Well I get $(sin20/sin^{2}(100))$*$(4\sqrt{3})*sin100+3)$  but I cant caculate that.. With a caculator it actually simplifies to $2\sqrt{3}$ but I cant use a caculator on these types of problmes..

i assumed A meant area and 2 means 2 times that area.

[quote="kohjhsd"]What does 2.A(DBE) mean?[/quote]

2 times area DBE + A(ABC)

and where is answer me@home?

[hide]Let $BH$ be the altitude of $\triangle ABC$. Then $\angle ABH = 90-80 = 10^\circ$

Let $F$ be the symmetric point of $A$ with respect to $H$.

Then $\triangle BAF$ is isosceles and similar to $\triangle BDE$. The ratio of the similarity is $2$, so the ratio of the areas is $4$.
Then $(ABH)=\frac{1}{2}(BAF) = 2(BDE)$

$(ABC)+2(BDE) = (ABC)+(ABH) = (BCH)$

But $\triangle BCH$ is a 30-60-90 triangle, so $(BCH) = \frac{\sqrt{3}}{{8}}BC^{2}=2\sqrt{3}$
[img]8427[/img]

[b]P.S.[/b]
$(X)$ denotes the area of $X$[/hide]

It is true pontios.
that is nearly same with yours.

[hide]
GDA and BDE equal and A(BDG)=A(GDA)
so A(ABC)+2A(BDE)=A(GBC)
[img]http://img50.imageshack.us/img50/9800/grvi6.png[/img][/hide]


ARML
email
geometry
AMC
AMC 12
AMC 12 B
AIME

I'm sure almost everybody had received an email containing following information.

[u]A Team[/u]
Cheng, Long
Chung, Hooyoung
Cui, He
Deng, Alan
Guirguis, Maria
Hong*, Gee-hoon
Hwang, Kevin
Kats, Stan
Naumov, Victor
Shin, Tai Ho
Sterling, Ethan
Uemura, Sho
Xu*, Yuetian
Yan, Hanpin
Zhu, Danny

[u]N Team[/u]
Bai, Yang
Cai, Francisco
Chiang, Zoe
Engelstein, Max
Lin*, Yuri
Ling, Xiao
Liu, Yitian
Som, Suraj
Spindler, Mark
Tang, Brady
Tao, Zuoyu
Weng*, Barry
Wu, Yi
Yang, Tao
Zverovich, Olga

[u]Y Team[/u]
Bin, Danny
Chang, Alex
Chen, Qin Feng
Deng, Tony
Ennever, Paul
Israel*, Evelyn
Kogan, Roman
Sung, Edward
Tsinis, Ilya
Wang, Sai Ke
Wong*, Tony
Yang, John
Zakharevich, Val
Zhang, Peisheng
Zhu, Wen Si

[u]C Team[/u]
Chien, Abraham
Guan*, Wilson
Hinchen, Naomi
Hui, Sam
Jang, Janice
Kwon, Kevin
Lee, Jonathan
Ma, Justin
Novodvorskiy, Yuriy
Pozin, Konstantin
Senie, Brian
Shen*, Diana
Teng, Steve
Zaslavskiy, Alex
Zhu, Weisha

[u]R Team[/u]
Choi*, Andrew
Ham, Dahye
Lipton, Ben
Meneses, Sergio
Mirzaee, Amin
Mui, Jason
Ye, Chun
Yu, Wensi
Yu*, Steven
Zhang, Hui Chuan
Zhao, Xiao Ting

i am the only one left a team,but i deserves it...*.*
anyway,i am quite sure A team is in div A,and
is C ,Y, N also on division A?

He and Tai Ho are the new additions, correct?  (Don't forget to tell them to do well, for the ice cream ... ;-) )

Hi Joel,
          Just wondering: For how many years has TJ taken first???And when they were not first, was NYC first???
                                                             Stan

TJ has won for the past 3 years.  New York City has won in the past -- six or seven years ago, NYC A, San Francisco A and Chicago A were all within 1 point of each other.  (I don't remember if NY and one of the others tied for second or if SF and Chicago tied for first.)  Some of the recent history is available on arml.com and older info is in the ARML/NYSML question books.  The teams that, in my experience and memory, most regularly place highly are:
Eastern Mass
San Francisco Area
Chicago
NYC
Thomas Jefferson
Mongomery County
and, perhaps less consistently or less impressively
Phillips Exeter, Texas, Minnesota, Georgia

Just for the record, TJ is totally beatable -- my senior year, without the relay the top three teams would have been
San Francisco 137
New York 132
TJ 131

You guys can do this :)

Yes, another monster has been added.

He Cui, Benjamin N. Cardozo High School.
NYCIML junior division starting member for Cardozo math team,
127.5 on AMC12B, 13 on AIME I 2005.
as far as I know, he missed MOSP by one point  :(

you SHALL win arml, you SHALL.  do it for the Gellerian Empire.


Putnam
function
trigonometry
integration
college contests

Suppose $ f(x)$ is a twice continuously differentiable real valued function defined for all real numbers $ x$ and satisfying

$ |f(x)|\leq1$ for all $ x$ and $ (f(0))^2\plus{}(f'(0))^2\equal{}4$.Prove that there exists a real number $ x_{0}$ such that

$ f(x_{0})\plus{}f''(x_{0})\equal{}0$

I don't have it yet. Here are some thoughts that don't go anywhere yet.

Suppose not. Assume that $ f(x)\plus{}f''(x)$ is never zero.

Let $ g(x)\equal{}f(0)\cos x\plus{}f'(0)\sin x.$ We know that we can write $ g(x)\equal{}2\cos(x\minus{}\delta)$ for some $ \delta$ and that $ g$ has a maximum of $ 2$ and a minimum of $ \minus{}2.$

Let $ h(x)\equal{}f(x)\minus{}g(x).$ Then $ h(x)\plus{}h''(x)\equal{}f(x)\plus{}f''(x),$ meaning that $ h(x)\plus{}h''(x)$ is never zero, so $ h(x)\plus{}h''(x)$ is always positive or always negative. Suppose WLOG that it's positive, so $ h''(0)>0.$ We also have that $ h(0)\equal{}h'(0)\equal{}0.$

But now what? Can $ h(x)$ ever be zero again? If not, we'd be done. Even showing that $ h(x)>\minus{}1$ would suffice. But I don't have anything here.

What if you assume $ h(x)$ to be $ h(x)\equal{}(f(x))^2\plus{}(f'(x))^2$

Well, if you do that, $ h'(x) = 2f'(x)(f(x) + f''(x)),$ so if $ f(x) + f''(x)$ is never zero, we have $ h'(x)$ changes sign only where $ f'(x)$ changes sign - but where does that argument go?

I would note that $ h'(0)\ne0,$ since we must have $ |f'(0)|\ge\sqrt {3}.$ So $ 4$ isn't the maximum of $ h$ ...

================================

Added in edit:

Here's another representation:

Let $ w(x)=f''(x)-f(x).$ We have that $ w(x)$ is never zero; WLOG we could say that $ w(x)>0.$ Then we can write $ f$ as follows:

$ f(x)=f(0)\cos x+f'(0)\sin x+\int_0^x\sin(x-y)w(y)\,dy$

$ =2\cos(x-\delta)+\int_0^x\sin(x-y)w(y)\,dy$

where either $ \frac{\pi}3\le\delta\le\frac{2\pi}3$ or $ \frac{4\pi}3\le\delta\le\frac{5\pi}3.$

The first interval for $ \delta.$ can be immediately dismissed, as $ f(x)\ge2\cos(x-\delta)$ for $ 0\le x\le \pi.$ The second interval is where the action lies.

Let $G(x)=(f(x))^2+(f'(x))^2$.Consider the interval $[-2,2]$.

For some $a\in[-2,0]$,we have by LMVT $\left|f'(a)\right|=\left|\frac{f(0)-f(-2)}{0-(-2)}\right|\leq\frac{|f(0)|+|f(-2)|}{2}\leq1$.

Similarly,for some  $b\in[0,2]$,we have by LMVT $\left|f'(b)\right|=\left|\frac{f(2)-f(-0)}{2-0}\right|\leq\frac{|f(2)|+|f(0)|}{2}\leq1$.

Now, $G(a)=(f(a))^2+(f'(a))^2\leq1+1=2$,i.e $G(a)\leq2$

Similarly, $G(b)\leq2$

But $G(0)=4$

Thus $G(x)$ must be having a point(say $x=x_{0}$) of global(absolute) maximum on $[-2,2]$ for which

$G'(x_{0})=0$

$2f'(x_{0})\left(f(x_{0})+f''(x_{0})\right)=0$.

Clearly,$f'(x_{0})\neq0$. Because if $f'(x_{0})=0$. then  $G(x_{0})\leq1$,but $G(x_{0})\geq4$

Therefore,there definitely exists some $x=x_{0}$ for which $f(x_{0})+f''(x_{0})=0$

This is certainly, JEE problem . Seems like they copied the problem from here !  :blush:  :w00tb:


Here is a question that I am not sure how to start off and finish.

It says at a school
90%take maths
85%take Science
80%take English
Then at least how many(%) has to take all three subjects?

I'll tell you if the answer is correct but please do explain.
Thanks

First, for the sake of clarity, draw a Venn diagram containing three intersecting circles and label these Math, Science and English.

Now, let $a=$ proportion of students taking all three subjects,
$b=$ proportion of students taking exactly Math and English,
$c=$ proportion of students taking exactly Math and Science, and
$d=$ proportion of students taking exactly Science and English.

So, proportion of students taking only Math $= 0.9-(a+b+c)$,
proportion of students taking only Science $= 0.85-(a+c+d)$, and,
proportion of students taking only English $= 0.8-(a+b+d)$.

Now, we must have $a+b+c+d+(0.9-(a+b+c))+(0.85-(a+c+d))+(0.8-(a+b+d)) = 1$
$\Rightarrow a = \frac{1.55-(b+c+d)}{2}$

But note that $b+c+d \le 1.$
So, $a \ge \frac{1.55-1}{2}= 27.5\%$, which is our answer.


calculus
integration
logarithms
limit
function
calculus computations

Compute the integral,
\[ \int_{0}^1 {\ln x \over 1\minus{}x^2}dx. \]
Thanks for help.

Use that $ \frac1{1 \minus{} x^2} \equal{} \sum_{k \equal{} 0}^{\infty}x^{2k}$ to turn this into

$ \sum_{k \equal{} 0}^{\infty}\int_0^1x^{2k}\ln x\,dx.$

Now do the usual integration by parts to find that

$ \int_0^1x^{2k}\ln x\,dx \equal{} \minus{} \frac {1}{(2k \plus{} 1)^2}.$

Hence, $ \int_0^1\frac {\ln x\,dx}{1 \minus{} x^2} \equal{} \minus{} \sum_{k \equal{} 0}^{\infty}\frac {1}{(2k \plus{} 1)^2}.$

Now, knowing that sum is equivalent to knowing $ \zeta(2).$ In particular, the answer is $ \minus{} \frac {\pi^2}{8}.$

Well, thanks for the natural proof!
My teacher said, since $ \lim_{x \to 1}{\ln x \over 1-x^2}=-\frac{1}{2}$, so he considered the function,
$ F(x)=\{\begin{array}{c}-\frac{1}{2}\quad{\rm if}\quad x=1 \ {\ln x \over 1-x^2}\quad{\rm if}\quad x \in [0,1).\end{array}$
Now, we would evaluate $ \int_{0}^{1} F(x)dx$ by substituting $ x=e^t$, them we compared it to other integral. I tried this but failed. Is there something wrong here?

That particular limit shows that the integral is not really improper at the $ 1$ end of the interval. It is improper as $ x\to0^ \plus{} ,$ but the singularity there is merely logarithmic, which is slow enough growth for the improper integral to converge.

The substitution has some merit, but cannot change the fact that the function is question does not have an elementary antiderivative. We are walking into the computation of $ \zeta(2).$ You will find that referenced elsewhere on this site, including multiple topics that contain links to a paper with more than a dozen computations of that.


counting
distinguishability
geometry
3D geometry
octahedron
rotation
symmetry

Eight congruent equilateral triangles, each of a different color, are used to construct a regular octahedron. How many distinguishable ways are there to construct the octahedron? (Two colored octahedrons are distinguishable if neither can be rotated to look just like the other.)

[b]$(\text{A})\,210\,\,\,\,\,(\text{B})\,560\,\,\,\,\,(\text{C})\,840\,\,\,\,\,(\text{D})\,1260\,\,\,\,\,(\text{E})\,1680$[/b]

[hide="Takin a shot"]
Fix one of the 8 faces as color #1.
Given this, there are $7!$ ways to apply colour to the 7 remaining faces.

However, as this face is adjacent to 3 different faces, we can rotate it 1/3 of a turn to reach another position, and another 1/3 of a turn to reach a third position, and another 1/3 of a turn to get back to where we started. This mean's there's 3-fold symmetry, and we must divide by 3:

$7!/3=1680$, $\boxed{E}$
 :maybe:
[/hide]

Fix one of the faces a certain color; then we can split the remaining faces into three categories; the face opposite the fixed one, the three faces in the same pyramid as the opposite faces, and the remaining three faces.

The face opposite can be any color, so there are $7$ combinations.

