Let's consider the equation
P(x)P(-x) = P(x^2-1)
Is well known that if P has only real roots then this roots can be decribed and we can describe all such polynomials. I thought on the question to find all solutions, but the success up to now is not so far. Anyone have idea about this problem?

Namdung

well this problem can be considered solved, since I have proposed almost  the same problem in Round 4, 2nd Edition of the MathLinks Contest. Only now I have seen this post :D:D

Valentin, this problem still open. Your problem in 4-round states that "The real polynomial f(x) has odd degree and co-prime with x^2-x-1 and f(x^2-1) = f(x)f(-x) (*) then f(x) has at least two complex roots". It well-known fact and we can see that if f(x) satisfy (*) and f has only real roots then
f(x) = x^k.(x+1)^k.(x-a)^p.(x-b)^q where a, b are two roots of x^2-x-1.

My question is: Find all real polynomial f(x) (f can has complex roots) satisfying (*).

For example, P(x) = x^6 + x^5 - 2x^4 - x^3 + x^2 + 1 is one of such solution.

So, dont hurry to move this one to solved problems section.

Namdung

oh you are right Namdung. I have misread the statement of the problem. I shall move it back to unsolved.

If it is an open question then shall we move it to Open Questions section?


Βρειτε συνάρτηση $f: \ \mathbb{N} \rightarrow \mathbb{N}$ τετοια ώστε

ι) $f(x+y)=f(x)+f(y)+2xy$   $\forall x,y \in \mathbb{N}$

ii) $f(x)$ τελειο τετράγωνο για καθε φυσικο!

 :)

Απλή:

  Δείχνω αρχικά επαγωγικά ότι f(kx)=kf(x) + k(k-1)* x^2 .
  Έπειτα για χ=1 έχω ότι f(k) = k( f(1) + k-1 ) . Θεωρώ k κάποιο πρώτο
  αριθμό, τότε αφού f(k) τετράγωνο έπεται ότι αφού k| f(k) τότε
  k| f(k)/k δηλαδή k| f(1) + k -1 => k| f(1) -1 για κάθε k πρώτο αριθμό.
  Άρα f(1)=1 οπότε  f(k)=k^2. Επαληθεύω και ισχύει!


Αν $ x,y,z$ μη αρνητικοί πραγματικοί αριθμοί αποδείξτε ότι \[ \frac{{{x^2} \minus{} yz}}{{x \plus{} y}} \plus{} \frac{{{y^2} \minus{} zx}}{{y \plus{} z}} \plus{} \frac{{{z^2} \minus{} xy}}{{z \plus{} x}} \ge 0\].

Πρέπει να δείξουμε ότι $ \frac {{x^2}}{{x \plus{} y}} \plus{} \frac {{y^2}}{{y \plus{} z}} \plus{} \frac {{z^2}}{{z \plus{} x}} \ge\frac {yz}{x \plus{} y} \plus{} \frac {xz}{y \plus{} z} \plus{} \frac {xy}{z \plus{} x}$
Εύκολα προκύπτει ότι $ LHS\equal{}\frac {{y^2}}{{x \plus{} y}} \plus{} \frac {{z^2}}{{y \plus{} z}} \plus{} \frac {{x^2}}{{z \plus{} x}}$ αφού αυτό ισοδυναμεί με $ (x \minus{} y) \plus{} (y \minus{} z) \plus{} (z \minus{} x) \equal{} 0$, που ισχύει.

Αρα τώρα αρκεί να δειχτεί ότι:
$ \frac {{x^2 \plus{} y^2}}{{x \plus{} y}} \plus{} \frac {{y^2 \plus{} z^2}}{{y \plus{} z}} \plus{} \frac {{z^2 \plus{} x^2}}{{z \plus{} x}} \ge\frac {2yz}{x \plus{} y} \plus{} \frac {2xz}{y \plus{} z} \plus{} \frac {2xy}{z \plus{} x}$
Oμως:
$ \frac {{x^2 \plus{} y^2}}{{x \plus{} y}} \equal{} x \plus{} y \minus{} \frac {2xy}{x \plus{} y}$ και τα λοιπά κι έτσι η ανισότητα γράφεται ως:
$ x \plus{} y \plus{} z\geq \frac {y(x \plus{} z)}{x \plus{} y} \plus{} \frac {z(x \plus{} y)}{y \plus{} z} \plus{} \frac {x(y \plus{} z)}{x \plus{} z}$

Τώρα έστω ότι $ x\geq y\geq z$ (για να είμαστα σωστοί επειδή η ανισότητα [i]δεν είναι συμμετρική[/i] πρέπει να πάρουμε και την περίπτωση $ x\geq z\geq y$ αλλά ευτυχώς είναι ίδια :P )
Eτσι τώρα έχουμε $ x(y \plus{} z)\geq y(x \plus{} z)\geq z(x \plus{} y)$ και $ \frac {1}{y \plus{} z}\geq \frac {1}{x \plus{} z}\geq \frac {1}{x \plus{} y}$.

Αρα από ανισότητα [b]αναδιάταξης[/b] $ RHS\leq \frac {x(y \plus{} z)}{z \plus{} y} \plus{} \frac {y(x \plus{} z)}{x \plus{} z} \plus{} \frac {z(x \plus{} y)}{x \plus{} y} \equal{} x \plus{} y \plus{} z$

[hide]υστερα απο λιγες πραξεις,
$ \sum x^3y \geq xyz(x \plus{} y \plus{} z) \leftrightarrow \sum \frac {x^2}{z} \geq \sum x$
αληθης αφου
-$ \sum \frac {x^2}{z} \geq \sum x$ κατευθειαν απο andreescu
ή,
-$ \frac {x^2}{z} \plus{} \frac {z^2}{y} \plus{} \frac {x^2}{z} \plus{} y \geq 4x$
κυκλικα και QED  :)[/hide]


consider the set $S=\{1,2,3,4,5\}$.
consider all n-digit numbers whose digits are chosen from S.
for any given n, how many of these numbers have the property that any adjacent digits differ by exactly 1?

(for example for n=2 you have:  12, 21, 32, 23, etc)

[hide]For any $n$ consecutive digits, there are $2$ such numbers ($1\ldots n$ and $n\ldots1$).

The number of $n$ consecutive digits in a set with $k$ consecutive digits is $k-n+1$.

Therefore, the number of $n$-digit numbers, all of whose adjacent digits differ by  exactly $1$, is $2(k-n+1)$.

In this case, $k=5$, so the expression becomes $2(5-n+1)$, or $12-2n$.[/hide]

[quote="Teki-Teki"]nice solution, but you make one assumption: the digits are distinct.  this makes it a lot nicer, however, b/c if two digits can be the same you get nasty stuff like complicated recursions and whatnot.  sorry to spoil a good soln.[/quote]
Then wouldn't it be $\infty$ because you could have numbers like $12121212\ldots$

I think "all $n$-digit numbers whose digits are chosen from $S$" means you select digits from $S$, but you cannot have more than $1$ of each digit because $S$ only contains one of each.

[quote="i_like_pie"]
Then wouldn't it be $\infty$ because you could have numbers like $12121212\ldots$
[/quote]

but if you only use n digits, you can't use that unless $n=\infty$

repetition of digit is allowed. for example if n=3, the number 343 is in the desired form.

no one can solve this?   :maybe:

[hide="possible hint"]
1)this problem has a non-combinatorial solution.
2) for any n, how many n-digit numbers that have the following property end in 1 or 5?

[/hide]

[quote="i_like_pie"][hide]If this helps at all, here's the first $9$:

$\begin{tabular}{|c|c|c|c|}\hline n & \# & 1's, 5's & 2's, 3's, 4's\\ \hline 1 & 5 & 2 & 3\\ \hline 2 & 8 & 2 & 6\\ \hline 3 & 14 & 4 & 10\\ \hline 4 & 24 & 6 & 18\\ \hline 5 & 42 & 12 & 30\\ \hline 6 & 72 & 18 & 54\\ \hline 7 & 126 & 36 & 90\\ \hline 8 & 216 & 54 & 162\\ \hline 9 & 378 & 108 & 270\\ \hline \end{tabular}$

The 1's and 5's column means how many numbers end in 1 or 5, and the same for the 2's, 3's, 4's column.[/hide][/quote]

[hide]
Well, that helps a lot.  Call the number that end in $1$ or $5$ $a_{n}$, the number that end in $2$ $3$ or $4$ $b_{n}$ and the total $c_{n}$.  Then we can guess from th table that

$c_{n}=a_{n-1}+2b_{n-1}$.

This makes sense, because if we have an $n-1$ digit number that ends in $2$ $3$ or $4$, we can add one of the digits on either side of the last digit, while for $1$ or $5$ there is only one adjacent digit.

Now, the problem is finding a closed form.

 [/hide]

1. Why are you quoting the post directly preceding yours?

2. The relation you have written down is wrong.  Even if it were right, you would need more relations than that to calculate the sequence we desire.

3. In fact, 2 and 4 are essentially equivalent in this problem, but they are not equivalent to 3.  So, it would be best to consider not your $a_{n}$ and $b_{n}$ but instead three sequences, $a_{n}$, $b_{n}$ and $c_{n}$, corresponding to those ending in 1 or 5, 2 or 4, and 3.

4. Another (actually easier) question we can ask: what if we consider the digits mod 5 (so that 1 and 5 are adjacent to each other)?

well if you are given an n-1 digit number that does not end in 1 or 5. then there is only two choices for the next digit that can be added to it.
but if the n-1 digit ends in 1 or 5 then there is only one digit that can be added.

(so in the table provided, if you add twice the "2,3,4" column and the "1 or 5" column. then you get the number of desired numbers for the next n)

i dont know if this is useful; but just a thought.

[quote="JBL"]1. Why are you quoting the post directly preceding yours?[/quote]
Why wouldn't I?  It shows that I was responding to it.

[quote="JBL"]2. The relation you have written down is wrong.  Even if it were right, you would need more relations than that to calculate the sequence we desire.[/quote]
I know, I edited it.

[quote="JBL"]3. In fact, 2 and 4 are essentially equivalent in this problem, but they are not equivalent to 3.  So, it would be best to consider not your $a_{n}$ and $b_{n}$ but instead three sequences, $a_{n}$, $b_{n}$ and $c_{n}$, corresponding to those ending in 1 or 5, 2 or 4, and 3.[/quote]
[hide]
Using that idea, we can see that $a_{n}=b_{n-1}$, $b_{n}=a_{n-1}+2c_{n-1}$, and $c_{n}=b_{n-1}$.

That is enough relations to calculate the sequence recursively.
[/hide]

Because in a format like this, the default assumption is always that you are refering to the thing that directly preceded you.  (For example, it is completely clear to whom this paragraph is addressed; I did not need to quote your question to answer it.)

Also, now you're making progress.  Your recursive relation for $b_{n}$ is off, though -- given a number ending in 3, we can extend it either by adding a 2 or a 4, so $b_{n}= a_{n-1}+2c_{n-1}$.  Now, using your relations we can get a recursion for $b_{n}$ and so also for $a_{n}$ and $c_{n}$.  Then we can solve these from their initial values and then add the results.

[hide="A few more details"]We have $b_{n}= 3b_{n-2}$ with $b_{1}= 2$ and $b_{2}= 4$.  This is a standard homogeneous linear recursion relation, to be solved in any of the standard ways.  We get $b_{n}= a\left(\sqrt{3}\right)^{n}+b\left(-\sqrt{3}\right)^{n}$ for some $a, b$, and with our initial data we can actually calculate $a, b$.  Then we know also the values for $a_{n}, c_{n}$ and so we can calculate $a_{n}+b_{n}+c_{n}= \frac{(12+7\sqrt{3})3^{n/2}+(12-7\sqrt{3})(-1)^{n}3^{n/2}}{9}$.[/hide]

Here is an algorithmic way of thinking about it:

consider a graph with nodes 1,2,3,4,5, and connections between numbers with adjacent numbers, we want the total number of paths, for n=1, we have
A=
[1,0,0,0,0]
[0,1,0,0,0]
[0,0,1,0,0]
[0,0,0,1,0]
[0,0,0,0,1]

for an adjacency matrix, we have
B=
[0,1,0,0,0]
[1,0,1,0,0]
[0,1,0,1,0]
[0,0,1,0,1]
[0,0,0,1,0]
where a_{i,j}=1 if |i-j|=1, else =0

we need the total possible numbers...i.e. the total paths length n on our graph...i.e.

the sum of the entries in B^{n-1}*A...A is the identity matrix... so just B^{n-1}

or basically just det( [1,1,1,1,1] *B^{n-1} * ( [1,1,1,1,1]^T) )

[T is transpose]

for n=1,2,...12 i get
5,8,14,24,42,72,126,216,378,648,1134,1944
finite differences
3,6,10,18,30,54,90,162,270,486,810

now this follows a bizarre pattern... ignoring the first pair of terms, it goes by multiplying by 5/3, 9/5

suppose the differences form $b_{n}$ where $b_{n}=a_{n+1}-a_{n}$, n=1,2,....
$b_{1}=3$, $b_{2}=6$, $b_{3}=10$,  $b_{n+2}=3b_{n}$...ignore the first term, this is a difference equation...

$b_{n}=\frac{5\sqrt{3}+9}{9}*(\sqrt{3})^{n}+\frac{9-5\sqrt{3}}{9}*(-\sqrt{3})^{n}$ n=2,3,...

now we have that $a_{n}=3+\sum_{k=1}^{n-1}b_{k}$...we apply the geometric series formula....

$a_{n}=3+\frac{7\sqrt{3}+12}{9}* (\sqrt{3})^{n}+\frac{4\sqrt{3}-7}{3}-\frac{7\sqrt{3}-12}{9}*(-\sqrt{3})^{n}$

ok yuck jbl is probably right then...wow wtf was i thinking that was unnecessary...

