An infinite sequence $a_0, a_1, a_2 \ldots$ of real numbers satisfies the condition $a_n = |a_{n+1} - a_{n+2}|$ for every $n \geq 0$, with $a_0, a_1$ distinct and positive. Can this sequence be bounded?

It can be proven that for such a sequence, the following assertions are equivalent

$\bullet$ the sequence is bounded.
$\bullet$ the function $f$, defined by $f(n,k) = a_na_k(a_n-a_k)$ for $n,k \geq 0$, is identically null.

In the case at hand, $f(0,1) = a_0a_1(a_0-a_1) \neq 0$, so it follows the sequence is unbounded. An extremely elegant proof, using a sequence $b_n = \max\{a_n,a_{n+1},a_{n+2}\}$ was provided during a 2005 Romanian Selection Test by Alexander Zamorzaev, from Moldova (but the solution is lost  :(

It immediately follows the next assertion is also equivalent with the previous two

$\bullet$ the sequence is of type $\ldots, 0,x,x,0,x,x,0, \ldots$, for some $x\geq 0$.

When the sequence is unbounded, we may either have $\lim_{n\to \infty} a_n = \infty$, e.g. for $a_{n+2} = F_na_0 + F_{n+1}a_1$ for $n\geq 0$ (where $F_k$ are the Fibonacci numbers), or a bounded subsequence may exist, e.g. for $a,b,b+a,a,b+2a,b+3a,a,b+4a,b+5a,a,\ldots$.

One proof can be seen [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=198930&sid=ce1fe403a564b98143cee5a93d3f28ad#p198930]here[/url]. You can also watch the discussion in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=152740&sid=ce1fe403a564b98143cee5a93d3f28ad#p152740]this post[/url].

I use one idea from that post, and then the idea of taking the sequence $b_n$. Of course, if the sequence has a zero then it has only two values, therefore, suppose the sequence is never zero. I want to prove that if the sequence has non-zero terms then it cannot get arbitrarily small. Argue by contradiction that there are arbitrarily small numbers in our sequence. Then, at one instance, there must be a term which is smaller than any term before it.

Suppose $a_i$ is the smallest term until $i$, i.e. $a_j>a_i$ for $j <i$. Suppose $i \geq 3$. Then $a_{i-1}>a_i$ so $a_{i-2}=a_{i-1}-a_i<a_{i-1}$. Then $a_{i-3}=a_{i-1}-a_{i-2}=a_i$. Contradiction. Therefore $i<3$. This means that for $i \geq 3$, $a_i$ cannot be the smallest term of the sequence. In conclusion, all the terms in the sequence are greater than the minimum value among $a_0,a_1,a_2$, which is non-zero by our assumption.

Take now the sequence $b_n=\max\{a_n,a_{n+1},a_{n+2}\}$. Using the recurrence relation we easily find that $b_n=\max\{a_{n+1},a_{n+2}\}\leq b_{n+1}$. Therefore $b_n$ is increasing. Moreover, we have
$b_{n+1}=\begin{cases} b_n \\
a_{n+3}=a_{n+2}+a_{n+1}=2b_n-|a_{n+2}-a_{n+1}|=2b_n-a_n=b_n+(b_n-a_n)
\end{cases}$

Now suppose $(a_n)$ is bounded. Then obviously $(b_n)$ is also bounded. A increasing and bounded sequence is convergent. Therefore $b_n \to L$. Then obviously $b_{n+1}-b_n \to 0$, and it can get smaller than any of the first three terms. If $b_{n+1}-b_n\neq 0$, then $b_{n+1}-b_n=b_n-a_n$ which can be:
1) $a_{n+1}-a_n=a_{i-1}$. Contradiction, because it is smaller than the first three terms.
2) $a_{n+2}-a_n=a_{i+1}$. Again contradiction for the same reason.

Therefore, $(b_n)$ is eventually constant. This means $\max\{a_{n},a_{n+1}\}=\max \{a_{n+1},a_{n+2}\}$ for very large $n$. If this max is $a_n$ or $a_{n+2}$ then we can find a zero in our sequence, and we are done.

( why? because either two consecutive terms are equal, either $a_n=a_{n+2}$. The $a_{n+1}$ can be $0$ or $2a_n$. In our assumptions it follows $a_{n+1}=2a_n$. From $a_{n+1}=|a_{n+2}-a_{n+3}|$ the only possible value for $a_{n+3}$ is $3a_n$. Then $b_n=2a_n$ and $b_{n+1}=3a_n$. $b_{n+1}=b_n$ yields $a_n=0$, contradiction)

 Else we have $a_{n+1}>a_{n+2}$, and we find a zero term from $\max\{a_{n+1},a_{n+2}\}=\max \{a_{n+2},a_{n+3}\}$.

I'm not sure this is the elegant proof, but it is a try.


Consider a non-empty set $S= \{1,2,3,...,n\}.$ Let us define a function $f(A)$ on a non-empty set $A$ of $S$ as follows: Arrange the elements of $A$ in a decreasing order, say, $a_{k},a_{k-1},a_{k-2},a_{k-3}...,a_{2},a_{1}$ where $1 \leq k \leq n.$ Then, $ f(A) = \frac{a_{k}}{a_{k-1}}\frac{a_{k-2}}{a_{k-3}}...$ [For example, $f(\{1,2,3\}) = \frac32 1,$ and $f(\{1,2,3,4,5,6\}) = \frac65 \frac43 \frac21.$]
Find $\frac1{n}\prod_{A \in S}f(A)$ where $A$ is a non-empty subset of $S.$

[Again, for those of you who think it's easy, just provide a hint to others, not the complete solution. Thanks!  :) ]

[hide="hint"] Factorial and Simplification if I understood the problem.[/hide]


geometry
trigonometry
search
geometry proposed

$ HA+HB+HC=2R(\cos A+\cos B+\cos C)=2R\left(4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}+1\right)=2r+2R\geq 6r$

[quote="chien than"]Let $ABC$ be an acute triangle with orthocenter $H$.Suppose that $r$ is radius of incircle of triangle $ABC$.Prove that $HA+HB+HC \geq 6r$[/quote]

I think that it is true for every point $P,$ inwardly to an acute triangle $\bigtriangleup ABC.$ $($ $PA+PB+PC \geq 6r$ $).$

I will search my archive and if I am not mistaken, I will post here a solution of this well known result.

Kostas vittas.

Kostas, this is also proven in http://www.mathlinks.ro/Forum/viewtopic.php?t=19316 .

  Darij


An isosceles trapezoid with base lengths 16 and 36 is circumscribed about a circle. Find the circumference of the circle.

use circumscribed quadrilateral rule that opposite sides sum up to the same.

[hide] Using Spectacular's hint, we can say that the non-parrellel sides are both x, so 16+36=x+x.  This means that 52=2x, so x=26.  Now, draw AX perpindicular to BC so that X is on BC.  Thus, we have that AB=26, AX=h, and BX=(36-16)/2=10.  So, we can use the Pythagorean Theorem to find that h=24.  Thus, the radius=h/2=12, which is our answer.  [/hide]


if$a+b+c\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ then prove it:
$a+b+c\ge \frac{3}{a+b+c}+\frac{2}{abc}$

Maybe this:

Multiply with $(a+b+c)abc$, and get \[(a+b+c)^{2}abc \geq 3abc+2(a+b+c)\]

But we know that \[(a+b+c)^{2}abc \geq (a+b+c)(ab+bc+ac)= 3abc+(a^{2}b+a^{2}c+b^{2}c+b^{2}a+c^{2}a+c^{2}b)\]
So we only have to prove that

\[(a^{2}b+a^{2}c+b^{2}c+b^{2}a+c^{2}a+c^{2}b) \geq 2(a+b+c)\]

Now, we know that \[(a^{2}b+a^{2}c+b^{2}c+b^{2}a+c^{2}a+c^{2}b) \geq \frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{a+b+c}.(a^{2}b+a^{2}c+b^{2}c+b^{2}a+c^{2}a+c^{2}b)\]

So we only have to prove that \[\sum_{cyc}{(ab+ac+bc.\frac{b+c}{a}+b^{2}+c^{2})}\geq 2(a+b+c)^{2}\]

This one is equavalent to \[\sum_{sym}{a^{3}b^{2}}\geq \sum_{sym}{a^{2}b^{2}c}\]

Which is obviously true.

I hope there are no mistakes, because I'm very tired when I'm writing this. Comments are always welcome


Jenny pulls two marbles out of a bag containing 4 red and 6 green marbles. Find the probability
that the two marbles are of the same color.

Either both marbles are red, or both marbles are green.

Probability (that both marbles are red) $ \plus{}$ Probability (that both marbles are green)

$ \frac{4}{10} \cdot \frac{3}{9} \plus{} \frac {6}{10} \cdot \frac {5}{9} \equal{}$

$ \frac{12}{90} \plus{} \frac{30}{90} \equal{}$

$ \frac {42}{90} \equal{}$

$ \frac {7}{15}$


geometry
3D geometry
sphere
geometric transformation
reflection
combinatorics unsolved
combinatorics

Through a point in the interior of a sphere we put three pairwise perpendicular planes. Those planes dissect the surface of the sphere in eight curvilinear triangles. Alternately the triangles are coloured black and wide to make the sphere surface look like a checkerboard. Prove that exactly half of the sphere's surface is coloured black.

You need to  reflect each plane in respect to the midpoint of the sphere. Then you have 26 areas of which many are equal. The problem reduces to counting the areas which are coloured white and black.

I guess the question talks about the curvilinear triangles which are on the outside. Otherwise the planes cut the sphere in 8 different entities each having 4 independent surfaces to be painted. But we are not really dissecting the sphere. So we only have to color the outer 8 curvilinear triangles which equally divide the outer surface of sphere in 8 parts. As we only have two colors 4 of them will definetely have the same color which means half of the sphere..
Could it be this simple...??
I dont know, but this is what I think is the Solution !!

"Half" here means "half of the surface area," not "half of the regions into which we've divided the surface."  There are many ways to divide a sphere into 8 regions, 4 white and 4 black, so that the area of the black regions does not equal to the area of the white regions.

The result is easy to verify in two dimensions. Now, move the initial point perpendicularly to one of the planes, say $ \alpha$. Then $ \alpha$ is going to move together with the initial point. The black area and the white are in each side of $ \alpha$ will equally increase/decrease because of the two dimensional case.


Determine the scalar equation of the plane through A(2,1,-5),perpendicualr to both 3x-2y+z=8 and 4x+6y-5z=10.

****if it help the answer is:4x+19y+26z+103=0*****i just don't understand how to get it!!!!!

hello
Let the eq. be $ ax \plus{}by \plus{}cz \equal{}d$
U just need to understand d fact that the normal of the above eq. is perpendicular to the normals of the other 2 planes .
So
taking the dot product
$ 3a\minus{}2b\plus{}c\equal{}0$
$ 4a\plus{}6b\minus{}5c\equal{}0$
Also
$ 2a\plus{}b\minus{}5c\equal{}d$
Solving the three
$ a \equal{}\frac{\minus{}4d}{103},b\equal{}\frac{\minus{}19d}{103},c\equal{}\frac{\minus{}26d}{103}$
thank u


geometry
rectangle
probability
analytic geometry
trigonometry
integration
conditional probability

Let C be the unit circle $x^2 + y^2 = 1$.  A point p is chosen randomly on the circumference C and another point q is chosen randomly from the interior of C (independently and randomly over their domains).  Let R be the rectangle with sides parallel to the x and y-axes with diagonal pq.  What is the probability that no point of R lies outside of C?

[hide="My solution"]Point $P$ has coordinates $(\cos\theta,\sin\theta)$ where $\theta$ is distributed uniformly on $[0,2\pi].$ Given that point $P$ has been chosen, then construct the rectangle $W$ inscribed in the circle with sides parallel to the axes with point $P$ as one vertex. If point $Q$ is in rectangle $W$, then the rectangle $R$ whose diagonal is $PQ$ is contained within $W$ and thus within the circle.  If point $Q$ lies outside of $W,$ then one of the edges of $R$ must have a side which extends one of the sides of $W$ outside the circle.

Hence the conditional probability that $R$ is contained in the circle is the area of $W$ divided by the area of the circle, that is

$P(R\subset C|\theta)=\frac4{\pi}|\cos\theta\sin\theta|$

Hence $P(R\subset C)=\frac1{2\pi}\int_0^{2\pi} \frac4{\pi}|\cos\theta\sin\theta|\,d\theta$

$=\frac2{\pi}\int_0^{\frac{\pi}2} \frac4{\pi}(\cos\theta\sin\theta)\,d\theta=\frac4{\pi^2}$[/hide]


trigonometry
geometry
circumcircle
geometric transformation
rotation
symmetry
geometry solved

Let $ABC$ be a right isosceles triangle with the right angle at $A$. Let $M$ and $N$ be points on the side $BC$ so that $N$ is inside the segment $MC$ and $BM^{2}-MN^{2}+NC^{2}=0$. Prove that $\measuredangle MAN=45^{\circ}$.

[i]Alternative formulation.[/i] Let $ABC$ be a triangle with $\measuredangle CAB=90^{\circ}$ and $AB=AC$. Let $M$ and $N$ be points on the side $BC$ so that $N$ is inside the segment $MC$ and $BM^{2}+CN^{2}=MN^{2}$. Prove that $\measuredangle MAN=45^{\circ}$.

Here's an idea : try  using the Law of Cosines

[color=blue][b]Problem.[/b] Let ABC be an isosceles right triangle with the right angle at A. Let M and N be two points on the hypotenuse BC such that the point N lies inside the segment MC and such that $BM^2-MN^2+CN^2=0$. Prove that < MAN = 45°.[/color]

[i]Solution.[/i] We will use directed segments on the line BC.

The equation $BM^2-MN^2+CN^2=0$ yields $BM^2+CN^2=MN^2$, and thus

$BM^2=MN^2-CN^2=\left(MN-CN\right)\cdot\left(MN+CN\right)$
$=MC\cdot\left(\left(MC+CN\right)+CN\right)=MC\cdot\left(MC+2\cdot CN\right)$.

On the other hand, let U be the point on the ray MC such that < MAU = 45°. Then, after http://www.mathlinks.ro/Forum/viewtopic.php?t=27981 post #4, we have $BM^2+CU^2=MU^2$. Similarly to how we inferred $BM^2=MC\cdot\left(MC+2\cdot CN\right)$ from $BM^2+CN^2=MN^2$, this equation will lead to $BM^2=MC\cdot\left(MC+2\cdot CU\right)$. Thus, we have $MC\cdot\left(MC+2\cdot CN\right)=MC\cdot\left(MC+2\cdot CU\right)$. Hence, CN = CU. Since the points N and U both lie on the line BC, and the segments are directed, it follows that the points N and U coincide. Hence, < MAU = 45° becomes < MAN = 45°. And the problem is solved.

[b]PS.[/b] This problem hasn't been solved on this forum for almost two years because it was moved into the solved section...

