If the square of Winslow's age is added to Abby's age, the sum is 209.  If the square of Abby's age is added to Winslow's ages, the sum is 183.  What is the sum of their ages?

Algebraic Number Theory sounds intimidating -- this problem is easy enough to solve via [i]Guess and Check[/i].   :)

I like to throw in advanced terminology once in a while with problems that are either solved by easier methods or solved in a manner that is ultimately described by the advanced terminology.  My hope is that this breaks students in to the terminology.  I try not to overdo it.

Since this is a sprint question, you are not allowed to use a calculator. Also guess and check method is not suggested for this question.

[hide="my simple solution"]

x^2+y=209             (1)
x+y^2=183             (2)

(1)-(2) we have (x-y)(x+y-1)=26 = 1*26=2*13.

let x-y=1 and x+y-1=26. Solve this system equation we find x=14 and y=13. This works. Then 14+13 = 27 is the solution.

x-y=2 and x+y-1=13 or other combinations are not working.

[/hide]

Nice work cj.  You are clearly prepared for the Intermediate Number Theory class (whenever it is that we should first offer it).

[quote="cj"]Since this is a sprint question, you are not allowed to use a calculator. Also guess and check method is not suggested for this question.[/quote]

Guess and check is very easy for this if you know your squares and inject a little logic into your guesses:
[hide]
W2  + A = 209 (1)
A2  + W = 183 (2)

it's clear that we need  W > A, and that W2 < 209 (so, W=15 is too large)
so try W = 13, then W2 = 169.  Then from (1) A would have to be 40 which is way too big for (2), so try a bigger W
try W=14, then  W2 = 196.  From (1) A would be 13.  Plug into (2) to test and lo and behold it works, so A+W = 27

[/hide]

cj's method is perfectly fine, and it wouldn't do any harm for Middle Schoolers to study it.  But, I'm going to weigh-in as preferring solafidefarms and mathmoms methods for most MATHCOUNTS-level students.

[hide="slight tweak to mathmoms method"]
Given the two equations:
\begin{eqnarray}
W^2+A=209 \\
A^2+W=183
\end{eqnarray}

As mathmom suggests, $W>A$, and equation (1) tells us $W<15$.

If we add the equations and rearrange:
\begin{eqnarray}
\refstepcounter{equation}
\refstepcounter{equation}
(W^2+W)+(A^2+A)=402
\end{eqnarray}

The large part plus the small part must equal 402.  Therefore,
\begin{eqnarray}
\refstepcounter{equation}
\refstepcounter{equation}
\refstepcounter{equation}
W^2+W>201 \\
A^2+A<201
\end{eqnarray}

Equation (4) tells us that $W>13$.  Now, there really isn't any guess and check, at all.  And there were no tricky factorizations.  $13<W<15$.  Then solve for A.
[/hide]

i think guess and check would be the best way since most of us haven't learned any complicated formulas that give us another alternative. But you have to guess logiclly and fast since it is a sprint question

[quote="GoBraves"]Guess and Check would probably be the easiest way. The closest perfect square to 209 is 196.  Thats 14 squared. 209-196=13.13 squared is 169. 169+14=183

14+13=27[/quote]

I agree with GoBraves (except the Red Sox are better than any NL team.. sry for that wild tangent).. I just thought of the two closest squares less than 200, and I got 13^2 and 14^2, 169 and 196 respectively. It just happened that those two numbers worked, and that was like number 2 on the sprint so you didn't really want to use any number theory or algebra or anything as number two is designed by nature to be easy.

[quote]I agree with GoBraves (except the Red Sox are better than any NL team.. sry for that wild tangent)..[/quote]

 :D You Fair Weather :D


Let $ ABC$ be a triangle , $ R$ the radius of circumcircle and $ S$ its area .

If $ a^2\plus{}b^2\plus{}c^2\equal{}4$ then prove that $ 6R^2\plus{}S^2\ge 3$ .

If you put $ x\equal{}a^2\plus{}b^2\minus{}c^2$ and the similars it becomes:

\[ \frac{xy\plus{}yz\plus{}zx}{4}\ge \frac{3xyz}{xy\plus{}yz\plus{}zx}\]

which is trivial.


symmetry geometry perimeter calculus trigonometry integration inequalities

We are given a finite collection of segments in the plane, of total length 1. Prove that there exists a line $ l$ such that the sum of the lengths of the projections of the given segments to the line $ l$ is less than $ \frac{2}{\pi}.$

The ideas of the two following proofs may apply to other related problems.

Choose an origin $O$ in the plane, and translate each of the $n$ segments so that its midpoint coincides with $O$. Denote the affixes of the $2n$ ends of these segments by $z_k$, $1\leq k \leq 2n$, say in trigonometric sense around $O$. Clearly $z_{n+k} = -z_k$ for $1\leq k \leq n$, so $\sum_{k=1}^{2n} z_k = 0$. Consider now the points of affixes $w_m =  \sum_{k=1}^{m} z_k$, $1\leq m \leq 2n$; they are the vertices of a convex, centrally symmetric polygon $\mathcal{P}$ with $2n$ sides, and symmetry centre $\Omega$. Take the circle of centre $\Omega$ and largest radius, inscribed in $\mathcal{P}$. The circle will be tangent to a pair of opposite sides, and each half of the polygon projects on the diameter $d$ of these points of tangency, covering it fully, with no overlaps, so the sum of the lengths of the projections is equal to $2d$.

But that circle is a smooth convex body, included in the convex polygon $\mathcal{P}$. The perimeter of $\mathcal{P}$ is the sum of the lengths of each half-segment from our collection, so its value is $1$, larger than the perimeter of the circle, which is $\pi d$. Thus the sum of the lengths of the projections, being $2d$, is less than $\frac {2} {\pi}$. This value can never be reached, but can be approached as closely as wanted, by taking large $n$, and the segments making angles of $\frac {2\pi} {n}$ around the origin.

A different approach is using calculus; consider a line of variable angle $\theta$, for which the length of the projection of a segment of length $\ell$ and angle $\varphi$ on it is $|\cos (\theta - \varphi)|\ell$. Then the average length of this projection is $\frac {1} {\pi} \int_{0}^{\pi} |\cos (\theta - \varphi)|\ell \textrm{d} \theta = \frac {2\ell} {\pi}$.
Therefore, the average sum of the lengths of the projections is $\frac {1} {\pi} \int_{0}^{\pi} \sum_{k=1}^n |\cos (\theta - \varphi_k)| \ell_k \textrm{d} \theta =$ $ \frac {1} {\pi} \sum_{k=1}^n \int_{0}^{\pi} |\cos (\theta - \varphi_k)|\ell_k \textrm{d} \theta =$ $  \frac {1} {\pi} \sum_{k=1}^n 2\ell_k = \frac {2} {\pi}$.
Consequently there will exist a direction $\theta$ for which the sum of the lengths of the projections is at most $\frac {2} {\pi}$ (and since this sum cannot be constant over all $\theta$, the inequality is strict).

Discussed also in here :

http://mzargar.wordpress.com/2009/04/08/integrals-combinatorics-and-geometry/

Is the problem related to [url=http://mathworld.wolfram.com/BuffonsNeedleProblem.html]Buffon's Needle Problem[/url]?

Probably, just by both using the calculus method described above; otherwise the statements seem quite unrelated.


geometry
circumcircle
trigonometry
vector
geometry proposed

Given a convex pentagon ABCDE with DC = DE and <BCD = <DEA = pi/2, let F be the point on segment AB such that AF/BF = AE/BC. Show that <FCE = <ADE and <FEC = <BDC.

Construct circumcircle $\omega$ of triangle $CDE$. Since $\angle AED=\angle CDE=\pi/2$, lines $EA$ and $CB$ intersect on $\omega$.
Let $P$ be point of intersection of $\omega$ and $DA$; $F'$ be point of intersection of $AB$ and $PC$. Let's calculate $AF':F'B$.
Indeed, using sine theorem, from triangles $APF'$ and $BCF'$ we obtain [b](*)[/b]
\[AF':F'B=\frac{AP}{BC}\cdot \frac{\sin APF'}{\sin BCF'}.\]
But $\angle APF'=\angle DPC=\angle EPD$ (since $ED=DC$). Thus $\angle APF'=\angle EPA$. Also it is easy to see $\angle BCF'=\angle PEA$. Applying sine theorem for triangle $EPA$, we obtain [b](**)[/b]
\[\frac{\sin APF'}{\sin BCF'}=\frac{\sin EPA}{\sin PEA}=\frac{EA}{AP}.\]
So from [b](*)[/b] and [b](**)[/b] we conclude $AF':F'B=AE:BC$. It means $F=F'$. But is clear $\angle ADE=\angle F'CE$.

That's all.

yes, this direct argument is good! But I like the following too, which reduces it to just one case:

For any such pentagon, keep all vertices except A fixed; let A vary on the line EA. Then we let U be the unit vector perpendicular to ED, so if AE is length v (this varies), and BC is length c (constant), and E is the origin, F = (cA + vB) / (c+v) = (cvU + vB)/(c+v) = (cU+B)*v/(c+v), where the vector is constant, so F lies on a line through E and thus <FEC is constant so we just need to show <FEC = <BDC for one position of A. Now, to keep things simple, we can just take A s.t. AE = BC and F is midpoint of AB, in which case the whole thing is symmetric over DF and we just have a little angle chasing: letting X = EC^DF, we have <FEC = pi/2 - <XFC = pi/2 - <DBC (FBDC cyclic) = <BDC, done.

If P=EA^CB, PCDE is cyclic. Let I on EA a part from A such that EI=BC. It holds that DEI is congruent to DCB. Therefore <DBC = <DIE which implies that PBDI is cyclic too. By Thales EF//IB so, if H=EF^DB, <EHD = <DBI = <DPI which means that H lies on c(PCDE). Then <BDC = <FEC = <QED.


Stanford
college
Stanford Math Tournament

I was wondering if anyone on AoPS was planning to go to the stanford math tournament, or the santa clara valley one the day before?

Santa Clara is March 3 (this Saturday), and SMT is Sunday.

I'll be there, as one of the organizers.  You might not see me until the end, though, as I plan on spending most of my afternoon grading the power contests.

Oh yeah I am going to Stanford, but not to the Santa Clara thing. BTW what is the power contest? Oh yeah and what subjects will you be taking? I am taking Geo.+Alg. (should have taken General instead considering I am in 8th grade)

Hopefully, I can make both, as my school is only 8 minutes from stanford.


Suppose x, y \geq 0 such that x2 + y3 \geq x3 + y4. Prove that x2 + y3 \leq 2.

I can solve it only for x^3+y^3 \leq 2. Are you sure the inequality is correct ?

That's the way it's stated at the kalva site :

http://www.kalva.demon.co.uk/russian/rus99.html

So I suppose this is correct ...

I think that there is also a mistake!! manlio is correct!!

[b]Solution:[/b]

we know that x^3+x>=2x^2 and y^4+y^2>=2y^3
and by adding them we get that: x^3+y^4+x+y^2>=2(x^2+y^3). we have that x^3+y^4<x^2+y^3 so we get that x+y^2>x^2+y^3.

On the other hand we have that x^2+1>=2x and y^4+1>=2y^2 and by adding them we have that:
2+x^2+y^4>=2(x+y^2)>2(x^2+y^3)>x^2+y^3+x^3+y^4
and we get the ineq!~!!!

cheers!!! :D :D :D

So is this the correct version of the problem ?

If x, y \geq 0 such that x2 + y3 \geq x3 + y4 then x3 + y3 \leq 2.

Maybe the other version is also correct but more difficult ? I don't know.
Can you give a counterexample to prove that the other version is false ?

Yeah, Manlio is right. It's problem 10.7 from the second day of the ALL-Russian olympiad 1999.
See also: http://www.mccme.ru/olympiads/vmo/25/vmo25.htm#kl10d2r

I'll send John Scholes a mail to tell him he's wrong :D

Of course it's difficult to prove a problem which is false ... :?

[quote="Ji Chen"]
In fact, we have  $x^{5}+y^{5}\leq 2$ for $x^{2}+y^{3}\geq x^{3}+y^{4}$.
[/quote]
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=105686
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=111000

Sort-of-ugly solution to the stronger one:
[hide]
We have
\[x^{2}+y^{2}\geq x^{3}+y^{4}-y^{3}+y^{2}\geq x^{3}+y^{3}\]
by the condition and AM-GM. We use this new condition to prove $x^{5}+y^{5}\leq 2$, which follows from
\[(x^{5}+y^{5})(x^{2}+y^{2})^{5}\leq 2(x^{3}+y^{3})^{5}\]
\[\iff (x^{4}-x^{3}y+x^{2}y^{2}-xy^{3}+y^{4})(x^{2}+y^{2})^{5}\leq 2(x^{2}+2xy+y^{2})^{2}(x^{2}-xy+y^{2})^{5}\]
Letting $u=\frac{x}{y}+\frac{y}{x}\geq 2$, this is equivalent to
\[u^{5}(u^{2}-u-1)\leq 2(u+2)^{2}(u-1)^{5}\]
Now the ugly part: multiplying everything out, we pull out [i]two[/i] (!) factors of $u-2$. It remains to prove
\[u^{5}+3u^{4}-3u^{3}-4u^{2}+6u-2\geq 0\]
But this is simple: $u\geq 2$ implies $3u^{4}\geq 6u^{3},\ u^{5}\geq 8u^{2},\ 6u\geq 12$.
[/hide]

My proof similar to your proof, scorpius119:
From $x^{2}+y^{2}\geq x^{3}+y^{3}$ we obtain if $x=ky$ then $y\leq\frac{k^{2}+1}{k^{3}+1}.$
Thus, it remains to prove that $k^{5}\left(\frac{k^{2}+1}{k^{3}+1}\right)^{5}+\left(\frac{k^{2}+1}{k^{3}+1}\right)^{5}\leq2,$ which equivalent to your inequality:
$(x^{5}+y^{5})(x^{2}+y^{2})^{5}\leq 2(x^{3}+y^{3})^{5}.$ :D
Practically the same. :wink:

[quote="arqady"] we obtain if $x=ky$ then $y\leq\frac{k^{2}+1}{k^{3}+1}.$[quote]
what is k? :huh:

[quote="arkhammedos"][quote="arqady"] we obtain if $x=ky$ then $y\leq\frac{k^{2}+1}{k^{3}+1}.$[/quote]
what is k? :huh:[/quote]
$k=\frac{x}{y}$ for $y\neq0.$:wink:

oops sorry for my bad question because it comes in my mind that $k$ is an integer
thanks.


