I solved IIT maths question today(I am in 12 now).It was quite easy wasnt it?I saw 6 problems which can be solved mentally only (seeing the options).........

might be for genious like u

but u sit in the jee exam hall and then see how tough the paper is !!

btw are u getting 160 to call it easy  :D

If six problems are easy, it doesn't mean the entire paper was easy.

[hide]If this be your question paper, I don't think you'd call it easy!

Q1) $ 1\plus{}2 \equal{} ?$
Q2) $ 2\times5\equal{}?$
Q3) $ 0.4/0.2 \equal{} ?$
[hide="Q4)"] Find the six thousandth derivative of $ \frac{\sin^{25} x\cos^{19} x \plus{} \sqrt{(1\minus{}\sec^{92}x\csc^{65}x)\left(4x^9\minus{}\frac{2}{x^6}\right)}}{x^3\cosh x \plus{} 15092\sin^2 x \cos^x}$
[/hide]

[hide]Hey you book worms out there, don't solve this and even if you do, I'm not interested in knowing the answer![/hide][/hide]

u call it easy??
i got 140 in phy,100 in chem
and 53 in maths :rotfl:

Me: Ph-:92, Math same as Riruraj and Chem (please don't ask). I never studided chem throughout the year. I never liked the boring stuff.

(Although I'm not competent enough, I hope I'm welcome in discussions here?)

[quote="preetham"]

but u sit in the jee exam hall and then see how tough the paper is !!

btw are u getting 160 to call it easy  :D[/quote]

sitting in the examination hall is different...I agree.
And I wanted to mean that IIT questions have tricky options(usually)...and they can rarely be solved by eliminating options..

Anyway  I didnt said only 6 problems are easy.Knowing some advanced maths make several questions easy.

[b]you may think taht I am a liar...But  one of my frnd probably you will see him in next years merit list   answered  almost all .  His name is Siddharatha Chatterjie.[/b]

Hey dude, you seem to be infuriated by our rematks! Don't take us seriously, we are just fooling around! No one meant to underestimate your problem-solving capabilities.

Your friend   Chatterjie    how much is he getting in maths murgi ie exact score out of 160 ?
 :oops:  (even i answered almost all but got few wrong)

i think he will appear in the exam next year??... :P

O.K now leave this discussion..I think you all gave IIT this year only.Please suggest a good book on physics containing very hard problems......I have Irodov,D.C pandey,200 challenging problems in physics Do you know anymore..............?

Anyway  arenot you all giving ISI admission test.

add it to it Krotov..

and i dont think DC pandey has more difficult questions than HCV or resnick.... :oops:

I don't know abt .C pandey new edition

but the D C Pandey old edition(i proudly own that book after $ n$ attempts where $ n$ tends to ur sig  :rotfl: )

D.C pandey (old,when IITJEE was subjective)is far far tougher than H C verma and in some topics,i'ld say even bettert than irodov(or sum ques are same  :rotfl:

Its my personal opinion and may vary 4m ppl to ppl

well i said that i hav never owned DC Pandey

i just saw 1-2 questions..adn i was able to solve it..so i thought no need to read another book

[hide]waise i bought HCV just 5 mnths before JEE and resnick 0.5 mnths before....never been interested in reading those thick books :rotfl:  :mad: [/hide]

what is the publication of krotov
No one amswered my 2nd question though??

@murgi krotov is probably in Science for everyone series(probably by CBS publishers)

@ ritu:even then u performed that well,it only shows ur class

yes
Krotov is by CBS publishers

Atleast i  m not giving ISI test :P

deeper understanding>>> solving million problems

solving good problems ==========> greater understanding
both of wich din not happen with me :D


function
real analysis
real analysis solved

Let $\alpha\in[0,1)$, $f:\mathbb{R}\to\mathbb{R}$ be a primitvable function and $F$ one of its primitives.
We have that $|f(x)\le\alpha|$ for all reals $x$.
Let $(x_n)_n$ be a sequence of real numbers so that $(x_{n+1}+F(x_n))_n$ is a convergent sequence.
Prove that $(x_n)_n$ is also convergent

Here’s a quick sketch:

Let $a_n=x_{n+1}-x_n$. We have $a_{n+1}+(F(x_{n+1})-F(x_n))\to 0$. However, $ F(x_{n+1})-F(x_n)=(x_{n+1}-x_n)\cdot F'(c_n)=a_n\cdot F'(c_n)$, where $c_n$ is between $x_{n+1}$ and $x_n$. This means that $a_{n+1}+a_n\cdot F'(c_n)\to 0$. It’s not hard to derive from here that $a_n\to 0$, so $x_n+F(x_n)=x_{n+1}+F(x_n)+x_n-x_{n+1}$ is also convergent.

We then show that $x+F(x)$ is bijective (and it is also continuous, of course), so from the convergence of $x_n+F(x_n)$ the convergence of $x_n$ follows easily.

[quote="grobber"]Let $a_n=x_{n+1}-x_n$. We have $a_{n+1}+(F(x_{n+1})-F(x_n))\to 0$.[/quote]
Are you sure? :?

It is time to know what is "Birch & Swinnerton Dyer"  :D  :D  :D  :first:

That's simply $x_{n+2}+F(x_{n+1})-(x_{n+1}+F(x_n))\to 0$, which holds because $x_{n+1}+F(x_n)$ is convergent.

Am I missing something? :?

Ops... Sorry. I was confused with the index. :blush:


ABC triangle and P1 and P2 are on the AC side.If P[hide="Index"]P1BP2[/hide]=AB+BC then find position of P1,P2


induction
combinatorics unsolved
combinatorics

Given $S_n=\{1;2;...;n\}$. Determine the least integer $k$ such that one can colour each subset of $S_n$ by one of $k$ colours and if $A$ and $B$ are the same colour then either $A\subset B$ or $B\subset A$

hello

i tried ur q and got a soln
but i dunno whether it is correct or not
it is a really very long

my answer is if n=2m then k=nCm    and if  n=2m+1   then k=nCm
nCm is n choose m

i mainly used induction in the entire q  :|

first i'm defining some symbols that i shall be using

1)A<B  for A belongs to B
2)[A] denotes cardinality of A

3)if a set A = {{a1},{a2}..........................{ar}}
   then (A,a) = {{a1,a},{a2,a}.....................{ar,a}}

clearly {ai}<{ai,a}
ie we can colour  A,(A,a)  by r colours     where  [A]=r..................................1

let Y(n) = {1,2,..............n}

first i am assuming that n is even = 2m

S(n) denote the set of all subsets of Y(n)
we arrange elements of S(n) in increasing order of cardinality

ie. s(i) = set of elements of S with cardinality i

s(1) = {{1},{2}.....................{2m}}
s(2) = {{1,2},{1,3}.................{1,2m}}
.
.
.
.
s(2m)={{1,2,3............2m}}

now starting from s(i) such that [s(i)] is the least( then initially i=2m) colour 1 element from each of the above sets s by a single colour so that

if Aj,Ak < s(j),s(k) resp then

Aj<Ak  for j less than k
ie. the condition in ur q
repeat this procedure with a diff color everytime till all elements of s(i) r consumed and then do the same for the left sets.
such a coloring always exists can be proven using induction

so finally we will reach a situation in which only some elements of s(m) r left

no. of elements left = 2mCm - 2mC(m-1) = t = [T]

and let the set of elements left = T

also no. of colors used till now = 2mC(m-1)

now we extend the process for S(n+1)

clearly S(n+1) = S(n) + (S(n),2m+1)

for A=(S(n),2m+1) we will follow exactly the same coloring procedure that we did in the previous case( as both the cases r exactly similar except that in the 2nd case {2m+1} has been added to every element of S(n))

so again for A no. of elements left = t
set of elements left = (T,2m+1)  as we followed the same procedure
no. of colors used = 2mC(m-1)

now colors reqd to color the remaining elements ie T,(T,2m+1) = [T] = t........from 1

hence total colors reqd to color elements of S(n+1) = 2mC(m-1) + 2mC(m-1) + t
                                                                          =(2m+1)Cm

but this only proves the upper bound that with 2m+1Cm colors we can certainly have a required coloring

to prove the lower bound consider all the subsets of Y(n+1) with cardinality m
no. of such subsets = 2m+1Cm

also we need different colors for these subsets as none of them contains all the elements of any other of them

ie. we need atleast 2m+1Cm colors
hence k min = 2m+1Cm

similarly we can go abt with odd n to prove for even nos.

plz do reply me back if u find some mistake, or some imp point missing :yup:

[quote="nthd"]My answer is $\binom{n}{[\frac{n}{2}]}$ ;)[/quote]

hi
that looks same as mine
any proof
 :?:

i wud love to see a shorter one


on augmented matrices, in what order should you get zeros under the main diagonal on a 5x4 matrix like this one:

w+x+y+z=3
-w+x-2y+z=-2
    2x-y+z=1
w     +y-z=2

i think the matrix should look something like this

?????
0????
00???
000??     i just dont know in what order to get the zeros...

i dont think it should matter, but maybe going from 2nd row to 4th row?
its easier to get multiple zeros on a row like that

i know this much of the order

?????
0????
20???
130??           but i dont know the rest, i think the rest MIGHT

                   be

?????
4????
26???
135??

Go one variable at a time using [url=http://en.wikipedia.org/wiki/Gauss%E2%80%93Jordan_elimination]the method of Gauss-Jordan elimination[/url]

Arrange your equations so that they are in some favorable order.

Now, make the w coefficient on the top line zero.

Now, add/subtract multiples of the first equation to/from the other equations to make all the other w coefficients zero

For example, if your equations are:
(1) w+x+y+z=3
(2) -w+x-2y+z=-2
(3) 2x-y+z=1
(4) w +y-z=2

Replace them with:
(1)
(2)+(1)
(3)-2*(1)
(4)-(1)

Now, do the same thing to eliminate x from all but the second line (which is now (2)+(1)).

This algorithmic approach works because the second equation has no w term, so you won't be undoing any previous work.


Find all 3-digit natural numbers $n$, which are the sum of all 2-digit natural numbers with digits only the digits of the number $n$

$\overline{abc}=\overline{aa}+\overline{bb}+\overline{cc}+\overline{ab}+\overline{bc}+\overline{ca}+\overline{ba}+\overline{cb}+\overline{ac}$


I have a solution (starts of like that :)). This problem is meant as a challenge to people in the begginer's corner.

[quote="Anh Cuong"]By B.C.S inequality,we have
Sumx_1/sqrt(1-x_1) \geq (x_1+...+x_n) 2 /sum(x_1*sqrt(1-x_1).
We have(by B.C.S again)
sqrt(x)*sqrt(x-x 2) \leq sqrt(1-sumx 2 )
By Chebysev,we have
n*sumx 2 \geq (sumx) 2 =1.
By all this ,we have the result we need.
 :D[/quote]
Sorry Anh Cuong but I found your solution extremely hard to read :( , I couldn't grasp the essential ideas from your solution (as a beginner's perspective). Here is my attempt to 'translate' his solution, which makes more sense to me, if this helps the beginners:

Using the Cauchy-Schwarz inequality on the numbers xi/ \sqrt (1-xi) and xi* \sqrt (1-xi) we have
( \sum (xi/ \sqrt (1-xi)))( \sum (xi* \sqrt (1-xi)))  \geq  ( \sum xi)2
So \sum (xi/ \sqrt (1-xi))  \geq  (( \sum xi)2) / ( \sum (xi* \sqrt (1-xi))) = 1 / ( \sum (xi* \sqrt (1-xi)))
On the other hand, by Cauchy-Schwarz again on the numbers  \sqrt xi and  \sqrt (xi(1-xi)),
( \sum (xi* \sqrt (1-xi)))2 \leq ( \sum xi) ( \sum (xi(1-xi)))
Now \sum (xi(1-xi))) =  \sum xi -  \sum xi2  \leq  1 - 1/n since by Chebyshev ineq., we can show that n \sum xi2 \geq ( \sum xi)2 = 1.
Therefore, 1 / ( \sum (xi* \sqrt (1-xi))) \geq  \sqrt (n/(n-1)) and we have the desired inequality. :cool:

3X.lich,you are right.My solution is really messy. :(
I'm sorry if it makes everybody tired


Find the value of $ \sqrt{12\minus{}\!\sqrt{12\minus{}\!\sqrt{12\minus{}\cdots}}}$.

This is what's known as a nested radical.

Let that whole radical be $ x$. So we then can say that $ \sqrt{12\minus{}x} \equal{} x$. Square both sides to get:

$ x^2 \equal{} 12 \minus{} x \implies x^2 \plus{}  x \minus{} 12 \equal{} 0$.

Factoring gives us $ x \equal{} \minus{}4$ and $ 3$. It can't be negative, so the answer is $ \boxed{3}$.


logarithms
function
number theory unsolved
number theory

we w'll have $a\gneqq 0$ the equation is equivalente to : $2^a\log 2= \ln a$
 we considere the function : $f\equiv {2^x\ log 2-\log x }$
 $\forall x\gneqq 0$ we have : $f'(x)={(\log 2)}^2 e^{x \log 2}-\frac{1}{x}$ we denote ${(log2)}^2= c$
 so $xf'(x)=cxe^{x \log 2}-1$ if we denot $k$ the unique solution $\gneqq 0$ of $cxe^{x \log 2}-1=o$
then $f$ has a minimum in $k$
$k$ must be $\lneqq 1$ so $f(k)=2^k\ log 2-\log k \gneqq o$
so the equation has no solution .


algorithm
combinatorics theorems
combinatorics

what is Robinson-Schensted correspondence?
I just know that  i can use it for counting standard tableaux

Also, if i want to count number of standard tableau, then i should use Robinson-Schensted correspondence or hook length formula?

Please answer specifically...

Have a nice day! :)

There is a nice informal description of it [url=http://math.mit.edu/~musiker/rstan7-8.pdf]here[/url].  If you want to count standard Young tableau of a particular shape, that's what the hook-length formula is for.  The RSK algorithm is better suited for proving statements that apply to tableau of many shapes, for example, the classic formula $ \sum_{\lambda} f_{\lambda}^2 \equal{} n!$.  It's possible that RSK can be used to prove the hook-length formula, although I have not seen such a proof.


What is the smallest possible average of four distinct
positive even integers?

The even integers would be 2,4,6, and 8, which have an average of 5.

it is not a positive integer or negative

Yeah 0 is not positive. But the actual answer is 5555, not just 5


complex analysis
function
complex analysis unsolved

1. If $ f$ is a holomorphic function on $ C$ such that the image of $ f$ does not intersect the region {$ z\in R|z \geq 5$}. Prove that $ f$ is constant.

Do you need elementary tools to solve this problem?

If not apply little picard theorem, if f entire avoid two values than f is constant.

