geometry
rectangle
combinatorics proposed
combinatorics

Let $n$ be an integer greater than $1$. For a positive integer $m$, let $S_{m}= \{ 1,2,\ldots, mn\}$. Suppose that there exists a $2n$-element set $T$ such that
(a) each element of $T$ is an $m$-element subset of $S_{m}$;
(b) each pair of elements of $T$ shares at most one common element;
and
(c) each element of $S_{m}$ is contained in exactly two elements of $T$.

Determine the maximum possible value of $m$ in terms of $n$.

Let the sets $ A_{1}, A_{2}, ..., A_{2n}$ be the elements of $ T$. Then for each $ n \in S$ there is exactly one, unique, pair of $ i, j$ such that $ A_{i}\cap A_{j}= \{n\}$ (second and third condition), there are possibly some empty intersections, so the number of intersection is not less then the number of elements of $ S$. In other words: $ \binom{2n}{2}\geq mn$ or $ m \leq 2n-1$. We will show that $ m=2n-1$ is attainable. We construct $ A_{1}, ..., A_{2n}$ as follows. $ A_{1}=\{1,2,...,2n-1\}$. If we have constructed $ A_{1}, ..., A_{k}$ then for $ A_{k+1}$ we take an element on the $ k$-th position from each of the builded sets and for the rest of elements we choose smallest, unused elements of $ S$. So it goes like:
$ A_{1}=\{1,2,...,2n-1\}$
$ A_{2}=\{1,2n,...,4n-3\}$
$ A_{3}=\{2,2n,4n-2...,6n-6\}$
$ A_{4}=\{3, 2n+1, 4n-2, 6n-5, ..., 8n-10\}$
...
$ A_{2n}=\{2n-1, 4n-3, 6n-6, 8n-10, ..., (2n-1)n\}$

It's easy to verify that such construction satisfies all required conditions.

Here's another way of thinking about the problem:
Consider a table with $ mn$ columns and $ 2n$ rows. Let each of the $ 2n$ rows represent each subset that is an element of $ T$, and each of the $ mn$ columns represent a subset; for each element that is an a certain subset of $ T$, in the row representing that subset, mark the relevant box. The given conditions tell us that: Each row has $ m$ marked squares, and each column has $ 2$ marked squares; and it's easy to see that each pair of elements in $ T$ shares at most one common element, iff there are no rectangles in the figure. So we have to find the largest $ m$, so that there are no rectangles.
Suppose $ m \geq 2n$. WLOG (for convenience) the first row in the figure has the first $ m$ squares marked. Also WLOG, that the 2nd row has the 1st square marked, the 3rd row has the 2nd square marked ... the 2n-th row has the $ 2n\minus{}1$-th square marked (to satisfy the condition that each column has 2 marked squares). Consider the $ 2n$-th row. We must have one marked square apart from the one already marked, but we easily see that marking any of the remaining of $ 2n\minus{}1$ squares will complete a rectangle - which gives us a desired contradiction.
When $ m \equal{} 2n\minus{}1$, greedy principle does the job.  Mark the first $ 2n\minus{}1$ squares of the first row. Starting from the $ 1$-st square in the $ 2$-nd row, tick the diagonal going down. Now mark, the next set of $ 2n\minus{}2$ squares, and mark the diagonal downwards again - keep doing this, and some simple computations show that when you finish you just finish filling the table.

Oh. This problem is very easy. If you have a graph with the vertexs is all sets. If two vertex are conected, we have two sets respect to two sets have common elements. So by the given conditions, we have each sides is respect to elements of the set
$S_m$.
Easily, we have $m_{max}=2n-1$

I got a solution similar to DangChien's, but the write-up is a bit more formal.

[hide="Solution"]
I claim the maximum value of $m$ is $2n-1$. First, we show that $m \le 2n-1$. By condition (b), there are at most $\binom{2n}{n}$ elements of $S_m$ which are in 2 sets in $T$. However, by condition (c) every element of $S_m$ is in 2 sets in $T$, so $\left|S_m\right| = mn \le \binom{2n}{n} = \frac{(2n)(2n-1)}{2}=n(2n-1)$, and $m \le 2n-1$.

Now, to see that $2n-1$ is achievable, consider a $K_{2n}$. Note that this $K_{2n}$ has $n(2n-1)$ edges, so we label the edges with distinct elements of $S_{2n-1}$. Now, let $T$ be the family of sets formed from taking each vertex and creating the set of labels of incident edges. Note that any two sets in $T$ have exactly one element in common -- the label of the edge between them -- so condition (b) is satisfied. Note also that each element of $S_m$ is in exactly two elements of $T$: the two endpoints of the edge with the given label. Thus condition (c) is satisfied. Condition (a) is also satisfied because every vertex in a $K_{2n}$ has degree $2n-1=m$, so every element of $T$ has $m$ elements. Finally, since a $K_{2n}$ has $2n$ vertices, $|T|=2n$, as desired.
[/hide]

[b]ANSWER:[/b] The maximum possible value of $m$ is $m=2n-1$.

[b]PROOF:[/b] Let $T=\{A_1,A_2, \dots ,A_{2n}\}$. Consider a $mn \times 2n$ matrix with all entries either $0$ or $1$, such that $a_ij=1$ iff $j \in A_i$. Call a triplet of the form $(i,A_j,A_k)$ [i]nice[/i] if $i \in A_j$ and $i \in A_k$. We count the number of [i]nice[/i] triplets in two different ways:-
[list][*] First fix an element $i$. Then, as $i$ is present in exactly two unique sets $A_j$ and $A_k$ according to problem condition $(c)$, so we get that $$\text{Total number of nice triplets}=\sum_{i=1}^{mn} \binom{2}{2} =mn$$
[*] This time, we choose two sets $A_j$ and $A_k$ in $\binom{2n}{2}$ ways. Then, according to condition $(b)$, these two sets have atmost one common element, which gives $$\text{Total number of nice triplets} \leq \binom{2n}{2}$$ [/list]
Thus, we get that $mn \leq \frac{2n(2n-1)}{2} \Rightarrow m \leq 2n-1$. Now, all that needs to be done is show that this bound is achievable.

Let us take $S=\{1,2, \dots ,n(2n-1)\}$ for $m=2n-1$. Then we define the elements of $T=\{A_1,A_2, \dots ,A_{2n}\}$ as follows:-
\[\displaystyle A_i = \left\{ \begin{array}{lr}  \{x:x=(p-1)(2n-1)+q \cup (n-1)(2n-1)+i \text{ for } 1 \leq p \leq n \text{ and } 1 \leq q \leq 2\} & \text{when}\ \ 1 \leq i \leq n-1 \\ \\ \{x:x=a(2n-1) \cup (n-1)(2n-1)+b \text{ for } 1 \leq a \leq n \text{ and } n \leq b \leq 2n-1\} & \text{when}\ \ i=n \\ \\ \{x:x=(i-n-1)(2n-1)+j \text{ for } 1 \leq j \leq 2n-1\} & \text{when}\ \ n+1 \leq i \leq 2n \end{array} \right.\]
Then one can easily see that this set $T$ satisfies all the three conditions given in the problem statement. Hence, done. $\blacksquare$

[hide=Solution]
By PIE, note that $|A_1 \cup A_2 \cup .... A_{2n}|=2mn-\binom{2n}{2}\leq mn$, so solving for $m$ in terms of $n$ we get that $m\leq 2n-1$.

To show that this is attainable, consider the following construction of $A_i$:

$A_1={1,2,3,...m}$
$A_2={m+1,m+2,...2m}$
....
$A_n={(n-1)m+1,(n-1)m+2,...mn}$
$A_{n+1}={1,m+1,2m+1,...(n-1)m+1}$
$A_{n+2}={2,m+2,2m+2....(n-1)m+2}$
...
$A_{2n}={m,2m,...mn}$

Clearly, the intersection between any two sets in this list is less than or equal to $1$, and each element occurs twice, so $m=2n+1$ is indeed attainable and is the desired maximum. $\square$

[/hide]

Easy problem. Solved with [b]geoking[/b] and myself.

The bound is pretty easy to get consider a $\binom{2n}{2} \text{x } mn$ incidence matrix, where the rows represent the pair of $A_{i}$ and the column $j$ represents the number $j$. If the number $j$ is in the i'th pair, we put 1 in the $(i,j)$ cell. Note that for each element, there is a pair of set and the intersection of any set is atmost 1. So we get $\binom{2n}{2} \geq mn$, by further simplifying one can get $m \leq 2n-1$. For construction, note that this incidence matrix is a square, and now diagonally fill $1$ to finish.


A point $M$ is taken on the perpendicular bisector of $AC$ in acute triangle $ABC$. $M,B$ are on the same side of $AC$. $\angle BAC=\angle MCB$, $\angle ABC+\angle MBC=180$. Find $\angle BAC$.

hmm..not sure??

[hide]
let $\alpha=\angle{BAC}$

$\angle{MCB}+\angle{ABC}= 180$
$\angle{MCB}+(180-2\alpha) = 180$
$\angle{MCB}=2\alpha$

EDIT: ya thats what i have so far..
EDIT2: I thought M was on $\overline{AC}$..so its completely wrong..just ignore
it might be wrong because my diagram couldve been wrong..but could someone please verify..
[/hide]

Here is my attempt at the diagram.

EDIT: By the way, why can't I ever get any of my diagram's to look nice? They always look so blurry.

k i c i didnt know $M$ could be any point on the perpendicular bisector...even though it says in the question  :blush:


I don't know if my team is going to make it.

The top how many teams advance?

The tope how many individuals not on those teams advance?

Please and Thank You for your time.

From mathcounts.org
[quote]The winning team will advance to the State Competition. Additionally, the two highest-ranking competitors not on the winning team (who may be registered as individuals or as members of a team) will advance to the State Competition. This is a minimum of six advancing Mathletes (assuming the winning team has four members). Additional teams and/or Mathletes also may progress at the discretion of the state coordinator.[/quote]

Here in PA, the top team and top four individuals all advance to the state competition.

To get to the nationals, you have to be in the top 4 as an individual in the state competition.

depends what state youre in i think

i live in pennsylvannia. we do the 1st place teams advance and the top 4 indiviuals or more advance. depends on the team. then the top scoring 2nd place teams advance.

EDIT: hey you live in pennsylvannia too

Well, at our school, top 5 teams and top 5 individual from each chapter go to state. (North Carolina)

In NE, I think it's the top 3 to 4 teams advance from Chapter and first 4 individuals. We're a state that likes 4. :lol:

I should have specified for chapter, so thank you all.

I need to practice more...


search
geometry
geometric transformation

[quote="maokid7"]My problem is that I have 2 arrays, $ A$ and $ B$. Lets say that one holds the values $ x$ while the other holds the values of $ f(x)$. so the first elements of array $ A$ hold the first $ x$ and the first element of array $ B$ is its corresponding $ f(x)$. I am performing a search on list $ B$ to see if it has a certain number (let us assume that it does). How would I be able to find out what element in array $ A$ it goes with?[/quote]

That would depend entirely on how the arrays are set up.

For example, say the arrays were

A[0]=1
A[1]=2
A[2]=3
A[3]=4

and

B[0]=1
B[1]=4
B[2]=9
B[3]=16

If you use this set up and know the "index" when you search then the A index will be identical.

You could also use a struct or a multi-dimensional array to solve the problem

Its better to just use the first idea mentioned above.

When you search through B, don't just store the maximum element, but the position of the maximum element as well. Whenever you find something bigger as you search, update both. Then at the end, you know the position, so just look that up in A.

[quote="maokid7"]ok here is an example

A={1,3,5,6,24}
B={2,5,6,12,0}.

so i search B to find its max element (12) and then need a way to show that this corresponds to 6 in A.

how would i use multi-dimensional arrays or structures to solve this?
I just read about how to make structures so i dont have much experience using them so this would be a good place to learn i suppose.[/quote]

Here's the code the maximum element

k=0;
for(i=1;i<N;i++)
  k=(b[i]>b[k])?i:k;
cout<<a[k];

[quote="maokid7"]thanks. but could you please explain the following line?

 k=(b[i]>b[k])?i:k;[/quote]

Basically, in C (and a few other languages C++, PHP, etc)

You can use an operator which has the effect of

(STATEMENT) ? RETURN IF TRUE : RETURN IF FALSE ;

It's the same as:

[code]
if (b[i] > b[k]) {
	k = i;
} else {
	k = k;
}
[/code]

(Yes, the else part is not necessary, this is just a literal translation of the line of code you asked about.)

[quote="maokid7"]thanks. but could you please explain the following line?

 k=(b[i]>b[k])?i:k;[/quote]

Its called ternary. The above ppl have explained it well enough but anyways, it can be used for simple if else jobs. The code is much more compact this way. Other than this, you are always free to use if else.


x^y>y^x if y>x>e, and x^(y^c)>(y^(x^c) if y>x>e,c>1. Therefore power expression increase when change (x,y) to down x<y. Therefore left power expession is maximum.


A ball is shot from a cannon. if velocity=v, height=h, time=t, and angle a ball is shot=$ \theta$, then write a formula where the time is the independent variable and the height is the dependent variable.

[hide="i am getting"]$ h\equal{}\minus{}4.9 \frac{t^2} {v}\plus{}t*\tan{\theta}$[/hide]

I think you're trying to make this too complicated.  Just take the regular equation,
$ d\equal{}v_{0}t\plus{}\frac{1}{2}at^{2}$
and modify the variables.  The initial velocity is $ v\sin{\theta}$ and the acceleration (or deceleration) is $ \minus{}g$ so plug it in:
$ h\equal{}v\sin{\theta}t\minus{}\frac{1}{2}gt^{2}$


function
calculus
limit
geometry
3D geometry
algebra
factorization

i cant seem to figure out this limit

lim x-> 0 [(1+x)^(1/3) - (1-x)^(1/3)] / x

by plugging in values I know the limit is 2/3 as x approaches 0, but I just cannot figure out how to modify the function to show such a value  :blush:

thanks

Wrong forum; next time, try the calculus section.

Anyway, we have $ \lim_{x\to0}\frac{(1\plus{}x)^\frac{1}{3}\minus{}(1\minus{}x)^\frac{1}{3}}{x}$. If we simply plug in $ x\equal{}0$, we get $ \frac{0}{0}$. This indicates that we can use L'Hopital's rule.

This means that the limit is the same as $ \lim_{x\to0}{\frac{1}{3}(1\plus{}x)^{\minus{}\frac{2}{3}}\plus{}\frac{1}{3}(1\minus{}x)^{\minus{}\frac{2}{3}}}$

Simply plugging in $ x\equal{}0$ gives $ \frac{1}{3}\plus{}\frac{1}{3}\equal{}\boxed{\frac{2}{3}}$.

is that the only way?

because i just started precal and ive never heard of that rule at all..

