factorial
induction
floor function
modular arithmetic
number theory

Prove that for every natural $ n$

$ n! | (2^n\minus{}2^0)(2^n\minus{}2)(2^n\minus{}2^2)\cdot \ldots \cdot (2^n\minus{}2^{n\minus{}1})$

This isn't quite HSB material.  The cleanest proof I know uses the fact that the RHS is the order of $ GL_n(\mathbb{F}_2)$ and that $ S_n$ is a subgroup.

Edit:  For what it's worth, the first time I saw this problem I computed the power of $ p$ dividing both sides for all $ p$, which wasn't too bad.  But really, who doesn't love $ GL_n(\mathbb{F}_2)$?  :)

[hide="The insufficient lacking non-Proof by Induction"]For $ n \equal{} 1$:

$ 1|2 \;\surd$

Hypothesis:

$ k!|(2^k \minus{} 2^0)(2^k \minus{} 2^1)\cdot \ldots \cdots (2^k \minus{} 2^{k \minus{} 1})$

$ \therefore (2^k \minus{} 2^0)(2^k \minus{} 2^1)\cdot \ldots \cdots (2^k \minus{} 2^{k \minus{} 1}) \equal{} C\cdot k!\;\surd$

Thesis:

$ (k \plus{} 1)!|(2^{k \plus{} 1} \minus{} 2^0)(2^{k \plus{} 1} \minus{} 2^1)\cdot \ldots\cdot (2^{k \plus{} 1} \minus{} 2^{k})(2^{k \plus{} 1} \minus{} 2^k)$

$ \therefore (k \plus{} 1)k!|2^k (2^{k \plus{} 1} \minus{} 2^0)(2^k \minus{} 2^0)(2^k \minus{} 2^1)\cdot \ldots\cdot (2^{k} \minus{} 2^{k \minus{} 2})(2^{k} \minus{} 2^{k \minus{} 1})$

$ \therefore (k \plus{} 1)k!|2^k(2^{k \plus{} 1} \minus{} 1)\cdot C\cdot k!\;\surd$

And the proof is complete, if I am not mistaken.[/hide]

[hide="But..."]
Your proof contains the unproven assumption that $ k\plus{}1|C \cdot 2^k(2^{k\plus{}1} \minus{} 1)$. Note that you need the $ C$, as $ k\plus{}1|2^k(2^{k\plus{}1} \minus{} 1)$ is not true for all $ k$.
[/hide]

True, I just proved that they have a common factor.

 I'll leave the work there in case someone wants to take it from there.

Thanks for the observation mano!

If it's not HSB material, mods please move it :)
[b]t0rajir0u[/b] i don't know what $ GL_n(\mathbb{F}_2)$ means...  :P

[hide="hint"]For all primes $ p>2$, if we choose any $ p$ contained in $ n!$, say in the term $ \ldots(kp-1)\cdot kp\cdot(kp+1)\ldots$, $ p$ is also contained in the term $ 2^{n-kp+k}(2^{k(p-1)}-1)=2^n-2^{n-kp+k}$.  If the term in the factorial contains $ p$ more than once, then so will the term, by Euler's Totient theorem.

If $ p=2$, then there are $ \frac{n(n-1)}$ instances of $ p$ in the $ 2^n$ product, because there are $ k$ 2's in $ 2^n-2^k$.  Then there are $ \lfloor{\frac{n}{2}}\rfloor+\lfloor{\frac{n}{4}}\rfloor+\ldots\< n$ 2's in the factorial.  This can only be more than $ \frac{n(n-1)}2$ if $ n<3$.  We have already checked from 1 to 3, and there are no exceptions.[/hide]

This is not a complete solution, because it fails to show that $ n-kp+k$ is always positive and less than $ n$, and other stuff...  I will leave this to the reader, and to show that this argument is in fact rigorous.

[hide="this is a pretty weak bound"] Call the right hand side $ f(n)$. Taking out factors of $ 2$ gives $ f(n) \equal{} 2^{n(n \minus{} 1)/2}(2^n \minus{} 1)(2^{n \minus{} 1} \minus{} 1)...(2 \minus{} 1)$. Denote the exponent of $ p$ in $ N$ as $ v_p(N)$. Note $ v_p(n) \equal{} \sum_{i \equal{} 0}^{\infty}\lfloor \frac {n}{p} \rfloor \leq \sum_{i \equal{} 0}^{\infty} \frac {n}{p} \equal{} \frac {n}{p \minus{} 1}$, which implies $ v_p(n) \leq \lfloor \frac {n}{p \minus{} 1} \rfloor$. Now, we will prove $ v_p(n!) \leq v_p(f(n))$ for all prime $ p$ after which we are done.

Case: $ p \equal{} 2$. We have $ v_2(n!) \leq n$ and $ v_2(f(n)) \equal{} \frac {n(n \minus{} 1)}{2}$. Then the claim holds true for $ n\geq 2$. For $ n \equal{} 1$ we have $ n! \equal{} 1$ which divides $ f(n) \equal{} 1$ and for $ n \equal{} 2$ we have $ n! \equal{} 2$ which divides $ f(n) \equal{} 6$.

Case: $ p$ is odd. We have $ v_p(n) \leq \lfloor \frac {n}{p \minus{} 1} \rfloor$. By Fermat's little theorem, $ 2^{p \minus{} 1}\equiv 1 \pmod p$. Thus, $ p|2^{k(p \minus{} 1)} \minus{} 1$. Then $ v_p(f(n))\geq \lfloor \frac {n}{p \minus{} 1} \rfloor \geq v_p(n!)$.

[/hide]


There are three allowable moves in the portion of the bee hive shown: from one cell to a cell directly to the right; from one cell to a bordering cell which is up and to the right; or from one cell to a bordering cell which is down and to the right.  How many distinct paths are there from cell A to cell B?

Blagh.. Casework sucks, is there a better way?

[hide="Answer"]
144
[/hide]

Think of block-walking.

and if you want to be more formal, its a recurrence.


Can anybody tell me what $ x\in (0,1)^2$ means?

I've never seen anything notated that way before... the closest thing I've seen is the following notation (which is fairly dissimilar, anyways)

$ (x,y) \in [0,1] \text{ x } [0,1]$
Which means an ordered pair $ (x,y)$ such that $ x \in [0,1] \text{ and } y \in [0,1]$

It probably means $ x$ is a vector $ (a, b)$ such that $ a \in (0, 1)$ and $ b \in (0, 1)$.

Yes, so to speak; this is known as the [url=http://en.wikipedia.org/wiki/Cartesian_product]Cartesian product[/url] of sets.

shudnt it be (x,y) belonging to (0,1)^2 ?? i mean i single variable cant be an element a AxB right?? has to be a ordered pair.

Yes, it must be a ordered pair, but probably the first poster wanted to concentrate on the $ (0,1)^2$ part.

[quote="Zed_darkx"] i mean i single variable cant be an element a AxB right??[/quote]
$ A \times B$ is a set like any other; $ x$ is an element of a set like any other.  Granted, this is somewhat confusing notation, but there's nothing formally wrong with it.  In other words, you don't have to specify a vector by its components.

It is perfectly legitimate to write $ x\equal{}(x_1,x_2)$. In fact, by writing $ x\in A\times B$, you are precisely specifying that $ x$ is an ordered pair.


Hi!

I've started a xanga that covers a few cool topics in number theory. It'll be updated once or twice a week, check it out!

http://www.xanga.com/learn_number_theory

Thanks!

bah. if someone actually wants to learn number theory, get a book from the library.

Gee... thanks for the encouragement buddy.

Instead of number theory, make me a xanga about polar.... :)


function
trigonometry
topology
advanced fields
advanced fields unsolved

Let $X,Y$ be metric spaces, with $Y$ complete, and let $A$ be a subset of $X$. Finally, let $f: A\to Y$ be a continuous map.

Prove that we can extend $f$ to a continuous map $g: B\to Y$, where $B$ is a $G_\delta$ subset of $X$ which contains $A$ and is contained in the closure $\overline A$ of $A$.

P.S.

This is somewhat related to the problem in [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=complete&t=103633]this[/url] topic, which, by the way, is [i]not[/i] solved correctly there :).

$B(x,r)$ is an open ball. For $n\ge 1$ let $U_{n}=\{x\in X\colon \exists r>0\ \mathrm{diam}\,(f(A\cap B(x,r)))<\frac{1}{n}\}$. Since each $U_{n}$ is open, the set $B=\overline{A}\cap\bigcap_{n}U_{n}$ is $G_{\delta}$. We can define $f(x)$ for $x\in B$ as the only element of $\bigcap_{r>0}\overline{f(A\cap B(x,r))}$. If $f$ is not continuous on $B$, then there exists $n\ge 1$ and a point $x\in B$ such that $\mathrm{diam}\,(f(B\cap B(x,r))>\frac{1}{n}$ for all $r>0$. But $\mathrm{diam}\,(f(B\cap B(x,r))=\mathrm{diam}\,(f(A\cap B(x,r))$, a contradiction.

Does this extension property characterize $G_{\delta}$ sets? I mean, if $A$ is $G_{\delta}$, does there exist a continuous $f\colon A\to \mathbb R$ that cannot be continuously extended to any larger subset of $\overline{A}$?

Yes, it does. Let $X$ be a metric space and let $A$ be a $G_\delta$ set. Replacing $X$ by $\overline A$, we see that we can always assume that $A$ is dense in $X$. We can represent $X\setminus A$ as a countable union of an increasing sequence of closed sets $F_{k}$. Suppose that, for each $k$, we can construct a function
$f_{k}: X\setminus F_{k}\to \mathbb R$ such that $|f_{k}|\le 1$, $f$ is continuous on $A$, and for all $y\in F_{k}$, we have $\liminf_{x\in A,x\to y}f_{k}=-1$, $\limsup_{x\in A,x\to y}f_{k}=1$. Then $f=\sum_{k}3^{-k}f_{k}$ is the function with the desired property. Now, how to construct $f_{k}$.