Picard's theorem is a huge overkill here. Liouville is more than enough ;)

I looked at Wikipedia that the Little PIcard Theorem states the converse of the statement. So I think we can't use it, and so as Liouville's theorem. Can anyone give another proof ? Or, why can we determine that $ f$ is bounded, because the Liouville states that "An entire function $ f$ that is bounded must be constant." Thanks in advance.

It's not directly Liouville, since we're only missing a pair of rays. To deal with that, transform the range: let $ g(z)\equal{}\sqrt{f(z)\plus{}5)}$ (the usual principal square root, putting the branch cut on the negative real axis). The range of $ g$ is contained in the right half-plane. Then $ h(z)\equal{}\frac{g(z)\minus{}1}{g(z)\plus{}1}$ is an entire function with range contained in the unit disk.


geometry
3D geometry
MATHCOUNTS
logarithms
ratio
geometric series

[quote="t0rajir0u"]Find $x$ such that

$3^{3x}+3^{2x+1}+3^{x+1}+1 = 1000$[/quote]

[hide="answer"]
$3^{3x-1}+3^{2x}+3^{x}= 333$

3x-1, 2x, and x obviously have to be less than 6. Testing x = 0 gives the wrong answer, testing x = 1 gives the wrong answer, but testing $\boxed{x=2}$ doesn't.
[/hide]

[hide]This is a cubic expansion which can be factored as $(3^{x}+1)^{3}=1000$. For all real solutions, we can cube root both sides and subtract one. By writing 9 as $3^{2}$, we notice that $x=2$.[/hide]

Wow I'm stupid <.< .

it's a mathcounts solution, stop talking

[quote="t0rajir0u"][quote="Treething"][hide="answer"]
$3^{3x-1}+3^{2x}+3^{x}= 333$

3x-1, 2x, and x obviously have to be less than 6. Testing x = 0 gives the wrong answer, testing x = 1 gives the wrong answer, but testing $\boxed{x=2}$ doesn't.
[/hide][/quote]

This solution works for this problem because the problem writer intentionally made the answer neat.  Of course, if I had rewritten it as

$3^{3x}+3^{2x+1}+3^{x+1}+1 = 2006$

You would have been out of luck.[/quote]

This can be written as $(3^{x}+1)^{3}=2006$

So $3^{x}+1=\sqrt[3]{2006}$

$3^{x}=\sqrt[3]{2006}-1$

$x=\log_{3}{(\sqrt[3]{2006}-1)}$

[quote="Treething"]it's a mathcounts solution[/quote]

The High School Basics forum should be the beginning of your growth out of MathCounts problem-solving.

[b]Another possible solution[/b]

The beginning of another elegant solution could be made by noticing that the proposed sum on the left is the sum of a geometric series whose common ratio is $x$.

Even if you meant $3^{-x}$, you would be incorrect.

[hide="solution"]We subtract 1 and factor:

$3^{x+1}\left(3^{2x-1}+3^{x}+1\right)=37*3^{3}$

By inspection, x=2.[/hide]


Euler
LaTeX
articles
number theory
prime numbers
geometric series

I was reading a book about math and came across this infinite product attributed to Euler:

product(1/(1-1/(p^2)))=(pi^2)/6

(where p stands for every prime number)

(ie, take the product of the reciprocals of one minus the reciprocal of the square of every prime number and you get pi squared divided by 6)

The implications of this formula (to me) seem pretty astonishing at first glance; it seems to be making a clear connection between discrete mathematics (prime numbers) and continuous mathematics (irrational numbers). However, I have been unable to find this formula on the internet.

I was interested in how Euler went about proving  a statement like this when we know so little about the distribution of prime numbers (the book I was reading said virtually nothing about the formula, other than that Euler "miracled it up" and that "we still hardly understand all it implies"). Can anyone here shed some insight into this problem? Is it significant that the product is equal to the sum of the reciprocals of all the perfect squares?

Learning to use [[Latex]] (see that link or http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690 ) would make your posts far more readable.

Actually, the connection between this product and the sum of the reciprocals of the squares is close to trivial:
\[ \prod_{i \geq 1} \frac {1}{1 \minus{} 1/p_i^2} \equal{} \prod_{i \geq 1} \sum_{k \geq 0} p_i^{ \minus{} 2k}
\]
is just the expansion of the inner term as a geometric series; expanding the resulting product gives exactly $ \sum_{n \geq 1} n^{ \minus{} 2}$ (just think about prime factorizations).

I suggest reading [url=http://en.wikipedia.org/wiki/Basel_problem]Wikipedia's article[/url].


algebra
polynomial
LaTeX
college contests

Algebraic problems this year are easy and not interesting (by me). For getting 1 rst prize, students have to make perfect score. I post one of problem:

1) It there exists polynomial P(x), for which P(x) >= P"(x) and P'(x) > P"(x) for all x from R?
2) Show that if P(x) >= P'(x) for all x from R then P(x) >=0 for all x from R.

Namdung

Hi Mr Nam Dung!
If I'm not wrong the answer of 1) is :There no exists P(x)
I don't know LaTeX,so I don't write the full solution for 1) and 2).Can you help me?Thx!

[quote="lexuanson"]
I don't know LaTeX.Can you help me?Thx![/quote]

On the top screen there is a circle with $\sum,\pi$ Latex


geometry
circumcircle
symmetry
perpendicular bisector

(a) Prove that the set of all points that are equidistant from two distinct points $ A,B$ in the plane is a line.
(b) Prove that the line is perpendicular to $ \overline{AB}$ and bisects it.

(c) Prove that the set of all points that are equidistant from three distinct points $ A,B,C$ in space is a line.
(d) Prove that the line is perpendicular to face $ ABC$ and intersects it at the circumcenter of $ \triangle ABC$.

[hide="(a) and (b)"]Suppose $ M$ is the midpoint of $ AB$. Consider two arbitrary points $ C$ and $ D$ in the set. We know that $ AC\equal{}BC$ and $ AD\equal{}BD$, so $ \triangle ABC$ and $ \triangle ABD$ are isosceles. They also share $ AB$, so $ CADB$ is a kite. It's a well-known property of kites that the diagonals intersect at right angles and that one diagonal (the one that isn't the axis of symmetry, not sure if this diagonal has a name), which in this case is $ AB$, is bisected by the other. This implies that the line $ CD$ is a perpendicular bisector of $ AB$, and therefore passes through $ M$. So $ C,M,D$ are collinear for any $ C$ and $ D$, forming a line.[/hide]

(c) and (d) should follow by something similar, I think.


calculus
integration
number theory
relatively prime
greatest common divisor

Hello,

Sorry this might be a silly question but I'm want to check this

Is it this true?

If $ (a,b)\equal{}x$ then $ (a^2,b^2)\equal{}x^2$

Thanks

[hide="Solution"]Yes. $ \frac{a}{x}$ and $ \frac{b}{x}$ are relatively prime and integral, so their squares are also relatively prime, so $ x^2$ is the greatest common divisor of $ a^2$ and $ b^2$.[/hide]

Thanks a lot but what do you mean by "integral"

Excuse my bad knowledge

Thanks a lot


2 days until October SAT's!

For us seniors it's one of the last chances we get to prove ourselves to the colleges that we can read, write, and do a little math. The rest of ya'll have a bit more time to get your scores settled out, but it's always nice to get it over with on the first try.

So what are your goals for the SAT? Trying to get that 2400? Or struggling to get over 600 on reading/writing?

Personally, my goal is to get 700+ on reading/writing, making my overall above 1500 and 2200 (I took it last year and got 690 reading, 670 writing, 790 math giving me a 1480/2150.. just shy of what I'd like although I am fairly content with the scores).

Well, just want to give a good luck shout out to all of you supporting our monopolistic collegeboard friends and find out who's gonna spend a fun 5 hours on Saturday filling in bubbles!

Trying to get an 800 on the SAT II Chemistry :!:.

[quote="aznphatso"]Taking Sat 2 Physics.
Just trying to get a third 800 to show --- also, University of California EECS departments like the SAT 2 Physics.[/quote]

Gosh... that was an ugly test. I got like four questions on bubbles... which I knew nothing about.

Taking them. Aiming for 2400. First time I'm taking the new one.

I took the SAT Reasoning test last year, and got a good score (2330).  Now, on to the subject tests...
Math IIC (easy)
Chemistry (half the material wasn't covered in my year-long chem class)
French (I'm a bit worried about this one, too, as I'm a little out of practice.)

Taking the SAT. I meant to but didn't get time to prepare. Oh, well.

Instead of the real one, i took a practice one at kaplan's prep class yesterday, hehe.

[quote="WindSlicer"]So... Results everyone?

Me: I completed all of my goals!!

Math: 800
Reading: 740
Writing: 700

Got the 1500+, 700+ on all, and 2200+!

Soo Happy! Yay![/quote]

Are you saying that the results are back?

A score of 2300 on the new SAT I is at the 99+ percentile: very good. I'm glad to say that one of the first guys on my math team, now a senior, just scored the same on the October SAT I.

[url=http://www.collegeboard.com/prod_downloads/highered/ra/sat/SATPercentileRanksCompositeCR_M_W.pdf#search=%22SAT%20Percentiles%20Total%20Group%20site%3Acollegeboard.com%20filetype%3Apdf%22]SAT Percentile Ranks for Males, Females, and Total Group[/url]

[quote="1234567890"]Nice job, paladin and Windslicer!
I think I'm done with standardized testing.  :lol:

800 Math
760 Writing
740 Critical Reading
2300 Composite[/quote]

Great job everyone, great job.

[quote="neelnal"]800 math
800 critical reading

and um... dissapointment

680 writing. apparently, i got a 7/12 on the essay but only one multiple choice wrong.

2280 total.

going to take it again in june to get perfect hopefully.[/quote]

great job. You'll definetely get it next year.


如果學生考試不及格，實在是不能夠怪他們啊！
因為一年只有365天。原因如下：

1。禮拜日
一年裡面有52天的禮拜日，扣除那些休息的日數，
平均起來，一年只剩下313天。

2。暑假
在一年內，大約有50天的天氣非常熱，而導致無法唸書，
因此，一年平均起來，只剩下263天。

3。每天8小時的睡眠時間
佔了一年的30天，平均起來，一年只剩下141天。

4 。每天1小時的遊戲時間
佔了一年的15天，平均起來，一年只剩下126天。

5。每天2小時的吃飯及吃零食的時間
佔了一年的30天，平均起來，一年只剩下96天。

6。每天花了1小時的時間溝通
佔了一年的15天，平均起來，一年只剩下81天。

7。考試及測驗至少佔了一年的35天，
所以平均起來，一年只剩下46天。

8。另外，包括了其他節日的假日，
就佔了一年的40天，平均起來一年只剩下6天。

9。預估一年生病3天，平均起來，一年只剩下3天。

10。去看電影或是有一些活動，佔了一年的2天，
平均起來，一年只剩下1天。

11。而這1天，就是我們自己的生日................

試問，這麼完全沒有時間可以唸書 ,
學生在考試又怎麼能夠及格呢？
所以啊，我們身為學生，考試不及格是有原因的... ( 同學們真辛苦 )

同学，那可是重复计算的啊……

[quote="dingdongdog"]同学，那可是重复计算的啊……[/quote]
Yes, i know.

[quote="ifai"]7。考試及測驗至少佔了一年的35天，
所以平均起來，一年只剩下46天。

[/quote]
这个太夸张了...

[quote="ifai"]如果學生考試不及格，實在是不能夠怪他們啊！
因為一年只有365天。原因如下：

1。禮拜日
一年裡面有52天的禮拜日，扣除那些休息的日數，
平均起來，一年只剩下313天。
2...

这是很好的借口呀！

用这样的理由，会被老师砸死的....  :ninja:

[quote="ifai"]
1。禮拜日
一年裡面有52天的禮拜日，扣除那些休息的日數，
平均起來，一年只剩下313天。

8。另外，包括了其他節日的假日，
就佔了一年的40天，平均起來一年只剩下6天。 [/quote]
在這些日子都可溫書呢!

我可不願考試不及格。看來我得不睡覺，好騰出一個月

[quote="modeler"]我可不願考試不及格。看來我得不睡覺，好騰出一個月[/quote]
相信談讀書沒有其他地方的人比澳門學生更輕鬆--以我這條大懶蟲為例, 即使不怎麼溫習也不會不及格(當然, 要取得優異的成績還得努力, 但就是勤力一點就可以了). 當我考大學時, 依然渾渾噩噩, 但最後我所報考的幾所大學(澳門的澳大, 理工; 大陸的南開和中山; 台灣的清華)都錄取了. 不幸的是, 我現在要休學...... :(

x,y,z>=0 satisfying x+y+z=1,
find the maxinum of $\sum{sqr{frac{1-x}{1+x}}}$

You mean $\sum{\sqrt{\frac{1-x}{1+x}}}$?
It is nothing to do with 考試不及格的原因  :D

这也太消极了， :lol:
人生还是挺美好的。。。。

for someppl, sleep averages about 12 hrs, so that's leaves only about 182.5 days.


A boy has forgotten the second and fourth digits of his four-digit locker combination. What is the largest number of combinations he must try to open the locker?

There are $ 10$ possible digits for both the second and fourth digit, making a total of $ 10\times10\equal{}\fbox{100}$ maximum combinations to try before succeeding.


Like $\angle OBC = 200^{\circ}\Rightarrow$ arc $BC = 200^{\circ}$ or something like that...

I've looked all over the internet, to no avail  :ninja:

Ah, but did you search this forum because it has been answered a number of times?
To save you looking, try \stackrel{\frown}{BC} as in $\stackrel{\frown}{BC}=200^{\circ}$
A possible alternative that some people use is ${\widehat{BC}}=200^{\circ}$

hmm... thanks.

They still don't look very good, though...

I agree with you there. I have just discovered a package called [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=arcs]arcs[/url] which MiKTeX will install for you and allows you to use \overarc{BC}. It does give a nicely rounded arc which you may prefer. It looks like the picture below:

I have MikTex and TexnicCenter, and got the package, and it still doesn't work... any suggestions?

Install? I extracted it...

And supposedly MikTex already includes it as well, but it says

"undefined control sequence"

"undefined control sequence" means the package still has to be installed.

The easiest way to install any package is to use the MikTex Package Manager, which should be on your Start menu somewhere. When it gives the list of packages, right click on arcs and choose install. That's it.