V is a vector space
T is a linear transfermation from V to V(operator)
suppose for any linear trasfermation S from V to V
The identity TS=ST holds

Prove that T is a multiple of the identity map

Suppose that $ v$ and $ Tv$ are linearly independent for some $ v\in V$. Choosing $ S$ so that $ Sv\equal{}v$ and $ S(Tv)\equal{}0$, we arrive at a contradiction: $ TSv\equal{}Tv\ne 0\equal{}STv$.

Therefore, $ Tv\equal{}\lambda_vv$ for all $ v\in V$. It remains to prove that the scalar $ \lambda_v$ does not depend on $ v$. If $ \lambda_v\ne \lambda_w$, then $ v$ and $ w$ must be linearly independent, but on the other hand
\[ \lambda_{v\plus{}w}(v\plus{}w)\equal{}T(v\plus{}w)\equal{}\lambda_vv\plus{}\lambda_ww\]
which contradicts the linear independence.


Could somebody be able to send me a previous Nationals competition copy (or file)?
I'm interested to see what a full competition at that level looks like.
I've only done up to States, and my school doesn't have any Nationals

I'd be much obliged if someone could send me one.

Thanks.

I think you should look at one of the stickies.

I don't have any PDFs, and I'm not sure if we're allowed to share those files without the permission of mathcounts.

Are you sure your coach doesn't have old nats?

Mathcounts national competitions are under strict copyrights, and if your coach doesn't have any old ones, you'll have to buy an old competition set from Mathcounts, which will include the national competition from that year.

Yeah, I've read the stickies, and they didn't make it too clear.
They did mention that the files were allowed to be shared
amongst people in private for 'educational purposes'.

I've asked my teacher. He doesn't have any Nationals, although he does have a bunch of States that I've already done.


induction
LaTeX
search
number theory
relatively prime

i am assuming that you want to say that $2^{3^k}+1$ is divisible by $3^k$

i guess that induction would work, but there is a better solution

$2^{3^k}+1 \mod{3^k}$

2 and 3 are relatively prime, so using an extension of fermat's little theorem, (note that $\phi(3^k)=2*3^{k-1}$

so $2^{3^k}+1 \equiv 2^{3*3^{k-1}}+1 \equiv 2^3+1\equiv 0 \mod{3^k}$

Nea , I would recommend to use Latex, it's very useful and easy to read.

[quote="Altheman"]...
2 and 3 are relatively prime, so using an extension of fermat's little theorem, (note that $\phi(3^k)=2*3^{k-1}$

so $2^{3^k}+1 \equiv 2^{3*3^{k-1}}+1 \equiv 2^3+1\equiv 0 \mod{3^k}$[/quote]

A typo, or something is missing here .. why $2^3+1 \equiv 0 \mod {3^k}$ )not to mention why of the previous step.. may like to edit it.
Nea - This problem  was posted recently here  so you may like to look it up with search  function..

[hide]yes, i made a mistake, and i see that induction would actually work better than i originally thought

$k=1$

$2^{3^1}+1=9$ which is divisible by $3^1$

then assume that $2^{3^k}+1 \equiv 0 \mod{3^k}$

using euler's extension of FLT ($\phi(3^k)=2*3^{k-1}$)

we have that $2^{3^k}+1 \equiv 2^{3^k+n*2*3^{k-1}}+1$

for any natural number n, those are equivalent, so letting n=3, we get

$\equiv 2^{3^k+3*2*3^{k-1}}+1 \equiv 2^{3^k+2*3^{k}}+1\equiv 2^{3^{k+1}}+1\equiv0$

so proven by induction[/hide]

Starting from the begining:
Altheman thanx.
tµtµ your induction solution is what i want but i don't understand why you write
$2^{3^{k+1}} + 1 = ((2^{3^{k}}+1) - 1)^3$
At my opinion there is a $+1$ missing at the right side of the formula. It will be great if you'll write step by step how you've past from $2^{3^{k+1}} + 1$ to $((2^{3^{k}}+1) - 1)^3$
Gyan i've been searching for a while but i haven't found it. Can you plz tell me the adress of the post?
Thanx to all

the link is here

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=2%5E%7B3%5En%7D&t=56978[/url]  ;)

Altheman  Also,  you were in rigth direction, when  you said
[quote]2 and 3 are relatively prime, so using an extension of fermat's little theorem, (note that $\phi(3^k)=2*3^{k-1}$[/quote]
For simplicity if we call $3^{k-1} = m$ so you said above  $2^{2m} \equiv 1 \bmod 3^k$
So $2^m \equiv \ either \ +1 \ or - 1 \bmod 3^k$
But if it were $+1 \bmod 3^k$ then it will be $+1 \bmod 3$  too.
But $2 \equiv (-1) \bmod 3$ and since $m$ is odd, so is $2^m$
So $2^{3m} \equiv -1 \bmod 3^k$ or $2^{3m} +1 \equiv 0 \bmod 3^k$

Hope this helps.
Added later:
With the above proof:  Actually ${2^{3^k}}+1$ is divisible by not only $3^k$but   $3^{k+1}$

[quote="Nea"]At my opinion there is a $+1$ missing at the right side of the formula.[/quote]
Yes it's a typo.


integration
logarithms
limit
real analysis
real analysis unsolved

Let $n,k$ be positive integers.

$(1)$　Evaluate $\int_{1}^{2^{n}}\log_{2}x dx.$

$(2)$　Let $a_{k},\ b_{k}$ be the integer part and fractional part respectively of $\log_{2}k.$ Find $\sum_{k=1}^{2^{n}}a_{k}.$

$(3)$　Find $\lim_{n \to \infty}\frac{1}{2^{n}}\sum_{k=1}^{2^{n}}b_{k}.$

[b]Solution to (1)[/b]: Note that $\int \log_{2}x \,dx = \frac{1}{\ln 2}\int \ln x \,dx = \frac{1}{\ln 2}(x \ln x-x).$

So, $\int_{1}^{2^{n}}\log_{2}x \,dx = \boxed{\frac{1}{\ln 2}\left( 2^{n}(n \ln 2-1)+1 \right)}.$

[b]Solution to (2)[/b]:

$\sum_{k=1}^{2^{n}}a_{k}= \sum_{k=2}^{2^{2}-1}a_{k}+\sum_{k=2^{2}}^{2^{3-1}}a_{k}+\ldots+\sum_{k=2^{n-1}}^{2^{n}-1}a_{k}+a_{2^{n}}$

$= 1\cdot 2+2\cdot 2^{2}+3\cdot 2^{3}+\ldots+(n-1)2^{n-1}+n = \sum_{k=1}^{n-1}k\cdot 2^{k}+n$

Now note that $\sum_{k=0}^{n-1}x^{k}= \frac{x^{n}-1}{x-1}$ for $x\not = 1.$

Differentiating both sides w.r.t. $x$ and rearranging terms, we get

$\sum_{k=0}^{n-1}k x^{k-1}= \frac{(n-1)x^{n}-n x^{n-1}+1}{(x-1)^{2}}$

$\Rightarrow \sum_{k=1}^{n-1}k x^{k}= x\cdot \frac{(n-1)x^{n}-n x^{n-1}+1}{(x-1)^{2}}$

Put $x=2$ in the above identity to get

$\sum_{k=1}^{n-1}k\cdot 2^{k}= 2 \left( (n-1)2^{n}-n\cdot 2^{n-1}+1 \right)$

$\Rightarrow \sum_{k=1}^{2^{n}}a_{k}= \boxed{2 \left( (n-1)2^{n}-n\cdot 2^{n-1}+1 \right)+n}.$

[b]Solution to 3[/b]: This is a bit tedious and I will do it later. We will, of course, have to use the result from (2) above and also somewhere in the solution use Stirling's asymptotic formula for $n!$.


algebra
function
domain
advanced fields
advanced fields theorems

Let $T$ be a bounded linear operator in a normed linear space $X$ ($T$ is not supposed to be compact, and $X$ is not necessarily a Banach space).

Let $\lambda$ be an element of the spectrum of $T$. The author of that book then says:

(a) the range of $\lambda Id - T$ is dense in $X$ and $(\lambda Id-T)^{-1}$ exists but is unbounded. We then say that $\lambda$ belongs to the continuous spectrum of $T$.

(b) $(\lambda Id-T)^{-1}$ exists, but its domain is not dense in $X$. We then say that $\lambda$ belongs to the residual spectrum of $T$.

(c) $\lambda Id - T$ does not have an inverse.

What I can't see is why in (a) the inverse is necessarily unbounded. Why couldn't the range of $\lambda Id - T$ be $\neq X$ and dense in $X$ and still $(\lambda Id - T)^{-1}$ is bounded on that range ? I must miss something easy.

Because a bounded operator defined on a dense subspace can be uniquely extended to a continuous operator on the whole space. In the situation you're describing, this would mean that you can define $(T-\lambda)^{-1}$ on the whole $X$, and then $\lambda$ wouldn't be in the spectrum, by definition.

I deleted my last reply; the question was obvious, thanks for your answer.


Is given  the triangle ABC where AB=10 , BC=8 , AC=9 .  The angle (B)  bisector  intersects the side AC in point P and devides it in 2 segments respecively AP= 5 and PC= 4 . Find the length of the bisector's segment BP.

See [[Stewart's Theorem]]; the answer is $ \sqrt{60}$.

I have 1 quesion :

Cevians are called only angle bisector , median and altitude ?

Cevian is generally any line connecting a vertex and opposite side of the triangle.

http://mathworld.wolfram.com/Cevian.html


function
complex numbers
real analysis
real analysis unsolved

Find a function $ f: I\subset \mathbb{R} \to \mathbb{R}$ such that $ f$ is differentiable on $ I$ and $ f(x\plus{}1)\equal{}f'(x)$.

A similar problem was solved here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=8145[/url].

The same method should work for your problem.  Namely, let $ c$ be a complex number such that $ e^c \equal{} c$.  Then let $ f(x)$ be the real part of $ e^{cx}$.


function
calculus
derivative
calculus computations

let f be a real function from $ (\minus{}1,1)$ to $ R$ withe $ lim_{x\to(\minus{}1)^{\plus{}}}f\equal{}lim_{x\to(1)^{\minus{}}}f\equal{}\plus{}\infty$,show that there existe a $ c$ wish belong to $ (\minus{}1,1)$ such that $ f'(c)\equal{}0$.

You left out a key word: differentiable.
There is a local minimum; we can find $ a<0$ and $ b>0$ so that $ f(a),f(b)>M>f(0)$, and then find the minimum on $ [a,b]$. The derivative there is zero.

I'd do it like this
We have that $ \forall\; M>0 \; \exists\; \delta>0$ such that if $ x\in V_{\delta}(1)$ or $ x\in V_{\delta}(\minus{}1)$ then $ f(x)>M$

Bolzano's Intermediate Value Theorem gives that $ \exists\; m,n$ with $ f(m)\equal{}f(n)\equal{}M$ We can now apply Rolle's theorem on $ [m,n]$


The players of a football team are to be given awards by their club directors. Including the reserves, the team of 15 players stand in a line along the pitch with one meter between them each. The manager will begin with any player and then continue presenting the awards to the others. If this procedure is to be completed using the shortest route, then beginning with player number 1 and continuing in line until player number 15 is the most appropriate. (In this case 14 meters would be covered). What will the total distance be if the same procedure is to be completed using the longest possible route?

The total distance would be at most $4(7+6+...+1) = 112$ if a closed loop were used, since this would be the length if we went from the centre to each point and back. The longest open path must be at least one less, or $111$.

It is easy to verify that such a path is possible.

I found 111 too. I think the anwers  is 111.


What is the arithmetic mean (average) of the areas of all noncongruent rectangles with integer side lengths and perimeter 8 units? Express your answer, in square units, as a decimal to the nearest tenth.

They are 1x3 and 2x2, and their average is 7/2=[b]3.5[/b]


algebra
polynomial
function
limit
rational function
complex analysis
complex analysis unsolved

prove that if the only singular point $ a$ on the circle of convergence of the power series $ f(z)\equal{}\sum_{n\equal{}0}^{\infty}a_n.z^n$ is a pole, then $ lim \frac{a_n}{a_{n\plus{}1}}\equal{}a$

We may assume that $ a\equal{}1$. Then $ f(z)\equal{}\frac{c}{(1\minus{}z)^k}\plus{}g(z)$, where $ c\ne 0$ and the coefficients of $ g$ tend to zero, because $ g$ is analytic in a disk of radius $ >1$. The conclusion follows.

That $ c$ is actually a polynomial in $ z$ of degree $ \le k\minus{}1$; you get an arbitrary polynomial in that spot rather than just a constant and a power of $ a\minus{}z$.

Indeed. Fortunately, the conclusion still follows. Or you can integrate $ f$ until the pole becomes a simple pole.

