Thanks for at least clicking on the link.

now onto the problems
P.S. if you could, show some work so i know how u did the problems, its for a test/Quiz make up and i have no clue to what im doing, tutoring isnt helping me :S

-----------------------------------------------------------
Test 1

-----------------------------------------------------------
1)
Prove the Identity:

$\frac{\tan^{2}\theta}{1+\tan^{2}\theta}= \sin^{2}\theta$

2)
Solve for $\theta$ on the interval $[0,2\pi)$

(A)
$\frac{2\sin \theta}{\cos \theta}= \sin \theta$

[edit jli]: i think i messed up texing this one because it doesnt make cosine is restricted between 1 and -1.  just press edit at the bottom of this post.  sorry for the confusion

(B)
$\sin^{2}\theta-\sin \theta = \cos^{2}\theta$

3)
Find the EXACT values for the Following:

(A)
$\tan 15^\circ$

(B)
$\sin 105^\circ$

4)
Given:
$0<\beta <\frac{\pi}{2}<\alpha<\pi$   with   $\sin \beta=\frac{24}{25}$ and $\sin \alpha=\frac{4}{5}$,
FIND $\sin (\alpha-\beta)$

-----------------------------------------------------------
Quiz 1
-----------------------------------------------------------
++++1)

Prove the following identities:

Tangent^2theta-Sine^2theta=Sine^2theta*tangent^2theta

++++2)

Cosine theta-1/1-Secant theta=Cosine theta+1/1+Secant theta

-----------------------------------------------------------
Quiz 2

-----------------------------------------------------------
++++1)

Consider the graph of the function y=sinx-sqrt(3)cosx

(A)Find the Amplitude and Period of the graph.
(B)Rewrite the Equation of this graph as a transformed Sin function.

++++2)

To the nearest degree, find the Acute angle between the lines  2x+y=10  and  x+2y=10

-----------------------------------------------------------
Test 2
-----------------------------------------------------------
++++1)

If Sinx+Bcosx=asin(x+3pi/4), Find the values of a and B.

++++2)

Find the Acute angle between the line 2x+3y=6 and its inverse.

++++3)

Solve for theta on [0,2pi)

(Cos theta)-(Cos 2theta)=1

++++4)

(A)
If Angle theta was any quadrant  2  angle, what quadrant(s) could angle 2theta lie in?

(B)
Given  (sin theta)=8/17, where pi/2<theta<pi, find the following:
-B1) Sin2theta
-B2) Cos2theta
-B3) the quadrant containing 2theta

(C)
Simplify the following:

(2SinxCosx)^2+(Cos^2x-Sin^2x)^2

-----------------------------------------------------------
I really appriciate if anyone even attempts this stuff :S i will be going in for more tutoring and hopefully will get this stuff. Thanks again guys and you really do help out alot, lots of props for helping me on my last problems...

Thanks again,
-Zabije Cie

you need to go through

[b]s.l.loney plane trignometry part 1[/b]

TEST 1
1)
$\frac{\frac{\sin^{2}\theta}{\cos^{2}\theta}}{1+\frac{\sin^{2}\theta}{\cos^{2}\theta}}=\sin^{2}\theta$

$\Rightarrow \frac{1}{\cos^{2}\theta}=\frac{\sin^{2}\theta+\cos^{2}\theta}{\cos^{2}\theta}$

$\Rightarrow \frac{1}{\cos^{2}\theta}=\frac{1}{\cos^{2}\theta}$

$Q.E.D.$

2)
a.I get $\cos \theta =2$ which is not possible

3)
a.Use the half angle formula for tangent:

$\tan \frac{a}{2}=\frac{1-\cos a}{\sin a}$

$\tan 15=\tan \frac{30}{2}=\frac{1-\cos 30}{\sin 30}=\frac{1-\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\boxed{2-\sqrt{3}}$

b.
Use the trigonometric sum formulas:

$\sin(a+b)=\sin a \cdot \cos b+\sin b\cdot \cos a$

$\sin 105=\sin(45+60)=\sin 45\cos 60+\sin 60\cos 45=\frac{\sqrt{2}}{2}\cdot \frac{1}{2}+\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{2}}{2}=\boxed{\frac{\sqrt{2}+\sqrt{6}}{4}}$

4)
$\sin(\alpha-\beta)=\sin \alpha \cos \beta-\cos \alpha \sin \beta$

$\cos \alpha=\frac{3}{5}$ and $\cos \beta=\frac{1}{25}$

$\sin(\alpha-\beta)=\frac{4}{5}\cdot \frac{1}{25}-\frac{3}{5}\cdot \frac{24}{25}=\boxed{\frac{-68}{125}}$

i think i made a mistake on this one :maybe:

Thanks guys for the help so far you are really helpful, this stuff is so hard to me.

again, THANKS!!!!

[quote="7h3.D3m0n.117"]mods please move this to intermediate

[hide]

$\tan 15 = 2-\sqrt{3}$

[/hide]

i did this yesterday for another Intermediate problem =)

hint: use half angle formula

@zabije cie, I'm really glad that even though you joined yesterday you find the time to learn LaTeX...it makes everyone's life easier.[/quote]

oh...that was me but whatever

didnt have enough time to latex them all though

the moving thing isnt working for me  :(

[quote="zabije cie"]thanks guys for the help so far, if anyone could, id really like it to be finished

btw THANKS!!! YOU GUYS ROCK![/quote]

ill finish over the weekend.  have a really big competition tommorow (i probably spelled that wrong)

[hide="2a"] 2a makes plenty of sense if you want to find the value of $\theta$.  jli proved that you you can't divide by $\sin\theta$ or else you get the impossible conclusion that $\cos\theta=2$.    That implies $\sin\theta=0\Rightarrow \theta=0$ [/hide]

[hide="2b"] For this one, I assume that you have to solve for $\theta$.  We have that \[\sin^{2}\theta-\sin\theta=\cos^{2}\Rightarrow 2\sin^{2}\theta-\sin\theta=\cos^{2}\theta+\sin^{2}\theta\Rightarrow 2\sin^{2}\theta-\sin\theta=1\Rightarrow 2\sin^{2}\theta-\sin\theta-1=0\Rightarrow (2\sin\theta+1)(\sin\theta-1)=0\Rightarrow \sin\theta=-\frac{1}{2}, \sin\theta=1\Rightarrow \theta=210,\theta=90\] [/hide]

[hide="4"] jli, you made a mistake.  $\cos\beta=\frac{7}{25}$, but there is still a mistake since the final answer should be more than 0 because $\alpha-\beta>0$[/hide]

[hide="Quiz 1, Problem 1."] We know that \[\sin^{2}\theta+\cos^{2}\theta=1\Rightarrow 1-\cos^{2}\theta=\sin^{2}\theta\]  Multiplying both sides by $\tan^{2}\theta$, we have that \[\tan^{2}\theta-\cos^{2}\theta\tan^{2}\theta=\sin^{2}\theta\tan^{2}\theta\]  Since $\tan^{2}\theta\cos^{2}\theta=\sin^{2}\theta$, we have that \[\tan^{2}\theta-\sin^{2}\theta=\sin^{2}\theta\tan^{2}\theta\][/hide]

[hide="Quiz 1, Problem 2."] We know that \[\cos\theta-\sec\theta+1-1=\cos\theta-\sec\theta+1-1\]  Now, we also know that $\sec\theta\cos\theta=1$, so \[\cos\theta-\sec\theta-\cos\theta\sec\theta+1=\cos\theta-\sec\theta+\cos\theta\sec\theta-1\]  Factoring, we obtain \[(\cos\theta-1)(\sec\theta+1)=(1-\sec\theta)(1+\cos\theta)\] Thus, \[\frac{(\cos\theta-1)}{(1-\sec\theta)}=\frac{(\cos\theta+1)}{(\sec\theta+1)}\] [/hide]


Given that the sum of the distances from point $ P$ in the interior of a convex quadrilateral $ ABCD$ to the sides $ AB, BC, CD, DA$ is a constant, prove that $ ABCD$ is a parallelogram.

Given that the sum of the distances from point  in the interior of a convex quadrilateral  to the sides  is a constant, prove that  is a parallelogram.

Cant we give a solution using complex numbers ? I do not know the latex code for it.


i have a german  question.
since i am not native german i can't understand what does this question means.
so here it is please
solve and tell me detail about.

i would really thankful in advance

ein kapital wächst bei 5,5% nach einem Jahr auf 3798 DM an. Anfrangskapital_

Vielen Danke...

what does Anfrangskapital mean i can solve it if you tell me what that means.

i think Anfangskapital means that you have some money and one year later there are 5,5% more. so the answer would be $\frac{3798 DM}{1,055}$.

Naphthalin

$3798 = a \cdot (1+\frac{5,5}{100})^{1}$
$\frac{3798}{(1+\frac{5,5}{100})^{1}} = a$
a is the "Anfangskapital" or the point of intersection with the axis of y

this is what i think it means
a principal increases with 5,5% after one year on 3798 DM. Anfrangskapital

ein kapital wächst bei 5,5% nach einem Jahr auf 3798 DM an. Anfrangskapital

One section increases from 5.5% to 3798 DM every year.    And I don't know what Anfangskapitel is, but you can google it and I bet you can get it.

Again: Kapital (asset; maybe also "capital", but I am a bit sceptical) is _not_ the same as Kapitel (chapter; maybe section, but the former is better).

Yeah I think kapitel is capital because if you use Google translate.

So the question in english is kind of like: Bob put money in the bank. He gets 5.5% interest. Now he has 3798 dollars. How much money did he put in the bank?
It's not the same words but it'd get the same answer.

It's something like that except the money is in for 1 year and it's annual interest.
So the full question is: Bob put some money in the bank. He invested it at 5.5% annual interest. After 1 year, he has 3798 DM (German Marks). How many marks did he invest in the first place? 1 DM=0.63 USD.

Es sagt:
At 5.5%, capital grows to DM 3,798 after one year. Initial capital


Suppose we have $ a \geq b$ and $ a \geq c$, what conditions are needed to have that $ b \geq c$ ?


quadratics
calculus
calculus computations

Here's a task i think i have the right formula to solve the problem with, but i can't make it work and getting the result i need. I have now encountered several types of this task and here's one of them.

$f(x)=x^2-8x+4$ has two tangents that goes through the spot (0,0).
Find the intersection between the tangents and the graph.

I use this formula:

$x=x1-\frac{f(x1)}{f'(x1)}$

$x=x1-\frac{x^2-8x+4}{2x-8}$

I believe this is right until now, but i'm not sure.

PS. Could you plz make a $x$ with a lower case $1$. The x1 just means that it is a different $x$ than the regular $x$.

That would be x_1: $x_1$.
The second equation boils down to $0=x_1-\frac{x_1^2-8x_1+4}{2x_1-8}$, since we're looking at tangents that pass through the origin. Bring that together into one fraction, and then solve the quadratic for $x_1$.

Could you show me how you would do it? I thought i knew what to do, but i can't make it give any sence. If you show me how to do this all the way i should be able to do any of the other tasks (i hope) in the future of this kind.

Could you plz show me how you would solve this? I don't really know how to do it...

That's something you can do, yes, i see that, but don't know what to do from here actually. Sorry if i might be stupid and can't see what to do from here, but i have a feeling i am suppose to split the top of the fraction into fractions:

$0=\frac{(x_1-2)(x_1+2)}{2x_1-8}$

But not sure this will help me.

The fraction is zero exactly when its numerator is zero.

So from now on i just forget everything about the denominator.

$x_1^2-4=0$

$x=-2...x=2$

So if i place two tangents, one where $x=-2$ and one where $x=2$ they should both go through $(0;0)$. If i am right on this then i think i understand it now, finally. Thx...

Let $Y-f(x)=f'(x)(X-x)$ be the equation of a tangent which passes through the origin and touches the graph in the point $T(x,f(x))$. Therfore, $X: =0,\ Y: =0$ $\Longrightarrow$ $f(x)=xf'(x)$, i.e. $x^2-8x+4=x(2x-8)$ $\Longrightarrow$ $x^2=4$ $\Longrightarrow$ $T_1(-2,24)\ ,\ T_2(2,-8)$.

Though i can't follow your way of thinking right now we've got the same results now, finally. Thx...

I have another problem now though. I have done a handful of simular tasks, but this one isn't that easy i think. Plz tell me where i do something wrong and what i should do instead:

$X=x-\frac{f(x)}{f'(x)}$

$0=x-\frac{x^{-1}}{(x^{-1})'}$

$0=x-\frac{x^{-1}}{-x^{-2}}$

$0=\frac{x-x^{-1}}{-x^{-2}}$

$0=-x^3+x$

This doesn't seem right to me. Could you plz guide me a step further and tell me if we have to do something entirely different from what we did with the last problem...?

Nevermind, have solved the last problem now...


For each problem substitute $ a\equal{}\frac{1}{x\plus{}1}$, with $ x\geq 0$ and the problems become trivial.

[quote="stergiu"]If $ a , b , c \in [0,1]$, prove that :

  [b] b)[/b] $ 2(a \plus{} b \plus{} bc) \leq ab \plus{} bc \plus{} ca \plus{} abc \plus{} 2$[/quote]
I think, it's wrong. Try $ a\equal{}c\equal{}\frac{1}{2}$ and $ b\equal{}1.$
If you mean to we need to prove that $ 2(a \plus{} b \plus{} c) \leq ab \plus{} bc \plus{} ca \plus{} abc \plus{} 2$
then it's still wrong: try $ a\equal{}b\equal{}c\equal{}\frac{1}{2}.$ :wink:

[quote="arqady"][quote="stergiu"]If $ a , b , c \in [0,1]$, prove that :

  [b] b)[/b] $ 2(a \plus{} b \plus{} bc) \leq ab \plus{} bc \plus{} ca \plus{} abc \plus{} 2$[/quote]
I think, it's wrong. Try $ a \equal{} c \equal{} \frac {1}{2}$ and $ b \equal{} 1.$
If you mean to we need to prove that $ 2(a \plus{} b \plus{} c) \leq ab \plus{} bc \plus{} ca \plus{} abc \plus{} 2$
then it's still wrong: try $ a \equal{} b \equal{} c \equal{} \frac {1}{2}.$ :wink:[/quote]

 Dear Arqady, thank you. I wiill tell your remmarks to the creator. By telephon he told me that he has the solution as in the previous post. I will check it again.

