[quote="catcurio"]Let $f$ be a function for which $f\left(\frac x3\right)=x^2+x+1$. Find the sum of all values of $z$ for which $f(3z)=7$.

$\text{(A)}\ -\frac13\qquad\text{(B)}\ -\frac19 \qquad\text{(C)}\ 0 \qquad\text{(D)}\ \frac59 \qquad\text{(E)}\ \frac53$[/quote]

[hide]Multiplying x/3 by 9 to get 3z is :
$(9z)^2+9z+1=81z^2+9z+1=7$
$81z^2+9z-6=0$
$27z^2+3z-2=0$
$(9z-2)(3z+1)=0$
$z_1=\frac{2}{9}, z_2=-\frac{1}{3}, \frac{2}{9}-\frac{1}{3}=\boxed{-\frac{1}{9}\text{(B)}}$[/hide]

Oh! i go confused witht the f(x/3)=x^2+x+1.

[quote="catcurio"]Let $f$ be a function for which $f\left(\frac{x}{3}\right)=x^{2}+x+1$. Find the sum of all values of $z$ for which $f(3z)=7$.[/quote]
[hide="Slightly quicker solution:"]
Put $x=9z$ and we get:
\[f(3z) = 81z^{2}+9z+1 =7 \]
Now, by Vieta's formula we get that:
\[\sum z = (-1)^{1}\cdot \frac{ 9}{81}= \boxed{-\frac{1}{9}\implies B}\]
(Note that this solution works for [i]any[/i] constant value of $f(3z)$, even if the polynomial is irreducible over the reals.)
[/hide]

Source: 2000 AMC 10/12

[hide="Different, nice solution"]
First we find the values of $x$ for which $f(x/3) = 7.$ Then we get $x^2+x+1=7,$ or $x^2+x-6=0$; thus by Vieta, the sum of these values is $-1.$ Now, note that the values of $z$ should be $1/9$ of the values of $x,$ since $3z = x/3$ gives $z = x/9.$ Therefore the sum of the values of $z$ is $\boxed{-1/9}.$
[/hide]

You guys could have just used viete’s formula.
Btw
Viete’s formula=
If you have the quadratic ax^2 + bx + c,

Sum of roots = -b/a
Product of roots = c/a
But the method you guys used was good too.

Find f(9z/3) and use viete's formula.Answer is -1/9

[hide=Solution]We know that $f\left(\frac{9z}{3}\right)=f(3z)=81z^2+9z+1=7$. By Vieta's Formulas, the sum of the values is $-\frac{9}{81} \implies -\frac{1}{9} \implies \boxed{\text{(B)}}$.[/hide]

[hide=u give a sol? i will too >:D]
We must find the value of $x$ for which $f\left(\frac{x}{3}\right)=f(3z)$, so $x=3z \cdot 3 = 9z$. from this we see that $81z^2+9z+1=7$, so by vietas our answer is $-\frac{1}{7}$
[/hide]

also imho this is basic basic msm :ooo

@OP

Substitude 9x instead of x, you will have f(3x)=81x^2+9x+1, we need 81x^2+9x+1=7, solving quadratic, x can be -1/3 and 2/9, suming these who we get -1/9 so the answer is B

[quote=peace09][hide=u give a sol? i will too >:D]
We must find the value of $x$ for which $f\left(\frac{x}{3}\right)=f(3z)$, so $x=3z \cdot 3 = 9z$. from this we see that $81z^2+9z+1=7$, so by vietas our answer is $-\frac{1}{7}$
[/hide]
[/quote]

you typoed

[hide=Solution]We have $$f(3z)=(9z)^2+(9z)+1=81z^2+9z+1=7\iff81z^2+9z-6=0\iff27z^2+3z-2=0\iff\text{sum of all values of }z=-\frac{3}{27}=\boxed{\textbf{(B) }-\frac19}$$ by Viete's Formulae, and clearly there is no double root for $z$ as the discriminant of the quadratic is $3^2-4(-2)(27)\neq0,$ so we can be sure our answer is correct.[/hide]

A one-liner. :)

Everyone's giving the same solution, why don't I join in? :)
[hide=Solution]$f(3z)=81z^2+9z+1=7\implies 81z^2+9z-6=0\implies 27z^2+3z-2=0$. The sum of the roots is just $-\frac{3}{27}=-\frac{1}{9}$.[/hide]

[hide=Solution]
Plugging in $x=9z$,
$$81z^2+9z+1=7 \implies 81z^2+9z-6=0 \implies 27z^2+3z-2=0$$By Vieta's the sum of the roots is $$-\frac{b}{a}=\boxed{-\frac{1}{9}}$$

[hide=Solution]We begin by finding the values of $z$ in which $f(3z)=7$.
$$f(3z)=81z^2+9z+1=7$$
$$81z^2+9z-6=0$$
By Vieta's Formulas, the sum of the values of $z$ that satisfy this equation is
$$-\frac{9}{81}=-\frac{1}{9}$$
Therefore, the correct answer is $\boxed{\text{(B)}\ -\frac19}$[/hide]

Yay! We have like 10 identical solutions now! :D

[quote=Lamboreghini][hide=Solution]We have $$f(3z)=(9z)^2+(9z)+1=81z^2+9z+1=7\iff81z^2+9z-6=0\iff27z^2+3z-2=0\iff\text{sum of all values of }z=-\frac{3}{27}=\boxed{\textbf{(B) }-\frac19}$$ by Viete's Formulae.[/hide]

A one-liner. :)[/quote]

If only one $z$ satisfies it, you haven't counted that case.

[hide = Solution]$f(3z) = f(\frac{9z}{3}) = (9z)^2 + (9z) + 1 = 7 \Rightarrow 81z^2 + 9z - 6 = 0 \Rightarrow 27z^2 + 3z - 2 = 0 \Rightarrow 27z^2 + 9z - 6z - 2 = 0 \Rightarrow 9z(3z + 1) - 2(3z + 1) = 0 \Rightarrow (9z - 2)(3z + 1) = 0 \Rightarrow z = \frac{2}{9}, -\frac{1}{3} \Rightarrow \text{Sum of all values of }z = \frac{2}{9} + \left(-\frac{1}{3}\right) = \frac{2}{9} - \frac{3}{9} = \boxed{\text{(B)}\ -\frac19}.$[/hide]

Didn't use vieta's formulas bcz I'm a rebel :evilgrin:

[size=50]and also there are many solutions above using it so I decided to go old-school :P[/size]

This is an AMC 10 Problem 24? Is it likely that the same problem on this year's exam would be a lot harder, or about the same?

@above a lot harder
-----
Plugging in $9z$, we need the sum of $z$ such that $$81z^2+9z-6=0,$$ which is clearly $\boxed{-\frac{1}{9}}$ via Vieta's since there is not a double root.

[quote=Taco12]@above a lot harder
-----
Plugging in $9z$, we need the sum of $z$ such that $$81z^2+9z-6=0,$$ which is clearly $\boxed{-\frac{1}{9}}$ via Vieta's.[/quote]

nice

Notice that $f(x)=9x^2+3x+1$. We need $f(3z)=7$. $9(3z)^2+3(3z)+1=7$. We don't even need to expand as we need the sum of all values, which by vietas its simply $-\frac{1}{9}$.


calculus
integration
trigonometry
function
calculus computations

Evaluate $ \int_0^1 \frac{x}{1\plus{}x}\sqrt{1\minus{}x^2}\ dx$.

The solution is very beautiful without trigonometry. Thanks again kunny; great problem.  :D [hide="My Method"]\[ I\equal{}\int^1_0 \frac{x}{x\plus{}1}\sqrt{1\minus{}x^2}\text{ d}x\]
$ \equal{}\int^1_0 \frac{(x\plus{}1)\minus{}1}{x\plus{}1}\sqrt{1\minus{}x^2}\text{ d}x$

$ \equal{}\int^1_0 \sqrt{1\minus{}x^2}\text{ d}x\minus{}\int^1_0 \frac{\sqrt{1\minus{}x^2}}{x\plus{}1}\text{ d}x$

$ \equal{}\frac{\pi}{4}\minus{}\int^1_0 \sqrt{\frac{1\minus{}x}{1\plus{}x}}\text{ d}x\equal{}\frac{\pi}{4}\minus{}\int^1_0 \frac{1\minus{}x}{\sqrt{1\minus{}x^2}}\text{ d}x$

$ \equal{}\frac{\pi}{4}\minus{}\int^1_0 \frac{1}{\sqrt{1\minus{}x^2}}\text{ d}x\plus{}\int^1_0 \frac{x}{\sqrt{1\minus{}x^2}}\text{ d}x$

$ \equal{}\minus{}\arcsin{x}\minus{}\sqrt{1\minus{}x^2}\bigg |^1_0\plus{}\frac{\pi}{4}\equal{}\boxed{1\minus{}\frac{\pi}{4}}$
[/hide]

That's right, I also had your solution.

Let me make a complementary explanation for Japanese high school students who don't learn arc $ sin x$.

$ \int_0^1 \frac {1}{\sqrt {1 \minus{} x^2}} dx\ [x \equal{} \sin \theta]$

$ \equal{} \int_0^{\frac {\pi}{2}} \frac {1}{\sqrt {1 \minus{} \sin ^ 2 \theta}}\cos \theta d\theta$

$ \equal{} \int_0^{\frac {\pi}{2}} \frac {1}{|\cos \theta|}\cdot \cos \theta d\theta$

$ \equal{} \int_0^{\frac {\pi}{2}} \frac {1}{\cos \theta}\cdot \cos \theta d\theta \left(\therefore 0\leq \theta \leq \frac {\pi}{2}\right)$.

$ \equal{} \frac {\pi}{2}$.

[quote="milin"][quote="kunny"]
Let me make a complementary explanation for Japanese high school students who don't learn arc $ sin x$.
[/quote]

Huh?  :huh: What do you mean?[/quote]

He simply evaluated the integral with trig substitution for any japanese students who may read it. Apparently, japanese schools do not cover arcsine integration. There is however a wonderful derivation of it by Louis Brand using inverses and the equation of a circle. A bit random, but I thought it might be worth mentioning.

I understood what he did, but the fact about Japanese students was the thing that got me.. So I was hopping for some more info.. I wasn't aware of this. I had an expression that they were somewhat a "world force" in the mathematics department, so being unfamiliar with inverse trigonometric functions is weird for my concept!


Our water polo team has 15 members. I want to choose a starting lineup consisting of 7 players, one of whom will be the goalie (the other six positions are interchangeable). In how many ways can I choose my starting lineup?

i think it might be 15C7 which is 6435 but i'm probably wrong

I... don't think so. I was trying to decipher the question. Is there anything special with water polo?

Pick the goalie first. There are 15 people for this. Then choose 6 for the rest. $ 15\cdot\binom{14}{6}\equal{}15\cdot3003\equal{}\boxed{45045}$.

Seems excessive but, whatever.


I am aware of the remainder theorem (stating that "if a polynomial function f(X) is divided by x - c, the remainder is f(c)"), but I am unaware of the reasoning behind this theorem.

I would greatly appreciate a thorough explanation of why this theorem is true.

Thank you to all who post and help explain

Let $ f(x)\equal{}(x\minus{}c)(P(x))\plus{}a$ (This is splitting up $ f(x)$ into a quotient and remainder.) Putting in $ x\equal{}c$ gives $ f(c)\equal{}a$, implying $ f(x)\equal{}(x\minus{}c)(P(x))\plus{}f(c))$.

Let $ P(x) \equal{} (x\minus{}c) \dot Q(x) \plus{} R$ be a polynomial function of degree 1 more than the degree of $ Q(x)$, and $ Q(x)$ is the quotient of $ f(x)$ when divided by $ x\minus{}c$ and $ R$ is the remainder (This is the Division Algorithm for Polynomials). Plugging $ c$ gives us $ P(c) \equal{} R$ which is the remainder.


$p$ is an odd prime number. Prove that all the prime divisors of $2^{p}+1$ (not including $3$) are in the form $2kp+1$.

Well known and already proven: the prime divisors of $x^{p-1}+...+x^{2}+x+1$, $p$ still prime, are all $\equiv 1 \mod p$ or $=p$. They are also clearly $\neq 2$, thus odd, this $\equiv 1 \mod 2$ and by this $\equiv 1 \mod 2p$ (or still equal to $p$). This is the statement if you set $x=-2$, because $2^{p}+1=(2+1)(2^{p-1}-2^{p-2}+-...+2^{2}-2+1)$.

Let q prime. suth that $q|(2^{p}+1)$.  Let $p\not |(q-1)$, then exist x, suth that $px=1+y(q-1)$. Therefore $2^{px}=2^{1+y(q-1)}=2*2^{(q-1)y}=2(mod q)$, but in other hand $2^{px}=(-1)^{x}(mod q)$. It give $2=1(mod q)$ (contradition) or $3=0(mod q)\Longrightarrow q=3$.

sorry guys, but i cant get it :(  can you explain more please............

If q>3 we get $p|(q-1)$, because p is odd $q=2kp+1$.
If k is odd $(\frac{2}{q})=-1$, therefore $k=p(mod 4)$.
If k is even, then $(\frac{2}{q})=1$. Therefore $4|k$.
It give $q=2kp+1, k=p(mod 4)$ or $q=8mp+1$.

let q a prime number divide 2^p +1.then ord(2)/2p .then ord(2)=2.p.2p.we check all cases we find that ord(2)=2p then 2p/q-1 =>q=2pk+1. :ninja:


geometry
circumcircle
trigonometry
angle bisector

Let $ ABC$ a triangle. The internal bisector of $ \hat{A}$ intercept $ BC$ at the point $ D$ and the circumcircle at the point $ E$. If $ \overline{AB}\equal{}8$, $ \overline{AC}\equal{}6$ and $ \overline{DE}\equal{}3$, find $ \overline{AD}$.

Let $ AD\equal{}k$ and $ \angle BAE\equal{}\angle EAC\equal{}\theta$.
$ BE\equal{}EC\implies (k\plus{}3)^2\plus{}8^2\minus{}2(k\plus{}3)(8)\cos{\theta}\equal{}(k\plus{}3)^2\plus{}6^2\minus{}2(k\plus{}3)(6)\cos{\theta}$$ \implies \cos{\theta}\equal{}\frac{7}{k\plus{}3}$
By angle bisector theorem we know $ \frac{BD}{DC}\equal{}\frac{4}{3}\implies \frac{BD^2}{DC^2}\equal{}\frac{16}{9}$
$ \frac{k^2\plus{}8^2\minus{}2(k)(8)\cos{\theta}}{k^2\plus{}6^2\minus{}2(k)(6)\cos{\theta}}\equal{}\frac{16}{9}\implies \cos{\theta}\equal{}\frac{7k}{48}$
$ \frac{7}{k\plus{}3}\equal{}\frac{7k}{48}\implies k^2\plus{}3k\minus{}48\equal{}0$, $ k\equal{}\frac{\sqrt{201}\minus{}3}{2}$

Let BD=4a and CD=3a.(Apply angle bisector theorem).By Stewart's theorem,
a(AD^2+xy)=b^2 x+c^2 y where a,b and c are the sides of triangle ABC opposite to angle BAC, angle ABC and angle ACB respectively.

