algebra
polynomial
calculus
derivative
quadratics
function
functional equation

Detetermine all polynominals $ P(x)$ (real coefficients, x real) such that:

$ 1\plus{}P(x)\equal{}\frac{P(x\minus{}1)\plus{}P(x\plus{}1)}{2}$

It is well-known that if $ P$ is a polynomial with degree $ n$ then $ Q(x) \equal{} P(x \plus{} 1) \minus{} P(x)$ is a polynomial with degree $ n \minus{} 1$.

Now, let $ P$ be a solution with degree $ n$, let $ Q$ defined as above, and $ R(x) \equal{} Q(x) \minus{} Q(x \minus{} 1)$.
The given functional equation leads to $ R(x) \equal{} 2$ for all $ x$.
Thus, $ R$ has degree $ 0$, so that $ Q$ has degree $ 1$, which forces $ P$ to have degree $ 2$.

Conversely, it is a straightforward verification that the only solutions are the polynomials of the form $ P(x) \equal{} x^2 \plus{} bx \plus{} c$, where $ b,c$ are any real constants.

Pierre.

[quote="Invariant"]Detetermine all polynominals $ P(x)$ (real coefficients, x real) such that:

$ 1 \plus{} P(x) \equal{} \frac {P(x \minus{} 1) \plus{} P(x \plus{} 1)}{2}$[/quote]

Let $ Q(x)\equal{}P'(x)$ be the derivative of $ P(x)$. We have $ Q(x\plus{}1)\equal{}2Q(x)\minus{}Q(x\minus{}1)$ and so :

$ Q(x\plus{}1)\minus{}Q(x)\equal{}Q(x)\minus{}Q(x\minus{}1)$ and so $ Q(x\plus{}1)\minus{}Q(x)$ is a periodic polynomial, so is a constant, and we have $ Q(x\plus{}1)\equal{}Q(x)\plus{}a$

So $ Q(x\plus{}1)\minus{}a(x\plus{}1)\equal{}Q(x)\minus{}ax$ and $ Q(x)\minus{}ax$ is a periodic polynomial, so is a constant, and we have $ Q(x)\equal{}ax\plus{}b$

So $ P(x)$ is a quadratic polynomial $ P(x)\equal{}ax^2\plus{}bx\plus{}c$ for some reals $ a,b,c$

Plugging back this expression in the original equation, we get $ a\equal{}1$ and so the general solution is $ \boxed{P(x)\equal{}x^2\plus{}ax\plus{}b}$

[quote="pbornsztein"]
Now, let $ P$ be a solution with degree $ n$, let $ Q$ defined as above, and $ R(x) \equal{} Q(x) \minus{} Q(x \minus{} 1)$.
The given functional equation leads to $ R(x) \equal{} 2$ for all $ x$.
Pierre.[/quote]

I do not understand. Why the given functional equation leads to $ R(x) \equal{} 2$ for all $ x$??

:wink: I am not asking about the denominator. $ R(x)\equal{}P(x \minus{} 1) \plus{} P(x \plus{} 1)$ and it has $ n\minus{}2$ degree. The lefi side of the equation is $ 1\plus{}P(x)$ and it has $ n$ degree. So:
$ 2\plus{}2P(x)\equal{}R(x)$ - and?

How is that $ R(x)\equal{}2$?? It means that $ P(x)\equal{}0$.... I do not understand

[quote="Invariant"]$ R(x) \equal{} P(x \minus{} 1) \plus{} P(x \plus{} 1)$[/quote]

No.
$ R(x)\equal{}Q(x)\minus{}Q(x\minus{}1) \equal{} (P(x\plus{}1)\minus{}P(x))\minus{}(P(x)\minus{}P(x\minus{}1))$
Thus $ R(x)\equal{}P(x\plus{}1)\minus{}2P(x)\plus{}P(x\minus{}1)\equal{}2.$

Pierre.

[hide="Solution"]Let $ P(x) \equal{} a_nx^n \plus{} \ldots \plus{} a_1x \plus{} a_0$ with $ a_n \neq 0$ and suppose $ n > 2$. Consider the coefficient of $ x^{n\minus{}2}$ in $ P(x\minus{}1) \plus{} P(x\plus{}1) \minus{} 2P(x)$. For $ P(x)$ to be a solution, this coefficient must be $ 0$. The binomial theorem tells us it is equal to $ 2 {n \choose 2} a_n$. Since $ a_n \neq 0$, this is never $ 0$, so contradiction. Therefore $ n \leq 2$. We easily see that $ n < 2$ gives $ P(x) \equal{} \frac{P(x\minus{}1) \plus{} P(x\plus{}1)}{2}$, also a contradiction. So $ n \equal{} 2$.

Let $ P(x) \equal{} ax^2 \plus{} bx \plus{} c$. Expand the given equation and simplify to show $ a \equal{} 1$ is necessary. It is easy to verify that $ P(x)$ being a solution implies that both $ P(x\minus{}r)$ and $ P(x) \plus{} s$ are solutions for any reals $ r,s$. So we get that all solutions are of the form $ P(x) \equal{} x^2 \plus{} bx \plus{} c$ for any reals $ b,c$.[/hide]

I am guessing the first paragraph of that solution is equivalent to pbornsztein's $ R(x)$ argument.

Something in the way of motivation (as well as an alternate solution) for the problem:

We shall write $ p_n$ to denote $ P(n)$.  Moreover, we will only consider the values of $ p_n$ at non-negative integers $ n$.  The given is equivalent to

$ p_{n \plus{} 2} \equal{} 2p_{n \plus{} 1} \minus{} p_{n} \plus{} 2$

Without the constant, this recursion has characteristic polynomial $ (x \minus{} 1)^2$, so we would expect it to have constant or polynomial solutions.  Let $ G(x) \equal{} \sum_{n \equal{} 0}^{\infty} p_n x^n$ be the generating function of $ p_n$.  Multiplying the above by $ x^n$ and summing across $ n \ge 0$, we obtain

$ \frac {G(x) \minus{} p_0 \minus{} p_1 x}{x^2} \equal{} 2 \cdot \frac {G(x) \minus{} p_0}{x} \minus{} G(x) \plus{} \frac {2}{1 \minus{} x} \Leftrightarrow$
$ G(x) \minus{} p_0 \minus{} p_1 x \equal{} 2x(G(x) \minus{} p_0) \minus{} G(x) x^2 \plus{} \frac {2x^2}{1 \minus{} x} \Leftrightarrow$
$ (x^2 \minus{} 2x \plus{} 1) G(x) \equal{} (p_1 \minus{} 2p_0)x \plus{} p_0 \plus{} \frac {2x^2}{1 \minus{} x} \Leftrightarrow$
$ G(x) \equal{} \frac {(p_1 \minus{} 2p_0)x \plus{} p_0}{(1 \minus{} x)^2} \plus{} \frac {2x^2}{(1 \minus{} x)^3} \Leftrightarrow$
$ G(x) \equal{} \left( (p_1 \minus{} 2p_0)x \plus{} p_0 \right) \left( \sum_{n \equal{} 0}^{\infty} (n \plus{} 1)x^n \right) \plus{} 2x^2 \left( \sum_{n \equal{} 0}^{\infty} \frac {(n \plus{} 2)(n \plus{} 1)}{2} x^n \right) \Leftrightarrow$
$ G(x) \equal{} \sum_{n \equal{} 0}^{\infty} \left( n^2 \plus{} (p_1 \minus{} 1 \minus{} p_0)n \plus{} p_0 \right) x^n \Leftrightarrow$
$ p_n \equal{} n^2 \plus{} (p_1 \minus{} 1 \minus{} p_0)n \plus{} p_0, n \ge 0$

If $ P(x)$ is a polynomial, then $ P(x) \minus{} x^2 \minus{} (p_1 \minus{} 1 \minus{} p_0)x \minus{} p_0 \equal{} 0$ for all integer $ x \ge 0$, hence $ P(x) \equal{} x^2 \plus{} (p_1 \minus{} 1 \minus{} p_0)x \plus{} p_0$ as desired.  :)

[b]N.B.:[/b]  Note that we only used the fact that $ P(x)$ is a polynomial in the last step.  I believe we can loosen the condition to "$ P(x)$ is $ C^{\infty}$," and perhaps further.

$ \forall n\in\mathbb{N}: \ p'(n) \equal{} \frac {p'(n \minus{} 1) \plus{} p'(n \plus{} 1)}{2}$ gives $ \exists (a,b)\in \mathbb{R},\forall n\in\mathbb{N}: \ p'(n) \equal{} 2an \plus{} b$
so $ \forall x\in\mathbb{R}: \ p'(x) \equal{} 2ax \plus{} b \Rightarrow p(x) \equal{} ax^2 \plus{} bx \plus{} c$
and $ 2 \plus{} 2p(x) \equal{} p(x \minus{} 1) \plus{} p(x \plus{} 1)$ gives $ a \equal{} 1$

we can solve any equation of the form $ Q(x)\plus{}P(x)\equal{}\frac{P(x\minus{}k)\plus{}P(x\plus{}k)}{2}$ for a given polynominal $ Q(x)$ and real $ k$ by just deriving it enough times. like pco did but more times if needed. in our case $ Q(x)\equal{}1$ and $ k\equal{}1$

[quote="Jure the frEEEk"]we can solve any equation of the form $ Q(x) \plus{} P(x) \equal{} \frac {P(x \minus{} k) \plus{} P(x \plus{} k)}{2}$ for a given polynominal $ Q(x)$ and real $ k$ by just deriving it enough times. like pco did but more times if needed. in our case $ Q(x) \equal{} 1$ and $ k \equal{} 1$[/quote]
yes $ Jure the frEEEk$ if $ deg(Q)\equal{}n$ we have $ \forall h\in\mathbb{N},p^{(n\plus{}1)}(kh)\equal{}\frac{p^{(n\plus{}1)}(k(h\minus{}1))\plus{}p^{(n\plus{}1)}(k(h\plus{}1))}{2}$ so $ \forall h\in\mathbb{N}: \ p^{(n\plus{}1)}(h)\equal{}ah\plus{}b$ ==> $ deg(p^{(n\plus{}1)})\le 1$ ==> $ deg(p)\le n\plus{}2$

Sorry to revive, but here is a link: https://artofproblemsolving.com/community/c6h159183p890647


At Makettle's, Jim bought 30 oranges at a price of three for 59 cents. At Paduka's, he then bought 20 oranges at a price of two for 41 cents. What was the number of cents in the average price of an orange?

Makketle => bought 30 oranges at prices of 3 for 59 cents = 590 cents
Paduka's => bought 20 oranges at prices of 2 for 41 cents = 410 cents

total = 1000 cents, 50 oranges.

so average price of an orange = 20 cents

answer ; 20


inequalities
geometry
perimeter
Pythagorean Theorem
IMO Shortlist
geometry solved

Let $ABC$ be a triangle with sidelengths $a$, $b$, $c$, and let $X$ be a point on its plane. Let $u$, $v$, $w$ be the distances of the point $X$ to the sides $BC$, $CA$, $AB$ of the triangle $ABC$. Prove the inequality:

$4\left(XA+XB+XC\right)^{2}\geq\left(a+b+c\right)^{2}+4\left(u+v+w\right)^{2}$.

Let U, V, W be the orthogonal projections of the point X on the sidelines BC, CA, AB of triangle ABC. Then, the distances u, v, w of the point X to these sidelines are u = XU, v = XV and w = XW. Also, BU + UC = BC = a, CV + VA = CA = b and AW + WB = AB = c (if some of the points U, V, W happen to lie outside the respective sides BC, CA, AB, then we have to consider the segments BU, UC, CV, VA, AW, WB as directed; note that the distances u, v, w remain non-directed). Now, by the Pythagorean theorem in the right-angled triangle BUX, we have $XB=\sqrt{BU^2+XU^2}=\sqrt{BU^2+u^2}$, while the Pythagorean theorem in the right-angled triangle CUX yields $XC=\sqrt{UC^2+XU^2}=\sqrt{UC^2+u^2}$; similarly, $XC=\sqrt{CV^2+v^2}$, $XA=\sqrt{VA^2+v^2}$, $XA=\sqrt{AW^2+w^2}$, $XB=\sqrt{WB^2+w^2}$. Now, by the Minkowski inequality (see, e. g., Lemma 2 in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=17384#p17384]http://www.mathlinks.ro/Forum/viewtopic.php?t=5355 post #2[/url], for n = 2 and u = 6), we have

$\sqrt{BU^2+u^2}+\sqrt{UC^2+u^2}+\sqrt{CV^2+v^2}+\sqrt{VA^2+v^2}$
$+\sqrt{AW^2+w^2}+\sqrt{WB^2+w^2}$
$\geq\sqrt{\left(BU+UC+CV+VA+AW+WB\right)^2+\left(u+u+v+v+w+w\right)^2}$.

Equivalently,

$XB+XC+XC+XA+XA+XB\geq\sqrt{\left(\left(BU+UC\right)+\left(CV+VA\right)+\left(AW+WB\right)\right)^2+4\left(u+v+w\right)^2}$,

what rewrites as

$2\left(XA+XB+XC\right)\geq\sqrt{\left(a+b+c\right)^2+4\left(u+v+w\right)^2}$.

Upon squaring, this becomes

$4\left(XA+XB+XC\right)^2\geq\left(a+b+c\right)^2+4\left(u+v+w\right)^2$,

and we are done.

