geometry
geometric transformation
reflection
articles

When writing geometry proofs,

1. Is it ever necessary to redraw your diagram if it increases clarity?
2. How do you show an angle chase on a diagram so a reader can follow it and tell the difference between given and derived information?
3. Where would be a good place to read about writing geometry proofs?

Geometry proofs somehow seem different than proofs for other types of problems. For example, to a show an algebraic manipulation, you might write out following lines of an equation. However, in geometry, would you ever rewrite your diagram with new information as you would rewrite a line of an equation?

If you draw any additional lines, it's a good idea to redraw the diagram so it looks like what you're working with in your proof, unless it's very simple and understandable. Also, if it's a very long proof, it might be better to draw occasional diagrams that reflect your progress, increasing readability. Otherwise, unless there's some other reason to redraw the diagram, it's not necessary.

Another option would be to utililze the two-column proof format (gasp). It's a nice way of organizing your thoughts and reasoning without all that clutter. If you're not a fan of the two column, try using it to organize your response before you put it in paragraph form. You can access more information on proofs in general (ones with long and tedious parts) [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_HowWriteOutline.php]here[/url], in the AoPS Resource Articles section.

Is the two-column proof usually frowned upon?

I noticed in the article you mentioned that the geometry proofs sometimes started with a diagram without most of the angles and lengths labeled. Is it better to start with a diagram that contains just the information given or should we start with a diagram that has all the lengths and angles that we found?

That depends.  If you have a mess of measurements, say, from an angle chasing problem, don't include them all.  But if it's not a lot, it couldn't hurt.  You should include lines and marks that demonstrate your proof, say, an altitude that you may drop in a triangle.

[quote="aufha"]Is the two-column proof usually frowned upon?[/quote]

It usually is if it's a final proof, but you can use two column as an organization aspect before you write your final proof.


Butandioic Acid has $ pK_a$ values $ 4.21$ and $ 5.64$ for the first and second dissociation respectively.

Selenous Acid $ \ce{H2SeO3}$ has $ pK_a$ values $ 2.62$ and $ 8.32$.

Explain why the values differed so greatly in Selenous Acid while they are quite similar for butandioic acid.

The relevant question is: why is $ \ce{HO2C\minus{}CH2CH2\minus{}CO2\minus{}}$ a better acid than $ \ce{(HO)SeO2\minus{}}$?

I am confused by how to go about explaining that would it just be that butandioate ion has more resonance structures than $ \ce{SeO3}^{ 2\minus{}}$ hence more stable?

Sounds like there is another explanation. But how can I come up with an explanation that accounts for the small $ pK_a$ variation for butandioic acid (also the large variation for selenous acid)?

$ \ce{CO2\minus{}}$ is an electronwithdrawing group.


Prove that there isn't a, b, c, d, n which are natural numbers
such that ad=bc, n^2<a<b<c<d<(n+1)^2

Try not to double post :)
$a,b,c,d,n \in \mathbb{N}, \ ad=bc, n^{2}<a<b<c<d<(n+1)^{2}$
Prove this is impossible :wink:
[hide="hint"]Just for fun relabel $a,b,c,d$ with $d-a,c-b,c+b,d+a$ and consider product fact in problem[/hide]

$(a+d-b-c)(a+b+c+d)=(a+d)^{2}-(b+c)^{2}=(d-a)^{2}-(c-b)^{2}<(d-a)^{2}\leq (2n)^{2}$
But, evident, $a+d>b+c$, hence $a+d-b-c\geq 1$; also $a+b+c+d>4n^{2}$. All this leds to contradiction with the first inequality.


it's very simple!
by holder
$\prod_{cyc}(a^{3}+b^{3}+1)\geq (a^{2}+b^{2}+c^{2})^{3}$
so we are left with $a^{2}+b^{2}+c^{2}\geq a+b+c$ which is obvious :wink:

Very nice!So we have three solutions for this ineq :wink:

Here is the first soln:By Cauchy we have: \[(a+b+c)\cdot\left(a+\frac{c^{2}}{b}+\frac{b^{2}}{c}\right)\geq\left(a+b+c\right)^{2}\] ,i.e \[\left(a+\frac{c^{2}}{b}+\frac{b^{2}}{c}\right)\geq(a+b+c)\] .By the same manner we get the other two cyclic ones.
But on the other hand we have: $a+\frac{b^{2}}{c}+\frac{c^{2}}{b}=\frac{1+b^{3}+c^{3}}{bc}$ :wink:

[quote="Albanian Eagle"]another nice solution:
$a^{3}+b^{3}\geq ab(a+b)$
so $a^{3}+b^{3}+1\geq ab(a+b+c)$ multiply these three inequalities :lol:

this problem is quite easy![/quote]
It's wrong: $a=-1,$ $b=-\frac{1}{2}$ and $c=2.$ :wink:
[quote="andyciup"]Let $a,b,c$ be real numbers with $a+b+c\geq 0$ and $abc=1$.Prove that the following inequality occurs: \[\left(a^{3}+b^{3}+1\right)\left(b^{3}+c^{3}+1\right)\left(c^{3}+a^{3}+1\right)\geq\left(a+b+c\right)^{3}\] [/quote]
It's wrong too.

@Albanian Eagle:That was also my second solution,but it works only on $\mathbb{R}_{+}$,since we need the condition $a+b\geq0$.Yesterday I asked [hide="here"] http://www.mathlinks.ro/Forum/viewtopic.php?t=112747 [/hide] if $a+b\geq0$ for all a and b and recieved a counter-example.

@Arqady:I was really amazed to see that counter-example,but I fail in finding the flaw in my proof..what is it?

[quote="andyciup"] I fail in finding the flaw in my proof..what is it?[/quote]
[quote="andyciup"]Here is the first soln:By Cauchy we have: \[(a+b+c)\cdot\left(a+\frac{c^{2}}{b}+\frac{b^{2}}{c}\right)\geq\left(a+b+c\right)^{2}\] [/quote]
The mistake here. :wink:

I didn't pay atention and I thought it was for positive variables :oops:


factorial
number theory
prime factorization

The number 15! is written in base 12. How many zeroes does it end with?

[hide="Solution"]Basically, we are trying to find the largest integer $ n$ for which $ 12^n\mid15!$.

$ 15!$ can be expressed as $ 2^{11}\cdot3^6\cdot5^3\cdot7^2\cdot11\cdot13$, and $ 12$ as $ 2^2\cdot3$. Thus, the largest integer $ n$ is $ \boxed{5}$.[/hide]

You aren't labeling your variables quite correctly.  What you want is the largest integer $ n$ such that $ 12^n | 15!$.  Note that to do this you don't need the part of the prime factorization past $ 3$.

Thanks for the correction (I meant what you said; I just didn't write it properly). I realized that the rest of the factorization wasn't necessary only after I had already figured it out, so I left it there anyway. :)


1) How many combinations of Baskin-Robbins ice cream and Heinz ketchup are there?

2) What is the probability of winning the lottery if you play in a pool of $1,000$ people every day for a year?

3) A point is chosen at random on a circle.  An arc of $pi$ radians is drawn off the perimeter of the same circle at random.  A chord connects the two ends of the arc.  What is the probability that the randomly selected point is in this region?  (Sorry that I couldn't word that any better!)

4) How many ways are there to give $7$ M&Ms to $49$ children, assuming not every child gets an M&M, and assuming that one child can get more than one M&M.

[hide]1) Depends on where you live :P
2) $/frac {73}{200}
3) Don't understand
4) I'm getting too large of a number--maybe I'm doing it wrong... :blush:[/hide]

3
[hide]$\pi$ radians is $180^{\circ}$ so the answer is $\frac 12$.[/hide]

I've got 4:

[hide]  If seven kids each get one M&M, then you can do that in 49!/(42!x7!) ways, which is 85900584.

Because one kid can be given more than one M&M, then he can get 2,3,4,5,6,or 7 M&Ms as well.

In the case of 2: You can only give the M&Ms to six kids. Thus, take 49!/(43!x6!), which is 13983816. You still have one last M&M to give away, so there are 6 ways to give it to one of the six kids. But it doesn't matter what order you gave the two M&Ms to the lucky kid who got 2, so you would divide by 2. This gives 13983816x6/2=41951448.

For 3: You can give the candy to five kids. 49!/(44!x5!)=1906884. Then you can give the remaining two M&Ms in five ways (since they all go to one kid). But it doesn't matter whether you gave one candy before the others, so you divide by 3 in the end. 1906884x5/3=3178140.

For 4: 49!/(45!x4!)=211876. Then we multiply this by four, but divide by 4. This gives 211876.

For 5: 49!/(46!x3!)=18424. Multiply by 3, but divide by 5. This gives approximately 11054

For 6: 49!/(47!x2!)=1176. Multiply this by 2, and divide by 6. This gives 392.

For 7: 49!/(48!x1!)=49. This one we don't multiply or divde, because we give all the M&Ms at once.

Adding these all up gives 131253543.

This might be way off, so don't take me for granted this is right. This problem might involve using the Principle of inclusion and exclusion, in which case the problem's going to get much more difficult than it already is.

Answer (hopefully):131253543[/hide]

4
[hide]Okay the real answer is the same thing as above except there's six ways to do it with six kids, five with five kids and so on until only one with one kid. There's only one way to do it with seven kids. So it's $nCr (49, 7 ) + nCr(49, 6) * 6 + nCr ( 49, 5) * 5 + nCr ( 49, 4 ) * 4 + nCr (49 , 3) * 3 + nCr ( 49, 2 ) * 2 + nCr ( 49, 1) = 180243077$[/hide]

[quote="tarquin"]4
[hide]There are several cases to consider when donig this problem. First, if you pick seven children out of the fourty-nine, then there's only one way to give all of them seven M&M's, and that is to give each child one M&M. Similarly, when choosing six children, there's seven ways to give all of them seven M&M's total and that is to give one of the six children two M&M's, and the other five one M&M. If we continue to write down these numbers, we get the following: Choosing 7, 1 way, 6, 6 ways, 5, 15 ways, 4, 20 ways, 3, 15 ways, 2, 6 ways, and 1, 1 way. The answer, therefore, is the number of ways you can choose seven kids out of fourty-nine, times 1, plus, then number of ways you can choose six children out of fourty-nine, times six, and so on. This is equal to $\frac {49!}{7!42!} + \frac {49!}{6!43!} * 6 + \frac {49!}{5!44!} * 15 + \frac {49!}{4!45!} * 20 + \frac {49!}{3!46!} * 15 + \frac {49!}{2!47!} * 6 + \frac {49!}{48!} = 202927725$.[/hide]
Oops I did it where more than one child can get more than one M&M...[/quote]

No you were right in the first place, Tarquin...sorry about the cofusing wording....

Edit: 10 secs. later:  Nobody got problem #1, the easiest one, (though it requires background knowledge, so you wouldn't see it on MathCounts.)

What is it, 31 flavors and 59 Heinz varieties?

I assumed not... but if they were I think that I'd have to multiply my answer by $7!$.

[quote="Magixmaniak"]For the last question, are the M&Ms distinguishible or not?[/quote]

no (bad question, i messed it up: the answer should have been much lower!)


How many numbers between 1200 and 3522 can you form with digits 1,2,3,4,5,6,7 without repeating digits?

I know that the answer is 305 but I dont understand how to calculate that, please help.

When solving a combinatorics problem, always keep in mind that dividing the problem into several easier ones might help.

[hide]In this case, we can classify our numbers into:

Numbers between 1200 and 2000: $ 6*5*4\equal{}120$ numbers.

Numbers between 2000 and 3000: $ 6*5*4\equal{}120$ numbers.

Numbers between 3000 and 3500: $ 3*5*4\equal{}60$ numbers.

Numbers between 3500 and 3522: $ \{12, 14, 16, 17, 21\}\equal{}5$ numbers.

$ \Longrightarrow$ There are $ 120\plus{}120\plus{}60\plus{}5\equal{}305$ numbers between 1200 and 3522.
[/hide]


inequalities
function
inequalities unsolved

Given $x,y,z>0$ prove that
$\sum\frac{x}{\sqrt{y^{2}+yz+z^{2}}}\ge \sqrt{3}$

Apply Cauchy-Schwarz we have :

$\sum \frac{x^{2}}{x \sqrt{y^{2}+yz+z^{2}}}\geq\ \frac{(x+y+z)^{2}}{ \sum x \sqrt{y^{2}+yz+z^{2}}}\geq\ \frac{(x+y+z)^{2}}{ \sqrt{(x+y+z)(xy(x+y)+yz(y+z)+zx(z+x)+3xyz)}}\geq\ \sqrt{3}$

[quote="Crux"]you can get first one be applying weighted jensen's inequality with function $\frac{1}{\sqrt{x}}$ and weighting factors $\frac{x}{x+y+z},\frac{y}{x+y+z},\frac{z}{x+y+z}$[/quote]

 can you show?

Thanks It is enough for my result, can it be made stronger
$\sum\sqrt[k]{\frac{x}{\sqrt{y^{2}+yz+z^{2}}}}\ge \frac{3}{\sqrt[2k]{3}}$
what's is the best constant $k$?

[quote="tanpham90"]
$\frac{(x+y+z)^{2}}{ \sum x \sqrt{y^{2}+yz+z^{2}}}\geq\ \frac{(x+y+z)^{2}}{ \sqrt{(x+y+z)(xy(x+y)+yz(y+z)+zx(z+x)+3xyz)}}\geq\ \sqrt{3}$[/quote]
I'm sorry, but I didn't quite catch it :blush:
Can anyone explain it to me?

I think he mean by Cauchy-Swart
$\sum (x\sqrt{y^{2}+yz+z^{2}})^{2}\le (x+y+z)(\sum(\sqrt{x(y^{2}+yz+z^{2})})^{2})$ and the last is Schur!


geometry
geometric transformation
reflection

Umm well not right now but would anyone want to duel? I mean not right now b/c i'd need more people. It'd be kind of like RAF but the proper way to duel online without wands. I'm really okay with any curses except for unforgivable ones. Anyone interested? I've got to go right now so..umm...i'll reply later...

[quote="mustafa"]Could you explain what dueling is? Is this a Harry Potter thing?[/quote]
Yes, read the second Harry Potter book. :D

I read it, it's just that nothing in the topic made it clear that it had anything to do with Harry Potter, except for a vague reference to unforgivable curses. What does RAF stand for?

Ready? Aim. Fire!(i think that's how you write it)
I'd make a list of curses allowed and I'd not make it 1 on 1 since that'd be too easy for people who know such spells as exanimo...or avada kedavra which will not be allowed...
you'd start with an amount of galleons and duel for more and then we'd end when someone has a set amount of galleons...
(the person would be able to pick things such as dodge, or expelliarmus(reflects spells) and it would be sort of like RPG...)

