I have my bachelor's in math from UNLV.
I got it in 2003, and have been working as a math tutor in the mean time.
My GPA was 3.59, in math it was 3.89.
I got a 1450 GRE score (800 in math), and a 880/900 on the math subject GRE (98th percentile).
I got 4 nontrivial Putnam scores, my highest was a 39, and that ranked me (barely) in the top 200 that year.
Anyway, I know I don't have the best stats in the world, but I thought I wouldn't have a problem getting into grad school. But I was turned down by all three schools I applied to: UW seattle, UC Berkeley, and MIT.
MIT is rediculously competitive, and I know UC Berkeley is up there too. UW is a good school, but I thought I was pretty sure to get in. I thought I had a decent chance at the other two, as well.
I know, I should have applied to more schools.
I plan to make the most of it and attend UNLV in the master's program for 1 year, and try to transfer to a PhD program somewhere else next year.
Anyway, the point of all this is to solicit advice. Is there anything in particular that you think hurt me? UNLV is a pretty mediocre school, to be nice. But my GRE and Putnam scores rank me against the whole country. Or is it that I just need to apply to some safer schools? Would being out of school for a couple years hurt me, even though my GRE subject score is recent?
I just know that something is hurting me, since I'm certain that I was pushed aside for someone with lower test scores (even MIT can't demand the top 2% in test scores, I should think). I just want to know why.

Have you gotten to know your current professors and shown them that you have research potential? What graduate schools are looking for, in any field, is ability to complete thesis research.

You should definitely read [url=http://www.amazon.com/gp/product/082183455X/]A Mathematician's Survival Guide: Graduate School and Early Career Development[/url] by Steven G. Krantz at your earliest opportunity, for more tips. Good luck.

when I apply next year, I will hopefully have been published. I've had professors offer to work with me, and I have started to try to contribute to their projects. It's good to know that that will make me look better.
Thank you.

It's been really very competitive to get into mathematics and computer science graduate schools in recent years.  I've seen undergraduate students who have papers submitted for publication, an REU or two, good test scores, etc. have troubles getting into grad school.
Berkeley and MIT are both in the top five, so it's no surprise you didn't get in either place (no matter how good you are, it's always a roll of the dice).
I don't know how difficult it is to get into UWash these days.

It's possible that the general GRE score is hurting you.  There are two layers of admissions, one in the department and one at the university's graduate school.  Some graduate schools basically expect all applicants to score close to the 99th percentile on the generals.

You say "(even MIT can't demand the top 2% in test scores, I should think)."  I went to Michigan, and in my class I was the only person who didn't score 900/900 on the Math subject GRE.  So yes, MIT can demand the top 2% in test scores.

It's possible that your essays hurt you.  (I have no idea how seriously they're taken.)

All that said, here's my best guess:  What's hurting you is in your recommendation letters.  All, or most, of the information in those letters is three years old, and you haven't done anything distinguishing in the years since graduation.  (Of course, that's an assumption I'm making; you just didn't mention anything other than math tutoring in your post.)  Thus, the recommenders aren't able to be as enthusiastic as possible.  Plus, you're being compared to current college seniors, most of whom have no time-gaps in their fantasticness.

My advice (this is for everyone applying to grad school, not just for you) is to get ahold of a ranked list of math departments (US News is the only place I know currently publishing them, though your library may have the Gourman Report which is old but good enough for this purpose) and ask a couple of professors to draw a couple of lines on it to indicate the range of schools they think are at your level.  Apply to a couple of schools above this level, 6 - 8 in the level, and a couple below the level for safety.  (For lots of people this is overkill, but better safe than sorry---better to have 5+ choices of places to go than 0.)

Wow, I thought only about 3000 people took the GRE math subject test each year, so top 2% would be about 150. I know that between MIT and Berkeley, there are at least 150 admissions. Maybe I'm wrong about some part of this, but it seems to me that even a top 5 school will almost HAVE to let in more than just the top 2%.

Anyway, I need to stop obsessing on the score. It's one of the best things the admissions coucil can see on my application.

I think you're right about the letters of reccommendation.
I'll be fresh from school next time, I'll have fresh reccommendations, and I'll hopefully have been published. I should look much better, plus I'll make sure to apply to some schools that I'll be sure to get into.
Thank you.


whatever happened to the trivia challenge thing?

it died down?

[hide]it DOES seem as though all the old ppl are gone... its kinda depressing...[i was looking at oooold posts that ppl like tare & mathfanatic put up....[/hide]
so many changes... *reminicense* [or however that's spelled]

What do you mean with old

if you either mean 21 or up, or having joined in 2003--->i am old!

i used to do it but then less and less people played so it wasn't fun anymore and then i stopped. i didn't like when they redid the format to where the question became  easy so it was a race to finish instead of the challenge that it is supposed to be.

[quote="Lucky707"]The AOPS Trivia Invitational is still played from time to time over AOL Instant Messenger. It's more of an impromptu thing.

If you leave your AIM here I can contact you when the impromptu trivia starts (it's totally random as to when it is)

Mine is: hezgotzbeenz, in case I don't see this thread ever again (I can be forgetful).[/quote]

I haven't seen one of those since last Spring. I always liked those. Please contact me when they start again. Mine's whitehorseking88, though I think I have you on my AIM contacts.


An urn contains n+m balls, of which n are red and m are black. They are withdrawn from the urn, one at a time and without replacement. Let X be the number of red balls removed before the first black ball is chosen. Find E[X]. To obtain this quantity, number the red balls from 1 to n. Now define the random variables $ X_i$ , i=1,….,n by

$ X_i \equal{} 1$    if red ball i is taken before any black ball is chosen

0 otherwise

a) Express X in terms of the $ X_i$
b) Find E[X]

As far as I can tell, a) is simply the sum of all of the X_i.

For b), my initial response was:
$ P(X_1 \equal{} 1) \equal{} \frac{n}{n\plus{}m}$

$ P(X_2 \equal{} 1) \equal{} \frac{(n)(n\minus{}1)}{(n\plus{}m)^2}$
...

and E[X] would be $ \sum(i)P_i$

However, this results in E[X] = $ m^2$, which is apparently not the correct answer. My TA's solution was:

$ \frac{\frac{(n\plus{}m)!}{(n\minus{}1)!(m\plus{}1)!}}{\frac{(n\plus{}m)!}{(n\minus{}1)!(m!)(1!)}}$

The denominator is the number of combinations of black balls, our red ball $ i$, and the other red balls. The numerator is the number of combinations of red and black balls if one of the red balls were treated as black. I'm having a hard time following the logic here; it seems not very intuitive. Is there a more intuitive way to do this?

[quote="sandlily"]For b), my initial response was:
$ P(X_1 \equal{} 1) \equal{} \frac {n}{n \plus{} m}$

$ P(X_2 \equal{} 1) \equal{} \frac {(n)(n \minus{} 1)}{(n \plus{} m)^2}$[/quote]  Neither of these is right.  You appear to be confusing "red ball number 1" (henceforth RB1) with "the first red ball drawn."  The probability that RB1 is drawn before any black ball is just $ \frac{1}{m \plus{} 1}$ -- this is your TA's number, but we can justify it with the simpler observation that if we ignore all red balls other than RB1, there are $ m \plus{} 1$ possible orders for RB1 and the $ m$ black balls, these are all equally likely, and in only one of them is RB1 before all the black balls.

Now observe that the exact same argument applies to RB2, etc.


quadratics
modular arithmetic
algebra
function
domain
polynomial
abstract algebra

[hide]
Lets work in mod 4 for now.

$a^2 + b^2$ must be $2 \mod 4$ because $8$ is $0 \mod 4$ and $6$ is $2 \mod 4$

That means $a^2 = 1 \mod 4$ and $b^2 = 1 \mod 4$

So $a$ and $b$ are odd.

Rewrite $a$ as $2p+1$ and $b$ as $2q+1$
$(2k+1)^2 + (2q+1)^2 - 8c = 6$
$4k^2 + 4k + 1 + 4q^2 + 4q + 1 - 8c = 6$
$4k^2 + 4k + 4q^2 + 4q - 8c = 4$
$k^2 + k + q^2 + q - 2c = 1$
$k(k+1) + q(q+1) - 2c = 1$

Now look at parity, even + even - even cannot be an odd.[/hide]

$a^2+b^2-6=8c$

$1+1-6 \equiv 0(mod 8)$ which is false..? :blush:

Um, the quadratic residues modulo 8 are 0, 1, and 4 and can never sum to 6 with only 2 summands.

[quote="chess64"]Show that there are no integers $a,b,c$ for which $a^2+b^2-8c=6$.[/quote]

[hide]
This is equivalent to showing that $a^2 + b^2 \equiv 6 \pmod{8}$ is not possible. The quadratic residues $\pmod{8}$ are $1,4,1,0,1,4,1,0$.
There's no way to add 2 of those to get $6$. QED. [/hide]

Aw, that's what i did too, so I don't get to post. :(

I guess it was a little obvious though.  :roll:

[quote="t0rajir0u"][hide="Solution using Gaussian integers"]
$a^2 + b^2 = (a + bi)(a - bi) = (1 + i)(1 - i)(4c + 3)$

But $4c + 3$ contains the product of an odd number of Gaussian primes; therefore, such a factorization is not possible. [/hide][/quote]

Yeah,I was thinking of that,saw it in a previous post of yours.
What is the criterion then?Something like what's posted in the number theory formulary thing?

[quote="shadysaysurspammed"]
What is the criterion then?Something like what's posted in the number theory formulary thing?[/quote]

The criterion for what?

[quote="t0rajir0u"][hide="Solution using Gaussian integers"]
$a^2 + b^2 = (a + bi)(a - bi) = (1 + i)(1 - i)(4c + 3)$

But $4c + 3$ contains the product of an odd number of Gaussian primes; therefore, such a factorization is not possible. [/hide][/quote]

Whoa can you explain this please? It sounds interesting. What's a Gaussian prime?

Gaussian Primes are $z=a+ib$ such that $a^2+b^2\in\mathbb{P}$. They are also $z=a+bi$ if $|a|,|b|\equiv 3\pmod 4$, where $|a|,|b|\in\mathbb{P}$.

Well lets define prime in general:
A number $p$ (whatever a 'number' is ;) ; well, it's an element from a ring here) is called "prime" if $p|ab$ implies that $p|a$ or $p|b$. Here $m|n$ means that there is a number $t$ from the same domain with $n=m \cdot t$.
A number $\epsilon$ is called a "unit" when $u|1$, so when there is $t$ with $\epsilon \cdot t=1$. For example, $-1$ is a unit in the integers, and $1,-1,i,-i$ are the units in gaussian integers (note that $1$ itself is a unit everytime by $1 \cdot 1 =1$).

Now when $p$ is a prime and $\epsilon$ a unit, then $p \cdot \epsilon$ will be prime also. Thus we consider all primes to be "the same up to units" that differ just by multiplication of a unit; as a consequence of this, we choose one (random?) special prime for all those that are considered "the same" by this (we often ignore it, but we already do this in the integers: we use the positive primes, but we always could use their negatives!).

This gives that the gaussian primes are (up to units again, so e.g. $1-i =-i \cdot(1+i)$ is prime, too, but not mentioned here):
a) $1+i$.
b) when $p \equiv 3 \mod 4$, then $p$ seen as $p+0i$ is still prime.
c) when $p \equiv 1 \mod 4$, there is (up to sign and changing $a \leftrightarrow b$; this up to stuff is exactly what the units and conjugation[see 2) below] do!) exactly one pair $(a,b)$ of integers with $p=a^2+b^2$. Then $\pi_1=a+bi$ and $\pi_2 =a-bi$ are two primes, and they are [u]not[/u] the same up to units (for example, $5=2^2+1^2$, thus $2+i$ and $2-i$ are primes in the Gaussian integers).

These are really all, and there are some nice things about them to know:

1) At first, every gaussian integer (and, since the normal integers are contained in them, especially the normal integers itself) is the product of such primes and a unit. But we have more:
Every gaussian integer $\neq 0$ can be expressed as product of primes and a unit in a unique way (don't forget that unique is still seen as "up to units", thus we have to take care which prime representing all those that are seen as "the same" we choose).
But this is nothing we are not used to: the same is the case in the normal integers where any number can be factored in a unique way up to sign (the sign equalls the units since all units in $\mathbb{Z}$ are $\pm 1$) as product of normal primes.

2) There is the conjugation map $c+di \mapsto \overline{c+di} = c-di$. This map preserves addition and multiplication as we know from complex numbers. Especially, this can be used to show that it also preserves units and primes, thus the conjugate of each of these has the same property again.

3) There is some mapping $N$ called the "norm" that sends $a+bi$ to $a^2+b^2=(a+bi) \cdot \overline{a+bi}$. This is nothing else than the complex modulus to the square, thus as some nice property, this map is "multiplicative", so $N(xy) = N(x)N(y)$ for all gaussian integers $x,y$.
When using $N$ to $\pi_1$ (or $\pi_2$, there is no difference), we get $N(\pi_1)= N(a+bi) =\pi_1 \overline{\pi_1} = \pi_1 \pi_2 =a^2+b^2=p$.

Well, one can ask: what is special about the Gaussian integers¿ The answer would be: not much.
One other useful set are the "Eisensteinian integers" of type $a+b\zeta_3$ with $\zeta_3 = e^{\frac{2 \pi i}{3}}$ being the "first" third root of unity.
There will be $6$ units then (ugly to work with ;) ), and there are also some different types of primes again. Even unique factorisation as product of primes , the conjugation map and the norm reappear.
Another example are numbers of type $a+ \sqrt{2}b$, where again all those properties hold, but: there are $\infty$ many units! (all units are given by $\pm (1+\sqrt{2})^k$ with $k$ any integer then; this corresponds to the Pell equation $x^2-2y^2=1$).