[b]EDIT:[/b] The problem is a particular case (for triangles) of the IMO Shortlist 1996 problem G9:

[color=blue]In the plane, consider a point $X$ and a polygon $\mathcal{F}$ (which is not necessarily convex). Let $p$ denote the perimeter of $\mathcal{F}$, let $d$ be the sum of the distances from the point $X$ to the vertices of $\mathcal{F}$, and let $h$ be the sum of the distances from the point $X$ to the sidelines of $\mathcal{F}$. Prove that $d^2-h^2\geq\frac{p^2}{4}$.[/color]

The solution of this problem is an obvious generalization of my above solution of its particular case.

  Darij


algebra
polynomial
system of equations
algebra proposed

Find the polynomial $P(x)$ with real coefficients such that $P(2)=12$ and $P(x^2)=x^2(x^2+1)P(x)$ for each $x\in\mathbb{R}$.

$P(x^2)=x^2(x^2+1)P(x)$
Firstly, it is easy to see that $P(0)=0$, $P(1)=0$ and $P(-1)=0$. If $P(x)$ has any real root other than $0$, $1$ and $-1$, say, $t$, then $P(t^2)=t^2(t^2+1)P(t)=0\Rightarrow$ $t^2$ is also a real root of $P(x)$, giving infinitely many solutions. Therefore, $P(x)$ has only three real roots $0$, $1$ and $-1$.
From the given condition we also know that $deg P(x)=4$
So $P(x)=a(x-1)^2x(x+1)$ or $a(x-1)x^2(x+1)$ or $a(x-1)x(x+1)^2$
Sub $x=2$, $12=P(2)=6a\text{ or } 12a\text{ or } 18a\Rightarrow a=2\text{ or } 1\text{ or }\frac{2}{3}$
Therefore $P(x)=2(x-1)^2x(x+1)\text{ or } (x-1)x^2(x+1)\text{ or } \frac{2}{3}(x-1)x(x+1)^2$.
By checking, only $P(x)=(x-1)x^2(x+1)$ is correct.
$\therefore P(x)=(x-1)x^2(x+1)$

Yes, now you are correct . This is also my solution with a little difference. thank you

It's easy to see that $ P(x) \equal{} P( \minus{} x)$, which means that $ P(x)$ is even. It's also easy to see that $ P(0) \equal{} P( \minus{} 1) \equal{} P(1) \equal{} 0$.
Since $ P(x)$ is even and $ P(0) \equal{} 0$, then $ x^2\mid P(x)$.
Therefore we can write $ P(x) \equal{} x^2(x^2 \minus{} 1)Q(x)$. Doing this, it's easy to see that $ x^4(x^4 \minus{} 1)Q(x^2) \equal{} x^4(x^4 \minus{} 1)Q(x)$, which means that $ Q(x^2) \equal{} Q(x)$, $ \forall x\in\mathbb{R}\setminus\left\{ \minus{} 1,0,1\right\}$.
It's also easy to see that $ Q(2) \equal{} 1$ and we can conclude that $ Q(x) \equal{} 1$, $ \forall x\in\mathbb{R}: x \equal{} 2^{2^a}$, with $ a\in\mathbb{Z}$.
This means that $ Q(x) \equal{} 1$ has infinite solutions, so $ Q(x)$ is constant, which means that $ Q(x) \equal{} 1$ and $ P(x) \equal{} x^4 \minus{} x^2$.

$ P(x^{2}) \equal{} x^{2}(x^{2} \plus{} 1)P(x)$
$ P(0) \equal{} 0$ then $ P(x) \equal{} xS(x)$

$ x^2S(x^2) \equal{} x^2(x^2 \plus{} 1)xS(x)$
$ S(x^2) \equal{} x(x^2 \plus{} 1)S(x)$
$ S(0) \equal{} 0$ then $ S(x) \equal{} xD(x)$

$ x^2D(x^2) \equal{} x(x^2 \plus{} 1)xD(x)$
$ D(x^2) \equal{} (x^2 \plus{} 1)D(x)$

Let $ ax^n$ be the leading term of $ D(x)$ then for the above expression to hold
$ a(x^2)^n \equal{} (ax^n)(x^2)$
$ x^{2n} \equal{} x^{n \plus{} 2}$ or $ n \equal{} 2$
Let $ D(x) \equal{} ax^2 \plus{} bx \plus{} c$
So $ ax^4 \plus{} bx^2 \plus{} c \equal{} ax^4 \plus{} ax^2 \plus{} bx^3 \plus{} bx \plus{} cx^2 \plus{} c$
$ \therefore b \equal{} 0$ and $ c \equal{} \minus{} a$ or $ D(x) \equal{} a(x \plus{} 1)(x \minus{} 1)$

Now $ P(x) \equal{} xS(x) \equal{} x^2D(x) \equal{} ax^2(x \plus{} 1)(x \minus{} 1)$
We are given $ P(2) \equal{} 4a(3)(1) \equal{} 12$ or $ a \equal{} 1$

$ \therefore$ $ P(x) \equal{} x^4 \minus{} x^2$

[quote="silouan"]Find the polynomial $ P(x)$ with real coefficients such that $ P(2) \equal{} 12$ and $ P(x^2) \equal{} x^2(x^2 \plus{} 1)P(x)$ for each $ x\in\mathbb{R}$.[/quote]
It's easy to see that $ deg\{P(x^2)\} \equal{} 2n$, and $ deg\{x^2(x^2 \plus{} 1)P(x)\} \equal{} n \plus{} 4$, so $ n \equal{} 4$ ($ n$ is degree of polynomial here). $ P(x)$ is even, so $ P(x) \equal{} ax^4 \plus{} bx^2 \plus{} c$. From $ P( \minus{} 1) \equal{} P(0) \equal{} 0$  and $ P(2)\equal{}12$ we will have system of equations, and $ a\equal{}1$, $ b\equal{}\minus{}1$, $ c\equal{}0$.

[quote="Kirill"]It's easy to see that $ deg\{P(x^2)\} \equal{} 2n$, and $ deg\{x^2(x^2 \plus{} 1)P(x)\} \equal{} n \plus{} 4$, so $ n \equal{} 4$ ($ n$ is degree of polynomial here). $ P(x)$ is even, so $ P(x) \equal{} ax^4 \plus{} bx^2 \plus{} c$. From $ P( \minus{} 1) \equal{} P(0) \equal{} 0$  and $ P(2) \equal{} 12$ we will have system of equations, and $ a \equal{} 1$, $ b \equal{} \minus{} 1$, $ c \equal{} 0$.[/quote]
Nice solution. :wink:
[quote="Complex_Ninja"]$ x^2S(x^2) \equal{} x^2(x^2 \plus{} 1)xS(x)$
$ S(x^2) \equal{} x(x^2 \plus{} 1)S(x)$
$ S(0) \equal{} 0$ then $ S(x) \equal{} xD(x)$[/quote]
The conclusion is true, but the argument is incorrect... You are dividing by 0...

[quote="Joao Pedro Santos"]
[quote="Complex_Ninja"]$ x^2S(x^2) \equal{} x^2(x^2 \plus{} 1)xS(x)$
$ S(x^2) \equal{} x(x^2 \plus{} 1)S(x)$
$ S(0) \equal{} 0$ then $ S(x) \equal{} xD(x)$[/quote]
The conclusion is true, but the argument is incorrect... You are dividing by 0...[/quote]
But aren't u just factoring an x...  Like isn't it basically taking $ x^2 \minus{} x$ and writing as $ x(x\minus{}1)$?

Let $\deg P=n$,  so we get that $\deg RHS=2n\text{ and }\deg RHS=n+4\text{ thus }2n=n+4\Longrightarrow n=4$

Furthermore letting $x\to0$ yields $P(0)=0$ so let $P(x)=ax^4+bx^3+cx^2+dx$
Now notice that $x\to1$ yields $P(1)=2P(1)\Longrightarrow P(1)=0\text{ so }a+b+c+d=0$
Moreover $x\to-1$ yields $P(1)=P(-1)\Longrightarrow P(-1)=0\text{ so }a-b+c-d=0\text{ thus }a+c=b+d$
Now using the fact that $a+b+c+d=0$ we get that $c=-a\text{ and }d=-b$
So we can rewrite $P$ as $P(x)=x(x^2-1)(ax+b)$
Letting $x\to2$ in yields $P(2)=6(2a+b)=12\Longrightarrow b=2(1-a)$ so $P(x)=x(x^2-1)(ax+2-2a)$ and by plugging this into our original equation we obtain $2x^2(x^4-1)(1-x)(1-a)=0$
Thus for $x\neq\pm1\text{ or }0$ we must have $a=1$ and so $P(x)=x^2(x^2-1), \forall x\neq\pm1\text{ or }0$ however since $P(0)=P(\pm1)=0$ we obtain $P(x)=x^2(x^2-1), \forall x\in\mathbb{R}$

To sum up $\boxed{P(x)=x^2(x^2-1), \forall x\in\mathbb{R}}$


Solve: \[\sin cos x = \cos \sin x \] $x$ in radians.

[hide="I have an idea..."]
We can set $x= \arctan(n)$, which gives the equivalent equality $\sin \left( \frac{1}{\sqrt{n^{2}+1}}\right) = \cos \left( \frac{n}{\sqrt{n^{2}+1}}\right) \Longrightarrow \frac{1}{\sqrt{n^{2}+1}}= \frac{\pi}{2}-\frac{n}{\sqrt{n^{2}+1}}$
$\Longrightarrow 0=\left(\frac{\pi^{2}}{4}-1\right) n^{2}-2n+\frac{\pi^{2}}{4}-1$

But the discriminant of this is $2 \pi^{2}-\frac{\pi^{4}}{4}< 0$; thus no $n$ exists, and there is $\boxed{ \text{No Solution!}}$.
[/hide]

[hide="solution"]
We know that $\sin(\frac{\pi}{2}-\theta) = \cos{\theta}$, so $\sin^{-1}\cos{x}= \frac{\pi}{2}-x$.

Therefore:

$\sin\cos x = \cos\sin x$
$\cos x = \frac{\pi}{2}-\sin x$
$\cos x+\sin x = \frac{\pi}{2}$
$\frac{1}{\sqrt{2}}\cos x+\frac{1}{\sqrt{2}}\sin x = \frac{\pi}{2\sqrt2}$
$\sin\frac{\pi}{4}\cos x+\cos\frac{\pi}{4}\sin x = \frac{\pi}{2\sqrt2}$
$\sin{\frac{\pi}{4}+x}= \frac{\pi}{2\sqrt2}$
$x = \sin^{-1}(\frac{\pi}{2\sqrt2})-\frac{\pi}{4}+2k\pi$.

Did I lose any solutions anywhere?

EDIT: Lol I just decided to check and $\frac{\pi}{2\sqrt2}> 1$, so no real solutions.
[/hide]

[quote="lotrgreengrapes7926"][quote="Hamster1800"]$\cos x+\sin x = \frac{\pi}{2}$[/quote]
$\cos x+\sin x\le\sqrt{2}<\frac{\pi}{2}$[/quote]

already realised that. Check my edit.

But if you allowed complex solutions, would it work?

EDIT: Yes, if you extend the trig functions to allow complex number solutions, my number is correct.

[quote="t0rajir0u"]The way this problem is normally phrased is to prove the inequality

$\cos \sin x > \sin \cos x$[/quote]

Would it be valid to solve for $\cos\sin x-\sin\cos x = 0$ and find no real solutions, then show that the difference is continuous and so it's always positive because it's positive at $x=0$?

Yes, but as I recall there was an easier way to show it.

Oh... I thought this was a little random problem of my own... never really seen it when I posted this :(


conics
ellipse
AMC
AIME
USA(J)MO
USAMO
analytic geometry

This is the thread where you'll discuss about AMC-12 B. However, this thread will be locked until AMC-12 B is done.

[rant]I'm really frustrated with this year's 12B.

After looking over most of the problems on the A, I was becoming pretty confident that I could get over 140, which would mean, at worst, 2 blank or 1 blank and 1 wrong. I could solve the hardest problems in five minutes or less, and I could solve every problem on the test. It's possible that this confidence negatively affected me going into the B, but even so, there's still a problem. The first part of the B test went pretty well, although I noticed a few that seemed more difficult than their A counterparts. Then I reached the last 5 problems.

I spent 30 minutes on #21 and 23-25, and didn't solve a single one of them. In fact, I didn't make any progress on any of them. I still have no idea how to do any of those four problems. Anyone else see something wrong there? I'm willing to put some of it down to time pressure, and some of it down to the fact that ellipses are probably my worst subject in math at the moment (#21). I don't really care about that. What I do care about is the fact that, based on everything else I've gone over, I should have been able to get over 140, possibly even a perfect score, on the AMC12. So, this is 10th grade, it only really matters for the prestige. But this means that the B was quite noticeably harder than the A. What happens to people who might have qualified for the AIME but didn't, because the B was harder than the A? What happens to juniors and seniors who could qualify for USAMO based on index, but can't, because they could only take the B test? The A and B are supposed to be equal in difficulty. Frankly, with the AMC being the biggest high school math competition in the country, I expected a little more. They claim that they make their best effort to make these two tests equal, and similarly with the AIME I and II. I don't think it's just coincidence that I honestly thought that the AIME II was supposed to be harder than the I. And now I'm noticing the same thing with the AMC. I'm very disappointed. I just hope that the AIME is a little more balanced this year.[/rant]

Yeah.   I thought the B was harder as well.  But the school I took the B at held the AMC 12B in the evening rather than in the morning.  I can't really think clearly in the morning,  so I actually improved on the B despite the increase in difficulty (122 --> 129)  :D

i salute u jesusfreak.
as a junior who had to take only the 12 B, i agree with you.
when i solved the 12 A for practice, i was at least able to do 2 questions after 20. but when i took the 12 B yesterday, when i solved the first 18 in 35 mins and looked at the problem afterward. it was just different. 18~20, it was doable. But after than it was just,, my mind just went blank looking at them. i believe i got about 10 points less than if i have took 12 A. now i refuse to look at the problems 20~ in 12 B.
i hope AIME B would not be harder as some people say because i would have that.

well, those people were trying to qualify for AIME and were on the borderline shouldn't have been hit hard

the first 15 problems were pretty normal, IMO

i have to admit though, the last 5 were pretty bad

Does anybody have #18?  I don't remember what the problem was, but I didn't get that one.  Same thing with number 22... I got really close but I think I misread the problem during the test.

