geometry
trapezoid
rectangle
Pythagorean Theorem

[quote="Cronus"]There is a trapezoid ABCD where AB||CD and AB=51,CD=91,AC=13,CD=37.
Find thae area of the trapezoid.[/quote]

So CD= 91 or CD= 37???  :huh:

[quote="ernie"]I don't think this is possible, because $ AC$ would have to be a diagonal and therefore larger than $ AB$.

The opposite vertices are:

$ A$ and $ C$

$ B$ and $ D$[/quote]

Not necessarily.  You can make a trapezoid with opposite angles obtuse, and then if you try hard enough, you can get a trapezoid for which AB is clearly longer than AC

but the 91=37 thing...

try,  instead of AB=51,CD=91,AC=13,CD=37, its AB=51,CD=91,BC=13,AD=37.
i solved it using guess and check but i might put a rigorous solution later

btw the height is 12 (you could figure that out using guess and check with pythagorean triples)
let the altitude from A to DC meet DC at E. similarly let the altitude from B to CD meet CD to F

let AE=w, ED=x, BF=y, FC=z

we have, by the pythagorean theorem $ x^2\plus{}w^2\equal{}37^2\equal{}1369, \text{and} y^2\plus{}z^2\equal{}13^2\equal{}169$

since AB and DC are parralel and angles AEC and BFD are right, ABFE is a rectangle so FE=51 and $ x\plus{}z\equal{}40, \text{and} w\equal{}y$

so we substitute y for w and 40-z for x

$ (40\minus{}z)^2\plus{}y^2\equal{}1369, \text{and} y^2\plus{}z^2\equal{}169$
subtracting the second equation from the first gives $ (40\minus{}z)^2\minus{}z^2\equal{}1200$ applying diff. of squares

$ 40(40\minus{}2z)\equal{}1200 \implies 40\minus{}2z\equal{}30 \implies z\equal{}5$

so z=5, x=35, and w=y=12 (which is the height) and the averages of the bases is 71 so the area is 71*12=852

Sorry guys.CD=91.The length of the not parallel sides are 13 and 37.(Stevenmeow guessed it right)I found it by guessing too.(at first)
[hide="Rigorous Solution"]Let the height be h.Draw 2 perpendicular lines AE & BF from A & B to CD.
Now in triangle BDF 169=h^2+x^2.
       in triangle ACE 37^2=h^2+(40-x)^2[from pythagoras And the fact that
                                                        AEFB is a rectangle]
From this we can find h=12 by substituting 169 for h^2+x^2 in the seconfd equation.So the area is 852. :D [/hide]

Do you mean BC=37 not another CD? I think that this trapezoid must be stretched out awkardly to make these measurements right. weird question................. :?: this is marked on my forehead


search
calculus
real analysis
real analysis unsolved

[quote="Lomisedu"]How to use the method of "increasing rational sequence" to construct the real number system?

I search many article and web page, but most of the article is using Dedekind's method or Cantor's method to construct the real number system.

thanks~[/quote]

We usually don't use that approach and rely on our old trusty friend Dedekind of Dedekind cut fame.

However, the irratioanl numbers can be seperated into two categroies; algebraic and transcendental.

http://en.wikipedia.org/wiki/Algebraic_number

" Even though the calculus can be developed over the computable numbers, the set of computable numbers is not closed under basic operations such as taking the supremum of a bounded sequence, so this set cannot be used as a replacement for the full set of real numbers in classical mathematics"

So you will have failed straight away [and that is nad news as can you imagine handling some transcedental numbers that are known to be non computable - that would be a bitch).

You will be interseted to read more about;

http://en.wikipedia.org/wiki/Dedekind_cut

What have algebraic, transcendental and computable numbers got to do with the question of constructing the real numbers?

Here's the reason using increasing subsequences of rationals is no fun: we would like just to define a real number to be a bounded, increasing subsequence of rationals.  But clearly this is not right, since every sequence converging to the same value should give us the same value.  So to do this you need to find an equivalence relation that means the same thing as "these two sequences converge to the same real value" that doesn't explicitly mention the value, and then define the real numbers to be the set of equivalence classes of this relation.  This is less pleasant to write down than, say, a Dedekind cut, though I suppose it makes defining arithmetic on $ \bf R$ a bit easier.  (You may have some freedom in choosing the definition, but the obvious thing to do is to try some souped-up version of the definition of a Cauchy sequence.)

Here's a try at the requisite equivalence relation: two strictly increasing sequences $ (x_n)$ and $ (y_n)$ are equivalent iff they intertwine. That is, for every $ n$ there exists $ k$ such that $ y_k>x_n$ and for ever $ n$ there exists $ k$ such that $ x_k>y_n.$

I think it can be made to work. I can see how to define order: if two sequences don't intertwine, then we should be able to decide which one is "larger."

We do also have $ \plus{}\infty$ appearing as an equivalence class; I think we want to define that away by restricting to bounded sequences.


number theory
greatest common divisor
logarithms
geometry
number theory proposed

For each $m\in \mathbb N$ we define $rad\ (m)=\prod p_i$, where $m=\prod p_i^{\alpha_i}$.

[b]abc Conjecture[/b]
Suppose $\epsilon >0$ is an arbitrary number, then there exist $K$ depinding on $\epsilon$ that for each 3 numbers $a,b,c\in\mathbb Z$ that $gcd (a,b)=1$ and $a+b=c$ then: \[ max\{|a|,|b|,|c|\}\leq K(rad\ (abc))^{1+\epsilon} \]

Now prove each of the following statements by using the $abc$ conjecture :
a) Fermat's last theorem for $n>N$ where $N$ is some natural number.
b) We call $n=\prod p_i^{\alpha_i}$ strong if and only $\alpha_i\geq 2$.
c) Prove that there are finitely many $n$ such that $n,\ n+1,\ n+2$ are strong.
d) Prove that there are finitely many rational numbers $\frac pq$ such that: \[ \Big| \sqrt[3]{2}-\frac pq \Big|<\frac{2^ {1384}}{q^3}  \]

ABC proves only an asymptotic version of Fermat conjecture, not the full Fermat's Last Theorem.

More:
e) prove an asymptotic version of Catalan's conjecture.
f) prove that there are infinitely many Wieferich primes.

I think that Omid took a mistake. part $a)$ must be : Prove Last fermat theorem for $n$ bigger than something

part a) let $K=f(\epsilon )$.suppose $x^n+y^n=z^n$ for large enough $n$. by abc Conjecture we have $f(\epsilon )(rad(xyz))^{1+\epsilon}\geq max\left \{ x^n,y^n,z^n \right \}\geq (xyz)^{n/3}$. We define $\pi (m)$ be the number of prime divisor of $m$. easy to see there exist at least one $q$ such that $q$ is prime and divides one of the $x,y,z$ and $\sqrt[\pi (x)+\pi (y)+\pi (z)]{k}\geq p_{i}^{n/3-1-\epsilon }$ . so $k\geq q^{(n/3-1-\epsilon).(\pi (x)+\pi (y)+\pi (z)) }\geq q^{(n/3-1-\epsilon )}\geq 2^{(n/3-1-\epsilon )}$but we let $n$ be large enough so we have $\log _{2}f(\epsilon )\geq \infty $ contraction.
the part d is not hard by Conjecture and Geometry of Numbers but my solution is very long and I had no time for write it.
I cant solve the part c can anybody help me about this part?

Does this work?
part c) First AFTSOC so there exist infinitely many $n$ s.t $n,n+1,n+2$ are strong and take $n$ sufficiently large.
Note that $(n^2)+(4(n+1))=(n+2)^2$
So by abc conjecture we have:
$$(n+2)^2\leq f(\epsilon)(rad((n^2 )(4n+4)((n+2)^2)))^{1+\epsilon}=f(\epsilon)(rad((n)(4n+4)(n+2)))^{1+\epsilon}$$ (*)
(note that by the condition non of the numbers $n,n+1,n+2$ are 2(mod 4) so $n+1$ must be a multiple of 4 thus $n,n+1,n+2$ are pairwise coprime)
$n+1$ is even so $rad(4n+4)=rad(n+1)$
We know that :
$rad(n)\leq \sqrt{n},rad(n+1)\leq \sqrt{n+1},rad(n+2)\leq \sqrt{n+2}$ (since $n,n+1,n+2$ are strong then)
so:
$$rad((n)(4n+4)(n+2))\leq ((n)(n+1)(n+2))^{\frac{1}{2}}$$ (**)
Now by (*) , (**) we have the following inequality which has to be tru for all positive $\epsilon$:
$$(n+2)^4\leq f(\epsilon)^2((n)(n+1)(n+2))^{1+\epsilon}$$
But $((n)(n+1)(n+2))^{1+\epsilon}\leq (n+2)^{3+3\epsilon}$ so finally we have:
$$n+2 \leq f(\epsilon)^2(n+2)^{3\epsilon}$$
but the last inequality is obviously wrong for $\epsilon=\frac{1}{4}$ and sufficiently large $n$ so we're done.


I'm not too sure about the mathematical environment in Ontario but are there any upcoming events that concerns with math in Toronto or Ontario?

Well, the National Math Camp began today in London, and Waterloo Seminar isin a weeks. But those are by invitation only.

Other than that, I don't think so, since it is near the end of the school year.

[quote="kool_dudy"]are you attendin math camp now?[/quote]

No, I went to National Camp last year. You can only go once, since they apparantly don't change the materials.

I will be at Waterloo Seminar though, and IMO training after that.

Oh...never knew that since I never attended...heh

Well, Good Luck


I think that $\frac{1}{\sqrt[3]{1+26x}}+\frac{1}{\sqrt[3]{1+26y}}+\frac{1}{\sqrt[3]{1+26z}}\geq1$ would look more beautifully. Moreover, 26 and 81 can hint the intended solution.


Ring Theory
inequalities
inequalities proposed

Let $ a,b,c,d$ be nonnegative numbers such that $ a\plus{}b\plus{}c\plus{}d\equal{}4$. Prove that

(a) $ a^3\plus{}b^3\plus{}c^3\plus{}d^3\plus{}32 \ge 6(a^2\plus{}b^2\plus{}c^2\plus{}d^2)$;

(b) $ 3(a^3\plus{}b^3\plus{}c^3\plus{}d^3)\plus{}32 \ge 10(a^2\plus{}b^2\plus{}c^2\plus{}d^2)$.

Nice, arqady. I think the first is equivalent to $ \sum abc\equal{}0$, too.
Nice, ehku. AC-Theorem is very useful for this inequality.

[quote="ehku"]Using the AC-theorem for the second.
We can also use the PID theorem.[/quote]

Can someone post the general form of PID theorem or give me a link?

thanks  :)

[quote="arqady"][quote="Vasc"]Let $ a,b,c,d$ be nonnegative numbers such that $ a \plus{} b \plus{} c \plus{} d \equal{} 4$. Prove that

(a) $ a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3 \plus{} 32 \ge 6(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)$;
[/quote]
It has turned out easily. Maybe I am wrong. But I don't see where.
Let $ f(a,b,c,d) \equal{} a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3 \plus{} 32 \minus{} 6(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2).$
Hence, $ f(a,b,c,d) \minus{} f(a,b \plus{} c \plus{} d,0,0) \equal{} 3a(bc \plus{} bd \plus{} cd) \plus{} 3bcd\geq0.$
But $ f(a,b,0,0) \equal{} 0,$ where $ a \plus{} b \equal{} 4$ and done.[/quote]

nice 2 :lol: .

[quote="arqady"][quote="Vasc"]Let $ a,b,c,d$ be nonnegative numbers such that $ a \plus{} b \plus{} c \plus{} d \equal{} 4$. Prove that

(a) $ a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3 \plus{} 32 \ge 6(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)$;
[/quote]
It has turned out easily. Maybe I am wrong. But I don't see where.
Let $ f(a,b,c,d) \equal{} a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3 \plus{} 32 \minus{} 6(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2).$
Hence, $ f(a,b,c,d) \minus{} f(a,b \plus{} c \plus{} d,0,0) \equal{} 3a(bc \plus{} bd \plus{} cd) \plus{} 3bcd\geq0.$
But $ f(a,b,0,0) \equal{} 0,$ where $ a \plus{} b \equal{} 4$ and done.[/quote]

arqady: if you have some similar inequalities(wich use the same method), please send me the links.

Thanks  :)

For example see here the fuzzylogic's problem:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=79554
[hide="The hint"] Let $ f(a,b,c) \equal{} a \plus{} b \plus{} c \plus{} 4abc \minus{} 2$ and $ f(a,b,c) \minus{} f\left(a \plus{} c \plus{} 2ac \plus{} \frac {ac}{b},b,0\right) \equal{} ...$  :wink:
But there are  ways.
[/hide]

$ f(a+c+2ac+\frac{ac}{b},b,0)\geq 0$ is true

but $ f(a,b,c,d)\geq f(a+c+2ac+\frac{ac}{b},b,0)$ is not always true:

It needs

$ ac(4ab-\frac{1}{b} - 2) \geq 0$
Or
$ 4ab\geq \frac{1}{b}+2$
Or
$ 4ab^{2}-2ab-1\geq 0$
Equivalent to
${ b\geq \frac{1+\sqrt{4a+1}}{4a}}$

which implies necessary $ 4ab-1 \geq 0$ but this is not always true.