We'll drop the index $k$ everywhere. Choose a sequence $\delta_{j}\to 0+$. Take $\delta_{1}$ and consider the maximal $\delta_{1}$-separated subset $S_{1}$ of $A\cap F(\delta_{1})$ where $F_\delta$ is  the $\delta$ neighborhood of $F$. It is easy to see that $S_{1}$ is a $\delta_{1}$-net in $F(\delta_{1})$ and $A\setminus S_{1}$ still has every point of $F$ as its accumulation point. Now choose the maximal $\delta_{2}$-separated set $S_{2}$ in $(A\setminus S_{1})\cap F(\delta_{2})$ and so on. Let $S=\cup_{j}S_{j}$. It is easy to see that each point of $X\setminus F$ has a neighborhood containing only finitely many points of $S$ and each point of $F$ is an accumulation point of $S$ and, moreover, for each point $y\in F$ and each $d>0$, the ball $B(y,d)$ intersects all but finitely many $S_{j}$. This allows us to construct a continuous function $f : X\setminus F\to \mathbb R$ such that $f=(-1)^{j}$ on $S_{j}$. The rest should be clear.

One question:

if $X=Y=R, A=(0,1), f=sin\frac{1}{x}sin\frac{1}{1-x}$, How to extend this map?

[b]Victor96[/b]: The claim is that $f$ can be extended to [b]some[/b] $G_{\delta}$ set, not to any $G_{\delta}$ set. If $A$ is itself $G_{\delta}$, as in your example, the statement is vacuous.

[b]fedja[/b]: Thanks for the explanation. I suppose $f_{k}(x)=\sin(\mathrm{dist}(x,F_{k})^{-1})$ would work, too? (OK, now I see that it wouldn't.) (But now I think that it would. :cool: )

Oh... I see, the $G_\delta$ set is not given.


trigonometry geometry calculus derivative Stanford college probability

has anyone taken algebra II? should i post these probs?

Well...some people might've skipped a couple of grades, so why not?

i know chinaboi and me are, but let me first do round 1

So so far three people skipped to 10th, correct?

...that might not be quite enough...oh well.

whats the difference in precalc and trig? are they the same thing

On a similar note, can anyone explain the difference between Algebra 1, Algebra 2, Geometry (several different topics from congruence to Trig) and Precalculus? (also the corresponding grades?)

Algebra I, Algebra II, and Precalc are basically 3 algebra courses ranging from very simple to pretty simple.

Geometry should be a course on geometry, but it's taught really badly, so for most people, it's a horrific course on memorizing random formulas.

in the New York school system (honors) for a regular public school Algebra 1 is 8th grade, Algebra 2 and Geometry are kind of mixed together into 9th and 10th grade, and precalculus is offered sometimes in 11th grade but other schools choose to go straight to AP. Precalculus (in my school anyhow) is really calculus in disguise, that's why most other schools (around here) don't bother with it. Because my school thinks we're all stupid, they believe we need one whole year on the differentiation stuff and one whole year on integration.

Of course, the NYS system does a horrible, boring, useless job of it all, so some of us choose to skip a whole lot of stuff, so the grade divisions kind of don't exist.

So I guess Massachusetts is the same as New York, probably. Fiertycoon and TheSouthpaw?

Here, the levels are as follows:

Algebra I in 8th grade
Algebra II in 9th
Geometry in 10th
Precalculus in 11th
and Calculus AB/BC in 12th.

What grade are you in, Tare? 8th?

yup, yet 10th. I think we even do a bit of Calculus AB in 11th Honors here, but I'm not sure.

My school:

7 - algebra I
8 - something weird
9 - geometry
10 - algebra II
11 - precalc
12 - calc AB/BC
discrete math

You can't skip subjects in out school :(, but many people take two maths their freshman year, so then everything would be bumped down a grade.

my school...

8-Algebra 1
9-Algebra 2
10-Geometry
11-Pre Calc (or Calc if you test out of precalc)
12-Nothing, only 4 math courses are required: Algebras 1&2, Geometry and calc/precalc

So Calculus isn't mandatory? (If you just take normal math classes, would you still learn Calculus?)

[quote="Chinaboy"]I think in our school its

7th - alg.I
8th - geom
9th - alg. II
10 - pre-calc / trig... not sure...
11- Calc. AB
12- Calc BC...
correct me if I'm wrong Neal.[/quote]

Yep, you're definately wrong.  You can't take both Calc AB and BC as BC includes all of the AB curiculum and then adds some.  BC is just a more advanced version of AB.  Since my school only offers AB, I took that and am taking Calc C online from Stanford EPGY.

[color=#1FE0FF]Actually, you guys are wrong about the calculus thing -- many schools have AB calculus and then BC calculus in sequence -- in fact, in some places it is the only way you can do it.  (This information is direct from two of my roommates, one from Ohio and the other from near Los Angeles.  The school which requires you to take AB before BC is a private school in his area.)

New York has what is known as an "integrated mathematics" curriculum.  That is, they don't break it down into algebra, geometry and algebra 2 but instead teach the information usually contained in those classes over a period of 3 years, but not in the same order necessarily.  To get a state high school diploma, you must pass exams in those subjects.  I believe that after that, it's up to the individual school district what happens next.  In Stuyvesant, for example, precalculus was always next, ususally followed by AP calculus (either AB or BC), although some students took an easier precalculus-like course and non-AP calculus.  I was actually a bit surprised to hear what Jen said about how they do it where she lives.   Where do you actually live, by the way?  Anyhow, not all students take the first course in the sequence (what would be algebra 1 elsewhere) in eigth grade, because at least in NYC not all middle schools offer it, although I think all ofthe honors-type programs do.[/color]

Well, it would be stupid to take both AB and BC as BC includes all of AB.

As for the EPGY, I cannot say much about it yet as I have yet to take a course.  However, I can give you the site: epgy.stanford.edu.  What courses did you take A+?  I'm taking Calc C.

correct, calculus isn't mandatory. I don't know why...

My school is strange. Honors students used to take Algebra II 9th, then Precalc, then Calc AB or BC. Senior year students could just not take a class, take Stat AP (though you dont have to be a senior for it) or they could go free to the local college and do Multivariate Calculus. Then my school district decided it was too expensive to pay for students going to the college. So they took out the trig from A2 and some of the precalc stuff and made a whole extra year from it called Trigonometry and Discrete Mathematics. Discrete involves some very very basic probability and combinatorics. It was really quite an annyoing class.

Hmm...

Um, so are you going to post the questions?
If not I'll just change the topic name to "Calculus and Stanford EGPY" and send it over to Other Problem Solving Topics.

[quote="DaSweepa"]As for the EPGY, I cannot say much about it yet as I have yet to take a course.  However, I can give you the site: epgy.stanford.edu.  What courses did you take A+?  I'm taking Calc C.[/quote]

  :mrgreen:

I did grades 5 and 6, prealgebra and beginning algebra.

In my school, a really smart kid takes AP Calc BC in his 10th grade. I take Algebra II Honors, but I will skip to AP Calc AB next year as a sophomore.

Mmm, same here. I plan to finish high school math by the end of my sophomore year, if I am able to stay here until that time. At our school: (All these courses are honor courses)
9th grade: Basic Geometric Proofs and some Algebra
10th grade: Precalculus (everything but calculus, includes some overlap with the 9th grade curriculum depending on the teacher)
11th grade: Calculus AB
12th grade: Calculus BC

anything below high school level is just the repetition of basic junk for [b]8 years[/b], talk about a waste of time :)


modular arithmetic
number theory
relatively prime

[url=http://www.artofproblemsolving.com/Wiki/index.php/Wilson%27s_Theorem]Proof of Wilson's Theorem on AoPS Wiki[/url]

"However, if $ {p}$ is prime, then each of the above integers are [[relatively prime]] to $ {p}$. So, for each of these integers a, there is another $ b$ such that $ ab \equiv 1 \pmod{p}$."

How do we prove the above statement (specifically how do we know $ b\in\{1,2,3,...,p\minus{}1\}$?)

Claim: Consider $ ax \pmod p$, where $ a$ is an integer relatively prime to $ p$. Each value of $ x$ from $ 0$ to $ p\minus{}1$ gives a unique value of $ ax \pmod p$.
Proof: Assume there are two distinct numbers $ r$ and $ s$ between $ 0$ and $ p\minus{}1$,inclusive such that $ ar\equiv as \pmod{p}$. Then we would have $ a(r\minus{}s)\equiv 0 \pmod p$. Since $ a$ is relatively prime to $ p$, this would mean that $ r\minus{}s\equiv 0 \iff r\equiv s$, a contradiction. So all the values of $ x$ give a unique value of $ ax$, and in particular, this implies that each number has a unique multiplicative inverse.

existence of modular multiplicative inverses - note that $ ax\equiv 1 \pmod{p}$ implies $ ax\minus{}1\equal{}kp$ for some $ k$, that is $ ax\plus{}(\minus{}k)p \equal{} 1$.  For fixed $ a$ and $ p$, this has solutions as long as $ gcd(a,p)\equal{}1$, and a solution can be found via the euclidean algorithm.

For uniqueness: note that if $ x_1$ and $ x_2$ are both inverse of $ a$, we have $ x_1 \equiv (x_2a)x_1 \equiv x_2(ax_1) \equiv x_2 \pmod{p}$.  Or alternatively we can note that if $ (x_0,k_0)$ is a solution to the above equation, then all solutions can be expressed as $ (x_0\plus{}np,k_0\plus{}na)$ for integer $ n$ [since $ lcm(a,p)\equal{}ap$].  And these solutions for $ x$ are all equivalent in mod $ p$.


Consider the concentration, C, (in mg/liter) of a drug in the blood as a function of the amount of drug given, x, and the time since the injection, t. For $ 0 \leq x \leq 4$ mg and $ t \geq 0$ hours, we have

$ C \equal{} f(x,t) \equal{} t e^{ \minus{} t(5 \minus{} x)}$

Find f(3,2). Give an exact answer.