Oh, sorry. L'Hopital's rule involves differential calculus, so you wouldn't have heard of it in precalc :wink:

It took me a while to mess around with the difference of cubes factorization, but I got something:

Recall that $ x^3\minus{}y^3\equal{}(x\minus{}y)(x^2\plus{}xy\plus{}y^2)$. If we let $ x\equal{}a^\frac{1}{3}$ and $ y\equal{}b^\frac{1}{3}$, we get:
\[ a\minus{}b\equal{}(a^\frac{1}{3}\minus{}b^\frac{1}{3})(a^\frac{2}{3}\plus{}a^\frac{1}{3}b^\frac{1}{3}\plus{}b^\frac{2}{3}).\]

If we let $ a\equal{}(1\plus{}x)$ and $ b\equal{}(1\minus{}x)$ we get:
\[ 1\plus{}x\minus{}1\plus{}x\equal{}2x\equal{}((1\plus{}x)^\frac{1}{3}\minus{}(1\minus{}x)^\frac{1}{3})((1\plus{}x)^\frac{2}{3}\plus{}(1\plus{}x)^\frac{1}{3}(1\minus{}x)^\frac{1}{3}\plus{}(1\minus{}x)^\frac{2}{3}).\]

Which means we can multiply our limit by $ \frac{(1\plus{}x)^\frac{2}{3}\plus{}(1\plus{}x)^\frac{1}{3}(1\minus{}x)^\frac{1}{3}\plus{}(1\minus{}x)^\frac{2}{3}}{(1\plus{}x)^\frac{2}{3}\plus{}(1\plus{}x)^\frac{1}{3}(1\minus{}x)^\frac{1}{3}\plus{}(1\minus{}x)^\frac{2}{3}}$ to get:

\[ \lim_{x\to0}\frac{2x}{x((1\plus{}x)^\frac{2}{3}\plus{}(1\plus{}x)^\frac{1}{3}(1\minus{}x)^\frac{1}{3}\plus{}(1\minus{}x)^\frac{2}{3})}.\]

Cancel out the $ x$ in the numerator and denominator, and plug in $ x\equal{}0$ to get $ \frac{2}{1\plus{}1\plus{}1}\equal{}\boxed{\frac{2}{3}}$.

Right, that's the smart way of doing it :)

Alternately, you can note that $ 1\plus{}x \approx \left( 1 \plus{} \frac{x}{3} \right)^3$ for small $ x$ (the terms of order higher than 1 become insignificant). You can show that this is the case more formally using calculus.


A player chooses one of the numbers $ 1$ through $ 4$. After the choice has been made, two regular four-sided (tetrahedral) dice are rolled, with the sides of the dice numbered $ 1$ through $ 4$. If the number chosen appears on the bottom of exactly one die after it is rolled, then the player wins $ \$1$. If the number chosen appears on the bottom of both of the dice, then the player wins $ \$2$. If the number chosen does not appear on the bottom of either of the dice, the player loses $ \$1$. What is the expected return to the player, in dollars, for one roll of the dice?

$ \textbf{(A)}\ \minus{}\frac{1}{8}\qquad
\textbf{(B)}\ \minus{}\frac{1}{16}\qquad
\textbf{(C)}\ 0\qquad
\textbf{(D)}\ \frac{1}{16}\qquad
\textbf{(E)}\ \frac{1}{8}$

[hide]There is $ \frac {1}{4}$ chance that one dice will roll chosen number, and $ \frac {1}{2}$ chance that one of two dice will roll chosen number earning $ 1\$$. There is $ \frac {1}{4}*\frac {1}{4} \equal{} \frac {1}{16}$ chance that both dice will roll chosen number earning $ 2\$$. For each dice there is $ \frac {3}{4}$ chance not to roll chosen number which makes total of $ \frac {3}{4}*\frac {3}{4} \equal{} \frac {9}{16}$ chance to loose $ 1\$$. Hence the expected return for the player is $ \frac {1}{2}*1\$ \plus{} \frac {1}{16}*2\$ \minus{} \frac {9}{16}*1\$ \equal{} \frac {1}{16}\$$. The answer is $ \boxed{\textbf{(D)}}$.[/hide]

[hide="Huh..."]
The probability of the number appearing $ 0$ times is $ P(0)\equal{}\frac{3}{4} \times \frac{3}{4}\equal{}\frac{9}{16}$.  The probability that it appears $ 1$ time is $ P(1)\equal{}2 \times \frac{1}{4} \times \frac{3}{4}\equal{}\frac{6}{16}$.  The probability that it appears $ 2$ times is $ P(2)\equal{}\frac{1}{4} \times \frac{1}{4}\equal{}\frac{1}{16}$.  So the expected return is $ P(0) \times (\minus{}1)\plus{}P(1) \times (1)\plus{}P(2) \times (2)\equal{}\frac{\minus{}9\plus{}6\plus{}2}{16}\equal{}\minus{}\frac{1}{16}$, or $ \boxed{\textbf{(B)}}$.
[/hide]

Here's how I did it:
There's 1/4 of chance on each dice that you get the designated #, 3/4 of chance on each dice that you do not get the designated #, so:
the expected return is:

(1/4)(1/4)*2+(1/4)(3/4)*2*1+(3/4)(3/4)*(-1)= -.0625

Or (B).

$\text{Please remember to hide your answers.}$


[hide]

We show $(n!)^2 > n^n$ by induction. The base case, for $n=3$ is true, since $36 > 27.$ Now we assume $(n!)^2 > n^n$. We show $((n+1)!)^2 > (n+1)^{n+1}$. But $((n+1)!)^2 = (n+1)^2(n!)^2 > (n+1)^2n^n$.

It suffices to show

$(n+1)^2n^n > (n+1)^{n+1}$ or
$(n+1)(n^n)>(n+1)^n$ or
$n+1 > (\frac{n+1}{n})^n$
$n+1 > (1+\frac{1}{n})^n$

The RHS is increasing and tends to $e$. The right hand side, with $n\geq 3$ is larger, and we are done.[/hide]

[quote="frt"]Prove:

If $n$ is a positive integer such that $n\ge 3$, then $(n!)^2>n^n$[/quote]

Induction

Initial case: $n= 3$
$(3!)^2 = 36$
$3^3 = 27$
So $(3!)^2> 3^3$

Induction Hypothesis:
Assume that this inequality holds for a fixed $k\ge 3$.
Induction step: $(k!)^2 > k^k$

We must show it holds for $k+1$
$(k+1)!^2 > (k+1)^(k+1)$
$((k+1)^2)((k!)^2)> ((k+1)^ k) (k+ 1)$
$ (k+1)((k!)^2) > (k+1)^ k$

Never mind. I am stuck. Anyone want to complete it for me? Or maybe it isn't simple induction at all.

Edit: nice solution, tetrahedon.

I used induction...
We assume that $(n!)^2>n^n$
Now to prove that $((n+1)!)^2>(n+1)^{n+1}$
It is sufficient to prove that $(n+1)^2 \ge (n+1)(\frac{n+1}{n})^n$
$n+1 \ge (\frac{n+1}{n})^n$
We are dealing with $n\ge3$.  However, the upper bound on $(\frac{n+1}{n})^n$ is $e$, and $e<3$.  Thus, the inequality is proven.
For a proof that $e<3$,
$(1+\frac{1}{n})^n=1+1+\frac{n(n-1)}{n^22}+\frac{n(n-1)(n-2)}{n^33!}+...+\frac{1}{n^n}<1+1+\frac{1}{2}+\frac{1}{3!}+\frac{1}{4!}+...$ too lazy to go on  :|
but you get the point.  At each step, the amount you must add will be greater than or equal to what you just added, but the things you add keep getting smaller.

Edit: Oh, my solution is identical to tetrahedr0n's :D .Looks like many induction approaches have been used.  You know, as they say, great minds think alike ;) and fools seldom differ  :P

[quote="K81o7"]Edit: Oh, my solution is identical to tetrahedr0n's  .Looks like many induction approaches have been used. You know, as they say, great minds think alike  and fools seldom differ [/quote]

Haha... I wonder if there is a simple non-induction solution. Maybe we should try proving the square of the $\gamma$ function is strictly greater than the $n^n$ function.


geometry
perimeter
trapezoid
Pythagorean Theorem


If $ p$ is a prime, prove that:

$ p\,|\,a^{p}+(p-1)!a$ and $ p\,|\,(p-1)!a^{p}+a$

for any integer $ a$

[quote="sludgethrower"]If $ p$ is a prime, prove that:

$ p\,|\,a^{p}+(p-1)!a$ [/quote]

In other words,
$ a^{p}+(p-1)!a\equiv 0 (\bmod p)$
By Fermat's theorem we got,
$ a+(p-1)!a\equiv 0 (\bmod p)$
Factor,
$ a[(p-1)!+1]\equiv 0 (\bmod p)$
But,
$ (p-1)!+1\equiv 0 (\bmod p)$ by Wilson's theorem.

For the second one, $ (p-1)!\cdot a^{p}\equiv-a\pmod{p}$ by Wilson's and Fermat's Theorems, the result follows.


Help!
Been set this one to look at:

Determine all pairs of positive integers (a,b) such that a^2/(2ab^2-b^3+1) is a positive integer.

After a lot of scratching around all i can establish is that b=2a and this generates a load..but can't see an easy way into it. Not sure our teacher can do it!

It certainly is a hard problem.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=262#p262]IMO 2003, problem 2[/url]


[b][u]Problem[/u][/b]: A bag contains some red balls and blue balls. If we remove 1 red ball from the bag, then $\frac{1}{7}$ of the remaining balls in the bag are red. If, instead of 1 red ball, we remove 2 blue balls from the ball, then $\frac{1}{5}$ of the remaining balls in the bag are red. How many balls are there in the bag at first?

[quote="mathisfun1"]x = # of red balls, y = # of blue balls.

$\frac{x-1}{x+y-1}= \frac{1}{7}, \ \frac{y-2}{x+y-2}= \frac{1}{5}$

This system has no integer solutions.[/quote]

It should be $\frac{y-2}{x+y-2}= \frac{4}{5}$ instead of $\frac{y-2}{x+y-2}= \frac{1}{5}$. And system has a solution $(x,y)=(4,18)$ so at first there was $22$ balls.


geometry
circumcircle
geometric transformation
homothety
geometry proposed

Ignore the circumcircle $\Omega$. Let E be any point on the line AD different from the points A, D, not necessarily lying on the circumcircle $\Omega$. Let $X \equiv BE \cap AC$ be the intersection of these 2 lines. Let the parallel to BC through the point X meet the line AB at a point Y'. The triangles $\triangle ABC \sim AY'X$ are centrally similar with similarity center A.

$\frac{AY'}{AB} = \frac{AX}{AC},\ \ \frac{AB + BY'}{AB} = \frac{AC + CX}{AC},\ \ 1 + \frac{BY'}{AB} = 1 + \frac{CX}{AC},\ \ \frac{BY'}{BA} = \frac{CX}{CA}$

Since D is the midpoint of the segment BC, the line AD intersects the segment XY' centrally similar to the segment BC at its midpoint Z, i.e., $\frac{ZX}{ZY'} = 1$. Consequently,

$1 = \frac{BY'}{BA}\cdot \frac{CA}{CX} = \frac{BY'}{BA}\cdot \frac{CA}{CX} \cdot \frac{ZX}{ZY'}$

By Ceva's theorem, the cevians AZ, XB, Y'C are concurrent. Since the line $AZ \equiv AD$ intersects the line $BX$ at the point E by the definition of the line BX, the 3 cevians concur at this point. As a result, the points Y', E, C are collinear, which means that the line CE intersects the line $AB \equiv AY'$ at the point Y'. Since $Y \equiv CE \cap AB$ by the definition of the point Y, the points $Y' \equiv Y$ are identical and the lines $XY \parallel BC$ are parallel.

Let A, B, C, D, E, F be some points which belong to the same circle. Prove that the following relations are equivalently:
1. AB.CD.EF=BC.DE.FA;
2. AC.BD.FE=BC.DF.AE;
3. AB.CE.DF=AC.BF.DE.

[quote="levi"]Let ABC be a triangle inscribed in the circle  $\Omega$ and the middlepoint D of the side [BC]. I note the intersections:
$E\in AD\cap \Omega , \ \  X\in BE\cap AC ,\ \ Y\in CE\cap AB$ Prove that XY || BC.[/quote]

The problem is absolutely easy. Like Yetti, I note that the circle $\Omega$ is unnecessary, so we will solve the following more general problem:

[color=blue][b]Problem.[/b] Let ABC be a triangle, and D the midpoint of its side BC. Let E be an arbitrary point on the line AD. Denote $X=BE\cap AC$ and $Y=CE\cap AB$. Prove that XY || BC.[/color]

[i]Solution.[/i] Let the parallel to the line BC through the point X meet the line AB at Y'. Then, XY' || BC, or, equivalently, Y'X || BC. Once we succeed to show that Y' = Y, we are done, since then, XY' || BC becomes XY || BC, and thus the problem is solved.

So it remains to show that Y' = Y. Well, let $E^{\prime}=BX\cap CY^{\prime}$, and let L be the midpoint of the segment XY'. Then, since XY' || BC and $BX\cap CY^{\prime}=E^{\prime}$, there exists a homothety with center E' mapping the point X to the point B and the point Y' to the point C. Hence, this homothety must map the midpoint L of the segment XY' to the midpoint D of the segment BC. It follows that the points L and D are collinear with the center E' of this homothety. In other words, the line LD passes through the point E'. On the other hand, since Y'X || BC and $BY^{\prime}\cap CX=A$, there exists a homothety with center A mapping the point Y' to the point B and the point X to the point C. Hence, this homothety must map the midpoint L of the segment Y'X to the midpoint D of the segment BC. It follows that the points L and D are collinear with the center A of this homothety. In other words, the line LD passes through the point A.

So, now, the line LD must pass through the points E' and A; thus, the points L, D, E' and A are collinear, and thus the point E' lies on the line AD. Therefore, $E^{\prime}=AD\cap BX$. But we know that $E=AD\cap BX$. Hence, E' = E. Consequently, since $Y^{\prime}=CE^{\prime}\cap AB$ and $Y=CE\cap AB$, we must have Y' = Y, and we are done.

  Darij

I choose a miserable particular case of a general problem. This general problem is placed at http://www.mathlinks.ro/Forum/viewtopic.php?t=43170


vector
linear algebra
linear algebra solved

I was hoping some one out there could provide a description and breakdown of the 8 axioms of vector spaces because I am having difficultly visualizing vector spaces.

Also visualizing a subspace.

Giving an example vector space problem carried out step by step through whether something is a subspace of the vector space and a step by step of how that was determined would be great.

Sorry for asking what seems like a maybe simple question, but I am having a lot of difficulty putting the words of vector spaces and subspaces to numbers and seeing it.

Thanks.

[url]http://en.wikipedia.org/wiki/Vector_space[/url] Wikipedia is always a good place to start. :wink:

you can go to [url]http://www.cs.ut.ee/~toomas_l/linalg/lin1/node5.html[/url]


So this is kinda confusing if you haven't reviewed exponents(especially negative ones) for a while
$ 8^{\minus{}1}\equal{}\frac18$
$ 4^{\minus{}1}\equal{}\frac14$
so $ \frac{\frac18}{\frac14}\equal{}\frac18*4\equal{}\frac12$
So $ a^{\minus{}1}\equal{}\minus{}\frac12$
now you know that $ a\equal{}\minus{}2$ (obvious)


Suppose each element $A_{k}$ in set $\{A_{k}\}_{k \in \mathbf{N}}$ is (Lebesgue) measurable and $\sum_{k=1}^{\infty}\mu(A_{k}) <+\infty$. Prove that $\mu(limsup_{k\to \infty}A_{k}) = 0$.

