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inequalities
inequalities unsolved

[quote="cocoowner"]$x_{1}, x_{2}, ... , x_{n}$ are positive real numbers, and $x_{1}x_{2}...x_{n}=1$ Prove that $\frac{1}{n-1+x_{1}}+\frac{1}{n-1+x_{2}}+...+\frac{1}{n-1+x_{n}}\leq 1$
I have written the same question in here, but that was algebra. @.@ My fault.[/quote]
This one is easy and I think it was proposed and solved in this forum,try to search!
Here is a link to a similar problem,but harder than this one! :P
[url]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=1999+romanian&t=64349[/url]

substitute each $x_{i}$ to exponential form and use Jensen inequality

For this problem, in the book [b]Vasile Cirtoaje (vasc)[/b], ''Algebraic Inequalities. Old and New Methods.'', GIL Publishing House, 2006 there are 4 solutions.

[quote="BoesFX"]It doesn't work since $f(x)=\frac{1}{n-1+e^{x}}$ is not concave for all real numbers $x$.

It is not that easy. :([/quote]
That's not really important. See [url=http://www.artofproblemsolving.com/Resources/Papers/MildorfInequalities.pdf]Mildorf's Olympiad Inequalities[/url] (it uses the Half Concave - Half Convex Theorem, as called by zhaobin).

PS: The problem is also in [b]Old and New Inequalities[/b].

That's why I'm saying it is not that easy especially for some that haven't known that technique :P

[b]Solution 1: [/b]

We clear denominators to have the inequality equivalent to \[ n(n-1)^{n-1} + \sum_{d=1}^{n-1} \sum_{1 \leq i_1 < \cdots < i_d \leq n} (n-d)(n-1)^{n-d-1} x_{i_1} \cdots x_{i_d} \leq 1 + \sum_{d=1}^{n-1} \sum_{1 \leq i_1 < \cdots < i_d \leq n} (n-1)^{n-d} x_{i_1} \cdots x_{i_d} + (n-1)^n \]

Note that it suffices to show that \[\sum_{1 \leq i_1 < \cdots < i_d \leq n} \left[ (n-1)^{n-d}  - (n-d)(n-1)^{n-d-1} \right] x_{i_1} \cdots x_{i_d} \geq n(n-1)^{n-1} - (n-1)^n -1 = (n-1)^{n-1} - 1 \] We have by AM-GM; \[\dbinom{n}{d} \left[(n-1)^{n-d} - (n-d)(n-1)^{n-d-1 } \right] \leq \sum_{1 \leq i_1 < \cdots < i_d \leq n} \left[ (n-1)^{n-d}  -(n-d)(n-1)^{n-d-1} \right] x_{i_1} \cdots x_{i_d}\] We then only need to show that \[ \sum_{d=1}^{n-1} \dbinom{n}{d} \left[(n-1)^{n-d} - (n-d)(n-1)^{n-d-1 } \right] = (n-1)^{n-1} + 1  \qquad (*) \] We  do that by binomial theorem by viewing $n^n = ( n-1 + 1)^n $ ;\[ \sum_{d=1}^{n-1} \dbinom{n}{d}(n-1)^{n-d} = n^n -1 - (n-1)^n \qquad (1)\]
And using the identity $ n \binom{n}{d-1} = (n-d) \binom{n}{d} $, we have again by binomial theorem;

\[
\begin{array}{lcl}
\sum_{d=1}^{n-1} (n-d) \dbinom{n}{d}  (n-1)^{n-d-1 } & = & n \sum_{d=1}^{n-1} \binom{n-1}{d}(n-1)^{n-1} \\
& = & n \left(\sum_{d=0}^{n-1} \binom{n-1}{d}(n-1)^{n-1} - (n-1)^{n-1} \right) \\
& = & n((n-1)^n - (n-1)^{n-1}) \qquad (2)
\end{array}
 \]
 Now subtracting $(2)$ from $(1)$ we obtain $(*)$ as desired

[b]Solution 2[/b]:

 We only try to clear denominators by Cauchy-Schwarz, but first we want to have our fractions on the right hand-side, we can do that by multiplying the inequality by $n-1$, to have an equaivalent

\[ 1 \leq \frac{x_1}{n-1 + x_1} + \cdots + \frac{x_n}{n-1 + x_n}  \qquad \qquad (*)\]

Now we note that \[n(n-1) + x_1 + \cdots + x_n = n(n-1) \sqrt{x_1^{n-1}  \cdots x_n^{n-1}} + x_1 + \cdots + x_n  \leq (\sqrt{x_1} + \cdots + \sqrt{x_n})^2      \qquad \qquad (1)\]

Now by Cauchy-Schwarz, \[ (\sqrt{x_1} + \cdots + \sqrt{x_n})^2 \leq \left( \frac{x_1}{n-1 + x_1} + \cdots + \frac{x_n}{n-1 + x_n} \right)(n(n-1) + x_1 + \cdots + x_n  ) \qquad \qquad (2) \]

Combining $(1)$ and $(2)$, we can obtain $(*)$.


Has your school ever won the states? Which state?

Nope, the best my school ever came (we are in Texas) was 8th place team last year.  The other years we got like 19th or 20th.

1st in NJ about five times.

Last year, we had two people from our school at Nats, both in the top 70 =D .

My team has won New Hamshire the last 4 years, and I'm hoping for #5 this year!!  I've had at least one kid make it to Nationals each of the 6 years of coached and had 2 kids make it twice.

Good Luck to all,

Josh Frost

My school won it last year and had one person go to nationals and this year I beat the only of the remainig members of that team each time we did a competion

Well, to be honest it is hard work.  For example, my to 6 kids and I practiced from 2pm until 7pm after school tonight and tomorrow(Saturday) we practice from 9:30am - 2:30pm.  None of my kids are past Algebra 1 yet, and never are, so we meet alot, take every old competition(including Team and Countdown), and I spend a ton of my own time collecting information and problem solving techniques from here :)

As I always say, Hard Work, Hard Work, Hard Work

That is a lot of time and effort.  Your kids and you deserve to win...

my school has never done better than 4th in chapter, hopefully that will change on Saturday

We've won the PA states 2 of the last 3 years. :D

my school got 2nd in chapter and 10th in CA last year, which was our best year.

My school didn't even get to chapter, if you know what I mean  :D


[b] Problem 5. [/b]Denote with $d(a,b)$ the numbers of the divisors of natural $a$, which are greater or equal to $b$. Find all natural $n$, for which
$d(3n+1,1)+d(3n+2,2)+\ldots+d(4n,n)=2006.$

[i]Ivan Landgev[/i]

$2006\geq n\geq \sqrt[3]{500^2}$
and check all the cases... :D

The sum equals the cardinal of the set $\{(x,y)\in\mathbb{N}^2\,|\,3n<xy\leq 4n,\, x\geq xy-3n\}$, but we can count how many elements have $y$ equal to $1, 2, 3,\dots$, rewriting the sum in the form of the telescoping series
\[ \left(\left[\frac{4n}1\right]-\left[\frac{3n}1\right]\right)+ \left(\left[\frac{4n}2\right]-\left[\frac{3n}2\right]\right)+ \left(\left[\frac{4n}3\right]-\left[\frac{3n}3\right]\right)+ \left(\left[\frac{4n}4\right]-\left[\frac{3n}4\right]\right)+ \left(\left[\frac{3n}4\right]-\left[\frac{3n}5\right]\right)+ \left(\left[\frac{3n}5\right]-\left[\frac{3n}6\right]\right)+ \left(\left[\frac{3n}6\right]-\left[\frac{3n}7\right]\right)+ \cdots  \]
All that remains is to solve the equation $3n-\left[\frac n2\right]+\left[\frac n3\right]=2006$, but this inmediately implies $\frac{2006-1}{\frac{17}6}<n<\frac{2006+1}{\frac{17}6}$, and so $n=708$.


ok the question is....
Suppose that a, b, c are positive intergers with
a+b+c=2006, and a!b!c!=m (times) 10(to the nth power), where m and n are integers and m is not divisible by 10, what is the smallest possible value of n???
A) 484
B) 492
C) 495
D) 498
E) 501

i know the answer but i need to know how to figure it out plz help...if u can email me the stuff at UNCshady107@gmail.com but i guess i can check back on here

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=75360 has already answered that topic.

Billy


Let $ (x_n)$ be a sequence of real numbers.

(a) Prove that $ (x_n)$ converges to a number $ A$ iff every subsequence of $ (x_n)$ has a subsequence which converges to $ A$.
(b) Does $ (x_n)$ converge if every subsequence of $ (x_n)$ has a subsequence which converges?

(b) is certainly false. For one counterexample, just use $ (\minus{}1)^n.$ Note that any subsequence of that has either infinitely many $ 1$s or infinitely many $ \minus{}1$s. In fact, every bounded sequence, no matter its other behavior, satisfies the condition of (b).

In fact, try proving this:

(b') Prove that $ (x_n)$ is bounded iff every subsequence of $ (x_n)$ has a subsequence which converges.

Proof of (b`):

1) Suppose sequence is bounded. Then every subsequence is also bounded and therefore has a convergent subsequence.

2) Suppose every subsequence has a subsequence which converges. Then if sequence is not bounded it is also not bounded eventually, therefore we can construct a subsequence a1 >= 1, a2 >= a1 + 1, ..., a(n) >= a(n-1) + 1, which does not have a convergent subsequence. Contradiction.

Finito :)

I've seen (a) under the name "Urysohn's principle". It's more general than just over real numbers, and very useful in some situations.


I made a mini AoPS puzzle hunt! Clues are scattered all over AoPS, and maybe some other places as well... (don't worry you won't have to leave your chair). Each clue will lead you to the next one in its own cryptic way .. good luck!

There is no prize at the end, except for maybe the satisfaction of figuring out my riddles.

[b]Clue #1[/b]
ION, who is a certain CON's direct opposite in one way, has a rather bizarre interest...

(BTW: Happy April Fool's Day, but this post is not a joke)

[quote="perfect628"]Could it be stopping early-onset nostalgia and Alzheimer's disease?[/quote]
It could possibly. (BTW, that part's not relevant to the puzzle. That's what happens when you involve outsiders in your setups, though the ION/CON thing works so beautifully)

Oh, to clarify, clue #1 is the only clue that you get for free. All of the other clues have already been planted in one way or another on AoPS (or somewhere else) You might have to think outside the box to reach some of them.

[quote="Nerd_of_the_Ages"]If we get to a clue, will we know it?

Like for example, let's say you are browsing through the forums and you happen to read where the 3rd clue is. Will you be able to tell it is a clue?[/quote]

I didn't make it that easy ... you will probably miss it, unless you are very observant. Don't trust them though ... I might have put a red herring or two in some of my posts.

Some of the clues will lead you to a particular post. The clue will be hidden in the post, sometimes rather well.

oh, gosh, oh gosh...

I just got a clue for free Because someone helped me without meaning to (you probably know who, randomdragoon)

I finally found the answer to the first clue, but i don't get how that is going to help me with my second clue. Is my second clue going to be around a place where they're talking about the first clue?

How far is this supposed to go?  I am not sure if where I got was the end.

You get to the end when you reach a page written by me that says "You win!"

"Red goes the moon" is a red herring. I put a couple of those around my posts. None of the clues lead to them. I put them there so you can't just search for all my posts and find clues that way. In other words, the phrase is not relevant to the puzzle at all.

As I said before, you'll have to think outside the box, but there isn't really anything that would require too much outside knowledge.

yay i won

time to get back on with my life

Weird...
that manage thing is so messed up...

[quote="CatalystOfNostalgia"]ION is my brother[/quote]

Who's ION?

my brother

obviously you didn't read my post very carefully

Dude, you basically just gave it away...
I thought there were no free clues...

[quote="randomdragoon's page"]
(No really, you win. Go gloat about it in the thread)
[/quote]

*gloats*

[quote]
You win! But was it really worth it? (No really, you win. Go gloat about it in the thread)
[/quote]

Omg, yes!
Finally.

Hey, I noticed that you only need the username to win, not the password...
Is this a bug?

[quote="bowei"]alright found the second clue?
weird
what the heck does y=2x-3 have to do with anything?[/quote]

The math problem isn't part of the puzzle. It's like one of those messages that mean one thing but when you look carefully it means something completely different

(the math problem is there so the post doesn't get rated as spam and deleted)

ZOMG I DON'T EVEN GET THE FIRST CLUE !!!1

Can you tell me what the first clue is supposed to lead to, because I didn't even understand it in the first place.


algebra
polynomial
group theory
abstract algebra
Galois Theory
superior algebra
superior algebra solved

Given a positive integer $n$, a field $F$, and a polynomial $P$ with degree $n$ in $F[X]$, one can show that a splitting field for $P$ has degree at most $n!$ over $F$.
Now given any positive integers $n,k$ (with $k \leq n!$), is it always possible to construct a field $F$ and a polynomial $P$ with degree $n$ in $F[X]$ such that a splitting field for $P$ has degree $k$ over $F$ ?

Surely not, the condition has to be $k|n!$ and not $k<n!$ (because the algebraic degrees multiply). Thus taking $k=n!-1$ will lead you to something impossible to do when $n\geq 3$.

To show the existence of such an extension for the possible cases:
Take a finite field and the $\frac{n}{k}$-th power of an irreducible polynomial of degree $k$. This works because all (finite) extensions of finite fields are normal.

For the general case:
Choose any irreducible polynomial with a Galois group that has a subgroup of order $k$.
More explicit example for $char(F)=0$:
Coose any $m$ such that the group $\mathbb{Z}/m\mathbb{Z}^*$ has a subgroup of order $k$, e.g. $m=2^k-1$.
Take $\mathbb{Q}$ and consider the field $H$ generated by the $m$-th roots of unity. There is also subgroup of order $k$ by the condition giving a fixed field $G$, thus there is an extension with $[H: G]=k$.
Let $a$ ba primitive element for that extension, then (a power of) the minimal polynomial of $a$ together with $G$ fulfills the property.


Wynona bought 4 video tapes and 2 video discs. The cost of the 4 tapes was the same as the cost of the 2 discs. If she spent a total of $ \$104$ and each tape was the same price, how many dollars did each tape cost?

