Vieta calculus derivative algebra polynomial function algorithm

By studying the derivative it has exactly one real zero. This is quite clearly not a triple zero, since the coefficients don't match (probably what you've already done, or just use binomial theorem). It cannot be a double root, since then the third root would be complex, but complex roots always come in pairs. So it is a simple zero. Then the other two roots are complex and they come as a conjugate pair.

[hide="how to do it with with just algebra"]
Suppose there is a double root, say $a$. Let $b$ be the other root, then by vieta's,
\[2a+b=1,\ 2ab+a^{2}=-1\]
Substituting $b=1-2a$ into the second equation,
\[2a(1-2a)+a^{2}+1=1+2a-3a^{2}=(1-a)(1+3a)=0\]
so $a$ is either $-\frac{1}{3}$ or $1$, but neither of these are roots of the polynomial, contradiction.
[/hide]

[quote="bpms"][quote="srulikbd"]it's should be easy  :maybe:  I tried to use vieta's formula and then assume that $a=b$ adn get contradiction. if $a=b=c$ I succeed in getting contradiction.[/quote]
[hide="hm"]Using the formula for the sum of a geometric series:
$\frac{x^{4}-1}{x-1}=0$
So the roots follow from the 4th roots of unity[/hide]
Or you could just plug in the values into the cubic equation if you are lazy and show that they are different.[/quote]

Umm... $x^{3}-x^{2}-x-1 \neq \frac{x^{4}-1}{x-1}$.

Sketch the graphs of $y=x^{3}-x^{2}-x$ and $y=1.$
You will see only one intersection point whose abcissa is greater than $\frac{5}{3}.$ Since the given cubic equation has real coefficients, the other roots are conjugated complex numbers.

[quote="scorpius119"][hide="how to do it with with just algebra"]
Suppose there is a double root, say $a$. Let $b$ be the other root, then by vieta's,
\[2a+b=1,\ 2ab+a^{2}=-1 \]
Substituting $b=1-2a$ into the second equation,
\[2a(1-2a)+a^{2}+1=1+2a-3a^{2}=(1-a)(1+3a)=0 \]
so $a$ is either $-\frac{1}{3}$ or $1$, but neither of these are roots of the polynomial, contradiction.
[/hide][/quote]
Note that $-\frac13$ and $1$ are exactly the extremal points of the given polynomial. If you think a little bit, the solution with 'calculus' and with vieta are exactly the same. I wouldn't call using derivatives calculus when it comes to polynomials. It seems stupid to artificially distinguish calculus and algebra when working with polynomials, since many of the algebraic properties of polynomials in $\mathbb{C}$ are closely connected to theorems about analytic functions.

[quote="scorpius119"][hide="how to do it with with just algebra"]
Suppose there is a double root, say $a$. Let $b$ be the other root, then by vieta's,
\[2a+b=1,\ 2ab+a^{2}=-1 \]
Substituting $b=1-2a$ into the second equation,
\[2a(1-2a)+a^{2}+1=1+2a-3a^{2}=(1-a)(1+3a)=0 \]
so $a$ is either $-\frac{1}{3}$ or $1$, but neither of these are roots of the polynomial, contradiction.
[/hide][/quote]
ty I knew it's should be easy...

[quote="srulikbd"]it's should be easy  :maybe:  I tried to use vieta's formula and then assume that $a=b$ adn get contradiction. if $a=b=c$ I succeed in getting contradiction.[/quote]

You then proved it!!! You sucessfully proved that if a=b, you get a conradiction, and you can apply the same logic to b=c and a=c.

[quote="srulikbd"]and the second part of the ques plz...any hint  :)[/quote]

those follow from simple factorizations...

[quote="srulikbd"]but $a,b,c$ are the roots of the equation from my first post...not just integers...[/quote]

ohhhh ok...

i guess i have a solution, it might not be exactly what you want, but anyway,

after expanding, it is clearly a symmetric polynomial in a,b,c

a lemma (from ACoPS 3.1.16e) states that any symmetric polynomial sum can be writen as a polynomial in the elementary symmetric functions (those are s_3=abc, s_2=ab+bc+ca, s_1=a+b+c), since those are integers, the result is an integer too

i was unable to prove this lemma, but of course, it is stuck in my mind...

[hide="solution"]
Use the division algorithm. There exist polynomials $Q$ and $R$ with integer coefficients such that
\[x^{1000}=(x^{3}-x^{2}-x-1)Q(x)+R(x)\]
for all $x$ and $R$ has degree at most two or equals 0. (Yes, the division algorithm works for integer coefficients, [b]if[/b] the divisor has leading coefficient 1, such as in this case.) So let
\[R(x)=ux^{2}+vx+w\]
where $u,v,w$ are integers. We obtain
\begin{eqnarray*}\sum_\text{cyc}\frac{a^{1000}-b^{1000}}{a-b}&=&\sum_\text{cyc}\frac{R(a)-R(b)}{a-b}\\ &=&u\left( \sum_\text{cyc}(a+b) \right)+v\left( \sum_\text{cyc}1\right) \\ &=& 2u+3v\\ \end{eqnarray*}
which is an integer.
[/hide]

thanks a lot!  :)  but how exactly Use the division algorithm. There exist polynomials...?

actually, it can be done without proving the division algorithm:
[hide]
Define
\[t_{n}=\frac{a^{n}-b^{n}}{a-b}+\frac{b^{n}-c^{n}}{b-c}+\frac{c^{n}-a^{n}}{c-a}\]
for nonnegative integers $n$. Then $t_{0}=0,\ t_{1}=3,\ t_{2}=2$, and the sequence satisfies the recurrence
\[t_{n+3}=t_{n+2}+t_{n+1}+t_{n}\]
so by strong induction, $t_{n}$ is an integer for all $n$; in particular, $t_{1000}$ is an integer.
[/hide]


Let$\ a,b,c$ are the measures of the sides of the triangle satisfying the condition$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3$ .
Find the least value of the expression
$\ A=\sqrt[4]{a+b-c}+\sqrt[4]{b+c-a}+\sqrt[4]{c+a-b}$

It's not open question. Similar problem proposed by Hungkhtn is in his book SangtaoBDT (I think you must know it, PhongVan). This problem solved case $\sqrt{a+b-c}$, not $\sqrt[4]{a+b-c}$, but two these problems can be solved by contradiction method.

To GacKiem:

I do not want to talk about things like this but this time I have to because you always annoy me.

You keep saying that pvthuan advertising his book. And you seem to be
envious about that. You yourself are saying too much about hungkhtn's book.

So it's not fair to do such things. You yourself should stop things like that. You keep saying that other should be POLITE, but you are not.


Points $P,Q$ do not lie on the diagonal $AC$ of parallelogram $ABCD$. $\angle ABP=\angle ADP, \angle CBQ = \angle CDQ$. Prove that $\angle PAQ = \angle PCQ$


Euler
calculus
integration
derivative
calculus computations

When estimating the value of $ \zeta(2)$, the normal technique is to split the infinite series into a partial summation, in which case you are left with a tail (a.k.a the error of the approximation). Note that this looks like:
$ \sum_{n \equal{} 1}^{\infty} \frac {1}{n^2} \equal{} \sum_{n \equal{} 1}^{N} \frac {1}{n^2} \plus{} \sum_{n \equal{} N \plus{} 1}^{\infty} \frac {1}{n^2}$

And note that this end tail summation can generally be regarded as $ O(1/N)$ (I believe this is correct. Correct me if it is not.)

However, can you find an even better estimation for the end tail summation?

[hide]
By better I mean if you sum the first 1000 terms you would get an error of about $ 10^{ \minus{} 3}$. However, with the refined estimation you would get an error of about $ 10^{ \minus{} 6}$.
[/hide]

Euler did this when approaching the Basel problem by using some transformations and estimations on a certain integral. If you study his work you can learn it. I'm sure someone here knows more and can give a more informed presentation.

http://www.math.nmsu.edu/~davidp/euler2k2.pdf

The "tail" is $ \sum_{n \equal{} N \plus{} 1}^{\infty}\frac1{n^2}.$ Compare this to the integral

$ \int_{N \plus{} \frac12}^{\infty}\frac1{x^2}\,dx \equal{} \sum_{n \equal{} N \plus{} 1}^{\infty} \int_{n \minus{} \frac12}^{n \plus{} \frac12}\frac1{x^2}\,dx.$

Note that we are essentially using the midpoint rule on each integral in that last sum; the error on doing so is proportional to the second derivative of $ \frac1{x^2},$ which is $ \frac {6}{x^4}.$ That means that if we add $ \int_{N \plus{} \frac12}^{\infty}\frac1{x^2}\,dx \equal{} \frac1{N \plus{} \frac12}$ to the partial sum, the remaining error will look like $ \sum_{n \equal{} N \plus{} 1}^{\infty} \frac {C}{n^4}\approx\frac {C}{N^3}.$

From here, there's an idea from numerical analysis that I'd like to exploit: the notion of a Richardson extrapolation table. To use a Richardson table, we must presume a series or asymptotic expansion, and we must know what exponents are present in that expansion.  In this case, let $ S(N) \equal{} \sum_{n \equal{} 1}^N\frac1{n^2}$ and let $ S \equal{} \zeta(2)$ be the whole sum. Let $ T(N) \equal{} S(N) \plus{} \frac1{N \plus{} \frac12}.$ Our presumption is that we can write

$ S \equal{} T(N) \plus{} \frac {A}{N^3} \plus{} \frac {B}{N^4} \plus{} \frac {C}{N^5} \plus{} \cdots.$

I'm going to create a table. My first column will be $ N,$ going by powers of $ 2.$ My second column will be $ S(N)$ and my third column will be $ T(N).$

The next column, call it $ T_1(N),$ will be $ \frac {8T(N) \minus{} T(N/2)}7.$

The column after that, call it $ T_2(N),$ will be $ \frac {16T_1(N) \minus{} T_1(N/2)}{15}.$

Then $ T_3(N) \equal{} \frac {32T_2(N) \minus{} T_2(N/2)}{31},$ and so on until we run out of data. Here's my table, going out to $ N \equal{} 32:$

$ \begin{tabular}{ccccccccc} N & S(N) & T(N) & & & & & & \\
1 & 1.000000 & 1.66666667 & & & & & \\
2 & 1.250000 & 1.65000000 & 1.64761905 & & & & \\
4 & 1.423611 & 1.64583333 & 1.6452381 & 1.64507937 & & & \\
8 & 1.527422 & 1.64506911 & 1.64495994 & 1.64494139 & 1.64493694 & & \\
16 & 1.584347 & 1.64495259 & 1.64493595 & 1.64493435 & 1.64493412 & 1.64493408 & \\
32 & 1.614167 & 1.64493649 & 1.64493419 & 1.64493408 & 1.64493407 & 1.64493407 & 1.64493407\end{tabular}$

I've been doing the computations on a spreadsheet, using floating-point arithmetic. I'm about at the point at which roundoff error makes it difficult to improve on the accuracy from what I already have - but, based on just 32 terms, I already have more accuracy than the displayed digits. The last two numbers on the bottom row have an accuracy of about $ 10^{ \minus{} 11}.$


calculus
integration
trigonometry
calculus computations

[quote="Fermat -Euler"]Find
${I_n = \int_0^\pi \frac{\cos (nt) dt}{1+\cos^2 t}}$[/quote]

Just an idea..

Do the substitution $t=\pi -x$...

hope it is working  :)

Should they be given recursively, or do you want an explicit formula?

$I_{n}\equiv0$ when n is odd, so let's define $J_{n}=I_{2n}$. we know $J_{n+2}+J_{n}=\int_{0}^{\pi}\frac{2 \cos\left(2nt+2t\right)\cos 2t}{1+\cos^2 t}dt$,
thus $J_{n+2}+6J_{n+1}+J_{n}=4\int_{0}^{\pi}\cos\left(2nt+2t\right)dt=0$, now everyone knows how do get $I_{2n}$.


Supposed that i already 'installed' a "LATEX" in my computer.How would i use the latex in another "computer cafe"
in the nearby neighborhood computer shop? Is latex 'downloadable?'

                                                      Respectfully,
                                                     A.Eddington

LaTeX has to be downloaded onto each individual computer that will be using it.  It is free to download.

Go to http://miktex.org for a commonly-used free download for Windows.


Give $ a,b,c \ge 0$.Prove that:
$ \sqrt{\frac{a(b\plus{}c)}{4b^2\plus{}10bc\plus{}4c^2}}\plus{} \sqrt{\frac{b(c\plus{}a)}{4c^2\plus{}10ca\plus{}4a^2}}\plus{} \sqrt{\frac{c(a\plus{}b)}{4a^2\plus{}10ab\plus{}4b^2}} \ge 1$
Equality occur if and only a=b=c;a=b,c=0 or any cyclic permution.
I have nice proof for this problem :)

If the above inequality is true,  the following must be valid

Let $ x,y,z >0$ then

\[ \sum_{cyc}\frac{x(y\plus{}z)}{4y^2\plus{}10yz\plus{}4z^2} \ge \frac{1}{3}\]

[quote="great math"]If the above inequality is true,  the following must be valid

Let $ x,y,z > 0$ then
\[ \sum_{cyc}\frac {x(y \plus{} z)}{4y^2 \plus{} 10yz \plus{} 4z^2} \ge \frac {1}{3}\]
[/quote]
Dear my friend,this ineq is obivious true by cauchy and schur :).The strongest ineq:
$ \sum\frac{x(y\plus{}z)}{y^2\plus{}yz\plus{}z^2} \ge 2$(vasc)

[quote="babyltv"]Give $ a,b,c \ge 0$.Prove that:
$ \sqrt {\frac {a(b \plus{} c)}{4b^2 \plus{} 10bc \plus{} 4c^2}} \plus{} \sqrt {\frac {b(c \plus{} a)}{4c^2 \plus{} 10ca \plus{} 4a^2}} \plus{} \sqrt {\frac {c(a \plus{} b)}{4a^2 \plus{} 10ab \plus{} 4b^2}} \ge 1$
Equality occur if and only a=b=c;a=b,c=0 or any cyclic permution.
I have nice proof for this problem :)[/quote]
By Holder we obtain:
$ \left(\sum_{cyc}\sqrt {\frac {a(b \plus{} c)}{4b^2 \plus{} 10bc \plus{} 4c^2}}\right)^2\sum_{cyc}\frac{a^2(4b^2 \plus{} 10bc \plus{} 4c^2)}{b\plus{}c}\geq(a\plus{}b\plus{}c)^3.$
Id est, it remains to prove that
$ (a\plus{}b\plus{}c)^3\geq\sum_{cyc}\frac{a^2(4b^2 \plus{} 10bc \plus{} 4c^2)}{b\plus{}c},$ which is easy SOS.

