floor function
logarithms
inequalities
limit
number theory
prime numbers
number theory open

Can anybody prove that the sequence n -> floor(ln(n)) contains infinitely many prime numbers? Or finitely many,if that is the case?

That sequence contains every non-negative integer, so...

let $ n \equal{} e^{x}$ with $ x > 0$. We thus have $ \lfloor \ln(e^{x}) \rfloor \equal{} \lfloor x\ln(e) \rfloor \equal{} \lfloor x \rfloor$

$ \lfloor x \rfloor$ produces any desired positive integer, and thus any desired prime as well.

Edit: sorry, we can't assume $ n \equal{} e^{x}$ since $ e^{x}$ is irrational while it's generally assumed $ n \in \mathbb{Z} > 0$

I'd say by intuition, you'd try to either show that you can satisfy $ \ln(n) \equal{} A \plus{} \epsilon$ for any non-negative integer $ A$ (since $ \lfloor A \plus{} B \rfloor \equal{} A$ so long as $ |B| < 1$), or you'd try to show that there exists suitable non-negative integer values $ C$ such that $ |x\ln(C) \minus{} x| < 1$.

Sorry, I misinterpreted the problem.  The idea is that once $ \ln (n\plus{}1) \minus{} \ln n$ grows sufficiently small the sequence will hit every integer greater than some constant $ M$.

Continuing from my original post, we have $ \lfloor \ln(n) \rfloor$ containing infinitely many prime numbers if it satisfies the following condition;

$ 0 < x\ln(A) \minus{} x < 1\;\;\; \forall A \in \mathbb{Z} > 0$

From the first inequality we get;

$ x\ln(A) \minus{} x > 0$

$ \ln(A) > 1$

$ e > A$

This is satisfied for $ A \equal{} \{ 1, 2 \}$

The second inequality yields;

$ x\ln(A) \minus{} x < 1$

$ \ln(A) < \frac {1 \plus{} x}{x}$

$ e^{\frac {1 \plus{} x}{x}} < A$

Taking the $ \lim_{x \rightarrow \infty}$ gives;

$ e < A$

This is satisfied when $ A \equal{} \{ 3, 4, 5, \cdots \}$

So our set of solutions is $ S \equal{} \{ 1, 2 \} \cup \{ 3, 4, 5, \cdots \} \equal{} \mathbb{Z} > 0$, which proves that our conditions on the value of $ A$ are satisfied.


ratio
geometry
circumcircle
trigonometry
angle bisector
geometry solved

Let $ABC$ be a triangle, let $D$ be a point on the side $BC$, and let $E$ be a point on the side $CA$ such that the lines $AD$ and $BE$ are the angle bisectors of the angles $CAB$ and $ABC$, respectively. If the line $DE$ bisects the angle $ADC$, find $\measuredangle BAC$.

Is it $A=120^0$ ?? ;)

Take a look at my approach:  :)

We have $\frac{AE}{EC}=\frac{AB}{BC}=\frac{c}{a}$ and

$\frac{AE}{EC}=\frac{AD}{DC}$ FROM THE ANGLE bisector theorem.

So $\frac{AD}{DC}=\frac{c}{a}$   $(1)$

$AD$ also bisects $A$ so again FROM THE ANGLE bisector theorem we get $\frac{DC}{DB}=\frac{AC}{AB}=\frac{b}{c}$ or $DC=\frac{ab}{b+c}$    $(2)$

Combining relations $(1)$ and $(2)$ we have $AD=\frac{c}{a} \cdot DC= \frac{c}{a} \cdot \frac{ab}{b+c}=\frac{bc}{b+c}$

However $AD=\frac{2\sqrt{bc}}{b+c} \cdot \sqrt{p(p-a)}$ where p denotes the semiperimeter of the triangle.

Finally we obtain $a^2=b^2+c^2+bc$ that is $A=120^0$

Your approach is correct, Socrates, but there is also a purely synthetic solution:

[quote="marko avila"]Let $D$ and $E$ be points on the sides $BC$ and $CA$ of a triangle $ABC$ such that the lines $AD$ and $BE$ are the angle bisectors of the angles $CAB$ and $ABC$, respectively. If the line $DE$ bisects the angle $ADC$, find $\measuredangle BAC$.[/quote]

We have to show that, if the line DE bisects the angle ADC, then A = 120°.

The angle bisector of an angle of a triangle divides the opposite side in the ratio of the adjacent sides. Applying this to the triangle ABC with its angle bisector BE, we get $\frac{AE}{EC}=\frac{c}{a}$. Applying this to the triangle ADC with its angle bisector DE, we get $\frac{AE}{EC}=\frac{AD}{CD}$. Thus, $\frac{AD}{CD}=\frac{c}{a}$. From now on, we can forget about the point E.