2 discs = 4 video tapes

8 video tapes = 104 dollars

104/8=13

answer : 13


Hello everybody!
Here there are some questions found by me in www but for which, unfortunately, i haven't found the solutions.
[color=red][b]1[/b].[/color]What limits how fast you can walk? How fast can a horse walk?

[color=red][b]2[/b].[/color]Atmospheric pressure supports a column of mercury 0.76m high, and a column of water 9m high (because water is 11 times less dense than mercury). How do you explain trees taller than 9m?

[color=red][b]3[/b].[/color]Into how many regions can n planes divide space? Find the maximum number (what conditions on the planes ensure that the number is a maximum?). For example, one plane divides space into two regions; two planes divide space into at most four regions (but only three if you are unlucky, and only two if you are really unlucky). Hint: Play with the one- and two-dimensional versions of this problem, and then try to generalize the patterns that you find.

[color=red][b]4[/b]. [/color]a) In the (infinite) two-dimensional lattice of 1 \Omega resistors shown below, what is the resistance between points A and B?
b)(Hard!) What is the resistance between points A and C?

Bye!! :D

[hide="number 1"] I guess it's many factors, like the length of your legs, the friction coefficient between your feet and the gronud, gravity, ... I don't feel like analysing it into depth;)[/hide]

[hide="number 2"] Well, I guess the trees don't have hoses inside and don't suck the water from the top. There are many processes that pump the water upwards (maybe you should ask in the biology forum;-), like osmosis, diffusion, and who knows what else. And, there is another important factor, the moisture in the air. The 100 m tall redwwods on California coast get almost ALL their water from the humid air...[/hide]

[hide="number 3"] $N$ lines divide a plane into $\frac{N(N+1)}{2}+1$ regions. You can prove this by induction easily, the (N+1)th line crosses all N already drawn lines, therefore crosses N+1 regions and therefore adds N+1 more regions, which maked after adding K lines 1+(1+2+3+...+K) regions.

Now, let's take planes. 0 planes make 1 region. Each next plane (let's have N planes placed already) crosses all placed planes and if we select a good location, all those planes cross the new plane in lines that intersect each other line the lines in the 2D problem. That makes $\frac{N(N+1)}{2}+1$ planar regions on the new plane. Each of those planar regions divide a spatial region formed by the old N planes, that means that (N+1)th plane adds $\frac{N(N+1)}{2}+1$ new regions. Nth plane adds $\frac{N(N-1)}{2}+1$ regions. Together we have

$R(N) = 1 + \sum_{i=1}^N \frac{i(i-1)}{2}+1$
$R(N) = N+1 + \sum_{i=1}^N \frac{i(i-1)}{2}$
$R(N) = N+1 + \frac{1}{2} \sum_{i=1}^N i^2 - \frac{1}{2} \sum_{i=1}^N i$
$R(N) = N+1 + \frac{N(N+1)(2N+1)}{12} - \frac{N(N+1)}{4}$
$R(N) = \frac{N(N^2-1)}{6} + N + 1$
[/hide]

[hide="number 4"]
a) If you connect point A to a voltage source U (0 is in infinity), and current I will flow into the network, due to symmetry, current I/4 will flow towards B. Similarly, when you only connect voltage -U to point B, current I will flow out of the network and I/4 will flow from A. Now, if you connect BOTH whose voltages, due to the superposition theorem (or however you call it), the situation will act as a sum of previous situations. Therefore, there will be total current I through the network and I/2 of this current will flow through the resistor AB. Since the same amount of the current flows throught the rest of the network, its resistance (without AB) is the same as the resistance of AB. Therefore the whole network acts as two parallel ABs between A and B, that is 0.5 Ohm.

I still have a small problem in b) part but I hope I'll solve it some time later;)

EDIT:

concerning b) part, it turns out to be something around 0,637 according to my computer simulation, but I still cannot solve it...
[/hide]

BTW, could you please specify, where did you get those problems at? I'm really interested.

regarding 2...
water moving up trees isnt solely caused by vacuums, it's also due to the cohesive property of water

[quote="Humpty_Dumpty"]2. Can the answer be connected with capillarity?[/quote]

Of course capilarity has something to do with it, but it is not the sole mechanism pushing the water upwards. Try posting in the biology forum;)

[quote="Rushil"]Has anyone got a solution for 4(b) ???[/quote]

I tried, but could not. My computer program says approximately 0.637 $\Omega$. I could solve it for example if I could find out how much current flows through C if you only connect 1V voltage to point A.

Do you know the solution for 4b, Humty_Dumpty?

kubus, unfourtunately the site where i found these exercises doesn't provide solutions nor answers...  :(  (p.s. the site is that specified by Rushil).


I still don't get how our teacher explained it.  Didn't he keep getting *one less than the answer*?  Do you know how to do it, or are you asking for help?

Hmm, I got 19 too. Anyone have the solution?

just do 119/8 and 119/6

so 14 7/8 and 19 5/6

19 20/24-14 21/24= 4 23/24 since 5 feet is less than 1/24 of a mile, you don't have to account for that.  so 4 23/24 miles*4 since it is a quarter mile track.

(4 23/24)*4 is 19 5/6 or round down to 19. So the solution is wrong.

Dude. $ 19 \frac{5}{6}$ rounds up to 20....

[quote="cognos599"][quote="mathcountsfreak"]7. Two runners began running in the same direction on a quarter-mile track. The first runner began five feet ahead of the second runner. The first runner's pace is 6 minutes per mile, while the second runner's pace is 8 minutes per mile. Both runners continue at their respective paces for 119 minutes. How many times will the first runner pass the second runner?

[hide="answer"]
20
[/hide]

The solutions probably simple, but no matter what, I always end up getting one less than the answer.[/quote]

omg we just did this problem in a team round in mathcounts practice today and i put 20 and one of my teammates caught my mistake and said the answer was 19 and that was the correct answer.[/quote]
So 19 was counted right?

[quote="darkmatter47"]Dude. $ 19 \frac {5}{6}$ rounds up to 20....[/quote]

If you read the question, to pass someone, you have to do a complete go around. You can't be 3/4 of the way toward the person and then say "I passed him/her." For this reason, you go to the greatest integer that is lower than what you have.

@mathcounts Yes, the answer is 19.

I got 20, but my method was pretty unorthodox :P .... I found that the first runner would run 19 5/6 miles plus the five feet, which doen't make a difference, and the second guy runs 24 7/8 miles... This way, the 1st guy is passed up by 5 miles, and b/c it is a quarter mile track, you get 4*5 = 20...... :lol:

um, the five feet actually I think does make a difference.  maybe not, eh, I forget

Based on trivea's info, the first runner goes $ 19\frac{5}{6} * 5280 \plus{} 5$ feet, so he runs 104,725 feet. The second runner runs $ 24\frac{7}{8} *5280$ feet, so he runs 131,340 feet. There is a difference of 26,615 feet. Since there are 1,320 feet per lap, the second runner passes the first $ \frac{26,615}{1,320}$ times, which simplifies to $ \frac{5323}{64}$ or roughly $ 20.1628...$. I get $ \boxed {20}$

The 5 feet don't make a difference.
The quarter-mile track is 1320 feet long...

trivea's info is partially wrong...it's 14 7/8, not 24 7/8.

But really....

[b]mathcountsfreak's "correct" answer is wrong.[/b]

I own the 1998-99 competition book, and my answer key clearly states [b]19 times[/b].

[quote="Ihatepie"][quote="mathcountsfreak"]Thanks for the help. I was pretty sure 19 was right (the 5 feet don't seem to make a big difference), but I was just wondering if there was something we were missing.

Then again, Dr. Chi also got 19, so I don't see why I should've doubted it :P[/quote]

well if you read the entire topic, it was resolved.[/quote]

Other people don't think so. I agree with you, I just wanted to point it out to other people a little bit clearly.


My Prediction-- (May Sound Highly Pro-Ligon)

[hide="prediction"]
1. Allen Yang (Ligon)

2. Calvin Deng (Ligon)

3. Some Sixth Grader (Some School)

4. Nick Tobey (Ligon)
[/hide]

This is because Allen, Calvin, and Nick were the the top non-eighth grade Mathletes last year.

1. Academy At Lincoln

2. Ligon

3. SMITH

4. Ligon

state results

Assume that 250 people go to state

1. James Yang--Martin 8
2. Sanjay ____ --Ligon 6

...
...

248. Allen Yang--Ligon 8
249. Nick Tobey--Ligon 8
250. Ivan Kutznesomething--Ligon 8
251. Calvin Deng--Ligon 7

[quote="Teki-Teki"]I think that my little bro will make nats this year.[/quote]

You have a younger brother?  I had not heard about him yet...

Yes, I do.  Two, in fact, although only one is still in middle school - the other's at high school with me.  Yes, that was a joke.  He's almost certainly not going to make the Ligon Mathcounts team, even if he tries.  I know that's kinda mean, but it is true.

If Damien made that prediction, I might believe him, however.

Neither are you, at least when it comes to reading.

[quote="Teki-Teki"]
I might believe him[/quote]

Key word is "might", mate.

[quote="dnkywin"]Well, Damien's brother isn't that good... (and he's going to a chess tournament right after individual round is over). Not to be mean or anything...[/quote]

hmm ok

thank you very much

I will be sure to forward this information to him

maybe he can make AIME this year... except he's sick

And the fact that he didn't even make AMC 10... (at our school)

um however he did? and is not taking it because he has the flu?

Guys let's avoid needlessly bashing people to their face for no reason. Do I need to lock this again???

In other news, I am an idiot. More on this later.

Oops. i take that back. However, Alan did not do well enough to qualify for AIME.

Correct. Calvin Deng, hi!  :P     Nick Tobey was first, not me. :furious:

Oops. I should have said (NOT IN THAT ORDER, NECCESARILY) !!!!

Hey, I owned him in chess at Calvin's birthday party.

Well, if 1 is 3, and 2 is 6, then 3 must be [size=200]1[/size][/hide]

[quote="not_trig"][quote="PI-Dimension"]Dude Alan Jiang is a beast.

He like, owned me in chess =(. In like, 20 moves =(.[/quote]

Dude I don't know "Alan Jiang."

Apparently the real Allan Jiang says he was screwing around while playing chess.[/quote]

Well I was just trying to congratulate him for something unlike the other 99.9999999999999% of the population.

Stop posting :spam: :spam: :spam: :spam:


A pool in the shape of a circle measures 10 feet across. One cubic yard of concrete is to be used to create a circular border of uniform width around the pool. If the border is to have a depth of 4 inches, how WIDE will the border be?

NOTE: 1 cubic yard = 27 cubic feet.

[quote="Interval"]I don't clearly understand the concept of cubics.

Can you explain what exactly is meant by cubic feet, cubic inches, cubic yards, etc?

Are we talking about volume here?[/quote]
Cubic means something multiplied three times. So three cubed means three multiplied three times. So cubic feet means that there was a number in feet that was multiplied three times. And yes this usually talks about volume because volume usually is something cubed.

[quote="Interval"]I don't clearly understand the concept of cubics.

Can you explain what exactly is meant by cubic feet, cubic inches, cubic yards, etc?

Are we talking about volume here?[/quote]
Yes, volume. Regular measures are used to measure length, width, height, etc. Square units are used to measure surface area. Cubic units are used for volume. A rectangular prism with sides 2 feet, 3 feet, and 5 feet has a volume of 30 cubic feet.


Let there be $ n \geq 2$ and $ x_{1},x_{2},..,x_{n} \in\mathbb{R}$ such that $ x_{1}\plus{}x_{2}\plus{}...\plus{}x_{n}\equal{}a$ and $ x_{1}^{2}\plus{}x_{2}^{2}\plus{}...\plus{}x_{n}^{2}\equal{}b$.Find the maximum value of $ x_{1}$.

wellllll if $ b>a^2$ then the obvious choice is:

$ x_1\equal{}\alpha , x_2 \equal{}x_3 \equal{}... \equal{} x_n \equal{} \frac{a\minus{}\alpha}{n\minus{}1}$

$ \alpha^2 \plus{}(n\minus{}1)\left(\frac{a\minus{}\alpha}{n\minus{}1}\right)^2\equal{}b$

$ \frac{(a\minus{}\alpha)^2\plus{}\alpha^2(n\minus{}1)}{n\minus{}1}\equal{}b$

$ a^2\minus{}2a\alpha\plus{}n\alpha^2\equal{}b(n\minus{}1)$

$ n\alpha^2\minus{}2a\alpha\plus{}a^2\minus{}b(n\minus{}1)\equal{}0$

$ \alpha\equal{}\frac{2a \pm \sqrt{4a^2\minus{}4n(a^2\minus{}b(n\minus{}1))}}{2n}\equal{} \frac{a \pm \sqrt{(b\minus{}a^2)(n\minus{}1)}}{n}$  and pick the plus value because its bigger.

if $ \frac{a^2}{n}<b<a^2$ then we pick

$ x_1\equal{}\sqrt {b \minus{} \delta}, x_2\equal{}x_3\equal{}...x_n\equal{} \frac{a\plus{} \minus{} \sqrt {b\minus{}\delta} }{n\minus{}1}$

and so $ (n\minus{}1)\delta \equal{} (a \plus{}\sqrt{b\minus{}\delta})^2\equal{}a^2\plus{}2a\sqrt{b\minus{}\delta}\plus{}b\minus{}\delta$

$ n\delta\minus{}a^2\minus{}b\equal{}2a\sqrt{b\minus{}\delta} \to n^2\delta^2 \minus{}2(b\plus{}a^2)n\delta\plus{}(a^2\plus{}b)^2\equal{}4a^2(b\minus{}\delta)$

more pleasantly:
$ n^2\delta^2\minus{}[2n(b\plus{}a^2)\minus{}4a^2]\delta\plus{}[(a^2\plus{}b)^2\minus{}4a^2b]\equal{}0$

so $ \delta \equal{} \frac{2n(b\plus{}a^2)\minus{}4a^2 \pm \sqrt{ [2n(b\plus{}a^2)\minus{}4a^2]^2\minus{}4n^2[(a^2\plus{}b)^2\minus{}4a^2b]}}{2n^2}\equal{}$

$ \frac{n(b\plus{}a^2)\minus{}2a^2 \pm \sqrt{ [n(b\plus{}a^2)\minus{}4a^2]^2\minus{}n^2[(a^2\plus{}b)^2\minus{}4a^2b]}}{n^2}\equal{}$

$ \frac{b\plus{}a^2}{n}\minus{}2\left(\frac{a}{n}\right)^2 \pm \frac{\sqrt{ [(b\plus{}a^2)\minus{}4a^2]^2\minus{}[(a^2\plus{}b)^2\minus{}4a^2b]}}{n}\equal{}$

$ \frac{b\plus{}a^2}{n}\minus{}2\left(\frac{a}{n}\right)^2 \pm \frac{2a\sqrt{2a^2 \minus{}b}}{n}\equal{}\delta$

so we pick the bigger one again.

maybe?


