limit
logarithms
ratio
inequalities
calculus
integration
geometric series

Let $(x_n)_{n \geq 1}$ be a sequence of real numbers s.t. $x_i > 1$, $\forall i \in \mathbb{N}$. Also consider $P_n=x_1 x_2 \ldots x_n$.

Is it true that $\lim_{n \to \infty} P_n = \infty$?

If it turns out not to be true, then a counter-example would be appreciated    :roll:  Or some sufficient conditions...

Again, sorry if this question is the dumbest ever posted here  :blush:

We have \[\lim_{n \rightarrow + \infty}\left(\prod_{i = 0}^n\left(1 + \frac{1}{2^{2^i}}\right)\right) = 2.\] So that's a nice counterexample.

Obviously we can also use the following example:

\[\lim_{n \rightarrow + \infty}\left(\prod_{i = 0}^n\ \sqrt[2^i]{2}\right) = 4\] but that one is less spectacular, in my opinion ;)

Intuitively it should be obvious that the limit doesn't need to be $+\infty$, I can easily explain why: define $y_i = \ln{x_i}$ for all $i$. Then \[\ln{P_n} = \sum_{i = 1}^ny_i\] but the only restrictions on the numbers $y_i$ is that they must be positive (since the $x_i$ are all bigger than 1). But it is obvious that an infinite sum of positive reals doesn't need to be infinite - think of geometric series with ratio smaller than 1, for example. So, take your sequence $\{y_i\}$ such that the limit $\sum_{i = 1}^{+ \infty}y_i$ is finite, then $\lim_{n \rightarrow + \infty}\ln{P_n}$ will be finite, and hence $\lim_{n \rightarrow + \infty}P_n$ will be finite too.

i thought about this while trying to solve a bloody (for me) problem (in which the product diverges).

i also tried (for that problem) taking logarithms but came up empty-handed  :blush:

The problem was:

Find \[ \lim_{n \to \infty} \prod_{k=1}^{n} \frac{8k+3}{8k-3} .\]

Well, in that case, the limit is $+ \infty$ indeed.

A proof? Well, if $a_1,a_2,...$ are positive reals then, for any $n$, \[\left(1 + a_1\right)\left(1 + a_2\right)\cdots\left(1 + a_n\right) > 1 + a_1 + a_2 + \dots + a_n.\]

Now, take $a_k = \frac{6}{8k - 3}$. Then the left hand side is exactly the given product.

Now, we get \[a_1 + a_2 + \dots + a_n > \frac{6}{8} + \frac{6}{16} + \dots + \frac{6}{8n}.\]

The latter sum is \[\frac{3}{4}\left(1 + \frac{1}{2} + \dots + \frac{1}{n}\right)\] and it is well known that the harmonic series diverges.

Hence, the sum must diverge as well, and hence the limit must be $+ \infty$.

[quote="Arne"]Well, in that case, the limit is $+ \infty$ indeed.[/quote]

I know that, but a proof would be more appreciated. I've been struggling with it for a few days  :P

[EDIT:] Thanks for the proof. It is very nice

My teacher solved it in two ways:

$1^{\circ}$ Very similar to yours; it start with the same trick and then used the following generalization: $\frac{1}{a} + \frac{1}{a+r} + \frac{1}{a+2r} + \frac{1}{a+3r} + \ldots$ diverges for $a,r>0$. The proof is a lot like yours: take $k \in \mathbb{N}$ s.t. $kr > a$. it is clear that $\frac{1}{a+tr} > \frac{1}{a+tr+kr-a} = r \cdot \frac{1}{t+k}$, for all $t \in \mathbb{N_0}$. Thus, our sum is $>r \left( \frac{1}{k}+\frac{1}{k+1}+\ldots \right)$, which famously diverges.

$2^{\circ}$ We have $\frac{8k+3}{8k-3} > \frac{8k}{8k-4} = \frac{2k}{2k-1} \Leftrightarrow -8k-12 > -24k \Leftrightarrow 16k > 12$, which is true for all $k \geqslant 1$. Therefore $\prod_{k=1}^{n} \frac{8k+3}{8k-3} > \prod_{k=1}^{n} \frac{2k}{2k-1}$.
The inequality $\frac{2k}{2k-1} > \frac{2k+1}{2k} \Leftrightarrow 0 > -1$ is true for all $k \geqslant 1$. Thus, $A = \prod_{k=1}^{n} \frac{2k}{2k-1} > \prod_{k=1}^{n} \frac{2k+1}{2k} = B$. Multiply this inequality by $A$ to obtain $A^2 > AB = \prod_{k=1}^{n} \frac{2k(2k+1)}{2k(2k-1)} = \prod_{k=1}^{n} \frac{2k+1}{2k-1} = 2n+1$, so $A > \sqrt{2n+1}$. In a similar manner, $A < 2 \sqrt{n}$.

My solution for the divergence of $A$:

A more natural integral based approach works for estimating $A$, but it's not that accurate.

We have \[ \ln A = \sum_{k=1}^n \left( \ln (2k) - \ln (2k-1) \right) = \sum_{k=1}^n \int_{2k-1}^{2k} \frac{1}{x} dx > \sum_{k=1}^n \frac{1}{2k} = \frac{1}{2} \left( \sum_{k=1}^n \frac{1}{k} \right) = \frac{1}{2} \left( \ln n + \gamma \right) = \ln \sqrt{n e^{\gamma}} > \ln \left( 1.334568 \cdot \sqrt{n} \right) .\]

We also have \[ \ln A = \sum_{k=1}^n \int_{2k-1}^{2k} \frac{1}{x} dx < \sum_{k=1}^n \frac{1}{2k-1} < 1+\ln n - \frac12 \ln(n+1) = \ln \frac{en}{\sqrt{n+1}} < \ln \left( e \cdot \sqrt{n+1} \right) .\] Therefore $1.334568 \cdot \sqrt{n} < A < e \cdot \sqrt{n+1}$. Can these estimates be sharpened? (the lower bound can as seen above, but what about the upper one)

Furthermore, can this approach be used for the problem with $\frac{8k+3}{8k-3}$ to find some bounds? I tried that before seeing any solution to this problem and I didn't succeed.

If the terms $b_k$ are all eventually of the same sign, then
\[\prod_{k=1}^{\infty}(1+b_k)\]
converges if and only if
\[\sum_{k=1}^{\infty}b_k\]
converges.

For a proof, take the logarithm of the product.

In your case, $\frac{8k+3}{8k-3}=1+\frac6{8k-3}$ and the series diverges by comparison to the harmonic series.

I think (but can't prove) that $\left( \frac{\displaystyle \prod_{k=1}^n \frac{2k}{2k-1}}{\sqrt{n}} \right) _{n \geq 1}$ is increasing.

This implies $\lim_{n \to \infty} \frac{\displaystyle \prod_{k=1}^n \frac{2k}{2k-1}}{\sqrt{n}} = c$, with $\sqrt{2}<c<2$.

Can a similar approximation be found for $\prod_{k=1}^n \frac{8k+3}{8k-3}$?

First approximation:

From $\sum_{k=1}^n\frac{c}k\approx c\ln n$, we should get, more or less,

$\prod_{k=1}^n\left(1+\frac{c}k\right)\approx e^{c\ln n}=n^c$

(in some loose sense of approximation).

So trying to compare $\prod_{k=1}^n\frac{2k}{2k-1}$ to $\sqrt{n}$ is the direction you should be looking.

Applying your approach to the other product would yield \[ \prod_{k=1}^n \frac{8k+3}{8k-3} = O \left( n^{\frac34} \right) . \] Checking with a C++ program, the constant seems to be $1.61 \ldots$.

I'm just wondering... Is what you wrote sufficient to prove the product grows in the same manner as $n^c$?

[quote="Kent Merryfield"]If the terms $b_k$ are all eventually of the same sign, then
\[\prod_{k=1}^{\infty}(1+b_k)\]
converges if and only if
\[\sum_{k=1}^{\infty}b_k\]
converges.

For a proof, take the logarithm of the product.
[/quote]

And inequalities $0\le \frac{b_k^2}{2}\le b_k-\frac{b_k^2}{2}\le ln(1+b_k)\le b_k$ for $k>M$  , right ?
(for $b_k\ge 0$)

[quote="Arne"]We have \[ \lim_{n \rightarrow + \infty}\left(\prod_{i = 0}^n\left(1 + \frac{1}{2^{2^i}}\right)\right) = 2. \] So that's a nice counterexample.[/quote]

how do you prove that? it looks very nice...

There are two ways I know of. The first one: expand the product, you get a 1 and a sum of terms of the form $\frac{1}{2^{2^{a_1} + 2^{a_2} + ... + 2^{a_j}}}$. Now think about binary representation of positive integers and conclude that the product is exactly $1 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + ... = 2$.

Alternatively, you could use the fact that \[ \displaystyle 1 + \frac{1}{2^{2^i}} = \frac{\displaystyle 1 - \frac{1}{2^{2^{i + 1}}}}{\displaystyle 1 - \frac{1}{2^{2^{i}}}}  \] and then you get a telescoping product.

both solutions are   :thumbup: . thanks Arne


function
modular arithmetic
Other problems

Let $ \{F_{n}\}$ be the sequence of Fibonacci numbers and let $ \varphi$ denote Euler's totient function. Prove that

$ \varphi(F_{n}) \equiv 0 \mod 4 \qquad$

if $ n \geq 5.$

P.S. Hope the proposal qualifies for the section of official [b]P-E-N[/b] problems!  :wink:

it is equivalent to show there is no $ 4k \plus{} 3$ prime in the Fibonacci sequence except $ 3$
It is really an interesting thing :)
When I found $ F_{22} \equal{} 17711 \equal{} 89*199$,I think I must miss something to prove the assertion.Can you post the solution?I am interested in it :)

$ \phi(n)\equiv 0\mod 4$ if $ n\not \equal{}1,2,4,p\equiv 3\mod 4$ p- prime.
Therefore sufficiently to show, that $ F_n\not \equal{}p\equiv 3\mod 4$.
$ F_n\equal{}\frac{q_1^n\minus{}q_2^n}{q_1\minus{}q_2},q_1\equal{}\frac{1\plus{}\sqrt 5}{2},q_2\equal{}\frac{1\minus{}\sqrt 5}{2}$. Therefore $ F_n|F_m$ if $ n|m$ and sufficiently consider case, when n - prime >3.
If n - odd $ L_n^2\minus{}5F_n^2\equal{}\minus{}4$, it mean $ (\frac{\minus{}1}{F_n})\equal{}1\to F_n\not \equal{}p\equiv 3\mod 4$.

[quote="Rust"]$ \phi(n)\equiv 0\mod 4$ if $ n\not \equal{} 1,2,4,p\equiv 3\mod 4$ p- prime.
Therefore sufficiently to show, that $ F_n\not \equal{} p\equiv 3\mod 4$.
$ F_n \equal{} \frac {q_1^n \minus{} q_2^n}{q_1 \minus{} q_2},q_1 \equal{} \frac {1 \plus{} \sqrt 5}{2},q_2 \equal{} \frac {1 \minus{} \sqrt 5}{2}$. Therefore $ F_n|F_m$ if $ n|m$ and sufficiently consider case, when n - prime >3.
If n - odd $ L_n^2 \minus{} 5F_n^2 \equal{} \minus{} 4$, it mean $ (\frac { \minus{} 1}{F_n}) \equal{} 1\to F_n\not \equal{} p\equiv 3\mod 4$.[/quote]
Nice work :lol:

I think we should also rule out the case $ F_n \equal{} 2^{\epsilon}p^k, \epsilon \equal{} 0,1, p \equiv 3 \pmod 4, p$ is a prime...

[quote="coquitao"]Wow! That was quick, [b]Rust[/b]...

There is an alternative solution. Does anyone feels like looking for it? :wink:

Best regards,[/quote]
Of course ,esteemed coquitao :)

$ F_{2k}\equal{}F_k*L_k$ is not prime if k>2.
$ F_{2k\plus{}1}\equal{}F_k^2\plus{}F_{k\plus{}1}^2$ can not divisors form $ 3\mod 4$.

TO RUST:

don't see how that relates to bambaman's objection...  :wacko:

$ F_{2k\plus{}1}\equal{}F_k^2\plus{}F_{k\plus{}1}^2$ and $ (F_k,F_{k\plus{}1}\equal{}1$, therefore $ p|F_{2k\plus{}1}\to p\equal{}2 or p\equal{}1\mod 4$.
Because for $ k>2$ $ F_{2k\plus{}1}>4$ we get $ 4|\phi(F_{2k\plus{}1}$.
If $ n\equal{}2^lk$, were $ l>0,k$ - odd, then $ 4|\phi(F_k)|\phi(F_n)$ if $ k>3$.
If $ n>4, k\equal{}1$, then $ F_8|F_n\to 4|12\equal{}\phi(F_8)|\phi(F_n)$,
if $ n>4,k\equal{}3$, then $ F_6|F_n\to 4\equal{}\phi(F_6)|\phi(F_n)$.


geometry
geometric transformation
reflection
combinatorics unsolved
combinatorics

Two players by turns draw diagonals in a regular $(2n+1)$-gon ($n>1$). It is forbidden to draw a diagonal, which was already drawn, or intersects an odd number of already drawn diagonals. The player, who has no legal move, loses. Who has a winning strategy?
[i]K. Sukhov[/i]

However, the answer to this problem is not what N.T.TUAN said.
If n is odd, then the second person has a winning stragety; If n is even, then the first one will win at last.

Let's note the number of legal moves a player can make by $k$. Clearly, first we have $k=(2*n+1)*(n-1)$. By each move, a player when draws a diagonal, it divides the poligon into 2 parts. These parts have $a$ and $b$ points, $2*n-1$ in total (without the points of the diagonal), and between them there are $a*b$ diagonals, which is therefore an even number. So when we draw a diagonal, we intersect an even number of drawn diagonals, but we intersect an even number of all ($a*b$) diagonals, so we intersect an even number of not drawn diagonals. From these $u1$ diagonals have an odd and $u2$ have an even number of intersections (with drawn diagonals) before we draw the diagonal ($u1+u2$ is even). After we draw it, $u1$ will have an even and $u2$ will have an odd number of intersections. $k$ will change to $k+u1-u2-1$ so it will change its parity($u1-u2$ is even). This means that a player has always an odd , while the other one an even number of options, so he's the only one who can and must lose regardless of his cleverness. If $n$ is even then the initial $k$ is odd, so the first player will win and if $n$ is odd the second one.

I think the solution has a problem or I didn't understand the last part correctly. Can you explain the point that you said k changes to k+u1-u2-1?

you said that the parity of intersections of u1 and u2 changes but we may draw a diagonal which doesn't cut a diagonal of u1 then for that diagonal the parity does not change

[quote=SOUMYASHANT NAYAK]the second player has a winning strategy.

suppose the first person draws a legitimate line AB.

strategy of the winning person:
choose the other adjacent side of B say 'BC'

draw its perpendicular bisector.
reflect the line drawn by first person on this perpendicular bisector.
(it will always be a line segment with its ends on the vertices of the polygon as it is regular polygon)
you always have such a line for the second person and such a trick leads to a unique line.(because there are odd no. of sides of polygon)

its not difficult to see that the 2nd person always has the [i]last laugh[/i]
 :lol:  :lol:[/quote]

[quote=rashah76]A simple solution when n is even.
When n is even 1 st person will draw the longest diagonal that passes through center (note that this will divide polygon into two equal halves).
Then next move B has to play is in one half only or else it will cross the longest diagonal.
Then A can play the same move in other half with corresponding vertices.
So if B has a move A also has a move and so A will win when n is even.[/quote]

I think there is a flaw in these solutions (which try to exploit symmetry). In particular, it might happen that if a player draws a diagonal $d_1$ which is symmetrical to $d_2$ (diagonal just drawn by his opponent), then it might happen that $d_1,d_2$ intersect.