          Therefore, AD^2=48-12a^2-------1
Using the fact that triangle ADB is similar to triangle CDE, we will get AD=4a^2--2
Square the second equation and equate it to the first equation to find a^2. The result follows.


come up with a clear dfn for a new operation which will be two for addition and subtraction in clock arithmetic, operating only on the numbers{1,2,3....12} The first argument should be current time(hour only) second argument should be elapsed time , and the output should be the new time (after the elapsed hours have passed) also on a 12hr clock. Make as much as normal addition/ subtraction  include properties(closure, associative, cummuative, identity, inverse) for your defn use a formula or words, etc.. Eplain how to calculate 3+5, 7+10, 3-5,5-3,7-10,10-7.  Now write a careful dfn of your operation ie describe operations explain why it has or does not have each properties.

This looks like you're asking us to "re-create" modular arithmetic, which already exists.

Modular addition and subtraction have all the properties listed, I believe.


trigonometry
inequalities
inequalities unsolved

[quote="mathmen"]:arrow:  :arrow:[/quote]

if $ x \ge \frac {\pi}{2}$, that we have $ \cos{x}\cos{y}\cos{z} \le 0$;

if $ x,y,z \in (0,\frac {\pi}{2})$, we have

$ \cos{x}\cos{y}\cos{z} \equal{} \frac12 (\cos{(x \plus{} y)} \plus{} \cos{(x \minus{} y)})\cos{z}\le \minus{} \frac12 \cos^2{z} \plus{} \frac12 \cos{z}$ $ \equal{} \minus{} \frac {1}{2}\left( {\cos z \minus{} \frac {1}{2}} \right)^2 \plus{} \frac {1}{8} \le \frac {1}{8}$

when $ x \equal{} y \equal{} z \equal{} \frac {\pi}{3}$ get "$ \equal{}$", so maximum is $ \frac18$.

it is very old and well known inequality  $ \cos x \cdot \cos y \cdot \cos z \le {1 \over 8}$  where  $ x,y,z$  are angles of triangle. Maximum achieved when triangle is equilateral.

See here http://reflections.awesomemath.org/ look for Trigonometry substitution.


Consider a bijection using base 2 and 3.


Hmm, i know this year, my friend in 8th grade got 6th... which i suppose is the cutoff (5th is still alternate)
so ill ask him what his scores were etc.

Lawl, I obviously didn't make that cutoff since I got 7th place after your friend  :P

No scores until after April though...

I was wondering about past years.

But it should be higher in past years since virginia like a couple of years ago was National Champion.

[quote="infinity4ever"]I know so many smart people from VA.... yet VA has like the 3rd most inactive forum in the US community  :rotfl:  :rotfl:  :rotfl:

Anyways, I was just wondering, what's the average cutoff to make Nats in MATHCOUNTS in VA?[/quote]

The cutoff for this year is ~35.


I still don't know how to do this. Help?

For all non-zero real numbers,
Prove that if a+b+c=0, then 3abc = a^3 + b^3 + c^3.
 :whistling:

[hide]
$a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

So, if $a+b+c=0$, the RHS equals 0 and the equality follows.[/hide]

[hide="Method 3"] $a^{3}+b^{3}+c^{3}= (a+b)(a^{2}-ab+b^{2})+c^{3}= (a+b)( (a+b)^{2}-3ab )+c^{3}=-c( c^{2}-3ab)+c^{3}= 3abc$ [/hide]
Of course, these are all variations on the same theme.

Only slightly harder:  Find necessary and sufficient conditions so that $3abc = a^{3}+b^{3}+c^{3}$.

[hide="Big hint"] $a^{3}+b^{3}+c^{3}-3abc = \frac{1}{2}(a+b+c)( (a-b)^{2}+(b-c)^{2}+(c-a)^{2})$ [/hide]


Let A={1,2,....,n} .prove or disprove  the followsing statement for all integers n>1 there exists function f:A_A and g:A_A which sastisfy f(f(k))=g(g(k))=k all k=1,2,...,n
                                                                   g(f(k))=k+1 all k=1,2,...,n-1


algebra
polynomial
geometry
geometric transformation
number theory
least common multiple
greatest common divisor

For each positive integer $ n$, let $ c(n)$ be the largest real number such that
\[ c(n) \le \left| \frac {f(a) \minus{} f(b)}{a \minus{} b}\right|\]
for all triples $ (f, a, b)$ such that

--$ f$ is a polynomial of degree $ n$ taking integers to integers, and
--$ a, b$ are integers with $ f(a) \neq f(b)$.

Find $ c(n)$.

[i]Shaunak Kishore.[/i]

When Shaunak posed this problem to me I suggested that $ c(n) \equal{} \frac {1}{n!}$, but I'm not 100% certain.  We can suppose WLOG that $ b \equal{} 0$ and $ f(0) \equal{} 0$ due to translation invariance, so write

$ f(x) \equal{} \sum_{i \equal{} 1}^{n} a_i {x \choose i}$

for some integers $ a_i$.  We want to minimize $ \left| \frac {f(x)}{x} \right|$ over all choices of $ a_i$ and $ x$ such that the quotient is well-defined and nonzero.

[b]Lemma 1:[/b]  For fixed $ x$, the minimum nonzero value of $ f(x)$ over all choices of $ a_i$ is $ \gcd \left( {x \choose 1}, {x \choose 2}, ... {x \choose n} \right)$.

[i]Proof.[/i]  This is [url=http://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity]Bezout's identity[/url].

[b]Lemma 2:[/b]  $ \gcd \left( {x \choose 1}, {x \choose 2}, ... {x \choose n} \right) \ge \frac {x}{n!}$.

[i]Proof.[/i]  Induction.  I don't know a nice way to prove this.

The conclusion follows; the equality case occurs whenever $ x$ is a multiple of $ n!$.

It looks like $ 1/(\text{lcm}\{1,2,\cdots,n\})$ to me. We have $ \binom{m}{k}\equal{}\frac{m}{k}\binom{m\minus{}1}{k\minus{}1}$, so $ \frac1{m}\binom{m}{k}\equal{}\frac1k\binom{m\minus{}1}{k\minus{}1}$
Multiply by that least common multiple $ L$, and all of the denominators are cleared. Also, if $ m$ is divisible by $ k$, $ \binom{m\minus{}1}{k\minus{}1}\equiv 1\mod k$. If $ m$ is a multiple of $ L$, we therefore have that the actual denominator of each $ \frac1k\binom{m\minus{}1}{k\minus{}1}$ is $ k$, and the common denominator is $ L$. That makes $ \frac{m}{L}$ the gcd of $ \{\binom{m}{1},\binom{m}{2},\cdots,\binom{m}{k}\}$ as desired for the equality case.

[quote="jmerry"]It looks like $ 1/(\text{lcm}\{1,2,\cdots,n\})$ to me. We have $ \binom{m}{k} \equal{} \frac {m}{k}\binom{m \minus{} 1}{k \minus{} 1}$, so $ \frac1{m}\binom{m}{k} \equal{} \frac1k\binom{m \minus{} 1}{k \minus{} 1}$
Multiply by that least common multiple $ L$, and all of the denominators are cleared. Also, if $ m$ is divisible by $ k$, $ \binom{m \minus{} 1}{k \minus{} 1}\equiv 1\mod k$. If $ m$ is a multiple of $ L$, we therefore have that the actual denominator of each $ \frac1k\binom{m \minus{} 1}{k \minus{} 1}$ is $ k$, and the common denominator is $ L$. That makes $ \frac {m}{L}$ the gcd of $ \{\binom{m}{1},\binom{m}{2},\cdots,\binom{m}{k}\}$ as desired for the equality case.[/quote]
when :$ c(n)\equal{}|\frac{f(a)\minus{}f(b)}{a\minus{}b}|$ where $ c(n)\equal{}\frac1{lcm(1,2,...,n)}$

Read the other post in the thread.

Also, don't quote the entire preceding post. It just wastes space.

Can anybody post a complete solution ? :maybe:

I've spent several days on that problem and I've really got much pleasure from it, even though I haven't got the full solution yet. Here's what I've done.
   Lemma:  Assume the polynomial $ P$ with degree $ n$ is taking integers to integers. If we write $ P(x) = \frac {a_{n}}{b_{n}}x^{n} + \frac {a_{n - 1}}{b_{n - 1}}x^{n - 1} + ... + \frac {a_{1}}{b_{1}}x + \frac {a_{0}}{b_{0}}$, where numbers $ a_{1},...,a_{n},b_{1},...,b_{n}$ are integers and $ (a_{i},b_{i}) = 1$ for any $ i$, then $ lcm(b_{1},...,b_{n})\leq{n!}.$
   Proof: Represent $ P$ is the following way:
     $ P(x) = L_{n}(x - 1)(x - 2)...(x - n) + L_{n - 1}(x - 1)(x - 2)...(x - (n - 1)) + ... + L_{2}(x - 1)(x - 2) + L_{1}(x - 1) + L_{0}$(Easy to notice that it can be done in an unique way).
      As $ P(m)$ is an integer for any integer value of $ m$, the numbers $ P(1),P(2),...,P(n)$ are all integers:
   $ P(1) = L_{0}$ is an integer. $ P(2) = L_1(2 - 1) + L_{0}$ is an integer. So, $ L_{1}$ is an integer. It's quite easy to prove bu induction that for any integer $ 0\leq{k}\leq{n - 1}$,  $ k!\cdot{L_{k}}$ is an integer number. Now consider $ P(n + 1)$:
 $ P(n + 1) = L_{n}\cdot{n!} + L_{n - 1}2\cdot{3}\cdots{n} + ... + L_{k}(n - k + 1)...(n - 1)n + ... + L_{1}n + L_{0}$, but easy to prove that for any positive integer $ 1\leq{k}\leq{n}$,  $ \frac {(n - k + 1)(n - k + 2)...n}{k!}$ is an integer (It's the same as $ \binom{n}{k}$). So, $ L_{k}(n - k + 1)(n - k + 2)...n)$ is an integer for any $ 0\leq{k}\leq{n - 1}$ but $ P(n + 1)$ is also an integer. It means that $ n!\cdot{L_{n}}$ is an integer too.  As $ k!\cdot{L_{k}}$ is an integer and $ n\geq{k}$, $ n!\cdot{L_{k}}$ is an integer. Easy to see that $ \frac {a_{i}}{b_{i}}$ can be represented as a linear combination of numbers $ L_{0},...,L{n}$. It means that $ n!a_{i}$ is divisible by $ b_{i}$,(As ${ n!\cdot{\frac {a_{i}}{b_{i}}}}$ must be integer) but $ (a_{i},b_{i}) = 1$. So, $ b_{i}|n!$, which means that  $ lcm(b_{0},...,b_{n})\leq{n!}$, Q.E.D.
      pent several days on that problem and I've really got much pleasure from it, even though I haven't got the full solution yet. Here's what I've done.
   Lemma:  Assume the polynomial $ P$ with degree $ n$ is taking integers to integers. If we write $ P(x) = \frac {a_{n}}{b_{n}}x^{n} + \frac {a_{n - 1}}{b_{n - 1}}x^{n - 1} + ... + \frac {a_{1}}{b_{1}}x + \frac {a_{0}}{b_{0}}$, where numbers $ a_{1},...,a_{n},b_{1},...,b_{n}$ are integers and $ (a_{i},b_{i}) = 1$ for any $ i$, then $ lcm(b_{1},...,b_{n})\leq{n!}.$
   Proof: Represent $ P$ is the following way:
     $ P(x) = L_{n}(x - 1)(x - 2)...(x - n) + L_{n - 1}(x - 1)(x - 2)...(x - (n - 1)) + ... + L_{2}(x - 1)(x - 2) + L_{1}(x - 1) + L_{0}$(Easy to notice that it can be done in an unique way).
      As $ P(m)$ is an integer for any integer value of $ m$, the numbers $ P(1),P(2),...,P(n)$ are all integers:
   $ P(1) = L_{0}$ is an integer. $ P(2) = L_1(2 - 1) + L_{0}$ is an integer. So, $ L_{1}$ is an integer. It's quite easy to prove bu induction that for any integer $ 0\leq{k}\leq{n - 1}$,  $ k!\cdot{L_{k}}$ is an integer number. Now consider $ P(n + 1)$:
 $ P(n + 1) = L_{n}\cdot{n!} + L_{n - 1}2\cdot{3}\cdots{n} + ... + L_{k}(n - k + 1)...(n - 1)n + ... + L_{1}n + L_{0}$, but easy to prove that for any positive integer $ 1\leq{k}\leq{n}$,  $ \frac {(n - k + 1)(n - k + 2)...n}{k!}$ is an integer (It's the same as $ \binom{n}{k}$). So, $ L_{k}(n - k + 1)(n - k + 2)...n)$ is an integer for any $ 0\leq{k}\leq{n - 1}$ but $ P(n + 1)$ is also an integer. It means that $ n!\cdot{L_{n}}$ is an integer too.  As $ k!\cdot{L_{k}}$ is an integer and $ n\geq{k}$, $ n!\cdot{L_{k}}$ is an integer. Easy to see that $ \frac {a_{i}}{b_{i}}$ can be represented as a linear combination of numbers $ L_{0},...,L{n}$. It means that $ n!a_{i}$ is divisible by $ b_{i}$,(As ${ n!\cdot{\frac {a_{i}}{b_{i}}}}$ must be integer) but $ (a_{i},b_{i}) = 1$. So, $ b_{i}|n!$, which means that  $ lcm(b_{0},...,b_{n})\leq{n!}$, Q.E.D.
     Easy to see also that any polynomial $ f$ taking integers to integers can be represented as $ f(x) = \frac {P(x)}{lcm(b_{0},...,b_{n})}$,where $ P$ is a polynomial with integer coefficients, but $ lcm(b_{0},...,b_{n})\leq{n!}$. So, $ c_{n}\geq{n!}$, because otherwise there would be a polynomial $ f$ and integer mu,bers $ a,b$ such that $ f(a)$ and $ f(b)$ are distinct and $ \frac {1}{n!} > \frac {P(a) - P(b)}{(a - b)lcm(b_{0}...,b_{n})}\geq{\frac {1}{n!}}$, which is obviously contradiction. (As $ P$ is having integer coefficients, $ \frac {P(a) - P(b)}{a - b}$ is an integer). Hence,  $ c_{n}$ can't be less than $ \frac {1}{n!}$.
    Now we are left to produce a polynomial $ P$ such that $ lcm(b_{0},...,b_{b}) = n!$ and there exist two distinct integers $ x,y$ with $ P(x) - P(y) = x - y$.
    This polynomial can be like a $ P(x) = \frac {(x + 1)(x + 2)...(x + n)}{n!}$, which is obviously taking integers to integers. Note that for $ n = 3$, the polynomial $ P(x) = \frac {(5x^{3} - 179x)}{6}$ is such that $ P(x) - P(y) = x - y$ holds for $ (x,y) = (6,0)$.

Well, i'm reviving this post, but to post a complete solution. In fact, we have $c(n)= \frac{1}{lcm(1,2,...,n)}$.