Babis

[quote="andyciup"]For each problem substitute $ a \equal{} \frac {1}{x \plus{} 1}$, with $ x\geq 0$ and the problems become trivial.[/quote]

nice idea  :)

[quote="stergiu"]If $ a , b , c \in [0,1]$, prove that :

  [b] a)[/b] $ ab \plus{} bc \plus{} ca \leq 2abc \plus{} 1$

  [b] b)[/b] $ 2(a \plus{} b \plus{} c) \leq ab \plus{} bc \plus{} ca \plus{} abc \plus{} 2$

 [color=red]Edited[/color][/quote]

 The second question, according Arqady's remark , is wrong. Only the first is true.Sorry.I had seen also only the  first part and the argument worked. But every problem is a different problem !

 Babis

[quote="stergiu"]If $ a , b , c \in [0,1]$, prove that :

  [b] a)[/b] $ ab \plus{} bc \plus{} ca \leq 2abc \plus{} 1$

 [color=red]Edited[/color][/quote]
We have:
$ RHS\minus{}LHS\equal{}z(1\minus{}x)(1\minus{}y)\plus{}(xy\minus{}1)(z\minus{}1) \geq 0$
This inequality is true because $ a , b , c \in [0,1]$ :wink:


summer program
PROMYS
Mathcamp
HCSSiM
Ross Mathematics Program
MIT
college

Any comments or suggestions?
I applied and got into these.  PROMYS was going to be my first choice, but I just need to make sure before I let them all know.  Which will be the most fun?  Where will I learn the most?

Any feedback would be appreciated.

I haven't gone to either of the three (I went to Ross last year) but one of my best friends went to Promys and said he had a great time/learned a lot.  He would definitely recommend going to Promys.

Hmm...I went to HCSSiM last summer and had a lot of fun, learned a lot. I haven't been to either of the others, but I seriously can't imagine the instructors being cooler than the ones at Hampshire. This summer should be even better (since it's #34!)

IIRC, PROMYS is more number theory-oriented than either HCSSiM or Mathcamp, so if you're really into NT, that might be the best for you. Of course, on the flip side, if you don't really like number theory, you might not want to do PROMYS.

I have a good friend who went to Mathcamp last summer, and he said there was a lot of work, but he had a good time nevertheless. Other than that, I don't know too much about it.

Hope (against hope) this helps!

for this year, it's hands down HCSSiM.  This is the 34th HCSSiM program and an absolutely amazing program is getting arranged.

http://www.hcssim.org/ypmd.html

Plenary Speakers

David Kelly (Hampshire College) will speak on The Mathematical and Social History of 17
Eric Lander (Broad Institute at MIT)
Susan Landau (Sun Microsystems)
Erik Winfree (Caltech)
Dana Randall (Georgia Tech)
Jim Propp (University of Wisconsin-Madison)

dont know about hampshire, iwent to mathcamp past two years and had a blast but im going to promys this year, mathcamp has very little structure so if you are motivated or really interested in a topic (or likely to find an interesting topic) it can be really good because you can do as much work related to whatever you want  as you watn on the other hand if you don't do work without being told no one will tell you and you probably wont learn as much (you will still learn) as at the other programs


If $ x$ is an element of the set $ \{ \minus{}1, 1, 2 \}$ and $ y$ is an element of $ \{ \minus{}2, \minus{}1, 0, 1, 2 \}$, how many distinct values of $ x^y$ are positive?

The only way to do this is to test. For -1; -2, -1, and 2 works. For 1, everything works, and for 2, everything works. But we are asked for the number of distinct values. Counting gives $ 1, \frac {1}{4}, 2, \frac {1}{2}, 4$, so $ \boxed{5}$ is the answer.


calculus
integration
logarithms
calculus computations

$ \int_{0}^{1}\frac {x^{b} \minus{} 1}{ln x}dx$

Thanks for any help.

this  has  already been  done  , see old posts.

sorry b555 i am unable to find or dont know exactly where and how to find.

Using Leibnitz formula, differentiate:

$ y\equal{} \int_0^1\frac {x^b \minus{} 1}{\ln x}\,dx \\ \\
\frac{dy}{db} \equal{} \int_0^1 x^b dx \equal{} \frac{1}{1\plus{}b} \\ \\
\int dy \equal{} \int \frac{db}{1\plus{}b} \\ \\
y \equal{} \ln(1\plus{}b) \plus{}c \\ \\
f(0) \equal{} 0, \; \; \therefore c\equal{}0 \\ \\
\therefore \boxed{\int_0^1\frac {x^b \minus{} 1}{\ln x}\,dx \equal{} \ln (1\plus{}b)}$

Also, see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1371319#1371319]this[/url].

hello
i couldn't find that post for u but ya i can give u the solution.
[hide="solution"]
$ I(b)\equal{}    \int_{0}^{1}\frac {x^{b} \minus{} 1}{ln x}dx$
$ I'(b)\equal{}\int_{0}^{1}\frac{x^{b}ln x}{ln x} dx$
$ I'(b)\equal{}\frac{1}{b\plus{}1}$
integrating this
$ I(b)\equal{}ln(b\plus{}1)\plus{}C$
$ I(0)\equal{}0$
thus $ c\equal{}0$
[/hide]
thank u

[quote="kabi"]
i couldn't find that post for u but ya i can give u the solution.
[/quote]

I have posted the link in the post above your post!


probability
inequalities
conditional probability
calculus
calculus computations

I wasn't sure where to post this, but can anybody help me with the following problems:

1. Let A,B be two events with P(B)>0 Show that
a) If B ⊂ A, then P(A|B)=1,
b) If A ⊂ B, then P(A|B)=P(A)/P(B).

2. Three biased coins C1, C2, C3 lie on a table. Their respective probabilitues of falling heads when tossed are 1/3, 2/3, and 1. A coin is picked at random, tossed, and observed to fall heads. Calculate the probability that it is Ck, for each k= 1, 2, 3.
Given that a coin has been tossed once and observed to fall heads, calculate the probability that a 2nd throw of the same coin will also produce heads.
The experiment is begun again with the same three coins. This time the coin selected is tossed twice and observed to fall heads both times. Calculate the probability that it is Ck, for each k= 1, 2, 3.
Gicen that a coin has been tossed twice and observed to fall heads both times, calculate the probability that a third throw of the same coin will also produce heads.

3. Let A1, A2, ...., An be independent events. Show that the probability  that none of the events A1, A2, ...., An occur is less than ${e^{-\sum_{i=1}^{n}P \left(Ai \right) }}$

For 1a) I got that P($A\cap B$)=1 and P(${A}^{c}\cap B$)=0. So P(B)= P($A\cap B$) + P(${A}^{c}\cap B$) =1 Then P(A|B)=P($A\cap B$)/P(B)=1 Is this right? I don't think so since on part B If B ⊂ A then P($A\cap B$) would equal 1 as well.

[quote="223rem"]I wasn't sure where to post this, but can anybody help me with the following problems:

1. Let A,B be two events with P(B)>0 Show that
a) If B ⊂ A, then P(A|B)=1,
b) If A ⊂ B, then P(A|B)=P(A)/P(B).

For 1a) I got that P($A\cap B$)=1 and P(${A}^{c}\cap B$)=0. So P(B)= P($A\cap B$) + P(${A}^{c}\cap B$) =1 ... Is this right? [/quote]

No, it isn't. You cannot find $P(B)$ from the conditions of the problem. On the other hand, you don't need to: the conditional probability $P(A|B)$ is just $P(A\cap B)/P(B)$. In the case (a), we have $A\cap B=B$, so we get $P(B)/P(B)=1$. In the case (b), we have $A\cap B=A$, so we get $P(A)/P(B)$.

For the third problem, write the exact expression for the probability you need and use the inequality $1-x\le e^{-x}$.

Number 2 is a longer story, so I'll leave it to somebody else. It is also all about conditional probabilities.

The first problem is trivial;
(a) $A,B,\mathbb{P}(B)>0,B \subset A$ then, $\mathbb{P}(A|B)=\frac{\mathbb{P}(A\bigcap B)}{\mathbb{P}(B)}=\frac{\mathbb{P}(B)}{\mathbb{P}(B)}=1$
(b)The same.

Edit:Sorry, i didnt see someone posted it. :blush:


do u mean $n=2002$ is the unique solution

[hide="This is what i have.."]
$= 2^{1994}(1+2^{4}+2^{5}+2^{6}+2^{8}+2^{n-1994})$
$= 2^{1994}(369+2^{n-1994})$

So if $368+2^{n-1994}) =$ a perfect square then the whole thing is a perfect square since $2^{1}994$ is a perfect square..also $n > 1994$..if i'm not mistaken..so n = 2002 does work..how can u prove thats the only one if it is...
[/hide]

Note that this problem is very similar to an IMO Shortlist problem, 1979, proposed by the US and A. ( :P )

The Shortlist problem (it was much easier than the problem in this topic) was this:

Find all natural numbers $n$ so that
\[2^{8}+2^{11}+2^{n}\]
is a perfect square.

The only answer is indeed $12$.
You can see IMO Shortlist problems were very easy once  :)

ya i can see why n = 12 is the only answer..nothing else seems possible..indeed they are pretty easy..


U.S. all the way!

U.S. dominates in track and field, but I'm thinking we can count off a medal for basketball.  Wow, the "Dream Team" really stinks this year.

Wow, USA just lost there first game for basketball.  That is horrible.  It will be really hard to come back and win a medal now.

Yeah, the US will do much better in Track/Field.

The U.S. can be counted out in the equestrian events too for medals.  I don't think much for the team this year though some are okay.  The British team will dominate the equestrian events.  Australie seems to be right there with the U.S. in swimming so who knows who will get the most there.  I don't know if the U.S. will get the most.  It will be close.

The US Dream Team stinks. I can't believe they have Iverson and they're still losing incredibly to Puerto Rico, whos basketball team I haven't even heared of. The United States has won only one gold so far, compared to China's 5. So, I think China will definitely be in the top 3 gold and total medal contenders. The United States is good at Track and Field - that's correct, but they do need to make up ground for the events in which they haven't performed as good as they would have liked to such as Men's Gymnastics and other obscure events(In my opinion). I was really hoping the basketball team would do better...

US can be lucky Vatican City doesn't participate. The pope would've taken all golds :D

[quote="white_horse_king88"]The US Dream Team stinks. I can't believe they have Iverson and they're still losing incredibly to Puerto Rico, whos basketball team I haven't even heared of ... [/quote]are they loosing ... or did they actually lose?  :maybe:

I'm going for U.S., but China comes pretty close. Yeah, U.S. bastketball sucks. They should have put garnett on it or something.

Speaking of basketball, go NZ! What a nearly-great comeback..

Just to remember Sydney 2000  :D
1. USA 40 gold medals;
2. Russia 32;
3. China 28.
BTW I've just voted for Russia. I believe that their nice mascot Cheburashka will help them  :)

[quote="Syntax Error"][url=http://sports.yahoo.com/oly/news;_ylc=X3oDMTBpcnQzN2d2BF9TAzk2NjczMTUwBHNlYwN0bQ--?slug=ap-us-puertoricobko&prov=ap&type=lgns]Puerto Rico tops US by 19[/url]

::sigh::

[url=http://sports.yahoo.com/oly/news;_ylc=X3oDMTBpbXVpdmtqBF9TAzk2NjczMDkwBHNlYwN0bQ--?slug=ap-mens400freerelayswm&prov=ap&type=lgns]Phelps Messes Up[/url]

Yeah...I embellished that last headline a bit, but yeah.[/quote]

How'd Phelps mess up? I mean, yeah, he can't win 8 golds now, but you really can't say it's his fault. The first swimmer had a false start that really cost them.

So, the US only has one gold so far, and that's from Phelps. Jeez...

bah...i mind myself cheering for china...lol...

btw, did anyone watch men's synch diving this afternoon?...shocking messup by the chinese guys and they were up by like 15 points too *sigh*...there goes another medal...

I think that was probly one of the most interesting 5th round dives ever...the three best teams in the world failed on their last dive...

Yes, it was unbelievable! I watched this Live on Eurosport.

In 5th round one of Chinese (they where on the 1st place by that moment) lost orientation (!), so China got 0s for this round. After that one of Russians flying touched the tramp by leg and also nearly lost orientation, so Russia got nearly 0s for technique. At last,as I remember,one of the americans, who where 2nd to the moment, just before the water nearly lost orientation too!!!

So, if you reverse the table of results, you get:
1.China;2.Russia;3.USA. And actually it was 6.USA;7.Russia;8.China.
Now gess who got the gold medals? GREEKS!!!!! And this was their first gold medal in this olympiad!!!

I ll tell you what I have heard on tv.  The Korean guy should have gotten .1 more points for difficulty.  But one guy also said that the judges also missed that he had 4 holds in his routine when 3 is the max.  This should have been a .2 point reduction.  So I wouldn't argue over this too much.