  Darij

Let O be the midpoint of BC and R OA = OB = OC the circumradius of the right isosceles $\triangle ABC.$

$MN^{2}= (2R-BM-CN)^{2}= 4R^{2}+BM^{2}+CN^{2}-4R(BM+CN)+2 BM \cdot CN$

$2R^{2}-2R(BM+CN)+BM \cdot CN = 0$

$BM = \frac{2R (R-CN)}{2R-CN}= \frac{2R \cdot ON}{BN}$

$MN = BN-BM = \frac{BN^{2}-2R \cdot ON}{BN}=$

$= \frac{(R+ON)^{2}-2R \cdot ON}{BN}= \frac{R^{2}+ON^{2}}{BN}= \frac{AN^{2}}{BN}$

$\frac{MN}{MB}\cdot \frac{ON}{OB}= \frac{AN^{2}}{2R \cdot ON}\cdot \frac{ON}{R}= \frac{AN^{2}}{AB^{2}}\ \Longrightarrow$

AM, AO are isogonals with respect to $\angle BAN,$ hence $\angle MAN = \angle BAO = 45^\circ.$

A rotation of 90 degrees about A yields an elegant [synthetic] solution...I believe I have posted this already...but I cannot find it now...?

The foll. solution I posted be4, but I am not able to find it now:
Take a quarter of a circle $ C(A, AM)$ and $ P$ a point on the arc $ BC$, draw then the bisectors of the angles $ \angle BAP$ and $ \angle CAP$; these angle bisectors will intersect $ BC$ at M and N respectively (by symmetry $ \angle APM\equal{} \angle ABM \equal{}45^\circ$ and $ \angle APN \equal{} \angle ACN \equal{} 45^\circ$), hence $ \angle MPN \equal{} 90^\circ$ and $ BM^2\plus{}CN^2\equal{}MN^2$ (since $ BM\equal{}MP$ and $ CN\equal{}NP$); consequently $ \angle MAN\equal{}\frac{\angle BAP\plus{}\angle PAC}{2}\equal{}\frac{\angle BAC}{2}\equal{}45^\circ$.

Best regards,
sunken rock


$ 1)$ Να βρειτε τους $ a,b,q,p \in R$ ετσι ωστε:
$ (2x \minus{} 1)^{20} \minus{} (ax \plus{} b)^{20} \equal{} (x^2 \plus{} px \plus{} q)^{10}$

$ 2)$ Eστω $ a,b$ δυο διαφορετικες ριζες της $ x^4 \plus{} x^3 \equal{} 1$
Nδο οτι το γινομενο $ ab$ ειναι ριζα της $ x^6 \plus{} x^4 \plus{} x^3 \minus{} x^2 \minus{} 1 \equal{} 0$

$ 3)$ Να βρειτε ενα πολυονυμο $ F(x)$ του μικροτερου δυνατου βαθμου ωστε διαιρουμενο με τα $ (x \minus{} 1)^2$ και $ (x \minus{} 2)^3$ να αφηνει υπολοιπο $ 2x$ και $ 3x$ αντοιστιχα

Καλημέρα

2)Θέτω p=ab,s=a+b Έχω $ 1\plus{}a\equal{}1/a^3 , 1\plus{}b\equal{}1/b^3, a\ne b$ αφαιρώντας κατά μέλη ...$ p\minus{}p^3\equal{}s^2$
Ακόμη $ a^4\minus{}b^4\equal{}b^3\minus{}a^3$... αρα $ (s\plus{}1)(s^2\minus{}p)\equal{}ps$ τότε $ \minus{}p^2((1\plus{}s)\equal{}s$ συνεπώς $ p\minus{}p^3\equal{}p^4/(1\plus{}p^2)^2$ που μετα τις πράξεις είναι η αποδεικτέα
Μια καλή συλλογή στα πολυωνυμα είναι στο συνημμένο (Δες Γ16,Θ11)

[quote="Dimitris X"]
$ 2)$ Eστω $ a,b$ δυο διαφορετικες ριζες της $ x^4 \plus{} x^3 \equal{} 1$
Nδο οτι το γινομενο $ ab$ ειναι ριζα της $ x^6 \plus{} x^4 \plus{} x^3 \minus{} x^2 \minus{} 1 \equal{} 0$
[/quote]

Den eimai sigouros ean einai sosti mia perergi lusi pou mou irthe molis twra
Here it goes:
$ a^4 \plus{} a^3 \equal{} 1$
$ b^4 \plus{} b^3 \equal{} 1$
Etsi $ (ab)^4 \plus{} a^3b^3(a \plus{} b) \plus{} (ab)^3 \equal{} 1$ $ \boxed{1}$
Exoume na deiksoume oti:
$ (ab)^6 \plus{} (ab)^4 \plus{} (ab)^3 \minus{} (ab)^2 \minus{} 1 \equal{} 0$

Apo $ \boxed{1}$:
$ (ab)^6 \plus{} 1 \minus{} a^3b^3(a \plus{} b) \minus{} (ab)^2 \minus{} 1 \equal{} 0 \Leftrightarrow (ab)^6 \minus{} a^3b^3(a \plus{} b) \minus{} (ab)^2 \equal{} 0 \Leftrightarrow a^2b^2(a^4b^4 \minus{} ab(a \plus{} b) \minus{} 1) \equal{} 0$
Diakrinoume loipom duo periptoseis:
$ ab \equal{} 0$ h:
$ (a^4b^4 \minus{} ab(a \plus{} b) \minus{} 1) \equal{} 0$,gi na apodiksoume to zitoumeno arkei ndo to deutero einai adunato.
Theorouyme poluonomo $ F(x)$ ws pros $ a$.
$ F(x) \equal{} x^4b^4 \minus{} x^2b \minus{} xb^2 \minus{} 1 \equal{} 0$ kai arkei na deiksoume oti den exei zeros:
Gia na exei rational zeros prepei eite:

$ 1)$
$ F(1) \equal{} 0\Leftrightarrow b^4 \minus{} b^2 \minus{} b \minus{} 1 \equal{} 0 \Leftrightarrow b^4 \minus{} b(b \plus{} 1) \minus{} 1 \equal{} 0\Leftrightarrow b^4 \minus{} \frac {1}{b^2} \minus{} 1 \equal{} 0 \Leftrightarrow b^6 \minus{} b^2 \minus{} 1 \equal{} 0 \Leftrightarrow (b^2)^3 \minus{} b^2 \minus{} 1 \equal{} 0$,pou den exei rationals zeros

$ 2)$
$ F( \minus{} 1) \equal{} 0\Leftrightarrow b^4 \plus{} b^2 \minus{} b \minus{} 1 \equal{} 0 \Leftrightarrow (b \minus{} 1)(b^3 \plus{} b^2 \plus{} b \plus{} 1) \plus{} b(b \minus{} 1) \equal{} 0 \Leftrightarrow (b \minus{} 1)(b^3 \plus{} (b \plus{} 1)^2) \equal{} 0$
To $ b \equal{} 1$ aporiptetai ara prepei:
$ b^3 \plus{} (b \plus{} 1)^2 \equal{} 0$,Alla $ b^3 \equal{} \frac {1}{b \plus{} 1}$, ara exoume ndo:
$ (b \plus{} 1)^3 \plus{} 1 \equal{} 0 \Leftrightarrow (b \plus{} 2)[(b \plus{} 1)^2 \minus{} (b \plus{} 1) \plus{} 1] \equal{} 0 \Leftrightarrow b \equal{} \minus{} 2$ pou aporiptetai kai auto......
Den eimai sigouros ean einai swsti gi auto parakalo na tin eleksete :roll:


Hi, I'm having trouble solving these problems with variables on both sides

Example:

2z - 9 < 3z + 9

Can someone please post a solution? The math book doesnt explain it.

I think this is too easy to this forum. If you add $-2z-9$ to the both sides of inequality you get $2z-9+(-2z-9)<3z+9+(-2z-9)$ which is equivalent to $-18<z$ or $z>-18$.


Factor: 16a^4-16a^2+4

My work:

4(4a^4 - 4a^2 + 1)

Where do I go from there?

If I let x = a^2, how do I plug and chug to complete the factoring?

Letting $x=a^{2}$ means that we will have a binomial. So plugging that in, we will have this equation $16x^{2}-16x+4$. I'm sure you can do the rest.


symmetry inequalities counting distinguishability search

[quote="Elemennop"]It's not quite nesbitt's.[/quote]

Yes, not Nesbitt's.
[hide="hint"]a) mixing...
b) for you[/hide]

$(\frac{a}{a+b}+\frac{b}{a+b})^2=1$---(1)
$(\frac{b}{b+c}+\frac{c}{b+c})^2=1$---(2)
$(\frac{c}{c+a}+\frac{a}{c+a})^2=1$---(3)

(1)+(2)+(3), then expand LHS and redistribute terms:

$U + V + (\frac{2ab}{(a+b)^2}+\frac{2bc}{(b+c)^2}+\frac{2ca}{(c+a)^2})=3$,
where $U=\frac{a^2}{(a+b)^2}+\frac{b^2}{(b+c)^2}+\frac{c^2}{(c+a)^2}$ and         $V=\frac{b^2}{(a+b)^2}+\frac{c^2}{(b+c)^2}+\frac{a^2}{(c+a)^2}$.
But $\frac{2ab}{(a+b)^2}+\frac{2bc}{(b+c)^2}+\frac{2ca}{(c+a)^2} \leq \frac12+\frac12+\frac12 = \frac 32$ by AM-GM inequality.
This implies $U+V\geq 3-\frac32=\frac32$. ---(*)
By the symmetry of the roles of a, b and c in U and V, the minimum values of U and V must be the same. Denote this commom value by M.
(*) implies $2M\geq \frac32$  and therefore $M\geq \frac34$.
Finally, $U\geq M\geq \frac34$. The equality sign can be attained when $a=b=c$.
$Q.E.D.$

Remark. I guess the inequality can be generalized to $\frac{a^n}{(a+b)^n}+\frac{b^n}{(b+c)^n}+\frac{c^n}{(c+a)^n}\geq \frac{3}{2^n}$ for  all positive integers n > 1. Can anyone give a proof or disproof ?

yes, the way I solve without thinking is:
we can conceptually see that equality is obtained when $a=b=c$. Thus we show:
$(\frac{1}{2})^2+(\frac{1}{2})^2+(\frac{1}{2})^2=\frac{3}{4}$.
done.

oops, forgot the inequality part.
now we just notice that if we keep a=b but make c>a, we get:
$(\frac12)^2+(\frac{a}{a+c})^2+(\frac{c}{c+a})^2$
now we just show that $(\frac{a}{a+c})^2+(\frac{c}{c+a})^2 > \frac12$ when $a>c$.
since $a$ and $c$ are indestinguishable, this is the same as saying $a\not=c$.
so we show that the above equation is really: $\frac{a^2+c^2}{(a+c)^2} = 1 - \frac{2ac}{(a+c)^2}$
WLOG, we can write this as: $1 - \frac{2a^2 + 2an}{(2a+n)^2} > \frac12$ which can be shown conceptually.
Then if all the logical magically actually works the problem is solved.
whatever...

[quote="me@home"]oh and by the way, what is Nesbitt's?[/quote]

$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}$

For the proof search  [color=orange]Nesbitt  [/color]

[quote="luimichael"]Hi! Lovasz,
Would you please explain more for the last step of your proof?
 $\frac{1}{(1+a)^2}+\frac{2}{(1+\frac{1}{\sqrt{a}})^2} \ge \frac34 \Leftrightarrow (a-1)^2 \ge 0$[/quote]

Ok,
\[ \frac{1}{(1+a)^2}+\frac{2}{(1+\frac{1}{\sqrt{a}})^2} \ge \frac34 \Leftrightarrow (a-1)^2 \ge 0  \]
By BSC inequality:
\[ \frac{2}{(1+\frac{1}{\sqrt{a}})^2} = \frac{2a}{(\sqrt{a}+1)^2} \ge \frac{a}{a+1}  \]
Now, expanding the ineq:
\[ \frac{1}{(1+a)^2}+\frac{a}{a+1} \ge \frac34  \]
\[ \Leftrightarrow (a-1)^2 \ge 0  \]
we get the result. :)

[quote="luimichael"]Dear Lovasz,
Thanks a lot for the explanation! :)
But I can't imagine how you figure out the crucial step $f(a,b,c)\geq f(a,\frac{b+c}2, \frac{b+c}2)$. :idea:
I have applied your nice method to solve two similar problems.
Best Regards,
Michael[/quote]

luimichael, that's mixing variables, a famous method. Can anybody give him a link? because I don't have time now!

[quote="Svejk"]$\sum \left( \frac{a}{b+c}\right)^{2}\geq ...$[/quote]
But Svejk, it's $\sum \left( \frac{a}{a+b}\right)^{2}\geq...$  :wink:

Stuff on mixing variable for luimichael, enjoy!
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=113248[/url]

Note that the problem is a special case of :
 Let $a,b,c,d>0$ with $abcd=1$  . Then

$\sum\frac{1}{(1+a)^{2}}\geq 1$

[quote=Lovasz]Let $a,b,c>0$. Prove:
$$(\frac{a}{a+b})^2+(\frac{b}{b+c})^2+(\frac{c}{c+a})^2\ge \frac34$$[/quote]
[url=https://artofproblemsolving.com/community/c6h2376928p19458801]2008 Indonesia[/url]
Let $a,b,c$ be positive real numbers such that $a+b+c=1$. [url=https://artofproblemsolving.com/community/c6h2013496p14121239]Prove that [/url]$$\frac{a^2}{(b+c)^3}+\frac{b^2}{(c+a)^3}+\frac{c^2}{(a+b)^3}\geq \frac98$$
[b]Belarus 2014[/b]
[url=https://artofproblemsolving.com/community/c6h112469p638944]h[/url] [url=https://artofproblemsolving.com/community/c6h303978p1644305]h[/url] [url=https://artofproblemsolving.com/community/c6h506415p2844417]h[/url] [url=https://artofproblemsolving.com/community/c6h1519087p9390576]h[/url]


trigonometry
calculus
calculus computations

Prove that sequence $\{n\sin{n}\}_{n\geq 1}$ is not bounded.

Consider the interval $I_{k}=\left[2\pi k+\frac{\pi}6,2\pi k+\frac{5\pi}6\right].$ Since the length of $I_{k}$ is greater than one (in fact, it's greater than 2), for each $k,$ there is an integer $n_{k}\in I_{k}.$ Then $n_{k}>2\pi k$ and $\sin(n_{k})\ge\frac12,$ so $n_{k}\sin(n_{k})>\pi k.$

Nice solution! Thank you very much  :wink:

It's a cheap and crude solution for a rather weak result; unbounded is easy. One could ask tighter questions that would require more careful arguments. For instance: for each $\epsilon>0,$ is it true that $n\sin n>n-\epsilon$ for infinitely many $n?$


After seeing Silverfalcon's 15 AMC problems posted on the forum, I have decided to dedicate a thread to AMC-12 problems. Please do not post such large quantities of problems like that in the future.