Could anyone explain me what AC-theorem and PIC theorem are? Thanks!

[quote="FOURRIER"]$ f(a \plus{} c \plus{} 2ac \plus{} \frac {ac}{b},b,0)\geq 0$ is true

but $ f(a,b,c,d)\geq f(a \plus{} c \plus{} 2ac \plus{} \frac {ac}{b},b,0)$ is not always true:

It needs

$ ac(4ab \minus{} \frac {1}{b} \minus{} 2) \geq 0$
[/quote]
No! It gives $ \frac{ac(4b^2\minus{}2b\minus{}1)}{b}$ and two cases: $ b\geq\frac{1\plus{}\sqrt5}{4}$ and $ 0<b<\frac{1\plus{}\sqrt5}{4}.$ :wink:

[quote="arqady"]
No! It gives $ \frac {ac(4b^2 \minus{} 2b \minus{} 1)}{b}$ and two cases: $ b\geq\frac {1 \plus{} \sqrt5}{4}$ and $ 0 < b < \frac {1 \plus{} \sqrt5}{4}.$ :wink:[/quote]

Sorry I had a small mistake in my computation

Ok, so for case $ 0 < b < \frac {1 \plus{} \sqrt5}{4}.$ it's not true, we should prove the inequality for this case using another method.

[quote="FOURRIER"]
Ok, so for case $ 0 < b < \frac {1 \plus{} \sqrt5}{4}.$ it's not true, we should prove the inequality for this case using another method.[/quote]
Sure, but we should prove the inequality for this case using the same mathematic.  :)


I know, this is really strange to grasp. But try to imagine this:

Problem:

Two trains are driving on the same rail. Assume a straight rail with no curves.

Train T1 starts from the left at position P1 = 0. Train T2 starts from the right at position P2 = 100.

On top of T1 is another Rail which extends out infinitely. Please imagine. Not realistic. T2 is driving on top of that Rail.

For every step T1 moves forwards towards P2, T2 moves 2.5 steps towards P1 on that top-rail. As T1 moves forward, it pushes T2 backwards by that same amount of movement.

How can it be calculated at which point on the bottom Rail (the Rail T1 is driving on) the trains would theoretically crash into eachother (assuming an collision when T2 reaches T1 even if on different rails).

I asked several math genius and none of them were able to help me here. I hope one of you knows how to solve this. I would also like to understand a little bit the formula to learn math better. Well I think I need linear algebra but I'm not sure how to set this up. I was never really good at maths :(

Welcome to AoPS! :) I see you are new.

[hide="Solution"]
For every foot $ T_1$ moves, $ T_2$ moves $ 2.5\minus{}1\equal{}1.5$ feet. So every step, they cover a total of $ 1\plus{}1.5\equal{}2.5$ feet. That means that after $ 40$ steps, they will "collide" since $ 40\times2.5\equal{}100.$ After $ 40$ steps, $ T_1$ has traveled $ 40\times1\equal{}\boxed{40}$ feet along $ R_1.$[/hide]

Hope that was what you were looking for. :)

Thanks, that looks promising. Can you explain this calculation in more detail? That would be great. I try to wrap my head around it, but it's really hard.

Think about it like this.

$ \dots \dots \dots\dots\dots T_2$
$ T_1\dots \dots \dots\dots\dots$
(note that the dots are the rails)

Alright. So whenever $ T_1$ moves forward 1 step (or for practical purposes 1 foot) forward, $ T_2$ moves back one foot, right? But we also know that for every foot $ T_1$ moves forward, $ T_2$ moves 2.5 feet forward. So for every foot $ T_1$ moves forward, $ T_2$ moves 1 foot backwards and 2.5 feet forward, or just 1.5 feet forward. So whenever $ T_1$ takes one foot forward, the total distance covered by BOTH of them is 1+1.5=2.5, agreed? Now how does that help? Well, we know that the distance between the two trains is 100 feet, right? So that means if they cover 2.5 feet every movement, then in $ \frac {100}{2.5} \equal{} 40$ movements they will collide, right? So what does that tell us? That means since $ T_1$ moves forty times, it will be $ \boxed{40}$ feet along $ R_1$, and since $ T_2$ moves 40 times as well, it will be 40x1.5=$ \boxed{60}$ feet along $ R_2$.

In 40 movements, train 1 will have moved 40 feet, and train 2 will have moved 40x1.5=60 feet. That is a nice little way to double check our work, since 40+60=100.


can someone give me a nice explanation for this problem?

A fair coin is tossed repeatedly. Find the probability of obtaining five consecutive heads before two consecutive tails.

I recommend this article [url]http://mathworld.wolfram.com/CoinTossing.html[/url] and to find others like it. Coin flipping was never so complicated.

[hide]Let $P$ be the desired probability, $P_1$ be probability given first flip is heads, $P_2$ be probability given first flip is tails.

$P=\frac{P_1+P_2}{2}$ by adding cases where first flip is heads or tails.

$P_2=\frac 12 P_1$ because either the second flip is tails (failure) or heads (begins a new run with heads). So $P=\frac 34 P_1$.

Now. if the first flip is heads, then either:

- Four heads follow (success) with probability $\frac{1}{16}$.

- Up to three heads follow, but then tails. In order to have a chance at success, the flip after tails must be heads. If it is, then a new run begins with heads, which is neutral. (That means success is independent of these flips.)

The combined probability of TH, HTH, HHTH, HHHTH is $\frac 14+\frac 18 +\frac{1}{16}+\frac{1}{32}=\frac{15}{32}$.

Then $P_1=\frac{1}{16}+\frac{15}{32}P_1\Rightarrow \frac{17}{32}P_1=\frac{1}{16}\Rightarrow P_1=\frac{2}{17}$.

Multiplying by $\frac 34$, $P=\boxed{\frac{3}{34}}$.[/hide]

[hide="another solution"]first, start with heads, then including that, HT,HHT,HHHT,HHHHT can occur to delay the pattern, and consider the expansion (HT+HHT+HHHT+HHHHT)^2, this covers all possible ways of having 2 delaying events, we can have from 0 to infinity delaying events, so
also the sum of those probabilities is $\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}=\frac{15}{32}$, also we multiply this by $\frac{1}{32}$ because we have to take into account the probability of teh 5 heads at the end
$\frac{1}{32}*\sum_{k=0}^{\infty} \left(\frac{15}{32}\right)^k=\frac{1}{17}$

then if it is tails first, then it is just $\frac{1}{2}*\frac{1}{17}$ because after the first tail, the sequence is forced back to a head again, so our answer is $\frac{1}{17}+\frac{1}{34}=\frac{3}{34}$[/hide]

[hide="to scorpius"]does this problem seem familiar...lol[/hide]

Oh the horror.

[hide]

Obviously, the string of H's and T's could be of infinite length, but is there a way to condense this length and represent it as a probability of key components?

Yes...there is.

Let a successful string be one that encounters a run of 5 heads before it encounters 2 tails.

Let $p$ be the probability that a successful string is created if our sequence starts with a heads; H.

Let $t$ be the probability that a successful string is created if our sequence starts with a tails; T.  In this case, the T MUST be followed by an H.  Therefore, since there is a $\frac{1}{2}$ probability of selecting T first, then this case is essentially T followed by a successful string that starts with H.  Hence, $t = \frac{1}{2}*p$.

So adding up these two probabilities gives our answer.

Note that our H strings can be interrupted in the following ways:

HT...

HHT...

HHHT...

HHHHT...

In each of these, a successful string is assumed to have followed after the T.  If a successful string does not follow, then we would essentially keep repeating portions of the above strings, but the probabilities of those will have been accounted for.

HT... has an initial probability of occurence $\frac{1}{2}*t$ since HT is followed by a successful string in T, and H has a 1/2 probability of occuring.

Likewise, HHT has a probability of $\frac{1}{4}*t$.

Therefore,

$p = t * (\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16}) + \frac{1}{32}$

since we must also consider the string HHHHH of occuring initially (probability 1/32).

Since $t = \frac{p}{2}$, it follows from substituting in and solving the equation above that:

$\frac{17}{32}*p = \frac{1}{32}$

$\Rightarrow p = \frac{1}{17}$

Therefore, our desired probability is:

1/17 + 1/34 = $\frac{3}{34}$[/hide]


limit
real analysis
real analysis unsolved

Let $ a_1,a_2,...,a_k$ be numbers so that $ \sum_{j\equal{}1}^k a_j \equal{}0$ Find if the limit:
$ \lim_{n\to\infty} \sum_{j\equal{}1}^k a_j\sqrt{(n\plus{}j)}$ exists and if it does, find the value of it.

I have no idea. Any tips/solutions are welcome!

Write $ \sqrt{n\plus{}j}\equal{}\sqrt{n}\plus{}f_n(j)$. $ \sum_{j\equal{}1}^k a_j\sqrt{n}\equal{}0$, so we're looking at $ \lim_{n\to\infty}\sum_{j\equal{}1}^k a_j f_n(j)$. Now, how big can the values of $ f_n$ get? Show that they go to zero, and estimate the worst-case size of that sum.

I'm not sure I followed completly, but thanks! :)


Bob can run 1 mile in the same time that Gill can run $ \frac56$ of a mile. Cindy can run $ \frac34$ of a mile in the same time that Gill can run 1 mile. If each person runs at those rates, how many miles will Cindy run in the time it takes Bob to complete an 8-mile run?

Let's say Bob runs at $ 1$ mile per hour (dang he's even slower than I am).  Then, Gill runs at $ \frac{5}{6}$ miles per hour and Cindy runs at $ \frac{5}{6}\cdot \frac{3}{4} \equal{}\frac{5}{8}$ miles per hour.

It takes Bob $ 8$ hours to run $ 8$ miles, so in that time Cindy runs $ \frac{5}{8}\cdot 8\equal{}\boxed{5}$ miles.


Fie $ABCD$ un patrulater inscriptibil. Pe semidreptele $(AB$ şi $(AD$ se consideră punctele $P$, respectiv $Q$, astfel încât $AP=CD$ şi $AQ=BC$. Să se arate că $PM=MQ$, unde $M=(PQ)\cap (AC)$.

Intrebare de bun simt: Cine e M ???  :D  :D

Edit: Multumesc de modificare...

Este eleganta solutia ta. Mi-a placut; sincer nu m-am gandit la metoda inversiunii.
Eu am gasit o solutie fara inversiune.

Fie $T$ punctul de intersectie al paralelei la $AD$ prin $P$ cu $AC$; $\alpha = \angle BAC$, $\beta = \angle CAD$.

Iese ca $AQ = BC = 2R \sin \alpha$ si $AP = CD = 2R \sin \beta$. Avem $\angle ATP = \beta$, deci, din teorema sinusurilor, $\frac{AP}{PT} = \frac{\sin \beta}{\sin \alpha}$, adica $PT = 2R \sin \alpha = AQ$.

In concluzie, $APTQ$ este paralelogram si $PM = MQ$.

E buna solutia?

Probabil trigonometria putea fi evitata, dar in cazul de fata nu duce la calcule grele.

Excelenta solutie! Folosirea trigonometriei nu este un defect in cazul asta.
Problema a fost data la $ONM 1984$, iar solutia pe care am gasit-o eu este identica cu cea a autorului (nu foloseste trigonometrie).
Enuntul original mai dadea si $N$ mijlocul lui $(BD)$ si se cerea sa se arate ca $PM=QM=CN$.
Oricum, felicitari pentru solutia frumoasa pe care ai dat-o!

Multumesc pentru aprecieri... Mi-a luat destul de mult s-o rezolv si m-am mirat ca iese asa de simplu.

Ultima parte iese din $\triangle DBC \equiv \triangle ATP$. Acum sa vad care e solutia ta.

Fie $P'$ simetricul lui $P$ fata de $A$ si $N'$ mijlocul segmentului $P'Q$.
Evident triunghiurile $CBD$ si $AQP'$ sunt congruente (prin ipoteza de inscriptibilitate a lui $ABCD$). Atunci si medianele lor $[CN]$ si $[AN']$ sunt congruente. Cum $[AN']$ este linie mijlocie in triunghiul $P'PQ$ si $AMQN'$ este paralelogram ($\angle BAC \equiv \angle CDB \equiv \angle AP'Q \Rightarrow AM\|N'Q$), rezulta ca $PM=MQ=AN'=CN$ si solutia se incheie.
[img]http://i19.photobucket.com/albums/b200/mariusdamian/geometrieONM.gif[/img]


Find the number of terminating zeros in 36!-24!

[hide]It's just the number of zeroes at the end of 24!, which is 4.[/hide]


[quote="nodet"]ba arghame 3va5 chandadad mitvan tolid kard betori ke majmue arghame har adad told shode 500 bashad?[/quote]

ba rahe hal lotfan :oops:

$3x+5y = 5 * 100 \Rightarrow  x=5k \Rightarrow  3k+y = 100$
$k = 33,y = 1 \Rightarrow  x=165, tedade-adad : \binom{166}{1}$
$k = 32,y = 4 \Rightarrow  x=160, tedade-adad : \binom{164}{4}$
..
$k = 0,y = 100 \Rightarrow  x=0, tedade-adad : \binom{100}{100}$

pas tedade kolle adade tolid shode barbare $\sum_{i=0}^{33}{\binom{2(83-i)}{3i+1}}$


number theory
greatest common divisor
Euler
function
totient function

Be $m=2^4.3^3.5^2$ find how many positives integers numbers $\leq$$m$  are primes with $m$.