[quote="santosguzella"]$a$, $b$, $c$ are whole numbers such $100a+10b+c$ is divisible by $109$. Prove that $(9a-c)^{2}+9b^{2}$ is divisible by $109$.[/quote]

This is easy. From hypothesis we have $(9a-c)^{2}+9b^{2}=(9a+100a+10b)^{2}+9b^{2}=(10b)^{2}+9b^{2}=109b^{2}=0(\mod 109)$


hello, we have $ x\equal{}2$, $ x\equal{}4$ and $ x\equal{}\minus{}\frac{2\text{LambertW}\left(\frac{1}{2}\ln(2)\right)}{\ln(2)}$.
Sonnhard.

let f(x)=x^2-2^x. This is a smooth function.
f'''(x)= - (ln2)^3 2^x <0, therefore f has at most three roots.
two of them are 2, 4; both of them are positive.
f(0)=-1<0, f(-1)=1/2>0. therefore there exists -1<u<0 such that  f(u)=0
Therefore 2,4,u are all the solutions.


inequalities
induction
inequalities unsolved

I know that this inequality was mentioned a few times on this forum, but I haven't found any solution of this problem, which is:

\[\left(\frac{a}{a+b}\right)^{3}+\left(\frac{b}{b+c}\right)^{3}+\left(\frac{c}{c+a}\right)^{3}\geq \frac{3}{8}\]

for any positive reals $a$, $b$, $c$.

As I found on mathlinks, on China TST there was a problem

\[\left(\frac{a}{a+b}\right)^{2}+\left(\frac{b}{b+c}\right)^{2}+\left(\frac{c}{c+a}\right)^{2}\geq \frac{3}{4}\]

Which also hasn't been solved. Hope that you'll help :)

[quote] I found on mathlinks, on China TST there was a problem

$\left(\frac{a}{a+b}\right)^{2}+\left(\frac{b}{b+c}\right)^{2}+\left(\frac{c}{c+a}\right)^{2}\geq \frac{3}{4}$
[/quote]

The second isn't hard. Here is my proof.
Let $x= \frac{a}{a+b}, y=\frac{b}{b+c}, z= \frac{c}{c+a}$
Then $x^{2}+y^{2}+z^{2}+2xyz-1= \frac{a^{4}b^{2}+a^{2}c^{4}+b^{4}c^{2}-3a^{2}b^{2}c^{2}}{(a+b)^{2}(b+c)^{2}(c+a)^{2}}\geq 0$
so $x^{2}+y^{2}+z^{2}\geq 1-2xyz$
Now we have to prove $1-2xyz \geq \frac34$
or $\frac14 \geq 2xyz$
or $1\geq 8xyz$
But $8xyz= \frac{8abc}{(a+b)(b+c)(c+a)}\leq 1$
 :)

[quote="robpal"][quote="The soul of rock"]Oh, It's quite easy
Since
$\frac{1}{(1+a')^{2}}+\frac{1}{(1+b')^{2}}+\frac{1}{(1+c')^{2}}+\frac{1}{(1+1)^{2}}\ge 1$
[/quote]

How did you get that?[/quote]
Above problem is a problem in China TST 2004  :D Solution used behind inequality:
$\frac{1}{(a+1)^{2}}+\frac{1}{(b+1)^{2}}\ge \frac{1}{1+ab}\\ \to (ab+1)(a-b)^{2}\ge 0$
It alway true  :) And if you beware be above inequality is Nesbitt'inequality open:
 $(\frac{a}{a+b})^{2}+(\frac{b}{b+c})^{2}+(\frac{c}{c+d})^{2}+(\frac{d}{d+a})^{2}\ge 1$
For problem 1 we can solve it by Mixing Variable  :) And generalization problem is very very difficult. Name of it is "Thách Thức" problem or "defiant problem" and author of problem is Pham Kim Hung (my brother  :D )

[quote="zaizai-hoang"]Note: solution of generalization problem used I.G.I theorem or  induction generalization method.
This problem is:
Let $a_{1},a_{2},..,a_{n}\ge 0$ such that: $a_{1}a_{2}...a_{n}=1$. Find the best k constant to behind inequality alway right:
$\frac{1}{(1+a_{1})^{k}}+\frac{1}{(1+a_{2})^{k}}+...+\frac{1}{(1+a_{n})^{k}}\ge \frac{n}{2^{k}}$[/quote]
ehhh..sorry, but can you post the solution?


Could someone explain to me basically what SSS, SAS, ASA is with some examples ( if you feel it explains more.) Thanx!

Two triangles that have all three corresponding sides congruent must be congruent.
Two triangles that have two corrseponding sides and the angle in between these sides congruent must be congruent.
Two triangles that have all corresponding angles congruent and one pair of corresponding sides congruent must be congruent.

I'm kinda overwhelmed by what my teacher told me...is there a way that you memorized this, or used some example to find it easier to remember what corresponded to where?

Well, I just look at the letters:

SSS-Side Side Side- All three sides are the same
SAS-Side Angle Side- Two sides and the angle between them are the same
ASA-Angle Side Angle- Two angles and the side between them are the same

If any of them apply, the two triangles are congruent

No problem. Is this for some sort of a test?

yeah.....chapter 4 test in the Glencoe Geometry 1998 version... I'm taking a 6 week Geometry course that covers all of the 13 chapters

Well, you could practice on [url=http://www.glencoe.com/sites/michigan/student/mathematics/index.html]The glencoe website.[/url]

Specifically, the 1988 Geometry book tests are [url=http://www.glencoe.com/sec/math/geometry/geo/index.htm]here[/url], although I can't find much useful in that last one.


function
induction
number theory
prime numbers
number theory solved

Prove that  $\pi(n) > log_2 n$ for $n \geq 5$. Do not use any theorems you can not prove by yourself or that are not elementary.

good to know that \pi (n) denotes the number of prime numbers smaller than n. :D - i think it's more maths for college than high-school, mindspa :)

He wants an elementary solution, so I think that this problem can sit comfortably in this section.

Trivial using Chebyshev's Theorem and induction.
(I have proved Chebyshev's Theorem in [url=http://www.mathlinks.ro/viewtopic.php?t=4130]this thread[/url].)

(By Chebyshev's theorem we have [tex]\pi(2n)\geq\pi(n)+1[/tex].)

We know that max power of p which divides n! is kp=[n/p]+[n/p2]+..., so kp < n/p+n/p2+... = n/(p-1). It follows that $n! < (\prod p_i^{1/(p_i-1)})^n$ where pi is prime and  pi <= n, i=1,2,...,$\pi(n)$.
Since $x^{1/(x-1)}$ is decreasing for x>1 we have $p_i^{1/(p_i-1)}\leq 2$. Thus $n!<2^{n\pi(n)}$.
Now we use n!>(n/4)n (by induction) and obtain $\pi(n) > log_2 n - 2$.
You may slightly modify this proof for n>5 and get $\pi(n) > log_2 n$.


modular arithmetic
algebra
polynomial
number theory unsolved
number theory

If $ a,b,c,d,m$ are integers so that $ 7|am^3\plus{}bm^2\plus{}cm\plus{}d$ and $ m$ is not divisible by $ 7$ then prove that there is an integer $ n$ so that $ 7|dn^3\plus{}cn^2\plus{}bn\plus{}a$.

[hide="Constructive solution"]
For convenience, let $ f(x)=ax^3+bx^2+cx+d$ and $ g(x)=dx^3+cx^2+bx+a$.  Since $ m \not\equiv 0 \pmod{7}$, it has an inverse $ m^{-1} \pmod{7}$.
Then,
\begin{align*}
m^3g(m^{-1})&=m^3(dm^{-3}+cm^{-2}+bm^{-1}+a)\\
&= am^3+bm^2+cm+d\\
&= f(m)\\
&\equiv 0 \pmod{7}
\end{align*}
Thus, $ 7\mid m^3g(m^{-1})$.  Because $ 7\nmid m$ and 7 is a prime, we can conclude that $ 7\mid g(m^{-1})$, so any integer congruent to $ m^{-1} \pmod{7}$ works.
[/hide]

[hide]
Lemma: In a field, $ P(x)$ is reducible iff $ x^dP(1/x)$ is reducible where $ d\equal{}\deg P$.
Proof: If $ P(x)\equal{}\prod_{g\text{ irreducible}}g$, then $ x^dP(1/x)\equal{}\prod_{g\text{ irreducible}}g(1/x)x^{d_g}$ where $ d_g\equal{}\deg g$, as desired.
The first statement is equivalent to saying that the polynomial $ P(x)\equal{}ax^3\plus{}bx^2\plus{}cx\plus{}d$ has a root in $ F_7$ (which denotes the field $ \mathbb{Z/}7\mathbb{Z}$).  Applying the above lemma will give that $ x^3P(1/x)$ also has a root, as desired.[/hide]


number theory
prime numbers
number theory unsolved

prove that any prime no.  $ 2{}^2{}^n+$ $ 1$ cannot be represented as a difference of 2 fifth powers of integers

Actually the question asked was $ 2^{2^{n}} \plus{} 1$ not $ 2^{2n} \plus{} 1$ ,anyway it doesnt matter.
Suppose that there exists such integers $ a,b$ ..
$ 2^{2n} \plus{} 1 \equal{} a^{5} \minus{} b^{5} \equal{} (a \minus{} b)(a^{4}\plus{} a^{3}b \plus{} a^{2}b^{2} \plus{} ab^{3} \plus{} b^{4})$ .
We have $ \text{L.H.S}$ is a prime ,its only factors being $ 1$ and itself ,its trivial from now on taking cases in the R.H.S .

srinath, all numbers of the form $ 2^{2^{n}}\plus{}1$ are not primes.
they are a class of numbers called Fermat Numbers.
in fact, $ 2^{2^{5}}\plus{}1$, the fifth in the series itself is not a prime.
and so your proof is incorrect.
though computed proofs shall exist, we should find a theoretical proof.

@phymax....
Sorry...I think it was given in the problem that $ 2^{2^{n}}$ or $ 2^{2n}$ are prime numbers , I never said all are primes........I know that $ 641 \mid 2^{2^{5}} \plus{} 1$....I think you overlooked the condition in the problem that $ 2^{2^{n}} \plus{} 1$ (this was only asked in the exam,anyway it doesnt matter whether the power is $ 2n$ or $ 2^{2^{n}}$) or $ 2^{2n} \plus{} 1$ is a prime number.

What if $ a \minus{} b \equal{} 1$? I think your proof is incomplete.

I never gave a complete proof :wink:...I just gave the required hint...from there isnt it easy ?...substitute the value of $ a\equal{}b\plus{}1$ in the other equation ,you will get a contradiction...also checking for other cases is also easy...


[quote="jhredsox"]What is the hardest sprint question problem 1?
-jorian[/quote]
Why does it matter? I could understand hardest problem, but why question number 1?

[quote="mathgeniuse^ln(x)"]He is thinking that the harder the first question the harder the test, and wants to know...[/quote]
Generally not true, that's like saying the harder the last question the harder the test (though that was true for 05 :D). I'd say 05 has the hardest first question...

[quote="mathgeniuse^ln(x)"]He is thinking that the harder the first question the harder the test, and wants to know...[/quote]

But he already created a forum that has people posting the hardest mathcounts years so I doubt that's why he wants to know the hardest first questions.

I think this thread isn't going anywhere, knowing the first question really tells you nothing about the test in general.  Locked.


inequalities
trigonometry
inequalities unsolved


arithmetic sequence
number theory
relatively prime
number theory unsolved

Given n integers which are the n consecutive tems of an arithmetic progression, whose common difference is relatively prime to n!.
Prove that the product of the n integers is divisible by n!

Assume $a,a+d,a+2d,...,a+(n-1)d$  the $n$ consecutive tems of an arithmetic progression, where $\gcd (d,n!)=1$. For every prime $p$ and positive integer $k$, have at least $[\frac{n}{p^{k}}]$ elements of $\{a,a+d,...,a+(n-1)d\}$ divisble by $p^{k}$. We're done!.

Note: In $\{a,a+d,...,a+(p^{k}-1)d\}$ has at least one element divisble by $p^{k}$, so in $\{a+p^{k}d,...,a+(2p^{k}-1)d\}$, etc. And if $p^{r}\parallel n!$ then $r=\sum_{i=1}^{\infty}[\frac{n}{p^{i}}]$.


calculus
derivative
calculus computations

Find the equations of the lines tangent to the curve x^2- xy+y^2=9 at the two points where the curve intersects the x-axis.

A swimming pool is 60 ft long and 25 ft wide. Its depth varies uniformly from 3 ft at the shallow end to 15 ft at the deep end. Suppose the pool is being filled with water at the rate of 800 ft^3/min. At what rate is the depth of water increasing at the deep end when it is 5 ft deep at that end?

You're curve intersects at $(3,0)$ and $(-3,0)$. The derivative is $2x-y-x\frac{dy}{dx}+2y\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=\frac{y-2x}{2y-x}$. So evaluated at $(3,0)$ you have $\frac{dy}{dx}=3$ and at $(-3,0)$ you have $\frac{dy}{dx}=-3$, so the equations are $y=3(x-3)$ and $y=-3(x+3)$, as long as none of my signs are backwards (did the actual calculations in my head).


find all the x,y,z, in N

 so:     *x|yz-1

           *y|xz-1

           * z|xy-1  :!:

hi
the solution was easy and it is here:
 each of the (x,y,z) divides xy+yz+xz-1.(this is a result of the main problem)and we have:(x,y)=(x,z)=(y,z)=1 because if (x,y)=d then d divides dk-1 so d is equal to 1.
so because each of (x,y,z)divides xy+xz+yz-1 and they have no common divisior ,xyz divides xy+xz+yz-1.but its easy to prove that for natural numbers of x,y,z we have this inequlity that : xy+yz+xz-1 is bigger than xyz and smaller than 2xyz(the proof is by induction)so we can easily find the answers which are the answers that they turn the above inequlity to equality.(I dont want to find them here) :D

Isn't this from Romaian Junior TST or something?


ARML AMC AIME Stanford college geometry email

When do the lectures start? The post on Yahoo groups is somewhat confusing.

huh? link if applicable, plz
i wasnt aware that arml gave lectures
 :huh:

I have no idea what that post meant either, but I'll just follow the crowd when the date comes near.

I think the first practice is tentatively scheduled for April 6th.  We're still working on room reservations, so it probably won't be officially announced until then.

Just a random-ish question, what level/type of math is ARML?

Hold on, confirmation: ARML practice is at Stanford?

Yes. As far as I know, it should be in the same room as Stanford Math Circle (C380 or C308, something like that).

In the past, it's been 380-380C.  That may change; don't assume anything until you get official word.  I'll post something here when we know details.

[quote="serialk11r"]Hmmm I sorta need to know really soon to make Chinese class plans (I have private small chinese class where there's some flexibility for rescheduling...).[/quote]
thanks for letting us know; we really care

Wow, that sounder meaner than I think you intended it to be junggi...

Anyways, from the flyer that they handed out at BAMO Awards Ceremony:

"ARML practices for the San Francisco Bay Area Teams will begin at Stanford University Sunday, April 6 2 PM- 4:45 PM"

Join the message group at: http://groups.yahoo.com/group/SFBA-ARML for more information as it becomes available."

If that's the case then I will not be able to make the first meeting  :(
Shouldn't the more information be coming about now?

Wait holy crap TWO???? Are you sure about this? That means I'm gonna be late to pretty much all of them...

[quote="Ubemaya"]Which J-PLL... not saying much. It is the one that looks like an L or a backwards L. Plus, there are several algorithms for those...

I can do >10 turns a second, but only on stuff like R U R' U' x 6.[/quote]

Well so can I on those since only 2 sides are rotated... it's other turns that make it harder.