Feel free to post problems and solutions here.  :)

Edit: Yeah, I really found that kind of strange.

[quote="JesusFreak197"]Umm... for one thing, Silverfalcon is a mod, so he's not going to be doing something like this for no reason, and for another, he's posting them one at a time to add them to the directory of Olympiad/Pre-Olympiad problems being created on AoPS.[/quote]

Bingo.

Don't worry. My posts look like spams but they're not. There are some special moderators who were given permission to post problems from contest and add to Olympiad. I'm one of them and along with Miths (and joml88 soon I'm pretty sure), we take problems on USA problems.

No need to panic.  :)

Since this topic is done, I'm going to lock it.


[url=http://bildr.no/view/510515][img]http://bildr.no/thumb/510515.jpeg[/img][/url]

In ii , why is $ E_f$ a interval as $ E_f \subset f([0,1])$ ?
thanks

[quote="mavropnevma"]They just put the cart in front of the horses - the conclusion before the proof.

At point i) it was proven that $ f([0,1]) \subseteq E_f$.
At point ii) they prove that $ E_f \subseteq f([0,1])$ (see the last line there).

Only now, that we know that $ E_f \equal{} f([0,1])$, can we apply the knowledge that $ f([0,1])$ is an interval, because $ f$ is continuous.[/quote]

Thanks ,
Is my question stupid?


function
integration
trigonometry
real analysis
real analysis unsolved

Let $f: R\rightarrow R$ be a continuous function such that

$\int_{0}^{x}f(x-y)\sin y \,dy=0$, for all $x \in R$.

Prove that $f=0$.

This makes a lot more sense if you choose $f$ supported on $[0,\infty)$ and $x>0$. $x<0$ just gives a mirror image depending on values of $f$ for $x<0$, which doesn't allow anything new.

Once we reduce to $x>0$, Laplace transforms look useful. The transform of $\sin$ is never zero, so the transform of $f$ is zero wherever it's defined. Unfortunately, this doesn't help if $f$ grows too fast; we need a different approach to handle that case.

Rewrite the relation as $\int_{0}^{x}f(y)\sin(x-y)\,dy=0$ and differentiate twice:
$\int_{0}^{x}f(y)\cos(x-y)\,dy=0$
$f(x)-\int_{0}^{x}f(y)\sin(x-y)\,dy=0$
By the last line, $f(x)=0$. Since $x$ was generic, $f$ is identically zero.

My solution  is almost the same and  I think that $\sin$ can be replaced by $\sinh$ and the result still holds.


I'm working on a problem and am caught on the last step...I believe that in order to solve it, I have to prove that:

Sum of kS(n, k) = B(n+1)-B(n)

where S(n, k) is a Stirling number of the second kind.

Of course, if this is wrong then I guess I'm on the wrong track.

Thanks!

Thanks. I'm working on the RHS of that last equation and have put B(n+1) and B(n) into their Stirling number equivalents and then combined them into one summation but am having trouble finishing it. Frankly, I just don't know what can be cancelled out/changed into a different form to get it to match the LHS.

[quote="hsiljak"]It is true.

It's known that $ \sum_{k \equal{} 0}^nS(k,n) \equal{} B(n)$, and also that $ S(n,k) \equal{} S(n \minus{} 1,k \minus{} 1) \plus{} kS(n \minus{} 1,k)$. These facts are sufficient to prove your claim that $ \sum_{k \equal{} 0}^nkS(k,n) \equal{} B(n \plus{} 1) \minus{} B(n)$[/quote]

Just to finish this up:

$ S(n \plus{} 1,k) \equal{} S(n,k \minus{} 1) \plus{} kS(n,k)$

If we let $ k \equal{} 1\ldots n$ and sum the equations, we have:

$ \sum_{k \equal{} 1}^nkS(n,k) \equal{} \sum_{k \equal{} 1}^nS(n \plus{} 1,k) \minus{} \sum_{k \equal{} 0}^{n \minus{} 1}S(n,k)$

noting that $ S(n,0) \equal{} 0$ for nonzero $ n$, and $ S(n,n) \equal{} 1$ yields

$ \sum_{k \equal{} 0}^nkS(n,k) \equal{} \sum_{k \equal{} 0}^{n \plus{} 1}S(n \plus{} 1,k) \minus{} \sum_{k \equal{} 0}^nS(n,k)$

$ \sum_{k \equal{} 0}^nkS(n,k) \equal{} B(n \plus{} 1) \minus{} B(n)$


geometry
circumcircle
power of a point
radical axis
geometry proposed

Let $D$ be inside $\triangle ABC$ and $E$ on $AD$ different to $D$. Let $\omega_1$ and $\omega_2$ be the circumscribed circles of $\triangle BDE$ and $\triangle CDE$ respectively. $\omega_1$ and $\omega_2$ intersect $BC$ in the interior points $F$ and $G$ respectively. Let $X$ be the intersection between $DG$ and $AB$ and $Y$ the intersection between $DF$ and $AC$. Show that $XY$ is $\|$ to $BC$.

Let $P=DE\cap BC$. By Menelaus in $ABP,ACP$ with transversals $DG$ and $DF$ respectively, we see that $\frac{XB}{XA}=\frac{DP}{DA}\cdot\frac{GB}{GP}$, and $\frac{YC}{YA}=\frac{DP}{DA}\cdot\frac{FC}{FP}$. Proving that the two LHS's in the expressions above are equal is then equivalent to proving that $\frac{GB}{GP}=\frac{FC}{FP}$, which, in turn, is equivalent to $PF\cdot PB=PG\cdot PC$. This is just another way of saying that the powers of $P$ wrt $\omega_1,\omega_2$ are equal, which is clear because $P$ belongs to the radical axis $DE$ of the two circles.

P.S.

You said $\omega_1$ is the circumcircle of $BDC$. I'm pretty sure, however, that you meant $BDE$, right?

I'm sorry you are right it's $\triangle BDE$  :D

And it was also my solution with menelaos  :roll:


[quote="NguyenDungTN"]$ 8(x^3 \plus{} y^3 \plus{} z^3)^2 \geq 9(x^2 \plus{} yz)(y^2 \plus{} zx)(z^2 \plus{} xy)$ ??????????????[/quote]

Yes, Ive edited it

[quote="FOURRIER"]Prove that:
$ 8(x^3 \plus{} y^3 \plus{} z^3)^2 \geq 9(x^2 \plus{} yz)(y^2 \plus{} zx)(z^2 \plus{} xy)$

Where $ x,y,z$ are arbitary positive reals[/quote]
Very easy and not strong! :(
$ 8(x^3 \plus{} y^3 \plus{} z^3)^2\geq\frac{8}{3}\cdot(x^2\plus{}y^2\plus{}z^2)^3\geq$
$ \geq\frac{1}{3}\cdot(x^2\plus{}y^2\plus{}z^2\plus{}xy\plus{}xz\plus{}yz)^3\geq9(x^2\plus{}yz)(y^2\plus{}xz)(z^2\plus{}xy).$

[quote="arqady"][quote="FOURRIER"]Prove that:
$ 8(x^3 \plus{} y^3 \plus{} z^3)^2 \geq 9(x^2 \plus{} yz)(y^2 \plus{} zx)(z^2 \plus{} xy)$

Where $ x,y,z$ are arbitary positive reals[/quote]
Very easy and not strong! :(
$ 8(x^3 \plus{} y^3 \plus{} z^3)^2\geq\frac {8}{3}\cdot(x^2 \plus{} y^2 \plus{} z^2)^3\geq$
$ \geq\frac {1}{3}\cdot(x^2 \plus{} y^2 \plus{} z^2 \plus{} xy \plus{} xz \plus{} yz)^3\geq9(x^2 \plus{} yz)(y^2 \plus{} xz)(z^2 \plus{} xy).$[/quote]

It was my solution too  :ninja:


Let:$ a;b;c>0;a^3b\plus{}b^3c\plus{}c^3a\equal{}3$.Prove that:
$ \frac{(a\plus{}b)(b\plus{}c)(c\plus{}a)}{a\plus{}b\plus{}c}\plus{}\frac{24}{(a^2b\plus{}b^2c\plus{}c^2a)^2} \ge \frac{16}{3}$

[quote="caubetoanhoc94"]Let:$ a;b;c > 0;a^3b \plus{} b^3c \plus{} c^3a \equal{} 3$.Prove that:
$ \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{a \plus{} b \plus{} c} \plus{} \frac {24}{(a^2b \plus{} b^2c \plus{} c^2a)^2} \ge \frac {16}{3}$[/quote]
We have $ (a^2b \plus{} b^2c \plus{} c^2a)^2 \leq (a^3b \plus{} b^3c \plus{} c^3a)(ab \plus{} bc \plus{} ca)$
and $ 9(a \plus{} b)(b \plus{} c)(c \plus{} a) \geq 8(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)$  :P


Let M be a metric space. Show that M is compact $ \Longleftrightarrow$ to every family (Fx) of closed substes of M in which $ \cap$ Fx = $ \emptyset$, exists a finite number of elements of the family Fx[size=59]1[/size],.., Fx[size=59]k[/size] in which $ \cap$ Fx[size=59]1[/size] $ \cap$ Fx[size=59]2[/size] $ \cap$ ... $ \cap$ Fx[size=59]k[/size] = $ \emptyset$.


Prove or disprove:

$a,b,c,t>0$

$\frac{a+tb}{at+c}+\frac{b+tc}{bt+a}+\frac{c+ta}{ct+b}\ge3$

[quote="blahblahblah"]Please do not spam the forum with so many of these topics (especially some that are obviously false!).  If you want to know whether a certain inequality is true, that is fine, but this is overkill.[/quote]

I don't spam the forum. I just think they are interesting problems.. Some of them are of the same kind, but I think they contains new ideas...

Anyway I have no reasons to spam the forum; and if you think I do so, then sorry.

[quote="spanferkel"]Just put $a+tb=x$ etc... ;) $\Leftrightarrow \frac xz+\frac yx+\frac zy\ge 3$ which is trivial.[/quote]

Yes spanferkel, is very easy!! :P


One of the topic suggestions up for John Conway's lecture is "Angels and Demons."  What the heck is that about?  Dan Brown's novel?

Google is your friend. Googling "Conway angel demon" (no quotes) gave:
http://en.wikipedia.org/wiki/Angel_problem
as the first result. (Well, actually I already knew what the suggestion meant, but I googled anyway to make sure it was easy to find)

Hmm.  It wasn't too easy for me though.   Google didn't show much for "angels and demons," "angels and demons math problem," or "angels and demons mathematics."

That's because it's actually called the "Angles and [i]Devils[/i] game"  :wink:

Lol well it's listed as Angels and Demons, not my fault :roll:


algebra
polynomial
number theory proposed
number theory

If p is a prime number, and d is a positive integer greater than 1,
prove that x^d = a (mod p) has more than one solution for some integer a (we don't have a = 1 (mod p)), if and only if d is a divisor of p-1.

Hint: The reverse direction of this proof is fairly simple, but the forward direction is somewhat difficult.

I think the statement needs to be corrected:
the beginning is fine until ... if and only if gcd(d,p-1) is greater than 1

In other words, the map $ x \mapsto x^d$ is one-to-one iff $ \gcd(d, p\minus{}1) \equal{} 1$.  This is trivial using primitive roots.

t0rajir0u,

could u pls. show details of the demonstration?

Something more general holds:

Let $ G$ be a finite group. Then $ x \mapsto x^d$ is injective/surjective/bijective iff $ \gcd(|G|,d)\equal{}1$.

Thanks.
If f(x) is a polynomial of degree n, prove that f(x)= 0(mod p) has exactly n solutions if and only if f(x) divides x^p -x(mod p).


probability
analytic geometry
geometry
floor function

1. Let $\overline{CD}$ be a line segment of length 6. A point $P$ is chosen at random on $\overline{CD}$. What is the probability that the distance from $P$ to $C$ is smaller than the square of the distance from $P$ to $D$?

2. Let $P$ be a point chosen at random on the line segment between the points $(0,1)$ and $(3,4)$ on the coordinate plane. What is the probability that the area of the triangle with vertices $(0,0)$, $(3,0)$, and $P$ is greater than 2?

3. A real number $y$ is chosen at random such that $0\leq y\leq 100$. What is the probability that $y-\lfloor y\rfloor\geq \frac{1}{3}$?

[hide="1"]

$PC=x$
$PD=6-x$
$PD=x^{2}-12x+36$

$PC<PD^{2}$
$PD^{2}-PC>0$
$x^{2}-13x+36>0$
$(x-9)(x-4)>0$

The root is 4. (9 doesn't matter since it isn't the interval [0,6])the equation goes below 0 at 4. Thus the probablility is $\frac{4}{6}=\frac{2}{3}$

[/hide]

[hide="2"]
The y-value of the point on the line from $(0,1)$ to $(3,4)$ has to be greater than $\frac43$ (base times height must be greater than  2).

$\boxed{\frac89}$

[/hide]

[hide="3"]
You can break it up into 100 identical cases.

$\boxed{\frac23}$
[/hide]


limit
symmetry
real analysis
real analysis unsolved

Let $ y$ and $ z$ tend to $ 1$, then the limit becomes

$ \lim_{x\to 1}\frac{(x+2)^{3}-27x}{x^{3}+2-3x}=\lim_{x\to 1}\frac{x^{3}+6x^{2}-15x+8}{x^{3}-3x+2}$ which is indeterminate so L´Hopital it.

$ =\lim_{x\to 1}\frac{3x^{2}+12x-15}{3x^{2}-3}=\lim_{x\to 1}\frac{6x+12}{6x}=3$

I did a second L´Hopital, but decided not to break the flow, and I don´t know how to do the darn $ =$ with the H over it.

You assumed that $ \lim_{(x,y,z)\rightarrow (1,1,1)}f(x,y,z)=\lim_{x\rightarrow 1}\lim_{y\rightarrow 1}\lim_{z\rightarrow 1}f(x,y,z)$ , and that doesn't have to be true in every situation.

I did know that, but I purposely didn´t state it knowing that my reason for thinking that wasn´t sound. I just went with the guess that it was true since due to the symmetry, any order of taking the limits one at a time wouldn´t change the limit. I know, though, that this doesn´t necessarily mean that the limit even exists. Sorry.

The limit is indeed 3. Some algebra:
We write the fraction as $ 1+\frac{3(x(y-z)^{2}+y(x-z)^{2}+z(x-y)^{2})}{\frac12(x+y+z) ((y-z)^{2}+(x-z)^{2}+(x-y)^{2}}=1+2\frac{3}{x+y+z}\cdot \frac{x(y-z)^{2}+y(x-z)^{2}+z(x-y)^{2}}{(y-z)^{2}+(x-z)^{2}+(x-y)^{2}}$

The last term is a weighted average of $ x,y,z$ with positive weights, so it lies somewhere between the minimum of the three and the maximum. If we take $ x\le y\le z$ WLOG, we get $ 1+\frac{6x}{x+y+z}\le\frac{(x+y+z)^{3}-27xyz}{x^{3}+y^{3}+z^{3}-3xyz}\le 1+\frac{6z}{x+y+z}$. The outer quantities clearly tend to 3 as $ x,y,z\to 1$, and we are done.


function
Support
real analysis
real analysis unsolved

Let $H=L^2[0,1]$ and $L(H,H)$ be the space of linear continious operators from $H$ to $H$. $F_g$ maps $f(x)$ to $f(x+g)$ for any real $g$.
1). Prove that if $g_1 \neq g_2$, then $||F_{g_1}-F_{g_2}|| \geq \sqrt{2}$.
2).Prove that $L(H,H)$ is not separable.

