probability
calculus
analytic geometry
geometry

pick any three points on a circle and form a triangle.  Find the probability that the triangle is an obtuse triangle?

You can solve this straightforwardly by integrating over the possible arclength of the third point as the second point varies (taking the first point as fixed): this is essentially what someone posted in another forum.  Please don't post your questions in more than one place.  Also, what is the source?  It didn't just come to you in a dream, did it?  (I think there is a nice non-calculus approach to this problem as well, but I'm not sure.)

Regarding a non-clculus approach: assume one of the points ($A$) is fixed. Now consider $\angle ABC$ and $\angle ACB$ as variables $x,y$, i.e. the coordinates of a point in the plane. They can represent any point in the triangular region $x,y>0,x+y<\pi$, and the region corresponding to the case we're interested in consists of three smaller triangles determined by the additional conditions:  a) $x+y<\frac\pi 2$, b) $x>\frac \pi 2$, c) $y>\frac \pi 2$. If you make a drawing you'll see that the area of the region we're interested in is $\frac 34\cdot$ the total area, so the probability is $\frac 34$.


A farmer grows 252 kilograms of apples. He sells them to a grocer who divides them into 5 kilogram and 2 kilogram bags. If the grocer uses the same number of 5 kg bags as 2kg bags, then how many bags did he use in all?

[hide]Since they're both the same amount of bags, let that be x.

5x+2x=252.
7x=252
x=36.

But he used two types of bags, 36*2=72. [/hide]

[hide]
He uses the same amount of bags for 5 and 2.
Let the number of bags be x.
5x+2x=252
7x=252
x=36
Since there are 2 kinds of bags, the answer 36*2=72[/hide]


What is the reciprocal of $ 0.0\overline{2}$ ? Express your answer in
simplest form.


[quote="zaizai-hoang"]$\bigstar$ Let $a,b,c$ are positive real numbers such that $a+b+c=1$. Prove that:
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge \frac{25}{1+48abc}$[/quote]

posted recently!

Dear e.lopes  :D Can you send to me the link of above problem?  :blush:
I have proved it by S.O.S method but I think my solution is not nice.

it was posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=145628]here[/url]

Thank you,  scorpius119  . I like your solution.  :wink:

The following inequality is true

For all positive real numbers $a,b,c,d$ with sum $1$ then
\[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\ge \frac{63}{2(1+47abcd)}. \]


integration
complex numbers
calculus
calculus computations

Observe that ${(1-\frac{1}{\sqrt{2}})^{\frac{1}{\sqrt{2}}}= \sum_{k = 0}^{\infty}{-\frac{1}{\sqrt{2}}\choose k}1^{r-k}(-\frac{1}{\sqrt{2}})^{k}\quad\quad\quad}$

${-\frac{1}{\sqrt{2}}\choose k}\&{}={1 \over k!}\prod_{n=0}^{k-1}(-\frac{1}{\sqrt{2}}-n)=\frac{-\frac{1}{\sqrt{2}}(-\frac{1}{\sqrt{2}}-1)(-\frac{1}{\sqrt{2}}-2)\cdots(-\frac{1}{\sqrt{2}}-(k-1))}{k!}.$
Plugging in the first few terms should yield the result we are after.

if i [u]didn't [/u] give you the answer how would you have proceeded ?  :roll:

Finding the closed form of a series is generally harder than showing that it is equal something for it often requires coming up with some tricks.  Do I have a trick for finding the closed form of the series in question? No, I don't. But someone might have already done so and is sitting somewhere in the library.

Online reverse decimal number looker-upper :D

One classic trick is to introduce a variable: instead of $\sum a_{n}$, look at $f(x)=\sum a_{n}x^{n}$. What does that do here? We have $a_{0}=\frac12$ and $a_{n+1}=-\frac{(4n+1)(4n+3)}{(4n+4)(4n+6)}a_{n}$, or $(2n+\frac12)(2n+\frac32)a_{n}=-(2n+2)(2n+3)a_{n+1}$. To deal with the index-doubling, we look instead at $g(x)=\sum a_{n}x^{2n}$; this leads to a differential equation. In particular, $\sum (2n+2)(2n+3)a_{n+1}x^{2n+1}=\frac{d^{2}}{dx^{2}}(xg(x))$ and $(2n+\frac12)(2n+\frac32)a_{n}x^{2n-\frac12}= \frac{d^{2}}{dx^{2}}(x^{\frac32}g(x))$. Now
$\frac{d^{2}}{dx^{2}}(xg(x)=-x^{\frac32}\frac{d^{2}}{dx^{2}}\left(x^{\frac32}g(x)\right)$
$\frac{d}{dx}(g(x)+xg'(x))=-x^{\frac32}\frac{d}{dx}\left(\frac32x^{\frac12}g(x)+x^{\frac32}g'(x)\right)$
$2g'(x)+xg''(x)=-\frac34xg(x)-3x^{2}g'(x)-x^{3}g''(x)$
$(x^{3}+x)g''(x)+(3x^{2}+2)g'(x)+\frac34xg(x)=0$.
We also have the initial conditions $g(0)=\frac12,g'(0)=0$. The solution is rather difficult in this form.

Fortunately, there's another trick: we can interpolate some missing terms, and look at $h(x)=\sum_{n}b_{n}x^{n}$, where $b_{n}=(i^{n})\frac{1\cdot 3\cdots (2n-1)}{2\cdot 4\cdots (2n+2)}$; $g(x)=\frac12(h(x)+h(-x))$ is the real part of $h$. This gives the relation $(n+2)b_{n+1)=i(n+\frac12)b_{n}}$ and the differential equation
$\frac{d}{dx}(xh(x))= ix^{\frac32}\frac{d}{dx}\left(x^{\frac12}h(x)\right)$
$h(x)+xh'(x)=\frac{i}{2}xh(x)+ix^{2}h'(x)$
$(x-ix^{2})h'(x)+(1-\frac{i}{2}x)h(x)=0$
$\frac{h'(x)}{h(x)}=\frac{\frac i2x-1}{x-ix^{2}}$
$h(x)=\exp\left(\int \frac{\frac i2x-1}{x-ix^{2}}\right)$
This is an ugly but routine exercise in partial fractions. I don't feel like doing it right now.
The initial condition $h(0)=\frac12$ will come in the exact choice of antiderivative.

By the way, akech is wrong.

holy smokes, that's deep !   :o

i will continue from the partial fractions and see what happens ! :ninja:


algebra
polynomial
Galois Theory
superior algebra
superior algebra solved

Show that if E is a finite field and F is a subfield then the norm homomorphism from E* to F* is surjective.

Nice  :)
But my solution is not  :(

Let $q$ be the order of $F$ and $n$ the degree of $E$ over $F$.
The (cyclic) Galois group of $E$ over $F$ is generated by $x \mapsto x^q$. Since the norm is the product of all conjugates, the norm is given by $N(x) = x^{1+q+q^2+...+q^{n-1}}$. But for any $a \in E^*$, the polynomial $N(x)-a$ has at most $1+q+q^2+...+q^{n-1}$ roots (finite fields are perfect), and $N(x)=0 \iff x=0$.
Let $Q \subset F^*$ be all norms that are archieved. Then $|Q| \geq \frac{E^*}{1+q+q^2+...+q^{n-1}} = \frac{q^n-1}{1+q+q^2+...+q^{n-1}}=q-1=|F^*|$, so $Q=F$.


Let $ ABC$  be triangle,with $ \angle A\equal{}30$ and $ BC\equal{}d$.Let $ \omega_A(A,\frac{d}{3})$,$ \omega_B(B,\frac{d}{3})$,$ \omega_C(C,\frac{d}{3})$, be a circles.Do there exist points $ A_1,A_2,A_3$,inside or one these circles,such that $ A_1A_2\equal{}A_3A_1\equal{}A_3A_2$,and $ A_1\in\omega_A$,$ A_2\in\omega_B$,$ A_3\in\omega_C$

It is one month left and still nobody solved this beautiful problem,so i'll post a solution:
Answer:Yes and there is unique such triangle.
Solution:WLOG assume that $ d\equal{}3$.
Let $ \zeta$ be the circumcircle of $ \triangle ABC$.Then its radius obviously equal to $ 3$.
It is well-known that there exist circle $ \omega(O,d_1)$,which is externally tangent to each of $ \omega_A,\omega_B,\omega_C$.
It is obviously that $ O$ coincides with circumcenter of $ \triangle ABC$ and hence $ d_1\equal{}3\minus{}1\equal{}2$.
Let us consider the locus $ P$ of the points,such that for every point $ X\in P$,there exist $ Y\in \omega_B$ and $ Z\in \omega_C$,such that $ XYZ$ is equaterial triangle.
At first we will turn every point of $ \omega_B$ to $ 60$ degree in a clockwise direction,w.r.t to every point of $ \omega_C$.
If we turn point $ B$ w.r.t  to every point inside of $ \omega$ to $ 60$ degree in a clockwise direction,then we get another circle $ \omega '$,which is equal to $ \omega_C$,and if its center $ O'$,then $ \angle O'BC\equal{}60$.So if we replace every point of $ \omega '$ with the circle,equal to $ \omega '$,we obtain $ \omega$.
So $ P\equiv\omega$.But $ \omega_A$ and $ \omega$ have a common point,moreover they have exactly one common point,so we are done.

[quote="Virgil Nicula"][color=darkblue]See the my lower topic for you, [b]Erken[/b] ![/color]

http://www.mathlinks.ro/Forum/viewtopic.php?p=1015349#1015349[/quote]
Oh,thank you much,if i have enough time i'll try to solve them.


Last year, each of Big Al's five brothers gave him a gift of money to Big Al. The dollar amounts were consecutive integers, and their sum was a perfect cube. If the brothers decide to give Big Al cash gifts this year and next year, and this years sum is larger than last years, and next year's sum still later, what is the least possible dollar amount Big Al could get next year from his 5 brothers combined?

[hide="Solution"]The sum of five consecutive integers must be divisible by $ 5$. Actually, any integer divisible by $ 5$ can be expressed as the sum of five consecutive integers. The first cube is $ 5^3$. The next year, he receives $ 6^3$ dollars, and the year after, he receives $ 7^3 \equal{} \boxed{343}$ dollars.

The above is assuming the condition does not hold for the next years. If the condition does hold, then the solution would be $ 15^3 \equal{} \boxed{3375}$ dollars.[/hide]

I think this was #6 on the recent California Math League (CAML) test.

LOL this is from NEAML. I did this 5 days ago  :P .

OK.

This is how I went about this problem:
(a) The dollar amounts were consecutive integers and there are 5 brothers. This is expressed as:                  (x-2)+(x-1)+x+(x+1)+(x+2) which simplifies to 5x.
(b) The sum has to be a cube. Knowing that 5x has to equal a cube for some "x" we try to dina a pattern.,
    (1) We know that 5x either ends in a 0 or 5. Hence, the cube must end in 0 or 5.
    (2) The least possible cube with these characteristics is 5^3 (woah..coincidence? :D) or 125. This is how   much Big Al got the previous year.
(c) We want to know the least dollar amount Big Al gets next year; so we can either do trial and error or just find a pattern.
     (1) The cubes of all multiples of 5 can fit into the divisible by 5x category.
(d) The above information leaves us with the least dollar amount Big Al can receive this year  is (10^3) and next year is (15^3) or:

                                   $ \boxed{3375 DOLLARS}$

[hide=" :>("]
BLEH beaten![/hide]

lolwut, this was also from the Ohio Math League #6. Apparently, separate competitions seem to share the same problems.

Solution is basically the same as others.

Google Math League. The same competition is held in different states/regions, but is named, scored, and compared among regions. I bet my friends in Maryland also encountered this question.


advanced fields
advanced fields unsolved

1. Brooks' theorem states that if $G$ is a connected graph then $\chi(G)\leq \triangle (G)$ unless $G$ is complete or is an odd cycle. Prove that theorem for 3-connected graph.

2. Let $G$ be a graph, and let $d_{1}+d_{2}=\triangle (G)-1$. By considering a partition $V_{1},V_{2}$ of $V(G)$ that minimizes the quantity $d_{2}e(G[V_{1}])+d_{1}e(G[V_{1}])$, show that there is a partition with $\triangle(G[V_{i}])\leq d_{i},i=1,2$.

3. By taking $d_{1}=3$, show that if a graph $G$ contains no $K_{4}$ then $\chi(G)\leq \frac{3}{4}\triangle(G)+\frac{3}{2}$

4. Where can I found a complete proof of the Brooks' theorem.

I'm new in graph theory. It would be nice if someone will say if the following proof for #3 is correct.

Suppose $G$ is a $K_{4}$-free graph, with max degree $D(G)$, and chromatic number $c(G)$. We claim that $c(G) \leq 3/4 \cdot D(G)+1$. We know that $G$ has an induced subgraph $H$, such that $D(H) \leq 3$, and $D(G\H) \leq D(G)-4$. Since $G$ is $K_{4}$-free, Brooks' theorem implies that $c(H)\leq 3$. So colour $H$ with $3$ colours.

It remains for us to colour $G\setminus H$. Note that $c(G\setminus H) \leq D(G)-4$ (by Brooks' theorem). But we can repeat the above argument on $G\setminus H$ : after using another $3$ (new) colours, we are left with a graph with chromatic number at most $D(G)-8$.

Repeat until we are left with a graph with chromatic number at most $d$, where $d\leq 4$. Finally, just use $d$ new colours to finish off.

This process will use $\frac{3}{4}(D(G)-d)+d$ colours, which is at most $\frac{3}{4}\cdot D(G)+1$. Thus $c(G) \leq \frac{3}{4}\cdot D(G)+1 < \frac{3}{4}\cdot [D(G)+2]$.


calculus
integration
geometry
derivative
function
calculus computations

The line $ y \equal{} mx$ has three intersection points with the curve $ y \equal{} |x(x \minus{} 1)|.$ Find the value of $ m$ such that the areas of two parts bounded by the line and the curve are equal.

The points of intersection are:

$ \plus{}x (x \minus{} 1) \equal{} m x$ $ \Longrightarrow$ $ A \equal{} (1\minus{}m,m(1\minus{}m) )$
$ \minus{}x (x \minus{} 1) \equal{} m x$ $ \Longrightarrow$ $ B \equal{} (1\plus{}m,m(1\plus{}m) )$

The 2 ares desired are:

$ S_1\equal{}\int_0^{1\minus{}m} \minus{}x (x\minus{}1) \, dx\minus{}\int_0^{1\minus{}m} m x \, dx\equal{}\minus{}\frac{1}{6} (m\minus{}1)^3$
$ S_2\equal{}\minus{}\int_{1\minus{}m}^1 \minus{}x (x\minus{}1) \, dx\minus{}\int_1^{m\plus{}1} x (x\minus{}1) \, dx\plus{}\int_{1\minus{}m}^{m\plus{}1} m x \, dx\equal{}m^2$

The areas are equal if and only if $ \minus{}\frac{1}{6} (m\minus{}1)^3\equal{}m^2$

By the substituition $ m\equal{} \minus{}(n\plus{}1)$ the equation simplifies to $ n^3 \equal{} \minus{}2$

So the Real solution for $ m$ is $ m\equal{} \minus{}1\plus{}\sqrt[3]{2}$

What's the range of $ m$ for which two curves have three intersection points? :wink:

This is very easy my friend.