Yo lewis, my mom says you were reading a book during chadbourne safety patrol. Man, chadbourne really needs another set of safety patrol people to take care of the safety patrol people.

LOL WTF? NO COMMENT.

 :rotfl:  :rotfl:  :rotfl:

WHAT HAPPENED AT ARML

I'm so proud I live in Fremont:
[youtube]vSb-nV8l2QY[/youtube]

wow

_________________
Procrastination is the FASTEST way to get things done.
Talent=Apparent Skill/Work Ethic
If you like chess, go to http://www.artofproblemsolving.com/Forum/viewtopic.php?t=193031 and post if interested in the 5 minute BLITZ TOURNAMENT!

Why exactly am I watching this right now?

Although I'll admit, that was the most pwnage video I've seen in a while.

[quote="Alumen"]That was disturbing[/quote]
Not everyone agrees with you; that video got a TON of awards. Btw, did you know that that video was done entirely by hand (as in no computer editing)? Dude, that video wasn't actually even ever posted on the internet by Bitter Films, but now it has like, what, 3 mil views?

EDIT: My bad, 1.7 mil views. While your at it, here's one of his first videos.
NOTE: May contain material offensive to girls.
[youtube]kYqKucJTn2c[/youtube]

[quote="hunter34"]When I was watching that, I thought I was on acid.[/quote]
Yep, same here. "I live in a giant bucket." :rotfl:

Without the pyschos in that youtube video, the world would be in a wonderous turmoil.

To me the world is a wonderous turmoil...

then again, its probably the acid..


function
inequalities unsolved
inequalities

[quote="thanhnam2902"]Let $ x \plus{} y\ge 0$. Find maximum of funtion: $ f(x,y) \equal{} \frac {1 \plus{} 2^{x \plus{} y}}{1 \plus{} 4^x} \plus{} \frac {1 \plus{} 2^{x \plus{} y}}{1 \plus{} 4^y}$.[/quote]
$ f(x,y) \minus{} 2 \equal{} \frac {(2^x \minus{} 2^y)^2(2^{x \plus{} y} \minus{} 1)}{(4^x \plus{} 1)(4^y \plus{} 1)}\geq 0$
i think that,  $ x \plus{} y \leq 0$ then  $ Max f(x,y)\equal{}2$
 if  following  condition of this problem then  $ Min f(x,y) \equal{} 2$
 occur equal  if and only if : $ x \equal{} y$or $ x \plus{} y \equal{} 0$


Let $ a,b,c$ be positive real number
Prove that

$ \frac{a}{\sqrt{\minus{}a\plus{}b\plus{}c}}\plus{}\frac{b}{\sqrt{a\minus{}b\plus{}c}}\plus{}\frac{c}{\sqrt{a\plus{}b\minus{}c}} \ge \sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}$.

I think that$ a, b$ and $ c$ must be sides of a triangle , so let $ b \plus{} c \minus{} a \equal{} x,a \plus{} c \minus{} b \equal{} y$ and $ a \plus{} b \minus{} c \equal{} z$ .
$ \Longleftrightarrow \sum \frac {y \plus{} z}{2\sqrt {x}} \ge \sum \sqrt {\frac {x \plus{} y}{2}}$
by chebychev's theorem  $ \sum \frac {y \plus{} z}{2\sqrt {x}} \ge \frac {1}{3}(\sum \frac {1}{\sqrt {x}}) \sum \frac {y \plus{} z}{2} \ge$
$ \frac {1}{9}(\sum \frac {1}{\sqrt {x}})(\sqrt {\frac {y \plus{} z}{2}} \plus{} \sqrt {\frac {x \plus{} z}{2}} \plus{} \sqrt {\frac {y \plus{} x}{2}})^2$
$ \ge \frac {1}{9}(\sum \frac {1}{\sqrt {x}})(\sum \sqrt {x})(\sqrt {\frac {y \plus{} z}{2}} \plus{} \sqrt {\frac {x \plus{} z}{2}} \plus{} \sqrt {\frac {y \plus{} x}{2}})$
$ \ge \sqrt {\frac {y \plus{} z}{2}} \plus{} \sqrt {\frac {x \plus{} z}{2}} \plus{} \sqrt {\frac {y \plus{} x}{2}}$.

I think that it's false for $ a \equal{} 4,b \equal{}$1 and $ c \equal{} 2.$
but in this case $ a,b$ and $ c$ are not sides of triangle.

Let $ a,b,c$ be positive real number
Prove that

$ \frac{a^2}{(\minus{}a\plus{}b\plus{}c)^4}\plus{}\frac{b^2}{(a\minus{}b\plus{}c)^4}\plus{}\frac{c^2}{(a\plus{}b\minus{}c)^4} \ge \frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^2}$.

[quote="soruz"]In any triangle prove that

$ \frac {a^2}{( \minus{} a \plus{} b \plus{} c)^3} \plus{} \frac {b^2}{(a \minus{} b \plus{} c)^3} \plus{} \frac {c^2}{(a \plus{} b \minus{} c)^3} \ge \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}$.[/quote]
Yes, it's true! SOS kills it easy enough.
[quote="soruz"]Let $ a,b,c$ be positive real number
Prove that

$ \frac {a^2}{( \minus{} a \plus{} b \plus{} c)^4} \plus{} \frac {b^2}{(a \minus{} b \plus{} c)^4} \plus{} \frac {c^2}{(a \plus{} b \minus{} c)^4} \ge \frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}$.[/quote]
Try $ a\rightarrow0^\plus{}.$ :wink:

[quote="soruz"]tchebytchev wrote:
$ \frac {1}{9}(\sum \frac {1}{\sqrt {x}})(\sqrt {\frac {y \plus{} z}{2}} \plus{} \sqrt {\frac {x \plus{} z}{2}} \plus{} \sqrt {\frac {y \plus{} x}{2}})^2$
$ \ge \frac {1}{9}(\sum \frac {1}{\sqrt {x}})(\sum \sqrt {x})(\sqrt {\frac {y \plus{} z}{2}} \plus{} \sqrt {\frac {x \plus{} z}{2}} \plus{} \sqrt {\frac {y \plus{} x}{2}})$
Can someone detail please!!!Thanx![/quote]
$ (\sqrt {\frac {y \plus{} z}{2}} \plus{} \sqrt {\frac {x \plus{} z}{2}} \plus{} \sqrt {\frac {y \plus{} x}{2}})^2$
$ \equal{}(\sqrt {\frac {y \plus{} z}{2}} \plus{} \sqrt {\frac {x \plus{} z}{2}} \plus{} \sqrt {\frac {y \plus{} x}{2}})(\sqrt {\frac {y \plus{} z}{2}} \plus{} \sqrt {\frac {x \plus{} z}{2}} \plus{} \sqrt {\frac {y \plus{} x}{2}}).$
and since $ \sqrt {\frac {p^2 \plus{} q^2}{2}} \ge  \frac {p \plus{} q}{2}$. we have:
 $ \sqrt {\frac {y \plus{} z}{2}} \plus{} \sqrt {\frac {x \plus{} z}{2}} \plus{} \sqrt {\frac {y \plus{} x}{2}} \ge \sqrt{x}\plus{}\sqrt{y}\plus{}\sqrt{z}.$

In any triangle prove that
$ \frac{\minus{}a^3\plus{}b^3\plus{}c^3}{\minus{}a\plus{}b\plus{}c}\plus{}\frac{a^3\minus{}b^3\plus{}c^3}{a\minus{}b\plus{}c}\plus{}\frac{a^3\plus{}b^3\minus{}c^3}{a\plus{}b\minus{}c} \le ab\plus{}bc\plus{}ca$.

[quote="soruz"]In any triangle prove that
$ \frac { \minus{} a^3 \plus{} b^3 \plus{} c^3}{ \minus{} a \plus{} b \plus{} c} \plus{} \frac {a^3 \minus{} b^3 \plus{} c^3}{a \minus{} b \plus{} c} \plus{} \frac {a^3 \plus{} b^3 \minus{} c^3}{a \plus{} b \minus{} c} \le ab \plus{} bc \plus{} ca$.[/quote]
$ \sum_{cyc}\frac {a^3 \plus{} b^3 \minus{} c^3}{a \plus{} b \minus{} c}\leq ab\plus{}ac\plus{}bc\Leftrightarrow\sum_{cyc}(a^5\minus{}a^4b\minus{}a^4c\plus{}a^3bc)\geq0.$

[quote="soruz"]In any triangle prove that
$ a^2( \minus{} a \plus{} b \plus{} c) \plus{} b^2(a \minus{} b \plus{} c) \plus{} c^2(a \plus{} b \minus{} c) \le 3abc$.[/quote]
We have:
$ \sum_{cyc}(b\minus{}c)^2(b\plus{}c\minus{}a) \geq 0$
<=>$ \sum_{cyc}(b\minus{}c)^2(b\plus{}c)\minus{}\sum_{cyc}a(b\minus{}c)^2 \geq 0$
<=>$ \sum_{cyc}(b^2\minus{}c^2)(b\minus{}c)\minus{}\sum_{cyc} a(b\minus{}c)^2 \geq 0$
<=>$ 6abc\minus{}2\sum_{cyc}a^2(b\plus{}c\minus{}a) \geq 0$
<=>$ \sum_{cyc} a^2(b\plus{}c\minus{}a) \leq 3abc$
Done!

[quote="soruz"]In any triangle prove that
$ a^2( \minus{} a \plus{} b \plus{} c) \plus{} b^2(a \minus{} b \plus{} c) \plus{} c^2(a \plus{} b \minus{} c) \le 3abc$.[/quote]

This is Schur inequality, and we can prove it without the condition a,b,c are three length sides of a trangle.

[quote="soruz"]In any triangle prove that
$ a^2(\minus{}a\plus{}b\plus{}c)\plus{}b^2(a\minus{}b\plus{}c)\plus{}c^2(a\plus{}b\minus{}c) \le 3abc$.[/quote]
$ a^2(\minus{}a\plus{}b\plus{}c)\plus{}b^2(a\minus{}b\plus{}c)\plus{}c^2(a\plus{}b\minus{}c) \le 3abc$([url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=346917&sid=97ce8668540a928189b121cb6b617957#p346917]IMO1964[/url])
$\Leftrightarrow abc\ge(b+c-a)(c+a-b)(a+b-c)$
$\Rightarrow\sqrt{\frac{a}{b+c-a}}+\sqrt{\frac{b}{c+a-b}}+\sqrt{\frac{c}{a+b-c}}\ge 3$
([b]Sqing[/b]:Mathematics teaching (China Shanghai)No.5(2001) problem543 )
$\Rightarrow\sqrt{\frac{a}{b+c-a}}+\sqrt{\frac{b}{c+a-b}}+\sqrt{\frac{c}{a+b-c}}>2\sqrt{2}$
$\Leftrightarrow \sqrt{\frac{a}{p-a}}+\sqrt{\frac{b}{p-b}}+\sqrt{\frac{c}{p-c}}>4$.
(Mihaela Berindean: GMB 9/2013, pag.406,Q26737)

[quote="soruz"]In any triangle prove that
$ a^2(\minus{}a\plus{}b\plus{}c)\plus{}b^2(a\minus{}b\plus{}c)\plus{}c^2(a\plus{}b\minus{}c) \le 3abc$.[/quote]

In any triangle prove that
\[3abc \geq a^2(\minus{}a\plus{}b\plus{}c)\plus{}b^2(a\minus{}b\plus{}c)\plus{}c^2(a\plus{}b\minus{}c) \ge \frac{8}{\sqrt{3}}w_aw_bw_c\]

[quote=sqing]Let $a,b,c$ be three sides of a triangle with circumradius  $R$ and inradius  $ r $ .Prove that
$$ \sqrt{\frac{2R}{r}}+1\leq\sqrt{\frac{a}{b+c-a}}+\sqrt{\frac{b}{c+a-b}}+\sqrt{\frac{c}{a+b-c}}\leq\frac{R}{r}+1$$[/quote]

Long ago I solved the following one equality:
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For $ x, y, z > 0 $ real numbers, we have:

$ \sqrt{\frac{y+z}{x}} + \sqrt{\frac{x+z}{y}} + \sqrt{\frac{x+y}{z}} = \sqrt{\frac{(x+y)(y+z)(z+x)}{xyz}} + \frac{2(x+y+z)}{ \sqrt{x(y+z)} + \sqrt{y(x+z)} + \sqrt{z(x+y)} } \ \ ; $

Author: Lorian Saceanu
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It will help here! Greetings dear teacher Sqing!

Let $a,b,c$ be three sides of a triangle with circumradius $R$ and inradius $ r $ .Prove or disprove
$$ \sqrt{\frac{2R}{r}}+1\leq\frac{a}{\sqrt{b(c+a-b)}}+\frac{b}{\sqrt{c(a+b-c)}}+\frac{c}{\sqrt{a(b+c-a)}}\leq\frac{R}{r}+1$$
Let $a,b,c$ be three sides of a triangle . Prove  that
$$\frac{a}{\sqrt{b(c+a-b)}}+\frac{b}{\sqrt{c(a+b-c)}}+\frac{c}{\sqrt{a(b+c-a)}}\geq 3$$

[quote=Lonesan]We can improve a little the right part of inequality as follows:
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$ \frac{2(x+y+z)}{ \sqrt{x(y+z)} + \sqrt{y(x+z)} + \sqrt{z(x+y)} } \le \sqrt{\frac{(y+z)(z+x)(x+y)}{4xyz}} = \sqrt{\frac{R}{r}} \ \ ; $
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So our inequality will become:
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Let $a,b,c$ be three sides of a triangle with circumradius $R$ and inradius $ r $ .Prove that:
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$\sqrt{\frac{2R}{r}}+1 \le \sqrt{\frac{a}{b+c-a}}+\sqrt{\frac{b}{c+a-b}}+\sqrt{\frac{c}{a+b-c}}\leq 3\sqrt{\frac{R}{2r}} \ \ ; $
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Greetings from Lorian Saceanu![/quote]

${\it q492}=3\,{\frac { \left( x+y+z \right) \sqrt {2}xyz}{\sqrt {x
 \left( y+z \right) }+\sqrt {y \left( z+x \right) }+\sqrt {z \left( x+
y \right) }}}
$

$q492$ is here:

\[[492, [402, 103, 341], [19, 386]]\]

$[402,103,341]=$
$[\sqrt {3}\sqrt {2} \left( \sqrt {{\frac {{x}^{3}}{
 \left( y+z \right)  \left( {x}^{2}+2\,yz \right) }}}+\sqrt {{\frac {{
y}^{3}}{ \left( z+x \right)  \left( {y}^{2}+2\,zx \right) }}}+\sqrt {{
\frac {{z}^{3}}{ \left( x+y \right)  \left( {z}^{2}+2\,xy \right) }}}
 \right) xyz,$
$1/2\, \left( \sqrt {yz \left( y+z \right) }+\sqrt {zx
 \left( z+x \right) }+\sqrt {xy \left( x+y \right) } \right) \sqrt {xy
z}\sqrt {2},1/2\,{\frac {yz \left( x+y \right)  \left( z+x \right) }{y
+z}}+1/2\,{\frac {x \left( x+y \right) z \left( y+z \right) }{z+x}}+1/
2\,{\frac {xy \left( y+z \right)  \left( z+x \right) }{x+y}}]$

$[19,386]=$
$[ \left( 2\,\sqrt {{\frac {y+z}{6\,x+y+z}}}+2\,\sqrt {{\frac
{z+x}{6\,y+x+z}}}+2\,\sqrt {{\frac {x+y}{6\,z+x+y}}} \right) xyz,
\sqrt {3} \left( \sqrt {{\frac {{x}^{2}}{xy+zx+yz}}}+\sqrt {{\frac {{y
}^{2}}{xy+zx+yz}}}+\sqrt {{\frac {{z}^{2}}{xy+zx+yz}}} \right) xyz]
$

choice  nice ineq ,as

\[\frac{1}{6}\sum{\sqrt {{\frac {y+z}{x}}}}\geq {\frac {x+y+z}{\sqrt {x \left( y+z \right) }+
\sqrt {y \left( z+x \right) }+\sqrt {z \left( x+y \right) }}}\geq \frac{\sqrt{2}}{3}\sum{\sqrt {{\frac {y+z}{6\,x+y+z}}}}\]

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

\[\sqrt {2}\sqrt {{\frac { \left( xy+zx+yz \right)  \left( x+y+z
 \right) }{xyz}}- \left( 27-18\,\sqrt {2} \right)  \left( \frac{1}{9}-{\frac {
xyz}{ \left( xy+zx+yz \right)  \left( x+y+z \right) }} \right) }\geq
\sqrt {{\frac {y+z}{x}}}+\sqrt {{\frac {z+x}{y}}}+\sqrt {{\frac {x+y}{
z}}}\]

[quote=xzlbq]
choice  nice ineq ,as

\[\frac{1}{6}\sum{\sqrt {{\frac {y+z}{x}}}}\geq {\frac {x+y+z}{\sqrt {x \left( y+z \right) }+
\sqrt {y \left( z+x \right) }+\sqrt {z \left( x+y \right) }}}\geq \frac{\sqrt{2}}{3}\sum{\sqrt {{\frac {y+z}{6\,x+y+z}}}}\][/quote]

indeed nice and hard!

