For a function f: Z-->R, the following statement is true:

$ f(z) \equal{} z \minus{} 10$   for $ z > 100$
      and   $ f(z)\equal{}f(f(z \plus{} 11))$ for $ z\le11$

Prove that for all $ z\le100, f(z) \equal{} 91$.

[quote="Smartguy"]For a function f: Z-->R, the following statement is true:

$ f(z) \equal{} z \minus{} 10$   for $ z > 100$
      and   $ f(f(z \plus{} 11))$ for $ z\le11$

Prove that for all $ z\le100, f(z) \equal{} 91$.[/quote]
could you please double-check your problem  :idea:

well i copied it exactly from a source--> Bratislava Correspondence Seminar, Fall 1998 3rd series

$ \blacktriangleright$ Did you copy this from 8.28 on page 8 of [url=http://mathcircle.berkeley.edu/recur.pdf]this presentation handout[/url]?
$ \blacktriangleright$ Perhaps the source meant $ f(z) \equal{} f(f(z \plus{} 11))$ for $ z \leq 101$?

[b]Objection to problem as-is:[/b]

Consider $ f(z) \equal{} 1 \; \; \forall \; \; z \leq 22$. This is internally consistent because:

1. $ z \leq 11 \implies z \leq 22 \implies f(z) \equal{} 1$
2. $ z \leq 11 \implies z \plus{} 11 \leq 22 \implies f(z \plus{} 11) \equal{} 1 \implies f(f(z \plus{} 11)) \equal{} f(1) \equal{} 1$

[quote="TZF"]$ \blacktriangleright$ Did you copy this from 8.28 on page 8 of [/quote]

Indeed, I did.

Well, I emailed the guy that made those notes -- let's see if he says anything (or if someone else figures this out).

So I emailed the author of the lecture notes and he gave me the link to the actual competition problems. Indeed, there was a typo.

[url=http://kms.sk/bkms/starezadania/9798/az979813.html]See here[/url].

The question should read:

For a function $ f \; : \; \mathbb{Z} \to \mathbb{R}$, the following statement holds:

$ f(z) \equal{} z\minus{}10$, for $ z \geq 101$
$ f(z) \equal{} f(f(z\plus{}11))$, for $ z \leq 100$

Prove that for all $ z \leq 100$, $ f(z) \equal{} 91$.


superior algebra
superior algebra unsolved

<Q,+> is not isomorphic to <Q*, x>
<R,+> is not isomorphic to <R*, x>

Thanks

$ \mathbb{Q}^\times\le\mathbb{R}^\times$ contain the element $ \minus{}1$ of finite order $ 2$.
But an element $ x$ of $ \mathbb{Q}$ or $ \mathbb{R}$ of order $ 2$ would satisfy $ 2x\equal{}0\not\equal{}x$.


Of the following five statements, $I$ to $V$, about the binary operation of averaging(arithmetic mean),

I. Averaging is associative
II. Averaging is commutative
II. Averaging distributes over addition
IV. Addition distributes over averaging
V. Averaging has an identity element

those of which are always true are

(A) All
(B) I and II only
(C) II and III only
(D) II and IV only
(E) II and V only

Let's denote $\frac{x+y}{2}=x*y$
$I$ is definetly false.
$II$ is true.
$III$ is false $x*(a+b)=\frac{x+a+b}{2}$ while $x*a+x*b=x+\frac{a+b}{2}$.
$IV$ is true $x+(a*b)=x+\frac{a+b}{2}$ and $(x+a)*(x+b)=\frac{2x+a+b}{2}=x+\frac{a+b}{2}$.
$V$ is false , if $x<y$ $x*e=x\to x=e$ and $y*e=y\to e=y$ thus contradiction ($x=y$).
So the answer is $(D)$.


For $ a,b,c,d >0$ and $ abcd\equal{}1$, we define a real function $ f$ as $ f(x)\equal{}a^x\plus{}b^x\plus{}c^x\plus{}d^x$

(a). For two positive reals $ x$ and $ y$, prove that $ x\ge y$ implies $ f(x)\ge f(y)$

(b). Prove or disprove the converse statement.

$ ya^x + (x - y).1\geq xa^y$
$ \Rightarrow\sum ya^x\geq\sum xa^y - 4(x - y) = y\sum a^y + (x - y)(%Error. "suma" is a bad command.
^y - 4)$
And $ \sum a^y\geq 4\Rightarrow f(x)\geq f(y)$

I don't understand your solution..

Part (b), please ? I got stuck..

My solution for part (a) is inspired by [b]nayel[/b]..

because $ x\ge y$, we may set $ x \minus{} y \equal{} k,k\ge 0$

Using Chebyshev, we see that :

$ a^x \plus{} b^x \plus{} c^x \plus{} d^x \ge \frac {(a^k \plus{} b^k \plus{} c^k \plus{} d^k)(a^y \plus{} b^y \plus{} c^y \plus{} d^y)}{4} \ge a^y \plus{} b^y \plus{} c^y \plus{} d^y$

Because $ a^k \plus{} b^k \plus{} c^k \plus{} d^k\ge 4\sqrt [4]{(abcd)^k} \equal{} 4$

[quote="Ronald Widjojo"]For $ a,b,c,d > 0$ and $ abcd = 1$, we define a real function $ f$ as $ f(x) = a^x + b^x + c^x + d^x$

(a). For two positive reals $ x$ and $ y$, prove that $ x\ge y$ implies $ f(x)\ge f(y)$

(b). Prove or disprove the converse statement.[/quote]

We have $ f(x) = a^x + b^x + c^x + d^x$

$ \rightarrow f '(x) = a^x.lnx + b^x.lnb + c^x.lnc + d^x.lnd$

Follow Chebyshev we have

$ f '(x) \ge\ \frac {1}{3}(a^x + b^x + c^x + d^x)(lna + lnb + lnc + lnd)
= \frac {1}{3}(a^x + b^x + c^x + d^x)ln(abcd) = 0$

So with $ x \ge\ y$ we have $ f(x) \ge\ f(y)$

Yeah, the (a) part is easy, but for part (b) ?

I can't solve it, though i can make it..  :oops:

We have $ \frac{f(x)\minus{}f(y)}{x\minus{}y}\equal{}f'(\xi)\ge0$ for a certain $ \xi\in[y,x]$ (mean value theorem). So it only remains to show $ f(x)\equal{}f(y)\Rightarrow x\equal{}y$ which is easy again.


Prove that for any four real numbers $a$, $b$, $c$, $d$, the inequality
\[ \left(a-b\right)\left(b-c\right)\left(c-d\right)\left(d-a\right)+\left(a-c\right)^2\left(b-d\right)^2\geq 0  \]
holds.

[hide="comment"]This is inequality (350) in: Mihai Onucu Drimbe, [i]Inegalitati, idei si metode[/i], Zalau: Gil, 2003.

Posted here only for the sake of completeness; in fact, it is more or less the same as http://www.mathlinks.ro/Forum/viewtopic.php?t=3152 .[/hide]

  Darij

We need to prove : $-xyz(x+y+z)+(x+z)^2(y+z)^2 \ge 0$

$\Leftrightarrow (z^2+xy+yz+zx)^2 \ge xyz(x+y+z)=xy[z(x+y)+z^2]$

Which this one is obvious true!

And I think It's still true:

$2(x-y)(y-z)(z-t)(t-x)+(x-z)^2(y-t)^2 \ge 0$


i was solving some problems the other day and saw this thing called
Descartes Theorum
something to do with simplifying polymonials
the book did a horrible job of explaining it
and i have no idea what to do

can someone explain Descartes Theorum to me?
in easy words?

Are you saying about Decartes sign rule?

I'd say he is, but it's definitely not going to be used in MOEMS :wink:
[url]http://www.purplemath.com/modules/drofsign.htm[/url]


rewrite as
 $  \frac{dy}{dx} = \frac{(3\frac{x}{y}+2\frac{y}{x}-1)}{(3\frac{x}{y}+2\frac{y}{x}+1)}$
make the subtitution  $y = vx$  ..change of variable
so  $ \frac{dy}{dx}=v+x\frac{dv}{dx}$

from then on it easy to seperate it and integrate to find v as a function of x and then multiply by x to find y the orginal function.

hello, with $ y\equal{}xv$ i have got
$ \frac{3\plus{}2v^2\plus{}v}{3\plus{}v^2\minus{}4v\minus{}2v^3}\,dv\equal{}\frac{dx}{x}$
Sonnhard.


find the product
catechol + chloroform(KOH) ->A + $ \ce{CH2I2} / \ce{KOH}$ ->B

obviously you know the first part( how cho came to the para position). in the second you dont get a carbene as it is unstable. (it is not a dihalocarbene) you get a carbocation(CH2I+)( also bcoz I- is a better leaving group as compared to cl- so removal of I- takes place b4 removal of h) which is attacked by oh. similarly the other oh attacks replacing i giving piperonal.
alternative mechanism:
CH2I2+base->CH2(OH)2->HCHO which on reaction with substituted catechol gives piperonal

Thanks how -CHO came to the para posistion?

Doesn't your book explain that? Does your book also says that the yield of A is only about 11%, and so that reaction product is just a curiosity?

no Carcul please explain the entire thing

The only explanation that comes to my mind is a statistical reason: most of the product obtained from catechol comes from the normal Reimer-Tiemann reaction*, with formyl in position 3, since this is the most favoured position (probably due to some kind of complexation); however, some dichlorocarbene may also attack position 4, since it is also electronically favoured.

* Actually, the Reimer-Tiemann reaction cannot in general be made on catechols because of the ease with which they are oxidized, specially under the basic condictions of the reaction. As far as I know, the biggest yield of 4-formylcatechol from catechol with the Reimer-Tiemann reaction that was obtained was a 29% (and only with a special technique).


The cost of five pencils and one pen is $ \$2.50$, and the cost of one
pencil and two pens is $ \$1.85$. What is the cost of two pencils and
one pen?

Let a pencil value=x and a pen=y.
Add the 2 equations together and 6x+3y=4.35.
Divide by 3 and get 2x+y=1.45.


Let K include {Rational Numbeers}  /\ (0,1)
Prove if Points of K partition (0,1) into adjacent open intervals
[b]==> K is CLOSE [/b]

Hmm... The phrase  "Points of $K$ partition $(0,1)$ into adjacent open intervals" may only mean that $(0,1)\setminus K$ is a union of open intervals and, therefore, open. Thus, $K$ is relatively closed in $(0,1)$ (but not necessarily closed: take the sequence $1,\frac12,\frac13,\dots$; it partitions $(0,1)$ into adjacent open intervals in any reasonable sense but doesn't form a closed set). In any case, the problem, as stated, is either very simple or wrong. Perhaps, I misunderstand something :?


If I want to make a Latex file (I am later going to convert it to PDF) and I want to make a nice PDF, how do I skip multiple lines between the title and the first line? The \par command only allows me to skip one line, no matter how many times I type it. Is there something I can do to skip multiple lines?

The command \vspace allows you to skip space (v=vertical, \hspace skips horizontal space). For example, \vspace{2.3in} skips 2.3 inches. You can also use other units, such as cm, pt, etc. A particularly useful one to know is \baselineskip, which is the height of a line of text. So \vspace{2\baselineskip} skips 2 lines.

Warning: don't overuse it! The automatic formatting on LaTeX is designed to be aesthetically pleasing, and artificial formatting commands such as this can make the document look less appealing. I usually only use this when positioning a picture on the page or in similar circumstances.

Also, if you want to end a page and have the text stop until the next page, you can use the \pagebreak command.


[hide]if you mean the percent chance that you will lose 75 times in a row, it's 1/655^75[/hide]

no i meant what is the chance i will lose at all in the 75 times i play

would it be

(1/655)^75

to get around a 10% chance?????

I you mean the probabiliity you'll lose only once, then
[hide]$\frac{1}{655}$ chance you'll lose once, and then to win 74 times, the probability is $(\frac{654}{655})^{74}$. However, there are 75 places to put this one loss.
So the total is $\frac{1}{655}\cdot(\frac{654}{655})^{74}\cdot75\approx.102$[/hide]

[quote="G-UNIT"][quote="mathbeginner"]Hey this is a fairly simple question I think but I don't know how to do it...

If I'm playing a game and I have a 1/655 chance to lose, what is the % chance that I will lose if I play 75 times?

Thanks.

P.S.  At first I thought to just say 75/655, but that is wrong because if that were true then if I played 655 games I would have a 100% chance to lose, which is obviously incorrect...[/quote]

  The probability that you will lose at least once, while playing 75 times is:

$1-(\frac{654}{655})^{75}= \approx.01$[/quote]Explanation of his solution:
Note that if you have two probabilities $P(A)$ and $P(\text{not }A)$ then $P(A)+P(\text{not }A)=1$.

In this case $P(A)$ is the probability that lose at least once
So $P(\text{not }A)$ is the probability that you don't lose at least once, in other words the probability that you win every single game.

So if you find the probability that he wins all $75$ games: it is $\left(\frac{654}{655}\right)^{75}$  In this case basically you just found $P(\text{not }
A)$.  So to find $P(A)$ you want $1-\left(\frac{654}{655}\right)^{75}$.

OK its done, i obviously am not good with words and its hard to explain but i figured that its going to be near 10% and thats all i wanted to know, thx for the help everyone

I think my answer comes out to be about 10 percent, saying that you will lose once and [i]only[/i] once.

[quote="mathbeginner"]Hey this is a fairly simple question I think but I don't know how to do it...

If I'm playing a game and I have a 1/655 chance to lose, what is the % chance that I will lose if I play 75 times?

Thanks.

P.S.  At first I thought to just say 75/655, but that is wrong because if that were true then if I played 655 games I would have a 100% chance to lose, which is obviously incorrect...[/quote]

[hide]
1:655::x:75
655x=75
x=75/655=15/131
15/31|65
15/31*1/65=15/2015=3/403$\approx$.7%?
[/hide]

whoa there's been about 5 answers all different, which one is correct?

I guess the wording needs to be improved.


从本站下载的 MathLinks Contest 上的题应该用什么“打开方式”来阅读？

那是 pdf 檔，你要用 Adobe Reader 來打開。 ;)
你可到以下網址下載。
[url=http://www.chinese-s.adobe.com/main.html]http://www.chinese-s.adobe.com/main.html[/url]


algebra
polynomial
modular arithmetic
coefficients
IMO Shortlist

For a polynomial $ P$ of degree 2000 with distinct real coefficients let $ M(P)$ be the set of all polynomials that can be produced from $ P$ by permutation of its coefficients. A polynomial $ P$ will be called [b]$ n$-independent[/b] if $ P(n) \equal{} 0$ and we can get from any $ Q \in M(P)$ a polynomial $ Q_1$ such that $ Q_1(n) \equal{} 0$ by interchanging at most one pair of coefficients of $ Q.$ Find all integers $ n$ for which $ n$-independent polynomials exist.

First we can remove the condition that $P(n)=0$ (it's superfluous).

Let $d = 2000$. Clearly $0,1$-independent polynomials of degree $d$ exist, so suppose $n\ne 0,1$ and an $n$-independent $P$of degree $d$ exists, with coefficients $a_0<a_1<\cdots<a_d$.