One intersection is trivial $ (0,0)$

The other only can be on $ 0<x<1$ with the curve $ \minus{}x(x\minus{}1)$.

With that $ m$ cannot be negative, only positive. $ m>0$, because for $ m\equal{}0$ it would touch the graph at $ (1,0)$.

Now, the derivative of the function at $ 0$ need to be equal to maximum $ m$.

Why? I first made the limit of a point in the curve tending to $ (1,0)$ and now the limit of this point to $ (0,0)$.

I don't know how to express it with betters words, Is like those thinks you think in one second but to explain is more difficult.


Let  $ a,b,c,d$  positive integer numbers  such that$ ax_1 \plus{} b \equal{} cx_2 \plus{} d$.
Find minimum of  expression $ P \equal{} x_1 \plus{} x_2$ (  note :  where $ x_1,x_2$  are natural  roots)

For some a,b,c,d we do not have solutions

Such as  a=2, b=1, c=6, d=4


How did you guys do????!
I watched CD in my teachers room on her comp, and that CA kid was crazy, and i dont mean crazy smart, crazy rude. I don't like him. I laughed the entire CD bcuz of him... :P

I am mad at whoever graded Tonys test for grading it wrong!  :mad:

How was the team round?

Did anything hilarious or notable happen?

TELL ME EVERYTHING.

well ur going to have to wait 2 days cuz their going to disney...   i want to go to disney

btw did u really get a gift card for winning state cd?

Well, all of the top 12 made usamo, so they were all pretty smart. Mathcounts also has a lot to do with luck and how you feel on the day. Hope AL is at least in top 20.

New Mexico got most improved. :( :( :( If I remember correctly, they had a 6th grader named Joshua Brakenseik who made top 57 last year, I bet on him to make CD this year. Last year their team was 3 8th graders and a 6th grader... wonder what grades ppl were on there this year?

@yaofan yes i got 50 dollars to best buy for winning state cd last year. did you get anything this year other than a trophy?

?!?!?!?!?!?

that's unfair. i didn't get anything... who gave it to you?

btw thank you for taking out tony for me. i wish cd was official...

and i'm watching the webcast and cubex why are u running so fast? it's not really necessary... tony can't really keep up...

i don't remember what all of our scores were but,

Me-32 43rd place
rickie-22ish idk place
john-15-22 idk place
ash-15 like 180somthing place

we got a 6 on the team round and 32th...

wow gj chestnut...  can i have a copy of the test?

what place did u get last year?

good job on the big improvement from last year.

I don't think he got to keep a copy of the test.

edit: watching cd right now - that guest speaker had a pretty good speech.

I will shout at Ash accordingly (jk). Wow, good job Rickie. Awesome for a 6th grader. And Hooray Tony!!!! Top 57!!!

CONGRATULATIONS team AL. 11 place improvement is awesome.

Tony got (i think) a 32 and either 42-43rd place.
John got a 23 and I got a 22. We were in the 120's.
And Ash got a 19 (i think) and 160th place..idk

I got to hold the banner! Yay!  :D  :D  :D

Hooray... I was able to solve all of the problems Ash remembered :( I can't wait to take it when someone gets it....

i got 182 and DID NOT GET A 19.
they said i got a 15 but after i checked answers i knew i had at least a 17.  :(
the test was somewhat easy this year and i had a horrible day
i have looked all over the internet but cannot find a copy of the 2009 national mathcounts test

dynamo, this year's test was really easy compared to last year; I should've gotten at least 5-7 more points from careless mistakes (8 times 7 is 64 :blush: )

@ ash-Pokemon
Ash u know how I feel towards you right now. Haha jk u will do better next year if u rnt a state failure like yours truly.

@ chestnut
Grr... I know I would have done at least OK, ash remembered a spectacularly low amount of problems and I got them all... Don't beat yourself up about it, u got to nats. also, you did the best any AL person has done in a long time. and, u have all of high school to out-shine me even more :P

If any of u nats ppl remember questions from nats post them here with the answers hidden pleeeeease its the least you could do for failing old me. Why am I the only person who seems to remember every question on a test? :( ASH haha jk

[quote="dynamo729"]Why am I the only person who seems to remember every question on a test?[/quote]

wow. I could remember about 2 -3 questions an hour after taking mathcounts test. I suppose my memory is pretty bad.

and im really being serious and not exaggerating :(

edit: dynamo or ash, have either of you attended arml practices? How have they been? Hoover High school's school day ends at 3:27 for some reason, and I also do not have the ability to teleport, so I don't think I can make the practices.

Hah, no I haven't bee to an arml practice. Yeah Huntsville is out 27th because we missed two days because Madison pol had swine flu so we were out for two days.


logarithms
integration
trigonometry
real analysis
real analysis unsolved

$ \frac {( \minus{} 1)^n}{n(n^2 \plus{} 1)} \equal{} ( \minus{} 1)^n \left[ \frac1n \minus{} \frac {n}{n^2 \plus{} 1} \right]$

So your sum is equal to $ \minus{}\ln 2 \minus{} \sum\limits_{n \equal{} 1}^\infty \frac {( \minus{} 1)^n n}{n^2 \plus{} 1}$.

I have not been able to evaluate that second sum.

(edit: minus sign)

hello, Mathematica gives
$ \minus{}\log (2)\plus{}\frac{1}{4} \left(\minus{}\psi ^{(0)}\left(\frac{1}{2}\minus{}\frac{i}{2}\right)\minus{}\psi ^{(0)}\left(\frac{1}{2}\plus{}\frac{i}{2}\right)\plus{}\psi
   ^{(0)}\left(1\minus{}\frac{i}{2}\right)\plus{}\psi ^{(0)}\left(1\plus{}\frac{i}{2}\right)\right)$
Sonnhard.

$ \sum_{n\equal{}1}^{\plus{} \infty} \frac{(\minus{}1)^n n}{n^2\plus{}1} \equal{} \minus{} \int_0^{\plus{} \infty} \frac{\cos x}{1 \plus{} e^x}\,dx$.


number theory
greatest common divisor
relatively prime

prove that any two numbers of the following sequence are relatively prime:

2 + 1, 22 + 1, 24 + 1, 28 + 1, ... , 22^n + 1, ...

show that this result proves that there are an infinite number of primes.

I don't think I made any mistakes....but who knows.  My congruance skills are far from perfect.  I feel like something isn't right, so if someone would check this that'd be great.

[hide]

Suppose that there are two numbers 2^(2^a)+1=k*m and 2^(2^b)+1=k*n.

Obviously k is odd, because 2^(2^b)+1 is odd.

In modulo notation, this is:

2^(2^a) :equiv: -1 (mod k)

2^(2^b) :equiv: -1 (mod k)

Therefore:

2^(2^a) :equiv: 2^(2^b) (mod k)

2^(2^a-2^b) :equiv: 1 (mod k)

(2^(2^a))^(2^(b-a)-1) :equiv: 1 (mod k)

But since we know that 2^(2^a) :equiv: -1 (mod k), and GCD (2^(2^a),k)=1 because k is odd, we can change that last statement to

(-1)^(2^(b-a)-1) :equiv: 1 (mod k)

But since 2^(b-a)-1 is odd, this statement becomes

-1 :equiv: 1 (mod k)

Contradiction.

[/hide]

These are called the Fermat numbers, so let's denote them by Fn. Here's an easy way to prove that property:

Try to show (really easy) that F1*F2*..*Fn=Fn+1-2, and this shows that Fn+1 is coprime with Fi for all i from 1 to n, which is enough to get what we want.

I just realized I forgot part two of the question, about infinite primes.

Suppose that there are a finite number of primes, call this number n.
Take n+1 numbers in the form 2^(2^a)+1.  There must be at least n+1 prime factors here because they are all mutually prime.  We found an extra prime outside that set; contradiction.

Bump.  Will someone confirm my proof?  I just want to make sure it's all legit.

just a note, you didn't contradict anything, -1 :equiv: 1 (mod k) is only true for k = 1 or 2, and since you said k was odd, k = 1, and specifically say that k = the gcd of the two numbers.

Yes, as you say, Magnara's argument was a little incomplete near the end, but other than that, it looks like a sound proof.  In effect, he showed that 2^(2^b) - 1 is a multiple of 2^(2^a) + 1 for b > a.


induction
inequalities
algebra solved
algebra

Prove that the sequence (an)n>=1 given by a_(n+1)=n/an+1; (n>=1) and a1=1 is increasing.

cheers! :D :D

a2=1/a1+1 =1/2<a1!!
is it (an)+1 in the denominator?

hello!

a2=2>a1;
an+1 is not in the denominator, only "an"!!!
a_(n+1)=n/an+1; (n>=1) and a1=1

it is correct like I posted! :D :D

cheers! :D :D

I hope tha the following solution is correct. We will prove by induction that sqrt(n)<=a_n<sqrt(n)+1 for all n. For n=1 it is true. Suppose it is true for n. Proving that a_n+1<sqrt(n+1)+1 reduces to n/a_n+1<sqrt(n+1), or a_n>n/sqrt(n+1), which is true, since a_n>=sqrt(n)>n/sqrt(n+1). On the other hand, proving that a_n+1>sqrt(n+1) reduces to
n/a_n>= sqrt(n+1)-1=n/(1+sqrt(n+1)), which is true because a_n<1+sqrt(n)<1+sqrt(n+1). Hence the induction works and a_n<1+sqrt(n) for all n.
  But a_n+1>a_n is equivalent to 1+ n/a_n>a_n, or a_n<(1+sqrt(1+8n))/2, which is true for n>1 because a_n<1+sqrt(n)<(1+sqrt(1+8n))/2.

Little mistake, harazi.
Inequality an+1 > an is equivalent to an < (1+ \sqrt (1+4n) )/2 and  (1+ \sqrt (1+4n) )/2 < \sqrt n + 1.

Yes, you' re wright. Fortunately, as you have seen, the solution is not wrong.

Oh, boy, in fact my solution is quite wrong. I have completely lost my minds.

[quote="Lagrangia"]
a2=2>a1;
an+1 is not in the denominator, only "an"!!!
a_(n+1)=n/an+1; (n>=1) and a1=1
[/quote]

rather you should have written it as a_{n+1}=1+n/a_n.
and also since a_2=a_3=2 the sequence cannot be increasing.

Maybe you wanted it to be non-decreasing (the definition of an increasing sequence is a_{n+1} > a_n , and a non-decreasing sequence is a sequence for which a_{n+1} \geq a_n ).

The problem wants to show that a_(n+1)>=an.
but isn't it also like so: incresing means a_(n+1)>=an and strictly increasing:a_(n+1)>an?

cheers! :D

no. :) only if you directly (mot-a-mot in french) translate the wording into
Romanian.

and by the way, I looked at the problem, and did not find it that hard. I have written a complete solution, which you can access here: http://www.mathlinks.ro/diff/lagrangia%20soln.pdf

:D


OK, the problem is, I want to know much more about LaTeX packaging, but the Not So Short Introduction to LaTeX does not feature much about the topic.  Neither does the page "LaTeX" on AoPS, so I was wondering, where can I find a really detailed packaging pdf on LaTeX?\

No, I do not need help on how to enter a package into the next MiKTeX edition, just how to make one and all the commands necessary.

Writing packages is non-trivial and requires a good knowledge of TeX, not just LaTeX so you should read Donald Knuth's [url=http://www.amazon.co.uk/TeXbook-Donald-E-Knuth/dp/0201134489/sr=8-2/qid=1159825979/ref=sr_1_2/202-0137577-3200634?ie=UTF8&s=books]The TeXBook[/url] first and [url=http://www.amazon.co.uk/LaTeX-Companion-Frank-Mittelbach/dp/0201362996/sr=1-1/qid=1159826014/ref=sr_1_1/202-0137577-3200634?ie=UTF8&s=books]The Latex Companion[/url] (probably the best up-to-date book for advanced LaTeX users) has details about writing packages and classes.

Other places to look are:
[url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=writecls]Learning to write LaTeX classes and packages[/url].
You will also need to know how to prepare the distribution files .dtx and .ins and you can get help on these at:
[url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=dtxtut]How to Package Your LaTeX Package[/url]
and [url=http://www.tex.ac.uk/tex-archive/support/makedtx/manual.html]makedtx html version[/url] or [url=http://www.tex.ac.uk/tex-archive/support/makedtx/makedtx.pdf]makedtx pdf version[/url]

Not a task to be undertaken lightly.


Let $\{x_n\}$ be a sequence of natural numbers such that \[(a) 1 = x_1 < x_2 < x_3 < \ldots; \quad (b) x_{2n+1} \leq 2n \quad \forall n.\] Prove that, for every natural number $k$, there exist terms $x_r$ and $x_s$ such that $x_r - x_s = k.$

Conditions a) and b) cannot hold at the same time.

Accordingly to my IMO Problems book, 21-st IMO was in 1979 and 22-nd IMO was in 1981  :!:

Anyway, I think (b) is $x_{n+1}\leq 2n$. Then solution is the following.
Suppose the contrary. Consider numbers $x_{1}<x_{2}<...<x_{k+1}\leq 2k$. Suppose $x_{i}<k$ and $x_{i+1}\geq k$. Numbers $x_{1},...,x_{k+1}, x_{1}+k,...,x_{i}+k,x_{i+2}-k,...,x_{k+1}-k$ are pairwise distinct, they are in $\{1,2,...,2k\}$ and their count is $2k+1$. Contradiction.


geometry
trapezoid
quadratics
area of a triangle
Heron\u0027s formula

A convex pentagon inscribed in a circle has side length of 16, 16, 16, 16, and 6. What is the area of the pentagon?

