Prove that, for every positive real numbers $ a,b,c$, the following inequality always holds: $ \frac{a^2b^2}{a\plus{}b}\plus{}\frac{b^2c^2}{b\plus{}c}\plus{}\frac{c^2a^2}{c\plus{}a}\le\frac{a^3\plus{}b^3\plus{}c^3}2.$

[quote="Johan Gunardi"]Prove that, for every positive real numbers $ a,b,c$, the following inequality always holds: $ \frac {a^2b^2}{a \plus{} b} \plus{} \frac {b^2c^2}{b \plus{} c} \plus{} \frac {c^2a^2}{c \plus{} a}\le\frac {a^3 \plus{} b^3 \plus{} c^3}2.$[/quote]
Note that $ ab \ge \frac {{4a^2 b^2 }}{{\left( {a \plus{} b} \right)^2 }}$, hence we have
\[ \sum {\frac {{a^2 b^2 }}{{a \plus{} b}}} \le \sum {\frac {{ab\left( {a \plus{} b} \right)}}{4}} \le \frac {{a^3 \plus{} b^3 \plus{} c^3 }}{2}
\]

we have $ \frac {a^2b^2}{a\plus{} b} \leq \frac {a^3 \plus{} b^3}{4}$
then sum these inequalities.

The Idea is Holder Inequality :

$ (a\plus{}b)(a^3\plus{}b^3) \ge (a^2\plus{}b^2)^2 \ge 4a^2b^2$


Let $ K$ be a finit field such that there exists a nonzero element $ u$ in $ K$ with the property that $ u^{3}\equal{}1\plus{}1$. Prove that for every element $ a$ in $ K$ there exist $ x,y$ in $ K$ such that $ a\equal{}x^{3}\plus{}y^{3}$.

We can actually prove something a little more specific. Let $q=|K|.$

i) If $q=3n,$ for some integer $n,$ then $a=a^q=(a^{n})^3$ for all $a \in K.$

ii) If $q=3n+2,$ for some integer $n,$ then $a=a^{2-q}=(a^{-n})^3$ for all $0 \ne a \in K,$ and clearly $0=0^3.$

So, in both cases i), ii), for every $a \in K,$ there exists $x \in K$ such that $a=x^3.$

iii) Now suppose that $q=3n+1,$ for some integer $n$ and let
$$A:=\{x^3: \ \ \ x \in K\}.$$
Let $K^{\times}$ be the multiplicative group of $K.$ We know $K^{\times}$ is cyclic and so $K^{\times}=\langle c \rangle.$ We have
$$\{c^{3m}: \ \ 1 \le m \le n\} \cup \{0\} \subseteq A$$
(in fact the two sets are equal) and so $|A| \ge n+1.$ Let $a \in K$ and consider the following sets
$$B:=\{a-x^3: \ \ \ x \in K\}, \ \ \ C:=\{x^3-a: \ \ \ x \in K\}.$$
Clearly $|A|=|B|=|C| \ge n+1$ and so $|A|+ |B|+ |C| \ge 3n+3 > |K|.$ So at least two of the sets $A,B,C$ have non-empty intersection.
If $A \cap B \neq \emptyset$ or $A \cap C \neq \emptyset,$ then $a=x^3+y^3$ for some $x,y \in K$ and we are done.
Now suppose that $B \cap C \ne \emptyset.$ So there exist $x,y \in K$ such that $a-x^3=y^3-a$ and so
$$2a=x^3+y^3.$$
Since, as given in the problem, $u^3=2$ for some $u \ne 0,$ we have $2 \ne 0$ and $2^{-1}=u^{-3}.$ Hence
$$a=u^{-3}(x^3+y^3)=(u^{-1}x)^3+(u^{-1}y)^3$$
and that completes the solution.


function
induction
logarithms
algebra unsolved
algebra

Find all functions $f: \mathbb{Z}^+\to \mathbb{R}$, which satisfies $f(n+1)\geq f(n)$ for all $n\geq 1$ and $f(mn)=f(m)f(n)$ for all $(m,n)=1$.

By induction $f$ is nondecreasing.  As $f(1)=f(1)^2$, $f(1)$ is either 0 or 1.  If $f(1)=0$ then $f(n)=0$ for all $n$.  So suppose $f(1)\neq 0$; then $f(1)=1$ and $f(n)\geq 1$ for all $n$.

Suppose there exist integers $a<b$ such that $f(a)=f(b)=k$.  Then $k=\frac{f(p)}{f(q)}$ for all $a\leq\frac{p}{q}\leq b$ such that $(a,q)=(b,q)=1$ (since $f(q)f(a)=f(qa)\leq f(p) \leq f(qb)=f(q)f(b)$.)  If we then construct an increasing sequence of rational numbers ${1, \frac{p_1}{q_1}, \frac{p_2}{q_2}, \ldots \frac{p_l}{q_l}, a}$ all coprime to $b$, $a$ and each other, and each no more than $\frac{b}{a}$ times the last, then \[ \frac{p_i}{q_i}<\frac{p_{i+1}}{q_{i+1}}<\frac{bp_i}{aq_i} \]
so
\[ ap_iq_{i+1}\leq ap_{i+1}q_i\leq bp_iq_{i+1} \]
so
\[ kf(p_iq_{i+1})=f(ap_iq_{i+1})\leq f(ap_{i+1}q_i)\leq f(bp_iq_{i+1})=kf(p_iq_{i+1}) \]
so $f(ap_iq_{i+1})=f(ap_{i+1}q_i)$ and $\frac{f(p_i)}{f(q_i)}=\frac{f(p_{i+1})}{f(q_{i+1})}$.  Continuing down the list, we find by induction that $1=f(1)=f(a)$, by extension that $f(n)=1$ for all $n\leq b$, and by a similar construction that $f(n)=1$ for all $n$.

Otherwise we can assume the function is strictly increasing, and in this case I don't think there's much we can say:  just define on all prime powers in increasing order (choose $f(2)$ then $f(3)$ then $f(4)$ then $f(5)$ then $f(7)$, etc.) so that each time the value chosen is within the acceptable interval for the function to be increasing.  I haven't checked it rigorously but I think that this can be done pretty arbitrarily, and the function need not be particularly nice.

you are wrong
  we choose$f(2)=2 ,f(3)=5$ ...then$f(4)=5<f(3)$

I thing the root is
         $f(n)=1 \forall n \in N*$
   OR  $f(n)= n^a \forall n \in N*$ and with $a>0$

[quote="ZetaX"]How do you get $f(4)=5$¿[/quote]
 I am very sorry
  $f(4)=f(2)f(2)=4< f(3)=5$ but  we have f(n+1)>=f(n)

[quote="ZetaX"]Note that $\gcd(2,2)=2$, so you cannot claim $f(4)=f(2)f(2)$![/quote]
   sorry .I  don't read it careful but I still thing the solution of Agrippina is wrong
   choose $f(2)=2 ,f(3)=5 ,f(5)=6, f(7)=8$
             then we claim$f(6)= 10>f(7)$
  and my root can be wrong

[quote="tranngoctrung"]
   choose $f(2)=2 ,f(3)=5 ,f(5)=6, f(7)=8$
             then we claim$f(6)= 10>f(7)$[/quote]

What I meant in the condition in my earlier solution ("so that each time the value chosen is within the acceptable interval for the function to be increasing") is that when you pick the value of the function for a prime power $p^k$ you must ensure that the value you pick is strictly between $\frac{f(m_1)}{f(n_1)}$ and $\frac{f(m_2)}{f(n_2)}$, where $\frac{m_1}{n_1}$ and $\frac{m_2}{n_2}$ are the closest rational approximations from above and below to $p^k$ for coprime $m$ and $n$ whose only prime power factors are less than $p^k$.

This avoids the sort of contradiction you've been producing, but I don't think it necessarily yields a "nice" solution like $f(n)= n^a$.

Erdos proved that if $f$ is multiplicative (i.e $f(mn)=f(m)f(n)$ when $(m,n)=1$), positive, and monotonic then there is a constant $c$ such that $f(n)=n^c$. Thus since $f(1)$ can be only $0$ or $1$, the only possible solutions are $f(n)=n^c$ for some constant $c$ or $f(n)=0$. I don't think it's all right to put such problems on an Olympiad :(.

f(1)=1,

so f(n)>=0

so, f(2)>=1

suppose f(2)=a, then, we could know that f(mn)>=af(m).

like this, a is the smallest value of f(n) except f(1), we also could know that  a^n+1>f(2^n)>=a^n.

like this, at any p(p is prime number), we can find the value like f(p)=a_p

a_2=<a_3=<a_3.......=<a_p.

 like above, f(p^n)>=(a_p)^n

and, gernerally, we can find that  f(m)>=f([m/p]x p)

thus we can reason that f(p^n)=(a_p)^n

f shoul satisfy this condition--->f(mn)=f(m)f(n)

thus, f(p)=p^n(n is one of real number)

Kimjiman and {x}, I wonder whether you are taking the range of $f(n)$ to be the integers rather than the reals?  In particular, Kimjiman, if I understand you correctly you deduce from $a^n+1>f(2^n)\geq a^n$ that $f(2^n)=a^n$, which you can only do if you know that $f(n)$'s range is restricted to $\mathbb{Z}$.

When you do restrict $f(n)$ to taking only integer values, I think you are right:  the only solutions are indeed the "nice" ones.  I'm afraid I can't follow your proof, though!

I don't have the reference I used yesterday right now, but I am almost 100 % sure that Erdos's prove does concern multiplicative function from $\mathbb{Z^+}$ to $\mathbb{R}$

You're right; Google turned up this reference:

http://alumnus.caltech.edu/~however/papers/paper01.html

to "Erdos' theorem" that "every increasing multiplicative function from the positive integers to the reals is of the form $f(n)=n^a$ for some real $a>1$".  When I've convinced myself of it I'll let you know.  Kimjiman, I don't understand the step in your proof when you go from $f(mn)\geq af(n)$ for $m$ coprime to $n$ and $\geq 2$, to $a^n+1>f(2^n)\geq a^n$.  Does anyone else have a proof?

Anyway, I agree; a classic theorem seems an odd choice of problem for an Olympiad.

Apologies for my earlier skepticism: [x] and Erdos were right, and my "counterexample" was based on fallacious reasoning.  Here is a proof from the [i]American Mathematical Monthly[/i], October 1986.

$f(p^{n+1})=f(p)f(p^n)$ for all naturals $n$ and primes $p$.  For \[ \frac{f(xp-1)}{f(x)}\leq \frac{f(p^{n+1})}{f(p^n)}\leq \frac{f(xp+1)}{f(x)} \] for all $(x,p)=1$, and so
\[ f(p)\frac{f(x-p)}{f(x)}\leq \frac{f(p^{n+1})}{f(p^n)}\leq f(p)\frac{f(x+p)}{f(x)} \]
for all $(x,p)=1$.  If $\frac{f(p^{n+1})}{f(p)f(p^n)}=r$ then $f(x+kp)\geq r^kf(x)$ for all naturals $k$; for some sufficiently large $x$, $f(2)f(x)=f(2x)>f(x+kp)$ as the function is increasing, so $f(2)>r^k$ for all integers $k$, and so $r\leq 1$.  Likewise $r \geq 1$, so $\frac{f(p^{n+1})}{f(p)f(p^n)}=r=1$ for all $p$, $n$.

Therefore $f(p^n)=f(p)^n$ for all $p$, $n$ and so $f$ is strictly multiplicative as well as increasing.

Then if $f(n)$ is not of the form $n^a$ for some constant $a$, then there would exist integers $l$, $m$ so that $\frac{\ln l}{\ln m}\neq \frac{\ln f(l)}{\ln f(m)}$.  If the LHS were greater, there is a rational number $\frac{\ln l}{\ln m}>\frac{s}{t}>\frac{\ln f(l)}{\ln f(m)}$, so that $l^t>m^s$ but $f(m^s)=f(m)^s>f(l)^t=f(l^t)$, a contradiction; similarly, there is a contradiction if the RHS is greater.

Differentiating functions defined on $\mathbb Z^{+}$ only ¿

hi
i can recommend you something about the problem though i cannot solve this problem. For example, you can look at IMO 1994/5 and 1997/4 in this step


algebra
polynomial
superior algebra
superior algebra unsolved

How would you go about describing the splitting field of x^q - a over K, where a is in K and char K= q?

If $ K$ is perfect, then the polynomial has only one prime factor in an algebraic closure of $ K$, so in fact the polynomial is reducible. Over a imperfect field, the extension is purely inseparable in $ a$ is not a $ q$th root (and otherwise it's of course just $ K$). I'm not entirely sure otherwise. In general, in $ \mathbb{F}_q(t)$, we can add in a root of $ x^q \minus{} t$ to get an extension of degree $ q$. I'm not sure about cases in general.


What is the least common multiple of 12, 18 and 30?

The lcm of 12,18, and 30 is $ 2\cdot2\cdot3\cdot3\cdot5\equal{}4\cdot9\cdot5\equal{}20\cdot9\equal{}\boxed{180}$.