[quote="arqady"]
Id est, it remains to prove that
$ (a \plus{} b \plus{} c)^3\geq\sum_{cyc}\frac {a^2(4b^2 \plus{} 10bc \plus{} 4c^2)}{b \plus{} c},$ which is easy SOS.[/quote]
Yes, you are right. It's can be write into:
$ \sum\ a^3\plus{}6abc \geq\ \sum\ ab(a\plus{}b)\plus{}2abc\sum\ \frac{a}{b\plus{}c}$
It's easy with SOS  :)


Suppose $ (u_n)_{n\in \mathbb{N}^*}$ is a increase sequence such that $ lim_{n\to + \infty}{(u_n)} = 1$ and $ \forall n\in \mathbb{N}^*, u_n>0$

Prove that $ {\lim }\limits_{n \to \infty } \sqrt [n]{{\sum\limits_{i = 1}^n {u_i^n} }} = 1.$

$ n^{\frac {1}{n}}u_n \ge \sqrt [n]{\sum u_i^n} \ge u_n$
$ \Rightarrow \lim \equal{} 1$

I think the first condition isnot nessesary
$ \forall \epsilon > 0 :$exist $ K \in N^*: 1 \minus{} \epsilon \le u_n < 1 \plus{} \epsilon \forall n > K$
$ \Rightarrow \sum^{n}_{i \equal{} 1} u_i^n \equal{} C \plus{} \sum^n_{i \equal{} K \plus{} 1} u_i^n ( C \equal{} \sum^{K}_{i \equal{} 1} u_i^n)$
$ \Rightarrow \sqrt [n]{ C \plus{} (n \minus{} K)(1 \minus{} \epsilon)^n} < \sqrt [n]{\sum^{n}_{i \equal{} 1} u_i^n} < \sqrt [n]{ C \plus{} (n \minus{} K)(1 \plus{} \epsilon)^n}$
$ \Rightarrow 1 \le \lim_{n\rightarrow \infty} \sqrt [n]{\sum^{n}_{i \equal{} 1} u_i^n} \le 1 \plus{} \epsilon \forall \epsilon > 0$
$ \Rightarrow \lim_{n\rightarrow \infty} \sqrt [n]{\sum^{n}_{i \equal{} 1} u_i^n} \equal{} 1$


lemma 1: if we have x1,x2,x3,x4 are integer numbers and:
x1+x2=x3+x4  (mod 3)
x1x2= x3x4     (mod 3)
then we have {x1,x2}={x3,x4} ( mod 3)

lemma 2: (a1,a2,...,a9) is a root of this equation.
Then all permutations of that set that a1,a4,a7 or a2,a3 or a5,a6 or a8,a9 are permuted are all also roots of this equation.
We can say that a1,a4,a7 are same rolls.
                        a2,a3      are same rolls.
                        a5,a6 are same rolls.
                        a8,a9 are same rolls.

We easily have that: a1+a4+a7 mod 3= 0
                               (a1)^2+(a4)^2+(a7)^2 mod 3 =0
hence, there are always at least 2 of these 3 numbers are the same in mod 3.
Using lemma 2 ,we can suppose a1=a7=a; a4= b (certainly in mod 3)

if a+ba1+a2+a3+a4=a4+a5+a6+a7 and (a1)^2+(a2)^2+(a3)^2+(a4)^2= (a4)^2+(a5)^2+(a6)^2+(a7)^2 show that: a2+a3=a5+a6 (mod 3) and a2a3=a5a6 (mod 3)
hence, {a2,a3}={a5,a6} => {a2,a3}={b,c}
So, {a8,a9}={a,c} => b=c (contradic)
So, a=b
We can easily do the next step since we have a1=a4=a7 (mod 3)

I was wondering is there any other way to solving this?
For example how they did for the official APMO solution.  But I think the official solution was slightly unclear...at least to me.
could anyone try and explain the question or provide a similar one which is simple please?  or if better how to solve all these kinds of problems in general.

Could anyone make suggestions/point out errors in this solution?
[hide="Solution"]
WLOG, let $a_1 < a_4 < a_7$. Also, WLOG let $a_{3i-1} < a_{3i}$ for $i = 1, 2, 3$.
Adding the three expressions in the 1st equality yields $2a_1 + a_2 + a_3 + 2a_4 + a_5 + a_6 + 2a_7 + a_8 + a_9 = 45 + a_1 + a_4 + a_7$.
Thus, since the three expressions were equal, their sum is divisible by 3. Thus, $45 + a_1 + a_4 + a_7$ is divisible by 3.
This implies that $3\mid a_1 + a_4 + a_7$. Doing the same for the 2nd equality yields $3\mid 285 + a_1^2 + a_4^2 + a_7^2$, or $3\mid a_1^2 + a_4^2 + a_7^2.$
The only solutions to these equations are $a_1 \equiv a_4 \equiv a_7 (mod 3)$. We now have three cases:

Case 1: The common modular residue is 1.
From the WLOG, $a_1 = 1$, $a_4 = 4$, and $a_7 = 7$. Then, $a_1^2 + a_2^2 + a_3^2 + a_4^2 = (285 + a_1^2 + a_4^2 + a_7^2)/3 = (285+66)/3 = 117$.
Thus, $a_2^2 + a_3^2 = 117 - a_1^2 - a_4^2 \implies a_2^2 + a_3^2 = 100$, which only has the solution $a_2 = 6$, and $a_3 = 8$. But then, $a_5^2 + a_6^2 = 52$, and $a_8^2 + a_9^2 = 67$. Since $81>67, 81>52$, each of $a_5, a_6, a_8, a_9$ is less than 9. But this is contradiction, since one of these must be 9(9 has not appeared in the permutation yet). Thus, this case is impossible.

Case 2: The common modular residue is 2.
Following a similar procedure to case 1, $a_1 = 2$, $a_4 = 5$, $a_7 = 8$, and $a_1^2 + a_2^2 + a_3^2 + a_4^2 = 126$. This yields that $a_2^2 + a_3^2 = 97 \implies a_2 = 4, a_3 = 9$. Similarly, $a_5^2 + a_6^2 = 37 \implies a_5 = 1, a_6 = 6$, and $a_8^2 + a_9^2 = 58 \implies a_8 = 3, a_9 = 7$.
Since $2 + 4 + 9 + 5 = 5 + 1 + 6 + 8 = 8 + 3 + 7 + 2 = 20$, this permutation does indeed satisfy the conditions. However, removing the WLOG restriction, swapping $a_{3i}$ and $a_{3i-1}$ keeps the validity of the permutation, for $i = 1,2,3$. Permuting around the sets $ \{a_1, a_2, a_3, a_4\}$, $\{a_4, a_5, a_6, a_7\}$, $\{a_7, a_8, a_9, a_1\}$ (for instance, permuting $\{a_1, a_2, a_3, a_4\}$, $\{a_4, a_5, a_6, a_7\}$, $\{a_7, a_8, a_9, a_1\}$ from {2, 4, 9, 5}, {5, 1, 6, 8}, {8, 3, 7, 2} to {8, 3, 7, 2}, {5, 1, 6, 8}, {2, 4, 9, 5}.)

Case 3: The common modular residue is 3.
Following the same procedure yields $a_2^2 + a_3^2 = 137 - 45 = 92$. However, this has no integer solutions. Thus, this case has no solutions.

Thus, the permutations that satisfy the given equalities are found from $(2, 9, 4, 5, 1, 6, 8, 3, 7)$, and others found by swapping $a_{3i}$ and $a_{3i-1}$ for $i = 1,2,3$, or by permuting the sets $ \{a_1, a_2, a_3, a_4\}$, $\{a_4, a_5, a_6, a_7\}$, $\{a_7, a_8, a_9, a_1\}$  as described above in case 2.
[/hide]

@cpma213 can you more explain why does 45+a_1+a_2+a_3 is dibisible by 3?thanks

There are 4 even integers to use. Any of them can be counted in one or two of the three equal expressions, and each of the expressions must have the same number of even integers (consider the squares mod 4). The total number of times even integers will appear in the three sums is between 4 and 8, but must be divisible by three, so each each sum has exactly two even integers, and two of the four even integers are counted twice, so they are among $a_1, a_4, a_7$.  Now note:

$$3(a_1+a_2+a_3+a_4) = (a_1+a_2+a_3+a_4) + (a_4+a_5+a_6+a_7) + (a_7+a_8+a_9+a_1)= 45 + a_1 + a_4 + a_7$$

Thus $3|(a_1+a_4+a_7)$ so either all are congruent mod 3 or all are distinct. The latter cannot be true because all of the sums must have the same number of multiples of three (consider the squares mod 3). Thus the only option is $2,5,8$ since two of these must be even.

Now the sum of all three sums is $45+2+5+8=60$ so each sum is $20$. Whichever sum has $2,5$ in it must have either $9,4$ or $7,6$ as the other two. We can check each of these, filling in the rest with the fact that each side has one multiple of three and two even numbers, to find that

$$\{2,9,4,5\} \{5,1,6,8\} \{8,3,7,2\}$$

and its associated permutations are the only solutions.

it can be easily done by cases on congruence of x1 x4 x7 MOD 3

Notice $45 + a_1 + a_4 + a_7 =2a_1 + a_2 + a_3 + 2a_4 + a_5 + a_6 + 2a_7 + a_8 + a_9 = 3(a_1 + a_2 + a_3 + a_4)$; in particular, a_1,a_4,a_7 are either all congruent or all distinct mod 3. In the same way we get 285+a_1^2+a_4^2+a_7^2=3(a_1^2+a_2^2+a_3^2+a_4^2); in particular, if they're distinct mod 3 the RHS is 0 mod 3 but LHS is not 0 mod 3.

Case 1. a_1,a_4,a_7=1,4,7. Then a_2+a_3=6+a_5+a_6, a_7+a_8=3+a_5+a_6, so it follows by adding them onto a_5+a_6 that a_5+a_6=8, so a_5,a_6=2,6 or 3,5; keep caseworking like this until the end (im lazy to  type this up but its easy casework)
Case 2 and 3 do the same thing

solutions are (2,9,4,5,1,6,8,3,7) and permutations by cyclically switching them by 3 spots.


geometry
parallelogram
trigonometry
algebra
polynomial
complex numbers
geometry proposed

a) Given a convex hexagon ABCDEF with all the equal angles.
   Prove that |AB|-|DE| = |EF|-|BC| = |CD|-|FA|.
b) The opposite problem: Prove that it is possible to build a convex
   hexagon with equal angles of six segments a_1, a_2, ... , a_6, whose
   lengths satisfy the condition a_1 - a_4 = a_5 - a_2 = a_3 - a_6.

Well, there's a well-known and elegant solution of this one. We can construct an equilateral triangle by drawing parallels to the sides of the hexagon and forming some parallelograms, but here's another one:

We can use complex numbers. Let $\omega=\cos\frac \pi 3+i\sin\frac\pi 3$. Such a hexagon exists (with sides $a_1,a_2,a_3,a_4,a_5,a_6$ in this order) iff $a_1+a_2\omega+a_3\omega^2+a_4\omega^3+a_5\omega^4+a_6\omega^5=0$, which is equivalent to $(a_1-a_4)+(a_2-a_5)\omega+(a_3-a_6)\omega^2=0$, and since the minimal polynomial of $\omega$ over $\mathbb R$ is $1-\omega+\omega^2=0$, we get the conclusion (in both a) and b)).

Damn, that is the coolest solution...indeed i didnt want people to post the equilateral triangle solution but this one....this is much more nice than the other and also shows a nice aplication of complex numbers and polinomials.

Doing some tidy-up work on my new laptop, I have just unearthed a file with the solution of this problem I thought to have lost. I bet this won't interest anybody, but I post it since I don't know what else I should do with the solution...
_________________________________________________________________________________________

I slightly modify the notations of the problem:

[i][b]a)[/b] Given a convex hexagon ABCDEF whose angles are all equal.
Prove that AB - DE = CD - FA = EF - BC.

[b]b)[/b] Given six segments a, b, c, d, e, f such that a - d = c - f = e - b.
Prove that there exists a convex hexagon ABCDEF whose angles are all equal and whose sides are AB = a, BC = b, CD = c, DE = d, EF = e and FA = a.[/i]

Actually, there are two synthetic solutions, and I find both of them not less cool than the one using complex numbers... probably that's because I am always in war with polynomials... anyway, that's a matter of taste...

So here are the two synthetic solutions, for those who share my taste:

[i]First solution.[/i] [b]a)[/b] This is the proposed solution of problem [b]a)[/b] when it was posed on the Allrussian Olympiad 1964 for the 9th class:

Since the angles of the hexagon ABCDEF are all equal, while their sum is 4 $\cdot$ 180° = 720°, these angles are all equal to 120°. Thus, < (AB; BC) = 120°, < (BC; CD) = 120° and < (CD; DE) = 120°, with angles directed modulo 180°. Hence, < (AB; DE) = < (AB; BC) + < (BC; CD) + < (CD; DE) = 120° + 120° + 120° = 360° = 0°, so that AB || DE. By analogy, BC || EF and CD || FA.

Now, let x be the parallel to the lines BC and EF through the point A, let y be the parallel to the lines AB and DE through the point C, and let z be the parallel to the lines CD and FA through the point E. The lines x and y meet at U, the lines y and z meet at V, and the lines z and x meet at W. Then,

  < UVW = < (y; z) = < (DE; CD)          (since y || DE and z || CD)
        = - < (CD; DE) = - 120° = 60°.

Similarly, < VWU = 60° and < WUV = 60°. Hence, the triangle UVW is equilateral, and UV = VW = WU. But since AU || BC and CU || AB, the quadrilateral ABCU is a parallelogram, and we have CU = AB. Similarly, CV = DE. Thus, UV = CV - CU = DE - AB. Similarly, VW = FA - CD and WU = BC - EF. Hence, we have DE - AB = FA - CD = BC - EF. In other words, AB - DE = CD - FA = EF - BC. Solution of [b]a)[/b] complete.

[b]b)[/b] I work with directed angles modulo 180° again.

WLOG assume that a - d = c - f = e - b $\leq$ 0 (else, replace a, b, c, d, e, f by b, c, d, e, f, a). Then, we have d - a = f - c = b - e $\geq$ 0. Now construct an equilateral triangle UVW with sidelength VW = WU = UV = d - a = f - c = b - e. We can WLOG assume that the orientation of this equilateral triangle UVW is clockwise, so that we have < WUV = < UVW = < VWU = + 60°.

Let E be a point on the ray WV such that WE = f. Then, VE = WE - VW = f - (f - c) = c. Similarly, define a point A on the ray UW such that UA = b and obtain WA = e, and define a point C on the ray VU such that VC = d and obtain UC = a. Now, let the parallels to the lines CU and AU through the points A and C meet each other at the point B; then, the quadrilateral ABCU is a parallelogram, and thus we have AB = UC = a and BC = UA = b. Similarly, define two points D and F such that the quadrilaterals CDEV and EFAW are parallelograms, and you will see that CD = c, DE = d, EF = e and FA = f.

Now, we have found a hexagon ABCDEF with sidelengths AB = a, BC = b, CD = c, DE = d, EF = e and FA = f. The convexity of this hexagon is clear. Remains to show that all angles of this hexagon are 120°.

