modular arithmetic
quadratics
number theory
relatively prime

Hello,

$ i)$ Prove that , $ \forall n \in \mathbb{N}$ : $ \sqrt{\frac{n}{n\plus{}2}}\not\in \mathbb{Q}^*$

$ ii)$Prove that , $ \forall a,b \in \mathbb{N}^*$, with $ a>b$ : $ \frac{a^2\plus{}b^2}{a^2\minus{}b^2}\not\in \mathbb{N}$

$ iii)$ Let $ a,b \in \mathbb{N}$, such that : $ \alpha^2\plus{}2(a\plus{}b)\alpha \minus{} (a^2\plus{}b^2)\equal{}0$ , Prove that :     $ \alpha \not\in\mathbb{Q}$

[hide="#1"]$ \frac{n}{n\plus{}2}\equal{}1\minus{}\frac{2}{n\plus{}2}$ $ \implies$ $ \sqrt{\frac{n}{n\plus{}2}}$ is Irational number.[/hide]

[quote="enndb0x"][hide="#1"]$ \frac {n}{n \plus{} 2} \equal{} 1 \minus{} \frac {2}{n \plus{} 2}$ $ \implies$ $ \sqrt {\frac {n}{n \plus{} 2}}$ is Irational number.[/hide][/quote]

why ? it's not clear

Note: If  $ \sqrt[k]{a}$ is natural number then $ a\equal{}t^k$ for some integer $ t$ and $ k$.

$ 1\minus{}\frac{2}{n\plus{}2}$ is not perfect square since $ \forall n\in \mathbb{N}$  the number $ 1\minus{}\frac{2}{n\plus{}2}< 1$ so it can not be perfect square.

The fact that is it less than one doesn't mean that it can't be a perfect square. Try $ \sqrt {\frac {9}{16}}$.

EDIT:

[hide="#1"]
We have $ \sqrt {\frac {n}{n \plus{} 2}} \equal{} \frac {\sqrt {n}}{\sqrt {n \plus{} 2}}$. We know that $ \gcd{(n,n \plus{} 2)} \equal{} 2$, so we write $ n \equal{} 2q$ for $ q \in \mathbb{N}$. In order for the fraction to be rational, the numerator and denominator must be rational. Therefore, $ q \equal{} j^2$ and $ q \plus{} 2 \equal{} k^2$ for $ j,k \in \mathbb{N}$. We have $ k^2 \minus{} j^2 \equal{} 2$. We know that the difference between two perfect squares is odd, thus the expression is irrational for all $ n \in \mathbb{N}$. $ \Box$
[/hide]

Uhh... a couple of flaws. First, for odd $ n$, $ \gcd(n,n \plus{} 2) \equal{} 1$.

[hide="#1"]
We have two cases: $ 2|n$ or $ 2\not|n$.

Case 1: $ 2 |n$. We have $ \gcd(n,n \plus{} 2) \equal{} 2$, so substituting $ n \equal{} 2k$  we have $ \frac n{n \plus{} 2} \equal{} \frac k{k \plus{} 1}$. Since $ \gcd(k,k \plus{} 1) \equal{} 1$, we have that $ \sqrt {\frac k{k \plus{} 1}} \in \mathbb Q$ if and only if both $ k$ and $ k \plus{} 1$ are perfect squares. Let $ k \equal{} a^2$ and $ k \plus{} 1 \equal{} b^2$. We have $ b^2 \minus{} a^2 \equal{} (b \minus{} a)(b \plus{} a) \equal{} 1 \implies b \minus{} a \equal{} 1\text{ and }b \plus{} a \equal{} 1$ so $ a \equal{} 0$, a contradiction.

Case 2: $ 2 \not| n$. We have $ \gcd(n,n \plus{} 2) \equal{} 1$, so $ \sqrt {\frac n{n \plus{} 2}}\in \mathbb Q$ if and only if both $ n$ and $ n \plus{} 2$ are perfect squares. Let $ n \equal{} a^2$ and $ n \plus{} 2 \equal{} b^2$. We have $ b^2 \minus{} a^2 \equal{} (b \minus{} a)(b \plus{} a) \equal{} 2$. However, by parity, $ (b \minus{} a)(b \plus{} a) \not\equiv 2 \pmod 4$, a contradiction.[/hide]

[quote]$ (n,n \plus{} 2) \equal{} 2$[/quote]

For even $ n$.

[quote="sunehra"]
In order for the fraction to be rational, the numerator and denominator must be rational. [/quote]

Only if it's in lowest terms.

[quote]We know that the difference between two perfect squares is odd[/quote]

Consecutive squares.

[hide="Solution"]Obviously $ (n,n \plus{} 2)\le 2$; if it's equal to $ 1$, then the two have to be $ 2$ apart $ \pmod{4}$; contradiction by quadratic residues in that mod. If it's $ 2$, then it's $ \sqrt {k/(k \plus{} 1)}$; clearly these are relatively prime, hence both numerator and denominator are rational, hence $ k \equal{} 0\Rightarrow n \equal{} 0\not\in \mathbb{N}$.[/hide]

EDIT: Beaten by Yongyi. He managed to stretch the same solution out into around four times as many lines though.  :P

Temperal: Your solution is more handwaving :P

[hide="#2"]
Assume that $ a^2\plus{}b^2\equal{}k(a^2\minus{}b^2)$ for some $ k\in \mathbb N$. This equation transforms into $ (k\minus{}1)a^2\minus{}(k\plus{}1)b^2\equal{}0 \implies (k\minus{}1)a^2\equal{}(k\plus{}1)b^2 \implies \frac{k\plus{}1}{k\minus{}1}\equal{}\left(\frac ab\right)^2$, which is proven false by (1).
[/hide]

[hide="#2"]
Suppose by way of contradiction that $ \frac {a^{2} \plus{} b^{2}}{a^{2} \minus{} b^{2}}\in\mathbb{N}$ for some $ a, b$. Without loss of generality we can assume that $ gcd(a, b) \equal{} 1$ since we can divide the numerator and denominator by their common factor. Since there are no two positive squares which differ by 1, $ a^2 \minus{} b^2 > 1$, so there exists a prime $ p$ such that $ p|a^2 \minus{} b^2$, so $ p|a^2 \plus{} b^2$ as well. Thus, $ p|2a^2$ and $ p|2b^2$, so $ p \equal{} 2$ since $ gcd(a, b) \equal{} 1$. However, for $ a^2 \minus{} b^2$ to be even, it must be divisible by 4, but then $ 4|a^2 \plus{} b^2$ and both $ a, b$ must be even since squares are 0 or 1 mod 4, contradicting the fact that $ gcd(a, b) \equal{} 1$. Thus, no such $ a, b$ can exist.
[/hide]

"Handwaving"? It was exactly the same as yours, logically, as I noted.

[hide="3"]
For $ \alpha$ to be rational, the discriminant $ \sqrt{b^2\minus{}4ac}\equal{}\sqrt{4(a\plus{}b)^2\plus{}4(a^2\plus{}b^2)}\equal{}2\sqrt{2(a^2\plus{}ab\plus{}b^2)}$ needs to be rational. But for the expression in the parentheses to have an odd power of two, $ ab$ needs to be $ 2\pmod{4}$; but then the expression is odd; contradiction.
[/hide]


inequalities
induction
rearrangement inequality

How do you prove the rearrangement inequality by induction?

Anyone?

If not by induction what other method of proof could be used?

Here's a heuristic argument (that is, not actually a proof :wink: ):

You're in front of a cash register, full of 1, 5, 10, 20 and 100 dollar bills.  You only have enough time to grab 3 of one type of bill, 5 of another, 7 of another, 9 of another and 11 of the last type of bill before the shopkeeper returns.  In order to maximize your profit, what do you do?

The full proof follows from answering the question, "How much do I actually lose if I make a single 'sub-optimal' choice in grabbing the money?" where by "sub-optimal" I mean relative to what the rearrangement inequality tells you is optimal.  If you calculate this difference and it's positive (i.e. you actually do lose money) then you can induct.


geometry
function
trigonometry
analytic geometry

In the cartesian plane, consider the curves $x^2+y^2=r^2$ and $(xy)^2=1$. Call $F_r$ the convex polygon with vertices the points of intersection of these 2 curves. (if they exist)

(a) Find the area of the polygon as a function of $r$.
(b) For which values of $r$ do we have a regular polygon?

due to symettry we only consider the first quadrant.

We have a quarter circle and the graph of y = 1/x.  It either intersects in one point (r=1, a square) or two points (where we must have an octahedron.)

The points of intersection between y = 1/x and $x^2 + y^2 = r^2$ are when $x^4 -r^2x^2 + 1 = 0$ so that $x = \sqrt{ \dfrac{r^2 \pm \sqrt{r^2 - 4} }{2} }$.  The larger of those x values must be $r \cos \pi /8$.  We solve, and get $r^2 + \sqrt{r^2 - 4} = 2r^2\cos \pi /8$.  Now $r^4(2\cos \pi/8 -1)^2 = r^2 - 4$.  Finally, from cos(x/2) = sqrt( (cosx + 1) /2 ), we have $r^4(2 + \sqrt{2} + 1 - 2\sqrt{2 + \sqrt{2}}) - r^2 + 4 = 0$.

Now it is possible to solve for r^2 and hence for r.


Given that $ a_{1}, a_{2}, ..., a_{29}$ is a strictly monotonic increasing sequence of integers such that $ a_{k\plus{}7} \minus{} a_{k} \leq 13$ for every integer $ 1 \leq k \leq 22$, prove that there exist indices $ i$ and $ j$ such that $ a_{i} \minus{} a_{j} \equal{} 4$.

we suppose that $ c_1,c_2,...,c_{29}$ is a strictly monotonic increasing sequence of integers such that $ c_{k\plus{}7}\minus{}c_k\le 13$
we take $ a_i\equal{}c_i\minus{}c_1$ then $ 0\equal{}a_1<a_2<....<a_{29}$
we have $ a_{29}\equal{}\sum_{k\equal{}0}^{3}a_{7k\plus{}8}\minus{}a_{7k\plus{}1}\le 13\times 4$
so $ \{a_1,a_2,...,a_{29}\}\subset\{0,1,2,...,13\times 4\}$ gives $ \exists\ a_{b_1}<a_{b_2}<..<a_{b_{8}}$ such that $ a_{b_{i}}\equiv a_{b_j}(mod\ 4)$ we take ($ a_{b_{i\plus{}1}}\minus{}a_{b_i}\equal{}4q_i$)
and we have $ \sum_{i\equal{}1}^7 a_{b_{i\plus{}1}}\minus{}a_{b_{i}}\equal{}a_{b_8}\minus{}a_{b_1}\le 13\times 4$
so $ \sum_{i\equal{}1}^7 q_i\le 13$ so $ \exists i\in\{1,2,..,7\}: \ q_i\equal{}1$
then $ c_{b_{i\plus{}1}}\minus{}c_{b_i}\equal{}a_{b_{i\plus{}1}}\minus{}a_{b_i}\equal{}4$

[hide="Solution"]We can add a constant to all terms without changing anything, so WLOG $ a_{1} \equal{} 0$.  Then using $ a_{k \plus{} 7} \minus{} a_{k}\le 13$ four times, we get $ a_{29}\le 52$.  Assume no terms differ by 4.  We put the integers from 0 to 52 in sets based on their residues modulo 4.  There are 13 integers in each set except the 0 modulo 4 set, which has 14 integers.  We can take at most 7 integers from each set, but since there are 28 integers total, we must take exactly 7 from each set.  There is only one possible selection of integers, $ 0,1,2,3,8,9,10,11,..., 48,49,50,51$.  But here $ a_{9} \minus{} a_{2} \equal{} 16 \minus{} 1 > 13$, a contradiction.  Therefore we can always find two terms with difference 4.[/hide]


algebra
polynomial
Putnam
induction
strong induction

Prove that for $m = 0, 1, 2, ...$

$S_{m}(n) = 1^{2m+1}+2^{2m+1}+...+n^{2m+1}$

Is a polynomial in $n(n+1)$.

(This may actually be just Olympiad level, since most of the problems in this book are quite tough, and some are obviously Putnam-like.)

I got inductive proof, though I guess more concise approaches could be found.

[hide]We'll induct over $m$. Denote $S_{k}=1^{k}+\dots+n^{k}$.

\begin{eqnarray*}(j+1)^{2m+1}&=& j^{2m+1}+{2m+1\choose 1}j^{2m}+{2m+1\choose 2}j^{2m-1}+\dots+{2m+1\choose 2m}j+1\\(j-1)^{2m+1}&=& j^{2m+1}-{2m+1\choose 1}j^{2m}+{2m+1\choose 2}j^{2m-1}+\dots+{2m+1\choose 2m}j-1\end{eqnarray*}

Adding those two and taking $\sum_{j=1}^{n}$, we get

\begin{eqnarray*}S_{2m+1}+(n+1)^{2m+1}-1+S_{2m+1}-n^{2m+1}=2S_{2m+1}+2{2m+1\choose 2}S_{2m-1}+\dots+2{2m+1\choose 2m}S_{1}\end{eqnarray*}

or

$(n+1)^{2m+1}-n^{2m+1}-1=2{2m+1\choose 2}S_{2m-1}+\dots+2{2m+1\choose 2m}S_{1}$

Obviously, if we prove that $(n+1)^{2m+1}-n^{2m+1}$ is a polynomial in $n(n+1)$, then from $S_{1}={1\over 2}n(n+1)$, by strong induction, it follows that $S_{2m-1}$ is a polynomial in $n(n+1)$.

Put $P_{2m+1}=(n+1)^{2m+1}-n^{2m+1}$. Then $P_{1}=1, P_{3}=3n(n+1)+1$, both being polynomials in $n(n+1)$. Also

$[(n+1)^{2}+n^{2}]P_{2m+1}=(n+1)^{2m+3}-n^{2m+3}+n^{2}(n+1)^{2}[(n+1)^{2m-1}-n^{2m-1}]$, giving

$P_{2m+3}=[2n(n+1)+1]P_{2m+1}-[n(n+1)]^{2}P_{2m-1}$

Hence by induction we get that $P_{2m+3}$ is a polynomial in $n(n+1)$, which in turn proves the initial statement.[/hide]


trigonometry geometry calculus integration

Find the area of the region bounded by y=cos x and y= 2cos x, where -pi/2<=x<=pi/2.