The three faces next the opposite face can be colored in $3!$ ways, but we have to account for cyclic permutations, so it's $3!/3=2$ ways, but we have to choose the colors, so the total combinations is $\binom{6}{3}2!$.

Finally, the last three faces can be colored any way with the remaining colors, so $3!$.

Multiply it all together to get $7\cdot\binom{6}{3}\cdot2\cdot3!=1680$.

[hide="Symmetries"]For each distinct coloring we can generate a total of $24$ which are non-distinct (by fixing the first color to any of 8 triangles and picking any of its 3 vertices as the "top apex", which automatically determines the colors of the remaining 7). Therefore each coloring is represented by $24$ which fall into the same class. Hence the total number is ${8!\over 24}={7!\over 3}=1680$[/hide]

[hide="But"]If the reflections w.r.t. either of 3 symmetrial planes of an octahedron are also considered the same, then the real number is ${1680\over 3}=560$[/hide]

How can 8 equilateral triangles form an octagon?
EDIT: Nevermind I get it... :wallbash:

[hide]The first triangle can be in any of 8 spots, but these are all indistinguishable. It touches three triangles, arranged in a circle, which has a $\frac{7\cdot6\cdot5}{3}$ possibilities. Any arrangement of the last 4 triangles is unique, $4!$.
$\frac{7\cdot6\cdot5}{3}4!=1680$.[/hide]


\[ \begin{gathered}
  {\text{Found }}\left( {f^{ \minus{} 1} } \right)\left( 6 \right){\text{ if }}f\left( x \right) \equal{} 2x^3  \plus{} 3x^5  \plus{} x \hfill \\
  thanks \hfill \\
\end{gathered}\]

Note that:
 $ f^{\minus{}1}(6) \equal{} a \iff f(a) \equal{} 6$
$ \implies 2a^3 \plus{} 3a^5 \plus{} a \equal{} 6 \implies a^5 \plus{} \frac{2}{3}a^3 \plus{} \frac{1}{3}a \minus{} 2 \equal{} 0$

I get $ a \equal{} 1$.
Hence $ f^{\minus{}1}(6) \equal{} 1$


An apple is in the shape of a ball of radius 31 mm. A worm gets into the apple and digs a tunnel of total length 61 mm, and then leaves the apple. (The tunnel need not be a straight line.) Prove that one can cut the apple with a straight slice through the center so that one of the two halves is not rotten.

?  Why would the apple necessarily be rotten?

i think he means wormified

and well if the worm's 16 mm in diameter then it's kinda difficult to do that


I have a test coming up tomorrow and I am still having trouble factoring certain polynomials. For example, 2p^3(p+2)+(p+2) now I know that the factor is (p+2)[2p^3+1] but why is that the factor?

Maybe it might help if you thought of a more "concrete" example. Lets use 5 and 7.
So we have 5*7 + 7. You can see that since multiplication is repeated addition, there is "one more" 7, so 5*7 + 7 = 6 * 7. So in your example, $ 2p^3(p \plus{} 2) \plus{} (p \plus{} 2)$, using your reasoning you can see that there is one more (p + 2). Also, you can just rewrite as $ 2p^3(p \plus{} 2) \plus{} 1(p \plus{} 2)$ and see, using the distributive property, except backwards, that this is equal to what you said. Sorry if that didn't really help. I usually don't post in intermed, usually too hard for me

[quote="krak"]why is that the factor?[/quote]

It's the distributive rule in reverse:  $ a(b \plus{} c) \equal{} ab \plus{} ac$.  In this case, you have $ a \equal{} p \plus{} 2$, $ b \equal{} 2p^3$, $ c \equal{} 1$.


Prove that among the positive numbers $ x, 2x, 3x, 4x, ..., (n \minus{} 1)x$ there is at least one which differs from an integer by at most $ \frac {1}{n}$.

[quote="AndrewTom"] which differs form an integer by at most $ \frac {1}{n}$.[/quote]

Do you mean: which differs "from" an integer by at most $ \frac {1}{n}$. If not please be more specific ?

Oops! Sorry Obel1x, it was a typo and should, of course, read "from". Thanks, I've edited it.

For real $ r$, denote by $ \{r\} \equal{} r \minus{} \lfloor r \rfloor$ its fractional part ($ 0 \leq \{r\} < 1$). Consider the intervals $ I_k \equal{} \left [ \frac {k} {n}, \frac {k \plus{} 1} {n} \right )$, for $ k \equal{} 0,1,\ldots,n \minus{} 1$, and look at the values $ \{\ell x\}$ for $ \ell \equal{} 1,2,\ldots,n \minus{} 1$.

If any falls in $ I_0$ or $ I_{n \minus{} 1}$, we are done; if not, there will be two of them, say $ \{ix\}$ and $ \{jx\}$ with $ 1\leq i < j\leq n \minus{} 1$, falling within the same $ I_k$, for some $ 1\leq k \leq n \minus{} 2$ (the Dirichlet part...).

Then $ (j \minus{} i)x$ is the one sought-after.


quadratics
number theory unsolved
number theory

Let n be a positive integer.Show that there are infinitely many prime p such that the smallest positive primitive root of p is greater than n.

Take $ n$ such that $ 1,2,...,n$ are quadratic residues. You will need Dirichlet's theorem (but the easy case $ 1 \mod m$) and quadratic reciprocity for this.


The problem from my frind :
When $ 10^{n} \plus{} 1$ becomes a prime number?
$ n\equal{}1,2,3,4,5, \cdots$

well, if $ n>1$ has an odd factor greater than 1, then $ 10^n\plus{}1$ factors into a product of numbers greater than 1 and therefore can't be prime: if $ n$ is odd, then $ 10^n\plus{}1$ is divisible by 11, and if $ n$ is even and has an odd factor, then $ n$ must be composite, so $ n\equal{}pq$, where $ p$ is odd and $ q>1$, so $ 10^q\plus{}1|10^n\plus{}1$, where $ 1<10^q\plus{}1<10^n\plus{}1$. so $ 10^n\plus{}1$ can only be prime if $ n$ is a power of 2. other than "only if" statements like this, we can't really say anything about when $ 10^n\plus{}1$ is prime.


geometry
perimeter
trigonometry
trig identities
Law of Sines

The perimeter of a triangle ABC is 6 times the arithmetic mean of the sines of its angles. If side $ a \equal{} 1$, find angle A.

[hide]
$ \sin(A)\plus{}\sin(B)\plus{}\sin(C) \equal{} \frac{2S}{bc}\plus{}\frac{2S}{ac}\plus{}\frac{2S}{ab} \equal{} \frac{2S(a\plus{}b\plus{}c)}{abc}$

So if $ \frac{\sin(A)\plus{}\sin(B)\plus{}\sin(C) }{3} \equal{} \frac{a\plus{}b\plus{}c}{6}$

$ \therefore 4S\equal{}abc\equal{}bc$

Hence, $ sin(A) \equal{} \frac{2S}{bc} \equal{} \frac{2S}{4S} \equal{} \frac{1}{2}$[/hide]

By the Law of Sines,

$ \frac {a}{\sin A} = \frac {b}{\sin B} = \frac {c}{\sin C}
= \frac {a+b+c}{\sin A+\sin B+\sin C}........(1)$

we are given that

$ a + b + c$ = $ 6 \frac {\sin A+\sin B+\sin C}{3}
= 2 (\sin A+\sin B+\sin C)$

substituting this into (1),

$ \frac {a}{\sin A} = 2$

since a = 1, this implies

$ \sin A = \frac {1}{2}$

$ \implies A = \frac {\pi}{3}$ or $ \frac {2\pi}{3}$


Show that for a set $ \mathbb{S} \equal{} \{1,2,...,n\}$ the sums of elements in the subsets of $ \mathbb{S}$ can produce any element of the set $ \{0,1,2,3,...,\frac {n(n \plus{} 1)}{2}\}$.

[hide="Induction"]
Ok clearly its true for n=0.

Assume its true for n. Then we must show it for n+1.

We can just take the subsets of (0,1,2...,n) and make any of the numbers from 0 to n(n+1)/2. We then need to make the numbers (n^2+n+2)/2 to (n+1)(n+2)/2. Note that by subtracting n+1 from these numbers, we get (n^2-n)/2 through n(n+1)/2, all of which were included in the original set.[/hide]


geometry
geometric transformation
national olympiad

Anyone post Russian Math Olympiad 2009? I think it has already been held.

Dear greentreeroad and mathlinkers,

You can find in the attachment the problems of All-Russian Mathematical Olympiad 2009 in Vietnamese provided by [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=273]Tran Nam Dung[/url]. I've already posted some problems on MathLinks, and hope some other members will post the rest soon.

Regards,
April

[quote="April"]I am sorry, I am not very good at English, so I am afraid that I can make some mistakes when translating the problems. I really appreciate if there are some Vietnamese members can translate and post the other problems. I know that there are two Vietnamese moderators [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=4576]Pham Van Thuan[/url] and [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=5820]Nguyen Trung Tuan[/url], they are really good at Mathematical English, so I hope they can come here and help us.

Here are the problems that I posted:
Problem 9.2 and 10.2: http://www.mathlinks.ro/viewtopic.php?t=276014
Problem 9.5: http://www.mathlinks.ro/viewtopic.php?t=276020
Problem 10.7: http://www.mathlinks.ro/viewtopic.php?t=276013
Problem 11.5: http://www.mathlinks.ro/viewtopic.php?t=276018
Problem 11.7: http://www.mathlinks.ro/viewtopic.php?t=276011

Enjoy!!!

[color=red][b]EDIT.[/b][/color]
Problem 9.4: http://www.mathlinks.ro/viewtopic.php?t=274838
Problem 10.1: http://www.mathlinks.ro/viewtopic.php?t=276091[/quote]

Very good, thanks. Would you also translate 11.3, 11.4, 11.8?

[quote="April"][b]Day 1[/b]

[b]Grade 9:[/b]
Problem 9.1: http://www.mathlinks.ro/viewtopic.php?t=276878
Problem 9.2: http://www.mathlinks.ro/viewtopic.php?t=276014
Problem 9.3: http://www.mathlinks.ro/viewtopic.php?t=276876
Problem 9.4: http://www.mathlinks.ro/viewtopic.php?t=274838

[b]Grade 10:[/b]
Problem 10.1: http://www.mathlinks.ro/viewtopic.php?t=276091
Problem 10.2: http://www.mathlinks.ro/viewtopic.php?t=276014
Problem 10.3: http://www.mathlinks.ro/viewtopic.php?t=276374
Problem 10.4: http://www.mathlinks.ro/viewtopic.php?t=276875

[b]Grade 11:[/b]
Problem 11.1: http://www.mathlinks.ro/viewtopic.php?t=276766
Problem 11.2: http://www.mathlinks.ro/viewtopic.php?t=276376
Problem 11.3: http://www.mathlinks.ro/viewtopic.php?t=277238
Problem 11.4: http://www.mathlinks.ro/viewtopic.php?t=277244

[b]Day 2[/b]

[b]Grade 9:[/b]
Problem 9.5: http://www.mathlinks.ro/viewtopic.php?t=276020
Problem 9.6: http://www.mathlinks.ro/viewtopic.php?t=277245
Problem 9.7: http://www.mathlinks.ro/viewtopic.php?t=276765
Problem 9.8: http://www.mathlinks.ro/viewtopic.php?t=277239
[b]
Grade 10:[/b]
Problem 10.5: http://www.mathlinks.ro/viewtopic.php?t=277241
Problem 10.6: http://www.mathlinks.ro/viewtopic.php?t=277243
Problem 10.7: http://www.mathlinks.ro/viewtopic.php?t=276013
Problem 10.8: http://www.mathlinks.ro/viewtopic.php?t=276340

[b]Grade 11:[/b]
Problem 11.5: http://www.mathlinks.ro/viewtopic.php?t=276018
Problem 11.6: Will be posted soon
Problem 11.7: http://www.mathlinks.ro/viewtopic.php?t=276011
Problem 11.8: http://www.mathlinks.ro/viewtopic.php?t=276340

My thanks to [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=3272]Andrei Frimu[/url], [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=58637]Christian Aravanis[/url] for correcting my translations and to [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=270]Fedor Petrov[/url] for providing the translations of problems 10.5, 10.6 and 11.4  :first:[/quote]

Many thanks to all of you. Now let's make it to the resource section.

As far as i can see from IMO's official page, there are three 15-aged student in Russian team, isn't it kind of interesting? Anybody has information about whether it is true or not, because I have seen many mistakes in this site, especially about ages. This may be another example.

No, this easily may be true. Even though they are 15, they are already 10 graders, and have in mind, that there are only 11 grades in Russia.