[b]Case 1: $|n| \ge 2$.[/b] Call a permutation $\pi$ [i]good[/i] if $\sum_{k=0}^{d} a_{\pi(k)} n^k = 0$, and [i]bad[/i] otherwise.

By rational approximation, we can find pairwise distinct integers $b_0<b_1<\cdots<b_d$ with $\sum b_{\pi(k)} n^k = 0$ if and only if $\sum a_{\pi(k)} n^k = 0$. Indeed, by $d+1$-dimensional Kronecker's theorem, we can find, for any $\epsilon>0$, arbitrarily large positive integers $N$ with $\| N a_i \| \le \epsilon$ for all $i$. Then we can take $b_i = [Na_i]$ (where $[x] = x - \|x\|$) if $\epsilon |1+n+\cdots+n^d| < 1$, $N>\frac{2}{\min|a_i - a_j|\setminus\{0\}}$, and $N>\frac{1}{\min|\sum a_{\pi(k)} n^k| \setminus\{0\}}$. (The second guarantees the $b_i$ are pairwise distinct, while the first and third guarantee $\sum b_{\pi(k)} n^k$, which is an integer, vanishes only when $\pi$ is good---this follows from the triangle inequality.)

[b]Lemma.[/b] Suppose there are $r_i$ indices $j$ with $b_j \equiv i \pmod{n}$. Then $\sum_{k=0}^{d} b_{\pi(k)} n^k = C$ has at most $r_0!\cdots r_{n-1}!$ solutions $\pi$. Furthermore, if equality holds, then for each $j$ with $b_j \equiv C \pmod{n}$, $\sum_{k=0}^{d-1} b_{\sigma(k)} n^k = \frac{C-b_j}{n}$ must have exactly $r_0!\cdots (r_C-1)! \cdots r_{n-1}!$ solutions $\sigma$ (with $\sigma$ restricted to the indices $k\ne j$); in particular, there must be at least one index $k\ne j$ with $b_k \equiv \frac{C-b_j}{n} \pmod{n}$.

[i]Proof.[/i] Induct on $r_0 + \cdots + r_{n-1} = d+1 \ge 1$ using the requirement $a_{\pi(0)} \equiv C \pmod{n}$. $\blacksquare$

WLOG assume $\gcd(b_0,b_1,\ldots,b_d) = 1$, so $0<r_0<d+1$, and there exists $1\le t\le n-1$ with $r_t>0$; WLOG $b_d \equiv t \pmod{n}$. Now focus on the $\pi$ with $\pi(0) = d$: they must all be bad. On the other hand, any good $\pi$ "saves" (via a single transposition) exactly one bad $\pi'$ with $\pi'(0) = d$. But each bad $\pi'$ (with $\pi'(0) = d$) is saved by at least one good $\pi$, so we must have at least $d!$ good $\pi$. Yet by the lemma and the fact that $a! b! < (a+1)! (b-1)!$ whenever $a\ge b\ge 1$, we have at most $d! 1!$ good $\pi$, with equality only when $\{r_0,r_t\} = \{1,d\}$.

Now we will invoke the equality clause of the lemma to derive a contradiction. If $r_0 = 1$, then WLOG $b_0 \equiv 0\pmod{n}$ and $b_1,\ldots,b_d \equiv t\pmod{n}$. It follows that $\sum_{k=0}^{d-1} b_{\pi(k)} n^k = \frac{-b_0}{n}$ for all $\pi$, from which we easily obtain $b_1 = \cdots = b_d$ (e.g. consider the effect of a transposition), contradiction.

Otherwise, $r_t = 1$, so $b_0,\ldots,b_{d-1}\equiv0\pmod{n}$. Take a good $\pi$ and suppose $\pi(M) = d$. If $M\le d-2$, let $\{u,v\} = \{0,1,\ldots,d\}\setminus\{\pi(0),\ldots,\pi(d-2)\}$. Then by repeated application of the equality clause, $b_{\sigma(d-1)}+nb_{\sigma(d)} = -\frac{\sum_{k=0}^{d-2} b_{\pi(k)} n^k}{n^{d-1}}$ has $2!$ solutions $\sigma$ (where $\{\sigma(d-1),\sigma(d)\} = \{u,v\}$), since $b_u,b_v \equiv 0\pmod{n}$; hence $b_u = b_v$, which is impossible.

Thus $M\ge d-1$ for any $\pi$, whence the equality clause gives us $n^k \mid b_0,\ldots,b_{d-1}$ for $k=1,2,\ldots,d-2$ (just induct on $k$). Now fix some $\pi$; the equality clause tells us there exists $\pi' \ne \pi$ with $\pi'(i) = \pi(i)$ for $i=0,1,\ldots,d-3$. We cannot have $\pi'(d-2) = d$ (by the previous paragraph) or $\pi'(M) = \pi(M) = d$ ($\pi',\pi$ can't agree in any more places), so WLOG $M = d-1\implies \pi(d-1) = d$ and $\pi'(d) = d$. Thus
\begin{align*}
0 &= \sum (b_{\pi(k)} - b_{\pi'(k)}) n^k \\
&= (b_{\pi(d-2)} - b_{\pi'(d-2)})n^{d-2} + (b_d - b_{\pi'(d-1)})n^{d-1} + (b_{\pi(d)} - b_d)n^d,
\end{align*}which is impossible mod $n^d$ (we have $n^2\mid n^{d-2} \mid b_{\pi(d-2)},b_{\pi'(d-2)},b_{\pi'(d-1)}$, but $n\nmid b_d$).

[b]Comment:[/b] The official solution uses $|n|^d > |n|^{d-1}+\cdots+1$ instead to show $a_d n^d + \cdots \ne a_0 n^d + \cdots$, whence one finds an inductive bound of $|\{\sum a_{\pi(k)} n^k\}| \ge 2^{d-1}$ (to finish, we have $\{\sum a_{\pi(k)} n^k\} \subseteq \{(a_{\pi(j)} - a_{\pi(i)})(n^j - n^i)\}$, and the latter is too small). However, only the above method generalizes if we replace the "polynomial basis" $(n^0,\ldots,n^d)$ with $(n_0,\ldots,n_d)$, where for each $k$, $\frac{n_k}{n^k}$ is an integer relatively prime to $n$.

[b]Case 2: $n=-1$.[/b] Then for any indices $i_1<i_2<\cdots<i_{1000}$, there exist indices $j,k$ with
\[ a_{i_1}+\cdots+a_{i_{1000}} = C + (a_{i_j} - a_k), \]where $2C = a_0+\cdots+a_{2000}$. This already tells us that
\[ (a_{1001}+\cdots+a_{2000}) - (a_0+\cdots+a_{999}) \le 2(a_{2000}-a_0), \]which although "unlikely," still needs a small improvement. Now note that $\{a_0+a_1+a_{i_3}+\cdots+a_{i_{1000}}\}_{1< i_3<\cdots<i_{1000}<1999}$ [url=https://artofproblemsolving.com/community/c6h79397p454617]has at least[/url] $1+998\cdot999$ elements, one of which must, by pigeonhole, take the form $C+a_u - a_v$ for $u,v\ne 0,1,1999,2000$. But for this choice of $(i_j)$, we then have
\begin{align*}
(a_{i_3}+\cdots+a_{i_{1000}}+a_{1999}+a_{2000}) - (a_0+a_1+a_{i_3}+\cdots+a_{i_{1000}})
&\le (C+a_{2000}-a_0) - (C+a_u-a_v) \\
&\le (a_{2000}-a_0) + (a_{1998} - a_2),
\end{align*}which is absurd.


[quote="harazi"]I don't even know I am posting it here (probably just to add it to the list of topics with no answer), but here it goes:
  For any $x_1,...,x_n>0$ we have the inequality:
   $\frac{x_1}{\sqrt{x_1+x_2}}+\frac{x_2}{\sqrt{x_2+x_3}}+...+\frac{x_n}{\sqrt{x_n+x_1}}\leq\sqrt{n-\frac{23}{16}}$ if $x_1+x_2+...+x_n=1$.
    I have no idea whether we can find a constant $k<1$ which does not depend on $n$ such that $\frac{x_1}{\sqrt{x_1+x_2}}+\frac{x_2}{\sqrt{x_2+x_3}}+...+\frac{x_n}{\sqrt{x_n+x_1}}\leq\ k\sqrt{n}$.[/quote]

 :)

I think I have a kind of solution.

It's clear that out sum does not exceed $\sum_{i=1}^n \sqrt{x_i}$. If one of the $x_i$ exceeds $\frac 14$, $\sum_{i=1}^n \sqrt{x_i}$ will be at most $\frac 12+\sqrt{\frac 34 (n-1)}<\sqrt{n}-\frac 12$ for $n>50$. By the methods below, replacing 3), we can prove that our sum doesn't exceed $\sqrt n-\frac 12$ for $n\geq 7$. Now suppose $n>7$. Then we may assume none of the numbers exceed $\frac 14$.
We need some preliminary results:
1)Let $x_1,x_2,\ldots,x_n$ be real numbers. Then $\sum_{i,j}(x_i-x_j)^2\geq \frac{n+5}8 \sum_{i=1}^n (x_i-x_{i+1})^2$ (subscripts are taken modulo $n$).

Proof: let $\delta_k=\sum_{i=1}^n (x_{i+k}-x_i)^2$. By using the inequality $x^2+y^2\geq \frac{(x+y)^2}2$ we get $(x_{i+k}-x_i)^2+(x_{i+k+1}-x_i)^2\geq \frac{(x_{i+k+1}-x_{i+k})^2}2$. Now summing over all $i$ we get $\delta_k+\delta_{k+1}\geq \frac{\delta_1}2$. From here, we easily derive that our sum is $\delta_1+\frac{\delta_2+\delta_3+\ldots+\delta_{n-2}}2\geq \frac{n+5}8 \delta_1$, as desired.

2)Denote by $s=\sum_{i=1}^n \frac{(x_{i+1}-x_i)^2}{x_i+x_{i+1}}^2$. Then $\sum_{i=1}^n \sqrt{x_i} \leq \sqrt{n-\frac{n+5}{16}s}$.

Proof: we have $n(\sum_{i=1}^n x_i)-(\sum_{i=1}^n \sqrt{x_i})^2=\sum_{i,j}(\sqrt{x_i}-\sqrt{x_j})^2\geq \frac{n+5}8 \sum_{i=1}^n (\sqrt{x_i}-\sqrt{x_{i+1}})^2$, according to (1). Finally, $(\sqrt{x_i}-\sqrt{x_{i+1}})^2=\frac{(x_i-x_{i+1})^2}{(\sqrt{x_i}+\sqrt{x_{i+1}})^2}\geq \frac{(x_i-x_{i+1})^2}{2(x_i+x_{i+1})}$. From here we deduce the conclusion.

3)We have $\sum_{i=1}^n \frac{x_i}{\sqrt{x_i+x_{i+1}}}\leq \sum_{i=1}^n \sqrt{x_i}- \frac{x_ix_{i+1}}{x_i+x_{i+1}}$.

Proof: We have $\frac{x_i}{\sqrt{x_i+x_{i+1}}}=\sqrt{x_i}-\frac{\sqrt{x_i}(\sqrt{x_i+x_{i+1}}-\sqrt{x_i})}{\sqrt{x_i+x_{i+1}}}=\sqrt{x_i}-\frac{x_{i+1}\sqrt{x_i}}{\sqrt{x_i+x_{i+1}} (\sqrt{x_i+x_{i+1}}+\sqrt{x_i})}$. We can deduce $\frac{\sqrt{x_i}}{(\sqrt{x_i+x_{i+1}}+\sqrt{x_i})(\sqrt{x_i+x_{i+1}})}>\frac{x_i}{x_i+x_{i+1}}$ for $x_i<\frac 14$, so our term is at most $\sqrt{x_i}-\frac{x_ix_{i+1}}{(x_i+x_{i+1})}$. By summing over all (i), we get our conclusion.

Now, let's finish the problem. Set $s$ as in 2). Then $\sum_{i=1}^n \frac{x_ix_{i+1}}{x_i+x_{i+1}}=\sum_{i=1}^n \frac{x_i+x_{i+1}}4-\frac{(x_i-x_{i+1})^2}{4(x_i+x_{i+1})}=\frac 12-\frac s4$.

$\sum_{i=1}^n \frac{x_i}{\sqrt{x_i+x_{i+1}}}\leq \sum_{i=1}^n \sqrt{x_i}- \frac{x_ix_{i+1}}{x_i+x_{i+1}}\geq \sqrt{n-\frac{n+5}8s}-\frac 12+\frac s4$. However, it's clear that $\sqrt{n-\frac{n+5}8s}<\sqrt n-\frac{n+5}{16\sqrt n} s<\sqrt n-\frac s4$ for $n\geq 7$. So for $n\geq 7$ our sum is at most $\sqrt n-\frac 12$, which is more than we wanted.

By replacing $\frac 14$ by $a=O(\sqrt[3]{n})$, we can deduce that the sum is at most $\sqrt n- \lambda \sqrt[6] n$, for some constant $\lambda$.


inequalities
number theory unsolved
number theory

Prove that n!m!(m+n)!|(2n)!(2m)! where n, m are natural numbers.

Posted before .
Use the inequalities :
$ [2x]\plus{}[2y]\geq [x]\plus{}[y]\plus{}[x\plus{}y],\forall x,y\geq 0$

if $ p^a$ devides $ n!$ and $ a$ is the largest integer such that this holds, than $ a\equal{}\sum_{i\equal{}1}^{\infty}[\frac{n}{p^i}]$

we have to prove that the exponent of every prime number $ p$ is smaler in integer factorization of $ m!n!(m\plus{}n)!$ than in $ (2n)!(2m)!$.
so we have to prove the following inequality:
$ \sum_{i\equal{}1}^{\infty}[\frac{n}{p^i}] \plus{} \sum_{i\equal{}1}^{\infty}[\frac{m}{p^i}] \plus{} \sum_{i\equal{}1}^{\infty}[\frac{n\plus{}m}{p^i}] \leq \sum_{i\equal{}1}^{\infty}[\frac{2n}{p^i}]\plus{}\sum_{i\equal{}1}^{\infty}[\frac{2m}{p^i}]$

which is true because of the inequality TTsphn mentioned above, when used for every $ i$


I've got a question that I'm having some issues with if someone can give me some guidance, it would be appreciated.

If $ y$ is given by a composite function of $ x$, use the Chain Rule to show that:

\[ \frac{d^{2}y}{dx^{2}}=\frac{d^{2}y}{du^{2}}\cdot\left(\frac{du}{dx}\right)^{2}+\frac{dy}{du}\cdot\frac{d^{2}u}{dx^{2}}\]

Thanks in advance.

I think this works...

$ \frac{d^{2}y}{dx^{2}}= \frac{d}{dx}\left(\frac{dy}{dx}\right)\\ = \frac{d}{dx}\left(\frac{dy}{du}\cdot\frac{du}{dx}\right)\\ = \frac{d}{dx}\left(\frac{dy}{du}\right)\cdot\frac{du}{dx}+\frac{d^{2}u}{dx^{2}}\cdot\frac{dy}{du}\\ = \frac{d^{2}y}{du^{2}}\cdot\left(\frac{du}{dx}\right)^{2}+\frac{dy}{du}\cdot\frac{d^{2}u}{dx^{2}}$


ARML is putting together their next book of contests, and they need the full list of winners from 2004 and 2005.  They have first place, but are looking for the rest.  If you have that information, can you post it here?

don't they already have the results?  After all, they are the ones that make the contest!

