limit Euler integration calculus quadratics logarithms geometry

Let $a_0=1$, $a_{k+1}=a_k+\frac{1}{a_k}$, $\forall k \in \mathbb{N}_0$.

[b]a)[/b] Prove that $\lim_{k \to \infty} a_k=\infty$. (there are at least three methods to prove this)

(the part in which i'm more interested) [b]b)[/b] Find some good bounds for $a_k$ (actually two of the solutions i know use some weak lower / upper bound for $a_k$ - let's see if you can do better)

Here's the cheapest answer to part a): This sequence is clearly increasing, hence it either converges or tends to infinity. If it converges to a limit $x,$ then $x$ satisfies the equation $x+\frac1x=x.$ Since that has no real solution, convergence is not possible; thus $\lim_{k\to\infty}a_k=\infty.$

Not very satisfying, is it? In particular, this proof leaves us absolutely no insight for part b) - yes, the sequence goes to infinity, but we don't know how fast.

For a different intuitive approach, consider this:
\[a_{k+1}-a_k=\frac1{a_k}\]
is a difference equation that approximates (by the Euler method of approximation) the differential equation
\[y'=\frac1y.\]
This differential equation is readily solved by separation of variables; the solutions are $y=\sqrt{2x+C}.$ Hence we should look for estimates that say that for large $k,\,a_k$ behaves asymptotically like $\sqrt{2k}.$

That's not very rigorous, so far. I invite others to sharpen it.

Here's another step in the process, albeit still a long ways from a complete answer.

Suppose $a_k=\sqrt{2k+b_k}.$

In other words, define $b_k=a_k^2-2k.$

Algebraic manipulation of the recursive formula gives us

$b_{k+1}=b_k+\frac1{a_k^2},$ or $b_{k+1}=b_k+\frac1{b_k+2k}.$

This is another strictly increasing sequence. Hence we have the following lower bound:

For $k>2,\,a_k>\sqrt{2k}.$

I believe that $b_k\to\infty,$ at a rate that is nearly logarithmic. Nailing that down would sharpen our estimates.

[quote="Kent Merryfield"]For a different intuitive approach, consider this:
\[a_{k+1}-a_k=\frac1{a_k}\]
is a difference equation that approximates (by the Euler method of approximation) the differential equation
\[y'=\frac1y.\][/quote]

i have a question about differential equations (separation method)

i searched on the net, and here's how i've understood it:

(1) $\frac{dy}{dx}=\frac{1}{y}$

(2) $y dy = 1 dx$

(3) $\int y dy = \int 1 dx$

(4) $\frac12 y^2 = x+c $

(5) $y = \sqrt{2x+2c}$

why are the steps $(2)$ and $(3)$ valid operations? i'm new to calculus, so forgive me if it's a dumb question

as for the original problem, your attempts are welcome

[quote="perfect_radio"]why are the steps (2) and (3) valid operations?[/quote]
You may not like it, but the attitude I usually take in teaching it is this: just play along - let the notation be what it wants to be. There is a very big theorem that we can call on in the end. (My running joke is that it's the only theorem in an entire semester of a differential equations course, and everything else is a corollary.) That theorem is the existence-uniqueness theorem for differential equations.

What does that - in particular, the uniqueness - get us? Suppose that by some method - some structured guess, some symbolic manipulation (even if the notation for that manipulation leaves you a little queasy) - we find a family of solutions. If these are indeed solutions (which can be checked easily enough) and if we can solve for all possible initial conditions, then we have found all solutions.

In your solution, there is one side note on the step from (4) to (5). The quadratic equation in (4) does have two different possible roots for $y.$ However, in our case, we have the additional information that the initial condition is going to have $y>0.$ Since the differential equation isn't valid or well-formed for $y=0$, the solutions are not going to cross the axis and will always have $y>0.$

(As for the fact that you wrote $y=\sqrt{2x+2c}$ and I wrote $y=\sqrt{2x+c}$ : twice an arbitrary constant is just some other arbitrary constant.)

--

A numerical fiddle related to my previous post: running it out past $k=1000,$ it looks like $b_k\approx \frac12\ln(ck),$ where $c$ is a constant a little bit bigger than $\frac12.$

Yes, by using your previous recurrence, we can show that $b_k - \frac{1}{2} \ln k$ approaches a constant.

By the way, at the very beginning, it helps to square the original recurrence:
\[
  a_{k+1}^2 = a_k^2 + \frac{1}{a_k^2} + 2 \, .
\]
If we let $s_k = a_k^2$, we get the nice recurrence
\[
  s_{k + 1} = s_k + 2 + \frac{1}{s_k} \, .
\]
That's a quick way to see that $s_k \sim 2k$.

[quote="Ravi B"]Yes, by using your previous recurrence, we can show that $b_k - \frac{1}{2} \ln k$ approaches a constant.[/quote]

how did you show that? first of all, why doesn't $b_k - \frac12 \ln k \to \infty$? (i get some very ugly relations  :blush:  )

Inspired by Kent's recurrence for $b_k$, let's define a simpler recurrence
\[
  c_{k + 1} = c_k + \frac{1}{2k+1} \, ,
\]
with $c_0 = 1$.  That recurrence has solution
\[
  c_k = 1 + \sum_{i = 0}^{k-1} \frac{1}{2i+1} \, .
\]
From results about the harmonic series, it is known that $c_k - \frac{1}{2} \ln k$ approaches a constant (involving Euler's constant).

By comparing the recurrences for $b_k$ and $c_k$, we see that $b_k \le c_k$.  Let's examine the difference $d_k = c_k - b_k$.  By subtracting the recurrences for $b_k$ and $c_k$, we get the recurrence
\begin{eqnarray*}
d_{k+1}
  & = & d_k + \frac{1}{2k+1} - \frac{1}{2k + b_k} \\
  & = & d_k + \frac{b_k - 1}{(2k+1)(2k+b_k)} \, .
\end{eqnarray*}
From that recurrence, we see that $d_k$ is an increasing sequence.  By summing up, we get the equation
\[
  d_k = \sum_{i = 0}^{k-1} \frac{b_i - 1}{(2i+1)(2i + b_i)} \, .
\]
Because $1 \le b_i \le c_i = O(\ln i)$, we get
\[
  d_k = \sum_{i = 0}^{k-1} \frac{O(\ln i)}{(2i+1)^2} = O(1) \, .
\]
In other words, $d_k$ is bounded.  A bounded, increasing sequence must approach a limiting constant.  That's what we wanted to show.

[quote="Ravi B"]\[
  c_k = 1 + \sum_{i = 0}^{k-1} \frac{1}{2i+1} \, .
\]
From results about the harmonic series, it is known that $c_k - \frac{1}{2} \ln k$ approaches a constant (involving Euler's constant).[/quote]

how do you prove it?

i have a found a result which was similar to yours (but not the same), and the constant was called Euler-Mascheroni. i don't how to prove that either

Define the usual harmonic series
\[
  H_n = \sum_{i = 1}^n \frac{1}{i} \, .
\]
It is known that $H_n - \ln n$ approaches a constant, the Euler (or Euler-Mascheroni) constant.  You can prove that result by comparing the harmonic series to the integral
\[
  \int_1^n \frac{1}{x} \, dx  = \ln n \, .
\]
I'm not going to give a proof here; it's a classical result.

Anyway, we can express the sum of odd reciprocals in terms of harmonic numbers:
\[
  \sum_{i=0}^{k - 1} \frac{1}{2i+1} = H_{2k} - \frac{1}{2} H_k \, .
\]
So from the estimate on harmonic numbers, we get the estimate for our sum.

I am really dumb... I knew this result, $\ln (n+1) < H_{n}< 1+\ln n$, and also how to prove it (considering some rectangles + integrals).

I have found that on the net that the existence of the constant can also be proven in the same way. (actually it gave the same hint as Ravi B + a nice drawing).

So, draw the graph of $x \mapsto \frac{1}{x}$. Also draw the "big" and "small" rectangles coresponding to every pair $\{n,n+1\}$. Let $D_{i}$ be the "difference" rectangles between the big and the small ones. Then we have $\sum_{i=1}^{n}\frac{1}{i(i+1)}= \sum_{i=1}^{n}\left( \frac{1}{i}-\frac{1}{i+1}\right) = 1-\frac{1}{n+1}$, so $\textrm{Area of all }D_{i}\textrm{ rectangles}= \sum_{i=1}^{\infty}\frac{1}{i(i+1)}= 1$.

Consider the sets $F_{i}$ formed by the $D_{i}$ rectangles and the graph of $x \mapsto \frac{1}{x}$ (the lower side). Thus, $\sum_{i=1}^{n}F_{i}= 1+\ln n-H_{n}$. This sum increases at each step, and is bounded by $1$, thus, it is convergent...

i know the presentation is not the most fortunate, but, did I got it right?

[quote="Kent Merryfield"]Here's the cheapest answer to part a): This sequence is clearly increasing, hence it either converges or tends to infinity. If it converges to a limit $x,$ then $x$ satisfies the equation $x+\frac1x=x.$ Since that has no real solution, convergence is not possible; thus $\lim_{k\to\infty}a_k=\infty.$

Not very satisfying, is it? In particular, this proof leaves us absolutely no insight for part b) - yes, the sequence goes to infinity, but we don't know how fast.
[/quote]
i believe the ratio test is much more convinient, not sure if it's cheaper though.
 :oops:
btw, im new here, seems like a good maths community.

[quote="perfect_radio"]i know the presentation is not the most fortunate, but, did I got it right?[/quote]Yes, your method is fine.


function
inequalities
algebra proposed
algebra

Let $ a,b,a',b'$ be real numbers such that $ a\neq a'$ and $ f: \mathbb R\to\mathbb R$ be a function satisfying
\[ f(x)\plus{}f(a\minus{}x)\equal{}b\qquad\text{and}\qquad f(x)\plus{}f(a'\minus{}x)\equal{}b'\]
for all $ x\in\mathbb R$. Show that there exist real numbers $ k,c,C$ such that
\[ |kx\plus{}c\minus{}f(x)|\le C\]
holds for all $ x\in\mathbb R$.

we suppose that $ h\equal{}a\minus{}a'>0,g\equal{}b\minus{}b'$; $ f(x\plus{}h)\minus{}f(x)\equal{}f(x\plus{}h)\plus{}f(a\minus{}x\minus{}h)\minus{}(f(x)\plus{}f(a'\minus{}x))\equal{}b\minus{}b'\equal{}g$
now for $ r\in[0,h[: \ \forall x\in S_r\equal{}\{nh\plus{}r: \ n\in \mathbb{N}\}: \ f(x)\equal{}kx\plus{}m(r)$ where $ k\equal{}g/h$ and $ m: [0,h[\to \mathbb{R}$

$ f(x)\plus{}f(a\minus{}x)\equal{}b\equal{}kx\plus{}m(r)\plus{}k(a\minus{}x)\plus{}m(r')$ where $ x\equal{}nh\plus{}r,a\minus{}x\equal{}n'h\plus{}r'$ and $ r,r'\in[0,h[$
NOW we must pove that $ |m|<C$

[quote="nayel"]I'm really really sorry I forgot to mention that $ f$ is continuous in $ \mathbb R$. :([/quote]
so $ m$ is continue in $ [0,h[$ so $ M$ existe $ m<M$


probability
AMC
AIME
MATHCOUNTS
number theory

hm that's a tough problem i wonder how to do it jkjk i'm too lazy

i just made that question up out of no where cuz im bored and just want to get my posts up

is there a reason i'm doing the same thing??? lol

[quote]it's not wise to post just to get your post count up, but at least it has some meaning

I believe this problem was posted already. I'll try to find the thread[/quote]

That's why he's the wiseidiot.

[quote="wiseidiot"][quote="Ihatepie"][quote="wiseidiot"]whats the probability that a randomly selected interger from 1 to 100 can be expressed as the sum of two squares?[/quote]

I'm pretty sure this was in AOPS Counting and Probability book. I'm too lazy to check  :lol: I also think that it was asking for positive perfect squares.

Well anyways, list all squares: 1,4,9,16,25,36,49,64,81,100

The sum of two squares cannot be the same as the sum of two others.

So total ways to be added are 10C2 or 45, BUT, some additions like 100+81 don't work since they add to over 100. For this, you can easily see that 100 won't work. so 9C2 or 36 so far. 81+64,81+49,81+36,81+25 don't work
64+49 doesn't work

so 36-5 or 31.[/quote]

You forgot 0 is a perfect square.......[/quote]

Err... the two numbers can be the same. Anyway, here's my solution.

$ 0 \qquad 1 \qquad 4 \qquad 9 \qquad 16 \qquad 25 \qquad 36 \qquad 49 \qquad 64 \qquad 81 \qquad 100$.

If the smaller number is $ 0$, we have 10 choices (since $ 0 \plus{} 0$ is not in the range specified by the problem).
If the smaller number is $ 1$, we have 9 choices ($ 1^2$ to $ 9^2$).
If the smaller number is $ 4$, we have 8 choices.
If the smaller number is $ 9$, we have 6 choices (since the sum $ 25$ has already been counted by $ 0^2 \plus{} 5^2$, we don't need to add it again).
If the smaller number is $ 16$, we have 5 choices (since the sum $ 65$ has already been counted by $ 1^2 \plus{} 8^2$, we don't need to add it again).
If the smaller number is $ 25$, we have 4 choices ($ 5^2$ to $ 8^2$).
If the smaller number is $ 36$, we have 1 choice (since the sums $ 85$ and $ 100$ have already been counted by $ 2^2 \plus{} 9^2$ and $ 0^2 \plus{} 10^2$, we don't need to add them again).
If the smaller number is $ 49$, we have 1 choice ($ 7^2 \plus{} 7^2$).
If the smaller number is $ 64$, we have 0 choices.
If the smaller number is $ 81$, we have 0 choices.
If the smaller number is $ 100$, we have 0 choices.

Adding them all up, we find that $ 44$ out of the $ 100$ numbers can be expressed as the sum of two squares. Therefore, the required probability is $ \boxed{\frac {11}{25}}$.

[b]Edit[/b]: Added a $ 1$ in my list of squares! Also removed remaining duplicates.

Yongyi, you forgot to put a 1 in your list of squares. Just saying.

[quote="alanchou"][quote="the future"][quote="wiseidiot"]whats the probability that a randomly selected interger from 1 to 100 can be expressed as the sum of two squares?[/quote]

change it to "whats the probability that a randomly selected interger from 1 to 100 can be expressed as the [i]difference[/i] of two squares?" and youll get a more number theory based question that doenst require brute force[/quote]

heh, the only way it can't be is if there's only 1 factor of 2 in the number

as in, 2 isn't the difference of 2 squares, 6 isn't, 10 isn't, 18 isn't, etc.[/quote]

yea lol it was an old AIME number 1, probalby mathcounts level

its pretty easy to see by a pattern but its provable with some easy NT.

[quote="wiseidiot"][quote="Ihatepie"][quote="wiseidiot"]whats the probability that a randomly selected interger from 1 to 100 can be expressed as the sum of two squares?[/quote]

I'm pretty sure this was in AOPS Counting and Probability book. I'm too lazy to check  :lol: I also think that it was asking for positive perfect squares.

