trigonometry quadratics algebra polynomial

Prove that  Tan70 =Tan20 +2Tan40 +4Tan10

I think you meant Tan 70o &c. The symbol o in plain text is coded [code]o[/code].

Denote $x = \tan 20^\circ$. Using the formula $x = \tan 20^\circ = \frac{2\tan 10^\circ}{1 - \tan^2 10^\circ}$ and solving this quadratic equation for the unknown $\tan 10^\circ$,

$x(1 - \tan^2 10^\circ) = 2 \tan 10^\circ,\ \ x \tan^2 10^\circ + 2 \tan 10^\circ - x = 0$

$\tan 10^\circ = \frac{-2 \pm \sqrt{4 + 4x^2}}{2x} = \frac{-1 \pm \sqrt{1 + x^2}}{x}$

and we have to take the positive root (with the plus sign), because both $x = \tan 20^\circ > 0$ and $\tan 10^\circ > 0$. Using this and the formulas

$\tan 70^\circ = \frac{1}{\tan 20^\circ} = \frac 1 x$

$\tan 40^\circ = \frac{2 \tan 20^\circ}{1 - \tan^2 20^\circ} = \frac{2x}{1 - x^2}$

the equation in question can be written as

$(?)\ \ \ \frac 1 x = x + \frac{4x}{1 - x^2} + \frac{4(-1 + \sqrt{1 + x^2})}{x}$

$(?)\ \ \ \frac{1 - x^2}{x(1 - x^2)} = \frac{x^2(1 - x^2)}{x(1 - x^2)} + \frac{4x^2}{x(1 - x^2)} - \frac{4(1 - x^2)}{x(1 - x^2)} + \frac{4(1 - x^2)\sqrt{1 + x^2}}{x(1 - x^2)}$

$(?)\ \ \ 1 - x^2 = x^2(1 - x^2) + 4x^2 - 4(1 - x^2) + 4(1 - x^2)\sqrt{1 + x^2}$

$(?)\ \ \ x^4 - 10x^2 + 5 = 4(1 - x^2)\sqrt{1 + x^2}$

$(?)\ \ \ x^8 + 100x^4 + 25 - 20x^6 + 10x^4 - 100x^2 = 16(1 - 2x^2 + x^4)(1 + x^2) =$

[color=white]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .[/color] $= 16(1 + x^2 - 2x^2 - 2x^4 + x^4 + x^6)$

$(?)\ \ \ x^8 - 20x^6 + 110x^4 - 100x^2 + 25 = 16(x^6 - x^4 - x^2 + 1)$

$(?)\ \ \ x^8 - 36x^6 + 126x^4 - 84x^2 + 9 = 0$

If this 8th degree polynomial has the root x, it also has the root -x, because it is a 4th degree polynomial in $x^2$. Thus we have to show that $x = \pm \tan 20^\circ$ are roots of the above 8th degree polynomial. Using the formulas for trisection of the angle 60o and dividing them,

$\frac{\sqrt 3}{2} = \sin 60^\circ = 3 \sin 20^\circ - 4 \sin^3 20^\circ$

$\frac 1 2 = \cos 60^\circ = 4 \cos^3 20^\circ - 3 \cos 20^\circ$

$\sqrt 3 = \frac{3 \sin 20^\circ - 4 \sin^3 20^\circ}{4 \cos^3 20^\circ - 3 \cos 20^\circ} = \frac{\sin 20^\circ}{\cos 20^\circ} \cdot \frac{3 - 4 \sin^2 20^\circ}{4 \cos^2 20^\circ - 3}$

Substituting $x = \tan 20^\circ = \frac{\sin 20^\circ}{\cos 20^\circ}$ and

$\sin^2 20^\circ = \frac{\tan^2 20^\circ}{1 + \tan^2 20^\circ} = \frac{x^2}{1 + x^2}$

$\cos^2 20^\circ = \frac{1}{1 + \tan^2 20^\circ} = \frac{1}{1 + x^2}$

and squaring the result, we obtain a 6th degree polynomial, which definitely has the roots $x = \pm \tan 20^\circ$.

$\sqrt 3 = x\ \frac{3(1 + x^2) - 4x^2}{4 - 3(1 + x^2)} = x\ \frac{3 - x^2}{1 - 3x^2}$

$3(1 - 3x^2)^2 = x^2(3 - x^2)^2$

$3(1 - 6x^2 + 9x^4) = x^2(9 - 6x^2 + x^4)$

$x^6 - 33x^4 + 27x^2 - 3 = 0$

Now we take the 8th degree polynomial in question and divide it by the above 6th degree polynomial using the long division. The result is

$x^8 - 36x^6 + 126x^4 - 84x^2 + 9 = (x^2 - 3)(x^6 - 33x^4 + 27x^2 - 3)$

which can be easily verified by multiplying out the factors on the right side. Thus we see that the 8th degree polynomial in question has 2 roots $x = \pm \sqrt 3 = \pm \tan 60^\circ$ (we are not interested in these), 2 roots $x = \pm \tan 20^\circ$ and 4 other roots that we are not interested in (without a proof, these are $x = \pm \tan 40^\circ$ and $x = \pm \tan 80^\circ$). But this is exactly what we wanted to prove.


[color=darkblue] Let $ m,n \in \mathbb{N}$, $ m\le 2000$, and $ k \equal{} 3 \minus{} \dfrac{m}{n}$. Determine the minimum possible value for $ k$[/color]

hi ;) !!!

$ n \in \mathbb{N}^*$

$ k\equal{}3 \minus{} \frac{m}{n}$

 $ k \geq 3\minus{}2000 \equal{} \minus{}1997$ than $ min(k)\equal{}\minus{}1997$

....


I created some simple cryptograms!!!
Tell me if you know how to solve them.
1. DBSPIMSBKOTMDFIZOIFFIPSPOSIFFIPRQWRFTFQFTSHNDRRTVNQHZPIZYTYFCM SBWTHDFTSHRDRRQBTHAFNDFHSOIFFIPTRP IKODMIZJTFNTFRIOYDHZTRFNQRNDPZIPFSRSOLIFNDHDMDIRDPMTKNIPOIFFIPYPIGQIHMTIRZTAPDBRDHZ FPTAPDBRMDHNIOKCSQZIFIPBTHIJNTMNOIFFIPRFDHZRYSPJNTMNSFNIPOIFFIPQOFTBDFIOCJIATLICSQFJIHFCRILIHYSQPFNRDRFNIDHRJIP
2.
APTRICVNLOEVOSLNTYARSNOPISITNOONSTBUTSTITUOITNEHAWTYFOGIRUTEIHO STUSIOTXEMANITEEHRFQEEUCNDYSIRTBITUOIANDNONITECHTTATISIXERTMELEC YOLESOTLPIATNXEATISTTRUSNUOETCACHAHARTCREFOHTSIETTXAWTSARSNOPES WDTITHEHNONEXETTIOWTIHHCAMEKFSRO SARURPSINILGUYRNAEADLBMESEASEGINENETNEIPSIHTSELOTUOITNTOIHPSRAATSLPORAETSI BAAEELICHPREABESODCNAHTPRENOOETFEHOBKOHTMENAHWLOVODENOYLUNMBRES
3.
NHPSSSTNOTBETTHA
TTIMIEACOSHMRORW
ASSOIICIENWUTTTI
SOIEEBAVHOYBBLLW
ISRWCTRHTDEIXHRH
ESNOAQOQEARSEEOA
TTVEFOUDXHLESPIE
SUIDNEEFURLYNWGA
SIOERONRRWPRUESEX
HSNMRMDROTTXTSOT
HYRUIAESOEGHCRNI
SNOTEAOLEWEOTTOD
OEFEESUNNUYHNOLT
EITAODLEOEYMUNOT
HAFRRQOETCIACADY
SESONMDFTHRSIOEE Mr. Green

DO NOT ANSWER THIS THREAD!
These are problems from the 2005 Collaborative problem solving exam, which starts today and ends on April 20th.

[quote="weiyangmiami"]I created some simple cryptograms!!!
Tell me if you know how to solve them.[/quote]
Why did you say you created these cryptograms when you didn't?


geometry
number theory
number theory unsolved

Can we find all [u]real[/u] numbers $x,y,z$ which satisfy $3^x+4^y=5^z$
What about $3^x+4^x=5^x$
The same question with $5^x+7^y=13^z$

As the term 'real number' says, this is not number theory  :(

Take any $x,y$ and solve for $z$, there will be a solution everytime.

Number theory, Algebra, Geometry it is all maths. Can you solve this: $3^x+4^x=5^x$ in real numbers

Do you think the division into different themes here is just there to neglect it¿ No! When I look into Number Theory, I want to see Number Theory!

And for the $3^x+4^x=5^x$ question: some analysis shows that there is no solution except $2$ (find the one and only maximum and take limits to both directions).


analytic geometry
symmetry
absolute value
combinatorics proposed
combinatorics

In a $ 19\times 19$ board, a piece called [i]dragon[/i] moves as follows: It travels by four squares (either horizontally or vertically) and then it moves one square more in a direction perpendicular to its previous direction. It is known that, moving so, a dragon can reach every square of the board.

The [i]draconian distance[/i] between two squares is defined as the least number of moves a dragon needs to move from one square to the other.

Let $ C$ be a corner square, and $ V$ the square neighbor of $ C$ that has only a point in common with $ C$. Show that there exists a square $ X$ of the board, such that the draconian distance between $ C$ and $ X$ is greater than the draconian distance between $ C$ and $ V$.

Consider a set of coordinates so that corner $ C$ is $ (0,0)$, the opposite corner of the $ 19\times19$ grid is $ (18,18)$, $ V$ is obviously $ (1,1)$ and wlog because of symmetry the first coordinate is horizontal and the second coordinate is vertical.  We will call "horizontal" a move that changes the horizontal coordinate by 4 and the vertical coordinate by 1; "vertical" moves are defined reciprocally.

Since from $ (0,0)$ to $ (1,1)$, the horizontal coordinate changes by 1, the number of vertical moves must be odd.  Likewise, the number of horizontal moves must be odd.  Assume that the dragon may move from $ (0,0)$ to $ (1,1)$ with less than 8 moves.  Then it must be able to do it with 6 or less, i.e., at most 3 moves in each direction.  So, the net change introduced by horizontal moves in the vertical coordinate is at most 3 (at most 3 times 1), while the change introduced by vertical moves in the vertical direction is at least 4 (odd number of times 4).  Since the net change in the vertical coordinate is $ \plus{}1$, we would necessarily need two "upward" vertical moves, one "downward" vertical move and three horizontal moves so that in the three of them the change in the vertical coordinate is downward.  Similarly, we would need two "rightbound" horizontal moves, one "leftbound" horizontal move and three vertical moves so that in the three of them the change in the horizontal coordinate is to the left (assuming 0 is the leftmost column, 18 the rightmost column).  But this is impossible, since the first move must be horizontal righbound and up, or vertical upward and rightbound.  So at least 8 moves are necessary.  With 8 moves it is possible: $ (0,0)\rightarrow(4,1)\rightarrow(8,0)\rightarrow(7,4)\rightarrow(6,8)\rightarrow(2,7)\rightarrow(1,3)\rightarrow(5,2)\rightarrow(1,1)$.

So we need to find a square such that the dragon takes at least 9 moves to get to it, and one such square is $ (18,17)$.  Since the change in vertical coordinates is odd, an odd number of horizontal moves is needed; similarly, an even number of vertical moves is needed.  So if it is possible to get from $ (0,0)$ to $ (18,17)$ in 8 moves or less, it is possible to do it in at most 7 moves.  If the change in both coordinates is added in absolute value, the net change in the sum of coordinates is at most 5, so 7 moves may bring a sum of changes in both coordinates of at most 35.  But the net change in coordinates from $ (0,0)$ to $ (18,17)$ is exactly 35, so all changes in the horizontal direction (be it through horizontal or vertical moves) must be of the same sign, and likewise in the vertical direction.  Or, if we have exactly $ h$ horizontal moves and $ v$ vertical moves, it must hold that $ 18 \equal{} 4h\plus{}v$ and $ 17 \equal{} h\plus{}4v$, yielding $ 1 \equal{} 3(h\minus{}v)$, absurd.  So at least 9 moves are needed to get from $ (0,0)$ to $ (18,17)$.  Such a sequence of 9 moves is $ (0,0)\rightarrow(4,1)\rightarrow(8,0)\rightarrow(9,4)\rightarrow(10,8)\rightarrow(11,12)\rightarrow(12,16)\rightarrow(13,12)\rightarrow(14,16)\rightarrow(18,17)$.

I just realized that I have actually gone beyond the scope of the problem, since it was enough to show that it is not possible to get from $ (0,0)$ to $ (18,17)$ in less than 9 moves (I have shown that it is actually possible to do it in exactly 9 moves) and that it is possible to get from $ (0,0)$ to $ (1,1)$ in no more than 8 moves (I have shown that 8 is the minimum number of moves required)...  :wink:

same solution and i like this question. good number 4 i think.

Yep, numbers 1 and 4 problems in Ibero are typically problems where no "heavy equipment" is necessary at all, but if you fail to see the "essence" of the problem (in this case I would say parity is this "essence"), you bang your head against the wall to no avail...   :(   Overall very simple (to IMO standards) problems, but good fun nonetheless   :)


analytic geometry
geometry
graphing lines
slope

If a Cartesian coordinate grid were placed on top of the map of Math City, Algebra
Lane would go through the points (1, 3) and (1, 1). Geometry Boulevard runs
perpendicular to Algebra Lane and passes through the points (2, 2) and (2, y ). What
is the value of y ?

[quote="anirudh"]If a Cartesian coordinate grid were placed on top of the map of Math City, Algebra
Lane would go through the points (-1, 3) and (1, -1). Geometry Boulevard runs
perpendicular to Algebra Lane and passes through the points (2, 2) and (-2, y ). What
is the value of y ?[/quote]

[hide]i assume you mean negative signs because there are boxes on my computer....