[quote=Lonesan]The left part of inequality (using my equality from above) becomes:
\\
\\
$ \frac{1}{6}\sum{\sqrt {{\frac {y+z}{x}}}}\geq {\frac {x+y+z}{\sqrt {x \left( y+z \right) }+
\sqrt {y \left( z+x \right) }+\sqrt {z \left( x+y \right) }}}  \iff  \frac{1}{3}\sum{\sqrt {{\frac {y+z}{x}}}}\geq {\frac {2(x+y+z)}{\sqrt {x \left( y+z \right) }+
\sqrt {y \left( z+x \right) }+\sqrt {z \left( x+y \right) }}} \iff  \frac{1}{3}\sum{\sqrt {{\frac {y+z}{x}}}}\geq \sqrt{\frac{y+z}{x}} + \sqrt{\frac{x+z}{y}} + \sqrt{\frac{x+y}{z}} - \sqrt{\frac{(x+y)(y+z)(z+x)}{xyz}} \iff \frac{3}{2} \ge \sum \sin \frac{A}{2} \ \ ; true $
\\
\\
Greetings from Lorian Saceanu![/quote]

full do it.

[quote=Lonesan]\\
Let $a,b,c$ be three sides of a triangle with circumradius $R$ and inradius $ r $ .Prove that:
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$\sqrt{\frac{2R}{r}}+1 \le \sqrt{\frac{a}{b+c-a}}+\sqrt{\frac{b}{c+a-b}}+\sqrt{\frac{c}{a+b-c}}\leq \sqrt{\frac{4R+r}{r}} \le 3\sqrt{\frac{R}{2r}} \ \ ; $
\\
\\
Greetings from Lorian Saceanu![/quote]
In #24,
Equivalent transformation process，A little more detail。

And an old one:
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If $ \triangle ABC $ is an acute angled one:
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$ \sum \sqrt{\frac{a}{s-a}} \ge \frac{5R+2r}{2\sqrt{Rr}} \ \ ; $
\\
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Greetings from Lorian Saceanu!


can u tell in details what gaussian pairing actually is? plz? :D

[quote="mathwizarddude"]Prove that
$ 1 \plus{} 2\cos(2\pi/5) \plus{} 3\cos(4\pi/5) \plus{} 4\cos(6\pi/5) \plus{} 5\cos(8\pi/5) \equal{} \minus{} \frac52$
and
$ \sec^2(\pi/7) \plus{} \sec^2(2\pi/7) \plus{} \sec^2(3\pi/7) \equal{} 24$.[/quote]

[hide="solution of the first one"]
For complex number $ z\ne 1$,we have the identity:

$ z\plus{}z^2\plus{}z^3\plus{}z^4\plus{}z^5\equal{}\frac{z^6\minus{}z}{z\minus{}1}$,and

$ \frac{d}{dz}(z\plus{}z^2\plus{}z^3\plus{}z^4\plus{}z^5)\equal{}\frac{d}{dz}(\frac{z^6\minus{}z}{z\minus{}1})$,thus we have a identity:

$ 1\plus{}2z\plus{}3z^2\plus{}4z^3\plus{}5z^4\equal{}\frac{(6z^5\minus{}1)(z\minus{}1)\minus{}z(z^5\minus{}1)}{(z\minus{}1)^2}$

when $ z\equal{}\cos(2\pi/5)\plus{}i\sin(2\pi/5)$,

we have the LHS of given $ \equal{}Re(1\plus{}2z\plus{}3z^2\plus{}4z^3\plus{}5z^4)\equal{}Re(\frac{(6z^5\minus{}1)(z\minus{}1)\minus{}z(z^5\minus{}1)}{(z\minus{}1)^2})$

$ \equal{}Re(\frac{5}{(z\minus{}1)})$ (because $ z^5\equal{}1)$

$ \equal{}Re(\frac{5}{(\cos(2\pi/5)\plus{}i\sin(2\pi/5)\minus{}1})\equal{}Re(\frac{5(\cos(2\pi/5)\minus{}1\minus{}i\sin(2\pi/5))}{(\cos(2\pi/5)\minus{}1)^2\plus{}\sin^2(2\pi/5))})$

$ \equal{}\frac{5(\cos(2\pi/5)\minus{}1)}{2(1\minus{}\cos(2\pi/5))}\equal{}\minus{}\frac{5}{2}$　,it's done.[/hide]


Prove that for every value of $ n >10000$ we can find two numbers $ a,b /in N$ such that
                    $ n<a^2\plus{}b^2<n\plus{}3n^{1/4}$.

Let $ a \equal{} [\sqrt {n \plus{} 1}]$, then $ a^2\le n \plus{} 1 \le a^2 \plus{} 2a$. If $ a^2\ge n$ we take $ b \equal{} 1$. It give $ n < a^2 \plus{} b^2\le n \plus{} 2 < n \plus{} 3\sqrt [4]{n}$.
If $ a^2 < n$, then take $ b \equal{} 1 \plus{} k\ge 1, k \equal{} [\sqrt {n \minus{} a^2}]$. Then $ n < a^2 \plus{} b^2\le a^2 \plus{} k^2 \plus{} 2k < n \plus{} 2k$ or $ a^2 \plus{} b^2 < 2\sqrt 2 \sqrt a < 2\sqrt 2 \sqrt [4] n$.
It is true for any $ n\in N$ and exactly (exist $ a,b\in N\equal{} \{1,2,...\}$), suth that $ n<a^2\plus{}b^2<2\sqrt 2 \sqrt[4] n.$


Find all complex ordered triplets $(x,y,z)$ such that

$x+y+z=x^{2}+y^{2}+z^{2}=x^{3}+y^{3}+z^{3}=3$.

[quote="mathnerd314"][quote="Allan Z"]$(x,y,z)=\boxed{(1,1,1)}$[/quote]

Of course, that follows by inspection, from the fact that $1+1+1=1^{2}+1^{2}+1^{2}=1^{3}+1^{3}+1^{3}=3$.

But I think he's asking for complex solutions.[/quote]

Nowhere did Allan Z assume that the roots were real, so his solution is fine.


AMC
AMC 10
AMC 12
USA(J)MO
USAMO
AIME
probability

Dear AMC,

I know you are very busy this week, and I appreciate all that everyone at AMC does to make this excellent program available.

However, I do not understand why the AMC website says:

[quote]          7. The selection process is designed to favor students who take the more mathematically comprehensive AMC 12A and AMC 12B contests.

    * In advising students who desire to be selected for the USAMO whether to take the AMC 12 level contests or the AMC 10 level contests, it will be to their advantage to take the AMC 12 level contests. [/quote]

Source:  http://www.unl.edu/amc/e-exams/e8-usamo/usamo.html

I believe this statement was true in some earlier years (e.g., 2000), but it appears to be no longer true.

Clarification would be appreciated, so that future students facing the choice between AMC10 and AMC12 can make better informed decisions.

Thanks

Edited:  I did not intend for this to be a letter of complaint--and I realize that the decision about cutoff levels must be made by more than one person, not just the director.   I believe that this language first appeared when the AMC introduced the 10 and 12 split, in 2000.  The committee did have lower index cutoffs that year for young students who took the AMC12 rather than the AMC10.

Since then I have wondered whether the statement above still represents current policy and I think clarification would be helpful.

[quote="Marc982"]Dear Prof. Dunbar,

I know you are very busy this week, and I appreciate all that you and the other folks at AMC do to make this excellent program available.

However, I do not understand why the website says:

[quote]          7. The selection process is designed to favor students who take the more mathematically comprehensive AMC 12A and AMC 12B contests.

    * In advising students who desire to be selected for the USAMO whether to take the AMC 12 level contests or the AMC 10 level contests, it will be to their advantage to take the AMC 12 level contests. [/quote]

Source:  http://www.unl.edu/amc/e-exams/e8-usamo/usamo.html

I believe this statement was true in some earlier years (e.g., 2000), but it appears to be no longer true.

Clarification would be appreciated, so that future students facing the choice between AMC10 and AMC12 can make better informed decisions.

Thanks[/quote]

I really don't think this was necesarry.

First of all, if you have a complaint to make to Prof. Dunbar, it would make more sense and be more polite and likely to be listened to if you e-mail him instead of posting your complaint on a public forum.

Secondly, he doesn't want people e-mailing him because he's busy COMPILING THE LIST OF USAMO QUALIFIERS. If he doesn't want you e-mailing him, and e-mailing is better than posting on a public forum, then he certainly doesn't want you posting complaints to him on a public forum.

Also, you wouldn't even know about the whole floor/index/AMC10-12/whatever if he hadn't been kind enough to post the information on AoPS. So in return for his consideration he gets yelled at. That doesn't seem very fair either.

On another note, thank you very much to Professor Dunbar for providing us with the Floor/Index information.

[quote="Phelpedo"]Obviously the AMC 12 is advantageous under certain circumstances. Mathematically, they select more people who take the AMC 12 than the 10, and you can qualify pretty much based on a high AMC 12 score and a mediocre AIME score, unlike the 10.

Bottom line is, you know the selection process- it's quite explicitly stated on the site- so you can make your own decision about which is better to take.[/quote]

Yep, it explicitely recommends the amc 12, and they count equally (NOT mentioned)!  And the 10 is obviously easier.

[quote="filletwho"]It was in the handbook. It was mentioned that they take the index for tenth graders and below.[/quote]

Care to give a link?

What it seems to boil down to is that if you have a legitimate shot at qualifying for the USAMO, and you are young enough to take the AMC10, then you should take the AMC10.  If you merely want to qualify for the AIME, then you should take the AMC12 instead.  This is because it is easier to qualify for the AIME via the AMC12, but easier to qualify for the USAMO via the easier AMC10 (as long as you can qualify for the AIME first!)

An excerpt from a message from Prof. Dunbar:

[quote]Here is a quote from the notes that I made in the selection process last year (March 25, 2005):

   "Double-checking:  In this group of 241 are 196 with a USAMO index of 233.0 or greater, among whom are 31 10-takers, so I have selected 165 12-takers with the highest USAMO index.   Everybody with a lower index was a 10th grader or below with a 9 or better on the AIME."

Among those 10th graders I made no distinction about which contest they took, AMC 10 or AMC 12, and that is corroborated in my notes elsewhere.

We will change the written protocol to reflect the reality.

As for advising your student, the latest figures I have handy indicate  that in 2004, 1596 10th graders and below qualified for AIME by virtue of their score on the AMC 12.  On the other hand 1309 10th graders and below qualified for AIME on the basis of their AMC 10 score.  So it appears that chances are better for qualifying for AIME by way of the AMC 12 for which the door is open wider.
[/quote]

[quote="probability1.01"][quote="Phelpedo"]Obviously the AMC 12 is advantageous under certain circumstances. Mathematically, they select more people who take the AMC 12 than the 10, and you can qualify pretty much based on a high AMC 12 score and a mediocre AIME score, unlike the 10.

Bottom line is, you know the selection process- it's quite explicitly stated on the site- so you can make your own decision about which is better to take.[/quote]

That doesn't make sense. If you're talking about the selection before the floor score selection, the AMC 12 only gives you that extremely tiny opportunity of setting the floor yourself. So yeah, I have no idea what you're talking about (btw, the selection process in the event of a floor score tie thing is not explicitly stated at all, unless something was added relatively recently).[/quote]

If you calculate the probability of setting the floor as 1/240, then yeah, there is practically no chance. I'm looking at it like this: say you are good at speed, and generally taking speed tests like the AMC. But you are not good at the AIME ( not good meaning that you can't qualify by matching the floor). Then your only chance is to have a good enough AMC12 score and a good enough AMC12 index to make the initial top 240 selection. i.e., on this test there would be no way to make the USAMO with an AIME of 7 if you took the AMC10, but it would be possible if you got a high score on the AMC12.

Let's generously say that the chance of setting the floor value is 1/50. Something tells me that improving your chances by 1/50 and jeopardizing your opportunity in the event of a floor tie (very likely, especially with expansion) is not worth it. Notice how setting the floor could only possibly allow you to have an AIME score of one lower, which is not much help even in such rare circumstances, meaning that you had a good chance of making the floor in the first place.

[quote="seesaw"]What some people tend to forget is that one who takes the AMC10 must get a 120  to get into the AIME, even though it is easier.  About 1/5 of the people who took the AMC12 got into the AIME, compared to only the 2% of those who took the AMC10.[/quote]

Yes, but this topic refers to about usamo qualification, but whatever this thread is getting pointless.
Moral of Story: tough luck, life isn't fair, so live with it and study more

Yes, but the website says that the AMC12 is better to get into the [b]USAMO[/b]. Anyway, the people who are  complaining wouldn't have a problem making the 120 AMC10 cutoff. [/hide]

[quote="Phelpedo"]Yes, but I'm not saying you should try to be that one person to set the floor. Obviously, it's not a good decision to go and try to set the floor, and throwing away your index score. But if you know that you can't get a high enough AIME score, your only hope is of obtaining a high AMC12 score to be in the initial selection.[/quote]

Well, then it still makes no sense. The only time AMC 12 benefits over AMC 10 is when you yourself set the floor score. Otherwise, even if your AMC 12 + AIME qualified through the initial selection, your AIME score was already high enough to qualify by the floor score. So yeah... I'm not going to discuss this anymore, I think you can figure it out.

The reason I'm closing this thread is explained at:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=467288#p467288.


Express this sum as a common fraction:

\[ \sqrt{0.16} \plus{} \sqrt[3]{\minus{}0.027}\]

hello, we have $ \sqrt{0.16}\plus{}\sqrt[3]{\minus{}0.027}\equal{}0.4\minus{}0.3\equal{}0.1\equal{}\frac{1}{10}$.
Sonnhard.


linear algebra
matrix
linear algebra unsolved

$E$ a $\mathbb{C}$ vectorial space.
If $M$ is a matrix, $C(M)$ is the subspace of matrices that commute with $M$.
1/ if $H$ is an hyperplan of $M_n(\mathbb{C})$, does there always exist a matrix $M$ such that
$H$ contains $C(M)$ ?
2/ what is the minimal number $k < n^2$ such that for any vectorial space $V$ of dimension
$k$, there exists a matrix $M$ such that $V$ contains $C(M)$ ?