[quote="mlok"]Or you can integrate $ f$ until the pole becomes a simple pole.[/quote]
Erm... How would you integrate $ \frac 1{z^2}\plus{}\frac 1z$? :?

mlok - how does the conclusion follow? can you please explain..i dont understand. if the pole is simple , then all is fine.

The Taylor series at $ 0$ of the rational function $ \frac{q(z)}{(1\minus{}z)^k}$ has coefficients $ a_n\equal{}p(n)$ (at least for large $ n$), where $ p$ is a polynomial. I'll leave the details to an interested reader.  :)

How about a simple case first:  Assume $ f(z)$ has a pole on the unit circle at $ z=a$.  Then $ f(z)$ can be written as:

$ f(z)=\frac{\phi(z)}{(z-a)^k}$

where $ \phi(z)$ is analytic and non-zero at $ z=a$.  Here's the simple part:  Let's assume for starters that $ \phi(z)=c$:

Now, the Taylor series for $ f(z)$ at $ z=0$ then is:

$ f(z)=\sum_{n=0}^{\infty}\frac{1}{n!}\frac{d^n}{dz^n}
\left\{\frac{c}{(z-a)^k}\right\}\Biggr\vert_{z=0} z^n$

and therefore:

$ \frac{b_n}{b_{n+1}}=
\frac{(-1)^n ck(k+1)...(k+n-1)(-a)^{-(k+n)}}{n!}
\frac{(n+1)!}{(-1)^{n+1}ck(k+1)...(k+n)(-a)^{-(k+n+1)}}$

which if you simplify that, it reduces to just $ a$.

therefore:

$ \mathop\lim\limits_{n\to\infty}\frac{b_n}{b_{n+1}}=a$

Looks so anyway.  Pretty new to me.


symmetry
geometry
geometric transformation
rotation

What is symmetry and how do you use it to solve problems?

Thank you

rts2007

Dude.
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=267768
Seriously, it's like 5 topics below this one.

But anyway, symmetry is useful when you have congruent/similar "stuff" and you prove something that can be applied to other things that are symmetric.

Perhaps as a stupid example, suppose you have a square with a diagonal drawn in, creating 2 congruent triangles. You've proven the triangles congruent, and found the area of one triangle. Now, because of symmetry across the diagonals, you can say "By symmetry, triangle ABC (or whatever) has the same area as triangle DEF (or whatever), so the area of the square is..."

Clearly I just made that example up, so it's perhaps not so helpful, but looking at other problems that involve symmetry (usually something like regular polygons, etc) might help.

Symmetry also works in counting problems, like when you're counting the number of ways to arrange beads on a bracelet such that flipping the bracelet doesn't change the arrangements of the beads. It's also conjuncted (dunno if that's a word) with rotations in terms of combinatorics, like when arranging things in a circle such that rotating objects is considered indistinguishable.

In terms of combinatorics, you generally use symmetry just for the sake of correcting for overcounting, which is covered in Chapter 3 of [i]Introduction to Counting and Probability.[/i]


How do you solve this?

Link to problem : http://books.google.com/books?id=MI6c6nZKpjMC&pg=PA48&lpg=PA46&ots=EoMf3lQBM2&dq=2005+amc+10b+%238&output=html

(It's not on AOPS wiki)

On the other hand, it [b]is[/b] in the Contests Section.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1376098#1376098]Here[/url] is the link to the thread.


This is Jeopardy!!!! This has three different players each time. It is a trivial like game. There are seven categories, each with five questions in them. The first person chooses the category, then the point amount. The person to correctly answer the question first, will get the point amount, and be allowed to choose the next category and point amount. I will start as soon as I have the three contestants.

join....
BUTLER, YOU CANNOT MOD 3 GAMES AT THE SAME TIME.

It only says RPG's, and this is a very educational knowledge game![/hide]

[quote="andrewjjiang97"][quote="randomguy64"]I do not join.

How do you prevent people from simply looking it up?

In addition, you will have timezone issues. For example, if your timezone is GMT-6, and the first guy's timezone is also GMT-6, and the other two people's timezones are both GMT+6, the first guy obviously has an advantage.[/quote]

Good point. This is basically just an active test.

\quits[/quote]

very good point.
/quits

Hmmm..... Okay, let's have 4 people play. :maybe:

k00lpersoon will choose from the categories:

AoPS   History    Beginning and ending with "M"  Science   Movie stars   Music   Animals
100       100                 100                            100           100            100       100
200       200                  200                           200            200             200     200
300       300                 300                           300             300            300       300
400       400                 400                           400             400            400        400
500       500                 500                           500              500            500        500

NO REALLY, IT'S SUPER HARD.
See, no one is answering it.

OMG I think its science...
no i'm sure
really... :rotfl:

oh yeah!
it should be science because there is obviously lots of experiements here and stuff about plants and stuff like that.

but it's definitely not math...

I never do any math on this website, just science...

not only that but i don't think there even is MATH on the website!
I mean if you click bookstore, geometry isn't math!
And also latex doesn't sound familiar.
and plus the side bar has science stuff on it.

Yeah. I mean like geometry deals with dissecting frogs... DUH!!

Dude the answer is pretty obviously the hidden "E. none of the above".

OMG OMG OMG!!!!!! WILL YOU GUYS STOP BUGGING ME ON THIS!!!???????????????? IF YOU WANT TO BUG SOMEONE GO DO IT SOMWHERE ELSE, BUT NOT HERE!!!!!!!!!!!!!!!!! ANSWER ME THIS!!!!!!!!!!!!!!

Would you like people continuously BUGGING you on the subject on the difficulty of a question???????

Just look. I already have a hard enough life without people criticising me on what I do here too.

what the heck butler,
we really don't know!
we are discussing what the answer would be.

And by the way, I'm serious.  AoPS has science and social studies forums, and math sometimes requires critical thinking.

K, it's my question.
hey butler it's math

Wow! You are so smart!

I never could have figured that out!

it really is math?? i would have never guessed that...

[quote="piegraph"]
BUTLER, YOU CANNOT MOD 3 GAMES AT THE SAME TIME.[/quote]

Now you can't anymore. Remember the reason why we had a policy about RPGs anyways?

Also, since you got annoyed at the people playing this, I locked it.


Find an integer $n$, where $100 \leq n \leq 1997$, such that \[\frac{2^{n}+2}{n}\] is also an integer.

Let $ T_p$ is half period, suth that $ p|2^p \plus{} 1$ and $ p|2^n \plus{} 1\to n \equal{} mT_p,m \equal{} 1\mod 2$.
If $ n|2^n \plus{} 2$ then for all odd prime divisor $ p|n$ we get $ \frac {n \minus{} 1}{T_p}$ is odd integer. Therefore $ v_2(T_p)$ is same for all odd prime divisors.
If $ n \equal{} p$ (is prime) we get $ p|4\to p \equal{} 2$. Obviosly $ v_2(n)\le 1$. If $ n \equal{} 2p$, then $ p|2^2 \plus{} 2 \equal{} 6\to n \equal{} 6$.
From conditions we get, that n had odd prime divisor $ p\le 43$ and $ T_3 \equal{} 1,T_5 \equal{} 2$, $ T_7$ is not defined, because period is 3 and $ 7\not |2^k \plus{} 1$. $ T_{11} \equal{} 5, T_{13} \equal{} 6,T_{17} \equal{} 4, T_{19} \equal{} 9,T_{23} \equal{} 11,T_{29} \equal{} 14,$ $ T_{31}$ is not defined.
$ T_{37} \equal{} 18,T_{41} \equal{} 10,T_{43} \equal{} 7$.
If $ p|n$ and $ T_p$ is even, then $ p \equal{} 1\mod 4$, therefore $ n \minus{} 1$ must be even and $ n \equal{} pk, \frac {k \minus{} 1}{T_p} \minus{} \ odd$. It mean, that $ v_2(k \minus{} 1) \equal{} v_2(T_p) < v_2(p \minus{} 1)$, therefore k had prime divisor q, suth that $ v_2(q \minus{} 1)\le v_2(T_p)$ and $ v_2(T_q) < v_2(T_p)$ - contradition.
We get, that all odd prime divisors n had odd $ T_p$, therefore n=2m, m- odd. Therefore one one prime divisor of n $ \le 31$.
It is $ p \equal{} 3$ or $ p \equal{} 19$ or $ p \equal{} 23$. If $ n \equal{} 2*23*k$, then $ 11|2k \minus{} 1$. But $ k \equal{} 17,39$ not work.
If $ n \equal{} 2*19k$, then $ 9|2k \minus{} 1, k \equal{} 5,23,41,59$ not work. Therefore $ n \equal{} 2*3^l*k$. Obviosly $ 2^n\equiv 1\mod 3^{l\plus{}1}$, therefore $ l$ can not be $ >1$. Therefore $ n\equal{}6k$, were $ k$ odd prime $ > 43$ or $ k \equal{} 1$, because prime divisors of k must be $ >43$ If k is prime $ k|2^6\plus{}2$. If $ k$ is odd prime $ T_k|6\minus{}1 \equal{} 5\to k\equal{}11$.
Therefore all solutions $ n|2^n \plus{} 2,n\le 1997$ are $ n \equal{} 1,2,6,66.$

[quote="Rust"]Let $ T_p$ is half period, suth that $ p|2^p \plus{} 1$ and $ p|2^n \plus{} 1\to n \equal{} mT_p,m \equal{} 1\mod 2$.[/quote]Heh?  :|

And it's wrong anyhow, 946 is a solution.


algebra
function
domain
geometry
3D geometry
polynomial

$ \text{Find all} \ x \ \text{such that} \ \minus{}\pi < \frac{1}{\sqrt[3]{x}} < \pi$.

Please write it out step by step algebraically. I know the answer, but I did it analytically with no algebra.

Take $ \frac {1}{\sqrt [3]{x}} > \minus{} \pi$, multiply both sides by $ \sqrt [3]{x}$, and then divide by $ \minus{} \pi$ to get $ \sqrt [3]{x} > \minus{} \pi \implies x > \minus{} \pi ^3$.

Now take $ \frac {1}{\sqrt [3]{x}} < \pi$, multiply both sides by $ \sqrt [3]{x}$, and then divide by $ \minus{} \pi$ to get $ \sqrt [3]{x} < \pi \implies x < \pi ^3$.

Thus, $ \{ x \ | \ \minus{} \pi ^3 < x < \pi ^3 \}$.

EDIT: Ah ha! I knew I messed up somewhere. Thanks, Farenhajt.

[quote="isabella2296"]Take $ \frac {1}{\sqrt [3]{x}} > \minus{} \pi$, multiply both sides by $ \sqrt [3]{x}$, and then divide by $ \minus{} \pi$ to get $ \sqrt [3]{x} > \minus{} \pi \implies x > \minus{} \pi ^3$.

Now take $ \frac {1}{\sqrt [3]{x}} < \pi$, multiply both sides by $ \sqrt [3]{x}$, and then divide by $ \minus{} \pi$ to get $ \sqrt [3]{x} < \pi \implies x < \pi ^3$.

Thus, $ \{ x \ | \ \minus{} \pi ^3 < x < \pi ^3 \}$.[/quote]

So, you're including $ 0$ in your solution, in spite of the expression not being defined for $ x\equal{}0$? :)

[hide="Solution"]$ \left|{1\over\sqrt[3]x}\right|<\pi$

$ \left|{1\over x}\right|<\pi^3$

$ |x|>{1\over\pi^3}$

$ x<\minus{}{1\over\pi^3}$ or $ x>{1\over\pi^3}$[/hide]

hello, i have only $ \frac{1}{\pi^3}<x$ because of the range of definition of $ \sqrt[3]{x}$ is here $ x>0$.
Sonnhard.

Thanks. $ \sqrt[3]{x}$ can be negative so both answers should work right?

The domain of $ \sqrt[2n]x$ is $ [0,\plus{}\infty)$

The domain of $ \sqrt[2n\plus{}1]x$ is $ (\minus{}\infty,\plus{}\infty)$

[quote="Dr Sonnhard Graubner"]hello, see here for the domain of the cube root
http://mathworld.wolfram.com/CubeRoot.html
Sonnhard.[/quote]

Quote from the page given: "The schoolbook definition of the cube root of a negative number is $ ( \minus{} x)^{1/3} \equal{} \minus{} \left(x^{1/3}\right)$" - and in this problem we're absolutely not interested in definition of the cube root function over complex numbers.

By the way, Sonnhard, I've seen you were using Maple to write some paper in geometry (about some "special" triangle). Then you're obviously using Maple to find your polynomial factorizations too? Naughty, naughty...

hello, ok, if you use this definition we have additionally $ x<\minus{}\frac{1}{\pi^3}$.
Sonnhard.


probability
analytic geometry
integration
calculus

A point $(x,y,z)$ is randomly chosen in 3-dimensional space such that $0 \leq x \leq 3, \ 0 \leq y \leq 4 , \ 0 \leq z \leq 5$.
 What is the probability that none of the three coordinates (i.e. $x$, $y$, or $z$) vary by more than one unit from any of the other two?

the probability can be constructed by volumes.

you have a Parallelepiped of volume $3.4.5=60$

And your set problem is just a simple Box. As I understand.

But it is can be just constructed in a box of $3^{3}$ of volume. because the coordinates

think about this.

I think It helps

[quote="Eternica"]Uh. I find something different.

[hide]
You have a couple of cases:

$0,1$ are the only numbers (not counting $1,1,1$).

$1,2$ are the only numbers (not counting $2,2,2$).

$2,3$ are the only numbers (not counting $3,3,3$).

$3,4$ are the only numbers

For the former three cases, there are $2^{3}-1=7$ choices.
For the latter case, there are $4$ choices.

