higher than the floor on the AIME and you are under 10th grade, that still doesnt mean you will qualify for the AIME (theoritically last year, you couldve gotten 2 higher than the floor on the AIME as a 10th grader and below and still not have qualified)

No, I *believe* they use the AIME qualifying score... of course, you don't have to worry about this if you get like a 10 on AIME. So do that :)

If you were able to get a 103 on the AMC 12, that would be monstrous.

Also, the 2005 floor was 9 (it was a particularly easy year), but it will probably be near 6 in the following years due to the fact that the number of USAMO qualifiers doubled in 2 years. But, your time would be better spent trying to maximize your score on the AIME (or whatever test you are preparing for), than trying to find out what you should get on the AIME. What you get is what you're going to get, and they'll tell you if you made the USAMO or not.

Read the selection process step carefully from the 2007 USAMO Teachers' Manual at:
http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2007-ua/2007usamoqual.shtml
The process will be the same this year.

Steve Dunbar
MAA Director, American Mathematics Competitions


Yup 8 is Roman Catholic because thats the religion most everyone was then.

Protestant?

Almost no one in America was Roman Catholic, relatively speaking.

what's cool is both senators from masatchusets are catholic that is john kerry and edward kennedy.

also gray davis was catholic and i think bustamante is catholic (99% of mexicans are roman catholic except for in utah).
i can be considered mormon which i think isn't part of protestantism or maybe it is

[quote="churchilljrhigh"]scout guy-kit carson[/quote]

Kevin is right.  My hometown (well, my current town of residence) is named after Kit Carson.

1. Lansford Hastings?

I think if you REALLY desired the answers, they're easy enough to find on the internet.


calculus
integration
limit
function
calculus computations

lim(h->0) (1/h)* [(integral from 2 to 2+h) root(1+t^3) dt]

is this as simple as solving the integral on the given interval, then solving the limit for h? If not, any help would be much appreciated... thanks

[quote]is this as simple as solving the integral on the given interval, then solving the limit for h?[/quote]
No, it's simpler than that. This is good thing, because you're not going to be able to "solve the integral".

Look at the problem this way. We have:

$\lim_{h\to0}\frac1h\int_2^{2+h}f(t)\,dt$

where $f$ is a function that is continuous at $2.$

This is essentially the fundamental theorem of calculus; the only things that matter about $f$ are that it is continuous and what its value is at $2.$

Ah, think about the definition of derivative.

$\lim_{\Delta x \to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$

Here we have $f(x)=\int^{x}_{2}\sqrt{1+t^{3}}dt$

Then you'll reason with $f(2)$ and then $f(2+h)$


trigonometry
trig identities
Law of Cosines
circumcircle

[b]Show that for any quadrilateral inscribed in a circle of radius 1, the length of the shortest side is less than or equal to $ \sqrt{2}$[/b]

[hide="my approach"]imagine such a quadrilateral. the shortest side must have a central angle of corresponding chord that is in the range of $ 0<\theta<\frac{\pi}{2}$. use law of cosines on the max and min value of $ \theta$ to get the shortest side $ x$ is in the range $ 0<x<\sqrt{2}$
is that ok?[/hide]

Does it lead you to a correct proof?  If so, then your approach works.

Would this work?

The shape with the greatest four sides lengths would be the square with side length $ \sqrt{2}$, so it follows that any other quadrilateral has at least one side less than $ \sqrt{2}$?

[quote="mihail911"][b]Show that for any quadrilateral inscribed in a circle of radius 1, the length of the shortest side is less than or equal to $ \sqrt {2}$[/b]

[hide="my approach"]imagine such a quadrilateral. the shortest side must have a central angle of corresponding chord that is in the range of $ 0 < \theta < \frac {\pi}{2}$. use law of cosines on the max and min value of $ \theta$ to get the shortest side $ x$ is in the range $ 0 < x < \sqrt {2}$
is that ok?[/hide][/quote]
by ptolemy's theorem if $ a,b,c,d$ were the sides and $ e,f$ the diagonals then
$ ac\plus{}bd \equal{}ef \leq 2.2\equal{}4$---------------*
also let $ x\equal{}min[a,b,c,d]$ thus we have from * $ 2x^{2} \leq ac\plus{}bd \leq 4 \Rightarrow x \leq \sqrt{2}$

[b]Similar question[/b]
$ ABCD$ is a quadrilateral inscribed in a circle of unit radius.if $ AB.BC.CD.DA \geq 4$ .Prove that $ ABCD$ is a square.