Well anyways, list all squares: 1,4,9,16,25,36,49,64,81,100

The sum of two squares cannot be the same as the sum of two others.

So total ways to be added are 10C2 or 45, BUT, some additions like 100+81 don't work since they add to over 100. For this, you can easily see that 100 won't work. so 9C2 or 36 so far. 81+64,81+49,81+36,81+25 don't work
64+49 doesn't work

so 36-5 or 31.[/quote]

You forgot 0 is a perfect square.......[/quote]

[quote="Ihatepie"]I also think that it was asking for positive perfect squares.
[/quote]

well...


Paula can have pennies, nickels, dimes and quarters in her purse.  The average value of the coins in her purse is 20 cents.  If she adds one more quarter to her purse, the average value of all of the coins will be 21 cents. How many nickels does she have in her purse?

Assume that the value of every con is 20c. Now we add a quarter. The quarter is worth 25c, and "loses" 4c. Thus 4c is added to every other coin, and there are 4 coins. These 4 coins are worth 80c, so we have 3 quarters and $ \boxed{1}$ nickel.


hello, a day has $ 24 h\equal{}24\cdot60min\equal{}1440min$ so we get $ 3600: 1440\equal{}2.5$ the answer is $ 2.5$ days.
Sonnhard.

[quote="GameBot"]How many days are there in 3600 minutes? Express your answer as a decimal.[/quote]

Another way of looking at this is to note that $ 3600\equal{}60^2$, which is 60 hours. Since we have 24 hours in a day, 2.5 days.


Are there integers $ a,b,c,d,x,y,z,t$ such that each of the numbers:

$ |ay\minus{}bx|,|az\minus{}cx|,|at\minus{}dx|,|bz\minus{}cy|,|bt\minus{}dy|,|ct\minus{}dz|$

equals either $ 1$ or $ 2005$?

Suppose the opposite case. Take the sets $ A\equal{}(a,b,c,d)$ and $ B\equal{}(x,y,z,t)$. In each one of this ones, there is at most one even number, otherwise one of the modulus would be even. If all the elements in some set are odd, there would be a even modulus made by two odd elements from the other set. So each set has just one even element. But, in this case, is easy to see that some modulus will be even (you can suppose WLOG that some element of a set is even and divide in some cases), contradiction. It finishes the problem.


Let   $AM, BN , CP$   three concurrent cevians in the triangle  $ABC$. The parallel from $N$  to $AB$  meets $PM$at $E$ and the parallel from $M$  to $AB$ meets $PN$ at $F$ . Prove that the straight lines $MN , EF$ and $PC$  are concurrent.

[u] Babis[/u]

Let $X$ be the intersection of $EN$ with $PC$
$MN,EF,PC$ are concurrent iff
$\frac{EM}{MP}\frac{PF}{FN}\frac{NX}{XE}=1$
Since $MF \parallel EN \parallel AB$, then $\triangle PMF$ is similar to $\triangle PEN$ so $\frac{PM}{ME}=\frac{PF}{FN}$
and the problem is equivalent to proving
$\frac{NX}{XE}=1$
This equation doesn't involve the construction of $F$, and according to a few of my diagrams, it isn't true. Did I make a mistake? Am I misunderstanding the problem?

NO ! Yes , there is really a problem , probably some typo in the book. I have pleased Mr Enescu in Romany to help us.
 Really sorry!I thought that for junior level  it could not be difficult and I wrote it in it's initial form without to solve it. This was my panishment !!!
 I apologize.

 [u]babis[/u]

 [color=red]Virgil can help us , as well ![/color]
[i]( Dear schulmannerism, you must inverse the relation from the similarity-in fact this relation is thales and not from the similar triangles(so in my figure!). But the method is the best for these problems!)[/i]

Suppose the parallel to $AB$ through $C$ intersects $PN,PM$ at $U,V$, respectively. Then triangles $NCU,NAP$  are similar, hence $\frac{CU}{AP}=\frac{NC}{NA}$. Similarly, $\frac{CV}{PB}=\frac{MC}{MB}$, consequently,
\[ \frac{CU}{CV}=\frac{NC}{NA}\cdot \frac{MB}{MC}\cdot\frac{PA}{PB}=1. \]
Since $CU=CV$, the claim follows easily from the well known fact that in a trapezoid the midpoints of the basis, the point of intersection of the non-parallel sides and the point of intersection of the diagonals are concurrent.
I do not know at this moment the author's solution. Perhaps it does not use Menelaus theorem.


Find all functions : $ f:\mathbb{R}\to\mathbb{R}$ so that

$ (f(x\plus{}y))^{2}\equal{} f(x)f(x\plus{}2y)\plus{}yf(y)$ :)

[hide="Solution"]Replace $ x$ with $ x\plus{}2y$ and $ y$ with $ \minus{}y$.

$ f(x\plus{}y)^{2}\equal{} f(x\plus{}2y)f(x)\minus{}yf(\minus{}y)$
Subtract from the original equation to get
$ yf(y) \equal{}\minus{}yf(\minus{}y)$, so $ f(x)$ is odd for nonzero $ x$.

$ x \equal{}\minus{}y\neq 0$:
$ f(0)^{2}\equal{} f(\minus{}y)f(y)\plus{}yf(y)$
$ f(0)^{2}\plus{}f(y)^{2}\equal{} yf(y)$

$ x \equal{}\minus{}2y\neq 0$
$ f(\minus{}y)^{2}\equal{} f(\minus{}2y)f(0)\plus{}yf(y)$
$ f(y)^{2}\plus{}f(2y)f(0) \equal{} yf(y)$

Transitively, $ f(0)^{2}\plus{}f(y)^{2}\equal{} f(2y)f(0)\plus{}f(y)^{2}$ so
$ f(0)^{2}\equal{} f(2y)f(0)$
Then we either have $ f(2y) \equal{} f(0)$ or $ f(0) \equal{} 0$.

If $ f(2y) \equal{} f(0)$, then $ f(x) \equal{} c$. Plug that in to the original equation to get $ c^{2}\equal{} c^{2}\plus{}yc$, so $ c \equal{} 0$. $ f(x) \equal{} 0$ is valid.

If $ f(0) \equal{} 0$, then $ x \equal{} 0$ gives $ f(y)^{2}\equal{} yf(y)$, so either $ f(x) \equal{} 0$ (already got that) or $ f(x) \equal{} x$, also valid.

$ \boxed{f(x) \equal{} 0}$ or $ \boxed{f(x) \equal{} x}$[/hide]

I think your solution have a problem  :) You can read my solution, then , you will see this problem :wink:

Another solution .
[hide]Let $ f(0) \equal{} a$
$ x \equal{}\minus{}y$  we have : $ a^{2}\equal{} f(y).f(\minus{}y)\plus{}yf(y),\forall y\in\mathbb{R}$(*)
In (*) , replace $ y$ by $ \minus{}y$, we have $ yf(y) \equal{} yf(\minus{}y),\forall y\in\mathbb{R}$
Then , for all $ y\not \equal{} 0$, $ f(y) \equal{}\minus{}f(\minus{}y)$.
For all $ y\not \equal{} 0$, $ a^{2}\equal{}\minus{}f^{2}(y)\plus{}yf(y)\leftrightarrow [f(y)\minus{}\frac{y}{2}]^{2}\plus{}a^{2}\minus{}\frac{y^{2}}{4}\equal{}0,\forall y\not \equal{} 0$
So :
$ a^{2}\le\frac{y^{2}}{4},\forall y\not \equal{} 0\Rightarrow a \equal{} 0$.

Hence : $ f(x)[f(x)\minus{}x] \equal{} 0$

If there exists $ a,b\not \equal{} 0$ so that : $ f(a) \equal{} 0,f(b) \equal{} b$
-$ x \equal{} a,y \equal{} b$, we have $ f^{2}(a\plus{}b) \equal{} b^{2}\not \equal{} 0$ so $ f(a\plus{}b) \equal{} a\plus{}b\leftrightarrow (a\plus{}b)^{2}\equal{} b^{2}$
Because $ a\not \equal{} 0$ then $ a \equal{}\minus{}2b$.
Similary, $ f(a\plus{}b) \equal{} a\plus{}b$ and $ f(a) \equal{} 0\Rightarrow a \equal{}\minus{}2(a\plus{}b)$
Contracdiction .Thus we have $ f(x) \equal{} 0,\forall x\in\mathbb{R}$ or $ f(x) \equal{} x,\forall x\in\mathbb{R}$ :) [/hide]

Based on your comments, $ f(0)^{2}\equal{} f(2y)f(0)$ implies $ f(0) \equal{} f(2y)$ or $ f(0) \equal{} 0$ is the only thing I can think of that you're saying is flawed. But I think it's ok, since $ f(0)$ is constant. If i$ f(0)\neq 0$, then clearly $ f(2y) \equal{} f(0)$. Can you be more specific?

But if there exists $ a,b$ (follow my solution $ f(a) \equal{} 0,f(b) \equal{} b$) then your solution is wrong.You must show that it is contracdiction :)

[quote="MellowMelon"]Oh woops, I missed that last line. :( You are correct, I missed that possibility.[/quote]

You can use my solution to proof it  :lol:


function
limit
Putnam
quadratics
algebra
algebra unsolved

Let $f$ be a continuous function , $f : R \to R$ and $0<a,b<1/2$ such that:
$f(f(x))=af(x)+bx$ $\forall x \in R$
Prove that there is  $k \in R$ such that $f(x)=kx$ $\forall x \in R$

[quote="cuopbientcd"]Use equal $x^{2}-ax-b=0$ and properties of limit.(From IMO)[/quote]

Can you post solution?

uhm, can you post your solution, cuopbientcd?

[color=darkblue]If you live in Romania, then seek for the book [b]Analiza matematica (clasa a XI - a), Editura TEORA, colectia Lyceum, Bucuresti, 1999 [/b] and read at the page 328 down, Example 18, where this exercise is solved completely.[/color]

[quote="Virgil Nicula"]If you live in Romania[/quote]
But we aren't live in Romania  :D

Then soon I"ll write here a solution of this problem.

[quote="Bazooka"][color=red]Let $f: \mathbb R\rightarrow \mathbb R$ be a continuous function and $\{a,b\}\subset\left(0,\frac{1}{2}\right)$

such that $f(f(x))=af(x)+bx$ $\forall x \in R$.

Prove that there is  $k \in R$ such that $f(x)=kx$ $\forall x \in R$.[/color][/quote]
[color=darkred][b][u]A helpful problem.[/u][/b] Let $f: \mathbb R\rightarrow\mathbb R$ be a continous function

and $a<1\ ,\ b\ne 0\ ,\ a+b\ne 1$ for which $f(f(x)=af(x)+bx\ ,\ \left(\forall\right) x\in\mathbb R$.

Prove that the function $f$ has an alone fixed point,

i.e. $x_{f}\in\mathbb R$ so that $f(x_{f})=x_{f}$.[/color]

[color=darkblue][b][u]Proof.[/u][/b]

$1.\blacktriangleright$ The function $f$ is injection. Indeed, $f(x_{1})=f(x_{2})$ $\Longrightarrow$

$f(f(x_{1}))=f(f(x_{2}))$ $\Longrightarrow$ $af(x_{1})+bx_{1}=af(x_{2})+bx_{2}$ $\Longrightarrow$

$\left\{\begin{array}{c}b(x_{1}-x_{2})=0\\\ b\ne 0\end{array}\right\|$ $\Longrightarrow$ $x_{1}=x_{2}$, i.e. the function $f$ is strict monotonously.

$2.\blacktriangleright$ [b][u]Existence of the fixed point $x_{f}$.[/u][/b]

Denote $X\ .s.s.\ Y$ ("[b][u]s[/u][/b]ame [b][u]s[/u][/b]ign") $\Longleftrightarrow X=Y=0\ \vee\ XY>0$.

Appear two cases :

$2.1.$ The function $f$ is strict increasing ($\nearrow$).

In this case $f(0)=f(0)-0\ s.s.\ f(f(0))-f(0)=(a-1)f(0)$ $\Longrightarrow$

$f(0)\ .s.s.\ (a-1)f(0)$ $\Longrightarrow$ $\left\{\begin{array}{c}(a-1)f^{2}(0)\ge 0\\\ a-1< 0\end{array}\right\|$ $\Longrightarrow f(0)=0$ $\Longrightarrow$ $x_{f}=0$.

$2.2.$ The function $f$ is decreasing ($f\searrow$).

In this case suppose that $f(x)\ne x$ for any $x\in\mathbb R$. Then :

$f(x)<x$ for any $x\in \mathbb R$ $\Longrightarrow \lim_{x\to-\infty}f(x)=-\infty$- false, because $f\searrow$.

$f(x)>x$ for any $x\in\mathbb R$ $\implies$ $\lim_{x\to\infty}f(x)=\infty$- false, because $f\searrow$.

Therefore, exists $x_{f}\in\mathbb R$ so that $f(x_{f})=x_{f}$.

[u]Otherwise[/u], the function $g(x)=f(x)-x$ is decreasing

and the equation $g(x)=0$ has at most (being injection) and at least a zero

because $\lim_{x\to-\infty}g(x)=\infty$ and $\lim_{x\to \infty}g(x)=-\infty$.

$3.\blacktriangleright$ [b][u]Oneness of the fixed point $x_{f}$[/u][/b]. Indeed,

$f(x_{f})=x_{f}\Longrightarrow f(f(x_{f}))=f(x_{f})\implies af(x_{f})+bx_{f}=f(x_{f})\implies$

$ax_{f}+bx_{f}=x_{f}$ $\implies$ $\left\{\begin{array}{c}(a+b-1)x_{f}=0\\\ a+b\ne 1\end{array}\right\|$ $\implies$ $x_{f}=0$.

In conclusion, the function $f$ has a fixed point $x_{f}$ and it is uniquely.

[/color]

[quote="arkhammedos"]see page 14 in this book
http://www.animath.fr/cours/eqfonc.pdf[/quote]

That is not this problem  :D

[quote="N.T.TUAN"]this problem posted here http://www.mathlinks.ro/Forum/viewtopic.php?t=47940[/quote]
This proof is truly awesome.
Just a few remarks.
First it holds for all $a\not =0,b>0,a+b\not =1$!

In order to show, that $f$ is bijective, one needs not only $f$ continuous, injective and therefore strictly monotonic,
but also that $f$ is not bounded from above or below. But otherwise $\lim_{x\rightarrow \infty}f(x)$ or $\lim_{x\rightarrow-\infty}f(x)$ exists in $\mathbb{R}$.
Say $\lim_{x\rightarrow t}f(x)=c\in\mathbb{R}$ for $t=\infty$ or $-\infty$. But then
$\lim_{x\rightarrow t}x=\lim_{x\rightarrow t}{f(f(x))-af(x)\over b}={f(c)-ac\over b}$ exists in $\mathbb{R}$ for $t=\infty$ or $-\infty$, as $f$ is continuous and $b>0$. Contradiction!