1/ take a funtion $G$ with support $[0; \frac{|g_2-g_1|}{3}]$ such that $|G|_2=1$. Then $|(f_{g_1}-f_{g_2})(G)|_2=\sqrt{2}$
2/ follows directly from 1/ .


find all integer solutions to the following:

$ x^{2}+y^{2}= z^{2}+t^{2}$

i have the exhaustive solution set,but i just have it! i mean i don't know how to Derive it...somewhere i read this can be found in "Diophantine Equations" by Mordell ,but,that is somewhat Advanced number theory...
you know,finding the solutions to $ x^{2}+y^{2}= z^{2}$ is not that advanced; i need something like that...
anyways,any effort is appreciated...

P.S: inform me if it's better be moved to Olympiad Section...

[hide="Lemma 1"] Any odd prime $ \equiv 1 \bmod 4$ can be written as a sum of squares in exactly one way. [/hide]
[hide="Lemma 2"] $ (a^{2}+b^{2})(c^{2}+d^{2}) = (ac \pm bd)^{2}+(ad \mp bc)^{2}$ [/hide]

Thanks guys,but i want to know [b]How to Derive Them.[/b]
besides,the exhaustive solution set is not exactly what you believe...:

$ {x}= \frac{{ac}+{bd}}{2}$
$ {y}= \frac{{ad}-{bc}}{2}$
$ {z}= \frac{{ac}-{bd}}{2}$
$ {t}= \frac{{ad}+{bc}}{2}$

still want answers... :)

sorry for my bad english , :blush: you want to know how we obtain these expressions of $ x,y,z,t$?

[quote="matt276eagles"][quote="K81o7"]The number $ P*Q$ can be written as a sum of two squares in $ 4\tau(P)$, where Q is all of the 3mod4 primestuff and P is all of the 1mod4 primestuff and 2-stuff...[/quote]

I have no idea what you just said.  Could you explain?[/quote]

Sorry, I had a typo there.  The correct statement is below.  The idea is to use the nice behavior of the Gaussian integers and the norm function in that ring.

Ie, if $ n=2^{k}p_{1}^{a_{1}}p_{2}^{a_{2}}...p_{x}^{a_{x}}q_{1}^{b_{1}}q_{2}^{b_{2}}...q_{y}^{b_{y}}$, where all $ p_{i}$ are $ 1\mod{4}$ primes, and all $ q_{j}$ are $ 3\mod{4}$ primes, then the number of ordered pairs $ (c, d)$ of integers such that $ c^{2}+d^{2}=n$ is $ 4\tau(p_{1}^{a_{1}}p_{2}^{a_{2}}...p_{x}^{a_{x}})$.

The simplest way to see this is through the Gaussian integers, $ \mathbb{Z}[i]$.  We want the number of $ c+di$ with norm $ n$ (where the norm of $ c+di$ is $ c^{2}+d^{2}$, ie the number times its conjugate).  First, we have to show that, given $ n$ has norm $ ef$ for some rational integers $ e, f$, there exist $ n_{0}, n_{1}$ with norms $ e, f$ respectively such that $ n_{0}n_{1}=n$.  That means that each Gaussian integer $ c+di$ with norm $ n$ can be factored into Gaussian integers with norms equal to the prime factors.

Remember that if a Gaussian integer has prime norm, then it is a prime Gaussian integer (because norm is multiplicative).  Then, we know if an integer prime $ p$ is factorable in the Gaussian integers, it will be factored into two prime Gaussian integers, each of norm $ p$.  Moreover, once we prove UFD, we will know this factorization is unique.  Clearly, if $ p \equiv 3\mod{4}$, then $ p$ is a prime in the Gaussian integers, because $ -1$ is a quadratic nonresidue mod p.  We can show that any prime $ p \equiv 1\mod{4}$ can be expressed as a sum of two squares.  Down to multiplication by units, there are exactly two numbers then with norm $ p$ in the Gaussian integers.  Then there are $ a_{i}+1$ different Gaussian integers with norm $ p_{i}^{a_{i}}$.  The result then follows (we have to multiply by $ 4$ because we only considered one Gaussian itneger from each possible associate class).

I skimmed over the proof very briefly there...but I wrote a paper on it at PROMYS.  I can send that to you if you want.

[quote="K81o7"]the number of ordered pairs $ (c, d)$ of integers such that $ c^{2}+d^{2}=n$ is $ 4\tau(p_{1}^{a_{1}}p_{2}^{a_{2}}...p_{x}^{a_{x}})$.[/quote]

To clarify, $ \tau(n)$ is the number of divisors of $ n$, which can also be written $ d(n)$ or $ \sigma_{0}(n)$.

$ x^{2}+y^{2}=z^{2}+t^{2}$ $ \Rightarrow $ $ x^{2}-z^{2}=t^{2}-y^{2}$ which is  written $ (x-z)(x+z)=(t+y)(t-y)$

Lemma All the interger solutions of the equation $ xy=zt$ are given by $ x=ac,y=bd,z=ad,t=bc$ where $ a,b,c,d$ are intergers
to prove this just take $ (x,z)=a$ and use simple divisibiliity.

so we have $ x-z=ac,x+z=bd,t-y=ad,t-y=bc$ so we get the expressions of $ x,y,z,t$

WOW,that was Miraculous dear Kostas "the Sorcerer"!! :D
i even did'nt think this could be this easily done...
however,where did you find that lemma?!

thanks anyone...

Well you can find it in some books for elementary number theory.
Moreover we can prove these :

All the interger solutions of the equation $ xy=z^{2}$ are fiven by this : $ x=u^{2}t, y=v^{2}t, z=uvt$ where $ u,v,t$ are arbitrary intergers.

All the interger solutions to the equation $ xy=z^{3}$ are given by $ x=uv^{2}t^{3}, y=u^{2}vw^{3}, z=uvtw$ $ u,v,t,w$ are arbitrary intergers
if you want any explanation let me know it  :)

thanks,i know how to prove the first one...but for the second one i don't have a clue...


how did you think of using the weak bounds like that?

[quote="Altheman"]you can find that the answer is what he gave by partial derivatives....then you just have to reverse the process[/quote]

How to do it?


function
limit
induction
inequalities
algebra proposed
algebra

[quote="Diogene"]I denote : $f(u-) = \lim_{(t\rightarrow u)\& (t<u)}f(t) ,\ f(u+)= \lim_{(t\rightarrow u)\& (t>u)}f(t)$
$f$ is increasing , so $f(0+),f(x-),f(x+)$ exist for all $x\in ]0,+\infty[$
You make $x\rightarrow 0 , y\rightarrow 0, x > y$ in the definition relation ship, and you obtain : $f(0+)-f(0+) = 4\sqrt{f^2(0+)}$, so $f(0+)=0$ .
You make $y\rightarrow 0$ in the definition relation ship, and you obtain :
$f(x+)-f(x-) = 4\sqrt{f(x+)f(0+)} =0$, so $f(x+)=f(x-)$, so $f$ is continous because $f$ is increasing.
By induction you prove $f(nx)=n^2f(x)$ where $n$ is natural.
From $m^2f(1)=f(m)=f(n\frac mn)=n^2f(\frac mn)$ you obtain $f(\frac mn)=\frac{m^2}{n^2}f(1)$ for all $\frac mn \in Q^+$(positive rational numbers)
Finally $f(x)=x^2f(1)$, because of $f$ is continous .
Do you follow me ?:cool:[/quote]
How do you guess to get it in $lim$ or you do it by setting something inseat of $x$ there is some problems that are solved by $log$ or$ln$ or $lg$so i am stopped there I understood but i think i cannot  do in new problems , can you give some advices.
       Abdurashid

[quote="bilarev"]Let $f:\mathbb{R}^+\to\mathbb{R}^+$ be a function that satisfies for all $x>y>0$
\[f(x+y)-f(x-y)=4\sqrt{f(x)f(y)}\]
a) Prove that $f(2x)=4f(x)$ for all $x>0$;
b) Find all such functions.

[i]Nikolai Nikolov, Oleg Mushkarov [/i][/quote]

a) Let $f\left( x \right) = a,f\left( {2x} \right) = b$.
We have $f\left( {3x} \right) - f\left( x \right) = 4\sqrt {f\left( {2x} \right)f\left( x \right)}  \Rightarrow f\left( {3x} \right) = a + 4\sqrt {ab} $
$f\left( {4x} \right) - f\left( {2x} \right) = 4\sqrt {f\left( {3x} \right)f\left( x \right)}  \Rightarrow f\left( {4x} \right) = b + 4\sqrt {a\left( {a + 4\sqrt {ab} } \right)} $
$f\left( {5x} \right) - f\left( x \right) = 4\sqrt {f\left( {3x} \right)f\left( {2x} \right)}  \Rightarrow f\left( {5x} \right) = a + 4\sqrt {b\left( {a + 4\sqrt {ab} } \right)} $
$f\left( {5x} \right) - f\left( {3x} \right) = 4\sqrt {f\left( {4x} \right)f\left( x \right)}  \Rightarrow f\left( {5x} \right) = a + 4\sqrt {ab}  + 4\sqrt {a\left( {b + 4\sqrt {a\left( {a + 4\sqrt {ab} } \right)} } \right)} $
So we have
$\begin{array}{l}
a + 4\sqrt {b\left( {a + 4\sqrt {ab} } \right)}  = a + 4\sqrt {ab}  + 4\sqrt {a\left( {b + 4\sqrt {a\left( {a + 4\sqrt {ab} } \right)} } \right)} \\
 \Leftrightarrow \sqrt {b\left( {a + 4\sqrt {ab} } \right)}  = \sqrt {ab}  + \sqrt {a\left( {b + 4\sqrt {a\left( {a + 4\sqrt {ab} } \right)} } \right)}
\end{array}$
$\begin{array}{l}
 \Leftrightarrow {\left( {\sqrt {b\left( {a + 4\sqrt {ab} } \right)} } \right)^2} = {\left( {\sqrt {ab}  + \sqrt {a\left( {b + 4\sqrt {a\left( {a + 4\sqrt {ab} } \right)} } \right)} } \right)^2}\\
 \Leftrightarrow b\left( {a + 4\sqrt {ab} } \right) = ab + a\left( {b + 4\sqrt {a\left( {a + 4\sqrt {ab} } \right)} } \right) + 2\sqrt {ab} .\sqrt {a\left( {b + 4\sqrt {a\left( {a + 4\sqrt {ab} } \right)} } \right)} \\
 \Leftrightarrow 4b\sqrt b  = b\sqrt a  + 4a\sqrt {a + 4\sqrt {ab} }  + 2\sqrt b .\sqrt {a\left( {b + 4\sqrt {a\left( {a + 4\sqrt {ab} } \right)} } \right)}
\end{array}$
Use assess inequality
If $a > \frac{1}{4}b$ right-hand side > left - hand side, and If $a < \frac{1}{4}b$ right-hand side > left - hand side.
So we have $a = \frac{1}{4}b \Rightarrow f\left( {2x} \right) = 4f\left( x \right)$


Phenolic derivatives containing $ \minus{}\text{CHO}$ group are unaffected in Clemmensen's Reduction.
Carcul did u get my Pm

The explanation of that is perhaps related to the fact that the carbonyl carbon is less positive in those kind of aldehydes.


superior algebra
superior algebra unsolved

1)  Consider the set Z of integers, and let R denote the subset all multiples of 4.  Define addition as ordinary addition in Z, and define multiplication * in R by a*b = ab/4

a.	Show that (R, +, *) is a ring with unity (what is the unity of R?)

b.	Show that the mapping Ø: R → Z defined by Ø(x) = x/4 is an isomorphism

(R,+,*) is a set with two binary operations. b is totally trivial (look at it, put in the definitions), and a follows from b.


Prove that for all $a,b,c$ are non-negative reals and $abc=1$ then the following inequality is true

\[ \frac{1}{(1+a)^3}+\frac{1}{(1+b)^3}+\frac{1}{(1+c)^3}+\frac{5}{(1+a)(1+b)(1+c)} \ge 1  \]

Also, $k=5$ is the best constant for this inequality

\[ \frac{1}{(1+a)^3}+\frac{1}{(1+b)^3}+\frac{1}{(1+c)^3}+\frac{k}{(1+a)(1+b)(1+c)} \ge \frac{k+5}{8}  \]

No, a bit wrong. The idea to my problem is very nice, not as a enconter as you.

Now, let's try to solve the first one. I don't understand what's PVT want to do? :D

The first inequality seems to be nice and hard!It's equivalent to:If $xyz=(1-x)(1-y)(1-z)$ then :$x^3+y^3+z^3+7xyz\geq 1$
  When does the equality occur here,Hungkhtn?

[quote="pvthuan"]Hung, when does equality hold?

When we have imposed a condition, it is easy to think of a number of inequalities but it is not that easy to solve them. Let me give an analogue.

Let $a,b,c$ be positive real numbers such that $abc=1$, prove that
\[ \frac1{(1+a)^2}+\frac1{(1+b)^2}+\frac1{(1+c)^2}+\frac2{(1+a)(1+b)(1+c)}\geq1.  \]

So, as you can see, mine is NOT less beautiful than yours, Hung. :P[/quote]
this is nice, but easier because equality holds when a=b=c. In fact, if i am not wrong, it is easy to prove that if $ab \ge 1$,
$\frac1{(1+a)^2}+\frac1{(1+b)^2}+\frac1{(1+c)^2}+\frac2{(1+a)(1+b)(1+c)} \ge \frac{1}{(1+ab)} + \frac{1}{(1+c)^2} + \frac{2}{(1+\sqrt{ab})^2(1+c)}$
Then it's easy to prove because it is a one-variable inequality.

To keira_khtn, I think the equality holds when a=b -> infinity, c ->0

[quote="hungkhtn"]Also, the solution is very nice.  ;)[/quote]
Yes, we want to know how nice it is. Hung post your solution.

Oh, you should think about it one time, don't always see only solution. I see you post problem here only the result you guess and most if it is the wrong problem which must be repair many times.

OK, I can see that the idea of hung is quite quite different. The inequality looks symmetric but the equality does not hold when $a=b=c=1$. That is hard.

