geometry
geometric transformation
rotation

[hide="Let..."]
...the boys be called $A$, $B$, $C$ and $D$.
Between $A$ and $B$ there are $a$ girls, between $B$ and $C$ there are $b$ girls, between $C$ and $D$ there are $c$ girls and between $D$ and $A$ there are $d$ girls.

$a,b,c,d$ are positive integers with $a+b+c+d=10$. Assuming we count permutations $(a,b,c,d)$ of each other as the same solution, there are [b]nine[/b] ways this can be done: $(7,1,1,1)$, $(6,2,1,1)$, $(5,3,1,1)$, $(5,2,2,1)$, $(4,4,1,1)$, $(4,3,2,1)$, $(4,2,2,2)$, $(3,3,3,1)$ or $(3,3,2,2)$.

 [/hide]

[quote="tim1234133"][hide="Let..."]
...the boys be called $A$, $B$, $C$ and $D$.
Between $A$ and $B$ there are $a$ girls, between $B$ and $C$ there are $b$ girls, between $C$ and $D$ there are $c$ girls and between $D$ and $A$ there are $d$ girls.
$a,b,c,d$ are positive integers with $a+b+c+d=10$. [b]Assuming we count permutations $(a,b,c,d)$ of each other as the same solution[/b], there are [b]nine[/b] ways this can be done: $(7,1,1,1)$, $(6,2,1,1)$, $(5,3,1,1)$, $(5,2,2,1)$, $(4,4,1,1)$, $(4,3,2,1)$, $(4,2,2,2)$, $(3,3,3,1)$ or $(3,3,2,2)$.[/hide][/quote]
[quote="Treething"]I don't think you're taking into consideration that there are multiple ways to get each of your nine combinations.[/quote]
He assumed they are all the same...

[hide="Solution"] To begin the simplification, imagine that we are simply looking at a line of people instead of a circle. One starting condition is that no two boys can sit next to each other, meaning there must be at least one girl between each boy. To simplify this, remove one girl from in between each boy, taking 4 girls total. Then we are just considering how to place 6 girls and 4 boys around a table with no restrictions. Since all the children are unique and special, this is just $10!$ ways. Now we could choose 4 of the 10 girls to have been taken out for simplification (with the other 6 girls accounted for above) [b]in a particular order[/b] in $\frac{10!}{6!}$ ways. And lastly, our initial simplification of a line contains 14 different arrangements for every 1 arrangement in a circle (rotations of the circle), so we must divide our number of possible lines by 14 to get the number of possible circles. Then unless I made a mistake somewhere, our total answer is:
$(10!)(\frac{10!}{6!})/14 = 1306368000$[/hide]

[quote="pipi"][quote="boxedexe"]There are 10 girls and 4 boys.  In how many ways can these girls and boys sit around a circular table such that no boys are next to each other.[/quote]
I consider this question as follows:
Arrange the 10 girls in circular table, there $9!\/$ ways. There are 10 spaces between girls. Choose 4 of them and place one boy for each, the 4 boys can interchange among them. This can be done by $\binom{10}{4}4!\/$ ways.

Hence the answer is $\binom{10}{4}4!9! \/$.[/quote]

This assumes the order that the girls sit in is important. I think we only want the number of ways the boys can sit.

It says "In how many ways can these girls and boys sit around a circular table". I think this means that the way the girls sit in matters.


rotation
trigonometry
LaTeX
Pythagorean Theorem
geometry
trig identities
Law of Cosines

Isosceles $\triangle ABC$ has a right angle at $C$.  Point $P$ is in $\triangle ABC$ such that $PA=11$, $PB=7$, and $PC=6$.  Legs $\overline{AC}$ and $\overline{BC}$ have length $s=\sqrt{a+b\sqrt{2}}$, where $a$ and $b$ are positive integers.  What is $a+b$?

[hide="Hint:"]
Rotation.  :wink: [/hide]

[hide="Solution"]
Rotate the triangle $90^\circ$ clockwise about $C$ to get the triangle $A'B'C'$ with point $P'$ within the triangle. Then, we have that $\angle PCB+\angle PCA = \angle P'C'B'+\angle PCA$, thus $\triangle PCP'$ is right and $PP'=6 \sqrt 2$. Then, we have a few different ways of finding $BC$, call it $x$.

[img]http://i61.photobucket.com/albums/h70/cincodemayo5590/rotatedtriangle.png[/img]

[u]Method 1 (longer):[/u]
Label $\angle BPC=\alpha,\ \ \angle APC=\beta, \ \angle APB=\gamma$. Then look at the quadrilateral $PAP'C$; we have that the angle $\angle PAP'=360-90-\alpha-\beta=270-\alpha-\beta$ by rotation. But, $\gamma=360-\alpha-\beta$, thus $\gamma = \angle PAP'+90^\circ$.
Now, we can find the cosine of $\angle PAP'$ and use that to find the hypotenuse $AB$ and thus the side length $x$.
By cosine law on $\triangle PAP'$, (or by simply noticing, as have [b]Elemennop[/b] and [b]scorpius119[/b], that $\triangle PAP'$ is right), we have that $\cos \angle PAP' = \frac{7}{11}$. Then applying cosine law to $\triangle PAB$, with $\cos \gamma =-\sin \angle PAP'$ we get that $(x \sqrt 2)^{2}= 170+84 \sqrt2 \implies x=\sqrt{ 85+42 \sqrt{2}}\implies 85+42= \boxed{127}$.

[u]
Method Two:[/u]
Notice that $B$, $P$, and $P'$ are colinear. See [b]scorpius119[/b]'s post (#5) for an explanation. Then apply cosine law to $\triangle BP'C$ to get that $x=\sqrt{ 85+42 \sqrt{2}}\implies 85+42= \boxed{127}$.[/hide]

EDITs: Edited a bit based on comments from [b]Elemennop[/b] and [b]scorpius119[/b]. Thank you. :)

[hide="Bigger Hint: Method of Solution"] Rotate the triangle counterclockwise around B to $BC$, and multiply it by a factor of $\sqrt{2}$. Then you have an isosceles right triangle.  The rest is cosine law.[/hide]

cincodemayo ~ Your first (longer) method is shortened considerably (at least getting rid of an unneccesary cosine law application) if you notice PAP' is a right triangle.

[quote="scorpius119"][quote="cincodemayo5590"]
Notice that $B$, $P$, and $P^\prime$ are collinear. (I'm not sure why, can someone explain?)
[/quote]

[hide]We find $\angle PP^\prime A$ to be $90^\circ$ by the Pythagorean Theorem (actually, its converse, but whatever). Then notice that $\angle BPC$ and $\angle CP^\prime A$ must be congruent because the first triangle is rotated to get the second one. It follows that $\angle BPC=135^\circ$, $\angle BPP^\prime =180^\circ$. yay![/hide][/quote]

Ah, great. Thanks!

[quote="Altheman"][quote="cincodemayo5590"]
Notice that $B$, $P$, and $P'$ are concurrent. [/quote]

points are not concurrent

also you can just be like PP'A=90 by law of cosines, then AC can be found using LoC also

note that the best method is to find cos PP'A regardless[/quote]

Blah, colinear. You know what I mean.  :P

Sorry, this Latex program is too big for my pc.
[hide] draw a perpendicular from CA to P and call it h. Let the length of the line from C to the foot of this perpendicular be x, and also the length of the line from A to the same point be y.
So, (1) x + y = s
Draw another perpendicular from CB to P. This will have length of x. The length of the line from C to the foot of this perpendicular will be h. Let the length of the line from B to the foot of this perpendicular be k.
So, (2) k + h = s

(3) k^2 + x^2 = 49
(4) x^2 + h^2 = 36
(5) y^2 + h^2 = 121

(3) - (2): k^2 - h^2 = 13
(k + h)(k - h) = 13
s(k - (s - k)) = 13
2k - s = 13/s
k = (s^2 + 13)/2s

(5) - (4): y^2 - x^2 = 85
(y + x)(y - x) = 85
s(y - (s - y)) = 85
y = (s^2 + 85)/2s
y + x = (s^2 + 85)/2s + x
s = (s^2 + 85)/2s + x
x = s - (s^2 + 85)/2s = (s^2 - 85)/2s

(3): ((s^2 + 13)/2s)^2 + ((s^2 - 85)/2s)^2 = 49
(s^2 + 13)^2 + (s^2 - 85)^2 = 196s^2
s^4 + 26s^2 + 169 + s^4 - 170s^2 + 7225 = 196s^2
2s^4 - 340s^2 + 7394 = 0
s^4 - 170s^2 + 3697 = 0
(s^2 - 85)^2 - 7225 + 3697 = 0
(s^2 - 85)^2 = 3528
s^2 = 85 +- sqrt 3528
discard the solution 85 - sqrt 3528 as this will result in negative length sides
s^2 = 85 + sqrt 3528
s = sqrt(85 + sqrt 3528) = sqrt (85 + (sqrt 1764)(sqrt 2))
= sqrt(85 + 42 sqrt 2)

[/hide]

There is no program required when you want to post on the internet.  :wink:

boxedexe

Thanks, didn't know that. I'll look into it.


calculus
integration
calculus computations


[hide]Suppose the common difference ($=a_2 - a_1$, $a_3 - a_2,$ etc.) is $d$.  Then $a_4 = a_7 - 3d$ and $a_{10} = a_7 + 3d$, so the first equation means $3a_7 = 17$, or $a = \frac{17}{3}$.  From similar reasoning, $a_4 + _5 + \ldots + a_{10} = 7a_7$.  Using this and $a_{11} = a_7 + 4d$, etc., yields the following version of the second equation:

$7a_7 + (a_7 + 4d) + (a_7 + 5d) + (a_7 + 6d) + (a_7 + 7d) = 77$
$11a_7 + 22d = 77$
$22d = \frac{44}{3}$
$d = \frac{2}{3}$

$13 - \frac{17}{3} = \frac{22}{3}$, so $a_k = 11d + a_7$, i.e. $k = 18$.  (B).

(Edited to fix LaTeX problems)[/hide]

[hide]The sum of arithmetic sequence with [b]odd term [/b]is [b]the middle term times the number of term[/b].

First condition:$3a_7=17\Longleftrightarrow a_7=\frac{17}{3}$.

Second condition:$11a_9=77\Longleftrightarrow a_9=7$

We have $a_n=\frac{7-\frac{17}{3}}{9-7}(n-9)+7=\frac{2}{3}n+1$.

Solving $a_k=13$,the answer is $k=18$[/hide]


I'm attempting to close the gap between me and the top ten when I go to states on saturday by doing a few problems from the All Time Greatest Mathcounts Problems, but some of the problems seem too easy. Are the old problems easier, or am I just getting lucky and picking the easy ones?

[quote="stevenmeow"]The way i see the all-time ggreatest problems is where MATHCOUNTS messed up. some of the porblems * maybe most or all* were made for the purpose of being easy problems. But, some porblems hatched huge protests because of there ambiguoug wording.
     Example: (may not be exact)
there are 190 marbles. what is the most number of bags that u can have such that the # of marbles in each bag is different. all bags have at least 1 marble

[hide="listed ans/solution"]
19...
the question WANTED you to find a number n such that 1+2+3+4...+n=190
i will not solve this but u can on your own
[/hide]
[hide="protested ans/solution"]
the PROTESTED answer is 190 because the problem never said the bags couldnt be inside one another
therefore, one bag could have 1 marble and then be places in a slightly bigger bag. then another marble would be put into the bigger bag (not in the 1st bag) and THEN put into a 3rd bag...
and so on...
[/hide]

in my opinion, the all-time greatest problems are more for entertainment than practice. in some cases, you MAY be able to pick up techniques although this should not be you MAIN source of practice.[/quote]

Dude, the "protested" solution is based off of twisting the words (to an unreasonable extent).

My math teacher once said, "If there is ambiguity and one solution is much easier than the other, pick the harder solution."

[quote="stevenmeow"]The way i see the all-time ggreatest problems is where MATHCOUNTS messed up. some of the porblems * maybe most or all* were made for the purpose of being easy problems. But, some porblems hatched huge protests because of there ambiguoug wording.
     Example: (may not be exact)
there are 190 marbles. what is the most number of bags that u can have such that the # of marbles in each bag is different. all bags have at least 1 marble

[hide="listed ans/solution"]
19...
the question WANTED you to find a number n such that 1+2+3+4...+n=190
i will not solve this but u can on your own
[/hide]
[hide="protested ans/solution"]
the PROTESTED answer is 190 because the problem never said the bags couldnt be inside one another
therefore, one bag could have 1 marble and then be places in a slightly bigger bag. then another marble would be put into the bigger bag (not in the 1st bag) and THEN put into a 3rd bag...
and so on...
[/hide]

in my opinion, the all-time greatest problems are more for entertainment than practice. in some cases, you MAY be able to pick up techniques although this should not be you MAIN source of practice.[/quote]

eh, the only reason I don't really find them the greatest problems is because I've seen so many of those types of problems the solution seems kinda, like, plain.

I actually think the old problems were easier.  remember, if there's 100 or so questions, some will probably be very easy.  it also depends on if you're looking in the countdown section or not :P


polynomial
algebra
Irreducible
factoring polynomials
IMO
IMO 1993
irreducible polynomial

well i think is an easy one...lets say that $g(x)=x^g+a_{g-1}x^{g-1}+...+a_{0}$ and $q(x)=x^q+b_{q-1}x^{q-1}+...+b_{0}$ then $a_{0}b_{0}=3$ so WLOG $b_{0}$ is multiple of 3 so lets prove by induction that $b_{n}$ is multiple of 3
first lets see that $b_{0}$ is and we know that the coefficient of $x^1$ in $p(x)$ is 0 so $b_{0}a_{1}+b_{1}a_{0}=0$ or $b_{0}a_{1}=-b_{1}a_{0]}$ and $a_{0}$ is not multiple of 3 so $b_{1}$ must be and that way you can get the result.

that is looking for the coefficient 0 of $x^{n}$ in $p(x)$ and having that $b_{i}=3k$ for $i=0,1,...,n-1$ all that is sufficient if the coefficient is 0 in $x^{q}$ and doesnt occur if $q=n-1$ if that is true then $g(x)=x+g$ with g not multiple of 3. so then $g^n+5g^{n-1}=g^{n-1}(g+5)=-3$ for some integrer g, but then $g^{n-1}$ divides 3, so $g=1$ or $g=+/-3,n=2$ and both are wrong :!:  :D

is it right? :roll:  :lol: ?