I'll need people willing to do it by PM...

Oooo... sounds really really fun.  [i]Accio Wand[/i].

what are we supposed to sign up?
then Im in or something

I'll give it a try. It does seem quite a bit like Ready, Aim, Fire, but more complex.

Count me in, I hope I have time.  (I used to be a total freak about HP, but it took too much time  :blush: .)

[i]Accio your wand[/i]  :P

accio everybody's wand.... Do I win now? :D

Not really, I can just deflect the spell with a flick of my own wand.

[i]Sectumsempra![/i] (hint, start bleeding to death now).

Protego! Your spell bounces off and hits you!

Sounds fun!
[i]levicorpus[/i]! And you don't even see it coming.. :wink:

Impedimenta! Did both daine and pieterminate attack me? I might have to use my ninja-like powers to dodge your spells :ninja:

no you are still playing because dueling doesn't end when you lose!!! ok first person to 7 points wins!!!

ok i am pming daine but beware of my lateness everyone :P

umm is that about me or you :o wtf i never get it

ok you and mustafa can start and junggi can wait and if no one shows up kyyaunmathcount can play

yeah kyyaunmathcounts can umm play srry about my spelling....i will make this better to see...

just here to say i will def. have written up my 2 complete duels by the new year :lol:

I keep using [i]reducio[/i] for 14 years until you are less than 1 planck length at which point your existence is meaningless in physics so I win (kind of)

also why is this tagged as "geometry" and "geometric transformation" and "reflection"

wow how did you bump a 14 year old thread??? what a legend

anyways... [i]Geminio[/i] 10 times and multiply myself to all duel against you

I already won lol, we can start a new one though

Here are the scores in a more organized format:
[hide=Scores]
[list=1]
[*]mustafa: 2 :first:
[*]kyyuanmathcount: 1 :)
[*]thatin: 1 :)
[*]junggi: 1 :)
[*]john0512: 1 :)
[*]jackdillon: 0 :wallbash_red:
[*]daine: 0 :wallbash_red:
[/hide]


topology
function
parameterization
advanced fields
advanced fields solved

I'm beginning to study homotopy theory. How I can show that
$\{(x,0),x\in\mathbb{R}\}\cup \{(0,y),y\in\mathbb{R}\}\cup \{x+y=1,0\leq x\leq 1\}\simeq \mathbb{S}^{1}$.

I'd do it in stages. First, show that you can pull the spikes on the axes in, making it homotopic to the triangle. Then expand the triangle radially around an interior point until it's a circle.

I was thinking something similar. I know for example that I can map the real line to the unit interval by using the function arctan and then an affine mapping. But if I take only $x$ to my parameter, I will map the $y$-axis to a constant point, which is something I don't want. It looks like I have to use maybe some function like arctan x + arctan y. Well, I will go and think more.

Use a different formula for each ray, and leave the triangle itself fixed. Don't try to make the whole line into a point.


Hey, i am new in this site and my english are bad(i apologise for this). My teacher gave  me this  problem and he
claimed that it is easy. I have tried to solved it but i couldn't.I ask for the sollution but he insist
 that the sollution is easy.So any help;

Problem:

Let S(n) be the sum of the first n primes ( S(1)=2,S(2)=2+3,S(3)=2+3+5 ...)
Prove that  in  the interval [S(n),S(n+1)] exist a square number for all natural n.

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?t=40296

Pierre.


Hello Everyone,

Can anyone please explain the uses for arithmetic mean, geometric mean, and harmonic mean (and any other mean)? I mean, I know their respective formulas, but I don't understand their uses. Which one should I use in what situation?

Er, I'm not sure what you mean. Most often in Mathcounts, questions will directly ask for the arithmetic/geometric mean. Arithmetic means definitely occur the most, I've seen a few geometric means, and I've NEVER seen a harmonic mean. If you want something past Mathcounts, look at the Arithmetic Mean-Geometric Mean Inequality, or even Root Mean Square-Arithmetic Mean-Geometric Mean-Harmonic Mean, abbreviated as RMS-AM-GM-HM, or the Power Mean Inequality. Too lazy to actually link to them.

The most useful application of harmonic mean for me is for those kind of problems that say if "Mr. A drives 20 mph, he arrives 5 minutes late. If he drives 40 mph, he arrives 5 minutes early. What speed should he go at to arrive perfectly on time", or something like that. If he arrives equally late and early, the answer is just the harmonic mean of the two speeds.

Arithmetic means: Obvious use most of the time; too general to discuss specifically.
Geometric means: There are a few uses in geometry. For example, take a right triangle, and drop the altitude from the right angle to the hypotenuse. Then the geometric mean of the two lengths you broke the hypotenuse into is equal to the length of the height. There are also many other geometric mean relations in this figure (try to find them).
Harmonic means: Refer to xpmath's second paragraph.

But as xpmath said, there's not many situations in which finding some mean will give you the key insight to a problem. In some of the instances described above, they simply make calculations quicker.


integration
logarithms
induction
inequalities unsolved
inequalities

Your sum < $\int_{n}^{2n}\frac{1}{x}=\ln 2n-\ln n=\ln 2<0.7$

Misha

another proof.
we  can prove  1/(n + 1) + 1/(n + 2) + 1/(n + 3) + ... + 1/(2n)<7/10-1/4n  by induction
for $n$ enough large. :)

$zhaoli$ your name are rather like mine.we look like brothers. :D
and this problem comes from the magzine of《chinese mathematics olympiad（high school）》（数学竞赛之窗（高中版）》，right?
and thank you(include everyone) for supporting our magzine. :)

I'm only a person who colleges information for the magazine sometimes.
and I'm sorry I'm not able to make the web site more stable because I don't manage the website.but I'll say to the manager of the website.thank you :)

[quote="wenxin"][quote="siuhochung"]"our magazine"?
You are one of the committees who published it?
I have an opinion, can you make the site of the magazine more stable? the asp scripts there often encounter errors.[/quote]

If not all right, have a try of this

http://www.mathfan.com/Default2.html[/quote]

asp error encounters immediately when i log in....  :(  :(
[quote]“/?应用程?中的?务器错误。
--------------------------------------------------------------------------------

无法找到资?。
说明: HTTP 404。您正在查找的资?(或者它的一个?赖项)?能已被移除，或其??称已更改，或暂时??用。请检查以下 URL 并确?其拼写正确。

请求的 URL: /Home.aspx

--------------------------------------------------------------------------------
版本信?: Microsoft .NET Framework 版本:1.1.4322.2032; ASP.NET 版本:1.1.4322.2032 [/quote]

[quote="zhaobin"][quote="siuhochung"][quote="zhaobin"]are you sure your net link is public net.if your net link is edcuted net will not able to enter.I think :)[/quote]

may i know what's edcuted net?[/quote]
my fault.forget it.your must be educated net.because you can enter mathlink.[/quote]

i still dun understand.
wt're public net & educated net and wt're their differences?
maybe we talk about this on private msg system.

[quote="zhaobin"]maybe I am not able to express.for example,my school link is educated link.so in my school I can't enter mathlink.now I can enter the mathfan.I don't know why you can't ;)[/quote]

In Hong Kong very few schools have dormitory. Most students access the internet at home, like me.


geometry
projective geometry
geometry proposed

I know this is maybe the wrong section to post this but I really do not know where to post it otherwise...

Most of you guys probably know Kiran Kedlaya's notes on Euclidean geometry... well, there is a few problems in there which are obviously incorrectly stated. I was just wondering whether anyone knows the correct version...

*Problem 3.3.3
Let $C_1, C_2, C_3$ be circles in the plane. $C_1, C_2$ intersect at $A, B$; $C_2, C_3$ intersect at $C, D$; $C_1, C_3$ intersect at $E, F$. Let  $P_1$ be an arbitrary point on $C_1$ and define points $P_2$, ...$P_7$ as follows: $P_2 = C_2 \cap P_1A$, $P_3 = C_3 \cap P_2C$, $P_4 = C_1 \cap P_3E$, $P_5 = C_2 \cap P_1B$, $P_6 = C_3 \cap P_2D$ and $P_7 = C_1 \cap P_3F$. Prove that $P_1 = P_7$.

*Problem 8.6.4
$A,B,C,D$ lie on a straight line, in this order. A circle $k$ passes through $B$ and $C$, and $AM$, $AN$, $DK$, $DL$ are tangent to $k$ at $M$, $N$, $K$ and $L$. Lines $MN$ and $KL$ intersect $BC$ at $P$ and $Q$.
Prove that $P$ and $Q$ do not depend on $k$.
If $AD = a$, $BC = b$ and segment $BC$ moves along $AD$ find the minimum length of $PQ$.

It is useless to disturb Kedlaya :
In Problem 3.3.3, some little typos : P[5] = C[2]^P[4]B, P[6] = C[3]^P[5]D, P[7] = C[1]^P[6]F.
I suggest to use the fact that, if two circles (o), (o') intersect at P, and Q, the direct similitude with center Q mapping (o) to (o') maps any point M of (o) to the second common point of (o') with the line MP.
Here, we have 6 such similitudes and their product is identity.
Problem 8.6.4 is quite correct : As MN is the polar line of A wrt k, P is the harmonic conjugate of A wrt (B,C); same way, Q is the harmonic conjugate of D wrt (B,C).
PQ will be minimal when AB = CD, in which case PQ = b^2/a.
Kind regards.

[quote="Arne"]*Problem 3.3.3
Let $C_{1}, C_{2}, C_{3}$ be circles in the plane. $C_{1}, C_{2}$ intersect at $A, B$; $C_{2}, C_{3}$ intersect at $C, D$; $C_{1}, C_{3}$ intersect at $E, F$. Let  $P_{1}$ be an arbitrary point on $C_{1}$ and define points $P_{2}$, ...$P_{7}$ as follows: $P_{2}= C_{2}\cap P_{1}A$, $P_{3}= C_{3}\cap P_{2}C$, $P_{4}= C_{1}\cap P_{3}E$, $P_{5}= C_{2}\cap P_{1}B$, $P_{6}= C_{3}\cap P_{2}D$ and $P_{7}= C_{1}\cap P_{3}F$. Prove that $P_{1}= P_{7}$.[/quote]

In my opinion, the definitions $P_{5}= C_{2}\cap P_{1}B$, $P_{6}= C_{3}\cap P_{2}D$ and $P_{7}= C_{1}\cap P_{3}F$ should be replaced by $P_{5}= C_{2}\cap P_{4}B$, $P_{6}= C_{3}\cap P_{5}D$ and $P_{7}= C_{1}\cap P_{6}F$. Actually, the problem is an easy angle-chasing exercise.

[quote="Arne"]*Problem 8.6.4
$A,B,C,D$ lie on a straight line, in this order. A circle $k$ passes through $B$ and $C$, and $AM$, $AN$, $DK$, $DL$ are tangent to $k$ at $M$, $N$, $K$ and $L$. Lines $MN$ and $KL$ intersect $BC$ at $P$ and $Q$.
Prove that $P$ and $Q$ do not depend on $k$.
If $AD = a$, $BC = b$ and segment $BC$ moves along $AD$ find the minimum length of $PQ$.[/quote]

I don't see what is wrong about this problem. See http://www.mathlinks.ro/Forum/viewtopic.php?t=3482 for this problem.

  darij

Oh, I tried to make an accurate diagram for #2 few times and that didn't work, but I probably just misread something.


FAU has a math day, and this is their latest individual competition:

http://www.math.fau.edu/web/mathday/mday08individual.pdf

Could you guys look at it and give me an idea of the difficulty, in your opinion? I didn't think any were particularly tough, save one or two, and if someone could compare it to the AMC 12, that would be great.

I'd say that they're hard AMC 10 - medium AMC 12 and maybe even EXTREMELY EXTREMELY easy AIME.

Um, I'd say it's actually a pretty good mock AMC 10.

I agree with AIME15 on this one. Some of the questions might be easier-AMC 12 problems, but the AMC 10 and AMC 12 difficulty overlap  a bit so it's hard to say.


Frankly it seems quite inactive. It is nice, with many resources, but most of those can be better found with Wikipedia or Wolfram's MathWorld. The highlight would then be the forums, which have many sections with no posts. But still, it is a very nice resource and it seems highly developed. It problem is its community base. If you got more people (and more professionals), then maybe it could be something like AoPS...

[quote="yppetrov"]What do you think about the new math site?
[url=http://www.math10.com]Math[/url][/quote]

I looked through some of the files it seems pretty nice. Although more information can be brought from Wolfram MathWorld, math10 does feature a nice features such as drawing graphs (under logarithms). It is quick and to the point.


In a quadrilateral ABCD, AB is parallel to CD, and is perpendicular to AD;AB=3CD.A circle is drawn touching the sides of the quadrilateral. If the area of the quadrilateral is 4 sq. units, find the radius of the circle:

This is more appropriate for Intermediate.

[hide="Solution"]If $CD=b$, then $AB=3b$. Put $h=AD, y=BC$. Then

$h+y=3b+b=4b$
$h^{2}+(2b)^{2}=y^{2}\iff (y-h)(y+h)=4b^{2}\implies y+h=b$

From there $y={5b\over 2},h={3b\over 2}$. Since $[ABCD]={b+3b\over 2}h=3b^{2}$, we get $b={2\sqrt{3}\over 3}\implies h=\sqrt{3}\implies r={h\over 2}={\sqrt{3}\over 2}$[/hide]


calculus
integration
calculus computations

hi all, sorry if this has been posted before

i really want to evaluate a series of the form

$ \sum_{k=1}^\infty \alpha^n n^p$ for a fixed $ 0<\alpha <1, 2p\in \mathbb c{Z}$ is negative,

anyone got any ideas?  I got something of an interated integral formula for when p is a negative integer itself, but im hoping for an all encompassing solution if there is one.
thanks

This is the definition of the polylogarithm (modulo mixing up $ k$ and $ n$) -- since it has a name, it's safe to assume that it isn't nice in general but might be nice in some special cases.  Just knowing that $ p$ is a half-integer doesn't seem to be enough to do anything with it (or at least, Mathematica won't express it in a closed form).


Suppose that each box of cereal contains or not exactly one of $ r$ different coupons. Each coupon is equally likely to be placed in any given box and placing the coupons into boxes is independent. Determine the probability that a person who buys $ n$ boxes of cereals $ (n \geq r)$ will obtain a complete set of $ r$ different coupons.

inclusion-exclusion:

the number of ways that we can miss $ k$ particular coupons, $ 1\leq k \leq r$ is $ (r\minus{}k)^n$. so by PIE, the number of ways to miss at least $ 1$ coupon is $ S\equal{}\sum_{k\equal{}1}^r\binom{r}{k}(r\minus{}k)^n$. so the probability is $ 1\minus{}\frac{S}{r^n}$.


algebra
system of equations
algebra proposed

Solve the system of equations $ \begin{cases} x^3 \plus{} x(y \minus{} z)^2 \equal{} 2\\ y^3 \plus{} y(z \minus{} x)^2 \equal{} 30\\ z^3 \plus{} z(x \minus{} y)^2 \equal{} 16\end{cases}$.