The olympics would be terrible if you could debate everything a day or two after the event has taken place.  I guarantee you that if this were the case we would see plenty more medals being switched after they are given out.  There would be terrible debates.  Countries would hate each other and it might even turn into war.  What would the point of having judges there be if we just argued everything the day after? There wouldn't be a point.  You have to respect the call of the judges and abide by the rules.  This isn't the first time a call has gone the wrong way( and this is not nearly as bad as some other calls have been).

War is a bit of an extreme situation over the Olympics.  In ancient times, I presume it's happened before, but I doubt it's a major likelihood today.

I can understand that the judging process is a lot more complicated than the way it is said to be, and that debate is frequently pointless on the matter, but the point I was trying to make is that judgings are a great deal less subjective when there is already an existing control value.

Well, it seems that the U.S. tops the competition with 103 medals, 35 of them golds. Russia is second with 92 medals and 27 golds, and China is third with 63 medals and 32 golds. You can see the full results at [url]http://sports.yahoo.com/oly/medals[/url]

And Hamm refuses to give up his medal. I still think him foolish!

I want to mention that if the medals would be summed of the former Soviet Union they would have an overwhelming lead with 162.

[quote="yif man12"]Anyone see the marathon? I only heard about it but it sounds awful, especially since it might have cost De Lima a gold.[/quote]Yes, I've also heard that some public interventions along the way helped modify the final rankings, but I do not know more ...  :maybe:

[quote="yif man12"]Anyone see the marathon? I only heard about it but it sounds awful, especially since it might have cost De Lima a gold.[/quote]

I really wanted to see the marathon but we got hit by a tropical storm so all we had was storm coverage the whole day.  I heard about the marathon though.  I m glad the American got second but I also think it is unfortunate what happened to De Lima.

For the people who think that Hamm should return his gold medal, do you think the person who won the marathon should give his gold to Lima? If Lima hand't been attacked, he would have probably won gold.

The man claimed he touched him to "prepare for the second coming of Christ."

Oh, did anyone see the volleyball gold medal match between China and Russia? Great match (of course China came out on top even though they were 2 sets down)

And China won Olympic gold in Women's doubles in Tennis.

And the track and field hurdler, ya.

Diving was so great, too bad they failed big time in men springboard synchro... gymnastics was a DISASTER.

Hopefully, Chinese atheletes will not succumb to pressure in the 2008 Beijing Games.

[quote="RobinHood3000"]It's a difficult question, but I think probably not.  de Lima only might have won, and so it is impossible to judge the actual outcome of the race, had de Lima not been attacked.[/quote]
True, I don't think the runner who won gold should give it up, but there may be a chance that the olympics give him another gold, like what happened in the winter games' figure skating fiasco.

Although, I think it was a good move giving him the special award.

So how many of the poll votes about which country would get the most medals came on the last day of the poll?  ;)

An astute user of this site I know noticed that polls can be set to run specific numbers of days, so maybe next time (Winter Olympics?) the poll should close before the end result is known.

Congratulations to the United States.

Eum... no tokenadult, US was leading the race all the way from the start,
I think there will be very few extra votes to it
(last time I checked, sum week before the end, there were 33 US votes yet...)

Um it's us by far

They will keep on winning


inequalities
trigonometry
geometry proposed
geometry

Prove that $ \triangle ABC\ \implies\ \boxed {\ 1\ \le\ \frac {1}{a\plus{}b\plus{}c}\cdot\sum\ \frac {a^2}{b \plus{} c \minus{} a}\ \le\ \frac {2R}{r} \minus{} 3\ }$ (standard notations).

\[ b\plus{}c\equal{}2R(\sin B\plus{}\sin C)\equal{}4R\cos\frac{A}{2}\cos\frac{B\minus{}C}{2}\le 4R\cos\frac{A}{2}\]

\[ \frac{r}{b\plus{}c\minus{}a}\equal{}\frac{r}{2(s\minus{}a)}\equal{}\frac{1}{2}\tan\frac{A}{2}\]

So

\[ \frac{r(b\plus{}c)^2}{b\plus{}c\minus{}a}\le 16R^2\cos^2\frac{A}{2}\frac{1}{2}\tan\frac{A}{2}\equal{}8R^2\sin\frac{A}{2}\cos\frac{A}{2}\equal{}4R^2\sin A\equal{}2Ra\]

So

\[ \frac{r(b\plus{}c)^2}{b\plus{}c\minus{}a}\plus{}\frac{r(c\plus{}a)^2}{c\plus{}a\minus{}b}\plus{}\frac{r(a\plus{}b)^2}{a\plus{}b\minus{}c}\le 2R(a\plus{}b\plus{}c)\]

[quote="Virgil Nicula"] [color=darkred]Prove that $ \triangle ABC\ \implies\ \boxed {\ 1\ \le\ \frac {1}{a \plus{} b \plus{} c}\cdot\sum\ \frac {a^2}{b \plus{} c \minus{} a}\ \le\ \frac {2R}{r} \minus{} 3\ }$ (standard notations).[/color] [/quote]
[color=darkblue]Hence my inequality is no difficulty. An its equivalent form is $ 2p\ \le\ \sum\frac {a^2}{b \plus{} c \minus{} a}\ \le \frac {2R \minus{} 3r}{r}\cdot 2p$ .

The left inequality is easily, with [b]C.B.S.[/b] I found the right inequality starting from the following [b]interesting

identity[/b] in any triangle $ ABC$ : $ \boxed {\ \frac {(b \plus{} c)^2}{p(p \minus{} a)} \plus{} \frac {(b \minus{} c)^2}{(p \minus{} b)(p \minus{} c)} \equal{} \frac {4aR}{S}\ }$ . Thus, $ \frac {(b \plus{} c)^2}{p(p \minus{} a)}\ \le\ \frac {4aR}{S}$ a.s.o.[/color]

hello, a nice inequality! Lets solve it. By substituting $ R=\frac{abc}{4A}$ and $ r=\frac{A}{s}$ with the well-known formula $ A=\sqrt{s(s-a)(s-b)(s-c)}$
with $ s=\frac{a+b+c}{2}$ we have to prove that
${ -\frac{(b+c)^2}{a+b-c}-\frac{(a+c)^2}{a-b+c}-\frac{(b+c)^2}{-a+b+c}+\frac{abc(a+b+c)}{2\left(-a+\frac{a+b+c)}{2}\right)\left(-b+\frac{(a+b+c)}{2}\right)\left(-c+\frac{a+b+c}{2}\right)}}\geq0$.
Building the common denominator and factoring the inequality is equivalent with:
$ \frac{2(a+b+c)(a^3+b^3+c^3+3abc-a^2b-ab^2-a^2c-ac^2-b^2c-bc^2)}{(-a+b+c)(a-b+c)(a+b-c)}\geq0$. Now we have to show that
$ a^3+b^3+c^3+3abc \geq a^2b+ab^2+a^2c+ac^2+b^2c+bc^2$.
By substituting $ a=y+z,b=x+z,c=x+y$ we have:
$ x^2y+xy^2+x^2z+y^2z+xz^2+yz^2\geq6xyz$ which is true by AM-GM.
Sonnhard.

Nice other corollary

\[ \frac{4R}{r}\ge\frac{(b\plus{}c)(c\plus{}a)(a\plus{}b)}{abc}\]


If $\ a,b,c>0$, $\ n\in\mathbb{N},n\geq\ 2$ then
$\ \sum\sqrt[n]{4bc(b+c)}\leq\ 3\sqrt[n]{(a+b)(b+c)(c+a)}$

Very nice inequality!We have the inequality $(a+b)(b+c)(c+a)\geq\frac{8(\sum a)(\sum ab)}{9}$ so we have to prove: \[\sum\sqrt[n]{9\left(\frac{bc}{\sum bc}\right)\cdot \left(\frac{b+c}{2\sum a}\right)}\leq 3\] .We use now AM-GM for these two fractions from one sqrt and $\frac{1}{3}$ taken $(n-2)$ times and we get the result. :)

How on earth did you find this inequality...
$\ (a+b)(b+c)(c+a)\geq\frac{8(a+b+c)(ab+bc+ca)}{9}$?

Well, it's kind of "well-known". To solve it, you can use another "well-known" identity: $\prod (b+c) = (a+b+c) (bc+ca+ab)-abc$.

Yes.For an aplication of this usefull inequality look here:
http://www.mathlinks.ro/Forum/viewtopic.php?p=232780#p232780

[quote="stancioiu sorin"]If $\ a,b,c>0$, $\ n\in\mathbb{N},n\geq\ 2$ then
$\ \sum\sqrt[n]{4bc(b+c)}\leq\ 3\sqrt[n]{(a+b)(b+c)(c+a)}\quad (*)$[/quote]
My solution: \[{(*) \iff \sum_{\rm cyc}\sqrt[n]{4bc \over (a+b)(a+c)}\leqslant 3 \\ \iff \sum_{\rm cyc}\sqrt[n]{\underbrace{1\cdot 1\cdots 1}_{n-2\ \{\rm times}}\cdot{4bc \over (a+b)(a+c)}}\le \sum_{cyc}{1 \over n}\left(n-2+{2b \over a+b}+{2c \over a+c}\right) = 3.\] AM-GM only...


Hi,
I was wondering if anyone on this forum has some advice on thoroughly preparing for the AIME... like any good books/other materials to study, or um.. just general advice. I'd like to qualify for USAMO next year, so I'm aiming for 8ish on the AIME. (is that an appropriate goal...?) I'm more of a self-studying person, but I just don't know how to create some sort of plan/what to do/etc for the AIME. So yeah.. I'd appreciate any help. Thanks! :)

This is the right site to get it all.

Go to the bookstore and get Volume 1 and 2, as well as the Intro/Interm books for the topics you need, and sign up for classes. Maybe also some contest books

This topic has been asked million times. No joke. Use Search button.

But, seeing that you're new, I'm going to send you PM to answer your question (I've heard them so much that I can recite them off my head).


[quote="max1546"]Hi my apologies if this is in the wrong category section.  I'm new here :)

Put brackets in the following statement so that it is now true:
3+4-5-3-1-5=12

I think it needs to have brackets in a way that they multiply.
Thanks[/quote]
number of odd numbers (3,5,3,1,5) is 5 - odd number. Terefore for all ways put brackets result is odd.

This is from a olympiad practise test so it will surely work with multiplications?
maybe put brackets between signs to make them even or positive?

Whatever you do, you'll always get that $1\equiv-1\mod{2}$, so $3+4-5-3-1-5$ will always equal $3+4+5+3+1+5\equiv 1$ modulo 2, as Rust pointed out.

Just placing brackets will only affect the sign ($+$ or $-$), but "$+\equiv-\mod{2}$". You can't multiply numbers just by placing some brackets in this case. Therefore, it's impossible.

I finally solved it.
I am not sure of any shortcuts, but I did them one by one and there are three solutions.
I only know one of them,
It is:
3 +(4 - 5)(- 3 - 1 - 5)=12

Could anyone tell me more? And how to do them; not by doing them one by one.


limit
integration
function
calculus
calculus computations

Find:
$\mathop {\lim }\limits_{n \to \infty } \left( {2n + 1} \right)\int\limits_0^1 {x^n e^x dx} $

[hide="Comment."]Although convenient for explicit calculations, if you want those, the function $e^x$ could be replaced with by other continuous function having the same value at $1$ without changing the answer.[/hide]

Suppose $f(x)$ is continuous on $[0,1]$.

[b]Proposition 1[/b]. $\lim_{n\to\infty} \int_0^1 x^nf(x)dx = 0$.
The proof is easy, just using the boundness of $f(x)$.

[b]Proposition 2[/b]. $\lim_{n\to\infty} n\int_0^1 x^nf(x)dx = f(1)$.
Since $\lim_{n\to\infty} n\int_0^1 x^n dx = \lim_{n\to\infty} \frac{n}{n+1} = 1,$
it suffices to show that $\lim_{n\to\infty} n\int_0^1 x^n (f(x) - f(1))dx = 0$.

For any given $\varepsilon>0$, since $f(x)$ is continuous at $x=1$, we have $\delta>0$ such that $|f(x)-f(1)|<\varepsilon/2$ for $x\in[1-\delta,1]$. Thus $\left|\int_0^1 nx^n(f(x)-f(1))dx\right| \leq \left|\int_0^{1-\delta} nx^n(f(x)-f(1))dx\right| + \int_{1-\delta}^1 nx^n|f(x)-f(1)|dx$
Since $nx^n$ is increasing, $n(1-\delta)^n\to 0$, and $f(x)-f(1)$ is bounded (recall that $f$ is continous), we can find $N$ large enough such that the first integral above is less than $\frac{\varepsilon}{2}$ for all $n\geq N$.
The second integral is less than $\epsilon\int_{1-\delta}^1 nx^n dx\leq \epsilon\int_0^1 nx^n dx = \frac{n}{n+1}\cdot\frac{\varepsilon}{2} < \frac{\varepsilon}{2}$
Thus $n\int_0^1 x^n (f(x) - f(1))dx<\epsilon$ for $n\geq N$.

=======================================

This problem must have been discussed before...


function
induction
algebra
functional equation
algebra unsolved

Find every function $ f: \mathbb{N}\to\mathbb{N}$ such that $ f\left(f\left(n\right)\right)\equal{}n\plus{}2,\forall n\in\mathbb{N}$

$ f(n\plus{}2)\equal{}f(f(f(n)))\equal{}f(n)\plus{}2$. Therefore if $ f(1)\equal{}a,f(2)\equal{}b$, then $ f(n)\equal{}n\plus{}a\minus{}1$ if n -odd, $ f(n)\equal{}n\plus{}b\minus{}2$ if n even.
Because $ f(a)\equal{}3,f(b)\equal{}4$ may be only $ a\equal{}2,b\equal{}3$ or $ f(n)\equal{}n\plus{}1$.