Thus, we have $3*7+4=25$ total out of $4*5*6=120$ choices and so your probability should be $\frac{25}{120}=\frac{5}{24}$.[/hide][/quote]

But the variables aren't restricted to integers.   :)

Hope you had fun with that calculus  :lol:
In general, though, restrict calculus to the calculus forum; these problems are non-calculus based.

Well, I couldn't find a geometric solution, because the region seems completely intractable :) Prove me wrong. I know I would have finished this calculus solution eons faster than a geometric one. Obviously in hindsight you can try to reconstruct the region by considering the six triple integrals in the final computation, but even then I wish you good luck.


AoPSwiki
articles
modular arithmetic
number theory
LaTeX

I was trying to format a series of equations such that each equal sign was directly underneath the other. I can't figure out how to do this, can someone please help??

Use the align environment as described in AoPSWiki [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Math#Aligning_Equations_.28align.29]Aligning Equations (align)[/url].

i'm sorry I'm reviving this topic but that didn't work.

The equal signs were still not lined up

It should work. Can you show us an example that doesn't work for you?
Check that this minimal document
[code]\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\begin{document}
\begin{align*}
1 + 1 &= 2 \\
11 + 22 &= 33
\end{align*}
\end{document}[/code]gives you this
\begin{align*} 1 + 1 & = 2 \\
11 + 22 & = 33 \end{align*}

Oh I was putting the ampersand in the wrong places oops... thank you!!

EDIT: Does this work on the forum because my code didn't work. Is my code wrong?[code]\begin{align*}
3^{6n}-2^{6n} & = (3^6)^n - (2^6)^n\\
&= (729)^n - (64)^n
\end{align*}
[/code]

[code]\begin{align*}
(729)^n - (64)^n &\equiv (29)^n - (29)^n \pmod{35}\\
&\equiv 0 \pmod{35}
\end{align*}[/code]

btw this align environment works for modular arithmetic right?

You don't say what "didn't work" means? What happens? Are there any errors or warnings?
The code works fine in a document and here:

\begin{align*}
3^{6n}-2^{6n} & = (3^6)^n - (2^6)^n\\
&= (729)^n - (64)^n
\end{align*}

\begin{align*}
(729)^n - (64)^n &\equiv (29)^n - (29)^n \pmod{35}\\
&\equiv 0 \pmod{35}
\end{align*}

Yes there's an error message and it doesn't show up on the final pdf file

So what is the error message???!!
Did you get the same error message with the minimal document I asked you to try? If not, give your own [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]minimal document[/url] which shows the error.
You can't expect to get help unless you give us the information to work with.

the error message is

file ended while scanning use of  \end{align*}

ANd as for the minimal document, the file doesn't show up.

IN other words, a pdf file is not produced

That error message usually means there's a mistake in the code eg a missing } or the like. Can you post the code you are using - the complete document to check for errors?
Do you get the same message with my minimal document where the pdf isn't produced? Can you post the log file here (if your file is called something.tex then the log will be called something.log and will be in the same directory)
What happens with an even more minimal document
[code]\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\begin{document}
Sample text $1+1=2$
\end{document}[/code]


Solve equation:

$ \sqrt[3]{3x\plus{}2}\plus{}\sqrt{3x\minus{}5}\equal{}x^2\minus{}1$

$ \sqrt [3]{3x \plus{} 2} \plus{} \sqrt {3x \minus{} 5}\leq \frac {7}{4}x \minus{} \frac {1}{2}\leq 4x \minus{} 5\leq x^2 \minus{} 1$ for $ x\geq \frac {5}{3}$.

Actually, for $ x \in \left[\dfrac{5}{3},2\right)$ we have $ \frac {7}{4}x \minus{} \frac {1}{2} > 4x \minus{} 5$


$abc=2$
prove that:
$a^3+b^3+c^3 \geq a\sqrt{b+c}+b\sqrt{a+c}+c\sqrt{a+b}$

The inequality is reversed and it has also been discussed. The idea is to apply Cauchy-Schwarz in the RHS and then Chebyshev inequality and at last AM-GM.

Anyway,i posted a better one a month ago, I think, without any condition.The inequality is:

     $a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{c+a}\leq\sqrt{2(a+b+c)(ab+bc+ca)}$ ;)

  and it follows imediately from Cauchy-Schwarz. ;)

Excuse me,Harazi,but maybe I didn't make myself clear here.I said that we can use the inequality ,I posted to solve
this "difficult" inequality from the junior balkan shortlist.Then, after what it remains we can surely use the condition given. ;)

we have
   ${x^3 +y^3 +z^3 }\geq $
 other hands ,
${x^3 +y^3 +z^3 }$( \sum{ \frac {a}{b+c }}$ \geq A

[quote="tranthanhnam"]we have
   ${x^3 +y^3 +z^3 } \geq {\sum{ \frac{a}{b+c }}}$
 other hands ,
${x^3 +y^3 +z^3 }( \sum{ \frac{a}{b+c }} \geq A$

Okie, Here my solution.
Set $P = a^3 + b^3 + c^3$.
Lemma: If $x,y>0$ then: $x^3 + y^3 \geq xy(x+y)$.
Proof: Using AMGM, we have:
$a^3 + b^3+c^3+c^3 \geq ab(a+b) + (c^3 + c^3) \geq 2\sqrt{2c^3ab(a+b)} = 4c\sqrt{a+b}$.
=> $P + c^3 \geq 4c\sqrt{a+b}$.
Oh, $P + b^3 \geq 4b\sqrt{a+c}$.
      $P + a^3 \geq 4a\sqrt{b+c}$.
=> $P \geq a\sqrt{b+c} + b\sqrt{a+c} + c\sqrt{a+b}$ (yeah).
Equality holds if $a=b=c = \sqrt[3]{2}$.


Find $ a, b, c, d, e, f, g$ where:

$ 1$. All the seven are positive integers
$ 2$. $ \frac{1}{a} \plus{} \frac{1}{b} \plus{} \frac{1}{c} \plus{} \frac{1}{d} \plus{} \frac{1}{e} \plus{} \frac{1}{f} \plus{} \frac{1}{g} \equal{} 1$
$ 3$. $ a \times b \equal{} c$ and  $ a \times c \equal{} d$ and  $ a \times d \equal{} e$ and $ d \times f \equal{} g$

[hide="Answer"]We need a = 2 otherwise if a is any higher we don't get to 1 fast enough. b can't be 3, since 1/2 + 1/3 + 1/6 = 1 already, so a reasonable guess would be to have a = 2, b = 4 ... e = 32.

This leaves $ \frac{1}{f} \plus{} \frac{1}{16f} \equal{} \frac{1}{32}$, and it turns out that $ f\equal{}34$ works $ \implies g \equal{} 16 \cdot 34 \equal{} 544$

Hence, $ (a,b,c,d,e,f,g) \equal{} (2,4,8,16,32,34,544)$[/hide]


AMC
USA(J)MO
USAMO
AIME
calculus
trigonometry
AMC 10

I'm still holding out hope that there will be a Mock USAMO for those who qualified through the Mock AMC A/B and Mock AIME A. Is there any new information on this?

If we do plan to keep this thing alive(yet I don't know how possible it'll be when everyone is in school, sports, extracirriculars, etc.) then I have a plan.
Since the USAMO is like a culmination, we could average out everyone's performance on all of the AIME's and then take their highest AMC score ever and then the top ten indices would make USAMO. This is due to the fact that good OLympiad problems are much harder to come up with than the average AMC question.

Maybe we can ask some of the advanced users if they are interested in writing/grading this? Not all of them visit the AMC forum.

Yeah, I definitely need a lot more practice AIME. I am kind of depressed that I only got a 3 on the Mock AIME! I

Can non-moderators host the mock contests also?

Yeah, Mildorf is not a moderator. It's funny that people who are hosting Mock AMCs are all intermediate forum moderators.

Yeah, I think anyone can make it.  Mildorf isn't a moderator but he made the Mock AIME.  The more contests the better!

If we're going to have two AMC's a month, make one about AMC 10 level, and the other a AMC 12 level.  I would like to see at one test in which the last 10 problems are at least doable for me.

I'll co-write a Mock AMC 10 test...I have a bunch of easy problems all around my room.  But I'd like it if someone would look it over, to make sure it isn't too easy.

I'd be willing to co-write an AMC 10 one day... key word: co-write, since I don't have enough problems to write one all by myself

I can co-write the Mock AMC 10 - no problem. I have tons of spare problems lying around.

I am a moderator and I would be glad to write a contest.  I was thinking about doing a contest all on one subject so that it would fit into my forum.  I could do some cool calculus problems but that might anger AoPS because you guys have a problem with calculus (JK, but you know what I mean).  I was also thinking about writing contests about a single subject, like 10 challenging problems about imaginary numbers, or trigonometry, or something like that...an in depth sort of contest instead of a general level like "mock AMC 10" But then again, my contest would not be a "mock" anything =)  Interested?

Hello,

I have written a mock AMC 12. It is currently being edited by someone who will not be taking it. I will start a new thread later where I will talk more about it. May I call it Mock AMC D, or is that name already taken?

[quote="JS1527"]I could do some cool calculus problems but that might anger AoPS because you guys have a problem with calculus (JK, but you know what I mean).  [/quote]
I $\heartsuit$ the Calculus.

I am working on a Mock AMC.  I guess mine could be the E.  It will take a few weeks before I will be able to administer it.

I will write down any problems that I think of and eventually pool 15 of them together as Mock AIME 3. This will not be for at least a month or so though.


Find all $p$ is a prime such that $p^2-p+1=a^3$ where $a$ is integer


another one

for every positive composite number the following is true:-

nl(n-1)! where n>4
get the formal proof please

Since $ n$ is composite, let $ n \equal{} xy$, where $ 1 < x\le y < n$. First assume that $ x < y$. Since $ x\ge 2$, we must have $ y\le n/2\le n \minus{} 1$. Thus $ x$ and $ y$ are both in $ \{1,2,\dots,n \minus{} 1\}$ and therefore $ n \equal{} xy\mid (n \minus{} 1)!$.

If $ x \equal{} y$ then $ n \equal{} x^2$ and we have $ n^2 \minus{} 1\ge 2n$, which implies $ x^2\mid (2x)!\mid (x^2 \minus{} 1)!$. Hence in all cases $ n\mid (n \minus{} 1)!$.

This should go into the Number Theory fora...


is $ a^*b^* \cap c^*d^* \equal{} \oslash$

Well, can you think of any string of length 0 or more which consists of 0 or more a's followed by 0 or more b's, and is also 0 or more c's followed by 0 or more d's?


calculus
integration
trigonometry
logarithms
real analysis
real analysis unsolved

Calculate: 1. $ \int\frac {dx}{e^x \plus{} 1}$
2. $ \int\frac {1 \minus{} 2\sin^2x}{2 \plus{} \sin 2x}dx$

1.  $ \int \left( 1 \minus{} \frac{e^x}{e^x \plus{} 1} \right) \, dx \equal{} x \minus{} \ln (e^x \plus{} 1) \plus{} C$.

2.  $ \int \frac{\cos 2x \, dx}{2 \plus{} \sin 2x} \equal{} \frac{1}{2} \ln (2 \plus{} \sin 2x)$.


search
inequalities
inequalities unsolved

[quote="S_Alex"]a1, a2...,a10 >0 a1+a2+...+a10=1. Find the maximum of
E= \sqrt (a1*a2) +  \sqrt (a2*a3) + ... +  \sqrt (a9*a10)[/quote]

I'm sorry. The source is not Arad 99, it's Iasi 98. And by the way the answer is cos pi/11 but i bon't know how to prove this.

[quote="harazi"]I am astonished by the beauty of this problem. I think I would have never solved it. Some ideas of how to get the maximum:
  We search for x_1,...,x_9 and y_1,...,y_9 such that we could apply (x_1a_1-y_1a_2) 2 +...+(x_9a_9-y_9a_10) 2  \geq 0. This can be written (a_1) 2 (x_1) 2 +(a_2) 2 ((x_2) 2 +(y_1) 2 )+...+(a_10) 2 (y_9) 2  \geq 2(x_1y_1a_1a_2+...+x_9y_9a_9a_10). So, it would be better to have x_1y_1=...=x_9y_9=1 and (x_1) 2 =(x_2) 2 +(y_1) 2 =...=(y_9) 2 . If we put (x_1) 2 =b_1 and so on we have to solve the system b_1=1/b_1+b_2=1/b_2+b_3=...=1/b_9. Unfortunately, I cannot solve this system. Help![/quote]
The ideea is good and quite natural only that you should replace a_i with  \sqrt a_i. I've found the official solution to this problem in one of my books. i don't have time know to write it , and i don't want to spoil the fun for those of you who are trying to solve it.

well,admittedly this is the most unnatural (is there such a word) solution i have ever given.and i saw later that my idea was the same as Harazi's.but with a subtle difference :D.he should have tried taking
x_1=y_9=1
(x_1) 2  =(x_2) 2  +(y_1) 2 2 =...=(y_9) 2
and  x_1y_1=...=x_9y_9
this is actually more symmetric and reduces the number of variables very fast (to 8) as we can see x_9=y_1 ; x_8=y_2  and so on  (my variable were different,so i might have made a mistake here,just chek)
then then aftere the elimination (it's easy :) ) the equation for the maximum is t^5-t^4-4t^3+3t^2+3t-1=0 where the maximum value of the given expression is t/2
i checkd by the calculator that t= 2cos( \pi /11) is a root of the above but i dont what abt the other roots :?
i am working the stuff out BTW

I also got that equation, but the question is: how on earth could anyone see that the solution is 2cospi/11, during a contest?

i agree ,harazi.that's what makes this problem soo beautiful.
anyways,let's keep trying the problem (not the equation),mate
well if one tried lagrange multipliers ,he'd be kicking his own ass

I'm really not going to think on how to solve the equation. There must be some other way to solve the problem, but I don't feel it. It is an extraordinary beautiful problem.