[quote="zaya_yc"]
$ x\equal{}\minus{}\frac{d}{2}$ and $ d\equal{}0$

$ d\equal{}0$----> $ x\equal{}y\equal{}z\equal{}\sqrt[3]{2}$

$ x\equal{}\minus{}\frac{d}{2}$-----> $ d\equal{}2\sqrt[3]{2}$----> $ x\equal{}\minus{}\sqrt[3]{2}$ ; $ y\equal{}\sqrt[3]{2}$ ; $ z\equal{}3\sqrt[3]{2}$[/quote]
I don't know how he obtained the above solutions.
It is clearly that $(x,y,z)=(1,3,2)$ is a solution.
Anyone help me?
P.S. I had solved it myself.

[quote="zaya_yc"]$ \begin{cases} x^3\plus{}xd^2\equal{}2 \\
(x\plus{}2d)^3\plus{}(x\plus{}d)d^2\equal{}x^{3}\plus{}6x^{2}d\plus{}11xd^{2}\plus{}9d^{3}\equal{}30 \\
(x\plus{}d)^3 \plus{} (x\plus{}d)4d^2\equal{}x^{3}\plus{}3x^{2}d\plus{}7xd^{2}\plus{}5d^{3}\equal{}16\end{cases}$[/quote]

His problem lies here.  It should read

$ \begin{cases} x^3\plus{}xd^2\equal{}2 \\
(x\plus{}2d)^3\plus{}(x\plus{}2d)d^2\equal{}x^{3}\plus{}6x^{2}d\plus{}13xd^{2}\plus{}10d^{3}\equal{}30 \\
(x\plus{}d)^3 \plus{} (x\plus{}d)4d^2\equal{}x^{3}\plus{}3x^{2}d\plus{}7xd^{2}\plus{}5d^{3}\equal{}16\end{cases}$.

With this modification, the equations we get are direct multiples of each other, so we need a new workaround.

Your solution is probably correct; could you post it?

[quote="tenniskidperson3"]
Your solution is probably correct; could you post it?[/quote]
Yes, of course. This is my solution, I found it while I didn't known the official solution yet.
$ \begin{cases} x^3 \plus{} x(y \minus{} z)^2 \equal{} 2\\ y^3 \plus{} y(z \minus{} x)^2 \equal{} 30\\ z^3 \plus{} z(x \minus{} y)^2 \equal{} 16\end{cases}$
$\Leftrightarrow \begin{cases}x^3+xy^2+xz^2=2xyz+2,\\y^3+yz^2+yx^2=2xyz+30,\\z^3+zx^2+zy^2=2xyz+16.\end{cases}$
Let $u=x^2+y^2+z^2, v=xyz$ then $(x,y,z)=(\frac{2v+2}{u},\frac{2v+30}{u},\frac{2v+16}{u})$.
Now, since $x^2+y^2+z^2=u$ and $xyz=v$,
$\frac{(2v+2)^2}{u^2}+\frac{(2v+30)^2}{u^2}+\frac{(2v+16)^2}{u^2}=u$,
and $\frac{(2v+2)(2v+30)(2v+16)}{u^3}=v$.
These ones give $(2v+2)(2v+30)(2v+16)=v((2v+2)^2+(2v+30)^2+(2v+16)^2)$
or, $v^3+4v-240=0$. Or $(v-6)(v^2+6v+40)=0$.
Thus $v=6$, and $u=14$, and $(x,y,z)=(1,3,2)$.


geometry
rectangle
3D geometry
pyramid
circumcircle
sphere
tetrahedron

The rectangle $ABCD$ below has dimensions $AB = 12 \sqrt{3}$ and $BC = 13 \sqrt{3}$.  Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at $P$.  If triangle $ABP$ is cut out and removed, edges $\overline{AP}$ and $\overline{BP}$ are joined, and the figure is then creased along segments $\overline{CP}$ and $\overline{DP}$, we obtain a triangular pyramid, all four of whose faces are isosceles triangles.  Find the volume of this pyramid.

[asy]
pair D=origin, A=(13,0), B=(13,12), C=(0,12), P=(6.5, 6);
draw(B--C--P--D--C^^D--A);
filldraw(A--P--B--cycle, gray, black);
label("$A$", A, SE);
label("$B$", B, NE);
label("$C$", C, NW);
label("$D$", D, SW);
label("$P$", P, N);
label("$13\sqrt{3}$", A--D, S);
label("$12\sqrt{3}$", A--B, E);[/asy]

somebody should try this, it is pretty routine, the only thing to note is that
[hide="hint"]the altitide to the base B(A)CD is the circumcenter of that triangle[/hide]

[quote="Zellex"]Altheman, does that apply to any trangular pyramid?[/quote]

[hide="..."]what do you mean by trianglular pyramid? the reason is that all the lateral edges are equal, so you can think of a sphere with the upper vertex of the pyramid as the center, and the verticies of the base on the sphere, then the plane of the base determines a circle with the sphere, being the circumcircle of the base, furthermore, the altitude from the center of a sphere to a crossection obviously hits the center of the crossection circle[/hide]

Oh...I asked because I didn't realize that the lateral edges were equal  :blush: :/. Sorry.

With the strong hint by Altheman (takes me 5 min to figure out), I will type up the mechanical work:

[hide="solution"]
first we find the length of the diagonal.  So $144*3 + 169*3 = a^2$, and $a = \sqrt{939}$, consequently $\frac{a}{2} = \frac{\sqrt{939}}{2}$.  The area of the triangle base is $\frac{bh}{2} = \sqrt{399} * 6\sqrt{3} = 18\sqrt{133}$.  also, $\frac{abc}{4R} = 18\sqrt{133}$, so $R = \frac{168\sqrt{3}}{2\sqrt{133}}$.  finally we get the altitude of the pyramid, which is $\frac{\sqrt{40215}}{2\sqrt{133}}$.  Applying the pyramid formula we get $3\sqrt{40215}$
[/hide]

That is nasty...(wonder how to do all that stuff accurately on AIME...)

Kalva says the answer is 594...you answer seems to be 601.616033...

[hide]

When folded, let the pyramid's points A and B coincide at E

By pythagorean, EP is sqrt(13^2*3+12^2*3)/2 = sqrt(939)/2

Since ECD is iscosceles (sp), so the height EF is sqrt(13^2*3-6^2*3)=sqrt(399)

Since triangle PCD is also iscosceles (sp), the height PF is sqrt(939/4-6^2*3)=13sqrt(3)/2

Now look at triangle FPE

Let the altitude of the tetrahedron meet ECD at G

By law of cosines, FG is equal to ((PF)^2+(FE)^2-(PE)^2)/(2(PF)(EF))=291/(39sqrt(133))*sqrt(507)/2

by pythagorean theorem (PF)^2-(FG)^2=(PG)^2

sqrt(507/4(1-291^2/39^2/133))=(PG)^2

after some awful arithmetic you get PG=99/sqrt(133)

multiply that by the area of ECD which is 6sqrt(3)*sqrt(399) you get 1782

divide by 3 to get the volume and you get 594[/hide]

These types of problems can always be solved by using ANALYTIC GEOMETRY.

I'm too tired to post my solution now, but here is a hint.
[hide="hint"]
Just set the vertex as $(x, y, z)$ and use the distance formula. Eventually we get $z$, which is the altitude and multiply $\frac{1}{3}$ the base by it. [/hide]

I haven't really seen a way with analytical geometry (which is probably becuase of my own stupidity), but distance formula seems to be an idea.

[hide]Since the 3 edges connecting the base are all equal, we first find the circumcircle of the base.

herons formula is equivalent to sqrt(4a^2b^2-(a^2+b^2-c^2)^2)/4

Using 13sqrt(3), 13sqrt(3) and 12sqrt(3) we have 18 sqrt(133) as the area

the circumcircle is abc/(4ABC) where ABC denotes the area of abc

circumcircle=169sqrt(3)/2/sqrt(133)

The three edges connect p to the base are all sqrt((13^2*3+12^2*3))/2= sqr(939)/2 because of pythagorean theorem

now we use pythagorean theorem again:  939/4-169^2*3/4/133=(939*133-169^2*3)/532=3*(313*133-169^2)/133/4=

9801/133=99/sqrt(133)

multiply this by the area of the base =18*sqrt(133)*99/sqrt(133)/3=594[/hide]

Here is my solution.

[hide="Solution using analytic geometry"]
Let $\triangle{ABC}$ (or the triangle with sides $12\sqrt{3}$, $13\sqrt{3}$, $13\sqrt{3}$) be the base of our tetrahedron. We set points $B$ and $C$ as $(6\sqrt{3}, 0, 0)$ and $(-6\sqrt{3}, 0, 0)$, respectively. Using Pythagoras, we find $A$ as $(0, \sqrt{399}, 0)$. We know that the vertex of the tetrahedron ($D$) has to be of the form $(x, y, z)$, where $z$ is the altitude of the tetrahedron. Since the distance from $D$ to points $A$, $B$, and $C$ is $\frac{\sqrt{939}}{2}$, we can write three equations using the distance formula:

$x^{2}+(y-\sqrt{399})^{2}+z^{2}=\frac{939}{4}$

$(x-6\sqrt{3})^{2}+y^{2}+z^{2}=\frac{939}{4}$

$(x+6\sqrt{3})^{2}+y^{2}+z^{2}=\frac{939}{4}$

Solving these equations with a bit of effort, we eventually get $x=0$, $y=\frac{291}{2\sqrt{399}}$, $z=\frac{99}{\sqrt{133}}$.
Since the Area of a triangle is $\frac{1}{2}\cdot bh$, we have the base area as $18\sqrt{133}$. Thus, the volume is $V=\frac{1}{3}\cdot18\sqrt{133}\cdot\frac{99}{\sqrt{133}}=6\cdot99=594$.

[/hide]

I do not understand why the altitude to the base goes through the circumcenter.


topology
real analysis
real analysis unsolved

Prove or disprove: for any set $A$ of real numbers with measure $0$ and for any $\epsilon$ one can cover $A$ with a countable family of open intervals $I_{n}$ such that $I_{n}$ has length smaller than $\frac{\epsilon}{2^{n}}$.

If such a covering exists, then for any $s>0$ the $s$-dimensional Hausdorff measure of $A$ is at most $\sum_{n\ge 0}\left(\frac{\epsilon}{2^{n}}\right)^{s}=\frac{\epsilon^{s}}{1-2^{-s}}$. Since $\epsilon$ is arbitrary, we have $\mathcal H^{s}(A)=0$, which implies that the Hausdorff dimension of $A$ is zero.

There are sets of zero measure with positive Hausdorff dimension, e.g. standard Cantor set.


trigonometry
inequalities
triangle inequality

Multiply the second equation by $i$ to get that:

$\sum \text{cis}\, x_{i}=n*\left(\frac{\sqrt{6}+\sqrt{2}}{4}+\frac{\sqrt{6}-\sqrt{2}}{4}i\right)=n*\text{cis}\, \frac{\pi}{12}$

If we use the triangle inequality, we have that:

$n=\left|\sum\text{cis}\, x_{i}\right|\le \sum|\text{cis}\, x_{i}|=n$

Hence $\text{cis}\, x_{i}$ all have the same direction. Hence we have that $x_{i}=\frac{\pi}{12}+2k_{i}\pi$ where $k_{i}$ are integers.

Nice problem.

Just so other people do not get confused, Altheman meant to write $n=\left|\sum\text{cis}\, x_{i}\right|\le \sum|\text{cis}\, x_{i}|=n$.  My solutions is identical to yours.


inequalities
induction
function
calculus
derivative
search
inequalities proposed

Let $ a>0$ and $ b>0$  such that $ a^{(2n \plus{} 1)^2} \plus{} b^{(2n \plus{} 1)^2} \equal{} 2,$ where $ n\in\mathbb N.$ Prove that
\[ \frac {a^{n \plus{} 1}}{b^n} \plus{} \frac {b^{n \plus{} 1}}{a^n}\geq2
\]

Answer of Reynaldo Gil Pons:
First Solution:
$ \frac {a^{n + 1}}{b^n} + \frac {b^{n + 1}}{a^n}\geq2$
$ \Leftrightarrow a^{2n + 1} + b^{2n + 1}\geq2a^nb^n$
Let $ \frac {b}{a} = c$
Then $ a^{(2n + 1)^2} + b^{(2n + 1)^2} = 2$ implies that:
$ a^{(2n + 1)^2}(1 + c^{(2n + 1)^2}) = 2\hspace{1cm}(1)$
And:
$ a(1 + c^{2n + 1})\geq2c^n\hspace{1cm}(2)$
Replacing a from $ (1)$ at $ (2)$:
$ (\frac {1 + c^{2n + 1}}{2})^{(2n + 1)^2}\geq\frac {c^{n(2n + 1)^2} + c^{(n + 1)(2n + 1)^2}}{2}$
Let $ c^{2n + 1} = d$. Then we have to prove that:
$ (\frac {1 + d}{2})^{(2n + 1)^2}\geq\frac {d^{n(2n + 1)} + d^{(n + 1)(2n + 1)}}{2}$
$ \Leftrightarrow (\frac {1 + \frac {1}{d}}{2})^{(2n + 1)^2}\geq\frac {d^{-n(2n + 1)} + d^{-(n + 1)(2n + 1)}}{2}$
(Thanks for arqady, I fix the mistake)
Then, multipliplying the two previous inequalities:
$ \Leftrightarrow (\frac {2 + \frac {1}{d} + d}{4})^{(2n + 1)^2}\geq\frac {d^{ - (2n + 1)} + d^{(2n + 1)} + 2}{4}$
Let $ d = x^2$:
$ \Leftrightarrow (\frac {(\frac {1}{x} + x)^2}{4})^{(2n + 1)^2}\geq\frac {(x^{ - (2n + 1)} + x^{(2n + 1)})^2}{4}$
$ \Leftrightarrow (\frac {\frac {1}{x} + x}{2})^{(2n + 1)^2}\geq\frac {x^{ - (2n + 1)} + x^{(2n + 1)}}{2}\hspace{1cm}(3)$

Let`sprove the following inequality by induction in $ i$.
$ (\frac {\frac {1}{x} + x}{2})^{8i}\frac {x^{ - (2i - 1)} + x^{(2i - 1)}}{2}\geq\frac {x^{ - (2i + 1)} + x^{(2i + 1)}}{2}\hspace{1cm}(4)$
It easy to prove the inequality for $ i = 1$
Suppose that the inequality holds for $ i = k$. We will demostrate that the inequality hold for $ i = k + 1$

Now, we will prove that the following inequality holds for every natural number $ j$.
$ (\frac {\frac {1}{x} + x}{2})^{8}(\frac {x^{ - (2j + 1)} + x^{(2j + 1)}}{2})^2\geq\frac {x^{ - (2j - 1)} + x^{(2j - 1)}}{2}\frac {x^{ - (2j + 3)} + x^{(2j + 3)}}{2}\hspace{1cm}(5)$
$ \Leftrightarrow(\frac {\frac {1}{x} + x}{2})^{8}(\frac {x^{ - (2j + 1)} + x^{(2j + 1)}}{2})^2\geq\frac {x^{ - 2(2j + 1)} + x^{2(2j + 1) + x^{ - 2} + x^{2}}{4}$
$ \Leftrightarrow(\frac {\frac {1}{x} + x}{2})^{8}(\frac {x^{ - (2j + 1)} + x^{(2j + 1)}}{2})^2\geq(\frac {x^{ - (2j + 1)} + x^{(2j + 1)}}{2})^2 + (\frac {x^{ - 1} + x}{2})^2$
$ \Leftrightarrow((\frac {\frac {1}{x} + x}{2})^{8} - 1)(\frac {x^{ - (2j + 1)} + x^{(2j + 1)}}{2})^2\geq(\frac {x^{ - 1} + x}{2})^2$
But it is easy to prove that $ (\frac {\frac {1}{x} + x}{2})^{8} - 1\geq(\frac {x^{ - 2} + x^2}{2})^{2} - 1 = (\frac {x^{ - 1} + x}{2})^2$
$ \Leftrightarrow(\frac {x^{ - (2j + 1)} + x^{(2j + 1)}}{2})^2\geq1$
The previous inequality is easy to prove by $ MA%Error. "geqMG" is a bad command.
$. Then the inequality $ (4)$ holds for every natural number $ j$.