To do it yourself you need to:
1. extract the files and compile them as the Readme says
2. then you put the resulting arcs.sty file in the right place eg C:\texmf\tex\latex\arcs
3. then you must tell MiKTeX where it is, by running MiKTeX Options, File name database, then Refresh Now

It's a lot easier to let MiKTeX do it all for you as in the first paragraph and only takes a few minutes :)


is this correct?    :(

[hide]minimum: $ \frac {13 \minus{} \sqrt {133}}{2}$
maximum: $ \frac {13 \plus{} \sqrt {133}}{2}$[/hide]

[quote="enndb0x"]Let $ a,b,c$ be three positive real number such that :
\[ \left(a \plus{} b \plus{} c\right)\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \right) \equal{} 16\]
Find the minimum and maximum value of :
\[ P \equal{} \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\]
[/quote]
if we can prove that
 Let $ a,b,c$ be three positive real number.k is constant positive such that :
\[ \left(a \plus{} b \plus{} c\right)\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \right) \equal{} k\]
Find the minimum and maximum value of :
\[
P = \frac {a}{b} + \frac {b}{c} + \frac {c}{a}

[quote="Pain rinnegan"][quote="Obel1x"]$ \left(a + b + c\right)\left(\frac {1}{a} + \frac {1}{b} + \frac {1}{c}\right) = 16$

${{ (\frac {a}{b} + \frac {b}{c} + \frac {c}{a}) + (\frac {a}{c} + \frac {b}a} + \frac {c}{b}) = 13 \implies \frac {a}{c} + \frac {b}a} + \frac {c}{b} = 13 - (\frac {a}{b} + \frac {b}{c} + \frac {c}{a}) \ge 3$

$ \implies \frac {a}{b} + \frac {b}{c} + \frac {c}{a} \leq 10$ ?[/quote]

Your solution is obviously wrong .[/quote]

why :(

[quote="Obel1x"]
why :([/quote]

When you need to find a maximum (minimum) be sure to check the equality cases each time ;)

Let $ x\equal{}\frac{a}{b}$, $ y,z$.

Then we are looking at $ x\plus{}y\plus{}z$ under $ x\plus{}y\plus{}z\plus{}xy\plus{}yz\plus{}zx\plus{}3\equal{}16$ and $ xyz\equal{}1$ so I think we can proceed by using a cubic ...?


geometry
circumcircle
trigonometry
geometry proposed

let ABC be a scalene triangle. Denote by P,Q the midpoints of the altitudes BB' and CC' respectively. Prove that MN bisects PQ.(M is the midpoint of BC and N the circumcenter of triangle MB'C').

Let $ X$ be the intersection of $ MP$ and $ AB$. Since $ MP\parallel AC$, $ X$ is the midpoint of $ AB$. It is well known that $ C',X,B',M$ lie on the nine-point circle. Note that $ BM$ is a median of $ \triangle BB'C$, so $ B'M\equal{}BM\equal{}MC$. Thus $ \angle BB'M\equal{}\angle B'BM\equal{}\pi/2\minus{}C$. Similarly, $ \angle XB'B\equal{}\pi/2\minus{}A$. Therefore $ \angle XB'M\equal{}\pi\minus{}A\minus{}C\equal{}B$. Hence $ \angle XNM\equal{}2B$ and $ \angle PMN\equal{}\pi/2\minus{}B$. Therefore \[ [PNM]\equal{}\frac12PM\cdot MN\cdot\sin\angle PMN\equal{}\frac12\cdot\frac12 B'C\cdot r\cdot\cos B\equal{}\frac14\cdot BC\cos C\cdot r\cdot\cos B\equal{}\frac14\cdot ar\cos B\cos C\] Similarly we get that $ [QNM]\equal{}\frac14\cdot ar\cos B\cos C$, thus $ [QNM]\equal{}[PNM]$ and the conclusion follows.

Applying Menelaus's theorem we have $ LP\equal{}LQ$ iff $ \frac{NP}{NB'}\equal{}\frac{MQ}{MB'}\equal{}\frac{BC'}{BA}\equal{}cos B$
But $ \frac{NP}{NB'}\equal{}\frac{h_b\minus{}R}{R}\equal{}\frac{OB'}{OA}\equal{}cos B$
$ \Rightarrow$ QED

[quote="livetolove212"]Applying Menelaus's theorem we have $ LP \equal{} LQ$ iff $ \frac {NP}{NB'} \equal{} \frac {MQ}{MB'} \equal{} \frac {BC'}{BA} \equal{} cos B$
But $ \frac {NP}{NB'} \equal{} \frac {h_b \minus{} R}{R} \equal{} \frac {OB'}{OA} \equal{} cos B$
$ \Rightarrow$ QED[/quote]Why is $ \frac{BC'}{BA}\equal{} \cos B$? Did you assume that $ BA\equal{}BC$? Take note that scalene $ \ne$ isosceles.

Sorry I don't know the meaning of scalene  :blush:
This is my proof:
Let $ MN\cap B'C' \equal{} \{J\}$ then J is the midpoint of B'C' (Because $ MB' \equal{} MC'$)
$ JP// \equal{} QM$ so JM passes through the midpoint of PQ.
Is it too easy??   :D


See the image.  We begin with a cube of side length 1, and then repeat 3 times a process which I will describe as "cutting out the front corners of a solid."  Given that the cut will divide the edges it is cutting in half, if we repeat this process n times, what will the volume of the solid left over be?  What will its surface area be?  What will the sum of the lengths of its edges be?  What will the limit of these three values as [tex]\displaystyle{n\rightarrow\infty}[/tex]?  What if the cut only takes out one third of the edge?

A little clarification, just in case: by "cutting the edges in half," I wish to indicate that in the second picture from the left, exactly one eigth of the cube would have been removed.  For the latter question, I wish to change it so that exactly 1/27th is removed.

Thanks for help, rcv!

Many would be quick to assume that as n approaches infinity, the corner would flatten into a triangle. This isn't the case. Instead it will look like the sierpinski triangle with points popping out in the center and every other triangle.

The volume: The volume loses 1/8 the first time, then 3/64, then 9/512. It forms a geometric sum (which will be subtracted from the initial volume, 1. The ratio is 3/8, so g1/(1-r) 1/8//5/8=1/5. So as n-> :inf: , there will be 4/5 of the cube's volume left.

The surface area: It will stay the same throughout, because each time you're basically pushing in a corner. I can see a little flash animation for it. Look at the colored pictures, and imagine all the blue areas being pushed up, the red and white areas being pushed left in right. It all fits back together into a perfect cube.

The edges: I don't know, i've tried a buncha times. I'll leave this for someone else.

Oh yeah, and how did you make the pic?

And anyone here in 3d engineering? I would love to see a pic of a physical cube thats been cut out 100 times and see what it looks like.

Now its 1/3 instead of 1/2? Interesting.

Volume: Like last time, you have 1/27 of the cube removed the first time. Next, you have 3 cubes each 1/729 of the cube (!!!) or 1/243. so the sequence is slightly easier to figure: 1/3^3, 1/3^5, 1^3^7, etc. Once again, geometric sum is g1/(1-r) ... 1/27/(1-1/9) = 1/27//8/9 = 1*9/27*8 = 1*1/3*8 = 1/24. So using the 1/3 method, there will be 23/24 left of the volume.

Surface area: is still the same. Read last post.

Edge length: (This is easier to figure out, 1/3) Each time you cut out a cube, you have basically 3 sides bent into another shape, and created 3 segments which add up to one line. Next one, each segment created is 1/9 a length. You have 9 of these, so it adds one more. The sequence is divergent because you add a constant, so there is no limit. But there is a formula: With n cuts of the cube, S.L. = 12+n.

[quote="Hexaditidom"]Oh yeah, and how did you make the pic?[/quote]

Geometer's Sketchpad, copied into Paint and saved as a .gif.

Here's a 7th iterate that I made with mathematica:

BTW, this is a fractal, right? If so, what is its dimention? How do we calculate fractal dimension?

Well, it's definitely a fractal, since it is just like Serpinski's triangle.  I have no idea what fractal dimension is, however.

There are actually several different notions of "fractal" dimension. The most common notion is Hausdorff Dimension. The basic idea of Hausdorff Dimension is that just like a line has no area in 2 dimensions, and a plane has no volume in 3 dimensions - it's the infimum of all the "dimensions" in which the factal has no "volume" (actually, Hausdorff measure).

It's a little bit difficult to define formally, but that's the basic idea.


Find integers $ m,\ n$ such that $ m\leq \log_3 1000 < m \plus{} 1$ and $ n\leq \log_2 \frac {1}{200} < n \plus{} 1$.


calculus
integration
search
symmetry
real analysis
real analysis unsolved

compute it for any power of the tangent, the result is the same and it's $ \frac\pi4.$

you can search around the forum and you'll find it.

Making a search with keywords 'putnam, symmetry, integral' gives us

http://www.mathlinks.ro/viewtopic.php?t=103824
http://www.mathlinks.ro/viewtopic.php?t=78817

Note the interesting pattern in topic naming :D


isnt this just cauchy's in engel form?

edit:  yea...just another name for it apparently.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15125[/url]

I believe there is another proof of it by a simple substitution into cauchy...I may be wrong though.

You can also view the lemma as a special case of Holder's inequality, with exponents $\frac{1}{2}$ and $-1$.  (Because one of the exponents is negative, the sign of Holder's inequality reverses.)

n=2 is true by the trivial inequality

then use induction (using the two variable case as the induction step)

Yes I know, but there is a way of showing that this is equivalent to Cauchy (there is a substitution to go back and forth)- I think.

Multiply both sides through by the denominator, and I think you can straightforward Cauchy-Schwartz from there.


$ A(x_0,y_0)$ is a fixed point on the plane. $ BC$ is a line and $ A$ lies on it. $ BC$ meets $ Ox$ at $ C$, meets $ Oy$ at $ B$

Then find the minimum of $ |OB|\plus{}|OC|\plus{}|BC|$ and prove it

Let $ \overline{OC}=p, \overline{OB}=q$, the equation of $ BC$ is $ l: \frac{x}{p}+\frac{y}{q}=1$.

$ (x_0,\ y_0)\in{l}\Longleftrightarrow \frac{x_0}{p}+\frac{y_0}{q}=1$.

$ %Error. "thereorefore" is a bad command.
L=p+q+\sqrt{p^2+q^2}\geq p+q+\frac{p+q}{\sqrt{2}}=\frac{2+\sqrt{2}}{2}(p+q)$.

Now $ (p+q)\left(\frac{x_0}{p}+\frac{y_0}{q}\right)\geq (\sqrt{x_0}+\sqrt{y_0})^2$, yielding
$ p+q\geq (\sqrt{x_0}+\sqrt{y_0})^2$. $ \therefore L_{min}=\frac{2+\sqrt{2}}{2}(\sqrt{x_0}+\sqrt{y_0})^2$.

[quote="kunny"]Let $ \overline{OC} = p, \overline{OB} = q$, the equation of $ BC$ is $ l: \frac {x}{p} + \frac {y}{q} = 1$.

$ (x_0,\ y_0)\in{l}\Longleftrightarrow \frac {x_0}{p} + \frac {y_0}{q} = 1$.

$ %Error. "thereorefore" is a bad command.
L = p + q + \sqrt {p^2 + q^2}\geq p + q + \frac {p + q}{\sqrt {2}} = \frac {2 + \sqrt {2}}{2}(p + q)$.

Now $ (p + q)\left(\frac {x_0}{p} + \frac {y_0}{q}\right)\geq (\sqrt {x_0} + \sqrt {y_0})^2$, yielding
$ p + q\geq (\sqrt {x_0} + \sqrt {y_0})^2$. $ \therefore L_{min} = \frac {2 + \sqrt {2}}{2}(\sqrt {x_0} + \sqrt {y_0})^2$.[/quote]

I'm sorry that your answer is wrong.

In the first inequality, the $ ''=''$ holds iff $ p=q$, but in the second one which using $ Cauchy-Schwarz$ Inequality, the $ ''=''$ holds iff $ \frac{x_0}{p^2}=\frac{y_0}{q^2}$


geometry
3D geometry
AMC
AIME
algebra
polynomial

[b]Find the smallest positive integer whose cube ends in 888.[/b]

[hide="!?!?!?"]I caved.

for(w,0,999)
$ 1000$fPart$ ((10^{ \minus{} 1}n)^3)$$ \rightarrow$N
if $ N \equal{} 888$:Disp $ N$
End:End[/hide]
Hint please?

EDIT: That really shows why calculators aren't allowed on AIME.  :maybe:

Start from the end: which digit should the number end with? Then, once you figure that out, try to find the previous digit, and so on.

This is the wiki solution for[url=http://www.artofproblemsolving.com/Wiki/index.php/1988_AIME_Problems/Problem_9]1988 AIME #9[/url] (wow ... I just randomly guessed that the aime would be 1988, and it turned out to be there).

As fedja said, establish the units digit of the number, than the tens, and so forth; don't be afraid to use a bit of casework.

[quote="t0rajir0u"]See http://www.artofproblemsolving.com/Forum/viewtopic.php?p=457062#457062 and http://www.artofproblemsolving.com/Forum/viewtopic.php?t=75142 .

A very general approach:  solve $ \bmod 8, 125$ using [url=http://planetmath.org/encyclopedia/HenselsLemma.html]Hensel's lemma[/url] and combine results with the Chinese Remainder Theorem.[/quote]
So we can use Hensel's Lemma to solve nth degree equations in moduli?
Can you explain how this works?

Hensel's Lemma gives us a way to solve a polynomial $ f(x) \equiv 0 \bmod p^{k\plus{}1}$ given a solution $ \bmod p^k$.  Without $ p$-adic notation, if there exists $ a$ such that $ f(a) \equiv 0 \bmod p^k$ and $ f'(a) \not \equiv 0 \bmod p$, then there exists $ a'$ such that $ f(a') \equiv 0 \bmod p^{k\plus{}1}$.

The elementary version of this is solving for the last digit, then the second-to-last digit, and so forth.  I mention Hensel's Lemma only as a generalization.

[quote="t0rajir0u"]Hensel's Lemma gives us a way to solve a polynomial $ f(x) \equiv 0 \bmod p^{k \plus{} 1}$ given a solution $ \bmod p^k$.  Without $ p$-adic notation, if there exists $ a$ such that $ f(a) \equiv 0 \bmod p^k$ and $ f'(a) \not \equiv 0 \bmod p$, then there exists $ a'$ such that $ f(a') \equiv 0 \bmod p^{k \plus{} 1}$.

The elementary version of this is solving for the last digit, then the second-to-last digit, and so forth.  I mention Hensel's Lemma only as a generalization.[/quote]
How do we use the knowledge of the existence of a solution to find it though (or does hensel's lemma state this)?

DO some proof mining: by showing the lemma, you already construct the solution explicitely.