#18, you had to realize that it could only end up on coordinates (a,b), where a+b is even

this is one of the more basic examples of an invariant

there are 11 "rings" you can end up on, 1 origin, those in 2 steps going striaght out, 4 steps, etc till 10 steps

the number of points on these rings is 121

I thought the intermediate problems, those numbered in the range of about 10-20, were easier than their counterparts on the AMC 12A. The geometry problems in that range were much more straightforward, in my opinion.

The last five were not actually too much harder than the last five on the 12A, either. If you've been keeping up with the Mock AMC's, you should have realized that #23 was very similar to a problem in the mock AMC B by Rep123max and joml88. The main difference for me was that the last five on the previous round had more problems in algebra, a subject with which I am more familiar. Furthermore, #24, $(x+y+z)^{2006}+(x-y-z)^{2006}$, was easily done by an estimate through expansion of the smaller powers, especially with the convenience of a TI-89.

All said and done, I ended up with 131, only six points lower than my score on the 12A. But yes, I agree that the 12B was harder than the AMC 12's given in recent years. I would expect the AIME to be pretty difficult too. It looks like a trend this year.

[quote="AMC website"]The top one percent of students on the AMC 10 A will be invited to take the AIME. The top one percent of students on the AMC 10 B will be invited to take the AIME. The top 5 percent of students on the AMC 12 A or students scoring better than 100 on the AMC 12 A will be invited to take the AIME. The top 5 percent of students on the AMC 12 B or students scoring better than 100 on the AMC 12 B will be invited to take the AIME.[/quote]

[quote="AMC website"]The USAMO index cutoff for participants was 233 if you took the AIME I, or 220.5 if you took the AIME II.[/quote]

can someone please direct to me to this year's amc12B contest, i would much appreciate that.
thanks.

we'll be posting all the problems soon, if you are looking for the answers:

http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2006-12a/06AMC12AB-answers.html

look in the Contests/Resources section soon, it'll have an index of all the AMC B problems

I would agree in saying these last few (and comparably some of the first few) were by far more difficult than the A.  If it wasn't for careless errors I would have had a perfect on the A, yet i didn't get a single one of these last 5 (6, actually.. altho I suppose I could have gotten 22 if I had given it some thought).  Maybe it would have been more equal if this was a no calculator contest, but A was pretty brutable.

For me, $\Delta Score = -14$

I thought it was fairly easy. Maybe I had high overexpectations, but I didn't find it that difficult.

Like, the 19 about the ages was really, really, easy for a number 19 in my opinion.

I was so close to making the AIME!!!  :read:  :wallbash_red:  :wallbash:  :(  :mad:  I took the AMC 12B I guess I shouldnt be mad cause I'm only in 5th/6th grade, but if I had gotten 2 more probs right I would have gone to the AIME.  I am so close!!!  What is your best advice, if you are as close as I was to making the AIME, how do you get TO making the AIME?  Any specific classes/books?  Once your sure you'll make the AIME hwo do you prepare for it?

The AMC 8 (i got a 21 or 22, i'm not sure, my coach never toald me, i just know i was second in my school  :winner_second: ) and AMC12 (got a 89 cause I guessed trying to get into AIME, since i wasn't taking it with my school, otherwise would have got a 94) are way easier than the AMC 10.

[quote="fuzzylogic"][quote="The Zuton Force"]Does anyone know a good way of doing #24?

I sketched it out quickly and estimated C but I donno how to do it properly.

[/quote]

try multiplying $\sin{x}+\sin{y}$ on both sides. Then it becomes, $\sin{3x}+\sin{3y}\ge0$[/quote]

Oh nice it reduces to $\sin{3x}$ formula.  I applied quadratic formula and it does work(but I never finished because I was short on time)

[quote="frt"][quote="fuzzylogic"][quote="beta"]
#23 would be difficult if you have never seen the equilateral triangle version.[/quote]
Using rotation is pretty nice, it turns out $11,7,6\sqrt{2}$ is a right triangle.

If you have enough time left (like 10 min),  it can be cracked by brute-force (use coordinates). It turns out it isn't that bad. The square of the leg satifies a quadratic equation.[/quote]
I noticed that rotation method would work, but I didn't see that $11$, $7$, $6\sqrt{2}$ was a right triangle :wallbash:

Another method for #23 is to use reflection... But either way, you get a $11$-$7$-$6\sqrt{2}$ triangle and you need to notice that it's a right triangle...[/quote]

hm, even if it wasn't right, you can get the angle with law of cosines, then add 45 to it and law of cosine again to get a leg

[quote]From that point on, all you need is common sense (however inelegant and unmathematical it may be). The last two digits are Mr. Jones's age. He cannot possibly be 0 years old, and think about what would happen (i.e. social implications, etc) if a nationwide math contest ever suggested the possibility of a 90-year old guy with 8 kids under the age of 10.[/quote]

I found the introduction of this clue somewhat odd... certainly it was easy enough to figure out the correct multiple-choice answer without knowing what the four-digit number was, but anyone trying to brute-force the problem would have benefited from knowing or agreeing with the socially normal age to have children.  There tend to be cultural/economic differences in the minimum "acceptable" age, for instance; I'd assume that the AMC would put the father between 30 and 50, for instance, because I was actually considering his age at the oldest daughter's birth during the test, but not everyone would either think that way or make the same assumption.

Obviously this isn't that important, but there's so rarely an instance in which cultural knowledge becomes an issue that this question stood out.

[quote="frt"][quote="fuzzylogic"][quote="beta"]
#23 would be difficult if you have never seen the equilateral triangle version.[/quote]
Using rotation is pretty nice, it turns out $11,7,6\sqrt{2}$ is a right triangle.

If you have enough time left (like 10 min),  it can be cracked by brute-force (use coordinates). It turns out it isn't that bad. The square of the leg satifies a quadratic equation.[/quote]
I noticed that rotation method would work, but I didn't see that $11$, $7$, $6\sqrt{2}$ was a right triangle :wallbash:

Another method for #23 is to use reflection... But either way, you get a $11$-$7$-$6\sqrt{2}$ triangle and you need to notice that it's a right triangle...[/quote]

my first instinct was also rotation...but i drew my diagram so terribly that i thought the right angle was between the $11$ and the $6\sqrt{2}$.  so i wasted a few minutes wondering why $193$ is not a perfect square.  and then time was called :(

For some reason, my score on the A (I took it at home) was 12 points lower than my B (the one I officially took).  I admit, though, the last 5 on the B were much harder than the last 5 on the A.  Weird.

The B was definitely more difficult in terms of trying to get a perfect score. After a 144 on the A I was ready to ace the B, but alas I cannot guess that well (I guessed on three problems, got one right, but missed another one stupidly). So I ended up with 132. Oh well.

Well I got the same score on B as I did on A. And both scores were worse than what I got last year on amc 12 b.

I relied on calculator too much

omg ppl were complaining about how hard last questions on 12 A was after they took the test. i thought 12 B couldn't be worse than that.. but oh well it was..!!!!!!!ahhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh :(

Ooph, they're gettin good at writing up those last couple questions.

#23 -- I immediately thought about the beginnning problem in Mathematical Olympiad Challenges, where you rotate it, but never got around to doing it.

#24 -- I got that $\frac{\sin^3 x+\sin^3 y}{\sin x+\sin y}\le\frac{3}{4}$ but somehow I didn't think about cross multiplying and getting the simplified version of this. Oh well.

#22 -- Any way to do this other than guess that the partition is 499,499,1008?

[quote="beta"]I thought it was one of the hardest AMC in the past few years.[/quote]

I agree with that statement, so much for the AMC A and B are equally hard. In my opinion #19 was rather simple for a problem that late in the test (but I missed it due to a very stupid error). The score I got was on the lower end of my practice AMC test scores.

[quote="alan"][quote="Elemennop"]#22 -- Any way to do this other than guess that the partition is 499,499,1008?[/quote]

guessing that the partition is 624,624,758 :P[/quote]

lol

Anybody got tips for guessing numbers really well? haha

Maybe realizing that $624=5^3-1$, so it has fewer $5$'s. ;)

Ahh, I got 21 right and then erased it because I thought I made an algebraical error.
So is 150 on the 10 better than 120 on the 12, or does it not matter since it's not aime?

Oh right, was the rotation thing for 23?  How does that work?


Solve $\frac{sin10 +sin30 +sin50}{cos10 +cos30 +cos50}$.

No Calculators.

Hint: Multiply the numerator by $-2sen(10)$ and the denominator by $-2cos(10)$.

[quote="FMako"]No need to spam.  The hint wasn't necessarily for you, it could be for anyone wanting a slight edge in solving it.[/quote]

Ditto.

I deleted all of the extra posts by citistarz.  Either he accidentally kept clicking submit or he was just being obnoxious.  I'll give him the benefit of the doubt  :D


Dude ccy what are you talking about? 11:09 is not morning. That's much too early to be waking up.

[quote="ccy"]yayyy good MORNING![/quote]

you mean "WHY AM I AWAKE ALREADY?"


trigonometry
Trigonometric Equations
geometry
IMO
IMO 1966

Let $a,b,c$ be the lengths of the sides of a triangle, and $\alpha, \beta, \gamma$ respectively, the angles opposite these sides. Prove that if \[ a+b=\tan{\frac{\gamma}{2}}(a\tan{\alpha}+b\tan{\beta}) \] the triangle is isosceles.

Let ABC be a triangle, which biggest (in other case is same) angle is <ACB. Let CH is a perpendicular to AB, H lie on the side AB. Let <ACH=ф. In triangle AHC we have tga=cotgф =>atgatgф=a and in triangle BCH btgbtg(y-ф)=b => a+b=atgatgф+btgbtg(y-ф). In this broblem we have a+b=tg(y/2)(atga+btgb) => tg(y/2)(atga+btgb)=atgatgф+btgbtg(y-ф).  atga(tgф-tg(y/2))=btgb(tg(y/2)-tg(y-ф)). We use that tgx-tgy=(sin(x-y))/cosxcosy => atgasin(ф-y/2)/cosфcos(y/2)=btgbsin(ф-y/2)/cos(y-ф)cos(y/2) => atga/cosф=btgb/cos(y-ф). But in triangle AHC cosф=sina and cos(y-ф)=sinb. We use sinus law: a/sina=b/sinb and we have tga=tgb. Is it OK?

$\left(A+\frac C2\right)+\left(B+\frac C2\right)=180^{\circ}\Longrightarrow \cos \left(A+\frac C2\right)=-\cos \left(B+\frac C2\right).$
$a+b=\tan \frac C2\cdot (a\tan A+b\tan B)\Longleftrightarrow a\left( \cot \frac C2-\tan A\right)=b\left(\tan B-\cot \frac C2\right)\Longleftrightarrow$
$\frac{a\cos \left(A+\frac C2\right)}{\sin \frac C2\cos A}=\frac{b\cos \left(B+\frac C2\right)}{\sin \frac C2\cos B}\Longleftrightarrow$$\cos \left(A+\frac C2\right)=0\ \ \vee\ a\cos B=b\cos A\Longleftrightarrow$
$A+\frac C2=B+\frac C2\ \vee\ \tan A=\tan B\Longleftrightarrow A=B.$

[quote="DPopov"]Let $ a,b,c$ be the lengths of the sides of a triangle, and $ \alpha, \beta, \gamma$ respectively, the angles opposite these sides. Prove that if
\[ a \plus{} b \equal{} \tan{\frac {\gamma}{2}}(a\tan{\alpha} \plus{} b\tan{\beta})\]
the triangle is isosceles.[/quote]
The way I did it was this:
Let $ x \equal{} \tan \frac {\alpha}{2}, y \equal{} \tan \frac {\beta}{2}$.
Then: $ a \equal{} 2R \cdot \sin a \equal{} 2R \frac {2x}{1 \plus{} x^2}, b \equal{} 2R \frac {2y}{1 \plus{} y^2}$
$ \tan \alpha \equal{} \frac {2x}{1 \minus{} x^2}, \tan \beta \equal{} \frac {2y}{1 \minus{} y^2}$
$ \tan \frac {\gamma}{2} \equal{} \frac {1 \minus{} xy}{x \plus{} y}$
Then it is equivalent to:
$ (x \plus{} y)^2(1 \minus{} x^2)(1 \minus{} y^2) \equal{} 2(1 \minus{} xy)^2(x^2 \plus{} y^2)$
And:
$ (x \plus{} y)^2 \le 2(x^2 \plus{} y^2)$ with equality iff $ x \equal{} y$
$ (1 \minus{} x^2)(1 \minus{} y^2) \le (1 \minus{} xy)^2$ with equality iff $ x \equal{} y$.
So: $ RHS \ge LHS$ with equality iff $ \tan \frac {\alpha}{2} \equal{} \tan \frac {\beta}{2} \iff \alpha \equal{} \beta \iff a \equal{} b$. Since there is equality we can conclude that the triangle is isosceles.

my solution =
let angles of triangle be $A,B,C$ respective to side lengths $a,b,c$
wlog let $\angle A\ge \angle B$ and hence $a\ge b$
so by thchebytcheff we have

$2(a+b)= 2\tan {\frac{C}{2}}(a\tan A +b\tan B)\ge \tan {\frac{C}{2}}(a+b)(\tan A+\tan B)$
so $2\ge \tan {\frac{C}{2}}(\tan A+\tan B)$
or $2cosAcosB\ge \tan {\frac{C}{2}}(sin(A+B))=\tan {\frac{C}{2}}(sinC)$
or we get,

$cos(A-B)-cosC\ge 2(\sin\frac{C}{2})^2\leftrightarrow cos(A-B)\ge 1$
and thus $\angle A=\angle B$

so $ABC$ is isosceles triangle.
we are done :D


Eh, you know what I do?
I just go into the hall, sneak into a computer lab, and do somethings on the computer.
Nah, not really.
But yeah, I do go on the computer during class every now and then.

