[quote="djimenez"]Someone asked me this one yesterday, I thought it could be nice to post it in this forum.

Prove that if $\alpha>0$ is a real number, then $\arctan\alpha+\arctan\alpha^{-1}=\frac\pi2$. I found three different solutions to it and at least two of them are completely basic, just trigonometry, the other needs a simple fact from calculus.

Enjoy :)[/quote]

[hide="Proof"]

Suppose $\exists\beta\ni\,\arctan\beta+\arctan\beta^{-1}\neq\frac{\pi}{2}$ then

$\arctan\beta^{-1}\neq\frac{\pi}{2}-\arctan\beta\\
\\
\Rightarrow \tan(\arctan\beta^{-1})\neq\tan(\frac{\pi}{2}-\arctan\beta)\\
\\
\Rightarrow \beta^{-1}\neq\frac{\sin(\frac{\pi}{2}-\arctan\beta)}{\cos(\frac{\pi}{2}-\arctan\beta)}\\
\\
\Rightarrow \beta^{-1}\neq\frac{\sin(\frac{\pi}{2})\cos(\arctan\beta)-\cos(\frac{\pi}{2})\sin(\arctan\beta)}{\cos(\frac{\pi}{2})\cos(\arctan\beta)+\sin(\frac{\pi}{2})\sin(\arctan\beta)}\\
\\
\Rightarrow \beta^{-1}\neq\frac{\cos(\arctan\beta)}{\sin(\arctan\beta)}\\
\\
\Rightarrow \beta^{-1}\neq\frac{1}{\tan(\arctan\beta)}\\
\\
\Rightarrow \beta^{-1}\neq\frac{1}{\beta}\\
\\
\Rightarrow \beta\cdot\beta^{-1}\neq\beta\cdot\frac{1}{\beta}\\
\\
\Rightarrow 1\neq 1 $

Our supposition has lead us to a contradiction; it follows that our supposition is false.

That our supposition is false implies: $\forall\alpha\in\textbf{R}\,\arctan\alpha+\arctan\alpha^{-1}=\frac{\pi}{2}\;\square\\
\\$
[/hide]

[quote="Dr. No"][hide="Proof"]Suppose $\exists\beta\ni\,\arctan\beta+\arctan\beta^{-1}\neq\frac{\pi}{2}$ then

$\arctan\beta^{-1}\neq\frac{\pi}{2}-\arctan\beta\\
\\
\Rightarrow \tan(\arctan\beta^{-1})\neq\tan(\frac{\pi}{2}-\arctan\beta)\\
\\
\Rightarrow \beta^{-1}\neq\frac{\sin(\frac{\pi}{2}-\arctan\beta)}{\cos(\frac{\pi}{2}-\arctan\beta)}\\
\\
\Rightarrow \beta^{-1}\neq\frac{\sin(\frac{\pi}{2})\cos(\arctan\beta)-\cos(\frac{\pi}{2})\sin(\arctan\beta)}{\cos(\frac{\pi}{2})\cos(\arctan\beta)+\sin(\frac{\pi}{2})\sin(\arctan\beta)}\\
\\
\Rightarrow \beta^{-1}\neq\frac{\cos(\arctan\beta)}{\sin(\arctan\beta)}\\
\\
\Rightarrow \beta^{-1}\neq\frac{1}{\tan(\arctan\beta)}\\
\\
\Rightarrow \beta^{-1}\neq\frac{1}{\beta}\\
\\
\Rightarrow \beta\cdot\beta^{-1}\neq\beta\cdot\frac{1}{\beta}\\
\\
\Rightarrow 1\neq 1 $

Our supposition has lead us to a contradiction; it follows that our supposition is false.

That our supposition is false implies: $\forall\alpha\in\textbf{R}\,\arctan\alpha+\arctan\alpha^{-1}=\frac{\pi}{2}\;\square\\
\\$
[/hide][/quote]The proof is basically right... and understandable even when... you never used the fact that $\alpha>0$... and that $\tan x$ is not injective!

We can say that you used it ($\alpha>0$) but you didn't explicitely mention it. Can you tell in what step you used it strongly without mention it? ;)

[quote="djimenez"]The proof is basically right... and understandable even when... you never used the fact that $\alpha>0$... and that $\tan x$ is not injective!

We can say that you used it ($\alpha>0$) but you didn't explicitely mention it. Can you tell in what step you used it strongly without mention it? ;)[/quote]

David,

I don't believe the constriant $\alpha>0$ is necessary.

Both $\sin x$ and $\tan x$ are odd functions; as such they have the same parity as their arguments. Therefore, the proof is valid regardless of the parity of $\alpha$, or rather $\beta$, in this case.

[quote="djimenez"][quote="Dr. No"]I don't believe the constriant $\alpha>0$ is necessary.[/quote] Of course it is... note that

$(-1)^{-1}=-1$ and $\arctan(-1)+\arctan(-1)=-\frac\pi2\neq\frac\pi2$ ;)[/quote]

Ah, but    $-\frac{\pi}{2}\equiv \frac{\pi}{2}\bmod{\pi}$.

Further, $\pi$ is the period of $\tan x$

Ergo, any answer resulting from not supplying the constraint will be likewise congruent to $\frac{\pi}{2} \bmod{\pi}$.  This suggests a more general statement of the problem is:

$\forall\alpha\in\textbf{R}\;\arctan\alpha + \arctan\alpha^{-1}\equiv \frac{\pi}{2}\bmod{\pi}$

 :lol:

[quote="djimenez"]Someone asked me this one yesterday, I thought it could be nice to post it in this forum.

Prove that if $\alpha>0$ is a real number, then $\arctan\alpha+\arctan\alpha^{-1}=\frac\pi2$. I found three different solutions to it and at least two of them are completely basic, just trigonometry, the other needs a simple fact from calculus.

Enjoy :)[/quote]

I had this problem in the semestrial math paper. We had to calculate for some particularily $\alpha$'s. But I told one of my classmates that this is true for any $\alpha>0$.

[color=blue]The problem
If $x\in (0,+\infty)$ show that: $\arctan x+\arctan\frac{1}{x}=\frac{\pi}{2}.$[/color]

[i]The proof[/i]
It's clearly that $\tan(\arctan x)=x$. Because $\tan(\frac{\pi}{2}-m)=\cot m$ results that ${\tan(\frac{\pi}{2}- \arctan\frac{1}{x})=\cot(\arctan\frac{1}{x})=\frac{1}{\tan(\arctan\frac{1}{x}})}=\frac{1}{\frac{1}{x}}=x$
So we found out that $\arctan x=\frac{\pi}{2}-\arctan\frac{1}{x}+k\pi$ with $k\in Z$ or else written it's:
$\arctan x+\arctan\frac{1}{x}=\frac{\pi}{2}+k\pi, k\in Z$.
But it's said that $x>0$ so $\arctan x\in(0,\frac{\pi}{2})$ and  $\arctan\frac{1}{x}\in(0,\frac{\pi}{2})$. It fallows that $\arctan x+\arctan\frac{1}{x}\in (0,\pi)$. Finally that's possible when $k=0$. And that's all  :D

[quote="djimenez"][quote="amirhtlusa"]isn't that obvious?? ;)[/quote]Well, it could be for you, but given that I have seen you in the College fora pretty often I assume you are not at HS level and you are not a beginner learning from more experienced problems solvers, and this forum is thought to be! ;)[/quote]
i'm a Junior in HS :D

, my proof is more like "AntonioMainenti"'s

[quote="AntonioMainenti"][hide="what is needed to make this sufficiently rigorous.."]Suppose $\alpha$ is the tangent of the angle $\theta$. Then $\tan\theta=\alpha\Rightarrow\frac{\sin\theta}{\cos\theta}=\alpha\Rightarrow\alpha^{-1}=\frac{\cos\theta}{\sin\theta}=\frac{\sin\left(\frac{\pi}{2}-\theta\right)}{\cos\left(\frac{\pi}{2}-\theta\right)}=\tan\left(\frac{\pi}{2}-\theta\right),$ so $\arctan\alpha+\arctan\alpha^{-1}=\theta+\frac{\pi}{2}-\theta=\frac{\pi}{2}.$[/hide][/quote]

Antonio,

The problem is sufficiently rigorous as stated. A proof wherein $\alpha > 0$ was assumed has already been given.

The question at hand is whether this restriction is necessary. David has proved that it is by counterexample -- I shant quibble with the obvious.  For me, the thing that remains is whether this is the most general form of the problem that can be given. If it is then the resulting theorem is of little value since it is merely an abstruse restatement of an obvious fact pointed out by another poster earlier in the thread: $\tan (\frac{\pi}{2}-m) =\cot m$.

[quote]Antonio,

The problem is sufficiently rigorous as stated. A proof wherein $\alpha > 0$ was assumed has already been given.

The question at hand is whether this restriction is necessary. David has proved that it is by counterexample -- I shant quibble with the obvious.  For me, the thing that remains is whether this is the most general form of the problem that can be given. If it is then the resulting theorem is of little value since it is merely an abstruse restatement of an obvious fact pointed out by another poster earlier in the thread: $\tan (\frac{\pi}{2}-m) =\cot m$.[/quote]Ah I see. It seems like the sum would be $\frac{\pi}{2}$ if $\alpha>0$, undefined for $\alpha=0$, and $-\frac{\pi}{2}$ if $\alpha<0$. I'm sure you already realize this though (assuming it's right). As for generalizing further.. I have no idea.


vector
function
group theory
abstract algebra
induction
superior algebra
superior algebra solved

let (G.*) a croupe such that for every element X    X*X=e          prove that exist k \ card(G)=2^k :?:

Let $x,y\in G$. Then $e=x^{2}y^{2}=(xy)^{2}$ which leads to $xy=yx$, so $G$ is abelian. From now on, denote $xy$ by $x+y$.
With this in view, it should be easy to see that $G$ has a natural structure of $\mathbb Z/2\mathbb Z$ vector space.
Choose $B$ a basis for $G$ as a vector space. Then we can identify any element of $G$ with a finite subset of $B$, because all the coefficients that appear are either 0 or 1 and there are only finitely many 1's, so we can look at this as of a characteristic function of a subset of $B$ with finite support.

If $B$ is finite, which I think is what you care for, then if $|B|=k$, then $|G|=|\mathcal P(B)|=2^{k}$.
If $k$ is not a finite, then the cardinal of finite subsets of $B$ is the same as the cardinal of $B$, so $|G|=|B|$.

But maybe by "croupe" you mean "finite group"  :D

[quote="-oo-"]alternative: the assumption tells us that every element has order $1$ or $2$. so by cauchy's theorem, the order of $G$ has no odd prime divisor.[/quote]

Alternative to the finite setting you mean.

the problem - as stated - is false.
you actually need finiteness hypothesis, since there are at least two ways to find a counterexample: either $(\overline{\mathbb{F}}_{2},+)$ and $(\mathbb{Z}/2\mathbb{Z})^\omega_{0}$ (i.e. definitively null sequences in $(\mathbb{Z}/2\mathbb{Z})^\omega$) are countable.

ma_go the problem is not false becouse we have card(G)=card(H)card(G\H).H={e.x} is subgroup of G. and we prove that G\H is abelien .and done by induction in n. :ninja:

mine was just an answer to amfulger.

as i said, the statement needs finiteness hypothesis: you can't drop it, since i found two [isomorphic] infinite counterexamples.

of course, if you ask for your group to be finite, there are at least three different proofs of it.

Seems I forgot to end my argument there. I proved that if $G$ is infinite, then $|G|$ is the same as the cardinal of a basis $B$ for the $\mathbb Z/2\mathbb Z$ vector space $G$.

What I forgot to mention and what ma-ga completed, is that there exist cardinals that are not $2^{\mbox{some other cardinal numer}}$, for example $\aleph_{0}=|\mathbb N|$.


Hi all,

I had a question about the rule of signs that wasn't covered in my algebra 2 or precalc course.

How do we determine the number of zeros there are if we are missing a term, such as x^3+6x+20 or x^5-4x^4+2x^2-4

You can ignore that term since 0 is not considered a sign change.

[quote="SuperMathBoy"]Do you mean that, for example: in the polynomial, x^2-4, we can just skip the x term and look at the sign change between x^2 and -4?[/quote]That's right. As you can see, there is only 1 sign change so you should expect one 1 positive root and 1 negative root or two complex roots (if the roots are not real).


I use UTC+0 (= GMT) timezone and it is showing as an hour behind. I am posting this at 23:57 UTC but the forum is showing the time as 22:57. When the clocks go forward in a week's time to UTC+1 (= BST, British Summer Time) it will be still be an hour behind if I use the summer time option.

This seems global. My UTC-8+DST (Pacific time zone in the USA) is displaying one hour too early, and I just saw the same complaint from someone two hours east of me.

That would be me I guess. I made a topic a few hours earlier but put in in the wrong forum. It's here: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=195533[/url], and I also experimented with turning DST off.

same here too the clocks are slow by an hour

Yeah, we know.  It should be fixed by tomorrow.


calculus
integration
trigonometry
logarithms
geometry
geometric transformation
reflection

The problem was didn't make sense, so that was deleted.

but when $x=1$ , the expression inside the integral is undefined(reaching infinity)? So we cannot perfome the integration ?

Sorry for that , you are right. I will edit.

kunny

Could you give a hint please? I've tried all the tricks in my book with no success.

Kalle, sorry for my delaying reply to you.This problem is very trciky. Remark that $\sin^{2}\theta+\cos^{2}\theta =1.$

kunny

Kunny do you know solution. I tried very much but i can not solve . As if it is a mount.

Hello, everyone.

I'm very sorry that this problem is mistaken and taking troubles you to take much time. I should have examined in detail. I see to it that afterwards.