I'm not sure, but I think this is more calculus than trig...  :?:

[hide="calculus way"]$ A \equal{} \displaystyle\int^{\pi/2}_{\minus{}\pi/2} {2\cos{x}\minus{}\cos{x}}\,dx \equal{} \displaystyle\int^{\pi/2}_{\minus{}\pi/2} {\cos{x}}\,dx \equal{} \sin{(\pi/2)}\minus{}\sin{(\minus{}\pi/2)} \equal{} \boxed{2}$[/hide]

Integral $ cos(x) dx \equal{} sin(x)$
Integral $ 2cos(x) dx \equal{} 2sin(x)$
It is definite.
$ 2sin(\frac {\pi}2)\minus{}2sin(\minus{}\frac {\pi}2)\minus{}sin(\frac {\pi}2)\plus{}sin(\minus{}\frac {\pi}2)\equal{}4\minus{}2\equal{}2$

LOOL: Beaten talking on MSN


function
induction
algebra
functional equation

Find all functions $f : \mathbb{Z} \rightarrow \mathbb{Z}$ such that for all integers $m,n$,
\[f(m - n + f(n)) = f(m) + f(n).\]

$f(m) = 0$ and $f(m) = 2m$

My solution:

$f(m-n + f(n)) = f(m) + f(n)$

$m = n = 0 \ yields \ f(f(0)) = 2f(0)$.

$m = n \ yields \ f(f(m)) = 2f(m)$.

Put $m = m+n-f(n)$ to obtain $f(m) = f(m+n-f(n)) + f(n)$.
Setting $n=m$ yields $f(m) = f(2m-f(m)) + f(m)$ and $f(2m-f(m)) = 0$. Either $f(m) = 0$ for all m or $f$ is nonconstant and $2m - f(m) = b$ where b is a zero of $f$. Substituting $f(m) = 2m-b$ into the original equation, we conclude that $b=0$ and hence $f(m) = 2m$.

I need to account for the possibility that $f$ has infinitely many zeros, thus, that $2m-f(m)$ is not constant.

i think this solution is not ok because you can't assume that $2m-f(m)=k$ has solution for every k if it is not costant.

 [hide="this is my solution"] $f(f(m))=2f(m)$
suppose $m\neq n$ but $f(m)=f(n)$

than

$f(a-n+f(n))=f(a-m+f(n)$  $\forall\ a\in \mathbb{Z}$
if $a-n+f(n)=b$ than $f(b)=f(b+n-m)$ than the function is periodic but this is not possible because $\exists m: f(m)\neq 0$ and so the serie $f(m);\ f(f(m)); \ \dots$ has infinitely many different value

than f is iniective...

than $m-n+f(n)=n-m+f(m)$ than $f(m)-2m=c$ and substituting we have $c=0$

[/hide]

sorry for my english

Showing injectivity was the key; I did  not know how to do so. Could you please explain why $f$ cannot be periodic?

Because, if $f: \mathbb{Z}\to\mathbb{Z}$ is periodic, then its range is finite. Thus you can not get infinitely many different values in the sequence $f(m), f(f(m)), \ldots$.

My solution does not use injectivity.
[hide] Use the fact that $f(f(n)) = 2f(n)$ for all $n\in\mathbb{Z}$ and set $n: =f(n)$ in the functional equation

$f(m-n + f(n)) = f(m) + f(n)$

to get

$f(m + f(n)) = f(m) + 2f(n)$ for all $m, n \in \mathbb{Z}$.

On the other hand, if we let $m: =m+n$ (again) in the original equation, we get

$f(m + f(n)) = f(m+n) + f(n)$ for all $m, n \in \mathbb{Z}$.

The last two equations imply $f(m+n) = f(m) + f(n)$ for all $m, n \in \mathbb{Z}$. Hence $f(n) = kn$ for some constant $k\in\mathbb{Z}$ ($k = f(1)$). Then one easily gets $k=0$, or $k=2$.[/hide]

I did not read the previous solutions because they're boring.  :P
Here goes a simple one.

If $f(n)=0$ for every $n$, then the equation is satisfied.
Otherwise, $f(a)\neq 0$ for some $a$. Plugging $m=n=a$ yields to $f(f(a))=2f(a)$, and by induction $f^{(k)}(a)=2^{k-1}f(a)$, so $f$ has infinite image.
$f(n)=0$ for some fixed $n$ implies $n=0$, otherwise $f(m-n)=f(m)$ for every $m$, and the image of $f$ would have cardinality at most $n$.
Now plug $m=2n-f(n)$ to get $f(2n-f(n))=0$, yielding $2n-f(n)=0$ by the above argument.
So $f(n)=2n$ for every $n$, which indeed satisfies the equation.

$m=n,f(f(n))=2f(n)$
Replace $n$ by $f(n)$ in the original equation,$f(m-f(n)+f(f(n)))=f(m)+f(f(n))$
$f(m+f(n))=f(m)+2f(n)$
Replace $m$ by $m+n$ in the original equation,$f(m+f(n))=f(m+n)+f(n)$
So $f(m+n)=f(m)+f(n)$ that's Cauchy equation.
So $f(x)=cx$,then easy to know $c=0$ or 2.

Hi [b]jin[/b] and [b]MindFlyer[/b],

Next time please read the postings before yours. This could have saved you time writing the same stuff again ;)

Take it easy!

[quote="Azarus"]Find all functions $ f : \mathbb{Z} \rightarrow \mathbb{Z}$ such that

$ f(m \minus{} n \plus{} f(n)) \equal{} f(m) \plus{} f(n)$

for all $ m,n \in \mathbb{Z}$[/quote]
I do not believe that the following solution has been posted yet:
[hide="Solution"]
Consider $ g(m) \equal{} f(m) \minus{} m$.  Then, the condition becomes $ g(m \minus{} n \plus{} n \plus{} g(n)) \plus{} m \minus{} n \plus{} f(n) \equal{} f(n) \plus{} g(m) \plus{} m$.  This means that $ g(m \plus{} g(n)) \equal{} n \plus{} g(m)$.  Plugging in $ n \equal{} 0$, we have that $ g(m \plus{} g(0)) \equal{} g(m)$, meaning that $ g(x)$ must be periodic with period $ g(0)$ or $ g(0) \equal{} 0$.  In the former case, $ g(x)$ is bounded (since it may only take on values $ g(1), g(2), g(3),\cdots , g(g(0))$).  Hence, $ n \equal{} g(m \plus{} g(n)) \minus{} g(m)$ is bounded, which is clearly false since $ n$ may go arbitrarily high.  Thus, $ g(0) \equal{} 0$.  It follows that, after plugging in $ m \equal{} 0$, we have that $ g(g(n)) \equal{} n$.  Plug in $ g(n)$ for $ n$, so $ g(m \plus{} n) \equal{} g(m \plus{} g(g(n))) \equal{} g(m) \plus{} g(n)$, and, by Cauchy's function, we have that $ g(m) \equal{} cm$ for a constant, $ c$.  Plugging it back in the original equation, we get that $ cm \plus{} c^2n \equal{} n \plus{} cm$, so $ c \equal{} \pm 1$.  This gives that $ f(x) \equal{} 0, 2x$. [/hide]

Let $P(m,n)$ be $f(m-n+f(n))=f(m)+f(n)$.

$P(2n-f(n),n)\implies f(2n-f(n))=0$

[b]Case1:[/b]$f\equiv 0$
This satisfies the condition.

[b]Case2:[/b]$\exists \alpha\in \mathbb Z$ s.t. $f(\alpha)\neq 0$
If $\exists n_0\in \mathbb Z$ s.t. $\beta =2n_0-f(n_0)\neq 0$.
$P(m,\beta)\implies f(m-\beta)=f(m)$.
This implies that $f$ is periodic.Thus $f$ is bounded.On the other hand,
$P(m,m)\implies f(f(m))=2f(m)$.Then
$f^{k}(\alpha)=2^{k-1}f(\alpha)$.Hence $|f|$ is unbounded which is absurd.Thus
$\forall n\in \mathbb Z:f(n)=2n$ which satisfies the condition.

From case1,2,the answer is
$\boxed{f\equiv 0 \ \text{or} f(n)=2n(\forall n\in \mathbb Z)}\blacksquare$ :agent:

[quote=IstekOlympiadTeam]$P(m,n)$ be the assertion of $f(m - n + f(n)) = f(m) + f(n)$

$P(0,0) \to f(f(0))=2f(0)$

$P(0,f(0)) \to f(f(0))=f(0)+2f(0)=2f(0) \to f(0)=0$

$P(0,n) \to f(f(n)-n)=f(n)$ Then $P(m,f(n)-n) \to f(m+n)=f(m)+f(n)$ It proves $f(x)=cx$ where $c \in \mathbb{Z}$

 Plugging back we get $c=0$ or $c=2$[/quote]

I admire your solution, which doesn't use injectivity or periodicity. However, is there a way to prove injectivity directly (for $f(n)=2n$ )?

$P(0,0): f(f(0))=2f(0)$.

$P(0,f(0)): f(f(0))=f(0)+2f(0)\implies f(0)=0$.

$P(0,m): f(f(m)-m)=f(m)$.

$P(m,f(n)-n): f(m+n)=f(m)+f(n)\implies f(x)=kx$ for some constant $k$.

Plugging this back gives $km-kn+k^2n=km+kn\implies k^2n=2kn\implies k=\{0,2\}$.

Thus, the only solutions are $\boxed{f\equiv0}$ and $\boxed{f(x)=2x\forall x\in\mathbb{R}}$, which work.

$P(n,n)\Rightarrow f(f(n))=2f(n)$
$P(m,f(n))\Rightarrow f(m+f(n))=f(m)+2f(n)$
$P(m+n,n)\Rightarrow f(m+n)=f(m)+f(n)$ so by Cauchy's FE $f(n)=cn$. Testing, we have $\boxed{f(n)=0}$ and $\boxed{f(n)=2n}$ as the only solutions.

$P(m+2n-f(n),n) : f(m+n) = f(m+2n-f(n)) + f(n) (1)$.
$P(m,f(n)-n) : f(m+n) = f(m) + f(f(n)-n) (2)$.
Now applying $m=n$ in $(1),(2)$ $\implies$ $f(3n-f(n)) = f(f(n) - n)$.

Claim: $f$ is injective.
Proof : Assume $f(a_1) = f(a_2)$. Now we have $f(z-a_1+f(a_1)) = f(z-a_2+f(a_2)$ for every $\ z\in \mathbb{Z}$ so if $k = z-a_1+f(a_1)$ then $f(b) = f(a_1 + (a_1-a_2))$ so $f$ is periodic. Now we have 2 cases, either $\forall x:$ $f(x) = 0$ or $\exists x: f(x)\neq 0$. first case is one of the answers, second case gives contradiction with $f$ being periodic so $z - a_1 + f(a_1) = z - a_2 + f(a_2)$ so $a_1 = a_2$.

Now that $f$ is injective we have $f(3n-f(n)) = f(f(n) - n) \implies 2f(n) = 4n \implies f(n) = 2n$.
Answers : $f(n) = 0,f(n) = 2n$.


trigonometry
inequalities unsolved
inequalities

Prove that for x in (2,oo) we have sin(pi/x)>3/sqrt(x^2+9)

[quote="Lagrangia"]Prove that for $x \in [2,+\infty[$ we have $\sin\left(\frac{\pi}{x}\right)>\frac{3}{\sqrt{x^2+9}}$[/quote]This is equivalent to  $x^2\sin^2\left(\frac{\pi}{x}\right) > 9-9\sin^2\left(\frac{\pi}{x}\right)$ or  $\tan\left(\frac{\pi}{x}\right) > \frac3x$, for $x>2$.

Substituting $y=\frac{\pi}x $ we get: $\tan (y) > \left(\frac{3}{\pi}\right)y$ for $0<y<\frac{\pi}2$

Which is trivial since $\tan(x)>x$. :)

[quote="Peter VDD"]This is equivalent to  $x^2\sin^2\left(\frac{\pi}{x}\right) > 3-3\sin^2\left(\frac{\pi}{x}\right)$[/quote]

Are you sure? :)


floor function
logarithms
number theory proposed
number theory

Let $ (a,b,c) \in \mathbb{N}^3$ be a triple of [i]positive[/i] fixed integers such that $ \min\{a,b,c\} > 1$. Show that exist infinitely many $ n \in \mathbb{N}$ such that the greatest prime factor of $ \displaystyle \left\lfloor \frac {a^n \minus{} 1}{a^b \minus{} 1} \right\rfloor$ is greater than $ \log_a(n^{cn})$.  :D

[i](Paolo Leonetti and Salvatore Tringali)[/i]

[size=75]Edit @Zetax: you're right: it had no effects, d=1 is sufficient[/size]  :wink:

What's the point in having $ d$¿ (And you maybe want that $ b|n$.)


limit
inequalities
quadratics
trigonometry
triangle inequality
algebra
calculus

Prove that $ \lim_{x\to1}(2x+1)=3.$

[hide="Proof"]
We have $ |(2x+1)-3|<\epsilon\implies|x-1|<\frac\epsilon2.$ So from here we take $ \delta=\frac\epsilon2.$

Since $ 0<|x-1|<\delta\implies2|x-1|<2\delta\implies|(2x+1)-3|<\epsilon.$ It follows that by definition $ \lim_{x\to1}(2x+1)=3.\quad\blacksquare$[/hide]

But when I have to prove that $ \lim_{x\to6}x^2=36$ I got lost; since we have $ |x^2-36|<\epsilon\implies|x+6\parallel{}x-6|<\epsilon,$ then from here what do I do to choose my $ \delta$?

[hide="Similar case"]
When I gotta prove that $ \lim_{x\to2}\frac{2x-5}{x+3}=-\frac15,$ so

\begin{eqnarray*}
   \left| \frac{2x-5}{x+3}+\frac{1}{5} \right|&<&\epsilon \\
  \left| \frac{10x-25+x+3}{5(x+3)} \right|&<&\epsilon \\
  \frac{11}{5}\left| \frac{x-2}{x+3} \right|&<&\epsilon,
\end{eqnarray*}

and I dunno what to do from there. :([/hide]

We're looking at $ x\to 6$, so $ |x\minus{}6|$ will be really small. As long as the other term $ |x\plus{}6|$ isn't too big, we can handle this. How big can $ |x\plus{}6|$ get? If we first assume that $ |x\minus{}6|\le 1$, or $ 5\le x\le 7$, we have $ |x\plus{}6|\le 13$. Now, if we also choose $ |x\minus{}6|<\frac{\epsilon}{13}$, we get $ |x\minus{}6\parallel{}x\plus{}6|<\epsilon$ as desired. That condition is $ |x\minus{}6|<1$ and $ |x\minus{}6|<\frac{\epsilon}{13}$, or equivalently $ |x\minus{}6|<\min\{1,\frac{\epsilon}{13}\}$.