I make up my own math problems and solve them. It's quite fun, actually. :P I have a bunch of random problems written all over my binders now.

[quote="crazychinesemaniac"]So people what do you do when you're in a boring class and there's still about another 45 minutes left in the class? Me sometimes I read, a kid I know yesterday was clipping his fingernails during class. It was pretty funny when the teacher busted him. Yes it was a guy  :rotfl:[/quote]

clip finger nails...great idea! thanx zo much lolz

Sit at the back of the classroom, and either sleep or do some math. lol :sleeping:

Eh most of my classes only last 40 minutes anyway, and the only one that ever goes an hour is math. So yeah. I rarely get bored during class.

[quote="ZhangPeijin"]meh, all my classes are 45 minutes, and I usually fall asleep after the 10 minute mark.
5 minutes if its French.
As soon as I get in if I can get away with it in Math.[/quote]

aren't they 49 minutes?

today, I was pulling out AIME problems whenever I could...

Yeah that's what I did today.
I put my cd of all the AIME problems in my pocket.
During first block I had to endure the eardrum shattering 'music'.
Second block I left the class and spent around 5 minutes around the printer waiting for the AIME problems I printed to come out.
Spent most of the day looking at them and got yelled at by my english teacher for not doing english in english class.

[quote="aanjohri"]ahg. my school uses block scheduling so i only have 4 out of 8 classes a day,so each class is 90 min. long. in a way its gud, so u can manage hw and stuff, but sometimes it gets soooo boring :([/quote]

block scheduling is much better, longer classes, more flexible hw with two days, less hw in general, and we get the time to do stuff like labs that we couldn't do in 50 minutes.

though it can get boring...

With the way this year is going, I think I might start photocopying pages of Latin to translate in Physics, English, Bio, etc. The book is too bulky to open discreetly.


The sum of all positive integer factors of a number is 28. What is the number?

i don't know a formula for this, but with some brute force, you can get....

$ 12$ has the factors: $ 1, 2, 3, 4, 6, 12$
$ 1 \plus{} 2 \plus{} 3 \plus{} 4 \plus{} 6 \plus{} 12 \equal{} 28$

so, our answer is $ \boxed{12}$

The formula for the sum of all positive integer factors of a number with prime factorization \[ p_{1}^{e_1}p_{2}^{e_2}p_{3}^{e_3}\cdots p_{n}^{e_{n}}\]is \[ (1\plus{}p_1\plus{}p_1^2\plus{}p_1^3\plus{}...\plus{}p_1^{e_1})(1\plus{}p_2\plus{}p_2^2\plus{}p_2^3\plus{}...\plus{}p_{2}^{e_2})\cdots (1\plus{}p_{n}^{1}\plus{}p_{n}^2\plus{}p_{n}^3\plus{}...\plus{}p_{n}^{e_{n}})\]
Or, if you know the product and sum things, this is
\[ \prod_{i\equal{}1}^{n}\left(\sum_{j\equal{}0}^{e_{i}}p_i^{j}\right)\]

we then note that
1+2+4=7
1+3=4
7(4)=28
So the number is $ 2^2\cdot 3\equal{}12$.

Summation notation rhymes and sounds scary. It is scary.


algebra polynomial function LaTeX factorial

Hmm, I used finite differences, although I'm reallly new to those so I probably screwed up. Something interesting was that every single time you took the difference, the sequence of numbers is symmetric. I had to use it 22 times before it came constant at 1 1. I'm not sure what to do right now though.

There is always a 23rd degree polynomial (there are 24 terms there) that fits these numbers for $ x \equal{} 0, 1, 2, \dots, 24$. :D

And what do you mean "formula" for a "finite sequence"?

A function $ f(n)$? We could plug in $ n > 24$ and add more terms... unless this sequence is special...

this sequence is very special. it ends in $ 58$ no more no less.
can you please tell me the 23rd degree polynomial for it and how you got it?
i would be so happy if u explain step by step,as it is very important for me to know it.
thank you

Hi binomial_4eva! I use difference to done your problem but it's very long and very complicate. I give you a question: Have you got a solution of this problem. If you have, Can you post it?

[quote="binomial_4eva"]this sequence is very special. it ends in $ 58$ no more no less. [/quote]
It would help to know the motivation and source for this problem; otherwise, there's no particular reason to prefer one kind of sequence rule or another.  I could just tell you that $ f(n) \equal{} \text{that sequence}$, which a perfectly valid (but completely useless) function rule.  What exactly are you looking for?

[quote="binomial_4eva"]can you please tell me the 23rd degree polynomial for it and how you got it?[/quote]
[url=http://mathworld.wolfram.com/LagrangeInterpolatingPolynomial.html]Substitute[/url].  This is, in all likelihood, not the answer you actually want (if you're actually looking for a sensible answer to some actual problem).

there are loads of reasons that i am not sayin the exact problem here:

1: i cannot put tables like i want in latex. nor can i list

2: the problem is very long, its only a bit of it i am writing here, so if i write one part i have to explain all bits here, with proofs

3:it is a complete new system i have invented for solving a problem, so it might need loads of explanations if i write it here,

4:this is the most important, my idea is new and i think its right.
if i tell you here ,you might disagree with me in the whole thing, like you did with this post that you locked http://www.artofproblemsolving.com/Forum/viewtopic.php?t=184288, and that upsets and discourages me of working on it more.

so please just help me with questions i am asking.
i need the 23th degree polynomial for it,please find that for me if u can.

tor :why do u say it wont be the answer i am looking for ?

It gives no information.  To an arbitrary finite sequence of length $ d$ one can associate a polynomial of degree at most $ d\minus{}1$ that outputs that sequence, and you have learned nothing about the sequence (unless it turns out that the degree of the interpolating polynomial is much less than $ d$).

so what kind of information do you need to find the exact sequence rule for it ?i can answers your questions about it.ask i will answer

The point I'm trying to make is that without knowing where these numbers come from, there is no well-defined notion of "[b]the[/b] exact sequence rule."  I don't know whether to prefer any given rule over any other given rule.  So it'd be nice to know whether these numbers are the result of a computation, an experiment, some counting problem...

ok how do i put tables in latex , like tables in excel that u can write inside it? like a table of numbers
 like this http://www.techonthenet.com/excel/formulas/images/cell001.png

ok here is the source of my question t0rajir0u. i hope it helps
[geogebra]0ca645de2b8daa97cb0f716d2b6a3ffa2ae7de61[/geogebra]

this is the one example of the system i created and i need your help in this:
i call the number of elements (the $ a,b,c$ )used in that, $ n$
and i call how many each separated as $ b$ for example in that table the $ n \equal{} 3$ and $ b \equal{} 3$
because i have used $ a,b,c$ to make the whole table from, thats why $ n \equal{} 3$
and because each row is made of three of number or in other words its separated as $ abc$
thats why $ b \equal{} 3$.
[size=150]
important[/size] remember i choose the $ n \equal{} 3$ and $ b \equal{} 3$ because it will give me an easy small table. if i used big $ b$ and $ n$ the table would have been huge enormous which would be so dificult to list.
i know the formula for how many it would have been but any way.i will leave how many and what i know about them for now i mean the formulas.

that table is only one example out of the infinity, because its up to us to change the $ n$ and $ b$
but $ n,b\in \mathbb{N}$

i have found loads of formulas about that kind of tables.but i am stuck in one thing, although i have found a formula for that as well but not an explicit one. i have a formula that always works but i cannot write it fully in mathematical form. the formula i was asking for that sequence is somehow related to that problem.
i have spent like about 2 years on this system.and have found loads of formulas but not the one i am looking for.

so i will ask that question now and you can use the table for help.but i will draw another table with $ n \equal{} 4$ and $ b \equal{} 3$ which will be easier because $ n$ and $ b$ are different.

how do i find the place of the none-repeated ones: by none-repeated i mean like $ acb$ and $ bca$ not like $ aab$ or $ bbc$ because as you can see b is repeated twice in $ bbc$.

by place i mean like for example  in that table $ abc$ is first place, and $ acb$ is the second place for the none-repeats. i know the place formula for repeat one. i dont know the none-repeat one.

say we didnt have the table or the table was too big. how would we know what place is $ cba$ is?
its in the 6th place because we can see the table. but if the number of $ n$ and $ b$ was different how would we find it ? please your formula must contain $ b$ and $ n$

That's just a list of base-$ 3$ representations of the integers.  There are lots of questions you can ask about such lists (how often a given digit is used, etc.) and most of them have pretty straightforward answers.

yes and that is what $ b$ stand for. $ b$ stands for base.
i need help in this someone please help me.

anyone plz helpppppppppppp

here is a table with $ b\equal{}4$ and $ n\equal{}3$ i think this helps more for finding that formula
 [geogebra]58d324535a08aac3bfacf0f9b1803b7721724cc5[/geogebra]

i am looking for a formula like this
this formula is for the repeated ones only
 $ P(e)\equal{}(L_1\minus{}1)b^{n\minus{}1}\plus{}(L_2\minus{}1)b^{n\minus{}2}\plus{}(L_3\minus{}1)b^{n\minus{}3}\plus{}\dots\plus{}(L_i\minus{}1)b^{n\minus{}i}\plus{}\dots\plus{}L_n$

say we wanted to find what the place of $ bad$ in the table with $ b\equal{}4$ and $ n\equal{}3$
we can use the formula to find it.

$ P(bad)\equal{}(2\minus{}1)4^{2}\plus{}(1\minus{}1)4^1\plus{}4\equal{}4^2\plus{}4\equal{}16\plus{}4\equal{}20th$ place
which you can see in the table $ bad$ is in the 20th place in that table above.

i am stuck on this:
how do we find what place for the non-repeat ones for example  $ dbc$  without looking at the table, but only using the information $ b\equal{}4$ and $ n\equal{}3$.

can we find a formula like the one for the repeat ones?

in the table we can see $ dbc$ is the $ 22th$ place from the non-repeats but how do we find this without looking at the table?

this might help as will the number of none-repeats in the tables are equal to the permutation $ n$ and $ b$
number of none-repeats$ \equal{}\frac{b!}{(b\minus{}n)!}$ which i call $ P$ for permutation

number of repeats$ \equal{}b^n\minus{}\left(\frac{b!}{(b\minus{}n)!}\right)\equal{}b^n\minus{}P$

Unsolved Questions:
1:because the formula for the repeats are a polynomial and the degree is $ n\minus{}1$
does it mean that all polynomials has a table like that?

2:is roots of all polynomials a base? like the $ b$ in those table?

3:if yes then can we have imaginary,irrational and negative bases as well ?

4:how do we show a negative base in that kind of table:

and many more i wont make the post long and boring for reading.

[quote="azjps"][quote="binomial_4eva"]i am stuck on this:
how do we find what place for the non-repeat ones for example  $ dbc$  without looking at the table, but only using the information $ b \equal{} 4$ and $ n \equal{} 3$.[/quote]

Try fixing the digits from the left side one at a time; let our number be $ k \equal{} \overline{a_1 a_2 a_3 \cdots a_{n_b}}, 0 \le a_i < b$. Now, a non-repeating number smaller than $ k$ will have a first digit $ c_1$ such that $ c_1 \le a_1$. If $ c_1 < a_1$, then there are $ a_1$ ways to pick $ c_1$ (counting $ 0$), and there are $ \frac {(b \minus{} 1)!}{(b \minus{} n)!}$ ways to fix the rest of the number; if $ c_1 \equal{} a_1$, then we proceed to check if $ c_2 \le a_2$, and so forth.

Then, $ P(\overline{a_1 a_2 a_3 \cdots a_{n_b}}) \equal{} \left(\sum_{i \equal{} 1}^n a_i \cdot (b \minus{} i)! \right) \left/ (b \minus{} n)! \right.$ Eg, for your example, $ P(312_4) \equal{} [3 \cdot (4 \minus{} 1)! \plus{} 1 \cdot (3 \minus{} 1)! \plus{} 2 \cdot (2 \minus{} 1)!]/1! \equal{} 22$.[/quote]

Thankx azjps but $ dbc$ is $ 423$ not $ 312$
using your formula
$ P(dbc_4)\equal{}P(423_4)\equal{}[4\cdot(4\minus{}1)!\plus{}2\cdot(4\minus{}2)!\plus{}3\cdot(4\minus{}3)!]/1!\equal{}[24\plus{}4\plus{}3]/1\equal{}31$
not $ 22$
have i misundrestood anything ?

