geometry perimeter incenter circumcircle AMC AIME search

1.Three parallel chords of a circle have lengths 2,3,4 and subtend angles x,y,x+y at the center (x+y <180). Find cos x

2.A hexgon is inscribed in a circle. Five sides have length 81, and the other side length 31.  Find the sum of the three diagonals form a vertex on the short side

3.The triangle ABC has AB=1995, bc=1993, CA=1994. CX is an altitude. Find the distance between the points at which the incircles of ACX and BCX touch CX

4. The incircle of ABC touches AB at P and has radius 21. If AP=23 and PB=27 find the perimeter of ABC

5. A circle radius r is inscribed in ABCD. It touches AM at P and CD at Q. AP=19, PB=26, CQ=37,QD=23. Find r.

6. A triangle has sides 20,21,22. The line through the incenter parallel to the shortest side meets the other two sides at X and Y. Find XY

7. The triangle ABC has AB=24. The median CE is extended to meet te circumcircle at F. CE=27, and the median AD=18. Find area of ABF

a couple of these problems have been posted before on this forum:

for example, heres 5

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=385895#p385895

EDIT: Actually, now that i think of it, these are like all AIME problems....use the search function or check Contests


logarithms
floor function
algebra proposed
algebra

Powers of $2$ in base $10$ start with $3$ or $4$ more frequently? What is their state in base $3$? First write down an exact form of the question.

You have to find whether the density of those $n$ for which $\left\{n\log_{10}2\right\}\in(\log_{10}3,\log_{10}4)$ (the necessary and sufficient condition for $2^n$ to begin with $3$) is larger than that of those $n$ for which $\left\{\log_{10}2\right\}\in(\log_{10}4,\log_{10}5)$ (the necessary and sufficient condition for $2^n$ to begin with $4$). Use Weyl's Theorem: the first density is $\log_{10}\frac 43$, while the second is $\log_{10}\frac 54$, smaller than the first. In conclusion, more powers of $2$ begin with $3$ than with $4$.

I'm not sure what the second question means though. In base $3$, which numbers begin with $3$ in base $3$? Maybe the question turns into: are there more numbers of the form $2^n$ that begin with $10$ or with $11$ in base $3$? Anyway, an analysis like that above can be performed.

[quote="grobber"]the density of those $n$ for which $\{\left\lfloor\log_{10}2\right\rfloor\}\in(\log_{10}3,\log_{10}4)$ (the necessary and sufficient condition for $2^n$ to begin with $3$)[/quote]
I think you mistyped something. The expression $\{\left\lfloor\log_{10}2\right\rfloor\}\in(\log_{10}3,\log_{10}4)$ doesn't make sense! (There is no $n$ in it and the left side is always $0$)


The following inequality is still easy.
Let $ a,$ $ b,$ $ c,$ and $ d$ are positive numbers such that $ a^3 \plus{} b^2 \plus{} c^3 \plus{} d^3 \equal{} 4.$ Prove that
\[ \frac {a^3}{b^2 \plus{} c^2} \plus{} \frac {b^3}{c^2 \plus{} d^2} \plus{} \frac {c^3}{d^2 \plus{} a^2} \plus{} \frac {d^3}{a^2 \plus{} b^2} \ge\ 2
\]
What about the following? :)
Let $ a,$ $ b,$ $ c,$ and $ d$ are positive numbers such that $ a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4 \equal{} 4.$ Prove that
\[ \frac {a^3}{b^2 \plus{} c^2} \plus{} \frac {b^3}{c^2 \plus{} d^2} \plus{} \frac {c^3}{d^2 \plus{} a^2} \plus{} \frac {d^3}{a^2 \plus{} b^2} \ge\ 2
\]

OK. But with $ a^4\plus{}b^4\plus{}c^4\plus{}d^4\equal{}4$ it's still unsolving.

[quote="arqady"]OK. But with $ a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4 \equal{} 4$ it's still unsolving.[/quote]
 why？i  think the proof is right   :maybe:

[quote="arqady"][quote="ye109"]
and $ \sum_{cyc}{a^6(b^2 \plus{} c^2)}\leq\ \frac {1}{2}(\sum{a^4})^2$
[/quote]
Why it's true?[/quote]
 :wink: i find it's wrong,i think another proof.


In triangle $ABC$, prove that: \[\cot\frac{A}{2}+\cot\frac{B}{2}+\cot\frac{C}{2}\ge 3\left(\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}\right)\]

[hide]we have $\tan{\frac{A}{2}}=\frac{2r}{b+c-a}$ and so on.
So the inequality becomes
$\frac{b+c-a}{2r}+\frac{a+c-b}{2r}+\frac{a+b-c}{2r}\geq3\left(\frac{2r}{b+c-a}+\frac{2r}{a+c-b}+\frac{2r}{a+b-c}\right)$
$\frac{a+b+c}{12r^{2}}\geq\frac{1}{b+c-a}+\frac{1}{a+c-b}+\frac{1}{a+b-c}$

but we have $r=\sqrt{\frac{(p-a)(p-b)(p-c)}{p}}$,so the inequality becomes
$\frac{(a+b+c)^{2}}{3(a+b-c)(a+c-b)(b+c-a)}\geq\frac{1}{b+c-a}+\frac{1}{a+c-b}+\frac{1}{a+b-c}$

and with the ravi substitution
$b+c-a=x$
$a+c-b=y$
$a+b-c=z$
it becomes

$\frac{(x+y+z)^{2}}{3xyz}\geq\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$
$(x+y+z)^{2}\geq3(xy+yz+zx)$
$x^{2}+y^{2}+z^{2}\geq xy+yz+zx$
that is the simpler of the inequality :lol: [/hide]

bye

[hide="alternatively you could"]
substitute $x=\tan \frac{A}{2}$ etc. so $x,y,z$ positive. Then use trig addition to prove $xy+yz+zx=1$.
[/hide]

Wow yeah, since I stink at trig, this is what I think he's saying.

[hide]Since for triangle $ABC$, $\cot{A/2}\cot{B/2}\cot{C/2}= \cot{A/2}+\cot{B/2}+\cot{C/2}$, we can see that for our equation, equating
$tan{A/2}= x$...etc as Scorpius said,

we have $\frac{1}{xyz}= \frac{1}{x}+\frac{1}{y}+\frac{1}{z}= \frac{xy+yz+zx}{xyz}$...

Edit: I think this works.

Edit: Sigh... does not work... what's the next step?[/hide]

[hide="finishing the solution"]
So after the substitution, the inequality becomes
\[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq 3(x+y+z)\]
which simplifies to
\[(xy+yz+zx)^{2}\geq 3xyz(x+y+z)\]
\[\iff x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}\geq x^{2}yz+xy^{2}z+xyz^{2}\]
This is true just by AM-GM's such as:
\[\frac{x^{2}y^{2}+x^{2}z^{2}}{2}\geq x^{2}yz\]
[/hide]


When placing each of the digits 2, 4, 6, 8 and 9 in exactly one
of the boxes in the addition problem shown, what is the greatest
possible sum?

[asy]size(100);
path box = (0,0)--(1,0)--(1,1)--(0,1)--cycle; real shifter = 1 + .2; draw(box ^^ shift(shifter*right)*box ^^ shift(2shifter*right)*box ^^ shift(shifter*down + shifter*right)*box ^^ shift(shifter*down + 2shifter*right)*box); label("$+$",shift(shifter*down + shifter*left)*(.5,.5));
draw(shift(1.15shifter*down)*((-1,0)--(3shifter,0)));[/asy]

We want to have 9xx added to 8x, in order to have the biggest sum.

$ 942 \plus{} 86 \equal{} 1010$

Let's see if we can get anything bigger.

$ 962 \plus{} 84 \equal{} \boxed{1046}$

We want the largest numbers in the largest places, so we have a 9 in the hundreds place.  We have the 8 and the 6 in the tens place, and 4 and 2 in the ones place.  Adding yields 9*100+14*10+6=1046.

Also, the diagram doesn't show up in the game.


Here's a good problem I wanna share with you;
Let $S$ be the set of real numbers, $x\in [0,1]$, such that for all $\epsilon>0$ there exists some rational number $\frac{p}{q}\ \text{st}\ |x-\frac{p}{q}|<\frac{\epsilon}{q^{3}}$. Prove that $S$ is infinite and that it does not contain any interval. What more can you say about $S$?

Let $x_{n}= \sum_{k=1}^{n}a^{-b^{k}}$ for $a\geq 2$ and $b \geq 4$ and $x$ be the limit of $(x_{n})$. Then

\begin{eqnarray*}x-x_{n}& = & a^{-b^{n+1}}+a^{-b^{n+2}}+a^{-b^{n+3}}+\cdots\\ & \leq & a^{-b^{n+1}}+a^{-b^{n+1}(1+(b-1))}+a^{-b^{n+1}(1+2(b-1))}+\cdots\\ & = & a^{-b^{n+1}}\sum_{k = 1}^{\infty}\left( a^{-b^{n+1}(b-1)}\right)^{k-1}\\ & = & a^{-b^{n+1}}\frac{1}{1-a^{-b^{n+1}(b-1)}}\\ & < & 2\cdot a^{-b^{n+1}}\\ & = & 2\cdot \left( a^{-b^{n}}\right)^{3}a^{-(b-3)b^{n}}\\ \end{eqnarray*}

And for any $\epsilon$, we can choose $n$ such that $2\cdot a^{-(b-3)b^{n}}< \epsilon$. So reversing this process would show that $x$ is in $S$. But there are infinitely many choice of $(a, b)$, $S$ is infinite.

You can use Liouville's theorem to show that $\frac{r}{\sqrt{2}}$ for nonzero rational $r$ is always not in $S$, so for any two distinct $x, y \in S$, we can always find a number $z\notin S$ that lies between $x$ and $y$. So any interval (except for an empty set or a singleton) cannot be a subset of $S$.

If you don't require that $|x-\frac{p}{q}| > 0$ then $S$ is infinite just because it contains all rational numbers (between 0 and 1).  :)

Now I will give an alternative proof that $S$ doesn't contain any interval. It doesn't use Liouville's theorem.
Let us compute the measure of the set $S$. The set $S$ is covered by intervals $(\frac{p}{q}-\frac\varepsilon{q^{3}}, \frac{p}{q}+\frac\varepsilon{q^{3}})$. The total measure of all intervals is
\[\sum_{q=1}^\infty \sum_{p=0}^{q}\frac{2\varepsilon}{q^{3}}= 2\varepsilon \cdot \sum_{q=1}^\infty \frac{q+1}{q^{3}}\leq C \varepsilon. \]
Since $\varepsilon$ can be arbitrarily small, $S$ has measure $0$. Therefore, $S$ does not contain any interval.


integration
real analysis
calculus
algebra
function
domain
trigonometry

Friends....
This question may be silly for many of you.
We define $ X$ to be integrable if $ E|X|$ is finite.
My question is, does $ E(X)$ finite imply $ X$ is integrable?  Because I think $ E(X)$ could be finite when  $ E|X|$ is infinite.

NO: $ E[X]$ finite does not imply $ E[|X|] < \infty$. Try to find a counter example when $ X$ oscilate a lot.

But for my question in Analysis forum, Professor Kent Merryfield gave the following reply. That time I accepted, but now I am realizing.

The hypothesis was that $ \int_Xf\ d\mu$ is finite. This only makes sense as a Lebesgue integral if $ \int_X|f|\ d\mu$ is finite.

And I'm sticking with that answer.

alekk, the problem with what you just said is that "oscillate" implies that you're taking the domain in some particular order. This can be true for integrals over $ \mathbb{R}.$ For instance, we can say that $ \int_0^{\infty}\frac{\sin x}{x}\,dx$ converges as an improper integral despite not being absolutely integrable - but we can [i]only[/i] say that because we've agreed to take the real line in a particular order - and that's because $ \mathbb{R}$ (or $ [0,\infty)$) has some structure [i]other[/i] that that of a measure space.

What about a probability sample space? The basic issue is that the [i]only[/i] structure such a space has is that of a measure space - and in a measure space, and in the definition of the Lebesgue integral, we must be free to select and rearrange measurable subsets in any order whatsoever. Hence the definition of the Lebesgue integral - and the Lebesgue integral is the only definition we have that makes any sense for what $ E(X)$ is - must be insensitive to rearrangements, from which we arrive at the requirement for absolute integrability.

Example: Suppose $ U$ is uniformly distributed on $ \left(\minus{}\frac{\pi}2,\frac{\pi}2\right)$ and let $ X\equal{}\tan U.$ This gives a reasonably well-known distribution on the line. We can say that $ X$ has a probability density function of $ f(x)\equal{}\frac{1}{\pi(1\plus{}x^2)}$ on $ \mathbb{R}.$ Despite the symmetry of this distribution, despite the fact that $ 0$ is clearly the median (and mode) of this distribution, we must say that $ E(X)$ is undefined.

Yes sir, you are right.  We define $ EX\equal{}EX^\plus{}\minus{}EX^\minus{}$. So if $ EX$ is finite then both $ EX^\plus{}$ and $ EX^\minus{}$ have to be finite and hence $ E|X|\equal{}EX^\plus{}\plus{}EX^\minus{}$ is also finite. Am I right sir?


calculus
derivative
logarithms
function
integration
calculus computations

If $ y \equal{} x^{\ln x}$, then $ y'$ is _____.

I got $ \frac {x^{\ln x} \ln x}{x}$, but the answer key says $ \frac {2x^{\ln x} \ln x}{x}$.  Could someone please explain what I did wrong?  Thanks in advance!

check again, actually your mistake is quite obvious, your derivative for (ln x)^2 is wrong...

