calculus
integration
inequalities
real analysis
real analysis unsolved

[color=green] [b] If  $[ \cdot ]$denotes integral part  and  $A_n: = \displaystyle \frac{ 2^{2n} (n!)^2}{ (2n)!}$ investigate if     equation    \[ \; \; \displaystyle \left [\frac{ A_n^{2}}{\pi} \right]= 2006\; , \; \; n\in {\mathbb N}^* ,  \]
has solution(s) .
[/b] [/color]

If $a_n = \int_0^{\pi/2} sin^n x dx$ then $\frac{2n}{2} \leq \frac{A_n^2}{\pi} = \frac{2n+1}{2}\frac{a_{2n+1}}{a_{2n}} \leq \frac{2n+1}{2} < n + 1$
For the first inequality we use: $\frac{n}{n+1} \leq \frac{a_{n+1}}{a_n} = \frac{n}{n+1}\frac{a_{n-1}}{a_n}$
So, $[\frac{A_n^2}{\pi}] = n$, so equation has unique solution n = 2006  :blush:


trigonometry
inequalities proposed
inequalities

Prove that for every $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n}$ in the interval $(0,\pi/2)$
\[\left({1\over \sin \alpha_{1}}+{1\over \sin \alpha_{2}}+\ldots+{1\over \sin \alpha_{n}}\right) \left({1\over \cos \alpha_{1}}+{1\over \cos \alpha_{2}}+\ldots+{1\over \cos \alpha_{n}}\right) \leq\]
\[\leq 2 \left({1\over \sin 2\alpha_{1}}+{1\over \sin 2\alpha_{2}}+\ldots+{1\over \sin 2\alpha_{n}}\right)^{2}.\]

[i]Proposed by A. Khrabrov[/i]

[url=https://artofproblemsolving.com/community/c6h2076541p14910549]See here for an alternative solution by me.[/url]


If $ \log_6 x\equal{}2.5$, the value of $ x$ is:
$ \textbf{(A)}\ 90 \qquad\textbf{(B)}\ 36 \qquad\textbf{(C)}\ 36\sqrt{6} \qquad\textbf{(D)}\ 0.5 \qquad\textbf{(E)}\ \text{none of these}$

Solving the equation,

$ \log_6 x\equal{}2.5$
$ x\equal{}6^{2.5}$ (the inverse to a log is using the log base as a base)
$ x\equal{}6^{\frac{5}{2}}$
$ x\equal{}\sqrt{6^5}$ (the denominator of a rational exponent represents a root)
$ x\equal{}36\sqrt6$

The answer is $ \fbox{C}$.


Hi!

Let $x,y,z$ be real numbers such that $x^{2}+y^{2}+z^{2}=9$. Prove that :
\[{{1+xyz}\over{3}}\geq \min (x,y,z).\]

[quote="yuyang"]Let $x,y,z$ be real numbers such that $x^{2}+y^{2}+z^{2}=9$. Prove that :
\[{{1+xyz}\over{3}}\geq \min \{x,y,z\}. \]
[/quote]
See: http://www.mathlinks.ro/Forum/viewtopic.php?t=141298
Do you have another solution?  :wink:


search
function
combinatorics proposed
combinatorics

A graph with $n$ vertices contains no $p$-clique ($p\geq 2$).

Show that it has at most $(1-\frac{1}{p-1})\frac{n^{2}}{2}$ edges

Am I wrong or this is Turan's theorem?

Pierre.

The bound indeed follows directly from Turan's theorem.

Thanks, but how to prove Turan's?

(I have used the search function, but it seems there are no proofs)

I don't understand the part that proved the edges in $B$ by induction...


Are there any numbers n such that the sum of the digits of 2n (in base 10) equals the sum of the digits of 2n + 1?

Find a base (other than 10) in which such powers of 2 [i]do[/i] exist.
Find a base (other than 10) in which such powers of 2 [i]do not[/i] exist.

[hide]Not in base 10.

Define S(x) = sum of digits of x. In any base k, S(a) = S(b) --> a :equiv: b mod k-1. Brief proof:

Let nk = apkp + ap-1kp-1 + ... + a1k + a0, where ai's are digits of n in base k.

k :equiv: 1 mod k-1

apkp + ap-1kp-1 + ... + a1 + a0 :equiv: ap + ap-1 + ... + a1k + a0 = S(n) mod k-1

So if S(2n) = S(2n+1), then 2n :equiv: 2n+1 mod 9, which doesn't happen. I think the only way it happens is if for some n, 2n :equiv: 0 mod k-1. So if k = 2n + 1, we should be able to find such numbers.[/hide]

Mmmm, yummy.  Your generalization looks good to me.


Show that there are no integers $ a,b,c$ for which $ a^{2}\plus{}b^{2}\minus{}8c\equal{}6$

[hide]$ a^{2}\plus{}b^{2}\minus{}8c \equal{} 6\Rightarrow a^{2}\plus{}b^{2}\equiv 6\pmod{8}$ but $ x^{2}\equiv 0,1,4\pmod{8}$[/hide]


AMC AIME algebra polynomial ratio geometry rectangle

This is where you'll vote for your scores from AMC-10 B this year.

I am quite pleased to have answered 21 problems correctly :D unless I mis-bubbled or the machine misscores. Yay! AIME!

20 right, 2 wrong, 3 omit.

Made stupid mistakes  :(

Ok nevermind, I saw the answers.

And I am so POed at myself right now >_>

I made the stupidest mistake that anyone could ever make... on number 16... I did that problem in 3 seconds and got the right answer, well that's dandy, right? Well, with 2 minutes left in the AMC, I was checking some answers to see if they were right, and when I checked number 16, I thought my first answer was wrong, so I changed it from E to D (only one year).

That said, I got a 117. If I hadn't changed that answer, I would have gotten a 123 and got to take the AIME. Now, I wouldn't be as annoyed right now had I gotten another question wrong... but no. I just had to change my answer... >_<;;

(U*#(*U (UJJMNTO IUO*P(%

Okay, I'm done.

If it is okay with the AMC folks, could someone post the answers to the AMC 10 B test.  We got our test booklets back but not the answers so we cannot score them.
Thanks

Where are the answers?  How come everyone else knows what they got except for me?  :(  :D

OK, I got 131: 21 right, 2 omitted, and 2 wrong.  :(  :(

EDIT: Actually 137: 22 right, 2 omitted, and 1 wrong.

I can't believe I actually took it, but I decided to last moment.

I think 150. We'll see if I bubbled correctly.

Sigh... still doesn't make that miserable AMC 12 score go away. :(

I'm new here. Got 20 right, 1 wrong, 4 omitted. Hooraz.

[quote="biffanddoc"]grr...
i think i got 22 right 1 wrong 2 blank

for that weird polynomial one, 2+2+.5=9/2 not 5/2  :wallbash_red:  :wallbash_red:  :wallbash_red:

For the triangle one, that was a bit weird? I don't see how to get the answer.
For the circle one, I got that the radius of the quarter-circle was $2-\sqrt{2}$.
All of the choices had $\sqrt{3}$ :?[/quote]

That's not really a quarter circle, the arc is the arc of the large circle of radius 2, not it's arc.

Anyways I got 140.5, did 24 and missed 1 due to carlessness. Got the same score on the A, sadly enough.

I answered all questions, but got the last one wrong. So my score would be 144, assuming I did the bubbling correctly.

I loved #23. It was a pretty problem, and yet it was hard. Not many students from my school (Stuy) got it right. I think it is the hardest problem on this year's 10B.

Also, I think #24 should not have been #24. It was just way too easy.

Edit:
OOPS! I posted in the wrong thread.

The way they presented the solution to number 23 on the solutions packet was a little overdone. LOL.

It was just two ratios with two variables, the classic algebra in geometry problem.

*shoots myself for missing #5*. I'll never forget how stupid I was. :(

yay 133.5 (21r 1w 3b) -- i'm probably gonna fail the aime

hmm... i probably should be taking the 12 instead of the 10 (im a sophomore).. heh..

i'm glad they have those MathJam transcripts, they helped alot when trying to figure out how to do #23  :D

[quote="joml88"][quote="white_horse_king88"]I think he means that a you will get a 146.5 or a 20% chance at a 150, and maybe better than 20% depending on if you can eliminate choices. For 150, You get a picture on the website and niceness.[/quote]

Additionally...if you're taking the 10 then you don't have to worry about how much you qualify by but only that you qualify.  After the AMC, all you care about is the floor...right?[/quote]

True

Thankfully I have two more years to try to get a perfect on the AMC 10

I got 117-ish.

Assuming I'm taking 9th grade math and am in 8th grade, do you think i'll make it next year?

[quote="X_wing1redleader"]I got 118 but i dont see how that is possible I answered 20... or do they round your score?[/quote]

From this, it seems like you answered 18 correctly, left 4 blank, and answered three incorrectly.

They don't round your score

I got 130, 20 right, 1 wrong, and 4 omits.

Gave up on #19 cause I used degrees  :blush:

150.

A nice score since it's the first time I'm sitting for the AMC 10.  :)  By the way, it helps if you do problems from other mathematical competitions such as the Australian Mathematics Competition because some of the non-routine problems test the same concepts. But of course, the Australian competition is not so well known to AOPSers and the problems and solutions are not extensively publicized as well. However their books are worth buying.

And to me the score doesn't really matter - it's the participation. I'm always proud to have participated in an international math competition.  :)

138  :D  23 right and 2 wrong
Pretty good score for being an 8th grader.

121.5 19 right, 3 wrong 3 omit!!! WOOT BARLEY made AIME, i'm in 8th grade....... i hope i didn't bubble wrong, i already screwed up in mathcounts

126.5! 19 right, 1 wrong (extremely stupid mistake on #4) and 4 blank. But I guess if I visited AOPS like my friend tells me to, I wouldn't have such a close score. I'm going to prepare for the AIME this time. Wait, so once you're in 11th grade do you have to take AMC12? I'm going to be taking Calculus anyway but I'd like to take the easier one if possible.

Hah 126.5 isn't close, you can miss bubble one and still make it, I GOT A 121.5 if i make the slittest bubbling error i'm screwed

I am in 8th grade, and I got either a 112 or a 118, I don't remember whether I got 30 right.   :( Compared to your scores, it looks like mine isn't so great


a,b,c,d are real numbers such that a 2 +b 2 +c 2 +d 2 =1 and ac+bd=0. Determine the value of a+b+c+d. Determine the value of ab+cd.

Yes, this sounds a bit incorrect :
Let $a=r\cdot \cos (\theta)$ and $b = r \cdot \sin (\theta )$, where $r \in [0,1]$ and $\theta \in [0, 2 \pi[.$

Then, the first equation is equivalent to $c = \sqrt {1-r^2} \cos ( \phi )$ and $d = \sqrt {1-r^2} \sin ( \phi )$. Therefore, the second is equivalent to ${\phi = \theta + \frac {\pi} 2} $, so that I cannot think you could determine the values you are wanting to.
Or did you mean the range?

Pierre.

Sorry! The question will be like that:
a,b,c,d are real numbers such that a 2  + b 2  = c 2 + d 2  = 1 and ac+bd=0. Determine the value of a+b+c+d. Determine the value of ab+cd.

Let $a= \cos (\theta), b= \sin (\theta), c = \cos ( \varphi), d= \sin (\varphi).$
Then, from $ac+bd = 0$ you deduce that $\theta = \phi \pm \frac {\pi} 2.$
It means that either $a=-d,b=c$ or $a=d,b=-c$.
In the first case $a+b+c+d = 2c$ and in the second case $a+b+c+d = 2d.$
But, in both cases $ab+cd = 0.$

Pierre.

If I don't make mistakes (is it english ?! :)), you can't find $a+b+c+d$ be a constant. Take $a=b=c=-d=\frac{1}{\sqrt{2}}$ and $a=d=1, a=c=0$


[quote="theax"]Isn't that question (in my subject line) true? I am not sure if my reasoning is correct, so could anyone help by verifying my results and responding by posting whether or not my reasoning is correct???????

Take for example - 1 inch = 4 wows (I am simply making up a unit to illustrate my point). So 4 inches = 16 wows. Square rooting, we get 2 inches = 4 wows. But it doesn't make sense. And then picking other units and square rooting them will make other different square roots values. Wait... [b]Is it true that you can't square root conversion factors ( as a unit - say inches - squared is inches squared but the square root of inches is...) ?????????? So therefore the numbers obtained from square rooting would be off, too, right?????????[/b]

So is it also true that when in a problem - you are given a length, call it X, and you need to construct the square root of X. If you are not given a unit length, then there are infinitely many solutions (using the reasoning shown above) - as you don't know the unit of X. Right????????????????

Thanks in advance!  :D[/quote]

We can't sqrt different lengths measures (wows and inches aren't the same)

When you square root both sides of the equation,you have to square root the units as well. Therefore, although meaningless, $ 4$ inches=16 $ wows$ would turn into $ 2\sqrt{inches}\equal{}4\sqrt{wows}$. Now, $ \sqrt{inches}$ and $ \sqrt{wows}$ are meaningless, however if we have something like $ 1$ (inch)^2=$ 6.25$ (cm)^2, then we could square root both sides to get the true statement $ 1$ inch=$ 2.5$ cm

[quote="rofler"]When you square root both sides of the equation,you have to square root the units as well. Therefore, although meaningless, $ 4$ inches=16 $ wows$ would turn into $ 2\sqrt {inches} \equal{} 4\sqrt {wows}$. Now, $ \sqrt {inches}$ and $ \sqrt {wows}$ are meaningless, however if we have something like $ 1$ (inch)^2=$ 6.25$ (cm)^2, then we could square root both sides to get the true statement $ 1$ inch=$ 2.5$ cm[/quote]

yes, exactly.  I can't tell you how many times I've seen somebody on the the internet claim to prove that $ 10 {cents} \equal{} 1 {dollar}$ by taking the square root of the equation $ 100 {cents} \equal{} 1 {dollar}$.


modular arithmetic
number theory unsolved
number theory

How to prove: Let $ n \equal{} h\cdot 2^k \plus{} 1, h < 2^k, 2\nmid h$. Then $ n$ is prime if and only if $ a^ {(n \minus{} 1)/2}\equiv \minus{} 1\pmod n$ for some integer $ a$.