So lets take a step further: what do the numbers we look at have in common¿
The most important thing would be: they are "algebraic integers", thus are the roots of monic polynomials with integer coefficients.
Or seen another way (these are for what one often looks at equivalent, but the proof of this is somewhat hard and deep):
there are finetely many (complex) numbers $a_1, ..., a_m$ that are also "algebraic integers" such that all our numbers are of type $n_1 a_1 + ... + n_m a_m$ with all $n_i$ normal integers.
Then one can ask some natural questions like whether there is still unique factorisation. But there is not; it's replaced by something called "unique prime ideal decomposition/factorisation" (well, Dedekindian rings only).
And the norm is the product of the so-called "conjugates" then, not meaning just complex conjugation.
Or another question: how do the units look like¿ This is answered by Dirichlets Theorem on units (for those who know the terms: the unit build up a group that is isomorphic to $\mathbb{Z}^{r+s-1} \times U$ with $U$ being the roots of unity contained and $r,s$ some integers, more exactly the number of real and conjugated complex "embeddings").

All these stuff is what Algebraic Number Theory concerns with, and even more (e.g. valuation theory). There are a lot of nice theorem with consequences for the normal integers, too, e.g. Dirichlets Theorem on units generalizes Pells equation (which is a very simple corollary of it). Another one is Fermats Last Theorem, for which these techniques give partial solutions, or Catalans Conjecture, or an answer to the question when a number is a $n$-th power residue $\mod$ some prime (generalizing the law of quadratic reciprocity).
But I will end this now...

[quote="shadysaysurspammed"]So in this question,who said $a^2+b^2$ is a prime? :?  And torajirou,I  also don't understand the part where you said something about $4c+3$ that it contains product of odd number of gaussian primes....[/quote]

1.  Masoud said that $a + bi$ is a Gaussian prime iff $a^2 + b^2$ is an integer prime.  His definition works (I think) because the norm (length) of a Gaussian integer (complex number) is multiplicative, so if you had a non-prime (say, $3 + i$, with norm $3^2 + 1^2 = 10$) norm, then you would be able to decompose it into a product of two other Gaussian integers.  In fact, $3 + i = (2 + i)(1 - i)$, with norms $5$ and $2$.

2.  Well, consider it as an integer.  $4c + 3$ is odd, so its prime factorization contains only odd numbers.  They are either $\equiv 4k + 1$ or $\equiv 4k + 3$.  There must be an odd number of factors of the form $4k + 3$ or else the whole thing $\bmod 4$ would not be $3$.  Each of these is either a Gaussian prime or a product of an odd number of Gaussian primes.  Hence $4c + 3$ contains the product of an odd number of Gaussian primes.

That means

$(a + bi)(a - bi) = (1 + i)(1 - i)(4c + 3)$

Has an even number of Gaussian primes in the LHS (since any Gaussian prime that divides the LHS divides either $a + bi, a - bi$ and then must divide the other) but an odd number in the RHS - contradiction.

Thanks a lot toirajrou(kinda weird username  :D )

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.CMO1969Problem7

Vo Duc Dien

Please do not post the same thing that people posted two years ago


Let $ a_i, b_i$ be positive real numbers, $ i\equal{}1,2,\ldots,n$, $ n\geq 2$, such that $ a_i<b_i$, for all $ i$, and also \[ b_1\plus{}b_2\plus{}\cdots \plus{} b_n < 1 \plus{} a_1\plus{}\cdots \plus{} a_n.\] Prove that there exists a $ c\in\mathbb R$ such that for all $ i\equal{}1,2,\ldots,n$, and $ k\in\mathbb Z$ we have \[ (a_i\plus{}c\plus{}k)(b_i\plus{}c\plus{}k) > 0.\]

For each i=1,.., n. There exist an integer k_i such that

di=[k_i - bi, k_i  -ai ] is a sub close interval of [0,1]

The sum of those n segments is less than 1, so we could find a real number c that does not belong to any of di

This number c satisfies ........

I dont quite understand your solution... Can some1 post another solution or can u help me understand yours? You say there exists $ k_i$ such that your $ d_i$ is in $ [0,1]$. But if thats the case, the task is solved,, we just let $ c\equal{}0$... Maybe im getting something wrong. My idea is something similar, but I didnt manage to complete it.

The author of this problem is prof. Vasile Pop, from Cluj.

I'll post here the official solution.

[hide=" Solution"] The condition is equivalent to no integer $ k$ to be found in any interval $ [a_i\plus{}c;b_i\plus{}c]$ , i.e. in the translates by $ c$ of $ [a_i;b_i]$. Moreover, the sum of the lengths of these intervals is known to be less than 1, since $ \sum_{i\equal{}1}^n(b_i\minus{}a_i)<1.$

We are thus asked to show that the union $ S$ of a (finite) set of non-degenerate intervals, of total length less than 1, may be translated such that it becomes disjunct of $ \mathbb{Z}$. For all $ k \in \mathbb{Z}$, consider the sets $ S_k\equal{}S \cap [k;k\plus{}1)$ and their translates $ S_k\minus{}k \subset [0;1)$. Due to possible overlaps, the total length of the union $ \cap \{S_k; k \in \mathbb{Z} \}\equal{}S$, hence less than $ 1$. Therefore, there exists $ \minus{}c \in [0;1) \cup \{ S_k\minus{}k; k \in \mathbb{Z} \}$, hence $ (S\plus{}c) \cap \mathbb{Z}\equal{} \emptyset$.  [/hide]

Note that the problem is equivalent to finding some $c$ so that when we translate the intervals $(a_i,b_i)$ to $(a_i+c,b_i+c)$ none of the new intervals doesn't contain an integer. Now we can divide our intervals to intervals that already don't contain an integer and the ones that do.Let the first ones be called good and the others be bad. So let the first intervals $WLOG$ be $(c_i,d_i)$ such that $0<c_i<d_i<1$ and the others be $(1-a_i,1+b_i)$. Now note that all the bad intervals actually form one interval $(1-a,1+b)$ such that $a=max(a_1, \dots , a_n)$ and $b=max(b_1, \dots , b_n)$.Now note that we can rearrange good intervals to some new intervals $(c_i,d_i)$ such that they don't overlap. Now we can suposse that none of the good intervals overlaps with our bad interval(because if they do we can just make one bad interval out of those 2 intervals). Let our good intervals be $(c_1,d_1),(c_2,d_2), \dots ,(c_n,d_n)$ such that $c_1<d_1<c_2<d_2< \dots <c_n<d_n$. It also holds that $d_n<1-a$ so $d_n+a<1$ so we can choose some $\epsilon$ such that $d_n+a+\epsilon<1$ and $a+b+\epsilon<1$. Now by taking $c=a+\epsilon$ we see that our bad interval goes to good interval and that $(c_n,d_n)$ stays subset of $(0,1)$ so the other good intervals also stays subsets of $(0,1)$. We see that after translating for such chosen $c$ all of the intervals are good, so $c$ exists.

similar to official solution, just more expanded upon:

Consider the quadratic polynomial $p_i(x)=(x+a_i)(x+b_i).$ We see that the roots of $p_i$ are $-a_i$ and $-b_i$, so that $p_i(c+k)\le 0$ iff $-b_i-c\le k\le -a_i-c$ for all integers $k$. Therefore, in order to show that there exists some $c\in\mathbb{R}$ such that $p_i(c+k)>0$ for all $i\in[n]$ and $k\in\mathbb{Z}$, it suffices to find some $c$ such that the interval $[-b_i-c,-a_i-c]$ and equivalently $$[a_i+c,b_i+c]$$ contains no integers for all $i\in[n].$

Now, define $$S=\bigcup_{i=1}^n[a_i,b_i].$$ Letting $l$ be the total length that $S$ encloses, we find that
\begin{align*}
l&=\sum_{i=1}^n(b_i-a_i)\\
&=\sum_{i=1}^nb_i-\sum_{i-1}^na_i\\
&<1,
\end{align*}
as given in the problem statement. Define the set $S'$ to contain the fractional parts of $S$, so that $x\in S$ iff $x-\lfloor x\rfloor\in S'.$ Clearly, $S'$ is a subset of $[0,1)$ and encloses a length of at most $l<1$, implying that $S'$ is a proper subset of $[0,1).$ Hence, there must exist some $c'\in[0,1)$ such that $c'\notin S'.$

But this means that $[a_i-c',b_i-c']$ cannot contain any integers for all $i\in[n]$, as otherwise there would be some $k'\in\mathbb{Z}$ such that $k'+c'\in[a_i,b_i]\subseteq S$, so that
\begin{align*}
c'&=(k'+c')-k'\\
&=(k'+c')-\lfloor k'+c'\rfloor\\
&\in S',
\end{align*}
a contradiction. Hence, setting $c'=-c$ implies the result, so we are done. $\blacksquare$


Find all positive integers $k$ such that the product of the digits of $k$, in decimal notation, equals \[\frac{25}{8}k-211\]

The product of the digits of $k$ is an integer so $8 \mid k, k=8n$, and the product of the digits is $25n-211$. Now since $8 \mid k$ the rightmost digit is even, so $25n-211$ is even; then $25n$ is odd, so $n$ is odd.

Also the product of the digits of $k$ is at most $k$ (trivial to show) so $25n-211 \leq 8n, 17n \leq 211, n \leq 12$. On the other hand, it is a non-negative number so $25n-211 \geq 0, 25n \geq 211, n \geq 9$. Now we just have to check $n=9, 11$ because it is odd, that is, $k=72, k=88$. Inspection shows both work.


if $ p$  and $ p^2\plus{}2$  are numbers prime   then  $ p^3\plus{}2$  also prime number
  caculate   sum of sequence :
 $ 1\plus{}2^2x\plus{}3^2x^2 \plus{} ...\plus{}n^2x^n$

for the first one p can only be 3.
Because if it's not 3, then it is not a multiple of 3, then $ p^2\plus{}2$ can be divided by 3. So $ p^3\plus{}2$=29 is prime

[quote="Hong Quy"]caculate   sum of sequence :
 $ 1 \plus{} 2^2x \plus{} 3^2x^2 \plus{} ... \plus{} n^2x^n$[/quote]
The general term doesn't seem right.
I assume Hong Quy wanted to calculate
$ S(x)\equal{}1\plus{}2^2x\plus{}3^2x^2\plus{}\dots\plus{}(n\plus{}1)^2x^n$
$ S(x)\equal{}(2\cdot 1\plus{}3\cdot 2x\plus{}4\cdot 3x^2\plus{}\dots\plus{}(n\plus{}2)(n\plus{}1)x^n)\minus{}(1\plus{}2x\plus{}3x^2\plus{}\dots\plus{}(n\plus{}1)x^n)$
$ S(x)\equal{}\frac{d}{dx}\frac{d}{dx}(1\plus{}x\plus{}x^2\plus{}\dots\plus{}x^{n\plus{}2})\minus{}\frac{d}{dx}(1\plus{}x\plus{}x^2\plus{}\dots\plus{}x^{n\plus{}1})$
$ S(x)\equal{}\frac{d}{dx}\frac{d}{dx}\left(\frac{x^{n\plus{}3}\minus{}1}{x\minus{}1}\right)\minus{}\frac{d}{dx}\left(\frac{x^{n\plus{}2}\minus{}1}{x\minus{}1}\right)$
The final expression isn't that nice, though.

We have two cases.

For $ x \equal{} 1$. the desired sum: $ S_n(x) \equal{} \sum_{k \equal{} 1}^n k^2 \equal{} \frac {1}{6}n(n \plus{} 1)(2n \plus{} 1)$.

For $ x\neq 1$, let $ f(k) \equal{} \frac{1}{1\minus{}x}k^2\plus{}\frac {2}{(1\minus{}x)^2}k\plus{}\frac{1\plus{}x}{(1\minus{}x)^3}$,

$ S_n(x) \equal{}\sum_{k \equal{} 1}^{n\plus{}1} k^2x^{k\minus{}1}\equal{} \sum_{k \equal{} 1}^{n\plus{}1} \{f(k \minus{} 1) \minus{} f(k)\} \equal{} f(0) \minus{} f(n)$
$ \equal{} \frac{1\plus{}x}{(1\minus{}x)^3}\minus{}\left\{\frac{1}{1\minus{}x}n^2\plus{}\frac{2}{(1\minus{}x)^2}n\plus{}\frac{1\plus{}x}{(1\minus{}x)^3}\right\}$.

But

$ f(k \minus{} 1) \minus{} f(k) \neq k^2 x^k$

I got:

$ f(k \minus{} 1) \minus{} f(k) \equal{} \frac {(3 \minus{} 2k)x}{(1 \minus{} x)^2}$

Also, we're not looking for:

$ \sum^{n}_{k \equal{} 1}k^2 x^k$

But for:

$ S_n(x)\equal{}\sum^{n}_{k \equal{} 1}k^2 x^{k \minus{} 1}$ (I think)

The series it written incorrectly because the left side of "..." is of the form $ k^2 x^{k \minus{} 1}$ and the right is $ k^2 x^k$.

Am I doing something wrong?

Also, I got:

$ S_n(x) \equal{} n^2 \frac {x^n}{x \minus{} 1} \minus{} 2n\frac {x^n}{(x \minus{} 1)^2} \plus{} (x \plus{} 1)\frac {x^n \minus{} 1}{(x \minus{} 1)^3}$


geometry solved
USA(J)MO
USAMO
geometry
geometric transformation
homothety
parallelogram

Let $ABC$ be a triangle and let $\omega$ be its incircle. Denote by $D_1$ and $E_1$ the points where $\omega$ is tangent to sides $BC$ and $AC$, respectively. Denote by $D_2$ and $E_2$ the points on sides $BC$ and $AC$, respectively, such that $CD_2=BD_1$ and $CE_2=AE_1$, and denote by $P$ the point of intersection of segments $AD_2$ and $BE_2$. Circle $\omega$ intersects segment $AD_2$ at two points, the closer of which to the vertex $A$ is denoted by $Q$. Prove that $AQ=D_2P$.

When replying to the problem, I ask that you make posts for solutions and submit comments, jokes, smilies, etc. separately. Furthermore, please do not "hide" any portion of the solution. Please use LaTeX for posting solutions. Thanks.