I actually think the ARML winners can be found at the ARML website.

They have full team results, but not individuals; that was settled in the tiebreaker, and apparently their records got lost. The web site has the individual top ten for 2006 and later only.

Here's what I can reconstruct from old threads here:

2004:
1. Aaron Pixton, Upstate New York
2. Jongmin Baek, SFBA A
3. Anders Kaseorg, North Carolina
Other perfect scores:
Ricky Biggs, TJHSST A
Brian Lawrence, Montgomery County A
Eric Price, TJHSST A

Sevens included:
Darren Yin, SFBA A
Adam Hesterberg, Washington A
Alison Miller, Upstate NY
Dmitriy Karabash, NYC A
Yan Zhang (compucomp), Lehigh Valley?
(rand%(0)), Missouri?

There were about 10-15 more sevens, and they may not have been in the tiebreaker at all. In fact, the individual results seem to have been announced differently at the three sites, with only the Penn State site hearing about all of the perfect scores.

For 2005, I can't find any individual results at all.

I have to agree with abcak; it strikes me as extremely strange that such a prominent national, well-organized competition would not have the results of their competition from just a few years ago...

The 7's from Iowa in 2005 were Zach Abel and one other student whose name I don't remember (he was on the Chicago team, and I seem to recall that he might not have been on Chicago A).  I think Zach placed second or third; the other student did not place.  I'm getting into pretty wild guesses here, but I think Ryan Ko won that year (or if not, he was second and Zach was third).

That was also the first year site awards were given out, so that other student would have been the Iowa site winner.  If anybody out there thinks they might recognize him, I've got his picture and can send it to you.

The following is all from what I remember from actually being at ARML 4 years ago. There was only one 7 at the San Jose ARML site in 2005, and that was Curtis Liu from SFBA. I believe it was a 3-way tiebreaker over the different sites to determine first place.

The SoCal team in 2005 had three people who scored 6, and I think I remember them taking part in some form of tiebreaker - although I'm pretty sure that would have been for site honors, not national honors. What "Myself" is saying is consistent with what I remember, but I don't remember anything about the result of the national tiebreaker.

We had one person with a 6 in 2004; again, that didn't put him in the national tiebreaker.

According to Wikipedia, Ryan Ko won in 2005.


Perform the addition or subtraction or operation and write the result in standard form

-(3/4 + 7/5i) - (5/6 -1/6i)

(3 + sqrt of -5)(7 - sqrt of -10)

This is exersise for ... well, I'll better silent.
In any case it is not good for this section.


assume that $ a\sqrt{6} \plus{} b\sqrt{10} \plus{} c\sqrt{15}\in\mathbb{Q}$ for some $ a,b,c\in\mathbb{Z}$,

does it imply that $ a\equal{}b\equal{}c\equal{}0$ ?

(and why, if so....)

A bit of theory:  $ \mathbb{Q}[ \sqrt{2}, \sqrt{3}, \sqrt{5} ]$ has dimension $ 8$ over $ \mathbb{Q}$ and can be given a basis $ \{ 1, \sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{6}, \sqrt{10}, \sqrt{15}, \sqrt{30} \}$.  This can be proven by adjoining $ \sqrt{2}, \sqrt{3}, \sqrt{5}$ separately, and generalizes to the square root of every prime.


linear algebra
matrix
algebra
polynomial
search

How can we find all eigenvalues of nxn all-ones matrix?

In other words, the matrix known as $ J.$

This is really pretty simple.

Direct multiplication gives us that $ J^2\equal{}nJ.$ That means that $ J$ satisfies the polynomial $ x^2\minus{}nx$ (in fact, that's its minimal polynomial), which means that any eigenvalues of $ J$ must also satisfy $ x^2\minus{}nx\equal{}0.$ So the only possible eigenvalues are $ 0$ and $ n.$

I'll let you finish it: how many of each of those must we have? (That is, I'm asking about the algebraic multiplicity of each of those two eigenvalues.)

Thx, I have gotten the answer.
There are (n-1) 0s and 1 n
Actually, the full question is that
[img]http://img131.imageshack.us/img131/9812/12842740.jpg[/img]

I am afraid I get the wrong eigenvectors
Here is my work
[img]http://img131.imageshack.us/img131/949/ans1.jpg[/img]

It is for eigenvalue 0

There are (n-1) eigenvectors like that column vector.

However, when I want to search for the answer form the internet.
I find that the answer is one -1  and one 1 in the column vector.

for the eigenvalues of n, the answer is all 1. I can't find this answer. Is it right? I am afraid of the wrong eigenvectors will get the wrong matrix P.

for the following part. I think I can try to cope with it.
In a iii , P should be the component of corresponding eigenvectors.

a iv Form the "trend" of the eigenvectors value, i actually guess that P may be an orthogonal matrix.. Otherwise, I think it is quite difficult to obtain it.

It seems that I can think of where the problems lie on. I can solve the whole question. Thankyou.


[quote="pascal_1623"]Tamyra is making four cookies and has exactly four chocolate chips. If she distributes the chips randomly into the four cookies, what is the probability that there are no more than two chips in any one cookie? Express your answer as a common fraction.[/quote]

[hide]I got $\frac{19}{35}$...don't know if it's right[/hide]

[quote="pascal_1623"]Tamyra is making four cookies and has exactly four chocolate chips. If she distributes the chips randomly into the four cookies, what is the probability that there are no more than two chips in any one cookie? Express your answer as a common fraction.[/quote]

Two questions:

Are the cookies distinct?
Are the chips distinct?

I remember the answer being something different but then again, I did it over a year ago.

[hide="Solution"]We will find the probability that there are [i]more[/i] than two chips in some of the cookies and subtract that from 1. There are two cases:

[i]Case 1:[/i] There are 3 chips in one cookie and 1 chip in another. There are a total of 4 x 3 ways to select the cookies and 4 ways to choose the chips. Thus there are a total of 48 ways for this case.

[i]Case 2:[/i] There are 4 chips in one cookie. There are just 4 ways to select the cookie.

Now we add up the total number of ways that satisfy either of these cases and divide by $4^4$, the total number of ways to put the chips in the cookies (each of 4 chips can go into one of 4 cookies), and the probability of either case 1 or 2 occurring is $\frac{52}{256}=\frac{13}{64}$, so our answer is $1-\frac{13}{64}=\boxed{\frac{51}{64}}$.[/hide]

[quote="Max300"][hide]We need to consider the number of partitions of the 4 chocolate chips.

There are 4 partitions if we put all 4 chips into one cookie.
There are 12 partitions if we put 3 chips in one cookie and 1 in another.
There are 6 partitions if we put 2 chips into one cookie and 2 into another.
There are 12 partitions if we put 2 chips into one cookie and put the other two in different cookies.
And finally there is 1 partition if we put each chip into a different cookie.

There are a total of 35 partitions with 19 of them having no more than two chips in a cookie.

Thus our answer is $\frac{19}{35}$[/hide][/quote]

[hide="Comment"]
The answer is incorrect since  you are distinguishing among the cookies but not among the chips.

You have to distinguish among both.[/hide]

[quote="SplashD"][quote="pascal_1623"]Tamyra is making four cookies and has exactly four chocolate chips. If she distributes the chips randomly into the four cookies, what is the probability that there are no more than two chips in any one cookie? Express your answer as a common fraction.[/quote]

Two questions:

Are the cookies distinct?
Are the chips distinct?
[/quote]

Am I supposed to consider my questions answered?

I think nebula's answer is right, I remember the answer being fifty-something over 64

[quote="SplashD"][quote="SplashD"]
Two questions:

Are the cookies distinct?
Are the chips distinct?
[/quote]

Am I supposed to consider my questions answered?

I think nebula's answer is right, I remember the answer being fifty-something over 64[/quote]

Yes I think I sort of answered it in my previous post.

They both are distinct.

I got something along those lines, and basically did the same thing as [b]Max300[/b]...i think my answer is a little different

[hide="something along the lines of Max300"]
$9/17$

here's the work...("o"s are chips, "x"s are cookies, not that it matters)

[b]o o o o

x x x x[/b]

4 chips per cookie

4000
0400
0040
0004

3 chips per cookie + 1 chip per cookie

1300
1030
1003-3
3010
3100
3001-6
0031
0013
0103-9
0301
0130
0310-12

2 chips per cookie + 2 chips per cookie

2020
2200
2002-3
0022
0202
0220-6

2 chips per cookie + 2 chips in different cookies

2110
2011
1021-3
1012
1102
1120-6
1210
2011
2101-9
0112
0121
0211-12

12+6+12+4=34
6+12=18

9/17[/hide]

[quote="bpms"][quote="Ignite168"][quote="rruszyk"]I get the same answer.[/quote]

You're probably going to be banned for impersonation, but before you are, who are you?[/quote]
Hey, we don't have to click on your sig to quote anymore.[/quote]

That's because you need a double hide tag to hide the quote button.

51/64 is correct. I have seen the answer key.


1.) Is the set D = {(1,1,2),(5,-2,3)} a spanning set?

2.) If the vector(1,0,0) is added to set D, will the enlarged set be a spanning set?

[quote="Kent Merryfield"]Spanning set of what? Unless you specify what space or subspace you are trying to span, the question cannot be interpreted.[/quote]

I guess there is something wrong with the book. I suppose, it is 3-space.

All right, we'll assume you're talking about spanning $ \mathbb{R}^3.$

Two vectors cannot possible span $ \mathbb{R}^3.$ An spanning set for a three-dimensional space must have at least 3 elements.

Now if we add that particular third vector, would the resulting set span $ \mathbb{R}3?$ There's a way to find out. Write these vectors as columns, and stack them together as a matrix:
\[ \begin{bmatrix}1&1&5\\0&1&\minus{}2\\0&2&3\end{bmatrix}\]
Now, perform Gaussian elimination. When you reach row echelon form, is there a pivot in every row? If there is if pivot in every row, then the set does span $ \mathbb{R}^3.$ Otherwise, it doesn't.

That's also why the two-element set cannot span $ \mathbb{R}^3.$ Take the $ 3\times 2$ matrix you get and row eliminate it. There can be at most two pivots, and with three rows, there can't be a pivot in every row.


trigonometry
function
algebra
functional equation
system of equations
algebra unsolved

Determine all functionf $f : \mathbb{R} \mapsto \mathbb{R}$ such that

\[ f(x+y) = f(x)f(y) - c \sin{x} \sin{y}  \] for all reals $x,y$ where $c> 1$ is a given constant.

First put in $y=0$. You get $f(x)=0 \; \forall x$ or $f(0)=0$. So $f(0)=0$, because the first function doesn't work.
Now put $x=t,y=\pi$ for any real $t$. You get $f(\pi + t)=f(\pi )f(t)$. For $x=\frac{\pi }{2}+t,y=\frac{\pi }{2}$ you get $f(\pi +t)=f(\frac{\pi }{2})f(\frac{\pi }{2} + t) - c \cos t$. Finally for $x=\frac{\pi }{2}, y=t$ you get $f(\frac{\pi }{2} + t)=f(\frac{\pi }{2})f(t)-c \sin t$. Now you have a system of equations (if you put in the original functional equation $x=y=a$ ($a=f(\frac{\pi }{2})$) you get $f(\pi )=a^2-c$) and from it you get $f(t)=a \sin t + \cos t$. To get $a$ you put $t=\pi$ and from that $a_1= \sqrt{c-1}$ and $a_2=- \sqrt{c-1}$.


Using two distinct members of the set $ \{0, 1, 2, 3, 4\}$, how many different sums can be attained?

The least sum is $ 0\plus{}1\equal{}1$ and the greatest is $ 4\plus{}3\equal{}7$.  Clearly all sums in between can be attained.  Then we have $ 7\minus{}1\plus{}1\equal{}\boxed{7}$ different sums.


Provide a synthesis of acetophenone from ethylbenzene.

I am not sure if this works (just starting to learn about Organic Synthesis). Please guide me along where I am going wrong, thanks.

I would start off by oxidising the ethylbenzene by a powerful oxidant (e.g. hot, concentrated, acidified $ KMnO_4$):

$ C_6H_5C_2H_5 \plus{} 6[O] \rightarrow C_6H_5COOH \plus{} 2H_2O \plus{} CO_2$

I am thinking along the lines of reducing the carboxyl group to a primary alcohol (e.g. $ LiALH_4$) and maybe oxidising that to an aldehyde.

The aldehyde can then be converted to a secondary alcohol using the Grignard Reagent ($ CH_3MgI$) and then oxidised to the ketone desired.

Are the reductants that can convert carboxylic acids directly to aldehydes?

[quote="BanishedTraitor"]Please guide me along where I am going wrong, thanks.[/quote]

In organic synthesis it is common practice to not show balanced chemical reactions: we are just to show the relevant information: starting material(s), product(s), and reaction conditions.

Your synthesis is correct, good job. However, it uses too many steps: two are enough.

[quote="BanishedTraitor"]Are the reductants that can convert carboxylic acids directly to aldehydes?[/quote]

Yes there are (via the formation of some intermediate or by some indirect route), but the common practice (specially in organic chemistry books) is to first convert the acid into an acid chloride, esther, or cyanide, and then reduce one of these with a reducing agent not so powerfull as LiAlH4 (for example, DIBALH). Here's an indirect route that I particularly like:

1) Treat the acid with two equivs. of butyl lithium;
2) Treat with one equiv. of ethyl formate.
3) Make an acidic workup and heat the mixture.

The explanation of these steps will be left as an additional exercise.

Hmmm...... I really need to learn more organic reactions.

P.S. What are some useful chain lengthening and chain degradation process?

[quote="BanishedTraitor"]What are some useful chain lengthening and chain degradation process?[/quote]

In this particular case you don't need that (just oxidation reactions). However:

1) Chain lengthening reactions: aldol, SN with alkyne anions and organometalics, Fridel-Crafts, Wittig, Michael, Diels-Alder.

2) Chain degradation reactions: oxidative cleavage of alkenes and alkynes (ozonolysis), oxidation of side chains in aromatic compounds.

Well I tried working backward but I always get stuck at some point when I need the branch to be halegen substitued but I only know using $ hf$ and $ X_2$ (radical reaction).

That's very close. Here are some of my approaches:

[hide="Two step synthesis"]1. $ C_6H_5CH_2CH_3 \plus{} Br_2/light \longrightarrow C_6H_5CHBrCH_3$

2. $ C_6H_5CHBrCH_3 \plus{} DMSO \longrightarrow C_6H_5COCH_3$.[/hide]

[hide="Three step synthesis"]1. $ C_6H_5CH_2CH_3 \plus{} KMnO_4/HO^\minus{}/heat \longrightarrow C_6H_5CO_2H$

2. $ C_6H_5CO_2H \plus{} 2\,CH_3Li \longrightarrow C_6H_5C(CH_3)(O^\minus{})_2$

3. $ C_6H_5C(CH_3)(O^\minus{})_2 \plus{} H_2O \longrightarrow C_6H_5COCH_3$.[/hide]

[hide="One step synthesis"]$ C_6H_5CH_2CH_3 \plus{} SeO_2 \longrightarrow C_6H_5COCH_3$.