The slope of Algebra lane is -2. That means the slope of the geometry Blvd is 1/2.

y=mx+b
2=1+b
b=1

y=-1+1
=0. [/hide]

[hide="click"]
so the slope of the first line is 2. since geometry blvd is perpindicular to algebra lane then it would have a slope of one-half.  so to find the slope you do (x-x)/(y-y) so then it becomes (2-y)/4=1/2. So y =0[/hide]

[hide]

The slope of the first line is -2, and since the second line is perpendicular to the first, the second line's slope is $\frac{1}{2}$.
So the equation to find y is $\frac{1}{2} = \frac{y - 2}{-4}$
Simplify, and you get $y = \framebox{0}$[/hide]


Compute $ 9^3 \plus{} 3\times 9^2 \plus{} 3\times 9 \plus{} 1$.

Note that this is just $ (9\plus{}1)^{3}\equal{}10^{3}\equal{}1000$.

More detailed solution:

Expanding $(a+b)^3$ gives us $a^3+3a^2b+3ab^2+b^3$. Note that $9^3 + 3\times 9^2 + 3\times 9 + 1$ can be rewritten as $9^3 + 3\times 9^2\times 1 + 3\times 9 \times 1^2+ 1^3$. We see this is of the form $(a+b)^3$ with $a=9$ and $b=1$. Thus, we have $9^3 + 3\times 9^2 + 3\times 9 + 1 = (9+1)^{3}=10^{3}=1000$


Inequality
three variable inequality
calculus
maximization
IMO
IMO 1984

forgot to add in:

$x,y,z \ge 0$, $x+y+z =1$

Without loss of generality, suppose $z\leq \frac{1}{2}$. Thus write $x+y=\frac{1}{2}+t,z=\frac{1}{2}-t$, where $0\leq t\leq \frac{1}{2}$.

It is easy to see that $xy+yz+zx-2xyz=xy(1-2z)+(x+y)z\geq 0$.

$xy+yz+zx-2xyz=xy(1-2z)+(x+y)z\leq \left(\frac{x+y}{2}\right)^2(1-2z)+(x+y)z = 2t\left(\frac 14+\frac t2\right)^2+\frac{1}{4}-t^2 = \frac{4t^3-4t^2+t-2}{8}$, denote it by $f(t)$.

Let $f'(t)=0$ and we get $t=\frac{1}{6}$ and $t=\frac{1}{2}$. It is easy to verify that $f\left(\frac{1}{6}\right)=\frac{7}{27}$ is maximal.

Hence $xy+yz+zx-2xyz\leq \frac{7}{27}$

Ok, I'll give another solution:
Let $W(t) = (t-x)(t-y)(t-z)$ then:
$W(t) = t^3 - t^2 + qt + r$
We have to prove that $q + 2r \leqslant \frac{7}{27}$
$W(\frac{1}{2}) = -\frac{1}{8} + \frac{1}{2}q + r = -\frac{1}{8} + \frac{q+2r}{2}$,
So we must show that $W(\frac{1}{2}) \leqslant \frac{1}{216}$
When one of x,y,z, is bigger or equal $\frac{1}{2}$  then $W(\frac{1}{2}) \leqslant 0$
So let $x,y,z < \frac{1}{2}$ We have got:
$W(\frac{1}{2}) = (\frac{1}{2}-x)(\frac{1}{2}-y)(\frac{1}{2}-z) \leqslant (\frac{\frac{1}{2}-x +\frac{1}{2}-x+\frac{1}{2}-x}{3})^3 = \frac{1}{216}$
(I used $G_k \leqslant A_k$)

The left inequality is trivial. And the right inequality follows from
\[
(1-2x)(1-2y)(1-2z)\leq\frac{1}{27},
\]
which is true by AM-GM inequality and that at most one term among $1-2x,1-2y$ and $1-2z$ would be negative.

homogenize everything and kill it with schur and muirhead.  :D

[quote="xxreddevilzxx"]what was muirhead again???[/quote]

it's also called bunching too.

If you really want to see the proof using Schur and Muirhead, then look at paragraph 2.4 of Hojoo Lee's "Topics in Inequalities", which you can find at http://my.netian.com/~ideahitme/eng.html (the proof doesn't explicitely use Schur or Muirhead, but a theorem is applied which follows from them). But this is an absolutely stupid way to solve this inequality; any of the proofs you gave above in this topic, or Kalva's proof ( http://www.kalva.demon.co.uk/imo/isoln/isoln841.html ) is better.

  Darij

Just a notice that this problem looks similar to BMO1999: If $p,q,r$ are nonnegative and $p+q+r=1$, then
\[
7(pq+qr+rp)\leq2+9pqr.
\]
Also, this can be solved using the same argument, that is it follows from
\[
(7-9p)(7-9q)(7-9r)\leq64,
\]
which is true by AM-GM inequality and that at most one among $7-9p,7-9q$, and $7-9r$ would be negative. :lol:

here is a proof with Schur and Muirhead, which I tend to think is not so ugly:

We have $\sum xy\cdot \sum x=\sum x^2y+3xyz\geq 2xyz$

This establishes the leftmost inequality

Similarly, we need to show

$27\sum x^2y+27xyz\leq 7(x+y+z)^3$

This is just

$27\sum x^2y+27xyz\leq 7\sum x^3+21\sum x^2y+42xyz$

By Schur,

$7\sum x^3+21xyz\geq 7\sum x^2y$, so

$7\sum x^3+21\sum x^2y+42xyz\geq 28\sum x^2y+21xyz$

and since $\sum x^2y\geq 6xyz$, we're done.

Another solution :

We set $ p = x+y+z = 1$, $ q = xy+xz+yz$, $ r = xyz$

The inequality rewrite :

$ 0\leq q-2r\leq\frac{7}{27}$

The Schur's inequality is:

$ x^{t}(x-y)(x-z)+y^{t}(y-x)(y-z)+z^{t}(z-x)(z-y)\geq0$ for all real t and positive reals x, y,z

With $ t = 1$ we have : $ x(x-y)(x-z)+y(y-x)(y-z)+z(z-x)(z-y)\geq0\Leftrightarrow pq-9r\geq0$

With $ t = 2$ we have : $ x^{2}(x-y)(x-z)+y^{2}(y-x)(y-z)+z^{2}(z-x)(z-y)\geq0\Leftrightarrow p^{3}-4pq+9r\geq0$

From these 2 inequalities, we deduce a framing for q and r :

$ \frac{4pq-p^{3}}{9}=\frac{4q-1}{9}\leq r\leq\frac{pq}{9}$ (1)

$ \frac{9r}{p}=9r\leq q\leq\frac{p^{3}+9r}{4p}$ (2)

From (2) we have $ q\geq9r$ and because $ 9r\geq2r$ then by transitivity, $ q\geq2r\Leftrightarrow q-2r\geq0$

The inequality to show rewrite $ 27q-54r-7\leq0$

From (1) we have $ 27q-54r-7\leq27q+(6-24q)-7=3q-1$

But it's known that $ 3q\leq p^{2}=1$ indeed $ 3q-1\leq0$ and by transitivity, $ 27q-54r-7\leq0$

Finally, we showed $ 0\leq q-2r\leq\frac{7}{27}$

If we have elementary symmetric sums in three variables, $ \sigma_{i}$ for $ i=1,2,3$, then the inequality becomes:

$ 0\le \sigma_{2}\sigma_{1}-2\sigma_{3}\le \frac{7}{27}(\sigma_{1})^{3}$.

We can apply the substitutions: $ \sigma_{1}=s$, $ \sigma_{2}=4rR+r^{2}$, $ \sigma_{3}=sr^{2}$ [for a triangle].

$ 0\le (s)(4rR+r^{2})-2sr^{2}\le \frac{7}{27}s^{3}\iff 0\le 4rR-2r^{2}\le \frac{7}{27}s^{2}$.

The left inequality becomes $ R\ge 2r$ which is euler's inequality. For the right side, note that $ 9GI^{2}\ge 0\implies s^{2}\ge 16rR-5r^{2}$.

Then it suffices to show that $ \frac{7}{27}*(16rR-5r^{2})\ge 4rR-2r^{2}\iff\frac{19}{27}r^{2}+\frac{4}{27}rR\ge 0$ which is obvious.

my soln

let $ {a\plus{}b\plus{}c\equal{}p\equal{}1}$
$ {ab\plus{}bc\plus{}ac\equal{}q}$
$ {abc\equal{}r}$

now we have to prove $ {q<\equal{} 7/27 \plus{} 2r}$

but from first degree schur we know that

$ {p^3 \plus{} 9r >\equal{} 4pq}$

putting p=1

we know $ {1\plus{}9r>\equal{}4q}$

we know $ {r/4<\equal{}1/27.4}$  (am,gm)

hence $ {q<\equal{}1/4\plus{}2r \plus{} r/4 <\equal{}2r\plus{}1/27.4\plus{}1/4}$

hence
$ {q<\equal{}2r\plus{}7/27}$

qed

Let's use this substitutes: $ x=\frac{a}{a+b+c}$, $y=\frac{b}{a+b+c}$, $z=\frac{c}{a+b+c}$ ($a,b,c \ge 0$, and they are not 0 simultaneously). The left inequality is true by AM-GM Inequality. The right one is equivalent with $27(a+b+c)(ab+bc+ca)-54abc \le 7(a+b+c)^3$. From Schur, we have: $7(a+b+c)^3 \ge 28(a+b+c)(ab+bc+ca)-63abc$, so it is enough to prove $(a+b+c)(ab+bc+ca) \ge 9abc$, which is obviously true.

This is a solution for right side of inequality

taking $x=\frac{a+b}{2}$ , $y=\frac{b+c}{2}$ and $z=\frac{c+a}{2}$

then we have to prove that for real numbers $a,b,c$ such that $a+b+c=1$ and $a+b,b+c,c+a$ are non-negative hold

$abc\leq\frac{1}{27}$ or $\sqrt[3]{abc}\leq \frac{1}{3}=\frac{a+b+c}{3}$ which is equivalent to

$\frac{1}{2}(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c})(\sum_{cyc} (\sqrt[3]{a}-\sqrt[3]{b})^2)\geq 0$

which is true since $\sum_{cyc} (\sqrt[3]{a}-\sqrt[3]{b})^2\geq 0$ and $\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\geq 0$

last inequality is true since $a+b\geq0\Rightarrow a\geq -b\Rightarrow \sqrt[3]{a}\geq -\sqrt[3]{b}\Rightarrow \sqrt[3]{a}+\sqrt[3]{b}\geq 0$

similar we have $\sqrt[3]{b}+\sqrt[3]{c}\geq 0$ and $\sqrt[3]{c}+\sqrt[3]{a}\geq 0$ adding this three inequality we have desired result .

It is obviously true that more than one variable is lower than $\frac{1}{2}$

suppose that $a\leq \frac{1}{2}$ $\space$, $\{x, y, z\}=\{a, b, c\}$

without loss of generality, let $f(a, b, c)=ab+bc+ca-2abc=yz+zx+xy-2xyz$

Prove $f(a, \frac{b+c}{2}, \frac{b+c}{2}) \geq f(a, b, c) $
we get
$$f(a, \frac{b+c}{2}, \frac{b+c}{2})=a(\frac{b+c}{2})+(\frac{b+c}{2})^2+a(\frac{b+c}{2})-2a(\frac{b+c}{2})^2 \geq ab+bc+ca-2abc=f(a, b, c)$$
equivalent with
$$\frac{(b+c)^2}{4}-bc \geq 2a(\frac{(b+c)^2}{4}-bc)$$
which is true because it equals with $$1 \geq 2a$$
Therefore $$f(x, y, z)=f(a, b, c) \leq f(a. \frac{b+c}{2}. \frac{b+c}{2}) \leq . . . \leq f(\frac{a+b+c}{3}. \frac{a+b+c}{3}. \frac{a+b+c}{3})=\frac{7}{27}$$

Solution for the right inequality:

Without loss of generality, we may assume that $x=min(x,y,z) \Rightarrow x \le \frac{1}{2}$

Let $yz=t$ .We have $0 \le \ t \le \frac{(y+z)^2}{4}=\frac{(1-x)^2}{4}$.

We need to prove that $F(t)=t(1-2x)+x(1-x) \le \frac{2}{27}$. But $F(t)$ is an increasing linear function of $t$, so is enough to prove that $F(\frac{(1-x)^2}{4})=\frac{(1-x)(2x^2+x+1)}{4} \ge \ 0$, which is obviously true. Done!

[quote=xxreddevilzxx]Prove:
$0 \le yz+zx+xy - 2xyz \le \frac{7}{27}$[/quote]

Prove the left side inequality,
Since $x+y+z=1 \implies 0\leq x, y,z\leq 1 \implies  xy\geq xyz , yz\geq xyz \implies xy+yz+ zx\geq2xyz$

the second inequality,
Case (1)
Since $ x+y+z=1$ , at most one of three variables is greater  than $\frac 12$,
without loss of generosity, suppose $x> \frac 12 ;  y,z\leq \frac 12$
$(1-2x)(1-2y)(1-2z)=1-2(x+y+z)+4(xy+yz+zx)-8xyz=-1+4(xy+yz+zx-2xyz)$

$ \implies xy+yz+zx-2xyz= \frac {(1-2x)(1-2y)(1-2z)}{4}+\frac 14\leq \frac 14=\frac {7}{28}< \frac {7}{27}$,

because $(1-2x)(1-2y)(1-2z)\leq 0$.

Case (2)
When the three variables $x,y,z$ are all less than $\frac 12$.
Then

$(1-2x)(1-2y)(1-2z)\leq \left (\frac{1-2x+1-2y+1-2x}{3}\right)^3=\frac 1{27}$,  by AM_GM.

$xy+yz+zx-2xyz= \frac {(1-2x)(1-2y)(1-2z)}{4}+\frac 14\leq \frac 14\left(\frac 1{27}\right)+\frac 14=\frac 7{27}$

Hence, we are done.


limit
logarithms
calculus
calculus computations

Find \[ \lim_{n \to \infty} \left( \frac{e}{\left( 1+\frac{1}{n} \right)^n } \right) ^n .  \]

Multiple Taylor expansions using the development of $ln$, $exp$, $x \rightarrow \frac{1}{1+x}$, and $ln$ again. Eventually you find $\sqrt{e}$.