In order to prove this, we first prove that, for every polynomial $f$ taking integers to integers, we have $M (\frac{f(a)-f(b)}{a-b}) \in \mathbb{Z}$, where $M=lcm(1,2,...,n)$ and $a>b$ are integers. Let $g(x)=f(x+b)-f(b)$. Clearly, $g$ take intergers to integers, and $\frac{f(a)-f(b)}{a-b} = \frac{g(\delta)}{\delta}$, where $\delta = a-b$. Also, note that we can write $g(x)=\sum_{k = 1}^{n}a_i{x\choose k}$, with $a_i \in \mathbb{Z}$ and $a_0=0$, since $g(0)=0$. So, we have:

$M \frac{g(\delta)}{\delta}= \frac{M}{\delta} ( \sum_{k = 1}^{n}a_k{ \delta \choose k}) = \frac{M}{\delta} ( \sum_{k = 1}^{n}a_k \frac{ \delta}{k}{ \delta - 1 \choose k -1}) =\sum_{k = 1}^{n}a_k \frac{M}{k}{ \delta - 1 \choose k -1}$, which is an integer, since $a_k , \frac{M}{k}, { \delta - 1 \choose k -1} \in \mathbb{Z}, \forall k=1, \dots , n; \delta \in \mathbb{Z}$.

So, we proved that $M (\frac{f(a)-f(b)}{a-b}) \in \mathbb{Z}$, which implies $\left|\frac{f(a)-f(b)}{a-b}\right| \ge \frac{1}{M} = \frac{1}{lcm(1,2,...,n)}$ when $f(a)\ne f(b)$. Thus, we have $c(n) \ge \frac{1}{lcm(1,2,...,n)}$, since $c(n)$ is the largest.

Now we only need to prove that $c(n)= \frac{1}{M}$ works. To do this, we want to find a polynomial $f_0(x)= \sum_{k = 1}^{n}a_i{x\choose k}$, with $a_i \in \mathbb{Z}$, such that $f_0(M)=1$. If that happens, we'll have $\left|\frac{f_0(M)-f_0(0)}{M-0}\right| = \frac{1}{M}$, showing that $c(n)= \frac{1}{M}$.

Let $d=gcd({M \choose 1},{M \choose 2}, \dots, {M \choose n})$. If there's a prime $p$ such that $p|d$, then $p| {M \choose 1}$, since $d|{M \choose 1}$. Thus, $p|M$, which implies $p \le n$. Let $\alpha \in \mathbb{N}$ such that $p^{\alpha} \le n <p^{ \alpha +1}$. We have $d|{M \choose p^{\alpha}} \Rightarrow p|{M \choose p^{\alpha}}$, but this is a contradiction, since $M=M_0  . p^{\alpha}$, where $M_0$ isn't divisible by $p$ (you can see quickly that this is a contradiction using Lucas Theorem).

So, $gcd({M \choose 1},{M \choose 2}, \dots, {M \choose n})=1$, and Bézout Theorem states that there exists $a_1,a_2, \dots, a_n \in \mathbb{Z}$ such that $\sum_{k = 1}^{n}a_i{x\choose k}=1$. This implies $f_0(M)=1$, which finishes the proof that $c(n)= \frac{1}{M}$ works.

Let $L_n=\text{lcm} (1,2,3,...,n)$. We claim that $c(n)=\frac{1}{L_n}$.

Note that any polynomial $f(X)$ that maps the integers to the integers has a representation $f(X)=c_0+c_1\binom{X}{1}+c_2\binom{X}{2}+...+c_n\binom{X}{n}$.

Lemma 1: $L_n\cdot\frac{\binom{a}{n}-\binom{b}{n}}{a-b}\in\mathbb{Z}$.

Proof: Write $g(X)=\binom{X+b}{n}-\binom{b}{n}$. Note that $g(X)=d_1\binom{X}{1}+...+d_{n}\binom{X}{n}$. Now, $\frac{1}{X}\binom{X}{n}=\frac{1}{n}\binom{X-1}{n-1}$ and so the denominator of $\frac{g(X)}{X}$ must have size at most $L_n$; thus $L_n\cdot\frac{g(X)}{X}\in\mathbb{Z}$ as desired. $\blacksquare$

Now, write $T=\frac{f(a)-f(b)}{a-b}=\sum_{k=0}^n c_k\frac{\binom{a}{k}-\binom{b}{k}}{a-b}$. In particular, $v_p\left(c_k\frac{\binom{a}{k}-\binom{b}{k}}{a-b}\right)\ge -v_p(L_k)\ge -v_p(L_n)$, and thus $v_p(T)\ge -v_p(L_n)$, so $T\cdot L_n\in\mathbb{Z}$. Thus if $T\neq 0$, then $T\ge\frac{1}{L_n}$, so we established a lower bound on $c(n)$.

We show that this lower bound is attainable. We claim that, for a suitable choice of $c_i$, $\frac{f(N!)-f(0)}{N!}=\frac{1}{L_n}$ for large $N$.

First, note that $\frac{\binom{N!}{k}-\binom{0}{k}}{N!-0}=\frac{\binom{N!}{k}}{N!}=\frac{\binom{N!-1}{k-1}}{k}$. Moreover, no prime less than or equal to $k$ divides $\binom{N!-1}{k-1}$, as the expression can be written as $\prod_{i=1}^{k-1}\frac{N!-i}{i}$ and note that $\gcd\left(\frac{N!-i}{i}, L_k\right)=1$ for large $N$ and $k\le n$. In particular, this means that $\frac{f(N!)-f(0)}{N!-0}=\sum_{k=0}^{n} \frac{c_k t_k}{k}$ for $\gcd(t_k,k)=1$ fixed and some $c_k$. By Bezout, we can choose a suitable choice of $c_i$ such that the expression equals $\frac{1}{L_n}$, so we are done. $\square$

Let $g(x)=f(x+b)-f(b)$ and $m=a-b$. Then, $g(x)$ is an integer-valued polynomial with $g(0)=0$, and we want to find the largest $c(n)$ such that \[c(n)\le\left|\frac{g(m)}{m}\right|\] for all $m$.

As $g$ is integer-valued with $g(0)=0$, we can use Newton interpolation and find integers $c_1,c_2,\ldots,c_n$ such that \[g(x)=c_1\binom x1+c_2\binom x2+\ldots+c_n\binom xn\] so \[\frac{g(m)}{m}=\frac{c_1\binom m1+c_2\binom m2+\ldots+c_n\binom mn}{m}=\frac{c_1\binom{m-1}0}1+\frac{c_2\binom{m-1}1}2+\ldots+\frac{c_n\binom{m-1}{n-1}}n\] which must be some integer divided by $\text{lcm}(1,2,\ldots,n)$, so $c(n)\ge\frac1{\text{lcm}(1,2,\ldots,n)}$. To show that $c(n)\le \frac1{\text{lcm}(1,2,\ldots,n)}$, note that a construction is to set $m=1$ so it suffices to find integers $c_1,c_2,\ldots,c_n$ such that $\frac{c_1}1+\frac{c_2}2+\ldots+\frac{c_n}n=\frac1{\text{lcm}(1,2,\ldots,n)}$, which is an easy exercise in induction and Bezout's lemma.

Let $L = \operatorname{lcm}(1,\dots,n)$. We claim the answer is $c(n) = 1/L$.

By shifting suitably, we seek \[ c(n) = \min_{\substack{f(0)=0 \\ \deg f = n}} 	\min_{\substack{x \\ f(x) \neq 0}} 	\left\lvert \frac{f(x)}{x} \right\rvert. \] Taking the basis $\tbinom x1$, $\tbinom x2$, ..., $\tbinom xn$ for ${\mathbb Z} \to {\mathbb Z}$ polynomials, we may write $f(x) = a_1 \tbinom x1 + a_2 \tbinom x2 + \dots + a_n \tbinom xn$. Now for a [i]fixed[/i] $x$ the smallest nonzero value of $f(x)$ across all choices of $f$ is exactly $\gcd_{1 \le k \le n} \tbinom xk$, by using Euclidean algorithm on $a_k$. So we now interchange the order of the $\min$ operators to obtain \[ 	\boxed{c(n) = \min_{x \neq 0} \frac{\left\lvert G(x) \right\rvert}{x}} 	\quad\text{where}\quad 	\boxed{G(x) \overset{\text{def}}{=} 	\gcd\left( \tbinom x1, \tbinom x2, \dots, \tbinom xn \right)}. \] First, we note $G(L) = 1$, since if $p^e$ exactly divides $L$ (even if $e=0$) then $p \nmid \tbinom{L}{p^e}$ by Lucas theorem, say. Thus $c(n) \le 1/L$.

On the other hand since  \[ L \cdot \binom xk = L \cdot \frac{x}{k} \binom{x-1}{k-1} 	= x \cdot \frac Lk \binom{x-1}{k-1} \in x \cdot {\mathbb Z} \] we see that $L \cdot G(x) \ge x$ for any $x \in {\mathbb Z}$. Hence $c(n) \ge 1/L$, as desired.

The following problem appears as an exercise in chapter 10 of PftB. It is cited as "MOSP 2001."
[quote]
Let $f$ be a polynomial with rational coefficients such that $f(n) \in \mathbb{Z}$ for all $n \in \mathbb{Z}$. Prove that for any integers $m$, $n$ the number
\[\mathrm{lcm}[1, 2, \dots, \deg(f)] \cdot \frac{f(m) - f(n)}{m - n}\]
is an integer.

The answer is $c(n)=1/d_n$ where $d_n:=\mathrm{lcm}(1,\ldots,n)$.

We first show that $c(n)\le 1/d_n$. To do this, we need a lemma.

[u]Lemma 1:[/u] $\gcd\left[\binom{d_n}{1},\ldots,\binom{d_n}{n}\right]=1$.

[u]Proof of Lemma 1:[/u] Suppose that there was a common prime factor, call it $p$. Now, let $p^e$ be the largest power of $p$ that is at most $n$. We see that $e=\nu_p(d_n)$ (this is the key fact of the lcm we are using). Now, we claim that $\nu_p\left(\binom{n}{p^e}\right)=1$, which would be a contradiction.

We have that $d_n=kp^e$ for $\gcd(k,p)=1$. For $0\le a<p^e$, we have $\nu_p(kp^e-a)=\nu_p(p^e-a)$ (if $a=0$ then this is obvious, and if not, both quantities are equal to $\nu_p(a)$). Thus,
\[\nu_p((kp^e)(kp^e-1)\cdots(kp^e-p^e+1))=\nu_p((p^e)(p^e-1)\cdots(p^2-p^e+1)),\]
so $\nu_p\left(\binom{n}{p^e}\right)=1$, which is the desired contradiction. $\blacksquare$

Now, by induction with Bezout's lemma as a base case, we have that there are integers $a_1,\ldots,a_n$ such that
\[a_1\binom{d_n}{1}+\cdots+a_n\binom{d_n}{n}=1.\]
Let $f(X)=a_1\binom{X}{1}+\cdots+a_n\binom{X}{n}$. Then,
\[\frac{f(d_n)-f(0)}{d_n-0}=\frac{1}{d_n},\]
so we must have have $c(n)\le 1/d_n$.

We'll now show that $c(n)\ge 1/d_n$, which is the same as showing that
\[\left|\frac{f(a)-f(b)}{a-b}\right|\ge\frac{1}{d_n}.\]
We have a lemma.

[u]Lemma 2:[/u] $d_n\frac{f(a)-f(b)}{a-b}\in\mathbb{Z}$.

[u]Proof of Lemma 2:[/u] Let $\delta=a-b$. We see that
\[g(\delta):=f(b+\delta)-f(b)\]
is a polynomial in $\delta$ with degree at most $n$ that outputs integral values, and $g(0)=0$. Expand $g(X)$ in the binomial basis, so
\[f(X)=a_1\binom{X}{1}+\cdots+a_n\binom{X}{n}\]
for some rational numbers $a_1,\ldots,a_n$. Since $g(1)\in\mathbb{Z}$, we have $a_1\in\mathbb{Z}$. Now, since $g(2)\in\mathbb{Z}$, we deduce $a_2\in\mathbb{Z}$, and so on. Therefore, all the $a_i$ are integers. Now, $\frac{1}{X}\binom{X}{p}=\frac{1}{p}\binom{X-1}{p-1}$, so
\[f(X)/X=\frac{a_1}{1}\binom{X-1}{0}+\frac{a_2}{2}\binom{X-1}{1}+\cdots+\frac{a_n}{n}\binom{X-1}{n-1}.\]
In particular then, we see that
\[\frac{f(a)-f(b)}{a-b}=g(\delta)/\delta=\frac{A}{d_n}\]
for some integer $A$. Therefore, the lemma follows. $\blacksquare$

Now, since $\frac{f(a)-f(b)}{a-b}\ne 0$, and its an integer over $d_n$, its smallest possible absolute value is $1/d_n$, so $c(n)\ge 1/d_n$, as desired.

Combining the two parts, we have $c(n)=1/d_n$. $\blacksquare$


function
integration
algebra
functional equation
real analysis
real analysis unsolved

$ \textbf{1.}$ Evaluate:

$ \sum^{\infty}_{m\equal{}1}\sum^{\infty}_{n\equal{}1}\frac{m^{2}n}{3^{m}\left(n3^{m}\plus{}m3^{n}\right)}$

$ \textbf{2.}$  $ x$ is a nonzero real and $ x<1$. Prove that there exist numbers $ c_{1}$ and $ c_{2}$ such that:

$ \frac{c_{2}}{\sqrt{1\minus{}x}}\leq\prod^{\infty}_{k\equal{}0}\left(1\plus{}x^{4^{k}}\right)\leq\frac{c_{1}}{\sqrt{1\minus{}x}}$

$ \textbf{3.}$ Prove that $ \sqrt{2}$ is the limit of a sequence of numbers in the form $ \sqrt[3]{n}\minus{}\sqrt[3]{m}$

$ \textbf{4.}$ Find all continous functions $ f: \mathbb{R}^{\plus{}}\rightarrow\mathbb{R}$ such that for all positive reals $ x$ :

$ f\left(x^{2}\right)\equal{}\frac{f(x)}{x}$

[hide="My Solution"]if we put $ \sqrt{x}$ instead of $ x$ in the equation we'll have:
$ f\left(x\right)\equal{}\frac{f\left(\sqrt{x}\right)}{\sqrt{x}}$
as a consequence $ f\left(x^{2}\right)\equal{}\frac{f\left(\sqrt{x}\right)}{x\sqrt{x}}\equal{}\frac{f\left(x^{\frac{1}{2}}\right)}{x^{1\plus{}\frac{1}{2}}}$

then we can easily prove that for all $ n\in\mathbb{N}$, $ f\left(x^{2}\right)\equal{}\frac{f\left(x^{\frac{1}{2^{n}}}\right)}{x^{1\plus{}\frac{1}{2}\plus{}\cdots\plus{}\frac{1}{2^{n}}}}$

now if $ n\rightarrow\infty$ we have $ \frac{1}{2^{n}}\rightarrow0$ and $ 1\plus{}\frac{1}{2}\plus{}\cdots\plus{}\frac{1}{2^{n}}\rightarrow2$

so the last equation will be in the form: $ f\left(x^{2}\right)\equal{}\frac{f(1)}{x^{2}}\equal{}\frac{k}{x^{2}}$

but as $ x^{2}$ is a positive real so we can say that the solution for this functional equation will be $ f(x)\equal{}\frac{k}{x}$ for constant $ k$.
[/hide]

$ \textbf{5.}$ Let $ f$ be a $ n\plus{}1$ times differentiable function on interval $ [a,b]$. Also for $ k\equal{}0,1,\cdots,n$ we have $ f^{(k)}(a)\equal{}f^{(k)}(b)\equal{}0$. Prove that there exist a number $ c$ in the interval $ (a,b)$ such that $ f^{(n\plus{}1)}(c)\equal{}f(c)$.
 :D