Other oxidizing agents include CrO3/acetic acid, Jones reagent, and PCC.[/hide]


I started to write mock SAT. For this summer, SAT is all I'm doing and I thought that I would score better if I can think as a ETS's view as well. There isn't official date yet but I wanted to see how many people will be interested on taking this mock SAT.

By the way, math section will be harder than the usual so you can even use this for AMC practice haha.

Anyone?  :)

Sure I'll take it but I'll probably end up skipping the essay because its so boring and I hate writing.

[quote="SplashD"]I 'll do it but what are you gonna do with the essay section?[/quote]

For essay section, I'll grade as follows:

I'll take all answers and go through them once for 3 minute. This will be the first score. Then after I went through them once, I'll take about 5-6 minute detailed examination to give the second score. So, your score will be holistic, just like in reality (except I play two roles).

I would recommend you to do essays because if you don't, your writing score will be relatively low.

count me in, but I probably won't turn in the essay. I will try it of course, so it won't go to waste.

I will try to see if I have time to do it.

So I guess I'm in. :)

[quote="SplashD"]Are you certified to holistically grade an essay?

Can we not include the math section? Everybody knows over 90% of AoPsers can get a perfect 800 on it, so it would be pointless.[/quote]

I'm not certified. I'm just junior student who just wants to feel like a test maker. If you don't like my grading, fine. But I'm going to do my best to give a fair score.

Math section has to be included. I know that 90% of AoPSers can get perfect score. However, that doesn't mean that all of them will get perfect scores. In math section, perfect score depends on not making careless mistakes, which are also important on the real one.

Plus, I'm going to try to make it harder than usual SAT style.

OK, I'll try, if you will be so kind as to remind me by PM before the test. :)

[quote="SplashD"][quote="Silverfalcon"]
Plus, I'm going to try to make it harder than usual SAT style.[/quote]

The math section or the whole test?

Just out of curiosity what was your highest SAT score so far?[/quote]

Nothing is certain yet.

My SAT score is pretty low (I rather not say the number but I'm still practicing). As I said, this is not anything that's goign to happen very soon. I'm going to do lots of practices to see the styles of questions and create (and borrow some problems from books) a test that will be very similar to real one.

[quote="Silverfalcon"][quote="SplashD"]Would you be able to finish it before school starts?[/quote]

Yup.

Moreover, for all students who submit answers, they're going to receive full solution and for essay section, your feedback and sample essays.[/quote]

Are you going to write the sample essay? By the way, are you going to make up the topic for the essay or take an existing one?

The topic should involve penguins.

And count me in.

bump

also, is this going to happen seeing as how summer's almost over?

[quote="moogra"]if it is harder, then how is a mock sat i? it should be mock sat 2? and also where are teh problems[/quote]
Problems haven't been posted yet... :huh:


function
algebra
polynomial
ratio
rational function
functional equation
number theory

Find all functions such that f (3x + 2y) = f (2x) + f (3y) + 6xy
If there does not exist such the function justify your answer.
x,y are rationals and f (x) is a rational function

[hide="Some thoughts/questions related to this problem and other functional equations in general"]I was wondering is it possible to assume f(x) = (a (x^3)) + (b (x^2)) + (c x) + d and solve for f (x) ? Even if it does give a solution would it be possible to find any other functions?  Also if it is the only function, how would you prove it?[/hide]

Regarding the functional equation
[hide]
$ f(3x \plus{} 2y) \equal{} f(2x) \plus{} f(3y) \plus{} 6xy$; $ x$, $ y$ rational; $ f$ rational-valued.

Assume such a function $ f$ exists.
Putting y=0 in the original equation,
$ f(3x) \equal{} f(2x) \plus{} f(0)$.
Putting x=0 in the original equation,
$ f(2y) \equal{} f(3y) \plus{} f(0)$.
Putting x=y and adding these two equations,
$ f(0) \equal{} 0$.
Thus
$ f(3x \plus{} 2y) \equal{} f(2x) \plus{} f (3y) \plus{} 6xy \equal{} f(3x) \plus{} f(2y) \plus{} 6xy \equal{} f(2x \plus{} 3y)$.

Now put $ x \equal{} \minus{} \frac {3}{2}y$ above and get

$ f( \minus{} \frac {5}{2}y) \equal{} f(0)$
so $ f$ is constant, but that would mean $ 6xy$ is constant, which is false.

Thus such a function does not exist.
[/hide]
Regarding your question, [hide]
you may indeed do that to find a particular solution. I don't think you get marks "for the answer."
However, first you might try to prove that $ f$ takes a particular form. I do not see this so often; instead usually you just use the properties in the question to construct the answer (perhaps with some guessing in the scrap work).
If you don't show that f is a particular kind of function (for example, polynomials $ F$would arise under certain conditions; and linear functions arise in $ F(m \plus{} n) \equal{} F(m) \plus{} F(n)$ for $ m$ and $ n$ rational; and multiplicative arithmetical functions arise in $ F(ab) \equal{} F(a)F(b)$ for $ a$ and $ b$ relatively prime), but to show that the solution $ F$ to a functional equation is unique, perhaps you may
"Let $ F$ and $ G$ be solutions to --(some given functional equation)--" and then show that $ F(t) \minus{} G(t) \equal{} 0$ or $ F(t)/G(t) \equal{} 1$ everywhere, or even "suppose there exists some $ t$ for which $ F(t)$ > $ G(t)$" (and prove by contradiction) etc. I don't know what to say except that the conclusion you will be looking for will be similar to proving to real numbers are equal, or that two sequences are identical.
Instead what we usually do is just ... find the function directly. I suppose if you use that method to find a particular solution, and if you are "confident" that there are no others, you may use facts about your particular solution as hints.
It becomes much different if $ F$ is known to be continuous! Then you can consider small changes. But then, when you know more facts about the function you would just be using those facts...[/hide]

I thought "rational function" usually means "ratio of polynomials with integer coefficients." Correct me.


Find all pairs of non-negative integers $ (m, n)$ such that $ m^{n}\equal{}n^{m}$.

[quote="AndrewTom"]Find all pairs of non-negative integers $ (m, n)$ such that $ m^{n} \equal{} n^{m}$.[/quote]
Did you mean to place the restriction that m doesn't equal n?

[quote="varunrocks"]>> (0,0) << and (2,4)[/quote]

I say you can't use $ (0, 0)$ because $ 0^0$ is indeterminate
to begin with.

Taking the ln of both sides, $ \frac {ln (m)}{m} \equal{} \frac {ln (n)}{n}$. But $ f(x) \equal{} \frac {ln (x)}{x}$ has a maximum at $ x \equal{} e$, so if we don't have $ m \equal{} n$, we must have one of $ m$,$ n$ be either $ 1$ or $ 2$. There are clearly no solutions for $ m \equal{} 1$ and as said before, we have the solutions $ (2,4)$ and $ (4,2)$.


The sum of a negative integer, $ N$, and its square is 6. What is
the value of $ N$?

$ N^2\plus{}N\minus{}6\equal{}0$
Solving quadratics, we get $ N\equal{}\minus{}3$, because $ N<0$


Hi all

I have done 5 out of 6 given questions in INMO09. What are my chances????   :maybe: :huh:

Do keep in mind i am in XII

Do Reply to
  :arrow:  http://www.mathlinks.ro/Forum/viewtopic.php?t=252391

You have the best chances provided u have written all of them very properly. I suppose u didnt solve the combi. :) {An INidan would be so often}.
12thers here say they have got only 3 or 3 and a half. So obvioulsy u have great chances.
Congrats. I fpossible pl tel, how others {11th and below} did in ur centre. And continue this in the INMO Discussion Thrad.

ya mostly in my center were saying they did 2 -3 questions.

the one quest i m not sure is 4th.

& last question i did by geometry.

Very good Naveen.
Yeah a nice solution exists by geometry.
And also, 4th sum is misleading wit its geometric progression.

Anyone getting 4 or 5 correct and is from class 12  :?:

and 12thers what are ur scores in Fiitjee PT3?

well i cracked 4 out of 6 wat are my chances. and for naveen great if i get through we shall surely meet at the IMOTC. best of luck

Hello naveen and inabash.

Last year the cut-offs were 40 for everyone not in 12th standard and 50 for those students in 12th standard.
Literally 50 translates to just getting three problems correct. :D However finally it just boils down to whether the solutions are "rigorous". Yeah last year a 12thie from my center got through with 4 questions right. The cutoffs never go above 50 or 55 for 12thies. So all the best and hope u make it

[quote="Akashnil"]Well there is some1 in the name of Sanjit in the list .. hope he is naveenkr, coz his email address is Sanchitml@gmail.com . ch and j ... or maybe his name is sth else ...[/quote]
Actually, I think Sanchit and Naveen are best friends and use the same ID?  :rotfl:
But I don't think either qualified coz they are in 12th to the best of my knowledge.  :huh:


$ \lim_{p \to \infty} \left( \frac{ (px\plus{}1)^e}{p^{e\minus{}1} \sqrt[p]{e^{px\plus{}1}}} \right) \minus{} \lim_{p \to \infty} \left( \frac{px^e}{e^x} \right) \equal{}$

NB : $ x$ is a variable, and $ e \approx 2.71828...$

Hmm, no one?
Big hint : Let $ f(x) \equal{} \frac{x^e}{e^x}$

The correct statement should be :  $ \lim_{p \to \infty} \left( \frac{ (px\plus{}1)^e}{p^{e\minus{}1} \sqrt[p]{e^{px\plus{}1}}}  \minus{}  \frac{px^e}{e^x}\right)$

From $ \frac{ (px\plus{}1)^e}{p^{e\minus{}1} \sqrt[p]{e^{px\plus{}1}}}  \minus{}  \frac{px^e}{e^x} \equal{} \frac{ (x\plus{}1/p)^ee^{\minus{}1/p} \minus{} x^e}{1/p}$ it's easy to see that the required limit is $ f'(0)$ where $ f(t)\equal{}(x\plus{}t)^ee^{\minus{}t}$ .
 :cool:


geometry
circumcircle
trigonometry
geometric transformation
reflection
symmetry
Miquel point

Triangle $ ABC$ is inscribed in circle $ \omega$. The tangent lines to $ \omega$ at $ B$ and $ C$ meet at $ T$. Point $ S$ lies on ray $ BC$ such that $ AS \perp AT$. Points $ B_1$ and $ C_1$ lie on ray $ ST$ (with $ C_1$ in between $ B_1$ and $ S$) such that $ B_1T \equal{} BT \equal{} C_1T$. Prove that triangles $ ABC$ and $ AB_1C_1$ are similar to each other.

Nice problem!  What number was it?

Unfourtunately, I have a very long and ugly proof for this, but I hope it is worth mentioning.  A shorter proof would definitely be more appropriate and appreciated:
[hide="Solution"]
Allow us to take points $ B'$ and $ C'$ so that $ AT$ bisects $ \angle BAC'$, $ AT$ bisects $ \angle B'AC$, and $ TB \equal{} TC \equal{} TB' \equal{} TC'$ (so that $ B'$ is closer to $ B_1$ than $ C_1$ and $ C'$ is closer to $ C_1$ than $ B_1$).  Also, allow the intersection between $ B'C'$ and $ Bc$ to be $ S'$.  Now, we call $ \angle BAB' \equal{} \alpha$ and $ \angle B'AT \equal{} \beta\implies \angle TAC \equal{} \beta$ and $ \angle CAC' \equal{} \alpha$.

[b]Lemma 1: $ B'$, $ T$, and $ C'$ are collinear[/b]
[i]Proof:[/i] We notice that $ \angle BAC \equal{} \angle TBC \equal{} \angle TCB$ and that $ \angle BAC \equal{} 2\beta \plus{} \alpha$.  Thus, $ \angle BTC \equal{} 180 \minus{} 2\angle TBC \equal{} 180 \minus{} 4\beta \minus{} 2\alpha$.  Now, we have that $ AT$ bisects $ \angle B'AC$ and $ B'C \equal{} TC$, so $ ACTB'$ is cyclic.  Similarly, $ BAC'T$ is cyclic.  Thus, $ \angle CTB' \equal{} 180 \minus{} \angle B'AC \equal{} 180 \minus{} 2\beta$ and $ \angle BTC' \equal{} 180 \minus{} \angle BAC' \equal{} 180 \minus{} 2\alpha \minus{} 2\beta$.  So, $ \angle BTB' \equal{} \angle B'TC \minus{} \angle BTC \equal{} 180 \minus{} 2\beta \minus{} (180 \minus{} 4\beta \minus{} 2\alpha) \equal{} 2\alpha \plus{} 2\beta$.  We can also find that $ \angle CTC' \equal{} \angle BTC' \minus{} \angle BTC \equal{} 180 \minus{} 2 \minus{} \alpha \minus{} 2\beta \minus{} (180 \minus{} 4\beta \minus{} 2\alpha) \equal{} 2\beta$.  Thus, we find that $ \angle B'TC' \equal{} \angle BTB' \plus{} \angle BTC \plus{} \angle CTC' \equal{} 2\beta \plus{} 180 \minus{} 4\beta \minus{} 2\alpha \plus{} 2\beta \plus{} 2\alpha \equal{} 180$, and thus $ B'$, $ T$, and $ C'$ are collinear and Lemma 1 is proven.

[b]Lemma 2: $ AS'\perp AT$[/b]
[i]Proof:[/i] We notice that $ C'T \equal{} TB' \equal{} TC \equal{} TB$ and thus $ CBB'C$ is cyclic since it has center $ T$.  Furthermore, $ C'$, $ T$, and $ B$ are collinear, and thus $ C'B'$ is the diameter of the circumcircle of $ CB'BC'$.  Thus, we find that $ \angle SCB' \equal{} \angle BCB' \equal{} \frac {\angle BTB'}{2} \equal{} \beta$.  Also, we  have isoceles triangle $ CTB'$ and $ \angle CTB' \equal{} 180 \minus{} 2\alpha \minus{} 2\beta$.  Thus, $ \angle CB'T \equal{} \alpha \plus{} \beta$.  Now, we notice that
\[ \angle CSB' \plus{} \angle SCB' \equal{} \angle CB'T\implies \beta \plus{} \angle CSB' \equal{} \alpha \plus{} \beta\implies \angle CSB' \equal{} \alpha \equal{} \angle BAB'
\]
and thus $ \angle BSB' \equal{} \angle BAB'$ thus giving us cyclic quaderilateral $ ASBB'$.  Now, this implies that
\[ \angle B'AS \equal{} \angle B'BS \equal{} 180 \minus{} \angle B'BC \equal{} \angle TC'C \equal{} \frac {180 \minus{} \angle CTC'}{2} \equal{} \frac {180 \minus{} (2\alpha \plus{} 2\beta)}{2} \equal{} 90 \minus{} \alpha \minus{} \beta
\]
Thus, we see that $ \angle TAS \equal{} \angle TAB \plus{} \angle BAB' \plus{} \angle B'AS \equal{} \beta \plus{} \alpha \plus{} 90 \minus{} \alpha \minus{} \beta \equal{} 90\implies TA\perp AS'$.