We will show this now: Since AB || CU and CB || AU, we have < ABC = < (AB; CB) = < (CU; AU) = < VUW = - < WUV = - 60° = 120°. Hence, we have showed < ABC = 120°; similarly, < CDE = 120° and < EFA = 120°. Now, just as we saw AB || CU, we also have AF || EW, and thus < FAB = < (AF; AB) = < (EW; CU) = < WVU = - < UVW = - 60° = 120°. Therefore, we have < FAB = 120°. Similarly, < BCD = 120° and < DEF = 120°. Together with < ABC = 120°, < CDE = 120° and < EFA = 120°, this entails that all angles of the hexagon ABCDEF are 120°, and the proof is done.

[i]Second solution.[/i] [b]a)[/b] We work with non-directed angles this time...

Let the lines BC and DE meet at X, let the lines DE and FA meet at Y, and let the lines FA and BC meet at Z. As in the first solution, we can easily see that all angles of the hexagon ABCDEF are 120°; in particular, < ABC = 120°. Thus, < ZBA = 180° - < ABC = 180° - 120° = 60°. Similarly, < ZAB = 60°. Hence, the triangle ZAB is equilateral, and it follows that ZA = ZB = AB and < AZB = 60°. Similarly, the triangle XCD is equilateral, XC = XD = CD and < CXD = 60°, and the triangle YEF is equilateral, YE = YF = EF and < EYF = 60°. Since < YZX = < AZB = 60° and < ZXY = < CXD = 60°, the triangle XYZ is also equilateral, and thus YZ = ZX = XY.

Now, YZ = YF + FA + ZA = EF + FA + AB and ZX = ZB + BC + XC = AB + BC + CD. Since YZ = ZX, it follows that EF + FA + AB = AB + BC + CD, so that EF + FA = BC + CD, and EF - BC = CD - FA. Similarly, CD - FA = AB - DE. Hence, AB - DE = CD - FA = EF - BC. This completes the proof of [b]a)[/b].

[b]b)[/b] This is a sketch of a solution:

We have a - d = c - f = e - b. Subtracting a + c + e and multiplying with -1, we get c + d + e = e + f + a = a + b + c.

So let XYZ be an equilateral triangle with sidelength YZ = ZX = XY = c + d + e = e + f + a = a + b + c. Construct two points F and A on its side YZ such that YF = e and ZA = a; then, FA = YZ - YF - ZA = (e + f + a) - e - a = f. Similarly construct two points B and C on the side ZX with ZB = a and XC = c, and two points D and E on the side XY with XD = c and YE = e; the triangles XCD, YEF and ZAB are easily seen to be equilateral, ... you know how to continue.

[b]EDIT:[/b] The following is not part of the solution anymore; it's a note on how to compute the area of the hexagon ABCDEF knowing its sidelengths AB, BC, CD, DE, EF, FA.

In fact, using the notations of the second solution, part [b]a)[/b], it is clear that, if we denote by $\left|P_1P_2...P_n\right|$ the non-directed area of a polygon $P_1P_2...P_n$, then

| ABCDEF | = | XYZ | - | XCD | - | YEF | - | ZAB |.

But the triangles XYZ, XCD, YEF and ZAB are equilateral and have the sidelengths YZ = EF + FA + AB, CD, EF and AB, respectively; hence, their areas are $\left|XYZ\right|=\frac{\sqrt3}{4}\cdot\left(EF+FA+AB\right)^2$, $\left|XCD\right|=\frac{\sqrt3}{4}\cdot CD^2$, $\left|YEF\right|=\frac{\sqrt3}{4}\cdot EF^2$ and $\left|ZAB\right|=\frac{\sqrt3}{4}\cdot AB^2$, so that

$\left|ABCDEF\right|=\left|XYZ\right|-\left|XCD\right|-\left|YEF\right|-\left|ZAB\right|$
$=\frac{\sqrt3}{4}\cdot\left(EF+FA+AB\right)^2-\frac{\sqrt3}{4}\cdot CD^2-\frac{\sqrt3}{4}\cdot EF^2-\frac{\sqrt3}{4}\cdot AB^2$
$=\frac{\sqrt3}{4}\cdot\left(\left(EF+FA+AB\right)^2-CD^2-EF^2-AB^2\right)$.

This expression is asymmetric on the first sight, but actually, since AB - DE = CD - FA = EF - BC, it can be rewritten in a more symmetric form, e. g. as follows:

$\left|ABCDEF\right|=\frac{\sqrt3}{4}\cdot\left(AB\cdot CD+BC\cdot DE+CD\cdot EF+DE\cdot FA+EF\cdot AB+FA\cdot BC-2K^2\right)$,

where K = AB - DE = CD - FA = EF - BC.

  Darij

[quote=grobber]........ a hexagon exists (with sides $a_1,a_2,a_3,a_4,a_5,a_6$ in this order) iff $a_1+a_2\omega+a_3\omega^2+a_4\omega^3+a_5\omega^4+a_6\omega^5=0$,...[/quote]
Could you explain this part in a more detailed way?


[hide]I don't know where to post this because there appears to be no such forum for such questions.Anyways -----[/hide]

Can somebody recommend some excellent books on [b]Inequalitites[/b]. I am looking for books which cover all [b]basic ones as well the advanced ones [/b]which can be useful in exams of all kinds. Also, such books should contain sufficient amount of [b]practice and problems of all difficulty level[/b].I would appreciate if you can tell me about it!!!

Also I would like to ask the above questions regarding some sites or webpages or other resources.Do you know of any?

Can you help me out???? :)  :?  :(  :roll:  :thumbup:

Perhaps do a search...[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=16022]here[/url] is one past topic where this has been discussed (I'm not sure if there are other topics though)

That particular forum has suggested only one book and the others are in Romanian and the rest goes on about how to order that book and delivery woes.Can somebody plz post here other books and web resources on inequalitites which can be useful???!!!! :lol:  :roll:

Somebody suggested to me Hardy's Classic . How good is it> Is it too archaic??

i wanted to ask such a question too !! but i wanted to say that if you found it tell me too !!
thanks a lot !! :)


I am really very stupid!!!

I have made a very very big mistake when I copied the text.
The correct text is

Let 1=a_1<a_2<a_3<.....<a_1993=1993 real numbers. Prove that

  \sum (k=1,..,1992)(a_(k+1)-a_k)/(1+(a_k)^2) \geq 3/4

Thank you very much to Harazi to have found this error.

Sorry again!!

I guess I shouldn't say this but: stop saying this. Old members know why.

The following is the ugliest possible solution to an inequality, but it's I have. We have sum (a_k+1-a_k)/(1+(a_k)^2)>sum (a_k+1-a_k)/(1+a_k*a_k+1). Now put a_i=tg x_i, where 0<x_1<...<x_1993<pi/2. We have to prove that sum tg(x_k+1-x_k)>3/4. But sum tg(x_k+1-x_k)>sum (x_k+1-x_k)=x_1993-x_1. But x_1993-x_1=arctg1993-arctg1=arctg1992/1994>3/4.


Ten points in space, no three of which are collinear, are connected, each one to all the others, by a total of 45 line segments. The resulting framework F will be “disconnected” into two disjoint nonempty parts by the removal of one point from the interior of each of the 9 segments emanating from any one vertex of f . Prove that F cannot be similarly disconnected by the removal of only 8 points from the interiors of the 45 segments.

[hide="Solution"]Consider two arbitrary points $ A$ and $ B$. Note that we can go from $ A$ to $ B$ in 9 different ways without any overlap of segments used: going straight from $ A$ to $ B$, or going from $ A$ to any of the 8 other vertices and then to $ B$. It follows that no two points can be disconnected by less than 9 breaks in the segments, and the conclusion follows.[/hide]


Positive real number $ k$ is given. Find the maximum of $ \lambda$ such that for all $ a \ge b \ge c \ge d \ge 0$ and $ a \plus{} b \plus{} c \plus{} d \equal{} 1$ the following inequality is true.

$ \frac {a \minus{} b}{k \plus{} a \plus{} b} \plus{} \frac {b \minus{} c}{k \plus{} b \plus{} c} \plus{} \frac {c \minus{} d}{k \plus{} c \plus{} d} \plus{} \frac {d \minus{} a}{k \plus{} d \plus{} a} \ge \lambda (a \minus{} b)(b \minus{} c)(c \minus{} d)$

You're right, it's not a easy one at all,
but I think a nice solution exists.
Actually it's one of my inspiration.

Take $ a\equal{}\frac{1}{4}\plus{}3\epsilon, b\equal{}\frac{1}{4}\plus{}\epsilon, c\equal{}\frac{1}{4}\minus{}\epsilon, d\equal{}\frac{1}{4}\minus{}3\epsilon$ where $ \epsilon \rightarrow 0^{\plus{}}$, one can get that the inequality is equivalent to

$ 2\epsilon^3 \left( \frac{64}{(2k\plus{}1)^3\minus{}64(2k\plus{}1)\epsilon^2}\minus{}\lambda \right) \ge 0$

Observe that $ \epsilon\rightarrow 0^{\plus{}}$ we obtain

$ \lambda \le \frac{64}{(2k\plus{}1)^3}$

On the other hand, denote $ x\equal{}a\minus{}b, y\equal{}b\minus{}c, z\equal{}c\minus{}d$ we have

$ \quad\frac {a \minus{} b}{k \plus{} a \plus{} b} \plus{} \frac {b \minus{} c}{k \plus{} b \plus{} c} \plus{} \frac {c \minus{} d}{k \plus{} c \plus{} d} \plus{} \frac {d \minus{} a}{k \plus{} d \plus{} a}\\
\equal{}\left( \frac {a \minus{} b}{k \plus{} a \plus{} b}\plus{}\frac {c \minus{} d}{k \plus{} c \plus{} d} \right)\plus{} \left( \frac{b\minus{}c}{k\plus{}b\plus{}c}\plus{} \frac{d\minus{}a}{k\plus{}d\plus{}a} \right)\\
\equal{}\frac{k(a\minus{}b\plus{}c\minus{}d)\plus{}(a^2\minus{}b^2\plus{}c^2\minus{}d^2)}{(k\plus{}a\plus{}b)(k\plus{}c\plus{}d)}\plus{} \frac{k(b\minus{}c\plus{}d\minus{}a)\plus{}(b^2\minus{}c^2\plus{}d^2\minus{}a^2)}{(k\plus{}b\plus{}c)(k\plus{}d\plus{}a)}\\
\equal{}(k(a\minus{}b\plus{}c\minus{}d)\plus{}(a^2\minus{}b^2\plus{}c^2\minus{}d^2))\\
\quad \cdot\left( \frac{1}{(k\plus{}a\plus{}b)(k\plus{}c\plus{}d)}\minus{} \frac{1}{(k\plus{}b\plus{}c)(k\plus{}c\plus{}d)} \right)\\
\equal{}\frac{(k(a\minus{}b\plus{}c\minus{}d)\plus{}(a^2\minus{}b^2\plus{}c^2\minus{}d^2))((k\plus{}b\plus{}c)(k\plus{}d\plus{}a)\minus{}(k\plus{}a\plus{}b)(k\plus{}c\plus{}d))}{(k\plus{}a\plus{}b)(k\plus{}b\plus{}c)(k\plus{}c\plus{}d)(k\plus{}d\plus{}a)}\\
\equal{}\frac{((k\plus{}a\plus{}b)x\plus{}(k\plus{}c\plus{}d)z)(x\plus{}y)(y\plus{}z)}{(k\plus{}a\plus{}b)(k\plus{}b\plus{}c)(k\plus{}c\plus{}d)(k\plus{}d\plus{}a)}\\
\ge\frac{8xyz\sqrt{(k\plus{}a\plus{}b)(k\plus{}c\plus{}d)}}{(k\plus{}a\plus{}b)(k\plus{}b\plus{}c)(k\plus{}c\plus{}d)(k\plus{}d\plus{}a)}\\
\equal{}\frac{8xyz}{\sqrt{(k\plus{}a\plus{}b)(k\plus{}c\plus{}d)(k\plus{}b\plus{}c)^2(k\plus{}d\plus{}a)^2}}\\
\ge \frac{64}{(2k\plus{}1)^3}xyz$

Where the last inequality is valid using AM-GM and by the fact $ a\plus{}b\plus{}c\plus{}d\equal{}1$.

Therefore we get the result $ \lambda\equal{}\frac{64}{(2k\plus{}1)^3}$  :wink:

Typo: the original inequality should be

$ \frac {a \minus{} b}{k \plus{} c \plus{} d} \plus{} \frac {b \minus{} c}{k \plus{} d \plus{} a} \plus{} \frac {c \minus{} d}{k \plus{} a \plus{} b} \plus{} \frac {d \minus{} a}{k \plus{} b \plus{} c} \ge \lambda (a \minus{} b)(b \minus{} c)(c \minus{} d)$


calculus
integration
function
probability
binomial coefficients
geometric series
combinatorics solved

Hey
here's a problem that i wud like u ppl to solve (and this time around i'm sure wot i'm posting is correct :D  :D )
\[ \sum_{k=0}^{n} \binom{n}{k}^{-1} = \frac{n+1}{2^{n+1}} \sum_{i=1}^{n+1} \frac{2^i}{i} \]
i have a solution which uses calculus (but it's pretty simple), though i'd luv to better amd more olympiad-like solutions
(i hope the question is clear, $\binom{n}{k}$ denote  the binomial coefficients)
cheers

As I recall, the problem can be done pretty simply by proving that both sides satisfy the same recurrence.  But I 'd like to see your calc proof.

well,i'll prove a lemma first
(all definite integrals in the following para are from 0 to 1)
LEMMA:
 \int x^k*(1-x)^(n-k) dx= 1/[(n+1)(nCk)]
one can easily prove this by considering the integral  \int (tx+1-x)^(n) dx
.u compute it in two ways and compare the coefficients of t^k
or better still if one knows the beta function ,this can be proved pretty easily.

now since we want the sum of the inverses of the binomial coefficients.consider the integral which represents the sum
(n+1)*\int [\sum  x^k*(1-x)^(n-k) ]dx where the sum ranges from k=0 to n.sum up the geometric series inside and perform the substitution
2x-1=t.change the limits to -1 to 1.and the integral we are left with is
[ (n+1)/2^(n+1)]  \int [{(1+t)^(n+1) - (1-t)^(n+1)}/2t] dt.
now if we prove that the integral is the same as the following integral
 \int [(1+t)^n - 1)/t)] dt (limit -1 to 1) ,then our job will be done as the preceding integral  represents  \sum  i=1 to (n+1) {2^i}/i .
now just proceed to show that the difference of the two integrals is an odd function so its definite integral from -1 to 1 will be zero.
QED (hope it's correct :D  :D )
CHEERS!!