[quote="MithsApprentice"]Let $ABC$ be a triangle and let $\omega$ be its incircle. Denote by $D_{1}$ and $E_{1}$ the points where $\omega$ is tangent to sides $BC$ and $AC$, respectively. Denote by $D_{2}$ and $E_{2}$ the points on sides $BC$ and $AC$, respectively, such that $CD_{2}=BD_{1}$ and $CE_{2}=AE_{1}$, and denote by $P$ the point of intersection of segments $AD_{2}$ and $BE_{2}$. Circle $\omega$ intersects segment $AD_{2}$ at two points, the closer of which to the vertex $A$ is denoted by $Q$. Prove that $AQ=D_{2}P$.[/quote]

[b][u]SOLUTION[/u][/b]
[color=indigo]
Because of the homothety and the equality of segments (isotomic points and tangents), we get:

$\frac{AQ}{QD_{2}}=\frac{AE_{1}}{E_{1}Z}=\frac{CE_{2}}{E_{1}C+CD_{2}}=\frac{CE_{2}}{CD_{1}+BD_{1}}=\frac{CE_{2}}{BC}$ $(1)$

Now we apply [i]Menelaus's Theorem[/i] to the triangle $AD_{2}C$: $\frac{PD_{2}}{PA}=\frac{BD_{2}\cdot E_{2}C}{BC \cdot E_{2}A}=\frac{CE_{2}}{BC}\cdot \frac{BD_{2}}{E_{2}A}=\frac{CE_{2}}{BC}$ $(2)$

From $(1)$ and $(2) \Longrightarrow q.e.d.$

Nick Rapanos[/color]

I'm sorry for reviving an old thread, but I have an alternate solution (though not as short as the above one) to this problem.

It suffices to prove that $\frac{AQ}{AD_2}=\frac{PD_2}{AD_2}$. We will find the values of these fractions separately, and prove that they are equal.

Before we begin, for the sake of convenience, let us have $AE_1=a$, $BD_1=b$, and $D_1D_2=c$.

First, we find the value of $\frac{AQ}{AD_2}.$
Since $BD_1=CD_2$, the $A$-excircle meets $BC$ at $D_2$, and so $Q$ is diametrically opposite to $D_1$. Hence, the line through $Q$ parallel to $BC$ is tangent to the incircle. Furthermore, let the homothety centered at $A$ that takes $Q$ to $D_2$ take $B$ to $T$.

We have $\frac{AQ}{AD_2}=\frac{AB}{AT}=\frac{a(a+b)}{a^2+3ab+ac+2b^2+bc}$; the length of $AT$ can be found easily using the relation $\frac{a}{a+b}=\frac{a+2b+c}{AT}$.

Now, we find the value of $\frac{PD_2}{AD_2}$.
Assign a mass of $a, b, b+c$ to points $A, B, C$, respectively. Then $P$ has a mass of $a+2b+c$. This implies $\frac{AP}{PD_2}=\frac{2b+c}{a}$.
Adding one to each sides, $\frac{AD_2}{PD_2}=\frac{a+2b+c}{a}$.
Reciprocating both sides, we have $\frac{PD_2}{AD_2}=\frac{a}{a+2b+c}$.

Simple algebra reveals that $\frac{a(a+b)}{a^2+3ab+ac+2b^2+bc}=\frac{a}{a+2b+c}$, and we have $AQ=D_2P$, as desired. $\blacksquare$

[quote][color=darkred]Let $ABC$ be a triangle and let $w(I)$ be its incircle. Denote by $D_{1}$ and $E_{1}$ the points where $\omega$ is tangent to sides $BC$ , $AC$ respectively. Denote by $D_{2}$ , $E_{2}$ the points on sides $BC$ and $AC$ respectively, such that $CD_{2}=BD_{1}$ ,  $CE_{2}=AE_{1}$ and denote by $P$ the point of intersection of segments $AD_{2}$ and $BE_{2}$ . Circle $w$ intersects segment $AD_{2}$ at two points, the closer of which to the vertex $A$ is denoted by $Q$ . Prove that $AQ=PD_{2}$ .[/color][/quote]

[b][u]Proof[/u].[/b] Is well-known property $Q\in ID_1\ \ (*)$ . Therefore, $\frac {AQ}{AD_2}=\frac {h_a-2r}{h_a}=\frac {ah_a-2ar}{ah_a}=\frac {2pr-2ar}{2pr}=\frac {p-a}{a}\ \ (1)$ .

 Apply [u][b]Menelaus[/b]' theorem[/u] to $\overline {BPE_2}/AD_2C\ \ :\ \ \frac {BD_2}{BC}\cdot\frac {E_2C}{E_2A}\cdot\frac {PA}{PD_2}=1$ $\implies$ $\frac {p-b}{a}\cdot\frac {p-a}{p-c}\cdot\frac {PA}{PD_2}=1$ $\implies$

$\frac {PA}{p}=\frac {PD_2}{p-a}=\frac {AD_2}{a}$ $\implies$ $\frac {PD_2}{AD_2}=\frac {p-a}{a}\ \ (2)$ . From the relations $(1)$ , $(2)$ obtain $AQ=PD_2$ .

==================================

$(*)\ \blacktriangleright$ Denote $\{D_1,Q_1\}=w\cap D_1I$ .Prove easily that $\frac {AD}{Q_1D_1}=\frac {h_a}{2r}=\frac {ah_a}{2ar}=\frac {2pr}{2ar}=\frac pa$ and

$\frac {DD_2}{D_1D_2}=\frac {p|b-c|}{a}\cdot \frac {1}{|b-c|}=\frac pa$ . In conclusion, $\frac {AD}{Q_1D_1}=\frac {DD_2}{D_1D_2}$ , i.e. $Q_1\in w\cap AD_2$ $\implies$ $Q_1\equiv Q$ .

In my solution, I used also the fact that $Q\in D_1I$ which is new for me, I just remarqued it from the diagram and proved it, and I think if we knew this result, the problem becomes easy.
[hide=" My solution"]
Start by proving the following lemma.
[b]Lemma.[/b] [i]the incircle $\omega$ of triangle $\triangle{ABC}$ touch side $BC$ at $D$, the incenter of $\triangle{ABC}$ is $I$, denote $Q\equiv (\omega)\cap(DI)$ and $R\equiv (BC)\cap(AQ)$, then $BR=CD$.[/i]

[b]Proof.[/b] Apply Menelaus on triangle $\triangle{PBD}$ that is cutted by the line $(AQR)$, we get then: $\frac{BR}{DR}=\frac{PQ}{DQ}.\frac{BA}{PA}=\frac{1}{t}$  then  $BR=\frac{BD}{1+t}$. Project $A$ on $BC$ at a point $H$, then $(DI)\parallel (AH)$. We have  $\frac{DQ}{PQ}=\frac{DQ}{DP-DQ}$  and   $DP=\frac{AH}{BH}.BD$  thus    $\frac{DQ}{PQ}=\frac{DQ.BH}{AH.BD-BH.QD}$, and we have   $\frac{PA}{BA}=1-\frac{BP}{BA}=\frac{BH-BD}{BH}$,  then    $t+1=\frac{BD(AH-QD)}{AH.BD-BH.QD}$,  therefor:    $\frac{1}{BR}=\frac{t+1}{BD}=\frac{AH-QD}{AH.BD-BH.QD}$   call this result $(*)$. Put $AB=c$ , $AC=b$ and $BC=a$ and $p=\frac{a+b+c}{2}$ and $S$ the area of $\triangle{ABC}$ and $B=\angle{ABC}$. Then: $AH=c.sin{B}$,   $QD=2r=\frac{2S}{p}=\frac{ac.sin(B)}{p}$,   $BD=\frac{a+c-b}{2}$ and  $BH=c.cos(B)$, replacing on $(*)$ we get that   $\frac{1}{BR}=\frac{2(p-a)}{p(a+c-b)-2ac.cos(B)}$ From the law of cosines, $2ac.cos(B)= a^2+c^2-b^2$ replacing on the previous result we get: $\frac{1}{BR}=\frac{2}{a+b-c}$ from which we deduct that: $BR=CD$. That end the proof of the lemma.

Return to our problem. Let $I$ the incenter of $\triangle{ABC}$ and $H$ the projection of $A$ on $BC$. Obvious that $D_2B=E_2A$, then by Menelaus applied on $\triangle{AD_2C}$ we have $\frac{AP}{D_2P}.\frac{CE_2}{CB}=1$ then $\frac{PA}{PD_2}=\frac{b+c-a}{2a} \Longrightarrow \frac{AD_2}{AP}=\frac{p}{a} \Longrightarrow AP=\frac{a.AD_2}{p}$. From the lemma above, points $D_1$ , $I$ and $Q$ are collinear, then $\frac{AP}{D_2Q}=\frac{AD_2}{QD_2}.\frac{a}{p}=\frac{a.AH}{2p.r}=1$ then $AP=D_2Q$ from which follow that $AQ=D_2P$.
[/hide]

[quote="jayme"]Dear mathlinkers,
this problem come from Lalesco a geometer of the beginning of the XX centuries...
Why P which is the Nagel point is not used in order to have a synthetic proof?
Sincerely
Jean-Louis[/quote]

I am trying to get a proof using the fact that $P$ is the nagel point but I am not able to complete it. May be someone can.

From an homothety with centre A and ratio $\frac{s-a}{s},$ the point $D_2$ maps to $Q$ and hence $\frac{AD_2}{AQ}=\frac{s}{s-a}$ and hence $\frac{AP}{AQ}+\frac{D_2P}{AQ}=1+\frac{a}{s-a}$ It suffices to prove that $\frac{AP}{AQ}=\frac{a}{s-a}\Longleftrightarrow \frac{AP}{AD_2}=\frac{a}{s}$ and from here, I am not able to complete the proof. If someone knows the length of the nagel point from the vertex...

Nagel point $P$ of $\triangle ABC$ is the incenter of its antimedial triangle $\triangle A_0B_0C_0$ and orthogonal projections $M,M_0$ of $A,A_0$ on $BC$ are symmetric about the midpoint of $BC.$ Let the incircle $(P,2r)$ of $\triangle A_0B_0C_0$ touch its side $B_0C_0$ at $D_0$ and let $R \equiv AP \cap A_0M_0.$ Since $A_0D_1$ is the $A_0$- Nagel ray of $\triangle A_0B_0C_0,$ it follows that $A_0D_1 \parallel AR$ $\Longrightarrow$ quadrilateral $QRA_0D_1$ is a parallelogram, hence  $A_0R=QD_1=PD_0=2r.$ $T$ is the orthogonal projection of $P$ onto $BC,$ since $D_0T=A_0M_0.$ Then $PT=RM_0$ $\Longrightarrow$ $\overline{D_2T}=\overline{D_2M_0}=\overline{D_1M}.$ If the projections of  $\overline{PD_2},\overline{AQ}$ on $BC$ are equal, then $PD_2=AQ,$ as desired.

Here is a rather nice solution using barycentric coordinates:

Let $A$ be $(1,0,0)$, $B$ be $(0,1,0)$, and $C$ be $(0,0,1)$. Let the side lengths of the triangle be $a,b,c$ and the semi-perimeter $s$.

Now, $CD_1=s-c$, $BD_1=s-b$, $AE_1=s-a$, $CE_1=s-c$.

Thus, $CD_2=s-b$, $BD_2=s-c$, $AE_2=s-c$, $CE_2=s-a$.

Therefore, the non-normalized coordinates of $D_2$ are $(0,s-b,s-c)$ and of $E_2$, $(s-a,0,s-c)$.

Clearly then, the non-normalized coordinates of $P$ are $(s-a,s-b,s-c)$.

Normalizing, we have that $D_2$ is at $\left(0,\frac{s-b}{a},\frac{s-c}{a}\right)$, $E_2$ at $\left(\frac{s-a}{b},0,\frac{s-c}{b}\right)$, and $P$ is at $\left(\frac{s-a}{s},\frac{s-b}{s},\frac{s-c}{s}\right)$

Now, we find the point $Q'$ inside the triangle on the line $AD_2$ such that $AQ'=D_2P$. It is then sufficient to show that this point lies on the incircle.

$P$ is the fraction $\frac{s-a}{s}$ of the way "up" the line segment from $D_2$ to $A$. Thus, we are looking for the point that is $\frac{s-a}{s}$ of the way "down" the line segment from $A$ to $D_2$, or, the fraction $1-\frac{s-a}{s}$ of the way "up".

Thus, $Q'$ has normalized $x$-coordinate $1-\frac{s-a}{s}=\frac{a}{s}$.

As the line $AD_2$ has equation $(s-c)y=(s-b)z$, it can easily be found that $Q'$ lies at $\left(\frac{a^2}{as}, \frac{(s-a)(s-b)}{as}, \frac{(s-a)(s-c}{as}\right)$, or in non-normalized coordinates with sum $as$, at $(a^2,(s-a)(s-b),(s-a)(s-c)$.

Recalling that the equation of the incircle is $a^2yz+b^2xz+c^2xy+(x+y+z)[(s-a)^2x+(s-b)^2y+(s-c)^2z]=0$, we must show that this equation is true for $Q'$'s values of $x,y,z$.

Plugging in our values, this means showing that

$a^2(s-a)^2(s-b)(s-c)+a^2(s-a)[b^2(s-c)+c^2(s-b)]+as[a^2(s-a)^2+(s-a)(s-b)^3+(s-a)(s-c)^3]=0$.

Dividing by $a(s-a)$, this is just

$a(s-a)(s-b)(s-c)+a[b^2(s-c)+c^2(s-b)]+a^2s(s-a)+s(s-b)^3+s(s-c)^3=0$.

Plugging in the value of $s$:

$\frac{a(-a+b+c)(a-b+c)(a+b-c)}{8}+\frac{ab^2(a+b-c)}{2}+\frac{ac^2(a-b+c)}{2}+\frac{a^2(a+b+c)(-a+b+c)}{4}+\frac{(a+b+c)(a-b+c)^3}{16}+\frac{(a+b+c)(a+b-c)^3}{16}=0$.

$2a(-a+b+c)(a-b+c)(a+b-c)+8ab^2(a+b-c)+8ac^2(a-b+c)+4a^2(a+b+c)(-a+b+c)+(a+b+c)(a-b+c)^3+(a+b+c)(a+b-c)^3=0$

$2a[(-a+b+c)(a-b+c)(a+b-c)+4b^2(a+b-c)+4c^2(a-b+c)]+(a+b+c)[4a^2(-a+b+c)+(a-b+c)^3+(a+b-c)^3]=0$

The first bracket is just:

$-a^3-b^3-c^3+a^2b+a^2c+b^2c+b^2a+c^2a+c^2b-2abc+4ab^2+4b^3-4b^2c+4ac^2-4bc^2+4c^3=-a^3+3b^3+3c^3+a^2b+a^2c-3b^2c+5b^2a+5c^2a-3c^2b-2abc$

and the second bracket is $-2a^3+4a^2b+4a^2c+6ab^2+6ac^2-12abc$.