Wow Really nice solution :) can anyone solve $ 1$? i spent 2 days on it, i'm really confused!  :(

for $ 3$, you can replace $ x^3$ by any function function $ f$ such that $ f(x) \to \infty$ and $ f(n\plus{}1)\minus{}f(n) \to 0$

for 2/, here is an approach (already discussed recently). You have to show that if $ f(x) \equal{} \sum_{k\equal{}0}^{\infty} \ln(1\plus{}x^{4^y})$ then $ f(x) \minus{} \frac{|\ln(1\minus{}x)|}{2}$ stays uniformly bounded on $ [0;1[$. By a simple serie/integral comparison (i.e. $ \ln(1\plus{}x^{4^k}) \approx \int_{k}^{k\plus{}1} \ln(1\plus{}x^{4^y}) dy$, one easily obtains that $ f(x) \equal{} \int_0^{\infty} \ln(1\plus{}x^{4^y}) dy \plus{} O(1)$ where $ \int_0^{\infty} \ln(1\plus{}x^{4^y}) dy \equal{} \frac{1}{\ln(4)} \int_{|\ln(x)|}^{\infty} \frac{\ln(1\plus{}\exp(\minus{} \theta))}{\theta} d \theta \equal{} C \plus{} \frac{|\ln(1\minus{}x)|}{2} \plus{} o(1)$. Hence the conclusion

[hide="1"]
Let $ f(m,n)$ be the function under the summation and $ S=\sum_{m,n\geq 0} f(m,n)$. (Any ordering of the summation is valid since the terms go to 0 faster than $ 2^\max \{ m,n\}$.) Then also $ S=\sum_{m,n\geq 0} f(n,m)$ so our sum is
\[ \frac{1}{2}\sum_{m,n\geq 0} \frac{m^2n}{3^m(3^mn+3^nm)}+\frac{mn^2}{3^n(3^nm+3^mn)}\]
\[ =\frac{1}{2}\sum_{m,n\geq 0} \frac{mn(3^nm+3^mn)}{3^m3^n(3^mn+3^nm)}\]
\[ =\frac{1}{2}\sum_{m,n\geq 0} \frac{m}{3^m}\cdot \frac{n}{3^n}=\frac{1}{2}\left(\sum_{m=0}^\infty \frac{m}{3^m}\right)^2\]
The sum on the right can easily be evaluated to $ \frac{3}{4}$, making the answer $ \boxed{\frac{9}{32}}$.
[/hide]

[quote="alekk"]for $ 3$, you can replace $ x^3$ by any function function $ f$ such that $ f(x) \to \infty$ and $ f(n \plus{} 1) \minus{} f(n) \to 0$[/quote]

$ (x \plus{} 1)^3 \minus{} x^3 \overset{x \to \infty}{\longrightarrow} \infty$, but luckily $ (x \plus{} 1)^{\frac{1}{3}} \minus{} x^{\frac{1}{3}} \overset{x \to \infty}{\longrightarrow} 0$ ;)

I of course meant that $ f(x)\equal{}x^{\frac{1}{3}}$ satifies $ f(x) \to \infty$ and $ f(n\plus{}1)\minus{}f(n) \to 0$  :blush:


function
Euler
trigonometry
search
real analysis
logarithms
calculus

Solving ${f}$ from

${f(x)=f(P+x)}$

To Simplify

${\frac{f(P+x)}{f(x)}=1}$

Assuming that the form of ${f}$ is exponential, because of the fraction of the function and that function "slided" is constant.

${f(u)=\alpha e^{u \beta }}$

Substituing

${\frac{f(P+x)}{f(x)}=e^{P \beta }=1}$

From [b]Euler[/b]:

${c_{1}\in \mathbb{Z}}$
${e^{2 \pi i c_{1}}=1}$

Substituing

${e^{P \beta }=e^{2 \pi i c_{1}}}$
${P \beta =2 \pi i c_{1}}$
${\beta =\frac{2 \pi i c_{1}}{P}}$

Partial Solution

${f(u)=e^{\frac{2 i \pi u c_{1}}{P}}\alpha }$

By a basic proof, the sum of two solution is a solution too.
And a solution times constant is a solution too.
${f_{1}}$ and ${f_{2}}$ are solution.

${f(u)=f_{1}(u)+f_{2}(u)}$

${f(x)=f(P+x)}$
${f_{1}(x)+f_{2}(x)=f_{1}(P+x)+f_{2}(P+x)}$
${f_{1}(x)-f_{1}(P+x)=-\left(f_{2}(x)-f_{2}(P+x)\right)}$

${f_{1}(x)-f_{1}(P+x)=0}$
${f_{2}(x)-f_{2}(P+x)=0}$

So ...

${f(u)=\sum_{n=0}^{\infty }e^{\frac{2 i \pi u n}{P}}\alpha_{n}}$

From Euler Again

${e^{\frac{2 i \pi u n}{P}}\alpha_{n}=\cos \left(\frac{2 n \pi u}{P}\right) \alpha_{n}+i \sin \left(\frac{2 n \pi u}{P}\right) \alpha_{n}}$

Simplifying, because ${\alpha}$ can be complex.
So a solution times constant is a solution too.

${\cos \left(\frac{2 n \pi u}{P}\right) \alpha_{n}+\sin \left(\frac{2 n \pi u}{P}\right) \beta_{n}}$

And Finally the Fourier Series

${\sum_{n=0}^{\infty }\left(\cos \left(\frac{2 n \pi u}{P}\right) \alpha_{n}+\sin \left(\frac{2 n \pi u}{P}\right) \beta_{n}\right)}$

Now, if someone interested

Solve:

${\alpha f(x)+\chi f(x+\delta )+\beta f(x+\epsilon )=0}$

It's analogous, but much more harder

[hide]
Use the lagrange inversion
[/hide]

It's all nice but the hardest part is missing: one has to show that there are no other solutions. The question in what sense to understand the sum of the series (it may diverge) is another issue to be addressed. ;)

fedja
its nonsense proof this things

because fourier had alread proof this!!

but i search on all internet a deduction of fourier series and i didint find, so i put here my own deduction i made

the title is "Fourier Series Deduction"

not "Fourier Series Proof"

but tomorrow i will complet tis topic with this proof

im bad with this thing, if someone want to proof, write here

thanks

[quote="Thales418"]fedja
its nonsense proof this things
because fourier had alread proof this!!
[/quote]
I'm not so sure :). Kent and a few other people here know history much better than I but as far as I remember Fourier himself was very far from a rigorous proof that every reasonable periodic function can be represented by what we now call Fourier series (please, correct me if I'm wrong). The fact that all solutions of a convolution equation $f*\mu=0$ with compactly supported measure $\mu$ can be approximated by linear combinations of exponential solutions in many reasonable spaces of functions is definitely something proved much later. And the problem you proposed in the end of your post requires exactly this result though for quite special measures.

Anyway, I just wanted to attract your attention to the fact that while your "deduction" may clarify something to you (and, if it does, that is wonderful), without the "proofs" you omitted it isn't worth much: the main problem lies not in guessing that certain exponents fit and showing that the solutions form a linear space but in proving that in a certain sense, those are all possible solutions. What you "deducted" is that every Fourier series represents a periodic function; what one usually understands by "deduction of Fourier series" is showing that every reasonable periodic function can be represented by a Fourier series. See the difference?

As to posting proofs tomorrow, don't bother doing it for $f(x)=f(x+P)$ (those can be found in any textbook) but it would be really interesting to see what and how you suggest to prove for the equation you posted in the end. ;)

fedja

the last equation is not for me to solve, is a question for people in this forum

i thought that people will at least try to solve it

but lets the hints

${\alpha f(x)+\chi f(x+\delta )+\beta f(x+\epsilon )=0}$

Because of it was a linear recursive equation the solution is exponential. A simple book of recursive equation have a proof of this. This is why before i suposed it was exponential.

The solution can be   ${f(x)=a b^{x c}}$

But it' s better use  ${b=e}$

In this case  ${a e^{x c}}$

can be  ${c=\log (b) s}$

So  ${a e^{x c}=a b^{s x}}$

We eliminate a letter that is unecessary

Using

${f(x)=a e^{x b}}$
${\alpha f(x)+\chi f(x+\delta )+\beta f(x+\epsilon )=0}$
${a e^{b x}\alpha+a e^{b (x+\epsilon )}\beta+a e^{b (x+\delta )}\chi =0}$
${a e^{b x}\left(\alpha+e^{b \epsilon }\beta+e^{b \delta }\chi \right)=0}$

If we assume that  ${a\neq 0}$ and ${e^{b x}\neq 0}$ .
We get

${\alpha+e^{b \epsilon }\beta+e^{b \delta }\chi =0}$

Substituing

${b=\log (k)}$
${\alpha+e^{b \epsilon }\beta+e^{b \delta }\chi =0}$
${\chi k^{\delta }+\beta k^{\epsilon }+\alpha =0}$

Now we need to find ${k}$, but i don't know how to do that.
We can try use Lagrange Inversion. But I Dont know how to use it.
So. that was my Hint.

I don't know if fourier proof the equation that has his name. maybe euler the one who proof

heehhe

and at last but not the lest

i proof that every periodic function can be represented by a fourier series

the solution of a linear recursive equation is exponential, dont have another solution.

and i proof that the sum of a solution is a solution

so the sum of all possibel solutions is a solution too

so i put all the solutions

and put negative ${n}$ is nonsens, cause cosine and sine are even or odd.

you that dont understand that it was a proof

but find ${k}$ in my hint

[quote="Thales418"]
Because of it was a linear recursive equation the solution is exponential. A simple book of recursive equation have a proof of this.
[/quote]
This is exactly what I don't believe. If your $f$ were defined on integers only, I would agree that it would be a finite linear combination of "exponentials" but it is defined on reals so everything gets much more complicated. No simple book on recursive equations will contain a statement like the one you made. Also pure exponents may be not enough: one really needs to consider exponential polynomials. Even your equation with certain choice of coefficients (say, $f(x)-2f(x+1)+f(x+2)=0$ may have a solution ($f(x)=x$ in this case) that is not a sum of pure exponents. Lagrange inversion doesn't help much when you want to find $k$ (it can take infinitely many possible complex values). Actually, no explicit formula for the roots of a linear combination of more than two generic exponential functions exists.

[quote]
so the sum of all possibel solutions is a solution too
so i put all the solutions
[/quote]
Don't you see a logical fallacy here? You just got [b]very many[/b] solutions by adding exponents together but there is no guarantee that they are [b]all[/b] possible solutions.

Again lets explain....

Im borred

${-2 f(x+1)+f(x+2)+f(x)=0}$

More Generally

${\beta f(x+1)+\chi f(x+2)+\alpha f(x)=0}$

${f(x)=a b^{x}}$

${\beta f(x+1)+\chi f(x+2)+\alpha f(x)=0}$
${a \alpha b^{x}+a \beta b^{x+1}+a \chi b^{x+2}=0}$
${a b^{x}(\alpha+b (\beta+b \chi ))=0}$

${a\neq 0\land b^{x}\neq 0}$

${\alpha+b (\beta+b \chi )=0}$
${b=\frac{-\beta \pm \sqrt{\beta^{2}-4 \alpha \chi }}{2 \chi }}$

${f(x)=a_{2}\left(\frac{-\beta-\sqrt{\beta^{2}-4 \alpha \chi }}{2 \chi }\right)^{x}+\left(\frac{\sqrt{\beta^{2}-4 \alpha \chi }-\beta }{2 \chi }\right)^{x}a_{1}}$

But If:

${-2 f(x+1)+f(x+2)+f(x)=0}$
${f(x)=a b^{x}}$
${a b^{x}-2 a b^{x+1}+a b^{x+2}=0}$
${a (b-1)^{2}b^{x}=0}$
${b=1}$

What means that the characteristic equation have a double root.It means that the solution may be multiplied by x.
In the fourier deduction

${f(x)=f(P+x)}$
${f(x)=a b^{x c}}$
${a b^{c x}=a b^{c (P+x)}}$
${a b^{x}\left(b^{P}-1\right)=0}$

We dont have a double root. So the solution is exponential
Try solve the differential analogous

${f(x)-2 f'(x)+f''(x)=0}$

The solution is

${e^{x}c_{1}+e^{x}x c_{2}}$

Because of the double root

[quote]
Don't you see a logical fallacy here? You just got very many solutions by adding exponents together but there is no guarantee that they are all possible solutions.
[/quote]

I proof that one solution is solution.
but ${n}$ can be every integer possible.
And the irrational square root have infinity solution.

I made a proof of this.

please, read again

if dont understant i will make a less difficult proof


You're studying. You're absorbed in your work and having lots of fun.
A few hours later, you get up, and you realize your neck/back/arms hurt...

In your opinion, what is the best working position?

I have a (presumably bad) habit of leaning towards my work and curling myself so my is head directly above it. If my neck is extended as a line perpendicular to the chair, I won't be able to see the paper so obviously that's not completely right.

Against the side arms of the couch, knees bent up, math book and paper in lap. Always seems best for me.

Doing the work in a chair.

But I sometimes do it by laying on my back and my head hanging off the side of the bed working the problems.

Sometimes, unless you have a pillow propped up in the chair, the plastic/wood/metal will get to you, if you're sitting there for a long time. That's what I don't like  :wink:

[quote="7h3.D3m0n.117"][hide="The Most Comfortable"]
Sometimes I will lay on my bed and do some math. This hurts primarily my spine and elbows (laying on my stomach). I can usually relax and do some free math during this time.
[/hide]

[hide="The Better Suited"]
I think the best (and pain-free) position is to just sit at your desk. Don't cross your legs in an awkward position or anything, just a normal "sit-in-the-chair-properly." Ideal for important math or serious work.
[/hide][/quote]

D'OH. Well said. Ditto here. When I'm just relaxing I just work on my bed. Hurts you back after a while, though.

I work in a sort of squatting position with my chin resting on my knees.  This is actually comfortable and it is good for your legs and back.

We wasted a bunch of time in class talking about how you shouldn't do your work in bed because you're brain is usually trained to sleep on it, so you won't be able to concentrate as much.


geometry
geometric transformation
reflection

Please don't discuss the these problems now.  :lol:

So when's the deadline?

And what about the previous issues?  :)

1)At least 60 days.
2)You can find here http://www.nxbgd.com.vn/toanhoctuoitre/?p=8 .  :wink:

Someone in Mathlinks have posted M&Y problem before. If you see it , don't reply and call a MOD to lock this topic if it isn't deadline  :wink:

I think the problems in this magazine are really interesting, but it's such a shame that the solutions are written in Vietnamese. Can any of the Vietnamese members translate some of the problems already past the deadline, so that we can enjoy them here?

[quote="tuannguyen"]To massnet : Why don't you learn Vietnamese, that's a good idea.[/quote]

as long as you don't try to pronounce the words...

[quote="N.T.TUAN"]Yes, I can.  :wink:[/quote]
I can too :)

2010, issue 5: http://trungtuan.wordpress.com/2010/06/05/topic-101/ .