Why can't we just use the arithmetic mean, so that E \leq (a1+a2)/2 +(a2+a3)/2 +...+ (a9+a10)/2 = 1 with equality if a1=a2=...=a10?

[quote="pbornsztein"]Because in your rhs, the sum is not 1, since a_1 and a_10 appear only one time.
The inequality is not symmetric in each of the variables.

Pierre.[/quote]
Like duhhhhhh   :)


ratio
complex numbers
algebra unsolved
algebra

If z1,z2,z3 are complex numbers such that $ \frac {2}{z_1} \equal{} \frac {1}{z_2} \plus{} \frac {1}{z_3}$ show that points represented by z1,z2,z3 lie on a circle passing through origin

Take $ z_1 \equal{} 1$, $ z_2 \equal{} \minus{} 1$, $ z_3 \equal{} \frac {1}{3}$.  These numbers satisfy the equality and don't lie on any circle.  Did you mean something else?

We have $ (0,z_1,z_2,z_3) \equal{} (\infty, 2/z_1, 2/z_2, 2/z_3) \equal{} (\infty, 1/z_2 \plus{} 1/z_3, 2/z_2, 2/z_3)$. Letting $ w_i \equal{} 1/z_i$, we have that $ 2w_2, 2w_3,$ and $ w_2 \plus{} w_3$ are collinear, so this cross-ratio is real. Therefore $ 0,z_1,z_2,z_3$ lie on the same line-circle.

I did like this
Let angles at A and C be $ \alpha \ and \beta$.
and let us denote A=z1,B=z2,C=z3,D=origin

SO, $ \minus{}z_1\equal{}(z_2\minus{}z_1)e^{i \alpha},(z_2\minus{}z_3)\equal{}\minus{}z_3e^{i \beta}$
Multiplying both

$ z_1z_2\minus{}z_!z_3\equal{}(z_2z_3\minus{}z_1z_3)e^{i(\alpha \plus{}\beta )}$

Assuming that points are colcyclic
$ \equal{}> \alpha \plus{}\beta \equal{}\pi$
So,$ z_1z_2\plus{}z_2z_3\equal{}2z_1z_3$

Dividing whole equation by $ z_1z_2z_3$
we get the final answer


Amazing!!!! (or not?). The minimum to that simple expression is $21.601886904415354486926355909659$. I don't have the means to obtain more decimals right now.
Take the usual $a=y+z, b=x+z, c=y+x$. Now we can easily make $x+y+z=1$. More than that, make the notation $\frac {25(1-x)}{1+x}=u$ etc. Our inequality becomes $u+v+w\geq t$, with $x=\frac {25-u}{25+u}$ etc. We have $x+y+z=1$
We use Cauchy for the following:
\[1\cdot [(25-u)(25+u)+(16-u)(16+u)+(9-u)(9+u)]\geq (25+16+9-u-v-w)^2\Rightarrow 25^2+16^2+9^2-u^2-v^2-w^2\geq 2500+S^2-100S\], where $S=u+v+w$. Further more with $u^2+v^2+w^2\geq 1/3S^2$, we obtain the inequation $4S^2-300S+4614\leq 0$, which not only gives the lower bound but also gives the upper bound. The equality holds for $u=v=w$, and from here we [u]can[/u] obtain $x,y,z$, and from here we [u]can[/u] obtain $a,b,c$ for which the equality holds.

Mistakes up there :D

At LAST!!!  I realised where I am wrong. My Cauchy can't get to the equality case.
Yes, you're right Sasha. Just take care of the first sum and $y=3z$ cause you didn't edit them good.

Try! :D . $a,b,c$ be length of sides of triangle

[quote]But Sasha already solved it hxtung.....[/quote]
Note:$a,b,c $  be length of sides of triangle

[quote="hxtung"]Let $a,b,c$ be length of the sides of triangle $ABC$. Find the greatest real number $t$ such that:
$\displaystyle \frac {25a}{b+c}+\frac {16b}{c+a}+ \frac {9c}{a+b} \geq t$[/quote]

I proved that they also existed the upper bound. In fact, if $x,y,z$ are sides of triangle, we will have
\[ \displaystyle \frac {25x}{y+z}+\frac {16y}{z+x}+ \frac {9z}{x+y} \leqslant 41 \]

[quote=sqing]If $a,b,c>0$ , then\[ \frac {25a}{b+c}+\frac {16b}{c+a}+ \frac {9c}{a+b} \geq 22.\][/quote]
[url=http://www.artofproblemsolving.com/community/c6h493853p2771450]General Nesbitt inequality[/url]

[quote=virus_dien_khung]I proved that they also existed the upper bound. In fact, if $x,y,z$ are sides of triangle, we will have
\[ \displaystyle \frac {25x}{y+z}+\frac {16y}{z+x}+ \frac {9z}{x+y} \leqslant 41 \][/quote]
Which is clearly after using Ravi substitute.


AMC
AIME
AMC 8
AMC 10
AMC 12
MATHCOUNTS
USAMTS

1) AIME is the American Invitational Mathematics Examination. You can qualify through USAMTS (although not recommended), AMC10A/B (top 1%, generally around 120/150), or AMC12A/B (top 5%, generally 100/150). Most people qualify via AMC12A/B, I would think.

2) AMC is the American Mathematics Competition. Anyone can take it.

3) AMC8 is designed for 8th graders and below, AMC10 is designed for talented 10th graders and below, and AMC12 is designed for mathematically talented students who have had some decent experience with high school mathematics. No calculus is on any of the tests, although every once in awhile, there will be a problem with a valid calculus solution.

4-6) Check online for all kinds of regional competitions your state or county may have as well as other big national competitions.

[quote="Silverfalcon"]I heard a lot kind of competition and I kinda don't get it...
Number one question: What's AIME?
Number two question: What's AMC?
Number three question: What's difference between AMC8, AMC10, and AMC12?
Number four question: Are there any big math competition during summer that is big like Mathcounts?
Number five question: Are there any math competition where you can just sit in your computer and do it?
Number six question: Other than ones I mentioned, are there any good competition to enter?[/quote]

Rlee has given you some good answers already. The [url=http://www.unl.edu/amc/index.html]AMC Web site[/url] will answer your questions one, two, and three if you follow a lot of links and read it carefully. That's a good site to have bookmarked on your computer.

People in North America believe in taking the northern hemisphere summer off to have fun. For many of the people who visit this AoPS site, "fun" includes going to a summer math program. There is still (barely) time to sign up for one of the summer programs in math for summer 2004. See this site's list of [url=http://www.artofproblemsolving.com/Resources/AoPS_R_Programs.php]programs[/url] for more information. I am not aware of any North American competition programs that have meets over the summer, but there may be some southern hemisphere programs going on during what is winter there that you could participate in by postal mail or email.

I think there are some math competitions based on solving problems on your computer and emailing answers, but I have never signed up anyone on my local math team for something like that. Rather, I try to get the guys I coach OFF the computer  :)  so that they spend more time learning how to use their own brains.

Good competition programs, in the opinion of this site's owner, are listed on this site's [url=http://www.artofproblemsolving.com/Resources/AoPS_R_Contests.php]contests[/url] page. Anyone doing all of those contests would be awfully busy, but might have a lot of fun and would surely learn a fair amount of math.

Hope this helps!

[quote="Silverfalcon"]I heard a lot kind of competition and I kinda don't get it...
Number one question: What's AIME?[/quote]

The AIME is fairly difficult compared to MATHCOUNTS, and you have to qualify for it.

Quoted from the [url=http://www.unl.edu/amc/]AMC website[/url]:

"The AIME (American Invitational Mathematics Examination) is an intermediate examination between the AMC 10 or AMC 12 and the USAMO. All students who took the AMC 12 and achieved a score of 100 or more out of a possible 150 are invited to take the AIME. All students who took the AMC 10 and were in the top 1% also qualify for the AIME. The AIME is intended to provide further challenge and recognition, beyond that provided by the AMC 10 or AMC 12, to the many high school students in North America who have exceptional mathematical ability. The top scoring USA American Mathematics Competitions participants (based on a weighted average) are invited to take the USAMO.

The AIME is a 15 question, 3 hour examination in which each answer is an integer number from 0 to 999. The questions on the AIME are much more difficult and students are very unlikely to obtain the correct answer by guessing. As with the AMC 10 and AMC 12 (and the USAMO), all problems on the AIME can be solved by pre-calculus methods. The use of calculators is not allowed.

The AIME provides the exceptional students who are invited to take it with yet another opportunity to challenge their mathematical abilities. Like all examinations, it is but a means towards furthering mathematical development and interest. The real value of the examination is in the learning that can come from the preparation beforehand and from further thought and discussion of the solutions."

[quote]Number two question: What's AMC?[/quote]

AMC = American Mathematics Competitions, a 25 question, 75-minute math test with multiple-choice format.  Calculators are allowed.  See the [url=http://www.unl.edu/amc/]AMC website[/url].

[quote]Number three question: What's difference between AMC8, AMC10, and AMC12?[/quote]

The main difference is difficulty, although AMC8 is conceptually easier than the other two exams.  Again, see the AMC website.

[quote]Number four question: Are there any big math competition during summer that is big like Mathcounts?[/quote]

Not to my knowledge, although you might try asking that specific question in the "Other Contests..." forum.

[quote]Number five question: Are there any math competition where you can just sit in your computer and do it?[/quote]

The USAMTS (USA Mathematical Talent Search) is sort of like that.  Basically, there are five problems each round, and there are four rounds each year.  The fourth round was due March 14th, so it's too late for this year, but keep an eye out for it in the future.  The website is http://www.nsa.gov/usamts/.  (You will probably want to work them out on paper and send them in by writing up your solutions by hand, so it's not completely computer-based, but yes, all you need to participate is a computer with internet access (to download the problems from the website.)

[quote]Number six question: Other than ones I mentioned, are there any good competition to enter?[/quote]

[url=http://www.mandelbrot.org]Mandelbrot[/url], [url=http://www.olemiss.edu/mathed/contest/]Ole Miss Problem of the Week[/url], [url=http://mathforum.org/pow/]MathForum Problem of the Week[/url], [url=http://www.codybowl.com]Cody Bowl[/url].

P.S. This isn't really a question relating to the AIME - more about math contests in general.  As a result, I've renamed your topic.

It is sort of silly that we had three people answering the same question...


geometry
3D geometry
number theory unsolved
number theory

[quote="cnyd"]a,b is a positive integer 2a(a^2+3b^2) is a perfect cube a=b[/quote]

Are you saying that only $ a\equal{}b$ makes $ 2a(a^2\plus{}3b^2)$ a cube?

yes only a=b makes 2a(a^2+3b^2) perfect cube

Hello, we can rewrite the equation as
2a(a^2 + 3b^2) = (a+b)^3 + (a-b)^3. If a is greater then b , a-b is positive, based on the Fermat's Last Theorem we get no solution.
If b is greater then a, (b-a)^3 + 2a(a^2 + 3b^2) = (b+a)^3. Since b-a is positive , with the same reason we get no solution.
Finally, if a=b then, 2a(a^2 + 3b^2) = (2a)^3 which is a perfect square.

This problem was a step in the proof of Fermat's Last Theorem, case $ n\equal{}3$.

See: [url]http://fermatslasttheorem.blogspot.com/2005/05/fermats-last-theorem-proof-for-n3.html[/url]


[quote="djimenez"]Hello,

I read this and seems to me pretty amuzing, specially given the explanation. I hope you will enjoy it.

LETTER OF A DRIVER COUGHT SPEEDING AT 250 KM/H IN A STREET LIMITED TO 70 (in France)
([b]Translated[/b])
(Remark, 1km ~ 0.625miles)

"Your honor, I have perfectly seen the signal of [b]70[/b] in black with a red circle around, and without any other indication of units. You know that the law of July 4th 1,837 stablish as mandatory the use of the metric system in France, and that the law nº 65-501 of May 3rd, 1,961 modify it, definiing as units of legal base, the units of this metric system. You can verify all this in the webpage of the goverment. According to the International metric system, the length unit is the meter, and the time unit is the second, therefore it is obvius that the LEGAL speed unit is, consecuently, METERS PER SECOND. I cannot imagine that the Traffic Department is not applying the laws of the Republic.

That is why, 70 meters per second would correspond exactly to 252 kilometers per hour. The Police states that I have been cought at a speed of 250 Km/h, I do not doubt that. I was, thouth, 2km/h under the authorized limit. I will appreciate you will take this in consideration to give me back my driver's license and set me free."

 :D  :D  :D  :D[/quote]

lmao!

Btw- What happened after that?

Now that is a smart, creative (wo)man...

[quote="yif man12"]Here's another way.

Cop: Sir, you were speeding. License and Registration?
Driver: Sorry, I don't have one. I had it suspended when I got my 5th DUI.
Officer: May I see the registration for this vehicle?
Driver: It's not my car. I stole it.
Officer: The car is stolen?
Driver: That's right. But come to think of it, I think I saw the owner's card in the glove box when I was putting my gun in there.
Officer: There's a gun in the glove box?
Driver: Yes sir. That's where I put it after I shot and killed the lady who owns this car and stuffed her in the trunk
Officer: There's a BODY in the TRUNK?!?!?
Driver: Yes, sir.