I can't chat on ftw, unless i have a higher post rating...

So can someone please find it in their good hearts to do that.

This is helpful...3+1=4

I bumped you up with an excellent (6).  Now you can help me out whenever you see my posts.

You kinda need more than one post...  cough make multis cough...after all, thats what missouri is known for...

[quote="pianoforte"]Dang, I rated like 50 of your posts 1s and your rating is not going down.[/quote]
Darn I gave him 40 the other day...did you notice any effect? btw piano like my rating? :)

Dang, you're right, non-mods can't see ratings.

Thinking back, I am 90% sure rrusczyk modded me.

[quote="Mewto55555"][quote="007math"]We can't see your rating.  Only you can see yours, along with mods.

EDIT: pianoforte is a mod, isn't he. :wallbash_red:[/quote]

Yeah idk which n00b came up with THAT brilliant idea![/quote]
I wonder why you're saying that...seeing that you'd probably be pretty embarrassed from seeing how low your rating is compared to other people :O

I don't really see why people can't see other people's ratings. What good is a mod alone supposed to do with that knowledge?

It would be a better indicator of how spammy a person is.


For every $ x,y\in \Re ^{\plus{}}$, the function $ f: \Re ^{\plus{}} \to \Re$ satisfies the condition $ f\left(x\right)\plus{}f\left(y\right)\equal{}f\left(x\right)f\left(y\right)\plus{}1\minus{}\frac{1}{xy}$. If $ f\left(2\right)<1$, then $ f\left(3\right)$ will be

$\textbf{(A)}\  2/3 \\ \textbf{(B)}\  4/3 \\ \textbf{(C)}\ 1 \\ \textbf{(D)}\ \text{More information needed} \\ \textbf{(E)}\ \text{There is no }  f \text{ satisfying the condition above.}$

[hide]Putting $ y\equal{}1$, with the notation $ c : \equal{}f(1)$, we get

$ f(x)\plus{}c\equal{}cf(x)\plus{}1\minus{}{1\over x}\iff (c\minus{}1)f(x)\equal{}c\minus{}1\plus{}{1\over x}\iff f(x)\equal{}1\plus{}{1\over (c\minus{}1)x}$

Putting $ x\equal{}1$ into the last expression, we get $ c\equal{}1\plus{}{1\over c\minus{}1}\iff (c\minus{}1)^2\equal{}1\iff c\in\{0,2\}$

For $ c\equal{}2$ we get $ f(x)\equal{}1\plus{}{1\over x}$, but that function doesn't satisfy $ f(2)<1$

For $ c\equal{}0$ we get $ f(x)\equal{}1\minus{}{1\over x}$ and that's the desired function. Hence $ f(3)\equal{}{2\over 3}$ and the answer is $ A$.[/hide]


[quote="math1990"](b^n-1)/(b-1) is a perfect square then prove that b is divisible by 8[/quote]
wrong!conterexample:$ n\equal{}1,b\equal{}2$ or $ n\equal{}2,b\equal{}3$...

I'm sorry but I have seen that problem, may be I forget something.
may be n>2 or additionally b- is even


Prove that the sequence is the sum of two well known sequences.

[quote="Ubemaya"]Wow, how do you know that?[/quote][url=http://www.unl.edu/amc/a-activities/a7-problems/putnam/-pdf/1990.pdf]Putnam 1990[/url]

Well you could:

[hide="be cheap"]And notice that the ratio of consecutive terms seems to approach the sequence of natural numbers, subtract out $n!$, and then notice that the result is $2^{n}$. Proving that $T_{n}=n!+2^{n}$ is very easy by induction[/hide]

[hide="or"]
use a telescoping sum.
[/hide]


If $ ABC$ is a triangle with the sides $ a,b,c$ and $ 0< \alpha \le \beta <\frac{\pi}{2}$ then: $ b^2\plus{}c^2 \tan^2 \beta > a^2 \sin^2 \alpha$.


message from mathfanatic: I will return your post tomorrow.  Until then, it will sit in a word document on my computer.  I intend to enforce the no-discussion-of-scores-until tomorrow rule as strictly as possible.

We're locking/deleting the threads regarding discussing the test (even just general comments like difficulty) until tomorrow.   Tomorrow you will be free to discuss anything you like about the tests.


algebra
polynomial
inequalities open
inequalities

Hi every one! I put the following problem when try to generate another problem, and I think it is realy strange and interested :D

Find the largest const r such that for all $x,y,z \in [-r,r]$ and satisfy $x+y+z \ge 0$, we ussual have:
$(1+x+y^2)(1+y+z^2)(1+z+x^2) \ge 1$

longer a week but still have no idead ...

Ok, first at all, you should try to check some remarks:  ;)

1)the condition $x+y+z\ge 0$ can change to $x+y+z=0$

2)If $r\le \frac{1}{2}$ then the problem is correct!

3)r is not found when chose (x,y,z)=(r,-r,0) and cyclics  :D


The digits $ 1$, $ 2$, $ 3$, $ 4$, $ 5$, $ 6$, $ 7$, and $ 9$ are used to form four two-digit prime numbers, with each digit used exactly once. What is the sum of these four primes?

$ \text{(A)}\ 150 \qquad
\text{(B)}\ 160 \qquad
\text{(C)}\ 170 \qquad
\text{(D)}\ 180 \qquad
\text{(E)}\ 190$

[hide="Solution"]One example: 41+67+23+59=190 (E)

There are other combinations of numbers that give the same answer.[/hide]

[hide="Quickest solution"]1, 3, 7, and 9 must be the numbers in the units places as none of the others can finish a prime. So it's $ 20\plus{}40\plus{}50\plus{}60\plus{}1\plus{}3\plus{}7\plus{}9\equal{}190$.[/hide]


linear algebra
matrix
linear algebra solved

The linear transformation $f$ on the $x-y$ plane satisfies the following three conditions.

(1) A point $(1,0)$ is mapped to the inner part of the fourth quadrant by $f$.

(2) A point $(0,1)$ is mapped to the inner part of the secod quadrant by $f$.

(3) A point $(1,1)$ is mapped to the inner part of the first quadrant by $f$.

Then prove that there exist the inverse transformation for $f$ and if  the image of a point $P$ denoted by $f(P)$ lie in the inner part of the first quadrant, then the point  $P$ also lies in the inner part of the first quadrant.

Let's think matrix instead of linear transformation. Conditions (1) and (2) tell us that the matrix of this transformation can be written as

$A=\begin{bmatrix} a & -b \\ -c & d\end{bmatrix}$, where $a,b,c,$ and $d$ are all positive.

Condition (3) than tells us that $a>b$ and $d>c.$ From that we conclude that $ad>bc$ and hence $\det A = ad-bc>0.$

Thus $A$ is invertible. We can write its inverse as:

$A^{-1}=\frac1{ad-bc}\,\begin{bmatrix} d & b \\ c & a\end{bmatrix}$

Since all the entries in $A^{-1}$ are positive, it generates a linear transformation that maps the first quadrant into itself.

Thank you for your reply, Kent Merryfield. Your solution is the very model answer.

kunny


Let $ ABC$ be a triangle. Consider the circles with the centers $ A$ , $ B$ , $ C$ and the radius $ p \minus{} a$, $ p \minus{} b$ , $ p \minus{} c$ respectively, where $ 2p \equal{} a \plus{} b \plus{} c$ . These circles are exterior tangent to the circle $ C(X,x)$  and are interior tangent to the circle $ C(Y,y)$ . Ascertain $ x$ , $ y$ , $ IX$ , $ IY$ and prove that $ x(2 \plus{} \sqrt 3)\le r\le y(2 \minus{} \sqrt 3)$ , where $ C(I,r)$ is the incitcle  of $ \triangle ABC$ .

[hide="Answer."]
$ x \equal{} \frac {pr}{2p \plus{} (4R \plus{} r)}$ and $ y \equal{} \frac {pr}{2p \minus{} (4R \plus{} r)}$ . Observe that $ \frac 1x \plus{} \frac 1y \equal{} \frac 4r$ .[/hide]
See and [url=http://www.mathlinks.ro/viewtopic.php?t=199811][color=red][b]here[/b][/color][/url] .

x and y are the radius of the Soddy Circles and so

$ x \equal{} \frac {A}{2p \plus{} (4R \plus{} r)}$ and $ y \equal{} \frac {A}{2p \minus{} (4R \plus{} r)}$

Then the first and the second inequality become $ 4R\plus{}r \ge \sqrt{3}p$ wich is know

O.K., their names are the Soddy's circles, if you say.  But $ IX$ and $ IY$ ?


inequalities
trigonometry
inequalities unsolved

[quote="iptgut"][/quote]
the inequality is equivalent to $ \sum {\frac {1}{{1 \plus{} \frac {a}{{a \plus{} c \minus{} b}}}}} \le \frac {3}{2}$
By Cauchy we have :   $ \frac{{a}}{a \plus{} c \minus{} b} \plus{}1\ge 2\sqrt {\frac{{a}}{a\plus{}c\minus{}b}}$
So we need to prove $ \sum {\sqrt {\frac {{a \plus{} c \minus{} b}}{a}} } \le 3$
Because $ a,b,c$ be the sides of a triangle so we have:
                 $ \frac {{a \plus{} c \minus{} b}}{a} \equal{} \frac {{\sin A \plus{} \sin C \minus{} \sin B}}{{\sin A}} \equal{} \frac {{2\sin \frac {C}{2}\cos \frac {B}{2}}}{{\cos \frac {A}{2}}}$
it's following that
$ \sum {\sqrt {\frac {{a \plus{} c \minus{} b}}{a}} } \equal{} \frac {{\cos \frac {B}{2}\sqrt {\sin C} \plus{} \cos \frac {C}{2}\sqrt {\sin A} \plus{} \cos \frac {A}{2}\sqrt {\sin B} }}{{\sqrt {\cos \frac {A}{2}\cos \frac {B}{2}\cos \frac {C}{2}} }}$
$ \le \sqrt {\frac {{\left( {{{\cos }^2}\frac {B}{2} \plus{} {{\cos }^2}\frac {C}{2} \plus{} {{\cos }^2}\frac {A}{2}} \right)\left( {\sin C \plus{} \sin A \plus{} \sin B} \right)}}{{\cos \frac {A}{2}\cos \frac {B}{2}\cos \frac {C}{2}}}}$
$ \equal{} \sqrt {2\left( {\cos B \plus{} \cos C \plus{} \cos A} \right) \plus{} 6} \le \sqrt {2.\frac {3}{2} \plus{} 6} \equal{} 3$
Have done! :D

Put $ a\equal{}y\plus{}z,b\equal{}x\plus{}z,c\equal{}x\plus{}y$, where $ x,y,z\in \mathbb{R}^\plus{}$

Than the inequality is
$ \sum_{cyclic} \frac{y\plus{}z}{3y\plus{}z}\ge \frac32$

or

$ A\equal{}\sum_{cyclic} \frac{y}{3y\plus{}z}\le \frac34$

This can be easily proven.


suppose that S={1,2,3...3^k} and A is a subset of S with 2^k+1  element(k>2)
prove that there exists a,b,c such that
i, a,b,c belong to A
ii,a+b=2.c

[quote="comet_vn"]suppose that S={1,2,3...3^k} and A is a subset of S with [text]2^k[\text] element
prove that there exists a,b,c such that
i, a,b,c belong to A
ii,a+b=2.c[/quote]

This seems clearly wrong.
Let $k=2$, $S=\{1,2,3,4,5,6,7,8,9\}$, and $A=\{1,2,4,5\}$

$a$, $b$ and $c$ dont exist.

--
Patrick

[quote="comet_vn"]suppose that S={1,2,3...3^k} and A is a subset of S with [text]2^k[\text] element
prove that there exists a,b,c such that
i, a,b,c belong to A
ii,a+b=2.c[/quote]

This is quite false for every $k$. Even if $T=\left\{ 1,2,3,\cdots ,\frac{3^{k}+1}{2}\right\}$, then there exists a subset of $T$ with $2^{k}$ elements such that no three are in arithmetic progression. For example, we can take the numbers whose base 3 expressions contain no 2's and add 1 to each of them. And of course $T$ is contained in $S$.

[quote="Baronov"]I think that comet_vn mean this one:
Let A={1,2...3n} and B is subset of A with n elements then there are three numbers a,b,c in B such
that a+b=c.[/quote]

Not sure if this was intended.

take B = {2n+1, 2n+2, ..., 3n} has n elements and we can never have a+b=c in this set.

In fact, if we take a set with any 3n integers, there is some subset of more than n integers for which a+b=c is never satisfied. (A famous result of Paul Erdos on sum-free subsets of a given set).

Aryabhatta of course you are right this is exactly what i wanted to say.
Sorry for this stupid mistake.

[quote="Aryabhatta"][quote="comet_vn"]who can help me ?[/quote]

No one can, unless you specify what the problem is exactly. The original statement of the problem was shown to be false by scorpius119.

Also, please avoid using subject titles like "old?". Give some description. How would it be if there are a 100 topics all titled "old?", "easy", "hard"?[/quote]
I saw that problem when I read it in a notebook
and somebody asked me about it so I  think "old"


define positive integers   $ x, y$, such that      $ x^4  \minus{} 3y^4 \equal{} \minus{}2$

are we supposed to solve for $ x$ and $ y$ ??

x,y integers

$ x^2\equal{}a,y^2\equal{}b$

$ a^2\minus{}3b^2\equal{}\minus{}2$

hello, i have found $ x\equal{}\minus{}1,y\equal{}\minus{}1$ ; $ x\equal{}\minus{}1,y\equal{}1$ ; $ x\equal{}1,y\equal{}\minus{}1$ ; $ x\equal{}1,y\equal{}1$.
Sonnhard.