Dividing everything by $2a$ gives

$-a^3+3b^3+3c^3+a^2b+a^2c-3b^2c+5b^2a+5c^2a-3c^2b-2abc+(a+b+c)(-a^2+2ab+2ac+3b^2+3c^2-6bc)$, which is $0$, as desired.

As $Q'$ lies on the incircle and $AD_2$, $Q'=Q$, and our proof is complete.

Some days ago i saw another problem solved by using this Lemma but I couldn't solve.This one can be solved by following Lemma  too: :w00tb:

[color=#800000][b]Lemma[/b][/color]:
$I$(incenter) , $G$(centroid),$N$(Nagel point) lie in that order on a line (Known as the Nagel line) & $GN=2IG$
Proof:[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=50&t=518404[/url]
---------------------

Draw a line through $D_{1}$ & $G$ so it intersects$PQ$ at$ F$.Name midpoint of $BC$ : $D"$

As $PI$ bisects $QD_{1}$ & $G$ is on $PI$ such that $GP=2IG$(according to the[color=#800000][b]Lemma[/b][/color]) so $G$ is the centroid of $\vartriangle PQD_{1}$ so$F$ is the midpoint of $ QP$  (1)
In $AD_{1}D_{2}$: $AD"$ bisects $D_{1}D_{2}$ and $G$ is on it  such that $AG=AGD"$ so  $G$ is the centroid of$\vartriangle AD_{1}D_{2}$ and $F$ is the midpoint of $AD_{2}$.  (2)
By using (1) & (2) we can conclude  .

The above lemma is awesome.  See [url=http://www.cut-the-knot.org/Curriculum/Geometry/NagelLine.shtml]here[/url] and [url=http://www.cut-the-knot.org/Curriculum/Geometry/MedialNagel.shtml#NI]here[/url].

The key point is that the Nagel point of a triangle (in this case, $P$) is the incenter of its anticomplementary triangle.

[hide="Solution"]
Let $\triangle AFE$ be such that $B$ is the midpoint of $AF$ and $C$ is the midpoint of $AE$.  Let $D$ be the midpoint of $FE$.  (So $\triangle DBC$ is the medial triangle of $\triangle AFE$.)  Now let $P'$ be the Nagel point of $\triangle DCB$.  By lemma 2 [url=http://yufeizhao.com/olympiad/geolemmas.pdf]here[/url] (which is just an easy homothety between the incircle and excircle), $QD_1$ is a diameter of circle $\omega$.

Now $D_1$ and $D_2$ are also the tangency points of the $A$-excircle and $D$-excircle to side $BC$ with respect to $\triangle ABC$ and $\triangle DCB$ respectively.  Thus, since $ACDB$ is a parallelogram and $\angle CAD_2=\angle BDD_1$, we have that $AD_2 \parallel D_1D$.

Ok so let $I$ be the center of circle $\omega$, and let $AI \cap DD_1$ at $P'$.  By our lemma, Then $P'$ is the incenter of $\triangle AFE$ and the Nagel point of $\triangle DCB$.  Since $AQ \parallel D_1P''$ and $D_1I=QI_1$, we have that quadrilateral $AQP;_1$ is a parallelogram.  Thus $AQ=D_1P'=D_2P$.

[/hide]

[hide="Solution"]
Denote $a=BC$, $b=AC$, $c=AB$, $s$ as the semiperimeter, and $K$ as the area of $\triangle{ABC}$.

We claim that $D_1Q$ is a diameter of $\omega$. Since $CD_2=BD_1\implies CD_1=BD_2$, $AD_1$ is isotomic to $AD_2$. Since $D_1$ is the point of tangency of the incircle of $\triangle{ABC}$ with $BC$, $D_2$ must be the point of tangency of the $A$-excircle with $BC$ (the Gergonne and Nagel points are isotomic conjugates). Consider the homothety centered at $A$ taking $D_2$ to $Q$. Then this homothety takes line $BC$ to a line $l$ such that $Q$ passes through $l$ and $l$ is parallel to $BC$. But $BC$ is tangent to the $A$-excircle, so $l$ must be tangent to $\omega$. It follows that $D_1Q\perp BC$, so $D_1Q$ is a diameter of $\omega$.

Let $D$ be the projection of $A$ onto $BC$. We have $\angle{QD_2D_1}=\angle{AD_2D}$ and $\angle{QD_1D_2}=\angle{ADD_2}$, so $\triangle{QD_1D_2}\sim\triangle{ADD_2}$. We then can write $QD_2/QD_1=AD_2/AD$. But $QD_1=2r=2K/s$ and $AD=2K/a$, so
\[ \frac{QD_2}{2\frac{K}{s}}=\frac{AD_2}{2\frac{K}{a}}\implies\frac{QD_2}{a}=\frac{AD_2}{s}=\frac{AD_2-QD_2}{s-a}=\frac{AQ}{s-a},\]
or $\frac{AQ}{AD_2}=\frac{s-a}{s}$.

From $CE_2=AE_1$, we obtain $CE_1=AE_2$. Since $CE_1$ and $CD_1$ are tangents, we have $AE_2=CE_1=CD_1=BD_2$. Applying Law of Sines on $\triangle{BPD_2}$ and $\triangle{APE_2}$, we have
\[ \frac{D_2P}{\sin\angle{E_2BC}}=\frac{BD_2}{\sin\angle{D_2PB}}\text{ and }\frac{AP}{\sin\angle{AE_2B}}=\frac{AE_2}{\sin\angle{APE_2}},\]
respectively. But $\angle{D_2PB}=\angle{APE_2}$ and $\sin\angle{AE_2B}=\sin\angle{CE_2B}$, so
\[ \frac{D_2P}{\sin\angle{E_2BC}}=\frac{AP}{\sin\angle{CE_2B}}.\]
By Law of Sines on $\triangle{BCE_2}$, $CE_2/\sin\angle{E_2BC}=a/\sin\angle{CE_2B}$, so \[\frac{D_2P}{CE_2}=\frac{AP}{a}=\frac{AP+D_2P}{a+CE_2}=\frac{AD_2}{a+CE_2}.\]
Note that $CE_2=AE_1=AC-CE_1=b-(s-c)=b+c-s=s-a$. Therefore,
\[ \frac{D_2P}{s-a}=\frac{AD_2}{a+s-a}\implies \frac{D_2P}{AD_2}=\frac{s-a}{s}.\]
As a result, $AQ/AD_2=D_2P/AD_2$, so $AQ=D_2P$, as desired. // [/hide]

Let $\omega_a$ be the excircle of $\triangle{ABC}$ opposite to vertex $A.$

$\omega_a$ touches $BC$ at $D_2$

Let $\omega_a$ touches $AC$ at $H.$

Then $\angle{D_2HE}=\frac{\angle{C}}{2}$

It is well known lemma that $Q,I,D_1$ are collinear.

$\angle{E_1D_1C}=\angle{D_1E_1C}=\pi-\frac{\angle{C}}{2}$

So, $\angle{AE_1Q}=\frac{\angle{C}}{2}=\angle{D_2HE}$

Hence $D_2H || QE_1$

So $\frac{AQ}{AD_2}=\frac{AE_1}{AH}=\frac{s-a}{s}$

Applying menelus theorem ([color=darkred]$\triangle{AD_2C},\overline{E_2PB}$[/color]) it is easy to obtain $\frac{PD_2}{AD_2}=\frac{s-a}{s}$

Hence $AQ=D_2P$

Let $ M$ is the midpoint of $ BC \implies D_1M=D_2M $ , $D_1Q $ is the diameter and $ I $ is the center of $\omega$,

 implies $IM \parallel AD_2$ by using midpoint theorem in $ \triangle QD_1D_2 $ and $ 2IM=QD_2$

$ AD_2 \cap BE_2 =P$ (Nagel point ) $ AM \cap IP =G $(Centroid) and by using the fact that $ 2IG=GP$ we get that $AP=2IM=QD_2 \implies AQ=D_2P $

[b]Lemma:[/b] In $\triangle ABC$, $N$ is Nagel point, $I$ is incenter and $G$ is centroid. Prove that, $I, G, N$ are collinear and $GN = 2GI$.

[b]Proof:[/b]

[b]Claim 1:[/b] $IK \parallel AD_2$

[b]Proof:[/b] In $\triangle QD_1D_2$,
\[ID_1 = IQ\]
\[D_1K = D_2K\]
\[\text{So,} IK \parallel QD_2\]

$A, Q, D_2$ is collinear. So, $IK \parallel AD_2$.

[b]Claim 2:[/b] $I$ is the Nagel point of $\triangle KLM$.

[b]Proof:[/b]  $\triangle KLM$ is homothetically similar to $\triangle ABC$ and the center of homothety is $G$.

\[IK \parallel AD_2\]
Similarly, other $I$ to vertex going lines of $\triangle KLM$ are also parallel to Nagel point going lines of $\triangle ABC$. As they are homothetically similar and as Nagel point going lines of $\triangle ABC$ concur at $N$, so the $IK, IM, IL$ have to concur and that point will be the Nagel point of $\triangle KLM$.

So,
\[IK \parallel AD_2 \parallel AN\]

As the homothety ratio is $-\dfrac{1}{2}$,
\[IK = \dfrac{AN}{2}\]
\[IK \parallel AN \implies \angle GAD_2 = \angle GKI\]

\[AG = 2GK\]

So, $\triangle AGN \sim GIK$.
\[\therefore \angle IGK = \angle AGN\]
That implies, $I, G, N$ collinear.
\[\therefore GN = 2GI\]

Now we are gonna finish it with this magical theorem.

Here, $N = P$.
\[2IK = AP [\text{ Homothetical Ratio: }\frac{1}{2} \text{ }]\]
\[2IK = QD_2\]

So, $AP = QD_2 \implies AQ = PD_2$.

By the Diameter of the Incircle Lemma, $Q$ is the point diametrically opposite $D_1$ across $\omega$. Then, construct the points $I$, $M$ and $G$, which denote the incenter, midpoint of $BC$, and the centroid of $\triangle ABC$ respectively. Note that a homothety with ratio 2 centered at $D_1$ sends $IM$ to $QD_2$, so $QD_2=2IM$. Moreover, because $P$ is the Nagel point, we know that a homothety with ratio 2 centered at $G$ sends $IM$ to $PA$, so $PA=2IM$. Hence, $PA=2IM=QD_2$, so $AQ=D_2P$ as desired.

Let $I$ be the incenter of $ABC$, the midpoints of $BC, CA, AB$ be $M, N, R$ respectively, $CP$ meet $AB$ at $F_2$, and $A_1$ be the reflection of $A$ over $I$. Clearly, $P$ is the Nagel Point of $ABC$.

By the Diameter of the Incircle Lemma, we know $Q$ is the $D_1$-antipode wrt $\omega$, $AD_2 \parallel IM$, $BE_2 \parallel IN$, and $CF_2 \parallel IR$. Now, observe $IR$ is the $A$-midline of $ABA_1$, so $$A_1B \parallel IR \parallel CF_2.$$ Analogously, we have $$A_1C \parallel IN \parallel BE_2$$ so $BPCA_1$ is clearly a parallelogram. Hence, we know $A_1, P$ are symmetric about $M$.

It's easy to see $BD_1 = CD_2$ yields $MD_1 = MD_2$, so $A_1D_1PD_2$ is also a parallelogram. Now, this result combined with properties of point reflections yield $$AQ = A_1D_1 = D_2P$$ which finishes. $\blacksquare$

WLOG let $b>c$.

Let $D$ be the foot of the altitude from $A$ to $BC$ and let $D_3$ be the foot of the altitude from $P$ to $BC$. It suffices to show that $D_2D_1=D_3D=b-c$. By mass points we have $D_3D=D_2D\left(\frac{2a}{a+b+c}\right)$ and since $BD=\frac{c^2-b^2+a^2}{2a}$ we know that $D_2B=\frac{a+b-c}{2}$ we have that $D_2D=\frac{ab-ac-c^2+b^2}{2a}$ and $D_3D=b-c$, done.

[quote=rcorreaa]We use Barycentric Coordinates with $ABC$ being the reference triangle. Since $P$ is the Nagel Point, $$P= (s-a : s-b : s-c)= (\frac{b+c-a}{a+b+c},\frac{c+a-b}{a+b+c},\frac{a+b-c}{a+b+c}) \qquad (\star)$$

Let $I$ the incenter of $ABC$. It's well known that $I$ is the midpoint of $D_1Q$ and $I=(\frac{a}{a+b+c},\frac{b}{a+b+c},\frac{c}{a+b+c})$. Hence, $$Q=(\frac{2a}{a+b+c},\frac{2b}{a+b+c}-\frac{s-c}{2a},\frac{2c}{a+b+c}-\frac{s-b}{2a})$$
so $\frac{P+Q}{2}= (1,1-\frac{s-c}{2a},1-\frac{s-b}{2a})= \frac{A+D_2}{2}$, since $D_2=(0 : s-c : s-b)$.

Thus, the midpoint of $PQ$ coincides with the midpoint of$AD_2$, so $AQ=D_2P$, as desired.

$\blacksquare$[/quote]

Very ingenious solution.

Note that $P$ is the Nagel point and let $I$ and $G$ be the incenter and centroid, respectively. Then, $Q$ is the antipode of $D_1$ with respect to $\omega.$ Also, $\triangle GAP\sim\triangle GMI$ since $AG/GM=PG/GI=2.$ Hence, $QD_2=2IM=AP.$ $\square$

[b]This screams bary lmao. [/b]

Note that $Q$ is the antipode of $D_1$ on the incircle. We consider bary with $\triangle ABC$ as the reference triangle. Then, $I = (a:b:c), D_1 = (0:s-c: s-b)$, and thus, $Q= 2I - D = (\frac{2a}{a+b+c}, \frac{2b}{a+b+c} - \frac{s-c}{a}, \frac{2c}{a+b+c} - \frac{s-b}{a})$.