This is a standard technique- first restrict to "reasonably close", then restrict further.

[quote="|-.-|"]Prove that $ \lim_{x\to1}(2x + 1) = 3.$

We have $ |(2x + 1) - 3| < \epsilon\implies|x - 1| < \frac\epsilon2.$ So from here we take $ \delta = \frac\epsilon2.$ [/quote]

you can stop here, why are you stating the next line saying "Since $ 0 < |x - 1| < \delta \ldots$ "
you have already identified  $ \delta$ that fulfills the claim  $ \text{whenever} \qquad |x - 1| < \delta \qquad \text{then}\qquad |f(x) - 3| < \epsilon$

hence, u r done. no need to say anything else.

[quote="|-.-|"]
But when I have to prove that $ \lim_{x\to6}x^2 = 36$ I got lost; since we have $ |x^2 - 36| < \epsilon\implies|x + 6\parallel{}x - 6| < \epsilon,$ then from here what do I do to choose my $ \delta$?
[/quote]

isn't it obvious ?  $ |f(x) - 36| < \epsilon\implies |x^2 - 36 | < \epsilon\implies |(x - 6)(x + 6) | < \epsilon$.
now by hypothesis, $ |x - 6| < \delta$ , and so $ |x + 6| = |(x - 6) + 12 | < |x - 6| + 12 < \delta + 12$
(i used the triangle inequality)

so, pick a suitable $ \delta$ such that $ \delta\; (\delta + 12) = \epsilon$

you get a quadratic equation in $ \delta$, which gives $ 2$ solutions, of which you pick the positive one since by hypothesis, $ \delta > 0$, i.e. you pick  $ \delta = \sqrt {36 + \epsilon} - 6$

[quote="|-.-|"]

When I gotta prove that $ \lim_{x\to2}\frac {2x - 5}{x + 3} = - \frac15,$ so

\begin{eqnarray*} \left| \frac {2x - 5}{x + 3} + \frac {1}{5} \right| & < & \epsilon \\
\left| \frac {10x - 25 + x + 3}{5(x + 3)} \right| & < & \epsilon \\
\frac {11}{5}\left| \frac {x - 2}{x + 3} \right| & < & \epsilon, \end{eqnarray*}

and I dunno what to do from there. :([/quote]

again, by hypothesis, you chose $ |x - 2| < \delta$ , and so $ |x + 3| = | (x - 2) - (- 5) | > \bigg| |x - 2| - |-5| \bigg| < 5 - \delta\qquad (\because \delta > 0)$
(i used the "inverse" triangle inequality)

hence, $ |x + 3| > 5 - \delta$, from which you can see
$ \frac {11}{5}\;\frac {\delta}{5 - \delta} = \epsilon$

solve for $ \delta$ to show you need to pick $ \delta = \frac {25\epsilon}{5\epsilon + 11}$

so now, let me ask you the following $ 3$ Q's, let

$ f(x) \equal{} \frac {1}{\sqrt [3]{x}}$

give a $ \delta \minus{} \epsilon$ proof that $ f(x)$ has a limit as $ x\to 1$

 then, let $ g(x) \equal{} \frac {\sin\,x}{x}$

give a $ \delta \minus{} \epsilon$ proof that $ g(x)$ has a limit as $ x\to 0$

and then, let $ h(x) \equal{}\frac{\sqrt{x^3\minus{}2x^2\plus{}3x}}{\sqrt{x^3}\minus{}\sqrt{2x^2}\plus{}\sqrt{3x}}$

give a $ \delta \minus{} \epsilon$ proof that $ h(x)$ has a limit as $ x\to \infty$

Thanks guys, it's clearer now.

I'm afraid I won't be able to do your problems misan, I'm trying to understand the steps involved.


Asymptote
function
calculus
calculus computations

Find the vertical asymptotes of the graph of F(x) = (3 - x) / (x^2 - 16)

ok if i factor the denominator.. i find the vertical asymptotes to be x = 4, x = -4.

The 2nd part of the problem asks:
Describe the behavior of f(x) to the left and right of each vertical asymptote.. I'm not sure what i need to write for this.

They want you to investigate how the function behaves as it $x$ approaches $4$ or $-4$ from the right or left. You know that the function then approaches infinity, so basically what you need is to figure out whether it's $+\infty$ or $-\infty$. For example, when $x$ approaches $4$ from the left, $3-x$ is negative and $x^2-16$ is also negative, so here it's $+\infty$. And so on.


The Worldwide Youth in Science and Engineering (WYSE) state contest is coming up on this Thursday, April 17th.

Are there any fellow AoPSers going? And what are you participating in?

I myself will be competing in Computer Science and Chemistry.

By the way, why was New Trier conspicuously absent from the Vernon Hills Sectionals?

I would be going except for the fact that I don't like WYSE that much and I'm already missing 3 days of school this week.

I'd be doing math/physics.

I'll be there (taking math/physics). Due to a lack of participation from other schools, we automatically qualified for state.


[quote="gameworld7"]please delete this post mods.

mods...you guys have deleted 30 (exactly or almost exactly), 30 posts of mine in 2 days.  how could you guys do that? that's just evil.  i mean, i used to have like 420 posts.  now i only have 390.  could one of the mods explain to me how this "unusual" thing happened.  i need a good explanation. i'm pissed off right now... and...don't ever do that again. please. sorry about this post. the mods could delete this.  i just wanted to ask why...[/quote]

That's because they're mostly spam. If I'm not mistaken, just about all but 40 or so of your posts are in F&G. There's nothing wrong with that, but if most of the posts are spam, then that's a problem. Mods have a right to delete spam.

Also, why do you want to pay attention to your post count so much? It's the quality, not the quantity.

That's funny, because his post quality is at around 0, more on the negative side... 0*390=0!

[quote="gameworld7"]:(  :(  :(  :(  :(  :( does anyone in this site care (even a little) about me? if i died, would u guys feel sympathy for me?[/quote]I care about you ;) as about any other user on the site. And so does Richard, and Mathew and David.

But, there's a different issue here at stake. I know you sometimes feel bored, because nobody answers your posts. Try to relax. Watch some TV, go play a game. Try to solve a problem :)

Don't post two posts one after another. If it's really necessary you can edit your post, and say whatever you forgot to say. Even more, nobody will even see that you edited, because if your post is the last, it won't show up as edited.

Since this is F&G, your posts won't be touched as long as they respect some normal guidelines (which you have as I have seen :) ). But don't post twice in one topic without any response, because that is SPAM, and will be deleted.

the issue is not how much you post but what is in it...

you could make 100 posts a day, and as long as they serve some sort of purpose besides raising your post count, no one is gonna do anything about it. personally, i wouldn' be bothered even if you were just posting to raise your post count, however we want to keep meaningless

the problem occurs when you start making consequtive posts, posting on every single possible thread just to ask "what are we talking about" or something like that, posting "LOL", or things like that... If you'd like to know what the topic is about, go back to the front and read the thread like everyone else. If you'd like to say that someone's post was funny, or you agree, at least make some other sort of comment besides LOL or a smiley.

believe it or not, we've actually been rather lenient about this lately and not everything that should/could have been deleted has been. but obviously no one likes to have what they've written taken off the board, so just keep in mind what's been said here and it shouldn't be a problem. remember that it's actually work for us to delete posts, and we'd rather not do it as much as you would rather us not do it.

[edit] btw, you might also want to stay away from pointing and then asking a mod to delete the post...that generally doesn't sit well with too many people as being productive...

i doubt the real gameworld actually posted that cuz he gave his pass to so many ppl (like me)

Thread has been split. You can find the other portion in Questions, Suggestions, and Announcements.

hey wait a minute! who posted this topic??????? yeah it says gameworld7 but i didn't do it...oh yeah, 5 other people know my password. :D

for the people that know my password, don't make me look bad. please.

gameworld7, I have changed your password. use the standard procedure to retrieve it ...

Also, do not share your password with other people, no matter how deeply attached to them you are.

You did not indeed start the topic, but those are your posts, basically you started the topic in another topic, and I separated it.


Let $a_1$ be a positive integer.
Build the sequence $a_1 , a_2.....$ such that:
$a_{n+1}= 10a_n + i$
$i\in {0,1,2,...,8}$

Prove that there are $\infty$ composite numbers in the sequence.

Is $i$ arbitrary? I mean we can choose any $i$ from 0 to 8 for each $a_n$?

Yes....consider al th sequences that you can make choosing $i$. And prove there all these sequences have $\infty$ composite numbers.

For example a sequence could be so...

$k, 10k +3 , 10(10k+3)+5 , .....$

[hide]since the 10 shifts it place values, all the numbers are composed of digits 0 to 8, except the first string that could be anything

automatically, if we get an even number, it is composite, then we will be forced to get odds, and eventually, it will be divisible by 3...this argument could be strengthened, but this is the idea[/hide]

[quote="Altheman"]...and eventually, it will be divisible by 3...[/quote]

[hide="Counterargument"]Add $3$ or $5$ or $7$ first (whatever makes $a_2$ NOT divisible by $3$), and then keep adding $3$ - you'll never get a multiply of $3$.[/hide]

Assume the contary, that there is only a finite amount of composite numbers in it.
Therefore, above a certain point, there is only primes.

Now, after this point, we cannot attach any even numbers, otherwise it would be divisable by 2. Also, we cannot attach a 5, or else it would be divisable by 5.

So, from a point up, we can only attach 1's, 3's, or 7's.

Look at how the sequence changes (mod 3), if we attach some ones or sevens, we add 1 to the precedding term mod 3. Since this will yeild a multiple of 3 at some stage, we conculed that from a point onwards, we can only use the digit 3. So, our number looks something like: 1765453333333...., no matter how many 3's we add, its always prime.

Now, notice that $a_{n+3} \equiv a_{n}-1 \mod{37}$, so after a certain number of interations, our number will be divisable by 37. Contradiction. QED

i know that i did not consider that case, and that is why i said that it was not complete, but it was the idea anyway...good problem


function
superior algebra
superior algebra unsolved

[quote="peteryellow"]How can I show that the map, here Z= integers

\phi: Z/3Z /times Z/3Z ---> Z/3Z defined by phi(x,y) = xy is well-defined.[/quote]

well in $ \mathbb{Z}/3\mathbb{Z} \cong  \mathbb{Z}_3$ the function $ \phi$ is simply the product.

It's well defined if the product is not infuenced by the choice of the element in the congruence class.


inequalities
function
trigonometry
inequalities unsolved

$ 0\leq x,y,z<\frac {\pi}{6}\Rightarrow \sin x \plus{} \sin y \plus{} \sin z\geq\tan(x \plus{} y \plus{} z) \minus{} \frac {1}{2}\tan^3(x \plus{} y \plus{} z)$

Firstly, let's prove that:

$ sinx\plus{}siny\plus{}sinz \geq sin(x\plus{}y\plus{}z)$

$ \Longleftrightarrow \quad sinx\plus{}siny\plus{}sinz \geq sinxcos(y\plus{}z)\plus{}cosx(sinycosz\plus{}cosysinz)$

$ \Longleftrightarrow \quad sinx(1\minus{}cos(y\plus{}z))\plus{}siny(1\minus{}cosxcosz)\plus{}sinz(1\minus{}cosxcosy) \geq 0$ which is obviously true.

If we prove that $ sin\alpha \geq tan\alpha\minus{}\frac{tan^3\alpha}{2}$ where $ \alpha\equal{}x\plus{}y\plus{}z$ then the proof is done.

$ \Longleftrightarrow \quad sin\alpha(2cos\alpha\plus{}tan^2\alpha\minus{}2) \geq 0 \quad \Longleftrightarrow$

$ sin\alpha(1\minus{}cos\alpha)^2(2cos\alpha\plus{}1) \geq 0$ which is obviously true.  :wink:

Inequality holds if and only if $ x\equal{}y\equal{}z\equal{}0$


Hi, I have a simple factoring question

1. How do you factor x^8 + 64? Can you use roots of unity?

2. (Related) How do you take square roots of complex numbers?

Thanks.


geometry induction ratio algebra polynomial

Please help me, I have two questions,

1. Aside from mathematical induction, what are the other proofs of the formula n(n+1)(2n+1)/6. (Sum of the squares of numbers from 1 - n.)

 I need many proofs, Ill use this for training.

2. What theorem states that the diameter of a circle that can be inscribed in an equilateral triangle is 2/3 of the altitute of the triangle? Is this true for all triangles or just for equilateral triangle? Does it have to do with the centroid theorem?

Sorry but I have no time to research...

Thanks and God bless.

2. Just use the fact that the centroid and circumcentre and orthocentre coincide ina an equilateral triangle.Then use the fact that the centroid divides median in ratio 2:1. ANd, you are done.

Sanjith

First one:
One can count $\sum_{k=1}^{n} k^2$ using some useful tranformation. It is called propably Abel's transformation (at least in my country).
It states, that:
If $\sum_{i=1}^{k}b_i=B_k$ then:
$\sum_{k=1}^{n} a_kb_k=\sum_{k=1}^{n-1} B_k(a_k-a_{k+1})+a_nB_n$
To evaluate our sum just put $a_k=k^2$ and $b_1=b_2=\cdots b_n=1$ or inversely; I don't remember exactly.

Second way:

Let $a_k=\frac{(2k-1)^3}{24}$
Count $a_{k+1}-a_k$ and then use trivial formula:
$a_{n+1}-a_1=(a_2-a_1)+(a_3-a_2)+(a_4-a_3)+\cdots+(a_n-a_{n-1})+(a_{n-1}-a_n)$

two ways to do #1:

[hide="first"] $n^3=\sum_{k=1}^n k^3-(k-1)^3=\sum_{k=1}^n 3k^2-3k+1$. Since $\sum_{k=1}^n 1=n$ (obviously) and $\sum_{k=1}^n k=\frac{n(n+1)}{2}$ (can be shown by adding $1+2+3+\cdots +n$ to $n+(n-1)+(n-2)+\cdots +1$ and dividing by 2), you can solve for $\sum_{k=1}^n k^2$.[/hide]

[hide="second"]Use binomial coefficients. For any nonnegative integer $a$, $\sum_{k=1}^n \binom{k}{a}=\binom{n+1}{a+1}$ which can be shown using Pascal's Identity. Since $k^2=2\binom{k}{2}+\binom{k}{1}$, the sum is $2\binom{n+1}{3}+\binom{n+1}{2}$. Now just convert back to normal polynomial form.[/hide]

Thank you very much guys. By the way, I will appreciate it if you could post more solutions to number 1.