Denote the pentagon $ABCDE$, $a = AB = BC = CD = DE = 16$, $c = EA = 6$. The quadrilateral $BCDE$ is an isosceles trapezoid with equal diagonals $d = BD = CE$ and the base $b = BE$. The quadrilateral $ACDE$ is also cyclic and because the triangle $\triangle ACE$ is isosceles, $AC = CE = d$. Using Ptolemy theorem for the cyclic quadrilaterals $BCDE$ and $ABCE$,

$a^2 + ab = d^{2}$
$ac + ad = bd$

Calculating the base b from the 2nd equation and substituting to the 1st one,

$b = \frac{a}{d(c + d)}$

$\frac{a^2}{d(c + 2d} = d^2$
$d^3 - 2a^2d - ac^2 = 0$

Substituting $a = 16, c = 6$, we get the following cubic equation for the diagonal $d$:

$d^3 - 2^9 d - 3 \cdot 2^9 = 0$

A cubic equation has constructible roots iff one root is rational. Perhaps this cubic equation has a rational root - it is unlikely that a problem leading to a cubic equation with non-constructible roots would be given. If it does, then (because the leading coefficient is one) this root must be an integer of the type $2^n$ or $3 \cdot 2^n$, where $ \le n \le 9$. Trying $3 \cdot 2^n$:

$27 \cdot 2^{3n}- 3 \cdot 2^{n+9}d - 3 \cdot 2^9 = 0$

Notice that $27 - 3 \cdot 2^3 = 27 - 24 = 3$. It is clear that for $n = 3$, the equation becomes a numerical identity. Hence, the equation has an integer root equal to $3 \cdot 2^3 = 24$. Factoring it out:

$d^3 - 2^9d - 3 \cdot 2^9 = (d - 24)(d^2 + 24d + 2^6) = 0$

The roots of the remaining quadratic factor, equal to $-12 \pm 4 \sqrt 5$, are both negative, i.e., not acceptable.

Divide the pentagon $ABCDE$ into 3 non-overlaping isosceles triangles $\triangle ABC, \triangle CDE, \triangle ACE$ with known sides $(a, a, d), (a, a, d), (c, d, d)$ and calculate the area of each by Heron's formula:

$S(\triangle ABC) = S(\triangle CDE) = 48 \sqrt 7$
$S(\triangle ACE) = 27 \sqrt 7$
$S(ABCDE) = S(\triangle ABC) + S(\triangle CDE) + S(\triangle ACE) = 123 \sqrt 7$

That is correct, yetti. That is the method of solution that I have. Not that you needed it, but the length of the other diagonal (the one you called $b$) is 20.


Find all positive interger $ n$ so that $ n^3\minus{}5n^2\plus{}9n\minus{}6$ is perfect square number.

[quote="tdl"]Find all positive interger $ n$ so that $ n^3 \minus{} 5n^2 \plus{} 9n \minus{} 6$ is perfect square number.[/quote]

[b]Case 1:[/b] $ 3|n$, then $ x^2\equal{}n^3 \minus{} 5n^2 \plus{} 9n \minus{} 6\equal{}3(mod 9)$, impossible.
[b]Case 2:[/b] $ n\equal{}1(mod 3)$, then $ x^2\equal{}2(mod 3)$, impossible.

[b]Case 3:[/b] $ n\equal{}3k\plus{}2$ for some non-negative integers $ k$.

Then $ x^2\equal{}3k(9k^2\plus{}3k\plus{}1)$

For $ k \neq 0$,
as $ gcd(3k,9k^2\plus{}3k\plus{}1)\equal{}gcd(3k,1)\equal{}1$, we get $ 9k^2\plus{}3k\plus{}1 \equal{}y^2$ is a square.

$ (3k)^2<9k^2\plus{}3k\plus{}1<(3k\plus{}1)^2$, impossible.

For $ k\equal{}0$, the solution given is $ n\equal{}2$, which is a solution after checking.

[hide]
$ n^3\minus{}5n^2\plus{}9n\minus{}6 \equal{} (n\minus{}2)(n^2\minus{}3n\plus{}3)$

$ \gcd (n\minus{}2,n^2\minus{}3n\plus{}3) \equal{} \gcd ( n\minus{}2, (n\minus{}2)(n\minus{}1)\plus{}1) \equal{} \gcd ( n\minus{}2,1) \equal{} 1$

Therefore we can say that $ n\minus{}2 \equal{} a^2, n^2\minus{}3n\plus{}3 \equal{} b^2$

Therefore $ n \equal{} a^2 \plus{} 2$ and $ (a^2\plus{}2)^2 \minus{}3(a^2\plus{}2) \plus{} 3 \equal{} a^4 \plus{} a^2 \plus{} 1 \equal{} b^2$

But $ (a^2\plus{}1)^2 \ge b^2 > a^2$

Therefore we have the solution $ a \equal{} 0$, that is $ n\equal{}2$ and no other solutions
[/hide]

The answer is $n=2$ only. It is easy to check that this works.

	Now, $n^3-5n^2+9n-6=(n-2)(n^2-3n+3)$. By the euclidean algorithm we see that $\gcd(n-2,n^2-3n+3)=1$, so we must have $n-2=a^2$ and $n^2-3n+3=b^2$ for some positive integers $a$ and $b$. Plugging $n=a^2+2$ in the second equation, we obtain
	\begin{align*}
		a^4+a^2+1&=b^2\Longleftrightarrow\\
		4a^4+4a^2+1+3&=4b^2\Longleftrightarrow\\
		(2a^2+1)^2-4b^2&=-3\Longleftrightarrow\\
		(2a^2-2b+1)(2a^2+2b+1)&=-3
	\end{align*}
	From here it is standard, and we get $a=0$, i.e. $n=2$ as the only solution.


Let $ a,b,c$ be real constants such that $ x^2\plus{}x\plus{}2$ is a factor of $ ax^3\plus{}bx^2\plus{}cx\plus{}5$,and $ 2x\minus{}1$ is a factor of $ ax^3\plus{}bx^2\plus{}cx\minus{}\frac{25}{16}$. Find $ a\plus{}b\plus{}c$.

Let the three roots of $ ax^3 \plus{} bx^2 \plus{} cx \plus{} 5 \equal{} 0$ be $ r_1,r_2,r_3$ and the roots of $ x^2 \plus{} x \plus{} 2 \equal{} 0$ be $ r_1,r_2$. Then by Vieta's, $ r_1 \plus{} r_2 \equal{} \minus{} 1$, $ r_1r_2 \equal{} 2$, $ r_1 \plus{} r_2 \plus{} r_3 \equal{} \minus{} \frac ba\implies r_3 \equal{} 1 \minus{} \frac ba$, $ r_1r_2 \plus{} r_3(r_1 \plus{} r_2) \equal{} 2 \minus{} r_3 \equal{} \frac ca\implies r_3 \equal{} 2 \minus{} \frac ca$, and $ r_1r_2r_3 \equal{} \minus{} \frac 5a\implies r_3 \equal{} \minus{} \frac5{2a}$, so $ ar_3 \equal{} a \minus{} b \equal{} 2a \minus{} c \equal{} \minus{} \frac52\implies(a,b,c) \equal{} \left(a,a \plus{} \frac52,2a \plus{} \frac52\right)$
$ \implies a \plus{} b \plus{} c \equal{} 4a \plus{} 5$.

From the second condition,
$ \frac a8 \plus{} \frac b4 \plus{} \frac c2 \minus{} \frac {25}{16} \equal{} 0\implies2a \plus{} 4b \plus{} 8c \equal{} 2a \plus{} (4a \plus{} 10) \plus{} (16a \plus{} 20)$
$ \equal{} 22a \plus{} 30 \equal{} 25\implies a \equal{} \minus{} \frac5{22}\implies a \plus{} b \plus{} c \equal{} 4a \plus{} 5 \equal{} \boxed{\frac {45}{11}}$.


Let $a$,$b$,$c$,$d$ be positive real numbers with $a^{2}+b^{2}+c^{2}+d^{2}=4$
Prove that the following inequality occurs: \[\frac{a^{3}}{2+3bcd}+\frac{b^{3}}{2+3acd}+\frac{c^{3}}{2+3abd}+\frac{d^{3}}{2+3abc}\geq\frac45\] And there is also a generalization.

$a^{3}/(2+3bcd)=a^{4}/(2a+3abcd)$
and then if you use Cauchy inequality,you can get it!

Yes.I told you it was easy..and the generalization is solvable in the same manner: \[\frac{a^{3}}{t+pbcd}+\frac{b^{3}}{t+pacd}+\frac{c^{3}}{t+pabd}+\frac{d^{3}}{t+pabc}\geq\frac4{t+p}\]

I don`t think I know what you mean.The inequality occurs for all positive real numbers $t$,$p$.

obviously, i was talking what happens if p is negative.

Wlog $t=1$. (At least $t$ must be positive) :wink: Putting $a=0$ we find that $p\ge\sqrt{3}-2$. From $b^{2}+c^{2}+d^{2}\le4$ we get $bcd\le\frac8{3\sqrt{3}}$. So for $\sqrt{3}-2\le p\le0$ we have $|pbcd|<1$, and the solution with CS still applies.
This means in general the ineq holds for $p\ge(\sqrt{3}-2)t$.  :)


Here is a fantastic inequality:

\[ \sum\limits_{i \equal{} 1}^n {\sqrt {\left( {x_i^2  \plus{} y_i^2 } \right)} }  \ge n\sqrt {\left( {\frac{{\sum\limits_{i \equal{} 1}^n {x_i } }}{n}} \right)^2  \plus{} \left( {\frac{{\sum\limits_{i \equal{} 1}^n {y_i } }}{n}} \right)^2 }
\]

:spider: :surf:

Draw on the coordinate plane the points $ (0,0)$, $ (x_1,y_1)$, $ (x_1\plus{}x_2,y_1\plus{}y_2)$, ..., $ (x_1\plus{}x_2\plus{}...\plus{}x_n,y_1\plus{}y_2\plus{}...\plus{}y_n)$. Now broken line joining these consecutive points has $ \ge$ length than straight line going from $ (0,0)$ straight to $ (x_1\plus{}x_2\plus{}...\plus{}x_n,y_1\plus{}y_2\plus{}...\plus{}y_n)$.

i know that solution. give an algebraic solution.

i presume holder or minkowski would give the result but i just can't put it together

[quote="HTA"][quote="apratimdefermat"]Here is a fantastic inequality:
\[ \sum\limits_{i \equal{} 1}^n {\sqrt {\left( {x_i^2 \plus{} y_i^2 } \right)} } \ge n\sqrt {\left( {\frac {{\sum\limits_{i \equal{} 1}^n {x_i } }}{n}} \right)^2 \plus{} \left( {\frac {{\sum\limits_{i \equal{} 1}^n {y_i } }}{n}} \right)^2 }
\]
:spider: :surf:[/quote]
hmm... i'm gonna need to take some kinda eyes exam  :roll:
\[ \sum\limits_{i \equal{} 1}^n {\sqrt {\left( {x_i^2 \plus{} y_i^2 } \right)} } \ge n\sqrt {\left( {\frac {{\sum\limits_{i \equal{} 1}^n {x_i } }}{n}} \right)^2 \plus{} \left( {\frac {{\sum\limits_{i \equal{} 1}^n {y_i } }}{n}} \right)^2 }
\]
$ \leftrightarrow \sum\limits_{i \equal{} 1}^n {\sqrt {\left( {x_i^2 \plus{} y_i^2 } \right)} } \ge \sqrt {n^2 (\left( {\frac {{\sum\limits_{i \equal{} 1}^n {x_i } }}{n}} \right)^2 \plus{} \left( {\frac {{\sum\limits_{i \equal{} 1}^n {y_i } }}{n}} \right)^2) } \equal{} \sqrt {\left( \sum\limits_{i \equal{} 1}^n {x_i } \right)^2 \plus{} \left( \sum\limits_{i \equal{} 1}^n {y_i } \right)^2 }$ and it's Minkowski 's inequality[/quote]

u see u said u needed an eye exam . ...but see me i have got blind. ..all the same thanks for reopening my eyes!


function
real analysis
real analysis solved

Here is a nice problem :)

Let f:R-->(0,oo) a function with the property that lim(x-->0) (f(x)+1/f(x))=2.
Show that lim(x-->0)f(x)=1.

Unknown source

Of course, we can find e>0 such that when |x|<e we have f(x)<3. Let eps>0 and let delta>0 such that when |x|<delta we have |f(x)+1/f(x)-2|<(eps)^2/3. Then for |x|<delta we will have |f(x)-1|^2=|f(x)|*|f(x)+1/f(x)-2|<3*(eps)^2/3=(eps)^2, so f has limit 1 in 0.


inequalities
LaTeX
Cauchy Inequality
inequalities proposed

I purpose a nice inequality, in atachement.Have a nice day!

this is just cauchy...
by cauchy inequality we can prove that:

$ \frac{a_{1}^{2}}{b_{1}}\plus{}...\plus{}\frac{a_{n}^{2}}{b_{n}}\geq\frac{(a_{1}\plus{}...\plus{}a_{n})^{2}}{b_{1}\plus{}...\plus{}b_{n}}$

now let $ n \equal{} 3$ and:

$ a_{1}\equal{}\sqrt{x^{2}\plus{}t}$
$ a_{2}\equal{}\sqrt{y^{2}\plus{}u}$
$ a_{3}\equal{}\sqrt{z^{2}\plus{}v}$
$ b_{1}\equal{}\sqrt{y^{2}\plus{}u}$
$ b_{2}\equal{}\sqrt{z^{2}\plus{}v}$
$ b_{3}\equal{}\sqrt{x^{2}\plus{}t}$

we get your inequality

You could just post here isn't it?
I'll do it,because i think it will be better,to read it here,without downloading. :wink:
$ \frac{x^{2}\plus{}t}{\sqrt{y^{2}\plus{}u}}\plus{}\frac{y^{2}\plus{}u}{\sqrt{z^{2}\plus{}v}}\plus{}\frac{z^{2}\plus{}v}{\sqrt{x^{2}\plus{}t}}\geq\sqrt{x^{2}\plus{}u}\plus{}\sqrt{y^{2}\plus{}u}\plus{}\sqrt{z^{2}\plus{}v}$.
[hide="Solution"]One line with Cauchy-Schwarz.

[/hide]

Or just Holder (but actually it is the same thing). I will detail:

\[ \left(\sum_{cyc}{\frac{(\sqrt{x^{2}\plus{}t})^{2}}{\sqrt{y^{2}\plus{}u}}}\right)\left(\sum_{cyc}{\sqrt{x^{2}\plus{}t}}\right)\left(\sum_{cyc}{\sqrt{y^{2}\plus{}u}}\right)\geq\left(\sum_{cyc}{\sqrt{x^{2}\plus{}t}}\right)^{3}.\]

Here it is my solution:(Sorry but i can't handle Latex)

soruz,
You can learn $ \text{\LaTeX}$ here http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php .  :wink:


The heights of a triangle is 12, 15 and 20 units. Whats the area of the triangle?

It is well known (and if you don't know it, it's very easy to prove) that $\frac{1}{r}=\frac{1}{h_{a}}+\frac{1}{h_{b}}+\frac{1}{h_{c}}$, so it follows that $r=5$

Now, use Heroon's formula:

\[S=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{\frac{S}{5}\left(\frac{S}{5}-\frac{S}{6}\right)\left(\frac{S}{5}-\frac{2S}{15}\right)\left(\frac{S}{5}-\frac{S}{10}\right)}\]
So this gives us:
\[S=S^{2}\sqrt{\frac{1}{5.30.15.10}}\]
So finally $S=150$.