[quote="pvthuan"][quote="soarer"] this is nice, but easier because equality holds when a=b=c. In fact, if i am not wrong, it is easy to prove that if $ab \ge 1$,
$\frac1{(1+a)^2}+\frac1{(1+b)^2}+\frac1{(1+c)^2}+\frac2{(1+a)(1+b)(1+c)} \ge \frac{1}{(1+ab)} + \frac{1}{(1+c)^2} + \frac{2}{(1+\sqrt{ab})^2(1+c)}$
Then it's easy to prove because it is a one-variable inequality.

To keira_khtn, I think the equality holds when a=b -> infinity, c ->0[/quote]

Soarer, I'm afraid it is not true. You should reverse the sign in the second estimate. We should have $(1+a)(1+b)\geq (1+\sqrt{ab})^2$.[/quote]

I think you should check it carefully. I didn't use two estimates and sum them up. It is one inequality not two.

[quote="hungkhtn"]I've edit it. Sorry by my wrong menmory. $7$ replace by $5$.[/quote]
i think you should edit also the fraction $\frac{k+7}{8}$ to $\frac{k+3}{8}$  ;)

For $k=5$, we have to show that $xyz=(1-x)(1-y)(1-z)$ implies $x^3+y^3+z^3+5xyz \geq 1$, where $0<x,y,z<1$.
Let $S=x+y+z$ and $Q=xy+yz+zx$.
We have to show that $2xyz=1-S+Q$ implies $8xyz+ S^3-3SQ \geq 1$.
Substituting $xyz$, we have to show that
$S-1<Q \leq S^2/3$ implies $S^3-4S+3 \geq (3S-4)Q$.
Consider three cases.
(a) If $S \leq 1$, then $S^3-4S+3=(1-S)(3-S-S^2) \geq 0>(3S-4)Q$;
(b) If $1<S<4/3$, then $S^3-4S+3-(3S-4)Q>S^3-4S+3-(3S-4)(S-1)=(S-1)^3>0$;
(c) If $S\geq 4/3$, then $S^3-4S+3-(3S-4)Q \geq S^3-4S+3-(3S-4)S^2/3=(2S-3)^2/3 \geq 0$.
Very nice inequality, Hungkhtn! :lol:

Vasc got very nice and understandable solution, too.  :lol:

I think that the posted problem is easy, isn't it?  :lol:

[quote="silouan"]I think I used wrong my English sorry. Vasc could you post your solution for this

\[ \frac1{(1+a)^2}+\frac1{(1+b)^2}+\frac1{(1+c)^2}+\frac2{(1+a)(1+b)(1+c)}\geq1.  \][/quote]
If we follow Vasc's hint, we will reduce the inequality (by pvthuan) to $c+1\geq c(a+b)$, which (I think) is not true for all $a,b,c$, $abc=1$. The most natural proof is due to Arqady. In Arqady's proof, we just use some identities like
$(1+a)(1+b)(1+c)=2+a+b+c+ab+bc+ca$, and
$X^2+Y^2+Z^2=(X+Y+Z)^2+2(XY+YZ+ZX)$, then we reduce it to
\[ a^2+b^2+c^2\geq3.  \]

How did you solve that problem silouan ask you, Vasc?

Of course, this problem can be solved easier by the way I solve my problem, use substitite.

[quote="Vasc"][quote="pvthuan"] If we follow Vasc's hint, we will reduce the inequality (by pvthuan) to $c+1\geq c(a+b)$, which (I think) is not true for all $a,b,c$, $abc=1$. [/quote]
This last inequality is true. Who guess the hint?[/quote]

Oh, really,  :o Vasc, I tried $c+1\geq c(a+b)$ for one hour but I could not solve it.

The inequality $c+1\geq c(a+b)$ is equivalent to $(1-a)(1-b) \geq 0$.
We may always find two of the numbers $a,b,c$, let those $a$ and $b$, such that the both be either less or larger than 1. That's all.

Oh, that is really a good solution. I had come to $1+ab \geq a+b$, and then $(1-a)(1-b)\geq0$. Thank you Vasc for the line following the inequality.

Can you explain your post a little more? I'm confused what the graph represents exactly :maybe:

[quote="hungkhtn"]Prove that for all $ a,b,c$ are non-negative reals and $ abc \equal{} 1$ then the following inequality is true
\[ \frac {1}{(1 \plus{} a)^3} \plus{} \frac {1}{(1 \plus{} b)^3} \plus{} \frac {1}{(1 \plus{} c)^3} \plus{} \frac {5}{(1 \plus{} a)(1 \plus{} b)(1 \plus{} c)} \ge 1
\]
Also, $ k \equal{} 5$ is the best constant for this inequality
\[ \frac {1}{(1 \plus{} a)^3} \plus{} \frac {1}{(1 \plus{} b)^3} \plus{} \frac {1}{(1 \plus{} c)^3} \plus{} \frac {k}{(1 \plus{} a)(1 \plus{} b)(1 \plus{} c)} \ge \frac {k \plus{} 5}{8}
\]
[/quote]
Finally I got to know that this is TRIVIAL in the sense of the Schur inequality. Do you know how?

[quote="pvthuan"]Soarer, can you be more specific?

I see that you used
\[ \frac1{(1+a)^2}+\frac1{(1+b)^2}\geq\frac1{1+ab}, \;\;\;\;\; \frac1{(1+a)(1+b)}\geq \frac1{(1+\sqrt{ab})^2}.  \]
But the second one should be reversed. Otherwise, could you prove it in your way?

Thuan[/quote]

$\frac{1}{(1+a)^2} + \frac{1}{(1+b)^2} \ge \frac{1}{1+ab}$
$<=> (1+ab)[(1+a)^2+(1+b)^2] - (1+a)^2(1+b)^2\ge 0$
$<=>ab^3+a^3b +1 \ge  a^2b^2 + 2ab $
by AM-GM we have: $ab^3+a^3b \ge  2a^2b^2$
and $ab^3+a^3b + 2 \ge 4ab  $ -> end.

[quote="hungkhtn"]Prove that for all $a,b,c$ are non-negative reals and $abc=1$ then the following inequality is true

\[ \frac{1}{(1+a)^3}+\frac{1}{(1+b)^3}+\frac{1}{(1+c)^3}+\frac{5}{(1+a)(1+b)(1+c)} \ge 1  \]

[/quote]
Demonstracion :


quadratics
modular arithmetic
induction
floor function
binomial coefficients
algebra
special factorizations

Prove that for any positive integer $n$, the number
\[ S_n = {2n+1\choose 0}\cdot 2^{2n}+{2n+1\choose 2}\cdot 2^{2n-2}\cdot 3 +\cdots + {2n+1 \choose 2n}\cdot 3^n \] is the sum of two consecutive perfect squares.

[i]Dorin Andrica[/i]

Whenever I see a sum of alternate binomial coefficients I try to write it as $(X+Y)^n + (X-Y)^n$.  In this case I think we get $S_n = \frac14 [ (2+\sqrt3)^{2n+1} + (2-\sqrt3)^{2n+1} ]$.

Sketch of a solution that I have (I believe) worked out: Everything up to what rgep said is both natural and standard.  Now, write
\begin{eqnarray*}x^2 + (x + 1)^2 & = & \frac {(2 + \sqrt {3})^{2n + 1} + (2 - \sqrt {3})^{2n + 1}}{4} \\
& = & \frac {(1 + \sqrt {3})^{2(2n + 1)} + (1 - \sqrt {3})^{2(2n + 1)}}{2^{2n + 3}}\end{eqnarray*}
The discriminant of the quadratic equation is $ \Delta = \left(\frac {(1 + \sqrt {3})^{2n + 1} + (1 - \sqrt {3})^{2n + 1}}{2^{n}}\right)^2$.
Completing the square, we have $ x_{1,2} = \frac { - 2\pm\frac {(1 + \sqrt {3})^{2n + 1} + (1 - \sqrt {3})^{2n + 1}}{2^{n}}}{4}$ as the solutions to the above quadratic equation.  It suffices to prove that $ x_{1,2}\in\mathbb{Z}$, i.e. $ (1 + \sqrt {3})^{2n + 1} + (1 - \sqrt {3})^{2n + 1}\equiv 2^{n + 1}\pmod{2^{n + 2}}$.  You can prove (by induction for example) that
\begin{eqnarray*}(1 + \sqrt {3})^{2n + 1} + (1 - \sqrt {3})^{2n + 1} & \equiv & \sqrt {3}((1 + \sqrt {3})^{2n + 1} - (1 - \sqrt {3})^{2n + 1}) \\
& \equiv & 2^{n + 1}\pmod{2^{n + 2}}\end{eqnarray*}
The conclusion follows. $ \Box$

I hope that everything is correct. :)

Lemma. If $ n \in \mathbb{N}$ then exist $ (x_n,y_n) \in \mathbb{N}^2$ such that $ (2 \plus{} \sqrt {3})^{2n \plus{} 1} \equal{} 1 \plus{} x_n^2 \plus{} y_n\sqrt {3}$.
If it is true, then $ x_n$ is odd and since we have that $ S_n: \equal{} \frac {(2 \plus{} \sqrt {3})^{2n \plus{} 1} \plus{} (2 \minus{} \sqrt {3})^{2n \plus{} 1}}{4} \equal{} \frac {1 \plus{} x_n^2}{2} \equal{} z^2 \plus{} (z \plus{} 1)^2$ for $ x_n \equal{} 2z \plus{} 1$ then we have only to show that $ x_n$ is integer.
But $ x_0$ and $ x_1$ are integers and $ x_n \equal{} \frac {1 \plus{} \sqrt {3}}{2}(2 \plus{} \sqrt {3})^n \plus{} \frac {1 \minus{} \sqrt {3}}{2}(2 \minus{} \sqrt {3})^n$ satisfies a recurrence of second order in integers.

Dear rgep, boxedexe, bboypa , I'm sorry for my silly but I don' know how to get this :(
[quote="rgep"]$ S_n \equal{} \frac14 [ (2 \plus{} \sqrt3)^{2n \plus{} 1} \plus{} (2 \minus{} \sqrt3)^{2n \plus{} 1} ]$.[/quote]
Could you make it clearer ? Thanks :)

What happens if you multiply $ S_n$ by $ 2$ and treat $ 3^k$ as $ (\sqrt{3})^{2k}$? (Hint: Use Newton's Binomial Theorem).  You can also determine the original thought process by expanding $ \frac{1}4 [ (2\plus{}\sqrt3)^{2n\plus{}1}\plus{}(2\minus{}\sqrt3)^{2n\plus{}1}]$ using Newton's Binomial Theorem, and determining what cancels in the process and how only terms with $ \binom{2n\plus{}1}{2k}$ remain.  By going through this, you [i]may[/i] learn how to use roots of unity to determine sums with forms resembling $ \sum_{k\equal{}0}^{\lfloor\frac{n}{m}\rfloor}\binom{n}{km}x^ky^{n\minus{}km}$, $ n\geq m>0$.

[quote="forevercontinue"]No one help me ? :(  Just need to give me some hints or some links to further explain . Thanks.[/quote]

Let $ S_n\equal{}\sum_{k\equal{}0}^{n}\binom{2n\plus{}1}{2k}2^{2n\minus{}2k}3^k$

$ 2S_n\equal{}\sum_{k\equal{}0}^{n}\binom{2n\plus{}1}{2k}2^{2n\plus{}1\minus{}2k}(\sqrt 3)^{2k}$
Let $ T_n\equal{}\sum_{k\equal{}0}^{n}\binom{2n\plus{}1}{2k\plus{}1}2^{2n\plus{}1\minus{}(2k\plus{}1)}(\sqrt 3)^{2k\plus{}1}$

$ 2S_n\plus{}T_n\equal{}\sum_{k\equal{}0}^{2n\plus{}1}\binom{2n\plus{}1}{k}2^{2n\plus{}1\minus{}k}(\sqrt 3)^{k}$ $ \equal{}(2\plus{}\sqrt 3)^{2n\plus{}1}$
$ 2S_n\minus{}T_n\equal{}\sum_{k\equal{}0}^{2n\plus{}1}\binom{2n\plus{}1}{k}2^{2n\plus{}1\minus{}k}(\minus{}\sqrt 3)^{k}$ $ \equal{}(2\minus{}\sqrt 3)^{2n\plus{}1}$

And so $ S_n\equal{}\frac 14((2\plus{}\sqrt 3)^{2n\plus{}1}\plus{}(2\minus{}\sqrt 3)^{2n\plus{}1})$

Thanks boxedexe and pco too much . I appreciate your help.:clap2:


Prove that the root of x(x+1)(x+2)...(x+1981) = 1 is less than 1/1981! .

Suppose that x is a positive real root (otherwise there is nothing to do), then :
x = 1/[(x+1)(x+2)...(x+1981)] =< 1/1981!...

Pierre.


What is the least positive integer value for $ n$ such that $ 0.5^n<0.001$?

Let's first change this into fraction form:
\[ \left(\frac {1}{2}\right)^n < \frac {1}{1000}\]
We know that $ \left(\frac {1}{2}\right)^{10} \equal{} \frac {1}{1024} < \frac {1}{1000} < \frac {1}{512} \equal{} \left(\frac {1}{2}\right)^{9}$
Therefore, our answer is $ \boxed{10}$


im pretty sure that all numbers can be written in every other base. that is rational numbers, at least.

[quote="aufha"]Can we always determine if a fraction has a different base representation? If so, how? For example, how do we know that 1/10 cannot be expressed in binary?

I don't know the answer. Can someone explain to me, please?[/quote]
If a number is expressed in a base other than base ten, it is notated like this
$ \left(\frac{1}{10}\right)_{_}2$ for Binary (but that looks wierd)

Or maybe it would be like this: $ \frac{1_{2}}{10_{2}}$

[quote="aufha"]Can we always determine if a fraction has a different base representation? If so, how? For example, how do we know that 1/10 cannot be expressed in binary?

I don't know the answer. Can someone explain to me, please?[/quote]

Yes, all fractions have a base-representation.

Now, to your example, $ \frac{1}{10}\equal{}\frac{1}{2}\cdot\frac{1}{5}\equal{}0.1_{2}\cdot\frac{1}{5}$. Since 2 and 5 share no common factors, and since 5 will never divide into a power of 2, we know that 1/5 will be repeating, so we must divide/ Now we divide in binary $ 1_{2}/101_{2}$ as such:

[code]
     __.0011_________
101|1.00000000
      0.101
      -----
        .0110
        .0101
        -----
             1
[/code]
Since we have 1 again, we are repeating from here onwards and thus $ \frac{1}{10}\equal{}0.1_{2}\cdot0.\overline{0011}_{2}\equal{}0.0\overline{0011}$

This is often be the simplest way to find the decimal forms of such numbers (I could have made a computation error, but still thats the way), or try making an infinite geometric series to express the fraction. The point is, all rational numbers can be expressed as decimals.