If $ n$ is prime, then any nonsquare $ a$ will satisfy $ a^\frac{n\minus{}1}{2}\equiv \minus{}1\pmod{n}$. It remains to show the converse.

If $ a^\frac{n\minus{}1}{2}\equiv \minus{}1\pmod{n}$, then let $ p$ be a prime divisor of $ n$. We must have $ a^\frac{n\minus{}1}{2}\equiv \minus{}1\pmod{p}$ so the order of $ a$ modulo $ p$ must divide $ 2^k$ and be divisible by $ p\minus{}1$. This gives $ p\equiv 1\pmod{2^k}$.

Now $ p$ is of the form $ 1\plus{}a\cdot 2^k$. Also, $ \frac{n}{p}\equiv 1\pmod{2^k}$, so we have $ n\equal{}(1\plus{}a\cdot 2^k)(1\plus{}b\cdot 2^k)$ for some $ b$.

This means $ h\equal{}a\plus{}b\plus{}ab\cdot 2^k$. If $ b\equal{}0$, then $ n\equal{}p$, a prime. Suppose otherwise. Then $ b\geq 1$, and we cannot have $ a\equal{}b\equal{}1$ since $ h$ is odd. This means
\[ a\plus{}b\geq 3,ab\geq 2\Rightarrow h\geq 3\plus{}2^{k\plus{}1}\]
an absurd contradiction for $ h<2^k$. So $ n$ prime.

I think that there is a mistake in the proof:

When you have $a^{\frac{n-1}{2}} \equiv -1\pmod{p}$, you can deduce that
$\text{ord}_p(a)$ divides $n-1 = h2^k$, and
$\text{ord}_p(a)$ divides $p - 1$.

Nevertheless, you can't deduce than $\text{ord}_p(a)$ is divided by $p-1$, so you can't deduce that $p-1$ divides $\text{ord}_p(a)$ and, thus, divides $h2^k$.

I think that to aviod it we can deduce $\text{ord}_p(a) = 2^\alpha$, for any $\alpha \in {0,\ldots, k}$ and then write $p = a\cdot 2^\alpha +1$. If so,

\[N/p \equiv 1/(-1) \equiv -1 \pmod{2^\alpha} \Rightarrow N/p = b\cdot 2^\alpha +1\]

Then,

\[ h\cdot 2^k = N = (N/p)\cdot p = (a\cdot 2^\alpha + 1)\cdot (b\cdot 2^\alpha + 1) \]

I suppose that I should try to use that $h < 2^k$ so $h\cdot 2^k < 2^{2k}$, but I don't know how to continue...

PD. I know that this is an old post, but I'm trying to find a correct proof of Proth's Theorem and I can't find it anywhere!


[hide="1"]
We have the Binomial Theorem \[(x+y)^{n}= \sum_{k=0}^{n} \binom{n}{n-k}x^{n-k}y^{k}\] Substituting $x=1$ and $y=-1$ \[(1-1)^{n}= \binom{n}{n}-\binom{n}{n-1}+\cdots+(-1)^{n+1}\binom{n}{0}\] Using identity $\binom{n}{i}= \binom{n}{n-i}$, we can rewrite the above equation as \[0 = \binom{n}{0}-\binom{n}{1}+\cdots+(-1)^{n+1}\binom{n}{n}\] Since $\binom{n}{0}$ is equal to $1$, moving everything after that term on the RHS to the LHS yields the desired identity.
[/hide]

[hide="2"]By the identity $\binom{n}{r}=\binom{n}{n-r}$, we have

$\sum_{k=1}^{n}\binom{n}{k}\binom{n}{k-1}=\sum_{k=1}^{n}\binom{n}{k}\binom{n}{n-k+1}$.

Now we make a 1-1 correspondence between this last sum and $\binom{2n}{n+1}$:

Suppose we want to form a committee of $n+1$ people from $n$ men and $n$ women.  The number of committees with $k$ men is $\binom{n}{k}\binom{n}{n-k+1}$.  As $k$ ranges from  1 to $n$, we get all possible committees with $n+1$ members.  This is also known as Vandermonde's Identity.[/hide]

[hide="3"]The expression $\binom{n}{k}\binom{n-k}{m-k}$ gives the number of ways to pick $k$ leaders for a committee, and then pick $m-k$ other members for the committee, resulting in a committee with $m$ total members and $k$ leaders.  We could also form this committee by first choosing the $m$ members, and then choosing the $k$ leaders from those $m$ members.  Therefore, $\binom{n}{k}\binom{n-k}{m-k}=\binom{n}{m}\binom{m}{k}$.  Therefore, we have

$\sum_{k=0}^{m}\binom{n}{k}\binom{n-k}{m-k}=\binom{n}{m}\sum_{k=0}^{m}\binom{m}{k}$

But $\sum_{k=0}^{m}\binom{m}{k}=2^{m}$, which can be seen by expanding $(1+1)^{m}$.  Therefore,

$\sum_{k=0}^{m}\binom{n}{k}\binom{n-k}{m-k}=2^{m}\binom{n}{m}$[/hide]

[hide="1"]The sum equals the Binomial Theorem expansion of $(1-1)^{n}$, minus the first term, $(-1){n\choose 0}$.
$(1-1)^{n}-(-1){n\choose 0}=1$.[/hide]
[hide="2"]$\sum_{k=1}^{n}{n\choose k}{n\choose k-1}=\sum_{k=0}^{n-1}{n\choose k+1}{n\choose k}=\sum_{k=0}^{n-1}{n\choose n-1-k}{n\choose k}={2n\choose n-1}$, from Vandermonde's Indentity.[/hide]


Let p>3 be a prime. For each k belonging to {1,2........p-1}, choose unique x_k such that kx_k=1 ( mod p ). Define n_k  for 1 \leq k  \leq p-1 by setting kx_k= 1+pn_k. Prove that sum ( k=1, 2,  ... p-1) kn_k
 = (p-1)/2 ( mod p)

I think that the unicity of x_k supposed that you choose x_k in {1,2,...,p-1}}.

For each k, we have
k*n_k = (k 2 *x_k - k)/p.   (1)

In another hand, modulo p :
(p-k)*x_(p-k) = 1 = -k*x_(p-k) = k*(-x_(p-k)) = k*x_k
thus x_(p-k) = p - x_k.  (2)

From (1) and (2), we deduce that :
(p-k)*x_(p-k) = ( (p-k) 2 *x_(p-k) - (p-k) )/p = -k*n_k + k 2  + 1 mod[p].   (3)

Let p = 2q+1. Since p > 5 we deduce that q =/= 1 mod[3], then q(q+1) = 0 mod[6].

Then, from (3) :
 \sum {1 \leq k \leq p-1} k*n_k = \sum {1 \leq k \leq q} (k*n_k + (p-k)*n_(p-k) )
= \sum {1 \leq k \leq q} ( k 2 + 1) mod[p]
= q(q+1)(2q+1)/6 + q mod[p]
= p*q(q+1)/6 + (p-1)/2 mod[p]
= (p-1)/2 mod[p] since q(q+1)/6 is an integer.

Pierre.

Suppose $ i \in \{1,2,\cdots,p \minus{} 1\}$. Let $ \alpha_i \in \{1,2,\cdots,p \minus{} 1\}$ be such that $ i\cdot\alpha_i \equiv 1 (\text{ mod } p)$ for $ 1 \leq i \leq p \minus{} 1$.
It is clear that $ (p \minus{} i)(p \minus{} \alpha_i)\equiv 1 (\text{ mod } p)$ and $ i\cdot\alpha_i \equal{} p\cdot n_i \plus{} 1$ for $ 1 \leq i \leq p \minus{} 1$.
Therefore,
$ \sum_{i \equal{} 1}^{p \minus{} 1}i\cdot n_i \equal{} \sum_{i \equal{} 1}^{p \minus{} 1}i\cdot \left(\frac {i\cdot\alpha_i \minus{} 1}{p}\right) \equal{} \frac {1}{p}\sum_{i \equal{} 1}^{p \minus{} 1}(i^2\cdot\alpha_i \minus{} i) \equal{} \frac {1}{p}\sum_{i \equal{} 1}^{p \minus{} 1}i^2\cdot\alpha_i \minus{} \frac {1}{p}\sum_{i \equal{} 1}^{p \minus{} 1}i$
$ \equal{} \frac {1}{p}\sum_{i \equal{} 1}^{p \minus{} 1}i^2\cdot\alpha_i \minus{} \frac {p \minus{} 1}{2} \hspace{0.1in}\left(\because \sum_{i \equal{} 1}^{n}i \equal{} \frac {n(n \plus{} 1)}{2}\right)$
$ \equal{} \frac {1}{p}\sum_{i \equal{} 1}^{\frac {p \minus{} 1}{2}}\left(i^2\cdot\alpha_i \plus{} (p \minus{} i)^2\cdot(p \minus{} \alpha_i)\right) \minus{} \frac {p \minus{} 1}{2}$
$ \equal{} \sum_{i \equal{} 1}^{\frac {p \minus{} 1}{2}}\left(p^2 \minus{} (2i \plus{} \alpha_i)p \plus{} (i^2 \plus{} 2i\cdot \alpha_i) \right) \minus{} \frac {p \minus{} 1}{2} \equiv \sum_{i \equal{} 1}^{\frac {p \minus{} 1}{2}}(i^2 \plus{} 2i\cdot \alpha_i) \minus{} \frac {p \minus{} 1}{2} (\text{ mod } p)$
$ \equiv \sum_{i \equal{} 1}^{\frac {p \minus{} 1}{2}}(i^2 \plus{} 2) \minus{} \frac {p \minus{} 1}{2} (\text{ mod } p)\hspace{0.1in} \left(\because i\cdot\alpha_i \equiv 1 (\text{ mod } p)\right)$
$ \equiv \frac {\left(\frac {p \minus{} 1}{2}\right)\left(\frac {p \plus{} 1}{2}\right)p}{6} \plus{} 2\cdot\frac {(p \minus{} 1)}{2} \minus{} \frac {p \minus{} 1}{2} (\text{ mod } p)\hspace{0.1in} \left(\because \sum_{i \equal{} 1}^{n}i^2 \equal{} \frac {n(n \plus{} 1)(2n \plus{} 1)}{6}\right)$
$ \equiv \frac {p \minus{} 1}{2} (\text{ mod } p)\hspace{0.1in} \left(\text{ Note that } \frac {\left(\frac {p \minus{} 1}{2}\right)\left(\frac {p \plus{} 1}{2}\right)p}{6} \text{ is an integer and is } \equiv 0 (\text{ mod } p) \text{ as } p > 3 .\right)$


Hola,

Hace varios meses tuvimos aqui una pequenia diferencia de opiniones acerca de lo que era vocabulario aceptable y lo que no. A raiz de eso se me ha ocurrido: Por que no discutimos acerca de las differencias regionales en nuestro lenguaje, especificamente acerca de palabras que en algunos lugares son groceras o bulgares, y en otros no, o que tienen completamente significados. Se les puede agregar anecdotas, si desean. Asi que aqui empiezo yo

(1) En mi pais tenemos un postre al que llamamos [b]Cajeta[/b], hace como un anio organizamos un almuerzo en el que amigos de un monton de lugares (basicamente de todos los continentes) llevaban algo que compartir y todos comiamos algo que para nosotros lucia exotico. Bueno, para no hacer el cuento largo, cuando una amiga (argentina) nos pregunto el nombre del postre y le digimos... se sonrojo toda. Podran imaginarse que parte de la anatomia femenina es usualmente conocida como cajeta en Argentina.

(2) Hace posiblemente unos quince o dieciseis anios (tenia ocho o nueve anios de edad entonces) estaba en el deposito libre de Colon, en Panama, y resulta que en Costa Rica usamos la palabra [b]Chunche[/b] como sinonimo de cosa. Cuando no te viene a la mente cual es el nombre del artefacto al que te referis, decis chunche. Bueno, en ese momento una de las dependientas de la tienda en la que estaba tenia en su mano un walkman (si, en esos tiempos los diskmans no eran muy comunes que digamos) y le dije que [i]queria ver ese chunche[/i], acto seguido, recibi una bofetada. Resulta que en Panama Chunche tambien se refiere a la misma parte de la anatomia femenina que Cajeta en Argentina.

Alguna anecdota de parte de ustedes?

"Cajeta" me imagino es lo Mismo que "Buseta" que aqui simplemente es lo que en puertorico es una "guagua" chiquita. :D

Buseta, esa no la habia escuchado. Buseta en Costa Rica es un bus pequenio, digamos para unas quince o veinte personas.

[color=blue]Muchas de las palabras que utilizan en España son consideradas vulgaridades en Paraguay.
No hago problema cuando oigo alguna de ellas de parte de alguien que aprendió otro significado.
El único problema conmigo es que soy "purista", y no puedo pronunciar palabras que alguna persona en este mundo considera grosería.
Es por ese motivo que he aprendido a expresarme con un español más internacional (me lo hicieron notar en las olimpiadas…).[/color]


How many points must you get on the AMC12 (or 10 even) for there to be a lower than normal score on the AIME required to qualify for the USAMO.

There are some people who get a high score on the AMC12 but a score below average qualification on the AIME, but still make it to the USAMO.

I have to get a 150 on AMC 12 or 10 and then around a 7 on the AIME to qualify...that's pretty minimum...sometimes a 6 would do it with a perfect AMC.

I would estimate 140-145 on AMC for getting a 7 in AIME, or a perfect score on AMC plus a 6 on AIME, or something close.

[quote="justinian"]wait, i thought they made a floor score from qualifying people who took the amc 12's, and then anyone who scored above that floor score on the AIME that took the amc 10 got into the USAMO[/quote]

Basically. Nothing said contradicts that I dont think...