Now we can to prove the induction multiplicying $ (4)$ (when $ j = k$) and $ (5)$:
$ (\frac {\frac {1}{x} + x}{2})^{8(i + 1)}\frac {x^{ - (2i + 1)} + x^{(2i + 1)}}{2}\geq\frac {x^{ - (2i + 3)} + x^{(2i + 3)}}{2}\hspace{1cm}(3)$
Then, by the induction, we prove that $ (4)$ holds for every natural number $ i$.

Now, we are ready to prove $ (3)$ by induction in $ n$.
It is easy to see that $ (3)$ holds for $ n = 1$.
Suposse that $ (3)$ holds for $ n = p - 1$:
$ (\frac {\frac {1}{x} + x}{2})^{(2p - 1)^2}\geq\frac {x^{ - (2p - 1)} + x^{(2p - 1)}}{2}\hspace{1cm}(6)$
Then we can obtein that $ (3)$ holds too for $ n = p$ multiplicying $ (6)$ and $ (4)$ (when $ i = p$):
$ (\frac {\frac {1}{x} + x}{2})^{(2p + 1)^2}\geq\frac {x^{ - (2p + 1)} + x^{(2p + 1)}}{2}$
The equality holds iff $ n = 0$ or $ a = b = 1$. And the problem is solved.

Answer of Reynaldo Gil Pons:
Second Solution:

$ \frac {a^{n \plus{} 1}}{b^n} \plus{} \frac {b^{n \plus{} 1}}{a^n}\geq2$
$ \Leftrightarrow a^{2n \plus{} 1} \plus{} b^{2n \plus{} 1}\geq2a^nb^n$
Let $ \frac {b}{a} \equal{} c$
Then $ a^{(2n \plus{} 1)^2} \plus{} b^{(2n \plus{} 1)^2} \equal{} 2$ implies that:
$ a^{(2n \plus{} 1)^2}(1 \plus{} c^{(2n \plus{} 1)^2}) \equal{} 2\hspace{1cm}(1)$
And:
$ a(1 \plus{} c^{2n \plus{} 1})\geq2c^n\hspace{1cm}(2)$

Replacing a from (1) at (2):
$ (\frac {1 \plus{} c^{2n \plus{} 1}}{2})^{(2n \plus{} 1)^2}\geq\frac {c^{n(2n \plus{} 1)^2} \plus{} c^{(n \plus{} 1)(2n \plus{} 1)^2}}{2}$
Let $ c^{2n \plus{} 1} \equal{} d$. Then we have to prove that:
$ (\frac {1 \plus{} d}{2})^{(2n \plus{} 1)^2}\geq\frac {d^{n(2n \plus{} 1)} \plus{} d^{(n \plus{} 1)(2n \plus{} 1)}}{2}$
$ \Leftrightarrow (\frac {1 \plus{} \frac {1}{d}}{2})^{(2n \plus{} 1)^2}\geq\frac {d^{ \minus{} n(2n \plus{} 1)} \plus{} d^{\minus{}(n \plus{} 1)(2n \plus{} 1)}}{2}$

Then, multipliplying the two previous inequalities:
$ \Leftrightarrow (\frac {2 \plus{} \frac {1}{d} \plus{} d}{4})^{(2n \plus{} 1)^2}\geq\frac {d^{ \minus{} (2n \plus{} 1)} \plus{} d^{(2n \plus{} 1)} \plus{} 2}{4}$
Let $ d \equal{} x^2$:
$ \Leftrightarrow (\frac {(\frac {1}{x} \plus{} x)^2}{4})^{(2n \plus{} 1)^2}\geq\frac {(x^{ \minus{} (2n \plus{} 1)} \plus{} x^{(2n \plus{} 1)})^2}{4}$
$ \Leftrightarrow (\frac {\frac {1}{x} \plus{} x}{2})^{(2n \plus{} 1)^2}\geq\frac {x^{ \minus{} (2n \plus{} 1)} \plus{} x^{(2n \plus{} 1)}}{2}\hspace{1cm}(3)$
Suposse that:
Let $ f(k) \equal{} (x^{ \minus{} k} \plus{} x^{k})^{k^{ \minus{} 2}}$

Then the function $ f$ ($ k$ is a natural number) is decreasing:
$ f`(k) \equal{} (x^{ \minus{} k} \plus{} x^k)^{(k)^{ \minus{} 2}}ln(x^{ \minus{} k} \plus{} x^k)( \minus{} 2)k^{ \minus{} 3} \plus{} k^{ \minus{} 2}(x^{ \minus{} k} \plus{} x^k)^{k^{ \minus{} 2} \minus{} 1}(1 \minus{} x^{ \minus{} 2k})k^kln(x) < 0$
Then $ f(1)\geq f(2)\geq f(3)\geq...$
And let $ g(x) \equal{} 2^{ \minus{} x}$, that is a decreasing function.
Then $ h(x) \equal{} f(x)g(x)$ is a decreasing function too.
And the problem is solved again.

[quote="gilcu3"]Answer of Reynaldo Gil Pons:
First Solution:
$ \frac {a^{n \plus{} 1}}{b^n} \plus{} \frac {b^{n \plus{} 1}}{a^n}\geq2$
$ \Leftrightarrow a^{2n \plus{} 1} \plus{} b^{2n \plus{} 1}\geq2a^nb^n$
Let $ \frac {b}{a} \equal{} c$
Then $ a^{(2n \plus{} 1)^2} \plus{} b^{(2n \plus{} 1)^2} \equal{} 2$ implies that:
$ a^{(2n \plus{} 1)^2}(1 \plus{} c^{(2n \plus{} 1)^2}) \equal{} 2\hspace{1cm}(1)$
And:
$ a(1 \plus{} c^{2n \plus{} 1})\geq2c^n\hspace{1cm}(2)$
Replacing a from $ (1)$ at $ (2)$:
$ (\frac {1 \plus{} c^{2n \plus{} 1}}{2})^{(2n \plus{} 1)^2}\geq\frac {c^{n(2n \plus{} 1)^2} \plus{} c^{ \minus{} (n \plus{} 1)(2n \plus{} 1)^2}}{2}$
[/quote]
Explain the last step, please. :wink:

Replacing $ a$ from $ (1)$ at $ (2)$:
$ (\frac {1 \plus{} c^{2n \plus{} 1}}{2})^{(2n \plus{} 1)^2}\geq\frac {c^{n(2n \plus{} 1)^2} \plus{} c^{(n \plus{} 1)(2n \plus{} 1)^2}}{2}$
I raise the inequality $ (2)$ to the power of $ (2n \plus{} 1)^2$, and find the value of $ a$ at $ (1)$. Then the result comes from the replacement.

[quote="gilcu3"]Replacing $ a$ from $ (1)$ at $ (2)$:
$ (\frac {1 \plus{} c^{2n \plus{} 1}}{2})^{(2n \plus{} 1)^2}\geq\frac {c^{n(2n \plus{} 1)^2} \plus{} c^{ \minus{} (n \plus{} 1)(2n \plus{} 1)^2}}{2}$
I raise the inequality $ (2)$ to the power of $ (2n \plus{} 1)^2$, and find the value of $ a$ at $ (1)$. Then the result comes from the replacement.[/quote]
I don't see it. :maybe:

Yes, my prove is similar to Ibn Rochde's. I think there is also a proof using Chebychef. What is your proof, arqady?
 :)

[quote="arqady"][quote="Ibn Rochde"][quote="arqady"]Let $ a > 0$ and $ b > 0$  such that $ a^{(2n \plus{} 1)^2} \plus{} b^{(2n \plus{} 1)^2} \equal{} 2,$ where $ n\in\mathbb N.$ Prove that
\[ \frac {a^{n \plus{} 1}}{b^n} \plus{} \frac {b^{n \plus{} 1}}{a^n}\geq2
\]
[/quote]

we know that  $ a^{(2n \plus{} 1)^2} \plus{} b^{(2n \plus{} 1)^2} > 2(ab)^{2n \plus{} 1}$ ==> $ 1 > ab$ (1)[/quote]
OK. But why $ a^{(2n \plus{} 1)^2} \plus{} b^{(2n \plus{} 1)^2} > 2(ab)^{2n \plus{} 1}$ is true? :wink:[/quote]
With $ n\equal{}0$ it is not true
But with $ n\ge 1$ we have $ (2n\plus{}1)^2>2(2n\plus{}1)$ Then it is obovius to find it since AMGM

[quote="Allnames"][/quote][quote="arqady"][quote="Ibn Rochde"][quote="arqady"]Let $ a > 0$ and $ b > 0$  such that $ a^{(2n + 1)^2} + b^{(2n + 1)^2} = 2,$ where $ n\in\mathbb N.$ Prove that
\[ \frac {a^{n + 1}}{b^n} + \frac {b^{n + 1}}{a^n}\geq2
\]
[/quote]

we know that  $ a^{(2n + 1)^2} + b^{(2n + 1)^2} > 2(ab)^{2n + 1}$ ==> $ 1 > ab$ (1)[/quote]
OK. But why $ a^{(2n + 1)^2} + b^{(2n + 1)^2} > 2(ab)^{2n + 1}$ is true? :wink:[/quote][quote="Allnames"]
But with $ n\ge 1$ we have $ (2n + 1)^2 > 2(2n + 1)$ Then it is obovius to find it since AMGM[/quote]
Are you sure?

[quote="arqady"][quote="Ibn Rochde"][quote="arqady"]Let $ a > 0$ and $ b > 0$  such that $ a^{(2n \plus{} 1)^2} \plus{} b^{(2n \plus{} 1)^2} \equal{} 2,$ where $ n\in\mathbb N.$ Prove that
\[ \frac {a^{n \plus{} 1}}{b^n} \plus{} \frac {b^{n \plus{} 1}}{a^n}\geq2
\]
[/quote]

we know that  $ a^{(2n \plus{} 1)^2} \plus{} b^{(2n \plus{} 1)^2} > 2(ab)^{2n \plus{} 1}$ ==> $ 1 > ab$ (1)[/quote]
OK. But why $ a^{(2n \plus{} 1)^2} \plus{} b^{(2n \plus{} 1)^2} > 2(ab)^{2n \plus{} 1}$ is true? :wink:[/quote]

i think he used $ a^2\plus{}b^2 \geq 2ab$

[quote="arqady"][quote="Ibn Rochde"][quote="arqady"]Let $ a > 0$ and $ b > 0$  such that $ a^{(2n \plus{} 1)^2} \plus{} b^{(2n \plus{} 1)^2} \equal{} 2,$ where $ n\in\mathbb N.$ Prove that
\[ \frac {a^{n \plus{} 1}}{b^n} \plus{} \frac {b^{n \plus{} 1}}{a^n}\geq2
\]
[/quote]

we know that  $ a^{(2n \plus{} 1)^2} \plus{} b^{(2n \plus{} 1)^2} > 2(ab)^{2n \plus{} 1}$ ==> $ 1 > ab$ (1)[/quote]
OK. But why $ a^{(2n \plus{} 1)^2} \plus{} b^{(2n \plus{} 1)^2} > 2(ab)^{2n \plus{} 1}$ is true? :wink:[/quote]

 :huh:  sorry guys i did a mistake i taught $ (a^{2n \plus{} 1})^2$

i hope this is right $ a^{4n^2 \plus{} 4n \plus{} 1} \plus{} b^{4n^2 \plus{} 4n \plus{} 1} > \frac {1} {2} (a \plus{} b)(a^{4n^2 \plus{} 4n} \plus{} b^{4n^2 \plus{} 4n})$ (chebyshev)

so $ 4 > (a \plus{} b)(a^{4n^2 \plus{} 4n} \plus{} b^{4n^2 \plus{} 4n} )$ ==> $ 4 > 2(ab)^{\frac {1} {2} } *2ab^{2n^2 \plus{} 2n}$

so $ 1 > ab$  :|  (and maybe what allnames said is true  because $ max(a.b) >1$ so we have to try to  prouve his idea )

[quote="arqady"][quote="Allnames"]
OK. But why $ a^{(2n \plus{} 1)^2} \plus{} b^{(2n \plus{} 1)^2} > 2(ab)^{2n \plus{} 1}$ is true? :wink:[/quote]
But with $ n\ge 1$ we have $ (2n \plus{} 1)^2 > 2(2n \plus{} 1)$ Then it is obovius to find it since AMGM[/quote]
Are you sure?[/quote]
I cant understand your question $ n\ge 1$ then $ 2n \plus{} 1 > 2$ hence $ (2n \plus{} 1)^2 > 2(2n \plus{} 1)$ and we have that since $ x^2 \plus{} y^2\ge 2xy$ ($ a,b > 0$)

[quote="Allnames"]
I cant understand your question $ n\ge 1$ then $ 2n \plus{} 1 > 2$ hence $ (2n \plus{} 1)^2 > 2(2n \plus{} 1)$ and we have that since $ a^2 \plus{} b^2\ge 2ab$ ($ a,b > 0$)[/quote]
no it not $ a^2$ but it $ a^{4n^2\plus{}4n\plus{}1}$ this is the problem

[quote="Allnames"][quote="arqady"]
I cant understand your question $ n\ge 1$ then $ 2n \plus{} 1 > 2$ hence $ (2n \plus{} 1)^2 > 2(2n \plus{} 1)$ and we have that since $ a^2 \plus{} b^2\ge 2ab$ ($ a,b > 0$)[/quote][/quote]
Allnames, my question is why $ a^{(2n \plus{} 1)^2} \plus{} b^{(2n \plus{} 1)^2} \geq 2(ab)^{2n \plus{} 1}$ is true?  Explain it. Write a full reasoning.  :wink:

[quote="arqady"][quote="Allnames"]
Allnames, my question is why $ a^{(2n \plus{} 1)^2} \plus{} b^{(2n \plus{} 1)^2} \geq 2(ab)^{2n \plus{} 1}$ is true?  Explain it. Write a full reasoning.  :wink:[/quote][/quote]
Oh,sorry arqady,i am wrong,in the trulth,i only read your question to Ibn Rochde  (and not problem(because i think i cant solve it  :blush: ),and then,i forgot $ a,b$ are not greater than $ 1$ ,Really sorry, :blush: .