[b]Lemma 3: $ B' \equal{} B_1$, $ C' \equal{} C_1$, and $ S' \equal{} S$.  [/b]
[i]Proof: [/i] We know that $ S$ is on $ BC$ and $ SA\perp AT$ and we know that $ S'$ is on $ BC$ and that $ S'A\perp BC$.  Thus, we have that the intersection between the perpindicular to $ TA$ at $ A$ and $ BC$ is both $ S$ and $ S'$.  Two lines only meet at one point, so $ S \equal{} S'$.  Furthermore, $ B'$ and $ C'$ are located on $ ST$ so that $ B \equal{} B'$ and $ C \equal{} C'$.  Now, $ B_1$ and $ C_1$ therefore follow the same conditions as $ B'$ and $ C'$ respectively, so $ B_1 \equal{} B'$ and $ C_1 \equal{} C'$.

[b]Solution to Problem[/b]
From here, the problem finishes quite quickly.  We notice that $ \angle CAS \equal{} \angle TAS \plus{} \angle CAC' \plus{} \angle C'AT \equal{} 90 \plus{} \alpha \plus{} \beta$.  Also, we have already seen that $ \angle TC'C \equal{} 90 \minus{} \alpha \minus{} \beta$.  Thus, $ \angle SC'C \plus{} \angle CAS \equal{} 90 \plus{} \alpha \plus{} \beta \plus{} 90 \minus{} \alpha \minus{} \beta \equal{} 180$.  So, we obtain that $ CASC'$ is cyclic.  Thus, $ \angle ACB \equal{} \angle ACS \equal{} \angle AC'S \equal{} \angle AC'B'$.  Similarly, we can show that $ \angle ABC \equal{} \angle AB'C'$ and thus $ \triangle ABC\sim \triangle AB'C'$.  [/hide]
This is my 500th post  :D !

[quote="MellowMelon"]This was 5... so as long as you got the (apparently much easier) 1,2,4, you made the team by solving this one. I'm told the intended solution uses symmedians.[/quote]

Pretty much...I remember how distressed I was to find two geometry problems among the six total.

Ah..here we go...here's their solution:

[hide]
So $ AT$ is a symmedian of the triangle...this is well-known and can be easily proven (I can prove it if people want).

Now let $ M$ be the midpoint of $ BC$.  Given that, we have
$ m\angle{TBA}\equal{}180^{\circ}\minus{}m\angle{C}$
So, by the Law of Sines,
\[ \frac{BT}{AT}\equal{}\frac{\sin{\angle{BAT}}}{\sin{\angle{TBA}}}\equal{}\frac{\sin{\angle{CAM}}}{\sin{\angle{BCA}}}\]
\[ \equal{}\frac{MC}{AM}\]
But $ BT\equal{}TC_1$, so $ \frac{TC_1}{TA}\equal{}\frac{MC}{MA}$.
So because $ m\angle{TMS}\equal{}m\angle{TAS}\equal{}90^{\circ}$, quadrilateral $ TMAS$ is cyclic.  So $ m\angle{AMC}\equal{}m\angle{AMS}\equal{}m\angle{ATS}\equal{}m\angle{ATC_1}$.  By SAS similarity, then, we have $ \triangle{AMC} \sim \triangle{ATC_1}$.  Then because $ \frac{BC}{MC}\equal{}\frac{B_1C_1}{TC_1}\equal{}2$, we get again by $ SAS$ similarity that $ \triangle{ABC} \sim \triangle{AB_1C_1}$, as desired.
[/hide]

A very nice problem. This solution isn't nearly as nice as the official solution, it's a completely different approach:
Lemma $ 1$: Consider a circle centre $ O$, radius $ r$, and a point $ A$ outside the circle. Let a line $ l$, parallel to $ AO$, intersect the circle at $ D, E$. If the perpendicular from $ A$ to $ l$ meets $ l$ at a point $ B$, choose $ C$ so that $ A$ is the midpoint of $ BC$. Let $ AD$ meet the circle again at $ G$, and $ AE$ meet the circle again at $ F$. Then $ G, F, C$ are collinear.
Proof: Let's use inversion about a circle centre $ A$, radius $ \sqrt {AD.AG}$. Clearly, this inversion switches $ D,G$; $ F,E$, suppose $ C$ is taken to a point $ C'$. Proving that $ G, F,C$ are collinear reduces to proving that their images lie on a circle through $ A$, i.e. that $ DECC'$ is cyclic, or equivalantly $ BA*BC' \equal{} BD*BE$. But we note
$ BD*BE \equal{} BO^2 \minus{} r^2 \equal{} BA^2 \plus{} AO^2 \minus{} r^2 \equal{} BA^2 \plus{} AC*AC' \equal{} BA^2 \plus{} BA*AC' \equal{} BA*(BA \plus{} AC') \equal{} BA*BC'$,
as required. Here we use the fact that $ AC*AC' \equal{} AO^2 \minus{} r^2$, since both expressions are equal to the power of $ A$ with respect to the circle.
Now let's solve the problem:
Let the circle centre $ T$, radius $ TB \equal{} TC$ intersect $ AB$ again at $ B_{1}$ and $ AC$ again at $ C_{1}$. A quick angle chase tells us that $ B_{1}, C_{1}, T$ are collinear, and moreover, $ AB_{1}C_{1}$ is similar to triangle $ ACB$. Let $ B_{3}, C_{3}$ be the reflections of $ B_{1}, C_{1}$ about the line $ AT$ - by symmetry, we now have that $ AC_{3}B_{3}$ is similar to $ ABC$, $ T$ is the midpoint of $ B_{3}C_{3}$, and $ B_{1}C_{3}$, $ B_{3}C_{1}, AT$ are parallel. Let $ AB_{3}, AC_{3}$ intersect our circle (the circle centre $ T$, radius $ BT$), again at $ B'$ and $ C'$ respectively. Let the perpendicular from $ A$ to $ C_{1}B_{3}$ intersect $ C_{3}B_{1}$ at a point $ X$. From Lemma $ 1$ applied to this configuration, we have: $ B', C, X$ are collinear. But from Pascal applied to the $ 6$ points $ B_{3}, C_{3}, B, C, B_ {1}, B'$, we have that the $ 3$ points, $ BC \cap B_{3}C_{3}$, $ BB_{1} \cap B_{3}B'$, $ CB' \cap C_{3}B_{1}$, are collinear - i.e. $ A, X$ and $ B_{3}C_{3} \cap BC$ are collinear. If $ S'\equal{}B_{3}C_{3} \cap BC$ , we then have $ \angle TAS' \equal{} 90$. Since $ S'$ lies on $ BC$, going back to the problem statement, we have that, $ S' \equal{} S$. We also, the $ B_{3}$ in this diagram is $ B_{1}$ in the problem statement (not the $ B_{1}$ in my figure, sorry for the confusion.), and $ C_{3}$ in this diagram, is $ C_{1}$ in the problem statement. Since $ AB_{3}C_{3}$ is similar to $ ABC$ (already proven), the conclusion follows.

Nice problem! :blush:  Could anyone tell me the author of this problem?
Some more properties:
Let $ SA \cap B_1C\equal{}{K}$.Let K,A,B,C lie on a circle.

And a similar one:
Let $ BB_1 \cap CC_1\equal{}{L}$.Then L is lie on the symmedian wrt triangle ABC and on the circumcircle of triangle ABC.

Let $D$ and $E$ be on $AB$ and $AC$ such that $BCED$ is cyclic and $DE$ passes through $T$. By tangent angle theorem, $\angle{BCT}=\angle{CBT}=\angle{BAC}$. Hence $\angle{BDT}=\angle{ACB}=180^\circ - \angle{CBT} - \angle{ABC}=\angle{TBD}$. Hence $TD=TB=TC$. By the same argument, it follows that $TD=TB=TC=TE$. Let $M$ be the midpoint of $BC$. Since $BCED$ is cyclic, triangles $ABC$ and $AED$ are similar. Since $T$ is the midpoint of $DE$, it follows that $TB/TA=TD/TA=MC/MA$. Since $BTC$ is isosceles, $TM \perp BC$ which implies that $TMAS$ is cyclic. Hence $\angle{AMS}=\angle{ATS}$ which since $C_1 T/TA=BT/TA=MC/MA$ implies that triangles $ATC_1$ and $AMC$ are similar. By the same argument, triangles $ATB_1$ and $AMB$ are similar. This implies that $ABC$ and $AB_1 C_1$ are similar.

Dear Mathlinkers,
the circle centered at T passing through B1, C1, B and C leads to the Boutin's theorem... then we can think to the Liquel situation... in order to have a synthetic proof.
Sincerely
Jean-Louis

[quote="Zhero"]Let $\Gamma$ be the circle centered at $T$ with radius $TB$. Extend $AB$ and $AC$ to intersect $\Gamma$ at $B_2$ and $C_2$, respectively. It is clear that $\triangle ABC$ and $\triangle AC_2B_2$ are directly similar. Since $BC$ and $B_2C_2$ are antiparallel with respect to $AB$ and $AC$ and $AT$ is a symmedian of $\triangle ABC$, $AT$ is a median of $\triangle AC_2B_2$, i.e., $T$ is the midpoint of $B_2 C_2$.

Reflect $B_2$ and $C_2$ across $AT$ to get $C_1'$ and $B_1'$, respectively. Since this reflection maps $\Gamma$ to itself, $B_1'$ and $C_1'$ both lie on $\Gamma$, and $B_1'$, $T$, and $C_1'$ are collinear. Since $\triangle AC_2 B_2$ and $\triangle AB_1' C_1'$ are directly similar, $\triangle ABC$ and $\triangle AB_1' C_1'$ are directly similar.

Let $BC$ intersect $B_1' C_1'$ at $S'$ (which may be infinite.) Because $S' = BC \cap B_1' C_1'$, $T$ is the center of $BCC_1' B_1'$, and $A$ is the Miquel point of $BCC_1'B_1'$, we have $TA \perp AS'$. Hence, $S' = S$, whence $B_1 = B_1'$ and $C_1 = C_1$', so $\triangle ABC \sim \triangle AB_1C_1$, as desired.[/quote]

 how you prove $A$ is Miquel of $BCC_1B_1$ very nice.

If $K = \overline{BB_1} \cap \overline{CC_1}$ then $\angle BKC = \tfrac{\widehat{B_1C_1} - \widehat{BC}}{2} = \tfrac{1}{2} \left( 180 - \angle BTC \right) = \angle BAC$, so $B,K,A,C$ are concyclic.  So $A$ is the unique point with $AS \perp AT$ and $A \in (KBC)$; therefore $A$ is the Miquel point of $B_1BCC_1$, i.e. the center of the spiral similarity giving $\triangle ABC \sim \triangle AB_1C_1$.

[asy]/* DRAGON 0.0.9.6
Homemade Script by v_Enhance. */

import olympiad;
import cse5;
size(12cm);
real lsf=0.8000;
real lisf=2011.0;
defaultpen(fontsize(10pt));

/* Initialize Objects */
pair B_1 = (3.0, -3.0);
pair C_1 = (11.0, -3.0);
pair T = midpoint(B_1--C_1);
pair B = (6.451128850753037, 0.9621636086265188);
pair C = (10.082123836968163, -0.45038970554465996);
pair S = IntersectionPoint(Line(B,C,lisf),Line(B_1,C_1,lisf));
pair K = IntersectionPoint(Line(B,B_1,lisf),Line(C,C_1,lisf));
pair A = foot(T,K,S);

/* Draw objects */
draw(CirclebyPoint(T,B_1), rgb(0.0,0.6,0.6));
draw(circumcircle(A,B,C), rgb(0.0,0.6,0.8) + linetype("4 4"));
draw(T--B, rgb(0.0,0.8,0.8));
draw(T--C, rgb(0.0,0.8,0.8));
draw(K--S, rgb(0.0,0.6,0.8));
draw(circumcircle(K,B_1,C_1), rgb(0.0,0.8,0.8) + linewidth(1.0) + dotted);
draw(T--A, rgb(0.0,0.8,0.8) + linetype("4 4"));
draw((abs(dot(unit(T-A),unit(S-A))) < 1/2011) ? rightanglemark(T,A,S) : anglemark(T,A,S), rgb(0.0,0.8,0.8));
draw(B--B_1, rgb(1.0,0.6,0.0) + linewidth(1.0));
draw(B_1--C_1, rgb(1.0,0.6,0.0) + linewidth(1.0));
draw(C_1--C, rgb(1.0,0.6,0.0) + linewidth(1.0));
draw(C--B, rgb(1.0,0.6,0.0) + linewidth(1.0));
draw(K--B, rgb(0.0,0.8,0.8) + dashed);
draw(K--C, rgb(0.0,0.8,0.8) + dashed);
draw(C--S, rgb(0.0,0.8,0.8) + dashed);
draw(C_1--S, rgb(0.0,0.8,0.8) + dashed);

/* Place dots on each point */
dot(B_1);
dot(C_1);
dot(T);
dot(B);
dot(C);
dot(S);
dot(K);
dot(A);

/* Label points */
label("$B_1$", B_1, lsf * dir(225));
label("$C_1$", C_1, lsf * dir(-45));
label("$T$", T, lsf * dir(-90));
label("$B$", B, lsf * dir(140));
label("$C$", C, lsf * dir(70) * 1.414);
label("$S$", S, lsf * dir(-45));
label("$K$", K, lsf * dir(60));
label("$A$", A, lsf * dir(45));
[/asy]

But how do you know $K, A$ and $S$ are collinear?

Sorry, $TA \perp AS$ was meant rather than $TA \perp KS$. We don't need this collinearity at all since Miquel points does this all for us.

I will edit my post when I get home, seems I am unable to do so on my phone.

Let $X$ be the intersection of $\omega$ and $AT$ and let $BX$ intersect $ST$ at $Y$ and $CX$ intersect $ST$ at $Z$ and let the foot of perpendicular from $X$ on $BC$ be $R$.Not hard to observe that $\angle ABX =\angle ARC,, .ARXS$ is cyclic $\Rightarrow \angle ABX=\angle ATY \Rightarrow ABTY$ is cyclic.$\Rightarrow \angle XBT=\angle XAB=\angle BYT \Rightarrow Y \equiv C_1$ .Similarly $Z \equiv B_1.,, \Rightarrow \angle C_1BB_1=90 ,\angle SBB_1=90+\angle C_1BS=90+\angle XB_1C_1=90+\angle XAB_1=\angle B_1AS \Rightarrow ABB_1S$is cyclic.Similarly $ACC_1S$ is cyclic. $\Rightarrow \angle ABC=\angle AB_1C_1$ and $\angle ACB=\angle AC_1B_1$ and we are done.

Let $Q$ - Miquel point of $BCC_1B_1$. With $M,T$ being midpoints of $BC, B_1C_1$, by gliding principle we know that $Q\in (SMT)$. Also, with $T$ being circumcenter of $BCC_1B_1$ we know that $\angle SQT=90^\circ$.
Clearly $A\in (SMT)$ ($ST$ - diameter) and $\angle SAT=90^\circ$ by the statement. Hence $A\equiv Q$ and that is all we need.