[quote="JBL"][quote="314.math"]Could someone please explain what I did wrong?[/quote]  No, because you only showed what answer you got, not how you got it.  Show your workings, then we can discuss what's wrong with them.  (I personally would have approached this problem by rewriting $ x^{\ln x} \equal{} e^{(\ln x)^2}$ before taking the derivative, but probably you tried to use some formula or other instead.)[/quote]

Here's what I used:

$ \frac {d}{dx}a^u \equal{} a^u \ln a \frac {du}{dx} \Rightarrow \frac {d}{dx} \equal{} x^{\ln x} \cdot \ln x \cdot \frac {1}{x}$

hello, writing your term in the form
$ y\equal{}e^{x\ln(x)}$ we get for the first derivative
$ y'\equal{}e^{x\ln(x)}(\ln(x)\plus{}1)$
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, writing your term in the form
$ y \equal{} e^{x\ln(x)}$ we get for the first derivative
$ y' \equal{} e^{x\ln(x)}(\ln(x) \plus{} 1)$
Sonnhard.[/quote]

I don't get how you rewrote my original problem.  Where did the $ e$ come from?  Also, what is incorrect about my method?

Thanks!

hello, sorry, i have done a mistake, lets make it clear
$ y\equal{}x^{\ln(x)}$ taking the logarithm of both sides we get
$ \ln(y)\equal{}\ln(x)\ln(x)\equal{}(\ln(x))^2$
or
$ y\equal{}e^{(\ln(x))^2}$
from here we get
$ y'\equal{}e^{(\ln(x))^2}(2\ln(x)\cdot1/x)$
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, sorry, i have done a mistake, lets make it clear
$ y \equal{} x^{\ln(x)}$ taking the logarithm of both sides we get
$ \ln(y) \equal{} \ln(x)\ln(x) \equal{} (\ln(x))^2$
or
$ y \equal{} e^{(\ln(x))^2}$
from here we get
$ y' \equal{} e^{(\ln(x))^2}(2\ln(x)\cdot1/x)$
Sonnhard.[/quote]

Ok, I follow what you did, but why is my way wrong?  What is the motivation to take the ln of both sides?

[quote="314.math"]Here's what I used:

$ \frac {d}{dx}a^u \equal{} a^u \ln a \frac {du}{dx} \Rightarrow \frac {d}{dx} \equal{} x^{\ln x} \cdot \ln x \cdot \frac {1}{x}$[/quote]  Now, go look in your textbook or notes and find under what assumptions the rule $ \frac {d}{dx} a^x \equal{} a^x \ln a$ is valid.  (Hint: there is some restriction on $ a$.)

(I really like the way this rule applies to the function $ g(x) \equal{} x^2$, with $ a \equal{} x$ and $ u \equal{} 2$: we have $ \frac {d}{dx} x^2 \equal{} x^2 \ln x \cdot \frac {du}{dx} \equal{} 0$.   :wink:)

General principle: when you remember a theorem (one special case is a computational rule of this sort), you must remember not only the [i]conclusion[/i] of the theorem but also the [i]conditions under which the conclusion is valid[/i].  In this case, you are applying the theorem in a circumstance in which it is not valid, so the fact that  you get a wrong answer is not terribly surprising.

[color=green][Moderator says: please do not quote the preceding post for no good reason.][/color]

Hmmm... My sheet only says that $ \frac {d}{dx} (a^x) \equal{} a^x \ln a$.  The only restriction that I'm aware of is that for $ \int u^n du \equal{} \frac {u^{n \plus{} 1}}{n \plus{} 2} \plus{} C, n\neq \minus{} 1$.  Can you please tell me the restriction?  I know one exists, since $ \frac {d}{dx} x^2 \equal{} 2x$, not 0. :)

What JBL is trying to say it that $ \frac{d}{dx}a^x \equal{} a^x \ln a$ only if $ a$ is constant and does not depend on $ x$. Since this is not your case, you cannot use this formula. What you are dealing with is a non-standard function (non-standard is the sense that you can't find it in most derivation tables) so you need to use a somewhat more complicated method than simply reading it off from a table. In the same sense you could say that since $ \frac{d}{dx} x^n \equal{} n x^{n\minus{}1}$ then $ \frac{d}{dx}x^{\ln x} \equal{} \ln x \, x ^{\ln x \minus{} 1}$. This is, of course, wrong for the same reason - the formula holds only if $ n$ is constant / does not depend on $ x$.

[quote="milin"]What JBL is trying to say ...[/quote]  Well, yes and no: I was trying not to say it but instead to get 314.math to discover it for him or herself.   :)

I'm aware of that and would not have saved him the trouble had he not asked for it as directly as he did. :)


modular arithmetic
number theory
greatest common divisor
number theory unsolved

Find all integer solutions to $2 x^4 + 1 = y^2.$

For $|y|<2$ the solutions $(x,y)$ are $(0,-1),(0,1)$.
Let's assume $y>1$, the equation is equivalent to $(y+1)(y-1)=2x^4$. Because $y+1,y-1$ are coprime for odd divisors, then there exist such odd, coprime numbers $u,v$ that $x=2^tuv$ and $y-1=2^mu^4$ , $y+1=2^nv^4$, where $m,n$ are postive integers and $t$ is a non-negative positive integer.
Let's note that $m+n = 4t+1= 1$ $(\mod 4)$.
So we have $2^nv^4-2^mu^4=2$, so $min\{m,n\}=1$ and $max\{m,n\}>min\{m,n\}$:
1) $m=1$ thus $2^{n-1}v^4-u^4=1$ ($n>1$), but because $m+n = 1$ $(\mod 4)$ so $n-1=3$ $(\mod 4)$,so $n \geq 3$. However, $u^4+1=2$ $(\mod 4)$, so we have a contradiction because $8|2^{n-1}$ so $2^{n-1}=0$ $(\mod 4)$.
2)$n=1$ thus $v^4-2^{m-1}u^4=1$ ($n>1$), so $(v^2-1)(v^2+1)=2^{m-1}u^4$. Because $v^2+1,v^2-1$ ($v>1$) are coprime for odd divisors, then there exist such odd, coprime numbers $a,b$ that $u=ab$ and $v^2-1=2^{m-2}a^4$ , $v^2+1=2b^4$. Thus
$(v+2^{\frac{m-2}{2}}a^2)(v-2^{\frac{m-2}{2}}a^2)=1$ ( because $m+n=1$ $(\mod 4)$ and $n=1$ then $m-2$ is even), but that is impossible.
1) and 2) haven't given any solutions, so there aren't any solutions for $y>1$ and also for $y<-1$- the equality is symmetric, so the only solutions are: $(0,1),(0,-1)$.

if we don't consider trivial answers,then
no solution,
we know that $ x$ is even so let's replace $ x$ by $ 2x$ and we get,
$ 32x^4 \equal{} (y \minus{} 1)(y \plus{} 1)$
so we have 2 cases :
1-$ y \equiv \minus{} 1 \pmod {4}$ which implies
$ y \minus{} 1 \equal{} 2t^4,y \plus{} 1 \equal{} 16s^4,(t,s) \equal{} 1,ts \equal{} x$
so we get
$ 16s^4 \minus{} 2t^4 \equal{} 2 \implies 8s^4 \equal{} t^4 \plus{} 1$
but we obviously have
$ t^4 \plus{} 1 \equiv 2 \pmod {8}$.
so this case is ommitted.
2-$ y \equiv 1 \pmod {4}$ which implies
$ y \minus{} 1 \equal{} 16s^4,y \plus{} 1 \equal{} 2t^4,(t,s) \equal{} 1,ts \equal{} x$
so we get
$ 2t^4 \minus{} 16s^4 \equal{} 2 \implies t^4 \minus{} 8s^4 \equal{} 1$.
or in other words,
$ t^4 \minus{} 1^4 \equal{} 2(2s^2)^2$.
u can easily prove that the equation $ x^4 \minus{} y^4 \equal{} 2z^2$ has no integer solutions(by infinite descent).
so we r done!

[quote="skytin"]x^4 = 1 mod 5
2*x^4+1 = 3 mod 5
no y that y^2 = 3 mod 5[/quote]
what if $ 5 | x$ ?

We may first assume $ x,y \geq 0$.

$ 2x^{4}\equal{}y^{2}\minus{}1\equal{}(y\plus{}1)(y\minus{}1)$.
$ 2|(y\plus{}1)(y\minus{}1) \Rightarrow 4|(y\plus{}1)(y\minus{}1) \Rightarrow 2|x$.
Let $ x\equal{}2x_{1}$, then $ 32x_{1}^{4}\equal{}(y\plus{}1)(y\minus{}1) \Rightarrow y\equal{}16y_{1}\plus{}1$ or $ 16y_{1}\minus{}1$.
Then $ x_{1}^{4}\equal{}y_{1}(8y_{1}\plus{}1)$ or $ y_{1}(8y_{1}\minus{}1)$.
But since $ gcd(y_{1},8y_{1}\minus{}1)\equal{}gcd(y_{1},8y_{1}\plus{}1)\equal{}1$, both $ y_{1}$ and $ 8y_{1}\plus{}1$ (or $ \minus{}1$) must be $ 4^{th}$ powers.
But $ 8y_{1}\minus{}1$ cannot be a $ 4^{th}$ power, so we can only have $ x_{1}^{4}\equal{}y_{1}(8y_{1}\plus{}1)$ (i.e. $ y\equal{}16y_{1}\plus{}1$).

Note that $ y_{1},8y_{1}\plus{}1$ are still $ 4^{th}$ powers, so let $ y_{1}\equal{}a^{4}$, $ 8y_{1}\plus{}1\equal{}b^{4}$.
Then $ b^{4}\minus{}1\equal{}(b^{2}\plus{}1)(b\plus{}1)(b\minus{}1)\equal{}8a^{4}$.

Clearly b is odd.
Consider then the integers $ \frac{b^{2}\plus{}1}{2},\frac{b\plus{}1}{2},\frac{b\minus{}1}{2}$.
Taken pairwise, each gcd is 1 ($ \frac{1}{2}(b\plus{}1)(b\minus{}1)\equal{}\frac{b^2\minus{}1}{2}\equal{}\frac{b^2\plus{}1}{2}\minus{}1$).
However, their product = $ a^{4}$, so each of them must be a $ 4^{th}$ power.
In particular, $ \frac{b\plus{}1}{2},\frac{b\minus{}1}{2}$, 2 consecutive integers, must both be $ 4^{th}$ powers, possible only when $ \frac{b\minus{}1}{2}\equal{}0, \frac{b\plus{}1}{2}\equal{}1 \Rightarrow b\equal{}1 \Rightarrow y\equal{}16y_{1}\plus{}1\equal{}2(8y_{1}\plus{}1)\minus{}1\equal{}2b^{4}\minus{}1\equal{}1$, and $ x\equal{}0$.

$ \therefore (x,y)\equal{}(0,1),(0,\minus{}1)$ are the only solutions in integers.

im not sure if my ideas are correct cuz i wasnt sure how to proceed

but this is what i had:

$ 2x^4\equal{}(y\minus{}1)(y\plus{}1)$

we take this mod 4 and find that $ x$ is even and $ y\equal{}4k\plus{}1$ or $ y\equal{}4k\plus{}3$

with $ y\equal{}4k\plus{}3$, we find a contradiction. so, $ y\equal{}4k\plus{}1$

let $ x\equal{}2a$

$ 32a^4\plus{}1\equal{}(4k\plus{}1)^2$

then we have
$ 8a^4\equal{}(2k)(2k\plus{}1)$

i think we could use infinite descent as mentioned before to show that either a or k must be 0 but im not sure how to.  does anyone know how?

Here is my solution

I will assume $y$ is positive.
$2x^4+1=y^2\implies 2x^4=(y-1)(y+1)$, so replace $y-1$ with $2z$. This gives $2x^4=4z(z+1)$, and again we will replace $x$ with $2x_1$ to get $8x_1^4=z(z+1)$. Now $\gcd(z,z+1)=1$, hence either $z=8a^4$ and $z+1=b^4$ or that $z=a^4$ and $z+1=8b^4$  for some integers $a,b$ (since $y$ is positive- we can ignore the case $z=-8a^4$ and $z+1=-b^4$ etc).

[b]Case 1[/b]
$z=8a^4$ and $z+1=b^4$
Then $8a^4+1=b^4$ which rearranges to $8a^4=(b-1)(b+1)(b^2+1)$. Since $b$ must be odd, substitute $b-1=2c\implies b+1=2(c+1)$ and $b^2+1=2(2c^2+2c+1)$. Then $a^4=c(c+1)(2c^2+2c+1)$.

$\gcd(c,c+1)=1$
$\gcd(c, 2c^2+2c+1)=\gcd(c,(c+1)^2+c)=\gcd(c,(c+1)^2)=1$
$\gcd(c+1, 2c^2+2c+1)=\gcd(c+1,(c+1)^2+c)=\gcd(c+1,c)=1$

Hence all of $c$, $c+1$ and $2c^2+2c+1$ are fourth powers, i.e. $c$ and $c+1$ are consecutive squares. This implies $c=0\implies a=0 \implies z=0 \implies y=1$. This gives the solution $(0,1)$.

[b]Case 2[/b]
$z=a^4$ and $z+1=8b^4$
Then $a^4+1=8b^4$. Looking modulo $5$, the LHS is one of $1,2$ while the RHS is one of $0,3$, a contradiction.

Thus the only solution is $(0,1)$, however since we assumed $y$ is positive, there is also the solution $(0,-1)$.