Kunny

[quote="anih"][quote]but when x=1 , the expression inside the integral is undefined(reaching infinity)? So we cannot perfome the integration ?[/quote]

I don't get it:  $1 \notin [e^\frac{\pi}{4},e^\frac{\pi}{3}]$
Is that the correct version of the problem(after that remark)?[/quote]

As kunny said, the original post was edited to reflect shyong's complaint.  Mathematica (and also kunny, from his last post) thinks there is no elementary expression for this integral.

[quote="JBL"]Mathematica thinks there is no elementary expression for this integral.[/quote]
How do you tell Mathematica to investigate this?

Mathematica has built-in simplification algorithms in its integrator, so it usually doesn't spit out things like a hypergeometric function (as it does in this case) unless it is unable to find an elementary form.  There are also the functions Simplify and FullSimplify, which respectively try and try harder to make things less ugly.  It's not flawless, of course, (I'm always annoyed because it doesn't simplify $\sqrt{x^{2}}$) but for the vast majority of familiar integrals it will give you a nice answer when a nice answer exists.


[hide="hint"]vectors[/hide]
any other way?
[hide]it's actually easy to see that this is right if you look at this a,b c,d e,f g,h and put some of them negative, but what about a generalization a,b,c d,e,f..[/hide]

The sums naturally hint a vector solution.

Consider the plane vectors (a,b),(c,d),(e,f),(g,h).

The dot product between some two of these must form an angle less than 90 degrees.


[quote="physicsholic"]Weather right or not..I wrote a math paper(theme is secret for now :)) At least, I want to confirm it. Is there anyway?

One more thing..flag is S.Korea's, but I live in USA.[/quote]

Talk to some college professor, he'll know

[quote="José"]
Talk to some college professor, he'll know[/quote]

Or she'll know, as the case may be.

Whether you're a high-school student, college student, graduate student, or faculty member, the process of publishing original research is the same.  (Expository papers are different.)  You need to find an appropriate journal, and follow the submission instructions there.  Most journals require (or at least request) that papers be typeset in LaTeX.  What journals would be appropriate depends on the subfield of mathematics and on how significant the result is.  So, in order to get specific advice, you'll have to share your results.

In general, researchers do not keep their results secret.  If you're worried about having your results stolen, the thing to do is to submit your paper to the arXiv.  Then the results appear with your name on them and the date of submission, so you have evidence of when you produced the results.

[quote="physicsholic"]
One more thing..flag is S.Korea's, but I live in USA.[/quote]

Hm, you're the one who choose the flag when you created your account...identity crisis perhaps :D (just kidding with you).  Go to profile > country flag and change it to the usa flag.  Then just enter your password and hit submit.

[quote="physicsholic"]Weather right or not..I wrote a math paper(theme is secret for now :)) At least, I want to confirm it. Is there anyway?

One more thing..flag is S.Korea's, but I live in USA.[/quote]

Well, what have you done finally???


I feel this question is easy but I post it in High School forum and I feel it's too hard there. So I just get confused:
The question is:

Prove that:
$ \tan 35 \left(\sin 10 \plus{} \cos 10\right) \equal{} \cos 10 \minus{} \sin 10$

This is my first post in this hard forum be merciful in the solutions please don't put harsh stuff  :D

Thanking you

Take a $ \cos 10$ outside from both the sides. so we need to prove  $ \tan 35 ( 1 \plus{} \tan 10 ) \equal{} 1 \minus{} \tan 10$

$ \tan 35 \equal{} \tan ( 45 \minus{} 10 ) \equal{} \frac { 1 \minus{} \tan 10 }{ 1 \plus{} \tan 10 }$  as $ \tan 45 \equal{} 1$

Therefore $ \tan 35 ( 1 \plus{} \tan 10 ) \equal{} 1 \minus{} \tan 10$ which is the result we wanted

$ \tan 35(\sin10 \plus{} \cos 10) \equal{} \cos 10 \minus{} \sin 10$
$ \Leftrightarrow \tan 35 \equal{} \frac{1 \minus{} \sin 20}{\cos20}$
$ \Leftrightarrow \sin70\tan35 \plus{} \cos70 \equal{} 1$
$ \Leftrightarrow \sin^2(35) \plus{} \cos^2 (35) \equal{} 1.$

done :P

Please guys could you explain your computaions  :?:  :(

$ \tan 35 \equal{} (\cos 10 \minus{} \sin 10)\div (\cos 10 \plus{} \sin 10) \equal{} (\cos 10 \minus{} \sin 10)^2 \div (\cos^2 10 \minus{} \sin^2 10) \equal{} (1 \minus{} \sin 20) \div (\cos 20)$

use the fact that $ \cos 20 \equal{} \sin 70$ and $ \sin 70 \equal{} 2 sin 35 cos 35$ and $ \cos 70 \equal{} \cos^2 35 \minus{} sin^2 35$ then we get the desired result.

To prove $ \tan 35(\sin 10+\cos 10) =\cos 10-\sin 10$

Now take out   $ \cos 10$ on both sides so we have to prove $ \cos 10 \tan 35(\tan 10+1) =\cos 10(1-\tan 10$

Cancel the two $ \cos 10$ on both side leaving us to prove $ \tan 35(\tan (10) +1 ) = 1-\tan 10$

now $ \tan 35 =\tan ( 45-10 ) =\frac{ 1-\tan 10 }{ 1+\tan 10 }$

 $ [ \tan ( A-B) = \frac{ \tan A - \tan B }{ \tan A + \tan B }$ and $ \tan 45 = 1 ]$

Therefore $ \tan 35(\tan (10) +1 ) = 1-\tan 10$

Hence proved

Thanks a lot guys

The last two questions  :oops:

@zool007
Just explain this only  :)  $ \tan 35 \equal{} (\cos 10\minus{}\sin 10)\div (\cos 10\plus{}\sin 10)$
@"vijaymenon"
OK   :|  where is the $ \sin$

When iam taking a $ \cos 10$ outside then it becomes $ \frac{ \sin 10 }{\cos 10 } \equal{}  \tan 10$

If u have any doubt jus write $ \tan 10 \equal{}  \frac{ \sin 10 }{\cos 10 }$ and we will get back the original expression.

What do u do ?? in high school ??

it's something called working backwards.


Let $\left\{U_{1},....,U_{1985}\right\}$ and $\left\{K_{1},....,K_{1985}\right\}$ two partitios of the set $A=\left\{1,2,...,n\right\}$
,(i.e $U_{i}\cap U_{j}=\oslash$ for all $i\neq j$ and $U_{1}\cup ...U_{1985}=A$
and $K_{i}\cap K_{j}=\oslash$ for all $i\neq j$ and $K_{1}\cup ...K_{1985}=A$)

such that for all $1\leq i,j\leq 1985$ if $U_{i}\cap K_{j}=\oslash$ , then
$\left|U_{i}\right|+\left|K_{j}\right|\geq 1985$. Find the minimum value of n for which such partitions exist .

Hey I wonder about the source of this problem :blush:( [color=red]Armenia TST[/color]  :roll: ). From which TST is this problem? If it is from IMO team selection test, I can't figure out where did you find this problem?As far as I know there is no any book with Armenian TST,is there?
Please, don't leave me in the dark :D .

[quote="KMN"]Let $\left\{U_{1},....,U_{1985}\right\}$ and $\left\{K_{1},....,K_{1985}\right\}$ two partitios of the set
such that for all $1\leq i,j\leq 1985$ if $U_{i}\cap K_{j}=\oslash$ [/quote]
there existe $1\leq j\leq 1985$ such that $1\in k_{j}$
there existe $1\leq i\leq 1985$ such that $1\in U_{i}$
then $U_{i}\cap K_{j}\neq \oslash$
so, ((((((((((for all $1\leq i,j\leq 1985$,$U_{i}\cap K_{j}=\oslash$)))))))) is false :huh:

The assumption on $U, K$ is $\forall i\forall j (A(i,j) \Longrightarrow B(i, j))$
where $A(i, j)$ is $U_{i}\cap K_{j}= \emptyset$ and $B(i, j)$ is $|U_{i}|+|K_{j}| \geq 1985$.

thank you,
sorry becaue i am bad in englesh  :blush:


function
algebra
domain
induction
algebra solved

Find all functions f : Z (integers) \ {0} -> Q (rationals) such that
f((x+y)/3) = 1/2*(f(x)+f(y)) for all x,y in the domain of the function.

First of all we can observe that by taking x=1 and y=2 we get f(1)=f(2). We then take x=y=3 and we get f(2)=f(3). We now take x=2, y=4 to get f(1)=f(2)=f(3)=f(4). The next step is x=1, y=5 and we get f(1)=f(2)=..=f(5).

From now on we can use induction: Let's assume we've proven f(1)=f(2)=f(3)=..=f(3k-1) (let's call this P_k). We make x=3 and y=3k to get f(k+1)=(f(3)+f(3k))/2, meaning that f(3k)=f(1). Then x=2, y=3k+1 gives us f(3k+1)=f(1). The next step is x=1,y=3k+2. This gives f(3k+2)=f(1) so f(3(k+1)-1)=f(1), which is, in fact, P_(k+1), so induction works: P_k->P_(k+1).

This means that f is constant for the positive numbers (call its value a). By the same method we prove that it's constant for the negative numbers (call its value b). x=-1,y=4 gives us f(1)=(f(-1)+f(4))/2, so a=(a+b)/2, so a=b, so f is constant on Z\{0}.


How do you find the equation of the line tangent to the graph of  $ f$ at $ (1,1)$ where $ f$ is given by $ f(x)\equal{}3x^3\minus{}3x^2\plus{}1$.

$ y\minus{}y_{0}\equal{} (y_{0})^{\prime}(x\minus{}x_{0}),$ where $ (x_{0}, y_{0})\equal{}(1,1)$ and $ (y_{0})^{\prime}\equal{}\frac{dy}{dx}(x_{0}).$

Just express $ y$ in terms of $ x$ in the above equation.

Hmm... I am a little confused about how you solve this part: $ (y_{0})^{\prime} \equal{} \frac {dy}{dx}(x_{0})$

Is it $ {(f(x)}'\equal{}9x^2\minus{}6x)*(1)$?

Making it $ y\minus{}1\equal{}(9x^2\minus{}6x)(x\minus{}1)$ which would make $ y(x)\equal{}9x^3\minus{}15x^2\plus{}6x\plus{}1$?

Or am I way off and just confusing myself with my limited skills?

You should calculate $ f^{\prime}(x_{0})\equal{}f^{\prime}(1)\equal{}3$.

Now you get $ y\minus{}1\equal{}3(x\minus{}1)$. So $ y\equal{}3x\minus{}2.$


Let $ ABC$ be an acute triangle and $ H$ its orthocentre.

Show that: $ \sqrt{2}< \frac{AB\plus{}BC\plus{}CA}{AH\plus{}BH\plus{}CH}<2.$

Actually, I think your inequality is not very strong.
Try that proof:
$ \frac{AB\plus{}BC\plus{}CA}{AH\plus{}BH\plus{}CH}<2
\Leftrightarrow AB\plus{}BC\plus{}CA<2AH\plus{}2BH\plus{}2CH
\equal{}(AH\plus{}BH)\plus{}(BH\plus{}CH)\plus{}(CH\plus{}AH)$
Which is obviously true (triangle basic inequality)
One inequality's done
Now another:
Let $ N\in AB, M\in AC$ such that $ \widehat{NHB}\equal{}\widehat{MHC} \equal{}90^o$
 $ AM\plus{}AN\equal{}AN\plus{}NH>AH$
       $ NB>BH$
$ MC>HC$
$ \rightarrow AB\plus{}AC>HA\plus{}HB\plus{}HC$
Do the same work we have:
$ AB\plus{}AC>HA\plus{}HB\plus{}HC$
$ AB\plus{}BC>HA\plus{}HB\plus{}HC$
$ AC\plus{}BC>HA\plus{}HB\plus{}HC$
$ \rightarrow AB\plus{}BC\plus{}CA>\frac{3}{2}(HA\plus{}HB\plus{}HC)>\sqrt{2}(HA\plus{}HB\plus{}HC)$

[quote="Mateescu Constantin"]Let $ ABC$ be an acute triangle and $ H$ its orthocentre.

Show that: $ \sqrt {2} < \frac {AB \plus{} BC \plus{} CA}{AH \plus{} BH \plus{} CH} < 2.$[/quote]

the inequality is equivalent to

$ \sqrt {2}(R\plus{}r) < p < 2(R \plus{} r)$ Which is true and kind of weak


quadratics
LaTeX
algebra
linear equation

How to solve this?

2x + sqr{2x^2} = 16 + 16sqr{2}  ?

Thanks.

[hide="Hint"]Show that $ x$ has to be positive, and then take the root (making $ \sqrt 2x^2\equal{}x\sqrt2$) and solve the linear equation regularly.

[/hide]

[hide="Answer"]$ x\equal{}8 \sqrt 2$[/hide]

I dont know how to get rid of LHS root, could you please show me the procedure? Thanks.

2x + sqr{2x^2} = 16 + 16sqr{2}

in the first term 2x, look at 2 as (sqr{2})^2
and second term sqr{2x^2} = x(sqr{2})

then move everything to the right , and we have a quadratic equation

x (sqr{2})^2 + (x-16)(sqr{2}) -16 = 0
factor it out

( x(sqr{2}) - 16 )*(sqr{2} + 1 ) = 0
so
x(sqr{2}) - 16 must = 0

x(sqr{2}) - 16 =0
x(sqr{2})=16
x=16/sqr{2} = 16(sqr{2})/2 = 8(sqr{2})

(sorry, i dont know how to use LaTex, so this is kind of messy :oops: )

[/b]

I'll LaTeXify it for you.  :)

$ 2x \plus{} \sqrt{2x^2} \equal{} 16 \plus{} 16\sqrt{2}$

In the first ter $ 2x$, look at $ 2$ as $ \sqrt{2}^2$
and second term $ \sqrt{2x^2} \equal{} x\sqrt{2}$.

Then move everything to the right, and we have a quadratic equation.