Then, note that $P = (s-a: s-b: s-c)$ is the Nagel point. Then, it suffices to show that $P+Q = A+D_2$, and since they're collinear we check the first coordinate, so
\[\frac{-\frac12 a + \frac12 b + \frac12c}{\frac{a+b+c}{2}} + \frac{2a}{a+b+c} = 1 + 0\]
which is true, so $AQ = D_2P$ and we're done. $\blacksquare$.

[hide=Solution]Let $A=(1:0:0),B=(0:1:0),C=(0:0:1).$ Note that this implies $$E_1=(s-c:0:s-a),E_2=(s-a:0:s-c),D_1=(0:s-c:s-b),D_2=(0:s-b:s-c)\implies P=(s-a:s-b:s-c).$$ Thus, $P$ is the Nagel point of $ABC.$ This means that $$P=G-2I\implies p-D_1=G-2I-D.$$ Also, since $D_1Q$ is the diameter of the incircle, $$2I-D_1=Q\implies A-Q=A-2I+D_1.$$ It is clear that $$A-2I+D_1=G-2I-d_1,$$ since $$A+D_1+D_2=(1:1:1)=G.$$ Therefore, $D_2P=AQ_1,$ as desired.[/hide]

$AP=2IM=QD_2$, where $M$ is the midpoint of $BC$.

Note that $Q$ is the antipode of $D$ in the incircle. $I$ is the midpoint of $QD$ and $M$ the midpoint of $D_1D_2.$ Thus, $IM=\frac12 QD_2.$ Also, $I,G,P$ are collinear with $G$ one-third of the way from $I$ to $P$. Since $MI\parallel AP$, $\triangle MIG\sim \triangle APG\implies AP=2MI$ which implies result.

[hide=sketch]Apply barycentric coordinates on $\triangle ABC$. Note that $P=(s-a:s-b:s-c)$, as it is the Nagel Point. We also have the points $$I=(a:b:c), D_1=(0:s-c:s-b), D_2=(0 : s-c : s-b).$$ Some computation now gives $P+Q=A+D_2$, which finishes.
[/hide]

actually human solution:

[b]Claim: $Q$ is the reflection of $D_1$ over $I$[/b]

Consider the $A$-excircle, and let the tangents from $A$ to the $A$-excircle have length $x$, then considering the tangents from $B,C$ gives $x - b + x - c = a$, so $x = s$, then $D_2$ is the tangency of the $A$-excircle and $BC$ by length equivalence. Then by Fact 5, $A$, and the centers of the incircle and the $A$-excenter are collinear, so we consider the homothety mapping the incircle to the $A$-excircle, this also maps $Q$ to $D_2$, so we are done.

Now, we prove the result that $\frac{AQ}{QD_2} = \frac{D_2P}{PA}$, which finishes. Notice that $\frac{D_2P}{PA} = \frac{CE_2}{E_2A} \frac{BD_2}{BC} = \frac{s - a}{a}$ by ratio lemma. Then we use ratio lemma again, get $\frac{AQ}{QD_2} = \frac{AB}{BD_2} \frac{\sin ABQ}{\sin QBD_2}$, then drop perpendiculars $D_3, Q_2$ from $D_1,Q$ to $AB$, then we have $\frac{AB}{BD_2} \frac{\sin ABQ}{\sin QBD_2} = \frac{c}{s - c} \frac{QQ_2}{QD_1} $, now notice (where undefined points are to be defined analgously) that $2F_1F_2 = D_1D_3 + QQ_2$, so we can evaluate $QQ_2 = 2r - (s-b)\sin B$, so our desired equality is $\frac{s-a}{a} = \frac{c}{s-c} \frac{2r - (s-b) \sin B}{2r}$, which we can reduce to $\frac{b^2 - (a-c)^2}{4ac} = 1 - \frac{\sin B}{2 \tan \frac B2}$, which reduces further to $\frac{b^2-a^2-c^2}{4ac} = \frac{1}{2} - \cos^2 \frac B2$, which we can reduce further to $\frac{b^2 - a^2 - c^2}{4ac} = \frac 12 - \frac{1 + \cos B}{2}$, which rearranges to the Law of Cosines, so because all steps are reversible, we are done.

Very easy especially for U2.

By Menelaus on triangle $AD_2C$ with respect to $\overline{BPE_2}$, we obtain $\frac{AP}{PD_2} = \frac a{s-a}$. Also, $$\frac{D_2Q}{D_2A} = \frac{2r}{h_A} = \frac as,$$ so the result follows.

$Q$ is the antipode of $D_1$ with respect to the incircle, so the tangent to the incircle at $Q$ is parallel to $\overline{BC}$. If the triangle $\triangle$ formed by it and $\overline{AB},\overline{AC}$ has similarity ratio $k$ from $[ABC]$, then we have
$$k=\frac{AQ}{AD_2}=\sqrt{\frac{[\triangle]}{[ABC]}}=\sqrt{\frac{k(s-a)}{s}} \implies \frac{AQ}{AD_2}=\frac{s-a}{s}.$$
On the other hand, in barycentric coordinates is clear that $D_2=(0:s-b:s-c)=(0,\tfrac{s-b}{a},\tfrac{s-c}{a})$ and $P=(s-a:s-b:s-c)=(\tfrac{s-a}{s},\tfrac{s-b}{s},\tfrac{s-c}{a})$, hence $\frac{PD_2}{AD_2}=\frac{s-a}{s}$ as well, so we're done. $\blacksquare$


modular arithmetic
pigeonhole principle
induction
number theory proposed
number theory

Let $ p \not\equal{}2,3,5$ be a prime number. For any integer $ k\geq 1$, prove that there is a multiple of $ p^k$ with its sum of digits equal to $ p$.

[quote="pmvald"]Let $ p \not \equal{} 2,3,5$ be a prime number. For any integer $ k\geq 1$, prove that there is a multiple of $ p^k$ with its sum of digits equal to $ p$.[/quote]

Let $ P(k)$ be the proposition that "there is a multiple of $ p^k$ with its sum of digits equal to $ p$".

For $ k\equal{}1$ we have $ 1111...1 (p '1's)$ satisfies the condition.
Assume that $ P(m)$ is true for any $ m \in N$ then $ \exists$ positive integers $ a_1<a_2<...<a_p$ so that $ p^k|10^{a_1}\plus{}10^{a_2}\plus{}...\plus{}10^{a_p}$. let $ 10^{a_1}\plus{}10^{a_2}\plus{}...\plus{}10^{a_{p\minus{}1}}\equal{}K$

Now for $ P(m\plus{}1)$,

[b]Case 1:[/b] $ 10^{p\minus{}1} \neq 1 \pmod{p^{m\plus{}1}}$, then $ 10^{(p^m\minus{}p^{m\minus{}1})n} \neq 1 \pmod{p^{m\plus{}1}}$ but $ \equiv 1 \pmod{p^m}$  $ \forall n\equal{}1,2,...,p\minus{}1$

So $ K\plus{}10^{a_p}, K\plus{}10^{a_p\plus{}(p^m\minus{}p^{m\minus{}1})}, K\plus{}10^{a_p\plus{}2(p^m\minus{}p^{m\minus{}1})}, ... ,$
$ K\plus{}10^{a_p\plus{}(p\minus{}1)(p^m\minus{}p^{m\minus{}1})}$, are pairwisely distinct $ \pmod{p^{m\plus{}1}}$ and all of them are multiples of $ p^m$ , so by pigeonhole-principle we have one of the $ p$ numbers is divisible by $ p^{m\plus{}1}$ . As sum of digits $ \equal{}p$ for all these numbers , we have $ P(m\plus{}1)$ is true for this case.

[b]Case 2:[/b] $ 10^{p\minus{}1} \equiv 1 \pmod{p^{m\plus{}1}}$ , then $ 111...1 (p '1's) \equiv \frac{10^p\minus{}1}{9} \equiv 1 \pmod{p^{m\plus{}1}}$ (because $ 9|10^{p\minus{}1}\minus{}1$ ) . So $ 111...1$ satisfies the condition $ \Rightarrow P(m\plus{}1)$ is true for this case also.

So by induction $ P(k)$ is true for all $ k \in N$

[quote="stephencheng"]

For $ k \equal{} 1$ we have $ 1111...1 (p '1's)$ satisfies the condition.
[/quote]

this isn't true for any $ p$ except 3.

i didn't read stephencheng's solution but for the case $ k\equal{}1$ we can consider the number $ 1\plus{}10^{p\minus{}1}\plus{}10^{2(p\minus{}1)}\plus{}...\plus{}10^{(p\minus{}1)^2}$ which is obviously divisible by $ p$ and has the sum of digits equal to $ p$.

shoki, you usually don't consider trivial cases but now you only consider the trivial case! :D

yes but this time i wanted to complete stephencheng's solution(and i hope now it is correct!) :)

[quote]
stephencheng wrote:
For $ k\equal{}1$ we have $ 1111...1 (p '1's)$ satisfies the condition.
[/quote]

Well this small problem was addressed by shoki

[quote]
stephencheng wrote:
Assume that $ P(m)$ is true for any $ m \in N$
[/quote]

Better to say $ P(m)$ is true for some $ m \in N$

[quote]
stephencheng wrote:
... so that $ p^k|10^{a_1}\plus{}10^{a_2}\plus{}...\plus{}10^{a_p}$
[/quote]

Should say $ p^m|10^{a_1}\plus{}10^{a_2}\plus{}...\plus{}10^{a_p}$

[quote]
stephencheng wrote:
Case 1: $ 10^{p\minus{}1} \neq 1 \pmod{p^{m\plus{}1}}$, then $ 10^{(p^m\minus{}p^{m\minus{}1})n} \neq 1 \pmod{p^{m\plus{}1}}$
[/quote]

How do we conclude this?


This is A2 from ISL 03:
Find all non-decreasing functions f : R --> R such that:
(i) f(0)=0;f(1)=1
(ii) f(a)+f(b)-f(a)f(b)=f(a+b-ab) for all real numbers a, b s.t. a<1<b.

Let $g(x)=1-f(x),\ h(x)=g(1-x)$. We get $h(a)h(b)=h(a,b),\ \forall a<0<b$. Moeover, $h$ is non-decreasing (I'll just call it increasing, because that's what we call non-decreasing in Romanian; the notion of increasing is denoted by strictly increasing in Romanian). We also know that $h(0)=0,\ h(1)=1$.

Functions which are increasing, $h(1)=1$ and $h(x)=0,\ \forall x\le 0$ are clearly Ok, and if we assume our function does not have this form it's easy to see that $h(x)=0\iff x=0$. Another possible solution is $h(x)=k<0,\ \forall x<0,\ h(0)=0,\ h(x)=1,\ \forall x>0$, and if we rule this one out too we get $h(b)\ne 1,\ \forall b\ne 1$.

It is easy to show that $h(-1)=-1$, so $h(-x)=-h(x)$, so $h(ab)=h(a)h(b)$. This is the Cauchy equation, and since we have the extra condition that $h$ is increasing, we know that $h(x)=x^t,\ \forall x\ge 0$. Together with $h(-x)=-x^t,\ \forall x\ge 0$, this defines the function. We thus have these three types of solutions.

Is it correct?


The points (0,0), (1,0), (1,1) and (0,1) are the vertices of a square [i]S[/i].  Find an equation in [i]x[/i] and [i]y[/i] in whose graph in the [i]xy[/i] - plane is [i]S[/i].

I'm not sure what you mean. if the function is "in x and y" doesn't that mean the 'graph' is in R(3)?

That's what I was thinking, but I was confused.  How do you make a function in three-dimensional space turn at corners like that?

Alright I'm supposing the third component is just meant to be zero and you're mapping to the boundary of C=[0,1]x[0,1]x{0}.

In which case don't you just define a real valued function on the boundary of C with value 0 everywhere?

Yeah, that makes sense. I guess I was trying to get an equation that defined the edge of [i]S[/i].

It's wellknow curve of Peano.
http://www.cultinfo.ru/fulltext/1/001/009/001/235992157.jpg

I thought that might be it.  :blush:

I was thinking he wanted to traverse the square (the boundary) as opposed to the 1-cube. Comes from the ball/sphere distinction. Sorry!


[quote="richard"]I'm really confuse about what we can do by using affine transformations. For example: in order to prove Ceva's theorem, could we do the following?

- Make an affine transformation of the 3 vertices of the triangle to the 3 vertices of an equilateral triangle.
- Prove that the result is valid in the equilateral triangle.
- Say that, since it's valid after the transformation, it also was valid before the transformation.[/quote]

hmmm....

to prove Ceva theorem by affine tranformations we can say that:

each three ceva lines in a triangle, after using a paralell projection to another plane,

changes to: the three altitudes of a nother triangle.

then if we prove the concurrently of altitudes, every thing will be ok.

sorry, i think ure ideas dont work atlist for this problem :(

how ever, i can send the solution, later. ;)

another thing is that: affine transformations, fixes the relation betwin the lenghts.

foe example: if $AB$ is aline and $A'B'$ is its projective form, for each point $C$ on $AB$ ,and its projective

point ,$C'$,we have:

$\frac{AC}{CB}=\frac{A'C'}{B'C'}$

the ideas u said dont obay this rule. ;)

[quote="Centy"]Affine transformations preserve the ratios of distance along the same line. So if the cevian from $A$ meets $BC$ at $D$, we can affine transform $\triangle ABC$ to an equilateral triangle and $\frac{BD}{DC}$ is preserved. So if $\frac{BD.CE.AF}{DC.EA.FB} = 1$ for an equilateral triangle, it's true for all triangles.[/quote]

yes it is....

but, is it make the problem easier than that one?

[quote="Quick"][quote="Ashegh"]
yes it is....

but, is it make the problem easier than that one?[/quote]

That was not the question !

The question was whether this was a mathematically correct approach, and indeed it is...

Now, if this can lead to a proof is yet another question !   :lol:[/quote]

 i know it my dear.

i only wanted to say that the way which centy suggested , dont solve the problem easier than the first one.