Nice!  I think I have an alternate proof for your integral:

\int x^k*(1-x)^(n-k) dx= 1/[(n+1)(nCk)]

Now, the left hand side is the probability that if you randomly pick a real x_0 = x in [0, 1] and x_1, ... x_n in [0, 1] that x_1, ... x_k < x and x_(k+1)... x_n > x.  But because two of the x_i are equal with probability 0, we can assume that all the x_i are distinct, 0 <= i <= n.  But because we only care about order properties, the probability is the same as if the x_i are a permutation of {0, 1, ... n}.

So in other words, we want the probability that if x_0, x_1, ... x_n is a permutation of {0, 1, ... n}, x_0 = k, for 1 < i <=k, x_1 < k and for i>k x_i > k.  But the probability that x_0 = k is 1/(n+1), and given that x_0 = k, and the probability that 1 < i <=k, x_1 < k and for i>k x_i > k is 1/nCk.

So putting that together shows the integral evaluates to 1/[(n+1)(nCk)]


Vieta
algebra
polynomial
Rational Root Theorem

I've been hearing solutions with Vieta's formulas mentioned again and again, but when I looked it up, I didn't understand a thing. Can anyone please explain (in layman's terms) how Vieta's formulas work?

Also, if this is related (maybe), how do you find the roots of a polynomial with a degree higher than 2?

Your two questions are opposite in some sense.  Vieta's formulas tell you how to compute the coefficients of a polynomial given its roots, whereas solving a polynomial is the question of computing the roots given the coefficients.

Question 1:  Vieta's formulas are easiest to understand by looking at them in small cases.  Expand out $ (x \minus{} a)(x \minus{} b)$.  What are the coefficients?  Now do the same for $ (x \minus{} a)(x \minus{} b)(x \minus{} c)$.  Now try $ (x \minus{} a)(x \minus{} b)(x \minus{} c)(x \minus{} d)$.  Now...

Question 2:  There are formulas for degree $ 3$ and $ 4$.  They are very ugly; you will never use them.  What's rather curious is that there is [url=http://mathworld.wolfram.com/AbelsImpossibilityTheorem.html]no general formula[/url] in terms of roots for degree $ 5$ or higher.  In a competition setting, you'll mostly want to factor, possibly by using the [url=http://en.wikipedia.org/wiki/Rational_root_theorem]rational root theorem[/url].

I see the pattern, but how can this be applied?

Vieta's formulas are often used to answer questions like "what are the sums of the squares of the roots of this polynomial"?  We'll take the example of a cubic polynomial.  Write $ (x \minus{} a)(x \minus{} b)(x \minus{} c) \equal{} x^3 \minus{} r_2 x^2 \plus{} r_1 x \minus{} r_0$ where

$ r_0 \equal{} abc$
$ r_1 \equal{} ab \plus{} bc \plus{} ca$
$ r_2 \equal{} a \plus{} b \plus{} c$.

Then the sum of the squares is given by $ a^2 \plus{} b^2 \plus{} c^2 \equal{} (a \plus{} b \plus{} c)^2 \minus{} 2(ab \plus{} bc \plus{} ca) \equal{} r_2^2 \minus{} 2r_1$.  Usually you are given the coefficients $ r_1, r_2, r_3$ and are unable to compute the roots, so the key idea here is that you can compute $ a^2 \plus{} b^2 \plus{} c^2$ without computing $ a, b, c$.

boop
http://www.artofproblemsolving.com/Wiki/index.php/Vieta%27s_formulas
bop
http://www.artofproblemsolving.com/Wiki/index.php/Newton_sums

If you really want to learn them, I suggest doing lots of problems or else you'll forget. You'll also get to see how they're useful.

@differ- u stole mine and brutforc3's avatar! how did u modify it to say "avatar stealer"? this is more HSB level i think


prove that a polygon with $ n$ cotes has $ \frac{n(n\minus{}3)}{2}$ diagonales

$ \binom{n}{2}\minus{}n \equal{} \frac{n(n\minus{}1)}{2}\minus{}n \equal{} \frac{n(n\minus{}3)}{2}$

[quote="Kouichi Nakagawa"]$ \binom{n}{2} \minus{} n \equal{} \frac {n(n \minus{} 1)}{2} \minus{} n \equal{} \frac {n(n \minus{} 3)}{2}$[/quote]
hi Kouichi Nakagawa !!
i dont understand your solution
please to explain.

We put $ n$ point on the circumference of circle.
$ \binom{n}{2}$ : The number of the line joining the 2 points.
$ n$ : The number of the side of the polygon.


umm i've been told that there may have been a change and now canadians are able to qualify for official usamo testing from AIME and AMC?  anyone confirm?  what about MOP?

Impossible, for both, at least for now, since you have to be a U.S. citizen or is an immigrant with a green card. They'll only apply if an American is just visiting Canada. Besides I think the Canadians have their own Olympiad.

Besides it's USAMO not USA&C, MO :lol:

hey, dont diss the canadians...  i thought it was American MC, and we get to take AMC officially...  and American IME, and of course we get to write it officially as well.  so why not the USA MO??  what are you guys afraid of?  Canadians beating you?

The joke is, it contains the phrase "MO" which sounds funny if you say it outloud, like this: "Hey Jim, I made it to the USA and Canada, MO!" :lol:

its not really the us. its just among the permanent residents of the us.

[quote="jeffrey"]Maybe because its our one of the ways we select the IMO team and promote the growth of mathematics in the us. Administering the test to canadians doesn't help do either. Besides, you can get the questions online with solutions so you could just take it at home for fun...

And you do have the CMO...people in the US can't take that.[/quote]

a few years ago the canadians are allowed to participate in the USAMO officially but then i think it was some guy in canada who topped all the us contestants.  they did not like that so they made USAMO official to only those with us status.  but there may be some changes this year

my bad, i didnt think there was actually any reason to think it might happen...

if anyone cares, the last time canadians are allowed to compete in USAMO is 1991 (or 90?)   and JP Grossman had a perfect paper i think.  then washington did not like the best math student being canadian then we are not allowed to compete officially anymore.  but at the end of the day arent we all just trying to do some math??


calculus
integration
logarithms
derivative
trigonometry
function
calculus computations

[quote="prithwijit"]Evaluate $\int_{0}^{1}\ln(\frac{1+ax}{1-ax})\frac{dx}{x\sqrt{1-x^{2}}}$. $a \in (-1,1)$.[/quote]

Let the given integral be denoted by $I(a)$ so that (noting that differentiation under the integral sign is justified,) we have

$\frac{dI}{da}=\frac{d}{da}\int_{0}^{1}\ln\left(\frac{1+ax}{1-ax}\right)\frac{dx}{x\sqrt{1-x^{2}}}=\int_{0}^{1}\frac{\partial}{\partial a}\ln\left(\frac{1+ax}{1-ax}\right)\frac{dx}{x\sqrt{1-x^{2}}}$
$=\int_{0}^{1}\frac{1-ax}{1+ax}\frac{x(1-ax)+(1+ax)x}{(1-ax)^{2}}\frac{dx}{x\sqrt{1-x^{2}}}=\int_{0}^{1}\frac{2dx}{(1-a^{2}x^{2})\sqrt{1-x^{2}}}$

now let $u=\sin\theta\Rightarrow du=\cos\theta d\theta$ so that $0\leq x\leq 1 \Rightarrow 0\leq \theta\leq \frac{\pi}{2}$ and the integral becomes

$\frac{dI}{da}=2\int_{0}^{\frac{\pi}{2}}\frac{d\theta}{1-a^{2}\sin^{2}\theta}=2\int_{0}^{\frac{\pi}{2}}\sum_{k=0}^{\infty}a^{2k}\sin^{2k}\theta \, d\theta=2\sum_{k=0}^{\infty}a^{2k}\int_{0}^{\frac{\pi}{2}}\sin^{2k}\theta \, d\theta$
$=2\sum_{k=0}^{\infty}a^{2k}\frac{1}{2}\mbox{B}\left(k+\frac{1}{2}, \frac{1}{2}\right)=\sum_{k=0}^{\infty}a^{2k}\frac{\Gamma \left( \frac{1}{2}\right)\Gamma \left( k+\frac{1}{2}\right)}{\Gamma (k+1)}=\sum_{k=0}^{\infty}a^{2k}\frac{\Gamma\left( \frac{1}{2}\right)\Gamma\left( k+\frac{1}{2}\right)\Gamma\left( \frac{1}{2}-k\right) }{\Gamma\left( \frac{1}{2}-k\right) \Gamma (k+1)}$

recall that $\Gamma\left( k+\frac{1}{2}\right)\Gamma\left( \frac{1}{2}-k\right) = \frac{\pi}{\sin \pi\left(k+\frac{1}{2}\right)}= (-1)^{k}\pi$ so that

$\frac{dI}{da}=\pi\sum_{k=0}^{\infty}(-1)^{k}a^{2k}\frac{\Gamma\left( \frac{1}{2}\right) }{\Gamma\left( \frac{1}{2}-k\right) \Gamma (k+1)}=\pi\sum_{k=0}^{\infty}\left(\begin{array}{c}-\frac{1}{2}\\k\end{array}\right) (-a^{2})^{k}= \frac{\pi}{\sqrt{1-a^{2}}}$

this gives a equation for our integral, $\frac{dI}{da}=\frac{\pi}{\sqrt{1-a^{2}}}$ and hence

$I(a)=\int\frac{\pi da}{\sqrt{1-a^{2}}}=\pi \mbox{arcsin}(a)+C$

where C is a constant which is easily determined by evaluating I(0), viz.

$I(0)=\int_{0}^{1}\ln(1)\frac{dx}{x\sqrt{1-x^{2}}}=0=C$

thus

$\int_{0}^{1}\ln\left(\frac{1+ax}{1-ax}\right)\frac{dx}{x\sqrt{1-x^{2}}}=\pi \mbox{arcsin}(a)$

[quote="prithwijit"]Evaluate $\int_{0}^{1}\ln(\frac{1+ax}{1-ax})\frac{dx}{x\sqrt{1-x^{2}}}$. $a \in (-1,1)$.[/quote]
Here is an elementary solution that doesn't use advanced techniques -- mainly because I don't know any of the advanced techniques  :)

We know that if $|x| < 1,$ then
$\ln(1+x) = x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\frac{x^{5}}{5}-\cdots$, and
$\ln(1-x) =-x-\frac{x^{2}}{2}-\frac{x^{3}}{3}-\frac{x^{4}}{4}-\frac{x^{5}}{5}-\cdots$, so that
$\ln{(\frac{1+x}{1-x})}= 2x(1+\frac{x^{2}}{3}+\frac{x^{4}}{5}+\frac{x^{6}}{7}+\cdots)$

$\because |ax| < 1,$ we have
$\ln{(\frac{1+ax}{1-ax})}= 2(ax)(1+\frac{{(ax)}^{2}}{3}+\frac{{(ax)}^{4}}{5}+\frac{{(ax)}^{6}}{7}+\cdots)$

Making the substitution $x=\sin\theta$ in the original integral, we get
$I = 2a\int_{0}^{\frac{\pi}{2}}(1+\frac{{(a\sin\theta)}^{2}}{3}+\frac{{(a\sin\theta)}^{4}}{5}+\frac{{(a\sin\theta)}^{6}}{7}+\cdots){d\theta}$

Now, we use the identity $I_{2n}= \int_{0}^{\frac{\pi}{2}}{\sin}^{2n}x dx = \frac{2n-1}{2n}\cdot\frac{2n-3}{2n-2}\cdots\frac{1}{2}\cdot\frac{\pi}{2}$ to get
$I = \pi (a+\frac{a^{3}}{6}+\frac{3a^{5}}{40}+\frac{5a^{7}}{112}+\frac{35a^{9}}{1152}+\cdots) = \pi\arcsin{(a)}$.

[b][u]Note:[/u][/b] I kind of cheated on the last statement above because after integrating the factors term by term, I got $I = \pi (a+\frac{a^{3}}{6}+\frac{3a^{5}}{40}+\frac{5a^{7}}{112}+\frac{35a^{9}}{1152}+\cdots)$. I then compared this with known existing series and found it matched with $\pi\arcsin{(a)}.$ Maybe someone can come up with a better explanation or better way to reach the final conclusion.

I like FieryHydra's solution mainly because i was able to follow it.

however, i found benorin's solution to be interesting in his use of taking the derivative inside the integral buy once it got to the use of gama functions i was 100% lost.  :(
hopefully i will be able to understand it in a few years.  :)  :lol:

[quote="benorin"][quote="prithwijit"]Evaluate $\int_{0}^{1}\ln(\frac{1+ax}{1-ax})\frac{dx}{x\sqrt{1-x^{2}}}$. $a \in (-1,1)$.[/quote]

Let the given integral be denoted by $I(a)$ so that (noting that differentiation under the integral sign is justified,) we have

$\frac{dI}{da}=\frac{d}{da}\int_{0}^{1}\ln\left(\frac{1+ax}{1-ax}\right)\frac{dx}{x\sqrt{1-x^{2}}}=\int_{0}^{1}\frac{\partial}{\partial a}\ln\left(\frac{1+ax}{1-ax}\right)\frac{dx}{x\sqrt{1-x^{2}}}$
$=\int_{0}^{1}\frac{1-ax}{1+ax}\frac{x(1-ax)+(1+ax)x}{(1-ax)^{2}}\frac{dx}{x\sqrt{1-x^{2}}}=\int_{0}^{1}\frac{2dx}{(1-a^{2}x^{2})\sqrt{1-x^{2}}}$

now let $u=\sin\theta\Rightarrow du=\cos\theta d\theta$ so that $0\leq x\leq 1 \Rightarrow 0\leq \theta\leq \frac{\pi}{2}$ and the integral becomes

$\frac{dI}{da}=2\int_{0}^{\frac{\pi}{2}}\frac{d\theta}{1-a^{2}\sin^{2}\theta}$ [/quote]
I found out another way to circumvent the use of Gamma function. This one is better than my earlier solution because the last step of the latter was not very convincing. Here, I will continue from [b]benorin's[/b] last derived result above. The solution below may seem a little complicated but it is relatively elementary.