Anyway, to get as low a score as possible you just want to get 150. Though it is unlikely that you can get a 150 and only a 6 on the AIME, it is still possible. If "average" means the floor, last year it was 7. And the cutoff was 210. So you would need a 150 AMC12 score to get lower than "average."

Floor Score=Lowest AIME score of one of the top 150 people to qualify based on the system (10*AIME+AMC)

[quote="Derek"]The floor would be around 7, wouldn't it?[/quote]

That's what it was last year. But it has been 8 and 6 also.


[quote="jhcreinhardt"]Hint:

[hide]Geometric Probability, graph it on the x-y plane and find areas.[/hide][/quote]  ????

[hide]For the first one I got $\frac{299}{324}$...I have no clue how to do the second one...[/hide]

He's right, geometric probability is the best way to go here.  Usually these problems only involve 2 people, which allows you to graph on the x-y plane the two inequalities necessary.  For instance $x < y + 20$ and $x > y - 20$ would be the ones here, since he must arrive less than 20 minutes after or more than 20 minutes before to be there in time.  Then you divide the area between those 2 lines (in the appropriate constraints), and divide it by the total area of the appropriate constraints, such as $0 < x,y < 60$ (in terms of minutes).  However, there are 3 people here, so you have to find equations in terms of $x,y,z$, and find the VOLUME of the bounded area, and divide that by the volume of the total area.

I got the equations $z<y+15$, $z>y-20$, $x>z-25$, $x<z+20$, $y<x+25$, and $y>x-15$. But then when i bounded the equations...I couldnt find the area  :help:

[quote="mna851"]Hmmm, so what if there are four people do we lose hope of applying a geometric probability argument.[/quote]

i don't think so, it is more difficult, but you have to apply the inclusion-exclusion principle, (does that work instead of doing the 3-d way?)

[quote="aLPEREN"]3 people are to arrive at a certain location at some randomly chosen time between 1:00 PM and 2:00 PM, and each person will wait a certain amount of time before leaving.
Say, for example, with n = 3 people, one can wait for w1  and the other can wait for w2..... (both expressed as fractions of the total time interval).  What is the probability that they will meet? ( w1=w2=w3= 20 minutes)

or   w1= 15minutes  , w2= 20 minutes, w3= 30 minutes  What is the probability that they will meet?[/quote]

Does this mean all three will meet or that at least two of the three will meet?? :?:

I think the 3-d way would work.  The only obstacle is that it can be pretty hard to conceptualize to find the volume of the figure.  Especially in your second case where the ranges are all different.
If there are 4 people... I have no idea.  Maybe, if a generalization could be made, you could kinda pretend you're working in 4d.


What is the greatest common factor of 39 and 91?

39 = 13 *3
91 = 13 *7
thus the greatest factor that they share in common is 13


Gage skated 1 hr 15 min each day for 5 days and 1 hr 30 min each
day for 3 days.  How many minutes would he have to skate the ninth day in
order to average 85 minutes of skating each day for the entire time?

[quote="ernie"]75 mins x 5 days = 375 mins total
   90 mins x 3 days = 270 mins total
                8 days = 635 mins total

aim = 85 mins x 9 days = 765 mins

765-635 = [b]120[/b][/quote]

Acutally, 375+270 = 645. And 765-635 = 130, not 120. Just to avoid any confusion. :maybe:


Let $ u_n$ is a series with $ \begin {cases}a_1 \equal{} 0 \\
\mid a_2 \mid \equal{} \mid a_1 \plus{} 1 \mid \\
\mid a_3 \mid \equal{} \mid a_2 \plus{} 1 \mid \\
... \\
\mid a_n \mid \equal{} \mid a_{n \minus{} 1} \plus{} 1 \mid \end {cases}$

Find maximum of $ \boxed{S \equal{} \frac {a_1 \plus{} a_2 \plus{} a_3 \plus{} ... \plus{} a_n}{n}}$

Obviosly $ a_i\le i\minus{}1$, therefore $ S\le \frac{n\minus{}1}{2}$.
If $ a_i\equal{}i\minus{}1$, $ S\equal{}\frac{n\minus{}1}{2}.$


Let $ a_k$ be the coefficient of $ x^k$ in the expansion of $ (1 \plus{} 2x)^{100}$, where $ 0 \le k \le 100$. Find the number of integers $ r$: $ 0 \le r \le 99$ such that $ a_r < a_{r\plus{}1}$.

[quote="aidan"]$ r \equal{} 0$ works also right? That would give 67 values[/quote]
yes yes  :lol:

Sorry for bringing this up again, but does this solution imply that the coefficients increase monotonically from $ a_0$ to $ a_66$ and then decrease monotonically after that?

How do we know the coefficient sequence behaves as such? (i.e. could there be more than one maxima, etc)

[quote="aidan"]does this solution imply that the coefficients increase monotonically from $ a_0$ to $ a_{66}$ and then decrease monotonically after that?[/quote]  Yes.  (Well, it doesn't rule out possible equalities between terms, but whether those occur is obvious from what is written -- just look for equality conditions.)

[quote="aidan"]How do we know the coefficient sequence behaves as such?[/quote]  Because the given solution demonstrates that it does.  (And it's not too hard to tinker with the given solution to see that it works if we replace "2" and "100" with any positive real and positive integer, respectively.)


Let $a,b,c$ be positive real numbers and $abc=1$. Prove that
\[\frac{\sqrt{a^{2}+1}-\sqrt{2}a}{b+c}+\frac{\sqrt{b^{2}+1}-\sqrt{2}b}{c+a}+\frac{\sqrt{c^{2}+1}-\sqrt{2}c}{a+b}\le 0. \]
This was posted by Gackiem: http://www.mathlinks.ro/viewtopic.php?t=135860


Hint: $ n^4\plus{}4\equal{}(n^2\plus{}2n\plus{}2)(n^2\minus{}2n\plus{}2)$
Now generalize

An old IMO problem:  show that $ n^4 \plus{} 4^n$ is always composite, $ n \ge 2$.

Proof:  It's obvious when $ n$ is even.

[b]Sophie Germaine's Identity:[/b]  $ x^4 \plus{} 4y^4 \equal{} (x^2 \plus{} 2xy \plus{} 2y^2)(x^2 \minus{} 2xy \plus{} 2y^2)$

Now, when $ n \equal{} 2k\plus{}1$ is odd,

$ n^4 \plus{} 4^{2k\plus{}1} \equal{} n^4 \plus{} 4 \cdot \left( 2^k \right)^4$

which is composite by Sophie Germaine.  :)


[quote="alex2008"]If $ a,b,c > 0$ prove that :

$ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\ge \frac {4a}{2a^2 \plus{} b^2 \plus{} c^2} \plus{} \frac {4b}{2b^2 \plus{} c^2 \plus{} a^2} \plus{} \frac {4c}{2c^2 \plus{} a^2 \plus{} b^2}$[/quote]
Nice inequality
We have:$ \frac{4a}{2a^2\plus{}b^2\plus{}c^2} \le \frac{a}{a^2\plus{}b^2}\plus{}\frac{a}{a^2\plus{}c^2}$
Then $ RHS \le \sum \ \frac{a\plus{}b}{a^2\plus{}b^2} \le \sum \ \frac{2}{a\plus{}b} \le \sum \ (\frac{1}{2a}\plus{}\frac{1}{2b}) \equal{}LHS$
then,we have Q.E.D


I've been pondering this for a long time now and have finally given up.

http://www.geocities.com/jacobin2001/congr.jpg

source: unknown

We have that the least residue of $x$ modulo $m_{i}$ is $\equiv a_{i}N_{i}M_{i}\equiv a_{i}(mod m_{i})$, because all the other terms in the sum that gives $x$ ($x \equiv a_{i}N_{i}M_{i}+...+a_{k}N_{k}M_{k}(mod M)$) have the factors of $m_{i}$, therefore they don't contribute to the residue modulo $m_{i}$. And because $N_{i}M_{i}\equiv 1 (mod m_{i})$, the solution is unique modulo $M$.


Hello guys,
I am a junior who is currently in physics (1st year course, non ap) in high school. I was wondering if I should take the SAT II Physics for June. I already have taken 3 SAT II's (Math 2C, Chemistry, and Biology-M), but I am kind of leaning towards not taking it. On the other hand, although I am not exceptional at the subject, I think maybe I should give it a shot. Is there anyone here who has taken the SAT II Physics and is willing to share their thoughts about it? Thanks a lot!

The SAT II Physics Exam is about as comprehensive as the AP Physics B Exam. That is to say there is a LOT of material on the test. I got a 5 on my AP Physics B test, but I had to leave like 4-5 blank on the SAT II Physics because I hadn't learned about those topics (this got me a 790). The questions aren't excruciatingly difficult (like some of the AP ones) but they are in a wide range of topics, so you have to make sure you know all of them. If I recall correctly, there was a question on bubbles on my test. I was like :?. But anyway, if you plan on taking it, definitely review all the physics you have learned and check out Sparknotes or something. Good luck.

I heard Sparknotes' version of study guide has everything you want.... (heard from a guy who aced the test and couple others who came close) ---- personally, I just went over my notes (heh and i got 780) and did quite well. but like paladin said a lot of topics, so be careful.


inequalities
geometry
geometric inequality
inequalities unsolved

Without using a geometric proof, can anyone suggest an alternative approach to solving this inequality

sqrt(a^2+b^2+ab)+sqrt(c^2+b^2+cb)>=sqrt(a^2+c^2+ac)?

Thank you!

What about $a,b,c$? if $b=0$ and $a,c\ne0$ than it`s not true

I think it's supposed to be
\[\sqrt{a^{2}-ab+b^{2}}+\sqrt{b^{2}-bc+c^{2}}\geq \sqrt{a^{2}+ac+c^{2}}\]
But the geometric approach is so nice... of course, you can square, isolate radical, square again, and reduce to
\[(ab+bc-ac)^{2}\geq 0\]
but you must see this factorization. I only found it by using the equality case in the geometric inequality (where you have a triangle with sides $a,c$, included angle of 120 degrees bisected by segment of length $b$, then you can split up the area of the big triangle into the little triangles and use that to find $b=\frac{ac}{a+c}$)


ok so i understand how to do theses problems but for some reason i keep getting the wrong answer. If anyone could solve this step by step or at least solve into the correct form of a quadratic equation that would be great. Thanks

[img]http://webwork2.math.utah.edu/math1050fall2007-4/tmp/l2h/set2/13-19754108/19754108img1.gif[/img]

Answer

The first thing to notice is that $ x\neq\minus{}2$ and $ x\neq 1$.

Multiply throughout by $ (x\plus{}2)(x\minus{}1)$.

$ \Rightarrow (2x\minus{}1)(x\minus{}1)\minus{}(x\plus{}1)(x\plus{}2)\minus{}2(x\plus{}2)(x\minus{}1) \equal{} 0$

$ \Rightarrow 2x^{2}\minus{}3x\plus{}1\minus{}(x^{2}\plus{}3x\plus{}2)\minus{}2(x^{2}\plus{}x\minus{}2) \equal{} 0$

$ \Rightarrow 2x^{2}\minus{}3x\plus{}1\minus{}x^{2}\minus{}3x\minus{}2\minus{}2x^{2}\minus{}2x\plus{}4 \equal{} 0$

$ \Rightarrow\minus{}x^{2}\minus{}8x\plus{}3 \equal{} 0$

$ \Rightarrow x^{2}\plus{}8x\minus{}3 \equal{} 0$

$ \Rightarrow x \equal{}\frac{\minus{}8\pm\sqrt{64\plus{}12}}{2}\equal{}\frac{\minus{}8\pm\sqrt{76}}{2}\equal{}\frac{\minus{}8\pm 2\sqrt{19}}{2}\equal{}\minus{}4\pm\sqrt{19}$


Has any one heard about the profession of "Actuary".
 I don't know what a math major can do apart from becoming a professor but i want to do the math in the real world. I don't want to go towards academics(professor) ."Actuary" sounds like a dream career for all math lovers because this is what i came to know about the profession.

It is said that  best choice for people genuinely interested in math, who like to solve real world out of ordinary problems, and who love mathematical puzzles is becoming an "Actuary".

 To become a fully qualified Actuary though, you need to pass a series of about 9  rigorous tests conducted by the S.O.A.(Society of Actuaries).
These exams are very difficult though. But for a math lover, they are just merely math contests which can be won certainly with minimum required effort.

After passing all the 9 levels you will get a designation of F.S.A. (Fellow of Society of Actuaries).  This is a financiall rewarding profession based on merit.  Because your salary will be proportionally increasing with the number of exams you clear.

I love math, and i am seriously considering becoming an Actuary.
Any opinions on this Profession.