Hearing this, the officer immediately called his captain. The car was quickly surrounded by police, and the captain approached the driver to handle the tense situation:

Captain: Sir, can I see your license?
Driver: Sure. Here it is.
It was valid.
Captain: Who's car is this?
Driver: It's mine, officer. Here's the registration.
The driver owned the car.
Captain: Could you slowly open your glove box so I can see if there's a gun in it?
Driver: Gun? What gun ?? ...There's no gun in it.
Sure enough, there was nothing in the glove box.
Captain: Would you mind opening your trunk? I was told you said there's a body in it.
Driver: I said what ????
Trunk is opened; no body.
Captain: I don't understand it. The officer who stopped you said you told him you didn't have a license, stole the car, had a gun in the glove box, and that there was a dead body in the trunk.
Driver: Yeah, And I'll bet the lying son-of-a-***** told you I was speeding too![/quote]

rofl
both are really funny

[quote="sheepwarrior"]u could just kill the police officer and go to death row... i doubt theyd remember to give u a ticket  :D[/quote]

Of course, the idea is that you are making the officer look like a stupid and he would just not give you the ticket because of that.

[quote="mysmartmouth"]Lady To Cop: (After being stopped for speeding) Didn't I see you at the Grand Highwaypatrolmen's Ball?

Cop:  Highwaypatrolmen don't have balls.

When the cop realized what he said, he was too flustered to give a ticket....

Mods: Delete if innapropriate..[/quote]

sorta funny i guess...but pointless

[quote="maniacman384"]funny except the last ones that was just stupid. these are good jokes i hope to see mroe of em[/quote]

Well, the first one I got it from a non-reliable source, but any way, I wouldn't get surprised if it is true. I mean, those kind of things may happen. May be a french mathlinker could tell us if it is true! :D

[quote="Gyan"]Old story but a  physics graduate (foreign and very new in US)  was speeding  and I mean really speeding and was stopped by the police  officer. The officer, seeing he was a young and foreign and thinking he may not be familier with all the laws,  was a little friendly. So he remarkd:

"How fast do you think you can in your car around here?"

He looked at the speedometer, and in a tone only a true mathematician can say .."Up to 120?"  (
Yes his  maximum mark on his speedometer was 120).

He was let go with a warning. :)[/quote]

That one is really really funny!!, I could may be use it  :D  :D (Just kidding :P ).

Best,

Okay one more -Now from Florida.

A senior citizen in Florida bought a brand new Mercedes convertible. He tookoff down the road, flooring it to 80 mph and enjoying the wind blowing through what little hair he had left on his head.

"This is great," he thought as he roared down I-75. He pushed the pedal to the metal even more. Then he looked in his rear view mirror and saw a highway patrol trooper behind him, blue lights flashing and siren blaring.

"I can get away from him with no problem" thought the man and he tromped it some more and flew down the road at over 100 mph. Then 110, 120 mph!

Then he thought, "What am I doing? I'm too old for this kind of thing." He pulled over to the side of the road and waited for the trooper to catch up with him.

The trooper pulled in behind the Mercedes and walked up to the man "Sir," he said, looking at his watch. "My shift ends in 30 minutes and today is Friday. If you can give me a reason why you were speeding that I've never heard before, I'll let you go."

The man looked at the trooper and said, [hide]"Years ago my wife ran off with a Florida state trooper, and I thought you were bringing her back."
[/hide]
The trooper replied, "Sir, have a nice day."

[quote="mysmartmouth"]Lady To Cop: (After being stopped for speeding) Didn't I see you at the Grand Highwaypatrolmen's Ball?

Cop:  Highwaypatrolmen don't have balls.

When the cop realized what he said, he was too flustered to give a ticket....

Mods: Delete if innapropriate..[/quote]

lmao

this is another way to get out of a traffic ticket. (I had to revise it)

A police officer pulls a guy over for speeding and has the following exchange:

Officer: May I see your driver's license?
Driver: I don't have one. I had it suspended when I got my 5th DUI.

Officer: May I see the owner's card for this vehicle?
Driver: It's not my car. I stole it.

Officer: The car is stolen?
Driver: That's right. But come to think of it, I think I saw the owner's card in the glove box when I was putting my gun in there.

Officer: There's a gun in the glove box?
Driver: Yes sir. That's where I put it after I shot and killed the woman who owns this car and stuffed her in the trunk.

Officer: There's a BODY in the TRUNK?!?!?
Driver: Yes, sir.

Hearing this, the officer immediately called his captain. The car was quickly surrounded by police, and the captain approached the driver to handle the tense situation:

Captain: Sir, can I see your license?
Driver: Sure. Here it is.

It was valid.

Captain: Who's car is this?
Driver: It's mine, officer. Here's the owner's card.

The driver owned the car.

Captain: Could you slowly open your glove box so I can see if there's a gun in it?
Driver: Yes, sir, but there's no gun in it.

Sure enough, there was nothing in the glove box.

Captain: Would you mind opening your trunk? I was told you said there's a body in it.
Driver: No problem.

Trunk is opened; no body.

Captain: I don't understand it. The officer who stopped you said you told him you didn't have a license, stole the car, had a gun in the glovebox, and that there was a dead body in the trunk.

Driver: Yeah, I'll bet the guy told you I was speeding, too!

oh by the way i didn't steal yif man 12's way. I just got it in an email from my friend  :D

No offense, but I don't think this one can be topped.  It is one of the funniest things on the interweb.

[url=http://www.boreme.com/bm/APR04/a/t_speeding_offence/a.htm]www.boreme.com/bm/APR04/a/t_speeding_offence/a.htm[/url]

Make sure to go to the next two pages.  The first page alone isn't funny...

[quote="A+MATH"]Things not to do when a cop pulls you over:

Cop: Your eyes look red, have you been drinking.
You: Your eyes look glazed, have you been eating doughnuts[/quote]

Not bad!

That would sometimes have the opposite effect: get you another ticket for insulting an officer :D  :lol:

Did you learn this through personal experience? :D:D

elvenchamp777

Here's a joke:

A young man was speeding on the highway when he was pulled over by a traffic cop. "Look," said the cop, "you were doing fifteen over the limit. I'll have to write you a ticket."
"Sorry, mister," said the driver, "but you don't understand, I'm-"
"Kid, you broke the law, now you have to pay the consequences."
"But I can explain -"
"That's it! I've had enough of your bull. Now be quiet while I write this ticket."
"But officer -"
"All right! I'm taking you in."
And so the officer "accompanied" the young man to the police station where the young man was put in a temporary cell. After about an hour, the officer noticed the young man's troubled countenace and lightened up. "It's OK," said the officer, "the chief of police is a nice guy and he'll be in a good mood when he gets back, he's at his daughter's wedding."
"Don't count on it," said the young man, "I'm the groom."

Here's one that i heard a while ago and can't completely remember:

A cop pulls over a speeding car, and asks the driver, "are you wearing a seatbelt?"

driver: "yes."

cop: "congratulations! you just won 5,000 dollars! what will you spend it on?"

driver: "a new gun for robbing the bank."

passenger: "don't listen to him! he's drunk!"

other passenger: "Da*n it. I knew we wouldn't get far in a stolen car."

voice from the trunk: "are we past the border yet?"


A narcissistic number can be represented by taking its digits in the order given in combination with some mathematical operations. For example, $ 145\equal{}1\plus{}4!\plus{}5!$, and $ 81 \equal{} (8\plus{}1)^{2}$. Write 355 as a narcissistic number.

I don't think this should be in FTW, since i got this problem and put 3*5!-5, which was incorrect, and was the given answer.


inequalities
algebra
polynomial
three variable inequality

Here's an inequality by Vasile Cartoaje that appeared in Gazeta Matematica from Romania:

Let a,b,c>0 and x=a+1/b, y=b+1/c, z=c+1/a
              xy+yz+zx >= 2(x+y+z).

___________________
Gil Publishing House
gil1993@zalau.astral.ro
___________________

Very nice inequality. Have someone a very nice idea to solve it ?
I cannot solve it by the usual methods.

This problem is very nice.
I suppose the following way:

From 3 numbers, we can always choose two which either both  $\geq 2$, either both $\leq 2$. WLOG, suppose that are x,y. Then we have

\[ (x-2)(y-2)  \geq 0 \Rightarrow xy + 4  \geq 2(x+y) \quad (1) \]

In the other hand, we have
\[ z(x+y-2) = \left(c+\frac 1a\right)\left(a+\frac 1b+b+\frac 1c-2\right)  \geq \left(c+\frac 1a\right)\left(a+\frac 1c\right)  \geq 4 \Rightarrow  zx + zy \geq 4 + 2z \quad (2) \]
(By AM-GM inequality we have $ b+\frac 1b \geq 2$ and $(c+\frac 1a)(a+\frac 1c) \geq 4$)

From (1) and (2) we get the resulting inequality.

[quote="Mircea Lascu"]Let $ a,b,c>0$ and $ x\equal{}a\plus{}\frac{1}{b}, y\equal{}b\plus{}\frac{1}{c}, z\equal{}c\plus{}\frac{1}{a},$ prove that $ yz\plus{}zx\plus{}xy\geq 2(x\plus{}y\plus{}z).$[/quote]$ \sum{yz}\minus{}2\sum{x}$

$ \equal{}\sum_{cyc}\frac{bc(ca\minus{}1)^2\plus{}bc(a\minus{}1)^2(b\minus{}1)^2\plus{}(bc\plus{}1)(ca\plus{}ab\minus{}a\minus{}b)^2}{2abc(a\plus{}b\plus{}c)}\geq0.$

See also : http://www.mathlinks.ro/viewtopic.php?t=1739

By the way, the the polynomial

$ a(ab \plus{} 1 \minus{} b)(bc \plus{} 1 \minus{} c) \plus{} b(bc \plus{} 1 \minus{} c)(ca \plus{} 1 \minus{} a) \plus{} c(ca \plus{} 1 \minus{} a)(ab \plus{} 1 \minus{} b) \minus{} 3abc$

cannot be represented  as $ \sum{P(a,b,c)Q^2(a,b,c)}$,

where $ P(a,b,c)$ is polynomial with positive coefficients and $ Q(a,b,c)$ is polynomial with real coefficients.


Front side:10
Rear side:10
Right side:3
Left side: 3
Up side: 4
Down side: 4 (note that those 2 parts of a side of the cube give exactly one side)

Together: 34


Let f(1)=1 and f(n)=n+f(n-1) for n=2,3,4,... Find f(100).

I have some terms set up but I'm not sure what to do with them or how to find f(100).

[hide]
Find a pattern:

$ f(1)\equal{}1$
$ f(2)\equal{}2\plus{}f(1)\equal{}3$
$ f(3)\equal{}3\plus{}f(2)\equal{}6$
$ f(4)\equal{}4\plus{}f(3)\equal{}10$

Quickly one can see that $ f(n)$ gives the nth triangular number, or $ \frac{n(n\plus{}1)}{2}$.

So, $ f(100)\equal{}\frac{100*101}{2}\equal{}5050$
[/hide]

More generally:  $ f(n)$ is the sequence of partial sums of an arithmetic series.  In particular, $ f(n) \equal{} n \plus{} (n \minus{} 1) \plus{} (n \minus{} 2) \plus{} ... \plus{} 2 \plus{} 1$.  How do you find the sum of an arithmetic series?


有一个大圆;一个中圆和一个小圆分别与它内切;作中圆和小圆的两条内公切线,这两条内公切线与大圆交与四个点A.B.C.D;证明AB//CD
别以为这题简单哦!!希望有人能提供这题来源!谢了

这是上海叶中豪早期出的一道题。据作者透露，本题与两个小圆的两个极限点有关。

感觉这道题目有点问题，证明的应该是
ＡＢ，ＣＤ分别于两圆的两外公切线平行．．．． :)


$\text{Let}\; a,b,c,d\; \text{positive real numbers such that}\; a^2+b^2+c^2+d^2=1, \; \text{prove or disprove that}$
\[ \frac1{1-ab}+\frac1{1-bc}+\frac1{1-cd}+\frac1{1-da}+\frac1{1-bd}+\frac1{1-ca}\leq8.  \]

(In CRUX Math, Vasile Cirtoaje proposed the inequality
\[ \frac1{1-ab}+\frac1{1-bc}+\frac1{1-ca}\leq\frac92,  \]
where $a,b,c\geq0$ and $a^2+b^2+c^2=1$.)

First, i will prove the following assertion

For every positive real numbers $x,y$ such that $x,y \leq 1$, the following inequality holds.

$\frac{2}{1-xy} \leq \frac{1}{1-x^2} + \frac{1}{1-y^2}$

Proof :

Since $x , y =< 1$, we have $1-xy \geq 0$, so we can clear the denominator to obtain

$2(1-x^2)(1-y^2) \leq (1-x^2)(1-xy) + (1-y^2)(1-xy)$

$2 - 2x^2 - 2y^2 + 2x^2y^2 \leq 2 - x^2 - y^2 - 2xy + x^3y + xy^3$

$0 \leq x^2 + y^2 - 2xy + x^3y -2x^2y^2 + xy^3$

$0 \leq (x-y)^2 + xy(x-y)^2$

The last inequality is trivially true.

Now we can apply this inequality to both of the proposed inequality.

right?  :lol:

how are you going to prove the initial inequalities?

uuups sorry
my lemma can't be used

if you use my lemma, the resulting inequality will contradict AM-HM.

 :blush: sorry

it's hard

Pvthuan could you give a hint gor the ineq you proposed?

[quote="galeos"]judging from the phrase "prove or disprove", i don't think he has proven it to give a hint for the solution. :agent:[/quote]

No, I have not a solution for this one.


What is [i]N[/i]  :cap: [i]Z[/i]?

That might be too easy to post here if you've seen the notations before... oh well.

I'm going to step in on this one.  If you know the notations, the problem is easy.  If you don't know the notations, this problem is impossibly hard!

So let's let this be educational...