For all integers n /= 0, let the "superpower" of n be the number of times $ n $ is raised to the power of itself. For example:

$ 1^{super6} $ = ((((1^1)^1)^1)^1)^1 = 1

$ 2^{super4} $ = ((2^2)^2)^2) = 256

$ 3^{super5} $ = (((3^3)^3)^3)^3) = 58149737003040059690390169

(Just as $ x^1 $ will equal $ x $ , so will $ x^{super1} $)

Now for all positive integers n /= 0, let the "superfactorial" of n be the successive powers of $ n, n-1, n-2, ... 3, 2, 1 $ as shown:

$ 4super! $ = ((4^3)^2)^1 = 4096
$ 5super! $ = (((5^4)^3)^2)^1 = 59604644775390625

1. Calculate $ \frac{6super!}{3super! * 4super!} $

2. Find a way to express $ 100super! $ without using the superfactorial sign. (This should make #1 easier)

3. Find the last 3 digits of $ {-6}^{super4}. $

4. Find the first 3 digits of $ {3^{super4}} - {4^{super3}} $.

Latex is awesome by the way  :lol:

I immediately saw a few problems with my problems and was just on my way to change them... I didn't think someone would reply in the next 5 seconds  :blush:

Here are the new problems. Enjoy!

1. Express $ \frac{7^{super7}}{7^7} $ as a power of 7.

2. Express $ 529super! $ without the superfactorial.

3. Simplify $ \frac{24super!}{{12super!}^2} $

4. Find the last 3 digits of $ (-6)^{super5} $.

5. Find the last 3 digits of $ 3^{super8} - 8^{super3} $

A $\LaTeX$ advice for everyOne here :P use \cdot instead of * like $2\cdot 3$ and not $2 * 3$ ;)

You can use [url=http://mathworld.wolfram.com/ArrowNotation.html]arrow notation[/url] instead of 'superpowers'.

Corrtect me if im wrong, but i think the super factorial means

$\displaystyle n^{{n-1}^{{n-2}^{n-3}.....}}$

As opposed what everyone else is writing, same for the super powers...


If $ a$, $ b$, and $ c$ are positive integers, the radicals $ \sqrt{a\plus{}\frac{b}{c}}$ and $ a\sqrt{\frac{b}{c}}$ are equal when and only when:

$ \textbf{(A)}\ a\equal{}b\equal{}c\equal{}1 \qquad
\textbf{(B)}\ a\equal{}b\text{ and }c\equal{}a\equal{}1 \qquad
\textbf{(C)}\ c\equal{}\frac{b(a^2\minus{}1)}{2} \\
\textbf{(D)}\ a\equal{}b \text{ and }c\text{ is any value} \qquad
\textbf{(E)}\ a\equal{}b \text{ and }c\equal{}a\minus{}1$

hello, squaring both sides of the equation $ \sqrt{a\plus{}\frac{b}{a}}\equal{}a\sqrt{\frac{b}{a}}$ we get
$ a\plus{}\frac{b}{c}\equal{}a^2\frac{b}{c}$ multiplying by $ c$ we have $ ac\equal{}a^2b\minus{}b \Leftrightarrow c\equal{}\frac{b(a^2\minus{}1)}{a}$.
Sonnhard.

You all made typo mistakes.

[quote="AIME15"]If $ a$, $ b$, and $ c$ are positive integers, the radicals $ \sqrt {a \plus{} \frac {b}{c}}$ and $ a\sqrt {\frac {b}{c}}$ are equal when and only when:

$ \textbf{(A)}\ a \equal{} b \equal{} c \equal{} 1 \qquad \textbf{(B)}\ a \equal{} b\text{ and }c \equal{} a \equal{} 1 \qquad \textbf{(C)}\ c \equal{} \frac {b(a^2 \minus{} 1)}{2} \\
\textbf{(D)}\ a \equal{} b \text{ and }c\text{ is any value} \qquad \textbf{(E)}\ a \equal{} b \text{ and }c \equal{} a \minus{} 1$[/quote]

Answer $ \bf (C)$ should be $ c\equal{}\frac{b(a^2\minus{}1)}{a}$.

[quote="Dr Sonnhard Graubner"]hello, squaring both sides of the equation $ \sqrt {a \plus{} \frac {b}{a}} \equal{} a\sqrt {\frac {b}{a}}$ we get
$ a \plus{} \frac {b}{c} \equal{} a^2\frac {b}{c}$ multiplying by $ c$ we have $ ac \equal{} a^2b \minus{} b \Leftrightarrow c \equal{} \frac {b(a^2 \minus{} 1)}{a}$.
Sonnhard.[/quote]

Sonnhard's first equation should be $ \sqrt {a \plus{} \frac {b}{c}} \equal{} a\sqrt {\frac {b}{c}}$.

[quote="enndb0x"]There's none of the alternatives , but I think AIME15 made writing mistake at $ \boxed E$[/quote]

And yours should be pointing out $ \boxed C$ because there is nothing wrong with choice $ \bf (E)$.


geometry
parallelogram
perimeter
geometry unsolved

Suppose that parallelogram $ ABCD$ has $ \angle A \equal{} \angle C \equal{} 30^\circ$, $ \angle B \equal{} \angle D \equal{} 150^\circ$, and the shorter diagonal $ BD$ has length 2. If the height of the parallelogram is $ a$, find the perimeter of $ ABCD$ in terms of $ a$.

[quote="hlee.mathcountscoach"]Suppose that parallelogram $ ABCD$ has $ \angle A = \angle C = 30^\circ$, $ \angle B = \angle D = 150^\circ$, and the shorter diagonal $ BD$ has length 2. If the height of the parallelogram is $ a$, find the perimeter of $ ABCD$ in terms of $ a$.[/quote]

We denote $ AD = \lambda$ and $ AB = l$.Let $ H$ be the orthogonal projection of $ A$ at $ DC$ it follows that $ \sin(30^{\circ}) = \frac {a}{\lambda} \Rightarrow \lambda = 2a$ in $ \Delta ABD$ we have $ \cos(30^{\circ}) = \frac {l^2 + \lambda^2 - 4}{2.l.\lambda} \Rightarrow 4a^2 + l^2 - 2\sqrt {3}al = 4 \rightarrow l = \sqrt {8a^2 + 4} + 2\sqrt {3}.a$ and hence we deduce that $ p = 2(\lambda + l) = 2(2a + \sqrt {8a^2 + 4} + 2\sqrt {3}.a)$


In the following, the abbreviation $g \cap h$ will mean the point of intersection of two lines $g$ and $h$.

Let $ABCDE$ be a convex pentagon. Let $A^{\prime}=BD\cap CE$, $B^{\prime}=CE\cap DA$, $C^{\prime}=DA\cap EB$, $D^{\prime}=EB\cap AC$ and $E^{\prime}=AC\cap BD$. Furthermore, let $A^{\prime\prime}=AA^{\prime}\cap EB$, $B^{\prime\prime}=BB^{\prime}\cap AC$, $C^{\prime\prime}=CC^{\prime}\cap BD$, $D^{\prime\prime}=DD^{\prime}\cap CE$ and $E^{\prime\prime}=EE^{\prime}\cap DA$.

Prove that:
\[ \frac{EA^{\prime\prime}}{A^{\prime\prime}B}\cdot\frac{AB^{\prime\prime}}{B^{\prime\prime}C}\cdot\frac{BC^{\prime\prime}}{C^{\prime\prime}D}\cdot\frac{CD^{\prime\prime}}{D^{\prime\prime}E}\cdot\frac{DE^{\prime\prime}}{E^{\prime\prime}A}=1.  \]

  Darij

my solution is very ugly.
I used more than eight times the menelau's theorem  :(

anyone have one simple solution?

Take a look in there for additional identities.

  http://mathworld.wolfram.com/HoehnsTheorem.html

  Sorry Darij,  but they beat you to it.

  Was this problem the thing that inspired you for a construction problem you posted

  some time ago?

  Thank you.

    M.T.

[quote="armpist"]Take a look in there for additional identities.

  http://mathworld.wolfram.com/HoehnsTheorem.html[/quote]

Thanks for the link! In fact, these identities are what inspired me to create this problem, I just forgot their name (Hoehn's identities). As for the construction problem you are talking about, in case you meant http://www.mathlinks.ro/Forum/viewtopic.php?t=54608 : no, it was not inspired by any pentagram identities, and I actually don't see how to apply them in the problem. (It was inspired by the problem posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=38828 , but, unlike the latter one, [i]is[/i] solvable).

As for the solution, yes, the problem is really tricky if one tries to solve it by Menelaos/Ceva only (such a solution is possible, but if I would have written it down, it would probably take 4-5 pages).

Here is my "official" solution of the problem:

After the sine law in triangle $EA^{\prime \prime }A^{\prime }$, we have

$\frac{EA^{\prime \prime }}{A^{\prime }A^{\prime \prime }}=\frac{\sin\measuredangle EA^{\prime }A^{\prime \prime }}{\sin \measuredangle A^{\prime}EA^{\prime \prime }}=\frac{\sin \measuredangle EA^{\prime }A}{\sin\measuredangle BEC}$.

After the sine law in triangle $AA^{\prime }E$, we have

$\frac{\sin \measuredangle EA^{\prime }A}{\sin \measuredangle AEA^{\prime }}=\frac{EA}{AA^{\prime }}$, thus $\sin \measuredangle EA^{\prime }A=\sin \measuredangle AEA^{\prime }\cdot \frac{EA}{AA^{\prime }}=\sin \measuredangle AEC\cdot \frac{EA}{AA^{\prime }}$.

Hence,

$\frac{EA^{\prime \prime }}{A^{\prime }A^{\prime \prime }}=\frac{\sin\measuredangle EA^{\prime }A}{\sin \measuredangle BEC}=\frac{\sin\measuredangle AEC\cdot \frac{EA}{AA^{\prime }}}{\sin \measuredangle BEC}=\frac{EA\cdot \sin \measuredangle AEC}{AA^{\prime }\cdot \sin \measuredangle BEC}$.

Similarly,

$\frac{BA^{\prime \prime }}{A^{\prime }A^{\prime \prime }}=\frac{AB\cdot \sin\measuredangle DBA}{AA^{\prime }\cdot \sin \measuredangle DBE}$.

Thus,

$\frac{EA^{\prime \prime }}{BA^{\prime \prime }} =\frac{EA^{\prime \prime }}{A^{\prime }A^{\prime \prime }}: \frac{BA^{\prime \prime }}{A^{\prime}A^{\prime \prime }}=\frac{EA\cdot \sin \measuredangle AEC}{AA^{\prime}\cdot \sin \measuredangle BEC}: \frac{AB\cdot \sin \measuredangle DBA}{AA^{\prime }\cdot \sin \measuredangle DBE}$
$=\frac{EA}{AB}\cdot \frac{\sin \measuredangle AEC}{\sin \measuredangle DBA}\cdot \frac{\sin \measuredangle DBE}{\sin \measuredangle BEC}$.

Analogously,

$\frac{AB^{\prime \prime }}{CB^{\prime \prime }} =\frac{AB}{BC}\cdot \frac{\sin \measuredangle BAD}{\sin \measuredangle ECB}\cdot \frac{\sin\measuredangle ECA}{\sin \measuredangle CAD}$;
$\frac{BC^{\prime \prime }}{DC^{\prime \prime }} =\frac{BC}{CD}\cdot \frac{\sin \measuredangle CBE}{\sin \measuredangle ADC}\cdot \frac{\sin\measuredangle ADB}{\sin \measuredangle DBE}$;
$\frac{CD^{\prime \prime }}{ED^{\prime \prime }} =\frac{CD}{DE}\cdot \frac{\sin \measuredangle DCA}{\sin \measuredangle BED}\cdot \frac{\sin\measuredangle BEC}{\sin \measuredangle ECA}$;
$\frac{DE^{\prime \prime }}{AE^{\prime \prime }} =\frac{DE}{EA}\cdot \frac{\sin \measuredangle EDB}{\sin \measuredangle CAE}\cdot \frac{\sin\measuredangle CAD}{\sin \measuredangle ADB}$.

Therefore,

$\frac{EA^{\prime \prime }}{BA^{\prime \prime }}\cdot \frac{AB^{\prime\prime }}{CB^{\prime \prime }}\cdot \frac{BC^{\prime \prime }}{DC^{\prime\prime }}\cdot \frac{CD^{\prime \prime }}{ED^{\prime \prime }}\cdot \frac{DE^{\prime \prime }}{AE^{\prime \prime }}$
$=\left( \frac{EA}{AB}\cdot \frac{\sin \measuredangle AEC}{\sin\measuredangle DBA}\cdot \frac{\sin \measuredangle DBE}{\sin \measuredangle BEC}\right) \cdot \left( \frac{AB}{BC}\cdot \frac{\sin \measuredangle BAD}{\sin \measuredangle ECB}\cdot \frac{\sin \measuredangle ECA}{\sin \measuredangle CAD}\right)$
$\cdot \left( \frac{BC}{CD}\cdot \frac{\sin \measuredangle CBE}{\sin \measuredangle ADC}\cdot \frac{\sin \measuredangle ADB}{\sin \measuredangle DBE}\right) \cdot \left( \frac{CD}{DE}\cdot \frac{\sin \measuredangle DCA}{\sin \measuredangle BED}\cdot \frac{\sin \measuredangle BEC}{\sin \measuredangle ECA}\right)$
$\cdot \left( \frac{DE}{EA}\cdot \frac{\sin \measuredangle EDB}{\sin \measuredangle CAE}\cdot \frac{\sin \measuredangle CAD}{\sin \measuredangle ADB}\right)$.

Most of the fractions in this product cancel each other; what remains is:

$\frac{EA^{\prime \prime }}{BA^{\prime \prime }}\cdot \frac{AB^{\prime\prime }}{CB^{\prime \prime }}\cdot \frac{BC^{\prime \prime }}{DC^{\prime\prime }}\cdot \frac{CD^{\prime \prime }}{ED^{\prime \prime }}\cdot \frac{DE^{\prime \prime }}{AE^{\prime \prime }}$
$=\frac{\sin \measuredangle AEC}{\sin \measuredangle DBA}\cdot \frac{\sin \measuredangle BAD}{\sin \measuredangle ECB}\cdot \frac{\sin \measuredangle CBE}{\sin \measuredangle ADC}\cdot \frac{\sin \measuredangle DCA}{\sin \measuredangle BED}\cdot \frac{\sin \measuredangle EDB}{\sin \measuredangle CAE}$
$=\frac{\sin \measuredangle AEC}{\sin \measuredangle CAE}\cdot \frac{\sin\measuredangle BAD}{\sin \measuredangle DBA}\cdot \frac{\sin \measuredangle CBE}{\sin \measuredangle ECB}\cdot \frac{\sin \measuredangle DCA}{\sin \measuredangle ADC}\cdot \frac{\sin \measuredangle EDB}{\sin \measuredangle BED}$.

