Let $ABC$ be a right-angled triangle $(\angle A=90^\circ)$ and let $(I,\,r)$ be the incircle of its. Take the orthogonal projection $E$ of $A$ on th line $BI$ then the orthogomal projection $F$ of $A$ on the line $CE.$
Prove that: $EF=r$

[quote="delta"]They are close but no equal, so says my computer! http://mathsrv.ku-eichstaett.de/MGF/homes/grothmann/java/zirkel/doc_en/
or may be I missinturpreted the problem.[/quote]
Agree. I used the GSP and it also said that they are not the same.

They ARE equal!
Extend $AE, AF$ to intersect $BC$ at $M, N$ respectively; clearly $EF$ is midline of $\triangle AMN$ and for any $A-$right angled triangle there exists the equality $b+c=2(R+r)$. With $BM=AB=c, CN=AC=b$ we get $MN=2r\implies EF=r$

2nd proof:
Easy angle chase gives $\angle MAN=45^\circ$; by symmetry $I$ is the circumcenter of $\Delta MAN$. Let $G$ be the projection of $I$ onto $BC$, i.e. midpoint of $MN$; $I$ is orthocenter of $\Delta EFG$ with $\angle EGF=\angle MAN=45^\circ$ and, by property, $GI=EF$.

Best regards.
sunken rock

[quote="delta"]@sunken rock
Clearly you misread the problem $F$ is the projection of $A$ on line $CE$, not on line $CI$![/quote]

You are right, I misread it, but it should have been the way I took it, that's clear, the author mityped!

Best regards,
sunken rock


Find all positive integer solutions to
$ x^2 \minus{} y^2 \equal{} 20$
$ xy \plus{} 5x \plus{} 3y \equal{} 200$

[hide]im getting no solution[/hide]

I get the roots of $ y^4\plus{}10y^3\plus{}14y^2\minus{}1400y\plus{}39500$, can't be factored according to [url]http://mcis.jsu.edu/faculty/desmith/polyfactor-cgi.html[/url].

If you switch the $ 200$ and the $ 20$, the solution $ (\minus{}10,\minus{}10)$ pops out. SFFT should find them all fairly quickly.

[quote="abacadaea"]Find all positive integer solutions to
$ x^2 \plus{} y^2 \equal{} 20$
$ xy \plus{} 5x \plus{} 3y \equal{} 200$

[hide]im getting no solution[/hide][/quote]

doesn't x^2+y^2=20 only have solutions x=2, y=4, and x=4, y=2?

$ 200\equal{}xy\plus{}5x\plus{}3y \leq xy\plus{}5xy\plus{}3xy \equal{} 9xy \leq \frac{9}{2}(x^2\plus{}y^2) \equal{} 90$

sorry, it was supposed to be $ x^2\minus{}y^2$

Mellow-i used SFFT originally and got $ (x\plus{}5)(y\plus{}3)\equal{}215$. and then the solutions are (38,2) and (0,40), neither satisfy the first one.

Since $ 11^2\minus{}10^2\equal{}21$, the first equation means that $ x<11$, simce otherwise, even subtracting the next lower square would make $ x^2\minus{}y^2 \geq 21$

The max value of $ x$ is thus $ 10$, and the max value of $ y$ is thus $ 9$, and the max value of $ xy\plus{}5x\plus{}3y$ is $ 167<200$


Hi! As we know from Cauchy-Schwarz: $ (a^3 \plus{} b^3 \plus{} c^3)(a \plus{} b \plus{} c) \geq (a^2 \plus{} b^2 \plus{} c^2)^2$
We have the stronger inequality:

Given that $ a,b$ and $ c$ are positive reals, prove that:
\[ (a^3 \plus{} b^3 \plus{} c^3)(a \plus{} b \plus{} c) \geq (a^2 \plus{} b^2 \plus{} c^2)^2 \plus{} ac(a \minus{} c)^2
\]
:)

EDIT: It is not nice at all...  :blush:
It is ridiculously easy.

Let's figure it out!
After expanding and simplifying, we have
$ \sum_{cyc}{ab(a\minus{}b)^2} \ge ac(a\minus{}c)^2$
which is obvious.


group theory
superior algebra
superior algebra unsolved

Let $ p>3$ be prime.  Let $ \Omega \equal{}\{0,1,\ldots,p\}$ and let $ S$ be the symmetric group on $ \Omega$.  Let $ \alpha \equal{} (0,1)(2,3)\cdots (p\minus{}1,p)$ and let $ \beta$ be a $ p$-cycle.  Let $ G \equal{} <\alpha,\beta>$.

1.  Show that $ G$ acts transitively on the set $ \{(a,b) | (a,b)\in\Omega\times\Omega, a\neq b\}$.

2.  If $ p\equiv 1\mod 4$, then $ G$ is not simple.

1. If k denotes the unique fixed point of beta, then beta operates transitively on the set Omega\{k}. Since there is a unique point k' in Omega\{k}, which gets exchanged with k under operation of alpha, we conclude that G acts transitively on Omega. Now let (a,b) be an arbitrary element of T={ (c,d) | c<>d elements in Omega}. Exploiting the transitivity of G on Omega we can map (a,b) to some (k,b') in its G-orbit. Because b'<>k holds, we can find an appropriate power of beta which maps b' into k' and hence (k,b') is mapped into (k,k') proving the transitivity of G on T.

2. Let sgn: S->{-1,+1} be the group homorphism, which assigns to each permutation its signature and A = ker(sgn) be the alternating group of even permutations. We observe, that sgn(alpha)=(-1)^[(p+1)/2] holds, i.e. in case of p congruent to 1 modulo 4 we have sgn(alpha)=-1. This is equivalent to S = A*<alpha>=A*G. Since A is normal in S with index 2, the intersection A and G lies with index 2 in G and is normal as well.


Find all pairs of prime numbers $(p,q)$, such that the number is $(p+1)^q-1$ perfect power.

Is too easy to probe that $p=q$, so if $(p+1)^q-1=A^k$, with $A.\, k\in \mathbb{Z_+}$ then as $k>1$ then $q=p$ and $k=2$, so your equation is equivalent to $(p+1)^p-1$ is a perfect square, then $p$ must be an odd prime, so ... good luck!

Best regards.

RX the Crazy Math

[quote="Fati"]Why is it easy to prove that $p=q$ ?[/quote]
Because by congruence $(p+1)^q-1 =0 \ mod p$ and this number is a perfect power, namely $A^k$ then $p^k$ ($k>1$) divides $(p+1)^q-1$ and $q$ is a prime, so $p^2|(pq)$ then $p=q$.

RX the Crazy Math


algebra
polynomial
linear algebra
matrix
vector
limit
induction

Let we have a sequence $ a_{n\plus{}1}\equal{}a_{n}\minus{}\frac{1}{4}a_{n\minus{}1}$, where $ a_{0}\equal{}a_{1}\equal{}1$. Without finding out the formula for $ a_{n}$ show that the sequence is convergent.

The characteristic polynomial for this homogeneous linear recursion is $ x^2 \minus{} x \plus{} \frac14,$ which has $ x \equal{} \frac12$ as a double root. Without even looking at the initial conditions, we know that we can write $ a_n \equal{} (c_1 \plus{} c_2n)2^{ \minus{} n}$ for some constants $ c_1$ and $ c_2.$ Hence, $ a_n\to 0.$

We would draw the same conclusion for any homogeneous linear recursion whose characteristic polynomial has roots entirely in the interior of the unit disk.

The following is the same question:

Let $ A \equal{} \begin{bmatrix}0 & 1 \\
 \minus{} \frac14&1\end{bmatrix}.$ Show that if $ v$ is any column vector, then $ \lim_{n\to\infty}A^nv \equal{} 0.$ Or you could just say that $ \lim_{n\to\infty}A^n \equal{} 0.$

Here's a slight variation:

Define $ a_{n \plus{} 1} \equal{} a_n \minus{} \frac14\,a_{n \plus{} 1} \plus{} 3,$ and choose $ a_0$ and $ a_1$ arbitrarily. Prove that $ \lim_{n\to\infty}a_n$ exists and compute that limit.

Thank you, Kent.

Here is another way of solving the first task using just induction. We define a new sequence $ b_n\equal{}1\minus{}\frac{1}{4b_{n\minus{}1}}$, where $ b_{0}\equal{}1.$ Then we show with induction that $ \frac12 \leq b_{n} < b_{n\minus{}1} \leq 1$ and determine that $ lim_{n \to \infty}b_{n}\equal{}\frac12.$

Now we use $ \left\{ b_{n} \right\}$ to show that $ \left\{ a_{n} \right\}$ is descending and bounded, therefore it has a limit. Again induction $ 0 \leq b_{n\minus{}1}a_{n\minus{}1} < a_{n} \leq a_{n\minus{}1}$. So $ lim_{n \to \infty}{a_{n}}\equal{}lim_{n \to \infty}{a_{n\minus{}1}}\minus{}\frac14 lim_{n \to \infty}{a_{n\minus{}2}}.$ Therefore $ L\equal{}L\minus{}\frac14 L$ and $ L\equal{}0$.

Variation: $ a_{n \plus{} 1} \equal{} a_n \minus{} \frac14\,a_{n \plus{} 1} \plus{} 3$

This is a non-homogeneous equation but it could be transformed into homogeneous defining $ a_{n\plus{}1}\equal{}b_{n}\plus{}12$. We get $ b_{n\plus{}1}\equal{}b_{n}\minus{}\frac14 b_{n\minus{}1}$. Since $ lim_{n \to \infty}{b_{n}}\equal{}0 \rightarrow lim_{n \to \infty}{a_{n}}\equal{}12.$


Let triangle $ ABC$ have $ A\equal{}30^\circ$. Let $ S$ be  its circumcenter and $ I$ its incenter. Choose a point $ D$ on $ BA$ and a point E on CA such that $ BD\equal{}CE\equal{}BC$. Show that $ SI$ is perpendicular and equal to $ DE$
thx for your nice solution

See here : http://www.mathlinks.ro/resources.php?c=37&cid=47&year=1988&p=269054


What is the probability of two 1000 digits number have G.C.D is greater than a 700 digit number?

I think this involves finding the number of primes with more than 700 digits.And the number of their multiples that has 1000 digits or in other words in the interval [1000...,999...] and then the number of ways to choose 2 numbers from those 1000digit numbers and then division by 2.


Yuliya has a piece of meat which measures 5'' x 6'' x 8''. In order to make a stew, she would like to cut pieces which measure 2'' x 3'' x 4''. What is the maximum number of
such pieces she can cut from this piece of meat?

But $\frac52=2.5$, that gives you an extra 1/2 piece, which I would think would just go to waste. Wouldn't the answer be $\left\lfloor \frac52 \right\rfloor *\frac63 *\frac84=8$?

The funny thing is, I actually guessed 8.

Note that the question never said that the meat pieces had to be in the same orientation. Therefore, I think there is a way to cut the meat into 10 pieces (some pieces might be vertical, while others are horizontal).


When (a-b)^n, n>=2, is expanded by the binomial theorem, it is found that, when a=kb, where k is a postive integer, the sum of the 2nd and 3rd terms is zero. Find n in terms of k.

I never understood this: when a problem says the "2nd and 3rd terms", is it referring to a^(n-1)*-b  and a^(n-2)*b^2 or a*(-b)^(n-1) and a^2*(-b)^(n-2)?

Ok, lil' help would be appreciated here.

[hide]

(a-b)^n where a = kb

(kb-b)^n = ([k-1]b)^n = (k-1)^n * b^n

(k-1)^n = k^n - nC2 k^n-1 + nC3 k^n-2.....

so we have

nC3k^n-2 = nC2k^n-1

am i right so far?

[/hide]

2nd, 3rd referring to descending order of the first #, here meaning descending order of powers of a. THus, the answer to your question is the former choice.


Thanks.
It was great summer.
It couldn't have been any better.
I really had so much fun working with the greatest people on earth.

Thank you, thank you, and thank you!!

You juggle, sing and hold up boards for other people to break. You rock at math, play Canadian Fish, and everything else. What can't you do?

You've been great to work with and so has everyone else. It was so much fun that I almost feel guilty...

[quote="Geehoon"]Thank you, thank you, and thank you!![/quote]
No, thank you!

it has been a wonderful summer and i already miss it too much! hope that we could get together sometime for a TA gathering of some sort... and not geehoon, i see him too often already (jk)

[quote="Mariag"]not geehoon, i see him too often already (jk)[/quote]
so maria.. are you done with your bio pset?


Find all pairs of positive integers $ (x,y)$ such that both $ x^2 \plus{} 3y$ and $ y^2 \plus{} 3x$ are perfect squares.

wlog assume $ x\geq y$ than we have $ x^2 < x^2 \plus{} 3y < x^2 \plus{} 4x \plus{} 4 \equal{} (x \plus{} 2)^2$ so $ x^2 \plus{} 3y \equal{} (x \plus{} 1)^2$ and further $ 3y \equal{} 2x \plus{} 1$
by solving this linear diophan equation we get $ x \equal{} 1 \plus{} 2k$ and $ y \equal{} 1 \plus{} 3k$ for a non-negative integer $ k$
$ y^2 \plus{} 3x \equal{} (1 \plus{} 2k)^2 \plus{} 3(1 \plus{} 3k) \equal{} 4k^2 \plus{} 13k \plus{} 4$ but $ (2k \plus{} 3)^2 < 4k^2 \plus{} 13k \plus{} 4 < (2k \plus{} 4)^2$ for $ k > 5$ so we just need to manually check $ k$ up to five. it turns out to be that $ k \equal{} 5$ and $ k\equal{}0$ are the only solutions. for which we get $ x \equal{} 16$ and $ y \equal{} 11$ giving $ x^2 \plus{} 3y \equal{} 17^2$ and $ y^2 \plus{} 3x \equal{} 13^2$ in the first and $ y\equal{}x\equal{}1$ and $ x^2\plus{}3y\equal{}y^2\plus{}3x\equal{}4$ in the second case


do there exist 2 polygons, not necessarily convex, but necessarily non-selfintersecting, such that they share the same vertices and don't have any side in common ? :)

yes  :P
cute one

[code] AB
 CD
E  F
 GH
 IJ[/code]

ECADBFHJGIE and EABCDFJIHGE

it's difficult to understand your notation, i mean it's cruical where your put the vertices exactly. Please post it more understandable. And have you checked that your polygons are non-selfintersecting ?