[quote="Rust"]Because $ f(a) \equal{} 3,f(b) \equal{} 4$ may be only $ a \equal{} 2,b \equal{} 3$ or $ f(n) \equal{} n \plus{} 1$.[/quote]
I don't understand this.

Besides, my solution gives two solutions for $ f(n)$

$ f(n) \equal{} n \plus{} 1 \plus{} k( \minus{} 1)^{n \plus{} 1}\ \forall n$ where $ k$ is a constant $ \equal{} 0\text{ or }2$

That is $ (a,b) \equal{} (2,3)\text{ or }(4,1)$

[quote="karis"]why $ f(n \plus{} 2) \equal{} f(n) \plus{} 2$[/quote]
[hide]From the given equation:
$ f(f(n)) \equal{} n \plus{} 2$ for all $ n\in \mathbb N$

$ f(LHS) \equal{} f(RHS)$
Because $ LHS$ is the same natural number as $ RHS$ is.
[property of function: A function associates exactly one output to each input element drawn from a fixed set on which it is defined (here $ \mathbb N$, the set of natural numbers)]

So $ f(f(f(n))) \equal{} f(n \plus{} 2)$(1)

Now assume $ f(n) \equal{} m$

Since $ f(f(n)) \equal{} n \plus{} 2$ [b]for all[/b] $ n\in \mathbb N$
$ f(f(m)) \equal{} m \plus{} 2$ [$ m$ is a natural number]

Hence replacing $ f(n)$ for $ m$:
$ f(f(f(n))) \equal{} f(n) \plus{} 2$(2)

By the transitive property of equality (1) and (2) implies:
$ f(n \plus{} 2) \equal{} f(n) \plus{} 2$[/hide]

Ok,Let try this proof
Find the condition of $ b$ such that the Functional Equation has at least a solution
$ f: N\to N$
$ f(f(n))\equal{}n\plus{}b$ :lol:

Yeah, it is clearly that when $ b$ is an even number, there is a function $ f(n) \equal{} n \plus{} \frac {b}{2}$ satisfy the condition.

When $ b$ is an odd number, we will prove that there is no solution for the functional equation $ f(f(n)) \equal{} f(n) \plus{} b$.

Suppose function $ f: \mathbb{N}\to\mathbb{N}$ satisfies our conditon.

Replacing $ n$ by $ f(n)$ yields $ f(f(f(n))) \equal{} f(f(n)) \plus{} b$ or $ f(n \plus{} b) \equal{} f(n) \plus{} b,\forall n\in\mathbb{N}$

By induction, we obtain $ f(n \plus{} kb) \equal{} f(n) \plus{} kb,\forall n,k\in\mathbb{N}$, our mission now is to construct a map from $ \{0,1,...,b \minus{} 1\}\to\mathbb{N}$

Consider $ r\in\{0,...,b \minus{} 1\}$ and put $ f(r) \equal{} bk \plus{} s$ where $ s\in\{0,...,b \minus{} 1\}$ then:
$ f(f(r)) \equal{} f(bk \plus{} s) \equal{} f(s) \plus{} bk \equal{} r \plus{} b\Rightarrow r \equal{} f(s) \plus{} b(k \minus{} 1)$.

Since $ f(s)\in\mathbb{N}$ and $ r\leq b \minus{} 1$ then $ k \equal{} 0$ or $ k \equal{} 1$.

1. If $ k \equal{} 0$ then $ f(r) \equal{} s$ and $ f(s) \equal{} r \plus{} b$ for every pair of $ (r,s)$ where $ r,s\in\{0,...,b \minus{} 1\}$, which leads to contradiction because $ |\{0,...,b \minus{} 1\}| \equal{} b$, which is an odd number.

2. If $ k \equal{} 1$ then $ f(r) \equal{} s$ and $ f(s) \equal{} r$, similarly we got the contradiction, which end the proof.

In conclusion, the given functional equation has solution if and only if $ n$ is an even number.

Another problem: Let $ k\in\mathbb{N}$, how many functions $ f: \mathbb{N}\to\mathbb{N}$ are there that satisfy $ f(f(n)) \equal{} n \plus{} 2k$.

[quote="Mashimaru"]$ f(n \plus{} b) \equal{} f(n) \plus{} 2b,\forall n\in\mathbb{N}$[/quote]
How do you get this?
I'm getting $ f(n\plus{}b)\equal{}f(f(f(n)))\equal{}f(n)\plus{}b$
I don't have time now. Maybe I'll think about these later...

[quote="Mashimaru"]Yeah, it is clearly that when $ b$ is an even number, there is a function $ f(n) \equal{} n \plus{} \frac {b}{2}$ satisfy the condition.

When $ b$ is an odd number, we will prove that there is no solution for the functional equation $ f(f(n)) \equal{} f(n) \plus{} b$.

Suppose function $ f: \mathbb{N}\to\mathbb{N}$ satisfies our conditon.

Replacing $ n$ by $ f(n)$ yields $ f(f(f(n))) \equal{} f(f(n)) \plus{} b$ or $ f(n \plus{} b) \equal{} f(n) \plus{} 2b,\forall n\in\mathbb{N}$

By induction, we obtain $ f(n \plus{} kb) \equal{} f(n) \plus{} kb,\forall n,k\in\mathbb{N}$, our mission now is to construct a map from $ \{0,1,...,b \minus{} 1\}\to\mathbb{N}$

Consider $ r\in\{0,...,b \minus{} 1\}$ and put $ f(r) \equal{} bk \plus{} s$ where $ s\in\{0,...,b \minus{} 1\}$ then:
$ f(f(r)) \equal{} f(bk \plus{} s) \equal{} f(s) \plus{} bk \equal{} r \plus{} b\Rightarrow r \equal{} f(s) \plus{} b(k \minus{} 1)$.

Since $ f(s)\in\mathbb{N}$ and $ r\leq b \minus{} 1$ then $ k \equal{} 0$ or $ k \equal{} 1$.

1. If $ k \equal{} 0$ then $ f(r) \equal{} s$ and $ f(s) \equal{} r \plus{} b$ for every pair of $ (r,s)$ where $ r,s\in\{0,...,b \minus{} 1\}$, which leads to contradiction because $ |\{0,...,b \minus{} 1\}| \equal{} b$, which is an odd number.

2. If $ k \equal{} 1$ then $ f(r) \equal{} s$ and $ f(s) \equal{} r$, similarly we got the contradiction, which end the proof.

In conclusion, the given functional equation has solution if and only if $ n$ is an even number.

Another problem: Let $ k\in\mathbb{N}$, how many functions $ f: \mathbb{N}\to\mathbb{N}$ are there that satisfy $ f(f(n)) \equal{} n \plus{} 2k$.[/quote]
Very nice,mashimaru,I tried and found the result $ 2^k(2k \minus{} 1)(2k \minus{} 3)..5.3$,Perhaps,I am wrong
Can you show your result?

[quote="Akashnil"][quote="Mashimaru"]$ f(n \plus{} b) \equal{} f(n) \plus{} 2b,\forall n\in\mathbb{N}$[/quote]
How do you get this?
I'm getting $ f(n \plus{} b) \equal{} f(f(f(n))) \equal{} f(n) \plus{} b$
I don't have time now. Maybe I'll think about these later...[/quote]
you are right!mashiru is wrong!
but his proof seem to nice!I think it is good way to solve other function! :wink:

Greatly thanks to Akashnil and Math pro. I have edited my post.

@Allnames: I did not try to count how many function $ f$ are there but the hint of my teacher is to generate a sequence $ (u_k)$ where $ (u_k)$ denotes the number of functions $ f$ satisfy $ f(f(n))\equal{}n\plus{}2k$.

[quote="Mashimaru"]
@Allnames: I did not try to count how many function $ f$ are there but the hint of my teacher is to generate a sequence $ (u_k)$ where $ (u_k)$ denotes the number of functions $ f$ satisfy $ f(f(n)) \equal{} n \plus{} 2k$.[/quote]
Yes ,It is the way that I found my result.But ,my cacular is very bad,so I am not sure my one is true,Some one can help me?


inequalities
linear algebra
linear algebra unsolved

Let $A\in M_n(C)$ and $\lambda_1,...,\lambda_n$ its eigenvalues.
Show that:
$max_{i,j}|\lambda_i-\lambda_j|^2\leq 2(tr(A^*A) -\frac{1}{n}|tr(A)|^2)$.

what are those $z_i$ ? (there is no linear algebra here. This is just an inequality, isn't .. I'm sure I have misunderstood the pb. Could you clarify ?)

Also, what is $A*A$? Do you mean $A^2$ or $A^*A$?

I think I was drunk this morning when I posted this problem.Forget about those $z_k$.And... $A*A=A^* A$.It's not actually a problem of liniar algebra,it uses only a well known result about complex matrices but since it was with matrices i've considered it's better to put it here.

Hmm... sound like Schur's theorem, magistraly used by Myth to solve a difficult chinesse TST problem. Am I right?  ;)

Yes.I think those $z_k$ came from there. :)


geometry
circumcircle
geometric transformation
reflection
perpendicular bisector
geometry solved

:D

I would propose one problem. Please solve this.

Triangle ABC is on a circumcircle O

Bisector of angle ABC and ACB are relatively meeting AC and AB at D and E

Straight line DE is meeting O with M,N  (M is left, N is right)

If  I is the excenter of triangle ABC ( angle A) ,   Prove that IM=IN

Sorry, I'm poor at English  :(

Rewriting your problem:

[color=blue][b]Problem.[/b] Let $ I_{a}$ be the $ A$-excenter of a triangle $ ABC$. Let the angle bisectors of the angles ABC and BCA intersect the sides CA and AB at the points $ B^{\prime}$ and $ C^{\prime}$. Let the line $ B^{\prime}C^{\prime}$ intersect the circumcircle of triangle $ ABC$ at the points $ M$ and $ N$. Prove that $ I_{a}M \equal{} I_{a}N$.[/color]

[i]Solution.[/i] According to [url=http://www.cip.ifi.lmu.de/~grinberg/Schroeder/Schroeder.html#6][Schroeder6] in my Schröder points database[/url], we have $ B^{\prime}C^{\prime}\perp OI_{a}$, where $ O$ is the circumcenter of triangle $ ABC$. In other words, $ MN\perp OI_{a}$.

[b]EDIT:[/b] Two other proofs of $ MN\perp OI_a$ can be found in the [url=http://reflections.awesomemath.org/2007_4/MR_4_2007_solutions.pdf]solution of Mathematical Reflections problem S53[/url].

The segment $ MN$ is a chord of the circumcircle of triangle $ ABC$; thus, the perpendicular bisector of this segment $ MN$ passes through the center $ O$ of this circumcircle (since the perpendicular bisector of any chord of a circle passes through the center of the circle). In other words, the perpendicular bisector of the segment $ MN$ is the perpendicular to $ MN$ through $ O$. But the perpendicular to $ MN$ through $ O$ must be the line $ OI_{a}$, since $ MN\perp OI_{a}$. Hence, the perpendicular bisector of the segment $ MN$ is the line $ OI_{a}$. This means that the point $ I_{a}$ lies on the perpendicular bisector of the segment $ MN$. Thus, $ I_{a}M \equal{} I_{a}N$, completing the solution.

  Darij


function
integration
limit
real analysis
real analysis unsolved

Suppose $f$ is a real  continuous function over a closed interval $[0,T]. T>0$.
Prove that for every real number $t\in[0,T]$ we have
\[ \int_0^tf(u)du=\lim_{x\rightarrow +\infty}(\sum_{k=1}^{+\infty}\frac{(-1)^{k-1}}{k!}\int_0^Te^{kx(t-u)}f(u)du) \]
Apply the above result to prove that: if $g$ is a continuous function over $[0,T]$ such that:
\[ |\int_0^Te^{nt}g(t)dt|\geq M\ \ \ \forall n, \textrm{ and for some positive integer M} \]
then $g=0$


Nice!

All we need to do is to prove

$ 9a^2(a\minus{}1)^2 \le (a^3\minus{}1)^2$

It deduces to prove that

$ 9a^2 \le (a^2 \plus{} a\plus{}1)^2$ (*)

Moreover, we use the identity $ A^2\minus{}B^2 \equal{} (A\plus{}B)(A\minus{}B)$

(*) becomes $ (a\minus{}1)^2(a^2\plus{}4a\plus{}1) \ge 0$ which is trivial.
Finally, the equality cannot hold obviously so we have the strict inequality.
Done!


calculus
integration
function
real analysis
real analysis unsolved

Prove that if f is continuous on [a,b] and

$\int_a^b |f(x)|\,dx = 0$

then f(x) = 0 for all x in [a,b].

Thanks!

so I'll have to use an epsilon delta proof by contradiction here. I'll have to assume that there exists a c such that f(c) != 0 and for all x = f(c)/2, there exists a delta such that |f(x)-f(c)|< epsilon for |x-c| < delta. and then I should make |f(x)| > epsilon /2. This would contradict the original hypothesis...

Sometimes it's worth it to just indicate the main idea of a proof with a simple, easy-to-remember statement. For this proof it's this: "If it's not the zero function, then stick a box under the curve."

yur integrating over a function thats always positive or zero
if f(x)!=0 abs(f(x))>0 and integrating over any constant over any interval is the constant


inequalities
parameterization
inequalities proposed

[b]Prove that:[/b]$\underline {\overline {\left| \triangle ABC,\ 60^\circ \in \{A,B,C\}\Longrightarrow s^2\ge 27r^2+3(R-2r)^2\ \right| }}$

[b]Prove that:[/b]$\underline {\overline {\left| \triangle ABC,\ 45^\circ \in \{A,B,C\}\Longrightarrow s<2R+3r\ \right| }}$

Specifying an angle in the triangle should produce an equation between $s,r,R$, not only an inequality.   For fixed $r$ and $R$ there is a 1-parameter family of triangles, and a finite number of these triangles have a given angle at one of the vertices.


take a look at this:

[hide="it's a problem"]blah blah[/hide]

see the code doesn't work. but it appears when i remove the ' sign it works again.