Now, $\frac{\sin \measuredangle AEC}{\sin \measuredangle CAE}=\frac{AC}{CE}$ by the sine law in triangle $ACE$, and similarly $\frac{\sin \measuredangle BAD}{\sin \measuredangle DBA}=\frac{BD}{DA}$, $\frac{\sin \measuredangle CBE}{\sin \measuredangle ECB}=\frac{CE}{EB}$, $\frac{\sin \measuredangle DCA}{\sin \measuredangle ADC}=\frac{DA}{AC}$ and $\frac{\sin \measuredangle EDB}{\sin \measuredangle BED}=\frac{EB}{BD}$. Thus,

$\frac{EA^{\prime \prime }}{BA^{\prime \prime }}\cdot \frac{AB^{\prime \prime}}{CB^{\prime \prime }}\cdot \frac{BC^{\prime \prime }}{DC^{\prime \prime }}\cdot \frac{CD^{\prime \prime }}{ED^{\prime \prime }}\cdot \frac{DE^{\prime\prime }}{AE^{\prime \prime }}=\frac{AC}{CE}\cdot \frac{BD}{DA}\cdot \frac{CE}{EB}\cdot \frac{DA}{AC}\cdot \frac{EB}{BD}=1$,

and the problem is solved.

  Darij


That's Popa Calin.
[url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=92]pcalin[/url].

Sorry, but this is my inequality. It has been published in Gazeta Matematica three years ago.

Thanks!
Two years later, in 2004, I published in G.M. the following inequality

$5+\sqrt{2(x^2+y^2+z^2)(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2})-2} \geq (x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$.

Yes,Vacs.I have solved it with the same method :) .But,do you have the generalization?
:?

I found an alternative proof to the first problem: Let
$X=\sum_{cycl} a\sum\frac{1}{a}$
and
$Y=\sum_{cycl} a^2\sum\frac{1}{a^2}$
Then
$X = \sum a\sum\frac{1}{a} = \sqrt{ \left(\sum a^2+2\sum ab\right)\left(\sum\frac{1}{a^2}+2\sum\frac{1}{ab}\right) } \ge \sqrt{Y}+2\sqrt{X}$
Hence $(\sqrt{X}-1)^2\ge 1+\sqrt{Y}$, from which the result follows.

This is also Harazi's solution posted long time ago on this forum.


Just in case you're a nonbeliever. pssh.

http://www.sciencedaily.com/releases/2006/12/061224130514.htm

Sorry jo, but I am not an 'it's impossible' non-believer but an 'it's wrong' non-believer. :P

I never believed in Santa Clause.
I did believe in Sinterklaas.
"Santa Clause" is nothing more than a wrong pronunciation of the Dutch word "Sinterklaas". :)

[img]http://upload.wikimedia.org/wikipedia/en/thumb/d/d0/DutchSinterklaas2005.jpg/250px-DutchSinterklaas2005.jpg[/img]

A friend of mine, who is working in school with youngest children, asked their postman to bring them a greeting card from Santa Clause. She made a card, postman delivered it and children suddenly became half a dozen times better then they used to be.  :D
They are still wondering how Santa Clause knew their address. :lol:

Isn't Saturnalia where all the mods/admins lose their powers and the regular members become mods/admins :D  :lol:

maybe yay i am an admin since i am a particularly annoying user!!! JK though you might think otherwise

Actually, this is the link but it says it is "disputed":

http://en.wikipedia.org/wiki/Saturnalia#Saturnalia.27s_relation_to_Christmas

I am sure you can find better things than the wiki though.

face it guys there is no santa claus that lives in the north pole  :rotfl:

when you put out the presents you get an idea of that :lol:

[quote="moogra"][quote="Ignite168"][quote="purple plume"]yeah me 2. One year, no one in our family put santa claus presents, so I put a present of a giant lego set for my sister from santa claus. She believed it.

Tooth fairy's fake too if you don't know  :rotfl:

I got a chinese penny in 1st grade. !

I had recently gone to china[/quote]

Good luck ever finding a penny again. Are you sure that was a penny? How old are you? They stopped making pennies in 1962 and I think you may be confusing one fen with one jiao.[/quote]

hey it's possible. I have a bunch of 1 PENNIES (my dad collects coins). I know what 分 looks like. It's quite different from 角.[/quote]

So do i. Especially if your chinese  :rotfl:

Most of my family including my parents didn't even get to go to school until they were well into their teen years because of Wenhua Dageming so I doubt we have anything that old.

I'm just saying a jiao and a fen are very easy to confuse if you don't know Chinese


function
integration
real analysis
real analysis unsolved

Consider the function $ A(t)=\int_{R}{e^{itx+i\cdot\frac{x^3}{3}}}$. Prove that for $t$ going to $-\infty$ we have $ A(t)=\frac{2\sqrt{\pi}\cdot cos( \frac{2}{3}\cdot |t|^{\frac{3}{2}}-\frac{\pi}{4})}{\sqrt[4]{|t|}}+O( |t|^{-\frac{7}{4}})$.

see also

http://ocw.mit.edu/NR/rdonlyres/Mathematics/18-305Fall-2004/A05AA4A1-7D6B-4F4E-8FEC-DB582B41DC9C/0/hwV1.pdf


Can you help me? this appeared randomly out of the blue on my math homework. I think the answer is 720, but I'm not sure. Here's the question:

A twelve hour clock loses 1 minute every hour. Suppose it shows the correct time now. What is the least number of hours from now when it will again show the correct time?

[hide="Hint they gave:"]How many minutes will the slow clock lose before it shows the correct time again?[/hide]

If it were 12 hours, and it started at 12:00, for example, it would end at 11:48, not 12:00. [hide] I think it's $\boxed{720}$ hours. It would have to loose 12 hours to show the right time. And this is $12\cdot60=720$ hours later. [/hide]

Well, you want to find the lowest intersection of

$(x-\frac{1}{60}x)\mod12 \equiv (x)\mod12$

Where $x>0$ and $x\subseteq{Z}$

i don't really have ...anything close to a good solution but..
[hide]
say you start with 12PM
if it loses 1 minute every hour, then after 12 hours, it will be 11:48. and after another 48 hours, it will be 11AM. so after 60hours, you have 1 hour setback. then following this...
60 x 12
 to get to 12AM, but then times another 2 to get to 12PM...?
$1440 hours$?

..oh wait...is this hmm..maybe it's just 720 because 12oclock is just the same on an analog clock.
[/hide]

[quote="PPonte"][size=150][hide]1440 hours. (24 hours / 60^-1)[/hide][/size][/quote]
It's a 12 hour clock, no AM or PM. :D

that is correct since it's a 12-hour clock the correct answer should be $(12)(60)=720 %Error. "hours" is a bad command.
$

[size=150]Sorry. You are completly right.[/size]  :D

heh yeah..guess i shouldn't have thought that am and pm made a difference.


geometry
calculus
3D geometry
prism
trigonometry
inequalities
number theory

.iam currently in INDIA doing by 12th standard ie my last year of schooling and next year i have to join a college for my undergraduate study's .i am very very deeply interested in MATHEMATICS and its my passion.i want to be a mathematician,though i haven’t even passed the REGIONAL MATHEMATICS OLYMPIAD that's conducted in my country.but the problem is i have no idea as to what i should be working so that i can become a mathematician.i have no clue on where i have to join for my under- graduation or what course i have to go for. but the thing is that i very much love the subject and i would practically do anything to achieve my goal.i jus have a worry tah whether it is too late for me to improve on my mathematical abilities as iam already sixteen.my main areas of interest include some amt of elementary calculus,elemantary sequeces and series,and i also like properties of numbers and fascinated by them. please help me and advice me on which books i should study and which fields should i develop my knowledge more on.

Visit the following link for books and the path of a mathematician.

http://www.physicsforums.com/showthread.php?t=122924

For olympiad preparation  visit http://cmi.ac.in/~vipul/olymp_resources/

If want to do a undergraduate course in pure mathematics then I think
 ISI Bangalore and CMI are the best ones of India.

And one more thing,since u are in the 12th, getting selected in RMO is now even more difficult for u as only 6 12th graders are selected ffor RMO every year!So u have got to put in extra efforts.Here's some help:
•	What are the books available, and how do I get  them?
The following book treats the topics which are covered in the olympiads and also is a rich source of problems; (highly recommended)
•	V. Krishnamurthy, C. R. Pranesachar, K. N. Ranganathan and B. J. Venkatachala, Challenge and Thrill of Pre-College Mathematics, New Age International Publishers.
•	C. R. Pranesachar, S. A. Shirali, B. J. Venkatachala, and C. S. Yogananda, Mathematical Challenges from the Olympiads, Prism Books Pvt. Ltd. (Contains problems and solutions of International Mathematical Olympiad from 1986-1994).
•	C. R. Pranesachar, B. J. Venkatachala, and C. S. Yogananda, Problem Primer for the Olympiad, Prism Books Pvt. Ltd., #1865, 32nd. Cross, BSK II Stage, Bangalore 560 070. or 49, Sardar Sankar Road, Kolkata 700029. Phone: 24633890/24633944.
•	M. R. Modak, S. A. Katre, V. V. Acharya, An Excursion in Mathematics, Bhaskaracharya Pratisthan, 56/14 Erandavane, Damle Path, Pune 411 004.
The books listed below form the recommended reading for the various math competitions. Some are elementary, and some are not so elementary. As far as possible there are indicators to the type of the book but, of course, these can only be indicators....
•	GEOMETRY
1.	Durrel M. A., Modern Geometry, Macmillan & Co., London.
2.	H. S. M. Coxeter and S. L. Greitzer, Geometry Revisited, Mathematical Association of America.
3.	S. L. Loney, Plane Trigonometry, Macmillan & Co., London.
•	NUMBER THEORY
1.	I. Niven & H. S. Zuckerman, An Introduction to the Theory of Numbers, Wiley Eastern Ltd. New Delhi.
2.	David Burton, Elementary Number Theory, Universal Book Stall, New Delhi.
3.	G. H. Hardy & Wright, An introduction to the theory of numbers, Oxford University Publishers.
•	PROBLEM BOOKS
o	I M O Problem Collections
1.	S. L. Greitzer, International Mathematical Olympiad 1959-1977, MAA.
2.	M. S. Klamkin, International Mathematical Olympiad 1978-1985, MAA.
o	General
1.	M. S. Klamkin, USA Mathematical Olympiads 1972-1985, MAA.
2.	D. O. Shklyarshky, N. N. Chensov and I. M. Yaglom, Selected problems and Theorems in Elementary Mathematics.
3.	W. Sierpenski, 250 Problems in Elementary Number Theory, American Elsevier.
4.	I. R. Sharygin, Problems in Plane Geometry, MIR Publishers.
•	JOURNALS
1.	Samasya, journal devoted to problem solving, published by Leelavati Trust, Bangalore.
2.	Bona Mathematica, published by Bhaskaracharya Prathistana, Pune.
•	General Reading
1.	Arthur Engel, Problem-Solving Strategies, Springer.
2.	S. A. Shirali, A Primer On Number Sequences, University Press.
3.	S. A. Shirali, First Steps In Number Theory--- A Primer On Divisibility, University Press.
4.	B. J. Venkatachala, Functional Equations---A Problem Solving Approach, Prism Books Pvt. Ltd., #1865, 32nd. Cross, BSK II
Stage, Bangalore 560 070. or 49, Sardar Sankar Road, Kolkata 700029. Phone: 24633890/24633944.
5.	S. Barnard & J.M. Child, Higher Algebra, Macmillan & Co., London, 1939; reprinted Surjeet Publishers, Delhi, 1990
6.	W. S Burnside & A.W. Panton, The Theory of Equations, Vol. 1 (13th Edition), S. Chand & Co., New Delhi, 1990
7.	D. M. Burton, Elementary Number Theory, Second Edition, Universal Book Stall, New Delhi, 1991
8.	RA. Brualdi, Introductory Combinatorics, Elsevier, North-Holland, New York, 1977
9.	H.S.M. Coxeter & S.L. Greitzer, Geometry Revisited, New Mathematical Library 19, The Mathematical Association of America, New York, 1967
10.	C.V. Durell, Modern Geometry, Macmillan & Co., London, 1961
11.	D. Fomin, S. Genkin & 1. Itenberg, Mathematical Circles, First Reprinted Edition, University Press, New Delhi, 2000
12.	H.S. Hall & S.R Knight, Higher Algebra, Macmillan & Co., London; Metric Edition, New Delhi, 1983
13.	R Honsberger, Mathematical Gems, Part I (1973), Part II (1976), Part III (1985), The Mathematical Association of America, New York
14.	N.D. Kazarinoff, Geometric Inequalities, New Mathematical Library 4, Random House and The L.W. Singer Co., New York, 1961
15.	P.P. Korovkin, Inequalities, Little Mathematics Library, MIR Publishers, Moscow, 1975
16.	I. Niven, H.S. Zuckerman & H.L. Montgomery, An Introduction to the Theory of Numbers, Fifth Edition, Wiley Eastern, New Delhi, 2000
17.	A.W. Tucker, Applied Combinatorics, Second Edition, John Wiley & Sons, New York, 1984
18.	G.N. Yakovlev, High School Mathematics, Part II, MIR Publishers, Moscow, 1984
And one more thing,do selectively as there is always time shortage in the 12th standard!
Best of luck! :coolspeak:

You gotta be a millionaire to buy all those. :wink:

Mathlinks is the best I suppose, you might start with the questions from the previous years papers!


Hey, everyone! This problem's really bothering me...

Problem 4.23:

In poker, a 5-card hand is called a [i]three of a kind[/i] if there are three cards of one rank and two other cards which are not the same rank as each other or as the other three cards.  How many 5-card hands are three of a kind? (Assume we are using a standard 52-card deck with 13 ranks in each of 4 suits.)

[hide="Solution to 4.23"]The three cards of same rank can be any of the $ 13$ ranks, and the suits have $ \binom{4}{3}$ possible combinations.  The next card can be any of the $ 12$ remaining ranks and any of the $ 4$ suits.  The last card can be any of the $ 11$ remaining ranks and any of the $ 4$ suits.  However, the last two cards are interchangeable (their order doesn't matter), so we must divide by $ 2!$ to correct for overcounting.  Thus the number of 3-of-a-kind hands is $ \frac{(13\cdot4)\times(12\cdot4)\times(11\cdot4)}{2!}\equal{}\boxed{54912}$.[/hide]

OK: So first, why do we overcount only for the two other cards, not the group of cards in general? Shouldn't we have to divide by $ 5!$ instead, since we can rearrange all the cards and still have a 3-of-a-kind hand? Also, I don't understand why the $ (13\cdot4)$ accounts for all of the 3-of-a-kind possibilities. When I tried this problem, I got $ 13\cdot4\cdot\binom{4}{3}\equal{}208$ different ways to get just the 3 cards of the same rank. That's not counting the number of ways to get the other two cards.