Then there exists a $\gamma\in\mathbb{R}$ with $f(\gamma)=0$ and $f(0)=f(f(\gamma))=af(\gamma)+b\gamma=b\gamma$. And then $f(0)-0=b\gamma$ and $f(\gamma)-\gamma=-\gamma$ have different signs ($b>0$), and since $f$ is continuous, we must have some $\delta\in\mathbb{R}$ between $0$ and $\gamma$ with $f(\delta)-\delta=0$. But then $\delta=f(f(\delta))=af(\delta)+b\delta=(a+b)\delta$ and $\delta=0, f(0)=0$. ($a+b\not =1$)

And now $a^{2}+4b>0$, so $x^{2}-ax-b=0$ has $2$ real solutions $\lambda\not =\mu$,
with $f^{(n)}(x)={f(x)-\mu x\over\lambda-\mu}\lambda^{n}+{f(x)-\lambda x\over\mu-\lambda}\mu^{n}$ for all $n\in\mathbb{Z}$.
Also $\lambda \mu=-b<0$, so $\lambda ,\mu$ have different signs, say $\lambda >0>\mu$.
And $\lambda+\mu=a\not =0$, so $\lambda ,\mu$ have different absolute values.

Now the above proof goes through with only little more case work, to show $f(x)=\lambda x\text{ or }\mu x$ for all $x\in\mathbb{R}$:
 Assume $f(x)\not =\lambda x,\mu x$ for some $x\in\mathbb{R}\setminus\{0\}$.
 Then if $\lambda>|\mu|$, $f^{(n)}(x)$ have the same signs for $n>>0$ and alternating signs for $n<<0$,
 and if $\lambda<|\mu|$, $f^{(n)}(x)$ have alternating signs for $n>>0$ and the same signs for $n<<0$.
 But if $f$ is increasing, then $f^{(n)}(x)$ have the same signs for all $n\in\mathbb{Z}$,
 and if $f$ is decreasing, then $f^{(n)}(x)$ have alternating signs for all $n\in\mathbb{Z}$.
Contradiction!

Now $A: =\{x\in\mathbb{R}: f(x)=\lambda x\}$ and $B: =\{x\in\mathbb{R}: f(x)=\mu x\}$ are closed subsets of $\mathbb{R}$, as $f$ is continuous,
with $\mathbb{R}=A\cup B$ and $A\cap B =\{0\}$. And then $\mathbb{R}_{+},\mathbb{R}_{-}$ are connected subsets of $\mathbb{R}$
with $\mathbb{R}_{+}=(\mathbb{R}_{+}\cap A)\sqcup (\mathbb{R}_{+}\cap B)$ and $\mathbb{R}_{-}=(\mathbb{R}_{-}\cap A)\sqcup (\mathbb{R}_{-}\cap B)$.
But a connected topological space can't be the disjoint union of $2$ non-empty closed subsets.
So we must have $\mathbb{R}_{+}\subseteq A$ or $\mathbb{R}_{+}\subseteq B$, and $\mathbb{R}_{-}\subseteq A$ or $\mathbb{R}_{-}\subseteq B$.
We also have $\mu B=\{\mu x: x\in B\}\subseteq B$, because $f(\mu x)=f(f(x))=af(x)+bx=(a\mu+b)x=\mu^{2}x$ for all $x\in B$.
So in case $\mathbb{R}_{+}\subseteq B$ or $\mathbb{R}_{-}\subseteq B$ we have $\mathbb{R}=\{0\}\cup\mathbb{R}_{+}\cup \mu\mathbb{R}_{+}=\{0\}\cup\mathbb{R}_{-}\cup \mu\mathbb{R}_{-}\subseteq B$ (as $\mu <0$)
and otherwise $\mathbb{R}=\{0\}\cup\mathbb{R}_{+}\cup \mathbb{R}_{-}\subseteq A$.

This now gives $f(x)=\lambda x$ for all $x\in\mathbb{R}$, or $f(x)=\mu x$ for all $x\in\mathbb{R}$.

This is Putnam 2001 B5.

I did write a solution for my class in 2001; of course it's similar to what is above.

We can solve for $ x$: $ x \equal{} \frac {g(g(x)) \minus{} ag(x)}b .$ Given any $ x,$ consider $ y \equal{} \frac {g(x) \minus{} ax}b .$ Then if there is a $ w$ such that $ g(w) \equal{} x,$ then $ w \equal{} y.$ This means that $ g$ is invertible on its range, which means that it is one-to-one. A one-to-one continuous function is monotone, so $ g$ is monotone. However, $ g$ cannot be bounded, because if $ g$ were bounded, then $ g(g(x)) \minus{} ag(x)$ would be bounded and hence $ x$ would be bounded. So the range of $ g$ is unbounded either above or below.

Suppose $ g$ is increasing. If it were bounded above, then for $ x > 0$ it would be bounded below as well, and $ g(g(x)) \minus{} ag(x)$ would be bounded as $ x \to\infty,$ which can’t happen. If it were bounded below, then for $ x < 0$ it would be bounded above as well, and $ g(g(x)) \minus{} ag(x)$ would be bounded as $ x \to \minus{} \infty,$ which can't happen. Similar arguments can be made for g decreasing. So the range of $ g$ is $ ( \minus{} \infty,\infty),$ $ g$ is invertible, and $ g^{ \minus{} 1}(x) \equal{} \frac {g(x) \minus{} ax}b .$

Let $ x_{0}$  be an arbitrary nonzero number. For all $ n \ge 0,$ define $ x_{n \plus{} 1} \equal{} g(x_{n}).$ For all $ n \le 0,$ define $ x_{ \minus{} n \minus{} 1} \equal{} g^{ \minus{} 1}(x_{ \minus{} n}).$  The defining condition $ g(g(x)) \equal{} ag(x) \plus{} bx$ leads to $ x_{n \plus{} 2} \equal{} ax_{n \plus{} 1} \plus{} bx_{n}.$  This is a homogeneous constant coefficient second order difference equation. To find its general solution, assume $ x_{n} \equal{} \lambda n$ for some $ \lambda.$ This leads to the quadratic equation $ \lambda^{2} \minus{} a\lambda \minus{} b \equal{} 0.$  Let $ \alpha$ be the positive root of this equation and $ \beta$ the negative root. That is, $ \alpha \equal{} \frac {a \plus{} \sqrt {a^{2} \plus{} 4b}}2$   and $ b \equal{} \frac {a \minus{} \sqrt {a^{2} \plus{} 4b}}2 .$ Since $ a < \frac12 ,\ b < \frac12 ,$  $ \alpha < (1/2 \plus{} \sqrt {5/4})/2$ which is clearly less than $ 1.$ Since the sum of $ \alpha$ and $ \beta$ is positive, that means  $ \minus{} 1 < \beta < 0.$ Furthermore, $ |\beta| < |\alpha|.$ The general solution of the difference equations is   $ x_{n} \equal{} C_{1}\alpha^{n} \plus{} C_{2}\beta^{n}.$  Since $ x_{n}\to 0$ as $ n \to \infty$ and $ g$ is continuous, we must have that $ g(0) \equal{} 0.$ Consider $ h(x) \equal{} \frac {g(x)}x$   for $ x \ne 0.$ Since $ g$ is monotone and $ 0$ at $ 0,$ then $ h(x)$ is either always positive or always negative. For a given $ x_{0},$  we form the double-ended sequence $ x_{n}.$  Then $ h(x_{n}) \equal{} \frac {x_{n \plus{} 1}}{x_{n}} \equal{} \frac {C_{1}\alpha^{n \plus{} 1} \plus{} C_{2}\beta^{n \plus{} 1}}{C_{1}\alpha^{n} \plus{} C_{2}\beta^{n}}.$ If both $ C_{1}$ and $ C_{2}$  are nonzero, then  $ \lim_{n\to\infty}h(x_{n}) \equal{} \alpha > 0$ and  $ \lim_{n\to \minus{} \infty}h(x_{n}) \equal{} \beta < 0.$ Since $ h$ must be of constant sign, this cannot be. The only way to avoid this is to have either $ C_{1} \equal{} 0$  or $ C_{2} \equal{} 0.$ Furthermore, if $ C_{1} \equal{} 0$  starting at some $ x_{0},$ then the same must hold for all $ x_{0},$ and the same must be true if $ C_{2} \equal{} 0.$  If $ C_{1} \equal{} 0,$ then $ h(x_{0}) \equal{} \frac {x_{1}}{x_{0}} \equal{} \frac {C_{2}\beta}{C_{2}} \equal{} \beta,$ and thus $ h(x) \equal{} \beta$ for all $ x \ne 0,$ which means that $ g(x) \equal{} \beta x.$ Similarly, if $ C_{2} \equal{} 0$ for any $ x_{0},$ then $ g(x) \equal{} \alpha x$ for all $ x.$     [i]q.e.d.[/i]


function
integration
real analysis
real analysis solved

P9

f:[0, 1]->R is a continuous function s.t.  \int {0, 1} f(x)dx =  \pi/4. Prove that there is an x0 in (0, 1) s.t. 1/(1+x0) < f(x0) < 1/2x0.

If there exists a real number c in (0,1) satisfies :f(c) =1/(1+c 2 ) , the problem is obvious .
  Now , assume the contracdition.
   For f is a continous function we have
   f(x) > 1/(1+x 2 ) for all x in (0,1) ...........  (1)
   ( or f(x) < 1/(1+x  2  ) for all x in (0,1)......(2)  )
   So
       \int (0-->1) f(x) dx >  \int (0-->1) (1+x 2 )dx = \pi /4(in the case (1))
   (or \int (0-->1) f(x)dx <  \pi /4 in the case (2))

The condition can be written as \int_[0,1](f(x) - 1/(1+x^2))dx = 0
The function g(x) = f(x) - 1/(1+x^2) is continuous on [0,1]
Suppose that g > 0 on the compact [0,1] then \int_[0,1]g>0,
we also get a contradiction if we suppose that g < 0.
So there exist a \in [0,1] such that g(a) = 0
f(a)= 1/(1+a^2) from now it is obvious.


1994-1995 sprint question:

Suppose a coin purse has 30 coins in it, all of which are either nickels, dimes, or quarters. How many different combinations of these coins are there whose value is $5?

What is the most efficient way to solve this? Post your solution, not just answer, plz.

So would I. you know that there has to be several quarters since 30 dimes only make $3.

Are you kidding? An organized list would take forever. Well not really since there aren't that many combinations.
[hide]We know that $N + D + Q = 30$ and $5N + 10D + 25Q = 500$.
From this, we can form a bunch of equations from subtraction and stuff:
$D + 4Q = 70$

$3Q - N = 40$

Since the second equation has infinite solutions due to its positive slope, we look for solutions for the first equation such that $3Q > 40$ so $N$ won't be negative which are:

$(D, Q) = (2, 17), (6, 16), (10, 15), (14, 14)$ and so on.

We put the values for $Q$ in the second equation and find solutions for $(D, N, Q)$:

$(D, N, Q) = (17, 11, 2), (16, 8 , 6), (15, 5, 10), (14, 2, 14)$

Giving us a grand total of 4 different possibilities.
Of course we didn't really have to go through and find all of the acutal numbers... :)[/hide]

[quote="236factorial"]you can go try that yourself, because I didn't quite get it.[/quote]
You didn't understand his solution?

What he's saying is:
[hide]
\begin{eqnarray*}
N+D+Q&=&30\\
5N+10D+25Q&=&500\\
\end{eqnarray*}
So
\begin{eqnarray*}
5N+5D+5Q&=&150\\
5N+10D+25Q&=&500\\
5D + 20Q&=&350\\
D + 4Q&=&70\\
\end{eqnarray*}
And
\begin{eqnarray*}
10N+10D+10Q&=&300\\
5N+10D+25Q&=&500\\
-5N + 15Q&=&200\\
15Q&=&200+5N\\
3Q&=&40+N\\
\end{eqnarray*}
Now we have two equations:
\begin{eqnarray*}
3Q-N&=&40\\
D + 4Q&=&70\\
\end{eqnarray*}
The second one has positive slope so it has infinite solutions.
Yeah, tarquin has the rest...
[/hide]

I just basically made a list. The first time I did this problem, I just skipped it, but then I realized there weren't a lot of combinations at all (because of the 30-coin restriction). There's actually only four combos. Logic makes this listing much faster.


I seem to recall there being a post like this a few days before the AMC. I decided to start one earlier. So any tips on test taking strategies( such as how often u check, whether or not to bring food...) please share I'll  start with one: It's three hours, don't drink too much water. :D

[quote="h_s_potter2002"]You should guess on each problem that you don't know how to do (unlike the AMC) , and make sure you don't guess the same answer for each problem because it's unlikeley that all the answers will be the same.[/quote]

I'm gonna guess 025, 026, 027, 028, 029 for all the questions I can't do.(I hope I can at least solve 10 questions)

How much can pressure affect you on AIME?

I completely lost self-confidence in qualifying for USAMO so how to get self-confidence back?


calculus
integration
logarithms
function
absolute value
real analysis
calculus computations

Hey,

[b]1. The problem statement, all variables and given/known data.[/b]
Let,
$ {{y}, {{y}^{\prime}}, {{y}^{\prime\prime}},...,} = {{f(x)}, {{{f}^{\prime}}{(x)}}, {{{f}^{\prime\prime}}{(x)}},...,}$

Given the First-Order Linear Differential Equation (F-OLDE) in Standard Form (SF),
$ {{{f}^{\prime}(x)}+{{P(x)}{f(x)}}} = {Q(x)}$

Derive the General Solution (GS) to the First-Order Linear Differential Equation in Standard Form.

[b]2. Relevant equations.[/b]
Knowledge of Calculus and solving Differential Equations.
$ {{\int_{}^{}}{\frac{du}{u}}} = {{\ln{\lvert{u}\rvert}}+{C}}$

[b]3. The attempt at a solution.[/b]
[hide="My Attempt at a Solution."]
Given,
$ {{{f}^{\prime}(x)}+{{P(x)}{f(x)}}} = {Q(x)}$

Multiply both sides of the equation by some function, $ {s(x)}$.
$ {{s(x)}{\left[{{f}^{\prime}(x)}+{{P(x)}{f(x)}}\right]}} = {{\left[Q(x)\right]}{s(x)}}$
$ {{{s(x)}{{{f}^{\prime}}{(x)}}}+{{s(x)}{P(x)}{f(x)}}} = {{s(x)}{Q(x)}}$

Note that,
$ {{\frac{{d}^{}}{d{x}^{}}}{\bigg[{s(x)}{f(x)}\bigg]}} = {{{s(x)}{{f}^{\prime}(x)}}+{{f(x)}{{s}^{\prime}(x)}}}$

Letting,
$ {{{s(x)}{{{f}^{\prime}}{(x)}}}+{{s(x)}{P(x)}{f(x)}}} = {\bigg({{s(x)}{{f}^{\prime}(x)}}+{{f(x)}{{s}^{\prime}(x)}}\bigg)}$
We see that,
$ {{s(x)}{P(x)}} = {{s}^{\prime}(x)}$

Rearranging and Integrating leads to,
$ {{\int_{}^{}}{\left[P(x)\right]}{dx}} = {{\int_{}^{}}{\left[\frac{{s}^{\prime}(x)}{s(x)}\right]}{dx}}$
$ {{\int_{}^{}}{\left[P(x)\right]}{dx}} = {{\ln{\lvert{s(x)}\rvert}}+{{C}_{1}}}$
At this step, I believe the constant $ {{C}_{1}}$ can be disregarded, if that is the case what is the justification for disregarding the constant $ {{C}_{1}}$?