[color=darkblue]Let $ABCD$ be a square with $AB=a\ .$ Denote the intersection $O\in AC\cap BD\ .$ The circle $C_{1}=C(D,DO)$ meets the side $[CD]$ in the point $U\ .$ The circle $C_{2}=C(C,CO)$ meets the side $[BC]$ , $[CD]$ in the points $H$ , $T$ respectively. The circle $C_{3}(B)$ is tangent to the circle $C_{2}\ .$ The common tangent of the circles $C_{2}$ and $C_{3}$ cuts the sideline $BC$ in the point $M\ .$ Denote the point $R\in [BC]$ for which $UR\parallel BD\ .$ Prove that :
$1.\blacktriangleright\ BM=OA\ ;$
$2.\blacktriangleright\ OT\perp OR\ ;$
$3.\blacktriangleright$ The segment $[TU]$ is the side of the regular octagon inscribed in the given square.[/color]

[quote="Asha"][color=darkred]Let $ABCD$ be a square with $AB=a\ .$ Denote the intersection $O\in AC\cap BD\ .$ The circle $C_{1}=C(D,DO)$ meets the side $[CD]$ in the point $U\ .$ The circle $C_{2}=C(C,CO)$ meets the side $[BC]$ , $[CD]$ in the points $H$ , $T$ respectively. The circle $C_{3}(B)$ is tangent to the circle $C_{2}\ .$ The common tangent of the circles $C_{2}$ and $C_{3}$ cuts the sideline $BC$ in the point $M\ .$ Denote the point $R\in [BC]$ for which $UR\parallel BD\ .$ Prove that :
$1.\blacktriangleright\ BM=OA\ ;$
$2.\blacktriangleright\ OT\perp OR\ ;$
$3.\blacktriangleright$ The segment $[TU]$ is the side of the regular octagon inscribed in the given square.[/color][/quote]

[color=darkblue][b][u]Proof.[/u][/b] I"ll suppose w.l.o.g. that $OA=1\ (AB=\sqrt 2)\ .$ Therefore,

$DT=CU=BH=CR=\sqrt 2-1\ ,\ TU=HR=2-\sqrt 2\Longrightarrow \boxed{\ TU=\sqrt 2\cdot UC\ }\ \ (*)\ .$

$1.\blacktriangleright$ $\frac{MB}{MC}=\frac{HB}{HC}$ (the points $H$ , $M$ are harmonic conjugate w.r.t. $B$ , $C$) $\Longrightarrow$

$\frac{MB}{BC}=\frac{HB}{HC-HB}$ $\Longrightarrow$ $\frac{MB}{\sqrt 2}=\frac{\sqrt 2-1}{1-(\sqrt 2-1)}$ $\Longrightarrow$ $MB=1$ $\Longrightarrow$ $\boxed{\ MB=OA\ }\ .$

$2.\blacktriangleright$ $\{\begin{array}{c}DO=CO\\\ DT=CR\\\ \widehat{ODT}\equiv\widehat{OCR}\end{array}\ \ (S.A.S.)$ $\Longrightarrow$ $\triangle DOT\equiv\triangle COR$ $\Longrightarrow$

$\widehat{DOT}\equiv\widehat{COR}\ ,\ \boxed{\ OT=OR\ }$ $\Longrightarrow$ $m(\widehat{TOR})=90^{\circ}$ $\Longrightarrow$ $\boxed{\ OT\perp OR\ }\ .$

$3.\blacktriangleright$ From the relation $(*)$ obtain $TU^{2}=UC^{2}+UC^{2}=CU^{2}+CR^{2}$ $\Longrightarrow$

$TU^{2}=UR^{2}$ $\Longrightarrow$ $TU=UR=HR\ ,\ m(\widehat{TUR})=135^{\circ}$ $\Longrightarrow$ The segment $[TU]$

is the side of the regular octagon inscribed in the given square and $OU^{2}=UD\cdot UC\ .$

[b]Remark.[/b] If denote the intersection $N\in OR\cap DC$ , then $\boxed{\ CN=AB\ }\ .$ This nice

problem presents us [b][u]an interesting construction[/u][/b] of the regular octagon inscribed

in a given square $ABCD$ : $\{T,U\}\subset [BC]$ so that $DU=CT=OA$ !

[b][u]A small dictionary (at least for Asha).[/u][/b]

$S.A.S.$ means the congruence criterion [b]S[/b]ide - [b]A[/b]ngle - [b]S[/b]ide of two triangles.

$\mathrm{w.r.t.}$ means " [b]w[/b]ith [b]r[/b]espect [b]t[/b]o ".

$\mathrm{w.l.o.g.}$ means " [b]w[/b]ithout the [b]l[/b]oss [b]o[/b]ne's [b]g[/b]eneralization ".[/color]

Congratulations for the mode in which you present the solution of the problem and for the solution, Mr Virgil Nicula.

Thank you for your help, Mr. Nicula. Congratulations for the wonderfull solution!

I wish you a Happy New Year!

Teacher Asha

[color=darkblue]Thank you a lot, [b][u]Mychrom[/u][/b] and my mate [b][u]Asha[/u][/b] for your appreciations of the my reply.

I remember you that this nice problem was proposed by [b]Asha[/b] ![/color]

[size=134][color=darkred][b]A happy new year to all linkers ![/b][/color][/size]


The number of ways in which n distinct objects can be arranged in a circle is (n-1)!

Four men and four women are forming a circle for a folk dance.  In how many ways can this be done if we require that the men and women alternate positions.

If anyone can help with this problem, please do so.

The answer the the teacher has and even isn't sure of is 1152.  Is that right?

You shouldn't have bet!

http://www.mathlinks.ro/Forum/viewtopic.php?t=59927

You lost. :P What do I win for that? :)

This should be moved to the Getting Started forum, but here's your help.

[hide]We know that the order must be M,W,M,W,M,W,M,W or W,M,W,M,W,M,W,M. For the first man, we have 4 choices. For the first woman, 4 choices. For the second man, 3 choices, and so on. So our total number of combinations is: $2*4*4*3*3*2*2*1*1=1152$. [/hide]

I don't think that's right.

That's not taking into account that the placement of the first person is arbitrary.

The women can be placed in (N-1)! ways, leaving n slots for the men to be placed N! ways, making the answer (N-1)! x N! = 3! X 4! = 6 X 24 = 144.

Take one of the people and designate him to be first. Then there are 3! ways to arrange the other men and 4! ways to arrange the other women.

If one of the chairs was different then the rest, it would be 2*4! *4!


Let $ x,y,z > 0$,prove that
$ {\frac {\sqrt {y \left( x \plus{} y \right) \left( y \plus{} z \right) \left( x \plus{} y \plus{} z \right) }}{\left( 2\,y \plus{} x \plus{} z \right) y}} \plus{} {\frac {\sqrt {z \left( z \plus{} x \right) \left( y \plus{} z \right) \left( x \plus{} y \plus{} z \right) }}{\left( 2\,z \plus{} y \plus{} x \right) z}}\geq \frac{3\sqrt{3}}{4}{\frac {2\,x \plus{} z \plus{} y}{x \plus{} y \plus{} z}}$
BQ

background is:in triangle ABC
$ {\frac {{\it w_b}}{s \minus{} b}} \plus{} {\frac {{\it w_c}}{s \minus{} c}}\geq \frac {3\sqrt {3}}{2}\frac {b \plus{} c}{s}$

More strong is BF181(g):
$ \sqrt{\frac{w_bw_c}{(s\minus{}b)(s\minus{}c)}}\geq \frac{3\sqrt{3}}{2}(1\minus{}\frac{r}{w_a})$
BQ

[quote="hedeng123"][quote="xzlbq"]Let $ x,y,z > 0$,prove that
$ {\frac {\sqrt {y \left( x \plus{} y \right) \left( y \plus{} z \right) \left( x \plus{} y \plus{} z \right) }}{\left( 2\,y \plus{} x \plus{} z \right) y}} \plus{} {\frac {\sqrt {z \left( z \plus{} x \right) \left( y \plus{} z \right) \left( x \plus{} y \plus{} z \right) }}{\left( 2\,z \plus{} y \plus{} x \right) z}}\geq$ $ \frac {3\sqrt {3}}{4}{\frac {2\,x \plus{} z \plus{} y}{x \plus{} y \plus{} z}}$
BQ[/quote]

only prove

$ 2\sqrt {{\frac {\sqrt {y \left( x \plus{} y \right) \left( y \plus{} z \right) \left( x \plus{} y \plus{} z \right) }}{\left( 2\,y \plus{} x \plus{} z \right) y}} {\frac {\sqrt {z \left( z \plus{} x \right) \left( y \plus{} z \right) \left( x \plus{} y \plus{} z \right) }}{\left( 2\,z \plus{} y \plus{} x \right) z}}}\geq$ $ \frac {3\sqrt {3}}{4}{\frac {2\,x \plus{} z \plus{} y}{x \plus{} y \plus{} z}}$

$ \Longleftrightarrow \frac {16(x \plus{} y)(y \plus{} z)^2(x \plus{} y \plus{} z)^2(z \plus{} x)}{yz(2y \plus{} z \plus{} x)^2(2z \plus{} y \plus{} x)}\geq \frac {729(2x \plus{} y \plus{} z)^4}{256(x \plus{} y \plus{} z)^4}$

tsds  one  time[/quote]

Who is caushy?

[quote="xzlbq"]
Who is caushy?[/quote]

I believe that you think about [url]http://en.wikipedia.org/wiki/Augustin-Louis_Cauchy[/url]. In inequalities Cauchy is refered to [url]http://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality[/url]

I will read the page Carefully,thank you!
BQ


A white cylindrical silo has a diameter of 30 feet and a height of 80 feet. A red stripe with a horizontal width of 3 feet is painted on the silo, as shown, making two complete revolutions around it. What is the area of the stripe in square feet?

I'm having a hard time visualizing the stripe when it is unrapped. I know its a parallelogram, but I don't see how its hieght is 80. Help?

Hint:  Remember that similar problem on State Sprint?  Well, it looks like that.  Then, find the height of the parallelogram.

EDIT: The problem on the wiki has a picture.


Randy Pausch passed away today:

- http://www.cmu.edu/homepage/beyond/2008/summer/an-enduring-legacy.shtml
- http://download.srv.cs.cmu.edu/~pausch/news/index.html

He became famous for his positive life attitude as shown in his "Last Lecture":

- http://video.google.com/videoplay?docid=-5700431505846055184

"Randy Pausch Almost all of us have childhood dreams: for example, being an astronaut, or making movies or video games for a living. Sadly, most people don't achieve theirs, and I think that's a shame. I had several specific childhood dreams, and I've actually achieved most of them. More importantly, I have found ways, in particular the creation (with Don Marinelli), of CMU's Entertainment Technology Center (etc.cmu.edu), of helping many young people actually *achieve* their childhood dreams. This talk will discuss how I achieved my childhood dreams (being in zero gravity, designing theme park rides for Disney, and a few others), and will contain realistic advice on how *you* can live your life so that you can make your childhood dreams come true, too"

I heard the news from a girl I know who goes there. She didn't get to see his lectures live, but dropped by at his office.

I'm deeply saddened by the news, even though I don't know him. I think the world needs more people with an attitude like that.

He was one in a billion...Always optimistic, never a sad day in his life. RIP :(


Your book is correct.  Here's how it works.

Your velocity vector is given by $\left< 1.6 \cos \theta+v_{current}, 1.6 \sin \theta \right>$.  You want to arrive directly opposite the starting point, which means you want your $x$-velocity to be zero.

Now, when $\theta = 90^{\circ}$ (the initial case, when you were traveling straight across) your vector was

$\left< v_{current}, 1.6 \right>$

And you traveled a displacement of

$\left< 40, 80 \right>$

Therefore, $v_{current}= 0.8$.  Then we want

$1.6 \cos \theta+v_{current}= 0$
$1.6 \cos \theta =-0.8$
$\cos \theta =-\frac{1}{2}$
$\theta = 120^{\circ}$

So the swimmer travels a direction $30^{\circ}$ to the left of her initial direction.

[quote="t0rajir0u"]Your book is correct.  Here's how it works.

[/quote]
  Thanks t0rajir0u, I learnt something new today. :)

[quote="Gibbenergy"]I solved this problem but my result is different from the book. I can't see any mistakes from my arguments. Please help me!

   A swimmer heads directly across a river, swimming at 1.6 m/s relative to the water. She arrives at a point 40 m downstream from the point directly across the river, which is 80 m wide. In what direction should the swimmer head to arrive at the point directly opposite her starting point?

  My answer is 26.57 degree but the result (in the book) is 30 degree. So who is corrected?[/quote]
Gibb,

I believe that I understand how the author got $30^{\circ}$ as his answer.

If the water is still then the swimmer swims directly across at a speed of 1.6 m/s. It therefore takes her 50 seconds to make such a trip. However, in this same 50 seconds the river's current carries her 40 meters downstream. Thus the velocity of the current is $v_{\text{current}}=\frac{40\text{ m}}{50\text{ s}}= 0.8$ m/s.

Now, assuming 1.6 m/s is as fast as the swimmer can swim in whatever direction then the direction she must swim, $\psi$, in order to end exactly opposite her starting point is given by: $\psi+\sin^{-1}\left(\frac{0.8\text{ m/s}}{1.6\text{ m/s}}\right) = 0$

$\sin^{-1}\left(\frac{0.8\text{ m/s}}{1.6\text{ m/s}}\right) = 30^{\circ}$

$\boxed{\therefore \psi =-30^{\circ}}$


[size=150]Suppose $ \Delta ABC$ is a equilateral triangle, $ O$ is a  inner-point  of $ \Delta ABC$, and $ OD\perp BC$, $ OE\perp AC$, $ OF\perp AB$.
Prove: $ S_{\Delta OAF} \plus{} S_{\Delta OBD} \plus{} S_{\Delta OCE} \equal{} \dfrac{1}{2}S_{\Delta ABC}$.[/size]

draw parralel lines to sides of triangle from O and see the pic.

[quote="maomao13"]Is this a ......eh...... Olympiad Section Problem?[/quote]

I'd say it is - at least I found it really hard some time ago  :oops:
http://www.mathlinks.ro/Forum/viewtopic.php?t=32524

  darij


calculus
derivative
inequalities
algebra
function
domain
real analysis

How can we prove AM-GM inequality using partial derivative?

Let's try Lagrange multipliers, and I'll go for a weighted version of AM-GM.

Suppose $p_i>0$ for each $i$ and $\sum_{i=1}^np_i=1.$

Maximize $f(x_1,x_2,\dots,x_n)=\prod_{i=1}^nx_i^{p_i}$ subject to the constraint $\sum_{i=1}^np_ix_i=M.$ Restrict this to positive $x_i.$

Here are the Lagrange multiplier equations:

For each $j,$ $p_jx_j^{p_j-1}\prod_{i\ne j}x_i^{p_i}=\lambda p_j$

$\sum_{i=1}^np_ix_i=M.$

We get for each $j$ that $\lambda = \frac1{x_j}\prod_{i=1}^nx_i^{p_i}.$

We conclude that the maximum happens when $x_1=x_2=\cdots=x_n.$ By the last equality, that happens when $x_1=x_2=\cdots=x_n=M.$

But then, $f(M,M,\dots,M)=M^{p_1+p_2+\cdots+p_n}=M.$

Since that is the maximum, then for all (positive) values of the $x_i,$

$\prod_{i=1}^nx_i^{p_i}\le M = \sum_{i=1}^np_ix_i.$

The weighted geometric mean is less than or equal to the weighted arithmetic mean.

Why do we know that that is the maximum? Because it is the only critical point, and all around the boundaries of the bounded domain, as any $x_i\to 0,$ then $f$ tends to zero.