$a,b,c > 0$, prove that:

$\sum_{cyc}\frac{a^{2}}{ab+b^{2}}\geq \sum_{cyc}\frac{a}{\sqrt{ac}+c}\geq \frac{3}{2}$

no it's not a typo, the form between LHS and MHS is similar, but cyclic in the different direction :P

MHS="Middle Hand Side" ..  :rotfl:

For the LHS-MHS inequality consider \[\sum_{cyc}\frac{a^{2}\sqrt{a}(\sqrt{b}+\sqrt{a})-b^{2}(a+b)}{b\sqrt{a}(\sqrt{a}+\sqrt{b})(a+b)}\geq 0\] which follows from Rearrangement and $[5,1,0]\succ [4,2,0]$


modular arithmetic
AoPSwiki
induction
number theory
relatively prime

I'm having trouble with number theory.

What is the remainder when $ 103^{53} \plus{} 53^{103}$ is divided by 7?

Can you all give me some practice problems to try out (but not too hard please!)? Thanks!

$ \pmod{k}$

let's look at the numbers.

We have what's called Fermat's Little Theorem which states that $ a^{p\minus{}1}\cong 1 (\mod {p})$ where a is relatively prime to p and p is a prime. (the proof can be found on AoPSWiki)

Then, we apply this to the problem

We can reduce it to $ 103^5\plus{}53^1$ but it's still a mess.

Let's reduce 103 now. since 103 has a remainder of 5 when divided by 7, we can replace 103 with 5.

Also, we can replae 53 by its remainder when divided by 7, or 4.

so we get $ 5^5\plus{}4$. you can compute it now, but there is a slicky method;

we stretch the definitions of mod a bit and allow fractions into it!

then by FlT (fermat's little theorem), we replace $ 5^5$ by $ \frac{1}{5}$. Now we ask:"what number is congruent to 1 mod 7 when multiplied by 5?"

it can be trial-and-error ed that it is indeed 3. So now we have 3+4 and which has remainder 0 when divided by 7!

how about this one:

prove that $ 12^{2n\plus{}1}\plus{}11^{n\plus{}2}$ is divisible  by 133 for all positive integers n.

[hide="hint1"]Try induction[/hide]
[hide="hint2"]Note that $ 12^2\minus{}11\equal{}133$ and $ 11^2\plus{}12\equal{}133$[/hide]
[hide="hint3"]Try to use the squares and expand them. What do you see?[/hide]

[quote]We can reduce it to $ 103^5 \plus{} 53^1$ but it's still a mess.[/quote]

I don't understand how we reduce it.

From Fermat's Little Theorem $ 103^6 \equiv 1 \bmod{7} \Rightarrow 103^{54} \equiv 1 \bmod{7}$ Now $ 103 \times 53 \equiv \minus{}1 \bmod {7}$ and hence $ 103^{53} \times 103 \equiv 1 \bmod {7} \Rightarrow 103^{53} \equiv \minus{}53 \bmod {7}$

Likewise $ 53^{103} \equiv 53 \bmod{7}$ Hence $ 103^{53} \plus{} 53^{103} \equiv 0 \bmod{7}$

another one.....:D
(?) find remainder when 2222^5555 + 5555^2222 is divided by7 ? coincidence!!!  7 here too.....
please hide ur solutions if u answer :mad:
 EDIT:more probs:
simple ones:
[hide]1)give remainders: 81^119(mod 13), 13^216(mod19),7^1001(mod11)  and u can try other mods for same probs for practice[/hide]

q)remainder 1000! divided by 3^300?
q)prove xy divisible by 12 in equation:   x^2+y^2=z^2
                         hint
[hide]try remainders modulo 3 and 4 in squares[/hide]


Let's put it that way:

Find the flux of the electric field, produced by a point charge $ q$, in vacuum, through a circle of radios $ R$ in a plane perpendicular to the segment of length $ {h \over 2}$ with ends in the center and in the charge, with no use of integrals.

I cant visualize the situation :( Can someone draw  :roll:

It's just the flux in one of the cylinder base of the former problem.

@Conjurer: hopefully my drawing will make the situation clear.
@everyone: we discussed this problem [url=http://www.mathlinks.ro/viewtopic.php?t=256060&start=20]here[/url], and the answer was calculated ([b]using integrals[/b]) to be $ \phi_E \equal{} \frac {q}{2\epsilon_0}\left(1 \minus{} \frac {h}{\sqrt {4R^2 \plus{} h^2}}\right)$.

[url=http://imageshack.us][img]http://img6.imageshack.us/img6/85/mathllo1.jpg[/img][/url]

Well I dont know how we can do it without using any integration, but its easy enough with integration. :)

Of course it is. But for improving insight and experience in application of Gauss Theorem ( and for the matter already discussed in a similar subject), I think it's important to do with no use of integrals. Besides, it's an fine geometrical exercise...


Prove that $ f(n) \equal{} 1 \minus{} n$ is the only integer valued function defined on the integers that satisfies the following condition:

  $ f(f(n)) \equal{} n$ for all integers n;
  $ f(f(n \plus{} 2) \plus{} 2) \equal{} n$ for all integers n;
  $ f(0) \equal{} 1$

My proof:

   Let g be another function that satisfies the conditions, then g < 1 - n or > 1 - n. So there exist a real number q, q not zero, s.t. g(n) = (1 - n) + q.  Since g(0) = 1, then g(0) = 1 - 0 + q = q, which is a contradiction.

Is my proof right?

   Also, I read Kalva's solution; I am confused as to what he iis trying to prove there.

   Kalva states the problem as  "Let Z be the integers. Prove that if f : Z → Z satisifies f( f(n) ) = f( f(n+2) + 2 ) = n for all n, and f(0) = 1, then f(n) = 1 - n."

   But I thought the original problem was trying to prove uniqueness.

   Am I totally confused?

That's an easy problem.

1st step: You would show that $ f(n) \equal{} 1 \minus{} n$ satisfies the given condition.

2nd step: You would obtain $ f(n \plus{} 2) \minus{} f(n) \equal{} 2$ by the given condition. Solving the recurrence gives $ f(n) \equal{} 1 \minus{} n.$

[quote="sexynsmartjenny"]   Kalva states the problem as  "Let Z be the integers. Prove that if f : Z → Z satisifies f( f(n) ) = f( f(n+2) + 2 ) = n for all n, and f(0) = 1, then f(n) = 1 - n."

   But I thought the original problem was trying to prove uniqueness.[/quote]
This statement is the same. What he means is "If $ f$ satisfies those conditions, then $ f$ [i]must equal[/i] $ 1\minus{}n$". That's not the same at all as "then $ f$ [i]could equal[/i] $ 1\minus{}n$". If you prove what he wrote, uniqueness comes with that.

For your question about whether or not your proof was correct:
[quote="sexynsmartjenny"]Let g be another function that satisfies the conditions, then g < 1 - n or > 1 - n. So there exist a real number q, q not zero, s.t. g(n) = (1 - n) + q.[/quote]
Basically everything I've quoted is flawed.  If $ g$ is a function that satisfies the given condition then at any given value of $ n$ we could have any of $ g(n) < 1 \minus{} n$, $ g(n) > 1\minus{} n$ or $ g(n) \equal{} 1 \minus{} n$.  All we know from the fact that $ g(x)$ is not the function $ 1 \minus{} x$ is that there is [i]some value[/i] at which $ g(x) \neq 1\minus{} x$.  Or, in other words, there is some [i]function[/i] $ q$ such that $ g(n) \equal{} (1 \minus{} n) \plus{} q(n)$, and $ q(n)$ is not always zero.  However, "not always 0" potentially allows for "sometimes zero," or even "zero everywhere except one single point."  So we are not able to draw the contradiction you want from this fact alone.

[quote="Xevarion"][quote="sexynsmartjenny"]   Kalva states the problem as  "Let Z be the integers. Prove that if f : Z → Z satisifies f( f(n) ) = f( f(n+2) + 2 ) = n for all n, and f(0) = 1, then f(n) = 1 - n."

   But I thought the original problem was trying to prove uniqueness.[/quote]
This statement is the same. What he means is "If $ f$ satisfies those conditions, then $ f$ [i]must equal[/i] $ 1 \minus{} n$". That's not the same at all as "then $ f$ [i]could equal[/i] $ 1 \minus{} n$". If you prove what he wrote, uniqueness comes with that.[/quote]

Ok, I think the part I don't understand is why uniqueness follows from that.  I was able to derive at $ f\equal{}1\minus{}n$ when I was doing the problem, but I don't see why it must be unique.  Why?

The statement
(1) "if [conditions] then $ f \equal{} 1 \minus{} n$"
is logically equivalent to the statement
(2) "$ f \equal{} 1 \minus{} n$ is the unique function satisfying [conditions]."
The statement
(3) "$ f \equal{} 1 \minus{} n$ satisfies [conditions]"
is implied by both (1) and (2) but it implies neither of them.  When you say that you have derived $ f \equal{} 1 \minus{}n$, you appear to be claiming that you have proved either (1) or (2) (which are, again, equivalent), but you haven't written here a proof of either of these.  The fact that you don't see why it must be unique suggests that in fact you've proved (3), which does not move you very far towards proving either (1) or (2).

Kunny gave a hint about how to prove (1) or (2).  The key step is a clever substitution for $ n$ in the equation $ f(f(n \plus{} 2) \plus{} 2) \equal{} n$:
[hide]Substitute $ n \equal{} f(k) \minus{} 2$ to get $ f(f(f(k)) \plus{} 2) \equal{} f(k) \minus{} 2$, so by the other given condition $ f(k \plus{} 2) \equal{} f(k) \minus{} 2$ for any $ k$.  Then it suffices to be able to calculate $ f(n)$ for one odd and one even value of $ n$.[/hide]


integration
calculus
algebra
polynomial
number theory
least common multiple
calculus computations

\[ Evaluate : \int \frac{\left ( 4x \plus{} 5 \right )^{\frac{1}{3}}}{\left ( 4x \plus{} 5 \right )^{\frac{1}{4}} \plus{} 4} dx\]

[quote="tariq1440"]
Evaluate : $ \int \frac {\left ( 4x \plus{} 5 \right )^{\frac {1}{3}}}{\left ( 4x \plus{} 5 \right )^{\frac {1}{4}} \plus{} 4} dx$
[/quote]

[hide="Hint"]

Just 1 substitution

$ 4x \plus{} 5 \equal{} t^{12} \ \Rightarrow \ 4 dx \equal{} 12 t^{11} dt$

Integral reduces to

$ 3 \int \frac{t^{15}}{4 \plus{} t^3} dt$

Now, its simple .... you need to algebraically simplify it.

[/hide]

Let $ y \equal{} 4x \plus{} 5$.

Then $ dy \equal{} 4dx$ and

$ \int \frac {(4x \plus{} 5)^{\frac {1}{3}}}{(4x \plus{} 5)^\frac {1}{4} \plus{} 4}\, dx \equal{} \frac {1}{4}\int\frac {y^{\frac {1}{3}}}{y^\frac {1}{4} \plus{} 4}\,dy$

Then let $ z \equal{} y^\frac {1}{12}$, so that $ dz \equal{} \frac {1}{12}y^{\frac { \minus{} 11}{12}}\,dy$.

And so the integral becomes

$ 3\int \frac {z^{15}}{z^{3} \plus{} 4}\, dz$

This can now (I think) be dealt with by division to get a number of powers of $ z$, corresponding to powers of $ (4x \plus{} 5)^\frac {1}{12}$ and we should also get a term which integrates to $ ln$ and possibly an $ \arctan$.

Well done, Nora.91: You've not only posted your solution before mine but yours is a bit neater.

hello
for these types of question in which a polynomial like $ (ax\plus{}b)$   has different fraction powers $ ( \frac{p_1}{q_1},\frac{p_2}{q_2} ,.......)$ , it is advisable to put that $ (ax\plus{}b) \equal{} t ^m$ where $ m$ is LCM of denominators $ (q_1,q_2,....)$
thank u


Solve the equation  $ \sqrt {m{x^2}\minus{}30x\minus{}5} \equal{} \minus{}x^2\plus{}3x\minus{}2$

@zaya_yc: $ t \equal{} x\minus{}\frac{3}{x} \Rightarrow t^2 \equal{} x^2\minus{}6\plus{}\frac{9}{x^2}$.

Anyway...

Note that $ \minus{}x^2\plus{}3x\minus{}2 \equal{} \minus{}(x\minus{}1)(x\minus{}2)$ but the LHS must be nonnegative.  Hence $ x \in [1,2]$.

Setting $ x \equal{} 1$ gives $ \sqrt{m\minus{}35} \equal{} 0$.  Hence $ (x,m) \equal{} (1,35)$ is a solution.

Setting $ x \equal{} 2$ gives $ \sqrt{4m\minus{}65} \equal{} 0$.  Hence $ (x,m) \equal{} (2,16.25)$ is also a solution.

Do you want a general form for $ m$ in terms of $ x$?