Please help... :(

First, the 3-of-a-kind is distinct from the other two single cards, so is not interchangeable with them. In the expression $ 13\cdot 4$, $ 13$ indicates the choice of rank, and $ 4$ is same as $ \binom{4}{3}$, meaning the posssible combination of suits.

Let's try something different.

[hide]There are:

$ 13$ choices for three of a kind;
$ \binom{4}{3}$ ways to take the suits for the three of a kind;
$ \binom{12}{2}$ ways to take the other two cards;
$ 4$ choices for each suit of the 2 cards left.

Hence,

The result is given by

$ 13\times\binom{4}{3}\times\binom{12}{2}\times 4\times 4\equal{}54912.$[/hide]

Another way
[hide]
$ \binom{13}{3}$ ways to choose the three ranks
$ 3$ ways to choose the rank with 3-of-a-kind
$ \binom{4}{3}$ ways to choose the suit of the 3-of-a-kind
$ 4$ ways to choose each of the rest card
[/hide]


If $ \frac{\sin A}{\sin B} \equal{} \sqrt{3}$ and $ \frac{\tan A}{\tan B} \equal{} \sqrt{5},$ find $ A$ in radians where $ 0 < A < \frac{\pi}{2}$.

[quote="Silverfalcon"]If $ \frac {\sin A}{\sin B} \equal{} \sqrt {3}$ and $ \frac {\tan A}{\tan B} \equal{} \sqrt {5},$ find $ A$ in radians where $ 0 < A < \frac {\pi}{2}$.[/quote]

[hide="ugh messy"]
this implies $ \frac {\cos A}{\cos B} \equal{} \frac {\sqrt {15}}{5}$
or in other words $ \cos A \equal{} \frac {\cos B \sqrt {15}}{5}$
square both sides $ \cos^2 A \equal{} \frac {3\cos^2 B}{5}$
the first part implies $ \sin^2 A \equal{} 3 \sin^2 B$
add the equations to get $ \frac {1}{6} \equal{} \sin^2 B$
substitute into first equation again $ \sin A \equal{} \frac {\sqrt {2}}{2}$
so A=$ \frac {\pi}{4}$

something like that

[/hide]

[hide="Solution"]
$ \sin A \equal{} \sqrt {3} \cdot \sin B \; \; \implies \; \; \sin ^2 A \equal{} 3 \sin ^2 B \; \; \textbf{(i)}$
$ \tan A \equal{} \sqrt {5} \cdot \tan B \; \; \implies \; \; \tan ^2 A \equal{} 5 \tan ^2 B \; \; \textbf{(ii)}$

Dividing $ \textbf{(i)}$ by $ \textbf{(ii)}$ gives $ \cos^2 A \equal{} \frac {3}{5} \cos ^2 B \; \; \textbf{(iii)}$

Adding $ \textbf{(i)}$ and $ \textbf{(iii)}$ and using $ \sin ^2 \plus{} \cos ^2 \equal{} 1$ gives:
$ 1 \equal{} \frac {3}{5} \plus{} \frac {12}{5} \sin ^2 B \implies \frac {2}{5} \equal{} \frac {12}{5} \sin ^2 B \implies \frac {1}{6} \equal{} \sin ^2 B$

Using $ \textbf{(i)}$, we have $ \sin ^2 A \equal{} \frac {1}{2} \implies \sin A \equal{} \frac {1}{\sqrt {2}} \equal{} \frac {\sqrt {2}}{2}$

We recognize this as $ \sin \frac {\pi}{4}$, hence $ \boxed{a \equal{} \frac {\pi}{4}}$
[/hide]

[b]Edit:[/b] Looks like CircleSquared beat me to it

I did the same thing as both of you guys.

I just wondered this though... Is it possible to approach this problem with geometric solution (I think not because then I'm not good at geometry).

The geometric approach is a (slightly) less obvious path to take, but is cleaner.

[hide="Geometric Solution"]
Consider triangle ABC, and point D, the (perpendicular) projection of C onto AB.

Let $ AC \equal{} 1$ and $ BC \equal{} \sqrt {3}$, giving the first given equation by Law of Sines.

Let $ DA \equal{} a$, $ DB \equal{} b$ and $ DC \equal{} c$.

We have that $ b \equal{} a \sqrt {5}$ by the second given equation.

Then, $ 1 \equal{} a^2 \plus{} c^2$ and $ 3 \equal{} 5a^2 \plus{} c^2$, giving $ 1 \minus{} a^2 \equal{} 3 \minus{} 5a^2 \iff a \equal{} \frac {\sqrt {2}}{2}$, which gives $ \text{m} \angle A \equal{} \frac {\pi}{4}$
[/hide]


arkadaşlar bu sorunun daha önce forumda yer aldığını biliyorum ricam odur ki biri şu soruyu adam akıllı türkçe anlatabilir mi? forumdaki çözümde kafam karıştı soru ise şu
$\frac{2^{n}+2}{n}$ nin tamsayı olmasını sağlayan $100 \leq n \leq 1997$  aralığındaki n tamsayılarını bulun

evet, orada her sey dogrudur siz hangi noktayı anlamadınız?
hatta http://www.kalva.demon.co.uk -da bakabilirsiniz :-)

devran

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=1997+apmo&t=79669 adresindeki çözümün 4. satırındaki $d_{i}\ or \ d(p)$ gösterimi ile $d_{i}$ veya $d(p)$ ile $2^{d_{i}}\equiv-1\left(mod \ p_{i}\right)$ denkliğini sağlayan minimum $d_{i}$ değerini mi gösteriyoruz.Eğer böyleyse 6. satırda $ord_{2}\left(d_{k}\right)$ ile gösterdiği değerler nedir?
bunları cevaplarsanız anlamadığım yerleri yine sorabilirim herhalde :lol:  :lol:
He bu arada http://www.kalva.demon.co.uk/apmo/asoln/asol972.html deki çözümdede kafamda soru işaretleri oluştuğu için buradaki çözüme baktım...

o sorduğun yerleri tam olarak ben de anlamadım ama zaten buraya yazılan çözümler genelde tam olmaz. kalva da da deneyerek çözmüş ama sorunun kalıbı zaten bir tane değer istiyor. kalva da ilk sayımız asalsa demiş olmamış. sonra p.q ise demiş olmamış sonra 2.p.q demiş onda bir değer bulmuş yani bence anlaman gereken pek de bir şey yok sadece biraz deneme ile bir çözüm bulacaksın buradaki çözümde de kısa bir deneme yapmalıyız demiş sonra bütün asalları denemiş


calculus
integration
function
inequalities
number theory
prime factorization
Diophantine equation

I'm having a bit of trouble with diophantine equations.
I'm taking problems 46-50 of Engel of which he solves the (a) parts and for the (b) parts he just says to solve by yourself.
The problem is I can't solve either :(

For 47, he's like find all integral solutions to $ x^2\minus{}y^2\equal{}2xyz$.
I tried doing factoring by going like $ (x\plus{}y)(x\minus{}y)\equal{}2xyz$, but then couldn't do anything. Then I tried again with $ x^2\equal{}2xyz\plus{}y^2\equal{}y(2xz\plus{}y)$ but that didn't lead anywhere either.

For 48b, he's like find integral solutions to $ 2xy\plus{}3y^2\equal{}24$. So I did $ y(2x\plus{}3y)\equal{}24$ and $ 24\equal{}1*24, 2*12, 3*8, 4*6$ and solved, but that's lame. What if it were way larger? Like a huge number?

For 49b, it was like $ x^2\plus{}y^2\plus{}z^2\plus{}u^2\equal{}2xyzu$ and he just says to use infinite descent, but I have never done a proof like that before, so I'm not even sure how to roll one off like that.

Help?

[quote="#H34N1"]find all integral solutions to $ x^2 \minus{} y^2 \equal{} 2xyz$.[/quote]
The important thing to note is that this equation is homogeneous in $ x, y$, so you can divide out by $ \gcd(x, y)^2$ and assume without loss of generality that $ \gcd(x, y) \equal{} 1$.  Now we have to have $ x | y^2$ and $ y | x^2$.  When can this happen?

[quote="#H34N1"]find integral solutions to $ 2xy \plus{} 3y^2 \equal{} 24$. So I did $ y(2x \plus{} 3y) \equal{} 24$ and $ 24 \equal{} 1*24, 2*12, 3*8, 4*6$ and solved, but that's lame. What if it were way larger? Like a huge number?[/quote]
That's pretty much it.  If $ 24$ were much larger, there would be many more solutions depending on the prime factorization of whatever you replaced it with.  The question "find all integer solutions to $ xy \equal{} n$" is equivalent to finding all divisors of $ n$.

[quote="#H34N1"]$ x^2 \plus{} y^2 \plus{} z^2 \plus{} u^2 \equal{} 2xyzu$ and he just says to use infinite descent, but I have never done a proof like that before, so I'm not even sure how to roll one off like that.[/quote]
Did you follow the examples given in the chapter?  The standard example is the Diophantine equation $ x^2 \equal{} 2y^2$; the fact that this equation has no solutions implies that $ \sqrt{2}$ is irrational.  This proof is along the same lines, although it's likely you'll need to do some work modulo some small numbers to get the right descent.

i'm not sure if this helps, but notice that for the last problem you may assume that the four variables are all odd by considering it mod 4.

Does homogenous mean that even if you replace x and y it doens't make a difference? BEcause that's not necessarily true...

Also why can you divide thorugh by gcd(x,y)^2? Is there somewhere where I can learn these basics of NT?

[quote="#H34N1"]Does homogenous mean that even if you replace x and y it doens't make a difference? BEcause that's not necessarily true...[/quote]

A function $ f(x, y)$ is homogeneous of degree $ k$ if $ f(tx, ty) \equal{} t^k f(x, y)$.  This implies that the solutions to $ f(x, y) \equal{} 0$ are only determined up to multiplying $ x, y$ by the same constant.  In particular, this is true of the function $ x^2 \minus{} y^2 \minus{} 2xyz$.  (The word for what you are referring to is "symmetric.")  It helps to think of $ z$ as a constant.

Let $ d \equal{} \gcd(x, y)$.  Since $ d^2 | x^2, d^2 | y^2, d^2 | xy$, it divides every term on both sides.  This is a common tactic in solving both Diophantine equations and inequalities; it's just something I picked up on the forums myself, so I'm afraid I can't refer you to anything except possibly the AoPS Introduction to Number Theory.


geometry
circumcircle
geometry proposed
U sviti matematyky

The straight line $ l$ intersects the side $ BC$ of triangle $ ABC$ at point $ X$ and the straight lines $ AC,$ $ AB$ at points $ M,$ $ K$ respectively. Point $ N$ is chosen at the straight line $ l$ in such way that $ AN$ touches the circumcircle of triangle $ ABC.$ Let $ L$ be the intersection point of the circumcircles of triangles $ ABC$ and $ ANX,$   $ L\ne A.$ Prove that points $ A,M,L,K$ lie at a common circle.

we can prove L,C,M,X are concyclic  and K,L,X,B are concyclic
thus we are done


Five runners together complete a $ 100$-mile endurance race by running separate, non-overlapping portions of the course.  Runner B's portion is $ 1.5$ times the length of Runner A's portion.  The combined portion for Runners C and D is twice the length of the combined portion for Runners A and B.  Runner E then runs the last $ 10$ miles of the race.  How many miles did Runner B run?

somewhat quicker way:
E finished the last ten, so without him, A,B,C,and D ran 90. together, c and d ran twice as much as a and b, so c and d ran 60 and a and b ran 30.

$ b\equal{}1.5a$

$ a\plus{}1.5a\equal{}30$

$ 2.5a\equal{}30$

$ a\equal{}12$

$ \boxed {b\equal{}18}$


I got really bored, so I made up a problem.  It might be a little too easy for high school.  Please post your procedure and your answer.

What is the simplified value of:

$ \frac{1}{(cos X \minus{} sin X)^2 \plus{} (cos X \plus{} sin X)^2}$

Express as a common fraction.

And I just realized that this [b]should[/b] belong in High School Basics.  Please move it.

[hide]
First, expand the denominator to get $ \cos^2x\minus{}2\cos x \sin x \plus{}\sin^2 x \plus{}\cos^2x\plus{}2\cos x \sin x \plus{}\sin^2 x$. We know that $ \cos^2x\plus{}\sin^2x\equal{}1$, and that $ \minus{}2\cos x \sin x \plus{}2\cos x \sin x\equal{}0$. So, our expression becomes $ \frac{1}{2\plus{}0}\equal{}\boxed{\frac{1}{2}}$
[/hide]

[quote="lifeisacircle"][hide]
First, expand the denominator to get $ \cos^2x \minus{} 2\cos x \sin x \plus{} \sin^2 x \plus{} \cos^2x \plus{} 2\cos x \sin x \plus{} \sin^2 x$. We know that $ \cos^2x \plus{} \sin^2x \equal{} 1$, and that $ \minus{} 2\cos x \sin x \plus{} 2\cos x \sin x \equal{} 0$. So, our expression becomes $ \frac {1}{2 \plus{} 0} \equal{} \boxed{\frac {1}{2}}$
[/hide][/quote]

Yeah, pretty much.

Easy, huh?

Yup, that was probably one of the easiest trig problems ever.  I might soon make one that is [i]slightly[/i] more challenging.  Meh.

I found two trig problems, the hardest it goes in my geometry book.  Note:  These problems were [b]not[/b] created by ernie.

Prove these equations true:

$ tan^2 \theta \minus{} sin^2 \theta \equal{} tan^2 \theta \cdot sin^2 \theta$

$ \frac {sin X \minus{} cos X}{sin X \plus{} cos X} \equal{} \frac {2 sin^2 X \minus{} 1}{1 \plus{} 2 sin X \cdot cos X}$

#1

$ tan^2 \theta \minus{} sin^2 \theta \equal{} tan^2 \theta \cdot sin^2 \theta$

$ tan^2\theta \minus{} sin^2\theta \minus{} tan^2\theta\times sin^2\theta \equal{} 0$

$ tan^2\theta(1 \minus{} sin^2\theta) \minus{} sin^2\theta \equal{} 0$

$ 1 \minus{} sin^2\theta \equal{} cos^2\theta$

$ tan^2\theta\times cos^2\theta \minus{} sin^2\theta \equal{} 0$

$ sin^2\theta \minus{} sin^2\theta \equal{} 0$

Q.E.D.