I used the digits $ A,B \cdots \equal{} 0,1,\cdots,b \minus{} 1$ (the digits base $ b$). If instead we let $ A$ be $ 1$ etc, then there are only $ (a_i \minus{} 1)$ smaller digits for each $ i$, so the that summation would look like $ \left(\sum_{i \equal{} 1}^n (a_i \minus{} 1)(b \minus{} i)! \right) / (b \minus{} n)!$.

By the way, note that this sum is analogous to the polynomial you wrote above, except there is no longer $ b^{n \minus{} i}$ possible ways to select the remaining digits as they must be distinct.

it still dont work for some.
$ abc$ should equal to $ 1$

but according to the formula it is:
$ [(1 \minus{} 1)(4 \minus{} 1)! \plus{} (2 \minus{} 1)(4 \minus{} 2)! \plus{} (3 \minus{} 1)(4 \minus{} 3)!]/1! \equal{} [0(3)! \plus{} 1(2)! \plus{} 2(1)!]/1 \equal{} 4/1 \equal{} 4$

it should be $ 1$ but in the formula its $ 4$ and also for other bases it dont always work:

sorry am i miscalculating or  ? i dont know
i am confused

Hmm it appears that $ dbc$ appears in the $ 21$st spot or so  :oops:  You can fix for that by subtracting from each factorial by a function which describes the number of smaller digits before that particular digit, or something along those lines.

yes true
i wrote this earlier
[quote]i have a formula that always works but i cannot write it fully in mathematical form.[/quote]

this is the formula i was talking about i will write it here but its not fully mathematicalised because there are something in the formula i dont know how to write in mathematics.this formula involves the terms you said in your formula.

the formula is this and it always works
$ (L_1\minus{}1)\left(\frac{(b\minus{}1)!}{(b\minus{}n)!}\right)\plus{}\left((L_2\minus{}1)\minus{}(L_1\minus{}1)\left(\frac{(b\minus{}2)!}{(b\minus{}n)!}\right)\right)\plus{}\cdots\plus{}\left((L_i\minus{}1)\minus{}(L_{i\minus{}1}\minus{}1)\minus{}\cdots\left(\frac{(b\minus{}i)!}{(b\minus{}n)!}\right)\right)\plus{}\cdots\plus{}1$

the problem is in the $ (L_{i\minus{}1}\minus{}1)$ bits.for that bit you have to look at the letters it self to know what to put in the formula thats the problem with that formula.i will explain this later clearly, ihave to go now i will continue soon

sorry about that i was in a hurry, that formula is not exactly what i wanted to right,
i looked at my book and it is this:

$ (L_1 \minus{} \phi)\left(\frac {(b \minus{} 1)!}{(b \minus{} n)!}\right) \plus{} \left((L_2 \minus{}\phi_2)\left(\frac {(b \minus{} 2)!}{(b \minus{} n)!}\right)\right) \plus{} \cdots \plus{} \left((L_i \minus{}\phi_i)\left(\frac {(b \minus{} i)!}{(b \minus{} n)!}\right)\right) \plus{} \cdots \plus{} 1$

where $ \phi_i$ is the thing i am stuck on.i need to find the $ \phi_i$ in terms of $ b$ and $ n$.
i can explain what it is in english ,but i cannot put it in maths.

helppppppp someone please. i thought this would be solved in a day if i put it on Aops

[quote="azjps"] a function which describes the number of smaller digits before that particular digit[/quote]

Or, $ \phi_m \equal{} 1 \plus{} \sum_{i \equal{} 1}^{m\minus{}1} A_i$, where $ A_i \equal{} \begin{cases}1 &\quad a_i < a_m\\ 0 &\quad a_i > a_m\end{cases}$.

thank you  :lol: , i just dont get this, is $ A_i \equal{} a_i$? or are they different things.

can you apply it for $ bdca$ in the list with $ b \equal{} 5$ $ n \equal{} 4$? please
i want to completely understand your notation.

and can we write that in a closed form without the sigma you think?


inequalities
logarithms
calculus
function
inequalities proposed

You're faster than me,billzhao .But i have other proof  :D .We have :
[tex]\sum a\geq \sum\frac{2}{a+1}[/tex]
[tex]<=>(\sum a)(2+\sum abc+\sum ab +\sum a) \geq 2(\sum abc+2\sum ab+3\sum a +4)[/tex]
[tex]<=>\sum abc(a+b+c) +\sum ab(a+b) +\sum abc +\sum a^2 \geq 2\sum ab +4\sum a +8[/tex].
By AM-GM :
$a^2b+a^2c+a^2d+bcd \geq 4\sqrt[4]{a^6b^2c^2d^2}=4a $ ( By abcd=1)
Do similarly , we have : $ \sum ab(a+b) +\sum abc \geq 4\sum a$.
[tex] \sum a^2\geq 4[/tex] ( By AM-GM ) .We try to prove :
[tex] \sum abc(a+b+c) \geq 2\sum ab [/tex].
Since :
$a^2bc+ab^2d \geq 2\sqrt{a^3b^3cd}=2ab $ ( by abcd=1) .
.... [tex]=> \sum abc(a+b+c) \geq 2\sum ab[/tex]. :)

[quote="billzhao"]The inequality is equivalent to
\[
\sum_\mathrm{cyc} a + \frac{2a}{a+1} \geq 8
\]
Let $f(x) = e^x + \frac{2e^x}{e^x+1}$, then $f''(x) = e^x(e^{3x} + 3e^{2x} + e^x + 3)(e^x+1)^{-3} > 0$, so $f$ is convex. Thus by Jensen
\[
\sum_\mathrm{cyc} a + \frac{2a}{a+1}
= \sum_\mathrm{cyc} f(\ln a)
\geq 4 f\left( \frac{\ln a + \ln b + \ln c + \ln d}{4}\right)
= 4 f(0) = 8
\]
I am interested in seeing a non-calculus solution.[/quote]

Actually,
\[
\sum_\mathrm{cyc} a + \frac{2a}{a+1} \geq 8
\]
can be proved without calculus.
My proof:
\[
\begin{array}{l}
 \sum {a + \frac{{2a}}{{a + 1}}}  \ge 8 \\
  \Leftrightarrow \sum {\frac{{a + 1}}{2} + \frac{{a + 1}}{2} + \frac{{2a}}{{a + 1}}}  \ge 12 \\
 \end{array}
\]
Now,
\[
\begin{array}{l}
 \sum {\frac{{a + 1}}{2} + \frac{{a + 1}}{2} + \frac{{2a}}{{a + 1}}}  \\
  \ge 12\sqrt[{12}]{{\frac{{abcd(a + 1)(b + 1)(c + 1)(d + 1)}}{{16}}}} \\
  \ge 12\sqrt[{12}]{{\frac{{abcd(2\sqrt a )(2\sqrt b )(2\sqrt c )(2\sqrt d )}}{{16}}}} \\
  = 12 \\
 \end{array}
\]

It 's immediate prove by my generel inequal:

$ a_1+a_2+...+a_n \ge \frac{2}{a_1+1}+\frac{2}{a_2+1}+...+\frac{2}{a_n+1}$

for all positive $x_i$ and $x_1x_2...x_n=1$

Try to prove this!

Indeed,your Proof is very nice,Siuhochung.
Ok,It's isn't my trongest Inequality.I will post it when I find the number $K(n)$.
See  http://www.mathlinks.ro/Forum/viewtopic.php?t=30734

Now I just solve if $k(n)$ is a constant.
But I think the best $k(n)$ in a function of $n$.


Given a quadrilateral $ ABCD$, find a point $ P$ on segment $ BC$ such that $ \angle{BAP}\equal{}\angle{PDC}$.

It has been almost one year since this problem was posted. Does anybody get any idea :maybe: ?


find [b]a,b[/b] that [b](a 2+b)(a+b 2)=(a-b) 3                                  GET SET[/b]

The solution of this problem is given in the attachment (pdf-file):

oh, thanks  ;)  :lol: , it's for Mon Dec 20, 2004 12:14 pm


Hi every body,
here is music forum.
I love metallica group and their songs make me crazy!!
What about you? what's your opinion about $ Metallica$?? :D

Great group.  In no particular order, I would have to list among my favorite songs: Dyers Eve, Fight Fire with Fire, Fade to Black, One, Battery, Orion.  Of course this list is far from complete.

'enter sandman' is also pretty good, in my opinion  :)

I think it's somewhat overrated.  However, it is great when Mariano Rivera comes into a game. :)

'...And Justice For All' is a classic. Great band, great band. Anyone here like Beatallica?

Don't post illegal links to copyrighted material.

sorry.I do'nt post  download links.
thanks.

First 4 albums great.

Then I think they're pretty bad.
I think Megadeth is better :))))))

Um... I never got into them really.
They're okay.

Metallica's pretty good
I like rage against the machine, green day, and guns n roses better tho
but i like rap the best:
eminem
ludacris
lupe fiasco
jay-z
kanye west
b.o.b
50 cent
dr. dre
2pac
mos def
bustarhymes
andre 3000
The Notorious B.I.G

Now that I'm a little bit older, I realize that Metallica is not great, but the greatest.  :P

Well, [url=http://www.mediafire.com/?46dh3y65lp7nu79]Metallica's lyrics[/url] are great, and the melodies are excellent. I used to listen to rap songs, but I just found that James Hetfield can sing Eminem's songs (just listen to "Stone Cold Crazy" song in "Garage, Inc" album), however Eminem would never can James' riffs. :)

Metallica can never match great rappers' lyricism like eminem, 2pac, ludacris, etc.
you know all of those guys are professional poets as well which metallica will never be

[url=http://www.spinner.com/2007/03/27/the-worst-lyrics-ever-no-14/]Sure[/url]. :)

Let's leave this topic as we would never can say that Metal > Rap or Rap > Metal. Both metal/rock and rap have pretty nice/bad songs we can't deny.

1) The youtube link has been removed there, and I don't even find the name of "The Worst Metallica Song."
2) I'm sure you just found it in Google without checking it's availability.

[quote="fedja"][quote="amparvardi"]their songs make me crazy!!
[/quote]
OK, so what's your favorite Metallica lyrics? I clicked on a few songs at random and somehow I'm not impressed though I admit that my random selection may be non-representative at all.[/quote]
If you want good lyrics, search up eminem lose yourself lyrics (one of the best lyrical compositions ever)
or just any ludacris song for he is always consistent on his amazing lyrics

Do you mind dropping a link or two? My search was a complete failure :(.

sure i can give you some song lists by some artists but if u want links then pm me

ludacris:
bada boom
get back
my chick bad
act a fool

eminem:
not afraid
lose yourself
no love
the real slim shady
the way i am
stan
space bound
rock bottom

2pac:
hit em up
changes

btw i forgot to mention that all rappers started off as freestyle rappers and freestyle rap battlers here is a good example of a freestyle(non-battle type) by the best freestyler(mc jin)
http://www.youtube.com/watch?v=Xfi1-NugocM

[quote="fedja"]
OK, so what's your favorite Metallica lyrics? I clicked on a few songs at random and somehow I'm not impressed though I admit that my random selection may be non-representative at all.[/quote]

Sorry for the late reply, I was at school.

BTW, nice question!

We may divide Metallica songs into some groups (I just give few examples for each group. There are many more to see):

1) Social ones. Some songs which describe society's problems:

[list]
[*]"[url=http://en.wikipedia.org/wiki/Master_of_Puppets_%28song%29]Master of Puppets[/url]" deals pretty much with drugs. Listen to it and read the lyrics to find out how nice James describes addiction as a person who controls the addict person.

[*]"[url=http://en.wikipedia.org/wiki/Sad_But_True]Sad But True[/url]" speaks about how your bad-human-part tempts you and then "You're the one who's shamed...". The interesting thing about this song is that James Hetfield speaks as different persons.

[*]"[url=http://en.wikipedia.org/wiki/One_%28Metallica_song%29]One[/url]" is about the war and how it can destroy one's dreams (read wikipedia about the concept of this song).[/list]

2) Emotional ones. Non-metal-fans probably like this kind of songs much than others:

[list]
[*]"[url=http://en.wikipedia.org/wiki/Nothing_Else_Matters]Nothing Else Matters[/url]", the most popular Metallica song. A very nice melodic song which has been [url=http://en.wikipedia.org/wiki/Nothing_Else_Matters#Covers]covered[/url] by many other singers.

[*]"[url=http://en.wikipedia.org/wiki/The_Unforgiven_%28song%29]The Unforgiven I, II, and III[/url]". These three songs are about forgiving. I like The Unforgiven I the most because of its nice lyrics and music. The video clips of "Un I" and "Un 2" are really nice (but you are probably not able to realize what they mean when you see them for the first time. I highly suggest you seeing these clips).

[*]"[url=http://en.wikipedia.org/wiki/Until_It_Sleeps]Until It Sleeps[/url]": The lyrics of the song, written by James, address the loss of his father after a battle with cancer, similar to his mother's death. He was absolutely very depressed and you can really feel this if you scrutinize the lyrics. The video clip of this song is one of the most intelligent clips I've ever seen: every scene in the clip describes something really nice (I would first suggest you seeing the clip, and then I will write something about how strong it is and what anything means in the clip).

[*]"[url=http://en.wikipedia.org/wiki/Fade_to_Black_%28song%29]Fade to Black[/url]". The lyrics of this song are really sad - but poetic - and it has a very nice solo at beginning. One of my favorite songs.
[/list]

3) Instrumental ones. No vocals!

[list]
[*]"[url=http://www.youtube.com/watch?v=WuyZ4Bt73mg]Suicide and Redemption[/url]". Yes... instrumentals can be more powerful than words. I should explain what does this song mean. This place is not large enough to write!