How can one miss this solution?
Consider the pell's equation $y^2-2z^2=1$ . We need to find when $z$ is a perfect square. Note that $(y,z)=(3,2)$ is a solution and we have all solutions from this.
 $Y+\sqrt2 Z=(3+2\sqrt2)^N=(1+\sqrt2)^{2N}$ this gives $Z= 2^{N-1}+2^{N-2}+....+2+1=2^N-1$
and $2^N-1$ is a perfect square only when Z=0,1 so the solution are $(0,1) (0-1)$


geometry
3D geometry
ratio
rectangle
probability

I just did this sprint yesterday, so these answers are fresh in my mind...
8. For these problems, I always start by looking for a pattern. There are 64 unit cubes, so you get 64 there. Notice that the unit cubes are 1*1*1 and 64 is 4*4*4. There is also 1 4*4*4 cube, and there are 8 (2*2*2) 3*3*3 cubes. If you notice this pattern, you realize that you have to just sum up all of the cubes less than 4^3, and that gives you 100. THe reason this works is because the numbers corresponds to what I call its opposite cube.. at least for this figure. $64+27+8+1=100$ This problem is easier to do than to explain...

13. In this problem, the width of the paper is the one being folded, so I drew a diagram of an unfolded and folded paper. I labeled the width of the folded paper w, the length l, and the width of the unfolded paper 2w. I then set up the proportion $l/w=2w/l$ By cross multiplying, I got $2w^{2}=l^2$}. By finding the roots of these two, I got $w\sqrt{2}=l$. Which therefore gave me the ratio of $w/l = 1/\sqrt{2}$ (as in the equation, if the width was 1, then the width was sqrt(2). By simplifying that, you get the answer $\sqrt{2}/2$. Again, this problem sounds much more complex than it really is, as it is just a simple algebra problem.

21. For the last problem, you just multiplied wrong. 5/6*1/6*2 is 5/18 not 5/6... the way I did it is list all ten combinations (or just think them out) and put the denominator as 36 which also simplifies to 5/18.

[quote="SoccerBrainy40"]Wow, thank you! I totally understand the first one now, and silly me on the third one.

There's just one thing I don't understand on the second one...my friend was trying to explain it, and I don't completely understand him, neither.

But my question is, why is the porportion l/w = 2w/l instead of l/w = l/2w? I thought that since in the first ratio, you were comparing the length to the width, you had to do the same in the second ratio.

Thanks again! :)[/quote]

No prob... On th second one, you have to remember that the question states that the rectangles are similar, and if you fold the paper, then the [b]width gets smaller[/b], and the only way the rectangle can remain similar is if the length is proportional to the width in the same way the width was proportional to the length in the unfolded paper. Also, if you cross-multiply l/w=2l/w, then you get 2lw=lw, which will get you nowhere. Sprint problems are (usually) pretty straightforward... I think this and number 24 are the hardest if you don't know how to solve them.

The only one I can grasp is the first one, where cubes are same as squares. E.G. if you have an 8x8 checkerboard you will have 64+49+36+25 etc. squares, so cubes its the same.

[quote="SoccerBrainy40"]Hi everyone, I just practiced with the 2001 State Sprint Round, and there's a few I'm struggling with.

8. Sixty-four unit cubes are placed together to create a large cube. How many cubes with integer dimensions are in the 4x4x4 cube?

Answer: 100

I'm trying to figure out a way to do this that will work with all problems similar to this...I tried dividing 64 by 1, 8, 27, and 64 (four types of cubes that will fit), getting rid of the remainder, and adding the results together, but that came out to be 75.

13. What is the ratio of the width to length of a floded sheet of peper if the rectangle formed when the sheet is folded in half as shown is similar to the original rectangular shset? Express you answer as a common fraction in simplest radical form.

Answer: (:sqrt:2)/2

No idea how that came to be...

21. A standard die is tossed twice. What is the probability of obtaining exactly one 5? Express you answer as a common fraction.

Answer: 5/18

I got 5/6 by doing 1/6 (probability of getting a 5 on the first roll) times 5/6 (probability of gettin anything but 5 on the second roll). Then, I multiplied that by 2 because it could also come out vice versa (getting anything but a 5 and then getting a 5).

There are a few more, but I'll post them later. Can I just get help on these three for now? Thank you to whoever responds!  :)[/quote]

8.
find a pattern

1X1X1=1=1^3
2X2X2=9=1^3+2^3
3X3X3=36=1^3+2^3+3^3
4X4X4=100=1^3+2^3+3^3+4^3

so the generic formula for this is $1^3+2^3+3^3...n^3$

13.
I need the picture..I dont know what it looks like :blush:

21.
one case is rolling 5 on the first and then rolling anything other than a 5 the second time
another is rolling anything but a 5 and then rolling a 5

the probability of them happening for both is 5/36

5/36+5/36=10/36=5/18


Where did you get this problem?  It doesn't seem to have a nice answer.

$\frac{\tan {115}+\tan {55}+\tan {15}}{\tan {75} + \tan {70} + \tan {35}}$

The denominator equals $\tan 75\tan 70\tan 35$...if that helps any.

In other words, if $\alpha +\beta +\gamma =180^{\circ}$, then $\tan\alpha +\tan\beta +\tan\gamma=\tan\alpha\tan\beta\tan\gamma$.

Masoud Zargar


counting
derangement
combinatorics unsolved
combinatorics

How many biyectios from {1, 2, 3, 4...n } to {1, 2, 3, 4...n } are that follow that f( x) is different from x... for any value of x?

If you mean the number of bijections from $ \{1, 2, \ldots, n\}$ that have no fixed points, I believe that's just the number of [url=http://en.wikipedia.org/wiki/Derangement]derangements[/url]...


Two months ago, Javier Heredia found himself in a different country without any of the local money in his wallet. Not a penny, and he had some big purchases to make.

Heredia arranged to meet a currency broker at a busy city landmark, and they agreed that Heredia would pay $130 for the new money. As they met, sketchy characters with weapons gathered nearby. The broker handed Heredia the cash without problems.
But the smooth transaction didn't happen here on Earth - it took place in the fictional city of Freeport on the make-believe planet of Norrath, inside the Internet video game EverQuest. Heredia wasn't even himself at time; he was Gretk, a 5-foot-tall talking rat.
In fact, the only thing that was real was the money.
Video games are nothing new, but the latest generation of games, so-called MMORPGs, or massively multiplayer online role-playing games, has spawned an interesting new wrinkle in electronic entertainment: the virtual economy.
At auction houses and on eBay, players can use real money from the real world to buy game money, implements and weapons of their game's universe. Brokers take orders for fictional currency, make-believe swords, even entire characters (a Ranger Wood Elf male with very nice gear for $269).

The practice, while increasingly common as online games grow in popularity, raises interesting questions about intellectual property rights - shouldn't gamemakers get a percentage since they invented all this stuff? - and whether gamers like Heredia are being true to the rules of play.

Software and computer companies produce a wide variety of MMORPGs with an array of fantasy worlds to explore, from outer space to Tolkeinesque lands. Yet for all their differences these games share a form of meritocracy; through experience (and hours of play) gamers can build characters, called "avatars" among gamers, with ever greater strengths and ever greater goods, permitting an ever more interesting level of play.

Because the level of interaction in the games is so high, and because the social rewards can be so great, it has become attractive for players to pump the hard currency of the real world into these fake universes, all in the name of achieving status and prestige in the virtual society.

* * *

Edward Castronova is an economist at Indiana University who specializes in the video game industry. He estimates that revenues for online gaming were $1.9-billion in 2003 and will grow to $9.8-billion by 2009. He also figures that up to $100-million in real-world money is captured annually by dealers in virtual currency and goods, and that number will keep growing.
According to Castronova's survey of players, the average gamer is a man in his 20s, with a full-time job and disposable income. The average time spent within the virtual worlds is 20 hours per week, he said, adding that 20 percent of the EverQuest gamers he surveyed declared, I live in Norrath but I travel outside of it regularly.
In EverQuest, like other MMORPGs, players create a character, or avatar, that they will play each time they return to the game. This character has special powers, such as strength or agility, based on its race and class. An ogre, for instance, has more raw strength than a half-elf, and a barbarian is better in combat than a Vah Shir (a feline humanoid).
Avatars take on tasks in exchange for experience, game money or items. Castronova found that fighting magical beasts yielded friends, currency and prestige.
But in between killing magical beasts and accompanying lizards on adventures, Castronova noticed that Norrath's economy was booming.
He witnessed virtual bakers making virtual bread. Virtual merchants selling virtual boots and hats, along with weapons and armor. Even virtual homes. Castronova realized Norrath is a consumer society. Everyone wants more game money (platinum is the currency) to gain higher status and, well, amass more stuff.
He did some research, and the results were startling: The average player was generating 319 platinum pieces, about $3.42, for each hour spent "working" in the game. Castronova tallied the wealth all the players created in one year in Norrath: $2,266 per capita.

The exchange rate between Norrath's currency and the U.S. dollar is determined in a highly liquid (if illegal)
 currency market. According to the daily exchange rate published at [url=http://www.gameusd.com/everquest-plat.htm]Everquest Plat Research[/url]  ,its value exceeds that of the Japanese yen and the Italian lira. The creation of dollar-valued items in Norrath occurs at a rate such that Norrath's GNP per capita easily exceeds that of dozens of countries, including India and China.

Castronova discovered that EverQuest is the 77th richest country in th

i don't think you finished typing lol...

but i know Everquest is richer than ukraine is lol

it's kinda pathetic  :P

everquest is not richer than ukraine. it has a more affluent economy in terms of sales and trade, and has a higher ratio-ed income per capita.

if you dont believe that the virtual world has gone to hell yet, here's a little light reading

[url=http://www.msnbc.msn.com/id/7341316/]Gamer stabs gamer over virtual sword[/url]

[quote="L_Li"]everquest is not richer than ukraine. it has a more affluent economy in terms of sales and trade, and has a higher ratio-ed income per capita.

if you dont believe that the virtual world has gone to - yet, here's a little light reading

[url=http://www.msnbc.msn.com/id/7341316/]Gamer stabs gamer over virtual sword[/url][/quote]
Let me see if I can dig up the link where a duper made thousands of dollars...

MMO's are rediculous. Everyone and their mom want to make one, the whole industry is full of low quality clones, the players are all 12-year olds who curse at you, and the servers are always down.

I prefer something like diablo over a LAN.

One lady made 16 dollar bills and tried to use 10 of them at Target :o


Let $a,b,c>0$.Prove the following inequality:

            $\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\geq\frac{2a^2}{b+c+2a}+\frac{2b^2}{c+a+2b}+\frac{2c^2}{a+b+2c}$ ;)

The way to proceed is clear: $\frac{2a^2}{b+c+2a}\leq\frac{1}{2}(\frac{a^2}{b+a}+\frac{a^2}{c+a})$. Then it is Chebyshev inequality.


function
integration
geometry
abstract algebra
limit
calculus
inequalities

For open sets $ U \subset \mathbb{C}$, let $ L^p_a(U)$ be the space of all analytic functions $ f$ in $ U$ such that $ \int_U |f|^p \, dA < \infty$, where $ dA$ is the usual area measure.

i) Construct a set such that $ L^1_a(U)$ has finite (non-zero) dimension
ii) Can $ L^2_a(U)$ have finite (non-zero) dimension?

If $ U\subset V$ then $ L_a^p(V)\subset L_a^p(U)$. So we are looking for "big" open sets. The entire plane $ \mathbb C$ is, obviously, too big. How about $ \mathbb C$ minus a point? Nope, the spaces are still empty... Two points don't seem to help either... But without three points you have $ \frac{1}{(z\minus{}1)(z\minus{}2)(z\minus{}3)}$ which is in $ L^1$.

The problem with $ L_a^2$ seems that it does not allow the function to have poles...

For i), I can prove that $ f \in L^1_a(\mathbb{C} \setminus \{1,2,3\})$ has at most simple poles in 1, 2 and 3. But how do I finish to show that $ f$ is a multiple of the function mlok wrote down?

For ii), I have no idea. I tried to look at reproducing kernels in $ L^2_a(U)$, but there doesn't really seem to be a contradiction in $ L^2_a(U)$ being finite-dimensional there.

For i): Since $ f$ has to be $ L^1$, $ \lim_{z\to\infty} f(z) \equal{} 0$, thus $ f$ can be holomorphically continued to $ \infty$. To be in $ L^1$, $ f$ has to have a zero of order at least $ 3$ in $ z \equal{} \infty$.

For ii): Not yet a full solution but two seemingly crucial thoughts:
(1) If $ U^c \equal{} \mathbb{C}\setminus U$ has nonempty interior, then $ L^2_\alpha(U)$ has infinite dimension.
(2) If $ z\in U^c$ is an isolated point of $ U^c$, then $ L^2_\alpha(U) \cong L^2_\alpha(U\cup\{z\})$.

Ok, I can prove i) completely now. Although I don't feel I have a satisfactory proof of the fact that $ f$ can't be integrable around an essential singularity (in 0, let's say). My argument is to make the change of variable $ z \equal{} \frac {1}{w}$, so that we are considering the convergence of $ \int_{\mathbb{C}} \left | \frac{a_{\minus{}2}}{w^2} \plus{} \frac {a_{\minus{}1}}{w} \plus{} g(w) \right | \, dA$ where $ g$ is entire. But $ \int_{|w| < R} \left | \frac{a_{\minus{}2}}{w^2} \plus{} \frac {a_{\minus{}1}}{w} \right | \, dA$ grows like $ R$ and unless $ g \equal{} 0$ then $ \int_{|w| < R} | g(w) | \, dA$ grows faster than $ R^2$, by the subharmonicity of $ |g|$.. It just feels like I don't completely understand what is going on with this argument. How fast does the integral of a transcendental entire function really grow?