Every commutant contains $I$, but there are hyperplanes in $\mathcal M_n(\mathbb C)$ which do not contain $I$, so the answer to the first question is 'no'. This makes the second question pointless, so what did you really mean, alekk? :)

:blush:  all vetorial spaces here have to contain the identity matrix.


logarithms inequalities symmetry function calculus derivative parameterization

Set $ f(t) \equal{} \frac {1}{1 \plus{} e^t}$ and note that $ f(t) \plus{} f( \minus{} t) \equal{} 1$ fairly trivially.

Setting $ x \equal{} \log a, y \equal{} \log b, z \equal{} \log c, w \equal{} \log d$, the problem transforms into

For real numbers $ x\geq y \geq z \geq w$ such that $ x \plus{} y \plus{} z \plus{} w \equal{} 0$, show that
\[ \frac {f(x) \plus{} f(y) \plus{} f(z)}{3} \geq f\left(\frac {x \plus{} y \plus{} z}{3}\right)
\]
Noting that the case where all the numbers are equal is trivial, we can assume that $ w < 0 < x$.

Now, if $ x,y,z \geq 0$, the result easily follows from Jensen's Inequality (as $ f(t)$ is convex on $ t\geq 0$). We may therefore assume $ z<0$ also.

If $ y<0$ as well, then let us take $ p\equal{}\minus{}y, q\equal{}\minus{}z, r\equal{}\minus{}w$. By the symmetry of the $ f$ function, we have to show

\[ \frac {f(x) \plus{} 2 \minus{} f(p) \minus{} f(q)}{3} \geq f\left(\frac {r}{3}\right)\]
or
\[ f(p\plus{}q\plus{}r) \plus{} 2\geq 3f\left(\frac {r}{3}\right)\plus{}f(p)\plus{}f(q)\]

Let's transform again, setting $ g(x) \equal{} 2f(x) \minus{} 1$. This doesn't actually change the inequality, but we now have $ g(x) \equal{} 2f(x) \minus{} 1 \equal{} 1 \minus{} 2f( \minus{} x) \equal{} \minus{} g( \minus{} x)$. I.e. $ g$ is odd.
Thus
\[ \frac {g(x) \plus{} g(y) \plus{} g(z)}{3} \geq g\left(\frac {x \plus{} y \plus{} z}{3}\right)
\]
becomes, using the constraints and the fact that $ g$ is odd
\[ g(x) \plus{} g(y) \plus{} g(z) \plus{} 3g(\frac {w}{3}) \geq 0.
\]

Let $ f(x)\equal{}\frac{1}{1\plus{}e^x}$. Verify that $ f''(x)\equal{}\frac{e^x(e^x\minus{}1)}{(1\plus{}e^x)^3}\ge 0$. Hence $ f$ is convex in $ \mathbb R^\plus{}$. Therefore from Jensen's inequality we conclude that
\[ f(\ln a)\plus{}f(\ln b)\plus{}f(\ln c)\ge 3\cdot f\left(\frac{\ln a\plus{}\ln b\plus{}\ln c}{3}\right)\equal{}3\cdot f\left(\ln(abc)^{\frac 13}\right)\\
\iff \frac{1}{1\plus{}a}\plus{}\frac{1}{1\plus{}b}\plus{}\frac{1}{1\plus{}c}\ge \frac{3}{1\plus{}\sqrt[3]{abc}}\]

Form our condition abcd=1 we can see that variable d dont have place in inequality and we have to prove it for any a>=0 , b>=0 , c>=0 .
When we will reduce the left part of inequality to a common denominator we will achive :
(3+2a+2b+2c+ab+bc+ac) / (1+a+b+c+ab+bc+ac+abc) >= 3 / 1 + abc^(1/3)
If we will simplify it we will achive :
(abc)^(1/3) * (3+2a+2b+2c+ab+ac+bc) >= a+b+c+2ab+2bc+2ac+3abc
Lets simplify again :
(a+b+c)(2(abc)^(1/3)-1)+(ab+bc+ac)((abc)^(1/3)-2)+3((abc)^(1/3)-abc) >= 0
If we do the Cauchis theorem to (a+b+c) and (ab+bc+ac) we will achive :
a+b+c >= 3*(abc)^(1/3)
ab+bc+ac >= 3*(abc)^(2/3)
Lets put it into our inequality :
6*(abc)^(2/3) - 3*(abc)^(1/3) + 3abc - 6*(abc)^(2/3) + 3*(abc)^(1/3) - 3abc >= 0
0 >= 0
We achived the right inequality so the starting inequality was also right.
So we are done.

p.s n^k means n to deegre k

- consider fuction $ f(x) \equal{} \frac {1}{1 \plus{} e^x}$
- $ f''(x) \equal{} \frac {e^x(e^x \minus{} 1)}{(1 \plus{} e^x)^3}$ and $ f''(x)\geq 0$ when $ e^x\geq 1$ or in other words when $ x\geq 0$. function $ f$ is therefore convex for positive real numbers.
- we can conclude from the formulation of the problem that $ abc\geq 1$ and $ a\geq 1$
we have to deal with $ 3$ cases since we don't know the $ max(1,b)$ and $ min(1,c)$ and we need to consider all possibilities.

[b]*Case1. $ a\geq b\geq c\geq 1$[/b]
- denote $ a \equal{} e^y$,, $ b \equal{} e^z$,, $ c \equal{} e^w$. where $ y,z,w$ are some positive reals due to the definition of $ a,b,c$ in this case. we have:
$ \frac {1}{1 \plus{} a} \plus{} \frac {1}{1 \plus{} b} \plus{} \frac {1}{1 \plus{} c} \equal{} f(y) \plus{} f(z) \plus{} f(w)$
- and by Jensen Inequality for convex functions: $ f(y) \plus{} f(z) \plus{} f(w)\geq 3f(\frac {y \plus{} z \plus{} w}{3})$
- now we also have:
$ 3f(\frac {y \plus{} z \plus{} w}{3}) \equal{} \frac {3}{1 \plus{} e^{\frac {y \plus{} z \plus{} w}{3}}} \equal{} \frac {3}{1 \plus{} \sqrt [3]{e^y\cdot e^z\cdot e^w}} \equal{} \frac {3}{1 \plus{} \sqrt [3]{abc}}$
- therefore the inequalitiy holds

[b]*Case2. $ a\geq b\geq 1\geq c$[/b]
- denote $ c \equal{} \frac {1}{p}$. therefore $ p\geq 1$ and $ ab\geq p$
- the original inequality is therefore equivalent to:
$ \frac {1}{1 \plus{} a} \plus{} \frac {1}{1 \plus{} b} \plus{} \frac {p}{1 \plus{} p}\geq \frac {3}{1 \plus{} \sqrt [3]{\frac {ab}{p}}}$
and further equivalent to (since $ 1 \equal{} \frac {1}{x \plus{} 1} \plus{} \frac {x}{x \plus{} 1}$):
$ \frac {1}{1 \plus{} a} \plus{} \frac {1}{1 \plus{} b} \plus{} 1 \geq \frac {3}{1 \plus{} \sqrt [3]{\frac {ab}{p}}} \plus{} \frac {1}{1 \plus{} p}$
- now denote $ a \equal{} e^y$,, $ b \equal{} e^z$,, $ p \equal{} e^w$. where $ y,z,w$ are again positive reals and $ y\geq z$ and $ y \plus{} z\geq w$ due to the choice of $ a,b,c$ and the definition of $ p$ in this case. note also that $ 1 \equal{} \frac {1}{1 \plus{} e^0} \plus{} \frac {1}{1 \plus{} e^0}$
- the inequalitiy is therefore also equivalent to:
$ f(y) \plus{} f(z) \plus{} 2f(0)\geq 3f(\frac {y \plus{} z \minus{} w}{3}) \plus{} f(w)$ where $ y,z,w,y \plus{} z \minus{} w\geq 0$. this last inquality holds due to the Karamata Inequality for convex functions since:
$ y\geq \frac {y \plus{} z \minus{} w}{3}$
$ y \plus{} z\geq \frac {2(y \plus{} z \minus{} w)}{3}$
$ y \plus{} z \plus{} 0\geq \frac {3(y \plus{} z \minus{} w)}{3}$
and $ y \plus{} z \plus{} 0 \plus{} 0 \equal{} \frac {3(y \plus{} z \minus{} w)}{3} \plus{} w$

[b]*Case3. $ a\geq 1\geq b\geq c$[/b]
- denote $ c \equal{} \frac {1}{p}$ and $ b \equal{} \frac {1}{q}$ where $ p,q\geq 1$ and $ a\geq pq$
- the original inequality is therefore equivalent to:
$ \frac {1}{1 \plus{} a} \plus{} \frac {q}{1 \plus{} q} \plus{} \frac {p}{1 \plus{} p} \geq \frac {3}{1 \plus{} \sqrt [3]{\frac {a}{pq}}}$
and further equivalent to (since $ 1 \equal{} \frac {1}{x \plus{} 1} \plus{} \frac {x}{x \plus{} 1}$):
$ \frac {1}{1 \plus{} a} \plus{} 2 \geq \frac {3}{1 \plus{} \sqrt [3]{\frac {a}{pq}}} \plus{} \frac {1}{1 \plus{} q} \plus{} \frac {1}{1 \plus{} p}$
- now denote $ a \equal{} e^y$,, $ q \equal{} e^z$,, $ p \equal{} e^w$ where $ y,z,w$ are again positive reals and $ y\geq z \plus{} w$ due to the choice of $ a,b,c$ and the definition of $ p,q$ in this case. note also that $ 2 \equal{} \frac {4}{1 \plus{} e^0}$
- the inequalitiy is therefore also equivalent to:
$ f(y) \plus{} 4f(0)\geq 3f(\frac {y \minus{} z \minus{} w}{3}) \plus{} f(z) \plus{} f(w)$ where $ y,z,w,y \minus{} z \minus{} w\geq 0$. this last inquality holds due to the Karamata Inequality for convex functions since:
$ y\geq \frac {y \minus{} z \minus{} w}{3}$
$ y \plus{} 0\geq \frac {2(y \minus{} z \minus{} w)}{3}$
$ y \plus{} 0 \plus{} 0\geq \frac {3(y \minus{} z \minus{} w)}{3}$
$ y \plus{} 0 \plus{} 0 \plus{} 0\geq \frac {3(y \minus{} z \minus{} w)}{3} \plus{} z$
and $ y \plus{} 0 \plus{} 0 \plus{} 0 \plus{} 0 \equal{} \frac {3(y \minus{} z \minus{} w)}{3} \plus{} z \plus{} w$

- all three possible cases are therefore prooven. therefore
$ \frac {1}{1 \plus{} a} \plus{} \frac {1}{1 \plus{} b} \plus{} \frac {1}{1 \plus{} c}\geq \frac {3}{1 \plus{} \sqrt [3]{abc}}$ for all $ a\geq b\geq c$ such that $ abc\geq1$,, ($ abcd \equal{} 1$) Q.E.D.

We attack this problem with convexity and symmetry.

Let $ f$ be the real function $ v \mapsto \frac {1}{1 \plus{} \exp(v)}$.  Let us denote $ x \equal{} \log a$, $ y \equal{} \log b$, $ z \equal{} \log c$, and $ w \equal{} \log d$.  We know that $ x \plus{} y \plus{} z \plus{} w \equal{} 0$ and that $ x \ge y \ge z \ge w$; we wish to prove that
\[ \frac {f(x) \plus{} f(y) \plus{} f(z)}{3} \ge f \left( \frac {x \plus{} y \plus{} z}{3} \right) .
\]
[b]Lemma 1 (symmetry).[/b] For any real $ r$, $ f(r) \plus{} f( \minus{} r) \equal{} 1$.

[i]Proof.[/i]  We note that
\[ f(r) \plus{} f( \minus{} r) \equal{} \frac {1}{1 \plus{} e^r} \plus{} \frac {1}{1 \plus{} e^{ \minus{} r}} \equal{} \frac {e^{ \minus{} r/2}}{e^{ \minus{} r/2} \plus{} e^{r/2}} \plus{} \frac {e^{r/2}}{e^{r/2} \plus{} e^{ \minus{} r/2}} \equal{} 1. \qquad \blacksquare
\]
[b]Lemma 2 (convexity).[/b] The function $ f$ is convex on the interval $ [0,\infty)$ and concave on the interval $ ( \minus{} \infty,0]$.

[i]Proof.[/i]  The derivative of $ f$ is
\[ f'(x) \equal{} \minus{} \frac {e^x}{(1 \plus{} e^x)^2} \equal{} \frac {1}{(1 \plus{} e^x)^2} \minus{} \frac {1}{1 \plus{} e^x},
\]
so the second derivative of $ f$ is
\[ f''(x) \equal{} e^x \left( \frac { \minus{} 2}{(1 \plus{} e^{x})^3} \plus{} \frac {1}{(1 \plus{} e^{x})^2} \right) \equal{} e^x \cdot \frac {e^x \minus{} 1}{(1 \plus{} e^x)^3} ,
\]
which is evidently greater than or equal to zero on $ [0,\infty)$ and less than or equal to zero on $ ( \minus{} \infty,0]$.  $ \blacksquare$

[b]Lemma 3 (a bound).[/b] If $ r$ and $ s$ are reals such that $ r \ge \lvert s \rvert$, and $ \lambda_1, \lambda_2$ are positive reals such that $ \lambda_1 \plus{} \lambda_2 \equal{} 1$ and $ \lambda_1 r \plus{} \lambda_2 s \ge 0$, then
\[ \lambda_1 f(r) \plus{} \lambda_2 f(s) \ge f(\lambda_1 r \plus{} \lambda_2 s) .
\]
[i]Proof.[/i]  If $ s \ge 0$, then by Lemma 2, we simply apply Jensen's Inequality and we are done.  Hence we assume that $ s < 0$.

Let $ (t,\rho(t))$ be the parametrization of the line in $ \mathbb{R}^2$ passing the points $ (r,f(r))$ and $ ( \minus{} r,f( \minus{} r))$, and let $ (t,\sigma(t))$ be the parametrization of the line passing through the points $ (r,f(r))$ and $ (s,f(s))$.

By Lemma 1, the line $ (t,\rho(t))$ contains the point $ (0,1/2)$.  Since $ s \in [ \minus{} r,0]$, it follows from the concavity of $ f$ on $ ( \minus{} \infty, 0]$ that
\[ \sigma(s) \equal{} f(s) \ge \rho(s) .
\]
Since
\[ \sigma(r) \equal{} f(r) \equal{} \rho(r),
\]
it follows that for any real $ t \in [s,r]$, $ \sigma(t) \ge \rho(t)$.  In particular,
\[ \lambda_1 f(r) \plus{} \lambda_2 f(s) \equal{} \sigma (\lambda_1 r \plus{} \lambda_2 s) \ge \rho (\lambda_1 r \plus{} \lambda_2 s) . \qquad (*)
\]
And since $ \lambda_1 r \plus{} \lambda_2 s \in [0,r]$, $ f$ is convex on that interval, and $ \rho$ is the linear function connecting the endpoints,
\[ \rho(\lambda_1 r \plus{} \lambda_2 s) \ge f(\lambda_1 r \plus{} \lambda_2 s) . \qquad (**)
\]
We prove the lemma by combining $ (*)$ and $ (**)$.  $ \blacksquare$

The problem will now follow from repeated application of Lemma 3.