Okay, I tried

$ 2e^{ \minus{} 4}$ and $ \frac{2}{e^4}$ and both doesn't give me the correct answer... what am I doing wrong here?

Nothing. Your answer in either form is perfectly fine.

well.. I am using this web based online assignment thing and it always gives me a wrong result when I input those... any ideas?

Mistakes happen. Reporting mistakes can get them fixed.


function
limit
LaTeX
real analysis
real analysis unsolved

Let $ f: \mathbb{R} \rightarrow \mathbb{R}$ be a function with the property of Darboux. If the function $ g: \mathbb{R} \rightarrow \mathbb{R}, g(x)\equal{}f^2(x)\minus{}2f(x),$ has a limit in each point of it, then prove that $ f$ is continuous on $ \mathbb{R}$.

Ok.It's time to give a sketch of the  solution :) . Fix point $ x_0$ and suppose that $ \lim \limits {x\rightarrow x_0} g(x) \equal{} a$. If there does not exist $ \lim \limits {x\rightarrow x_0} f(x)$ then we choose two different sequences $ x_n$ and $ y_n$ which converge to $ x_0$, such that $ \lim \limits {y_n\rightarrow x_0} f(x) \equal{} b$ and
$ \lim \limits {x_n\rightarrow x_0} \equal{} c$ and $ b^2 \minus{} 2b \equal{} c^2 \minus{} 2c \equal{} a$ and if $ b\ne c$ applying Darboux condition we get a contadiction ( because we can take third sequence $ z_n$ such that $ \lim \limits {z_n\rightarrow x_0}$ $ f(x) \equal{} \frac {b \plus{} c}{2}$). So for every $ x_0\in\mathbb {R}$ there exist  $ \lim \limits {x\rightarrow x_0} f(x)$. Suppose that for some $ t\in\mathbb{R}$   $ \lim \limits {x\rightarrow t} f(x) \equal{} a > f(t)$. Then applying Darbou condition it's easy to construct a sequence $ t_n$ for which $ t_n\rightarrow t$ and $ f(t_n)\rightarrow \frac {f(t) \plus{} a}{2}$.

[quote="leshik"] If there does not exist $ \lim \limits {x\rightarrow x_0} f(x)$ then we choose two different sequences $ x_n$ and $ y_n$ which converge to $ x_0$, such that $ \lim \limits {y_n\rightarrow x_0} f(x) \equal{} b$ and
$ \lim \limits {x_n\rightarrow x_0} f(x)\equal{} c$ [/quote]
Why should this be true?

Latex tips: Use the following for limit:  \lim_ {y_n\to x_0} f(x)

Because there exist two sequences for which $ \lim_{y_n \to x_0}f(x)= \underline{\lim}\limits_{x\to x_0}f(x)=b$ and $ \lim_{x_n \to x_0}f(x)= \overline{\lim}\limits_{x\to x_0}f(x)=c$

[quote="leshik"] and if $ b\ne c$ applying Darboux condition we get a contadiction ( because we can take third sequence $ z_n$ such that $ \lim \limits {z_n\rightarrow x_0}$ $ f(x) \equal{} \frac {b \plus{} c}{2}$). So for every $ x_0\in\mathbb {R}$ there exist  $ \lim \limits {x\rightarrow x_0} f(x)$. [/quote]
Darboux property says only the following(not the limiting thing):
If $ f(a) < z < f(b)$, then there exists $ c$ s.t. $ f(c) \equal{} z$
Even if it so how can you conclude that $ \lim_ {x\to x_0} f(x)$ exists? because you have just two sequences $ x_n$ and $ y_n$ s.t. $ \lim_ {x_n\to x_0} f(x) \equal{} \lim_ {y_n\to x_0} f(x)$

Sorry if I m mistaken!

About your second question- if $ \underline{\lim}_{x\to x_0} f(x)\equal{}\overline {\lim}_{x\to x_o} f(x)$ then there exists $ \lim_{x\to x_0} f(x)$- this is standart fact from Calculus. :)  About Darboux property- yes you are right, but you can easily aplply this formulation to get appropriate sequence $ z_n$ :)

[quote="leshik"]About Darboux property- yes you are right, but you can easily aplply this formulation to get appropriate sequence $ z_n$ :)[/quote]

I'm not convinced.

Ok.I'll try to explain. Let us suppose that we have two sequences $ x_n$ , $ y_n$ s.t $ lim_{x_n\to x_o} f(x)\equal{}b$ and  $ lim_{y_n\to x_o} f(x)\equal{}c$ and $ b>c$.
Passing to the subsequences we can suppose that $ f(x_n)>b\minus{}\frac{b\minus{}c}{10}$ and $ f(y_n)<c\plus{}\frac{b\minus{}c}{10}$.Then because $ f(y_n)<c\plus{}\frac{b\minus{}c}{10}<\frac{b\plus{}c}{2}<f(x_n)$ we can choose $ z_n$ which lies on the segment with endpoints $ x_n,y_n$ for which $ f(z_n)\equal{}\frac{b\plus{}c}{2}$ (this follows from Darboux condition.) But then the sequence $ z_n$ tends to $ x_o$ and we are done :)


complex numbers
number theory unsolved
number theory

Consider sequence  $ a_{1},a_{2},..... \in N$ such that : $ a_{k+1}=a_{k-1}a_{k}+1 \forall k>1$
Prove that $ \forall k>9$ we have $ a_{k}-22$ is never a prime number.

[quote="Nbach"]Consider sequence  $ a_{1},a_{2},..... \in N$ such that : $ a_{k+1}=a_{k-1}a_{k}+1 \forall k>1$. Prove that $ \forall k>9$ we have $ a_{k}-22$ is never a prime number.[/quote]
Are the terms of the sequence $ a_{1}, a_{2}, \ldots$ supposed to be (possibly positive) integers? Rationals? Reals? Complex numbers? Hamilton's quaternions? What!? :maybe:

Ok,sorry ,indeed, $ a_{i}\in N$ $ \forall i \in N$


inequalities
geometric inequality
inequalities proposed

[color=purple]Let a triangle ABC with the side length a,b,c and the side of three median $m_{a},m_{b},m_{c}$.Prove that
  $\frac{a}{m_{b}m_{c}}+\frac{b}{m_{a}m_{c}}+\frac{c}{m_{b}m_{a}}\geq\frac{4(a+b+c)}{a^{2}+b^{2}+c^{2}}$
        Have fun !  :D [/color]

[color=purple]I have a nice solution  :blush: but it's a algebraic proff. :( If you have geometric inequality please post it  :D [/color]


How many ways can change for $ \$16.00$ be made using quarters and dimes if at least one of each type of coin must be used?

There must be an even number of quarters.
Start from 0 and go all the way up to 64.
This gives the sequence: 0, 2, 4, 6, ..... 60, 62, 64.
The number of terms is 31.

mathyoga, if the sequence is 0-64, then there would be 33 even terms. (64-0)/2 + 1 = 33, not 31

Well there's at least one of each type of coin. So, we have to subtract off the cases with only quarters and only dimes (2 cases). 33 - 2 = 31


> Three questions, though. (1) Will there be more detailed scoring (as in names, not just
> numbers) on the AMC site? (2) Will our papers be returned to us? (3) If Question 2 is true, will > our papers have comments on them, so that we can see our errors and improve next year?

In response to these and many similar questions:
(1) No we will not post names and scores together on the AMC site:  Basically the reason is this: The Family Educational Right to Privacy Act, (FERPA) also called the Buckeley Amendment, passed by Congress in 1984 forbids most educational institutions from releasing grades to anyone other than the grade recipient himself/herself unless authoized in writing to do so (for instance for a reference check, for a transcript, etc.)  As an educator, I have found that this law is basically a sound and reasonable policy.  The AMC is not an educational institution and is probably not bound by the law, but it makes a good foundation on which to base a policy.    Furthermore, it is a policy and a law which is familiar to most of the people working on our committees.  So what you see on the AMC website is what you get.  (We do bend the policy a bit in the AMC 10/AMC 12 Summary of Results where we list student names and Honor Roll scores,  that is by popular demand from teachers and has fostered no complaints yet.)

In fact, up until a few years ago, the only information that was returned was a SCORE INTERVAL.  That is, you only found out that your score was in a certain range, not what your exact score was.  When we started posting scores (anonymously by Student ID number) on the web, we did away with the intervals, and reported the scores.

(2) No.  You do have your originals (most of you anyway) since you faxed us a copy, and we worked from the faxed copies.  I checked with the long-time office staff and we have never,ever returned USAMO papers  Furthermore, we posted the solutions on our site on Thursday after the USAMO.   In years past, we sent a copy to the teacher weeks later.

(3) The comments on the papers (if any, and there are very few) would be useless to you.  I was present at the grading,  I had the other graders there in the same room with me, and it  takes detailed explanation and work to intepret a mark on a paper.  I suggest you compare your solution to the official solution and try to understand the differences and similarities.  It is a long-standing policy of the AMC that decisions of USAMO graders are final and unequivoal.

In addition, each paper is read twice and independently and  the two opinions have to be reconciled against a standard rubric and the graders must agree before a score is assigned.  Then all papers in the top 25% are read again and  the second reading must agree with the standard rubric.   Believe me, your score is correctly and fairly assigned.  Incidentally, all grading is blind grading, all we have is the student number, nothing else.

Steve Dunbar
AMC Director

Excellent piece of explanation.Hats off to you and your committe to have done such a wonderful job.

The description of IMO grading in Steve Olson's book Count Down is quite interesting, and moderately relevant to the discussion here.

Yeah, first-timers are usually surprised by how they do. By this I mean that learners taking their first math course demanding rigorous proofs are usually STUNNED at what low grades they get on their proofs at first. Learning proof writing takes plenty of practice, including plenty of reading good published proofs.