For $ 1\leq x\leq 2$, let $ t\equal{}x\minus{}\frac{3}{x}$ we can rewrite the given equation as $ m\equal{}t^2\minus{}6t\plus{}19$ for $ \minus{}2\leq t\leq \minus{}1$.

Answer:

1. $ m<26\cdots$ no solution

2. $ m\equal{}26\cdots x\equal{}2$

3. $ 26<m<35\cdots x\equal{}\frac{m\pm \sqrt{m^2\plus{}12}}{2}$

4. $ m\equal{}35\cdots x\equal{}1$

5. $ m>35\cdots$ no solution

Umm...something is obviously wrong there, because $ (x,m) \equal{} (2,16.25)$ is a solution.


function
limit
real analysis
real analysis unsolved

Let $ f: [a,b] \minus{} > R$ a incresing function s.t $ f(x\plus{}0) \plus{} f(x\minus{}0) \equal{} 2f(x)$ for all $ x \in (a,b)$ then f is continuos ?

[quote="Akashnil"]Please clearly define $ f_d(x)$ and $ f_s(x)$ and what is the meaning of the title?[/quote]
$ f_d(x_{0})\equal{} \lim_{x\to x_{0} , x \geq x_{0} }f(x)$

I think that:
$ f: [\minus{}4;4]\to{R}$,
$ f(x)$ =  $ 2x\plus{}4$ ,if $ x\in{[\minus{}4;0)}$
            $ 5$ if $ x\equal{}0$
            $ 2x\plus{}6$  if $ x\in{(0,4]}$
is a counterexample


we have:
$3(a^{2}+b^{2}+c^{2})\ge a^{2}+b^{2}+c^{2}+2ab+2bc+2ac = (a+b+c)^{2}=81\Leftrightarrow \sqrt{a^{2}+b^{2}+c^{2}}\ge \sqrt{27}$
$27= 3(a+b+c)\ge a+b+c+2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ac}=(\sqrt{a}+\sqrt{b}+\sqrt{c})^{2}\Leftrightarrow \sqrt{a}+\sqrt{b}+\sqrt{c}\le \sqrt{27}$
therefore $\sqrt{a^{2}+b^{2}+c^{2}}\ge 27\ge \sqrt{a}+\sqrt{b}+\sqrt{c}$ which proves the first part.

[quote="Fitim"]If $a,b,c>1$ and $a+b+c=9$; prove that:

$\sqrt{ab+ac+bc}\leq \sqrt{a}+\sqrt{b}+\sqrt{c}$.[/quote]

For argady!
$\sqrt{ab+ac+bc}\leq \sqrt{a}+\sqrt{b}+\sqrt{c}\Leftrightarrow\sqrt{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\leq \frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{ac}}+\frac{1}{\sqrt{bc}}$

If $a=\frac{1}{m^{2}},b=\frac{1}{n^{2}},c=\frac{1}{p^{2}}$, where $m,n,p$ are real positive number. Because $a,b,c>1$ and

\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}\geq \frac{4}{\sqrt{a}+\sqrt{b}}\geq \frac{4}{\sqrt{2(a+b)}= \frac{4}{\sqrt{2(9-c)}= \sqrt\frac{8}{9-c}>\frac{1}{\sqrt{c}}
Now we have $m+n>p$ and $m,n,p$ are sidelength of tringle.
Put $2m=x+y$ ,$2n=x+z$ ,$2p=z+y$    :wink:

[quote="Fitim"][quote="Fitim"]If $a,b,c>1$ and $a+b+c=9$; prove that:

$\sqrt{ab+ac+bc}\leq \sqrt{a}+\sqrt{b}+\sqrt{c}$.[/quote]

For argady!
$\sqrt{ab+ac+bc}\leq \sqrt{a}+\sqrt{b}+\sqrt{c}\Leftrightarrow\sqrt{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\leq \frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{ac}}+\frac{1}{\sqrt{bc}}$

If $a=\frac{1}{m^{2}},b=\frac{1}{n^{2}},c=\frac{1}{p^{2}}$, where $m,n,p$ are real positive number. Because $a,b,c>1$ and

$\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}\geq \frac{4}{\sqrt{a}+\sqrt{b}}\geq \frac{4}{\sqrt{2(a+b)}}=\frac{4}{\sqrt{2(9-c)}}= \sqrt\frac{8}{9-c}>\frac{1}{\sqrt{c}}$
Now we have $m+n>p$ and $m,n,p$ are sidelength of tringle.
Put $2m=x+y$ ,$2n=x+z$ ,$2p=z+y$    :wink:[/quote]
Id est, remain to prove that
 $(mn+mp+np)^{2}(m^{2}n^{2}+m^{2}p^{2}+n^{2}p^{2})\geq9(m^{2}+n^{2}+p^{2})m^{2}n^{2}n^{2},$
where $m,$ $n$ and $p$ side lengths of triangle.
It is true. I proved it! :D

For all ineq. (with three variables) can prove $F(a,b,c)\geq F(a, \frac{b+c}{2},\frac{b+c}{2})$????? :roll:
Or when we can use this hence???? :roll:

[quote="Fitim"]For all ineq. (with three variables) can prove $F(a,b,c)\geq F(a, \frac{b+c}{2},\frac{b+c}{2})$????? :roll:
[/quote]
Far not for everyone. As any other method.

[quote="Fitim"]If $a,b,c>1$ and $a+b+c=9$; prove that:

$\sqrt{ab+ac+bc}\leq \sqrt{a}+\sqrt{b}+\sqrt{c}\leq \sqrt{{a^{2}+b^{2}+c^{2}}}$.[/quote]

[size=200]12-th Macedonian Mathematical Olympiad 2005?[/size]

[quote="BttfUS"]Please write it in English or most of us will not understand!![/quote]
Long time no see
[hide]话说这说法居然成标准的了............给跪了
[/hide]

[quote="BttfUS"]Please write it in English or most of us will not understand!![/quote]

it's fine to read it even he write it in chinese. there are just few words on it.

[quote="Fitim"]If $a,b,c>1$ and $a+b+c=9$; prove that:

$\sqrt{ab+ac+bc}\leq \sqrt{a}+\sqrt{b}+\sqrt{c}\leq \sqrt{{a^{2}+b^{2}+c^{2}}}$.[/quote]
$\sqrt{ab+bc+ca} \le \sqrt{a} +\sqrt{b} + \sqrt{c}\iff (\sqrt{ab}-1)^2+(\sqrt{bc}-1)^2+(\sqrt{ca}-1)^2\leq 12.$

[quote=Fitim]If $a,b,c>1$ and $a+b+c=9$; prove that:

$\sqrt{ab+ac+bc}\leq \sqrt{a}+\sqrt{b}+\sqrt{c}\leq \sqrt{{a^{2}+b^{2}+c^{2}}}$.[/quote]
See also here:
https://math.stackexchange.com/questions/3373703/

[quote=Fitim]If $a,b,c>1$ and $a+b+c=9$; prove that:

$\sqrt{ab+ac+bc}\leq \sqrt{a}+\sqrt{b}+\sqrt{c}$.[/quote]
Here is another solution :
For $a\rightarrow a^2,b\rightarrow b^2$ and $c\rightarrow c^2$, the inequality is equivalent to $$(a+b+c)^2(a^2+b^2+c^2) \ge 9(a^2b^2+a^2c^2+b^2c^2)$$
$$\Leftrightarrow \sum (a^4+b^4-2a^2b^2) +4\sum (a^3b+b^3a-2a^2b^2)-2\sum (a^2b^2-2a^2bc+a^2c^2) \ge 0$$
$$\Leftrightarrow \sum (a-b)^2 \left[ (a+b)^2+4ab-2c^2 \right] \ge 0$$
WLOG, assume that $a \ge b \ge c$ and $\left\{\begin{matrix}
S_a=\left ( b+c \right )^2+4bc-2a^2\\
S_b=\left ( a+c \right )^2+4ac-2b^2\\
S_c=\left ( a+b \right )^2+4ab-2c^2
\end{matrix}\right.$ we have $S_a \le S_b \le S_c$.

Hence, It sufficient to show that $S_a+S_b \ge 0$ which transforms to $2c^2+6c(a+b)-(a^2+b^2) \ge 0$

Or $2(c-a)(c-b)+8c(a+b)-(a+b)^2 \ge 0$

Or $2(c-a)(c-b)+8(a+b)(c-1) \ge 0$ which is obvious.

[quote=Fitim]If $a,b,c>1$ and $a+b+c=9$; prove that:
$$\sqrt{ab+ac+bc}\leq \sqrt{a}+\sqrt{b}+\sqrt{c} $$.[/quote]
$ \Leftrightarrow ab+bc+ca \le 9+2( \sqrt{ab}+ \sqrt{bc}+ \sqrt{ca}) $
WLOG, assume $b \ge a \ge c$. From $a+b+c=9$ and $c>1$ we get $a+b <8$, implies $a<4$. So,
$( \sqrt{b} +\sqrt{c})^2=b+2 \sqrt{bc}+c> b+\sqrt{a} \cdot \sqrt{a} \cdot 1+c=9$
Now, by Rearrangement inequality we have:
$ \sqrt{ab}+\sqrt{ca}+\sqrt{bc} \ge a + 2 \sqrt{bc} =a+b+c- \frac{(b-c)^2}{(\sqrt{b}+\sqrt{c})^2} $
$ \ge 9- \frac{(b-c)^2}{8} $
It suffices to show that:
$$ bc+(b+c)(9-b-c) \le 27- \frac{(b-c)^2}{4}$$
$$ \Leftrightarrow (b+c-6)^2 \ge 0 $$
clearly true !

Let $a,b,c\ge m$ such that $a+b+c=9$. Find the smallest value of $m$ such that
$$\sqrt{a}+\sqrt{b}+\sqrt{c}\ge \sqrt{ab+bc+ca} .$$


Hello guys,

Every day I read how much unhealthy is to sleep less than 8 years. I usually study at night (21-22:30) and I don't sleep a lot - about 6 hours. What do you think? How much do you sleep? What is the best time for studying and how much sleep does a 16-year-old student need? Do you think I have to study earlier and how can i change it ?
thank you in advance,
Excellence

I think if you get to read everyday about how unhealthy it is too sleep too little, then you're probably subjected to a bit of fear propaganda. While it is unhealthy shortterm, and actually cuts away life span if you do it your whole life, I seriously doubt having messed up sleep-habits while you're 16 is going to do much to you. Or otherwise the constantly jetlagged partying 16-25 crowd will be in serious trouble.

I think that the best time to study is morning-noon, that's when most things happen at work places anyway. But I don't know anyone at university who actually does this. I like to study at night too, although typically much later than you.

I personally think (I surmise it has also been discussed among the medical society) : The quality of sleep matters much more than the quantity of the sleep; This is a subjective opinion, do what you feel is best for you.  :)

I usually start studying at 21:00, but I finish at about 12:00, sometimes earlier/later...
I have to get up at 7 o'clock because my school starts in 8.. :)

[quote="evilhamster"][quote="Kalle"]I seriously doubt having messed up sleep-habits while you're 16 is going to do much to you. Or otherwise the constantly jetlagged partying 16-25 crowd will be in serious trouble.[/quote]
Well, if everyone constantly parties, then there is no control of people who actually get sleep, and thus it's impossible to tell. I doubt you would be fine, because otherwise we wouldn't naturally want to go the sleep for eight hours a day. There's probably a reason for researchers saying children and teens need 8 hours of sleep each day (10 according to one of my teachers!).
In any case, if you don't get enough sleep, while your body might not complain, at some time in the day you'll nod off, often in a most inconvenient time.[/quote]
I didn't mean to imply that everyone "constantly parties", just that there is a large group of people who do. I think actually most people's day is longer than 24 hours. That is, if living in a place where you can't tell time, you have a longer cycle than 24 hours. Sleep-patterns are probably more contingent on when it's light outside than anything else.

It is [i]healthier[/i] to sleep in a regular manner, but there are also costs of doing this, in that you're not as socially flexible and may miss out on a lot. I doubt your teachers factor in this. It's as if teenagers are not real people yet, so the only goal should be to somehow maximize teenager's overall well-being; well-being in an adult sense measured in terms of success and virtues. If they really cared about how much you sleep, how about starting school later? But I don't know - it seems like a lot of american adolescents here are under a lot of pressure to act adult, for example the importance of having some kind of job. How much is the pay or experience of having some crappy job when you're 16 really going to matter in the end?

The point is that sleeping regularly, at "adult" times, has a huge moral value to it. A lot of people will think badly of you if you wake up at noon (unless of course you have a night job, that could excuse such decadent behaviour). That's why I think all advice on how important it is to sleep needs to taken with a grain of salt. (It's kind of a no-brainer that sleeping is good for you, but are you sure that getting your 8 hours between 10 pm - 6 AM is the best thing for you?)

Well, people do have a 25-hour sleep cycle, but more importantly, sleep is regulated by hormones that are triggered by darkness. So while it is not important to get sleep at night, it is definitely easier to do so. I know for a fact that it is possible to get eight hours of sleep each night, because I get nine hours usually (you think I'm lying, don't you?).
But I guess you may have a point about socially fitting in. I wouldn't know, though, because everyone knows me as really quiet and sort of an outcast(and I like it that way). So I can't offer any advice on that. But I really don't think that all the studies showing the importance of sleep can be tossed out so easily.

[quote="exceLLence"]I usually start studying at 21:00, but I finish at about 12:00, sometimes earlier/later...
I have to get up at 7 o'clock because my school starts in 8..[/quote]
Well, I have to get up at 7 as well, but i find it easy to get to bed on time. Even if my teachers were to stack on another hour of homework every night, I would still be able to get the eight hours of sleep I need and find time for extracurricular activities.