We want to prove that $ f(x)$ is irreducible over $ \mathbb{Z}$. Since $ f(x)$ largest degree's term is $ 1$, its factors' largest degree's terms are $ 1$. By the Extended Eisenstein Criterion, $ f(x)$ has an irreducible factor over $ \mathbb{Z}$ with degree larger that $ n\minus{}2$. So it has an irreducible factor over $ \mathbb{Z}$ with degree $ n\minus{}1$ or $ n$. Let's suppose $ f(x)$ has an irreducible factor $ \mathbb{Z}$ with degree $ n\minus{}1$. So it also has an irreducible factor with degree $ 1$, and since its largest degree's term is $ 1$, it has a zero over $ \mathbb{Z}$, which means $ f(x)$ also has a zero over $ \mathbb{Z}$. But $ f(x)$ is always odd, therefore it can't be $ 0$. Contradiction! So $ f(x)$ has a irreducible factor over $ \mathbb{Z}$ with degree $ n$, which is $ f(x)$, QED.

Just edited my post at http://www.mathlinks.ro/viewtopic.php?t=37593 to include this problem as well. Just take $ p \equal{} 3$ (which is prime) and $ q \equal{} 1$ (which is squarefree and not divisible by $ p$). And consider the case $ n\equal{}2$ separately.

  darij

It is a direct consequence of extended Eisenstein's criterion and rational root theorem

you can use peron criterion it's obvious with peron

He wants to say Perron. It's Theorem A in http://rms.unibuc.ro/bulletin/pdf/53-3/perron.pdf . Applying it here is pretty much overkill, though.

You could use perron's, but saying 5>4 is a little boring, don't you think?

Another way to do this problem is to reduce it mod 3. We have $x^n+5x^{n-1}+3 \equiv x^n+2x^{n-1}$ (mod 3). Simplifying gives $x^{n-1}(x-2)$. Suppose f(x)=g(x)h(x), for some $g, h \in \mathbb{Z}[x]$ nonconstant. Since x and x-2 are irreducible in $\mathbb{F}_3[x]$, we have (WLOG) $g(x)=x^k+3g_1(x)$ and $h(x)=x^{n-k-1}(n-2)+3h_1(x)$, for some integer polynomials $g_1$ and $h_1$. First we consider the case where k=0. This means that g is nonconstant, a contradiction. If k=n-1, then h(x) is a linear polynomial, so f must have an integer root. We can easily check that this is false. Finally, if 1<k<n-1, multiplying g and h gives $x^{n-1}(x-2)+3g_1(x)x^{n-k-1}(x-2)+3h_1(x)x^k+9g_1(x)h_1(x)$. Setting this equal to our original polynomial and plugging in x=0 gives $9g_1(0)h_1(0)=3$. Since $g_1, h_1 \in \mathbb{Z}[x]$, this is clearly a contradiction. Thus, our polynomial is irreducible in $\mathbb{Z}[x]$.

I think this problem could also be classified as number theory. Irreducible polynomials is one of the biggest overlaps of NT and algebra.

So, first, from $f(a_i) = a_i^n + 5a_i^{n-1} + 3 = 0$, you deduce $a_i^n + 5a_i^{n-1} = -3$, which is correct, only that you write $f(a_i) = a_i^n + 5a_i^{n-1} = -3$. Nice piece of algebra! All we have is  $f(a_i) -3 = a_i^n + 5a_i^{n-1} = -3$. And now we will have $g(a_i)h(a_i) -3 = -3$, i.e. $g(a_i)h(a_i) = 0$; obviously, since $g,h$ are the [b]factors[/b] of $f$.

Therefore $(g(a_i), h(a_i)) = (\pm 1, \mp 3)$ is clearly wrong, since at least one of them must be $0$. Moreover, in your mistaken way, you miss the possibility $(g(a_i), h(a_i)) = (\pm 3, \mp 1)$.

Nevermind, indeed under these wrong conclusions we would have $p(a_i) = g(a_i)+h(a_i) \in \{-2,2\}$, which you then write $p(a_i) =\pm 2$. True, $\deg p \leq n-1$. But you don't know that the $a_i$'s, $1\leq i \leq n$ are distinct, and even if they were, this does not contradict that Fundamental Theorem (rather the simple theorem that a polynomial cannot have more roots than its degree), since $\pm 2$ is not [b]just one value[/b]; for example for the polynomial $p(x) = x^2-3$ we have $p(\pm 1) = -2$ and $p(\pm\sqrt{5}) = 2$. So even if your argumentation up to this point were correct, you still have no contradiction. When only knowing $\deg p \leq n-1$, you need $p(x) = \pm 2$ for [b]at least[/b] $2n-1$ distinct arguments, in order to get a contradiction.

Suppose
f(x) = $(x^r + a_{r-1}x^{r-1} + ... + a_1x + 3)(x^s + b_{s-1}x^{s-1} + ... + b_1x + 1)$. We show that all the a's are divisible by 3 and using that we will establish a contradiction .[Here , we will show using only + ve sign of 3 , the other case is
f(x) = $(x^r + a_{r-1}x^{r-1} + ... + a_1x - 3)(x^s + b_{s-1}x^{s-1} + ... + b_1x - 1)$ , in which constant terms contains only - ve terms and the casee follows only the + ve case . So , we will show here only the first case] .
r and s must be greater than 1.Because
for if r = 1, then 3 is a root . Now we will make two cases :
case 1:
n is even, then it follows that 0 = $3^n +5\cdot$$3^{n- 1} + 3 = 3^{n-1}( 3 +5)+ 3$, which is false since 3+5 = 8.
case 2 :
if n is odd we would have 0 = $3^{n-1}(3 + 5) + 3$, which is false since 3 + 5 = 8.
If s = 1, then 1 is a root and we will argue by contradiction in the same way .
So r$\leq$n - 2, and hence the coefficients of x, $x^2, ... , x^r$ are all zero. Since the coefficient of x is zero, we have: $a_1+3b_1 = 0$, so $a_1$ is divisible by 3.
Now , we will use the induction hypothesis .
We can
now proceed by induction.
Assume $a_1, ... , a_t$ are all divisible by 3. Then
consider the coefficient of $x_{t+1}$. If s-1 $\geq $t+1, then $a_{t+1}$ = linear combination of $a_1,... , a_t + 3b_{t+1}$. If s-1 $\leq$ t+1, then $a_{t+1}$ = linear combination of some or all of $a_1,... , a_t.$
Either way, $a_{t+1}$ is divisible by 3.
Now we will consider the coefficients of x, $x^2$,
... , $x^{r-1}$ which gives us that all the a's are multiples of 3. Now consider the coefficientof $x^r$ which is also zero.
Now we will get , that is a sum of of terms which are multiples of 3 .
Then it does not becomes 0.
So , follows a contradiction !
So , the factorization is not possible .( proved ).
-NEELANJAN MONDAL.

[hide=Solution]
Assume FTSOC that there do exist $g(x), h(x)$ such that $f(x)=g(x)\cdot h(x)$. Observe that there are no integer roots of $f(x)$ from the Rational Root Theorem. Thus, $g(x), h(x)$ cannot be linear, and their degree is greater than or equal to $2$. We have $g(0)h(0)=3$. WLOG $g(0)=1$. Let $r_1, r_2, \dots r_j \in\mathbb{C}$ be the roots of $g(x)$. We have $r_1r_2\dots r_n=\pm 1$. Multiplying the equalities $r_i^{n-1}(r_i+5)=-3$ for all $1\leq i\leq j$, we obtain \[\vert g(-5)\vert = \vert (r_1+5)(r_2+5)\dots (r_m+5)\vert = 3^m.\] But $g(-5)f(-5)=3$, contradiction.
[/hide]

Extended Eisenstein $->$ RRT $->$ $\mod 2$ $->$ Done.

My point is, would your solution even count if you directly apply Eisenstein on this problem, because it completely trivialises the problem... (apparently any use of a theorem is prohibited if it trivialises the problem, and this theorem certainly does, right?)

It is a direct result of Perron's criterion. Alternatively, suppose $x^n+5x^{n-1}+3 = f(x)g(x)$. Reduce it modulo 3 to get $f(x)g(x) \equiv x^{n-1}(x+5)$, so we can assume that $f(x) = x^{p}(x+5) + 3F(x)$, $g(x) = x^{q} + 3G(x)$. Multiply them to get $x^qF(x)+x^p(x+5)G(x)+3F(x)G(x) = 1$. Plug in $x=0$, then $3F(0)G(0)=1$, clearly impossible.

[quote=pi_is_3.141][hide]Since $5>1+3$, it is true by Perron's Criterion[/hide][/quote]

oh lol someone is taking AMSP Alg 3...

[hide=Solution 1]Trivial by Perron's Criterion.[/hide]
[hide=Solution 2]
Assume for the sake of contradiction, let $f(x)$=$g(x)$$h(x)$, where $g$,$h$ are integer polynomials with degree greater than or equal to $2$. $g(0)$$h(0)=3$, so we can assume $g(0)$ to be $\pm1$. Let $y$ be the degree of $g(x)$. Let the roots be $a_{1}$,$a_{2}$,...,$a_{y}$. The product is $\pm1$. Plugging in $a_d$ in $f(x)$, you get $f(a_{d})=a_{d}^{n-1}(a_{d} +5)=-3$. If you multiply all of the inequalities for all of the roots of $g(x)$, we get $|g(-5)|=|(a_{1}+5)(a_{2}+5)...(a_{y}+5)|=3^y$.
But $f(-5)=g(-5)h(-5)=3$, which is a contradiction.[/hide]

[quote=fukano_2][hide=one liner]$f(x)=x^n+5x^{n-1}+3 \equiv x^n-2 = \overline{f(x)} \pmod 5$, and $x^n-2$ is irreducible, so $f(x)$ is irreducible[/hide]
[hide=-_- Remark]$f(x)$ reducible implies $\overline{f(x)}$ reducible, so by contrapositive, $\overline{f(x)}$ irreducible implies $f(x)$ irreducible[/hide][/quote]

I think this is wrong? $\overline{f(x)}$ only needs to be irreducible $\pmod 5$, not completely irreducible (for example, if I took $f(x)=x^2+2x+1$ and had $x^2+2x+1\equiv x^2+1\pmod 2$, $x^2+1$ being irreducible doesn't imply $x^2+2x+1$ is irreducible, since $x^2+1$ could still be reducible $\pmod 2$).

This can be done by using perrons criterian

By pure coincidence, I had been looking at irreducibility criterion when I encountered this problem. Perron's kills it.

Since $5>1+3$, $f$ is irreducible over $\mathbb Z[x]$. $\square$

What on earth?!

[hide=Solution]Since $5>1+3$, from Perron's Criterion this polynomial is irreducible over the integers as desired.[/hide]

[quote=orl]Let $n > 1$ be an integer and let $f(x) = x^n + 5 \cdot x^{n-1} + 3.$ Prove that there do not exist polynomials $g(x),h(x),$ each having integer coefficients and degree at least one, such that $f(x) = g(x) \cdot h(x).$[/quote]

1 liner by Perron's

By IG's inequality, $5$ is a greater number than $1+3=4$. Hence, we are done since Perron's finishes over $\mathbb{Z}\left[x\right]$ $\square$

Good thing i still have my irreducibility handouts :gleam:

it's also not too hard to prove perron's irreducibility criterion in olympiads, assume the assertion and suppose that $f(x)=g(x)h(x)$. This means that $f$ has only 1 root of modulus greater than 1, which means either $g$ or $h$ has roots not outside the unit circle. You'll reach a contradiction that $|g(0)| < 1$ yet $g(0)$ is, by definition, a nonnegative integer.

Everyone says "Perron's" but there's no explanation of what Perron's Criterion is, so I'll put the general statement here.
[quote]Let $f(X)=X^n+a_{n-1}X^{n-1}+\cdots+a_1X+a_0$ be a polynomial with integer coefficients such that $a_0\neq 0$ and \[|a_{n-1}|>1+|a_{n-2}+\cdots+|a_1|+|a_0|.\] Then $f(X)$ is irreducible in $\mathbb{Z}[X].$[/quote]
We come back to the original problem.
[quote]Prove that the polynomial $X^n + 5X^{n-1} + 3$ is irreducible in $\mathbb{Z}[X].$ [/quote]

Direct application of Perron's gives $5>1+3,$ so the polynomial is irreducible as required. Below is an alternate solution through Eisenstein's Criterion, of which here is the statement:
[quote]Let $f(X)=a_nX^n+a_{n-1}X^{n-1}+\cdots+a_1X+a_0$ be a polynomial with integer coefficients. Then $f(X)$ is irreducible in $\mathbb{Z}[X]$ if there exists a prime $p$ such that
(i) $p \mid a_0, p \mid a_1, \cdots, p \mid a_{n-1},$ and
(i) $p \nmid a_n, p^2 \nmid a_0.$ [/quote]

It is easy to verify by the rational root theorem that no linear factors divide $X^n+5X^{n-1}+3.$ Assume on the contrary that $P(X)=X^n+5X^{n-1}+3=f(X)g(X).$ Taking mod three, we get \[f(X)g(X)\equiv X^{n-1}(X-1)\pmod 3.\] Let $f(X)\equiv X^a\pmod 3,g(X)\pmod X^b(x-1)\pmod 3.$ Now, $a,b\neq 0$ since $f(X),g(X)$ both have degree at least two. Thus, we have $3\mid f(0)$ and $3\mid g(0)$ so $9\mid P(0)$ which is clearly false because $P(0)=3.$

Wow, I have finally done an irreducibility problem.  Let's not use Perron's Criterion.

Write $x^n + 5x^{n-1} + 3 = g(x) h(x)$ for some polynomials $g\in \mathbb Z[x]$ and $h\in \mathbb Z[x]$.  Plugging in $x = 0$ yields $g(0)h(0) = 3$, so without loss of generality $3\nmid h(0)$.  However, taking $f$ modulo $3$ yields
\[
x^n - x^{n-1} \equiv \bar g(x) \bar h(x)\pmod 3.
\]
Thus, $x$ cannot divide $\bar h(x)$.