[quote="murgi"]Cannt we get the previous issues
 magazines in english.[/quote]
 Mathematics and Youth Magazine, Problem in 411 Issue: [code]http://mathzine.com/2011/09/mathematics-and-youth-magazine-problem-in-411-issue/[/code]
Mathematics and Youth Magazine, Problem in 412 Issue: [code]http://mathzine.com/2011/10/mathematics-and-youth-magazine-problem-in-412-issue/[/code]

[quote="mathdoi"]
 Mathematics and Youth Magazine, Problem in 411 Issue: [code]http://mathzine.com/2011/09/mathematics-and-youth-magazine-problem-in-411-issue/[/code]
Mathematics and Youth Magazine, Problem in 412 Issue: [code]http://mathzine.com/2011/10/mathematics-and-youth-magazine-problem-in-412-issue/[/code][/quote]

Continue
Mathematics and Youth Magazine, Problem in 414 Issue: [code]http://mathzine.com/2011/12/mathematics-and-youth-magazine-problems-in-414-issue/[/code]

[quote="amparvardi"]Thank you, but can you please tell us the deadlines? It would be great to find the cheaters. :)[/quote]
20/2/2012

Find collection of problems from mathematics and youth magazine here https://artofproblemsolving.com/community/u324646h1377574p7625991

[quote=molympiad]Find collection of problems from mathematics and youth magazine here http://www.molympiad.ml/2017/02/mathematics-and-youth-magazine-problems/[/quote]

[quote=N.T.TUAN]2007, issue 9[/quote]

I wonder if the link of the material works well as the download page says "The page returned wrong reply.."

All problems from Mathematics and Youth Magazine in 2016

[quote=Skravin][quote=molympiad][quote=Skravin]

I wonder if the link of the material works well as the download page says "The page returned wrong reply.."[/quote]

Please find all MYM's problems here http://www.molympiad.ml/2017/02/mathematics-and-youth-magazine-problems/[/quote]

thanks

Is it possible to use those materials for non-commercial purposes?(and up to what year's material..)[/quote]

Yes, you can use those materials for non-commercial purpose, up to issue 473, november 2016. Thanks

issue 476, february 2017 http://www.molympiad.ml/2017/04/mathematics-and-youth-magazine-problems-feb-2017-issue-476/

how can we find solution to the problem 4 in 2007 Issue 9????

[quote=yangml]Cn you send all the mathmatical reflection problem to me ?[/quote]

You can get all mathematical reflection problems at http://molympiad.blogspot.com/2017/06/mathematical-reflections.html.

[url=http://molympiad.blogspot.com/2017/06/mathematics-and-youth-magazine-problems_1.html][b]MYM T5/480.[/b][/url] Suppose that $a,b,c$ are three positive numbers and given $\alpha,\beta,\lambda\geq2$. Prove that \[\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}+\frac{(a-b)^{2}}{\alpha(b+c)(c+a)}+\frac{(b-c)^{2}}{\beta(c+a)(a+b)}+\frac{(c-a)^{2}}{\lambda(a+b)(b+c)}.\]

Link to view mathematics and you magazine for anyone who are interested in
https://drive.google.com/drive/folders/0B8d9MNCjnXuoN2hSUVFGMjY3a0E


geometry
circumcircle
geometric transformation
reflection
parallelogram
perpendicular bisector

$ ABC$ is a triangle with orthocenter $ H$. The median from $ A$ meets the circumcircle again at $ A_1$, and $ A_2$ is the reflection of $ A_1$ in the midpoint of $ BC$. The points $ B_2$ and $ C_2$ are defined similarly. Show that $ H$, $ A_2$, $ B_2$ and $ C_2$ lie on a circle.

Let the reflection of $ H$ with respect to side $ BC$ be $ H_1$, and the reflection of $ H$ with respect to the midpoint $ M$ of $ BC$ be $ H_2$. Let the centroid of the triangle be $ G$ and the circumcentre be $ O$. Then:

It is known that $ H_1$ is on the circumcircle of $ \triangle{ABC}$. We can get that $ H_1H_2 \parallel BC$. Since $ O$ is the midpoint of $ AH_2$ and $ M$ is the midpoint of $ HH_2$ and $ A$, $ H$, $ H_2$ are on the same line, so we know that $ OG$ is the perpendicular bisector of $ H_1H_2$, which gives that $ H_2$ is also on the circumcircle of $ \triangle{ABC}$.

Hence, $ \angle{AA_1H_2}\equal{}90^o$. Because $ A_2HA_1H_2$ is a parallelogram, so $ HA_2 \perp AA_1$. Which gives that $ \angle{GA_2H}\equal{}90^o$. Similarly, we have that $ \angle{GB_2H}\equal{}\angle{GC_2H}\equal{}90^o$. So $ A_2$, $ B_2$ and $ C_2$ are on the same circle with diameter $ GH$.

Dear Mathlinkers,
with a generalization, you can see
http://jl.ayme.pagesperso-orange.fr/   vol. 5, le cercle de Hagge p. 43-47.

Sincerely
Jean-Louis


Let $ p,\ q$ be prime with $ p < q$.

(1) Prove that there don't exist integer $ r$ such that $ \frac {1}{p} \plus{} \frac {1}{q} \equal{} \frac {1}{r}$.

(2) Prove that it is limted to $ p \equal{} 2,\ q \equal{} 3$ such that there exists integer $ r$ such that $ \frac {1}{p} \minus{} \frac {1}{q} \equal{} \frac {1}{r}$.

[hide="(1)"]
$ \frac 1p \plus{} \frac 1q \equal{} \frac {p \plus{} q}{pq}$. Now, $ \gcd(p \plus{} q,pq) \equal{} 1 \iff \gcd(p,q) \equal{} 1$, which is true since $ p$ and $ q$ are different prime numbers. Therefore, it follows that the fraction is irreducible, and since $ p \plus{} q > 1$, it cannot be expressed in the form $ \frac 1r$.[/hide]

[hide="(2)"]
$ r\equal{}6$: $ \frac 12 \minus{}\frac 13 \equal{} \frac 16$, which is true, QED.[/hide]

OK.
[hide="(2)"]
$ \frac 1p \minus{} \frac 1q \equal{} \frac{q\minus{}p}{pq}$. Now, $ \gcd(q\minus{}p, pq)\equal{}1 \iff \gcd(p\plus{}q,pq)\equal{}1 \iff \gcd(p,q)\equal{}1$, which is true, so the fraction is irreducible. The only two values of $ p$ and $ q$ such that $ q\minus{}p\equal{}1$ are $ p\equal{}2$ and $ q\equal{}3$, which are the only two prime numbers which differ by 1, QED.[/hide]


[quote="rd5493"][quote="kohjhsd"][quote="happyme"]If the ratio of $y+5x$ to $x+2y$ is ${2}/{3}$, what is the ratio of $y$ to $x$?[/quote]

[hide]$3y+15x=2x+4y$
$13x=y$

Therefore $\frac{y}{x}=\boxed{13}$[/hide][/quote]

Actually, it would go like this.

If the ratio of y+5x to x+2y is 2/3, then we could say that (y+5x)/(x+2y)=2/3.  Through cross-multiplication we find that 3y+15x/2x+4y=1.  By rearranging the numbers we get 3y+15x/4y+2x=1.  We then see that 3/4y+15/2x=1.  Subtract 3/4y from each side.  15/2x=-3/4y.  Cross multiply again.  We get -60/6x=y.  Divide.  We get -10x=y.  So the ratio of y to x is -10.[/quote]
[hide]
Actually ,after cross multiplying, you get
$3y$ $+$ $15x$ $=$ $2x$ $+$ $4y$
$13x$ $=$ $y$
[/hide]

o, my bad. can u xplain how the 2 answers r set equal 2 each other.

what do you mean by set equal to each other?

[hide]if the ratio of y+5x to x+2y is 2/3, then (y+5x)/(x+2y)=2/3

after you cross multiply, you get 3(y+5x)=2(x+2y)

after distribution, it becomes 3y+15x=2x+4y

simplified, it becomes y=13x[/hide]


Discuss each equation below; that is, find the center, foci and vertices of each given ellipse.

(1) x^2 + 3y^2 - 12y + 9 = 0

(2) 4x^2 + 3y^2 + 8x - 6y = 5

Refer to my post here http://www.artofproblemsolving.com/Forum/viewtopic.php?t=190807 and try solving them.


prove the following stronger schur for nonnegative $a,b,c$
$\sum_{\substack{cyc}}a^{3}+3abc\ge \sum_{\substack{sym}}a^{2}b+2.\sum_{\substack{cyc}}(\sqrt{a}-\sqrt{c})^{2}(\sqrt{ab}(a-b)+\sqrt{bc}(c-b)+b\sqrt{ca})$

I didn't check is it right or wrong...but How do you make it? ($RHS$)

i will definitely tell u but sometime later and alongwith the generalization to n   terms. it is not a computer based question and is observed with hand only.

Nice!
[hide="hint"] :) Put $x=\sqrt{a}$ etc. and first show that $last\ sum\ of\ RHS=\sum_{cyc}x^{3}(y+z)(x-y)(x-z)$. ;) [/hide]


inequalities
quadratics
inequalities unsolved

Let $x_i,$ $1 \leq i \leq n$ be real numbers with $n \geq 3.$ Let $p$ and $q$ be their symmetric sum of degree $1$ and $2$ respectively. Prove that:

i) $p^2 \cdot \frac{n-1}{n}-2q \geq 0$

ii) $\left|x_i - \frac{p}{n}\right| \leq \sqrt{p^2 - \frac{2nq}{n-1}} \cdot \frac{n-1}{n}$ for every meaningful $i$.

i) is easy, ii) i still dont know how to deal with x - p/n.

here is part i

Let t = $\sum x_i^2$.

i)  Rewrite as

$(n-1)p^2 \geq 2qn$

$\Leftrightarrow (n-1)t \geq 2q$

But $t \geq q$ and $(n-1) \geq 2$ with equality when $n=3$ and $x_1 = x_2 = x_3$.

[quote="N.T.TUAN"][quote="orl"] Let $p$ and $q$ be their symmetric sum of degree $1$ and $2$ respectively. [/quote]

What are those?[/quote]

$\sum x_i, \sum x_ix_j$

[quote="warut_suk"]

$\sum x_ix_j$[/quote] $n!$ element in this sum?

Thank you very much.

ii) can also be proven if we look at it as quadratic inequality towards $x_i$ and show ( using i) ) that discriminant is negative iff $(n-1)p^2 - 2nq>0$ and if the last is $0$ then all $x_i$ are equal and LHS of ii) is 0


$ x^3\plus{}3367\equal{}2^n$    find natural $ x,n$

Surely, this is number theory... and it's an easy one, by the way:

Notice that the remainder of $ x^3$ modulo $ 7$ can be $ 0$, $ 1$ or $ \minus{}1$. Since $ 7|3367$ and the possible remainders of $ 2^n$ modulo $ 7$ are $ 2$, $ 4$ and $ 1$, we conclude that $ 3|n$. Let $ n\equal{}3n_1$. Then we have $ 3367\equal{}(2^{n_1}\minus{}x)(2^{2n_1}\plus{}2^{n_1}\cdot x\plus{}x^2)$, which should be easy enough to solve ($ 3367\equal{}7\cdot 13\cdot 37$). I don't have time right now, so I'll see to it later.

OK, it's simple after this:

$ 2^{n_1}\minus{}x$ is either equal to $ 1$, $ 7$, $ 13$, $ 37$, since $ 2^{2n_1}\plus{}2^{n_1}\cdot x\plus{}x^2$>$ 2^{n_1}\minus{}x$ and $ 2^{2n_1}\plus{}2^{n_1}\cdot x\plus{}x^2>0$ Let $ t\equal{}2^{n_1}\minus{}x$. Obviously, we have $ x\equal{}2^{n_1}\minus{}t$, which transforms our original equation to
$ 2^{3n_1}\minus{}3\cdot 2^{2n_1}\cdot t\plus{}3\cdot 2^{n_1}\cdot t^2\minus{}t^3\plus{}3367\equal{}2^n$, which is equivalent to
$ 3ts^2\minus{}3t^2s\plus{}(t^3\minus{}3367)\equal{}0$.
We can solve this quadratic equation by $ s$ ($ s\equal{}2^{n_1}$) for all $ t$. Since we have only four values for $ t$, it's not that hard.
It turns out that the only value for $ (n, x)$ is $ (12,9)$. I might have missed some, of course, but that doesn't really change the way we came to the solution.


search
inequalities unsolved
inequalities

For what pairwise different positive integers is the value of

$ \frac{x}{x\plus{}1}\plus{}\frac{y}{y\plus{}1}\plus{}\frac{z}{z\plus{}1}\plus{}\frac{k}{k\plus{}1}$

an integer?

[quote="ashrafmod"]For what pairwise different positive integers is the value of

$ \frac {x}{x \plus{} 1} \plus{} \frac {y}{y \plus{} 1} \plus{} \frac {z}{z \plus{} 1} \plus{} \frac {k}{k \plus{} 1}$

an integer?[/quote]

$ \frac {x}{x \plus{} 1} \plus{} \frac {y}{y \plus{} 1} \plus{} \frac {z}{z \plus{} 1} \plus{} \frac {k}{k \plus{} 1}$

Since each value is less than $ 1$, the sum must be $ <4$

If it was $ \leq 2$ then each term would have to have a mean of $ \frac{1}{2}$. This is impossible, hence we must search for solutions to

$ \frac {x}{x \plus{} 1} \plus{} \frac {y}{y \plus{} 1} \plus{} \frac {z}{z \plus{} 1} \plus{} \frac {k}{k \plus{} 1}\equal{}3$

We note that $ \frac{1}{2}\plus{}\frac{2}{3}\plus{}\frac{3}{4}\plus{}\frac{4}{5} < 3$
and $ \frac{2}{3}\plus{}\frac{3}{4}\plus{}\frac{4}{5}\plus{}\frac{5}{6} > 3$

We need to search for solutions in the range


Members of the Rockham Soccer League buy socks and T-shirts. Socks cost $ \$4$ per pair and each T-shirt costs $ \$5$ more than a pair of socks. Each member needs one pair of socks and a shirt for home games and another pair of socks and a shirt for away games. If the total cost is $ \$2366$, how many members are in the League?

$ \textbf{(A)}\ 77 \qquad
\textbf{(B)}\ 91 \qquad
\textbf{(C)}\ 143 \qquad
\textbf{(D)}\ 182 \qquad
\textbf{(E)}\ 286$

t-shirt is $ 4\plus{}5\equal{}\$9$so one member needs $ 2(4\plus{}9)\equal{}\$26$ so team has $ \frac{2366}{26}\equal{}91$ members

[quote="pouyanss26"]but this is not correct because you play with 91 teams so there is 92 teams !!!!!!![/quote]
There is not 91 teams, there are 91 members in a team. There does not mention anything about playing with other teams.

A pair of socks and a t-shirt sum to $13$, therefore each player needs $26$, so we have $\frac{2366}{26} \rightarrow \boxed{91}$


source: my own :D
prove if for real numbers $a_1,a_2,a_3,...,a_n$ we have
$a_0+\frac{a_1}{2}+\frac{a_2}{3}+...+\frac{a_n}{n+1}=0$
then we can find at least one $x_0\in [0,1]$ that:
$a_0+a_1x_0+....+a_nx_0^n=0$

[hide]A simple application of Roll's Theorem[/hide]

By the way, it's not original.

that's correct. :D
the actual problem was in a really old book from IRMO.!!

Yes, it's very old - it's in Rudin's Principles of Mathematical Analysis, which has first printing in 1953 (although I don't know if the problem appeared in that edition of the text)


function
integration
calculus
limit
inequalities
real analysis
real analysis unsolved

Characterize the class of functions $f:\mathbb{R}\to\mathbb{R}$ for which there exists a function $g:\mathbb{R}\to\mathbb{R}^+$ (yes, I intentially wrote $\mathbb{R}^+$) such that

$f\left(x\right)=\int_{-\infty}^{\infty}\left|x-t\right|g\left(t\right)dt$.