Problem 11.6 has already posted  :showoff:

[b]Day 1[/b]

[b]Grade 9:[/b]
Problem 9.1: http://www.mathlinks.ro/viewtopic.php?t=276878
Problem 9.2: http://www.mathlinks.ro/viewtopic.php?t=276014
Problem 9.3: http://www.mathlinks.ro/viewtopic.php?t=276876
Problem 9.4: http://www.mathlinks.ro/viewtopic.php?t=274838

[b]Grade 10:[/b]
Problem 10.1: http://www.mathlinks.ro/viewtopic.php?t=276091
Problem 10.2: http://www.mathlinks.ro/viewtopic.php?t=276014
Problem 10.3: http://www.mathlinks.ro/viewtopic.php?t=276374
Problem 10.4: http://www.mathlinks.ro/viewtopic.php?t=276875

[b]Grade 11:[/b]
Problem 11.1: http://www.mathlinks.ro/viewtopic.php?t=276766
Problem 11.2: http://www.mathlinks.ro/viewtopic.php?t=276376
Problem 11.3: http://www.mathlinks.ro/viewtopic.php?t=277238
Problem 11.4: http://www.mathlinks.ro/viewtopic.php?t=277244

[b]Day 2[/b]

[b]Grade 9:[/b]
Problem 9.5: http://www.mathlinks.ro/viewtopic.php?t=276020
Problem 9.6: http://www.mathlinks.ro/viewtopic.php?t=277245
Problem 9.7: http://www.mathlinks.ro/viewtopic.php?t=276765
Problem 9.8: http://www.mathlinks.ro/viewtopic.php?t=277239
[b]
Grade 10:[/b]
Problem 10.5: http://www.mathlinks.ro/viewtopic.php?t=277241
Problem 10.6: http://www.mathlinks.ro/viewtopic.php?t=277243
Problem 10.7: http://www.mathlinks.ro/viewtopic.php?t=276013
Problem 10.8: http://www.mathlinks.ro/viewtopic.php?t=276340

[b]Grade 11:[/b]
Problem 11.5: http://www.mathlinks.ro/viewtopic.php?t=276018
Problem 11.6: http://www.mathlinks.ro/viewtopic.php?t=277621
Problem 11.7: http://www.mathlinks.ro/viewtopic.php?t=276011
Problem 11.8: http://www.mathlinks.ro/viewtopic.php?t=276340

[quote="Umut Varolgunes"]Yes I know that. Yet, I thought a tenth grader is most possibly 16 years old, and wanted to check.[/quote]

well, that most probably because they went to school when they were 6 or 5, or alternatively if they skipped some grade, just like I did.

[quote]
Problem 11.6 has already posted Showoff [/quote]

Well done April!!! :D

[quote="Erken"]what about the results? I am really eager to know them.[/quote]

You can find the results [url=http://olympiads.mccme.ru/vmo/35/]here.[/url]


geometry
rectangle
analytic geometry
LaTeX
ratio
AMC

Rectangle $ ABCD$ has $ AB \equal{} 4$ and $ BC \equal{} 3$.  Segment $ EF$ is constructed through $ B$ so that $ EF$ is perpendicular to $ DB$, and $ A$ and $ C$ lie on $ DE$ and $ DF$, respectively.  What is $ EF$?

$ \textbf{(A)}\ 9\qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ \frac {125}{12}\qquad \textbf{(D)}\ \frac {103}{9}\qquad \textbf{(E)}\ 12$

Similar triangles. Since $ BD \perp EF$, we know that $ \triangle BDF \sim \triangle EDB \sim \triangle ECF$. Then you just find $ EB$ and $ BF$, and add them up.

yongyi, can you explain how  ECF is similar since I drew it out and it looks nothing like the other two.

uh, isn't the answer [hide]125/12 (C)?[/hide]

My math genius friend said he got something else, but I'm pretty sure I'm right.

This is a case of trying to visualize diagrams in my head :ewpu:

I meant $ \triangle EDF$.

To erw314: yes.

I actually used a coordinate-bash because I always mix up sides of similar triangles.  It's really not that bad with this problem.

[hide="Solution"]
I couldn't seem to get LaTeX to include my scanned image, but (as you probably already know) the figure is a right triangle with sides $ ED$, $ DF$, and $ EF$ and the rectangle is wedged into the bottom of the triangle with $ DC$ on $ DF$.  We know that the $ AB\equal{}DC\equal{}4$ and $ AD\equal{}BC\equal{}3$, so the rectangle's diagonal is $ DB\equal{}5$.  From here, you could look at $ \triangle DBF$ and notice that $ \triangle DBF \sim \triangle BCF \sim \triangle DCB$, use ratios of similarity to solve for $ CF$, and then do the same for $ \triangle EBF$.  I just used a theorem from geometry, which you can derive using similar triangles.  If we let $ CF\equal{}x$, then we have $ 3^2\equal{}4x \Longrightarrow x\equal{}\dfrac{9}{4}$, so $ DF\equal{}\dfrac{25}{4}$, and doing the same for $ \triangle EBD$ gives us $ EA\equal{}\dfrac{16}{3}$ and $ ED\equal{}\dfrac{25}{3}$.  Thus we have $ EF \equal{} \sqrt{\left(\dfrac{25}{3}\right)^2 \plus{} \left(\dfrac{25}{4}\right)^2} \equal{} 25\sqrt{\dfrac{1}{9}\plus{}\dfrac{1}{16}} \equal{} 25\sqrt{\dfrac{25}{144}} \equal{} \boxed{\dfrac{125}{12}}$.
[/hide]

I'm having a hard time visualizing this can someone please send me a picture?
sorry about reviving old topics!

Here's what the diagram looks like:
[geogebra]b34b7471b13f7842b04984269278fc9569a310d0[/geogebra]

Just do similar triangles. EF=4*(5/3)+3*(5/4)=125/12 which is answer C


I roll 100 standard dice.  I get one point for each die for which the number of dots on the top face is greater than two.  What is the probability that the number of points I get is...

a)   Even?
b)   Divisible by three?
c)   $\equiv 1\pmod{3}$?

[quote="hydro"]
[hide="a)"]If $p(i)$ is the probability to obtain $i$ dices greater than 2, than $p(i)=\binom{100}{i}\cdot \left( \frac{2}{3}\right)^{i}\cdot \left( \frac{1}{3}\right)^{100-i}$[/hide]
[/quote]

[hide="Hint"]$p(i)$ is a term of $\left(\frac{2}{3}+\frac{1}{3}\right)^{100}$.  How can we add up every other term?[/hide]

[quote="hydro"][hide="a)"]

the probability is the sum of the probabilities to obtain on 2 rolls a number of points greater than 2, on 4 rolls, on 6 and so on. If $p(i)$ is the probability to obtain $i$ dices greater than 2, than $p(i)=\binom{100}{i}\cdot \left( \frac{2}{3}\right)^{i}\cdot \left( \frac{1}{3}\right)^{100-i}$

So $p=\binom{100}{2}\cdot \left( \frac{2}{3}\right)^{2}\cdot \left( \frac{1}{3}\right)^{98}+\binom{100}{4}\cdot \left( \frac{2}{3}\right)^{4}\cdot \left( \frac{1}{3}\right)^{96}+$$...+\binom{100}{100}\cdot \left( \frac{2}{3}\right)^{100}\cdot \left( \frac{1}{3}\right)^{0}= \left( \frac{1}{3}\right)^{100}\cdot \sum_{i=1}^{50}\binom{100}{2i}\cdot 2^{2i}$, but I don't know how to calculate this sum...
[/hide][/quote]

In the expansion of \[ \left( \frac{2}{3}-\frac{1}{3}\right)^{100},\] every term where $\frac{1}{3}$ is raised to an odd power has a minus sign in front of it.

If we let $s$ denote the sum ${100\choose 1}\left(\frac{2}{3}\right)^{99}\left(\frac{1}{3}\right)+{100\choose 3}\left(\frac{2}{3}\right)^{97}\left(\frac{1}{3}\right)^{3}+\cdots+{100\choose 99}\left(\frac{2}{3}\right)\left(\frac{1}{3}\right)^{99}$, then we can write the equation \[ \left(\frac{2}{3}+\frac{1}{3}\right)^{100}= \left( \frac{2}{3}-\frac{1}{3}\right)^{100}+2s\] From this, we solve for $s$ and get \[s = \frac{1-(\frac{1}{3})^{100}}{2}\] The sum of ${100\choose 0}\left(\frac{1}{3}\right)^{100}+{100\choose 2}\left(\frac{2}{3}\right)^{2}\left(\frac{1}{3}\right)^{98}+\cdots{100\choose 100}\left( \frac{2}{3}\right)^{100}\left(\frac{1}{3}\right)^{0}$ is equal to $1-s$. The desired sum does not contain the term ${100\choose 0}\left(\frac{1}{3}\right)^{100}$, so we subtract that and get \[1-s-{100\choose 0}\left(\frac{1}{3}\right)^{100}\] [/hide]


Let x,y,z be real positive numbers.Let s=x+y+z, a=y+z,b=x+z,c=x+y. Prove that

1) (s^s)(x^x)(y^y)(z^z) \leq (a^a)(b^b)(c^c)

2) (s^ss)(x^xx)(y^yy)(z^zz) \geq (a^aa)(b^bb)(c^cc)

I think these are very nice twin inequalities.

Source: AMM   prob 10766 (1999) Proposed by Szilard Andras, Babes-
                      Bolyai University,Cluj-Napoca, Romania

I have the solution if somebody is interested.

Yes, I'm interested. I've tried last evening to prove it, but nothing. It is also very nice and I think the technique is nice.

1) Expanding both sides shows that (x+y+z)z \leq (y+z)(z+x) and therefore

     (x+y+z)^z*z^z \leq (y+z)^z*(z+x)^z

and similar

     (x+y+z)^x*x^x  \leq (x+y)^x*(z+x)^x

     (x+y+z)^y*y^y \leq (x+y)^y*(y+z)^y

Multiplying these three together gives the desidered inequality.

2) Recall Bernoully's inequality : If t \geq -1 and c \geq 1 then (1+t)^c \geq 1+ct. Assume that z>0 substitute y/(z+x) for t and (z+x)/z for c, raise both sides of the inequality to the power z and then multiply both sides by (z+x)^(z+x)*z^z to obtain

   (x+y+z)^(z+x)*z^z \geq (z+x)^(z+x)*(y+z)^z

Likewise we have

   (x+y+z)^(y+z)*z^z \geq (y+z)^(y+z)*(z+x)^z

Multiplying these two we obtain

(x+y+z)^(x+y+2z)*z^2z \geq (z+x)^(z+x)*(y+z)^(y+z)*((z+x)(y+z))^z \geq (z+x)^(z+x)*(y+z)^(y+z)*((x+y+z)z)^z

From this easily follows

  (x+y+z)^((x+y+z)z)*z^(zz) \geq (z+x)^((z+x)z)*(y+z)^((y+z)z)

Likewise (x+y+z)^((x+y+z)y)*y^(yy) \geq (x+y)^((x+y)y)*(y+z)^((y+z)y) and (x+y+z)^((x+y+z)x)*x^(xx) \geq (z+x)^((z+x)x)*(x+y)^((x+y)x).
Multiplying these three inequalities we obtain the desidered inequality.

This solution is by Ronald Kopas, Clarion University of Pennsylvania.

The solution is very ingenious and hard.


rotation
analytic geometry
function
trigonometry
geometry

A very long tube found in a vertical plane rotates around its end $ O$, with the constant speed $ \omega$. At $ t \equal{} 0$, the tube is in horizontal position. In its interior, a ball of mass $ m$ moves around. Initially, it rests in $ O$. The tube rotates counter-clockwise.

Describe the ball's motion (algebraically :().

I will assume that the movement of the ball has no effect on the tube since you said it has constant angular velocity $ \omega$
The simplest way is I think to observe the movement of the ball in the tubes rotating frame
first case: no friction.
This is the easier part.

So since no friction only the centrifugal force has an effect on the ball radially outwards so, the equation of motion in polar coordinates:

$ mr\omega^2 \equal{} m\ddot{r}$
That is:
$ \ddot r \minus{} r\omega^2 \equal{} 0$

This is a simple diff. equation.
The solution is with initial conditions $ r(0) \equal{} 0$ and $ \dot r \equal{} v_0$:
$ r \equal{} \frac {v_0}{\omega}\sinh{\omega t}$

Case 2, there is friction.

This is bit trickier, since now coriolis force comes into the game also :D
so the friction force:
$ F_f \equal{} \mu \sqrt {(mg)^2 \plus{} (2m\dot r \omega)^2}$
Here $ \mu$ is the coefficient of friction.

So the equation of motion:
$ \ddot r \plus{} \mu\sqrt {4\dot r^2 \omega^2 \plus{} g^2} \minus{} r\omega^2 \equal{} 0$

Now I tried to solve this, and I got as far that I think, that it cannot be given with elementary functions, I got elliptic integrals.

But maybe someone can find a nice solution? :D

Oops, now reading the problem again, I see that the tube moves in the vertical plane, and I solved it for the horizontal plane...

So here it is for the vertical plane:
similar just now we have to take gravity into the game, so the equation of motion:

$ \ddot r \minus{}r\omega^2 \equal{}g\sin(\omega t)$

Solution of this is simple, just add solution of homogenous part to particular solution, (It was fishy before about the initial conditions...but now its clear):
SO:
$ r(t)\equal{}\frac{g}{2\omega^2}(\sinh(\omega t)\minus{}\sin(\omega t))$

Oops... I forgot to mention that the motion is frictionless!

Anyway, this is from a "solved problems" book, and the solution given there involves energies and lagrangians :( (among others, the potential energy $ mgh$ also appears, leading to an extra term in the differential equation). Once we'll agree on this one (and hopefully I will have understood it), I'll show a similar problem, in which an extra twist that I also can't understand appears.

BTW, thank you for your efforts on my problems! This is a very dark area for me...