We can construct one easily.  Note that this is nothing more than an arithmetic sequence with $ a_{1}\equal{}28$, $ d\equal{}2$,  and $ a_{n}\equal{}98$.  The sum of an arithmetic series is given by
$ S_{n}\equal{}\frac{n(a_{1}\plus{}a_{n})}{2}$
Now, we must simply figure out just what $ n$ is.  Not too hard, use the formula:
$ a_{n}\equal{}a_{1}\plus{}(n\minus{}1)d\Rightarrow 98\equal{}28\plus{}(n\minus{}1)2\Rightarrow n\equal{}36$
Now go back to our first formula and we have
$ S_{36}\equal{}\frac{36(28\plus{}98)}{2}\equal{}18(126)\equal{}2268$.

Of course, there are other ways if you don't know all of those sequence formulae.  Try to find one.


quadratics
algebra
geometry solved
geometry

Let OU and OV be perpendicular rays in a circle center O and M the middle of chord UV. PQ is a chord through M. W is a pt on UV s.t. PW=PM and R is a pt on the ray (PW s.t. PR=MQ. Prove that MR=UV.

It is fact MP*MQ=MU*MV=R2/2

Let a=angle PMV. Since OM is perpendicular to UV:
R2=OP2=OM2+MP2-2OM*MP*cos(\pi/2+a)=R2/2+MP2+\sqrt 2*R*MP*sin a
R2=OQ2=OM2+MQ2-2OM*MQ*cos(\pi/2-a)=R2/2+MQ2-\sqrt 2*R*MQ*sin a

So MP and -MQ are roots of the quadratic equation x2+\sqrt 2 *R*x*sin-R2/2=0  ==> MP2+MQ2=(\sqrt 2*R*sin a)2+2*R2/2=2R2(sin a)2+R2.

Let us consider trianle MPR: angle MPR equals \pi-2a, PR=MQ ==>
MR2=MP2+MQ2-2MP*MQ*cos(\pi-2a)=2R2(sin a)2+R2+2MP*MQ*cos 2a=2R2(sin a)2+R2+R2(1-2(sin a)2)=2R2, i.e. MR=UV.

Not really a nice geom problem, huh? I also have a computational proof, using complex numbers. Bye! :D


Let $ a\;,\;b\;,\;c$ are the sides of a triangle .

So what is the range of $ \mathbf{I}=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}$

[hide="Answer"]$ \frac{9-8\sqrt{2}+5\sqrt{5}}{6}<\mathbf{I}<\frac{9+8\sqrt{2}-5\sqrt{5}}{6}$[/hide]

[hide="HINT"]
Similar triangles are insignificant in changing the expression's value.
[/hide]


[color=blue][u]If $ 0 < x < 1$, what is the maximum value of $ x \minus{} x^3$? Figure it out WITHOUT using calculus.[/u][/color]

uuuhhh.......idk? im just doing this because you said you needed someone posting in your topic in my game. (this post is gonna sooo be rated one star...)

I will post the solution after seeing what different people churn up.


trigonometry
geometry
trapezoid
circumcircle
parallelogram
geometry proposed

[quote="Virgil Nicula"][color=darkred]Let $ ABC$ be a triangle. Construct outside of it the right triangles $ ABM$ , $ ACN$ so that $ BA\perp BM$ , $ CA\perp CN$ and

$ ABM\sim ACN$ . Prove that for a point $ S\in (MN)$ exists the equivalence $ SB \equal{} SC\ \Longleftrightarrow\ SM \equal{} SN$ . In this case show

that $ S$ is the circumcenter of $ \triangle ABC$ $ \Longleftrightarrow$ $ m(\widehat {BAM}) \equal{} 90^{\circ} \minus{} A$ and if $ L\in MN\cap BC$ , then the ratio $ \frac {LM}{LN}$ is constant.[/color][/quote]
[hide="Proof with complex numbers."]
[color=darkblue][b][u]Proof.[/u][/b] Define $ X(x)$ - the point $ X$ with the affix $ x\in\mathbb C$ . Suppose w.l.o.g. that $ a \equal{} 0$ .

Denote the midpoint $ D(d)$ of the side $ [BC]$ , i.e.  $ d \equal{} \frac {b \plus{} c}{2}$ . Since $ ABM\sim ACN$ can write $ \frac {BM}{BA} \equal{} \frac {CN}{CA} \equal{} r$ .

Therefore, $ \left\|\begin{array}{c} m \equal{} b \plus{} r(a \minus{} b)i \\
\ n \equal{} c \plus{} r(c \minus{} a)i\end{array}\right\|$ and $ \boxed {\ SM \equal{} SN\ }$ $ \Longleftrightarrow$ $ s \equal{} \frac {m \plus{} n}{2}$ $ \Longleftrightarrow$ $ s \equal{} \frac {b \plus{} c}{2} \plus{} r\frac {c \minus{} b}{2}i$ $ \Longleftrightarrow$

$ s \equal{} \frac {b \plus{} c}{2} \plus{} ri\left(c \minus{} \frac {b \plus{} c}{2}\right)$ $ \Longleftrightarrow$ $ s \minus{} d \equal{} r(c \minus{} d)i$ $ \Longleftrightarrow$ $ DS\perp DC$ $ \Longleftrightarrow$ $ SD\perp BC$ $ \Longleftrightarrow$ $ \boxed {\ SB \equal{} SC\ }$ .

[b][u]Remark.[/u][/b] Observe that $ SD \equal{} r\cdot CD$ , i.e. $ \frac {DS}{DC} \equal{} \frac {BM}{BA} \equal{} \frac {CN}{CA}$ what means $ BDC\sim ABM\sim ACN$ .[/color][/hide]
[hide="A synthetical proof."]
[color=darkblue]Denote the projections $ X$ , $ Y$ of the points $ M$ , $ N$ on the sideline $ BC$ respectively and the middlepoint $ D$ of the side $ [BC]$ .

Observe that $ XB \equal{} YC \equal{} h_a\tan\phi$ , where $ \phi \equal{} m(\widehat {BAM}) \equal{} m(\widehat {CAN})$ . Thus, in the right trapezoid $ MXYN$ the point $ D$

is the midpoint of $ [XY]$ . Therefore, $ SM \equal{} SN$ $ \Longleftrightarrow$ $ SD\perp BC$ $ \Longleftrightarrow$ $ SB \equal{} SC$ . In this case, $ MXB\sim BTA$ and $ NYC\sim CTA$ ,

where $ T\in BC$ , $ AT\perp BC$ . Thus, $ \frac {MX}{XB} \equal{} \frac {BT}{TA}$ and $ \frac {NY}{YC} \equal{} \frac {CT}{TA}$ from which obtain $ \frac {MX}{NY} \equal{} \frac {TB}{TC}$ , i.e. $ \frac {LM}{LN} \equal{} \frac {TB}{TC}$ (constant).

Denote the circumcenter $ O$ of $ \triangle ABC$ . Then $ O\equiv S$ $ \Longleftrightarrow$ $ \widehat {OBC}\equiv \widehat {SBC}$ $ \Longleftrightarrow$ $ 90^{\circ} \minus{} A \equal{} \phi$ $ \Longleftrightarrow$ $ \phi \equal{} 90^{\circ} \minus{} A$ .[/color] [/hide]
[size=117][b][color=darkblue]See and [/color][/b][/size][url=http://www.mathlinks.ro/viewtopic.php?t=209609][color=red][b]here[/b][/color][/url] [size=117][color=darkblue][b].[/b][/color][/size]

A simple proof: Denote D,E be the orthogonal projections of M,N resp on BC. H is the orthocenter of triangle ABC. Easy to show that BD=CE, hence result.

Yes, it is near my solution.
E, F is are orthogonal projections of M, N on BC. We have MENF is a trapeze, and EB= CF( very simple) so if SM=SN then DB=DC, which is SB=SC ( D is orthogonal projection of S on BC).
Analogously if SB=SC.
Dear mr.danh: Are you Vo Quoc Ba Can' brother?
Sorry, I'm poor in English

[b]Edit :[/b] I added two new questions to the initial enuntiation.

Quite simple but very nice, Virgi Nicula. :D . I  indentify that you usually use trigonometric in geometry. I like it ( athough sometime they not very nice). :)

Equivalent wording, with straight forward solution:

"[i]Let $ABCD$ be a parallelogram, with $\angle A $ obtuse, $M \in BC$, $AM \perp BC$, $N \in CD$, $AN\perp CD$. Then:
   a) $O$, the midpoint of the diagonals is the circumcenter of the $\triangle AMN$
   b) for any $X\in BM$ and $Y\in ND$ such as $\triangle AMX \sim \triangle ANY$, the midpoint of $XY$ belongs to the segment joining the midpoints of $(BD)$ and $(MN)$."[/i]

I do not think there is necessary to show the proof of the above statements.

Best regards,
sunken rock


function
algebra
linear equation
algebra unsolved

[quote="tranthanhnam"]find all $f(x) : (0;+\infty )\to (0;+\infty )$ and $f(x)$increasing on ${[1;\infty)}$  and$\forall a,b,c \in R^{+}$ such that :
         $f(ab)f(bc)f(ca)=f(a^2b^2c^2)+f(a^2)+f(b^2) +f(c^2)$[/quote]
Let a=c=1/b=sqrt(x). We have f(1)=2,f(1/x)=f(x). Let a=x^2,b=c=1/x. We have f(x)^2=2+(2+f(x^4))/f(x^2)>3.
If y(n)=f(x^(2^n)), then y(n+1)=y(n)[y(n-1)^2-2]-2. If y(1)<=y(o)=f(x)<2 then exist n, with y(n)<sqrt(3) contradition.
Therefore f(x)=2 for all x.

[quote="harazi"]Unfortunately, there are many more solutions than the constant one.[/quote]
An example, harazi!!!!

[quote="harazi"]Try functions of the form $x^k+x^{-k}$.[/quote]
This function is not increase in [1, infinity]

[quote="jin"]I think it is increase.[/quote]
I am sorry, I think f(x) increase if  f(y)<=f(x) for y>=x, und don't undestoud.
Let a=sqrt(x)/y,b=sqrt(x)y,c=1/sqrt(x). We have f(x)[f(y)^2-2]=f(xy^2)+f(x/y^2). If z(k)=f(y^2k) we have linear equation
z(k+1)-c(y)z(k)+z(k-1)=0, c(y)=f(y)^2-2>=2. Therefore z(k)=d(y)^k+d(y)^(-k)=f(y^2k), becouse f increase f(x)=x^s+x^(-s), s=const.

[quote="Hong Quy"]I think the problem was from IMO shortlist 2003. Check it http://www.mathlinks.ro/Forum/post-119999.html#p119999[/quote]
It is sufficiently - Exict x0 not equal 1, were f(x) is continuosly. Then exist real const s, that f(x)=x^s+x^(-s).


Given point A on a line  r   and point B on a circle center   o

  Construct point X on the circle (point Y on a line) so that X,Y,B are collinear

  and triangle AXY has equal angles at X and Y.

  TY.

$ \bullet$ Let $ C$ be, the orthogonal projection of $ B,$ on the given line $ (r)$, through the fixed point $ A.$

The circumcircle $ (K)$ of the right triangle $ \bigtriangleup CAB$ with diameter $ AB,$ passes through the midpoint $ M$ of the segment $ XY,$ because of $ AM\perp MN.$

Based on the below [b][size=100]Lemma[/size][/b], we have that the point $ Y\equiv (r)\cap BX,$ when $ X$ is a mobile point on the given circle $ (O)$ and such that $ MY \equal{} MX,$ where $ M\equiv (K)\cap BX$, is the circle $ (O'),$ centered at the fixed point $ O',$ on the line segment $ OK$ such that $ KO \equal{} KO'$ and the solution of the proposed problem is completed.

$ \bullet$ The circle $ (O')$ intersects the given line $ (r)$ also at point $ Y'$, such that $ AY' \equal{} AX'$, where $ X'\equiv (O)\cap Y'B.$

[b][size=100][color=DarkBlue]LEMMA. - Three circles $ (O),\ (O_{1}),\ (O_{2})$ are given with the segment $ AB$ as their common cord and let $ C,\ D,\ E,$ their points of intersection, from an arbitrary line through the point $ A.$ Prove that $ \frac {CD}{DE} \equal{} \frac {OO_{1}}{O_{1}O_{2}}$, where $ O,\ O_{1},\ O_{3},$ are the centers of the given circles $ (O),\ (O_{1}),\ (O_{2}),$ respectively.[/color][/size][/b]

Kostas Vittas.

PS. I will post here later, the easy proof of the above [b][size=100]Lemma[/size][/b] I have in mind.

[quote="vittasko"][b][size=100][color=DarkBlue]LEMMA. - Three circles $ (O),\ (O_{1}),\ (O_{2})$ are given with the segment $ AB$ as their common cord and let $ C,\ D,\ E,$ their points of intersection, from an arbitrary line through the point $ A.$ Prove that $ \frac {CD}{DE} \equal{} \frac {OO_{1}}{O_{1}O_{2}}$, where $ O,\ O_{1},\ O_{3},$ are the centers of the given circles $ (O),\ (O_{1}),\ (O_{2}),$ respectively.[/color][/size][/b][/quote]
The line through the point $ B$ and perpendicular to $ AB,$ intersects the given circles $ (O),\ (O_{1}),\ (O_{2},)$ at points $ C',\ D',\ E'$ respectively and $ AC',\ AD',\ AE'$ are their diameters as well.

So, from $ \angle ACC' \equal{} \angle ADD' \equal{} \angle AEE' \equal{} 90^{o}$ $ \Longrightarrow$ $ CC'\parallel DD'\parallel EE'$.

Hence, based on the [b][size=100]Thales theorem[/size][/b], we conclude that $ \frac {CD}{DE} \equal{} \frac {C'D'}{D'E'} \equal{} \frac {OO_{1}}{O_{1}O_{2}}$ and the proof of the [b]Lemma[/b] is completed.

Kostas Vittas.


Of course it's valid!

In an isosceles right triangle, the hypotenuse is $\sqrt{2}$ times as long as a catet, so if the side lengths were integers, you could represent $\sqrt{2}$ as the ratio of two integers (namely the hypotenuse and the catet), what would mean that $\sqrt{2}$ is rational; but you know that $\sqrt{2}$ is not rational, and hence there is no isosceles right triangle with integer side lengths.