[*]"[url=http://www.youtube.com/watch?v=OzMJhOwBLqw]Orion[/url]" is what really reminds me of old Metallica, with the first - and IMHO, the best - bassist of the group.[/list]

4) Covered ones. Other groups' songs re-played by Metallica. Tracks of the second CD of "Garage, Inc" album go under this part. Some example are "Sabbra Cadabra" by [i]Black Sabbath[/i], "It's Electric" and "Am I Evil?" by [i]Diamond Head[/i], "Breadfan" by [i]Budgie[/i], etc.

I hope I find some time to write the explanations of some nice songs of Metallica. :)


Let .... Prove that there exists infintly many $n$, such that ....

Edited by pbornsztein, according to the above message from Laurent.

In practice, for organisation problems, the tournament of towns is not organized the same day in all countries. So the rule fixed by the turnir goradov team is to not publish the tournament till one month after the official competition time.
Please respect this rule.

Thank you,
Laurent


trigonometry
ratio
quadratics
geometry
angle bisector
algebra

Construct an isoscele triangle with angle measure: $36^o $-$72^o$-$72^o$. Draw the angle bisector of one of the $72^o$ angle. Then it will give us the Golden Ratio. Last use the Law of Cosine and obtain the answer.

[quote="shobber"][quote="Slizzel"]Find $\cos\frac{\pi}{5}$ :)[/quote]
Is it:
[hide]
$\frac{1+ \sqrt{5} }{4}$[/hide][/quote]

$\cos{\frac{\pi}{5}}$ is half the golden ratio?!  :!: there isn't a possible cosine value for the golden ratio though, because it's over 1.

Well, I mean the construction I made to get the answer involves the Golden Ratio. If you do the construction as I said in that post, you will get that the ratio of the base and side of the triangle is $\frac{\sqrt{5}-1}{2}$. Then use the Law of Cosine.

[quote="Slizzel"]Well I don't know what's the Golden Ratio :? Sorry...  :blush:[/quote]
Contruct $\triangle{ABC}$ with $\angle{A}=36^o$, $\angle{B}=\angle{C}=72^{o}$. Thus  $AB=AC$.
Now construct the angle bisector of $\angle{C}$ to meet $AB$ at $D$. So $\triangle{BCD} \sim \triangle{BAC}$.
Hence: $BC^2=BD \cdot BA= (BA-BD) \cdot AB= (BA-BC) \cdot AB$. Solve this equation we have:
\[\frac{BC}{AB}=\frac{\sqrt{5}-1}{2}\].

This is called the Golden Ratio.

So apply the law of cosine: $\cos{36^o}=\frac{\sqrt{5}+1}{4}$.

I think that Algebra is more complex than this one but the basic theory is the same: $36^o=2 \cdot (90^o-2\cdot 36^o)$

[quote="shobber"]$\frac{BC}{AB}=\frac{\sqrt{5}-1}{2}$.

This is called the Golden Ratio.
[/quote]

Ii thought it was $\frac{\sqrt{5}+1}{2}$....

To find the golden ratio, solve this equation and take the larger root (the smaller one is its reciprocal):

$x-\frac{1}{x}=1$.

Yeah, the golden ration [i]is[/i] $\displaystyle \frac{\sqrt{5}+1}{2}$.

It is an algebraic number, the root of the quadratic $x^2-x-1=0$ ,similar to what 236factorial wrote.

Isnt this well known ?

Let $y=\frac{\pi}{5}$ so the question is now find $\cos y$

$y=\frac{\pi}{5}$ :arrow:  $\pi-3y=2y$  :arrow: $\sin(\pi-3y)=\sin 2y$

:arrow: $3\sin y-4\sin^3y=2\sin y\cos y$
:arrow: $3-4\sin^2 y=2\cos y$     (since $\sin y\neq 0$ ,we can cancel off)
:arrow: $4\cos^2y-2\cos-1=0$
Solving quadratic equation and get

 $\cos y=\frac{1+\sqrt 5}{4}$   (we take positive sign since $\cos y$ is in first quadrant)


I was wondering what is they favorite sport of aops members?

really as you can see when you look at my avartar i hate tennis  :rotfl:

haha.
i'm into basketball, tennis, baseball, and i play football.
meh. i'm into track too.(i sprinted the 50 yrd dash in 6.8 seconds.)

oh yes, chess i a non-physical sport, i am pwning currently(on a big winning streak, something like 92)

Where do you play chess people? playchess.com FTW

Tennis, swimming, volleyball, chess

chess.com all the way. Even though the advertisements and the constant reminders ("Switch to Premium membership NOW!") are annoying, it's a good site.

I disagree that chess is a "sport", it's more of a "competition".

Also, I'm a tennis player

Chess is indeed a sport. A competition is when a bunch of people get together and compete. There are chess competitions, but chess is a sport.

I play on [hide="I"] internet[/hide] Chess Club.

Maybe we should make a topic about swimming styles :P

I am free-styler (not Boomfunk MC's)

@isabella
I thought a sport was a physical competition.

Don't get me wrong, I don't find anything wrong with chess, I play it too

I didn't know chess was a sport, oh well.

At least I learn other stuff other than math on AoPS. :)

Really?

I've never played Ultimate Frisbee before...

i put basketball because I love to play street basketball! But tied with that is track and field!


Can someone explain me about [url=http://en.wikipedia.org/wiki/Briggs-Rauscher_reaction]this[/url] reaction....
I need it urgenty tomorrow for a project tomorrow

Serves u right for choosing to quote such complex reactions :rotfl:  :rotfl: (SORRY!!!)

Anyway Rohit told me of BZ reaction in one of the older year projects of our school. I don't remeber of its contents. So probably rohit knows it. :maybe:  :noo:

dei it's not complex atleast the reaction it's understandable...we saw that in the KVPY camp...it was wonderful...just wanted a second thought from people like Carcul...we have our pracs tommoroww

I believe all the information required is on the wikipedia page. Otherwise, what are your doubts?

If you have doubts, then be more explicit about them.

[quote="pardesi"]yes they are there but i jsut wanted to know how i would explain...i am not pretty much confident with many terms of the page...what i know and probably understand is only the reactions taking place...which i read from somewhere else...[/quote]

Actually i was supposed to be doing this project  :D  (i.e. oscillations and clock rns) for my project but my teacher said it wont give a good result (even though i told her that i am fascinated by this rn and would like to do it) and hence dissuaded my group from taking the project  :(

Anyways what are ur doubts in the process  :?:


search
number theory unsolved
number theory

Find all $ (x,y,z) \in N$ such that $ 2^x\plus{}3^y\equal{}5^z$

already posted, have a look at http://www.mathlinks.ro/viewtopic.php?search_id=687566259&t=38955

I searched but found nothing, sorry...
However, I don`t understand the result of the problem... which are the solutions he found?


One out of every five Americans suffers from allergies. In a random sample of 250 Americans, how many would you expect to suffer from allergies?

This is simply a matter of ratio:

$ \frac{1}{5}\equal{}\frac{x}{250}$

Solving the above by multiplying $ 250$ on both sides, we find that our answer is $ \fbox{50}$.


¡Hola allí cada uno! ¡Alegre estar aquí en tu foro!
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If  want learn English, please to my blog, where I posted some website on English :)

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[quote="Ulanbek_Kyzylorda KTL"]sin(x)cos(x)cos(2x)cos(8x)=1/4sin(12x).
Find x.[/quote]

$ \sin(x)\cos(x)\cos(2x)\cos(8x)\equal{}\frac 14\sin(12x)$ $ \iff$ $ 2\sin(2x)\cos(2x)\cos(8x)\equal{}\sin(8x\plus{}4x)$

$ \iff$ $ \sin(4x)\cos(8x)\equal{}\sin(8x)\cos(4x)\plus{}\sin(4x)\cos(8x)$

$ \iff$ $ \sin(8x)\cos(4x)\equal{}0$

Hence the solutions : $ x\equal{}k\frac{\pi}8$

[quote="Ulanbek_Kyzylorda KTL"]sorry pco can you explain why x=kп/8[/quote]

$ \sin(8x)\cos(4x)\equal{}0$ $ \iff$ $ \sin(8x)\equal{}0$ or $ \cos(4x)\equal{}0$

$ \cos(4x)\equal{}0$ $ \implies$ $ \sin(8x)\equal{}2\sin(4x)\cos(4x)\equal{}0$ and so : $ \sin(8x)\cos(4x)\equal{}0$ $ \iff$ $ \sin(8x)\equal{}0$ $ \iff$ $ 8x\equal{}k\pi$ $ \iff$ $ x\equal{}\frac{k\pi}8$


[quote="cuenca"]Let $ABC$, a triangle such that $\angle A>45$, $\angle B>45$, If $PQRS$, is a square with $A,P,Q,B$ in a line in that order, $R$ in $BC$, $S$ in $AC$, Let $Q_{1}$ the foot of the perpendicular from $Q$ to $AC$, and $P_{1}$ the foot of the perpendicular from $P$ to $BC$, let $H$ the intersection of the lines $QQ_{1}$ and $PP_{1}$, prove that $CH$ is perpendicular to $AB$.[/quote]
You can easy see that as $\angle A>45$, $\angle B>45$=>$Q_{1}\in{[AS]}$ and $P_{1}\in{[BR]}$.We also have that $P,Q,P_{1},R,S,Q_{1}$ lie on the same circle, and even in this order :) .So let $H_{1}$ be the feet of perpendicular from $H$ to $AB$.Then \[\angle{QHH_{1}}=\angle{Q_{1}QR}=\angle{Q_{1}P_{1}R}=\angle{Q_{1}P_{1}C}=\angle{Q_{1}HC}\]
=>
\[\angle{QHH_{1}}=\angle{Q_{1}HC}\]
=>$C,H$ and $H_{1}$ are collinear :wink: .

interesting problem! actually, is also valid if $PQRS$ is a rectangle!

Let $E\neq C$ such that $E= (CPS)\cap (CRQ)$.

Notice that if $C, H$ and $E$ are collinear, then we know that:

Radical axis of $(PQRS)$ and $(CPSE)$ is $PS$
Radical axis of $(PQRS)$ and $(CRQE)$ is $RQ$
Radical axis of $(CPSE)$ and $(CQRE)$ is $CE$

So, $PQ, QR, CE$ concurr, but as $PS$ and $RQ$ concurr in the point at infinity, then the point of infinity of these lines is in $CE$, so $CE// PS$. Finaly, we would have that $CH//PS\perp AC\implies CH\perp AC$ as wanted.
[hide= Remark] There's a simpler way to prove that $C, H, E$ collinear implies directly that $CH\perp AC$ that a friend explained to me... Is just that the center of $(PSCE)$ is on the perpendicular bisector of $PS$. And the center of $(CRQE)$ is on the perpendicular bisector of $RQ$. More than that! The centers are in the same line, after all $PS= RQ$ and $PS//RQ$. So, it's known that the radical axis of two circles is a line that is perpendicular to the line that connects the two centers. Then, $CH$ is perpendicular to the line that connects the two centers, who is also the perpendicular bisector of $PS$, which is parallel to $AC$, and finished.[/hide]

Okay, now we just gotta prove that $C, H, E$ are collinear.
Notice that $\angle CEP= 180-\angle PSC= 90-A$ and similarly, $\angle QEC=90-B\implies \angle PEQ= C$. So, $HPQE$ is cyclic!

As $\angle CEP= 90-A$ and $\angle HQP= \angle Q_1QA= 90-A\implies \angle CEP= \angle HQP= \angle HEP\implies C, H$ and $E$ are collinear and doone!

Just to clearfy: $A= \angle CAB, B=\angle ABC, C=\angle BCA$


quadratics
inequalities
algebra
inequalities unsolved

Assume that the equation $ x^3 \plus{} pqx \plus{} r \equal{} 0$ has at least one solution $ x$ in reals. Prove that:

$ (\frac {pq}{\sqrt {2}})^2 \geq 2xr$

P.S. I don't know exactly how it states but it was something like this - maybe a little bit complicated

Please check the problem... Is the equation really $ x^3 \plus{} pq \plus{} r \equal{} 0$ ? Obviously, the equation $ x^3 \plus{} pq \plus{} r \equal{} 0$ always has at least one solution in reals.

  darij

please forgive me Darij  :)  - what a shame   :blush: - after all THERE WAS a mistake in this equation - I forgot about $x$ near $pq$ - now it's corrected. I'm really sorry ...

Hmm, that's nice, but... can't we treat pq as one variable now? Moreover, the problem as you posted it is VERY easy, so I think it's not what you wanted to post. I'm still confused...

  dg

it is EASY, so post your solution cos it's probably correct. Even pq are not variables but prior chosen numbers so the problem is really simply, but quite nice. So after you post your solution we can complicate it a little  :D

Okay, but I have said that the problem doesn't look authentic. In fact, we can replace pq by s and the problem becomes:

$ x^3 \plus{} sx \plus{} r \equal{} 0$ has at least one solution in reals, prove that:

$ \left(\frac {s}{\sqrt {2}}\right)^2 \geq 2xr$.

In fact, we can rewrite the equation $ x^3 \plus{} sx \plus{} r \equal{} 0$ as $ xx^2 \plus{} sx \plus{} r \equal{} 0$. If we now consider the quadratic equation $ xX^2 \plus{} sX \plus{} r \equal{} 0$, with X being a variable, then it follows that X = x is a solution of this equation. Hence, the discriminant $ s^2 \minus{} 4xr$ of this quadratic equation must be $ \geq 0$, and thus, we have $ s^2 \geq 4xr$, so $ \frac {s^2}{2} \geq 2xr$, i. e. $ \left(\frac {s}{\sqrt2}\right)^2 \geq 2xr$.