Contradiction also works: if the length of all four sides are greater than $ \sqrt {2}$, then all angles are greater than $ 90^\circ$, which is false. Thus the given statement is true.

Tim, why are all the angles greater than 90 degrees when the lengths of all four sides are greater than $ \sqrt{2}$?

The smallest angle must have measure at most 90, so the side opposite this angle can have length at most $ \sqrt{2}$.

Hmmm...is the shortest side in a quadrilateral always opposite the smallest angle?

[quote="10000th User"][quote="aufha"]Tim, why are all the angles greater than 90 degrees when the lengths of all four sides are greater than $ \sqrt {2}$?[/quote]Because the radius of the circumscribing circle is 1. Apply Law of Cosine, for example.[/quote]

I've read the problem, but I'm apparently missing something obvious. I drew a circle with radius one and a quadrilateral in it, but I still don't see what you mean. Why does that the radius of the circumscribed circle is 1 imply that all the angles greater than 90 degrees when the lengths of all four sides are greater than $ \sqrt {2}$? How exactly did you use the law of cosines?

hmmmmmm.......

Let $ w\ge x\ge y\gez>0$ be the measures of the central angles, to a smaller central angle opposes a smaller side...

We have that $ w+x+y+z=360\ge 4z$ so $ 90\ge z$ and the side oposite to "$ z$" is at most $ r\sqrt{2}$

(the fundament in this is semi evident, you can always use Pithagora's if you want)

 :)

(reply to aufha): Suppose $ O$ is the center of the circle and $ AB$ are two points on the circle. We have $ AO \equal{} BO \equal{} 1$. Suppose $ AB > \sqrt{2}$. By the law of cosines on $ \angle AOB$, $ 1^2 \plus{} 1^2 \minus{} 2 \cos( \angle AOB) \equal{} AB^2 > 2$. Canceling the $ 2$s gets $ \minus{}2 \cos( \angle AOB) > 0$, or $ \cos( \angle AOB ) < 0$. This is true only if $ \angle AOB$ is obtuse.


You have a equilateral triangle with area 4 rad(3). Then you have a square inscribed in it. What's the length of the square?

P packages of food is needed for two people to survive for 1 month.
You have Q packages. How long can C people survive? Express answer in terms of P, Q and C.

[hide="second question"]
$\frac{P}{2}$ packages are needed per person per month.
If we have $Q$ packages, then $\frac{2Q}{P}$ people can last a month. But we have $C$ people, so we need to divide by $C$ to get the number of months they can survive for:

$\frac{2Q}{PC}$ [/hide]

[b]First question.[/b] See the attached diagram.

From $\frac{a^{2}\sqrt{3}}{4}=4\sqrt{3}$ we get $a=4$

From $\triangle CUQ\sim\triangle CVA$ we get

\begin{eqnarray*}\frac{CU}{UQ}=\frac{CV}{VA}&\implies& \frac{h-x}{x/2}=\frac{h}{a/2}\\ &\implies& ah-ax=hx\\ &\implies& x(a+h)=ah\\ &\implies& x=\frac{ah}{a+h}\end{eqnarray*}

Now put $h=\frac{a\sqrt{3}}{2}$:

\begin{eqnarray*}x &=& \frac{\frac{a^{2}\sqrt{3}}{2}}{\frac{a(2+\sqrt{3})}{2}}\\ &=& \frac{a\sqrt{3}}{2+\sqrt{3}}\\ &=& a\sqrt{3}(2-\sqrt{3})\\ &=& a(2\sqrt{3}-3)\\ &=& 4(2\sqrt{3}-3)\end{eqnarray*}

And the perimeter of the square (if that's what's asked for) is $p=4x=16(2\sqrt{3}-3)$


Solve in $ \mathbb R$ the equation $ \sqrt{1\minus{}x}\equal{}2x^2\minus{}1\plus{}2x\sqrt{1\minus{}x^2}$.