[quote="archimedes1"]Do you mean expressed with a terminating decimal?[/quote]

[hide="If yes"] $ \frac{1}{p}$ can be expressed in base $ q$ with a terminating decimal if and only if all prime factors of $ p$ are also prime factors of $ q$.  For example, when $ q \equal{} 10$, $ p$ can only have the prime factors $ 2, 5$.  The "if" is fairly clear to see, since if the decimal terminates you can just multiply by $ q^{n}$ for large enough $ n$ and you'll get a fraction with $ q^{n}$ on the bottom which can be simplified.  The "only if" requires some fun number theory; see the other hide tag :) [/hide]
[hide="If no"] We are going to deal with a subcase that is pretty much equivalent to the general statement.

[b]Lemma:[/b]  If $ gcd(p, q) \equal{} 1$, then $ \frac{1}{p}$ has a repeating decimal expansion in base $ q$.

Proof:  By Euler's Totient Theorem, $ p | q^{\varphi(p) }\minus{}1$.  Let $ \frac{q^{\varphi(p)}\minus{}1}{p}\equal{} r$.  Then

$ \frac{1}{p}\equal{}\frac{r}{q^{\varphi(p) }\minus{}1}\equal{}\sum_{k\equal{}1}^{\infty}\frac{r}{q^{k\varphi(p)}}$

The last step is just the familiar sum of a geometric series.  This will make more sense with a concrete example, say $ q \equal{} 10$.  Let's pick $ p \equal{} 7$; that's a meaty example.  Euler's Totient Theorem tells us $ 7 | 10^{6}\minus{}1 \equal{} 999999$; in fact, $ \frac{999999}{7}\equal{} 142857$.  Then

$ \frac{1}{7}\equal{}\frac{142857}{999999}\equal{}\sum_{k\equal{}1}^{\infty}\frac{142857}{ 10^{6k}}\equal{} 0.\overline{142857}$

:D [/hide]

[quote="aufha"]I made a mistake- I wasn't clear about what I meant. On a worksheet for homework I got in computer science, I got this question:

"Can the decimal number 1/10 be expressed in the binary system?"

My teacher, Mr. Steuben, says it can't. I imagine he means a terminating decimal like archimedes1 mentioned. Can 1/10 be expressed as a fraction in a different base? How do I know 1/10 can't be expressed as a terminating decimal?[/quote]

It can be expressed in any base that's a multiple of 10, though that isn't really applicable to computer science.

Also, computers can't hold infinite precision ... it will have to truncate the repeating decimal somewhere. And, since computers work in binary, it means that 1/10 is not [i]exactly[/i] 1/10 when stored in a computer. So, in the context of your computer science class, your teacher is right.

Usually, computer systems are designed so it's not really a problem, but I remember there being some missile defense system or something where the round-off error accumulated from incrementing the time by 1/10 of a second each 1/10 second resulted in the complete system failing when it was actually needed.

Another example is old pocket calculators: evaluating out 1/3 is 0.333333333. if you do 1/3+1/3+1/3 (assuming it had order of operations) you might get 0.999999999 instead of the correct answer. Many new calculators fix this problem by storing some extra digits past the precision that is displayed on the screen, and then rounding the answer. By storing just one extra digit in the above example, the answer would be correctly rounded to 1.

The notation $ 0b$ denotes a binary number. $ \frac{1}{10}$ is $ \frac{0b1}{0b101}$. Dividing by a binary can be said to be a re-expression of the digits in the divisor through a form of the dividend. For instance, the number $ \frac{0b110}{0b10}$ from the denominator can be re-expressed as $ 0b01$ (make sure not to drop the 0's). Then take the first $ 1$ from the right in the divisor and re-express it as the dividend from right to left starting from the $ 1$ that you are evaluating. (In this case the $ 0b1\underline{10}$ is $ 0b01$ or just $ 0b1$). Do the same with all the other $ 1$'s, so you get $ 0b010$ or just $ 0b10$. Sum the values, $ 0b1\plus{}0b10\equal{}0b11$. Unfortunately, I don't know what happens when you get decimal binaries (like $ \frac{0b1}{0b10}$).

[quote="aufha"]How do I know 1/10 can't be expressed as a terminating decimal?[/quote]

I already answered this question in my previous post ;)

[quote="t0rajir0u"][quote="archimedes1"]Do you mean expressed with a terminating decimal?[/quote]

[hide="If yes"] $ \frac{1}{p}$ can be expressed in base $ q$ with a terminating decimal if and only if all prime factors of $ p$ are also prime factors of $ q$.  For example, when $ q \equal{} 10$, $ p$ can only have the prime factors $ 2, 5$.  The "if" is fairly clear to see, since if the decimal terminates you can just multiply by $ q^{n}$ for large enough $ n$ and you'll get a fraction with $ q^{n}$ on the bottom which can be simplified.  The "only if" requires some fun number theory; see the other hide tag :) [/hide][/quote]


logarithms
inequalities unsolved
inequalities

It's equivalent to $ S \equal{} \sum a_i \log a_i \geq \minus{} \log n$.

$ x \log x$ is convex, so $ S \geq n \left(\frac {\sum a_i}{n} \right) \log \left( \frac {\sum a_i}{n} \right) \equal{} n \cdot \frac {1}{n} \log \frac {1}{n} \equal{} \minus{} \log n$

Or use weighted GM-HM. (Or weighted power mean)
The result follows immediately by letting a_1, a_2 etc... be the weights.


analytic geometry
geometry
incenter
geometry theorems

Are there some properties of barycentric coordinate which about perpendicularity, angle, or circle?
If there aren't, how can we find Feuerbach point(or some special points)'s barycentric coordinate?

hihi
Mr. Shin uploaded some stuff related to bary coords at cafe.naver.com/manifold

Here I can give you a few barycentric coordinates:
$ G(\frac{1}{3},\frac{1}{3},\frac{1}{3})$-the centroid
$ I(\frac{a}{2p},\frac{b}{2p},\frac{c}{2p})$-the incenter(a,b,c sides,p-semiperimeter)
$ I_a(\frac{a}{2(p\minus{}a)},\frac{b}{2(p\minus{}a)},\frac{c}{2(p\minus{}a)})$ the exincenter
$ H(\cot C\cot B,\cot C\cot A,\cot A\cot B)$ the ortocenter....and many others...you cand also find many of these on wiki


Does anyone know what the qualifying score is for the AMC 10 invitation?

I think that it is open to everyone...

I remember that I got a 24 last year, but I don't remember getting an invitation, but I might have forgotten about it.

from what I've heard it is 120, and 119 or 119.5 if you are really lucky.

Wait, let me rephrase my quetion, What do you have to get on the AMC 8 to get an invitation to the AMC 10. I'[m sure u understood me though.

i would think any high score.

I took the amc 8 also, got a 23.

AMC 10 is open to everyone, but there's just this special invitation that high-scorers on the AMC 8 get.

[quote="Ceva Theorem"]To qualify for the AIME, you need at least a score of 120 or get in the top 1% for AMC 10.
AMC 12- get at least a score of 100 or get in the top 5%
They made it harder this year becuase they only give 1.5 points if you leave an answer blank, it used to be 2.5.... :ninja:[/quote]

Sorry for the offtopicness - I think it should be 1.5.  I remember people talking about just doing 11 problems and making AIME.

Yes, you do not need an invitation.  Just about 3 dollars will do.

Yeah, don't worry about the invitation; just take the AMC 10 anyhow. I think as a coach that all of my honor roll or merit roll awardees on the AMC 8 have been invited to take the AMC 10, and that is usually a score of 19 or higher.

just a 19? everyone above that invited to amc 10?

well.... woopty doo!

But, again, you should take the AMC 10 if you feel ready for it, whether you are invited to take it or not (indeed, whether you took the AMC 8 or not).

in my old middle school, only one person ( the top scorer out of all the participants) gets invited to take AMC10. whcih is odd ..... :o

ok. thanks. Yeah I think we should all take the AMC 10 even if it's just for experience.

[quote="mathgeniuse^ln(x)"]Well...I am not one of those people...

And no you don't need to have 3 dollars...if your school pays for it...

and yeah, take both...

Qualifying score for AMC 10...usually around 20/21 on the AMC 8...[/quote]
isn't just on a school-wide basis?
i got a 20 and nothing happened

oh well

should i call up the high school in my town and ask if i can take the amc 10? or how should i do it?'
-jorian

I thought someone cleared this up but some people are still talking about "qualifying scores." So, once and for all, [b]there is no qualifying score for the AMC 10[/b]. You can take it if you get a 0 on the AMC8.


[quote="warut_suk"]Is there an easy way to do
$\frac{x-1}{x-2} + \frac{x-2}{x-3} = \frac{x-3}{x-4} + \frac{x-4}{x-5}$[/quote]

If I didn't screw it too much, the solutions are $x=7\pm\sqrt{26}$, but I suppose you are more worried about the procedure than about the answer itself.

Note if $f(x)=1-\frac1x$, then, the equation that you have is $f(x-2)+f(x-3)=f(x-4)+f(x-5)$, but then $[1-f(x-2)]+[1-f(x-3)]=[1-f(x-4)]+[1-f(x-5)]$ (just pluggin), but this mean that

$\frac{1}{x-2} + \frac{1}{x-3} = \frac{1}{x-4} + \frac{1}{x-5}$

And this is quite easier to solve. According to my calculations (that I would verify in your possition, I am tired, just about going to bed, and I don't think clearly at midnight when I woke up at 5 am  :P ) the solutions are the ones I said.

Best regards (and good nights),


A quick idea: What if, instead of emails like

[quote]Hello darij grinberg,

You have received a new private message to your account on the MathLinks Forum and you have requested that you be notified on this event. You can view your new message by clicking on the following link:

http://www.mathlinks.ro/Forum/privmsg.php?folder=inbox

Remember that you can always choose not to be notified of new messages by changing the appropriate setting in your profile.

AoPS - MathLinks Administrators[/quote]

one would get a copy of the private message (the source code is enough, no need for LaTeX formulas as graphics) to one's email address when receiving a PM on MathLinks with the email notification option being set on? Would help keeping some order in one's tasks, having both PMs and regular emails in one (email) account.

(I didn't appreciate this until I changed to a Gmail account, which really allows to keep things in order...)

  Darij

Then what if someone repents of what they said and deletes the unread private message? then it's already at the recipient... I have done this before.

[quote="solafidefarms"]Then what if someone repents of what they said and deletes the unread private message? then it's already at the recipient...[/quote]

Actually I see no use in this... What if the recepient goes online quickly after the message has been sent and reads it though? ;)

  Darij

Sorry Darij, but for the reasons solafidefarms already explained this feature will not be implemented.


Din pacate, nu stiu cine iti va raspunde in urmatorii ani la intrebarea asta si chiar daca cineva va raspunde vreodata, cu siguranta nu vei intelege demonstratia. Doar ca amuzament, incearca sa intelegi demonstratia faptului ca in cazul in care exista o infinitate de numere prime gemene, atunci suma inverselor lor converge. Vei vedea ca nu e simpla!

[quote="harazi"] in cazul in care exista o infinitate de numere prime gemene, atunci suma inverselor lor converge [/quote]
nu doar in cazul in care exista o infinitate :P.

Fie pn sirul numerelor prime: p1=2,p2=3,..etc -in ordinea lor  crescatoare; atunci  pentru orice n natural>=1 avem
p1*p2*p3*....*pn-1 si p1*p2*p3*....*pn+1 sunt numere prime,formand o perche de forma (p,p+2) ;mai mult penru doua numere m<n naturale perechile (p1*p2*...*pm-1,p1*p2*....*pm+1) si (p1*p2*..*pn-1,p1*p2*...*pn+1) sunt distincte.Din cate am inteles asta trebuia demonstrat.

E clar ca nu ai inteles ceea ce trebuia :)

[quote="rockymarcianocosmin99"]Numim numere gemene o pereche de numere prime de forma(p,p+2).Aratati ca exista o infinitate de astfel de numere.[/quote]

Poti sa ne arati solutia ta?  :P  http://mathworld.wolfram.com/TwinPrimeConjecture.html

[quote="rockymarcianocosmin99"]Numim numere gemene o pereche de numere prime de forma(p,p+2).Aratati ca exista o infinitate de astfel de numere.[/quote]
     Daca vrei sa afli un raspuns, atunci invoca spiritul lui Riemann si poate ti-o face el.


inequalities
number theory unsolved
number theory

Let x and y be positive integers such that xy divides x^2 +y^2 +1.Prove that
                (x^2+y^2+1)/xy=3

See http://www.mathlinks.ro/Forum/viewtopic.php?t=77730
and http://www.mathlinks.ro/Forum/viewtopic.php?t=18491
for the method.

Thank you very much,I really appreciate your kindess

$x^{2}+y^{2}+1-kxy = 0$ where $k \ne 3$  and k is a positive integer .so $x \ne y$ .using AM-GM inequality it is obvious that $k > 3$  . take ${(X,Y})$ an answer to tour equation which $max(x,y)$ is the smallest.we can suppose $X>Y$ .but another answer to the above equation is $(\frac{Y^{2}+1}{X}),Y^{2})$  but $max(\frac{Y^{2}+1}{X}),Y^{2}) < max(X,Y)$ so the equation won't have an answer which $k \ne 3$

but i think that the equation have infinitly many solution in $x;y$. there is generalisation for this problem.called $markove equation$. here you are :
the eguation $a^{2}+b^{2}+c^{2}=Kabc$. has solution in natural numbers only for $K=1;3$.wich we can prove by infinit descente :roll:

Guys, I think this problem has to be named the most famous equation in ML :lol: . In fact, the links proposed by me@home are only a small part of the group of topics were this problem was posted :wink: .

Yes I am a relatively new user :oops: could you point me to some of those other topics?


Let ABC be a triangle. D, E, F lie on BC, CA, AB respectively, and AD, BE, CF are concurrent. M, N, P lie on EF, DF, DE respectively, and MD, NE, PF are concurrent.
Prove that AM, BN, CP are concurrent.
By using Ceva theorem, you can solve it. Am I right?

The all segments are orientated.

Denote $U\in BC\cap AM$, $V\in CA\cap BN$, $W\in AB\cap CP$.