[quote="Altheman"]now this follows a bizarre pattern... ignoring the first pair of terms, it goes by multiplying by 5/3, 9/5[/quote]

Yes.  In fact, the original sequence also follows such an odd pattern, alternating by $\frac{7}{4}$ and $\frac{12}{7}$ (excluding the first case).  (You did some extra work by going to the first differences and then coming back.)  This has to do with the nature of the recurrence relation, i.e. the sequence can most naturally be expressed in terms of the term two previous.  Thus the answer actually looks a lot nicer if you're willing to separate the cases $n = 1$, $n = 2m$ and $n = 2m+1$ for $m \geq 1$.


linear algebra
matrix
induction
vector
linear algebra unsolved

Two $ n\times n$ matrices $ A, B$ over a field $ F$ are said to be [i]p-similar[/i] if there exist a permutation matrix $ Q$ such that $ A\equal{}QBQ^{\minus{}1}$, and are said to be [i]p-equivalent[/i] if there exist permutation matrices $ P, Q$ such that $ A\equal{}PBQ$.

These clearly define equivalence relations on the set of $ n$ by $ n$ matrices. Find a set of canonical forms for each of these relations. Another related problem would be,
Given a matrix $ A$ find the set of matrices p-similar/p-equivalent to it.

PS. Let $ \sim$ be an equivalence relation on $ S$. A subset $ C\subset S$ is said to be a set of [i]canonical form[/i]s for $ S$ if each equivalence class contains exactly one member of $ C$.

(I am not sure if this is correct:) If $ Q$ is not the identity, $ B \mapsto QBQ^{ \minus{} 1}$ permutes the diagonal entries of $ B$, so one canonical form is the set of matrices with increasing (not necessarily strictly) diagonal entries $ b_{ii}$ with respect to some well-ordering (which may require the axiom of choice).  Even if this works, it is not likely to be a good solution.

Edit:  A solution along the same lines for the second part (and equally tenuous).  $ B \mapsto PBQ$ permutes the rows and then the columns of $ B$, so we can clearly pivot the smallest entry in $ B$ (again with respect to some well-ordering) into $ b_{11}$.  By induction, the $ k \times k$ minor containing $ b_{11}$ can be made to contain the $ k$ smallest entries in $ B$ in order.  In other words, the diagonal entries contain the $ n$ smallest entries in $ B$ in increasing order.

One problem with both of my above proofs is equal entries.  But note that permutation matrices do not affect the other elements in the same row and column of a given entry, so we can well-order the vector with $ 2k \minus{} 1$ entries containing the other elements.


Let $ A$ be a positive 16 digit integer.If you take out some consecutive digits integers among $ A$, prove that we can make the product of the numbers be square number. For example if some digit of $ A$ is 4, you may take out only the digit.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=161221]Try this.[/url]

The idea is very similar.


If the line $y=mx+1$ intersects the ellipse $x^2+4y^2=1$ exactly once, find $m^2$.

At the point of the intersection, $x^2+4(mx+1)^2=1\Rightarrow (4m^2+1)x^2+8mx+3=0$.
Since the line and the ellipse intersect only once, this quadratic must have discriminant of zero.
Hence, $(8m)^2-(4)(4m^2+1)*3=0$
$64m^2-48m^2=12\Rightarrow m^2=\frac{3}{4}$.


HAPPY NEW YEAR!!!

Alex

Don't Spend it on AoPS!  :rotfl:


geometry
incenter
angle bisector
projective geometry
geometry proposed

In ABC, the bisectors of the angles intersect the opposite sides in A', B' and C'. Through the intersection of B'C' and AA', a line parallel to BC is drawn, which intersects AB and AC in M and N. Prove that 2MN = BM + CN.

Right... my solution started nicely but then I had to prove something interesting (I won't spoil the fun by telling you what :)) about the position of points A', I, S, A (where I is the incentre and S is the intersection of B'C' and AA') which I could only do with barycentrics.

Any synthetic solution??  :(

op02 project, Bosnia problem 16. My solution is not really beautiful either, but here it is.

[hide="Solution"]
[color=blue][b]Problem.[/b] A triangle ABC is given in a plane. The internal angle bisectors of the angles A, B, C of this triangle ABC intersect the sides BC, CA, AB at A', B', C'. Let P be the point of intersection of the angle bisector of the angle A with the line B'C'. The parallel to the side BC through the point P intersects the sides AB and AC in the points M and N. Prove that $2\cdot MN = BM + CN$.[/color]

[i]Solution.[/i] Let the angle bisectors of the angles A, B, C of triangle ABC meet each other at I.

Since MN || BC, we have

  $\frac{AP}{AA^{\prime}}=\frac{AM}{AB}$;
  $\frac{AP}{AA^{\prime}}=\frac{AN}{AC}$;
  $\frac{AP}{AA^{\prime}}=\frac{MN}{BC}=\frac{2\cdot MN}{2\cdot BC}$.

Now remember the following algebraic rule: If the real numbers x, u, u', u", v, v', v" satisfy $x=\frac{u}{v}=\frac{u^{\prime}}{v^{\prime}}=\frac{u^{\prime\prime}}{v^{\prime\prime}}$, then $x=\frac{u+u^{\prime}+u^{\prime\prime}}{v+v^{\prime}+v^{\prime\prime}}$. Hence, we obtain

  $\frac{AP}{AA^{\prime}}=\frac{AM+AN+2\cdot MN}{AB+AC+2\cdot BC}$,

or, equivalently,

  $\frac{AP}{AA^{\prime}}=\frac{AM+AN+2\cdot MN}{b+c+2a}$.

Now, we have to compute $\frac{AP}{AA^{\prime}}$. Well, apply the Complete Quadrilateral Theorem to the complete quadrilateral formed by the lines BC, CA, AB and B'C'; then, you see that the points P and A' divide the segment AI harmonically. Hence, with directed segments,

  $\frac{AP}{PI}=-\frac{AA^{\prime}}{A^{\prime}I}=\frac{AA^{\prime}}{IA^{\prime}}$.

If the notation d(P; g) denotes the distance from a point P to a line g, then, by Thales, $\frac{AA^{\prime}}{IA^{\prime}}=\frac{d\left(A;\;BC\right)}{d\left(I;\;BC\right)}=\frac{h_a}{r}$, where $h_a$ is the A-altitude of triangle ABC, and r is the inradius. Now we know that $h_a = \frac{2S}{a}$ and $r=\frac{2S}{a+b+c}$, where S is the area of triangle ABC; therefore,

  $\frac{AA^{\prime}}{IA^{\prime}}=\frac{2S/a}{2S/\left(a+b+c\right)}=\frac{a+b+c}{a}$.

Consequently, $\frac{AP}{PI}=\frac{AA^{\prime}}{IA^{\prime}}=\frac{a+b+c}{a}$. Thus, $\frac{PI}{AP}=\frac{a}{a+b+c}$ and

  $\frac{AI}{AP}=\frac{AP+PI}{AP}=1+\frac{PI}{AP}=1+\frac{a}{a+b+c}=\frac{2a+b+c}{a+b+c}$,

so that

  $\frac{AP}{AI}=\frac{a+b+c}{2a+b+c}$.

On the other hand, from $\frac{AA^{\prime}}{IA^{\prime}}=\frac{a+b+c}{a}$, it follows that $\frac{IA^{\prime}}{AA^{\prime}}=\frac{a}{a+b+c}$, and thus

  $\frac{AI}{AA^{\prime}}=\frac{AA^{\prime}-IA^{\prime}}{AA^{\prime}}=1-\frac{IA^{\prime}}{AA^{\prime}}=1-\frac{a}{a+b+c}=\frac{b+c}{a+b+c}$.

Hence,

  $\frac{AP}{AA^{\prime}}=\frac{AP}{AI}\cdot\frac{AI}{AA^{\prime}}=\frac{a+b+c}{2a+b+c}\cdot\frac{b+c}{a+b+c}=\frac{b+c}{2a+b+c}$.

Thus, by what we have found above,

  $\frac{AM+AN+2\cdot MN}{b+c+2a}=\frac{b+c}{2a+b+c}$,

so that

  $AM + AN + 2\cdot MN = b+c$,

so that $b+c-\left(AM+AN\right)=2\cdot MN$, and thus

  $BM + CN = \left(c - AM\right) + \left(b - AN\right) = b+c - \left(AM + AN\right) = 2\cdot MN$,

qed..[/hide]

  darij

Right, so you got the harmonic division from the complete quadrilateral theorem. How could I miss that? Oh well. Not a very beautiful problem :)


no se como puedo contestar la pregunta necesito un  grafico de ello :huh: ????

puedes hacer la imagen digital y subirla a la pagina, o sino usar una buena notacion que puede describir Al principiod e la solucion...


Let $ O$ the circuncircle of a triangle $ ABC$($ \angle B > \angle C$,$ \angle B > \angle A$), $ T$ the circumcircle of $ AOC$, and $ M$ the midpoint of $ AC$, if $ D$ and $ E$ are points in $ AB$ and $ BC$ respectively such that $ \angle MDB \equal{} \angle MEB \equal{} \angle ABC$ then prove that $ BT$ and $ DE$ are perpendicular.


Let  x_i  be positive  numbers  with x_1*x_2*...x_n=1 . Prove  that
Σ[1/(n-1+xi) <=1
Thanks

Suppose that

Sum 1/(n-1+x_i) > 1
Then
  1/(n-1+x_1) > Sum (1/(n-1) - 1/(n-1+x_i)) = Sum (x_i/(n-1)(n-1+x_i)
                        i<>1                                     i<>1
>= (Product x_i/(n-1+x_i))^(1/n)
Similarly for j=2, 3, ..., n
Multiply these ineq. we get
    1 > x_1.x_2...x_n
Contradiction.

Namdung

If I'm not wrong, I also posted two solutions for this one. I think Arne posted this problem.

Don't you remember that you wrote me a private message asking to post that "splendid" solution with AM-GM? It seems that I am wrong.

harazi was correct :D

http://www.mathlinks.ro/phpBB/viewtopic.php?t=2286


function
geometry
3D geometry
sphere
trigonometry
analytic geometry

A power station and a factory are on opposite sides of a river 60m wide. A cable must be run from the power station to the factory. It costs 25 dollars per meter to run the cable in the river and 20 dollars per meter on land.

a. Express the total cost C as a function of x, the distance downstream form the power station to the point where the cable touches the land.
b. Find the minimum cost.

So directly across the river is 60m. Then the cable goes across diagonal forming a hypotenuse of the triangle. And the adjacent side is labeled x. Then on the side of with the factory the beach is 200m long but the diagonal cuts somewhere in those 200m.

Another question....

A cylinder is insribed in a sphere of radius 1.
a. Express the volume V of the cylinder as a function of the base radius r.
b. Find the radius and height of the cylinder having maximum volume.

Is this from the book "Algebra and Trigonometry with Analytic Geometry"? Authors are Swokowski & Cole?

I've had this exact problem for math class homework.

Is it chapter 3?


Can anyone post the problems of "baltic way 1990 and 1991" in this forum?

Thanks a lot!

what is the baltic way exactly? another sort of semi-international olympiad?

Yes, indeed  :D - this a contest for all the countries placed near the  Baltic Sea I suppose

yeah, but berlin seems to be too far away from the baltic sea so i cannot participate, only the really northern part of germany can participate which leads to really bad results for germany :(
anyway, i can't find these problems. on every site i visited there are always only the problems 1992 onwards.

Peter

Actually, I have seen a Korean(?) site with those problems... the only problem is that I don't read Korean...

Eh, Peter, I remember Christian Sattler has a booklet with the problems and solutions from all Baltic Ways (well, maybe not from all, but the 1990-91 ones are there)...

  Darij

The problems of Baltic Way 1990 and 1991 can be found in the following website.
The peoblems was written not in English, but in Estonian.
I can not read Estonian, can anyone help to translate it into English?

http://www.math.ut.ee/olympiaadid/ftp/bw90/bw90est.ps

http://www.math.ut.ee/olympiaadid/ftp/bw91/bw91est.ps

[quote="zhaoli"]
I can not read Estonian, can anyone help to translate it into English?
[/quote]

Well, I guess this is what you have been looking for:

http://math.ut.ee/~jan/olymp/BW90-93.ps

I must say I especially like problem number 20 of BW90 :)

Jan

Thank you very much, janwil.

Do you have the English version of solutions to Baltic Way 1994-1996?

I only have Estonian version solutions 1994-1996, some of them are difficult to understand.

Helo

I would be reale gratefull if you will send me solutions for baltic ways 93 and 94, if you had one

saruman@o2.pl

or pleasewrite here some links to the possible answers wchich are on the net, because i cannot find them

thank lot

saruman


Prove that $\forall x \in \mathbb{R}:$ \[\frac{x^{2}+2}{\sqrt{x^{2}+1}}\geq 2\]


function trigonometry logarithms limit algebra polynomial calculus

If I'm given a set of functions that are solutions to some differential equation, what process can I use to find the simplest differential equation that fits those functions? Constant coefficients are simpler than non-constant, lower order is simpler than higher order.

I really need some help on this so don't even worry about the "simplest" part. Just explain to me a process where I can backwards engineer a differential when I'm given a set of particular solutions.

For example, if I'm given the set of functions $ \{x^2\}$, one possible differential equation is $ y'''\equal{}0$, but the simplest differential equation is $ y'\minus{}\frac{2}{x}y\equal{}0$. Both of these equations are obvious fits because of such a simple set, but what about sets like $ \{x, e^x, sin(x)\}$ or even $ \{\ln(x)\}$?

[quote="epkid08"]the simplest differential equation is $ y' \minus{} \frac {2}{x}y \equal{} 0$[/quote]  Not $ y \minus{} x^2 \equal{} 0$?

I don't think there's very much to be said in general.  What's the motivation for this question?
For the example $ \{x, e^x, \sin x\}$ (note \sin, just like \ln, \lim, etc.), here are two possible ways to proceed:

1) $ e^x \equal{} e^{1 \cdot x}$; $ x \equal{} x \cdot e^{0 \cdot x}$, and $ \sin x$ comes as the imaginary part of $ e^{i x}$.  So, we need a polynomial with roots $ 0, 0, 1, i$.  This strongly suggests $ t^2 \cdot (t \minus{} 1) \cdot (t^2 \plus{} 1) \equal{} t^5 \minus{} t^4 \plus{} t^3 \minus{} t^2$.  So the equation we want is $ y^{(5)} \minus{} y^{(4)} \plus{} y^{(3)} \minus{} y'' \equal{} 0$.