Best wishes to all of this year's USAMO qualifiers who will be coming back next year to learn a lot about proof-writing in the next year.

Many thanks to Professor Dunbar for his great work in administering the AMC contests and putting up with the emotional reactions of teenage math-learners.

[quote="pieterminate"][quote="tokenadult"]
Yeah, first-timers are usually surprised by how they do. By this I mean that learners taking their first math course demanding rigorous proofs are usually STUNNED at what low grades they get on their proofs at first. Learning proof writing takes plenty of practice, including plenty of reading good published proofs.
[/quote]

What?  Is there a prejudice against first-timers?  :lol:  :roll:[/quote]

Yeah and the winter comes because it sees the leaves turn red :):)

Check the statement in Item 9, page 14 of the USAMO Teachers' Manual, the S-II Form, "Instructions to USAMO Participants"

"The USAMO is graded more strictly than most school examinations. so it is typical for students to overestimate the scores they will receive."

That statement has been on the form as long as I can remember.

The grading is truly blind.  With that many papers to look at, there is no time to analyze handwriting,  we focus on the mathematics, read for correct logic, and move to the next paper.

Steve Dunbar
AMC Director

[quote="MysticTerminator"][quote="AMCDirector"]Incidentally, all grading is blind grading, all we have is the student number, nothing else.

Steve Dunbar
AMC Director[/quote]

hmm . . . what if someone, hypothetically, writes really small? Would that be recognizable by the USAMO graders? (many of whom were MOP graders as well, I think)

seriously though, thanks for the explanation - I had no idea the papers were graded so meticulously (by $\ge$ three graders!)[/quote]

Ah yes, I remember your tiny handwriting. I could probably still recognize it.

[quote="Ravi12346"]Will we find out our scores for individual problems?[/quote]
As of current policy, no. We do not get any information about our actual test, except the final score.


Countdown Champ: Daesun Yim, NJ

Written Champ: Neal Wu, LA

Team Champ: Virginia

Congratulations!

I live in VA  :lol: Congratulations to all teams and to Daniel Li.

Yay Andrew Ardito!!! He's from my state (he won the countdown at my state of course), which is New York.

Could someone make a listing of the written ranking and scores of AoPSers and their countdown ranking if they made it to that?

Oh it's today?

I thought it was tomorrow :?

Oh well, CONGRATULATIONS

Did michigan or any michigan people make it?

Nope (pretty amazing huh?)

The top 12 written were:

1. Neal Wu
2. Daniel Li
3. Kevin Chen (aopser)
[b]4. Nathan Benjamin (wow surprising)[/b]
5. Daesun Yim (also aopser - treething)
6. Sam Keller (samath)
7. Brian Hamrick (hamster1800)
8. George Silvas
9. Rolland Wu
[b]10. Andrew Ardito (WOW)[/b]
11. George Yu
12. someone i forget who

Countdown results:
1. Daesun Yim
2. Andrew Ardito
[b]S. Neal Wu (lost to daesun)[/b]
S. Kevin Chen
Q. Nathan Benjamin
Q. Sam Keller
Q. Daniel Li
Q. George Silvas
P. Rolland Wu
P. George Yu
P. Brian Hamrick
P. Other guy

I probably messed up a couple written, but I am sure about countdown.

Teams:
1. Virginia
2. Washington
3. Indiana
4. Texas
5. Illinois
6. Massachusetts
7. Maryland
8. Pennsylvania
9. North Carolina
[b]10. California[/b]
Probably screwed up a few, but CA WAS 10th and the top 5 are right

Masters-still waiting

The test was hard, about 2004 nats level. I do know that there were a gazillion ties, and neal wu got 41, which was the highest.

#12 was 38

I got 36 or 35 depending on whether I got one of them right (I forget)

The team round was weird because there were 8 easy ones (everyone got them), but nearly all teams got #4 and #10 wrong. I dont think anyone got 10, but I know that VA got 1-9

Wow, this was really unexpected, all the more so because I thought the competition was next week or something! I'm pretty surprised that with all the hubbub about Michigan winning, they weren't even in the top 10 and had nobody in the countdown. And we still don't have a 2 time national champion!

[quote="rrusczyk"]Countdown runner-up: Andrew Ardito.

Written test runner-up: Daniel Li.

States 2nd and 3rd: Washington, Indiana.[/quote]

Congratulations Team Washington!

[quote="Drunken_Math"]Kevin Chen?

When I used to live in California (in 5th grade :P) I used to have a best friend called Kevin Chen.

Is he from California? o.O[/quote]
Kevin YANG is from California, I believe.

[quote="Ignite168"][quote="pgpatel"][quote="biffanddoc"]The person you really have to feel bad for is Eric McCabe. As 13th last year (omg that sucks), he was guaranteed countdown..... and didnt make it. If he got 13th, then if I were him I would shoot myself.

Also, I thought that countdown was really slow this year. Neal Wu wasnt that impressive at all, and I could answer most of the questions before the competitiors... especially the first 2...

And as a last note, they thought my girl teammate was a guy (roomed her with another guy) and both M and F were checked under gender for her........[/quote]
That's SO not true biff.
I was sitting next to you during countdown and you [b]DIDN'T[/b] answer most questions faster.
P.S. Why did you get banned treething?[/quote]

He posted a picture of something and said it was one of the admins.
If you're talking about questions where after one person got it wrong you got it faster than the other person, then you cannot compare. They are using their time to check the problem. I only got 3 or 4 questions faster than both of the competitors. (such as, what is the sum of the first 40 positive odd integers ;)) Eric McCabe got 22nd I believe.[/quote] What was it a picture of?

wow, lauren got last in the 14-way tiebreaker for a score of 30

[quote="Fanatic"]Indiana:

Nathan = 4th, 39 points.
John = 13th, 36 points (ouch, 13th)
Sachin = 39th, 31 points
Lauren = 57th, 30 points (16/14 split).

39th-57th place = At least 19 people separated by one point.[/quote]
I was 56th with a 16/14 split as well. 14 people got a score of 30. 58th place is walking away with a laptop, $\$$2000, and a lot more with a 21/8 split. I had 2 right in the last 10.

[quote="Fanatic"]Indiana:

Nathan = 4th, 39 points.
John = 13th, 36 points (ouch, 13th)
Sachin = 39th, 31 points
Lauren = 57th, 30 points (16/14 split).

39th-57th place = At least 19 people separated by one point.[/quote]
Make that 22. 36th had a 31 as well (23/4!).

Congratulations to all the Virginia team members and their coaches.  You guys did a great job at nationals and desreve all your kudos!!!

Also, congratulations to the Texas team for their 4th place showing.  4th in the Nation is a great job.  You should be proud of what you have done this year.
Matt, Kevin, Kevin and Pulak...thanks for all your effort in preparing for Nationals.

A special congratulations to Kevin Chen for his 3rd place finish in the nation as a 7th grader.  What an improvement...29th as a 6th grader and 3rd as a 7th grader...WOW!!

Jeff Boyd
Texas Coach

[quote="Phelpedo"]A woman named MCrawford :D  But let's leave that in the past.[/quote]

Huh?

[quote="Treething"]no... no... no... it's all in the past, let's be glad EVERYBODY HAS FORGOTTEN ABOUT IT AND WILL NOT TALK ABOUT IT ANYMORE.[/quote]

It doesn't sound at all like what happened, which is why I asked.

I think it's one of the older minor crimes I commited :p . Either way I'm not commiting any more crimes.

does anyone have an idea how New York came out?

I think they got in the 10th-15th place region.


rotation
integration
projective geometry
calculus
calculus computations

The region in the first quadrant bounded by the graphs of $y=x$ and $y=\frac{x^{2}}{2}$ is rotated around the line $y=x$. Find the centroid of the region and the volume of the solid of revolution.

[quote="liyi"]What is Pappus-Guirdin's theorem?[/quote]

[url=http://mathworld.wolfram.com/PappussCentroidTheorem.html]Pappus's Theorem[/url]

that's two of them anyhow...

[quote="kunny"]The volume is $\frac{\pi}{\sqrt{2}}\int_{0}^{2}\left(x-\frac{x^{2}}{2}\right)^{2}dx=\frac{2\sqrt{2}}{15}\pi.$

For the centroid, may be you are about to use Pappus-Guirdin's theorem,
This theorem is prohibitted in Japanese university entrance exam. :D

I will post the problems like this one later. :)[/quote]

Here is the more generalised problem.

Please take a look. [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=109728[/url].


calculus
integration
trigonometry
algebra
linear equation
calculus computations

Is there a reduction formula for cosec? This is for integration of $ cosec^3 x$.

hello
[hide="solution"]
$ I_n \equal{} \int cosec^{n}x dx$
      =$ \int cosec^{n \minus{} 2}cosec^2xdx$                      use integration by parts I m skipping some steps here.
      =$ \minus{} cosec^{n \minus{} 2}xcotx \minus{} \int(n \minus{} 2)cosec^{n \minus{} 2}x(cosec^2x \minus{} 1)dx$
      =$ \minus{} cosec^{n \minus{} 2}xcotx \minus{} (n \minus{} 2)\int(cosec^{n}x \minus{} cosec^{n \minus{} 2})dx$
      =$ \minus{} cosec^{n \minus{} 2}xcotx \minus{} (n \minus{} 2)I_n \plus{} (n \minus{} 2)I_{n \minus{} 2}$

$ (n \minus{} 1)I_n \equal{} \minus{} cosec^{n \minus{} 2}xcotx \plus{} (n \minus{} 2)I_{n \minus{} 2}$
$ I_n \equal{} \minus{} \frac {cosec^{n \minus{} 2}xcotx }{n \minus{} 1} \plus{} \frac {n \minus{} 2}{n \minus{} 1}I_{n \minus{} 2}$

  $ \int cosec^{n}xdx \equal{} \minus{} \frac {cosec^{n \minus{} 2}xcotx }{n \minus{} 1} \plus{} \frac {n \minus{} 2}{n \minus{} 1}\int cosec^{n \minus{} 2}xdx$

put  n=3 to find ur answer.