There are two cases to consider.  In the first case, $g(x) = x^{n-1} + 3p(x)$ and $h(x) = x - 1 + 3q(x)$ for some polynomials $p\in \mathbb Z[x]$ and $q\in\mathbb Z[x]$.  Since $\deg g \geq n - 1$ and $\deg h \geq 1$, these inequalities are actually equalities.  That is, $h$ is linear and $f$ has a rational root.  But the only rational roots $f$ can possibly have are $-1$ and $-3$ (RRT combined with the fact that $f$ has no positive roots), and both of these fail.

In the second case, $g(x) = x^n - x^{n-1} + 3p(x)$ and $h(x) = 1 + 3q(x)$ for some polynomials $p\in \mathbb Z[x]$ and $q\in\mathbb Z[x]$.  Comparing degrees in the same way implies that $h$ is a constant.  In turn, $f$ is irreducible.

[quote=orl]Let $n > 1$ be an integer and let $f(x) = x^n + 5 \cdot x^{n-1} + 3.$ Prove that there do not exist polynomials $g(x),h(x),$ each having integer coefficients and degree at least one, such that $f(x) = g(x) \cdot h(x).$[/quote]

Also [url=https://artofproblemsolving.com/community/c146h150952p850093]PEN Q7[/url]

For storage:\\

FTSOC assume the contrary and set $f(x)=h(x)\cdot g(x)$ where $h(x),g(x) \in \mathbb{Z}[x]$. plugging in $x=0$ yields $h(0)g(0)=3$ and plugging $x=-5$ we get $h(-5)g(-5)=3$.\\

now by RRT, we can clearly observe that there is no integer root of $f(x)$ hence $h(x)$ and $g(x)$ are not linear.\\

set $h(x)=(x-z_{1})(x-z_{2})\cdots (x-z_{r})$ where $\text{deg(h(x))}=r$ and $z_{i}$'s are root of $h(x)$ for $1\leqslant i \leqslant r$\\

now WLOG take $h(0)=\pm 1$ ( since $h(x) \in \mathbb{Z}[x]$)\\

we get $\left|\prod_{i=1}^{r} z_{i}\right|=1$\\

and $\left|\prod_{i=1}^{r}(z_{i}+5)\right|=|(z_{1}+5)(z_{2}+5)\cdots (z_{r}+5)|$\\

now notice for $z_{l} \in \{z_{1},z_{2},\cdots z_{r}\}$ we get $z_{l}^{n-1}(z_{l}+5)=-3$ hence we get:\\

$\left|\prod_{i=1}^{r}(z_{i}+5)\right|=\left|\prod_{i=1}^{r}\frac{-3}{z_{i}^{n-1}}\right|=3^{r}$ which gives that $|h(-5)|=3^{r}$\\

but also since $|h(-5)||3$ ( as $h(-5)g(-5)=3$ and $h(x)\in \mathbb{Z}[x]$) we have that $3^{r}|3$ for $r>1$ which gives a contradiction , and hence the result follows $\blacksquare$

Using Einstein's Criterion we take p=3, in this case 3 divides a0 , 3 dosent  divide an and 3^2 dosent divide 3 therefore by Einstein's criterion it implies that f(x) is irreducible.

@above: that doesn't work.  Eisenstein's Criterion requires $p = 3$ to divide [i]every coefficient except the leading one[/i].  Because $3$ does not divide $5$, you can't naively use Eisenstein here.


geometry
rectangle
combinatorics unsolved
combinatorics

given a rectangle with sides m;n. It is divided into mn squares . there are 3 colors to paint vertexss of squares, they are red, green,blue . it is known that red is paint around the boundary of rectangle. a square is called " properly colored" if there are exact 2 adjacent vertexs which is paint same color ! prove that the number of squares which is call "properly colored" is even!!


calculus
integration
logarithms
derivative
inequalities
calculus computations

I am to determine whether a given series converges or diverges

The Series is

$\sum_{n=1}^{\infty} \frac{\ln n}{n^2}$

I figure I am supposed to use the integral test (that's what section the problem is listed in) but I do not know how to integrate $f(x)$ if $f(x) = \frac{\ln x}{x^2}$

I know the derivative of $\ln x$ is $\frac{1}{x}$

but if I put $\ln x$ as $u$, then after converting $\frac{1}{x} \ dx$ to $du$, I am still left with a $x$ in the denominator.

I'd appreciate if someone could tell me what I'm missing, or show me the way....I feel the solution is simple, I'm just not seeing it...

$x=e^u$.

It's usually a good idea to write your substitutions backwards as well as forward.

[quote]I figure I am supposed to use the integral test...[/quote]
You've come to a site devoted to mathematical problem solving. The general principles that problem solvers live by include never assuming that we're "supposed" to work a problem by any particular method; we work the problem by whatever works.

This is a positive term series, so it either converges or tends to infinity. The issue is how big the terms are, "eventually", and we will decide the question by some sort of comparison. The integral test is one version of a comparison test but by no means the only one.

How does $\ln n$ behave as $n\to \infty?$ It goes to $\infty,$ but slowly - slower than any positive power of $n.$

So what about $\sum \frac{\ln n}{n^2}?$ Can we compare to $\sum \frac1{n^2}.$ No, that's a convergent series, but $\frac{\ln n}{n^2}$ is bigger than $\frac1{n^2}.$ Bigger than the terms of a convergent series doesn't tell us anything; that's an inequality in the wrong direction.

How about using that $\ln n<n$ to compare $\frac{\ln n}{n^2}$ to $\frac{n}{n^2}=\frac1n?$ That's no good, either , since $\sum\frac1n$ diverges; smaller than the terms of a divergent series also doesn't tell us anything.

But note that $\ln n$ is eventually smaller than [b]any[/b] positive power of $n.$ Hence for large enough $n,$ we have $\frac{\ln n}{n^2}<\frac{n^{1/2}}{n^2}=\frac1{n^{3/2}}$

That does it: $\sum\frac1{n^{3/2}}$ is a convergent series (a "$p$-series" with $p>1$). We have a convergent series.

---

OK, you could do it by the integral test. Integrate by parts, differentiating $\ln x$ and integrating $\frac 1{x^2}.$

$\int\frac{\ln x}{x^2}\,dx=-\frac{\ln x}x+\int\frac1{x^2}\,dx=-\frac{\ln x}x-\frac1x.$

---

But I actually saw the problem and immediately saw that it was a convergent series. How? By the thought processes in the first part of my post, only compressed into about 2% of the time it took to write them out.


inequalities
real analysis
real analysis unsolved

Let $ (a_{n})_{n\in\mathbb{N}}> 0$. Prove that then it's equivalent:
(a) $ \sum_{n\equal{}1}^{\infty}\frac{1}{a_{n}}<\infty$
(b) For any given $ (x_{n})_{n\in\mathbb{N}}\geq0$;
$ \sum_{n\equal{}1}^{\infty}a_{n}x_{n}^{2}<\infty$ $ \Rightarrow$ $ \sum_{n\equal{}1}^{\infty}x_{n}<\infty$

First, suppose $ \sum\frac{1}{a_{n}}<\infty.$

By Cauchy's inequality,

$ \left(\sum x_{n}\right)^{2}<\left(\sum\frac{1}{a_{n}}\right)\left(\sum a_{n}x_{n}^{2}\right)$

Hence, if $ \sum a_{n}x_{n}^{2}$ converges, then so also does $ \sum x_{n}.$

The other direction of the equivalence remains to be proved.


inequalities
induction
inequalities proposed

generally of Bernoulli's Inequality:
If $x_{1},x_{2},x_{n}>-1$ is all positive or all negative numbers.Then we have:
$(1+x_{1})(1+x_{2})\cdots(1+x_{n})\ge 1+x_{1}+x_{2}+\cdots+x_{n}$
It can be proved by induction,
And your inequality is a  special case :)

[quote="zhaobin"][quote="Fitim"]$(1+(-x_{1}))(1+(-x_{2}))\cdots(1+(-x_{n}))\geq 1-(x_{1}+x_{2}+\cdots+x_{n})$

 :wink:[/quote]
Please give your reason for it :wink:I don't think you can get any point if you write this in the IMO :P[/quote]

This solutions it's true!
Ok! Why not is true?


trigonometry
logarithms
calculus
calculus open

Let $f(x) = \prod_{n=1}^{\infty}\cos \left( \frac{x}{n \pi}\right)$ and $\sum_{n=0}^{\infty}a_{n}x^{n}$ be its maclaurin series.

Then $a_{n}= 0$ for all positive odd $n$, so it is rational. But what about for even $n$?

I conjectured that every $a_{n}$ is rational. For example, $a_{0}= 1$, $a_{2}=-\frac{1}{12}$, $a_{4}= \frac{11}{4320}$.

Is there someone who can prove or disprove this?

The conjecture is correct. All one needs to do to prove it is to note that $\ln\cos x=\sum_{k\ge 1}b_{k}x^{2k}$ with rational $b_{k}$ (this immediately follows from rationality of the coefficients of $\cos x-1$ and $\ln(1+y)$) and that $\sum_{n\ge 1}\frac 1{(\pi n)^{2k}}$ is rational (this is a well-known fact that can be easily derived from the rationality of the Taylor coefficients of $\tan z$ at $0$ and the residue theorem applied to $\frac 1{z^{2k}\tan z}$). Together, these two observations imply that $\ln$ of the product under consideration can be written as $\sum_{k\ge 1}c_{k}x^{2k}$ with rational $c_{k}$ and it remains to recall that the Taylor coefficients of $e^{z}$ are also rational.


calculus
integration
quadratics
modular arithmetic
number theory unsolved
number theory

Let $ S(n)$ be the sum of decimal digits of a natural number $ n$. Find the least value of $ S(m)$ if $ m$ is an integral multiple of $ 2003$.

[quote="mathangel"]$ \min(S(m)) \equal{} 2$
Choose $ m \equal{} 10^{1001} \plus{} 1 \vdots 2003$ because we have $ 2003|10^{2002} \minus{} 1$ and $ 2003 \not| 10^{1001} \minus{} 1$[/quote]
 :)
In fact,$ 10^{1001} \plus{} 1$ is not divisible by $ 2003$ and the number $ 10^{1001} \minus{} 1$ is.
I will provide my proof later.
[hide="answer:"]3[/hide]

[b]Solution:[/b]
Since there is no $ k$,such that $ 2003|10^{k}$,we may suppose that $ S(m)\geq 2$.

[b]-Claim 1:[/b]

$ 1001$ is the order of $ 10$ modulo $ 2003$.

[i]Proof:[/i]
Consider the set of all positive divisors of $ 1001$.There are exactly $ 8$ of them.After doing some calculations,one can ascertain that neither of these numbers,except $ 1001$, is an order modulo $ 2003$.
On the other hand $ 10^{1001}\equiv 1$ modulo $ 2003$.
[hide="Calculations"]The way to establish these relations could be described as follows:
$ 10^{4}\equiv \minus{} 15$ modulo $ 2003$,therefore, $ 15^{4}\equiv 550$ modulo $ 2003$,therefore
$ 10^{16}\equiv 550$ modulo $ 2003$.
$ 550^2\equiv 47$ modulo $ 2003$,so $ 10^{32}\equiv 47$ modulo $ 2003$.So $ 10^{64}\equiv 206$ modulo $ 2003$.
so $ 10^{96}\equiv 206\cdot 47\equiv 1670$ modulo $ 2003$.
Therefore,$ 10^{100}\equiv 1670\cdot ( \minus{} 15)\equiv 989$.
Now we only need to establish that $ 10^{200}\equiv 989^{2}\equiv 657$ and $ 100^{400}\equiv 1004$.So
$ 10^{800}\equiv 507$.
So $ 10^{1000}\equiv 657\cdot 507\equiv 601$.Therefore,
$ 10^{1001}\equiv 6010\equiv 1$ modulo $ 2003$.[/hide]
Suppose that $ r$ is the order of $ 10$ modulo $ 2003$,then $ r|1001$,but as we've already ascertained,this is impossible,as long as $ r\neq 1001$.

$ \square$

[b]-Claim 2:[/b]
There is no such number $ n$,with $ S(n) \equal{} 2$ and $ n$ is divisible by $ 2003$.

[i]Proof:[/i]

Assume the contrary,so there exist $ n$,of the form either $ 2\cdot 10^{k}$ or $ 10^{k} \plus{} 1$,that is divisible by $ 2003$.
The first case is clearly impossible.
According to the Claim no.1,if $ 10^{k}\equiv \minus{} 1$ mod $ 2003$,then $ 2k$ is divisible by $ 1001$.So $ k$ is divisible by $ 1001$,but then $ 10^{k}\equiv 1$ mod $ 2003$,contradiction.

$ \square$
[b]
-Claim 3:[/b]There exist some natural numbers $ a,b$,such that
$ 10^{a} \plus{} 10^{b} \plus{} 1$ is divisible by $ 2003$.

[i]Proof:[/i]
Assume the contrary.Consider the following sets, specifically reminders of elements modulo $ 2003$:
\[ A \equal{} \{10^{a}|a \equal{} 1,\dots,1001\}
\]

\[ B \equal{} \{ \minus{} 10^{b} \minus{} 1|b \equal{} 1,\dots,1001\}
\]
According to the Claim no.1, neither of the elements from one set,gives the same reminder.
The Claim no.2 yeilds that neither of these two sets contains $ 0$ and $ 2002$,so in common these two sets have at most $ 2001$ elements,but then $ A$ and $ B$ have some common element,it means
$ 10^{a}\equiv \minus{} 10^{b} \minus{} 1$ mod $ 2003$,or
$ 10^{a} \plus{} 10^{b} \plus{} 1$ mod $ 2003$.
And this number has sum of digits precisely equal to $ 3$.

$ \blacksquare$

P.S:
I wonder if it is possible to determine the exact values of $ a,b$?

[quote="greentreeroad"]This is the one of the coolest calculation problems in Olympiad I've ever seen. :wink:

Are students allowed to use calculator?[/quote]

I have the same question  :lol: .
I had to even use C++ to figure out whether my claims are really true or not,actually,mathangel's message was really helpful,by the use of his statement,I understood that,in fact,$ 1001$ is the order of $ 10$ modulo $ 2003$...I have no idea, how to guess any approach to solve it,without this "hint".