2) We have $ a y \plus{} b y' \plus{} cy'' \plus{} \ldots \equal{} 0$ for $ y \equal{} x$, $ y \equal{} e^x$ and $ y \equal{} \sin x$.  The first suggests $ ax \plus{} b \equal{} 0$, the second suggests $ a \plus{} b \plus{} \ldots \equal{} 0$ and the third suggests $ a \minus{} c \plus{} \ldots \equal{} 0$ and $ b \minus{} d \plus{} \ldots \equal{} 0$.  Now just make some simple choices.  How about $ a \equal{} 1$, $ b \equal{} \minus{} x$? Okay, then we can choose $ d \equal{} \minus{} x$ and we have to solve $ a \plus{} b \plus{} c \plus{} d \plus{} e \equal{} 0$ and $ a \minus{} c \plus{} e \equal{} 0$ for $ c$ and $ e$, and this leads to an equation with only four derivatives but nonconstant coefficients.

Note that the first method is restricted to very particular sets of functions: we can only work with exponentials, polynomials, sine and cosine, and products or sums of these things.

Thanks so much! Your first method will help a bunch for most of my problems, actually all but one.

[quote="JBL"]
2) We have $ a y \plus{} b y' \plus{} cy'' \plus{} \ldots \equal{} 0$ for $ y \equal{} x$, $ y \equal{} e^x$ and $ y \equal{} \sin x$.  The first suggests $ ax \plus{} b \equal{} 0$, the second suggests $ a \plus{} b \plus{} \ldots \equal{} 0$ and the third suggests $ a \minus{} c \plus{} \ldots \equal{} 0$ and $ b \minus{} d \plus{} \ldots \equal{} 0$.  Now just make some simple choices.  How about $ a \equal{} 1$, $ b \equal{} \minus{} x$? Okay, then we can choose $ d \equal{} \minus{} x$ and we have to solve $ a \plus{} b \plus{} c \plus{} d \plus{} e \equal{} 0$ and $ a \minus{} c \plus{} e \equal{} 0$ for $ c$ and $ e$, and this leads to an equation with only four derivatives but nonconstant coefficients.[/quote]

For my last problem, I need to use the set $ \{\ln(x)\}$, so $ a y \plus{} b y' \plus{} cy'' \plus{} \ldots \equal{} 0$, for $ y \equal{} \ln(x)$, but what equation of $ \{a, b,...,n\}$  does this "suggest?"

[quote="JBL"]the simplest differential equation is $ y' \minus{} \frac {2}{x}y \equal{} 0$ Not $ y \minus{} x^2 \equal{} 0$?
[/quote]

Unfortunately, I don't think my professor would be all too happy about me putting that type of equation down for every one of my answers. :D

Well, again, for sets of a single function, I personally like $ y \minus{} f(x) \equal{} 0$.  But, of course, $ y' \minus{} f'(x) \equal{} 0$ works fine, too, as do dozens of possible variations on this theme.  In the particular case of $ f(x) \equal{} \ln x$, you can get a simple-looking second-order equation $ a y \plus{} b y' \plus{} cy'' \equal{} 0$ where $ a, b, c$ are all polynomial.

[quote="t0rajir0u"]In particular you can get $ xy' \equal{} 1$.[/quote]

I'm not sure, but I think this is also against the rules. I don't have the paper with me, but I think it might have to be a homogeneous equation and can't be separable.

[quote="t0rajir0u"]So you can get $ (xy')' \equal{} xy'' \plus{} y' \equal{} 0$.  (One of your earlier examples is also separable, so I don't really see the point of that restriction.  This is a very artificial problem.)[/quote]

Alright, I'll give it a shot, thanks for the help.


answer 2^6/3^6  [i][Moderator's note: this is the answer to a different problem][/i]

Two parallel chords in a circle have lengths 10 and 14, and the distance between
them is 6. The chord parallel to these chords and midway between them is of length :sqrt: a
 find a.

I know the answer is [hide]184[/hide] because I saw this question before but I just couldn't understand the picture...

Beta, can you explain this question?

haha i remember that problem!!

wasn't it like #29?
i know it wasn't #30

I have split this thread from the "AMC Marathon" thread, because there was considerable discussion that did not involve answering the previous question and posting a new question.

From now, on, please try and limit discussions on these threads to answering and posting questions only.

[quote="Silverfalcon"]I know where the picture is and stuffs but I just don't understand it...[/quote]

Not elegant, but fairly elementary...  I put the diagram on a coordinate grid, and label the points I know.

I have 4 known points on the circle.  These gives me 4C2=6 known chords of the circle.  I choose 2 non-parallel chords.  The perpendicular bisectors (dashed red) of the chords of a circle pass through the center.  We find the coordinates of the center (0,1) at the intersection of two of these perpendicular bisectors.  Then we find the radius, using half of one known chord and the Pythagorean Theorem.  r=sqrt(1^2+7^2)=sqrt(50)

Then we find the half-length of the unknown chord using another application of the Pythagorean Theorem.  a=sqrt(50-2^2)=sqrt(46).

Chord length = 2a = sqrt(184).


[hide]you can simplify the expression too...
(6?3)+4-1=5
(6?3)+3=5
(6?3)=2
so you can see the ? is a "/"[/hide]

[quote="rrusczyk"]Which the arithmetic symbols x, +, -, or / should be placed where the ? is to make the following statement true:

(6 ? 3) + 4 - (2 - 1) = 5[/quote]

[hide="solution"]
(6 ? 3) + 4 - (1) = 5
(6 ? 3) + 3 = 5
(6 ? 3) = 2
6 [b]/[/b] 3 = 2
[/hide]

I hate arithmetic.....

well,

I'll just give you a hint..

[hide="hint"]remove the parenthesis and solve it backward ...you know what, I'm just going to solve it
6/3=2 got that? so it's division [/hide]

Which the arithmetic symbols x, +, -, or / should be placed where the ? is to make the following statement true:

(6 ? 3) + 4 - (2 - 1) = 5

[hide](6?3)=2
?=[hide]/[/hide][/hide]


Vieta
algebra
polynomial
number theory unsolved
number theory

hope it has not been posted before, if so please let me have the link.

prove that

\[ \sum_{k=1}^{p-1} k^{p-3}\equiv 0 \textrm{(mod} \ p\ \textrm{),} \]
where $p$ is a prime greater than $3$.  :D

A bit stronger statement can be proven
$V(n)= \sum_{k=1}^{p-1} k^n\equiv 0 (\mod p)$ for $p-1\not|n$ and
$V(n)\equiv -1 (\mod p)$ for $p-1|n$.
The second part follows imidiately from FLT. About the first part:
$P(x)=x^{p-1}-1$ has p-1 roots mod p, which are all the numbers from 1 to p-1. From Vieta's formulas we know that all symetrical polynomials of numbers 1 to p-1 are congrouent with coeficients in P(x) (with alternating signs). So we have the symetrical polynomials of degree<p-1 equivalent to 0, the last is -1 (by the way this is one implication of Willson's theorem). Because V(n) for n<p-1 can be expressed as o sum of products of symetrical polynomials degree at most p-2, V(n) is also 0 mod p.
We know $V(n)\equiv V(n+k(p-1)) (\mod p)$, so we're done.

Take a look at : http://www.mathlinks.ro/Forum/viewtopic.php?t=80845


(Japan) Determine the smallest integer n with the property that in every n-digit number,
there exists a string of consecutive digits whose product is a perfect square.

[hide="A Start"]
Anything containing $ 0,1,4,9$ is automatically has a string of length 1 that multiplies to a perfect square. So let's look at numbers that use only the digits $ 2,3,5,6,7$ ($ 8$ can be treated as a $ 2$, since it is $ 2$ times a perfect square, so we'll just not consider it).[/hide]

[hide="Solution"]
We prove that $ n \equal{} 16$.  Assume that $ n \equal{} 16$ fails.  Consider a $ 16$ digit number, $ a_1a_2a_3\cdots a_{16}$.  Now, we consider the products $ a_1, a_1a_2, a_1a_2a_3, \cdots, a_1a_2\cdots a_{16}$.  We notice that there are only $ 4$ primes under ten, namely $ 2$, $ 3$, $ 5$, and $ 7$.  Each of the $ 16$ numbers described can be written as $ 2^a3^b5^c7^d$ for $ a, b, c, d\ge 0$ and $ a \plus{} b \plus{} c \plus{} d\ge 1$.  Now, there are $ 2^4 \equal{} 16$ ways to order the parities of $ a$, $ b$, $ c$, and $ d$ (for example $ a$ and $ c$ can be odd and $ b$ and $ d$ can be even or $ a$ can be odd and $ b$, $ c$, and $ d$ could be even, etcetera.)  Yet, by our assumption, $ a, b, c, d\equiv 0\bmod 2$ cannot happen or else one of the described numbers is a square.  Thus, there are actually only $ 15$ parity orderings.  Hence, one of these is repeated, say $ a_1a_2\cdots a_i$ and $ a_1a_2\cdots a_j$ with $ i > j$.  Then, we can divide these to get $ a_{j \plus{} 1}a_{j \plus{} 2}\cdots a_i$ to get a perfect square.  This is a contradiction, and hence a perfect square exists.

EDIT: [b]MellowMellow[/b] showed the counterexample to 15, mine was wrong.  I believe the motivation is to list out the fifteen ways to order the parities of $ a$ and $ b$ and then find a number that works.
[/hide]

Counterexample is $ 232523272325232$. Notice the "ruler"-like structure, with a $ 7$ in the middle, a $ 5$ midway between it in edges, a $ 3$ midway between the edges and every other number...

EDIT: This extension is almost trivial if you've solved the original problem, but what if you're working in base $ b > 2$?

thank you soooo much quattomaster!!  :D  :D


The measure of the longer base of a trapezoid is $97$. The measure of the line segment joining the midpoints of the diagonals is $3$. Find the measure of the shorter base.

[hide]
Let the diagonals intersect at $O$.
Let the midpoints of the diagonals be $S$ and $T$.
Let the diagonal containing $S$ be have vertices $B$ (on the shorter side) and $D$, the diagonal containing $T$ have $C$ (longer side) and $A$.
The distance of the shorter side is d.

$DS = SO+ZB$
$SO/SD = 3/97 \implies 94 * SO = 3 * OB$ ; edit: woops, should be $SO/(SO+DS) = 3/97$
$\triangle AOB \sim \triangle OST \implies \frac{SO}{OB}= \frac{3}{d}\implies \boxed{d = 94}$ ; edit: which now gives $\boxed{d=91}$

Wow, i have to learn to be more concise with labelling..
[/hide]

I am not so sure if that is right.
[hide] Say the labels for each point and line is the same as JavaMan's.  Now, extend ST to hit AD and BC at M and N respectively.  Notice that Triangle ATM is similar to Triangle ACD and Triangle BMS is similar to Triangle BCD.  Since S is the midpoint of BD, and BS/BD=MS/CD, we have that CE=2SN.  With similar reasoning, 2TM=CD.  Notice that MN is also the median of Trapezoid ABCD.  This means that MN=(AB+CD)/2.  This implies that SN+TM=MN+ST=(AB+CD)/2+ST=CD/2+CD/2=CD.  This means that ST=(CD-AB)/2.  Plugging in all of the values, we get that AB=91. [/hide]  I am sorry, the solution probably made little sense.


Let $n\ge 2$ be a positive integer and let $a,b\in \mathbb {Z}$ such that $a\ne 0$. Consider the function $u:\mathbb {Z}\rightarrow \mathbb {Z}$, $u(x)=ax+b.$
Prove that if $a\ne 1$ then there exist infinitely many functions $f:\mathbb {Z}\rightarrow \mathbb {Z}$ such that $f^{(n)}=u$.
If $a=1$ prove that one can find $b$ such that there is no function   $f:\mathbb {Z}\rightarrow \mathbb {Z}$ such that $f^{(n)}=u$.
(Here, $f^{(n)}=\underset{n\text{ times}}{\underbrace{f\circ f\circ \ldots \circ f}}$)

This might be wrong, I'm not sure right now, but, at first sight, it looks like there are no such functions when $a=1$ and $n\not|\ b$ (this concerns the second question of the problem). We can always choose $b=1$, but I was trying to do more (find all such $b$ for given $n$, maybe).

It is so bad that nobody is interested in this problem, though we have seen quite lots of particular cases till now and we gave all our interest in them. As usual, nice problems remain in darkness. This one worths efforts...

Consider the equivalence relation $\sim_a$ defined on $\mathbb Z$ as follows: let $g(x)=ax+b,\ \forall x\in\mathbb Z$, and put $x\sim_ay$ iff there exists a natural $m\ge 0$ s.t. $y=g^{(m)}(x)$ or $x=g^{(m)}(y)$ (where $g^{(m)}$ is the $m$-fold composition of $g$ with itself, and $g^{(0)}$ is defined to be the identity function).

Suppose now we have shown that $\sim_a$ has infinitely many classes $C_x,\ x\in\mathbb Z$ ($C_x$ being the class of $x$). We can then find infinitely many permutations of the set of classes consisting of disjoint $n$-cycles. For each class pick an element as follows: if the class is finite take any element, and if it's infinite take the "smallest" element $x$, in the sense that there's no $y\ne x$ s.t. $x=ay+b$ (it's easy to check that such an element must exist in all infiniet classes of $\sim_a$ when $a\ne 1$). Now, for each $n$-cycle $(C_{x_1},C_{x_2},\ldots,C_{x_n})$ in our permutation, where $x_i$ are the chosen elements from $C_{x_i}$, put $f(x_i)=x_{i+1},\ \forall i\in\overline{1,n-1}$, and extend this function s.t. $f(ax+b)=af(x)+b,\ \forall x$.