[hide] What does remember $\varphi$?[/hide]

Let $f(n)$ the Euler's function. From the theorem of the $f(n)$ we have that

$f(2^4\cdot 3^3\cdot 5^2)=2^4\cdot 3^3\cdot 5^2(1-\dfrac{1}{2})(1-\dfrac{1}{3})(1-\dfrac{1}{5})=2880$ I think

[quote="felipesa"]$Silouan$
What exactly Eulers fuction did you use?
thanks for the help!!!
Felipe[/quote]
He was using Euler Totient Function and its property that:
$\varphi (n)=n\prod_{p|n}(1-\dfrac{1}{p})$


Find the largest value of $ x$ for which $ (x \plus{} 1)$ and $ (x \minus{} 1)$ are reciprocals.

$ \frac{x\plus{}1}{1}\equal{}\frac{1}{x\minus{}1} \implies x^2\minus{}1\equal{}1 \implies x^2\equal{}2 \implies x\equal{}\pm \sqrt{2}$ because we want the largest the answer is $ \sqrt{2}$.


Solve
\[ \left\{ \begin{array}{l}
 \log_2 x\plus{}\log_4 y\plus{}\log_4 z\equal{}2 \\
 \log_3 y\plus{}\log_9 z\plus{}\log_9 x\equal{}2 \\
 \log_4 z\plus{}\log_{16} x\plus{}\log_{16} y\equal{}2 \\
 \end{array} \right.\]

In the beginning the condition for which the logarithm can be defined is $ x > 0,\ y > 0,\ z > 0$. On the basis this condition, for $ a > 0,\ a\neq 1$,

Use $ \boxed{\log_{a^2} A \equal{} \frac {\log_a A}{\log_a a^2} \equal{} \frac {1}{2}\log_a A}$ to rewrite the given sytem of equation, we have

$ \log_2 x^2yz \equal{} 4,\ \log_3 y^2zx \equal{} 4,\ \log_4 z^2xy \equal{} 4$

$ \Longleftrightarrow x^2yz \equal{} 4^2\ \cdots [1],\ y^2zx \equal{} 9^2\ \cdots [2],\ z^2xy \equal{} 16^2\ \cdots [3]$

Multiplying both sides of these equations gives $ (xyz)^4 \equal{} (4\cdot 9\cdot 16)^2 \equal{} (2\cdot 3\cdot 4)^4$, since $ x > 0,\ y > 0,\ z > 0$, yielding $ xyz \equal{} 2\cdot 3\cdot 4\ \cdots [4]$.

Thus from $ [1],\ [4]]$, we have $ (2\cdot 3\cdot 4)x \equal{} 4^2\Longleftrightarrow x \equal{} \frac {2}{3}$, Analogously from $ [2],\ [4]$ and $ [3],\ [4]$,

we obtain the answer $ \boxed{x \equal{} \frac {2}{3},\ y \equal{} \frac {27}{8},\ z \equal{} \frac {32}{3}}$

[b]Little extra theory:[/b]
[b]Definition:[/b] $\boxed{\log_a{x}=y\Leftrightarrow x=a^y}$
[b]consequence 1:[/b] $ \boxed{\log_a{a^y}=y}$
[b]consequence 2:[/b] $\boxed{ a^{\log_a{x}}=x}$
[b]Property 1:[/b] $\boxed{\log_{a^k}{x}=\frac{1}{k}.\log_a{x}}$
[b]Proof:[/b]
$\log_{a^k}{x}=\log_{a^k}{a^{log_a{x}}}=\log_{a^k}{\left(a^k\right)^{\frac{1}{k}.\log_a{x}}}=\frac{1}{k}.\log_a{x}.$
[b]Property 2:[/b] $\boxed{\log_a{x^k}=k.\log_a{x}}$
[b]Proof:[/b]
$\log_a{x^k}=\log_a{\left({a^{log_a{x}}}\right)^k}=\log_a{{a^{k.log_a{x}}}}=k.\log_a{x}.$
[b]Consequence 3: [/b]$\boxed{\log_{a^q}{x^p}=\frac{p}{q}.log_a{x}}$

basically I 'm new to this site so I am not sure about the codes that denote symbols etc.
log_2X+ log_4Y+ log_4Z = 2 ; log_3Y +log_9Z+ log_9X =2 ; log_4Z+log_16x+log_16y = 2

log_2X+log_4Y+log_4Z =2
 (log_4x) log_2 4 +log_4Y+log_4z =2
   log_4 X^2YZ = 2 ; X^2YZ = 16 {as log_2 4 = 2}
similarly proceed for the next two equations and then u  would land up in this :
    Y^2XZ =81 & Z^2XY=256 .
 solving X^2YZ =16 & Y^2ZX= 81 & Z^2XY= 256
we get X= 2/3 & y= 27/8 & Z = 32/3

Using the logarithmic sum identity, we can rewrite the equations as so:
$$\log_{4}x^2yz=2$$
$$\log_{9}xy^2z=2$$
$$\log_{16}xyz^2=2$$

In exponential form, we have:
$$x^2yz=16$$
$$xy^2z=81$$
$$xyz^2=256$$

Multiplying the equations and taking the 5th root yields $xyz=24$, and accordingly (keeping $x,y,z$ positive) we have $\boxed{x=\frac{2}{3},y=\frac{27}{8},z=\frac{32}{3}}$.


$ (a_n)_{n \equal{} 0}^\infty$ is a sequence on integers. For every $ n \ge 0$, $ a_{n \plus{} 1} \equal{} a_n^3 \plus{} a_n^2$. The number of distinct residues of $ a_i$ in $ \pmod {11}$ can be at most?

$\textbf{(A)}\ 2  \qquad\textbf{(B)}\ 3  \qquad\textbf{(C)}\ 4  \qquad\textbf{(D)}\ 5  \qquad\textbf{(E)}\ 6$

[hide="answer"]
Answer is [b]B[/b]
Edit: Had the solution right but I wrote $ C$ instead of $ B$

First write all the congruences in mod 11:

$ 0,1,2,3,4,5,6,7,8,9,10\pmod{11}$

Now their squares, in mod 11:

$ 0,1,4,9,5,3,3,5,9,4,1\pmod{11}$

Now their cubes in mod 11:

$ 0,1,8,5,9,4,7,2,6,3,10\pmod{11}$

And the sum of their cubes and squares in mod 11:

$ 0,2,1,3,3,7,10,7,4,7,0$

Now that we have the congruences, we can work the cases:

If $ a_0\equiv 0\pmod{11}$ then $ a_{n \plus{} 1}\equiv 0\pmod{11}$, this only has one distinct residue.

If $ a_0\equiv 1\pmod{11}$ then $ a_{n \plus{} 1}\equiv 2\pmod{11}$, this only has two distinct residues (2 and 1).

If $ a_0\equiv 2\pmod{11}$ then $ a_{n \plus{} 1}\equiv 1\pmod{11}$, this only has two distinct residues (2 and 1).

If $ a_0\equiv 3\pmod{11}$ then $ a_{n \plus{} 1}\equiv 3\pmod{11}$, this only has one distinct residue.

If $ a_0\equiv 4\pmod{11}$ then $ a_{n \plus{} 1}\equiv 3\pmod{11}$, this only has two distinct residues (4 and 3).

If $ a_0\equiv 5\pmod{11}$ then $ a_{n \plus{} 1}\equiv 7\pmod{11}$, this only has two distinct residues (5 and 7).

If $ a_0\equiv 6\pmod{11}$ then $ a_{n \plus{} 1}\equiv 10\pmod{11}$, this has 3 distinct residues (6, 10 and 0).

If $ a_0\equiv 7\pmod{11}$ then $ a_{n \plus{} 1}\equiv 7\pmod{11}$, this only has one distinct residue.

If $ a_0\equiv 8\pmod{11}$ then $ a_{n \plus{} 1}\equiv 4\pmod{11}$, this has 3 distinct residues (8,4 and 3).

If $ a_0\equiv 9\pmod{11}$ then $ a_{n \plus{} 1}\equiv 7\pmod{11}$, this only has two distinct residues (7 and 9).

If $ a_0\equiv 10\pmod{11}$ then $ a_{n \plus{} 1}\equiv 0\pmod{11}$, this only has two distinct residues (10 and 0).

The greatest number of distinct residues is: $ \boxed{3}$
[/hide]

tony you got a couple of errors. Check the residues of the cubes.  I think the answer is $ \boxed{D}$.

Are you sure?, I just made a program to tell me the residues and all the cubes, and squares are correct.


Let $ X$ be a real number. Prove that among the numbers $ X,2X,\dots,(n\minus{}1)X$, there is one which differs from an integer by at most $ \frac1{n}$.

[hide]

AFTSOC: blah

Consider the fractional part of each of the multiples of X, if none are within 1/n of an integer than the fractional parts must each lie in one of the intervals  (2/n,3/n), (3/n,4/n),...,([n-2]/n,[n-1]/n).

Okay so thats n-2 intervals and n-1 multiples of X hmmm...

Blah pidgeonhole, JX and KX have fractional parts in the same interval. Wlog J<K so (K-J)X was one of these multiples and must have a fractional part less than 1/n.

[/hide]


$x\geq y\geq z>0$

Prove that

                        $\frac{x^2y}{z}+\frac{y^2z}{x}+\frac{z^2x}{y} \geq x^2+y^2+z^2$

:!:

Λάθος σόρρυ :blush:

[quote="silouan"]
$\frac{x^3y^2+y^3z^2+z^3x^2}{xyz} \geq \frac{(x^3+y^3+z^3)(x^2+y^2+z^2)}{3(xyz)}$
[/quote]

Για να χρησιμοποιήσεις Chebychev πρέπει οι δυο ακολουθίες $(x^3,y^3,z^3)$ και $(y^2,z^2,x^2)$ να είναι όμοια διατεταγμένες.Κάτι τέτοιο όμως στην περίπτωση αυτή δεν ισχύει.

Καμιά άλλη ιδέα;

Υ.Σ. Έχω βρει δυο τρόπους λύσεις (ο ένας συζητήσιμος(AM-GM)) ,αλλά επειδή έχουν αρκετή δουλειά, είπα μήπως υπάρχει καμιά πιο προφανής λύση!

Στέλιος... :!:

Στέλιο έχεις δίκιο έκανα χαζομάρα παραπάνω με τον Chebychev . Να μια άλλη λύση

Παρατηρούμε ότι
$\sum_{cycl}\frac{ab(a-b)}{c}=\frac{(a-b)(a-c)(b-c)(ab+bc+ca)}{abc} \ge 0$
άρα
$\sum_{cycl}\frac{a^2b}{c} \ge \sum_{cycl}\frac{ab^2}{c}$
και έτσι
$2\sum_{cycl}\frac{a^2b}{c}\ge \sum_{cycl}\frac{ab(a+b)}{c}$
Άρα μένει να δείξουμε ότι
$\sum_{cycl}\frac{ab(a+b)}{c} \ge 2(a^2+b^2+c^2)$
ή ισοδύναμα
$\sum_{cycl}a^2b^2(a+b) \ge 2abc(a^2+b^2+c^2)$
αλλά
$\sum_{cycl}a^2b^2(a+b) - 2abc(a^2+b^2+c^2)=\sum_{cycl}c^3(a-b)^2 \ge 0$

MMMMMmmm  μου  φαίνεται  οτι  η  άσκηση  βγαίνει  εύκολα  αν  θεωρήσεις  οτι  υπάρχουν $a> 0$ $b,c \geq 0$ωστε

       $z =a$, $y= a +b$ και  $x=a + b+ c$

 Έτσι  μετά  από πράξεις  καταλήγεις   οτι  $(a +b +c)^3(ab + b^2) + a^3c(a+b+c) \geq (a+b)^3(ab +ac)$
  οπου  προφανώς  ισχύει  κλασικότατη  ανισότητα
  Παιδιά  κοιτάξτε  λίγο  και  αυτήν  την  λύση   και  πείτε  μου  οκ...............

  Στέλιο  καλή  επιτυχιά  στην  βαλκανιάδα  αντε  και  με  ένα  ανευ  (χρυσό  μετάλλιο  πίσω)
  Σιλουανέ  θα  δω  και  την  δική  σου  λύση .....Βέβαια  ο  Σιλουανός  και  να  έχει  λάθος  στις  ανισότητες  είναι
  ατοπο  δεν  γίνεται ............ :D θα  τα  πούμε

[quote="silouan"]
Στέλιο για την παραπάνω δε'υτερη λύση τι λες ?Είναι σωστή ?
[/quote]
Ολόσωστη ,αλλά γιατί αντικατέστησες τα x,y,z με a,b,c; Σε βολεύουν περισσότερο;

[quote="gdeskos"]
Στέλιο καλή επιτυχιά στην βαλκανιάδα αντε και με ένα ανευ (χρυσό μετάλλιο πίσω)
[/quote]
Ευχαριστώ πολύ!

________

Βάζω και μια πιο εύκολη ανισότητα,που την ανακάλυψα λύνοντας μία άλλη:

$a,b,c>0$ τέτοια ώστε $abc=1$ νδό

$a^3+b^3+c^3\geq a+b+c$

(Υπάρχει και γενίκευση!!!)