Note that $\frac{dI}{da}= \int_{0}^{\frac{\pi}{2}}\left( \frac{1}{1-a\sin\theta}+\frac{1}{1+a\sin\theta}\right) d\theta = I_{1}+I_{2}$ (say)

Substituting $\sin\theta = \frac{2\tan{(\frac{\theta}{2})}}{1+{\tan}^{2}{\frac{\theta}{2}}}$ in $I_{1}$ and after some manipulation, we get

$I_{1}= \int_{0}^{\frac{\pi}{2}}\frac{{\sec}^{2}\frac{\theta}{2}d\theta}{{(\tan \frac{\theta}{2}-a)}^{2}+{(\sqrt{1-a^{2}})}^{2}}$.
Now again make the substitution $y = \tan{\frac{\theta}{2}}-a$ so that $I_{1}= 2\int_{-a}^{1-a}\frac{dy}{y^{2}+{(\sqrt{1-a^{2}})}^{2}}= \frac{2}{\sqrt{1-a^{2}}}{\tan}^{-1}\left( \frac{y}{\sqrt{1-a^{2}}}\right)\Big|_{-a}^{1-a}$
${= \frac{2}{\sqrt{1-a^{2}}}\Big[{\tan}^{-1}\left( \sqrt{\frac{1-a}{1+a}}\right)+{\tan}^{-1}\left( \frac{a}{\sqrt{1-a^{2}}}\right)}\Big]$

Again, in a similar way, we obtain ${I_{2}= \frac{2}{\sqrt{1-a^{2}}}\Big[{\tan}^{-1}\left( \sqrt{\frac{1+a}{1-a}}\right)-{\tan}^{-1}\left( \frac{a}{\sqrt{1-a^{2}}}\right)}\Big]$

$\therefore \frac{dI}{da}= I_{1}+I_{2}= \frac{2}{\sqrt{1-a^{2}}}\Big[{\tan}^{-1}\left( \sqrt{\frac{1-a}{1+a}}\right)+{\tan}^{-1}\left( \sqrt{\frac{1+a}{1-a}}\right)\Big]$
$\Rightarrow \frac{dI}{da}= \frac{2}{\sqrt{1-a^{2}}}\cdot \frac{\pi}{2}= \frac{\pi}{\sqrt{1-a^{2}}}$ ... $\Big[$ Use ${\tan}^{-1}x+{\tan}^{-1}y ={\tan}^{-1}\left( \frac{x+y}{1-xy}\right) \Big]$

Finally, now we get $I(a) = \pi{\sin}^{-1}(a)+C$. But, $C=0$ as shown in [b]benorin's[/b] solution earlier.

And, we are done.


Find all odd n such that $ n$ divides $ a^{n \minus{} 1} \plus{} 1$, a is an integer

It 's like a APMO problem.
n is even number.

Can you post a full solution?
I've only been able to show that any prime divisor of n is of the form 4k+1.
Or, can you tell me the year? I've searched through all the apmo problems and can't find it.

Let $ p$ is prime divisor n, suth that $ ord_2(p\minus{}1)$ is minimal. Then $ ord_2(n\minus{}1)\ge ord_2(p\minus{}1)$. Therefore $ a^{n\minus{}1}\not \equal{}\minus{}1 \mod p$. Contradition with $ n|a^{n\minus{}1}\plus{}1$.

the original problem is find 0<n<2000 such that $ 2^n\plus{}2$ is divided by n.

Rust, can you (or somebody else) please explain your solution better? I think you mean $ ord_{p\minus{}1}(2)$, as $ ord_{2}(p\minus{}1)$ doesn't exist since 2 divides p-1. Anyways, I remember that problem that math man just gave had a large solution, like 63 or something, so I don't think the solution can be that simple.

[quote="dgreenb801"]Rust, can you (or somebody else) please explain your solution better? I think you mean $ ord_{p \minus{} 1}(2)$, as $ ord_{2}(p \minus{} 1)$ doesn't exist since 2 divides p-1. Anyways, I remember that problem that math man just gave had a large solution, like 63 or something, so I don't think the solution can be that simple.[/quote]

Rust mean $ ord_{2}(p \minus{} 1)$ is the exponent of 2 in the decomposition of p-1 :wink:
by rust 's solution we have p must be 2 and choose a=(n-1) .


Does there exist a function $ f\colon \mathbb{R}\to\mathbb{R}$ such that for all x the following equality holds.
$ \lvert\ f(x\plus{}y)\plus{}f(y\plus{}z)\plus{}f(x\plus{}z)\minus{}f(x\plus{}y\plus{}z)\rvert>1$

_______________________________________
Azerbaijan Land of Fire  :lol:

yes, $ f(x)=2009$ for all $ x \in \mathbb{R}$.

[quote="Raja Oktovin"]yes, $ f(x) = 2009$ for all $ x \in \mathbb{R}$.[/quote]

:)

Pirkuliyev Rovsen's teacher gives rather strange problems for Olympiad preparation !!


real analysis
function
Support
search
real analysis unsolved

1)  Let $ A$ be a Lebesgue measurable subset of $ \mathbb{R}$ with $ m(A)>0$.  Is it necessarily true that there exist distinct points $ a,b\in A$ with $ |a\minus{}b|\in\mathbb{Q}$?

2)  Let $ A$ and $ B$ be Lebesgue measurable subsets of $ \mathbb{R}$ with $ m(A),m(B)>0$.  Is it necessarily true that there exist distinct points $ a\in A,b\in B$ with $ |a\minus{}b|\in\mathbb{Q}$?

I am rather unfamiliar with the finer points of measure theory, but if the following lemma holds:

[b]Lemma:[/b]  A Lebesgue measurable subset of $ \mathbb{R}$ with strictly positive measure contains an interval $ (x, y), y > x$.

then 1) follows trivially.  The above lemma seems true to me, but then again I can't think of an example of a subset of $ \mathbb{R}$ with positive Lebesgue measure that is not a countable union of disjoint intervals...

I'm sure the fact that $ A\plus{}B$ (here we're concerned with $ A\minus{}B$, but that makes no difference) contains a non-degenerate interval whenever $ A,B$ are measurable real sets with positive measure must have been discussed before around here more than once. Here's a proof I like:

Assume $ A,B$ are also bounded (there's no loss of generality here; simply replace $ A,B$ with bounded subsets having positive measure). Let $ \chi_{A},\chi_{B}$ be the characteristic functions of $ A$ and $ B$ respectively. Since $ \chi_{A},\chi_{B}$ are bounded and integrable, their convolution $ f \equal{}\chi_{A}*\chi_{B}$ is continuous, with support contained in $ A\plus{}B$. On the other hand $ f$ is non-zero (its Fourier transform is the product of the Fourier transforms of $ \chi_{A}$ and $ \chi_{B}$, and this product, when evaluated in $ 0$, is the positive product of the measures of $ A$ and $ B$). This means that $ f$ is non-zero on some open interval which must be contained in $ A\plus{}B$.

yes, it's been posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1271122204&t=19478]before[/url] :)

anyways it's a cute result.

There's another version [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=113431]here[/url], stated for $ \mathbb{R}^{n}$ and differences.

And finally, there's another (maybe simpler) proof of the result jmerry speaks of [url=http://en.wikipedia.org/wiki/Steinhaus_theorem]here[/url] :)

I guess we have everything taken care of if this topic ever comes up again!


logarithms
AMC
AIME
AMC 10
AMC 12
probability
expected value

So like yea, I just did 2002 12b for practice and I got only 124.5 because of a handicap of 10 oranges sitting next to me (for which I couldn't think about math while i ate) :( What makes me worried though is I randomly got rid of the logarithm on number 22 and got 1/2002, and I thought 9 was not a perfect square :O

SO GOOD LUCK GUYS DONT FAIL

139.5... with no stupid mistakes?
good job but I think 2002 was easies. The last problem was pretty cool I'd have to say though.

Blah, I just want to qualify for AIME with a decent score that will give me a shot at USAMO...

[quote="Brut3Forc3"]Hi Bay Area. I'm from Southern California. That means I hate you guys.
Eh, just kidding. Good luck to you guys! Spam!

I'm going for 150, by the way. :) (I'll probably get a 138 though.)[/quote]
Well you're pretty beast, so you'll probably achieve that goal. However my goal of 150/144 (10a/12b) probably won't happen, as with my 11 aime goal.

Good luck, everyone! Goal is 150, except I haven't prepared at all so expected value is more like...I dunno, something ugly. Whatever.

Haha, my AMC goal scores are lower than my scores last year =)  So hopefully I'll make them...
Um...out of the 11 practices amc 10s I took this weekend, my average was in the mid 140s and out of the 4 practice amc 12s I took, my average was in the lower-mid 130s (no thanks to 2001 =( )

2001? Hmm...I did all the 12as and 10as last year, so I probably don't remember anything. I think I can reliably say I will get >140 amc10, although nothing these days is sure.

12a/10b goal: 144/150

except I'll probably get 138-ish on the 12.*

*Although this is taking the AMC12 in about 60 minutes

Hmm...so I don't really remember how to do math anymore.

Me neither!
BTW Lynnelle WHY WHY WHY did you have to put that stupid mandelbrot problem on there??? I did not want to solve it, so I failed to come up with a solution on the spot when I had to explain it to them!

Huh? None of them were Mandelbrot problems. In fact, they were all (very low-numbered) old AMC-12/AHSME problems. Also, I gave you guys summaries of all the solutions. I'm confused.

What? The magic multiplicative square thingy??? And also, I didn't want to read the summary either because that problem was annoying. On Thursday we should bring some calculator bashable problems for them, because Ms. Franklin wants to do calculator bash lesson.

[quote="serialk11r"][quote="james4l"]Likely score: 138 on the AMC12.

Epic phail...

*hopes that MC and AMC10 will fare better*[/quote]
138 is decent though...[/quote]

blah, nosebleed. You should have seen my scratchpaper...

it's red... and that's purely natural dye!  :roll:

my goal is 144. I seem to be missing my goals by exactly one question, so I hereby declare:

I RAISE MY AMC10 GOAL TO 156.  :)

[quote="bookaholic"][quote="Ubemaya"][quote="bookaholic"]130.5 on the AMC-12. Even more epic phail.

@Damien: haha I got it down to (a-5)(b-5)(c-5)=25*79-125[/quote]

wait how did you get it to that

Also, flu/fever/headache on the AMC 12. Most epic phail.[/quote]
1/a+1/b+1/c=5
abc=5ab+5bc+5ac
abc-5ab-5bc-5ac=0
abc-5ab-5bc-5ac+5a+5b+5c-125=5a+5b+5c-125
(a-5)(b-5)(c-5)=5(a+b+c)-125=5*79-125[/quote]

hmmm, how did u get the second step?
should it be: 5abc=ab+bc+ac ?

[quote="DataBox"][quote]
wait how did you get it to that

Also, flu/fever/headache on the AMC 12. Most epic phail. [/quote]

can you quantify your fail ubeymaya?

lol the amc 10 was soo trivial this year[/quote]
Um, are you forgetting that it's really trivial every year? Don't let that 150 get to your head.

^Thanks for the advice, but if it were really that trivial every year, all of us should get 150s every time...

And homie, just wondering...what accomplishment of yours inspires [b]your[/b] cockiness?

Because it seems to me as if you "fail" everything of any significance....Enlighten me...

[quote="DiscreetFourierTransform"]#7 is a+b+c=79, 1/a+1/b+1/c=5. [/quote]
(Yay, interjecting again!)
I remember this problem from mandlebrot. :P
My solution was using $ \frac{1}{a}\equal{}\frac{1}{a\plus{}1}\plus{}\frac{1}{a(a\plus{}1)}$, we have $ \frac{1}{5}\equal{}\frac{1}{6}\plus{}\frac{1}{30}\equal{}\frac{1}{7}\plus{}\frac{1}{42}\plus{}\frac{1}{30}$. :)

[quote="DataBox"]^Thanks for the advice, but if it were really that trivial every year, all of us should get 150s every time...
[/quote]
uh homies people make silly mistakes on problems that they immediately know how to do?

I guess he "failed" at what he was trying to accomplish, which was a 150.

[quote="Brut3Forc3"][quote="DiscreetFourierTransform"]#7 is a+b+c=79, 1/a+1/b+1/c=5. [/quote]
(Yay, interjecting again!)
I remember this problem from mandlebrot. :P
My solution was using $ \frac {1}{a} \equal{} \frac {1}{a \plus{} 1} \plus{} \frac {1}{a(a \plus{} 1)}$, we have $ \frac {1}{5} \equal{} \frac {1}{6} \plus{} \frac {1}{30} \equal{} \frac {1}{7} \plus{} \frac {1}{42} \plus{} \frac {1}{30}$. :)[/quote]
Either the problem is wrong(it should be $ 1/a\plus{}1/b\plus{}1/c\equal{}1/5$) or your solution is wrong(doesn't match with the 2nd equation). I think the 1st assumption is true. Nice identity, Brut3Forc3 :).
My solution is different:
From the 2nd equation we have: $ (1\minus{}5a)(1\minus{}5b)(1\minus{}5c) \equal{} \minus{}376$(since $ a\plus{}b\plus{}c\equal{}79$). Now factorizing -376= -2^3.47 and solving some equations, we get the result as you mentioned above(note that the sum of $ (1\minus{}5a)\plus{}(1\minus{}5b)\plus{}(1\minus{}5c)\equal{} \minus{}392$, which is an even number. So it simplifies the cases much.
This problem appeared on the 1st Mock AIME of Mildorf. It's a great problem though.

-partial necropost-

wow... got a..
[hide="1"][hide="2"][hide="6"]
So shoot me, sue me, blah.
Well that is amazingly second place in MSJ. You probably won't guess who got first. (it's not Bowei or Marianna)
[/hide][/hide][/hide]

Who got first?  And how did Bowei and Marianna do?

121.5, 123 respectively... Lawrence Ku got first.

Dude so like today my math teacher walks into the room and tells us that the AMC answer forms (half of them for our school) showed up recently back at our school with something that said like, because of concerns about terrorism, packages over 13 oz. blah blah, even though she went to the post office and sent it. I hope my stuff wasn't in that half :O

^Wow, that really really sucks for the half that got their tests returned =(

America is far too paranoid for its own good...


Hi
Show that Tr[(AB)^n]=Tr[(BA)^n] n is a positive integer

We know that $ \text{Tr}(AB)\equal{}\text{Tr}(BA)$ for any $ A,B.$

Thus, $ \text{Tr}[(AB)^n)]\equal{}\text{Tr}[(ABA\cdots BA)B]\equal{}\text{Tr}[B(ABA\cdots BA)]\equal{}\text{Tr}[(BA)^n].$


who can share some ebook about the function equation ?
I hope someone can help me ... :(

Check out:
http://www.mathlinks.ro/index.php?f=319
http://www.mathlinks.ro/index.php?f=121


search
superior algebra
superior algebra unsolved

Let $ F$ be a finite field.Prove that every element of $ F$ can be written as sum of two perfect squares.
$ \forall a\in F$ there are $ x,y \in F$ such that:
$ a\equal{}x^2\plus{}y^2$

If $ char\ F\equal{}2$, all elements are squares, and we are done.  Otherwise:

Let $ S$ be the set of nonzero elements of $ F$ that are the sum of two squares.  If $ a,b\in S$, $ a\equal{}x^2\plus{}y^2$, $ b\equal{}u^2\plus{}v^2$, then $ ab\equal{}(xu\minus{}yv)^2\plus{}(xv\plus{}yu)^2  \in S$.

Since multiplication is injective, $ F$ is finite, and $ 1\in S$, it follows that $ S\leq F^*$ is a group under multiplication.

Note that there is some $ a$ such that $ a^2\plus{}1$ is not a square. (otherwise all elements are squares!)  So $ S$ contains at least one non-square, as well as all squares.  Since there are $ \frac{|F^*|}{2}$ nonzero squares, $ |S| > \frac{|F^*|}{2}$, so $ S\equal{}F^*$, and we are done.