All that's left is proving that if $a\ne 1$, then $\sim_a$ has infinitely many classes:

If $a=-1$, then all the classes have at most two elements, and we're done, while if $|a|\ge 2$, all the classes have density zero, so it's obvious that $\mathbb Z$ canot be covered by finitely many of them.

[quote="grobber"]Now, for each $n$-cycle $(C_{x_1},C_{x_2},\ldots,C_{x_n})$ in our permutation, where $x_i$ are the chosen elements from $C_{x_i}$, put $f(x_i)=x_{i+1},\ \forall i\in\overline{1,n-1}$, and extend this function s.t. $f(ax+b)=af(x)+b,\ \forall x$.[/quote]
And what do you put for $f(x_n)$? I suppose you mean $f(x_n)=ax_1+b$ but I think that won't work if there are both finite and infinite classes in the chosen $n$-cycle.

You're right, that's what I meant, but that can easily be fixed. If $|a|\ge 2$ there is at most one finite class (this happens when $1-a|b$, and it's the class of $\frac b{1-a}$, which has only one member); just keep that class fixed and perform the permutations on the other classes (and you can also work it out easily if $a=-1$). I hope I didn't miss anything else.


$ 4^{75}\equal{}2^{150}$
$ 8^{50}\equal{}2^{150}$
since these are equal, none of them can be largest, so we can eliminate them.
$ 3^{100}\equal{}9^{50}$
$ 9>6$
3^100 is biggest


[size=200]Let $ a,b,c$ be real positive numbers  and $ a\plus{}b\plus{}c\equal{}1$, Prove  that

$ 2 \geq \frac{(1\minus{}a)^3}{3a\plus{}b\plus{}c}\plus{}\frac{(1\minus{}b)^3}{3b\plus{}c\plus{}a}\plus{}\frac{(1\minus{}c)^3}{3c\plus{}a\plus{}b} \geq  \frac{8}{15}$
 BQ[/size]

Please use a smaller font size.

Anyway, it's very easy.

$ \bullet$ By Cauchy, we have
\[ \sum \frac{(b\plus{}c)^4}{(b\plus{}c)(b\plus{}c\plus{}3a)} \ge \frac{\left( \sum (b\plus{}c)^2 \right)^2}{\sum (b\plus{}c)(b\plus{}c\plus{}3a)} \ge \frac{15}{8}.\]
Denote $ \sum a^2\equal{}x$, $ \sum ab\equal{}y$. Then the inequality is equivalent to
\[ 9x^2\plus{}6xy \ge 17y^2,\]
which is obvious, as $ \sum a^2 \ge \sum ab$ with equality if and only if $ a\equal{}b\equal{}c$.

$ \bullet$ We have $ 3a\plus{}b\plus{}c \ge a\plus{}b\plus{}c$ and so on. Hence
\[ \sum \frac{(b\plus{}c)^3}{b\plus{}c\plus{}3a} \le \sum \frac{(b\plus{}c)^3}{a\plus{}b\plus{}c} \le 2 \\
\Leftrightarrow 3\sum (a^2b\plus{}ab^2) \plus{} 12abc \ge 0,\]
which is true. Iquality holds if and only if $ a\equal{}b\equal{}0$, $ c\equal{}1$ and permutations thereof.

[quote="xzlbq"][size=200]Let $ a,b,c$ be real positive numbers  and $ a \plus{} b \plus{} c \equal{} 1$, Prove  that

$ 2 \geq \frac {(1 \minus{} a)^3}{3a \plus{} b \plus{} c} \plus{} \frac {(1 \minus{} b)^3}{3b \plus{} c \plus{} a} \plus{} \frac {(1 \minus{} c)^3}{3c \plus{} a \plus{} b} \geq \frac {8}{15}$
 BQ[/size][/quote]

while $ x \in (0,1)$ we have

$ \frac {(1 \minus{} x)^3}{1 \plus{} 2x}\geq \frac {116}{225} \minus{} \frac {76x}{75}$

$ \Longleftrightarrow\frac {(109 \minus{} 25x)(1 \minus{} 3x)^2}{225(1 \plus{} 2x)}\geq 0$

Therefore

$ \frac {(1 \minus{} a)^3}{3a \plus{} b \plus{} c} \plus{} \frac {(1 \minus{} b)^3}{3b \plus{} c \plus{} a} \plus{} \frac {(1 \minus{} c)^3}{3c \plus{} a \plus{} b}$

$ \geq \frac {116}{75} \minus{} \frac {76(a \plus{} b \plus{} c)}{75} \equal{} \frac {8}{15}$

and

$ \frac {(1 \minus{} a)^3}{3a \plus{} b \plus{} c} \plus{} \frac {(1 \minus{} b)^3}{3b \plus{} c \plus{} a} \plus{} \frac {(1 \minus{} c)^3}{3c \plus{} a \plus{} b}$

$ < \frac {(1 \minus{} a)^3}{ b \plus{} c} \plus{} \frac {(1 \minus{} b)^3}{c \plus{} a} \plus{} \frac {(1 \minus{} c)^3}{a \plus{} b}$

$ \equal{} (b \plus{} c)^2 \plus{} (c \plus{} a)^2 \plus{} (a \plus{} b)^2$

we only prove $ (b \plus{} c)^2 \plus{} (c \plus{} a)^2 \plus{} (a \plus{} b)^2\le 2$

substitution $ a \equal{} \frac {x}{x \plus{} y \plus{} z},b \equal{} \frac {y}{x \plus{} y \plus{} z},c \equal{} \frac {z}{x \plus{} y \plus{} z}$

$ (b \plus{} c)^2 \plus{} (c \plus{} a)^2 \plus{} (a \plus{} b)^2\le 2\Longleftrightarrow\frac {2(xy \plus{} zx \plus{} yz)}{x^2 \plus{} y^2 \plus{} z^2}\geq 0.$

[quote="xzlbq"][size=200]Let $ a,b,c$ be real positive numbers  and $ a \plus{} b \plus{} c \equal{} 1$, Prove  that
>
> $ 2 \geq \frac {(1 \minus{} a)^3}{3a \plus{} b \plus{} c} \plus{} \frac {(1 \minus{} b)^3}{3b \plus{} c \plus{} a} \plus{} \frac {(1 \minus{} c)^3}{3c \plus{} a \plus{} b} \geq \frac {8}{15}$
>  BQ[/size][/quote]
>
> $ f(t) \geq \frac {(1 \minus{} a)^t}{ta \plus{} b \plus{} c} \plus{} \frac {(1 \minus{} b)^t}{tb \plus{} c \plus{} a} \plus{} \frac {(1 \minus{} c)^t}{tc \plus{} a \plus{} b} \geq g(t)$
> ?!


I rewrite the javascript code to generate the problem templates each time I upload a new set of problems (b/c I then delete the code), so it would certainly be nice.

But seeing as though there isn't much interest for uploading the problems to the wiki until around AMC time anyway, I don't think there would be much community interest for re-coding awupu.

:huh:
The message is too small. Please make the message longer before submitting.

[quote="#H34N1"]chess64 offered to re-code/re-upload awupu if there is "sufficient interest."

This is just a poll :D[/quote]
awupu = ?

[b]A[/b]oPS[b]W[/b]iki [b]U[/b]nofficial [b]P[/b]roblem [b]U[/b]ploader

Definitely, chess! AWUPU was so useful... I was considering writing my own version.

PHP > Perl!

Meh, I'll pin this topic...

Erm, alright, I guess that's sufficient agreement. I'll get on it, as soon as I'm finished a few other projects (you all should like my current project, which should be out within the next week, hopefully).


Use the distance formula to find the distance between the points (4,5) and (2,3)

[quote="236factorial"][hide]$2\sqrt{2}$ is what I got. [/hide][/quote]

You are correct. But how did you get that answer?

that is correct
yust use the following uhm formula  :D
distance between -> db

db[(a,b);(c,d)]=db[(0,0);(c-a,d-b)]
which is very obvious ; its just " dragging" the line to the origin without changing its form size or angle so the length (distance) is identical.

so distance of (4,5) and (2,3) is equivalent to  (0,0) and (2,2) and using pythagoras formula (c²=a²+b²) we get 2*root(2)

i hope i made it clear and not more complicated  :P

@math 92 : be careful root [(5-3)+(4-2)] = root[2+2]= root(4) = 2 which is false ! you forgot to square the diferences to obtain root[4+4]=root(8)=2 root(2).
c²=a²+b² so c is root(a²+b²) never forget it. funny mistake  :rotfl:
btw how do you use symbols on this forum like roots and stuff ??
im new here  :blush:

oops i forgot the squares.

You can make things look fancy by enclosing it in dollar signs for example if you put 5 in dollar signs it would be $5
$.  If you put \frac{5}{2} in dollar signs it would be  $\frac{5}{2}$  d=\sqrt{(x_{1}-x_{2})^2+(y_{1}-y_{2})^2}

would be $d=\sqrt{(x_{1}-x_{2})^2+(y_{1}-y_{2})^2}$

[quote="math92"]i think he used the formula: $d=\sqrt{(x_{1}-x_{2})+(y_{1}-y_{2})}$[/quote]

Thanks math92,

I was trying to 236factorial to show his (her?) work. It's one thing to be able to solve a problem; it's quite another to show how you did it so that others can learn from you.

I tried to show my work, except I got lazy with the latex, and they used to have an emoticon concerning the square root right?

And by the way, I'm a he (his in your case). It's shown right under my avartar and my status.

[quote="236factorial"]And by the way, I'm a he (his in your case). It's shown right under my avartar and my status.[/quote]

I meant no disrespect. Your gender doesn't show-up if you are offline.

[quote="236factorial"][quote="math92"]Well on my computer, if someone is offline, it shows their gender.[/quote]
It does on mine too, but of course, not everyone's computer is the same.[/quote]

Mine does at work but not at home... weird.

[quote="4everwise"]Yeah. The distance formula is a fancy way to show the pythagoren theorem, which i find interesting.[/quote]

To make things a little harder, can anyone show that? It isn't too hard, but it's a formula that mathcounts students should be able to see its sources.


Let triangle $ ABC$ be an equilateral triangle.  Two points $ D$ and $ E$ are chosen on segments $ AB$ and $ AC$, respectively, such that $ AD\equal{}CE$.  Let $ F$ be the intersection of segments $ BE$ and $ CD$.  The area of triangle $ ABC$ is $ 14$ and the area of triangle $ ACF$ is $ 3$.  If $ \frac{CE}{EA}\equal{}\frac{p\plus{}\sqrt{q}}{r}$ is in simplest form, find $ p\plus{}q\plus{}r$.

okay i set $ CE\equal{}x$ and i got something amazingly ugly

$ x\sqrt{x^2\plus{}\frac{9\sqrt3}{52}}\equal{}\frac{\sqrt{2783}}{3}\minus{}x\sqrt{\frac{2783}{156}}$

yeah not the best way i don't think  :P

[hide="not much but..."]
Let $ x\equal{}\frac{CE}{EA}\equal{}\frac{[CEF]}{[AEF]}\equal{}\frac{[BEC]}{[BEA]}$

$ [CEF]\plus{}[AEF]\equal{}3$
$ [BEF]\plus{}[BEA]\equal{}14$

Dividing, we have

$ \frac{[CEF]\plus{}[AEF]}{[BEF]\plus{}[BEA]}\equal{}\frac{3}{14}$
$ \Rightarrow***$
$ \frac{[AEF]}{[BEA]}\equal{}\frac{[CEF]}{[BEC]}\equal{}\frac{3}{14}$

I don't know if this will go anywhere. I can't find a answer unless I solve for the area of a particular triangle, which I haven't been able to do. I haven't used $ AD\equal{}BC$ yet, so maybe that's the key.

$ ***$
The reason this works:
$ \frac{a}{b}\equal{}\frac{c}{d}$
$ \frac{a\plus{}b}{2b}\equal{}\frac{c\plus{}d}{2d}$
$ \frac{a\plus{}b}{c\plus{}d}\equal{}\frac{b}{d}\equal{}\frac{a}{c}$
[/hide]


[color=blue]Rather preolympiad , but ...never mind.[/color]

Let $AB,AC$ the tangent segments from point $A$ to a circle and $P$ be a point on the line which  passes through the midpoints of segments $AB , AC$.If $PD$ is a tangent segment to the circle, prove that $PD = PA$

 [u] Babis[/u]

Let $AO$ meets the mid line of triangle $ABC$ and side $BC$ at point $H$ and $G$ respectively, then
$PA^{2}= AH^{2}+PH^{2}= AH^{2}+PO^{2}-HO^{2}= PO^{2}+(AH+OH)(AH-OH)$

Since $AH+OH=AO$, $AH-OH=HG-OH=-OG$, so we have:
$PA^{2}=PO^{2}-AO \cdot OG = PO^{2}-OB^{2}= PO^{2}-R^{2}= PD^{2}$. Therefore $PA=PD$.


Prove or disprove that if $ x,y,z>0$, then \[ x^3 y \plus{}y^3 z \plus{}z^3 x \geq (xy)^2 \plus{}(yz)^2 \plus{}(zx)^2.\]

Only $ \sum_{sym}x^3y \ge 2(x^2y^2\plus{}y^2z^2\plus{}z^2x^2)$ :wink:

We have Schur's ineq $ x^{m}(x^{n} \minus{} y^{n})(x^{n} \minus{} z^{n}) \plus{} y^{m}(y^{n} \minus{} z^{n})(y^{n} \minus{} x^{n}) \plus{} z^{m}(z^{n} \minus{} y^{n})(z^{n} \minus{} x^{n})$ Which is true for reals $ m$ and positive reals $ n,x,y,z$

Now, $ x^3y \minus{} x^2y^2 \equal{} x^2y(x \minus{} y) \equal{} xy(x^2 \minus{} xy)$ in the second equality we consider the term $ x^2 \minus{} xy$ which is appeared in Schur's ineq $ xy(x \minus{} y)(x \minus{} z)$ So, we may assume that the desired ineq can rewrite in the Schur's form.