  Darij

heh :P darij we don't need to help you getting started in geometry do we? :D

I didn't see the "very easy" and thought that the one who posted the problem didn't know the solution... I'm really missing this MathLinks classification in "Unsolved Problems" vs. "Proposed & Own Problems"...

  darij

LynnelleYe wrote:Prove that there is no isosceles right triangle with integer side lengths.
I made this problem myself, so there is a chance it isn't even valid, but I think it's okay.

[hide]All triangles have their angles measure add up to[tex]180^{\circ}[/tex]

All triangles have 3 angles

All right triangles have one angle that is [tex]90^{\circ}[/tex]

All isoceles triangles have two congruent sides.  Let's call them x and y.

[tex]{x} = {y}[/tex]

All isoceles triangles have two congruent angles.

In an isoceles triangle, the two congruent sides make up the the angle that is not congurnet with any of the other angles

The can not be two right angles in the triangle, since 90 and 90 = 180, and there are only 2 angles.

x and y make 90

Using the pythagorean theorom, we can get the equation,

[tex]{x^2}+{y^2} = {c^2}[/tex]

Since y = x :

[tex]{x^2} + {x^2} = {c^2}[/tex]

Let's substitue 1 for x.

[tex]{1^2} + {1^2} = {c^2}[/tex]

[tex]{1} + {1} = {c^2}[/tex]

[tex]{2} = {c^2}[/tex]

[tex]{c} = \sqrt{2}[/tex]

[tex]\sqrt{2}[/tex] is not rational.

All numbers are divisible by 1.

All right isoceles triangles have the same ratio as this one.

[tex]\sqrt{2}[/tex] multiplied by an integer can not be an integer.

So, whenever you multiply 1 by s, you have to multiply [tex]\sqrt{2}[/tex]s, causeing either 1, or [tex]\sqrt{2}[/tex], or both to be irrational, not an integer.

[/hide]

[quote="Peter VDD"]that's what darij just said :?[/quote]

How am I supposed to know what a catet is?

I think you can tell from the context of what he said that it is equivalent to a leg.  I think it s a German word... can anyone confirm this?

Well, I really don't know how this is called in English! In German, it is "Kathete". Indeed, if ABC is a right-angled triangle with the right angle at A, then the sides AB and AC are what I call the catets of the right-angled triangle ABC.

  Darij

[quote="darij grinberg"]Well, I really don't know how this is called in English! In German, it is "Kathete".[/quote]

This question came up in an earlier thread right after the merger. Yes, the two shorter sides of a right triangle (the two sides other than the hypotenuse) are called "legs" in English. Similarly, trapezoids have "bases" (two of them) and "legs" (the two nonparallel sides).

NightFlarer wrote:Couldn't you just have done:
[hide]Obviously if this proof works the legs are going to be the equal segments.  Call the legs x.  Thus, the hypotenuse is the diagonal of a square with side x.  Thus the 90 degree angle gets bisected and we have a 45-45-90 triangle.  So the hypotenuse is x root 2.  But isn't any integer times root 2 irrational?
Then the proof is proved false.[/hide]

That's basically what I said, except mine way was a whole lot more drawn out.


Forty cards are placed into a box, each bearing a number 1, 2, 3, 4, 5, 6,
7, 8, 9, or 10, with each number entered on four cards. Four cards are
drawn from the box at random and without replacement. Let $ p$ be the probability
that all four cards bear the same number. Let $ q$ be the probability that
three of the cards bear a number $ a$ and the other bears a number $ b$ that is not equal to $ a$.
What is the value of $ q/p$?

[hide]$ p\equal{}\frac{4}{40} \times \frac{3}{39} \times \frac{2}{38} \times \frac{1}{37}$

And:

$ q\equal{}\binom{4}{3} \times \frac{4}{40} \times \frac{3}{39} \times \frac{2}{38} \times \frac{36}{37} \equal{} 4(36)(p) \implies \boxed{144}$.[/hide]

Hi, nice solution, however I have a quick question.  I understand the first part of your solution from Alcumus's solution, and I understand all parts of the second part of your solution from Alcumus's solution, besides for the $\binom{4}{3}$.  Any help?  Thanks :).

That's because out of $4$ cards, you're choosing $3$ that bear $a$. (Alternatively, you could do $\binom{4}{1}$ to choose the $1$ card that bears $b$.)

The expression for $p$ carries an implicit $\binom{4}{4}=\binom{4}{0}$.

I think this way is easier:

to pick all numbers the same the probability is $\frac{4}{40}\times\frac{3}{39}\times\frac{2}{38}\times\frac{1}{37}\times10$

then the probability of choosing 3a's and 1b is the number of ways you can arrange aaab which is $4\times\frac{4}{40}\times\frac{3}{39}\times\frac{2}{38}\times\frac{36}{37}$

I got $\frac{36}{37}$  by saying that there were 36 other numbers other than a. then I just divided the top directions by the bottom directions! I think it's easier than the way the answer key showed. the answer is $\boxed{144}$

I think it would be easier to do:

$p= \frac{1}{\binom{40}{4}}$

$q= \frac{4 \cdot 36}{\binom{40}{4}}$

Which means $q/p= 4 \cdot 36 = \boxed{144}$

[hide=Solution]Note that denominators of the probabilities $p$ and $q$ are identical, so $p/q$ does not require solving for the full probability, and just the number of ways each of $p$ and $q$'s scenarios could happen.

Obviously, $p$ is just $10.$ Notice $q=10\cdot\dbinom{4}{3}\cdot9\cdot\dbinom{4}{1}.$

Thus, our final ratio is $\dfrac{10\cdot\dbinom{4}{3}\cdot9\cdot\dbinom{4}{1}}{10}=9\cdot4^2=\boxed{144}.$[/hide]

Sidenote: This is an AMC10 problem except slightly modified, just the value of $q.$

#nicestsolever

[hide=sol]

Let us say that we are zoning upon the card $t$which is the card that determines whether that set of cards is a $q$ or a $p$.  There is a $36:1$ ratio when we pick $t$ (assuming all others have already been picked), as there are $36$ cards that will give you $q$ and 1 that will give you $p$.  Since there are $4$ chances for this to happen, we multiply $36 \cdot 4 = 144$.

[hide=note]
IDK how this works.  A little bit.
[/hide]
[/hide]

[hide=sol]There are 4c4*10=10 ways to select 4 cards of same color, and 4c3*10*(40-4)=1440 ways to select 3 of one number and 1 of other number, so the answer is 1440/10=144 (dont feel like doing latex)[/hide]
I put 1/144 :wallbash:

[quote=vsar0406]I need some help understanding this part of the solution for this problem:
[quote = Alcumus Solution]Now we need to determine the number of ways that three cards can have a number $a$ and the other card have a number $b$, with $b\ne a$. There are $10\cdot 9 = 90$ ways to choose the distinct numbers $a$ and $b$. (Notice that the order in which we choose these two number matters, since we get 3 of $a$ and 1 of $b$.)[/quote]

Why would it not be $\binom{10}{2}$ ways to choose the two values for $a$ and $b$? I also did another alcumus problem exactly like this one but with the definition of $q$ changed so that $q$ is the probability that $2$ of the cards bear a number $a$ and the other $2$ bear a number $b$ such that $a \ne b.$ In the solution for that problem, there are $\binom{10}{2}$ ways to choose $a$ and $b$. But that's not true here. Why is that?[/quote]

For this problem there are $10\cdot9$ ways to choose $a$ and $b$ because the outcome will be $aaab$, which means that having $a=2$ and $b=3$ will not be the same as having $a=3$ and $b=2$ (the first case will give $2223$, while the second will give $3332$.) For the other question you had, that problem has the outcome as $aabb$, so there are equal amounts of $a$'s and $b$'s in your answers. In this case $a=2$ and $b=3$ will give the same outcome as $a=3$ and $b=2$ (the first case would give $2233$, while the second will give $3322$, which are the same case.) That's why you do $\binom{10}{2}$.

Does being able to distinguish 4 cards with the same numbers on them and not being able to change anything in this question?

i thought a and b were already chosen so i set a=1 and b=2 ut i was wrong :(

[quote=gary2003]What's wrong wit this solution?[hide=sol]
Assume the first card is arbitrary. The probability of daring 3 other of the same is $$\frac{3\cdot2\cdot1}{39\cdot38\cdot37}$$ Assume the first card you picked was the one of number $a$. Then the probability the next three are of $b$ is $$\frac{36\cdot3\cdot2}{39\cdot38\cdot37}$$ But the ratios of these are 36 not 144[/hide][/quote]

I also found that solution but it's wrong because you have to multiply a 4 choose 3 or 4 choose 1.

ohhhh this problem. this is literally the ONLY problem I have ever gotten wrong in Advanced Probability with Combinations :noo:

bruh i did stupid first answer and 2nd try i did 10*9*4*4/(10) but i thought 4*4=8 not 16  :(  :noo: :noo: :( :(

[quote=cothoop]who is game bot?[/quote]

Anyone can become gamebot. Users who complete problems and click discuss after become Gamebot! Zing!

[quote=cinnamon_e]Hello, can anyone explain the first line of the alcumus solution?
It says that the total number of ways that the numbers can be chosen is $\binom{40}{4}$, but I don't really get this, because I think that $\binom{n}{k}$ is only the number of ways to choose $k$ objects out of $n$ distinct objects.
[hide=example]you could have 10 identical candy bars and you want to pick 2 to eat, so obviously there's only one way, but if the choose thing is defined for indistinguishable objects, then there should be $\binom{10}{2}=45$ ways to choose your candy bars. something is wrong here :([/hide]
But we really only care about the numbers on our cards, so even if the actual cards were distinct (eg, colorwise), the numbers would still be the same.
Can anyone explain? Thanks in advance.  :jump:[/quote]

The probability of drawing one blue card from four distinct blue cards is the same as the probability of drawing one blue card from four indistinct blue cards.
If we begin by using $\binom{n}{k}$ to find the number of possible combinations, we can construct a probability by comparing it to the number of combinations that satisfy a condition. (Eg. $q$ and $p$)
You've already done the first step in your candy bar scenario. There are $\binom{10}{2}=45$ possible combinations.
[hide = second step]We know that ten of our candy bars are of kind $x$. The number of ways to pick two bars of kind $x$ from ten is $\binom{10}{2}=45$. Dividing, we find the probability that both of our candy bars are of kind $x$ is $45/45=1$.[/hide]

If I may modify the scenario a little:
[hide = example]You have ten candy bars in a bag. Five of them are white chocolate and five of them are milk chocolate. You randomly select two to eat. What is the probability that both of them are white chocolate?
There are $\binom{10}{2} = 45$ possible combinations when pulling two bars from a bag of ten.
The number of all-white combinations is equivalent to the number of ways to pick two white chocolate bars from five white chocolate bars: $\binom{5}{2} = 10$.
Dividing, we find the probability that both bars are white chocolate is $10/45 = 2/9$
There are always multiple roads to the same conclusion: The probability of pulling a white bar is $5/10 = 1/2$. The probability of then drawing another white bar is $4/9$. Multiplying, we find the probability of drawing a white bar and then drawing a white bar is $2/9$.[/hide]

Hope this clears things up!

[quote=programmeruser]Why do you have to do $\binom{4}{3}$ and $\binom{4}{1}$? Doesn't the first or the second one determine the position of the other one>?[/quote]

Infactual. You are misinterpreting what they are saying. They are saying that out of the $4$ cards that say $a$ (and the same for $b$), you have to choose which of those $4$ cards (in the box) that you will take out (same for $b$).


AMC
USA(J)MO
USAMO
number theory unsolved
number theory

if $a,b\in N$ and $P=\frac{b\sqrt\frac{2a-b}{2a+b}}{4}$ be a prime number find the maximum value of
$P$.

[hide="Click at your own risk"]First note that b cannot be odd. For if that is the case, then a trivial check will yield that the RHS of the equality in reduced form, has a denominator of 4. Assume b even. Now dividing both sides by b and squaring them gives:
16p^2/b^2 = (2a-b)/(2a+b). Rewriting it equivalently we have (2a+b)*16p^2 = b^2*(2a - b). These conditions imply that b|2a*16p^2. Also note, 2a>b>4p (this follows becase this relation always satisfies 0 < sqrt((2a-b)/(2a+b)) < 1 for positive a,b. Now several case arise:

so b = 16, 32 (since b > 4p >= 8) or (b,p) = p (b = 8p, 16p, 32p, p^2, 2p^2, 4p^2, 8p^2, 16p^2, 32p^2), or (b,p) = 1. For now assume p to be odd. we will consider the case p is even seperately.

If b = 32. Then p < 8. and we note that 2a = b(b^2 + p^2)/(b^2 – p^2) = 32(32^2 + p^2)/(32^2 – p^2) is not satisfied for naturals a. To see why, consider when p is not = 3, in that case the denominator is divisible by 3 but the numerator isn’t. When p = 3. One can check that a is not an integer.

If b = 16. p < 4. We use similar argument to conclude that no such p exists. (well I actually haven’t checked what happened when p = 3, but it doesn’t appear to be an integer from a preliminary glance)

If b = p. that forces p = 2 = b(since we assume b to be even). But this cant happen since b > 4p.

If b = p^2. we again get b = 4, p = 2. But this cant happen since b > 4p.

Now suppose b = kp. (assume p odd now) (with k being the naturals 8,16,32). Then 2a = kp(k^2 + 1)/(k^2 – 1). Since gcd((k/2)*(k^2+1),k^2-1) = 1 . We have k^2-1|p. implying p isn’t prime, for any even k >= 4.

Now suppose b = kp^2 (with k being 2,4,8,16,32, and p odd). Then kp^2(k^2p^2 + 1)/(k^2p^2 – 1) is certainly not an integer for any even k >= 2.

Now suppose (b,p) = 1, then b(2p^2)/(b^2 – p^2) is not an integer, and we are done.

Case p = 2 is trivial to check
I might have left some cases as i am pressed for time.
[/hide]

Too long for me to concentrate throughtout whole thing without having to write something down. NOOOOOO! I shouldn't have taken the risk. Can someone give me a breakdown? Thank you.

Well you didnt heed my warning :P

I terribly apologize for the ghastly incoherent proof i presented above. I am terrible at proof writing...and that costed me heavily during USAMO :(. So here is what i did, showed first that b is even, by a simple contradiction argument. the goal of my proof was to show that under those circumstances, a cannot be natural. So first i derived conditions on b. ie b must divide 32ap^2, and b must satisfy 2a>b>4p Then I considered two seperate cases (b,p) = p and the case (b,p) = 1 if p is odd, and the seperate case where p is even (ie: p = 2).