I know (online) an actuary, and he does indeed some to like his profession. You may want to look at the Department of Labor Web site for occupational outlook statistics to see what is predicted for demand for that occupation. I have also heard that biostatisticians will be in strong demand in the next few decades.


inequalities
vector
email
trigonometry
inequalities open

:D . JG's inequalities are interesting not just because of their attractive simplicity but also difficult ones. I'm going to introduce you a new solution to a familiar one
       M =\sum (cos((B-C)/2)) >= 2/ \sqrt3*( \sum sinA)
  I think that all of you know these inequalities easily proved by vectors:
      If XYZ is a triangle and x,y,z are real numbers then
1)  \sum (xy*cosZ) <=  1/2*(\sum (x ^2)).
2)  \sum (xy*(sinZ)^2)<=1/4*[( \sum x)^2).
      So my idea is how to replace sinA, sinB, sinC.
      We notice that for every (x,y,z) (x,y,z>0) there exists a triangle XYZ such that
       cotgX = x/( \sqrt (xy+yz+zx)), ...........
 (since we have  \sum (cotgA*cotgB) =1).
      so we get sinX =  [\sqrt ((xy+yz+zx)/(x+y)(x+z))],..........
                     cosX= x/[ \sqrt(x+y)(x+z)],..........
    Now by replacing x,y,z below by respectively \sqrt(yz/x),  \sqrt(zx/y),
  \sqrt(xy/z) then we have
         sinX =  \sqrt [(x*(x+y+z))/[(x+y)(x+z)]],...........(*)
         cosX =  \sqrt [(y*z)/[(x+y)(x+z)]],.............[color=blue]
     Let's come back to our problem , we replace x,y,z in (*) by sinA, sinB, sinC and we have

    sinX =  \sqrt (sinAcos(A/2)/[cos((A-B)/2)*cos((A-C)/2)]),.............
    It 's time we should use the above inequality we have
          M>=2* (\sqrt( \sum sinAcos(A/2) ))= N
    On the other hand, by using Bulhiacovski's inequality we have
     N>=2* \sqrt [( \sum sinAcos(A/2))*( \sum sin(A/2))*(2/3)]
        >=2/ \sqrt 3 *  (   \sum sinA).[/color]
       So we just need a few lines to write the solution.
      Another JG's inequality is
             \sum (tg^2(A/2))+8sin(A/2)sin(B/2)sin(C/2) >=2.
       With this repalcement, I have
         ( \sum tg(A/2))^2>=2*( \sum cosA)/[cos((B-C)/2)*os((C-A)/2)
*cos((A-B)/2)]  = K    (ABC is an acute triangle)
       But I get trouble with this equality
            K+8sin(A/2)sin(B/2)sin(C/2)>=2
       I think it's right because I have checked it many times. So I really look forward to your reply. My email address is TREEGONER@yahoo.com.
    Bye.

Treegoner,what is K?.
Let A=B=C then K=3.So I think you want to prove that
K/3+8sin(A/2)sin(B/2)sin(C/2) \geq 2.
or K=2*( \sum cosA)/( \sum cos((A-C)/2)) :?

AnhCuong, K here is equal to 2*( \sum cosA)/prod cos[(B-C)/2].
    At that time, I made such a stupid mistake when I wrote wrongly
       [color=blue] K + 8prodsin(A/2) >=4 (*)[/color] not 2. SORRY ! :(
   But luckily I finally have the result of this one with another simple direction.
    All we have to do is to prove
      (sum a^2)/( \sum ab)+8abc/prod(b+c)>=2 (a,b,c >0)
    <=>  (sum a)^2/( \sum bc) +8abc/prod(b+c) >=2.
     As we know, there exists a triangle ABC satisfying sinA =  \sqrt [ab+bc+ca)/(a+b)(a+c)].
       We apply this inequality  \sum xy(sinC)^2 <= 1/4(x+y+z)^2 for every x,y,z, we have
       LHS >= 4[\sum bc/(a+c)(a+b)] + 8abc/prod(a+b) = 4.  :D
    But I can't prove the (*) inequality. I think I should put it to Unsolved Problem. What a difficult one ! :?

How to prove this one ?

\[ \sum (xy \sin^2 z) \leq \frac 14 \left( \sum x\right)^2 \]

About Treegoner first inequalities (1) and (2) and to answer a question of Manlio : I don't know if these relations can be proved using vectors, but they both are a direct consequence of a more general inequality, which states :
If $A+B+C = p \pi$ and n is a positive integer and x,y,z are real numbers then,
\[ x ^2  + y ^2  + z ^2 \geq  2(-1)^{ np+1 }(xy\cos(nC) + yz\cos(nA) + zx\cos(nB)). \]

This inequality holds because it is equivalent to :
\[ (x + (-1) ^{np}(y\cos(nC) + z\cos(nB)) ^2 + (y\sin(nC) - z\sin(nB))^2 \geq 0 \]

The inequality (1) is the special case $n = p = 1.$
The inequality (2) is the special case n = 2 and p = 1 (using that cos(2t) = 1 - 2*sin 2 (t) ).

Pierre.

[quote="manlio"]How to prove this one ?

\[ \sum (xy \sin^2 C) \leq \frac 14 \left( \sum x\right)^2 \][/quote]

This came directly from the general inequality by taking M to be the circumcenter.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15558


If $ a, b, c$ distinte positive real numbers such that $ \sqrt{ab} \plus{} \sqrt{bc} \plus{} \sqrt{ca} \equal{} 1$, prove that

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

Well, I strongly recommend this method to solve this problem properly is using " separating ways"
First group, we have
$ \sum_{cyc}\frac {a^2}{a \plus{} b} \ge \frac {a \plus{} b \plus{} c}{2}$
Subsequently, we have
$ \sum_{cyc}\frac {a^2}{(a \plus{} b)(a \plus{} c)} \ge \frac{3}{4}$
Some readers should take these above exercises as a practice!
Good luck!

[b]Great Math [/b]I have understood few-little...


Is there any specific page on LaTex Page where I can learn how to make the title pages nicely?

I tried to look but I fell sleep when I was doing it :D (I guess I have to get some regular sleep).

I'm Texing some problems to give to my teacher for his nice work this year so...

Thank you!

[quote="DPatrick"]\title{Your title here}
\author{Your name here}
\date{The date here}
\maketitle[/quote]

In article mode, this just puts a title at the top of the first page.  Report and Book modes make a new title page.

[quote]In article mode, this just puts a title at the top of the first page. Report and Book modes make a new title page.[/quote]

if you want to have new title page in article using thoses command, you need to specify it:

\documentclass[12pt,titlepage]{article}

ledurt


Have anybody ever taken part in this program?I really need to listen your comment on it since I consider taking part in this this year. The main reason is it's free, not expensive as Math camp or Ross program but I will think to make a right decision! If you have experienced in it, please tell me how you feel, lecture, accomodation good/bad. I'm pleased to hear from all of your guys' idea! :)

First, the program is not free.  Applying is free, but it's not free to attend (though it is cheaper than CTY, which I attended before HCSSiM).  For that reason, I'd say definitely at least apply.  You get to take the Interesting Test if you apply, which is really a lot of fun, and taking it will hopefully increase your interest in the program.  I went there last summer, as did some other people on this forum, and had an amazing time.  The math was amazing, so much more interesting and challenging than what I'd done in school, and the entire environment (students, staff, yellow pigs, etc.) was just incredible.  I'd go back if I could, but I'm a senior, so I can't :( I can't speak of Math Camp or Ross as I've been to neither, but I believe that at least Math Camp is also supposed to be very good.  In short, definitely apply, and if you get in, strongly consider attending, it is an awesome experience.

Thanks for your advice,  Reynold! I saw your quote on the official website! But I'm more curious about the expense that I have to pay if I'm admitted to go there! I also don't see anything mentioned about the financial aid, which is usually populated in other similar programs.  And does it have any contest(I meant in there) to qualify for this program? Or we just need to exchange idea about Math, do as many problems as we can as well as new method trained to apply? I know every program like this gives me many precious experience! Wonderful environment is considered, in my opinion, as the most important element to perform the best!

Haha, thanks.  Financial aid is mentioned somewhere on there, I believe they provide some if necessary, but I didn't take any so I don't know.  The Interesting Test, which is a big part of your application, is emailed to you after you apply.

Wait! When did you submit your application form last year? Since I think it will affect a part of my admission, I intend to write it carefully before submitting!  But I don't think it's fair to everyone since they just can start doing problem set when submitting the application form. Anyway, I want to have a chance to improve my ability in Discrete, so I will finish writing application next week! Hopefully those problems are interesting as you introduced! :)

I submitted my application in early May.  Do spend time on it, and know that the interesting test won't be sent to you until March anyway, so there's not a huge rush.

Its really a good time.  If you are worried about money i think several people last year went for next to nothing.  Also they have no admission deadline.  I know because i submitted the interesting test i think 5 days before the program started last year.


[quote="Tiks"][quote="nayel"]$a,b,c$ are the side lengths of a triangle. prove that $\sum \frac{a}{\sqrt{a^{2}+2bc}}>\sqrt 2$[/quote]
As we have that $a^{2}+2bc\leq a^{2}+b^{2}+c^{2}$=>

$\sum \frac{a}{\sqrt{a^{2}+2bc}}\geq \frac{a+b+c}{\sqrt{a^{2}+b^{2}+c^{2}}}$

So we need to prove that $\frac{a+b+c}{\sqrt{a^{2}+b^{2}+c^{2}}}>\sqrt{2}$<=>

$(a+b+c)^{2}>2(a^{2}+b^{2}+c^{2})$<=>

$2(ab+bc+ac)>a^{2}+b^{2}+c^{2}$<=>

$a^{2}+b^{2}+c^{2}>(a-b)^{2}+(a-c)^{2}+(b-c)^{2}$

Which is obvious from the triangle's inequality! :wink:[/quote]

note that $2(ab+bc+ac)>a^{2}+b^{2}+c^{2}$ also follows from

$\sum{a(b+c-a)}>0$

[quote="ishfaq420haque"]my solution is by holder :D[/quote]
 I  too  use  Hoder's  Ineq:
 Rewrite  ineq  we have :
$ A^2 (a^3 \plus{} b^3 \plus{} c^3 \plus{} 6abc) \geq (a \plus{} b \plus{} c)^3 .$
Hence : $ A^2 \geq \frac {(a \plus{} b \plus{} c)^3}{a^3 \plus{} b^3 \plus{} c^3 \plus{} 6abc} \geq 2 \to \sum a^3 \plus{} 6abc \leq 3\sum a^2b$
By Other hand, $ a(a \minus{} b)^2 \leq ac^2$. hence :
 $ a^3 \plus{} b^3 \plus{} c^3 \leq 2\sum a^2b$
and by Cauchy's ineq $ \sum a^2b \geq 6abc.$


Determine all pairs of integers $(x,y)$ which satisfy the equaton
$6x^{2}-3xy-13x+5y=-11$

[hide="hint"]
Group the terms with $y$ together, factoring out the $y$. Then move all the other terms to the right side of the equation, isolate $y$ and use long division. That should work.
[/hide]


inequalities
geometry
trigonometry
trig identities
Law of Cosines
inequalities unsolved

Prove that if a triangle has sides $a,b,c$ has area $A$, and is acute, then

$27(b^2+c^2-a^2)^2(a^2+b^2-c^2)^2(a^2+c^2-b^2)^2 \leq (4A)^6$

Note that this is note true for obtuse triangles...

Use law of cosines $b^2+c^2-a^2=2bc\cos{A}$ and area formula $4A=2bc\sin{A}$,..., it reduces to

$(\tan{A}\tan{B}\tan{C})^2 \geq 27$

Since $\tan{x}$ is convex in interval $(0, \pi/2)$, we have

$\tan{A}\tan{B}\tan{C} = \tan{A}+\tan{B}+ \tan{C} \geq 3\sqrt{3}$


What Artist/Group do you listen to while doing MathCounts?

If you don't listen to anything then don't post
If you listen to Classical, don't post.

That's a bit biased there, don't ya think? But hmmm... I'd have to say EVERYTHING but country , classical, and metal. Not saying that I'm not a fan of those two, but I don't find it suits the occasion.

While doing mathcounts? Like at the competition or just practicing at school or practicing at home?

Are you allowed to listen to music? Would they think your music player is some sort of calculator in disguise?

I mean when you're practiceing at home.

And here are my favorite artists while doing MathCounts:
1 or 2 tracks of 2pac
Most of G-Unit
Black Eyed Peas
Linkin Park
All-American Rejects
Green Day
Eminem

[quote="lizcheang"]country , classical, and metal. Not saying that I'm not a fan of those two,[/quote]

That's three :D

I pretty much listen to everything too--except perhaps opera :P

At Aznwithabrain's request: I usually listen to the radio, so I don't really have any control over what artists I do listen to. And seeing how I can't post about Mozart and all of them... :mad: ;)

Please have future posts include your favorite artists/groups you listen to while doing MathCounts.

[quote="mathnerd314"][quote="lizcheang"]country , classical, and metal. Not saying that I'm not a fan of those two,[/quote]

That's three :D

I pretty much listen to everything too--except perhaps opera :P[/quote]

HAHAHA. sorry, I added metal in there and I didn't change "two." ahahaha. that's funny

but ummmm. let's see..... i'm a huge fan of the all american rejects, relient k, & taking back sunday.

listen to almost everything modern like relient k, chamilionare, etc
get sick of the old songs played over and over again
:)

[quote="Treething"]rhcp > you[/quote]
What's rhcp?

.... Red Hot Chili Peppers?

They're ok IMO, but I don't do MathCounts at home so I can't post in this topic I guess.

Asian Music is really vague, that's like, 50 genres right there.

[quote="Treething"]i don't know if any of you have heard my opinions on j-pop/k-pop but it deserves to burn[/quote]

"burn" as in "die"?
then I completely agree.

[quote="Ignite168"]Yes, the whole world should just listen to English songs  :roll:[/quote]

YES A GOOD IDEA

ALTHOUGH

MAYBE

IF THEY PEOPLE PLAYING THEM WERE GOOD THEY COULD HAVE INSTRUMENTS FROM OTHER COUNTRIES IN THE BACKGROUND

LIKE A HARP
OR ONE OF THOSE CHINESE INSTRUMENTS
OR THOSE KOREAN INSTRUMENTS
BUT NOT THOSE INDIAN THINGS

PINK FLOYD! I'll meet you on the dark side of the moon!*babbles incoherently* But most of my ipod is devoted to an almost scizophrenic brain...I have EVEY genre even unknown but mostly rap and classical.

I listen to mostly classic rock, Led Zeppelin, Jimi Hendrix, Pink Floyd....etc.

[quote="funcia"]I have EVEY genre even unknown but mostly rap and classical.[/quote]

Really? Do you have darkwave or industrial or synthpop? What bands?

Oh man...let me go get them...Really I was trying to say how my music varies not that I have classical,dark wave,dance,country(i'd never have it in my life!),pop, jazz....(really the list goes on and I haven't put 2 of my genres) My point was that most of us don't listen to the same music...but now that you mentioned it actually my favorite of those are Clan of Xymox No Human Can Drown...I kind of want it now...