We use inclusion-exclusion in the following way. (Observe that $ x^2 \minus{} (y \plus{} z)x \plus{} yz \equal{} (x \minus{} y)(x \minus{} z)$)
$ x^3y \minus{} x^2y^2 \equal{} xy(x^2 \minus{} xy \minus{} zx \plus{} yz) \plus{} xy.zx \minus{} xy.yz$
$ y^3z \minus{} y^2z^2 \equal{} yz(y^2 \minus{} yz \minus{} xy \plus{} zx) \plus{} yz.xy \minus{} yz.zx$
$ z^3x \minus{} z^2x^2 \equal{} zx(z^2 \minus{} zx \minus{} yz \plus{} xy) \plus{} zx.yz \minus{} zx.xy$

Summing up these ineq we get the desired ineq is equivalent to $ xy(x \minus{} y)(x \minus{} z) \plus{} yz(y \minus{} z)(y \minus{} x) \plus{} zx(z \minus{} x)(z \minus{} y) \geqslant 0$ Which is Schur's ineq.

By the way, since (3,1,0) majorizes (2,2,0) we get $ \sum_{sym}x^{3}y\ge \sum_{sym}x^2y^2$
expand the sum we get $ xy^3 \plus{} yz^3 \plus{} zx^3 \ge x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2$ Which is equivalent to $ xy(y \minus{} z)(y \minus{} x) \plus{} yz(z \minus{} x)(z \minus{} y) \plus{} zx(x \minus{} y)(x \minus{} z) \ge 0$ So, Mr.Dimitris's ineq is equivalent to $ \sum_{cyc}xy(x \minus{} y)(x \minus{} z) \plus{} \sum_{cyc}xy(y \minus{} z)(y \minus{} x) \ge 0$

I'm sorry if my ENG is not good  :(


Hi!
I've got a question...Where should I post topics concerning physics...in particular theoretical and mathematical physics?...I mean in which subforum are they more likely to be answered...and not to be deleted or closed...Could it be "Other Problem Solving Topics" or "The Round Table" or "Games & Fun Factory"?

Hm...Why don't you create such a subforum...devoted to physics?

Post questions about physics in Round Table if they are mostly about physics, or in Other Problem-Solving Topics if they have a problem-solving aspect to them.


Don't use dollar signs. It makes the LaTEX unhappy, which is why your post looks messed up.

[quote="Treething"]
[hide="Solution"]
She spent 108.24 dollars, so she must've bought three items that end in an 8.

She had to have bought a shirt and a drill.

25.78 + 30.98 = 56.76. She had to have bought one of the houseware items.

So her current total could be 68.14, 69.74, or 86.34 dollars. Which means she either needs 40.10, 38.50, or 21.90 more. It can't be 21.90, so it must be either 40.10 or 38.50. But it can't be either of those. I don't know.
[/hide][/quote]

That's what I got too.

[hide]Okay, you forget that more than three items can be bought. What if they buy three housewares and then one from each other department? That is legitimate.[/hide]

[quote="Treething"][quote="-Terminator-"]Mabel and her mom are going shopping on Saturday. They bought at least one item from each of the 3 departments that they visited. Mabel gave the clerk $120$ and she got back $11.76$ change. What items did they buy? Think about how much they spent. NO TAX TODAY!

HOUSEWARES
Dishtowels: $11.38$
Curtain Rods: $12.98$
Bath Mats: $29.58$

CLOTHING
Shirt: $30.98$
Dress: $49.90$
Slacks: $39.90$

TOOLS
Hammer: $17.90$
Saw: $23.90$
Drill: $25.78$[/quote]

[hide="Solution"]
She spent 108.24 dollars, so she must've bought three items that end in an 8.

She had to have bought a shirt and a drill.

25.78 + 30.98 = 56.76. She had to have bought one of the houseware items.

So her current total could be 68.14, 69.74, or 86.34 dollars. Which means she either needs 40.10, 38.50, or 21.90 more. It can't be 21.90, so it must be either 40.10 or 38.50. But it can't be either of those. I don't know.
[/hide][/quote]

[hide="Solution"]
yea, you guys forgot that part, so we try combonations of 2 housewares, guess and check, and we find that we get totals of 83.88, 65.68, and 67.28 when we try combos of 2 housewares, we find by guess and check again, that our answer is from 65.68,

$Curtain Rods, Bath Mats, Drill, and Slacks$
[/hide]

It'll be nice if the numbers weren't that huge...


[quote="risinglegend"]A shop owner sells a product and charges 4% GST of the selling price. He decides to give a 10% discount. He has 2 options:
(i) gives a 10% discount first and then charges a 4% GST or
(ii) charges a 4% GST first and then gives a 10% discount.
As a consumer, which option would you prefer? Show your calculations clearly.[/quote]
[hide="Hint"]
Take an example case, like if the product is $\$10$.[/hide]

[hide]using Drunken_Math's hint:
take 10 and multiply it by .04.  You then get 9.6.  Then multiply 9.6 by .1.  You get .96.  Subtact .96 from 9.6 and you get 8.64.

in the other case you take ten and multiply it by .9.  You get 9.  Then multiply .04 and 9 together and you get .36. .36 subtracted from 9 is 8.64.

so as a consumer they are both equal.[/hide]

A car travels x km/h for p km and makes another journey at y km/h for q km.
Find
(i) The total distance (r km) the car travels;
(ii) The formula for the average speed.[/b]


$ \boxed{1}$ Let n be a positive integer. PT there exist distinct positive integers $ x, y, z$ such that
$ x^{n \minus{} 1} \plus{} y^n \equal{} z^{n \plus{} 1}$
$ \boxed{2}$Let $ a, b, c\in\mathbb{N}$ such that a & b are relatively prime; and c is relatively prime to either a or b. PT there exists infinitely many triples $ (x,y,z)$ of distinct positive integers x, y, z such that $ x^a \plus{} y^b \equal{} z^c$
$ \boxed{3}$ let $ b, m, n\in\mathbb{N}$ such that $ b > 1$ and $ m,n$ are distinct. PT if $ b^m \minus{} 1$ and $ b^n \minus{} 1$ have the same prime divisors then $ b \plus{} 1$ is a power of $ 2$
thanks!! :)

See the resources section.
And note that this subforum is a bit inappropriate for newer IMO level problems. (And note that you should post every problem in a new topic, not 2 in one).

Oh, sorry! :blush:
Can you please give me a link so that I can get to the resources section? Thanks for your advice.

Just click the Contests button (if you're using AoPS skin) or the Resources button (if you're using ML skin) above (it will take you to http://www.mathlinks.ro/resources.php ), and there you'll find

http://www.mathlinks.ro/resources.php?c=1&cid=17&year=1997

http://www.mathlinks.ro/resources.php -> http://www.mathlinks.ro/resources.php?c=1&cid=17 -> http://www.mathlinks.ro/resources.php?c=1&cid=17&year=1997

Hello, where is the appropriate place to post these types of problems? :blush:  I have got all the resources thanks to Zetax and hsiljak. Thanks to both of you for your help. :)

These problems are number theory problems. If you have solved it, you can post it here: http://www.mathlinks.ro/index.php?f=57 . If you haven't solved it, post it here: http://www.mathlinks.ro/index.php?f=56 .

By the way, do not post them again, because these problems have been posted before.


Hi,

I'm writing an assignment on knot theory in LaTeX (this is my first major document using LaTeX btw) and need to insert diagrams of the knots.  I have a program that draws knots nicely and outputs them as .eps files and I can include these in my document.
What I would like to do is:
1) Label the diagrams with text in the same style as the rest of the document (ie. LaTeX text)
2) Add arrows and extra curves

What's the best way to do this?  (On Windows)
I've installed WinFig and read some of the documentation on XY-Pic but I'm not sure if these are the right tools.  In WinFig it doesn't display the original .eps picture when I open it; just the filename.

If anyone can help I'd be very grateful,

Ed

[quote="Eddie"]
2) Add arrows and extra curves
What's the best way to do this?  (On Windows)
I've installed WinFig and read some of the documentation on XY-Pic but I'm not sure if these are the right tools.  In WinFig it doesn't display the original .eps picture when I open it; just the filename.
[/quote]

For your first question, well, it has been around five years since the last time that I did something like that, but for the second, you cannot modify the eps itself with latex code. I mean, you have to edited by yourself (or at least that is what I think) On the other hand, if it works as you need it to work, then it is the right tool, may be not the best, but for sure the right one  ;)

I'm not sure to understand what you want exactly but to add a caption to a figure simply use the command

\caption{text}

in the figure environnement.  From what I can see in my documents, it's the same style of the rest of the document.

also,

http://www.iam.ubc.ca/~newbury/tex/figures.html

there is a small section on drawing lines and arrows in latex

hope it will help you

Cheers,  will check out the stuff on that link.
It wasn't captions I wanted to add but text over the top of the picture.  Looks like \put will do what I want.

E


[color=blue][size=167]If the equations x^2+ax+b=0 and x^2+px+q=0 have a common root, prove that (a-p)(bp-aq)=(b-q)^2.[/size][/color]

[quote="ifai"][hide="Solution"]Let $X$ be the common root of the equations, and $R, r$ the other roots of $x^2+ax+b=0, x^2+px+q=0$, respectively.
Because $R+X=-a, RX=b; r+X=-p, rX=q$. Thus $(a-p)(bp-aq)=(-R-X+r+X)(-RX(r+X)+(R+X)rX)=X^2(r-R)^2$
$(b-q)^2=(RX-rX)^2=X^2(r-R)^2$
Therefore, $(a-p)(bp-aq)=(b-q)^2$.
QED[/hide][/quote]

[color=indigo][size=150]actually I have a less complicated method :

[hide]

x^2+ax+b=0

x^2+px+q=0

By comparing the cofficient of x,
a=p

By comparing the constant terms,
b=q

L.H.S.= (a-p)(bp-aq)
= (a-a) (ab-ab)
= 0

R.H.S.= (b-q)^2
= (b-b)^2
= 0^2
= 0

therefore, (a-p)(bp-aq)=(b-q)^2 [/hide][/size][/color]

BEE, your solution is wrong because if the 2 equations only have ONE common root, then $a\neq p, b\neq q$.


Its just another variable, usually relating to $m_i$ for any i.

Used a lot in sequences.

I should be a little more specific...I remember seeing this on a topic and tried searching for it, but i couldn't find it. It was like $2_2$ has a value of something.

Used to write sequences or progressions.........

[quote="Jhonny Ruiz"]I think it's also used for identifying the base in which a number is written, for example,
$85_8$ is $85$ written in the base $8$,
$101001_2$ is $101001$ written in the base $2$, and that's how it works.[/quote]

$85$ is not written in base $8$. If it was, it'd be $105$

well yeah you may be right, I just wrote that to sort of show an example of how I remember about it, sorry guys

Johnny Ruiz is right, although his example doesn't really work, because for a number $m_n$ ($m$ in base $n$), no digit can be greater than $n$...

A better example would be to say $125_8$ is $85$...

And btw, $105_8$ is $69$.

[quote="easyas3.14159..."]correct me if i'm wrong, but doesnt something like $3_4$ mean $3^{3^{3^3}}$?[/quote]I don't think $3_4$ can mean that, but I think one notation for power towers is indeed $^43$, for example, in this case, so you were relatively close.

[quote="easyas3.14159..."]correct me if i'm wrong, but doesnt something like $3_4$ mean $3^{3^{3^3}}$?[/quote]

Don't think so, easy... Generally, you never use $3_4$, you use $3_a;3_b$ Maybe $3_4$ is the fourth term of a sequence previously known by you...... (stated in a problem, for example)

No, subscripts only mean different terms of a sequence for a variable, i.e. $n_4$ would mean the 4th term in a sequence of numbers...

When you see a constant with a subscript, it is for bases...

so heres my view:
if it is a variable with a subscript, it means the number is in a sequence of numbers

while
if it is a constant with a subscript, it means the number is in the value of the subscript's base

What is it from the thread with like the biggest number you can make using 1,2, and 3 or something?


Anyone know a way to enter safe mode during startup?

[quote="JBL"]You should have an opportunity to press a button -- probably F10, but wait until someone else who knowns something comes by -- during the start-up process.  If a DOS screen rolls by, do it then.  On XP, I think it's when you've got the blue bar scrolling at the bottom, it should say something like "F12 for setup" or something like that, and you can do it then.

You're probably better off waiting for advice from someone who actually knows this, however.[/quote]
It won't hurt to push buttons during the startup. I think it might be F8, but I'm not sure either. It probably depends on your version of Windows. I assume that if you know what safe mode is, you also know enough to know what the bios is, and know that if you accidentally get into the settings for that while pushing buttons as your computer boots, it would probably be best not to change anything. Unless, of course, the computer is broken and you know what you're doing in the bios.

i've noticed that its different for many computers/versions.. when the computer starts up and the black screen comes, just press randomly all of the high F's... from F7-F12. never fails.


[hide="Answer"]$5x=\frac{10}{17}\cdot n \Rightarrow x=\frac{2}{17}\cdot n$
$7x=42 \Rightarrow x=6$
$6=\frac{2}{17}\cdot n \Rightarrow n=\boxed{51}$[/hide]

Works for me.

No, she can't do half a word problem :) . How did you get that? I got a different answer.

Right answer JesusFreak197

I thought in two days, she did 42 problems, but this problem meant in 7 days, she read 42 books (5+2=7).

Obviously $42\cdot\frac{17}{14}=51$ :roll:

[quote="G-UNIT"][quote="236factorial"]I'm getting 357/2. I don't think she can do half a word problem.  :P[/quote]

  The reason you derived that answer is because of the sentence "after another two days she solved 42 problems".  You interpreted as "she solved 42 problems in 2 days" when the intended definition was: "after another 2 days she solves a [b]total[/b] of 42 problems".[/quote]

Yeah, that's what i said above. I guess the another meant nothing to me  :roll:

[quote="LadyKn1ght"]Amy solved an equal number of word problems each day. After five days, she had 7/17 of the word problems left to solve. After another two days she had solved 42 word problems. How many did she have at first?[/quote]
my answer
[hide]5x=10/17 so x=2/17                      x=days
         2x=4/17
         7x=42 so x=6
        2/17=6/y                                         y=problems she had at first
        therefore 2*3=6, so 17*3=51
        51 is the answer.[/hide]

[quote="ritchjp"][quote]Amy solved an equal number of word problems each day. After five days, she had 7/17 of the word problems left to solve. After another two days she had solved 42 word problems. How many did she have at first?[/quote]

[hide]
Let [tex]x[/tex] = the total number of problems

After five days she had [tex]\frac  {7}{17} \ x[/tex] of the problems left, which means that she worked [tex] \frac {10}{17} \ x[/tex] of the word problems.