$ x\sqrt{2}^2 \plus{} (x \minus{} 16)(\sqrt{2}) \minus{} 16 \equal{} 0$

Factor it out.

$ (x\sqrt{2} \minus{} 16)(\sqrt{2} \plus{} 1) \equal{} 0$

so

$ x\sqrt{2} \minus{} 16$ must be equal to $ 0$.

$ x\sqrt{2} \minus{} 16 \equal{} 0$

$ x\sqrt{2} \equal{} 16$

$ x \equal{} \frac{16}{\sqrt{2}} \equal{} \frac{16\sqrt{2}}{2} \equal{} \boxed{8\sqrt{2}}$

No real need for complicating it (btw, generally speaking, keep in mind that $ \sqrt x^2 \neq x$, it is $ |x|$)

so our equation could have been rewritten as

$ 2x \plus{} |x|\sqrt 2 \equal{} 16 \plus{} 16\sqrt 2$

Now, if we look at two cases, we have:

1. $ x < 0$, LHS is obviously negative (it becomes $ 2x \minus{} x\sqrt 2 \equal{} x\sqrt 2 (\sqrt 2 \minus{} 1))$, and as $ x$ is negative, it is all negative), and RHS is positive, so it is a contradicting case. No solutions.

2.$ x\geq 0$, our equation is now $ 2x \plus{} x\sqrt 2 \equal{} 16 \plus{} 16\sqrt 2$, or $ x\sqrt 2(\sqrt 2 \plus{} 1) \equal{} 16(\sqrt 2 \plus{} 1)$, i.e. $ x \equal{} 16/\sqrt 2 \equal{} 8\sqrt 2$.


[quote="mountain.dew"]Let $ z$ be a complex number and $ \bar{z}$ be its conjugate. What are the values of $ z$ for which $ z\bar{z}=5$ and $ z^{2}+\bar{z}^{2}=6$?[/quote]

If $ z\bar{z}=5$ and $ z^{2}+\bar{z}^{2}=6$ then $ (z+\bar{z})^{2}=6+2\cdot5=16$ and $ (z-\bar{z})^{2}=6-2\cdot5=-4$. Therefore $ z=\pm2\pm i$.


Let be $ a\in \mathbb{R}$ . Find $ x,y,z\in \mathbb{R}$ that satisfies $ x\plus{}y\plus{}z\equal{}4a$ and $ xy\plus{}yz\plus{}zx\equal{}7a^2\minus{}az$ .

[quote="alex2008"]Let be $ a\in \mathbb{R}$ . Find $ x,y,z\in \mathbb{R}$ that satisfies $ x \plus{} y \plus{} z \equal{} 4a$ and $ xy \plus{} yz \plus{} zx \equal{} 7a^2 \minus{} az$ .[/quote]
$ \Leftrightarrow 16xy \plus{}16 yz \plus{} 16zx \equal{} 7*16a^2 \minus{} 16az$
$ \Leftrightarrow 16xy \plus{}16 yz \plus{} 16zx \equal{} 7(x\plus{}y\plus{}z)^2 \minus{} 4(x\plus{}y\plus{}z)z$
$ \Leftrightarrow 16xy \plus{}16 yz \plus{} 16zx \equal{} 7(x\plus{}y\plus{}z)^2 \minus{} 4(x\plus{}y\plus{}z)z$
$ \Leftrightarrow 7x^2\plus{}7y^2\plus{}3z^2\minus{}2xy\minus{}6yz\minus{}6xz\equal{}0$
$ \Leftrightarrow (x\minus{}y)^2\plus{}\frac{3}{2}(z\minus{}2x)^2\plus{}\frac{3}{2}(z\minus{}2y)^2\equal{}0$
$ \Leftrightarrow \left\{\begin{array}{c} x\equal{}y\\z\equal{}2x\\z\equal{}2y\end{array}\right.$
Consider $ x\equal{}y\equal{}t \Rightarrow z\equal{}2t$
Since $ x\plus{}y\plus{}z\equal{}4a \Rightarrow 4t\equal{}4a \Rightarrow t\equal{}a \Rightarrow \left\{\begin{array}{c} x\equal{}y\equal{}a\\z\equal{}2a\end{array}\right.$


Can anyone post a solution to the following:
If $ x \plus{} y \equal{} 270$, $ 0 < x,y < 180$, and $ x,y \neq 90$, find the maximum value of $ 4sinx \plus{} 3siny$.

What I did was very non-legit. I said that since 4 and 3 are relatively close, just set both x,y =135 to get the answer of 4.95 when the actual answer is 5 (this wasnt a proof question, so it didnt really matter that I made this assumption since I rounded up to 5 anyways). I was wondering what the actual legit solution to this is.

Thanks.

[hide="solution"]$ 4\sin x \plus{} 3\sin y \equal{} 4\sin x \plus{} 3\cos (90 \minus{} y) \equal{} 4\sin x \plus{} 3\cos(x \minus{} 180) \equal{} 4\sin x \minus{} 3\cos x$

$ \equal{} 5\left(\frac {4}{5}\sin x \minus{} \frac {3}{5}\cos x\right) \equal{} 5\sin \left(x \minus{} \sin^{ \minus{} 1}\frac {3}{5}\right)\leq 5$.

Thus the answer is 5, for $ x \equal{} 90 \plus{} \sin^{ \minus{} 1} \frac {3}{5}$ and $ y \equal{} 270 \minus{} x$.[/hide]


geometry
similar triangles
arithmetic sequence
perpendicular bisector

Note to BHorseMath:  [hide]What you did was okay because it worked for this problem.  However you cannot concluded that x=2 and y=1.  Try doing it a little more algebra-ish by simplifying 2x + 2y = 6 so that the coefficents are relativley prime. [/hide]

[hide]2) the triangle is a 5-12-13, the altitude of the hypotenuse is 5 units long

[/hide]

Be more careful mathfiend -- look what it is asking you for and then check if that's what you found.

I believe it would depend on how one would draw the triangle. When I drew it, the altitude was [hide]5[/hide]. When drawing it another way, the altitude would be [hide]12[/hide].

[quote="Rep123max"]
2. A right triangle has an area of 30. If the length of the hypotenuse is 13, what is the length of the altitude [b]to the hypotenuse[/b]?
[/quote]

That means the one perpendicular to the hypotenuse.

An altitude is the line at right angles to the given side of a triangle through the opposite vertex.. its not actually one of the sides of a triangle. Maybe you are mixing it up with the word "adjacent"?

On number 3, why do you even need to know that [tex]1+3+5+7+9+ \ldots +n=(\frac{n+1}{2})^2[/tex]?

Mathfiend -- what is the hypotenuse of a right triangle?

Rep123max wrote:Exactly. Instead of just telling the formula for the sum of all even numbers, I told the formulas for the sum of all natural numbers and the sum of all the odd numbers.

I think Robert was trying to insinuate that:

[hide]The sum of the first n natural numbers multiplied by two gives you some even numbers [/hide]

[hide] 1. x+y+z=8, 2x+2y=6, this means the x+y=3, so z=5.

2. If it has an area of 30, then 13*altitude/2=30, so 13*a/2=30, 13*a=60, a=60/13=4 and 8/13.

3. This is just double n(n+1)/2, so it is n(n+1), where is half the last number in the sequence, so n=60. 60(60+1)=3600+60=3660.

4. Using a:^2:+b:^2:=c:^2:, then we know that the sum of the areas of the other two semicircles should equal the area of the semicircle drawn on the hypoteneuse, therefore the answer is 29.7. [/hide]

[quote="Ragingg"]By the way Robert, knowing both those formulas would be good as some of the terms are similar in both.  Suppose you subtracted them... what would you get?[/quote]

couldnt you do the formula 1+2+3... and multiply it by 2?

1. [hide]2x+2y=6, so x+y=3 and substitute that into x+y+z=8 to get z=5.[/hide]

2. [hide]It's a 5-12-13 triangle. Draw the altitude to the hypotenuse to form two triangles that are similar to the big one.  The altitude to the hypotenuse is one of the sides of the small triangles.  Using the similar triangles, you can then solve for the altitude. (it's 60/13?)[/hide]

3. [hide]2+4+6+...+120 has 60 terms, sum of arithmetic sequence is n(t1+tn)/2=60(2+120)/2=3660.[/hide]

4. [hide]The radius of each semicircle is 1/2 the length of the side it's on: the area of a semicircle is :pi:/2*r^2. Multiply the Pythagorean Thm. by :pi:, you can see that the area of the semicircle on the hypotenuse is equal to the sum of the areas of the other two semicircles.[/hide]

1.[hide]x+y = 8-z

2(x+y) = 2x+2y = 6 = 2(8-z)

6 = 16-2z

-10 = -2z

z = 5[/hide]

2.[hide]

I don't think this is right, since it can't be factored.

[/hide]

3.[hide]You just add 1 to (n+1) since 1+1 = 2, 3+1=4, and so on.

(120+2)2/22

1242/4

622

3844[/hide]

4.[hide]Let's first try to find out the length of the hypontenuse.

29.7*2 = 59.4

59.4/3.14 is about 18.9

The radius is  :sqrt: 18.9

The diameter, or hypotenuse is 2 :sqrt: 18.9

Squared, that is 75.6, or the other two sides' added together after being suqared.

I don't know where to go after here, since the number isn't really a number I fell confortably working with.  Did something go wrong here?[/hide]

h_s_potter2002 wrote:Am I supossed to notice that the triangle in number 2 is a [hide]5-12-13[/hide] triangle?  I didn't think I should have, and I complicated it I guess.

YOu don't have to notice that it's a 5-12-13 triangle. Hint: [hide]Hypotenuse [tex]\times[/tex] Altitude to the hypotenuse/2=the area of the triangle[/hide]

Help with 2.

[hide]It's a 5-12-13 triangle.  Doesn't the altitude to the hypotenuse cut the hypotenuse in half?

If it does, that makes the hypotenuse 6.5, and so 6.5 2 +x 2 =12 2 .  So then we get x 2 =101.75, and x=root101.75

Is this correct?[/hide]

[quote="NightFlarer"]Help with 2.
[spoiler]It's a 5-12-13 triangle.  Doesn't the altitude to the hypotenuse cut the hypotenuse in half?
[/quote]

Definitely no! A point lies on the perpendicular bisector of a segment iff it's equidistance from the endpoints. Since 5 is not equal to 12, it's impossible.

oh my god...that are problems we are doing in school...except that it could be that we are not given the formula for number 3 but have to derive them on our own...

Peter

Oh yeah, Peter, you really hit the point... and actually, even [i]my[/i] maths teacher, after I returned from the IMO, organized a little geometry contest in my class, and the problems given in that contest were much more "challenging" (oh dear, I hate this word) problems than the ones proposed in this thread...

But I can understand Rep123max's intention on posing those problems; he just wanted "[i]the people who only visit the non math forums[/i]" to post less spam and to learn the tiniest bit of maths...  :D

  Darij (yes I know I'm spamming again, I'll try to post some more mathematical comments the next days)

Do you guys have any respect for people no as smart, or as old as you?  There is a reason this part of the board is called Getting Started.  The people here aren't as smart as IMO patricipants.  A lot of people haven't even gone into Algebra yet, they can probably barely do the American qualifying rounds for their team.

If you don't want to try easy problems, then don't come to this part of the board, go to the Advanced or Undergraduate boards.  Not everyone can get into IMO, or even do well in the qualifying rounds since we're barely in our teens and haven't learned much maths yet.

So basically, don't make fun of other people just becuase they don't know as much maths as you do.

Maybe you didn't notice that these problems are posted in the Getting Started area.  Perhaps you should look up "getting started" in the dictionary.  That should help.  Or perhaps your English is poor and you think "Getting Started" means "Advanced."  Need I remind you that both of you were once beginners?  It's sad how so many people become cocky once they make it to the IMO.

And if I have misunderstood your messages, excuse me.

It  appears to be the German thing over again:

      " Deuchland Deuchland  uber alles ".

     You guys had to do much better at IMO to make your statements.

    Pestich.

I feel I'm getting mocked... :blush:

I agree with hspotter and nightflarer but Rep says this is for everyone to do.  Maybe you IMO people should just keep it low a little bit.

Actually, maybe it was about a misinterpretation of the Getting Started topic. As far as I understand, it is a topic aiming at school students who are already good at maths in school but want to learn some more maths. At least, this is the usual aim of such fores. Helping students who are not even on the school level in maths is usually the task of pedagogists and, on the internet, the task of some other fores, which are devoted to helping people with their homework. Maybe I just misunderstand the intention of this forum - then I will be glad if you clear me up about these things. But if I am right thinking that this forum is devoted for school students already being at school niveau and now wishing to increase their level beyound what is usually studied in school - then the "challenge" problems posted in these thread clearly miss the point.

Concerning the ages, both of us - Peter and me - are teens, too. Peter is 16, and I am 15. I am actually opposed to thinking that the mathematical abilities of someone are tied to his age - but even if you think otherwise, you will still agree that the main question is not the age.

So much concerning the posting of h_s_potter2002. Actually I am not sure about some of the points since I really have not seen any official and precise information what kind of students the Getting Started forum aims at, beyond the very laconic subtitle "For beginners to learn from more experienced problems solvers." leaving the question whether the writer meant "beginners" in mathematical olympiads (that is the way I understand the title) or "beginners" in school mathematics. I admit that I hadn't read the http://artofproblemsolving.com/Forum/viewtopic.php?t=10857 thread before, but now I have read it and it didn't answer most of my questions, but all discussions, including this one, make me think that the users of the Getting Started forum are participants of some mathematical olympiads, although not the IMO, and not just school students seeking help with their homework.

What concerns NightFlarer's post, the arguments were quite the same so I refer to above.