I've never met a person who would sleep 8 years... :huh: Or did you mean 8 years per life-time?!

If you meant 8 hours, instead of 8 years:

I doubt that less than 8 hours sleep could somehow damage you. As for me, I feel myself quite awake,even after 36 hours of not sleeping, however, I could easily doze off, while sitting in a lecture, even if I slept like a log that night.

I sleep approximately 8 hours and that seems OK for me...though now, I simultaneously study at school and university, so that might make sense to change my schedule...6 hours sleep would be normal for me...

Regarding your question:

You'd better spend more time sleeping than doing your assignments, homeworks or whatever.

The key is time management. Is it really worth it to stay up to do this assignment? Can I do this assignment during school the next day at lunch or whenever? How much will I actually accomplish by doing this assignment?

A lot of teachers assign "busywork." If you're homework is indeed busywork, then you shouldn't waste your time doing them. You can spend more time sleeping, or if you wish, actually [i]studying[/i] the course materials.

This week so far, I've gotten less than 8 hours everyday (due to a flood of labs due). This may sound normal for any hardworking high schooler, but I normally get 9 or 10 hours every night. So this transition was pretty hard on me; I was constantly tired throughout and I found it difficult to fully concentrate. (I said the mean of a set of numbers from 2 through 7 was 18  in my AP Statistics course :ninja: )

Speaking as a college student with a (at best) mediocre sleep schedule, this talk of "8 hours per night" seems quite quaint - I rarely get more than 6.5 nightly, and have gotten 8+ hours about five times this semester.  Last semester, I averaged about 4.8 hours per night (yes, I kept track).  I was perfectly fine after adjusting.  So speaking from personal experience, 8 hours of sleep per night is not at all necessary for health.

[quote="7h3.D3m0n.117"]The key is time management. [/quote]

Perfect. As a boarding school student, It's hard to study/do homework during the three hours given to the students because of the busy dorm life. I normally end up procrastinating and getting my work done at 12 or 1. and wake up at 8. (I have a free. thank goodness), but I decided one day, to try to manage my time. I got to my room at 7, and finished my work at 9:30. I went to bed at 11. simple.

Also. According to a video I watched in health class (yes. I actually remember things from health class), students are supposed to get 9.25 hours of sleep every night. What do you guys think about that? I normally get 7-8 hrs of sleep.

Time management- in two words described what could have been described in sentences. It's the best you can do.

Don't study late, study when you have time on [b]daylight[/b]. When you are bored, go study (sounds like a nerd). And homework, did you notice what kind of your teachers are? Do they all take notebooks and look for your homework? Don't do what you don't have. And of course, don't leave things for tomorrow if you can do them today.

Sleeping is very important. Have that in your mind.

The amount of sleep you need to perform optimally depends on you: age, how active you are, diet, etc.

Sleep affects your efficiency a lot.  That said, sometimes you just have so much to do and so little time that you need to make this tradeoff.  Caffeine actually can help counter the loss of alertness that comes with sleep deprivation.  This is all in the short run though, don't try doing this day after day...you'll become very unhealthy.  Being at college I've seen some pretty hard crashes.  I have some friends that overloaded their class load and ran a huge sleep deficit over the term.  One of them drove their immune system through the floor and got very sick and the other had a pretty epic caffeine crash one night ended him up at medical.

In short, realize that '8 hours a night' is a guideline.  I know people that need more and people that need less.  I feel that you can probably train your body to perform better on less sleep (at least I feel I have...I might be kidding myself, I dunno).  Know your own limits.  It's ok to go beyond them at times...just try not to be stupid about it ;)


Prove that:
$ \frac{1}{ab\plus{}2c^{2}\plus{}2c}\plus{}\frac{1}{bc\plus{}2a^{2}\plus{}2a}\plus{}\frac{1}{ca\plus{}2b^{2}\plus{}2b}\geq\frac{1}{ab\plus{}bc\plus{}ca}$
for all $ a,b,c>0$ satisfy $ a\plus{}b\plus{}c\equal{}1$

its 2007 not 2006...
[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=810857#p810857[/url]


group theory
abstract algebra
superior algebra
superior algebra unsolved

[quote="bike_me"]Let $ G$ be a finite group and let $ P$ be a normal $ p$-subgroup of $ G$. Show that $ P$ is contained in every Sylow $ p$-subgroup of $ G$.[/quote]

[hide]One way is to show that every $ p$-subgroup is contained within a Sylow $ p$-subgroup, and then use the fact that all Sylow $ p$-subgroups are conjugate.[/hide]

Or, more directly:

[hide]Let $ H$ be a Sylow $ p$-subgroup. Let $ |H| \equal{} p^n$ and $ |P| \equal{} p^a$. The coset $ HxP$ has $ \frac {(p^n)(p^a)}{|H \cap xPx^{ \minus{} 1}|} \equal{} \frac {(p^n)(p^a)}{|H \cap P|}$ elements since $ P$ is normal. We must have $ |H \cap P| \ge p^a$, or else $ p^{n \plus{} 1} \mid |G|$, as the double cosets partition $ G$. It follows that $ P \subseteq H$.[/hide]

For the first version I cannot seem to prove that every $ p$-subgroup is contained within a Sylow $ p$-subgroup. Any ideas?

people.brandeis.edu/~igusa/Math101aF07/Math101a_notesA6.pdf theorem 6.6 (1).


inequalities
algebra
function
domain
inequalities unsolved

\[ \begin{gathered}
   \hfill \\
  \sqrt {x^2  \minus{} 1}  \cdot \left( {ax^2  \minus{} x} \right) \geqslant 0{\text{ with a }} \in {\text{ }}\left] {{\text{0}}{\text{,1}}} \right[ \hfill \\
\end{gathered}
\]

[quote="sfsreporter"]domain of x?[/quote]
\[ x \in \mathbb{R}
\]


Express the area A of an isosceles right triangle as a function of x if the length is twice the width of a rectangle.

Does the base of an isosceles right traingle equal the length of a rectangle?

Does the height of an isosceles right triangle equal the width of a rectangle?

your first question makes absolutely no sense

and neither do the other two actually

[quote="sharkman"]In that case, how do I answer:

Express the area A of an isosceles right triangle as a function of x if the length is twice the width of a rectangle.

Thanks[/quote]

Wait, this isn't much better.

What rectangle? The length of what? "The width of a rectange"?

[quote="sharkman"]The answer is A(x) = (x^2/2).

How do I find it?[/quote]

I guess that means that both legs have length $x$?

[hide="hint"]Shark, let the rectangles length be the height of the isosceles. The width, the base. Then apply $A = \frac{1}{2}bh$

[/hide]

The last replied was very helpful and helped me to see why the answer is A(x) = (x^2/2).

Thanks


geometry
ratio
trapezoid
probability
number theory
relatively prime

From an ICTM State math contest; #20:

A circle with center O is inscribed in a quadrilateral ABCD with points of tangency at E (on AB), F (on BC), G (on CD), and H (on AD). Ab is parallel to DC, but AD is not parallel to BC. If AD = 337, BE = 62, FC = 225, and DG = AE + 175, then DO:AO = k:w, where k and w are two relatively prime positive integers. Find the value of k + w.

So... it's possible and impossible at the same time... credits to djshowdown for bringing it up at that particular moment in time.

It would appear possible at first glance because AB and DC are parallel, so GE goes through O and is perpendicular to both DC and AB, and because you know that c is supplementary to B..

So then you can easily find that the radius of the circle is (sqrt[287^2-163^2])/2
which is sqrt (13950).  Then you can find AD, which is sqrt(13950+175^2)...
From there, you just need to use the fact that it's a inscribed circle to find AE...then you can solve it...

go ahead and work it out if you would, I'm curious as to see what people get.  I think the diagram and lengths are possible to find, but using the answer presented in the contest solutions, I am pretty sure it is impossible.  In other words, I think the answer they gave is wrong, but maybe there's something I'm missing.  See what you get.

Well once you get the radius to be $\frac{\sqrt{273138}}{4}$,

you can easily find $AO$ and $DO$.  The ratio is $\frac{1961}{561}$

$1961+561=\boxed{2522}$

The provided solution was 25.  my schools' geometry teacher had a look at it, and figured that was from the ratio being 16/9, making those segments form a right angle.  I was playing around with it, but problem is, if that's the case, the radius becomes 144, diameter 288, but that one side is only 287.

I agree with the official answer.  I'm checking now for the possibility that the question is over-determined.

I don't see how that's possible.  You can easily find the side lengths of the quadrilateral, and even the lengths of the vertices to the tangency points.  I think that should pretty uniquely determine it...

So, based on an experiment with Geometer's Sketchpad, I believe with very high confidence that the problem is bad -- the quadrilateral described, a trapezoid with 4 sides that are (effectively) given, does not in fact have an incircle.  I would file an appeal with the contest administrators as soon as possible.  Your best bet for a formal demonstration is to take some subset of the givens (I started with the lengths of DE, AH, the relationship between DG and AE, the fact that an incircle exists, and the fact that AB and CD are parallel, not using the given length of CF) and prove that the final given cannot hold.  Via my method, I found that the length of CF would have to be roughly 335, not the 225 they want it to be.  Also, you can note that the diameter will be different if we calculate it on the left hand side or the right hand side.  One leads to nat mc's value for the radius, the other to a value of 144.

WRT what probability said before I finished that post, side lengths do not determine a quadrilateral.  However, side lengths and having two given sides parallel do.  In this case, those conditions preclude the existence of an incircle.

JBL's post overrides mine (seems to be in general too :P).


[quote="countdownmath"]Give an example of a divergent series whose partial sums are bounded.[/quote]

$\sum_{n=1}^{\infty}(-1)^{n}$  :rotfl:


Let $ x,y$ be non negative real numbers satisfy : $ x^6\plus{}y^5 \leq x^5\plus{}y^4$.
Prove that $ x^5\plus{}y^5 \leq 2$.

by Cauchy Schwarz ,$ ( x^5\plus{}y^4)^2\le(x^4\plus{}y^3)(x^6\plus{}y^5)\le(x^4\plus{}y^3)(x^5\plus{}y^4)$
so $ x^5\plus{}y^4 \le x^4\plus{}y^3$
similarly,$ x^4\plus{}y^3\le x^3\plus{}y^2 \le x^2\plus{}y \le x\plus{}1$
so $ x^6\plus{}y^5 \le x\plus{}1 \iff y^5\le x\plus{}1\minus{}x^6$
$ x^5\plus{}y^5\le 1\plus{}x\plus{}x^5\minus{}x^6\equal{}2\minus{}(x\minus{}1)(x^5\minus{}1)\equal{}2\minus{}(x\minus{}1)^2(x^4\plus{}x^3\plus{}x^2\plus{}x\plus{}1)\le 2$


Prove that if $m$ is not prime and $\phi(m) \mid m-1$, then $m$ has at least four distinct prime factors.

Note:  The proof that it has at least 3 distinct prime factors is good practice as well.

[hide="Through three"] $m$ has one prime factor:  if $m = p^{k}$ then $\varphi(m) = (p-1) p^{k-1}\not | p^{k}-1$, contradiction.

$m$ has two prime factors:  $m = p^{a}q^{b}$ and $\varphi(m) = (p-1) p^{a-1}(q-1) q^{b-1}| p^{a}q^{b}-1$.  But if $b > 1$ then $q^{b-1}|-1$, contradiction; and similarly for $a$.  Then $m = pq, \varphi(m) = (p-1)(q-1) \not | pq-1$, contradiction.

$m$ has three prime factors:  We know that they cannot have power greater than $1$; hence $m = pqr, \varphi(m) = (p-1)(q-1)(r-1) | pqr-1$.  But then we have the system

$p-1 | qr-1$
$q-1 | pr-1$
$r-1 | pq-1$

[b]Case:[/b]  WLOG $p = 2$.  Then $q-1 | 2r-1, r-1 | 2q-1$.  But WLOG $r > q$; then $r-1 \ge q$ so we must have $r-1 = 2q-1 \implies r = 2q$, contradiction.

[i]To be continued...[/i] [/hide]

Hint for method that helps solve the problem for at least 3 distinct prime factors and can be extended to 4.
[hide]$\frac{m-1}{\phi(m)}=\frac{p*q-1}{(p-1)(q-1)}$
Manipulate right side of equation
2nd hint (if needed):
[hide]$\frac{pq-p-q+1+p+q-2}{(p-1)(q-1)}$[/hide][/hide]

Solution to "at least 3 distinct prime factors":

[hide]Let $m=p*q$,
$\frac{m-1}{\phi(m)}=\frac{pq-1}{(p-1)(q-1)}$

Realize that $p*q-1=pq-p-q+1+p+q-2=(p-1)(q-1)+(p-1)+(q-1)$

$\frac{(p-1)(q-1)+(p-1)+(q-1)}{(p-1)(q-1)}=1+\frac{1}{q-1}+\frac{1}{p-1}$

$\frac{m-1}{\phi(m)}= 1+\frac{1}{q-1}+\frac{1}{p-1}$

Which implies that $0<\frac{1}{q-1}+\frac{1}{p-1}\leq 2$

The only possible integer solutions are when $q=p=2$ or $q=p=3$ we have a contradiction since the primes are not distinct.[/hide]

Solution to "at least 4 distince prime factors

[hide]Same methodology, just use the expression $\frac{pqr-1}{(p-1)(q-1)(r-1)}=\frac{pqr-(pq+pr+qr)+(p+q+r)-1+(p+q+r)-2(p+q+r)+(pq+pr+qr)+3-3}{(p-1)(q-1)(r-1)}=\frac{(p-1)(q-1)(r-1)+(p-1)+(q-1)+(r-1)+(p-1)(q-1)+(p-1)(r-1)+(q-1)(r-1)}{(p-1)(q-1)(r-1)}$
[/hide]


Let a,b,c be nonnegative numbers with a+b+c=3.Prove the inequality

a/b^2+1  + b/c^2+1  + c/a^2+1   >= 3/2

and this the doc format of the problem:

That is Cauchy
$ a\minus{} \frac{a}{b^2\plus{}1} \equal{} \frac{ab^2}{b^2\plus{}1} \le \frac{ab^2}{2b} \equal{} \frac{ab}{2}$

the inequality is true only if $ a,b,c \neq 0$ (take $ a\equal{}3$ and $ b\equal{}c\equal{}0$)
[b]great math[/b], let me complete your proof,
we have $ \frac{a}{b^2\plus{}1} \geq a\minus{}\frac{ab}{2}$
then $ S\equal{}\frac{a}{b^2\plus{}1} \plus{} \frac{b}{c^2\plus{}1} \plus{} \frac{c}{a^2\plus{}1} \geq 3\minus{} \frac{ab\plus{}bc\plus{}ca}{2}$
it's easy to prove that: $ ab\plus{}bc\plus{}ca \leq \frac{1}{3}(a\plus{}b\plus{}c)^2 \equal{}\frac{3}{2}$ thus $ S\geq 3\minus{}\frac{3}{2}\equal{}\frac{3}{2}$

That is absolutely fantastic!
The point of using Cauchy- Schwartz has been commented on many forums and its present usefulness is still highly recommended.