[quote="yfehehey"][quote="Jan"]It is well known (and if you don't know it, it's very easy to prove) that $\frac{1}{r}=\frac{1}{h_{a}}+\frac{1}{h_{b}}+\frac{1}{h_{c}}$  [/quote]
how do we call this formula (if it has a name) and How can we prove it  :ewpu:[/quote]

Each side is equal to Semiperimeter/Area.

lol.. from the number, i realized this right away.. the triangle is a right triangle with sides 15 20 and 25 haha

A=150!

[quote="yfehehey"][quote="Jan"]$\frac{1}{r}=\frac{1}{h_{a}}+\frac{1}{h_{b}}+\frac{1}{h_{c}}$  [/quote]
how do we call this formula (if it has a name) and How can we prove it  :ewpu:[/quote]

It's hasn't name. Prove by using area formula:
$s=p.r=\frac{1}{2}a.h_{a}=\frac{1}{2}b.h_{b}=\frac{1}{2}c.h_{c}$

[quote="hien"]Prove by using area formula:
$s=p.r=\frac{1}{2}a.h_{a}=\frac{1}{2}b.h_{b}=\frac{1}{2}c.h_{c}$[/quote]
I got $\frac{1}{r.(a+b+c)}=\frac{1}{a.h_{a}}=\frac{1}{b.h_{b}}=\frac{1}{c.h_{c}}=\frac{3}{a.h_{a}+b.h_{b}+c.h_{c}}$
then multiplied by $(a+b+c)$ but still can't get it

I think I should return to the primary school or sleep at home :sleeping:           :D


I am a lot interested in Mathematics and ready to dedicate my whole life to it.
I am from India and want to know which college would be the best for me.
But the problem is that I've always scored only in Mathematics and not overall so my results in high school
are not really good.
I have won a Mental Mathematics Competition conducted in my city and I am very good at these calculations.
I might try for International Mental Calculation World Cup next year.
I could not participate in the Regional Mathematics Olympiad (First Round to be cleared in India) and thus don't have
any Olympiad Experience though I am still a lot interested in it (but cannot take part as I am 20 years old now).
I still practice problems and continuously trying to improve my problem solving techniques.
I think of mathematics day and night and thus I am ready to devote my life to it.
I mean it when i say this.

Is there anyone who can guide as to what should I do ?

Thanks

Go to the best college you can get into (in terms of quality, cost, everything) and don't worry too much. If you study hard and passionately and look out for opportunities in college, you can go on into a good graduate program.


geometry circumcircle incenter function inequalities

$ \Delta$  $ PQR$  is acute ; S,I,O are respectively the circumcenter , incenter, And orthocenter.
$ < QSR \equal{} \alpha$ ; $ < QIR \equal{} \beta$ ; $ < QOR \equal{} \gamma$ all measured in degrees

prove
\[ \frac {1}{\alpha } \plus{} \frac {1}{\beta } \plus{} \frac {1}{\gamma } > 1/45
\]
[hide="Solution!"]
$ < QSR \equal{} 2P$
$ < QIR \equal{} Q \plus{} R$
$ < QOR \equal{} P \plus{} Q/2 \plus{} R/2$
\[ \frac {1}{\alpha } \plus{} \frac {1}{\beta } \plus{} \frac {1}{\gamma } \equal{} 1/2P \plus{} 1/(Q \plus{} R) \plus{} 1/(P \plus{} Q/2 \plus{} R/2)
\]
$ 1/x$ is a convex function so by Jensen,
$ 1/2P \plus{} 1/(Q \plus{} R) > 2/((2P \plus{} Q \plus{} R)/2) \equal{} 4/(2P \plus{} Q \plus{} R)$
therefore , $ 1/2P \plus{} 1/(Q \plus{} R) \plus{} 1/(P \plus{} Q/2 \plus{} R/2) > 4/(2P \plus{} Q \plus{} R) \plus{} 2/(2P \plus{} Q \plus{} R) \equal{} 6/(2P \plus{} Q \plus{} R) \equal{} 6/(P \plus{} 180) > 6/(90 \plus{} 180) \equal{} 6/(270) \equal{} 1/45$
 $ Q.E.D$
[/hide]

But I wud like a synthetic solution !

Can you please complete the question? I don't understand what we have to prove.....

@Potla Sorry for the typo :blush:  i have edited! thanx for noticing

[hide]r u by any chance a Bengali , the name seems so , a very common Bengali nickname! I apologize :D  if u r not a bengali......[/hide]


[quote="zaya_yc"][quote="aev5peru"]If $ a,b,c,d \in R^ \plus{}$, where$ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 1$.
Demostrar que:
=>$ \sqrt a \plus{} \sqrt b \plus{} \sqrt c \plus{} \sqrt d \leq \sqrt x \plus{} \sqrt y \plus{} \sqrt z \plus{} \sqrt w$, where $ x \equal{} 1 \minus{} a$, $ y \equal{} 1 \minus{} b$, $ z \equal{} 1 \minus{} c$, $ w \equal{} 1 \minus{} d$.
Suerte=good luck>!![/quote]

http://reflections.awesomemath.org/2008_2/MR_2_2008_solutions[1].pdf[/quote]
I think that it's solutions are incorrect, because WLG $ a \leq b \leq c \leq d$, moreover, $ a^2\plus{}b^2\plus{}c^2\plus{}d^2\equal{}1$, then $ 1/2 \leq d$...
Then , $ (1\minus{}2d)/ (\sqrt d \plus{} \sqrt w) \leq (1\minus{}2a)/ \sqrt 2$.
My question is the next, when $ a$ is changed by $ d$, in the first solution the affiration :
$ (1\minus{}2a)/ \sqrt 2 \leq (1\minus{}2d)/ (\sqrt d \plus{} \sqrt w)$, is false...then how prove that the line 4 is true...
In the solution 2 ...well, I have not had time for revised, because is very large and the IMO is the next week...
In the solution 3....in the line 8 is false, because LHS $ < 8$, in change the other side is more big that 8 when $ d\equal{}63/64$...

Bye...Suerte!!
Nos vemos en la IMO! :D...

You cand find [url=http://www.mathlinks.ro/viewtopic.php?t=194791]here[/url] another approach.


A car holds exactly six people, but only two of those six people can drive the car. What is the number of ways that the six people can be seated in the car on a drive?

There are 2 people that can sit in the driver's seat. There are 5 choices for the next seat, 4 choices, 3 choices,  2 choices, and 1 person left. There are 2x5x4x3x2=240


Is anyone going?

What's the competition going to be like?

I'm going.  It's a really hard competition for the time limit, so be prepared.

you must be a genius then.  for ordinary people it's pretty difficult.  unless perhaps you took the junior exam.

I took the senior, and it's not overwhelmingly hard; I think two or three people from Vestavia (Cicatriz one of them) could definitely get 30/32 or more in under an hour...

It's... not too hard. But honestly, what kind of test gives you bonus problems for which you show your work??? I mean on olympiads you prove things that are actually worth proving, but why show your work on more-or-less routine problems?

And what's ciphering? Is that like deciphering secret codes?

Er, you've never been to a contest with ciphering?  Basically at you have a team of 4 people.  Each team has one person go up at first.  They display a problem on an overhead.  Everyone works on it at the same time and the first person to answer correctly (by passing a slip of paper to a proctor right in front of them) gets the most points.  Second gets a little less points and so on.  After some number of people correctly answering the point value becomes constant...  I don't remember the exact numbers for the points but basically you want to be one of the first people to answer.  So each person on your team gets like 5 problems or something and then the next person on your team goes down etc.

The bonus problems are just for breaking ties.

Last year I think I missed one problem on the test and didn't get to one tie-breaker.  Luckily I didn't tie anyone...the next 3 people or so tied (they missed 2 problems) but they all got all of the tie-breakers.  So  I guess they just went with the person who can write best...

The test isn't extremely hard...but I think that less than an hour is really fast.

I meant to ask: what good schools will we be competing with? I went three years ago so I remember a little about what it's like. Are you going, jb05?

nope
"The message is too small. Please make the message longer before submitting"


Hello. Is there any one who can post problems from Bulgarian Mathematical Olympiad  2003 and team selection tests. THAN S

Hey, can you post bulgaria 2002,2003, india 1999->2003, mathematical olympiads, problems & team selection test? Thanhks :D

I can tell you the hardest problem at 2003 Bulgarian Olympiad:
Find all polynomial p(x) with integer coefficients such that for any positive integer n there exists a integer m such that p(m)=2^n


Fran bought a pair of roller blades for $ \$100$ and sold them to Toni for a $ 10\%$ profit. Toni then sold the roller blades back to Fran for a $ 10\%$ loss. Fran then sold the roller blades to Gary for $ \$100$. What percent profit did Fran earn on the original $ \$100$?

Fran starts out at a net profit of $ \minus{}\$100$.
First transaction: Fran sells it for $ \$100\cdot1.1\equal{}\$110$ (net profit of $ \$10$).
Second transaction: Toni sells it for $ \$110\cdot0.9\equal{}\$99$ (net profit of $ \minus{}\$89$).
Third transaction: Fran sells it for $ \$100$ (net profit of $ \$11$).

Thus, there is an overall profit of $ \boxed{11\%}$.


abstract algebra
algebra
function
domain
Ring Theory
superior algebra
superior algebra unsolved

Suppose that $A$ is a commutativ noetherian ring and $I$ is a proper ideal of $A$. The ideal $I^n$ is the ideal generated by the products of $n$ elements of $I$. Show that $\cap_{n \geq 1} I^n=\{0\}$.

Generally, this is not true. Just think about the ring $\mathbb F_2\times\mathbb F_2$, where $\mathbb F_2$ is the field with two elements. Or any Noetherian ring containing a non-zero non-invertible idempotent element for that matter. Maybe you wanted your ring to be a domain (no zero divisors)?

P.S.

It's also true for local Noetherian rings, if you take your $I$ to be the unique maximal ideal.

:blush: , of course I forgot to add the hypothesis that there are no zero divisors...


conics
ellipse
hyperbola
geometry
circumcircle
ratio
geometric transformation

Relabel the given point to construct another tangent from as $ J.$

Label the lines $ a, b, c, d, e$ and their intersections $ A \equiv a \cap b,$ $ B \equiv b \cap c,$ $ C \equiv c \cap d,$ $ D \equiv d \cap e,$ $ E \equiv e \cap a.$ Using Brianchon theorem for the pentagon $ ABCDE$ tangential to the conic, find the tangency points $ A_t, B_t, C_t, D_t, E_t$ with $ AB, BC, CD, DE, EA.$ That is, diagonals $ AC, EB$ from $ A, B$ meet at $ A'$ and $ DA'$ cuts $ AB$ at its tangency point $ A_t.$ The tangency points $ B_t, C_t, D_t, E_t$ are constructed in the same way. Given five points $ A_t, B_t, C_t, D_t, E_t$ of a conic (using the same notation), the conic tangents $ a, b, c, d, e$ at these points can be constructed similarly, using Pascal theorem.

If the pentagon $ ABCDE$ is convex, the inscribed conic is an ellipse. However, if all tangency points $ A_t, B_t, C_t, D_t$ lie on a semi-ellipse or on one hyperbola branch, the conic type may not be obvious. To decide, if the conic is an ellipse or a hyperbola, construct the circumcircle $ (O_t)$ of the $ \triangle A_tB_tC_t$ and find isogonal conjugate points $ D_t^*, E_t^*$ of $ D_t, E_t$ with respect to this triangle. If the line $ D_t^*E_t^*$ cuts $ (O_t)$ at two points, the conic has two points at infinity and it is a hyperbola. If $ D_t^*E_t^*$ does not cut $ (O_t),$ all points of the conic are finite and it is an ellipse. To construct just another tangent from $ J$ is a slim result, we can construct the conic main axes and foci. Construction of another conic tangent from the point $ J$ is then trivial.

[color=blue]Ellipse.[/color] Let $ A_m, B_m$ be midpoints of the ellipse chords $ E_tA_t, A_tB_t.$ The lines $ a_m \equiv AA_m, b_m \equiv BB_m$ meet at the ellipse center $ I,$ because the ellipse $ \mathcal K$ inscribed in $ ABCDE$ can always be projected to an incircle of $ A'B'C'D'E'$ by a parallel projection, which preserves segment ratios on the corresponding lines. Since $ a_m$ cuts the ellipse chord $ E_tA_t$ at its midpoint, it also cuts any other ellipse chord parallel to $ E_tA_t$ at its midpoint. Let $ a_i$ be parallel to $ E_tA_t$ through $ I.$ Then $ a_m, a_i$ are directions of two conjugate diameters of the ellipse.

Given the ellipse center $ I,$ the directions of two conjugate diameters $ a_m, a_i$ and an ellipse tangent $ a$ with its tangency points $ A_t,$ we can construct endpoints $ U_1, U_2$ and $ V_1, V_2$ of the conjugate diameters with these directions. The parallel to the conjugate diameter $ a_i$ through $ A_t$ (this is $ E_tA_t$) cuts $ AI$ at $ A_m$ and $ IU_1^2 \equal{} IA \cdot IA_m.$ Again, parallel projection of the ellipse to a circle makes $ E_t'A_t' \perp I'A',$ $ I'U_1'$ becomes radius of the projected circle, and $ A_m'$ is inversion of $ A'$ is this circle. The other end of the diameter with the direction $ a_m$ is the reflection $ U_2$ of $ U_1$ in $ I.$ In exactly the same way, we can construct endpoints $ V_1, V_2$ of the conjugate diameter with the direction $ a_i.$

Given two conjugate diameters $ U_1U_2, V_1V_2$ of the ellipse $ \mathcal K,$ we can construct both lengths and directions of its main axes $ p \equal{} P_1P_2, q \equal{} Q_1Q_2$: Let a perpendicular to the conjugate diameter $ V_1V_2$ at $ I$ cut the circle centered at $ I$ with radius $ IV_1 \equal{} IV_2$ at $ W_1, W_2,$ so that $ IW_1 \equal{} IW_2 \equal{} IV_1 \equal{} IV_2.$ Let $ M$ be the midpoint of $ U_1W_1,$ let $ \mathcal M$ be a circle with center $ M$ and radius $ MI,$ and let $ XY$ be its diameter through $ U_1, M, W_1.$ Then $ IX, IY$ are the directions of the ellipse major and minor axes $ P_1P_2, Q_1Q_2$ and $ \frac {_p}{^2} \equal{} XW_1 \equal{} YU_1,$ $ \frac {_q}{^2} \equal{} XU_1 \equal{} YW_1.$

[hide="Lengths of the main axes."] By definition, if $ u \equal{} U_1U_2$ is any ellipse diameter through its center $ I,$ any chord parallel to $ u \equal{} U_1U_2$ is cut in half by its conjugate diameter $ v \equal{} V_1V_2.$ When the ellipse is projected to a circle by parallel projection, the ellipse conjugate diameters are projected to perpendicular diameters of this circle. Let $ \mathcal K'$ be the ellipse pedal circle centered at $ I$ with radius $ \frac {_p}{^2}.$ Projecting $ U_1, V_1$ to $ U_p, V_p$ on the major axis and to $ U_1', V_1'$ on the pedal circle, circle, the right $ \triangle IU_pU_1' \cong \triangle V_1'V_pI$ with $ IU_1' \equal{} V_1'I = \frac {_p}{^2}$ and the corresponding sides perpendicular are congruent by ASA, hence $IU_p = V_1'V_p, IV_p = U_1'U_p$ and