Finally, about Pestich's reply, I think he gives just one reasonable argument or at least something that can be understood as an argument in the wider sense of the word. By this I mean the fact that we (Peter and me) were not the best scorers in this year's IMO. But actually we didn't claim anything like this. We just claimed that the problems posted in this thread were faaaar too easy, and in order to understand this it is not of much importance whether one has 42 points on the IMO or a honorable mention (many people who don't even have a HM on the IMO are far above this level of problems). This, BTW, has nothing to do with Germany. None of us spoke in the name of Germany, neither in the name of the German IMO team. Each of us spoke in his and only his name. Incidentally, I have emigrated from Russia, now living in Germany, my roots are Jewish and my mother was born in the Ukraine. I don't identify with any nationality just 'cuz I don't find this necessary (more to this point, I find this contraproductive)! I don't know what is Peter's opinion about this, but I can hardly imagine he spoke out of proudness for Germany.

  Darij[/i]

Ok.  I will try to outline what kinds of problems should and should not go in the getting started forum:

These problems should be 1-3 step problems.  By this I mean that all you have to do is apply a couple of formulas or draw in a few lines, etc. to finish the problem.

Anything with the word prove in it is 99% of the time not supposed to be in this forum.

Any kind of Olympiad problem regardless of country or how stupidly easy it was should go in another forum.

These problems should be from contests such as Mathcounts/AMC 8/AMC 10/ and easier AMC 12 problems.  If you are not familiar with these contests here is a link for where you can find AMC 12 problems.

[url]http://www.math.ksu.edu/main/events/hscomp/samples/amc12/sample.htm[/url]

I also advise some of you newer people to look at old topics.  You should get a sense of what this forum used to be.

Finally, if you are not sure what level a problem is put it in Intermediate Forum.  If it is too easy for that forum they can put it in here... if it is too hard they can advance it to a higher forum.

You IMO guys didn't just know how to do beginner stuff when you are born.  You shouldn't expect everyone on this forum to know as much as you(or even half as much for that matter).  This is the forum where people can acquire such knowledge.  It seems like you are just trying to do away with the "Getting Started" forum and create another forum for olympiad problems.  Maybe I should ve mentioned this earlier but this is not Getting Started to Olympiad problems it is Getting Started to problem solving in general.  I am sorry if you were confused about that. You guys already have more than 10+ forums/subforums for Olympiad level problems.  I can't imagine that would want more.

I am sorry if I came across rude but I think that we really need to reach an understanding on what difficulty this forum should be.

Thank you,

Joseph L.

Okay, looking at the problems posted here I find you are right. Thanks for giving a very precise definition of the aim of this forum, now it's clear.

  Darij

[hide]

1. If 2x + 2y  = 6, (x+y) = 3, and so z + 3 = 8, so z is 5.

2. Since it is a right triangle, the altitude must be half the hypotenuse, but (13*6.5)/2 =/= 30. I don't think it's possible for a right angled triangle to have an area dependant on anything else but the hypotenuse or the adjacent sides.

4. As the area of the semicircles are proportional to the squares of the lengths=, pythagoras will work here (hopefully). So the sum of the two semicircles on the legs will also be 29.7[/hide]

[quote="darij grinberg"]Actually, maybe it was about a misinterpretation of the Getting Started topic. As far as I understand, it is a topic aiming at school students who are already good at maths in school but want to learn some more maths.[/quote]

There has been a misinterpretation here about what "Getting Started" means. The Getting Started forum is for AoPS-Mathlinks site participants who have little or NO experience with mathematics competitions at any level. Anyone who has been to IMO, even once, is far beyond the intended level of the Getting Started forum and shouldn't be answering challenge problems in this forum.

In the United States, and in China (at least), mathematics competitions start at the late elementary school level and include problems that involve NO knowledge of algebra. Problems at that level are very appropriate for this forum, even if they look easy to people who have more experience in mathematics competitions. The solution for finding harder problems is to look for them in the higher level forums (Intermediate, Advanced, and College Playground). The persistent problem on most forums on AoPS has been problems that are too hard for their stated level, not problems that are too easy, so I as moderator often "promote" problems to a higher level if they are posted here but not really at a getting started level.

[quote="darij grinberg"]Helping students who are not even on the school level in maths is usually the task of pedagogists and, on the internet, the task of some other fores, which are devoted to helping people with their homework. Maybe I just misunderstand the intention of this forum - then I will be glad if you clear me up about these things.[/quote]

Yes, you are misunderstanding the intent of this forum. Thank you for asking to be cleared up. On this Getting Started forum, anyone is welcome to ask informational questions about mathematics at even the most elementary level. It is best to post questions with the title "Question: " so that everyone is clear what is a question and what is a challenge problem. Anyone is also welcome to post challenge problems here, but we moderators ask that only the BEGINNING problem-solvers post answers to challenge problems, until the true beginners have had a chance to answer. The experienced problem-solvers may then comment (preferably in spoiler text) on the solutions posted by the beginners. Please give the beginners a chance to practice answering easy challenge problems, and please remember that everyone has to start at the beginning.

[quote="darij grinberg"]Concerning the ages, both of us - Peter and me - are teens, too. Peter is 16, and I am 15.[/quote]

Several of the participants on this forum are eleven years old. In any case, some of the older participants have had very limited opportunities to learn mathematics well at school (not all schools in all places are very good) and have no math competition experience. This forum is designed to help the beginners, even if they are older than you are.

[quote="darij grinberg"]Actually I am not sure about some of the points since I really have not seen any official and precise information what kind of students the Getting Started forum aims at, beyond the very laconic subtitle "For beginners to learn from more experienced problems solvers." leaving the question whether the writer meant "beginners" in mathematical olympiads (that is the way I understand the title) or "beginners" in school mathematics.[/quote]

You admitted that you didn't read earlier threads on the intended level of this forum until AFTER you posted in this thread the first time. What you still need to think about is that the structure of mathematics competitions in different countries is not always the same. In China, for example, the term "olympiad" is used for mathematics competitions at all levels, but many of those competitions involved very elementary mathematics (although they are given to pupils of impressively young ages, by comparison to math contests in the United States). Much of the English-speaking world has math competitions for young people who are completely ignorant of algebra, and that is expecially the level that this Getting Started forum is intended for. If you have the experience level you indicate for yourself (and at the AGE you indicate for yourself, in any event), then you should be posting answers to challenge problems AT LEAST in the Intermediate forum, and probably mostly in the Advanced forum. The Getting Started forum is for younger people with less competition experience.

I hope that clears everything up. Different countries have different systems, but on this international Web site the Getting Started is consciously intended to serve up EASY problems for absolute beginners in any form of math competition.

Thanks a lot for explaining. I usually thought that "Getting Started" meant something like the former "Beginner's Corner" on MathLinks, but now I see it's a big difference...

  Darij

yeah, i agree with darij in any point. i thought as well this "getting started" would already deal with easy olympiad level problems, sorry for misunderstanding. i won't do such a post in the future.
and though i am actually a german(from the former gdr) and not immigrated like darij, i don't hold much onto the fact i am one and words like "deutschland deutschland ueber alles" would never come out of my mouth.
so, again sorry for that post.

Peter

some problems like that you are doing at school Peter? Lucky you. I'm as old as you're but I started in school first when I was seven.
I have to say some of that problems I would have to think some time but I would be able to solve them. And I'm a real unexperienced problem solver. But maybe it's just because of my maths teacher. Once he wasn't able to solve a problem into our math book 10 minutes later I solved it for him (and these are real easy ones). Later I told it to my boyfriend who was a member of the German IMO team 2003 and who had the same math teacher he just asked: What do you expect from a basketball player?
So maybe I have just really bad luck with this stupid teacher.


AMC
USA(J)MO
USAMO
geometry
similar triangles
perpendicular bisector

[quote]Let ${\cal C}_1$ and ${\cal C}_2$ be concentric circles, with ${\cal C}_2$ in the interior of ${\cal C}_1$. From a point $A$ on ${\cal C}_1$ one draws the tangent AB to ${\cal C}_2 (B\in {\cal C}_2)$. Let $C$ be the second point of intersection of $AB$ and ${\cal C}_1$, and let $D$ be the midpoint of $AB$. A line passing through $A$ intersects ${\cal C}_2$ at $E$ and $F$ in such a way that the perpendicular bisectors of $DE$ and $CF$ intersect at a point $M$ on $AB$. Find, with proof, the ratio $AM/MC$.[/quote]
I like this problem. You should too. It is very simple  :lol:

[quote="probability1.01"]Yeah. 5/3.[/quote]
what part of "with proof" do you not understand, the "with" or the "proof"?? JK, lol. proof, please  :D

[quote="TMill"][hide]WLOG, assume that length AB = 1
Let x = length AE

AE * AF = AB * AB by a certain geometry theorem with a name I don't remember
AF = 1/x

After this point, I assume E is between A and F rather than F between A and E.  If this assumption is not true, just switch all references to E and F past this point.
AE/AD = AC/AF, since x/(1/2) = 2/(1/x) = 2x
triangle ADE ~ triangle AFC since sides are proportional, and angle DAE is equal to itself
angle ADE = angle AFC by similar triangles
angle CDE is supplementary to ADE, and therefore supplementary to AFC
therefore C, F, E, and D are cyclic.
Since M is on the perpendicular bisector of CF and DE, it must be the center of this circle.

CM = MD = 1/2 * 3/4 = 3/8
AM = 3/8 + 1/4 = 5/8
AM/MC = 5/3[/hide]
My geometry is a bit rusty.[/quote]
very good  :lol: and welcome to AoPS


As you know, Cantor's diagonalization can be used to prove that EXPTIME problems exist, by listing out all the PTIME programs, and then listing out the decision problems each one solves, selecting all the values on the diagonal and flipping them, thus creating a new decision problem that is not solved by any program in PTIME.  However, why can't we use the same proof, but list out all possible turing machines instead of just the PTIME ones?   Then when we pick out the values along the diagonal and flip them, we get a decision problem that cannot be solved by any turing machine, thus its unsolvable.  But it can't be unsolvable because you can always generate the nth turing machine, calculate the output and flip it to output the answer.

There's a problem in the logic somewhere here, but I'm not sure where it is.

[i] But it can't be unsolvable because you can always generate the nth turing machine, calculate the output and flip it to output the answer.[/i]

The problem is in the "calculate the output" part. For an arbitrary Turing machine you can not calculate the output AND THEN flip it, because a Turing machine can reject a word by making an infinite computation that never reaches a final (accepting) state.

In other words, what you found is just a proof that there really are problems that are "not solvable", or, better phrased, you constructed a language that is not recursively enumerable. (Please note the difference between solvable=recursive and partially solvable=recursively enumerable problems/languages.)


algebra
polynomial
inequalities
complex numbers
advanced fields
advanced fields unsolved

This problem is difficult for me . Can anyone help me ? :)

Let $A$ be a commutative Banach Algebra with Unit . Suppose $x\in A$ and the set of all polynomials in $x$ is dense in $A$ .
Prove that the completion of the spectrum of $x$ (in $\mathbb{C}$) is connected .

What do you mean by "completion of the spectrum"? :? My first guess was that you meant "closure of the spectrum", but that would be pointless, since the spectrum is already closed (compact, actually).

Oh , I'm sorry .My English is terrible :(
Denote $e$ the unit of $A$
The spectrum of $x$ is the set of all complex numbers $\lambda$ such that $x-\lambda e$ is not invertible . We denote it $\sigma (x)$ .
The complement of $\sigma (x)$ is $\mathbb{C}\backslash \sigma (x)$ :)

You didn't need to actually define the terms, jusy mention that instead of "completion" you meant "complement" :).

Suppose $\mathbb C\setminus\sigma(x)$ has at least one bounded connected component $U$, and let $\lambda\in U$. Take some $\varepsilon>0$ (we'll determine later how small it should be), and a complex polynomial $P$ such that $\|P(x)-(x-\lambda e)^{-1}\|<\varepsilon\ (*)$. From the fact that every complex homomorphism of $A$ has norm at most $1$ and $(*)$ we deduce that $|P(z)-(z-\lambda)^{-1}|<\varepsilon,\ \forall z\in\sigma(x)$ ($\sigma(x)$ is the set of all images of $x$ through complex homomorphisms of $A$), which means that $|(z-\lambda)P(z)-1|<\varepsilon M,\ \forall z\in\sigma(x)\ (**)$, where $M=\max_{\omega\in\sigma(x)}|(\omega-\lambda)^{-1}|$. By the Maximum Modulus Principle, the inequality $(**)$ holds for all $z\in U$ as well, and by putting $z=\lambda$ we get $1<\varepsilon M$. Taking $\varepsilon<\frac 1M$ in the beginning gives the desired contradiction.


If someone could explain how to do the following problems, I'd greatly appreciate it.

$ \bold{1.}$ A town contains $ 4$ people who repair televisions.  If $ 4$ sets break down, what is the probability that exactly $ i$ of the repairers are called? Solve the problem for $ i\equal{}1,2,3,4$.  What assumptions are you making?

[hide="Answers"]$ P(i\equal{}1)\equal{}\frac{1}{64}$
$ P(i\equal{}2)\equal{}\frac{21}{64}$
$ P(i\equal{}3)\equal{}\frac{36}{64}$
$ P(i\equal{}4)\equal{}\frac{6}{64}$[/hide]

$ \bold{2.}$ The second Earl of Yarborough is reported to have bet at odds of $ 1000$ to $ 1$ that a bridge hand of $ 13$ cards would contain at least one card that is ten or higher.  (By [i]ten or higher[/i] we mean that it is either a ten, a jack, a queen, a king, or an ace.)  Nowadays, we call a hand that has no cards higher than $ 9$ a [i]Yarborough[/i].  What is the probability that a randomly selected bridge hand is a Yarborough?