Now, $ x \plus{} y \ge \frac {x \plus{} y \plus{} z \plus{} w}{2} \equal{} 0$; since $ x\ge y$ and $ x \plus{} y \ge 0$, it follows that $ x \ge \lvert y \rvert$.  Hence $ (r,s,\lambda_1,\lambda_2) \equal{} (x,y,1/2,1/2)$ satisfy the hypotheses of Lemma 3, so
\[ f(x) \plus{} f(y) \ge 2 f \left( \frac {x \plus{} y}{2} \right) .
\]
Also, $ \frac {x \plus{} y \plus{} z}{3} \ge \frac {x \plus{} y \plus{} z \plus{} w}{4} \equal{} 0$.  If $ \frac {x \plus{} y}{2} \plus{} z \le 0$, then $ \frac {x \plus{} y}{2} \plus{} w \ge 0$, so $ w \ge z$,  a contradiction.  Hence $ \frac {x \plus{} y}{2} \plus{} z \ge 0$.  Therefore
\[ (r,s,\lambda_1, \lambda_2) \equal{} \left( \frac {x \plus{} y}{2} , z, 2/3, 1/3 \right)
\]
satisfies the hypotheses of Lemma 3, so
\[ \frac {2}{3} f \left( \frac {x \plus{} y}{2} \right) \plus{} \frac {1}{3} f(z) \ge f \left( \frac {x \plus{} y \plus{} z}{3} \right).
\]
In summary,
\[ \frac {f(x) \plus{} f(y) \plus{} f(z)}{3} \ge \frac {2 f\bigl( (x \plus{} y)/2 \bigr) \plus{} f(z)}{3} \ge f \left( \frac {x \plus{} y \plus{} z}{3} \right),
\]
which is what we wanted to prove.  $ \blacksquare$

[b]Lemma: If $ n,m,p \in R^ \plus{}$, $ p\leq mn$.$ \equal{} > \frac {1}{p \plus{} m^2} \plus{} \frac {1}{p \plus{} n^2}\geq \frac {2}{p \plus{} mn}$[/b]
Dem:
$ \frac {1}{p \plus{} m^2} \plus{} \frac {1}{p \plus{} n^2}\geq \frac {2}{p \plus{} mn}$
$ < \equal{} > (m^2 \plus{} n^2 \plus{} 2p)(mn \plus{} p)\geq 2(m^2n^2 \plus{} n^2p \plus{} m^2p \plus{} p^2)$
$ < \equal{} > (m \minus{} n)^2(nm \minus{} p)\geq 0$, $ Q.E.D.$
------------------------------------
In the problem.$ a\geq b \geq c \geq d > 0$, and $ abcd \equal{} 1$...(1)
if: $ a \equal{} \frac {x^4}{w}$, $ b \equal{} \frac {y^4}{w}$, $ c \equal{} \frac {z^4}{w}$ $ \equal{} > d \equal{} w^3$, where $ x,y,z,w \in R^ \plus{}$
In (1): $ xyz \equal{} 1$, $ x\geq y \geq z \geq w$..(2),
From (1) and (2), $ 1 \equal{} abcd \geq d^4$ $ \equal{} > d \leq 1$, $ 1 \equal{} xyz\geq z^3$ $ \equal{} > z\leq 1$, but $ d \equal{} w^3$=> $ w\leq z \leq 1.$ ...(3).
Replacing in the inequality.
$ \frac 1{1 \plus{} a} \plus{} \frac 1{1 \plus{} b} \plus{} \frac 1{1 \plus{} c} \geq \frac {3}{1 \plus{} \sqrt [3]{abc}}.$
$ \minus{} >$ $ \frac 1{w \plus{} x^4} \plus{} \frac 1{w \plus{} y^4} \plus{} \frac 1{w \plus{} z^4} \geq \frac {3}{1 \plus{} w}$.
now,how $ z\leq 1$ $ \equal{} > xyz \leq xy$ $ \equal{} > xy\geq 1$, $ \equal{} > x^2y^2\geq xy \geq 1 \geq w$
$ \equal{} > x^2y^2\geq w$, Applying Lemma to $ x^2,y^2,w$.
$ \equal{} > \frac 1{w \plus{} x^4} \plus{} \frac 1{w \plus{} y^4}\geq \frac {2}{x^2y^2 \plus{} w}$, but $ xyz \equal{} 1$
$ \equal{} > \frac 1{w \plus{} x^4} \plus{} \frac 1{w \plus{} y^4}\geq \frac {2z^2}{1 \plus{} z^2w}$
$ \frac 1{w \plus{} x^4} \plus{} \frac 1{w \plus{} y^4} \plus{} \frac 1{w \plus{} z^4} \geq \frac {2z^2}{1 \plus{} z^2w} \plus{} \frac 1{w \plus{} z^4}$.
Now.
$ \frac {2z^2}{1 \plus{} z^2w} \plus{} \frac 1{w \plus{} z^4} \geq \frac {3}{1 \plus{} w}$...(4)
$ < \equal{} > (2z^6 \plus{} 3z^2w \plus{} 1)(w \plus{} 1) \geq 3(z^4 \plus{} w \plus{} w^6 \plus{} z^2w^2)$
$ < \equal{} > (z \minus{} 1)^2(2z^4 \plus{} 4z^3 \plus{} 3z^2 \plus{} 2z \plus{} 1 \minus{} w(z^4 \plus{} 2z^3 \plus{} 3z^2 \plus{} 4z \plus{} 2))\geq 0$
$ < \equal{} > (2z^4 \plus{} 4z^3 \plus{} 3z^2 \plus{} 2z \plus{} 1 \minus{} w(z^4 \plus{} 2z^3 \plus{} 3z^2 \plus{} 4z \plus{} 2)) \geq 0$...(*)
$ M \equal{} z^5 \minus{} z^3 \plus{} 2z^2 \minus{} 1 \equal{} (z \minus{} 1)(z^4 \plus{} z^3 \plus{} z \plus{} 1)\leq 0$
$ S \equal{} (2z^4 \plus{} 4z^3 \plus{} 3z^2 \plus{} 2z \plus{} 1 \minus{} w(z^4 \plus{} 2z^3 \plus{} 3z^2 \plus{} 4z \plus{} 2))$
$ S \plus{} M \equal{} (z \minus{} w)(z^4 \plus{} 2z^3 \plus{} 3z^2 \plus{} 4z \plus{} 2)\geq 0$ $ (z\geq w$
$ \equal{} > S \plus{} M\geq 0 \geq M$ $ \equal{} > S\geq 0$..=> (*) is true.=> (4) is true...
$ \equal{} > \frac 1{w \plus{} x^4} \plus{} \frac 1{w \plus{} y^4} \plus{} \frac 1{w \plus{} z^4} \geq \frac {2z^2}{1 \plus{} z^2w} \plus{} \frac 1{w \plus{} z^4} \geq \frac {3}{1 \plus{} w}$
 $ \frac 1{w \plus{} x^4} \plus{} \frac 1{w \plus{} y^4} \plus{} \frac 1{w \plus{} z^4} \geq \frac {3}{1 \plus{} w}$, $ Q.E.D.$

I misunderstand something  :!:  There are 4 variables but inequality only have 3 variables???

We put $ a \equal{} x^6$ , $ b \equal{} y^6$ , $ c \equal{} z^6$ and $ d \equal{} w^6$
From the condition $ x\geq y\geq z\geq w > 0$ and $ xyzw \equal{} 1$ and we easily obtain that
$ xy\geq 1$ , $ xyz\geq 1$ and $ xyz^2\geq 1$ (1)

(just replace at $ xyzw \equal{} 1$ the $ zw$ with $ xy$ , the $ w^3$ with $ xyz$ and the $ w$ with $ z$ )

We have that $ \frac {1}{1 \plus{} x^6} \plus{} \frac {1}{1 \plus{} y^6}\geq\frac {2}{1 \plus{} x^3y^3}$ which after the expanding is equivalent to

$ (x \minus{} y)^2(xy \minus{} 1)(x^2 \plus{} xy \plus{} y^2)^2(1 \plus{} xy \plus{} x^2y^2)\geq 0$

which is clearly true since $ xy\geq 1$

So we have to prove that

$ \frac {2}{1 \plus{} x^3y^3} \plus{} \frac {1}{1 \plus{} z^6}\geq\frac {3}{1 \plus{} x^2y^2z^2}$
which after expanding is equivalent to

$ (xy \minus{} z^2)^2(2x^2y^2z^4 \plus{} x^3y^3z^2 \minus{} 2xy \minus{} z^2)\geq 0$ so we have only to prove that

$ 2x^2y^2z^4 \plus{} x^3y^3z^2 \minus{} 2xy \minus{} z^2\geq 0$ .

We have from (1) that $ x^3y^3z^2\geq x^2y^2$ and $ x^2y^2z^4\geq z^2$ and also $ x^2y^2z^4\geq 1$ .

Adding these three we have that
$ 2x^2y^2z^4 \plus{} x^3y^3z^2\geq x^2y^2 \plus{} 1 \plus{} z^2$

so we have to prove that $ x^2y^2 \plus{} 1 \plus{} z^2 \minus{} 2xy \minus{} z^2\geq 0$ which is equivalent to

$ (xy \minus{} 1)^2\geq 0$ which is clearly true and we are done

We must prove that $ \frac 1{1\plus{}a} \plus{} \frac 1{1\plus{}b} \plus{} \frac 1{1\plus{}c} \geq \frac {3}{1\plus{}\sqrt[3]{abc}}$. Denote $ a\equal{}e^x,b\equal{}e^y,c\equal{}e^z$. Consider the fuction:
$ f(t)\equal{}\frac 1{1\plus{}e^t}$, we have
$ f"(t)\equal{}\frac{2e^t}{(1\plus{}e^t)^2}>0$.
By Jensen inequality, we have $ f(x) \plus{} f(y) \plus{} f(z) \geq 3f(\frac{x\plus{}y\plus{}z}{3})$. Hence, the solution is complete.

.

Since the harmonic mean is less than or equal to the geometric mean, let's switch the geometric mean in the denominator to a harmonic mean to make the right side larger. We find that the inequality still holds as follows, proving the desired result. Let $ P \equal{} abc$, $ S \equal{} a \plus{} b \plus{} c$, and $ S_2 \equal{} ab \plus{} bc \plus{} ac$.
\[ \frac1{1 \plus{} a} \plus{} \frac1{1 \plus{} b} \plus{} \frac1{1 \plus{} c} \geq \frac3{1 \plus{} \frac3{\frac1a \plus{} \frac1b \plus{} \frac1c}}
\]
After some algebra (I'm short on time), this leads to
\[ 9P \plus{} 6SP \plus{} 2PS_2 \geq 0
\]
, which is clearly true.

[quote=Valentin Vornicu]If $ a\geq  b\geq c\geq d > 0$ such that $ abcd\equal{}1$, then prove that  \[ \frac 1{1\plus{}a} \plus{} \frac 1{1\plus{}b} \plus{} \frac 1{1\plus{}c} \geq \frac {3}{1\plus{}\sqrt[3]{abc}}.\][/quote]
[url]https://artofproblemsolving.com/community/c6h433168p2447520[/url]
Let $a,b,c>0, \ \ abc\geq 1$. Prove that
\[ \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq\frac{3}{\sqrt[3]{abc}+1}\]


geometry
incenter
circumcircle
angle bisector
geometry proposed

Let ABC be a rectangular triangle with <C = 90°. Let R and S be the points where the incircle touch AC and BC respectively. Let K = AI^RS where I is the incenter of ABC. Let l be the parallel to BI through K and let J = l^IR. If D is the foot on AC of the angle bisector BI, prove that JD bisects AB.

[quote="pestich"]Lines parallel to angle bisectors go thru a fixed point on circumcircle right above the
   midpoint of AB.

Maj. Pestich[/quote]

yes! that's a rigth observation!
and it is more generally true, for a generic traingle ABC not only for rectangular triangle.


Ma perchè il sito di oliforum non si apre???? :?:  :maybe:

Uppo il topic xchè oggi ho scoperto che il sito e il forum nn si aprono... è così x tutti? Soprattutto, perchè?

:maybe: Stesso problema di sopra ... come mai? Sono l'unico?

P.S.: salve a tutti, comunque. :)

No beh succede a tutti... ormai è tradizione che verso natale non funziona mai.
Tradizione dovuta a cosa, non lo so! Sarà che vuole andare in vacanza anche lui,  o meglio, che vuole farci fare vacanze più "vacanze" a noi  :)

non si apre piùùùùùùùùùùùùùùùùùùùùùùùùù :mad:  :mad:


when will mr andreescu send out accpetance/rejection letters?

On this page (http://www.awesomemath.org/application.shtml) in the table, it says "Applicant informed of Admission Decision:  [b]April 14th, 2008[/b]."


A rectangle has a diagonal length of $ 2\sqrt{10}$ centimeters and an area of 12 square centimeters. What is the number of centimeters in the perimeter of the rectangle?

By trial and error, the side lengths are $ 6$ and $ 2$. Therefore, the area is $ 2(6\plus{}2)\equal{}\boxed{18}$

We are given that

$ x^2\plus{}y^2\equal{}40$

$ xy\equal{}12$

For some side lengths x,y.

Keep your eye on the ball! We only need $ 2(x\plus{}y)$.

Multiplying the second equation by 2 and adding gives

$ x^2\plus{}2xy\plus{}y^2\equal{}64$.

Thus,

$ (x\plus{}y)^2\equal{}64$
$ x\plus{}y\equal{}8$, since clearly sides can't be negative.

Thus, $ 2(x\plus{}y)\equal{}16$.

We didn't need to find x,y, just x+y!

[Regarding the previous post:
1. You meant perimeter, not area
2. $ 2(6\plus{}2)\equal{}18$?]


In how many ways $ n$ Queens can be placed on on a $ n*n$ chessboard so that none of them attack each other.
here is the value for some $ n$:
$ g(1) \equal{} 1$
$ g(2) \equal{} 0$
$ g(3) \equal{} 0$
$ g(4) \equal{} 2$
$ g(5) \equal{} 10$
$ g(6) \equal{} 4$
$ g(7) \equal{} 40$
$ g(8) \equal{} 92$
$ g(9) \equal{} 352$
$ g(10) \equal{} 724$
$ g(11) \equal{} 2680$
$ g(12) \equal{} 14200$
$ g(13) \equal{} 73712$

Here is a program that can compute $ g(n)$ (Source & EXE) :
Attention:Program Updated.Please Try New Version. :lol:

Then you should report [url=http://queens.inf.tu-dresden.de/]here[/url], because they have enormous problems with computing even $ Q(27)$.


linear algebra
matrix
function
calculus
inequalities
Functional Analysis
superior algebra

Hello,

This problem is something that I just thought some minutes ago.

Suposse that $n>m$ and that we have a matrix $A\in\mathcal{M}_{n\times m}$, such that $\mbox{Rank}(A)=m$ (this is equivalent to say that $A^TA$ is invertible, or $A$ as a linear function is injective or one-to-one).The problem is to determine if there exist a collection of integers $1\leq k_1 < k_2 < \cdots < k_m \leq n$, fixed, and a matrix $B\in\mathcal{M}_{m\times m}$, such that given any $x\in\mathbb{R}^m$, $Ax=y$, and $y=(y_1,y_2,\dots,y_n)^T$, then $B\tilde{y}=x$, where $\tilde{y}=(y_{k_1},y_{k_2},\dots,y_{k_m})^T$.

I hope you'll enjoy it. I will think about it later.

Best,

Hello Kent,

[quote="Kent Merryfield"]The first thing I'll do with this is to move it from "Calculus" to "Algebra".[/quote]

I disagree with you, because even when Linera Algebra has the word Algebra in its name, I am not totally convinsed it should be named that way. It, I think, should be called Linear Analisis. I think that the name comes from a more historical than applicational point of view. I think that Linar Algebra was introduced originally to solve explicitely systems of equations. The truth is that I have seen more applications and uses of "Linear Algebra" in an analytical fashion than in an algebraic fashion. For example, what is another name for Linear Algebra, specially when you are talking about infinite dimensional spaces: [b]Functional [u]Analysis[/u][/b].

Any way, if the moderator wants to move it to another forum in algebra, I don't really care, at least not too much!