[hide="its a problem"]blah blah[/hide]
why?

I don't think it ever worked with that character. Just don't use it in the hide lines.

An [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=31780]old thread[/url] with some discussion.

Characters that work: ! " .
Characters that don't work: ~ : ; ' < > , \ /

There are plenty of others that I didn't test.
These are disabled because they interact with other pieces of bbcode and smilies; trying to make them work would require chasing down a lot of bugs.

[quote="jmerry"]These are disabled because they interact with other pieces of bbcode and smilies; trying to make them work would require chasing down a lot of bugs.[/quote]Exactly.


counting
distinguishability
geometry
analytic geometry
graphing lines
slope
perpendicular bisector

1. hide]9!/2!2! = 9*8*7*6*5*4*3*2*1/4 = 9*8*7*6*5*3*2 = 1048320... hm... odd answer[/hide]

3. [hide]that's like placing two books on a shelf, so it's gonna be 2 ways?[/hide]

For 3 I would assume that you can shuffle the books around.

so it would be [hide]2*6*24=288[/hide]

[hide]CALABOOSE = 9 letters, 2 a's, 3 o's so 9!/(2!3!) which makes 9*8*7*6*5*2 = 30240.[/hide]

[hide](4,-2) and (-2,6)
The slope is equal to (y2 - y1)/(x2 - x1) = 8/-6. This makes the slope -4/3, and in standard form the first part of the equation would be 4x + 3y. Putting in points for x and y comes out 16 - 6 = 10 so 4x + 3y = 10. The x-intercept = 2 1/2 and the y-intercept = 3 1/3. The answer is x/(2 1/2) + y/(3 1/3) = 1.[/hide]

[hide]The geometry books can be permutated 24 ways and the algebra books 6 ways. For every permutation of geometry books, there are 6 permutations of Algebra books, and the two groups can be reversed. 24 * 6 * 2 = 288.[/hide]

The equation your actually finding is the perpendicular bisector, so you would have to use the mid-point formula for each point, and since the the line is perpendicular the slope would be negative reciprocial.

[hide="3"]
2 because either the algebra or the geometry books go first
3! because there are three algebra books (maybe four, I don't remember)
4! because there are four geometry books (maybe three, I don't remember)

2(3!)(4!)
2(6)(24)
288
[/hide]

[quote="deroundup378"]1. How many distinguishable linear  permutations can be made from the letters in the word calaboose?

2. Find the equation of the perpendicular bisector of the line segment whose endpoints are (4, -2) and (-2, 6). Write the equation in double-intercept form.

3. How many ways can 4 geometry books and 3 algebra books be arranged in a row if the geometry books must be kept together and the algebra books must be kept together?[/quote]

[hide="1"]
9!/(2!)(2!)
1048320
[/hide]

Please try to pay greater attention to the levels of problems posted in each forum.  I'm moving this one to The Basics.


trigonometry
geometry
3D geometry
tetrahedron
ratio

What is the cosine of the angle formed by two adjacent faces of a regular tetrahedron?

[hide]

Off the top of my head i think the answer is 1/3.

[/hide]

[hide]Just note that medians divide one another into segments of ratio 2:1, and you should get 1/3[/hide]


after going through some old tests, you kinda get the notion that the harder problems are in the back (duh), but what about the 1993-1994 national sprint? i thought the first couple of pages were actually harder than the last page. does anyone else notice that, or do i need to seek professional help?

Yeah.  Generally, the older tests were about doing quick arithmetic more than anything else.  They were also more confusing (but most MathCounts problems are, anyway :-) ).  That's why I like AMC/AIME/USAMO better.

[quote="bob123"]the older tests are harder because most of them didnt tell u how to express ur answers and wasnt descriptive enough about the problems...[/quote]

Yeah, I've noticed with a lot of math competitions that as they get older, the questions become clearer, and it's easier to understood what the question is asking (and I guess also, how you should solve it).  More recent Mathcounts do seem to be a little simpler.


[quote="t3hggnore"]Yeah my entire school will have no choice but to take the AIME B. Seeing as how the AIME A was so easy, the AIME B is bound to be way harder.[/quote]

Same here, as we're on spring break. Hopefully the AIME Bs aren't that much harder...

[quote="fourierseries"]I'm taking the alternate AIME.  Is anyone else taking it?

Is the index the same for both AIMEs?

Are we allowed to post alternate AIME problems?  I seem to see very few of those.[/quote]

The index may or may not be the same. For instance, last year there was a considerable difference. And there probably aren't many alternate AIME problems because it hasn't existed for very long yet.

last years aime2 was harder than the aime1 if i remember

[quote="t3hggnore"]Yeah my entire school will have no choice but to take the AIME B. Seeing as how the AIME A was so easy, the AIME B is bound to be way harder.[/quote]
WHich doesn't really help, because I'm probably gonna get a 1 or 2 >_<

[quote="tiredepartment"]I'm taking alternate AIME this year.  I heard that the AMC contest people purposely make the AIME Alternate slighly more difficult becuase of the time difference.  after all 2 weeks of studying can help if used wisely.[/quote]

I was born in November, does that give me a few extra months over the other people? I am, on average, 3.5 months older, than other people in my grade (assuming the birthday distribution is constant--which it probably isn't) :P


quadratics
algebra
polynomial
superior algebra
superior algebra unsolved

That's not the fundamental theorem of algebra for three reasons:
a) The fundamental theorem concerns fields, not groups.
b) In fact it only concerns one special field, $ \mathbb C$.
c) It doesn't state there are at most $ n$ solutions, but that there are ($ n$) solutions.

So nothing special on the counterexample $ (\mathbb Z/2\mathbb Z)^2$.

I've been saying this recently, but please TRY SMALL CASES.  There is one group of order 1, one of order 2, one of order 3, two of order 4... and already we have a counterexample.  The more you are willing to explore a problem to understand what is going on, the more you will not need people on the internet to solve your homework for you. :)

[quote="MaY"]Is it possible to for the quadratic equation x^2=e to have more than 2 solutions in a group with identity e?If it is so,will it contradict the fundamental theorem of algebra,i.e.,equations of power n can at most have n solutions?[/quote]

The power in groups is very different from the power in fields... Try to write the equation in additive notation...
The solutions of that equation are $ e$ and all the elements in the group with order $ 2$.

"Equation" means "polynomial," and polynomials as they are usually defined only make sense where you have both a multiplication [i]and[/i] an addition.  The result you quote is true fundamentally because if $ P(r) \equal{} 0$ then $ x \minus{} r$ is a factor of $ P$ over a field (the factor theorem), and there is no way to make sense of this statement in a multiplicative group because there is no addition.

(That's a small lie.  You can embed any group into a [url=http://en.wikipedia.org/wiki/Group_ring#Definition]group ring[/url] and define the "sum" of two elements in a multiplicative group as a formal sum, but in general this ring will not be a field and therefore the factor theorem will not hold.)


Give eight distinct positive integer $ a,b,c,d,e,f,g,h $. Find the least value of number $ n $ that satisfying:
$ n=ab+cd=ef+gh $.
Can you general it?. :)

That's from a Turkish Olympiad first round, see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=546593&p=3164379#p3164379]here[/url] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2815209&sid=cb30de0b42e364be74cc6d1524733aaa#p2815209]here[/url].
It's basically simple case analysis, I got $32$ some time ago but I may have been wrong.

It was a tedious round of trial and errors, although thanks for the link. :)


Vieta
real analysis
real analysis unsolved

consider the curve $(C):y=x^4 + ax^3 + 6x^2 + ax + 1$.
1/ find $a$ such that $(C)$ contains $5$ points on a line.
2/ find $a$ such that $(C)$ contains $4$ points on a line.

The first question is a joke, isn't it?  :D  Now, the second one... Suppose $ y=Ax+B$ is a line and the different point $A_i(x_i, y_i)$ are on the curve and line in the same time. Then the equation $ x^4+ax^3+6x^2+(a-A)x+1-B=0$ has four different real roots $x_i$ and thus $\sum{x_i}=-a, \sum{x_ix_j}=6$ (*). Reciprocal, if we have an $a$ such that there exist $x_i$ differents verifying the above equations, then it suffices to use Vieta's formulae to choose $A,B$ corresponding to the equation. Thus the question is under which condition for $a$ the system (*) has a solution in different real numbers. We have $\sum{x_i^2}=a^2-12$ and noting $ p=a^2-12$ we need to discuss $\sum{x_i}=-a, \sum{x_i^2}=p$. By Cauchy Schwarz, since the numbers are different we find $ 4p>a^2$, thus $a^2>16$. For the reciproque, suppose we have $ 4p>a^2$. It is immediate to see that for $\epsilon$ convenable there is a solution of the form $\frac{-a}{4}+\epsilon, \frac{-a}{4}+\epsilon, \frac{-a}{4}+2\cdot\epsilon, \frac{-a}{4}-2\cdot\epsilon$. Is it right? :?

and the last question was to find those $a$ such that there exists a unique line
that is tangent to $(C)$ in exactely $...$ points...


How do I make a matrix, with Latex. (and its determinant).

In a document you need \usepackage{amsmath} but on the forum you just need the code below.
[code]\begin{pmatrix}
1 & 2 & 3 \\
4 & 5 & 6\\
7 & 8 & 9
\end{pmatrix}
\begin{vmatrix}
1 & 2 & 3 \\
4 & 5 & 6\\
7 & 8 & 9
\end{vmatrix}[/code]gives
$ \begin{pmatrix}
1 & 2 & 3 \\
4 & 5 & 6\\
7 & 8 & 9
\end{pmatrix}
\begin{vmatrix}
1 & 2 & 3 \\
4 & 5 & 6\\
7 & 8 & 9
\end{vmatrix}$

Also try replacing pmatrix by bmatrix, Bmatrix, Vmatrix for other types of brackets.

$ \begin{Bmatrix}1&2&3\\4&5&6\\7&8&9\end{Bmatrix}$

That isn't what I expected, although I don't know what I was expecting. Now I know.

$ \begin{Vmatrix}1&2&3\\4&5&6\\7&8&9\end{Vmatrix}$

That is exactly what I was expecting.

$ \begin{pmatrix}1&2&3\\4&5&6\\7&8&9\end{pmatrix}$

Mix-and-match is usually bad notation and bad mathematics - but it doesn't seem to be illegal in $ \text{\LaTeX}.$ And stevem didn't mention using just {matrix} for no bracketing symbol at all.

[quote="Kent Merryfield"]Mix-and-match is usually bad notation and bad mathematics - but it doesn't seem to be illegal in $ \text{\LaTeX}.$[/quote]Actually it is, as you get the following error
[code]LaTeX Error: \begin{pmatrix} on input line 5 ended by \end{bmatrix}.

See the LaTeX manual or LaTeX Companion for explanation.
Type  H <return>  for immediate help.
 ...

l.5 ...in{pmatrix}1&2&3\\4&5&6\\7&8&9\end{bmatrix}
                                                  $
Your command was ignored.[/code]However, just as with misplaced dollar signs, LaTeX does its best to guess what you want. These error messages are ignored in the forum so you only see the output.


integration
calculus
invariant
quadratics
real analysis
real analysis unsolved

Calculate : $ \int\int\int_{x^2 \plus{} y^2 \plus{} z^2 \equal{} 1}{(xy \plus{} xz \plus{} yz)^mdxdydz}$
 With $ m \in \mathbb{N^{*}}$

The value of the integral is 0, since the set $ \{x^2\plus{}y^2\plus{}z^2\equal{}1\}$ has measure zero.

I think there must be $ x^{2}\plus{}y^{2}\plus{}z^{2} \leq 1$

Hmm, in this case one could apply an orthogonal transformation (leaving the unit ball invariant) to the bilinear form $ \beta$ whose quadratic form is the integrand (without the exponent). Reading off my scribblings, $ \beta$ has the eigenvalues $ 1, \minus{} \frac {1}{2}, \minus{} \frac {1}{2}$. After all it turns out to calculate the integral of $ (x^2 \minus{} \frac {1}{2}(y^2 \plus{} z^2))^m$ over the interior of the unit ball. By Cavalieri or Fubini, this should be easier than the original integral.


I'm taking Geometry at a high school first semester, so I was wondering if Introduction to Geometry will cover things not in a regular geometry curriculum and if it would still be a good idea for me to take it.

Thanks,

Nonie

I'm taking geometry at school too but I think I'll still take the Intro. to Geometry course. It depends on your school; I know that a whole lot of American high schools really neglect doing geometry. I think the AoPS class would teach more. Also, I think the kind of problems in the Intro. to Geometry class is going to be more "contest" centered as opposed to the stuff you do in class.

I don't think the classes at your high school will cover as much material as the AoPS classes will. I mean, AoPS classes are aimed for people getting ready for competitions or really enjoy math, so they should cover more ground than a regular geometry course.  :)

Thanks for your replies. :)

Oh, and my dad thinks I should take the class, so I probably will. :)

Trust me... if you spend time and are dedicated, you can probably do it  :wink:

[quote="nonie"][quote="Silverfalcon"]I'll drop my 2 cents here.

I took intro to geometry in my freshman year just like what you're planning to do. Instructor covered first two weeks of school geometry in one class. So yes, the course goes pretty fast but you have to know that they're oriented toward math olympiad.