#1
[hide]
The squares are: $1,4,9,16,25,36,49,64,81,100...$

The sum of the first $n$ terms is: $1,5,14,30,55,91,140,204,285,385...$

So...using my favorite method... you subtract consecutive terms until you hit a repeating digit.

$1,5,14,30,55,91,140,204,285,385\rightarrow 4,9,16,25,36,49,64,81,100\rightarrow 5,7,9,11,13,15,17,19\rightarrow2,2,2,2,2,2....$

We did this process $3$ times, so this sequence can be represented in the form $y=ax^3+bx^2+cx+d$. Taking the first four points, $(1,1);(2,5);(3,14);(4,30)$, we plug them in and solve for $a,b,c,d$.

We solve and get $a=\frac13,b=\frac12,c=\frac16,d=0$.

Therefore, we get $y=\frac13x^3+\frac12x^2+\frac16x\rightarrow \frac{x}{6}(2x^2+3x^1+1) \rightarrow \frac{x}{6}(2x+1)(x+1) \rightarrow \frac{(x)(x+1)(2x+1)}{6}$

[/hide]


Given that
$ \frac {x^2}{\mu^2}\plus{}\frac {y^2}{\mu^2\minus{}b^2}\plus{}\frac {z^2}{\mu^2\minus{}c^2}\equal{}1,$
$ \frac {x^2}{\nu^2}\plus{}\frac {y^2}{\nu^2\minus{}b^2}\plus{}\frac {z^2}{\nu^2\minus{}c^2}\equal{}1,$
$ \frac {x^2}{\rho^2}\plus{}\frac {y^2}{\rho^2\minus{}b^2}\plus{}\frac {z^2}{\rho^2\minus{}c^2}\equal{}1$.
Find the value of $ x^2\plus{}y^2\plus{}z^2$

[quote="hell_ever"]Given that
$ \frac {x^2}{\mu^2} + \frac {y^2}{\mu^2 - b^2} + \frac {z^2}{\mu^2 - c^2} = 1,$
$ \frac {x^2}{\nu^2} + \frac {y^2}{\nu^2 - b^2} + \frac {z^2}{\nu^2 - c^2} = 1,$
$ \frac {x^2}{\rho^2} + \frac {y^2}{\rho^2 - b^2} + \frac {z^2}{\rho^2 - c^2} = 1$.
Find the value of $ x^2 + y^2 + z^2$[/quote]The equation $ \frac {x^2}{k^2} + \frac {y^2}{k^2 - b^2} + \frac {z^2}{k^2 - c^2} = 1$ has 6 solutions in $ k$: $ \pm \mu, \pm \nu, \pm \rho$. The equation can be rearranged as \[ k^6-(b^2+c^2+x^2+y^2+z^2)k^4+(x^2b^2+x^2c^2+y^2c^2+z^2b^2+b^2c^2)k^2-x^2b^2c^2=0\] The coefficient of $ k^4$ is the sum of all possible products of roots taken two at a time: \begin{eqnarray*}-b^2-c^2-x^2-y^2-z^2&=&\mu(-\mu+\nu-\nu+\rho-\rho)-\mu(\nu-\nu+\rho-\rho)+\nu(-\nu+\rho-\rho)-\nu(\rho-\rho)+\rho(-\rho)\\
&=&-\mu^2-\nu^2-\rho^2\end{eqnarray*} $ \implies x^2+y^2+z^2=\mu^2+\nu^2+\rho^2-b^2-c^2$
See also: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=66517]1984 AIME 15[/url]


geometry
circumcircle
geometric transformation
geometry proposed

$ ABCD$ is a cyclic quadrilateral. $ E\equal{}AB\cap CD,F\equal{}BC\cap DA,G\equal{}AC\cap BD$. Two circumcircles $ (DAE),(DCF)$ intersect at $ H$. The bisector of angle $ AHB$ meets $ AB$ at $ I$, the bisector of angle $ DHC$ meets $ CD$ at $ J$. Show that $ I,G,J$ are collinear.

[quote="mr.danh"]$ ABCD$ is a cyclic quadrilateral. $ E \equal{} AB\cap CD,F \equal{} BC\cap DA,G \equal{} AC\cap BD$. Two circumcircles $ (DAE),(DCF)$ intersect at $ H$. The bisector of angle $ AHB$ meets $ AB$ at $ I$, the bisector of angle $ DHC$ meets $ CD$ at $ J$. Show that $ I,G,J$ are collinear.[/quote]
[hide="Solution"]
Notice that $ \angle ABF \equal{} \angle ABC \equal{} \angle ADE \equal{} \angle AHE$, so $ ABFH$ is cyclic.  Similarly, $ EBCH$ is cyclic.  Now, this gives that $ \angle ABH \equal{} \angle EBH \equal{} \angle ECH \equal{} \angle DCH$ and $ \angle HAB \equal{} 180 \minus{} \angle HFB \equal{} 180 \minus{} \angle HFC \equal{} \angle HDC$, so $ \triangle HAB\sim \triangle HDC$.  Thus, a spiral similarity maps $ \triangle HAD\sim \triangle HBC$.  This gives that $ \frac {AI}{IB} \equal{} \frac {HA}{HB} \equal{} \frac {HD}{HC} \equal{} \frac {DJ}{JC}$ since $ HI$ and $ HJ$ are bisectors of $ \angle AHB$ and $ \angle DHC$ respectively.  Furthermore, $ \frac {HA}{HB} \equal{} \frac {AD}{BC}$, so $ \frac {AI}{IB} \equal{} \frac {AD}{BC} \equal{} \frac {DJ}{JC}$.  Since $ ABCD$ is cyclic, we realize that $ \triangle AGD\sim \triangle BGC$, so $ \frac {AD}{BC} \equal{} \frac {GD}{GC} \equal{} \frac {GA}{GB}$.  It follows that $ \frac {GA}{GB} \equal{} \frac {AI}{IB}$, so $ IG$ bisects $ \angle AGB$.  Similarly, $ GJ$ bisects $ \angle DGC$, so $ \angle JGC \equal{} \frac {\angle DGC}{2} \equal{} \frac {\angle AGB}{2} \equal{} \angle AGI$, so $ I$, $ G$, and $ J$ are collinear. [/hide]


trigonometry
limit
logarithms
real analysis
real analysis unsolved

Since
\[ \left( f \right) \to \mathop {\lim }\limits_{n \to \infty } \frac {{\frac {{a_n }} {{\sqrt n }}}} {{a_n }} \equal{} \mathop {\lim }\limits_{n \to \infty } \frac {1} {{\sqrt n }} \equal{} 0\equal{} \mathop {\lim }\limits_{n \to \infty } \frac {1} {n} \equal{} \mathop {\lim }\limits_{n \to \infty } \frac {{\frac {{a_n }} {n}}} {{a_n }} \leftarrow \left( g \right)
\]
then (f) and (g) converge.

\[ \begin{gathered}
  \left( d \right) \to \frac{{a_n^2 }}
{{1 \plus{} a_n^2 }} \leqslant \frac{{a_n^2 }}
{{2a_n }} \equal{} a_n  \hfill \\
  \left( c \right) \to \frac{{a_n }}
{{1 \plus{} a_n }} \leqslant a_n . \hfill \\
\end{gathered}
\]

Yes, that's right, although the limit should be zero.

[hide="Another way for (f) and (g)"]We can use Abel's Test.[/hide]

\[ \left( a \right) \to a_n  \equal{} \frac{1}{{n^2 }}.
\]

\[ \left( b \right) \to a_n^2  \leqslant a_n \sum\limits_{n \equal{} 0}^{ \plus{} \infty } {a_n } .
\]
\[ \left( h \right) \to a_n  \equal{} \frac{1}
{{n^2 }}.
\]

convergent mean that $ a_{n} \minus{} > 0$ and for all $ N$ exist $ C$ such us
$ \sum_{n \equal{} 1}^{N}a_{n} < C$ .
for
(a) we can say nothing it convergent and divergent (example $ a_{n} \equal{} \frac {1}{n^{2}}$ and $ a_{n} \equal{} \frac {1}{n^{4}}$
for
(b) convergent  because exist $ N$ such as $ a_{N} < 1$ then $ a^{2}_{N} < < 1$ :)
for
(c) convergent because it's enough to write for big $ N$ $ \frac {a_{N}}{1 \plus{} a_{N}} \equal{} a_{N} \plus{} O(a^{2}_{N})$
for
(d) convergent it's easly seen if we use (c) and (b).
for
(e) convergent because ..
obviusly
for
 (f) convergent by Dirichlet pricipe:
$ \frac {1}{\sqrt {n}} \minus{} > 0$ monotony and serie $ \sum a_{n}$ convergent
 for
(g)
convergent because
$ \frac {a_{n}}{n} > a_{n}$
for
(h)
we can say nothing because
eample $ a_{n} \equal{} \frac {1}{n^{2}}$
and $ a_{n} \equal{} \frac {1}{n^{5}}$ :)
for
(i)
divergent because
$ \cos a_{n_{N}}$~$ 1 \minus{} O(a^{2}_{N})$
for
(j)
convergent because $ \sin x$~$ x$ for small $ x$
for
(k)
it's analogysli (j)
for
(l)
divergent because this sum is something like this $ N \minus{} C$ where $ C$ is some constant
for
(m)
convergent because
$ \cos \sqrt {a_{N}}$~$ 1 \minus{} \frac {a_{N}}{2}$ after is simple
for
(n)
hint : look (n)
for
(o) convergen use aproximation
for
(p)
divergent
also use aproximation
for
(q)
because for big $ N$ we have $ a_{N} > a_{N}^{3/2}$
for
(r) analogysli (q)

for (s) convergent it's wellknown
for
(t) convergent use aproximation
for
(u)divergent use $ \frac {1}{1 \plus{} a_{N}}$~$ 1 \minus{} a_{N}$
for
(v) convergent ba leibnits principe
for
(w) can convergent can divergent
example $ a_{n} \equal{} \frac {1}{n}$ and
$ a_{n} \equal{} \frac {1}{n^{2n}}$
for
(x) convergent is obviusly
for
(y) convergent (use $ \ln 1 \plus{} x$~$ x$ for small $ x$
for
(z)
 :bye:

\[ \left( x \right) \to \sum\limits_{k \equal{} 1}^n {\left( {a_k  \minus{} a_{k \plus{} 1} } \right)}  \plus{} \sum\limits_{k \equal{} 2}^n {a_k }  \equal{} a_1  \minus{} a_{n \plus{} 1} \xrightarrow{{n \to  \plus{} \infty }}a_1 .
\]

For (b), not only can we say that $ \sum a_n^2$ converges, we can even do so quantitatively: $ \sum a_n^2\le\left(\sum a_n\right)^2.$

Proof: Let $ A\equal{}\sum a_n.$ Let $ b_n\equal{}\frac{a_n}A.$ Note that $ b_n\le 1$ for each $ n,$ hence $ b_n^2\le b_n.$

It follows that $ \sum b_n^2\le\sum b_n\equal{}\frac1A\sum a_n\equal{}\frac{A}{A}\equal{}1.$

$ \frac1{A^2}\sum a_n^2\le 1$

$ \sum a_n^2\le A^2\equal{}\left(\sum a_n\right)^2.$

By the way, (e) converges.

[hide="Hint."]Cauchy-Schwarz[/hide]

\[ \left( e \right) \to \frac{{\sqrt {a_n } }}
{n} \leqslant \frac{1}
{2}\left( {a_n  \plus{} \frac{1}
{{n^2 }}} \right)
\]
by KM.

Very nice Extremal, but what are your answers for (q) and (r)?

[hide="Another hint for (e)"]AM-GM.[/hide]

\[ \left( s \right) \to \frac{{a_1 }}
{n} \leqslant \frac{{a_1  \plus{} ... \plus{} a_n }}
{n}.
\]

repeat:
 convergent of serie $ \sum a_{n}$ mean $ a_{n} \minus{} > 0$ last mean that for big $ N$ I can say $ a_{N} < 1$
therefore
$ a_{N}^{3/2} < a_{N}$ or $ a_{N} < 1$  :P
analogisly for (q)

\[ \begin{gathered}
  \left( q \right) \to a_n^{\frac{3}
{2}}  \leqslant a_n  \hfill \\
  \left( r \right) \to a_n^3  \leqslant a_n  \hfill \\
\end{gathered}
\]
for $ n$ large enough.

[quote="bookworm_vn"]
\[ \begin{gathered} \left( q \right) \to a_n^{\frac {3} {2}} \leqslant a_n \hfill \\
\left( r \right) \to a_n^3 \leqslant a_n \hfill \\
\end{gathered}
\]
for $ n$ large enough.[/quote]
The argument I gave in my previous post works in exactly the same way here:

For $ p> 1,\ \sum a_n^p\le\left(\sum a_n\right)^p.$

Furthermore, if we fix $ A\equal{}\sum a_n$ and $ p>1,$ I can find examples of strictly positive sequences $ (a_n)$ such that $ \sum a_n^p$ equals any specified value in the open interval $ (0,A^p).$


Two sides of a square are divided into fourths and another side of the square is trisected, as shown. A triangle is formed by connecting three of these points, as shown. What is the ratio of the area of the shaded triangle to the area of the square? Express your answer as a fraction.

[asy]draw((0,0)--(12,0)--(12,12)--(0,12)--cycle,linewidth(1));
fill((0,4)--(9,12)--(9,0)--cycle,gray(.7));
draw((0,4)--(9,12)--(9,0)--cycle,linewidth(1));
draw((-.5,4)--(.5,4),linewidth(1));
draw((-.5,8)--(.5,8),linewidth(1));

draw((3,-.5)--(3,.5),linewidth(1));
draw((6,-.5)--(6,.5),linewidth(1));
draw((9,-.5)--(9,.5),linewidth(1));

draw((3,11.5)--(3,12.5),linewidth(1));
draw((6,11.5)--(6,12.5),linewidth(1));
draw((9,11.5)--(9,12.5),linewidth(1));[/asy]

The sides are divided into $ s/3$ and $ s/4$ fragments. Therefore the area of the triangle is $ \frac{1}{2} \cdot \frac{3s}{4} \cdot s \equal{} \frac{3s^2}{8}$. The area of the square is $ s^2$, so the ratio is $ \frac{ \frac{3s^2}{8}}{s^2} \equal{} \boxed{\frac{3}{8}}$.