[hide]
Clearly $ \minus{}1\le x\le 1$.so we can let $ x\equal{}\cos{y}(0\le y\le \pi)$,then
$ \sqrt{1\minus{}x}\equal{}\sqrt{1\minus{}\cos{y}}\equal{}\sqrt{2\sin^2{y/2}}\equal{}\sqrt{2}\sin(y/2)$,
$ 2x^2\minus{}1\equal{}2\cos^2{y}\minus{}1\equal{}\cos{2y}$,and $ 2x\sqrt{1\minus{}x^2}\equal{}2\cos{y}\sin{y}\equal{}\sin{2y}$,

so we have the equation is equivalent to
$ \sqrt{2}\sin(y/2)\equal{}\cos{2y}\plus{}\sin{2y}\equal{}\sqrt{2}\sin(2y\plus{}\pi /4)$

$ \Longleftrightarrow \sin(y/2)\equal{}\sin(2y\plus{}\pi /4)$.
so $ 2y\plus{}\pi /4\equal{}y/2\plus{}2m\pi,2y\plus{}\pi /4\equal{}(\pi\minus{}y/2)\plus{}2m\pi(m\in \mathbb{Z})$,because $ 0\le y\le \pi$,we have $ y\equal{}3\pi /10$,

thusthe root is $ x\equal{}\cos(3\pi /10)\equal{}\frac{\sqrt{10\minus{}2\sqrt{5}}}{4}$.[/hide]


Let $e,\ p,\ k$ be positive numbers. $f\left(  x\right)  \symbol{126}x+ax^{p+1}$ as $x\rightarrow0^{+}.\ g\left(  x\right)  \symbol{126}ex^{k}$ as $x\rightarrow+\infty.\ f\left(  n,x\right)  $=$f\left(  ...f\left(  f\left(  x\right)  \right)  ...\right)  \ $ is the nth iteration of $f\left(  x\right).$

Find \[H\left(  a,p,e,k\right)  =\lim_{x\rightarrow0^{+}}f\left(  \left[  g\left(
1/x\right)  \right]  ,x\right)  /x.\]

As we all know, $\left[  x\right]  $ is the integer part of $x$.

[hide="Those who clicked here own me a proof"]
If $p>k$, $H\left(  a,p,e,k\right)  =1.$
If $p=k$ and $ape<1$, $H\left(  a,p,e,k\right)  =1/\sqrt[p]{1-ape}.$
If $p<k$ and $a<0$, $H\left(  a,p,e,k\right)  =0.$
The limit does not exist in all the other cases.
[/hide]

I didn't read so carefully, but if it is just the analysis of asymptotics of $f^{\circ n}$ then we discussed this type of problem here many times.  Search for "asymptotics" and my username (and maybe Myth or alekk?) to find the threads.


$a,b,x,y \in \mathbb{N}$ such that $a^2+b^2 \mid ax+by$. Prove that $gcd(x^2+y^2, a^2+b^2)=1$.

Set $a=x , b=y$ to see that the statement is wrong.


um...
i really don't understand what that is

[quote="Ineedhelp123"]how do u convert [i][size=150]([/size]b
                          ___
                          a [size=150])[/size] [i] into [i][size=150]([/size]_b__
                                                                 2a[size=150])[/size][/i] ????? :?:  :?:

sorry if u cant read this I havent realy found out how 2 use this yet
if u cant read this please post a reply telling me 2 make things clear
I would also like it if u could post a reply telling me how 2 use this topic posting thing  :lol:  :)  :ninja: [/u][/quote]

Do you mean convert $ \dfrac{a}{b}$ into $ \dfrac{b}{2a}$ ?

[quote="Ineedhelp123"]I would also like it if u could post a reply telling me how 2 use this topic posting thing  :lol:  :)  :ninja: [/u][/quote]

There is a forum instructing you how to use LATEX. You should go look at it :)

and yes it is $ \dfrac{a}{b}$ into $ \dfrac{b}{2a}$  :)

What do you mean convert $ \frac {a}{b}$ to $ \frac {b}{2a}$?

probably what do u need to multiply to $ \frac{a}{b}$ to get $ \frac{b}{2a}$

[quote="yaofan"]probably what do u need to multiply to $ \frac {a}{b}$ to get $ \frac {b}{2a}$[/quote]
You are just dividing by $ \frac{b}{2a}$
So we get:

$ \frac{a}{b}\cdot\frac{2a}{b}$

Multiply, then take the square root.