[/hide]
Thank u.

I just happen to have some edition or another of Stewart's calculus book lying near my computer. He puts a table of integrals in the back end papers, and there's a section just for reduction formulas:

73. $ \int\sin^nu\,du\equal{}\minus{}\frac1n\sin^{n\minus{}1}u\cos u\plus{}\frac{n\minus{}1}n\int\sin^{n\minus{}2}u\,du$

74. $ \int\cos^nu\,du\equal{}\frac1n\cos^{n\minus{}1}u\sin u\plus{}\frac{n\minus{}1}n\int\cos^{n\minus{}2}u\,du$

75. $ \int\tan^nu\,du\equal{}\frac1{n\minus{}1}\tan^{n\minus{}1}u\minus{} \int\tan^{n\minus{}2}u\,du$

76. $ \int\cot^nu\,du\equal{}\frac{\minus{}1}{n\minus{}1}\cot^{n\minus{}1}u\minus{} \int\cot^{n\minus{}2}u\,du$

77. $ \int\sec^nu\,du\equal{}\frac1{n\minus{}1}\tan u\sec^{n\minus{}2}u\plus{}\frac{n\minus{}2}{n\minus{}1}\int\sec^{n\minus{}2}u\,du$

78. $ \int\csc^nu\,du\equal{}\frac{\minus{}1}{n\minus{}1}\cot u\csc^{n\minus{}2}u\plus{}\frac{n\minus{}2}{n\minus{}1}\int\csc^{n\minus{}2}u\,du$

An formula 72 is exactly what you were looking for - the integral of $ \csc^3u.$

kabi: my only comment is that the name of the technique is [b]integration by parts,[/b] not partial fractions. All of these formulas can be re-derived without too much trouble; the derivation always has an integration by parts, a Pythagorean identity, and a step in which a linear equation is solved for the unknown integral. Remember that idea and you can recover them.

Hello Kent
Sorry, by mistake I wrote it partial fraction.I know that's integration by parts.I have edited that.
Thank u.

Hello
There are some more reduction formulas ( which I have derived during my Integration classes ) besides the basic ones given by Kent.
1.$ \int sin^{m}x cos^{n}x dx\equal{}\frac{sin^{m\plus{}1}xcos^{n\plus{}1}x}{m\plus{}n} \plus{} \frac{n\minus{}1}{m\plus{}n}\int sin^{m}x cos^{n\minus{}2}x dx$
2.$ \int cos^{m}x sinnx dx \equal{}\minus{}\frac{cos^{m}xcos nx}{m\plus{}n} \plus{}\frac{m}{m\plus{}n}\int cos^{m\minus{}1}xsin(n\minus{}1)x dx$
3$ \int cos^{m}x cos nx dx \equal{} \frac {cos^{m}x sin nx}{m\plus{}n} \plus{} \frac{m}{m\plus{}n}\int cos^{m\minus{}1}x cos(n\minus{}1)x dx.$
4.$ \int \frac{sin^{n}x}{cos^{m}x}\equal{}\frac{1}{m\minus{}1}\frac{sin^{n\minus{}1}x}{cos^{m\minus{}1}x} \minus{} \frac{n\minus{}1}{m\minus{}1}\int \frac{sin^{n\minus{}2}x}{cos^{m\minus{}2}}dx$

Thank u.


limit
superior algebra
superior algebra unsolved

Does the CRT still hold for infinitely many pairwise coprime ideals in a commutative ring?

E.g. there is no integer $ n$ such that $ n \equiv 1 \mod 2$ and $ n \equiv 0 \mod p$ for all other primes $ p$.

Aight. The exam question was. Given a set of primes and a map $ \pi: P \to \bf{Z}$ find a sequence $ a_n \in \bf{Z}$ such that $ a_n$ converges to $ \pi(p)$ in $ \bf{Z}_p$ (p-adic integers) for every $ p$. So I wanted to use something like CRT, but noticed a problem when $ P$ was infinite. Any other suggestions?

We can even make it a bit more general: for every prime $ p$ let $ c_p \in \mathbb Z_p$. Then there is a sequence $ a_n \in \mathbb Z$ converging in $ \mathbb Z_p$ to $ c_p$ for all primes $ p$.

Proof:
Let the primes be $ p_1, p_2, ...$ (in any ordering).
Take $ a_n$ such that $ a_n \equiv c_p \mod p_i^{n\minus{}i}$ for all $ i\equal{}1,2,...,n\minus{}1$ (exists by the CRT, the number of conditions is finite).

Then $ a_{n\plus{}k} \equiv c_{p_k} \mod p_k^n$, thus with $ n$ growing, the sequence $ a_{n\plus{}k}$ converges to $ c_{p_k}$ in $ \mathbb Z_{p_k}$. Thus the same holds for the sequence $ a_n$ itself. As this $ p_k$ was arbitrary, the result follows.

Note that this results says nothing else than $ \hat{\mathbb Z} \equal{} \prod_{p \text{ prime}} \mathbb Z_p \equal{} \lim_{\substack{\leftarrow \\ n}} \mathbb Z / n \mathbb Z$ is the completion of $ \mathbb Z$.


inequalities
LaTeX
inequalities unsolved

Jokeccil I hope you won't mind, but I have edited your messege in Latex, because instead of $\in$ there was a nice square! :D :D


probability
limit
function
probability and stats

Given $ n$ balls and $ n$ boxes. We put each ball in each box arbiterary, that is to say, suppose that each ball is put in some box with probability $ \frac {1}{n}$ independently. Let $ p_n$ be the probability such that only two boxes are empty for $ n\geq 3$.

(1) Find $ p_3,\ p_4$.

(2)  For $ n\geq 4$, find the probability $ q_n$ such that only two boxes are empty, three balls are put in a box and each ball is put in the rest of $ (n \minus{} 3)$ boxes.

(3) For $ n\geq 5$, find $ p_n$.

(4) For $ q_n$ found in (2), find $ \lim_{n\to\infty} \frac {q_n}{p_n}$.

Number the balls from 1 to n, and number the boxes from 1 to n.

Note there are $ \binom{n}{2}$ distinct ways to select two of the n boxes to be empty.  Among the remaining n-2 boxes, there are $ (n\minus{}2)^n$ ways to distribute the n balls.  The total number of arrangements of the n balls is $ n^n$, so the desired probability is

$ p_n \equal{} \left(\frac{n\minus{}2}{n}\right)^n \binom{n}{2}$.

Now, there are $ \binom{n}{3}$ ways to select 3 of the n boxes, and 3 ways to select one of the three boxes such that this box will contain three balls, and the other two will contain none.  Since exactly one ball must be placed in each of the remaining n-3 boxes, there are $ (n\minus{}3)!$ such arrangements of these balls.  Therefore,

$ q_n \equal{} 3\binom{n}{3} \frac{(n\minus{}3)!}{n^n} \equal{} \frac{n!}{2 n^n}$.

Hence

$ \frac{q_n}{p_n} \equal{} \frac{(n\minus{}2)!}{(n\minus{}2)^n}$,

and since we have

$ 0 < \frac{(n\minus{}2)!}{(n\minus{}2)^n} \le \frac{(n\minus{}2)^{n\minus{}2}}{(n\minus{}2)^n} \equal{} \frac{1}{(n\minus{}2)^2}$,

it follows that $ \lim_{n \rightarrow \infty} \frac{q_n}{p_n} \equal{} 0$.

[quote="atomicwedgie"]Note there are $ \binom{n}{2}$ distinct ways to select two of the n boxes to be empty.  Among the remaining n-2 boxes, there are $ (n \minus{} 2)^n$ ways to distribute the n balls.[/quote]  Surely not, if you expect this process to fill the other $ n \minus{} 2$ boxes!  (And this is not the unique error in this solution.)

What we really need is the number of surjective functions from an $ n$-set to an $ n \minus{} 2$-set.  There are two ways to do this: either one box gets three balls or two boxes get two.  For the first, there are $ (n \minus{} 2)\cdot \binom{n}{3} \cdot (n \minus{} 3)! \equal{} \frac {(n \minus{} 2)n!}{6}$ ways (box to hold three balls times the three it holds times arrangements for the leftovers), while for the second there are $ \binom{n \minus{} 2}{2} \cdot \binom{n}{2} \cdot \binom{n \minus{} 2}{2}\cdot (n \minus{} 4)! \equal{} \frac {(n \minus{} 2)(n \minus{} 3)n!}{8}$ (boxes to hold 2 balls times the balls in the first box times the balls in the second box times arrangements for the leftovers).  This gives us $ p_n \equal{} \binom{n}{2} \cdot \frac {(n \minus{} 2)(3n \minus{} 5)n!}{24n^n}$, and in particular $ p_3 \equal{} \frac {1}{9}$ and $ p_4 \equal{} \frac {21}{64}$.  This takes care of parts (1-3); for part (4), we're looking at the limit of $ \frac {\frac {(n \minus{} 2)n!}{6}}{\frac {(n \minus{} 2)n!}{6} \plus{} \frac {(n \minus{} 2)(n \minus{} 3)n!}{8}}$, and a small amount of simplification shows that this is, in fact, $ 0$.

Edit added later:
Kunny correctly points out via PM that I never actually calculated $ q_n$.  What I did calculate was that there are $ (n \minus{} 2)\cdot \binom{n}{3}\cdot (n \minus{} 3)! \equal{} \frac {(n \minus{} 2)n!}{6}$ ways to place $ n$ balls in $ n \minus{} 2$ boxes so that every box gets at least one ball and one box gets three balls.  However, I needed to calculate the number of ways to place $ n$ balls in $ n$ boxes so that two boxes get zero balls, one gets three and the rest get one.  In order to calculate this, I just need to multiply my previous result by $ \binom{n}{2}$ (choosing the two empty boxes) to get $ \frac {n(n \minus{} 1)(n \minus{} 2)n!}{12}$ and so an associated probability of $ q_n \equal{} \frac {n(n \minus{} 1)(n \minus{} 2)n!}{12n^n}$.

This was the portion of the combinatorial part where atomicwedgie was closest to correct: atomicwedgie's expression $ q_n \equal{} 3\binom{n}{3} \cdot \frac {(n \minus{} 3)!}{n^n}$ was only missing a factor of $ \binom{n}{3}$, accounting for the fact that we need to choose three of our $ n$ balls to go into a single box.

Kunny also asked if I could check that the problem was phrased in grammatically correct English.  I've rewritten it slightly here:[hide][quote]Given $ n$ balls and $ n$ boxes. We put each ball in a box arbitrarily, that is to say, suppose that each ball is put in some box with probability $ \frac {1}{n}$ independently. Let $ p_n$ be the probability such that only two boxes are empty for $ n\geq 3$.

(1) Find $ p_3,\ p_4$.

(2)  For $ n\geq 4$, find the probability $ q_n$ such that only two boxes are empty, three balls are put in a box and a ball is put in each of the remaining $ (n \minus{} 3)$ boxes.

(3) For $ n\geq 5$, find $ p_n$.

(4) For $ q_n$ found in (2), find $ \lim_{n\to\infty} \frac {q_n}{p_n}$.[/quote][/hide]


Let ABC be a acute trıangle such that the ınsrıbed cırcle meets AB ın poınt D and AC ın poınt E. We denote by X and Y the ıntersectıon poınt of the angles bısectors of the angle ACB and ABC wıth lıne DE and wıth Z the mıdpoınt of BC. Prove that the trıangle XYZ ıs equılateral ıf and only ıf angle A ıs equal  to 60 Degree.

It is Balkan MO 2005 Q1,you can solution in the resource forum


inequalities
calculus
function
inequalities solved

by the condition we have
x = tan A
y = tan B
z = tan C
where A+B+C= 180
Now the inequality becomes equivalent to proving
(sinA)^3*(sinB) 2 *sinC   \leq  27/(25 \sqrt 5)
now that's just what Bao posted :D  :D
anyways,i'm checking if AM-GM helps (and it seems that u have found a solution using it :lol:
best regards

i saw your proof of this inequality,Namdung and it used calculus.here's my different proof using calculus
first let's fix one of the angle,say B.Then to maximize (sinA)^3*(sinC)
we get the condition that tanA=3tanC
let tan C=y
AND tanA+tanB+tanC=tanA*tanB*tanC gives
tanB=4y/(3y 2 -1)
tanA=3y
and the function can be explicitly written as a function of y
f(y)=27*16*y^6/(9y 2 +1) 2 ( y 2 +1) \sqrt [(9y 2 +1)(y 2 +1)]
now we see by applying AM-GM to the quintuple to
(2y 2,2y 2,2y 2,2y 2,y 2 +1) that
(9y 2 +1)^(5/2)*(y 2 +1)^(3/2)  \geq 5*4 \sqrt (5)*(y^4)*(y 2 +1) 2
 another application of AM-GM and we have
that f(y) is atmost (27*16)/25*16 \sqrt 5
which is just what we wanted.
so if we could prove that (sinA)^3*(sinC) IS MAXIMUM when tanA=3tanC
WITHOUT CALCULUS,then we could have a complete proof without calculus
ideas ne1?? :D

Nice idea, Galois, I think I can prove your hypothese (to show tgA=3tgC without calculus). Now, let's try to use AM-GM to solve my variant. Note that (very useful information!) the equality holds when x=3, y=2, z=1.

Namdung

note that the problems admits a generalization.
(sinA)^k*(sinB)^l*(sinC)^m  \leq  (k^k)(l^l)(m^m)(S)^(S/2) *(Sk^2 + P)^(-k/2)*(Sl^2+P)^(-l/2)*(Sm^2+P)^(-m/2)
where S=k+l+m and P=klm
this can be simplified but i have tried to make it look nice
and that reminds that this problem was originally proposed by
Murray Klamkin (a noted problemist)
so what's the easiest proof u can come up with?? :D  :D


algebra polynomial search calculus derivative

enjoy  :lol:
and what does these notations means?

This was solved on this forum already, try to search for it (otherwise write $Q'(x)=P(x)+Q(x)$ and see what happens when $x$ is an extremal point of $Q$).

No calculus in intermediate?
I'm new here, but that seemed to be the rule in the sticky.
Or are derivatives not classified as calculus anymore?
Oh well. Sounds like a cool proof anyways.


[quote="DrKevin"]The MathZoom Summer Camp 2008 will be held at the University of California, Los Angeles, from July 13 through August 2, 2008.

Please visit the [url=http://www.artofproblemsolving.com/Forum/index.php?f=453]MathZoom[/url] forum or [url]http://www.mathzoom.org[/url] for details.[/quote]
That's not bay area but thanks anyways!


[b]25. John ordered several copies of his favorite rock star's latest album to give to his friends. He ordered 4 CD's and some additional tapes. The price of a CD was double the price of a tape. When the order arrived, the number of CDs had been interchanged with the number of tapes. The bill was 50% more than he expected. How many tapes did John order?[/b]

[b]27. Point P is on AB. Point A has coordinates (3,11) and Point B has coordinates (18,1). The ratio of AP:PB=2:3 What is the sum of the coordinates of point P?[/b]

[b]30. What integer n has the property that $ 5^{96}$ is greater than $ n^{72}$ and $ 5^{96}$ is less than $ (n\plus{}1)^{72}$?[/b]

[hide="Hint for 27."]Use a modified version of the midpoint formula [/hide]

25. [hide]formulate it as an algebra problem.[/hide]
30. [hide]5^96>n^72, so 5^(4/3)>n, and 5^(4/3)<n+1. So, n=8.[/hide]

[hide="'nother hint for 27"]Just look at the x-coordinate, and then you can find the y-coordinate.[/hide]

[hide="27"]$ P's$ x coordinate must lie between 3 and 18 such that $ 3(x\minus{}3)\equal{}2(18\minus{}x)$
$ 5x\equal{}45$
$ x\equal{}9$
$ y\equal{}7$ similarly.

Sum is 16[/hide]


Let $k$ be a positive integer. Prove that there exists an infinite  monotone increasing sequence of integers $\{a_{n}\}_{n \ge 1}$ such that \[a_{n}\; \text{divides}\; a_{n+1}^{2}+k \;\; \text{and}\;\; a_{n+1}\; \text{divides}\; a_{n}^{2}+k\] for all $n \in \mathbb{N}$.

$ ab|a^2\plus{}b^2\plus{}k$
Let $ a_0\equal{}1,b_0|k\plus{}1$ and and
Let $ k\plus{}1\equal{}mb_0$
$ d\equal{}\frac{mb\plus{}b^2}{b}\equal{}m\plus{}b_0$
Consider the equation:
$ a^2\plus{}b^2\minus{}dab\plus{}k\equal{}0$
$ \Delta\equal{}(d^2\minus{}4)b^2\minus{}4k$
Chose :
$ (d^2\minus{}a)b^2\minus{}4k\equal{}y^2$
$ \Leftrightarrow y^2\minus{}(d^2\minus{}4)b^2\equal{}\minus{}4k$
This equation has an solution $ (y,b)$ so it has infinite solution.

Tell me if it needs translate :)
http://www.irysc.com/forum/t10301/#post126935
I construct a sequence that satisfy $a_{n+2} . a_n = {a_{n+1}}^2 + k$. From this relation we conclude that \[ a_{n}\;\text{divides}\; a_{n+1}^{2}+k\;\;\text{and}\;\; a_{n+1}\;\text{divides}\; a_{n}^{2}+k \]

Define the sequence $a_{n}$ such that
$a_{1} = 1,a_{2} = k+1,a_{n+2} = (k+2)a_{n+1}-a_{n}.$

And we can prove the following equation  by induction.
$(a_{n+1})^{2}+(a_{n})^{2}-(k+2) a_{n+1}a_{n} = 0.$
So,we have done.


inequalities
geometry
circumcircle
geometry proposed

[color=purple]Let triangle ABC and M is the point in plane.$MX\perp BC,MY\perp CA,MZ\perp AB$.Let$R,R_{M}$ are the circumradius of ABC and XYZ.
        Prove that
 $2RR_{M}(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}})\ge 9$
  Have fun ! :D [/color]

[color=purple] No one like this problem ? :D [/color]

Use the fact that $R_{M}\geq r$. Actually this is true for any point $X, Y, Z$ on the sides of $ABC$. It is also clear that $\sum \frac{1}{bc}= \frac{1}{2Rr}$


geometry
geometric transformation
reflection

like i can see how this would be funny...like in 8th grade, i would be like OMG HAHA SUCK THAT PEOPLE WHO ARE UPTIGHT

but really...why should you be acting like this???  it says something about you that you have the desire to be obnoxious, like you might say that it is just for fun, but that is being shallow, there is more too it

there is a subtlety, even the fact that you consider doing this reflects something about your character, why do other people not post like this? they have restraint and they are considerate of others, what can i say about you? you take pleasure in being annoying?

if i may predict your reaction, this is funny to you; i am feeding the troll, but if that is the sad truth, then you just cannot accept your shortcomings, you use laughter to hide what you do not want to face

oh and i just take this tone for the sake of argument, i don't really care what you do, but i do ask that you reflect about why you do what you do


logarithms
calculus
derivative
real analysis
real analysis unsolved

For natural number $n,$ let $f_n(x)=x\ln \left(\frac{ne}{x}\right)\ (x>0).$

(1) Find the maximum value of $f_n(x)$ at $x>0.$

(2) Prove the following inequality.

\[ \frac{e}{1+\frac{1}{n}}<\left(1+\frac{1}{n}\right)^n<e. \]

the sign of $f_n'$ gives you an insight of what the graph of $f_n$ looks like. once you draw this graph, you cannot miss the maximum..


trigonometry
inequalities proposed
inequalities

Find the minimum value of the following expression:
\[ |1\plus{}2\cos{x}|\plus{}|1\plus{}2\sin{x}|\]

[quote="zaizai-hoang"]Find the minimum value of the following expression:
\[ |1 \plus{} 2\cos{x}| \plus{} |1 \plus{} 2\sin{x}|
\]
[/quote]

$ |1 \plus{} 2\cos{x}| \plus{} |1 \plus{} 2\sin{x}|\geq|1 \plus{} 2\cos{x} \plus{} 1 \plus{} 2\sin{x}|\equal{}|2 \plus{} 2(\cos{x} \plus{} \sin{x})|\equal{}|2 \plus{} 2\sqrt{2}\cos({45^{0}\plus{}x})|\geq |2\plus{}2\sqrt{2}*(\minus{}1)|\equal{}2\sqrt{2}\minus{}2$
$ x\equal{}225^{0}$

[quote="zaya_yc"][quote="zaizai-hoang"]Find the minimum value of the following expression:
\[ |1 \plus{} 2\cos{x}| \plus{} |1 \plus{} 2\sin{x}|
\]
[/quote]

$ |1 \plus{} 2\cos{x}| \plus{} |1 \plus{} 2\sin{x}|\geq|1 \plus{} 2\cos{x} \plus{} 1 \plus{} 2\sin{x}| \equal{} |2 \plus{} 2(\cos{x} \plus{} \sin{x})| \equal{} |2 \plus{} 2\sqrt {2}\cos({45^{0} \plus{} x})|\geq |2 \plus{} 2\sqrt {2}*( \minus{} 1)| \equal{} 2\sqrt {2} \minus{} 2$
$ x \equal{} 225^{0}$[/quote]

Sorry if I say that you are wrong  :(

where is a simple solution? argady... can you solve it simply?

sorry if I say you are wrong, argady  :(  The minimum value is $ \sqrt{3}\minus{}1$

[quote="zaizai-hoang"] [color=darkred]Find the [b]minimum[/b] value of the function $ f(x) \equal{} |1 \plus{} 2\cos x| \plus{} |1 \plus{} 2\sin x|\ ,\ x\in\mathcal R$ .[/color] [/quote]
[color=darkblue][b][u]Proof.[/u][/b] Observe that for any $ x\in\mathcal R$ we have $ f(x) > 0$ and

$ f^2(x) \equal{} 4(\sin x \plus{} \cos x) \plus{} 6 \plus{} 2\cdot\left|2(\sin x \plus{} \cos x)^2 \plus{} 2(\sin x \plus{} \cos x) \minus{} 1\right|$ .

I"ll use the substitution $ \boxed {\sin x \plus{} \cos x \equal{} t}\in\left[ \minus{} \sqrt 2,\sqrt 2\right]$ . Thus,

$ f(x)$ is [b]minimum[/b] on $ \mathcal R$ if and only if  $ h(t) \equal{} t \plus{} \left|t^2 \plus{} t \minus{} \frac 12\right|$ is [b]minimum[/b] on $ \left[ \minus{} \sqrt 2,\sqrt 2\right]$ .

In conclusion, $ \min_{x\in\mathcal R}f(x)\equal{}\sqrt {6\plus{}4\cdot\min_{|t|\le\sqrt 2}h(t)}$ a.s.o.[/color]

Yes I also got $ \sqrt3\minus{}1,$ but I don't see a nice solution.
We obtain it for $ \sin x\plus{}\cos x\equal{}\frac{\sqrt3\minus{}1}{2}$ or for $ \sin x\plus{}\cos x\equal{}\frac{\minus{}\sqrt3\minus{}1}{2}.$

[quote="Virgil Nicula"][quote="zaizai-hoang"] [color=darkred]Find the [b]minimum[/b] value of the function $ f(x) \equal{} |1 \plus{} 2\cos x| \plus{} |1 \plus{} 2\sin x|\ ,\ x\in\mathcal R$ .[/color] [/quote]
[color=darkblue][b][u]Proof.[/u][/b] Observe that for any $ x\in\mathcal R$ we have $ f(x) > 0$ and

$ f^2(x) \equal{} 4(\sin x \plus{} \cos x) \plus{} 6 \plus{} 2\cdot\left|2(\sin x \plus{} \cos x)^2 \plus{} 2(\sin x \plus{} \cos x) \minus{} 1\right|$ .

I"ll use the substitution $ \boxed {\sin x \plus{} \cos x \equal{} t}\in\left[ \minus{} \sqrt 2,\sqrt 2\right]$ . Thus,

$ f(x)$ is [b]minimum[/b] on $ \mathcal R$ if and only if  $ h(t) \equal{} t \plus{} \left|t^2 \plus{} t \minus{} \frac 12\right|$ is [b]minimum[/b] on $ \left[ \minus{} \sqrt 2,\sqrt 2\right]$ .

In conclusion, $ \min_{x\in\mathcal R}f(x) \equal{} \sqrt {6 \plus{} 4\cdot\min_{|t|\le\sqrt 2}h(t)}$ a.s.o.[/color][/quote]

Can you show me $ \cdot\min_{|t|\le\sqrt 2}h(t)$ ? Please complete your solution  :wink:

[color=darkblue][b]Zaizai-Hoang[/b], I see that you wish to make nothing !

In the my opinion, the finish is very, very simply.

[b]Answer[/b] : $ \min_{|t|\le\sqrt 2}h(t) \equal{} \minus{} \frac {1 \plus{} \sqrt 3}{2}$

and $ \min_{x\in\mathcal R}f(x) \equal{} \sqrt {6 \minus{} 2\left(1 \plus{} \sqrt 3\right)}$ $ \implies$

$ \boxed {\min_{x\in\mathcal R}f(x) \equal{} \sqrt 3 \minus{} 1}$ . [b]Is it O.K. ?![/b] [/color]

Your answer is correct, Virgil Nicula. :)

Here is my solution.

After all, this problem results in the condition for which the circle: $ (x \minus{} 1)^2 \plus{} (y \minus{} 1)^2 \equal{} 4$ intersects with the square $ |x| \plus{} |y| \equal{} k,$ yielding $ k_{min} \equal{} \sqrt {3} \minus{} 1.$
The equality is left for you.

[color=darkred][size=117][b]Very nice the your idea, Kunny ![/b][/size][/color] [color=darkblue][b]Using it, I obtain the following extension :[/b][/color]

[quote]$ \left\{\begin{array}{c}
0\le z\le\frac {\sqrt 2}{2}\\\\
x^2\plus{}y^2\equal{}1\end{array}\right\|$ $ \implies$ $ z\plus{}|x\plus{}z|\plus{}|y\plus{}z|\ge \sqrt {1\minus{}z^2}\ .$[/quote]
[color=darkblue][b]Particularly, for $ \left\|\begin{array}{c}
x: \equal{}\cos t\\\\
y: \equal{}\sin t\\\\
z: \equal{}\frac 12\le\frac {\sqrt 2}{2}\end{array}\right\|$ $ \implies$ $ \left|\cos t\plus{}\frac 12\right|\plus{}\left|\cos t\plus{}\frac 12\right|\ge \frac {\sqrt 3\minus{}1}{2}$ ,

i.e. the proposed problem : $ |2\cos t\plus{}1|\plus{}|2\sin t\plus{}1|\ge \sqrt 3\minus{}1$ for any $ t\in\mathcal R\ .$[/b][/color]

Thank you very much, Virgil Nicula.

Here is my final solution:

Let $ \left(\begin{array}{c} x \\
y\end{array}\right) \equal{}\left(\begin{array}{c} 1 \\
1\end{array}\right) \plus{} \cos \theta \left(\begin{array}{c} 2 \\ 0
\end{array}\right) \plus{} \sin \theta \left(\begin{array}{c} 0 \\
2\end{array}\right)\in{C: \ (x\minus{}1)^2 \plus{} (y\minus{}1)^2 \equal{} 4},\ C$ intersects with the Square $ |x| \plus{} |y| \equal{} k\Longrightarrow k\geq \sqrt {3}\minus{}1.$
The equality holds is left for you.

kunny


Good luck all!!

Same rule as other revival rounds.... have fun!!! :D

I'll start.

[b]25[/b]

hmm

i'll pretend izzy posted 3
[b]
0[/b]

YAY girls still win XD


[quote="Farnak"]Find all integer solutions (x, y, z) to $ x^2 \plus{} y^2 \equal{} 3z^2$[/quote]

[hide="Err...this work? =/"]The greatest $ m$ for which $ 3^m | x^2\plus{}y^2$ is even...and the greatest $ n$ for which $ 3^n | 3z^2$ is odd...so they can never be equal...?[/hide]

[quote="Farnak"]Find all integer solutions (x, y, z) to $ x^2 \plus{} y^2 \equal{} 3z^2$[/quote]

Consider mod 3, its quadratic residues are 0 and 1. The RHS is congruent to 0, so we need x and y to be divisible by 3, implying x^2 and y^2 are divisible by 9 and thus z is divisible by 3. Set x=3x_1, y=3y_1, z=3z_1, to get $ 9x_{1}^{2}\plus{}9y_{1}^{2}\equal{}27z_{1}^{2}$, divide out by 9, by infinite decent there are no solutions other than (0,0,0).

You sure?

[hide=" I think"] Let $ (x,y,z)$ be the smallest solution. Since $ 3|3z^{2} \Rightarrow 3|x^{2} \plus{} y^{2}$ but $ x^{2} \equiv 0,1 (mod3)$ and $ y^{2} \equiv 0,1 (mod3)$ so $ x^{2} \equiv y^{2} \equiv 0 (mod3)$. Hence let $ x \equal{} 3a$ and $ y \equal{} 3b$ then $ 3(a^{2} \plus{} b^{2}) \equal{} z^{2}$. Thus $ 3|z^{2}$ and let $ z \equal{} 3c \Rightarrow a^{2} \plus{} b^{2} \equal{} 3c^{2}$. We obtain solution $ (a,b,c)$ which is smaller than the original solution, a contradiction. Hence the only solution is $ (0,0,0)$.[/hide]

Edit: Oops didn't see 13375P34K43V312 has posted a solution. The 'you sure' was addressed at ch1n353ch3s54a1l

[hide]Using mod 3, both x and y must be divisible by 3, so $ \implies 3x_1^2\plus{}3y_2^2\equal{}z^2$. Again, using mod 3, z must be div by 3, $ \implies x_2^2\plus{}y_x^2\equal{}3z_1^2$. Infinite descent, only way out is (0,0,0).[/hide]

[hide="solution"]We take note of the obvious (0,0,0).

Now since the RHS is divisible by 3, the LHS must consist of squares which are multiples of 3. Therefore, x=3a and y=3b.

$ x^2\plus{}y^2\equal{}9a^2\plus{}9b^2\equal{}3z^2$

$ 3a^2\plus{}3b^3\equal{}z^2$

Therefore, z is also a multiple of 3.

$ z\equal{}3c\Rightarrow 9c^2\equal{}3a^2\plus{}3b^2$

$ a^2\plus{}b^2\equal{}3c^2$

Infinite descent, so (0,0,0) is the only solution.[/hide]

[quote="ch1n353ch3s54a1l"]The greatest $ m$ for which $ 3^m | x^2 \plus{} y^2$ is even[/quote]

This is true, but it needs to be proven.  [hide="In particular..."] $ gcd(x, y) \equal{} 1 \implies x^2 \plus{} y^2 \equiv 1, 2 \bmod 3$, hence if $ 3^m | x^2 \plus{} y^2$ then $ 3^m | gcd(x,y)^2$. [/hide]
By the way, I prefer this solution over infinite descent because it begins to introduce the concept of [url=http://mathworld.wolfram.com/p-adicNumber.html]the p-adics[/url].  ch1n353ch3s54a1l's solution simply demonstrates that the $ 3$-adic valuation of both sides is never equal.  ;)

We should consider the solution where $ |x| \plus{} |y| \plus{} |z|$ is minimal or something.

It's enough to consider the solution where $ z$ is minimal.  Equivalent condition requiring less work to demonstrate.

I guess that I was hinting at the idea that we want to do $ |z|$ or w/e because we can go negative...

But we're only dealing with squares here; we can assume that $ x, y, z$ are non-negative.


geometry
circumcircle
cyclic quadrilateral
geometry proposed

In an acute-angled triangle $ ABC$, $ H$ is the orthocenter. The feet of the perpendiculars from $ A$ to $ BC$ , from $ B$ to $ AC$ and from $ C$ to $ AB$ are $ A_{1}$ , $ B_{1}$ and $ C_{1}$ , respectively. $ K$ is a point on the small $ AB_{1}$ arc of the circle with diameter $ AB$ . $ \angle{HKB}$ = $ C_{1}KB$. Let $ KB$ and $ CC_{1}$ meet at $ L$, a circle whose radius with $ CL$ and whose center is $ C$ meets $ AA_{1}$ at $ M$ . A circle whose center is $ B$ and radius is $ BM$ intersects the line $ CC_{1}$ at $ P$ and $ Q$. Show that $ A$ , $ K$ , $ P$ and $ Q$ are concylic.

[quote="grupyorum"]In an acute-angled triangle $ ABC$, $ H$ is the orthocenter. The feet of the perpendiculars from $ A$ to $ BC$ , from $ B$ to $ AC$ and from $ C$ to $ AB$ are $ A_{1}$ , $ B_{1}$ and $ C_{1}$ , respectively. $ K$ is a point on the small $ AB_{1}$ arc of the circle with diameter $ AB$ . $ \angle{HKB}$ = $ \angle C_{1}KB$. Let $ KB$ and $ CC_{1}$ meet at $ L$, a circle whose radius with $ CL$ and whose center is $ C$ meets $ AA_{1}$ at $ M$ . A circle whose center is $ B$ and radius is $ BM$ intersects the line $ CC_{1}$ at $ P$ and $ Q$. Show that $ A,$ $ K,$ $ P,$ and $ Q$ are concylic.[/quote]

Here is my solution

Let $ W$ be the intersection from $ AK$ to $ CC_1$, then $ KA\perp KL$. But in the other hand, we have $ KL$ is the internal bisector of $ \angle C_1KA_2$. Thus, $ (KC_1,KH,KL,KW) \equal{} \minus{} 1$, which implies $ (C_1HLW) \equal{} \minus{} 1$. Since $ L$ is the orthocenter of $ \triangle WAB$. Hence, if we let $ A_2$ be the second intersection of $ WB$ to $ (AA_1B)$ $ \equiv$ $ (\omega)$, then $ A,$ $ L,$ $ A_2$ are collinear, which leads to the fact that $ A_2,$ $ H,$ $ K$ are collinear as well. Now due to $ (AC_1,AH,AL,AW) \equal{} \minus{} 1$ and $ B,$ $ A_1,$ $ A_2,$ $ K$ respectively are the intersections of $ AC_1,$ $ AH,$ $ AL,$ $ AW$, hence $ BA_2A_1K$ is [i]harmonic quadrilateral[/i]. Therefore, respectively, the tangents $ t_1$, $ t_2$ at $ A_2$, $ K$ of $ (\omega)$ and $ BA_1$ are concurrent, let call this point is $ C'$. The same arguments holds for $ t_1,$ $ t_2$ with $ AB_1$, let this point be $ C''$. As the consequence, $ C'$ must coincide with $ C''$ and coincide with $ C$, which implies $ C$ is the circumcenter of $ (KLW)$ $ \equiv$ $ (\Omega)$. Since $ B$ is the polar of $ AH$ with respect to $ (\Omega)$, hence if $ M$ is the intersection of $ (\Omega)$ with $ AA_1$ $ \equiv AH$, then $ BM$ is a tangents from $ B$ to $ (\Omega)$ $ \Longrightarrow$ $ BM^2$ $ \equal{}$ $ \overline {BA_2}.\overline {BW}$ $ \equal{}$ $ \overline {BC_1}.\overline {BA}$ $ \equal{}$ $ BP^2$, which implies $ \angle BPA \equal{} 90^{\circ}$. Similarly, $ \angle BQA \equal{} 90^{\circ}$. As the result, $ P,$ $ Q$ both lie on the circle with diameter $ AB$, which imples $ Q,$ $ A,$ $ K,$ $ P$ are concyclic. Our proof is completed then. $ \square$

Nice solution,here is mine. :)
$ AKLC_1$ is a concyclic quadrilateral.
Then,$ \angle C_1 AL = \angle C_1 KL = \angle BKH$
So,   $ AL \cap KH = X \\
\end{array}$ , with X is a point on the circle with AB is the diameter.
ANd we have, $ XH.HK = A_1HAH = C_1H.CH$
Therefore, $ C_1KCX$ is a concyclic quadrilateral.
$ \angle C_1KX = \angle C_1CX = \angle C_1OX$
therefor, CX,CK is the tangents of the circle with AB is the diameter.
then, $ \angle XLC = 90^0 - \angle BAX = \angle LXC$
Therefore, LC=XC=MC
We obtain
$ MC^2 = CP.CQ$
$ MC^2 = CA_1.CB$ (MC is the tagent to the circle with the center is B)
So, $ P,Q,A_1, B$ is concyclic.

Let $ N$ be the intersection of $ KH$ and $ (AB)$.Since $ AC_1LK$ is cyclic quadrilateral we get $ \angle BAN\equal{}\angle BKN\equal{}\angle C_1KN\equal{}\angle C_1AL$. Therefore $ A,L,N$ are collinear.
$ \angle B_1HC\equal{}\angle BAC\equal{}\angle LKB_1$ then $ LKB_1H$ is cyclic quadrilateral.
$ \Rightarrow \angle B_1LC\equal{}\angle HKB_1\equal{}\angle LAB_1 \Rightarrow CL^2\equal{}CA.CB_1$.
$ \Rightarrow CM^2\equal{}CA_1.CB \Rightarrow \angle BMC\equal{}90^o \Rightarrow BM^2\equal{}BA_1.BC$.
Let $ P',Q'$ be the intersection of $ (AB)$ and $ CC_1$.
Since $ \angle BP'A_1\equal{}\angle BAA_1\equal{}\angle BCP' \Rightarrow BP'^2\equal{}BA_1.BC$
$ \Rightarrow BQ'\equal{}BP'\equal{}BM \Rightarrow P'\equiv P,Q'\equiv Q$. We are done.


function
algebra
polynomial
algebra proposed

Find all monotonic  functions  $ f: \mathbb{R} \to \mathbb{R}$ which there is a positive integer $ n$ such that:
$ \forall \alpha \in \mathbb{R}$    $ f^n(\alpha)\equal{}\minus{}\alpha$
  $ (f^n\equal{}f\circ f\circ ...\circ f)$

Let $ a_{n}(x) \equal{} f \circ f \ldots \circ f (x).$ (n times)

Then we have $ a_{2n} \equal{} \minus{} a_n,$ whose characteristic polynomial is $ r^n \plus{} 1$, hence

$ a_n(x) \equal{} \sum_{j}{F_j (x)e^{\frac {2\pi \plus{} j}{n}i}}$

Real root of the characteristic polynomial will be if $ n$ is odd and $ r \equal{} \minus{} 1$.

Now we get $ f(x) \equal{} a_1(x) \equal{} a_0(x) \equal{} r x \equal{} \minus{} x$


In trapezoid $ABCD$, the diagonals intersect at $E$, the area of $\triangle ABE$ is 72 and the area of $\triangle CDE$ is 50. What is the area of trapezoid $ABCD$?

Let area of triangle $AED=x$,so area of $CEB=x$
$\frac{50}{x}=\frac{x}{72}$
$x^2=3600$
$x=60$
so area of$ABCD=242$