Or we can imagine all four sides separated into $4$ units.

Area of square: $4*4=16$

Area of triangle: $4*3/2=6$

Triangle/Square: $6/16=3/8$


limit
calculus
integration
logarithms
real analysis
calculus computations

\[ \lim_{n\rightarrow \infty}{\sum_{i\equal{}1}^n{\left(\frac{1}{(n\plus{}i\plus{}1)^2}\plus{}\frac{1}{(n\plus{}i\plus{}2)^2}\plus{}\cdots\plus{}\frac{1}{(n\plus{}i\plus{}n)^2}\right)}}\]

How to consider this one?

First rewrite it in terms of a double series, then bring the limit inside to the the general term of the inside series.

:blush: Can you give me more details....

That doesn't sound like a method I'd trust, at least without a lot more care.

At $ n$, this is a sum of $ n^2$ terms, each between $ \frac1{n^2}$ and $ \frac1{(3n)^2}$. The limit should be a number between $ 1$ and $ \frac19$, and the structure strongly suggests that we should turn this into an integral.

As suggested, write it as a double sum
$ S(n)\equal{}\sum_{i\equal{}1}^n\sum_{j\equal{}1}^n \frac1{(n\plus{}i\plus{}j)^2}$
$ S(n)\equal{}\sum_{i\equal{}1}^n\sum_{j\equal{}1}^n \frac1{n^2}\cdot\frac1{(1\plus{}\frac{i}{n}\plus{}\frac{j}{n})^2}$
That's a Riemann sum for a double integral; look at a grid with corners $ (x_i,y_j)\equal{}(\frac in,\frac jn)$.
$ \lim_{n\to\infty}S(n)\equal{}\int_0^1\int_0^1 \frac1{(1\plus{}x\plus{}y)^2}\,dy\,dx$
$ \equal{}\int_0^1 \left[\minus{}\frac1{1\plus{}x\plus{}y}\right]_{y\equal{}0}^{y\equal{}1}\,dx$
$ \equal{}\int_0^1 \frac1{x\plus{}1}\minus{}\frac1{x\plus{}2}\,dx$
$ \equal{}\left[\ln(x\plus{}1)\minus{}\ln(x\plus{}2)\right]_{x\equal{}0}^{x\equal{}1}$
$ \equal{}2\ln 2\minus{}\ln 3\equal{}\ln\left(\frac 43\right)$

Can you explain to me why you can't bring the limits inside to the general term?

If the limit was to any other number, you couldn't, but since it's to infinity, all the terms go to the same value of zero, regardless of the index variables, right...

The number of terms is not constant.

Your idea says the sum should go to zero. My crude initial estimate says it can't go to zero, since it's between $ \frac19$ and $ 1$ for all $ n$.

Oh, now I see, I for some reason didn't notice that the inner sum was an infinite sum by itself.

jmerry,you've made the same mistake with me.I got 0 point by using this method in the exam :blush: .Anyone have another thought?

the same mistake ?
Could    someone   tell   me    what   is     it?     I  think  jmerry  was  right...............

another  method

$ (n \plus{} i \plus{} 1)(n \plus{} i) < (n\plus{}i\plus{} 1)^2 <(n \plus{} i \plus{} 1)(n \plus{} i \plus{}2)$,

the answer is $ \frac {1}{2}\ln{\frac {4}{3}}$,as we should have $ i\leq j$.My opinion...

That's not what's printed. The inner sums all clearly go from $ 1$ to $ n$. My post is completely correct for the problem as you posted it.


In rectangle $ ABCD$, $ AB = 16$cm and $ AD = 5$cm. $ \overline{FG}$, $ \overline{BD}$, and $ \overline{AC}$ are concurrent at point O. What is the number of square centimeters in the area of the shaded region?
[asy]import olympiad; import geometry_dev; import graph; size(150); defaultpen(linewidth(0.9));
pair A = (0,3), B = (6,3), C = (6,0), D = origin, O = (3,3/2), F = (2,3), G = (4,0);
draw(A--B--C--D--cycle);
draw(A--C); draw(B--D);
draw(F--G);
filldraw(B--F--O--cycle); filldraw(C--G--O--cycle);
label("$A$", A, NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$O$",O,S); label("$F$",F,N); label("$G$",G,S);[/asy]

Let $ \{ ABCD \}$ be the area of figure $ ABCD$.

Note that $ \{ OGC \} = \{ AFO \}$, so
\begin{align*} \{ FBO \} + \{ OGC \} & = \{ FBO \} + \{ AFO \} \\
& = \{ ABO \} \\
& = \frac {1}{4} \cdot \{ ABCD \} \\
& = \frac {1}{4} \cdot 16\cdot 5 \\
& = \boxed{20} \end{align*}

You can just look at the problem and realize that the shaded area is just 1 fourth of the whole figure.
Thus: 5(16)/4=20.


Έστω ΑΒΓ τρίγωνο εγγεγραμμένο σε κύκλο και Κ, Λ, Μ τα μέσα των τόξων ΑΓ,ΑΒ,ΒΓ αντίστοιχα. Αν η ΚΜ τέμνει την ΒΓ στο Δ και η ΚΛ τέμνει την ΑΒ στο Ε, ν.δ.ο. η ΔΕ είναι παράλληλη στην ΑΓ.

έστω ότι οι διχοτόμοι ΑΜ,ΚΒ,ΛΓ τέμνουν τις ΚΛ,ΛΜ,ΜΚ αντίστοιχα στα Ρ,Σ,Τ και η ΚΛ τέμνει τις ΑΓ και ΑΒ αντίστοιχα στα ,Ε,Ε΄η ΛΜ τις ΑΒ , ΒΓ στα Ζ,Ζ΄και ΜΚ τις ΒΓ,ΓΑ στα Δ,Δ΄
Τότε η γωνία ΑΡΛ=90 διότι τα τόξα ΑΛ+ΓΚ+ΓΜ σχηματίζουν ημικύκλιο
Συνεπώς οι διχοτόμοι του ΑΒΓ είναι ύψη του ΚΛΜ και τ έκκεντρο Ι του ΑΒΓ ορθόκεντρο του ΚΛΜ
Απο τα ίσα τρίγωνα ΑΡΕ,ΑΡΕ΄και κυκλικά τα  Ρ,Σ,Τ είναι μέσα των ΕΕ΄,ΖΖ΄, ΔΔ΄
Απο το εγγραψιμμο ΙΡΚΤ οι γωνία ΙΡΤ=γωνία ΙΚΜ=Α/2 αρα ΡΤ//ΑΓ και λογω του προηγούμενου ΔΕ΄//ΑΓ

Πολύ καλή λύση. Παρόμοια είχα κι εγώ στο μυαλό μου.

Φερνουμε ΑΚ, ΒΚ, ΚΓ.  τοτε ΚΛ, ΚΜ διχοτομοι στα τριγωνα ΑΚΒ, ΒΚΓ αντιστοιχα. Αρα απο Θεωρημα Διχοτομων ΚΒ/ΑΚ = ΕΒ/ΕΑ (1), ΚΒ/ΚΓ = ΒΔ/ΔΓ (2), και απο την ισοτητα των χορδων ΑΚ,ΚΓ και τις (1),(2) λαμβανουμε: ΒΔ/ΔΓ = ΕΒ/ΕΑ, οποτε απο το θεωρημα του Θαλη οι ΕΔ και ΑΓ ειναι παραλληλες...


geometry
topology
advanced fields
advanced fields unsolved

Is any polyhedron a disjoint union of finitely many convex polyhedra?

I have seen this claim several places on the internet, (often used to calculate the volume of the polyhedron), and if it is true, there is probably a standard proof somewhere.

Does anybody know the proof or know a reference to such a proof?

Can you please give your definition of a Polyhedron¿

I'm not picky, so any "geometric" definition will do. This could be something along the lines of:

"In geometry, a polyhedron is traditionally a three-dimensional shape that is made up of a finite number of polygonal faces which are parts of planes; the faces meet in pairs along edges which are straight-line segments, and the edges meet in points called vertices. "

(Taken from wikipedia)

But if an advanced viewpoint (like something with simplices) will help the argument, then that's fine.

Draw all planes defining the faces. They split the entire space into finitely many convex regions. Now the interior of each region is either entirely contained in the polyhedron or is entirely outside it (otherwise there would be a boundary point of the original polyhedron that doesn't lie on any of the hyperplanes). The rest should be clear. As to the definition of a polyhedron, the one for which this proof works is "the closure of a bounded open set whose boundary is contained in the union of finitely many hyperplanes".


Ugh, I can't figure out C++ I/O. How does >> work with getline? How do you guys parse numbers along with strings? Do you use fin.getline and then convert? Or?

I've been spending an hour searching on Google trying to figure this out, but to no avail. Any help would be greatly appreciated. I'm a beginner at C++ so please explain the syntax. Thanks!

You don't combine >> with getline

The following is from the MSDN library.
int main()
{
   string s1;
   cout << "Enter a sentence (use <\n> as the delimiter): ";
   getline(cin,s1, '\n');
   cout << "You entered: " << s1 << endl;;
}
Input
test this
Sample Output
Enter a sentence (use <\n> as the delimiter): test this
You entered: test this

If you use the following form:

cin>>str;
cout<<str;   // output: test

>> uses <space> as the delimiter

What do you mean by parse numbers along with strings?

int temp;
cin>>temp;

can read an integer into "temp" var.

Also, MSDN is a great place search for functions, although you need to have some background in C/C++, else they may not make much sense

getline is very, very evil... My suggestion is: don't use it at all!

My old "hack" to it was (I think) calling getline twice where the problem exists. First time, it will read a empty string (clearing up the newline/space from last ">>" call) and second time read the "line" you want. :D

As matter in fact, you should try to avoid iostream in competition enviroments... Use "C-style" stdio instead. ;)

Why use C-style stdio?

Why avoid iostream in competition environment?

[quote="Weeblie"]
As matter in fact, you should try to avoid iostream in competition enviroments... Use "C-style" stdio instead. ;)[/quote]

Speed... stdio is at least 3 times faster than iostream.

Some tasks reads a lot of stuff (usually numbers) and cannot be done using iostream. :D

But, iostream is very nice to use, if you are going to read or write 64 bits integers though... :D (the scanf/printf "code" for 64 bits integers seems to be different for GCC and Visual C++ :P)

Just curious, can you give me an example problem that C stdio is easier/faster than iostream
?

I'm a novice, so haven't seen that type of problem yet

Thanks!
 :)

Hm... I think one of the tasks on USACO training page required it, but not 100% sure...

Anyway, I've got another nice little example. :)

"Clouds" in CEOI 2004 (http://www.oi.edu.pl/php/ceoi2004.php?module=show&file=tasks).

I'm getting these results on the biggest test case (biggest input file) using a Pentium Celeron D 2.54 ghz measured with "timeGetTime":

Solution found on that page: 913 ms
Solution modifed to use iostream: 2555 ms

The task says max-running time 3 secs, but keep in mind that the grading computer is only 1.6 ghz...

That's of cource an extreme example, 19 mb (!) big test files is not so very common... ;)

Wow... that's a huge difference...
Thanks for the tip!

 :D

Thanks for the tip! I'll check out stdio. What I meant about combining getline and >> is for instance, let's say we have the following thing to read:

5
somedude
34 23
blah

You could use >> for the first 5, but then when you use getline, it reads the 5 again... as either a char array or a string depending on which version you use (I'm still not familiar with pointers, but there was something in there with that also).

I was trying something like:

int num_of_people;
fin >> num_of_people;
string somedude;
getline(fin, somedude);
int first_num, second_num;
fin >> first_num >> second_num;

etc.etc. some dumb hack I was trying to do

and it was working horribly, and I couldn't even test since I'm in Florida and the Visual C++ on my laptop suddenly disappeared (I bet it was my sister).

The hardest part for me is learning C++ syntax and stuff. I was trying to figure out how to use an associative array on the Greedy Gift Givers problem (which I'm stuck on now...) and I ended up searching up maps on C++ and iterators........stuff that is too advanced for me lol.

If you want, this is the code I was trying to use, but I couldn't get I/O to work so... it wasn't working very well. I'm not even sure if the I/O was the original one I had. I was trying to hack it to work and it got messed up along the way.

I think that something seems to be causing a floating point error or something like that, but I'm not sure what: (I remembered in the original I was using short, but I couldn't find a function like atoi that converted to short, since I was messing around with getline again)

[hide]
[code]
//c**p program that should work if I have the I/O and syntax correct. Might not be very optimized though.

/*
ID:zyloch1
TASK:gift1
LANG:C++
*/
#include <iostream>
#include <fstream>
#include <string>
#include <map>

//Use namespace
using namespace std;

//Streams
ifstream fin("gift1.in");
ofstream fout("gift1.out");

//Total amount of people
int people_num;

//Our associative array of people and balances
map <string, int> people;

//Our temporary usage array to hold values before being converted to different type
char temp[5];

int main()
{
	//Store our comet and group names
	fin.getline(temp, 3);
	people_num = atoi(temp);
	delete temp;

	string name;
	int gifting, giving_num, giving_div;

	for (unsigned short i = 0;i < people_num;i++)
	{
		getline(fin, name); //get person's name
		people[name] = 0; //start person's balance at 0
	}

	for (unsigned short i = 0;i < people_num;i++)
	{
		getline(fin, name);

		//What they give
		fin.getline(temp, 5, ' ');
		gifting = atoi(temp);
		delete temp;

		fin.getline(temp, 3);
		giving_div = atoi(temp);
		delete temp;

		giving_div = gifting / giving_num;
		people[name] += (giving_div * giving_num); //update account of generous person

		for (unsigned short j = 0;j < giving_num;j++)
		{
			getline(fin, name);
			people[name] -= giving_div; //update account of beneficiary
		}
	}

	//Print out our results
	map <string, int> :: iterator i;
	for (i = people.begin();i != people.end();i++)
	{
		fout << (*i).first << " " << i->second << endl;
	}

	return 0;
}
[/code]
[/hide]

Yeah...it's pretty bad.

Just curious, you already know the which line is string and int, why are you using getline instead of >> ?

I only use getline when I want the whole sentence including the space, else >> works fine as well.

Really? >> works with strings also? I thought it was only char arrays.