[hide=fun stuff]That is what I originally wrote:
With $M$ - midpoint of $BC$, clearly $M\in (ATS)$. Gliding principle + $\angle TAS=90$ give us that $A$ - Miquel point of $BCC_1B_1$.

But then decided that some explanations won't hurt. Trying to find the right balance.[/hide]

[hide=an elementary proof]WLOG, $AB \ge AC$, so will now not worry about configuration issues. Let $M$ be the mid-point of side $BC$. Then $TM \bot SM$. So points $S,T,M,A$ lie on the circle with diameter $ST$. Note that it suffices to show $\triangle ACC_1 \sim \triangle ABB_1$. Let $\angle CTC_1 = 2 \alpha,\angle BTB_1 = 2 \beta$. Now, an easy angle chase yields $\angle ACC_1 = \angle ABB_1$. So it suffices to show $\frac{CC_1}{BB_1} = \frac{AC}{AB}$. Now, $\frac{CC_1}{BB_1} = \frac{\sin \alpha}{\sin \beta}$. Also, we have that
\begin{align*}
 & 2 \alpha = \angle STC = \angle STA - \angle CTA = \angle SMA - (\angle B - \angle CAT) = (\angle SMA - \angle B) + \angle CAT = \angle BAM + \angle CAT = 2 \angle BAM \\
& \implies 2 \beta = 2 \angle A - 2 \alpha = 2 \angle CAM
\end{align*}
So we have that
$$\frac{CC_1}{BB_1} = \frac{\sin \alpha}{\sin \beta} = \frac{\sin \angle BAM}{\sin \angle CAM} = \frac{MB \cdot \frac{\sin \angle AMB}{AB}}{MC \cdot \frac{\sin \angle AMC}{AC}} = \frac{AC}{AB} $$
as desired. $\blacksquare$[/hide]

It's easy to see $BCB_1C_1$ is cyclic with center $T$ and diameter $B_1C_1$. Now, let $BC_1 \cap CB_1 = A_1$. It suffices to show $A \in (A_1BC), (A_1B_1C_1) = \Omega$, as the Spiral Center Lemma would imply $ABC \sim AC_1B_1$.

Because $\angle B_1BC_1 = \angle C_1CB_1 = 90^{\circ}$, we know $B_1B \cap C_1C = H$ is the orthocenter of $A_1B_1C_1$.

[b][color=#f00]Claim:[/color][/b] $H$ lies on $\omega$.

[i]Proof.[/i] We know $T$ is the midpoint of $B_1C_1$ and $TB, TC$ are tangent to $\omega$. Because $B$ and $C$ are feet of altitudes of $A_1B_1C_1$, the Three Tangents Lemma applied to $A_1B_1C_1$ implies the desired result. $\square$

Thus, we know $A_1$ also lies on $\omega$.

Now, let the second intersection of $\omega$ and $\Omega$ be $A_2$. It suffices to show $A_2 = A$.

[b][color=#f00]Claim:[/color][/b] $A_1A_2$, $BC$, $B_1C_1$ concur at $S$.

[i]Proof.[/i] Applying the Radical Axis Theorem to $\omega$, $\Omega$, $(BCB_1C_1)$ yields the desired result. $\square$

Thus, $S, A_2, A_1$ are collinear. Because $AS \perp AT$, it suffices to show $\angle SA_2T = 90^{\circ}$ (which is equivalent to $\angle A_1A_2T = 90^{\circ}$ due to the collinearity).

Observe $\angle A_1AH = 90^{\circ}$, as $A \in \omega$ and $A_1H$ is a diameter by orthocenter configuration properties. Now, define $A_1'$ to be the antipode of $A_1$ wrt $\Omega$. It follows that $\angle A_1AA_1' = 90^{\circ}$, so $A, H, A_1'$ are collinear.

Because $H$ is the orthocenter of $A_1B_1C_1$ inscribed in $\Omega$, we know $T \in HA_1'$ by the Orthocenter Reflection Lemma. Thus $\angle A_1AT = \angle A_1AH = 90^{\circ}$ as desired. $\blacksquare$

[b]Disclaimer:[/b] My solution to this problem is horrendous. Do not attempt to replicate at home.

Unfortunately, I constructed $S$ on the incorrect ray, so part of my diagram is essentially flipped. As a result, the similarity in my diagram is actually $ABC \sim AC_1B_1$.

Also, my (incorrect) GGB can be found [url=https://www.geogebra.org/geometry/fbecrxas]here[/url]!

[asy] /* coordinates from geogebra */
import geometry; size(10cm); pair A,B,C,T,S,B_1,C_1,D; A=(3,6); B=(-5,-1); C=(4,-1); T=(-0.5,-7.9); S=(30.8,-1); B_1=(-8.6,-9.7); C_1=(7.6,-6.1); D=(-1.7,7.2); fill(A--B--C--cycle,purple+white+white+white); fill(B--C--C_1--B_1--cycle,blue+white+white+white+white); draw(circle(A,B,C),dashed);  draw(circle(B_1,C_1)); draw(B--C--C_1--B_1--cycle,blue); draw(A--B--C--cycle,red); draw(A--B_1--C_1--cycle,pink+red); draw(D--S); draw(S--C); draw(D--B); draw(D--C); draw(B--T); draw(C--T); draw(A--T); draw(S--C); dot(A^^B^^C^^T^^S^^B_1^^C_1^^D); label("$A$",A,NE); label("$B$",B,2W); label("$C$",C,NE); label("$T$",T,SW); label("$S$",S,2E); label("$B_1$",B_1,SW); label("$C_1$",C_1,SE); label("$D$",D,2N); [/asy]

Let $D=\overline{BB_1}\cap\overline{CC_1}.$ Notice that $$\measuredangle BDC=\pi+\measuredangle DB_1C_1+\measuredangle DC_1B_1=\frac{\pi+\measuredangle BTC}{2}=\measuredangle BAC$$ since $T$ is the center of $(BB_1C_1C).$ Hence, $ACBD$ is cyclic. Also, $\angle TAS=90,$ so $A$ it is the Miquel Point of $BB_1C_1C$ and therefore the center of a spiral similarity that maps $\overline{BC}$ and $\overline{B_1C_1}.$ $\square$

Notice that $BT=B_1T=CT=C_1T$ which implies $T$ is the circumcentre of $BCC_1B_1$
Let $BB_1 \cap CC_1=X$ and we know that $BC \cap B_1C_1=S$
By angle chasing,
Let  $\angle BB_1T=x$ and $\angle CC_1T=y$
$$\angle BTC=2x+2y-180 \implies \angle CBT=\angle BCT=180-(x+y)=\angle BAC=\angle BXC$$
where the last part comes from the cyclic quadrilateral.
and therefore $X \in \odot(ABC)$
Clubbing all the information it follows that, $A$ is the Miquel point of cyclic quadrilateral $BCC_1B_1$ and now taking spiral symmetry we get that $\triangle ABC \sim \triangle A_1B_1C_1$ as desired.

Easy one
Let $BC_1$ and $CB_1$ meet at $K$. $\angle BKC = \angle 180 - \angle BC_1T - \angle CB_1T = \angle TBC = \angle BAC$ so $BAKC$ is cyclic so $A$ is Miquel point of $BCB_1C_1$. Now the problem is almost Done the rest is just angle chasing...

Let $F=SA\cap(ABC)$, $L=AT\cap(ABC)$, $Q=FL\cap BC$ and $M=SL\cap (ABC)$
\\ Claim 1. M-Q-A
\\ Pf. Notice that $$-1=(A,L;B,C)\stackrel{F}{=}(S,Q;B,C)\stackrel{L}{=}(M,F;B,C)\stackrel{A}{=}(MA\cap BC,S;B,C)$$ which implies that $MA\cap BC=Q$

\\ See how $(M,F;B,C)=-1$ also shows that $T-M-F$, and now by looking at cyclic quadrilateral $(FALM)$  by Brocard we get that $ST$ is the polar of $Q$. Thus, if we let $O$ be the circumcenter of this circumference, then $OQ\perp ST$, but as $\angle LAF= 90$ we get that $FL\perp ST$. Deffine $N=FL\cap ST$,  $C_1*=FC\cap ST$, and $B_1*=C_1*L\cap ST$. By Brocard on $(BFCL)$ we conclude $B_1*-B-F$, and thus $L$ is the orthocenter of $FC_1*B_1*$, which implies $C_1*=C_1$ and $B_1*=B_1$.
\\ Now we're done as $L$ and $A$ are inverses with respect to $(C_1BCB_1)$, and now it is well-known that $A$ is the Miquel point that sends $BC$ to $B_1C_1$ which implies the desired result.

A quick angle chasing gives that $M=CC_1\cap BB_1$ lies on $(ABC)$. We will prove that if $N=CB_1\cap BC_1$, then $A, T, N$ are collinear, implying that $A$ is a Miquel point of $C_1CBB_1$.
Let $AB\cap ST=P, AT\cap (ABC)=N_1$ and  $BN_1\cap ST=C_0$. Then $-1=(B,C;A,N_1)\stackrel{B}{=} (T,S;P,C_0)$ and $\angle SAT=90^\circ$,so $AT$ is a bisector of $\angle BAC_0$. Then $\angle C_0AT=\angle TAB=\angle N_1BT=\angle C_0BT$, so $AC_0TB$ is cyclic, implying that $TC_0=TB$, so
$C_0\equiv C_1$.Similarly, $CB_1\cap AT\in (ABC)$, so $N=N_1$ and $N\in AT, N\in (ABC)$, so we are done.

Not hard at all.

Suppose that [i]B[size=50]2[/size][/i] and [i]C[size=50]2[/size][/i] are the intersection points of [i]ST[/i] and [i](SAB)[/i] and [i](SAC)[/i]. Let [i]M[/i] be the midpoint of [i]BC[/i]. Since that [i]TM[/i] is perpendicular to [i]BC[/i], we have that [i]ASTM[/i] is cyclic.

It's well-known that [i]AT[/i] is a symmedian of triangle [i]ABC[/i]. After that it's easy using just anglechase prove that [i]TC[size=50]2[/size]=BT=TB[size=50]2[/size][/i].

[asy] /* coordinates from geogebra */
import geometry; size(10cm); pair A,B,C,T,S,B_1,C_1,D; A=(-0.89,1.11); B=(1.96,-1.44); C=(-1.44,-1.06); T=(-0.1,-4.5); S=(-9.76,-0.13); B_1=(3.26,-6.02); C_1=(-3.47,-2.98); D=(1.16,1.4); fill(A--B--C--cycle,white+white); fill(B--C--C_1--B_1--cycle,white+white+white+white); draw(circle(A,B,C)); draw(circle(B_1,C_1)); draw(circle(A,C_1,B_1));draw(C_1--D--B_1--cycle,green); draw(A--B--C--cycle,red); draw(A--B_1--C_1--cycle,red); draw(D--S); draw(S--C); draw(D--B); draw(D--C); draw(B--T); draw(C--T); draw(A--T); draw(S--C);draw(S--C_1);dot(A^^B^^C^^T^^S^^B_1^^C_1^^D); label("$A$",A,NE); label("$B$",B,2W); label("$C$",C,NE); label("$T$",T,SW); label("$S$",S,2E); label("$B_1$",B_1,SW); label("$C_1$",C_1,SE); label("$D$",D,2N); [/asy]
$B_1B \cap C_1C= D$ . Chase, $\angle BDC= 180^{\circ}-(\angle DB_1T +DC_1T) = \frac{180^{\circ}- \angle BTC}{2}=\angle BCT$.Thus, $ACBD$ is cyclic. As $T$ is the center$(B_1C_1CB)$ , and as $A,D,S$ are collinear by radical axes ,we use [url=https://artofproblemsolving.com/community/c6h60787p366594]IMO 1985/5 [/url], which proves $A$ is the miquel point of the complete quad $B_1C_1CBDS $. The property that the miquel point is the centre of the spiral sim. which sends $BC$ to $B_1C_1$ finishes the proof.

Let $D = \overline{BB_1} \cap \overline{CC_1}$ and $E = \overline{B_1C} \cap \overline{BC_1}$. Note that $T$ is the center of $(BB_1C_1C)$ so $\overline{B_1B} \perp \overline{BC_1}$ and $\overline{B_1C} \perp \overline{CC_1}$. Hence $E$ is the orthocenter of $\triangle KB_1C_1$, and $\overline{TB}$ and $\overline{TC}$ are tangent to $(DBEC)$ by three tangents lemma, which is enough to imply that $(DBEC)$ and $(ABC)$ coincide.

Thus, we now know that $A$ is one of the intersections of $(ST)$ with $(DBEC)$; denote these two intersections as $A_1$ and $A_2$. Then it's well-known that $A_1$ and $A_2$ are just the $K$-Humpty and Queue points of $\triangle DB_1C_1$ (as $S$ is the $D$-ex point); WLOG assume that $A_1$ is the Humpty point and $A_2$ is the Queue point. If $A = A_1$, then $A$ lies on $(EB_1C_1)$ so $\triangle ABC \sim \triangle AC_1B_1$. If $A = A_2$, then $A$ lies on $(DB_1C_1)$ so $\triangle ABC \sim \triangle AB_1C_1$.

hello

Let $G=BB_1\cap CC_1$, notice that we have the good old orthocenter configuration hence $G\in \omega$. Let $A'$ be the foot of the altitude from $T$ to $SG$, it's well-known that $A'$ is the Miquel Point of $BCC_1B_1$. Hence $A'=\omega\cap (ST)$ which is also just $A$ so we're done??

Wait I think I fakesolved

Note that $BCB_1C_1$ is cyclic with circumcenter $T$. let $P$ be the intersection of $BC_1$ and $CB_1$.\\

It is well known that $TB$ and $TC$ are tangent to $(PBC)$ (this can be shown by angle chasing and showing that $\angle BTC=2\angle B_1+2\angle C_1-180$).  Hence, $P$ lies on $(ABC)$.\\

Thus, $A$ lies on the circle with diameter $ST$ as well as the $(PBC)$. Thus, it is the Miquel point, which establishes the spiral similarity, as desired.


ratio
complex analysis
complex analysis unsolved

Prove that $ \mathbb{C}\setminus \{ 0,1,2 \}$ is not conformally isomorphic to $ \mathbb{C}\setminus \{ 0,1,3 \}$

A conformal isomorphism is a 1-1 conformal map between sets; a conformal map is one that preserves angles between curves. This angle-preserving condition is equivalent to acting as multiplication by a nonzero complex number at a point (in $ \mathbb{R}^{2}$), so conformal maps in the plane are complex analytic functions.