The following two days, she worked an additional [tex]42[/tex] problems.

So in that seven day period she worked a total of [tex] \frac {10}{17} \ x + 42[/tex] problems.

So then we end up with the equation [tex]\frac {10}{17} \ x + 42 = x[/tex]

[tex] \frac {7}{17} \ x = 42[/tex] so [tex] x = 102[/tex][/hide][/quote]

But it never said that after doing 42 problems, she finished all the problems, thus the different answer.

[quote="LadyKn1ght"]Amy solved an equal number of word problems each day. After five days, she had 7/17 of the word problems left to solve. After another two days she had solved 42 word problems. How many did she have at first?[/quote]

5 days: 10/17 problems.

7 days: 14/17 word problems. 14x/17=42
14x=714

714/14=51

 :)


[quote="Go Around the Tree"]The radius of a circle is 5". Tangents drawn from an external point P form an angle of 120 degrees. How far is P from the center of the circle?[/quote]
[hide]Draw the segment from P to the center of the circle and you get a 30-60-90. So the length is $\frac{5 \sqrt{3}}{3}$[/hide]
EDIT: To k8reindeer
bmps :P ? That's a new one.
Anyways, perhaps I should clarify.
Draw the circle and the two tangents. Then connect the points that the tangets hit the circles and you have a 120-30-30 triangle. Draw the altitude/median/angle bisector of the 120 degree angle. Now you have two 30-60-90s from which you can evaluate to find the segment lengths.

bmps did you find the length of one of the tangents? it asks for the length of the segment from p to the center of the circle...
that or I'm just confused :ninja:


for the left one, it is true by
\[
\frac{{2a^2  + \frac{{a(b + c)}}{2}}}{{2(ab + bc + ca)}} \ge \frac{a}{{b + c}}
\]

The other inequality is stronger than Nesbitt, but I cannot solve it in a simple way.

I found a simple solution for the right one.
\[
\begin{array}{l}
 \frac{a}{{b + c}} + \frac{b}{{c + a}} + \frac{c}{{a + b}} \\
  \ge \frac{{\left( {a + b + c} \right)^2 }}{{2(ab + bc + ca)}} \\
  = \frac{{a^2  + b^2  + c^2 }}{{2(ab + bc + ca)}} + 1 \\
  = 1 + \frac{1}{2}\left( {\frac{{a^2  + b^2  + c^2 }}{{ab + bc + ca}}} \right) \\
  \ge 1 + \frac{1}{2}\left( {2 - \frac{{ab + bc + ca}}{{a^2  + b^2  + c^2 }}} \right) \\
  = 2 - \frac{{ab + bc + ca}}{{2(a^2  + b^2  + c^2 )}} \\
 \end{array}
\]
by cauchy schwarz and am-gm

[quote="manlio"]Let be given three positive real numbers $a,b,c$. Prove that

$ \frac{1}{2}+\frac{a^2+b^2+c^2}{ab+bc+ca} \geq \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geq 2 -\frac{1}{2}\frac{ab+bc+ca}{a^2+b^2+c^2}$[/quote]
for the right one:
We need prove:
$(a+b+c)(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}) -\frac{9}{2}$ $\le \frac{a^2+b^2+c^2}{ab+b+ca}-1$
$<->\sum {\frac{(a-b)^2}{(a+c)(b+c)}} \le \sum{\frac{(a-b)^2}{ab+bc+ca}}$
It's true because $(a+c)(b+c) \ge ab+bc+ca$

for the right,same way!
$2(a+b+c)(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}) -9 \ge \frac{ab+bc+ca}{a^2+b^2+c^2}-1$
$<->2 \sum\frac{(a-b)^2}{(a+c)(b+c)} \ge \sum{\frac{(a-b)^2}{a^2+b^2+c^2}}$
because
$2(a^2+b^2+c^2)-(a+c)(b+c)=c^2+2a^2+2b^2-ac-bc-ab>(c/2-a)^2+(c/2-b)^2+/frac{(a-b)^2}{2}$
Then we have result!

[quote="manlio"]Dear hungkhtn,

I don't understand this step

$\displaystyle 2(a+b+c)(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}) -9 = \sum\frac{(a-b)^2}{(a+c)(b+c)} $[/quote]

$2(a+b+c)(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}) -9 $
$= \sum{\frac{a+b}{a+c} +\frac{a+c}{a+b}-2}$
$=\sum{\frac{(a+c-a-b)^2}{(a+c)(a+b)}}$}

Is it ok?

Yes, now it is all Ok.

Thank you very much for this very nice proof  :)


Using unlimited pennis, nickels, dimes, quarters, and half-dollars how many ways are there to create 50 cents.

how much pennies/dimes/nickels=a dollar :blush:
i believe that 100 cents=1 dollar? :blush:  :blush:

A dollar is 100 cents.
Half a dollar is 50 cents.
Quarter is 25 cents.
Dime is 10 cents.
Nickel is 5 cents.
Penny is 1 cent.

[hide="hint"]we are to find the number of solutions in non-negative integers of the equation \[a+5b+10c+25d+50e=100\].[/hide]

[quote="1=2"]I think that he means that you can use any kind of coin(even multiple kinds) to create 50 cents.

[hide="solution"][b]Case 1:[/b] You use a 50-cent piece.
There's only 1 way that you can do that.

[b]Case 2:[/b] You don't use a 50-cent piece.
This case is much larger. We can divide it into sub-cases.

[i]Case 2I:[/i] You use 2 quarters.
There is only 1 way you can do that.

[i]Case 2II:[/i] You use 1 quarter.
[u]Case 2IIA:[/u] You use 2 dimes.
There are 2 ways you can do that: a nickel or 5 pennies make up the rest.

[u]Case2IIB:[/u] You use one dime.
Now you must use nickels and pennies to make 15 cents. There are four ways to do that.

[u]Case2IIC:[/u] You don't use any dimes.
There are a total of 6 ways to make 256 cents out of pennies and nickels.

[i]Case2III:[/i] You use no quarters.
[u]Case 2IIIA:[/u] You use 5 dimes.
Only 1 way to do that.

[u]Case2IIIB:[/u] You use 4 dimes
3 ways to do that.

Now we see a pattern; every dime we throw away, we get 2 more ways to do it.

1+(1+(2+4+6)+(1+3+5+7+9+11))=50 total ways to make 50 cents.[/hide]

Is there a better way than this???[/quote]

bascally what I did, ended up getting $\binom{7}{2}+\binom{6}{2}+2\binom{4}{2}+2$ which is 50 (which are the solutions for p+5n+10d+25q+50h=50, using balls and urns)

[quote]bascally what I did, ended up getting  which is 50 (which are the solutions for p+5n+10d+25q+50h=50, using balls and urns)
[/quote]

Art of Owna, can you please explain how to use balls and urns formula for p + 5n + 10d + 25q + 50h = 50?

I only understand the balls and urns formula if all the coefficient is 1 (such as p + n + d + q + h = x), but not with coefficients such as 5, 10, 25, and 50 in the LHS.

Yeah, I would like to see that as well please.

my balls and urns isnt really that..it does still involve setting h and q for numbers (but its easy because h cannot exceed 1 and q cannot exceed 2), and then you have to notice some parity/sum tricks.


Hi everybody,

I was wondering if anybody knows the proof for Descartes' Rule of Sighs.  In other words, I know it works but WHY does it work?

If anybody could post a proof, I'd greatly appreciate it.

Thanks in advance!

It's rule of SIGNS, not rule of Sighs.  For an elementary, albeit long proof, see [url=http://sepwww.stanford.edu/oldsep/stew/descartes.pdf]here[/url].

It's somewhere on here I think:
http://www.cut-the-knot.org/fta/ROS2.shtml

Google and Wikipedia are your friends.

Oops. Signs not sighs typo.:oops_sign:   lol.
thanks you guys!

ooo
i was looking for the same thing
thank you


AMC
AMC 12
USA(J)MO
USAMO
AIME
MATHCOUNTS
number theory

How'd you do? I know we don't actually know the answers till tomorrow, but still.
I think I got somewhere from 4-6. Dunno yet.
I answered #1-7 and 11, but definitely got #2 wrong because I misread the problem, and would've gotten #11 had I not forgotten to multiply by 2. Ah, stupid mistakes.  :roll:
Of course, I guessed the rest.

4. My team always reviews each others answers, so a four.

6

I think it was harder this year, I got an 8 on last year's II, and I've practiced a lot since last year.

Other people confirmed this (numerous low scores), although I unfortunately was not feeling good yesterday (not in school today), so I don't know. (Hmm I almost wonder if they'd let me take the alternate...)

I answered 11 of them
I'm not sure what I got

Guessed on 8,10,14,15
I probably made some mistakes on a few of them, but overall, I feel that I did well

I would've had 13, but I made stupid mistakes on #4, 5, 8, and 12.
Then #9 and #14 I didn't actually "solve" so I expected to get them wrong.

So I'm pretty sure of a 9.
My AMC12 was a 117.

I would be content, but I think I'm going to be the first person ever to have the highest index in their state, but not make USAMO, due to an underclassman making the floor. (he got 6 or 7)

I guess practicing for USAMO and not AIME was kindof a bad idea. :-/

I got an 8 or 9, depending on the answer to 9. Saiem got whichever one of those scores I didn't get [I think]. I thought it was easier, as I made stupid mistakes on #1 and #4 and solved 9 others, hopefully all right.

5. I'm very happy with this score, considering that

1) I had no prior preparation whatsoever.
2) My AMC 12B score, 99, doesn't even qualify me for AIME (I qualified through AMC 10A).

Answered 1, 2, 3, 4, and 5 correctly, by the way. Also attempted 6, 9, 12, and 13 but didn't answer them correctly.

7 and 13 were cool. So much that I might post some solutios to them!

I did like #7. Just intimidating notation, no?

Coming out of the exam, I thought I'd have either 14 or 15, but based on the solutions posted in this forum, I'm thinking a 12.  Once the official answers come out, we'll see.

Edit: And to tell people about the ones I messed up,

#15: 451, I probalby made a computational error.
#14: 225, I guessimated a little off, never occurred to me that the limit of that expression as you take later and later terms might be 225.
#10: 880, Overcomplicated and "solved" it in a very peculiar way, when I came back to check it, I never caught my mistake (which was blatant and stared me in the face)

But that's not as bad as the stupid mistakes I made last year...so I feel good.

BSed 3 of them.

[b]12[/b] legitimately solved [b]-6[/b] stupid mistakes = [b]6[/b]

Wow, there's something I need to work on...  :(

I think I got a 5 or 6, depending on the answer on #11, I misplaced my answer sheet.

I thought #7 was cool too...really simple problem once you figured out what they were asking.

I think it's interesting how several other people answered 1-7 and 11, just like me.

Anyway, missed #2 (misread) and #4 (stupid mistake), according to the answers calc rulz posted I miscounted on #5 and #6 though I haven't checked them yet, and I somehow guessed #8  :huh:

I also put 862 for 10 and the answer is 860...to think that I even considered putting 860 down, guessing 2 right when the odds are 1/1000 (though actually better, I made sure the answers were somewhat reasonable) would have been cool.

Out of curiosity, since I'm taking the II, would you say this AIME was harder than past ones?

I thought problem 1 was nice this year.

*shrugs*

It was nicer last year. I mean come on, you can't get any prettier than using the Pythagorean Theorem twice.

I think this is a wake-up call for the AMC people. Hopefully they'll take everyone's comments and make a better test next year (or maybe the AIME II is good this year).

[quote="tjhance"]I liked #11[/quote]
i liked it too, becasuse it was one of the few three i got right :P

1 guesstimate/almost there but stupid mistake, 8 legitimately solved -3 stupid mistakes (!) got #1 wrong >.<, so 5 :(
2, 3, 6, 7, 11? Something like that.  Dunno.  Got the three sequence problems on the first two pages right :P but...how did I do so badly on the first page >.<

[quote="indianamath"][quote="kyyuanmathcount"][quote="johnz"]15 aime :D[/quote]

Lol, good one there.  8th grader gets full score, but can't stop from flaming in the MATHCOUNTS forum.  Are we looking at a new Arnav?  no offense to anybody of course  :D

-_- seriously.[/quote]

lol he told me he was in 9th grade yet he said in the MATHCOUNTS forum that he made Nats. now he's saying he got 15...[/quote]

Stop bull shitiating

what bull?:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=138254

Doesn't my signature say enough about how badly I did?

Uh, I think those are his goals. Note the part in the beginning.

I do not think the LCM/GCF method works for all situations. (It just happened to work by chance for that problem) There is a really good method in the math jam, though. Basically, you calculate the angular velocities for each planet. Then, you set one of the planets to be your new reference, so it is remaining at rest. This way, we only have to concern ourselves with two of the planets because the third one is always at rest. As for those two planets, one is traveling at exactly double the velocity as the other planet, so it can be confirmed that when the slower planet makes it to the other side of the circle of rotation, the faster one will have returned to the starting point. And all three planets will be collinear.

[quote="bonnyjain"]I got a 7 (1, 2, 3, 6, 7, 8, 11).

I've solved #4 and #5 (SOOOO easy in hindsight) after coming home. I looked on the AoPS wiki for problem #4, and the reason given is that 105 is the gcf/lcd. Can anyone tell me why the gcf/lcd is the solution to the problem? I tried the gcf/lcd method for the same problem with different numbers, and it didn't work.[/quote]
It doesn't work.  The solution posted on the wiki was incorrect.  To see one reason why, suppose each planet takes twice as long to orbit the star.  Then it would be twice 105, or 210, years until they were next collinear, yes?  But least common multiple divided by greatest common factor would not change.  The author of the solution was merely lucky that the numbers worked out the way they did.  Thanks for pointing this out.