[hide] some how i think you'll believe me if i said i had all the harry potter tracks...[/hide]

Cool! That is farther from industrial than I usually listen to, but it's neat anyway. Regardless, my point was a lot of people say they "like all music" but probably they don't realize what's out there. If you don't believe me, I'll send you a song by the band Converter, which is a genre called "Power Noise", and then we'll talk again. ;)

edit: PS -- if you haven't heard of them already, you might like Covenant. they're awesome. Check it out! :D :D

I will...straying a bit from the genre(i believe) is Mogwai 1% of monster it gets boring but if you are reading...(or on AoPs..) I'm going to go find some of those clips(hopefully.)

krrish is ok, a little kiddish, and priyanka chopra is reallly bad inthi s movie.

golmaal, dus, main hoon na, bluffmaster, kal ho na ho, kuch kuch hota hai, Bunty aur Bubli, K3G, and lots more movies have really good soundtracks.  I only have Indian filmi on my ipod.

I just found out that my brother put my MP3 Player in the wash yesterday. IT BROKE!!! :cursing:  :furious:  :wallbash_red:
 :bomb:
I am so bad at him. [b]He forgot his I-POD charger at college and went to mow the lawn.[/b] He wanted to listen to music, so I gave him the mp3 Player to listen to while mowing the lawn. After he got back, I dunno what he did with it. He probably kept it in his pants or jacket pocket. When I asked him about it 20 minutes after he mowed the lawn, he told some stupid joke that was intended to be funny that implied that it was safe on some table inside the house. When I asked him again, he jokingly said, "I dunno," in another bad joke. This morning when I wake up, it s in the wash and it s ruined.

owned.

do what i'd do - buy an ipod and take the ipod out, put your ruined ipod inside, and return the ipod you bought


superior algebra
superior algebra theorems

Give a formal proof of the following:

(-1)x = -x ,for all ,x


A man walks $ x$ miles due west, turns $ 150^{\circ}$ to his left and walks $ 3$ miles in the new direction.  If he finishes at a point $ \sqrt3$ miles from his starting point, then $ x$ is

$ \textbf{(A)}\ \sqrt3 \qquad \textbf{(B)}\ 2\sqrt3 \qquad \textbf{(C)}\ \frac32 \qquad \textbf{(D)}\ 3\qquad \textbf{(E)}\ \text{not uniquely determined by the given information}$

When he turns, he makes a $ 180\minus{}150\equal{}30^{\circ}$ angle with the line that has length $ x$.  We know that the base subtending the $ 30$ degree angle has length $ 3$, so the hypotenuse, or $ x$, is just $ \frac{3}{\sqrt{3}}*2\equal{}\boxed{2\sqrt{3}}$.  The information about the distance from his starting point is irrelevant, although a quick check confirms that it does indeed satisfy the properties of a $ 30\minus{}60\minus{}90$ triangle.


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

I am currently a tenth grader.  In order to help me get into USAJMO or USAMO, should I take

AMC 10 twice
AMC 12 twice
AMC 10 and AMC 12

I'm fairly certain that I can make the AIME through both the AMC 10 and AMC 12.

Thanks a lot!

You should take both of them.  If you don't make it pass the amc10a/amc12a you can change which one you want to take on the b date.

If your only goal is the USAJMO, take AMC 10 twice. You can't qualify for the USAJMO through the AMC 12.
If your only goal is USAMO, then take AMC 12 twice because you can't qualify for the USAMO through the AMC 10 anymore. The one caveat is that if your AIME score is at least 11, you can qualify for the USAMo through the AMC 10.
If you want to maximize your chances of getting into either of them (like me), I would take the AMC 12 once and the AMC 10 once. That way, you can get into the USAMO if your AMC 12/AIME index is high enough or you score at least 11 on the AIME, and you can get into the USAJMO if your AMC 10/AIME is high enough.

[quote="batteredbutnotdefeated"]It depends on your own skills. If you think you can make USAMO, give taking the AMC12 twice some thought. Else, if you want to play it safe, take the AMC10 twice and qualify for USAJMO. If you're not sure if you'll do well or not, take both.(usually AMC10A and AMC12B)[/quote]

Thanks for your input.  Just wondering, why is it usually AMC10A and AMC12B?  Is this typically the easier combo?

[quote="Ihatepie"]Neither the A nor the B are any harder than each other. I think the reason that people say to take the 12 later is because the 12 is the harder test and you'll need more time to prepare.[/quote]

Just wondering, what are you planning to take?

I'm taking 12A and 10B because my school only supplies the 12 (and only on the A date.) I was able to find an alternate testing location for the B date, so I'm taking the B then. If I had a choice, I would do it the other way though.


algebra
polynomial
number theory solved
number theory

The real polynomial p(x) = ax3 + bx2 + cx + d is such that |p(x)| <= 1 for all x such that |x| <= 1. Show that |a| + |b| + |c| + |d| <= 7.

a solution at:
http://www.kalva.demon.co.uk/short/soln/sh965.html


[b]1.[/b]  Suppose that $ n,k\in\mathbb N$ satisfies $ \frac7{13} < \frac {n}{n \plus{} k} < \frac6{11}.$ Now for some given $ n$, there is unique $ k$ such that the above inequality holds. Find the maximum and the minimum of such $ n$.

[b]2.[/b]  Suppose that $ x, y\in\mathbb N\cup\{0\}$ satisfies $ 5|x \plus{} 2y$, $ 3|x \plus{} y$, and $ 2x \plus{} y\geq99$. Find the minimum of $ 7x \plus{} 5y$.

Hint for (1):
$ \frac{7}{13}<\frac{n}{n\plus{}k}<\frac{6}{11}\  \Rightarrow\  \frac{13}7>\frac{n\plus{}k}n>\frac{11}6 \ \Rightarrow \ \frac67>\frac kn>\frac56.$

For $ k$ to be unique we must have $ (k \plus{} 1)/n > 6/7,(k \minus{} 1)/n < 5/6\iff (7k \plus{} 7)/6 > n > (6k \minus{} 6)/5$
$ (7k \plus{} 7)/6 > (6k \minus{} 6)/5\iff71 > k$

so the maximum is $ \lfloor{\frac {7\cdot70 \plus{} 7}6}\rfloor \equal{} 82$, which works after substituting in the values.


geometric series
real analysis
real analysis unsolved

sigma j=0 to infinity nx^n converges to what?

The sum is acually from 1.Recall $\sum_{k=0}^{\infty}x^{k}$ derive an you should get your result. Your question isnt clear to me ,because you ask to what does it converge, but instead you should ask what is the radius of convergence.

I assume the thing to be computed is $\sum_{n=0}^{\infty}nx^{n}.$

(And yes, it doesn't matter whether that sum starts at 0 or at 1.)

This is a power series with radius of convergence 1. It converges only for $|x|<1.$ As for what it converges to: (For all of the following manipulations, assume that $|x|<1$ or if necessary, $|x|\le s$ for some arbitrary $s<1.$ The justifications are the basic theorems about power series proved in a beginning analysis course.)

Start with a geometric series:

$\frac1{1-x}=\sum_{n=0}^{\infty}x^{n}$

Differentiate:

$\frac1{(1-x)^{2}}=\sum_{n=1}^{\infty}nx^{n-1}$

Finally, multiply by $x$:

$\frac{x}{(1-x)^{2}}=\sum_{n=1}^{\infty}nx^{n}$


function
geometry
3D geometry
algebra
factorization
sum of cubes

Determine, with proof, the positive integer whose square is exactly equal to the number 1 + \sum 2001 i = 1  (4i-2) 3.

You should find the function f(i)=ai^4+bi^3+ci^2+di+e
to f(i+1)-f(i)=(4i-2)^3
Then sum=f(2002)-f(1)
Try it!!! :D

You still need to find the values of a,b,c,d,e  :(

Since this is solved, I'll post my solution using the known series  \sum {j=1 to n} j3 = (n2(n+1)2)/4 only. In full generality, I'll prove that p(n) is always a square for all natural numbers n, where p(n) = 1 + \sum {j=1 to n} (4j-2)3.

Consider \sum {j=1 to n} (2j-1)3. Since this sum is just the sum of cubes of odd integers up to 2n-1, this sum is just the same as the sum of 2n cubes minus the sum of even cubes up to 2n; in other words,
\sum {j=1 to n} (2j-1)3 = (\sum {j=1 to 2n} j3) - (\sum {j=1 to n} (2j)3) = ((2n)2(2n+1)2)/4 - 23(n2(n+1)2)/4 = ... = n2(2n2-1), after simplifications.
Therefore, p(n) = 1 + 23(n2(2n2-1)) = ((2n)2-1)2
 :cool:


Let the roots of  f(x)=x be p & s  where f(x) is  quadratic  polynomial ax^2 +bx+c.
P&q are also roots of f(f(x))=x .let the other two roots of  f(f(x))=x be r & s . then prove that
1.If  p&q  are real and unequal then  r&s are also real .
2. if p &q are imaginary then r & s are  also imaginary.
3. if p +  q= r + s, then p,q & r, s can not be real.
4. if p & q are real & equal then   p & s are imaginary.

Why have you posted this in the LaTeX forum? Is there a LaTeX question you wish to ask or should this go in another forum, if so which one?


geometry
circumcircle
rhombus
search
inequalities
national olympiad

Please discuss the problems and the solutions by following the link on the problem number.
The first two tests were given at the NMO, Bistrita, as follows:

Test 1, 31 March 2005

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=32119]Problem 1[/url]
Let $\mathcal{C}_1(O_1)$ and $\mathcal{C}_2(O_2)$ be two circles which intersect in the points $A$ and $B$. The tangent in $A$ at $\mathcal{C}_2$ intersects the circle $\mathcal{C}_1$ in $C$, and the tangent in $A$ at $\mathcal{C}_1$ intersects $\mathcal{C}_2$ in $D$. A ray starting from $A$ and lying inside the $\angle CAD$ intersects the circles $\mathcal{C}_1$, $\mathcal{C}_2$ in the points $M$ and $N$ respectively, and the circumcircle of $\triangle ACD$ in $P$.

Prove that $AM=NP$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=32124]Problem 2[/url]
Find the largest positive integer $n>10$ such that the residue of $n$ when divided by each perfect square between $2$ and $\dfrac n2$ is an odd number.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=32123]Problem 3[/url]
In a country 6 cities are connected two by two with round-trip air routes operated by exactly one of the two air companies in that country.

Prove that there exist 4 cities $A$, $B$, $C$ and $D$ such that each of the routes $A\leftrightarrow B$, $B\leftrightarrow C$, $C\leftrightarrow D$ and $D\leftrightarrow A$ are operated by the same company.

Test 2, 1 April 2005

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=32122]Problem 4[/url]
Let $a,b,c$ be positive numbers such that $a+b+c \geq \dfrac 1a + \dfrac 1b + \dfrac 1c$. Prove that
\[ a+b+c \geq \frac 3{abc}. \]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=32120]Problem 5[/url]
On the sides $AD$ and $BC$ of a rhombus $ABCD$ we consider the points $M$ and $N$ respectively. The line $MC$ intersects the segment $BD$ in the point $T$, and the line $MN$ intersects the segment $BD$ in the point $U$. We denote by $Q$ the intersection between the line $CU$ and the side $AB$ and with $P$ the intersection point between the line $QT$ and the side $CD$.

Prove that the triangles $QCP$ and $MCN$ have the same area.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=32121]Problem 6[/url]
Let $ABC$ be an equilateral triangle and $M$ be a point inside the triangle. We denote by $A'$, $B'$, $C'$ the projections of the point $M$ on the sides $BC$, $CA$ and $AB$ respectively. Prove that the lines $AA'$, $BB'$ and $CC'$ are concurrent if and only if $M$ belongs to an altitude of the triangle.

More tests will follow and they will be posted here.

[quote="Myth"]Who take part in JBMO?[/quote]What do you mean? The team is not yet established. The results are in this file: http://www.mathlinks.ro/Forum/download.php?id=1832

Is it so difficult to undertsand a such easy question? :huh:

I meant for what grades is JBMO aimed?

If it is not properly formulated, yes it is difficult :)
There are no grades at jBMO, but there is an $15\dfrac 12$ age limit (at the date of the competition) to be able to attend it.

[quote="Myth"][quote="Valentin Vornicu"]If it is not properly formulated, yes it is difficult :) [/quote]
So can you show how to properly formulate my question, great english speaker?[/quote]You did it yourself the second time ;)

They were given 4 hours each day. The jBMO will be slightly easier (I think) but with 4 problems in 4 hours, rather than 3.

in cat timp de la trimiterea mesajului pe adresa rgologan primesc un raspuns?

[quote="deedee"]sorry! how much time do i have to wait until i receive an answer from rgologan@...[/quote]We have no idea. Have patience :)

[quote="mmm"]why nobody post here the latest tests for the juniors?[/quote]Because I didn't have the time :P and I didn't consider it that important :D :D ... I will post them don't worry :)

[quote="Valentin Vornicu"]
More tests will follow and they will be posted here.[/quote]

Have the tests taken place?

Where can I find all junior TSTs Romania 2005??

Will they appear in resources project,too?

Yes they will be in the resources pages. The problems have been posted on the forum already. Just search for "JBMO 2005" and you will find them ;)

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=452801]Problem 7[/url]

A phone company starts a new type of service. A new customer can choose $k$ phone numbers in this network which are call-free, whether that number is calling or is being called. A group of $n$ students want to use the service.

(a) If $n\geq 2k+2$, show that there exist 2 students who will be charged when speaking.

(b) It $n=2k+1$, show that there is a way to arrange the free calls so that everybody can speak free to anybody else in the group.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=61543]Problem 8[/url]

Let $a, b, c>0$ such that $(a+b)(b+c)(c+a)=1.$ Prove that the following inequality holds:

                                                   \[ ab+bc+ca\leq\frac{3}{4}.  \]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=452802]Problem 9[/url]

In $\triangle ABC$, $BC>CA>AB$.