Same idea as julien, but with more details shown:

$\ln\left[\left(\frac{e}{\left(1+\frac1n\right)^n}\right)^n\right]= n\ln\left(\frac{e}{\left(1+\frac1n\right)^n}\right)$

$=n\left(1-n\ln\left(1+\frac1n\right)\right)$

$=n\left(1-n\left(\frac1n-\frac1{2n^2}+O(n^{-3})\right)\right)$

$=n\left(1-1+\frac1{2n}+O(n^{-2})\right)$

$=\frac12+O(n^{-1})\to\frac12$ as $n\to\infty$

So the answer is $e^{1/2}=\sqrt{e}.$


AMC
USA(J)MO
USAMO
AIME
probability
AMC 10
AMC 10 A

Vote for USAMO index. Also, what do you guys think the floor score will be?

I got 214... please less than 214...

I am pretty sure (just wanna let you know.. I am seriously bad at guessing).... that the index would be above 204

but not sure how high it would be....

and I LOVE!!!! AIME  #1 (I got it wrong haha..)

Lol these things are always hard to predict but probably a 206ish.

I'll wildly guess 208 index, and 7 floor with 214+ index for underclassmen.

I see you got a 198 index (i only got 199). I will join the group of praying people.

everybody Pray that the indexx is 197!!!

What's the probability that it will be higher than 227?

i got 202 last year and the cutoff was 202.5
I just don't want to get screwed over this year too.

i'm praying too...i completely FAILED FAIL FAIL FAILED the AMC 10A and i wasn't allowed to take the second one..i got 127.5 can you believe it i was so bad that day...i have 197.5...   :oops:  :o

Probably 4 people who just wanted to screw up the poll.

[quote="AIME15"][quote="johnk4521"]The index will be 197.5[/quote]

I pray you are right.[/quote]

haha... HA

I got a 196, according to the answers I've seen thrown around so far. A few stupid mistakes too many. And missing #1 on the AIME... is like missing a layup or the first question on millionaire.

[quote="Brut3Forc3"]... Who the heck picked >230? If 8 on the AIME is supposed to be a usually safe score, then 150+80 is... uh...[/quote]

That's exactly what I got o.0. This year was really easy, though, I got lucky.

And I have no idea how to prep for USAMO.

How are these polls turning out so nicely...

I hope that the AMC computers glitch and the index is 194.5. Or maybe I can bribe the committee with rice krispie squares.

Considering the AIME I, I'm guessing the index should be around 195-205

The poll is suppose to be what you think the index will be, not what you got.
I don't think there has ever been a 230+ index.

brightzhu, a couple years ago, there was an index of 233 or 233.5 (somewhere around there).  Of course, since the number of qualifiers has been upped to 500, I believe the two cutoffs were 197.5 and 202.5.

Anyone 220+ is very safe (safe enough that you should be looking at the olympiad forums than lurking here searching for cutoff predictions, at least).

I'm going to guess a cutoff of 6, from all the mistakes people made.

I believe.....(since last year's index was between 200 and 205)

this year's index shall be in the similar range.

Last year, I got  150+10*8 = 230

This year, I got  144(A stupid mistake in the AMC10A :(  ) + 10*11 = 254

I personally think that this year's AIME is somewhat more "confusing"

Floor score - 6, maybe 7.

[quote="Goldey"]I will join you guys in your prayer.

127.5 on AMC 10 cuz of stupid mistakes,
got a 7 instead of 9 on AIME because 9*4=36 and i didnt read a problem fully

So....
please be 197.5!!![/quote]

eh, 9*4 DOES equal 36...:P

1 in a 100 chance of the floor being 192.5 or lower.  Didn't that happen 2 years ago, though?

100th vote :D, I think everyone here is over thinking the index. Everyone who actually posts here are probably going to get higher scores than average ...

minumum index score=300 :P
ok seriously i think it will be close to 200


vector
graph theory
combinatorics unsolved
combinatorics

[quote="Zorro"]There is a graph with $2^n+1$ vertices and each edge is in one of $n$ colors. Prove that there is a cycle with odd lenght and in only one color[/quote]

For sure there is some missing hipothesis in your problem. Consider the following graph: $V=\{1,2,\dots,2^n+1\}$, and $E=\{\{x,y\}\subseteq V : |x-y|=1\}$, and color the edges as you want. You don't have a cylcle with odd lenght in only one color, because, of course, you don't have a single cycle in the graph.

He's talking about the complete graph with $t=2^n+1$ vertices. It has $\frac{t(t-1)}2$ edges, and lots and lots of cycles :).

yes, sorry
there is a edge between each two vertices
this is a generalisation, in original problem there is $n=9$

suppose not,i.e. suppose that there is a coloring of the edges of $K_{2^n+1}$ with the $n$ colors $1,2,\ldots,n$ such that there is no monochrome odd cycle.in other words $K_{2^n+1}$ can be decomposed into $n$ edge-disjoint bipartite graphs.For each color $i$ and the bipartite graph $G_i$ determined by the edges of color $i$,let $c_i$ be a proper 2 coloring of all the vertices. for each vertex,now consider the vector $\mathbf{v_c}=(c_1(v),c_2(v),\ldots,c_n(v))^{T}$. since every edge belongs to one of the color classes, for any vertices,$v\neq w$, there exists some $i$ such that for the graph $G_i, c_i(v)\neq c_i(w)$.hence,$\mathbf{v_c}\neq \mathbf{w_c}$.so the vectors $\mathbf{v_c}$ distinguish between all the vertices.but there are only $2^n$ such possible vectors where as there are $2^n+1$ vertices and that is a contradiction.

thank you
but could you explain to me what does "proper 2 coloring of all the vertices" mean??

a proper 2-coloring refers to coloring the vertices in such a way that adjacent vertices receive different colors.in a bipartite graph,since there are 2 color disjoint groups with the edges going only from one vertex group to the other we can give one color to the first group and another color to the second group thereby coloring the graph with just 2 colors and haveing adjacent vertices have different colors.


Given triangle $ ABC$ with $ \angle CAB\equal{}30^{\circ}, \angle ABC\equal{}27,5^{\circ}$. Point $ D$ is inside $ ABC$ such that $ \angle DAC \equal{}25^{\circ}, \angle DBC\equal{} 25^{\circ}$. Find $ \angle DCA$.

Let $ P$ a point inside $ \triangle ABC$. Let $ (A): (B) \equal{} (a_1, a_2, a_3): (b_1, b_2, b_3)$ where $ m(\angle BAP) \equal{} a_1, m(\angle ACP) \equal{} a_2, m(\angle CBP) \equal{} a_3$ and  $ m(\angle CAP) \equal{} b_1, m(\angle BCP) \equal{} b_2, m(\angle ABP) \equal{} b_3$.

It can easily be shown that elements of $ A$ and $ B$ are interchangeable in their own sets $ A$ and $ B$. e.g. if $ (A): (B) \equal{} (a_1, a_2, a_3): (b_1, b_2, b_3)$ is a model, then $ (A): (B) \equal{} (a_1, a_3, a_2): (b_1, b_2, b_3)$ is also a model.

 I have found five models.

1) $ (A): (B) \equal{} (t, 30^\circ, 90^\circ\minus{}3t): (2t, 30^\circ \minus{} t, 30^\circ\plus{}t)$

2) $ (A): (B) \equal{} (t, 30^\circ\minus{}2t, 90^\circ\plus{}t): (2t, 30^\circ \minus{} 2t, 30^\circ)$

3) $ (A): (B) \equal{} (t, 60^\circ\minus{}4t, 60^\circ\plus{}t): (3t, 30^\circ \minus{} 2t, 30^\circ\plus{}t)$

4) $ (A): (B) \equal{} (t, 30^\circ\minus{}t, 90^\circ\minus{}t): (2t, 30^\circ \minus{} t, 30^\circ)$

5) $ (A): (B) \equal{} (t, r, 90^\circ\minus{}t\minus{}r): (t, r, 90^\circ\minus{}t\minus{}r)$

Your problem is an element of the model type $ 2$, $ (A): (B) \equal{} (t, 30^\circ\minus{}2t, 90^\circ\plus{}t): (2t, 30^\circ \minus{} 2t, 30^\circ)$ where $ t\equal{}2.5^\circ$. $ (A): (B) \equal{} (2.5^\circ , 25^\circ , 92.5^\circ ): (5, 25^\circ, 30^\circ)$. So the desired angle is $ \boxed{30^\circ}$.

The problem is similar to the problem $ (A): (B) \equal{} (x, 2t, 30^\circ \minus{} 2t): (30^\circ\minus{}2t, t, y)$ where $ x$ and $ y$ are unknown and $ x$ is the desired angle.

(The configuration is according to my previous post - the where statement in the first sentence of the previous post)

Let $ A'$ be the symmetric of $ A$ according to $ CP$. $ \angle BCA' \equal{} t$, $ \angle BPA \equal{} 60 \minus{} t$, $ \angle BPA' \equal{} t$. So $ \triangle A'PA$ is equilateral and $ BPCA'$ is cyclic quadrelateral because $ \angle BPA' \equal{} \angle BCA' \equal{} t$. So $ \angle BA'P \equal{} t$ and $ \angle AA'B \equal{} 60\minus{}t$. This means $ B$ is the incenter of $ \triangle A'PA$. So $ \boxed {\angle BAP \equal{} x \equal{} 30^\circ}$

PS. all angles are in degrees.


factorial
LaTeX
number theory
prime factorization

Just wondering: Is there a nice way to do this without [hide=" this solution?"] Listing the prime factorization, taking out $ 2^7\cdot5^7$, and calculating it $ \bmod 10$?[/hide]
When $ 30!$ is computed, it ends in $ 7$ zeros. Find the digit that immediately precedes those zeroes.

LaTeX suggestions: \cdot looks much nicer than * for multiplication.  Also, the align* environment can be used to make the sorts of simplifications you just did look nicer:

\begin{align*}
\frac{30!}{10^7} & \equiv 3\cdot 7\cdot9\cdot2\cdot11\cdot12\cdot13\cdot14\cdot3\cdot16\cdot 17\cdot18\cdot19\cdot4\cdot21\cdot22\cdot23\cdot 24\cdot26\cdot27\cdot28\cdot29\cdot6\mod{10} \\
& \equiv 2^3 \cdot 3^4 \cdot 4^3 \cdot  6^3 \cdot 7^3 \cdot 8^2 \cdot  9^3 \mod{10} \\
& = 8 \cdot 1\cdot 4 \cdot 6 \cdot  3 \cdot 4 \cdot  9 \mod{10} \\
& = 6 \mod{10}
\end{align*}


A certain organism begins as three cells. Each cell splits and becomes two cells at the end of two days. At the end of another two days, every cell of the organism splits and becomes two cells. This process lasts for a total of 8 days, and no cells die during this time. How many cells are there at the end of the $ 8^\text{th}$ day?

This process repeats 4 times. Since after the first time the number of cells is $ 2\cdot 3$, and it doubles each time, after the fourth there are $ 3 \cdot 2^4\equal{}\boxed{48}$ cells.


email MATHCOUNTS inequalities geometry \/closed

Hi everyone,

We regret that some of the database got lost in the attack last week, including some of our moderators.

Therefore, we ask any person wishing to moderate a forum to please pm one of the board administrators.

Please include a short list of reasons why you should be a moderator, and the forums you'd like to moderate.

Thanks.

In particular, if you are interested in moderating a forum you can also email me directly at crawford@artofproblemsolving.com.  (Let Val sleep for now.)

I'm willing to volunteer for a few more boards if there's need of more volunteers. I definitely understand letting people who aren't mods have a chance though...it's just if there's a need.

[quote="tokenadult"]I notice that there is a new moderator on Getting Started, PavolWavol, and I haven't seen the new moderator's self-introduction. (It seems that this is a brand-new username, but possibly belonging to a long-time participant.) I always welcome new moderators, and it would be good for management to encourage new mods to introduce themselves to their boards.

AoPS forums home page is my browser home page, so as you can imagine I have been very eager to see the boards up and working again. Congratulations to management on the hard work involved in putting together the new and improved forums.[/quote]

PavolWavol has been released from his duties as of now :D because it was just a test user.

Finally someone's not complaining about the minor stuff, and looks on the bright side ...  :) you're welcome.

[quote="Myth"][quote="Valentin Vornicu"]Therefore, we ask any person wishing to moderate a forum to please pm one of the board administrators.[/quote]
Very strange policy.[/quote]Ok it may sound strange. Let's make it clear. Not necesarilly all people applying for this will become moderators, and even if they become, they are still in "probation" time.
That's all.

[quote="Myth"][quote="Valentin Vornicu"]Therefore, we ask any person wishing to moderate a forum to please pm one of the board administrators.[/quote]
Very strange policy.[/quote]

I think this an excellent policy. All people who are motivated and want to get involved and push the boards to new fronties are welcome. I do not like it if people dislike things but don't suggest better ideas.

[quote="Megus"]Well, Valentin made me a new moderator today so would you like me to introduce ? :D[/quote]

I listen...  ;)

[quote="Megus"]that  ;)  stress me a bit   :P[/quote]

You have curious habits.  At first you suggest it and then you don't take up the opportunity to introduce yourself.  :?

So I should introduce myself in the MathCounts forums?

If you want to it's ok, but it's not mandatory. Same goes for Megus.

Ok, I wanted to introduce myself [orl: that  ;)  just looks a bit strange while taking into account whole context, really   :) ]. and here it goes, so:

I live in Poland near Warsaw, capital city. I'm 16 and attend to Staszic High School in Warsaw - not yet participated in IMO. Obviously I love maths especially inequalities, though geometry is still my weakest area.
Any questions ?   :D

In that spirit, I'd also request like to request your billing address...for future reference of course...