Assume to the contrary that there exists a conformal isomorphism $ f:{\mathbb C}\setminus \{0,1,2\}\to{\mathbb C}\setminus \{0,1,3\}$. Note that each of the points $ 0, 1,2$ and $ \infty$ is either a simple pole or a removable singularity of $ f$. Therefore, $ f$ can be extended to a conformal automorphism of $ {\mathbb C}\cup \{\infty\}$. Hence $ f$ is a Möbius transformation. So it must preserve the cross ratio. Since $ f$ maps the set $ \{0, 1, 2, \infty\}$ to $ \{0, 1, 3, \infty\}$, the cross ratio of points in the former set equals the cross ratio of the points in the latter set (appropriately ordered).
Now the cross ratio of $ 0, 1, 2, \infty$ equals
\[ (0, 1; 2, \infty) = \frac{0-2}{1-2}= 2. \]
The cross ratio of points $ 0$, $ 1$, $ 3$, $ \infty$ equals (depending on the order):
\[ \frac{0-1}{0-3}= \frac13 \text{ or }\frac{0-3}{0-1}= 3 \text{ or }\frac{1-3}{1-0}=-2 \text{ or }-\frac{1}{2}\text{ or }\frac{3-0}{3-1}= \frac{3}{2}\text{ or }\frac{2}{3}. \]
None of these numbers equals $ 2$. We got a contradiction.


When writing a proof for an inequality on the USAMO, is it necessary to state when equality occurs even if the problem doesn't specifically ask to do so?

Thanks.

I don't think it does, but that is just my opinion. I think they care more about how you arrive at the inequality itself.

You are not required to write equality cases unless the problem asks for them. (The problem-writers and graders do not assume that the test-takers are psychic.)

Yeah, sometimes it's helpful to find the equality cases. But sometimes it's harder to find the equality cases than it is to prove that inequality holds.

[quote="Ravi B"]I agree with the responses.  Figuring out the equality cases, however, can be useful in trying to prove an inequality.  They can sometimes suggest promising approaches and rule out unpromising ones.  That's often one of the first things I do when facing an inequality.[/quote]

That was one of the things that threw me off with zscool's last posted inequality - I could have sworn that there were no equality conditions.

It's probably reasonably stylish to toss in the equality conditions at the end, but I wouldn't do that unless you'd already proven them inside your work -- no point to doing additional work that count for anything.


trigonometry
trig identities
Law of Cosines

I am interested in knowing the degree of curvature of the earth's
surface.  I am attempting to find a method which would allow me to
conceptualize and calculate the making of such a depiction on paper.

That is to say, if the earth curves from point A to point B, points
exactly 10 kilometers apart, if one were to draw a straight line
betwen the two points, then WHAT WOULD BE THE HEIGHT of the highest
point the curvature AB would be above the line AB?  Moreover, if the
point AB were now to be 20 kilometers apart, would the height be 2
times the height calculated earlier or would it be something
altogether different?

why is this here? do you want us to do it?

radius of the earth = 3960 mi.
set up an isosceles triangle with two legs = 3960 mi.
the third leg is 10 km (in miles)
use law of cosines to find the angle the two 3960 legs make with each other
Find 3960 - (height of the triangle)

[quote="Interval"]Law of Cosines is trigonometry.

I have not seen trig in over 25 years.

Can you show me how to use trig to answer this question?[/quote]

yeah, trig generally does not belong in this forum but whatever.

law of cosines: (i really hope you have gone to yahoo and google)

$a^{2}=b^{2}+c^{2}-2bc\cos A$ where $a,b,c$ are sides of a triangle and side $a$ is opposite of $\angle A$.

[quote="Interval"]Now that we got law of cosines here, can you show me how to use it to answer the given question?[/quote]

no, you will have to figure that out urself.  ask if you have any questions

[quote="Interval"]jli:

I appreciate your replies but make the hints a bit more easy for people who are not math-inclined.

Can you do that?

If not, then please do not answer my posts.

Thanks,
Interval[/quote]

all right.

ill do an example with my avatar an see if you can do the problem.

we are trying to find the value of $x$.  let $a=x$ so $b=3$ and $c=3$.  Also, the angle opposite of $a$ is $90$ degrees.

use the formula:

$x^{2}=3^{2}+3^{2}-2\cdot 3\cdot 3\cdot \cos 90=18$.

so $x=3\sqrt{2}$

get it?


trigonometry
geometry
inequalities
inequalities proposed

Prove that for x in (0, pi/2) we have sin(x)/x<3/(4-cos(x))<1

3<4-cosx easy to prove on )0,pi/2(

f(x)=3x-sinx(4-3cosx) study the variation on )0;pi/2(

Why did you put the trig inq in geometry Lagrangia ?

Well, I have put them because I though that trigonometry and geometry go together... I may be wrong.. :D :D

Would be nice to get a geometrical proof of this inequality

for sin(x)/x<1 it's easy to see a geometrical proof. but for sin(x)/x<3/(4-cos(x)) .. it's harder.. (x in (0,pi/2))


If 1,2, and 3 are solutions to the equation $ x^4 \plus{} ax^2 \plus{} bx \plus{} c \equal{} 0,$ then $ a\plus{}c$ equals

A. -12
B. 24
C. 35
D. -61
E. -63

[hide]We know that $ (x \minus{} 1)(x \minus{} 2)(x \minus{} 3)$ is a factor of our quartic because each individual factor also is.  This means that $ (x^2 \minus{} 3x \plus{} 2)(x \minus{} 3) \equal{} x^3 \minus{} 3x^2 \minus{} 3x^2 \plus{} 9x \plus{} 2x \minus{} 6 \equal{} x^3 \minus{} 6x^2 \plus{} 11x \minus{} 6$ is also a factor.

Let the fourth root be $ q$, we have $ (x \minus{} q)(x^3 \minus{} 6x^2 \plus{} 11x \minus{} 6) \equal{} x^4 \plus{} ax^2 \plus{} bx \plus{} c$.  Looking at the absent $ x^3$ term, we know that $ \minus{} 6x^3 \minus{} qx^3 \equal{} 0$ by finding ways to get $ x^3$ from distribution.  $ q \equal{} \minus{} 6$, so our polynomial in factored form is $ (x \minus{} 1)(x \minus{} 2)(x \minus{} 3)(x \plus{} 6)$.

Using Viete's formulae, we know that $ c \equal{} ( \minus{} 1)( \minus{} 2)( \minus{} 3)(6) \equal{} \minus{} 36$ and $ a \equal{} ( \minus{} 1)( \minus{} 2) \plus{} ( \minus{} 1)( \minus{} 3) \plus{} ( \minus{} 1)(6) \plus{} ( \minus{} 2)( \minus{} 3) \plus{} ( \minus{} 2)(6) \plus{} ( \minus{} 3)(6) \equal{} \minus{} 25$

$ a \plus{} c \equal{} \minus{} 36 \minus{} 25 \equal{} \minus{} 61 \equal{} > D$[/hide]


[quote="7h3.D3m0n.117"]$35y^{2}+2y-24$

factor the above...is it prime? i thought it was but then again working on AIM is harder than you think it is[/quote]

[hide]Here is the answer: (5y + 4)(7y - 6)[/hide]

OOPS  :blush: .........forgot to cross multiply something....thanks anyway!

pn12345, you made a mistake you wrote:
[quote]Here is the answer: (5y + 4)(7y - 6)[/quote]
(5y + 4)(7y - 6) = 35y^2 - 30y + 28y - 24 which is wrong.
Should be (5y - 4)(7y + 6)


function
calculus
derivative
vector
LaTeX
real analysis
real analysis unsolved

given acceleration $ a \equal{} 1 \plus{} ln x$
i can find that $ \Delta v^2 \equal{} 2xlnx$ and since it is given that when $ t \equal{} 0, x \equal{} 1,v \equal{} 0$
$ \therefore v^2 \equal{} 2xlnx$
however i have been asked to prove $ v > 0 \; when \;t > 0$ i have no idea how to explain it in mathematcial terms, can anyone please give any suggestion?
also howcome the symbols are not working

What is $ x$?  What is $ a$?  I'm going to take a guess and say that $ x$ is a function from $ \mathbb{R}$ to $ \mathbb{R}$, a twice-differentiable function in fact, and that $ a$ is the second derivative of $ x$.  Is that at all correct, or am I way off?

yes you are correct a=f(x) where a is accel and x is bounded vector displacement,you can use d(v^2/2)/dx=a
but the resulting v has power two, and i am asked to eliminate the negative sign, how do you do that mathematically
 and how come my latex codes are not appearing right

What is $ v$?  Also, I guessed that $ a \equal{} x''$, and you said I was right but then you said $ a \equal{} f(x)$, which seems to be different than what I said.  What is $ f$?

I don't know why your LaTeX wasn't showing up correctly.  Were you enclosing it in dollar signs?

v is the velocity, a is accel and x is bounded vector displacement

Suppose that $ x$ is a twice-differentiable function from the non-negative reals to the reals and that $ x''(t) \equal{} 1 \plus{} \ln(x(t))$ for all $ t \ge 0$.  Suppose also that $ x(0) \equal{} 1$ and $ x'(0) \equal{} 0$.  I'll outline a proof that if $ t > 0$, then $ x'(t) > 0$.

Assume the claim is false.  Let $ t_0 \equal{} min(\{t > 0 | x'(t) \equal{} 0 \})$.  (It takes a little work to prove that such a minimum exists.)  By Rolle's theorem, $ x''(c) \equal{} 0$ for some $ c$ between $ 0$ and $ t_0$.  Thus $ 1 \plus{} \ln(x(c)) \equal{} 0$, so $ x(c) \equal{} e^{\minus{}1} < 1$.  By the mean value theorem, there exists $ d$ between $ 0$ and $ c$ such that $ x'(d) < 0$.  By the intermediate value theorem, there exists $ w$ between $ 0$ and $ d$ such that $ x'(w) \equal{} 0$.  (It takes a little work to justify that step, but the idea is that $ x'$ is positive just to the right of $ 0$.)  This contradicts the definition of $ t_0$.

I wouldn't be surprised if there is some easier proof I haven't thought of.


My school finally did it this year. Are you doing it this year? Do you have any tips for beginners?

Hmm the spelling bee...I think I might actually do decently well (at least pwn everyone in my school), not like make it to nationals but possibly states. Well I don't have a whole lot of time to study, with tennis and piano and math.

I would guess learn your roots and the languages, and if according to Akeelah and the Bee, if you have no clue, spell it the way it sounds.

did it in elementary school. and got 2nd in 3rd and 2nd in4th grade school spelling bee.. yeah i suck.

the school i went to in 5th didnt do it.6th grade- didn't do it

7th- got 3rd in my school , missed some poetry word grr..... could have gotten 2nd easily

I NEVER studied for any spelling stuff. The prepared ones always win. so I would say, PREPARE

I don't care for spelling anyway, who cares if you know how to spell some odd word no one's ever going to use, and neither will you?

You need to get How to Spell Like a Champ, by Trinkle, Kimble, and Andrews, some stuff from http://hexco.com/nsb.htm, and the Paidiea or Spell-It, whatever they call the word list this year. After you've learned all the words in there get the official dictionary of the NSB (isn't that Webster's Unabridged?) and learn it all. But the important part is committing to study.

Meh, vocabulary: I'm working on studying my first actual unabridged dictionary. At my present rate it'll take 32 more hours of reading and marking words I don't know, and I'll end up with 16,000 words I don't know yet. And I intend to learn every one of them, because I love vocabulary :D.


Let $ 0<x_0,x_1, \dots , x_{669}<1$ be pairwise distinct real numbers. Show that there exists a pair $ (x_i,x_j)$ with
$ 0<x_ix_j(x_j\minus{}x_i)<\frac{1}{2007}$


can u ppl tell me your nsec scores.
i'm expecting somewhere around 210. is that gud enough???

plz reply.

Have you checked with the solns.
They are in the Indian forum so plz. check them.

Out of curiosity, just what is the NSEC?

NSEC - National Standard Examination in Chemistry.

Its the first round in India leading to the ICHO. I think this thread belongs in the Indian section.

Actually, I don't mind it's being here.  Since we have only one chemistry forum and no subforums, I feel that anything related to chemistry has a place here.

No reason to be US centric about these things.  That being said, an explanation of topics probably unfamiliar to someone from the US would be nice!


integration
calculus
function
calculus computations

[quote="vijaymenon"]If a>1 then evaluate $ \int_{0}^{1}\frac {ln\frac {1 \plus{} ax}{1 \minus{} ax} dx}{x\sqrt {1 \minus{} x^{2}}}$[/quote]
Is numerator ok?
(1 + ax/1 -ax)<0 when x belongs to interval (1/a,1),a>1 and we want to integrate at [0,1]

hello
[hide="Solution"]
Consider
 $ f(a) \equal{} \int_{0}^{1}\frac {ln\frac {1 \plus{} ax}{1 \minus{} ax}dx}{x\sqrt {1 \minus{} x^{2}}}$
$ f(a) \equal{} \int_{0}^{1}\frac {ln(1 \plus{} ax)dx}{x\sqrt {1 \minus{} x^{2}}} \minus{} \int_{0}^{1}\frac {ln (1 \minus{} ax)dx}{x\sqrt {1 \minus{} x^{2}}}$
Differentiating wrt $ a$ using $ Leibnitz Rule$
$ f' (a) \equal{} \int^{1}_{0} \frac {dx}{(1 \plus{} ax)\sqrt {1 \minus{} x^{2}}} \plus{} \frac {dx}{(1 \minus{} ax)\sqrt {1 \minus{} x^{2}}}$
$ f' (a) \equal{} \int ^{1}_{0} \frac {2 dx}{(1 \minus{} (ax)^2)\sqrt {1 \minus{} x^{2}}}$
 put $ x \equal{} \frac {1}{t}$
It will become
$ f'(a) \equal{} \int^{\infty}_{1} \frac {2t dt }{(t^2 \minus{} a^2)(\sqrt {t^{2} \minus{} 1})}$
Let $ t^2 \equal{} u$ and find the integral value .
and then integrate wrt to a.

[/hide]
If u find problem in solving further kindly tell me . I will solve it fully .
Thank u

Ya i almost forgot that it $ \ a$ can be taken a a  parameter.

Thanks KABI :)

the definite integral is not right. the argument of ln will be negative if x > 1/a. x varies from 0 to 1 and 1/a < 1. so, after x = 1/a, the function won't be defined.


modular arithmetic
number theory proposed
number theory

I had nothing to do so I thought I could post a nice and easy problem :):

If $m,n\in\mathbb N^*$, show that we can find integers $a_1,a_2,\ldots,a_n,b_1,b_2,\ldots,b_m$ s.t. $a_ib_j$ are all distinct $\pmod {mn}$.

[quote] I had nothing to do so I thought I could post a nice and easy problem

[/quote]

So, why don't you solve all the remaining unsolved problems? ;)

Pierre.


geometry
rectangle
analytic geometry
geometric transformation
homothety
circumcircle
ratio

Let $ ABCD$ be a rectangle with $ AB\equal{}b, AD\equal{}a$. A circle with center $ A$ and radius $ AC$ intersect line $ BD$ and $ E$ and $ F$. Bisector of angle $ ECF$ intersect chord $ EF$ at $ N$. Calculate $ \dfrac{EN}{NF}$.

first apply the angle bisector theorem and thentry inversion through  :)  $ C$ then the solution follows easily

Please post your solution if you have already solved it.