Given a point $P$ on the side $\left[AB\right]$ of a triangle $ABC$. Construct two lines through this point $P$ which divide the triangle $ABC$ into three parts of equal area.

1. First of all we bisect the area of $ABC$ (Figure 1). Construct point $M$ at the midpoint of $AB$ and draw a line thought $M$ parallel to $CP$. Suppose WLOG that it intersects side $BC$ at p.$Q$. Then $PQ$ bisects the area of $ABC$. (exercise!)

2. Construct $R\in QB$ s.t. $2QR=RB$. Then $[PRB]=\frac{1}{3}[ABC]$.

3. Construct $R'$ on ray $QC$ s.t. $QR'=QR$. There are two cases:
a) Figure 2. $R'\in QC$. Then $PR$ and $PR'$ are two required lines.
b) Figure 3. $R'\not\in QC$. Construct a line through $R'$ parallel to $CP$, which intersects $AC$ at point $R''$. Then $PR$ and $PR''$ are required lines (exercise!).


MATHCOUNTS
Alcumus
geometry
probability
FTW
AoPSwiki
complementary counting

I am posting this because I am going to be competing in the mathcounts competition!  :football:   So I just wanted to ask anybody who has previously competed for some advice! Hopefully I can get some help!!  :coolspeak:

[quote="celiasoccer"]Well i think maybe you should study your formulas so our team will win!!![/quote] I didn't mean replies from people on my team!

Okay guys, here we go-
1. Do old competitions.
This one is obvious, but make sure that, if you get one wrong or can't solve it, look at the solution and try to understand it.
2. Mathcounts marathon!
It has lots of problems, and lots of solutions.
3. Counting is a must!
Make sure you know combinations, permutations, complementary counting, ect. (hint:Alcumus)
4. The subjects:
You need to know some algebra 1 (for example, FOIL), some geometry (In my opinion, the hardest problems are geometry), counting and probability, number theory, a bunch of terminology (generally for geometry), and so on and so forth.
5. understanding good. memorizing... BAD!
If you understand a formula, you are more likely to remember it. for example:
$ 1\plus{}2\plus{}3\plus{}...\plus{}n\equal{}?$
$ 1\plus{}2\plus{}3\plus{}...\plus{}n\minus{}2\plus{}n\minus{}1\plus{}n\equal{}$
$ \frac{n}{2}*(n\plus{}1)\equal{}$
$ \frac{n(n\plus{}1)}{2}$
explanation:
every value can be paired up with another to create n/2 pairs of value n+1. Multiply these together to get total value.
While it may take longer to memorize, If you happen to forget the formula, you can derive it.
Though, if you can't just remember the Pythagorean theorem...
6. do problems.
7. do more problems.
8. do... I really need to say it?

I have done this before, I scored high on chapters.(will edit for score when is allowed.)
I am hoping I can do better on this competition than last year.
the biggest helper, though, is
9. experience
You aren't likely to win on your first try. I can just hope you aren't inn 8th grade...

extra tips-
play FTW!
post problems you don't understand on this forum.
AOPSwiki to find formulae you don't know.
look at my signature. It basically says that no matter how hard you train, somebody can still be better.
Thusly, practice does not make perfect.

Thanks!
I'm really nervous cause this is my first competition!!  :help:

Take a look at these websites-

[url]http://mathcounts.saab.org/mc.cgi[/url]

[url]http://www.mathcounts.org[/url]
 go to previous competitions.
The most important rounds in chapter are sprint and team.
Team is the hardest round, but sprint is key- It has the most problems, and because you don't have a calculator,
It is easy to make stupid mistakes.
don't really worry about memorizing pi, or the square roots, You will either:
1. be allowed to use a calculator
2. be asked to use simplest radical form(wiki if you don't know!) or in terms of pi.

Being nervous is okay! Just do your best, and if you win, YAY! If you lose, you will have that experience for next year.

A quick problem help (this kind stumped me last year)-
what is the last digit of $ 4^{201}$
This is easy! The last digit of the powers of 4 are 4, 6, 4, 6, 4, etc.
So, divide 201 by 2(the length of the pattern), find the remainder(1) and find that position in the pattern(the first number in the pattern is 4) so the answer is 4.

If you post the formulae you know, I can ad some more that may be of help.
Just do your best.
And If you do AWESOME (and I do too) I will see you in Orlando!


geometry
circumcircle
trigonometry
modular arithmetic
cyclic quadrilateral
geometry proposed

Let $ ABCD$ be a circumscriptible quadrilateral and let $ \gamma$ be an arbitrary line tangent to its incircle. Let $ A'$, $ B'$, $ C'$, $ D'$ be the orthogonal projections of $ A$, $ B$, $ C$, $ D$ on $ \gamma$, respectively. Prove that
\[ \frac{AA' \cdot CC'}{BB' \cdot DD'} \equal{} \frac{AO \cdot CO}{BO \cdot DO}.\]

[b]Lemma:[/b] $ ABCD$ is a cyclic quadrilateral and $ P$ is a point lying on its circumcircle $(O).$ Let $ X,Y,Z,V$ the orthogonal projections of $P$ onto sidelines $AD,AB,BC,CD.$ Then $ PX \cdot PZ = PY \cdot PV.$

$\frac {PX}{PY} = \frac {\sin \widehat{ PAX}}{\sin \widehat{PAB}} \ , \ \frac {PZ}{PV} = \frac {\sin \widehat{PCB}}{\sin \widehat{ PCD}}$

Since $ \angle PAB = \angle PCD$ and $ \angle PAX = \angle PCD$ $\pmod\pi,$ it follows that $ PX \cdot PZ = PY \cdot PV$ and the proof of the lemma is completed. $\square$
___________________________________________

Let  $ M,N,L,Q$ be the tangency points of  the incircle $(O)$ with $ DA,AB,CB,CD.$ Let $ X,Y,Z,V$ be the projections of $ P$ onto $ MN,NL,LQ,QM,$ where $ P \equiv   \gamma \cap (O).$ $\gamma$ is the polar of $ P$ WRT $ (O)$ and $ MN,NL,LQ,QM$ are the polars of vertices $ A,B,C,D,$ respectively. By Salmon's theorem, we deduce that

$ \frac {AA'}{AO} = \frac {PX}{R}  \ \ , \ \ \frac {CC'}{CO} = \frac {PZ}{R}$

$ \frac {BB'}{BO} = \frac {PY}{R} \ \ , \ \ \frac {DD'}{DO} = \frac {PV}{R}$

But according to the previous lemma, we have $ PX \cdot PZ = PY \cdot PV$

${\Longrightarrow  \frac {AA' \cdot CC'}{BB' \cdot DD'} = \frac {AO \cdot CO}{BO \cdot DO}}$

Very nice Luis, thanks. My proof was way uglier (and by this I mean really computational).

With the same conditions, but now $ ABCD$ is bicentric. Show that

$ \frac {AA' \cdot CC'}{BB' \cdot DD'} \equal{} \frac {[ABCD]}{BO \cdot DO} \minus{} 1 \ , \ \frac {BB' \cdot DD'}{AA' \cdot CC'} \equal{} \frac {[ABCD]}{AO \cdot OC} \minus{} 1$


Write $\frac{1}{x(x+1)}=\frac{1}{x}-\frac{1}{x+1}$  and telescope to have $\frac{1}{x}-\frac{1}{x+10}<\frac{1}{x-2}$ which is equivalent to $\frac{-20-x^{2}}{x(x+10)(x-2)}<0$
which is equivalent to $x(x+10)(x-2)>0$ which is true for
$x\in\left(-10,\ 0\right)\cup\left(2,\+\infty\right)$.

I hope it is true


Find the hundreds digit in the sum $\sum_{k=1}^{999}k*k!$

[hide]I will say 9.
$\sum_{k=1}^{999}k\cdot k! = \sum_{k=1}^{999}((k+1)!-k!)$

$\implies 1000! - 1!$

1000! clearly has 000 as the last three digits and when we minus 1, it will become 999 and which is why i think the hundreds digit is 9.[/hide]


inequalities function algebra domain probability calculus integration

$P(|x-\mu|\geq k\sigma) \leq \dfrac{1}{k^2}$

Can someone prove this without using 'integral'?  :)

(I know the proof using Markov's Ineq. I hope I can find a proof not using 'integral'. Please, help)

There is no way to avoid the generalized concept of integral.

A random variable is a (measurable) function whose domain is a probability sample space. A probability sample space is a measure space of total mass 1.

For any random variable, $E(Y)$ is the integral of $Y$ over the probability space, with respect to the probability as the measure. Hence every reference to an expected value is a reference to an integral.

Chebyshev's inequality has a hypothesis that cannot be ignored: we suppose that $E(X^2)<\infty.$ (In real analysis terms, this is the same as claiming that $X\in L^2.$) The statement of the theorem involves $\sigma,$ and $\sigma$ is defined by $\sigma^2=E(X^2)-E(X)^2.$ Since that involves expected values, it involves the concept of integration.

In a Lebesgue measure and integration course, the follwing inequality is proved: Suppose $f\in L^p$ for some $p<\infty$. Then for all $\alpha>0,$

$m(\{x: |f(x)|>\alpha\})\le\dfrac{||f||_p^p}{\alpha^p}$

You might as well call that Chebyshev's inequality; it really is the same thing, with the same argument.


I had been solving problems from the Contest section and I cannot verify if my answers are correct. So I think, it is also better if the moderators (or administrators) post  the answer keys. Of course, we cannot post all the problems here to verify the correct answers.

Thanks.

Click on the number of the problem. It will open up a seperate thread where usually the solution is discussed.

[quote="Knuth's_check"]I would rather have the answer key, so I can continue to work on the  problem rather than just skipping to the solution.  :D Also, some problems don't have the solutions posted yet. Anyway you can find solutions at the [url=http://www.unl.edu/amc/]AMC website[/url]. ;)[/quote]

Right... It is better if only correct letters are displayed so that if our answers are wrong we can go back to the problem and solve it again. But if we immediately proceed to the thread where the solutions are discussed, I think that's not problem solving anymore... :D


LaTeX
Asymptote
MATHCOUNTS
number theory
least common multiple
geometry
rectangle

[quote="levans"]I'm looking for some interesting math problems for younger students. These can be regular math problems, games, puzzles, etc. The criteria for these problems is that they must be pure text (no images, no LaTeX, no Asymptote) and should be suitable for kids somewhere in the age range of 6 and 11.

Please post any suggestions here.[/quote]

Is this a good one?:

A hose fills a pool in 3 hours. A drain empties the pool in 5 hours. If the hose fills the pool when the drain is open, how long will it take to fill the pool?

[hide="solution"]

fill will be $ \frac{1}{3}$ - $ \frac{1}{5}$

this will become $ \frac{2}{15}$

so $ \frac{15}{2}*60$

$ \boxed{450}$ minutes[/hide]

[quote="MCNational!"][hide="solution"]

fill will be $ \frac {1}{3}$ - $ \frac {1}{5}$

this will become $ \frac {2}{15}$

so $ \frac {15}{2}*60$

$ \boxed{450}$ minutes[/hide][/quote]

we're giving problems to levans, so he can use them for some unknown reason. We're not solving problems.
Aw man, there is no sad :P face. :noo:

Thanks, this is an appropriate problem. Also, others can feel free to post solutions.

The problems are for a gifted e-mail list.

I know 3 other good ones :)

[hide="1"] One light flashes every 5 minutes. Another flashes every 12 minutes. The third light flashes every 8 minutes. If they all flash together at the same time at 10:00 AM, when is the next time they will flash together?[/hide]

[hide="2"] John runs a mile in 6 minutes. Bob runs a mile in 9 minutes. Suppose they were on a circulur track and ran at the same time. In how many minutes will John meet up with Bob again after he passes Bob? [/hide]

[hide="3"] Refer to the problem with John and bob. edits:
a) suppose John gets a 30 second start. When will John meet up with Bob again after passing Bob TWICE?

b) suppose Bob gets a 45 second start. In how many minutes will John pass Bob after passing bob 3 times?[/hide]

[hide="Bonus"] Two trains head toward each other at the same speed: 45 mph. Train A starts at 10:00 AM to travel and Train B starts at 10:15 PM to travel. If the trains pass at 11:00 AM, how long is the path they will both travel?[/hide]

Is it possible to fill a 3x3 table with integers such that the sum of the numbers in each row is odd and the sum of the numbers in each column is even? (or any oddxodd)

Is it possible to tile an 8x8 chess board where opposite corners are removed with 1x2 tiles?