#2

$ \frac {sin X \minus{} cos X}{sin X \plus{} cos X} \equal{} \frac {2 sin^2 X \minus{} 1}{1 \plus{} 2 sin X \cdot cos X}$

$ sin2X \equal{} 2sinX cosX$

$ \frac {sin X \minus{} cos X}{sin X \plus{} cos X} \equal{} \frac {sin^2X \plus{} sin^2X \minus{} 1}{1 \plus{} 2sin2X}$

$ \frac {sin X \minus{} cos X}{sin X \plus{} cos X} \equal{} \frac {sin^2X \minus{} cos^2X}{1 \plus{} 2sin2X}$

$ \frac {sin X \minus{} cos X}{sin X \plus{} cos X} \equal{} \frac {(sinX \plus{} cosX)(sinX \minus{} cosX)}{1 \plus{} 2sin2X}$

$ \frac {1}{sinX \plus{} cosX} \equal{} \frac {sinX \plus{} cosX}{1 \plus{} 2sin2X}$

Cross Multiply.

$ sin^2X \plus{} 2sinXcosX \plus{} cos^2X \equal{} 1 \plus{} 2sin2X$

$ sin^2X \plus{} cos^2X \equal{} 1, 2sinXcosX \equal{} 2sin2X$

$ 1 \plus{} 2sin2X \equal{} 1 \plus{} 2sin2X$

Q.E.D.

Trig is fun!

lol, i was editting

First one gets a point!

$ sin^5 X cos^2 X \equal{} (cos^2 X \minus{} cos^4 X \plus{} cos^6 X) sin X$

Prove it's true.

It's easy.

[quote="Brut3Forc3"]I think it isn't true... must check though.[/quote]

It is true.  That's your hint.  I proved it.

phew solved it

$ sinX*sin^4X*cos^2X\equal{}sinX(cos^2X)(1\minus{}cos^2X\plus{}cos^4X)$

$ sin^4X\equal{}cos^4X\minus{}cos^2X\plus{}1\equal{}(cos^2X\minus{}1)^2$

$ cos^2X\minus{}1\equal{}sin^2X$

$ sin^4X\equal{}(sin^2X)^2$

Q.E.D.

Ummm... It's $ \sin X$, not $ sin X$.

Just correcting your LaTeX.

And DUDE! You wrote

$ \cos^2{X}\minus{}1\equal{}\sin^2{X}$

which is obviously false!

[quote="FantasyLover"]phew solved it

$ sinX*sin^4X*cos^2X \equal{} sinX(cos^2X)(1 \minus{} cos^2X \plus{} cos^4X)$

$ \boxed{sin^4X \equal{} cos^4X \minus{} cos^2X \plus{} 1 \equal{} (cos^2X \minus{} 1)^2}$

$ cos^2X \minus{} 1 \equal{} sin^2X$

$ sin^4X \equal{} (sin^2X)^2$

Q.E.D.[/quote]
I think $ (\cos^2 x \minus{}1)^2\equal{} \cos^4 x \minus{} 2 \cos^2 x \plus{}1$.

I am not sure if you can find the answer to this one.  I made it based on a problem in my book.

$ \frac{1\plus{}tan^2 X}{sin^2 X \plus{} cos^2 X} \equal{} \frac{1}{ax}$

A is a trigonometric function.  Find A.

I am not sure if finding this solution is possible or not.

DUDE!!! It's obviously the square of the cosine!

$ \frac {1 \plus{} \tan^2 x}{\sin^2 x \plus{} \cos^2 x} \equal{} \frac {\sec^2 x}{1} \equal{} \frac {1}{a(x)} \Rightarrow a(x) \equal{} \cos^2 x$
Hey, you beat me to it 'cause I have to LaTeX it! :P

lol im right then!

...to lazy to edit so w/e

ok since ernie isnt making prob for now

im gonna make a prob

<extremely easy>

What is $ \lim_{x\to0}\frac {1 \minus{} \cos{x}}{x^2}$?

________________________________________________

EDIT//

answer for ernie's prob

$ \frac{\pi}{3}, \frac{4\pi}{3}$

i dont think i need to explain this

[quote="FantasyLover"]lol im right then!

...to lazy to edit so w/e

ok since ernie isnt making prob for now

im gonna make a prob

<extremely easy>

What is $ \lim_{x\to0}\frac {1 \minus{} \cos{x}}{x^2}$?

________________________________________________

EDIT//

answer for ernie's prob

$ \frac {\pi}{3}, \frac {4\pi}{3}$

i dont think i need to explain this[/quote]
In [0,360)!!!!!!!!!

And also, $ \lim_{x\to0}\frac {1 \minus{} \cos{x}}{x^2}\equal{}\frac{1}{2}$

$ \lim_{x\to0}\frac {1 \minus{} \cos{x}}{x^2}$

$ \equal{}\lim_{x\to0}\frac {1 \minus{} (1 \minus{} \frac{x^2}{2!} \plus{} \frac{x^4}{4!} \minus{} \cdots)}{x^2}$

$ \equal{}\lim_{x\to0}\frac {\frac{x^2}{2!}\minus{}\frac{x^4}{4!}\plus{}\frac{x^6}{6!} \minus{} \cdots}{x^2}$

$ \equal{}\lim_{x\to0}(\frac{1}{2!}\minus{}\frac{x^2}{4!}\plus{}\frac{x^4}{6!} \minus{} \cdots)$

$ \equal{}\frac{1}{2!}\minus{}\frac{0}{4!}\plus{}\frac{0}{6!}\minus{}\cdots$

$ \equal{}\frac{1}{2!}$

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

[quote="FantasyLover"]lol im right then!

...to lazy to edit so w/e

ok since ernie isnt making prob for now

im gonna make a prob

<extremely easy>

What is $ \lim_{x\to0}\frac {1 \minus{} \cos{x}}{x^2}$?[/quote]
Whoa, didn't see ernie's problem.
$ \lim_{x\to 0}\frac {1 \minus{} \cos x}{x^2} \equal{} \lim_{x\to 0}\frac {\sin x}{2x} \equal{} \frac {1}{2}$? I still haven't learned calc yet.

EDIT: Btw, I think that problem doesn't belong in this forum.

meh its not really calculusy tho...

i didnt use l'hopital

its $ \lim_{x\to0}\frac{(1\minus{}\cos{x})(1\plus{}\cos{x})}{x^2(1\plus{}\cos{x})}$

which equals to $ \lim_{x\to0}\frac{\sin^2{x}}{x^2(1\plus{}\cos{x})}$

and its $ \lim_{x\to0}\frac{\sin^2{x}}{x^2}\times\frac{1}{1\plus{}\cos{x}}$

...which is just

$ \lim_{x\to0}\frac{1}{1\plus{}\cos{x}}$

$ \cos{0}$ is 1, so its

$ \frac{1}{2}$

Fantasy, you still haven't posted the correct answer. :P

Okay,

First of all, no marathons.

Second of all, this is HSB, not a college forum.

Locked.


calculus
inequalities
algebra unsolved
algebra

Suppose a and b are real numbers such that the roots of the cubic equation ax^3-x^2+bx-1 has positive real roots.javascript:emoticon(':(')

Prove That       0<3ab<=1 :(  :wink:

$ P(x)\equal{}ax^3\minus{}x^2\plus{}bx\minus{}1$ has 3 real roots so $ P'(x)\equal{}3ax^2\minus{}2x\plus{}b$ has real roots and its discriminant has to be non negative $ 1\minus{}3ab \geq 0\Longrightarrow  3ab \leq 1$
The roots $ x_1,x_2,x_3$ of $ P(x)$ are positive so  $ x_1x_2\plus{}x_2x_3\plus{}x_3x_1 \equal{} \frac ba > 0\Longrightarrow 0 <3ab$
 :cool:

I think Vìete theorem and AM-GM work in this problem

i also got it applying calculas i want to get it without using the slightest concept of calculas :)
can u please elaborate duongptnkvn :blush:

this is a solution which doesnt involve any calculus....

the given expression is $ ax^3 \minus{} x^2 \plus{} bx \minus{} 1 \equal{} 0$ which has positive real roots

let $ x$, $ y$, $ z$ be the roots.

we know that
$ x^2 \plus{} y^2 \plus{} z^2 \geq \ xy \plus{} yz \plus{} zx.$

i.e $ x^2 \plus{} y^2 \plus{} z^2 \plus{} 2xy \plus{} 2yz \plus{} 2zx \geq \ 3xy \plus{} 3yz \plus{} 3zx.$

i.e. $ \frac {( x \plus{} y \plus{} z )^2}{3} \geq \ xy \plus{} yz \plus{} zx$.

from the given expression, we know that
$ x \plus{} y \plus{} z \equal{} \frac {1}{a}$-(i)  and  $ xy \plus{} yz \plus{} zx \equal{} \frac {b}{a}$ -(ii).

so, the above inequality becomes,
$ \frac {1}{3a^2} \geq \ \frac {b}{a} \longrightarrow\ \frac {1}{3a} \geq \ b.$

which gives $ 1 \geq \ 3ab.$
by (i) and (ii) $ ab \geq \ 0.$

thus $ 1 \geq \ 3ab > 0.$ :D  :P  :lol:


AMC
AMC 12
AMC 12 A
AMC 10
AMC 10 B
AIME

Is it possible for the AMC12A cutoff to be a little lower than 100 this year??????? Also, can the cutoff for the AMC10B possibly be like 105 this year????

I haven't taken the test, so I don't know if you're question about the AMC 12A is true (which I doubt a litte), but I highly doubt that the AIME cutoff score for the AMC 10 will ever be even close to 105. Sorry  :(

Cutoff for AMC 10 are seldom lower than 110.
Cutoff score for AMC 12... well, most likely not be under 100.


Why just YouTube?  I tried to embed a Yahoo! Video into my post and ... well, I can't.  Yahoo! Video is much faster and with better resolution, and a cleaner video player.  You should find a way to allow Yahoo! Video and Dailymotion embedding.

[quote="minicon"]Or how about flash tags?
That would allow every video host.[/quote]Flash tags definitely not. Yahoo Video I might implement at some point in the future.

[quote="minicon"]Or how about flash tags?
That would allow every video host.[/quote]
do you know the security implications of that? any user with the slightest know-how could virtually destroy AoPS...


inequalities
trigonometry
induction
inequalities open

Here is a super-problem, comunicated to me by Popa Calin. It asks to find the minimal constant $ k$ such that for any $ a_1,...,a_n\geq 0 $ we have the inequality
   $ k(a_1^2+a_2^2+...+a_n^2-a_1a_2-a_2a_3-...-a_na_1)\geq a_1^2+a_2^2...+a_n^2-nG^2 $, where $G$ is the geometric mean of $ a_1,...,a_n $. I have found that $ k\geq\frac{1}{1-cos\frac{pi}{n}} $, but I dont know how to prove that this is the minimal constant. Can anyone prove or disprove this conjecture?

Subject to constraints:
$ \Sigma\(a_i^2=A $
$ \Sigma\(a_i a_{i+1}=B $
$ A>B>0 $.
We have to find minimum of   $p=a_1 a_2 ... a_n $.
For each $i=1,n$ exists $C_1,C_2$:
$ 2C_1 a_i^2+C_2 a_i (a_{i-1}+a_{i+1})+p=0 $
and:
$ 2C_1 A + 2C_2 B + np = 0 $
If we find the solutions of this system, vary $A,B$ in order to get that $k$.
(btw how did you manage $k>= \frac{1}{1-cos\frac{pi}{n}}$ ?)

I used the following result of Ozeki:
  $ max\frac{x_1\cdot x_2+x_2\cdot x_3+...+x_{n-1}\cdot x_n}{ x_1^2+x_2^2+...+x_n^2} =cos\frac{pi}{n} $. Old members probably remember a problem posted by S_Alex which reduces immediately to this one. But then I didnt know this and I struggled a lot to solve that problem.

...and how comes this result of Ozeki?
I don't find it trivial! Can you supply a proof for it, pls!

can you supply a proof or at least a link for that max?

Enjoy the proof!.
We have $C(x_1^2+\cdots x_n^2)-(x_1x_2+\cdots x_{n-1}x_n)=$
$=C_1(x_1-\frac{x_2}{2C_1})^2+C_2(x_2-\frac{x_3}{2C_2})^2+ \cdots$
$\cdots C_{n-1}(x_{n-1}-\frac{x_n}{2C_{n-1})^2+C_n x_n^2}$
Now the necessary and sufficient condition our expression to be positive is
all of
$C_i$ to be positive.
Now if we put $C_1=C=\cos \alpha$ one deduce by induction
$C_k=\frac{\sin {(k+1)\alpha}}{2 \sin {k\alpha}}.
Now we must have $0 \leq \alpha \leq \frac {\pi}{n+1}$ an the minimal $C$ is
$\cos {\frac{\pi}{n+1}}.
This implies what Harazi said.

Im sorry, there is a typo there. The maximum is $cos\frac{pi}{n+1} $.  Here is a hint for a possible proof. Consider $ t=\frac{pi}{n+1} $ and define $ u_1=2cost, u_k=2cost-\frac{1}{u_{k-1}} $. Observe that the sequence $ u_1\cdot u_2\cdot\cdot u_k $ verifies a second degree linear recurrence and hence his general term can be easily computed. This shows that we can also compute the general term of $ u_k $ . After that, we must verify that $ u_i>0 $ for $ i=1,n-1$ and $ u_{n-1}=\frac{1}{2cost} $. This is easy after we have the general formula. This shows that we have in fact
 $ u_1=u_2+\frac{1}{u_1}=u_3+\frac{1}{u_2}==u_{n-1}+\frac{1}{u_{n-2}}=\frac{1}{u_{n-1}}=2 cost $. Next, define $ a_i=\sqrt{u_i} $ and $ b_i=\frac{1}{a_i} $. Then it is easy to see that $ 2cost\cdot (x_1^2+...+x_n^2)-2(x_1\cdot x_2+...+x_{n-1}\cdot x_n)=(a_1\cdot x_1-b_1\cdot x_2)^2+...+(a_{n-1}\cdot x_{n-1}-b_{n-1}\cdot x_n)^2 \geq 0$. After that we can check the equality cases.

...and minimality of $k$ (in the original inequality) occurs when one of that $a_i$ is 0. Thanks!
I did not expect such a proof!

Sorry if I am a  bad mathematician to understand this step of Harazi reasoning.

A homogeneous linear 2nd degree recurrence relation with constant coefficient is of the form

$(*)$ $ u_{k+1} +Au_{k}+Bu_{k-1} =0$ with $ k=1,2,...$ and $u_0=C,u_1=D$ where $A;B;C;D$ are assigned constant.

Harazi relation is $ u_1=2cost, u_k=2cost-\frac{1}{u_{k-1}} $.

How do you obtain from this a relation like $(*)$ ?