For property (1), what did you have in mind? When I saw of your post I thought of the following: If $ a$ is in the interior of $ \mathbb{C} \setminus U$, then $ g(z) \equal{} \frac {1}{z \minus{} a}$ is analytic and bounded in $ U$. Hence if $ 0 \neq f \in L^2_a(U)$, then $ g^n f \in L^2_a(U)$ are linearly independent.

To estimate the integral of $ |f|$ near an isolated singular point $ 0$, just note that if $ f(z)\equal{}\sum_{k\equal{}\minus{}\infty}^\infty a_k z^k$, then the integral of $ f$ over the circumference of radius $ r>0$ is at least $ 2\pi |a_k|r^k$ for every $ k$. Integrating this inequality with respect to $ r$, we see that $ f\notin L^1_a$ unless $ a_k\equal{}0$ for $ k\le \minus{}2$. Similarly, $ f\notin L^2_a$ unless $ a_k\equal{}0$ for $ k\le \minus{}1$.

The $ L^2_a$ question is more interesting. I have an idea of how to solve it but I do not really like my approach. Let $ F\equal{}\mathbb C\setminus U$. Note that $ F$ does not contain isolated points (we can extend any function from $ L^2_a$ analitically to those points). Also, F cannot have bounded connected closed subsets (otherwise we can consider the conformal map $ g$ of the unbounded component of the complement of such a subset to the unit disk mapping $ \infty$ to $ 0$ and $ g^2,g^3,g^4,\dots$ will be in $ L^2_a$. In each bounded component, we can just take the identically $ 0$ function. This (if I do not mistake) should imply that connected components of $ F$ are just points. Now just apply the separation of singularities lemma to $ f$ to build functions whose singularities lie in arbitrarily small neighborhoods of $ n\plus{}1$ distinct points of $ f$. This will give you $ n\plus{}1$ linearly independent functions $ g_k$ that are as good as $ f$ locally but may decay like $ \frac 1z$ at $ \infty$. To improve the decay at infinity to $ z^{\minus{}2}$ just subtract $ g_1$ from $ g_k$ with an appropriate coefficients (assuming that it is $ g_1$ that fails to decay as $ z^{\minus{}2}$ at infinity from the very beginning).

What do you mean, exactly, by a singularity here? And what if the $ f$ you happen to pick only has finitely many?

By "singularity" I mean a point $ z$ to which the function cannot be analytically extended from $ U$ (i.e., there is no open $ V\supset U\cup\{z\}$ such that $ f$ is analytic in $ V$). Note that it is enough to consider the case when $ F$ has Lebesgue measure $ 0$ (otherwise we can create a bounded function decaying at infinity by just taking the Cauchy integral of the characteristic function of some bounded portion of $ F$), so such analytic extension, if it is there, is just the extension by continuity and we do not need to go into the tricky business of checking whether 2 different extensions agree here.

There cannot be finitely many singularities because then they would be isolated, which is impossible for an $ L^2_a$ function.

Anyway, I found a better idea that works in $ L^p_a$ with $ p>2$. If $ f$ is a not identically zero function in $ L^p_a$, then $ g(z)\equal{}\frac {f(z)\minus{}f(w)}{z\minus{}w}$ is another such function. It seems that applying this operation many times, we should get linearly independent functions unless $ f$ is rational. For $ L^2_a$, we can try $ g(z)\equal{}f(w_2)\frac {f(z)\minus{}f(w_1)}{z\minus{}w_1}\minus{}f(w_1)\frac {f(z)\minus{}f(w_2)}{z\minus{}w_2}$. See if you can make this trick work. :)

Well, as you indicated, the coefficients have to be adjusted so that $ f$ decays fast enough at infinity. It seems easiest to assume then that $ U$ contains a neighborhood of $ \infty$ (the change of variables $ z \equal{} \frac {1}{w}$ does not change the dimension).

But in fact, once you have a function $ f$ that decays like $ \frac {1}{z^n}$ at $ \infty$, it is easy to choose coefficients $ c_j$ so that $ \sum c_j \frac {f(z) \minus{} f(w_j)}{z \minus{} w_j}$ decays like $ \frac{1}{z^{n\plus{}1}}$. Just calculate the coefficients in the laurent expansion around $ \infty$ explicitly to get a system of linear equations that is not over-determined. So if $ f$ has infinitely many non-zero coefficients in its laurent expansion, then we can construct countably many linearly independent functions in $ L^2_a(U)$. But what if $ f$ does not have an infinite expansion?

About this:
[quote="fedja"] Note that it is enough to consider the case when $ F$ has Lebesgue measure $ 0$ (otherwise we can create a bounded function decaying at infinity by just taking the Cauchy integral of the characteristic function of some bounded portion of $ F$)[/quote]
Isn't such a Cauchy integral zero?

[quote="Kalle"]the change of variables $ z \equal{} \frac {1}{w}$ does not change the dimension[/quote]
Why?
[quote]
But what if $ f$ does not have an infinite expansion?
[/quote]
If we consider the case when $ F$ has measure $ 0$, then no rational function can be in $ L^p_a(U)$ with $ p\ge 2$.
[quote]
Isn't such a Cauchy integral zero?[/quote]
I mean the integral $ \int_S \frac {dm_2(w)}{z\minus{}w}$ where $ S$ is a bounded subset of $ F$ of positive measure, which behaves like $ \frac{m_2(S)}z$ at $ \infty$. ($ m_2$ is the Lebesgue measure on the plane, as usual).

[quote="fedja"][quote="Kalle"]the change of variables $ z \equal{} \frac {1}{w}$ does not change the dimension[/quote]
Why?
[/quote]
If $ V$ is the image of $ U$ under $ 1/z$, then $ f(z) \leftrightarrow \frac{f(w)}{w^2}$ is a unitary correspondence between $ L^2_a(U)$ and $ L^2_a(V)$.

[quote="fedja"][quote="Kalle"]But what if $ f$ does not have an infinite expansion?[/quote]
If we consider the case when $ F$ has measure $ 0$, then no rational function can be in $ L^p_a(U)$ with $ p\ge 2$.
[/quote]
Ooops. I am stupid  :P

Do you agree that the problem is solved now?

[quote="Kalle"]Do you agree that the problem is solved now?[/quote]
Seems like that :). Does it mean that it is time to post the next one? :P


geometry AMC AIME email ratio perimeter circumcircle

Has anyone of you cleared the Regional Mathematics Olympiad ???
Can you please help out, i am in 11th now and i would be appearing for the first time in RMO (2009). I don't want to miss this opportunity.can you please tell me , how much number thoery and geometry must i do...in order to be able to solve RMO Problems .....(actually i am able to solve many of the previous years' RMO Problems and very few INMO Problems ,but not all)

Please help me out

hey hi check this out [u]www.artofproblemsolving.com/forum/index.php?f=513[/u]

Hey !! Would it help if i solve AIME nad Junior Balkan Question Papers coz' i frequently solve them and am able to crack some of those problems.....????

how does one apply for RMO? im in 10th and wish to apply

hey its easy u just see se who ur regional coordintor is and then contact him:
check this http://www.iapt.in
btw where do u live?
 :D

through which medium should i contact? email, post, telephone....which one?

To Analytic:
You need to tell us where you live.
And you better hurry.
Enrollment might have been already over by now.

Here's some info:
http://www.isical.ac.in/~rmo/

In W.B. the procedure is:
http://www.isical.ac.in/~rmo/rmo08.html

Here's the list of regional coordinators:
http://olympiads.hbcse.tifr.res.in/enrollment/list-of-rmo-coordinators

well, i live in delhi. it will be sad if the enrollment has ended. i know that professor tripathi is the my regional co-ordinator and ways of contacting. normally, how do you contact your regional coordinator? phone or email?
i found email as a more convenient way but he has not responded.

this information seems to be useful, but i'll still confirm with the coordinator.but thanks :)

also send ur class tenth marksheet as a conformation and it should be [b]CERTIFIED[/b]

[quote="amangupta2"]C'mon you all don't panic, last date for registration is 24th october. so there is quite a lot of time you see!! :oops:

Registeration is quite easy:
1) Either the forms are made available by the school and sent to the cordinator
OR
2) You can individually register yourself by sending a DD of 50 rupees in the name of regional cordinator, delhi Region to Vasant Valley school , New delhi with the necesary info_ --> Name, Class, age, Address, Father's Name, School's Name and 2 passport size Photographs

I hope you would find this info useful!!!

              :idea: :wink:[/quote]

I am sorry but the last date was 26th septempber

oh!
noooooooooooooo  :(  :mad:
:'(
very bad.
darn darn!!!!

C'mon ths is not possible , coz' when i filed the form, it cleary mentioned that the registration has to be done before 24th pctober. and no markshet is required..(maybe it s required if you register individually)

Anyways i need not worry coz' i have already registered for RMO through my school(DPS Noida)  [And not markshet copy was required]

Hope that you are all able to register for it.(coz'there is still time)

ALL THE BEST !!

i duuno about other :)  regions but yes in uttar pradeshthe date of rmo is over!@

well, there is hope. today, professor just responded. so, maybe last date is 24th october.

hey delhi guys!!

RMO registration isn't over yet.
yes!! its 24 th oct....

i dunno why many of u have some confusoion

i had also sent a mail i got the reply.. i'm in XI as well

can the genius one who is able to solve past RMO problems let me which ones u r able to???let me try and see if a stupid one like me has any hope of even having a hope!!
reply fast dude!!!

hey solve these i was able to solve only three (with partial solution s............... :P )

1. Two boxes contain between them 65 balls of several different sizes. Each ball is white, black,
red or yellow. If you take any 5 balls of the same colour at least two of them will always be
of the same size (radius). Prove that there are at least 3 balls which lie in the same box have
the same colour and have the same size (radius).

.
3. A square sheet of paper ABCD is so folded that B falls on the mid-point M of CD. Prove
that the crease will divide BC in the ratio 5 : 3.
4. Find the remainder when 21990 is divided by 1990.
5. P is any point inside a triangle ABC. The perimeter of the triangle AB + BC + CA = 2s.
Prove that
s < AP + BP + CP < 2s.
6. N is a 50 digit number (in the decimal scale). All digits except the 26th digit (from the left)
are 1. If N is divisible by 13, find the 26th digit.
7. A censusman on duty visited a house which the lady inmates declined to reveal their individual
ages, but said — “we do not mind giving you the sum of the ages of any two ladies you may
choose”. Thereupon the censusman said — “In that case please give me the sum of the ages of
every possible pair of you”. The gave the sums as follows : 30, 33, 41, 58, 66, 69. The censusman
took these figures and happily went away. How did he calculate the individual ages of the ladies
from these figures.
8. If the circumcenter and centroid of a triangle coincide, prove that the triangle must be equilateral.
1

RMO Problems are quite easy , so even if i have solved may, it doesn't make me a genius anyways (as you mentioned)

actually i have solved many of them almost 50% of all RMO's Problems and about 10% of all INMO's Problems :

But the Problem is that I am still weak in Combinatorics and set theory and Algebra part

As far as Number Theory and Geometry and  General Questions (Tricky) are concerned I have ben able to solve most of them.....

Anyways!! Let's Look forward to RM0 2009..I would be appearing for RMO for the first time this year!!!
Bye the way which region are you all appearing from??, I am Appearing from Delhi Region.

ALL THE BEST  TO YOU ALL TOO !!!
] :P  :idea:  :roll:  :blush:  :wink:

[quote="amangupta2"]Have you People heard of Terence Tao ??? The Best Mathematician at Present . If not then go to google and Surery read about it !! It is truely Inspiring....[/quote]
Another senseless post...well,even if it had such great influence on you, it doesn't yet mean that everyone wants to hear or to read about it.If you really want to argue about this theme then go to the Round Table section or somewhere else and discuss who is the best mathematician in the world, as much as you wish.
P.S:
Furthermore I doubt that Terence Tao is the best mathematician...Perelman is much better, even Terence Tao admitted this fact.

Sir
I am really sorry , If I hurt you in anyways through that post.
Actually, my aim of creating that post was not to prove whether Terence Tao is the best mathematician or not  ,but i just wanted to let my Indian friends know about his childhood and his World Record at IMO  and that is all .

   :(  :huh:  :!:

same region.. let's see what happens....

the thing which made me call u a genius...that u cud spare time for atlest knowing what combi...  is all abt i dont know even that!!!lolzzz
so forget abt what i do..that will not pose any kind of trouble 2 u...
bye

ps: sorry didn't read the other page..

hey y so many processes
last yr there was no proof--roll no etc that the one appearing 4 RMO was me!!!! :rotfl:  :rotfl:


Was anyone at the SEIBA honor band tryouts this Saturday? I didn't even get recalled, but I swear there were like a 100 trumpets trying out for 20 spots...

Currently I'm working on the jazz band guitar part. As soon as my director gets an electric guitar for the school I can practice for the  jazz competition in March. I'm thinking about "borrowing" some Hendrix solos if I ever need a solo part...

No, I wasn't. I'm not in band, but I play piano for Ames Jazz II.

Did you tryout for all-state?

Penguin, are you a high school trumpeter?

I auditioned and made the band, again.  I've made it three of my possible four years.

[quote="PathofNeo"]Penguin, are you a high school trumpeter?
[/quote]
Yes (8th grade), though I suck. By suck I mean I can't even manage an A above the staff consistantly...

Oh, you've got plenty of time.  A good regiment of practice will straighten almost anything out.  If you want to increase your range, try ascending long tones (12-20 counts) until you reach your limit.  When you're there, hold it as long as you can while producing a good tone.  Then do it again.  And again.  And again. :lol:  (Hooray for being a trumpet major)


function
geometry
geometric transformation
reflection
quadratics
algebra
quadratic formula

Here's another problem i need help with, hope this is the right section for it

Find an equation that defines the inverse of function f(x)= 2x-3

[hide]
$f(x)=y=2x-3$.