Στέλιος... :!:

[quote="stedes"]
Βάζω και μια πιο εύκολη ανισότητα,που την ανακάλυψα λύνοντας μία άλλη:

$a,b,c>0$ τέτοια ώστε $abc=1$ νδό

$a^3+b^3+c^3\geq a+b+c$

(Υπάρχει και γενίκευση!!!)

Στέλιος... :!:[/quote]

Στελιο, λαμβανω την τιμη να απαντησω στο προβλημα σου!  :)

λοιπον, απο ανισοτητα μεσης τιμης δυναμεων έχουμε $\frac{a^3+b^3+c^3}{3}\geq (\frac{a+b+c}{3})^3$

oποτε μενει νδο $a+b+c\geq 3=3 \sqrt[3]{abc}$ (αληθες απο AM-GM)

H γενικευση ειναι:
"Αν $a_1,a_2,...,a_n>0$ με γινομενο τη μοναδα νδο $a_1^n+a_2^n+...+a_n^n\geq a_1+a_2+...+a_n$ οπου $n$ φυσικος"

Η αποδειξη αναλογη της παραπανω!!

Καλο πασχα σε ολους!
 :surf:

ΘΑ ΗΘΕΛΑ ΠΡΟΤΟΥ ΕΥΧΗΘΩ ΑΚΛΗ ΑΝΑΣΤΑΣΗ ΣΕ ΟΛΥΣ ΜΑΣ ΚΙ ΕΥΧΗΘΩ ΤΑ ΚΑΛΥΤΕΡΑ ΣΤΗΝ ΕΠΕΡΧΟΜΕΝΗ ΒΑΛΚΑΝΙΑΔΑ ΝΑ ΘΕΣΩ ΜΙΑ ΑΣΚΗΣΗ ΦΤΙΑΓΜΕΝΗ ΑΠΟ ΤΩΝ ΓΡΑΦΩΝ
ΝΑ ΔΕΙΞΕΤΕ ΟΤΙ ΜΕΣΑ ΣΤΟ ΕΠΙΠΕΔΟ ΧΡΩΜΑΤΙΣΜΕΝΟ ΑΠΟ Ν ΧΡΩΜΑΤΑ ΕΙΝΑΙ ΔΥΝΑΤΟΝ ΝΑ ΒΡΟΥΜΕ ΕΥΘΥΓΡΑΜΜΟ ΤΜΗΜΑ ΑΒ ΜΕ Α,Β ΙΔΙΟΥ ΧΡΩΜΑΤΟΣ ΚΑΙ ΑΒ=Γ ΟΠΟΥ Γ ΑΝΗΚΕΙ ΣΤΟΥΣ ΠΡΑΓΜΑΤΙΚΟΥΣ Υ.Γ. ΟΧΙ ΚΙ ΠΟΛΥ ΠΡΩΤΟΤΥΠΗ Ε?
ΚΑΛΟ ΠΑΣΧΑ ΛΟΙΠΟΝ

Θανάση ωραία η λύση σου  :) (εγώ την έλυσα διαφορετικά).

$a^3+b^3+c^3\geq a^2b+b^2c+c^2a=\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\geq \frac{a+b+c}{\sqrt[3]{abc}}=a+b+c$

[quote="socrates"]
H γενικευση ειναι:
"Αν $a_1,a_2,...,a_n>0$ με γινομενο τη μοναδα νδο $a_1^n+a_2^n+...+a_n^n\geq a_1+a_2+...+a_n$ οπου $n$ φυσικος"
[/quote]

Στο μυαλό μου είχα την εξής γενίκευση:

Αν $a_1,a_2,...,a_n>0$ και  $k,m \in R$ με $k\geq m$ ,τότε

$\frac{a_1^k+a_2^k+...+a_n^k}{\sqrt[n]{a_1^ka_2^k...a_n^k}}\geq \frac{a_1^m+a_2^m+...+a_n^m}{\sqrt[n]{a_1^ma_2^m...a_n^m}}$

Καλό Πάσχα και από μένα σε όλους!!!

Στέλιος... :cool:

Ορίστε και η δική μου λύση.

Από AM-GM ισχύει ότι a^3+2>=3a οπότε a^3+b^3+c^3>=a+b+c+2(a+b+c-3)>=a+b+c
αφού a+b+c>=3(abc)^(1/3)=3   :lol:

Με τη σειρά μου εύχομαι και εγώ καλό πάσχα σε όλους και πολλά (χρυσά) μετάλλια στη βαλκανική μας ομάδα  ;)

[quote="stedes"]
Στο μυαλό μου είχα την εξής γενίκευση:

Αν $a_1,a_2,...,a_n>0$ και  $k,m \in R$ με $k\geq m$ ,τότε

$\frac{a_1^k+a_2^k+...+a_n^k}{\sqrt[n]{a_1^ka_2^k...a_n^k}}\geq \frac{a_1^m+a_2^m+...+a_n^m}{\sqrt[n]{a_1^ma_2^m...a_n^m}}$

[/quote]

H αποδειξη??
 :)

Καλη επιτυχια!
 :lol:

[quote="socrates"][quote="stedes"]
Στο μυαλό μου είχα την εξής γενίκευση:

Αν $a_1,a_2,...,a_n>0$ και  $k,m \in R$ με $k\geq m$ ,τότε

$\frac{a_1^k+a_2^k+...+a_n^k}{\sqrt[n]{a_1^ka_2^k...a_n^k}}\geq \frac{a_1^m+a_2^m+...+a_n^m}{\sqrt[n]{a_1^ma_2^m...a_n^m}}$

[/quote]

H αποδειξη??
 :)
[/quote]
Έχω βρει τρεις διαφορετικές αποδείξεις,αλλά πριν γράψω τις λύσεις μου το αφήνω ως άσκηση για όποιον θέλει να ασχοληθεί (είναι εύκολο!).

[quote="ambros"]
ΝΑ ΔΕΙΞΕΤΕ ΟΤΙ ΜΕΣΑ ΣΤΟ ΕΠΙΠΕΔΟ ΧΡΩΜΑΤΙΣΜΕΝΟ ΑΠΟ Ν ΧΡΩΜΑΤΑ ΕΙΝΑΙ ΔΥΝΑΤΟΝ ΝΑ ΒΡΟΥΜΕ ΕΥΘΥΓΡΑΜΜΟ ΤΜΗΜΑ ΑΒ ΜΕ Α,Β ΙΔΙΟΥ ΧΡΩΜΑΤΟΣ ΚΑΙ ΑΒ=Γ ΟΠΟΥ Γ ΑΝΗΚΕΙ ΣΤΟΥΣ ΠΡΑΓΜΑΤΙΚΟΥΣ Υ.Γ. ΟΧΙ ΚΙ ΠΟΛΥ ΠΡΩΤΟΤΥΠΗ Ε?
[/quote]
Υποθέτω ότι το $\Gamma$ είναι σταθερό και δοσμένο.

Διαλέγουμε έναν αριθμό $d \in R$ έτσι ώστε να ισχύει $\frac{\Gamma}{2}<d<\frac{\Gamma\sqrt{2}}{2}$.
Με πλευρά $d$ κατασκευάζουμε $9$ τέτράγωνα ,ανά δύο διαφορετικού χρώματος.
Ενώνοντας αυτά τα $9$ μικρά τετράγωνα κατασκεύζουμε ένα μεγαλύτερο διαστάσεων $(3d)x(3d)$.
Καλύπτουμε όλο το επίπεδο με τετράγωνα διαστάσεων $(3d)x(3d)$ ,όμοια με αυτό που κατασκευάσαμε.

Οποιαδήποτε δύο σημεία ίδιου χρώματος:
1)αν βρίσκονται στο ίδιο μικρό τετράγωνο θα απέχουν απόσταση $v\leq \sqrt{2}d<\Gamma$ ,
2)ενώ αν βρίσκονται σε διαφορετικό τετράγωνο θα απέχουν απόσταση $v\geq 2d>\Gamma$ .

Συνεπώς αν χρωματιστεί το επίπεδο κατ'αυτό τον τρόπο,οποιαδήποτε δύο σημεία ίδιου χρώματος
θα απέχουν απόσταση $v\not=\Gamma$.

Μάλλον πρέπει να ισχύει μια επιπλέον συνθήκη για το $N$(π.χ. $N\leq 8$(??)).

Στέλιος... :pilot:

ΣΤΕΛΙΟ ΕΠΕΞΕΡΓΑΖΟΜΑΙ ΤΗΝ ΛΥΣΗ ΣΟΥ ΚΑΙ ΜΟΥ ΦΑΙΝΕΤΑΙ ΣΩΣΤΗ ΟΜΩΣ ΘΑ ΗΘΛΕΑ ΝΑ ΔΕΙΣ ΚΑ ΤΗΝ ΔΙΚΙΑ ΜΟΥ ΔΙΟΤΙ ΟΥΤΕ ΕΓΩ ΤΗΝ ΕΧΩ ΛΥΣΕΙ ΜΗΝ ΚΟΙΤΑΣ ΠΟΥ ΤΗΝ ΠΡΟΤΕΙΝΩ  :rotfl:  :rotfl:
ΛΟΙΠΟΝ ΘΕΩΡΟΥΜΕ ΜΙΑ ΚΙΤΡΙΝΗ ΕΥΘΕΙΑ  ΛΕΓΟΝΤΑΣ ΚΙΤΡΙΝΗ ΘΕΩΡΟΥΜΕ ΟΤΙ Η ΠΛΕΙΟΝΟΤΗΤΑ ΤΩΝ ΣΗΜΕΙΩΝ ΤΗΣ ΕΥΘΕΙΑΣ ΕΙΝΑΙ ΚΙΤΡΙΝΗ ΚΙ ΠΑΙΡΟΥΜΕ ΤΟ ΤΕΛΕΥΤΑΙΟ ΣΗΜΕΙΟ ΔΙΑΦΟΡΕΤΙΚΟΥ ΧΡΩΜΑΤΟΣ ΔΗΛΑΔΗ ΑΥΤΟ ΤΟ ΟΠΟΙΟ ΑΦΗΝΕΙ ΤΑ ΑΛΛΑ ΔΙΑΦΟΡΕΤΙΚΟΥ ΧΡΩΜΑΤΟΣ ΠΡΟΣ ΤΟ ΙΔΙΟ ΜΕΡΟΣ ΚΙ ΕΞΕΤΑΖΟΥΜΕ ΤΙΣ ΠΕΡΙΠΤΩΣΕΙΣ Δ<>=Γ,Γ/2 ΑΝΔΝ ΦΕΡΟΥΜΕ ΤΟΥΣ ΑΝΤΙΣΤΟΙΧΟΥΣ ΚΥΚΛΟΥΣ ΑΠΟ ΤΟ ΣΗΜΕΙΟ ΑΥΤΟ ΕΣΤΩ Ε  ΓΙΑ ΚΟΙΤΑ ΤΟ ΛΙΓΟ ΣΕ ΠΑΡΑΚΑΛΩ
ΚΑΛΗ ΑΝΑΣΤΑΣΗ

ΦΙΛΑΡΑΚΟ ΣΕ ΕΥΧΑΡΙΣΤΩ ΠΑΡΑ ΠΟΛΥ ΓΙΑ ΑΥΤΗΝ ΤΗΝ ΠΡΟΣΠΑΘΕΙΑ ΝΟΜΙΖΩ ΟΤΙ ΜΕ ΜΙΑ ΜΙΚΡΗ ΒΕΛΤΙΩΣΗ ΛΥΝΕΙΣ ΤΟ ΠΡΟΒΛΗΜΑ ΓΙΑ ΔΕΣ ΕΔΩ
ΠΡΑΓΜΑΤΟΠΟΙΗΣΕ ΤΗΝ ΙΔΙΑ ΚΑΤΑΣΚΕΥΗ ΜΕ ΤΑ ΤΕΤΡΑΓΩΝΑ ΜΕ  Γ/2<Δ<Γ  ΤΟΤΕ
ΟΠΟΙΑΔΗΠΟΤΕ 2 ΣΗΜΕΙΑ ΙΔΙΟΥ ΧΡΩΜΑΤΟΣ
Α ΑΝ ΒΡΙΣΚΟΝΤΑΙ ΣΤΟ ΙΔΙΟ ΤΕΤΡΑΓΩΝΟ ΘΑ ΙΣΧΥΕΙ Χ<=Δ*2^1/2 <Γ*2^1/2
Β ΑΝ ΒΡΙΣΚΟΝΤΑΙ ΣΕ ΔΙΑΦΟΡΕΤΙΚΑ  ΤΕΤΡΑΓΩΝΑ  ΘΑ ΙΣΧΥΕΙ Χ>= Δ*2^1/2 >(Γ*2^1/2)/2
ΤΩΡΑ ΑΝ    Δ*2^1/2=Γ ΠΟΥ ΜΠΟΡΕΙ ΝΑ ΣΥΜΒΕΙ ΑΥΤΟ ΤΟΤΕ ΑΠΛΑ ΤΟ Χ =ΘΑ ΕΙΝΑΙ Η ΔΙΑΓΩΝΙΟΣ ΕΝΟΣ ΤΕΤΡΑΓΩΝΟΥ ΑΛΛΙΩΣ ΣΕ ΜΙΑ ΑΠΟ ΤΙΣ ΔΥΟ ΠΕΡΙΠΤΩΣΕΙΣ ΕΙΝΑΙ ΔΥΝΑΤΟΝ ΝΑ ΥΠΑΡΧΕΙ ΕΥΘΥΓΡΑΜΜΟ ΤΜΗΜΑ Χ=Γ< Γ*2^1/2  (Α)  ‘Η  Χ=Γ>(Γ*2^1/2)/2 (Β)
ΓΙΑ ΔΕΣ ? ΤΩΡΑ ΓΙΑ ΤΟ Ν ΔΕΝ ΝΟΜΙΖΩ ΟΤΙ ΥΠΑΡΧΕΙ ΣΥΝΘΗΚΗ


geometry
3D geometry
linear algebra
matrix
linear algebra unsolved

Denote by $ H_n$ the linear space of $ n\times n$ self-adjoint complex matrices, and by $ P_n$ the cone of positive-semidefinite matrices in this space. Let us consider the usual inner product on $ H_n$
\[ \langle A,B\rangle \equal{} {\rm tr} AB\qquad (A,B\in H_n)\]
and its derived metric. Show that every $ \phi: P_n\to P_n$ isometry (that is a not necessarily surjective, distance preserving map with respect to the above metric) can be expressed as
\[ \phi(A) \equal{} UAU^* \plus{} X\qquad (A\in H_n)\]
or
\[ \phi(A) \equal{} UA^TU^* \plus{} X\qquad (A\in H_n)\]
where $ U$ is an $ n\times n$ unitary matrix, $ X$ is a positive-semidefinite matrix, and $ ^T$ and $ ^*$ denote taking the transpose and the adjoint, respectively.