Thanks guys. I've got them all down. If anyone wants to try solving them now, you are free to do so. If anyone has any more problems, also please post. I still need some for younger students, but can probably cull some from MOEMS or easy MATHCOUNTS stuff.

Ah.... I had a really really good logic problem but I can't find it now... maybe among my mountain of papers I'll find it.

there are fun math/logic games here [hide] http://www.mathplayground.com/[/hide]

Uhhh...I'd like to Answer some problems I guess.

Lights Problem:
To find when they will blink again find a multiple of 5,8, and 12.
The LCM of 5 and 8 is 40, the LCM of 12 and 40 is 120, so 120 minutes later is 12:00.

Bob and John:

The LCM of 6 and 9 is 18 minutes, at which point John will have run 3 miles and Bob 2.

I'll let someone else do a) and b) if anyone is going to..

I have a *REALLY* good crossword puzzle. I think it's probably too hard though :(

Here are some problems I made;

I went to the store with 20dollars and bought three apples that all cost the same amount of money. Then
I bought a banana for 4 dollars. When I spent all the money I was left with 7 dollars. What was the cost of one apple?

Five people are supposed to sit in a row of five chairs to have a picture of them taken. How many ways are there to arrange the five people?

A square has a side length of 10. It is broken into two equal rectangles. One of the rectangles is broken into two equal squares, one of which is removed. What is area of the remaining parts of the square?

How many positive factors does the number 24 have?

So, working backwards on the banana apple problem, you had 7 dollars left when you bought 3 apples and a banana for 4 bucks. That means 20-4-7= 9 dollars for the 3 apples, so each apple was ... 3 bucks!!

For a problem like the sitting in a row, use a factorial. 5 times 4 times 3 times 2 times 1 equals... 120! so there are 120 ways. :o weird huh?

Okay so, basically a square with an area of 100 was divided into three parts. One 50 piece rectangle and two 255 pieces. If one square was removed, 25+50 remains.

The area of the remaining pieces is 75.

Hmm... Is this becoming a marathon...  :?:  :!:

Sorry, I just now noticed the thread. Is it too late to suggest problems?

For it's original purpose, yes. But feel free to post more. I especially like interesting problems in which my wee little brain can solve :)


find all positive integers $ {x},{y}$ such that:

$ {x} \plus{} {x}^3 \plus{} {x}^5 \equal{} {y}^2$


superior algebra
superior algebra unsolved

Does an $ M>0$ exist, where $ |\alpha| \geq M$ for all algebraic integers $ \alpha$ of degree greater than 1?

No, $ (\sqrt{2}\minus{}1)^n\to 0$ for $ n\to\infty$.


Prove that if $\forall x \in R$ we have $acosx+bsinx+c cos 2x+d sin 2x \geq-\sqrt{c^{2}+d^{2}}$ then $a=b=0$

$r=\sqrt{c^{2}+d^{2}}$
Let $\phi$ such that $0\leq \phi<\pi$ and
$cos2\phi=\frac{c}{r}and sin2\phi=\frac{d}{r}$
$A=acos\phi+bsin\phi$ and $B=bcos\phi-asin\phi$
then I have
$rcos2x+Acosx Bsinx \geq-r$
replace $x$ on$x+\pi/2$then
$rcos2x-Bcosx+Asinx\leq r$ for all$x$
then QED.
Cuopbientcd.


group theory
abstract algebra
superior algebra
superior algebra solved

Prove that if N is a normal subgroup of G, and H is any subgroup of G, then H ∩ N is normal subgroup of H.

since $e\in H \cap N,$ and for any $a,b \in H \cap N, ab^{-}1 \in H \cap N$ hence $H \cap N$ is a subgroup.
for any $h\in H$, and for any $k\in H\cap N, hkh^{-}1 \in H$ and $hkh^{-}1 \in hNh^{-}1 = N$ as $N$ is normal in $G$. Then $hkh^{-}1 \in H\cap N$ implies that  $h(H\cap N)h^{-}1 \in H\cap N$, hence $h(H\cap N )h^{-}1 = H\cap N$. and so it is a normal subgroup of $H$.


Check out this prank from trigger happy tv

http://www.youtube.com/watch?v=KNPAtGQsxoI&mode=related&search=

no one thought it was funny? i was laughing so hard the first time i saw it  :rotfl:


Find all positive integers a,b,c such that a|(bc+1), b|(ca+1), c|(ab+1)

[quote="galois"]the only such set is {2,3,5}.i do have a proof of this one .it's kinda long .so i'll do the posting stuff later.k
 :D[/quote]

you think so? is 2.3+1 (seven thus) dividable by 5???

I think it's not.
I think only my solutions are correct :D

Those are actually not all the solutions: for instance, permutations of $(1,1,2)$ and $(2,3,7)$ work as well. I have a full solution posted [url=http://wangsblog.com/jeffrey/?p=85]here[/url].


θεωρούμε έναν 2x2 πίνακα Α.Δείξτε ότι υπάρχουν 2 πίνακες 2x2 Β και C ώστε Α=ΒC-CB αν και μόνο αν trA=0 (ίχνος του Α=0)
βασικά το αντίστροφο δεν μπορώ να αποδείξω,δηλαδή αν ένας πίνακας Α που έχει trA=0,γιατί να γράφεται έτσι, δηλαδη Α=BC-CB?

Νομίζω ότι θα έπρεπε Το γραμμικό σύστημα που προκύπτει με αγνώστους τα στοιχεία του Β και παραμέτρους τα στοιχεία του C να έχει πάντα μια τουλάχιστον λύση,δηλαδή να μην είναι αδύνατο, για οποιονδήποτε Α με tr(A)=0. Έτσι καταλήγεις στην διερεύνηση συστήματος 4x4 ...Το Mathematica δίνει αρκετή βοήθεια σε αυτό

Spinos, σκέψου τι σημαίνει να έχει ένας πίνακας μηδενικό ίχνος: το άθροισμα των ιδιοτιμών του είναι $ 0$.

Όμως, το ίχνος είναι επίσης ίσο εξ'ορισμού με το άθροισμα των διαγώνιων στοιχείων.

Άρα, αρκεί να μπορέσεις να γράψεις στη ζητούμενη μορφή τον πίνακα {{1,b},{c,-1}} ή τον {{0,b},{c,0}} :)

Cheerio,

Durandal 1707


Find all $ (l,k)$ with $ 2k > l > k$ ,such that -
$ (l \minus{} k)^3 \plus{} (l \plus{} k \minus{} 1)(l \minus{} k) \minus{} lk \equal{} 0$

Are you asking for integer solutions or real solutions?

if $ l$ and $ k$ are real number , then there are infinity solutions

i want intger solutions(if any) to the equation
@R.Maths
how is there infinite solutions if it is real?

how to solve this equation? it has been a long time and no-one has replied  :(


Dumb things that people do:

See this site:

http://www.dumbthings.com

:D:D:D

Those are so dumb that they're funny. :lol:

haha, I'll remember this when I have a bad day  :rotfl:

oy that's funny!!!! Yah i know this other site, about graphitti or something, it has some of that stuff on there.


Our club has 25 members, and wishes to pick a president, secretary, and treasurer.  In how many ways can we choose the officers, if individual members may hold all 3 offices?

Humph I'm not so sure but I believe it is $ 25^3$

25 ways for prez
25 for sec
25 for treas

[quote="antybash"]why didn't it accept my 25^3?[/quote]

I think you have to expand it, so the correct answer should have been $ 25^3\equal{}15625$... What did the answer say?

[quote="GameBot"]if individual members may hold all 3 offices?[/quote]

Yep, another vote for the system accepting $25^3$.

[quote=Cromptu]Wouldn't the answer be 13800? that is, 25* 24 * 23?
Because no person can hold more than one office, so for the president we have 25 choices, for the secretary we have 24 choices ( leaving out the one who is the president ) and for the treasurer we have 23 choices ( leaving out the 2 who are the president and secretary)[/quote]

People can hold more than one office. It says so in the problem


AMC AIME quadratics calculus derivative ratio geometry

Was it just me, or was the AIME II much easier than AIME I?  I was on spring break and took the AIME I as practice, and yesterday's test seemed much easier.  Are there different cutoffs for each test or are AIME I participants just at a disadvantage?

The problems were nicer. Especially 12, 13, 14, and 15. Maybe a more approachable last third and first third made it easier? Though
5, 6, and 7 were pretty hard if one didn't look for good solutions.

8 was the problem I felt was personally best on the test. Diophantine equations and Combinatorics rule.

The cutoff levels will depend solely on how well the people who took the test did.

I tried both AIMEs, and I thought each AIME this year was comparable to previous years’ difficulties, as they should.  Of course, both AIMEs featured different topics and questions, but both had their share of “easier” and harder questions (see the straightforward AIME I #3) -- neither test had significantly easier questions than the other.  (As PenguinIntegral posted, #5,6,7 on the AIME II “were pretty hard if one didn't look for good solutions”.)  People more prepared for certain types of problems no doubt will find those problems easier than other types; it depends on the individuals.  AIME I and II are of comparable difficulty, just hosted at different times.  Each year, I see personal bias on difficulty levels being expressed, but it does not change the fact that AIME I and II are of similar difficulty level; as the AMC Director said, “When the AIME Committee creates the AIME (or AIME I) and the AIME Alternate (or AIME II), two contests which are referred to as AIME Heads and AIME Tails are deliberately constructed to be balanced, in the experienced judgement of the committee, in content and difficulty.”  I think the AMC has successfully achieved this goal this year, as every year.

500 people total

I thought AIME II was nicer than AIME I, but a tad bit harder and easier to make bad mistakes with lots of crazy addition and a certain $x_{0}$ term. I really didn't like #1, especially as a 1. 3 was fun, 10, 11, 12, and 13 were really nice problems. The problem with this AIME is that so many problems took forever to understand like 1 and 8.

8 was really ambiguous. Even though it's kinda obvious, I think they should include stuff like "the lines have to be within the paper" (hey, someone could be drawing on the desk) and "the lines have to go from side to side." I thought this aime was about the same level, possibly a littler easier (though I suspect AIME I people will start complaining about how it was a lot easier).

[quote="rem"]Yeah I think the questions were nicer even though some were still just ugly bash.

Do they take top 500 ppl from both AIMEs or for each one there's a different cutoff?[/quote]
The number of people who qualify from each AIME is proportional to the number of participants for each AIME, with a grand total of as close to 500 as they can manage.

I felt that they were of even difficulty.

AIME II problems overall were generally more approachable in my opinion, but conceptwise, they were harder.

I don't see how 8 was Diophantine equations and combinatorics at all  :huh: I made a quadratic equation which I simply had to find the maximum integer of and then plug that number in.

5 was a very fun problem, 6 I misintepreted for 3 hours (kept thinking for a four digit number abcd, a<b>c<d was what they were asking). 7 was weird, I don't know about its difficulty really but I did get it right.

It was much less computation-heavy, although deeper... that actually made it much more fun lol

#8, I was lazy and took a derivative to maximize :rotfl: and actually did it right I think

OMG, i loved #5! #9 took me like an hour...cuz i did it three times, but i am sooo glad i got it right! i've looked at the previous tests (this is my first year doing it) and i thought the AIME was actually a bit easier (concepts)...

Either a miracle happened or yesterday's AIME II was one of the easiest ever, cause I went from last year's 7 to 14.

12, 13, and 15 were relatively easy. Especially 12.

It seems like the earlier and middle problems were harder than normal and the later questions were much easier and overall conceptually harder but also less error prone than the AIME I. This seems like it lead to people who were on the bubble for qualifying to get lowered scores (judging by very limited results from the polls so far) and people who usually do pretty well to do extremely well. I wonder how this will affect the floor/qualifying score.

The last problems of AIME II were definitely much easier....i did not like #6 (i had to do this one like 5 times). AIME I was relatively easy this year too. But, both were prone to stupid mistakes.

12 and 13 weren't too bad once you made some mental connections..

14 and 15 i thought were fairly difficult

haha i messed up on one and two though..... so much for easy first problems :P

[quote="Phelpedo"]8 was really ambiguous. Even though it's kinda obvious, I think they should include stuff like "the lines have to be within the paper" (hey, someone could be drawing on the desk) and "the lines have to go from side to side."[/quote]

Agreed. I feel stupid saying this but it took me forever to realize what #8 was asking because it took me forever to realize that the lines were [i]on the paper[/i] and went entirely across it.

i dunno i took the aime I for practice and got a 9 but i could have solved 13 but i took the aime II for real and got a 7 but could only solve 10-11 so iono i thought the aime I was full of doable questions just kinda hard to get started on...

I did feel the AIME II was easier. My score ratio (5/6) does not really fully reflect that; I made more dumb mistakes on the II.

I think the AoPS crowd has enough experience and whatnot to like (and do better at) the "pretty" problems, those that have relatively simple but hard-to-find solutions. In that regard, most of us are going to think the AIME II was easier. But I think that overall, they are going to come out to be approximately the same.

I wish I had taken the I - I took it in a competition setting two days later and got a 7. But I got a 5 on the II, so there went my shot at USAMO

I took the II and thought it was much easier.  I screwed 3 up stupidly and got a 9, which I wouldn't have gotten on the I even if I wasn't retarded on the 3 I missed.

Hopefully, it will be about even, since it will help me get into USAMO.


trigonometry
Pythagorean Theorem
geometry

I do not know the correct answer for this question but I think I'm right:

State the origin of the Pytheogrem Theorem in terms of other polygon.