Thanks!

$\limsup_{k\to\infty}A_{k}\subset \bigcup_{i=n}^{\infty}A_{k}$ for any $n$. And $\mu\left(\bigcup_{i=n}^{\infty}A_{k}\right)\le \sum_{i=n}^{\infty}\mu(A_{i})\to 0$ as $n\to\infty$.


calculus
integration
trigonometry
logarithms
calculus computations

solve :D

$ \int_0^\infty \; \frac{ \cos\; x \; \cdot \ln \; x} { 1\plus{}x^2} \; \; dx$

hello, Mathematica says the integral is
$ \frac{\pi\left(e^2ExpIntegralEi(\minus{}1)\minus{}ExpIntegralEi(1)\right)}{4e}$.
Sonnhard.


The deadline's past. So when do I know what I got???

I want to know :lol:

The results for Round 1 will not be available until early November.  It takes time to read this many solutions.  We haven't even filed all the papers yet, much less started to grade them.

dude do you have any idea how long it takes for 4-5 graders to read over a thousand solutions? In fact, 3 weeks is pretty quick in my opinion. Oh, not to mention that this contest is free.

You sit there for a few days anxious to know what you got, then you kind of forget about it. A few weeks later you get the pleasant surprise of seeing what you got when you're not even thinking about it anymore. That's the way it always works. You get used to it.

Maybe we should have some sort of sticky thread that everyone wondering how long it will take, freaking because they didn't send in the forms [i]before[/i] the solutions, etc. could read so they didn't have to ask every round or year...

I think they have more than 4-5 graders, but nonetheless there's a lot of filing of paperwork, distribution of solutions to graders, grading, collection of grades/comments, input of grades/comments into the database, and then finally you can see what you got.  Unlike USAMO, you actually get to see what you got on each problem and a little comment.  There's also more competitors than USAMO.  Though, the problems are easier to grade than USAMO problems, I'd think.


inequalities
algebra
polynomial
inequalities unsolved

let x,y,z positifs reals numbers such that 1/x+1/y+1/z=1 then de folloing inequality holds:
(1-1/(1+x²))(1-1/(1+y²))(1-1/(1+z²)) sup 1/2

[quote="diane"]$x,y,z \in \mathbb{R}| \frac1x+\frac1y+\frac1z=1$
Prove: $\left( 1-\frac{1}{1+x^{2}}\right)\left( 1-\frac{1}{1+y^{2}}\right)\left( 1-\frac{1}{1+z^{2}}\right)\geq \frac12$[/quote]
at least that is how I understand it...

Well at first question contains x,y,z are positive real number

but translater mistook condition of x,y,z ^^  :D

Oops my mistake!!! (sorry)
In the third line, you use $\prod (1-k)\geq 1-\sum k$. Does this inequality have a name, or is it just duh after multilying out?

That's very handy, how is it proven? Is it generalized polynomial work?

Nice proof on not enough sleep! It's probably just the textbook proof, but good job!

[quote="Albanian Eagle"]yes, it has a name.
[b]Bernoulli Inequality[/b]
let $x_{1},x_{2},..., x_{n}$ be real numbers either all in $[-1,0]$ or all in $[0,\infty)$ then:
$\prod_{i=1}^{n}(1+x_{i})\geq 1+\sum_{i=1}^{n}x_{i}$[/quote]

Hmm, it's  new for me to have this Bernoulli's. The Bernoulli I know is
$(1+x)^\alpha\geq 1+\alpha x$ if $x>-1$ and $\alpha<0$ or $1<\alpha$. If $0<\alpha<1$, then the sign is reversed.

The original inequality is equivalent to: \[(1+a^{2})(1+b^{2})(1+c^{2}) \leq 2 \big| a,b,c \geq 0,\ a+b+c=1.\] I think it's not very hard.

[quote="me@home"]That's what i was going to say when he called it bernoulii's...
anyway on your transformation lovasz can you prove that with only am-gm or am i mistaken?[/quote] \begin{eqnarray*}(1+a^{2})(1+b^{2})(1+c^{2})&=&(1+a^{2})(1+b^{2}+c^{2}+b^{2}c^{2}) \\& =& (1+a^{2})(a^{2}-2a+2+(bc)^{2}-2bc) \quad (bc \ge 0) \\ &\leq& (1+a^{2})(a^{2}-2a+2) \\ &=&(a-1)(a^{2}-a+2)+2 \\ &\leq& 2. \end{eqnarray*} P. complete.


integration
calculus
trigonometry
logarithms
calculus computations

I'm not sure if it's a good section to post these problems...

[color=green][The "Calculus Computations and Tutorials" forum would be the right place. Moved. - Moderator][/color]

[quote="tranthanhnam"]$\boxed{\ I_{2}=\int_{0}^{\infty }\frac{\arctan x}{(1+x^{2})^{2}}\ dx\ }\ .$[/quote]
$\left\{\begin{array}{c}I_{2}=\int_{0}^{\infty}\frac{\arctan x}{1+x^{2}}\cdot \phi '(x)\ dx\\\\ t=\phi (x)=\arctan x\ (x=\tan t)\ ,\ \phi '(x)=\frac{1}{1+x^{2}}\end{array}\right\|$ $\Longrightarrow$ $I_{2}=\int_{0}^{\frac{\pi}{2}}t\cos^{2}t\ dt\mathrm{\ ...\ }$


inequalities
geometry
angle bisector
inequalities proposed

Given acute triangle $ ABC$ and three altitudes $ h_{a},h_{b},h_{c}$ and medians $ m_{a},m_{b},m_{c}$,area $ S$ prove that
\[ m_{a}h_{b}+m_{b}h_{c}+m_{c}h_{a}\ge 3\sqrt{3}S \]
Compare with problem (not of me)
\[ l_{b}l_{c}+l_{c}l_{a}+l_{a}l_{b}\geq 3\sqrt{3}S \]
with angle bisector $ l_{a},l_{b},l_{c}$.

[color=purple]The 2nd problem was proposed in Crux 1996 vol 3.  :D.
   The following problem from Viet Nam  :D . Given two triangle and six angle bisectors of them $ l_{a},l_{b},l_{c},l_{x},l_{y},l_{z}$. Two triangles have areas S and S'. Prove that
   $ l_{a}(l_{y}+l_{z})+l_{b}(l_{z}+l_{x})+l_{c}(l_{x}+l_{y})\ge 6\sqrt{3SS'}$
 Have fun gemath !  :P Along time don't meet you  :) [/color]

[color=purple] About the 1st gemath i can prove more :
   $ xm_{a}h_{b}+ym_{b}h_{c}+zm_{c}h_{a}\ge 3\sqrt{xy+yz+zx}S$
  :D  :D  :P
   Have fun ! [/color]

[quote="evarist"]The 2nd problem was proposed in Crux 1996 vol 3.  :D.
   The following problem from Viet Nam  :D . Given two triangle and six angle bisectors of them $ l_{a},l_{b},l_{c},l_{x},l_{y},l_{z}$. Two triangles have areas S and S'. Prove that
   $ l_{a}(l_{y}+l_{z})+l_{b}(l_{z}+l_{x})+l_{c}(l_{x}+l_{y})\ge 6\sqrt{3SS'}$
 Have fun gemath !  :P Along time don't meet you  :)[/quote]
Nice problem but really of Viet Nam ?  :D it is corollary of two lemmas
$ \sum xa^{2}\ge\sqrt{xy+yz+zx}S$ and $ \sum l_{b}l_{c}\ge\sqrt{3}S$ (above) and
[quote="evarist"] About the 1st gemath i can prove more :
   $ xm_{a}h_{b}+ym_{b}h_{c}+zm_{c}h_{a}\ge 3\sqrt{xy+yz+zx}S$
  :D  :D  :P
   Have fun ! [/quote]
oh really a strong problem :D  try with this acute triangle
$ a=6,b=7,c=10,x=1,y=z=7$ goodluck  :D


[quote="IntrepidMath"]Solve:
Factor:
$x^2-3x+2$
[/quote]
[hide]
(x-2)(x-1)[/hide]

[quote="IntrepidMath"]Solve:

$3x^2-25x+\sqrt{42}=0$

Factor:
$x^2-3x+2$

Evaluate:
$\frac{5\sqrt{45}}{3\sqrt{42}\sqrt{63}}$[/quote]

[hide="Solve"]

Do we use the quadratic formulae for this?

[/hide]

[hide="factor"]$(x-2)(x-1)$[/hide]

[quote="anirudh"]
Do we use the quadratic formulae for this?
[/quote]

I would use the formula...i think that is the best approach. :)

[hide][quote="IntrepidMath"]Solve:

$3x^2-25x+\sqrt{42}=0$

Factor:
$x^2-3x+2$

Evaluate:
$\frac{5\sqrt{45}}{3\sqrt{42}\sqrt{63}}$[/quote]

for the first one, I got as far as

$ \frac{25\pm\sqrt{625-12\sqrt{42}}}{6} $

The second is $ (x-2)(x+1) $

The third is:

$ \frac{5\sqrt{45}}{3\sqrt{42}\sqrt{63}} = \frac{5\sqrt5}{21\sqrt6} $[/hide]


vector
Functional Analysis
real analysis
real analysis unsolved

Let $ M$ be a sub_vector_space of a normed space $ E$. Give that $ U=B'(0,r)$ and $ c\in (0,1)$. Suppose that $ U\subset M+cU$. Prove that $ U\subset\overline{M}$ and $ M$ is dense in $ E$

This is essentially the contrapositive of a basic result in functional analysis: if $ M\subset E$ is a proper closed subspace, then for any $ \epsilon>0$ there is a point in $ B(0,r)$ at distance $ >r-\epsilon$ from $ M$.

Form $ U\subset M\plus{}cU$, how can we show that $ U\subset M\plus{}c^{n}U,\forall n$.

$ U\subset M\plus{}cU\subset M\plus{}c(M\plus{}cU)\equal{}(M\plus{}cM)\plus{}c^{2}U\equal{}M\plus{}c^{2}U$ etc.


[quote="AndrewTom"]Show that there is an integer $ n$ with the following properties:

(i) the binary representation of $ n$ has exactly $ 2004$ $ \ 0s$ and $ 2004$ $ \ 1s$

(ii) $ 2004$ divides $ n$.[/quote]

$ a\equal{}70\times 2004\equal{}140280\equal{}\overline{100010001111111000}_2$ with exactly $ 9$ digits $ 1$ and $ 9$ digits $ 0$.
$ b\equal{}8\times 2004\equal{}16032\equal{}\overline{11111010100000}_2$ with exactly $ 7$ digits $ 1$ and $ 7$ digits $ 0$.

Let then $ n\equal{}a\plus{}b\sum_{k\equal{}0}^{284}2^{14k\plus{}18}$ $ \equal{}2004(70\plus{}8\sum_{k\equal{}0}^{284}2^{14k\plus{}18})$

$ n$ is a multiple of $ 2004$ and is the juxtaposition of $ 285$ sequences $ 11111010100000$ (from $ b$) and one sequence $ 100010001111111000$ (from $ a$)

So binary representation of $ n$ contains exactly $ 285\times 7\plus{}9\equal{}2004$ digits $ 1$ and $ 2004$ digits $ 0$.

Hence the result.


$n$ is integer $>0$ .let $A={1,2,...n}$.find $f(n)$ is integer min  such that every subset of $A$ include $f(n)$ elements  ,there exist three elements $x<y<z$ and $x+y\geq z$.

I think I have lot of problem form this
[i]trinhcongdien_511@yahoo.com[/i]

I don't understand why $f(n)=f(n-1)+f(n-2)+1$


Να δειχτεί ότι αν ισχύει $ |a_{n + 1} - a_{n}|\leq{\frac {1}{n^2}}$ τότε η $ a_n$ είναι ακολουθία του Cauchy.

Μάλλον τετριμμένο μου φαίνεται...

[hide]Χρησιμοποίησε τριγωνική ανισότητα και το γεγονός ότι η σειρά $ \sum_{n\equal{}1}^\infty \frac{1}{n^2}$ συγκλίνει.[/hide]
Cheerio,

Durandal 1707

Δεν είναι τετριμμένο γιατί για να δείξεις ότι είναι ακολουθία Cauchy αρκεί να δείξεις ότι \[ (\forall{\varepsilon}>0)(\exists{n_0}\in{\mathbb{N}})(\forall{n,m}\geq{n_0}) \Longrightarrow |a_n-a_m|<\varepsilon\]

Αρκει να δειξεις οτι η $ a_n$ συγκλινει.

[quote="Carl Friedrich Gauss"]Δεν είναι τετριμμένο γιατί για να δείξεις ότι είναι ακολουθία Cauchy αρκεί να δείξεις ότι
\[ (\forall{\varepsilon} > 0)(\exists{n_0}\in{\mathbb{N}})(\forall{n,m}\geq{n_0}) \Longrightarrow |a_n - a_m| < \varepsilon
\]
[/quote]

Έλα βρε Gauss :)

[hide]Για $ n\geq m$ έχουμε:
$ |a_{n + 1} - a_m| = |(a_{n + 1} - a_{n}) + \cdots + (a_{m + 1} - a_m)| \leq \sum_{k=m}^n|a_{k + 1} - a_{k}| \leq \sum_{k=m}^n \frac {1}{k^2}$

Όμως, αφού η σειρά $ \sum_{k = 1}^\infty \frac {1}{k^2}$ συγκλίνει, έπεται ότι η ακολουθία μερικών αθροισμάτων $ S_n = \sum_{k = 1}^n \frac {1}{k^2}$ θα είναι Cauchy και... τέλος :)
[/hide]
Cheerio,

Durandal 1707

Μια άλλη προσέγγιση
$ |a_{n \plus{}m\plus{} 1} \minus{} a_m| \equal{} |(a_{n \plus{}m\plus{} 1} \minus{} a_{n\plus{}m}) \plus{} \cdots \plus{} (a_{n \plus{} 1} \minus{} a_n)|\le \cdots \leq \sum_{k\equal{}0}^m\frac{1}{(n\plus{}k)^2}\equal{}\frac{1}{n}\frac{2\minus{}1}{n}\sum_{k\equal{}0}^m (\frac{1}{1\plus{}k \frac{2\minus{}1}{n}})^2$
όμως $ \frac{1}{n} \to 0 , \frac{2\minus{}1}{n}\sum_{k\equal{}0}^m (\frac{1}{1\plus{}k \frac{2\minus{}1}{n}})^2 \to\displaystyle\int^2_1 \frac{1}{x^2}\,dx$ άρα ...
(Ελπίζω να μην έκανα κάποιο λάθος γιατί την απόδειξη την έκανα πολύ βιαστικά. Πάντως συμφωνώ με την λύση του Παναγιώτη)

[quote="r_boris"]Μια άλλη προσέγγιση
$ |a_{n \plus{} m \plus{} 1} \minus{} a_m| \equal{} |(a_{n \plus{} m \plus{} 1} \minus{} a_{n \plus{} m}) \plus{} \cdots \plus{} (a_{n \plus{} 1} \minus{} a_n)|\le \cdots \leq \sum_{k \equal{} 0}^m\frac {1}{(n \plus{} k)^2} \equal{} \frac {1}{n}\frac {2 \minus{} 1}{n}\sum_{k \equal{} 0}^m (\frac {1}{1 \plus{} k \frac {2 \minus{} 1}{n}})^2$
όμως $ \frac {1}{n} \to 0 , \frac {2 \minus{} 1}{n}\sum_{k \equal{} 0}^m (\frac {1}{1 \plus{} k \frac {2 \minus{} 1}{n}})^2 \to\displaystyle\int^2_1 \frac {1}{x^2}\,dx$ άρα ...
(Ελπίζω να μην έκανα κάποιο λάθος γιατί την απόδειξη την έκανα πολύ βιαστικά. Πάντως συμφωνώ με την λύση του Παναγιώτη)[/quote]

Τέλεια ιδέα!