Now let Y and Z be the orthogonal projections of the point D on the sides CA and AB of triangle ABC. Since the point D lies on the angle bisector of the angle CAB, it is equidistant from its sides CA and AB; thus, DY = DZ.

Now, since < AYD = 90° and < AZD = 90°, the points Y and Z lie on the circle with diameter AD; in other words, the circumcircle of triangle AYZ has diameter AD. Since a sidelength of a triangle equals the diameter of the circumcircle multiplied with the sine of the opposite angle, we thus have $YZ=AD\cdot\sin\measuredangle YAZ$; in other words, $YZ=AD\cdot\sin A$. On the other hand, in the right-angled triangle CYD, we have $DY=CD\cdot\sin\measuredangle DCY=CD\cdot\sin C$. Finally, by the sine law in triangle ABC, we have $\frac{\sin A}{\sin C}=\frac{a}{c}$. Thus,

$\frac{YZ}{DY}=\frac{AD\cdot\sin A}{CD\cdot\sin C}=\frac{AD}{CD}\cdot\frac{\sin A}{\sin C}=\frac{c}{a}\cdot\frac{a}{c}=1$,

so that YZ = DY. Together with DY = DZ, this yields YZ = DY = DZ, and thus the triangle DYZ is equilateral, so that < YDZ = 60°. Since the points Y and Z lie on the circle with diameter AD, the quadrilateral AYDZ is cyclic, and thus < YAZ = 180° - < YDZ = 180° - 60° = 120°. In other words, A = 120°. And the problem is solved.

  Darij

$\frac{AB}{BC}=\frac{EA}{EC}=\frac{AD}{DC}\Longrightarrow \frac ca =\frac {l_a(b+c)}{ab}\Longrightarrow bc=l_a(b+c)\Longrightarrow bc=2bc\cos {\frac A2}$ $\Longrightarrow A=120.$

Nice solutions from both of you, too, darij and levi $!!!$ :lol:  :)

here is my solution:
in triangle $ADC$ point $E$ is excenter wrt $B$ therefore $AE$ bisects angle $DAX$ ($X$ is a point on line AB that A is between X and B)
hence $\angle BAC=120^{\circ}$

In Socrates's solution, I don't understand how you got (2).

How do you get $\frac{DC}{DB} = \frac{AC}{AB} = \frac{b}{c}$ to $DC = \frac{ab}{b+c}$

$DC+DB=a$ and $\frac{DC}{DB}=\frac{b}{c}$. Just solve this system.


quadratics
number theory proposed
number theory

Show that there do not exist any distinct natural numbers $ a$, $ b$, $ c$, $ d$ such that $ a^3\plus{}b^3\equal{}c^3\plus{}d^3$ and $ a\plus{}b\equal{}c\plus{}d$.

[quote="Leon"]Show that there do not exist any distinct natural numbers $ a$, $ b$, $ c$, $ d$ such that $ a^3 \plus{} b^3 \equal{} c^3 \plus{} d^3$ and $ a \plus{} b \equal{} c \plus{} d$.[/quote]

Let $ a\plus{}b\equal{}c\plus{}d\equal{}S\neq 0$. Then $ a^3\plus{}b^3\equal{}S^3\minus{}3abS$ $ \equal{}c^3\plus{}d^3\equal{}S^3\minus{}3cdS$ imply $ ab\equal{}cd\equal{}P$

So $ (a,b)$ and $ (c,d)$ both are the roots of the same quadratic $ X^2\minus{}SX\plus{}P\equal{}0$ and so either $ (a,b)\equal{}(c,d)$, either $ (a,b)\equal{}(d,c)$

Q.E.D.

If the both conditions are valid then $ a^2\minus{}ab\plus{}b^2\equal{}c^2\minus{}cd\plus{}d^2$, as $ a^3\plus{}b^3\equal{}(a\plus{}b)(a^2\minus{}ab\plus{}b^2)$, and as $ a^2\plus{}2ab\plus{}b^2\equal{}c^2\plus{}2cd\plus{}d^2$, we conclude $ ab\equal{}cd$

Now $ a^2\plus{}b^2\equal{}c^2\plus{}d^2$ from $ (a\plus{}b)^2\equal{}(c\plus{}d)^2$, so $ (a\minus{}b)^2\equal{}(c\minus{}d)^2$, from which we conclude easily with a+b=c+d that $ a\equal{}c$ or $ a\equal{}d$.