Does this class contain all convex functions, or at least the most important ones?

  Darij

It's not going to work, in general, because the integral may fail to converge.

Assume for the moment that the target convex function $f$ is smooth. Differentiate twice. We get:

$f'(x)=\int_{-\infty}^xg(t)dt-\int_x^{\infty}g(t)dt$

$f''(x)=2g(x).$

So then, if $f(x)=x^2$, we would need $g(x)=1.$ But then the integral would diverge for all $x.$

To have any hope, we'd at least need the following:

1. This would only work for $f$ such that $\lim_{x\to\infty}f'(x)$ and $\lim_{x\to -\infty}f'(x)$ both exist and are finite. That's what went wrong with the example.

2. We'd have to be allowed to add an arbitrary linear function to the integral.

3. We'd have to let $g$ be more general than a function: a distribution (generalized function) or maybe more particularly a positive measure might be required. In particular, if you want to recover $f(x)=|x|$, you'll have to let $g$ be the Dirac delta.

Thanks, Kent, for your reply. Actually the problem is that I really don't have a clue on calculus (I don't even know a really strict definition of the integral). When I invented the question above, I was trying to get a freaky proof of the Karamata inequality and I wrote everything out of intuition.

Yes, both allowing to add an arbitrary linear function to the integral and letting g be a distribution would be okay for me; as for the existence of $\lim_{x\to\infty}f'(x)$ and $\lim_{x\to -\infty}f'(x)$, I am not really conscious of how strong these conditions are and how much is still left of the original question after this all. Maybe we should better restrict everything to an interval rather than the whole real axis, and then transform the question into:

[i]Let [a; b] be an interval on the real number axis. Characterize the class of functions $f:\left[a;\;b\right]\to\mathbb{R}$ for which there exists a function $g:\left[a;\;b\right]\to\mathbb{R}^+_0$[/i] (yes, this should be $\mathbb{R}^+_0$ - in my first post, I accidentally wrote $\mathbb{R}$, sorry) [i]such that

$f\left(x\right)=\int_{a}^{b}\left|x-t\right|g\left(t\right)dt$.[/i]

I hope that this version of the problem avoids, at least, the troubles with divergent integrals. Anyway, lots of thanks for making the things clear.

  Darij


algebra
function
domain
superior algebra
superior algebra solved

Let $A$ be a ring for which exists $k$ such that $x^{2k+1}=x,\forall x\in A$. Prove that :$(xy)^{2k}=(yx)^{2k}$

[quote="DiaDi"]I know it must be easy, but I just can't seem to figure it out right now. how did you prove step 3?
If the ring were a domain of integrity, or finite, the conclusion would be obvious, but as it is...
I would appreciate your help :)[/quote]

you only need to use (1) and (2) to prove (3).

I have a question, how did you find (2)? (I do not mean prove it, I mean how you discover it)

anyway, another way to proceed after step 3. Notice $xy = 0$ implies $yx = 0$, then by
$(xy)^{2k+1} = x^{2k+1}y$, $xyx((yx)^{2k-1}y - x^{2k-1}) = 0$, $x^2y((yx)^{2k-1}y-x^{2k-1}) = 0$. Multiply both sides by $x^{2k-1}$, $xy((yx)^{2k-1}y-x^{2k-1}) = 0$, $x((yx)^{2k-1}y-x^{2k-1})y = 0$ or $(xy)^{2k} = x^{2k}y$
By (3), $(xy)^{2k} = x^{2k}y = y^{2k}x = (yx)^{2k}$

:blush: I was just looking at the condition given by the problem, not at what was already proven. my mistake  :oops:


function
parameterization
limit
algebra proposed
algebra
functional equation

Find all continuous functions $f:\mathbb R\to \mathbb R$ such that $f(2x)=2f^{2}(x)-a$ for all $x \in \mathbb R$ and a fixed real $a$ (example: $a=1$ and $f(x)=\cos x$).

[quote="Rust"]Find continiosly function $R\to R$, satisfyed $f(2x)=2f^{2}(x)-a$, a fixed real parametr (for example a=1 f(x)=cosx).[/quote]

Hello Rust!
The answer to this problem is strongly linked to the value of $a$ (in fact to the existence of fixed points for $g(x)=2x^{2}-a$).

1) First it is obvious that we can choose $f(x)$ on $[1,2]$, for example, with the only constraints of continuity and $f(2)=2f^{2}(1)-a$ and then we can define $f(x)$ on $[1,+\infty)$ without difficulty. Same thing for an arbitrary definition on $[-2,-1]$, with continuity and $f(-2)=2f^{2}(-1)-a$, and then complete definition on $(-\infty,-1]$.

2) the problem is to be able to build $f(x)$ on $[0,1)$ and to have the continuity in $0$.

2.1) if $a<-\frac{1}{8}$, the function $g(x)=2x^{2}-a$ has no fixed point and it is rather easy to see that it is impossible to define $f(x)$ about $0$ $\Rightarrow $ no solution.

2.2) if $-\frac{1}{8}\leq a\leq 0$, it is generally possible to build $f(x)$ on $(0,1]$ by using the formula $f(x)=\sqrt{\frac{f(2x)+a}{2}}$.  We must take the positive value of the square root and not the negative one.
One condition for this to work is that the reference value ($f(x)$ in $[1,2]$ or $[-2,-1]$) must be either the lowest fixed point of $g(x)$ (and the function is constant), either $>$ to this lowest fixed point and then $f(0)=$ the highest fixed point of $g(x)$.

2.3) if $0<a<\frac{3}{8}$. it is again possible to build $f(x)$ on $(0,1]$ by using the formula $f(x)=\sqrt{\frac{f(2x)+a}{2}}$.  One condition for this to work is that the reference value ($f(x)$ in $[1,2]$ or $[-2,-1]$) must be $>-a$ and then $f(0)=$ the highest fixed point of $g(x)$.
But it is then possible to use the negative value of the square ($f(x)=-\sqrt{\frac{f(2x)+a}{2}}$ in some conditions). Due to the fact that $g'(x)>-1$ on the lowest fixed point, it is possible to have this lowest fixed point as a value for $f(0)$.

2.4) if $\frac{3}{8}\leq a$, it is impossible (except $f(x)$ is constant) to have $f(0)=$ lowest fixed point. So $f(0)=$ highest fixed point.
The reference value ($f(x)$ in $[1,2]$ or $[-2,-1]$) must be $>-a$.
It is possible during some steps to use $f(x)=-\sqrt{\frac{f(2x)+a}{2}}$ but, at the end, we must use $f(x)=\sqrt{\frac{f(2x)+a}{2}}$

As a conclusion :
if $a<-\frac{1}{8}$ : no solution
if $-\frac{1}{8}\leq a\leq 0$ : infinite many solutions (starting wih values $>$ lowest fixed point), all non constant solutions having $f(0)=$ highest fixed point.
if $0<a<\frac{3}{8}$ : infinite many solutions and two possible values for $f(0)$.
if $\frac{3}{8}\leq a$ : infinite many solutions and all non constant solutions having $f(0)=$ highest fixed point.

--
Patrick

[quote="Rust"]Ok!
Had these equation nonconstant continiosly periodic solution for $a\ge \frac{3}{8}$?
Is these periodic solution with period $2\pi$ unique (a=1, f(x)=cos x)?
Obviosly, if f(x) is solution, then f(bx) is solution. Therefore I interested about minimal period $2\pi$.[/quote]

For the moment, I can show that if $f(x)$ is continuous, non constant and periodic, then $a=1$ :

We have $f(2x)=g(f(x))$ with $g(x)=2x^{2}-a$ and $f(x)$ periodic (period $T$).

Let $p=\frac{1-\sqrt{1+8a}}{4}$ and $q=\frac{1+\sqrt{1+8a}}{4}$ the two fixed points of $g(x)$ (which exist since $a\geq\frac{3}{8}>-\frac{1}{8}$).

Since $f(x)$ is continuous and periodic, it exists lower and upper bounds for $f(x)$ : $A\leq f(x)\leq B$ $\forall x\in \mathbb{R}$. But, if $|f(x_{0})|>q$ for some $x_{0}$ it is easy to see that $\lim_{n\rightarrow+\infty}f(2^{n}x_{0})=+\infty$ and so we can conclude $|f(x)|\leq q$ $\forall x\in \mathbb{R}$

I have shown in the previous post that $f(0)=q$.

If $f(\frac{kT}{2^{n}})=q$ $\forall k,n\in\mathbb{N}$, it is immediate (since f is continuous), that $f(x)=q$ $\forall x$). So, since $f(x)$ is supposed periodic non constant, it exist $k$ and $n$ such that $f(\frac{kT}{2^{n}})\neq q$. But $f(kT)=f(0)=q$. So it exist in $[1,n]$ an integer $i$ such that $f(\frac{kT}{2^{i}})= q$. and $f(\frac{kT}{2^{i+1}})\neq q$. But, if $f(x)=q$, and since $q=2q^{2}-a$, $f(\frac{x}{2})=q$ or $f(\frac{x}{2})=-q$. So $f(\frac{kT}{2^{i+1}})=-q$.

So (remember that $|f(x)|\leq q$ $\forall x\in \mathbb{R}$) lower and upper bounds of $f(x)$ are $-q$ and $+q$ and these values may be reached.
Since $f(x)$ is continuous with values in $[-q,+q]$, it exist $x_{1}$ such that $f(x_{1})=0$. Then $f(2x_{1})=-a$ and so $-a\in[-q,+q]$, so $a\leq q$
But, since $f(x)=2f^{2}(\frac{x}{2})-a$, we have $f(x)\geq-a$ $\forall x$, and so $-q\geq-a$ and $q\leq a$

So $q=a$, and, since $q=\frac{1+\sqrt{1+8a}}{4}$, we can conclude $a=1$

So $a=1$

--
Patrick


1)Εστω ακολουθια $x_{n}$με $x_{1}=1$ και $x_{n+1}=x^{n}+nx_{n}$ με $x\in \mathbb R$ και $n\geq 1$ ν.δ.ο $\prod_{n=1}^\infty (1-\frac{x^{n}}{x_{n+1}})=\frac{1}{e^{x}}$
2)Εστω ακεραιος $n>1$.ν.δ.ο δεν υπαρχει αρρητος αριθμος $a$ ωστε $\sqrt[n]{a+\sqrt{a^{2}-1}}+\sqrt[n]{a-\sqrt{a^{2}-1}}$να ειναι ρητος!

Eιναι λιγο καλυτερες απο τις προηγουμενες.....

ΡΑΛΛΗΣ :D

γεια σε ολους....ραλλη σιγουρα γραφεις x στην πρωτη ασκηση? η δευτερη ειναι ευκολη απλα θετεις α τη μια και β την αλλη ριζα?

Ναι καλύτερα θα ήταν αν η σταθερά στο πρώτο είχε άλλο γράμμα, πχ $a$.

Ελπίζω να μην έχω κάποιο λάθος:

Στο ζητούμενο γινόμενο παρατηρούμε ότι "φεύγουνε" όλοι οι όροι εκτός απο τα $n$ και τα $x_{1}$ και $x_{n+1}$ οπότε αρκεί να δείξουμε ότι:

$\lim \frac{x_{n+1}}{n!}= e^{a}$

Έστω $g_{n}(a) = a^{n}$.

Παρατηρούμε ότι $x_{n+1}= a^{n}+nx_{n}= a^{n}+n(a^{n-1}+(n-1)x_{n_{1}}) = a^{n}+na^{n-1}+n(n-1)x_{n-1}+\ldots$

Επαγωγικά βγάζουμε: $x_{n+1}= a^{n}+na^{n-1}+\ldots+n(n-1) \cdots 3 \cdot 2 \cdot a+n(n-1) \cdots 3 \cdot 2 \cdot 1$, δηλαδή:

$x_{n+1}= g_{n}(a)+g_{n}^{'}+\ldots+g_{n}^{(n-1)}(a)+g_{n}^{(n)}(a) = \sum_{k=0}^{n}g_{n}^{(k)}(a)$

Άρα το ζητούμενo κλάσμα γίνεται (όπου $\lim$ εννοείται $\lim$ του $n$ στο $+\infty$):

$\lim \frac{\sum_{k=0}^{n}g_{n}^{(k)}(a)}{n!}= \lim \sum_{k=0}^{n}\frac{1}{n!}\cdot g_{n}^{(k)}(a) = \lim \left( \frac{1}{n!}\cdot a^{n}+\frac{1}{n!}\cdot na^{n-1}+\frac{1}{n!}\cdot n(n-1)a^{n-2}+\ldots \right) = \lim \left( \frac{1}{n!}\cdot a^{n}+\frac{1}{(n-1)!}\cdot a^{n-1}+\frac{1}{(n-2)!}\cdot a^{n-2}+\ldots \right) = \lim \sum_{k=0}^{n}\frac{1}{k!}\cdot a^{k}= e^{a}$

H τελευταία ισότητα ισχύει λόγω της σειράς Maclaurin του $e^{a}$.

Μπραβο cmad!Αυτη ακριβως ειναι η λυση...(η καλυτερα ετσι την ελυσα εγω)+μου φαινεται οτι ειναι και η μοναδικη....μακαρι να με διαψευσει καποιος...
Οντως θα ηταν καλυτερα αν η σταθερα ηταν καποιο αλλο γραμμα αλλα
ετσι την βρηκα και για λογους πνευματικων δικαιωματων και ιδιωτικων απορητων :gathering:  αναγκαστηκα να την  βαλω ετσι +εγω....

ΡΑΛΛΗΣ :D


geometry
incenter
circumcircle
trigonometry
ratio
angle bisector
trig identities

Very cool problem. I don't have a solution yet, so let me spam a bit:

- The point N lies on the angle bisector of the angle CAB, i. e. on the line AI, where I is the incenter of triangle ABC. [Trivial.]
- The line KL coincides with the perpendicular to the line AI at the point I. [Not trivial, au contraire: Long ago a rather tricky proof was posted on MathLinks.]
- The line MN is nothing directly known to me, but the line AM passes through the external center of similitude of the incircle and the circumcircle of triangle ABC. [This follows from Monge's theorem.]
- The line BC is a sideline of a triangle ABC. [ :D ]

  Darij

Now I have a solution. It is elementary and it won't use my remarks above.

I will rewrite the problem:

Let ABC be a triangle, and let v be a circle touching the segments AB and AC and the circumcircle of a triangle ABC. The circle v touches the segments AB and AC at the points K and L and the circumcircle of triangle ABC at the point M. Further, let N be the midpoint of the arc BC not containing A on the circumcircle of triangle ABC.

Then, show that the lines KL, BC and MN concur.

[i]Solution.[/i]

We will use [i]non-directed distances[/i].

If the lines BM and CM meet the circle v at the points P and Q (apart from M), then after the theorem about the angle between a tangent and a chord, both of the angles < MPQ and < MBC are equal to the angle between the line MQ = MC and the common tangent to the circle v and the circumcircle of triangle ABC at the point M. Therefore, < MPQ = < MBC, and PQ || BC. It follows that $\displaystyle \frac{BM}{CM}=\frac{BP}{CQ}$. Hence,

  $\displaystyle \frac{BM\cdot BP}{CM\cdot CQ}=\frac{BM}{CM}\cdot \frac{BP}{CQ}=\frac{BM}{CM}\cdot \frac{BM}{CM}=\frac{BM^{2}}{CM^{2}}$.

But $BM\cdot BP=BK^{2}$ and $CM\cdot CQ=CL^{2}$ (intersecting secant and tangent theorem), so we have $\displaystyle \frac{BK^{2}}{CL^{2}}=\frac{BM^{2}}{CM^{2}}$. This entails

  $\displaystyle \frac{BK}{CL}=\frac{BM}{CM}$.

Now, if the line MN meets the line BC at X, then we have to prove that the point X lies on the line KL.

By the Sine Law in triangles BMX and CMX, we have

  $\displaystyle \frac{BX}{CX}=\frac{\sin \measuredangle BMX\cdot XM:\sin \measuredangle XBM}{\sin \measuredangle CMX\cdot XM:\sin \measuredangle XCM}=\frac{\sin \measuredangle BMX:\sin \measuredangle XBM}{\sin \measuredangle CMX:\sin \measuredangle XCM}$
  $\displaystyle =\frac{\sin \measuredangle BMN:\sin \measuredangle CBM}{\sin \measuredangle CMN:\sin \measuredangle BCM}$

(since either < BMX = < BMN or < BMX = 180 - < BMN, so in any case sin < BMX = sin < BMN, and similarly sin < XBM = sin < CBM, sin < CMX = sin < CMN and sin < XCM = sin < BCM). Now, if R is the circumradius of triangle ABC, then we can apply the extended law of sines, and get

  $\displaystyle \frac{BX}{CX}=\frac{\sin \measuredangle BMN:\sin \measuredangle CBM}{\sin \measuredangle CMN:\sin \measuredangle BCM}=\frac{\frac{BN}{2R}:\frac{CM}{2R}}{\frac{CN}{2R}:\frac{BM}{2R}}=\frac{BN:CM}{CN:BM}=\frac{BN}{CN}\cdot \frac{BM}{CM}$
  $\displaystyle =\frac{BM}{CM}\text{\ \ \ \ \ \ \ \ \ \ (because }BN=CN,\text{ since }N\text{ is the midpoint of the arc }BC$
  $\displaystyle =\frac{BK}{CL}=\frac{BK}{AK}\cdot \frac{AL}{CL}\text{\ \ \ \ \ \ \ \ \ \ (since }AK=AL\text{)}$,

and thus

  $\displaystyle \frac{BX}{CX}\cdot \frac{CL}{AL}\cdot \frac{AK}{BK}=1$.

By Menelaos, it follows that the points X, L and K are collinear, i. e. the point X lies on the line KL. Problem solved.

The circle v is one of the three [i]mixtilinear incircles[/i] of triangle ABC. Each of these three mixtilinear incircles touches two sides of the triangle and the circumcircle. We can also define the three [i]mixtilinear excircles[/i] of the triangle; each of them touches the extensions of two sides and the circumcircle. These circles have many properties and it seems that they are often used in olympiads. Especially the Iranian TSTs have many problems on them.

  Darij

I think I have another solution without any calculations.
First, [tex]MK[/tex] and [tex]ML[/tex] meet [tex]w[/tex] at the midpoints [tex]C_1[/tex] and [tex]B_1[/tex] of arcs [tex]AB[/tex] and [tex]AC[/tex] respectively.
Let [tex]I[/tex] be the incenter of [tex]ABC[/tex], then [tex]CC_1[/tex] meet [tex]BB_1[/tex] at [tex]I.[/tex] So by Pascal thm for [tex]ACC_1MB_1B[/tex]   I lies on [tex]KL.[/tex] Since [tex]K[/tex] and [tex]L[/tex] are symmetric across the bisector of [tex]\angle BAC,[/tex] [tex]I[/tex] is the midpoint of [tex]KL[/tex]. Let [tex]P[/tex] be the midpoint of arc [tex]CAB[/tex] (i.e. [tex]NP[/tex] is a diameter of [tex]w[/tex]). We have [tex]\angle LIC = \angle AIC - \pi/2 = \angle ABC/2 = \angle LMC.[/tex] This means, [tex]LIMC[/tex] is cyclic and [tex]\angle CMI = \angle ILA = (\pi -\angle BAC)/2 = \angle CMP[/tex] so [tex]M,I,P[/tex] are collinear. Let [tex]Q[/tex] be the midpoint of [tex]BC.[/tex] Then triangles [tex]NQB[/tex] and [tex]NBP[/tex] are similar which implies [tex]NB^2 = NQ\cdot NP.[/tex] Since [tex]NB=NC=NI,[/tex] [tex]NI^2 = NQ\cdot NP[/tex] and triangles [tex]NIQ[/tex] and [tex]NPI[/tex] are similar. Then [tex]\angle NIQ = \angle NPI[/tex]
Let [tex]MN[/tex] intersect [tex]BC[/tex] at [tex]R.[/tex] Then [tex]\angle NRQ = \widehat{MN}/2 = \angle MPN = \angle NPI = \angle NIQ.[/tex] Hence, [tex]NQIR[/tex] is cyclic and [tex]\angle NIR = \angle NQR = \pi/2.[/tex] But since [tex]NI \bot KL,[/tex] [tex]KL[/tex] passes through [tex]R.[/tex]

[quote="grobber"]I used the easy-to-show fact that $MK,ML$ are the angle bisectors of $\angle AMB,\angle AMC$ respectively, which means that if $T=KL\cap BC$ then , by Menelaus, $\frac {TB}{TC}=\frac {BK}{KA}\cdot \frac {AL}{LC}=\frac {BM}{MC}$, which means that $MT$ is the external angle bisector of $\angle BMC$, which is also $MN$, and we're done.[/quote]

Grobber, your proof is really marvellous, so let me write it up in a more complete form in order to make it understandable not for insiders only ;-)

[b]Lemma 1.[/b] Let a circle k touch the circumcircle of a triangle ABC at its vertex B and touch the side CA at a point T. Then the point T lies on the angle bisector of the angle ABC.

[i]Proof.[/i] If M is the center of k, then MB = MT, so that the triangle BMT is isosceles. Thus, < MBT = < MTB. Now let H be the foot of the B-altitude of triangle ABC. Then, since the lines MT and BH are both perpendicular to CA, they are mutually parallel; thus we have < MTB = < HBT. Consequently, < MBT = < HBT. If O is the circumcenter of triangle ABC, then the point B lies on the line MO (since the circle k touches the circumcircle of triangle ABC at the point B); therefore, < MBT = < OBT, and we get < OBT = < HBT.

Now, on the one hand, in the right-angled triangle AHB we have < HBA = 90 - < BAH = 90 - < BAC; on the other hand, since O is the circumcenter of triangle ABC, we have < BOC = 2 < BAC, and the triangle BOC is isosceles, so that

  < OBC = (180 - < BOC) / 2 = (180 - 2 < BAC) / 2 = 90 - < BAC,

and finally < OBC = < HBA. Therefore,

  < CBT = < OBC + < OBT = < HBA + < HBT = < ABT.

Therefore, the point T lies on the angle bisector of the angle ABC, what proves Lemma 1.

Now to the problem: The circle v touches the circumcircle of the triangle AMB at the vertex M, and it touches the side AB at the point K. Hence, after Lemma 1, the point K lies on the angle bisector of the angle AMB. Since an angle bisector in a triangle divides the opposite side in the ratio of the adjacent sides, we have $\displaystyle \frac{AK}{KB} = \frac{AM}{BM}$. Similarly, the point L lies on the angle bisector of the angle CMA, and $\displaystyle \frac{CL}{LA} = \frac{CM}{AM}$.

Finally, we have < CAN = < BAN, since the point N is the midpoint of the arc BC on the circumcircle of triangle ABC. Therefore,

  < CMN = < CAN       (cyclic)
        = < BAN = 180 - < BMN        (cyclic again),

from what it follows that the line MN is the exterior angle bisector of the angle BMC. Now, if X is the point of intersection of this line MN with the line BC, then, by the fact that an exterior angle bisector in a triangle divides the opposite side externally in the ratio of the adjacent sides, we get $\displaystyle \frac{BX}{XC} = - \frac{BM}{CM}$. Therefore,

  $\displaystyle \frac{BX}{XC} \cdot \frac{CL}{LA} \cdot \frac{AK}{KB} = - \frac{BM}{CM} \cdot \frac{CM}{AM} \cdot \frac{AM}{BM} = -1$,

and by Menelaos, it follows that the points X, L, K lie on one line. This means that the line KL passes through X, and hence the lines KL, BC and MN concur at the point X.

The problem is solved, but we have also learnt a very nice application of the above reasoning:

[b]Proposition 2.[/b] If D is an arbitrary point in the plane of triangle ABC, then the point of intersection of the external angle bisector of the angle BDC with the line BC, and the points of intersection of the internal angle bisectors of the angles CDA and ADB with the lines CA and AB, respectively, are collinear.

  Darij

Dear Mathlinkers,
a synthetic proof can be obtain in this way:*
let I be the incenter of ABC, L' the second meet point of RL with the circumcircle of ABC and U the meet point of BC and MN.
According to Pascal's theorem applied to NACBL'MN,   I, L and U are collinear.
We know that K, I and L are collinear...
Sincerely
Jean-Louis

I've got a generation of this problem.

$ \odot \Omega$ passes through the vertex $ B,C$ of $ \triangle ABC$, $ \odot \omega$ inscribe $ \odot \Omega$ at $ T$, and touches $ AB,AC$ at $ P,Q$ respectively. $ M$ is the midpoint of arc $ BC$(contain $ T$)

Then we have

  $ PQ,BC,MT$ are concurrent.

suppose $ PQ\cap BC \equal{} K$,$ MT\cap BC \equal{} K'$

Apply the [b]Menelaus Theorem [/b]to $ \triangle ABC$ and $ PQK$, we have

$ \frac {AP}{PB}\cdot\frac {BK}{KC}\cdot\frac {CQ}{QA} \equal{} 1$

$ \Rightarrow$ $ \frac {BK}{KC} \equal{} \frac {BP}{CQ}$

Let $ R$ be the radii of $ \odot \Omega$ and $ r$ be the radii of $ \odot\omega$,

easyly, we can get

$ \frac {BP^2}{BT^2} \equal{} \frac {2R \minus{} 2r}{2r} \equal{} \frac {CQ^2}{CT^2}$

$ \Rightarrow$ $ \frac {BP}{CQ} \equal{} \frac {BT}{CT}$

Obviously, $ MT$ is the ex-angle bisector of $ \angle MTC$

we have

$ \frac {BT}{CT} \equal{} \frac {BK'}{K'C}$

$ \Rightarrow$ $ \frac {BK}{KC} \equal{}\frac{BK'}{K'C}$

so $ K \equal{} K'$

Hence, $ PQ,BC,MT$ are concurrent.

When $ A$ lies on $ \odot \Omega$, it becomes the problem which [b]andrew[/b] proposed.

Dear [b]tobeno_1[/b], your generalization is a problem from MOSP Tests 2007. I've presented my solution in: http://www.mathlinks.ro/viewtopic.php?t=154186  :P


trigonometry
inequalities
inequalities proposed

For $ 0\leq a,\ b,\ c,\ d\leq \pi$,
Prove that $ \sin a \plus{} \sin b \plus{} \sin c \plus{} \sin d \plus{} \left|a \minus{} \frac {\pi}{2}\right| \plus{} \left|b \minus{} \frac {\pi}{2}\right| \plus{} \left|c \minus{} \frac {\pi}{2}\right| \plus{} \left|d \minus{} \frac {\pi}{2}\right|\leq 2\pi$.

If we put here $ a\equal{}b\equal{}c\equal{}d\equal{}\frac{\pi}{2}$, we will get: $ 4\geq 2\pi$  :maybe:

Sorry for that I made a typo. I have just edited it.

kunny

Let`s prove that: $ \sin a \plus{} \left|\frac{\pi}{2}\minus{}a\right|\leq \frac{\pi}{2}$
This means: $ \sin a \minus{}\frac{\pi}{2}\leq \frac{\pi}{2}\minus{}a\leq \frac{\pi}{2}\minus{}\sin a$

Fist meas: $ a\plus{}\sin a \leq \pi$
The second: $ \sin a\leq a$
They have simple proof.

Now if we take four such inequalities and add them we will get the answer :)

That's right. I have no comment to add your solution more than this.


What is the measure of any exterior angle of a regular decagon?

[asy]for(int i = 0; i<10; ++i)
{
	draw(dir(18+36i)--dir(18+36(i+1)));
}

draw(interp(dir(90),dir(126),2)--dir(90));

pair A,B,C;

A = dir(126)+.3*dir(198);
B = dir(126)+.3*dir(216);
C = dir(126)+.3*dir(234);

draw(A..B..C);[/asy]

For a regular polygon with n number of sides, the measure of the exterior angle is 360/n, so the exterior angle is 360/10=36 degrees

Each angle of a decagon is 180*8/10=144.
Since an exterior+interior angle=180, 180-144=36.
Or,
if an angle of a decagon is 180*8/10, an exterior angle equals 180*(1-8/10)=180*2/10=180*1/5=36.


In triangle $ ABC$ with $ \left|BC\right|>\left|BA\right|$, $ D$ is a point inside the triangle such that $ \angle ABD\equal{}\angle DBC$, $ \angle BDC\equal{}150{}^\circ$ and $ \angle DAC\equal{}60{}^\circ$. What is the measure of $ \angle BAD$?

$\textbf{(A)}\ 45  \qquad\textbf{(B)}\ 50  \qquad\textbf{(C)}\ 60  \qquad\textbf{(D)}\ 75  \qquad\textbf{(E)}\ 80$

[hide="Solution"]If $ D$ is angle bisector of $ \triangle ABC$, $ \angle BDC \equal{} 90^\circ \plus{} \frac {\angle A} {2} \equal{} 90^\circ \plus{} \frac {120^\circ} {2} \equal{} 150^\circ$. Wow! So it is.

Anyway, Let $ E$ be a point on $ [BA$ such that $ BE \equal{} BC$. So $ \triangle EDC$ is an isosceles triangle. $ \angle BDC \equal{} \angle BDE \equal{} 150^\circ$.
$ \angle EDC \equal{} 360^\circ \minus{} 2\cdot150^\circ \equal{} 60^\circ$. So  $ \triangle EDC$ is equilateral. $ \angle DEC \equal{} \angle DAC \equal{} 60^\circ$. So $ A,D,C,E$ are concylic.
Because $ ADCE$ is cyclic, $ \angle DCE \equal{} \angle BAD \equal{} 60^\circ$.
$ \boxed C$.[/hide]


inequalities
real analysis
real analysis unsolved

Let a>o, and 0<b<2

Sum_k [min(1, |ka|^2 )/(k^(1+b))] =< C*(a^b)

where C is a constant that can depend on b.

Thank you,

gibraltor

That sum there says Sum_k, but you don't specify what $k$ starts and ends as.  Is it $\sum_{k=1}^{\infty}$?

I am sorry,

its over all integers k, except 0.

Is this what you mean?

[quote]Let $\displaystyle b \in \left( 0,2 \right)$.

Prove that there is a constant $\displaystyle C$ that depends on $\displaystyle b$ such that
\[ \displaystyle \sum_{k \in \mathbb Z \setminus \{ 0 \}} \frac{\min \left\{ 1, \left| ka \right|^2 \right\}}{k^{1+b}} \leq C a^b , \, \forall a > 0 . \][/quote]

If no, what's the correct form of the problem?


The product of the digits of a nine-digit palindrome is 144. How many such palindromes exist?

[hide="solution.."]I think its $62$. but I have to go to sleep..

Just consider $a^{2}b^{2}c^{2}d^{2}e=144$,

e must be a perfect square so there are only $3$ cases of $e=1,4,9$

then $abcd=12$-40 solutions
$abcd=6$-16 solutions
$abcd=4$-10 solutions

the first case has $4$ extra solutions since all must be single digits..

If you do the three cases-4 (66-4) it should be $62$..[/hide]


Please have your sound turned on.

http://www.birthdaycards.com/pickup?ID=A222-RHRA-7C76-SRS

[hide="Happy Thanksgiving!"]
Since there are at least two flamingos, two turkeys, and two pigs we have that $ 14$ legs are already accounted for, now we need to count how many turkeys can fit with $ 7$ legs left. The maximum amount of turkeys that can fit with $ 7$ legs is $ 3$ and we have one extra flamingo.

So there can be a maximum of $ 5$ turkeys inside the fence.
Edit: Whoops thanks!
[/hide]

[quote="tonypr"][hide="Happy Thanksgiving!"]
Since there are at least two flamingos, two turkeys, and two pigs we have that $ 14$ legs are already accounted for, now we need to count how many turkeys can fit with $ 7$ legs left. The maximum amount of turkeys that can fit with $ 7$ legs is $ 3$ and we have one extra flamingo.

So there can be a maximum of $ 5$ flamingos inside the fence.
[/hide][/quote]

tonypr,

you meant to type this final answer instead:
[hide]$ 5$ turkeys inside the fence[/hide]


[quote="solafidefarms"]In a certain group started in June, there were 9 messages posted in June, 54 messages posted in July, and 324 posted in August. Given that this geometry progression continues, how many messages will be posted in September?[/quote]

[hide]It is 1944, because each time, it gets multiplied by 6...
I have a more complicated, faster method, but its not MOEMS level[/hide]

[quote="bpms"][quote="anirudh"][quote="solafidefarms"]In a certain group started in June, there were 9 messages posted in June, 54 messages posted in July, and 324 posted in August. Given that this geometry progression continues, how many messages will be posted in September?[/quote]

[hide]It is 9244, because each time, it gets multiplied by 6...
I have a more complicated, faster method, but its not MOEMS level[/hide][/quote]
That is wrong ;)[hide]The answer is 324*6=1944, since september is one month after august, and after each month the value is multiplied by six. There isn't any way faster... Even if it were more months...[/hide][/quote]

I edited it. :blush: Silly computational error :blush:

[quote="anirudh"][/quote][quote="bpms"][quote="anirudh"][quote="solafidefarms"]In a certain group started in June, there were 9 messages posted in June, 54 messages posted in July, and 324 posted in August. Given that this geometry progression continues, how many messages will be posted in September?[/quote]

[hide]It is 9244, because each time, it gets multiplied by 6...
I have a more complicated, faster method, but its not MOEMS level[/hide][/quote]
That is wrong ;)[hide]The answer is 324*6=1944, since september is one month after august, and after each month the value is multiplied by six. There isn't any way faster... Even if it were more months...[/hide][/quote][quote="anirudh"]

I edited it. :blush: Silly computational error :blush:[/quote]
Two things. One, don't edit your posts when you make a mathematical mistake or say something that is later incorrect (typos or latex errors are ok for editing...). People who come to this website may think that the people here never make mistakes, a false notion. You consistently edit your posts when you shouldn't...
Two, I strongly doubt there is a faster  method for the problem. Can you please post your method...

If you are weird, like me, just take all the ratios like so: 9:54:324 and it ends up like this if you simplify:  1/36:1/6:1.  So the next month will be 1/216:1/36:1/6:1, and because 9 is 1/216, you multiply 216 and 9 and get 1944


let $x_{n}$ be sequence difined by:
$x_{0}=1$
$x_{1}=2$
$x_{n+1}=\sqrt{x_{n}+6\sqrt{x_{n-1}}}$
find relation between $x_{n}$ and $n$.

How did you find out this ?
[quote="Rust"] $x_{n}=4*2^{-2^{-(n-1)}}$.[/quote]

[quote="Nameless"][quote="boxedexe"]Rust used the principle of mathematical induction.[/quote]
I don't ask how to prove it. I ask why did Rust find out $x_{n}=4*2^{-2^{-(n-1)}}$ if he hasn't done this problem before?[/quote]

By trying out the first few values and finding a pattern? :wink:


The product $2^2\cdot 3^3 \cdot 5^7$ can be rewritten in the form $a^2\cdot b^3\cdot c$ with positive integers $a$, $b$ and $c$. What is the greatest possible value of $a\cdot b -c$?

[quote="Captain Sugar"][hide]-78119 because a^2 has to be 2^2, because the powers of 3^3 and 5^7 are not divisible by 2.  b^3 has to be 3^3 because the power of 5^7 isn't divisible by 3. so c must be 5^7.
6-5^7=-78119.[/hide][/quote]

You can take out powers...for example $3^3$ can be $3^2\cdot 3$...

[quote="Captain Sugar"][hide]-78119 because a^2 has to be 2^2, because the powers of 3^3 and 5^7 are not divisible by 2.  b^3 has to be 3^3 because the power of 5^7 isn't divisible by 3. so c must be 5^7.
6-5^7=-78119.[/hide][/quote] What is the GREATEST possible value. Oh, wait. Yeah, Captain Sugar, it doesn't have to be divisible. [hide] $5^7=25^2*5^3$ :D [/hide]

[quote="pgpatel"][quote="Captain Sugar"]don't blame me, i did the problem at 11 p.m.! :blush:
oops, i forgot... wow. now that i look at it, it's really simple. boo hoo. :([/quote]
I did too... But I was only 3 off :D.[/quote]

Actually you were 2 off, but if you insist whatever.

749-747=2

[hide]
pft this is wrong
6*25-15=135[/hide]


In the decomposition of $ KClO_3$ with $ MnO_2$ as catalyst, $ MnO_2$ increases the rate of chemical reaction.But, why do catalysts behave the way they do?What is the working principle?

Thanks.

A catalyst find an alternative pathway for the reaction, a pathway which requires a lower activation energy.

Thank you.
It seems to understand this, I need to have a greater knowledge of chemistry.


geometry
integration
perimeter
quadratics
calculus
calculus computations

Find the area of this surface when the cylinder Y^2+Z^2=1 intersects the cylinder X^2+Z^2=1.

I was stuck on this question for a while. Any help is highly appreciated.

Surface area = $2\pi \int^{b}_{a} f(x)\sqrt{1+f'(x)^2} dx$.
From the first equation, we have $z^2 = 1-y^2$.
From the second equation, we have $z^2=1-x^2$.
So if $1-y^2=1-x^2$ then $y^2=x^2$.
I'm not too sure if this is right, but I think $f(x) =\sqrt{x}$, and $a=0$ and $b=1$.
If this is right, then the surface area is about $5.5156$.

If you inscribe a sfere in the solid made from the intersection of two cylinder, we can write the proportion:
 Surface area: Area sfere = Perimiter quadratic section: Perimiter circular section.
Then the surface area A = $ \frac {4 \pi r^2\cdot 8r} {2 \pi r}=16r^2$. But r=1 and A = 16 qu.
If you want to use the integral calculus, you must calculate the double integral:
\[
A = 4 \int \int \sqrt {1+ \frac {z^2} {1-z^2}} \ dydz
\]
in the set $ 0 \leq y \leq 1 ; 0 \leq z \leq \sqrt {1-y^2} $.


:D I almost missed the sign up  :oops:

Oh, and I'm loving the difficulty of the math section of the SAT  :rotfl: I did about 100 questions the other night and didn't get any wrong... but sadly, I get slaughtered on the writing and reading sections

Yeah.

I'm not sure if I'll have time to "study". Which, theoretically, could cost me thousands of dollars in scholarship money. Meh.


superior algebra
superior algebra solved

If $S=\{2,4\}$ and $R=\mathbb{Z}_{6}$ then show that $S^{-1}R$ is isomorphic to the field $\mathbb{Z}_{3}$.
PS Solving this problem is as easy as ABC; (one can even consider all cases! Does any one know a better way, an extension or sth with proof? :D) [hide="This special case"]$0\sim\frac32\sim\frac34,\;1\sim\frac14\sim\frac52,\; 2\sim\frac12\sim\frac54$ is easily seen to be isomorphic with $\mathbb{Z}_{3}.$[/hide]

Let $K,L$ be fields. Consider $R=K\times L$ and consider the multiplicative system $S=K^{*}\times\{0\}$.
Then $S^{-1}R\simeq K$.

I leave the proof to you. I am not entirely sure this is true, because I just guessed it :)
The connection to the previous problem is $\mathbb Z_{6}\simeq \mathbb Z_{2}\times\mathbb Z_{3}$.


Prove that
\[ a \plus{} n \plus{} x \equal{} \sqrt {{{(a \plus{} n)}^2} \plus{} ax \plus{} x\sqrt {{{(a \plus{} n)}^2} \plus{} a(n \plus{} x) \plus{} (n \plus{} x)\sqrt {{{(a \plus{} n)}^2} \plus{} a(2n \plus{} x) \plus{} (2n \plus{} x)\sqrt {...} } } }\]

[quote="Rose_joker"]just only an out line for the solution...we see that
$ (a \plus{} n \plus{} x)^2 \equal{} (a \plus{} n)^2 \plus{} 2x(a \plus{} n) \plus{} x^2 \equal{} (a \plus{} n)^2 \plus{} ax \plus{} x(a \plus{} 2n \plus{} x) \equal{} (a \plus{} n)^2 \plus{} ax \plus{} x\sqrt {(a \plus{} 2n \plus{} x)^2}$
and in the same way we have
$ (a \plus{} 2n \plus{} x)^2 \equal{} (a \plus{} n)^2 \plus{} 2(a \plus{} n)(x \plus{} n) \plus{} (x \plus{} n)^2 \equal{} (a \plus{} n)^2 \plus{} a(x \plus{} n) \plus{} (x \plus{} n)\sqrt {(a \plus{} 3n \plus{} x)^2}$
:)[/quote]

But how do we know it's not tend to infinity ??? Because the last term is larger .


logarithms
function
limit
algebra unsolved
algebra

Find $ x \in R$ of this equation: $ \log_3{(x\plus{}1)}\plus{}\log_5{(2x\plus{}1)}\equal{}2$

[b]Solution:[/b]
If a>1, then $ (x_1 > x_2 > 0 \Rightarrow log_a{x_1} > log_a{x_2}).$
[b]1).[/b] x=2 solution!
[b]2).[/b] if $ \minus{} \frac {1}{2} < x < 2 \Rightarrow log_3(x \plus{} 1) < log_3{3} \equal{} 1$ and $ log_5(2x \plus{} 1) < log_5{5} \equal{} 1 \Rightarrow log_3(x \plus{} 1) \plus{} log_5(2x \plus{} 1) < 2$;
[b]3).[/b] if $ x > 2 \Rightarrow log_3(x \plus{} 1) > log_3{3} \equal{} 1$ and $ log_5(2x \plus{} 1) > log_5{5} \equal{} 1 \Rightarrow log_3(x \plus{} 1) \plus{} log_5(2x \plus{} 1) > 2$.
[b]2).[/b] and [b]3).[/b] $ \Rightarrow x\ne 2$ non solution!

hello, by defining the function $ f(x)=\frac{\ln(x+1)}{\ln(3)}+\frac{\ln(2x+1)}{\ln(5)}-2$ we get $ f^'(x)=\frac{1}{(x+1)\ln(3)}+\frac{2}{(2x+1)\ln(5)}>0$ and from here follows, that $ f(x)$ is strictly monotonously increasing. Further we have $ \lim_{x\to-1/2+0}f(x)=-\infty,\lim_{x\to \infty}f(x)=+\infty$. Hence we get the only solution $ x=2$.
Sonnhard.


$p$ prime number,
$A,B \in M_{n}(Z)$ such that $AB=p(A+B)$

Prove that $det(A-pI_n)-det(B-pI_n)$ is divisible by $p$ or $p^2-1$

We have $(A-p)(B-p)=p^2$, so $\det(A-p)\det(B-p)=p^{2n}$ and $\det(A-p)-\det(B-p)=\pm p^a(p^{2n-2a}-1)$ for some $a$ with $0\le a\le n.$ Actually, the second factor is always divisible by $p^2-1$.


I don't have a specific example, but many counting problems ask you to find the distinct possible orders of a number (possibly a four digit number), where two of the digits are the same. I know the rule, take the factorial of the number of digits, then divide by the number of same digits (in this case, 2). But why?

Why is the following logic wrong?

[hide]
For a four digit number with two digits indistinct,
The first digit has 3 distinct choices, the second also has 3, the third has 2, and the last has one. 3 x 3 x 2 x 1= 18.
The actual answer is 4!/2 = 24/2 =12.
[/hide]

[quote="darkmatter47"]The first digit has 3 distinct choices, the second also has 3[/quote]

Not if you picked one of the digits that isn't doubled!  Then there are only two.  You have to split into two cases depending on whether you picked the doubled digit or not, and then you'll get the right answer.

Counting is subtle like that.  :)

[quote="darkmatter47"]I know the rule, take the factorial of the number of digits, then divide by the number of same digits (in this case, 2). But why? [/quote]

Ah, I hadn't noticed you'd asked this question.  There's a very good tutorial about counting written by a friend of mine at http://brainjammer.com/math/combinatorics/ .


[quote="i_like_pie"][quote="Ignite168"][quote="i_like_pie"]$17^{3}>2^{11},$[/quote]

How did you calculate this?[/quote]
$17^{3}>16^{3}=2^{12}>2^{11}$

Treething beat me.[/quote]

What was the point of you posting then if only to offer the exact same solution?

[quote="Ignite168"][/quote][quote="i_like_pie"][quote="Ignite168"][quote="i_like_pie"]$17^{3}>2^{11},$[/quote]

How did you calculate this?[/quote]
$17^{3}>16^{3}=2^{12}>2^{11}$

Treething beat me.[/quote][quote="Ignite168"]

What was the point of you posting then if only to offer the exact same solution?[/quote]
Your question was directed at me, so I answered it.

[quote="i_like_pie"][/quote][quote="Ignite168"][/quote][quote="i_like_pie"][quote="Ignite168"][quote="i_like_pie"]$17^{3}>2^{11},$[/quote]

How did you calculate this?[/quote]
$17^{3}>16^{3}=2^{12}>2^{11}$

Treething beat me.[/quote][quote="Ignite168"]

What was the point of you posting then if only to offer the exact same solution?[/quote][quote="i_like_pie"]
Your question was directed at me, so I answered it.[/quote]

I'm talking about your solution, not that question.

[quote="Ignite168"][quote="i_like_pie"]Your question was directed at me, so I answered it.[/quote]

I'm talking about your solution, not that question.[/quote]
I didn't read your solution until now.

[quote="mathnerd314"]Of course, if you don't have a calculator...[/quote]

Yes, what then?

[quote="mathnerd314"][hide]$11 \log 31$ or $14 \log 17$? 16.4 or 17.2? I'd definitely say $17^{14}$ is larger.

Of course, if you don't have a calculator...[/hide][/quote]

My 1st solution is easier with a calculator. Does not require knowledge of logs.