$ \frac {u^2 \plus{} v^2}{4} \equal{} IU_1^2 \plus{} IV_1^2 \equal{} IU_p^2 \plus{} U_1U_p^2 \plus{} IV_p^2 \plus{} V_1V_p^2 \equal{}$

$ \equal{} IU_p^2 \plus{} \frac {q^2}{p^2}\ U_1'U_p^2 \plus{} IV_p^2 \plus{} \frac {q^2}{p^2}\ V_1'V_p^2 \equal{}$

$ \equal{} \left(1 \plus{} \frac {q^2}{p^2}\right) IU_p^2 \plus{} \left(1 \plus{} \frac {q^2}{p^2}\right) U_1'U_p^2 \equal{}$

$ \equal{} \left(1 \plus{} \frac {q^2}{p^2}\right) IU_1'^2 \equal{} \frac {p^2 \plus{} q^2}{4}$

When projecting a circle into ellipse by a parallel projection, any square circumscribed around the circle is projected to a parallelogram circumcsribed around the ellipse, with a pair of conjugate diameters as its midlines. Areas of all these parallelograms are equal, $ uv\ \sin \widehat{uv} \equal{} pq.$ By the cosine theorem for the $ \triangle IU_1W_1,$

$ U_1W_1^2 \equal{} IU_1^2 \plus{} IW_1^2 \minus{} 2\ IU_1 \cdot IW_1\ \cos \widehat{U_1IW_1} \equal{}$

$ \equal{} \frac {u^2 \plus{} v^2 \mp 2uv\ \sin \widehat{uv}}{4} \equal{} \frac {p^2 \plus{} q^2 \mp 2pq}{4}$

$ U_1W_1 \equal{} \frac {p \mp q}{2}$

The upper resp. lower sign is for an acute resp. obtuse $ \angle U_1IW_1.$ $ IM \parallel W_1U_2$ is midline of the $\triangle U_1W_1U_2$ $\Longrightarrow$

$ IM \equal{} \frac{U_1W_2}{2} \equal{} \frac {p \pm q}{4}.$[/hide]

[hide="Directions of the main axes."]Lengths $ p, q$ of the ellipse main axes are already known. We have to show that $ U_1, V_1$ lie on an ellipse with center $ I$ and main axes in the directions $ IX, IY.$ The $ \triangle IMX$ is isosceles with $ IM \equal{} XM,$ hence parallels to $ IY, IX$ through $ M$ bisect the angle $ \angle IMX$ internally resp externally. If we extend $ IM$ by the distance $ MW_1$ into $ IU_1',$ then the quadrilateral $ IXU_1'W_1$ is an isosceles trapezoid with equal diagonals $ IU_1' \equal{} XW_1 \equal{} \frac {_p}{^2}$ and $ W_1U_1' \parallel IX.$ Perpendicular to the conjugate diameter $ V_1V_2$ (parallel to $ IW_1$) through $ U_1$ cuts $ IM$ at $ K$ and the parallel to $ U_1W_1$ through $ I$ at $ L.$ $ IU_1KW_1$ is a parallelogram, because $ U_1K \parallel IW_1$ and the diagonal $ IK$ cuts the other dioagonal $ U_1W_1$ in half at $ M.$ $ MU_1 \parallel IL$ is a midline of the $ \triangle IKL,$ therefore, $ U_1$ is the midpoint of $ KL$ and $ U_1L \equal{} U_1K \equal{} IW_1.$ Parallel to $ IX$ through $ L$ cuts $ IM$ extended at $ N.$ Since the line $ IY \perp LN$ bisects the angle $ \angle LIN,$ it cuts $ LN$ at its midpoint $ O.$ Then $ OU_1 \parallel (NK \equiv IU_1')$ is a midline of the $ \triangle KLN.$ Consequently, the $ \triangle IW_1U_1' \cong \triangle U_1LO$ are congruent, having the corresponding sides parallel and $ U_1L \equal{} IW_1.$ This means that $ OU_1U_1'I$ is a parallelogram, $ U_1U_1' \parallel (OI \equiv IY)$ and in addition, $ IU_1' \equal{} XW_1 \equal{} \frac {_p}{^2}.$ The ellipse conjugate radius $ IU_1$ therefore projects to the pedal circle radius $ IU_1'$ in the direction $ IY.$

Let $ V_1'$ be projection of the conjugate diameter enpoint $ V_1$ to the pedal circle $ \mathcal K'.$ Then $ IV_1' \equal{} IU_1' \equal{} \frac {_p}{^2},$ $ IV_1 \equal{} IW_1 \equal{} \frac {_v}{^2},$ $ IV_1 \perp IW_1,$ $ V_1V_1' \perp W_1U_1',$ so that the $ \triangle IV_1V_1' \cong IW_1U_1'$ are congruent by Ssa and $ IU_1' \perp IV_1'.$ As a result, the ellipse conjugate radii $ IU_1, IV_1$ project to the pedal circle perpendicular radii $ IU_1' \perp IV_1'$ in the direction parallel to $ IY,$ which means that $ IX, IY$ are the directions of the ellipse major and minor axes. Obviously,

$ XW_1 \equal{} IM \plus{} MW_1 \equal{} IM \plus{} \frac {U_1W_1}{2} \equal{} \frac {p \pm q}{4} \plus{} \frac {p \mp q}{4} \equal{} \frac {p}{2}$

$ XU_1 \equal{} |IM \minus{} MU_1| \equal{} \left|IM \minus{} \frac {U_1W_1}{2}\right| \equal{} \left|\frac {p \pm q}{4} \minus{} \frac {p \mp q}{4}\right| \equal{} \frac {q}{2}$
[/hide]

I may do a hyperbola later, the construction up to a pair of conjugate diameters is similar and the rest is even simpler than for an ellipse.

[color=blue]Hyperbola.[/color] Let $ A_m, B_m$ be midpoints of the hyprtbola chords $ E_tA_t, A_tB_t.$ The lines $ a_m \equiv AA_m, b_m \equiv BB_m$ meet at the hyperbola center $ I,$ because the $ \triangle IE_tA_t$ or the $ \triangle IA_tB_t$ can alway be projected to an isosceles $ \triangle IE_t'A_t'$ or $ \triangle IA_t''B_t''$ by parallel projections, which preserve segment ratios on the corresponding lines. By symmetry, $ A'A_m'$ then passes through $ I,$ hence $ AA_m$ also passes through $ I.$ Since $ a_m$ cuts the hyperbola chord $ E_tA_t$ at its midpoint, it also cuts any other hyperbola chord parallel to $ E_tA_t$ at its midpoint. Let $ a_i$ be parallel to $ E_tA_t$ through $ I.$ Then $ a_m, a_i$ are directions of two conjugate diameters of the hyperbola. Only one of the lines $ a_m, a_i$ can intersect the given hyperbola $ \mathcal K$ (with the tangents $ a, b, c, d, e$) at the enpoints of one conjugate diameter, the other line intersects its conjugate hyperbola $ \mathcal K^*$, i.e., the hyperbola with the major and minor axes exchanged.

Given the hyperbola center $ I,$ the directions of two conjugate diameters $ a_m, a_i$ and a hyperbola tangent $ a$ with its tangency points $ A_t,$ we can construct endpoints $ U_1, U_2$ and $ V_1, V_2$ of the conjugate diameters with these directions. Assume that $ a_m$ cuts the given hyperbola $ \mathcal K$ and $ a_i$ its conjugate hyperbola $ \mathcal K^*.$ The parallel to the conjugate diameter $ a_i$ through $ A_t$ (this is $ E_tA_t$) cuts $ AI$ at $ A_m$ and $ IU_1^2 \equal{} IA \cdot IA_m.$ Again, the $ \triangle IE_tA_t$ can be projected to an isosceles $ \triangle IE_t'A_t'$ by a parallel projection, which projects its conjugate diameter $ U_1U_2$ in the direction $ a_i$ into the major axis $ U_1'U_2'$ of the projected hyperbola. If equation of the projected hyperbola is $ \frac {x^2}{p^2} \minus{} \frac {y^2}{q^2} \equal{} 1,$ the equations of its tangents at $ E_t' \equal{} (x_{et}, y_{et}), A_t' \equal{} (x_{at}, y_{at})$ are $ \frac {x_{et}x}{p^2} \minus{} \frac {y_{et}y}{q^2} \equal{} 1,$ $ \frac {x_{at}x}{p^2} \minus{} \frac {y_{at}y}{q^2} \equal{} 1.$ By symmetry, $ x_{et} \equal{} x_{at}, y_{et} \equal{} \minus{} y_{at}$ and $ A_m' \equal{} (x_{at}, 0).$ Solving for the x-coordinate $ x_a$ of their intersection $ A'$, $ \frac {x_{at}x_a}{p^2} \equal{} 1$ or $ IA' \cdot IA_m' \equal{} IU_1'^2.$ The other end of the diameter with the direction $ a_m$ is the reflection $ U_2$ of $ U_1$ in $ I.$ In exactly the same way, we can construct endpoints $ V_1, V_2$ of the conjugate diameter with the direction $ a_i.$

Given two conjugate diameters $ U_1U_2, V_1V_2$ of the hyperbola $ \mathcal K,$ we can construct both lengths and directions of its main axes $ p \equal{} P_1P_2, q \equal{} Q_1Q_2$: Let $ u_1, u_2$ be parallels to the conjugate diameter $ V_1V_2$ through $ U_1, U_2$ and let $ v_1, v_2$ be parallels to the conjugate diameter $ U_1U_2$ through $ V_1, V_2.$ The lines $ u_1, v_1, u_2, v_2$ form a parallelogram $ KLMN$ with $ K \equiv v_2 \cap u_1,$ $ L \equiv u_1 \cap v_1,$ $ M \equiv v_1 \cap u_2,$ $ N \equiv u_2 \cap v_2$ with the conjugate diameters $ U_1U_2, V_1V_2$ as its midlines, i.e., $ U_1, V_2, U_2, V_2$ are midpoints of $ KL, LM, MN, NK.$ Since $ u_1, u_2$ are tangents of the given hyperbola $ \mathcal K$ at $ U_1, U_2,$ while $ v_1, v_2$ are tangents to the conjugate hyperbola $ \mathcal K^*$ at $ V_1, V_2,$ the diagonals $ KM, LN$ of the parallelogram $ KLMN$ are their common asymptotes. The hyperbola main axes bisect the angles between its asymptotes. Any tangent to the given hyperbola $ \mathcal K$ has its tangency point in the plane quadrant bisected by the major axis. The hyperbola tangent $ a$ with its tangency point $ A_t$ is the internal bisector of the angle $ \angle F_2A_tF_1,$ where $ F_1, F_2$ are the hyperbola foci. Thus $ a$ intersects the perpendicular bisector of $ F_1F_2,$ the hyperbola minor axis, at $ X_a$ on the circumcircle $ \mathcal O_a$ of the $ \triangle A_tF_1F_2.$ The perpendicular bisector of $ A_tX_a$ cuts the minor axis at the circumcenter $ O_a$ and a circle centered at $ O_a$ with radius $ O_aA_t \equal{} O_aX_a$ cuts the hyperbola major axis at $ F_1, F_2.$ Length of the hyperbola major axis is given by $ p \equal{} P_1P_2 \equal{} |P_1F_2 \minus{} P_1F_1| \equal{} |A_tF_2 \minus{} A_tF_2|.$

Here is another construction that works for any conic.

Denote by $a,b,c,d,e$ the given tangents. $a,b,c$ cut $d$ at $A,B,C,$ respectively and $a,b,c$ cut $e$ at $A',B',C',$ respectively. $\{A,B,C \} \mapsto \{A',B',C' \}$ is then a homography between $d$ and $e$ inducing a homography betwen the superposed pencils $P(A,B,C)$ and $P(A',B',C').$ Hence tangents from $P$ to the desired conic are the double rays in the referred homography. That is, draw arbitrary circle $\omega$ through $P$ intersecting the pencils again at two circular series, their homography axis cuts $\omega$ at $U,V$ $\Longrightarrow$ lines $PU,PV$ are the desired tangents.


Eight women of different heights are at a party. Each woman decides to only participate in a handshake with women shorter than herself. How many handshakes take place?

There's no way for a woman to shake hands with someone shorter than herself, because one person will always be taller. Therefore, our answer is $ \boxed{0}$.

This has always been one of my favorite problems. Seen it lots of times on FTW. I always get it correct in 1 second, but unfortunately there are always ppl in my game who get it quicker.

Think of it this way, who will the shortest person shake hands with? This thought step makes it much easier to understand.


inequalities
number theory unsolved
number theory

Let n be a positive integer >1. let d(n) be the sum of the divisors of n.

prove there do not exist a rational q such that d(n)=n^q.

I'm not sure I understand this: what if $p_2^{e_2}|\frac{p_1^{e_1+1}-1}{p_1-1}$, for example?

[quote="jpsteenstra"]The sum of the divisors of $n$ can be written as
$\displaystyle d(n)=(p_1^{e_1}+p_1^{e_1-1}+\cdots+1)(p_2^{e_2}+p_2^{e_2-1}+\cdots+1)\cdots(p_k^{e_k}+p_k^{e_k-1}+\cdots+1)$
$\displaystyle=\left(\frac{p_1^{e_1+1}-1}{p_1-1}\right)\left(\frac{p_2^{e_2+1}-1}{p_2-1}\right)\cdots\left(\frac{p_k^{e_k+1}-1}{p_k-1}\right)$.
Since $d(n)$ contains the factor $p_k^{e_k}$, there exist no power of $n$ is equal to $d(n)$,[/quote]

This is clearly wrong - it works only for $k$ at most $2$, but it is not true when we have at least $3$ distinct prime factors! Interesting I put myself the same problem few months ago, and I did not know that it was a problem given to some contest.

[quote="Myth"]I am really confused. :oops:
Suppose $q=b/a$, then $n=p_1^{k_1a}...p_l^{k_la}$, so
\[d(n)=\frac{p_1^{k_1a+1}-1}{p-1}...\frac{p_l^{k_la+1}-1}{p-1},\]
but $\frac{p_1^{k_1a+1}-1}{p-1}<p_1^{k_1a+1}\leq p_1^{k_1b}$, since undoubtedly $b>a$. Therefore $d(n)<n^{b/a}$.