As soon as I saw the problem, I thought 2 is the answer. However, this requires $ 10^a\plus{}1$ to be a multiple of 2003 for some a. I tries the case a=1001 using calculator but got the opposite result... Then I guess the answer is 3. However, this means 1001 needs to be an order. I gave up the problem before the terrible amount of checking :oops:

P.S. Is there a good way to test whether a number is a primitive root for a given prime?

[quote="greentreeroad"]
P.S. Is there a good way to test whether a number is a primitive root for a given prime?[/quote]
Calculating the order of $ a$ modulo $ p$ is not a computationally hard problem as these things are usually measured, but in a problem solving context there are various shortcuts that are worthwhile remembering. For example, if $ a$ is a quadratic residue modulo $ p$ then $ a$ is [i]not [/i]a primitive root, and this can often be checked easily using the Legendre symbol.

As an example, in this problem you get a very quick proof that the order of $ 10$ modulo $ 2003$ divides $ 1001$. The Legendre symbol can be calculated literally in your head:

$ \left(\frac{10}{2003}\right) \equal{} \left(\frac{2}{2003}\right) \left(\frac{5}{2003}\right) \equal{} \minus{}\left(\frac{5}{2003}\right) \equal{}
 \minus{}\left(\frac{2003}{5}\right) \equal{} \minus{}\left(\frac{3}{5}\right) \equal{} 1$
where we've used the multiplicative property, then the quadratic character of $ 2$ ($ 2$ is a QR modulo $ p$ iff $ p \equiv \pm 1 \pmod{8}$), then quadratic reciprocity, and finally the trivial fact that $ 3$ is not a square modulo $ 5$.

This shows, very quickly, that $ 10^{1001} \equiv 1 \pmod{2003}$. Checking whether $ 10^k \equiv 1 \pmod{2003}$ for some divisor of $ 1001$ requires more calculation, although of course the computation of powers of $ a$ mod $ p$ can be done relatively fast.

Are the below two statements equivalent?

a)a is a primitive root for a given prime p
b)$ a^{\frac {p \minus{} 1}{2}} \plus{} 1$ is divisible by p

a) leads to b) as $ a^{p \minus{} 1}\equiv 1\bmod p\iff (a^{\frac {p \minus{} 1}{2}} \minus{} 1)(a^{\frac {p \minus{} 1}{2}} \plus{} 1)\equiv 0\bmod p$, so $ a^{\frac {p \minus{} 1}{2}}\equiv 1\bmod p$ (which is impossible since $ p \minus{} 1$ is the order of $ a\bmod p$) or $ a^{\frac {p \minus{} 1}{2}}\equiv \minus{} 1\bmod p$.  The second must hold, so a) leads to b).

b) does not lead to a).  Consider $ p \equal{} 13$ and $ a \equal{} 5$.  We have that $ a^2\equiv 25\equiv \minus{} 1\bmod 13$, so $ a^{3\cdot 2}\equiv ( \minus{} 1)^3\equiv \minus{} 1\bmod 13$.  Then, $ a^6\equiv \minus{} 1\bmod 13$, but $ a^4\equiv 1\bmod 13$.

[quote="mathangel"]My propositions:

1/ (A piece of cake) For any $ m,n \in \mathbb{N}$, if $ m|n$ then there exists a number whose decimal representation consists of $ S(n)$ ones and the rest zeros that is divisible by $ m$.

2/ For any odd prime $ p$, $ \min S(pk | k \in \mathbb{N}) \leq 3$.[/quote]
It is easy to show? that 1) is true.
But 2) is not true. For example consider $ p\equal{}41$, then $ 10^k\equal{}1\mod 41,k\equal{}0\mod 5$,
$ 10^k\equal{}10\mod 41, k\equal{}1\mod 5$,  $ 10^k\equal{}18\mod 41,k\equal{}2\mod 5$, $ 10^k\equal{}16\mod 41,k\equal{}3\mod 4$, $ 10^k\equal{}37\mod 41,k\equal{}4\mod 5$. Therefore $ 41|n\to 41|S_0(n)\plus{}10S_1(n)\plus{}18S_2(n)\plus{}16S_3(n)\minus{}4S_4(n)$, were $ S_i(n)$ is summ of digits for $ 10^k, k\equal{}i\mod 5$.
It is easy to chek, that for $ S(n)\equal{}S_0(n)\plus{}S_1(n)\plus{}S_2(n)\plus{}S_3(n)\plus{}S_4(n)\le 3$ $ 41\not |n$.


Can someone solve the following recursion for the recurrence equation using generating functions?
$p_n= \frac{1}{2} - \frac{1}{2}p_{n-1}$ where $p_1=0$

Why use generating functions? Characteristic equations do the job easily here.

Can someone post a solution using either characteristic equations or generating functions?

Ok, I don't know much about characteristic equations so I will show how I would do it with gf's.

First we define our generating function:

\[G(x)=p_1+p_2x+p_3x^2+\cdots.\]

Now, we look to express the problem in terms of $G(x)$.  We do this by multiply the whole thing by $x^n$ and summing over all values of $x$ in the domain.  Here is what it looks like (I've rearranged a little but you don't have to have it in any particular form):

\[2\sum_{n=2}^\infty x^np_n+\sum_{n=2}^\infty x^np_{n-1}=\sum_{n=2}^\infty x^n.\]

Now we evaluate each part separately:

\begin{eqnarray*}
2\sum_{n=2}^\infty x^np_n &=& 2(x^2p_2+x^3p_3+\cdots)\\
&=& 2(xG(x)-p_1x)\\
&=& 2xG(x)
\end{eqnarray*}

\begin{eqnarray*}
\sum_{n=2}^\infty x^np_{n-1} &=& x^2p_1+x^3p_2+\cdots\\
&=& x^2(p_1+p_2x+\cdots)\\
&=& x^2G(x)
\end{eqnarray*}

\[\begin{eqnarray*}
\sum_{n=2}^\infty x^n &=& x^2+x^3+\cdots\\
&=& x^2(1+x+x^2+\cdots)\\
&=& \frac{x^2}{1-x}\]

Putting this together we get:

\[2xG(x)+x^2G(x)=\frac{x^2}{1-x}\Rightarrow G(x)=\frac{x}{(2+x)(1-x)}.\]

Now we use partial fraction decomposition:

\[G(x)=\frac{x}{(2+x)(1-x)} = \frac 13\left(\frac{1}{1-x}\right)-\frac 13\left(\frac{1}{1+\frac x2}\right).\]

We can expand those expressions in the parentheses:

\begin{eqnarray*}
G(x) &=& \frac 13(1+x+x^2+\cdots)-\frac 13\left(1-\frac 12x+\frac 14x^2-\cdots\right)\\
 &=&  (1-1) +\frac 13\left(1+\frac 12\right)x+\cdots+\left(1+\frac{(-1)^n}{2^{n-1}}\right)x^n
\end{eqnarray*}

So $p_n=\frac 13 \left(1+\frac{(-1)^n}{2^{n-1}}\right).$

PHEW!!!...I really hope this helps  :)

Also, if we define $q_n = p_n - \frac{1}{3}$, we get the recurrence $q_n = -\frac{1}{2} q_{n-1}$.  That recurrence is easier to solve, because we have eliminated the constant.


probability
complementary counting
expected value

Three standard six-sided dice are tossed.  Calculate the number of 5-spots that you expect to appear, and also find the probability of seeing at least one 5-spot.

What does it mean by "5-spot"?

I think is the face of dice with 5 points.

i think by "appear", it means everything except the side facing down? that is, everything you can see.

....now could someone solve this?

No, usually it means that the side facing up is a 5 (and hence the probability is $ \frac{1}{6}$). However, the problem is solvable either way.

Answer:

In this question I assume that by appear we mean the number on the top face.

In this answer the three dice are not interchangable. This means that  two combinations will be the same if and only if the same numbers are shown on the same dice. For instance the combination: die one shows 1, die two shows 3, die three shows 5, is a different combination to: die one shows 1, die two shows 5, die three shows 3.

I will first take the situation when only one die shows a five:

If the first die shows a five then are 5 different possibilities for the second die and 5 different possibilities for the third die. This means that there are 25 different combinations for the first die to show a five. Since there are three dice then there are 75 ways for one die to show a five.

Lets take the case when only two dice show a five:

If both the first and second dice show a five then there are 5 possibilites for the third die. Since there are 3 different ways in choosing a pair of dice from a set of 3, then there are 15 different ways for only two dice to show a five.

There is only one way for all three dice to show a five.

In total there are $ 6\cdot 6\cdot 6 = 216$ different combinations (since the three dice are treated as being distinct).

If we roll all combinations we will see $ 75\cdot 1+15\cdot2+1\cdot 3 = 108$ fives appear.

We can find the density of fives to be $ \frac{108}{216}= \frac{1}{2}$

But this does not give an indication on how many fives we expect to see for only one throw. Hence we must turn to the probabilities of each of the four cases depending on the number of fives present.

For 1 five: chance $ = \frac{75}{216}$
For 2 fives: chance $ = \frac{15}{216}$
For 3 fives: chance $ = \frac{1}{216}$

We can answer now the second part of the question: the probability of at least 1 five $ = \frac{75+15+1}{216}= \frac{91}{216}$

And for the first part: Chance of finding no fives $ = \frac{216-91}{216}= \frac{125}{216}$

Hence I would be expecting no fives to turn up.

Alternatively, on the second part, you could use complementary counting: The probability of no fives is $ (\frac{5}{6})^{3}=\frac{125}{216}$; therefore the probability that there is at least one five is $ 1-\frac{125}{216}-\frac{91}{216}$ (do you see why?)

sludgethrower, your solution to part 1 is incorrect: The expected value is [b]not[/b] the outcome that is most likely to turn up, but the number (of fives) that turn up on average. As a simplified example, when you roll a single die, the expected number of dots is 3.5, even though it is clearly impossible to get this number.

mathnerd314...

[quote]sludgethrower, your solution to part 1 is incorrect: The expected value is not the outcome that is most likely to turn up, but the number (of fives) that turn up on average. As a simplified example, when you roll a single die, the expected number of dots is 3.5, even though it is clearly impossible to get this number.[/quote]

Well I was working by the dictionary definition of expected:

[quote]to look forward to; regard as likely to happen; anticipate the occurrence or the coming of[/quote]

It is extremely unlikely that we will get 0.5 fives. But the mathematical description ,I "expect", is different and more rigorous. When I first worked through the problem I used your definition but I began to question myself :). I have no idea what complementary counting is however! In Scotland probability and statistics is not taught beyond the age of 14...

According to the book [i]Statistics: Decisions and Applications in Business and Economics[/i], by Moshe Ben-Horim and Haim Levy:

[quote]Expected value: $ \mu_{X}= E(X) = \Sigma x P(x)$[/quote]

In other words, the expected value is defined as the sum of all the outcomes, multiplied by the probability of each outcome.

Complementary counting is a nice technique that is often used in American math competitions such as Mathcounts. Basically, the idea is that counting the [b]opposite[/b] of what you want is often simpler.


I think we should set up a general time for FTW users to log onto FTW so that everybody can play a huge game  :D  (the more the merrier  :roll: )

when can you guys be on?

i'm generally on ftw 24/7 except on mondays thursdays and fridays at 7:30-9:00 (i know not exactly 24/7  :| )

5:00 to 6:00-ish...
i think the general "ftw rush-hour" is between 8 and 10 eastern time... though i may be wrong...

[quote="hunter34"]Meh I am on 4:30-5:30.[/quote]

I think it might help if people stated their time zones  :wink: .

we tried to do something like this before and it didn't work out so well.

but i hope it'll work this time...

I'm on everyday after 4 p.m.

I'm in Midwest so the time zone would be CST.

5:00-6:00 and 7:00-8:00 unless it's Friday.
EST.


inequalities
function
inequalities proposed

[quote="arqady"]onlylove_math, Jensen works here too. :wink:[/quote]
Thanks Arqady , we also solve it by only AM-GM:

$ 1 \plus{} \frac {n^2}{n^2(a_1)^2}\ge\ (n^2 \plus{} 1)\frac {1}{\sqrt [n^2 \plus{} 1]{n^{2n}.(a_1)^2n}}$

$ \rightarrow LHS \ge\ (n^2 \plus{} 1)^n\frac {1}{\sqrt [n^2 \plus{} 1]{n^{2n^2}.(a_1.a_2...a_n)^{2n}}}$

$ \rightarrow LHS \ge\ (n^2 \plus{} 1)^n.\frac {1}{\sqrt [n^2 \plus{} 1]{n^{2n^2}((\frac {a_1 \plus{} a_2 \plus{} ... \plus{} a_n}{n})^n)^{2n}}} \ge\ (n^2 \plus{} 1)^n$

And with Jensen we have

Let $ f(t) \equal{} ln(1 \plus{} \frac {1}{(a_1)^2}) \rightarrow f"(t) \equal{} \frac {2(3t^2 \plus{} 1)}{(t^3 \plus{} t)^2} > 0$

$ \rightarrow f(t)$ is convex fuction

Follow Jensen we have

$ f(a_1) \plus{} f(a_2) \plus{} ... \plus{} f(a_n) \ge\ n f(\frac {a_1 \plus{} a_2 \plus{} ... \plus{} a_n}{n})$

$ \leftrightarrow ln(1 \plus{} \frac {1}{(a_1)^2}) \plus{} ln(1 \plus{} \frac {1}{(a_1)^2}) \plus{} ... \plus{} ln(1 \plus{} \frac {1}{(a_1)^2}) \ge\ nln(1 \plus{} \frac {1}{(\frac {a_1 \plus{} a_2 \plus{} ... \plus{} a_n}{n})^2}) \equal{} ln(n^2 \plus{} 1)^n$

$ \rightarrow \left(1 \plus{} \frac {1}{a_1^2}\right)\left(1 \plus{} \frac {1}{a_1^2}\right)\cdot...\cdot\left(1 \plus{} \frac {1}{a_1^2}\right)\geq\left(1 \plus{} n^2\right)^n$

We have done! :)


What's your favorite subject if math is not an option?

lol i wonder if english will get any votes...

essays are evil *shudder*

physics!!! the other sciences are okay, but physics is my favorite by far  :)

I love science, especially chemistry (probably because I have an awesome teacher).  I very much dislike history (mostly because I am bad at it...can't remember jack).  I like to read novels and such but essays kill me.  I'm not bad at them but they are just like pulling teeth to me.  I really like economics  but unfortunately AP didn't make at my school so I'm stuck in CP which honestly means I get only half a year, and much less insight into economics :(

MIne is science. :) I had the best science taeacher last year. I hope I have a good one this year too...