It seems like pell equation......Does anyone have a solution for this problem??? :?:


Say you choose a random permutation of 5 digits from the set {1,2,3,4,5}, where each digit is used exactly once. What's the probability that you choose a permutation that changes from increasing to decreasing, or decreasing to increasing at most once? That is, strings like 12345, 54321, 12543, and 53214 are acceptable, but strings like 13245 or 53241 are not.

I had this on a quiz earlier but had no success in finding a nice solution. Can anyone solve this to ease my mind?

How many permutations go up-up-up-up?  How many go up-up-up-down?  up-up-down-down? etc.

A different approach: consider first the case where we have ups followed by downs.  Let $ S \subseteq \{1, 2, 3, 4\}$.  How many such permutations are there if the elements of $ S$ are exactly the set of values preceding $ 5$ in the permutation?

(This is a rather silly question: there are few enough of these permutations that you can actually list them out by hand in only a minute or two.)

Sorry, I've been fiddling with it but I'm still drawing a blank. I don't understand how to systematically count all the possibilities using permutations and combinations.

I've proposed two different methods.  Have you tried either one?  If so, what have you accomplished?  Where are you stuck?

I've been trying your alternative approach. If I understand correctly, am I supposed to find all increasing sequences that can be made from the non empty (improper) subsets of $ \{1,2,3,4\}$?

I couldn't think of a way to do it systematically though! I just brute forced it as
1234 for four elements
123 134 124 234 for three elements
12 13 14 23 24 34 for two elements
1 2 3 4 for one element.

for a total of 15 increasing sequences that could come before 5. I suppose it follows that the decreasing sequence following 5 would then already be determined, as there would only be one possibility of the remaining unused elements. Thanks for your help and patience JBL. How would I do this more efficiently though? It seems I've only found the set of sequences that change from increasing to decreasing at 5.

This is a good time for you to learn this, if you haven't before: how many subsets (counting the empty set and the whole set) does a one-element set have?  A two element set?  A three element set?  Guess the pattern?  Prove it?

[quote="Mr Fubini"]It seems I've only found the set of sequences that change from increasing to decreasing at 5.[/quote]  At what other value could it change from increasing to decreasing?  (By the way, there was no reason to exclude the empty set, which gives rise to the permutation 54321.)

Then you must make a similar computation for permutations that go down first, then up (here you choose a subset of $ \{2, 3, 4, 5\}$ to precede the 1), and then you need to subtract 2 because you've counted the permutations 12345 and 54321 twice (once in each case).

Ah, now it's a little clearer. Thanks to the both of you, tancm and JBL.

Oh, and for your first question, a set of n elements has $ 2^n$ subsets, correct? As for each element in the set, we have the choice of including it or excluding it when building a subset, and that's 2 choices n times.

Right: your 15 was $ 2^4 \minus{} 1$ (with the $ \minus{}1$ because you excluded the empty set).


There are infinitely many  purses. One of them contains KM coins, the other are empty.
Every second the purses undergo the following procedure: One takes a coin from each non-empty purse and puts these coins together into an empty purse.
At some moment there are K purses holding M coins.
For which K, M is it possible?

These are cool problems, but there's always something unclear, isn't there? :)

When we take $5$ coins (for example) in one move, do we put them all in the same non-empty purse?

I failed to find any pairs except for $K=1,\ M\in \mathbb N^*$ and $(K,M)\in \{(2,1),(2,2)\}$. Maybe I missed some of them?

I think I've noticed something nice, but I'm not sure it's true: if $KM$ isn't a triangular number, then the sequence of configurations becomes periodic exactly after $KM$ steps (i.e. after the $KM$'th move we reach a configuration we have reached before, and this happens for the first time).

yes this is the answer. I don't know about your observation.
I tried to prove the sequence will become periodic after at most KM steps, but then I found the solution and stopped thinking..
Anyway, try to solve it.
It's not so hard, I wanted to put it into Solved Problems Section, but I don't know how I put it here

Well, it was in the "Solved Problems" section, but I think the moderators thought you wanted to put it here. Anyway, if you have the solution, you should post it in the "Proposed Problems" section.

The idea is that during the first $KM$ moves (after the first move), until we finish off the first purse, the other purses have a configuration of the type $1,2,3,\ldots,k,k+2,k+3,k+4,\ldots,t$ (it could also be $2,3,4,\ldots,t$). It's easy to show this (I didn't do it rigorously, but once you realize what's going on it becomes sort of boring to try to write it all down in a rigorous manner :)). For given $KM$, there is an $n$ s.t. $1+2+\ldots+n\le KM<1+2+\ldots+n+(n+1)$. In this case there is a $t\in [0,n]$ s.t. $KM=1+2+\ldots+n+t$. Then it's not that hard to show that the sequence which will repeat itself first is $1,2,\ldots,n-t,n-t+2,n-t+3,\ldots,n,n+1$.

A corollary of all of this is that the purse with the smallest number of coins always has $1$ or $2$ coins, so $M\le 2$. Now, if $M=1$ it means that at some point we will have $K$ purses each with $1$ coin, and at the next step we will have a purse with $K$ coins, so from the corollary we find that $K\le 2$. We do something similar if $M=2$.

That stuff I said in the previous paragraph can (I think) show us that for $KM$ not triangular the sequence gets periodic after $KM$ steps (until we finish off the first purse), and if $KM$ is triangular then it becomes periodic after $KM-n$ steps or something like that (maybe $KM-n+1$, I'm too lazy to check right now), where $KM=1+2+\ldots+n$.

Nice soln, grobber! Thanks!
Neither me nor the author didn't observe that law of yours.
Your solution is very nice and even explains more about how do the coins behave

I have to stress again upon the fact that it's not really a solution :D.


The probability that it will rain on any given day in April is $ \dfrac{2}{3}$. compute the probability that it will rain at least two days out of three consecutive April days.
This problem has something to do with the Bernoulli Experiment. But I don't know what the Bernoulli Experiment is or how to use it. Could someone please explain this to me.

I don't know what they are either, so...

[hide="Solution"]
The probability that it rains two days exactly is $ \binom32\left(\frac23\right)^2\left(\frac13\right) \equal{} \frac49$, and the probability that it rains all three days is $ \left(\frac23\right)^3 \equal{} \frac8{27}$, so our total probability is $ \frac49 \plus{} \frac8{27} \equal{} \boxed{\frac {20}{27}}$.
[/hide]

That's exactly what you used here, Bernoulli Experiment.


limit
number theory
relatively prime
number theory proposed

Let $ a,b$ and $ k$ be positive integers, and let $ p_k(a,b)$ be the perid of the sequence $ \{ a^n \text{mod} b^k\}^{\infty}_{n\equal{}1}$. Find the $ \lim_{k\rightarrow \infty} \frac{p_{k\plus{}1}(a,b)}{p_k(a,b)}$.

It is not hard $ \lim_{k\rightarrow \infty} \frac {p_{k \plus{} 1}(a,b)}{p_k(a,b)}\equal{}b$.

Nice problem!
Denote the limit as $ P(a,b)$
Let us first suppose without loss of generality that $ (a,b) \equal{} 1$ as $ P(a,b) \equal{} P(a,\frac {b}{(a,b)})$.
note that $ p_{k \plus{} 1}$ is divisible by $ p_k$ so put $ \frac {p_{k \plus{} 1}}{p_k} \equal{} q_k$ . where $ p_k \equal{} p_k(a,b)$ is the order of $ a (\mod b)$
so $ a^{p_k} \equal{} 1 \plus{} r_k b^k$ and $ (1 \plus{} r_k b^k)^{q_k}\equiv 1 \plus{} q_k r_k b^k (\mod b^{k \plus{} 1})$ so $ q_k \equal{} \frac {b}{(b,r_k)}$
as for large enough $ k$ $ (b,r_k) \equal{} 1$
we get that $ P \equal{} b$  :)

Edit: Rust was faster though...:D

Rust attention! the answer is actually $ \frac{b}{(b,a)}$  :wink:  see my post above

I consider only case $ (a,b)\equal{}1$.
[quote="Albanian Eagle"]Rust attention! the answer is actually $ \frac {b}{(b,a)}$  :wink:  see my post above[/quote]
If $ (a,b)\not \equal{}1$ , then your answer is wrong. Consider $ a\equal{}2,b\equal{}12$.
Let $ b\equal{}b_1p_1^{k_1}..p_m^{k_m}$ and $ p_i|a$ ,for $ i\equal{}1,...,m$ and $ (b_1,a)\equal{}1$.
Then $ \frac{p_{k\plus{}1}(a,b)}{p_k(a,b)}\equal{}b_1 \ \forall k>N$.

ooops! yes,
anyways thats what I meant (and my solution too)
so the answer should be the biggest divisor of b which is relatively prime to a. (I was stupid enough to call this b\(a,b))  :oops:


greatest common divisor
floor function
number theory
least common multiple
videos
MATHCOUNTS

How many of the first 500 positive integers are divisible by 3, 4
and 5?

$ \text{gcf}(3,4,5)\equal{}60$

There are $ \lfloor\frac{500}{60}\rfloor$ or $ \boxed{8}$ numbers that work.

[quote="ernie"]$ \text{gcf}(3,4,5)\equal{}60$[/quote]

I think you mean $\text{lcm}(3,4,5)\equal{}60$ up there, since $\text{gcf}$ usually stands for "Greatest Common Factor", and here we're looking for the least common multiple.

I used the same approach ernie did.

This problem is very similar to the second of the "Monkeys on a Number Line" problems that Richard talked about in his [url=http://www.artofproblemsolving.com/Resources/video.php?video_id=115&mc=1]November 2009 MathCounts Mini video[/url].

Tim

No, 0 is not a [b]positive[/b] integer, so the correct answer is 8.

[quote=tylia]isn't that 0 is also an integer and dividable by any number ?

if so isn't the answer supposed to be 9 ?[/quote]

Also, 0 is not divisible by everything.

[quote=tylia]isn't that 0 is also an integer and dividable by any number ?

if so isn't the answer supposed to be 9 ?[/quote]

0 is neither positive nor negative


Εστω ότι υπάρχει μια συνάρτηση f, επί, με f:$ M_{n}\rightarrow {{0,1,2,...,n}}$, όπου $ M_{n}$΄είναι το σύνολο των nxn πινάκων με πραγματικούς όρους

α) Να αποδειχθεί ότι για κάθε αντιστρέψιμο πίνακα Χ, $ f(X)\equal{}f(I_n)$ όπου $ I_n$ είναι ο μοναδιαίος
    πίνακας.

β) Για κάθε Χ αντιστρέψιμο και για κάθε Y ανήκει στο $ M_n$ να αποδειχθεί ότι
             $ f(XY)\equal{}f(YX)\equal{}f(Y)$

γ) Να αποδειχθεί ότι η f είναι η μοναδική και μάλιστα f=rank

[quote="spinos"]Εστω ότι υπάρχει μια συνάρτηση f, επί, με f:${ M_{n}\rightarrow \{0,1,2,...,n\}}$, όπου $ M_{n}$΄είναι το σύνολο των nxn πινάκων με πραγματικούς όρους

α) Να αποδειχθεί ότι για κάθε αντιστρέψιμο πίνακα Χ, $ f(X) = f(I_n)$ όπου $ I_n$ είναι ο μοναδιαίος
    πίνακας.

β) Για κάθε Χ αντιστρέψιμο και για κάθε Y ανήκει στο $ M_n$ να αποδειχθεί ότι
             $ f(XY) = f(YX) = f(Y)$

γ) Να αποδειχθεί ότι η f είναι η μοναδική και μάλιστα f=rank[/quote]

Spinos, θέλει λίγο φτιάξιμο η διατύπωση, δε βγαίνει ιδιαίτερο νόημα έτσι...

Cheerio,

Durandal 1707


A biologist wants to calculate the number of fish in a lake. On May 1 she catches a random sample of 60 fish, tags them, and releases them. On September 1 she catches a random sample of 70 fish and finds that 3 of them are tagged. To calculate the number of fish in the lake on May 1, she assumes that 25% of these fish are no longer in the lake on September 1 (because of death and emigrations), that 40% of the fish were not in the lake May 1 (because of births and immigrations), and that the number of untagged fish and tagged fish in the September 1 sample are representative of the total population. What does the biologist calculate for the number of fish in the lake on May 1?


Why do some forums for the states and regions say "Only users granted special acsess (sorry for mispelling) can see this forum" or something?  Because I live in Southern California and I can't see "my" on forum.

And how do you get that special acsess (sorry again)?

PM a moderator for that specific forum, and he will grant you access to it.

I'm not sure why there are private forums though...


Let $ f(x)\equal{}x^{2m}\minus{}x^{2m\minus{}1}\plus{}x^{2m\minus{}2}\minus{}...\plus{}x^2\minus{}x\plus{}1$.

Then we have to show that $ f(x_1)\plus{}f(x_2)\plus{}...\plus{}f(x_n) \le n$

$ f(x)\equal{}\frac{x^{2m\plus{}1}\plus{}1}{x\plus{}1}\le \frac{x\plus{}1}{x\plus{}1}\equal{}1$ So the result immediately follows.

Hope I did not make any mistakes.

Cheers,

Rofler

[quote="xzlbq"]Teacher csl:
Inequality In the opposite direction，you do not say....
BQ[/quote]

Form is beautiful, but prove too simple,see #2.


function
induction
algebra unsolved
algebra

$ f(x)$ is a function defined on $ [0,1]$ such that
1) $ f(1)\equal{}1$,
2) $ f(x)\equal{}\frac{1}{3}[f(\frac{x}{3})\plus{}f(\frac{x\plus{}1}{3})\plus{}f(\frac{x\plus{}2}{3})]$ for all $ x \in [0,1]$, and
3) For every $ \epsilon>0$, there exists a positive $ \delta$ such that: for all $ x \in [0;1]$ and $ |x\minus{}y|<\delta$, we have
$$ |f(x)\minus{}f(y)|<\epsilon.$$

Prove that $ f(x)\equal{}1$ for all $ x \in [0,1]$.