I highly recommend this class even for people who don't compete in math olympiad. Geometry plays a big portion in standard tests such as ACTs and SATs and one year of geometry is probably not "sufficient" to most people. Plus, it's a great branch of mathematics.[/quote]

Actually I'm going into 8th grade.  2 weeks in one class! wow

As long as the time works out for me, I'll do the class. :)[/quote]

I'd like to say the quality of what you are taught in AoPS is far better than that of school.  Not only does the course go faster, it goes more in depth.  With just school instruction, most competition problems are still impossible.  However, Intro to Geometry should cover all you need for MC - AMC10.

[quote="kyyuanmathcount"][quote="EmoryABrown"][quote="daermon"]is the geometry class reeeeealy hard?  cuz im probably going to take it and I am in 7th and geometry is not my specialty :maybe:[/quote]

I am taking the course and am in seventh also :D .  I came to ask the same question, but thanks for the answer Karth.[/quote]

Yeah, it should work well.  I took a normal 7th grade math class in 6th grade, then selfstudied a year during my 7th grade year.  When I arrived in 8th grade, I tested out of both Algebra I and Geometry.  You should be fine :).[/quote]

My school class is Algebra 1, but I finished the book after 4 weeks of school, so I pass the tests while focusing on geometry.  I want to skip the school geometry and get on my brother's old math trak if possible

Just watch out for the [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Calculus.php]Calculus Trap[/url]


Triangle ABC has sides with lengths 10, 23 and 27 units. The perimeter of a particular square is 60% of triangle ABC’s perimeter. What is the area of the square?

The perimeter of the triangle is $ 10\plus{}23\plus{}27\equal{}60$ units. The perimeter of the square is then $ .6 \cdot 60\equal{}36$ units. Since a square has three equal sides, each side has length $ \frac{36}{4}\equal{}9$. Finally, the area is $ 9^2\equal{} \boxed{81}$.

[quote="PowerOfPi"]The perimeter of the triangle is $ 10 \plus{} 23 \plus{} 27 \equal{} 60$ units. The perimeter of the square is then $ .6 \cdot 60 \equal{} 36$ units. Since a square has three equal sides, each side has length $ \frac {36}{4} \equal{} 9$. Finally, the area is $ 9^2 \equal{} \boxed{81}$.[/quote]

wow... makes sense, man

Just set up the equation 4x the perimeter of the square=60*.6 the perimeter of the triangle and multiplied by .6

You should get 4x=36  solve for x and get 9


Round 3 problems are now available on the [url=http://www.usamts.org]USAMTS website[/url].  The due date is January 12.

Please note new web submission deadline: [b]3 PM Eastern / Noon Pacific[/b] on the due date.

On Problem 2, does the last player to have a turn HAVE to take 2 stones if there are 2 stones remaining?

@gauss1181:

Be careful not to talk about any problems until this round is over. Email usamts if you have questions.

Xantos is correct - do not discuss the problems publicly.


Show that $\frac{1}{\pi} \arccos \left( \frac{1}{\sqrt{2003}} \right)$ is irrational.

Assume the contrary, namely $ \mathrm{arccos} \left( \frac{1}{\sqrt{2003}} \right) \equal{} \frac{\pi r}{n}$, for some $ r, n \in \mathbb{N}$, $ (r, n) \equal{} 1$ and $ r < \frac{n}{2}$. Taking the cosine of the both sides and, then, squaring yields $ \frac{1}{2003} \equal{} \cos^2 \frac{\pi r}{n} \equal{} \frac{1}{2} \cdot \left( \cos \frac{2 \pi r}{n} \plus{} 1 \right)$. Hence $ \cos \frac{2\pi r}{n} \equal{} \minus{} \frac{2001}{2003} \in \mathbb{Q}$, and by a well-known result $ n \in \{1, 2, 3, 4, 6\}$, however it is a check by hand that in no such case $ \cos \frac{2\pi r}{n} \equal{} \minus{} \frac{2001}{2003}$, the contradiction needed.

The well known result is http://www.mathlinks.ro/Forum/viewtopic.php?t=150579 .


For how many integers $ n$ is $ \frac{n}{20\minus{}n}$ the square of an integer?

Well $ n<20$ since otherwise we get negative numbers.  We also need $ n\geq 20\minus{}n\Rightarrow n\geq 10$ in order to possibly get an integer (although $ n\equal{}0$ works, too).  If $ n\equal{}10$, we get 1, which is fine.  If we set $ n\equal{}16$, we get 4, which is fine as well.  Also, if we set $ n\equal{}18$, we get 9, which works.  So $ n\equal{}0,10,16,18$.

$ 20\minus{}n | n$ and  $ 20\minus{}n | 20\minus{}n \Rightarrow 20\minus{}n | 20$ $ \Rightarrow 20\minus{}n \in \{ 1,2,4,5,10,20 \} \Rightarrow n \in \{ 0,10,15,16,18,19 \}$

Denote $ M\equal{}\frac{n}{20\minus{}n} \Rightarrow$

[b][color=darkblue][u]Case 1[/u]: [/color][/b]$ n\equal{}0 \Rightarrow M\equal{}0$
[b][color=darkblue][u]Case 2[/u]:[/color] [/b]$ n\equal{}10 \Rightarrow M\equal{}1$
[b][color=darkblue][u]Case 3[/u]:[/color][/b] $ n\equal{}15 \Rightarrow M\equal{}3$
[b][color=darkblue][u]Case 4[/u]:[/color][/b] $ n\equal{}16 \Rightarrow M\equal{}4$
[b][color=darkblue][u]Case 5[/u]: [/color][/b]$ n\equal{}18 \Rightarrow M\equal{}9$
[b][color=darkblue][u]Case 6[/u]:[/color] [/b]$ n\equal{}19 \Rightarrow M\equal{}19$

Finally, $ x \in \{0,10,16,18 \}$


Hi everyone,

I finished writing mock Mathcounts Sprint, Target, and Team rounds on Wednesday and the local middle school took Sprint today. I have been away from the Mathcounts arena for a long period of time, so apparently I cannot fairly judge difficulty levels; the highest sprint score was around 19/30. The tests are also noticeably deficient in geometry problems and the wording on some questions is awkward. Finally, I have not proofread the questions, so I cannot guarantee the validity of the answers. Five questions have already been corrected on Sprint. Nevertheless, if practice is desired, I shall post the tests sometime over the weekend (no earlier than Saturday evening), and post the statistics of all feedback that I receive on Monday evening.

oooooo post it we could all do with the practice

me too! (this is message is now not too small)

I will do it.
ill also post mine after I finish the last couple question.

I'll do it... if I can access the computer tomorrow (highly probable).

I'll take it, if I have time and remember...  :D

I'll take it and hope that I don't fail.

Count me in this competition
When will it be posted?

The Sprint and Target Rounds shall be posted momentarily.

he said momentarily 7 hours ago so i'm not sure

Here it is: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=76817 .


calculus
integration
function
algebra
polynomial
real analysis
real analysis unsolved

let f from [0,1] to R be continuous function.if(riemann) integral f(x)x^n on [0,1]=0 for all nonnegative integers n, prove f(x)=0 for all x in [0,1]. :huh:

[hide="Hint."]The Weierstrass Approximation Theorem says that a continuous function can be approximated abritrarily well in the uniform norm by a polynomial.[/hide]

Let $P_{n}$ a series of polynomials that converge uniformly to f. (we can find such a sequence by Weierstrass Theorem) Then , using the hypothesis, we get that $\int_{0}^{1}f(t)P_{n}(t)dt$ is 0 $\forall n$. Taking th limit, we get that $\int_{0}^{1}f(t)^{2}dt=0$. As a result f is constant 0.


Is there a simple proof to the following?
$\left(  \sum_{k=1}^{n}{k}\right)^{2}=  \sum_{k=1}^{n}{k^{3}}$

[hide="Induction"]
Well... if we can prove that this case works from $k=1$, and then prove that if it works for $k$, it will work for $k+1$, we can prove it.

So, if $k=1$, we get, $1^{2}=1^{3}$. That is true. Now...

The LHS of the equation can be written as $(k+k-1+k-2...+k-k)^{2}$ which is equal to
\[(\frac{(k)(k+1)}{2})^{2}= \]

\[\frac{k*k*(k+1)(k+1)}{4}= \]

\[\frac{k^{4}+2k^{3}+k^{2}}{4}\]
Adding $(k+1)^{3}=k^{3}+3k^{2}+3k+1=\frac{4(k^{3}+3k^{2}+3k+1)}{4}$ to this, we get
\[\frac{k^{4}+6k^{3}+13k^{2}+13k^{2}+4}{4}\]
which can be factored to
\[\frac{(k+1)(k+1)(k+2)(k+2)}{4}\]
This shows that your equation must be true for all integers equal or greater than 1.

EDIT: I should have started the induction with 0, but I will quickly show that it works for 0 too.
$0^{2}=0^{3}$. Therefore, your equation must be true for all integers equal or greater than 0.
[/hide]

Well, this is a forward-direction version of what is more naturally a go-backwards proof:

[hide]
$\sum_{k=1}^{n}2k-1=n^{2}$ $\implies \sum_{k=1}^{n}(k)+\sum_{k=1}^{n}(k)-n=n^{2}$ $\implies \sum_{k=1}^{n}k+\sum_{k=1}^{n-1}k = n^{2}$ $\implies n \left( \sum_{k=1}^{n}k+\sum_{k=1}^{n-1}k \right) = n^{3}$ $\implies \left( \sum_{k=1}^{n}k-\sum_{k=1}^{n-1}k \right) \left( \sum_{k=1}^{n}k+\sum_{k=1}^{n-1}k \right) = n^{3}$ $\implies \left( \sum_{k=1}^{n}k \right)^{2}-\left( \sum_{k=1}^{n-1}k \right)^{2}=n^{3}$

Since it works for the case $n=1$, repeatedly adding in the above identity evaluated at successive values of $n$ telescopes to $\left( \sum_{k=1}^{n}{k}\right)^{2}= \sum_{k=1}^{n}{k^{3}}$
[/hide]

I would be interested in seeing a geometric interpretation of some sort for this

(the proof without words that occasionally shows up in the left nav doesn't count!)

[b]Edit:[/b] See above, essentially the same.

[hide="Direct proof"] Verify the identity

$k^{3}= 6{k \choose 3}+6{k \choose 2}+{k \choose 1}$

Then

$\begin{eqnarray*}\sum_{k=1}^{n}k^{3}&=& \sum_{k=1}^{n}\left( 6{k \choose 3}+6{k \choose 2}+{k \choose 1}\right) \\ &=& 6{k+1 \choose 4}+6{k+1 \choose 3}+{k+1 \choose 2}\\ &=& \frac{(k+1)k(k-1)(k-2)}{4}+\frac{4(k+1)k(k-1)}{4}+\frac{2(k+1)k}{4}\\ &=& \frac{(k+1)k(k-1)(k+2)}{4}+\frac{2(k+1)k}{4}\\ &=& \frac{(k+1)k(k^{2}+k)}{4}\\ &=& \left( \frac{(k+1)k}{2}\right)^{2}$

By Hockeystick, as desired.  This method can be used to find close forms for $\sum_{k=1}^{n}k^{r}$, although the coefficients of the combinations don't follow a nice pattern. [/hide]
[b]Follow-up:[/b]  Find all positive integer sequences $a_{k}$ such that

$\left( \sum_{k=1}^{n}a_{k}\right)^{2}= \sum_{k=1}^{n}a_{k}^{3}$

[quote="The Zuton Force"]I would be interested in seeing a geometric interpretation of some sort for this.[/quote]I agree that a geometric proof would be very interesting.  Does anybody have one?


There are two natural ways to inscribe a square in a given isosceles right triangle. If it is done as the leftmost triangle below, then one find that the area of the square is $441 \text{cm}^2$. What is the area (in $\text{cm}^2$) of the square inscribed in the same $\triangle ABC$ as shown in the rightmost triangle below?

$\text{(A)} \ 378 \qquad \text{(B)} \ 392 \qquad \text{(C)} \ 400 \qquad \text{(D)} \ 441 \qquad \text{(E)} \ 484$

[hide="Answer"]With the original square, we have side length $21$. Also, since it is a square and the triangle is right isosceles, then one vertex of the square is at the midpoint of the hypotenuse of the triangle. Therefore, the leg length of the triangle is $42$, and the length of the hypotenuse is $42\sqrt{2}$. Now, I'm not sure how to prove this next part, but I kind of know that the two vertices of the square which are on the hypotenuse in the second image trisect the hypotenuse, so the side length of this square is $14\sqrt{2}$, and the area is $392\Rightarrow \boxed{B}$.

I guess I can prove that by assuming it, and then showing that it works? Not a very strong proof, I know...[/hide]

Hmm...
[url=http://www.kalva.demon.co.uk/aime/aime87.html]This[/url] makes me think the answer should be 440... :huh:

You should be able to assume that the right square trisects the hypotenuse, because it seems to be marked on the problem.

Besides, it must be true since the triangle is isosceles and the square is at a 45 degree angle from vertical.

Okay, I wasn't sure if that was marked, or if it was a weird thing with the image. :P

[quote="Elemennop"]That isn't an isosceles right triangle, chess64, so I don't believe you can assume that.[/quote]

Oh, oops. :blush:

[quote="dasherm"]You should be able to assume that the right square trisects the hypotenuse, because it seems to be marked on the problem.[/quote]

Heh, that's an effect of the image compression that the forum does. Click on the image to bring up the full size of it, and you'll see it isn't marked.

Anyways, the smaller right triangle is similar to the full right triangle, so you can angle chase and show that the square trisects it.

Elemennop is correct.

Even though what JF did worked out, it's best to make sure that it works (rather than assuming that it does). Do a simple angle chasing starting on one acute triangle. It should become pretty obvious.