Hope I helped!

@yellow duck: Your answer is the reciprocal of the actual one.

$ \frac {a}{b} \cdot \frac {2a}{b}$ $ \Rightarrow$ $ \frac {2a^2}{b^2}$ and $ \sqrt {\frac {2a^2}{b^2}}$ $ \neq \frac {2a}{b}$.  So I really don't see the point of that because it's not even the reciprocal.

Math Champion's answer is correct.

@Math Champion: Don't worry about it. I misread the question.
But why is it $ \frac{b^2}{2a^2}$?

Cause if you multiply that with the original, you get the wanted result.

So say for example a/b=2. Then to get b/2a, you take the reciprocal, getting b/a=1/2, and then divide it by two, getting b/2a=1/4.

He asked how to convert a/b into b/2a. He wasn't explicitly clear in his wording, so he could mean that if he had a/b=x, what would b/2a equal.

That didn't really have anything to do with quadratics...The only similarity is the $ \minus{}\frac{b}{2a}$ in the quadratic formula.

Let $ x$ be the thing we multiply $ \frac {a}{b}$ by to get $ \frac {b}{2a}$:

$ \frac {a}{b} \cdot x \equal{} \frac {b}{2a} \\
 \\
\frac {b}{a} \cdot \frac {a}{b} \cdot x \equal{} \frac {b} {a} \cdot \frac {b}{2a} \\
 \\
\boxed{x \equal{} \frac {b^2}{2a^2}}$

EDIT: 1000th post!

about the answer to the problem. What else would it be?

hi people ive got a new problem that i need help with :D    :lol:

If you look at his username, he created his account so that he could ask for help.  :rotfl:

if A 5/12 = b x 50% then what is A?  :ninja:  :)

do you mean $ A\cdot\frac {5}{12} \equal{} b\cdot50\%$?

There is no way to numerically solve for $ a,$ but you can solve for $ a$ in terms of $ b.$
\[ \begin{align*}\frac{5a}{12}&=\frac{b}{2}\\a&=\frac{12}{5}\cdot\frac{b}{2}\\a&=\boxed{\frac{6b}{5}}\end{align*}.\]

Lol, we didn't even get confirmation that that was what he meant.


What is the exact definition for a relation to be an equivalence relation?

Well i know that the conditions required are that it shd be reflexive symmetric and transitive :) But should it have elements that necessary display these properties?

This question aroused my doubts:
Consider the relation R = { (1,1) , (2,2) , (3,3) }
Is this relation an equivalence relation?  :)   :maybe:

I mean, the absence of any element in the relation which obeys symmetric n transition nature has aroused the doubt that whether this relation is an equivalence relation or not

Thanks for the help in advance

This might be interesting - http://en.wikipedia.org/wiki/Antisymmetric_relation.

IMO, I think it is an equivalence relation because [b]IF[/b] (a,b) belongs to R then (b,a) belongs to R as well. We have nothing against this condition and maybe if we take (a,b) = (1,1),it turns out to be true.

[quote="madness"]

I mean, the absence of any element in the relation which obeys symmetric n transition nature has arosed the doubt that whether this relation is an equivalence relation or not
[/quote]

Don't have any such doubts. The rules have to be directly applied. The relation is an equivalent relation.

This should help you : a void relation is symmetric and transitive but not reflexive.

Yes, this is an equivalence relation called "equality".

[quote="bruteforce"][quote="madness"]

I mean, the absence of any element in the relation which obeys symmetric n transition nature has arosed the doubt that whether this relation is an equivalence relation or not
[/quote]

Don't have any such doubts. The rules have to be directly applied. The relation is an equivalent relation.

[b][u]This should help you : a void relation is symmetric and transitive but not reflexive[/u][/b].[/quote]

Why is the underlined statement true? :)

Just saw that in wikipedia :)

its called a partial equivalence relation :)

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

The above relation will not be an equivalence relation if the set A on which it is defined happens to be A={1,2,3,4} in which case it will lose the reflexive property.