[quote="robbenmaths"]$ f(x)$ is a function defined on $ [0;1]$ such that
1) $ f(1) \equal{} 1$
2) $ f(x) \equal{} \frac {1}{3}[f(\frac {x}{3}) \plus{} f(\frac {x \plus{} 1}{3}) \plus{} f(\frac {x \plus{} 2}{3})]$ for all $ x \in [0;1]$
3) For every $ \epsilon > 0$, there exists a positive $ \delta$ such that: for all $ x \in [0;1]$ and $ |x \minus{} y| < \delta$, we have
$ |f(x) \minus{} f(y)| < \epsilon$.

Prove that $ f(x) \equal{} 1$ for all $ x \in [0;1]$[/quote]

The point $ 3)$ implies $ f(x)$ is continuous, so is bounded on $ [0,1]$. Let $ m$ the minimum value.

So it exists $ x_0$ such that $ f(x)\geq f(x_0)\equal{}m$ $ \forall x\in [0,1]$.
Then $ f(x) \equal{} \frac {1}{3}[f(\frac {x}{3}) \plus{} f(\frac {x \plus{} 1}{3}) \plus{} f(\frac {x \plus{} 2}{3})]$ implies $ f(x_0) \equal{} f(\frac {x_0}{3}) \equal{} f(\frac {x_0 \plus{} 1}{3}) \equal{} f(\frac {x_0 \plus{} 2}{3})\equal{}m$
Then, applying the same demo with $ \frac{x_0}{3}$,  $ \frac{x_0\plus{}1}{3}$ and  $ \frac{_0x\plus{}2}{3}$, we get $ f(\frac{x_0\plus{}k}{9})\equal{}m$ $ \forall k\equal{}0,1,2,...,8$.
And, with induction : $ f(\frac{x_0\plus{}k}{3^p})\equal{}m$ $ \forall k\equal{}0,1,2,...,3^p\minus{}1$
And, since $ f(x)$ is continuous, $ f(x)\equal{}m$ $ \forall x\in[0,1]$

And so, since $ f(1)\equal{}1$,  $ f(x)\equal{}1$ $ \forall x\in[0,1]$

we take for $ e>0$: $ E(e)\equal{}\{h\in[0,1]: \ \forall x,y\in[0,1],|x\minus{}y|<h\Rightarrow |f(x)\minus{}f(y)|<e\}$
and we take $ \delta\equal{}sup(E(e))$
if $ \delta <1$
we see that for $ h\in[\delta,min(3\delta,1)[$
if $ |x\minus{}y|\le h$ then for $ k\in\{0,1,2\}: \ |\frac{x\plus{}k}{3}\minus{}\frac{y\plus{}k}{3}|<\frac{h}{3}<\delta$
so $ \forall k\in\{0,1,2\}: \ |f(\frac{x\plus{}k}{3})\minus{}f(\frac{y\plus{}k}{3})|<e$
then $ |f(x)\minus{}f(y)|\equal{}\frac{1}{3}|\sum_{k\equal{}0}^{2}f(\frac{x\plus{}k}{3})\minus{}f(\frac{y\plus{}k}{3})|<e$
then $ min(3\delta,1) \in E(e)$ gives $ min(3\delta,1)\le sup(E(e))\equal{}\delta$ (impossible)

then $ \forall e>0: \ sup(E(e))\equal{}1$
so $ \forall e>0,\forall x,y\in]0,1]$ we have $ |x\minus{}y|<1$ then $ |f(x)\minus{}f(y)|<e$ then $ f(x)\equal{}f(y)\equal{}f(1)\equal{}1$
so $ \forall x\in]0,1]: \ f(x)\equal{}1$ and $ f(0)\equal{}\frac{1}{3}(f(0)\plus{}f(1/3)\plus{}f(2/3))\equal{}\frac{1}{3}(f(0)\plus{}2)$ gives $ f(0)\equal{}1$


Consider a $ 2^{1999}$ by $ 2^{1999}$ square, with a single 1 x 1 square removed. Show that no matter where the small square was removed, it is possible to tile this "giant square minus tiny square" with ells (An "ell" is made up of 3 1x1 tiles to form a shape in that resembles letter 'L').

Have fun & good luck.

The ell can tessellate. 4 ells can tessellate into an ell that is 4 times bigger. In turn, these ells can tessellate into ells that are 16 times bigger than the original. In fact, these ells can tessellate into any form that is a power of 2.

Take the 2^1999 by 2^1999 square, and put an ell in there of that magnitude, with the empty quadrant being the one with the removed square. Continue doing this until you cover everything besides the missing square.

Nice try. Your solution does not show anything, and I'm not sure how you arrived at your conclusion but your solution is not a "solution" but rather some type of conceptual blurb.

Induction on squares of the size $ 2^{n}* 2^{n}$
Assume it can be done for squares of the size $ 2^{k}*2^{k}$, with the addition that no matter where the missing hole is it can be tiled.

Now look at a square of size $ 2^{k\plus{}1}* 2^{k\plus{}1}$

Break it up into 4 $ 2^{k}*2^{k}$ squares. Now the one that has the missing square square in can be tiled by the induction hyphotehes. As for the other 3, tile them so that 3 squares in the middle (where the 4 $ 2^{k}*2^{k}$ squares meet) are missing which can again be done by the induction hyphotehes. Now we can cover those 3 squares with an extra ell piece, so now the whole thing is covered apart from the one missing square at the start. Induction complete.

Or course we still need to check the base case, a 2*2 square, which can be done.

You should also induct to show that the ells can create bigger ells, or how do we know that those three big squares can be covered?

This is straightforward with the following configuration:
[asy]size(50); draw((0,0)--(2,0)--(2,1)--(1,1)--(1,2)--(0,2)--cycle); draw((2,0)--(4,0)--(4,2)--(3,2)--(3,1)--(2,1)--cycle); draw((1,1)--(3,1)--(3,2)--(2,2)--(2,3)--(1,3)--cycle); draw((0,2)--(0,4)--(2,4)--(2,3)--(1,3)--(1,2)--cycle);[/asy]

[quote="Hamster1800"]You should also induct to show that the ells can create bigger ells, or how do we know that those three big squares can be covered?[/quote]
His inductive hypothesis is strong enough to do this.
[quote="seamusoboyle"]with the addition that no matter where the missing hole is it can be tiled.[/quote]

[quote="seamusoboyle"]tile them so that 3 squares in the middle (where the 4 $ 2^{k}*2^{k}$ squares meet) are missing which can again be done by the induction hyphotehes.[/quote]

Ah I missed that part. Your solution is complete :)


inequalities
function
logarithms
algebra proposed
algebra

For $a,b \in [1;\infty)$ show that \[ab\leq e^{a-1}+b\ln b\]

If $a=1$ then trivial else denote $f(b)=ab-b\ln{b}$ for $b\in[1,\infty)$, then $f'(b)=a-\ln{b}-1=0\Longleftrightarrow b=e^{a-1}\in (1,\infty)$ and $f(e^{a-1})$ is maximum of $f$. But $f(e^{a-1})=ae^{a-1}-(a-1)e^{a-1}=e^{a-1}$. Done!


Does anybody know what the 2000 USAMO cutoff score was? wikipedia says 164, but I'm highly doubtful.

[quote]Well, I looked on http://www.unl.edu/amc/e-exams/e8-usamo/archiveusamo.html and dug up the good stuff:
For 2000: 9th 164 (w/ 7 AIME) 10th 174 (w/ 8 AIME), 11th 204, 12th 212
[/quote]

I don't know how do this, but the quote is from mathfanatic.

What do you mean you don't know how to do this?

By the way, thanks for finding it. Where was that from?

Here's a list of the actual indexes [url]http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2000-ua/00usamoqualindex.shtml#n[/url]

I don't know how to say who the quote is by (ive seen people do it before).

This is from an old post: http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1081029061&t=6687

Tip: you can find a lot of things by just using the search button  :roll:

The syntax is (quote="name") instead of (quote).

That's interesting stuff. I would be interested in 1997 and 1998 by index myself, but they don't have those on the official site.


If $ 3x \equal{} 8y$, $ 12y \equal{} 7z$ and $ kz \equal{} 72x$, then what is the value of $ k$?

$ 3x\equal{}8y \implies 9x\equal{}24y$, and $ 12y\equal{}7z \implies 24y\equal{}14z$.  Using the transitive property, we get $ 9x\equal{}14z \implies 72x\equal{}\boxed{112}z$.


I am in need of some help with these problems, such as the one that follows:

1/
In one year there were 84 employees in a company. A dinner was held and the total number of people present was 90, 35 of whom were employees that had brought a partner.

The number of employees that didn’t attend the dance was?

2/
In one year the number of people at the dinner was 112. The number of employees who had brought a partner was 3 times the number of those who attended but did not bring a partner.

The number of employees of the company who attended the dinner but didn’t bring a partner was?

Am I supposed to use algebra to solve these or what is the best way?

and where can I find more of these to practice on?

Thank you!

[hide="My guess on the first problem"]$84$ employees
$90$ camed to the dinner
$35$ camed with a partner

$35 * 2 = 70$
$90-70 = 20$
$20$ employees camed without a partner

It was totally $20+35 = 55$ employees
And there was $84-55 = \boxed{29}$ employees who sat home and ate pizza instead[/hide]

I'm not sure if it's the correct answer.

For number two, you need algebra.

[hide="Hint"]Start with: "Let x represent the number of employees who attended without a partner" and see if you can represent the total number of people at the party in terms of x[/hide]

[hide]2) you have to figure that for the partners, they count as 2 people. We have 3 "2 peoples" for every loner. That means you have 7x=112, solve for x by dividing by 7 and get 16[/hide]

[quote="learningfish"]One year the number of employees of the company who attended the dinner with a partner was the same as the number who attended without a partner but more than the number who did not attend.

The total number of people who attended the dinner was

A/ Less than the number of employees of the company
B/ The same as the number of employees of the company
C/ Greater than the number of employees of the company
D/ any of A B or C depending on the number of employees of the company.

Is the answer C and is algebra required?[/quote]
The answer is C.  I am not sure if Algebra is required, but it certainly helps
[hide]  Say that there are x employees who came to the party with a partner.  Thus, x employeese came to the party without a partner.  Now, say that y employees did not come to the party.  From the given information, we know that x>y.

We also know that if x employees came with a partner then there are x pairs of 2 who came.  Thus, 2x people came from the employees with a partner.  However, we forgot about the employees who came, but without a partner.  This is simply x.  Thus, our total is 2x+x=3x.

Now, since we know that x>y, we know that x+2x>y+2x, so 3x>2x+y.  Notice that 3x is the number of people who came and 2x+y is the number of employees.  Thus, the number of people who came is greater than the number of employees. [/hide]

Thank you.

Do you know where I can find more of these word problems, if that's what they're called


function inequalities trigonometry calculus limit ratio

Find a function $f(x)$ such that $f(x)$ is discontinuous everywhere except for $x = a \in A$, where $A$ is (a) $\{0\}$, (b) $\{a_{1}, a_{2}, \ldots, a_{n}\}$ for reals $a_{i}$, (c) $\{k \pi | k \in \mathbb{Z}\}$.

if you define a piecewise function, this is kind of pointless

otherwise if f(x) is a composition of elementary functions, it is more interesting, assuming that we must have real outputs, the trivial inequality gives the following:

a) $\sqrt{-x^{2}}$

b) $\sqrt{-\prod(x-a_{n})^{2}}$

c) $\sqrt{-\sin^{2}x}$

This would probably be more interesting if we added the condition $f(x)$ is defined on the reals.

[quote="Farenhajt"][quote="Altheman"]a) $\sqrt{-x^{2}}$

b) $\sqrt{-\prod(x-a_{n})^{2}}$

c) $\sqrt{-\sin^{2}x}$[/quote]
Do we consider "undefined at $x$" and "discontinuous at $x$" to be interchangeable qualifications?[/quote]

my calculus book says that $f(x)$ is continuous at $a$ iff $\lim_{h\rightarrow 0}f(a+h)=f(a)$
that definition implies that $f(a)$ is defined, but it is not, so the function is not continuous...

[quote="Farenhajt"]Sure, of course, but creating a function which is NOT DEFINED everywhere except in zero seems a bit... I don't know, "ugly"? or "too easy"... though by the definition it's legitimate.

I do like aviateurpilot's examples better, if only for aesthetical reasons.[/quote]

lol ok...i suppose the problem boils down to finding a function that is discontinuous over reals, and then you can use piecewise to finish,

another function i propose to be discontinuous,

$A\subset \mathbb{R}$ the ratio of number of 1's to the number of 0's in the binary representation is defined

$f(x)=$
$0$ for $x\in A$
$1$ for $x\in\mathbb{R}-A$

but your example
f is continuous everywhere except for $x=a\in A$


I am attracting the bronze medal with my mind  :D

Unfortunately I lost my camera during that Balkans and so I don't have photos taken by me ...


modular arithmetic
number theory unsolved
number theory

Find all the values of $ p$, $ q$, $ r$ and $ n$ so that $ p^n\plus{}q^2\equal{}r^2$. ($ p$, $ q$ and $ r$ are positive primes and $ n\in\mathbb{Z}^\plus{}$.)

Note that one of $ p,q,r$ must be even, or more precisely, exactly one of $ p,q$ must equal $ 2$.

We have $ p^n \equal{} (r \plus{} q)(r \minus{} q)$. Hence $ r \plus{} q \equal{} p^a,r \minus{} q \equal{} p^b$. Thus
\[ r \equal{} \frac {p^b(p^{a \minus{} b} \plus{} 1)}{2},q \equal{} \frac {p^b(p^{a \minus{} b} \minus{} 1)}{2}\,\,(1)
\]
First let $ b > 0$.