Music is my thing, this is my 8th year of piano and my 4th year of french horn :) I am going to have an awesome science class this year with experiments that could blow my science teachers roof of or set it on fire if we are not careful.

hey, where is history? That should be on the list

I would definitely choose music, even if you had put math in the polling booth. For it is food for the soul...

I guess I could add it, but I though History could be part of Social Studies.

next to math, i love science, especially mechanics.


On a coordinate plane, Bob wants to travel from the point (0,0) to the point (5,5) without traversing through the points (1,2) and (3,4). Assuming he must either go up or right, how many ways can he do this?

[hide="Hint"] Inclusion-Exclusion [/hide]

I think
[hide]The number of ways to get to (5,5) minus the number of ways to get to (1,2) and (3,4) plus the number of ways to get from (1,2) to (3,4) which is

$\frac {10!}{5! \ 5!} - \frac {3!}{2!} - \frac {7!}{4!} + \frac {3!}{2!} = 252 - 35 = 217$[/hide]

There are 10 choose 5 (252) ways total to reach (5,5).  However, there are 3 choose 1 ways to reach (1,2) and then 7 choose 3 ways to reach (5,5) from there. 3 choose 1 times 7 choose 3 is 105, and then there are 7 choose 3 times 3 choose 1 ways to reach it going through (3,4).  252-105-105=42 ways for "Bob" to reach his destination.l

Hi Topper! It's been a while.

Anyways, both answers are not quite correct.

Remember that you are counting the ways to get from (0,0) to (1,2) to (3,4) to (5,5) twice in that arrangement, so you must account for the 'overcount'.

I'm not quite sure if that method would work tarquin... Haven't tried it.

[hide]
10C5, or 252 ways to get to (5,5) from (0,0).

3C1 ways to get to (1,2), and 7C4 ways to get to (5,5) from (1,2), so 3x35=105 ways to go from (0,0) to (5,5) that go through (1,2).

7C3 ways to get to (3,4), and 3C2 ways to get to (5,5) from (3,4), so again, we have 105 ways to go from (0,0) to (5,5) going through (3,4).

4C2 ways to get to (3,4) from (1,2), so there are 3*(4C2)*3= 54 paths from (0,0) to (5,5) that go through (1,2) and (3,4).

252-105-105+54=96 [/hide]

I used Assassin of Death's method.  I came out with the same answer, but I think he made a few addition errors.


Is there any one played game "Harvest moon"?Come in and show me,please?
I love this game,very,very,very,......I hope you also love it like me!  :P  :lol:

http://en.wikipedia.org/wiki/Harvest_Moon_%28series%29

Haven't played it, heard good things about it.

Does anyone ever not use wikipedia as a prime source of information?


Find all positive integers $ n$ such that there exists a permutation $ (a_1,a_2,...,a_n)$ of $ (1,2,...,n)$ satisfying
$ \sqrt{a_1 \plus{} \sqrt{a_2 \plus{} \sqrt{... \plus{} \sqrt{a_n}}}}$ is a rational number


function
calculus
integration
real analysis
real analysis solved

Let f a nonnegative ,continuous and periodical function defined on the reals, such that the arithmetic mean of the numbers f(1), f(2),...f(n) tends to zero when n tends to infinity. Prove that f(k)=0 for any natural number k.

Well, it seems that everyone consider this problem or very simple, or extremely difficult. I assure you that it's not impossible, but it's not trivial at all.

I kind of have a solution, but I used Weyl's theorem:

If a_n={n*i} with i being an irrational positive number then H_n(a,b)/n -> b-a as n->00, where H_n(a,b) = the number of terms from the set {a_1,.., a_n} which are in the interval (a,b), where 0<=a<b<=1.

I considered 2 cases:

Case 1: f has a rational period p/q. Then p is also a period for f and for and the sequence (f(1)+..+f(n))/n, and the conclusion is obvious because  f(k) >= 0 for all real k.

Case 2: f has an irrational period. Let's assume f(1)=k>0. Then for any t>0 there exists c>1 s.t. f(x) > k-t for every x in (1,c). Let the period of f be T, which is irrational. We define the function g(x) in a simmilar way to {x}: it's the fractional part with respect to T, meaning that g(x) is x-nT, where n is the largest natural s.t. nT<=x. Since Weyl's result is also valid here, this means that H(1,c)/n -> (c-1)/T as n->00, where H_n(1,c) is defined as above, only with g(x) instead of the regular {x}. This means that the fraction of numbers from the set {f(1),..,f(n)} which are > k-t becomes >= (c-1)/T- \varepsilon for large enough values of n (and a positive  \varepsilon). This yields the conclusion that (f(1)+..+f(n))/n becomes >= (k-t)((c-1)/T- \varepsilon ) for all large enough values of n, which is a contradiction to the fact that that specific sequence ->0 as n->00, because we can choose t and  \varepsilon as small as we want.

I hope people can understand all of this... :D

Well done, grobber. It's exactly the solution I gave to this problem.

When I solved this, I didn't even know Weyl's thm existed, but that thing it states looked sensible, so I assumed it was true. It wasn't until this morning that I read an article concerning this beautiful result. I don't have a proof of it though.

You know what, there is another solution, but which uses a theorem far less known than that you mentioned. It is due to Weil and says that if x is irrational and f is continuous then  the sequence 1/n* (f({x})+f({2x})+...+f({nx})) tens to the integral of f between 0 and 1. So, the integral must be zero and f must be 0 is it's period is irrational. This was the second solution I gave to this problem.

I think that the topic called " a difficult example" is a much more difficult problem. Have you thought about it?

Hmmm... Not yet. thx for showing me this second soln. It also seems sensible, but, i think, quite difficult to prove.


I am interested in awards at IMC.Are there gold,silver,bronze medals?
 :?:

As far as i know-No, there are the following awards: Grand Grand First Prize, Grand First Prize, First Prize, Second Prize and Third Prize

And how many people get corresponding prizes?

More people get prizes than in IMO ... but I don't know if there is an exact rule as to how the prizes are awarded :?

[quote="cezar lupu"]how can a student participate at IMC? I'm interrested. From what I know so far it is a contest for students from
 universities. can someone give me more details?[/quote]It's a contest for all students, even high-school ones (Kappa participated last year and he was 11th grade). He will probably be there this year too.

For more details you can check out the page http://www.imc-math.org

[quote="Valentin Vornicu"]Yes, if you do not hold a bachelor's diploma in mathematics :)[/quote]What if you're not studying mathematics? :? Then you can participate all your life? :P

Anyway, apparently our holiday plans changed yesterday and it seems I will be going to IMC this year :) Who from mathlinks is going to this competition? ( anyone planning to leave from munchen airport, germany? )

[quote="Peter VDD"][quote="Valentin Vornicu"]Yes, if you do not hold a bachelor's diploma in mathematics :)[/quote]What if you're not studying mathematics? :? Then you can participate all your life? :P[/quote]I think that after a while you will get bored :D

our university wants to save some money so we are going by train :(
~40 hours by train... damn... :mad:

[quote="makc"]our university wants to save some money so we are going by train :(
~40 hours by train... -... :mad:[/quote]Your university at least pays your stay. We have to pay everything on our own...

but we pay for visa ourselves (and btw as we are going by train we have to pay for transit romanian visa)  :?
i'd better pay for plane tickets than spend 40 ours in train...

[quote="Peter VDD"]can't you choose to pay your own plane ticket then? :?[/quote]
i think i can but than i'll have to go alone without our leader and without rest of our team :( and i don't even know where exactly are we going :D

We are going to get Sofia in the same train as you team,Makc; the only difference that we are starting from a day earlier


function
linear algebra
matrix
trigonometry
linear algebra unsolved

Suppose that $\|.\|$ is a norm on $\mathbb{R}^{n}$ and $\|M\|=\sup_{\|v\|=1}\|M.v\|$ the corresponding norm on $M_{n}(\mathbb R)$. Is it true that the function $M \to \| \exp(M) \|$ is convex ?

Breaking news: the exponential function is not convex!  :lol: Specifically, let $M=\begin{pmatrix}0 & t \\-t & 0\end{pmatrix}$. Then $\exp(M)=\begin{pmatrix}\cos t & \sin t \\-\sin t & \cos t\end{pmatrix}$. For any non-Euclidean norm on $\mathbb R^{2}$ the matrix norm $\|\exp(M)\|$ is a nonconstant periodic function of $t$.


solve this equation
$\large\sqrt[3]{1+x}-\sqrt[3]{1-x}=\sqrt[6]{1-x^2}$

[hide]
Let $a=\sqrt[3]{1+x}$ , $b=\sqrt[3]{1-x}$ where $a\geq b$ we get

$a-b=\sqrt{ab}$

$\implies (a-b)^3=ab\sqrt{ab}$

$\implies a^3-b^3-3ab(a-b)=a^3-b^3-3ab\sqrt{ab}=ab\sqrt{ab}$

Hence $2x=4\sqrt{(ab)^3}$ $\implies x=2\sqrt{1-x^2}$ which is $x=\frac{2\sqrt{5}}{5}$ (take only positive because $a\geq b$)
 ;)
[/hide]


ratio
factorial
calculus
integration
induction
function
binomial coefficients

The question did not say, but I assume $n,m$ are positive integers.

$n>m$
\[ S=\frac{m}{n}+\frac{m(m-1)}{n(n-1)}+\cdots+\frac{m!}{n(n-1)\cdots(n-m+1)}\]

Prove: $S= \frac{m}{n-m+1}$

I saw a way of proving such equalities by integrals, but it is long, tiresome and not natural
:)

I think some weird, twisted sort of induction works here:

Let's denote $k=n-m$. We can bring all the fractions to the same denominator and after some calculations we get \[(m+k-1)\cdots(k+1)+ (m-1)(m+k-2)\cdots(k+1)+\cdots+(m-1)! = (m+k)(m+k-1)\cdots(k+2).\] We have to prove this is true. Call this $P(m, k)$ (the sentence that states this is true).

Let's assume $P(m, 1), P(m, 2),.., P(m, k)$ are true for any natural $m$, so we have to prove $P(m, k+1)$ for every $m$. I'm not going to write it all, you can check the calculations. $P(m, k+1)$ true iff $P(m, k+1)-P(m, k)$ true. If we calculate $P(m, k+1)-P(m, k)$ and we divide it by $m-1$ we get $P(m-1, k+1)$, which is true iff $P(m-1, k+1)-P(m-1, k)$ true, since $P(m, k)$ is true for any $m$. We then repeat the process, until we have to prove that $P(1, k+1)$ is true, which is obvious (we just replace in the original sum and we find $1/n=1/n$ which is true). So we proved $P(m, k+1)$. If we proceed like this we can prove that $P(m, k)$ true for any natural $m, k$.

Please check it and tell me if I have made any mistakes. Bye!

the given sum is \[\sum \frac{\binom{m}{k}}{\binom{n}{k}}\] where $\binom{n}{k}$ denotes the binomial coefficients and the sum is from $k=1$ to $m$. Let S(m,n) denote the sum.

Then after fiddling with the sum you can get
\[ S(m,n) = \frac{m}{n}(1 + S(m-1,n-1)) \]
Now you can compute the nested sum. It seems imposing but it is not. Each denominator has $n-m+1$ and everything comes out to be good
If you didn't get something, give me a sign. I'll write it out more clearly
cheers

actually one can prove a stronger result
 \sum  (mCk)/[nC(r+k)]
with the summation variable 'k' ranging from 0 to m, r+m \leq n
the sum is (n+1)/[(n+1-m){(n-m)Cr}]
for ur problem .take r=0 (and subtract 1 ans summation variable ranged from 1 to m,instead of 0)
the proof uses vandermonde's identity and generating function
if u want i'll post it as well
cheers

is this the vandermonde's identity?

[img]
http://invis.free.anonymizer.com/http://mathworld.wolfram.com/c1img2968.gif
[/img]


I obtain this symbol $ n \in I$ when reading Series with Non-Negative terms. May I know how to read the symbol, especially the $ I$?

Thank You.

$ I$ is just the letter, and it stands for an index set of arbitrary size. Note that either all but countably many entries in the sum are zero, or the sum diverges.


geometry
inequalities
inequalities solved

Let Ai be the angles of a triangle , ai the lengths of the sides
of the triangle,and the k the area of the triangle(i=1,2,3).Prove:

(a)

Multiplying 4R^2, we get a1^2 + a2^2 + a3^2 =  \sum 2a1a2 cosA3

It's trivial by 2a1a2 cosA3 = a1^2 + a2^2 - a3^2.

(b)

In the triangle 0 <= A1,A2,A3 <=pi. so sin Ai >=0, and

sinA1sinA2 >=sinA1sinA2cosA3. And similarly,

sinA2sinA3 >=sinA2sinA3cosA1

sinA3sinA1 >=sinA3sinA1cosA2

Adding all the inequalities and multiplying 2, we get

2 \sum sinA1sinA2 >=  2 \sum sinA1sinA2cosA3= \sum sin^2 A1

by (a), as desired.

(equality holds when cosA1=cosA2=cosA3=1 : impossible.)

(c) : Counterexample : When B goes to pi, LHS <=8+8+8=24

But RHS goes 00, Am I confused?

(d)

By Jensen's inequality, (3 \sqrt 3)/2 >= sinA1 + sinA2 + sinA3

And by Cauchy, 3(a1^2 + a2^2 + a3^2)>=(a1+a2+a3)^2

Multiplying two inequalities, We get left inequlity.

(a1^2 + a2^2 + a3^2)(sinA1 + sinA2 + sinA3)

=(a1^2 + a2^2 + a3^2)(a1+a2+a3)/2R

>=27abc/2R

=54 (abc/4R)=54K, and we obtain right, as desired.


25^x - 5^(x+1) = 500

solve for x.

i know you need to make the bases the same but im not sure how.


blogs Asymptote USAMTS geometry email MATHCOUNTS \/closed

Hi all,

We just wanted to let you know that we have moved the site to a new server configuration (a more powerful, faster one) so some stuff might be broken (although we've taken steps to prevent that, we might have missed some of it).

So if you notice anything wrong with the site just let us know.

Also, if you can't log in, we've made it so that for a limited time you can post as a guest in this thread.