I'm looking forward to the complication...  ;)

  Darij

solution is absolutely correct - it was an old problem from the first stage of Polish Olympiad  :) - give me one day and I complicate it "a biiit"   :D.

Ok I've posted complicated version in "complicated short and nice" topic in "inequalities proposed and own problems" section - good luck  :cool: I hope you'll like it  :D


How $ m^3$ is $ 4.6 \ \text{cm}^3$?  I'm getting thrown off since the units is cubed.  :oops:

Thanks in advance!

$ 1\mbox{m^{3}}=100\text{cm}\times 100\text{cm}\times 100\text{cm}$

That should help you.

Just for future reference, if you ever need squared or cubic conversion factors, you can always use the conversion factors you already know to derive new ones. $ 1\text{m}\equal{}100\text{cm}$. If you cube both sides, you will find that $ 1\text{m}^{3}\equal{}1 000 000 \text{cm}^{3}$.


combinatorial geometry
combinatorics
polygon
Extremal combinatorics
IMO Shortlist

Given is a convex polygon $ P$ with $ n$ vertices. Triangle whose vertices lie on vertices of $ P$ is called [i]good [/i] if all its sides are unit length. Prove that there are at most $ \frac {2n}{3}$ [i]good[/i] triangles.

[i]Author: Vyacheslav Yasinskiy, Ukraine[/i]

[quote="delegat"]Given is a convex polygon $ P$ with $ n$ vertices. Triangle whose vertices lie on vertices of $ P$ is called [i]good [/i] if all its sides are equal in length. Prove that there are at most $ \frac {2n}{3}$ [i]good[/i] triangles.[/quote]

That's a demanding, but well-known problem.
A solution can be found for instance in:

---Janos Pach, Rom Pinchasi
---How many unit equilateral triangles can be generated by N points in convex position?
---American Mathematical Monthly 110(5): 400-406 (2003)

Hmmm, I think it is also true that there are at most $ \frac{n}{3}$ squares among the vertices of a convex $ n$-gon.

What is the maximum number of times that the unit distance can occur among the distances between n points in the plane?

Nice problem!! Here is my solution:

To each vertex $V$, assign at most two good triangles in the following way: Consider all the good triangles which have a vertex $V$, assign to $V$ the leftmost clockwise and the rightmost clockwise (this is possible because all good triangles lie on a half-plane determined by some line through $V$).

Now take $ABC$ a good triangle (those vertices are in clockwise order). I prove that it is either assigned to $B$ as the leftmost clockwise triangle, or assigned to $C$ as the rightmost clockwise triangle. To see this, assume not, then there are good triangles $BB'B''$ and $CC'C''$ more to the right clockwise and more to the left clockwise than $ABC$, respectively, such that $B'$ and $C$ are on different sides of $AB$, and $C'$ and $B$ are on different sides of $AC$. Then $\angle B'AC' > 180$ but $\angle BAC < 60$, contradiction since the polygon is convex.

From this, we see that every good triangle is assigned at least 3 times. Since each vertex is assigned at most 2 triangles, we easily finish and therefore there are at most $2n/3$ good triangles. $\blacksquare$

Consider triples $(T,v,\varepsilon)$ where $T$ is a good triangle, $v$ is one of its vertices, and $\varepsilon=1$ if there is no other good triangle with vertex $v$ that is $T$ rotated counterclockwise by some angle less than $\pi$, and $\varepsilon=-1$ if the previous statement holds with clockwise. Note that for a good triangle $T$ and one of its vertices $v$, there may not always exist an appropriate $\varepsilon$. Call such triples [i]extremal[/i].

We first claim that there exactly $2n$ extremal triples. Indeed, for each vertex $v$ of $P$, there is exactly one ``most clockwise'' good triangle, and one ``most counterclockwise'' good triangle. Note that these may be the same, but still count towards different triples since $\varepsilon$ is different. Thus, for each $v$, there are exactly $2$ extremal triples, so we have total $2n$ such triples.

The key claim is the following.

[b]Claim:[/b]  Consider a good triangle $T=abc$. Then, at least three of the following triples are extremal: \[(T,a,1),(T,a,-1),(T,b,1),(T,b,-1),(T,c,1),(T,c,-1)\]

[i]Proof:[/i]  Suppose WLOG that $abc$ is oriented counterclockwise. We claim that one of $(T,b,1)$ and $(T,c,-1)$ is extremal, and by symmetry thus one of $(T,c,1)$ and $(T,a,-1)$, and one of $(T,a,1)$ and $(T,b,-1)$ are extremal. This would complete the proof.

Suppose for the sake of contradiction that neither $(T,b,1)$ nor $(T,c,-1)$ are extremal. Let $\ell$ be the line through $a$ parallel to $bc$. It is easy to see then that one of the vertices of a good triangle with vertex $b$ which is $T$ rotated counterclockwise is above or on $\ell$, and similarly one of the vertices of a good triangle with vertex $c$ which is $T$ rotated clockwise is above or on $\ell$, so two vertices of $P$ are above or on $\ell$. Furthermore, these two are on opposite sides of the perpendicular line to $\ell$ through $a$, so $a$ is actually inside or on the convex hull of these two vertices along with $b$ and $c$, which violates the convexity of $P$. This completes the proof of the claim. $\blacksquare$

This now finishes, as we have at least $3X$ extremal triples, where $X$ is the number of good triangles, so $3X\le 2n$, or $X\le\frac{2n}{3}$, as desired.


Find maximum value of the expression:
\[ 2\sqrt{5-\frac{(2k+1)^2}{k^2+1}}+\sqrt{5-\frac{(2k-1)^2}{k^2+1}}.  \]
where k is real number.

This simplifies to $f(k)=\frac{2|k-2|-|k+2|}{\sqrt{k^2+1}}$, and then there are three branches to examine for extremes: $(-\infty,-2),[-2,2),[2,\infty)$. Will leave that to someone else.


trigonometry
inequalities
trig identities
Law of Cosines
triangle inequality

We know that given the sysem:
$ ay\plus{}bx\equal{}c\\
 az\plus{}cx\equal{}b \\
bz\plus{}cy\equal{}a$
for $ a,b,$ and $ c$ representing non-zero constants, we can apply a very creative and slick geometric approach (in addition to algebra-bashing) to solve for the variables $ (x,y,z)$ in terms of $ a,b,c$.
Solving, you will find that the solution set is very similar in form to the cosine in the law of cosines (e.g. $ \cos B\equal{}\frac{a^2\plus{}c^2\minus{}b^2}{2ac}$) We know this to work because if you consider an arbitrary triangle ABC, you can extend altitudes and express each side length as a product of the cosines of each angle adjacent to the one opposite the side in question and the adjacent side lengths.

My question is: what happens to this geometric approach to solving the system of the constants $ a,b,c$ do not satisfy the triangle inequality. This makes it so you can not form a triangle. Does this mean that the system has no solution, or does the geometric approach just fall apart at this point?

anybody have any kind words to contribute, or is my question just too vague?

$ x\equal{}\frac{b^2\plus{}c^2\minus{}a^2}{2bc}$ and symmetrical others is always true (for $ abc\ne 0$).

I don't know what else you are saying, but if $ a,b,c$ don't satisfy the triangle inequality, then you are going to get $ x\not\in [\minus{}1,1]$...

So basically:

You know there is a solution. The geometric way leads you to the explicit form for the solution to the system. That form works whether or not the a,b,c form a triangle.


After a decent amount of searching on the web, I couldn't find the dates for the 2008 PuMaC. Since it's coming up, I was wondering if the dates have been decided, and if so, when?

-diophantient

See the topic 3 below this

Quote: I heard it was ttoward the end of January...

I'm not sure that's accurate though

Official word on this would be much appreciated.

It's at the end of January -- we thought there would be fewer conflicts around that time. I'll post an official announcement soon.


Has anyone on this site heard of or have used Grammatix? It claims to have real SAT strategies that WORK. People on the College Confidential Forum said Grammatix really WORK because they have tried it. I am not sure if I should believe this. Has anyone heard of this "product"?

I've not heard of this particular product. Googling it, I see the product's Web site. At least it isn't quite as outrageously expensive as some other products I have seen, but I would still start with [url=http://www.amazon.com/exec/obidos/tg/detail/-/0874477182/learninfreed]The Official SAT Study Guide: For the New SAT[/url] published by the College Board before using any other test preparation books.

Are you a native speaker of another language besides English? That might raise some other issues about how you should best prepare.

What does your question "Are you a native speaker of another language besides English?" mean?

I do speak chinese at home but I was born in NY if thats what you mean.


function
integration
real analysis
real analysis unsolved

If $ \chi$ means the indicator function and $ \to$ means the convergence in $ \mathcal D'(\mathbb R)$ (or $ \mathcal S'(\mathbb R)$), then the limit is Dirac's delta function.

I guess the limit is 0, but I have not known how to prove it???

[quote="zhoraster"]If $ \to$ means pointwise convergence, then the limit is zero. Just try to prove that any point $ x$ does not belong to $ (0,1/n)$ for $ n$ big enough.[/quote]

Yes, this is the main point I am fighting. But I have not found any clues until now!

[quote="man.maiko"]$ n\chi_{(0,\frac {1}{n})}(x)\to ?$ as $ n\to \infty$?[/quote]

$ \exists \varepsilon>0, n_0\in N:$ For $ n>n_0$ \[ n\chi_{(0,1/n)}(x)\ge \varepsilon\] for every $ x\in R.$ Then \[ 1\equal{}\int_{R}n\chi_{(0,1/n)}(x)dx\ge \int_{R}\varepsilon dx\equal{}\infty.\] This contradiction yields \[ n\chi_{(0,\frac {1}{n})}(x)\to 0\] a.e. in $ R$, $ n\in N.$ :roll:

No, [b]livenju[/b], your argument is not correct, because $ \varepsilon$ might depend on $ x$. Also you cannot claim that $ f_n(x)>\varepsilon(x)$ for [i]all[/i] indices $ n$ large enough, you can claim this just for infinite amount of indices.

[quote="zhoraster"]No, [b]livenju[/b], your argument is not correct, because $ \varepsilon$ might depend on $ x$. Also you cannot claim that $ f_n(x) > \varepsilon(x)$ for [i]all[/i] indices $ n$ large enough, you can claim this just for infinite amount of indices.[/quote]

Thanks, zhoraster!

[quote="zhoraster"]For $ x > 0$ take $ n\ge 1/x$.[/quote]

For $ x>0,\exists n_0: \forall n>n_0, 1/n\to 0$ and $ x>1/n_0$. Then $ x>1/n$, so $ f_n(x)\equal{}n\chi_{(0,1/n)}(x)\equal{}0.$ We get $ f_n\to 0$ almost uniformly. Ok?

[quote="zhoraster"]Something like this, though very messy.[/quote]

Thank [b]zhoraster[/b]! Ha ha.


Yes, you got it right.  But aren't there two correct answer choices?
[hide="Specifically"][b]D[/b] is the answer mentioned in the solutions booklet and the AMC website, but isn't [b]A[/b] correct as well?  I'm guessing that [b]A[/b] was supposed to be $N < 1$.[/hide]
I looked at the official test booklet, and I have copied the problem exactly as written.

Does anyone know whether both correct answers were given credit on the 2002 AMC 12A?

If people don't understand logarithms, finding a log base 2002 might be a little difficult though.

I agree that the answer choices are odd ... $N<1$ would make much more sense for A.

Most likely a typo.  I would've said D.  But I would've also spoken with the proctor about it.

Most tests normally want you to give the "most correct" answer, which, I think, would be D in this case.

i recall that the answer for A was less than 1

Maybe.  All I can say is that choice [b]A[/b] says $N > 1$ in the test booklet I purchased from the AMC, in the "21st Century" CD I purchased from the AMC, in Doug Faires's book, and on the KSU website.

Ah yes.

In the CD, there may be a mistake, but on the pamphlet it is the real deal and that is <1.

[quote="mathgeniuse^ln(x)"]In the CD, there may be a mistake, but on the pamphlet it is the real deal and that is <1.[/quote]
That's interesting.  In the test pamphlet I purchased from the AMC about 2 years ago, choice A says $N > 1$.  So your test pamphlet from the AMC says $N < 1$?

[quote="Boy Soprano II"][quote="life=tennis+math"][quote="Art of Owna"]This question is kind of stupid because anyone can plug that into their calculator and do it in liek 5 secs.[/quote]

Can someone show me how to put the bases stuff on the calculator (for example, this one has a base of 2002). While I'm somewhat confident with logs, it's always nice to get the problems done quicker to be able to attempt the other problems.[/quote]
$a^{\log_{a}b \cdot \log_{b}c}= (b)^{\log_{b}c}= c \Longrightarrow \log_{a}b \cdot \log_{b}c = \log_{a}c \Longrightarrow \frac{\log_{a}c}{\log_{a}b}= \log_{b}c$ :!:
[hide="hint"]Try $a \in \{ 10, e \}$ (the latter is preferred)[/hide][/quote]

That's what I was saying about calculators on this problem; if people don't understand logarithms well enough, they won't know to do that, so they can't just plug it into their calculators.

yea...but like...all you need to know for amc logs is change of base then...


A point is chosen at random on the number line between 0 and 1, and the point is colored green. Then, another point is chosen at random on the number line between 0 and 1, and this point is colored purple. What is the probability that the number of the purple point is greater than the number of the green point, but less than twice the number of the green point?