The symbol $\cap$ is used to denote the intersection of two sets.  [The elements common to both sets.]  For example:
  $\{\hbox{Fred, Betty, Barney}\}\cap\{\hbox{Mr.Slate, Fred, Dino}\}=\{\hbox{Fred}\}$

The symbol $\cup$ is used to denote the union of two sets.  [The elements that occur in one or both sets.]  For example:
$\{\hbox{Red, Green}\}\cup\{\hbox{Yellow, Red}\}=\{\hbox{Red, Green, Yellow}\}$

The above notations *might* occur in advanced MATHCOUNTS material, so they would be good to know.

The following notations are probably beyond the scope of MATHCOUNTS.  I am quite confident that if MATHCOUNTS problems ever used the following "doublestruck" letters, they would define what they meant...

Very often, the symbol $\mathbb{N}$ is used to indicate the set of natural numbers.  $\{1, 2, 3, \ldots\}$

Very often, the symbol $\mathbb{Z}$ is used to indicate the set of all integers.  $\{\ldots, -3, -2, -1, 0, 1, 2, 3, \ldots\}$

So the intersection of these two sets would be the set of natural numbers.

Any deeper discussion on these "doublestruck" notations should probably be carried to the the "Intermediate" Forum.

[quote="09husbri"]For some reason I think the intersection/union symbols have been used on chapter or state tests sometime during the past three years. Could be hallucinating, though.[/quote]

They have.  That's why I posted this here in the first place....

Actually, it would be
[hide]{all natural numbers}[/hide]
wouldn't it?

No because  :cap: means intersection.   The only way it could work was if $N=Z$.

Well since rcv just said that Z is all integers and N is all natural numbers, it must be all natural numbers.

Sorry, didn't read the last part of his post. :blush:

I don't know how many of you use :cap: and  :cup: in your classes on a regular basis, but we use it a lot in our books (the Elements of Math series by IMACS). They are pretty handy symbols to know, and I think rcv did a great job of explaining it. :lol:

Have any of you heard of symetric difference, denoted by a triangle? It says that

a symetric difference b = (a :cup: b) - (a :cap: b)

[quote="09husbri"]I've never heard of that. Sounds like one of those bogus "operations" the MathCounts people make up some problems with... :D  What is it used for?[/quote]

MathCounts doesn't make up bogus operations, unless you mean things like #, which are totally random, such as #$\#(x)=7x-3$

[quote="09husbri"]That's what I mean. They'll say something like "Denote the operation & such that a & b is equal to 3a - 2b." :roll:[/quote]
no, this ones a real one. Other wise they would define it in a form of a function, and all the probs that I've seen that contain it never define  :cap: and :cup:

I know that union and intersection are real operations; I've used 'em before. I was talking about symetric difference, which I'm sure is useful but seems like one of those bogus complex operations.

And for the Union and intersection, they don't have to define them! :P

Sorry... I didn't mean to confuse people. Symetric difference doesn't occur in MATHCOUNTS that I know of. Seeing the :cup: and :cap: made me think of it.

In our math class we use the Elements of Math books by IMACS. [url]http://imacs.org/IMACSWeb/default.aspx[/url] (For their website, although I don't know anything about the online classes. We use their textbooks, although I don't know that those are available to any individual who wants to buy them. We get them from our teacher.)

Symetric difference is the notation they use to stand for a :cup: b - a :cap: b, where a and b are both sets.

I guess no one else has heard of it.... Oh well.  :)
 :)  :)  :)  :lol:

Could you plesae explain to us how it is used? I know it isn't important, but I'm curious.

Symmetric difference is just an operation that you can perform on a set.

Umm... let's see. Hold on... Let me get my old math book and I can get some specific problems that they had. On second thought it's 10:45, parents are asleep, and walking down the hallway wakes them up :lol: .

So, here we go. Say you have
a={1, 2, 3, Billy Bob, spaceship} and  b={1, 7, 8, Billy Bob, airplane}.

By symetric difference (which is denoted by a triangle, but, I don't have anything fancy to make it so...) You would have

 a :cup: b - a :cap: b

so

{1, 2, 3, 7, 8, Billy Bob, spaceship, airplane} - {1, Billy Bob} =
{2, 3, 7, 8, spaceship, airplane}
 :)  :)  :)  :lol:

Hope this helps!!!

And for those people who still think I'm crazy, here's a website that tells a little about it. Or you can just look it up on Google. [url]http://mathworld.wolfram.com/SymmetricDifference.html[/url]

Oooh I get it, it's basically the opposite of intersection; it's all the ones the sets [i]don't[/i] hold in common!

symetric difference:

Imagine you have a Venn diagram:

Symetric difference is whatever is in the two outer regions, but not in the middle region.

Okay, I got it. Thanks, that's pretty interesting. :)

[quote="tarquin"][hide]{}[/hide]
If you've ever taken a TMSCA number sense test these questions are on every test...[/quote]

tarquin, remember that TMSCA tests are only for people living in Texas so theres no way anyone would know what they are.


linear algebra
matrix
inequalities
combinatorics proposed
combinatorics

We select $16$ cells on an $8\times 8$ chessboard. What is the minimal number of pairs of selected cells in the same row or column?

I seem not to understand your question... it seems to me that if you take the upper 2 rows then the lower ones will not have pairs... and I cannot seem to find any problem making sense in this way... please clarify :)

If you take two upper rows, there are $8+2C_8^2=64$ such pairs.

Ah in that way... still, doesn't really make sense to me either, since I would then think the number is trivially 16:
Here an example:
x o o o o o o x
x x o o o o o o
o x x o o o o o
o o x x o o o o
o o o x x o o o
o o o o x x o o
o o o o o x x o
o o o o o o x x

Here a proof (if one can call this a proof):
If the amonts on the rows are {a1,a2,...,a8}, then we have sum a_i=16, so the minimum sum of $C^2_i$ is when they're all equal, that is possible here: 2 on each row.
If the amounts on the columns are {b1...b8}, same stuff.
So if we can find an example that fulfills both optimal conditions (i.e. everywhere 2), then it is certainly minimal in the sum of those. Such a configuration exists, see example.

Or did I still misget it? :)

Sixteen rooks are placed on the cells of an $8\times 8$ chessboard. What is the minimum number of pairs of rooks attacking each other?
An easy proof would be considering a $8\times 8$ matrix with entries $(0,1)$ and total sum $16$. Then, the required value is $\sum \binom{r_i}{2}+\sum \binom{c_j}{2}$ where $r_i$ and $c_j$ sums of $i$ row and $j$ column respectively.
By convexity of $\binom{x}{2}$ we can use Jensen inequality, and we have $\sum \binom{r_i}{2}+\sum \binom{c_j}{2}\geq 8\binom{2}{2}+8\binom{2}{2}=16$.  The example of $16$ pairs is constructed by placing the rooks on 2 diagonals.

[quote=Agr_94_Math]The problem was rephrased as rooks attacking each other and 16 rooks in [url]http://www.imomath.com/index.php?options=mbb|Rus|f|MO|05&p=0[/url].
(Though essentially it is the same, it was a nice way of rephrasing it).

The proof can also be obtained by going for this kind of a coloring.

2 3 4 0 1 2 3 4
1 2 3 4 0 1 2 3
0 1 2 3 4 0 1 2
4 0 1 2 3 4 0 1
3 4 0 1 2 3 4 0
2 3 4 0 1 2 3 4
1 2 3 4 0 1 2 3
0 1 2 3 4 0 1 2[/quote]

BRO CAN YOU PLEASE EXPLAIN HOW THIS COLORING WILL FINISH THE PROBLEM?

We can find the lower bound (to be rigorous a bit) and then show that bound holds true

Assume a bipartite graph between the $8$ rows and the $8$ columns. For each selected cell in the $i$th row and $j$th column, we draw an edge between the $i$th row and $j$th column in the bipartite graph. Then, we just wanna minimize the sum of each node's degree choose $2$. Since the sum of the degrees of all the nodes is $32$, the answer is clearly $16\binom{2}{2}=16$ with the construction below.

[code]
BBWWWWWW
BBWWWWWW
WWBBWWWW
WWBBWWWW
WWWWBBWW
WWWWBBWW
WWWWWWBB
WWWWWWBB
[/code]

where B represents selected and W represents not selected.


Let be $ t\in \mathbb{Q}_ \plus{}$ and $ m(t) \equal{} \min \left(t,\frac {t \plus{} 2}{t \plus{} 1}\right)$ and $ M(t) \equal{} \max \left(t,\frac {t \plus{} 2}{t \plus{} 1}\right)$ .

i)Prove that $ m(t) < \sqrt {2} < M(t)\ , \ (\forall) t\in \mathbb{Q}_ \plus{}$

ii)Compare $ \sqrt {2} \minus{} m(t)$ and $ M(t) \minus{} \sqrt {2}$

hello to i) we have $ \min\left(t,\frac{t\plus{}2}{t\plus{}1}\right)\equal{}t$ for $ t\le \sqrt{2}$ or  $ \min\left(t,\frac{t\plus{}2}{t\plus{}1}\right)\equal{}\frac{t\plus{}2}{t\plus{}1}$ for $ \sqrt{2}<t$.
In the first case is $ m(t)\le \sqrt{2}$, in the second one we get $ \frac{t\plus{}2}{t\plus{}1}<\sqrt{2} \Leftrightarrow \sqrt{2}<t$.
Sonnhard.

hello, could it be that you have made a typo and it is $ M(t)\equal{}\max\left(t,\frac{t\plus{}2}{t\plus{}1}\right)$?
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, could it be that you have made a typo and it is $ M(t) \equal{} \max\left(t,\frac {t \plus{} 2}{t \plus{} 1}\right)$?
Sonnhard.[/quote]

Yes , you're right . I don't know why everytime i make this typo . I'll try to be more careful  :oops:  .

hello, we have $ M(t)\equal{}\left(t,\frac{t\plus{}2}{t\plus{}1}\right)\equal{}\frac{t\plus{}2}{t\plus{}1}$ if $ t \le \sqrt{2}$ or $ M(t)\equal{}\left(t,\frac{t\plus{}2}{t\plus{}1}\right)\equal{}t$ if $ \sqrt{2}<t$ and we have $ \sqrt{2}<\frac{t\plus{}2}{t\plus{}1} \Leftrightarrow t<\sqrt{2}$.
Sonnhard.


$x-5\sqrt{x} +6 = 0$

$\LaTeX$ed by a mod

Treat this as a quadratic. If you are familiar with factoring, this turns into

$(\sqrt{x})^2-5\sqrt{x}+6=(\sqrt{x}-2)(\sqrt{x}-3)$

$\sqrt{x}$=2 or 3

x = 4 or 9

[quote="fourierseries"]Or just isolate your square root term then square both sides.[/quote]

to get $25x=x^2+12x+36$, or $x^2-13x+36=0$.

$(x-4)(x-9)=0$

$x=4,9$

Please try to limit the amount of homework problems that you post in this forum.

$\lim_{hwk\to post\#5}single~user=Unhappy~Mods$

extenstion problem : write out an algorithm for to solve this


In a EQUILATERAL triangle  ABC  a point X  is taken such that  area of ABX=ACX=BCX  HOW MANY WAYS WE CAN TAKE A POINT X .pl draw a figure

at first' let's find all points $ x$ such as $ S_{ABX} \equal{} S_{BCX}$ where by $ S_{ABX}$ I denote area of triangle $ ABX$
since $ S_{ABX} \equal{} \frac {|AB|*|XH|}{2}$ where $ XH$ is perpendicular line to $ AB$ thus it's easy to see that all such point's  lay on the bisectors  of the angle which is constructed by the lines $ AB$ and $ BC$ on the picture this is red lines .
similary for other triangles . finally , needed points will be intersection of red, blue and brown lines .
I'm not good painter  , but on my picture this is red points.
also ,one more red point which was not located on my picture top


Prove that the equation $ x^3\plus{}y^3\equal{}7z^3$ has infinitely many solutions with $ (x,y,z)\equal{}1$

Let $ (c,d) \in \mathbb{Q}^2$ a solution of the elliptic curve $ c^3\plus{}d^3\equal{}7$, so it exists $ A,B,C \in \mathbb{Z}$ s.t.:
i)$ c\equal{}\frac{A}{C}$
ii)$ d\equal{}\frac{B}{C}$
iii)$ (A,C)\equal{}(B,C)\equal{}(A,B)\equal{}1$

So, we starts from the equation $ {A^3}\plus{}B^3\equal{}7C^3$ and in particular $ C | A^3\plus{}B^3$.
If $ c\equal{}\frac{A}{C}$ is a solution, then $ c'\equal{}\frac{A'}{C'}$ $ \equal{}\frac{A}{C} \frac{2B^3\plus{}A^3}{A^3\minus{}B^3}$ is too.
( So we are consideing the trasformation $ (c,d) \to (c',d')\equal{}(\frac{2cd^3\plus{}c^4}{c^3\minus{}d^3}, \frac{2c^3d\plus{}d^4}{d^3\minus{}c^3})$)

It is easy to demostrate that $ A'>A$ and $ C'>C$ and so all the $ c$-solution are distinct (analogue for $ d$), in fact:
i)$ (A,C)\equal{}1$
ii)$ (A,A^3\minus{}B^3)\equal{}(A,B)\equal{}1$
iii)$ (2B^3\plus{}A^3,A^3\minus{}B^3)\equal{}1 \text{ or } 3$
iv)$ (C,2B^3\plus{}A^3)\equal{}(C,B^3)\equal{}1$ (we remember that $ C | A^3\plus{}B^3$)

so, at every step, the "size" of c and d, when they are in the least terms, grows up very very quickly  :D  :D


How do I use the unit circle to find sin and cos of an angle?
Thanks,

If the line segment (radius) joining the point $ P(x, y)$ on the circle to the origin (the centre of the circle) makes an angle of $ \theta$ with the positive direction of the $ x\minus{}$axis, then $ x\equal{} \cos \theta$ and $ y\equal{} \sin \theta$, by Pythagoras.