Please prove this inequality if it's true
Let a+b+c=3. Prove that
$ \sqrt{a\sqrt{a}}\plus{}\sqrt{b\sqrt{b}}\plus{}\sqrt{c\sqrt{c}}\geq\sqrt{3(ab\plus{}bc\plus{}ac)}$

[quote="hoangnhatvn"]Please prove this inequality if it's true
Let a+b+c=3. Prove that
$ \sqrt {a\sqrt {a}} \plus{} \sqrt {b\sqrt {b}} \plus{} \sqrt {c\sqrt {c}}\geq\sqrt {3(ab \plus{} bc \plus{} ac)}$[/quote]
Let $ a \equal{} x^4,$ $ b \equal{} y^4$ and $ c \equal{} z^4,$ where $ x,$ $ y$ and $ z$ are non-negative numbers.
Then $ x^4 \plus{} y^4 \plus{} z^4 \equal{} 3$ and $ \sqrt {a\sqrt {a}} \plus{} \sqrt {b\sqrt {b}} \plus{} \sqrt {c\sqrt {c}}\geq\sqrt {3(ab \plus{} bc \plus{} ac)}\Leftrightarrow$
$ \Leftrightarrow x^3 \plus{} y^3 \plus{} z^3\geq\sqrt {3(x^4y^4 \plus{} x^4z^4 \plus{} y^4z^4)}\Leftrightarrow$
$ \Leftrightarrow(x^4 \plus{} y^4 \plus{} z^4)(x^3 \plus{} y^3 \plus{} z^3)^4\geq27(x^4y^4 \plus{} x^4z^4 \plus{} y^4z^4)^2\Leftrightarrow$
$ \Leftrightarrow\left(\frac {x^4 \plus{} y^4 \plus{} z^4}{3}\right)^3\left(\frac {x^3 \plus{} y^3 \plus{} z^3}{3}\right)^{12}\geq\left(\frac {x^4y^4 \plus{} x^4z^4 \plus{} y^4z^4}{3}\right)^6,$ which is true
because $ \left(\frac {x^4 \plus{} y^4 \plus{} z^4}{3}\right)^3\geq\left(\frac {x^3 \plus{} y^3 \plus{} z^3}{3}\right)^{4}$ is true
and $ \left(\frac {x^3 \plus{} y^3 \plus{} z^3}{3}\right)^{16}\geq\left(\frac {x^4y^4 \plus{} x^4z^4 \plus{} y^4z^4}{3}\right)^6$ is true.
A hint for a proof of the last inequality see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=351034#351034

Sorry that I can't see the proof of the last inequality.
In your link I only found these inequalities:
$ a^3\plus{}b^3\plus{}c^3\equal{}3,  x^4y^4\plus{}x^4z^4\plus{}y^4z^4\le3$
and $ a\plus{}b\plus{}c\equal{}3, a^n\plus{}b^n\plus{}c^n \ge ab\plus{}ac\plus{}bc$

[quote="hoangnhatvn"]Sorry that I can't see the proof of the last inequality.
In your link I only found these inequalities:
$ a^3 \plus{} b^3 \plus{} c^3 \equal{} 3, x^4y^4 \plus{} x^4z^4 \plus{} y^4z^4\le3$
and $ a \plus{} b \plus{} c \equal{} 3, a^n \plus{} b^n \plus{} c^n \ge ab \plus{} ac \plus{} bc$[/quote]
But $ x^4y^4 \plus{} x^4z^4 \plus{} y^4z^4\le3\Leftrightarrow\left(\frac{x^4y^4 \plus{} x^4z^4 \plus{} y^4z^4}{3}\right)^3\leq1\Leftrightarrow$
$ \Leftrightarrow\left(\frac{a^3\plus{}b^3\plus{}c^3}{3}\right)^8\geq\left(\frac{x^4y^4 \plus{} x^4z^4 \plus{} y^4z^4}{3}\right)^3.$ :wink:

Sorry. I think you got a mistake.  :(
we have $ x^4\plus{}y^4\plus{}z^4\equal{}3$, not $ x^3\plus{}y^3\plus{}z^3\equal{}3$.
Is it different?

PS: Anyway, thanks for your nice solution  :)

Arquady is true because the last step can be killed by this inequality:
$ \left(\frac{a^3\plus{}b^3\plus{}c^3}{3}\right)^4 \le \left(\frac{a^4\plus{}b^4\plus{}c^4}{3}\right)^3 \Leftrightarrow a^3\plus{}b^3\plus{}c^3 \leq 3$:)

[quote="great math"]Arquady is true because the last step can be killed by this inequality:
$ \left(\frac {a^3 \plus{} b^3 \plus{} c^3}{3}\right)^4 \le \left(\frac {a^4 \plus{} b^4 \plus{} c^4}{3}\right)^3 \Leftrightarrow a^3 \plus{} b^3 \plus{} c^3 \leq 3$:)[/quote]
Of cause! See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1231215#1231215
 :wink:

Yeah, this inequality $ \left(\frac {x^4 \plus{} y^4 \plus{} z^4}{3}\right)^3\geq\left(\frac {x^3 \plus{} y^3 \plus{} z^3}{3}\right)^{4}$ is true  :lol:
But what about this?  :(
$ \left(\frac {a^3 \plus{} b^3 \plus{} c^3}{3}\right)^8\geq\left(\frac {x^4y^4 \plus{} x^4z^4 \plus{} y^4z^4}{3}\right)^3$
This inequality is only true when $ a^3 \plus{} b^3 \plus{} c^3 \equal{} 3$.
The condition of a,b,c we have is $ a^4 \plus{} b^4 \plus{} c^4 \equal{} 3$
so	$ \left(\frac {a^3 \plus{} b^3 \plus{} c^3}{3}\right)^4 \le \left(\frac {a^4 \plus{} b^4 \plus{} c^4}{3}\right)^3 \Leftrightarrow a^3 \plus{} b^3 \plus{} c^3 \leq 3$ but we can't assume that $ x^4y^4 \plus{} x^4z^4 \plus{} y^4z^4\le3$  :maybe:

PS: If you think you're right, please post your entire solution. Thanks  :)

[quote="hoangnhatvn"]
But what about this?  :(
$ \left(\frac {a^3 \plus{} b^3 \plus{} c^3}{3}\right)^8\geq\left(\frac {x^4y^4 \plus{} x^4z^4 \plus{} y^4z^4}{3}\right)^3$
[/quote]
I mean $ \left(\frac {x^3 \plus{} y^3 \plus{} z^3}{3}\right)^8\geq\left(\frac {x^4y^4 \plus{} x^4z^4 \plus{} y^4z^4}{3}\right)^3.$
This inequality is homogeneous. Thus, it is true for all non-negative $ x,$ $ y$ and $ z.$ :wink:
[quote="hoangnhatvn"]PS: If you think you're right, please post your entire solution. Thanks  :)[/quote]
Last time. OK? If still you'll have a question then I'll post a full proof.

oh, I understand  now :D but I can't prove the last inequality  :(
Please prove it

We'll prove that $ \left(\frac {x^3 \plus{} y^3 \plus{} z^3}{3}\right)^8\geq\left(\frac {x^4y^4 \plus{} x^4z^4 \plus{} y^4z^4}{3}\right)^3$ for non-negative $ x,$ $ y$ and $ z.$
Let $ x^3\plus{}y^3\plus{}z^3\equal{}3.$ Then, we need to prove that $ x^4y^4 \plus{} x^4z^4 \plus{} y^4z^4\leq3.$
By AM-GM we obtain: $ 1\plus{}x^3\plus{}y^3\geq3xy.$
Hence, $ \sum_{cyc}x^4y^4\leq\frac{1}{3}\sum_{cyc}x^3y^3(1\plus{}x^3\plus{}y^3).$
Thus, it remains to prove that $ \sum_{cyc}x^3y^3(1\plus{}x^3\plus{}y^3)\leq9.$
Let $ x^3\equal{}a,$ $ y^3\equal{}b$ and $ z^3\equal{}c.$
Hence, $ a\plus{}b\plus{}c\equal{}3$ and $ \sum_{cyc}x^3y^3(1\plus{}x^3\plus{}y^3)\leq9\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}ab(a\plus{}b\plus{}1)\leq9\Leftrightarrow\sum_{cyc}ab(4a\plus{}4b\plus{}c)\leq(a\plus{}b\plus{}c)^3\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^3\minus{}a^2b\minus{}a^2c\plus{}abc)\geq0,$  which is Schur.


I believe I found an easier solution for one of the problems in Intermediate Algebra. Should I post it at all, and if so, where should I post/PM/email it?

I think you're supposed to e-mail it to books@artofproblemsolving.com

If you're taking the Algebra 3 class, then you can post it on the message board.


$ a)$ Prove that for any strictly real $ x$, the following equality is true:

$ 1\plus{}\frac{1}{x^2}\plus{}\frac{1}{(1\plus{}x)^2}\equal{}\left ( 1\plus{}\frac{1}{x}\minus{}\frac{1}{1\plus{}x} \right ) ^2$

$ b)$ Prove that the following inequality is true:

$ \sqrt{1\plus{}\frac{1}{2^2}\plus{}\frac{1}{3^2}}$+$ \sqrt{1\plus{}\frac{1}{4^2}\plus{}\frac{1}{5^2}}$+$ \sqrt{1\plus{}\frac{1}{6^2}\plus{}\frac{1}{7^2}}$+...+$ \sqrt{1\plus{}\frac{1}{2008^2}\plus{}\frac{1}{2009^2}}$$ <\frac{1}{2} \left ( 2009\minus{}\frac{1}{2009} \right )$

Very nice your inequality, moldovan!  :lol:
But I think, it's better without $ a).$ :wink:


[quote="math92"]In 1969 the price of 5 kilograms of flour was $0.75. In 1970 the price was increased 15 percent. In 1971, the 1970 price was decreased by 5 percent. What was the price of 5 kilograms of flour in 1971?[/quote]
[hide]
Price in:

1969: 75 cents
1970: 75*1.15=86.25 cents
1971: 86.25-86.25*0.05=81.9375 or about 82 cents.[/hide]

[quote="math92"]In 1969 the price of 5 kilograms of flour was $0.75. In 1970 the price was increased 15 percent. In 1971, the 1970 price was decreased by 5 percent. What was the price of 5 kilograms of flour in 1971?[/quote]
[hide]
0.75x1.09= about 0.82
[/hide]

[color=darkred][size=75][Edited by 236factorial to fix latex viewing problem][/size][/color]


AMC
AIME
geometry
USA(J)MO
USAMO
similar triangles
Pythagorean Theorem

Wait.. so im guessing the cutoff is what? 6,7,8?  what do you guys think? Vote in the poll.  Was this years harder or was last years harder?  :?:

I think that this was right on par with some of the previous years... Although #2 was a bugger!

AIME floor will be 7???????????
NO!!!!!!!!!!!!!!!!!!!!!!!!!
i got a 6!!!!!!!!!!!!

Darn it.
If only i hadn't screwed up #4 and #10
darn darn darn....

[quote="ComplexZeta"]I predict AIME floor will be 7.[/quote]\

I say 8.  but then again, you never know.

This AIME seemed easier to me than last year's, but few seem to agree. I think what happened was the exact opposite of what happened with the AMC 12; on the AMC 12 most questions were easy but the hardest ones were brutal, but on this AIME most questions were harder, but there were no questions that really stood out as being remarkably difficult.

Anyone agree with that assessment? Or do I have no idea what I'm talking about?

As for the cutoff... I have no idea.

Assuming that a 210 will qualify (hopefully it does), I know that the floor will be at most a 7, since I know some people that got ~140 / 7.

Eugevir, I agree completely.  Last year was easy in a good way -- this was easy in a not-so-good way.

However, I've noticed in myself a tendency to grumble about tests not being hard in the right way, so that just might be me.

It was harder.

Knowing that geometry is the most underdeveloped branch of mathematics in the standard American math curriculum, the test authors screw people over by writing a load of geometric problems. How nice and thoughtful of them. :D

Yes, and they *really* liked similarity.  I think what they were trying to do was keep people from getting ahead on AIME geometry by memorizing a bunch of fancy theorems and instead doing stuff that pretty much didn't require more than similar triangles, the Pythagorean theorem, and ingenuity.

[quote="tAsDoOm"]
Knowing that geometry is the most underdeveloped branch of mathematics in the standard American math curriculum, the test authors screw people over by writing a load of geometric problems. How nice and thoughtful of them. :D[/quote]

It seems in the discussion that your grade level in school affects qualification for USAMO in some way. Can someone elaborate on this?


Find $ Min (Max|\sqrt {x} \plus{} ax \plus{} b|)$, where $ a,b$ are 2 real numbers and $ x \in [0;16]$

To mod: Please lock this topic because it is a contest problem in Mathematics And Youth Magazine of Viet Nam. The dealine is 15-04-2008!


algorithm
induction
algebra unsolved
algebra

This question deals with the greedy algoriithm, making the best choice at the time (not thinking about any others) hoping it leads to an optimal solution.  I'm working on a problem from my text book which deals with a graph:

So say fred wants to go to the store, and every street in his neighbourhood is a toll road.  This can be viewed as a graph with intersections and values assigned.  When he comes to an instersection (vertex), he gets a choice of which road (edge) to take next, and each road is labeled with it's destination and toll amount.  He decides to use a greedy algorithm, where he takes the cheapest toll road, unless it leads back to an intersection he's already been to.