That aside, assuming the constant $ {{C}_{1}}$ can be disregarded we have,
$ {{\int_{}^{}}{\left[P(x)\right]}{dx}} = {{\ln{\lvert{s(x)}\rvert}}}$
$ {{e}^{{\int_{}^{}}{\left[P(x)\right]}{dx}} = {{e}^{\ln{\lvert{s(x)}\rvert}}}}$
$ {{e}^{{\int_{}^{}}{\left[P(x)\right]}{dx}} = {\lvert{s(x)}\rvert}}$

From this step I believe the absolute value function can be disregarded, how I would justify that the absolute value function can be disregarded?
In other words, how do I justify going from,
$ {\lvert{s(x)}\rvert} = {{e}^{{\int_{}^{}}{\left[P(x)\right]}{dx}}}$
to
$ {s(x)} = {{e}^{{\int_{}^{}}{\left[P(x)\right]}{dx}}}$
[/hide]
Thanks,

-[u][b][url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=33715]APSStudent[/url][/b][/u]

[quote="APSStudent"]
$ {{\int_{}^{}}{\left[P(x)\right]}{dx}} = {{\ln{\lvert{s(x)}\rvert}} + {{C}_{1}}}$
At this step, I believe the constant $ {{C}_{1}}$ can be disregarded, if that is the case what is the justification for disregarding the constant $ {{C}_{1}}$?[/quote]
The justification is that you haven't defined what $ s$ is yet, so you can "absorb" the constant into $ s$ to agree with your initial conditions.

[quote="APSStudent"]how do I justify going from,
$ {\lvert{s(x)}\rvert} = {{e}^{{\int_{}^{}}{\left[P(x)\right]}{dx}}}$
to
$ {s(x)} = {{e}^{{\int_{}^{}}{\left[P(x)\right]}{dx}}}$[/quote]
$ e^x$ is always positive.  By the way, what you really want to say is that

$ s(x) = e^{\int_a^x P(t) \, dt}$

for some unspecified $ a$ or else this expression isn't particularly well-defined; this is what I meant when I said that you can absorb the constant into $ s$.

Hey,
[quote="t0rajir0u"]
The justification is that you haven't defined what $ s$ is yet, so you can "absorb" the constant into $ s$ to agree with your initial conditions.
[/quote]
I see. Another justification I've heard is,
[i]Only one $ {s(x)}$ is needed, therefore no arbitrary constant (as a result of the integration) is needed.[/i]
is the above correct?

[quote="t0rajir0u"]
$ e^x$ is always positive.  By the way, what you really want to say is that

$ s(x) \equal{} e^{\int_a^x P(t) \, dt}$

for some unspecified $ a$ or else this expression isn't particularly well-defined; this is what I meant when I said that you can absorb the constant into $ s$.
[/quote]
Ok.
Let me ask this then, why is it that we want a well-defined expression?
Since we're trying to find the general solution to this DE, isn't it better to find the general solution by using an indefinite integral as opposed to a definite integral since in the end we want a more general solution as opposed to a specific solution (to the DE)?

Thanks,

-[u][b][url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=33715]APSStudent[/url][/b][/u]

[quote="APSStudent"]Another justification I've heard is,
[i]Only one $ {s(x)}$ is needed, therefore there is no need for the arbitrary constant as a result of the integration.[/i][/quote]
I suppose this is also true; since we're just multiplying the original equation by $ s$ to get it into a form where we can integrate correctly, the actual constant comes afterwards, not here.

[quote="APSStudent"]isn't it better to find the general solution by using an indefinite integral as opposed to a definite integral since in the end we want a more general solution as opposed to a specific solution (to the DE)?[/quote]
See the above.  We're just multiplying by $ s$ to simplify, not to solve, so we don't need to worry about constants yet.  As an analogy, to solve an equation like

$ \frac{1}{1 \minus{} x} \plus{} \frac{1}{1 \plus{} x} \equal{} x$

you would multiply by $ (1 \minus{} x)(1 \plus{} x)$ to clear the denominators.  Does it matter whether you multiply by $ (1 \minus{} x)(1 \plus{} x)$ or by $ 3(1 \minus{} x)(1 \plus{} x)$?  No.

Hey,
[quote="t0rajir0u"]
I suppose this is also true; since we're just multiplying the original equation by $ s$ to get it into a form where we can integrate correctly, the actual constant comes afterwards, not here.
[/quote]
I see, thanks.

[quote="t0rajir0u"]
See the above. We're just multiplying by $ s$ to simplify, not to solve, so we don't need to worry about constants yet.
[/quote]
Ahh, ok. Well, I know you mentioned that we would like the expression to be well-defined, such as,
$ s(x) = e^{\int_a^x P(t) \, dt}$
however, is it technically incorrect then to proceed solving for the general solution without using a "well-defined" expression?

In other words, is it incorrect to keep solving for the general solution by using a "general" expression, such as,
$ {s(x)} = {{e}^{{\int_{}^{}}{P(x)}{dx}}}$

Thanks,

-[u][b][url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=33715]APSStudent[/url][/b][/u]

Nope; the two are pretty much equivalent since it doesn't matter what $ a$ is.  It's just a good habit not to write indefinite integrals if you can; if you ever take a real analysis course you'll learn that the actual theory here is completely built up in terms of the definite integral, and the indefinite integral just refers to exactly the expression I wrote down.


calculus
integration
function
analytic geometry
advanced fields
advanced fields theorems

currently i am studying integration of differential forms on chains.
however , i do not understand the idea behind integrating differential forms..
usually one integrates real valued functions, right?
so intuitively, what is the integral of a differential form and why would one wish to compute it?
could anyone please explain?
thanks in advance   :huh:

ps hope this is the right forum to post this topic...

Do you remember line and surface integrals from that initial multivariable class? Integrating differential forms is the generalization of that idea, and the only sensible way to handle integration when we don't have canonical coordinates.
Now, if you do have a standard volume form on the set you're integrating over, you can convert the integral of any other form into a volume integral of some real-valued function, with volume defined in terms of the standard form. If you want to compute things, you're probably going to pick coordinates and do exactly this with respect to the standard volume form $ dx_1\wedge dx_2\wedge\cdots\wedge dx_n$.


geometry
Fixed point
circles
parallelogram
conics
IMO
IMO 1979

Two circles in a plane intersect. $A$ is one of the points of intersection. Starting simultaneously from $A$ two points move with constant speed, each travelling along its own circle in the same sense. The two points return to $A$ simultaneously after one revolution. Prove that there is a fixed point $P$ in the plane such that the two points are always equidistant from $P.$

Let $O_1, O_2$ be the centers of the 2 circles and let the points B, C move on the circles $(O_1), (O_2)$, recpectively. Let A' be the other intersection of the 2 circles different from the point A. Since the angles $\angle BO_1A = \angle CO_2A$ are equal, the isosceles triangles $\triangle AO_1B \sim \triangle AO_2C$ are similar. Consequently, the angles $\angle BAO_1 = \angle CAO_2$ are equal and the angles

$\angle BAC = BAO_1 + \angle O_1AO_2 + \angle O_2AC =$

$= BAO_1 + \angle O_1AO_2 - \angle CAO_2 = \angle O_1AO_2$

are also equal. The angles $\angle A'BA = \frac{\angle A'O_1A}{2} = \angle O_2O_1A$ are equal and the angles $\angle A'CA = \frac{\angle A'O_2A}{2} = \angle O_1O_2A$ are also equal. As a result, the point A' is on the side BC of the triangle $\triangle ABC$, which is always similar to the fixed triangle $\triangle O_1AO_2$.

[b]Lemma[/b] (Yaglom's Geometric Transformations II, pp 68-69). Assume that a figure F' (the triangle $\triangle ABC$) moves in such a way that it remains at all times similar to a fixed figure F (the triangle $\triangle O_1AO_2$) and in addition, some point of the figure F' does not move at all (the point A). If a point B of the figure F' describes a curve $\Gamma$ (the circle $(O_1)$), then any other point M of the figure F' describes a curve $\Gamma'$ similar to the curve $\Gamma$.

Consequently, the midpoint M of the segment BC decribes a circle (Q). When the points B, C, become identical with the point A, the midpoint M of BC also becomes identical with A. Hence, the circle (Q) passes through the point A. The line BC always passes through A'. Just after the start of motion counter-clockwise, the points B, C, A' follow on the line BC in this order. The points B, C both pass through the point A' one after another and just before the end of one period of motion, the points A', B, C follow on the line BC  In this order. From the continuity principle, the midpoint M of BC also passes through the point A' and consequently, the circle (Q), the locus of the midpoints M, also passes through the point A'. Thus the circle (Q) belongs to the same pencil with the circles $(O_1), (O_2)$ and it is centered on their center line $O_1O_2$. Let p be the perpendicular bisector of the segment BC intersectong the circle (Q) at a point P different from the point M. The point A' and the circle (Q) passing through this point are fixed and the angle $\angle A'MP = 90^\circ$ is right. Therefore, A'P is a fixed diameter of the circle (Q), which means that the point P is also fixed. Q.E.D.

When the points B, C are both diametrally opposite to the point A on their respective circles, the segment $O_1O_2$ is the midline of the triangle $\triangle ABC$, i.e., $BC \parallel O_1O_2$, $BC \perp AA'$. The cyclic quadrilateral AA'MP inscribed in the circle (Q) is then a rectangle. Its circumcenter Q is the intersection of its diagonals AM, A'P intersecting at their midpoint and the midpoint of AM is identical with the midpoint of $O_1O_2$. Thus the fixed point P is the reflection of the intersection A' of the circles $(O_1), (O_2)$ in the midpoint Q of the segment $O_1O_2$.

[quote="yetti"]
 $\angle BAC = BAO_1 + \angle O_1AO_2 - \angle O_2AC = \angle O_1AO_2$ .[/quote]
I'm sorry, and maybe I mistake something.
but why is this true?
$O,A,C$ can not be on a line. :(

By the problem conditions, the angles $\angle AO_1B = \angle AO_2C$ are equal, because the points A, B travel on their respective circles with constant velocity and the same period, hence, with the same angular velocity. The triangles $\triangle s AO_1B, \triangle AO_2C$ are both isosceles, hence all their angles are equal and $\angle BAO_1 = \angle CAO_2$. The angle $\angle BAC$ can be expressed as a sum of directed angles

$\angle BAC = \angle BAO_1 + \am\angle O_1AO_2 + \angle O_2AC =$

$= \angle BAO_1 + \angle O_1AO_2 - \angle CAO_2 = \angle O_1AO_2$

The equation you refer to has a wrong sign (tx, fixed).

ok,thanks .
I haven't notice the condition which is "each travelling along its own circle in the same sense."
but if "each travelling along its own circle in the different sense"
can the conclusion also be true?

[quote="yetti"]The conclusion is true even if the 2 points B, C travel on the circles $(O_1), (O_2)$ in the opposite directions. The fixed point of the perpendicular bisector of the segment BC is then the reflection of the intersection A in the midpoint Q of the segment $O_1O_2$, but I do not have a proof yet.[/quote]
sorry.I don't know which is the fixed point you wrote because my weak english.
and I guess the fixed point is
let $AO_1$ intersect   circle $(O_1)$ at $M$,$AO_2$ intersect   circle $(O_2)$ at $N$,
and I think the fixed point is the midpoint of $MN$.
but no proof,either?
is your fixed point the same as me :) ?

[quote="zhaobin"]... is your fixed point the same as me :) ?[/quote]

Certainly. With your notation, the points M, N are diametrally opposite to the point A on their respective circles $(O_1), (O_2)$, so that $O_1O_2$ is the midline of the triangle $\triangle AMN$ parallel to the side MN. If P is the midpoint of MN, AP is the A-median of the triangle $\triangle AMN$, which cuts the midline segment $O_1O_2$ at its midpoint Q and AQ = PQ, so that P is a reflection of A in Q.

Let $B_0, C_0$ be the points on the circles $(O_1), (O_2)$ diametrally opposite to their intersection A and let A' be the other intersection of these 2 circles. Let B, C be the points on the circles $(O_1), (O_2)$ traveling in the same sense counter-clockwise. Denote $\alpha = \angle BAC = \angle B_0AC_0 = \angle O_1AO_2$, $\beta = \angle ABC = \angle AB_0C_0 = \angle AO_1O_2$, $\gamma = \angle ACB = \angle AC_0B_0 = \angle AO_2O_1$ the angles of the similar triangles $\triangle ABC \sim \triangle AB_0C_0 \sim \triangle AO_1O_2$. We know that $A' \equiv (O_1) \cap (O_2) \equiv B_0C_0 \cap BC$. Furthermore, denote $\phi = \angle BAO_1 = \angle CAO_2$ the base angles of the similar isosceles triangles $\triangle AO_1B \sim \triangle AO_2C$ and $k = \frac{BA}{O_1A} = \frac{CA}{O_2A}$ the ratio of their base to the shoulder side. Let $M_0$ be the midpoint of $B_0C_0$. Since $O_1O_2$ is a midline of the triangle $\triangle AB_0C_0$, the point $M_0$ is a reflection of the point A in the midpoint Q of the segment $O_1O_2$. Let C' be a point  traveling in the opposite sense on the circle $(O_2)$ than the the point B on the circle $(O_1)$. The point C' is obviously a reflection of the point C in the diameter $AC_0$ of the circle $(O_2)$. We will show that the triangles $\triangle C'BC \sim \triangle C'M_0C_0$ are similar. The angle $\angle BCC' \equiv \angle A'CC'$ and the angle $\angle M_0C_0C' \equiv \angle A'C_0C'$ spanning the same arc A'C' of the circle $(O_2)$ are equal. The isosceles triangles $\triangle ACC' \sim \triangle O_2C_0C'$ have the angles at the vertices $A, O_2$ equal, hence, they are similar and $\frac{CC'}{C_0C'} = \frac{CA}{O_2C_0} = \frac{CA}{O_2A} = k$. From similarity of the triangle $\triangle s ABC \sim \triangle AO_1O_2$ with similarity coefficient $\frac{BC}{O_1O_2} = \frac{CA}{O_2A} = k$ and from the parallelogram $O_1M_0C_0O_2$, we have $\frac{BC}{M_0C_0} = \frac{BC}{O_1O_2} = k$. Thus the similarity of the triangles $\triangle C'BC \sim \triangle C'M_0C_0$ by SAS is proven. But then the angles $\angle C'BA' \equiv \angle C'BC = \angle C'M_0C_0 \equiv \angle C'M_0A'$ are equal, which means that the quadrilateral $BC'A'M_0$ is always cyclic, regardles of the position of the point B on the circle $(O_1)$ and the corresponding position of the point C' on the circle $(O_2)$. From the quadrilateral ABA'C (concave in the figure below), the angle

$\angle BA'C' = \angle ABA' + \angle AC'A' + \angle BAC' =$

$= \angle AB_0A' + \angle AC_0A' + \angle B_0AC_0 - 2 \phi = \beta + \gamma + \alpha - 2 \phi = 180^\circ - 2 \phi$.