[hide="Note"]I found a solution (trying to search to avoid posting on here) [url]http://www.physicsforums.com/showthread.php?t=199138[/url] at that link, but I cannot make sense of what their final answer is and how they arrived at it.[/hide]

No one?  I also have another problem I'm interested in:

$ \bold{3.}$ An urn contains $ n$ red and $ m$ blue balls.  They are withdrawn one at a time until a total of $ r$, $ r \leq n$, red balls have been withdrawn.  Find the probability that a total of $ k$ balls are withdrawn.
[i]Hint:[/i] A total of $ k$ balls with be withdrawn if there are $ r \minus{} 1$ red balls in the first $ k \minus{} 1$ withdrawals and the $ k^{\text{th}}$ withdrawal is a red ball.

[quote="skimnc"]No one?[/quote]

Be patient -- the members of this forum are not an on-call service to help you with your work on your schedule :)

For your second question, post 2 is the only worthwhile portion of it.

bump

I figured out $ \bold{1.}$ and $ \bold{2.}$, but if anyone could help with $ \bold{3.}$, I would greatly appreciate it.


1. So far a student's math exam grades are 90, 96, 82, and 88. It takes a grade of 70 or better to pass and all grades are whole numbers. If the student receives a passing grade on each of the next two exams, what is the probability that the student will have an average of at least 90? Assume that any score is equally likely to occur and that all exams have 100 possible points.

2. A tank in the form of a rectangular solid with base area of 12 square feet has 8 inches of water in it. If a 1 foot gold cube is placed in the tank, by how many inches will the water rise?

[hide]I did these problems and got 153/961 and 1 respectively. But appearantly the answers are 133/961 and 8/11. Can someone point out where (if) I might be going wrong?[/hide]

Okay, I'm kinda new at this so try and bear with me (my Algebra is a little rusty). In order to find the first problem I put X in for the two unknown grades, and solved it so that the answer was 90(the minumum grade). X comes out to 92 then. That means there are 9 different grades that can make the average grade at least 90. That also means that there are 22 grades that would create an average of below 90%. So there fore the ratio is 9/22.
    I Thought the second problem was easy, but I came up with the same answer( 1 inch).
!!!!![b]Can someone pleaase help me [/b]!!!!

In the second problem, the cube is 12x12x12 so it has a volume
$12*12*12 = 1728\mbox{ in^3}$
So when the cube is placed in the water, it displaces that much water.  The volume of one inch of water is $12\mbox{ in^3}$ so the water rises $\frac{1728}{12} = 144\mbox{ in.}$

note that for number 2, not all of the cube is submerged in water..since the cube is 12 inches high and the water is only 8.
also, as more of the cube goes into the tank, the higher the water level is, thus the more of the cube is in the water, and thus the water level gets higher still...
this looks like calc to me....

ah oops!
yes it does look like calc now that i think about it.. i also didnt see that the area of the base is 12 sq ft for some reason

No, it's not really calc, but it has a trick. Have in mind that the total volume of water is the same, but without the cube its base is the rectangular of the pool, and with the cube its the rectangular minus square of the cube... now it should be easy.

the first one i got 483/1010 which i used my calculator... but for the second one i got 12... notice it says inches while a 144 in^3 cube is dropped. so basically,

8*12=96 and 12*12=144
so the water is 96 in^3 and the cube is 144 in^3
and since it is already 8 inches high, it rises 144/12 inches or 12 inches

remember that gold is denser then water... so it [u]will[/u] sink... and the base of the cube dropped epuals the base of the container (tank) sorry  :lol:  :lol:


Can somebody explain a few approaches and differences for the following problems?

Let a, b, c, d, and e be integers whose sum is 150. How many possible combinations are there for the set {a,b,c,d,e}?

Let a,b,c,d and e be integers whose sum is 150 where a<b<c<d<e. How many possible combinations are there for the set {a,b,c,d,e}?

[quote]
Let a, b, c, d, and e be integers whose sum is 150. How many possible combinations are there for the set {a,b,c,d,e}?
[/quote]

Okay, well think about it this way:

150 sticks. we just need to somehow separate these sticks into 5 groups. the number of sticks in each group is a,b,c,d,e. we need to put in 4 separators. the sticks are identical, so the total number of arrangements is just 154C4.

Generalization: for x_1+....x_r=n, there are (n+r-1)C(r-1) ways to select the set.

Come to think of it though, maybe you shouldnt call it the "set" {a,b,c,d,e}. We should call it an 5-tuple.

When I use a different way like this, I don't get that answer. Can you tell me what I am doing wrong?

Somebody posted this strategy earlier, but I'm going to modify it for this problem:
Let a, b, c, d, and e be integers whose sum is 150. How many possible combinations are there for the set {a,b,c,d,e}?

I let v =a,  w=a+b,  x=a+b+c,  y=a+b+c+d,  z=a+b+c+d+e=150
Since, z must equal 150, and there are as many 5-tuples of (v,w,x,y,z) as (a,b,c,d,e), then we only need to find the number of possible combinations for v,w,y and z. I then get 149C4 as the answer.

edit: wait I just realized something. I forgot to consider that values of a,b,c,d, and e can be 0

Actually, your problem doesn't make it clear -- as far as the problem is concerned, they could be negative, too.  You might want to decide whether they are integers, positive integers, or non-negative integers.

Yea, sorry about the confusion. I wanted it to be only postive integers originally, but I somehow forgot to include that. I see my mistake now.

A good way to do it is just to make it a combinatorics problems. Imagine you have 150 marbles, labeled 1,2,3,4,........150. If we know four of the numbers, we don't need to know the fifth number. We can just pick the sums. call a=S_1, a+b=S_2, a+b+c=S_3, a+b+c+d=S_4.Note that there is only one way you can arrange S_1,S_2,S_3,S_4. so the answer is 149C4.


Find the number of primes$ p$, such that $ x^{3} \minus{}5x^{2} \minus{}22x\plus{}56\equiv 0\, \, \left(mod\, p\right)$ has no three distinct integer roots in $ \left[0,\left. p\right)\right.$.

$\textbf{(A)}\ 1  \qquad\textbf{(B)}\ 2  \qquad\textbf{(C)}\ 3  \qquad\textbf{(D)}\ 4  \qquad\textbf{(E)}\ \text{None}$

$ (x\minus{}2)(x\minus{}7)(x\plus{}4) \equiv 0 \pmod p$. So $ {2,7,\minus{}4}$ are the solution set.
$ 2 \equiv 7 \pmod p \Rightarrow p \equal{} 5$.
$ 2 \equiv \minus{}4 \pmod p \Rightarrow p \equal{} 2$ and $ p \equal{} 3$.
$ 7 \equiv \minus{}4 \pmod p \Rightarrow p \equal{} 11$.
So there are $ 4$ prime numbers. $ D$.


induction
modular arithmetic
number theory

Chairs are equally spaced around a table and numbered from $ 1$ to $ 2n$.  Prove that if every odd chair is directly opposite another odd numbered chair, then the number of chairs is a multiple of $ 4$.

Prove that $ n^{5}\minus{}n$ is divisible by $ 10$ for all integers $ n$.

[hide="Second Proof"]
Induction

For $ n=1$
$ 1-1 = 0 \equiv 0 \mod 10$.

For $ n=k$
$ k^{5} - k \equiv 0 \mod 10$.

For $ n= k+1$
\begin{eqnarray*} \left( k+1 \right) ^{5} -k -1 &=& k^{5} + 5k^{4} + 10 k^{3} +10 k^{2} + 5k + 1 - k -1 \\ &=& k^{5} - k + 10 \left( k^{3} + k^{2} \right) + 5k^{4} + 5k \\ &\equiv & 5k^{4} + 5k \mod 10 \end{eqnarray*}.
Then
$ k^4 + k \equiv 0 \mod 2$ because if $ k$ is odd or even the expression is even. $ \blacksquare$.
[/hide]

[hide="Direct noninductive proof for 2"]$ n^5 \equiv n \pmod{5}$ by FLT, so the given expression is divisible by $ 5$. $ n^2 \equiv n \pmod{2}$ by FLT. Therefore by multiplying by varying powers of $ n$ and applying the transitive property of equivalence $ n^5 \equiv n^4 \equiv n^3 \equiv n^2 \equiv n pmod{2}$, so the given expression is divisible by $ 2$.[/hide]

Find the largest positive integer $ N$ for which $ n^5\minus{}n$ is divisible by $ N$, for all integers $ n$, and $ N>10$ ;)

[quote="314.math"]For the second proof, I have two questions.

First, is there a way to find a proof without using $ \mod$?

Secondly, what's FLT?

Thanks.[/quote]There is a proof without mod, I'm sure someone will post that. FLT stands for "Fermat's Little Theorem" in here.

[quote="madness"]Actually i think the stricter divisibility is that $ n^5 \minus{} n \equiv 0 \mod 30$[/quote]Is that the largest positive integer answering my question posted above? Prove it :P

that can be easily proved by factorisation with cases :)

[hide="1"]
If all odd numbers are across from another, then they come in pairs, so there's an even number of odd numbers.
1 to 2n-1 are the odds, there being n of them. From above 2 divides n, so n=2k, k integer, then number of integers = 2n = 2(2k) = 4k.
[/hide]

[hide="2"]
n^5 - n = n(n-1)(n+1)(n^2+2), one of n and n-1 are even, one of n, n-1, n+1 is divisible by 3, by FLT, 5|n^5-n, so 30|n^5-n for all n, and for n=2, n^5-n=30, so 30 is the largest number that divides all n^5-n
[/hide]

I think this might be the solution for second problem,

10 | n^5 - n  , that means n^5 - n = 10*k

that is true if we prove that the last digit of n^5 = n for all integers,
the last digit in n^x , repeates periodically, so all that is left is to show that
last digit of n^5 = n, for numbers from 1 to 9.

i am sure there must be more elegant way but i did it with putting everything on paper(i have written just the last digits to show that they occur periodically):

1 1 1 1 1
2 4 8 6 2
3 9 7 1 3
4 6 4 6 4
5 5 5 5 5
6 6 6 6 6
7 9 3 1 7
8 4 2 6 8
9 1 9 1 9

Can please someone comment that solution tell if it's even suitable?

I got the first problem now, but am still confused about the second one.  Could someone help me, please?  Thanks!

[hide="Painless solution to number 2"]
Note that $ s(n) \equal{} n^5 \minus{} n \equal{} (n \minus{} 1)n(n \plus{} 1)(n^2 \plus{} 1)$, and it suffices to prove divisibility by 2 and 5:

$ 2|n(n \plus{} 1) \implies s(n) \equiv 0 \pmod{2}$

$ n^2 \plus{} 1 \equiv (n \minus{} 2)(n \plus{} 2) \pmod{5}$
$ \implies s(n) \equiv (n \minus{} 2)(n \minus{} 1)n(n \plus{} 1)(n \plus{} 2) \equiv 0 \pmod{5}$[/hide]

[quote="Gonky"]...that is true if we prove that the last digit of n^5 = n for all integers,
the last digit in n^x , repeates periodically, so all that is left is to show that
last digit of n^5 = n, for numbers from 1 to 9.
...
Can please someone comment that solution tell if it's even suitable?[/quote]

It sure is.

Is there anything wrong with my proof? I know it is proves the statements but it is strong enough? Can it be taken as less "acceptable" than the others?

The inductive proof is fine, although you could write it up more cleanly.

Thanks t0rajir0u.  :)
How could I present it more cleanly?

[quote="t0rajir0u"]You don't make it clear that $ k^5 \equiv k \bmod 10$ is your inductive step; it looks like you're assuming the result to prove it.[/quote]

Thanks.  :)

[quote="Gonky"]i am really quite new to this thing ...  but what is mod ? I have some clue what it does, but what is the definition?[/quote]

[b]Definition:[/b]  We say that $ a \equiv b \bmod m$ (read: $ a$ is congruent to $ b$ modulo $ m$) if and only if $ m | a \minus{} b$.  In other words, $ a, b$ give the same remainder upon division by $ m$.

[b]Examples:[/b]

$ 12 \equiv 5 \bmod 7$
$ 22 \equiv \minus{}11 \bmod 11$
$ 87 \equiv \minus{}13 \bmod 25$

[b]Properties:[/b]

$ a \equiv a \bmod m$
$ a \equiv b \bmod m \Leftrightarrow b \equiv a \bmod m$
$ a \equiv b \bmod m, b \equiv c \bmod m \implies a \equiv c \bmod m$
$ a \equiv b \bmod m \implies a \plus{} c \equiv b \plus{} c \bmod m$
$ a \equiv b \bmod m \implies ac \equiv bc \bmod m$

One useful way to think about mods (especially $ \bmod 10$) is that the last digit in base $ m$ must be equal.  So $ k^5 \equiv k \bmod 10$ mean that $ k^5, k$ have the same last digit in base $ 10$.  For example,

$ 2^5 \equal{} 32$
$ 3^5 \equal{} 243$
$ 4^5 \equal{} 1024$
$ 5^5 \equal{} 3125$

and so forth.


I keep looking, and I can't find a way.... Is there a way to display all the threads to which one is subscribed, and/or all the subscribed threads with new posts?

Thanks!

One thing you can do instead of subscribing is 'add the topic to favorites' (there's a little link for it near the top when you are looking at the topic).  You can then click the favorites button up top to see all your favorites.

There will be a watched topics list again (it used to be on the old ml board), but I need to debug it a little bit, since some of the structural matters of the forum have changed. Until then you can use Favorite topics :)

That is one option, but I get automatically subscribed to threads to which I post, so using the "subscribed" threads is a more appealing option, if it is possible.  I'm surprised this feature isn't build into the bb code, as it's one I've seen on many other boards.

Thanks for the update, Valentin, I'll watch for it :)

[quote="mathmom"]I'm surprised this feature isn't build into the bb code, as it's one I've seen on many other boards.[/quote]You shouldn't be. First of all this option is available default only on paid boards (such as invision/vbulletin). Second of all, as you might have seen this is not a regular bboard, it is very "customized" :D

[quote="Valentin Vornicu"] Second of all, as you might have seen this is not a regular bboard, it is very "customized" :D[/quote]

Yes, but it seems "better" than most other boards where I have seen that feature, so that is why I was surprised. :)

[quote="mathmom"]Yes, but it seems "better" than most other boards where I have seen that feature, so that is why I was surprised. :)[/quote]Thanks! :D


function
induction
algebra solved
algebra

Let $f$ be a function defined on the positive integers such that \[ f(1) = 2,\;f(2) = 1,\;f(3n) = 3f(n), \\ f(3n + 1) = 3f(n) + 2,\;f(3n + 2) = 3f(n) + 1.\]
How many $n$ satisfy $n \leq 2004$ and $f(n) = 2n$?