At first, if $ \phi(0)\equal{}X_0$, then $ \phi\minus{}X_0$ is also isometry of $ P_n$. Indeed, if $ \phi(X)\minus{}X_0\notin P_n$ for some $ X\in P_n$, then $ \phi(tX)\notin P_n$ for large $ t$. So, without loss of generality $ \phi(0)\equal{}0$. Next, consider extreme rays of $ P_n$. They are rank-1 nonnegative-definite operators . They should be moved by $ \phi$ to extreme rays again. The next is standard, since the angle in $ H_n$ between two such rays equals the angle between there images as lines in $ \mathbb{C}^n$. so our map on extreme rays is less or more the same as isometry of $ \mathbb{C}$.


Let x_n is the only positive real root of polynomial P_n(x) = x^n - x - 1 (n>=2). Find all real numbers  a  such that  lim (n^a)(x_n-x_(n+1)) exists, finite and not equal zero.

[quote]
Namdung Posted: Sun Nov 09, 2003 8:57 am    Post subject: Polynomial roots

--------------------------------------------------------------------------------

Let x_n is the only positive real root of polynomial
$P_n(x) = x^n - x - 1$($\ n\geq\ 2$). Find all real numbers a such that
$lim(n^a)(x_n-x_(n+1))$ exists, finite and not equal zero.
Why a=2? :(

[/quote]


How much money can you make teaching high school math?  With a master's degree?  Public vs. private?  If you have a strong math background, is it possible to get a better salary?  (I've heard you can make as much as $\$$70,000 a year in California public high schools, but I don't know how rare that is or how many years of experience that requires...and I've heard that salaries at private high schools are much worse.)

Teaching is really fun, but I'm afraid teaching high school would be draining...and the whole certification thing sounds like a waste of time...and man, you have to wake up so early!

Teacher's salaries vary very widely by location and by school type.  70K in public schools (once you have experience) is not unheard-of, but you'll likely start much lower.  Small non-super-rich private schools will pay much less than public schools, but you'll have much less crap to deal with.  In the public school system, a higher degree will translate to a higher salary.

Teaching is fun, and is draining.  I only made it for a semester in the regular classroom before I tried something else.

I will note something one of my favorite high school teachers said: Those that go into teaching for the money are almost always awful teachers.  This is probably true of any profession in which the professionals are helping people.  If you are very good at math, but are choosing a profession for financial reasons, pick something besides teaching.  But if you love teaching, we could always use more teachers who actually like math.  I fear there are far too many who do not.

I'm a freshman at Brown, thinking about going into teaching.
Last year, I actually did a small paper on teacher salaries in my high school, but the surveys returned were not nearly enough to make any generalizations. It did seem that those with higher degrees and taught longer had the higher salaries. If I remember correctly, salaries ranged from less than $50,000 to over $70,000 a year.
Some teachers also complemented their incomes by advising certain clubs, coaching, and tutoring privately.
I also had the opportunity to teach a calculus class for a week, while a proper (certified) substitute teacher was being recruited. I don't know why it was, but like rrusczyk said, it was both physically and mentally draining. For my services, I received a gift from the school for $250, which translates to about $12000 a year. :P
I recently saw my an old teacher of mine, and he could barely stay awake during our conversation. Between yawns, he told me that he put it on himself to carefully go over some of the students' homework and write down exactly what was done wrong.  Then again, I've also had teachers who collect homework once and then never look at a student's work between exams, so the amount of time and energy invested in teaching is mostly up to the teacher.

Well whatev... teaching salaries at my school are as follows

starting: 80K+
47% of teachers make 100K+
around 12 teachers make 120K+

these figures [i] are [/i] indeed accurate; in fact, there was a Westchester section NYTimes article about it a while ago.

oooh yeah and my school is public

keep in mind though, most teachers teach for a few years at other schools before coming here, and you need at least a master's degree to teach in new york (idk about other states)

but whatev... so if you randomly get married and you both become teachers, you can bring home like a quarter of a million each year, which sounds like plenty of money to me... plus you get summers off.

Yeah, but keep in mind not everyone lives in Scratchsdale err I mean Scarsdale, where property tax gives the school like a billion dollars.

I think the average salaries at my school (Public school in Texas) are around 40-45 K per year.

I think that's about typical for my region.

Do also keep in mind that teaching takes many forms.  In major cities, tutors can be very highly paid.  Also, in the coming years there will be more and more ways to teach online.  There also will (hopefully) be many more alternatives to regular public school.  For example, in Asia, as I understand, there are many what we would call after-school programs that are designed for kids to excel (as opposed to remedial programs, which dominate the market in the US).  This will likely become more common in the US as No Child Left Behind leaves all our eager students far, far behind their potential (not to mention far behind their competitors in other countries).

[quote]Well whatev... teaching salaries at my school are as follows

starting: 80K+
47% of teachers make 100K+
around 12 teachers make 120K+[/quote]

Anyone know if they have places like this in California?

What's the most that a high school teacher can make at a private school?  There is a high school in LA where the tuition is $\$$100,000 a year, but I don't know if that means their teachers are paid especially well.

If you have a master's in math, is teaching in high school considered to be a good background for becoming a community college professor?  For example, if you were applying to a full-time community college position, and had taught high school for a few years, how would you do against applicants who had been "freeway flyers" for a few years?  (A "freeway flyer" is someone who teaches part-time at several different community colleges...you have to drive back and forth, hence the name.)

Thanks for all the info.

By the way...  I've heard that $\$$100,000 tuition thing from a few different sources, but now that I think about it again it just seems ridiculously high... But in any case, there are certainly schools in LA where the tuition is $\$$25,000 per year.

There are a handful of private schools in the LA area that charge high tuition. A few (but not most) of the students on our ARML team are from schools like that. I don't actually know what salaries they pay teachers, but I'm extremely skeptical of the numbers that have been thrown around here. I thinks it's far more likely that you can't afford to live in the community the school is located in if you earn the salary of a teacher at that school.

The religious schools, for the most part, pay less than the public schools.

Meanwhile, back at the public schools: the number of new openings for math teachers each year exceeds the number of math majors in the state intending to go into teaching who get bachelors degrees each year - I'm sure by at least a factor of two. Part of the reason there are so many openings is that a lot of teachers quit within their first few years. It is, as someone noted above, a profession that requires a lot of mental and physical energy. There's also the distressingly common practice of saddling new teachers with the worst classes to teach: Pre-algebra for 9th or 10th graders, Algebra I spread over two years, "consumer math," and so on.

The principals of the public schools are desperate for warm bodies to put into classrooms. As to the ways you can get certified as "subject matter competent" - I rather not even get into that.

Dr. Merryfield is certainly correct, you can't afford to live in Scarsdale on 100K/year if you have a family and all, but you can certainly live comfortably in a nearby suburb.  But I'm sure the same goes for most good schools and their surrounding communities.  I'm sure my figures are accurate; if you subscribe to TimesSelect you can do an easy search for "Teacher Salaries" in the Westchester section and read the article at http://www.nytimes.com; otherwise, you'll just have to take what I said on faith.  But whatev.  It is sort of fool's good, as the bagels here cost 10 anyway, so you might as well just earn 40K/year and move somewhere where the bagels cost 4 dollars.

[quote="Aunt Sally"][quote]The principals of the public schools are desperate for warm bodies to put into classrooms.[/quote]

Is it possible to teach part-time at a public high school, for half the full-time salary?

I'm fairly young and I don't plan to get married or have a family any time soon.  I don't care much about money, as long as I can get by; what I'd really like is a lot of time and freedom to read and learn about whatever I want.  (And I don't think professors at research universities struggling to get tenure necessarily have that kind of freedom.)[/quote]

I would imagine this is hard, but not impossible.  You might consider a place like Kaplan, or any of the other test-prep outfits, if your goal is to teach part-time and still be able to have enough money to do wild and crazy things like, say, eat.

[quote="Aunt Sally"]
If you have a master's in math, is teaching in high school considered to be a good background for becoming a community college professor?  For example, if you were applying to a full-time community college position, and had taught high school for a few years, how would you do against applicants who had been "freeway flyers" for a few years?  [/quote]

I think that being a freeway flyer would be seen as a better qualification for a full-time community college position.  (So would a Ph.D., especially in California.)  The exception would be if you were going to teach mathematics education classes rather than mathematics content classes.

[quote="Aunt Sally"]
Is it possible to teach part-time at a public high school, for half the full-time salary?
...what I'd really like is a lot of time and freedom to read and learn about whatever I want.
[/quote]

I suspect there are very few part-time positions offered.  More importantly, they're likely to come without benefits (such as health insurance).  High-school teaching is very time-intensive not only because of the number of classes one must teach and amount of homework one usually grades, but because of the amount of time one must spend in communication with parents.

If you'd like lots and lots of time, probably teaching is not the way to go.  Even part-time teaching will not leave you a lot of time, unless you choose to do less than the best job you can do.  Teaching tends to consume mental energy even when you're not at work or not focusing on it.  You'd be better in a 9-to-5 job which can be left both mentally and physically at the work site.

Um, what would be the starting salary for an inner-city school, if you have a mster's degree or a PhD?  And would the PhD make any difference?

Instead of pouring all this money into accountability and extra help programs and new methods of teaching, why not increase teacher salaries, attracting better teachers, thus giving students a better education?

I don't think anyone can argue that many smart people out there shy away from teaching because of the money.  It's not a bad profession, but we tend to think of it as a lower-paying job, even though it's perhaps the most important job in America.  If we starting pouring our money into teachers, instead of new football equiptment, think about how much better teachers we would have.  For example, I bet even some of the AoPS instructors would have thought twice about quitting teaching if they were getting paid as much as some sinecure executive at Walt Disney who does next to nothing.

Even if people think it wouldn't work, I believe it's a worthy experiment to try in some under-performing school district.  Start paying teachers upwards of $\$$200,000, and just see what happens.  I realize that it would be impractical to pay that much on a regular basis, but in the interests of science, lets do it in one place, and see what happens.

What I can tell you: in France, with a teaching salary, you've got just enough to survive ... but there's nothing to complain about, it comes with a package of personal advantages.

It is good in USA that average teaching salary is about 50,000$.
It is maximum 200$ in Tajikistan. I think you must thank GOD that you are living in USA.

I've been looking at teaching at the college or university level. Do you know how much people make on average at city colleges and universities?

I guess particularly in California at the jc, csu and uc.

[url=http://www.cpec.ca.gov/higheredupdates/update2000/up00-1.asp]This link[/url] contains some information about salaries in my system, the CSU, as well as the University of California. I imagine that if you search around, you might find something similar for community colleges.

Teacher is a precious career. If you want to be rich, don't be a teacher!

[quote="tokenadult"]Teacher salaries are high in some places. But union work rules and the lack of competing providers mean that the learners don't necessarily get good teaching that way.

The best education reform would be to give learners more power to shop. That provides incentives for school administrators to do whatever it takes--which might include paying teachers more, or perhaps hiring different teachers--to make schools better.

See

http://learninfreedom.org/school_state.html

for more background reading on this subject.[/quote]

i agree with this notion. there was a report on abc a few months ago about making the education system more privatized, which in turn increases competition, and thus quality of education.

Dunno if this is appropriate, but how much do the AOPS guys make? (rrusyck, nsato, etc.)

Does anyone have an idea how much University Professors make?

Gib Z: look a few post up, to a link I posted.

I can add to that the statistics for Michigan State University mathematics department: the average full professor salary is about $ \$$107,000. After taxes, retirement plans and other such things it gives you 5,000-6,000 dollars a month in free money. Not very much, of course, but quite enough to live on :D.


let $\{a_{n}\}_{n=1}^{\infty}$ be a sequence and  $\lim_{n\rightarrow \infty}(a_{n+1}-\frac{1}{2}a_{n})=0$ prove : $\lim_{n\rightarrow \infty}a_{n}=0$

[quote="Rust"]Exist N, suth that $|a_{n+1}|<\frac{3}{4}|a_{n}| \ \forall n>N$[/quote]
Why?  What if, for example, $a_{n}= \frac{1}{n}$ for all $n$?