[quote="benjamin"][quote="richard"]I'm really confuse about what we can do by using affine transformations. For example: in order to prove Ceva's theorem, could we do the following?

- Make an affine transformation of the 3 vertices of the triangle to the 3 vertices of an equilateral triangle.
- Prove that the result is valid in the equilateral triangle.
- Say that, since it's valid after the transformation, it also was valid before the transformation.[/quote]

I think it’s even better to apply a central projection !
Indeed let A, B, C, D be 4 points in a plane P, and let A’, B’, C’, D’ be 4 points in a plane P’. Then you can always place the planes P and P’ so that there exists a central projection from P to P’ which carries ABCD into a quadrilateral similar to A’B’C’D’. You can find proofs in Yaglom’s book “Geometric transformations 3”, but it’s intuitively obvious.
Less obvious is the fact that the ratio DB/DC . EC/EA . FA/FB remains unchanged through a central projection (the preservation of the cross-ratio is a special case of this property).
Thus, you can find a central projection which maps A, B, C, and the intersection of AD and BE onto an equilateral triangle and its center ! Then Ceva’s theorem is trivial.

Benjamin[/quote]

good.

this solution also works.

do u know any projection solution, with out central projection?

i mean , for exampla paralell projectios.

[quote="Ashegh"]do u know any projection solution, with out central projection?[/quote]

I just read one !

Let a triangle ABC and points M, N, P on its sides be given, such that AM/MB . BN/NC . CP/PA = 1
We first prove that if A'N', B'P' and C'N' are the altitudes of a triangle A'B'C' (Fig. 134a), then A'M'/M’B’ . B'N'/N’C’ . C'P'/P’A’ = 1
Indeed, the similarity of the right triangles A'N'C' and B'P'C' implies that C'P'/N'C' = a/b where a = B'C' and b = A'C' In much the same way we can show that B'N'/M'B' = c/a, A'M'/P'A' = b/c where c = A'B'. These three equalities imply that :
A'M'/M’B’ . B'N'/N’C’ . C'P'/P’A’ = A'M'/P’A’ . B'N'/M’B’ . C'P'/N’C’ = b/c . c/a . a/b = 1
Now we construct a triangle A'B'C' such that the feet N' M' and P' of its altitudes divide the sides of the triangle in the given ratios :
A'M'/M’B’ = AM/MB ; B'N'/N’C’ = BN/NC ; C'P'/P’A’ = CP/PA
We divide an arbitrary segment B'C' in the ratio B'N'/N'C' = BN/NC.
At N' we erect a perpendicular to B'C'. Then we divide segment C'N' in the ratio C'Q/QN' = CP/PA and erect at Q a perpendicular to C’B’ (Fig. 134b). Let P' be the point where this perpendicular meets the semicircle constructed on the segment C'B' as diameter, and let A' be the point where the line C'P' meets the perpendicular to C'B' at N'. We claim that A'BC' is the required triangle. Indeed A'N' and B'P' are two altitudes of this triangle ; let C'M' be its third altitude.
Since B'N'/N'C' = BN/NC and C'P'/P'A' = C'Q/QN' = CP/PA , and since A’M’/M’B’ . B’N’/N’C’ . C’P’/P’A’ = 1 = AM/MB . BN/NC . CP/PA, it follows that A'M'/M’B' = AM/MB.
We now map ABC by a parallel projection onto a triangle similar to A'B'C'. In view of the property that a parallel projection preserves the ratio of the lenghts of 2 collinear segments, points N, P, M are mapped onto the feet of the altitudes of the image of ABC and the lines AN, BP, and CM onto its altitudes. Since the altitudes of a triangle are concurrent, the same is true of the lines AN, BP and CM.

Benjamin


Carrying out the operations, we have $ 625x^{4}\equal{}1000x^{3}\Rightarrow x\equal{}\frac{1000}{625}\equal{}\frac{8}{5}$.


symmetry
integration
geometry
calculus
analytic geometry
LaTeX

A uniform circular plate of radius 2R has a circular hole of radius R cut into it. The centre of the smaller circle is a distance 0.80 R from the centre of the larger circle. What is the position of the centre of mass of the plate?

Do you have any reason to believe that this comes out as something nice? What page in Giancoli? I have the book at home, I would like to see the problem and it's context.

It is clear by symmetry that y = 0, and $x = \frac{{ \int_A x f(x)\;dx}}{M}$ where f(x) is the difference between the circles and M is the area between them. I found this very difficult to calculate. It might be possible to do it as a line integral + green's theorem.

As for its context, I don't believe it should be very difficult or should require calculus. It's on page 206, problem 51.("Physics: Principles with applications", 5th ed.)

Oh and there appears to be a hint too "[Hint: Try subtraction]".

[quote="shobber"]Is there a non-calculus way to solve this problem? Also, is there a genenral way to find the centre of mass of an object?[/quote]

[b]There definitely is a NON-CALCULUS method for solving center of mass, moment of inertia and other prolems relating to properties of matter[/b].The name and theoretical aspects may [b]startle[/b] you but it is an [b]extremely simple [/b]method with wide applicability and with which [b]even the toughest problems can be solved without use of any calculus.[/b] :lol: Here's how it goes.It's called the [b]negative mass [/b] method.I'll illustrate it in the above question(the general case can be easily understood by extrapolation!!)

Consider a Circular disc of Radius = 2R and having density = X (SI units).Now imagine another circular disk of radius R made of  a material having hypothetical density = -X (don't worry about true science , just imagine!).Now superimpose these 2 bodies with the smaller dish 0.8R from center of larger disk. Consider them to be bodies centered at their respective centres.Now we can easily calculate the CM of the combined body using the 2 body formula.Imagine a coordinate system at centre of original disk and x axis along the line joining the centres of the 2 disks.Let M be the effective total mass of the combined body.\[\overline X =\ \frac { \varrho \times \pi (\ 2R)^2\ \times 0\ +\ (\ -\varrho)\ \times \pi R^2\ \times (\0.8R)\ }{M} =\ \frac {-8}{30} R \ (\ on\ simplification\ )\ \] Regarding the Y coordinate it will be zero by symmetry(actually i've become a bit lazy after writing that Latex code)

I hope you realise the immense use of this wonderful method and ude it wherever you find those useless CM questions.If my explanation isn't clear to you, I'll be happy to explain it further.

A simple way that actually uses "true science" would be to consider the original body A and the small circle B that combine to form the large circle C. You can get equations for the center of mass of C in terms of those of A, B which can be solved for the center of mass of A.

In other words, you can calculate the CM of the disk with the hollow like the CM of two particles that are in the center of the circles, one of which has a negative mass (the hollow). I have used this in many problems, it's a wonderful way to avoid integrations... Obviously you have to think about symmetry.


In my chapter competition on Saturday, I found that there could be two very frustrating cases for an individual:

1) I came in third place individually in my chapter competition, yet, because I am on my school's B team, I may not even get a chance to go to my state competition (second place, who finished one point ahead of me, was also on the B team).
2) Another kid there, who finished in 6th individually, was the only strong member on his team and will thus not be competing in state.

My question is: Do you think there should be a threshold in which high-scoring individuals can compete in their state's competition as an individual?

I don't know that it will help for this year, but I highly recommend writing MATHCOUNTS with this explanation.  Something similar happened in the San Diego Chapter this year, as well, and I'm not sure how it will be resolved.  In my 7th grade year (1984, gasp), I came in 2nd in chapter and 3rd place team.  My chapter sent the top two teams and the top individual (who was on one of those teams, but they still didn't go down to the next individual, me).  I imagine many other students around the country got hosed similarly - I didn't get to go to state that year.  We wrote MATHCOUNTS and they did change the rules for the following year.  Find out as soon as you can if you are going to state, and if you are not, start writing - write the state coordinator and the national office.

In my chapter (Santa Clara), they allowed 4 individuals and 4 teams . That's mainly because there were 167 participants and 20 teams. Metsfan001, you probably have a small chapter.

By size, the smallest in America. The same can not be said, however, about population. . . .

The first year that the "Individual" was introduced in MATHCOUNTS 4 years ago, my team members finished 1,2,3,4,  and my top individual took 6th.  At that time our chapter rule was the top 3 teams and any kid in  the top 4 not on one of those teams advanced to states.  My 6th place kid beat all but one kid from the other schools and didn't make it to states.

I contacted the state coordinator and the next year the rule was changed so that the top 4 kids not on one of those teams advance.  It's a good thing since 3 of my individuals beat all the kids on the 3rd place team, and one even finished 5th, beating all but my 4 team members who took 1st - 4th.

I think your best bet is to contact the state coordinator since I think they have the power to decide who advances to states.

Good luck,

Josh Frost

So this means I have to be in the top 2 on sprint/target to go to state?! :(  :?

I am not so sure about this anymore...

Billy

These rules vary widely from state to state, and even from chapter to chapter within a state. One key point is how much space there is in the place where the state meet is held, which sets a hard limit on how many mathletes can be invited to go there. Fortunately, in City of Lakes South Chapter in Minnesota, as in all Minnesota chapters, the rule is that some number of teams go on to state (based on how many teams registered in each chapter) and some number of individuals go (based on how many individuals registered in each chapter). We do NOT have a rule that goes further down the list of individuals if some of the top individuals have already won a spot at state by being members of a top team--which is a rule that I think is more fair.

At least in two chapters in KS, the top 6 individuals and top 3 teams qualify.  Usually there won't be more than one or two who are not in the top 3 teams.  This is a good way of not dropping off a strong mathlete.

This is from the MATHCounts website:

[quote]LOCAL competitions: Held between February 5 and February 27, 2005, the Chapter Competition consists of the Sprint, Target and Team rounds. The Countdown Round (official or just for fun) may or may not be included. The chapter and state coordinators determine the date and administration of the local competition in accordance with established national procedures and rules. Winning teams and students will receive recognition. The winning team will advance to the State Competition. Additionally, the two highest-scoring competitors not on the winning team (who may be registered as individuals or as members of a team) will advance to the State Competition. This is a minimum of six advancing Mathletes (assuming the winning team has four members). Additional teams and/or Mathletes also may progress at the discretion of the state coordinator, provided that the policy for progression is consistent for all chapters within a state.[/quote]  from [url=http://205.252.29.34/webarticles/anmviewer.asp?a=98]205.252.29.34/webarticles/anmviewer.asp?a=98[/url]

This seems to say that the top team + top 2 NOT on team go to state as a MINIMUM.
Interesting.

Billy

I'm not very familar with mathcount and the stuff about individual advancing or not and that stuff because I've never been to a math count thing before, but I think they should be allow from I've heard of the rules... it's just more fair... can someone explain more about the individual and team stuff? is it that if u qualified from your school top 4, you are on the team, if you are 5-8, then you compete as individual? if u are individual, do u get a different placement thing or something? i'm confused

If anyone does write the MATHCOUNTS coordinators, it might be reasonable to suggest that any student in the top 4 in any chapter be allowed to advance to the state tournament.  After all, the fourth student in a chapter might be the fourth student in a state.  During my 8th grade year this was the case for the Alabama team.  While these four students were all from 2 teams that advanced to state regardless, it would have been a shame had one of those students not advanced.

here's the way chapter works:
its predetermined (based upon chapter participation... number of people there) how many teams/individuals not an advancing team... for our chapter Preston Trail (TX) which is next week... there are 25 teams and 5 teams and 5 individuals not a winning team go to state.
now: the teams that advance to state: normally anywhere between 1 (small chapter) to generally at max. 6 (large chapter)... then the top individuals who ARE NOT PART OF THE TEAMS ADVANCING TO STATE go as individuals.
This means: say you placed 3rd as did your team. (not going in your case), the person who placed first: on a winning team. therefore you and the person placed before you should go... try asking your state coordinater.

I have twice had a kid finish 4th at chapter and then make it to Nationals(2nd and 3rd at State) and 3 times the kid who was 3rd made it to nationals(2nd, 2nd, 3rd at State) Why my kids keep doing better at State than they do at Chapter I do not know :)

In my chapter, top 28 individuals and top 6 teams advance. But the Maryland Potomac chapter is tough, so it balances it out.

Yup, it does vary a lot from chapter to chapter. For example, my chapter (Hartford) has, I believe, one of the better systems. Here, we take the top 6 or so teams out of 20-30, and also any individuals in the top 30 or so to go to states. I think this would be a good system to fix the problem of individuals not getting an opportunity to go to states.

yea im a competitor and Jason goes to my school and is in my grade demoting me to 2nd best on the team, yea it is a bit scary because he only missed 1 sprint problem at chapter and he got a perfect target, good luck to you too at states

Hehe, he and I watched Hannibal at nats last year. Pretty funny experience.

Hmm... i must have a really easy chapter:

Top 4 teams (MAYBE up to six!)
Top 12 Individuals (MAYBE up to 16!)

I'm in an Indiana Chapter.

[size=75]of course...it would be [/size][size=100]ridiculously [/size][size=150]outragiosly [/size][size=200]ginormously [/size][size=75]ufair if it wasn't[/size]

[quote="metsfan001"][quote="hockeyadi23"]lol.... man how'd u get over 1000 posts in Yifan? I just got started today... but still....[/quote]

He started a year before you.[/quote]
;) Of course...  ;)

Yah, my chapter in Southeast Nebraska takes the top two teams and also the top 4 individuals that aren't on the top teams.

I would know because that's how I got to state last year as a 7th grader.

:(  :(

Apparently New York is above the law, and I'm stuck.

 :(  :(

Definitely, we need more individuals going.

Of course there has to be some elimination in the chapter competition, but besides having enough space, there is nothing to lose by allowing 25 individuals to go to States. States is a harder test. If you are good enough to get to nationals, then you will get there, regardless of how many other students are competing against you. If you are in the top 25, you have a chance of making to nationals if you practice and get lucky by getting questions you can handle.

our chapter takes the top 4 team and the top 4 individual who's not on the team

[quote="uberl33tmage"]baseball freak, what school do you go to? And do you know two ppl in your chapter named Patricia and Maryanna?
BTW my names Kevin.[/quote]

>[ marianna not maryanna grr!