[quote="Math pro"]http://www.mathlinks.ro/viewtopic.php?t=218216&start=20
see my solution here![/quote]
 :rotfl:  :rotfl:  :rotfl:
Are you sure,with me (a stupid student),i cant see what are similar features in your solution(and i dont think it is a full solution because at least.can you write your derivative,) Arqady's is a hard problem!and Having mistakes is popular
I think,like zaizai,you have never solved before you post,any your post are some comment ,
[hide]see here http://www.mathlinks.ro/search.php?search_author=Math+pro[/hide]

[quote="gilcu3"]$ \Leftrightarrow (\frac {\frac {1}{x} \plus{} x}{2})^{(2n \plus{} 1)^2}\geq\frac {x^{ \minus{} (2n \plus{} 1)} \plus{} x^{(2n \plus{} 1)}}{2}\hspace{1cm}(3)$.[/quote]
We can reformulate and generalize this inequality as follows:

Let $ x,y$ be positive real numbers such that $ xy\equal{}1$. If $ k$ is a natural number, then

$ (\frac {x\plus{}y}{2})^{k^2} \ge \frac {x^k\plus{}y^k}{2}$.

Moreover, I think the following generalization holds:

Let $ x_1,x_2,...,x_n$ be positive real numbers such that $ x_1x_2...x_n\equal{}1$. If $ k$ is a natural number, then

$ (\frac {x_1\plus{}x_2\plus{}...\plus{}x_n}{n})^{\frac{k(kn\minus{}n\plus{}2)}{2}} \ge \frac {x_1^k\plus{}x_2^k\plus{}...\plus{}x_n^k}{n}$.


AMC
AIME
AMC 12
AMC 10
geometry
function
probability

Here are a few questions I have related to the AMC 10 and AMC 12:

1) Are the qualification scores the same for the AIME on both the AMC 10 and the AMC 12? i.e. Would it be required to score a 120 on the AMC 12 as in the AMC 10 even though it is relatively harder?

2) Was there a 2000AMC10B? Was there a 2001AMC10B?

3) Also, does the 'Math Club Book' for $\$$22 have problems from the AMC 10's or are they different?

Thank you very much.

Anirudh

1. No, of course not. A 100 on the AMC12 or top 5% is needed to qualify. A 120 on the AMC10 or top 1% is needed to qualify.

2. It's possible that they didn't start having two test dates until 2002.

There was only one test for 2000 and 2001.

Yay. Then I got all of them free of cost. :)

My coach has the Math Club Book.  I think the problems were original, but I'm not sure.  However, they were grouped by subject, so that made it a really good resource.  (I skipped the geometry ones  :wink: )

EDIT: I was wrong, I guess they are from old AMC 10s.  If I practiced more I might have known that... :oops:

[quote="mysmartmouth"]My coach has the Math Club Book.  I think the problems were original, but I'm not sure.  However, they were grouped by subject, so that made it a really good resource.  (I skipped the geometry one  :wink: )[/quote]

Well, my coach has the AMC 8 one and he copied the entire book for us. Every single problem was from a past AMC 8.

EDIT: How many pages does it have? The AMC 10 one?

anirudh. Where can you see 2004 to 2006 grouped in a category?

and did you buy those quetsions from 2002 and 2003 ? :blush:


[quote="nguoigo"]For $ t,n \in Z+$ such that $ t^{n}+\frac{1}{t^{n}}\in Z. Prothat: t^{n}+t^{2}n+t-n=2$[/quote]
We have $ \frac{1}{t^{n}}\in\mathbb N$, so $ t=1$ and we're done! :P

If so, $ t^{n}$ and $ \frac{1}{t^{n}}$ are rational roots of integer equation $ X^{2}-(t^{n}+\frac{1}{t^{n}})X+1=0$, therefore , they are integer and we're done , too :P

[quote="N.T.TUAN"]If so, $ t^{n}$ and $ \frac{1}{t^{n}}$ are rational roots of integer equation $ X^{2}-(t^{n}+\frac{1}{t^{n}})X+1=0$...[/quote]

Obviously it is in $ \mathbb{Q}+$ since $ t \in \mathbb{Q}+$.  But, why is $ t^{n}+1/t^{n}\in \mathbb{Z}$?  Thanks.

The hypotesis is that $ t^{n}+\frac 1{t^{n}}\in \mathbb{Z}$.

Oh!! woops, so $ t\in \mathbb{Q}+$ and $ t^{n}+1/t^{n}\in \mathbb{Z}$.  Thanks!!


[hide]
Dot product: $ a.b \equal{} |a|.|b| \cos{\theta}$

Now consider a unit circle with points $ P(\cos(x), \sin(x))$ and $ Q(\cos(y), \sin(y))$ where $ x>y$

In vector form
$ \vec{OP}\equal{}\cos(x)i \plus{} \sin(x)j$ and $ |\vec{OP}| \equal{} 1$
$ \vec{OQ} \equal{}\cos(y)i\plus{} \sin(y)j$ and $ |\vec{OQ}| \equal{} 1$

So the ange between the two vectors is
$ \vec{OP}.\vec{OQ}\equal{} |\vec{OP}| .|\vec{OQ}| \cos{x\minus{}y}$

$ \cos(x) \cos(y) \plus{} \sin(x) \sin(y) \equal{} \cos(x\minus{}y)$
Q.E.D.

$ 50^{th}$ POST!! :)
[/hide]


let P be an interior point in regular equilateral triangle ABC of side length 1
show PA*PB+PA*PC+PB*PC>=1

Okay.. lets take the general case of this problem!!! :D

Lets denote PA=x, PB=y, PC=z, BC=a, CA=b, AC=c.
Now we have to prove that (xy+yx+zx)(x+y+z)>=a^2x+b^y+c^z.

[b]Solution [/b]: we regroup and have x(y^2+z^2-a^2)+y(x^2+z^2-b^2)+z(x^2+y^2-c^2)+3xyz>=o which is written now like so:
sum ((y^2+z^2-a^2)/(2yz)) +3/2>=0 or:
cos(BPC)+cos(CPA)+cos(APB)+3/2>=0  [b] (1) [/b].
lets denote (for easyer computation) m(BPC)=alpha, m(CPA)=betha, m(APB)=gamma, alpha+betha+gamma=2* \pi
so (1) becomes cos(alpha)+cos(betha)+cos(gamma)+3/2>=0 which is:
2cos((alpha+betha)/2)cos((alpha-betha)/2)+cos(alpha+betha)+1+1/2>=0
2cos((alpha+betha)/2)[cos((alpha-betha)/2)+cos((alpha+betha)/2)]+1/2>=0

4cos^2((alpha+betha)/2)+4cos((alpha+betha)/2)cos((alpha-betha)/2)+cos^2((alpha-betha)/2)+sin^2((alpha-betha)/2)>=0 which is :

[2cos(gamma/2)-cos((alpha-betha)/2)]^2+sin^2[(alpha-betha)/2]>=0
relation which is obius true!! :D
we will have equality iff sin[(alpha-betha)/2]=0 and cos (gamma/2)=1/2 because alpha-betha is (-pi, pi) and gamma/2 is (o,pi) we have that alpha=betha=gamma=2pi/3 .

and for your case we consider a=b=c=1 and it's done! :D
cheers!!!


It is only week to go, and the IMO'2009 webpages at
http://www.imo2009.de/imo/
are still in VERY bad shape..

Even the most basic informations on the contest are missing:
http://www.imo2009.de/imo/index.php?option=com_content&view=article&id=25&Itemid=33&lang=en
For instance, are students allowed to use rulers, color-pens, etc?

The only pages that really seem to be complete are the pages with pictures of the organizers:
http://www.imo2009.de/imo/index.php?option=com_content&view=article&id=22&Itemid=31&lang=en
http://www.imo2009.de/imo/index.php?option=com_content&view=article&id=37&Itemid=44&lang=en
http://www.imo2009.de/imo/index.php?option=com_content&view=article&id=24&Itemid=32&lang=en
Admittedly, these picture pages indeed are very nice indeed.

Uhm... a guide should be able to more than just survive in a German city. I don't think translating those infos into English would add any substantial candidates ;)

EDIT: regulations about taking compasses and similar stuff into the test are there:
http://www.imo2009.de/imo/index.php?option=com_content&view=article&id=14&Itemid=6&lang=en

  darij

[quote="orl"]Look at [url=http://www.imo2009.de/imo/index.php?option=com_content&view=article&id=24&Itemid=32&lang=en]this page[/url] you linked to in your previous post. You can see junior organizer and it support [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=17461]Naphthalin[/url] right here on the forum!  :wink:

I also appreciate how they cannot be bothered to translate the information for potential [url=http://www.imo2009.de/imo/index.php?option=com_content&view=article&id=17%3Aguides&catid=1%3Ainhalte&lang=de&Itemid=32]guides[/url] into English.[/quote]

just 2 simple things:

1. if you visit the website as a contestant, you won't find this guide page.
2. as darij already said: the guides should understand german well, so if they don't understand this they can't hel really much ;)

Naphthalin

Yeah but this way the contestants do not know what they can expect from their guides!  :wink: How about a news section on the main webpage to keep people updated?


inequalities
number theory
greatest common divisor
modular arithmetic
number theory unsolved

For any finite set $Z$ we define $\gcd(Z)$ as the greatest common divider of all elements of this set, $\gcd(\{a\})=a$, $\gcd\left(\emptyset\right)=1$.
Let $n$ be a positive integer. Prove that for any numbers $N_{0},\ldots,N_{n-1}\in\mathbb{N}^{+}$ we have: \[\prod_{Z\subset\{0,\ldots,n-1\}\atop |Z|\equiv 1\pmod{2}}\bigl(\gcd\left(\{N_{i}: i\in Z\}\right)\bigr)^{|Z|}\geqslant \prod_{Z\subset\{0,\ldots,n-1\}\atop |Z|\equiv 0\pmod{2}}\bigl(\gcd\left(\{N_{i}: i\in Z\}\right)\bigr)^{|Z|}.\]

Is it to hard? Or maybe I didn't post it in apropriate forum?
Edit: there are very many interesting generalisations.

Moved from Intermediate, but not sure of the exact difficulty. Moderators of this forum, free feel to move it.

It is unsolved on this forum for more than one month. :(  :(  :(
Don't be afraid of the form of this inequality. It seems hard, but [b]it is really simple[/b]. If someone doesn't understand the notation, I will explain it.  :)

Take any prime $p$. Suppose that $p^{q_{i}}\| N_{i}$. Now WLOG assume that $q_{0}\geq q_{1}\geq \ldots \geq q_{n-1}$. It shouldn't be too hard, I think, but it's pretty tedious.


[quote="nurul_ashida"]3. Let $a,b,c$ be the sides  of a triangle ABC. Prove that
$\sqrt{a+b-c}+\sqrt{b+c-a}+\sqrt{c+a-b}\ge\sqrt{a}+\sqrt{b}+\sqrt{c}$

4. Let $\frac{p}{q}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...-\frac{1}{1318}+\frac{1}{1319}$ with $(p,q)=1$ . Show that $p$ divides $1979$.

5.  If integers a,b satisfy $a=a^{2}+b^{2}-8b-2ab+16$. Show that $a^{2}$ is a perfect square.

6. Let $x_{1},x_{2},..,x_{n}$ be positive real numbers and let $y_{1}, y_{2},...,y_{n}$ be their permutation. Prove that

$\frac{(x_{1})^{2}}{y_{1}}+\frac{(x_{2})^{2}}{y_{2}}+...+\frac{(x_{n})^{2}}{y_{n}}\ge x_{1}+x_{2}+...+x_{n}$.

7. Show that $3^{21}-2^{24}-6^{8}-1$ is divisible by 1930.

8. Let $f(x)=x^{2}/1+x^{2}$, $x\in\mathbf{N}$. Show that $f(\frac{1}{2000})+f(\frac{2}{2000})+...+f(\frac{1999}{2000})+f(\frac{2000}{2000})+f(\frac{2000}{1999})+...+f(\frac{2000}{1})=\sqrt{1999}$.[/quote]
Can this be moved to the appropriate forum.
There must be typos.
There was a missing square root in 3.
Number 5 is trivial, unless it was written wrong.
I think there is a typo here: $f(x)=x^{2}/1+x^{2}$

This should be moved the the intermediate forum.

[hide="number 3:"]

$2s=a+b+c$ so $a+b-c=2(s-c)$ and like wise for the other cases
($s$ is the semiperimeter of $ABC$)

now we have

$\frac{\sqrt{2}}{3}(\sqrt{s-a}+\sqrt{s-b}+\sqrt{s-c}) \geq \frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{3}$

now we apply AMGM to both sides and show that GM for the LHS is larger than that of the RHS thus the original LHS must be greater than the original RHS.
doing so we have

${{\sqrt{2}(\sqrt[3]{\sqrt{(s-a)(s-b)(s-c)}}})\geq \sqrt[3]{\sqrt{abc}}}$

we know that $(s-a)>a$ by the triangle inequality so $(s-a)(s-b)(s-c)>abc$ and we are done
[/hide]

*please let me know if i made a mistake somewhere  :oops: *

[hide="number 6:"]

just use cauchy's.

multiply both sides by $y_{1}+y_{2}+\dots+y_{n}$ and we get that the

$LHS\geq(x_{1}+x_{2}+\dots+x_{n})^{2}$ and the $RHS$ is equal to $(x_{1}+x_{2}+\dots+x_{n})^{2}$ since $y_{1},y_{2},\dots,y_{n}$ is a permutaion of the sequence of $x_{i}$
[/hide]


group theory
abstract algebra
Galois Theory
superior algebra
superior algebra unsolved

We have a finite algebraic extension L over the rationals, with [L:Q] = 4.  Is it always true that there exists a subfield K with Q < K < L with [L:K] = 2?