 :lol:

[quote="orl"][quote="pbornsztein"]You may laugh Myth, but look at you : you are now a green man here  :D  :P
Welcome to our new green friend! :)

Pierre.[/quote]

Finally he became a moderator (which should have taken place much earlier). :) Now we can ask him for his credit card too...  ;)[/quote]And with him becoming a moderator this Moderator topic ends. For now, no other moderator positions are open, so the topic is closed.


Is there a way to solve an equation (for $x$) in the form

$ab^x + cd^x = e$, where $a$, $b$, $c$, $d$, and $e$ are constants?

for example

$3(4)^x + 5(9)^x = 56.41$

Is there an algebraic way to solve that?

If $b=d$ than it is straightforward. If $1<b<d$ you can transform the equation in $a+c \left (\frac{d}{b}\right )^x=e\left (\frac{1}{b}\right ) ^x$ on the left side you have an increasing function while on the right side you have a decreasing one so it has an unique solution. The other cases are treated simillar.

Pay attention, this is a nice technique

Suppose that $ab^x+cd^x=e$. Raise $10$ to each side:
$10^{ab^x}10^{cd^x}=10^e$ The properties of exponents let us break down the barriers in this problem as we rewrite this as
$10^{(bd)^{x}}=10^{e-a-c}$. Then $(bd)^x={e-a-c}$ so that $x=\log_{bd}(e-a-c)$

[quote="drunner2007"]
$10^{ab^x}10^{cd^x}=10^e$ The properties of exponents let us break down the barriers in this problem as we rewrite this as
$10^{(bd)^{x}}=10^{e-a-c}$.[/quote]

How does this step work?

It doesn't.

For example, suppose we wanted to solve the equation (a=1,b=1,c=1,d=2,e=5) 2^x+1=5.  The answer is 2, of course, but drunner's method suggests that it should be ln(3)/ln(2).

It doesn't work, coz it's wrong :)

Namely, $10^{ab^x}=10^e$ doesn't imply $10^{b^x}=10^{e-a}$, rather $10^{b^x}=10^{e/a}$

But then you can't pull roots separately - $10^{ab^x}10^{cd^x}$ means that you have to pull $a$-th root from the first factor and $c$-th root from the second factor in order to simplify.

I'm beginning to have VERY serious doubts about drunner being a math teacher...

My bad man; people make mistakes. Who said I was going to be a math teacher?


Can someone please show methods of solving related rates and detailed examples. I tried learning it but i couldnt figure it out. thanks alot.
here are few examples:
Ex. 1: A snowball of spherical shape is melting. how fast is the radius changing when it is 2 inches and the snow is melting at a rate of 10 cubic inches a minute?       Ex. 2: Grain is being poured on a pile in the shape of a circular cone. Radius and height of that cone are equal at all times. At what rate is the height changing when the pile is 5 meters high and grain is being poured at a rate of 10 cubic meters a minute?                                                                                Ex. 3: Leibniz is running along the positive x-axis towards the origin. Newton is trying to hide from Leibniz, positioning himself on the y-axis so that he is always hidden behind the trunk of an apple tree at the point (9,18). How fast is newton running when leibniz is at the point (36,0), running at a speed of 9 meters per second?

Here is how I do related rates:

first list everything you know about the problem.

let x be the radius of the sphere, y is the volume, and t is time (we are assuming that the snowball will stay a sphere even though it probably wouldn't in real life.)

we are trying to find dx/dt when x=2.

dy/dt=-10 when x=2

now you should have enough info to solve the problem

y=4/3(pi)x^3
dy/dt=4(pi)x^2 (dx/dt)
-10=4(pi)(4)(dx/dt)
-5/8(pi)=dx/dt

and the problem is solved...now try the rest on your own


Hey guys, for my homework, I have to do some problems involving Sine, Cosine, Tangent on right triangles. If you had some time could u plz help me explain the functions and stuff.

1) We'll just use this as an example

You have a triangle with sides 4.5, 9, and 10.1. the hypotenuse is facing to the top right.. Angle A is not 90 degrees and is at the top between side 4.5 and side 10.1. Angle B is at the bottom between sides 10.1 and 9. Coul l du plz help to explain step by step through the problems

TY :D  :D

Wait, do you need to find the angles?
If so, just figure out $ \sin{a}$ and $ \sin{b}$, and take Arcsin (your calculator should have this as $ sin^{\minus{}1}$.


Έστω ισοσκελές τρίγωνο $ ABC$ με $ AB\equal{}AC$ και $ \hat{A}$ γωνία ίση με $ 20^{o}$ μοίρες. Θεωρούμε σημείο $ D$ πάνω στην πλευρά $ AC$ τέτοιο ώστε $ AD\equal{}BC$. Να υπολογιστεί η γωνία $ \hat{BDC}$.

[hide]Είναι μία καλή ευκαιρία να δει κανείς τη λύση, μέσα από το σχήμα ενός κανονικού [b][size=100]9-γώνου[/size][/b]. Προκύπτει εύκολα ότι $ \angle BDC \equal{} 30^{o}$.[/hide]

Κώστας Βήττας.

Απ το σημείο $ D$ φέρνουμε ευθεία τέτοια ώστε το τρίγωνο $ EAB$ να είναι ισόπλευρο. Τότε έχουμε ότι $ EA \equal{} ED \equal{} EB \equal{} AB$. Άρα απο αυτή την ισότητα προκύπτει ότι $ \angle EAD \equal{} \angle ADE \equal{} 80^{o}$ και $ \angle EBD \equal{} \angle EDB \equal{} 70^{o}$. Άρα $ \angle BDC \equal{} 30^{o}$

Καμιά άλλη όμορφη λύση??? :)


\[ \vert (a\plus{}b)(c\plus{}d)(ab(a\minus{}b)\plus{}bc(b\minus{}c)\plus{}cd(c\minus{}d)\plus{}da(d\minus{}a))\vert \leq A(a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}d^{2})^{5/2}\]

Find the least value A can have while the inequality makes sense for any real number \[ a,b,c,d\].

[quote="kyungissac"]
\[ \vert (a \plus{} b)(c \plus{} d)(ab(a \minus{} b) \plus{} bc(b \minus{} c) \plus{} cd(c \minus{} d) \plus{} da(d \minus{} a))\vert \leq A(a^{2} \plus{} b^{2} \plus{} c^{2} \plus{} d^{2})^{5/2}\]
Find the least value A can have while the inequality makes sense for any real number
\[ a,b,c,d\]
.[/quote]
$ A\equal{}\frac{\sqrt{2}}{8},a\equal{}d\equal{}0,b\equal{}\minus{}c$
it that right ?

I will first prove the inequality when $ A\equal{}\frac{32}{25\sqrt 5}$. Let
$ f(a,b,c,d)\equal{}1024(a^2\plus{}b^2\plus{}c^2\plus{}d^2)^5$ $ \minus{}3125((a\minus{}b)(a\minus{}c)(ab(a\minus{}b)\plus{}bc(b\minus{}c)\plus{}cd(c\minus{}d)\plus{}da(d\minus{}a)))^2$.
Then
$ f(a\plus{}d,b\plus{}c,\minus{}b\plus{}c,\minus{}a\plus{}d)\equal{}1024(x\plus{}y\plus{}p)^5\minus{}12500y(x^2\minus{}px\plus{}q)^2$
where $ p\equal{}2(c^2\plus{}d^2)$, $ q\equal{}(c^2\minus{}d^2)^2$, $ x\equal{}(a\plus{}b)^2$, $ y\equal{}(a\minus{}b)^2$. Let $ g(x,y,p,q)\equal{}f(a\plus{}d,b\plus{}c,\minus{}b\plus{}c,\minus{}a\plus{}d)$. Then
$ \frac{\partial g}{\partial p}(x,y,p,q)\equal{}5120(x\plus{}y\plus{}p)^4\plus{}25000xy(x^2\minus{}px\plus{}q)$ $ \geq 61440 x^2yp\minus{}25000x^2yp \geq 0$.
Therefore, we only need to prove the inequality when $ p^2\equal{}4q$. Then
$ \frac{\partial g}{\partial p}(x,y,2p,p^2)\equal{}10240(x\plus{}y\plus{}2p)^4$ $ \minus{}50000(p\minus{}x)^3y\geq y(40960(x\plus{}2p)^3\minus{}50000(p\minus{}x)^3)$ $ \equal{}240y(1157p^3\plus{}2673p^2x\plus{}399px^2\plus{}379x^3) \geq 0$.
Also, $ f(x,y,0,0)\equal{}4(x\minus{}4y)^2(256x^3\plus{}203x^2y\plus{}88xy^2\plus{}16y^3)\geq 0$. Therefore, we prove the inequality when $ A\equal{}\frac{32}{25\sqrt 5}$.
If $ A<\frac{32}{25\sqrt 5}$, then $ a\equal{}3$, $ b\equal{}1$, $ c\equal{}\minus{}1$, $ d\equal{}\minus{}3$ is a counterexample.


what is the gcd of two negative numbers?

and what is the gcd of a negative number and a positive number?

if a problem says that the gcd of two integers is 1, does that mean that they are both positive? or can they be negative?

Consider for example, -4 and -12.

The GCD of these numbers logically would be 4, just as if the numbers were both positive.

Another example, -6, -28

The GCD of these numbers is 2 as it is greater than all of the negative factors and 1.

-3, 6:

The GCD is 3 because it evenly divides -3 (by -1) and 6 (by 2). As we see, really, the GCD is is the same as if both numbers were positive.

In problems in which it tells you the GCD is 1, it is usually implied that the numbers are both positive. It would be unusual and tricky for them to include negative numbers as well. But logically, yes, I guess both numbers could be negative :)


number theory
greatest common divisor
induction
number theory solved

After, the problem GCD by Alekk, let's give another one with the same taste :
Let n,k be two nonnegative integers, with n \geq k. Dtermine the gcd of the numbers C(n,k), C(n+1,k),..., C(n+k,k).

Pierre.

Denote by di GCD( C(n,i),...,C(n+i,i)) for i=0,k
It can easily be proved by induction from k to 1 that
dk | dk-1 | ... | d1|d0=1, so dk=1.


Let $ F_0\equal{}0, F_1\equal{}1, F_k\equal{}F_{k\minus{}1}\plus{}F_{k\minus{}2},\, k\ge2,$ be the Fibonacci sequence. Prove that for every positive integers $ k\ge n$ the number $ 2^kF_{3n\plus{}2k}\plus{}F_{k\minus{}n}$ is divisible by $ 5.$

For $ n \equal{} 1$ and $ k \equal{} 2$, $ 2^k F_{3n \plus{} 2k} \plus{} F_{k \minus{} n} \equal{} 2^2 F_7 \plus{} F_1 \equal{} 4 \cdot 13 \plus{} 1 \equal{} 53$.

[quote="rogue"]Let $ F_0 \equal{} 0, F_1 \equal{} 1, F_k \equal{} F_{k \minus{} 1} \plus{} F_{k \minus{} 2},\, k\ge2,$ be the Fibonacci sequence. Prove that for every positive integers $ k\ge n$ the number $ 2^kF_{3n \plus{} 2k} \plus{} F_{k \minus{} n}$ is divisible by $ 5.$[/quote]

[quote="nsato"]For $ n \equal{} 1$ and $ k \equal{} 2$, $ 2^k F_{3n \plus{} 2k} \plus{} F_{k \minus{} n} \equal{} 2^2 F_7 \plus{} F_1 \equal{} 4 \cdot 13 \plus{} 1 \equal{} 53$.[/quote]

Sorry :blush:  Correct version is:

[quote=" "]Let $ F_0 \equal{} 0, F_1 \equal{} 1, F_k \equal{} F_{k \minus{} 1} \plus{} F_{k \minus{} 2},\, k\ge2,$ be the Fibonacci sequence. Prove that for every positive integers $ k\ge n$ the number $ 2^kF_{3n \plus{} 2k} \plus{} 3F_{k \minus{} n}$ is divisible by $ 5.$[/quote]


The question is simple: how could we know whether a problem which used to be open is solved or not?

I believe that there must be a site listing at least solved problems from very famous books, like for example Guy's "Unsolved problems in number theory" and Shanks's "Solved and unsolved problems in number theory". Does anybody know something like that?

Well it's not like Wiles's proving Fermats last would have gone unnoticed. Just Google it I guess :lol:


Harvard college calculus AMC USA(J)MO USAMO AIME

I want to have cool skills, like bowhunting skills, nun-chuk skills:
Alright, I am going off to Harvard this fall, and I will probably go for econ.  By the time I get there, I want to have International Math Olympiad skills.  I will study however much and whatever you tell me.

Has anyone been a part of any country's International Math Olympiad Team?  If so, tell me how I can study.  I can study up to four hours a day if it comes down to it.  I'm a beginner in a Calculus BC Class.

This would probably be better in the IMO - related boards.

4 hours a day will not be nearly enough to make the IMO team for the US if you've never done problem solving. It takes years and years to develop the skills and experience necessary even to make USAMO, much less IMO.

What if I'm at the level where I can solve about 4 AIME questions

I'm a rather quick learner, and I got this far by myself, I never had any special classes or extra studying

I'm also going to college but I want to be able to solve all 15 AIME questions

What would you recommend?

Put lots of time into it.  There are no true shortcuts to becoming good at math.  If you want to solve all 15 AIME problems, you need to a) be able to solve all of them at an intellectual level, b) have the experience necessary to catch all of your mistakes.

If I were to write the AIME today, I'd probably score about an 11 (more some years, less in others).  I've been able to score 4 on an AIME since I was in Grade 10 (three years ago) and I've done a lot of math since then (not necessarily problem solving related).  So you're looking at roughly a thousand hours of math.