I give contradition for case, when don't exist $\lim_{n\to \infty}a_{n}$ or $\lim_{n\to \infty}a_{n}\not =0.$
Second way. Let $\epsilon >0$, exist N, suth that $|a_{n+1}-a_{n}/2|<\epsilon /2$. Therefore $\frac{a_{n-1}-\epsilon }{2}<a_{n}<\frac{a_{n}+\epsilon }{2}\Longrightarrow \frac{a_{n-2}+\epsilon }{4}-\epsilon <a_{n}<\frac{a_{n-2}-\epsilon}{4}+\epsilon ...$ It give
$\frac{a_{N}+\epsilon }{2^{n-N}}-\epsilon <a_{n}<\frac{a_{N}-\epsilon }{2^{n-N}}+\epsilon$, therefore $\lim_{n\to \infty }a_{n}=0.$


My TV have 3 plugs, one for left audio channel, other for the right and the other for the video. I don't know the name of the cable to plug on it, sorry. The ouput of my camera (not webcam) cam be pluged on this TV using this cable.

I bought a new headphone that can be pluged on the computer (that small plug) and can be plugged on a old radio (the same plug of the TV). (It has a "conversor" of the big plug to the small (When we don't know the name of things to examplain is really a pain in the butt)).

So I came up with the ideia, why not plug my camera in the audio input of my computer?

The range of the ouput voltage could've (don't know) the same.

So If I plug it in and record it as a music. How could've I possible decode it in a video? Because a music is a plain list, but a video is a sucession of arrays (images).

So my question is: This stupid method would fail? If not, How can I decode It.

Note: I really intend to try it, sounds really fun and maybe do work. So why bother buy a expensive TV card if you can use your old sound card?  :D

Thanks.

Huh, sounds interesting :). But I seriously doubt that the video output will be indeed in the same range, and even if it is then your sound card probably won't be able to record it in high enough resolution. I'll ask my dad, he should know more about it since he studied electrical engineering and he usually knows everything about any electrical device which was used 20-30 years ago... I'll let you know if I find out something. Btw your connection is called "composite audio / video" (or AV).

I don't know how it is multiplexed, butthe frequency is really very high, It has much more informating per sec than audio, don't if the sound card will rec it all.

Supposing everything works, we still have the problem of deconding the signal. :P

Yeah... should have been quite obvious :P. The average sound card has a recording frequency of 44 kHz but apparently PAL video uses frequency of about 4.4 MHz, you'll never be able to record that with your sound card. It also uses QAM modulation which would require even greater resolution to correctly decode, I think? I don't remember this stuff anymore :blush:.

EDIT: But to answer the second part of your question, PAL encoding seems to be explained well enough for example on Wikipedia.


AMC
MATHCOUNTS
articles
summer program
Gmaas

Be creative.
In this game, I will start off if...
You respond with then...

OK...
If rrusczyk ran for president 08, then...

I'll run for president 09 (or is it 13? not sure)

If John Lennon were alive...

chaos would reign for all eternity.

If Elvis was alive, then

we would still live like it was the 80s *shivers*
If AoPS never existed then..

chaos would reign for all eternity.

If chaos reigns for all eternity, then

Crazychinesemaniac would be very happy because he would lead the chaos
If life were simple then...

NoTA would be content.

Wait a minute....

crazychinesemaniac = happy if chaos reigned for all eternity
chaos reigns for all eternity -> AoPS does not exist
-> crazychinesemaniac would be happy if AoPS did not exist

HOW DARE YOU!

Anyway...

If 42 were the answer to life, the universe, and everything (which it is)....

then life would continue as normal.

if 43 was the answer to life the universe and everything then...

then the entire foundation of mathematics would collapse, because it's actually 42.

If AoPS_RULZ (a username, look her up in the member list) sucks at math (which is also kind of true)....

then it's definitely true.

If every human was a Christian, then

Technology would recess to the point at which it was in the 1500's.

If I was president, then

I would hope it weren't president of the U.S.

If I weren't perfect(I know, a stretch, but try)...

We would use calculators, not nuclear weapons.  (Intelligence War)

Okay.

If everyone in the world was atheist or agnostic, then....

I would be picked and eaten

If we were all vegan...

I'd hate life.

If we started scoring f- on tests...

I would cry (What? F-'s???)

If you were a mod...

i would ban some people

If you were actually a turtle...

then I'll sleep all day in my shell.

If you were a toddler...

[quote=Mathbee2020]then I'll sleep all day.[/quote]

if you were 100 years old...

I would be in retirement

If you only ate pure sugar or salt for a day..... [size=50]Oh dear...[/size]

my cells would shrivel up like raisins in the sun due to the osmotic properties of cell membranes.

If you give a pig a pancake...

he would throw it back at your face.

If 1=2=3=4=5=6=7=????

The world would be in a GREAT chaos...

If you lived on Mars...

Then I would be a Martian.

If you used actual mice with your computer/laptop...

I would have lots of mice

if the sun was a meatball...

I'd eat it with the international date line, which in this case would be a long line of spaghetti.

If $e = \alpha \beta^2$...

then $\alpha \verb# and # \beta$ = $m \verb# and # c$

If life was made of variables...

then we would all be a huge math equation
if if...


function
algebra
domain
real analysis
real analysis unsolved

I was just on a different forum when someone asked a harmless-looking question.  I haven't had too much experience with the axiom of choice, but it seems to me that the proof requires it.  Am I right or can it be proven without the AoC?

Given a function $ f: A \rightarrow B$, we say it has a right inverse if there is a function $ g: B \rightarrow A$ such that $ \forall y \in B, f(g(y)) \equal{} y$.

Now the statement:
If a function $ f: A \rightarrow B$ is surjective, then $ f$ has a right inverse.

Thanks, but something else stands out then.  Do you actually mean $ B$?  Or did you mean $ A$?

When I showed it (using the axiom of choice), I thought I applied the axiom to $ A$...

It's about the domain of the choice function you're trying to construct- that domain is $ B$.

[quote="jmerry"]If $ B$ is infinite, this requires the axiom of choice- it basically is the axiom for general $ B$.[/quote]
More precisely, suppose that the statement is true. Take any set $ X$. Consider $ A$ to be the set of all pairs $ (y,Y)$ where $ \varnothing\ne Y\subset X$ and $ y\in Y$, and take $ B\equal{}P(X)\setminus\{\varnothing\}$. Let $ f(y,Y)\equal{}Y$. Then the right inverse mapping will, in effect, give you a choosing function.


geometry
geometric transformation
reflection
angle bisector
geometry unsolved

let $ABC$ be triangle $(AB=AC)$   and $\angle ABC=36^o$
          let $D$  be on $AC$  and satisfyind conditions:$ABD=DBC=18^o$, let$N$be on side$BD$such that : $NCB=12^o$.
 prove that :
                 $AC=CN$

BD is the bisector of the angle $\angle ABC = 36^\circ$. Let P be a point on the ray (BA, such that $\angle BCP = 72^\circ$. Then $\angle BPC = 180^\circ - (\angle BCP + \angle ABC) = 72^\circ$, which means that the triangle $\triangle BCP$ is isosceles with BC = BP. Hence, the bisector BD of the angle $\angle PBC \equiv \angle ABC$ is perpendicular to CP and cuts the segment CP at its midpoint M. The triangle $\triangle ACP$ is also isosceles with CP = CA, because $\angle APC \equiv \angle BPC = 72^\circ$ and $\angle PAC = \angle ABC + \angle ACB = 2 \cdot 36^\circ = 72^\circ$. Let the circle (C) centered at the point C and with radius CP = CA meet the angle bisector BD at a point N'. Then also CN' = CP = CA. Since $BD \equiv BM \perp CP$ is perpendicular to the radius of this circle at its midpoint M, the triangle $\triangle CPN'$ is equilateral and $\angle PCN' = 60^\circ,$ $\angle N'CB = \angle PCB - \angle PCN' = 72^\circ - 60^\circ = 12^\circ$. Consequently, the points $N' \equiv N$ are identical and $CN \equiv CN' = CA.$

See that $\widehat{CND}=30^\circ$ and the reflection of $C$ in $BD$ is $P$, on $AB$, with $\Delta PNC$ equilateral, so $PC=CN \ (\ 1\ )$. As $\widehat{PCA} = 60^\circ - 24^\circ = 36^\circ$ and $\widehat{PAC} = 72^\circ$, triangle $\triangle PAC$ is isosceles, with $PC = AC \ (\ 2\ )$, (1) and (2) giving $AC = CN$.

Best regards,
sunken rock


I recently encountered a problem, and I think that the catalan numbers are the solution. I've checked mathworld and read about them, but could someone maybe explain some more about them like interesting properties or what other problems they solved or what else they can be used for.
thanks

Here is an article you might find interesting.

http://www.geometer.org/mathcircles/catalan.pdf

Catalan numbers turn up in the solution to a number of different problems in combinatorics. The Mathworld article that you refer to mentions the following problems :-

- the number of ways a regular polygon can be divided into  triangles if different orientations are counted separately
- the number of binary bracketings of n letters
- the solution to the ballot problem
- the number of trivalent planted planar trees
- the number of states possible in an n-flexagon
- the number of different diagonals possible in a frieze pattern with n+1 rows
- the number of ways of forming an n-fold exponential
- the number of rooted planar binary trees with n internal nodes
- the number of rooted plane bushes with n graph edges
- the number of extended binary trees with n internal nodes
- the number of mountains which can be drawn with n upstrokes and n downstrokes
- the number of noncrossing handshakes possible across a round table between n pairs of people
- the number of sequences with nonnegative partial sums which can be formed from n 1s and n -1s

Most of these problems are described either in the Catalan number article itself or elsewhere in Mathworld.

The fact that Catalan numbers turn up in solutions to all these different problems indicates that there might be some link between them. In Conway and Guy's "Book of Numbers", they build explicit analogies between six or seven of these problems. These analogies show how a solution to one problem translates into a solution to another problem - in effect they are establishing an explicit 1-1 relationship between the sets of solutions to different problems.

Some of the analogies are quite natural, but others are very ingenious and not at all obvious !


Stanford
college
geometry
geometric transformation
reflection
percent
3D geometry

Scientists and researchers who are potentially finding cures for diseases and helping mankind directly are paid minimal salaries. Athletes, who simply serve a purpose of entertainment in society, are paid an exponential salary in comparison to these scientists.

Is this right? Should there be a set limit on salaries of professional athletes in America? What should it be?

Yeah, athletes are paid a lot because people are willing to shell out lots of money to see big events (this is not just a USA topic, footballers like Ronaldinho and Zidane are the highest paid athletes in the world iirc). If you don't like how much athletes get paid, the problem is in one of the following:
- The free market system is bad (i.e., just because someone does something that other people will pay for doesn't mean that the someone should get rich from it)
- You think that world interest in professional sports is way higher than it ought to be

I agree that it seems bad that people who provide services that we value highly (teaching, research, technology) get paid mediocre to good salaries whereas people who provide services that we think are not that important (pro sports) get paid vast fortunes, but in my opinion we'd either have to have a weird exception to our general rule or sacrifice something of value (free market system, people's freedom to entertain themselves) in order to rectify it.

[quote="math92"]But I'd say the average athlete gets paid more than the average scientist.[/quote]

[quote="Ignite168"]That is not true at all...[/quote]

Average salary (annual I assume) of a medical scientist is $\$$33,644 (USD).  This is per http://tms.ecol.net/jobs/asc2.htm
Other sites put scientists salaries around $\$$50,000 USD annually, on average.  This is purely research, not teaching.

The average salary of a professional athlete in the US is about $\$$27,000 USD annually.  A professional athlete is someone who does athletics for a living, whether it be minor or major.  This is per http://swz.salary.com/salarywizard/layouthtmls/swzl_compresult_national_AR17000014.html

But the more distrubing fact for me is that the average Major League Baseball player's salary was $\$$1.9 million, a National Basketball Association player's was $\$$2.6 million per http://www.megaessays.com/viewpaper/32426.html

So even though your average athlete for a living doesn't make more than your average medical scientists, high-profile, national calliber athletes do make a lot more.  I think that is what math92 meant for the most part.

I agree with Nathan that something must be sacrificed if you want professional athletes not to earn such quantities of money. Of course I'm pretty sure what is it that I would sacrifice, namely free market. This could turn out to be a very long post for me if I'm going to sustain carefully that opinion so I won't do that now.

One thing I do want to say is that free market makes it pretty logical Ronaldinho or Zidane earning lots of money. For both of them have proven to be unique soccer players in the world. What Ronaldinho or Zidane offer(offered in Zidane's case) to the world is something only them can give. To state it simply, there's the whole world demanding something only one man can give, therefore the price of it rises accordingly. What I think is this must be regulated, in science there are aldo people of unique genius, you surely know of many examples. What Einstein gave to this world is arguably far more important than what Ronaldinho or Zidane ever will. And it seems pretty obvious that at his time, only he could. The point is that the closer the enterprise is to the public, the more paraphernalia you can build around it, particularly the entertainment industry is so close to the public that they are allowed to raise Ronaldinho or Zidane to the state of human gods while Einstein remains a a wise, bit of a crazy old man. Free market is at the basis of all this, I'm not concerned about Ronaldinho getting more money for his exceptional talent, I'm concerned about the whole system that implements this and why.