[quote="metsfan001"] I came in third place individually in my chapter competition, yet, because I am on my school's B team, I may not even get a chance to go to my state competition[/quote]

It was ten years ago, but when I was a seventh grader I finished [b]first[/b] in my chapter (with a 44, highest in the organizers' memory). Since I was on my school's B team, I didn't go to state. The next year, I made the California team and finished fifteenth nationally.

I didn't even know I was very good at math contests until that experience.

By b team you all mean individual but not team right? Our coach ALWAYS no matter what unless the sevys are really dumb or something puts two 7th adn two 8th on the team so that there always is experience... and all the individuals are 8th graders unless someone in seventh is phenominal. Anyways if you came third in your chapter its safe to say you probably would have gone since a person on the top team probably was in the top two, but sometimes that doesn't happen.


inequalities
geometry
circumcircle
inequalities proposed

Prove that

$ \frac{a^2b^2\plus{}b^2c^2\plus{}c^2a^2}{a^2\plus{}b^2\plus{}c^2}\leq3R^2$

[quote="mathmen"][quote="Ligouras"]Prove that

$ \frac {a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2}{a^2 \plus{} b^2 \plus{} c^2}\leq3R^2$[/quote]

a,b,c $ \in$???[/quote]

sides of triangle $ ABC$

[quote="Ligouras"]Prove that

$ \frac {a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2}{a^2 \plus{} b^2 \plus{} c^2}\leq3R^2$[/quote]

 The following inequality is equivalent to

$ 3R^2 \minus{} \frac {s^4 \minus{} 8Rrs^2 \plus{} 2r^2s^2 \plus{} 16R^2r^2 \plus{} 8Rr^3 \plus{} r^4}{2(s^2 \minus{} 4Rr \minus{} r^2)}\geq 0$

$ \Longleftrightarrow \minus{} s^4 \plus{} (6R^2 \plus{} 8Rr \minus{} 2r^2)s^2 \minus{} 24R^3r \minus{} 22R^2r^2 \minus{} r^4 \minus{} 8Rr^3\geq 0$

$ \Longleftrightarrow \minus{} s^4 \plus{} (4R^2 \plus{} 20Rr \minus{} 2r^2)s^2 \minus{} r(4R \plus{} r)^3 \plus{} 2R((R \minus{} 6r)s^2 \plus{} r(4R \plus{} r)(2r \plus{} 5R))\geq 0$

can prove  $ \minus{} s^4 \plus{} (4R^2 \plus{} 20Rr \minus{} 2r^2)s^2 \minus{} r(4R \plus{} r)^3\geq 0$

and

$ (R \minus{} 6r)s^2 \plus{} r(4R \plus{} r)(2r \plus{} 5R)\geq \minus{} 4rs^2 \plus{} r(4R \plus{} r)(2r \plus{} 5R)$

$ \equal{} r( \minus{} 4s^2 \plus{} (4R \plus{} r)(2r \plus{} 5R))\geq r( \minus{} 4(4R^2 \plus{} 4Rr \plus{} 3r^2) \plus{} (4R \plus{} r)(2r \plus{} 5R))$

$ \equal{} r(5r \plus{} 4R)(R \minus{} 2r)\geq 0$

[quote="Ligouras"]Prove that

$ \frac {a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2}{a^2 \plus{} b^2 \plus{} c^2}\leq3R^2$[/quote]

$ LHS \le \frac{1}{3}(a^2\plus{}b^2\plus{}c^2) \le 3R^2$

$ OG^2 \equal{} R^2 \minus{}\frac{1}{9}(a^2\plus{}b^2\plus{}c^2)$ ($ O\minus{} circumcenter$ , $ G\minus{}triangle centroid$)


inequalities
algebra
function
domain
algebra solved

I will propose now an extremely difficult inequality:
  Prove that for any positive a,b,c with sum 1 we have the inequality:
  (ab+bc+ca)(a/(b^2+b)+b/(c^2+c)+c/(a^2+a))>=3/4.
   I gave this problem to lots of people very well prepared on this domain and they couldn't solve it. I have a very short and easy solution, but I would like to know if there aren't other solutions. I really love this problem.

Come on, noone tries this inequality? Is it so ugly?

Harazi, it is a beautiful inequality but extremely difficult to solve in a simple way.
It is like a CRUX solved inequality

 (ab+bc+ca)[1/(a+b) 2 +1/(b+c) 2 +1/(c+a) 2 ] \geq 9/4

but the same method of solution doesn't work.
I wait for your solution!!
Thank you.

I will let everyone try it for a week and after that I will post a solution. I'm interested if anyone had seen it somewhere, because I hadn't.

Still no solution? Boy, this one is tuff!

You can find my solution at :

http://www.iandrei.go.ro/ineq_harazi.pdf

I hope that it's correct!

Hmmm..., your solution is absolutely amazing. I haven't check if it is correct yet, but I believe it is. I will post now my initial solution (meanwhile I found a very simple solution, which I will post after a month-don't ask me why). So, my solution was:
  Take 0<=x<=1. From Cauchy we have:
  a/(x+b)^2 +b/(x+c)^2+c/(x+a)^2>=(a/(x+b)+b/(x+c)+c/(x+a))^2>=
(1/(x+ab+bc+a))^2. Now all we have to do is to integrate this inequality between 0 and 1 and use the fact that ab+bc+ca<=1/3. This one is more elegant, but allmost impossible to find. Anyway, I believe that this is really a super-inequality.

this problem was a the first problem(I think) of iranian team selection for year
2001 (1380 in iranian year).

Too bad! Just when I was proud I have invented a beautiful and difficult inequality. I said I will post a new solution after a month. Because I'm very sad, I will post it now. All we have to do is to write:
 sum a/(b^2+b)= sum (a/b)^2/(a+a/b) and apply Cauchy. You will obtain that sum a/(b^2+b)>=(sum a/b)^2/ (1+sum a/b), which is greater than
3/4(a/b+b/c+c/a). Now we have to prove that a/b+b/c+c/a>=1/(ab+bc+ca), which is again Cauchy: a/b+b/c+c/a=a^2/ab+b^2/bc+c^2/ca>=(a+b+c)^2/(ab+bc+ca)=1/(ab+bc+ca).

[quote="harazi"]Prove that for any positive $x,y,z$ with sum $1$ we have the inequality:
\[(xy+yz+zx)\left(\frac{x}{y^{2}+y}+\frac{y}{z^{2}+z}+\frac{z}{x^{2}+x}\right)\geq\frac{3}{4}.\][/quote]
$\left(\sum_{cyc}x\right)^{2}\sum_{cyc}xy\sum_{cyc}\frac{x}{y(2y+z+x)}\geq\frac{3}{4}\sum_{cyc}x(2y+z+x)\sum_{cyc}xy\sum_{cyc}\frac{x}{y(2y+z+x)}$
$\geq\frac{3}{4}\left(\sum_{cyc}x\right)^{3}$.


[hide]Mr. P is not the science, and has more experience than someone, so Mr. P is not the history one, so Mr. P is the math guy.

Mr. Eye is I believe not the sci from the second part, so perhaps Mr. Love is the science teacher.[/hide]

Billy

[hide] It is Mr. Eye because he is not an only child, and Mr. Problems is not the history teacher because he has more experience than the science teacher. Then Mr. Love is the history teacher.[/hide]

why do you need all the enpeirience parts?

[hide]The experience thing tells you that problems is a math teacher. That last sentence also tells you that Eye is not the history (as he has a sister) teacher and therefore is the science teacher. In that part, you get the answer to the question, Mr. Eye. [/hide]

[hide](a)At Agnesi Middle School, Mr. Eye, Mr. Love and Mr. Problems teach science, mathematics, and history -- but not necessarily in that order.(b) The history teacher, who was an only child, has the least experience. (c)Mr. Problems, who married Mr. Eye's sister, has more experience than the science teacher. (d)What teaches science?

i numbered the sentences in the question so it's easier to explain

from b and c, you can tell that Mr.Problems is not the history teacher.... and then the part about Mr.Eye's sister is married to Mr. Problems tells you that he's not the history teacher either... and so Mr. Love is the history teacher.... and then c tells u that Mr. Problems is nto the science teacher, so Mr. Eye is the science teacher. [/hide]

[hide]without the experience part, then you can't finish the problem since it tells you that mr.p is the math teacher and so on.[/hide]

[quote="dengmi"][hide]Mr. Love teaches history.  Since Eyes is not an only child, he can't teach history.  The next statement tells us that Problems couldn't have teached science.  But since we know the social studies teacher was the least experienced, and Problems was more experienced than the science teacher, he must have teached math.  Therefore, Eyes and Love didn't teach math so Eyes is the Science teacher, making Love the history teacher.[/hide]

Who picked the names?  I noticed that you could make two statements out of it.  Eye love problems as in your eyes like looking at problems and Eye (it really should be I) love problems.[/quote]

Just a quick note, you didn't answer the question: Who was the science teacher? Sry i just noticed that as I went over the forum.

[hide]Yeah, I'm guessing Mr. Eye since he isn't the history teacher and Mr. Problems isn't the science teacher. :lol: [/hide]

[quote="MCrawford"]At Agnesi Middle School, Mr. Eye, Mr. Love and Mr. Problems teach science, mathematics, and history -- but not necessarily in that order.  The history teacher, who was an only child, has the least experience.  Mr. Problems, who married Mr. Eye's sister, has more experience than the science teacher.  What teaches science?[/quote]

[hide]
[b][i]What teaches science?[/i][/b]
A teacher! :P

Ok then.
If the history teachers got least EXP. and Mr problems has more EXP. than the Sci teacher than he teaches Math.
If Mr Eye has a sister than hes not an only child so hes not the history teacher and he teaches sci.[/hide]

[quote="dengmi"][hide]Mr. Love teaches history.  Since Eyes is not an only child, he can't teach history.  The next statement tells us that Problems couldn't have teached science.  But since we know the social studies teacher was the least experienced, and Problems was more experienced than the science teacher, he must have teached math.  Therefore, Eyes and Love didn't teach math so Eyes is the Science teacher, making Love the history teacher.[/hide]

Who picked the names?  I noticed that you could make two statements out of it.  Eye love problems as in your eyes like looking at problems and Eye (it really should be I) love problems.[/quote]

About who wrote the problems... It's in the mathcounts school handbook. 1999-2000.


The perimeter of an isosceles triangle is 143.  Two of the sides are in a ratio 3:5.  Find the two possible lengths of the base.

Since it isosceles the third side of this triangle is also in proportion to the third side either 3:5 or 5:3

So if all sides are in proportion 3:3:5 (The base is the one not congruent to the other sides)

We have $ 3x \plus{} 3x \plus{} 5x \equal{} 143$ and $ x \equal{} 13$

So the length of the base here is $ 5 \cdot 13 \equal{} 65$

The second possibility we have the sides in proportion 5:5:3

So we have $ 5x \plus{} 5x \plus{} 3x \equal{} 143$ and $ x \equal{} 11$

So the length of the base is $ 11 \cdot 3 \equal{} 33$

and the possible lengths of the bases are $ x \equal{} 65, 33$


Hello,

Try this one.

Find all the positive integers $x$ and $y$ such that
\[y^x = x^{50}\]

Best regards,

Can this be solved w/o using any logs or calculators or any complicated maths?

SnowStorm, you're almost there, try to use variables in your solution and make a little extension to your solution and then show that there are no other values, if you think you found them all.

This is interesting.

Let's see: For all $y^x=x^{50}$.

If you take both sides to the $\frac{1}{x}$ power, you get:

$y=x^{\frac{50}{x}}$ power.

Notice that this works for all $x$ where $x$ is a factor of $50$.

So, $50 = 2 * 5^2$. This gives $2*3=6$ values for $x$ so far.

Now take $x=100$. I just noticed that it worked, though I'm still trying to figure out a formal proof as to why.

I can see that for $x$ more than $100$, $x^{\frac{50}{x}}$ converges to $1$, but I guess it can easily be done using limits.

In fact, that would be the proof now that I think a little.

However, limits are a little beyond getting started. Eh, is there another way to do this that's simpler?

[quote="SnowStorm"]Hmm
[hide]
simplifying:
$y^{\frac{x}{50}} = x$
So x must be divisible by 50?
Plugging 50 in
$y^1 = 50$
x = 50, y = 50
100:
$y^2 = 100$
x = 100, y = 10

I think these might be the only ones that work.
[/hide][/quote] assume x is divisible by 50.
all of those numbers work
1^50=1^1
2^50=(2^25)^2
so on for all divisible by 50
now we work on those above 40.
they must be multiples of 50.
x In y = 50 In x.
x must be a square (if x=100) cube (if x=150) or a nth power, if 50n is an nth power. 50n=y^n. 2 works (obviously). 50=5^2*2*n=y^n. so n = 5^(kn-2)*2^(bn-1) with some k and b. and then, there should only be one solution for this, because it increases exponentially.
so what am i missing?


inequalities
algebra
polynomial
limit
induction
function
inequalities unsolved

[quote="Golden Rule"]Prove that each of the following inequalities holds for all n∈ N
a) n2 ≤ 2n + 1
b) n3 ≤ 3n[/quote]
Are you sure? :P

I'm sorry the powers in the problem did not transfer from word.
The problem is:
Prove that each of the following inequalities holds for all n∈ N
a) n^2 ≤ 2^n + 1
b) n^3 ≤ 3^n

It quite easy you can Induction.
More general we a result on Analysis
$ a > 1$
$ P(x)$ is a polynomial ,least coffience >0
Then
$ \lim_{x\to\infty}\frac{a^{x}}{P(x)}\equal{}\plus{}\infty$

[quote="Golden Rule"]
a) n^2 ≤ 2^n + 1 (1)
b) n^3 ≤ 3^n[/quote]

a, Induction :

-$ n \equal{} 1,(1)\Leftrightarrow 1\le 2\plus{}1$

-If for each natural number $ i\le k$ then $ i^{2}\le 2^{i}\plus{}1$

We have $ k\ge 2$ then $ 2k\plus{}1\le k^{2}\minus{}1$ so:

$ (k\plus{}1)^{2}\equal{} k^{2}\plus{}2k\plus{}1\le 2k^{2}\minus{}1\le 2^{k\plus{}1}\plus{}1\Box$

b, Because the function $ f(n) \equal{}\sqrt [n]{n}\searrow,\forall n\in [3,n]$ and $ \sqrt{2}<\sqrt [3]{3}$  :)

Let's use induction: When $ n \equal{} 1$, $ n^{2}< 2^{n}\plus{}1$. Let, $ k^{2}\leq 2^{k}\plus{}1$ for any positive integer $ k$ and let's prove, that $ (k\plus{}1)^{2}\leq 2^{k\plus{}1}\plus{}1$.
$ 2^{k\plus{}1}\plus{}1\geq 2(k^{2}\minus{}1)\plus{}1 \equal{} 2k^{2}\minus{}1$. So, if we manage to prove that $ 2k^{2}\minus{}1\geq k^{2}\plus{}2k\plus{}1$, we are done. $ RHS\minus{}LHS \equal{} (k\minus{}1)^{2}\minus{}3\geq 0$,for any positive integer $ k$ greater then $ 2$. If $ k\leq 2$, we see that the given inequality holds.
Again induction method: If $ n \equal{} 1$,inequality holds. Suppose, $ k$ is positive integer and $ k^{3}\leq 3^{k}$ holds. Let's prove, that $ (k\plus{}1)^{3}\leq 3^{k\plus{}1}$.
So, if we manage to prove, that $ (k\plus{}1)^{3}\leq 3k^{3}$, we are done. $ RHS\minus{}LHS \equal{} (k\minus{}2)(k\plus{}1)(2k\minus{}1)\minus{}3\geq 0$, for $ k\geq 3$. If $ k\leq 3$, we see that inequality is right.
We are done.  :)


function
integration
real analysis
real analysis unsolved

a monotone function on[a,b] is called singular if f'(x)(derivative)=o a.e. show-
 1.let f be non-decrasing monotone fun.then f has following property-given c,d positive real nos. there is finite collection [yk,xk] of non overlapping intervals s.t./x1-y1/+...+/xn-yn/ < d and summation f(xk)-f(yk) >f(b)-f(a)-c.
2.conversely if f is non-decreasing function satisfying property in 1. then it's singular.