Best,

The point is to collect similar problems together. If you examine the problems that do get posted in "Superior Algebra", you'll see plenty of matrix problems. We may have momentary trends, like a succession of group or ring problems, but it does seem that this is the home of matrix problems.

Your particular problem does not involve limits, topological concepts, or inequalities, and although you specify real entries for the matrix, the problem should be the same over any field. It does involve the concepts of rank and linear independence - in fact, simply finding $m$ independent rows of $A$ should be most of the answer.

I would suggest that a problem, such as a functional analysis problem, that truly needs both analytic and algebraic concepts would most comfortably fit into "advanced fields."

[quote="Kent Merryfield"]... in fact, simply finding $m$ independent rows of $A$ should be most of the answer...[/quote]

Yes, you are right, it is that easy. Just pick $v_{k_1},v_{k_2},\dots,v_{k_m}$ independent rows of $A$. Then, define the matrix $B=[v_{k_1},\dots,v_{k_m}]$. $B$ is invertible and if $Ax=(y_1,\dots,y_n)^T$, then $Bx=(y_{k_1},\dots,y_{k_m})^T$.

Thank you. Best regards,


Given an odd integer $n>3$, let $k$ and $t$ be the smallest positive integers such that both $kn+1$ and $tn$ are squares. Prove that $n$ is prime if and only if both $k$ and $t$ are greater than $\frac{n}{4}$

If $ n$ is prime, we need $ n|t$ as $ tn$ has to contain the prime factor $ n$ at least twice. But this gives $ t\geq n > \frac {n}{4}$. Additionally, $ x^{2}\equiv 1\mod p$ has only the solutions $ \pm 1\mod p$, thus for $ kn \plus{} 1$ to be a square $ a^{2}$, we need $ a\equiv\pm 1\mod p$.  From $ a > 1$ we conclude $ a\geq n \minus{} 1$, yielding $ kn \plus{} 1\geq (n \minus{} 1)^{2}\iff k\geq n \minus{} 2$. But from $ n > 3$ we get $ n \minus{} 2 > \frac {n}{4}$, thus $ k > \frac {n}{4}$.

If $ n$ is composite, we consider two cases:
[b]Case 1:[/b] $ n$ has at least two different prime divisors $ p\neq q$, let $ p^{a}$ be the highest power of $ p$ dividing $ n \equal{} p^{a}m$.
We have $ m > 2$ (remember that $ m$ is odd) and $ \gcd(p^{a},m) \equal{} 1$, thus by the Chinese Remainder Theorem, we find an $ y$ such that $ y\equiv 1\mod p^{a}$ and $ y\equiv \minus{} 1\mod m$. We can assume that $ |y| < \frac {n}{2}$, too, giving $ y^{2} < \frac {n^{2}}{4}$.
From $ y^{2}\equiv 1\mod p^{a}$ and $ \mod m$ we get $ y^{2}\equiv 1\mod n$, from $ y\not\equiv 1\mod m$ we get $ y\not\equiv 1\mod n$ and $ y\not\equiv \minus{} 1\mod p^{a}$ gives $ y\not\equiv \minus{} 1\mod n$. Thus if we set $ k \equal{} \frac {y^{2} \minus{} 1}{n}$, this is a positive (as $ y\neq\pm 1$) integer and it fulfills $ k \equal{} \frac {y^{2} \minus{} 1}{n} < \frac {y^{2}}{n} < \frac {\frac {n^{2}}{4}}{n} \equal{} \frac {n}{4}$.
[b]Case 2:[/b] $ n$ has only one prime divisor, thus $ n \equal{} p^{a}$ for a prime $ p$.
If $ a$ is even, we can take $ t \equal{} 1 < \frac {n}{4}$ (as $ n\geq 5$ by assumption). If $ a$ is odd, it is $ \geq 3$ and we can take $ t \equal{} p < p\cdot\frac {p^{2}}{4} \equal{} \frac {p^{3}}{4}\leq\frac {n}{4}$.

In both cases, at least one of $ k,t$ can be choosen such that $ kn \plus{} 1$ or $ tn$ is a square, but being lower than $ \frac {n}{4}$, in other words, we are done.


number theory
least common multiple
MATHCOUNTS
AMC
AIME
USA(J)MO
USAMO

Consider the following sum:

$\frac{1}{1*2}+\frac{1}{2*3}+\frac{1}{3*4}+\frac{1}{4*5}+\frac{1}{5*6}$

Express the sum as a simple froaction in lowest terms.

[hide] yea everything would cancel out except the first and last terms, making the answer 1-1/6= [b] 5/6[/b] [/hide]

It is called a sum at Mengoli way (or a telescopic sum)

But usually telescopic sum problems don't just end after 5 terms, they go like 100 terms.

But the sum does telescope :wink:

The advantage of using the telescoping method for this problem is that a) it's much faster and b) you can generalize.

consider this problem
1/10+1/100+1/1000+1/10000
its
how many ones there are in this case there is 4 so put 4 ones
1111
now the last denominator which in this case is 10000
so the answer is 1111/10000

That is because $ 10\plus{}100\plus{}1000\plus{}10000\equal{}11110/100000 \equal{} 1111/10000$
Just use the unit fraction rule

Lol. yeah. That wasn't exactly a telescoping problem.

Better,more MC-y way.

The sum of the first 1 terms is 1/2

2 terms is 2/3

3 terms is 3/4

Clearly because this is mathcounts, the sum of the first n terms is n/n+1

You can't be sure. The pattern might change. I've made that mistake on a MC test before.

To do well at MC, you must balance the need for rigor with the need for speed. I weigh heavily toward the latter.

Agree. Speed in MC is more important than knowing all the concepts. For AIME and AMC and USAMO, on the other hand, the former is more important.

[quote="Math Champion"]Agree. Speed in MC is more important than knowing all the concepts. [/quote]

Bleah :|

yes, i diasgree a lot with this type of problem given in a olympic contest

I think this topic is resolved and Mods can lock... i mean we solved the problem


modular arithmetic
quadratics
algebra
polynomial
calculus
integration
quadratic formula

1. Find all solutions to the congruence $x^3 - 4x^2 + 33 \equiv 0$ (mod 125)

2. Determine the number of solutins to $x^{14} + x^7 + 1 \equiv 0$ (mod 61)

edit: Disregard what I said.

While I don't like your presentation with having $-3$ under a surd at all, it does work if you instead perform the solution by completing the square. Sorry about that.

According to the answers, there are solutions, although I don't know how to find them. Isn't there some theorem that can be used?  :P

For 1, before I solve your problem, I would like to give a general way to solve this type of congruence equation.

Let $f(x)$ be a polynomial in $x$ with integral coefficients and $p$ be a prime. In general, if we want to solve the congruence equation\[ f(x)\equiv 0\pmod{p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}} \],
by the Chinese Remainder Theorem, we can consider\[ f(x)\equiv 0\pmod{p^\alpha}\cdots (1) \]instead.
The last equation can be simplified to the congruence equation\[ f(x)\equiv 0\pmod{p}\cdots (2) \]
Every solution to $(1)$ is also a solution to $(2)$. Suppose\[ x\equiv x_1\pmod{p} \]is a solution to $(2)$, then $x=x_1+pt_1$ where $t_1$ is an integer. We put this into\[ f(x_1+pt_1)\equiv 0\pmod{p^2} \]and expand the left hand side using Taylor's expansion we have\[ f(x_1)+pt_1f'(x_1)\equiv 0\pmod{p} \]If $p\not| f'(x_1)$ then we have\[ \frac{f(x_1)}{p}+t_1f'(x_1)\equiv 0\pmod{p} \]and we can solve for $t_1$ modulo $p$. There is a unique solution $t_1\equiv t_1'\pmod{p}$. In this way we have found a solution to\[ f(x)\equiv 0\pmod{p^2} \] from a solution to $f(x)\equiv 0\pmod{p}$.
Similarly we can solve for higher degree of the prime modulo.

Let's go back to your problem: $x^3-4x^2+33\equiv 0\pmod{125}$.
Let $f(x)=x^3-4x^2+33$ and $f'(x)=3x^2-8x$. First consider the equation\[ f(x)\equiv 0\pmod{5} \]By inspection, $x\equiv 1\pmod{5}$ and $x\equiv 2\pmod{5}$ are solutions. Then we let $x=1+5t_1$ or $x=2+5t_1$. But for the first case, $f'(x)\equiv f'(1)\equiv 0\pmod{5}$ is divisible by $5$ so it is not a solution to the original equation. Now we want to solve\[ f(2)+5t_1f'(2)=25-20t_1\equiv 5t_1\equiv 0\pmod{25} \]This gives $t_1\equiv 0\pmod{5}$. Hence the solution to $x^3-4x^2+33\equiv 0\pmod{25}$ is $x\equiv 2\pmod{25}$.
Now let $x=2+25t_2$. We want to solve\[ f(2)+25t_2f'(2)=25-100t_2\equiv 25+25t_2\equiv 0\pmod{125} \]
This gives the solution $t_2\equiv 4\pmod{5}$. Hence we have $x\equiv 2+25t_2\equiv 102\pmod{125}$.

For 2, $x^{14}+x^7+1\equiv 0\pmod{61}\cdots (1)$.
Put $y=x^7$, we have
$y^2+y+1\equiv 0\pmod{61}\cdots (2)$
$4y^2+4y+1\equiv 58\pmod{61}$
$(2y+1)^2\equiv 58\pmod{61}$
Now put $z=2y+1$. We want to determine the number of solutions to
$z^2\equiv 58\pmod{61}\cdots (3)$.
We use the Legendre symbol to help us find out this.
$(\frac{58}{61})=(\frac{2}{61})(\frac{29}{61})=(-1)^{\frac{61^2-1}{8}}(\frac{61}{29})=-(\frac{29\times 2+3}{29})=-(\frac{3}{29})=-(\frac{29}{3})=-(\frac{2}{3})=-(-1)=1$
so there are two solutions to $(3)$, and hence two solutions to $(2)$.
Let the two solutions be $y\equiv a\pmod{61}$ and $y\equiv b\pmod{61}$. Now we want to find out the number of solutions to
$x^7\equiv a\pmod{61}$ or $x^7\equiv b\pmod{61}$
and I have to think more about it... :D

[quote="chess64"][quote="chess64"][hide="2"]
Put $y=x^7$. Now $y^2+y+1 \equiv 0 \pmod {61}$. By the quadratic formula, we have
\begin{eqnarray*} y &\equiv & \frac{-1 \pm \sqrt{-3}}{2} \pmod {61}\\ y &\equiv & \frac{-1 \pm \sqrt{-3+61 \cdot 19}}{2} \pmod {61}\\ y &\equiv & \frac{-1 \pm 34}{2} \pmod {61}\\ y &\equiv & \frac{-1 \pm 95}{2} \pmod {61}\\ y &\equiv & 47,-48 \pmod {61}\\ x^7 &\equiv & 47,-48 \pmod {61} \end{eqnarray*}

I don't think there are [i]any[/i] solutions...
[/hide][/quote]

Chess64, you're doing it wrong. :D You're supposed to complete the square! :roll:

[hide]
\begin{eqnarray*} y^2+y+1 &\equiv & 0 \pmod {61}\\ y^2+62y &\equiv & -1 \pmod {61}\\ (y+31)^2 &\equiv & 289 \pmod {61}\\ (y+31) &\equiv & \pm 17 \pmod {61}\\ y &\equiv & -14,-48 \pmod {61}\\ x^7 &\equiv & 47,-48 \pmod {61}\\ \end{eqnarray*}
[/hide]

Hmm...that's weird. I got the same answers as chess64 even by completing the square :huh:[/quote]That's just hilarious.  :rotfl:

[hide="2"]The congruence $x^{14}+x^7+1\equiv 0\pmod{61}$ does not have $x\equiv 0$ as a solution, so by Fermat's Little Theorem, $x^{60}\equiv 1$.

Multiplying the congruence by $x^{53}$, $x^7+1+x^{53}\equiv 0$.

Subtract this from the original to get $x^{53}\equiv x^{14}$. Since $x\not \equiv 0$, $x^{14}\not \equiv 0\Rightarrow x^{39}\equiv 1$.

From the original congruence $x^{14}+x^7+1\equiv 0$, multiply by $x^7-1$ to get $x^{21}\equiv 1$.

Then $x^{21}$ is congruent to $x^{39}\Rightarrow x^{18}\equiv 1\Rightarrow x^{21}\equiv x^{18} \Rightarrow x^3\equiv 1$.

Then any solution to $x^{14}+x^7+1\equiv 0$ will satisfy $x^3\equiv 1$. $x\equiv 1$ is not a solution by inspection, so this reduces to $x^2+x+1\equiv 0$.

Using the quadratic formula, $x\equiv \frac{-1\pm \sqrt{-3}}{2}$. A square root of -3 is 95, so $x\equiv \frac{-1\pm 95}{2}\equiv 47,\ -48\equiv 13,\ -14$.

Since $13^2+13+1=(-14)^2+(-14)+1=183=3*61\equiv 0$, both solutions check. Therefore there are $\boxed{2}$ distinct solutions, namely 13 and -14.[/hide]

[quote="4everwise"]That's just hilarious.  :rotfl:[/quote]
Surprisingly, chess64 also got a right answer in another problem using his "hilarious" method.
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=398860#p398860]http://www.mathlinks.ro/Forum/viewtopic.php?p=398860#p398860[/url]
Perhaps chess64's method is not a joke; I think his method agrees with completing the square, but his way of presenting things might be too strange.
Is this way of solving common in America and Canada? (I suspect this after reading scorpius119's post.)

[quote="Elemennop"]Chess64's method is correct; It's just a more direct solution that isn't quite as pretty as completing the square (though I prefer Chess64's method). Both of them are just as legit.[/quote]
Wait, but didn't chess64 get a different answer?
And in his first solution, how did he know to add 61*19?

:D same as in the other thread, guess and check.

Oh, and leepakhin, I think 4everwise was saying that me talking to myself was hilarious, not the method. I learned the method from Mildorf's NT lecture.

1234567890: see the last two lines of scorpius's solution.

the way that he does it works, but he does not explain why...

you cannot have roots, imaginary numbers, etc, but the places that he is placing in the extra numbers in is allowed because that is the result that would occur if he added numbers in the original equation, this is a working backwards method, but if you were to present a solution, you would need to go forwards i.e. make the additions before you use the quadratic formula


geometry
3D geometry
function
trigonometry
conics
parabola
inequalities

This is COP Shorlist-2. Post here if you have answer (even other people posted it) and use this as "Tool" for the COP, not just a post where you answer few questions.

There are 20 questions. Wording has been changed to prevent possible law problem even though it is posed up on the web.

1. In a cube, the space diagonal (longest diagonal) is defined as:

$\sqrt {s^2 + s^2 + s^2}$

Let's say that the space diagonal of cube J is 9. Find the volume of this cube J.

2. Simplify:

$(3 + k^{-1})^{-1}$

3. For all integers $s$ such that it lies in $(3, \infty)$, simplify the expression below so that it is true for all integers $s$.

$\frac {(s-3)!s!}{(s-2)!(s-1)!}$

4. In math, it's usually not right to have square root on denominator on your answer. So, you RATIONALIZE them. Rationalize the denominator and simplify for: $\frac {2 \sqrt 6 + \sqrt 3}{\sqrt 6 - \sqrt 3}$

5. Define $a(b) = b^2 - 2$. Then what is $a(x+2)$?

6. Function $u(x)$ is linear function such that $u(1) = 2$ and $u(3) = 8$. Find the value for $u(5)$.

7. Simplify the following trig equation:

$\sqrt {\sec^2 \alpha + \cos^2 \alpha + \csc^2 \alpha + \sin^2 \alpha - (\tan^2 \alpha + \cot^2 \alpha)}$

8. Let $4^{\frac {m}{n}} = 256$ and $3^{m} = \frac {1}{27}$. Compute the value of $n$.

9. Two distinct circles with radius 1 are tangent to each other. If we're to draw a circle with radius 2 that are tangent to both of the circles, then how many such circles can we draw?