From my understanding, char array and string are basically the same

only difference, is that String is a class because C++ is an OOP language which builds on C

In C, there is no String class, so you need to use char array.

I've rarely used any C-style syntax as I'm not too familiar with them except Char array.

Wow! You're right. I'm an idiot lol

I'm interested in <stdio.h>. I checked it out, and I found some good functions, although they all seem to take a *char parameter. Do we have to cast everything we get to int/short, or is there a better way?

Well, stdio works a little bit different... It works with "codes"... :D

For example:

[code]

int int_number;
unsigned int unsigned_int_number;
double double_number;
float float_number;
char char_array[100 + 1]; // maximum capacity is "100", not 101! (don't forget ending NULL character)

FILE* f = fopen("test.in", "r"); // "r" tells fopen to open the file for "r"-eading
fscanf(f, "%d", &int_number); // reads a number, must send the variable to store to as pointer
fscanf(f, "%u", &unsigned_int_number); // reads a unsigned number, must send the variable to store to as pointer
fscanf(f, "%lf", &double_number); // reads a "double", must send the variable to store to as pointer
fscanf(f, "%f", &float_number); // reads a "float", must send the variable to store to as pointer
fscanf(f, "%s", char_array); // reads a character array, notice that char_array is a "array", i.e. already a pointer... no & is needed. :)

fscanf(f, "%d %u %lf %f %s", &int_number, &unsigned_int_number, &double_number, &float_number, char_array); // first read a int, then a unsigned int, then a double, then a float, then a char_array... all of them separaterd by space, tab or newline.

fscanf(f, "This is a number: %d", &int_number); // For the string (without "-character): "This is a number: 12", int_number will be set to 12.

FILE* fout = fopen("test.out", "w"); // open/create for "w"-riting

fprintf(fout, "%d", int_number); // writes a number, notice that we don't have to send it as pointer now (but for gods sake, be sure that int_number is a int, and not a char or some other thing!) :D

fprintf(fout, "My name is %s and I'm %d years old.\n", "James Bond", "46"); // It will print (without "-character): "My name is James Bond and I'm 46 years old.", followed by a newline.

fclose(f); // don't forget to close file... not 100% necessary in competitions but a "must be done" in real programs
fclose(fout); // this file too...

[/code]

stdio is a little bit complicated to start with, but it's very useful after you've learned it... :)

Other than it's faster, how is it useful, in comparison to <iostream> ?

 :(

Thanks! I'm familiar with fopen modes and fscanf reminds me a bit of sprintf in PHP, so I should be fine. I'm not too familiar with pointers and their uses yet, but I'm learning. Thanks again :)

Hey!
C++ is very good for I/O.

I use evil Microsoft Visual Studio .NET.

A sample program that runs is as follows:
[code]
#include <iostream>
#incude <string>

using namespace std;

int main(){
string name;
int age, number;

cout << "What is your AoPS username";
cin >> name;

cout << "What is your age";
cin >> age;

number = age * age + age;

cout << "Your AoPS username is " << name << " , your age is " << age << " and your lucky number is " << " number."
return 0;
}

[/code]
Hope that helped!!!


What percentile does a 115 on the AMC 12 put you at (out of all the AMC 12 participants)
---approximately of course

At the very least the top 5%, I think...

Go here for last year's breakdown,

http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2009-12a/09AMC1012-scoredist.pdf

and here for more archives.

http://www.unl.edu/amc/e-exams/e6-amc12/archive12.shtml

Looks like for 2009 AMC12 A, that is 1.21 percentile, and 2.65 for the B.


Find all points like $M$ in the plane of the triangle $ABC$ so that $MA+MB+MC<AB+AC$.

It is well known that if $BC < AB < AC,$ then for any point $M,$ in the interior of a given triangle$\bigtriangleup ABC,$ we have that $MA+MB+MC < AB+AC.$

You can see at http://www.mathlinks.ro/Forum/viewtopic.php?t=77839

Kostas Vittas.

[quote="gemath"]I don't think it is easy as that. Becasue If I don't understand wrong then this problem is find locus of $M$ such that $MA+MB+MC<AB+AC$ with any triangle $ABC$ . In your problem you only show that when $M$ is inde triangle and $a<c<b$ that the inequality is satified! It is n't enough![/quote]

You are right dear gemath. But I think that when a locus is the wanted, it is better to be defined more conditions, because of we will be confused from the several and different each other configurations.

I am waiting to tell us Mathematica, what he is thinking about.

Kostas Vittas.

There was no specific information about the lengths of the triangle. :wink:


[quote="shaaam"]Hi. I want you to prove the following identity.
Let's first remember one of the famous infinite sum for $ pi^2$

$ pi^2/12 \equal{} 1/1^2 \plus{} 1/2^2 \plus{} 1/3^2 \plus{} 1/4^2 \plus{} ......$

Now i want you to prove that,

$ pi^19/2791563947 \equal{} 1/1^19 \plus{} 1/2^19 \plus{} 1/3^19 \plus{} 1/4^19 \plus{} .......$

I guess , it is really hard!  :wink:[/quote]

sorry , in the second one, it is pi to the power of 19. And all the numbers in the denominators are to the power of 19.
$ pi^(19)/2791563947 \equal{} 1/1^(19) \plus{} 1/2^(19) \plus{} 1/3^(19) \plus{} 1/4^(19) \plus{} .......$

The first given sum is off by a factor of 2: $ \zeta(2) \equal{} \sum_{n \equal{} 1}^{\infty}\frac {1}{n^2} \equal{} 1 \plus{} \frac {1}{2^2} \plus{} \frac {1}{3^2} \plus{} ... \equal{} \frac {\pi^2}{6}\ne\frac {\pi^2}{12}$

As for the given expression you ask us to prove:
\[ \zeta(19) \equal{} \sum_{n \equal{} 1}^{\infty}\frac {1}{n^{19}}\stackrel{?}{ \equal{} }\frac {\pi^{19}}{2791563947}
\]
this, along with other expressions of the form $ \zeta(2n \plus{} 1)$, doesn't have a known closed-form yet and is an open problem (unless this given statement is proven true for the specific case of $ \zeta(19)$).

It's also wrong: the expression involving $ \pi$ begins $ 1.0000000009\ldots$ while $ 2^{\minus{}19} \equal{} 0.0000019\ldots$, so the sum is strictly larger than the pi expression.

Sorry guys.. I am wrong then. But let me tell you how did i find the 2791563947.
I found a direct answer from the program that i made in my graphing calculator...
 :blush:


geometry MATHCOUNTS trigonometry calculus AMC AIME USA(J)MO

I'm wondering how much geometry is known to people who frequent this Getting Started Forum. If you have read any threads on this forum in the past two months, I'd like to hear from you how much geometry background  you have. Where did you learn your geometry? How do you keep it up?

I'll post an example answer, about myself, to the thread in a little while, but feel free to jump in right away with your description of your geometry background.

I've had no real geometry  course, but I have taught myself a lot of geometry from books I have.

I ve had a geometry course but it was nothing to brag about.  I have learned the basics from that class and a lot more from math team and competitions.  I would have to say that I have learned the most from AoPS volume one though.  I hope to learn more from Volume 2.

What's "college-level geometry"?  Like Differential Geometry that Simon takes?

I have taken geometry but I have forgotten a lot of the theorems and stuff...  :|

i'm just taking geometry online right now. i've learned some of it just for stuff like preparing for mathcounts and the AMC tests, but this year i'm actually formally studying it.

Very few high schools in America offer a second advanced geometry course, so it would be unusual for students to ever take two years of geometry in high school. In fact, I actually find more and more advanced students skipping high school geometry classes all together and just picking up bits and pieces on their own. But keep in mind that "geometry" has a very broad definition. Technically, a lot of trigonometry and calculus is also ctually geometry.

From my own personal experience, I never took geometry in high school. I got a little bit (basically SAT I geometry) in middle school, and then learned the rest in bits and pieces on my own. But I never learned any "advanced" geometry (like Ceva's Theorem or Menelaus' Theorem) in high school. Although I did learn Descartes' Circle Therem and a little about inversion (both of which I thought were cool, but never totally got until I learned them in more advanced settings in college).

Never learning these more advanced theorems may have hurt me competition wise, but I found that it never hurt me in the long run because if you know the basic skills of geometry (labeling, angle chasing, similarity, analytic techniques, using complex numbers, etc) you can more or less derive the more advanced theorems as need be. It may just take a lot (in some cases a [i]whole lot[/i]) longer. But speed was never a concern for me in high school because I wasn't much interested in competitions. For example, I would never learn the formula for Stewart's Theorem, I just knew it as, "the law of cosines twice."

In my opinion though, geometry in particular is best learned through problem solving. That's why the AoPS books are especially good for their Geometry sections. As are the AMC-Competitions (AMC, AIME, and USAMO) and the ARML contests. The more geometry problems you solve the better you will get at them.

Speaking of geometry problems, [url=http://155.225.48.46/statemat/stuf2004/Geoprobs.htm]here[/url] is a collection of  problems that I posted for practice for our ARML Team. A lot of them came from this site, so if you do a search for any particular one you can probably find a solution to it here.

[quote="ComplexZeta"]Most college geometry classes will have very little to do with competitions. I took the undergraduate differential geometry class, and I don't imagine that any of it could be used for contests. There are other sorts of geometry offered in college: we have a graduate sequence in algebraic geometry, an undergraduate sequence on noneuclidean geometry, and some Riemannian geometry (which I guess is more noneuclidean stuff) in the graduate differential geometry sequence. I'm probably forgetting about a few others.[/quote]

Where does projective geometry fit in there? Also, what is projective geometry? I've read a short article about it and it was pretty interesting, but it didn't explain much. Plus the name is pretty cool  8-)

Some colleges also offer classes in advanced Euclidean Geometry, Classical Alternative Geometries, and Foundations of Geometry (sometimes as different courses and sometimes all studied in one course).

The geometry that's learned in high school and that's found on high school competitions is basically the same geometry that was known to the Greeks some 2000 years ago (without the trigonometry - that came later). But most of the geometry covered in college is more modern. It's based on ideas from basically the last 200 years (especially Differential Geometry). So geometry classes in college are extremely advanced (both fundamentally and chronologically) in relation to the kind of geometry that is learned in high school.

Classical Projective Geometry, if it were studied anywhere in college, would probably be taught in a course on Classical Alternative Geometries or Foundations of Geometry. But ideas from Classical Projective Geometry show up in lots of courses, like Complex Analysis, Algebraic Number Theory, and Algebraic Geometry.

With one year of geometry(and a bit trig), a lot of AIME geometry problems are doable. But what I hate is that we will never cover some advanced theorems like Ceva's Theorem and Stewart's Theorem, which are also sometimes useful.


advanced fields
advanced fields theorems

Silly yes or no question:

The definition $ Bd A \equal{} \overline{A} \cap \left( \overline{X \minus{} A} \right)$ is equivalent to the definition $ Bd A \equal{} \overline{A} \minus{} Int A$ right?

\overline means the closure. Int means the interior.

That is true. Use De Morgan's rule and the facts that $ \overline{A} \equal{} A\cup\operatorname{Bd}A$, $ \operatorname{int}A \equal{} X\setminus \overline{X\setminus A}$, $ X\minus{}\operatorname{Bd}A\equal{}\operatorname{int }A\cup\operatorname{ext}A$


:(  alguien me puede decir como resolver sucesiones recurrentes???? :?por favor
 :help:

Tal vez [url=http://www.fing.edu.uy/~elvio/RELREC.pdf]esto[/url] te sea útil. Podrías encontrar más información buscando "Relaciones de recurrencia" en internet también.


How does the Hybridization change around the boron atom when a boron halide forms an adduct with a base like pyridine ($ \ce{C5H5N}$)? Is this more favourable for $ \ce{BF3}, \ce{BCl3} , \ce{BBr3}$? Hence deduce the relative Lewis acidity of the three.

[hide="Answer"]In a boron trihalide the geometry is planar trigonal, the hybridization is sp2, and the boron has an empty 2p orbital. Therefore, as the halogens have lone pairs of electrons, they can act as intramolecular Lewis bases (or pi donors in this context) and form a pi bond with the boron empty p orbital. As fluorine is about the same size as boron, and the halogen size increases down the group, this pi bond formation is most favorable for BF3 and least for BBr3. When a boron trihalide forms a complex, the donor atom of the base must first "displace" such a boron-halogen pi bond (with the boron's hybridization changing from sp2 to sp3, just like a nucleophilic addition to a carbonyl) - as BBr3 as the weakest of such pi bonds, BBr3 is the best Lewis acid of the three. As BF3 as the strongest pi bond, it is the weakest Lewis acid from the three.[/hide]


http://www.cs.umd.edu/class/sum2003/cmsc311/Notes/Data/fracToBaseK.html

This link will explain how to get from 10 to another base.  Volume 2 of AoPS treats this topic pretty well (chapter 16 I think?).

This might help: [url]http://www.cut-the-knot.org/blue/frac_conv.shtml[/url]


Prove that every positive integer $n$ satisfies
$\sum_{k=1}^{2n-1}\ (-1)^{k-1}\binom{2n}{k}^{-1} = \frac{1}{n+1}$

Yes!!! At last, I've got it (it makes me a long time before thinking at dominos)!
The key is to observe that :
$k!(2n-k)! = \frac 1 {2(n+1)} (U_k + U_{k+1})$, where $U_k = k!(2n-k+1)!$.

Thus the desired sum simplifies by dominos :
$ S = \sum_{k=1}^{2n-1} \frac {(-1)^{k-1} } {C_{2n}^k } = \frac 1 {(2n)!} \sum_{k=1}^{2n-1} (-1)^{k-1}k!(2n-k)! $
$= \frac 1 {(2n+2) \cdot (2n)!} \sum_{k=1}^{2n-1} (-1)^{k-1}(U_k + U_{k+1})$
$= \frac {U_1+U_{2n}} {(2n+2) \cdot (2n)!} $
$= \frac 1 {n+1}$

Pierre.

I solved it too, but I had some help. In a book I saw the following problem:

1) Show that $\frac{2n-k+1}{\binom{2n}k}=\frac{2n+1}{\binom{2n+1}k}$;

Let $\displaystyle S=\sum_{k=0}^{2n}(-1)^k\frac 1{\binom{2n}k},\ S_1=\sum_{k=0}^{2n}(-1)^k\frac {2n-k}{\binom{2n}k}$. Show that:

2) $S_1+S=2n+1$;

3) $S_1=nS$;

From here it's easy to find $S=\frac{2n+1}{n+1}$ and this solves our problem.


show that 3 is a primitive root modulo p for every Fermat prime Fn with n > 0.