I'm pretty sure the answer to this question is (in spoiler):

[hide]unknown.[/hide]

That's my best understanding of the issue, in terms of historical research in Babylonian, Chinese, and Greek sources. The earliest written documents that survive to the present with clear evidence of an understanding of the Pythagorean theorem are lists of Pythagorean triples on Babylonian cuneiform clay tablets that date far, far earlier than any Greek or Chinese source mentioning the theorem.

LOL.  I think that silverfalcon phrased the question sortof badly.  What is the question exactly as it comes from the textbook (or wherever you got it)?

Not really I got from the text but I think I made the question badly...

Um... I just thought answer could be like this:

Nvr mind.. wrong question.. it should be 45-45-90 triangle's hypotenuse, not the p. theorem...

And the correct answer I hoped was:

[hide]I think 45-45-90 triangle's hypotenuse would came from square's diagonal since if diagonal is crossed in square and made square into two pieces,  the result polygon is two 45-45-90 triangles. And hypotenuse of following triangle is same as diagonal of the square....[/hide]

Now, here's my question, I need to ask because I'm curious..

Did triangle found first, or did square did?

sure? (no offense) but can you like gives story or any background on it?
Because in universe, there is usually cause-and-effect in exception of which nobody's sure on which came first such as egg and chicken

[quote="Silverfalcon"]sure? (no offense) but can you like gives story or any background on it?[/quote]

There are a lot of historical questions (and you are asking a historical question) for which the answers are uncertain, because the observations people made at the time weren't written down nor preserved in artifacts, of if they were written down and preserved in artifacts, the writings and artifacts were later destroyed. The great library of Alexandria was burned up, for example. (Indeed, I have read of at least THREE instances in different centuries of the library in Alexandria being burned up.) Similarly, there was a huge book-burning in China at the time of the first Qin dynasty emperor. A lot of ancient literature is just plain lost, and a lot of interesting things that people did in the old days were never learned about by the rare literate people who could write them down. Much of the early history of mathematics is lost because there are no sources from which it could be reconstructed.

I'm told that the preferred textbook on the history of mathematics for today's reader is [url=http://www.amazon.com/exec/obidos/tg/detail/-/0321016181/learninfreed]A History of Mathematics: An Introduction[/url] (2nd Edition) by Victor J. Katz, but I haven't had a chance to read that book yet. But history isn't proved by textbooks, but by primary sources. And all the textbooks agree that there just aren't enough primary sources extant today to determine when people first learned about the Pythagorean theorem.

Hope this helps!

[quote="yif man12"]since we're on the subject of things coming first, here's my question. Which came first, the chicken or the egg?[/quote]
If you believe in god, I guess the chicken came first since god put chickens in the garden of eden...

If you believe in evolution, then you can't tell--the changes would be too gradual to be noticed.

But I guess the egg came first... I suppose you could get a chicken from another egg or something...

I agree with tokenadult that no one really knows.  They were probably found or defined around the same time because of their close relationship relationship.  If I had to guess I would say that the triangle was discovered first but I do not have a good basis to give any evidence for this and I'm pretty sure that no one can really tell which came first since there are no records.

[quote="joml88"] no records.[/quote]

That's the main point


What is $7\frac{5}{7}$% of $17.7-2\frac{1}{9}+2.3$? Express your answer as a decimal to the nearest hundredth.

Oh yeah, no calculators  :D

what's that bracket under the 5 in your fraction percent?

I hate when I don't hold down the shift button long enough :wallbash:

i got 199/700. can that be simplified any further?

I don't think that's the correct answer...  :(

yikes! :oops: i made a lot of mistakes in the beginning. now i got 1.38.


Of the 36 students in Richelle's class, 12 prefer chocolate pie,
8 prefer apple, and 6 prefer blueberry.  Half of the remaining
students prefer cherry pie and half prefer lemon.  For Richelle's pie graph
showing this data, how many degrees should she use for cherry pie?

There are 36-12-8-6=10 students who prefer either cherry or lemon, so 5 prefer cherry. Because there are 36 students, each one is represented by 10 degrees, so cherry pie has 50 degrees.

[quote="KingSmasher3"][hide="SOLUTION"]Subtracting $ 12$, $ 8$, and $ 6$ from $ 36$. We have $ 5$ for cherry and $ 5$ for lemon. Thus, if the number of degrees we are asked to find is $ \text{N}$, then $ \frac {5}{36} \equal{} \frac {\text{N}}{360}$$ \implies$ $ \text{N} \equal{} \boxed{50}$, thus our answer.[/hide]

The previous user submitted a few seconds before me. Mine uses LaTeX, so I spent a little more time on it.  :wallbash_red:[/quote]
I love how your emote ties the post together  :thumbup:


If the alphabet were printed in simple block printing, which capital letters would have BOTH vertical and horizontal symmetry?

isnt that just
[hide]
$\boxed{\text{I}}$ $\boxed{\text{H}}$[/hide]

Well because the letter I looks like this:

[size=200]I [/size]

Clearly it has both horizontal and vertical symmetry and if you rotate the I to an H you get the same result..there migth be more examples of letters, just write them out in block form and you will see them.


[quote="BillyBobGoo"]Grrr, ok I've completed every other problem in this section correctly and maybe I'm tired and burnt out, but I'm just not getting how the answer supplied was arrived at...

If someone please could walk me thru solving this problem you automatically go to heaven!  Haha

h
_____
h/w - h

"h" divided by "h" over "w" minus "h"

It's such a seemingly simple problem I'm down on myself for not getting it.

Thanks in advance for your help![/quote]

multiply by w in the numerator and denominator

you get
hw/(h-hw)

cancel h from numerator and denominator( provided that h is not zero)

your answer will be
w/(1-w)

KeepSmiling
Malay

That's what I had, grr, I guess the book's wrong, thanks! : D

And  I think I might have posted this in the wrong section, sowwy


For every point $ M$ inside triangle $ ABC$ we define a triplet $ (d_1, d_2, d_3)$,
where $ d_1, d_2, d_3$ are distances from $ M$ to the sides $ BC,AC,AB$. Prove that the set of point $ M$, satisfing condition $ d_1.d_2.d_3\geq r^3$, where $ r$ is the incircle’s radius, lies inside the circle with center $ O$ and radius $ OI$.

Observe that $ OM\le OI$ $ \Longleftrightarrow$ $ 2pr^2\le \sum ad_2d_3$ . I don't know to go on ...


algebra
polynomial
Vieta
algebra unsolved

Show that the product of the two real roots of $x^4+x^3-1$ is a root of $x^6+x^4+x^3-x^2-1=0$.

[quote="jin"]Suppose the four roots of $x^4+x^3-1$ is $a,b,c,d$
Consider $f(x)=(x-ab)(x-ac)(x-ad)(x-bc)(x-bd)(x-cd)$
We can soluted $f(x)=x^6+x^4+x^3-x^2-1$
So it's easy.[/quote]

How did you get from line 2 to line 3? :huh:

By Vieta on the first one you know
$abcd=-1$
$abc + abd + acd + bcd = 0$
$ab+ac+ad+bc+bd+cd = 0$
$a+b+c+d=-1$

Use Vieta on the other one, then get $ab+ac+ad+bc+bd+cd$ as the coefficient of $x^5$, but this is $0$ by the above.
The others work like that, too.

OK, that's what my solution was, but I thought there might be a better way than expanding all those symmetric sums. :D

I can't find such easy one.So I omited the long solution.


function
limit
real analysis
advanced fields
advanced fields unsolved

Give an example of a decreasing sequence $ (\mu_{n})$ of measures on a measurable space such that the set function $ \mu$ defined by $ \mu(E)\equal{}\lim \mu_{n}(E)$ is not a measure.

Say, $ X\equal{}\{1,2,3,\dots\}$ and $ \mu_n$ is the counting measure restricted to $ \{n,n\plus{}1,n\plus{}2,\dots\}$. Then $ \mu\equal{}\lim \mu_n$ vanishes on all finite subsets. But $ \mu(X)\equal{}\infty$; so $ \sigma$-additivity fails.

Thanks!
I had something different.
Let $ m$ be the Lebesgue measure on $ \mathbb{R}$ and let $ \mu_{n}(E)\equal{}m(E)2^{\minus{}n}$. Since $ \mathbb{R}\equal{}\bigcup_{k \in \mathbb{Z}} [k,k\plus{}1)$, $ \mu(\bigcup [k,k\plus{}1))\equal{}\lim \mu_{n}(\bigcup [k,k\plus{}1))\equal{}\lim m(\mathbb{R})2^{\minus{}n}\equal{}\infty$ but $ \sum_{k} \mu([k,k\plus{}1))\equal{}\sum_{k} \lim_{n} \mu_{n}([k,k\plus{}1))\equal{}\sum_{k} \lim_{n} 2^{\minus{}n}\equal{}0$. Hence $ \mu$ is not a measure.

But if the measures $ \mu_n$ are finite...

Oh? Each $ \mu_{n}$ is a constant multiple of Lebesgue measure on $ \mathbb{R}$ so I think they shouldn't be finite.

I meant to ask: if the measures $ \mu_n$ were assumed to be finite, would it follow that $ \mu\equal{}\lim_{n\to \infty}\mu_n$ is a measure?

If they were finite, then it seems that $ \mu$ would be a measure from the usual sort of arguments.


[quote="Protonios"]How did you create it if you cannot solve it? I mean how do you know it holds? :o[/quote]
I created this one when I tried to broaden a known inequality (I cannot post it here) and I have used computer to check that it is true but I cannot find any solutions for it. :(

[quote=can_hang2007]Let $ a,b,c$ be nonnegative real numbers. Prove that
\[ \frac{a}{\sqrt{(a^2\plus{}2b^2)(2b^2\plus{}c^2)}}\plus{}\frac{b}{\sqrt{(b^2\plus{}2c^2)(2c^2\plus{}a^2)}}\plus{}\frac{c}{\sqrt{(c^2\plus{}2a^2)(2a^2\plus{}b^2)}} \ge \frac{3}{\sqrt{2(a^2\plus{}b^2\plus{}c^2)\plus{}ab\plus{}bc\plus{}ca}}\]
Equality holds if and only if $ a\equal{}b\equal{}c.$[/quote]

After removing the sqrt sign, the SHED technique (buffalo way) works.  However, there are very long expressions and one needs math software.

[quote=xzlbq][quote=can_hang2007]Let $ a,b,c$ be nonnegative real numbers. Prove that
\[ \frac{a}{\sqrt{(a^2\plus{}2b^2)(2b^2\plus{}c^2)}}\plus{}\frac{b}{\sqrt{(b^2\plus{}2c^2)(2c^2\plus{}a^2)}}\plus{}\frac{c}{\sqrt{(c^2\plus{}2a^2)(2a^2\plus{}b^2)}} \ge \frac{3}{\sqrt{2(a^2\plus{}b^2\plus{}c^2)\plus{}ab\plus{}bc\plus{}ca}}\]
Equality holds if and only if $ a\equal{}b\equal{}c.$[/quote]
I had prove it.[/quote]

I have two solutions, one is degree 8

[quote=can_hang2007]Let $ a,b,c$ be nonnegative real numbers. Prove that
\[ \frac{a}{\sqrt{(a^2\plus{}2b^2)(2b^2\plus{}c^2)}}\plus{}\frac{b}{\sqrt{(b^2\plus{}2c^2)(2c^2\plus{}a^2)}}\plus{}\frac{c}{\sqrt{(c^2\plus{}2a^2)(2a^2\plus{}b^2)}} \ge \frac{3}{\sqrt{2(a^2\plus{}b^2\plus{}c^2)\plus{}ab\plus{}bc\plus{}ca}}\]
Equality holds if and only if $ a\equal{}b\equal{}c.$[/quote]
By Holder
$$\left(\sum_{cyc}\frac{a}{\sqrt{(a^2+2b^2)(2b^2+c^2)}}\right)^2\sum_{cyc}a(a^2+2b^2)(2b^2+c^2)(a+c)^3\geq\left(\sum_{cyc}(a^2+ab)\right)^3.$$
Thus, it's enough to prove that
$$\left(\sum_{cyc}(a^2+ab)\right)^3\sum_{cyc}(2a^2+ab)\geq9\sum_{cyc}a(a^2+2b^2)(2b^2+c^2)(a+c)^3$$ or
$$\sum_{cyc}(2a^8+7a^7b+7a^7c-a^6b^2+8a^6c^2+17a^5b^3-a^5c^3-32a^4b^4+25a^6bc+36a^5c^2b+40a^4b^3c-32a^4c^3b-129a^4b^2c^2+53a^3b^3c^2)\geq0,$$
which is true by Schur and AM-GM.

[quote=MathPassionForever]@above exactly how do you do such stuff? Like what was the motivation behind this? Infact I gave up trying to prove the last line just by seeing the length of that thing.[/quote]
The motivation it's just very interesting to solve  Can's problems. The last inequality is not hard. If you have motivation to prove it, so I am sure that you'll make it. Good luck!


search
algebra
polynomial
inequalities
Gauss
number theory unsolved
number theory

Determine whether $\sqrt{1001^2+1}+\sqrt{1002^2+1}+\cdots+\sqrt{2000^2+1}$ is rational.

It can be proved (by Pierre, for example  :D ) that if $\sum{\sqrt{a_i}}$ is rational and $a_i$ are integers, then all $a_i$ are squares. So...

Decent. Could you please give me a link to this proof?

Try to use search button with topic "Irrationality of polyquadratic numbers" , author vess. There you will find a link to Pierre's proof.

One can prove that the sum above is not an integer and that it is a zero of a monic polynomial => the sum is irrational.