My guess would be no, since there exist quartics with A4 or S4 as the Galois group, but I'm having trouble thinking about this precisely.

If $ \alpha$ is the root of a quartic with Galois group $ A_4$, and $ L\equal{}\mathbb{Q}(\alpha)$, then such a field $ K$ would be an intermediate field $ \mathbb{Q}\subset K \subset M$, where $ M$ is the splitting field of $ m_\alpha$.

This would correspond to a subgroup of $ A_4$ of index $ 2$, which does not exist.

Ah, and for the same reason, we can say the same thing about $ S_4$.  That is, no subgroup of index $ 4$ is contained in a subgroup of index $ 2$.

In summary: if $ \alpha$ is the root of an irreducible quartic, then there exists some intermediate field $ \mathbb{Q}\subsetneq K\subsetneq \mathbb{Q}(\alpha)$ iff the Galois group of $ \alpha$ is not either $ S_4$ or $ A_4$.

(for the other possible Galois groups, namely $ D_8$, $ V_4$, and $ Z_4$, it is easy to check that every subgroup of index $ 4$ is contained in a subgroup of index $ 2$)

How much of this has to change over a field that is not perfect?

Great!  That's what I thought.  Thanks for your help.


How many distinct three-digit positive integers have only odd digits?

To solve this problem we use the Fundamental Counting Principle, and multiply the number of choices for the digit of the hundreds, tens, and units digit of the three digit number. A quick breakdown for each digit:

Hundreds Digit:
This must be odd, so it must be amongst 1, 3, 5, 7, or 9. This allows for [b]5[/b] choices

Tens Digit:
Similar Case. 5 choices.

Units Digit:
Similar Case. 5 choices.

Thus we would have $ 5(5)(5)\equal{}5^3\equal{} 125$

I thought when it said "distinct" that the digits had to be distinct (so I thought the answer should be $5\cdot 4\cdot 3=60$. This problem might need to be reworded better.


When i type some words in between dollar signs is there a way I can make them not affected so they appear like normal text?

Use \text{...} but please use very sparingly on the forum and only when essential; it saves processing on the server. In documents LaTeX is quite happy to skip in and out of math mode, so again only use where you have to.

\sin x=0.5 \text{ and we know that } 0 \leq x \leq \pi
gives
$\sin x=0.5 \text{ and we know that }0 \leq x \leq \pi$
but is better written in 2 pieces as
$\sin x=0.5$ and we know that $0 \leq x \leq \pi$

Note that we have to use \text{ and we know that } rather than \text{and we know that} as spaces are eaten up outside \text so you'd get
$\sin x=0.5 \text{and we know that}0 \leq x \leq \pi$

oh ok thanks i was going to use it for texnic center anyway.


Can someone define the meaning of the pigeon hole theorem in an elimentary way . At the same time, pls show me some basic question using the pigeon hole theorem . Thanks!

Say you have more boxes then pigeons. You want to put all the pigeons into the boxes so that no box in empty. The pigeon hole says that there is at least one box with at least two pigeons.

A concrete example:
3 boxes.
5 pigeons.
Put all the pigeons in the boxes. Each box must contain a pigeon.
Pigeon 1 goes in Box 1. Pigeon 2 goes in Box 2. Pigeon 3 goes in Box 3.
Where does Pigeon 4 and Pigeon 5 go? In some of the three boxes.
So we know that in at least one box, there are at least two pigeons.

When using Pigeonhole theorem, the hardest is to identify what the pigeons are and what the boxes are.
Try yourself on these question:

(i). A pine tree can have at most 700.000 needles.
Show that there exists at least two pine trees with the same number of needles in the world.

(ii). If 9 people are seated in a row of 12 chairs, then some consecutive set of 3 chairs are filled with people.

About that seat question. so 3 other people will have to sit in 3 of the 12 chairs?

[quote="dran"]Say you have more boxes then pigeons. You want to put all the pigeons into the boxes so that no box in empty. The pigeon hole says that there is at least one box with at least two pigeons.
[/quote]

That should be more pigeons than boxes.

by the way, proof questions don't really belong in classroom math


OK. So here it goes again.

Gale buys 12 bags of mulch. Each bag contains 2 cubic feet of mulch. What part of a cubic yard are the twelve bags of mulch in total? Express your answer as a common fraction.

Answer: 8/9

My answer:

$ \frac {12 \times 2}{36^3} = \frac{1}{1944}$

What's wrong with my answer?

ok, and here is my other question (i'm sorry i have so much):
A piece of lumber 5 1/4 feet long is cut into 9 pieces. The waste per cut is 1/8 inch. How many inches of lumber are wasted?

Answer: 1

My answer: 9/8.

What's wrong with this one?
Thanks,
LHM

I havent taken a look at the first one, but in the second one, there are only 8 cuts to seperate them into 9 pieces, not 9. (so its 8/8 or 1)

For number 1, it should be
$ \frac{12\times2}{3^3}$, not $ 36^3$, since there are 3 feet in a yard.
So the answer is
$ \frac{24}{27}\equal{}\frac{8}{9}$


I think I saw this sequence first in 2nd grade, and I have no clue how I remember it.  I won't give it away, but I usually see it started with a 0 as the first term...

[hide]The sequence is all of the numbers that have an even number of letters when written out.[/hide]

wow lol at first i thought it had something to do with adding a pattern of 1s and 4s with the exeption of 9 to 11 which was 2.  wow i was really off!!! buts it is an interesting sequence.


trigonometry
calculus
function
induction
derivative
geometry
geometric transformation

The people in Calc at the high school are all talking about $ \minus{}e^{i\pi}\equal{}1$

Well, I am a 7th grader taking Alg II Honors, and I want a proof as to why this is true. How did anyone derive this??? It seems like someone was pushing random buttons on their TI-84 and found this out.

If you know, please try to make the proof as simple as possible. (I understand if Calc is involved.)

Do you know Euler's formula ; $ e^{ix}\equal{}\cos x\plus{}i\sin x$?

In the most straightforward sense, this is not a statement to be proved so much as a consequence of a definition.  In particular, we [b]define[/b] the exponential

$ e^x \equal{} \sum_{k \equal{} 0}^{\infty} \frac {x^k}{k!}$

for all complex numbers $ x$.  Euler's formula is then a direct consequence (because we define $ \cos x, \sin x$ in terms of $ e^x$ and then prove they have their familiar properties), as is this identity.

The key problem here is that if you have only seen $ e^x$ in real contexts, there is no clear way to define it for complex values of $ x$.  The tools for doing so belong to the realm of [url=http://en.wikipedia.org/wiki/Complex_analysis]complex analysis[/url].

It's impossible to prove this without using calculus, because it requires calculus to even define the objects involved.  Of course, it's also true that you need calculus to rigorously define trig functions, so I'm sort of dodging the question.  The best answer I can give, is that from an Alg II point of view (assuming you also know some trig and complex numbers), it's essentially a definition:

When $ x$ is a real number, $ e^{ix}$ may be *defined* by Euler's formula $ e^{ix} \equal{} \cos x \plus{} i\sin x$.  Now just plug in $ x \equal{} \pi$.

So the interesting question now becomes:  Why on earth is this definition a reasonable one?  One part of the answer is that this notation obeys the usual rules of exponentiation.  For example, for any natural number $ n$, $ (e^{ix})^n \equal{} e^{inx}$.  This translates into De Moivre's formula:

$ (\cos x \plus{} i\sin x)^n \equal{} \cos nx \plus{} i\sin nx$

I would argue that (from Alg II point of view), $ e^{i\pi} \equal{} \minus{} 1$ is just some fun notation, but the mathematical meat is in De Moivre's formula, which can be proved by induction.  (On the other hand, once you have the tools of calculus, you can establish Euler's formula as an honest-to-goodness theorem rather than definition, and then you could *prove* De Moivre without using the ugly induction argument.)

My way of phrasing the "mathematical meat," as you put it:

If we define $ e^z$ by the power series as in t0rajir0u's post, then we can prove (using the power series and nothing else) that $ e^{z\plus{}w}\equal{}e^ze^w.$ From this we can derive all of the other usual laws of exponents - which include DeMoivre's theorem as a special case. That justifies us calling it an exponential.

yenlee gave the more elementary idea of just defining $ e^{ix}\equal{}\cos x\plus{}i\sin x.$ OK, then start there and prove that $ e^{i(x\plus{}y)}\equal{}e^{ix}e^{iy}.$ (Some classic trigonometric identities that every student should be familiar with.) It's all part of the same package of ideas.

$ f(x)\equal{}\cos x\plus{}i\sin x\Longrightarrow f'(x)\equal{}if(x)\Longrightarrow f(x)\equal{}f(0)e^{ix}\equal{}e^{ix}$.

Of course, until you've developed the theory of complex differentiation, it's not clear at all that the derivative of $ f(x)$ is well-defined.

Yes it is, he's just differentiating it in a real variable, which is no different than differentiating a function $ \mathbb{R} \to \mathbb{R}^2$. Although I'm not sure about the last implication, but it's a good motivation for defining $ e^{ix}$ that way.

The actual "complex derivate" is quite a different thing. Formally it looks the same, but basically a complex differentiable function is a differentiable map $ \mathbb{R}^2 \to \mathbb{R}^2$ such that the Jacobian is a rotation in every point.

Thanks, guys. I don't understand most of it, but I expected that. My math teacher said the proof was pretty intense.

Consider, why $ e^{ix}$ can be described by $ \cos x,\ \sin x$?

Keyword: Wronskian

Regarding Merryfield's post:  I don't know why I emphasized the formula $ (e^{ix})^n=e^{inx}$.  The formula $ e^{i(x+y)=e^{ix}e^{iy}}$ is clearly more fundamental.

Recapping my previous post, I contend that a good way to understand the formula, armed only with knowledge of trig and complex numbers, is to *define* $ e^{ix}=\cos x+i\sin x$, and then observe that this notation is consistent with all the desirable properties of exponentiation, and therefore $ e^{i\pi}=-1$ is a reasonable thing to say (as opposed to arbitrarily defining $ e^{i\pi}=-1$ for no good reason).  As pointed out by others, there are many ways to understand the formula when armed with a calculus background.

Kalle: You are mostly correct, although usually one uses the words "holomorphic" or "analytic" rather than "complex differentiable," and you don't want to say that the Jacobian is a rotation at every point, but rather that the Jacobian is a *scalar multiple* of a rotation at every point.  (Rotations preserve lengths, angles, and orientation.  Scalar multiples of rotations only preserve angles and orientations.  In terms of plane geometry, they're similarities but not congruences.)  Exercise: What can you say about a function whose Jacobian is an honest-to-goodness rotation at every point?  (It's a much stronger condition than holomorphic.)

You don't have to correct me yenlee.  :P I know a lot of complex analysis (I mean as an undergrad student. I've taken corresponding to a full semester of courses in complex analysis, factorization theory and hardy spaces). I was just trying to quickly describe the fundamental difference between the two notions of derivatives.

I also like to inform beginners of the viewpoint that it's just a differentiable map whose jacobian is a rotation (times a dilation). Usually I think the C-R equations are observed as if it was some kind of coincidental relationship, with two somewhat weird calculations. Instead, I like to make a point that if a function is complex differentiable in a point $ z$ with derivative $ a \plus{} ib$ it means the function can be locally approximated as $ f(z\plus{}w) \minus{} f(z) \equal{} (a\plus{}ib)w$, and the action of multiplying by $ a\plus{}ib$ is exactly that of a rotation (times a dilation).

That's exactly the point that a student lecturer made during MIT's Campus Preview Weekend (hmm...).  I thought it was a great intuition for why the notion of complex differentiability is much stronger than the notion of real differentiability, and yes, it certainly puts the Cauchy-Riemann equations within the proper context.  It's the main thing I took away from the lecture :)

Kalle, I didn't mean to imply that you don't know what you're talking about; I made the correction because I wouldn't want other readers to get the wrong idea.

Incidentally, I very much like the way you described complex differentiation in your latest post.  This point is lost in most introductory calculus classes, but the fundamental idea of differentiation is that of linear approximation.  So for example, a real-valued (or real vector-valued) function of n real variables is differentiable at a point iff it can be approximated near that point by a real (affine-)linear function (i.e. a constant plus a real linear transformation).  If you think in this way, it is evident that the "right" definition of "complex differentiability" at a point, for a complex-valued (or complex vector-valued) function of n complex variables, is that the function should be approximated near that point by a complex (affine-)linear function (i.e. a constant plus a complex linear transformation).

One may regard it as a low-dimensional coincidence that if you identify $ \mathbb{C}$ with the $ \mathbb{R}^2$, then complex linear transformations from $ \mathbb{C}$ to $ \mathbb{C}$ are precisely scalar multiples of rotations of $ \mathbb{R}^2$.  (A coincidence in the sense that complex linear transformations from $ \mathbb{C}^n$ to $ \mathbb{C}^n$ are not, in general, scalar multiples of rotations of $ \mathbb{R}^{2n}$.)

Sorry for getting so far away from the original topic of this thread.


function
number theory unsolved
number theory

Tank you ZetaX,
et voila:

Let p an odd prime and let $f_p$ the following function:

$f_p$:N$\longrightarrow$N, where N is the set of nonnegative integers defined by $f_p$(n)=$\frac{p^n-1}{2}$.

I denote $f_p^{(s)}$(n) = $f_p$($f_p$(..(n))), s times.

Exist p and n such that n divides $f_p$(n), $f_p$(n) devides $f_p$($f_p$(n)),... $f_p ^{(s)}$(n) divides $f_p^{(s+1)}$(n), for all s?

bye, Uber

Well, at the moment it's as good as unreadable...
Please read http://www.mathlinks.ro/Forum/viewtopic.php?t=5261 and repair it :)
Especially the header is not necessary.


What are your suggestions for a good read in thermodynamics involving minimum grad level stuff and understandable by a grade 11 student(for eg, it shouldgive reasons for assumptions and the like)??

[url=http://www.ciao.co.uk/Physical_Chemistry_by_P_W_Atkins__21396]this one[/url] is nice

Thanks I'll try to get it!(And where exactly can u get in south india??)