I have a really ugly solution using coordinates:  :oops:

[hide="My Solution"]From the angle bisector theorem, $ \frac {EN}{NF} = \frac {CE}{CF}$. Consider the homothety that sends $ A$ to $ A'$ such that $ \overrightarrow{CA'} = 2\overrightarrow{CA}$. This sends rectangle $ ABCD$ to rectangle $ A'B'C'D'$ and we can restate our problem as follows: $ A'B'CD'$ is a rectangle and $ A,B,D$ are the midpoints of $ CA',CB'$ and $ CD'$ respectively. $ BD$ intersects the circumcircle of $ A'B'CD'$ at $ E$ and $ F$. Then we need to find $ \frac {CE}{CF}$.

Consider a coordinate system with $ A$ as the origin, and the axes along $ AD$ and $ AB$. Then $ D = (a,0),B = (0, - b), C = (a, - b)$. Hence the equation of $ BD$ (or $ EF$) is $ bx - ay = ab$ and the circumcircle of $ A'B'C'D'$ is $ x^2 + y^2 = a^2 + b^2$. From these we find that $ E = \left(\frac {a(b^2 + \sqrt {D})}{a^2 + b^2},\frac {b( - a^2 + \sqrt {D})}{a^2 + b^2}\right)$ and $ F = \left(\frac {a(b^2 - \sqrt {D})}{a^2 + b^2},\frac {b( - a^2 - \sqrt {D})}{a^2 + b^2}\right)$. Here $ D = a^4 + a^2b^2 + b^4$.

Thus (after simplification) $ CE = \frac {\sqrt {b^2(b^2 + \sqrt D)^2 + a^2(a^2 - \sqrt D)^2}}{a^2 + b^2}$ and ${ CF = \frac {\sqrt {a^2(a^2 + \sqrt D)^2 + b^2(b^2 - \sqrt D)^2}}{a^2 + b^2}}$. And finally $ \frac {CE}{CF} = \sqrt {\frac {b^2(b^2 + \sqrt D)^2 + a^2(a^2 - \sqrt D)^2}{{a^2(a^2 + \sqrt D)^2 + b^2(b^2 - \sqrt D)^2}}}$.

I haven't found any nice expressions. :blush: [/hide]

I don't think that my ratio is quite well. But at least it is a ratio with $ a$ and $ b$.

[color=red]Here goes my solution,[/color]
Consider a dilatation with center $ C$ and ratio $ 2$ which maps the rectangle $ ABCD$  to $ A'B'CD'$.
Now using angle bisector theorem we get, ratio,$ k \equal{} \frac {EN}{NF} \equal{} \frac {EC}{CF} \equal{} \frac {\sin \angle {EFC}}{\sin \angle{CEF}} \equal{} \frac {\sin \angle {EA'C}}{\sin \angle{CA'F}}$

$ B'EFD'$ is an isosceles trapezoid (as it is a cyclic one). So, $ B'E \equal{} D'F$ and $ \angle B'A'E \equal{} \angle FA'D'$

Let $ \angle BB'A \equal{} B$ and $ \angle ACD \equal{} C$

By easy angle chasing we get that, $ \angle EA'C \equal{} C \minus{} B/2$ and $ \angle CA'F \equal{} \pi/2 \minus{} C \minus{} B/2$
so,
$ \frac {\sin \angle {EA'C}}{\sin \angle{CA'F}} \equal{} \frac {\sin (C \minus{} B/2)}{\cos (C \plus{} B/2)} \equal{} \frac {\sin C.\cos{B/2} \minus{} \cos C.\sin{B/2}}{\cos C.\cos{B/2} \minus{} \sin C.\sin{B/2}}$

Let $ AC \equal{} r \equal{} \sqrt {a^2 \plus{} b^2}$
So,
$ \sin {B/2} \equal{} \sqrt {\frac {r \minus{} a}{2r}}$ and $ \cos B/2 \equal{} \sqrt {\frac {r \plus{} a}{2r}}$ and $ \sin C \equal{} \frac {a}{r}$ and $ \cos C \equal{} \frac {b}{r}$
so,
$ \frac {\sin C.\cos{B/2} \minus{} \cos C.\sin{B/2}}{\cos C.\cos{B/2} \minus{} \sin C.\sin{B/2}} \equal{} \frac {{\frac {a}{r}}\sqrt {\frac {r \plus{} a}{2r}} \minus{} {\frac {b}{r}}\sqrt {\frac {r \minus{} a}{2r}}}{{\frac {b}{r}}\sqrt {\frac {r \plus{} a}{2r}} \minus{} {\frac {a}{r}}\sqrt {\frac {r \minus{} a}{2r}}}$
After cancelling common terms we get,
$ k \equal{} \frac {a\sqrt {r \plus{} a} \minus{} b\sqrt {r \minus{} a}}{b\sqrt {r \plus{} a} \minus{} a\sqrt {r \minus{} a}}$

$ Q.E.D$

p.s. If the solution looks unclear please have a look at the diagram.

[url=http://img146.imageshack.us/my.php?image=figure2dm8.png][img]http://img146.imageshack.us/img146/6676/figure2dm8.th.png[/img][/url]

I have another trignometric solution (wasnt as trivial as I thought first)
let the diagonals meet each other at  $ K$
now $ \frac {EN}{NF} \equal{} \frac {CE}{CF}$ and $ K$ bisects $ AC$ let angle subtended by $ CE$ be $ x$ and that by  $ CF$ be $ y$ therefore $ \frac{CE}{CF}\equal{} \frac{ \sin{x}}{ \sin{y}}$
by simple angle chasing angle $ EAC \equal{} 2y$ and angle $ ACE \equal{} 90^{0} \minus{} y$
as $ EK$ is a median to triangle $ CEA$ we get $ \cos{x \plus{} y} \equal{} 2 \sin{x} \sin{y}$
which implies $ \cos{x}\cos{y} \equal{} 3 \sin{x} \sin{y}$ which implies $ 8 \sin^{2}{x} \sin^{2}{y} \plus{} sin^{2}x \plus{} sin^{2}y \equal{} 1$  __________________________$ eq^{n} 1$
also $ AC \cos{x \plus{} y} \equal{} c \cos{x \plus{} y} \equal{} \frac {ab}{c}$
this implies  $ \frac {ab}{c^{2}} \equal{} \sin{x} \sin{y}$
dividing $ eq^{n} 1$ by the one found now and substituting $ \frac {ab}{c^{2}} \equal{} \sin{x} \sin{y}$
you get a quadratic in the ratio required

i can't see anything...... :(  please correct your latex code

[quote="gauravpatil"]I have another trignometric solution (wasnt as trivial as I thought first)
let the diagonals meet each other at  $ K$
now $ \frac {EN}{NF} \equal{} \frac {CE}{CF}$ and $ K$ bisects $ AC$ let angle subtended by $ CE$ be $ x$ and that by  $ CF$ be $ y$ therefore $ \frac {CE}{CF} \equal{} \frac {\sin{x}}{\sin{y}}$
by simple angle chasing angle $ EAC \equal{} 2y$ and angle $ ACE \equal{} 90^{0} \minus{} y$
as $ EK$ is a median to triangle $ CEA$ we get $ \cos{x \plus{} y} \equal{} 2 \sin{x} \sin{y}$
which implies $ \cos{x}\cos{y} \equal{} 3 \sin{x} \sin{y}$ which implies $ 8 \sin^{2}{x} \sin^{2}{y} \plus{} sin^{2}x \plus{} sin^{2}y \equal{} 1$  __________________________$ eq^{n} 1$
also $ AC \cos{x \plus{} y} \equal{} c \cos{x \plus{} y} \equal{} \frac {ab}{c}$
this implies  $ \frac {ab}{c^{2}} \equal{} \sin{x} \sin{y}$
dividing $ eq^{n} 1$ by the one found now and substituting $ \frac {ab}{c^{2}} \equal{} \sin{x} \sin{y}$
you get a quadratic in the ratio required[/quote]

Could you please write a bit clearly. most probably I am missing something or you have missed something. :maybe:

I :)  have modified it a bit
let the diagonals meet each other at  $ K$ let $ c$ be third side
now $ \frac {EN}{NF} \equal{} \frac {CE}{CF}$ and $ K$ bisects $ AC$ let angle subtended by $ CE$ be $ x$ and that by  $ CF$ be $ y$ therefore $ \frac {CE}{CF} \equal{} \frac {\sin{x}}{\sin{y}}$ by sine rule
by simple angle chasing angle $ EAC \equal{} 2y$ and angle $ ACE \equal{} 90^{0} \minus{} y$
as $ EK$ is a median to triangle $ CEA$ we get $ \cos{x \plus{} y} \equal{} 2 \sin{x} \sin{y}$ substituting $ \cos{x \plus{} y} \equal{} \cos{x} \cos{y} \minus{} \sin{x} \sin{y}$
which implies $ \cos{x}\cos{y} \equal{} 3 \sin{x} \sin{y}$ which implies $ 8 \sin^{2}{x} \sin^{2}{y} \plus{} sin^{2}x \plus{} sin^{2}y \equal{} 1$  ______$ eq^{n} 1$
now we calculate distance between a and bd in 2 ways $ AE \cos{x \plus{} y} \equal{} c \cos{x \plus{} y} \equal{} b \sin{ABD} \equal{} \frac {ab}{c}$
this implies  $ \frac {ab}{c^{2}} \equal{} \cos{x \plus{} y} \equal{} \sin{x} \sin{y}$
dividing $ eq^{n} 1$ by the one found now and substituting $ \frac {ab}{c^{2}} \equal{} \sin{x} \sin{y}$
you get a quadratic in the ratio required or rather an equation of this form $ ratio \plus{} \frac {1}{ratio} \equal{} r$ where $ r$ is a known constant in $ a,b$ and $ c$
atleast i see nothing missing

[size=117][color=darkblue][b]Very nice application of the [u]Generalized Pythagoras' Theorem[/u] (G.P.T.) ![/b][/color][/size]

[quote="Ahiles"] [color=darkred]Let $ ABCD$ be a rectangle with $ AD = a$ , $ AB = b$ . The circle $ w = C(A,r)$ ,

where $ r = \sqrt {a^2 + b^2}$ , intersects $ BD$ in $ E$ and $ F$ so that $ B\in (ED)$ . Calculate $ \frac {CE}{CF}$ .[/color] [/quote]
[color=darkblue][b][u]Proof.[/u][/b] Denote $ \|\begin{array}{c} BE = x \\
 \\
DF = y\end{array}$ . Observe that $ \|\begin{array}{c} \cos\widehat {ABD} = \frac br \\
 \\
\cos\widehat {ADB} = \frac ar\end{array}$ . Apply [b]G.P.T.[/b] to :

$ \|\begin{array}{cccc} \triangle\ ABE\ : & r^2 = x^2 + b^2 + 2bx\cdot \frac br & \implies & rx^2 + 2b^2x - ra^2 = 0 \\
 \\
\triangle\ ADF\ : & r^2 = y^2 + a^2 + 2ay\cdot\frac ar & \implies & ry^2 + 2a^2y - rb^2 = 0\end{array}\|$ $ \implies$ $ \|\begin{array}{c} rx = \sqrt {\Delta} - b^2 \\
 \\
ry = \sqrt {\Delta} - a^2\end{array}\|\ \ (1)$

where $ \boxed {\ \Delta = a^4 + a^2b^2 + b^4\ }$ . Observe that $ \|\begin{array}{c} \cos\widehat {CBD} = \frac ar \\
 \\
\cos\widehat {CDB} = \frac br\end{array}$ . Apply [b]G.P.T.[/b] to :

$ \|\begin{array}{cc} \triangle\ BCE\ : & CE^2 = x^2 + a^2 + 2ax\cdot\frac ar \\
 \\
\triangle\ DCF\ : & CF^2 = y^2 + b^2 + 2by\cdot\frac br\end{array}\|$ $ \implies$ $ (\frac {CE}{CF})^2 = \frac {rx^2 + 2a^2x + ra^2}{ry^2 + 2b^2y + rb^2}\stackrel {(*)}{\ = \ }$ $ \frac {(ra^2 - 2b^2x) + 2a^2x + ra^2}{(rb^2 - 2a^2y) + 2b^2y + rb^2} =$

$ \frac {ra^2 + (a^2 - b^2)x}{rb^2 - (a^2 - b^2)y} =$ $ \frac {r^2a^2 + (a^2 - b^2)\cdot rx}{r^2b^2 - (a^2 - b^2)\cdot ry} =$ $ \frac {r^2a^2 + (a^2 - b^2)(\sqrt {\Delta } - b^2)}{r^2b^2 - (a^2 - b^2)(\sqrt {\Delta} - a^2)} =$ $ \frac {a^4 + b^4 + (a^2 - b^2)\sqrt {\Delta}}{a^4 + b^4 - (a^2 - b^2)\sqrt {\Delta}} =$

$ \frac {[(a^4 + b^4) + (a^2 - b^2)\sqrt {\Delta}]^2}{(a^4 + b^4)^2 - (a^2 - b^2)^2(a^4 + a^2b^2 + b^4)} =$ $ \frac {[(a^4 + b^4) + (a^2 - b^2)\sqrt {\Delta}]^2}{(a^4 + b^4)^2 - (a^2 - b^2)(a^6 - b^6)} =$ $ \frac {[(a^4 + b^4) + (a^2 - b^2)\sqrt {\Delta}]^2}{a^2b^2(a^2 + b^2)^2}$ .

In conclusion, $ \boxed {\ \frac {CE}{CF} = \frac {|a^4 + b^4 + (a^2 - b^2)\sqrt {\Delta}|}{ab(a^2 + b^2)}\ }$ . Remark that $ x - y = \frac {a^2 - b^2}{\sqrt {a^2 + b^2}}$ .[/color]


Five positive integers less than 100 have 12 positive integer factors. Which of these five integers is not divisible by 12?

[hide="Like this"]$12=2^2\cdot 3$, and the number of positive divisors that a number $n=p_1^ap_2^b...$ is $(a+1)(b+1)...$, where $p_1,p_2...$ are distinct primes. So, we take all possible combinations of the prime factorization of $12$, and find numbers $k=p_1^ap_2^bp_3^c$ so that $(a+1)(b+1)(c+1)=12$ (the largest possible number of primes is $3$ so that we can make $12$) and $k<100$. One of these numbers will not be divisible by $12$. That number is $90=2\cdot 3^2\cdot 5$, since it does not have two $2$'s in it.[/hide]

If you didn't understand that (I couldn't get everything), I'll put it in hopefully simpler terms.

[hide]Like Jesusfreak said, the number of positive factors a number has is when you take it into prime factorization, the powers of its prime factors + 1 all multiplied by each other. (72 is $2^3 * 3^2$, so it has (3+1)(2+1) factors, or 12 factors)

So for a number to have 12 factors, $(a+1)(b+1)$... must equal 12.

All the ways to make 12 $(2*2*3)$

$2*6, 3*4, 2*2*3.$

so our numbers can be either $a*b^5$ (only 96 works, which is divisible by 12),$a^2*b^3$ (only 72 works, which is divisible by 12) , or  $a*b*c^2$ (must be).

and this can't be divisible by $2^2*3.$

try numbers, with c not being 2.

$2*3*5^2$ is too high, so c has to be 3.

$2*5*3^2$ works. Our answer is 90.

Sorry if this was a bit too long, but hopefully it was clear enough.[/hide]