Look at the 2nd triangle. Ther vertex of square lying on BC be M, and let square side length be x.

BM = xsqrt2 / 2  MC = 2xsqrt2 /2

as determined previously BC = 42 = BM + MC = 3xsqrt2/2

solve for x we get 28/sqrt2

x^2 = 28*14 = 392


search
geometry
cyclic quadrilateral
circumcircle

After a long search on the net i found that there isn't a great lot in one place to refer when it comes to angle chasing...so this thread...the objective is to get as many results as possible that can be used as a refernce for angle chasing...

Guidelines
Classify your result under categories - circles, triangls, lines, polygons, etc(can't think of more!)
here are a few from me...

Polygons : The external angles of a polygon sum up to 360 degrees

Circles : Angles inscribed in the same arc are equal

[quote="fizsle"]Polygons : The external angles of a polygon sum up to 360 degrees[/quote]

sum(internal angles) = (n-2) 180

internal angleave = (n-2)180/n

So,

180 - (n-2)180/n = (180n-180n + 360)/n = 360/n

sum (360/n) = 360

The opposite sides of a cyclic quadrilateral add up to 180

Lines:
Alternating and corresponding angles are equal when working with parallel lines.
Co-interior angles add up to 180.

Complementary angles add to 90*.
Supplementary add to 180*.

If a+b = c = 90
and x+y = z = 90,
a and b are complementary, x and y and complementary, and z and c are supplementary.

This is used to prove near all theorems. No need for the rest.

For example, it is used for proving the 180* of a triangle, which is then used to prove the (n-2)180 theorem, etc.

Triangles: An external angle is equal to the 2 remote interior angles.

Parallel lines: Corresponding angles are equal. Alternate interior and exterior angles are congruent. Same side interior and exterior angles are supplementary.

Lines: Vertical angles are equal.

Cyclic Quadrilater = Quadrilateral Inscribed in a Circle

Opposite angles add up to 180 degrees.

Draw a cyclic quadrilateral with diagonals drawn in.  Form 4 pairs of equal angles.  Angles that subtend the same arc of the circumscribed circle are equal in measure.

Circles: A tangent to a circle is perpendicular to the radius drawn to the intersection.


For what kind of triangle $ ABC$ that the following relation among its sides and its angles holds
$ \frac{bc}{b\plus{}c} (cosA\plus{}1)\plus{}\frac{ca}{c\plus{}a}(1\plus{}cosB)\plus{}\frac{ab}{a\plus{}b}(1\plus{}cosC)\equal{}\frac{3}{16}(a\plus{}b\plus{}c)^2\plus{}cos^2A\plus{}cos^2B\plus{}cos^2C$

[quote="pro_math"]For what kind of triangle $ ABC$ that the following relation among its sides and its angles holds
$ \frac {bc}{b \plus{} c} (cosA \plus{} 1) \plus{} \frac {ca}{c \plus{} a}(1 \plus{} cosB) \plus{} \frac {ab}{a \plus{} b}(1 \plus{} cosC) \equal{} \frac {3}{16}(a \plus{} b \plus{} c)^2 \plus{} cos^2A \plus{} cos^2B \plus{} cos^2C$[/quote]
Who can help me?

[quote="pro_math"]For what kind of triangle $ ABC$ that the following relation among its sides and its angles holds
$ \frac {bc}{b \plus{} c} (cosA \plus{} 1) \plus{} \frac {ca}{c \plus{} a}(1 \plus{} cosB) \plus{} \frac {ab}{a \plus{} b}(1 \plus{} cosC) \equal{} \frac {3}{16}(a \plus{} b \plus{} c)^2 \plus{} cos^2A \plus{} cos^2B \plus{} cos^2C$[/quote]
I think we should use $ cosA\equal{}\frac{b^2\plus{}c^2\minus{}a^2}{2bc}$


for each natural number N>1, find some real number a s.t. for all integers p,q with $ 1\le q < N$, we have $ |aq \minus{} p|\ge\frac1N$

I am really frustrated.  It's so hard to study by myself.  Please help :(

ok, but can you explain it a little bit?  i mean by golly how did you get to 1/N?  I can barely understand the problem


Show that if $x + y + z = 0$ then
\[x^3 + y^3 + z^3 = 3xyz\]

This is basically how I've been trying to do it:
From the first equation we get:
$x = -(y + z)$
$y = -(x + z)$
$z = -(y + x)$
(However, these statements are 'empty' in the sense that they simplify to x = x, y = y, z = z. Something is strange).

Then you get that
\[3xyz = -3(y + z)(x + z)(x + y) = -3(2xyz + x^2y + x^2z + y^2x + y^2z + z^2x + z^2y) \]
At this point I'm kind of confused, as you get that 3xyz is equal to an expression which contains xyz, which I guess is a result of the 'empty' statements.

Then if you raise both sides of $x + y + z = 0$ to the power of 3 and solve for $x^3 + y^3 + z^3$ you get that
\[x^3 + y^3 + z^3 = -3(2xyz + x^2y + x^2z + y^2x + y^2z + z^2x + z^2y) = 3xyz \]

So, in the end it seems to work out (although I am unsure whether it is completely correct). I guess my question deals with what exactly I am doing. I don't get the whole deal with the empty statements, and 3xyz being equal to an expression with xyz in it. This is probably trivial to most of you, but there's just something I'm not getting. Any help is appreciated.

edit: I don't even know what I am asking. There is something I feel I don't understand in the method of the solution, but I can't pinpoint just what.

hey cant u do directly by using the identity

$a^3+b^3+c^3-3abc= (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

but $a+b+c=0$

hence
$a^3+b^3+c^3-3abc=0$

$a^3+b^3+c^3=3abc$

i know my method's lame :blush:  but weeell ......is it true? :oops:

[quote="Arne"]Kalle, there is a very quick way - just substitute all $z$ by $-(x + y)$ everywhere but keep the rest unchanged, and see what you get.

Need more help? :)[/quote]
Oh. Yeah. Showing that the two equations are consistent for all $x, y, z$ is enough, right?

I guess my solution is pretty strange. I was so stuck in this thought of trying to substitute all variables at once, but I see that simply substituting z is a much better solution. I feel dumb. Thanks.


Prove that the Cantor set does not contain any interval.

If the Cantor set is the middle-third, it suffices to observe that it has Lebesgue measure zero which in turn may be proved by observing that the complement has measure 1 (easy, it it is the union of disjoint intervals).

Thanks, but can we solve this problem without using measure theory?? I don't know measure theory...

The points of the Cantor Middle-third are identified by the expansion $ \sum_{k\equal{}0}^\infty {a_k\over 3^k}$ where $ a_k\equal{}0,2.$ It sufficies to prove that the complement is dense. To write the expansion of a point of the complement you need  to change the value of at least one $ a_k$ from 0 or 2 to 1 and this may be done choosing the index $ k$ as large as you need....

I hope you can continue by yourself.

Let's suppose $ x \equal{} \sum_{k \equal{} 1}^{\infty}\frac {a_k}{3^k}$($ a_k \equal{} 0,2$) is an element of the Cantor set and $ J$ is an interval which contains $ x$. Then, we can find a natural number $ K$ such that $ y \equal{} \left(\sum_{k \equal{} 1}^{K \minus{} 1}\frac {a_k}{3^k}\right) \plus{} \frac {b_K}{3^K} \plus{} \left(\sum_{k \equal{} K \plus{} 1}^{\infty}\frac {a_k}{3^k}\right)\in J$($ b_K \equal{} 0$ if $ a_K \equal{} 2$, $ 2$ if $ a_K \equal{} 0$). We changed one digit of the ternary representation of $ x$. If not(if we can't find such $ K$), $ J$ is not an 'interval', but just a set which contains one element. Let's think about $ z \equal{} \left(\sum_{k \equal{} 1}^{K \minus{} 1}\frac {a_k}{3^k}\right) \plus{} \frac {1}{3^K} \plus{} \left(\sum_{k \equal{} K \plus{} 1}^{\infty}\frac {a_k}{3^k}\right)$. It is obvious that $ x < z < y$ when $ a_K \equal{} 0$ and $ y < z < x$ when $ a_K \equal{} 2$. Since $ x,y\in J$ and $ z\notin J$, contradiction to $ J$ is an interval.

Please, check the error if there is.

I think $ J$ open should be added. Otherwise you cannot be sure that choosing $ K$ large enough, $ x$ stiil belongs to $ J$ regardless the value of $ a_K.$ It
sufficies to take $ x\in \partial J$ if $ J$ is closed at right/left or both.

The rest is O.K.

[quote="J.Y.Choi"]Let's suppose[/quote]

No, the condition that $ J$ is open is needless. Sorry, but I can't understnad what your assertion is. Please, tell me more minutely. Whether $ J$ is closed or not, for any $ x\in J$, we can find $ K\in\mathbb N$ such that $ y(\neq x)\in J$ because $ \mathbb N$ is not bounded above(definition of $ y$ is same as I mentioned above).

Following your argument, it seems that you first choose $ x\in C$ (Cantor set) and make the supposition that the interval $ J$ contains
$ x.$ Now if  $ J$ is closed and  $ x$ is one of the two endpoints (the right one for instance) and  $ a_k \equal{} 0$ from some $ k_0$ on, you have $ b_K \equal{} 2$ for some $ k \equal{} K$ and then $ y\not \in J$ since $ y > x.$ I think you couldn't continue  your argument.

Indeed the right approach to the problem, I think, is to prove that the complement of $ C$ is dense. So let's suppose  the contrary: there exists an interval  $ J\subset C$ (closed or open) such that $ J\cap ({\rm complement\ of}\ C) \equal{} \emptyset.$ You choose $ x$ in the open part of $ J$ and from now on your argument is o.k.

don't you agree?

[quote="J.Y.Choi"]Let's suppose $ x \equal{} \sum_{k \equal{} 1}^{\infty}\frac {a_k}{3^k}$($ a_k \equal{} 0,2$) ....
[/quote]

Here's another approach, which is pretty short. By construction, $ K\equal{}\bigcap_{n\equal{}0}^{\infty}K_n,$ where $ K_n$ is the union of $ 2^n$ disjoint closed intervals, each of length $ 3^{\minus{}n}.$

Suppose $ (a,b)\subset K.$ Then for every $ n,\ (a,b)\subset K_n.$ However, the longest interval that is a subset of $ K_n$ has length $ 3^{\minus{}n}.$ Hence $ b\minus{}a\le 3^{\minus{}n}$ for all $ n,$ which means that $ b\minus{}a\equal{}0.$ (And it made no difference to the argument whether I used $ (a,b),$ or $ [a,b].$)


function
inequalities
algebra unsolved
algebra

Find all real solutions to the following system of equations.  Carefully justify your answer.
\[ \left\{ \begin{array}{c} \displaystyle\frac{4x^2}{1+4x^2} = y \\ \\ \displaystyle\frac{4y^2}{1+4y^2} = z \\ \\ \displaystyle\frac{4z^2}{1+4z^2} = x \end{array} \right. \]

This problem is simple. I think, without visiting kalva... :) , hope it's true:
the first condition$\iff 4+\frac{1}{4x^2}=\frac{1}{y}$,...

More generally, if $f$ is an increasing function, then the solutions of the system
\[ \left\{ \begin{array}{c} f(x) = y \\ \\ f(y) = z \\ \\ f(z) = x \end{array} \right.  \]
satisfy $x=y=z$. Clearly, we have $x,y,z\ge 0$ and $f(x)=\frac{4x^2}{1+4x^2}$ is increasing on the interval $[0,+\infty)$.

Clearly, x,y,z >0 so we multiply all three eqns together to get AmGM inequality

Bomb

First, show that $x,y,z$ are non-negative.
Then, show that if one of x,y,z is negative, then all of x, y, z are negative. This means that to find other solutions, assume x,y,z are strictly positive.
Then, compare x and $\frac{4x^2}{1+4x^2} = y$, to get $x\geq \frac{4x^2}{1+4x^2}$ so then $x\geq y$. The problem is symmetric, so then $x\geq y\geq z\geq x$ which means $x = y =z $. Then solve  $\frac{4x^2}{1+4x^2} = x$ to get $(x,y,z)=(0,0,0), (\frac{1}{2},\frac{1}{2},\frac{1}{2})$.