$\blacktriangleright\ 1^{\circ}.\ AD\cap BE\cap CF\not =\emptyset\Longleftrightarrow \frac{DC}{DB}\cdot \frac{EA}{EC}\cdot \frac{FB}{FA}=-1$

$\blacktriangleright\ 2^{\circ}.\ DM\cap EN\cap FP\not =\emptyset\Longleftrightarrow \frac{MF}{ME}\cdot \frac{ND}{NF}\cdot \frac{PE}{PD}=-1$

$\blacktriangleright\ 3^{\circ}$. Lemma from http://www.mathlinks.ro/Forum/viewtopic.php?t=71064 $\Longrightarrow$

$\blacktriangleright\ 3.1.\ \frac{UB}{UC}=-\frac{AB}{AC}\cdot \frac{MF}{ME}\cdot \frac{AE}{AF}$

$\blacktriangleright\ 3.2.\ \frac{VC}{VA}=-\frac{BC}{BA}\cdot \frac{ND}{NF}\cdot \frac{BF}{BD}$

$\blacktriangleright\ 3.3.\ \frac{WA}{WB}=-\frac{CA}{CB}\cdot \frac{PE}{PD}\cdot \frac{CD}{CE}$

From the product of the above five relations results

$\frac{UB}{UC}\cdot \frac{VC}{VA}\cdot \frac{WA}{WB}=-1$, i.e. $AM\cap BN\cap CP\not =\emptyset.$

[quote="phuchung"]Let ABC be a triangle. D, E, F lie on BC, CA, AB respectively, and AD, BE, CF are concurrent. M, N, P lie on EF, DF, DE respectively, and MD, NE, PF are concurrent.
Prove that AM, BN, CP are concurrent.[/quote]

See
http://www.mathlinks.ro/Forum/viewtopic.php?t=71071 post #2
and http://www.mathlinks.ro/Forum/viewtopic.php?t=6579 .

  darij


Find $ f : (0,\plus{} \infty )  \to R$ satisfying
$ \large f(\sqrt[k]{\frac{x^{k}\plus{}y^{k}}{2}})\plus{}\large f(\sqrt[k]{\frac{2}{\frac{1}{x^{k}}\plus{}\frac{1}{y^{k}}}})\equal{}f(x)\plus{}f(y)$ $ \forall x,y >0$ where $ k >1$

It has property continous ?
If has only condition then it has many solution.

If f is continiosly, then it is easy to show, that $ f\equiv const$.

[quote="nedo477"]I've seen it before, f is a continuous function.[/quote]

Where is it ,nedo477? Please don't tell me it from VMF or MNF  :D


Prove that if $ x\plus{}y\plus{}z\equal{}6$ and $ xy\plus{}yz\plus{}zx\equal{}9$ ($ x,y,z\in\mathbb{R}$) then $ 0\le x,y,z\le 4$.

This problem is not true. Although we can edit it.
Let $ m \equal{} xyz$
$ x,y,z$ is 3 root of the equation :
$ f(x) \equal{} x^3 \minus{} 6x^2 \plus{} 9x \minus{} m$
$ f'(x) \equal{} 3(x \minus{} 1)(x \minus{} 4)$
Consider sign of $ f'(x)$

TTsphn, your factorization is wrong. $ f'(x) \equal{} 3(x\minus{}1)(x\minus{}3)$. The two equality cases do in fact use the values $ 1$ and $ 3$: two variables $ 3$ and the other $ 0$ give the minimum, and two variables $ 1$ and the other $ 4$ give the maximum.


modular arithmetic
superior algebra
superior algebra unsolved

Prove that the ring Z/2^nZ has no idempotents except [0] and [1]. (hint: for integers b, consider when b^2-b can be divisible by 2^n.)

If $ x^\equiv x\pmod 2^n$, then $ 2^n|x(x\minus{}1)$,  because $ gcd(x,x\minus{}1)\equal{}1$, so $ 2^n|x$ or $ 2^n|x\minus{}1$. So $ x\in\{0,1\}$


geometry
incenter
projective geometry
circumcircle
search
function

[quote="nguyentrang"]Given a triangle $ ABC$, let $ k$ be its circumcircle. Let $ AD$ be the bisector of angle $ BAC$ ($ D\in BC$). Let $ l$ be the circle which is internally tangnet to $ k$ and tangent to segments $ AD,BD$. If $ E$ is the point of tangency between $ l$ and $ AD$, prove that $ E$ is the incenter of triangle $ ABC$.[/quote]
[size=134][color=green]Lyness open[/color][/size] theorem:
$ ABCD$ is a quadrilateral incribed $ (O)$,Let $ (K)$ is internally tangency with $ AC,BD,(O)$ at $ F,E,M$.$ I_1;I_2$ is incenter of triangle $ ADC,BDC$
so $ E,F \in EF$
[img]http://i209.photobucket.com/albums/bb172/Ineq/mathlinksro.jpg[/img]
Apply we have design  :D

[quote="nguyentrang"]Given a triangle $ ABC$, let $ k$ be its circumcircle. Let $ AD$ be the bisector of angle $ BAC$ ($ D\in BC$). Let $ l$ be the circle which is internally tangnet to $ k$ and tangent to segments $ AD,BD$. If $ E$ is the point of tangency between $ l$ and $ AD$, prove that $ E$ is the incenter of triangle $ ABC$.[/quote]
See here: [url]http://www.mathlinks.ro/viewtopic.php?p=831617#831617[/url].

[quote="Clark Kent"][size=134][color=green]Lyness open[/color][/size] theorem:
$ ABCD$ is a quadrilateral incribed $ (O)$,Let $ (K)$ is internally tangency with $ AC,BD,(O)$ at $ F,E,M$.$ I_1;I_2$ is incenter of triangle $ ADC,BDC$
so $ E,F \in EF$[/quote]
This is far from being an open problem at the moment.
See Lemma 1 from Ayme's beautiful proof of Thebault's theorem: [url=http://forumgeom.fau.edu/FG2003volume3/FG200325index.html]Jean-Louis Ayme, Sawayama and Thébault's Theorem, Forum Geometricorum 3 (2003) 225--229.[/url]

[quote="pohoatza"]
[quote="Clark Kent"][size=134][color=green]Lyness open[/color][/size] theorem:
$ ABCD$ is a quadrilateral incribed $ (O)$,Let $ (K)$ is internally tangency with $ AC,BD,(O)$ at $ F,E,M$.$ I_1;I_2$ is incenter of triangle $ ADC,BDC$
so $ E,F \in EF$[/quote]
This is far from being an open problem at the moment.
See Lemma 1 from Ayme's beautiful proof of Thebault's theorem: [url=http://forumgeom.fau.edu/FG2003volume3/FG200325index.html]Jean-Louis Ayme, Sawayama and Thébault's Theorem, Forum Geometricorum 3 (2003) 225--229.[/url][/quote]
This problem was posted by me at this forum several months ago and i know at least four solutions,one of them was found by April, i found 2 solutions(one of them with Pascal theorem) and another one was given in Prasolov book. :)
I will try to find the link...

[quote="nguyentrang"]How to prove your statement?

And how to use it with my problem?[/quote]
[img]http://i209.photobucket.com/albums/bb172/Ineq/hay-8.jpg[/img]
$ AD\cap k\equal{}F$
$ E,M$ is tangency of $ l$ with $ AD,BD$.Follow Lyness open theorem we have incenter of $ ABC$ lie on $ EM$,and it lie on $ AD$ too
So $ E$ is incenter of $ ABC$
[hide]@nguyentrang:Còn về lời giải của định lý Lyness mở rộng bạn có thể đọc toán tuổi thơ 2 số 42 bài viết của thầy Minh Hà  :D  [/hide]

[quote="pohoatza"]
See Lemma 1 from Ayme's beautiful proof of Thebault's theorem: [url=http://forumgeom.fau.edu/FG2003volume3/FG200325index.html]Jean-Louis Ayme, Sawayama and Thébault's Theorem, Forum Geometricorum 3 (2003) 225--229.[/url][/quote]

I don't understand how they apply (the converse of) Pappus at the very last step. Anybody please help...

[quote="Erken"][quote="hữu quân"]we have the problem :

let triangle $ ABC$ .$ (O)$ be circumcenter,$ D$ is point on segmen$ BC$ .$ (O1)$ tangent $ [BC] ,[AD]$ and $ (O) .(O2)$ tangent $ [DC] ,[AD]$ and $ (O)$ . prove that $ O1,O2,I$ colinear[/quote]
Please post your problem in separate topic.[/quote]
Don't post this. This is Thebault's theorem. It has been posted many times before. Please use search function (and also read Ayme's paper).

@nguyentrang: There are some small typos in Ayme's article: It should be $ GF \| DP$, and $ DQ \| EH$ (Apparently the points $ F$, and $ H$ are mixed up in the third section of the paper). In this case, the conclusion follows from the converse of Pappus' theorem, applied in $ PEIGQD$, because $ PE \cap GQ \equal{} \infty$, $ EI \cap QD \equal{} \infty$, and $ IG \cap DP \equal{} \infty$, these three "intersections" being collinear on the line at infinity).

Since we are still talking about Thebault's theorem, try to prove the particular result from: [url]http://www.mathlinks.ro/viewtopic.php?p=1110205#1110205[/url].

Nice problem!  Using Clark Kent's diagram...

Let this circle be tangent to the circumcircle of $ \triangle{ABC}$ at $ P$.  The homothecy about $ P$ which maps the circle in question to the circumcircle maps $ M$ to $ F$.  Therefore, $ P$, $ M$, and $ F$ are collinear.

Now $ m\angle{MBF}\equal{}m\angle{BPF}$, so $ \triangle{FBM} \sim \triangle{FPB}$.  Thus, $ FM \cdot FP\equal{}FB^2$, so $ FB\equal{}FE$, by Power of a Point.  But $ FB\equal{}FI$, so we're done.


prove :D

$ \sum_{k=1}^\infty\;\frac{\zeta{(2k+1)}}{\left(2k+1\right)\cdot 2^{2k+1}}\;=\;\boxed{\frac{1}{2}\left(\ln2-\gamma\right)}$

If my calculation is correct, then

$ \sum_{n=1}^{\infty}\frac{\zeta(2n+1)}{2n+1}x^{2n+1}\; = \; \sum_{n=1}^{\infty}\left[ \frac{1}{2}\left( \ln(k+x)-\ln(k-x) \right)-\frac{x}{k}\right].$

Plugging $ x=\frac{1}{2}$ gives the answer.


geometric inequality
geometry
circumcircle
inradius
trigonometry

An acute triangle $ABC$ has circumradius $R$, inradius $r$. $A$ is the biggest angle among $A,B,C$. Let $M$ be the midpoint of $BC$, and $X$ be the intersection of two lines  that touches circumcircle of $ABC$ and goes through $B,C$ respectively. Prove the following inequality : $ \frac{r}{R} \geq \frac{AM}{AX}$.

It's well-known that $AX$ is the symmedian of $ABC$. It's easy to show by an angle chase that if we construct the circle centered at $X$ of radius $XB=XC$ then it cuts the rays $(AB,\ (AC$ in points $B',\ C'$ s.t. $X\in [B'C']$ and $B'C'$ is antiparallel to $BC$ wrt the angle $BAC$. In this case the triangles $ABC,\ AC'B'$ are similar (in the specified prder of vertices) and $\frac{AM}{AX}=\frac{BC}{C'B'}=\frac{BC}{2XB}=\cos{\angle XBC}=\cos A$.

I didn't continue, but this certainly looks friendlier: we have to show that $\frac rR \geq \min (\cos A,\ \cos B,\ \cos C)$.

Proceed as Grobber does to reduce the inequality to $\frac rR \geq \min\left(\cos A, \cos B, \cos C\right)$.
Let $D, E$ and $F$ be the feet of the perpendiculars from $O$ to $BC, CA$ and $AB$ respectively.  If $\min\left(OD, OE, OF\right) > r$, then

$2\left[ABC\right]=OD\cdot BC+OE\cdot CA+OF\cdot AB \geq \min\left(OD, OE, OF\right) \left(AB+BC+CA\right) > r\left(AB+BC+CA\right)=2\left[ABC\right]$,

a contradiction.  Therefore, $\min\left(\cos A, \cos B, \cos C\right) = \frac{\min\left(OD, OE, OF\right)}{R} \leq \frac rR$, proving the desired inequality. Equality holds iff $O=I$ or equilvanetly iff $ABC$ is equilateral.

Note: all arguments are valid since ABC is acute and $O$ lies in the interior of $ABC$

[quote="grobber"]It's well-known that $AX$ is the symmedian of $ABC$.[/quote]

I'm in trouble with proving that. Maybe you can help me ?

[quote="grobber"]Let $BY$ and $CZ$ the tangents in $B,C$ to the circumcircle. Just calculate the ratios in which $BY$, $CZ$ divide the sides $CA,\ AB$ respectively and then use Ceva's thm for $AX,\ BY,\ CZ$. It's really easy.[/quote]

Then i'm really stupid  :( becouse I still don't get it. Maybe you could explain a bit more ? I am able to get $\frac{BN}{CN}=\frac{AB^2}{BC^2}$, where N is the point where AX meets BC, but I don't know how to go further.

That's it then: that's the ratio in which the symmedian divides the side opposite to the vertex from where the symmedian starts.

[quote=vinoth_90_2004]a much easier way after proving its sufficient to prove r/R >= cos A, where
A >= B >= C, so  would be: since r/R = cos A + cos B + cos C - 1, so we want to prove cos B + cos C >= 1, where B+C<120. which follows from jensen.  :P[/quote]

I don't think the jensen's inequality really works in this case, since cosx is not convex on (0,π/2).

Let $(O)$ be the circumcircle of $\triangle ABC.$ The inversion about $(O)$ fixes $A$ and sends $M \mapsto X.$ Hence, the inversive distance formula yields \[\frac{AM}{AX} = \frac{OA \cdot OM}{R^2} = \frac{OM}{R}.\] Now, recall that $OM = 2R\cos A.$ Thus, by Euler's Inequality, $R \ge 2r$, it is sufficient to show that $\cos A \ge 1 / 2.$ Indeed, the assumption that $\angle A = \max\{\angle A, \angle B, \angle C\}$ entails $\pi / 2 > A \ge \pi / 3$ and the desired result follows.


trigonometry
algebra
Trigonometric Identities
Triangle
IMO Shortlist
IMO Longlist

Show that a triangle whose angles $A$, $B$, $C$ satisfy the equality
\[ \frac{\sin^2 A + \sin^2 B + \sin^2 C}{\cos^2 A + \cos^2 B + \cos^2 C} = 2 \]
is a rectangular triangle.

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

[quote="orl"]Show that the triangle whose angles $A$, $B$, $C$ satisfy the equality
\[ \frac{\sin^2 A + \sin^2 B + \sin^2 C}{\cos^2 A + \cos^2 B + \cos^2 C} = 2  \]
is a rectangular triangle.[/quote]

The op02 problem Germany 6 = geo problem 56 states even more:

[i]Prove that a triangle ABC is right-angled if and only if

$\frac{\sin^2 A + \sin^2 B + \sin^2 C}{\cos^2 A + \cos^2 B + \cos^2 C} = 2$.[/i]

The solution is simple once one knows the identity

$\sin^2 A + \sin^2 B + \sin^2 C = 2 \left(\cos A \cos B \cos C + 1\right)$,

which holds for every triangle ABC. This identity shows that, if we denote $S = \sin^2 A + \sin^2 B + \sin^2 C$, then we have S = 2 if and only if 2 (cos A cos B cos C + 1) = 2, i. e. if cos A cos B cos C = 0, i. e. if triangle ABC is right-angled.

Now, obviously, $\cos^2 x = 1 - \sin^2 x$ yields

$\cos^2 A + \cos^2 B + \cos^2 C = \left(1-\sin^2 A\right)+\left(1-\sin^2 B\right)+\left(1-\sin^2 C\right)$
$= 3-\left(\sin^2 A + \sin^2 B + \sin^2 C\right) = 3 - S$.