To find $f^{-1}(x)$, we swap the roles of $x$ and $y$. Note that this reflects the graph through the line $y=x$.

$x=2y-3$.

Solving for $y$, we get $y=\frac{x+3}{2}$.

This is the inverse of the original function.

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

now try the inverse of the function $f(x)=(2x-3)^{2}$
$\textbf{Hint}:$ Its similar to yours.

Would you do the same thing?
[hide]$f(x)=y=(2x-3)^{2}$

Switch x and y:

$x=(2y-3)^{2}$
$x=4y^{2}-12y+9$
$4y^{2}-12y+(9-x)=0$

Now, use the quadratic formula:

$y=\frac{12\pm\sqrt{144-16(9-x)}}{8}$
$y=\frac{3\pm\sqrt{9-9+x}}{2}$

$\boxed{y=\frac{3\pm\sqrt{x}}{2}}$

Now that I've done all that work, I realize I could have just taken the square root in the first step. :mad: [/hide]


[quote="socrates"]Try this problem!

      Find all polynomials f(x) such that f(x)+f(x+2)=2f(x+1)
      for all real x.

I don't find it difficult, nevertheless interesting! :)[/quote]

I love math problems in the F&G forum.  :D

[hide]f(x)+f(x+2)=2f(x+1)
f(x+1)=(f(x)+f(x+2))/2
This means f(x+1) is the arithmetic mean of f(x) and f(x+2), which indicates that f(x) is a linear function.
So, any linear function would satisfy f(x)[/hide]

[quote="socrates"]Very well!
Please tell me the general rule for this.! :)[/quote]

Well, you posted it in at least 2 other forums, check there! ;)  :rotfl:  :rotfl:

[quote="socrates"]Very well!
Please tell me the general rule for this.! :)[/quote]The rule is simple: you don't post serious math problems in the Games & Fun Factory forums, and you don't post a problem twice on the forum!

Discussion about the problem is carried on in this topic:
http://www.mathlinks.ro/Forum/viewtopic.php?p=271412


The origin is the limiting point of a system of co-axial circles of which $ x^2 \plus{} y^2 \plus{} 2gx \plus{} 2fy \plus{} c \equal{} 0$ is a member. The equation of the orthogonal system is given by-
A] $ (x^2 \plus{} y^2)(g \plus{} \lambda.f) \plus{} c(x \plus{} \lambda y) \equal{} 0$
B] $ (x^2 \plus{} y^2)(g \plus{} \lambda.f) \plus{} c(\lambda x \plus{} y) \equal{} 0$
C] $ (x^2 \plus{} y^2)(\lambda g \plus{} f) \plus{} c(x \plus{} \lambda y) \equal{} 0$
D] none of these

where $ \lambda$ belongs to R

Ans-A

IIT jee math syllabus includes the following for circles:

Equation of a circle in various forms, equations of tangent, normal and chord.

Parametric equations of a circle, intersection of a circle with a straight line or a circle, equation of a circle through the points of intersection of two circles and those of a circle and a straight line.

I dont think coaxial and other things are necessary.:) :blush:
May come in comprehension though but I am ready to risk it. :D

i am ready to risk everything in co- geo. :rotfl:

[quote="achu_182"]what do u mean by
1. coaxial circles (any 2 circles can be said to be coaxial, if axis is considered to be a diameter)
2. limiting point
??[/quote]

1. same radical axis.
2. members of the family with zero radius.

method... x2+y2=0 is also amember. s1+ts2=0. diffrntiate. replacce dy/dx by -dx/dy. int.


Let $ x,y,z>0$ and $ x\plus{}y\plus{}z\equal{}\sqrt{3}$. Find maximum of $ P\equal{}\sqrt{x^2\plus{}xy\plus{}y^2}\plus{}\sqrt{y^2\plus{}yz\plus{}z^2}\plus{}\sqrt{z^2\plus{}zx\plus{}x^2}$

Is the maximum achievable, or is it an upper bound?

Edit: I agree  :) , I thought it was weird though that $ P$ was defined, which at first glance suggested that the value would be attainable.

Upper bound -- it occurs at $ (\sqrt{3},0,0)$ and permutations, which are not in the domain space.

On the other hand, the minimum [i]is[/i] attainable.


find the product and suggest a mechanism

Isn't the product just 1,1-dimethylcyclohexane?

Interesting. So, what is the major product?

Is the major product 3,3,5-trimethylcyclopentanone? I am surprised that you have to study such a specific issue.

yes that is the right answer and the mechanism?

Well, to your information the mechanism of the Clemmensen reduction is not fully understood at present, as far as I know: the best we (or I) have is one or another [i]proposal[/i] for the mechanism. Also, I see no interest in knowing the mechanism of such a transformation - there are plenty of mechanisms much more important than this in an organic chemistry course. However, I will leave you the proposal I know - you can then figure out by yourself the mechanism of the particular Clemmensen reduction of 1,3-diketones.

The reaction of a carbonyl compound RCOR' occurs at the surface of the zinc catalyst as follows:

Step 1. The pi-bond of the C=O linkage undergoes homolitic cleavage to form the radical species $ Zn\minus{}O\minus{}C(RR')\cdot$.

Step 2. There is elimination of zinc oxide to form $ Zn\equal{}CRR'$.

Step 3. Addition of $ H^{\plus{}}$ to form $ Zn^{\plus{}}\minus{}CHRR'$.

Step 4. Further addition of $ H^{\plus{}}$ to give the final product, $ RCH_{2}R'$ and $ Zn^{2\plus{}}$.


Solve the system equation: $ \{\begin{array}{l}x^2 + y^2 = 1 \\
e^\frac{1}{e}(e^{x\ln x}+e^{y\ln y})=(x+y)^2 \\
\end{array} $

Nobody like this problem?
[hide="Hint"]$ x\ln x \geq \minus{} \frac {1}{e}$[/hide]


[quote="eureka-123"]a physical quantity  is related to four variables as
             Ω =( a^3.b^2 ) / √y . n

the percentage errors in measurement of a,b,y,n are 1%,3%,4% and 2% respectively. if value of Ω is calculated as 3.763 unit from the given values...to what value should you round off the result ??[/quote]
[hide]
Assuming you meant $ \Omega \equal{}\frac{a^3b^2}{\sqrt{y}\cdot n}$

then $ \ln\Omega \equal{} 3\ln a \plus{} 2\ln b \minus{}\frac{1}{2}\ln y \minus{}\ln n$.

Differentiating the above expression:

\[ \frac{d\Omega}{\Omega} \equal{} 3\frac{d a}{a} \plus{} 2\frac{d b}{b} \minus{}\frac{1}{2}\frac{d y}{y} \minus{}\frac{d n}{n}\]

Therefore,

\[ \frac{\sigma_{\Omega}}{\bar{\Omega}} \equal{} 3\frac{\sigma_a}{\bar{a}} \plus{} 2\frac{\sigma_b}{\bar{b}} \plus{}\frac{1}{2}\frac{\sigma_y}{\bar{y}} \plus{}\frac{\sigma_n}{\bar{n}}.\]

This implies $ \frac{\sigma_{\Omega}}{\bar{\Omega}} \equal{} 3(1\%) \plus{} 2(3\%) \plus{}\frac{1}{2}(4\%) \plus{} 2\% \equal{} 13\%$

\[ \sigma_{\Omega} \equal{} 0.13\cdot \bar{\Omega} \\ \\ \equal{} 0.13(3.763) \equal{} 0.49\quad \text{to two significant figures.}\]

\[ \therefore \Omega \in [3.3, 4.3)\][/hide]


Find all real functions f if f(xy+((f(x))^2)=f(f(x))f(x+y).
   Harazi

Just the beginning:

We put x=0 and find f(f 2 (0))=f(f(0))f(y) for all real y.

Case 1: f is constant. We put f(x)=k for all x, so k 2 =k, so f(x)=1 for all x or f(x)=0 for all x. Both of these verify the equation, so they're both solutions.

Case 2: f isn't constant. From f(f 2 (0))=f(f(0))f(y) we find f(f(0))=0. In the initial eqn we make x=f(0), so we get f(yf(0))=0 for all reals y, so f(0)=0, because otherwise yf(0) would cover all reals, so f would be constant.

Let's take an a s.t. f(a)=0. We put x=a and find f(ay)=0 for all reals y, so a=0 (by the same argument as above). This means that the only soln of f(x)=0 is x=0.

We take y=-x, so f(f 2 (x)-x 2 )=0, so, by the previous observation, f 2 (x)-x 2, which means that for every real x we have f(x)=x or f(x)=-x.

I haven't continued.

Another step:

If we make y=0 we get f(x2)=f(f2(x))=f(f(x))*f(x) (*)

We can write the initial relation like this:

f(x(x+y))=f(f(x))f(x+y). Let a=x and b=x+y. We have f(ab)=f(f(a))f(b) for all real a and b. Since f(ab) is symmetric in a and b we also have f(ab)=f(a)f(f(b)). We multiply these and we combine the result with (*) to get (ab)2 = f(a2)*f(b2), so f(x)/x is constant for all positive x, which means that we either have f(x)=x for all x>0 or f(x)=-x for all x>0.

I'm doing this step by step :D

Let's assume f(x)=x for all x>0. Let's take a,b<0. We have f(f(a))f(b)=f(ab)=ab, so we either have f(a)/a=f(b)/b=1 or f(a)/a=f(b)/b=-1, so for all x<= f(x)/x is constant, so we either have f(x)=x for all x<= or f(x)=-x for all x<0. Since we assumed f(x)=x for x>0, we have 2 cases:

f(x)=x if x>0 and -x if x<0
f(x)=x for all reals x.

I think that if we assume f(x)=-x for all x<0 we get the other 2 choices:

f(x)=-x if x>0 and f(x)=x if x<0
f(x)=-x for all reals x.

I think these 4 and the constant solutions f(x)=1 for all x and f(x)=0 for all x are the solutions of the problem. Is it right? I haven't checked the case f(x)=-x for all x>0, so I might be wrong.

Yes, grobber, you are right., but the function f(x)=-x if x>0 and x if x<0 is not solution.  I think that this problem solves itself, you have just to work on it. Let's remember the IMO 2002 function, another example of this type.


algebra
polynomial
function
real analysis
real analysis unsolved

let $f(x,y)$ be a continuous function from $R^2$
to $R$ such that:
*for all $x_0 \in R$, $y -> f(x_0, y)$ is a polynomial
*for all $y_0 \in R$, $x -> f(x , y_0)$ is a polynomial

prove that $f \in R[X,Y]$.

It has been proved many times on forum, Alekk. By the way, as far as I remember, the problem works without the assumption of continuity. I remember anyway to have seen it at ENS without this condition. I will check.

It's true, we don't need the hypothesis of continuity. I do believe it's been discussed before, and I sort of remember a proof by Myth, but I'm definitely not sure, and I can't find it.. :?

I will post a solution tomorow, I remember to have solved this one some time ago. I will try to rediscover the idea.

Here is what I did. Hope it's good. First of all, by the fact the the reals are not countable, one can find two numbers $m,n$ and two infinite sets $A,B$ such that there exist functions $ g_0,..., g_n$ from $A$ to the reals and functions $ h_0,...,h_m$ from $B$ to the reals such that $ f(x,y)=g_0(x)+g_1(x)y+...+g_n(x) y^n$ for all $x$ in $A$ and real $y$ and $ f(x,y)=h_0(y)+...+h_m(y) x^m$ for all $y$ in $B$ and real $x$ (I just wrote that there exist $m,n$ such that for infinitely many reals $x$ the function which takes $y$ to $ f(x,y)$ is a polynomial of degree at most $n$ and the same for $m$). Now, take $ y_0,...,y_n$ different from $B$ and write the condition $ h_0(y)+...+h_m(y)x^m=g_0(x)+...+g_n(x) y^n$ for $y_0,...,y_n$ and all $x$ in $A$. It will give you a linear system in the unknowns $ g_0(x),..., g_n(x)$ and you will easily see that we can find $P_0,...,P_n$ polynomials such that $ g_i(x)=P_i(x)$ when $x\in A$.  Thus we can find a two-variable polynomial $P_1(X,Y)$ such that $ f(x,y)=P_1(x,y)$ whenever $x\in A$ and $ y\in R$. The same argument shows that we can find a two variable polynomial $ P_2(X,Y)$ such that $ f(x,y)=P_2(X,Y)$ whenever $x\in R$ and $ y\in B$. It follows that $ P_1-P_2$ vanishes on the infinite set $AxB$ and thus it vanishes everywhere, hence $ f(x,y)=P_1(x,y)$ whenever $x$ is in $A$ or $y$ is in $B$. Now, take two reals $x,y$. I want to show that $ f(x,y)=P_1(x,y)$. If $x\in A$ it's clear. Assume the contrary. The function $ t(x)= f(x,y)$ is a polynomial $ S(x)$. Moreover $ S(a)=P_1(a,y)$ whenever $a\in A$, which is infinite, thus in fact the polynomials are identical and thus $ f(x,y)=S(x)=P_1(x,y)$.

very nice  :)
The next question asked if the conclusion holds if we only suppose that $f(x_0,y)$
and $f(x,y_0)$ are polynomial when $x_0 \in Q$ or $y_0 \in Q$ ?

This one is easier :), and I definitely remember it being solved by Myth. The answer is no. I don't know if this was the exact same example that Myth probided, but it was along the same lines:

We take $f(x,y)=\sum_{n\ge 1}\alpha_n(x-q_1)(y-q_1)\ldots(x-q_n)(y-q_n)$, where $(q_n)_n$ is an enumeration of the rationals, and $\alpha_n$ are rationals chosen to be so small that the series converges absolutely for all reals $x,y$.