By the way, I heard somewhere that particular examples apart, Ronaldinho, Zidane etc, the average salary of players on any of the four main major sports leagues in the USA, that would be baseball, basketball, american football and hockey, is greater than the average salary of the players in the european soccer leagues.

Ursula

I tend to stay away from discussions like this one because I believe they violate the fundamentals of a capitalist economy. When we start saying "athletes should only be paid x million per year" or "they should only charge us x dollars for a courtside seat at the NBA Finals", the position being advocated is the same one that Communism subscribes to - but we, being Americans (most of us anyway), can't use [i]that[/i] word now can we?

What is it that you really want when you rein in someone's salary? "Fairness"? "Equality"? "But they don't deserve..."? That's exactly why state planning is so popular in certain places. People start to feel that others earn too much for the skills and talents they possess. They think it's not fair. They think that they themselves would an absolutely fair judge at exactly what someone else should be able to earn from their abilities - or, if not themselves, then surely the men down in DC. And it comes from the same vein as absurd minimum wages (Chicago anyone?), abused welfare, and resource allocation by state initiative (food stamps?).

Am I taking this too far? Making presumptions? Yes. I think. And no. I am not against having a minimum wage or a welfare system. But I am certainly against the emotions of the public, sparked by feelings of injustice, dictating a "maximum" reward for the talents and skills of someone they can't begin to empathize with. A line should a drawn for this sort of intervention. A line must be drawn if we believe in capitalism.

Few people complain about Larry and Sergie being worth 11 billion apiece for their Stanford experiment. Mostly, they marvel. The same goes for Bill and Paul and Steve - admiration. The Waltons? Same deal. But what's the difference between these people, who in their moments of creativity came up with hundred-billion dollar ideas and, well, athletes who exhaust themselves in gyms and on courts and fields daily?

I think we tend to underestimate the dedication of athletes and  what is required to become Kobe or MJ or Shaq. It's easy to hear about them being paid 400k per game and forget that they are simply the best in the world and their performance is being watched by tens of millions. And frankly, for being the best in the world in such a huge market, it's not an exorbitant salary at all.

The best athletes aren't billionaires, are they? But the best businessmen are. The best lawyers, real estate agents, doctors. Authors. Candy makers (think Mars). Investors. Inventors. Stradivarius, if he were alive today, would probably find himself in this category. Einstein's [i]image[/i] earns a ten-digit (including cents  :) ) salary every year. The creators of YouTube - honestly, was that idea worth 1.6 billion? Was it more deserving of a 1.6 billion price tag than athletes' performances?

Tiger Woods gets paid millions for swinging a golf club. Tom Cruise for smiling (pretending no less) to a camera. Nicole Kidman for shooting an ad. Gisele Bundchen for wearing 10 million-dollar lingerie and making seductive strolls down a catwalk.

The truth is, it's incredibly tough to be the best. But if you are...If you make the necessary sacrifices and have the proper talent, you'll get there. And when you do, you'll earn a hellofalot for being the brightest bulb in that particular chandelier. No matter which chandelier you're affixed to.

Frankly, it's sad that we use academia as an example of people who "should" earn more but they don't, as if there were no rich men and women in academia. There are. Look for patent holders or surgeons who have invented procedures or lawyers who have furthered theories through litigation (or just bother to practice part-time) or mathematicians who consult for the NSA. Perhaps academia in itself doesn't pay well, but those who go through the rigors of advanced education and choose to do something with it in the marketplace will certainly be rewarded by Adam Smith's hand.

After all, Charles Darrow was a millionaire (for inventing Monopoly). Joe Jamail banked more than 400 million for convincing a jury that handshakes could constitute a contract. Hymen (often misspelled Hyman) Lipman put an eraser at the end of the pencil, got a patent, and got rich. So why not athletes? Would you say their contribution to society is more trivial than the eraser on the end of your pencil?

Additionally, I want to add that you're making huge assumptions when you say that scientists' work is more important than athletes' and actors' (or actresses) performances.  You're assuming that science is more good than entertainment.  Certainly there are intriguing reasons to believe this but there are also reasons to refute it as well.  For example, entertainment by some can be considered an intrinsic good and science an instrumental good.  In other words, entertainment is good in itself but science is good because it brings you other things which are good.

This is, in my opinion, a highly philosophical matter.  It's just always good to consider all of your assumptions with a grain of salt.  There seems to be a large assumption in society that work is "better" than play.  Well, the assumption might seem stronger to me as I'm at a school full of hard-workers and geniuses, etc.  But even in high school where a lot of students would do other things instead of their homework, for example, I could sense that even though they didn't act upon the principle that school is more important than play, they held that principle as their main belief and just did not have the discipline or motivation to act in accordance with it.

Honestly, I don't see any reason to object to a society that neglects science and simply just enjoys pleasure as long as they do so in a morally acceptable way.  Of course, keeping things morally acceptable is practically impossible.  Additionally, our intelligence prevents us from indulging in simple pleasures while ignoring science.

I'll bring up one last point.  We don't know the meaning of life and I doubt we will ever come close to figuring it out.  For this reason I feel that many, if not all, moral issues are only resolvable to a certain point before assumptions have to be made.

And a disclaimer: the arguments that I don't [i]necessarily[/i] reflect my personal beliefs.  For example, saying that entertainment is better than science is something I don't fully believe myself.  However, fully dismissing it should be an interesting task :D

[quote="MithsApprentice"]I tend to stay away from discussions like this one because I believe they violate the fundamentals of a capitalist economy...[/quote]

Yeah indeed, I'm against the fundamentals of a capitalist economy. The examples you put, I agree of course with them, however I point out the same argument in my post, which is, salaries as they are today make perfect sense in a free market capitalist economy.

The word communism is satanized in the US. In Europe in the other hand, there exist plenty of parties with the word communist in their names that get quite a few percent of the votes in the general elections. The so called communist regimes once obviously did horrible things to humanity though, so don't think I don't find logical for the US, being the other faction back then, to demonize the word communism. What I mean is that many people with pretty different ideas than those of Stalin or Mao have called themselves communists and still do. What they all coincide in is taking marxism as a basis but it seems apparent that there's a good spectrum still to differ.

To be frank, the problem with the salaries is something that doesn't concerns me much. I don't care if someone is getting a lot of money for his/her skills. What I see is this happening now because of the way the society is built. I'm not sure this would happen, at least, not so disproportionately, if the society is built from other principles I can think of. Capitalist economy as it is today is I think, unsustainable for the world. So either it changes -unlikely- or we will see, among other things, a lot of more wars in the upcoming years -we are seing some already-. Also you ask if I want "fairness", "equality", "but they don't deserve", well yeah except for what I think it means to be fair with everyone or that there is equality is something almost all countries have agreed already. I'm thinking for instance in the Declaration of Human Rights. The mere existence of such a document means people have agreed there must be some absolute concept of fairness and that all men should be equal up to some point. Far, very far away we are from achieving this simple one-would-think-obvious equality.

[quote="joml88"]Additionally, I want to add that you're making huge assumptions when you say that scientists' work is more important than athletes' and actors' (or actresses) performances...[/quote]

Actually I'm not making any huge assumptions. In fact, I think society should be built to value leisure. If you look at my post, I said [i]arguably[/i] Einstein's contributions are more important. And that is because although Einstein himself surely wasn't thinking about it, and though the purpose of scientists might be the pure generation of knowledge or the description of nature or whatever, science is useful to society because of its practical applications. This practical applications, [i]arguably[/i] contribute to our happiness, and particularly to entertaiment. If you don't have a TV set you wouldn't see very regularly Kobe Bryant, and of course the quantum mechanics used in a TV set can be traced back to Einstein. If Einstein wouldn't have been there, surely we would have TV sets nevertheless but if you take him out and say the ten other guys that were doing the stuff maybe the TVs would have come ten years later. In the case of general relativity, no one else was thinking about it at his time, so if Eisntein wasn't there GPS's -which rely heavily on general relativity corrections- might have come ten years later. (this argument is not very good because the appearance of GPS's is retarded fundamentally because of technical problems and not theoretical ones, nevertheless you get my point)

You can build upon the argument that one does not do science for these reasons and there are actually many discoveries in science that it is dubious will contrbute to our happiness in any way. Gödel's incompleteness theorem seems , for instance pretty far from applications in real life. The fact is that science is useful to society, and along the years we have achieved an amazingly efficient mechanism of producing scientific knowledge which includes the production of huge quantities of '[i]useless[/i]' knowledge which in fact it is not useless at all because it adds to our comprehension of the world and the more we understand the world the more capable we are to produce '[i]useful[/i]' things.

[quote="jomll88"]I'll bring up one last point. We don't know the meaning of life and... [/quote]

This one is obvious. It's 42.

Ursula

[quote="joml88"]First of all, I wasn't addressing "you" ursula.  I was making a generalized statement.  I was addressing anyone who held this belief.  I didn't even pay any attention to your post when I wrote mine.  I was addressing those who take it as fact that science is inherently better than entertainment (or some other entity in the world).  My claim was that "science is inherently good" cannot be used as a premise for one's argument unless adequate justification is given, of course.[/quote]

OK OK, since Miths message came after mine and seems to address it and your message starts with the word 'additionally' I assumed you were talking to me. No need to get mad at me :)

Ursula

[quote="7h3.D3m0n.117"]It's unfair for the jocks at school to laugh at you because they get more pay while we're saving <input high number here> people from cancer  :lol: .[/quote]

I think what you're addressing here is a completely different issue: the social interactions of different stereotyped groups in public high schools. It's certainly something that should be discussed, but I don't think this is the best context for that.

There is absolutely no law that says if you study hard in high school one day you will be financially better off than your peers who get drunk and fail classes. Perhaps there is a noteworthy statistical correlation between hard work, acceptance to good colleges, and high salaries later in life, but that doesn't make it an all-exclusive law.

But on the other hand, if someone does find a cure to cancer  I assure you that individual would find himself with a net worth higher than the combined net worths of all his (or her) former high school classmates, assuming he/she enforces the patent and doesn't get it stolen by a pharmaceutical giant. Governments protect the creator's interests in advances of math, science, and art for the very purpose of granting incentive for these kinds of discoveries.

@ursula: Joe and I did not intend to pick on you at all. In fact, both of our responses addressed the direction of the discussion in general.

@ignite: I'll keep deleting your posts if you can't respectfully disagree.

[quote="7h3.D3m0n.117"]As Bill Gates said:

"Be nice to nerds, chances are, you'll end up working for one someday." or something like that.

It's unfair for the jocks at school to laugh at you because they get more pay while we're saving <input high number here> people from cancer  :lol: .[/quote]

First of all, pro athletes do not on average get paid more than scientists.

Second, what's unfair about that? The majority of people have judged that entertainment is more important than researching, so there's nothing unfair.

[quote="Ignite168"][quote="math92"][quote="PenguinIntegral"]Why shouldn't we let the free market decide?

Scientists get payed loads of money too. It's not like you get an advanced degree then starve. :roll:[/quote]
But I'd say the average athlete gets paid more than the average scientist.[/quote]

That is not true at all, for ever star like DAISUKE MATSUZAKA W00T or TIKI BARBER GO GIANTS, you have tens and hundreds in the minors or even less known leagues that no one has heard of.

Also, the big names in sports are harder to come by than smart scientists and their contributions are worth more than the average scientists' contribution (as decided by consumers) so they should be paid more.[/quote]
I bet there are only about 1 or 2 famous scientists that the general public knows. Einstein and maybe Pythagoras (srry if spelled wrong) and Fibbonacci. People on this site might know alot but the people here are at the top of their math class. I can say that most people here can only name about 5-10 mathmaticians. If you think about sports, a person who is really into sports can name 100s of players, some can name who their wifes are, and most can name what position and what team they r on. I bet no one here knows that much about 100 mathmaticians.

dang, mithsapprentice, that would be a great essay for a college application

[quote="Ihatepie"]I bet no one here knows that much about 100 mathmaticians.[/quote]
You don't want to know how much I know about mathematicians and their personal history. I don't think the difference is that sports fans are generally more hardcore than math fans. It's just that there are way more people who are into sports total, so the number of fanatics is large; that's not evidence that the proportion of fanatics among people into math is not smaller.

I agree with Ignite. It's all about the dollar vote. When it comes down to it, people use their freedom to spend their money to "vote" about what they consider valuable. People consider the fruits of research valuable, so a lot of money gets to researchers, only indirectly through the people who do stuff with the research (such as medical technology). People also consider entertainment valuable so a lot of money flows to people who provide entertainment.

A large part of the reason I am not against capitalism is this: may people view the freedom in capitalism from the side of the earner. Entrepreneurs have the freedom to do what they want and the system rewards them if they provide something that's in demand. But you can also look at it in the way I just mentioned -- the freedom of the buyer. I am delighted by the fact that my value judgments can have some impact on the world through my ability to spend money according to those values, with fairly minimal interference.

In fact, you don't even need a dollar to vote.

When you watch that Super Bowl game, or the GIANTS LOSE A TWENTY ONE POINT LEAD TO THE TITANS IN THE FOURTH QUARTER, you're voting with what you watch and the advitisers capitalize that. Why do you think they try to get the best players? To draw the most people in.