Thus, the condition given in the problem,

$\frac{\sin^2 A + \sin^2 B + \sin^2 C}{\cos^2 A + \cos^2 B + \cos^2 C} = 2$,

is equivalent to $\frac{S}{3-S} = 2$. But this is a linear equation for S with the only solution S = 2. Hence, we have

$\frac{\sin^2 A + \sin^2 B + \sin^2 C}{\cos^2 A + \cos^2 B + \cos^2 C} = 2$

if and only if S = 2. But we know that S = 2 holds if and only if triangle ABC is right-angled. Thus, we have

$\frac{\sin^2 A + \sin^2 B + \sin^2 C}{\cos^2 A + \cos^2 B + \cos^2 C} = 2$

if and only if the triangle ABC is right-angled, and the problem is solved.

  Darij

[b]Solution by[/b] [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=11748]campos:[/url]

Multiplying out the fraction and adding $\cos^2A+\cos^2B+\cos^2C$ to both sides
gives
$3=3(\cos^2A+\cos^2B+\cos^2C) \Rightarrow 1=\cos^2A+\cos^2B+\cos^2C$.
Finally, using the trigonometry identity
$1=\cos^2A+\cos^2B+\cos^2C+2\cos{A}\cos{B}\cos{C}$ we have
$\cos{A}\cos{B}\cos{C}=0$, so it must be a rectangular triangle.


In quadrilateral $ABCD$, $\angle A=\angle C=90^\circ$, $AB+AD=7$ and $BC-CD=3$. Find the area of quadrilateral $ABCD$.

[hide="Solution"]Denote $x=AB$ and $y=BC$. Then $AD=7-x$ and $CD=y-3$

From Pythagoras we have

\begin{eqnarray*}x^2+(7-x)^2=y^2+(y-3)^2 &\implies& 2x^2-2y^2=14x-6y-40\\ &\implies& x^2-y^2=7x-3y-20\qquad(*)\end{eqnarray*}

From the formula for the area of the right triangle, we have that the quadrilateral area is

\begin{eqnarray*} A &=& \frac{x(7-x)}{2}+\frac{y(y-3)}{2}\\ &=& \frac{7x-3y-x^2+y^2}{2}\\ &=& \frac{7x-3y-(x^2-y^2)}{2}\end{eqnarray*}

Using $(*)$ we get $A=\frac{7x-3y-(7x-3y-20)}{2}=10$[/hide]


geometry
circumcircle
incenter
geometry unsolved

Dear Mathlinkers,
let ABC be a triangle rectangular at A with AB<AC, (0) the circumcircle of ABC,
P, Q two points on the segment BC so that  BQ=BA, CP=CA
B', C' the midpoints of AQ, AP,
D the second point of intersection of AQ with (0),
(1) the circle passing through A, P, D and
E the second point of intersection of BC with (1).
Prove : B'C' = CE.
Sincerely
Jean-Louis

Dear Mathlinkers,
the question can be also:
prove that B'C'CE is a parallelogram.
Sincerely
Jean-Louis

[hide]But dear [b][size=100]Jean Louis[/size][/b], $ B'C' \parallel \equal{} \frac {PQ}{2} \equal{} CE$ $ \Longrightarrow$ $ B'C'\parallel \equal{} CE$  :)  [/hide]
Kostas Vittas.

Dear Kostas and Mathlinkers,
I think that my point of view was not enough clear…Sorry
Now, my metric proof instead a proof without calculus.

Let BC = a, AC = c, AB = c.
We have : BP = a-c, CQ = a-b, PQ = a-(a-c)-(a-b) = b+c-a.
According to Haruki's theorem,  BP.CQ = PQ.CE  i.e.  (a-c).(a-b) = (b+c-a).CE
i.e.      a² -ab-ac+bc = (b+c-a).CE
i.e.    2a²-2ab-2ac+2bc = 2.(b+c-a).CE
According th Pythagoras"s theorem:  a² = b²+c²
then,          a²+b²+c² -2ab-2ac+2bc = 2.(b+c-a).CE
i.e.            (b+c-a)²=2.(b+c-a).CE
by simplification,    b+c-a = 2.CE
i.e.   PQ = 2.CE  and we are done.

Sincerely
Jean-Louis

Note taht the incenter $I$ is the circumcenter of $\triangle APQ$ and the circumcenter of $\triangle APD$ is the midpoint of $IC$ .


trigonometry
function
algorithm
linear algebra
matrix
calculus
calculus computations

Let $ y(x)$ satisfy the inhomogeneous equation
\[ y'' \minus{} 2x^{ \minus{} 1}y' \plus{} 2x^{ \minus{} 2} y \equal{} f(x). \qquad \qquad (*)\]
Set
\[ {y \choose y'} \equal{} u(x){y_1 \choose {y_1}'} \plus{} v(x){y_2 \choose y_2'},\]
where $ y_1(x)$ and $ y_2(x)$ are two independent solutions of $ (*)$ when $ f(x) \equal{} 0$, and $ u(x)$ and $ v(x)$ are unknown functions. Obtain first order differential equations for $ u(x)$ and $ v(x)$, and hence find the most general solution of $ (*)$ in the case $ f(x) \equal{} x \sin x$. Are the functions $ u(x)$ and $ v(x)$ completely determined?

hello, in the case that $ f(x)\equal{}0$ holds we get the solution
$ y(x)\equal{}C_1x\plus{}C_2x^2$
Sonnhard.

I also get that result; how do you get first-order DEs for $ u(x)$ and $ v(x)$?

You'll have been shown in lectures how to use phase space to derive the equation:

$ \left( {\begin{array}{cc} y_1 & y_2 \\
y_1' & y_2' \\
\end{array} } \right) \left( {\begin{array}{c} u' \\
v' \\
\end{array} } \right) \equal{} \left( {\begin{array}{c} 0 \\
f \\
\end{array} } \right)$

From this you can get first order DEs for u and v.

I obtained $ v'(x) \equal{} \minus{}f(x)$ and $ u'(x) \equal{} \frac{1}{x} f(x)$ according to that "algorithm" we learnt - I still don't understand everything about phase space though, so that's something to ask my supervisor soon!

[quote="aidan"]I obtained $ v'(x) = - f(x)$ and $ u'(x) = \frac {1}{x} f(x)$ according to that "algorithm" we learnt - I still don't understand everything about phase space though, so that's something to ask my supervisor soon![/quote]

What happened to y_1 and y_2? Look at the equation in my previous post. We want u' = something, v' = something. From that equation, we can either multiply out and solve the equations for u' and v', or write

$ \left({\begin{array}{c}u'\\ v'\\ \end{array}}\right) =
\frac{1}{W}
 \left({\begin{array}{cc}y_{2}'&- y_{2}\\ -y_{1}' & y_{1}'\\ \end{array}}\right) \left({\begin{array}{c}0\\ f\\ \end{array}}\right)$

From which we get $ u' = \frac{-y_2 f}{W}, \ v' = \frac{y_1 f}{W}$

Where W is the Wronskian. Now you have two first order DEs for u' and v' :)

Oh I solved for $ u'$ and $ v'$ using normal simultaneous methods without using the Wronskian and matrix inverse method taught in lectures, I think I got the correct answer though (unless there have been careless mistakes)  :maybe:


integration
logarithms
real analysis
real analysis unsolved

solve $ y' \equal{} exp(x \plus{} y) \plus{} exp(x)$

hello, we have $ y'\equal{}e^x(e^y\plus{}1)$ or
$ \int\frac{1}{e^y\plus{}1}\,dy\equal{}x\plus{}C$
$ \ln\left(\frac{e^{y}}{e^{y}\plus{}1}\right)\equal{}e^{x}\plus{}1$.
Where are the other diff equations from you?
Sonnhard.


integration
calculus
calculus computations

use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the y-axis. sketch the region and a typical shell.

y=4(x-2)^2 , y=x^2 -4x+7

i did the graph and all that, but i need help on doing the problem and the answer is 16 pie, but i keep on getting 27 pie

Is this your set-up? I'll work it through if it is, otherwise this is just the problem.

$2\pi\int_{1}^{3}x\cdot\left(x^{2}-4x+7-4(x-2)^{2}\right)dx$

Edit: Perfect, see one of your other shell questions. http://www.artofproblemsolving.com/Forum/viewtopic.php?t=149364 Scorpius119 has explained how to come up with that set-up.


Find $ a,b,c \in N$ such that $ ab$ divides $ a^2\plus{}b^2\plus{}1$.

[quote="SUPERMAN2"]Find $ a,b,c \in N$ such that $ ab$ divides $ a^2 \plus{} b^2 \plus{} 1$.[/quote]

[hide="My solution"]
$ a^2\plus{}b^2\plus{}1\equal{}kab$

If $ a\equal{}b$, we get $ 2a^2\plus{}1\equal{}ka^2$ and so $ k\equal{}3$ and $ a\equal{}b\equal{}1$
If $ a\neq b$, WLOG say $ a>b$ then $ 0<b^2\plus{}1<a^2$, so $ 0<kab\minus{}a^2<a^2$ and $ 0<kb\minus{}a<a$

But we have $ (kb\minus{}a)^2\plus{}b^2\plus{}1\equal{}k(kb\minus{}a)b$
So, from $ (a,b,k)$ with $ a>b$, we get $ (b,kb\minus{}a,k)$ with $ 0<kb\minus{}a<a)$

So any solution $ (a,b,k)$ gives, after some iterations, the solution $ (1,1,3)$

So the only solutions are $ (1,1,3)$ and all the subsequent transformations $ (a,b,3)\to(3a\minus{}b,a,3)$

So the only solutions are $ u_{n\plus{}1}^2\plus{}u_n^2\plus{}1\equal{}3u_nu_{n\plus{}1}$ $ \forall n>0$ with the sequence $ u_n$ defined as : $ u_1\equal{}u_2\equal{}1$ and $ u_{n\plus{}2}\equal{}3u_{n\plus{}1}\minus{}u_n$

[/hide]


just preorder it and send it to awesomemath with your name on it

I don't see the problem. Can't you just walk/talk the bus to the nearest bookstore at UTexas? Or maybe http://utdallas1.bkstore.com/bkstore/content might even hold it.

I don't see a problem with pre-ordering it to the camp. You'll get the address with your precamp packet, or if you need to do it sooner let me know.

[quote="13375P34K43V312"]So Harry Potter 7 is coming out July 21 (I told you it was lame). Is it possible that we can reserve through the amp people and the order comes in so we can read during camp?

because i do not like it when there's a harry potter book that someone has read and i have not[/quote]
can i borrow the book after you read it =D

[quote="Altheman"]i ono if ap english has made me elitist this year or w/e...but harry potter does not really have much substance...it is a "page-turner"...which has some value...but i don't see a reason to make such a huge effort for it...yea[/quote]

hey now

[quote="Aryth"]Hey, any other 8th graders other than 13375P34K43V312 and fanatic going that are here?[/quote]
me, b-flat, and splashD (I think)
Are you going?

[quote="kyyuanmathcount"]*ah-em*  I'm going!

imo, I would rather not get into a book during a [b]math[/b] camp.[/quote]
hehe, forgot about you. :D
Also, why not? Don't you get bored of math after 8 or 9 hours?
I know I do, but you probably have more stamina than me.

[quote="kyyuanmathcount"]*ah-em*  I'm going!

imo, I would rather not get into a book during a [b]math[/b] camp.[/quote]

Good point.

However, this will mean that I won't be able to go to the party and read the entire thing the day it comes out (assuming that it is short enough to read in 1 day) :(

Anyway, my mom booked 2 copies back in January, so I somehow doubt that she would get another one sent to texas for me...

dude arnav a lot of vietnamese people speak english.  just give one of them US$\$$1 and they'll read it to you.  okay, this is kind of mean and probably ineffective, but you should try it  :rotfl:

EDIT: stop complaining almost all of these people wish they had your problem.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=144187

there we go

[quote="Altheman"]i ono if ap english has made me elitist this year or w/e...but harry potter does not really have much substance...[/quote]

neither does junk food. Does that stop us?

[quote="billzhao"][quote="MysticTerminator"]OH SNAP IT'S COMING OUT DURING THE IMO NOW I'LL HAVE TO READ IT IN VIETNAMESE S_()UIFPOW@#":<>?SD<FJR*(SJD:FKLJ)#$*U%IP
$JT[/quote]

What's wrong with that? HP6 came out while we were in Mexico during IMO 2005 and we managed to get our hands on an American copy (note: American, not English). Just ask Brian or Sherry and they'll tell you about it :P[/quote]We all remember me making fun of you guys for reading that book during the hurricane!  :rotfl:


I want to move all of my LaTeX-related files to a new folder, but when I do so and try to compile a file, it doesn't work.  What do I need to do to fix this?

We'll need a little more information.  What exactly happens when you try to complie?  Have you copied every single file from one folder to the next (are you sure you didn't miss a .sty file you need or something like that)?

I have a big folder with all LaTeX-related stuff in it, and I moved the entire folder to a new location, so I'm sure I didn't miss any files.
When I try to compile, I get the following message (and nothing else):

C:\windows\...(OLD location of files)...\MiKTeX\MiKTeX\bin\pdflatex.exe Cannot execute the command.  Error: The system cannot find the file specified.

I checked, and the file pdflatex.exe still exists, and it's in the MiKTeX\MiKTeX\bin folder, but this folder is in a new location.  My guess is that I need to somehow tell TeXnicCenter that the files have moved.
I tried going to "define output profiles" and changing the file paths of the compiler, "BiBTex executable", and "MakeIndex executable" to the new locations.  Upon compiling, it got further than it had before, but the results showed (among other things that seemed inconsequential):

Error: PDFLATEX (file pdftex.cfg): Cannot open config file
==>Fatal error occurred, the output PDF file is not finished!

The pdftex.cfg file does exist and (as far as I know) has not been modified.


superior algebra
superior algebra solved

let G be a group with 2n elements in which exactly n have the property that x2 = 1. Prove that n is an odd number.

The number of elements that satisfy x^2 = 1 and are not equal to 1 is even (because every element pairs off with its inverse). Hence, in total (that is if we add 1), we have (even number) + 1 = n. Hence, n is odd.

Cheers,
Thanos

!! Warning Thanos, in general, the number of elements that satisfy "x^2 = 1 and are not equal to 1" could be odd .. !!

thanos' solutions is indeed not correct because  in my initial formulation of the problem was not so inspirate cuz I was quick to write it ... I meant if all elements of order 2 are n (that is not 1) not elements for which x2=1 ... my mistake

the correct solution is that each number for which x2 is different from 1, can be paired up with its inverse, and thus making them an even number of elements. But they are exactly n-1 such elements, therefore n is odd.

The initial problem is not too bad, but I prefer a more clever version :
Let G be a finite group and let n be the number of all the elements x having the property : (x <> 1)&(x^2 = 1).  Prove : " if n is a non-zero number, then it is odd " .

Still about the V.V initial problem " G is a group with 2*n elements in which exactly n have the property that (x<>1)&(x^2 = 1)". For this case, there is only one value for n : n=1  .. See you ..