Now for all positive n we have n < sqrt(n^2 + 1) < n + 1/n. The last inequality follows by squaring. So we have that the sum is of the form:
1001 + b_1 + 1002 + b_2 + ... + 2000 + b_1000, where 0 < b_i < 1/1001. Then we have 0 < b_1 + b_2 + ... + b_1000 < 1 and hence the sum is not an integer.

Now a more general proof: sqrt(n_1) + sqrt(n_2) + ... + sqrt(n_k) is a root of a monic polynomial, where n_i is not a square.
We know that sqrt(n_1) is a root of p(x) = x^2 - n1, so it's true for one term. Assume now that sqrt(n_1) + ... + sqrt(n_k) is a root of the monic polynomial p(x) = x^a + b_(a-1)x^(a-1) + ... + b_0. Now I'll prove that y = sqrt(n_(k+1)) + sqrt(n_k) + ... + sqrt(n_1) is a root of a monic polynomial. So we have p(y - sqrt(n_(k+1))) = 0 = (y - sqrt(n_(k+1)))^a + b_(a-1)*(y-sqrt(n_(k+1)))^(a-1) + ... + a_0 = y^a  + P(y) + sqrt(n_(k+1))*Q(y), where P(y) and Q(y) are polynomial with integer coefficients and degree at most a-1. So y^r + P(y) = -sqrt(n_(k+1))*Q(y) => y^2r + 2*y^r*P(y) + P(y)^2 = Q(y)^2 * n_(k+1) => y^2r + 2*y^r*P(y) + P(y)^2 - n_(k+1)*Q(y)^2 = 0, so we have a monic polynomial of degree 2r with y as a zero.

My solution was a little bit easier. Clearly, the numbers $\sqrt{k^2+1}$ are algebraic integers, so our sum is also an algebraic integer (since this set is a ring) and since the only rational numbers which are also algebraic integers are the integers, it suffices to show that our sum is not integer. This reduces to proving that $\sum{\frac{1}{\sqrt{k^2+1}+k}}$ is not integer, which is obvious since it is $>0$ and $<1$.

Hi Harazi,

can you give me a proof that the sum of algebric integer is also an algebric integer

Thanks!

Ok, here it goes. Suppose $a,b$ are algebraic integers and denote $a_1=a,a_2,...,a_n$ and $b=b_1,b_2,...,b_m$ all conjugates of $a,b$ (that is we regard the minimal polynomial of $a$ and $b$ over $Q$-I will let you prove that they have integer coefficients and are irreducible in $Q[X]$ easy consequence of Gauss lemma- and then consider $a_i$ and $b_j$ their roots). Define the polynomial $ F(x)=\prod_{1\leq i\leq n, 1\leq j\leq m}{(x-a_i-b_j)}$. Observe that its coefficients are symmetric in $a_i$ and $b_j$ thus they can be written as polynomial with integer coefficients in the symmetric fundamental sums of $a_i$ and respectively $b_j$. Since the symmetric sums in $a_i$ and $b_j$ are integer numbers (use Viete and the observation made in the beggining of the solution about minimal polynomials), it follows that $F$ has integer coefficients. It is obviously monic and has root $a+b$ thus conclusion.


geometry
3D geometry
algebra
factorization
sum of cubes

[quote="frt"][hide="answer"]
$(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$

I just knew the answer because I memorized it :roll:
[/hide][/quote]

Just as you....I memorize that answer, too.   But I wonder how this was factorized. I've tried about sum of cubes but still couldn't figure it out, though.

A method I have seen before is to use the 'zero term' factorization. What this means is that you supply terms whose sum is zero but allows you factor. The well-known example is the Sophie-Germain identity:

$a^4+4b^4=a^4+4a^2b^2+4b^4-4a^2b^2=(a^2-2ab+2b^2)(a^2+2ab+2b^2)$

For this problem, there are a lot of 'zero terms' and it can be very difficult to choose the correct terms.

it should look like that because x,y,z are interchangeable

This question alredy appeared here :) :
e.g. : http://www.mathlinks.ro/Forum/viewtopic.php?t=67280

And here's an "application exercise" :
http://www.mathlinks.ro/Forum/viewtopic.php?t=27112

You may want to try the general :
Solve in positive integers :
$x^3+y^3+z^3 = xyz + k$ $k \in \mathbb{N}$


limit
probability
calculus
real analysis
real analysis unsolved

Find the value of:\[
\mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\frac{{C_{2i}^i }}{{(2i - 1)4^i }}}
\]

Proposed by prowler on Olympiad section

Flip a coin repeatedly and count the heads and tails. The sum $\sum_{n=0}^\infty \frac{{2n\choose n}}{(2n+1)2^{2n+1}}$ is the probability that your counts of heads and tails will ever agree, and it's equal to 1.

This isn't exactly the problem asked, but it's nice too. Why $2i-1$ in the denominator? It's just a couple of calculus steps away, but $2i+1$ would be much more natural.


1.Find all integers a,b,c with 1<a<b<c such that the number

(a-1)(b-1)(c-1) is a divisor of abc-1

2.Find all pair of integer (x,y) such that

x^6+3x^3+1=y^4

3.solve the following equation in nonzero integers x,y

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

thank you

[quote="poonprang"]
2.Find all pair of integer (x,y) such that

x^6+3x^3+1=y^4

thank you[/quote]

We have that for $ x > 0$: $ (x^3 \plus{} 1)^2 < x^6 \plus{} 3x^3 \plus{} 1 < (x^3 \plus{} 2)^2$, so $ (x^3 \plus{} 1)^2 < (y^2)^2 < (x^3 \plus{} 2)^2$. Contradiction.
For $ x < 0$ and $ x\neq \minus{}1$: $ (x^3 \plus{} 2)^2 < x^6 \plus{} 3x^3 \plus{} 1 < (x^3 \plus{} 1)^2$, and the situation is the same. Contradiction.
For $ x\equal{}\minus{}1$ there is no solution.
For $ x \equal{} 0$ we have $ y \equal{} 1$ and $ y \equal{} \minus{} 1$.

a solution for the third one....
we have to solve the equation $ 2y^2\plus{}y(x^2\minus{}3x)\plus{}3x^2\plus{}x\equal{}0$ with descriminant $ D\equal{}x^4\minus{}6x^3\minus{}15x^2\minus{}8x\equal{}a^2$
Consequently $ a\equal{}kx$ and now we have $ \minus{}8\equal{}x(k^2\plus{}15\plus{}6x\minus{}x^2)$,so $ x|8$ and that shows that $ |x|\equal{}1,2,4,8$
By checking these cases we find the solutions...
I hope it is ok!!!! :)

I have a solution for first one. Let f(a,b,c) = (abc-1)/(a-1)(b-1)(c-1). Write it as 1+ 1/(a-1) + 1/(b-1) + 1/(c-1) + 1/(a-1)(b-1) + 1/(b-1)(c-1) + 1/(c-1)(a-1). Now it is only dealing with the cases. If a is more than or equal to 4, you can see that b is more than equal to 5 and c is more than equal to 6. Now try to get bounds and you will see that f(a,b,c) is less than equal to 119/60 which is smaller than 2. Checking yields no solution. We have 2 cases. a=2 or a=3. Now as usual, you can bound f(a,b,c) in order to solve it. The only thing you need is solving some diophantine equations.


Circles $O$ and $P$ have radii $16$ and $9$ respectively, and are tangent externally at point $A$.  Point $B$ is chosen on circle $O$ such that $AB=8$, and $BC$ is drawn tangent to circle $P$ at point $C$.  Compute $BC$.

[hide]Draw diameters of the circles through $A$ and extend $AB$ beyond $A$ to hit $P$ on a certain point $K$. The obtained triangle are similar for AAA and so $\frac{32}{8}= \frac{18}{AK}\Rightarrow AK = \frac{9}{2}$. On the other hand $BC^{2}= AB \cdot BK = 8\left(8+\frac{9}{2}\right) = 100 \Rightarrow BC = 10$.[/hide]

Sorry, but I don't see how the  corresponding angles are equal.  Could you explain this more in depth?

$\measuredangle ABX = \measuredangle AKY = 90^{\circ}$ and $\measuredangle BAX = KAY$. See the diagram, hope now it's clear.


I was wondering, how many countries impose limits on the number of international olympiads one can participate in each year?    I'm asking because our education ministry is planning to impose a 1 olympiad per year only rule this year (which of course a lot of people are unhappy about, including me), and I was wondering whether we're the only one doing this or if it's a common practice.

Most of the olympiads occur at the same time (e.g. IMO and IPhO), so there's a practical limit on how many one can do; usually, the practical limit is 2, I think, as IOI occurs later than most of the other ones.

They probably want you to focus most of your energy on one Olympiad so you can do very well in that one and not just okay in multiple ones.

[quote="MysticTerminator"]that seems silly as the smartest people in one of the mathematical sciences will probably excel as well in others

more likely it is simply the logistics

is your avatar a go game[/quote]

Maybe they want to give more people a chance? And my avatar is an upsidedown dog.

[quote="JellyWorld"]They didnt announce this rule until very very late, if they decide to go on with it they'll disqualify half the IOI team, since 1 is going to IMO and the other is going to ICHO, which means I might get to go to IOI instead of them.  Still, I'd rather they not have the rule, because I'm *hoping* to make it to both the IMO and IOI teams next year, and I won't be able to do that due to the rule.

I'm thinking of petitioning them if they really decide to push through the rule, I'm pretty sure a lot of people would sign.[/quote]

I think you should get started on a petition right now, show them the people that would be ineligible because of this and how that wuold basically demolish the team.

That seems like a very stupid rule.  It discourages the smartest people from competing?? WTF??  What country is that?

That's crazy.  While I see the idea behind it (excell at one subject, practice at it, do good at it, etc.) it still seems wrong to me.  Maybe it's my American set of values (earn it and it's yours and no one can take it away from you).

In Flanders (my half of Belgium) the finals of Chemistry, Physics and Biology are on the same day.
But it is not dependant of the math olympiad results.
I know a guy who went IMO 2003 and also went to the IPHO that year.
I have even heard of a Belgian who went to the IPHO 1992 in Finland, didn't stick around for the ceremony and went straight from Finland to Moscow instead, to participate in the IMO, and he actually obtained medals in both.

I think it not racist, just something worth looking into : I noticed in IMO 2002 that countries like Canada, USA and Australia have a rather high percentage of Asians.  Actually, there was one of Cantonese descent in my team, and last year we had a chinese in the team as well (and I think Belgium has a rather low percentage of Asians compared to USA for instance).

So what is it?  I heard some people saying : culture.  The motivation to do very well.  But let us not forget the average immigrant in country A from country B does not represent the entire population of country B : not everyone in country B can or wants to go to country A.

We had a girl in Iran that participated in IMO and IOI in 1994. Her name was Roya Beheshti and got Silver in IMO and Bronze in IOI.

This year Nima Ahmadipour participated in IMO (2006) and will partipate in IOI.

[url]http://imo2006.dmfa.si/participants/IRN.html[/url]

Omid I don't understand, you mean she is in the IMO 2006 team?

I think that students should be allowed to participate in multiple olympiads, iff they are willing to work hard in all olympiads they wish to participate in. Some students can handle intense Math and Physics at the same time. If they can, nothing should stop them.

I meant "she" for the girl in IMO 1994. Her name was "Roya Beheshti" and got Silver Medal in IMO and Bronze medal in IOI.

This year in 2006 Nima Ahmadipour (IRN1, Of Course he is a boy) participateded in IMO and WILL PARTICIPATE IN IOI 2006.


BTW, I don't think there exist real $x$, $y$, $z$ which satisfy given conditions.


Prove that if $a,b,c$ are positive integers such that $c^2 = a^2+b^2$, then both $c^2+ab$ and $c^2-ab$ are also expressible as the sums of squares of two positive integers.

It's easy to prove that if (a,b,c) is a "Pythagorian triple" then $a+b-c > 0$ (triangle ineq), $a+b+c$ and $a+b-c$ are even.

Now $c^2 + ab = ((a+b+c)/2)^2 + ((a+b-c)/2)^2$

I let you search the other one   ;)

Or just plainly use the known structure of Pythagorean triples $c=d(m^2+n^2)$, $\{a,b\} = \{2dmn, d(m^2-n^2)\}$.
Then $c^2 \pm ab =$ $ d^2\left ((m^2+n^2)^2 \pm 2mn(m^2-n^2)\right ) =$  $  (d(m^2 \pm mn))^2 + (d(n^2 \mp mn))^2$.


how can i solve the follow one?
x^14+y^11=1
x^7+y^17=1.

i also would like to help me to find (with steps if possible) the discriminant of a third degree polynomial
thnx


floor function
IMO Shortlist
geometry proposed
geometry

Let $k$ be the number of right angles of a given $n$-gon.
1) Prove that $k<\frac{2}{3}(n+2)$.
2) Assume that $n$ is divisible by 3. Prove that the case $k=\frac{2}{3}n$ is possible.

And for 2, do we have a triangle with two right angles? I translated the problem from the Finnish version of the paper and II'm also wondering what do they want.

The question about the maximal number of right angles in a n-gon was posed in the IMO Shortlist 2003 as problem C3 (though it is, indeed, much older - I remember seeing it discussed in a paper in a German magazine on mathematical education - probably "mathematik lehren"). It was posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=5785 , but until a solution is given, I can just refer you to the official solutions which are downloadable at http://www.mathlinks.ro/Forum/viewtopic.php?t=15621 , where also the correct answer is given (it is 1 right angle for n = 3, 4 right angles for n = 4, 3 for n = 5 and $\left\lfloor\frac{2n}{3}\right\rfloor+1$ for $n\geq 6$. (I will not post my solution, since it is way too ugly.)

  Darij