[quote="Indian"]I want to have cool skills, like bowhunting skills, nun-chuk skills
Alright, I am going off to Harvard this fall, and I will probably go for econ.  [/quote]

Well Indian, I am very bad with the bow (basically because I have used it just a few times in my life and I don't practice), but I am pretty good with numchakus. It is not hard. You know, if you practice you become very good quickly. You know, it is a discipline that has its own punishment. If you skrew up the technique, probably some place of your own body will receive a strike  :D , what is pretty anoying, therefore you will learn quickly how not to do that. Also, in the club where I practice, you'll receive 25 push-ups if your numchacku flies our of your hands, and 100 if you hit someone! that is how you lear fast  :D . Any way, any of those is not hard to learn, but there is no substitute to the regular practice and hard work.

For sure in Harvard there must be a couple of clubs where you can practice any of those.

[quote="Indian"]By the time I get there, I want to have International Math Olympiad skills.  I will study however much and whatever you tell me.[/quote]

What for? I mean, I participated in a few of those kind of contests (well, always at a level lower than the IMO) and I loved it while it lasted. Now, a few years after that, I discovered that I enjoy a lot more to solve the problems without the presure that I have just sixty or ninty minutes to solve it or I will skrew it up. Now, I can sit calmed, study the problem, see where it come from, interpret it in different ways, and enjoy it piece by piece, not having to swallow it all at once.

If you are going to be in Harvard, well I am sure that you can find some nice elementary but challenging courses in Olympiad related subjects. Also, you can work problems by your own, or even you can do your minor in Math if you like. But as MithsApprentice says, that is not something that you are going to learn from now to September, if you have never done problem solving before, not matter if you spend your last vacations before college doing only math problems! I don't want to discourage you, but even when there is no substitute for the hard work, there is also no substitute for the experience, and if you don't have it, well, it is the first thing you have to look for.

[quote="Indian"]Has anyone been a part of any country's International Math Olympiad Team?  If so, tell me how I can study.  I can study up to four hours a day if it comes down to it.  I'm a beginner in a Calculus BC Class.[/quote]

Calculus and Olympiads are totally different. Any way, I am sure you will find hundreds of people here that have gone to IMO at least once! :D .

Wait...you're going to Harvard next fall? And you generally score a 4 on AIMEs?

Well assuming you didn't score an 8 or above on this year's AIME from what you've told us, then there's really no way you could make IMO. IMO is only for high school students and you must qualify for USAMO first. Then you have to make a great score on USAMO to be selected as a Winner, make a good score on the TST, go through MOP, and then hit IMO. However, since this is your last year of high school that will be impossible. Sorry buddy.

However, I (and I'm sure everyone else as well) do recommend you to practice for the Putnam.

Thanks for the responses!  First of all, I have only posted on the original post.  Someone else seems to have acted like they continued my post below, but I only was the OP...so this is the only screen name I use.  The other person had valid questions too, but they are not me :)

I now understand that this can be more difficult than I thought.  Let me just put it this way: I want to go in the direction of Math Olympiad kids, although I know that at this point, I cannot get close to as far as they did.  What direction do I go in?  I am not doing this for competitions, I am doing this because I want a better grasp of math.  I can buy books, and at this point, what is the best way to advance my skills with only four hours of time every day?

My expertise is geography and government, so math hasn't come into play that much.  I am just now starting to realize the value of math.

Really I think that you do not comprehend what the level of preparing an IMO requires. It is impossible for you at this age to become in less than half a year (ie till fall) and IMO level student (or at least that is what I think given your background you presented here). Even if you work for 8, not 4 hours a-day, it would still be almost impossible to reach that level.

It requires time to process all the informations, and a lot of practice (and I mean tons :) ). Furthermore, I don't think it will be of much use to you now (you are already, as you say, accepted to Harvard) so I think you should spend those 4 free hours you have doing something more usefull for your future economical carrer :)

[quote="Valentin Vornicu"]I think you should spend those 4 free hours you have doing something more usefull for your future economical carrer :)[/quote]

Practice those ultimate frisbee moves.

[quote="Indian"]
I now understand that this can be more difficult than I thought.  Let me just put it this way: I want to go in the direction of Math Olympiad kids, although I know that at this point, I cannot get close to as far as they did.  What direction do I go in?  I am not doing this for competitions, I am doing this because I want a better grasp of math.  I can buy books, and at this point, what is the best way to advance my skills with only four hours of time every day?
[/quote]

You are underestimating the difficulty level of IMO. Consider the USA selection process, first there are around 100,000 High School students take the AMC, around 5% of those people make it to the second level(called AIME), then out of all the AIME qualifiers, only around 250 will qualify for USA Math Olympiads(which is a similar test to IMO). And then they select the top 12 scorers on the USAMO to take the Team Selection Test, and and they take the top 6 for the IMO team. Well the amount of work it takes.... after 4 years of practice you might have a slight chance, I'm not sure.

[quote="beta"]Consider the USA selection process, first there are around 100,000 High [/quote]

Actually I believe it's more like a quarter million. But regardless, working for 4 hours per day every day for 4 years...you might have a chance if you're also extremely talented at problem solving. Not all problem solving can be trained.

Once again thank you for the suggestions, and I apologize if I come across as underestimating the IMO.  I am not very familiar with the competition (as you can tell), so I had no idea how difficult it was.

OK, so now I know that that is out :).  But, now that we are on the topic, what can I do for four hours every day to improve my general math skills?  I mean, even for those of you who have participated in IMO's, what advice would you give in terms of any books that work well in particular.

Thanks!  And BTW, I think my Ultimate skills are nearly as hopeless as my bowhunting ones.  Hey, well at least I got a nice bike.

Vote Pedro!

Well, i think that i am hopeless even to come to the CMO. But i love math, i study math after school by myself but sometimes i really don't get the way. I don't know what and how should i study. I do not expect to be able to solve the IMO problems but i will try to appraoch it. So can you tell me some of the hints of studying for Math Competition?

There have been numerous discussions about this topic. Use the [url=http://www.mathlinks.ro/Forum/search.php]search[/url] function :P :)


summer program
PROMYS
quadratics
AMC
AIME
number theory

I'm thinking of applying for the PROMYS summer program. I'm a little skeptical about the mere 1 hour a day of instruction. Also, I've heard that it is geared towards deriving some elementary number theory results. I guess I would like to hear from alumni/counselors or other people that know something about it. More specifically, would it be worth it for a person who has a little background, i.e. knowledge of Euler's totient theorem, quadratic reciprocity, etc.

I got an application thingy in the mail too, let's just say I didn't like the application questions too much. I didn't feel any of the questions were particularly new appealing or elegant and I wasn't planning on going to a math summercamp in the first place. However, a plus is that their advanced seminars seem sufficiently enough challenging if not too much, they have a good reputation, and they said it was because of my AIME score, not my AMC score which gives me more faith in the talent pool of applicants.

I noticed that you commented on my PROMYS post in the Washington thread. I'll give my thoughts on the 1 hour of instruction per day here.

In reality, the instructive portion is far less important than the problem sets. Learning number theory at PROMYS is about doing the problem sets; you'll realize that the morning lectures are a sort of introduction and that the problem sets really push you to your limits. Having other students to work with on the problem sets is one of the biggest benefits of PROMYS, and you'll learn loads of stuff as well as how to really rigorize your proofs. I definitely would not worry about the amount of teaching not being enough.

If you have a pretty good number theory background, like knowing the totient theorem and quadratic reciprocity, the number theory stuff alone probably won't be quite worth it. But of course that's what the advanced seminars are for, and those are REALLY challenging from what I saw. The topics are also quite interesting, and personally I would go for it if you know a solid amount of NT.

And if you think you will have too much extra time since you only have a single hour of instruction, don't be fooled. The NT problem sets themselves can easily take an entire day, and you'll want time to actually sleep/shower/relax a little. And again, the advanced seminar problem sets are pretty ridiculously tough.

@Ignite168: Personally, I found the application problems somewhat cool, but PROMYS itself was way cooler. It brings back good memories :P.

I went knowing no number theory whatsoever, and I had a blast.  The morning lectures weren't really a time to learn a great deal: it was more like a prompt, or going over what came a few days before.  But the problem sets were extremely instructive, and took a great deal of thought.  Not to mention you learn rigor.  The group-theoretic approach to number theory was nice as well.

Yet I can definitely vouch taht the advanced seminars are, indeed, advanced, and will challenge you.  I sat in on one seminar: Combinatorics, and I can say the problem sets were challenging indeed, taking mroe time than the number theory.  Although it seems like you'll have lots of free time, I don't think you'll ever find yourself bored.  I can vouch that after the first two weeks, I was up past midnight working on problem sets every night (though maybe it was because I had such a bothersome roommate who brought such bothersome friends :))

Not to mention it's really FUN.  You have complete freedom to explore Boston, there are activities, you have lots of other kids and an entire lounge every day, great food, and lots of math.  What more could you ask for?  (also know that there is what is called "Exploration Lab" for first years: you get a general class of open-ended problems, and you join with a couple other students to work on it and develop a lecture and paper on your findings)

[quote="Phelpedo"]I was under impression from an older thread in this forum that first years were not allowed to attend the advanced seminars? I wouldn't really want to go to relearn basic number theory. Also, what kind of questions do the problem sets contain? Are these problems I couldn't get just by going through Engel or 104 Number Theory Problems?

@ignite: that's funny because I received an invite before I took the AIME...[/quote]

I just received it yesterday and it said congratulations high scoring AIMEr or something.

On the application it said something like sufficiently advanced students and returning students and then the blurb about the Advanced Lectures. I'm sure if you're excelling in basic NT and you say something, they'll listen as they 1) want you to go back and 2) want to challenge and motivate you.

[quote="Phelpedo"]I was under impression from an older thread in this forum that first years were not allowed to attend the advanced seminars? I wouldn't really want to go to relearn basic number theory. Also, what kind of questions do the problem sets contain? Are these problems I couldn't get just by going through Engel or 104 Number Theory Problems?

@ignite: that's funny because I received an invite before I took the AIME...[/quote]

That's vaguely true. There's really nothing barring first years from the advanced seminars, but a [b]LOT[/b] of them (definitely more than 3/4 of them) are busy enough with the NT sets and possibly an Exploration Lab. Even the best first years can only handle one advanced seminar, and the mere thought of a first year staying in two advanced seminars for more than a week is ridiculous.

The NT is more than what I would think of as "basic", since it covers quadratic residues and even a bit of stuff related to abstract algebra. Only a small portion of it is covered in AoPS Volume 2 or Engel, so it's extremely rare for a first year to already know all of the material (I think that there were only 2 last year and 0 the year before that. At least that's how many were making a big deal about it.)


1. For triangle ABC with AB=4, AC=5, and median AD=3, construct the triangle.

2. For triangle ABC with angle A=33, angle B=55, and the angle bisector of angle C with length 5, construct this triangle.

[hide="1,calculating way"]

$ AD\equal{}\frac{\sqrt{2b^2\plus{}2c^2\minus{}a^2}}{2}$

$ a\equal{}\sqrt{46}$, now you know all 3 sides.
[/hide]

[hide="2"]

Construct any triangle with those angles, draw that bisector, and then construct your triangle using similarity. [/hide]

[quote="Math Champion"]You can't draw the bisector. I forgot to say that you only have a compass and a straightedge.[/quote]

Why not? When you have a triangle, I think you can construct it with a compass and a straightedge. I said to construct any triangle with these angles, and in it construct the bisector, and compare with 5, do some similarity, and that's it.

Wow! Thanks a lot! That site is awesome.


geometry
geometric transformation
rotation
analytic geometry
parallelogram
complex numbers
geometry unsolved

Let $ ABCD$ and $ BNMK$ be two non overlapping squares and let $ E$ be the mid-point of $ AN$.If the point $ F$ is the foot of perpendicular from $ B$ to the line $ CK$.Prove that $ E$,$ B$ and $ F$ are collinear.

[hide="Using complex numbers"]
Consider complex plane with origin at $ F$, $ FK$ be the real axis and $ FB$ be the imaginary axis.Let $ c,k,bi$ be the complex coordinates. The rotation of $ C$ with $ B$ as centre gives $ A$ so $ A$ has coordinates.$ a\equal{}b(1\minus{}i)\plus{}ci$ and similarly for $ N$ we have the coordinates $ n\equal{}b(1\plus{}i)\minus{}ki$.
The mid point of AN is $ e\equal{}\frac{a\plus{}n}{2}$ so $ e\equal{}b\plus{}\frac{(c\minus{}k)i}{2}$ which gives the result.
[/hide]

[geogebra]4ac5d57e585ffad830c579716ede503f5b1ea9e5[/geogebra]

Let $ \angle{CBK}\equal{}x$. Construct parallelogram $ ABNX$, with $ E$ as the intersection of the diagonals. Note that $ \angle{ABN}\equal{}180\minus{}x$ and $ \angle{BAX}\equal{}x$. By SAS congruency, $ \triangle{BCK}$ is congruent to $ \triangle{BAX}$. Thus,

$ \angle{FBE}\equal{}\angle{FBC}\plus{}\angle{ABC}\plus{}\angle{ABX}\equal{}90\minus{}\angle{BCK}\plus{}90\plus{}\angle{BCK}\equal{}180$.

$ QED$.


Could someone explain this to me or indicate some material to what look for?

$ x \equal{} \sqrt{n \plus{} \sqrt{n \plus{} \sqrt {n} \plus{} ... }} \equal{} \sqrt {x \plus{} n}$

if x is a limit?

$ x \equal{} \sqrt {n \plus{} \sqrt {n \plus{} \sqrt {n} \plus{} ... }} \equal{} \sqrt {x \plus{} n}$
$ x^{2} \equal{} n \plus{} \sqrt {n \plus{} \sqrt {n \plus{} \sqrt {n} \plus{} ... }} \equal{} n \plus{} x \iff x \equal{} \sqrt {x \plus{} n}$

LoL
I see, how idiot of me, thanks a lot.


$ \boxed{1}$ For $ x \equal{} \frac {1}{\sqrt {2 \minus{} \sqrt {3}}},\ y \equal{} \frac {1}{\sqrt {2 \plus{} \sqrt {3}}}$, find the value of $ (x \minus{} y)^2$.

$ \boxed{2}$ For $ x \equal{} \sqrt {3 \minus{} \sqrt {5}},\ y \equal{} \sqrt {3 \plus{} \sqrt {5}}$, find the value of $ x \minus{} y \plus{} 2x^2y \plus{} 2xy^2 \minus{} x^4y \plus{} xy^4$.

$ \boxed{3}$ Let $ abc\neq 0,\ a \plus{} b \plus{} c \equal{} 0,\ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$, find the values of the following expressions.

(1) $ bc \plus{} ca \plus{} ab$

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

(3) $ a\left(\frac {1}{b} \plus{} \frac {1}{c}\right) \plus{} b\left(\frac {1}{c} \plus{} \frac {1}{a}\right) \plus{} c\left(\frac {1}{a} \plus{} \frac {1}{b}\right)$

[quote="kunny"]$ \boxed{1}$ For $ x \equal{} \frac {1}{\sqrt {2 \minus{} \sqrt {3}}},\ y \equal{} \frac {1}{\sqrt {2 \plus{} \sqrt {3}}}$, find he values of $ (x \minus{} y)^2$.

$ \boxed{2}$ For $ x \equal{} \sqrt {3 \minus{} \sqrt {5}},\ y \equal{} \sqrt {3 \plus{} \sqrt {5}}$, find the value of $ x \minus{} y \plus{} 2x^2y \plus{} 2xy^2 \minus{} x^4y \plus{} xy^4$.

$ \boxed{3}$ Let $ abc\neq 0,\ a \plus{} b \plus{} c \equal{} 0,\ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$, find the values of the following expressions.

(1) $ bc \plus{} ca \plus{} ab$

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

(3) $ a\left(\frac {1}{b} \plus{} \frac {1}{c}\right) \plus{} b\left(\frac {1}{c} \plus{} \frac {1}{a}\right) \plus{} c\left(\frac {1}{a} \plus{} \frac {1}{b}\right)$[/quote]
[color=blue][b]All these problems are very easy, I am very lazy to find all the results of them. I, however, find out the result of the problem 1 be 2  :rotfl: . The others are just for high school students  :rotfl:  :rotfl:  :rotfl:  :rotfl: [/b][/color]

Maybe you are an adult, but you are rude. I don't post them only for you.
Don't cheat us! :mad:

Japanese Communities Moderator

kunny

[quote="kunny"]$ \boxed{1}$ For $ x \equal{} \frac {1}{\sqrt {2 \minus{} \sqrt {3}}},\ y \equal{} \frac {1}{\sqrt {2 \plus{} \sqrt {3}}}$, find he values of $ (x \minus{} y)^2$.[/quote]

[hide="Solution for the first problem"]
$ x\minus{}y\equal{}\frac{1}{\sqrt{2\minus{}\sqrt{3}}}\minus{}\frac{1}{\sqrt{2\plus{}\sqrt{3}}}\equal{}\frac{\sqrt{2\plus{}\sqrt{3}}\minus{}\sqrt{2\minus{}\sqrt{3}}}{(\sqrt{2\minus{}\sqrt{3}})(\sqrt{2\plus{}\sqrt{3}})}\equal{}\frac{\sqrt{2\plus{}\sqrt{3}}\minus{}\sqrt{2\minus{}\sqrt{3}}}{\sqrt{(2\plus{}\sqrt{3})(2\minus{}\sqrt{3})}}\equal{}\frac{\sqrt{2\plus{}\sqrt{3}}\minus{}\sqrt{2\minus{}\sqrt{3}}}{\sqrt{4\minus{}3}}\equal{}\sqrt{2\plus{}\sqrt{3}}\minus{}\sqrt{2\minus{}\sqrt{3}}$

$ (x\minus{}y)^2 \equal{} (\sqrt{2\plus{}\sqrt{3}}\minus{}\sqrt{2\minus{}\sqrt{3}})^2 \equal{} (2\plus{}\sqrt{3})\plus{}(2\minus{}\sqrt{3})\minus{}2\sqrt{(2\plus{}\sqrt{3})(2\minus{}\sqrt{3})}\equal{}4\minus{}2\sqrt{1}\equal{}2$

 :D
[/hide]

That's right. :lol:

$ (x\minus{}y)^2\equal{}x^2\plus{}y^2\minus{}2xy$ directly leads you rapid calculation.  :)

2.

We have $ x^2\plus{}y^2\equal{}6$
$ (x\minus{}y)^2\equal{}x^2\plus{}y^2\minus{}2xy\equal{}6\minus{}2.2\equal{}2$
$ x\minus{}y\equal{}\minus{}\sqrt{2}$, because $ x<y$

$ (x\plus{}y)^2\equal{}x^2\plus{}y^2\plus{}2xy\equal{}6\plus{}2.2\equal{}10$
$ x\plus{}y\equal{}\sqrt{10}$

$ x^2\plus{}y^2\plus{}xy\equal{}6\plus{}2\equal{}8$

So, $ x\minus{}y\plus{}2x^2y\plus{}2xy^2\minus{}x^4y\plus{}xy^4\equal{}\minus{}\sqrt{2}\plus{}2.2.\sqrt{10}\minus{}2.\minus{}\sqrt{2}.8\equal{}4\sqrt{10}\plus{}15\sqrt{2}$

Hope, there's no miscalculating :)

3.

(1)

$ ab\plus{}bc\plus{}ac\equal{}\frac{(a\plus{}b\plus{}c)^2\minus{}(a^2\plus{}b^2\plus{}c^2)}{2}\equal{}\minus{}\frac{1}{2}$

(2)

$ a^4\plus{}b^4\plus{}c^4\equal{}(a^2\plus{}b^2\plus{}c^2)^2\minus{}2(a^2b^2\plus{}a^2c^2\plus{}b^2c^2)$
$ 1\minus{}2[(ab\plus{}bc\plus{}ac)^2\minus{}2abc(a\plus{}b\plus{}c)]\equal{}1\minus{}\frac{1}{2}\equal{}\frac{1}{2}$

(3)

The problem is equivalent with

$ \frac{a^2b\plus{}ab^2\plus{}ac^2\plus{}a^2c\plus{}bc^2\plus{}b^2c\plus{}abc}{abc}\equal{}\frac{(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ac)\minus{}2abc}{abc}\equal{}\minus{}2$
 :)

That's correct. :lol:

Here is my solution:

Alternative Solution 1:

$ a\plus{}b\plus{}c\equal{}0\Longrightarrow a\plus{}b\equal{}\minus{}c,\ b\plus{}c\equal{}\minus{}a,\ c\plus{}a\equal{}\minus{}b$, we have

$ a\left(\frac{1}{b}\plus{}\frac{1}{c}\right)\plus{}b\left(\frac{1}{c}\plus{}\frac{1}{a}\right)\plus{}c\left(\frac{1}{a}\plus{}\frac{1}{b}\right)\equal{}\frac{b\plus{}c}{a}\plus{}\frac{c\plus{}a}{b}\plus{}\frac{a\plus{}b}{c}\equal{}\minus{}3$.

Alternative Solution 2:

$ a\left(\frac{1}{b}\plus{}\frac{1}{c}\right)\plus{}b\left(\frac{1}{c}\plus{}\frac{1}{a}\right)\plus{}c\left(\frac{1}{a}\plus{}\frac{1}{b}\right)\equal{}(a\plus{}b\plus{}c)\left(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\right)\minus{}3\equal{}\minus{}3$.

[quote="Arrange your tan"]2.

We have $ x^2 \plus{} y^2 \equal{} 6$
$ (x \minus{} y)^2 \equal{} x^2 \plus{} y^2 \minus{} 2xy \equal{} 6 \minus{} 2\cdot2 \equal{} 2$
$ x \minus{} y \equal{} \minus{} \sqrt {2}$, because $ x < y$

$ (x \plus{} y)^2 \equal{} x^2 \plus{} y^2 \plus{} 2xy \equal{} 6 \plus{} 2\cdot2 \equal{} 10$
$ x \plus{} y \equal{} \sqrt {10}$

$ x^2 \plus{} y^2 \plus{} xy \equal{} 6 \plus{} 2 \equal{} 8$

So, $ x \minus{} y \plus{} 2x^2y \plus{} 2xy^2 \minus{} x^4y \plus{} xy^4 \equal{} \minus{} \sqrt {2} \plus{} 2\cdot2\sqrt {10} \minus{} 2 \minus{} \sqrt {2}(8) \equal{} 4\sqrt {10} \plus{} 15\sqrt {2}$

Hope, there's no miscalculating :)[/quote]

IW@IT,

do not try to use a period for multiplications
as you did in your post.  Instead, use a
multiplication dot, asterisk, sets of parentheses,
or other grouping symbols as I have done in making
changes to your prior post at the top of this post.

I was told that "\cdot" while in Latex mode makes
the multiplication dot, for instance.

[quote="kunny"]$ a \plus{} b \plus{} c \equal{} 0\Longrightarrow a \plus{} b \equal{} \minus{} c,\ b \plus{} c \equal{} \minus{} a,\ c \plus{} a \equal{} \minus{} b$, we have

$ a\left(\frac {1}{b} \plus{} \frac {1}{c}\right) \plus{} b\left(\frac {1}{c} \plus{} \frac {1}{a}\right) \plus{} c\left(\frac {1}{a} \plus{} \frac {1}{b}\right) \equal{} \frac {b \plus{} c}{a} \plus{} \frac {c \plus{} a}{b} \plus{} \frac {a \plus{} b}{c} \equal{} \minus{} 3$.[/quote]

The expression is equivalent to the symmetric sum
$ A\equal{}\sum \frac{a(b\plus{}c)}{bc}$

$ \minus{}a\equal{}b\plus{}c\implies a^2\equal{}b^2\plus{}c^2\plus{}2bc\equal{}1\minus{}a^2\plus{}2bc\implies bc\equal{}(2a^2\minus{}1)/2$

$ A\equal{}\sum 2a(b\plus{}c)/(2a^2\minus{}1)$

But according to your solution we have $ A\equal{}\sum \frac {b \plus{} c}{a} \implies 2a/(2a^2\minus{}1)\equal{}1/a \implies 2a^2\equal{}2a^2\minus{}1$
 :huh:  :huh:  :huh:


$abc=1,a,b,c>0$ prove that:
\[\sum_{cyc}{(a+\frac{1}{2a})}\geq 3\sum_{cyc}\sqrt[3]{\frac{c}{4(a+1)b}}\]
I just want to know how hard it is?I create it but I don't know if it is hard.
as far as I can see ,it is hard
...atually,I don't think this inequality is nice........

QUite nice solution.As you found, it is not strong.When I found it ,I have applied AM-GM twice too.


Did anyone get number #24?

Was this how you found the answer: $\left(\frac{2006 + 2}{2}\right)^2$, so $1008016$? I'm not really sure which letter it was.

Yes, i believe that was the answer...what was the question again?

it was like (a+b+c)^2006+(a-b-c)^2006
i expanded both with binomial, after grouping as: (a+(b+c))^2006+(a-(b+c))^2006

then a bunch of terms cancel, and yeah...

I'm really bad when there's 3 variables, so i let y = b+c, and then used binomial formula to expand it. Then after adding the two sums and cancelling out every other term, I realized that it was just the sum of the odds from 1 to 2007.

Oh yeah...cookies to kids problem, but kid $x$ must get an even number of cookies....so it becomes $2007+2005+...+1=\frac{2008(1004)}{2}=1008016$

i was talking to people after the test,,, most people will scared away by the large coefficients of terms,
but if you notice that niether the coefficients nor the powers of $x$ matters, the problems is very easy...

in general for any power of $n=2k$ there will be $(k+1)^2$ terms remaining.


calculus
integration
derivative
calculus computations

Hello,
Can someone explain to me how to get the antiderivative of a function. I know what it does. Pretty much, when you differentiate the antiderivative, you get the original function. But, How do you get the antiderivative. I know some stuff, such as x^n --> (x^(n+1))/(n+1).

What else?

Or maybe a good antiderivative link?

Thanks a lot in advance.

There's pretty much a whole course devoted to it. You will take Calc 2 at some point, and you will learn many methods of finding antiderivatives.

perhaps a dumb question: antiderivatives is the same as integration right?

They are often used interchangeably. Definite integrals are a little different, and don't always require an antiderivative to evaluate.

[url=http://en.wikipedia.org/wiki/Lists_of_integrals]To start you off...[/url]

You should be aware that unlike differentiation, it is not always possible to find the antiderivative of even a relatively simple-looking function.  In particular, there is no (direct) analogue of the product rule or the chain rule.


The harmonic mean, $ m$ , of $ a$ and $ b$ is given by $ \dfrac{2}{\frac{1}{a} \plus{} \frac{1}{b}}$.

What is the harmonic mean of $ 3$ and $ 6$ ?


Can anyone give me their [b]Combinatorial Arguements[/b] [notes from AMSP particularly at the start because I didn't know what I was doing when I was in that class and took really crappy notes so if anyone could post some of their notes form the beginning of the course that would be great.


Let $ n\geq 2$ be a positive integer. Consider a closed broken line $ P_1P_2...P_{2n}P_1$. What is the maximum number of self-intersections?

srikanth gave a partially correct solution (matching TZF's bound) [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1523385322&t=176965]here[/url]; the problem was posted with source and answer (but no solution) even earlier, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1523385322&t=50620]here[/url].

[quote="The Zuton Force"][hide="An Upper Bound"]Each sub-segment [ss] can, at most, intersect every other ss except itself and the two adjacent ss. For each of the n ss's, there are n-3 ss's it can intersect. Because there is pairwise symmetry, we must divide by 2. The maximum number is n(n-3)/2[/hide][/quote]

I suppose your $ n$ is my $ 2n$?

Even then, try to give an example. You won't find any  :P

No, actually that seems to work for odd numbers (ie my n is your 2k+1).

Simple examples include the triangle (n=3), the five-point star (n=5), the 7-point star, and generally, the 2k+1-point star.

While the particular construction is not possible for n=2k (pick a segment, consecutive points must be on opposite sides, and this puts the first point on both sides of the segment), and the upper bound I set is unattainable, it is still an upper bound of some sort  :P

[quote="The Zuton Force"]

While the particular construction is not possible for n=2k (pick a segment, consecutive points must be on opposite sides, and this puts the first point on both sides of the segment), and the upper bound I set is unattainable, it is still an upper bound of some sort  :P[/quote]

Yep  :D

Hint: The real upper bound here is $ n(2n\minus{}4)\plus{}1$.


[i]If (a + b)^2 + (b + c)^2 + (c + d)^2 = 4(ab + bc + cd), prove that a = b = c = d.[/i]

Thanks with this, i just can get it. I would really appreciate all your help.

sub tract 4ab from $ (a\plus{}b)^2$
you get $ (a\minus{}b)^2$
do this with the other 2 terms
the equation becomes $ (a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}d)^2\equal{}0$
assuming all numbers are real...
$ a\minus{}b\equal{}0, b\minus{}c\equal{}0, c\minus{}d\equal{}0, a\equal{}b\equal{}c\equal{}d$


geometry
perimeter
calculus
integration
search
LaTeX
congruent triangles

How many different (i.e. pairwise non-congruent) triangles are there with integer sides and with perimeter $ 1994$?

[hide="Solution :)"]Let $ x, y, z$ be the three integral sides then as given
$ x \plus{} y \plus{} z \equal{} 1994$
Also any of the three sides cannot be more than $ \frac{1994}{2} \equal{} 997$. So the maximum length of a side of the triangle is $ 996$.
Remember all lengths less than $ 997$ can form a triangle.
Now suppose I am giving a length of $ 996$ to each sides so that sum of length is exceeding the perimeter by $ (3)(996) \minus{} 1994 \equal{} 994$.
Hence now I need to remove $ 994$ units from three sides to restore the perimeter to be 1994.
Thus the number of positive integral solutions of $ x \plus{} y \plus{} z \equal{} 1994$ are equivalent to number of non- negative integral solutions of $ x' \plus{} y' \plus{} z' \equal{} 994$ which is nothing but $ (994 \plus{} 3 \minus{} 1, 3 \minus{} 1) \equal{} (996, 2) \equal{} (498)(995)$.

Now this result includes some congruent triangles. So to find the unordered solutions, first find the number of solutions in which all the three variables are same - $ None$ because $ 994$ is not a multiple of $ 3$.

Number of solutions in which any two of the three variables are same - $ 498$ - they are counted thrice in the above solution.

Number of solutions in which all the three variables are different  - they are counted six times in the solution  - $ \frac{(498)(995) \minus{} (3)(498)}{6} \equal{} 82336$

Hence the total number of pairwise non-congruent triangles are $ 0 \plus{} 498 \plus{} 82336 \equal{} 82834$[/hide]

The associated sequence is [url=http://www.research.att.com/~njas/sequences/A005044]A005044[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url]; it satisfies a simple linear recurrence.  Here's one past discussion on the forum: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=275216 (and there are doubtless many others -- try a search!).

Thanks.

kamallohia,  I don't get what you mean by $ (994 \plus{} 3 \minus{} 1, 3 \minus{} 1) \equal{} (996, 2) \equal{} (498)(995)$.

It's a binomial coefficient -- the correct LaTeX would have been \binom{996}{2}, giving $ \binom{996}{2}$

Thanks JBL  :)
That's what i was referring to.

Andrew  :)
I hope you have got your answer. If doubt persists do ask me. I'll try to share to best of my knowledge.


pigeonhole principle
calculus
calculus computations

Find a injective mapping  which maps an interval in $\mathbb{R}$ to an infinite number of points on $\mathbb{R}$

[quote="jmerry"]This question doesn't make sense. What do you really mean?[/quote]
Yes it does. The points should be different, they should not be connected.

[quote="Kent Merryfield"]For an injective mapping, the cardinality of the codomain must be at least as large as the cardinality of the domain. An interval has the (uncountable) cardinality of the continuum. Any injective image of it must also be uncountable.

[quote]The points should be different, they should not be connected.[/quote]
That's a very poorly worded sentence, and I really don't know what you mean by it. If you mean a discrete set (with no limit points), then that cannot be uncountable.[/quote]

Ok :( , but perhaps I could give an example and you could help me formulating the question :) .
Convert each number in a given interval, say [0,1] (suppose these are in 10-digit system) to its binary equivalent. Then consider these binary numbers as 10-digit-nubers again on the real line. These will be uncountable set of (dicrete) points distributed on R.

[quote="Kent Merryfield"]It's possible that the word you want is "totally disconnected" rather than "discrete." That would take a while to explain - do you know what "connected" means? And if so, "connected component"?[/quote]

Ok, so you mean that they will necessarily not be single point, but
short intervals?
I have just been introduced to connected set and components.

[quote="Kent Merryfield"]It's possible that the word you want is "totally disconnected" rather than "discrete." That would take a while to explain - do you know what "connected" means? And if so, "connected component"?[/quote]

Could you please explain the differences?
Or at least give a proof why we do not get points.

[quote="JBL"][quote="color"]These will be uncountable set of (dicrete) points distributed on R.[/quote]

No they won't: they have lots and lots of accumulation points.[/quote]
What do you mean by accumulation points? Could you please
explain why in a bit more formal way? Thanks.

A subset $S$ of $\bf R$ has an accumulation point $p$ if and only if every neighborhood of $p$ contains at least one point of $S$.  So, for example, the set of accumulation points of the rationals is the reals, and the set of accumulation points of $\{\frac{1}{n}\mid n \in{\bf Z}_{>0}\}$ is just 0 (plus perhaps the set itself; it doesn't really matter very much, but I think many authors may define it so that in general the set of accumulation points does not include the set, via the use of punctured neighborhoods.  This isn't very important).

We can divide the reals into the countably many intervals $[n, n+1]$ for $n \in \bf Z$.  (Overlap is okay.)  We have uncountably many points in the interval $[0, 1]$, so if we map it injectively into $\bf R$, there must be one of these intervals which contains uncountably many image points.  (This is the pigeonhole principle with infinite cardinalities.)  Since that interval is a compact subset of the reals, any infinite subset has an accumulation point (and since the subset is actually uncountable, we should expect many accumulation points, I believe).

Note that having an accumulation point is different than "being an interval."  Every neighborhood around an accumulation point might include both points from the set (that's required by the definition) and points not from the set (the definition doesn't say anything either way), whereas in an interval all small-enough neighborhoods contain only other points of the interval, except at the two endpoints.


[b]Prove $ 2\cos 20$ satisfies the unfactorable equation $ x^3\minus{}3x\minus{}1\equal{}0$[/b]

[hide="hint"]Let $ \cos 20$ equal $ x$, and use the triple-angle formula for $ \cos 3\theta$ to give you an expression for $ \cos 60$.[/hide]


Hi,

I came across this other problem in the book AoPS vol 1 in which the solution doesn't make sense to me. The problem is:

A crew of $ 30$ people can build a certain road in $ 60$ days. After the tenth day the plans are changed; the company wants the road built in $ 30$, not $ 60$, days. How many more people must be hired?

So I did:

$ 30$ people $ \rightarrow$ $ 60$ days;

Therefore, $ 30(2) \rightarrow 60/2 \Rightarrow 60$ people $ \rightarrow$ $ 30$ days;

Or:

$ 30$ people $ \rightarrow$ did $ \frac{1}{6}$ $ \rightarrow$ in $ 10$ days

Therefore, $ 30$ people $ \rightarrow$ did $ \frac{1}{6}(3)$ $ \rightarrow$ in $ 10(3)$ days

$ \Rightarrow$ $ 30$ people $ \rightarrow$ did $ \frac{1}{2}$ $ \rightarrow$ in $ 30$ days

$ \Rightarrow$ $ 30(2)$ people $ \rightarrow$ did $ \frac{1}{2}(2)$ $ \rightarrow$ in $ 30$ days

$ \Rightarrow$ $ 60$ people $ \rightarrow$ did $ 1$ $ \rightarrow$ in $ 30$ days.

So, we need $ 60\minus{}30\equal{}30$ [b]more[/b] people.

But according to the book, we need $ 45$ more people. Is there something wrong with my logic?

Thank you for your help.

Regards,
limac

Well, you know that each day $ \frac{1}{60}$ of the work is done by $ 30$ people.  So, every person does $ \frac{1}{1800}$ of the work every day.
After $ 10$ days, $ \frac{1}{6}$ of the work has been done.  So, $ \frac{5}{6}$ of the work remains.
So, we have that $ \frac{5}{6}\equal{}(20)(\frac{n\plus{}30}{1800})$, yielding $ n\equal{}\boxed{45}$.

Your problem is that you forgot the fact that $ 10$ days have already gone by.  In order to finish in $ 30$ days, you only have $ 20$ days remaining, not $ 30$.  So, although your logic was correct, the fact that you only have $ 20$ days left and not $ 30$ makes your answer incorrect.

nittanylion, thank you very much for your reply. I see what I was doing wrong. So, my way was basically doing everything for day 0, so therefore it was wrong.

Thank you for giving me that realization! :D

Regarda,
limac


Εστω ενα συνολο Α με n στοιχεια και εστω $S_{A}$ το αθροισμα των στοιχειων του συνολου Α.Θα ονομαζω $S_{k}$ το αθροισμα ολων των στοιχειων ολων των υποσυνολων του συνολου Α που το καθενα απο αυτα εχει κ στοιχεια. Νδο:
1) $S_{k}$/$S_{A}=$${n-1 \choose{k-1}}$
2)Για καθε τριγωνικο αριθμο χ υπαρχει  θετ ακερ κ ωστε  ο κχ να γραφεται ως διαφορα 2 τριγωνικων.
3)Λυσε την εξισωση Σ(Σκ)=$x^{2}$ (Tωρα θεωρησε ως συνολο Α το αρχικο αποκομμα (1,2,....n))

Ειναι καλη παρατηρηση και απλη.Προσπαθηστε την.Tην εφτιαξα με αφορμη μια παρατηρηση που ειχα κανει στην β λυκειου.Ετσι την ξεθαψα..... :lol:


group theory
abstract algebra
superior algebra
superior algebra unsolved

is the center of An included (as a setwise sense) in center of Sn for all n?

i.e. center of Sn is trivial => center of An is trivial?

ALSO! WHY isn't anyone responding to my previous question namely

Prove that every finite group G contains a nilpotent subgroup H whose normal closure H^G=<H^g:g in G> is G.

!!

The center of $S_{n}$ is trivial for $n\ge 3$. Indeed, $\sigma\in S_{n}$ commutes with the transposition $(ij)$ iff $\{\sigma(i),\sigma(j)\}=\{i,j\}$.
Since $A_{3}$ is abelian, the answer to your question is no.

Your previous question seems interesting, and I considered it for a while, but did not come up with anything worth posting.

i should have said n for greater than or equal to 4...

If $\sigma\in A_{n}$ commutes with the $3$-cycle $(ijk)$, then $\{\sigma(i),\sigma(j),\sigma(j)\}=\{i,j,k\}$. Hence $\sigma(i)$ is contained in the intersection of all $3$-point sets containing $i$, which is exactly $\{i\}$ provided that $n\ge 4$. Thus, the center of $A_{n}$ is trivial for $n\ge 4$.


[quote=Inequalities Master]Let $ a,b,c,x,y,z$ be real numbers. Prove that:
$ 4(a^2\plus{}x^2)(b^2\plus{}y^2)(c^2\plus{}z^2)\geq 3(bcx\plus{}cay\plus{}abz)^2$.[/quote]
Let $x=\frac{au}{\sqrt2}$, $y=\frac{bv}{\sqrt2}$ and $z=\frac{cw}{\sqrt2}$.
Hence, we need to prove that $(u^2+2)(v^2+2)(w^2+2)\geq3(u+v+w)^2$, which is very known.

[quote=arqady]
Let $x=\frac{au}{\sqrt2}$, $y=\frac{bv}{\sqrt2}$ and $z=\frac{cw}{\sqrt2}$.
Hence, we need to prove that $(u^2+2)(v^2+2)(w^2+2)\geq3(u+v+w)^2$, which is very known.[/quote]
There is a nice proof by Cauchy Schwarz inequality.
\[3(bcx+cay+abz)^2\leq 3(b^2+y^2)\left[(cx+az)^2+c^2a^2\right]\]
So it suffice to show that
\[4(a^2+x^2)(c^2+z^2)\geq 3\left[(cx+az)^2+c^2a^2\right]\]
which is just
\[(ac-2xz)^2+(az-cx)^2\geq 0\]
So the inequality is proved. From the proof, it is very easy to find the equal condition.

Let $ x,y,z$ be positive real numbers and $ a,b,c$ be real numbers. Prove that:
$$x^3(b^2+c^2)+y^3(c^2+a^2)+z ^3(a^2+b^2)\geq 2xyz(ab+bc+ca).$$
[size=50](ZXSXJXCK  No.7(2019))
[/size]


search AMC USA(J)MO USAMO ARML AIME USAMTS

How do you guys pronounce it? I spell the acronym but I've heard others do it phonetically as well. Obviously, most of the time when I talk about it it's online, so it isn't pronounced at all.

I alternate... usually I pronounce it, then when someone asks me what I said, I spell it, and then when they ask me what it is, I explain it. Then I get funny looks ;)

I spell it out for most people, but say the phonetic name around people who I know will know it.

There's some more discussion in this thread: http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=596257613&t=197459.

I always spell out U-S-A-M-O; I've gotten so used to spelling it that it sounds funny to me when I say you-sam-o.

I've only heard USAMO pronounced once in my life, and it was u. s. a. m. o. pronunciation.

I  personally think "Use a Moe" sounds kinsa unusua.

i use you ess a em oh
my friend uses you sam oh

Aye-Im-Cee
Aim!
U.S.A.M.O.
Mop. Mosp.
I.M.O.

Ewe-sam-ites (thank you Lynnelle :))
eye-Test
ewe-say-co?

I pronounce AIME as Amy, AMC as A.M.C, MOP as mop, IMO as I.M.O.