What is wrong?  :blush:[/quote]

Nothing is wrong!  :)  This is the short proof of this fact, I gave the same solution some time ago...
Apparently, you proved that $d(n)<n^{b/a}$ for any positive integers $n,a,b$, $1\leq a<b$, $GCD(a,b)=1$, and "the contradiction" seems to arise when $b/a$ decreases to $1$...
Indeed if $d(n)=n^{b/a}$ then $n^{b/a}$ is an integer and $n$ must be of the form $n=m^a$ for some integer $m$.
As you showed the following inequality:
$d(m^a)<m^b$
holds for any positive integers $m>1$, $a,b$, $1\leq a<b$, $GCD(a,b)=1$.
This is true all the time! No matter if we increase $a$ as much as we want and even we consider $b=a+1$!
Indeed for any $p$-prime, $a,b,k$ positive integers, $1\leq a<b$:
$\frac{p^{ka+1}-1}{p-1}\leq p^{ka+1}-1<p^{ka+1} \leq p^{k(a+1)} \leq p^{kb}$,
since $ka+1 \leq k(a+1) \leq kb$.
We have:
$d(m^a)<m^{a+1} \leq m^b$
You don't have to be "confused" anymore!  :)

[quote="heartwork"][quote="Myth"]I am really confused. :oops:
Suppose $q=b/a$, then $n=p_1^{k_1a}...p_l^{k_la}$, so
\[d(n)=\frac{p_1^{k_1a+1}-1}{p-1}...\frac{p_l^{k_la+1}-1}{p-1},\]
but $\frac{p_1^{k_1a+1}-1}{p-1}<p_1^{k_1a+1}\leq p_1^{k_1b}$, since undoubtedly $b>a$. Therefore $d(n)<n^{b/a}$.

What is wrong?  :blush:[/quote]

Apparently, you proved that $d(n)<n^{b/a}$ for any positive integers $n,a,b$, $1\leq a<b$, $GCD(a,b)=1$[/quote]
Just remark: I proved this inequality in only case when $n$ is an $a$-th power. For other cases we will encounter some problems. ;)

[quote="Myth"]Just remark: I proved this inequality in only case when $n$ is an $a$-th power. For other cases we will encounter some problems. ;)[/quote]

Since $n^{b/a}$ is an integer for some positive integers $n>1$, $a,b$, $1\leq a<b$, $GCD(a,b)=1$ then $n$ is an integer at power $a$, so I don't need to prove the inequality for $n$ not a power of $a$.
For other cases something similar does not hold!
Counterexample: $n=2$, $a=2$, $b=3$, $d(2)=3$ but $3 > 2^{3/2}$.
I might take some $n_1$ instead of $n$ which is not a power of $a$, for some fixed $a,b$, $1\leq a<b$, $GCD(a,b)=1$, $n_1$ large enough in order to have the reverse: $d(n_1) > {n_1}^{b/a}$.
Just think deeper a little bit! That's it!

An interesting question and somehow related to Ramanujan "superabundant numbers" is:
For what pairs of positive integers $a,b$, $1\leq a<b$, $GCD(a,b)=1$ we have:
$d(n)<n^{b/a}$,
for all positive integers $n>1$ or equivalent to find the supremum of:
$log_n {d(n)}$, for any positive integer $n>1$!

I think it works for all $a,b$, with $1\leq a<b$ and $(log_2 {3}) a < b$.


The Fredholm alternative states that if $T$ is a compact operator and $\lambda \ne 0$, then
(a) $\lambda I-T$ is injective if and only if $\overline{\lambda}I-T^{*}$ is injective
(b) $\lambda I-T$ is injective if and only if $\lambda I-T$ is surjective.

This is useful: note that [b]grobber[/b] used the result [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=fredholm&t=89706]here[/url] some time ago to good effect.

Anyone care to prove it? :D

Here's a generalization:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=operator+compact&t=100570[/url]


Let $ P$ be a polynomial with positive coefficients. Prove that if $ P\left(\frac{1}{x}\right)\geq \frac{1}{P(x)}$ holds for $ x\equal{}1$, then it holds for every $ x>0$.

See http://www.artofproblemsolving.com/Forum/viewtopic.php?p=17618#17618. This is an old problem that has been posted several times.


Let $x;y;z > 0$ and $xy+xz+yz=1$
Prove
$\frac{27}{4}(x+y)(y+z)(z+x)\geq (\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x})^{2}\geq 6\sqrt{3}$

LHS:

By Cauchy, we have $(\sqrt{x+y}+\sqrt{x+z}+\sqrt{z+y})^{2}\leq 6(x+y+z) \leq 6(x+y+z)(xy+yz+zx) = 6\sum{x^{2}y}+2\sum{xyz}$

by bunching, this is $\leq \frac{27}{4}(\sum{x^{2}y}+\frac{9}{4}\sum{xyz})$

RHS:

$(\sqrt{x+y}+\sqrt{x+z}+\sqrt{z+y})^{2}$
$\geq 3(\sqrt{(x+y)(x+z)}+\sqrt{(x+y)(y+z)}+\sqrt{(z+y)(x+z)})$
$= 3\sum{\sqrt{x^{2}+1}}$

now, use $x = cotg(a)$ and the reciprocals, and use jensen!

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=508186#p508186]see here[/url]  :D


Let $ O$ be the point of intersection of the diagonals $ AC$ and $ BD$ of the quadrilateral $ ABCD$ with $ AB \equal{} BC$ and $ CD \equal{} DA$. Again, let $ N$ and $ K$ be the feet of perpendiculars from $ D$ and $ B$ to $ AB$ and $ CD$, respectively.
Prove that the points $ N$, $ O$, and $ K$ are collinear.

$ AB\equal{}BC$, $ CD\equal{}DA \Longrightarrow AC \perp BD$. So $ D$, $ O$, $ A$ and $ N$ are concyclic. Hence $ \angle{DON}\equal{}\angle{DAN}$.

Similarly $ B$, $ O$, $ C$ and $ K$ are concyclic $ \Longrightarrow \angle{BOK}\equal{}\angle{BCK}$.

Since $ \triangle{ABD} \cong \triangle{CBD} \Longrightarrow \angle{BAD}\equal{}\angle{BCD} \Longrightarrow \angle{DAN}\equal{}\angle{BCK}$, so $ \angle{DON}\equal{}\angle{BOK} \Longrightarrow N, O, K$ are collinear.

I have almost same solution, so I am not going to post it. BTW the problem statement was not correct (Now it is OK)
Here goes the diagram.


$AB$ is a chord of a circle with center $O$, $M$ is the midpoint of $AB$. A non-diameter chord is drawn through $M$ and intersects the circle at $C$ and $D$. The tangents of the circle from points $C$ and $D$ intersect line $AB$ at $P$ and $Q$, respectively. Prove that $PA$ = $QB$.

Let $X =DQ\cap CP$ and $Y= AB\cap BC$. Then line $XY$ is polar to $M$, so $XY\bot OM$. Since $OM\bot PQ$ we have $XY||PQ$.
Let $S=DC\cap XY$. Now:
$(Q,P;M,\infty)=(XQ,XP;XM,XY)=(XD,XC;XM,XY)=(D,C;M,S) =(YD,YC;YM,YS)=
(YA,YB;YM,YS)= (A,B;M,\infty)=-1$
So $M$ on the middle of segment $PQ$ and conclusion folows.

Zhaoli could you post a solution with a picture of the problem please

Why? A solution has been posted already, and it's not hard to draw the picture yourself.

[quote="silouan"]Please maybe zaoli has another solution[/quote]

it is easy to draw a figure by yourself!

Solution 1 (From zhaobin)
Since $D, O, M, Q$ are concyclic, So $DQ = \frac{OD}{\sin \angle OMD} \cdot \sin \angle DMQ.$ (sine law)
Similarly, $CP = \frac{OC}{\sin \angle OMC} \cdot \sin \angle CMP.$
While $\angle OMD + \angle OMC = \pi$, $\angle DMQ = \angle CMP$, $OC = OD$, so $CP = DQ.$
$PB \cdot PA = CP^2 = DQ^2 = QA \cdot QB.$
While $PB \cdot PA = (PM - MB)(PM + MB) = PM^2 - MB^2, QB \cdot QA = (QM + MA)(QM - MA) = QM^2 - MA^2.$
So $PM = QM$, which means $QA = PB$.

Solution 2 (From lin)
Since $CD$ passes through $M$, so $C, D$ lie on the opposite sides of the diameter through $OM$. So $P, Q$ lie on opposite direction outside segment $AB$.
Since $OM$ is perpendicular to $AB$, $OC$ is perpendicular to $PC$, so $O, M, C, P$ concyclic, $\angle OPM = \angle OCM.$
Similarly, $\angle OQM = \angle ODM$, triangles $OCD$ and $OPQ$ are similar.
But $OC = OD$, $\angle OCM = \angle ODM$, so $\angle OPM = \angle OQM$, $OP = OQ$, result in $PM = QM$, $PA = QB$.

Since $\angle{QMO}=\angle{QDO}=90^o$, so $Q, D, M, O$ are cyclic. Hence $\angle{QOD}=\angle{QMD}$.

Similar, we have $C, P, M O$ are cyclic. Therefore $\angle{COP}=\angle{CMP}$.

Because $\angle{CMP}=\angle{DMQ}$, so $\angle{QOD}=\angle{COP}$. Along with $OD=OC$, $\angle{ODQ}=\angle{OCP}$, we will have: $\triangle{ODQ}\cong\triangle{OCP}$, which gives $OQ=OP$. Therefore $\triangle{OPB} \cong \triangle{OQA}$, hence $QA=PB$.

P.S. Hope this picture will help you, silouan.


limit
calculus
derivative
calculus computations

First evaluate the derivative, which reduces nicely to
$ \frac{1\plus{}x}{(1\minus{}x)^{3}}$
Now, when we approach 1 from the left side, the expression tends to $ \infty$ but as we approach from the right side, it goes to $ \minus{}\infty$ so the limit does not exist.


category theory
superior algebra
superior algebra unsolved

Let $C$ be a category, $S,T : C \to Set$ two functors and $\tau : S \to T$ a natural transformation. Then:

$\tau$ is monic/epi in $Set^{C}$ iff $\tau_{c}: S_{c}\to T_{c}$ is monic/epi in $Set$ for all $c \in C$

Well, $\Leftarrow$ is obvious. And $\Rightarrow$ follows in the case of monics with the Yoneda Lemma. But what about the case of epis?!

A bit of a revive but why not. Suppose $\tau:S\rightarrow T$ is natural between functors $S,T:\mathcal C\to \textbf{Set}$ and that there exists some $c\in \text{Ob}(\mathcal C)$ such that $\tau_c:S(c)\to T(c)$ is [i]not[/i] epic. One claims that there exist two distinct natural transformations $\eta,\eta':T\to T$ such that $\eta\circ \tau=\eta'\circ \tau$. Indeed, take $\eta=\text{id}$, and observe that there exists some $\kappa\in c$ such that $\kappa\not\in\text{im}(\tau_c)$ and furthermore some $\kappa'\in c\setminus \{\kappa\}$. Define $\eta'$ to agree with $\eta$ everywhere with the exception of $\eta'_c(\kappa)$ being $\kappa'$ and not $\kappa$. Then $\eta'$ can be seen to be natural (any composition series passing through $\eta'_c$ is "blind" to $\kappa$) and the claimed equality of compositions to hold, as desired $\blacksquare$


Okay. We have two equations:

$x^2-px+q = 0$ with roots $\tan \alpha$ $\text {and}$ $\tan \beta$

$x^2-rx+s = 0$ with roots $\cot \alpha$ $\text {and}$ $\cot \beta$

How can $rs$ be discussed in terms of $p$ and $q$? Explain.

One equation has reciprocal roots of the other. This is discussed in the polynomial chapter of AoPS Volume 2.

I don't have AoPS Volume 2 so I don't know it appeared there.

 :D

But you have not yet answered the question. ;)

Let $\tan\alpha=x$ and $\tan\beta=y$
We have $p=-(x+y)$ and $q=xy$
$\cot\alpha=\frac{1}{x}$ and $\cot\beta=\frac{1}{y}$
Thus, $r=-\frac{x+y}{xy}$ and $s=\frac{1}{xy}$
$rs=-\frac{x+y}{x^2y^2}$
after substitution, we have
$rs=\frac{p}{q^2}$


I have been doing IMO related stuff for a while - I've basically learnt all the basic theory for Olympiads in different fields -- simple and basic and necessary theorems, results, etc.. I have also some experience in solving Olympiad problems of different levels etc...

However, I feel my prep is somewhat corrupted due to reasons like impatience to see solution, etc.. I am willing to spend a month or so devoted to Olympiad prep starting from scratch -- not scratch in terms of theory, but rather scratch in terms of problem solving, mental appraoch, practice etc.

What schedule, books, etc. do you recommond for this endeavor of mine? Please suggest fast the type of  problems,. theory, mind frame, books etc, I should do in lieu of my above position
Thanks
Rushil.

Hello Rushil,
    Here are some of the books which could prove useful.
1) Challenges and Thrills of Pre College Mathematics-a standard easily available one-very good for Geometry.
2) Problem Solving Strategies-Arthur Engel by Dover Publishers
You would do well to cover these two in depth .You could also look up at IMO Problems.
These should be enough.
PM me if you need more clarifications.
Sathej

Sathej, Thanks!
I believe (1) is a bit too elementary at this stage! I've read it and found that its good if you get it in IX and use it extend your interest in Math(which  I could not get --- whuaaa!! (crying!) )... but Now its not much useful --- I hope you agree!? --- please comment on this!

(2) - I have that and it truly the best!!!

Please , all of you, give some other recommendations fast regarding books, schedule etc... also , What do you guys think about Larson's " Problem solving through problems" --- is it good for Olympiads?? Does somebody(form India!) know whether this book is available in India? I've searched for it practically everywhere~~!!

Hello,
    Well,Yes (1) is elementary. But, I reiterate the Geometry part is rather good and it would do well to learn all of it from there before doing anything else. Especially, the section 'Geometric Gems' contains results which are used even in some IMOTC problems.
 Actually, there are several books which are very good. But not easily available.You could order it from Prism Book House.
 A Path to Combinatorics for Undergraduates by Andreescu and Feng. It isreally exhaustive and has a huge collection of problems.
 Mathematical Olympiad Challenges by Andreescu and Eneescu
 Challenging Problems with Elementary Solutions-Books I and II by Yaglom and Yaglom.

Sathej

Whats 'Prism Book House'??? Where is it?

[quote="http://www.verveonline.com/others/india-stores.shtml"]....In Bangalore:
...
7.  	Bharathi News Agency, Mini Magazine Centre, Bhatt’s Book Bureau, Prism Book House  	Jayanagar Complex ....[/quote]
 :)

Hello Rushil,
    Actually, all the books I mentioned are by Dover or Birkhauser Publishers.Infact, I have already mentioned these in the India Forum under Good books thread. One Prism Book House with HQ in Bangalore,one branch here in Chennai, has many of them.Firstly, have you got the book Functional Equations by BJ Venkatachala? That is extremely self sufficient for functional equations. This is published by Prism themselves.
Sathej


analytic geometry
graphing lines
slope
trigonometry

Point P is on the y-axis. From P, two lines are drawn to be tangent to $x^2-6x+y^2+6y+14=0$ at Q and R. If Q is $(4, -3+\sqrt3)$, find the length QR.