Στα χαλάει λίγο το ότι το άθροισμα Riemann είναι μέχρι το $ m$ και άρα δε λειτουργεί το τελευταίο βήμα... ωστόσο, αντί για το $ \int_1^2\cdots$, αρκεί να συγκρίνεις με το πεπερασμένο $ \int_1^\infty\frac {1}{x^2}dx$ και είμαστε ζάχαρη :)

Cheerio,

Durandal 1707

[quote="r_boris"]Ναι απροσεξία. Εννοείς$ \displaystyle\int^{\infty}_1 \frac {1}{x^2}\,dx$?[/quote]

Ναι, το διόρθωσα :)

Πολύ σωστά παιδιά και οι δύο..μπράβο!!


symmetry
geometry
perpendicular bisector
power of a point
radical axis
geometry solved

In general, let $X'$ be the symmetric of $X$ wrt the perpendicular bisector of $BC$. If we show that $A'$ lies on the circle $(APQ)$, then we're done, because $A,A'$ also lie on $(ABC)$, so the line of the centers of $(APQ),(ABC)$ is $\perp AA'$, which is $\|BC$ (I'm assuming $AB\ne AC$, the other case is trivial).

We have $\angle PQ'Q=\pi-\angle QQ'C=\pi-\angle A$, so $Q'$ lies on $(APQ)$. In basicly the same way we show that $P'\in(APQ)$. Because of the symmetry of the situation, we also know that $Q\in(A'P'Q')$. We have thus shown that both $A,A'$ lie on $(PQQ')$, and this proves that $APQA'$ is cyclic.

[quote="orl"]Let the triangle $ABC$ with $BC$ the smallest side. Let $P$ on ($AB$) such that angle $PCB$ equals angle $BAC$. and $Q$ on side ($AC$) such that angle $QBC$ equals angle $BAC$. Show that the line passing through the circumenters of triangles $ABC$ and $APQ$ is perpendicular on $BC$.[/quote]

Another solution:

In the following, the abbreviation "circle M(r)" will mean the circle with the center M and the radius r, and the abbreviation "circle $P_1P_2P_3$" will mean the circle through three given points $P_1$, $P_2$, $P_3$.

We have to show that the line joining the centers of the circles ABC and APQ is perpendicular to the line BC.

Since < QBC = < BAC and < QCB = < BCA, the triangles QBC and BAC are similar. Thus, CA : CB = CB : CQ. In other words, $CA\cdot CQ=CB^2$. Since the point Q lies on the ray CA, we can therefore conclude that the points A and Q are inverse to each other with respect to the circle C(CB). Now, if a circle u passes through two points which are inverse to each other with respect to another circle k, then this circle u is orthogonal to the circle k. Hence, the circle APQ, passing through the points A and Q which are inverse to each other with respect to the circle C(CB), must be orthogonal to this circle C(CB). Similarly, the circle APQ is also orthogonal to the circle B(BC). But the centers of all circles orthogonal to two given circles lie on the radical axis of these two circles. Hence, since the circle APQ is orthogonal to the two circles B(BC) and C(CB), its center lies on the radical axis of these two circles B(BC) and C(CB). Now, clearly, these two circles B(BC) and C(CB) have equal radius, and since the radical axis of two circles with equal radius is the perpendicular bisector of the segment joining their centers, it follows that the radical axis of the circles B(BC) and C(CB) is the perpendicular bisector of the segment BC. Hence, we obtain that the center of the circle APQ lies on the perpendicular bisector of the segment BC. On the other hand, it is trivial that the center of the circle ABC also lies on the perpendicular bisector of the segment BC. Hence, the line joining the centers of the circles ABC and APQ is the perpendicular bisector of the segment BC, and thus it is perpendicular to the line BC. Problem solved.

  Darij

Let $K$ be the intersection of $CP$ and $(APQ)$ (other than $P$). It's easy to see that $KQ \parallel BC$, because $\angle QKC= \angle BAC= \angle BCK$.
We have $\angle KCB = \angle QBC$, so $BCQK$ is a isosceles trapezoid.
$I$, $Q$ lies on the perpendicular bisector of $KQ$, $BC$ respectively. But $KQ$, $BC$ have the same perpendicular bisector, so we are done.


[b]No Problem !!![/b]
Suggest:
Let $ P(x) \equal{} det(B \plus{} x.C) \equal{} det(B) \plus{} .... \plus{} det(C).x^{118}$
Then apply lagaranger Formula for $ x_{1} \equal{} 1890 ,x_{2} \equal{} 1891,....,x_{119} \equal{} 2008$
[b]to change !!![/b] Oke !!
[b]"..I remember Uncle ,I remember Uncle ,I remember Uncle .."[/b]
 :rotfl:  :rotfl:


Conside a sequence $ a_{n}$ ,x is a real number
$ a_{1}= 1$
$ a_{n+1}= x^{n}+n.a{n}$
Find
$ \large\prod\limits_{i = 1}^{\infty}\ (1-\frac{x^{n}}{a_{n}})$

[quote="TTsphn"]
$ a_{n\plus{}1}\equal{} x^{n}\plus{}nx_{n}$
[/quote]

Who is $ x_{n}$ ?


let f : ]0;+infini[ x ]0;+infini[ to [4;+infini[

  f((x;y))= (x+y) (1/x+1/y)

prove that this function is surjectif.

Is it injective?

Obviously surjective, Obviously not injective  :arrow:

can you tell me why ie it surjectif
a detailled solution please  :blush:  :blush:

could  you  find solution without continious

$ f(x,y)\equal{}2\plus{}\frac{x}{y}\plus{}\frac{y}{x}$
Let $ t\equal{}\frac{x}{y}$
$ f(x,y)\equal{}f(t)\equal{}2\plus{}t\plus{}\frac{1}{t}$
You can prove that the equation $ k\equal{}\frac{1}{t}\plus{}t$ has solution if and only if $ k\geq 2$
Because $ \Delta\equal{}k^2\minus{}4$
But I think use continous is the best solution :wink:


number theory
relatively prime
number theory unsolved

If $m,n\in\mathbb{N}$ where $n>1$ are relatively prime . Also , $q$ is the smallest positive divisor in $n$ such that

$m^n\equiv 1(\mod q)$ . Is it true that this equation have solution only when $m=1$ ?

[hide="remark"]Im not sure whether it is true or not , cause we have $m^{q-1}\equiv 1(\mod q)$ and we can deduce $(n,q-1)=1$ so I guess there is no answer from here other than m=1 . [/hide]

As I understand the following works:
Take $n=q(q-1)$ and every $m$ satisfies $m^n \equiv 1 \mod q$.
Especially $m=3,n=2$ works ;)

ZetaX:  but $q$ is supposed to be the smallest positive divisor with this property so you should also prove that some smaller than $q$ divisor of $n$ doesn't satisfy this (or is it obvious and I just don't see it?  :oops: )

Well, the existence of one such number implies the existence of a smallest such number, doesn't it¿

Of course it does, but I meant something different - forget about it  :) - but is it so easy to find a counterexample when it is demanded $q$ a proper divisor ?

Well , I also just figure out $m=3,n=2$ works because of catalan conjecture . But what about for n>2 ? Is there still any solution ? I only manage to reduce to $m\equiv 1(\mod q)$ ....

Can you describe your problem again¿
I think we speak of two different problems or so...   :?

[quote="shyong"]If $m,n\in\mathbb{N}$ where $n>1$ are relatively prime . Also , $q$ is the smallest positive divisor in $n$ such that

$m^n\equiv 1(\mod q)$ . Is it true that this equation have solution only when $m=1$ ?

[/quote]

Well , the question is above but since there are counter example for m=3 , n=2 , so im wondering whether we can prove that there are no other solution for $m^n\equiv 1(\mod q)$ other than $m=1$ and $m=3,n=2$ .

[quote="ZetaX"]I gave infinetely many examples above, didn't I¿  :?[/quote]

Well , you said that for every $m$ satisfying $m^n\equiv 1(\mod q)$ where $n=q(q-1)$ works but

1）$q$ is the smallest divisor (prime divisor actually) so $n=q(q-1)$ can only be $n=2$ . or else the prime divisor in $q-1$ would be smaller than $q$ and contradict

2) does $m^n\equiv 1(\mod q)$ have another solutions ? :huh:

[quote="shyong"][quote="ZetaX"]I gave infinetely many examples above, didn't I¿  :?[/quote]
Well , you said that for every $m$ satisfying $m^n\equiv 1(\mod q)$ where $n=q(q-1)$ works but

1）$q$ is the smallest divisor (prime divisor actually) so $n=q(q-1)$ can only be $n=2$ . or else the prime divisor in $q-1$ would be smaller than $q$ and contradict [/quote]
That was the reason why I asked you to state the problem correctly since that is not what is stated in the first post.

Well, every even $n$ works with every odd $m$...


Let v: [0,1] => R^2 be a continuous curve st |v'(t)| <=1 for all t in [0,1], and v(0) is a point inside a square of side 8. Find a continuous curve u: [0,1] => R^2 such that u(0) is the center of the square and u intersects all curves v that satisfies the previous conditions (i.e. there exists t0 st u(t0)=v(t0)).

Peano Curve is an answer but I don't know how to show this.

You seem to misinterpret the answer: the curve in question is a Peano curve, all right, but not every Peano curve has the desired property ;).


[quote="vishalarul"]Let $a$, $b$, $c$, and $d$ be nonnegative integers. Prove that

$\frac{1}{a}+\frac{1}{b}+\frac{4}{c}+\frac{16}{d}\geq \frac{64}{a+b+c+d}$[/quote][hide]By the [b]Cauchy-Schwartz Inequality[/b], \[\left(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}+\frac{16}{d}\right)\left(\frac{a}{64}+\frac{b}{64}+\frac{c}{64}+\frac{d}{64}\right)\ge\left(\frac{1}{8}+\frac{1}{8}+\frac{1}{4}+\frac{1}{2}\right)^{2},\]  \[\left(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}+\frac{16}{d}\right)\left(\frac{a+b+c+d}{64}\right)\ge1.\] Since $a+b+c+d$ is positive, then $\frac{a+b+c+d}{64}$ is positive. Hence, divide both sides by this term, to get \[\frac{1}{a}+\frac{1}{b}+\frac{4}{c}+\frac{16}{d}\geq \frac{64}{a+b+c+d},\] as desired.[/hide]

[quote="vishalarul"]Let $a$, $b$, $c$, and $d$ be nonnegative integers. Prove that

$\frac{1}{a}+\frac{1}{b}+\frac{4}{c}+\frac{16}{d}\geq \frac{64}{a+b+c+d}$[/quote]
[b]towersfreak2006[/b], you missed a nice (and simple) solution.  :wink:

Using Cauchy in Engel form,
$\frac{1}{a}+\frac{1}{b}+\frac{4}{c}+\frac{16}{d}\geq \frac{{(1+1+2+4)}^{2}}{a+b+c+d}= \frac{64}{a+b+c+d}$.

Well, technically, in order to have the inequality hold, you must not have $a=b=c=d=0$...

For the inequality to [i]make sense[/i] you must have none of $a, b, c, d$ equal to 0.


So this is the general form of this
$ abc \geq (a+b-c)(b+c-a)(a+c-b)$

I think we can use the similar method for it
$ (S-a_{1})...(S-a_{n-1}) \leq \frac{((n-1)S-(a_{1}+...a_{n-1}))^{n-1}}{(n-1)^{n-1}}=(\frac{a_{n}}{n-1})^{n-1}$

Do the same thing with the others

 :huh: But I have some troubles while multiplying all of them
I mean the factors aren't all positive

Oh, I see it now

WLOG we can assume that $ a_{1}\geq a_{2}\geq ...\geq a_{n}$

So $ (S-a_{2}),..., (S-a_{n})$ are positive

If $ (S-a_{1})$ is positive we can multiply it all

Otherwise the inequality is trivial

[quote="humanbeing"]
WLOG we can assume that $ a_{1}\geq a_{2}\geq ...\geq a_{n}$

So $ (S-a_{2}),..., (S-a_{n})$ are positive

[/quote]

This is not true :o Sorry for my mistake
But I think maybe we can use the idea
When $ \prod(S-a_{i})\leq0$ the inequality is trivial

When $ (S-a_{i})$ aren't all positive and $ \prod(S-a_{i})\geq0$, the inequality is not always true

For example $ n=4$; $ a_{1}=1$; $ a_{2}=2$; $ a_{3}=5$; $ a_{4}=6$

I'm not so sure, you may try again if the example is true.


Before practicing his routine on the rings, a 64.1 kg gymnast hangs motionless, with one hand grasping each ring and his feet touching the ground. Both arms make an angle of 25.9° with the vertical. If the force exerted by the rings on each arm has a magnitude of 284 N, what is the magnitude of the force exerted by the floor on his feet?

[hide]

$\Sigma F_{y}: 2(284)\sin\left(25.9^{\circ}\right)-64.1\cdot g+N = 0$

$\implies N = 381\text{ Newtons}$
[/hide]

[quote="sks119"]could you please elaborate on how you got that answer...i am still a little confused. thank you.[/quote]
Sure, it's a static problem. No acceleration. This implies the sum of forces must equal to zero.

Using that fact, I merely summed the vertical components of force on the gymnast and set the sum equal to zero. Each arm is under a tension of 284 N @ 25.9 degrees. The vertical component of this tension is 284 x sin(25.9). There are two such components, both directed upward. The gymnast's own weight pulls him downward (64.1 kg x 9.81 m/s^2). Finally the normal force, N, of the floor pushes up on the gymnast's feet; this is what you're trying to find.

Edit: Also it would help if I weren't dyslexic. The real answer is appearently $N = 381\text{ Newtons}$


[quote="rassvet"]There was a problem some decades ago(Jury, France, 82)

a1, a2,..., an >=0
Prove
a1 + (a1a2)^1/2 + (a1a2a3)^1/3 + ... + (a1a2..an)^1/n <= e(a1+a2+...+an)[/quote]
There was a problem 8 decades ago ( it's problem from G. Polya ). :P


HI
See this problem and aend your solutions here:

this problem comes from iranian Mosp ,Is there any solver? :roll:


AMC USA(J)MO USAMO geometry circumcircle

[b]The altitude CE of acute-angled triangle ABC crosses the circle with diameter AB at M and N, and the altitude BD crosses the circle with diameter AC at P and Q. Prove that M, N, P, and Q are concyclic[/b]

Denote the circumcenter by $ H$. Let the third altutde be $ AF$.