Make a graph with y-axis green and x-axis purple. Then mark in .25,.5,.75 and 1. Then test each one of those for purple and find the wanted region. You'll see a pattern. The shaded region should be 1/4 of all possible values.


trigonometry
exterior angle
geometry proposed
geometry

[color=darkblue]I"ll present you the following problem inspired from the K®izalid's problem 3 :[/color] http://www.mathlinks.ro/Forum/viewtopic.php?t=145944

[color=darkred]If, in a $\triangle ABC$, median $\overline{BD}$ is such that $A=2C$, and $m(\widehat{ADB })= 45^{\circ}$, then prove that $C=\frac{\pi}{12}$.[/color]

Trigonometrically: Denote $\angle{C}=x$, therefore $\angle{A}=2x, \angle{ABD}=135-2x, \angle{DBC}=45-x$. Now we have from the sin law: $\frac{CD}{\sin{(45-x)}}=\frac{BD}{\sin{x}}$ and $\frac{AD}{\sin{(45+2x)}}=\frac{BD}{\sin{2x}}$, and dividing them we obtain:
$\sin{x}\cdot \sin{(45+2x)}=\sin{2x}\cdot \sin{(45-x)}$
$\iff 2\cos{(45+x)}=\cos{(45-3x)}+\cos{(45+3x)}= 2\cos{45}\cdot \cos{3x}$, therefore:
$\cos45 \cdot \cos{x}-\sin{45}\sin{x}= \cos{45}\cdot \cos{3x}$
$\iff \sin{x}=\cos{x}-\cos{3x}=2\sin{x}\sin{2x}$, and now because $x$ is not $0$, we have $2\sin{2x}=1$, therefore $x=15^\circ$ , because it is acute beeing $<45^\circ$ from the triangle $BDC$ which has the exterior angle $\angle{BDA}=45^\circ$.

Let $\angle C=x$ then $\angle A=2x,\,\angle B=180^\circ-3x$
We have:
$\frac{AB}{AD}=\frac{\sin\widehat{ADB}}{\sin\widehat{ABD}}\Longleftrightarrow\frac{2c}{b}=\frac{\sin 45^\circ}{\sin (45^\circ+2x)}\Longleftrightarrow\frac{2\sin x}{\sin 3x}=\frac{1}{\sin 2x+\cos 2x}$

$\Longleftrightarrow 2\sin x(\sin 2x+\cos 2x)=3\sin x-4\sin^{3}x$ $\Longleftrightarrow 2\sin 2x+2(1-2\sin^{2}x)=3-4\sin^{2}$
$\Longleftrightarrow\sin 2x=\frac{1}{2}\Longleftrightarrow x=\frac{\pi}{12}$ because $x<\frac{\pi}{4}$

Nice proofs  :lol:

Thanks Virgil for inspirate on the problems  :ninja:

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=146184]By the way[/url]


Let be $ a,b,c>0$ such that $ \frac{1}{a^2\plus{}1}\plus{}\frac{1}{b^2\plus{}1}\plus{}\frac{1}{c^2\plus{}1}\equal{}2$ . Prove that :

$ ab\plus{}bc\plus{}ca\le \frac{3}{2}$

[quote="alex2008"]Let be $ a,b,c > 0$ such that $ \frac {1}{a^2 \plus{} 1} \plus{} \frac {1}{b^2 \plus{} 1} \plus{} \frac {1}{c^2 \plus{} 1} \equal{} 2$ . Prove that :

$ ab \plus{} bc \plus{} ca\le \frac {3}{2}$[/quote]

See http://www.mathlinks.ro/viewtopic.php?p=974624#974624


abstract algebra
algebra
polynomial
linear algebra
matrix
number theory
relatively prime

Let $R$ be a finite associative ring in which there is exactly one non-zero element that is neither a left nor a right divisor of 0. Prove that $R$ is boolean.

Denote the element, which is not a zero-divisor from both sides
by $ e $. Then $ -e , e^{2} $ are also not a zero divisors,
therefore because of uniqueness $ e = -e = e^{2} $ so $ 2e = 0 $,
therefore $ 2ex = 0 $, for every $ x \in R $. Also, the
homomorphism $ R \rightarrow R $, $ x \mapsto ex $ doesn't have
kernel, so it is a bijection. Therefore $ 2x=0 $ for every $ x $.
Now, for every $ x \in R $ we have $ e(ex-x) = e^{2}x-ex = ex-ex=0
$, therefore $ ex=x $. By analogy, $ xe=x $. So we get that $ e $
is a unit. Now, because of $ 2 = 0 $ in $ R $, then $ R $ is an $
\mathbb{Z}_{2} $ -algebra. Note also that if some element $ x $
has left and right inverse, it must be $ e $. Therefore there are
no nilpotents except $ x=0 $, because if $ x^{n} = 0 $, then $
(e-x)(e+x+x^{2} + \ldots + x^{n-1}) = (e+x+x^{2} + \ldots +
x^{n-1})(e-x) = e-x^{n} = e $, so $ e-x $ is invertible, so $ e-x
= e $, so $ x = 0 $. Take some element $ x \in R $, such that $ x
\neq 0,e $. Denote by $ P(t) \in \mathbb{Z}_{2}[t] $ the minimal
polynomial of $ x $. First of all, $ P(0) = P(1) = 0 $, because if
$ P(0) = 1 $, then $ x $ will be invertible, and if $ P(1) = 1 $,
then $ x-1 $ will be invertible. Then we can write $ P(t) =
t^{n}(t-1)^{m}Q(x) $, where $ n,m \geqslant 1 $ and $ Q(0)=Q(1)=1
$. Then element $ y = x(x-1)Q(x) $ is nilpotent, so $ y = 0 $,
therefore because of minimality of $ P $, we get that $ n=m=1 $.
Now, let us prove that $ deg Q = 0 $. Indeed, if it is not the
case, then consider $ T(t) = (t-1) + tQ(t) $. Then $ deg(T) =
deg(P) - 1 $ and it is easy to see that $ T(t) $ is relatively
prime to $ P(t) $. Therefore there exists a polynomial $ S(t) $,
such that $ T(t)S(t) \equiv 1 ( mod \ P(t) ) $. Therefore $
T(x)S(x) = S(x)T(x) = e $, so $ T(x) $ is invertible, therefore $
T(x) = e $, then $ P_{1}(t) = T(t) - 1 $ vanish $ x $, which
contradicts to minimality of $ P $. So we get that $ Q(t) \equiv 1
$, so $ P(t) = t(t-1) $, therefore $ x = x^{2} $. So we get that
for every $ x \neq 0,e $ we have $ x^{2} = x $, and obviously $
0^{2} = 0 $ and $ e^{2} = e $, so $ R $ is a boolean ring.

I hope it is correct.

I'm proud, since I also found this solution an believe me, there are not many problems in this field to which I find solutions.  :(  Anyway, I think it is the most natural approach.

It suffices to assume $R$ is left (or right) artinian instead of finite. We can prove just as Leva did that the ring has an identity $1$ (which is that unique non-divisor of zero element). Now we prove that the intersection of all the maximal left ideals is zero. This follows from the fact that $x$ belongs to this intersection (the Jacobson radical) iff $1-rx$ is invertible for all $r\in R$, which, in turn, means that $rx=0,\ \forall r\in R$. In particular, $x=1\cdot x=0$, so we're done.

Our ring is now left artinian and semiprimitive (i.e. its Jacobson radical is zero), so the [url=http://planetmath.org/encyclopedia/WedderburnArtinTheorem.html]Wedderburn-Artin Theorem[/url] states it must be isomorphic to a finite product of matrix rings over division rings, i.e. $R\cong\prod_{i=1}^n\mathcal M_{n_i}(D_i)$, where $D_i$ are division rings. If $n_1$, say, is $>1$ we get a contradiction, since it's easy to find an element $a$ in $\mathcal M_{n_1}(D_1)$ different from the identity which is not a divisor of zero, and then we take $x\in R$ to be $(a,1,\ldots,1)$, which is a non-divisor of zero different from $1$. This means that $n_i=1,\ \forall i$, so $R$ is a direct product of division rings $D_i$. Next, if, for example, $D_1$ has more than two elements, then we can pick $1_{D_1}\ne a\in D_1$ and the element $x\in R,x=(a,1,\ldots,1)$ is, again, a non-divisor of zero, so we have a contradiction again. This means that all the division rings $D_i$ are in fact fields with two elements, and the conclusion follows.


On the horizontal line $ y\equal{}5$, there are two points exactly 5 units from the point $ (7,1)$. What is the sum of the $ x$-coordinates of these two points?

The x-coordinates will average to 7, so the sum is 14.

I just added $7$ (first x-coordinate) and $5$: $7+5=12$ and then did $7-5=2$. $12+2=\boxed{14}$


A Ferris wheel of diameter 40m has its centre 21m above the ground. The wheel rotates once every 30s. Draw a graph to show a rider's height above the ground during a 4 min ride, starting at the lowest position. Write and equation for the graph.

I wasn't quite sure how to do this questions when it was given on my test, I only got 2/6 marks so can someone help me on this question?

[hide]I think that the equation, on plane $Oty$, can be $y=-20\cos (\omega t) +1$, and since $T=\frac{2 \pi}{\omega}=30 s$, $\omega=\frac{1}{15}\pi \frac{rad}{s}$, so it becomes $y=-20\cos \left( \frac{1}{15}\pi t \right) +1$.
The fast way to find this is thinking that at $t=0$ the rider must be at 1m above the ground, and you are searching for a periodic function that has its maximum at $\frac{T}{2}$. This function must be $-\cos$.
The long way is geometric, with harmonic motion. [/hide]

hydro, your formula has it's sinusoidal axis at 1, it needs to be at 21.  it should be y=-20cos(pi/15*t)+21.


calculus
integration
trigonometry
function
real analysis
complex numbers
complex analysis

I'm currently reading an introductory text in complex analysis. An exercise given there is to compute \[ \int_0^{\pi} e^{a \cos t} \cos ( a \sin t ) dt  \] by using tools specific to complex analysis. Can you help me here?

By the way, is there an elementary solution to this one? (by elementary I mean the use of real analysis only)

Maybe "elementary" isn't quite the word, but here's a calculus-using solution.

First, note the the function $e^{a\cos t}\cos(a\sin t)$ is even, so we can write the problem as

$\frac12\int_{-\pi}^{\pi}e^{a\cos t}\cos(a\sin t)\,dt=\text{Re }\left[\frac12\int_0^{2\pi} e^{ae^{it}}\,dt\right].$

Now define $f(a)=\frac12\int_0^{2\pi} e^{ae^{it}}\,dt.$ Differentiate it.

$f'(a)=\frac12\int_0^{2\pi} e^{ae^{it}}e^{it}\,dt= \left.\frac1{2ai}e^{ae^{it}}\right|_0^{2\pi}=0$

Since $f(a)$ is constant, we have $f(a)=f(0)=\frac12\int_0^{2\pi}1\,dt=\pi.$

And the real part of that is also $\pi.$

Caveat: I was assuming that $a\in\mathbb{R}.$

Let $z(t)=ae^{it}$, and make the change that Kent did to an interval of length $2\pi$.. We're trying to find the real part of $\frac12\int_0^{2\pi} e^{z(t)}\,dt=\frac12\int_0^{2\pi}\frac{e^{z(t)}}{z'(t)}\cdot z'(t)\,dt=\frac12\int_0^{2\pi}\frac{e^{z(t)}}{iz(t)}\cdot z'(t)\,dt$

That last integral is by definition the contour integral $\frac1{2i}\int_{|z|=a}\frac{e^z}{z}\,dz$. You should have the tools to do this.

Shouldn't that be $z(t)=ae^{it}$ and hence an integral on the circle $|z|=|a|?$

Oops. Fixed now. The other constants don't change, as $\frac{z'(t)}{z(t)}$ is still just $i$.

Kent's trick is nice. (In what cases can you differentiate under an integral, wrt another variable?)

Thanks for the help.


What integer, when added to both the numerator and the denominator of $\frac{6}{7}$, will give a value of 0.8?

[hide]
using guess and check i get $\boxed{-2}$[/hide]


[hide]
If all $a_{i}$ are 0, the inequality is trivial. Otherwise, make the substitution
\[x_{i}=a_{i}\sqrt[k]{\frac{n}{a_{1}^{k}+a_{2}^{k}+\cdots+a_{n}^{k}}}\]
so that $x_{1}^{k}+x_{2}^{k}+\cdots+x_{n}^{k}=n$ and our inequality is equivalent to
\[x_{1}+x_{2}+\cdots+x_{n}\leq n\]
after dividing both sides by $a_{1}^{k}+a_{2}^{k}+\cdots+a_{n}^{k}$. But this follows from applying the AM-GM inequality $x_{i}\leq \frac{x_{i}^{k}+k-1}{k}$ to each $i$ and adding up the inequalities.
[/hide]

Couldn't you just take the kth root of both sides and then you'd have the $RMS\ge AM$ inequality?

i seriously doubt that power mean is an appropriate way to prove power mean... unless you're like

circular logic works because\/
^-<-<-<-<-<-<-<-<-<-<-<-<-
 :P
[hide="but you could also use"]
induction! The expression
\[n(a_{1}^{k+1}+a_{2}^{k+1}+\cdots+a_{n}^{k+1})-(a_{1}+a_{2}+\cdots+a_{n})(a_{1}^{k}+a_{2}^{k}+\cdots+a_{n}^{k}) \]
can actually be multiplied out rather easily and simplifies to
\[\sum_{1\leq i<j\leq n}(a_{i}-a_{j})(a_{i}^{k}-a_{j}^{k}) \]
which is nonnegative.
[/hide]

It is trivial by Jensen inequality for convex function $f(x)= x^{k}$.