I got 8.  Don't remember which ones I got right.  This AIME had a lot of problems with obvious brute-force solutions and way less obvious elegant solutions that I could never find.


[quote="arqady"]Let $a,$ $b$ and $c$ are  sides lengths of triangle. Prove that
$\frac{(a+b)(a+c)(b+c)}{8}\geq\frac{(2a+b)(2b+c)(2c+a)}{27}.$[/quote]
now do you  have a nice solution in mind?


Could someone solve this equation for me (with working out) please.

Solve the equation z^4 + 1 = 0 for complex z. Give your answers in the rectangular form.

This website is not so other people can do your homework for you.  This is a place for students of mathematics to learn.  If you wanted to learn about complex numbers, post a more general question about concepts that people could reply to with concepts, problem approaches, etc.

[hide="Approach"]
Any complex number can be represented in the form $re^{i \theta}$, where $e^{i \theta} = \cos \theta + i \sin \theta$ (in radians).  The former is polar representation, and the latter is rectangular representation.  The preceding equation, often referred to as Euler's formula, can be used to derive the following:

$z^n = \left( re^{i \theta} \right)^n = r^n e^{i n \theta} = r^n \left( \cos n\theta + i \sin n\theta \right)$

(This result is referred to as deMoivre's theorem/formula.)

Taking $n = \frac{1}{4}$, for example, allows you to find all the fourth roots of a number $z$.  (Remember that $\theta = \theta + 2\pi k$ and so there are four possible values of $\frac{\theta}{4}$ that can be used in the RHS.)
[/hide]

[quote="coolio2005"]And you couldnt have replied with atleast a method on how to do  it? Why are you even a user of the forum?[/quote]

I'm a member of this forum because I enjoy solving, posing and discussing math problems with other similarly inclined individuals.  Things that annoy me about this community are people who come here in order to have other people do problems for them, the solutions of which they will then pass off as their own.  This behavior is obnoxious, dishonest, and devalues the efforts of people who do their own work.  If someone wants to ask people for help with concepts or for methods for a family of questions, that's fine by me, but if someone simply post one's assigned problems verbatim, it offends me.

Two further notes:  First, the use of the word "urgent" in the title is both a dead giveaway and additionally insulting -- why should I solve problems on your schedule?
Second, these are all my personal opinions -- others on this forum obviously don't share them, since some of them seem happy to do your work for you.  And as I am not an administrator, I of course have no power over your behavior on this site, except to berate you.

First of all, if you don't like questions that are posted in this forum, you just not bother to reply to them, it's pretty damn simple (give it a go!!!).
Secondly, i'm not too sure how well you know Australian Universities (considering i dont even know where you are from), but we are given tutorial questions each week for the particular subject, and in this case, a maths subject. I have a maths test next week, which will involve complex numbers and other topics. Of course, the tutorials are NOT compulsory to do or hand in, so no, i'm not planning on using someones answers and submitting them. But it's kind of hard to "learn" on a forum, as people aren't necessarily on their pc replying back every time a user writes something (this isn't chat!). So instead of me spending an entire day on 1 complex number question, i ask for assistance by showing me how it's done. If i was to ask someone a question, they'd tell me how to do it, then i'd attempt it, then i'd post it and wait for a reply, my test will be over already! I don't appreciate false accusations and what not, especially when you don't know the story behind it. And why on earth a person would write something in a forum topic they dont like, well god knows!!!

Observe that

  Z^4 + 1 = [ Z^2 - (Sqrt2).Z + 1 ].[Z^2 + (Sqrt2).Z +1] ,so the problem is now reduced to the solution of two quadratic equations and the use of DeMoivre's Theorem has been avoided altogether.

In using De Moivre's Theorem note ,that  for r an integer or zero

1 = Cos2r(pi) + iSin2r(pi)   and -1 =Cos(2r+1)pi + i Sin(2r+1)pi

Solutions of Z^4=-1 are therefore   Cos(2r+1)pi/4 +i Sin(2r+1)pi/4 where r takes the values 0,12,3.

I hope this helps!

[quote="coolio2005"]First of all, if you don't like questions that are posted in this forum, you just not bother to reply to them, it's pretty - simple (give it a go!!!).[/quote]

Keep in mind that JBL is a moderator.  The impression was that this was a homework problem to be simply done for you, and that is not appreciated on this forum.

That is beside the point. He jumped to conclusions claiming i was using it for homework or whatever, when clearly it is NOT the case. I have a test tomorrow and i'm studying tutorials, and i needed help.

Coolio2005 is right!

  Everybody can ask what he wants and if someone wants to answer , he can do it.How can we know the reason , for which somebody needs help ? If the problem is homework and someone wants help , he deserves to have it ,as well ! There are not rules in the forum for this subject. The only rule is that we must be kind to each other!

  [u]Babis[/u]

 [i](It is clear that pupils   must first try to solve their problems and then post them. To this point I agree with JBL.But ,if they write , they must have an answer. ) [/i]

Where did you see the sarcasm ? I'm trying to see  this point. Probably my bad english , but ... sarcasm , never. I have never done this in all my life(I'm 48 and mathteacher , so this is long away from my methods).SORRY.
I hope you 'll read my words again.

   [u]babis[/u]

[hide="Here is the roots of the biquadratic equation with real coefficients and the negative discrimination."][color=darkred] Given are the real numbers $a,b$ for which $b>a^2$ and the equation $f(z)=z^4+2az^2+b=0$. We observe that $b>0$ and $-\sqrt b< a<\sqrt b$. Thus,
$\sqrt b -a>0$ and $f(z)=\left(z^2+\sqrt b\right)^2-2\left(\sqrt b-a\right)\cdot z^2=$ $\left(z^2+\sqrt b\right)^2-\left[\sqrt{2\left(\sqrt b-a\right)}\cdot z\right]^2=$
$\left[z^2+\sqrt{2\left(\sqrt b-a\right)}\cdot z+\sqrt b\right]\cdot \left[z^2-\sqrt{2\left(\sqrt b-a\right)}\cdot z+\sqrt b\right]$. Thus, $a+\sqrt b>0$ and the discrimination is $\Delta =\left[\sqrt{2\left(\sqrt b-a\right)}\right]^2-4\sqrt b=-2\left(a+\sqrt b\right)<0$. Therefore,
\[ \blacktriangleright\ \boxed {a,b\in R,\ a^2<b,\ z^4+2az^2+b=0\Longrightarrow z_{1,2,3,4}=\pm\sqrt 2\left(\sqrt{\sqrt b-a}\pm i\sqrt{\sqrt b+a}\right)}\ \blacktriangleleft \][/color][/hide]


A parabola is the set of points in the plane whose distance from a fixed point $F$, called [i]focus[/i], is equal to their distance from a fixed line, called [i]directrix[/i].

How do I find the focus or directrix of any parabola?

use the definition of the focus and directrix to derive the arbitrary equation of a parabola in terms of the location of the focus and equation of the directrix. then for a particular quadratic equation, factor it to get it in the appropriate form, and it will make the location of the focus and equation of the directrix evident.  :)


$a_i\in R$ find minimum of $S$:
$S=\sum_{i=1}^n\sum_{j=1}^n\frac{a_ia_j}{i+j}$


inequalities
algorithm
algebra
polynomial
inequalities proposed

[quote="imedeen"]Let $ a,b,c > 0$ such that $ a \plus{} b \plus{} c \plus{} 1 \equal{} 4abc$. Prove that:
\[ \frac {1}{{{a^4} \plus{} b \plus{} c}} \plus{} \frac {1}{{{b^4} \plus{} c \plus{} a}} \plus{} \frac {1}{{{c^4} \plus{} a \plus{} b}} \le \frac {3}{{a \plus{} b \plus{} c}}\]
[/quote]

any one any idea pease  :maybe:

[quote="imedeen"][quote="imedeen"]Let $ a,b,c > 0$ such that $ a + b + c + 1 = 4abc$. Prove that:
\[ \frac {1}{{{a^4} + b + c}} + \frac {1}{{{b^4} + c + a}} + \frac {1}{{{c^4} + a + b}} \le \frac {3}{{a + b + c}}\]
[/quote]

any one any idea pease  :maybe:[/quote]

$ {a + b + c + 1 = 4abc} {\Longrightarrow}{abc\ge{1}},$

thus we  can  suppose

   $ a=(1+k){\frac{x}{y},b=(1+k){\frac{y}{z},c=(1+k){\frac{z}{x},{k\ge0}.}}}$

So, use a new mechanical proving algorithm which can be read for a kind of polynomial

inequality,we can  prove the inequality is true.

My Solution  :)

From Holder's Inequality, we have:

$ \left( {{a^4} \plus{} b \plus{} c} \right)\left( {\dfrac{1}{a} \plus{} b \plus{} c} \right)\left( {1 \plus{} b \plus{} c} \right) \ge {\left( {a \plus{} b \plus{} c} \right)^3}$

$ \Rightarrow \dfrac{{\left( {\dfrac{1}{a} \plus{} b \plus{} c} \right)\left( {1 \plus{} b \plus{} c} \right)}}{{{{\left( {a \plus{} b \plus{} c} \right)}^3}}} \ge \dfrac{1}{{{a^4} \plus{} b \plus{} c}}$

Analogously, we have:

$ \dfrac{{\left( {\frac {1}{b} \plus{} c \plus{} a} \right)\left( {1 \plus{} c \plus{} a} \right)}}{{{{\left( {a \plus{} b \plus{} c} \right)}^3}}} \ge \dfrac{1}{{{b^4} \plus{} c \plus{} a}},\dfrac{{\left( {\dfrac{1}{c} \plus{} a \plus{} b} \right)\left( {1 \plus{} a \plus{} b} \right)}}{{{{\left( {a \plus{} b \plus{} c} \right)}^3}}}\ge \dfrac{1}{{{c^4} \plus{} a \plus{} b}}$

Therefore, we need to prove that:

$ \dfrac{{\left( {\frac {1}{a} \plus{} b \plus{} c} \right)\left( {1 \plus{} b \plus{} c} \right) \plus{} \left( {\dfrac{1}{b} \plus{} c \plus{} a} \right)\left( {1 \plus{} c \plus{} a} \right) \plus{} \left( {\dfrac{1}{c} \plus{} a \plus{} b} \right)\left( {1 \plus{} a \plus{} b} \right)}}{{{{\left( {a \plus{} b \plus{} c} \right)}^3}}}$
$ \le \dfrac{3}{{a \plus{} b \plus{} c}}$

$ \Leftrightarrow \left( {\dfrac{1}{a} \plus{} b \plus{} c} \right)\left( {1 \plus{} b \plus{} c} \right) \plus{} \left( {\dfrac{1}{b} \plus{} c \plus{} a} \right)\left( {1 \plus{} c \plus{} a} \right) \plus{} \left( {\dfrac{1}{c} \plus{} a \plus{} b} \right)\left( {1 \plus{} a \plus{} b} \right)$

$ \le 3{\left( {a \plus{} b \plus{} c} \right)^2}$

From the condition:

 $ a \plus{} b \plus{} c \plus{} 1 \equal{} 4abc$

  $ \Rightarrow b \plus{} c \plus{} 1 \equal{} a\left( {4bc \minus{} 1} \right)$

  $ \Rightarrow \left( {\frac {1}{a} \plus{} b \plus{} c} \right)\left( {1 \plus{} b \plus{} c} \right) \equal{} \left( {\frac {1}{a} \plus{} b \plus{} c} \right)a\left( {4bc \minus{} 1} \right) \equal{} \left( {1 \plus{} ab \plus{} ac} \right)\left( {4bc \minus{} 1} \right)$

 $ \Rightarrow \displaystyle\sum {\left( {\frac {1}{a} \plus{} b \plus{} c} \right)\left( {1 \plus{} b \plus{} c} \right)} \equal{} \displaystyle\sum {\left( {1 \plus{} ab \plus{} ac} \right)\left( {4bc \minus{} 1} \right)}$

  $ \equal{} 2\left( {ab \plus{} bc \plus{} ca} \right) \minus{} 3 \plus{} 8abc\left( {a \plus{} b \plus{} c} \right)$

  $ \equal{} 2\left( {ab \plus{} bc \plus{} ca} \right) \minus{} 3 \plus{} 2\left( {4abc} \right)\left( {a \plus{} b \plus{} c} \right)$

  $ \equal{} 2\left( {ab \plus{} bc \plus{} ca} \right) \minus{} 3 \plus{} 2\left( {a \plus{} b \plus{} c \plus{} 1} \right)\left( {a \plus{} b \plus{} c} \right)$

  $ \equal{} 2\left( {ab \plus{} bc \plus{} ca} \right) \minus{} 3 \plus{} 2{\left( {a \plus{} b \plus{} c} \right)^2} \plus{} 2\left( {a \plus{} b \plus{} c} \right)$

  $ \equal{} \left( {{a^2} \plus{} {b^2} \plus{} {c^2} \plus{} 2ab \plus{} 2bc \plus{} 2ca} \right) \minus{} 3 \plus{} 2{\left( {a \plus{} b \plus{} c} \right)^2} \plus{} 2\left( {a \plus{} b \plus{} c} \right) \minus{} \left( {{a^2} \plus{} {b^2} \plus{} {c^2}} \right)$

  $ \equal{} 3{\left( {a \plus{} b \plus{} c} \right)^2} \minus{} \left( {{a^2} \minus{} 2a \plus{} 1} \right) \minus{} \left( {{b^2} \minus{} 2b \plus{} 1} \right) \minus{} \left( {{c^2} \minus{} 2c \plus{} 1} \right)$

  $ \equal{} 3{\left( {a \plus{} b \plus{} c} \right)^2} \minus{} {\left( {a \minus{} 1} \right)^2} \minus{} {\left( {b \minus{} 1} \right)^2} \minus{} {\left( {c \minus{} 1} \right)^2}$

  $ \le 3{\left( {a \plus{} b \plus{} c} \right)^2}$

The proof is completed. Equality holds if and only if $ a \equal{} b \equal{} c \equal{} 1$

[quote="imedeen"]Let $ a,b,c > 0$ such that $ a \plus{} b \plus{} c \plus{} 1 \equal{} 4abc$. Prove that:
\[ \frac {1}{{{a^4} \plus{} b \plus{} c}} \plus{} \frac {1}{{{b^4} \plus{} c \plus{} a}} \plus{} \frac {1}{{{c^4} \plus{} a \plus{} b}} \le \frac {3}{{a \plus{} b \plus{} c}}\]
[/quote]
The following inequality inequality is also true.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a\plus{}b\plus{}c\geq3.$ Prove that:\[ \frac {1}{{{a^4} \plus{} b \plus{} c}} \plus{} \frac {1}{{{b^4} \plus{} c \plus{} a}} \plus{} \frac {1}{{{c^4} \plus{} a \plus{} b}} \le \frac {3}{{a \plus{} b \plus{} c}}\]

[quote="arqady"]
The following inequality inequality is also true.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a \plus{} b \plus{} c\geq3.$ Prove that:
\[ \frac {1}{{{a^4} \plus{} b \plus{} c}} \plus{} \frac {1}{{{b^4} \plus{} c \plus{} a}} \plus{} \frac {1}{{{c^4} \plus{} a \plus{} b}} \le \frac {3}{{a \plus{} b \plus{} c}}\]
[/quote]
[quote="zhu rh"]It had been posted before and it can be done by CS like the Serbia's one in 2007[/quote]
What is your proof? Thank you!