Case 1: $ p \equal{} 2$. Then we have $ r \equal{} 2^{b \minus{} 1}(2^{a \minus{} b} \plus{} 1)$ and $ q \equal{} 2^{b \minus{} 1}(2^{a \minus{} b} \minus{} 1)$. As $ q,r$ are odd primes, we get $ b \equal{} 1$ which implies $ r \equal{} 2^{a \minus{} 1} \plus{} 1,q \equal{} 2^{a \minus{} 1} \minus{} 1\implies a > 1$. Since $ q \equal{} 2^{a \minus{} 1} \minus{} 1$ is prime, $ a \minus{} 1$ is prime. Thus $ a\ge 3$. If $ a \minus{} 1$ is odd, we have $ r\equiv 0\pmod 3\implies r \equal{} 3,a \equal{} 2\not\ge 3$, contradiction. Thus $ a \minus{} 1$ is even $ \implies a \equal{} 3, q \equal{} 3,r \equal{} 5,p \equal{} 2,n \equal{} a \plus{} b \equal{} 4$.

Case 2: $ p$ is odd. Then we have that $ 2\mid p^{a \minus{} b}\pm 1$. Thus from $ (1)$ $ p^b\mid q,p^b\mid r\implies q \equal{} p \equal{} r$, which doesn't yield any solutions.

Now assume that $ b \equal{} 0$. Then from $ (1)$, $ r \equal{} \frac{p^a \plus{} 1}{2},q \equal{} \frac{p^a \minus{} 1}{2}\implies r\minus{}q\equal{}1$. Hence $ q\equal{}2,r\equal{}3,p\equal{}5,a\equal{}1$.

Thus the solutions are $ (p,q,r,n) \equal{} (2,3,5,4),(5,2,3,1)$.

EDIT: Fixed mistake.

I have a very similar solution but i find that $ p \equal{} 5,n \equal{} 1,q \equal{} 2,r \equal{} 3$ is also solution where $ b\equal{}0$ in your proof .

Yes I made a stupid mistake :wallbash:. I fixed it. Thank you tchebytchev. :)


geometry
trigonometry
circumcircle
analytic geometry
incenter
geometry proposed

[quote="stergiu"]The incircle of a triangle $ ABC$ with $ AB\plus{}AC \equal{} 2BC$ touches side $ AB$ at $ D$ and side $ AC$ at $ E$.Prove that segment $ DE$ bisects the median $ AM$ of triangle $ ABC$.

[b] NOTE[/b]

 Does anyone have a collection with the most but the best properties of this kind of triangles ($ AB\plus{}AC \equal{} 2BC$)? I think , I have seen some other problems on this triangle , but unfortunately I did not collected them. Tnks.[/quote]
Let $ P,Q$ be the midpoint of $ AB,AC$. We can prove $ A,P,I,Q$ is cyclic.Therefore, projection of $ I$ on $ AP,PQ,AQ$ lie on a line. Since $ \angle{IAP}\equal{}\angle{IAQ}$,projection of I on PQ is midpoint of $ PQ$. Therefore, midpoint of $ PQ$ lie on $ DE$ but midpoint of $ PQ$ is midpoint of $ AM$. Threrefore, $ DE$ pass midpoint of $ AM$.

[quote="Leonhard Euler"][quote="stergiu"]The incircle of a triangle $ ABC$ with $ AB\plus{}AC \equal{} 2BC$ touches side $ AB$ at $ D$ and side $ AC$ at $ E$.Prove that segment $ DE$ bisects the median $ AM$ of triangle $ ABC$.

[b] NOTE[/b]

 Does anyone have a collection with the most but the best properties of this kind of triangles ($ AB\plus{}AC \equal{} 2BC$)? I think , I have seen some other problems on this triangle , but unfortunately I did not collected them. Tnks.[/quote]
Let $ P,Q$ be the midpoint of $ AB,AC$. We can prove $ A,P,I,Q$ is cyclic.Therefore, projection of $ I$ on $ AP,PQ,AQ$ lie on a line [color=red][b]( The simpson line ). [/b][/color]Since $ \angle{IAP}\equal{}\angle{IAQ}$,projection of I on PQ is midpoint of $ PQ$. Therefore, midpoint of $ PQ$ lie on $ DE$ but midpoint of $ PQ$ is midpoint of $ AM$. Threrefore, $ DE$ pass midpoint of $ AM$.[/quote]

 Mmm! This solution seems nice , but :
  a) How do you prove that $ A,P,I,Q$ is cyclic and where do you use the condition $ AB\plus{}AC \equal{} 2BC$?
  b) Why is $ K$ , the projection of $ I$ on $ PQ$ the midpoint of $ PQ$ ?( OK ! Now I see this :$ IKPD , IKEQ$ are incribed, so angles $ IPK, IDK, IEK,IQK$ are equal and so $ \triangle IPE$ is isosceles.).
 Well , only the first question arises !
 If you explain this , then this solution is a little better and shorter that the solution I have in my notes. I will send it , if asked !
[u] Babis[/u]

Let the incircle touch $ BC$ at $ F$. Quite simple calculations solve the problem using the well-known property that $ AM$, $ FI$ and $ DE$ are concurrent (this holds in any triangle, not only when $ b\plus{}c\equal{}2a$):
Let $ AA'$ be the altitude from $ A$, $ A'\in BC$. The problem then reduces (from the above-stated property) to proving that $ MA_{1}\equal{}2MF$.
Let $ s$ denote the semiperimeter. From $ AF\equal{}s\minus{}b$, we get $ MF\equal{}\frac{a}{2}\minus{}(s\minus{}b)\equal{}\frac{b\minus{}c}{2}$. And $ MA_{1}\equal{}\frac{a}{2}\minus{}c\cos B$.
Note that $ \cos B\equal{}\frac{a^{2}\plus{}c^{2}\minus{}b^{2}}{2ac}\equal{}\frac{\frac{1}4(c\plus{}b)^{2}\plus{}(c\plus{}b)(c\minus{}b)}{(c\plus{}b)c}$. So $ c\cos B\equal{}\frac{b\plus{}c}{4}\plus{}c\minus{}b$. Then $ MA_{1}\equal{}2MF\Leftrightarrow\frac{a}{2}\minus{}\left(\frac{b\plus{}c}{4}\plus{}c\minus{}b\right)\equal{}b\minus{}c$, which is true since $ \frac{a}{2}\equal{}\frac{b\plus{}c}{4}$.

[color=darkred][b]Lemma I[/b] ([u]an well-known remarkable relation[/u]). $ \{\begin{array}{c}\boxed{\triangle ABC}\\ \\ X\in (AC)\ ,\ Y\in (AB)\\ \\ R\in (XY)\ ,\ Z\in AR\cap BC\end{array}\|$ $ \implies$ ${ \frac{YB}{YA}\cdot ZC+\frac{XC}{XA}\cdot ZB =\frac{RZ}{RA}\cdot BC}$.

[b]Lemma II.[/b] Let $ C(I,r)$ be the incircle of $ \triangle ABC$ and denote the its tangent points $ E\in (AC)$, $ F\in (AB)$ and the midpoint $ M$ of the side $ [BC]$.
Then the line $ AM$, $ ID$, $ EF$ are concurrently in a point $ R$ for which $ \frac{RF}{RE}=\frac{b}{c}$ and $ \frac{RA}{RM}=\frac{b+c-a}{a}$.[/color]
[color=darkblue][b]Proof.[/b]

$ 1\blacktriangleright$ Denote $ R\in EF\cap AM$ and apply the [b]upper lemma [/b]for $ \{\begin{array}{c}\boxed{\triangle ABC}\\ \\ E\in (AC)\ ,\ F\in (AB)\\ \\ R\in (EF)\ ,\ M\in AR\cap BC\end{array}\|$ $ \implies$ ${ \frac{FB}{FA}\cdot MC+\frac{EC}{EA}\cdot MB =\frac{RM}{RA}\cdot BC}$,

i.e. $ \frac{p-b}{p-a}\cdot\frac{a}{2}+\frac{p-c}{p-a}\cdot\frac{a}{2}=\frac{RM}{RA}\cdot a$ $ \implies$ $ \frac{RA}{RM}=\frac{b+c-a}{a}$, more exactely $ \frac{RA}{b+c-a}=\frac{RM}{a}=\frac{AM}{b+c}$.

$ 2\blacktriangleright$ Denote the point $ D$ where the incircle $ \mathcal C(I,r)$ touches the side $ [BC]$, the projection $ S$ of the vertex $ A$ on the sideline $ BC$

and the intersection $ R'\in ID\cap AM$. The relations $ DM =\frac{1}{2}\cdot |b-c|$, $ DS =\frac{p-a}{a}\cdot |b-c|$ are well-known (you can prove it easily).

Therefore, $ \frac{R'A}{R'M}=\frac{DS}{DM}=\frac{b+c-a}{a}=\frac{RA}{RM}$ $ \implies$ $ \frac{R'A}{R'M}=\frac{RA}{RM}$ $ \implies$ $ R'\equiv R\in EF\cap ID\cap AM$ and $ \boxed{\frac{RA}{RM}=\frac{b+c-a}{a}}$.

From $ \{\begin{array}{c}AE = AF\implies\frac{RF}{RE}=\frac{\sin\widehat{BAM}}{\sin\widehat{CAM}}\\ \\ MB = MC\implies\frac{c}{b}\cdot\frac{\sin\widehat{BAM}}{\sin\widehat{CAM}}= 1\end{array}\|$ $ \implies$ $ \boxed{\ \frac{RF}{RE}=\frac{b}{c}\ }$, more precisely $ \frac{RF}{b}=\frac{RE}{c}=\frac{EF}{b+c}$.

[b]Remark I.[/b] The relation $ IA^{2}=\frac{bc(p-a)}{p}$ is well-known. Therefore, $ IO\perp IA$ $ \Longleftrightarrow$ the power of $ I$ w.r.t. the circumcircle $ w$ is $ p_{w}(I) = IA^{2}$ $ \Longleftrightarrow$

$ 2Rr =\frac{bc(p-a)}{p}$ $ \Longleftrightarrow$ $ 4Rpr = 2bc(p-a)$ $ \Longleftrightarrow$ $ abc = 2bc(p-a)$ $ \Longleftrightarrow$ $ b+c = 2a$. Thus, $ \boxed{\ IO\perp IA\Longleftrightarrow b+c = 2a\ }$.

[b]Remark II.[/b] $ b+c = 2a$ $ \Longleftrightarrow$ $ b+c-a = a$ $ \Longleftrightarrow$ $ RA = RM$, i.e. $ \boxed{\ b+c = 2a\Longleftrightarrow\mathrm{the\ line\ EF\ bisects\ the\ median\ [AM]}\ }\ \ (*)$.

[b]Remark III.[/b] Denote $ \{\begin{array}{c}N\in (AC)\ ,\ NA = NC\\ \ P\in (AB)\ ,\ PA = PB\end{array}$ and the second intersection $ A'$ of the circumcircle $ w$ with the line $ AI$.
Then the quad. $ APIN$ is cyclically $ \Longleftrightarrow$ $ m(\widehat{API}) = B+\frac{A}{2}$ $ \Longleftrightarrow$ $ \widehat{API}\equiv\widehat{ABA'}$ $ \Longleftrightarrow$ $ PI\parallel BA'$ $ \Longleftrightarrow$

$ IA = IA'$ $ \Longleftrightarrow$ $ IO\perp IA$. In conclusion, $ \boxed{\ \mathrm{the\ quad.\ APIN\ is\ cyclically\ }\Longleftrightarrow IO\perp IA\ }$. [/color]

[quote="freemind"]Let the incircle touch $ BC$ at $ F$. Quite simple calculations solve the problem using the well-known property that $ AM$, $ FI$ and $ DE$ are concurrent (this holds in any triangle, not only when $ b\plus{}c \equal{} 2a$):[/quote]

[quote="TomciO"] The fact that quadrilateral $ APIQ$ is cyclic can be proved in the such way:
The condition $ AB\plus{}AC \equal{} 2BC$ is equivalent to $ \angle AIO \equal{} 90^{\circ}$ where $ O$ is the circumcentre of the triangle $ ABC$. Indeed, let $ T$ be the midpoint of the arc $ BC$. We wish to prove that $ I$ is the midpoint of the segment $ AT$. But from the Ptolemy Theorem:
$ AT\cdot BC \equal{} TB\cdot AC\plus{}TC\cdot AB \equal{} TI(AC\plus{}AB) \equal{} 2TI\cdot BC$ (it's well known that $ TB \equal{} TC \equal{} TI$) so indeed $ TI \equal{}\frac{1}{2}AT$.
Now we see that points $ A, P, I, O, Q$ all lie on the circle with the diameter $ AO$.[/quote].

[b] NOTE[/b]

 After the above remark of Freeminf and your nice result , in triangle $ AMT$ segmants $ IK, TM$ are parallel and since $ I$ is the midpoint of $ AT$ , $ K$ will be the midpoint of $ AM$.

     All this is a nice and rich work :lol:  out of the problem. If somebody has any suggestion to my second part in the initial mepsage , please write here.

[quote="Virgil Nicula"][color=darkblue][b]See[/b] upper [b]Lemma I[/b] and [b]Lemma II[/b] ...................................................................................................................

<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<
$ \mathrm{-------------------------------------}$ $ \mathrm{-----------------}$
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>

[b][u]Remark I.[/u][/b] The relation $ IA^{2}=\frac{bc(p-a)}{p}$ is well-known. Therefore, $ IO\perp IA$ $ \Longleftrightarrow$ the power of $ I$ w.r.t. the circumcircle $ w$ is $ p_{w}(I) = IA^{2}$ $ \Longleftrightarrow$

$ 2Rr =\frac{bc(p-a)}{p}$ $ \Longleftrightarrow$ $ 4Rpr = 2bc(p-a)$ $ \Longleftrightarrow$ $ abc = 2bc(p-a)$ $ \Longleftrightarrow$ $ b+c = 2a$. Thus, $ \boxed{\ IO\perp IA\Longleftrightarrow b+c = 2a\ }$.

[b][u]Remark II.[/u][/b] $ b+c = 2a$ $ \Longleftrightarrow$ $ b+c-a = a$ $ \Longleftrightarrow$ $ RA = RM$, i.e. $ \boxed{\ b+c = 2a\Longleftrightarrow\mathrm{line\ EF\ bisects\ median\ [AM]}\ }\ \ (*)$.