[quote="harazi"] Consider $ t=\frac{pi}{n+1} $ and define $ u_1=2cost, u_k=2cost-\frac{1}{u_{k-1}} $. Observe that the sequence $ u_1\cdot u_2\cdot\cdot u_k $ verifies a second degree linear recurrence and hence his general term can be easily computed. This shows that we can also compute the general term of $ u_k $ . After that, we must verify that $ u_i>0 $ for $ i=1,n-1$ and $ u_{n-1}=\frac{1}{2cost} $. This is easy after we have the general formula. [/quote]

Thank you very much.

Just multiply the recurrence relation by $ u_1 u_2...u_{k-1} $.

I think that the initial inequality (as a super-problem noted) is still open.
Did you really proved that $ k_n = \frac {1}{1-cos \frac{\pi}{n} } ?$

No, I could not prove that this is the optimal constant. So, did you find this problem somewhere? And it is open? Then, there is no reason to think on it anymore... :(


I was looking over the usamo qualifiers, and I saw someone is grade 13?

How?

Maybe they're doing a post-graduate year at a high school.


hι παιδες!
ειμαι καινουρις στο forum και το βρισκω πολυ ωραιο και φιλοξενο...
τεσπα στο θεμα μου...αρχισα να ασχολουμαι με mathemaica...αλλα οπως πολλοι ξερετε windows sux...και η για την εκδοση για Linux δεν εχω ακουσει και τα καλυτερα...μηπως ξερετε κανενα εξισου απλο περιβαλλον ωστε να μπορω να κανω 2σδυαστατα plots(ε και κανενα 3d δε θα μ επειραζε αλλα δεν επειγει... :P )...α και αν ηταν opensource δε θα με πειρζε..αν και Linux που να μην ειναι δεν το βλεπω ιδιεταιρα πιθανο... :D

ευχαριστω προκαταβολικα
darksaga

Vasika den katalava an theleis gia Windows h linux kai ti.
To mathematica egw to douleuw se arch linux kai einai mia xara.
(kai to matlab alla telospantwn)
Gia apla plot synarthsewn mporeis na dikimaseis to gnuplot (Windows & Linux).

Steile mou to e-mail sou na sto steilw an thes (peripou 2MB) :wink:

@offtopic:mimmis έχεις pm (για να τραβήξω την προσοχή σου & δεν είμαι σίγουρος ότι στάλθηκε)


Find a formula for the sum:

$\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+ ... + \frac{1}{(2n+1)(2n+3)}$

And that form simplies nicely into
[hide]
$\frac{n}{6n+9}$
[/hide]

[quote="José"][quote="Adunakhor"][hide]
$S = \displaystyle \frac{1}{2} \displaystyle \left(\displaystyle \frac{1}{3}-\displaystyle \frac{1}{2n+3} \right)$

The expression telescopes nicely :)

[/hide][/quote]

Have you got a proof?? maybe it is using induction...

EDIT: amcavoy posted before me...[/quote]

there are a bunch of problem like this, the task of telescoping these kind of small sums kinda becomes innate

a more general approach is trying to think like this:
a fraction $\frac{1}{pq}$ where p,q are two liniar expressions
and p is close to q(so that the simplification of the telescopant
series will be possible) can be broken up into $\frac{a}{p} , \frac{b}{q}$
so if we consider $\frac{1}{pq}=\frac{a}{p}-\frac{b}{q}=\frac{qa-pb}{pq}$
with the restriction $bp-aq=1$
so here we :
$p= 2n+1$
$q= 2n+3$
so $bp-aq=1 \Rightarrow 2nb+b-2na-3a=1 \Rightarrow  n(2b-2a)+b-3a-1=0$
so for this to happen for every $n$ we need that both $2b-2a=0$ and
$b-3a-1=0$ so we have $b=a$ and $a=-1/2$ so $b=-1/2$ so we now know that
$\frac{1}{(2n+1)(2n+3)}= a\cdot \frac{1}{p} - b \cdot \frac{1}{q}=$
$=\frac{1}{2} \cdot \frac{1}{2n+1} - \frac{1}{2} \cdot \frac{1}{2n+3}$

from now on it is easy to see why if we sum up terms allot of them will simplify,its just
the fact that for this case $p,q$ are close enough

[quote="Chocoboom"]telescoping means that it certain parts cancel out nicely, but i dont get how you got this $\sum_{k=1}^{n}\frac{1}{\left(2n+1\right)\left(2n+3\right)}=\frac{1}{2}\sum_{k=1}^{n}\frac{1}{2n+1}-\frac{1}{2}\sum_{k=1}^{n}\frac{1}{2n+3}$
sorry im not good with sigma, is that some kinda identity or something?[/quote]
I used a simple partial fraction decomposition.  That is the only step you really need to take (the rest is obvious because every term in the middle cancels).

that is what im explaining in my post amcavoy
a simple fraction decomposition

Oh I'm sorry, I just hit the "quote" button on the poster's question without reading your post.


For what values of $ k$ does the graph of
\[ \frac{(x\minus{}2k)^2}{1}\minus{}\frac{(y\minus{}3k)^2}{3}\equal{}1\]
pass through the origin?

Thanks in advance!

We plug in the point $ (0,0)$ in our equation (since the hyperbola must pass throught the point). Thus,

$ \frac{(0\minus{}2k)^2}{1}\minus{}\frac{(0\minus{}3k)^2}{3}\equal{}1$

$ \implies 4k^2\minus{}\frac{9k^2}{3}\equal{}1$

$ \implies 4k^2\minus{}3k^2\equal{}1 \implies k^2(4\minus{}3)\equal{}1 \implies k^2\equal{}1 \implies k\equal{}\boxed{\pm1}$.

I get your solution, but I have a question.  Don't the $ \minus{}2k$ and $ \minus{}3k$ represent horizontal and vertical shifts of the function?  So why I couldn't I say that the graph passes through the origin when $ k\equal{}0$?

Thanks!

If you had k = 0, then the center of the hyperbola would be be at the origin. However, the hyperbola itself will not pass through the origin. The two portions will originate from vertices that are at (1,0) and (-1.0). It is also obvious that plugging in 0 for x when k is 0, will not satisfy the equation.  Check out some more information about hyperbolas on the internet to clear up your doubts.

Hi elcric.

See here: http://en.wikipedia.org/wiki/Hyperbola

or here: http://mathworld.wolfram.com/RectangularHyperbola.html

or here: http://xahlee.org/SpecialPlaneCurves_dir/Hyperbola_dir/hyperbola.html

or here: http://www.algebralab.org/lessons/lesson.aspx?file=algebra_conics_hyperbola.xml

Hope this helps you.


Is there a polynomial $f(x,y,z)$ with real number coefficients such that $f(x,y,z)$ is positive iff $|x|, |y|, |z|$ are sides of a triangle?

Yes, sorry, that is what I intended to write. Nice question: what is the minimal degree of such a polynomial?


Solve the equation
$18x^{2}-18x\sqrt{x}-17x-8\sqrt{x}-2 = 0$.

[quote]

${u (3 u-4)=2}$

[/quote]

Why do you have this agrument?

hmm....

i really dont know

hauhua

sorry


What type of hybridization would be found in the central atom of sulfur dioxide? Of carbon disulfide?

Is it generally true that if the central atom has one "group" attached, it is sp hybridized; if it has two "groups" attached it is sp2 hybridized; if it has 3 groups attached it is sp3 hybridized? For example, the carbon in methane has 3 "groups" (i.e. hydrogens) attached and is sp3 hybridized.

[quote="Aspiringchemlete"]Is it generally true that if the central atom has one "group" attached, it is sp hybridized; if it has two "groups" attached it is sp2 hybridized; if it has 3 groups attached it is sp3 hybridized?[/quote]

It would be more correct to say: if the central atom is attached to only two, three, or four other atoms, then it has hybridization sp, sp2, and sp3, respectively. However, in each case there can be no unshared pairs on the central atom.

[quote="Aspiringchemlete"]What type of hybridization would be found in the central atom of sulfur dioxide? Of carbon disulfide?[/quote]

In carbon dissulfide, the central atom is carbon, it has no unshared pairs, and is attached to only two atoms: therefore, it is sp hybridized. For sulfur dioxide, begin by writing its Lewis structure.

How about if a central atom is attached to 5 and 6 other atoms?

[quote="Carcul"]

How about if a central atom is attached to 5 and 6 other atoms?[/quote]

Would that be when we take hybridization with $ d$ orbitals into account (i.e. $ dsp^{2},dsp^{3},d^{2}sp^{3}$).

Yes, but I believe dsp2 hybridization does not exist.

I am pretty sure it does. Example: $ Ni(CN)_4^{2 \minus{} }$.

Yes, you are right BanishedTraitor, I was only thinking in tetrahedal geometry and in organic compounds.

On the note of Hybridization. I want to ask Carcul, how come the bond angle of $ H_2S$ is ideally $ 105^{\circ}$ according to hybridization but $ 92^{\circ}$ in actual fact? $ 92^{\circ}$ seems to be closer to $ 90^{\circ }$ which is the position of two $ p$ orbitals. So is hybridization actually present in $ H_2S$?

Hybridization is perhaps always present when elements form compounds. It is a known fact that the bond angles of the hydrogen compounds of the elements of the pnictides and chalcogens decrease down each group. Compare the following values:

NH3 (107.8º), PH3 (93.5º), AsH3 (91.8º), SbH3 (91.7), BiH3 (90.48º).

H2O (104.5º), H2S (92.1º), H2Se (91º), H2Te (89º).

This trend is generally atributed to more s character of the orbitals containing the lone pairs, and to more p character of the orbitals that combine with the H 1s orbitals. Therefore, as the lone pairs are more close to the central atom (compared with ammonia and water), then the H-X-H angle must be reduced in order to minimize electric repulsions.


Let $a,b,c\ge 0$ such that $a^{2}+b^{2}+c^{2}+d^{2}=4$ prove that:
$2(a^{3}+b^{3}+c^{3}+d^{3})\ge 2+\frac{3}{ \sqrt{2}}\sqrt{ab+bc+cd+da+ac+bd+2}$

My solution use AM-GM (second solution) and UC method (first solution) :D  and you ? :wink:  :P

Yes agrady  my solution use too equality
$2+ab+bc+cd+da+ac+bd= \frac{(a+b+c+d)^{2}}{2}$
And I use UC Method (Undetermined Coefficient Method).
we have inequality equavalent:
$2(a^{3}+b^{3}+c^{3}+d^{3})\ge \frac{3}{2}(a+b+c+d)+2$
now we will prove that inequality:
$2a^{3}\ge \alpha(a-1)(a+1)+\frac{1}{2}(3a+1)$
with  :alpha is a constant
Eassy by [b]UC method [/b] we can find  $\alpha = \frac{9}{4}$
And now we have:
$2a^{3}\ge \frac{9}{4}a^{2}+\frac{3}{2}a-\frac{7}{4}\\ \Leftrightarrow 8a^{3}-9a^{2}-6a+7\ge 0 \\ \Leftrightarrow (a-1)^{2}(8a+7)\ge 0$
Similarly with remain inequality we have:
$2(a^{3}+b^{3}+c^{3}+d^{3})\ge \frac{3}{2}(a+b+c+d)+2$
I have 3 solution. It is use Holder inequality and use only AM-GM inequality :D

we have got (a+b+c+d)(a^3+b^3+c^3+d^3)>=(a^2+b^2+c^2+d^2)^2
=16 by cauchy
after that we must show that 32>=2(a+b+c+d)+(3/2)(a+b+c+d)^2
and this s obvius because (a+b+c+d)<=4 by am qm

UC method can prove many problem. Ex: USA MO 2000, JAPAN MO 1997, IMO 2000... now I will use UC method to prove that one problem in Vasc'book (it's in hungkhtn'book, too)
[u][b]Problem:[/b][/u]
Let $a,b,c$are positive real number. Prove that:
 $\sqrt[3]{ \frac{a^{3}}{a^{3}+(b+c)^{3}}}+\sqrt[3]{ \frac{b^{3}}{b^{3}+(a+c)^{3}}}+\sqrt[3]{ \frac{c^{3}}{c^{3}+(b+a)^{3}}}\geq 1$
Now we want prove the follow inequality:
$\sqrt[3]{ \frac{a^{3}}{a^{3}+(b+c)^{3}}}\ge \frac{a^{p}}{a^{p}+b^{p}+c^{p}}$
by UC method we easily find $p=2$
continoues I will post old problem by UC method can prove:
 [u][b]Problem:[/b][/u]
Let $a,b,c$are positive real number. Prove that:
$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{b+a}\ge \frac{21}{16}+\frac{27(a^{3}+b^{3}+c^{3})}{16(a+b+c)^{3}}$
Solution:
Since the inequality is homogeneous, we may normalize to $a+b+c=3$.
from the assumption $a+b+c=3$ we have:
$\frac{a}{3-a}+\frac{b}{3-b}+\frac{c}{3-c}\ge \frac{21}{16}+\frac{27(a^{3}+b^{3}+c^{3})}{16}$
Now we want prove that the follow inequality:
$\frac{a}{3-a}\ge \alpha(a-1)+\frac{1}{16}(a^{3}+7)$
By [b]UC method[/b] we cant find  $\alpha = \frac{9}{16}$
And we have:
$\frac{a}{3-a}\ge \frac{9a+a^{3}-2}{16}\\ (a-1)^{2}(a^{2}-a+6)($
It true :D
[b]Problem of (Moldova TST 2005)[/b], we can use [b]UC method [/b]to prove... :D
And have many, many problem can prove that by [b]UC method[/b]

[quote="zaizai-hoang"]UC method can prove many problem. Ex: USA MO 2000, JAPAN MO 1997, IMO 2000... now I will use UC method to prove that one problem in Vasc'book (it's in hungkhtn'book, too)
[u][b]Problem:[/b][/u]
Let $ a,b,c$are positive real number. Prove that:
 $ \sqrt [3]{\frac {a^{3}}{a^{3} \plus{} (b \plus{} c)^{3}}} \plus{} \sqrt [3]{\frac {b^{3}}{b^{3} \plus{} (a \plus{} c)^{3}}} \plus{} \sqrt [3]{\frac {c^{3}}{c^{3} \plus{} (b \plus{} a)^{3}}}\geq 1$
[/quote]
By AM-GM we have:
$ \sqrt{1\plus{}x^3} \leq \frac{1\plus{}x\plus{}1\minus{}x\plus{}x^2}{2}\equal{}1\plus{}\frac{x^2}{2}$
So $ \sqrt[3]{\frac{a^3}{a^3\plus{}(b\plus{}c)^3}} \equal{}\frac{1}{\sqrt{1\plus{}(\frac{b\plus{}c}{a})^3}} \geq \frac{1}{1\plus{}\frac{1}{2}(\frac{b\plus{}c}{a})^2} \geq \frac{1}{1\plus{}\frac{b^2\plus{}c^2}{a^2}}\equal{}\frac{a^2}{a^2\plus{}b^2\plus{}c^2}$
Done!