[quote="zhu rh"]Can anyone tell me how to prove arqady's one?[/quote]
The hint see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=329526


does anybody know how to make a degree symbol in $LATEX$?  I can't find it in the guide  :(

I think you have to type this ^\circ

so it comes like this $^\circ$

For example $47^\circ$

oh I thought it said to just do \circ and nothing was happening  :lol:

[quote="bpms"][quote="mathgeniuse^ln(x)"]you need a carat symbol (^)

However, if you don't it just becomes \circ which turns to $\circ$

If you do 47\circ it becomes $47\circ$

Hence you should do 47^\circ to get $47^\circ$[/quote]
The carrot symbol is for putting the circle in as an exponent. $90^{0}$(90^0) also works but it isn't as good.[/quote]

Yes, true, but $90^{0}=1$, so, to avoid confusion, use $90^\circ$. (See the difference: $90^{0}\ne 90^\circ$).

does this thread belong here? I think you're supposed to post latex questions in the questions suggestions and announcements forum.


What is the value of $ n$ in the equation $ n \plus{} (n \plus{} 1) \plus{} (n \plus{} 2) \equal{} 9$?

$ n \plus{} (n\plus{}1) \plus{} (n\plus{}2) \equal{} 9$

$ 3n \plus{} 3 \equal{} 9$

$ 3n \equal{} 6$

$ \boxed{n \equal{} 2}$

Alternately, you could also deduce that $ n \plus{} 1$ is the average of the numbers.  So that means that $ n \plus{} 1 \equal{} \frac {9}{3} \equal{} 3\implies n \equal{} 2$


Evaluate 1+2+4+7+... is there e formula for this?  :?

(Add 1,2,3...n to the previos term)

Let's call this function f(n) where n is the number of terms.

The sequence of the sum of the first n integers is 1 3 6 10 15 21... and can be found by the formula n(n+1)/2.

1 2 4 7 11 16 22 is the same thing, with one extra term at the beginning and one added to each term.

1 + 1+1 + 1+1+2 + 1+1+2+3 + 1+1+2+3+4 + 1+1+2+3+4+5 + 1+1+2+3+4+5+6...

First lets get rid of the extra 1's. 1 for each term = n.
f(n) = n

0 + 1 + 1+2 + 1+2+3 + 1+2+3+4 + 1+2+3+4+5...
Each one is an n(n+1)/2.

n + 0 + 1(2)/2 + 2(3)/2 + 3(4)/2 + 4(5)/2 + 5(6)/2 ...
$ n + 0 + \frac{1(2) + 2(3) + 3(4) + 4(5)...}{2} $

The sequence created is 2 + 6 + 12 + 20 + 30... when divided by 2 is 1 + 3 + 6 + 10 which is the first sequence we mentioned! There's a way to sum this sequence, but I don't know what it is and I don't have time to finish this. Anyone else?

So you are stuck where you want to find the sum:

\[(1)(2)+(2)(3)+\cdots+(n-1)(n)\]

right?

Here are some hints and then a final answer:

[hide="hint 1"]
You can express it in sigma notation as:

\[\sum_{k=1}^{n-1}k(k+1)\][/hide]

[hide="hint 2"]
This equals

\[\sum_{k=1}^{n-1}{k^2}+\sum_{k=1}^{n-1}{k}\][/hide]

[hide="hint 3"]
The sum of the first $n$ squares is:

\[\frac{n(n+1)(2n+1)}{6}\].[/hide]

[hide="solution"]
Picking up where Hexaditidom left we evaluate the sum:

\begin{eqnarray*}
(1)(2)+\cdots+(n-1)(n) &=& \sum_{k=1}^{n-1}{(k^2+k)}\\
&=& \sum_{k=1}^{n-1}{k^2}+\sum_{k=1}^{n-1}{k}\\
&=& \frac{(n-1)n(2n-1)}{6}+\frac{(n-1)(n)}{2}
\end{eqnarray*}

Thus our desired formula is:

\[n+\frac 12\left[ \frac{(n-1)n(2n-1)}{6}+\frac{(n-1)n}{2}\right].\]

You can clean that up or multiply it out...but I'm gonna leave it like that ;) .[/hide]

...or to get harder. Sum the expresion $\frac{1}{n}+\frac{2}{n}+\frac{4}{n}+\frac{7}{n}...$  :)

[quote="Slizzel"]So how can I sum 1+2+4+7+11...?  :huh:[/quote]

I believe that I gave you the formula...[hide]\[n+\frac 12\left[ \frac{(n-1)n(2n-1)}{6}+\frac{(n-1)n}{2}\right].\][/hide]

Is that not what you asked for?  If you are asking for the infinite sum then the answer is pretty easy...it diverges.  And for you second question just factor out a $\frac 1n$ and it's essentially the same thing.


[hide="1"]$ 3^{2x}\minus{}6*3^{x}\plus{}8 \equal{} 0$...so if $ z \equal{} 3^{x}$, then
$ z^{2}\minus{}6z\plus{}8 \equal{} 0$, so factoring gives us
$ z \equal{} 2$ or $ z \equal{} 4$.
So...$ x \equal{}\log_{3}4$ or $ x \equal{}\log_{3}2$.
Not sure how to simplify further. Oh wait, I do....the question wants the sums...so adding those gives $ log_{3}8$[/hide]

[hide="2"]The formula for the number of diagonals is $ \frac{(n)(n\minus{}3)}{2}$ if I remember correctly, so plugging in 20 for $ n$ (Number of sides) gives...170.[/hide]

[hide="4"]First...$ x^{2}\plus{}y^{2}\equal{} (x\plus{}y)^{2}\minus{}2xy$, so with substitution we get that $ xy \equal{}\frac{\minus{}3}{2}$...then....
$ (x^{3}\plus{}y^{3}) \equal{} (x\plus{}y)(x^{2}\minus{}xy\plus{}y^{2})$, so with substitution...$ (x^{3}\plus{}y^{3}) \equal{} (1)(4\plus{}3)$, or $ 7$[/hide]

I don't know enough about probability to do 3 and 5, which is pitiful.... :(

[hide="(1)"]

$ 3^{2x}\minus{}6\cdot3^{x}\plus{}8 \equal{} 0$

By Vieta's formulas, we have that

$ 3^{x_{1}}\cdot 3^{x_{2}}\equal{} 8$

$ 3^{x_{1}\plus{}x_{2}}\equal{} 8$

$ x_{1}\plus{}x_{2}\equal{}\log_{3}8\approx 1.893$

Or if we substitute $ z \equal{} 3^{x}$ and actually find the roots from above, we just add $ \log_{3}2\plus{}\log_{3}4 \equal{}\log_{3}8$
[/hide]

[hide="(3)"]
The number of different ways we can roll two 1s in 5 rolls is $ {5\choose2}5^{3}$. This is because we need to order the positions of the 2 ones, and then there are 5 ways to pick the numbers of each of the remaining dies. There are a total of $ 6^{5}$ possible ways of rolling. Hence, our probability is $ \frac{5^{4}}{2^{4}3^{5}}$.
[/hide]

[hide="(5)"]
The girls must be all clustered together in the middle of the line, and so the boys must be placed at the ends. At the beginnning of the line there must be between $ 1$ and $ 5$ boys; hence there are 5 possible ways to order the lines just considering gender.

We can order the girls in $ 5!$ ways, and the boys in $ 6!$ ways, hence the total number of ways is $ 5\cdot 5! 6! \equal{} 432000$.
[/hide]

I was thinking along those lines for 3, but wasn't sure about the specifics, and I didn't really look at 5 at all. Stupid mistake on 1, didn't read the problem carefully enough. Oh well.

I didn't know you could apply vieta's to exponents too...that's interesting.

[quote="Icy"]I didn't know you could apply vieta's to exponents too...that's interesting.[/quote]

Misses the point.  azjps is applying Vieta's to the variable $ y \equal{} 3^{x}$.

[quote="boo"]1. The sum of the solutions of the equation $ 9^{x}\minus{}6\cdot 3^{x}\plus{}8 \equal{} 0$ is what?

2. How many diagonals does a regular 20-gon have?

3. When Five ordinary six sided dice are tossed simultaneously, the probability that a 1 shows on the top face of exactly two of the dice is what?

4. If $ x\plus{}y \equal{} 1$ and $ x^{2}\plus{}y^{2}\equal{} 4$, then what does $ x^{3}\plus{}y^{3}$ equal?

5. In how many ways can six boys and 5 girls stand in a row if all the girls are to stand together but the boys cannot all stand together?[/quote]

[hide="4"]
We have that
\[ xy\equal{}\frac{(x\plus{}y)^{2}\minus{}x^{2}\minus{}y^{2}}{2}\equal{}\frac{1\minus{}4}{2}\equal{}\minus{}1.5\]
We also have that
\[ x^{3}\plus{}y^{3}\equal{}(x\plus{}y)(x^{2}\plus{}y^{2}\minus{}xy)\equal{}1(4\minus{}xy)\equal{}4\minus{}(\minus{}1.5)\equal{}\boxed{5.5}\]
[/hide]

Ugh...I subtracted $ 2xy$ instead of just $ xy$...Can't get anything right ever.

[hide="1"]
$ 3^{2x}\minus{}6\cdot3^{x}\plus{}8\equal{}0\implies y^{2}\minus{}6y\plus{}8\equal{}0\implies (y\minus{}4)(y\minus{}2) \equal{} 0\implies y\equal{}4,2\implies 3^{x}\equal{}4,2\implies\log_{3}4\plus{}\log_{3}2\equal{}\boxed{\log_{3}8}$[/hide]
[hide="2"]
$ \frac{20\cdot17}{2}\equal{}\boxed{170}$[/hide]
[hide="3"]
$ \binom{5}{2}\cdot\frac{1}{6}^{2}\cdot\frac{5}{6}^{3}\equal{}\boxed{\frac{625}{3888}}$[/hide]
[hide="4"]
$ (x\plus{}y)(x\plus{}y)\minus{}2xy\equal{}4\implies (1)(1)\minus{}2(xy)\equal{}4\implies xy\equal{}\frac{\minus{}3}{2}\implies (x\plus{}y)(x^{2}\minus{}xy\plus{}y^{2})\implies (1)(5.5)\equal{}\boxed{5.5}$[/hide]
[hide="5"]
$ 5\cdot5!\cdot6!\equal{}\boxed{432000}$[/hide]


Let ABCD is a convex quadrilateral, AC=2a,BD=2b.Prove that at least one side of ABCD is not less than $ sqrt {a^2 \plus{} b^2 }$

[quote="TRAN THAI HUNG"]Let ABCD is a convex quadrilateral, AC=2a,BD=2b.Prove that at least one side of ABCD is not less than $ \sqrt {a^2 \plus{} b^2 }$[/quote]
If there is not such side then $ 4a^{2}\plus{}4b^{2} > AB^{2}\plus{}BC^{2}\plus{}CD^{2}\plus{}DA^{2} \geq AC^{2}\plus{}BD^{2}\equal{} (2a)^{2}\plus{}(2b)^{2}$


geometry
3D geometry
pyramid
sphere
tetrahedron
inradius
ratio

If a triangular pyramid has both an inscribed and circumscribed sphere with radii r and R, respectively. Find min(R/r) in terms of the lengths of the edges of the pyramid.

What do you mean by max?  A regular tetrahedron has a fixed inradius and circumradius.

[quote="al.M.V."]If a triangular pyramid has both an inscribed and circumscribed sphere with radii r and R, respectively. Find max(R/r) in terms of the lengths of the edges of the pyramid.[/quote]

Sorry, but I'm still not sure what you mean.  Given the lengths of the edges, R and r can be determined, so what are we maximizing over?

well, i was just trying to generalize the R/r for circles in triangles to spheres in pyramids. Supposedly as pyramids take different forms, then the ratio would change, so what's the maximum. This is a problem I made up myself, so it might not make much sense, in which case it should be ignored... :D

So I think what you mean is find the maximum of R/r over [b]all[/b] tetrahedrons.  In this case, it was posted here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=27187]www.mathlinks.ro/Forum/viewtopic.php?t=27187[/url]

oh thanks! that's exactly what i wanted!  :)

[quote]Slight correction: That post answers the question on the minimum value of R/r. There is no maximum value.[/quote]

Yeah, now I think about it, of course.  :D Just make the tetrahedron infinitely "thin" and "long". :lol: