Let $ ABC$ be a given triangle for which we denote by $ h_{a}$, $ h_{b}$, $ h_{c}$ the lengths of its altitudes, and by $ r_{a}$, $ r_{b}$, $ r_{c}$ the lengths of its exradii. Prove that
\[ \frac {h_{a}r_{a}}{r_{a}^{2} \plus{} 4h_{a}^{2}} \plus{} \frac {h_{b}r_{b}}{r_{b}^{2} \plus{} 4h_{b}^{2}} \plus{} \frac {h_{c}r_{c}}{r_{c}^{2} \plus{} 4h_{c}^{2}}\leq \frac {3}{5}.
\]

EDIT: Indeed, they were the exradii, thanks.

hello, do you meant with $ r_a$, $ r_b$ and $ r_c$ the lengths of the exradii?
Sonnhard.

[quote="pohoatza"]Let $ ABC$ be a given triangle for which we denote by $ h_{a}$, $ h_{b}$, $ h_{c}$ the lengths of its altitudes, and by $ r_{a}$, $ r_{b}$, $ r_{c}$ the lengths of its inradii. Prove that
\[ \frac {h_{a}r_{a}}{r_{a}^{2} \plus{} 4h_{a}^{2}} \plus{} \frac {h_{b}r_{b}}{r_{b}^{2} \plus{} 4h_{b}^{2}} \plus{} \frac {h_{c}r_{c}}{r_{c}^{2} \plus{} 4h_{c}^{2}}\leq \frac {3}{5}.
\]
[/quote]
The following inequality is true too.
\[ \frac {h_{a}r_{a}}{8r_{a}^{2} \plus{} h_{a}^{2}} \plus{} \frac {h_{b}r_{b}}{8r_{b}^{2} \plus{} h_{b}^{2}} \plus{} \frac {h_{c}r_{c}}{8r_{c}^{2} \plus{} h_{c}^{2}}\leq \frac {1}{3}
\]
By the way, $ \frac {h_{a}r_{a}}{9r_{a}^{2} \plus{} h_{a}^{2}} \plus{} \frac {h_{b}r_{b}}{9r_{b}^{2} \plus{} h_{b}^{2}} \plus{} \frac {h_{c}r_{c}}{9r_{c}^{2} \plus{} h_{c}^{2}}\leq \frac {3}{10}$ is wrong already. :wink:

Do you have any nice approaches, arqady? My solution is purely algebraic (and very computational).

[quote="pohoatza"]Do you have any nice approaches, arqady? [/quote]
No. It is an expanding and SOS.


We are considering offering a 1-day seminar for the AIME on March 20th.  The class would probably be 4.5 hours of class spread over around 6 hours (to allow for some breaks).  The problems covered will be different from those in the current AMC/AIME Problem Series.  The cost would be around $20.

If you are interested in the course, please let us know here your optimal timing structure.

I'm interested, though I probably won't sign up unless I actually make the AIME.

As early in the morning as possible is best for me...

Some guidelines on who might benefit from the class would be helpful.  Maybe if I have no idea what the AIME is, should be a big enough hint!  I'll look around for links on what it is.

Excuse my ignorance.

AIME is the second stage in the selection of the U.S. Mathematical Olympiad team and it stands for "American Invitational Mathematical Examination"

Reference: [url=http://www.unl.edu/amc/e-exams/e7-aime/aime.html]Official website[/url]

I'd like it if it was in the PM (after noon) but it doesn't have to be.

[quote="MarkSmith888"]Yea, that would be great.  If I make AIME, I'd like to take the cram session.  Last year I made AIME and got totally lost on just about all questions past that first page, so yea I could use some help.[/quote]

Well, if that is true, you're probably not very badly off. The first page has 9 out of the 15 problems, heh. :D

Ha ha, yet understanding is the first part, doing the question is even harder.  I think I finally figured out how to do like 5 on the whole test in 3 hours worth of thinking time.  Anyway I need some help since AIME is so different from every other competition we take (not to mention a lot harder).  First things first though, got to qualify for it first.

I would be interested, I believe. Times...I'll deal with them, whatever they are, though it wouldn't be hard to select something reasonable for all time zones in the US. I assume the course will cover a lot of problems in this time?

Yea time would be an issue since I'm in the Central Time Zone and I know the instructuors are in the Pacific Time Zone.  I could start anytime basically, preferably as early as possible.  Since Pacific is behind every other time zone in the Continental US, I suppose starting too early wouldn't be too much of an issue.

Yeah it might be a good idea to write what time zone we're in. I'm in Eastern by the way so that's +3 from Pacific and +1 from Central so even if we start at 9 AM Pacific I's still have no problem...

[quote="MithsApprentice"]I'm definitely interested...would it be possible to have two of these?...one before each AIME exam...[/quote]

Unless we had an absolutely enormous number of students register it is not likely that we would hold this session twice [this year].

I will, without a doubt, register for the AIME class.  Thanks for holding one!

[quote="ml2"]What would be discusssed in this class: different AIME problems and how to aprroach them?[/quote]

Yes.  Particularly how to approach them.

Unless something gets in the way, we are going to hold the cram session.  Details are not entirely worked out yet, but when they are we will make a general announcement and post the course for registration.

I would definitely be interested in a cram session and would probably attend. In terms of time, I'm part of CST so I guess sometime in the morning would be fine.


function
real analysis
real analysis unsolved

This is elementary but I still don't get it. I found this proof that the sum of two continuous functions is continuous.

The sum of continuous functions is continuous follows directly from the triangle inequality. Take any  e> 0. There exists d1 > 0 such that whenever

| x - c | < d1
we know that
| f(x) - f(c) | < e
(because f is continuous at c). There also exists  d2 > 0 such that whenever
| x - c | < d2
we know that
| g(x) - g(c) | < e
(because g is continuous at c). But then, if we let  d = min(d1 , d2), we have: if
| x - c | < d
then
| (f(x) + g(x)) - (f(c) + g(c)) | ≤
      | f(x) - f(c) | + | g(x) - g(c) | <
       e + e = 2e
That finishes the proof. (That we don't get a simple e should not bother us any more).

My question is: why are we not bothered by the fact that we get 2e instead of e?

... Because you can go back and replace every instance of $ \epsilon$ with $ \frac{\epsilon}{2}$ if you feel like it. It's just that that's never the way you do it the first time, because it's not very natural to predict how many pieces you'll end up estimating.


Which natural numbers cannot be presented in that way: $[n+\sqrt{n}+\frac{1}{2}]$, $n\in\mathbb{N}$
$[y]$ is the greatest integer function.

$a_n = [n+\sqrt{n}+\frac{1}{2}]$

An integer $A$ is missed iff $a_n < A < a_{n+1}$ for some $n$ since $a_n$ is strictly increasing.
using $x \leq [x] < x+1$ this gives after some rewritting $A = (A-n)^2$; i.e $A$ is a square.

$a_{n^2} = n^2 + n$ which is precisely the $n^2$th non square, so none are missed.

To sum up : $a_n$ takes all non-square values.

I have a small question. Greatest integer function is the greatest integer number less or equal to this number, so it should be:
$x-1<[x]\leq{x}$

Usually [x]  is defined as $[x] \leq x < [x]+1$ (i.e the greastest integer not bigger than x) but I may be wrong


Would someone be willing to give me a site where I can find past Mathcounts tests? Thanks.

http://www.4shared.com/dir/257226/be0f945c/Mathcounts_States.html

http://www.4shared.com/dir/268475/413f2797/sharing.html

The second one doesn't work right now, I dunno why...but it has nats

I will try to get you some tests in a little bit, still trying to figure out the file usage on AoPS.  Try [url=http://mathcounts.saab.org/mc.cgi]mathcounts.saab.org/mc.cgi[/url]   for past mathcounts problems

http://www.4shared.com/dir/268475/413f2797/sharing.html

is it just my computer, or does that link not work?

No, it didn't work for me earlier either.  :?  Don't those files get deleted after a certain period of time?

:P too arrogant to do state, really I don't want to revisit memories of the gaping holes I left in the state competitions, especially in the target round.

Another website.

http://goodlymaths.0catch.com/contests/mathcounts/mathcounts_archive.htm


inequalities
geometry
geometric transformation
reflection
inequalities proposed

:( , no solution ?
For inequality 2, can you  prove for $k=\frac{1}{4}$ ?

[quote="ehku"]$\sqrt{\frac{a^{2}}{a^{2}+\frac{ab}{4}+b^{2}}}+\sqrt{\frac{b^{2}}{b^{2}+\frac{bc}{4}+c^{2}}}+\sqrt{\frac{c^{2}}{c^{2}+\frac{ca}{4}+a^{2}}}\le 2$
Nice  :D[/quote]
This inequality is posted by Bin Zhao in Mathematical Reflections, last issue (no. 6). See
http://reflections.awesomemath.org/2006_6/2006_6_problems.pdf


limit
integration
logarithms
trigonometry
real analysis
real analysis unsolved

Hi all,
This is an interesting question that came up in our exam today. Might be too easy for all the brainy people here but still. Here it goes --

Evaluate
$ \lim_{R \to \infty}$ $ R^{\alpha} \displaystyle\int^{\frac {\pi}{2}}_{{0}} e^{ \minus{} Rsin\theta}\,d\theta$

$ \alpha$ $ <$ $ 1$.

I would love to see if there are any shorter solutions than what I did.

Thanks.

$ 0< R^{\alpha}\int^{\frac{\pi}{2}}_{{0}}e^{\minus{}Rsin\theta}\,d\theta \equal{}  \int^{\frac{\pi}{2}}_{{0}}R^{\alpha}e^{\minus{}Rsin\theta}\,d\theta \equal{}  \int^{\frac{1}{R}}_{{0}}R^{\alpha}e^{\minus{}Rsin\theta}\,d\theta  \plus{}  \int^{\frac{\pi}{2}}_\frac{1}{R} R^{\alpha}e^{\minus{}Rsin\theta}\,d\theta < \int^{\frac{1}{R}}_{{0}}R^{\alpha}\,d\theta  \plus{}  \int^{\frac{\pi}{2}}_\frac{1}{R} R^{\alpha}\frac{1}{1\plus{}Rsin\theta}\,d\theta \equal{} R^{\alpha\minus{}1}  \plus{}  \int^{\frac{\pi}{2}}_\frac{1}{R} R^{\alpha}\frac{1}{1\plus{}Rsin\theta}\,d\theta \le R^{\alpha\minus{}1}  \plus{}  \int^{\frac{\pi}{2}}_\frac{1}{R} R^{\alpha}\frac{1}{Rsin\theta}\,d\theta \le R^{\alpha\minus{}1}  \plus{}  R^{\alpha\minus{}1} \int^{\frac{\pi}{2}}_\frac{1}{R} \frac{1}{sin\theta}\,d\theta \equal{}   R^{\alpha\minus{}1}  \plus{}  R^{\alpha\minus{}1} \ln\tan\frac{\theta}{2} |^{\frac{\pi}{2}}_\frac{1}{R}  \equal{}  R^{\alpha\minus{}1} \minus{} R^{\alpha\minus{}1} \ln \tan \frac{1}{2R}$.
But $ \tan \frac{1}{2R} < \frac{1}{R}$ for big $ R$, and thus $ |\ln \tan \frac{1}{2R}|<|\ln \frac{1}{R}| \equal{} \ln R$.
And $ R^{\alpha\minus{}1} \minus{} R^{\alpha\minus{}1} \ln \tan \frac{1}{2R} < R^{\alpha\minus{}1} \minus{} \frac{R^{\alpha} \ln R}{R}\to 0.$

Hi,
nice solution. What I did is the following -

Use mean value theorem.
$ \exists$ $ \phi$ $ \epsilon$ $ [0,\frac{\pi}{2}]$ such that,

$ \displaystyle\int^{\frac {\pi}{2}}_{{0}} e^{ \minus{} Rsin\theta}\,d\theta$ = $ \frac{\pi}{2} e^{\minus{}Rsin\phi}$

Hence, the expression to be evaluated reduces to  $ \frac{\pi}{2}R^{\alpha} e^{\minus{}Rsin\phi}$

Clearly, for $ \alpha$ $ <$ $ 1$ this goes to zero as $ R$ goes to $ \infty$.

and what is you mark for exam ?  :blush:

Not out yet. Why? Is there any flaw in my solution?

[quote="avkuvalekar"]
Use mean value theorem.
$ \exists$ $ \phi$ $ \epsilon$ $ [0,\frac {\pi}{2}]$ such that...
[/quote]

why $ \phi$ is the same for different $ R$ ???

actually you have $ \phi(R)$ and $ \frac{\pi}{2}R^{\alpha}e^{\minus{}Rsin\phi(R)}$  :(

Well, still it will go to zero. I am going to be penalised heavily for this but I guess I can still use the fact that that there exists such $ \phi(R)$.
Now, $ R^{\alpha}$ / $ e^{Rsin(\phi(R))}$ will also go to zero as $ R$ goes to $ \infty$

But still, I am sc***d!  :(

[quote="anik_andrew"][quote="avkuvalekar"]
Now, $ R^{\alpha}$ / $ e^{Rsin(\phi(R))}$ will also go to zero as $ R$ goes to $ \infty$
[/quote]

depend on $ \phi (R)$...[/quote]

I guess not. I know that $ \phi(R)$ is not zero. Now, use L'Hospital's rule to find the limit knowing that $ \phi'(R)$ is also not equal to zero.

$ \lim_{R \to \infty} R^{\alpha}/e^{Rsin\phi(R)} \equal{} \frac {\alpha}{ R^{1 \minus{} \alpha} e^{Rsin\phi(R)}(sin\phi(R) \plus{} Rcos\phi(R) \phi'(R))}$
Now, the numerator is finite and the denominator goes to $ \infty$ as $ R$ goes to $ \infty$ because if $ sin\phi(R)$ goes to zero then cos goes to 1 and hence the term within brackets goes to $ \infty$

This is getting slightly messy now, I don't know. But I feel that $ sin\phi$ can not go to 1 as $ \phi$ reduces as R increases.

but if $ \phi(R)\to0$ and $ \phi'(R)\to 0$ then we cant say that denominator is goes to infty.

Yeah! I am sc***d basically. :(
I am just wondering if the intermediate value theorem can somehow be used though.

There exists $ c>0$ such that $ \frac{sin\theta}{\theta} > c$  on $ [0,\pi/2]$. Therefore $ e^{\minus{}Rsin\theta} \leq  e^{\minus{}Rc\theta}$ on $ [0,\pi/2]$. Everything is now easy, and your limit is $ 0$.


Inequality quadratics algebra polynomial

Here is my soln to this problem.
The above expression can be written like
${{(a+b)^2 - c^2}>2ab + 6(c-a)(c-b)},$
${(a+b)\left(a+b - \frac{c^2}{a+b}\right)>2ab + 6(c-a)(c-b)}.$
But using AM-GM we know that
$ {a+b +\frac{c^2}{a+b}>2c.}$
So we have that
$ {\left(a+b - \frac{c^2}{a+b}\right)\ge2\cdot\left(c - \frac{c^2}{a+b}\right).}$
Now, if we are able to prove that
$ {2\left(c - \frac{c^2}{a+b}\right)>2ab + 6(c-a)(c-b)},$
Or
$ {(a+b)\left(c - \frac{c^2}{a+b}\right)>ab + 3(c-a)(c-b)},$
then we will be done.
But on simplifying this we have
$ {c(a+b-c)>4ab -3c[a+b-c]},$
Or
$ {4c(a+b-c)>4ab},$
Or
$ {a+b>c+\frac{ab}{c}}$
Now here, set  ${c^3 = a^3 + b^3},$ and we are done.
[color=#0040FF][size=85][Mod edit: Properly LaTeXed][/size][/color]

[hide]By AM-GM $ \dfrac{bc^2\plus{}cb^2\plus{}b^3\plus{}ac^2\plus{}ca^2\plus{}a^3}{6}\ge abc$.
with equality when $ bc^2\equal{}cb^2\equal{}b^3\equal{}ac^2\equal{}ca^2\equal{}a^3$. But if there is equality, $ b\equal{}c$, and since $ a^3\plus{}b^3\equal{}c^3$, $ a^3\equal{}0$, and $ a\equal{}0$. That contradicts the condition in the problem that $ a$ is positive. Thus there is no equality case.
$ \dfrac{bc^2\plus{}cb^2\plus{}b^3\plus{}ac^2\plus{}ca^2\plus{}a^3}{6}>abc$
$ bc^2\plus{}cb^2\plus{}b^3\plus{}ac^2\plus{}ca^2\plus{}a^3>6abc$
Dividing by $ ab$ we get
$ \dfrac{c^2\plus{}cb\plus{}b^2}{a}\plus{}\dfrac{c^2\plus{}ca\plus{}a^2}{b}>6c$.
Multiplying by $ (c\minus{}a)(c\minus{}b)$ we get
$ \dfrac{(c\minus{}a)(c^3\minus{}b^3)}{a}\plus{}\dfrac{(c\minus{}b)(c^3\minus{}a^3)}{b}>6c(c\minus{}a)(c\minus{}b)$.
Since $ a^3\plus{}b^3\equal{}c^3$, we can substitute $ c^3\minus{}b^3$ with $ a^3$ and $ c^3\minus{}a^3$ with $ b^3$.
$ \dfrac{a^3(c\minus{}a)}{a}\plus{}\dfrac{b^3(c\minus{}b)}{b}>6c(c\minus{}a)(c\minus{}b)$
$ a^2(c\minus{}a)\plus{}b^2(c\minus{}b)>6c(c\minus{}a)(c\minus{}b)$
$ a^2c\minus{}a^3\plus{}b^2c\minus{}b^3>6c(c\minus{}a)(c\minus{}b)$
$ a^2\plus{}b^2\minus{}\dfrac{a^3\plus{}b^3}{c}>6(c\minus{}a)(c\minus{}b)$
Since $ c^2\equal{}\dfrac{a^3\plus{}b^3}{c}$,
$ a^2\plus{}b^2\minus{}c^2>6(c\minus{}a)(c\minus{}b)$.[/hide]

man what a soln MIJ i am glad.............

perfect soln .....

minblowing stuff

no words to praise it..................

 :omighty:

WLOG we can set c =1. So $ a^3\plus{}b^3 \equal{} 1$

So we have to prove that $ a^2\plus{}b^2\minus{}1 > 6(1\plus{}ab\minus{}a\minus{}b)$. Now if we set p = a+b and q = ab, the inequality becomes

$ p^2\plus{}6p\minus{}7>8q$ with $ p(p^2\minus{}3q) \equal{} 1$

Obviously p>1 and from Power Mean Inequality, $ p< 2^{\frac{2}{3}}$

We have $ 3q \equal{} p^2 \minus{} \frac{1}{p}$. So the inequality reduces to

$ 3p(p\plus{}7)>8(p^2\plus{}p\plus{}1)$ or proving that $ 5p^2 \minus{} 13p \plus{} 8<0$ which occurs when p lies between the roots 1 and 8/5.

We have already established p>1. Now if its true that $ 2^{\frac{2}{3}} < 8/5$ we are done. But this is equivalent to 125<128 and the inequality is proved

Obviously, $ c > b$ and $ c > a$. If $ 2c > \equal{} a$ and $ 2c > \equal{} b$, than $ c >\equal{} a \plus{} b$ and $ a^{3} \plus{} b^{3} > \equal{} (a \plus{} b)^{3}$, which is impossible. WLOG assume that $ a > \equal{} b$. Than, $ a > 2c$. Let's consider $ a^{2} \plus{} b^{2} \minus{} c^2 \minus{} 6(c \minus{} a)(c \minus{} b) \equal{} b^{2} \plus{} 6b(c \minus{} a) \plus{} (a \minus{} c)(a \plus{} 7c)$ as a quadratic polynomial on $ b$. As the coeeficient with $ b^{2}$ is $ 1$, if $ D < 0$, than the polynomial is always positive. $ D \equal{} 32(c \minus{} a)(2c \minus{} a) < 0$, which implies that the polynomial is positive, as desired.    Lasha Lakirbaia.

We need to show : $ a^2 \plus{} b^2 \minus{} c^2 > 6(c\minus{}a)(c\minus{}b)$
We are given $ a^3 \plus{} b^3 \equal{} c^3$

We use the following factorization: $ p^3 \plus{} q^3 \plus{} r^3 \minus{}3pqr \equal{} (p \plus{} q \plus{} r)(p^2 \plus{} q^2 \plus{} r^2 \minus{}pq \minus{}qr \minus{}rp)$

Substituting $ p \equal{} a$, $ q \equal{} b$ and $ r \equal{} \minus{}c$, we get   $ \;\;3abc \equal{} (a \plus{} b \minus{}c)(a^2 \plus{} b^2 \plus{} c^2 \minus{}ab \plus{} bc \plus{} ca)$

In other words,
$ a^2 \plus{} b^2 \plus{} c^2 \minus{}ab \plus{} bc \plus{} ca \equal{}  \frac{3abc}{a\plus{}b\minus{}c}$

$ a^2 \plus{} b^2 \minus{} c^2 \minus{}(c\minus{}a)(c\minus{}b) \equal{}  \frac{3abc}{a\plus{}b\minus{}c} \minus{}3c^2$

$ a^2 \plus{} b^2 \minus{} c^2 \minus{}(c\minus{}a)(c\minus{}b) \equal{}  \frac{3c(c\minus{}a)(c\minus{}b)}{a\plus{}b\minus{}c}$

Enough to show: $ \frac{3c(c\minus{}a)(c\minus{}b)}{a\plus{}b\minus{}c} > 5(c\minus{}a)(c\minus{}b)$

Enough to show: $ \frac{3c}{a\plus{}b\minus{}c} > 5$  (NOTE: $ a\plus{}b >c$)

Enough to show: $ a \plus{} b < \frac{8}{5}c$

We know that $ \left(\frac{a\plus{}b}{2}\right)^3 \leq \frac{a^3 \plus{}b^3}{2} \equal{} \frac{c^3}{2}$

This implies $ a \plus{} b \leq 4^{\frac{1}{3}} c < \frac{8}{5}c$.  Hence proved.

Jaswinder,

  Sorry to point out that your solution is incorrect.

You last line reads $ a \plus{} b > c \plus{} \frac{ab}{c}$

This is the same as $ 0 > c \minus{} a \minus{}b \plus{} \frac{ab}{c}$

which is the same as  $ 0 > \frac{c^2 \minus{} c(a \plus{} b) \plus{} ab}{c}$

In other words, you claim $ 0 > \frac{(c \minus{} a)(c \minus{} b)}{c}$ which is not true.

[quote="lasha"]Obviously, $ c > b$ and $ c > a$. If $ 2c > \equal{} a$ and $ 2c > \equal{} b$, than $ c > \equal{} a \plus{} b$ and $ a^{3} \plus{} b^{3} > \equal{} (a \plus{} b)^{3}$, which is impossible. WLOG assume that $ a > \equal{} b$. Than, $ a > 2c$. [/quote]

if $ 2c \minus{} a < 0, a > 2c > 2a \iff 0 > a ???$

i don't know if this is the same as the official solution because I can't open that .doc file... anyway here it goes...

note that $ c \minus{} a \equal{} \dfrac{b^3}{c^2 \plus{} ac \plus{} a^2}$, so we have to prove that

$ (a^2 \plus{} b^2 \minus{} c^2)(c^2 \plus{} ac \plus{} a^2)(c^2 \plus{} bc \plus{} b^2) > 6a^3b^3$...

after multiplying the terms and using that $ c^3 \equal{} a^3 \plus{} b^3$ a couple of times, we are left to show that $ c^2 \plus{} c(a \plus{} b) \plus{} a^2 \plus{} b^2 > 6ab$ (a $ a^2b^2$ factor eliminates on both sides)...

from am-gm and cauchy-schwarz we have that $ c^2 \plus{} c(a \plus{} b)\geq 2\sqrt {(a \plus{} b)(a^3 \plus{} b^3)}\geq 2(a^2 \plus{} b^2)$, and from this it's very clear how to conclude, and we're done  :D

we just have to show that equality cases are not possible, in fact, we can't have equality at the am-gm since it's obvious that $ a\plus{}b>c$...

We must take two cases:

First : c<a & c<b

Second : c>a & c>b

And we complete

[quote="Mij"][hide]By AM-GM $ \dfrac{bc^2 \plus{} cb^2 \plus{} b^3 \plus{} ac^2 \plus{} ca^2 \plus{} a^3}{6}\ge abc$.
with equality when $ bc^2 \equal{} cb^2 \equal{} b^3 \equal{} ac^2 \equal{} ca^2 \equal{} a^3$. But if there is equality, $ b \equal{} c$, and since $ a^3 \plus{} b^3 \equal{} c^3$, $ a^3 \equal{} 0$, and $ a \equal{} 0$. That contradicts the condition in the problem that $ a$ is positive. Thus there is no equality case.
$ \dfrac{bc^2 \plus{} cb^2 \plus{} b^3 \plus{} ac^2 \plus{} ca^2 \plus{} a^3}{6} > abc$
$ bc^2 \plus{} cb^2 \plus{} b^3 \plus{} ac^2 \plus{} ca^2 \plus{} a^3 > 6abc$
Dividing by $ ab$ we get
$ \dfrac{c^2 \plus{} cb \plus{} b^2}{a} \plus{} \dfrac{c^2 \plus{} ca \plus{} a^2}{b} > 6c$.
Multiplying by $ (c \minus{} a)(c \minus{} b)$ we get
$ \dfrac{(c \minus{} a)(c^3 \minus{} b^3)}{a} \plus{} \dfrac{(c \minus{} b)(c^3 \minus{} a^3)}{b} > 6c(c \minus{} a)(c \minus{} b)$.
Since $ a^3 \plus{} b^3 \equal{} c^3$, we can substitute $ c^3 \minus{} b^3$ with $ a^3$ and $ c^3 \minus{} a^3$ with $ b^3$.
$ \dfrac{a^3(c \minus{} a)}{a} \plus{} \dfrac{b^3(c \minus{} b)}{b} > 6c(c \minus{} a)(c \minus{} b)$
$ a^2(c \minus{} a) \plus{} b^2(c \minus{} b) > 6c(c \minus{} a)(c \minus{} b)$
$ a^2c \minus{} a^3 \plus{} b^2c \minus{} b^3 > 6c(c \minus{} a)(c \minus{} b)$
$ a^2 \plus{} b^2 \minus{} \dfrac{a^3 \plus{} b^3}{c} > 6(c \minus{} a)(c \minus{} b)$
Since $ c^2 \equal{} \dfrac{a^3 \plus{} b^3}{c}$,
$ a^2 \plus{} b^2 \minus{} c^2 > 6(c \minus{} a)(c \minus{} b)$.[/hide][/quote]

This is, in my opinion , the best and the easiest solution of this problem...... :wink:

Thanks, Mij, for your tremendously awesome and breathtaking but elementary solution....... :lol:

[quote="Aravind Srinivas L"]No there is a beutiful solution given by Mayank Magoo from India.

Let a,b,c be positive real numbers such that $ a^3 \plus{} b^3 \equal{} c^3$. Prove that
$ a^2 \plus{} b^2 \minus{} c^2 > 6(c \minus{} a)(c \minus{} b)$
Let $ x \equal{} \frac {a}{c}$, $ y \equal{} \frac {b}{c}$
Thus given $ x^3 \plus{} y^3 \equal{} 1$, $ x,y > 0$, prove that $ x^2 \plus{} y^2 \minus{} 1 > 6(1 \minus{} x)(1 \minus{} y)$
LHS =  $ x^2 \plus{} y^2 \minus{} 1 \equal{} x^2 \minus{} x^3 \plus{} y^2 \minus{} y^3 \equal{} x^2(1 \minus{} x) \plus{} y^2(1 \minus{} y) \geq 2xy\sqrt {(1 \minus{} x)(1 \minus{} y)}$.

Also $ x^3 \equal{} 1 \minus{} y^3 \equal{} (1 \minus{} y)(1 \plus{} y \plus{} y^2) \geq 3y(1 \minus{} y)$
Similarly $ y^3 \geq 3x(1 \minus{} x)$. Thus multiplying these two inequalities we get
$ x^2y^2 > 9(1 \minus{} x)(1 \minus{} y)$ or $ xy > 3\sqrt {(1 \minus{} x)(1 \minus{} y)}$ (we get strict inequality because both equalities cannot happen simultaneously)
Therefore
$ 2xy\sqrt {(1 \minus{} x)(1 \minus{} y)} > 6(1 \minus{} x)(1 \minus{} y)$
and we are done........................... :P   :D .[/quote]
Wow, I did not know so many solutions existed.... :D  :rotfl:
Thanks.

[quote="skand"]Let $ a,b,c$ be positive real numbers such that $ a^3 \plus{} b^3 \equal{} c^3$.Prove that:

$ a^2 \plus{} b^2 \minus{} c^2 > 6(c \minus{} a)(c \minus{} b)$.[/quote]
Assume wlog $ c\equal{}1$ and then use Langrange Multipliers:
$ F(a,b,c,\lambda) \equal{} a^2\plus{}b^2\plus{}6a\plus{}6b\plus{}6ab\minus{}7 \minus{} \lambda(a^3\plus{}b^3\minus{}1) \ge 0$
$ F_a \equal{} 2a\plus{}6\plus{}6b \minus{} 3 \lambda a^2 \equal{} 0$
$ F_b \equal{} 2b\plus{}6\plus{}6a \minus{} 3 \lambda b^2 \equal{} 0$

Then $ F_a\equal{}0, F_b \equal{} 0 \Rightarrow \minus{}4(a\minus{}b) \equal{} 3\lambda(a\minus{}b)(a\plus{}b)$ So there is a extremum when $ a\equal{}b$. Now assume $ a \neq b$:
$ 3 \lambda \equal{} \frac{\minus{}4}{a\plus{}b}$. Then:
$ F_a \equal{} 2a\plus{}6b\plus{}6b \plus{} \frac{4}{a\plus{}b} a^2 > 0$. So there is only extremum when $ a\equal{}b$:
Then $ a\equal{}b\equal{}\sqrt[3]{\frac{1}{2}}$:
$ a^2\plus{}b^2\plus{}6a\plus{}6b\plus{}6ab\minus{}7 > 0 \iff$
$ 8 \cdot \sqrt[3]{\frac{1}{4}} \plus{} 12 \cdot \sqrt[3]{\frac{1}{2}} \minus{} 7 > 0$ which is obvious.
Then we just have to consider the points $ a\equal{}1,b\equal{}0$ and $ a\equal{}0,b\equal{}1$ in which there are equality.

[quote="Agr_94_Math"]Yeah I too found the same mistake in his solution but waited for him to know it himself.
Jaswinder one advice: Giving a solution for any inequality problem is not important as anyone can do that.
Giving a proper and perfect solution only matters.[/quote]

trying to become bond........appreciate the fact that he tried..........
and he wasnt very wrong as well....

ps..nothing personal

 :ninja:  :noo:  :alien:

[quote="lasha"]Obviously, $ c > b$ and $ c > a$. If $ 2c > \equal{} a$ and $ 2c > \equal{} b$, than $ c > \equal{} a \plus{} b$ and $ a^{3} \plus{} b^{3} > \equal{} (a \plus{} b)^{3}$, which is impossible. WLOG assume that $ a > \equal{} b$. Than, $ a > 2c$. Let's consider $ a^{2} \plus{} b^{2} \minus{} c^2 \minus{} 6(c \minus{} a)(c \minus{} b) \equal{} b^{2} \plus{} 6b(c \minus{} a) \plus{} (a \minus{} c)(a \plus{} 7c)$ as a quadratic polynomial on $ b$. As the coeeficient with $ b^{2}$ is $ 1$, if $ D < 0$, than the polynomial is always positive. $ D \equal{} 32(c \minus{} a)(2c \minus{} a) < 0$, which implies that the polynomial is positive, as desired.    Lasha Lakirbaia.[/quote]
I'm afraid...but this solution has many errors in it....
[u]Ist error[/u](Also pointed out by LEE TAE HUN)
[quote]If $ 2c > \equal{} a$ and $ 2c > \equal{} b$, than $ c > \equal{} a \plus{} b$ and $ a^{3} \plus{} b^{3} > \equal{} (a \plus{} b)^{3}$, which is impossible \cdots \cdots Than, $ a > 2c$[/quote]
This is false....actually it should have been If $ c \ge 2a$ and $ c \ge 2b$, than $ c > \equal{} a \plus{} b$ and $ a^{3} \plus{} b^{3} > \equal{} (a \plus{} b)^{3}$, which is impossible...note the changes made in the inequality.
[u]IInd error[/u]
[quote]WLOG assume that $ a \ge b$[/quote]
Sorry, but the polynomial isn't symmetric in the three variables.....so this isn't allowed
Hope I'm clear......

Also [quote="Ali3985"]We must take two cases:

First : $ c < a , c < b$

Second : $ c > a , c > b$

And we complete[/quote]
@ali3985 - The first case is not valid at all.....The second case is the only one that holds....

@Rijul saini: of course it is allowed to assume $ a \ge b$.  :wink:

@Mathias DK : Why we can assume wlog c=1 ?

@enndb0x

take a=> a.k  b=> b.k c=>c.k where k is a positive real number so inequality satisfies again.So you can make c=1

Yes , I know about homogeneity ,but I didn't think we can change something in the condition  when using Lagrange Multipliers , but that's ok now -thanks.

And of course we can assume that $ a\ge b$ because we know for a fact that c is the greatest variable.  The inequailty might not be symmetric in a,b,c, but it IS symmetric in a,b.

[quote="sqing"][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=3105746#p3105746]Strengthen of India 2009:[/url]
Let $ a,b,c$ be positive real numbers such that $ a^3 \plus{} b^3 \equal{} c^3$. Prove that\[a^2 \plus{} b^2 \minus{} c^2 \ge\left(2+\sqrt[3]{4}+2\sqrt[3]{2}\right)(c \minus{} a)(c \minus{} b).\][/quote]
We'll find the maximal $k$ for which the inequality $ a^2 \plus{} b^2 \minus{} c^2 \geq k(c \minus{} a)(c \minus{} b)$
is true for all positives $a$, $b$ and $c$ such that $a^3 \plus{} b^3 \equal{} c^3$.
$a^2 \plus{} b^2 \minus{} c^2 \geq k(c \minus{} a)(c \minus{} b)\Leftrightarrow a^2+b^2-\frac{a^3+b^3}{c}\geq k(c-a)(c-b)\Leftrightarrow$
$\Leftrightarrow(c-a)a^2+(c-b)b^2\geq kc(c-a)(c-b)\Leftrightarrow$
$\Leftrightarrow(c-a)\cdot\frac{c^3-b^3}{a}+(c-b)\cdot\frac{c^3-a^3}{b}\geq kc(c-a)(c-b)\Leftrightarrow$
$\Leftrightarrow(c^2+cb+b^2)b+(c^2+ca+a^2)a\geq kabc\Leftrightarrow$
$\Leftrightarrow c^2(a+b)+c(a^2+b^2)+a^3+b^3\geq kabc\Leftrightarrow$
$\Leftrightarrow c(a+b)+a^2+b^2+c^2\geq kab\Leftrightarrow$
$\Leftrightarrow (a+b)\sqrt[3]{a^3+b^3}+a^2+b^2+\left(\sqrt[3]{a^3+b^3}\right)^2\geq kab$.
Now by AM-GM we obtain:
$(a+b)\sqrt[3]{a^3+b^3}+a^2+b^2+\left(\sqrt[3]{a^3+b^3}\right)^2\geq(2\sqrt[3]2+2+\sqrt[3]4)ab$.
The equality occurs for $a=b$.
Hence, the answer is indeed $2\sqrt[3]2+2+\sqrt[3]4$.

Given
$a^3+b^3=c^3$ eq1
To prove :$a^2+b^2-c^2>6(c-a)(c-b)$
Proof: let a/c=x,b/c=y
From above Eq 1
$x^3+y^3=1$
$x^2+y^2-1>6(1-x)(1-y)$eq2
Now substitute $x^3+y^3=1$ in eq2
We get $x^2(1-x)+y^2(1-y)>6(1-x)(1-y)$eq3
Now eq3 could be written as
$x^2/(1-y)+y^2/(1-x)>6$
Now substitute $x^2/(1-y)=1+y+y^2/x $from Eq 1
Similarly with other one also
Now we need to prove $(1+y+y^2/x)+(1+x+x^2)/y > 6$
This is true by simple AM-GMinequality

The crux to this problem is the identity $(a+b-c)^3=a^3+b^3-c^3+3(a+b)(c-a)(c-b)=3(a+b)(c-a)(c-b)$

Rewrite the ineq in the form $$(a+b-c)^2>8(c-a)(c-b)$$
Since $a+b>c$ the ineq is equivalent to
$$(a+b-c)^3>8(c-a)(c-b)(a+b-c)$$
by using the identity, we need to prove $$3(a+b)>8(a+b-c)$$
This is same as $8c>5(a+b)$. Using $\dfrac{a+b}{2} \le \sqrt[3]{\dfrac{a^3+b^3}{2}}=\dfrac{c}{\sqrt[3]{2}}$,  we need to show $4\sqrt[3]{2}>5$ which is true bcoz $128>125$.

The same proof actually yields a better constant $\dfrac{2+\sqrt[3]{4}}{\sqrt[3]{4}-1}\approx 6.10724$ in place of $6$

Let $ a,b,c$ be positive real numbers such that $ a^3 + b^3 = c^3$.Prove that
$$3a^2+3b^2-2c^2\geq 2(7+6\sqrt[3]{2}+4\sqrt[3]{4})(c-a)(c-b)$$
$$3a^2+3b^2-c^2\geq (22+18\sqrt[3]{2}+13\sqrt[3]{4})(c-a)(c-b)$$
[url=https://artofproblemsolving.com/community/c6h299899p1832596]h[/url] [url=https://math.stackexchange.com/questions/2673543/if-a3b3-c3-prove-that?noredirect=1]h[/url]


logarithms calculus integration geometry rectangle

I have two questions:
1) d/dx(ln t)^2 = ?  (ln is natural log)

2) If n is any positive integer which is larger: ln(n) or the sum 1/2 + 1/3 + 1/4 + ... + 1/n ?

Please show all your work (thanks so much)

Do you know the chain rule?

\[\frac{d((ln x)^2)}{dx} = \frac{d((\ln x)^2)}{d(\ln x)}\cdot \frac{d(\ln x)}{dx} = 2\ln x \cdot \frac{1}{x}\]

1) i assume you mean d/dx(ln x)^2
this is a simple application of the chain rule.
(2ln x)(1/x)

2) ln x is the integral from 1 -> infinity of 1/x
since 1/x is decreasing over that interval, the sum
1/2 + 1/3... is less than ln x (think of approximating the integral with rectangles measured from the right endpoint.)

I hope this makes sense.

I don't want to use integration for (2) (although it's much easier  :D ).

It is easy to show that $(1+\frac{1}{n})^{n+1}$ is decreasing, approaching to $e$.

Thus $e < (1+\frac{1}{n})^{n+1}$
Take logarithm, we have $\frac{1}{n+1} < \ln (1+\frac{1}{n}) = \ln (n+1) - \ln n$
Let $n=1,2,\dots,n-1$ we have
$\frac{1}{2} <\ln 2 - \ln 1$
$\frac{1}{3} <\ln 3 - \ln 2$
...
$\frac{1}{n} <\ln n - \ln (n-1)$
add them up
we get $\frac{1}{2} + \cdots + \frac{1}{n} < \ln n$.


inequalities
triangle inequality
real analysis
real analysis theorems

Let $ x\equal{}x_{1}x_{2}......x_{n}$ and similarly $ y\equal{}y_{1}y_{2}.....y_{n}$ and $ z\equal{}z_{1}z_{2}....z_{n}$ be n-tuples of real values. Define $ d(x,y)$ to be the number of positions in which each tuple vary. Someone help me to prove the triangle inequality please! :(

If $ x_i \neq z_i$, then at least one of this is true: $ x_i \neq y_i$, $ y_i \neq z_i$.

Suppose that $ x$ and $ z$ differ at the $ i$th position, so that $ x_{i}\neq z_{i}$.  What happens if $ x$ and $ y$ disagree at the $ i$th position?  What can you say about $ y$ and $ z$?


calculus
derivative
function
linear algebra
matrix
real analysis
real analysis unsolved

What is the physical interpretation of df/dxdy or df/dydx for a function f(x,y)?

That should perhaps depend on what the function f(x,y) represents.

first of all, a second-order mixed derivative is written as $ \frac{\partial^{2}f}{\partial x\;\partial y}$ or $ \frac{\partial^{2}f}{\partial y\;\partial x}$ (and not $ \frac{\partial f}{\partial x\;\partial y}$ or $ \frac{\partial f}{\partial y\;\partial x}$). [u]FYI[/u]: mixed derivatives are the entries off of the main diagonal in the Hessian matrix, do you know what that is?

second, what you are asking for is a [b]geometric [/b]interpretation and not a physical interpretation. if you are asking for a physical interpretation, you would first have to state in your question what physical property $ f(x,y)$ represents and what $ x$ and $ y$ are, i.e. whether $ f(x,y)$ represents the temperature of a rectangular plate where $ x$ and $ y$ represent a pair of orthogonal co-ordinates. only then can i give a physical interpretation.

and third, $ f(x,y)$ can be geometrically visualized as a surface in 3 dimensions, so a [b]geometric [/b]interpretation of $ \frac{\partial^{2}f}{\partial y\;\partial x}$ would be the [b]twist [/b]of that surface. for example, $ \frac{\partial f}{\partial x}$ tells us how fast $ f(x,y)$ changes with respect to changes in $ x$, and doing a $ \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right)\;\;\left(\;\equiv\;\frac{\partial^{2}f}{\partial y\;\partial x}\right)$ then tells us how fast $ \frac{\partial f}{\partial x}$ changes with respect to changes in $ y$, which in plain english is referred to as the twist of that surface.


[quote="academicsrule"]I saw this question somewhere, and it only gave the answer. I disagreed with the answer...

Basically, there is a trapezoid of sides 3, 3, 6, and 2, where the parallel sides are of lengths 3 and 6. It says to find the area.

Their answer: [hide]4sqrt5[/hide]

How?[/quote]
Well this isn't really MOEMS, but this is how you do it.
Set up an equation, with y being the altitude, and x being the leg of the triangle with 3 as the hypotenuse.
$ x^{2}\plus{}y^{2}\equal{}9$
$ (3\minus{}x)^{2}\plus{}y^{2}\equal{}4$
$ 9\minus{}6x\plus{}x^{2}\plus{}y^{2}\equal{}4$
$ 6x\minus{}9\equal{}5$
$ 6x\equal{}14$
$ x\equal{}\frac{7}{3}$
$ \frac{49}{9}\plus{}y^2\equal{}9$
$ y^2\equal{}\frac{32}{9}$
$ y\equal{}\frac{4\sqrt{2}}{3}$
$ \frac{3\plus{}6}{2}\cdot\frac{4\sqrt{2}}{3}\equal{}\frac{36\sqrt{2}}{6}\equal{}6\sqrt{2}$
Eh I probably did something wrong, it's late and I'm tired as it is.

That's what I got. So, I can safely say their answer was wrong?

I did it the same way as you did.

Eh...
I still think I did something wrong though...

I checked over your work, Zhang, and it seems right to me. I'm not the best person to check though.

everything seems right to me as well, but I didn't spend that long

Eh I guess the answer key is wrong then...
Not the first time I suppose.

Just curious, but what was the answer to the question?  I got the same answer as you, too, and your method is seems good for the question.

umm...are you asking what the answer sheet answer was because ZhangPeijins answer is $ 6\sqrt{2}$ and the book answe is $ 4\sqrt{5}$.

It's kinda implied that I would like to know the book answer, cuz I already mentioned that I got the same answer as Zhang :P .  Anyways, $ 4 \sqrt {5}$?  That's weird, though the two numbers are close in value...  Hmm...


Has anyone ever noticed their # of posts goes up when they press search over the one in the window of information on the left?...

[quote="eggylv999"]...over the one in the window of information on the left?...
[/quote]

What are you talking about?


rrusczyk wrote:Prove that if M is inside triangle ABC such that the areas of ABM and ACM are equal, then M lies on the median from A to the midpoint of BC.

[hide]Let [X] represent the area of region X.  Since [ABM] = [ACM], and both triangles have the same base AM, the altitudes from C to AM and B to AM must be equal.  Now extend AM to the point D on BC.  MD is clearly parallel to AM, the altitudes from B and C to AM are perpendicular to AM, so these lines are also perpendicular to MD.  Thus, they are the altitudes to MD in triangles BMD and CMD.  Since BMD and CMD have the same base MD and the same height, their areas are equal.  Thus [BMD] = [CMD], so [ABD] = [ABM] + [BMD] = [ACM] + [CMD] = [ACD].  Since ABD and ACD have the same altitude to BC, the bases corresponding to that altitude must be equal, i.e. BD = DC.  Thus D is the midpoint of BC, AD is the median to BC, and M is thus on the median to BC.[/hide]

[quote="mathfanatic"][quote="blahblahblah"]Mathfanatic, you are mostly correct, but you proved the wrong direction.

You proved that if M lies on the median from A to the midpoint of BC, then [ABM]=[ACM].  Now see if you can fix this.[/quote]

No, I think I proved the correct direction.  I definitely started from [ABM] = [ACM], so if I proved [ABM] = [ACM], then all that stuff above is both circular and pointless.[/quote]

Yeah, I read what you did wrong.  I actually started with the observation that if we pick a point on the median and consider the altitudes to BC, then [ABM] and [ACM] will be equal.  From there it's not difficult to show that picking a point which does NOT lie on the median will cause us to lose equality.  My mistake.


Let $a_0,a_1$ are positive real numbers. Let's define the sequence $a_n$ as follow :
$a_{n+1}=\dfrac{2}{a_n+a_{n-1}}$ $\forall n\ge 1$
Find $\lim_{n\rightarrow \infty} a_n$


a) Solve the following system of linear equations in under 15 seconds. (Of course actually timing yourself is pointless. Just follow the spirit of the rule--no calculators, try to do it entirely in your head, etc.)
61x+67y=4096
67x+61y=4096

b) Prove that the method you used above works for all systems of linear equations of the form
ax+by=c
bx+ay=c
(Provided of course that a and b are not equal.)

[hide]a) by elimination we get -6x+6y=0

y=x

128x=4096

x=32

and y=32

b) ax+by=c

bx+ay=c

ax-bx+by-ay=0

x(a-b)-y(-b+a)=0

x(a-b)=y(a-b)

x=y

ax+bx=c

(a+b)x=c

x=c/(a+b)

and since x=y, y=c/(a+b)[/hide]

Hello, LynnelleYe!

Quote:a) Solve the following system of linear equations in under 15 seconds.
61x + 67y = 4096
67x + 61y = 4096

b) Prove that the method you used above works for all systems of linear equations of the form:
ax + by = c
bx + ay = c
(Provided of course that a and b are not equal.)

[hide]

(a) Add the equations: 128x + 128y = 8192

I recognized the coefficients: 27(x + y) = 213

Hence: x + y = 26 = 64

From the symmetry of the equations: x = y

Therefore: x = y = 32.

(b) Add the equations: (a + b)x + (a + b)y = 2c

Then: x + y = 2c/(a + b)

From the symmetry: x = y.

Therefore x = c/(a + b), y = c/(a + b)

[/hide]


[quote="greatestmaths"]let  the group  A ={1;1/2;1/3;.....;1/n}  such that n>3

let the function   f: [0;1] ----->[0;1/4]

such that f(x) is the distance between x and the nearer member from A to x.

1) is this function injective ?

2) is this function surjective?[/quote]

$ f(\frac{2}{3})\equal{}f(\frac{5}{6})\equal{}\frac{1}{6}$ and so f is not injective.
$ f(1\minus{}a)\equal{}a$ $ \forall a\in[0,\frac{1}{4}]$ and so f is surjective.

[quote="greatestmaths"]no you are wrong in the second statemant because if wa take a = 1/4

we found f(x)=1/4   x=1- 1/4 = 3/4
that mean f(3/4) = 1/4 but that wrong because f( 3/4)=1/2[/quote]

I'm sorry, but you're wrong.

The elements of $ A$ which are the nearest from $ \frac{3}{4}$ are $ 1$ and $ \frac{1}{2}$ and so $ f(\frac{3}{4})\equal{}1\minus{}\frac{3}{4}\equal{}\frac{3}{4}\minus{}\frac{1}{2}\equal{}\frac{1}{4}$

[quote="greatestmaths"]yes but if we take x = 2/3

f(2/3) = 1/6

because 2/3+1/6 = 1/2

      but  2/3+1/3=1

so 1-1/6 =5/6 isn't 2/3

sorry but you're wrong[/quote]

What sort of game are you playing ?

First you say that my first statement is true and my second is wrong.
Then you seem to say my second statement is true and my first is wrong !

Obviously $ \frac{5}{6}\neq\frac{2}{3}$ !
What i wrote is $ f(\frac{5}{6})\equal{}f(\frac{2}{3})$ :
The element of $ A$ which is the nearest of $ \frac{5}{6}$ is $ 1$ so $ f(\frac{5}{6})\equal{}1\minus{}\frac{5}{6}\equal{}\frac{1}{6}$

The element of $ A$ which is the nearest of $ \frac{2}{3}$ is $ \frac{1}{2}$ so $ f(\frac{2}{3})\equal{}\frac{2}{3}\minus{}\frac{1}{2}\equal{}\frac{1}{6}$


[quote="aev5peru"][quote="aev5peru"]demostrar que al colorear los numero naturales de dos colores se puede encontrar 3 del mismo color que esten en progresion aritmetica.
esto se me ocurrio resolviendo un problema, porfavor espero soluciones.[/quote]
creo que este es un caso particular,  :o  y se generalisa para n colores y se llama teorema de van der waerden, y uno puede encontrar progresiones tan largas como quiera.
chau.[/quote]

Eso si es asi, aunque el enunciado de Van der Waerden tiene un enunciado un poco diferente. Es así:

Para todos $n$ y $k$ enteros positivos existe un número mínimo $W(n,k)$ tal que si se colorean los numeros desde $1$ hasta $W(n,k)$ con $n$ colores, existe una progresión aritmética monocromática de longitud $k$.

La prueba que conozco es una inducción rarisima. Es muy larga. No se si alguien tenga una solución corta de este teorema, pero es bastante complejo.


Objective: Dodge as many spikes as you can. My record is 417.

[url=http://www.freewebarcade.com/game/ragdoll-avalanche-2/]http://www.freewebarcade.com/game/ragdoll-avalanche-2/[/url]

i got 519
The message is too small. Please make the message longer before submitting.
 ok


No idea of how difficult this is, theres problem some theorems about logic that make it trivial but anyway: can you prove (or disprove!) that you cannot express a AND b using only => (implies).
(for example, how do you know something like ((a=>b)=>(b=>a))=>a is not the same as a AND b).

Cute problem.  Hint in spoiler: [hide]In any "implies" expression, set the rightmost variable to true.[/hide]

It is well known that:
[tex]\neg \left( {\neg p} \right) \equiv p[/tex]

[tex]\neg \left( {p \wedge q} \right) \equiv \left( {\neg p} \right) \vee \left( {\neg q} \right)[/tex]

[tex]\neg \left( {p \vee q} \right) \equiv \left( {\neg p} \right) \wedge \left( {\neg q} \right)[/tex]

otherwise let us denote by:

[tex]\overline {\overline p }  \equiv p[/tex]

[tex]\overline {p \wedge q}  \equiv \overline p  \vee \overline q [/tex]

[tex]\overline {p \vee q}  \equiv \overline p  \wedge \overline q [/tex]
(Moivre)

Then
[tex]p \to q \mathop \equiv  \overline p  \vee q[/tex]

[tex]p \to q\mathop  \equiv  \neg p \vee q[/tex]

whence

[tex]p \to q \equiv \overline p  \vee q \equiv \overline {p \wedge \overline q } [/tex]

[tex]p \to \overline q  \equiv \overline {p \wedge q} [/tex]

[b]Conclusion:[/b]
[tex]p \wedge q \equiv \overline {p \to \overline q }  \equiv \neg \left( {p \to \neg q} \right)[/tex]

This is going to Other Problem Solving Topics...

[quote="metru"][b]Conclusion:[/b]
[tex]p \wedge q \equiv \overline {p \to \overline q }  \equiv \neg \left( {p \to \neg q} \right)[/tex][/quote]
I think TripleM was asking whether it was possible using just implication, not allowing negation.  With just implication, it is impossible.  My hint supplies an approach for proving it.  Namely, by fixing one variable, we can make any implication expression true, but you can't set an AND to be true by fixing just one variable.

Ah that looks as if it works, Ravi. (yes I did mean just using implies, not "not" :)) Are there any generalisations to this? What other truth tables with 2 variables are impossible? More variables?

[hide]

Given (a => b), there are no restrictions stating that "b" cannot exist without "a".  Such a restriction would require an AND to include the condition (b => a).

Breaking the implications into separate parts would require an AND to make them all true, such as:

(a => b) AND ((a => b) => (b => a)).

If you use "implies", then:

(a => b) => ((a => b) => (b => a))

is equivalent to

(a => b) => (b => a)

Which is an illogical statement as stated above.

[/hide]


algebra
polynomial
complex numbers
algebra theorems

Let $ \omega \equal{} e^{\frac {2\pi}{p}}$ where $ p$ is a prime number and let $ P(x) \equal{} A_0 \plus{} A_1x \plus{} A_2x^2 \plus{} ... \plus{} A_{p \minus{} 1}x^{p \minus{} 1}$ where $ A_0, A_1, ..., A_{p \minus{} 1}$ are [b]rational[/b].
If we know that $ P(\omega) \equal{} 0$, then prove that $ A_0 \equal{} A_1 \equal{} ... \equal{} A_{p \minus{} 1}$.

Consider $ F(x) \equal{} \sum_{i \equal{} 0}^{p \minus{} 1} x^i$.
Let $ G(x)$ be the minimal polynomial of $ \omega$ - the monic polynomial of least degree such that $ \omega$ is its zero.
Such polynomial exists, because $ F(\omega) \equal{} 0$ and $ F$ has rational coefficients. It is unique, because if $ G_{1,2}$ are 2 polynomials that satisfy this property, their difference is also a polynomial with rational coefficient, but its degree is smaller (and we can normalize it so that it will be monic).

[u]Fact 1:[/u] $ F$ is irreducible - it is a known fact about Cyclotomic Polynomials, proofs can be found anywhere. A easy proof using Eisenstein's criterion can be found here:
[url]http://en.wikipedia.org/wiki/Eisenstein%27s_criterion[/url]

[u]Fact 2:[/u] $ G$ divides any polynomial with rational coefficients such that $ \omega$ is its zero.
Proof: assume otherwise. Divide those 2 polynomials, and call the residue $ R(x)$. R's degree is less than G's degree and $ \omega$ is its zero. Normalize it (possible because it is non-zero) and you get a contradiction to minimality.

Combine those 2 facts and get that $ F$ is the minimal polynomial: $ F \equal{} G$.

If $ A_0 \equal{} 0$, $ P$ must be the zero polynomial (its degree is smaller than F's degree). Thus $ A_0 \equal{} A_1 \equal{} \cdots \equal{} A_{p \minus{} 1} \equal{} 0$.
Otherwise, $ F | \frac {P}{A_0}, deg(F) \equal{} deg(P) \implies P(x) \equal{} A_0F(x)$ and the conclusion follows.

This is an interesting theorem when stated geometrically:  it states that an equiangular polygon with a prime number of sides, all of which have rational length, must in fact be equilateral, or stated perhaps more naturally, an equiangular polygon with a prime number of sides which is not equilateral has at least one irrational side.  See, for example, the discussion [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=152452]here[/url].

I would also like to mention that this is not only useful geometrically but it also has applications in combinatorics.  See, for example, the book [i]Complex Numbers from A to...Z[/i] (english version by Titu Andreescu and Dorin Andrica).

Thankyou. I love simple little things to like this.


topology
real analysis
real analysis unsolved

What is the simpliest way to prove that every infinite set has a countable dense subset?
and is there a set A such that the intersection of A with the rational numbers is not dense in A?

[quote="perfect_radio"]
This is not true. $\mathbb Z$ is infinite, but has no dense subset.[/quote]
What do you mean? $\mathbb{Z}$ is dense in $\mathbb{Z}$ and $\mathbb{Z}$ is infinite. In general, we often lose separability condition if we take the product of separable spaces over the index set which cardinal is bigger than continuum.

If it makes you feel any better, perfect_radio, I had the exact same thought - it's just that I saw the catch before I had gotten around to posting it.

I assume that you mean "every infinite subset of $\mathbb{R}$" where you wrote "every infinite set". There are metric spaces in which this property fails blatantly.

Let $X$ be a subset of $\mathbb{R}$. We want to find $Y\subset X$ countable so that $Y$ is dense in $X$.
Fix $n$ and find a countable set $S_{n}$ of real numbers such that all real numbers are within $\frac1n$ of some element of $S_{n}$. We can choose the concrete example $S_{n}=\{\frac{k}{n}: k\in \mathbb{Z}\}$.
For each $p\in S_{n}$, let $U_{p}$ be the set $\{q: |q-p|<\frac1n\}$. By our choice of $S_{n}$, the $U_{p}$ cover $\mathbb{R}$. Now we choose a countable set $Y_{n}\subset X$ as follows:
For each $p\in S_{n}$, include one element of $U_{p}\cap X$ if $U_{p}\cap X$ is nonempty. Include nothing otherwise.
$Y_{n}$ has the property that for any $x\in X$, there exists some $y\in Y_{n}$ such that $|x-y|<\frac2n$ (since each such $x$ is in some $U_{p}\cap X$, which also contains an element of $Y_{n}$).
Repeat this process for all positive integers $n$, and let $Y=\cup Y_{n}$.

This proof generalizes easily to show that any subset of a separable metric space is separable.

[quote="kgt1c"]and is there a set A such that the intersection of A with the rational numbers is not dense in A?[/quote]
Yes, for example the set of all irrational numbers.


The administrators come together and create a NewAoPS. Barely an hour after they set up the site, 237factorial joins on October 1st. Then 237factorial tells math93 about NewAoPS, and he joins on October 5th.

If 237factorial consistently posts 17 times a day, and math93 consistently posts 22 times a day, when will math93 catch up with 237factorial?

Show your work.

(Sorry I picked on you, 236factorial and math92, but you were on the top 50 list and were some of the only members with numbers in their names.)

haha nice problem  :P I have to say, I don't know what math92 will do....

I'd have to say that I found the content of this problem amusing.  :lol:

:rotfl:  :rotfl:
very nice problem!!!
i'm going to try and find the answer soon!

Here's my solution,

[hide]
Let x=days that math93 has been on the forums; x+4=days that 237factorial has been online.

$17(x+4)=22x$
$17x+68=22x$
$x=13.6\rightarrow x=boxed{14}$

$\therefore$ It'll take math93's posts 13.6 days$\rightarrow$14 days to catch up to 237factorial's number of posts.
[/hide]

Masoud Zargar

really amuzing problem and really nice solution for it.you know i think all of iranian are cool in math.(even you boxedexe,i was kidding you are so cool.age torshi nakhori ye chizi mishi ha.) :D

are u kidding mathnerd 314 :(

this is not the way of posting problems...

i think the problem is fine....no body was offended by it...


The fractions $ \frac{ax\plus{}b}{cx\plus{}d}$ and $ \frac{b}{d}$ are unequal if:

$ \textbf{(A)}\ a\equal{}c\equal{}1, x\neq 0 \qquad
\textbf{(B)}\ a\equal{}b\equal{}0 \qquad
\textbf{(C)}\ a\equal{}c\equal{}0 \\
\textbf{(D)}\ x\equal{}0 \qquad
\textbf{(E)}\ ad\equal{}bc$

[hide]
If $ \frac{ax\plus{}b}{cx\plus{}d} \equal{} \frac{b}{d} \Longrightarrow ad\equal{}bc$ So it's not $ E$.
$ B,C,D$ also turn out to make the expressions equal

So $ A$ is the only possible answer. Checking this
$ \frac{x\plus{}b}{x\plus{}d} \equal{} \frac{b}{d} \Rightarrow b\equal{}d$

...Well It doesn't really specify that $ d\neq b$ but I'll assume that it has been
So answer is $ \boxed{A}$
[/hide]


The sum of three numbers is 4, the sum of their squares is 10, the sum of their cubes is 22. What is the sum of their fourth powers?

java has solved it but i could not understand

[quote="yagaron"]kunny, define $a b c$ it's a little bit trouble.[/quote]

You can use the identity $a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$.

yes we could, i know that

and we need to define $a b+a c+b c$ too.

so the final formula is long.

i make that and simplificate. to become more compact and simple to calculate

:)


[quote="Mircea Lascu"]If $x,y,z>0$, then
\[ (x+y+z)\left[\frac{4yz}{(y+z)^{2}}\cdot\frac{1}{x}+\frac{4zx}{(z+x)^{2}}\cdot\frac{1}{y}+\frac{4xy}{(x+y)^{2}}\cdot\frac{1}{z}\right]\geq 9.  \][/quote]

This is just a reformulation of Iran'96. The inequality rewrites as

\[ \frac{1}{(x(y+z))^{2}}+\frac{1}{(y(z+x))^{2}}+\frac{1}{(z(x+y))^{2}}\geq\frac{9}{4xyz(x+y+z)}.  \]

Now, setting $yz=a, zx=b$ and $xy=c$ the inequality becomes Iran'96 just like I said:

\[ \frac{1}{(a+b)^{2}}+\frac{1}{(b+c)^{2}}+\frac{1}{(c+a)^{2}}\geq\frac{9}{4(ab+bc+ca)}.  \]

[quote="silouan"]By Chebychev AM-GM and then Iran 1996 inequality we  have
$(x+y+z)\left[\frac{4yz}{(y+z)^{2}}\cdot\frac{1}{x}+\frac{4zx}{(z+x)^{2}}\cdot\frac{1}{y}+\frac{4xy}{(x+y)^{2}}\cdot\frac{1}{z}\right]\geq (\frac{4}{(y+z)^2}+\frac{4}{(z+x)^2}+\frac{4}{(x+y)^2})\frac{(x+y+z)^2}{3}\geq (\frac{4}{(y+z)^2}+\frac{4}{(z+x)^2}+\frac{4}{(x+y)^2})(xy+yz+zx)\geq 9$[/quote]

This is wrong, Silouan. When you apply Chebyshef, the inequality is reversed.
WLOG, $x\leq y\leq z$. From here it clear that $x^{2}\leq y^{2}\leq z^{2}$ and it is also clear that

\[ y+z\geq z+x\geq x+y  \] and \[ (y+z)^{2}\geq (z+x)^{2}\geq (x+y)^{2}.  \]

In the same time it is obvious that \[ \frac{1}{(y+z)^{2}}\leq\frac{1}{(z+x)^{2}}\leq\frac{1}{(x+y)^{2}}  \]
and \[ \frac{yz}{x}\geq\frac{zx}{y}\geq\frac{xy}{z}.  \]

So, supposing that you have applied Chebyshef to the following two triplets \[ \left(\frac{yz}{x},\frac{zx}{y},\frac{xy}{z}\right)  \] and \[ \left(\frac{1}{(y+z)^{2}},\frac{1}{(z+x)^{2}},\frac{1}{(x+y)^{2}}\right),  \]

the inequality is reversed. :)

Yes  :blush: . Thank you for pointing out .
Your solution is nice......


$ \mbox{Given that :  cosec}\Theta \minus{} \mbox{sin}\Theta \equal{} m \ \ \mbox{and} \ \mbox{sec}\Theta \minus{} \mbox{cos}\Theta \equal{} n \\
\mbox{Show that : } \ \ (m^2n)^{\frac {2}{3}} \plus{} (mn^2)^{\frac {2}{3}} \equal{} 1$

Solution

[hide]$ m=\dfrac{1-\sin^2 \theta}{\sin \theta}, n=\dfrac{1-\cos^2 \theta}{\cos \theta}$.

$ m^2n=\dfrac{\cos^4 \theta}{\cos \theta}= \cos^3 \theta$.

$ n^2m=\dfrac{\sin^4 \theta}{\sin \theta}= \sin^3 \theta$.

$ (\cos^3 \theta)^\dfrac{2}{3}+(\sin^3 \theta)^\dfrac{2}{3}=1$.

$ \cos^2 \theta +\sin^2 \theta=1$.[/hide]


[quote=yiz76][quote=blondu]- Master, may I ask you a question?
- Yes
- I asked you.
- I answered.
- What was your answer?
- Is this another question?... :D[/quote]

 :rotfl:  :rotfl:  :rotfl:[/quote]

Agreed


Fiind all $ x,y,k,\alpha\in\mathbb{N}$ and the prime number $ p$ which satisfy the follow equation: \[ xy^{k}\minus{}yx^{k}\equal{}p^{\alpha}.\]  ( $ \mathbb{N}$ it`s the set of natural numbers)

If $ k \equal{} 0$, then we have $ x \minus{} y \equal{} p^a$, giving infinite solutions here.
If $ k > 0$, then we have $ xy(y^{k \minus{} 1}\minus{}x^{k \minus{} 1}) \equal{} p^a$, so $ x$ and $ y$ are both powers of $ p$. So we have
$ x \equal{} p^m, y \equal{} p^n$
giving $ p^{m \plus{} kn} \minus{} p^{n \plus{} km} \equal{} p^a$ but then we can factor out a common factor of $ p$ until one of the terms is 1, but the only solution here will be $ 2^1 \minus{} 1 \equal{} 2^0$, giving $ x \equal{} 2, y \equal{} 1, k \equal{} 0,p \equal{} 2,a \equal{} 0$.

But the equation is in naturals, so no solution.

Numbers $ xy^{k}$ and $ yx^{k}$ are both odd or both even, then $ p \equal{} 2$
From here it is Easily to find that :$ x \equal{} 2^{a}$ and $ y \equal{} 2^{b}$.
 We have $ xy^{k} \plus{} yx^{k} \equal{} p^{\alpha} \Rightarrow xy(y^{k \minus{} 1} \minus{} x^{k \minus{} 1}) \equal{} 2^{\alpha}$
From $ x \equal{} 2^{a}$ and $ y \equal{} 2^{b}$ we have that : $ y^{k \minus{} 1} \minus{} x^{k \minus{} 1} \equal{} 2^{b(k \minus{} 1)} \minus{} 2^{a(k \minus{} 1)} \equal{} 2^{a(k \minus{} 1)}(2^{(b \minus{} a)(k \minus{} 1)} \minus{} 1) \equal{} 2^{\alpha}$
From here we take the equation : $ 2^{(b \minus{} a)(k \minus{} 1)} \minus{} 1 \equal{} 1 \Rightarrow 2^{(b \minus{} a)(k \minus{} 1)} \equal{} 2^{1} \Longrightarrow b \minus{} a \equal{} 1$ and $ k \minus{} 1 \equal{} 1 , k \equal{} 2$
From here we can easily find that $ y \equal{} 2x$ and $ x \equal{} 1$ so $ y \equal{} 2$ then $ \alpha \equal{} 1$
$ (x,y,p,\alpha) \equal{} (1,2,2,1)$


It is a particular case for $AD, BE, CF$ are concurrent at a point $P$.

It follows from the same lemma that was used to prove problem 2 from 2nd China TST 2006.

Can someone bash this problem with complex numbers?

it s special case of hagge circles when $P$ is an infinity point

[quote=Electron_Madnesss]Actually, this problem was there in [u]Evan's[/u] textbook and my solution to it was (surprisingly!) the same as his, so I am posting it here.

$\rightarrow$ We $(ABC)$ to be the unit circle with $a=1,b=-1 $ and $k=\dfrac{-1}{2}$.
Also, let $s,t \in (ABC)$.
We claim that $K$ is the center of $(STUH)$.

Notice that $x = \dfrac{1}{2}(s+t-1+\dfrac{s}{t}) \implies 2\cdot(2\text{Re}x +1 ) = s+t+\dfrac{1}{s} +\dfrac{1}{t} + \dfrac{s}{t} + \dfrac{t}{s}$, which does not depend on $X$.

Also, $|k+\dfrac{s+t}{2}|^2 = 3+ 2\cdot(2Rex+2) \implies \dfrac{s+t}{2} $ has a fixed distance with $k$ as desired$\blacksquare$[/quote]

I do not understand.
You're not allowed to put $a=1,b=-1$ since $ABC$ is not a right triangle. And what is $X$??

[quote=Electron_Madnesss]Actually, this problem was there in [u]Evan's[/u] textbook and my solution to it was (surprisingly!) the same as his, so I am posting it here.

$\rightarrow$ We $(ABC)$ to be the unit circle with $a=1,b=-1 $ and $k=\dfrac{-1}{2}$.
Also, let $s,t \in (ABC)$.
We claim that $K$ is the center of $(STUH)$.

Notice that $x = \dfrac{1}{2}(s+t-1+\dfrac{s}{t}) \implies 2\cdot(2\text{Re}x +1 ) = s+t+\dfrac{1}{s} +\dfrac{1}{t} + \dfrac{s}{t} + \dfrac{t}{s}$, which does not depend on $X$.

Also, $|k+\dfrac{s+t}{2}|^2 = 3+ 2\cdot(2Rex+2) \implies \dfrac{s+t}{2} $ has a fixed distance with $k$ as desired$\blacksquare$[/quote]

Is this a solution to USAMO 2015 P2 or am I wrong?

Oh Darn, I copied the wrong solution from my notebook.
@above you are correct this the solution of USAMO 2015-2

[hide=Solution][quote=China TST 2006 D7 P1]$H$ is the orthocentre of $\triangle{ABC}$. $D$, $E$, $F$ are on the circumcircle of $\triangle{ABC}$ such that $AD \parallel BE \parallel CF$. $S$, $T$, $U$ are the semetrical points of $D$, $E$, $F$ with respect to $BC$, $CA$, $AB$. Show that $S, T, U, H$ lie on the same circle.[/quote]
[b][color=#000]Solution:[/color][/b] We assume, $D$ $\in$ arc $AC$ not containing $B$. Rename, $S,T,U$ as $D',E',F'$. Clearly, $DD'$, $EE'$, $FF'$ concur on $\odot (ABC)$, say at $P$. Also, it follows from Simson Lines, that $AF,AE'$ are Isogonal WRT $\angle BAC$. We'll now show $PD'CE'$ is cyclic.
\begin{align*} \angle PD'C=\angle PDC=\angle PEC=\angle EE'C \end{align*}
Hence, $PD'CE'$ is cyclic as desired! Similarly, $PF'AE'$ and $BF'D'P$ are cyclic.
\begin{align*} \angle HD'E'=\angle HD'P+\angle PD'E'=\angle ACD+\angle E'CP=\angle BCP \end{align*}
\begin{align*} \angle HF'E'=\angle HF'P -\angle E'F'P=\angle CBF-\angle E'AP=\angle FAP+\angle CAE'=\angle BAP=\angle BCP \end{align*}
These two angle equalities imply, $HD'E'F'$  cyclic $\qquad \blacksquare$
-------------------------------------------------
[hide=Excuse]I know, I should have attached a diagram with this to avoid any misunderstanding due to configuration issue. [/hide]
[/hide]

$a,b,c,d$, with the circumcircle as the unit circle, will be our free variables. Note that due to parallel conditions we have that $e=\frac{ad}{b}$ and $f=\frac{ad}{c}$

Then,
\[s=b+c-bc\overline{d}=b+c-\frac{bc}{d}\]
\[t=a+c-ac\overline{e} = a+c - ac \cdot \frac{b}{ad}=a+c-\frac{bc}{d}\]
\[u=a+b-ab\overline{f} = a+b - ab\cdot \frac{c}{ad}=a+b-\frac{bc}{d}\]
\[h =a+b+c\]
Now, we translate by $\frac{bc}{d}-a-b-c$ to get
\[s'=-a,t'=-b,u'=-c,h'=\frac{bc}{d}\]
These 4 points are clearly all on the unit circle, so we have shown that $S,T,U,H$ can be translated to a unit circle, and are therefore concyclic
$\blacksquare$

Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/vol5.html   then P-hagge circle,  p. 44-48.

Sincerely
Jean-Louis

We use complex number. $(ABC)$ is our unit circle. Consider $A=a,B=b,C=c$. So $h=a+b+c$. And suppose $D$ is any point on the unit circle. From the parallel condition we get a relation between points $a,b,c,d,e,f$ that
\begin{eqnarray*}
ad=be=cf
\end{eqnarray*}
The co-ordinate of $s=b+c-\frac{bc}{d}$. Similarly $t=a+c-\frac{ac}{e},u=a+b-\frac{ab}{f}$. For being $S,T,U,H$ concyclic, the quantity must be
\begin{eqnarray*}
\frac{s-u}{t-u}\div \frac{s-h}{t-h}&=&\frac{\frac{abd-bcf}{df}+c-a}{\frac{abe-acf}{ef}+c-b}\div \frac{c+d}{c+e},\hspace{2em}[abd=bcf, abe=acf]\\
&=&\frac{c-a}{c-b}\cdot \frac{c+e}{c+d}\in \mathbb{R}
\end{eqnarray*}
Substituting the conjugates of all points (1 minute computation) we get the same quantity. Therefore it must be a real number and we are done. $\square$

Sketch :Note that $S\in \odot(\triangle BHC)$ and similarly other cases hold.
Prove that $DS,ET,FU$ are concurrent and they lie on $\odot(\triangle ABC)$
If these 3 lines concurr at $G$ prove that $AGTU$ and similarly other symnetrical quadrilaterals are cyclic.
Now use this cyclicities to prove the final concyclicity.
All of these can be proved by ang[b]e[/b]ls $\blacksquare$

[quote=shobber]$H$ is the orthocentre of $\triangle{ABC}$. $D$, $E$, $F$ are on the circumcircle of $\triangle{ABC}$ such that $AD \parallel BE \parallel CF$. $S$, $T$, $U$ are the semetrical points of $D$, $E$, $F$ with respect to $BC$, $CA$, $AB$. Show that $S, T, U, H$ lie on the same circle.[/quote]

Cited in Titu's book as "MOP 2006"  :maybe:

The parallel condition gives us $ad = be = cf = z$ for some complex number $z$ of magnitude $1$, so $d = \tfrac{z}{a}$, $e = \tfrac{z}{b}$, $f = \tfrac{z}{c}$. Then, using the complex foot formula, we can find that $$\frac{\tfrac{z}{a} + s}{2} = \frac{1}{2}\left(b + c + \frac{z}{a} - \frac{abc}{z}\right).$$
This means $s = b + c - \tfrac{abc}{z}$, and we similarly find that $t = a + c - \tfrac{abc}{z}$, $u = a + b - \tfrac{abc}{z}$. Now, to show that $STUH$ is concyclic, it suffices to prove that $\tfrac{h - s}{u - s} \div \tfrac{h - t}{u - t}$ is real, or it equals its conjugate. First, we evaluate the quantity as it is; noting that $h = a + b + c$, we get
$$\begin{aligned} \frac{h - s}{u - s} \div \frac{h - t}{u - t} &= \frac{a + \tfrac{abc}{z}}{a - c} \div \frac{b + \tfrac{abc}{z}}{b - c} \\ &= \frac{a(z + bc)(b - c)}{b(z + ac)(a - c)}.\end{aligned}$$
The conjugate of this is
$$\begin{aligned} \frac{\tfrac{1}{a}(\tfrac{1}{z} + \tfrac{1}{bc})(\tfrac{1}{b} - \tfrac{1}{z})}{\tfrac{1}{b}(\tfrac{1}{c} + \tfrac{1}{ac})(\tfrac{1}{a} - \tfrac{1}{c})} &= \frac{a(z + bc)(b - c)}{b(z + ac)(a - c)} \\ &= \frac{h - s}{u - s} \div \frac{h - t}{u - t}, \end{aligned}$$
so we are done.

WLOG, let $(ABC)$ be the unit circle and rotate $ABC$ so that $AD$ is parallel to the $y$-axis. Therefore, we have that $d=\frac{1}{a}$, $e=\frac{1}{b}$, and $f=\frac{1}{c}$. Using reflection formulas, we find that $s=b+c-abc$, $t=c+a-abc$, and $u=a+b-abc$. \\

From here, we find that since $\frac{s-t}{h-t}*\frac{h-u}{s-u}$ is equal to its conjugate, meaning that it's real, $S$, $T$, $H$, and $U$ are concyclic, and we are done.

Let the real line be the line through $O$ perpendicular to all of $AD,BE,CF$. Thus, we have $d=\frac{1}{a}$ and so on. We have that $$D'=b+c-\frac{bc}{d}=b+c-abc,$$ and similarly $$E'=c+a-abc,F'=a+b-abc.$$ We wish to show that these are concyclic with $a+b+c$. Of course, shift by $-a-b-c+abc$, so we wish to show $$-a,-b,-c,abc$$ are concyclic. Thus, it suffices to show that $$\frac{c(b-a)(1+ab)}{b(c-a)(1+ac)}$$ is real. This is just because $$\frac{\frac{1}{c}(\frac{1}{b}-\frac{1}{a})(1+\frac{1}{ab})}{\frac{1}{b}(\frac{1}{c}-\frac{1}{a})(1+\frac{1}{ac})}$$ $$=\frac{b(ac-bc)(abc+c)}{c(ab-bc)(abc+b)}=\frac{c(a-b)(ab+1)}{b(a-c)(ac+1)},$$ as desired.

I'm so confused, why did everyone using complex finish by checking angle conditions for concyclicity?

Set $ABC$ on the unit circle and rotate so that $AD$, $BE$, and $CF$ are vertical lines. It follows that $(d, e, f) = (\overline{a}, \overline{b}, \overline{c})$ and $(x, y, z) = (b+c-abc, c+a-abc, a+b-abc)$.

Recalling $h = a+b+c$, quadrilateral $HXYZ$ has circumcenter $\omega = a+b+c-abc$ because \[ x-\omega = -a, \quad y-\omega = -b, \quad z-\omega = -c, \quad h-\omega = abc, \] and all four of these differences have magnitude $1$. $\blacksquare$

Sorry for bumping but I wonder if there exists a synthetic solution.


Three fair, standard six-faced dice of different colors are rolled.
In how many ways can the dice be rolled such that the sum of
the numbers rolled is 10?

Possible combinations:
1-3-6
1-4-5
2-2-6
2-3-5
2-4-4
3-3-4

Each combination has 6 different possibilities, since all 3 dice are distinct. Therefore, the final answer is $ 36$.


Suppose N = {0, 1, 2, 3, ...}. Let f: N -> N be a bijection.
Prove that there exist nonnegative integers a, b and c in N such that
1) a < b < c
2) f(a), f(b), f(c) are in arithmetic progression.

A try for this nice problem.

Let g=f^-1.

The conditions of the problem can be rewritten to : a<b<c and f(b) = (f(a)+f(c))/2
Or, equivalently, we have to find x,y,z, with a=g(x),b=g(y),c=g(z), with(x+z)/2€IN, such as g(x) < g((x+z)/2) < g(z).

Let suppose it's not possible to find such x, z. Then g(n) < g(0) or g(n) > g(2*n) for all n>0.
Since there are only a finite number of number < g(0) and g is injective, for n >= N0 we have g(n) > g(2n) > g(4n) >.... which is clearly impossible since { g(2^kn) } is an infinite set of integer >= 0 and hence not bounded. So x,y,z exist and hence a,b,c.

Sounds correct ?


Here's what I've got:
[hide]\[
4-\frac{3}{4-\frac{3}{4-\frac{3}{\ldots}}} =x \\\\
\frac{3}{4-\frac{3}{4-\frac{3}{\ldots}}}=4-x \\\\
\frac{3}{x}=4-x \\\\
-x^2+4x-3=0\\\\
x=3\text{ or }x=1
\]
[/hide]
I'm not sure which one to keep. Can someone explain where I went wrong, or which answer to keep if I was right?

To determine which answer to keep you could try computing the first couple partial sums.  Im pretty sure the answer is [hide] $x=3$[/hide]

We need rigorous argumentation here.

Denote $s_n=4-\frac{3}{4-\frac{3}{4-...}}$ ($n$ fractions). Then $s_0=4$, $s_{n+1}=4-\frac{3}{s_n}$. And we are to prove $\lim_{n\to\infty}s_n$ exists and find it.

Indeed, suppose $s_n> 3$. Then $s_{n+1}=4-\frac{3}{s_n}>4-1=3$. So $s_n> 3$ for all $n$. Moreover, consider $s_{n+1}-s_n=\frac{4s_n-s_n^2-3}{s_n}=-\frac{(s_n-3)(s_n-1)}{s_n}<0$. It follows that $s_n$ is a decreasing bounded sequence. Due to Wierstrass theorem it has limit $x$.
Then $x=4-\frac{3}{x}$. Therefore $x=1$ or $x=3$, but $s_n>3$ and it means $x=\lim s_n\geq 3$.
Thus $x=3$.

If, however, you defined the sequence so that $s_0 = 4 - 3, s_{n+1} = 4 - \frac{3}{s_n}$, you will get a sequence which converges to 1.  (In fact, you get a sequence which is always 1.)  The partial sums of this sequence have as their first operation subtraction; Myth's has as its first operation division.  His is the way that one normally interprets such an expression (so don't do it my way).  However, interpreting it my way shows why there is that second root.  Also, if you start with $s_0 = k$ for any positive k (maybe negative, too -- just not 0 :) ) the sequence will converge to 3, with the exception of 1.  If you start with k = 1 (as in my example), it sticks at 1.  So, 3 is an "attractor," I think the word is, and 1 isn't.


Putnam probability geometry calculus integration trigonometry email

I did not solve this problem (I put 3/5, but made all sorts of independence and weird assumptions thta are almost certainly false).

However! I've thought of this problem before now (in about October), and almost put it on the forums to ask if anyone had the answer. So now I really wish I'd done so, because the people in charge of the Putnam would have FREAKED OUT.

The problem is well-known; that doesn't mean that the answer is, or even less a proof.

This problem is reduced to finding the Expected Area of a triangle formed by selecting 3 random points in the circle.

Clearly the 4th point selected will create a convex polygon if the point is outside the triangle.

No, not simply "outside the triangle." Extend the sides of the triangle in both directions. This cuts the disk into 7 regions, one of which is the triangle itself, and the other 6 of which lie outside. The area you want is of the sum of the three outside regions which border the faces of the triangle.

There goes my 120!

Ahh yes this is obviously correct. I rushed into finding the average value of the triangle and missed the easy conceptual portion.

Actually it does indeed reduce to finding the average area of a triangle. Iff ABCD is not a convex quadrilateral, exactly one of the triangles formed by 3 of the points contains the fourth one. Hence the answer is 1-4P(triangle ABC contains D).

Great observation - Does anyone have an efficient manner to calculate the average area of the triangle?

I tried to use what I knew about point picking in a disk, is there a more direct method?

I guess you could take the centre of the circle, and sum the areas of the three triangles formed (being careful with orientation), and that should give you a bunch of ugly integrals which I really don't want to do. (Didn't have enough time left to do then, and with finals starting, no time to try it now, though I still have time to waste on the internet, apparently.)

The expacted area of the triangle given by three random points in the unit circle is 35/(48 pi).  The proof I have is not pretty and I wonder if
anybody knows a nice solution to the problem.

Yossarian, marcin:

At this point, I'll settle for ugly. What do you have?

I've been following a different path, based on asking whether two segments $AC$ and $BD$ intersect. This leads to the triple integral $\frac{3}{\pi^{3}}\int_{0}^{1}\int_{0}^{2\pi}\int_{0}^{2\pi}2s(1-s^{2})^{2}\sqrt{1-s^{2}\sin^{2}\theta}\sqrt{1-s^{2}\sin^{2}\phi}|\sin(\theta-\phi)|\,d\theta\,d\phi\,ds$ for the probability of a convex quadrilateral. If I can evaluate it, I'll post the derivation and we'll have an alternate solution.

The solution is in the attachement.
Just found that other solutions are at
 http://www.unl.edu/amc/a-activities/a7-problems/putnamindex.shtml

I could download the attachment. It looks like it was generated with some form of $\text{\TeX}.$ Marcin - could you copy the body of it and post it here in "regular" fashion? It wouldn't be too long a post. Either that, or email me the $\text{\TeX}$ code and I'll post it for you.

Well, here's the next best thing (click on it for better quality):

[img]6129[/img]

[quote="marcin"][b]Problem A6:[/b]

Four points are chosen uniformly and independently at random in the interior of a given circle.
Find the probability that they are the vertices of a convex quadrilateral.

[b]Solution:[/b]

Consider the set
\[\{ S=(\alpha,r,a,b,x,y): \alpha\in [0,2\pi),\ 0\leq r\leq 1,-\sqrt{1-r^{2}}\leq a,b\leq \ \sqrt{1-r^{2}},\]
\[\ x^{2}+y^{2}\leq 1 \}.\]
The map
\[F: (\alpha,r,a,b,x,y)\mapsto\]
\[(a\cos\alpha+r\sin\alpha,-a\sin\alpha+r\cos\alpha, b\cos\alpha+r\sin\alpha,-b\sin\alpha+r\cos\alpha,\]
\[x\cos\alpha+y\sin\alpha,-x\sin\alpha+y\cos\alpha)\]
is a parametrization of the product $T$ of three unit circles (the inverse map corresponds to rotating the triangle with vertices $A_{1}, A_{2}, A_{3}$ inside the unit circle  to a triangle $A_{1}^{'}A_{2}^{'}A_{3}^{'}$ such that the line $A_{1}^{'}A_{2}^{'}$ is parallel to the x-axis and above it; here  $A_{1}^{'}=(a,r)$, $A_{2}^{'}=(b,r)$, $A_{3}^{'}=(x,y)$; $\alpha$ is the angle of rotation).

For a point $t=(A_{1},A_{2}, A_{3})$ of $T$, let $A(t)$ be the area of
$A_{1}A_{2}A_{3}$. Thus the expected area of a random triangle in the unit circle is
\[E=\pi^{-3}\int_{T}A= \pi^{-3}\int_{S}A\circ F |JF| \]
where $JF$ is the Jacobian of $F$. Note that $A(F(\alpha,r,a,b,x,y))=|a-b\parallel y-r|/2$. Also it is not hard to see that $|JF|=|a-b|$ (this requires computation of a $6\times 6$ determinant, but it is not hard).

Thus
\[E=\pi^{-3}\frac{1}{2}\int_{0}^{2\pi}\int_{0}^{1}\int_{D}\int_{-\sqrt{1-r^{2}}}^{\sqrt{1-r^{2}}}\int_{-\sqrt{1-r^{2}}}^{\sqrt{1-r^{2}}}(a-b)^{2}|r-y|da\,db\,dx\,dy\,dr\,d\alpha \]
where $D=\{x^{2}+y^{2}\leq 1\}$ is the unit circle.

Since the angle is not involved in the function under integral,  after evaluation the first two most-right integrals we get
\[E=\frac{8}{3}\pi^{-2}\int_{0}^{1}\int_{D}|r-y|(1-r^{2})^{2}\,dx\,dy\,dr. \]
Now
\[\int_{D}|r-y|\,dx\,dy= \int_{-1}^{1}|r-y|\int_{-\sqrt{1-y^{2}}}^{\sqrt{1-y^{2}}}\,dx\,dy= 2\int_{-1}^{1}|r-y|\sqrt{1-y^{2}}\,dy=\]
\[2[\int_{-1}^{r}(r-y)\sqrt{1-y^{2}}\,dy+\int_{r}^{1}(y-r)\sqrt{1-y^{2}}\,dy]=\]
\[4r\int_{0}^{r}\sqrt{1-y^{2}}\,dy+\frac{4}{3}(1-r^{2})^{3/2}.\]

Thus
\[E=\frac{8}{3}\pi^{-2}\int_{0}^{1}(1-r^{2})^{2}[4r\int_{0}^{r}\sqrt{1-y^{2}}\,dy+\frac{4}{3}(1-r^{2})^{3/2}].\]
Since the derivative of $\int_{0}^{r}\sqrt{1-y^{2}}\,dy$ is $\sqrt{1-r^{2}}$ and
$\frac{d}{dr}(1-r^{2})^{3}=-6r(1-r^{2})^{2}$, integration by parts yields
\[\int_{0}^{1}(1-r^{2})^{2}4r\int_{0}^{r}\sqrt{1-y^{2}}\,dy =\frac{2}{3}\int_{0}^{1}(1-r^{2})^{7/2}\,dr.\]
Thus
\[E=\frac{16}{3}\pi^{-2}\int_{0}^{1}(1-r^{2})^{7/2}dr =\frac{35}{48\pi}. \][/quote]
I did a few format tweaks there. marcin - proofread it, please, see if there's anything wrong. If I get confirmation that this works, I'll probably delete joml88's post.

Here are my solutions to the Putnam problems
http://www.math.binghamton.edu/mazur/teach/putnam06.pdf

[quote="sexynsmartjenny"]What is the probability that the first three chosen points in this circle are linear?  Zero?[/quote]

Yes. Imagine you've already chosen the first two points. These determine a line segment, a chord of the circle. The probability that the third point is on the line is the area of the line divided by the area of the circle. The area of the circle is greater than 0. But a line segment is 1-dimensional so its area is 0.

I see that this problem amounts to finding the expected value of the area of a triangle inside the circle.  We divide this into three cases: points A, B, and C do not lie on a common semicircle; points A, B, and C do lie on a common semicircle in that order with point B outside of triangle OAC; points A, B, and C do lie on a common semicircle in that order with point B inside triangle OAC. Of course O is the center of the circle.  In the latter two cases, we define Theta1 as the angle formed by the line segments OA and OB, Theta2 as the angle formed by the line segments OB and OC, and Theta = Theta1 + Theta2.  Theta takes values on the interval [0, pi] and its probability density function is given by f(Theta) = 2Theta/(pi^2).  Does anyone understand how to determine that this is the probability density function for Theta?

I have just thought of a way to answer my own question.  We wish to find P(THETA <= theta).  We then take the derivative of P(THETA <= theta) with respect to theta, and this gives us the probability density function for THETA.  We know that THETA = THETA1 + THETA2.  So P(THETA <= theta) = P(THETA1 + THETA2 <= theta).

Before we find P(THETA1 + THETA2 <= theta) it makes sense to try and find P(THETA1 + THETA2 <= theta | THETA1 = theta1).  This is just P(THETA1 + THETA2 <= theta | THETA1 = theta1) = P(THETA2 <= theta - theta1).  THETA1 + THETA2 takes values on the interval [0, pi].  Therefore it makes sense to say that if THETA1 = theta1 then THETA2 takes values on the interval [0, pi - theta1].  In particular, the distribution for THETA2 is uniform on [0, pi - theta1] so its pdf is given by 1/(pi - theta).  Therefore P(THETA2 <= theta - theta1) = integral from 0 to theta - theta1 of 1/(pi - theta1) dx = (theta - theta1)/(pi - theta1).

What we really want to find is P(THETA1 + THETA2 <= theta).  Now that we know what P(THETA1 + THETA2 <= theta | THETA1 = theta1) is, we can use the laws of probability to say that P(THETA1 + THETA2 <= theta) = sum over all theta1 on the interval [0, theta] of P(THETA1 + THETA2 <= theta | THETA1 = theta1)P(THETA1 = theta1) = {(theta - theta1)/(pi - theta1)}P(THETA1 = theta1).  Of course it does not make sense to add over all theta1 on the interval [0, theta], but this does mean something intuitively.  It obviously must be some sort of an integral since THETA is a continuous random variable.  Let's approximate the area under the curve g(x) = (theta - x)/(pi - x) from x = 0 to x = theta using k rectangles where k is some positive integer.  Each rectangle has width theta/k.  The approximation is (theta/k){g(theta/k) + g(2theta/k) + ... + g(theta)}.  We can think of the width of each rectangle divided by pi as theta/k = P(0 < THETA1 < theta/k) = P(theta/k < THETA1 < 2theta/k) = ... = P((k-1)theta/k < THETA1 < theta).  If we let k approach infinity then the intervals become infinitesimally small so then it makes sense to speak of something like P(THETA1 = theta1).  So it follows that P(THETA1 + THETA2 <= theta) is the area under the curve g(x) from x = 0 to x = theta divided by pi.  This comes out to be theta/pi + (theta - pi)(ln(pi) - ln(pi - theta))/pi.

The derivative of P(THETA <= theta) = theta/pi + (theta - pi)(ln(pi) - ln(pi - theta))/pi with respect to theta is (1/pi)ln(pi/(pi - theta)).  This is not in agreement with what I found on http://www.unl.edu/amc/a-activities/a7-problems/putnam/-pdf/2006s.pdf which says that the pdf for THETA is actually 2theta/(pi^2) on [0, pi].  Can anyone tell me where I went wrong in my reasoning?  In particular, I'm not sure if my assumption that the distribution for THETA2 is uniform on [0, pi - theta1] is correct.

I have now figured out the answer to my own question.  THETA takes values on the interval [0, pi].  So we can imagine a line segment of length pi.  Then to find P(THETA <= theta) we need to place two stones on the interval [0, theta].  This happens with probability (theta/pi)(theta/pi) = (theta^2)/(pi^2).  So P(THETA <= theta) = (theta^2/pi^2) and the probability density function for THETA is the derivative of P(THETA <= theta) with respect to theta which is 2theta/(pi^2).

Anyway, in the case that has 3 stones on a common semicircle, we suppose that points A, B, and C lie in that order.  O is the center of the circle.  Either B lies inside triangle OAB, or it lies outside it.  If B lies outside triangle OAC, then [ABC] = [OAB] + [OBC] - [OAC].  If B lies inside triangle OAC then [ABC] = [OAC] - [OAB] - [OBC].  When we compute E([ABC]) I am not sure how to account for when B is inside the triangle and when B is outside the triangle.  Does anyone understand how to account for this?  I have tried reading the solution online, but it is too difficult to understand.

I understand most of this problem.  Unfortunately, I have hit a dead-end that I can't seem to get past.  Can someone take a look at what I have typed in the attachment please?  It is kind of long (a little more than 11 pages).  However, a lot of it is just tedioius math.  There are several pictures in the attachment that take up space.  And a lot of it is just straightforward explanation so that one can easily absorb the information.  Hence it is not as bad as it might at first seem.  So to whomever has any time to kill, can you take a look and answer my question that is described in the last 3 and a half pages?

Suppose that [i]x[/i], [i]z[/i], and theta are constants that do not depend on theta_{1}.  How does one evaluate the integral from 0 to theta of 1/{[i]x[/i]sin(theta_{1}) + [i]z[/i]sin(theta - theta_{1}}  [i]d[/i]theta_{1}?   It comes up when you suppose that three points have already been chosen and they lie on a common semicircle with no point lying inside the triangle formed by the other two points and the circle's center.

If this helps, I suppose that points [i]A[/i] and [i]C[/i] have already been chosen and that point [i]B[/i] is to be chosen somewhere in the circle above the line segment [i]AC[/i].  I am supposing that [i]OA[/i] has length [i]x[/i], [i]OC[/i] has length [i]z[/i], and that theta is the measure of the angle [i]AOC[/i].  Theta_{1} is the random variable that is the measure of [i]AOB[/i] and therefore takes values on the interval (0, theta).

Can I ask the motivation for most of the above solutions -- I have a hard time dealing with continuous probability problems like this.

In most problems involving continuous probability (and this is one of the problems in the handout -- and it is an a6..), we usually want to consider what are the degree of freedoms of the problems, and maybe somehow tweak them or find a perspective such that it is much doable to solve the problem (say Putnam 2021 B1 where you have degree of freedom: center of square and orientation of square); However in this problem, I think we have four degree of freedom (all four points independently), and I have no idea how to visualize or where to start other than if you fix three points, then the area of the circle minus the area of the triangle is where the last point should be; but somehow "integrating this area" with respect to any possible triangle $XYZ$ inside the circle doesn't seem intuitive, or seem pleasant to do. Can anyone share their thoughts?


[quote="Albanian Eagle"]$(1+a^{2})(1+b^{2})(1+c^{2})(1+d^{2})=(\sum a-\sum bcd)^{2}+(1+abcd-\sum ab)^{2}$
 :wink:[/quote]Yes, for an easy proof of that see
http://www.mathlinks.ro/Forum/viewtopic.php?t=82723


geometry
parallelogram
inequalities
ratio
geometric transformation
rotation
algebra

Hello. Here is problem 8 from the recent Canadian Math Olympiad Qualifying Repechage. It was the hardest problem on the test. Can anyone find a solution that is almost entirely based on construction?

[b]Problem 8[/b]
Consider three parallelograms $ P_1, P_2, P_3$. Parallelogram $ P_3$ is inside parallelogram $ P_2$, and the vertices of $ P_3$ are on the edges of $ P_2$. Parallelogram $ P_2$ is inside parallelogram $ P_1$, and the vertices of $ P_2$ are on the edges of $ P_1$. The sides of $ P_3$ are parallel to the sides of $ P_1$. Prove that one side of $ P_3$ has length at least half the length of the parallel side of $ P_1$.

[hide="My Kind of Synthetic Solution"]
[b] Please see the attached PNG for a diagram. [/b]

[b]Lemma 1: [/b] At least one of the numbers $ a\plus{}b$ and $ \frac{1}{a}\plus{}\frac{1}{b}$ is at least $ 2$ where $ a>0$ and $ b>0$.

First note that if $ a\plus{}b \ge 2$, then the claim holds true. Now take the second case where $ a\plus{}b<2$.

Note that for any $ a$, it follows that
\[ (a\minus{}1)^{2} \ge 2 \indent \Rightarrow \indent 1 \ge 2a\minus{}a^{2} \indent \Rightarrow \indent \frac{2}{a(2\minus{}a)} \ge 2\]
Decomposing into partial fractions yields that
\[ \frac{1}{2\minus{}a}\plus{}\frac{1}{a} \ge 2\]
Now since $ a\plus{}b<2$, then $ \frac{1}{b}>\frac{1}{2\minus{}a}$. Hence
\[ \frac{1}{a}\plus{}\frac{1}{b}>\frac{1}{2\minus{}a}\plus{}\frac{1}{a} \ge 2 \indent \Rightarrow \indent \frac{1}{a}\plus{}\frac{1}{b} > 2\]
Hence the lemma is proven.

-End of Lemma 1-

[b]Lemma 2: [/b] $ \triangle{AKL} \cong \triangle{CMN}$, $ \triangle{DNK} \cong \triangle{BLM}$, $ \triangle{NSR} \cong \triangle{LQP}$, and $ \triangle{KPS} \cong \triangle{MRQ}$.

First note that since $ KLMN$ and $ ABCD$ are parallelograms, that $ AB \| CD$, $ BC \| DA$ and $ KL \| MN$. Hence $ \angle{CNM}\equal{}\angle{ALK}$ and $ \angle{CMN}\equal{}\angle{AKL}$. Also since $ KLMN$ is a parallelogram, $ KL\equal{}MN$. Hence by ASA congruency, $ \triangle{AKL} \cong \triangle{CMN}$. By the same argument it follows that $ \triangle{DNK} \cong \triangle{BLM}$. Also, in examining $ KLMN$ instead of $ ABCD$ and $ PQRS$ instead of $ KLMN$ the second set of results is achieved by the same argument. Therefore $ \triangle{NSR} \cong \triangle{LQP}$ and $ \triangle{KPS} \cong \triangle{MRQ}$, and the lemma has been proven.

-End of Lemma 2-

[b]Lemma 3:[/b] $ HSWD \cong AUPT \cong VBXQ \cong RYCZ$.

First note that $ AB \| TX \| HY \| DC$ and $ AD \| UW \| VZ \| BC$ because $ ABCD$ and $ PQRS$ are parallelograms with parallel sides. Therefore it follows that $ HSWD$ and $ VBXQ$ are parallelograms. Also, it follows that $ \angle{KSH}\equal{}\angle{MQX}$. By Lemma 2, since $ \triangle{DNK} \cong \triangle{BLM}$ then $ \angle{SKH}\equal{}\angle{QMX}$. Also by Lemma 2, $ \triangle{KPS} \cong \triangle{MRQ}$ and $ KS\equal{}MQ$. By ASA congruence, $ \triangle{KSH} \cong \triangle{MQX}$ and therefore $ HS\equal{}QX$. By the same argument it follows that $ SW\equal{}VQ$. Also, since $ ABCD$ is a parallelogram, $ \angle{ADC}\equal{}\angle{ABC}$. Therefore since parallelograms $ HSWD$ and $ VBXQ$ have the same side lengths and angles, they are congruent.

By the same argument, it follows that $ AUPT \cong RYCZ$. Now note that since $ TPSH$ is a parallelogram because of the parallel lines established above, $ SH\equal{}PT$. Similarly, because $ UPQV$ is a parallelogram, them $ SW\equal{}VQ\equal{}UP$. Also since $ DW \| TP$ then $ \angle{DWS}\equal{}\angle{TPU}$. Therefore since parallelograms $ HSWD$ and $ AUPT$ have the same side lengths and angles, they are congruent. It then follows that $ HSWD \cong AUPT \cong VBXQ \cong RYCZ$ and hence the lemma has been proven.

-End of Lemma 3-

Now note that since $ SW \| KH$ and $ HS \| WN$, then $ \triangle{KSH} \sim \triangle{SNW}$. Also, since $ RY \| NZ$ and $ RZ \| MY$, then $ \triangle{RNZ} \sim \triangle{MRY}$. Therefore,
\[ \frac{WN}{HS}\equal{}\frac{NS}{SK} \indent \frac{NZ}{RY}\equal{}\frac{NR}{RM}\]
By Lemma 3, $ HSWD \cong RYCZ$ and therefore $ HS\equal{}RY$. Substituting this into the above equations and adding yields that
\[ \frac{WN\plus{}NZ}{HS}\equal{}\frac{NS}{SK}\plus{}\frac{NR}{RM}\]
Now note that $ WN\plus{}NZ\equal{}WZ\equal{}SR$ because $ SRZW$ is a parallelogram. Also since $ HSWD$ and $ RYCZ$ are parallelograms, $ HS\equal{}DW\equal{}ZC$. Therefore $ CD\equal{}DW\plus{}WZ\plus{}ZC\equal{}SR\plus{}2(HS) \Rightarrow 2(HS)\equal{}CD\minus{}SR$. Dividing the equation by $ 2$ and substituting yields that
\[ \frac{WN\plus{}NZ}{2(HS)}\equal{}\frac{SR}{CD\minus{}SR}\equal{}\frac{1}{2}(\frac{NS}{SK}\plus{}\frac{NR}{RM})\]
By the same argument it follows that
\[ \frac{QR}{BC\minus{}QR}\equal{}\frac{1}{2}(\frac{MQ}{QL}\plus{}\frac{RM}{NR})\]
By Lemma 2, $ \triangle{NSR} \cong \triangle{LQP}$ and $ \triangle{KPS} \cong \triangle{MRQ}$. Therefore $ MQ\equal{}SK$ and $ NS\equal{}QL$. Substituting this into the above equation yields that
\[ \frac{QR}{BC\minus{}QR}\equal{}\frac{1}{2}(\frac{SK}{NS}\plus{}\frac{RM}{NR})\]
Now let $ a\equal{}\frac{NS}{SK}$ and $ b\equal{}\frac{NR}{RM}$. Substituting yields that
\[ \frac{SR}{CD\minus{}SR}\equal{}\frac{1}{2}(a\plus{}b) \indent \frac{QR}{BC\minus{}QR}\equal{}\frac{1}{2} (\frac{1}{a}\plus{}\frac{1}{b})\]
By Lemma 1, it follows that one of $ a\plus{}b$ and $ \frac{1}{a}\plus{}\frac{1}{b}$ is at least $ 2$. WLOG this element is $ a\plus{}b$. It then follows that
\[ \frac{SR}{CD\minus{}SR} \ge 1 \indent \Rightarrow \indent SR \ge \frac{CD}{2}\]
Similarly, if this were not the case then $ QR \ge BC/2$. Therefore at least one side of $ PQRS$ which is $ P_3$ is greater than or equal to half the length of the parallel side of $ ABCD$ which is $ P_1$.
[/hide]

Did anyone find any helpful constructions?

Affine transformation yields an elegant solution.

First, note that the problem is affine, and that the 3 parallelograms have a common center. Affine transform $ P_1$ to be a square.

Consider the midpoint parallelogram $ L'M'N'K'$, with inscribed square $ P_4$,$ P'Q'R'S'$, both in the same orientation as $ P_2$ and $ P_3$. Then by the problems condition, and the fact that $ P_3$ is centered at the same point as $ P_4$, with sides parallel, and all sides strictly smaller, we have that $ P_3$ lies strictly inside $ P_4$. This implies:

$ \angle PLQ \le \angle PL'Q < \angle P'L'Q' \equal{} 90$, the first inequality using the fact that of all triangles with vertices P,Q, and a vertex on AB, the midpoint will make the smallest triangle, and the second following from the fact that $ P_3$ is strictly contained in $ P_4$. A similar argument works for the other 3 angles of $ P_2$, so the sum of its angles is strictly less than 360, contradiction!

Wow. That is a lot shorter than my solution. Even though this is a fairly noobie question, when you say that they have the same center, do you mean intersection of the diagonals?

Ina parallelogram, it is the intersection of the diagonals, and is also the point to which is has 180 degree rotational symmetry. When one of the parallelograms is inside the other, imagine rotating the entire figure by 180 degrees about the center of the outside one. Because the inner parallelogram divides the sides in the opposite ratio on opposite sides (which you proved in lemma 2 I believe), the inner parallelogram will be transformed to itself, so therefore the center of the larger one is also the center of the smaller one.

Cheers,

Rofler

BTW, this works equally well with parallelograms, the exact same arguments apply, no need for affine transformations at all!

Thank you. That helps a lot. I did not realize the rotation thing.


calculus
integration
algebra
polynomial
complex analysis
complex analysis unsolved

1. Suppose that $ f(z)$ is analytic and that $ f^{'}(z)$ is continuous in a region that contains the closed curve $ \gamma$. Show that $ \int_{\gamma}\overline{f(z)}f^{'}(z) dz$ is purely imaginary.
I know that if the Taylor series expansion of $ f$ has only real coefficients, $ f$ is essentially a polynomial of real coefficients, then we have $ \overline{f(z)}\equal{}f(\overline{z})$. Then we could use integration by parts formula to find the sum of this identity and this identity's conjugate is zero. Then we can say this identity is purely imaginary. However, I don't know how to do this for a general $ f$.

2. Show that if $ |a|<r<|b|$, then $ \int_{\gamma}\frac{1}{(z\minus{}a)(z\minus{}b)} dz \equal{} \frac{2\pi{i}}{a\minus{}b}.$ $ \gamma$ denotes the circle centered at the origin, of radius $ r$, with the positive orientation.
Is there a way of doing this just with Cauchy Theorem, not using Residue Formula?

[hide="Hints:"]1 - Differentiate $ h\equal{}\mid f\mid^2\circ\gamma$.

2 - Can you find constants $ \alpha$ and $ \beta$ such that \[ \frac{1}{(z\minus{}a)(z\minus{}b)}\equal{}\frac{\alpha}{z\minus{}a}\plus{}\frac{\beta}{z\minus{}b}\,?\][/hide]

From the  #2, you have another proof for Liouville's theorem

You don't even need Cauchy's Theorem for #2.

[hide="hint"]

Expand into infinite series.

[/hide]


Using three straight lines, waht is the maximum number of regions into which the enclosed region can be divided?

I can't draw the region, but it was a horseshoe type of region.
[/img]

Well I'm guessing 12 (3+4+5; The maximum area you can make with one is 3 I'm guessing so you can create that +1 with your second line and that +2 with the third so you get 3+3+1+3+1+1=3+4+5=12) but I'm not sure, what did you guess?

how can you cut it into 3 sections with 1 line?

tare's answer was too much

but anyways... how u cut it into 3 parts with one line,

just cut one down through the open section... if that's understandable...

what i did was just draw it...

too bad i got 1 less at competition...

[quote="white_horse_king88"]I can't draw the region, but it was a [b]horseshoe[/b] type of region.
[/img][/quote]
Well, I got 9 as the asnwer

that's what i got at nats, but the answer was...[hide] 10 [/hide]:(

Oh shoot I was imagining a cowboy boots sort of thing :lol: guess I'd better go and watch more Lucky Charms commercials...

It's like 9 or 10 depending on the actual horseshow region.

[quote="Tare"]It's like 9 or 10 depending on the actual horseshow region.[/quote]
Why would the region of the horseshoe matter?  As long as its a horseshoe.

I guess some horseshoes have this curvy thing sticking out of the ends.

o gosh...we did this one yesterday. I know how to do it. It just would be so much easier with the picture. Ok, I'll try to explain it.

First, draw a line straight down. Then, draw one slanted to the left, but starts tothe right of the first line. The third line is the key. If you draw it right, you have several small triangles that make so the horseshoe is divided into 10 regions.

Heres my way.  You know how to divide a rectangle into 7 regions with 3 lines.  Make another rectangle next to (but not touching) the first.  Connect them using a half circle on the bottom.  Make sure both lines pass through the space between the rectangles.

Can anyone give more info on this problem?  I've been assigned it & can only seem to get nine.

Editted: I figured it out.  Thanks.


limit
logarithms
real analysis
real analysis unsolved

I  think it's $1$:

${\frac{\sqrt{(n+1){\sqrt{(n+2){\sqrt{(n+3)\cdots{\sqrt{n+n}}}}}}}}{n}}=(1+\frac{1}{n})^{\frac{1}{2}}(1+\frac{2}{n})^{\frac{1}{4}}\cdots(1+\frac{n}{n})^{\frac{1}{2^n}}\cdot\frac{1}{n^\frac{1}{2^{n}}}$

Next take ln and see that it goes to $0$ so the original expression goes to ${e^{0}=1}$

$\begin{displaymath}\bigg(\frac{1}{2}\ln(1+\frac{1}{n})+\frac{1}{4}\ln(1+\frac{2}{n})+\cdots+\frac{1}{2^{n}}\ln(1+\frac{n}{n})\bigg)+\ln(\frac{1}{n^\frac{1}{2^{n}}})=\newline\frac{1}{n}\cdot\bigg(\frac{1}{2}\cdot\frac{\ln(1+\frac{1}{n})}{\frac{1}{n}}+\frac{2}{4}\cdot\frac{\ln(1+\frac{2}{n})}{\frac{2}{n}} +\cdots+\frac{n}{2^{n}}\cdot\frac{\ln(1+\frac{n}{n})}{\frac{n}{n}}\bigg)+\ln(\frac{1}{2^\frac{1}{n^{n}}})\leq\newline\frac{1}{n}\sum_{k=1}^{n}\frac{k}{2^k}+\ln(\frac{1}{n^\frac{1}{2^{n}}})\rightarrow 0 \end{displaymath}$

(since the last sum is not infinite (easily checked with d'Alembert's(?) criterion)), and since the original ln expression is clearly $\geq 0$ it goes to 0 and we're done.
We've also been using the fact that $\ln(x+1)\leq x$

Maybe...Maybe not
 :roll:


[quote="DonkeyKong"]What is the least perfect square which is the sum of the three different non-zero perfect squares?

[hide]49[/hide]

Is there a way without guessing and checking?[/quote]

Maybe you could find the squares by sum of odds and evens.....(dumb idea)...but the number wasn't that low anyway...

I think finding it without guess and check would involve some difficult modular aritmetic.

Not exactly so difficult...use the fact that all odd squares are $ 1\pmod4$ and all even squares are $ 0\pmod4$.

In order for the sum of the three distinct, non-zero perfect squares to be a perfect square, we can have one odd and two evens adding up to an odd square, or all evens adding up to an even square, which might reduce casework somewhat. However, this observation is not necessary, as the numbers are so small.

If perfect squares can be similiar then answer is $ 3^2\equal{}1^2\plus{}2^2\plus{}2^2$, in this situation guessing and checking is the best way to solve the problem.

If perfect squares must be different, then we have one odd square and two even squares as math154 wrote. All squares cannot be even, because the answer wouldn't be the smallest:

$ (2a)^2\plus{}(2b)^2\plus{}(2c)^2\equal{}(2d)^2$
$ a^2\plus{}b^2\plus{}c^2\equal{}d^2$

5 cannot be the answer:

$ 5^2\neq1^2\plus{}2^2\plus{}4^2$
$ 5^2\neq3^2\plus{}2^2\plus{}4^2$

7 is the answer: $ 7^2\equal{}2^2\plus{}3^2\plus{}6^2$

$ 2^2, 3^2$, and $ 6^2$ [i]are[/i] three different perfect squares.

[quote="007math"][quote="DonkeyKong"]What is the least perfect square which is the sum of the three [b][size=200]different [/size][/b]non-zero perfect squares?

[hide]49[/hide]

Is there a way without guessing and checking?[/quote][/quote]

misread  :oops:

For a sprint question, this is an awfully long solution

Well, for each sprint round question you have a minute and twenty seconds, and I don't think I could come up with a solution like that in a minute and twenty seconds.

But doesn't it make up for the #1's and #2's?

Math154 did the best- he showed that there are at least 2 even of 3 (20 seconds), so try 2,4 2,6 or 4,6 and find the 49 (1 min).


Determine all integers $a$ and $b$ such that $ab$ divide $c(c^2 - c + 1)$ and $c^2 + 1$ divide $a + b$ for $c \in \mathbb Z$


Given: m>0; a,b,c satisfying: a/(m+2)+b/(m+1)+c/m=0

Prove that  ax^2 + bx+c= 0 has root in  (0,1)

Have A Great Day!

[hide="Solution"] Let $ f(x) \equal{} ax^2 \plus{} bx \plus{} c$.  We are given that $ \int_{0}^{1} f(x) x^{m\minus{}1} \, dx \equal{} 0$, hence $ f(x) x^{m\minus{}1}$ changes sign or is constantly zero in that interval.  The conclusion follows. [/hide]

[quote="t0rajir0u"][hide="Solution"] Let $ f(x) \equal{} ax^2 \plus{} bx \plus{} c$.  We are given that $ \int_{0}^{1} f(x) x^{m \minus{} 1} \, dx \equal{} 0$, hence $ f(x) x^{m \minus{} 1}$ changes sign or is constantly zero in that interval.  The conclusion follows. [/hide][/quote]

Well-done! Can anyone solve it without using integration?

Have A Great Day!


an 8x8 array consists of  the numbers 1,2,3,4.....,64. Consecutive numbers are adjacent along a  row or coloumn.what is the minimum vale of sum of numbers along a diagonal.?

[hide="Idea for proof"]
Every numbering of our grid can clearly be completed by beginning with 1 at some point and then counting up to 64 by moving to an adjacent square on the grid; in other words, a path like the one above. Now, note that before all the diagonal elements are taken up, at least one half of the grid (a large 8x8 right triangle) needs to be covered. This triangle has 36 squares, so a lower bound to the value of the last square to be reached in the triangle is 36. Now, the minimums for the other diagonal squares are clearly 1,3,5,...,13. This gives us a lower bound of 85 for our minimum.

The rest is just discarding the possibilities of 85, 86, and 87 by showing that at least 3 squares of the complement of our 8x8 triangle need to be reached in order to complete the 8x8 square. I hope this is right, I don't have time to verify it now.
[/hide]


algebra
polynomial
linear algebra
matrix
system of equations
number theory unsolved
number theory

$P$ is a polynomial with degree $2$.For every $i\in Z$,$P(i)$ is square number.

Prove $P(x)=Q(x)^{2}$,for $Q(x)$ is a polynomial with degree $1$

Actually P can be of any even degree.  Clearing denominator, I'm going to assume that P is in Z[x].  clearly the leading term is positive.  Since the condition is unaltered by replacing P(x) with P(x+k) for any positive integer k, we can assume the coefficients of P(x) are all positive.

so for every integer x, P(x) = (r_x)^2 for some integer r_x.

if P(x) = ax^2 + bx + c then since P(x)/(r_x)^2 = 1 for every integer x (provided of course r_x != 0), (x/r_x)^2 goes to 1/sqrt(a) as x -> infinity.

Claim is that a is a perfect square.  If it isn't, then :

P(x+1) - P(x) = (r_(x+1))^2 - (r_x)^2 = (r_(x+1) - r_x)*(r_(x+1) + r_(x)).

and (P(x+1) - P(x))/x = (r_(x+1) - r_x)*(r_(x+1) + r_x)/x

But (r_x)/x and (r_(x+1))/x goes to 1/sqrt(a) as x -> infty, and r_(x+1) - r_x is an integer.  And the LHS goes to 2a, which is integer.  So 1/sqrt(a)) must be rational, i.e. a is a perfect square.

From now, just take g = vx + w so that g^2 = v^2x^2 + 2vwx + w^2.  Choose v,w such that v^2 = a, v > 0, and 2vw <= b <= 2v(w+1).  Then for sufficiently large x,

(g(x))^2 <= P(x) <= (g(x) + 1)^2 .

Since there is no integer square between n^2 and (n+1)^2, P(x) is either ((g(x))^2 or (g(x) + 1)^2 for infinitely many values of x.  Since two polynomials agreeing on infinitely many distinct values -> same as polynomials, you get P(x) is a square of a polynomial.

Here's a problem related to this (much harder, IMO):

suppose f,g are in Q[x] with f(Q) = g(Q) (as sets).  Show that not only deg f = deg g, but also there exist a,b rational numbers with a != 0 such that f(x) = g(ax + b) as polynomials.  Converse of this is clear.

[quote="sheaf"]Actually P can be of any even degree.  Clearing denominator, I'm going to assume that P is in Z[x]. [/quote]

Is it obvious that $P(x) \in \mathbb{Z}\forall x \in \mathbb{Z}\implies P(x) \in \mathbb{Q}[x]$?

Yes.  Vandermonde matrix is invertible.  So you got a system of equations P(n) = a_n for some integer a_n.  Then just invert the matrix to get the coefficients of P in terms of a_n, i.e. they are rational


In ham yek soale zibaye dige

Farz konid $ x,y\ge 0$ va $ x^3\plus{}y^3\equal{}1$ maximome ebarate zire ra bedast avarid
\[ \sqrt{x}\plus{}2\sqrt{y}\]

Salam

eyval halle khoby bod Ramtin Madani