Prove that
$\angle GCA+\angle GBC<\angle BAC<\angle GAC+\angle GBA$

($G$ is the centroid of $\triangle ABC$)

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=452917]Problem 10[/url]

Let $k,r\in N^*$ and let $x\in (0,1)$ be a rational number given in decimal representation
$x = 0.a_1a_2a_3a_4...$

Show that if the decimals $a_k, a_{k+r}, a_{k+2r},...$ are canceled, the new number obtained is still rational.


ARML
geometry
calculus
integration
trigonometry
3D geometry
Pythagorean Theorem

1.
In right triangle $ABC$, $\angle C = 90^\circ$, $c=3$, and $a+b=\sqrt{17}$.  Compute the area of the triangle.

2.
If $(a+b):(b+c):(c+a) = 6:7:8$, and $a+b+c=14$, compute the value of $c$.

3.
Let $z$ be a root of $z^5-1=0$, with $z \not= 1$.  Compute the value of

\[z^{15} + z^{16} + z^{17} + \cdots + z^{50}. \]

4.
A regular polygon of $2n$ sides is inscribed in a circle of radius 5.  The sum of the squares of the distances from any point on the circle to the vertices of the polygon is 600.  Compute the value of the integer $n$.

5.
The circles, each of radius 5, lie in perpendicular planes.  They intersect in two points, $X$ and $Y$, thus determining a common chord $\overline{XY}$.  If $XY = 8$, compute the distance between the centers of the circles.

6.
How many positive integers less than 54 are equal to the product of their proper divisors?

7.
Find all positive integers $n$, less than 17, for which $n! + (n+1)! + (n+2)!$ is an integral multiple of 49.

8.
In triangle $ABC$, $a \ge b \ge c$.  If $\displaystyle \frac{a^3+b^3+c^3}{\sin^3{A} + \sin^3{B} + \sin^3{C}} = 7$, compute the maximum possible value of $a$.

[quote="MCrawford"]1.
In right triangle $ABC$, $\angle C = 90^\circ$, $c=3$, and $a+b=\sqrt{17}$.  Compute the area of the triangle.

2.
If $(a+b): (b+c): (c+a) = 6: 7: 8$, and $a+b+c=14$, compute the value of $c$.

3.
Let $z$ be a root of $z^{5}-1=0$, with $z \not= 1$.  Compute the value of
\[z^{15}+z^{16}+z^{17}+\cdots+z^{50}. \]
4.
A regular polygon of $2n$ sides is inscribed in a circle of radius 5.  The sum of the squares of the distances from any point on the circle to the vertices of the polygon is 600.  Compute the value of the integer $n$.

5.
The circles, each of radius 5, lie in perpendicular planes.  They intersect in two points, $X$ and $Y$, thus determining a common chord $\overline{XY}$.  If $XY = 8$, compute the distance between the centers of the circles.

6.
How many positive integers less than 54 are equal to the product of their proper divisors?

7.
Find all positive integers $n$, less than 17, for which $n!+(n+1)!+(n+2)!$ is an integral multiple of 49.

8.
In triangle $ABC$, $a \ge b \ge c$.  If $ \frac{a^{3}+b^{3}+c^{3}}{\sin^{3}{A}+\sin^{3}{B}+\sin^{3}{C}}= 7$, compute the maximum possible value of $a$.[/quote]

[hide="1"]$a^{2}+b^{2}+2ab=17$
$a^{2}+b^{2}=9$

$ab=4$
$ab/2=Area=2$[/hide]

[hide="2."]2(a+b+c)=21x
a+b+c=10.5x

14=10.5x=21x/2

28=21x

x=4/3

a+b=8

14-8=c=6[/hide]

[hide="3."]That equals

$1+z+z^{2}+z^{3}+z^{4}+1+z+z^{2}+z^{3}+z^{4}+1+z+z^{2}+z^{3}+z^{4}+1+z+z^{2}+z^{3}+z^{4}+1=1$[/hide]

[hide="4."]$600/25=2n=24$

$n=12$[/hide]

[hide="5."]The chord is 3 away from the radius(from the pythagorean theorem and the bisection of the cord), so we can use the pythagorean theorem again to get $3\sqrt{2}$[/hide]

[hide="6."]With 1 positive divisor, it is obviously true. For 2, nope. For 3, nope. For 4, cube of a prime or product of two primes, it works. For 5, nope. For 6, nope. So on.

Numbers that have 1 divisor: 1
Numbers below 54 that have 4 divisors: 39

39+1=40[/hide]

[hide="7."]That equals:

$n!(1+n+1+n^{2}+3n+2)$, and that's $n!(n+2)^{2}$. Either n>13 or n=5mod7.

5,12,14,15,16[/hide]

[hide="8"]Observe that by the law of sines, $\sin{A}=\frac{a}{2R}$, and similar for the other angles*.

So we have $\frac{a^{3}+b^{3}+c^{3}}{\frac{a^{3}}{8R^{3}}+\frac{b^{3}}{8R^{3}}+\frac{c^{3}}{8R^{3}}}=8R^{3}=7$, where $a^{3}+b^{3}+c^{3}$ cancels.

Thus, $R=\sqrt[3]{7}/2$, and the largest possible side (a) is the length of the diameter of the circle, or
$\sqrt[3]{7}$.

*to see this for angle A, circumscribe triangle ABC with a circle, and draw diameter BE, then drawing EC. Since $\angle{A}$ and $\angle{BEC}$ intercept the same arc, they are equal. Also, $\angle{ECB}$ is a right because it intercepts arc BC, a diameter.
We then have $\sin{A}=\sin{BEC}=\frac{BC}{BE}=\frac{a}{2R}$. [/hide]


linear algebra
matrix
algebra
polynomial
Diophantine equation
linear algebra unsolved

Find $n$ such that the ecuation $A^2=\left ( \begin{array}{cccc}1&2&{...}&n\\1&2&{...}&n\\{...}&{...}&{...}&{...}\\1&2&{...}&n\end{array}\right )$ has solutions in $\mathcal l M_n(\mathcal l Q)$.
[hide="question"] I got an infinity of solutions through a Pell equation ... could it be correct?[/hide] :?

Well, can you please post a complete solution?  :)

It would be my pleasure cause i enjoyed solving it.
  Let`s denote $X= ( \begin{array}{cccc}1&2&{...}&n\\1&2&{...}&n\\{...}&{...}&{...}&{...}\\1&2&{...}&n\end{array} )$.
   Now we call $(\lambda , V)$ ,$V= (\begin{array}{c}x_1\\x_2\\{...}\\x_n\end{array} )$  a pair of eigenvalue,eigenvector of the matrix $X$ iff $XV=\lambda V$. Now let`s determine the eigenvalues of $X$.
We have that $XV=\lambda V$ and from here we get the system $ \{ \begin{array}{ccccc} {\lambda x_1=}&{x_1+}&{2x_2+}&{...+}&{nx_n}\\ {\lambda x_2=}&{x_1+}&{2x_2+}&{...+}&{nx_n}\\ {...}&{...}&{...}&{...}&{...}\\ {\lambda x_n=}&{x_1+}&{2x_2+}&{...+}&{nx_n} \end{array}$ so $x_1=x_2={...}=x_n$. So assuming $V$ not to be null, we get that $\lambda =\frac{n(n+1)}{2}$. So the eigenvalues of $X$ cand be  either $\frac{n(n+1)}{2}$ or $0$. But $Tr(X)=\lambda_1+\lambda_2+...\lambda_n$ and $Tr(X)=\frac{n(n+1)}{2}$ so we can conclude by saying that the eigenvalues of $X$ are $\lambda_1=\lambda_2=...=\lambda_{n-1}=0,\lambda_n=\frac{n(n+1)}{2}$.
  Then the eigenvalues of $A$ are $\alpha_1=\alpha_2=...=\alpha_{n-1}=0,\alpha_n=\sqrt{\frac{n(n+1)}{2}}$. The caracteristic polynomial of $A$ , $P_A\in \mathcal l Q[X]$ so should it have a root of the form $a+\sqrt{b}$ then $a-\sqrt{b}$ is also a root, so judging by the fact that $\alpha_n$ is the only non-null root it should be rationall more that it should be natural because $2|n(n+1)$. So we get the equation $\sqrt{\frac{n(n+1)}{2}}=k \rightarrow n^2+n-2k^2=0 \rightarrow n=\frac{\sqrt{1+8k^2}-1}{2}$. So $\sqrt{1+8k^2}$ shoul be a postive integer $1+8k^2=l^2\rightarrow l^2-8k^2=1$ which is a Pell equation. Observe that $l=3,k=1$ is a solution . We can get the general solution from the matrix $ ( \begin{array}{cc} l&8k\\k&l \end{array}  )$. Thus there are a infinity of $n$ for which the equation $A^2=X$ has solution in $\mathcal l M_n(\mathcal l Q)$.

Remarks: X^2=n(n+1)/2.X and rank(X)=1.
What follows is a generalization:
Let X be in Mn(Q) s.t. X^2=aX (a<>0) and rank(X) is odd.
Then X is a square in Mn(Q) iff a is a square in Q.
proof:  let d be a square root of a (in C).
- If a is a square  then 1/d.X is in Mn(Q).
- If X=A^2 with A in Mn(Q), and a is not a square then
A^2(a.I-A^2)=0 and a-x^2 is irreducible over Q. The characteristic polynomial of A is A^p.(a-x^2)^k and A is similar (in C) to:
diag(0..0,J..J,d.Ik,-d.Ik) where J is the 2 by 2 nilpotent Jordan matrix.
Thus X is similar to diag(0..0,0..0,a.Ik,a.Ik) and rank(X) is even!!
Remark: this result holds with Q a field ( and C his algebraic closure).


algebra
polynomial
vector
superior algebra
superior algebra theorems

How can I show that every element in the extension field GF(p^L) is algebraic over GF(p)(existence of minimal polynomial)? I've just started with abstract algebra so I would enjoy a proof that doesnt rely on too many other results.

finite extensions (in this case of degree $L$) are always algebraic.

Let $L/K$ be a finite field extension.  In order to prove that $L/K$ is an algebraic extension, we need to show that any element $\alpha\in L$ is algebraic, i.e., there exists a non-zero polynomial $p(x)\in K[x]$ such that $p(\alpha)=0$.

Recall that $L/K$ is a finite extension of fields, by definition, it means that $L$ is a finite dimensional vector space over $K$. Let the dimension be $[L\colon K]=n$ for some $n\in \mathbb{N}$. Consider the following set of ``vectors'' in $L$:
\[\mathcal{S}=\{ 1, \alpha, \alpha^{2},\alpha^{3},\ldots,\alpha^{n}\}\]
Note that the cardinality of $S$ is $n+1$, one more than the dimension of the vector space. Therefore, the elements of $S$ must be linearly dependent over $K$, otherwise the dimension of $S$ would be greater than $n$. Hence, there exist $k_{i}\in K,\ 0\leq i \leq n$, not all zero, such that
\[k_{0}+k_{1}\alpha+k_{2}\alpha^{2}+k_{3}\alpha^{3}+\ldots+k_{n}\alpha^{n}=0 \]
Thus, if we define
\[p(X)=k_{0}+k_{1}X+k_{2}X^{2}+k_{3}X^{3}+\ldots+k_{n}X^{n}\]
then $p(X)\in K[X]$ and $p(\alpha)=0$, as desired.


calculus
derivative
logarithms
algebra
polynomial
calculus computations

find the absolute minimum and absolute maximum value of h(x) on the closed interval 0.5<x<1.  h(x)=e^x-lnx

Take the derivative and set it to 0:
$ h'(x)\equal{}e^{x}\minus{}\frac{1}{x}\equal{}0\Rightarrow e^{x}\equal{}\frac{1}{x}$
Now we can use logs to get
$ \ln{1/x}\equal{}x\equal{}\ln{1}\minus{}\ln{x}\equal{}\minus{}\ln{x}\equal{}x$.  This has no solutions so there are no critical values.  If you look at a graph, you'll see this is indeed the case.

But since global extreme values are guaranteed, this would mean it takes its minima and maxima on the boundary. I'm not so sure it's correct that it has no zeros though, in fact I'm pretty sure JRav made a mistake.

$ e^x\equal{}\frac1x$ does indeed have a solution in the interval $ (.5,1).$ This number cannot be expressed in closed form but is about $ .56714.$ Since $ h''(x)>0$ everywhere, this is the location of an absolute minimum of $ h.$

JRav might want to redraw that graph and reconsider what he thought he saw in it.

On the interval $ [.5,1],$ $ h$ has a global (absolute) minimum at $ x\approx.56714.$ That minimum value is approximately $ 2.33037.$ It has an absolute maximum at $ x \equal{} 1$ of $ e\approx 2.71828.$ It's value at $ x \equal{} .5$ of approximately $ 2.34187$ is a local endpoint maximum but not a global maximum.

In this context, "absolute" and "global" are synonyms. American calculus books usually say "absolute."

Kent, but how did you measure the max absolute value in that interval?

I would stick in a calculator.  Or you could Taylor Expansion and try to get it as a polynomial and solve it that way.

"Stick it in a calculator" is close if not quite precise.

I got the root of $ h'(x)\equal{}0$ by using Newton's method. Since I was sitting at a computer, I used a spreadsheet - think of it as the big calculator.


pigeonhole principle
number theory unsolved
number theory

Let A be a set of 10 positive integers all less than 100. Show that there are two disjoint subsets of A with equal sums of members.

There are $ 2^{10}\minus{}1$ subsets of A, which can assume its sum of elements from $ 1$ to $ 90\plus{}...\plus{}99\equal{}945$.So there will be two subsets with equal sum, because $ 945$ is less than $ 2^{10}\minus{}1\equal{}1023$. And now, with these two subsets, take out their intersection of elements, that we'll have two disjoint subsets with equal sum.


A standard clock (not digital) has a second hand, minute hand, hour hand. The clock does not tick- it runs continuously. Even when the second hand moves 1/4 of a second, the minute hand moves 1/240, and the hour hand moves 1/14400.