We have $ MH \times HN \equal{} AH \times HF \equal{} PH \times HQ$, and we are done.

Can someone draw a diagram? I don't see how an altitude can cross the circle twice.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=356630#356630[/url]

Please just use the contests section next time. All the problems there have been discussed, and you can see the thread by clicking on the little problem number. Also, there's no need to post problems in bold.


$w_B$ and $w_C$ are excircles of a triangle $ABC$. The circle $w_B'$ is symmetric to $w_B$ with respect to the midpoint of $AC$, the circle $w_C'$ is symmetric to $w_C$ with respect to the midpoint of $AB$. Prove that the radical axis of $w_B'$ and $w_C'$ halves the perimeter of $ABC$.

Denote the triangle angles, sides and semiperimeter in the usual manner: $\alpha = \angle A, \beta = \angle B, \gamma = \angle C$, $a = BC, b = CA, c = AB$, $s = \frac{a + b + c}{2}$. Let $M, N$ be the midpoints of the segments $CA, AB$ and $E_B, E_C, E_B', E_C'$ centers of the circles $w_B, w_C, w_B', w_C'$. Since the circle $w_B$ is tangent to the line $CA$ and since both the circles $w_B, w_B'$ and the points $C, A$ are symmetrical WRT the midpoint $M$ of the segment $CA$, the circle $w_B'$ is also tangent to the line $CA$. For similar reasons, the circle $w_C'$ is tangent to the line $AB$. Let $S, S'$ be the tangency points of the circles $w_B, w_B'$ with the line $CA$ and $T, T'$ the tangency points of the circles $w_C, w_C'$ with the line $AB$. Then $AS' = CS = s - a$ and $AT' = BT = s - a$. Thus the tangent lengths from the vertex $A$ to the circles $w_B', w_C'$ are equal, which means that the vertex $A$ lies on the radical axis of these 2 circles.

Let $\triangle A'B'C'$ be a triangle having the original triangle $\triangle ABC$ as its medial triangle, such that $A \in B'C', B \in C'A', C \in A'B'$. The line $B'C'$ is symmetrical to the line $BC$ WRT both the midpoints $M, N$ of the segments $CA, AB$. Since the circles $w_B, w_C$ are both tangent to the line $BC$, the circles $w_B', w_C'$ are both tangent to the line $B'C'$. The line $A'B'$ is symmetrical to the line $AB$ WRT the midpoint $M$. Since the circle $w_B$ is tangent to the line $AB$, the circle $w_B'$ is tangent to the line $A'B'$. The line $C'A'$ is symmetrical to the line $CA$ WRT the midpoint $N$. Since the circle $w_C$ is tangent to the line $CA$, the circle $w_C'$ is tangent to the line $C'A'$. Hence, the centers $E_B', E_C'$ of the circles $w_B', w_C'$ lie on the internal bisectors of the angles $\angle B' = \beta, \angle C' = \gamma$ of the triangle $\triangle A'B'C'$ intersecting at its incenter $I'$. The incenter $I'$ of the triangle $\triangle A'B'C'$ is identical with the Nagel point of its medial triangle $\triangle ABC$.

Let $r_B, r_C$ be the radii of the excircles $w_B, w_C$ of the triangle $\triangle ABC$, identical with the radii of their reflections $w_B', w_C'$, and $r' = 2r$ the radii of the incircles of the triangles $\triangle A'B'C', \triangle ABC$. Then

$r_B = s \tan{\frac \beta 2},\ \ r_C = s \tan{\frac \gamma 2},\ \ r' = 2r = 2(s - b) \tan{\frac \beta 2} = 2(s - c) \tan{\frac \gamma 2}$

$B'I' = 2 \sqrt{ca \tan{\frac \gamma 2} \tan{\frac \alpha 2}},\ \ C'I' =2 \sqrt{ab \tan{\frac \alpha 2} \tan{\frac \beta2}}$

The vertex $B'$ is the external homothety center of the circle $w_B'$ and the incircle $(I')$ and the vertex $C'$ is the external homothety center of the circle $w_C'$ and the incircle $(I')$. Hence

$B'E_B' = B'I'\ \frac{r_B}{r'} = 2 \sqrt{ca \tan{\frac \gamma 2} \tan{\frac \alpha 2}} \cdot \frac{s \tan{\frac \beta 2}}{2(s - b) \tan{\frac \beta 2}} = \frac{s}{s - b} \sqrt{ca \tan{\frac \gamma 2} \tan{\frac \alpha 2}}$

$I'E_B' = B'E_B' - B'I' = \left(\frac{s}{s - b} - 2\right) \sqrt{ca \tan{\frac \gamma 2} \tan{\frac \alpha 2}} = \frac{2b - s}{s - b} \sqrt{ca \tan{\frac \gamma 2} \tan{\frac \alpha 2}}$

The power of the point $I'$ to the circle $w_B'$ is

$(r_B - I'E_B')(r_B + IE_B') = r_B^2 - I'E_B'^2 = s^2 \tan^2{\frac \beta 2} -  \frac{(2b - s)^2}{(s - b)^2}\ ca \tan{\frac \gamma 2} \tan{\frac \alpha 2} =$

$= \frac{s^2 r^2}{(s - b)^2} -  \frac{(2b - s)^2}{(s - b)^2} \cdot \frac{ca r^2}{(s - c)(s - a)} =$

$= \frac{r^2}{(s - b)^2} \frac{s^2(s-a)(s-c) - ac(2b - s)^2}{(s-a)(s - c)} =$

$= \frac{r^2}{(s - b)^2(s - a)(s - c)} \left[s^2 \left(s^2 - s(a + c) + ac \right) - ac(4b^2 - 4sb + s^2)\right] =$

$= \frac{r^2}{(s - b)^2(s - a)(s - c)} \left[s^3(b - s) - 4abc(b - s)\right] =$

$= \frac{r^2}{(s - b)(s - a)(s - c)} (4abc - s^3) = \frac{sr^2}{\triangle^2}(4abc - s^3) = \frac{4abc}{s} - s^2$

where $\triangle$ is the area of the triangle $\triangle ABC$. Since this expression is symmetrical WRT to the cyclic exchange of the sides $a, b, c$, it follows that the same expression could be obtained for the power of the point $(I')$ to the circle $w_C'$. Hence, the incenter of the triangle $\triangle A'B'C'$, identical with the Nagel point of its medial triangle $\triangle ABC$, lies on the radical axis of the circles $w_B', w_C'$. Since by definition, the Nagel point of a triangle is the intersection of the perimeter splitters, the radical axis $AI'$ of the circles $w_B', w_C'$ cuts in half the perimeter of the triangle $\triangle ABC$.

First of all, $A$ clearly lies on the radical axis of the two circles, since it's power wrt both of them is equal to $AE^2$, where $DEF$ is the intouch triangle of $ABC$. We now need to prove that the point $D'$, where $\omega_A$ touches $BC$, also lies on the radical axis.

Let $S,T$ be the points on the parallel through $A$ to $BC$ which satisfy $AS=AT=AE=AF$, s.t. $FS$ crosses $AC$, for example. Also, let $P=TE\cap SF,T'=TE\cap BC,S'=SF\cap BC$. Since $TFES$ is cyclic, and $F,S$ lie on $\omega_C'$, while $T,E$ lie on $\omega_B'$, it means that $P$ has equal powers wrt these two circles, so it lies on the radical axis. It means that the intersection between the radical axis and $BC$ is $AP\cap BC$, or, in other words, the midpoint of $S'T'$. It's easy now to perform the necessaru computations (which are not ugly at all) and find this point to be the tangency point between the excircle $\omega_A$ and $BC$, as we wanted.

Denote by $ I$, $ I_a$, $ I_b$, $ I_c$ the centers of the incircle, $ A$-excircle, $ B$-excircle and $ C$-excircle of triangle $ ABC$, respectively. Call $ M$ the mid-point of the side $ BC$ and $ D$, $ E$ the tangency points of the incircle and $ A$-excircle of triangle $ ABC$ with $ BC$, respectively.

We know that the line segment $ AE$ passes through the antipodal point of $ D$ w.r.t. the incircle $ (I)$, so we deduce that $ IM$ is parallel to $ AE$ $ (1)$.

Now, let us call $ I'_b$, $ I'_c$ the centers of the circles $ \omega_B'$ and $ \omega_C'$, respectively, and $ J$ the midpoint of arc $ BC$ not containing $ A$ of the circumcircle of triangle $ ABC$. We will prove that triangles $ IMJ$ and $ I'_bDC$ are similar.

We can easily to see that $ MJ\perp DC$ and $ IJ\perp I'_bC$, it implies $ \angle IJM\equal{}\angle I'_bCD$. Thus, it suffices to show that $ \frac{JM}{JI}\equal{}\frac{CD}{CI'_b}$, i.e. $ \frac{JB}{JM}\equal{}\frac{AI_b}{AF}$, where $ T$ is the tangency point of $ \omega_B$ with $ CA$. From the similarity of triangles $ JBM$ and $ AI_bF$, we obtain $ \frac{JB}{JM}\equal{}\frac{AI_b}{AF}$, thus $ \triangle IMJ\sim\triangle I'_bDC$, and hence we conclude that $ IM\perp I'_bD$ $ (2)$.

Combining $ (1)$ and $ (2)$, we have $ AE\perp I_b'D$. Similarly, $ AE\perp I_c'D$. It means $ I_b'$, $ I_c'$ and $ D$ lie on line that is perpendicular to $ AE$.

MP=MQ=MD=MD'=1/2(b-c)  =>  P,D,Q,D'  are concyclic
denote N is the intersection point of incircle and AD'，∠AND=90  which indicates N is on circleM，∠D1NP=∠PDB notice PD'//BI   =>  ∠D1NP=∠PDB=90-1/2∠B=∠AT3P  =>  A，T3，N，P are concyclic  thus T3,N,D are collinear
analogusly，T4,N,D are collinear  =>  T3T4⊥AD'
we already have T3T4⊥ST and  A,S,T are collinear because A is on radical axis of circle T3,T4 =>  T3T4⊥AT
therefore A,S,D',T  are collinear

[quote="April"]We will prove that triangles $ IMJ$ and $ I'_bDC$ are similar.[/quote]

It isn't true.

Dear [b]ArbelosYS[/b], could you please point out my mistake?

BTW, there was a typo in my solution, I just wanted to say $ \color{red} F$ is the tangency point of $ \omega_B$ with $ CA$, not $ T$ as listed below:

[quote="April"]... We will prove that triangles $ IMJ$ and $ I'_bDC$ are similar.

We can easily to see that $ MJ\perp DC$ and $ IJ\perp I'_bC$, it implies $ \angle IJM \equal{} \angle I'_bCD$. Thus, it suffices to show that $ \frac {JM}{JI} \equal{} \frac {CD}{CI'_b}$, i.e. $ \frac {JB}{JM} \equal{} \frac {AI_b}{AF}$, where $ T$ is the tangency point of $ \omega_B$ with $ CA$.[/quote]

Let the incircle of $\triangle ABC$ be tangent to $BC$, $CA$, and $AB$, at $D$, $E$, and $F$, respectively, and let $D'$ be the reflection of $D$ across $BC$. It is well-known that $AD'$ passes through the antipode $P$ of $D$.

Let $\ell$ be the line through $A$ parallel to $BC$. $\omega_C'$ must be the circle tangent to $AB$ at $F$ and tangent to $\ell$, and $\omega_B'$ must be the circle tangent to $AC$ at $E$ and tangent to $\ell$. Let $EP$ and $FP$ intersect $\ell$ at $U$ and $V$, respectively. Note that $\angle FAV = \angle FAC + \angle CAV = \angle A + \angle C = 180^{\circ} - \angle B$, while $\angle VFA = \angle PFA = \angle FPD = 90^{\circ} - \angle FDB = \frac{\angle B}{2},$ so $AF = AV$. By symmetry, $AE = AU$ as well. Hence, $U$ and $V$ must be the tangency points of $\omega_C'$ and $\omega_B'$ with $\ell$, respectively. Furthermore, $\angle UVF = \angle AVF = \angle VFA = \angle PEF = \angle UEF$, so $UVEF$ is cyclic, so $(PU)(PE) = (PV)(PF)$. But $P$ lies on chord $UE$ of $\omega_C'$ and chord $FV$ of $\omega_B'$, so $P$ lies on the radical axis of $\omega_B'$ and $\omega_C'$. Since $AF = AE$, $A$ must lie on the radical axis of $\omega_B'$ and $\omega_C'$ as well, so $AP$ is the radical axis of $\omega_B'$ and $\omega_C'$, as desired. Since $D'$ lies on line $AP$, and $AB + BD' = AB + \frac{CA+CB-AB}{2} = \frac{AB+BC+CA}{2}$, line $AD'$, i.e., the radical axis of $\omega_B'$ and $\omega_C'$ halves the perimeter of $\triangle ABC$.

My solution:

Let $ B'=w_B \cap BC , C'=w_C \cap BC $ .
Let $ B'' $ be the reflection of $ B' $ in the midpoint of $ AC $ .
Let $ C'' $ be the reflection of $ C' $ in the midpoint of $ AB $ .
Let $ l $ be a line passing through $ A $ and parallel to $ BC $ .

Easy to see $ B'', C'' \in l $
Since $ AB''=CB'=BC'=AC'' $ and $ l $ is the common tangent of $ w_B', w_C' $ ,
so $ A $ lie on the radical axis of $ w_B' $ and $ w_C' $ .

It's suffices to prove the radical axis of $ w_B' $ and $ w_C' $ is $ A $ -Nagel line of $ \triangle ABC $ .

Since the center $ I_b', I_c' $ of $ w_B', w_C' $ is the orthocenter of $ \triangle ICA, \triangle IAB $ , respectively ,
so the pole $ M $ of $ I_b'I_c' $ with respect to the incircle $ (I) $ of $ \triangle ABC $ is the midpoint of $ BC $ .
Since $ I $ is the Nagel point of the medial triangle of $ \triangle ABC $ ,
so $ MI $ is parallel to the $ A $ -Nagel line of $ \triangle ABC $ ,
Since the radical axis of $ w_B', w_C' $ is perpendicular to $ I_b'I_c' $ and $ I_b'I_c' \perp IM $ ,
so we get the radical axis of $ w_B', w_C' $ is $ A $ -Nagel line of $ \triangle ABC $ .

Q.E.D

$N$ is foot of nagel cevian, $M$ is midpoint of $BC$, $I$ is incentre.

Obviously $A$ lies on the radical axis. Further, the centres of $w_B', w_C'$ are the orthocentres of $IAC, IAB$. Since the line through these orthocentres is perpendicular to $IM$, and $AN \parallel IM \implies AN$ is the radical axis.