From the equation $x^2-6x+y^2+6y+14=0$,

$(x - 3)^2 + (y + 3)^2 = 4$

This is a circle centered at the point $C= (3, -3)$ and with radius $r = 2$. Is the point $Q = (x_Q, y_Q) = (4, -3 + \sqrt 3)$ on this circle ?

$(4 - 3)^2 + (-3 + \sqrt 3 + 3)^2 = 1 + 3 = 4$

Yes, it is. The equation of the line $CQ$ is

$y - y_C = \frac{y_Q - y_C}{x_Q - x_C} (x - x_C)$

$y + 3 = \frac{-3 + \sqrt 3 + 3}{4 - 3} (x - 3)$

$y = x \sqrt 3 - 3(1 + \sqrt 3) = m x + b$

The tangent $PQ$ passes through the point $Q$ and has the slope $-\frac 1 m = -\frac{1}{\sqrt 3} = -\frac{\sqrt 3}{3}$.

$y - y_Q = -\frac 1 m (x - x_Q)$

$y + 3 - \sqrt 3 = -\frac{\sqrt 3}{3} (x - 4)$

$y = -\frac{\sqrt 3}{3} x + \frac{7\sqrt 3}{3} - 3$

This tangent intersects the y-axis in the point $P = (0, y_P),\ y_P = \frac{7\sqrt 3}{3} - 3$ and the segment length $PQ$ is equal to

$PQ = \sqrt{(x_Q - x_P) ^2 + (y_Q - y_P)^2} = \sqrt{16 + \frac{16}{3}} = \sqrt{\frac{64}{3}} = \frac{8 \sqrt 3}{3}$

The angle $\angle PCQ$ is given by

$\tan{\widehat{PCQ}} = \frac{CQ}{PQ} = \frac{6}{8 \sqrt 3} = \frac{\sqrt 3}{4}$

$\sin{\widehat{PCQ}} = \frac{\tan{\widehat{PCQ}}}{\sqrt{1 + \tan^2{\widehat{PCQ}}}} = \frac{\frac{\sqrt 3}{4}}{\sqrt{1 + \frac{3}{16}}} = \sqrt{\frac{3}{19}}$

The segment length $QR$ is then

$QR = 2 PQ \sin{\widehat{PCQ}} =  \frac{16 \sqrt 3}{3} \cdot \sqrt{\frac{3}{19}} = \frac{16}{\sqrt{19}} = \frac{16 \sqrt{19}}{19}$


ok i found this problem in The Art and Craft of Problem Solving.

Sal the Magician ask you to pick any five cards from a standard deck.  you do so, and then show them to Sal's assistant Pat,  who places one of the five cards back in the deck and then puts the the remaining four cards into a pile. Sal is blindfolded, and does not witness any of this. then Sal takes off the blindfold, takes the pile of four cards, reads the four cards that Pat has arranged, and is able to find the fifth card in the deck (even if your shuffle the deck after Pat puts the card in the deck).  Assume that neither Sal nor Pat has supernatural powers, and that the deck of cards is not marked. How is the trick done?
Harder version: you pick which of the fie cards goes back into the deck (instead of Pat).

could someone please tell me how this is done. i have been thinking about these cases for about a week and have gotten nowhere. :(

the easier version involves pigeonhole
[hide]5 cards, 4 suits: 2 cards of a suit.
13 within a suit: cards always within $\pm 6$ units of one another.

so choose one of the doubles of a suit to return to deck.
the other stays to indicate the suit of the card.
the other 3 cards can be arranged in 6 ways.  each of these arrangements indicates incrementing the number on teh card that stays by 1-6.[/hide]

not sure about harder one...

[b]Edit[/b] Someone already solved it, so you can ignore this post.
I'm guessing that the four cards in the pile are arranged to point to what the fifth card is.
It might help to make a table of possible indexes for the cards and the different ways to indicate the value of a given index. Do all of the cards in the pile of 4 have to be face down and in the same direction?
I came up with the possible indexes of suit, rank, parity (for numbers), red/black, face card
And to indicate, position in pile, face up/down, turned vertical/horizontal.
This might go out of the boundary of the problem though. Which problem number is it?

is problems $6.2.8$ on page 222.
and alan i understand how pigeonhole  works for the suits but how dose it work for the card value.


calculus integration geometry perimeter inequalities

An isosceles triangle with integral side lengths has perimeter 87. Find the maximum length of the non-congruent side.

Remember, this problem isn't actually that hard, I thought the height had to be an integer too so i pulled out my $2rs,r^{2}-s^{2},r^{2}+s^{2}$ beastliness

[hide="Solution"] Say the sides are $a$, $a$, and $b$.  We have that $2a+b=87$ and $2a>b$.  Thus, we have that b+b<87, so 2b<87, so b<43.5.  Thus, the highest possible value for b is $\boxed{43}$.  [/hide]

How can that be when we want to find the maximum length of the [b]non[/b] congruent side? Isn't it
[hide]
85?
[/hide]

xscapezar: triangle inequality states that a + b > c, 1 + 1 is NOT greater than 85.

[hide]43 is correct, I believe :)[/hide]

[quote="magixter"]Who is Mr. Frost and where are his problems?[/quote]
Mr. Frost is the coach of MA this year.


function
modular arithmetic
algebra unsolved
algebra

Find the minimum positive integer $k$ such that there exists a function $f$ from the set $\Bbb{Z}$ of all integers to $\{1, 2, \ldots k\}$ with the property that $f(x) \neq f(y)$ whenever $|x-y| \in \{5, 7, 12\}$.

[quote="Megus"]I think the following works. It's easy to see that $k\geq3$ (Consider values $0,5,12$). Now our function will be:

for $x \equiv 0,1,2,3,4 \mod 15$ put $f(x)=1$
for $x \equiv 5,6,7,8,9 \mod 15$ put $f(x)=2$
for $x \equiv 10,11,12,13,14 \mod 15$ put $f(x)=3$

It works  :)[/quote]

I don't think so... $f(21) = f(9)$ and $21-9 = 12$.

EDIT: I think if you change it to $\mod 17$ you can get $k = 4$, which I think is the minimum.

$ k=1 $ is not a solution.
$ k=2 $ is not a solution because if a is red,a+7 is blue then a+12=(a+7)+5 can't be red or blue false.
$ k=3 $ isn't a solution because if a is red,a+5 blue,a+12 green then a+7 can't be green or red so is blue. so a+5 and a+7 are of the same colour so the even numbers are of a colour and the odd of the other false.
$ k=4 $ is the solution, the even numbers take 2 numbers and the odd other 2.
the numbers congruent with 0,2,4,6,8,10 mod 24 take a value and the numbers congruent with 12,14,16,18,20,22 take the other value.
for the odd numbers we take a value for the numbers congruent with 1,3,5,7,11 and the other for the numbers congruent with 13,15,17,19,21,23 mod 24.


calculus integration induction function factorial search blogs

The integral from 0 to infinity of (x^n)(e^(-x)) = n!.

Can some one explain this to me?
*how to derive it
*some practice problems
*a link to a lesson on this topic, if possible

You assert $ \int_0^\infty x^n e^{\minus{}x} \, dx \equal{} n!$.

Do you know what induction is? As well as integration by parts?

That is the correct integral.

I know parts.
I'm not sure about induction.
Our professor did this in class, but I don't remember everything.
I don't think he used induction.

Well, the function closely related to this integral is the Gamma function defined as $ \Gamma(z) \equal{} \int_0^\infty t^{z\minus{}1} e^{\minus{}t} \, dt$, where in our case we would be dealing with $ \Gamma(n\plus{}1)$.

Basically, you want to deduce that $ \Gamma(n\plus{}1) \equal{} n!$ from the fact that you know that $ \Gamma(n) \equal{} (n\minus{}1)!$. You do this by integrating by parts.

We have that $ \Gamma(1) \equal{} \int_0^\infty e^{\minus{}x} \, dx \equal{} 1$.

Now, if we know that $ \Gamma(n) \equal{} (n\minus{}1)!$, we can write (by integrating by parts):

$ \Gamma(n\plus{}1) \equal{} \int_0^\infty x^n e^{\minus{}x} \, dx \equal{} \minus{} x^n e^{\minus{}x} |_0^\infty \plus{} \int_0^\infty n x^{n\minus{}1} e^{\minus{}x} \, dx$

On the right side of this equation, you should see $ \Gamma(n)$ somewhere. Can you finish the proof?

It might be useful to read [url=http://en.wikipedia.org/wiki/Gamma_function]Gamma function wiki[/url], as well as [url=http://en.wikipedia.org/wiki/Mathematical_induction]Induction wiki[/url], but more importantly your class' notes and/or book.

Yeah this does not even resemble anything we did in class that day. And we have not did any proofs. Proof writing is another class. What is that little half-T thing? What he did was showed us some different values for the function and then from the pattern made the formula. I was just wondering how exactly that went though...

That half-T is the capital Gamma (from the Greek alphabet), created by typing \$\Gamma\$ (i.e. $ \Gamma$).

Well, first looking at some different values is of course a good idea to form an idea of what the general expression would be, but for any proof you would need induction in some form I think.

So, what do you want out of this thread? Do you want me to explain induction in more detail?

An example problem:

Integral from 0 to infinity of the function (x^6)(e^(-3x)).

I did it before but now I don't remember how to.
Is there a place that has a lesson on how to do these problems and offers more practice?
I know this problem can be done by parts, but I need to do it the "factorial" way.
And the gamma symbol never popped up once in the lecture.
And don't worry about induction since it is not pertinent to my class.

Right, so you want to reduce your integral ($ \int_0^\infty x^6 e^{\minus{} 3 x} \, dx$) to something where you can use the fact that $ \int_0^\infty x^n e^{\minus{}x} \, dx \equal{} n!$?

Well, the gamma function is probably a bit too advanced I guess. Knowing about it also isn't necessary to solve the problem you posed, it just makes it easier.

Try the substitution $ u \equal{} 3x$. I can help you, but I'm not going to do all the work for you.

Regarding the place you are searching for... most introductory calculus / real analysis books have a section about induction as well as one on integration by parts. You could also try to use the search function of this forum or just browse this subsection looking for problems.

Thanks, yeah the substitution rings a bell.
BTW you mentioned the gamma function is a bit too advanced, what level is that usually introduced (what age or year in college)

Also to be a math major what math should you be taking each year at college? And does it hinder my chances since I've never did any of the contest math?

Year 1: ?
Year 2: ?
Year 3: ?
Year 4: ?

Well, I'm not from America (where I presume you are from) so I don't really know how you can/should choose your courses (we have a different system here). Most universities have a standard course-package I would guess? And surely there are people there to help you choose. Just some basic analysis, algebra and such.

Hinder your chances on what? I had done some of it, but did not attain any noteworthy results. Of course experience always never hurts, but I wouldn't say contest math is compulsory to be succesful. You can check out Terence Tao's blog [url=http://terrytao.wordpress.com/career-advice/]here[/url] for more of this kind of talk.

The gamma function (for positive real numbers) was introduced at my university in the first year during a probability course.

We are kind of derailing the thread. I hope you can solve the problem you started the thread with now.


Drew, Shilen, and Rahul found a dollar and one of them dropped the dollar. One of them is a habitual liar, one ocasionally lies, and the other always tells the truth.
If:
Shilen said: “The dollar is mine!”
Rahul said: “Shilen is telling the truth.”
Drew said: “Its Rahul’s dollar.”
and if the dollar belongs to the habitual liar, then whose dollar is it?

[hide]If the habitual liar owns the dollar, then they cannot say that they own the dollar (they'd be lying). Therefore, Shilen is not the habitual liar. If Shilen always told the truth, he would still have the dollar (which he doesn't), so Shilen is the one that occasionally lies.

Since Shilen is lying, however, Rahul is also lying and Drew is telling the truth. Therefore, the dollar belongs to Rahul.[/hide]

[hide]Rahul's dollar--two people lied so one told the truth.  The first two statements would have to be true together, but they can't be because there have to be two people that lied.[/hide]

nice names you chose, PressAltF4

now where have i heard of them before?

[hide="My solution"]
There are 3 possibilities:

If Shilen is the habitual liar, then Rahul is also lying while Drew is telling the truth. The dollar must belong to Rahul.

If Rahul is the habitual liar, then Shilen is also lying while Drew is telling the truth. The dollar must belong to Rahul, who is the habitual liar.

If Drew is the habitual liar, then Shilen and Rahul are both telling the truth. The dollar must belong to Shilen.

The answer is therefore [b]Rahul[/b].
[/hide]

i think
[hide]its rahuls dollar. i  cant come up with a decent explanation.[/hide]

[hide]I saw something like this in the chapter countdown. Anyways it's my dollar!!! jk it's Rahul's dollar.[/hide]


calculus
integration
function
algebra
domain
complex analysis
complex analysis unsolved

[quote]Let $ f$ be analytic in the disk $ B(a,R)$ and suppose that $ \gamma$ is a closed rectifiable curve in $ B(a,r)$. Then $ \int_{\gamma} f \equal{} 0$.[/quote]

Show that the same holds if we replace the disk by a half plane

I suppose the main thing is that the domain you are working with is simply connected, and both the ball/half plane satisfies this.

[hide]
they gave as hint: Use a Mobius transformation
[/hide]
I don't see how.

Using a biconformal map between the disc and the half plane you make a change of variable to turn the integral over an arc in the right half plane to the integral over an arc in the disc.


Euler
topology
invariant
geometry
geometric transformation
advanced fields
advanced fields unsolved

Let $ T$ be the $ n$-torus. Suppose $ T$ acts on some (smooth) manifold $ M$. Show that $ \chi(M^{T})\equal{}\chi(M)$, where $ \chi$ is the Euler characteristic.

[quote="ma_go"]uhm.. what's $ M^T$? fixed points?[/quote]

Right.

There is a $ T$-equivariant cell decomposition by cells $ T/H \times D^k$. $ T/H$ is a torus by pr0ntryagin duality, so $ \chi(T/H \times D^k) \equal{} 0$ except when $ T \equal{} H$. From this it is obvious.

I'm not sure I follow your argument--did you implicitly apply cellular approximation?

I'll assume that $ M$ is compact, in order to make sure you have finite-dimensional cohomology groups.