PS If you have been redirected to http://209.41.180.112/**** address, it means that you cache hasn't yet fully updated. Give it a couple more hours.

Excellent! I can feel it!

I thought my Inet connection was slow, but now I know I was wrong.

Thanks Valentin!

Blog text got screwed up, I think. At least in the default theme. Everything else I looked at seems okay, but I didn't really investigate the whole site. Looking forward to upgraded classroom software... ;)

We have to copy the css files. We're doing it right now. The new classroom is still in the works, and it might take some time.

:pilot:

I was having problems today and I thought your server was down, but this one is so fast!

I think my post count went down... (I remember it being over 100)  :|  :|

[quote="7h3.D3m0n.117"]It's because the Community Forums were taken off the topic count list  :) .

Possible bug: When I'm logged onto AoPS and go to the Wiki, I am no longer logged in.[/quote]Delete your cookies and try again.

I can't access the my classes, whenever I do, it logs me out.

Are you using the ip address (look at your browsers' URL address bar)?

[quote="Valentin Vornicu"]The new classroom is still in the works, and it might take some time.[/quote]

zomg awesome! xDD

Is it an update or a complete new software?

[quote="Anaxerzia"]zomg awesome! xDD
Is it an update or a complete new software?[/quote]It's a complete new one that I'm writing from scratch. That's why it will take some time.

when i tried to go to the classroom, it would always pop open a new window/tab and I would be logged out.  after i logged in again and went to the classroom i got logged out again!

another problem is that when i DID get to the classroom, instead of "Classroom Loaded 100%" like it usually says in the white space on the page, it said "nulljava.lang.NullPointerException".  All three browsers said the same thing, and Java is otherwise still functioning on my computer.  Restarting it had no effect.

[quote="#H34N1"]It happened again. My account got automatically de-activated and I had to reset my password before I could log in again. I did not forget my password.[/quote]That cannot happen. Just email me when you think your account in deactivated before resetting the password.

during class last time, i typed something and i got logged out of class and had to reset my password

Make sure you don't have stuff affecting your keyboard (caps lock on, key loggers, etc.) I'm always typing my password in, it's long has letters and characters and still I never had to reset it. I'm thinking you must do something wrong somewhere (or rather your computer).

Maybe its a friend, who gave me a couple of virus applications he wrote for me to test out. I'll try reinstalling xp.

my computer (desktop) stilll freezes currently, but in the summer during Mathcounts class, it never did.... what's going on??!!?!?

My computer just kicked me off AoPS for 4 days straight what should I do

I'm not sure if this bug is related to the server move, but when you click a "Download PDF Version" link on a resources page, any images don't show up, they just appear as:
"[img]6611[/img]" or something similar

I believe that problem is unrelated to the server move. The contest section had a few pdfs that had that same problem.

[quote="minicon"]I don't know if this is related, but the avatars in the classroom are huge again.  I just went to test a new browser and it occured.  Just thought I'd let you know before classes start.
:)
EDIT: It seems to only happen with the user's avatar.  My Guitar Hero Avatar [i]only[/i], not the other faces.[/quote]

It seems to be happening to several avatars, not just one.  I noticed 3 that it happened to (including mine) last WOOT class.  Mine worked fine before the server change...

[quote="Nerd_of_the_Ages"]That happened a long time ago, it was happening in the summer classes. So I'm pretty sure that doesn't have to do with the server change.[/quote]

This was the first class that that happened in; my avatar worked fine in all of the previous classes.

It happened in a summer class for me.  Mine and several others' avatars were as big as on the forum.


modular arithmetic
induction
calculus
derivative
integration
geometry
number theory

1. Prove that $n^4 + 4^n$ is always composite for $n>1$.

2. (Easy) Prove that the square of every prime, $p^2 \equiv 1 \pmod {12}$ , $p > 3$

3. Let $x_1+ x_2 + x_3 +.......x_n = 2005$, with $x_1, x_2,x_3.....x_n > 0$, and x is is a positive integer. Find the maximum possible value of $x_1 * x_2 * x_3.....x_n$.

4. (Easy) Prove $n^3 - n$ is divisible by $3$ and $n^5 - n$ is divisible by 5 for the same set of $n$, $n>1$.
[hide]Bonus: For 4, don't use Fermat's Little Theorem for extra credit. Cool way.    [/hide]
5. [b]Poke-fun[/b]
$n$ pokemon are arranged in a circle, with each pokemon holding $1$ badge. When a pokemon has no badges left, he has to return to Ash's pokeball. The $n$ pokemon play the following game. The first pokemon passes $1$ badge to the second, the second passes $2$ badges to the third, the third passes $1$ badge to the fourth, the fourth passes $2$ badge to the fifth, and so on until the last player passes his badge to the first remaing pokemon. Find the values of $n$ for which the game will end with only $1$ pokemon having all the badges.

I've solved all but #3 (well, actually I got the answer, but can't prove it), so forgive me if I am not sure with your proof of 3.

The first one is the identity of Sohpie Germain for odd n, which it has to be or else the sum is even and greater than 2.

[hide="4."]
$n^3-n = n(n^2-1) = (n-1)n(n+1)$. Obviously one of them is a multiple of $3$.

$n^5-n = (n^4-1) = (n-1)n(n+1)(n^2+1)$. It's easy to check that for $n \equiv 0,1,2,3,4 \mod{5}$ one of those is a multiple of $5$.[/hide]

#3's been posted before, use as many 3's as possible and then the rest is 2's.  there's a 'proof' in "the art and craft of problem solving"'s section on eulerian mathematics that is interesting.  can be posted later if you want.

$a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)$ which is composite.

Hence for odd $n$ we let $n=2k+1$. Then $n^4+4^n=n^4+4 \cdot (2^k)^4$ which has the above form.

For number 4, i went for an unusual proof. You can do it by Fermats theorem, induction, or factoring, but by far the coolest, (which generalises to any prime, and thus another way of proving Fermat...) is this:

Consider a necklace with different beads (not circular). If there are 5 beads, of n different colours, there are $n^5$ different chains. If we now throw away the chains where all the colours are the same, we of course have $n^5-n$ different chains, with evey possibility but the chains we just threw out.

Now, if we take all of the necklaces, and make $n^5-n$ differnet circles with them by joining their two ends together, we are going to get the same chains 5 different times, for the five different orientations (kinda crap explination). Therefore $n^5-n$ is divisable by 5. QED

[quote="DPopov"]Prove that all primes greater than three are in the form $6k \pm 1$[/quote]

Well they are all $6k+a$ where $a < 6$. If $a$ is even, $6k+a$ is divisible by $2$ and if $a=3$, $6k+a$ is divisible by $3$. So the only possible values of $a$ are $1$ and $5$. And $6k+5$ is the same as $6k-1$.

Dpopov:

I don't get your proof for 1, nor do I get how to solve that problem.
 :|

cats: would you post it?

sure thing chief.  first i should say what "eulerian mathematics" is in case you're not familiar.  I dont really know but to the best of my knowledge its using calculus in places that you shouldn't to get some results.

the problem in the book is the same as your problem, but instead of 2005 he uses 1976.

"Once again, we shall inappropriately apply calculus to a discrete problem.  It makes intuitive sense for the numbers whose sum is 1976 to be equal (by idea of AM-GM).  But how large should these parts be?  Consider the optimization question of finding the max of $f(x):=(S/x)^x$, where $S$ is a positive constant.  An exercise in logarithmic differentiation shows that $S/x=e$.  Thus, if the sum is $S$ each part should equal $e$ and there should be $S/e$ parts.  Now this really makes no sense if $S$=1976 and the parts must be integers, and having a non-integral number of parts makes even less sense.  But it at least focuses our attention on parts whose size is close to $e=2.71828...$  Once we start looking at parts of size 2 and 3, the problem is close to solution."  (Zeitz, 313).

What don't you get about it? Any number in the form of $a^4+4b^4$ is composite since it can be factored as

\[a^4+4b^4=a^4+4a^2b^2+4b^4+4b^4-4a^2b^2\]
\[ \Rightarrow a^4+4a^2b^2+4b^4+4b^4-4a^2b^2=(a^2+2ab)^2-(2ab)^2\]
\[ \Rightarrow (a^2+2ab)^2-(2ab)^2=(a^2+2b^2+2ab)(a^2+2b^2-2ab)\]

Now for odd $n$, $n=2k+1$, I showed that you can factor $n^4+4^n$ as

\[n^4+4^n=n^4+4 \cdot 4^{2k}\]
\[\Rightarrow n^4+4 \cdot 4^{2k}=n^4+ 4\cdot (2^2)^{2k}\]
\[\Rightarrow n^4+ 4\cdot (2^2)^{2k}=n^4+4 \cdot 2^{4k}\]
\[\Rightarrow n^4+4 \cdot 2^{4k}=n^4+4 \cdot (2^k)^4\]

which has the form $a^4+4b^4$ which I showed is composite; therefore $n^4+4^n$ for odd $n$ is composite.

1234567890, you could have just asked our group. We got all five, and we were first on the first four.

Tetrahed0n: Well, actually, I solved all of them (except 3, which I solved, but couldn't prove) later, and some of them I think our group turned in 3rd. Just thought I'd share them with our fellow AoPS members.  :)

More geometry would be welcome in the Daily Challenges (we solved the trapexoid problem first!)..... I am not too good at counting or number theory. Your group probably will win or David Stiebel's group. We have no chance.....

DPopov, thanks for the explanation. When I was referring to the that problem, I was referring to the problem you posted.

Oh, no problem. I will leave it up a little longer before I post a solution.

For 3, a little number called e is in use

Hm? $x$ is a positive integer. So for $n>4$ we observe that $3(n-3)>n$ and each side has the same sum and the problem becomes trivial. If $x \in \Re$ then you would be correct about using $e$.

No I did! This is where I first learned the application of this partition into the greatest product.

[quote="DPopov"]Show that $n^5+n^4+1$ is composite for $n>1$[/quote]

we have, with $w^3=1$, that $w^2+w+1=0$

Putting this in, we get $w^5+w^4+1=w^2+w+1$

Therefore $n^5+n^4+1$ is divisable by $n^2+n+1$


how do u solve mod word problems?
ex: Several thieves found a package containing some dollar bills. They decided to share the money. They found that when they tried to give 3 dollars to each thug, they ran out of money and one thug received nothing. When each thug took 2 dollars, they had 1 dollar left over. What is the least possible sum of the number of dollars and the number of thieves?  :|

I think that problem was discussed here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1067966115&t=202077

[quote="toadoncart"]srry to dig up an old post, but found a new problem:
"How many positive integers divisible by 4 can be formed using the digits 1,2,3, and 4, each at most once for each integer?" (2001 State Sprint)

is there a quick way to do this w/o slow casework?  :| (the 1,2,3,and 4 makes me feel kinda suspicious)[/quote]\

this one's not too hard. the divisibility rule for 4 is that the last 2 digits of a number have to be divisible by 4. so just find how many numbers you can make that have those last two digits.

Yeah, use the fact that the last two digits must be divisible by four, and that the last digit can't be odd. Also, don't forget to count one digit, two digit, three digit, and four digits numbers, since I always end up forgetting to found the one digit ones.


geometry
circumcircle
trigonometry
symmetry
trapezoid
perpendicular bisector

Equilateral triangles ABD, BCE, ACF are erected outwards from the sides of the triangle ABC.
(i) it is a well known result that the circumcircles of the three equilateral triangles pass through a single point P. Prove it. (However, if you know this well enough, you may move on to the 2nd part, which should be just as easy :D )
(ii) Let DP = x, EP = y, FP = z. Express the area of the triangle ABC in terms of x,y,z.

I shall try to prove it, though it seems hard for me.
And I shall add aonther problem related to that:
[color=blue]If the centres of the 3 triangles are $O_1$, $O_2$,$O_3$, then $\triangle{O_1O_2O_3}$ is also an equilateral triangle.[/color]

[color=blue][b](1)[/b][/color] Let $(O_1), (O_2), (O_3)$ be the circumcircles of the equilateral traingles $\triangle ABD, \triangle BCE, \triangle CAF$ and let the circles $(O_1), (O_2)$ intersect at a point P. Since the angles $\angle ADB = \angle BEC = 60^\circ$ are spanning the major arcs $AB, BC$ of the circles $(O_1), (O_2)$, the angles $\angle APB = \angle BPC = 180^\circ - 60^\circ = 120^\circ$ are equal. Hence, the angle $\angle CPA = 360^\circ - (\angle APB + \angle BPC) = 360^\circ - 2 \cdot 120^\circ = 120^\circ$ is also equal to these 2 angles. Since the angle $\angle CFA = 60^\circ$ spans the major arc CA of the circle $(O_3)$, the minor arc CA of this circle is the locus of points Q, such that the angle $\angle CQA = 180^\circ - \angle CFA = 120^\circ$. Since the point P is such a point, the circle $(O_3)$ passes through the point P. The point P is called the 1st Fermat point of the triangle $\triangle ABC$.

[color=blue][b](2)[/b][/color] Let $P'$ be the intersection of the lines $AE, BF$. The triangles $\triangle CAE \cong \triangle CFB$ are congruent by SAS, because $CE = BC,\ CF = CB$ and the angles $\angle ACB = \angle C + 60^o = \angle BCF$ are equal. Hence, $AE = BF$ and the angles $\angle P'EC \equiv \angle AEC = \angle FBC \equiv \angle P'BC$ are equal. This means that the quadrilateral $P'BEC$ is cyclic and its circumcircle is the circumcircle $(O_2)$ of the equilateral triangle $\triangle BCE$. Similarly, the quadrilateral $P'AFC$ is also cyclic with the circumcircle $(O_3)$ and the point $P' \equiv P$ is identical with the intersection $P$ of these 2 circumcircles. In exactly the same way, we can show that $BF = CD$ and that their intersection is the point $P$.