Does the same stay true if we replace polynomials by rational functions?


In the following figure, $CD$ is a half chord perpendicular to the diameter $AB$ of the semicircle with center $O$. A circle with center $P$ is inscribed, toucher $AB$ at $E$ and arc $BD$ at $F$. Prove that $\Delta EDA$ is isosceles.

I've got a proof using trigometry and bisector theorem to show $\measuredangle CDE= \measuredangle EDB$. If you want I can post it. :D

[quote="M4RI0"]This is gonna be a little long :P

$WLOG$:$OP=1$
Also:$\measuredangle POE = \alpha$

$PE=CE=sin \alpha$, $OF=1+sin\alpha$
$\Rightarrow OC=cos \alpha-sin \alpha$

$(DA)^{2}=(AC)^{2}+(OD)^{2}-(OC)^{2}$
$(DA)^{2}=(1+cos\alpha)^{2}+(1+sin\alpha)^{2}-(cos\alpha-sin\alpha)^{2}$
$(DA)^{2}= 2(1+sin\alpha)(1+cos\alpha)=(1+sin\alpha+cos\alpha)^{2}$ $\Rightarrow$ $DA=1+sin\alpha+cos\alpha$

$AE=1+sin\alpha+cos\alpha$

$\Rightarrow$ $\boxed{DA=AE}$  $\mathbb{QED}$

Wow, I found a better solution :!: .[/quote]

Nice. See if you can find a way with massive Pythagoras.

Maybe if someone could show in a geometric way that $\measuredangle CDE=\measuredangle EDB$, it would be really nice.


[quote="myc"]The intergers from 1 to 10 inclusive, are partitioned at random into two sets of five elements each.  What is the probability that 1 and 2 are in the same set?

my method seems right, but I'm getting the wrong answer.  Would anyone mind checking my work?  Thank you:)
[hide]
There are $\frac{10!}{5!5!}=252$ ways to choose the elements in the two sets.  There are $\frac{8!}{3!5!}=56$ ways to choose the other three elements in set 1( the set that 1 and 2 are in).  so the my answer is $\frac{56}{252}$.  Do I have to multiply $\frac{56}{252}$ by 2, since there are two sets, 1 and 2 can be in set 1 or set 2.  [/hide][/hide][/quote]
Actually, you do have to multiply by 2, and then you can simplify $2 \times \frac{56}{252}$ to $\frac{4}{9}$, which I believe is the answer.

Another way to do the problem:
1 is in one of the sets. 2 can go in any of the other 9 spaces, 4 of which are in the same set as 1, 5 of which are not. Each space has equal probability. Thus our probability is $\frac49$.

[quote="lotrgreengrapes7926"][hide] This is my solution:
There are $\frac{10!}{2^{5}}$ ways to create the 5 sets.
Think of it as ordering the numbers in a list, and every 2 are in a set. For example, $abcdefghij$ would have the sets $\{a,b\}$, $\{c,d\}$, etc. Each pair of 2 could also be flipped. For example, $abcdefghij\equiv bacdefghij$. There are $10!$ ways to list the numbers and $2^{5}$ ways to represent the same 5-set solution in a list.

There $\frac{8!}{2^{4}}$ ways to create the 5 sets, given that one of the sets is $\{1,2\}$, in the same way as above.
Thus, the answer is $\frac{\frac{8!}{2^{4}}}{\frac{10!}{2^{5}}}=\frac{2}{10\cdot9}=\boxed{\frac{1}{45}}$ [/hide]
EDIT: Oops, I thought it was 5 sets of 2 elements. :blush:
Is my answer correct, though?[/quote]
Well, actually, if that had been the problem, the answer would be $\frac19$ because 1 is in one of the sets, and 2 has to be placed in one particular spot out of 9 possible ones.

Your solution was almost correct. You just forgot to multiply by 5 because {1,2} could be any of {a,b}, {c,d}, etc.

[quote="nebula42"]Well, actually, if that had been the problem, the answer would be $\frac19$ because 1 is in one of the sets, and 2 has to be placed in one particular spot out of 9 possible ones.[/quote] That was easy.  :|

yeah, this problem is incredibly easy if you think about it - the 1 has to be in a certain spot, and there are 9 spots for the 2 to be in, and only four work, making the probability $\frac49$.


inequalities
function
trigonometry
logarithms
real analysis
real analysis unsolved

Prove that $\sqrt[2006]{\cos x}\leq \cos \frac{x}{2006}$ for $0\leq x \leq \frac{\pi}{2}.$

Let $f(t)=t\ln \cos \frac{x}t,t \ge 1$
$f'(t)=\ln \cos \frac{x}t+\frac{\sin \frac{x}{t}}{\cos \frac{x}{t}}\frac{x}{t}$
Let:
$g(a)=ln \cos a+a\tan a,0<a<\frac{\pi}2$
We have $g'(a)=\tan a+\frac{a}{\cos^{2}a}-tan a>0$
and $g(0)=0$
then we have:$f'(t)=g(\frac{x}{t})>0$
so $f(2006)>f(1)$
Which comes the desired inequality :)


geometry
geometric transformation
reflection
projective geometry
geometry open

Could anybody please show me some results about the antipole and the perspector PLEASE  :(

Friendly

José Carlos

Antipole? Never heard about it. If you wanted to say "antipode", then the [i]antipode[/i] of a point P with respect to a circle k is the reflection of this point P in the center of the circle k. Usually, this term is only used if the point P lies on the circle k; then, the antipode of P is also called the [i]point diametrically opposite[/i] to the point P on the circle k.

As for the perspector, there is really not much to say. If $A_1A_2...A_n$ and $B_1B_2...B_n$ are two polygons such that the lines $A_1B_1$, $A_2B_2$, ..., $A_nB_n$ all concur at one point, then this point is called the [i]perspector[/i] of the polygons $A_1A_2...A_n$ and $B_1B_2...B_n$, and the polygons themselves are called [i]perspective[/i] (or [i]in perspective[/i]). The most important result on perspectivity is [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=30323]the Desargues theorem[/url].

  Darij

[quote="darij grinberg"]As for the perspector, there is really not much to say. If $A_1A_2...A_n$ and $B_1B_2...B_n$ are two polygons such that the lines $A_1B_1$, $A_2B_2$, ..., $A_nB_n$ all concur at one point, then this point is called the [i]perspector[/i] of the polygons $A_1A_2...A_n$ and $B_1B_2...B_n$, and the polygons themselves are called [i]perspective[/i] (or [i]in perspective[/i]). The most important result on perspectivity is [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=30323]the Desargues theorem[/url].

  Darij[/quote]

 :?  I see... let me ask you something... If you talk about two polygons such that the lines join the homologous vertices are concurrent at $O$, then we say that the polygons are in [i]perspective from [/i]$O$... the dual defintion says that if two polygons such that the intersetion points of the homologous sides lie in $l$, then we say that the polygons are in [i]perspective from [/i]$l$... What about the poligons (even more the n-line complete) wich are perspective from a point and from a line?? (I think it could be better if it's possible to prove that if two $n$-line complete or $n$-point complete are perspective from a line, then they are perspective from a point, because... What's the definition of perspective... How can u identify if two figures are in perspective froma  point and from a line... It have to use perspective froma  point and perspective from a line??...)

[b]Theorem. [/b](Desargues) Two triangles are in perspective from a point iff they are in perspective from a line.

So it follows in...

[b]Theorem.[/b] Given two complete quadrangles such that 5 points of intersection of their correspondent sides lie in a line, the the 6 intersection points of the correspondent sides lie in a line and the two quadrangles are perspective from a point and from a line.

PD. Have you took [b]Measure Theory??[/b]... I need some concepts about areas... :?:


An edgy spider walks only along the edges from $A$ to $B$ of the dodecahedron formed by the net shown. What is the number of edges in the shortest path that the spider could take?

[img]7605[/img]

I think my answer might be right, but I visualized it, so I also wouldn't be surprised if it's wrong.

[quote="jhredsox"]I think the answer is [hide]5[/hide] edges.

Jorian[/quote]

no, youre considering him walking on the net, instead of the dodecahedron...

"An edgy spider walks only along the edges from A  to B of the [b]dodecahedron formed by the net shown. [/b]

[hide]It's 5 on the net, but when it becomes a dodecahedron, it's less since the edges come closer and new connections are made.

I visualized it and got 3... not sure though, I suck at visualizing things :P
[/hide]

try putting some of the net together...

I still got my answer.

[hide="Solution"]By putting the net together, you get a dodecahedron. The "V" shaped gaps between the A and the dot next to it become one spot. Thus, the answer becomes $\boxed{3\ \text{edges}}$, which is our answer.[/hide]

[quote="mwpl11"]dodecahedron means twelve sides i think (?)[/quote]

no, 12 faces.

oh yeah. oops :blush:

I forgot it was 3D. my bad.

[quote=".:Logic:."][quote="bpms"]Uh, what net?[/quote]

That shape in the figure is supposed to be a net of a dodecahedron. Am I missing something here?  :maybe:[/quote]
It wasn't there when I posted.

[quote=".:Logic:."][quote="jhredsox"]How does [i]that[/i] make a dodecahedron?

-jorian[/quote]
Dodeca = 12 right?

That net forms a 12 faced polyhedron.[/quote]

oh, i thought you guys meant putting lines between each of the sides

whoops

jorian

Could someone post a solution and explain this? I don't get it at all.

It's more visualization than "solution".... :P

Here's how I did it

[img]http://www.artofproblemsolving.com/Forum/album.php?mode=attach&id=7638[/img]

I kind of saw that was connected

so that's the path (correct me if it isn't)

-jorian

edit: why isn't the pic showing??!!

[img]7638[/img]

that's what I did.

(picture jhredosx)

Just a note: that was the picture I made, anirudh just posted it because I had trouble posting it.

So yah, that's how I did it.

-jorian

Please do not call me by my name; rather #H34N1 is fine :)

Sorry #H34N1, I'm so used to anirudh, plus I can't remember #H34N1. Please don't reply as this topic has served its purpose (unless you have a question) and we're getting off-track.

-Jorian


sorry i know i shouldnt have posted this here, but..bcz.. i dont know where to post this question

can someone tell me this , does ti 89 titanium solve questions in step by step or does it just give the answer?
 thanks, adm u can delete this after the answr sorry

Just the answer, but you can easily program the calculator to show at least a few steps with a basic knowledge of the TI programming language.

[quote="vishalarul"][quote="binomial_4eva"]any one knows the calculus made easy key for ti 89 titanium?[/quote]

Um... You have to buy it to get the key.[/quote]

Buy what? :huh:

Do you guys mean an app? B/c I know one app for the calculus is one you must buy. But if by "key"  you mean something else...please explain.

Anyway since it is sort of trivial in my mind to make an 89 show some steps for you, has anyone done this and be willing to post the program?

Also can you do both dec->frac and frac->dec?


The distances from a point on the circumcircle of an equilateral triangle to the two nearest vertices are 3 and 6. Find the distance from the point to the furthest vertex. (Please explain.)

Thank you.

Let $ABC$ be the equilateral triangle and $P$ given point in the circumcircle. Let $PA=3$ and $PB=6$ Now angle $APB=120^\circ$ since $ABCD$ is cyclic and $C=60^\circ$. Now $AB^2=3^2+6^2-2\cdot 3\cdot 6\cos 60^\circ=45-18=27$ so $AB=3\sqrt{3}$. It follows from Ptolemy's theorem that $3\sqrt{3}\cdot 3+3\sqrt{3}\cdot 6=PC\cdot 3\sqrt{3}$ so that $PC=3+6=9$.

thank you, but can u please tell me what Ptolemys Theorem says.

Mathworld.Wolfram is a really good site for this. [url]http://mathworld.wolfram.com/PtolemysTheorem.html[/url]

Aside: does anyone know why it doesn't find the server unless you include the /index.html? I get a server not found when just typing in mathworld.wolfram.com.

http://www.mathlinks.ro/Forum/viewtopic.php?t=34094 ;)

[quote="Hokkage"]Aside: does anyone know why it doesn't find the server unless you include the /index.html? I get a server not found when just typing in mathworld.wolfram.com.[/quote]
I don't... which browser are you using?


real analysis
topology
real analysis unsolved

Suppose continuum hypothesis is wrong.
Let S be a set of real numbers and |N| < |S| < |R|. Is it true mes(S) = 0?

Sure. The measurable set of positive measure contains a closed subset of positive measure. But the closed set is either not-more-than-countable or have capacity continuum (it's classical result of, I think, Lindelof. It may be proved as follows: 1)call the point a from the set A to be a point of condensation, if any neighborhood of a contains more than countably many points of A. 2) All points of A except countably many are points of condensation. 3) The set of points of condensation of closed set A is perfect set (i.e. the closed set, which contains all its limit points) 4) not-empty perfect set has capacity of continuum; it's easy to construct injection from the set of sequences {a1,a2,...}, a[i]=0 or 1, to it, as it is made for Cantor set: choose a complementary interval, and consider points left to it, if a[1]=0 and right to it, if a[1]=1. Then choose a complementary interval left (or right) to the first choosen and do the same.By a principle of embeeded segments, you get a point of A from a sequence and it's clearly an injection.

You must not suppose S is measurable. Maybe there exists nonmeasurable set S with property |N| < |S| < |R|.

[quote]perfect set (i.e. the closed set, which contains all its limit points)[/quote]
Is it exact definition of perfect set? I mean closed set always contains its limit points.

Ops, sorry. The perfect set is such a close set, that each its point is a limit point (i.e. the set coinciding with its limit set).

As for non-measurable set, I do not know indeed, what to do.