The angles $\angle BM_0C' = \angle BA'C' = 180^\circ - 2 \phi$ are equal, because the quadrilateral $BC'A'M_0$ is cyclic. For the same reason, the angles $\angle BC'M_0 = \angle BA'M_0$ are also equal. But the angle $\angle BC'M_0 = \angle CA'C_0 = \angle CAC_0 \equiv \angle O_2AC_0 = \phi$. As a result, the remaining angle of the triangle $\triangle BCM_0$ is equal to

$\angle C'BM_0 = 180^\circ - (\angle BM_0C' + \angle BC'M_0) = 180^\circ - (180^\circ - 2 \phi + \phi) = \phi$

This means that the triangle $\triangle BC'M_0$ is always isosceles, i.e., $BM_0 = C'M_0$, regardles of the position of the point B on the circle $(O_1)$ and the corresponding position of the point C' on the circle $(O_2)$. The point $M_0$ is fixed and the perpendicular bisector of the segment BC' always passes through this point.

The locus of midpoints N of the segments BC'  is an ellipse centered at the midpoint Q of the seqment $O_1O_2$ and passing through the points $A, M_0$ (no proof).

As above, $O_1$ and $O_2$ are the centers of the 2 circles, $A$ and $A'$ being their common points; also, as already shown, $B, C, A'$ are collinear. When $BC$ becomes parallel to $O_1O_2$, their position become $D$, respectively $E$, diametrically opposite to $A$ on the 2 circles; let’s call $F$ the midpoint of $DE$.
It’s easy to see that always $\Delta ABC \sim \Delta AO_1O_2$, having $\hat B=\hat O_1$ and $\hat C=\hat O_2$, so always $\widehat{AQC}=\widehat{AOO_2}=\widehat{AFA'}$ ($O$-midpoint of $O_1O_2$).
Now let’s take $P'$ the symmetrical of $A$ w.r.t perpendicular bisector of $O_1O_2$, and we claim that this is the required fixed point. $AP'O_1O_2$ is isosceles trapezoid, hence $O_1P'=O_2A=O_2E \ (\ 1\ )$ and $O_2P'=O_1A=O_1D \ (\ 2\ )$ and $\widehat{P'O_1A}=\widehat {P'O_2A}$, resulting that $\Delta DO_1P'=\Delta P'O_2E$, i.e. $P'D=P'E$, hence $P'F \perp DE$ or $AP'FA'$ is a rectangle.
Let’s call $M$ the intersection of $BO_1$ and $O_2C$; as $\widehat{AO_1B}=\widehat{AO_2C}$, $AO_1MO_2$ is cyclic, or  $\widehat {BO_1P'}=\widehat{P'O_2C}$; with (1) and (2) we get $\Delta BO'P'=\Delta P'O_2C$, hence $P'B=P'C$, and $P'$ is always on the perpendicular bisector of $BC$, and $P'\equiv P$, done.

Notes: 1) There is not necessary for $DE$! We get easily the result from the other triangles and parallelogram.
2) When the points move in opposite direction, the result is even more obvious (of course, once you know the required position).

Best regards,
sunken rock

Let $O1$ and $O2$ be the centers of the two circles in the problem, denoted Γ and Φ. Using $O1$ as the center draw another circle, denoted Π, that has the same radius as the radius of Φ. Now use $O2$ as the center draw another circle, denoted Δ, that has the same radius as the radius of Γ. There are now two sets of concentric circles.
Now let Π intersect Δ at P with $P$ closer to $A$ than to $B$. This is the point $P$ in the problem where the two points $E$ and $F$ are equidistant from. Indeed, with $PE = PF$, it’s easily seen that ∠$AO1E$ = ∠$AO2F$ = α.

Let $X,Y$ be the two points, $\omega_X$, and $\omega_Y$ be the corresponding circles, and $B$ be the other intersection. It is simple to check that $BXY$ is always a line.

Now, invert at $B$ with radius $1$ so that $\omega_X$ and $\omega_Y$ map to lines $\ell_X$ and $\ell_Y$ through $A'$, and let $C$ be the point on $BX'Y'$ such that $(B,C;X',Y')=-1$. By well known facts about projective, $A'C$ is a fixed line - let $T$ be the foot from $B$ to this line.

Now, if we invert back, $\angle BC'T'=90^{\circ}$, and $C'$ is the midpoint of $XY$, so we are done as $T'$ is the fixed point.

Edit: Holy I'm bad at math, here's a quicker sol:

Let $X,Y$ be the two points, $\omega_X$, and $\omega_Y$ be the corresponding circles, and $B$ be the other intersection. It is simple to check that $BXY$ is always a line.

Now, if $M$ is the midpoint of $XY$, since the ratios of the powers of $M$ to $\omega_X$ to $\omega_Y$ is $-1$, by the forgotten coaxiality lemma, $M$ lies on a fixed circle through $A$ and $B$, and now the fixed point is just the $B$-antipode.

Solution from [i]Twitch Solves ISL[/i]:

Let $B$ and $C$ be the antipodes of $A$ on the two circles and let $D$ be the foot of the altitude from $A$, which is the other intersection point of the two circles. Also, let $M$ be the midpoint of $\overline{BC}$, and construct rectangle $ADMZ$. Our claim is that $Z$ is the fixed point.
We let $X$ and $Y$ be the two points; by the condition the angles are the same. So we have a spiral similarity \[ \triangle AXY \sim \triangle ABC. \]
[asy]size(10cm); pair A = dir(110); pair B = dir(210); pair C = dir(330); pair D = foot(A, B, C); pair M = midpoint(B--C); pair Z = A+M-D;
pair X = midpoint(A--B)+dir(200)*abs(A-B)/2; pair Y = -D+2*foot(midpoint(A--C), X, D);
draw(circumcircle(A, B, D), blue); draw(circumcircle(A, C, D), blue); draw(circumcircle(A, D, M), dashed+deepgreen); pair N = midpoint(X--Y); filldraw(A--X--Y--cycle, invisible, purple); filldraw(A--B--C--cycle, invisible, blue); draw(A--D, blue); draw(A--Z--M, deepgreen);
dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$M$", M, dir(M)); dot("$Z$", Z, dir(Z)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(Y)); dot("$N$", N, dir(N));
/* TSQ Source:
A = dir 110 B = dir 210 C = dir 330 D = foot A B C M = midpoint B--C Z = A+M-D
X = midpoint(A--B)+dir(200)*abs(A-B)/2 Y = -D+2*foot midpoint A--C X D
circumcircle A B D blue circumcircle A C D blue circumcircle A D M dashed deepgreen N = midpoint X--Y A--X--Y--cycle 0.1 red / purple A--B--C--cycle 0.1 yellow / blue A--D blue A--Z--M deepgreen
*/ [/asy]
Now let $N$ be the midpoint of $\overline{XY}$. By spiral similarity, since $N$ maps to $M$, it follows $M$, $N$, $A$, and $D$ are cyclic too. So actually $N$ lies on the circumcircle of rectangle $ADMZ$, meaning $\overline{ZN} \perp \overline{XY}$, hence $ZX = ZY$ as needed.

[b][color=red]Remark:[/color][/b]  	The special point $Z$ can be identified 	by selecting the special case $X \to A$, $Y \to A$ 	and $X=B$, $Y=C$.

Let the points on the circles be $E$ and $F$. Then, let $B$ the the other intersection of the two circles, and $C$ and $D$ be the centers of the circles containing $E$ and $F$, respectively. WLOG, let us name $E$ and $F$ such that $CA>DA$. Let $O$ be the midpoint of $CD$, and the antipode of $B$ with respect to the circle centered at $O$ with radius $OA$ be $P$. Let $M$ be where this such circle hits $EF$. We first note that $$\angle ADF = 2\angle ABF = 2 \angle APM = \angle AOM$$ and also $$2 \angle APM = 360^{\circ} - 2\angle MBA = 360^{\circ} - 2\angle EBA = \angle ACE$$ meaning that since $AD=DF$, $AO=OM$, and $AC=CE$, then SAS similarity gets us that $\triangle ADF \sim \triangle AOM \sim \triangle ACE$. This means we have $\frac{AO}{AM} = \frac{AC}{AE}$ so $\frac{AO}{AC} = \frac{AM}{AE}$ and because $\angle MAE = \angle OAC$, then $\triangle AOC \sim \triangle AME$ and similarly, we also have $\triangle AOD \sim \triangle AMF$. This means that $$\frac{EM}{MA} = \frac{OC}{OA} = \frac{DO}{OA} = \frac{FM}{MA} \to EM=FM$$ meaning that $M$ is the midpoint of $EF$. Now, since $P$ is the antipode of $B$, then $\angle BMP = 90^{\circ} = \angle EMP = \angle FMP$ so $EP=FP$ for all $E$ and $F$ for some fixed $P$ and we are done.

Let $P_1, P_2$ be the moving points (pun intended). Denote by $M$ the midpoint of $P_1P_2$.

[b][color=#f00]Claim.[/b][/color] The second intersection of the circles $B$ lies on $P_1P_2$.

Denote by $P_1', P_2'$ the $P_1, P_2$-antipodes. Evidently a spiral similarity at $A$ still sends $\overline{P_1P_1'} \to \overline{P_2P_2'}$. Thus we are done by spiral similarity lemma. $\blacksquare$

[b][color=#f00]Claim.[/b][/color] $M$ moves along a fixed circle through $A, B$.

Observe that $\angle AMB$ is constant because a spiral similarity at $A$ takes $\overline{O_1O_2} \to \overline{P_1P_2}$, and hence $$\angle AMP_1 = \angle ANO_1$$ which is constant where $N$ is the midpoint of $\overline{O_1O_2}$. $\blacksquare$

Therefore the $B$-antipode in the fixed circle $(AMB)$ will always satisfy $\angle PMB = 90^{\circ}$, and thus lies on the perpendicular bisector.

Let \(O_1\) and \(O_2\) be the centers of the two circles, and let \(P\) be the point such that \(PAO_1O_2\) is an isosceles trapezoid with \(\overline{PA}\parallel\overline{O_1O_2}\). I claim \(P\) is the fixed point equidistant from the two particles.

Let the particle on \((O_1)\) be \(C\) and  let the particle on \((O_2)\) be \(D\). Then \(\triangle AO_1C\sim\triangle AO_2D\), so since spiral similarites come in pairs, we have \(\triangle ACD\sim\triangle AO_1O_2\). Now let \(P'\) be the point such that \(\triangle ACD\cup P'\sim\triangle AO_1O_2\cup P\).

Since \(\triangle AO_1C\sim\triangle APP'\), we have \(PA=PP'\). But \(P'ACD\) is an isosceles trapezoid, so the perpendicular bisectors of \(\overline{CD}\) and \(\overline{AP'}\) coincide. Thus \(PC=PD\), as needed.

[i]There is no need to identify the point $P$, we can just prove its [b]existence[/b] as follows:[/i]

Let $\omega_1,\omega_2$ be the two circles with centers $O_1,O_2$ respectively and $B = \omega_1 \cap \omega_2 \ne A$. Let $X,Y$ be the points which move and $Z$ be the midpoint of segment $XY$ ($Z$ is also moving).

[b][color=#f00]Claim: [/color][/b] As $X,Y,Z$ move, $B$ always lies on line $XY$ and there is a fixed circle $\Gamma$ which passes through $A,B$ and $Z$.
[b][color=#00f]Proof: [/color][/b] If we consider two such positions $(X_1,Y_1,Z_1)$ and $(X_2,Y_2,Z_2)$. It suffices to show $B = \overline{X_1Y_1} \cap \overline{X_2Y_2}$ and points $A,B,Z_1,Z_2$ are concylic. Now it is not hard to note by $SAS$ criterion that
$$\triangle AX_1Y_1 \sim \triangle AO_1O_2 \sim \triangle AX_2Y_2$$
So $A$ is the center of spiral sim $\tau$ sending $\overline{X_1Y_1}$ to $\overline{X_2Y_2}$, which implies
$$\overline{X_1Y_1} \cap \overline{X_2Y_2} = \odot(AX_1X_2) \cap \overline{AX_2Y_2} \ne A = \omega_1 \cap \omega_2 \ne A = B$$
As $\tau$ sends $Z_1$ to $Z_2$, so it follows that points $A,B,Z_1,Z_2$ are concyclic. This proves our claim. $\square$

Now just note that $P$ as the antipode of $B$ wrt $\Gamma$ works. This completes the proof of the problem. $\blacksquare$

Let $X$ and $Y$ be the particles and $O_1$ and $O_2$ be the circumcenters. First, we claim that $X$, $Y$, and $B$ are always colinear. Indeed, notice that \[ \angle ABX = \dfrac{1}{2} \angle AO_1X = \dfrac{1}{2} \angle AO_2Y = 180^{\circ}-\angle ABY \] as desired. Thus, by spiral similarity, $\triangle XYA$ remains similar to every other instance of $\triangle XYA$. If $M$ is the midpoint of $XY$ then $M$ moves along a circle passing through $A$ and $B$ since $\angle AMX$ and $\angle AMY$ are fixed. So, the perpendicular bisector of $XY$ always passes through the antipode of $B$ in $(ABM)$ which we just established as fixed. $\blacksquare$

Let $X$ and $Y$ be the moving points on $\omega_1$ and $\omega_2$, respectively. Let $O_1$ and $O_2$ be the centers of $\omega_1$ and $\omega_2$, respectively and let $A_1$ and $A_2$ be the $A$ antipode with respect to $\omega_1$ and $\omega_2$, respectively. Let $B$ be the other intersection of $\omega_1$ and $\omega_2$.

[b][color=lightblue]Claim:[/b] $\overline{XY}$ passes through $B$.
[i]Proof.[/i] Since $X$ and $Y$ have equal angular velocities, \[\angle AA_1X=\tfrac{1}{2}\angle AO_1X=\tfrac{1}{2}\angle AO_2Y=\angle AA_2Y\] so $A$ is the center of the spiral similarity $\overline{A_1X}\mapsto\overline{A_2Y}$. Hence, $A$ is the Miquel point of $A_1XYA_2$ so $\overline{XY}\cap\overline{A_1A_2}$ lies on $(AA_1X)$ and $(AA_2Y)$. $\blacksquare$

Let $M$ and $O$ be the midpoints of $\overline{XY}$ and $\overline{O_1O_2}$, respectively.

[b][color=lightblue]Claim:[/b] $M$ lies on $\gamma$, the circle with center $O$ passing through $A$ and $B$.
[i]Proof.[/i] Notice $A$ is the center of the spiral similarity $\overline{OO_1}\mapsto\overline{XM}$ so \[\angle AMX=\angle AOO_1=\tfrac{1}{2}\angle AOB\] and $O$ lies on the perpendicular bisector of $\overline{AB}$. $\blacksquare$

Since the perpendicular bisector of $\overline{XY}$ is the line perpendicular to $\overline{BM}$ that passes through $M$, it must always pass through the antipode of $B$ with respect to $\gamma$. Since $\gamma$ and $B$ are fixed we are done. $\square$

Denote the two circles as $\omega_1, \omega_2$. Let $P_i$ and $O_i$ denote the moving point and center of $\omega_i$. Additionally, let $B \neq A$ be the second intersection of the two circles.