[color=blue][b](3)[/b][/color] The angles $\angle BPE = \angle CPE = 60^\circ$ spanning the congruent major arcs $BE, CE$ of the circumcircle $(O_2$) of the equilateral triangle $\triangle BCE$ are equal, hence, the line $AE$ bisects the angle $\angle BPC = 120^o$ and similarly, the lines $BF, CD$ bisect the angles $\angle CPA = \angle APB = 120^\circ$. Let the line forming an angle of $60^\circ$ with the line $AE$ meet the line $CD$ at a point $Q$. The triangle $\triangle APQ$ is equilateral, because $\angle PAQ = \angle APQ \equiv \angle APD = 60^\circ$. The triangles $\triangle APB \cong \triangle AQD$ by SAS, because $AP = AQ,\ AB = AD$ and $\angle DAQ = \angle DAP - 60^\circ AP = \angle BAP$ are equal. Hence, $DQ = PB$ and

$PA + PB + PC = PQ + DQ + PC = CD = BF = AE$

[color=blue][b](4)[/b][/color] Denote $x' = CP,\ y' = AP,\ z' = BP$ and $x = DP,\ y = EP,\ z = FP$.

[color=red][b](4a)[/b][/color] Assume that the point P lies inside of the triangle $\triangle ABC$, i.e., that the largest angle of this triangle is less than or equal to $120^\circ$. If this is the case, then

$\triangle_{ABC} = \triangle_{APB} + \triangle_{BPC} + \triangle_{CPA} =$

$= \frac{y'z'}{2} \sin 120^\circ + \frac{z'x'}{2} \sin 120^\circ + \frac{x'y'}{2} \sin 120^\circ = \frac{\sqrt 3}{4} (x'y' + y'z' + z'x')$

and we proved that

$x' + y' + z' = x + x' = y + y' = z + z'$

$x = y' + z',\ \ y = z' + x',\ \ z = x' + y'$

Solving this set of 3 equations for x', y', z' in terms of x, y, z, we get

$x' = \frac{y + z - x}{2},\ \ y' = \frac{z + x - y}{2},\ \ z' = \frac{x + y - z}{2}$

Substituting the expressions for x', y', z' leads to

$\triangle_{ABC} = \frac{\sqrt 3}{16} \left[(y + z - x)(z + x - y) + (z + x - y)(x + y - z) + (x + y - z)(y + z - x)\right] =$

$= \frac{\sqrt 3}{16} \left[z^2 - (x - y)^2 + x^2 - (y - z)^2 + y^2 - (z - x)^2\right] =$

$= \frac{\sqrt 3}{16} \left[x^2 + y^2 + z^2 - (x - y)^2 - (y - z)^2 - (z - x)^2\right] =$

$= \frac{\sqrt 3}{16} \left[2(xy + yz + zx) - (x^2 + y^2 + z^2)\right]$

[color=red][b](4b)[/b][/color] Assume that the point P lies outside of the triangle $\triangle ABC$ and for example, the angle $\angle C > 120^\circ$ is the largest angle of this triangle. If this is the case, then

$\triangle_{ABC} = \triangle_{APB} - \left(\triangle_{BPC} + \triangle_{CPA}\right) =$

$= \frac{y'z'}{2} \sin 120^\circ - \frac{z'x'}{2} \sin 60^\circ - \frac{x'y'}{2} \sin 60^\circ = \frac{\sqrt 3}{4} \left[y'z' - (z'x' + x'y')\right]$

and we proved that

$y' + z' - x' = y + y' = z + z' = x - x'$

$y = z' - x',\ \ z = y' - x',\ \ x = y' + z'$

Solving this set of 3 equations for x', y', z' in terms of x, y, z, we get

$x' = \frac{x - (y + z)}{2},\ \ y' = \frac{z + x - y}{2},\ \ z' = \frac{y + x - z}{2}$

Substituting the expressions for x', y', z' leads to

$\triangle_{ABC} = \frac{\sqrt 3}{16} \left[(z + x - y)(y + x - z) - (y + x - z)(x - y - z) - (x - y - z)(z + x - y)\right] =$

$= \frac{\sqrt 3}{16} \left[x^2 - (y - z)^2 - (z - x)^2 + y^2 - (x - y) + z^2\right] =$

$= \frac{\sqrt 3}{16} \left[x^2 + y^2 + z^2 - (y - z)^2 - (z - x)^2 - (x - y)^2\right] =$

$= \frac{\sqrt 3}{16} \left[2(xy + yz + zx) - (x^2 + y^2 + z^2)\right]$

This is exactly the same formula as for the triangle $\triangle ABC$ with the largest angle less than or equal to $120^\circ$.

[quote="shobber"][color=blue]If the centres of the 3 triangles are $O_1$, $O_2$,$O_3$, then $\triangle{O_1O_2O_3}$ is also an equilateral triangle.[/color][/quote]
The sidelines $O_1O_2, O_2O_3, O_3O_1$ of the triangle $\triangle O_1O_2O_3$ are the perpendicular bisectors of the segments AP, BP, CP. Since the lines $AP \equiv AE, BP \equiv BF, CP \equiv CD$ concurring at the 1st Fermat point P form pairwise angles $\angle APF = \angle BPD = \angle CPE = 60^\circ$, the lines $O_1O_2, O_2O_3, O_3O_1$ perpendicular to the lines AE, BF, CD, respectively, also form pairwise angles $60^\circ$. Since the lines $O_1O_2, O_2O_3, O_3O_1$ pairwise intersecting at the 3 different points $O_1, O_2, O_3$ are obviously not concurrent, the triangle $\triangle O_1O_2O_3$ formed by these 3 lines has all internal angles equal to $60^\circ$, hence, it is equilateral.

Wow, yetti, I'm amazed and impressed!

I can't believe you decided to type all that up! I guess I should have posted this in the olympiad section.. :first:  :10:

Sorry for the double post, but another problem could be added here, I think.

Draw the circumcircle of triangle DEF, and extend EA, DC, FB to meet the circle at A', B' and C', respectively.
Prove that CD=AA'+BB'+CC'.

I'm quite sure this works, but it hasn't been tested yet, so if you're going to prove it, be sure to test it first!

[quote="georgesand"]Draw the circumcircle of triangle DEF, and extend EA, DC, FB to meet the circle at A', B' and C', respectively.
Prove that CD=AA'+BB'+CC'.

I'm quite sure this works, but it hasn't been tested yet, so if you're going to prove it, be sure to test it first![/quote]
You have a typo in the problem, it should be: "extend EA, FB, DC to meet the circle at A', B' and C', respectively." Then it works, but it is a consequence of a simpler problem:

[color=blue][b]Lemma:[/b][/color] Let P be the 1st Fermat point of the triangle $\triangle ABC$ and (O) its circumcircle. Let the concurrent cevians AP, BP, CP intersect the circumcircle (O) at points $A_1, B_1, C_1$ different from the triangle vertices A, B, C, thus forming another triangle $\triangle A_1B_1C_1$ with the same 1st Fermat point and the same circumcircle as the triangle $\triangle ABC$ and with the point triples $(A, P, A_1),\ (B, P, B_1),\ (C, P, C_1)$ collinear. Then $A_1P + B_1P + C_1P = AP + BP + CP$.

[color=blue][b]Proof:[/b][/color] WLOG, assume that the angles $\angle A > \angle B > \angle C$. Since the angles $\angle A_1PB_1 = \angle APB = 120^\circ$, $\angle B_1PC_1 = \angle BPC = 120^\circ$, $\angle C_1PA_1 = \angle CPA = 120^\circ$, the point P indeed is the 1st Fermat point of the triangle $\triangle A_1B_1C_1$. Let $F, F_1$ be vertices of the equilateral triangles $\triangle CAF, \triangle C_1A_1F_1$ constructed on the bases $CA, C_1A_1$, such that the points $F, F_1$ lie on the opposite sides of the lines $CA, C_1A_1$ than the vertices $B, B_1$, respectively. The points $B, P, B_1$ are collinear by the definition of the triangle $\triangle A_1B_1C_1$, the points $B, P, F$ are collinear because P is the 1st Fermat point of the triangle $\triangle ABC$ and the points $B_1, P, F_1$ are collinear because P is the 1st Fermat point of the triangle $\triangle A_1B_1C_1$. Hence the 5 points $F, B_1, P, B, F_1$ are all collinear in this order. Let U, V be the intersection of the sides CF, AF of the equilateral triangle $\triangle CAF$ with the circumcircle (O) of the triangle $\triangle ABC$. Since the points O, F both lie on the perpendicular bisector of the side CA, by symmetry, the lines $UV \parallel CA$ are parallel and the triangle $\triangle UVF$ is also equilateral. The angle

$\angle UCC_1 = \angle UA + \angle ACC_1 \equiv \angle FCA + \angle ACP = 60^\circ + \angle ACP$

is spanning the [color=darkred][u]major[/u][/color] arc $UC_1$ of the circumcircle (O). The angle

$\angle VAA_1 = \angle VAC + \angle CAA_1 \equiv \angle FAC + \angle CAP = 60^\circ + \angle CAP$

is spanning the [color=darkred][u]minor[/u][/color] arc $VA_1$ of the circumcircle (O).

The sum of these 2 angles is

$\angle VCC_1 + \angle UAA_1 = 120^\circ + \angle ACP + \angle CAP = 120^\circ + (180^\circ - \angle CPA) =$

$= 120^\circ + (180^\circ - 120^\circ) = 180^\circ$.

Consequently, the angles spanned by the [color=darkred][u]major[/u][/color] arcs $UC_1, VA_1$ of the same circle (O) are equal, which means that the chords $UC_1 = VA_1$ are equal. This means that the quadrilateral $UVA_1C_1$ is an isosceles trapezoid, the lines $UA_1 \parallel VC_1$ are parallel and $UV = C_1A_1$. It follows that the equilateral triangles $\triangle UVF \cong \triangle C'A'F'$ are congruent and the line $BB_1 \equiv F_1F$ is parallel to the lines $UA_1 \parallel VC_1$. This in turn implies that the quadrilaterals $UA_1BB_1$ and $VC_1BB_1$ are cyclic trapezoids, hence, they are isosceles and $A_1B = UB_1, C_1B = VB_1$. Thus the triangles $\triangle A_1BC_1 \cong UB_1V$ are congruent by SSS and consequently, the triangle pairs $\triangle A_1F_1B \cong \triangle UFB_1,\ \triangle C_1F_1B \cong \triangle VFB_1$ are both congruent by SAS. As a result, we obtained $BF_1 = B_1F$ and

$A_1P + B_1P + C_1P = B_1F_1 = B_1B + BF_1 = B_1B + B_1F = BF = AP + BP + CP$

Now we apply this lemma to the original problem. Denote

$(1)\ \ \ d = AP + BP + CP = AE = BF = CD$

$(2)\ \ \ EP + FP + DP = (AE - AP) + (BF - BP) + (CD - CP) =$

[color=white]. . . . .[/color] $= (AE + BF + CD) - (AP + BP + CP) = 3d - d = 2d$

Since the triangles $\triangle EFD, \triangle A'B'C'$ have the 1st Fermat point P common and a common circumcircle and the point triples $(E, P, A'),\ (F, P, B'),\ (D, P, C')$ collinear, just like the triangles $\triangle ABC, \triangle A_1B_1C_1$, we also have

$(3)\ \ \ A'P + B'P + C'P = EP + FP + DP = 2d$

$(4)\ \ \ AA' + BB' + CC' = (A'P - AP) + (B'P - BP) + (C'P - CP) =$

[color=white]. . . . .[/color] $= (A'P + B'P + C'P) - (AP + BP + CP) = 2d - d = d$

where $d = AE = BF = CD$.

THE PROBLEM IS EASY.
FIRST,YOU CAN PROVE THAT,∠APB=∠BPC=∠CPA=120°
AND P IS THE FERMART POINT OF ΔABC
SECOND,YOU CAN PROVE THAT O1O2⊥CP,O2O3⊥AP,O1O3⊥BP
SO ∠O2O1O3=180°-∠BPC=60°
LIKE THIS,YOU CAN PROVE THAT ∠O1O2O3=∠O2O1O3=∠O2O3O1=60°


limit
logarithms
calculus
calculus computations

I searched but couldn't find these so bear with me and point to the forum links if they exist:

Show that if $ a_{n \plus{} 1} \minus{} a_n \to k$ then $ \frac {{a_n }}{n} \to k$

I also saw the following lemma on this forum but I can't remember what it said or what the preconditions were so I'll put it loosely and if you know of stronger conditions give them in your proof. (I'd like to see a different proof of the following)

Show that if $ a_n$is a positive sequence then $ \lim \frac {{a_{n \plus{} 1} }}{{a_n }} \equal{} \lim a_n ^{\frac {1}{n}}$if the former limits exists.

[quote="§outh§tar"]
Show that if $ a_n$is a positive sequence then $ \lim \frac {{a_{n \plus{} 1} }}{{a_n }} \equal{} \lim a_n ^{\frac {1}{n}}$if one of the limits exists.[/quote]
I think it is wrong,try:
$ a_n \equal{} 2 \plus{} ( \minus{} 1)^n$,then the second one limit exists,but the first one doesn't exist :)

And the right should be,if the former one exists,then the latter one also exists.

Assume that $ \lim\frac{a_{n\plus{}1}}{a_{n}}\equal{}l$. Let $ b_{n}$ be the sequence defined by:

$ b_{n}\equal{}\frac{a_{n\plus{}1}}{a_{n}}$.

Since $ b_{n}$ is a positive convergent sequence it is known that

the geometric mean of $ b_{n}$ converges to the same limit. Now,

$ \sqrt[n]{b_{1}b_{2}\cdots b_{n}}\equal{}\sqrt[n]{\frac{a_{n\plus{}1}}{a_{1}}}\rightarrow l$

and the problem is solved.

The first part of your problem is a simple application of Cesaro-Stolz lemma.

For the second- take logarithms to reduce to the first. The converse fails- there are no reasonable conditions to make it work.

Thanks to all. didilica the geometric mean application is a result of cesaro-stolz. jmerry you are the most intelligent person on this forum!

For reference:
http://www.math.ksu.edu/~nagy/snippets/stolz-cesaro.pdf


The problem's inspired from the other, but this one is unsolved now. It looks very simple, but really difficult!

Prove that for all $a,b,c$ are non-negative reals we have the inequality :

$3(a+b+c) \ge 2(\sqrt{a^2+bc}+\sqrt{b^2+ac}+\sqrt{c^2+ab})$

I hope I 'll see a nice solution for it!

Good luck every one!

To Vasc : I think You 've thought about it? Do you have a solution?