If (b,p) = p(which is odd) then b must be 8p,16p,32p, 2p^2, 4p^2, 8p^2, 16p^2 or 32p^2. And for each case i proved it seperately. Of course much after the proof i realized that i uselessly made the proof much longer than it really needed to be.  I uselessly considered numerous cases where i could have simply just considered b = kp for even k>4 or b = kp^2 with even k>=2. Each case can be easily proven by the same argument of substituting b and showing a cannot be an integer.

if (b,p(which is odd)) = 1 then b = 16, 32 (b cant = 2,4,8 since a p is alteast 2 and so 4p is alteast 8) and for any other possible b takes that satisfy the 2a>b>4p, i showed(or did I?) that a cannot be an integer. Which now i realize was superflous as i didnt need to seperately consider b = 16, or 32, as long as b>4*p>=4*3=12.

if p is 2, then we can easily check to see that a cannot be an integer.

Hope this helps.

Thank you. Much better. I understand exactly what you were trying to do before.

[quote=RealZeta]
If (b,p) = p(which is odd) then b must be 8p,16p,32p, 2p^2, 4p^2, 8p^2, 16p^2 or 32p^2. [/quote]
I don't think so. Indeed, $b=30, p=5$ is a solution.


function
real analysis
real analysis solved

[u][b]The author of this posting is : xxxxtt[/b][/u]
____________________________________________________________________

Give an example of a bijective function f:R->R s.t f(0)=0, f continuous in 0, but its inverse not continuous in 0.

On $[-1,1]$ define the function as follows: $f(x)=\bigg\{\begin{array}{c}x,\ x\ne\frac 1n,\ n\in\mathbb Z\\\frac1{2n},\ x=\frac 1n,\ n\in\mathbb Z\end{array}$. On $\mathbb R\setminus[-1,1]$ we don't actually need to define it. Just take $f=g$, where $g$ is a bijection from $\mathbb R\setminus[-1,1]$ to $\mathbb R\setminus[-1,1]\cup\{\frac 1{2n+1}|n\in\mathbb Z\}$.

I hope it works.


[quote="Tiago Guerreiro"]Hello  :)

Find the values of $a$ and $b$ such that:

The remainder of the division of $2x^3-ax^2+5x+b$ by $x-1$ and $x+2$ be $3$ and $0$, respectively[/quote]

[hide]The remainder theorem is very useful here ;)

We have two equations: $2-a+5+b=3$ and $-16-4a-10+b=0$

You can solve from there[/hide]

[quote="surge"]Remainder theorem is...

If polynomial $P(x)$ is divided by $x-p$ the remainder is equivalent to $P(p)$... right?[/quote]

yes

[quote="surge"]Remainder theorem is...

If polynomial $P(x)$ is divided by $x-p$ the remainder is equivalent to $P(p)$... right?[/quote]

Yes and factor theorem is sort of like the remainder theorem except it states that if f(a)=0, then f(x) has a factor of x-a.


MATHCOUNTS
geometry
3D geometry
sphere
topology
analytic geometry
linear algebra

Yes, but this shouldn't be discussed in the MC forum.

Posts on Mathcounts forums = Posts about Mathcounts problems and topics

4 dimensions $ \neq$ Mathcounts problem or topic

Because of Masters 05 you can talk about multiple dimensions in the mathcounts forum. Carl, would you like to post a summary of that Masters problem?

[quote="AstroPhys"]The regular three dimensions looked at over time, is four dimensional.[/quote]

Not quite. Technically, since you can't go backwards in time (at least not yet!), time isn't a dimension in the sense of the three spatial dimensions.

I guess the main issue here is what exactly you mean by "is". (Bear with me.) You might as well say, "is there such a thing as a circle?" because it's impossible to draw a perfect circle, at least not in the world that humans experience every day. Similarly, since humans usually only experience three spatial dimensions, it seems counter-intuitive to talk about a fourth dimension.

But in a mathematical sense, the 4th dimension exists, and so does the 5th, and the 6th, and the 42nd, and the 65537th. (I think I've heard a little bit about infinite-dimensional configuration spaces too, but I don't know anything about them, so I should probably keep silent.) This could get pretty philosophical, but basically you can define a four-dimensional space very easily: each point in that space corresponds to an ordered 4-tuple $ (a,b,c,d)$ (my choice of letters is completely arbitrary), just like the Cartesian plane with its $ (x,y)$ coordinates.

And then you can do geometry in such a space. I didn't go to Mathcounts 2005, but from what I gathered, the Masters Round was about properties of polyhedra in higher dimensions. (Correct me if I'm wrong.)

I hope this post makes at least a little bit of sense and I'm not just spouting nonsense like I usually do.

Einstein sad "Time is the 4th-dimension"

It is a long phrase of him about that.

This is not the right place.Is it ? :D

[quote="kyyuanmathcount"]Here is an interesting MATHCOUNTS problem that can be generalized:

If on a plane, 6 lines are drawn, what is the maximum number of regions formed?  What if we had n regions?

Now, put this into the 3rd dimension:  If we had a sphere, and had 6 great circles (these are "circumfrances," or circular disks that go through the center), what is the maximum number of regions formed?  What about n-disks?

How do we generalize the previous two problems into higher dimensions?[/quote]

Way to steal stuff from WOOT...

Also, this obviously isn't Mathcounts, but in linear algebra, analysis, topology, and alot of other stuff, 4, 5, n, or even infinitely many dimensions play a large role.

[quote="13375P34K43V312"]

Way to steal stuff from WOOT...

[/quote]

It was in ACoPS too - pretty standard generalization to 3-d.  Can everyone get back to mathcounts topics in this thread?

[quote="kyyuanmathcount"]Here is an interesting MATHCOUNTS problem that can be generalized:

If on a plane, 6 lines are drawn, what is the maximum number of regions formed?  What if we had n regions?

Now, put this into the 3rd dimension:  If we had a sphere, and had 6 great circles (these are "circumfrances," or circular disks that go through the center), what is the maximum number of regions formed?  What about n-disks?

How do we generalize the previous two problems into higher dimensions?[/quote]

Technically, your second question is two-dimensional too, except that you're using spherical geometry instead of Euclidean geometry now. If you want to generalize the first question, you'd have to use 6 planes separating 3-space into regions.

But yes, this is straying off-topic...

@Elio (n): To borrow a quote that has been used by numerous people, Einstein said that time is [u]a[/u] 4th dimension, but it is not [u]the only[/u] 4th dimension.

I've been poking around on Google and I discovered this link: http://www.geocities.com/CapeCanaveral/7997/index.html Hopefully it will help explain better than I can.


a side of 1 equi. triangle is congruent to altitude of anothe equi. triangle. what is the ratio of the areas of the triangle?

The area of the smaller triangle is $\frac{s^{2}\sqrt{3}}{4}$ and the area of the bigger one would be $\frac{s^{2}}{\sqrt{3}}$. So the ratio would be $\frac{3}{4}$.

[img]http://i96.photobucket.com/albums/l167/adasarathy/triangle_ratio.gif[/img]

So, you'd use the formula for the area of an equilateral with side $s$, $\frac{s^{2}\sqrt{3}}{4}$. Your small triangle's area would be $\frac{\sqrt{3}}{4}$. The side of the big one, $s_{b}$ would be, by the pythagorean:

$s_{b}^{2}+1^{2}=\left( \frac{s_{b}}{2}\right)^{2}$

$s_{b}^{2}-\left( \frac{s_{b}^{2}}{4}\right)=1$

$\frac{3s^{2}_{b}}{4}=1$

$3s^{2}_{b}=4$

$s_{b}=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3}$

Then, using the base times height formula, $\frac{2\sqrt{3}}{3}\times 1\times \frac{1}{2}=\frac{2\sqrt{3}}{4}$

So, $s: s_{b}=\frac{\sqrt{3}}{4}: \frac{2\sqrt{3}}{4}$

So, the ratio would be:

$\sqrt{3}: 2\sqrt{3}$, which is equal to $\boxed{1: 2}$

[quote="Ubemaya"]I think the answer's 3/4...[/quote]

And your justification...

the length of the median of an equilateral triangle is $\frac{s}{2}$ (half the side length) $\times \sqrt{3}$ (by 30-60-90 triangles). The square of that (since we're calculating area) is $\frac{3}{4}$.


Can anyone help me with this? i have no idea how to approach this problem:

Let $ W$ be a subspace of $ K^n$ of dimension $ k$ whose intersection with the subspace spanned by $ e_{k\plus{}1},...,e_n$ is $ \{0\}$.
(a) Show how $ W$ can be uniquely specified by a $ k$ by $ n\minus{}k$ matrix.
(b) Show that every $ k$-dimensional subspace of $ K^n$ can be described by a set of $ k$ integers $ 1\leq i_i < ... < i_k \leq n$, and an array of $ k(n\minus{}k) \minus{} \sum^k_{a\equal{}1} (i_a \minus{} a)$ numbers in $ K$.

Start with a basis for the subspace written as row vectors. Put these vectors into a $ k\times n$ matrix. What happens when you row-reduce this matrix?

you will get a matrix in the row echelon form, but the last (k+1)th to nth column will be all zero entries?
Is that right?

No. Those columns could be anything- but none of them can be pivots.

This is also useful in the general case.

I think it is equivalent to showing this:

If F is a vield and W is a subspace of the linear space F(n) with dim W<=W, then there is one and only one row reduced echelon form m x n matrix whose row space is W.

It is a little tedious to prove on computer but if you have access to the book by hoffman and kunze, there is a proof there.

As for the last part of b I'm not really sure what it means

I think that helps, thanks!
I'm not sure what part b want either, so I guess I'll just leave it

A hint for $ b$: the $ i_j$ are the locations of the pivot elements.


[quote="myc"]A secretatry writes letteres to 8 different people and addresses 8 envelopes with the people's addresses.  He randomly puts the letters in the envelopes.  What is the probability that he gets exactly 6 letters in the correct envelopes?

Can someone explain to me why my method is wrong?
Thanks:)

$\frac{1}{8}$= probability of picking the right envelope for letter # 1.
$\frac{1}{7}$= probability of picking the right envelope for letter # 2, since there are now 7 envelopes to choose from.
and so on...
$\frac{1}{8}*\frac{1}{7}*\frac{1}{6}*\frac{1}{5}*\frac{1}{4}*\frac{1}{3}*1*1=\frac{1}{20160}$

i understand the method of finding out the number of ways to choose the two wrong envelopes... but I don't understand why my method is wrong.[/quote]
Who says that the correct envelopes have to be 1-6 ;)?

[hide]There are $8\choose6$=28 ways to choose which letters are correctly addresed, then for the other two envelopes, there is only one arrangement that will make them both wrong. Thus the probability is $\frac{28}{8!}=\frac{1}{1440}$[/hide]

[quote="myc"]A secretatry writes letteres to 8 different people and addresses 8 envelopes with the people's addresses.  He randomly puts the letters in the envelopes.  What is the probability that he gets exactly 6 letters in the correct envelopes?

Can someone explain to me why my method is wrong?
Thanks:)

$\frac{1}{8}$= probability of picking the right envelope for letter # 1.
$\frac{1}{7}$= probability of picking the right envelope for letter # 2, since there are now 7 envelopes to choose from.
and so on...
$\frac{1}{8}*\frac{1}{7}*\frac{1}{6}*\frac{1}{5}*\frac{1}{4}*\frac{1}{3}*1*1=\frac{1}{20160}$

i understand the method of finding out the number of ways to choose the two wrong envelopes... but I don't understand why my method is wrong.[/quote]

Also, the second to last letter has to be in the wrong envelope, which is only a $\frac{1}{2}$ probability, not $1$.


ratio
trigonometry
LaTeX
geometry
similar triangles

http://i3.photobucket.com/albums/y92/musicis4fools2/similartriangles1.jpg

By hypothesis, triangle QRS is similar to triangle RTS. Show that x/(x-1) = 1/x

Usually in similarity questions you should look for 2 triangles which have all 3 angles equal or equilangular

We shall prove that triangle $ QRT /// QRS$

$ < Q$ is common to both triangles

$ < RTQ \equal{} < QRS$ (equal angles are opposite equal sides)

hence $ < RTQ \equal{} < RSQ$ ( < sum of triangle = $ 180^o$)

Hence $ QRT /// QRS$ (equilangular)
I dont know how to do 2nd part.

Similar triangles have therr ratios the same.

But which sides are in same ratio?

QRS is obviously isoscoles (sp? >.<), so draw a height from Q down to RS. To find x, we have sin18 = x/2/1, so x = 2sin18=0.618 approx. Blah. ANyway, since you gave us QRS ~ RTS, we know that x/TS = 1/x =(?) x/(1-x). IF this is true, then x^2 = 1-x. Moving everything together onto one side gets us x^2+x-1 = 0. Solving for x leads us to [-1 +- sqrt(5)]/2. x is obviously positive, so it equals (-1 + sqrt5)/2, or 0.618 approx. This matches up with the x above, so the equation is true, x^2 = 1-x. Therefore, x/(1-x) = 1/x.

I don't know how to do it w/o (a tiny bit of) trig, but you can just plug it into your handy dandy calculator.

edit: ... omg, wow. I can't believe I didn't see that the first time around. >.<

[quote="superwizard188"]QRS is obviously isoscoles (sp? >.<), so draw a height from $ Q$ down to $ RS$.

To find $ x$, we have $ sin18^o = x/2/1,$
$ x = 2sin18 = 0.618 approx$

since $ QRS ~ RTS$, we know that $ x/TS = 1/x = (?) x/(1 - x)$.

IF this is true, then $ x^2 = 1 - x$. Moving everything together onto one side gets us

$ x^2 + x - 1 = 0$. Solving for

$ x$ leads us to $ \frac { - 1 \pm\sqrt(5)}{2}.$

x is obviously positive, so it equals $ { - 1 + sqrt\5}{2}$, or $ 0.618$ approx. This matches up with the x above, so the equation is true, $ x^2 = 1 - x$. Therefore, $ \frac {x}{1 - x} = \frac {1}{x}$.

I don't know how to do it w/o (a tiny bit of) trig, but you can just plug it into your handy dandy calculator.[/quote]

If your gonna write a solution write it neatly. No one can read this....and make the effort to learn latex  :mad:

http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideSym.php
http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideCommands.php
i tried to clean it up a bit

...-_-...Thanks for the valuable lesson.

[quote="superwizard188"] ...IF this is true, then x^2 = 1-x. Moving everything together onto one side gets us x^2+x-1 = 0...[/quote]

I understand your method there, but the question says to PROVE \frac{x}{1-x} = \frac{1}{x}. When you do a mathematical proof, you're proving one side of the equation is equal to the other, i.e. LHS = RHS. I.e. you can't combine the two sides into the equation x^2+x-1 = 0

The proof would also be more elegant if trig wasn't involved, but I'm not finding an easy proof without trig...

P.S How do I write in LaTeX, cuz it obvioiusly didn't work here :|

Properties of similar triangles (the ratios of side lengths and such) can be used to prove $ \frac {x}{1 \minus{} x} \equal{} \frac {1}{x}$. No trigonometry is required, and I'm pretty sure the solution without it is simpler.

[quote="Mo R."]How do I write in LaTeX, cuz it obvioiusly didn't work here :|[/quote]
Your syntax is correct, but you have to place dollar signs around your code to render it into an image:

[code]$\frac{x}{1-x}=\frac{1}{x}$[/code]
Welcome to AoPS! :welcomeani:

[quote="Mo R."]

The proof would also be more elegant if trig wasn't involved, but I'm not finding an easy proof without trig...|[/quote]
Hint:
[hide]Do you see any isosceles triangles?[/hide]


*sigh*
I'm trying to improve the competition, so please state your comments here.
Also, if you think this is hopeless (there is no hope for salvation), please say so too.
*sigh*

lol. I think this should be deleted though...seeing as the poll is slightly biased.

[quote="Tare"]really?[/quote]
well, now that you've edited it, i guess it's not that biased.
still, there aren't any reasons corresponding with "yes"

Well if you like it, then that means there's no place to improve upon. If you don't, then there's a reason why, and that needs to be fixed.

[quote="Fierytycoon"]Well, the contest is for MATHCOUNTS people, so difficulty probably won't factor in too much. Though I think perhaps the title "AoPS member of the month" is not too attractive...perhaps something like "Problemsolving Machine" or "Problem magnet" would be better..I dunno, but a cool title.  :D[/quote]

The prize could be a date with me or Fierytycoon?   :wink:

[quote]The prize could be a date with me or Fierytycoon? [/quote]

After that is announced, Tare wouldn't have time to grade all the solutions that flood in.  8-)

[quote="Tare"]I decided to up the bet, and the questions. The last round's going to be MUFASA hard :) , but for a prize, you'll get your own personalized avatar (such as personalized pear, your favorite TV character, etc.), as long as the material's appropriate and the request isn't too hard.[/quote]
great. give us the last round then. in fact, forget about making it MUFASA hard, just make it WHURM!! hard. ;)

Patience, young grasshopper.

[size=59]...I only have two of the problems and I haven't finished either one :) [/size]

Can only people in grade 8 or lower participate, or are there no grade level limits to this?

no grade limit. you can still join now, since the due's at September 30.


Show that for any triangle $ABC$, the following inequality is true:
\[ a^2 + b^2 +c^2 > \sqrt{3} max \{ |a^2 - b^2|, |b^2 -c^2|, |c^2 -a^2| \} . \]

Suppose a>b>c. We can scale c = 1.

Then we want to show $a^2 + b^2 + c^2 > \sqrt{3}(a^2 - c^2)$

The most straightfoward solution is $b^2 > a^2 + c^2 - 2ac$ from the triangle inequality.  Then $b^2 > a^2 -2a + 1$.

Finally, we want to show $a^2 + (a-1)^2 + 1 > \sqrt{3}(a^2 - 1)$

It becomes $a^2(2-\sqrt{3}) - 2a + (2+\sqrt{3}) > 0$.  Examining the quadratic in a, reveals that it has one root, $4+\sqrt{3}$, which is clearly too big for any value of b to make (a,b,c) a triangle.

[b]@Singular[/b]

it is possible for any particular two lengths to constitute the sides of infinite number of triangles. hence your last argument is wrong.
how it should have been done is

$ a^2 \plus{} b^2 \plus{} c^2 > a^2 \plus{} (a \minus{} 1)^2 \plus{} 1$

and

$ a^2(2 \minus{} \sqrt {3}) \minus{} 2a \plus{} (2 \plus{} \sqrt {3}) > 0 \forall a \in R$

hence this is true for infinite values of a. hence we can have infinite b's with

$ b^2 > (a \minus{} 1)^2$

hence it is true for triangle with any side lengths.

WLOG assume $c^2-a^2$ is max. Also $b^2>a^2+c^2-2ac$
Substituting, $2a^2+2c^2-2ac>\sqrt{3}(c^2-a^2)$ or$(2+\sqrt{3})a^2+(2-\sqrt{3})c^2>2ac$
Since this is true by AM GM we are done


I remember that I read somewhere that the statement $ \phi(n) | n\minus{}1$ implies $ n$ is prime is a more than less known conjecture. I would like to read more about that problem, so if anyone knows the name of some mathematicians who had worked on that problem, or the conjectures's name it would be very useful.

:)

It's called the [url=http://mathworld.wolfram.com/LehmersTotientProblem.html]Lehmer totient problem[/url], and there are references available in the article.


Let $ A \in M_n{\mathbb{(C)}}$ such that $ A^{2} \plus{} I \equal{}0$
Prove that $ \det(A\overline{A} \plus{} I) \geq 0$
 Here $ \overline{A}$ Is  conjugate complex matrix of $ A$
 $ I$ is identity matrix

For $ B \equal{} A\overline{A}$ we have $ B\overline{B} \equal{} E$ (*). Thus it is easy to show that $ B$ and $ \overline{B}$ commute, thus they are simultaneously triangonalizable. Since $ B$ is similar to $ D \plus{} N$ with a diagonal matrix $ D$ having the eigenvalues of $ B$ on the diagonal and a strictly upper triagonal matrix $ N$, the relation (*) gives that $ D\overline{D} \equal{} E$, i.e. the set of eigenvalues $ \mathcal E$ of $ B$ is contained in $ S^1$. $ \mathcal E$ is preserved by conjugation, since the eigenvalues of $ B \equal{} A\overline{A}$ and $ \overline{B} \equal{} \overline{A}A$ coincide.

Then the determinant of $ E \plus{} B$ can be presented as the product of non-negative factors $ |1 \plus{} \lambda|^2 \equal{} (1 \plus{} \lambda)(1 \plus{} \overline{\lambda})|$ for non-real eigenvalues $ \lambda\in\mathcal E$ and $ 1 \plus{} \lambda$ for the real eigenvalues $ \lambda\in\mathcal E\cap\mathbb R\subset\{1, \minus{} 1\}$.


A guy walks into a 7-11 store and selects four items to buy.  The clerk at the counter informs the gentleman that the total cost of the four items is $ \$$7.11. He was completely surprised that the cost was the same as the name of the store. The clerk informed the man that he simply multiplied the cost of each item and arrived at the total. The customer calmly informed the clerk that the items should be added and not multiplied. The clerk carefully added the items together and informed the customer that the total was still exactly $ \$$7.11. (There was no sales tax involved)  Your task is to determine the exact costs of each item?

Is this even possible within the set of $ \mathbb{Z}$? The only factors of 711 are 1, 3, and 79...yet no combination of these 4 can equal 711 when multiplied and added...but is this solvable outside of $ \mathbb{Z}$? I tried some things, but nothing gets me anywhere D:

It is possible

[hide]
$ x.y.z.w\equal{}711000000$
$ x\plus{}y\plus{}z\plus{}w\equal{}711$[/hide]

I don't follow your logic...if the product and sum are equal...how can their values be $ 711000000$ and $ 711$?

Furthermore, if we assume the numbers are in terms of cents, then there have to be integer solutions to the equation, since you can't exactly have .5 cents in your pocket...

Which is where the multiplying dollars aspect confuses me...What does it mean to have $ \$1.50\cdot\$.50$?

[quote="Icy"]I don't follow your logic...if the product and sum are equal...how can their values be $ 711000000$ and $ 711$?

quote]
Because they have been converted to cents.  Consider this: if you have 0.10 dollars and you multiply it by 0.10 dollars, what do you get? 0.01.  That's one cent.  But try converting it to cents first.  10 cents multiplies by ten cents equals 100 cents, which is 1 dollar.
Think about why that happens.

I've seen this problem on AoPS before.
http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=711&t=44345

This concept is really strange to me....
Assuming that $ x, y, z, w$ are in cents, and 7.11 in dollars, then...

What I was doing before was:
$ (10^{ \minus{} 2})(x \plus{} y \plus{} z \plus{} w) \equal{} 7.11$ Which works, since you can add cents directly...but...
$ (10^{ \minus{} 2})xyzw \equal{} 7.11$, doesn't work. I just realized that that only converts one of the cent values to dollars, not all 4.

So...if the values are in cents...to convert them to dollars would be...

$ 10^{ \minus{} 2}(x \plus{} y \plus{} z \plus{} w) \equal{} 7.11$ and
$ (10^{ \minus{} 2}x)(10^{ \minus{} 2}y)(10^{ \minus{} 2}z)(10^{ \minus{} 2}w) \equal{} 7.11\Leftrightarrow$
$ (10^{ \minus{} 8})xyzw \equal{} 7.11\Leftrightarrow$
$ xyzw \equal{} 711000000$
Is that it?

Then the rest is just really intellegent algebra, too advanced for me. :(

Thanks for the help Archimedes.


algebra
polynomial
trigonometry
algebra unsolved

$P(x)$ is a polynomial of degree $3n$ such that
\begin{eqnarray*} P(0) = P(3) = \cdots &=& P(3n) = 2, \\ P(1) = P(4) = \cdots &=& P(3n-2) = 1, \\ P(2) = P(5) = \cdots &=& P(3n-1) = 0, \quad\text{ and }\\ && P(3n+1) = 730.\end{eqnarray*}
Determine $n$.

Eek... but if I may offer a possible simplification of the problem:  Consider $Q(x) = P(x) - 1$.  Then the problem can be rewritten

$Q(3k) = 1, k \le n$
$Q(3k-2) = 0, k \le n$
$Q(3k-1) = -1, k \le n$
$Q(3n+1) = 729 = 3^6 = 3 \times 3^2 \times 3^3$

Hope that leads someone to a solution 'cause I'm stuck and I have homework to do  :)   (Personally, I think $n = 3$ or $n = 6$)

[quote="t0rajir0u"]Eek... but if I may offer a possible simplification of the problem:  Consider $Q(x) = P(x) - 1$.  Then the problem can be rewritten

$Q(3k) = 1, k \le n$
$Q(3k-2) = 0, k \le n$
$Q(3k-1) = -1, k \le n$

Hope that leads someone to a solution 'cause I'm stuck and I have homework to do  :)   (Personally, I think $n = 3$ or $n = 6$)[/quote]

So we can get $f(3n+1)=(-1)^n\frac{2}{3}(\sqrt{3})^{3n+2}\cos\frac{3n+2}{6}\pi$ by Langrange Interpolation.
So $n=4$.

Using Lagrange Interpolation I get that $P(3n+1) = \sum_{k=0}^{3n}(-1)^{3n-k}\binom{3n+1}{k}f(k)$.
Now you could use the old trick with the roots of unity, but I don't see any way of keeping the computations short.

Before wriring down the proof(asuuming it to be correct) an intutive note since i was doing complex no.s these days period three reminds me of  $\omega$

[i]Proof[/i]
a bit guess work and manipulation may lead u to conclude that

$P(k)-1$ $=$ $\frac{2Ime^{\frac{\pi i}{3}}\omega^{k}}{\sqrt 3}$ for the given values of $k$

claim is

$P(x)=1+\frac{2\sum_{k=0}^{3n}\binom{x}{k}Ime^{\frac{pi i}{3}}(\omega-1)^{k}}{\sqrt 3}$
clearly  $P(j)=1+\frac{2Ime^{\frac{\pi i}{3}}\omega^{k}}{\sqrt 3}$ for $j=0,1,2\dots 3n$
so the polynomial cannot be anything but the above one
a bit of manipulation leads to
$P(3n+1)=\frac{-2 Ime^{\frac{\pi i}{3}}(\omega-1)^{k}}{\sqrt 3}$
note $\omega-1=1\sqrt 3 e^{\frac{\pi i}{3}}$
we get by De-moivre's Theorem $P(3n+1)=730=3^{6}+1$
this gives $n=4$

hey, there is a solution using difference operator by Dr Kin Yi Li in his journal "Mathematical Excalibur", which I think is more succinct than lagrange interpolation. here is the link:

http://www.math.ust.hk/excalibur/v11_n5.pdf


limit
calculus
derivative
logarithms
function
calculus computations

1 a) Calculate $ \lim_{h \rightarrow 0} (\ln(1 \plus{} h)/h)$ by identifying the limit as the derivative of $ \ln (1\plus{}x)$ at $ x\equal{}0$.

b) Use the result of part (a) to show that $ \lim_{h \rightarrow 0} (1 \plus{} h)^{1/h} \equal{} e$.

c) Use the result of part (b) to calculate the related limit, $ \lim_{n \rightarrow \infty} (1 \plus{} 1/n)^n$.

Some hints:
Okay part (a) is straight forward for you could also do it by using L'Hopital and the answer to part (a) is 1.
For part (b), apply $ e$ to both sides.
For part (c), just transform you funtion into something that look like the one in part (b).


[quote="jli"][quote="now a ranger"]A rhombus has interior angles of 30, 150, 30, and 150 degrees, and an area of 3. Find its perimeter.[/quote]

[hide="solution"]Let the side be $s$.  The height of the rhombus is $\frac{s}{2}$ so the area is $\frac{s^{2}}{4}$.  So $s=\sqrt{12}=2\sqrt{3}$ so the perimeter is $\boxed{8\sqrt{3}}$[/hide][/quote]

If the height is $\frac{s}{2}$, then the area is $\frac{s^{2}}{2}$, I believe.

[hide]oh yeah...i was thinking of a triangle

so it should be $8\sqrt{6}$[/hide]

So many different answers... I suggest you guys actually draw the diagram and do the math.

[hide="Correct answer is"]Why are you looking here? Do the problem :D[/hide]


suppose x=x0+tv and y=y0+sw are two parametric representations of the same line l in r^n
a. show that there are scalars t0 and s0 such that y0=x0+t0v and x0=y0+s0w
b. show that v and w are parallel

[hide="Sketchy"]a. What are $ x_0$ and $ y_0$? Are they on $ \ell$? If yes, then there must be a scalar, so...

b. Use part a. Perform dot product. Recall that two vectors are parallel when their dot product is _____ [/hide]
BIG EDIT: you seem desperate in seeking a solution and looks like a homework. Hide edited to help guiding you instead of providing you the solution.


Find all positive integers $x,y$ satisfying the equation \[ 9(x^2+y^2+1) + 2(3xy+2) = 2005 . \]

The equation is equivalent to :$9[(x+y)^2-2xy+1]+2(3xy+2)=2005$

Now we put $x+y=k$ $xy=l$ and this is equivalent $9k^2-18l+9+6l+4=2005$ <=>

$3k^2=4(l+166)$ => $k=even$. The above is equivalent to

$2k^2+k^2-4l-664=0$ (1) . The number $k^2-4l$ is the discriminant of $a^2-ka+l=0$

and to have roots it should be $\geq 0$. So $k^2-4l\geq 0$. So from (1) we have that

$2k^2\leq 664$ <=> $k\leq 18,..$ so $k\leq 18$ because $k$ is an integer.

Now if $k=18$ then from (1) we have that $l=77$ and $x=7$ or $11$ or $y=7$ or $11$.

For $k=16$ we have $l=1044$ and discriminant$<0$ ,no roots

If $k\leq 14$ , $3k^2<664$ and so $l<0$ absurb.

So the only roots are $x=7$ or $11$,$y=7$ or $11$

The given equation is equivalent to
$ 2(x+y)^2+(x-y)^2=664 $ Since $ x+y $ and $ x-y $ have the same parity they must be even. Now let $ x+y=2a $ and $ x-y=2b $. We have
$ 16a^2+4b^2=664, 2a^2+b^2=166, b=2c, a^2+2c^2=83 $. Since $ a^2<83, a<9 $. Now it is easy to check that the solutions are $ (x,y)=(11,7) and (7,11) $

Another way of proving this problem is considering the equation as quadratic $ 3x^2+xy2+3y^2-664=0 $, so the discriminant  must be perfect square, but doing it this way takes too much time.

Simplifying the equation we get
$ 3x^2 + 3y^2 + 2xy = 664 $ or $3(x-y)^2 = 664-8xy = 8(83-xy)$
implying that $x$ and $y$ are odd. Letting $x=2a+1$ and $y = 2b+1$ we get
$3(a-b)^2 = 2(83-xy) < 2\cdot 72$ or $(a-b)^2 < 46$
leaving $(a-b) = 6,4,2$ since $a$ and $b$ have the same parity. Now plugging these  three cases in the equation we get $(x,y) \in \{(7,11),(11,7)\}$ .

How about AM-GM and common divisors:
By AM-GM
$2005 = 9(x^2+y^2+1) + 2(3xy+2) \ge 9(2xy+1) +2(3xy+2) =24xy+13 $ which gives $xy \le 83$

Through letting $x=ad;y=bd, (a,b)=1 $ we see that $d=1,2$

If $d=2$ then $166 = 3a^2 +3b^2 +8ab \ge 14 ab \rightarrow ab \le 11$ and checking cases no solution.

Thus $d=1$.

Now letting $y=x+a$ we see that $8x^2+8ax+3a^2=664$ which implies $a$ is a multiple of four.

We now check coprime odd numbers that differ by a multiple of four and multiply to less than $83$, and we will obtain the result.

Super-Ugly and bashy yet ...

We can re-write the equation as:

$ 3x^2 + y^2 + 2(3x)(y) + 8y^2 + 9 + 4 = 2005$

or $ (3x + y)^2 = 4(498 - 2y^2)$

The above equation tells us that $(498 - 2y^2)$ is a perfect square.
Since $498 - 2y^2 \ge 0$. this implies that $y \le 15$

Also, taking $mod 3$ on both sides we see that $y$ cannot be a multiple of $3$. Also, note that $249 - y^2$ has to be even since $(498 - 2y^2) = 2(249 - y^2)$ is a perfect square.
So, $y^2$ cannot be even, implying that $y$ is odd.

So we have only $\{1, 5, 7, 11, 13\}$ to consider for $y$.

Trying above 5 values for $y$ we find that $y = 7, 11$ result in perfect squares.

Thus, we have $2$ cases to check:

$Case 1: y = 7$
$ (3x + 7)^2 = 4(498 - 2(7^2))$
$ =>  (3x + 7)^2 = 4(400)$
$ => x = 11$

$Case 2: y = 11$
$ (3x + 11)^2 = 4(498 - 2(11^2))$
$ => (3x + 11)^2 = 4(256)$
$ => x = 7$

Thus all solutions are $(7, 11)$ and $(11, 7)$.

Ok so a lil bit bashy  but I bet it will be quick and won't be rigorous at all.
Write the given equation as 3$(x-y)^{2}$ + 6$(x+y)^{2}$ = 1992
Now just put x-y as a and x+y as b.
So bound like 1992>6$b^{2}$. and then check , there wont be that much
You would be lucky if you start from the last like the max value b can take is 18, So you get the solns in this manner.

Solution expand and get
$3x^2+3y^2+2xy=664$
Then make quad in $x$
Thus $7968-32y^2=k^2$
Thus $249-y^2=2a^2$
Now we know y is odd and less than $15$ so we can take $y=1,3,5,7,9,11,13,15 8$ cases
And we get $y$ is $7$ or $11$ putting in original
When $y =7$ Solution expand get $x =11$ we know qtn statement is symmetric thus
WLOG if $y =11$ then $x=7$ done
$(7,11)(11,7)$


Let $ S$ be the area of the triangle $ ABC$. Suppose that we have constructed the point $ P$

[img]15916[/img]

The quadrilateral $ AFPE$ has area $ \frac {1}{3}S$
If we move the point $ P$ parallel to $ FE$, the area of the triangle $ FEP$ remains constant, so the area of $ AFPE$ remains constant, which is $ \frac {1}{3}S$

The (dashed) line $ IJ$ has this property, so the point $ P$ we are searching for should belong on $ IJ$

The triangle $ AEI$ has area equal to $ \frac {1}{3}S$. So it's easy to find the point $ I$. It should be $ AI \equal{} \frac {1}{3}\frac {AB\cdot AC}{AE}$

So we perform the following construction:

Let $ C_1\in AC$ such that $ AC_1 \equal{} \frac {1}{3}AC$
[img]15919[/img]
Let $ I \in AB$ such that $ IC_1 \parallel BE$

$ \frac {AI}{AB} \equal{} \frac {AC_1}{AE} \equal{} \frac {AC}{3AE} \Rightarrow AI \equal{} \frac {AB\cdot AC}{3AE}$

Similarly, we find points $ I',J'$ on the sides $ AC,BC$ respectively, with the corresponding attributes

Then $ (AFPE) \equal{} (CEPD) \equal{} \frac {1}{3}S$ and consequently $ (BDPF) \equal{} \frac {1}{3}S$

Sorry, I forgot to write. We construct $ I'J'$ with the same way as $ IJ$. Then $ P$ is on the intersection of $ IJ$ and $ I'J'$

Well, $ IJ$ has the properties : $ I\in AB, \ J\in AC, \ IJ \parallel FE$ and $ (AIE) \equal{} \frac{1}{3}S$

We construct $ I'J'$ so that it has the same properties with $ IJ$.

That is, $ I'\in CA, \ J' \in CB$ such that $ \ I'J' \parallel ED$ and $ (CI'D) \equal{} \frac{1}{3}S$

[i]Sorry, I could not find a simpler construction  :([/i]

We draw three lines:

$ a_1\parallel EF, \ A \in a_1$
$ b_1\parallel FD, \ B \in b_1$
$ c_1\parallel DE, \ C \in c_1$

We have $ \boxed{(AFE) \plus{} (PFE) \equal{} (BFD) \plus{} (PFD) \equal{} (CDE) \plus{} (PDE) \equal{} \frac {1}{3}S} \ \ (1)$

It is $ (AFE) \plus{} (PFE) \equal{} \frac {1}{2}(u_1 \plus{} u_2)\cdot FE$, where $ u_1,u_2$ are the altitudes respectively of $ \triangle AFE$ and $ \triangle PFE$ corresponding to the side $ FE$

If we set $ u_1 \plus{} u_2 \equal{} h_a$ then we simply have $ (AFE) \plus{} (PFE) \equal{} \frac {1}{2}h_a\cdot FE$

In other words, $ h_a$ is the distance from $ P$ to the line $ a_1$

Similarly, set $ h_b \equal{} d(P,b_1)$ and $ h_c \equal{} d(P,c_1)$

Now $ (1)$ becomes $ \boxed{EF\cdot h_a \equal{} FD \cdot h_b \equal{} DE \cdot h_c} \ \ (2)$

The formula $ (2)$ would be enough to find point $ P$ (we could draw lines parallel to $ a_1$ and $ b_1$ in the appropriate distances and find their intersection)

But there is one more elegant way...

The lines $ a_1,b_1,c_1$ are forming a triangle $ D'E'F'$, which is similar to $ \triangle DEF$

Then from $ (2)$ we get $ \frac {1}{2}E'F'\cdot h_a \equal{} \frac {1}{2}F'D' \cdot h_b \equal{} \frac {1}{2}D'E' \cdot h_c$

which means that $ (PE'F') \equal{} (PF'D') \equal{} (PD'E')$. So, $ P$ divides the triangle $ D'E'F'$ into 3 equal areas

So $ P$ is the centroid of $ \triangle D'E'F'$


OK, since I don't actually have a copy of the problems, here's the problem as I remember it:

There are two beads, each of mass $ m$, strung on a circular hoop of mass $ m_h$ that is positioned vertically on a flat surface.  The beads are released from the top of the hoop with negligible velocity (so that one bead goes down each side of the hoop).  What is the minimum value (or was it the maximum value?) of $ m/m_h$ so that the hoop doesn't leave the surface?  Neglect friction. (throughout the entire process, the hoop remains perfectly vertical)

EDIT: Oops, I'm stupid.  I meant "maximum value" - otherwise we could have $ m \equal{} 0$, obviously.

[quote="genericme"] so that the hoop doesn't leave the surface?
[/quote]

Mind explaining this line again?

I mean, so that the hoop doesn't jump up or become airborne.

And [b]oh crap[/b], I just realized that we probably aren't allowed to discuss these problems until March 10 or something...

But I guess it won't hurt to read the problem, since everyone who took the quarterfinal exam (which is now over) would know of it anyway.

That is a very interesting question, but perhaps the mod should lock this temporarily.

Just curious; exactly at what level is the quarter final? - does it correspond to AIME or USAMO ?

This corresponds to the AIME.

...but I think it was more a speed test. We'll see how it turns out.
Oh, and also, yeah, we are not supposed to talk about the problems until after March 10th.

I hope it's fine if we discuss this now

[hide]
Assume that the hoop (radius $ R$) never leaves the surface.

Suppose the beads have fallen through an angle of $ \theta$ (with the verticle)
Velocity of beads $ v\equal{}\sqrt{2gR(1\minus{}\cos\theta)}$
Therefore, the normal force at the hoop-bead contact is
$ N\equal{}\frac{mv^2}{R}\equal{}2mg(1\minus{}\cos\theta)$

Hence, net vertical, upwards force on the hoop is $ F\equal{}2N\cos\theta\equal{}4mg\cos\theta(1\minus{}\cos\theta)$

As the hoop never leaves the surface, $ m_hg\geq F$ at all $ \theta$, i.e., $ m_hg$ is not less than the maximum of $ F$

Maximum of $ F$ is at $ \theta\equal{}\tan^{\minus{}1}2$
Hence, $ F_{\text{max}}\equal{}4mg\left( \frac{\sqrt{5}\minus{}1}{\sqrt{5}}\right)$
Hence, $ m_hg\geq 4mg\left( \frac{\sqrt{5}\minus{}1}{\sqrt{5}}\right)$ or
$ \frac{m}{m_h}\leq \frac{\sqrt{5}}{4(\sqrt{5}\minus{}1)}$
[/hide]

That isn't quite right, the solutions are up on the AAPT site right now so you can just check there.


Alguien sabe exactamente por qué se le da una rotación de 60 grados???? (apart del triándulo equilátero)

Gracias, apreciare mucho la ayuda

Hola, trataría de ayudarte si explicases mejor lo que te inquieta. A que te refieres con rotación de 60 grados?, en que caso particular?. En el caso de la determinación del punto de Fermat del triángulo?. De ser el caso, puede verse el siguient link para una demostración [url]http://centros5.pntic.mec.es/~marque12/matem/fermat.htm[/url]


There are $ 50$ people on two dimensional plane. At the same time, every people shot the people nearest from him (if there are two people have the same distance from him, he can choose one people)
Find a configuration such that only $ 10$ people die

if A shot B and B shot A at the same time...who died?[considering of course that A is the nearest person for B and B is the nearest person for A]

In clockwise sense, let $ ABCDEF$ be a regular convex hexagon, with center $ O_1$, and $ AGHIO_1C$ be a regular convex hexagon with center $ B$. Thus, both hexagon have the same side length.

Now, each of $ A,B,C,D,E,F$ shots on $ O_1$, each of $ G,H,I,O_1$ shots on $ B$.
In that case,we have found a group of $ 10$ people where only two die.

Then, use $ 5$ similar groups, sufficientely far from each other so that each people from a group shots in the same group.

Pierre.


algebra
polynomial
quadratics
function
system of equations

Factor the following over polynomials with integer coefficients:

$x^4 - 4x^3 +14x^2 -4x + 13$

[hide]Write 14x^2 as 13x^2+x^2. Now the polynomial looks like

Which can be factored as

And...um...I think that's it.[/hide]

Quote:Factor the following over polynomials with integer coefficients:
[hide]

Another grouping approach . . .

We have: x4 + 14x2 + 13 - 4x3 - 4x

= (x2 + 1)(x2 + 13) - 4x(x2 + 1)

= (x2 + 1)(x2 + 13 - 4x)

= (x2 + 1)(x2 - 4x + 13)[/hide]

demji wrote:How would you do this problem anyway?  Could you explain.  Why do you have to break it up?

You want to make it so that [hide]two or more groups of terms have a common factor greater than 1. In this case, I noticed that the x^3 term had the same coefficient as the x term, so if there were two x^2 terms, x^4 had the same coefficient as one of them and and the other had the same coefficient as the x^0 term, then they'd have a common polynomial of degree 2. This means that one of the x^2 terms has to have a coefficient of 1, and the other a coefficient of 13. Luckily, they add up to 14 like I hoped. Then it's easy.[/hide]

(Or if you can't follow that, which is pretty likely considering how jumbled that was even to me, sufficient to say that it was really a matter of instinct. You get used to noticing patterns after a bit.)