Give that, can all the hands be exactly 120 degrees from each other?

I dont think so, if we set $\theta$ equal to the angle the hour hand makes with the verticle (from 12) then the minute hand is $60 \theta-360m$ which must also be equal to $\theta +- 120$ Doing the same for the second hand and playing round for a while we get a contradiction....

Seamus, you want to use "\theta," not "\theata."


Let $ S$ be the set of permutations of the sequence $ 1, 2, 3, 4, 5$ for which the first term is not $ 1$. A permutation is chosen randomly from $ S$. The probability that the second term is $ 2$, in lowest terms, is $ a/b$. What is $ a \plus{} b$?

$ \textbf{(A)}\ 5 \qquad
\textbf{(B)}\ 6 \qquad
\textbf{(C)}\ 11 \qquad
\textbf{(D)}\ 16 \qquad
\textbf{(E)}\ 19$

[hide]There is a 1/4 chance that the first number is 2, and a 3/4 chance that the first number is not 2
Given that the first number is not 2, there is a 1/4 chance that the next number is 2.
The probability is thus (3/4)(1/4)=3/16, and the answer is [b](E) 19[/b].[/hide]

[hide]There are $ 5!$ total permutations of the set.  Of those, $ 4!$ start with $ 1$. Therefore, there are $ 4 \bullet 4!$ permutations starting without a 1. The number of ways we can have the second number 2 is $ 3 \bullet 1 \bullet 3!$.  Hence, the probability that the set is such is given by: $ \frac{18}{96}\equal{} \frac{3}{16}$.  The sum of the numerator is thus $ 3\plus{}16\equal{}19$, or $ \boxed E$[/hide]

[quote="AlanLiangTheAsian"][quote="123456789"][hide]If the first term, it can be one of 2, 3, 4, or 5, and there are $ 4!$ ways to arrange the other 4 digits, so there are $ 4 \cdot 4!$ elements in $ S$.

If we want the second term to be 2, then there are $ 3$ ways to choose a first digit, and $ 3!$ ways to arrange the other 3 digits, so there are $ 3 \cdot 3!$ elements in this case.

The final probability is
\[ \frac {3 \cdot 3!}{4 \cdot 4!} \equal{} \frac{3}{16} \rightarrow \boxed{\textbf{E}}.
\]
[/hide][/quote]

Why is the total possible elements 4 * 4! and not 5!? Isn't 5! the total possible permutations of 5 objects? Or is it because the total possibilities MUST include the requirement of 1 not being the first term?[/quote]

We have four possibilities for the first number in the sequence, every one except for 1, and then we have 4 possibilities for the second number, all five except the one we used for the first term, then 3, then 2, then 1.

We have 4x4x3x2x1=4x4!

5! does not account for the fact that 1 cannot be the first term.


Εστω μια συναρτηση $ f: [0, 1] \rightarrow R$ με συνεχη 1η παραγωγο και $ \int_{0}^1{f(x)dx} \equal{} 0$

Δειξτε οτι για καθε $ a \in (0, 1)$ ισχυει:

$ \left|\int_{0}^a{f(x)dx}\right| \leq \frac{1}{8}|\max_{x \in [0, 1]}{[f'(x)]}|$

Θα ηθελα (αν γινεται) μια λυση που να μπορει να γινει κατανοητη απο καποιον που δεν εχει περισσοτερες γνωσεις σε ολοκληρωματα απο αυτες του Λυκειου. Επισεις, θα ηθελα να μου πειτε αν τα βιβλια του μαθηματος Αναλυση 1 (η Απειροστικος Λογισμος 1) του 1ου εξαμηνου στα μαθηματικα τμηματα, παρεχουν στον αναγνωστη τις απαραιτητες γνωσεις ωστε να μπορεσει να ασχοληθει με δυσκολα προβληματα με ολοκληρωματα οπως το παραπανω. Ακομα, αν οι γνωσεις που παρεχει το μαθημα Αναλυση 1 δεν ειναι αρκετες για την επιλυση τετοιου ειδους προβληματων, θα ηθελα να μου προτεινετε καποιο καλο βιβλιο το οποιο να ασχολειτε με προβληματα οπως το παραπανω και να παρεχει στον αναγνωστη τις γνωσεις που απαιτουνται για να λυθει ενα προβλημα οπως το παραπανω.

Ευχαριστω

Δεν είναι δύσκολο... Draw a picture. Δε θέλει "εξειδικευμένες" γνώσεις.

δες το συνημμένο


Let $ ABC$ be a triangle. $ D$, $ E$ and $ F$ points on $ BC$, $ AC$ and $ AB$.  $ AD$, $ BE$ and $ CF$ concurr at a point $ P$ inside the triangle.

prove that:

i) $ \frac{AD}{PD}+\frac{BE}{PE}+\frac{CF}{PF}\ge9$

ii) $ \frac{AD}{AP}+\frac{BE}{BP}+\frac{CF}{CP}\ge\frac{9}{2}$

iii) $ \frac{PD}{AP}+\frac{PE}{BP}+\frac{FP}{CP}\ge\frac{3}{2}$

iv) $ \frac{\sin{A}}{\sin{2A}}+\frac{\sin{B}}{\sin{2B}}+\frac{\sin{C}}{\sin{2C}}\ge3$

Try not to use any of them to prove the other so we could get different and may be interesting solutions.

 :)

[quote="ElChapin"]i) $ \frac{AD}{PD}+\frac{BE}{PE}+\frac{CF}{PF}\ge9$[/quote][hide="Solution"]Let $ [XYZ]$ be the area of $ \triangle XYZ$. Then $ \frac{[BPC]}{[BAC]}=\frac{PD}{AD}$, $ \frac{[APC]}{[ABC]}=\frac{PE}{BE}$, and $ \frac{[APB]}{[ACB]}=\frac{PF}{CF}$.
But we know $ \frac{[BPC]}{[BAC]}+\frac{[APC]}{[ABC]}+\frac{[APB]}{[ACB]}=1$ because $ [BAC]=[ABC]=[ACB]$.
So $ \frac{PD}{AD}+\frac{PE}{BE}+\frac{PF}{CF}=1$.
We can now use Cauchy-Schwarz: $ \left(\frac{PD}{AD}+\frac{PE}{BE}+\frac{PF}{CF}\right)\left(\frac{AD}{PD}+\frac{BE}{PE}+\frac{CF}{PF}\right)\geq (1+1+1)^{2}$
So $ \frac{AD}{PD}+\frac{BE}{PE}+\frac{CF}{PF}\geq 9$.
(we can similarly use this idea in parts ii, by noticing that $ \frac{AP}{AD}+\frac{BP}{BE}+\frac{CP}{CF}=1-\frac{PD}{AD}+1-\frac{PE}{BE}+1-\frac{PF}{CF}=2$)[/hide]


ARML
geometry
geometric transformation
reflection

Mr. Sigur is building a new Georgia ARML website [url=http://paideiaschool.org/TeacherPages/Steve_Sigur/resources/GAARML/GEORGIAARML/Welcome.html]here[/url].

I know that people on the mailing list have already been notified of it, but there are probably people interested in ARML who are not (yet) on the mailing list.

There's a different URL now.  I edited my first post to reflect that.


Suppose that a blackbody spectrum is specified by Spectral Radiancy $ R_{T} (\nu) d\nu$ and Energy Density $ \rho_{T} (\nu) d\nu$ then show that
$ R_{T} (\nu) d\nu \equal{} \frac {c}{4}\cdot \rho_{T} (\nu) d\nu$


Points $ M$, $ N$ and $ P$ are chosen on sides $ AB$, $ BC$ and $ CA$ of triangle $ \triangle ABC$ such that $ AM\equal{}BN\equal{}CP$. If $ \angle ACM \equal{} \angle BAN \equal{} \angle CBP$ prove that $ \triangle ABC$ is equilateral.


algebra
polynomial
algorithm
algebra unsolved

Find all Polynomials $ P(x)$ satisfying $ P(x)^2-P(x^2) =2x^4$.

Clearly the degree of $ P(x)$ is $ \geq2$.

If $ P(x)\equal{}a_2x^2\plus{}a_1x\plus{}a_0$, then $ a_2^2\minus{}a_2\equal{}2$ (1), $ a_1a_2\equal{}a_1a_0\equal{}a_0^2\minus{}a_0\equal{}0$ (2), and $ a_1^2\plus{}2a_0a_2\minus{}a_1\equal{}0$ (3).  Clearly, either $ a_0\equal{}1$ or $ a_0\equal{}0$ for (2).  In the first case, $ a_1\equal{}0$ and $ a_2\equal{}0$ for (2) and (3), inconsistent with (1), so $ a_0\equal{}0$, hence $ a_1a_2\equal{}0$; if $ a_2\equal{}0$, again (1) is impossible, so $ a_1\equal{}1$, and $ a_2$ is one of the roots of $ x^2\minus{}x\minus{}2\equal{}(x\minus{}2)(x\plus{}1)$, for $ P(x)\equal{}2x^2$ and $ P(x)\equal{}\minus{}x^2$.

If $ P(x)$ is a monomial, then $ P(x)\equal{}a_nx^n$, reverting to the previous case for $ a_0\equal{}a_1\equal{}0$, since clearly $ n\equal{}2$.

If $ P(x)$ is not a monomial and its degree exceeds $ 2$, then it may be written as $ P(x)\equal{}a_nx^n\plus{}a_mx^m\plus{}Q(x)$ for $ n\geq3$ and $ a_na_m\neq0$, $ m<n$, and degree of $ Q(x)$ less than $ m$ ($ Q(x)$ possibly zero).  Insertion in the condition yields either $ 2a_na_m\equal{}0$ (impossible) or $ 2a_na_m\equal{}2$ with $ m\plus{}n\equal{}4$, yielding $ P(x)\equal{}a_3x^3\plus{}\frac{x}{a_3}$ for any nonzero real $ a_3$, or $ P(x)\equal{}a_4x^4\plus{}\frac{1}{a_4}$ for any nonzero real $ a_4$.

How about this generalized version: Find all polynomials $ P(x)$ such that $ [P(x)]^2 \minus{} P(x^2) \equal{} \alpha x^{n}$ ?

I 've found sort of [b] algorithm [/b] to solve this. But I haven't seen any pattern.

First things first, I noticed a stupid mistake in my previous solution, since once polynomials of the form $ P(x) \equal{} a_3x^3 \plus{} \frac {1}{a_3}$ and $ P(x) \equal{} a_4x^4 \plus{} \frac {1}{a_4}$ are found to be the only ones that COULD be a solution, plugging them into the equation yields conditions $ a_3^2 \minus{} a_3 \equal{} 0$ and $ a_4^2 \minus{} a_4 \equal{} 0$, and since $ a_3,a_4\neq0$, $ a_3 \equal{} a_4 \equal{} 1$ follows, or the polynomials posted before my solution include all possible answers.  My bad, I apologize for it... :blush:

\medskip

Now, for the generalized version, we need to consider two cases, whether $ n \equal{} 2m \plus{} 1$ is odd, in which case the solution would always need to be of the form $ P(x) \equal{} a_ m x^m \plus{} Q(x)$, with $ m > \frac {n}{2}$ and $ m$ exceeding the degree of $ Q(x)$, or whether $ n \equal{} 2m$ is even, in which case there could be an additional solution of the form $ P(x) \equal{} a_mx^m \plus{} Q(x)$ with degree of $ Q(x)$ less than $ m$.

In the second case, clearly the cross-product term between the highest degree term in $ Q(x)$ and $ a_mx^m$ needs to be zero, since it would appear in $ P(x)^2$ but not in $ P(x^2)$, or $ Q(x) \equal{} 0$, yielding condition $ a_m^2 \minus{} a_m \equal{} \alpha$, where real solutions would exist iff the discriminant $ 1 \plus{} 4\alpha$ is non-negative, ie, two solutions of this form would exist for any $ \alpha > \minus{} \frac {1}{4}$, where $ a_m \equal{} \frac {1\pm\sqrt {1 \plus{} 4\alpha}}{2}$, and exactly one solution $ P(x) \equal{} \frac {x^m}{2}$ would exist for $ \alpha \equal{} \minus{} \frac {1}{4}$.

In both cases, consider a solution that is at least a trinomial of degree larger than $ \frac {n}{2}$, ie $ P(x) \equal{} a_mx^m \plus{} Q(x)x^\ell \plus{} a_kx^k$, where $ m \minus{} \ell$ exceeds the degree of $ Q(x)$, $ m > \frac {n}{2}$, $ \ell > k\geq0$, $ a_m,a_k\neq0$ and $ Q(x)$ is nonzero.  Clearly, the cross-product terms arising in $ P(x)^2$ mixing $ a_mx^m$ and the highest degree term in $ Q(x)x^\ell$, and mixing $ a_kx^k$ and the lowest degree term in $ x^\ell Q(x)$, cannot be cancelled by any other term in $ P(x)^2$ or in $ P(x^2)$, or at least one of them must be zero, the other one being possibly $ \alpha x^n$.  But this is clearly impossible, since then either $ a_m \equal{} 0$, or $ a_k \equal{} 0$, or $ Q(x) \equal{} 0$.  Contradiction, hence solutions exist that are only monomials or binomials.  Solutions that are monomials are only possible when $ n$ is even, reverting to the previous case, and solutions that are binomials of the form $ P(x) \equal{} a_mx^m \plus{} a_kx^k$ produce $ (a_m^2 \minus{} a_m)x^{2m} \plus{} 2a_ma_kx^{m \plus{} k} \plus{} (a_k^2 \minus{} a_k)x^{2m} \equal{} \alpha x^n$ with $ a_ma_k\neq0$, or $ a_m^2 \minus{} a_m \equal{} a_k^2 \minus{} a_k \equal{} 0$, yielding $ a_k \equal{} a_m \equal{} 1$, and $ \frac {\alpha}{2} \equal{} a_ma_k \equal{} 1$, impossible unless $ \alpha \equal{} 2$.  Hence solutions of this form may exist only when $ \alpha \equal{} 2$, and all of them are of the form $ P(x) \equal{} x^m \plus{} x^{n \minus{} m}$.