Where is this problem from? It may appear to be hopeless.

I made lectures about continuum hipothesis (cardinal numbers, ordinal numbers, Erdos theorem and so on). It is very interesting theme. And now I have some questions related to it. So this is my problem and don't know solution at the moment.

If we assume Martin's Axiom (MA) and the negation of Continuum Hypothesis (CH) then every set of cardinality less than continuum has measure zero. The proof of this fact is easy, and I can post it tomorrow if anybody is interested. (And it is known that MA + $\neg$ CH is consistent with ZFC.)

The other direction is trickier. It is consistent with ZFC that there is a set of cardinality less than continuum that is not a null set. However, I do not know a simple argument for this.  :(  If I don't come up with a short proof, I'll find a reference and post it here.

We will prove that MA implies that every set of cardinality less than continuum is a null set.

Consider a set $A\subset [0,1]$ s.t. $|A| < |{\mathbb R}|$. We are going to cover $A$ with null sets $S_{k}$, $k\in \omega$.

We consider sentences of two types
1) "$x\in \mathbf{S_{k}}$" for $x\in A$, $k\in\omega$. Here $\mathbf{S_{k}}$ is a variable name;
2) "$[s, t] \not\subset \mathbf{S_{k}}$" for $s,t \in{\mathbb Q}$ ($s<t$), $k\in \omega$.

A set of sentences is consistent if it doesn't contain a pair of sentences "$x\in \mathbf{S_{k}}$" and "$[s, t] \not\subset \mathbf{S_{k}}$" for $x\in [s, t]$. A condition $p$ is a finite consistent set of sentences.  Denote the set of all conditions by $P$. We say that a condition $p$ is stronger than a condition $q$ if $p\supset q$; denote this by $p\leq q$. Then $(P,\leq)$ is a partially ordered set. Two conditions $p$ and $q$ are consistent if $p\cup q$ is consistent.

We say that a set $D\subset P$ is dense, if for every condition $p\in P$ there exists $q\in D$, s.t. $q\leq p$.

We will now construct many dense sets. Roughly speaking each of them encodes a property we want to have. Then MA will ensure that all these properties can be simultaneously satisfied.

First, let $D_{x}= \{p\in P : \exists k \ (x\in \mathbf{S_{k}}) \in p\}$ for $x\in A$. We prove that the set $D_{x}$ is a dense set: Take an arbitrary condition $p$. Let $\mathbf{S_{k}}$ be a name that is not mentioned in $p$ (since $p$ is a finite set, only a finite number of $\mathbf{S_{k}}$ appear in it). Then $p \cup \{(x\in S_{k})\}$ is a valid condition that is stronger than $p$. Now, for $k\in \omega$, and $\varepsilon > 0$ let \[D_{k, \varepsilon}= \{p\in P: \exists \text{ intervals }[s_{1},t_{1}], \dots, [s_{k}, t_{k}] \text{ s.t. their union has measure at least }1-\varepsilon \text{ and }([s_{i}, t_{i}]\not\subset S_{k})\in p\}.\] The set $D_{k, \varepsilon}$ is dense: Let $p$ be a condition. The set $F$ of $x$ mentioned in $p$ is finite. Therefore, we can pick (finitely many) disjoint intervals in the complement of $F$ whose total length is greater than $1-\varepsilon$.

Now we prove that $P$ satisfies the countable chain condition. That is, every antichain (a set of pairwise inconsistent conditions) is at most countable. Observe that if two conditions $p$ and $q$ have the same sentences of the second type, then they are consistent. But there are only countably many different combinations of sentences of the second type. Hence every antichain is countable.

We are now ready to use MA. Recall that Martin's Axiom states that given
1) a partial ordered set $P$, that satisfies the countable chain condition,
2) a family of dense sets $D_\alpha$; the cardinality of the family is less than continuum,
there exists a ("generic") set $G$ s.t. $G$ is consistent, and $G$ intersects every $D_\alpha$.
We apply MA to our set $P$ and dense sets $D_{x}$ (for $x\in A$) and $D_{k, \varepsilon}$ ($k\in \omega$, $\varepsilon>0$, $\varepsilon\in{\mathbb Q}$). Define \[S_{k}= \{x: (x\in \mathbf{S_{k}}) \in p, p\in G\}.\] We prove that $A\subset\bigcup_{k\in \omega}S_{k}$. Let $x\in A$. Since $D_{x}$ is dense, $D_{x}\cap G \neq \emptyset$. Let $p\in D_{x}\cap G$. By the definition of $D_{x}$, there exists $k$ s.t. $(x\in \mathbf{S_{k}}) \in p$. Since $p\in G$, $x\in S_{k}$.
Finally, we prove that $\mu(S_{k}) = 0$. Let $p\in D_{k,1/2^{m}}\cap G$. The condition $p$ contains sentences $[s_{i}, t_{i}]\not\subset \mathbf{S_{k}}$, s.t. $\mu(\bigcup_{i=1}^{m}[S_{i}, t_{i}]) \geq 1-1/2^{m}$. Since $p\in G$, and all conditions in $G$ are consistent, no condition $q\in G$ contains "$(x\in \mathbf{S_{k}})$" for $x\in[s_{i}, t_{i}]$. We conclude $\bigcup_{i=1}^{m}[s_{i}, t_{i}] \cap S_{k}= \emptyset$. So $\mu(S_{m}) \leq 1/2^{m}$ for every $m$.

We have proved that $A \subset \bigcup_{k\in\omega}S_{k}$, where each $S_{k}$ is a null set.

Here is a reference to a good article on the subject:

H. Judah and S. Shelah, "The Kunen-Miller Chart (Lebesgue Measure, the Baire Property, Laver Reals and Preservation Theorems for Forcing)", The Journal of Symbolic Logic, Vol. 55, N 3, Sept. 1990.

Thanks. It should be studied. Too far away from clarity :)

But will the more general conclusion be true:
If $|A|<|B|$,then $m^{*}(A)\le m^{*}(B)$ :?:

Yes,that's what I mean,And sorry I forgot to type ^.
And if this is true,the original problem will be easy to solve,consider the Cantor-Set.

If there exists a set $S$ s.t. $|S| < \mathbf{c}$ and $\mu^{*}(S) > 0$, then the answer to your question is "no": when $A = S$, $B$ is the Cantor set, we have $|A|< |B|$, but $\mu^{*}(A) >\mu^{*}(B)$.

On the other hand, if for every set $S$ s.t. $|S| < \mathbf{c}$, $\mu^{*}(S) = 0$, then the answer is trivially "yes".

So your question is equivalent to Myth's original question. Hence the answer to it is also independent of ZFC.


logarithms
calculus
calculus computations

I need to find out if the following series convergers or divergers to infinity. Please show work

(cos (n*pi))/(1+sqrt n)    n runs from 1 to infinity

1/(ln (n!)                       n runs from 2 to infinity

Need some help fast

For 1 note: $\sum_{n=1}^\infty \frac{\cos(n\pi)}{1+\sqrt{n}}= \sum_{n=1}^\infty \frac{(-1)^{n}}{1+\sqrt{n}}$. Do you have it now?

My gut says diverge for 2, but don't see a way to show it right off. Maybe compare to $\sum\frac{1}{n\ln n}$ ?

Edit: Yeah, just tried it. $\sum_{n=2}^\infty \frac{1}{\ln n!}> \sum_{n=2}^\infty \frac{1}{n\ln n}$ and the latter diverges. Not sure what I was thinking since $n\ln n=\ln n^{n}$ and $n^{n}$ grows faster than $n!$... Oh well.

Thank you very much. And yes it helped. I see now that cos n*pi is just a funny way to write (-1)^n  :)

$\sum_{2}^{n}\frac{1}{k\ln{k}}=\ln{\ln{n}+C'+o(1).}$
where $C'$ is constant
 :wink:


χτες πήγαμε με το Ράλλη (RAKAR) να παίξουμε σκάκι...
καθώς τοποθετούσαμε τα πούλια είχα την εξής ιδέα:
πoια η πιθανότητα να τραβήξουμε τα πούλια από μια αδιαφανής σακούλα με μια καθορισμένη σειρά.
Μετά από κάποια ώρα αρχίσαμε να παίζουμε σκάκι... και στο τέλος αφού είχαμε μείνει με 1Κ και ένα πιόνι ο καθένας, βγάλαμε 2 Q (ταυτόχρονα!) και νίκησε ο Ράλλης. Εξετάστε επίσης ποιες διατάξεις Κ-Q επιτρέπουν τη νίκη του ενός παίκτη οπωσδήποτε και ποια η πιθανότητα να συμβεί αυτό.  :P

[color=#f8f8f8]PS: 2 ill S6[/color]:stretcher:

Μην δειτε το hint(ειναι η λυση...)πριν το προσπαθησετε....
[hide]Επ.....γιατι βλεπεις το hint?Ελα τωρα..,αφου δεν την εχεις προσπαθησει καθολου...Λοιπον ελα ,κλειστο και συνεχισε....[/hide]

[hide]Καλα παλι τα ιδια??Προσπαθησε ντε....[/hide]

[hide]Ε,δεν αντεχεσαι ρε παιδι μου....Τελος παντων,επιδη δεν βαζεις μυαλο θα σου πω το αποτελεσμα... (2(8!)^2)/31!  Και τι εγινε τωρα ευχαριστηθηκες??[/hide]

 :rotfl:  :rotfl:  :rotfl:.......

Πολύ ωραίο κι χιουμοριστικό το παιχνιδάκι σου αγαπητέ Ράλλη
με μεγάλη  δόση χιούμορ
αυτο δειχνει καλό χαρακτηρα
αλλά κι μεγάλο μυαλο  :D
Χαιρετώ


If a! divided by b! is a multiple of four but not a multiple of 8, then what is the maximum value of a-b?

I think it's 3. Any group of four consecutive integers, when multiplied together, will be divisible by 8.

This is because you have a multiple of $ 2$ and a multiple of $ 4$, and $ 2 \times 4 \equal{} 8$.


ratio
analytic geometry
arithmetic sequence

So, that name doesn't really mean anything, but here's the question:

Consider the infinite grid in the plane of points arranged in a square lattice with each point a distance 1 from its 4 nearest neighbors.  Say we tile it with a finite number $n$ of other square lattices such that the set of vertices of the original grid is equal to the disjoint union of the vertices of the $n$ new grids.  If the shortest distance between two points in the $i^{th}$ of the $n$ grids is $l_n$, prove that $\frac{1}{l_1^2}+\frac{1}{l_2^2}+\ldots+\frac{1}{l_n^2} = 1$.  Also prove that this does not necessarily hold for a tiling of a square lattice with infinitely many other such lattices.

Maybe I'm just tired.. I'm not understanding the problem  :blush:  :blush:

Do you mind clarifying it a bit? (Maybe a diagram would help)

What does your bathroom floor look like anyways?  :P

I can give you a brief description of one example, to which I will add a drawing tonight when I'm not running to class:

It's easy to dissect a grid into two grids: take the two new grids at 45 degree angles to the old grid with side lengths 1/2.  Or, in other words, on the cartesian plane take the set {(a, b) s.t. a + b odd} and the set {(a, b) s.t. a + b even}.  Each of these is a square lattice.  Their disjoint (meaning no over-lap) union is all the lattice points in the plane.  We can do this lots of other ways, as well.

A cute problem!  The underlying idea is very easy, but it's expressed in an interesting way.

Here's a generalization: Let $S$ denote the set of lattice points, which is partitioned into the sets $S_1, S_2, \dots, S_n$.  Each set $S_i$ is periodic, that is, there exist non-zero integers $e_i$, $f_i$, $g_i$, and $h_i$, such that if $(x,y) \in S_i$, then $(x + e_i, y + f_i) \in S_i$ and $(x + g_i, y + h_i) \in S_i$.

Then each set $S_i$ has a well-defined density $d_i$.  What does that mean?  Suppose we take a very large region $R$ in the plane; this will contain many lattice points, and many points in $S_i$.  Then the ratio of points in $S_i$ in $R$ to lattice points in $R$ goes to a limit as the region $R$ increases unboundedly.  This limit is the density $d_i$.

Now, since the $S_i$ partition $S$, the sum of the densities must be 1.  This is just the analogy of the result that if we have a number of mutually exclusive events that cover all possibilities, then the sum of their probabilities is 1.

In JBL's result, the density of the $i$th grid is $1/l_i^2$, and the result follows.

This idea also solves the following problem: Let $S$ denote the set of integer, which is partitioned into sets $S_1, S_2, \dots, S_n$, such that each $S_i$ is an arithmetic progression with common difference $d_i$.  Prove that
\[\frac{1}{d_1} + \frac{1}{d_2} + \dots + \frac{1}{d_n} = 1.\]


Hi this was shown in my book, but I do not understand how they deduced it.  Can someone please explain this to me?  Thanks.

[b]My book said this:[/b]
From the equations
$ XdX\minus{}YdY\plus{}cdc\equal{}0$
$ YdX\plus{}XdY\minus{}bdc\equal{}0$,
Hence,
$ \displaystyle\frac{dc}{X^2\plus{}Y^2}\equal{}\frac{dX}{bY\minus{}cX}\equal{}\frac{Xdc\minus{}cdX}{X(X^2\plus{}c^2)}$.  How did they come up with this? :huh:

Rewrite the condition as $ (X, \minus{} Y,\ c)\cdot (dX,\ dY,\ dc) \equal{} 0$ and $ (Y,\ X, \minus{} b)\cdot (dX,\ dY,\ dc) \equal{} 0$,

$ (dX,\ dY,\ dc)$ is perpendicular to two vectors $ (X, \minus{} Y,\ c),\ (Y,\ X,\ \minus{}b).$

$ \therefore (dX,\ dY,\ dc)\parallel{}(bY \minus{} cX,\ cY \plus{} bX,\ X^2 \plus{} Y^2).$

Okay kunny, thank you very much for the explanation.