10. $4z3 + 134 = 5q7$ and $5q7$ is divisible by 7. Find $z+q$.

11. Simplify:

$\frac {-4(k+3) + k^2}{(k+3)(k+2)}$

12. Let $COP = 2001$ such that $C,O,P$ are all positive integers. If 5 DIFFERENT digit are used to write $C,O,P$, what is the largest value of three integers?

13. Let $c_1$ be the center of the circle $x^2 = 9 - (y - 3)^2$ and $v_1$ be the vertex of the parabola $y = x^2 - 4x + 3$. Find the distance between $c_1$ and $v_1$.

14. Given that $i = \sqrt {-1}$, what is $z_9$ if $z_1 = i, z_2 = -2, z_3 = -4i$?

15. We call the number is $\text {abundant}$ if it is less than the sum of its positive proper divisors (including 1 but not including itself). For instance, 4 is not abundant number because 1+2 < 4. How many abundant number are less than 20?

16. For how many values of $y$ is the inequality $|y^2 - 8| < 4$ holds true?

17. $\mathbb {WOW} = 32^{- (\sqrt 5)^{-2}}$. Find the value of $\mathbb {WOW}$.

18. Let $p_s$ equals the sum of the coefficents in the expansion of the $(\frac {1}{2} - 4x + 4x^3)^{274} (6x^4 - 6x + 2)^{275}$. Find $p_s$

19. What is the product of the solutions of $x^{\log_{10} x} = \frac {10000}{x^3}$?

20. Exactly two integers between 75 and 85 are divisors of $3^{32} - 1$. What is the product of those integers?

I'm posting the rest now. But since I want to give much time as possible, you may start solving from 1-12.

So...
[hide]
1.
[hide] The space diagonal is $s\sqrt{3}=9\Rightarrow s=3\sqrt{3}$.  $V=81\sqrt{3}$[/hide]

2.
[hide] $\frac{k}{3k+1}$  :? [/hide]

3.
[hide] $\frac{(s-3)!\cdot s!}{(s-2)!(s-1)!}=\frac{(s-3)!\cdot (s-1)!\cdot s}{(s-3)!(s-2)(s-1)!}=\frac{s}{s-2}$[/hide]

4.
[hide] $\frac{2\sqrt{6}+\sqrt{3}}{\sqrt{6}-\sqrt{3}}=5+3\sqrt{2}$.[/hide]

5.
[hide] $a(x+2)=x^2+4x+2$. Is a 2nd grade function. :?[/hide]

6.
[hide] $u(5)=32$[/hide]

8.
[hide] $3^m=3^-3\Rightarrow m=-3. 4^\frac{-3}{n}=4^4\Rightarrow \frac{-3}{n}=4\Rightarrow n=\frac{-3}{4}$ [/hide]

10.
[hide] $5q7$ is divisible by 7 and $q$ is an pozitive integer results $q=6$ $4z3=567-134=433\Rightarrow z=3$. And $z+q=9$ [/hide]

11.
[hide] $\frac{-4(k+3)+k^2}{(k+3)(k+2)}=\frac{(k-2)^2-4^2}{(k+3)(k+2)}=\frac{(k-6)(k+2)}{(k+3)(k+2)}=\frac{(k-6)}{k+3}$
[/hide]
[/hide]

My englis is not so god  :blush:  What is $sec$, $csc$ and $cot$ .. Sorry for this stupid question  :D

Here is the version in Romanian for that one.

I do not know how to say sine and cotangent in Romanian.

$\sqrt {secant^2 x + cosinus^2 x + cosecanta^2 x + sin^2 x - (tangenta^2 x + cotangent^2 x)}$

$\sin x = cosinus (90 - x)$

$cotangent x = \frac {1}{tangenta x}$

Hope this helps!

P.S. Perhaps I can use little bit of help, Valentine? :D

[quote="Slizzel"]I've never heard of $secant, cosecant$ :huh:[/quote]

$\sec x = \frac{1}{\cos x}$, $\csc x = \frac{1}{\sin x}$

that might help if you really do not know what secant or cosecant means. Or you will have to relearn your trigonometry.

The answer must be [hide]$\sqrt{3}$[/hide]  :P

1.[hide]$9=\sqrt{3s^{2}}$
$27=s^{2}$
$s^{3}=81\sqrt{3}$[/hide]
2.[hide]$\frac{1}{3+k^{-1}}=\frac{1}{3+\frac{1}{k}}=\frac{k}{3k+1}$[/hide]
3.[hide]The factorials will cancel out leaving $\frac{s}{s-2}$[/hide]
4.[hide]Multiply the bottom by the conjugate and simplify.  $\frac{(2+\sqrt{6}+\sqrt{3})(\sqrt{6}+\sqrt{3})}{3}=\frac{15+9\sqrt{2}}{3}=5+3\sqrt{2}$[/hide]
5.[hide]$(x+2)^{2}-2=x^{2}+4x+2$[/hide]
6.[hide]Since it's linear, the slope remains constant.  The function increases by six from 1 to 3, so it increases by six from 3 to 5, so the answer is $14$.[/hide]
7.[hide]Ooh, a trig problem.  I need practice with those.
$\sqrt{\sec^{2}x+\cos^{2}x+\csc^{2}x+\sin^{2}x-\tan^{2}x-\cos^{2}x}$
$\sqrt{(\sec^{2}x-\tan^{2}x)+(\csc^{2}x-\cos^{2}x)+\cos^{2}x+\sin^{2}x)}$
$\sqrt{(1)+(1)+(1)}$
$\sqrt{3}$[/hide]
8. [hide]We can see that $m=-3$.  So we can write $4^{\frac{-3}{n}}=256$, so $\frac{-3}{n}=4$ and $n=-\frac{3}{4}$.[/hide]
9.dunno...
10.[hide]We find that 567 evenly divides 7, so $q=6$.  $567-134=433$, so $z=3$ and $q+z=9$.[/hide]
11.[hide]Expanding the numerator we find the expression is equal to $\frac{k^{2}-4k-12}{(k+3)(k+2)}=\frac{(k-6)(k+2)}{(k+3)(k+2)}=\frac{k-6}{k+3}$.[/hide]
12.[hide]Well, I found the prime factorization of 2001 is $3\times29\times23$, but they share the same digits, and they're all prime.  Either I'm doing something wrong or not understanding the question correctly.[/hide]

All 20 questions are finally up!

Sorry for the delays. :blush:

13
[hide]Formula for circle is $(x - h)^2 + (y - k )^2 = r^2$ where $(h, \ k)$ is the center of the circle. The center of this circle is then $(0, \ 3)$ and using $\frac {-b}{2a}$ the vertex of that porabola is $(2, \-1)$. Using the distance formula, their distance is $\sqrt {20} = 2 \sqrt {5}$.[/hide]

16
[hide]Notice that $y^2$ must be between 4 and 12 in order for $|y^2 - 8|$ to be less than 4. $2 < y < 2 \sqrt 3 $.[/hide]

[hide="To tarquin"]

[quote="tarquin"]$\sqrt {20} = 2 \sqrt {10}$.[/quote]

$\sqrt {20} \neq 2 \sqrt {10}$[/hide]

I'm only going to do a couple right now -

1.
[hide]$\sqrt{s^2+s^2+s^2} = 9 \\
3s^2 = 81 \\
s^2 = 27 \\
s = 3\sqrt{3} \\$

and since we need the volume -

$s^3 = 81\sqrt{3}$[/hide]

2.
[hide]${(3+k^{-1})}^{-1} = \frac{1}{3+\frac{1}{k}}$[/hide]

3.
[hide]$\frac{(s-3)!s!}{(s-2)!(s-1)!} = \frac{(s-3)!}{(s-2)!} \cdot \frac{s!}{(s-1)!} \\
= \frac{1}{s-2} \cdot s \\
= \frac{s}{s-2}$[/hide]

5.
[hide]$a(x+2) = (x+2)^2 - 2 \\
= x^2+4x+4-2 = x^2+4x+2$[/hide]

6.
[hide]Since it's linear, $u(3)-u(1) = u(5)-u(3)$.  Since we know what $u(1)$ and what $u(3)$ are, we can just ubsitute them in, and solve for $u(5)$.

$8-2 = u(5)-8$
$u(5) = 14$[/hide]

8.
[hide]$m = -3$, since $3^{-3} = \frac{1}{27}$.
$\frac{m}{n} = 4$, since $4^4 = 256$

$\frac{-3}{n} = 4 \\
-3 = 4n \\
n = \frac{-3}{4}$[/hide]

9.
[hide]3, one where the two circles are internally tangent to the larger circle, and two where they are externally tangent.[/hide]

A few of them for now...

1) [hide]Work backwards. We have $\sqrt{3s^2}=s\sqrt{3}=9$ so $s=\frac{9}{\sqrt{3}}=3\sqrt{3}$. Cubing this to find the volume we get $\boxed{81\sqrt{3}}$.[/hide]

2) [hide]$(3 + k^{-1})^{-1}=\frac{1}{(3+\frac{1}{k})}=\frac{1}{\frac{3k+1}{k}}=\boxed{\frac{k}{3k+1}}$[/hide]

3) [hide]We have $\frac {(s-3)!s!}{(s-2)!(s-1)!}$. Cancelling $(s-3)!$ from $(s-2)!$ leaves us with $s-2$ and cancelling $(s-1)!$ from $s$ leaves us with $s$. Therefore we have $\boxed{\frac{s}{s-2}}$.

By the way, shouldn't the interval be $[3,\infty)$ since $0! = 1$[/hide]

4) [hide]Multiply denominator and numerator by conjugate $\sqrt{6}+\sqrt{3}$, then simplify and we get $\boxed{5+3\sqrt{2}}$.[/hide]

5) [hide]Subsitute $(x+2)$ in for $b$ and simplifying, we have $\boxed{x^2+4x+2}$[/hide]

6) [hide]Since the function is linear and increases at constant rate, the difference between $u(1)$ and $u(3)$ is 6, therefore difference between $u(3)$ and $u(5)$ must also be 6, so therefore $\boxed{u(5)=14}$.[/hide]

7) [hide]Simplifying inside the radical, we first know that $\cos^2 \alpha + \sin^2 \alpha = 1$. We can also rewrite $\sec^2 \alpha = \tan^2 + 1$ and $\csc^2 \alpha = \cot^2 + 1$. Then, we can easily simplify this statement to $\boxed{\sqrt{3}}$[/hide]

8) [hide]We see $m=-3$ and that $\frac{m}{n}=4$. Therefore, $\boxed{n=-\frac{3}{4}}$[/hide]

9) [hide]If the two tangent circles are placed horizontally to each other. There can be one circle above, and one circle below. There can also be two circles that are internally tangent to one circle and externally tangent to the other. Finally, the circle can be internally tangent to both small circles. Therefore, there are a total of $\boxed{5}$ possible circles.[/hide]

10) [hide]We look for 3 digit numbers beginning with 5, ending with 7, and divisible by 7. The only one that fits that description is $567$. Subtracting 134 and we end up with $433$ so $z=3$ and $q=6$ and $\boxed{z+q=9}$[/hide]

11) [hide]Distributing and we get the numerator as a quadratic $k^2-4k-12=(k+2)(k-6)$. The $(k+2)$ cancels and we have $\boxed{\frac{k-6}{k+3}}$[/hide]

12) [hide]$2001=3\cdot 23 \cdot 29$. $C,O,P$ cannot be these three because a digit is used twice. However, if we multiply $3\cdot29$ we can have the numbers $1,23,87$ and our largest value is $\boxed{87}$ (is that what you're looking for?)[/hide]

13) [hide]We can move $(y-3)^2$ over to the left side to get $x^2+(y-3)^2=9$. This is in standard form for circle so $c_1=(0,3)$. The vertex of $y = f(x) = x^2 - 4x + 3$ is $\left(\frac{-b}{2a},f\left(\frac{-b}{2a}\right)\right)$. Therefore the $v_1=(2,-1)$. Using distance formula, we get distance as $\boxed{2\sqrt{5}}$.[/hide]

14) [hide]This is a geometric sequence with common ratio $2i$. We can use the formula, or we can calculate up to $z_9$ to get $\boxed{256i}$.[/hide]

15) [hide]The only two such numbers are $12$ and $18$ with some simple trial and error so there are $\boxed{2}$ such numbers under 20.[/hide]

16) [hide]$y^2-8$ must be between $4$ and $-4$. This means that $4<y^2<12$. This means that $\boxed{y\in(-2\sqrt{3},-2) \cup (2,2\sqrt{3})}$

(I just wrote the range since I wasn't sure what the question was asking.[/hide]

17) [hide]Working backwards, we end up with $\mathbb {WOW} = \boxed{\frac{1}{2}}$[/hide]

18) [hide]Because the variables do not contribute to the answer of this question, we can just use the numbers. They simplify to $\left(\frac{1}{2}\right)^{274}\cdot2^{275}=\boxed{2}$[/hide]

19) [hide]I liked this question. We have $x^{\log_{10} x} = \frac {10000}{x^3}$. We know that for some $n$, $x^n=10000$, therefore $n=\log_{x} 10^4$. Just looking at the exponents, we realize that $\log_{10} x = \log_{x} 10^4 - 3$.

I'm not too good with logs and log rules and such so I did it this way: I assumed that $x=10^y$ for some $y$. Then, we can rewrite as $\log_{10} 10^y = y = \log_{10^y} 10^4 - 3 \Rightarrow y = \frac{4}{y} - 3 \Rightarrow y^2 + 3y - 4 = (y+4)(y-1) = 0$ so $y=1$ or $y=-4$ and $x=10$ or $x=\frac{1}{10^4}$. Multiplying them together gives our final answer as $10^{-3}=\boxed{\frac{1}{1000}}$.[/hide]

20) [hide]We use difference of two squares over and over again until we get two factors of $3^{32}-1$ as $3^4+1$ and $3^4-1$ which are $82$ and $80$. Their product is $\boxed{6560}$[/hide]

18
[hide]All of the coefficients kill off to $\left(\frac 12\right)^{274}$ and $2^{275}$. The answer is then $2$.[/hide]

Alright, I'll give the rest of them a try:
13.[hide]We can see that the coordinates of the center of the circle are $(0,3)$.  We find the x-coordinate of the vertex is $2$.  Substituting this in we get the y-coordinate if $-1$. So the coordinate of the vertex is $(2,-1)$.  Using the distance formula we get $\sqrt{20}=2\sqrt{5}$[/hide]
14.[hide]It's like a geometric sequence with first term $i$ and common ratio $2i$.  So $i\times(2i)^{8}=256i$[/hide]
15.[hide]I just tested out numbers and $12$ was the only one that worked.  So there is only $1$.[/hide]
16.[hide]Uh...isn't this one is there are limitless values.[/hide]
17.[hide]$32^{-(\sqrt{5}^{-2})}=32^{-\frac{1}{5}}=\frac{1}{2}$[/hide]
18.[hide]We just plug in $1$ to find the sum of the coefficients. $\left(\frac{1}{2}-4+4\right)^{274}(6-6+2)^{275}=2^{275}\left(\frac{1}{2}\right)^{274}$.  We can rewrite this as $2\left(\frac{1}{2}\times2\right)^{274}$.  So this simplifies to $2$.[/hide]
19.don't know yet...I found one answer through guess and check, but don't know if there are others.
20.[hide]Using difference of squares repeatedly we can write $3^{32}-1$ as $(3^{16}+1)(3^{8}+1)(3^{4}+1)(3^{4}-1)$.  We see that $3^{4}+1=82$, and $3^{4}-1=80$, so the answer is $82\times80=6560$.[/hide]