$F_n | 3^{\frac{F_n -1}{2}}+1$ (3 is nonresidue mod $F_n$) and as order of 3 must be power of two it must be $F_n -1$ ie 3 is primitive root modulo $F_n$


This one is easier than the last, except I'm doing it wrong, so can someone post a solution and an answer?

 :?:: The sequence shown was formed by writing the first letter of the alphabet and continuin the pattern by writing one more letter of the alphabet each time. Continuing this pattern, what letter is the 280th letter in this sequence?

A, A, B, A, B, C, A, B, C, D, A, B, C, D, E, A, ...

[hide=" Answer"]
D
[/hide]

[hide]
The way I would do it is that every time you complete a sequence, there is one more a than b, 1 more b than c.

so say for example, 8 sequences.

the sequences have 8 a's, 7 b's, 6 c's,etc.

so, it is just 1+2+3+4+5+6.....

It helps to memorize the answer to 1+2+3.....+n, you need to use it a lot.

So I know 1-20 is 210

u have to find the 280th

210+21+22+23=276. Then count the first 4 letters. 277 is a, 278 is b, 279 is c,

and 280 is

"D"

[/hide]

[hide]So, like Ihatepie pointed out, it's just like $ 1\plus{}2\plus{}...\plus{}n$.We could either find the greatest number less than 280, or the smallest greater than 280. If we do the former, we have if n=23, the sum is 276. So it's A,B,C,D, and it's D.[/hide]


trigonometry calculus integration algebra polynomial

Evaluate $\sum_{k=1}^{n-1}\sin^2{\frac{k\pi}{n}}$ in terms of $n$.

very good! i could not think about how to deal with the squared part...i should have thought of the double angle forumla, that is a very standard technique dealing with even power integrals that i should have seen

Once you stumble over it, you never forget it  :lol:

I think there is a easier method using a polynomial with roots $sin^2(\frac{k\pi}{n})$

\[ r\in R^*,\ a_{k+1}=a_k+r,\ k\in N^*\Longrightarrow \] \[ \boxed {\ S_n\equiv \sum\limits_{k=1}^n \sin a_k=\frac{\sin \frac{a_1+a_n}{2}\sin \frac{nr}{2}}{\sin \frac r2}\ ;\ C_n\equiv\sum\limits_{k=1}^n \cos a_k=\frac{\cos \frac{a_1+a_n}{2}\sin \frac{nr}{2}}{\sin \frac r2}\ }\ . \]
Particularly, $A_n\equiv\sum\limits_{k=1}^{n-1} \sin ^2\frac{k\pi}{n}\Longrightarrow$ $2A_n=\sum\limits_{k=1}^{n-1} \left(1-\cos \frac{2k\pi}{n}\right)=\ldots$ a.s.o.


calculus
integration
calculus computations

One approach:

Let $ f(x)=\frac{x^{2}}{(x^{2}+1)(x^{2}+4)(x^{2}+9)(x^{2}+16)}$

Then:

$ \text{Residue}(f,i)=\frac{i}{720}$
$ \text{Residue}(f,2i)=\frac{-i}{180}$
$ \text{Residue}(f,3i)=\frac{3i}{560}$
$ \text{Residue}(f,4i)=\frac{-i}{630}$

(contour integral arguments left for the reader)

$ 2I=2\pi i (\frac{i}{720}-\frac{i}{180}+\frac{3i}{560}-\frac{i}{630})$

which gives:

$ I=\frac{\pi}{2520}$

Or just do partial fractions...


AMC
AMC 12
AMC 12 B
AIME
AMC 12 A
AMC 10
AIME I

AMC Stats are out today.... there is one thing i dont get...

for AMC12B, 8th grade and below average was higher than 9th 10th or 11th grade average... what the heck?? :?:

The top 1% cutoff for AMC 10 is high this year. There are more perfect papers on 10 this year.

And one seventh grade got perfect score on it.

[quote]for AMC12B, 8th grade and below average was higher than 9th 10th or 11th grade average... what the heck?? Question[/quote]
This is a consequence of self-selection. The AMC-12 is administered by high schools. If you're a 9th grader or older, you're already there at the high school, so you don't have to take any special trips. Maybe you're a member of the high school math club; maybe your teacher told you to sign up for it. The point is that there are a lot of ordinary ways to wind up taking the AMC12.

But how would you take it if you're an 8th grader or younger? It might take special arrangement. You probably had to deliberately seek it out. You might have needed a special trip to the high school just for the AMC. It could be that you're a middle schooler who travels to the high school every day anyway because you're taking a high school math course. But all of these scenarios are at least a little bit unusual, and the students who would fall into these categories are likely to be unusually well-qualified.

interestingly enough, my sophomore score of 131 on the AMC 12B does not seem to show up in the distribution by grade... maybe i bubbled in the wrong grade...

zabelman was the only perfect on the AMC 12A.

rickards got 2nd in the nation on the AMC12A.

we're an underfunded public school.

pwnd.

In the middle of Hickville, Florida too.

Whereas my high school only has one AIME qualifier, w00t w00t.  Oh yeah, AIME I conflicts with FCAT (standardized testing) for me but I'm still taking the AIME I.

Wow, qualification for AIME via AMC12A was huge; it wasn't even close to 5% qualification -- 9000+ qualified out of 85000, which is over 10% :o Same thing for the AMC12B -- 5000+ out of 45000.

Hehe, I guess that's good for me, I was something like 10 points away from the cutoff...if they raised it... :( Oh well, the AIME will filter out the unworthy ones (myself included, most likely).

No offense, but ... what the heck? 57 people got 144 on AMC 10A? That's downright amazing. Are there so many geniuses out there, or is the AMC 10 just really easy this year?


In a plane, take a point $ P$ inside of an acute angled triangle $ \triangle{ABC}$, not containing line segments and vertices. Let denote $ A'$ be the center of a circle passing through $ B,\ C,\ P$, similarly $ B'$ passing through $ C,\ A,\ P$ and $ C'$ passing through $ A,\ B,\ P$. Prove that the necessary and sufficient condition for which $ A,\ B,\ C,\ A',\ B',\ C'$ are on a same circle is that $ P$ coincides with the incenter of $ \triangle{ABC}$.

$ A$, $ B$, $ C$, $ A'$, $ B'$, $ C'$ are concyclic iff $ \angle BA'C\equal{}180^{\circ}\minus{}\alpha$, $ \angle CB'A\equal{}180^{\circ}\minus{}\beta$, $ \angle AC'B\equal{}180^{\circ}\minus{}\gamma$ ($ \alpha\equal{}\angle CAB$, $ \beta\equal{}\angle ABC$, $ \gamma\equal{}\angle BCA$). By Central Angle Theorem that is equivalent to $ \angle CPB\equal{}90^{\circ}\plus{}\alpha/2$, $ \angle APC\equal{}90^{\circ}\plus{}\beta/2$ and $ \angle BPA\equal{}90^{\circ}\plus{}\gamma/2$. Since that conditions are only fulfilled for one point, the incenter $ I$, we are finished.

I heard that the similar problem has been posed in USAMO.

Could anyone know the problem? I would like to look at the original problem.


1)
In problem 1, do any two sets in F have [b]exactly one[/b] common element or [b]at least one[/b] common element?

2)
In problem 2, shouldn't it be "infinely many multiples", instead of "infinity of multiples"?

1) when i read it, I thought that the "a common element" means "at least one element". but now i'm not that sure

2) i think the two formulations are equivalent

3) who proposed the inequality? (it's just a curiosity)

1) When I want to say [i]exactly[/i] I state it. Otherwise it has the other meaning ;)
2) Well it's a matter of English. I belive both statements mean the same thing :)
3) I cannot tell you what is the sources of the problems of this round it until the round is over.


Πως μπορουμε να αποδειξουμε οτι η διαφορα ενος μετρησιμου συνολου απο ενα μη μετρησιμο ειναι μη μετρησιμο συνολο???

Na se rotiso kati?? legontas metrisimo enois ari8misimo..etsi?An nai..nomizo ine poli aplo na to apodiksis..skeyu mono oti i enosi ari8misimu pli8us ari8misimon sinolon ine to poli ari8misimo sinolo... :wink:

 :D

nai..pos mporeis na to apodeikseis mathimatika omos??

to $\mathbb{N}^{2}$ einai arithmisimo kai mporeis epomenws na arithmiseis ta stoixeia ths enwshs metrisimou plithous metrisimwn sunolwn... :ninja:

Katarxas ti ennoeis me ton oro metrisimo giati nomizw exei ginei ena mikro berdema edw pera :) .I ennia tou metrisimou den exei sxesi me tin enia tou arithmisimou...I ennia tou metrisimou epivalei na exeis kapoio "metro" orismeno...Kai afto to "metro" prepei na ikanopoiei kapoies idiotites

opws eipe kai o rakar
A-M=B
tote BUM=A
an B arithm tote A arithm atopo qed

an omws metr<>arithm tote...

Paidia to eipa kai pio pano alla mallon prepei na to ksanapw...I enia tou metrisimou sinolou den exei sxesi me tin enia tou arithmisimou sinolou..To mono pragma pou isws mporoume na poume pws ta sindeei einai oti kathe arithmisimo sinolo einai kai metrisimo kata Lebesque , me metro Lebesque =0


Call a $ 3$-digit number [i]geometric[/i] if it has $ 3$ distinct digits which, when read from left to right, form a geometric sequence.  Find the difference between the largest and smallest geometric numbers.

[hide]First find the biggest and smallest.
Biggest: We start with a 9. The next 2 digits must be smaller than 9, so the common ratio must be a fraction with denominator 3. The biggest one that's less than 1 is $ \frac23$, so the number is 964.
Smallest: We start with 1. The next 2 digits must be bigger than 1, so the ratio must be a fraction greater than 1 with denominator 1; the smallest such fraction is $ \frac21$, so the number is 124.
$ 964\minus{}124\equal{} \boxed{840}$[/hide]

[hide]124 is the smallest and 964 is the largest. Hence the answer is $ \boxed{840}$.

The proof of 124 is obvious. The proof of 964 is as follows:

99-: not allowed
98-: not possible
97-: not possible
96-: 964 is possible
[/hide]

[hide="solution"] We assume the smallest starts with one. Since digits must be whole numbers and are distinct, the common ratio must be at least two, giving us 124.

Now, we assume the largest starts with nine. Once again using the fact that digits must be whole numbers and distinct, we let the common ratio be $ \frac{a}{b}$. Clearly b must divide 9, so it must be 3 or 9. After some quick checking, we see the largest common ratio is $ \frac{a}{b}\equal{}\frac{2}{3}$ which works and gives 964.

Our answer is 964-124=840.[/hide]

[hide="Solution"]So basically, were trying to find the largest in the smallest.

For the smallest, we have to start off with a 1 in the hundreds. We can see that 124 is the smallest. a=1, r=2 in the geometric sequence.

For the largest number, we have to start the hundreds with a 9. Only 964 works. a=9 and r=2/3 in the geometric sequence.

$ 964\minus{}124\equal{}$ $ \fbox{840}$[/hide]

I failed. When I looked I saw the largest geometric number, not the largest 3-digit geometric number.

Hm, well the largest geometric number is 8421, so $ 8421\minus{}124$ is not a valid answer on the AIME, so you would've caught that anyway.

[quote="Ihatepie"]I only took it unofficially, but why would that be an invalid answer?[/quote]
All answers on the AIME must be integers between 0 and 999.

Isn't 1,1,1 a "geometric" sequence however?

Yes, but it does specify the digits are distinct.

*sighs*
hopefully i don't overlook that on the actually AMC wow

[hide = Sol]
So to find the smallest is pretty simple. We want the first digit to be $1$ which means that the other digits are in the form of $r$ and $r^2$. The smallest positive integer that can do this is $2$, so we have the number $124$. For the largest number, we want the first digit to be 9. Then we want the ratio to be $\frac{x}{y}$. So the 2nd digit will be in the form $9\frac{x}{y}$ and the $3$rd digit will be in the form $9\frac{x^2}{y^2}$. We want the denominators to cancel and the only factor of $9$ besides $9$ is $3$.  So$ y =3$. It follows that $x$ has to be $2$ or $1$ but we want the largest so we take $2$. This leads the number to be $964$. So we have $964 - 124 = 840.$
[/hide]

[quote=sugar_rush][quote=RedFireTruck][hide=Solution]
We see that the smallest number should start with $1$.

We see that the smallest number should have a common ratio of $2$.

We see that the smallest number is $124$.

We see that the largest number should start with $9$.

We see that the largest number should have a common ratio of $\frac23$.

We see that the largest number is $964$.

We see that the difference between the largest and smallest is $964-124=\boxed{840}$.
[/hide]

Note: if you want to be double sure about largest common ratio, you can test 8/9 and 7/9 to see they both fail[/quote]

lol all of the lines of this solution start with "We see that…"[/quote]

thats because i read somewhere that you should always say WE

[quote=RedFireTruck][quote=asdf334]hi gollum[/quote]

? i am confoozled. i am not gollum.[/quote]

maybe he toking to gollum not u

p1: Our number is in the form $100a+10ar+ar^2$, with the digits being $a, ar, ar^2$. First we maximize. Clearly we want to maximize $a$, attained at $a=9$, so $r = \frac{2}{3}$ and the number is $964$. The minimum is obviously at $a = 1$, so $r = 2$ and the number is $124$. Subtracting gives $\boxed{840}$.

[color=#38761D][b]BJUMPS AIME GRINDS #9[/b][/color]
To get the desired smallest three digit geometric number we want the Hundreds digit to be equal to $1$ and the smallest non 1 ratio that we can use is two so are smallest number is $124$ For our largest number we need a $9$ in the hundreds digit by the fundamental theorem of logic we get the largest number to be $964$. Finding the difference $964-124$ we achieve the final result of $\boxed{840}$

The smallest, is obviously, $124,$ and the greatest, is hundreds digit $9,$ so $9ab$, now we do $\frac{2}{3}$, ratio, so $964$, hence, $964-124=\boxed{840}.$

[size=50][sub][sub][hide=.]
[quote=TheEvilMathematician]The largest number would be $964$, and smallest is $124$ by common sense, Ans is just difference or $\boxed{840}$[/quote]
WoW So FunI hahaha i forgot to laugh

[/hide][sub][sub][/size]

Let's start with finding the smallest geometric number. Clearly the first digit is $1$, after which the smallest digit is $2$, forcing us to end with $4$; i. e. $124$ is the smallest geometric number.
Now for the largest one: The first digit is $9$. After that the largest common ratio possible is $\frac{2}{3}$, gives us $964$.
The answer is the difference of these, which is $964 - 124 = \boxed{840}$.