1. Let $ E \equal{} \{x: f'(x) \equal{} 0\}$ then $ m(E) \equal{} b \minus{} a$, for each $ x \in E$, since $ f'(x) \equal{} 0$, for $ \eta \equal{} \epsilon/(b \minus{} a)$, there are intervals $ [x, x \plus{} h]$ for abritraty small $ h$ such that $ f(x \plus{} h) \minus{} f(x) < \eta h$. Hence all the intervals forms a cover of $ E$ in the sense of Vitali. So that there is a finite collection of intervals $ \{[x_k, y_k]\}$ such that
\[ \sum{y_k \minus{} x_k} > m(E) \minus{} \delta
\]
and
\[ \sum{f(y_k) \minus{} f(x_k)} < \eta m(E) < \epsilon
\]
so that let the final interval be $ [a,x_1], [y_1, x_2], \ldots, [y_n, b]$, so that the total length of the intervals is
\[ x_1 \minus{} a \plus{} x_2 \minus{} y_1 \plus{} \cdots \plus{} b \minus{} y_n \equal{} b \minus{} a \minus{} \sum{y_k \minus{} x_k} < \delta
\]
and the total sum is
\[ f(x_1) \minus{} f(a) \plus{} f(x_2) \minus{} f(y_1) \plus{} \cdots \plus{} f(b) \minus{} f(y_n) \equal{} f(b) \minus{} f(a) \minus{} \sum{f(y_k) \minus{} f(x_k)} > f(b) \minus{} f(a) \minus{} \epsilon.
\]

2. let $ g \equal{} \int_a^x{f'(x)} \plus{} f(a)$, $ h \equal{} f \minus{} g$. Hence $ g$ is absolutely continuous and $ h$ is singular. Let $ \delta$ and $ \epsilon$ be in absolutely continuous condition of $ g$. By the above proposition, there exists the collections $ \{[y_k, x_k]\}$, so that
\[ \sum{h(x_k) \minus{} h(y_k)} \ge h(b) \minus{} h(a) \minus{} \epsilon,
\]
But
\[ \sum{h(x_k) \minus{} h(y_k)} \equal{} \sum{f(x_k) \minus{} f(y_k)} \minus{} \sum{g(x_k) \minus{} g(y_k)} \le f(b) \minus{} f(a) \plus{} \epsilon \equal{} h(b) \minus{} h(a) \plus{} \epsilon \minus{} (g(b) \minus{} g(a)).
\]
Hence $ g(b) \minus{} g(a) \le 2\epsilon$, so that $ g(b) \equal{} g(a) \equal{} g(x) \equal{} 0$, $ f \equal{} h$ and $ f$ is singular.


algebra
polynomial
inequalities
factoring polynomials
rearrangement inequality
AMC

ok...this mite be the wrong forum but can anyone factor these polynomials:
a^10+a^5+1
prove that if a^2+ab+b^2 + 0 then a=0 and b=0

will learning how to factor polynomials with higher exponents be of any use on the AMC? (just to make it relevant)

I do not think there are many / any worthwhile techniques that can be specifically applied to factoring polynomials with large exponents. On the other hand, most such problems (and indeed the two you provided) can be approached using familiar methods. The first can be factored as $\frac{a^{15}-1}{a^{5}-1}$, since it is a geometric series. Whether or not this is any easier to deal with depends on the problem. The second is probably not easiest to attack by factoring. Recall from the rearrangement inequality that $a^{2}+b^{2}\geq 2 a b$. But both $a^{2}$ and $b^{2}$ are greater than 0, so we must have $|a^{2}+b^{2}| = |a b| \Rightarrow 2 a b = a b \Rightarrow a = 0$ or $b = 0$, which immediately gives the desired result (note that the factoring solution might be simpler to write, but this method can be worked out about as fast as you can read the problem, and using this type of logic is a very general way to approach such problems).

And yes, I think this is the wrong forum. :)

Well, you could think of the first as $X^{2}+X+1$ where $X = x^{5}$... though that's not really factoring  :P However, you could use this technique to solve for valuable roots, which would LEAD to factoring (of course, then again, not all terms in the factorization may necessarily be integers...).


geometry
trapezoid
circumcircle
ratio
trigonometry
power of a point
radical axis

Let a trapezoid ABCD ($ AB\parallel CD$). $ O\equal{}AC\cap BD$. M is the midpoint of side CD. The circumcircles $ (AOD),(BOC)$ intersect at K. Prove that $ \widehat{KOC} \equal{}\widehat{MOD}$

$ \odot (AOD), \odot (BOC)$ cut $ CD$ again at $ E, F.$ Powers of $ C, D$ to these circles are in the ratio

$ \frac{\overline{CD} \cdot \overline{CE}}{\overline{DC} \cdot \overline{DF}} \equal{} \frac{\overline{CO} \cdot \overline{CA}}{\overline{DO} \cdot \overline{DB}} \equal{} \frac{\overline{CO}^2}{\overline{DO}^2}\ \Longrightarrow\ \frac{\overline{CE}}{\overline{DF}} \equal{} \minus{}\frac{\overline{CO}^2}{\overline{DO}^2}$

Radical axis $ OK$ cuts $ CD$ at $ L,$

$ \overline{LC} \cdot \overline{LF} \equal{} \overline{LD} \cdot \overline{LE}\ \Longrightarrow\ \frac{\overline{LC}}{\overline{LD}} \equal{} \frac{\overline{LE}}{\overline{LF}} \equal{} \frac{\overline{CL} \plus{} \overline{LE}}{\overline{DL} \plus{} \overline{LF}} \equal{} \frac{\overline{CE}}{\overline{DF}} \equal{} \minus{}\frac{\overline{CO}^2}{\overline{DO}^2}$

$ OK \equiv OL$ is O-symmedian and $ OM$ the O-median of the $ \triangle COD.$

[b]Second proof :[/b]
Let the point E such that M is the midpoint of OE, then OCED is a parallelogram.
$ \widehat{OAK}\equal{}\widehat{ODK}, \widehat{OBK}\equal{}\widehat{OCK}\Rightarrow\triangle AKC\sim\triangle DKB$
$ \Rightarrow \frac{AK}{DK}\equal{}\frac{AC}{BD}$
Because $ AB\parallel CD$, then $ \frac{OC}{OD}\equal{}\frac{OA}{OB}\equal{}\frac{OA\plus{}OC}{OB\plus{}OD}\equal{}\frac{AC}{BD}$
So, $ \frac{AK}{DK}\equal{}\frac{OC}{OD}\equal{}\frac{OC}{CE}$
$ \widehat{AKD}\equal{}\widehat{AOD}\equal{}\widehat{OCE}$ because $ OD\parallel CE$.
hence, $ \triangle OCE\sim\triangle AKD\Rightarrow \widehat{DOK}\equal{}\widehat{COE}\equal{}\widehat{COM}$
$ \Rightarrow \widehat{KOC}\equal{}\widehat{MOD}$

[quote="Virgil Nicula"] [color=darkred]Let $ ABCD$ be a trapezoid , $ AB\parallel CD$ . Denote $ O\in AC\cap BD$ .The circumcircles of the triangles

$ AOD$ , $ BOC$ intersect again at $ K$ . Denote $ S\in KO\cap CD$ . Prove that $ \frac {SC}{SD} \equal{} \left(\frac {OC}{OD}\right)^2$ .[/color] [/quote]
[color=darkblue][b][u]Proof.[/u][/b]

$ \blacktriangleright\ \left\|\begin{array}{ccc} \widehat {KAC}\equiv\widehat {KAO}\equiv\widehat {KDO}\equiv\widehat {KDB} & \implies & \widehat {KAC}\equiv\widehat {KDB} \\
 \\
\widehat {KCA}\equiv\widehat {KCO}\equiv\widehat {KBO}\equiv\widehat {KBD} & \implies & \widehat {KCA}\equiv\widehat {KBD} \end{array}\right\|$ $ \implies$

$ KAC\sim KDB$ $ \implies$ $ \frac {KA}{KD} \equal{} \frac {AC}{DB}$ $ \implies$ $ \boxed {\ \frac {KA}{KD} \equal{} \frac {OC}{OD}\ }\ (1)$ .

$ \blacktriangleright$ Denote $ S\in OK\cap CD$ . Apply the [b]Sinus' theorem[/b] in the triangle $ KAD\ \ :$

$ \frac {KA}{KD} \equal{} \frac {\sin\widehat {KDA}}{\sin\widehat {KAD}} \equal{} \frac {\sin\widehat {KOA}}{\sin\widehat {SOD}} \equal{} \frac {\sin\widehat {SOC}}{\sin\widehat {SOD}}$ , i.e. $ \boxed {\ \frac {KA}{KD} \equal{} \frac {\sin\widehat {SOC}}{\sin\widehat {SOD}}\ }\ (2)$ .

$ \blacktriangleright$ In the triangle $ COD$ apply an well-known property : $ \boxed {\ \frac {SC}{SD} \equal{} \frac {OC}{OD}\cdot\frac {\sin\widehat {SOC}}{\sin\widehat {SOD}}\ }\ (3)$ .

From the relations $ (1)$ , $ (2)$ , $ (3)$ obtain : $ \frac {SC}{SD} \equal{} \left(\frac {OC}{OD}\right)^2$ , i.e. the rays $ [OS$ is [b]symmedian[/b] in the triangle $ COD$ .

[b][u]Remark 1.[/u][/b] If denote $ R\in OK\cap AB$ , then the ray $ [OR$ is symmedian in the triangle $ AOB$ .

[b][u]Remark 2.[/u][/b] Generally, in any convex quadrilateral $ ABCD$ we have $ \frac {SC}{SD} \equal{} \frac {CO\cdot CA}{DO\cdot DB}\ !$[/color]

[quote="mr.danh"]Let a trapezoid ABCD ($ AB\parallel CD$). $ O \equal{} AC\cap BD$. M is the midpoint of side CD. The circumcircles $ (AOD),(BOC)$ intersect at K. Prove that $ \widehat{KOC} \equal{} \widehat{MOD}$[/quote]
Nice problem!
[hide="Solution"]
Extend $ AD$ and $ BC$ to meet at $ P$.  It is well known that $ OP$ passes thorugh $ M$.  Now, invert around $ O$ with arbitrary radius.  Let it map any point $ X$ to $ X'$.  Let $ K'O$ meet $ D'C'$ at $ Q$.  Now, since the circumcircles of $ \triangle AOD$ and $ \triangle BOC$ meet at $ K$, we have that $ A'D'$ and $ B'C'$ meet at $ K'$.  Also, the inversion combined with the fact that $ AB\parallel DC\implies \triangle AOB\sim \triangle COD$ gives that $ \frac{A'O}{B'O}\equal{}\frac{BO}{AO}\equal{}\frac{DO}{CO}\equal{}\frac{C'O}{D'O}$, which implies that $ A'B'\parallel D'C'$.  Now, this means that $ OK'$ passes through the midpoint of $ D'C'$, so $ Q$ is the midpoint of $ D'C'$.  The inversion gives that $ \triangle OD'C'\sim \triangle OCD$, so $ \triangle OD'Q\sim \triangle OCM$, which implies that $ \angle MOC\equal{}\angle D'OQ\equal{}\angle DOK$, so $ \angle MOD\equal{}\angle KOC$, as desired.  [/hide]

Let $L\in AD\cap BC, L, O, M$ are collinear. From $OBCK, AOKD$ cyclic quads we get $\angle CBK=\angle COK=\angle ADK \  (\ 1\ )$ and $\angle BCK=\angle KOD=\angle KAD \ (\ 2\ )$, from (1) and (2) we get $\Delta BCK\sim \Delta DAK$, hence their altitudes from $K$ are proportional to their bases $BC$ and $AD$, but $\frac{BC}{AD}=\frac{CL}{DL}$, i.e. $KL$ is symmedian in $\Delta CDL$, or $\angle CLK=\angle MLD \ (\ 3\ )$. But (1) implies $LBKD$ cyclic, or $\angle LKB=\angle LDB\ (\ 4\ )$, from (3) & (4) getting $\angle CBK=\angle MOD\ (\ 5\ )$, as sum of the 2 equal angles; (5) & (1) give the relation to be proven.

Best regards,
sunken rock