I think this : $\ 3(a+b+c) \ge 2(\sqrt{a^2+bc}+\sqrt{b^2+ac}+\sqrt{c^2+ab})$ is the same as to prove one of this two $\ 3(a+b+c) \ge 2(\sqrt{a^2+ab}+\sqrt{b^2+bc}+\sqrt{c^2+ca})$ and/or $\ 3(a+b+c) \ge 2(\sqrt{a^2+ac}+\sqrt{b^2+ab}+\sqrt{c^2+bc})$
Since $\ 2(\sqrt{a^2+ac}+\sqrt{b^2+ab}+\sqrt{c^2+bc}) * 2(\sqrt{a^2+ab}+\sqrt{b^2+bc}+\sqrt{c^2+ca}) \geq (2(\sqrt{a^2+bc}+\sqrt{b^2+ac}+\sqrt{c^2+ab}))^2$ I could be wrong  :D
And give this at first might be a good idea : Assume $\ a\geq b\geq c \geq 0$

Or we may use you hint, Vasc, applying AM-GM twice, which leads to an really ugly equation, dunno what's worng about using it for the second time :(
Frank Morgan

To be honest, I thounght about it too. :)

[quote="Vasc"][quote="hungkhtn"]The problem's inspired from the other, but this one is unsolved now. It looks very simple, but really difficult!
To Vasc : I think You 've thought about it? Do you have a solution?[/quote] A hint for a solution to this inequality is the following:
Assume that $a \geq b \ge c$ and use the inequality $\sqrt{b^2+ca}+\sqrt{c^2+ab} \leq 2\sqrt{\frac{b^2+c^2+a(b+c)}{2}}$.[/quote]

It seem very obvious? can you post the more details?

[quote="Vasc"]The next step is squaring.
Cezar, can you continue the proof ?[/quote]

I can't believe that It can be solved by the method!  :) It's supricing!

I can prove a weaker inequality without equality and replacing $2/3$ with $\sqrt{3}$ using PM-AM  :lol: I'll have to think more to get the stricter inequality...

[quote="hungkhtn"] I can't believe that It can be solved by the method!  :) It's supricing![/quote]It can.
 By squaring, the inequality turns out into
$(a-2s)^2+20bc \geq 12(a+2s)(\sqrt{a^2+bc}-a)$, where $s=(b+c)/2$. Since
$\sqrt{a^2+bc}-a=\frac{bc}{\sqrt{a^2+bc}+a} \leq \frac{bc}{2a}$, the inequality reduces to
$(a-2s)^2+2bc+\frac{12(a-s)bc}{a} \geq 0$. ;)

[quote="hungkhtn"]Prove that for all $ x,y,z$ are nonnegative reals we have the inequality :

$ 3(x \plus{} y \plus{} z)\geq 2\left(\sqrt {x^2 \plus{} yz} \plus{} \sqrt {y^2 \plus{} zx} \plus{} \sqrt {z^2 \plus{} xy}\right).$[/quote]$ \sum{2\sqrt {x^2 \plus{} yz}}\leq \sum{\frac {11x^3 \plus{} 28x^2(y \plus{} z) \plus{} 7x\left(y^2 \plus{} 6yz \plus{} z^2\right) \plus{} 2(y \plus{} z)\left(y^2 \plus{} 10yz \plus{} z^2\right)}{5\left(x^2 \plus{} y^2 \plus{} z^2\right) \plus{} 14(yz \plus{} zx \plus{} xy)}}$

$ \equal{} 3(x \plus{} y \plus{} z).$

[hide="Proof."]$ \left[11x^3 \plus{} 28x^2(y \plus{} z) \plus{} 7x\left(y^2 \plus{} 6yz \plus{} z^2\right) \plus{} 2(y \plus{} z)\left(y^2 \plus{} 10yz \plus{} z^2\right)\right]^2$

$ \minus{} 4\left(x^2 \plus{} yz\right)\left[5\left(x^2 \plus{} y^2 \plus{} z^2\right) \plus{} 14(yz \plus{} zx \plus{} xy)\right]^2$

$ \equal{} \frac {\left[63x^3 \plus{} 84x^2(y \plus{} z) \minus{} x\left(125y^2 \minus{} 284yz \plus{} 125z^2\right) \minus{} 22(y \plus{} z)(y \minus{} z)^2\right]^2}{189}$

$ \plus{} \frac {3472x^3yz(y \plus{} z)}{9} \plus{} \frac {4x^2yz\left(36204y^2 \plus{} 27095yz \plus{} 36204z^2\right)}{189}$

$ \plus{} \frac {16x(y \plus{} z)(y \minus{} z)^2(y \plus{} z \minus{} x)^2}{9} \plus{} \frac {608xy^2z^2(y \minus{} z)^2}{189(y \plus{} z)}$

$ \plus{} \frac {4(y \minus{} z)^2\left[x^2(y \plus{} z)\left(68y^2 \plus{} 15yz \plus{} 68z^2\right) \minus{} 2x(y \minus{} z)^2\left(68y^2 \plus{} 117yz \plus{} 68z^2\right) \plus{} (y \plus{} z)(y \minus{} z)^2\left(68y^2 \minus{} 53yz \plus{} 68z^2\right)\right]}{189(y \plus{} z)}$

$ \geq 0,$

the last term is positive semi-definite for the discriminant

$ \Delta_{x} \equal{} (y \minus{} z)^4\left(68y^2 \plus{} 117yz \plus{} 68z^2\right)^2$

$ \minus{} (y \minus{} z)^2(y \plus{} z)^2\left(68y^2 \plus{} 15yz \plus{} 68z^2\right)\left(68y^2 \minus{} 53yz \plus{} 68z^2\right)$

$ \equal{} \minus{} 4y^2z^2(y \minus{} z)^2\left(3043y^2 \plus{} 6447yz \plus{} 3043z^2\right)\leq 0$.[/hide]
Another proof see : http://www.mathlinks.ro/Forum/viewtopic.php?t=124812

Analogic inequality see : http://www.mathlinks.ro/Forum/viewtopic.php?t=124116

[quote="Vasc"][quote="hungkhtn"] I can't believe that It can be solved by the method!  :) It's supricing![/quote]It can.
 By squaring, the inequality turns out into
$ (a \minus{} 2s)^2 \plus{} 20bc \geq 12(a \plus{} 2s)(\sqrt {a^2 \plus{} bc} \minus{} a)$, where $ s \equal{} (b \plus{} c)/2$. Since
$ \sqrt {a^2 \plus{} bc} \minus{} a \equal{} \frac {bc}{\sqrt {a^2 \plus{} bc} \plus{} a} \leq \frac {bc}{2a}$, the inequality reduces to
$ (a \minus{} 2s)^2 \plus{} 2bc \plus{} \frac {12(a \minus{} s)bc}{a} \geq 0$. ;)[/quote]
How can you get this part, Vasc? I also apply Cauchy-Schwartz and squaring exactly same as you did but I just get the following ugly inequality:
a^2+b^2+14ab+5c^2+10(ac+bc)>=8.square root(2)[(c^2+ab)(a^2+b^2+ac+bc)]
Sorry for my stupidity but how can you transform this inequality into the form above?

[quote=hungkhtn]The problem's inspired from the other, but this one is unsolved now. It looks very simple, but really difficult!

Prove that for all $a,b,c$ are non-negative reals we have the inequality :

$3(a+b+c) \ge 2(\sqrt{a^2+bc}+\sqrt{b^2+ac}+\sqrt{c^2+ab})$

[color=#f00]I hope I 'll see a nice solution for it![/color]

Good luck every one!

To Vasc : I think You 've thought about it? Do you have a solution?[/quote]
[url=https://artofproblemsolving.com/community/c6h1954963p13504395]anhduy98:[/url]
Suppose : $  c \ge a,b. $ Use  (AM-GM) we have :
$$\bullet \sqrt{a^2+bc} \le \frac{1}{2}\left( \frac{a^2+bc}{a+b}+a+b\right)=\frac{2a^2+b^2+2ab+bc}{2(a+b)}$$
$$\bullet \sqrt{b^2+ca} \le \frac{1}{2}\left( \frac{b^2+ca}{a+b}+a+b\right)=\frac{a^2+2b^2+2ab+ca}{2(a+b)}$$
$$\bullet \sqrt{c^2+ab}=\sqrt{\left(c+ \frac{ab}{a+b}\right)^2-\left(\frac{ab}{a+b}\right)^2-\frac{ab(2c-a-b)}{a+b}}\le c+\frac{ab}{a+b}=\frac{2ab+2bc+2ca}{2(a+b)}$$
$$\rightarrow LHS \le \frac{3a^2+3b^2+6ab+3bc+3ca}{2(a+b)}=\frac{3(a+b+c)}{2}.$$

[quote=Crazy_LittleBoy][quote=xzlbq]If this is hard?

$x,y,z>0$,prove that:

\[\frac{3}{2}\,\sqrt { \left( x+y+z \right) ^{2}-9\,{\frac { \left( xy+zx+yz
 \right) xyz}{ \left( x+y+z \right) ^{3}}}}\geq \sqrt {{x}^{2}+yz}+\sqrt {
{y}^{2}+zx}+\sqrt {{z}^{2}+xy}\][/quote]

Easy to prove:$$\sqrt {{x}^{2}+yz}+\sqrt {{y}^{2}+zx}+\sqrt {{z}^{2}+xy}\leq 1/2\,
\sqrt {9\, \left( x+y+z \right) ^{2}-81\,{\frac {xyz \left( xy+zx+yz
 \right) }{ \left( x+y+z \right) ^{3}}}-48\,{\frac { \left( x-y
 \right) ^{2} \left( y-z \right) ^{2} \left( z-x \right) ^{2}}{
 \left( x+y+z \right) ^{4}}}}
$$[/quote]

Could you please show the details of the derivation?


inequalities
geometry
circumcircle
trigonometry
power of a point
inequalities unsolved

Fuzzylogic used this ineq to solve a medians ineq

 $\displaystyle m_a \geq \frac{b^2+c^2}{4R}$

I would like to see a geometric and a trigonometric solution.

Thank you very much.

The proof I knew is a nice geometric proof using power of a point. I will leave this to interested forum members.

I used this inequality for the proofs of:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=16012

and

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15979

Fuzzylogic, can you show me your powerpoint proof? I'm very interested.

Here is my trigonometric proof:

Let D be the midpoint of the side BC of triangle ABC, and O the circumcenter of triangle ABC. Then $AD = m_a$, AO = R and OD = R cos A. Now, after the Cosine Law in triangle OAD, we have

$OD^{2}=AD^{2}+AO^{2}-2\cdot AD\cdot AO\cdot \cos \measuredangle DAO$,

so

$2\cdot AD\cdot AO\cdot \cos \measuredangle DAO=AD^{2}+AO^{2}-OD^{2}$.

In other words,

$2m_{a}R\cdot \cos \measuredangle DAO=m_{a}^{2}+R^{2}-\left( R\cos A\right) ^{2}$.

Now, since for every real x we have $1\geq \cos x$, we have

$2m_{a}R\geq 2m_{a}R\cdot \cos \measuredangle DAO=m_{a}^{2}+R^{2}-\left( R\cos A\right) ^{2}$
$=m_{a}^{2}+R^{2}-R^{2}\cos ^{2}A=m_{a}^{2}+R^{2}\left( 1-\cos ^{2}A\right) =m_{a}^{2}+R^{2}\sin ^{2}A$
$=m_{a}^{2}+\left( R\sin A\right) ^{2}=m_{a}^{2}+\left( \frac{1}{2}a\right) ^{2}$

since a = 2R sin A. Now use the formula $m_{a}^{2}=\frac{1}{4}\left( 2b^{2}+2c^{2}-a^{2}\right) $ to get

$2m_{a}R\geq m_{a}^{2}+\left( \frac{1}{2}a\right) ^{2}=\frac{1}{4}\left( 2b^{2}+2c^{2}-a^{2}\right) +\left( \frac{1}{2}a\right) ^{2}$
$=\frac{1}{4}\left( 2b^{2}+2c^{2}-a^{2}\right) +\frac{1}{4}a^{2}=\frac{1}{4}\left( 2b^{2}+2c^{2}\right) =\frac{1}{2}\left( b^{2}+c^{2}\right) $,

and $4m_{a}R\geq b^{2}+c^{2}$, thus $m_{a}\geq \frac{b^{2}+c^{2}}{4R}$.

  Darij

[quote="fuzzylogic"]Let M be the midpoint of BC, Let AM intersects the circumcircle at M', by power of a point: $AM\cdot MM'=BM\cdot MC$ So $MM' = a^2/(4m_a)$. Hence

$2R \geq AM^{\prime} = AM + MM^{\prime} = m_a + \frac{a^2}{4m_a} = \frac{4m_a^2 + a^2}{4m_a} = \frac{2b^2+2c^2}{4m_a} = \frac{b^2+c^2}{2m_a}$[/quote]

Very nice proof!

One final thing: Both of our proofs shows that in the inequality $m_a \geq \frac{b^2+c^2}{4R}$, equality holds if and only if b = c.

  Darij

hello, by substituting
$m_a^2=\frac{b^2}{2}+\frac{c^2}{2}-\frac{a^2}{4}$ and $R=\frac{abc}{4A}$ with $A=\sqrt{s(s-a)(s-b)(s-c)}$ and $s=\frac{a+b+c}{2}$
we get after some calculations from
$m_a \ge \frac{b^2+c^2}{4R}$
the following
$\frac{1}{16}(b-c)^2(b+c)^2(b^2+c^2-a^2)^2\ge0$.
Sonnhard.

Here is another way using Ptolemy Theorem and without using the formula for the median.

Let M be the midpoint of BC, Let AM intersects the circumcircle at M'. By Ptolemy Theorem,
 $AM'\cdot a = M'C\cdot c + M'B\cdot b$ ---------- (*)

Since $\Delta{M'CM}$ is similar to $\Delta{BAM}$, we have M'C/BA = MC/MA. That is, $M'C = ac/(2m_a)$. Similarly, $M'B = ab/(2m_a)$.

Hence, from (*) we deduce that $2R \geq AM^{\prime} = \frac{b^2+c^2}{2m_a}$.


Hi all,
In the equation a= b (over ) ed, the SI units of the quantity of b are Kg m^3 s^-2
the SI units of quantity e are Kg and the SI units of quantity d are m s^-2

What are the correct SI units for the quantity a  ?

Please help thx trigger.[/u]

Don't post the same question multiple times in different forums.  I answered your question in the other thread.