Posted three times before, take a search next time.

If $ a^2\plus{}1$ allways is a square, why does this imply that all elements are squares?

[quote="Dennen"]If $ a^2 \plus{} 1$ allways is a square, why does this imply that all elements are squares?[/quote]

I wrote that hastily; it's not as obvious as I thought. (I was thinking of $ \mathbb{F}_p$)

It would, however, imply that the number of squares is divisible by $ char\ F$.  Since the number of squares is $ \frac{|F|\plus{}1}{2}$, this is impossible.


[quote="thepunisher123"]somebody can say to me a site of theoremes for inequality?[/quote]
See here my friend  http://www.artofproblemsolving.com/Forum/index.php?f=55 :wink:
Dimitris


algebra function domain conics parabola calculus inequalities

Let $ f(x)\equal{}\sqrt{ax^2\plus{}bx}$. For how many real values of $ a$ is there at least one positive value of $ b$ for which the domain of $ f$ and the range of $ f$ are the same set?

$ \textbf{(A)}\ 0 \qquad
\textbf{(B)}\ 1 \qquad
\textbf{(C)}\ 2 \qquad
\textbf{(D)}\ 3 \qquad
\textbf{(E)}\ \text{infinitely many}$

I think it's a=-4 and a=0, here is my solution:
Assume the parabola $ ax^2\plus{}bx$ is facing up, then it will be positive for infinitely many negative x which will be in the domain of f but not the range, hence the parabola faces down, or a=0, which works if b is positive. Since the parabola faces down, it will be greater than 0 from 0 to $ \minus{}\frac{b}{a}$, the domain of f. The parabola is at its highest at the vertex when $ x\equal{}\minus{}\frac{b}{2a}$, $ y\equal{}\minus{}\frac{b^2}{4a}$, and it crosses through (0,0) so subbing into f(x) we will have $ \minus{}\frac{b}{a}\equal{}\sqrt{\minus{}\frac{b^2}{4a}} \implies a\equal{}\minus{}4$. Is this correct?

I don't understand how to do this problem. The answer is 2 with the values -4 and 0. Can someone please explain this without Calculus or any crazy lemmas, but simply an approach that has casework with $a>0$, $a = 0$, and $a<0$ and explanations on why I would get $a = 0$ and $a= -4$ to work.

I found $a=0$ to work but couldnt find $a=-4$. Is there an easy way to do this?

@TheMaskedMagician, any $b$ should satisfy $a=-4$.

[hide="Solution"]If $a>0$ then the domain is all $x$ where $ax^2+bx \geq 0$ or $x \geq 0 \cup x\leq -\dfrac{b}{a}$. From the last inequality, $x$ can take on negative values, but $f(x)$ is strictly nonnegative, thus $a \leq 0$.

Suppose $a=0$. Then $f(x) = \sqrt{bx}$ in which $x \geq 0$ which is also the range of $f(x)$. Thus any positive $b$ would satisfy the problem statement.

Finally, consider $a<0$. Then $ax^2+bx \geq 0 \rightarrow 0 \leq x \leq \dfrac{b}{a}$. Because the domain and range should be the same, $0 \leq f(x) \leq \dfrac{b}{a}$ as well. The left hand side is obvious, so now we find $b$ such that the right hand side is satisfied. Notice that $f(x)$ is the square root of a parabola with vertex or maximum at $x=-\dfrac{b}{2a} \rightarrow \max (f(x))=\sqrt{-\dfrac{b^2}{4a}}$. Then we want the maximum equal to $\dfrac{b}{a}$ in which $a=-4$ for any positive $b$.

$a=0,-4$ gives $\boxed{\textbf{(C)}\ 2}$ as the answer.[/hide]


[b]f(x)=arctg(x)
Calculate n-th derivate on f(x)[/b]

$ f(x)\equal{}\arctan x\equal{}\frac1{2i}(\log(1\plus{}ix)\minus{}\log(1\minus{}ix))$

$ f'(x)\equal{}\frac1{2i}\left(\frac{i}{1\plus{}ix}\plus{}\frac{i}{1\minus{}ix}\right)\equal{}\frac1{1\plus{}x^2}$

$ f^{[n]}(x)\equal{}\frac{(n\minus{}1)!(\minus{}i)^{n\minus{}1}}2 \left(\frac1{(1\plus{}ix)^n}\plus{}\frac{(\minus{}1)^{n\minus{}1}}{(1\minus{}ix)^n}\right)$

That does turn out to always be real for $ x$ real. And it's a lot easier to evaluate at $ x\equal{}0.$


Prove that the polynomials $ P(x) \equal{} 1 \minus{} \frac {x}{1} \plus{} \frac {x(x \minus{} 1)}{1.2} \minus{} \frac {x(x \minus{} 1)(x \minus{} 2)}{1.2.3} \plus{} ... \plus{} ( \minus{} 1)^n\frac {x(x \minus{} 1)...(x \minus{} n \plus{} 1)}{1.2.3...n}$ and $ Q(x) \equal{} \frac {( \minus{} 1)^n}{n!} (x \minus{} 1)(x \minus{} 2)...(x \minus{} n)$ are identical.

Nice solution by induction!

Another way is:

$ P(x) \equal{} 1 \plus{} \sum_{i\equal{}1}^n (\minus{}1)^i \binom {x}{i} \equal{} 1 \plus{} \sum_{i\equal{}1}^n (\minus{}1)^i \left[ \binom {x\minus{}1}{i} \plus{} \binom {x\minus{}1}{i\minus{}1} \right] \equal{}  \\ \equal{} 1 \plus{} \sum_{i\equal{}1}^n (\minus{}1)^i \binom {x\minus{}1}{i} \minus{} (\minus{}1)^{i\minus{}1} \binom {x\minus{}1}{i\minus{}1} \equal{} 1 \plus{} \sum_{i\equal{}1}^n f(i) \minus{} f(i\minus{}1)$

The telescopic series is clear.

$ P(x) \equal{} 1 \plus{} f(n) \minus{} f(0) \equal{} 1 \plus{} (\minus{}1)^n \binom {x\minus{}1}{n} \minus{} (\minus{}1)^0 \binom {x\minus{}1}{0} \equal{} \frac {(x\minus{}1)!}{n!(x\minus{}n\minus{}1)!} \Rightarrow \\ P(x)\equal{}Q(x)$


integration
limit
calculus
calculus computations

By $\frac{1}{k+1}=\int_0^1 x^k dx$, the answer is $\lim_{n\to\infty}\frac{1}{n+1}=0$


What is it? How do you prove it? Where can it be used?

lmaoo WOW

Two tangent theorem: Choosing any point in space not on or in a given circle, the two tangent lines from this point to the circle or equal; proven using simple congruency theorems


trigonometry
inequalities
inequalities unsolved

exactly are you summing in this problem?

[quote="e.lopes"]i have one solution using chebyschev and jensen, but i'm not certain about it.[/quote]
please post it :) thanks

to maokid7:
yes.

o oops sorry but in my last post i left out a very important word.

what i meant to say was, exactly [u][b]what[/b][/u] are you summing.  :oops:  sorry about the confusion.

oh,I am sorry,too ;)
I mean:
$\frac{\cos^2{2A}}{\sin^2 A}+\frac{\cos^2{2B}}{\sin^2 B} +\frac{\cos^2{2C}}{\sin^2 C} \geq 1$  :)

thanks.
maybe ill give the problem a shot now that i know what i am trying to solve.

[quote="zhaobin"][quote="hardsoul"][quote="zhaobin"]$A,B,C$ are three angle in a triangle.
prove:$\sum \frac{\cos^2{2A}}{\sin^2 A} \geq 1$
thanks[/quote]

I think that the problem is equivalent to that one http://www.mathlinks.ro/Forum/viewtopic.php?highlight=hard+for+beginners&t=4575[/quote]
sorry.I can't see it.can you help more :)[/quote]

In http://www.mathlinks.ro/Forum/viewtopic.php?highlight=hard+for+beginners&t=4575 it is proved that
Given $x,y,z >0$, $xy+yz+zx=1$ then
$\sum_{cyc} \frac{(1-x^2)^2}{1+x^2} \ge 1$ Now put $x=tgA,y=tgB,z=tgC$

$\sum_{cyc} \frac{(1-x^2)^2}{1+x^2} =\sum \frac{\cos^2{2A}}{\sin^2 A} \geq 1$

[quote="hardsoul"][/quote][quote="zhaobin"][quote="hardsoul"][quote="zhaobin"]$A,B,C$ are three angle in a triangle.
prove:$\sum \frac{\cos^2{2A}}{\sin^2 A} \geq 1$
thanks[/quote]

I think that the problem is equivalent to that one http://www.mathlinks.ro/Forum/viewtopic.php?highlight=hard+for+beginners&t=4575[/quote]
sorry.I can't see it.can you help more :)[/quote][quote="hardsoul"]

In http://www.mathlinks.ro/Forum/viewtopic.php?highlight=hard+for+beginners&t=4575 it is proved that
Given $x,y,z >0$, $xy+yz+zx=1$ then
$\sum_{cyc} \frac{(1-x^2)^2}{1+x^2} \ge 1$ Now put $x=tgA,y=tgB,z=tgC$

$\sum_{cyc} \frac{(1-x^2)^2}{1+x^2} =\sum \frac{\cos^2{2A}}{\sin^2 A} \geq 1$[/quote]
sorry,
but $x=tgA tgB+tgB tgC+tgC tg A=1$?????

[quote="hardsoul"][/quote][quote="zhaobin"][quote="hardsoul"][quote="zhaobin"]$A,B,C$ are three angle in a triangle.
prove:$\sum \frac{\cos^2{2A}}{\sin^2 A} \geq 1$
thanks[/quote]

I think that the problem is equivalent to that one http://www.mathlinks.ro/Forum/viewtopic.php?highlight=hard+for+beginners&t=4575[/quote]
sorry.I can't see it.can you help more :)[/quote][quote="hardsoul"]

In http://www.mathlinks.ro/Forum/viewtopic.php?highlight=hard+for+beginners&t=4575 it is proved that
Given $x,y,z >0$, $xy+yz+zx=1$ then
$\sum_{cyc} \frac{(1-x^2)^2}{1+x^2} \ge 1$ Now put $x=tgA,y=tgB,z=tgC$

$\sum_{cyc} \frac{(1-x^2)^2}{1+x^2} =\sum \frac{\cos^2{2A}}{\sin^2 A} \geq 1$[/quote]
sorry.you are right
but I think put $x=1/tgA,y=1/tgB,z=1/tgC$
thanks


$ a,b,c$ are all integers between $ \minus{}2007$ and $ 2007$, inclusive.  $ a,b,c$ are chosen so that the equation $ ax^{2}\plus{}bx\plus{}c\equal{}0$ has no real solutions.  Find the sum of all possible values of $ b$.

i dont have a exaclly proof. but we can use one thing.

Suppose $ b$ can assume different values, we just need to proof that number of the positive values that it can assume is equal to the number of negative.

so the sum will be zero.

$ b^{2}> 4 a c$

this is due this fact.

Actually, this holds for $ \minus{}s < a,b,c < s$

i believe that ur last statement wasn't true lord.

for example assume $ b\equal{}\minus{}2004$, $ a\equal{}\minus{}2006$, and $ b\equal{}2005$ in this case $ b^{2}<4ac$ thus disproving your last statement.

er!?

actually $ b^{2}> 4 a c$ need to be true. otherwise the solution of the quadratic equation would be real

and the problem is only for complexes ones

Dump

wait...i was confused. the solution to this problem is when $ b^{2}<4ac$ because its asking for complex solutions. right?

my algebra software rotates the signal < 180 degrees!!!

oh ...

yes, it's <

my error


You are given an arbitrary sequence $ u_{1}, u_{2}, \dots, u_{2^{n}}$, where each of the $ u_{i}$'s is either -1 or +1. You construct the new sequence $ v_{1}, v_{2}, \dots, v_{2^{n}}$ where $ v_{1} \equal{} u_{1}u_{2}$, $ v_{2} \equal{} u_{2}u_{3}$, $ \dots$, $ v_{2^{n}\minus{}1} \equal{} u_{2^{n}\minus{}1}u_{2^{n}}$ and $ v_{2^{n}} \equal{} u_{2^{n}}u_{1}$. Continue to construct new sequences successively, using the same rules. Show that after at most $ 2^{n}$ steps the resulting sequence will consist entirely of 1's.

[hide]Let

$ a_i \equal{} 0 \text{ if } u_i \equal{} 1$ and $ a_i \equal{} 1 \text{ if } u_i \equal{} \minus{} 1$

It is easy to see that the question is equivalent to proving that after at most $ 2^n$ steps of the transformation $ T: a_i\to a_i \plus{} a_{i \plus{} 1}$, the sequence is all even.

So, let the generating function of the sequence be $ f(x) \equal{} \cdots \plus{} a_{2^n \minus{} 2}x^{ \minus{} 2} \plus{} a_{2^n \minus{} 1}x^{ \minus{} 1} \plus{} a_{2^n} \plus{} a_1x^1 \plus{} a_2x^2 \plus{} \cdots$, where the coefficients repeat with period $ 2^n$.
After a transformation of $ T$, for any generating function $ g(x)$, $ T$ sends $ g(x)$ to $ (x \plus{} 1)g(x)$.  So, after $ 2^n$ transformations of $ T$, we get the function

$ h(x) \equal{} (x \plus{} 1)^{2^n}f(x) \equal{} \left(x^{2^n} \plus{} \binom{2^n}{1}x^{2^n \minus{} 1} \plus{} \cdots \plus{} \binom{2^n}{1}x \plus{} 1\right)f(x)$

Since we are only concerned with the parity, observe that

$ h(x)\equiv(x^{2^n} \plus{} 1)f(x)\pmod{2}$

So, if $ h(x) \equal{} \cdots \plus{} b_{2^n \minus{} 2}x^{ \minus{} 2} \plus{} b_{2^n \minus{} 1}x^{ \minus{} 1} \plus{} b_{2^n} \plus{} a_1x^1 \plus{} b_2x^2 \plus{} \cdots$, then for all $ i\equiv j\pmod{2^n}$, we have $ b_i\equiv a_j \plus{} a_{j \minus{} 2^n}\pmod{2}$.  But, since $ a_j \equal{} a_{j \minus{} 2^n}$, we have $ b_i\equiv0\pmod{2}$ for all $ i$.

[/hide]


trigonometry
geometry
area of a triangle
Heron\u0027s formula

Looking at the #4 solution, I understand everything except how we went from tan(a+b) = -tan(theta) to 4r^2 = (6-x)*x.  I must be missing something.  Help?

Thanks.

It's not an implication.  We are using the facts $\tan(\alpha + \beta) = -\tan(\theta)$, and a seperate fact $4r^2 = (6-x)(x)$.