I think it's difficult to argue that professional athletes are not overpaid, in the sense that many professional sports leagues have a great number of teams losing money (see: profit sharing in baseball, the lockout in the NHL, the generally mediocre state of affairs in the NBA).  Football is an exception, between the massive TV deals and general level of popularity that the NFL currently enjoys.

Professional sports owners frequently pay players far more than the market demands (see: Rodriguez, Alex, this MLB offseason, any of forty names I could easily list in the NBA) and turn around and pass the prices onto fans.  There's a big difference between a (quantitatively) fair market value for players and what various owners with too much money are willing to pay for players.  These foolish signings then drive up the contracts of everyone else.

[quote="calc rulz"]I think that it's really messed up that athletes get paid that much money in comparison to academic professionals. Yeah, we may live in a capitalist society, but I simply don't think it makes any sense for an athlete to get enormous amounts of money, much more than they really need, simply for being an athlete, while teachers and scientists get the short end of the stick (and in particular, I think teachers and scientists especially need more money because better quality teaching and scientific advancement is very important). Thus, I believe we do need to have some kind of limit to how far this can go.[/quote] I agree with you. Athletes are just one successful person. Teachers help millions succeed and keep most people from having a bad job with no pay. Scientists save lives, and help us learn more about the world. All athletes do is be good at sports. I believe in our society and that the consumers should decide but I think there is a way of going to far.

Small Athletes get paid less than Small Academic people
Big Athletes still get paid less than Big Academic people (Shaq vs Steve Jobs; think about it)

So what's the problem here?

A large amount of athletes make our lives more comfortable. So do scientists.
So?

And
Should education be our sole focus of life? Should that be the only sphere of capitalistic activity? Should we impose an artificial restriction that allows brains to be everything, and those born without it to have no chance?

And lastly
What would cutting athlete salary do for us? Give us less entertainment, shoot down advertising, lower the employment salaries or rates of companies involved in private market sectors which expound on the advertising (including, but not limited to academic corporations like microsoft or ATT)?

And if sports figures should be paid less for entertainment vs realwordimpacts, then theoretical mathematicians shouldn't be paid at all.

(Agreed, however, that people who do public services like teachers and army soldiers should be payed more. That doesn't mean that athletes should be paid any less; they probably trained 10 hours every day doing sports, if not more.)

well, given salary caps, it would seem that they are not paid enough....

but really, the only positive effect I can think of off the top of my head that pro-athletes have is to encourage kids to pursue sports and therefor become more fit and get less sick and cost less to fix up.

Basically, I should say that I don't think being a pro-athlete should be a viable career at this time.

and that is why we don't have many theoretical mathematicians!
But we have tons of athletes.

[quote="Scrambled"]In that case, theoretical mathematicians should never be paid.[/quote]

ugh your avatar gives me a headache lol

but mathematicians do get hired by the highest orders in the world (governments, space administrations, etc.)

Athletes have a purpose, still; making our lives more comfortable, just like most people in the private sector. Theoretical mathematicians have no purpose.
What you, d3m0n, are referring to is applied mathematicians.


In parallelogram $ ABCD$, the measure of angle $ ABC$ is 3 times the measure of angle $ BCD$. How many degrees are in the measure of angle $ ADC$?

knowing that it is a parallelogram, we know that consecutive angles add up to $ 90$
let n = degree measure of $ ADC$. since $ ABCD$ is a parallelogram, $ ADC$ is the same as $ ABC$. then $ n \plus{} 3n \equal{} 90$ solving for $ n$ we get $ n \equal{} 22.5$

Consecutive angles of a parallelogram actually add up to $180$.

Let $x=m\angle BCD$ and $3x=m\angle ABC$.

$3x+x=180^\circ$

$3x=135^\circ$

Opposite angles of a parallelogram are congruent. Hence the measure of angle $ADC$ is $\boxed{135}$


search
analytic geometry
geometry open
geometry

Can you tell a way to find a distance between two point $ X$and $ Y$ in the plain of any triangle $ ABC$ and please demonstrate when $ X\equiv O,I,G,H$ (and so $ Y$) $ X\not\equiv Y$ (something like complex number or other method)

I am sending attached a pdf file with an old paper by Viorel Bandila in which many distances between important points of the triangle are computed. The paper is in spanish, I had translated for my students, but there are only a few words and a lot of formulae.

Ok thanks but can you give me a link or something for some method

The paper shows applications of a theorem of Leibniz about the subject. Search in Google "Leibnitz theorem" and surely you will find it

If $M(x,y,z)$ and $N(u,v,w)$ are two points with the normalized barycentrical coordinates w.r.t. $\triangle ABC$ ,

then $\boxed{MN^2=-\left[(y-v)(z-w)\cdot a^2+(z-w)(x-u)\cdot b^2+(x-u)(y-v)\cdot c^2\right]}$ .


Diophantine equation
number theory proposed
number theory

Prove that there exist infinitely many pairs of integers $ (a;b)$ such that $ a^{2006}\plus{}1$ is divisible by $ b$ and $ b^{2006}\plus{}1$ is divisible by $ a$.

The fisrt we proof a litle problem
There exist infty (x,y) such that
$ y|x^{2}\plus{}1$ and $ x|y^{2}\plus{}1$
This solution use Lucas sequence
But $ x^{2}\plus{}1|x^{2006}\plus{}1 and y^{2}\plus{}1|y^{2006}\plus{}1$  so we have full solution

[quote="TTsphn"]The fisrt we proof a litle problem
There exist infty (x,y) such that
$ y|x^{2}\plus{}1$ and $ x|y^{2}\plus{}1$[/quote]
$ (x,y)\equal{}(1,1)$is such a pair .suppose that $ (x,y)$is such pair with $ x\le y$ and $ gcd(x,y)\equal{}1$ and consider $ (y,z)$ where $ y^{2}\plus{}1\equal{}xz$ then $ gcd(z,y)\equal{}1$.
$ x^{2}(z^{2}\plus{}1)\equal{}(y^{2}\plus{}1)^{2}\plus{}x^{2}\equal{}y^{4}\plus{}2y^{2}\plus{}(1\plus{}x^{2})$ is divisible by $ y$ since $ gcd(x,y)\equal{}1, z^{2}\plus{}1$ is divisible by $ y$ .so$ (y,z)$ is another such pair with $ y\le\frac{y^{2}}{x}<z$.this can be repeated infinitely many times.

An other solution
Easy to check that
$ xy|x^{2}\plus{}y^{2}\plus{}1$
Then
$ x^{2}\plus{}y^{2}\plus{}1 \equal{} 3xy$
Use the Pell equation we can find all solution of the equation.
An example
$ F_{2n\plus{}1}^{2}\plus{}F_{2n\plus{}1}^{2}\plus{}1 \equal{} 3F_{2n\plus{}1}F_{2n\minus{}1}$
Where $ F(n)$ is Fibonacci number.


In our school this year, there are 44 participants. What about your school?

I would say about thirty, probably 4-6 of which are likely to make AIME.  There were 30-something people taking it at my school last year, very few of which were 12s.

Nobody in my school's history has ever qualified for USAMO :(

Right around 70. Expecting 25 or so AIME qualifiers. Its impossible to tell whether or not anyone will make the USAMO - as only one former qualifier hasn't graduated yet.

At my old school, there are usually about 1300 to 1500 people who take the AMC, roughly 200-300 of whom move on to the AIME.  (Go Stuy!  Whee!)

in my school, all the kids get really high GPA's but only about 20 or so kids are taking the AMC's. And most of them are freshman like me. I assmue the AMC 10/12 are optional in high school, is that so? At my middle school, however, you had to take the AMC 8.

I don't think I've realized how bad my school is untill now.

I think there's only one person in both the junior high and high school combined (about 3,000 people) that will take the AMC 10.  I think you can all guess who that person is.

[quote="h_s_potter2002"]I don't think I've realized how bad my school is untill now.

I think there's only one person in both the junior high and high school combined (about 3,000 people) that will take the Mock AMC 10.  I think you can all guess who that person is.[/quote]

Well, considering that only about 20 to 30 ppl take *mock* amc's on aveage, that isn' bad  :D  :D .  You meant the AMC 10 not the mock AMC 10, right?

At my school we ordered 20 AMC 10's and 20 AMC 12's.  So we'll probably get a little less than 40 to take it.  I expect, o, about 5 to 8 AIME qualifiers.  As for USAMO, I have a little bit of a chance and if some of the other kids do better than they usually do, then they'll have decent shots also.

I go to a magnet school - everyone takes AMC A, math team takes AMC B also.

[quote="Zratonx"]more than 150 but only about 20-40 ish make AIME[/quote]

Only.  In my school more than 150 took the AMC, but only about 1ish made the AIME.

O wow AMC is more popular than I thought(at least according to the poll)...

Well about 5-7 people in our school will make AIME, maybe one of the seniors in our school will make USAMO  (I probably won't make it :( )

We had about 100 or so last year, but i think around 200 this year. last year about 15 made aime. usamo = 0 :(. this year its hoptefully at least 1 =P

You guys must go to pretty special schools. In my school district, with two middle schools and one high school, NO ONE takes the AMCs. That's even with me offering to pay for their registrations, just so that I can have a testing site for my homeschooling group. Yeesh. The latest word I've heard about our joint homeschooling group is that ten kids will take the 10B (including one AoPSer who is very likely to qualify for AIME), but only four kids will take any test on the "A" day. We are running one testing location on the A day, under one homeschooling group's name, and one location on the B, under a different group's name. Since my son is going to test on both dates (12A and 10B), this is going to royally goof up the AMC's system if he qualifies for AIME twice. (We haven't decided which location to use for the AIME; maybe we will poll all the qualifiers. We had three qualifiers last year.)

100+ taking AMC10.
Probably 120+ taking AMC12.

Last year we had 10 or so make the AIME.

No one in my school, it doesn't offer it.   Already too late for registration, I'll just try next year.

based on the AMC website, you can still register. AMC A registration is over but AMC B is still going on.
[url=http://www.unl.edu/amc/b-registration/registration.html]www.unl.edu/amc/b-registration/registration.html[/url]

we actually  have 300 ppl signed up this year... pretty amazing. but the showing up rate is usually only around 70% because 30% of the ppl a) don't know that they're signed up cuz teachers do it or b) don't bother coming or c) forget.

I have two others taking the test with me.  We are all hoping on making the second level.  And im hoping on making the third.
I got the school to register, so its in.  and the results are going to my dad's email address.

[quote]We usually have around 30 students who end up taking the AMC, with 7-8 moving on to AIME. We had one USAMO qualifier last year, who I believe was the first in the nation's history. This year I would guess that we'll have 0-3 qualifiers.[/quote]

You mean first in the school's history?

About 70-ish people taking the AMCs.
About 25 people will make it to the AIME.

The faculty supposedly chose the "Top 100" at our school to take it... I don't really believe in the accuracy of their selection.

My school has ~300 students; 30 take AMC
Last year, 3 qualified for AIME.
None made USAMO.

this is the first year my school's offering AMCs..
(very mathematically unenthusiastic school, unfortunately)
We only ordered 10 tests for each 10 and 12 (both A), so I guess 20 people will take it.
we have two people who got second place in the Mexican Math Olympiad, so they might make it into AIME.
As for me... probably not :(  but who knows what will happen.

My school has a large number of people taking AMC 10/12 - probably between 100-200, unfortuantely the school is only offering the A test. Last year about 25 made AIME, and no one has made USAMO yet.


[quote="boxedexe"]It seems that I haven't been able to solve this (I believe it is due to my small frontal lobe...always had trouble with language :P).  Here it is:

The supplement of an angle is 72 degrees less than twice the supplement of the complement of the angle.  Find the measure of the angle.[/quote]

[hide]
Supplement $= 180-x$
Complement $= 90-x$
Supplement of Complement $= 180-(90-x) = 90+x$

We have $180-x = 2(90+x)-72 \Rightarrow 72 = 3x \Rightarrow x = 24$.[/hide]

[hide]
Let the angle be $x$
$\pi - x=2(\pi -(\pi /2 - x)) -72$
$\Rightarrow \pi - x=\pi + 2x-72$
$\Rightarrow x=24$ degrees
[/hide]

Isthe above way of writing correct?
using $\pi$ radian and 72 degrees in the solution together.

What I want to say that should we use only one type of units in a solution or we can use both

Uuh, I find it really obvious that for any meaningful calculation you must use either radians or degrees, not both. Also, radians are unitless, which is why they are so good.


How many triples of positive integers a, b, c with a > b > c satisfy:  $a+b+c = 352$

Fix C at one.
Thus, we have a bunch of pairs (350 pairs) of a's and b's.
Fix C at two.
Nothing can use a one or a two, and our target is 350 with two numbers. There are 348 pairs.
all the way down:
Fix C at 116. We MUST have b = 117 and A = 119

Try and add up the pairs.

First, use balls and urns to find the number of permutations. Call this X. Find the number of (a,b,c) with a=b=c (0), then the number of (a,b,c) with a=b, b=c, or c=a. Divide this by 3. Call this Y.

Do X-3Y. Let this be Z. Divide this by 6, add the new number to Y to get the answer.

Sounds like a federal tax return form...

^ That too was my solution. Are there any other (easier) ways?  :)