It looks to me (it doesn't have to be correct, though :D) that $f(n)$ is obtained from $n$ by writing it in base $3$ and keeping all its $0$'s while changing $1$'s to $2$'s and vice-versa. I also believe that this means that the numbers for which $f(n)=2n$ are precisely those which have no $2$'s in their ternary representation.

Is this even remotely accurate,Arne? :)

Well it's quite correct Grobber ;) Of course there is still a proof to be given (induction :D)
The answer should be 127 :)

You don't really need an induction of any kind.

Since for $n \geq 2$, $\{f(i)\}_{i=1}^n$ uniquely determines $f(n+1)$, according to the given relations, it is sufficient to find a single function $f$ satisfying these conditions.
Ala grobber, we observe that $f(n)=$ $n$'s ternary representation with 1's and 2's switched trivially satisfies the given equations implying that it is valid and, by our above remarks, is the unique $f$ satisfying these conditions.
Then obviously $f(n) = 2n$ iff $n$ contains no $2$'s in its ternary representation.  The largest such $n \leq 2004$ is $1093=1111111_3$.  Reading this number in binary implies that there are indeed the $127$ different $n$ meeting the requirements.


Find $ 2 \times 2$ matrices $ A, B$ such that $ AB \equal{} 0$ but $ BA \neq 0.$

How did you construct this, and is this the only one?

it's not the only one , to construct such matrix , I fix a matrix $ A$ like in my previous post , for $ B$ set  $ B \equal{} (b_{i,j})$ with $ b_{1,1} \equal{}x, b_{2,1} \equal{} z , b_{1,2} \equal{} y$ and $ b_{2,2} \equal{} t$.solve the system $ AB\equal{}0$ to find that we must have $ x\equal{}z$ and $ y\equal{}t$ , and for $ BA \neq 0$ we choose $ x\minus{}y \neq0$.

General solution:
$ AB\equal{}0$ implies that $ A,B$ are simultaneously triangularizable.
In this case $ A\equal{}\begin{pmatrix}a&b\\0&c\end{pmatrix},B\equal{}\begin{pmatrix}d&e\\0&f\end{pmatrix}$
with $ ad\equal{}cf\equal{}0,ae\plus{}bf\equal{}0,db\plus{}ec\not\equal{}0$.
Then the general solution is:
$ A\equal{}P\begin{pmatrix}0&b\\0&c\end{pmatrix}P^{\minus{}1},B\equal{}P\begin{pmatrix}d&e\\0&0\end{pmatrix}P^{\minus{}1}$ with $ db\plus{}ec\not\equal{}0$ and $ P$ invertible.

Actually,a more general result holds: We can find two matrices $ A,B \in\mathcal{M}_n(\mathbb{R})$ such that $ AB\equal{}0$ and $ BA\neq 0$, where $ n\geq 2$.


trigonometry
analytic geometry
graphing lines
slope

[quote="chetjan"]How many points of intersection do the graphs of $ y \equal{} cos(\frac {\pi}{x})$ and $ y \equal{} 100x \plus{} 2$ have?[/quote]
[hide="Hint on what x can be"]$ \cos(\frac{\pi}{x})\equal{}100x\plus{}2$. cosine is between $ \minus{}1$ and $ 1$.
$ \minus{}1\le 100x\plus{}2\le 1$
$ \minus{}\frac{3}{100}\le x\le \minus{}\frac{1}{100}$[/hide]

[quote="thejackchaner"]Looks interesting. Can you give us a hint of how big the number will be?[/quote]
[hide="General Hint"]The number of intersection will be less than 100
[hide="Specific Hint"]The number of intersections will either be 67 or 68[/hide][/hide]

[hide="Answer"]67
[hide="Solution"] all you need to know is y(x) and y ' (x) for the given x.

When x=-0.03, the wavy line is at y=-0.5 and heading upwards steeply. The straight line is at y=-1.
The wavy line will then hit 1 when x=-1/34. It hasn't intersected yet.
By the time x reaches -1/36, when the wavy line is again at 1, it will have intersected the straight line twice.
By the time x reaches -1/38, it will have intersected it twice more.
And so on...
Between -1/98 and -1/100, it will intersect twice, and then intersect once more at the endpoint (-1/100, 1).

There are 33 even numbers from 36 to 100 inclusive. Double that to get 66. And add the last intersection to get 67.

To see that there are two intersections between -1/98 and -1/100, remember that, at (-1/100,1) the slope of the straight line is 100, and the slope of the wavy line is 0. So the straight line is under the wavy line for a brief moment beforehand.[/hide][/hide]


Mr, Mendez awards extra credit on quizzes to his students with quiz grades that exceed the class mean. Given that 107 students take the same quiz, what is the largest number of students who can be awarded extra credit?

Let's say one person gets 0 and everyone else (106 people) gets 100.  The mean will then be less than 100.  Then the total number of people who can get extra credit is $ \boxed{106}$.


AMC
AIME
trigonometry
geometry
geometric transformation
reflection
AoPSwiki

Hello. I was just wondering if you guys wouldn't mind helping me come up with as many non-trig solutions to this problem as possible. I pressed the link to the old thread and there was only one solution not involving trigonometry (construction of equilateral triangle $ CMN$ such that $ N$ is the reflection of $ M$ across median $ CD$). Anyways, I have included the problem as well as my solution with cyclic quadrilaterals. Thanks for your help.

Triangle $ ABC$ is isosceles with $ AC\equal{}BC$ and $ \angle{ACB}\equal{}106^{\circ}$. Point $ M$ is in the interior of the triangle so that $ \angle{MAC}\equal{}7^{\circ}$ and $ \angle{MCA}\equal{}23^{\circ}$. Find $ \angle{CMB}$.

[hide="Cyclic Quadrilateral Solution"]
Let $ D$ be the midpoint of $ AB$. Let $ E$ be the intersection of $ AB$ and $ CM$, and let $ H$ be the intersection of $ AM$ and $ CD$. By angle chasing one can determine that $ \angle{AEC}\equal{}120$, $ \angle{AHC}\equal{}120$, $ \angle{CED}\equal{}60$ and $ \angle{DAH}\equal{}30$. Because $ DH$ is the perpendicular bisector of $ AB$, then $ AH\equal{}BH$ and $ \angle{HAB}\equal{}\angle{HBA}\equal{}30$. Therefore, $ \angle{AHB}\equal{}120$ and since $ \angle{MEB}\plus{}\angle{BHA}\equal{}60\plus{}120\equal{}180$, $ MHBA$ is cyclic. Also, because $ \angle{AHC}\equal{}\angle{AEC}\equal{}120$, $ HEAC$ is cyclic. Therefore, by arc-angle theorem $ \angle{ACE}\equal{}\angle{AHE}\equal{}\angle{MBE}\equal{}23$. By pure angle chasing, one can obtain that $ \angle{CMB}\equal{}83$
[/hide]

You don't really need trigonometry - you are correct.

Here is a non trigonometric solution from the AoPSWiki:

[img]http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/0/a/c/0acadfa5d916afcaa6d423b731b121aa2a373bf2.png[/img]
[img]http://alt2.artofproblemsolving.com/Forum/latexrender/pictures/a/9/3/a93a41f2e73a496bb66b6bc1a1372ef1bb7ae0bb.png[/img]
Take point $ N$ inside $ \triangle ABC$ such that $ \angle CBN \equal{} 7^\circ$ and $ \angle BCN \equal{} 23^\circ.$

$ \angle MCN \equal{} 106^\circ \minus{} 2\cdot 23^\circ \equal{} 60^\circ.$ Also, since $ \triangle AMC$ and $ \triangle BNC$ are congruent (by ASA), $ CM \equal{} CN.$ Hence $ \triangle CMN$ is an equilateral triangle, so $ \angle CNM \equal{} 60^\circ.$

Then $ \angle MNB \equal{} 360^\circ \minus{} \angle CNM \minus{} \angle CNB \equal{} 360^\circ \minus{} 60^\circ \minus{} 150^\circ \equal{} 150^\circ.$ We now see that $ \triangle MNB$ and $ \triangle CNB$ are congruent. Therefore, $ CB \equal{} MB,$ so $ \angle CMB \equal{} \angle MCB \equal{} \boxed{083^\circ}.$


I'm wondering if it's about time we split the two help forums. There are many users asking questions about Asymptote so it would be nicer to have it in a separate forum.

What do you guys think?

The vote was 8-1 so topic closed, Asymptote forum created. Please place all your Asymptote related questions in that forum.


What is $1!+2!!+!3+4!+5!!+!6$?

[b]NOTE:[/b]$x!!\ne (x!)!$ (It's called a double factorial...)

Edited: Before, it ended up being an insanely large number, so I changed it so it would be smaller.
Edited again: Too big to find out

derangements

if you take $(1,2,3)$ it is all the possible ways such that none of the numbers are in their original place like $(2,3,1)$ or $(3,1,2)$

[quote="stupidityismygam"]
derangements

if you take $(1,2,3)$ it is all the possible ways such that [b]none [/b]of the numbers are in their original place like $(2,3,1)$ or $(3,1,2)$[/quote]
[quote="i_like_pie"]Then isn't $!n$ just $n!-1$? :maybe:[/quote]

[quote="i_like_pie"]Then isn't $!n$ just $n!-1$? :maybe:[/quote]
No.

For example, $!4=9\ne24-1$

[quote="i_like_pie"]Then isn't $!n$ just $n!-1$? :maybe:[/quote]
$!n=n!\cdot\sum_{k=0}^{n}\frac{(-1)^{k}}{k!}$


geometry
rectangle
probability
analytic geometry

A point $ (x,y)$ is randomly picked from inside the rectangle with vertices $ (0,0)$, $ (4,0)$, $ (4,1)$, and $ (0,1)$. What is the probability that $ x<y$?

$ \textbf{(A)}\ \frac{1}{8} \qquad
\textbf{(B)}\ \frac{1}{4} \qquad
\textbf{(C)}\ \frac{3}{8} \qquad
\textbf{(D)}\ \frac{1}{2} \qquad
\textbf{(E)}\ \frac{3}{4}$

[hide]
Since y can be at most 1, we know that for all points with an x coordinate greater than 1, x > y. Now we just need to consider the square formed by the points (0, 0), (0, 1), (1, 0), and (1, 1). Exactly 1/2 satisfies x > y, since it is random and no one has higher probability than the other. Or, you can see that for a point (x, y) within the square to have x < y, the x coordinate must be on the line formed by f(x) = y, where y is the y - coordinate, but behind f(x) = x, where x is the x - coordinate. If the length of the square increases, the width increases the same amount, so the possible region is a triangle formed by the points (0, 0), (1, 1), and (0, 1). This square is 1/4 of the total figure, so the probability is 1/4 * 1/2 = 1/8.

$ \boxed{A}$[/hide]
Sorry if I made a careless mistake if I wasn't clear.

[hide] We can solve this by considering the intersection of the line y=x with the rectangle, then taking only the region in the rectangle above the line. The intersection points are (0,0) and (1, 1), so our desired region is a triangle with vertices (0, 0), (0, 1), and (1, 1). The area is hence $ \frac{1}{2}$. The area of the rectangle is 4, so our answer is $ \textbf{(A)}\ \frac{1}{8}$.[/hide]


probability
ARML
geometric sequence
arithmetic sequence

One tries to pick 3 distinct numbers from {1,...34} so that their sum is divisible by 3. how many ways can this be done.

When a certain biased coin is flipped 5 times the probability of getting heads exactly once is the same as of getting heads twice. What is the probability that the coin comes up heads 3 times out of 5.

Arml How many ways can you pick 3 numbers from the set {2^1,...2^2002} so that they form a geometric progression?

[hide="1"]
This was the tie-breaker for last year's ARML, wasn't it?

There are 11 multiples of 3, 11 numbers 1 less than a multiple of 3, and 12 numbers 1 more than a multiple of 3.

Possibilities:
All 3 are multiples of 3: $(11)(10)(9)/3!=165$
All 3 are 1 less than multiples of 3: $(11)(10)(9)/3!=165$
All 3 are 1 more than multiples of 3: $(12)(11)(10)/3!=220$
One of each: $(12)(11)(11)=1452$

Total: $165+165+220+1452=\boxed{2002}$
[/hide]

[hide="2"]
Let $h$ be the probability of flipping a heads.
Then, $1-h$ is the probability of flipping a tails.

We have $C_{1}^{5}h(1-h)^{4}=C_{2}^{5}h^{2}(1-h)^{3}$
$(1-h)=2h$
$h=\frac13$

The probability of flipping 3 heads is:
$C_{3}^{5}h^{3}(1-h)^{2}=(10)\frac{1}{3^{3}}\frac{2^{2}}{3^{2}}=\boxed{\frac{40}{243}}$

[/hide]

[hide="3"]
I will assume that this implies 3 different numbers.

This is the same as asking how many ways you can pick an arithmetic progression of lenth 3 out of the set {1,2,3,...,2002}.

There is one such arithmetic progression for any pair of starting and ending values that differ by a mulitple of 2.

For first term 1, there are (2001-1)/2 = 1000 such values for the third term.
For 2, there are (2002-2)/2 = 1000 such values.
For 3, there is 1 less value than there is for 1.
For 4, there is 1 less value than there is for 2.
etc.
For 1999, there is (2001-1999)/2 = 1 value
For 2000, there is (2002-2000)/2 = 1 value.