This problems deals with Graphs, how can I prove that Fred may not make it to the store?  Also if he does make it, how can
I prove or disprove that he didn't take the cheapest route?  Am I supposed to use induction?  Or give graph examples that wouldn't work and use them to back my position up? Thanks for any help!

Here you go:

[img]http://img168.imageshack.us/img168/5545/friedrichnietzscheeh7.jpg[/img]

The left graph is a counter-example for your first question (Must Fred reach the store?), proving the answer is no and the right graph is a counter-example for your second question (If he reaches it, must he always pay the minimum amount of money getting there?) proving, again, the answer is no.

Maybe I have incorrectly understood this problem, but I think this is it.

More generally, using counter-examples is, in the great majority of problems, the best way of proving the answer is no. If you were to prove the answer to "Will always..." is yes, you can not use counter-esamples - you must rely on some other strategies.


Very Self Explanatory... the point is to get from 1 to 1000 with these pictures

[img]http://www.doubletheodds.co.uk/images/Number%201.jpg[/img]

Be creative!

[url=http://www.youtube.com/watch?v=Yu_moia-oVI]Click this link to measure your luck![/url]

...what the heck?

[img]http://www.uscg.mil/d14/cmd/img/D14%20logo%20(hi-res).jpg[/img]

RADIO FTW! Well actually iPods FTW! Heh.
[img]http://www.riseministries.com/resources/logoImages/kiss-107-fm20080507125627.gif[/img]

[img]http://www.arago.si.edu/media/000/028/397/28397_lg.jpg[/img]
this is the perfect image!
Ohio!!!
lulz.
except....that Cleveland..ya but w/e

[img]http://kara.allthingsd.com/files/2009/07/110percent.png[/img]

lol for some reason i get all the multiples of 5 from 80

I hope this will make the message longer.


calculus
integration
algebra
polynomial
function
geometry
calculus computations

[quote="kunny"]A eightic funtion $ y \equal{} ax^8 \plus{} bx^7 \plus{} cx^6 \plus{} dx^5 \plus{} ex^4 \plus{} fx^3 \plus{} gx^2\plus{}hx\plus{}i\ (a\neq 0)$ touches the line $ y \equal{} px \plus{} q$ at $ x \equal{} \alpha ,\ \beta ,\ \gamma ,\ \delta \  (\alpha < \beta < \gamma <\delta).$
Find the area of the region bounded by these graphs in terms of $ a,\ \alpha ,\ \beta ,\gamma ,\ \delta .$[/quote]

Find the area of the region bounded by these graphs in terms of $ a,\ \alpha ,\ \beta ,\gamma ,\ \delta $, [i]without[/i] $b, c, d, e, f, g, h, i $?

Well, subtracting the eight-degree polynomial by the straight line equation, we get $a(x-\alpha)^2(x-\beta)^2(x-\gamma)^2(x-\delta)^2$.  But the idea of integrating this thing terrifies me.

[quote="kunny"]Yes, we can find the area without using $b,\ c,\ d,\ e,\ f,\ g,\ h,\ i$.[/quote]

A eightic funtion  touches the line, Ok. But I find the area of the region bounded by these graphs, by the inteqral, using $x \in (\alpha ; \beta) , (\beta ; \gamma) , (\gamma ; \delta) $??? Very complex, maybe. Can U give to me your Sol.  Thank Mr.

[quote="kunny"]An eighth degree polynomial funtion $ y \equal{} ax^8 \plus{} bx^7 \plus{} cx^6 \plus{} dx^5 \plus{} ex^4 \plus{} fx^3 \plus{} gx^2\plus{}hx\plus{}i\ (a\neq 0)$ touches the line $ y \equal{} px \plus{} q$ at $ x \equal{} \alpha ,\ \beta ,\ \gamma ,\ \delta \  (\alpha < \beta < \gamma <\delta).$
Find the area of the region bounded by these graphs in terms of $ a,\ \alpha ,\ \beta ,\gamma ,\ \delta .$[/quote]
Can U give to me experimental method about Find the area of the region bounded by these graphs in terms of $ a,\ \alpha ,\ \beta ,\gamma ,\ \delta .$, please.


Find all natural numbers $ a,b$ and prime number $ p$ satisfying $ 2^a\plus{}p^b\equal{}19^a.$

$ p\equal{}17$ obviously
now use the equality $ b\equal{}v_{17}(19^a\minus{}2^a)\equal{}v_{17}(17)\plus{}v_{17}(a)\equal{}1\plus{}v_{17}(a)$ so we get $ 17^{b\minus{}1} | a$.
now use inequality.


The three-digit number $ 2a3$ is added to the number $ 326$ to give the three-digit number $ 5b9$. If $ 5b9$ is divisible by $ 9$, then $ a \plus{} b$ equals:

$ \textbf{(A)}\ 2 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ 9$

5+b+9=14+b must be divisible by 9. 18 is divisible by 9, so b=4. a+2=b, so a+2=4 giving a=2. a+b=2+4=6, so the answer is C.

[quote="isabella2296"][hide="Click for solution"]
$ 5b9$ is divisible by 9, so it follows that $ b \equal{} 4$. $ 549 \minus{} 326 \equal{} 223$, so $ a \equal{} 2$. $ a \plus{} b \equal{} 2 \plus{} 4 \equal{} 6$, so our answer is $ \boxed{\textbf{(C)}}$.
[/hide]

EDIT: Beaten.[/quote]
hmmm is it really that necessary to be asking a question and answering it yourself, izzy? if not, was that the actual published solution in the AHSME answer key?

They're posting the problems so it can be added to the resource section.

[quote="gauss1181"] was that the actual published solution in the AHSME answer key?[/quote]

No it was not.  It was similar by following the same method, but not word for word.


geometry
3D geometry
sphere
tetrahedron
ratio
pyramid

[quote="sonny"][quote="bpms"][quote="sonny"]Two spheres are inscribed in two cubes. The lengths of the sides of the cubes are 10 and 30. In the larger sphere another tetrahedron is inscribed. What is the ratio of the volumes of the triangular pyramid to the sphere with radius 10?[/quote]
[hide]With the 30-150 formula the edge of the tetrahedron is $30\sqrt{3}$. The volume of the tetra is $6750 \sqrt{3}$. The volume of the smaller sphere is $\frac{4000 \pi}{3}$. The ratio is $\frac{81 \sqrt{3}}{16}$[/hide][/quote]

Check the volume of the sphere again.[/quote]
What?
[hide]$\frac{4 \pi r^{3}}{3}$?[/hide]
Oh, I forgot pi in my answer so just divide my answer by pi.

[quote="bpms"][/quote][quote="sonny"][quote="bpms"][quote="sonny"]Two spheres are inscribed in two cubes. The lengths of the sides of the cubes are 10 and 30. In the larger sphere another tetrahedron is inscribed. What is the ratio of the volumes of the triangular pyramid to the sphere with radius 10?[/quote]
[hide]With the 30-150 formula the edge of the tetrahedron is $30\sqrt{3}$. The volume of the tetra is $6750 \sqrt{3}$. The volume of the smaller sphere is $\frac{4000 \pi}{3}$. The ratio is $\frac{81 \sqrt{3}}{16}$[/hide][/quote]

Check the volume of the sphere again.[/quote][quote="bpms"]
What?
[hide]$\frac{4 \pi r^{3}}{3}$?[/hide]
Oh, I forgot pi in my answer so just divide my answer by pi.[/quote]

The radius is 5 not 10.

[quote="sonny"][/quote][quote="bpms"][/quote][quote="sonny"][quote="bpms"][quote="sonny"]Two spheres are inscribed in two cubes. The lengths of the sides of the cubes are 10 and 30. In the larger sphere another tetrahedron is inscribed. What is the ratio of the volumes of the triangular pyramid to the sphere with radius 10?[/quote]
[hide]With the 30-150 formula the edge of the tetrahedron is $30\sqrt{3}$. The volume of the tetra is $6750 \sqrt{3}$. The volume of the smaller sphere is $\frac{4000 \pi}{3}$. The ratio is $\frac{81 \sqrt{3}}{16}$[/hide][/quote]

Check the volume of the sphere again.[/quote][quote="bpms"]
What?
[hide]$\frac{4 \pi r^{3}}{3}$?[/hide]
Oh, I forgot pi in my answer so just divide my answer by pi.[/quote][quote="sonny"]

The radius is 5 not 10.[/quote]
Heh, I feel really stupid now...
Good problem by the way :)
Well then my revised answer is
[hide]$\frac{81 \sqrt{3}}{2 \pi}$[/hide]
Unless I'm being stupid again.


How does your mathcounts coach run the meetings? How often are they, and what are normal procedures (no detail necessary)?

Here's how our meetings are ran:

First we get in the classroom after school, once a week (maybe two near competitions). we have some time-wasting people in our club, so the meetings usually start 15 minutes late. We then go over announcements, which takes us to the half-hour point out of 80 minutes. We work on a page in the handbook, which takes us 20-30 minutes, and then explain answers. By then, we would have 10 minutes left. We may do a few countdown problems (which are no fun since I take everyone of them  :roll: ). Then we leave.

That's probably the basic schedule. I'm wondering what you guys do.

Well we have 1.5hr meetings once or twice a week for a month before the state comp (im from DE we dont have chapters :D  :D ). First our coach talks for a while then we do half of a school round comp (either sprint/target or team/cd) and the other another week. Then we eat pizza and he gives us handbook problems to do for practice at home. done

I am one of the coaches, so I should definately know all of this. :D (The real coach asked me and another alum to come back and help teach.)

The school has expanded day, so each kid has to choose MC as their "class" for that day.  So, each week we have 45 min sessions starting from the first week of school and ending after the first semester.  Then we have about 2 practices a week leading up to chapters.

In a typical meeting, we will go over questions to the previous week's homework, give out new homework, and teach new concepts occasionally.  We are about to start giving practice tests.

50 minute meetings (too short), once a week every monday after school (3:00-3:50).

Check Problem of the week, MATHCOUTNS. there might be a short lesson on something, then we do a MATHCOUNTS warmup or a workout sheet. She posts 2 Manhattan Math Olympiad questions while we're finishing up the warmup/workout. We work on that if we're done. Go over answers to warmup/workout. Discuss solutions to M.M. Olympiad questions. Pass out new MATHCOUNTS Problem of the Week. Leave.

My dad is our coach.  We start at about 2:10 and end at various times.  Most people leave at 3:30, but for the 6th graders, it ends at 4:30 (I of course stay the entire time).  Generally, we do some kind of worksheet about new topics like sums of sequences.  I mostly help the other students.  Then, we learn how to do those problems.  After we finish that, we do some sort of competition (practice AMC, Mathcounts rounds...).  We explain the problems and by then, it is usually over.  In the extra time, we do stuff from the workbook and some coutdown round.  We practice once a week first semester and twice near the competitions.

[quote="236factorial"][quote="biffanddoc"] Then we eat pizza  [/quote]

WHAT?!?!?!?!?!?!?!?!?!?!?!  :mad:  That's not fair!  :P

I see some of you have a little bit of "homework" to do, we don't have any homework here (but I want some....  :D )[/quote]Oh yeah and soda too  :lol: (free)

In my mathcounts class, we do a quick worksheet and then check it.  THen our teacher teaches a new concept or lesson, and then we do another worksheet.  We do all this while eating free pizza and soda. :D

No fair, you guys get pizza and soda.
 :P

HOw long are your guy's classes?  Mines half an hour to an hour.

Mine's a hour and a half (but really an hour because we eat pizza for a half hour :) )

Mine:

Starts after school at 3:00. One warm-up completed, checked, and discussed before 3:30. One work-out completed, checked, and discussed before 4:00. Ends at 4:00.

However, we're going toward working less on the handbook and more on mock competitions. w00t!

We have a mock competition tomorrow (against another school). Do any of your schools do that, and if so, how often?

no. We start in February. seriously :mad: Im gonna talk to the coach

The only MathCounts that we do in our school is Homework.... We look over the answers once a week (Tuesdays) and get another sheet of Mathcounts to do at home. We work on mostly TMSCA stuff.


1. Determine all functions $f: \mathbb R \to \mathbb R$ such that \[ f(x\plus{}yf(x))\equal{}f(x)\plus{}xf(y)\] for all reals $x$ and $y$.
2. Determine all functions $ f: \mathbb R^* \to \mathbb R^*$ which satisfy \[ f(x^2\plus{}y)\equal{}f(f(x))\plus{}\frac{f(xy)}{f(x)}\] for $x$ and $y$ such that $x,y,x^2\plus{}y \in \mathbb R^*$.

[quote="KDS"][b]Problem 1[/b]:Determine all functions $ f: R \to R$ such that $ f(x \plus{} yf(x)) \equal{} f(x) \plus{} xf(y)$[/quote]

Let $ P(x,y)$ be the assertion $ f(x \plus{} yf(x)) \equal{} f(x) \plus{} xf(y)$

$ P(1,0)$ $ \implies$ $ f(0) \equal{} 0$

If $ f(1)\neq 1$, then $ P(1,\frac 1{1 \minus{} f(1)})$ $ \implies$ $ f(\frac 1{1 \minus{} f(1)}) \equal{} f(1) \plus{} f(\frac 1{1 \minus{} f(1)})$ and so $ f(1) \equal{} 0$. Then $ P(1,x)$ $ \implies$ $ f(x) \equal{} 0$ $ \forall x$ and we got one solution.