We have y = 4x^2/(1 + 4x^2)

Or, x>=y

Simmilarly, y>=z and z>=x

So, x = y = z

Now we will find (2x - 1)^2 = 0

So x=y=z=0 or 1/2

Multiply all three inequalities together to obtain \[\left(\frac{4x^2}{1+4x^2}\right) \left(\frac{4y^2}{1+4y^2}\right) \left(\frac{4z^2}{1+4z^2}\right) = xyz.\]

Assume none of $x, y, z = 0$. If one of $x, y, z$ is $0$, then we obtain the trivial solution $\boxed{(x, y, z) = (0, 0, 0)}$. Then we must have
\begin{align*}
(1+4x^2)(1+4y^2)(1+4z^2) &= 64xyz \\
\implies (1/x + 4x)(1/y + 4y)(1/z + 4z) &= 64.
\end{align*}

However $1/x + 4x \geq 4$ by the arithmetic-geometric mean inequality, so that we have equality only when $x = \pm 1/2$. However if $x<0$ we obviously obtain a contradiction. Thus $x = 1/2$, and similarly $y = 1/2, z = 1/2$, so that $\boxed{(x, y, z) = (1/2, 1/2, 1/2)}$ is the only other solution. $\blacksquare$


function
logarithms
calculus
calculus computations

For $ n\in\mathbb{N}$, prove: $ \left(1 \plus{} \frac {1}{n \plus{} 1}\right)^{n \plus{} 2} < \left(1 \plus{} \frac {1}{n}\right)^{n \plus{} 1}$

Hello

 For $ x \in [1, \plus{} \infty [$

let $ f(x) \equal{} exp((x \plus{} 1)ln(1 \plus{} \frac {1}{x}))$

$ f'(x) \equal{} (ln(1 \plus{} \frac {1}{x}) \minus{} \frac {1}{x})exp((x \plus{} 1)ln(1 \plus{} \frac {1}{x})) < 0$

because $ ( \forall x \in ]0,1]) ln(x \plus{} 1) \minus{} x < 0$

therefore  $ f$ is a strictly monotonous decreasing function

Thus , For $ n\in\mathbb{N}^{\large\ast}, \left(1 \plus{} \frac {1}{n \plus{} 1}\right)^{n \plus{} 2} < \left(1 \plus{} \frac {1}{n}\right)^{n \plus{} 1}$

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

For $ n\geq 1,$ we can readily show via expansion that

$ \left( 1\minus{}\frac{1}{(n\plus{}1)^{2}}\right) ^{n\plus{}1}\left( 1\plus{}\frac{1}{n\plus{}1}\right) \leq 1\minus{}\frac{1}{2\left( n\plus{}1\right) ^{2}}\minus{}\frac{1}{2\left( n\plus{}1\right) ^{4}}<1$

Hence,

$ \left( 1\plus{}\frac{1}{n\plus{}1}\right) ^{n\plus{}2}<\left( 1\plus{}\frac{1}{n}\right) ^{n\plus{}1}$

Set $ f_{t} \equal{} \left(1 \plus{} \frac{1}{t} \right)^{t \plus{} 1}$ and $ h(t) \equal{} \log f_{t}$ and show $ h(t)' <0$.

If Calculus approach is not permitted, then we can induct on n.


Hello.
I was wondering how can i calcualte the error factor when i am doing measurments.
Ex, mass of object a is 32.01 g.
so the error factor is 0.01 +/-
right?
well if so, and i got a bunch of these objects, and i decided to do some devision multiplicataion addition etc......
a/w
how do i calcualted the new error factor after lets say....i divide 32.01 by 15.02?
Thx Alot.

I dont know how this fits here -- might probably be better in some other forum....Coming to the error factor anyway , it depends on the type of device that you're using -- If you have a device that measures the weight and gives you the weight as either 10kg or 11kg and nothing inbetween , you say that the weight of the body is (10kg $\pm$ 0.5 kg )if the device reads 10kg.
I'm not sure if I've got your question completely right -- If you have any doubt with any further problem regarding the error factor , go ahead and post your doubt...

all right
first of all the error factor in that case would be 1 kg. since we donnon where the rounding is going.(9.5->10.4)
but if i get a bunhc of 1.kg error factor, then what would be the erro factor in the end of something like this:
mass 1: 10 kg +- 1
mass 2: 50 kb +-5
if i divide mass 2 by mass 1 what will the new error factor be?
Thx


modular arithmetic
geometry
geometric transformation
reflection
linear algebra
matrix
rotation

Find the number of permutations $ \sigma$ of $ S\equal{}\{1,2,...,n\}$ such that $ \forall i\equal{}\overline{1,n}, \sigma (i)\plus{} \sigma^{\minus{}1}(i)\equal{}n\plus{}1$.

Please give your post more informative titles!

Substituting $ \sigma(i)$ for $ i$, we get $ i \plus{} \sigma(\sigma(i)) \equal{} n \plus{} 1$ for all $ i$, or equivalently
\[ \sigma^2 \equal{} n(n \minus{} 1)(n \minus{} 2)\cdots21 \equal{} (1n)(2(n \minus{} 1))(3(n \minus{} 2))\cdots \equal{} : w.
\]
This shows immediately that there are no solutions if $ n \equiv 2, 3 \pmod 4$, since in this case we have that the square of our permutations must be equal to an odd permutation.  So, we are left with the cases $ n \equiv 0, 1 \pmod 4$.

If $ n \equal{} 4m \plus{} 1$ then $ w$ has the unique fixed point $ 2m \plus{} 1$, so $ \sigma$ must have $ 2m \plus{} 1$ as a fixed point and otherwise must look exactly like one of the solutions of length $ 4m$.

Now suppose $ n \equal{} 4m$.  It's easy to see that the only way $ \sigma$ can have the desired property is if its cycle structure consists of $ m$ 4-cycles, each of which is a (cyclically ordered) union of two of the cycles of $ w$.  Thus, we need to first pair up the $ 2m$ cycles of $ w$, a task we can perform in $ (2m \minus{} 1)!! \equal{} \frac {(2m)!}{2^m m!}$ ways.  Now, for each pair, there are two ways we can combine the transpositions into a valid cycle: $ (ab), (cd)$ can be recovered as the square of either $ (acbd)$ or $ (adbc)$.  Thus we have $ 2^m$ additional choices, for a final answer of $ \frac {(2m)!}{m!}$.  This answer is so nice that it suggests some other simple combinatorial proofs must be available.

The associated sequences in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url] are [url=http://www.research.att.com/~njas/sequences/A122670]A122670[/url] and [url=http://www.research.att.com/~njas/sequences/A001813]A001813[/url].

Note that this problem could have easily been stated in terms of symmetries of the permutation matrix.  In particular, it asks for the number of permutation matrices that are preserved by transposition followed by a reflection in the horizontal axis.  This is equivalent to asking that the matrix be preserved under a 90 degree rotation, which is an alternative explanation of the cycle structure of the solutions.


geometry
analytic geometry
USAMTS
conics
ellipse
trigonometry
circumcircle

I thought number 5 was easy, just common sense.  Numbers 1 and 2 were the usual difficulty, and number 4 was interesting.  I used a computer geometry program to experimentally find what the shape was, and then I proved it using Ptolemy's theorem (long proof, but no coordinate systems lol).

Number 3 was so annoying!  It took me about 10 minutes to write a computer program proving the answer was 300, but I made myself crank out a proof.  It took a lot of thinking, but it was annoying thinking.

I think numbers 1, 2, and 3 are pieces of cake...as long as I hadn't made stupid mistakes for #1 and #3 :(

#4 and 5 were interesting but I had brutish forceish solutions which I can't tell are right or wrong.

Overall, I think this round is the hardest of the three rounds we've had so far. (Not really, but maybe a LITTLE easier than R1, R2 was easiest)

i wonder if my solution for number 4 is considered coordinate bash... i did geometry, but two of the lengths of line segments i found happened to represent the x- and y- coordinates  :D
yeah this round was easy except i carelessed the first two question  :rotfl:
round two was killer (actullay cuz i spent no time on it and therefore made many stupids)

I thought round 2 was the easiest so far...which isn't good, because that's the one where solutions were lost, and I didn't have a total copy of my solutions. Argh.

Yeah, 1 was basically either 4 or 5 if you were organized; 2 was easy after listing a bunch of the first few numbers. 3 was nice, but I failed it. I also failed 4. Really liked 5.

[quote="Temperal"]#1 was pretty much guaranteed 4 or 5 - the obvious solution involved some brute force which was easy to mess up at the end, hence the 4 I got.

#2 was trivial.

#3 was probably easy, but I suck at scum - er, number theory, so somehow I messed it up.

#4 was clever if you didn't coordinate bash (coordinate bashing is basically bailing out).

#5 was intuitive, but not that easy in my opinion. I messed up the end slightly.[/quote]

Uh...coordinate bashing isn't un-clever or "bailing out" at all.  It's a legit way to solve a problem (and I've used it for the last three USAMTS rounds), and anyways, in round 3's case, I suck at working with ellipses outside of the coordinate plane.  Maybe that says something about my failure. ;\

[quote="Math Geek"]It's a legit way to solve a problem[/quote]

I'm sorry you feel that way.

I stewart/angle bisector theorem bashed 4.

Not that much better than coordinates, but less messy.

[quote="buzzer11"]I had literally three days to do this round because I was so busy, haha.

#1, 2, 3 I got.

#4, 5, i didn't. 5 felt very intuitive, since 0<a<(b+1)/b, and the case where 0<a<=1 was trivial. but i didn't get any further.[/quote]

I know this is nitpick, but 0<a<b/(b-1)

[quote="abacadaea"][quote="Temperal"]
#4 was clever if you didn't coordinate bash (coordinate bashing is basically bailing out).
[/quote]

i bet you were thinking of me when you wrote that. :P[/quote]
Not you specifically; I was just sick of all the people who say "ooh, look at my awesome and clever solution to #4" when they just coordinate bashed.

I found that I could use the property of cyclic quadrilaterals, and then use the point A to prove that $ BA\plus{}BC$ was constant, but in order to do that, I had to find the x-coordinate of $ A$. Then it was just simpler to say that $ x\equal{}\sqrt{2}c\sin\theta$ and $ y\equal{}c\sin\theta$ are the parametric equations of an ellipse.

[quote="Heero Kirashami"]For #4, I just drew a really good picture and then took a ruler and tried to see if anything popped out (which it did).[/quote]

But there were so many cases... how did you draw the "good picture"?

What cases? There were not that many choices for $ D$ if you fixed BC... it's pretty obvious (after staring at it for at least half an hour) that $ D$ must be on the perpendicular bisector of BC, and even then, the distance from $ BC$ can only range from $ \minus{} \frac 12 BC$ to $ \frac 12 BC$.

ya i proved by contradiction that on the circumcircle of abc d must lie at the point such that the tangent line to d was parallel to BC; otherwise there was always more than one point  :lol:

here's how i did:

#1: Wasted like 2 weeks trying to find a good way to count the number of 11-multiples in S, but ended up brute-forcing it (which is what the official solution did, ironically)

#2: Didn't get it (no time)

#3: Easy crap, though it took me a while to prove that it was 300, and only half-rigorously

#4: Just plain easy crap, got it in like 10 mins

#5: no time

the no times were because i only spent 2 weeks & like 10 mins a day

#4, I used Ptolemy's theorem and similar triangles (no coordinates) to show that AB + AC was always equal to BC*sqrt(2).


function
limit
calculus
calculus computations

[quote="N.T.TUAN"]Let function $f: \mathbb{R}\to\mathbb{R}$ satisfying
a)$f(x+1)=f(x)\forall x\in\mathbb{R}$
b) Exists $f'(x)$ for all $x\in\mathbb{R}$, and
c)$|f'(x)|<1\forall x\in\mathbb{R}$.
Prove that exist $\lim_{n\to+\infty}\frac{p_{0}(x)+p_{1}{(x)}+...+p_{n-1}(x)}{n^{2}}$ and it independent for $x$. Where $p(x)=x+f(x)\forall x\in\mathbb{R}$ and $p_{0}(x)=x,p_{n+1}(x)=p(p_{n}(x))(n=0,1,2,...)$.[/quote]

Lets denote L that limit. Then by Cezaro-Stolz:
$L = \lim_{n\to+\infty}\frac{p_{n}(x)}{2n+1}$
By iterating $p_{n}(x) = x+\sum_{k=0}^{n-1}f(p_{n}(x)) \leq x+n(f(0)+1)$ (since $|f'(x)| \leq 1$ and f(x) is periodic).

[quote="N.T.TUAN"]You have only $\frac{p_{n}(x)}{2n+1}\leq\frac{x}{2n+1}+\frac{n(f(0)+1)}{2n+1}$  :o[/quote]

Ops, sorry for stupid post  :oops:
Forgot n in numerator (thinked that limit is 0)...


it's just a classical example of thermodynamic versus kinetic equilibrium..
the activation energy for physical equilibrium is quite low as the bonds formed are very weak so the energy barrier is less similarly for the reverse reaction also that is bond breang it is less but may not be as less as bond breaking so at low temp it is favourable as the molecules can have activation enrgy for forward reaction(bond making)  but not backward(bond breaking)
in case of chemiosorption since actual stronger bonds are formed the activation enrgy for forward reaction is far more than in physiorption so not all molecules can climb this energy barrier at low temperature but at high temeprature it is possible also they acheie such a large stability that teh $ E_{act}$ for backward(bond breaking) is very high even at high temperature

On the other hand at high temperature thoug the forward reaction in physiorption is easy so is the backward of physiorption so on an average the molecules are not physioadsorbed

u are almost right!!!!!!

@pardesi- change this to question 59

[quote]Pardesi
On the other hand at high temperature thoug the forward reaction in physiorption is easy so is the backward of physiorption so on an average the molecules are not physioadsorbed[/quote]

ur are wrong.. at higher temperatures physissoprtion is not favoured... only chem is favoured....

this is beacuse-- solid+Gas----->  Solid/Gas + Heat
                                     <-----
SO increase in temperature favours the backward reaction and favours desorption.

that's wrong how come a reaction that is favoured that is has an $ E_{Act}$ which can be got thru at low temperature be not croosed at a higher temperature???
actually what happens is the $ E_{act}$ for both forward and backward reactions of physiorptions is so low that molecules have sufficient energy to undergo both so as a whole it 'looks' as if there has been no physiorption :wink:

how abt this explanation:
using dG=dH-TdS
for phsi adsorption dH is not very negative, whereas dS is negative, so -TdS is positive so at high temp, dG is positive, but in case of chemisorption, dG is negative even at high temp as dH is negative. :huh:

yes that's what i explained but more in intutiveness.....inspired by m&b :D

@Varun
as pardesi said, we can speak only kinetics for chemisorption....and thermodynamics for physissorption...
but ur explaination for physissorption should be modified....
dH<0 and dS<0 and dH is highly negative than TdS , so physissorption is facoured at low temp... since as temp increases dG becomes less negative and becomes positive at some temp... so desorption is favoured at hich temp..

Ranga .... overa schooliku pattikareee..... Challakarre Ranga