It's useful for finding values like $ \sin0^{\circ}$, $ \sin90^\circ$, $ \sin180^\circ$, and $ \sin270^\circ$ (and cos, tan, etc. of those angles too).

For example, for $ \sin90^\circ$, you would look at a $ 90^\circ$ angle and see that it goes through the point (0,1) on the unit circle. The y-coordinate is 1, so $ \sin90^\circ\equal{}1$.

For values like $ \sin30^\circ$, $ \sin45^\circ$, and $ \sin60^\circ$, you would draw a right triangle in the unit circle that goes from (0,0) to (x,y) to (x,0).
The hypotenuse is the radius of the unit circle, so it is 1.

For finding $ \sin30^\circ$ or $ \sin60^\circ$, you would use the relationships from a 30-60-90 triangle to find the side lengths of the right triangle you drew.
With $ 30^\circ$, you would have a triangle with hypotenuse 1, adjacent side $ \sqrt{3}/2$, and opposite side 1/2. Then $ \sin30^\circ\equal{}1/2$, and $ \cos30^\circ\equal{}\sqrt{3}/2$.
For finding $ \sin45^\circ$, you would use the relationships from a 45-45-90 triangle.

So to find sin and cos of an odd angle. say like 37 degrees, i would just have to graph it?

Actually, 37 and 53 degrees are two more special ones. Although it is not exact, sin37 is approximately 3/5 and cos 37 is approximately 4/5.

You can't graph a triangle by hand and find the sinusoidal functions accurately.

You can also use the [url=http://www.themathpage.com/atrig/sum-proof.htm]Sum-Difference formulas[/url], which say that $ \sin(\alpha \plus{} \beta) \equal{} \sin \alpha \cos \beta \plus{} \sin \beta \cos \alpha$, and that $ \cos(\alpha \plus{} \beta) \equal{} \cos \alpha \cos \beta \minus{} \sin \alpha \sin \beta$.  The link provided shows a proof.


group theory
abstract algebra
induction
number theory
relatively prime
Galois Theory
superior algebra

Hello!  I was asked to solve a problem, but I don't really even know how to start, so I'll appreciate any help.   The problem is this:

Let $ G$ be a group and let $ x,y \in G$.  If the orders of $ x, y$ and $ xy$ are pairwise relatively prime, then $ G$ is nonsolvable.

THANK YOU!

Let $ G$ be a minimal solvable counterexample with some $ (x,y)\not\equal{}(1,1)$, s.t. $ |\langle x\rangle |,|\langle y\rangle |,|\langle xy\rangle |$ pairwise relatively prime.
Then $ G\not\equal{}1$ has some minimal normal subgroup $ 1\not\equal{}N\unlhd G$ and $ xN,yN,(xN)(yN)\equal{}xyN$ have relatively prime orders in $ G/N$.
By induction we have $ xN\equal{}N\equal{}yN$ and $ x,y\in N$, and since $ N$ is an elementary abelian $ p$-group for some prime $ p$,
we get $ x,y,xy$ $ p$-elements of relatively prime orders. So two of them must be $ \equal{}1$, which gives the contradiction $ x\equal{}y\equal{}1$.

Interestingly, this is actually a characterization of solvable groups. In other words, the converse works as well: every non-solvable group contains two elements $ x,y$ such that the orders of $ x,y,xy$ are pairwise relatively prime. This is much harder and requires a sizable piece of the Classification theorem; it can be found as Corollary 3 in [url=http://www.ams.org/bull/1968-74-03/S0002-9904-1968-11953-6/S0002-9904-1968-11953-6.pdf]this[/url] paper by J. G. Thompson.

The easier direction which olorin just proved is actually also a published result of P. Hall.

[quote="olorin"]Let $ G$ be a minimal solvable counterexample with some $ (x,y)\not \equal{} (1,1)$, s.t. $ |\langle x\rangle |,|\langle y\rangle |,|\langle xy\rangle |$ pairwise relatively prime.
Then $ G\not \equal{} 1$ has some minimal normal subgroup $ 1\not \equal{} N\unlhd G$ and $ xN,yN,(xN)(yN) \equal{} xyN$ have relatively prime orders in $ G/N$.
By induction we have $ xN \equal{} N \equal{} yN$ and $ x,y\in N$, and since $ N$ is an elementary abelian $ p$-group for some prime $ p$,
we get $ x,y,xy$ $ p$-elements of relatively prime orders. So two of them must be $ \equal{} 1$, which gives the contradiction $ x \equal{} y \equal{} 1$.[/quote]

I have a couple of questions:

How do you know such a minimal normal subgroup  $ N$ exists?,
how did you use induction? (I do not see where you used it because I do not see any natural number)

Thanks in advance

[quote="Adriana"]
How do you know such a minimal normal subgroup  $ N$ exists?,
[/quote]

A solvable group by definition contains a non-trivial normal subgroup (EDIT: unless of course it's cyclic of prime order, which is a trivial case in this problem), and one (or more) of these normal subgroups must be minimal in the sense that it contains no others - this just follows from $ G$ being finite.

However, olorin also assumes that you are aware of the basic structure theorem for minimal normal subgroups, namely, that they must be elementary abelian $ p$-groups. This isn't too hard to prove and I can give an outline if you need it. I think that one should not expect to solve this problem in reasonable time without seeing some of those results first.

[quote]
how did you use induction? (I do not see where you used it because I do not see any natural number)
[/quote]

This is a different way of looking at proofs by induction: consider a minimal counterexample. Minimality means that every smaller case is necessarily not a counterexample, so we can assume the theorem does hold for that smaller case.

[quote="randomgraph"]it can be found as Corollary 3 in [url=http://www.ams.org/bull/1968-74-03/S0002-9904-1968-11953-6/S0002-9904-1968-11953-6.pdf]this[/url] paper by J. G. Thompson.[/quote]
Awesome, I didn't know that. :)
But it needs a little more work to complete the argument.

Let $ G$ be a minimal non-solvable counterexample, which doesn't contain $ x,y,xy\in G\setminus \{1\}$ with relatively prime orders.
Then every proper subgroup of $ G$ is solvable, and let $ 1\not\equal{}N\unlhd G$ be a minimal normal subgroup of $ G$.

If $ N\equal{}G$, then $ G$ is simple and must in fact be one of the minimal simple groups found by Thompson.
So this case can be checked explicitly.

And if $ N<G$, then $ N$ is a solvable minimal normal subgroup of $ G$.
So $ N$ must be an elementary abelian $ p$-group for some prime $ p$ and $ |N|\equal{}p^k$ for some $ k\in \mathbb{N}$.
Then $ G/N$ is non-solvable and by induction we find $ xN,yN,xyN\in G/N\setminus \{N\}$ with relatively prime orders $ l,m,n$.
At most one of those orders is divisible by $ p$.
If $ p|l$ we can replace $ (x,y,xy)$ by $ (xy,y^{\minus{}1},x)$.
If $ p|m$ we can replace $ (x,y,xy)$ by $ (x^{\minus{}1},xy,y)$.
So we can assume $ p\nmid l$ and $ p\nmid m$.

But then using the following lemma we can assume the orders of $ x,y\in G$ divide $ l^r,m^s$ for some $ r,s\in\mathbb{N}$.
But the order of $ xy\in G$ divides $ |\langle xy\rangle N|\equal{}|\langle xyN\rangle |\cdot |N|\equal{}np^k$
and $ l^r,m^s,np^k$ are pairwise relatively prime.
So the orders of $ x,y,xy\in G\setminus\{1\}$ are pairwise relatively prime. Contradiction!

[u][b]Lemma:[/b][/u]
For any $ N\unlhd G$ and $ xN\in G/N$ of order $ n\in \mathbb{N}$ we find $ y\in xN$ of order dividing $ n^k$ for some $ k\in\mathbb{N}$.
[u][b]Proof[/b][/u]
Let $ l$ be the order of $ x\in G$ and $ m|l$ maximal, which is coprime to $ n$.
Then all prime factors of $ {l\over m}$ must divide $ n$, so $ {l\over m}|n^k$ for some $ k\in\mathbb{N}$.
Also $ un\plus{}vm\equal{}1$ for some $ u,v\in\mathbb{Z}$ and $ N\equal{}(xN)^n\equal{}x^nN$ gives $ x^n\in N$
and $ y: \equal{}x^{1\minus{}un}\equal{}x(x^n)^{\minus{}u}\in xN$ and $ y\equal{}x^{vm}$ has order dividing $ {l\over m}|n^k$.
[u][b]q.e.d.[/b][/u]


Are there any books with past AIME questions 1988+ (similar to the ones for the AMC-12)? The kalva site doesn't have answers to some of the questions, and some of the answers that it does have are soometimes too brief to understand.

No books, only individual past tests that you can order from AMC.


Solve the ES:

$ \left\{\begin{array}{l} \sqrt {x \plus{} 1} \plus{} \sqrt {y \plus{} 1} \equal{} 4 \\
x \plus{} y \equal{} 6 \\
\end{array} \right.$

Note:

We don't solve: $ x \equal{} 6 \minus{} y$ so $ \sqrt {6 \minus{} y \plus{} 1} \plus{} \sqrt {y \plus{} 1} \equal{} 4$ and cyc  :D  :D

This is problem in University Exam 2006 of Vietnam. Out of normal solution, we have a great solution!! :D

[hide="Solution"]
Square the first equation, giving $ x \plus{} 1 \plus{} y \plus{} 1 \plus{} 2\sqrt {(x \plus{} 1)(y \plus{} 1)} \equal{} 16$. Substituting $ x \plus{} y \equal{} 6$, moving terms, and dividing gives $ \sqrt {(x \plus{} 1)(y \plus{} 1)} \equal{} 4 \Rightarrow (x \plus{} 1)(y \plus{} 1) \equal{} 16$. Expanding this gives $ xy \plus{} x \plus{} y \plus{} 1 \equal{} 16 \Rightarrow xy \equal{} 9$. By Vieta's, we have $ (3,3)$ as the only solution.[/hide]

Let $ X\equal{}\sqrt{x\plus{}1}$ and $ Y\equal{}\sqrt{y\plus{}1}$.  Then the equations become:
\[ X\plus{}Y\equal{}4
\]
and
\[ X^2\plus{}Y^2\equal{}8
\]
Then square the first equation and subtract the second to get $ XY\equal{}4$.  Then we see by vieta's that $ X\equal{}Y\equal{}2$ so $ \boxed{x\equal{}y\equal{}3}$

I understand that part of it because earlier in the solutions we had $ x \plus{} y$ and $ xy$.  :P
How does Vieta's directly give the solutions?  If it does not, then why state it since $ x\plus{}y$ and $ xy$ are already directly known?

For my solution note that
\[ \frac {X \plus{} Y}{2} \equal{} \sqrt {XY}
\]
so by AM-GM, this is true when $ X \equal{} Y$.

This also works for George's solution.

[quote="azjps"][quote="AstroPhys"]Each solution makes sense up to the using Vieta's part.  How do you use Vieta's in this?[/quote]

$ x,y$ are the solutions to $ (p \minus{} x)(p \minus{} y) \equal{} p^2 \minus{} 6p \plus{} 9 \equal{} 0$.   :wink:

[hide]
By AM-QM, $ 4 \equal{} \sqrt {x \plus{} 1} \plus{} \sqrt {y \plus{} 1} \le 2\sqrt {\frac {x \plus{} y \plus{} 2}{2}} \equal{} 4$ with equality iff $ x \equal{} y \equal{} 3$.
[/hide]

Edit: That is the application of Vieta's, as then $ (p \minus{} x)(p \minus{} y) \equal{} p^2 \minus{} 6p \plus{} 9 \equal{} (p \minus{} 3)(p \minus{} 3) \Longrightarrow x \equal{} y \equal{} 3$. The answer is pretty obvious here anyway, but, for example, it could be used for when more variables exist if all the [[elementary symmetric sum]]s are known.[/quote]

That's very good! We use Cauchy-Schwarz Inequality to solve that equation system! :)  :)

Well if we have $ X^2\plus{}Y^2$ and $ XY$ we can use $ X\plus{}Y\equal{}\sqrt{X^2\plus{}2XY\plus{}Y^2}$ and $ X\minus{}Y\equal{}\sqrt{X^2\minus{}2XY\plus{}Y^2}$ so we can find $ X$ and $ Y$.

All solution is true. That's my solution:

Condition: $ \left\{ \begin{array}{l}
 x \ge  \minus{} 1 \\
 y \ge  \minus{} 1 \\
 \end{array} \right.$

use Cauchy-Schwarz inequality to equation (1), we have:

$ 4^2  \equal{} (\sqrt {x \plus{} 1}  \plus{} \sqrt {y \plus{} 1} )^2  \le 2.(x \plus{} y \plus{} 2) \leftrightarrow x \plus{} y \ge 6$

"=" is happended so $ x\equal{}y\equal{}3$.

You can go to [url=http://www.mathlinks.ro/viewtopic.php?p=1240382#1240382][b]here[/b][/url] to see another equation system. That's very nice!!!  :)  :)

Dude, just square the first one to get

$ \sqrt{(x\plus{}1)(y\plus{}1)}\equal{}4$ or $ (x\plus{}1)(y\plus{}1)\equal{}16$

Now, $ x\plus{}1\plus{}y\plus{}1\equal{}8$ so x+1 and y+1 are solutions to the quadratic $ t^2\minus{}8t\plus{}16\equal{}0$ <--> $ (t\minus{}4)^2\equal{}0$ hence both x+1 and y+1 must be equal to 4, so x=y=3. Checking, we see that it works.