First, endow $ M$ with a $ T$-invariant Riemann metric. This makes it clear that $ M^T$ is itself a smooth manifold. Let $ r>0$ be some small positive integer, and denote by $ A,\ B$ and $ C$ the sets of points at distance $ \le r,\ r$ and respectively $ \ge r$ from $ M^T$. They're all manifolds, $ A$ and $ C$ with common boundary $ B$. We have $ \chi(M)\equal{}\chi(A)\minus{}\chi(B)\plus{}\chi(C)\ (*)$. If $ r$ is small enough, $ A$ deformation retracts onto $ M^T$, so $ \chi(A)\equal{}\chi(M^T)$. Also, $ B$ and $ C$ are compact manifolds on which $ T$ acts without fixed points. Hence, they both admit self-maps with no fixed points and homotopic to the identity (namely the translation by some topological generator of $ T$). By the [url=http://en.wikipedia.org/wiki/Lefschetz_fixed-point_theorem]Lefschetz Fixed Point Theorem[/url], they must both have Euler characteristic zero. Plugging all of this into $ (*)$ gives the desired result.


Βρειτε μια ορθοκανονικη βαση του $ E^{\bot}$, οπου $ E=span\lbrace [10-12]^T,[-1110]^T\rbrace$

Ας γραψω την λυση που βρηκα.

Τα διανυσματα $ [10-12]^T,[-1110]^T$ ειναι μια βαση του $ E$.

Αν το διανυσμα $ [xyzw]^T$ $ \in E^{\bot}$  τοτε

$ [xyzw]^T\circ [10-12]^T=0\Rightarrow x-z+2w=0$

$ [xyzw]^T\circ [-1110]^T=0\Rightarrow -x+y+z=0$

Λυνοντας το συστημα βρισκουμε $ x=z-2w$ και $ y=-2w$

Αρα $ [xyzw]^T=[z-2w-2wzw]^T=z[1010]^T+w[-2-201]^T$

Ετσι τα διανυσματα $ k_1=[1010]^T$ και $ k_2=[-2-201]^T$ αποτελουν μια βαση του $ E^{\bot}$

Με την μεθοδο ορθογωνοποιησης Gram-Schmidt βρισκουμε τα διανυσματα

$ v=\frac{\sqrt{2}}{2}[1010]^T$ και $ u=\frac{\sqrt{7}}{7}[-1-211]^T$

τα οποια ειναι ορθοκανονικη βαση του $ E^{\bot}$


group theory
abstract algebra
superior algebra
superior algebra solved

$G$ is a group. The normal subgroup of $G$ are only $\{e\}$ and $G$ itself.
$G$ is not commutative.
Show that $G$ is isomorphic to $Inn(G)$, where $Inn(G)$ is the group formed by all inner-automorphisms of $G$.

$Inn(G)$ is isomorphic to $G/Z(G)$, and, since $Z(G)$ is normal in $G$ and $Z(G)\ne G$, we get $Z(G)=\{e\}$, meaning that $Inn(G)\cong G/{e}\cong G$.

The isomorphism from $Inn(G)$ to $G/Z(G)$ can be constructed like this: $f_g\mapsto \hat g$, where $f_g:G\to G,\ f_g(x)=gxg^{-1},\ \forall x\in G$. It's easy to see that this really is an isomorphism.

Let $f:G\to Inn(G)$, $f(g)(x)=gxg^{-1}$.
It is easy to see that f is a group morphism.
$G/\ker(f)\simeq Inn(G)$ because f is obviously surjective.
$\ker(f)$ is a normal subgroup of G so it is $e$ or $G$. If it is $e$ then it's ok. If not then $f(g)=id_G\ \forall g\in G$. So $gxg^{-1}=x\ \forall x,g\in G$, so G is commutative. This is a contradiction.


function
induction
number theory unsolved
number theory

find all sequences $ (a_n)$ of nonnegative integers such that

i) $ a_0 \leqslant a_1 \leqslant \ldots$
ii) $ a_{m^2\plus{}n^2}\equal{}a_m^2 \plus{} a_n^2$

[quote="MS87"]find all sequences $ (a_n)$ of nonnegative integers such that

i) $ a_0 \leqslant a_1 \leqslant \ldots$
ii) $ a_{m^2 \plus{} n^2} \equal{} a_m^2 \plus{} a_n^2$[/quote]
So we are looking for a non decreasing function $ f(x) : \mathbb N_0\to\mathbb N_0$ such that the following assertion is true :
$ P(x,y)$ : $ f(x^2 \plus{} y^2) \equal{} f(x)^2 \plus{} f(y)^2$ $ \forall x,y\in\mathbb N_0$

$ P(0,0)$ $ \implies$ $ f(0) \equal{} 2f(0)^2$ and so $ f(0) \equal{} 0$ (since $ f(0)\in\mathbb N_0$)
$ P(1,0)$ $ \implies$ $ f(1) \equal{} f(1)^2$ and so either $ f(1) \equal{} 0$, either $ f(1) \equal{} 1$

1) If $ f(1) \equal{} 0$
Then $ P(1,1)$ $ \implies$ $ f(2) \equal{} 0$
$ P(x,0)$ $ \implies$ $ f(x^2) \equal{} f(x)^2$ and so we get $ f(2^{2^n}) \equal{} 0$ and so $ f(x) \equal{} 0$ $ \forall x$ since $ f(x)$ is non decreasing

2) If $ f(1) \equal{} 1$. Then :
$ P(1,1)$ $ \implies$ $ f(2) \equal{} 2$
$ P(0,2)$ $ \implies$ $ f(4) \equal{} 4$
$ P(1,2)$ $ \implies$ $ f(5) \equal{} 5$
$ P(2,2)$ $ \implies$ $ f(8) \equal{} 8$
$ P(0,4)$ $ \implies$ $ f(16) \equal{} 16$

$ f(2) \equal{} 2\le f(3)\le f(4) \equal{} 4$ and so $ f(3)\in\{2,3,4\}$
If $ f(3) \equal{} 2$, then $ P(0,3)$ $ \implies$ $ f(9) \equal{} 4 < f(8)$, impossible since $ f(x)$ is non decreasing
If $ f(3) \equal{} 4$, then $ P(1,3)$ $ \implies$ $ f(10) \equal{} 17 > f(16)$, impossible since $ f(x)$ is non decreasing
So $ f(3) \equal{} 3$

$ P(3,5)$ $ \implies$ $ f(34) \equal{} 34$
$ P(4,5)$ $ \implies$ $ f(41) \equal{} 41$

$ f(5) \equal{} 5\le f(6)\le f(8) \equal{} 8$ and so $ f(6)\in\{5,6,7,8\}$
If $ f(6) \equal{} 5$, then $ P(0,6)$ $ \implies$ $ f(36) \equal{} 25 < f(34)$, impossible since $ f(x)$ is non decreasing
If $ f(6) \equal{} 7$, then $ P(0,6)$ $ \implies$ $ f(36) \equal{} 49 > f(41)$, impossible since $ f(x)$ is non decreasing
If $ f(6) \equal{} 8$, then $ P(0,6)$ $ \implies$ $ f(36) \equal{} 64 > f(41)$, impossible since $ f(x)$ is non decreasing
So $ f(6) \equal{} 6$

So $ f(x) \equal{} x$ $ \forall x\in\{0,1,2,3,4,5,6\}$

Then, using the identities :
$ (2u \plus{} 1)^2 \plus{} (u \minus{} 2)^2 \equal{} (2u \minus{} 1)^2 \plus{} (u \plus{} 2)^2$
$ (2u \plus{} 2)^2 \plus{} (u \minus{} 4)^2 \equal{} (2u \minus{} 2)^2 \plus{} (u \plus{} 4)^2$

We can deduce :
$ P(2u \plus{} 1,u \minus{} 2)$ $ \implies$ $ f(2u \plus{} 1)^2 \equal{} f(2u \minus{} 1)^2 \plus{} f(u \plus{} 2)^2 \minus{} f(u \minus{} 2)^2$
$ P(2u \plus{} 2,|u \minus{} 4|)$ $ \implies$ $ f(2u \plus{} 2)^2 \equal{} f(2u \minus{} 2)^2 \plus{} f(u \plus{} 4)^2 \minus{} f(|u \minus{} 4|)^2$

And, applying this equalities from $ u \equal{} 3$, it's immediate to show thru induction that $ f(x) \equal{} x$ $ \forall x$

Hence the two sequences solution of the problem :
$ a_n \equal{} 0$ $ \forall n\ge 0$
$ a_n \equal{} n$ $ \forall n\ge 0$


geometry
circumcircle
incenter
trigonometry
geometry solved

Given a triangle ABC, let CH be an altitude, and let cevians CM and CN bisect angles ACH and BCH, respectively. The circumcenter of CMN is also the incenter of ABC. Prove that $[ABC] = \frac{AN \cdot BM}{2}$.

In $\triangle{CMN}$ we have: $r= CI\cdot{\cos{\frac{C}{2}}}$.
Hence $\angle{ACB}=90^\cdot$.
But $\angle{ACN}={ANC}$ and $\angle{BCM}=\angle{MBC}$.
Hence $AC\cdot{BC}=AN\cdot{BM}$ and we are done.

Sorry Pestich, when I read the problem I didn't see your post!


$ \theta _{1}, \cdots ,\theta _{n} , R$ are variables and $ d_{1}, \cdots ,d_{n}$ are given.
We have following equations. Can we find a formula for $ \theta _{1}, \cdots ,\theta _{n} , R$?
(All variables are positive.)

$ sin\frac{\theta_{1}}{2} \equal{} \frac{d_{1}}{2R}, \cdots ,sin\frac{\theta_{n}}{2} \equal{} \frac{d_{n}}{2R}, \theta_{1}\plus{}\cdots \plus{} \theta_{n}\equal{}2\pi$

[quote="paracompact"]$ \theta _{1}, \cdots ,\theta _{n} , R$ are variables and $ d_{1}, \cdots ,d_{n}$ are given.
We have following equations. Can we find a formula for $ \theta _{1}, \cdots ,\theta _{n} , R$?
(All variables are positive.)

$ sin\frac {\theta_{1}}{2} \equal{} \frac {d_{1}}{2R}, \cdots ,sin\frac {\theta_{n}}{2} \equal{} \frac {d_{n}}{2R}, \theta_{1} \plus{} \cdots \plus{} \theta_{n} \equal{} 2\pi$[/quote]

These are the equations used to inscribe a polygon whose sides are known in a circle. They may be written :

$ \theta_k\equal{}2\arcsin(\frac{d_k}{2R})$ and $ \sum_{k\equal{}1}^{n}\arcsin(\frac{d_k}{2R})\equal{}\pi$

Let  $ d_M\equal{}\max(d_k)$. The last equation has a solution as soon as $ \sum_{k\equal{}1}^{n}\arcsin(\frac{d_k}{d_M})\geq\pi$ but no explicit formula for this solution.


function
topology
real analysis
real analysis unsolved

1. Suppose X=(R,p) is a metric space and p is the discrete metric .Which of the followings function is continuous?
a.f:R^2->X , f(x,y)=x+y
b.g:X->R, g(x)=1,if x<0 g(x)=0 otherwise
c.h:R^2->X  h(x,y)=x
only one is the right answer

2.X  is a complete metric space<=>
a.every sequence in X has a subsequence converges in X.
b.every sequence in X has a cauchy subsequence
c.every convergent subsequence in X is cauchy
d. for every degressive sequence A(n),n=1,2,...non emptysubsets of X with limd(A(n))=o then intersectionclosedA(n)is no empty
only  one is the right answer

3.(X,p) is a metric space and x(n) a sequence in X.Then x(n) is caouchy <=>
a.for every e>0and x e X there is a m e N  so x(n) e S(x,e)for every n>=m.
b.x(n) converges in a x e X.
c.there is a subsequence x(k)(n) of x(n) which coverges in  a x e X.
d.limd(x(m):m>=n)=0(d is the diameter )
only one is the correct answer

Who in their right mind would create multiple-choice analysis questions?

These are quite easy- the second and third are merely testing definitions.

For the first: A continuous function from a connected set to a discrete set must be constant. Any function from a discrete set is continuous.


limit
logarithms
real analysis
real analysis unsolved

Evaluate

\[{ \lim_{n\to\infty}( (1+\frac{1}{n} )^{\frac{1}{2}}+(1+\frac{1}{n} )^{\frac{2}{3}}}+...+(1+\frac{1}{n} )^{\frac{n}{n+1}}-n )\]

[hide] Is the answer $ \boxed{1} ?$ .Is there any solution without Taylor expansion ?[/hide]

Let

$ x_{n}\equal{}\sum\limits_{k\equal{}1}^{n}\left(\left(1\plus{}\frac{1}{n}\right)^{k/(k\plus{}1)}\minus{}1\right)$.

A calculation, based on $ a^{k}\minus{}b^{k}$ formula, shows that

$ \left(1\plus{}1/n\right)^{k/(k\plus{}1)}\minus{}1\equal{}\frac{1}{n}\left(1\minus{}\frac{(1\plus{}1/n)^{k/(k\plus{}1)}}{(1\plus{}1/n)^{k/(k\plus{}1)}\plus{}(1\plus{}1/n)^{(k\minus{}1)/(k\plus{}1)}\plus{}\cdots\plus{}(1\plus{}1/n)^{1/(k\plus{}1)}\plus{}1}\right)$

$ \equal{}\frac{1}{n}(1\minus{}b_{k})$.

We get that $ x_{n}\equal{}1\minus{}\frac{1}{n}\sum\limits_{k\equal{}1}^{n}b_{k}$.

We have, since $ (1\plus{}1/n)^{j/(j\plus{}1)}\geq 1$ for all $ j\equal{}0\cdots k$ that $ b_{k}\leq\frac{1}{k\plus{}1}\left(1\plus{}1/n\right)^{k/(k\plus{}1)}$, and hence

$ \frac{1}{n}\sum\limits_{k\equal{}1}^{n}b_{k}\leq \frac{1}{n}\sum\limits_{k\equal{}1}^{n}\frac{1}{k\plus{}1}\cdot\left(1\plus{}1/n\right)^{k/(k\plus{}1)}\leq \frac{1}{n}\sum\limits_{k\equal{}1}^{n}\frac{1}{k\plus{}1}2^{k/(k\plus{}1)}\rightarrow 0$ by Cesaro Stoltz.

Thus $ \lim x_{n}\equal{}1$.

I hope there are no errors.

problem is simplest if you change every terms by approximations :
really you will have :
$ (1\plus{}\frac{1}{n})^{\frac{j}{j\plus{}1}}\approx 1\plus{}\frac{j}{n(j\plus{}1)}$
hence
$ \sum ^{n}(1\plus{}\frac{1}{n})^{\frac{j}{j\plus{}1}}\minus{}n\approx \sum^{n} \frac{j}{n(j\plus{}1)}\approx \frac{n\minus{}\ln n }{n}\minus{}>1$
and what is RMT?

Yes Extremal ,I also managed to prove this problem using almost the same idea after expanding with Taylor.

[quote="Extremal"]

and what is RMT?[/quote]

RMT is a Mathematical Magazine from Timisoara , but in Romanian we say '[b]R[/b]evista de [b]M[/b]atematica din [b]T[/b]imisoara'