What's Ewe-sam-ites?  Don't tell me that's how you pronounce USAMTS or something.

[quote="RisingStar"]You Ess Ey Mo is how some people up here say USAMO.[/quote]

OMG!  That's how I used to say it until someone corrected me!  THe surprising thing is that I'm from GA...

don't most ppl pronouce AIME as ay-mie, or do ppl say ay eye em ee?

My friends and I call it A.I.M.E. -- I know too many Amy's  :)

Outside, I hear people (my math teacher, much of my ARML team, etc) call it Amy.

[quote="Zmastr"]I thought that the AIME was pronounced aim as a freshman, and then a senior corrected me, saying that it was pronounced amy, and that people would think I'm an idiot if I didn't say it that way.   :roll:  I say amy, and U S A M O, like the letters.  you-sam-oh is just too weird for me  :)[/quote]

Same heh.

[quote="K81o7"][quote="worthawholebean"]Everyone calls it the "aim" at my school. IMO it sounds much better than "amy."[/quote]

But then it can be mistaken for AIM, or aim.  Amy is much harder to mistake for something else...[/quote]I've never had that issue. Plus, it's much more fun to pun with.

Everyone's spelling it you sam oh.
I say it you-same-oh, as a way to sort of have the spelling in there (the long 'a') but I'm kinda saying it too fast... and everyone I've ever heard calls it the same.
did you guys mean for 'sam' to have a long 'a' or do you really say it with a short 'a'?

[quote="tenniskidperson3"]I was kind of wondering about that, because at my ARML practice, the coach was talking about the techniques we can use, and he said you-sam-oh, and I was like, hmm, that's not what it is.  By the way, do you pronounce ARML as Ay Are Em El, or Arml?[/quote]

Are-mul

Yeah ARML is almost unanimously pronounced as "Are-mul" as I have heard it.

are-mul, you-say-moe, aye-me

How do you guys pronounce IMO?

eye-moh or eye-em-oh

aymee

you-ess-ay-em-oh

eye-em-oh

are-mul

[quote="Zmastr"]I wonder how the good people over at the AMC office pronounce these tests themselves?[/quote]

I will not say these are "official" pronunciations, but these are the pronunciations I hear most often.
AMC is spelled out.
AIME is like Amy.
USAMO is spelled out.
IMO is spelled out.

Around here we say ARML like arm with an l tacked on the end, but they're a different contest, so that's definitely not official. ;)

I had never heard "You-suh-mo" until I attended NCTM two weeks ago and some others attendees there called the USAMO that.

So there's the officially unofficial pronunciation guide. ;)

Elizabeth Claassen
Senior Accounting Clerk
(and resident number cruncher)

everyone i know says you saa mo. There is still the question over the a sound: some people say aaaaa and some say ahhhhhhh.

[quote="cyberspace"]some people say aaaaa and some say ahhhhhhh.[/quote]I would interpret those as the same sound: like the "a" in "father".

I usually spell everything out except ARML, which I pronounce "are-mul" (if I am interpreting this spelling correctly).

[quote="33286"][quote="cyberspace"]some people say aaaaa and some say ahhhhhhh.[/quote]I would interpret those as the same sound: like the "a" in "father". [/quote]

I meant aaaaaaa as in the sound in bat and yes ahhh like in father.

EYE-EM-OH

ARE-MUL

AY-EM-SEE

AY-ME

YOU-SAY-MOE

IMO= I.M.O

USAMO= U.S.A.M.O

AIME= Amy

AMC= A.M.C

ARML= are mul

MOP= mop (not M.O.P.)

With AIME ARML and MOP, if someone looks at me weird when I say it, I say the letters/acronym, but they usually still look at me strangely, and then I explain it and they keep on looking at me strangely. :wink:

The majority of those threads are on the pronunciation of AIME, which is not the point of this topic.
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=596257613&t=197459.


If you have a 3x3 matrix, and you switch two rows/columns of it, do you multiply the determinant of the original matrix by -1?
If so, if you switch two rows/columns and another two rows/columns does the determinant stay the same?
Ex. Switch rows 1 and 3, then switch rows 2 and 3.(Does the determinant stay the same since you switched twice?)

Yes, Ckck, that's right. We can consider any matrix $ A \in \mathcal{M} _{n} (\mathbb{K})$, and it is enough to consider rows, since $ \det(A) \equal{} \det(A^T)$. If B is a matrix resulting from A by k row exchanges, then $ \det(A) \equal{} (\minus{}1)^k \det(B)$.


The largest and smallest of three consecutive terms in an arithmetic sequence differ by 14. Half of the smallest term is added to each term and the sum of the resulting three numbers is 120. What is the value of the original smallest term?

Common ratio is $ r$, $ i$th term is $ a_i$.  $ a_3\minus{}a_1\equal{}14 \implies r\equal{}7$, and $ a_1\plus{}\frac{1}{2}a_1\plus{}a_1\plus{}7\plus{}\frac{1}{2}a_1\plus{}a_1\plus{}14\plus{}\frac{1}{2}a_1\equal{}120 \implies \frac{9}{2}a_1\equal{}99 \implies a_1\equal{}\boxed{22}$.

The answer is correct, but I just feel like pointing out that it's a common [b]difference[/b], not ratio :wink:


Each cube has an edge of one.
You can fit 9 cubes on the length, 3 on the depth and 3 on the height.  Multiply the numbers and you get 81.

You could also do $ (9 \cdot 3 \cdot 3) \div 1^3 \equal{} \boxed{81}$.


[hide]4 are multiples of 12, need 2^2*3 at least One uses no other factor, one uses the last 3, the last 5 and both[/hide] :ninja:

[quote="236factorial"]How many positive factors of $2^2\cdot3^2\cdot5$ are multiples of 12?[/quote]

[hide]
SInce we need to find factors of $12=2^2\cdot3$, we divide $2^2\cdot3^2\cdot5$ by it to get
$3\cdot5$
So now we need to find how many factors $3\cdot5$ has.  Since we have 2 choices for the power of each 3 and 5 (0 and 1), we have $2\cdot2 = \boxed{4}$ choices of factors.
[/hide]

[hide="Some explanation"]Basically, you're just factoring out the $12$ and taking a $15$. Then, you use the formula $a^xb^y... \Rightarrow (x+1)(y+1)...$ to find out how many distinct positive integer factors a given integer has, since you know that each factor is multiplied by $12$.[/hide]


calculus
integration
function
real analysis
real analysis unsolved

If f and g are continuous functions, and  $ \int_{0}^{1} |f(t) \minus{} g(t)|\not \equal{} 0$ show that  $ f(x)\not \equal{} g(x)$ for $ a < x < b$ where  $ (a,b)\subset [0,1]$

Intuitively it is really obvious, but I'm not sure how to do this in a formal setting.

Work with the difference $ h\equal{}|f\minus{}g|$. If it's nonzero somewhere, use continuity to show that it's nonzero in an interval around that point. You can even "put a box under the graph", and thus show that the integral is positive..


John attended a three-country multicultural festival. At the first exhibit, John paid a $ \$3$ admission fee. He then spent half the money he had left on souvenirs. Finally he spent $ \$3$ on food and left. At each of the second and third exhibits, he again paid a $ \$3$ admission fee, then spent half of the money he had left for souvenirs, then spent $ \$3$ on food, and left. When he left the third exhibit, he had no money remaining. How many dollars did he spend at the festival?

Working backwards yields $ 3(2)\plus{}3\equal{}9$ before the 3rd exhibit, $ 12(2)\plus{}3\equal{}27$ before the 2nd exhibit, and $ 30(2)\plus{}3\equal{}63$ before the first exhibit.  Thus, he had $ \boxed{\$63}$ at the start, and since he spent all of it, that is our answer.


geometry
parallelogram
geometric transformation
homothety
trigonometry
rotation

[quote="mathgeniuse^ln(x)"][quote="rnwang2"]$ABCD$ is a square. $DEFG$ is a square. $I$ is the center of $ABCD$ and $J$ is the center of $DEFG$. $M$ and $N$ are midpoints. Prove that $MJNI$ is a square.[/quote]

Wow, Hard Question.

Any hints by any chance.

Are the squares congruent.[/quote]

No, the squares are not necessarily congruent.

[hide="Hint"] Parallelograms![/hide]

Only [u]very basic[/u] proofs belong in HSB.

Request to move to Intermediate.

[quote="daermon"][hide="partial proof"]if we draw AC and GE, we find that MJNI is formed by connecting the midpoints of quadrilateral AGEC and is thus a parallelogram.  not sure how to extend to squares yet... :maybe: [/hide]
EDIT: [hide]maybe prove that AGEC is cyclic?[/hide][/quote]
Actually, I don't think it's necessarily cyclic.

And that was not the parallelogram I was hinting at.

oops.  i meant MJNI :blush:  :blush:
and now that i think about it, that actually doesn't seem like a good method anyways :D

[quote="SM4RT"]does this help extending $\overline{GA}$ and $\overline{EC}$ to meet at some point $Q$...[/quote]
Follow the hints...  :wink:

$MJ=IN=\frac{1}{2}AE$ and $MJ\parallel IN\parallel AE$
$MI=JN=\frac{1}{2}CG$ and $MI\parallel JN\parallel CG$

$AE=CG$ by cosines law or whatever you like

$CG$ is a 90 degree rotation of $AE$ about $d$

thus the sides of $MINJ$ are equal and perpendicular to each other.

That is clever.

But how did you get AE=CG, by cosine law.

And how do you know that CG is a 90 degree rotation about AE.

Hopefully the diagram is self-explanatory.

[quote="risinglegend"]how do u post a pic?[/quote]
You add it as an attachment.


Is it going to be on this year? I haven't heard of any air date.

I was told by a very reliable source (Mr. Rusczyk) a month or two ago that he did not think ESPN was going to air it this year.  Unless someone's heard something about it in the recent past, I'm pretty sure they won't air it.

[quote="Fanatic"]Yhey don't like Lockheed Martin.[/quote]

Are you kidding?  :huh: I hope so.

I imagine that it doesn't bring much money either.

I just emailed someone that works for mathcounts and he said that matcounts would not be broadcasting on espn.  Heres his message:

Unfortunately ESPN is not broadcasting a program on the National Competition this year.  We would have liked to work with them, but they decided it was not something that was going to fit their plans for the year.  Hopefully we can entice another network to broadcast this next year.


A typical compact disc plays 60 minutes of music and costs $ \$15.00$. A typical long-playing record plays 40 minutes of music. Assuming that the money spent of a recording is paying only for the amount of music, how many dollars should a record cost?

The problem states that money spent is based on amount of music, so if it's 2/3 the music, it's 2/3 the cost.

$ 15*\frac23\equal{}\boxed{10}$


complex analysis
algebra unsolved
algebra

can anyone give e the proof of this thorem?  its not there in wikipedia also.

Reason 1 you couldn't find one is that this question belongs in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=347]complex analysis section[/url]. This is not an elementary theorem by any means.


trigonometry
function
real analysis
real analysis solved

Moubinool, are you sure that you want the coefficient at $y$ to be $3+4x^2$ and not $2+4x^2$? :?

Well, I am teaching a course on differential equations, and the topic for next week is power series. So ...

$y=c_1\left(1-\frac{3x^2}2+\frac{25x^4}{4!}-\frac{331x^6}{6!}+ \frac{5937x^8}{8!}+\cdots\right)$

$+c_2\left(x-\frac{7x^3}{3!}+\frac{81x^5}{5!}-\frac{1303x^7}{7!}+ \frac{26785x^9}{9!}+\cdots\right).$

For power series $\sum a_nx^n$ the recursion for the coefficients is

$a_{n+2}=-\frac{(4n+3)a_n+4a_{n-2}}{(n+2)(n+1)}$ and the radius of convergence is $\infty.$

If there's more of a pattern than that, I haven't spotted it yet.

The answer I'm getting is $y=(C_1\cos{(x)}+C_2\sin{(x)})e^{-x^2}$.  (I got this from my calculator).  I can't figure out how to solve this by hand, but if we were given $y_1=e^{-x^2}sin{(x)}$, then it would be easy to show that the general solution is $y=(C_1\cos{(x)}+C_2\sin{(x)})e^{-x^2}$.

Jimmy is right, of course.

Let $y=e^{-x^2}u$. Then

$y'=e^{-x^2}u'-2xe^{-x^2}u$
$y''=e^{-x^2}u''-4xe^{-x^2}u'+(4x^2-2)e^{-x^2}u$

Put that in the original differential equation:

$e^{-x^2}\left[u''-4xu'+(4x^2-2)u+4x(u'-2xu)+(4x^2+3)u\right]=0$

Which simplifies to $e^{-x^2}(u''+u)=0$ or just $u''+u=0.$

[quote="Jimmy"]The only thing that bugs me is how did you get $e^{-x^2}$.  I know that after solving several differential equations, you can kind of get a feel of what the solution should look like, but is there a more rigorous way of getting $e^{-x^2}$?[/quote]
Yes. Read some stuff about asymptotic analysis.
Example:
$y^{\prime\prime}+p\left( x\right) y^{\prime }+q\left( x\right) y=0.\ $ Let's suppose $y\left( x\right) \ $ can be written as $e^{A\left( x\right) }$, where $A\left( x\right) $ is a so-called slowly-varying function when $x\rightarrow \infty $. Thus $A^{\prime }\left( x\right) \gg A^{\prime \prime}\left( x\right) \gg A^{\prime \prime \prime }\left( x\right) \gg \cdots\cdots .$ $y^{\left( n\right) }$~$\left( A^{\prime }\left( x\right)\right) ^{n}e^{A\left( x\right) },$ so $\left( A^{\prime }\right)^{2}+pA^{\prime }+q$~$0$ to the first order, $A^{\prime }$~$\frac{-p\pm \sqrt{p^{2}-4q}}{2}.$

In our case, $p=4x,q=3+4x\ $, so $A^{\prime }$~$-2x,\ A$~$-x^{2}.$

Thanks Buffalo
I was wondering, is asymptotic analysis covered in a Real Analysis Course?