EDIT: Ahh, this is the same as Telv's, yes by polarity I proved the orthocentre thing but it is pretty well known too :)

Obviously, $A$ lies on the radical axis of $\omega_B'$ and $\omega_C'$. Let $E$ be the touch point of the $A$ excircle with $BC,$ $I$ be the incenter of $ABC$ and $M$ be the midpoint of $BC$. From [url=http://yufeizhao.com/olympiad/geolemmas.pdf]Lemma 2,[/url] we get $AE \parallel IM$. [url=https://www.artofproblemsolving.com/community/c6h215424p1191663]Time Travelling to Tuymaada 2008,[/url] we see that $M$ lies on the polar of $I_B'$ and $I_C'$ wrt the incircle $\Longrightarrow M \, \text{is the pole of line} \, I_B'I_C' \Longrightarrow IM \perp I_B'I_C'.$ Thus, $AE$ is the radical axis of $\omega_B'$ and $\omega_C'$ and it bisects the perimeter of triangle $ABC$ as desired.

[quote=enhanced][quote=Taha1381][quote=TelvCohl]

Since the center $ I_b', I_c' $ of $ w_B', w_C' $ is the orthocenter of $ \triangle ICA, \triangle IAB $ , respectively ,
so the pole $ M $ of $ I_b'I_c' $ with respect to the incircle $ (I) $ of $ \triangle ABC $ is the midpoint of $ BC $ .

[/quote]

Why is this true?[/quote]

It is a particular case of [url=http://artofproblemsolving.com/community/c6h612004p3639577]this[/url][/quote]

Thanks.

Let $I_b',I_c'$ be the centers of $w_B'$ and $w_C'$ respectively, let $D$ be the $A$-extouch point and let $E,F$ be the $B$-intouch and $C$-intouch points, respectively. Note that $A$ lies on the radical axis of $w_B$ and $w_C$ since $AC$ and $AB$ are both tangent to the incircle and $w'_B$ and $w'_C$, respectively. Then it suffices to show that $D$ lies on the radical axis, or that $AD \perp I_b'I_c'$. By the perpendicular lemma, this is equivalent to $AI_b'^2 - AI_c'^2 = DI_b'^2-DI_c'^2$. Let $r_b$ and $r_c$ be the the radius of $\omega_B$ and $\omega_C$, respectively. Then $AI_b'^2-AI_c'^2=r_b^2-r_c^2$.

Let $X$ be the projection of $D$ onto $FB$. Then \[DI_c'^2 = (FB-YB)^2+(FI_c'^2-YD)^2 = ((s-b)-(s-c)\cos(B))^2+(r_c-(s-c)\sin(B))^2 \]
and expanding yields $r_c^2+(s-c)^2+(s-b)^2 -2(s-b)(s-c)\cos(B) - 2[ABC]\sin(B) $ where $[ABC]=(s-c) r_c$. Since $DI_b'^2$ is symmetric in $b,c$, we have \[DI_b'^2-DI_c'^2=r_b^2-r_c^2+2(s-b)(s-c) ( \cos(B) - \cos(C)) + 2[ABC](\sin(B) - \sin(C))\]
so it suffices to show that $(s-b)(s-c) ( \cos(B) - \cos(C)) + [ABC](\sin(B) - \sin(C)) = 0$ which follows immediately after expanding and using $\sin(B) = \tfrac{2[ABC]}{ac}$ and $\sin(C) = \tfrac{2[ABC]}{ab}$. $\square$

[b]s[/b]olved with Courageous Kanav Talwar

[hide = solution]
We advertise the use of the so called "lin pop trix".

In what follows, we will denote by $F(P, \omega)$ the power of point $P$ wrt circle $\omega$. Define $M_B, M_C$ as the midpoints of sides $AC, AB$ respectively. Let the incircle $\omega$ of $\triangle ABC$ meet sides $BC, CA, AB$ at points $D, E, F$ respectively. Define $D'$ as the $A$-extouch point.

Observe that $AE, AF$ are tangent to circles $w'_B, w'_C$ respectively and so $A$ has same power wrt $w'_B, w'_C$.  Therefore it suffices to show that $D'$ lies on the radical axis as well, since $AD'$ bisects the perimter of $\triangle ABC$.

We first aim to compute $F(B,w_B')$. Define $G(P)$ for a point $P$ as \[G(P) = F(P, w_B) - PM^2\]
By aforementioned tricx, we know that $G$ is linear. Let $X$ denote the reflection of $B$ in $M_B$. We have \[\frac{(c-a)^2}{4} = G(M) = \frac{G(X)+G(B)}{2} = \frac{G(X)+s^2-BM^2}{2}\] from where we deduce (after Apollonius drama) that \[F(B, w_B') = F(X, w_B) = \frac{3(a^2+c^2)-b^2-2ac-2s^2}{2}\]
Define a function $H(P)$ as \[H(P) = F(P, w_B')-F(P, w_C')\] over all points $P$. Remark that we've to prove $H(D') = 0$. Since $H$ is linear, we need only show that \[(s-b)H(B)+(s-c)H(C) = 0\]
Observe \[H(B) = F(B, w_B')-F(B, w_C') = \frac{3(a^2+c^2)-b^2-2ac-2s^2-2(s-b)^2}{2} = (a-c)^2-b^2 = -4(s-c)(s-a)\]
Similarily we'd have $H(C) = -4(s-a)(s-b)$ right? Nope.

Observe that $H$ was defined as $w_B'$ power minus $w_C'$ power, so we need to flip the sign for $H(C)$ to make sense. Thus \[H(C) = 4(s-a)(s-b)\] and we win for obvious reasons. Hence the radixal axis of $w_B'$ and $w_C'$ halves the perimeter of $\triangle ABC ~$ $\square$.
[/hide]


For fixed $n$ we have the numbers $x_1, \ldots, x_n$ satisfying  \[\sum^n_{k=1} x^3_k = \sum^n_{k=1} x_k = 0.\] How many sets of two numbers $x_k$ and $x_i,$ $i \neq k,$ can be found whose sum equals 0.

you mean at most? If not I don't get what to do, if yes that sounds incredibly simple:

It's just when the sets are all: a,-a,a,-a, etc. and for odd n add a zero.

So the answer is: n/4 (being  $\displaystyle\binom{\frac n2}1\cdot\binom{\frac n2}1$) for even n, and logically (n-1)/4 for odd n.

Yes, it shoud mean at most. And the numbers $x_i$ are provided. You cannot choose them as you prefer.  :)

Uh, you can't say both of those.. if you are going to say "at most" then you are going to have to let something be variable :P. If everything in that expression is fixed, how are you meant to maximise something?

So I'm not really sure what the actual question is..

Hmm... I think he means that the requirement is enough to figure out what numbers they are exactly. So in that case the answer would even be more simple:

$\frac n2$ for even $n$; $\frac{n-1}2$ for odd $n$

since the thingies are a permutation of a, -a, b, -b, c, -c, etc. and for odd n add a ", 0" to that.


algebra
polynomial
geometry
3D geometry
binomial coefficients
AMC

What is

 :Sigma: n= 1 to 1000 of 3n:^2: - 3n +1

The general method to derive a formula for the sum

1^k + 2^k + 3^k + ... + n^k :

Determine a polynomial of degree k + 1 (with constant term 0) such that for all x :

f(x) - f(x - 1) = x^k.

Then just telescope.
For example, k = 2 :

We are looking for a polynomial f(x) = ax + bx + cx such that
f(x) - f(x - 1) = x.
Now you can work this out (do it) and compare coefficients.
Then use a telescoping sum. Nice method huh :)

Arne, perhaps you can show how your method works by going through an entire example... say 1^3+2^3+...+n^3.

That k=2 example does not really make sense to me :\

[color=#00FF00]There is actually a much better way to do that particular problem.

3n^2 - 3n + 1 = n^3 - (n - 1)^3

So it just telescopes.  Much less work, yes?[/color]

aww c'mon Joel
now ur making me feel bad
how did u see that factoring?

Joel, I know that proof, and that proves 1^2+2^2+3^2... not 1^3+2^3+3^3...

Also, I wanted to see Arne's method work, if it will work for all powers k.

[color=#00F000]No, no -- I was explaining how to do the original problem SouthPaw asked, not the harder, more general question you guys are dealing with.

The numbers 3, 3, and 1 never come in that order unless there are cubes involved -- at least, if you see something that vaguely resembles binomial coefficients, it can be helpful to check out why they are there.  You don't get binomial coefficients without there being a good reason for them.[/color]

[color=#00FF0F]Besides which, that is exactly what Arne's method relies on -- he just takes as a given the (provable) statement
1^n + 2^n + 3^n + . . . + m^n is a polynomial of degree (n+1) of m.  Then he solves equations for the coefficients.[/color]

Sorry, the polynomial shouldn't be monic ...

An example for k = 2.

Suppose f(x) = ax + bx + cx and f(x) - f(x - 1) = x.
Now x = f(x) - f(x - 1)
= ax + bx + cx - a(x - 1) - b(x - 1) - c(x - 1)
= 3ax + (2b - 3a)x + (a - b + c)
Comparing coefficients we see that
3a = 1 => a = 1/3
2b - 3a = 0 => b = 1/2
a - b + c = 0 => c = 1/6.
So f(x) = x/3 + x/2 + x/6 = x(x + 1)(2x + 1)/6.
Now note that
n + (n - 1) + ... + 2 + 1
= (f(n) - f(n - 1)) + (f(n - 1) - f(n - 2)) + ... + (f(2) - f(1)) + (f(1) - f(0))
= f(n) - f(0) = n(n + 1)(2n + 1)/6.

However this becomes quite difficult if k becomes large.
Joel's recursion - style solution is an alternative but I think it's also difficult to apply his idea for large k.


linear algebra
matrix
vector
linear algebra unsolved

Suppose A and B are all n*k matrices with rank = k, here (n>k). So the columns of A and B are linearly independent vectors in R^n. Prove that there exists a n*n invertible matrix C such that C*A=B

Thx

This is basically just Gaussian elimination. Each matrix is row equivalent to a matrix in reduced row echelon form, and knowing that they are of maximal rank uniquely specifies for us what that reduced row echelon form is. And row equivalency is given by left-multiplication by an invertible matrix.

$ \exists\, n\times n$ invertible matrices $ P$ and $ Q$ such that $ PA\equal{}\begin{pmatrix}I\\0\end{pmatrix}$ and $ QB\equal{}\begin{pmatrix}I\\0\end{pmatrix}.$

Then all we have to do is to note that $ Q^{\minus{}1}PA\equal{}Q^{\minus{}1}QB\equal{}B.$ Your $ C\equal{}Q^{\minus{}1}P.$


Find k such that the equation $x^{2}-kx+4 = 0$ has a repeated real solution.

MY WORK:

I understand that I must find a value for k, right?

What is a repeated real solution?

How do I find k?

[quote="sharkman"]Find k such that the equation x^2 -kx + 4 = 0 has a repeated real solution.

MY WORK:

I understand that I must find a value for k, right?

What is a repeated real solution?

How do I find k?[/quote]

[hide]For a repeated solution, we must have the discriminant equal to $0$:

$0 = D = k^{2}-16$

$k^{2}= 16$

$k = \pm 4$[/hide]

The form must be a perfect square, so $(x+a)^{2}=x^{2}-kx+4\implies a=\pm 2$ etc.

I think a repeated solution is one that happens twice (basically, the discriminant is zero), for quadratics only in perfect square trinomials ( $x^{2}+2xy+y^{2}$).

If you're familiar with factoring these, just eyeball it and you can see its $\pm4$

[quote="sharkman"]bpms,

What's that weird (but interesting) looking symbol you presented here?[/quote]

That would be a sigma. Read about it [url=http://www.artofproblemsolving.com/Wiki/index.php/Sigma_notation]here[/url].

[quote="13375P34K43V312"][quote="bpms"]Try applying a full blooded substitution.
Like $x=\sum^{1}_{0}(.5)^{k+1}n^{k}$.
Then everything cancels out.[/quote]

LOL nice
way to try to scare him away

i have to try that...[/quote]
What do you mean?
That substitution is a cool one, it trivializes the problem.

[quote="sharkman"]Find k such that the equation $ x^{2}-kx+4 = 0$ has a repeated real solution.

MY WORK:

I understand that I must find a value for k, right?

What is a repeated real solution?

How do I find k?[/quote]

[hide]repeated real is when discriminent is 0. Therefore, k^2-16=0

k^2=16. k=+/-4[/hide]


geometry
function
calculus
integration
symmetry
calculus computations

Find the area between the curves $ y\equal{}x^{2}\plus{}1$ and $ y\equal{}11\minus{}\frac{9}{x^{2}}$.

Not possible to calculate the area
Because X = -1 and X = 1 points of intersection between the curves
And X = 0 between them and the second function is not defined when X = 0

i don't think so...we can integrate this of course...look at the graph below...the points of intersection are $ (1,2)$, $ (\minus{}1,2)$, $ (3,10)$ and $ (\minus{}3,10)$..the area would be twice the integral of, say $ f(x) \minus{} g(x)$, from $ x\equal{}1$ to $ x\equal{}3$, where $ f(x) \equal{} 11\minus{}9/x^2$ and $ g(x) \equal{} x^2\plus{}1$.

I'm really wrong, I did not realize
Question intelligent and deceptive

Sorry, I should have specified that I meant in the first quadrant, that is, between the points of intersection $ (1, 2)$ and $ (3, 10)$.

actually, i've  also encountered a similar problem before in which we are asked to find the area between blah3...and at first I thought that finding the area literally between the curves...so what I did was to look for the axis of symmetry but I couldn't get the area..so I asked our teacher to have some clarification then I realized what he meant by "between" was actually to simply get the area bounded by the given curves...then everything went well..

So I understand if anybody could mistakenly understood a problem...anyhow, we always make mistakes but during exams, as much as possible, we should be careful though...c",)


geometry
geometric transformation
rotation
linear algebra
matrix

Choose $ \theta$ such that $ 0 \leq \theta \leq \pi$. Show that counterclockwise rotation in $ \mathbb{R}^2$ by $ \theta$ is an orthogonal transformation.

I think I need to show that the matrix representation for this rotation with respect to the standard basis is orthogonal...not entirely sure how to do that though...?

What's the definition of an orthogonal matrix?  What is the matrix for rotation by $ \theta$?  (If you don't know how to do this, you can figure it out either by figuring out the images of $ (1, 0)$ and $ (0, 1)$ or by using polar coordinates.)


function
algebra
polynomial
abstract algebra

[b]Part One.[/b]  Find families of polynomials $f, g$ such that $f(g(x)) = g(f(x))$.

[b]Part Two.[/b]  Find families of functions $f, g$ such that $f(g(x)) = g(f(x))$.

Reminiscient of a WOOT problem...

(you might want to post that and its solution)

[quote="Thales418"]one possible solution is ${f}$ be the inverse of ${g}$[/quote]

Boring  :P

What I'm interested in part one is examples where $\text{deg }f, g > 1$, since I haven't come up with an easy way to generate such polynomials and am not sure if they even exist.  (Anyone want to venture a proof?)

And part two is basically asking for abelian function groups under composition.  There's a very nice example of such a group :)

Abelian means commutative, and it describes the group, not the functions.  Sorry for the terminology.


vector
linear algebra
linear algebra unsolved

Let $S$ and $T$ be disjoint affine subspaces of $\mathbb{A}^n$, having dimensions $s$ and $t$ respectively.