So [b](dX, dY, dc)[/b] is perpendicular to [b](X, -Y, c)[/b] and [b](Y, X, -b)[/b] which implies [b](dX, dY, dc)[/b] is a scalar multiple of [b](X, -Y, c)[/b] x [b](Y, X, -b)[/b].


trigonometry
inequalities proposed
inequalities

Let $a;b;c>0$ satisfying $a+b+c=1$.Prove that:
$\frac{a}{a+bc}+\frac{b}{b+ca}+\frac{\sqrt{abc}}{c+ab}\leq 1+\frac{3\sqrt{3}}{4}$

[hide="A hint"]Put $x=ab,y=bc,z=ca$ then $a=\tan{\frac{X}{2}}$, e.t.c. Where $X,Y,Z$ are angles of a triangle.[/hide]


The reaction; 4Ag(s) + O2 (g) yields 2Ag2O (s) is exothermic.  Which statement about the reaction is correct?

A. It is spontaneous at all temperatures.
B. It is spontaneous only at low temperatures.
C. It is spontaneous only at high temperatures
D. It is non-spontaneous at all temperatures.

Can someone answer this question and explain why the correct answer is the correct answer?
thanks![/code]

The criteria of spontaneity at constant T and P is $ \Delta G \equal{} \Delta H \minus{} T \Delta S \leq 0$. For this reaction, since it is exothermic, we already have $ \Delta H < 0$; however, a solid and [i]a gas[/i] are converted into [i]a solid[/i]: so $ \Delta S$ must be negative. Since the (thermodynamic) temperature T is always positive, the free energy variation will be negative only if $ |T \Delta S| < |\Delta H|$. Therefore, the correct answer seems to be B.


integration
trigonometry
limit
calculus
calculus computations

I was solving some sample problems for final exam, and i saw these three , they are really cute :D :

$1.$ Evaluate:

$I=\int xe^{2x}\cos3xdx$

$2.$ Prove that:

$\sqrt{3}\int^{\frac{1}{\sqrt{3}}}_{0}x\arctan xdx=\sum^{\infty}_{n=1}\frac{(-1)^{n-1}}{(4n^{2}-1)3^{n}}$

and then evaluate the sum of the series $\sum^{\infty}_{n=1}\frac{(-1)^{n-1}}{(4n^{2}-1)3^{n}}$

$3.$ For $n>1$ prove that:

$I=\int^{\infty}_{0}\frac{dx}{\left(x+\sqrt{1+x^{2}}\right)^{n}}=\frac{n}{n^{2}-1}$

For the first one

$\int x e^{ax}\cos bx \cdot dx = \int x e^{ax}\frac{e^{ibx}+e^{-ibx}}{2}\cdot dx$

and integrate by parts


function
trigonometry
calculus
vector
real analysis
real analysis unsolved

Let $ f,g: \mathbb{R} \to \mathbb{R}$ satisfy: for all $ x,y \in \mathbb{R}$
1- $ f(x \plus{} y) \equal{} f(x)f(y) \minus{} g(x)g(y)$.
2- $ g(x \plus{} y) \equal{} g(x)f(y) \plus{} g(y)f(x)$.
3- $ g'(0) \equal{} 0$.
Prove that $ f^2(x) \plus{} g^2(x) \equal{} 1$.

You mean, besides the trivial case $ f(x)\equiv g(x)\equiv 0$?

So.. it's pretty clear these are just the sine and cosine functions, but are we supposed to prove that those are the only ones satisfying this?

i dont think this problem goes here since it has to do with derivatives. i think it means that g(x) is tangent to the x-axis at the origin but it says that there shouldnt be calculus probs here. anyways...how do u solve this?

The problem is incorrect as stated. Something is wrong - probably the initial condition (3). Perhaps that was supposed to be $ g'(0)\equal{}1?$

It's clear to me that it is incorrect as stated because everything in statements (1), (2), and (3) is linear homogeneous, meaning that the set of all functions satisfying it is a vector space, and we could multiply any solutions by a fixed scalar to have another solution. Given that, there's no way to draw the nonhomogeneous conclusion $ f^2(x)\plus{}g^2(x)\equal{}1.$

Farenhajt's objection is really part of this observation.


Does anyone know of some good english sites besides amazon to get books?

are you asking for math books? then there is [url=higherpreception.com]this[/url]

That url doesn't seem to exist...
(at least for me) :)

Me neither...

[url=http://www.artofproblemsolving.com/Books/AoPS_B_Texts.php]This[/url] is much much better...

Is that a european site?  Because americans will have to pay uber-shipping for those...

Apparently we interpreted "english" differently...i'm very very sorry...

Jeez...

[quote="Poincare"]he asked for a european site dood! !!! ! ! ! ! ![/quote]
No amazon doesn't have some books like olympiad geometry and an introduction to diophantine equations do you know any north american sites that sell these?


In how many different ways can change be made for $ \$10$ if only quarters or dimes are used?

We want $ 25q \plus{} 10d \equal{} 1000\implies 5q \plus{} 2d \equal{} 200\implies q\equiv0\pmod2$
$ \implies q \equal{} 2m,m\in\mathbb Z\ge0\implies d \equal{} 100 \minus{} 5m$. Also, $ 2m,100 \minus{} 5m\ge0\implies0\le m\le20\implies\boxed{21}$.

Or, we know we can make it with 100 dimes, and we can substitute 5 dimes for 2 quarters, 100/5+1=21.

Another way:

Let's just figure out how to make each half dollar. Each half dollar can be 2 quarters or 5 dimes. So there are 2 ways to make each half dollar.

We are simply choosing 20 objects, the half dollars, from two possible choices (quarters or dimes).

Let the 0 be a half-dollar. We have 00000000000000000000. We add in a |. All 0's to the left of this divider will be made from quarters, and all 0's to the right will be made of dimes. There are now 21 symbols, 20 0's and 1 |. There are therefore $ \binom{21}{1}\equal{}21$ ways to do this.

Third way, sort of similar to AIME's.

we see that 10 dollars is 40 quarters. every two quarters can be replaced by dimes, so we have 40/2=20. but, we're forgetting the scenario where all of them are dimes, so $ 20\plus{}1\equal{}\boxed{21}$


Given $\triangle ABC$ with incenter $I$,$d$ is line through $I$ and cuts two segment $AB,AC$, with any $M,N \in d$ prove that
$b.S_{BMN}+c.S_{CMN}=a.S_{AMN}$

Let MN meet AB and AC at P and Q. Then $\frac{\triangle{BMN}}{\triangle{AMN}}=\frac{BF}{AF}$ and $\frac{\triangle{CMN}}{\triangle{AMN}}=\frac{CE}{AE}$, thus the equality becomes theorem 1 of: http://www.mathlinks.ro/Forum/viewtopic.php?t=100964

yes you are right, if I change incenter $I$ by excenter $J_{a}$ of angle $A$and $d$ isline thruogh $J_{a}$ and cuts ray $\vec{AB},\vec{AC}$ this problem is still true!
And if I change $I$ by $L$ Lemoine,$d$is line through $L$,with any $M,N\in d$ then in three number $a^{2}S_{AMN},b^{2}S_{BMN},c^{2}S_{CMN}$ there is one is sum other!
Can it be done by your menthod ? :)  :)


linear algebra
matrix
algorithm
ratio
superior algebra
superior algebra solved

If we start with a matrix with integer entries,of type (2,2),wich has the determinant 1, and make transformations by adding a line(colon) multiplied by an integer to other line(colon)in all possible ways, can we obtain all the matrixes
with integer entries wich have also the determinant 1?

Take A=(a b
             c d)
detA=1, so ad-bc=1. So all pairs (a,c), (a,b), (b,d), (c,d) are coprime.
Apply Euclid's algorithm to the lines 1 and 2. For instance suppose a=cq+r with 0<=r<|c|. So add line 2 multiplied by -q to line 1. Notice that the new matrix is of the form
(r b'
 c d)
Now repeat this (possible by Euclid's algorithm) until a matrix of the form
(1 x
 0 y)
The new matrix has y for it's deterninant, but the transformations we have made don't change the det, so y=1. Now add line 2 multiplied by -x to line 1 and get (1 0
                         0 1)
So we can bring the matrix A to I_2=(1 0
                                                       0 1)
Now I must prove the converse: we can turn I_2 into any matrix A with determinant 1. Get back to it later.

Sorry for the delay. I almost thought I missed this problem.
So back to my proof. I must prove that the matrix I_2 can be turned into any matrix with integer entries
a b
c d
of determinant 1 using only the following operation:
add a line (column) multiplied by an integer to the other line (column).
It is obvious that a,b are coprime or else det(a,b;c,d) is divisible by their gcd.

I now work only with lines and show that I can obtain a matrix of the form
a x
c y. For now I don't care about the elements x,y. Practicaly I want to reach from (1,0) to (a,c) only by one of the following operation repeated a certain number of times:
(q,w) -> (q+kw) or (q,w)->(kq+w) for some integer k.

To do this I walk back the path of Euclid's algorithm.
(a,c)<-(r1,c)<-(r1,r2)<-(r3,r2)<-...<-(1,0) where
a=cq1+r1
c=r1q2+r2
...
I think this part of the solution is complete.

I turned I_2 into
a x
c y
Because the transformations used don't modify the det ay-cx=1, but ad-bc=1 and by Chinese Remainder Theorem or smth simpler we get the existance of an integer m such that y=d+mc and x=b+ma.
Add column 1 multiplied by -m to column 2 and get
a b
c d

Hope you don't ask me to write my soln again.

P.S. There was one (among others  :D ) grammar mistake in my previous proof:
"Now repeat this (possible by Euclid's algorithm) until a matrix of the form
(1 x
0 y)"
And it stops here. I should have ended it in "is obtained.

P.P.S. There is a big grammar error in the problem text: xxxxtt writes colon instead of column.
From Collins English Dictionary:
colon1(`k?l?n) (pl, -lons) the punctuation mark : , usually preceding an explanation or an example of what has gone before, a list, or an extended quotation(pl, -lons) this mark used for certain other purposes, such as expressions of time, as in 2:45 p.m., or when a ratio is given in figures, as in 5:3(pl. -la (-l?)) (in classical prosody) a part of a rhythmical period with two to six feet and one principal accent or ictus[C16: from Latin, from Greek kolon limb, hence part of a strophe, clause of a sentence] colon2(`k?l?n) , pl  -lons  or  -la  (-l?)the part of the large intestine between the caecum and the rectum[C16: from Latin: large intestine, from Greek kolon] colon3(k?`ln; French kl) a colonial farmer or plantation owner, esp. in a French colony[French: colonist, from Latin colonus, from colere to till, inhabit] coln (k?`l?n; Spanish ko`lon) , pl  -lons  or  -lones  (Spanish -`lones)the standard monetary unit of Costa Rica, divided into 100 cntimosthe standard monetary unit of El Salvador, divided into 100 centavos[C19: American Spanish, from Spanish, after Cristbal Coln Christopher Columbus] Coln (k`ln; Spanish ko`ln) a port in Panama, at the Caribbean entrance to the Panama Canal. Chief Caribbean port. Pop.: 68 688 (1987). Former name: AspinwallArchipilago de  (arti`pjelao e). the official name of the Galpagos Islands.

I found out it was easier to prove the second part.
The transformations allowed act as the multiplication of A to the left or right by a matrix of the form:
1 0,  1 k, where k is an integer.
k 1,  0 1,
Notice that these matrices are invertible in Z and their inverses are
1 -k and 1 0. They are still allowed transformations.
0 1       -k 1
If in the first I proved that ant A can be transformed into I_2, I proved that there exist E_1, E_2...E_n and F_1...F_m matrices of order 2 which are allowed tranformations such that E_1*...E_n*A*F_1*...*F_m=I_2.
All we have to do now to go from I_2 to A is to simplify the matrices E_i and F_j by their inverses and get A=E_n'...E_1'I_2F_m'...F_1'.


geometry
analytic geometry
rectangle
combinatorics proposed
combinatorics

Given 3n points in the plane such that distance between any two point is not greater than 1 and no three are collinear. Prove that we can select n triangles satisfy:
1. Each of 3n points is a vertex of one triangle of n triangles.
2. Any two of n triangles have no common point in their area.
3. Sum of area of n triangles is less than 1/2
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[quote="pbornsztein"]Well, 3) is incorrect because it does not holds for $n=1$ and the three vertices of an equilateral triangle with side-length 1.

In another hand, since there is a finite number of points, we may find an orthogonal system of coordinates such that the $n$ points have pairwise distinct x-coordinate.
Let $A_1,...,A_{3n}$ be the points, where the ordering follows the ordering of their x-coordinate. Now choose the triangles $A_{3i+1}A_{3i+2}A_{3i}$ for $i=0,...,n-1$ and you will have 1) and 2).

Pierre.[/quote]
you re wrong. Because n=1, three points are $A_1 ,A_2 ,A_3$, so the area of triangle $A_1 A_2 A_3$ is ${\sqrt3 / 4} < 1/2$

We can find two perpendicular directions s.t. none of the lines determined by two points among the $3n$ has one of the two directions. Choose a rectangle having sides in the two directions s.t. it's minimal. Then both its sides are $\le 1$. We choose triangles as Pierre said, according to one of the directions. The triangles are covered by a set of disjoint rectangles (a rectangle for each triangle), and these rectangles are covered by the large rectangle, which has area $\le 1$. On the other hand, a triangle covered by a rectangle has area $\le \frac 12\times$(the area of the rectangle), so we can now obtain our conclusion.