Consider now $ f(1) \equal{} 1$

$ P(1,x)$ $ \implies$ $ f(x \plus{} 1) \equal{} f(x) \plus{} 1$
Suppose now $ f(a) \equal{} 0$. Then  $ P(a,1)$ $ \implies$ $ a \equal{} 0$ and so $ f(x) \equal{} 0\iff x \equal{} 0$

Let then $ y\neq 0$ : $ f(y)\neq 0$ and

$ P(y,\frac {x \minus{} y}{f(y)})$ $ \implies$ $ f(x) \equal{} f(y) \plus{} yf(\frac {x \minus{} y}{f(y)})$

$ P(y,\frac {x \minus{} y}{f(y)} \plus{} 1)$ $ \implies$ $ f(x \plus{} f(y)) \equal{} f(y) \plus{} yf(\frac {x \minus{} u}{f(u)} \plus{} 1)$ $ \equal{} f(y) \plus{} yf(\frac {x \minus{} u}{f(u)}) \plus{} y$

Comparing these two lines, we get $ f(x \plus{} f(y)) \equal{} f(x) \plus{} y$ $ \forall x,\forall y\neq 0$ and, since this equality is true also for $ y \equal{} 0$ :

New assertion $ Q(x,y)$ : $ f(x \plus{} f(y)) \equal{} f(x) \plus{} y$ $ \forall x,y$

$ Q(0,x)$ $ \implies$ $ f(f(x)) \equal{} x$ and so $ f(x)$ is a bijection.
Then $ Q(x,f(y))$ $ \implies$ $ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$
Then $ P(x,y)$ becomes $ f(x) \plus{} f(yf(x)) \equal{} f(x) \plus{} xf(y)$ and so $ f(yf(x)) \equal{} xf(y)$ and so $ f(xy) \equal{} f(x)f(y)$

So we have the classical equation :
$ f(1) \equal{} 1$
$ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$
$ f(xy) \equal{} f(x)f(y)$

Whose unique solution is $ f(x) \equal{} x$
[hide="See quick demo"]
$ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$ and $ f(1) \equal{} 1$ imply $ f(x) \equal{} x$ $ \forall x\in\mathbb Q$
$ f(xy) \equal{} f(x)f(y)$ implies $ f(x^2)\geq 0$ and so $ f(x)\geq 0$ $ \forall x > 0$ and so, using $ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$, $ f(x)$ is monotonous.
$ f(x) \equal{} x$ $ \forall x\in\mathbb Q$ and $ f(x)$  monotonous imply $ f(x) \equal{} x$ $ \forall x\in\mathbb R$
[/hide]

And so the two solutions for this equation :
$ f(x) \equal{} 0$ $ \forall x$
$ f(x) \equal{} x$ $ \forall x$

[quote="KDS"][b]Problem 2[/b]:Determine all functions $ f: R^* \to R^*$ such that $ f(x^2 \plus{} y) \equal{} f(f(x)) \plus{} \frac {f(xy)}{f(x)}$   $ \forall x,y,x^2 \plus{} y \in R^*$

$ R^*$ is the set of nonzero real numbers.[/quote]

Let $ P(x,y)$ be the assertion $ f(x^2\plus{}y)\equal{}f(f(x))\plus{}\frac{f(xy)}{f(x)}$

Suppose $ \exists a\in\mathbb R^*$ such that $ f(a)\neq a^2$. Then :

$ P(a,f(a)\minus{}a^2)$ $ \implies$ $ f(f(a))\equal{}f(f(a))\plus{}\frac{f(a(f(a)\minus{}a^2))}{f(a)}$ and so $ f(a(f(a)\minus{}a^2))\equal{}0$ which is impossible.

So $ f(x)\equal{}x^2$ $ \forall x\in\mathbb R^*$

But then $ f(x^2\plus{}y)\equal{}x^4\plus{}y^2\plus{}2x^2y$ while $ f(f(x))\plus{}\frac{f(xy)}{f(x)}\equal{}x^4\plus{}y^2$ and so $ P(x,y)$ is wrong.

So no solution.

Dear pco,i think KDS give us a fail question.It must be:
$ f(x^{2}+y)=f^2(x)+\frac{f(xy)}{f(x)} $
It a problem from Bulgaria
I have a similar problem from my school test selection team to VMO 2009
#2 $ f(x^{3}+y)=f^3(x)+\frac{f(xy)}{f(x)} $
The difficult is finding  $f(1)$...After have $f(1)=1$ we have new problems:
$f^n(x)=f(x^n)$ and
$f(x+n)=f(x)+n$
$\to$ become trivial
[hide]Find the fomula for $f$ in $\mathbb Q$ and prove that $f$ is inscreasing so take the limit to get fomula of $f$  over $\mathbb R$[/hide]
The step find $f(1)$ not very difficult,we can solve with a equation degree 9 for #2 of $f(1)$ and the only root is $\boxed1$ :wink: Try it.
[hide]Brute force :P [/hide]


[quote="Captain Sugar"][quote="lotrgreengrapes7926"][quote="Captain Sugar"]i have ten pairs of socks in my drawer. if I pick a random number of socks from my drawer, what is the chance that i will get a match?

and part 2, the hard part:
what is the probability that every sock i pick out will match another sock i picked out if i pick a random number of socks?[/quote]
Is the random number between 0 and 10?[/quote]
between 0 and 20, because there are 10 pairs of socks.[/quote]

You can't pick out 20? This is sorta expected value I guess..

what i really meant was between 1 and 20, inclusive.


This belongs here ???????

http://www.mathlinks.ro/Forum/latexrender/pictures/10f7967de5dd00b1f224345aee11b8c3.gif

Anyone can help plzz

what is there to solve? you posted $4y^{4}$..

:mad:  :spam:

PLEASE don't spam

SOME people actually use these forums to do math; you're not helping.

I believe the intended question was to solve : $x^{4}+8xy+4y^{4}-4=0$

[hide="From where Zuton left off..."]$\left(x^{2}+2y^{2}-2xy+2\right)\left(x^{2}+2y^{2}+2xy-2\right)=0$

So either one of those or both of those are 0.

The roots of x^2+2y^2-2xy+2 are not available. Please hang up and dial again.

Well, so much for the easy way. *throws phone over shoulder*

Let's find the roots of x^2+2y^2+2xy-2. (0,1) works, so we divide by something..... Can you divide????[/hide]

[hide="A little more..."] $x^{2}+2y^{2}-2xy+2 = (x-y)^{2}+y^{2}+2 \ge 2$ has no real roots.

The other factor can be written

$(x+y)^{2}+y^{2}= 2$

Which is an oblique (?) ellipse.  We can parametrize $x+y = \sqrt{2}\cos \theta, y = \sqrt{2}\sin \theta$ and this gives the solution set

$(x, y) = \boxed{ \left( \sqrt{2}(\cos \theta-\sin \theta), \sqrt{2}\sin \theta \right) }$

Unless you wanted [i]integer[/i] roots; the only integer roots occur when $x+y, y = \pm 1$ which gives the four solutions

$(x, y) = \boxed{ (0, 1), (2,-1), (-2, 1), (0,-1) }$ [/hide]

[quote="binomial_4eva"]i think that link contains virus.[/quote]
How do you figure?

So? I can create something that doesn't look right that isn't a virus.

david you never been hacked thats y, hackers do alot of things

for christ's sake it's a .gif on the aops site, it's not a virus

How can you make te trig substitution, since we do not know that x+y E[-1,1]?

It's not.  But the solution set to $a^{2}+b^{2}= r^{2}$ is given parametrically by $a = r \cos \theta, b = r \sin \theta$.  This is just a conversion from rectangular coordinates to polar coordinates, if you want to think about it that way.

1. Let's not get any religion involved.

2. The equation that he meant to post was $x^{4}+8xy+4y^{4}-4$

3. Since it was posted on Art of Problem Solving, either

(i). There has been a bug unleashed, which is unlikely, but since the universe might be infinite, there is a possibility.

(ii). Typo. Nothin' to worry about.

tnx guys for help and sry for mistake i clicked copy image location but it copied just 4y^2.


Hi all, Im working on a problem from a school contest. The problem is as follows:

Find all triples of numbers x, y and z such that when any one number is added to the product of the other two numbers the sum is 2.

I'm not sure if the work I am doing is correct, but this is what I have:

z + xy = 2
y + xz = 2
x + yz = 2

These three equations only satisfy one number added to the product of the other two so for ANY one number to be added to the product of the other two, it would only occur where the three equations intersect, right?

I add the three equations and get
x + y + z + xy + xz + yz = 6

from the original three equations any one variable can be eliminated by substitution.
i.e. x = 2 - yz
so if i plug all of those in for x I get:
3y + 3z - (y^2)z - y(z^2) = 4

So I can say that all points included in the graph of that function would yield a solution that satisfies the triples of numbers condition.
Is this correct?  Or am I making incorrect assumptions?  Any help is appreciated.. Thank You  :)

$z+xy=y+zx \iff (z-y)=x(z-y) \iff (z-y)(1-x)=0$.  Argue similarly for the other equations.  What is the conclusion?

By the way, this is in the wrong section.

Sorry for posting in the wrong section, I wasn't sure where post it.  I'm a little confused by what you're doing.  You are substituting y + xz for 2 correct?  I am confused since substuting that for 2, would take away the signifigance of 2.  The substitution would then work for any constant wouldn't it?
For example I can then say the same for:

z + xy = 3
y + xz = 3
And wind up with the same equation....unless I'm missing something?

x + yz = 2
If x = 2, and if y = 0, then z can be pretty much anything.

Yes, but only for that equation. The problem is to find all triples of #'s x, y , and z such that when [b]any[/b] number is added to the product of the other two the sum is 2.  For example if x=y=z=1, then it would satisfy the problem for all circumstances.
1 + 1*1 = 2 , for all of the equations.


Let a < b. If $a, b, 2a, 3a, b+2,$ and $b+7$ are all roots for x of the equation $x^{4}+cx^{3}+dx^{2}+ex+f = 0$ where $c, d, e,$ and $f$ are all integers, find the sum of all distinct possible values of f.

The first line of the solution says:
"The given information $a < b$ implies that all roots are real and thus there are four distinct roots among the six listed."

I don't get why... help please?

$<$ is only defined on real numbers.  Moreover, a quartic equation can only have up to four distinct roots.  Since $a < b$ we can't have $a = b$, so there is just some annoying casework to work through (which two pairs of the roots are equal).


calculus
integration
trigonometry
function
complex numbers
calculus computations

The integral equals $\infty$.

To see this use the substitutions $\tan(x)=y$ and $y^{2}=a$ and you get that

$\int_{0}^{\pi/2}\frac{dx}{\tan(x))^{\sqrt{2}}}=\frac{1}{2}\int_{0}^{\infty}\frac{a^{-1/2-\sqrt{2}/2}}{a+1}da=\beta(1/2-\sqrt{2}/2,\sqrt{2}/2-1/2)=\infty$,
since the beta function is defined for all complex numbers whose real parts are positive reals.

[quote="Carcul"]Are you sure you are not looking for $\int^{\pi/2}_{0}\frac{1}{1+(\tan x)^{\sqrt{2}}}\,dx$?[/quote]
What's the solution of this problem?!

Hi Payman, :)  if you're looking for the discussions about Elmo Sanaat's University Mathematics Competition i have posted them completely here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=151928


LaTeX geometry counting distinguishability

Are we allowed to use colored text (e.g. blue or red) in $\LaTeX$ solutions?

Wasn't this question just asked?  Anyway, I believe DPatrick answered that yes as long as you don't fax them (the fax wont show color...).

No, little bit different.

pkerichang asked whether he could use colors in his pictures. ZennyK is asking whether he can use color pencils in the text, not pictures.

I believe the answer will be similar, however. If you're faxing, it doesn't make a difference whether your color is black or blue. If you're faxing, I guess it's OK as long as you don't select fonts as yellow or light green which are hard to see.

Thanks, guys.
I was specifically asking about $\LaTeX$ solutions via e-mail, not fax. I'm guessing they probably print these solutions in B&W anyway, though, so it might not matter. I think I'll just stick to plain, boring old black. ;)

I was asking because often it is easier to see what's going on in big, ugly algebraic manipulations if you use different colors for separate "chunks" of an expression.

[quote="joml88"]O yeah, that's what it was.  Hmm, I would keep your text in black.  It would most likely annoy the graders to have to read something in multiple colors.  Also, it would be less professional.[/quote]
I disagree, but alright. :roll:

[quote="ZennyK"]
I was asking because often it is easier to see what's going on in big, ugly algebraic manipulations if you use different colors for separate "chunks" of an expression.
[/quote]
I've never seen that done before, and probably it's not a good idea, especially because you don't know whether the graders will even get color copies of the solution.  Probably a better way to separate parts of an expression would be to use parentheses (with \left and \right as necessary) to make reading it all easier on the graders' eyes.

[quote="ZennyK"]I was asking because often it is easier to see what's going on in big, ugly algebraic manipulations if you use different colors for separate "chunks" of an expression.[/quote]

If you're having trouble expressing yourself, your first instinct should be to ask, how can I re-organize what I am saying to make it more clear, instead of thinking of artificial methods like coloured text.

When dealing with large algebraic expressions, one thing that may help is to break up your expressing into smaller pieces (there may be a natural way to do this), do your manipulation on each piece, and then re-assemble them.