To discover the latter, consider ABC.  From pythagoras, we must have $(2r)^2 + (6-x)^2 + (2r)^2 + x^2 = (6)^2$, which simplifies to $4r^2 = (6-x)(x)$.

To use both facts to arrive at $a=2$, lets use $\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}$, $\tan\alpha = \frac{6-x}{r}$, $\tan\beta = \frac{a}{r}$, $\tan\theta = \frac{x}{r}$.  Then the first fact implies $\frac{-x}{r} = \frac{\frac{6-x}{r} + \frac{a}{r}}{1 - \frac{(6-x)(a)}{r^2}}$, simplifying to $(6-x)ax - xr^2 = 6r^2 + ar^2$ and then to $[(6-x)x - r^2]a = 6r^2$.

Also let me point out that the first solution is needlessly complicated - Heron's formula approach works very quick:

$\sqrt{(6-x)ax(6+a)} = (6+a)r$, where we square both sides, substitute $r^2 = \frac{(6-x)(x)}{4}$, cancel some terms, and we get $4a = a+6$ implying $a = 2$ immediately.


trigonometry
geometry
circumcircle
geometry proposed

Let $ P$ be a point in triangle $ ABC$, and define $ \alpha,\beta,\gamma$ as follows: \[ \alpha\equal{}\angle BPC\minus{}\angle BAC, \quad \beta\equal{}\angle CPA\minus{}\angle \angle CBA, \quad \gamma\equal{}\angle APB\minus{}\angle ACB.\] Prove that \[ PA\dfrac{\sin \angle BAC}{\sin \alpha}\equal{}PB\dfrac{\sin \angle CBA}{\sin \beta}\equal{}PC\dfrac{\sin \angle ACB}{\sin \gamma}.\]

Let $ A',B',C'$ be the second intersections of  rays $ AP,BP,CP$ with the circumcircle $ (O)$ of $ \triangle ABC.$ Angle chase gives

$ \angle B'CP \equal{} \angle BPC \minus{} \angle BB'C \equal{} \angle BPC \minus{} \angle BAC \equal{} \alpha$

Similarly, $ \angle C'AP \equal{} \beta$ and $ \angle A'BP \equal{} \gamma.$

By sine law in the triangles $ \triangle PCB',\triangle PAC'$ and $ \triangle PBA'$ we obtain

$ \frac {\sin\widehat{BAC}}{\sin \alpha} \equal{} \frac {PC}{PA} \ , \ \frac {\sin\widehat{CBA}}{\sin \beta} \equal{} \frac {PA}{PC'} \ , \ \frac {\sin\widehat{ACB}}{\sin \gamma} \equal{} \frac {PB}{PA'}$

From power of $ P$ WRT $ (O) \ , $ $PA \cdot PA' \equal{} PB \cdot PB' \equal{} PC \cdot PC'$ we get

$ PA \cdot \frac {\sin\widehat{BAC}}{\sin \alpha} \equal{} PB \cdot \frac {\sin\widehat{CBA}}{\sin \beta} \equal{} PC \cdot \frac {\sin\widehat{ACB}}{\sin \gamma}$


I think that's what it's called...

In a Menelaus figure (not sure how to explain, but I think it's pretty clear), there are a total of four triangles.  Prove that the circumcircles of all 4 triangles intersect at a common point.

In a pentagon, extend all five sides to their intersections to form a star.  There are five Menelaus figures in this star.  Prove that the five miguel points  are concyclic.

I believe this is called Miquel's theorem... you could look it up, or if you have Geometry Revisited it's in there.

[hide="approach"]You put the 5 points in groups of 4.  So by proving all 5 groups are concyclic, all 5 poitns must lie on the same circle.  Because we want to show opposite angles are supplementary, our motivation is to use the fact that the angles of a triangle add up to 180 degrees.  And we can see that its true through similar triangles.[/hide]

Miquel Point : [b]Let A,B,C,D four points , ABCD convex and E is the intersection of AB with CD, and F the intersection of BC with AD.The circumcircles of the triangles ABF, ADE,CFD,ECB intersect at a comon point. [/b]
I have a clasical demonstration. <CBA=x and <CDA=y.Let M be the comon point for the circumcircles of the triangles CEB and CFG. \[ <FAB \plus{} <FMB \equal{} <DAE \plus{} y \plus{} <CMB\equal{} <DAE \plus{} <ADE \plus{} <AED\equal{}180\] and results that AFMB is inscriptibile. $ <DAE \plus{} <DME\equal{}<DAE\plus{}<AFB\plus{}x\equal{}180$ and results that ABMF is inscriptible. With these informations we conclude that M is the comon point for the 4 circles mentioned.

PS:sorry if i made mistakes with the angles...try to correct me if I am wrong(sorry for my english:D)

[b]Miquel's star : Prove  that if you circumscribe circles to the triangles formed by each side of a pentagon and the extensions of the adjacents sides, the 5 points where the adjoining circles meet are concyclic.[/b]
Let ABCD be the pentagon and ABF,BCG,CDH,DEI,EAK the five triangles formed by each side of the pentagon and the extensions of the adjacents sides. and L,M,N,P,Q the points of meeting of the circumcircles of the triangles mentioned.The circles C(ABF), C(BCG), C(FCI) from the complet quadrilateral ABCIGF are passing thru the Miquel's point,L. Alsaw the circles C(CDH), C(EDI),C(FCI) from the complet quadrilateral CDEFIH are passing thru the point N. Therefore the circumcircle  of the triangle FCI passes thru L and N.The circles C(ABF),C(FCI) and C(LNQ) have a comon point L. FI passes thru the intersection point F of the first two circles, and it results that if we unite point A with I with the intersections Q and N of these 2 circles with the third, QA and IN, extended,are meeting in a point R from the circle LNQ..If we consider the points Q,N,E on the sides of the triangle ARI,the circles QRN,NEI,QEA are passing thru the same point P, the circles NEI and QEA are the same with the circles DEI and AEK. Therefore Q,L,P,N are concyclic.In the same way we prove that M is on this circle .[/b]


Ồ bài này trên báo Toán học và tuổi trẻ mà!

Please use English on this part of the message boards.

[quote="manutd"]It's so hard, isn't it?[/quote]
Not really, it's just annoying to solve something not that 'beautiful' like this

[hide="Click here if anyone is interested"]Let $p=5x-1$, $q=9-x$, $g=2x^2+3x-1$
Original equation is $\sqrt{p}+\sqrt[3]{q}=g$

$\sqrt[3]{q}=g-\sqrt{p}$

$q=(g-\sqrt{p})^3$

$g^3+3gp-q=(3g^2+p)\sqrt{p}$

$(g^3+3gp-q)^2=p(3g^2+p)^2$

This is a degree 9 polynomial.[/hide]

If this is what you define hard then, take a look at this :D

[hide="Ugly One"]Solve $\sqrt[2005]{x^{1001}-509x^{19}+1}-\sqrt[1729]{x^{826}+x^2+1999}=x^{666}+666x+999$[/hide]

[quote="Altheman"]$x=1$
once you show that this is true, we note that a similiar function to this equation increases indefinately, so there are no other roots[/quote]
are you sure?can you explain more? :!:

Let $f(x)=2\2x^{2}+3x-1-\sqrt{5x-1} - \sqrt[3]{9-x}$

one can show that this function is increasing in the interval $[\frac{1}{5},\infty)$,
and it starts below the x-axis, and then it is over the x-axis later,
and since it is continious in that interval, there is one root (by the intermediate value theorem),
but since the function is increasing, it never comes back down again for another root, so we have found all roots


calculus
derivative
function
calculus computations

Consider the equation

$ a^2\plus{}b^2\equal{}c^2$.

Differenciating both sides gives

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

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

$ 2a\plus{}2b\equal{}2c \implies a\plus{}b\equal{}c$.

Which can't be right, just consider (3,4,5).

What am I misunderstanding? THe steps seem correct to me.

You need to specify which variable you are differentiatng with respect to. In the third line, you seem to be differentiating each term with respect to a different variable, which is not correct.

Also, the equation has to be true everywhere for differentiating both sides to be valid.

For example, the equation $ x^2\equal{}1$ is not true for all $ x$, and differentiating gives $ 2x\equal{}0\implies x\equal{}0$, which is obviously false. The derivative gives the instantaneous rate of change, and if this is not the same as the rate of change of the other side, you can't expect solutions of the differentiated equation to satisfy the original equation.

However, something like $ e^x\equal{}\sum \frac{x^n}{n!}$ is valid for all $ x$ and you can differentiate both sides because the functions are exactly the same (and so their rates of change will be exactly the same everywhere).

[quote="BOGTRO"]Consider the equation

$ a^2 \plus{} b^2 \equal{} c^2$.

Differenciating both sides gives

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

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

[/quote]

$ 2ada \plus{} 2bdb \equal{} 2cdc$ would be correct.

Yeah, you want to find where the functions intersect, not where the tangent lines have the same slopes.

basically, once you have $ 2ada+2bdb = 2cdc$, plug in values to get $ 2(3)(0)+2(4)(0) = 2(5)(0)$, ie $ 0+0=0$;the derivative of a constant is zero
this equation is used if you have two of the sides of a triangle and their respective rate of change and then working backwards to find the missing value
$ %Error. "pretty" is a bad command.
(Q.E.D.)$


let $ a$ , $ b$ , $ c$ $ \in ]0, \plus{} \infty[$ , prove that :

$ \frac {a}{bc} \plus{} \frac {b}{ca} \plus{} \frac {c}{ab} \plus{} \frac {bc}{a} \plus{} \frac {ca}{b} \plus{} \frac {ab}{c}  \geq a \plus{} b \plus{} c \plus{} \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}$

${ LHS=\sum {\frac{a}{bc}}+\sum {\frac{ab}{c}}=\frac{1}{abc}(\sum {a^2+(ab)^2})\geq \frac{1}{abc}\sum {ab+a^2bc}}=RHS$

thank's
nice solution

this is my prof

$ a^2\plus{}b^2\plus{}c^2\geq ab\plus{}bc\plus{}ca$

$ \frac{a^2\plus{}b^2\plus{}c^2}{abc} \geq \frac{ab\plus{}bc\plus{}ca}{abc}$

$ \frac{a}{bc}\plus{}\frac{b}{ca}\plus{}\frac{c}{ab} \geq \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}$

and we now that :

$ \frac{bc}{a}\plus{}\frac{ca}{b}\plus{}\frac{ab}{c} \geq a\plus{}b\plus{}c$


[quote="GameBot"]A bag contains 4 balls. Each ball is a single color, and 3 distinct
colors are represented in the bag. What is the probability that
the first 2 balls selected without replacement are the same color?
Express your answer as a common fraction.[/quote]

There are 2 balls that have the same color, so our answer is $ \frac{2}{4} \cdot \frac{1}{3} \equal{} \frac{1}{6}$.


For what value(s) of $k$ will the graph of $y=kx-6$ be perpendicular to the graph of $y=\frac{4}{7}x+5$?

[i]No Calculators!!![/i]
[i]Solutions Appreciated[/i]

[i]Solutions are Appreciated :wink: [/i]

[hide]The perpendicular line to $y = mx+b$ is $y =-\frac{1}{m}x+b$. In this case, $m = \frac{4}{7}$, so we take the negative reciprocal of that and get $m =-\frac{7}{4}$.[/hide]

thanks
(this sentence is here just to make this message longer)

[hide]
If two lines are perpendicular, then when the slopes are multiplied, the product is $-1$
$\frac{-1}{\frac{4}{7}}=\boxed{-\frac{7}{4}}$[/hide]

[quote="i_like_pie"]The perpendicular line to $y = mx+b$ is $y =-\frac{1}{m}x+b$. In this case, $m = \frac{4}{7}$, so we take the negative reciprocal of that and get $m =-\frac{7}{4}$.[/quote]
Hide your solution...


inequalities
inequalities proposed
algebra

Let $ a,b,c$ be nonegative numbers,no two of which are zero.Prove that:
$ \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}\plus{}\frac{3\sqrt[3]{a^2b^2c^2}}{2(ab\plus{}bc\plus{}ca)} \ge 2$

[quote="quykhtn-qa1"]Let $ a,b,c$ be nonegative numbers,no two of which are zero.Prove that:
$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {3\sqrt [3]{a^2b^2c^2}}{2(ab \plus{} bc \plus{} ca)} \ge 2$[/quote]
Consider $ \sqrt[3]{a}\equal{}x;\sqrt[3]{b}\equal{}y;\sqrt[3]{c}\equal{}z$
Rewrite the ineq. as:
$ \left(\sum \frac{x^3}{y^3\plus{}z^3}\minus{}\frac{3}{2} \right)\minus{} \left(\frac{1}{2}\minus{}\frac{3x^2y^2z^2}{2(x^3y^3\plus{}y^3z^3\plus{}z^3x^3)}\right) \ge 0$
$ \Leftrightarrow \sum \frac{(x^3\minus{}y^3)^2}{2(x^3\plus{}z^3)(y^3\plus{}z^3)}\minus{}(xy\plus{}yz\plus{}zx) \sum \frac{(zx\minus{}yz)^2}{4(x^3y^3\plus{}y^3z^3\plus{}z^3x^3)} \ge 0$
$ \Leftrightarrow \sum \frac{(x\minus{}y)^2(x^2\plus{}y^2\plus{}xy)^2}{2(x^3\plus{}z^3)(y^3\plus{}z^3)}\minus{}(xy\plus{}yz\plus{}zx) \sum \frac{z^2(x\minus{}y)^2}{4(x^3y^3\plus{}y^3z^3\plus{}z^3x^3)} \ge 0$
$ \Leftrightarrow \sum (x\minus{}y)^2\left( \frac{(x^2\plus{}y^2\plus{}xy)^2}{2(x^3\plus{}z^3)(y^3\plus{}z^3)}\minus{} \frac{z^2(xy\plus{}yz\plus{}zx)}{4(x^3y^3\plus{}y^3z^3\plus{}z^3x^3)}\right) \ge 0$
what can we continue??