[b][color=#f00]Claim:[/color][/b] $B$ lies on $P_1P_2$
[b][color=#f00]Proof:[/color][/b] Let $P_1B$ intersect $\omega_2$ again at $P_2'$. Since $\measuredangle ABP_2' = \measuredangle ABP_1$, we can see that the directed arcs $\widehat{AP_1}$ of $\omega_1$ and $\widehat{AP_2'}$ of $\omega_2$ have equal measures, which implies that $P_2' = P_2$, as desired.

Now, let $M$ be the midpoint of $\overline{P_1P_2}$ and $O$ be the midpoint of $\overline{O_1O_2}$.

[b][color=#f00]Claim:[/color][/b] $M$ always lies on the circle centered at $O$ through $A$ and $B$
[b][color=#f00]Proof:[/color][/b] Note that since $\triangle AO_1P_1 \overset{+}{\sim} \triangle AO_2P_2$, $A$ is the center of the spiral similarity sending $\overline{O_1P_1}$ to $\overline{O_2P_2}$. Hence $A$ is also the center of the spiral similarity which takes $\overline{O_1O_2}$ to $\overline{P_1P_2}$, which also sends $O$ to $M$. Thus $\triangle AOM$ is similar to both $\triangle AO_1P_1$ and $\triangle AO_2P_2$, so $OM = OA = OB$, which proves the claim.

So, if we let $B'$ be the antipode of $B$ wrt the circle from the last claim (which is fixed), then we always have $\overline{B'M} \perp \overline{P_1P_2}$, and thus $B'P_1 = B'P_2$, so we are done.

Suppose the second intersection of the two given circles is $B$. Denote segments $P_1Q_1$ and $P_2Q_2$, with the intersection of their perpendicular bisectors as $X$, and $P_1P_2 \cap Q_1Q_2 = C$. Let $O_P$, $O_Q$, $O_1$, and $O_2$ be the circumcenters of $\triangle AP_1P_2$, $\triangle AQ_1Q_2$, $\triangle CP_1Q_1$, and $\triangle CP_2Q_2$, respectively. By Miquel point properties, it is well known that $AO_1O_2O_PO_Q$ is cyclic (and fixed), so proving that $X$ also lies on the circle suffices.

It's easy to show $B$ will lie on $P_1Q_1$ and $P_2Q_2$ through the angle condition implied by the problem. Spiral similarity at $A$ gives
\[\triangle AO_1O_2 \sim \triangle AP_1P_2 \sim \triangle AQ_1Q_2\]
\[\implies \angle O_1AO_2 = \angle Q_1AQ_2 = \angle Q_1BQ_2 = \angle O_1XO_2.\]

Thus $X$ lies on the fixed circle as a fixed point, completing the proof. $\blacksquare$

[asy]
size(250); defaultpen(linewidth(.4));

pair A,B,C,P1,P2,Q1,Q2,O1,O2,OP,OQ,X;
A = (0,24);
B = (0,-24);
P1 = 40dir(200)-32;
P2 = 40dir(260)-32;
Q1 = 51dir(315)+45;
Q2 = 51dir(15)+45;
OP = (-32,0);
OQ = (45,0);
C = extension(P1,P2,Q1,Q2);
O1 = circumcenter(C,P1,Q1);
O2 = circumcenter(C,P2,Q2);
X = extension(O1,foot(O1,P1,Q1),O2,foot(O2,P2,Q2));

filldraw(A--P1--P2--cycle^^A--O1--O2--cycle^^A--Q1--Q2--cycle, palegreen);
draw(A--P1--C--Q2--cycle^^Q1--P2--A--Q1--P1--P2--Q2^^A--O1--O2--cycle);
draw(circle(OP,40)^^circle(OQ,51));
draw(circumcircle(A,O1,OQ),blue+linewidth(.6)+dashed);
draw(O1--X--O2, blue+linewidth(.6));

dot("$A$", A, dir(90));
dot("$B$", B, dir(135));
dot("$C$", C, dir(270));
dot("$P_1$", P1, dir(180));
dot("$P_2$", P2, dir(225));
dot("$Q_1$", Q1, dir(0));
dot("$Q_2$", Q2, dir(20));
dot("$O_1$", O1, dir(270));
dot("$O_2$", O2, dir(0));
dot("$X$", X, dir(90));
dot(OP);
dot(OQ);
[/asy]


calculus
integration
number theory
prime numbers

Let $ a$ = the sum of the first five non-negative integral powers of 2. Let $ b$ = the sum of the first three prime numbers. Let $ c$ = the fifth prime number. Find $ abc$

a = the sum of the first five non-negative integral powers of 2 = 1+2+4+8+16=31

b = the sum of the first three prime numbers = 2+3+5=10

c = the fifth prime number = 11

abc=31*10*11=3140

answer : 3140

i came up with the same answer but on ftw it said the ans was 7040

I got 3140 as well...
How did they get 7040???????


A faster solution is with contrarmonical AM-GM :wink:

[quote="kritikos"]contrarmonical AM-GM[/quote]

Please could you explain.

Contrarmonical is this inequality

$ \dfrac{a_1^2+a_2^2+....+a_n^2}{a_1+a_2+.....+a_n}\geq\sqrt{\dfrac{a_1^2+a_2^2+....+a_n^2}{n}}$ and it continues exactly the same as AM-GM the first fragment is the contrarmonical.

[quote="kritikos"]Contrarmonical is this inequality

$ \dfrac{a_1^2 + a_2^2 + .... + a_n^2}{a_1 + a_2 + ..... + a_n}\geq\sqrt {\dfrac{a_1^2 + a_2^2 + .... + a_n^2}{n}}$ and it continues exactly the same as AM-GM the first fragment is the contrarmonical.[/quote]
I see! This is nothing but QM-AM (or RMS-AM) inequality :wink:


Find all primes $ p,q$ such that $ p|q\plus{}6$ and $ q|p\plus{}7$

[hide]
Okay, so we will deal with a few cases.  Firstly, let us look at the problem if p=q+6.  So, q is a factor of p+7, which is q+13.  Thus, q is a factor of q+13, so q is a factor of 13.  13 is the only prime factor of 13.  So, q is 13.  Thus, p must be 19.
Not, if 2p=q+6, then $ p \equal{} \frac {q \plus{} 6}{2}$.  So, q is a factor of $ \frac {q \plus{} 6}{2} \plus{} 7 \equal{} \frac {q \plus{} 20}{2}$.  So, $ q\le \frac {q \plus{} 20}{2}$ and therefore, $ p\le 20$.  If 3p=q+6, then $ q\le\frac {q \plus{} 27}{3}$.  Thus, $ q\le 13.5$.  So, it is obvious that when kp=q+6 and k increases, then q's upper bound decreases.  Thus, q is always less than 20 (it can't be equal to 20 because 20 isn't prime).
Now, let us try it when q is equal to 2 or 3.
Firstly, we will deal with 2.  If, q is two, then p must be a factor of 2+6=8.  Thus, we have that the only prime factor of 8 is 2.  So, p=2.  Thus, q must be a factor of p+7.  Thus, 2 is a factor of 9, which is false.  So, q is not equal to 2.
Now, with three.  So, we have that p is a factor of 3+6=9 and the only prime factor of 9 is 3.  So, p=3.  Thus, q is a factor of 3+7=10, which is also false.  So, q is not three.
Now, note that in all other possiblities of q, q must be congruent to 1 or -1 mod 6.  So, q+6 is congruent to plus or minus 1 mod 6.  Thus, p is a factor of something congruent to 1 or -1 mod 6.  So, something congruent to 1 or -1 mod 6 must be composite (since p is a factor of it and we have showed that the only possibilty when p=q+6 is when (p,q)=(19,13).)  The lowest value of q+6 is 25.  So, q=19.  But the only prime factor of 25 is 5.  So, p=5, thus, q is a factor of p+7=12.  Thus, 19 is a factor of 12, which is false.
Any other q would be higher than 20, which we know is impossible.  So, our only (p,q) is $ \boxed{(19, 13)}$[/hide]

From the system

\begin{eqnarray*}q+6 &=& ap\\ p+7 &=& bq\end{eqnarray*}

where $ a,b$ are natural, we get

$ p={6b+7\over ab-1},q={7a+6\over ab-1}$

From $ p\geqslant 2$ we get $ a\leqslant 3+{9\over 2b}$, hence $ a\in\{1,\dots,7\}$. By quick inspection of the possibilites, we find that only for $ a=1$ we have a solution, which is $ (19,13)$.


$ \frac{x\minus{}7}{ x\minus{}5}\plus{}\frac{x\minus{}8}{ x\minus{}4}\plus{}\frac{x\minus{}9}{ x\minus{}3}\plus{}\frac{x\minus{}10}{ x\minus{}2}\equal{}x\minus{}6$

$ \frac {x \minus{} 7}{ x \minus{} 5} \plus{} \frac {x \minus{} 8}{ x \minus{} 4} \plus{} \frac {x \minus{} 9}{ x \minus{} 3} \plus{} \frac {x \minus{} 10}{ x \minus{} 2} \equal{} x \minus{} 6$ with $ (x\minus{}5)(x\minus{}4)(x\minus{}3)(x\minus{}2)\neq 0$

$ \Longleftrightarrow 1\minus{}\frac{2}{x\minus{}5}\plus{}1\minus{}\frac{4}{x\minus{}4}\plus{}1\minus{}\frac{6}{x\minus{}3}\plus{}1\minus{}\frac{8}{x\minus{}2}\equal{}x\minus{}6$

$ \Longleftrightarrow 2\left(\frac{1}{x\minus{}5}\plus{}\frac{2}{x\minus{}4}\plus{}\frac{3}{x\minus{}3}\plus{}\frac{4}{x\minus{}2}\right)\equal{}10\minus{}x$

[quote="kunny"]$ \frac {x \minus{} 7}{ x \minus{} 5} \plus{} \frac {x \minus{} 8}{ x \minus{} 4} \plus{} \frac {x \minus{} 9}{ x \minus{} 3} \plus{} \frac {x \minus{} 10}{ x \minus{} 2} \equal{} x \minus{} 6$ with $ (x \minus{} 5)(x \minus{} 4)(x \minus{} 3)(x \minus{} 2)\neq 0$

$ \Longleftrightarrow 1 \minus{} \frac {2}{x \minus{} 5} \plus{} 1 \minus{} \frac {4}{x \minus{} 4} \plus{} 1 \minus{} \frac {6}{x \minus{} 3} \plus{} 1 \minus{} \frac {8}{x \minus{} 2} \equal{} x \minus{} 6$

$ \Longleftrightarrow 2\left(\frac {1}{x \minus{} 5} \plus{} \frac {2}{x \minus{} 4} \plus{} \frac {3}{x \minus{} 3} \plus{} \frac {4}{x \minus{} 2}\right) \equal{} 10 \minus{} x$[/quote]

With all due respect i don't think that this is a complete solution. What step follows next mr Kunny?

as far as i know it can't be simplified much farther unless you add x, and subtract $ 2\left(\frac{1}{x\minus{}5}\plus{}\frac{2}{x\minus{}4}\plus{}\frac{3}{x\minus{}3}\plus{}\frac{4}{x\minus{}2}\right)$ which doesn't help much
 that does however leave you with $ x\equal{}10\minus{}2\left(\frac{1}{x\minus{}5}\plus{}\frac{2}{x\minus{}4}\plus{}\frac{3}{x\minus{}3}\plus{}\frac{4}{x\minus{}2}\right)$
why don't you ask the original poster to elaborate?

hello, calculating the left-hand side minus the right-hand side of our equation and factoring we get
$ \frac{\minus{}2088\plus{}2440x\plus{}x^5\plus{}231x^3\minus{}1084x^2\minus{}24x^4}{(x\minus{}5)(x\minus{}4)(x\minus{}3)(x\minus{}2)}\equal{}0$
Sonnhard.

wow...that's a beautiful factorization

so is the expression true for all x that doesn't make one of the denominators 0?
or are there any specific values of x?


inequalities
inequalities proposed
algebra

[b]IF[/b] $ a$ , $ b$ , $ c$ are positive, [b]THEN[/b] there are the following inequalities :

$ 1\ \blacktriangleright\ \sum\frac {b^2 \plus{} c^2}{b \plus{} c}\ \ge\ \sqrt {3\left(a^2 \plus{} b^2 \plus{} c^2\right)}$ ([u]well-known[/u]).

$ 2\ \blacktriangleright\ \boxed {\ \sum\frac {b^2 \plus{} c^2}{b \plus{} c}\ \ge\ \sqrt {3\left(a^2 \plus{} b^2 \plus{} c^2\right) \plus{} f(a,b,c)\cdot\sum (b \minus{} c)^2}\ }$  ( [b][u]strongest[/u][/b]),

where $ f(a,b,c) \equal{} \frac {abc}{a^2 \plus{} b^2 \plus{} c^2}\cdot\sum \frac {b \plus{} c \minus{} 2a}{a(b \plus{} c)}\ \ge\ 0$ .

[quote="Virgil Nicula"]$ \sum\frac {b^2 + c^2}{b + c}\ \ge\ \sqrt {3\left(a^2 + b^2 + c^2\right) + 2abc\cdot\sum \frac {b + c - 2a}{a(b + c)}}$ Is it strongest ?![/quote]
[quote="nguoivn"]
$ \sum\ \frac {a^2 + b^2}{a + b} \geq\ \sqrt {4\sum\ a^2 - 2\sum\ ab + 4\sum\ \frac {a^2b^2}{(a + b)^2}}$
[/quote]

We can't compare 2 results above with another. (So, I think we can't say which one is strongest)
But I solved both of them by Am-Gm. How is your proof for your ineq ?  :)

[b][u]Generalization[/u].[/b] $ \lambda\in (0,2]\ \implies\  \sum\frac {b^2+c^2}{b+c}\ \ge\ \sqrt {3\left(a^2+b^2+c^2\right)+(2-\lambda )\cdot\sum \frac {bc(b+c-2a)}{b+c}}$

Yes, but we only need to prove with $ k\equal{}2$ because $ \sum \frac {bc(b\plus{}c\minus{}2a)}{b\plus{}c} \geq\ 0$
 :)

$ \sum\frac {b^2 + c^2}{b + c}\ \ge\ \sqrt {3\left(a^2 + b^2 + c^2\right) + (2 - \lambda )abc\cdot\sum \frac {b + c - 2a}{a(b + c)}}$ and for  $ \lambda : = \frac {2(ab + bc + ca)}{a^2 + b^2 + c^2}\le 2\ \Longrightarrow$

$ \sum\frac {b^2 + c^2}{b + c}\ \ge\ \sqrt {3\left(a^2 + b^2 + c^2\right) + f(a,b,c)\cdot\sum (b - c)^2}$ , where $ f(a,b,c) = \frac {abc}{a^2 + b^2 + c^2}\cdot\sum \frac {b + c - 2a}{a(b + c)}\ \ge\ 0\ .$

[quote][color=darkred][b]IF[/b] $ a$ , $ b$ , $ c$ are positive, [b]THEN[/b] there is the following inequality :

$ 1\ \blacktriangleright\ \sum\frac {b^2 + c^2}{b + c}\ \ge\ \sqrt {3\left(a^2 + b^2 + c^2\right)}$ ([u]well-known[/u]).[/color][/quote]

[color=darkblue][b][u]A nice and short proof[/u][/b] of [u]my friend[/u] [b]Marius Mainea[/b].

$ \left\|\ \boxed {\frac {b^2 + c^2}{b + c} - \frac {b + c}{2}} = \frac {(b - c)^2}{2(b + c)} = \frac {(b - c)^2}{2(a + b + c)} + \frac {a(b - c)^2}{2(b + c)(a + b + c)}\ge\boxed {\frac {(b - c)^2}{2(a + b + c)}}\ \right\|\ \sum\ \implies$

$ \sum\frac {b^2 + c^2}{b + c} - (a + b + c)\ge\frac {\sum(b - c)^2}{2(a + b + c)} =$ $ \frac {\sum(b - c)^2}{(a + b + c) + (a + b + c)}\ge$ $ \frac {3\left(a^2 + b^2 + c^2\right) - \left(a + b + c\right)^2}{\sqrt {3\left(a^2 + b^2 + c^2\right)} + (a + b + c)} =$

$ \sqrt {3\left(a^2 + b^2 + c^2\right)} - (a + b + c)\ \implies\ \sum\frac {b^2 + c^2}{b + c}\ \ge\ \sqrt {3\left(a^2 + b^2 + c^2\right)}\ .$

[b][u]My proof[/u].[/b] $ \blacktriangleright\ \sum\frac {b^2 + c^2}{b + c}\ \stackrel{(1)}{\ \ge\ }\ \frac {2\left(a^2 + b^2 + c^2\right) + (ab + bc + ca)}{a + b + c}\ \stackrel{(2)}{\ \ge\ }\ \sqrt {3\left(a^2 + b^2 + c^2\right)}\ .$

Denote $ \left\|\ \begin{array}{c} a^2 + b^2 + c^2 = 1 \\
 \\
ab + bc + ca = t \\
 \\
s = a + b + c = \sqrt {1 + 2t}\end{array}\ \right\|\ .$ Observe that  $ ab + bc + ca\le a^2 + b^2 + c^2\ \implies\ t\in (0,1]\ .$

$ \boxed 1\ \ \sum\frac {b^2 + c^2}{b + c}\ \ge\ (a + b + c) + \frac {\sum (b - c)^2}{2(a + b + c)} = s + \frac {2 - 2t}{2s} = \frac {2s^2 + 2 - 2t}{2s} =$

$ \frac {s^2 + 1 - t}{2s} = \frac {2 + t}{s} = \frac {2\left(a^2 + b^2 + c^2\right) + (ab + bc + ca)}{a + b + c}\ .$

$ \boxed 2\ \ \frac {2 + t}{\sqrt {1 + 2t}}\ge \sqrt 3\ \Longleftrightarrow\ (2+t)^2\ \ge\ 3(1+2t)\ \Longleftrightarrow\ (t - 1)^2\ge 0\ .$[/color]

[quote="Virgil Nicula"]
[quote][color=darkred][b]IF[/b] $ a$ , $ b$ , $ c$ are positive, [b]THEN[/b] there is the following inequality :

$ 1\ \blacktriangleright\ \sum\frac {b^2 + c^2}{b + c}\ \ge\ \sqrt {3\left(a^2 + b^2 + c^2\right)}$ ([u]well-known[/u]).[/color][/quote]

[color=darkblue][b][u]A nice and short proof[/u][/b] of [u]my friend[/u] [b]Marius Mainea[/b].

$ \left\|\ \boxed {\frac {b^2 + c^2}{b + c} - \frac {b + c}{2}} = \frac {(b - c)^2}{2(b + c)} = \frac {(b - c)^2}{2(a + b + c)} + \frac {a(b - c)^2}{2(b + c)(a + b + c)}\ge\boxed {\frac {(b - c)^2}{2(a + b + c)}}\ \right\|\ \sum\ \implies$

$ \sum\frac {b^2 + c^2}{b + c} - (a + b + c)\ge\frac {\sum(b - c)^2}{2(a + b + c)} =$ $ \frac {\sum(b - c)^2}{(a + b + c) + (a + b + c)}\ge$ $ \frac {3\left(a^2 + b^2 + c^2\right) - \left(a + b + c\right)^2}{\sqrt {3\left(a^2 + b^2 + c^2\right)} + (a + b + c)} =$

$ \sqrt {3\left(a^2 + b^2 + c^2\right)} - (a + b + c)\ \implies\ \sum\frac {b^2 + c^2}{b + c}\ \ge\ \sqrt {3\left(a^2 + b^2 + c^2\right)}\ .$

[b][u]My proof[/u].[/b] $ \blacktriangleright\ \sum\frac {b^2 + c^2}{b + c}\ \stackrel{(1)}{\ \ge\ }\ \frac {2\left(a^2 + b^2 + c^2\right) + (ab + bc + ca)}{a + b + c}\ \stackrel{(2)}{\ \ge\ }\ \sqrt {3\left(a^2 + b^2 + c^2\right)}\ .$

Denote $ \left\|\ \begin{array}{c} a^2 + b^2 + c^2 = 1 \\
 \\
ab + bc + ca = t \\
 \\
s = a + b + c = \sqrt {1 + 2t}\end{array}\ \right\|\ .$ Observe that  $ ab + bc + ca\le a^2 + b^2 + c^2\ \implies\ t\in (0,1]\ .$

$ \boxed 1\ \ \sum\frac {b^2 + c^2}{b + c}\ \ge\ (a + b + c) + \frac {\sum (b - c)^2}{2(a + b + c)} = s + \frac {2 - 2t}{2s} = \frac {2s^2 + 2 - 2t}{2s} =$

$ \frac {s^2 + 1 - t}{2s} = \frac {2 + t}{s} = \frac {2\left(a^2 + b^2 + c^2\right) + (ab + bc + ca)}{a + b + c}\ .$

$ \boxed 2\ \ \frac {2 + t}{\sqrt {1 + 2t}}\ge \sqrt 3\ \Longleftrightarrow\ (2 + t)^2\ \ge\ 3(1 + 2t)\ \Longleftrightarrow\ (t - 1)^2\ge 0\ .$[/color][/quote]

Nice proofs, my friends. My proof for this weaker is not longer than you (it's based on Am-Gm+only 2 lines).
See it here (#8):
http://www.maths.vn/forums/showthread.php?t=17775
I only take care of the strongers. Using my hint, we are easily to prove my stronger:
[quote="nguoivn"]
$ \sum\ \frac {a^2 + b^2}{a + b} \geq\ \sqrt {4\sum\ a^2 - 2\sum\ ab + 4\sum\ \frac {a^2b^2}{(a + b)^2}}$
[/quote]
With your stronger, Am-Gm solved it again (of course with different way):
[quote="Virgil Nicula"]$ \sum\frac {b^2 + c^2}{b + c}\ \ge\ \sqrt {3\left(a^2 + b^2 + c^2\right) + 2abc\cdot\sum \frac {b + c - 2a}{a(b + c)}}$ Is it strongest ?![/quote]
 :)

[quote="nuoivn"] [color=darkred]I only take care of the strongers. Using my hint, we are easily to prove my stronger.

With your stronger, Am-Gm solved it again (of course with different way)

$ 1.\blacktriangleright \sum\ \frac {b^2 + c^2}{b + c} \geq\ \sqrt {4\sum\ a^2 - 2\sum\ bc + 4\sum\ \frac {b^2c^2}{(b + c)^2}}$ ([b]@ [u]nuoivn[/u][/b]).

$ 2.\blacktriangleright\ \sum\frac {b^2 + c^2}{b + c}\ \ge\ \sqrt {3\sum a^2 + 2abc\cdot\sum \frac {b + c - 2a}{a(b + c)}}$ ([b]@ [u]Virgil Nicula[/u][/b]).[/color] [/quote]

$ \Delta\equiv\ \left[4\sum\ a^2 - 2\sum\ bc + 4\sum\ \frac {b^2c^2}{(b + c)^2}\right] -$ $ \left[3\sum a^2 + 2abc\cdot\sum \frac {b + c - 2a}{a(b + c)}\right] =$

$ \sum a^2 - \sum 2bc + \sum\frac {4b^2c^2}{(b + c)^2} - \sum\frac {2bc(b + c)(b + c - 2a)}{b + c} =$

$ \sum \frac {a^2(b + c)^2 - 2bc(b + c)^2 + 4b^2c^2 - 2bc(b + c)(b + c - 2a)}{(b + c)^2} =$
$ \sum\frac {a^2(b + c)^2 + 4abc(b + c) + 4b^2c^2 - 4bc(b + c)^2}{(b + c)^2} =$ $ \sum\frac {\left[a(b + c) + 2bc\right]^2 - \left[2(b + c)\sqrt {bc}\right]^2}{(b + c)^2}\ \implies$

$ \Delta= \sum\frac {\left[a(b+c)-2bc\right]^2+4bc\left[a(b+c)-\left(b^2+c^2\right)\right] }{(b + c)^2}$ [b]OR[/b]

$ \Delta=\sum\frac {\left[a(b + c) + 2bc - 2(b + c)\sqrt {bc}\right]\cdot\left[a(b + c) + 2bc + 2(b + c)\sqrt {bc}\right] }{(b+c)^2}\ .$

[color=darkblue]Observe that $ a(b + c) + 2bc - 2(b + c)\sqrt {bc} = (b + c)\left(a - \sqrt {bc}\right) - \sqrt {bc}\cdot\left(\sqrt b - \sqrt c\right)^2$

[size=134][b]What is more strong ? I don't know ![/b][/size] [/color]

[url=http://www.mathlinks.ro/viewtopic.php?p=1390271#1390271][color=red][b]Here[/b][/color][/url] [color=darkblue]I presented a nice and very strong own extension of $ \sum\frac {b^2 + c^2}{b + c}\ge\ \sqrt {3(a^2 + b^2 + c^2)}\ .$[/color]

[quote="Virgil Nicula"]
[size=134][b]What is more strong ? I don't know ![/b][/size] [/color]

[url=http://www.mathlinks.ro/viewtopic.php?p=1390271#1390271][color=red][b]Here[/b][/color][/url] [color=darkblue]I presented a nice and very strong own extension of $ \sum\frac {b^2 \plus{} c^2}{b \plus{} c}\ge\ \sqrt {3(a^2 \plus{} b^2 \plus{} c^2)}\ .$[/color][/quote]
I have said that:
[quote="nguoivn"]
We can't compare 2 results above with another. (So, I think we can't say which one is strongest)
But I solved both of them by Am-Gm.   :)[/quote]
[quote="Virgil Nicula"]
I presented a nice and very strong own extension of $ \sum\frac {b^2 \plus{} c^2}{b \plus{} c}\ge\ \sqrt {3(a^2 \plus{} b^2 \plus{} c^2)}\ .$[/color][/quote]
I saw it. It's also strong but it can be solved easily by SOS  :)
And here is my proof for your ineq in this topic:

Dear nguoivn, sorry  :blush: but I don't understand in your proof why Virgil Nicula stronger inequality is equivalent with

$ \sum a^2\plus{}4\sum \frac{a^2b^2}{(a\plus{}b)^2} \plus{}\frac{16abc\sum ab}{(a\plus{}b)(b\plus{}c)(c\plus{}a)} \geq 2\sum ab \plus{} 4abc\sum \frac{1}{a\plus{}b}$

Thank you very much. :)

$ (\sum\ \frac{a^2\plus{}b^2}{a\plus{}b})^2 \geq\ 3\sum a^2 \plus{} 2abc\cdot\sum \frac {b \plus{} c \minus{} 2a}{a(b \plus{} c)}$
<=> $ (\sum\ 2(a \plus{} b \plus{} c) \minus{} 2\sum\ \frac {ab}{a \plus{} b})^2 \geq\ 3\sum a^2 \plus{} 2abc\cdot\sum \frac {b \plus{} c \minus{} 2a}{a(b \plus{} c)}$
Now, square, we obtain that ineq.You can check it  :)

[quote="nguoivn"]$ (\sum\ \frac {a^2 \plus{} b^2}{a \plus{} b})^2 \geq\ 3\sum a^2 \plus{} 2abc\cdot\sum \frac {b \plus{} c \minus{} 2a}{a(b \plus{} c)}$
<=> $ (\sum\ 2(a \plus{} b \plus{} c) \minus{} 2\sum\ \frac {ab}{a \plus{} b})^2 \geq\ 3\sum a^2 \plus{} 2abc\cdot\sum \frac {b \plus{} c \minus{} 2a}{a(b \plus{} c)}$
Now, square, we obtain that ineq.You can check it  :)[/quote]

Thank you very much, nguoivn. :)

[quote="Virgil Nicula"][b]IF[/b] $ a$ , $ b$ , $ c$ are positive, [b]THEN[/b] there are the following inequalities :$ 1\ \blacktriangleright\ \sum\frac {b^2 + c^2}{b + c}\ \ge\ \sqrt {3\left(a^2 + b^2 + c^2\right)}$ ([u]well-known[/u]).[/quote]
A nice  proof :Because  $\sqrt{1+x} \le1+\frac{x}{2},$
so  \[\sqrt{3(a^2+b^2+c^2)}=\sqrt{(a+b+c)^2+(b-c)^2+(c-a)^2+(a-b)^2)}\]
\[\le{(a+b+c)+\frac{(b-c)^2+(c-a)^2+(a-b)^2}{2(a+b+c)}=\frac{b^2+c^2}{b+c} +\frac{c^2+a^2}{c+a} +\frac{a^2+b^2}{a+b}.}\]

[quote=nguoivn]It isn't the strongest. I don't think it's stronger than the following ineq:
$ \sum\ \frac {a^2 \plus{} b^2}{a \plus{} b} \geq\ \sqrt {4\sum\ a^2 \minus{} 2\sum\ ab \plus{} 4\sum\ \frac {a^2b^2}{(a \plus{} b)^2}}$
Dear my friend, I think you shouldn't analyse your original ineq to SOS form because we'll have to expand it again  :wink:
Please rewrite it.[/quote]

Using SOS :D
[hide] [img]https://i.imgur.com/MAsCJqS.png[/img] [/hide]
[quote=Virgil Nicula]$ \sum\frac {b^2 + c^2}{b + c}\ \ge\ \sqrt {3\left(a^2 + b^2 + c^2\right) + 2abc\cdot\sum \frac {b + c - 2a}{a(b + c)}}$ Is it strongest ?![/quote]

SOS kills it again and again :)
[img]https://i.imgur.com/7lc6gqg.png[/img]


Albatross runs a lottery where there are 100 balls, labeled 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, and 100. But the numbers are weighted, so basically, a certain number is more likely to pop up than another. It just so happens that the most likely ball to be picked is 35.

Frankinfueter decided to buy 1,000 lottery tickets. Every single time, he picked the same number, 100. After he spent 1,000 dollars and only won once, instead of ten times, as he expected, Frankinfueter decided to beat the crap out of Albatross. He figured out that the probability graph was a logarithmic function, peaking at 35 and decreasing rather rapidly away from it. He proceeded to beat the crap out of Albatross once he realized that the chance of getting 100 picked was rather close to 0.

After beating Albatross senseless, Frankinfueter looked at his data from the 1,000 trials, and he realized that 683 of the balls were numbered between 24 and 46. 133 numbers were between 47 and 57, and 135 were between 13 and 23. He pulled out his graphing calculator (a TI-83 he stole from Albatross) and graphed a logarithmic regression function for the number of times each lottery ball was picked. Each tick on the horizontal scale represents 10.

[img]http://img.photobucket.com/albums/v259/drspyder/calcu.jpg[/img]

[b]How many balls should he expect to be numbered greater than 67 if he buys another 1,000 lottery tickets?[/b]

I think I know how to do this, but it's [b]extremely[/b] complicated...

It's not that bad if you do it the special way... then again I might have more experience in statistics than you guys, but I don't know how much you've been exposed to it...


Some Problems from the 2004 Universtiy of Maryland competition:

1. What is the last digit in the decimal expansion of 7^(7^7).
a)1  b)3   c)5   d)7   e)9

2. Two hikers started walking at noon, the 1st from A to B, the 2nd from B to A to A along the same path. They each walked at a constant speed. They met at 3 pm. The first came to B 2.5 hours before the 2nd got to A. When did the second hiker get to A?
a)6:00   b)6:30   c)7:00   d)7:30   e)8:00

3. Peter Rabbit starts hopping at the point (0,0) and finishes at (6,0). Each hop is from a point of the form (x,y) to one of the points (x+1, y :pm: 1). However, he is not allwed to jump on a point (x,y) with y<0. How many paths are there from (0,0) to (6,0)?
a)5   b)6   c)7   d)8   e)9

4. Let k be the largest positive integer, not a multiple of 10, such that removing the last 2 digits of k 2 leaves a perfect sqare. Then:
a)0<k<20   b)20<k<50   c)50<k<100   d)100<k<200   e)200<k

5. How many values of A are there such that the sytem of equations
x2 = y2, (x-A)2+y2=1 has exactly 3 solutions (x,y)
a)0   b)1   c)2   d)3   e)4

I m moving this to Intermediate because no one has answered and it also seems better suited for that forum.

5.

[hide] x and y can be either -1, 0 or 1

thus we have 5 ordered pairs that work

they are

(0, 0) with A = 1

(-1, -1) with A = -1

(1, 1) with A = 1

(1, -1) with A = 1

(-1, 1) with A = -1

But there can only be 3 ordered pairs (x, y) that work.  Right now A is either -1 or 1, and since it is

-1 twice, getting rid of -1 will give 3 ordered pairs (x, y).  So A = 1 is the solution and the answer is

B) 1

[/hide]

2)  Let T be the total distance between A and B, and let x,y be the speeds of the person starting from A and the person starting from B respectively. From the given information we have 3(x+y) = T and T/y-T/x=5/2. Combining we get x/y-y/x=5/6 so the ratio of x:y is 3:2. If we split AB into 5 intervals then after 3 hours the first hiker has traveled 3 intervals while the second has traveled 2, which means in another 2 hours the first hiker will have reached point B. Since this is 2.5 hours earlier than when the first hiker finishes, the second hiker will have used 2+2.5=4.5 hours since 3pm to reach point A. Therefore the second hiker should reach point A at 7:30

Danbert wrote:[hide]
1)  The final digit of  for  is , cycling every four terms.
.
823543 mod 4=3
The third term of the sequence is 3.
The final digit of  is 3.
[/hide]

7^7 = 3^7 = (3^2)^3*3 = (9^3)*3 = (1^3)*3 = 3 mod 4, for those of us who don't have calculators.

Nightflarer and I solved #4 together.  Here is what we did.

The k that we are looking for satisfy  $k^2=100a^2+b$ where $a\geq 0$, $0\leq b<100$ and $k>0$ and they re all integers.  That was probably the crux move right there.

Now we can rearrange it:  $k^2-100a^2=b\Rightarrow (k-10a)(k+10a)=b$.  The part of the problem that states that k isnt a multiple of 10 makes it so $b\neq 0$ and if b did equal 0 there wouldn't be a highest.  It would just be infinity.

But anyway k is maximized if $k-10a=1$ and $k+10a=b$ which gives $2k=b+1$.  Let b=99 and we have k=50.  This is the upper bound.  Now we should just fine a number in the interval (20,50) which works.   We can see that 41 works AWD.

2) [hide] Peter Rabbit must jump to (x+1,y+1) 3 times and (x+1,y-1) 3 times. Denote the first type of jump by U and the second type of jump by D. We can thus represent Peter's actions by a six letter string of Us and Ds. The restriction that he cannot be on a point for which y<0 means that we cannot have more D's in our sequence before U's (e.g. we can't write DDUUUD). So from here, it's pretty easy to set up a little tree chart to count the number of acceptable sequences. (Start with U, it can be followed by U or D, each of those can be followed by etc.) This count gives 5 possible paths for Peter. However, it would be more cumbersome to count this if he had a longer way to go. There's probably a better way to count than to make a chart.[/hide]

Also, no one mentioned that Euler's Generalization of Fermat's Little Theorem (a^phi(n) = 1 (mod n) for (a,n) = 1) also makes 7^(7^7) (mod 10) fall out nicely.

Hello, EFuzzy!

[quote]5. How many values of A are there such that the system of equations
$.\quad x^2 = y^2,\; (x-A)^2 +  y^2 = 1$ has exactly 3 solutions (x,y)?
$.\qquad a)\;0\quad b)\;1\quad c)\;2\quad d)\;3\quad e)\;4$[/quote]

$y = \pm x$ is a pair of lines bisecting the quadrants.

$(x - A)^2 + y^2 = 1$ is a unit circle with its center on the x-axis.

To have exactly 3 intersection, the circle must pass through the Origin.
That is: $A = \pm 1.$

Therefore, answer choice [b](c) 2[/b].


[quote="lightrhee"]Korean team for IMO2005 is announced.

Kang Hyung Jun
Kang Hwan
Nam Ju Kang
Park Do Sung
Jung Young Hun
Jung Won Ho

I don't know how they spell they name, so I just took the spelling that I thought to be the best.[/quote]

Kang Hwan   :arrow:     Ghang Whan

:D Small correction there. But this should be announced in the dedicated topic: http://www.mathlinks.ro/Forum/viewtopic.php?t=24485, therefore topic will be closed down.


logarithms
real analysis
real analysis unsolved

if a positive nonincreasing sequence $\{a_n\}$ satifies $\sum_n a_n=\infty$,

Prove or disprove that $\frac{1}{a_n}=O(n)$? (i.e. $\frac{1}{a_n}<An$ for some fixed $A$)

Or, in other words, you want $a_n>\frac Cn$. This is, of course, not true. Take $a_n=\frac 1{n\ln n}$.


[color=blue][b](i)[/b][/color] Prove that $n\in N$ can be written in form $a^2+b^2$ where $(a,b)=1 \Longleftrightarrow n$ isn't divisible by 4 or by any $4k+3 \in N$
[color=blue][b](ii)[/b][/color]  Use [b][color=blue](i)[/color][/b] to deduce that if a prime $p=4k+3$ divides $a^2+b^2$, then $p|a$ and $p|b$

Well, it's trivial when using unique prime factorisation in gaussian integers.


so, basically, you just try to count using four fours.
an example is here:

1=
4/4x4/4

you can use any math operation possible as lon as it doesnt include an extra number suchas cubic roots...

so yeah.

1=
4/4 x 4/4

2= 4/4 + 4/4

Doubt we'll get any farther than that  :P

hmm

$ 3 \equal{}\frac{4}{4}\plus{}\frac{4}{4}\plus{}\frac{4}{4}$

this is ridiculous.

[quote="speedcuber"]so, basically, you just try to count using four fours.
[/quote]

So 3 should be:

$ 3 \equal{}\frac{(4*4\minus{}4)}{4}$

[quote="speedcuber"]yeah.

4= (4*4)/(4*4)

sorry, i dont have LaTex.
is it compatible with vista?[/quote]

4 should be:

$ 4\equal{}4\plus{}4*(4\minus{}4)$

If you are having trouble with LaTeX, start a post in the LaTeX to see if anyone can help.
http://www.mathlinks.ro/index.php?f=123

yeah, im not paying attention..
sorry...

anyways,
6=

4+4 -4 plus sqrt of four.

$ 11 \equal{} !4\plus{}\frac{4\plus{}4}{4}$

See [url=http://en.wikipedia.org/wiki/Subfactorial]subfactorial[/url].  :)

$ 105\equal{}4!*4\plus{}(!4: :!4)$

i keep on forgetting to attach my signature...

[quote]i keep on forgetting to attach my signature...[/quote]

you can set it to defalt in your profile

$ 107\equal{}!4\cdot!4\plus{}4!\plus{}\sqrt{4}$


The vertices of a regular 7-gon are colored black and white. Prove that there are three vertices of the same color forming an isosceles triangle. More generally, for which regular n-gons is the assertion valid?

When I was small child ( :P ) I was solving this problem for $n=5$ and $n=25$ :)

[hide="what i have so far"]looks pidgeon holey to me. given any point, there are 3 ways to choose an isosceles triangle with a total of 6 points. If that one is painted black, then the only way to have 4 blacks without that one being a vertice of an isosceles is to have all on one side be black. But with that, you can easilly make an isosceles. Same argument for white [/hide]


The ratio of the interior angles of two regular polygons is 3:2. How many such pairs are there?

[quote="arjunlandes"][hide]Let $n$ and $m$ be the number of sides of the two polygons. WLOG, let $n<m$ so that $3\cdot\frac{180(n-2)}{n}=2\cdot\frac{180(m-2)}{m}$. Simplifying, and multipying through by $nm$, we get $nm-6m+4n=0$. Factoring the LHS, we get $(6-n)(m+4)=24$. We know that $n$ and $m$ are natural numbers, and $n<6$ to satisfy the last equality. Looking at factors of $24$ less than $6$, there are $4$, so $\framebox{4}$ pairs of regular polygons work.[/hide][/quote]
There are no $1-gons$ or $2-gons$.

[quote="arjunlandes"]we get $nm-6m+4n=0$. Factoring the LHS, we get $(6-n)(m+4)=24$. ...[/quote]
 That means to find all [i]unordered[/i] pairs of positive integers $(a, b)$ such that $ab = 24$ and one of them is less than six and the other greater than 4. Thus we have the following ordered pairs:
$(1, 24)$, $(2, 12)$, $(3, 8)$.
The corresponding values of $n$ and $m$ are: $(5, 10)$, $(4, 8)$, and $(3, 4)$. Therefore there are $\boxed{3}$ pairs.


probability
modular arithmetic
LaTeX
number theory

4)What is the remainder when 5 to the power of 30 is divided by 7?
I don't want to write out 5 to the power of 30 and divide it by 7...

5)A bag contains 7 white, 9 blue, and 4 red marbles. IF the three marbles are pulled from the bag, what is the probability that two are blue and one is red? Express your answer as a common fraction.
I got 1/18....

6)At how many points do the graphs of the equations y=x to the power of 3 and y=x to the power of 2+4x intersect?
Dunno how to do coordinates...

7)What is the number of distinct ways of arranging the letters in the word AVERAGE?

I'll give you some hints for a couple:

4) use modular arithmetic

7) use overcounting and then correct for it.

5 and 6... I can't really give "hints"...

Hint for the first one: $ 5^6$ is congruent to $ 1 (mod 7)$

For number 7, what I got was since there's 7 numbers it's 7! but then you have to divide by 2! twice because a and e are repeated once. $ \dfrac{7!}{2! \times 2!} = \dfrac{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1 \times 2 \times 1} = 1,260.$ I think I got something wrong though?

Helpful tip: learn [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]LaTeX.[/url]

[hide="4)"]
By a [url=http://en.wikipedia.org/wiki/Fermat's_little_theorem]variation of Fermat's Little Theorem,[/url] $ 5^6\equiv5^{12}\equiv5^{18}\equiv5^{24}\equiv5^{30}\equiv1\pmod{7}$ so the remainder is $ \boxed{1}.$[/hide]

[hide="5)"]
There are $ \binom{20}{3}=1140$ ways to choose the 3 marbles.
There are $ \binom{9}{2}=36$ ways to choose 2 blue marbles.
There are $ \binom{4}{1}=4$ ways to choose 1 red marble.
Therefore, the probability is $ \frac{4\cdot36}{1140}=\boxed{\frac{12}{95}}.$[/hide]

[hide="6)"]
We basically need to find the number of solutions to the equation \[ x=x^2+4x.\] Obviously, there are 2 solutions, which are \begin{align*}x&=0\\&=3.\end{align*} Therefore, there are $ \boxed{2}$ intersection points.[/hide]

[hide="7)"]
Normally, there would be $ 7!$ ways to arrange the letters, but there are 2 e's and 2 a's, so we must divide for overcounting. Therefore, there are \[ \frac{7!}{2!2!}=\frac{5040}{4}=\boxed{1260}\] ways to arrange the letters.[/hide]

In 6, where do you get x^2+4x from?

It's merely 4x+2=3 which has 1 solution.

I thought the question for #6 is, "At how many points do the graphs of $ y \equal{} x^3$ and $ y \equal{} x^{2\plus{}4x}$ intersect?"

If that is indeed the case, then there should be three points of intersection.

$ 1,1 \rightarrow 1 \equal{} 1^3$ and $ 1 \equal{} 1^6$
$ 0,0 \rightarrow 0 \equal{} 0^3$ and $ 0 \equal{} 0^2$
$ \frac{1}{4},\frac{1}{64} \rightarrow .015625 \equal{} .25^3$ and $ .015625 \equal{} .25 ^ 3$

(If I misinterpreted your question due to lack of LaTeX, I am sorry. You should try to format your posts.)

that is what I was thinking as well... Maybe we should wait for K00l to reply, eh?

Well, he doesn't seem to be replying hmm?

So then all problems have been solved right? So mods can lock this thread?

There is no need to lock a thread with answered questions. Locking is necessary when the discussion is off topic.

Well since 5^6 equas 1(mod 7) then 5^30 equals (5^6)^5 (mod 7) then 5^6 equals 1 mod 7 so (5^6)^5  equals (1)^5 mod 7 and 1^5 equals 1 so 5^30 mod 7 equals 1.

um, i pretty sure someone else already posted that solution.
Next time, pls read all above posts.  :)

You've gotta stop this cgyao15. Too many of your posts are meant to "remind" others of forum etiquette. I also see you being impatient with others a lot:
[quote="cgyao15"]dude you might have not noticed the [b] AND[/b] between the 3 and 4....[/quote] They might be new or not have known.

If you really have a problem with someone else's post, pm them instead of posting.