[b][u]Remark III.[/u][/b] Denote $ \{\begin{array}{c}N\in (AC)\ ,\ NA = NC\\ \ P\in (AB)\ ,\ PA = PB\end{array}$ and the second intersection $ A'$ of the circumcircle $ w$ with the line $ AI$.
Then the quad. $ APIN$ is cyclically $ \Longleftrightarrow$ $ m(\widehat{API}) = B+\frac{A}{2}$ $ \Longleftrightarrow$ $ \widehat{API}\equiv\widehat{ABA'}$ $ \Longleftrightarrow$ $ PI\parallel BA'$ $ \Longleftrightarrow$

$ IA = IA'$ $ \Longleftrightarrow$ $ IO\perp IA$. In conclusion, $ \boxed{\ \mathrm{\ APIN\ is\ cyclically\ }\Longleftrightarrow IO\perp IA\ }$.

[b][u]Remark IV.[/u][/b] $ \boxed{\ IG\parallel BC\Longleftrightarrow b+c = 2a\ }$, where $ G$ is the centroid of $ \triangle ABC$. Indeed, $ IG\parallel BC$ $ \Longleftrightarrow$ $ 3r = h_{a}$ $ \Longleftrightarrow$

$ \Longleftrightarrow$ $ 3ar = 2S$ $ \Longleftrightarrow$ $ 3ar = 2pr$ $ \Longleftrightarrow$ $ 3a = a+b+c$ $ \Longleftrightarrow$ $ b+c = 2a$ (try show $ IG\perp BC\Longleftrightarrow$ $ b+c = 3a$ or $ b = c$).

[b][u]Remark V.[/u][/b] $ \boxed{\ b+c = 2a\Longleftrightarrow\cos\frac{B-C}{2}=\sqrt{\frac{2r}{R}}\ }$. Generally, $ \cos\frac{B-C}{2}\ge\sqrt{\frac{2r}{R}}$. Indeed, $ \cos\frac{B-C}{2}\ge\sqrt{\frac{2r}{R}}$ $ \Longleftrightarrow$

$ 2\sin\frac{B+C}{2}\cos\frac{B-C}{2}\ge 2\cos\frac{A}{2}\cdot\sqrt{\frac{2r}{R}}$ $ \Longleftrightarrow$ $ \sin B+\sin C\ge\sqrt{\frac{8rp(p-a)}{Rbc}}$ $ \Longleftrightarrow$ $ b+c\ge\sqrt{\frac{32Rrp(p-a)}{bc}}$ $ \Longleftrightarrow$

$ b+c\ge\sqrt{{\frac{8abc(p-a)}{bc}}}$ $ \Longleftrightarrow$ $ (b+c)^{2}\ge 4a(b+c-a)$ $ \Longleftrightarrow$ $ (b+c-a)^{2}\ge 0$, with equality if and only if $ b+c = 2a$.

[b][u]Remark VI.[/u][/b] $ \boxed{b+c = 2a\Longleftrightarrow p^{2}+9r^{2}= 18Rr\Longleftrightarrow\sin\frac{A}{2}=\sqrt{\frac{2r}{R}}\Longleftrightarrow\cos A = 1-\frac{r}{R}\Longleftrightarrow\cos B+\cos C =\frac{2r}{R}}$. Indeed,

$ IO\perp IA$ $ \Longleftrightarrow$ $ AO^{2}= IA^{2}+IO^{2}$ $ \Longleftrightarrow$ $ R^{2}=\frac{bc(p-a)}{p}+R^{2}-2Rr$ $ \Longleftrightarrow$ $ 2Rpr = bc(p-a)$ $ \Longleftrightarrow$ $ abc = 2bc(p-a)$ $ \Longleftrightarrow$ $ b+c = 2a$.

[b][u]Remark VII[/u][/b] (easily). If $ L\in AI\cap BC$, then $ b+c = 2a$ $ \Longleftrightarrow$ $ \frac{LA}{LA'}= 3$ $ \Longleftrightarrow$ $ 2h_{a}+3r = 3r_{a}$ $ \Longleftrightarrow$

$ r_{a}= 3r$ $ \Longleftrightarrow$ $ h_{a}= 3r$ $ \Longleftrightarrow$ $ \boxed{\ h_{a}= r_{a}\ }$ ($ r_{a}$ is $ A$- exinradius and $ h_{a}$ is $ A$- altitude).

[b][u]Remark VIII.[/u][/b] IF the point $ P$ has the barycentrical coordinates $ (x,y,z)$, $ x+y+z = 1$ w.r.t. $ \triangle ABC$, then the scalar product

$ \overrightarrow{IO}\circ\overrightarrow{IP}=-3Rr+\frac{1}{2}\cdot(x\cdot bc+y\cdot ca+z\cdot ab)$. In consequence, $ \boxed{\ IP\perp IO\Longleftrightarrow x\cdot bc+y\cdot ca+z\cdot ab = 6Rr\ }$.

Generally, for any point $ P(x,y,z)$, $ x+y+z = 1$ exists the equivalence $ x\cdot bc+y\cdot ca+z\cdot ab\ge 6Rr\Longleftrightarrow$ $ m(\widehat{OIP})\le90^{\circ}$.

[b]I"ll return here soon ![/b][/color][/quote]

 Well,  :lol: Virgil , this is nice work and organation ! Thank you very much!
[u] Babis[/u]

Sorry for doing this but I was tired to read all the above solutions . Maybe mine is somewhat different maybe not .
Let $ I,\ O,\ ,T,\ Q,\ F$ the incenter, the circumcenter, the midpoint of the arc BC , the point of intersection of DE and AM , and the point of tangency of the incircle with BC respectively .

It is well known that the lines IF,AM,DE are concurrent . Also from the hypothesis we have that $ AD\equal{}MC$ so the triangles $ TMC$ and $ ADI$ are congruent
so $ TM\equal{}ID\equal{}IF$ and $ TC\equal{}AI$ . But $ TC\equal{}TI$  so $ AI\equal{}IT$ so $ I$ is the midpoint of $ AT$ . Finally IQ //TM so the segment IQ connects the midpoints at the triangle ATM so M is the midpoint of AM and we are done . :wink:

[quote="silouan"]Sorry for doing this but I was tired to read all the above solutions . Maybe mine is somewhat different maybe not .
Let $ I,\ O,\ ,T,\ Q,\ F$ the incenter, the circumcenter, the midpoint of the arc BC , the point of intersection of DE and AM , and the point of tangency of the incircle with BC respectively .

It is well known that the lines IF,AM,DE are concurrent . Also from the hypothesis we have that $ AD \equal{} MC$ so the triangles $ TMC$ and $ ADI$ are congruent
so $ TM \equal{} ID \equal{} IF$ and $ TC \equal{} AI$ . But $ TC \equal{} TI$  so $ AI \equal{} IT$ so $ I$ is the midpoint of $ AT$ . Finally IQ //TM so the segment IQ connects the midpoints at the triangle ATM so M is the midpoint of AM and we are done . :wink:[/quote]

[b] Yes Silouan !It is a little different and ...more elementary[/b]

 You  do not use to prove Ptolemy to prove that $ AI\equal{}IT$ , but the fact that $ AD \equal{}\tau\minus{}\alpha \equal{} ... \equal{} CM$.

I gave this nice problem to my classmate [b][i]Kapshuk Alexander[/i][/b] (ML ID [b][i]plectrus[/i][/b]) and after discussing some related properties, he found the following simple beautiful synthetic proof:

Again, $ IF$, $ DE$ and $ AM$ are concurrent at $ T$. Draw through $ T$ a parallel to $ BC$. Let it intersect $ AB$ and $ AC$ at $ K,L$ respectively. It is well-known that $ KD\equal{}EL$ (this holds for any triangle, [i]not[/i] only when $ b\plus{}c\equal{}2a$). To see this, note that $ KTID$ and $ ITEL$ are cyclic (angles $ \angle{ITK},\angle{IDK},\angle{ITC},\angle{IEC}$ are all right). Then $ \angle{DIK}\equal{}\angle{DTK}\equal{}\angle{ETC}\equal{}\angle{EIC}\equal{}u$, hence $ KD\equal{}EC\equal{}r\sin u$.

Returning to the problem, we then have: $ AK\plus{}AL\equal{}AD\minus{}KD\plus{}AE\plus{}EL\equal{}2AD\equal{}a$, because $ AD\equal{}AE\equal{}s\minus{}a\equal{}\frac{a}{2}$, where we've used that $ b\plus{}c\equal{}2a$. But the midpoints $ M,N$ of $ AB$ and $ AC$ respectively; also satisfy $ MN\|BC$ and $ AM\plus{}AN\equal{}\frac{b\plus{}c}{2}\equal{}a$, thus $ M\equal{}K$ and $ N\equal{}L$, so $ T\in KL$ lies on the middle line of $ ABC$, hence $ AT\equal{}TM$.


function
calculus
derivative
linear algebra

How would I prove that preimages of points are isolated w/o using the inverse function thm.

What are the conditions on your function? You haven't asked a sufficiently specific question.

For a function with an invertible derivative, the inverse function theorem is the best way to do it, and the result that $f$ is injective on small enough balls is an intermediate step in the proof.


Find all values of x that satisfy the equation:

$x=\sqrt{1-\frac{1}{x}}+\sqrt{x-\frac{1}{x}}$

There's actually a mad easy way to do this....But, your way works also  :lol:

[quote="Altheman"]conjugate it...[/quote]
I don't know if that would really help.  It simplifies the right side, but you're still left with basically the same mess on the right.

$x=\frac{x-1}{\sqrt{x-1/x}-\sqrt{1-1/x}}$

$\sqrt{x-1/x}-\sqrt{1-1/x}=1-\frac{1}{x}$

blah...nevermind...you can solve for the radicals, but that still gives the ugly polynomial

$2\sqrt{x-1/x}=1+x-\frac{1}{x}$


Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that:
\[ \frac {2a \plus{} b}{a \plus{} c \plus{} 2b} \plus{} \frac {2b \plus{} c}{b \plus{} a \plus{} 2c} \plus{} \frac {2c \plus{} a}{c \plus{} b \plus{} 2a}\geq\frac {9}{4}
\]
There is a nice proof.

Johan Gunardi, my proof is similar:
$ \sum_{cyc}\frac {2a \plus{} b}{a \plus{} c \plus{} 2b} \equal{}\sum_{cyc} \frac {(2a\plus{}b)^2}{(2a\plus{}b)(a\plus{}c\plus{}2b)}\geq\frac{9(a\plus{}b\plus{}c)^2}{\sum(2a\plus{}b)(a\plus{}c\plus{}2b)}\equal{}\frac {9}{4}.$


geometry
trigonometry
function
incenter
trig identities
Law of Sines

In a triangle $ABC$, let $BD$ and $CE$($D$ on $AC$ and $E$ on $AB$) be the angle bisectors. If $\measuredangle BDE = 24^\circ$ and $\measuredangle CED = 18^\circ$. Find the measure of $\measuredangle ABC$?

Say <DBC = b and <BAC=a

Then, draw a line connecting E to D. The intersecting lines of the bisectors and ED form a triangle EDG. Below that is another triangle GBC.

<EGD=180-24-18=138=<BGC due to opposite angles.

Now, <ECB = c= 180-138-b = 42-b

Look at triangle ABC, 180-a+2b=2c=(42-b)2

180-a+2b=(42-b)2

Solving for a, we get a=96, so <BAC=96.

Now, we can set up a system of equations.

Say <ADE=d and <AED=e

d+24+(180-21-.5b)=180
2d+366-b=360
2d-b=-6

e+(180-42-b)+18=180
2e-2b=48

d+e+96=180
d+e=84

Combine 2e-2b=48 and 2d-b=-6 to get:

2e+2d-3b=42
2(84)-3b=42
b=42

So our answer is 2b=84

<ABC=84*.

Voila!

That took a long time to type guys. I hope you appreciate.

[quote="i_like_pie"][quote="Virgil Nicula"][quote="neelnal"][color=red][size=134].... <BAC=96. .... <ABC=84*. .... Voila! I hope you appreciate.[/size][/color][/quote]
[color=darkblue][size=117]$\implies m(\angle ACB)=0^{\circ}\ .$ I appreciate very much your effort.[/size][/color][/quote]
How can $m\angle ACB=0^{\circ}?$[/quote][b]I_like_pie[/b], at least you and I see that in the [b][u]Neelnal's space and geometry[/u][/b] it is perhaps !

O darn. Forgot to recheck my work. Well, then can we get a solution?

Here is the solution:

I made a silly mistake and it would be nice not to poke fun at me.

d+e=84

180-b-42+18+e=180
2b+84=36+e

180-21-.5b-42+24+d=180
b+63=24+2d

Combine the second and third to get

60+2(e+d)=b+63
e+d=84, so
solving, b=27

<ABC=54*. Voila!

Sorry,Virgil Nicula I made a silly mistake. You don't have to be so arrogant until you solve the problem yourself.

[quote="M4RI0"][quote="neelnal"]
180-b-42+18+e=180
2b+84=36+e
[/quote]

 :?:[/quote]

<BAC=96.

Now, we can set up a system of equations.

Say <ADE=d and <AED=e  and <DBC=b

From triangle ADE,
d+e=84

From point of intersection D on line AC,

180-21-.5b-42+24+d=180

From point of intersection D on line AC,

180-b-42+18+e=180

All I did was put e and d's on one side and b's on the other and add the equations.

[quote="neelnal"][color=darkred]... 60+2(e+d)=b+63 , e+d=84, so solving, b=27 , <ABC=54*.[/color] Voila!
[color=darkblue]Sorry,Virgil Nicula ! I made a silly mistake. [b]You don't have to be so arrogant until you solve the problem yourself.[/b][/color][/quote]
[color=darkblue]Sorry for that words ! Don't know you when is in jokingly ? See above. Is it all right ? I tried to solve and I failed to do this problem.

I obtained just this : if $m(\widehat{DBC})=x$, then exists the relation $2\sin 24^{\circ}\cos x=\cos (42^{\circ}-x)\ ,$ i.e. $\tan x=\frac{2\sin 24^{\circ}-\sin 48^{\circ}}{\cos 48^{\circ}}\ .$

Again, sorry for that words ![/color]

Ok sorry I accused you.

Anyways, the answer is <ABC=54. I went back and triple checked my work. It is a 96-54-30 triangle.