Thus there are a total of $\frac{(1000)(1001)}{2}+\frac{(1000)(1001)}{2}=\boxed{1001000}$  geometric progressions in this set.
[/hide]


I need help with these:

1) standard deviation = 15.6, 10% are above 17.2, find the mean

2) mean = 88, 2.5% are below 50, find the standard deviation.

3) While only 5% of babies have learned to walk by the age of 10 months, 75% are walking by 13 months of age. If the age at which babies develop the ability to walk can be described by a normal model, find the mean and standard deviation

Assuming these are all normal distributions:
[hide="1."]
The Z score is given by $Z = \frac{x-\mu}{\sigma}$. A table of Z-scores gives the probabilities of values less than or equal to Z.  We want the value greater than Z so we have
$1-\frac{17.2-\mu}{15.6}= .1$ The answer follows.  [/hide]

[hide="2."]
Assuming normal distribution, this uses the [url=http://www.isixsigma.com/dictionary/Empirical_Rule-467.htm]Empirical Rule[/url].  So that $95\%$ of the values are in $88 \pm 2\sigma$ or $5\%$ of the values are greater OR less than $88\pm2\sigma$.  Symmetry gives that $2.5\%$ are below $2\sigma$ so that $88-2\sigma = 50 \implies \sigma = 19$.  [/hide]

[hide="3."]Use the Z-scores again to get a system of equations.
$\frac{10-\mu}{\sigma}=.05$
$\frac{13-\mu}{\sigma}=.75$
Solve the first for $\mu$ and then substitute into the second:
$\frac{13+.05\sigma-10}{\sigma}=.75\sigma$ Solving for $\sigma$ gives $\sigma \approx 4.286$.  Then go back and solve for $\mu$.[/hide]


calculus
integration
logarithms
calculus computations


geometry
incenter
inradius
Pythagorean Theorem
geometry proposed

Let $\triangle ABC$ be a right triangle with hypotenuse $[BC]$. Let $[AH]$ be the altitude. Incircle of $\triangle ABH$ touches $[AB]$ at $D$. Incircle of $\triangle ACH$ touches $[AC]$ at $E$. Let $E'$ be a point on $[AB]$ such that $|AE|=|AE'|$. If $M$ is the midpoint of $[AH]$, and $I$ is the incenter $\triangle ABH$, prove that $\widehat{EE'M}= \widehat{IHD}$.

i apologize to everyone for this question. I had forgotten the point $E'$ in the problem. forgive me!

AE' = AE, $\triangle AEE'$ is isosceles right, $\angle AE'E = 45^\circ.$ HI bisects the right angle $\angle AHB,$ $\angle AHI = 45^\circ.$ $\angle DHI = \angle AHI-\angle AHD = 45^\circ-\angle AHD$ and $\angle ME'E = \angle AE'E-\angle AE'M = 45^\circ-\angle AE'M.$ To show $\angle DHI = \angle ME'E,$ it is sufficient to show that $\triangle AHD \sim \triangle AE'M$ with the common angle $\angle DAH = \angle MAE'$ are similar. Denote a = BC, b = CA, c = AB, h = AH and $s_{b},\ s_{c}$ the semiperimeters of $\triangle ABH,\ \triangle ACH.$ As $\angle A$ is right and AH the A-altitude

$h = AH = \frac{bc}{a},\ BH = \frac{c^{2}}{a},\ CH = \frac{b^{2}}{a}$

$AD = s_{b}-BH = \frac{h+c-BH}{2}= \frac{c(a+b-c)}{2a}$

$AE' = AE = s_{c}-CH = \frac{h+b-CH}{2}= \frac{b(c+a-b)}{2a}$

$\frac{AD}{AH}= \frac{AM}{AE'}\ \Longleftrightarrow\ AD \cdot AE' = AM \cdot AH\ \Longleftrightarrow$

$\frac{bc(a+b-c)(c+a-b)}{4a^{2}}= \frac{h^{2}}{2}= \frac{b^{2}c^{2}}{2a^{2}}\ \Longleftrightarrow\ a^{2}-(b-c)^{2}= 2bc$

which is true because of Pythagorean theorem $a^{2}= b^{2}+c^{2}.$

in my solution, the first part is same
[b]lemma:[/b]
Let $\triangle ABC$ be a right triangle with $\angle B = 90^\circ$. The incircle touches the hypotenuse at $D$. The area of the triangle is equal to $AD\cdot CD$.
[hide="Proof:"]
Let $s$ be the semiperimeter, $r$ be the inradius and $a,b,c$ be sides ($c$ is the hypotenuse). We have $r = s-b$. So
$[ABC]=s\cdot r= s(s-b)= \sqrt{s(s-a)(s-b)(s-c)}\Rightarrow$  $s(s-b)=(s-a)(s-c)=[ABC]$
[/hide]

From the lemma we know that $\frac{AH\cdot BH}{2}=AD\cdot BD$. From the similarity $\frac{BD}{AE}=\frac{BH}{AH}$. Multiply both equation. So we get $\frac{AH^{2}}{2}=AD\cdot AE$. We have $\frac{AH^{2}}{2}=AM\cdot AH$. So $AE\cdot AD=AE'\cdot AD=AM\cdot AH$ implies $E',M,H,D$ are concyclic. $\angle EE'M = |\angle AE'E-\angle AE'M|$. Similarly $\angle DHI = |\angle IHA-\angle DHA|$. We know $\angle AE'M = \angle AHD$ and $\angle AE'E = \angle IHA = 45^\circ$. So $\angle EE'M = \angle IHD$. Q.E.D


[quote="Ligouras"]Prove that

$ \frac {a}{a^{2} \plus{} 4w^{2}_{a}} \plus{} \frac {b}{b^{2} \plus{} 4w^{2}_{b}} \plus{} \frac {c}{c^{2} \plus{} 4w^{2}_{c}}\geq\frac {9}{8s}$[/quote]

Algebraization

Can be shown using the Muirhead theorem.

 :lol:


algebra
polynomial
function
geometry
geometric transformation
number theory
number theory unsolved

Let $ P(x) \in Z[x]$. Suppose that $ P(n)$ is a kth power for all $ n \in N$. Then, there exists a polynomial $ Q(X) \in Z[x]$ such that $ P(x)\equal{}Q(x)^k$.

CHECK OUT
George P´olya and G´abor Szeg˝o, Problems and Theorems in Analysis, vol. 2; Theory of functions, zeros,polynomials, determinants, number theory, geometry; revised and enlarged translation by C. E. Billigheimer of the fourth German edition, Springer-Verlag, New York, 1976. PROBLEM: VIII-114.
to prove the case whwn k=2, check http://www.mathlinks.ro/viewtopic.php?t=241684

[quote="conan_naruto236"]Let $ P(x) \in Z[x]$. Suppose that $ P(n)$ is a kth power for all $ n \in N$. Then, there exists a polynomial $ Q(X) \in Z[x]$ such that $ P(x) \equal{} Q(x)^k$.[/quote]
Please help me :(


Sabet konid agar $ (xy\plus{}1)(xz\plus{}1)(yz\plus{}1)$ Morabbae kamel bashad angah tamame adade xy+1,xz+1,yz+1 morabbae kameland.

kasai ke mikhan hallesho bebinan mitoonan be [url=http://www.mathlinks.ro/viewtopic.php?t=150369]inja[/url] moraje'e konan.


a,b, and c are the lengths of the three sides of a right triangle. a+b+c=22 and a^2 + b^2 + c^2 = 200. what is the number of square units in the area of the triangle. please help

Is there a theorem that you could relate to this problem that involves squares of sides of a right triangle?

Hint: It begins with a P...was a name of a person...he had a cult...golden ratio...ring a bell? :lol:

oh hehehe, i misread the reply as involving the shape of squares and triangles. i got it...PYTHAGOREAN!!! so could you explain further because i tried using it.

Um, just substitute the a^2 + b^2 with c^2 and go from there.

3 variables, 3 equations, one destiny :lol:

use the pythagorean theorem, and a simple manipulation will yield the result.

[quote="eqn1190"]i did substitute so i got a=6, b=8, and c=10 but it doesnt add up to a+b+c=22. but it works for a^2+b^2+c^2=200. where did i go wrong?[/quote]

where did you get a=6, b=8 from?

am i correct about c=10? then i can explain further.

c=10 is correct, but nothing in the question said a and b had to be integers.

(You really don't have to solve for a and b. Some simple algebraic manipulations can easily yield a relationship between a and b that would give away the area of the triangle.)

how do we make the third equation, and how do we solve when are equations arent equal (^2 and ^1)

Okay, full solution then...

[hide]WLOG assume a :le: b < c. Then, a2 + b2 = c2. Substituting that into a2 + b2 + c2 = 200 yields c2 = 100 and thus c = 10 because c > 0.

Therefore a + b = 22 - 10 = 12. Squaring this yields a2 + 2ab + b2 = 144; subtract a2 + b2 = c2 = 100 to get 2ab = 44, so ab/2 = 11 is the desired area.[/hide]

For the curious, a = [hide]6 - :sqrt:14[/hide] and b = [hide]6 + :sqrt:14[/hide].

yup that's what I got. remember that you don't always have to solve for specific variables, and that you can square expressions, etc. (in general, manipulate them) to get what you want.

"without loss of generality"

I don't think Osiris needed to put that in there, but it looks a whole lot cooler now.

Well you do need it actually.. nothing in the question said that c had to be the longest side. So you can't just assume that without saying WLOG :)

thanks a lot, i understand now. im not really good at simple manipulations like that because i dont really see it. are there any more problems i can try to solve that are like the one i posted?

4a + b = 1/(2a) + 2/b.  If a and b are positive, compute the value of ab.

If  r = 20/(r + 5) and  s = -5 + 20/s, where r and s are two distinct real numbers, compute r + s.

If r:^2: + br + 4 = 0  and  s + r = -b, what is the value of s:^2: + bs?


calculus
integration
least common multiple
superior algebra
superior algebra unsolved

Let $ x_i, y_i$ be integers for $ i \equal{} 1,2,\cdots, s$. Is there a solution of the following equation system on $ l$?

\[ \left\{\begin{array}{ll} \frac {x_1}{y_1}[\frac {y_1}{x_1},\frac {y_2}{x_2},\cdots,\frac {y_s}{x_s}]\cdot l\equiv 1(\bmod   x_1) \\
\frac {x_2}{y_2}[\frac {y_1}{x_1},\frac {y_2}{x_2},\cdots,\frac {y_s}{x_s}]\cdot l\equiv 1(\bmod  x_2) \\
\cdots\cdots\cdots \\
\frac {x_s}{y_s}[\frac {y_1}{x_1},\frac {y_2}{x_2},\cdots,\frac {y_s}{x_s}]\cdot l\equiv 1(\bmod  x_s) \\
\end{array}\right.\]

Not before you explain what the $ \left[...\right]$ are supposed to mean.

  darij

$ [n_1,n_2,\cdots,n_s]$ means least common multiple of numbers $ n_1,n_2,\cdots,n_s$.

What is the least common multiple of rational numbers supposed to mean? Maximal power of each prime? Is $ l$ to be choosen in a way that makes the left hand sides integral?

  darij

sorry, where $ \frac{y_i}{x_i}$ with $ i\equal{}1,2,\cdots,s$ and $ l$ are all integers.

The question still doesn't make much sense to me: take $ s\equal{}2$, $ x_1\equal{}2$, $ y_1\equal{}4$, $ x_2\equal{}1$, $ y_2\equal{}8$. Then your first equation spells $ \frac12\cdot 8\cdot l\equiv 1\mod 2$, impossible. On the other hand, there are many cases where a solution exists. I doubt they can be classified.

  darij


geometry
3D geometry
tetrahedron
analytic geometry
IMO
IMO 1964

In tetrahedron $ABCD$, vertex $D$ is connected with $D_0$, the centrod if $\triangle ABC$. Line parallel to $DD_0$ are drawn through $A,B$ and $C$. These lines intersect the planes $BCD, CAD$ and $ABD$ in points $A_2, B_1,$ and $C_1$, respectively. Prove that the volume of $ABCD$ is one third the volume of $A_1B_1C_1D_0$. Is the result if point $D_o$ is selected anywhere within $\triangle ABC$?

In tetrahedron $ ABCD $, vertex $ D $ is connected with $ D_{0} $, the centrod in $ \triangle ABC $. Line parallel to $ DD_{0} $ are drawn through $ A,B $ and $C$. These lines intersect the planes $ BCD, CAD $ and $ ABD $ in points $ A_{1}, B_{1} $ and $ C_{1} $, respectively. Prove that the volume of $ ABCD $ is one third the volume of $ A_{1}B_{1}C_{1}D_{0} $. Is the result if point $ D_{o} $ is selected anywhere within $ \triangle ABC $?

 [size=150]Part 1: Construction of the points $ A_{1}, B_{1}, C_{1} $[/size]

Given a tetrahedron $ ABCD $ with vertex $ D $,see picture.
Lines $d1$ and $d2$ are bisectors of $ \triangle ABC $. The crossing of $d1$ and $d2$ is the point $ D_{0} $.We draw the line $ D D_{0} $ and also the line $z$, parallel to $ D D_{0}$.
In the plane $ (ABC) $, we draw the line $ p $, parallel to $ BC $.
In this plane $ (ABC) $, the line $ p $ is crossing the 2 bisectors $d1$ and $d2$ in 2 points, where $b1$ and $b2$ are parallel to $ D D_{0}$.
In the plane $ (b1, d1,D D_{0},C) $, the lines $ d1 $ and $ CD $ are crossing in the point $A1$.
In the plane $ (b2, d2,D D_{0},B) $, the lines $ d2 $ and $ BD $ are crossing in the point $A2$.
In the plane $ (p, b1, b2,z,A1, A2) $, the lines $ A1A2 $ and $ z $ are crossing in the point $A'$; plane $ (BCD)  \cap  z $ = $A'$.
That point $A'$ is the asked point $ A_{1}$.
The same construction can be used for the points $ B_{1} $ and $ C_{1} $.
If the tetrahedron is regular, then the points $ A_{1}, B_{1}, C_{1} $ form a plane, parallel to the plane $ (ABC)$, see part 2.