Another possibility is to do your manipulations one basic step at a time, and then explain each step in detail, so that you're not doing a lot at once.

As others have pointed out, colour doesn't really belong in a mathematical proof, except perhaps in a geometry diagram.  In a proof, your words are your primary means of communication, so your priority should be to choose them well.

Assume that anything that is sent electronically will be printed in black and white.

If you mail it, the grader will see it with color.  If it is emailed or faxed, it will be black and white.

[quote="pkerichang"]uhh....so should I have coloured pictures??? because I have some 3D pictures, and without colors it looks ungly and indistinguishable (is that a word???)...[/quote]

Try experimenting with grey-scale; adjust colors and print it out in black and white until you're satisfied that it would be easily interpreted by the grader.


probability
LaTeX
trigonometry
geometry
trig identities
Law of Cosines
area of a triangle

An object moves $ 8$ cm in a straight line from $ A$ to $ B$, turns at an angle $ \alpha$, measured in radians and chosen at random from the interval $ (0,\pi)$, and moves $ 5$ cm in a straight line to $ C$.  What is the probability that $ AC < 7$?

$ \textbf{(A)}\ \frac {1}{6} \qquad \textbf{(B)}\ \frac {1}{5} \qquad \textbf{(C)}\ \frac {1}{4} \qquad \textbf{(D)}\ \frac {1}{3} \qquad \textbf{(E)}\ \frac {1}{2}$

[color=green][size=75]-LaTeX EDITED-[/size][/color]

[quote="314.math"]An object moves $ 8$ cm in a straight line from $ A$ to $ B$, turns at an angle $ \alpha$, measured in radians and chosen at random from the interval $ (0,\pi)$, and moves $ 5$ cm in a straight line to $ C$.  What is the probability that $ AC < 7$?

$ \mathrm{(A)}\ \frac {1}{6} \qquad\mathrm{(B)}\ \frac {1}{5} \qquad\mathrm{(C)}\ \frac {1}{4} \qquad\mathrm{(D)}\ \frac {1}{3} \qquad\mathrm{(E)}\ \frac {1}{2}$[/quote]

One of the brackets in your $ \text{\LaTeX}$ code is $ ]$ when it should be $ \}$.

[hide="Solution."]
=====
As $ \alpha$ ranges from $ 0$ to $ \pi$, the distance $ AC$ is decreasing, so we need only find what $ \alpha$ causes $ AC = 7$ and then deduce that any $ \alpha$ greater than this will have $ AC < 7$. For this we use the Law of Cosines, knowing that the angle subtending $ AC$ is $ \pi - \alpha$:
\[ AC^2 = AB^2 + BC^2 - 2\cdot AB\cdot BC\cos (\pi - \alpha)
\]

\[ 7^2 = 8^2 + 5^2 - 2\cdot 5\cdot 8\cos (\pi - \alpha)
\]
Solving this gives you $ \cos (\pi - \alpha) = \frac {1}{2}$, and since our angles are all in $ [0,\pi]$, we have $ \pi - \alpha = \frac {\pi}{3} \implies \alpha = \frac {2\pi}{3}$, so any $ \alpha$ in the interval $ \left(\frac {2\pi}{3},\pi\right)$ will have $ AC < 7$, which is $ \frac {1}{3}$ of the interval in which $ \alpha$ is chosen. $ \boxed{D}$
=====
[/hide]

For an alternate solution

[hide="Solution"]
Use two different areas of the triangle (Heron's and Sine Formula):
$ \dfrac{ab sinc}{2} \equal{} \sqrt {(s)(s\minus{}a)(s\minus{}b)(s\minus{}c)}$
$ 20sinC \equal{} 10\sqrt{3}$
$ c\equal{}\pi/3$
If the angle gets smaller, then $ AC$ decreases so you have $ \dfrac{\dfrac{\pi}{3}}{\pi} \equal{} \dfrac{1}{3}$[/hide]

Also Jvenezuela, there's a problem with your solution. As $ \alpha$ increases, $ AC$ increases not decreases. Kind of hard to explain why but say you have two sticks of fixed length $ AB$ and $ BC$ forming a triangle with a flat surface. If you increase the angle between AB and BC then the length of the surface that is part of the triangle increases not decreases.

[quote="crazypianist1116"]
Also Jvenezuela, there's a problem with your solution. As $ \alpha$ increases, $ AC$ increases not decreases. Kind of hard to explain why but say you have two sticks of fixed length $ AB$ and $ BC$ forming a triangle with a flat surface. If you increase the angle between AB and BC then the length of the surface that is part of the triangle increases not decreases.[/quote]

No, I'm right: $ \alpha$ is not the angle between $ AB$ and $ BC$, $ \pi \minus{} \alpha$ is. So as $ \alpha$ increases, $ \pi \minus{} \alpha$ decreases, and $ AC$ decreases, as I said.

[quote="314.math"][quote="crazypianist1116"]$ \dfrac{ab sinc}{2} \equal{} \sqrt {(s)(s \minus{} a)(s \minus{} b)(s \minus{} c)}$
$ 20sinC \equal{} 10\sqrt {3}$[/quote]

I get the left hand side, but for Heron's Formula, what are the three triangle side lengths?  I know there's 8 and 5...[/quote]

The other length is 7 which is the desired length of $ AC$.

[quote]No, I'm right: $ \alpha$ is not the angle between $ AB$ and $ BC$, $ \pi \minus{} \alpha$ is. So as $ \alpha$ increases, $ \pi \minus{} \alpha$ decreases, and $ AC$ decreases, as I said.[/quote]
I c wut u did thar \internet même

[quote="crazypianist1116"][quote="314.math"][quote="crazypianist1116"]$ \dfrac{ab sinc}{2} \equal{} \sqrt {(s)(s \minus{} a)(s \minus{} b)(s \minus{} c)}$
$ 20sinC \equal{} 10\sqrt {3}$[/quote]

I get the left hand side, but for Heron's Formula, what are the three triangle side lengths?  I know there's 8 and 5...[/quote]

The other length is 7 which is the desired length of $ AC$.[/quote]

Isn't $ AC$ supposed to be less than $ 7$?

Could you please explain this part:

[quote="crazypianist1116"]If the angle gets smaller, then $ AC$ decreases so you have $ \dfrac{\dfrac{\pi}{3}}{\pi} \equal{} \dfrac{1}{3}$[/quote]

May that'll help.

Thanks!

Well, the idea is that: If you find the angle for which AC = 7 ( pi/3), you know that AC > 7 when the angle is greater than pi/3. So you could do this

Probability that angle is greater than 7 = 2/3, so the probability that the angle is less than 7 is 1 - 2/3 = 1/3.

Is there any solution without trigonometry.

1.3e721122003

Yes, there is - and it's actually not that bad.  Looking at the answers, you know the angle ranges are going to go from $0$ to $\frac{\pi}{n}$ for $n = 6, 5, 4, 3, 2$, so the only angle that's not nice is $\frac{\pi}{5}$...so it's pretty likely that a non-trig solution is possible, since the angle is bound to be a "nice" (ie 30-60-90 or 45-45-90) angle.

[hide="Solution"]Draw an $5, 7, 8$ triangle with $AB= 8, BC = 5, AC = 7$, and drop an altitude to the side with length $8$.  You now have two right triangles.  Label the point on $AB$ where the altitude hits with an $H$.

Let $BH = x$, so that $HA = 8 - x$.  Let $CH = h$.  From the two right triangles we have:

$x^2 + a^2 = 25$

$a^2 + (8 - x)^2 = 49$

Solving gives $a = 2.5\sqrt{3}, x = 2.5$.  So $\triangle BHC$ is a right triangle with sides $2.5, 2.5\sqrt{3}, 5$.  Therefore, it's a 30-60-90 triangle, with the angle opposite the altitude measuring $\frac{\pi}{3}$ radians.

Thus, from $0$ until $\frac{\pi}{3}$, third side is smaller than $7$, and past $\frac{\pi}{3}$ to $\pi$, the third side is bigger than $7$.  Therefore, the answer is $\boxed{D}$.[/hide]


number theory
greatest common divisor
puzzles

Determine the smallest number N constituted entirely by 3’s and 7’s with each of the digits 3 and 7 occurring at least once such that both N and the sum of digits of N are divisible separately by 3 and 7.

[hide]
To be divisible by 3, the digits must add up to a multiple 3. If you take away all the 3's, the remaining digits must add up to a multiple of three because adding 3 doesn't change the divisibility by three. Since, there is at least one 7, and one 3, we test...

3777
7377
7737
7773

And find none of them are divisible by 7. So we move to 5 digits.

33777
37377
37737 <-- Bingo.
so... 37737?[/hide]

[hide] For the N to be divisible by 3, the sum of its digits must be congruent to 0 mod 3. Thus 7's must come in groups of three, since 3(7) = 0 mod 3. By having a number that consists of 7's in multiples of three and any number of 3's, we've satisfied the constraint of divisibility by 3. 7 is a little trickier. I start by looking at the sum of the digits being divisible by 7. SInce 7's won't effect this type of divisibility, 3's have to come in groups of 7 (gcd (3,7) = 1). I will now look at the possible arrangements of three digits of 7's and seven digits of3's to find one which is divisible by both, i.e. 21. We already have divisibilty by three due to the test, so what were really looking for is divisibilty by 7. in doing so, we can cast replace all 7's with 0's. (and eliminate ending zeros, they dont affect divisibilty by 7 either). Starting from the smallest, I'll check for divisibility by 7's

3333333: Not Divisible (ND)
33333303: ND
33333033: ND
33330333: ND
33303333: DIVISIBLE!

Thus N=3337333377 is the smallest N satisfying the given properties.[/hide]

Previous answer doesn't work: Needs sum of digits divisible by 7


vector
algebra
system of equations
linear algebra

Let $ u \equal{} (1,2,3)$, $ v \equal{} (2, \minus{} 3, 1)$, and $ w \equal{} (3, 2, \minus{}1)$ be vectors. Find $ c_1$, $ c_2$ and $ c_3$ such that $ c_1u \plus{} c_2v \plus{} c_3w \equal{} (6, 4, \minus{}2)$

[quote="Carcul"]Just solve by using, for example, determinants.[/quote]

I don't think that was of help. ;)

But the topic starter could clarify what exactly his difficulties are. $ c_1u \plus{} c_2v \plus{} c_3w \equal{} (6, 4, \minus{} 2)$ is equivalent to the system of equations

$ 1c_1 \plus{} 2c_2 \plus{} 3c_3 \equal{} 6$;
$ 2c_1 \plus{} \left( \minus{} 3\right)c_2 \plus{} 2c_3 \equal{} 4$;
$ 3c_1 \plus{} 1c_2 \plus{} \left( \minus{} 1\right)c_3 \equal{} \minus{} 2$.

You should know some tactics to solve such systems. One of them uses determinants, but look up "Gaussian elimination" first for a more elementary tactic.

  darij


calculus
derivative
function
linear algebra
linear algebra unsolved

Let T be a linear operator: V -> V, where V is the space of all infinitely differentiable functions.

Find all real eigenvalues and a real eigenfunction for each eigenvalue.

 T(f(x)) = f ' ' ' ' (x)  [4th derivative]

 with e^(4throot(lambda)x), you get any lambda > 0.

 But now I must find eigenvalues < 0.. I have a feeling that you cannot find such eigenvalues, it seems like you could only use e^x function for something like this, and it wouldn't work for e^x.. Don't know how to prove it though..

 any ideas?

thanks a lot

Let $ \lambda$ be a complex eigenvalue and let $ \mu$ be a complex number s.t. $ \lambda\equal{}\mu^4$. If $ \lambda\not\equal{}0$ then the solutions of $ y^{(4)}\equal{}\lambda{y}$ are $ y\equal{}ae^{\mu{x}}\plus{}be^{i\mu{x}}\plus{}ce^{\minus{}\mu{x}}\plus{}de^{\minus{}i\mu{x}}$. If $ \lambda\equal{}0$ then the solutions are $ y\equal{}ax^3\plus{}bx^2\plus{}cx\plus{}d$.
If $ \lambda>0$ then we can take $ \mu>0$ and the real associated eigenvectors are $ y\equal{}ae^{\mu{x}}\plus{}be^{\minus{}\mu{x}}\plus{}ccos(\mu{x})\plus{}dsin(\mu{x})$.
If $ \lambda<0$, for instance $ \lambda\equal{}\minus{}4$ then we can take $ \mu\equal{}1\plus{}i$. The eigenvectors are $ y\equal{}ae^{(1\plus{}i)x}\plus{}be^{(i\minus{}1)x}\plus{}ce^{(\minus{}1\minus{}i)x}\plus{}de^{(1\minus{}i)x}$. The real form is: $ e^x(acos(x)\plus{}bsin(x))\plus{}e^{\minus{}x}(ccos(x)\plus{}dsin(x))$.