Hi, i want to hear your ideas on these NFC games.

First of all, i want to point out that the Vikings vs Saints game was just terrible-- for minnesota.
They for sure were gonna win but they just didn't do anything good!
Really, the fumbles were outragous, you gotta hold on to the ball, Adreann and seriously Brett, stop playing hero.
I want to hear your thoughts and ideas too!

The saints cheated in that game... something that was not known when this was posted...

Please don't revive a 5 year old thread.

[quote=Sullymath]revival is the best[/quote]

No, it's the worst


N.δ.ο. για $ a,b,c>0$ ισχυει

$ \frac{bcd}{(1/3)acd\plus{}(2/3)abd\plus{}abc}\plus{}\frac{adc}{(1/3)abd\plus{}(2/3)abc\plus{}bcd}\plus{}\frac{abd}{(1/3)abc\plus{}(2/3)dbc\plus{}adc}\plus{}\frac{abc}{(1/3)bcd\plus{}(2/3)acd\plus{}abd}\geq{2}$

[quote="xp"]N.δ.ο. για $ a,b,c > 0$ ισχυει

$ \frac {bcd}{(1/3)acd \plus{} (2/3)abd \plus{} abc} \plus{} \frac {adc}{(1/3)abd \plus{} (2/3)abc \plus{} bcd} \plus{} \frac {abd}{(1/3)abc \plus{} (2/3)dbc \plus{} adc} \plus{} \frac {abc}{(1/3)bcd \plus{} (2/3)acd \plus{} abd}\geq{2}$[/quote]

Ωραια Ασκηση, η λυση μου σε hide:

[hide]Η Ανισοτητα ειναι ομογενης οποτε αρκει να την αποδειξουμε για $ abcd \equal{} 1$

Για $ abcd \equal{} 1$ η ανισοτητα γραφεται:

$ \frac {\frac {1}{a}}{(1/3)\frac {1}{b} \plus{} (2/3)\frac {1}{c} \plus{} \frac {1}{d}} \plus{} \frac {\frac {1}{b}}{(1/3)\frac {1}{c} \plus{} (2/3)\frac {1}{d} \plus{} \frac {1}{a}} \plus{} \frac {\frac {1}{c}}{(1/3)\frac {1}{d} \plus{} (2/3)\frac {1}{a} \plus{} \frac {1}{b}} \plus{} \frac {\frac {1}{d}}{(1/3)\frac {1}{a} \plus{} (2/3)\frac {1}{b} \plus{} \frac {1}{c}}\geq{2}$

Θετουμε $ \frac {1}{a} \equal{} x, \frac {1}{b} \equal{} y, \frac {1}{c} \equal{} z, \frac {1}{d} \equal{} w$, και η ανισοτητα γραφεται:

$ \frac {x}{(1/3)y \plus{} (2/3)z \plus{} w} \plus{} \frac {y}{(1/3)z \plus{} (2/3)w \plus{} x} \plus{} \frac {z}{(1/3)w \plus{} (2/3)x \plus{} y} \plus{} \frac {w}{(1/3)x \plus{} (2/3)y \plus{} z}\geq{2} \leftrightarrow \frac {3x}{y \plus{} 2z \plus{} 3w} \plus{} \frac {3y}{z \plus{} 2w \plus{} 3x} \plus{} \frac {3z}{w \plus{} 2x \plus{} 3y} \plus{} \frac {3w}{x \plus{} 2y \plus{} 3z}\geq{2} \leftrightarrow \frac {x}{y \plus{} 2z \plus{} 3w} \plus{} \frac {y}{z \plus{} 2w \plus{} 3x} \plus{} \frac {z}{w \plus{} 2x \plus{} 3y} \plus{} \frac {w}{x \plus{} 2y \plus{} 3z}\geq{\frac {2}{3}} \leftrightarrow \frac {x^2}{yx \plus{} 2zx \plus{} 3wx} \plus{} \frac {y^2}{zy \plus{} 2wy \plus{} 3xy} \plus{} \frac {z^2}{wz \plus{} 2xz \plus{} 3yz} \plus{} \frac {w^2}{xw \plus{} 2yw \plus{} 3zw}\geq{\frac {2}{3}}$

Απο την ανισοτητα Andreescu εχουμε:

$ LHS \geq \frac {(x \plus{} y \plus{} z \plus{} w)^2}{4(xy \plus{} xz \plus{} xw \plus{} zw \plus{} yz \plus{} yw)}$

Οποτε αρκει ν.δ.ο

$ \frac {(x \plus{} y \plus{} z \plus{} w)^2}{4(xy \plus{} xz \plus{} xw \plus{} zw \plus{} yz \plus{} yw)}\geq{\frac {2}{3}} \leftrightarrow 3(x \plus{} y \plus{} z \plus{} w)^2 \geq 8(xy \plus{} xz \plus{} xw \plus{} yz \plus{} yw \plus{} zw) \leftrightarrow 3((x \plus{} y)^2 \plus{} (z \plus{} w)^2 \plus{} 2(x \plus{} y)(z \plus{} w)) \geq 8(xy \plus{} xz \plus{} xw \plus{} yz \plus{} yw \plus{} zw) \leftrightarrow 3(x^2 \plus{} y^2 \plus{} z^2 \plus{} w^2 \plus{} 2xy \plus{} 2zw \plus{} 2xz \plus{} 2yz \plus{} 2xw \plus{} 2yw) \geq 8(xy \plus{} xz \plus{} xw \plus{} yz \plus{} yw \plus{} zw) \leftrightarrow 3(x^2 \plus{} y^2 \plus{} z^2 \plus{} w^2) \geq 2(xy \plus{} xz \plus{} xw \plus{} yz \plus{} yw \plus{} zw) \leftrightarrow (x^2 \plus{} y^2 \minus{} 2xy) \plus{} (x^2 \plus{} z^2 \minus{} 2xz) \plus{} (x^2 \plus{} w^2 \minus{} 2xw) \plus{} (y^2 \plus{} z^2 \minus{} 2yz) \plus{} (y^2 \plus{} w^2 \minus{} 2yw) \plus{} (z^2 \plus{} w^2 \minus{} 2zw) \geq 0 \leftrightarrow (x \minus{} y)^2 \plus{} (x \minus{} z)^2 \plus{} (x \minus{} w)^2 \plus{} (y \minus{} z)^2 \plus{} (y \minus{} w)^2 \plus{} (z \minus{} w)^2 \geq 0$, το οποιο ισχυει.

Αρα ισχυει και η αρχικη.

Ισοτητα εχουμε μονο οταν: $ x \equal{} y \equal{} z \equal{} w \leftrightarrow a \equal{} b \equal{} c \equal{} d$.[/hide]

[quote="xp"]Σωστος....Κριμα που δεν ασχοληθηκες απο παλαιοτερα με τους Μαθηματικους διαγωνισμους αν καταλαβα καλα απο καποια Post εισαι Γ λυκειου....[/quote]

Ναι πραγματι, φετος εμαθα για ολα αυτα λιγο πριν το Θαλη :(

Για την ασκηση, τωρα που την ξανακοιταω δεν χρειαζεται να χρησημοποιησουμε το οτι η ανισοτητα ειναι ομογενης, μια απλη αλλαγη μεταβλητων και μετα Andreescu ειναι αρκετα... απλα στην αρχη επηδη την ειδα πολυ μεγαλη το πρωτο που σκευτηκα ηταν να την απλοποιησω...


Given that a vowel cannot be the first or last letter, how many different arrangements are possible using all the letters in FIVE?

I and E are in the middle, so there are 4 ways (FIEV,FEIV,VEIF,VIEF).

Instead of listing, there are $ 2$ ways to arrange the first and last letter and $ 2$ ways to arrange the middle two, so there are $ 2 \times 2\equal{}\boxed{4}$ ways.

Or, there are $ 24$ ways to arrange without the restriction.  The probability that the middle two are vowels is $ \frac{1}{2} \times \frac{1}{3}\equal{}\frac{1}{6}$, so $ 24 \times \frac{1}{6}\equal{}\boxed{4}$.


Three positive number $x,y,z$,such that:$xyz(x+y+z)=1$ and $xy,yz,zx\le 1$,Prove:
\[ \left(\frac{1}{(x+y)(y+z)}\right)^2+\left(\frac{1}{(y+z)(z+x)}\right)^2+\left(\frac{1}{(z+x)(x+y)}\right)^2 \ge \frac 1 2 \]
Find when does the equality hold.
Hope it is nice,Enjoy!
Good luck.

Nice inequality. Here is my solution:
[hide]
Let's put $a=xy$ $b=yz$ $c=zx$
$ab+bc+ca=1$, $a,b,c\leq 1$
The inequality will become:
$\frac{1}{(a+b+c+\frac{ab}{c})^2}+\frac{1}{(a+b+c+\frac{bc}{a})^2}+\frac{1}{(a+b+c+\frac{ca}{b})^2}\geq \frac{1}{2}$
$a+b+c+\frac{ab}{c} = \frac{ab+bc+ca+c^2}{c} = \frac{1+c^2}{c}$
So the inequality is:
$\frac{a^2}{(1+a^2)^2}+\frac{b^2}{(1+b^2)^2}+\frac{c^2}{(1+c^2)^2}\geq \frac{1}{2}$
If we put $a=tg\alpha$, $b=tg\beta$, $c=tg\gamma$, where $\alpha,\beta,\gamma\leq \frac{\pi}{4}$, and
$\alpha+\beta+\gamma=\frac{\pi}{2}$
$\frac{a^2}{(1+a^2)^2}=sin^2{\alpha}cos^2{\alpha}$
The inequality will become:
$sin^2(2\alpha)+sin^2(2\beta)+sin^2(2\gamma)\geq 2$
So we have to prove the inequality:
$sin^2{A}+sin^2{B}+sin^2{C}\geq 2$
where $A,B,C$ are the angles of acute triangle.
It is equal to
$cos^2{A}+cos^2{B}+cos^2{C}\leq 1$, but
$cos^2{A}+cos^2{B}+cos^2{C}=1-2cos{A}cos{B}cos{C}\leq 1$
Equality holds if one of the angles equal to $\frac{\pi}{2}$
$A=\frac{\pi}{2}$

$\alpha=\frac{\pi}{4}$
$a=1$
$xy=1$
So equality holds if $xy=1$(or $yz=1$ or $zx=1$) :)
[/hide]
Sorry if solution is ugly. :roll:

Thanks very much,In fact your solution is nicer than me. :)


Time To Weigh the Hippos
Martha is the chief hippopotamus caretaker at the Wild Animal Park in San Diego,
California. She has just arrived at the cargo dock in the downtown harbor in order to pick
up four members of the endangered species hippopotamus mathematicus recently rescued
from African poachers. In order to complete the paperwork, she needs to weigh them, but
the only scale big enough to weigh a hippo is a truck scale that starts at 350 kilograms – far
more than any of the hippos weighs. Martha is puzzled for a few minutes, then gets the idea
of weighing them in pairs, thinking that if she gets the weight of every possible pair, she can
later figure out the weights of the individual hippos. She measures the weights pair by pair,
getting 362, 516, 406, 428, and 494. As she tries to weigh the heaviest pair of hippos, the
scale breaks.
1. What is the weight of the heaviest pair of hippos?
2. What are the weights of all the individual hippos?

[hide="1."]Let $ a,b,c,d$ be the weights of each individual hippo.

Without loss of generality, assume $ a\le b\le c\le d$

Now we can set the pairs in the following order.

$ a \plus{} b \equal{} 362$, $ b \plus{} c \equal{} 428$, $ c \plus{} d \equal{} ?$, $ d \plus{} a \equal{} 494$, $ b \plus{} d \equal{} 516$, $ a \plus{} c \equal{} 406$.

We need $ c \plus{} d$, we can find this with the following.

$ c \plus{} a \equal{} 406$
$ a \plus{} b \equal{} 362$
$ b \plus{} d \equal{} 516$

By adding these up we get $ (c \plus{} d) \plus{} 2(a \plus{} b) \equal{} 1284$. WE substitute $ 362$ for $ a \plus{} b$
$ c \plus{} d \plus{} 724 \equal{} 1284$, so $ c \plus{} d \equal{} 560$. $ Q.E.D$[/hide]

for the second one, i think there are different ways to do it but this is the first think I thought about.

[hide="Here we go"]

$ a \plus{} b \equal{} 362$, $ b \plus{} c \equal{} 428$, $ c \plus{} a \equal{} 406$.

Adding these up we get, $ 2a \plus{} 2(b \plus{} c) \equal{} 1196$. We substitute $ 428$ for $ b \plus{} c$

$ 2a \plus{} 856 \equal{} 1196$, therefore $ \boxed{a \equal{} 170}$, $ b \equal{} 362 \minus{} 170 \equal{} 192$, so $ \boxed{b \equal{} 192}$. $ c \equal{} 428 \minus{} 192 \equal{} 236$, therefore $ \boxed{c \equal{} 236}$. and finally
$ d \equal{} 516 \minus{} 192 \equal{} 324$, thus, $ \boxed{d \equal{} 324}$

So we are done. $ Q.E.D$[/hide]

[quote="jimi"]Jhonny

How do you know $ b \plus{} c \equal{} 428$ and not $ d \plus{} a$?[/quote]

I don't know really, that is one of the reasons why I said that there is more than one way to do it. But, I wonder if it matters...

I think this problem has more than 1 solution. Jhonny's and:
[hide]
(1) $ a \plus{} b \equal{} 362$
(2) $ a \plus{} c \equal{} 406$
(3) $ b \plus{} d \equal{} 516$

From here, there are 2 possibilities: either $ b \plus{} c \equal{} 428$ or $ a \plus{} d \equal{} 428$
If $ b \plus{} c \equal{} 428$ then you get Johnny's solution.
If (4) $ a \plus{} d \equal{} 428$ then:
(3) - (4) = (5) $ b \minus{} a \equal{} 88$
(1) + (5) $ 2b \equal{} 450$ so $ b \equal{} 225$ and so on until you finally get:
$ a \equal{} 137$
$ b \equal{} 225$
$ c \equal{} 269$
$ d \equal{} 291$
[/hide]

I agree, Even though I didn't expect more than one answer