[size=150]Part 2: Calculation of the coordinates of the points $ A_{1}, B_{1}, C_{1} $[/size]

Given a regular tetrahedron $ ABCD $ with vertex $ D $; each side has length 1.
	\[A(0,0,0), B(\frac{\sqrt{3}}{2},\frac{1}{2},0), C(0,1,0)\]
	\[D_{0}(\frac{\sqrt{3}}{6},\frac{1}{2},0), D(\frac{\sqrt{3}}{6},\frac{1}{2},\frac{\sqrt{6}}{3})\]
In this regular tetrahedron: $ D D_{0} \bot (ABC)$, hence $ D D_{0} \bot (x\;y)$.
The equation of a plane $(PQR)$ with
	\[P(x_{1},y_{1},z_{1}),Q(x_{2},y_{2},z_{2}),R((x_{3},y_{3},z_{3})\]
is given by:
\[ \left|
\begin{array}{ccc}
x- x_{1}& y- y_{1}& z- z_{1}\\
x_{2}- x_{1}& y_{2}- y_{1}& z_{2}- z_{1} \\
x_{3}- x_{1}& y_{3}- y_{1}& z_{3}- z_{1} \end{array}
\right|=0\]
a) equation of the plane $(DBC)$:
\[ \left |
\begin{array}{ccc}
x& y- 1& z\\
\frac{\sqrt{3}}{2}& - \frac{1}{2}& 0 \\
\frac{\sqrt{3}}{6}& - \frac{1}{2}& \frac{\sqrt{6}}{3} \end{array}
\right|=0\]
	\[\frac{\sqrt{6}}{6}x+\frac{\sqrt{2}}{2}(y-1)+\frac{\sqrt{3}}{6}z=0\]
Through point $A$, we are drawing a parallel with $D D_{0}$, i.e. this parallel $\bot (x\;y)$
The equation of this parallel is:
	\[ \left\{\begin{array}{ll} x=0\\ y=0\end{array}\right.\]
This parallel through $A$ $ \cap (DBC) = A'$:
	\[ \left\{\begin{array}{lll} \frac{\sqrt{6}}{6}x+\frac{\sqrt{2}}{2}(y-1)+\frac{\sqrt{3}}{6}z=0\\x=0\\ y=0\end{array}\right.\]
	\[A'(0,0,\sqrt{6})=A_{1}(0,0,\sqrt{6})\]
b) equation of the plane $(ADB)$:
\[ \left|
\begin{array}{ccc}
x& y& z\\
\frac{\sqrt{3}}{2}&  \frac{1}{2}& 0 \\
\frac{\sqrt{3}}{6}&  \frac{1}{2}& \frac{\sqrt{6}}{3} \end{array}
\right|=0\]
	\[\frac{\sqrt{6}}{6}x-\frac{\sqrt{2}}{2}y+\frac{\sqrt{3}}{6}z=0\]
Through point $C$, we are drawing a parallel with $D D_{0}$, i.e. this parallel $\bot (x\;y)$
The equation of this parallel is:
	\[ \left\{\begin{array}{ll} x=0\\ y=1\end{array}\right.\]
This parallel through $C$ $ \cap (ADB) = C'$:
	\[ \left\{\begin{array}{lll} \frac{\sqrt{6}}{6}x-\frac{\sqrt{2}}{2}y+\frac{\sqrt{3}}{6}z=0\\x=0\\ y=1\end{array}\right.\]
	\[C'(0,1,\sqrt{6})\]
c) equation of the plane $(ADC)$:
\[ \left|
\begin{array}{ccc}
x& y& z\\
0& 1& 0\\
\frac{\sqrt{3}}{6}&  \frac{1}{2}& \frac{\sqrt{6}}{3} \end{array}
\right|=0\]
	\[\frac{\sqrt{6}}{3}x-\frac{\sqrt{3}}{6}z=0\]
Through point $B$, we are drawing a parallel with $D D_{0}$, i.e. this parallel $\bot (x\;y)$
The equation of this parallel is:
	\[ \left\{\begin{array}{ll} x=\frac{\sqrt{3}}{2}\\ y=\frac{1}{2}\end{array}\right.\]
This parallel through $B$ $ \cap (ADC) = B'$:
	\[ \left\{\begin{array}{lll} \frac{\sqrt{6}}{3}x-\frac{\sqrt{3}}{6}z=0\\x=\frac{\sqrt{3}}{2}\\ y=\frac{1}{2}\end{array}\right.\]
	\[B'(\frac{\sqrt{3}}{2},\frac{1}{2},\sqrt{6})\]
Because the value of the z-coordinate of $A',B'$ and $C'$ is $\sqrt{6}$, are the surface of $\triangle ABC$ and the surface of $\triangle A'B'C'$ equal.
Because the height of the tetrahedron $ ABCD $ equals $\frac{\sqrt{6}}{3}$ (the z-coordinate of the point $D$) and the height of the tetrahedron $ A'B'C'D_{0} $ equals $\sqrt{6}$, is the volume of $ ABCD $ indeed one third the volume of $ A_{1}B_{1}C_{1}D_{0} $.

[size=150]Part 3: Calculation of another tetrahedron[/size]

a) Given a arbitrary tetrahedron $ ABCD $ with vertex $ D $.
	\[A(0,0,0), B(a,b,0), C(0,c,0)\]
Point $D_{0}$ is the crossing point of the bisectors of $\triangle ABC$.
$D D_{0}$ is perpendicular on the plane $(ABC)$, points $D$ and $ D_{0}$ have the same x-coordinate and y-coordinate; the z-coordinate of $D$ is 1.
The same calculation as in Part 2 is made with result: the volume of $ ABCD $ is one third the volume of $ A_{1}B_{1}C_{1}D_{0} $.
b) Given a regular tetrahedron $ ABCD $ with vertex $ D $; each side has length 1.
	\[A(0,0,0), B(\frac{\sqrt{3}}{2},\frac{1}{2},0), C(0,1,0), D(\frac{\sqrt{3}}{6},\frac{1}{2},\frac{\sqrt{6}}{3})\]
The point $ D_{o} $ is selected anywhere within $ \triangle ABC $:
\[D_{0}(k,l,0)\] with $k \neq \frac{\sqrt{3}}{6}$ \ and/or \ $l \neq \frac{1}{2}$.
The same calculation as in Part 2 is made with result: the volume of $ ABCD $ is NEVER one third the volume of $ A_{1}B_{1}C_{1}D_{0} $.


Is there another way to align equations other than eqnarray*?
Thanks!
minicon

Eqnarray is obsolete - see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]Latex FAQ 5[/url]
Make sure you have loaded amsmath package in your document and use align
\begin{align}
2x+y&=3 \\
x+2y&=5
\end{align}
[code]\begin{align}
2x+y&=3 \\
x+2y&=5
\end{align}[/code]


Is it possible to use \input on a file that's not in the same directory as your .tex file? I'm trying it now by including the full pathname in single quotation marks but it's not sure it this will work...

Difficult to diagnose the problem without the details. But remember
1. the path must be [i]relative[/i] to the directory of the tex file;
2. always avoid directories with spaces in them; you [i]may[/i] be able to use the DOS version of windows paths with spaces in them;
3. the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=import]import package[/url] may be useful.


[code][(x - x1) / a]^2 + [(y - y1) / b]^2 = 1
[(x - x2) / c]^2 + [(y - y2) / d]^2 = 1[/code]

Those are my two equations.

All I need is for someone to simplify these into a quartic. I don't have access to Mathematica right now, but I think this can easily be solved. I just need a quartic, I don't need the steps involved (I'm pretty sure that simplifies to a quartic).

Eliminating y from both equations gives this (if it helps) :
[b]sqrt {b^2 - b^2[(x - x1) / a]^2} + y1 = sqrt {d^2 - d^2[(x - x2) / c]^2} + y2[/b]

Obviously, if it's easier to solve for x than that would be just as good.

Anything is really appreciated and thank you for reading!!

I'll put what you posted into latex. I'm not going to try solving it now, as it's late and I need to do something else. Maybe tomorrow.

$\left(\frac{x-x_{1}}{a}\right)^{2}+\left(\frac{y-y_{1}}{b}\right)^{2}= 1$
$\left(\frac{x-x_{2}}{c}\right)^{2}+\left(\frac{y-y_{2}}{d}\right)^{2}= 1$

And you think it becomes
$\sqrt{b^{2}-b^{2}\left(\frac{x-x_{1}}{a}\right)^{2}}+y_{1}= \sqrt{d^{2}-d^{2}\left(\frac{x-x2}{c}\right)^{2}}+y_{2}$

Is this right?
----------------------------------
The only way to get a quartic is to multiply the first equation by the second, and that would be a lot of messy algebra. It's doable, but, as you said, mathematica would be nicer.

I have a copy at school, I'll burn this out for you tomorrow if nodoby answers sooner.

By the way, what are you solving for, and why do you want a quartic? That adds to the complexity. As it is, those equations are fairly nice looking and simple.

[quote="DiscreetFourierTransform"]The only way to get a quartic is to multiply the first equation by the second[/quote]

I beg to differ.

[hide="The method that I am too lazy to finish"] Let $x = a \cos \theta+x_{1}, y = b \sin \theta+y_{1}$.  Then let $t = \tan \frac{\theta}{2}$.

The first equation is automatically satisfied; and because we have the relations

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

The second equation gives us a quartic in $t$, which we can solve to calculate $x, y$. [/hide]
Edit:

[quote="DiscreetFourierTransform"]By the way, what are you solving for[/quote]

Looks like the intersections of two ellipses to me.

Yeah, I saw the elipses thingy too.
However, I think that other quartic method is nice too.
Tor, you pwn me on every math post I make  :oops:  :P
How long have you been doing comp. math? I just started this year, and I'm still kind of intimidated by the stuff in Pre-Olympiad...

Since my join date, more or less.  Read other people's solutions - you'll get a better flavor for the types of logic you need to solve interesting, non-school kinds of problems.

Thanks for the replies!! I actually found my solution. Someone used Maple from another math forum to get it.

[quote]Looks like the intersections of two ellipses to me.[/quote]
Indeed, you are right! :wink:

If anyone is interested in the horrible quartic that results, [url=http://mymathforum.com/viewtopic.php?p=2800#2800]here[/url] you go!


I think this is one of the things like $\infty!=\sqrt{2\pi}$.
It's clear to see it's not technically true since the sequence contains only positive numbers. :)

well I know that it is a special thing and not exactly equal. One person put a little a above the equals sig to show that they weren't really equal. I was just wodnering where that came from.

It's the Riemann Zeta Function at -1.  You can search here or look it up on mathworld for more.

What I find most fascinating about this summation is its applications in the real world.  The computation of certain physical quantities often involves summing this (divergent) series to give a small, negative number.  Quite interesting.  :)


Released today.

FIRST PRIZE
[b]Yufei Zhao[/b], Don Mills Collegiate Institute, Don Mills, Ontario. :)

SECOND PRIZE
[b]Jacob Tsimerman,[/b] University of Toronto Schools, Toronto, Ontario.

THIRD PRIZE
[b]David Rhee[/b], McNally School, Edmonton, Alberta.

HONOURABLE MENTIONS
[b]Boris Braverman[/b], Simon Fraser Junior High, Calgary, Alberta
[b]Dennis Chuang[/b], Strathcona-Tweedsmuir School, Okotoks, Alberta
[b]Gabriel Gauthier-Shalom[/b], Marianopolis College, Montreal, Quebec
[b]Oleg Ivrii[/b], Don Mills Collegiate Institute, Don Mills, Ontario
[b]Jnos Kramr[/b], University of Toronto Schools, Toronto, Ontario
[b]Andrew Mao[/b], A.B. Lucas Secondary School, London, Ontario
[b]Richard Peng[/b], Vaughan Road Academy, Toronto, Ontario
[b]Peng Shi[/b], Sir John A. MacDonald Collegiate Institute, Agincourt, Ontario.

Source: [url]http://www.cms.math.ca/MediaReleases/2004/cmo.html[/url]

Can someone please post the problems?
The CanMO website doesn't give them.

Yey, congrats Billzhao! :D I suppose you are admitted in the Canada IMO Team? :D

Could you please post the questions also? :)

[quote="billzhao"]The problems are posted [url=http://www.mathlinks.ro/viewtopic.php?t=5547]here[/url].

Btw, on the front page, it should spell Can[b]a[/b]da, not Canda[/quote] sorry, I will fix it  :blush:  :blush:  :blush:


[quote="Klaas"]Let f(x)= x^2 + x, prove that there doesn't exist such integers a,b for which 4f(a)=f(b).
 :blush:[/quote]

That's wrong :
$ 4f(0)\equal{}f(0)$
$ 4f(0)\equal{}f(\minus{}1)$
$ 4f(\minus{}1)\equal{}f(0)$
$ 4f(\minus{}1)\equal{}f(\minus{}1)$

In fact :

$ 4f(a)\equal{}f(b)$ $ \Leftrightarrow$ $ 4a^2\plus{}4a\equal{}b^2\plus{}b$ $ \Leftrightarrow$ $ (2a\plus{}1)^2\equal{}b^2\plus{}b\plus{}1$

Then, if $ b>0$, we have $ b^2<b^2\plus{}b\plus{}1<(b\plus{}1)^2$ and $ b^2\plus{}b\plus{}1$ can't be a perfect square.
Same, if $ b<\minus{}1$, we have $ (b\plus{}1)^2<b^2\plus{}b\plus{}1<b^2$ and $ b^2\plus{}b\plus{}1$ can't be a perfect square.

So $ 4f(a)\equal{}f(b)$ is impossible except in the four cases explicitely given above.