Let $ (u_n) $ be defined : $ u_1=2 , u_2=7 $
                                        $ u_{n+2}=[\frac{u_{n+1}^2}{u_n}+\frac{1}{2}] $
  a) Find the general formula of $ u_n $
  b) Prove that : $ |u_n^2-u_{n-1}u_{n+1}|=2^{n-2} $
  $ [x] $ is the integer part of $ x $

This question was posted (with an outline of a solution) [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=48609]here[/url].


trigonometry
calculus
calculus computations

Ok so I post a question similar to this one on another thread, and the guys help me a lot with that. but I really still confused on how to go on this one

[img]http://img39.imageshack.us/img39/562/85487490.jpg[/img]

$ \sin x  \equal{} \sum_{k\equal{}0}^n \frac{(\minus{}1)^k\cos(0)}{2k\plus{}1!}x^{2k\plus{}1} \plus{} \frac{(\minus{}1)^{n\plus{}1}\cos(c)}{2n\plus{}3!}x^{2n\plus{}3}$

$ |\sin x \minus{}  \sum_{k\equal{}0}^n \frac{(\minus{}1)^k\cos(0)}{2k\plus{}1!}x^{2k\plus{}1}| \le |\frac{\cos(c)}{2n\plus{}3!}x^{2n\plus{}3}| \le \frac{1}{2n\plus{}3!}$

so for $ 2n\plus{}3! \ge 10^5$ so $ 8<2n\plus{}3 \to \frac{5}{2}<n \to n\ge 3$

so we need a term for 2(3)+1 = 7 degree polynomial.


geometry
3D geometry
polynomial
number theory
number theory unsolved

The problem is actually from the Russian Math Olympiad this year. Using very advanced number theory, I managed to prove that the following equations do not have integer solutions. Unfortunately, I would truly appreciate an elegant or elementary solution to the problem:
Show that the equations 2x^2+1=y^15 and 2x^2-1=y^15 do not have non-trivial solutions.
I tried factoring the expressions y^15+-1 but the problems reduce to Pelle-like equations which should have common solutions. After that I couldn't do anything more.
I hope someone manages to help me solve these problems.

[b]alishkov[/b]: can you show your proof please ?

[b]spx2[/b]: what is the meaning of your post ? If it was supposed to be a solution of a given equation, it is wrong.

It depends how advanced you want to get.  $2x^2+1 = z^3$ can be factorised in the ring $Z[\sqrt{-2}]$, which is a UFD.  Since $z$ must be odd, the factors $1 \pm x\sqrt{-2}$ are coprime and so each is a cube times a unit, but the units are $\pm 1$, both cubes, so we can write $1 + x \sqrt{-2} = (u+v\sqrt{-2})^3 = u(u^2-6v^2) + v(3u^3 - 2v^2)\sqrt{-2}$.  Equating coefficients, $u = \pm 1$, $u^2-6v^2 = 1$, $v=0$ and there is only the trivial solution $x=0, z=1$.  Finally write $y^5 = z$.

For the other equation $2x^2-1 = z^3$ one could factorise in $Z[\sqrt2]$, which is harder to deal with because of the non-trivial unit, or write $2x^2 = z^3 + 1$ and factorise in the Eisenstein integers.

i really dont understand anything of this
i'm at the beggining


geometry
circumcircle
perpendicular bisector
geometry unsolved

Note that in a given triangle $ ABC$ with $ X_{a}$ as the tangency point of the incircle with $ BC$ and with $ Y_{a}$ as the tangency point of the $ A$-excircle with $ BC$, the line $ AY_{a}$ passes through the antipode of $ X_{a}$ wrt. the incircle of $ ABC$. Let this point be $ X_{a}'$. Thus, the construction goes as follows: Draw the perpendicular in $ X_{a}$ to $ X_{a}Y_{a}$ (note that this is the line $ BC$) and intersect it with $ AY_{a}$. This is $ X_{a}'$, and so, the midpoint $ I$ of $ X_{a}X_{a}'$ is the incircle of $ ABC$. On the other hand, the points $ X_{a}$, $ Y_{a}$ are isotomic wrt. the segment $ BC$, and so the midpoint $ M$ of $ X_{a}Y_{a}$ coincides with the midpoint of $ BC$. Now, draw the perpendicular in $ M$ to the line $ X_{a}Y_{a}$ (which is in fact the perpendicular bisector of $ BC$) and denote by $ M'$ the intersection of this line with $ AI$. This new point $ M'$ is the midpoint of the arc $ BC$, not containing the vertex $ A$. But we know that this is the circumcenter of triangle $ IBC$; hence, draw the circle $ \Gamma$ centered at $ M'$ with radius $ M'I$ and intersect it with the line $ X_{a}Y_{a}$ to obtain the vertices $ B$ and $ C$.


\[\iff \frac{16}{\sum a }\leq 2(\sum \frac{1}{a})-(\sum \sqrt[3]{\frac{\frac{1}{a^{3}}+\frac{1}{b^{3}}}{2}})\]
\[\iff (\sum \frac{\sqrt[3]{\frac{a^{3}+b^{3}}{2}}}{ab}+\frac{16}{\sum a }\leq 2(\sum \frac{1}{a})\]

More later sorry for not putting up anything... :(


One of the sides of a triangle is divided into segments of 6 and 8 units by the
point of tangency of the inscribed circle. If the radius of the circle is 4, then
the length of the shortest side is

(A) 12 units (B) 13 units (C) 14 units (D) 15 units (E) 16 units

By a well known theorem, K = rs (this is not difficult to prove, try it).

So: r = 2K/(a+b+c)

let x be the length of the other tangent segment:

Then 4(28+2x) = 2/4sqrt[(28+2x-14)(28+2x-(8+x))(28+2x-(6+x)(28+2x)]

8(28+2x) = sqrt[(28+2x-2*14)(28+2x-2(8+x))(28+2x-2(6+x)(28+2x)]

64(28+2x)^2 = (2x)(12)(26)(28+2x)
--> x=7

So the shortest side has length 13.


geometry
geometric transformation
homothety

In the figure the two small circles are tangent to the big circle internally at $A,D$ and between them externally at point $C$. Line $AC$ meets the outer circle at $B$ . Prove that the segments $CD,DB$ are perpendicular.

[u]  Babis[/u]

[quote="silouan"]It's equivalent to this one
 :wink:
http://www.mathlinks.ro/Forum/viewtopic.php?t=147868[/quote]

Please , fill the topic with your elementary solutions. If you can not find a solution, then you can see the link sent by Silouan :) .Try to manage the solutions in your own way!

  Babis

I have not looked at Silouan's link.
[hide][asy]size(200); pair A, D, Z, O, M, N, C, B, E; A = (-50,50); D = (50,50); Z = (0,length((50,50))); O = circumcenter(A,D,Z); draw(circumcircle(A,D,Z)); draw(A--O--D); M = point(A--O,arctime(A--O,0.25*arclength(A--O))); draw(circle(M,arclength(M--A))); N = point(D--O,arctime(D--O,0.6*arclength(D--O))); draw(circle(N,arclength(N--D))); draw(M--N); C = intersectionpoint(M--N, circle(N,arclength(N--D))); draw(D--C--A); path j = circle(O,arclength(O--D)); pair E=intersectionpoint(C--interp(D,C,50),j); pair B=intersectionpoint(C--interp(A,C,50),j); draw(E--C--B--cycle);draw(B--D); label("A",A,NW);label("D",D,NE);label("C",C,SE);label("E",E,W);label("B",B,SE);label("O",O,SW);label("M",M,SW);label("N",N,SE);dot(A);dot(B);dot(C);dot(D);dot(E);dot(M);dot(N);dot(O);[/asy]
Let $M$ be the center of the small circle with tangent point $A$, let $N$ be the center of the other small circle which has tangent point $D$, let $O$ be the center of the large circle.
Extend $AC$ and $CD$ past $C$ to meet circle $O$ at points $B$ and $E$, respectively. Then, since $MC$ is a radius of circle $M$ and $OB$ is a radius of circle $O$, $\angle AMC = \angle AOB$, so $MC\parallel OB$. Similarly $CN$ is a radius of circle $N$ and $EO$ is a radius of circle $O$ so $EO\parallel CN$. but that means $EO\parallel OB$, so $EB$ is a diameter, and $\angle EDB$ is a right angle, so $ED\perp DB$.[/hide]

[quote="minsoens"]I have not looked at Silouan's link.
[hide]Let $M$ be the center of the small circle with tangent point $A$, let $N$ be the center of the other small circle which has tangent point $D$, let $O$ be the center of the large circle.
Extend $AC$ and $CD$ past $C$ to meet circle $O$ at points $B$ and $E$, respectively. Then, since $MC$ is a radius of circle $M$ and $OB$ is a radius of circle $O$, $\angle AMC = \angle AOB$, so $MC\parallel OB$. Similarly $CN$ is a radius of circle $N$ and $EO$ is a radius of circle $O$ so $EO\parallel CN$. but that means $EO\parallel OB$, so $EB$ is a diameter, and $\angle EDB$ is a right angle, so $ED\perp DB$.[/hide][/quote]

 [u]Yes ! [/u]

 You get a good note  :) for this nice solution ! Now you can read and study Silouan's solution!Of course , if you know transformations , then you can find this way in your own !

Babis

let $M$, $N$ be the centers of the left and right circles, $O$ the center of the larger circle, extend $DC$ to hit the larger circle again at $E$, we know that circle $M$ maps to circle $O$ based on a homothety centered at $A$, so $MC$ // $OM$, likewise, $NC$ // $OE$, but since $M$, $C$, $N$ are collinear [the circles are tangent], $E$, $O$, $B$ are collinear, hence $\angle EDB$ subtends a semicircle, and has a degree measure of $90$, as desired


One of Euler's conjectures was disproved in the $1980$s by three American Mathematicians when they showed that there is a positive integer $n$ such that \[n^{5}= 133^{5}+110^{5}+84^{5}+27^{5}.\] Find the value of $n$.

As 110 and 84 are even, and 27 and 133 are odd, $2|n^5$.

Both 84 and 27 are divisible by 3, and $3|(133+110)|133^5+110^5$, so $3|n^5$.

By FLT $n^5 \equiv n \mod 5$, and by repeated use and addition, $n^5 \equiv 4 \mod 5$ and so $n \equiv 4 \mod 5$.

Both 133 and 84 are divisible by 7, $27^5 \equiv -1 \mod 7$, and $110^5 \equiv -32 \equiv 3 \mod 7$. Hence $n^5 \equiv 2 \mod 7$. We can then get $n \equiv 4 \mod 7$.

By CRT, $n \equiv 144 \mod 210$. By some really weak bounds, we can say that $n=144$ specifically.

[quote="Moonmathpi496"]n is obviously even.Now check the last digits of rhs(mod 10).
Rhs=3+0+4+7==4
so last digit of n will be 4.(bcz 2^5==2)
not we set a lower bound.Observe that rhs==0 mod 3.
k>133.
So possibilities are 144,174.
Now 174^5=(133+41)^5.Using binomial theorem,we may show that it is >rhs.
So 144 is the answer.

Moon[/quote]

Довольно интересно, что если мы захотим искать решения в целых комплексных числах, то можно даже получить формулу решения уравнения такого вида. И число слагаемых можно уменьшить.
Там опечатка сейчас только заметил. Конечно
 $b=p^2+2ps-s^2$

Answer is 144 which we get by applying modular arithmetic

Note that $2\mid n$ and $3\mid n,$ also it is clear that $n\equiv 4\pmod 5.$ So $n>133,$ which means $n=144$ or $n=174,$ it is not hard to disregard the latter.


geometry
parallelogram
geometric transformation
rotation
geometry unsolved

Convex pentagon $ABCDE$ satisfies:
\[BC=CD \quad DE=EA \quad \angle{BCD}=\angle{DEA}=\frac{\pi}{2}. \]
Prove, that using the segments $AC,CE,EB$ one can build the triangle and calculate its angles knowing, that $\angle{ACE}=\alpha$ and $\angle{BEC}=\beta$.

[color=blue]Lemma: ${R_{S}}^{-90}\circ{R_{T}}^{-90}=S$, where $S$ is a halfturn.[/color]

Let ${R_{c}}^{90}(E)=F$

\[{R_{E}}^{-90}\circ{R_{C}}^{-90}(F)=E\]
\[{R_{E}}^{-90}\circ{R_{C}}^{-90}(B)=A\]

From the lemma we have, that segment $FB$ is parallel to $AE$ and they are equal. Thus $FBEA$ is parallelogram $\Rightarrow AF=EB$ and $AF \parallel EB$. We see, that triangle $AFC$ is a triangle we are looking for. From trivial angle chasing we conclude, that its angles are: $\frac{\pi}{2}-\alpha , \frac{\pi}{2}-\beta, \alpha+\beta$

Hey, please explain what $R_{E}^{-90}$ etc. mean.

It's rotation with centre $E$ and angle $-90^{\circ}$


Hi, I saw a two day online seminar for AIME.
Does anyone know how it works? Is it helpful?
Thank you!

Information about the Special AIME Seminar can be found [url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=8]here[/url].

since the seminar uses stuff from AIME A, and since I took AIME A already, then the seminar won't be really useful?

The material in the Special AIME Seminar will have a very large overlap with the AIME A Problem Series.  So, if you took the AIME A class, and paid close attention, then you won't see a great deal of new material in this class.


I have seen an angle set equal to two capital letters with an arc over them. This seems to indicate that the angle is equal to an angle at the centre of a circle which, in radians, is equal to the length of the arc subtended by that angle. Is this correct? I have never seen a definition for this notation.

what you're referring to is, for example, $ m \angle ABC \equal{} m \stackrel\frown{AC}$, where the $ m$ means "measure of." but in practice the $ m$ is sometimes dropped for simplicity, making it look like the angle equals the arc. technically, of course, the two aren't equal per se, so this notation is loose, though acceptable. and yes, the measure of an arc is defined as the measure of the central angle subtending the arc.

Thanks pluristiq. I guess the circle has a radius of $ 1$ so that arc length $ \equal{}r \theta \equal{} \theta$?

Nothing to do with the length of the radius.  The measure of an arc is an angular measure; it's [i]not[/i] the same as the length of the arc. [quote="pluristiq"]the measure of an arc is defined as the measure of the central angle subtending the arc.[/quote]


Given [x]^2 - x[x]+3<=0 for positive real x. Prove that x>=4,75.

x>= n +3/n where n=[x].

but n+1>x>=n so that n+1>(n+3/n) or that n>3. or that n>=4.
in case of n=4 we get x>=4+3/4.
-a


geometry
analytic geometry
vector
geometry proposed

Prove that the centroid $ G$ of a triangle $ ABC$ is the unique point that minimizes the sum of its squared distances from the vertice.