[quote="Mr.Bikrone"][quote="quykhtn-qa1"]Let $ a,b,c$ be nonegative numbers,no two of which are zero.Prove that:
$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {3\sqrt [3]{a^2b^2c^2}}{2(ab \plus{} bc \plus{} ca)} \ge 2$[/quote]
Consider $ \sqrt [3]{a} \equal{} x;\sqrt [3]{b} \equal{} y;\sqrt [3]{c} \equal{} z$
Rewrite the ineq. as:
$ \left(\sum \frac {x^3}{y^3 \plus{} z^3} \minus{} \frac {3}{2} \right) \minus{} \left(\frac {1}{2} \minus{} \frac {3x^2y^2z^2}{2(x^3y^3 \plus{} y^3z^3 \plus{} z^3x^3)}\right) \ge 0$
$ \Leftrightarrow \sum \frac {(x^3 \minus{} y^3)^2}{2(x^3 \plus{} z^3)(y^3 \plus{} z^3)} \minus{} (xy \plus{} yz \plus{} zx) \sum \frac {(zx \minus{} yz)^2}{4(x^3y^3 \plus{} y^3z^3 \plus{} z^3x^3)} \ge 0$
$ \Leftrightarrow \sum \frac {(x \minus{} y)^2(x^2 \plus{} y^2 \plus{} xy)^2}{2(x^3 \plus{} z^3)(y^3 \plus{} z^3)} \minus{} (xy \plus{} yz \plus{} zx) \sum \frac {z^2(x \minus{} y)^2}{4(x^3y^3 \plus{} y^3z^3 \plus{} z^3x^3)} \ge 0$
$ \Leftrightarrow \sum (x \minus{} y)^2\left( \frac {(x^2 \plus{} y^2 \plus{} xy)^2}{2(x^3 \plus{} z^3)(y^3 \plus{} z^3)} \minus{} \frac {z^2(xy \plus{} yz \plus{} zx)}{4(x^3y^3 \plus{} y^3z^3 \plus{} z^3x^3)}\right) \ge 0$
what can we continue??[/quote]
My solution very nice :) .Not use SOS  :wink:  :)

Does the following inequality help?
$ 3 \sqrt[3]{a^2b^2c^2} \ge 4(ab\plus{}bc\plus{}ca) \minus{} (a\plus{}b\plus{}c)^2$

[quote="great math"]Does the following inequality help?
$ 3 \sqrt [3]{a^2b^2c^2} \ge 4(ab \plus{} bc \plus{} ca) \minus{} (a \plus{} b \plus{} c)^2$[/quote]
Yes :)

I got it :)
$ \sum_{cyc}\frac{a}{b\plus{}c} \ge \frac{(a\plus{}b\plus{}c)^2}{2(ab\plus{}bc\plus{}ca)}$

[quote="Johan Gunardi"][quote="great math"]Does the following inequality help?
$ 3 \sqrt [3]{a^2b^2c^2} \ge 4(ab \plus{} bc \plus{} ca) \minus{} (a \plus{} b \plus{} c)^2$[/quote]How to prove this?[/quote]
$ 3\sqrt{a^2b^2c^2}\ge \frac{9abc}{a\plus{}b\plus{}c} \ge RHS$
It 's shur 'inequality

[quote="quykhtn-qa1"]Let $ a,b,c$ be nonegative numbers,no two of which are zero.Prove that:
$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {3\sqrt [3]{a^2b^2c^2}}{2(ab \plus{} bc \plus{} ca)} \ge 2$[/quote]
$ LHS \geq\ \sum\ \frac{a}{b\plus{}c} \plus{} \frac{4abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)} \geq\ 2$ (obviously trues by Schur)  :)


inequalities
algebra
system of equations
algebra proposed

Suppose that $x_{1},x_{2},..., x_{n}\in \mathbb{R}$ and $\lambda \in \mathbb{R}-\{0\}$. Also for every $i \in \{1,2,...,k\}$ we have $\lambda^{i}=\sum_{j=1}^{n}{x_{j}}^{i}$. Find the smallest amount of $k$ such that from these $k$ equations it could be deduced that precisely $n-1$ of the $x_{i}$s are zero.

Here's an outline of a solution:
[hide]
Using the below facts prove that at most one of the $x_{j}$s is non-zero if k=3 (It can easily be proved that there exist infinite sets of $n$ non-zero real numbers that are  solutions to the system of equations in the case k=2):

1.$\sum_{1\leq i,j\leq n}{x_{i}}^{2}x_{j}=0$ since  $\lambda^{3}=\lambda(\lambda^{2})$ and so $\sum_{j=1}^{n}{x_{j}}^{3}=\left(\sum_{j=1}^{n}x_{j}\right) \left(\sum_{j=1}^{n}{x_{j}}^{2}\right)$.

2.$\left(\sum_{j=1}^{n}x_{j}\right) \left(\sum_{j=1}^{n}{x_{j}}^{3}\right)= \left(\sum_{j=1}^{n}{x_{j}}^{2}\right)^{2}$ because $(\lambda^{3})(\lambda )=(\lambda^{2})^{2}$. However, by the Cauchy-Schwarz inequality we have $\left(\sum_{j=1}^{n}x_{j}\right) \left(\sum_{j=1}^{n}{x_{j}}^{3}\right)\geq \left(\sum_{j=1}^{n}{x_{j}}^{2}\right)^{2}$, therefore you can apply the equality condition of this inequality here.[/hide]

[quote="mathematica"]Here's an outline of a solution:
However, by the Cauchy-Schwarz inequality we have $\left(\sum_{j=1}^{n}x_{j}\right) \left(\sum_{j=1}^{n}{x_{j}}^{3}\right)\geq \left(\sum_{j=1}^{n}{x_{j}}^{2}\right)^{2}$, therefore you can apply the equality condition of this inequality here.[/quote]
Only if the $x$'s are nonnegative; it's easy to find a counterexample when at least one is negative (for example, when $a=\pi,b=-e$, we get $(a+b)(a^{3}+b^{3})<(a^{2}+b^{2})^{2}$.

I believe that a necessary and sufficient condition is $k\geq n$. $k=n$ is certainly sufficient, since we can use the equations to arrive at
\[\prod_{j=1}^{n}x_{j}=0\]
and then use induction.

[quote="scorpius119"]
Only if the $x$'s are nonnegative; it's easy to find a counterexample when at least one is negative (for example, when $a=\pi,b=-e$, we get $(a+b)(a^{3}+b^{3})<(a^{2}+b^{2})^{2}$.
[/quote]
Yes, you are right. I've asked one of the moderators to change it.


23.) The average of 12 different positive integers is 12. What is the greatest possible value of any one of these numbers?

[quote="Ignite168"][hide]
The total of the numbers will be 12*12 or 144.
The least positive integer is 1. We will assume 11 of the numbers are 1 and subtract it from 144 giving you 133. Therefore, the greatest possible value one of the numbers would be 133.[/hide][/quote]

It said [b]different[/b] integers.

[hide]So, 1+2+3+...+11=66, the sum is 144, so the largest is 78.[/hide]

Woops  :blush:
[hide]
The first 11 numbers should be the arithmatic sequence $1+2+3....11$ as those are the 11 smallest integers. Apply formula for arithmatic sequence to get $\frac{11(12)}{2}$ which equals $66$. Finally, do $144-66$ to get $\boxed {78}$[/hide]

[hide]The sum would be 12*12=144. Letting 11 numbers be the smallest positive numbers, 1,2,...11, whose sum is 66, that means 144-66=78 is our answer.[/hide]


Here's how it will work. We'll start the game with 8 people (number subject to change), who are in a specified order. Rules of the game:

1. The player who leads gives a letter, say N.
2. The next player follows with another, say A. It must be possible that these two letters start a word, such as NAsty or NAstiness.
3. After this, each following player adds on to the chain of letters. If a player completes a word (say, the first four players say N, A, S, T in succession and the fifth says Y), that player recieves a letter, in the word Ghost. Once a player accumulates five letters, to spell the word ghost, they are eliminated from the game.
4. A word only counts as a word if it has three letters or more.
5. If it is a player's turn and believes there is no such word beginning with the given letters (say, PSEUD), s/he can challenge the preceding player. The preceding player must give a word starting with those letters (pseudonym), as a result the challenger recieves a letter. If the preceding player fails to do so, that person gets the letter.
6. If someone takes 24 hours or more for their turn (either giving a letter or answering a challenge), s/he forfeits their turn, takes the letter, and play continues with the same word.
7. The use of dictionaries is not allowed.
8. Once a cycle ends (a successful challenge or the completion of a word), the player who lost leads the next word. If that player was eliminated, the next player in the order leads.

I am the moderator of this game, I will also be playing (don't worry, I don't have an advantage and I won't cheat).
Sign up here.

Players:

1. 13375P34K43V312
2. jli
3. cincodemayo5590
4. hunter34
5. junggi
6. miyomiyo

I'll play; this is a popular game in the math team for my school.  :)

darn i'm too late to be signed in
oh well

[quote="Ignite168"]8's a bit much, try 6[/quote]

hmm, fine, good idea. I'm starting a second game for different people.
But let this game begin. I'm leading for the first word.

S

lol game b's 3 letters ahead of game a...

I want to be in a more active game.  Can
(A) jli be replaced
(B) jli get a letter right now and game continues
(C) move me to game B? :lol:

Dang people. After 24 hours, the next person can go. Read the rules. Seriously. Golly gosh gee wilikers.

sed[b]i[/b]

oops I didn't know we had to sign up

You have to sign up to play.

I'll go SED[b]I[/b], though.


Set A has 16 elements and set B has 37 elements. The union of
sets A and B has 43 elements. How many elements are in the
intersection of sets A and B?


The first term of a geometric sequence is 7, and the 7th term is 5103. What is the 5th term?

Since the first term is 7 and the 7th term is 5103, the common ratio is $ r$, so $ 7r^6\equal{}5103$, $ r^6\equal{}729$, $ r\equal{}3$.  So the 5th term is $ \frac{5103}{3^2}\equal{}\frac{5103}{9}\equal{}\boxed{567}$


geometry
geometric transformation
reflection
circumcircle
geometry proposed

Let $ABC$ be acute-angle triangle. $B'$ is the reflection of $B$ across $AC$, and $C'$ is the reflection of $C$ across $AB$. The circumcircles of $ABB', ACC'$ intersect at $P$. Show that $AP$ passes through the circum center of $ABC$.

Argh... Synthetic solution.

Let $D$ be the other intersection point of $AB$ and circumcicle of $\triangle ACC'$. Since $AD$ bisect $\angle CAC'$, $AD$ is the diameter of the circle.
Since $P$ and $C$ is passed through the circle, $\angle APD = \angle ACD = 90^\circ$

Let $E$ be the other intersection point of $AC$ and circumcicle of $\triangle ABB'$. Since $AE$ bisect $\angle BAB'$, $AE$ is the diameter of the circle.
Since $P$ and $B$ is passed through the circle, $\angle APE = \angle ABE = 90^\circ$

Since $\angle APE = \angle APD = 90^\circ$; Points $D$, $P$, and $E$ are colinear.

Let $F$ be the intersection point of $BE$ and $CD$ then $AF$ is the diameter of the circumcenter of $\triangle ABC$ i.e. $AF$ passes through the circumcenter of $\triangle ABC$.

But $F$ is the orthocenter of $\triangle ADE$ and $P$ is the foot of the altitude $AF$.
Hence $AP$ passes through the circumcenter of $\triangle ABC$.


The file is the the result of IMO 2008

Including Rank of country

Are you sure that those are the final results? On the website it says that results may not be complete.

The contents are all from the website  http://www.imo-2008.es/results.html

I think the resules are almost reliable because they're posted after the jury meeting.


Find all infinite sequences $ a_1, a_2, a_3, ...$ , each term $ 1$ or $ \minus{}1$, such that no three consecutive terms are the same and $ a_{mn} \equal{} a_ma_n$ for all $ m, n$.

No such sequence exists.

First, we claim that $ a_2=-1$ must hold. Otherwise assume $ a_2=1$. Now, $ (a_2,a_3,a_4)$ are given by $ (1,a_3,1)$, i.e. $ a_3=-1$. Checking $ (a_7,a_8,a_9)$ (resp. $ (a_8,a_9,a_{10})$) represented by $ (a_7,1,1)$ (resp. $ (1,1,a_5)$) yields $ a_5=a_7=-1$. But this means $ a_5=a_6=a_7=-1$, a contradiction. So from now on let $ a_2=-1$.

Now we claim, that $ a_3=-1$ must hold. Otherwise assume $ a_3=1$. Considering $ (a_3,a_4,a_5)$ we conclude $ a_5=-1$. Checking $ (a_5,a_6,a_7)$ yields $ a_7=1$ and checking $ (a_9,a_{10},a_{11})$ yields $ a_{11}=-1$. Substitution into ${ (a_{63},a_{64},a_{65}})$ then yields $ (1,1,1)$, a contradiction. So from now on let $ a_3=-1$.

Checking $ (a_4,a_5,a_6)$ determines $ a_5=-1$ and checking $ (a_9,a_{10},a_{11})$ requires $ a_{11}=-1$. Consideration of $ (a_{11},a_{12},a_{13})$ yields $ a_{13}=1$. Substitution into $ (a_{13},a_{14},a_{15})$ defines $ a_7=1$. Again, checking ${ (a_{63},a_{64},a_{65}})$ yields $ (1,1,1)$, a contradiction.

[quote="hjbrasch"]Now we claim, that $ a_3 = - 1$ must hold. Otherwise assume $ a_3 = 1$. Considering $ (a_3,a_4,a_5)$ we conclude $ a_5 = - 1$. Checking $ (a_5,a_6,a_7)$ yields $ a_7 = 1$ and checking $ (a_9,a_{10},a_{11})$ yields $ a_{11} = - 1$. Substitution into ${ (a_{63},a_{64},a_{65}})$ then yields $ (1,1,1)$, a contradiction. So from now on let $ a_3 = - 1$
[/quote]
Why $ a_{65} = 1$?
I think: $ a_{14}=a_2a_7=-1,a_{15}=a_3a_5=-1$ then $ a_{13}=1$, hence $ a_{65}=a_5a_{13}=-1$

you are right, seems :blush:  that I have to check again (maybe even change my working hypothesis)


Find all positive integer solutions of equation:$ x^2\plus{}7\equal{}2^n$

x^2+7=2^n    . . . (1)

  x^2 -1 +8=2^n

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

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

From (1) it's easy to see that  x is odd number ,so :
 x-1 =2^n-3 -1    or
 x+1=2^n-3  -1
    if and only if : 2^n-3 -1=0  or  2^n-3 -1=1

so we have the solution 1. 2^n-3 -1=0 =>n=3  and x=1
                                 2. 2^n-3 -1=1 =>n=4 , but there's no integer solution of x !

 The only solution is (x,n)=(1,3)

efejzullahu, your solution is not correct (try $ x \equal{} 3$ or $ 11$). One of the problems is because you are assuming that one of $ x \minus{} 1$ or $ x \plus{} 1$ do not have $ 2$ in its factorization. You forgot some steps. And you are assuming that $ 2^{n\minus{}3} \minus{}1$ is a prime number.

See PEN:
[url]http://www.mathlinks.ro/viewtopic.php?t=150658[/url]

Doesn't appear an elementary one  :o

[quote="erudito"]efejzullahu, your solution is not correct (try $ x \equal{} 3$ or $ 11$). One of the problems is because you are assuming that one of $ x \minus{} 1$ or $ x \plus{} 1$ do not have $ 2$ in its factorization. You forgot some steps. And you are assuming that $ 2^{n \minus{} 3} \minus{} 1$ is a prime number.[/quote]

hmm ...  : from (1) we can see that X is odd number so
                             X-1 and X+1  don't have 2 in their factorization

ohh sorry  :blush:
  it seems I'm drunk  :(  :(

No problem! Mistakes can happen sometimes.