Find the dimension of the smallest affine subspace of $\mathbb{A}^n$ containing both $S$ and $T$.

What is $\mathbb{A}^n$? It's not a standard notation afaik...

That's just the $n$-dimensional affine space :)

Hmm.. that's weird. I don't think it only depends on $s,t$... just take 2 coplanar/noncoplanar lines.

[I think the most used international notation is $AG(n,K)$ with $n$ with $n$ the dimension and $K$ the field.]

It is ok to write $\mathbb{A}^n (K)$ imo, but you should always state the field in which you're working, even if it's arbitrary.

[are you even working over a field? maybe that explains my silly answer...]

I'm working over $\mathbb{R}$ in this case.

Anyways, the answer is simple: $s + t + 1 - \dim(V \cap W)$.

And what are $V$ and $W$ supposed to mean?

$V$ and $W$ are just the underlying vector spaces, sorry :)

Hmm, with underlying vectorspaces you mean $span(S,\vec{o})$, right? But then your answer is still wrong, no?
- You need a 3-dimensional space to contain 2 noncoplanar lines (while your formula gives 2)
- You need a 2-dimensional space to contain 2 coplanar lines (while your formula gives sometimes 1 and sometimes 2)
- ...

And this regarless of the field you're working in, finite, real, complex, ...

[quote="Peter VDD"]
- You need a 3-dimensional space to contain 2 noncoplanar lines (while your formula gives 2)
- You need a 2-dimensional space to contain 2 coplanar lines (while your formula gives sometimes 1 and sometimes 2)
- ...
[/quote]

No, my formula gives the right results.

With the underlying vector space I mean: we can write $S = p + V$, where $p$ is a point and $V$ is a vector space.

Ah, in that way... yes, that's the same as my $\dim(span(V,W)) +1$ then...


Suppose we fill numbers 0-8 into a 3x3 matrix. Then we can swap as many times as we want. The rule is we can only swap 0 with its adjacent neighbors(at most 4 neighbors). How many permutations can we get?

if i'm understanding correctly...

the rule on 0 doesn't restrict anything. it can end up anywhere we want. likewise, any othere element can end up anywhere we want. so every permutation is attainable, and the number of all permutations is easy to count...

Think again.

For 2x2 case, suppose the original permutation is (0,1,2,3)(left to right, top to bottom), then you can never get permutation (0,1,3,2).

how so? just swap 2 and 3. your question implies you can swap any two elements. unless you'd like to clarify it for us...

Sorry for the obscurity. What I meant is you can only swap 0 and its neighbors. Swapping two neighbors which do not include 0 is illegal.

Probably half of them, check [url]http://en.wikipedia.org/wiki/Fifteen_puzzle[/url]


function
real analysis
inequalities
calculus
integration
derivative
real analysis unsolved

Prove that if $f:I\to\mathbb{R}$, (where $I$ is and interval in $\mathbb{R}$ )is a primitivable function and satisfies the relation
$f\left(\frac{x+y}2\right)\le\frac{f(x)+f(y)}2$ for any $x,y\in I$ then it is continuous.
[i]Note[/i]: A function $f:A\to\mathbb{R}$ is called primitivable if there exists a function $F:A\to\mathbb{R}$, $F$ derivable s.t. $F'=f$.

I assume you mean $f$ is "primitivable", not $I.$ I also assume you mean for $I$ to be a bounded interval.

Could you please define exactly what you mean by "primitivable"? Does it mean Riemann integrable? Lebesgue integrable? Lebesgue measurable?

I vaguely remember there being several topics a few months ago about convex ("Jensen-convexe") functions.

[quote="grobber"]Oh, there's no Lebesgue integration in high school contests here :). I'm sure it just means that there is a function $F$ s.t. $F'=f$: $f$ is a derivative.[/quote]
Actually you have to know Lebesgue criterion for integration for olympiads... is in that list with requirements grobber (I'm sure you remember)

[quote]I'm sure it just means that there is a function $F$ s.t. $F'=f$[/quote]
Everywhere? What about a function like $f(x)=\mathrm{sgn}(x)$ which is the derivative of $F(x)=|x|$ everywhere except the origin. (Note that this function- or any function with a jump discontinuity - doesn't satisfy the midpoint-convex condition, so it doesn't matter to this problem.)

hmm... the function $f(x)=sgn(x)$ is not primitivable.. exactly because there are no functions $F$ with the property that $F'=f$ EVERYWHERE
for a function to be primitivable it is necessary(not always sufficient) to has Darboux Property (it transoforms and interval into an interval)
the definition is with $f:A\to\mathbb{R}$ where (I think I omitted it) $A$ is an interval from $\mathbb{R}$.


Source: AHSME 1973 #16

If the sum of all the angles except one of a convex polygon is $2190^\circ$, then the number of sides of the polygon must be
$\text{(A)} \ 13 \qquad \text{(B)} \ 15 \qquad \text{(C)} \ 17 \qquad \text{(D)} \ 19 \qquad \text{(E)} \ 21$

[hide]Is it $13$?
$2190=180(n-2), n-2\approx12, n-2\approx 14. 14\times 180=2520$ so i rounded down.... and each angle is less than 180.  [/hide]

[hide]
\begin{eqnarray*}(n-2)\cdot 180=2190+\gamma\\(n-2)\cdot 180=12\cdot 180+30+\gamma\end{eqnarray*}

LHS is integer divisible by $180$ so must be RHS, and since $0<\gamma<180$ it's clear that $\gamma=150$.

So we have $(n-2)\cdot 180=13\cdot 180\implies n-2=13\implies n=15$
[/hide]


Find the mean of all numbers ordered from 0 to 10000 in decimal notation in which $ 1$ doesn't appear .

[hide]
Consider the following  ____   ____   ____   ____ .
Total number of $ 1$,$ 2$,$ 3$ and $ 4$ digit numbers which do not contain $ 1$ are the total number of ways in which we can fill the four blanks with the numbers $ 0$,$ 2$,$ 3$,$ .....$,$ 8$,$ 9$ with repetitions. (This generates $ 1$ digit numbers when the first three blanks are filled with $ 0$, two digit numbers when the first two blanks are filled with $ 0$. This also includes the number $ 0000$ which is equivalent to $ 0$
So, the total number of such numbers (including $ 0000$) is $ 9^{4}$
Now ,we list the numbers out:

$ 0 0 0 0$

$ 0 0 0 1$

$ 0 0 0 2$
...
...
$ 0 1 6 7$

$ 0 1 6 8$
...
...
$ 9 9 9 9$

We columnwise add them. Each column contains equal numbers of the numbers $ 0$,$ 2$,$ 3$...$ 9$. We are adding $ 0000$ but that does not change the sum.

sum of each column = $ 9^{3}\cdot (0 \plus{} 2 \plus{} 3 \plus{} 4...9)$
                              = $ 32076$

Now putting proper value for the units,tens and hundreds place
we get sum of all the numbers = $ 35636436$ [hide="like this"]
$ 32076 \plus{} 320760 \plus{} 3207600 \plus{} 32076000$[/hide]

Total number of such numbers = $ 9^{4}$.

Average = $ 5431.55$

[/hide]

Great solution graffiti, but wouldn't the total number of such numbers actually be $ 9^4$, since the problem asked: Find the mean of all numbers ordered from 0 to 10000 in decimal notation in which 1 doesn't appear. From 0 to 10000 means that you actually DO count 0, correct?


probability
combinatorics proposed
combinatorics

We form a decimal code of $21$ digits. the code may start with $0$. Determine the probability that the fragment $0123456789$ appears in the code.

I'v found $p = \frac {12 \cdot 10^{11} - 30} {10^{21}}$.
Is that right?

Pierre.

That's kind  :blush:

Ok, I write my solution then :

There are exactly $10^{21}$ codes.
Consider $0123456789$ as one digit in a code of $12$ digits. There are $12$ positions for this block. For each of these positions, we have to fulfill the others which gives $10^{11}$ choices, and then a total of $12 \cdot 10^{11}$ dispositions.

But some of these are counted twice, since two blocks may occur in the same code. There can be more than two blocks....and a code with two blocks is a code with three digits (the two blocks and the 21-th number). There are three ways to place the two blocks, and for each 10 ways to choose the last number, which give a total of 30 codes with two blocks.
Thus the desired probability is : $p = \frac {12 \cdot 10^{11} - 30} {10^{21}}$, as claimed (if we omit the typo that I have edited :D).

Pierre.

sorry pbornsztein. I don't what you understood from my reply. It was only a confirmation, I didn't wanted it to sound ironic. Sorry again.

Yes I think that's the way he understood it, and he just said thank you. :)

Exactly. I took that as a compliment which made me shy ;)

Pierre.

Ok first I understood that you took it as a non-credibility remark so you posted the solution. As a matter of fact I must admit that it wasn't that much a compliment as it was a confirmation after I solved it myself.


Solve in integer numbers the equation:
[i]y[/i] 2 =1 + [i]x[/i] + [i]x[/i] :^2: + [i]x[/i] :^3: + [i]x[/i] :^4:

Take it easy!

Here is one more similar to yours(Just as easy and interesting):
y 2 +y=x+x 2 +x 3+x 4

Solve in integer numbers the equation: $y^2 = 1 + x + x^2 + x^3 + x^4$

Here is my solution:

$y^2 = 1 + x + x^2 + x^3 + x^4 \Leftrightarrow y^2 = 1 + x ( 1 + x^2 ) + x^2 (
1 + x^2 ) \Leftrightarrow ( y - 1 ) ( y + 1 ) = ( x + x^2 ) ( 1 + x^2 ) \Leftrightarrow \left\{\begin{array}{l}
  y - 1 = x + x^2\\
  y + 1 = 1 + x^2
\end{array}\right. \vee \left\{\begin{array}{l}
  y - 1 = 1 + x^2\\
  y + 1 = x + x^2
\end{array}\right. \Leftrightarrow \left\{\begin{array}{l}
  x = - 1\\
  y = 1
\end{array}\right. \vee \left\{\begin{array}{l}
  x = 3\\
  y = 11
\end{array}\right.$

You are wrong, Maks.
We cant's state $(y-1)(y+1)=(x^2+1)(x^2+x)$ implies that $y-1=x^2+1$ and $y+1=x^2+x$ OR $y+1=x^2+1$ and $y-1=x^2+x$.

OK, I corrected myself:

$y^2 = 1 + x + x^2 + x^3 + x^4 \Leftrightarrow y^2 = 1 + x ( 1 + x^2 ) + x^2 (
1 + x^2 )
\Leftrightarrow ( y - 1 ) ( y + 1 ) = x ( 1 + x ) ( 1 + x^2 )
\Leftrightarrow \left\{\begin{array}{l}
  y - 1 = x ( 1 + x )\\
  y + 1 = 1 + x^2
\end{array}\right. \vee \left\{\begin{array}{l}
  y - 1 = 1 + x^2\\
  y + 1 = x ( 1 + x^{} )
\end{array}\right. \vee \left\{\begin{array}{l}
  y - 1 = x ( 1 + x^2 )\\
  y + 1 = 1 + x
\end{array}\right. \vee \left\{\begin{array}{l}
  y - 1 = x^{} + 1\\
  y + 1 = x ( 1 + x^2 )
\end{array}\right. \vee \left\{\begin{array}{l}
  y - 1 = ( 1 + x ) ( 1 + x^2 )\\
  y + 1 = x
\end{array}\right. \vee \left\{\begin{array}{l}
  y - 1 = x\\
  y + 1 = ( 1 + x ) ( 1 + x^2 )
\end{array}\right. $

Which gives me the integer solutions:

$\left\{\begin{array}{l}
  x = - 1\\
  y = 1
\end{array}\right. \vee \left\{\begin{array}{l}
  x = 3\\
  y = 11
\end{array}\right. \vee \left\{\begin{array}{l}
  x = - 1\\
  y = - 1
\end{array}\right.$

I hope that's finally right... (omitting the fact that it's horribly ugly,
ineffective, ...)  :blush:

[quote="Maks"]I hope that's finally right... (omitting the fact that it's horribly ugly,
ineffective, ...)  :blush:[/quote]
I don't want to disappoint you, but you are wrong again.
For example, take $x=12$, then $x(x+1)(x^2+1)$ can be expressed as product of two numbers in many ways. And you pointed only some of them.

3x.lich has just removed reply where he stated that $x=3$ and $y=11$ is not a solution. :lol:

You're right. I will stop fighting mathematics now :)
Anyway thanks for pointing this out to me ...

P.S: And the second time I will try to think more before sending a "solution"


Suppose a bounded sequence $\{a_{n}\}_{n\geq1}$ satisfies that
\[a_{n}<\sum^{2n+2006}_{k=n}{\frac{a_{k}}{k+1}}+\frac{1}{2n+2007},n=1,2,3,....\]
Prove that $a_{n}<\frac{1}{n},\forall n \in N$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=130463]Whole paper[/url]


[color=darkblue]Let $ u_n\equal{}\displaystyle \left ( 1\plus{}\frac{1}{2^2}\plus{}\frac{1}{3^2}\plus{}\frac{1}{4^2}\plus{}...\plus{}\frac{1}{n^2} \right)$.
[size=150]Find[/size] $ lim u_n$ ?[/color]

This should go in the Calculus forum.

EDIT Hint was regarding [i]if[/i] it converged or not.


if $ a\equal{}log2$ when which of the following represents $ log_5{2}$ in terms of $ a$

[hide="Faster"]
Notice that $ 2 \cdot 5 \equal{} 10$, therefore $ \log 2 \plus{} \log 5 \equal{} 1$. So, $ \log 5 \equal{} 1 \minus{} \log 2$.

Next, the answer we're looking for is $ \log_5 2$, which is $ \frac{\log 2}{\log 5}$. Substituting $ \log 5 \equal{} 1 \minus{} \log 2$:

$ \frac{\log 2}{\log 5} \equal{} \frac{\log 2}{1 \minus{} \log 2} \equal{} \frac a{1\minus{}a}$.
[/hide]

how does $ log2 \plus{} log5 \equal{} 1$? i mean i see that this is faster, buy why would $ log2 \plus{} log5$ not equal 10?

EDIT: Never mind, just realized log10 =1 srry DISREGARD POST  :blush:


Find all $a,m,n \in Z+$ such that: $(a+1)^{m}=a^{n}+1$

see http://www.mathlinks.ro/Forum/viewtopic.php?t=129357  :wink:

for instance suppose $m,n>1$... then $a^{n}=a(af(a)+1)\implies a^{n-1}\equiv 1 (a)$ for the messy $f(a)$ in terms of combinatorical sums.
Thus $n-1=0$, so $a+1=(a+1)^{m}$ so $m=1$, which is satisfied for all a.

EDIT: sorry didn't see post above mine

Hmm :oops: can you please point me to my mistake?