[color=darkblue]You can use the [b]Lagrange[/b]'s identity : if the point $ P$ has the barycentrical coordinates $ (x,y,z)$ , $ x\plus{}y\plus{}z\equal{}1$ w.r.t. $ \triangle ABC$ ,

then for any $ M$ we have $ \boxed {\ \sum x\cdot MA^2\equal{}MP^2\plus{}\sum x\cdot PA^2\ }\ge \sum x\cdot PA^2$ . For example,

$ I\left(\frac {a}{2p},\frac {b}{2p},\frac {c}{2p}\right)$ $ \implies$ $ \sum a\cdot MA^2\equal{}2p\cdot IA^2\plus{}\sum a\cdot IA^2\ge \sum a\cdot IA^2$ . Observe that $ \sum a\cdot IA^2\equal{}abc$ .

For $ G\left(\frac 13,\frac 13,\frac 13\right)$ obtain the proposed problem : $ \boxed {\ \sum MA^2 \equal{} 3\cdot MG^2 \plus{} \sum GA^2\ }\ge \sum GA^2$ . [/color]

Wow, wonderful solutions!
I've proved the property using vectors, but your solutions are better.
Thanks so much!

[quote="yetti"]1. Symmedian point $ G$ of a $ \triangle PQR$ minimizes the sum of squares of distances from the triangle sides,
see [url]http://www.mathlinks.ro/viewtopic.php?t=44102[/url].
2. Symmedian point $ G$ of the $ \triangle PQR$ is the unique point, such that $ G$ is centroid of its pedal $ \triangle ABC,$
see [url]http://www.mathlinks.ro/viewtopic.php?t=209057[/url].[/quote]

Sorry, Yetti, but what do you prove? If triangle $ ABC$ is constant, the anti-pedal triangle $ PQR$ of the point needs not be constant, or do I misunderstand you?

PS. See also http://www.mathlinks.ro/Forum/viewtopic.php?t=15558 for a generalization of the problem.

  darij

This is what I mean: $ G$ is centroid of a $ \triangle ABC.$ $ \triangle PQR$ is antipedal triangle of $ G$ WRT the $ \triangle ABC$ ($ A \in QR, B \in RP, C \in PQ$ and $ GA \perp QR, GB \perp RP, GC \perp PQ$) $ \Longrightarrow$ $ G$ is symmedian point of the antipedal $ \triangle PQR.$ $ X \not\equiv G$ is arbitrary point, $ D, E, F$ are feet of normals from $ X$ to $ QR, RP, PQ.$

$ XA^2 \plus{} XB^2 \plus{} XC^2 \equal{} XD^2 \plus{} AD^2 \plus{} XE^2 \plus{} BE^2 \plus{} XF^2 \plus{} CF^2 >$

$ > XD^2 \plus{} XE^2 \plus{} XF^2 > GA^2 \plus{} GB^2 \plus{} GC^2.$


Let P be an interior point of a triangle ABC and AP, BP, CP meet the sides BC, CA, AB in D, E, F respectively. Show that:
$\frac{AP}{PD}=\frac{AF}{FB}+\frac{AE}{EC}$

It's Van Aubel's theorem...rather well known  :)


Let f(x) = (x+1)(x-1). Clearly the solutions of f(x) = 0 are -1 and 1. Suppose we want to find the solutions of f(x) = 0 using Newton-Raphson iterations.

What map do we iterate when doing Newton-Raphson iterations? (I am not sure what is meant by this question, can someone explains what is a map in this context?)

A map or mapping is a function. The variant language is used in order to get you thinking in certain ways, but in the end, that's all it is.

In general the method is this: We're trying to solve $ f(x)\equal{}0$ for a differentiable function $ f.$ Once we have a guess $ x_n$, we proceed to the next guess by setting $ x_{n\plus{}1}\equal{}x_n\minus{}\frac{f(x_n)}{f'(x_n)}.$

Hence the mapping that is being iterated is $ g(x)\equal{}x\minus{}\frac{f(x)}{f'(x)}.$

In the case of $ f(x)\equal{}x^2\minus{}1,$ we iterate $ g(x)\equal{}x\minus{}\frac{x^2\minus{}1}{2x}\equal{}\frac{x^2\plus{}1}{2x}.$

Note that our first guess must not be equal to zero, and having a first guess be very close to zero would be terribly inefficient. Those features are normal for the method.


arithmetic sequence
number theory unsolved
number theory

Let a(1)=0, a(2)=1, and a(n+2)=a(n+1)+a(n) for all positive integers n. Show that there exists an increasing infinite arithmetic progression of integers, which has no number in common in the sequence {a(n)} n>=0.

We can consider the sequence $ a_n \mod 8$, and we will find that there is no Fibonacci numbers of form $ 8k \plus{} 4$ and $ 8k \plus{} 6$.


Show for all integers $ n \ge 2005$ the following chaine of inequalities:
$ (n\plus{}830)^{2005}<n(n\plus{}1)\dots(n\plus{}2004)<(n\plus{}1002)^{2005}$

Hmm... got the upper bound but I wonder if we can get the lower bound in a similar manner?

We can first apply AM-GM to the $2005$ terms $n, n+1, \cdots, n+2004$:

$$\frac{n+(n+1)+(n+2) \cdots + (n+2004)}{2005} \ge \sqrt[2005]{ n(n+1)(n+2) \cdots (n+2004) }$$

$$\frac{2005n + 1002(2005)}{2005} \ge \sqrt[2005]{ n(n+1)(n+2)\cdots(n+2004)}$$

$$n + 1002 \ge \sqrt[2005]{n(n+1)(n+2)\cdots(n+2004) }.$$

Now we claim that equality is never achievable. For the sake of contradiction, let's assume that $n+1002$ may be reached. Then, by the equality condition of AM-GM, $n=n+1=n+2=\cdots=n+2004$.  There are obviously no solutions to this case. Thus the quantity is strictly always less than $n+1002$.


I Have This Equation And I don't Know How To Solve It:
$x^4-x^3-10x^2+2x+4=0$


[quote="british8985"]NEED ADVICES.....
coz I'm repeatin' because of my English problem
and I don't wanna WASTE TIME
taking AP courses is one of the reason too

I don't mean MATH or CALCULUS....
coz I don't care how hard is it I WILL TAKE FOR SURE
it's an advantage that math12 is seperated from calculus......
and  math is my interest TOO
I mean science courses

do you guys recommend me to take those ??

I've heard ppl who got pretty good marks on gr11 courses but BAD in AP
not BAD...PRETTY BAD

so????what do u think??[/quote]

AP courses are a breeze.

My classes are more difficult to get an A in than to get a 5 on the AP test.
Yes i recommend you take them. It shows the colleges that you challenge yourself.

it really depends on ur teacher. some teachers r harder than others. my ap calc teacher turned 34 math newbs to all 5s on ap calc, except 2 got 4s. he was rather difficult. don't judge by other students, it depends totally on ur ability. i find a lot of ppl like to learn from reading, but that doesn't work, do problems!!! if u think u can take it, go for it, most teachers do mark adjustments to compensate for the difficulty. i heard waterloo is thinking of giving preference to ap students, i dunno if thats true, but i heard talks about it.

i thought ap was a good intro to what u would likely to expect in university per course.

AP courses are (relatively) easy.  For example, fully 40% of students score a 5 on the AP Calculus BC exam.  Bad teaching and whatnot can complicate things, but for the most part it is not difficult to score well on AP exams.

[quote="Singular"]Wow.  I can't read that.[/quote]
Me agree... :(
Plus she's from my high school...
Who's your english teacher??? :?
Ms Grenier??? :?


Points M and N are taken on the hypotenuse of a right triangle ABC so that BC=BM and AC=AN. Prove that the angle MCN is equal to 45 degrees. (tournament of towns 1993)

Which is the hypotenuse? AB, AC, or BC? And please don't answer "yes". ;)

Actually the hypotenuse has to be $AB$. Otherwise one of the points ($M$, $N$) would not be on the hypotenuse (unless hypotenuse is extended).

I have done this proof before, so I'll leave this to other people :)

$\angle BCN = \theta$, $\angle MCN = x$, $\angle MCA = \gamma$
$\theta + x = \alpha + \gamma$
$\gamma = \beta + \theta - x$
$2x = \alpha + \beta$
$x = 45^o$


linear algebra
matrix
combinatorics unsolved
combinatorics

The number $ 1$ or $ \minus{}1$ is written in each cell of a $ 2000 \times 2000$ array. The sum of all the numbers in the array is positive. Show that there are $ 1000$ rows and $ 1000$ columns such that the sum of the numbers at their intersections is at least $ 1000$. [/list]

[quote="Superficial"]The number $ 1$ or $ \minus{} 1$ is written in each cell of a $ 2000 \times 2000$ array. The sum of all the numbers in the array is positive. [/quote]

Do you mean a $ 2000\times 2000$ matrix like a $ 8\times 8$ chessboard ? In that case it is trivial. We have atleast $ 2.10^6\plus{}1$ 1's so as to make the sum of the matrix(your array) positive. Since every cell in the matrix can be uniquely identified by the ordered pair $ (r,c)$ (where $ r$ denotes the row number and $ c$ denotes the column number) we can atleast $ 2.10^6\plus{}1$ ordered pairs giving the sum $ 2.10^6\plus{}1$.

Please Clarify. OR do you mean to say each row and each column in the matrix has a positive sum ?

Srikanth, you cannot just choose ordered pairs like that.

You have to choose $ 1000$ rows and $ 1000$ columns, look at the intersection (so there will be $ 10^6$ cells), and prove that the sum of these numbers is at least $ 1000$.

Anyway, thanks for your interest of the problem  :)

Thanks a lot, Ehsan2004.

Maybe you want to try some more Russian problems I posted in this forum.  :)


How many seconds longer is $ 2\%$ of an hour than $ 30\%$ of a minute?

An hour is 60 minutes, or 3600 seconds. 2% of that is 72. A minute is 60 seconds, of which 30% is 18. Therefore, we have $ 72 \minus{} 18 \equal{} 54$.


[quote="linzk3"]The lines described by 2ax+7y=4 and 3bx-4y=5 are parallel.
Determine the ratio a:b.

i have it down to  -8ax=21ax so would the ratio be -8x:21x
or have i done something wrong or can you go farther..?
please and thanks =][/quote]
The x cancels.

[quote="linzk3"][quote="linzk3"]The lines described by 2ax+7y=4 and 3bx-4y=5 are parallel.
Determine the ratio a:b.

i have it down to  -8ax=21bx so would the ratio be -8x:21x
or have i done something wrong or can you go farther..?
please and thanks =][/quote]

so it would be -8a:21b[/quote]

$-8ax=21bx \implies-8a=21b \implies a: b=\frac{a}{b}=\frac{-21}{8}$


algebra
polynomial
Diophantine Equations

Find all cubic polynomials $x^3 +ax^2 +bx+c$ admitting the rational numbers $a$, $b$ and $c$ as roots.

De Viete gives us $ 2a\plus{}b\plus{}c \equal{} 0$, $ c \equal{}\minus{}abc$ and $ b \equal{} ab\plus{}bc\plus{}ca$.

[list][*]If $ c \equal{} 0$: if also $ b \equal{} 0$ then $ f(x) \equal{} x^{3}$, else $ a \equal{} 1$, so $ b \equal{}\minus{}2$, and $ f(x) \equal{} x^{3}\plus{}x^{2}\minus{}2x$ is ok.
[*]If $ c\not \equal{} 0$ then $ ab \equal{}\minus{}1$ and thus $ a,b\not \equal{} 0$. Mutliply the last equation with $ a$ to get $ ab \equal{} a^{2}b\plus{}abc\plus{}ca^{2}$ or $ \minus{}1 \equal{}\minus{}a\plus{}c\plus{}ca^{2}$, yielding $ c \equal{}\frac{a\minus{}1}{a^{2}\plus{}1}$. On the other hand is $ a(\minus{}2a\minus{}c) \equal{}\minus{}1$, so $ c \equal{}\frac{1\minus{}2a^{2}}{a}$. Since $ a(a\minus{}1) \equal{} (1\plus{}a^{2})(1\minus{}2a^{2})\Leftrightarrow 1\plus{}a\minus{}2a^{2}\minus{}2a^{4}\equal{} 0$ has no rational roots, no solutions here.[/list]
So $ f(x)\in\{x^{3},x^{3}\plus{}x^{2}\minus{}2x\}$.


A triangle of area 1/2 lies in a unit square. Prove that at least two of its vertices are also vertices of the square.

[hide]Suppose at least one of its vertices was on the inside of the unit square. If this is the case, it could be bounded by a smaller rectangle, and we know that the area of the triangle is at most 1/2 that of its bounding rectangle. If the triangle is obtuse, it could be even less than this. Let $ k$ be the area of the rectangle. We know that $ k=\frac{1}{2}$, but if $ k$ were inside a rectangle that was smaller than the unit square, then $ k\lte \frac{1}{2}\cdot1$ which is impossible.

Now we know that the vertices must all be on the edges of the unit square. With this in mind, pick two vertices and call there distance $ d$. $ d$ can be anything from 0 to $ \sqrt{2}$. Clearly, if $ d$ is anything from 0 to 1, the two points (which we will call $ A$ and $ B$ must be on the same side. Now, if the last point (point $ C$) is on the opposite side, the area would be $ \frac{1}{2}\cdot1\cdot d=\frac{1}{2}$. Meaning that $ d=1$ so that $ A$ and $ B$ are both vertices of the triangle. So for there to be any hope of the statement being false, $ 1>d>\sqrt{2}$. $ d$ cannot be $ \sqrt{2}$ because it it was, they would be on opposite vertices. So now we have all three vertices on different sides, and none of which are on a vertex. We can shrink the side that has no vertices on it and move it in until moving it any farther would move one of the vertices outside the triangle. The triangle is now bounded by a rectangle with an area smaller than 1, so the triangle's area must be smaller than $ \frac{1}{2}$.

QED[/hide]


linear algebra
matrix
linear algebra unsolved

For any $ A \in M_n(C)$ prove that there exist $ B \in M_n(C)$ that statisfies $ AB\equal{}0$ and $ rank(A)\plus{}rank(B)\equal{}n$. Thanks in advance.

Suppose $ A$ as a linear operator and if $ \mbox{rank A}\equal{}k$ then $ \mbox{ker A} \equal{}n\minus{}k$ and then there exist $ n\minus{}k$ columns which are linearly independent ,like $ X_1,...,X_{n\minus{}k}$ and then make a matrix $ B$ with all of these columns by repeating.