[hide]Assume it factors, at the very least, into two quadratic terms (take a look at the graph of the function and you'll kind of see why).  So (x 2 +ax+b)(x 2 +cx+d)=the original thing.  Now a, b, c, and d are integers and bd=13, so (WLOG) b=1 and d=13 (or b=-1 and d=-13, but check that case later if this one doesn't work).  Now you can just expand and set up a system of equations to find a and c.[/hide]

[quote="demji"]i think i understand.  you want to get it into (xe(something))(whatever)(sameaspreviouswhatever)
=(xe(something)+1)(whatever)
sorry if its confusing but it just want to keep tems of x out[/quote]
Yeah. Just don't forget that the 'sameaspreviouswhatever' could be multiplied by something other than 1.


For $ x,y,z$ positive reals prove that

$ \sum x^2 \plus{}xyz\plus{}2 \geq \sum x \plus{} \sum xy$

It is just

\[ x^2\plus{}y^2\plus{}z^2\plus{}2xyz\plus{}1\ge 2(xy\plus{}yz\plus{}zx)\]

and

\[ x^2\plus{}y^2\plus{}z^2\plus{}3\ge 2(x\plus{}y\plus{}z)\]

i have a different proof... rewrite the inequality as $ f(x)\equal{}x^2\plus{}x(yz\minus{}y\minus{}z\minus{}1)\plus{}(2\plus{}y^2\plus{}z^2\minus{}y\minus{}z\minus{}yz)\geq 0$...

if $ yz\minus{}y\minus{}z\minus{}1\geq 0$, then $ f(x)$ is increasing for nonnegative real numbers, and $ f(x)\geq f(0)\geq 1$...

if $ yz\minus{}y\minus{}z\minus{}1\leq 0$, then it's enough to prove that the discriminant is less than or equal to $ 0$... we have to prove that
$ (yz\minus{}y\minus{}z\minus{}1)^2\minus{} 4(2\plus{}y^2\plus{}z^2\minus{}y\minus{}z\minus{}yz)\leq 0$....

if $ (y\minus{}1)(z\minus{}1)\geq 0$, then, $ 2\geq 1\plus{}y\plus{}z\minus{}yz>0$, which implies that $ 4(2\plus{}y^2\plus{}z^2\minus{}y\minus{}z\minus{}yz)\geq 4\geq(yz\minus{}y\minus{}z\minus{}1)^2$, and we're done...

now suppose that $ (y\minus{}1)(z\minus{}1)\leq 0$... expanding the expression for the discriminant we have to prove that
$ 8\plus{}3y^2\plus{}3z^2\plus{}2y^2z\plus{}2yz^2\geq y^2z^2\plus{}6y\plus{}6z\plus{}4yz$...

this follows from $ 6\plus{}3y^2\plus{}3z^2\geq 6y\plus{}6z$, $ 1\plus{}y^2z\plus{}yz^2\geq 3yz$ and $ yz(y\plus{}z)\geq yz(yz\plus{}1)$, and we're done  :D


calculus
derivative
logarithms
function
calculus computations

So today was the first day of school for me, and for extra credit my Calculus AB teacher assigned this problem:

$ {{{{e^x}^x}^x}^{...}}\equal{}10$

I have had no previous encounter with any of the calculus problem solving techniques other than finding simple derivatives,(my teacher claims that this problem does NOt require the use of calculus, but rather the manipulation of properties of logarithms), but a friend who's in Calculus BC says that I might need to use the taylor series to solve...

Thanks in advance =D

No, you won't need Taylor series here (as a matter of fact, I do not see how they would help you anyway). What you do need is to understand how the function $ f(x)=x^{x^{x^{x^\dots}}}$ behaves. It is not obvious why it is well-defined at all but, assuming that it is, we have $ f(x)=x^{f(x)}$, so if $ f(x)=y$, then $ x=\dots$ . Well, enough said. ;)

so then we would have $ ln10 \equal{} x^{ln10}$
therefore, $ x \equal{} \sqrt [{ln10}]{ln10}$
right?


Define
$ g(n\plus{}2)\equal{}g(n)\plus{}g(n\plus{}1)\plus{}1,n \in \mathbb{N}$
$ g(1)\equal{}0,g(2)\equal{}1$
prove that for all prime $ p>5$ we have $ p | g(p)(g(p)\plus{}1)$

A first step
[hide]
Clearly we can see that g(n) = F(n) - 1 . where F(1) = 1 and F(2) = 2
[/hide]

I think $ g(n)\equal{}F(n\plus{}1)\minus{}1$ as$ \sum F(i)\equal{}g(n\plus{}1)\equal{}F(n\plus{}2)\minus{}1$

If $ \mathrm{f}(1)\equal{}0,\ \mathrm{f}(2)\equal{}1,\ \mathrm{f}(n\plus{}2)\equal{}\mathrm{f}(n\plus{}1)\plus{}\mathrm{f}(n)$, then

$ \mathrm{g}(n)\ \equal{}\ \sum_{k\,\equal{}\,1}^n{\mathrm{f}(k)}$

And there is already a nice handy formula for the Fibonacci sequence. :wink:

[quote="madness"]No according to the way in which i defined it , g(n) = F(n) - 1 only  :huh:

Yes Jane_Bennet but we arent going to expand the set of radicals are we?  :P
I think we have to deal with integer expressions only so that we can easily finish it off by modulo theory  :D[/quote]

i'm sorry..i didn;t notice that F(2)=2.... :oops:

that's right but the crunch now is to use Binet's Formulae to prove the divisibility
source ISL/ILL 1988 ireland

Phew I thought I was really losing it all .
Thank Goodness I never new Binet's
 :P   :rotfl:   :D

Pardesi could you tell me Binet's formula and explain how to do this sum with it.
Too lazy to search  :P  :D

ok here goes the Binet's Formulae
if u define the fibbonaci numners as hell_ever did then
$ F_{n}\equal{}\frac{\alpha^{n} \minus{} \beta^{n}}{\alpha \minus{} \beta}$
where $ \alpha \equal{} \frac{1\plus{}\sqrt{5}}{2}$ $ \beta\equal{}\frac{1\minus{}\sqrt{5}}{2}$

Well it seems that i knew the formula but not the name  :rotfl:
Do you mean to expand the whole thing?
I think there must be a better method ( Probably modular )  :maybe:


inequalities
geometry
geometric transformation
reflection

Let $a,\, b,\, c$ be positive real numbers. Prove that \[\frac{a}{b(b+c)^{2}}+\frac{b}{c(c+a)^{2}}+\frac{c}{a(a+b)^{2}}\ge\frac{9}{4(ab+bc+ca)}\]

Well, you should not post your solution, April. The deadline for Reflections Issue 2 is not over yet. And this topic should have been locked.

Anyway, no matter. It's an easy problem, if not trivial.  :D

[quote="April"]It's really easy problem  :D
Applying [b]Cauchy-Schwarz Inequality[/b], we have:
\[\left(\sum \frac{a}{b(b+c)^{2}}\right)\left(\sum ab\right)\ge\left(\frac{a}{b+c}\right)^{2}\ge\frac{9}{4}\]
[/quote]
You mean \[\left(\sum \frac{a}{b(b+c)^{2}}\right)\left(\sum ab\right)\ge\left(\sum\frac{a}{b+c}\right)^{2}\ge\frac{9}{4}\]


vector
topology
real analysis
real analysis unsolved

Let E be a normad vector space and $ C \subset E$ a convex subset. Let $ x \in C$ and $ y \in C^{o}$. Show that $ tx \plus{} (1\minus{}t)y \in C^{o}$ for all $ t \in [0, 1[$.

$ C^{o}$ is the interior of $ C$.

May someone help me?

C convex means$ \forall x,y \in C, \forall t\in [0,1)$

$ P(x,y,t)\equal{}tx\plus{}(1\minus{}t)y \in C$

so since $ y \in C^o$

$ \exists \epsilon > 0$ s.t. $ B(y, 2\epsilon) \subset C$

let $ B\equal{}B(y,\epsilon ) \subset C$

then $ \forall z \in B, P(x,z,t) \in C, \forall t$ by convexity

so for $ P(x,y,t)$ we can see that it is obviously contained in an open set in C
for one example,

$ B( P(x,y,t), \frac{1\minus{}t}{2}\epsilon )$


Let $x,$ $y$ and $z$ are non-negative numbers such that $x+y+z=1.$
Prove that $xy\sqrt{1-z}+xz\sqrt{1-y}+yz\sqrt{1-x}\leq\sqrt{\frac{2}{27}}.$
Good luck!

We have $(\sum_{cyc}xy\sqrt{1-z})^{2}\leq (\sum_{cyc}xy)(\sum_{cyc}xy-3xyz)$(CBS).Now we have $ (\frac{2}{3}\sum_{cyc}xy)(\sum_{cyc}xy-3xyz)\leq\frac{1}{4}(\frac{5}{3}\sum_{cyc}xy-3xyz)^{2}$ so we have to prove $\frac{5}{3}\sum_{cyc}xy\leq 3xyz+\frac{4}{9}$.Now we have from Schur $ 3xyz+\frac{1}{3}\geq \frac{4\sum_{cyc}xy}{3}$ so it remains to prove $\sum xy\leq \frac{1}{3}$(QED)

The following inequality is true
[quote="mateivld"]We have $(\sum_{cyc}xy\sqrt{1-z})^{2}\leq (\sum_{cyc}xy)(\sum_{cyc}xy-3xyz)$(CBS)[/quote]
and the end! Thank you. :lol:


function
limit
ratio
calculus
derivative
real analysis
real analysis unsolved

Let $ f$ be a diferentiable function on $ (0,\infty)$ and let $ \omega$ be a positive real number.

Prove that if $ \lim\limits_{n\rightarrow \infty}(f'(x) \plus{} \omega f(x)) \equal{} A$ then $ \lim\limits_{n\rightarrow \infty}f(x) \equal{} A/\omega$.

Since $ \lim_{x\to\infty} \frac{(f(x)e^{\omega x})'}{(e^{\omega x})'} \equal{} \frac{A}{\omega}$, by L'hospital rule, we have $ \lim_{x\to\infty} f(x) \equal{} \frac{A}{\omega}$.

[quote="Kalle"]Why $ f(x) e^{\omega x} \to \pm \infty$?[/quote]

Oh, I was unaware of that.

If $ A \neq 0$, it is easy to prove that $ |f(x)| \to \infty$ as $ x \to \infty$ using $ \epsilon \minus{} \delta$. But I'm not able to handle the case $ A \equal{} 0$ yet.

For the L'Hospital rule, you do not need to check that both functions tend to infinity. The fact that the denominator does and that the ratio of the derivatives has a limit is enough (just examine the proof to see it).

I have never seen the proof of L'hospital, or seen it stated anywhere except on the internet actually. Anyway, my complaint was not very substantial anyway, since the case $ A \equal{} 0$ is easily reduced to $ A \equal{} 1$ by adding $ 1/\omega$ to $ f$. I just wanted to point out that there are some conditions and stuff to L'hopital, since Kent Merryfield has conditioned me into disliking it. Although I don't know why.

This is indeed a very nice problem, and it surprises me that this condition ensures that $ f'$ converges. What is the intuition here? I don't get it

I do not like l'Hospital too  :)  just solve the differential equation $ f'(x)\plus{}\omega f(x) \equal{} A(x)$ where $ A(x) \to A$ with the variation of constant method. It will give you the result for almost nothing.

Yes, that is it, as alekk pointed out, I think that the problem can be solved, another method, by the simple application of $ \epsilon\minus{}\delta$ definition of limits and then by using the monotonicity of the left hand and right hand side functions.


[quote="kunny"]For $x>0,\ y>0$, find the minimum value of $x+y+\frac{2}{x+y}+\frac{1}{2xy}$.[/quote]
$x+y+\frac{2}{x+y}+\frac{1}{2xy}$

$\geq4\times\frac{x+y}{4}+2\times\frac{1}{x+y}+\frac{2}{(x+y)^{2}}$

$\geq(4+2+1)\sqrt[7]{\left(\frac{x+y}{4}\right)^{4}\left(\frac{1}{x+y}\right)^{2}\frac{2}{(x+y)^{2}}}=\frac{7}{2}$,

with equality if and only if $x=y=1$.

That's right. :) I haven't noticed in using $\frac{a+b+c+d+e+f+g}{7}\geq \sqrt[7]{abcdefg}\ (a,\ b,\ c,\ d,\ e,\ f,\ g>0)$.


What do you think about this new website ? It looks pretty but the menu/navigations are sometimes confusing and not helpful in finding desired information.

I think that Mathcounts website should at least keep most important historic information, like pictures of White House Visits, etc. New design only contains White House visit for 2006. Where are all the links for pictures in 2005 and 2007 ?

Type in your search in the Search function so it links to the archived content.

If you have trouble finding things (I did at first)  use the site map and it shows where everything is.


Find the best constant k (smallest :)) so that this inequality is true with x, y, z >0 and xyz=1
$ (x^{2}+y^{2}+z^{2})^{k}\ge x^{3}z+y^{3}x+z^{3}y$

Sorry, i edited above.

[quote="mathgeniuse^ln(x)"][quote="bbbbbbbbbb"]Find the best constant k (smallest :)) so that this inequality is true with x, y, z >0.
$ (x^{2}+y^{2}+z^{2})^{k}\ge x^{3}z+y^{3}x+z^{3}y$[/quote]

[hide]I think the answer is
2

WLOG, let $ x \ge y \ge z >0$.

Then, by rearrangement, $ x^{4}+y^{4}+z^{4}\ge x^{3}z+y^{3}x+z^{3}y$.

So, then we see that k=2, works, but k=1 doesn't.

I don't know how to show that fractional values are feasible.[/hide][/quote]
I don't think you can WLOG a cyclic but nonsymmetric inequality

[quote="mathgeniuse^ln(x)"]Wait, why can't you do WLOG in the inequality, because it is symmetric, and x, y, and z will have to have some order, right, but I don't know.[/quote]
what happens to the wlog when x>z>y
(the RHS isn't symmetric...symmetric would be like $ x^{3}z+xz^{3}+y^{3}x+x^{3}y+z^{3}y+yz^{3}$ )