Tell me why Zsigmondy's theorem is not elementary...  :P

Very sketchy (and only for $b=1$, but the general case is not harder, just "homogenize"):

Let $\Phi_{n}(x)$ be the $n$-th cyclotomic polynomial.
Known facts (solve them as problems if you want):

a) $\gcd(a^{m}-1,a^{n}-1)=a^{\gcd(m,n)}-1$.

b) $x^{n}-1 = \prod_{d|n}\Phi_{d}(x)$.

c) For $n\geq 3$, $k$ any integer and primes $p|n$, we have that $p^{2}\nmid \Phi_{n}(k)$.

d) Let $k,m,n$ be integers with $m|n$ and $m,n>0$. Then for any prime $p$, $\Phi_{m}(x)$ and $\Phi_{n}(x)$ can only have a common root $\mod p$ if $p|n$. In other words, if there is an integer $k$ and a prime $p$ with $p|\Phi_{m}(k), \Phi_{n}(k)$, then $p|n$.

Now to Zsigmondy:
- handle the case $n=2$, so assume $n\geq 3$.
- see that it suffices to show: $\Phi_{n}(a)$ has a prime divisor $p$ not dividing any $\Phi_{m}(a)$ with $m<n$.
- reduce it, using a), to the case $m|n$, $m \neq n$.
- write $\Phi_{n}(a)=uv$ with $u$ being the "bad" (already occured before) prime factors and $v$ being coprime to all those; all we want to show is $v>1$.
- use d) to get that all prime divisors that can be "bad" divide $n$.
- c) gives that $u \leq n$.
- show that $\Phi_{n}(a)>n$ except some small cases (this is where the 2^6-1 exception ocures).
- having $\Phi_{n}(a)>n$, the result follows, since then $v>1$.

[quote="Bazooka"]oh, what about  m=2,a=2,n=3?  :wink:
To NTT: I didn't see any real solution there  :D  Have you got any solution using elementary number theory?[/quote]
see http://www.kalva.demon.co.uk/ibero/isoln/isol003.html  :D

I have some similar problems :
 Problem 1:"Find all triples of positive integer $(a,m,n)$ such that $(a+1)^{n}$ is divisible by $a^{m}+1$".
 Problem 2:"Find all triples of positive integer $(a,m,n)$ such that $(a-1)^{n}$ is divisible by $a^{m}-1$".
 I have solved thems, they are not easy!!!

I'm sorry about my poor English.

Zsigmondy strikes again ;)
Why do you all dislike it :(

because we left a lot of time to solve this problem and your zsigmondy make it very easy;($a^{n}+b^{n}=c^{n}$)3 :P


[quote=Vasc]If $ a,b,c$ are nonnegative real numbers, then

$ (b) \ \ \ \  12\sum \frac {ab}{(a\plus{}b)^2} \le 5\plus{}\frac {4(ab\plus{}bc\plus{}ca)}{a^2\plus{}b^2\plus{}c^2},$

with equality for $ a\equal{}b\equal{}c$, and again for $ \ \frac a{2}\equal{}b\equal{}c$, or $ \ \frac b{2}\equal{}c\equal{}a$, or $ \ \frac c{2}\equal{}a\equal{}b$.[/quote]
We'll prove that this inequality is true for all reals $a$, $b$ and $c$ such that $\prod_{cyc}(a+b)\neq0$.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, our inequality it's $\frac{12\sum\limits_{cyc}ab(c^2+3v^2)^2}{(9uv^2-w^3)^2}\leq5+\frac{4v^2}{3u^2-2v^2}$ or
$\frac{4(3w^6+A(u,v^2)w^3+B(u,v^2))}{w^6-18uv^2w^3+81u^2v^4}\leq\frac{5u^2-2v^2}{3u^2-2v^2}$ or
 $(36u^2-24v^2-5u^2+2v^2)w^6+E(u,v^2)w^3+F(u,v^2)\leq0$ or $f(w^3)\leq0$,
where $f$ is a convex function  because $f(x)=(31u^2-22v^2)w^6+E(u,v^2)w^3+F(u,v^2)$  and $31u^2-22v^2\geq0$.
Thus, by [url=http://www.artofproblemsolving.com/community/c6h278791]uvw[/url] it remains to check one case only:  $b=c=1$,
which gives $(a-1)^2(a-2)^2\geq0$. Done!

[quote=Vasc]If $ a,b,c$ are nonnegative real numbers, then

$ (a) \ \ \ \  \sum \frac {a(ab\minus{}2bc\plus{}ca)}{ka^2\plus{}b^2\plus{}c^2} \ge 0,$

where $ k\equal{}1\plus{}2\sqrt 2$, with equality for $ a\equal{}b\equal{}c$, and again for $ \frac a{\sqrt 2}\equal{}b\equal{}c$, or $ \frac b{\sqrt 2}\equal{}c\equal{}a$, or $ \frac c{\sqrt 2}\equal{}a\equal{}b$;

$ (b) \ \ \ \  12\sum \frac {ab}{(a\plus{}b)^2} \le 5\plus{}\frac {4(ab\plus{}bc\plus{}ca)}{a^2\plus{}b^2\plus{}c^2},$

with equality for $ a\equal{}b\equal{}c$, and again for $ \ \frac a{2}\equal{}b\equal{}c$, or $ \ \frac b{2}\equal{}c\equal{}a$, or $ \ \frac c{2}\equal{}a\equal{}b$.[/quote]

szl...,please do it to sos.
BQ

[quote=Vasc]If $ a,b,c$ are nonnegative real numbers, then
$ (b) \ \ \ \  12\sum \frac {ab}{(a\plus{}b)^2} \le 5\plus{}\frac {4(ab\plus{}bc\plus{}ca)}{a^2\plus{}b^2\plus{}c^2},$

with equality for $ a\equal{}b\equal{}c$, and again for $ \ \frac a{2}\equal{}b\equal{}c$, or $ \ \frac b{2}\equal{}c\equal{}a$, or $ \ \frac c{2}\equal{}a\equal{}b$.[/quote]

We have
\[RHS-LHS=\frac{4}{a^2+b^2+c^2}\sum{\frac{(a-b)^2(ab-2c^2)^2}{(b+c)^2(c+a)^2}}+\frac{(a^2+4ab+4ac+b^2+4bc+c^2)(a-b)^2(a-c)^2(b-c)^2}{(a^2+b^2+c^2)(a+b)^2(b+c)^2(c+a)^2}\ge{0}\]

[quote=arqady]$(a)$ we can prove by $uvw$ technique, but it's very ugly.[/quote]

SOS is not hard to

[quote=szl6208][quote=Vasc]If $ a,b,c$ are nonnegative real numbers, then

$ (a) \ \ \ \  \sum \frac {a(ab\minus{}2bc\plus{}ca)}{ka^2\plus{}b^2\plus{}c^2} \ge 0,$

where $ k\equal{}1\plus{}2\sqrt 2$, with equality for $ a\equal{}b\equal{}c$, and again for $ \frac a{\sqrt 2}\equal{}b\equal{}c$, or $ \frac b{\sqrt 2}\equal{}c\equal{}a$, or $ \frac c{\sqrt 2}\equal{}a\equal{}b$;
[/quote]

We have
\[LHS=\frac{(8+4\sqrt{2})\sum{c[ab-(2-\sqrt{2})(bc+ca)]^2(a-b)^2}+\sum{c[a^2+b^2+(6\sqrt{2}-7)c^2][a+b-(1+\sqrt{2})c]^2(a-b)^2}}{\prod{[(1+2\sqrt{2})a^2+b^2+c^2]}}\ge{0}\][/quote]

Dear,put .txt here.

BQ

[quote=arqady]Dear szl6208. Can you show us a SOS solution for all reals $a$, $b$ and $c$ such that $\prod_{cyc}(a+b)\neq0$? Thank you!
By the way, in your solution something is not true. :maybe:[/quote]
[quote=Vasc]If $ a,b,c$ are nonnegative real numbers, then
$ (b) \ \ \ \  12\sum \frac {ab}{(a\plus{}b)^2} \le 5\plus{}\frac {4(ab\plus{}bc\plus{}ca)}{a^2\plus{}b^2\plus{}c^2},$

with equality for $ a\equal{}b\equal{}c$, and again for $ \ \frac a{2}\equal{}b\equal{}c$, or $ \ \frac b{2}\equal{}c\equal{}a$, or $ \ \frac c{2}\equal{}a\equal{}b$.[/quote]
Dear arqady,see following:
We have
\[RHS-LHS=\frac{1}{3(a^2+b^2+c^2)}\sum{\frac{(a^2b-2a^2c+ab^2+3abc-ac^2-2b^2c-bc^2+c^3)^2}{(b+c)^2(c+a)^2}}+
\frac{3}{a^2+b^2+c^2}\sum{\frac{(a-b)^2(ab-2c^2)^2}{(b+c)^2(c+a)^2}}\ge{0}\]

[quote=Vasc]If $ a,b,c$ are nonnegative real numbers, then

$ (a) \ \ \ \  \sum \frac {a(ab\minus{}2bc\plus{}ca)}{ka^2\plus{}b^2\plus{}c^2} \ge 0,$

where $ k\equal{}1\plus{}2\sqrt 2$, with equality for $ a\equal{}b\equal{}c$, and again for $ \frac a{\sqrt 2}\equal{}b\equal{}c$, or $ \frac b{\sqrt 2}\equal{}c\equal{}a$, or $ \frac c{\sqrt 2}\equal{}a\equal{}b$;

[/quote]
This inequality is equivalent to $f(w^3)\geq0$, where $f$ is a decreasing function, which says that it's enough to assume $b=c=1$.

[quote=Vasc]If $ a,b,c$ are nonnegative real numbers, then

$ (a) \ \ \ \  \sum \frac {a(ab\minus{}2bc\plus{}ca)}{ka^2\plus{}b^2\plus{}c^2} \ge 0,$

where $ k\equal{}1\plus{}2\sqrt 2$, with equality for $ a\equal{}b\equal{}c$, and again for $ \frac a{\sqrt 2}\equal{}b\equal{}c$, or $ \frac b{\sqrt 2}\equal{}c\equal{}a$, or $ \frac c{\sqrt 2}\equal{}a\equal{}b$;

$ (b) \ \ \ \  12\sum \frac {ab}{(a\plus{}b)^2} \le 5\plus{}\frac {4(ab\plus{}bc\plus{}ca)}{a^2\plus{}b^2\plus{}c^2},$

with equality for $ a\equal{}b\equal{}c$, and again for $ \ \frac a{2}\equal{}b\equal{}c$, or $ \ \frac b{2}\equal{}c\equal{}a$, or $ \ \frac c{2}\equal{}a\equal{}b$.[/quote]

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

\[-{\frac {15600883}{13824000}}+{\frac {67}{1536000}}\,\sqrt {797543121}
+{\frac { \left( {\frac {1960987}{1536000}}-{\frac {67}{1536000}}\,
\sqrt {797543121} \right)  \left( acd+abd+abc+bcd \right) }{{a}^{3}+{c
}^{3}+{b}^{3}+{d}^{3}}}\]
\[\geq {\frac {abc}{ \left( a+b+c \right) ^{3}}}+{
\frac {bcd}{ \left( b+c+d \right) ^{3}}}+{\frac {acd}{ \left( a+c+d
 \right) ^{3}}}+{\frac {abd}{ \left( d+a+b \right) ^{3}}}\]

equality hold when $a=b=c=(-1/20+{\frac {1}{60}}\,\sqrt {1929})d$.

[quote=Vasc]If $ a,b,c$ are nonnegative real numbers, then

$ (a) \ \ \ \  \sum \frac {a(ab\minus{}2bc\plus{}ca)}{ka^2\plus{}b^2\plus{}c^2} \ge 0,$

where $ k\equal{}1\plus{}2\sqrt 2$, with equality for $ a\equal{}b\equal{}c$, and again for $ \frac a{\sqrt 2}\equal{}b\equal{}c$, or $ \frac b{\sqrt 2}\equal{}c\equal{}a$, or $ \frac c{\sqrt 2}\equal{}a\equal{}b$;

$ (b) \ \ \ \  12\sum \frac {ab}{(a\plus{}b)^2} \le 5\plus{}\frac {4(ab\plus{}bc\plus{}ca)}{a^2\plus{}b^2\plus{}c^2},$

with equality for $ a\equal{}b\equal{}c$, and again for $ \ \frac a{2}\equal{}b\equal{}c$, or $ \ \frac b{2}\equal{}c\equal{}a$, or $ \ \frac c{2}\equal{}a\equal{}b$.[/quote]

Dear Vasc,very nice this:

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

\[{\frac {19}{21}}+\frac{2}{7}\,{\frac {ab+bc+cd+ad+ac+bd}{{a}^{2}+{c}^{2}+{b}^{
2}+{d}^{2}}}\geq{\frac {ab+bc+ac}{ \left( a+b+c \right) ^{2}}}+{\frac {bc
+cd+bd}{ \left( b+c+d \right) ^{2}}}+{\frac {cd+ad+ac}{ \left( a+c+d
 \right) ^{2}}}+{\frac {ad+ab+bd}{ \left( d+a+b \right) ^{2}}}\]

equality hold when $a=b=c=\frac{2}{3}d$


analytic geometry
combinatorics proposed
combinatorics

Find the smallest natural number $n$ that the following statement holds :
Let $A$ be a finite subset of $\mathbb R^{2}$. For each $n$ points in $A$ there are two lines including these $n$ points. All of the points lie on two lines.

Old but nice.

The smallest desired number is $6$.

Let $E$ be the set of the $6$ points with coordinates $(0,0),(2,0),(2,2),(0,2),(0,1),(1,1).$
It is easy to verify that any subset of $5$ points from $E$ can be covered by two lines, but $E$ cannot be covered by two lines (since there is no four collinear points in $E$, any covering of $E$ by two lines would separate the six points into three points of collinear points. The unique triple of collinear points from $E$ which contain $(0,1)$ is $(0,0),(0,1),(0,2)$. Therefore the second line has to cover the three other points which are not collinear...).

Now, we prove that $n=6$ has the desired property.
Let $E$ be any set of at least $6$ points in the plane, such that any subset of $6$ points from $E$ can be covered by two lines.
Let F be any subset of $E$ with $6$ points. From the pigeon-hole principle, at least three points from $F$, and then from $E$, must be collinear. Say $A,B,C$ are collinear, and belong to the line $\Delta$.
If $E$ contains at most two points which are not on $\Delta$, we are done. Otherwise, let $X,Y$ be two points from $E$ which are not on $\Delta$, and let $\Delta '$ be the line $(XY)$.
For any point $M$ in $E$ which is not $X$ or $Y$ and which does not belong to $\Delta$, we know that the set formed by $A,B,C,X,Y,M$ can be covered by two lines.
If none of these lines is $\Delta$ then the line which covers $A$ cannot cover $B$ or $C$. Thus the other line must cover $B$ and $C$ so that this is $\Delta$, a contradiction.
Therefore, one of the line is $\Delta$. It follows that the other line must cover $X,Y,M$, which forces this second line to be $\Delta '$ and $M \in \Delta '$.
Thus, any point from $E$ belongs either to $\Delta$ or $\Delta '$, which proves that $E$ can be covered by two lines, and we are done.

Pierre.


In a class a guy assigns an integer value for the prettiness of a girl. If three girls' prettiness are three consecutive integers, then he calls them triadic. What is the maximum number of triadic?

I don't think this question is well-posed.