In summary, if $ \alpha \equal{} 2$, solutions of the form $ P(x) \equal{} x^m \plus{} x^{n \minus{} m}$ exist for all $ n\geq m > \frac {n}{2}$.  There are no other solutions if $ n$ is odd, but there are additional solutions $ P(x) \equal{} 2x^\frac {n}{2}$ and $ P(x) \equal{} \minus{} x^\frac {n}{2}$ if $ n$ is even.  No other solutions exist for $ \alpha \equal{} 2$.  If $ \alpha\neq2$, solutions exist only if $ n$ is even, and they are of the form $ P(x) \equal{} a_mx^\frac {n}{2}$, where $ a_m$ is one of the roots of $ \rho^2 \minus{} \rho \minus{} \alpha \equal{} 0$.  Again, no solutions at all may exist if $ \alpha < \minus{} \frac {1}{4}$, exactly one solution $ P(x) \equal{} \frac {x^\frac {n}{2}}{2}$ exists for $ \alpha \equal{} \minus{} \frac {1}{4}$, and exactly two solutions exist for any $ \alpha > \minus{} \frac {1}{4}$, $ \alpha\neq2$, $ n$ even.

the above polynomial should be even or odd ,so if p(0)=0 then its odd otherwise its even


geometry
geometric transformation
homothety
vector
percent
similar triangles
3D geometry
\/closed

The Olympiad Problem Series: Geometry class will be held on Mondays from 7:30 PM - 9:30 PM EST (4:30 PM - 6:30 PM PST).  The class will run from February 16 through May 3.  Note that this class is 2 hours, instead of our usual 90 minutes.  The longer time is due to the breadth and depth of material we will cover in the course.  Topics will include similar triangles, cyclic quadrilaterals, power of a point, construction, locus, transformations, inversion, homothecy, 3D geometry, vectors, and a little more.  Most classes will start with a lecture on one of these topics, then continue with a series of problems relating to that area of geometry.  Numerous additional problems will be provided on the class message board.

This class is appropriate for anyone seeking a deep understanding of Euclidean geometry and a knowledge of those tools which are most useful for tackling Olympiad problems.  Not all problems will be Olympiad-level difficulty, as we will introduce concepts with more basic problems.  However, many problems both in class and on the message board will come from various national and international Olympiads.  The best suitability guide is probably your attitude towards AMC->12 geometry problems; if these are relatively easy for you, then you're ready for this class.  If these are still pretty challenging for you, then you should wait for our introductory geometry course which we hope to offer by this summer.

If you have any questions about the course, please post them here.

I would like to do this, but Im not sure if I am ready. Could you perhaps make some sort of pretest for it? Ive never actually taken an amc-12 (took the 10 last year, and have a problem book of 83-88 ashme's) and arent familiar with the difficulty level very much of those geometry problems.

Grab your AoPS Volume 1 - flip to the potpourri of geometry section.

If the following statements are true, then this class won't be too advanced for you:

1) Nearly all of the first 30 problems are very easy for you.

2) You can handle most of the last 10 problems, even if they aren't completely trivial for you.

If either of these statements isn't true, you should probably consider the introductory Geometry course we hope to offer in the spring/summer.

Just a question....  What is the percent of correct answer in the potpourri section one must gets for the course? For few problems in that chapter, I got them incorrect and although I can perfectly understand the approaches, I would not have thought of them. Would you recommend me the olympiad course or the introductory course for this matter?

would you have gotten the answer if you tried for a bit longer? although Mr. R said they should be fairly easy, but its just a few problems. oh well, just wait for what Mr. Rusczyk says

Of those first 30 problems, you should be able to do 25+ of them without thinking too hard.  For the last 10, you should be able to do at least 6-7 of them, though they don't need to be as obvious as the first 30.

It also depends on how willing you are to take a class in which some of the material requires you to ask many questions (either of yourself or of an instructor).  If you are the tenacious sort, then you should be comfortable taking the class.  Most of the ideas in the class will be initially presented with a few simple problems, then we'll dive into harder ones.  If you are the sort who, when starting to get lost, clams up and doesn't ask questions, then you'll want to wait until you are more certain of the fundamentals.  I obviously highly recommend training yourself to become the tenacious sort.  I was the latter when in high school - I'd know way more math now had I been the former, at least when I was at MOP.

Why is the price increased more for this class as opposed to the other classes in 2004?

2 hours per class compared to 1.5 i think

That's right - the class is 1/3 more than our other courses, which is the reason for the increase in price.

I cringe when people ask "price" of the AOPS classes. Today's mathematical education is showing a very clear intellectual class division line: the olympiad level elitism vs. "water downed"school math. A good amount of "in-between" bright kids are desperately looking for help to give a shot in the elite competitions. Yes, " We are all in the gutter, but some of us are looking at the stars.  "With this understanding in mind, I can only say "thanks" to Mr. R. and Mr. C., as far as price is concerned, I will work my rear off to pay for it without asking any questions. Yes, there is something money cannot buy, -----one is the "heart" of forming this AOPS community. I was looking at the stars one time, that was a long, long time ago. Back then, I wish I had this kind of class to join. Oh! how I wish.

I agree. And really, when you think about it, these courses are at an extremely reasonable price. I'm going to take an E.P.G.Y course this summer which will probably cost around $600...I understand it will be a more difficult course, but still, for most people AoPS courses are a great deal.

Just to clarify regarding the difficulty of EPGY's courses relative to ours - it really depends on which course of theirs you're comparing to ours.  I'd wager there aren't many (if any) EPGY courses that have problems as difficult as some of the ones we will dissect in this class.  (Not that all of the problems we tackle in this class are extremely hard, but we will have many very challenging problems.)  The EPGY course is likely more time-consuming than most of our courses, but I doubt it's more challenging.  Again, this depends on what you're comparing it to - I don't think we have a great deal of overlap with EPGY.

Ack, I meant more lengthy...I had a friend that took an EPGY course in 8th grade and it took up tons of time. I'm going to be taking the EPGY Pre-Cal course this summer because it's the only course that our school accepts for honors' credit. But I understand, yours are more difficult (and will probably be more enjoyable...;))


nice problem ☺
i'll try to figure it out, i wonder if there is a system to follow...

[quote="xpmath"][hide]161 squares?[/hide][/quote]

how did you get that?

Yup, made it myself for our local math league tryout and 0 out of 50 kids got it right.(of course I have seen enough problems that maybe I saw it somewhere else awhile ago and it came to me subconscientiously)

None of the given answers are right so far, although who knows maybe I am still wrong :)

[quote="frost13"]Ok, so how many squares are in this diagram?  It's my problem and it took me three attempts to get it right :blush:[/quote]

it's definitely wrong, but i got [hide]108[/hide]

jorian

edit:
[hide]
i got 91 regular
17 abstract  :P [/hide]

[hide]130 perhaps?[/hide]
I salute anyone brave enough to type up a full solution if they get it right.  ("But we have overcounted the 2 semi-semi large squares that are not quite in the upper-right corner, and the size 2 square that is 2 units away from the right side of the previous square, but the 3.5 by 3.5 square we have counted twice..."  *shudder*)

Ok, still no right answers..

I will say it is between 120 and 160 which eliminates most answers so far.

ok, my second guess is 126

hope it's correct

jorian

127? Somewhere in the 120s
Edit: Overcounted one, I think it is 126.

Sorry, but you are still undercounting.

frost

Hmm, this problem seems familiar.  :?:

[hide]I got 138 (Did I get it right this time?). If we imagine that the extra lines didn't exist, we'd have 91 squares.  Then consider the small one in the middle, there are 5 squares there.  For the next largest, we have 3X4+1=13 squares.  Finally, for the last, we have 5X4+1=21 squares.  Now we add.  91+8+5+13+21=138.  ( If I wanted, I could have just tried to find my copy and add the 8 I knew I missed, but I can't find it.)[/hide]

[hide]$91$ (original squares)
+
$4 \cdot (6+4+2)$ (added squares, when viewed from each corners)
-
$3 \cdot 3$ (the center ones of the added squares  - I overcounted it from all 4 corners)
+
$8$ (between the outermost added line and middle added line - you have 2 squares each sides)
=
$138$ is my 1st guess.[/hide]

kohjhsd wins!!!  alanchou gets honorable mention for getting it first, but since he's on my team and saw it already he had an advantage!

Frost

oh, ok. i missed the yellow and green in the top left, and two of the ones in the 2nd row to the right (the blank ones).

jorian


geometry
rectangle
algebra
system of equations

[quote="Treething"]Mr. Sam has a pet rectangle. It is twice as long as it is wide, and its area is 72 cubic centimeters. What is its length?[/quote]

PET RECTANGLE?! Anyways,
[hide]
$2w=l$

$w\cdot l=72 cm^3$

$w\cdot 2w=72$

$2w^2=72$

$w^2=36$

$w=6$

$l=2\cdot 6=12$

 :D [/hide]

[quote="Treething"]Mr. Sam has a pet rectangle. It is twice as long as it is wide, and its area is 72 cubic centimeters. What is its length?[/quote]

  I think you mean 72 centimetres squared.

[hide="w00t"]
$\text{i set up a system of equations}\\\\2w=l\\\\w \cdot l=72\\\\\text{plug in 2w for l in teh second equation}\\\\w \cdot 2w=72\\\\2w^2=72\\\\w^2=36\\\\w=6\\\\6 \cdot 2=12\\\\\boxed{w=6}\\\\\boxed{l=12}$[/hide]

[quote="Treething"]Mr. Sam has a pet rectangle. It is twice as long as it is wide, and its area is 72 cubic centimeters. What is its length?[/quote]

[hide]$x(2x)=72$

$2x^2=72$

$x^2=36$

$x=\pm6$

$x=6$

Length is 12.[/hide]

[hide]

Well, what I did was list all of the factors of 72:

1*72
2*36
3*24
4*18
6*12
8*9
The only factor pair that has one number double the other is 6*12

So the answer is $  \framebox{12}  $[/hide]


function
inequalities
induction
algebra proposed
algebra

should use induction
Note that we want to show that  f(x^n)^(n-1) =< [f(x) f(x^2)... f(x^n-1)] [ f(x)^(n-1) .... f(x^n-1)^ 1/(n-1)] = A . B
Now we can write the B as:

[f(x)] [f(x) f(x^2)^(1/2)] [ f(x) f(x^2)^1/2 f(x^3)^1/3] ... [f(x) f(x^2)^1/2 ... f(x^n-1)^1/n-1] which by induction hypothesis is greater than or equal to  f(x) f(x^2)...f(x^n-1) = A.

So A . B >=  A^2  = [f(x) f(x^n-1)]^2 [f(x^2)f(x^n-2)]^2 ... >= f(x^n)^ (n-1)

So the result follows.
pretty simple so i just came up with this. takes time to see what i did because of the bad write up.  ;)

[quote="silouan"]Let a function  $f: (0,+\infty)\rightarrow(0,+\infty)$   such that for every  $x,y>0$ we have that  $f(xy)\leq f(x)f(y).$

For each  $x>0$ and for each  $n\in\mathbb{N^*},$  prove that :
$f(x^n)\leq f(x)\cdot(f(x^2))^{\frac{1}{2}}\cdots(f(x^n))^{\frac{1}{n}}.$[/quote]

I don't understand.  Isn't it obvious if we take both sides to the power of $n!$?

Every thing is obvious!!!  :)

But to the extent i know ur obvious is not clear...

(Edited; ignore, misread an inequality sign.)

i don't see how that helps.
 u maybe making a mistake.

Oh.  I see.  The signs on the inequality don't work out  :blush:


For $n$ integer, $n\ge2$ prove that:

$1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}>n\left(\sqrt[n]{n+1}-1\right)$

(I have a solution)

Rewrite as

$1 + \frac{1 + \frac{1}{2} + ... + \frac{1}{n} }{n} > \sqrt[n]{n+1}$

Strongly suggesting AM-GM.  Indeed,

$\frac{n + 1 + ... + \frac{1}{n} }{n} = \frac{(1 + 1) + (1 + \frac{1}{2}) + ... + (1 + \frac{1}{n} ) }{n} = \frac{ \frac{2}{1} + \frac{3}{2} + ... + \frac{n+1}{n} }{n} > \sqrt[n]{n+1}$

(The equality case is impossible.  Clearly the terms are not equal.)  QED.


Prove that

\[ 2^{n}\equal{}\dbinom{n}{0}\plus{}\dbinom{n}{1}\plus{}...\plus{}\dbinom{n}{n\minus{}1}\plus{}\dbinom{n}{n}\]

[quote="13375P34K43V312"]consider the set (1,2,...,n), we want to find the number of subsets, there are 2^n because each element is either included or not included, but there are also the RHS many because we can look at 0 element subsets, 1 element subsets, etc.

alternatively consider the binomial expansion of (1+1)^n[/quote]

What is the RHS?

You can also use induction and $ \binom{n}{r}\equal{}\binom{n\minus{}1}{r}\plus{}\binom{n\minus{}1}{r\minus{}1}$.

[hide="Generating function"]
$ (1\plus{}x)^{n}\equal{}{n\choose 0}\plus{}{n\choose 1}x\plus{}...\plus{}{n\choose n}x^{n}$
Let $ x\equal{}1$.
[/hide]


$ ABCDEF$ is cyclic hexagon with $ BD$ be perpendicular to $ CF$ and $ AC$ parallel to $ DE$. Let $ BD$ and $ CF$ cut at $ G$, $ AC$ and $ BE$ at $ H$, and $ AD$ and $ CE$ at $ I$. Suppose further that $ CH \equal{} AH \plus{} DE$.

Prove that $ GH\cdot BD \equal{} BC\cdot DE$

Dear jbmorgan,
Download the following file to see a solution.