[quote="loup blanc"]Let $ \lambda$ be a complex eigenvalue and let $ \mu$ be a complex number s.t. $ \lambda \equal{} \mu^4$. If $ \lambda\not \equal{} 0$ then the solutions of $ y^{(4)} \equal{} \lambda{y}$ are $ y \equal{} ae^{\mu{x}} \plus{} be^{i\mu{x}} \plus{} ce^{ \minus{} \mu{x}} \plus{} de^{ \minus{} i\mu{x}}$. If $ \lambda \equal{} 0$ then the solutions are $ y \equal{} ax^3 \plus{} bx^2 \plus{} cx \plus{} d$.
If $ \lambda > 0$ then we can take $ \mu > 0$ and the real associated eigenvectors are $ y \equal{} ae^{\mu{x}} \plus{} be^{ \minus{} \mu{x}} \plus{} ccos(\mu{x}) \plus{} dsin(\mu{x})$.
If $ \lambda < 0$, for instance $ \lambda \equal{} \minus{} 4$ then we can take $ \mu \equal{} 1 \plus{} i$. The eigenvectors are $ y \equal{} ae^{(1 \plus{} i)x} \plus{} be^{(i \minus{} 1)x} \plus{} ce^{( \minus{} 1 \minus{} i)x} \plus{} de^{(1 \minus{} i)x}$. The real form is: $ e^x(acos(x) \plus{} bsin(x)) \plus{} e^{ \minus{} x}(ccos(x) \plus{} dsin(x))$.[/quote]

 Hi, I appreciate your solution, but I haven't studied many of your concepts there (the real and imaginary forms). I think there must be a more elementary solution ( for lambda > 0, e^(4throot(lambda)x) ), I need to think of one for < 0..

thanks a lot


function
limit
real analysis
real analysis unsolved

[quote="yagaron"]$f$ is a real function bounded above and below.
$g$ is a real function such as ${\lim_{x\to 0}\, g(x)=0}$

So, can we say that ${\lim_{x\to 0}\, g(x) f(x)=0}$ ?[/quote]

Let say $f(x)$ is a function defined on an open interval containing zero and it bounded by $K>0$.
That is, $|f(x)|\leq K$ on this open interval.

Then,
$|f(x)g(x)|\leq K\cdot |g(x)|$

But,
$\lim_{x\to 0}K\cdot |g(x)| = K\cdot 0 =0$.

By the squeeze principle the rest is finished.


Handy Aaron helped a neighbor $ 1\frac{1}{4}$ hours on Monday, 50 minutes on
Tuesday, from 8:20 to 10:45 on Wednesday morning, and a half-hour on Friday.
He is paid $ \$3$ per hour. How many dollars did he earn for the week?

First of all, rewrite everything into minutes.
Monday: 1 1/4hr = 60+15 = 75 mins
Tuesday: 50 mins
Wednesday: 2 hrs 25 mins = 145 mins
Friday: 30 mins

Add them all up, and you get 300 mins, or 6 hrs.
If he is paid $ 3 per hour, he earned$18.

300 minutes is $ 5$ Hours not $ 6$ :wink: . $ 300/60\equal{}5$ I'm sure you just made a small error. :wink:

So the answer is 5*3=$ 15$


inequalities
arithmetic sequence
geometric sequence
algebra unsolved
algebra

An arithmetical progression has $ n$ members from which we can choose $ k$ members, which are members of geometric progression.
Prove that $ n\geq2^{k\minus{}1}$.

Let $ h$ is step of arithmetic progression and $ x_1,x_2,...,x_k$ is geometric progression, then $ x_2\minus{}x_1\equal{}y_1h,...,x_k\minus{}x_{k\minus{}1}\equal{}y_{k\minus{}1}h$ give geometric progression with natural numbres (I consider increasing $ x_i$, else we can defined $ y_i\equal{}(x_{i\plus{}1}\minus{}x_i)/h$).
If $ q\equal{}\frac{a}{b}, (a,b)\equal{}1$, then $ n\ge |x_{k}\minus{}x_1|/h \plus{}1\equal{}c|a^{k\minus{}1}\minus{}b^{k\minus{}1}\plus{}1\ge 2^{k\minus{}1}.$

[quote="Rust"]If $ q \equal{} \frac {a}{b}, (a,b) \equal{} 1$, then $ n\ge |x_{k} \minus{} x_1|/h \plus{} 1 \equal{} c|a^{k \minus{} 1} \minus{} b^{k \minus{} 1} \plus{} 1\ge 2^{k \minus{} 1}.$[/quote]
What number is $ c$?And how did you got that inequality?
Thank you!

[quote="Rust"]$ c \equal{} \frac {y_1}{b^{k \minus{} 2}}\in N$[/quote]
How do you know that $ c$ is a natural number.


integration trigonometry quadratics algebra function domain calculus

Can someone help with this one?

$\int \frac{dx}{1+\sin^{2}(x)}$

in my solutions I have something like this: $t=\tan(x)$

But I cant understand how they reached $\sin(x) =\frac{t}{\sqrt{1+t^{2}}}$.

Can anyone offer any easy, understandable solution for this one?  :)

I put it in the Integrator and its not pretty at all

$t = \tan \left(\frac{x}{2}\right)$ gives $dx = \frac{2dt}{1+t^{2}}$, and $\sin x = \frac{2t}{1+t^{2}}$. It is straightforward to check this.

Some calculation shows that the substitution gives
$\int \frac{2(1+t^{2})}{1+6t^{2}+t^{4}}dt$

This shows that the problem was not taylored to be nice, and the best approach would be with contour integration.

Anyway, if we want to continue we notice $(1+6t^{2}+t^{4}) = (t^{2}+3+2\sqrt{2})(t^{2}+3-2\sqrt{2})$ (just solve the quadratic in $t^{2}$).

So, we do the whole partial fraction business and get
$\int \frac{2(1+t^{2})}{1+6t^{2}+t^{4}}dt = \int \frac{1+\frac{\sqrt{2}}{2}}{t^{2}+3+2\sqrt{2}}+\frac{1-\frac{\sqrt{2}}{2}}{t^{2}+3-2\sqrt{2}}dt$

It is not so hard to realize that the numerators will be constants in this partial fractionation, so it's not really that hard even though the constants are ugly.

Now, I hope you know how to deal with the remaining two terms (complete square and make a linear variable substitution).

We get
$\int \frac{1+\frac{\sqrt{2}}{2}}{t^{2}+3+2\sqrt{2}}+\frac{1-\frac{\sqrt{2}}{2}}{t^{2}+3-2\sqrt{2}}dt = \frac{\sqrt{2}}{2}\left( \arctan \left({\frac{t}{\sqrt{2}-1}}\right)+\arctan \left({\frac{t}{\sqrt{2}+1}}\right) \right)$

That is,

$\int \frac{dx}{1+\sin^{2}x}= \frac{\sqrt{2}}{2}\left( \arctan \left({\frac{\tan \left(\frac{x}{2}\right)}{\sqrt{2}-1}}\right)+\arctan \left({\frac{\tan \left(\frac{x}{2}\right)}{\sqrt{2}+1}}\right) \right)$

Hopefully it is right. There will be some restrictions to the domain of this expression, but it can be fixed by 'continuing' $\arctan$. I.e. branch different inverses to $\tan$ together to one differentiable function... However, I have written so much already on this problem.

By the way, the substitution I made here works to transform trigonometric integrands to rational functions. In it it is also good to know that $\cos x = \frac{1-t^{2}}{1+t^{2}}$.

I think there may be a simpler way to do it:

$\int \frac{dx}{1+\sin^{2}x}= \int \frac{\sec^{2}x dx}{\sec^{2}x+\tan^{2}x}=\int \frac{\sec^{2}x dx}{1+2\tan^{2}x}$.

Now substitute $u=\sqrt{2}\tan(x)$.

Yes, you are right - that substitution is more effective in this particular case. But the method I gave works in very general situations.

[quote="236factorial"][quote="RiemannZetaFunction"]I put it in the Integrator [/quote]

Is that a website, or a function on your calculator?[/quote]
http://integrals.wolfram.com/index.jsp

[quote="Kalle"]This shows that the problem was not taylored to be nice...[/quote]

Hehehe.  Perhaps "tailored" instead?  But I do like it this way ...

one should adopt the most efficent solution. but must also keep a general approch in mind :yoda:  :yoda: :yoda:


for ever countable and dense subset of R {x(n):n=1,2,...}then
a.the sequence (x(n)) is Cauchy?
b.the sets {x(2n)}and {x(2n+1)} are dense too in R?
c.{x(n)}intersectionQ=non empty?
d.{x(n):n>and =m}is dense in R for every m=1,2,..
only one is the correct answer

uhm... just (d) is correct.
for any other there's a counterexample:
(a) you can number rationals, and of course you can get no real limit (so the sequence can't be a cauchy sequence, otherwise it'd be definitely limited.
(b) again, you can number the rational in such a way that $x_{2n} <0$ and $x_{2n+1} > 0$, and none of this two sequences is dense onto the reals.
(c) just number rationals and add $\sqrt2$ to each... none of this numbers is rational.
just in order to prove (d), you can suppose that's not true. let's call $(a_n)$ our sequence. now there's $x \in \mathbb{R}$ such that there exist $\epsilon > 0$ such that there's no $a_n \in (x-\epsilon, x+\epsilon)$. but for hypothesis $\forall \epsilon > 0 \exists k | \exists n > k, a_n \in (x-\epsilon, x+\epsilon)$, and you get contraddiction...


Let $n$ $=$ $4^{k}$ $+$ $1$ such that:
     $3^{\frac{n-1}{2}}$ +$1$ is divisible by $n$
Prove that $n$ is a prime number

I had edited the problem,sorry for the topo

If $3^\frac{n-1}{2}$ is divisible by $n$ then the only prime factor of $n$ is three. Then we have $3^{l}=4^{k}+1$. But Mihailescu's theorem says that the only integer solutions for $x^{n}-y^{m}=1$ is $x=3,y=2,n=2,m=3$ if $n,m\geq 2$. Thus we must have $2k=3$ or $l, k\in \{0,1\}$. Thus we have no such $n$.

Can u tell me the proof of Mihailescu's theorem ??  :o
Really thx for that...

[quote="bui_haiha2000"]Let $n$ $=$ $4^{k}$ $+$ $1$ such that:
     $3^{\frac{n-1}{2}}$ +$1$ is divisible by $n$
Prove that $n$ is a prime number
[/quote]

This is the true problem

$3^{n-1}\equiv 1 (mod \ n) \Rightarrow ord_{n}(3) \mid (n-1 = 2^{2k}), \\ but \ ord_{n}(3) \nmid 2^{2k-1}\Rightarrow ord_{n}(3) = 2^{2k}= n-1. \\ ord_{n}(3) \mid \varphi (n) \Rightarrow 2^{2k}\mid \varphi (2^{2k}+1) \ so \ \varphi (n) = n-1, so \ n \ prime.$


analytic geometry
graphing lines
slope
quadratics
calculus
derivative
function

1. Given that A(0,-3) and B(-10,-28) are two points that lie on a straight line in x and y space, find:
(a)	The slope of the line
(b)	The intercept
(c)	The equation for the straight line
(d)	 Sketch the line

2.  Use the quadratic formula to find the solutions (or roots) of :

   3x*3x + 45x = 48

3. If the following system of linear equations has infinitely many solutions, find the value of M:

  -12x + 3y = 4

  +4x - y = M

4. Find the first derivatives,   dy   of :
                                        --
                                        dx

  y= -6x (-squared)

5.  A firm has a demand function  P = 32 – Q
At what quantity is the total revenue maximised?
[/i]

[quote="HELLLP"]1. Given that A(0,-3) and B(-10,-28) are two points that lie on a straight line in x and y space, find:
(a)	The slope of the line
(b)	The intercept
(c)	The equation for the straight line
(d)	 Sketch the line
[/quote]

$ y \plus{} y_0 \equal{} \frac {y_1 \minus{} y_0}{x_1 \minus{} x_2}(x \minus{} x_0)$

[quote="HELLLP"]
2.  Use the quadratic formula to find the solutions (or roots) of :

   3x*3x + 45x = 48
[/quote]

$ 3^2x^2 \plus{} 5 \cdotp 3^2x \minus{} 2^4 \cdotp 3 \equal{} 0$

$ x_{0,1} \equal{} \frac { \minus{} 5 \cdotp 3^2 \pm \sqrt {5^2 \cdotp 3^4 \plus{} 2^6 \cdotp 3^4}}{18} \equal{} \frac { \minus{} 5 \cdotp 3^2 \pm 3^2\sqrt {5^2 \plus{} 2^6}}{18} \equal{}$

$ \equal{} \frac { \minus{} 5 \pm \sqrt {5^2 \plus{} 2^6}}{2} \equal{} \frac { \minus{} 5 \pm \sqrt {89}}{2}\ \mathellipsis$

[quote="HELLLP"]
3. If the following system of linear equations has infinitely many solutions, find the value of M:

  -12x + 3y = 4

  +4x - y = M

[/quote]

$ y \equal{} 4x \plus{} \frac43$

$ y \equal{} 4x \minus{} M$

$ M \equal{} \minus{} \frac43$

[quote="HELLLP"]
4. Find the first derivatives,   dy   of :
                                        --
                                        dx

  y= -6x (-squared)
[/quote]

$ f(x) \equal{} \minus{} 6x^{ \minus{} 2}$

$ f'(x) \equal{} 12x^{ \minus{} 3}$


[url]http://reflections.awesomemath.org/2008_1/unexpected_ineq.pdf[/url]
First problem in 5th page. :mad:

Try to change $ a\rightarrow\frac {1}{a},$ $ b\rightarrow\frac {1}{b}$ and $ c\rightarrow\frac {1}{c}.$ :wink:
Thank you for the link. I didn't know about it!