[quote="neelnal"][color=darkred]... Anyways, the answer is <ABC=54. I went back and triple checked my work. [b][u]It is a 96-54-30 triangle.[/u][/b][/color][/quote]

[color=darkblue]If the my relation $\boxed{\ 2\sin 24^{\circ}\cos x=\cos (42^{\circ}-x)\ }$ is truly, then $B\ne 54^{\circ}\ !$

Indeed, $x: =27^{\circ}\Longrightarrow \cos 15^{\circ}=2\sin 24^{\circ}\cos 27^{\circ}<2\sin 30^{\circ}\cos 27^{\circ}=\cos 27^{\circ}$ $\implies$ $\cos 15^{\circ}<\cos 27^{\circ}\ ,$ what is falsely ![/color]

Virgil, calculator says that $\arctan\frac{2\sin 24^\circ-\sin 48^\circ}{\cos 48^\circ}=6^\circ$

[color=darkblue][b][u]Thanks, Farenhajt ![/u][/b] This means that the my relation is correctly (at a pinch, I can post its proof here) and $B=12^{\circ}$.
It is interestingly. How we solve this problem (now we know and answer) without calculator ?
Or how from the my relation we'll find $x=6^{\circ}\ ?$ Maybe says [b][u]M4RIO[/u][/b]. Please, M4RIO, do you know a proof ?[/color]

[color=red]Ready ![/color][color=darkblue] I finished with this problem. Let $f: (0,42^{\circ})\rightarrow R$, where $f(x)=2\sin 24^{\circ}\cos x-\cos (42^{\circ}-x)\ .$ Observe that the this function is strict decreasing and $f(6^{\circ})=2\sin 24^{\circ}\cos 6^{\circ}-\cos 36^{\circ}=$ $\sin 30^{\circ}+\sin 18^{\circ}-\cos 36^{\circ}=\frac{1}{2}+\frac{-1+\sqrt 5}{4}-\frac{1+\sqrt 5}{4}=0\ .$ In conclusion, $f(x)=0\implies\ x=6^{\circ}\ ,$ i.e. $\boxed{\ B=12\ }\ .$ Again, [b][u]thank you very much, Farenhajt ![/u][/b]

[b]P.S.[/b] The my final answer for [b][u]Neelnal[/u][/b] : it is $96^{\circ}-12^{\circ}-72^{\circ}$ triangle. [/color]

Let $\angle ABD=x$. (The incenter is $I$)

Then we obtain through the law of sines on $\triangle DIE$, $\triangle EIB$, $\triangle DIC$, and $\triangle CBI$:

$\frac{\sin18}{\sin24}=\frac{DI}{EI}= \frac{ \frac{CI*\sin42-x}{\sin 96+x }}{ \frac{BI*\sin x}{\sin 138-x}}= \frac{\sin138-x}{\sin96+x}$

$\cos 42-x=2\sin 24\cos x$

$\tan x=\frac{2\sin 24(1-\cos 24)}{\cos 48}$

i cannot solve this without a calculator...

In general, for a quadrilateral $BCDE$ with $CE\cap BD=I$, we can find $\frac{EI\cdot CI}{DI\cdot BI}$ in two ways using the law of sines.


inequalities
linear algebra
matrix
inequalities proposed

We have seen and applied too much times the following interesting inequality of VasC:
If $ a,b,c$ are real numbers, then
\[ (a^2 \plus{} b^2 \plus{} c^2)^2 \ge 3(a^3b \plus{} b^3c \plus{} c^3a).
\]
This inequality is so interesting that there are many solvers try to find their own proof for it. In here, I will present an identity which proves this nice inequality :)
Let $ A \equal{} \sum a^3b,B \equal{} \sum ab^3,C \equal{} \sum a^2b^2, D \equal{} abc\sum a$, then we have
\[ 4(C \minus{} D)[(a^2 \plus{} b^2 \plus{} c^2)^2 \minus{} 3(a^3b \plus{} b^3c \plus{} c^3a)] \equal{} (A \plus{} B \plus{} D \minus{} 3C)^2 \plus{}3 (B \plus{} C \minus{} A \minus{} D)^2 \ge 0
\]
Hope you all will enjoy it! :)

P/s: I established this identity according to $ pqr$ technique. ;) :)

This identity is very nice, Can_hang ! :lol:

Sorry all, I forgot the number $ 3$ at the second term, I've edited it. Very sorry! :(

[quote="can_hang2007"]
Hope you all will enjoy it! :)
[/quote]

I enjoyed it!  :)  i think we can sharpen VasC's inequality with using this identity. I will think about this.  :|

This kind of identity can be obtained by matrix i think.
Try google about the 17th problem of Hilbert.


The larger of two consecutive odd integers is three times the smaller. What is their sum?

$ \textbf{(A)}\ 4 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 20$

[hide="Solution"]The numbers must be 1 and 3, as they are the only consecutive odd integers which are a factor of three apart. Their sum is $\boxed{4}$, or choice (A) [/hide]

[hide]$x+2=3x$

$x=1$

$1+3=4\Rightarrow\boxed{\text{A}}$[/hide]

Pretty simple

1 3 5 7 .........
we quickly see that 3 x 1 = 3
so 3+1=4
Answer [b]4[/b]


Bob and a def other kids were playing cards when Bob, posed a question. "In an ordinary deck of playing cards, 2 of the Jacks have 2 eyes and 2 of the jacks have 1 eye. How many eyes are on the 4 Jack cards?

Spoiler Ahead -- "Hide" Not in Effect
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As I see it, your given statement is wrong...

         Because there are two Jack faces appearing on each of the four Jack cards, that means that half of them, or four Jacks apiece, would have two eyes, and the other half, or four Jacks apiece, would have one eye.

So your statement "...2 of the Jacks have 2 eyes and 2 of the jacks have 1 eye." is not accurate.

From my breakdown, 4*2 + 4*1 = 8 + 4 = 12.  Then there would be a total of twelve eyes among the Jacks on the four Jack cards.


Suppose $a_1,a_2,...a_n$ are positive real numbers. Let $e_{j,k}=n-1$ if $j=k$ and $e_{j,k}=n-2$ otherwise. Let $d_{j,k}=0$ if $j=k$ and $d_{j,k}=1$ otherwise. Prove that\[\sum_{j=1}^n\prod_{k=1}^n e_{j,k}a_k^2\geq \prod_{j=1}^n\left(\sum_{k=1}^n d_{j,k}a_k\right)^2\]

Let r be the product of all $a_i$, s the sum.  For convenience, denote $K = (n-2)^{n-1}(n-1)n$.

$LHS = Kr^2$

$RHS = \prod (s - a_j)^2$

Remains to show $K \geq \prod (\frac{s}{a_i} - 1)^2$

It is not necessarily true:  take $a_1$ arbitrarily large, then cannot be true.  But we have only rearranged, and not invoked any ineq's yet.  So it is false.


Type:

[code];$code here$[/code]

Or if you want the "teacher" type with the text then the formulas in the middle:

[code]; Text here $$Code here$$ More text (which is optional)[/code]


function
calculus
integration
complex analysis
complex analysis unsolved

Show that if $ f: \bf C \to C$ is a continuous function such that $ f$ is analytic off $ [\minus{}1,1]$, then f is an entire function (analytic on $ \bf C$).

It highly suggests to use Morera's theorem, but I don't see how :(

You want to show that the integral of $ f$ over any triangle is zero. If a triangle does not cross the real axis, this is clear. If one of its sides lies on the real axis, you can translate the triangle by $ \pm \epsilon i$, find that $ f$ integrates to zero over the translated triangle, and let $ \epsilon\to 0$. If the real axis cuts the triangle into two parts, deal with each part separately: their common boundary is traversed twice in opposite directions and therefore its contibution cancels out.

Is there a different idea? I don't like morera's theorem

Let $ C\equal{}\{z\colon |z|\equal{}2\}$ and $ D\equal{}\{z\colon |z|<2\}$. For $ z\in D$ define $ g(z)\equal{}\frac{1}{2\pi i}\int_C \frac{f(\zeta)}{\zeta\minus{}z}d\zeta$. Since $ g$ is holomorphic in $ D$, it suffices to prove that $ f(z)\equal{}g(z)$ for $ z\in D\setminus [\minus{}1,1]$ (the equality extends to $ [\minus{}1,1]$ by continuity). By Cauchy's integral formula we have $ f(z)\equal{}g(z)\plus{}\frac{1}{2\pi i}\int_{C_{\epsilon}} \frac{f(\zeta)}{\zeta\minus{}z}d\zeta$, where $ C_{\epsilon}$ is the boundary of $ \epsilon$-neighborhood of $ [\minus{}1,1]$ oriented clockwise, and $ \epsilon$ is sufficiently small. Letting $ \epsilon\to 0$ and using the continuity of $ f$, we obtain that $ f(z)\equal{}g(z)$.

Thanks mlok. I realize now that the argument with morera is quite beautiful and I do like it :) So this is also true if we replace $ [\minus{}1,1]$ with $ \mathbb{R}$ right? Maybe I am overlooking something (I am super-tired, I've been up for 24 hours)

Yes, the first proof works for $ \mathbb R$. The second doesn't.


Solve the equation
$4x^2-4x-10= \sqrt{8x^2-6x-10}$

It should be $8x^2-6x-10\geq 0$ so  $x\in R-\left(\frac{3-\sqrt{89}}{8},\frac{3+\sqrt{89}}{8}\right)$

Now square and you have to solve the equation $16x^4-32x^3-72x^2+86x+110$
Easily we find two roots of this which are $x=-1$ and $x=\frac{5}{2}$  and then we have only to solve a quadratic equation .But we should reject some roots and the finally roots are  $x=\frac{5}{2}$ and $x=\frac{1-3\sqrt 5}{4}$


quadratics
number theory solved
number theory

Let l be a quadric residue modulo p and k a non quadric residue modulo p . Then
               p / ka^2+lb^2 => p/a and p/b  , a,b are positive integers .
// "Gazeta Matematica" (a math magazine from Romania ) number
// 9/2000 page 332

I assume $l\neq 0 (mod\  p)$
$p|k\cdot a^2 +l\cdot b^2  \Leftrightarrow k\cdot a^2 \equiv -l\cdot b^2 (mod\  p).$
If p|a, then $p|l\cdot b^2 \Rightarrow p|b$.
If p does not divide a, then p does not divide b. Then a is invertible mod p and we get $k \equiv -l\cdot c^2 (mod\  p)$. This is a contradiction iff p=1(mod 4).

For all $p\equiv -1(mod\  4)$ take a=b=1, k=-1 and l=1. Notice that we don't have p|a and p|b.

If l=0, we can have p|a and p does not divide b.

So what is the full problem text?

{       Let l be a quadric residue modulo p and k a non quadric residue modulo p . Then
p | ka^2+lb^2 => p|a and p|b , a,b are positive integers .
// "Gazeta Matematica" (a math magazine from Romania ) number
// 9/2000 page 332 }

 I think the problem is false as amfulger showed!But we can "change" the problem like asthe following (sorry i haven't got the journal issue):
 Problem:
       Find all prime p so that for any l be a quadric residue modulo p and k a non quadric residue modulo p we have
           " if  a,b are positive integers and $p| k\cdot a^2+l\cdot b^2$ then p|a and p|b , a,b are positive integers "

As amfulger  showed we have p=4t+1

And last,just note that when we say k is a (non) quadric residue  we mean gcd(k,p)=1

[quote="pluricomplex"]
Problem: Find all prime p so that for any l be a quadric residue modulo p and k a non quadric residue modulo p we have
           " if  a,b are positive integers and $p| k\cdot a^2+l\cdot b^2$ then p|a and p|b"

As amfulger  showed we have p=4t+1
And last,just note that when we say k is a (non) quadric residue  we mean gcd(k,p)=1[/quote]

Let $p\equiv 1(mod\ 4)$. Then -1 is a quadratic residue.
Assume we have k,l,a,b like in the problem text such as $p\nmid a$. Then $k\cdot l^{-1}$ is a non-quadratic residue and a is invertible mod p. So WLOG we can assume l=a=1. We thus have $p|k+b^2 \Leftrightarrow k\equiv -b^2 (mod\ p)$.  Since -1 is a quadratic residue, we get that k is a quadratic residue (a contradiction).

So p|a and from this p|b.


Agar $ x,y,z>0$ bashand sabet konid
\[ \sum_{cyc}\frac{1}{x^2\plus{}xy\plus{}y^2} \geq\frac{9}{(x\plus{}y\plus{}z)^2}\]

khob bazam ye namosavie maroofe dge (@Toxic frog:manzooram in nis ke hame didanesh! :D)

khob namosavimoon hamgene,pas mishe farz kard $ x + y + z = 1$ ,khob hala deghat konid ke:

\begin{eqnarray*}1 = (x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx) = x^2 + y^2 + xy + (\underbrace{z^2 + zx + yz}_{  z(x + y + z) = z}) + (xy + yz + zx)\end{eqnarray*}

pas darim:

$ x^2 + xy + y^2 = 1 - z - (xy + yz + zx)$

khob hala gharar bedid $ S = 1 - (xy + yz + zx)$ pas:

$ x^2 + xy + y^2 = S - z$

pas namosavimoon be shekle zir dar miad:

$ \sum\frac {1}{S - x}\geq 9$

inam man tahala tanha rahi ke vasash didam ineke hamaro too ham zarb konimo bad az shur estefade konim,yeni:

$ \sum\frac 1{S - x}\geq 9$

$ \iff 9S^3 - 21S^2 + 12S\leq 9xyz + 1$

$ \iff S(12 - 9S)(1 - S)\leq 9xyz + 1$

$ \iff (12S - 9S^2)(1 - S)\leq 9xyz + 1$

$ \iff \left(4 - (3S - 2)^2\right)(1 - S)\leq 9xyz + 1$

pas kafie sabet konim:

$ 4(1 - S)\leq 9xyz + 1$

$ \iff 4(xy + yz + zx)\leq 9xyz + 1$

ke inam tebghe shur doroste cherake tebghe shur midoonim:

$ \sum x(x - y)(x - z)\geq 0$

$ \iff (x + y + z)^3 + 9xyz\geq 4(x + y + z)(xy + yz + zx)$

$ \iff 1 + 9xyz\geq 4(xy + yz + zx)$.