the equation satisfies for only similar integers and the question itself contains the verbal   :)
$ a+b=c+d$ as well as $ a^3+b^3 =c^3+d^3$
impossible ! for distinct integers


geometry
algebra
polynomial
integration
calculus
calculus computations

for $ p\in (0;\infty)$ calculate  area between $ y^{2} \equal{} 4px$ and $ 16py^{2} \equal{} 5(x \minus{} p)^{3}$

by descartes rule of signs, for each fixed positive p the polynomial

$ q(x)\equal{}x^3\minus{}3px^2\minus{}\frac{49}{5}p^2 x \minus{}p^3$ has exactly one positive root, which is $ 5p$

thus, the area between the two curves is

$ A(p)\equal{}\int_{\minus{}2p\sqrt 5}^{2p\sqrt 5} p\plus{}\left(\frac{16py^2}{5}\right)^{\frac{1}{3}}\minus{}\frac{y^2}{4p} dy$

$ \equal{}2\int_0^{2p\sqrt 5} p\plus{} \left(\frac{16p}{5}\right)^{\frac{1}{3}}y^{\frac{2}{3}}\minus{}\frac{1}{4p} y^2 dy$

$ \equal{}2 p(2p\sqrt 5) \plus{} \frac{6}{5}\left(\frac{16p}{5}\right)^{\frac{1}{3}}(2p\sqrt 5)^{\frac{5}{3}}\minus{}\frac{1}{6p}(2p\sqrt 5)^3$

$ \equal{}\frac{104p^2}{3\sqrt 5}$


Is there a formula for finding x^0 + x^2 + x^4.....?

[quote="Fierytycoon"]Its just a geometric series, the sum of which is

S = 1/(1-x^2).[/quote]

But only if -1 < x < 1. Otherwise it diverges.

Hm... I remember you can still use tht formula, because that is just the simplified form a a more general formula. I think it was something like 1- x^(n+1)/ 1 - x for x^0+x^1....+x^n. Since for that geometric sequence you factor out x^2, you can still get an answer... though if x is too large or small it becomes difficult.

When you learn complex analysis, you'll say that 1/(1-x^2) is an analytic continuation of 1+x2+x4+...

That's be a 1 at the beginning, not an x.


probability
Alcumus
expected value
Support

I just got a question:

Suppose you have a weighted coin in which heads comes up with probability $ \frac{3}{4}$ and tails with probability $ \frac{1}{4}$. If you flip heads, you win $ 2$ dollars, but if you flip tails, you lose $ 1$ dollar. What is the expected value of a coin flip? Express your answer as a decimal.

I entered the answer $ \frac{5}{4}$ and Alcumus said, "Way to Go!"  Yet, it said "Express your answer as a decimal".  What happened?

Alcumus actually accepts fractional answers as decimal, and vice versa.

In fact, Alcumus even accepts unsimplified fractions, like, for this case, $ \frac{25}{20}$.


A positive three-digit integer is formed by choosing three different digits from the set $ \{0,2,4,6,8\}$. The first digit cannot be 0. What is the probability that the number is divisible by 3? Express your answer as a common fraction.

Assume that $ abc$ is the $ 3$ digit number.

Case 1: $ ab \equiv 0 \pmod{3}$

The possible numbers are:

$ 24$, $ 42$, $ 48$, $ 60$, $ 84$

This chance is $ \frac{5}{16}$.

Case 2: $ ab \equiv 1 \pmod{3}$

The possible numbers are:

$ 28$, $ 40$, $ 46$, $ 64$, $ 82$

This chance is also $ \frac{5}{16}$.

Case 3: $ ab \equiv 2 \pmod{3}$

The possible numbers are:

$ 20$, $ 26$, $ 62$, $ 68$, $ 80$, $ 86$

This chance is $ \frac38$.

For Case 1: We need $ c \equiv 0 \pmod{3}$.

Two chances are possible, $ 0$ or $ 6$.  Thus, this probability is $ \frac{5}{16} \times \frac{2}{5}$, or $ \frac18$.

For Case 2: We need $ c \equiv 2 \pmod{3}$.

The two chances possible are $ 2$ and $ 8$.  Thus, the probabilitiy of this is $ \frac{5}{16} \times \frac25$, or $ \frac18$.

For Case 3: We need $ c \equiv 1 \pmod{3}$.

The only possible digit for this is $ 4$.  Thus, this probability is $ \frac{3}{8} \times \frac15$, or $ \frac{3}{40}$.

Thus, the answer is $ \frac18 \plus{} \frac18 \plus{} \frac{3}{40}$, or $ \boxed{\frac{13}{40}}$.

Does anyone listen to me? Ever? http://www.artofproblemsolving.com/Forum/viewtopic.php?t=229641 Although that's not the answer I got.

I just asked this problem, and got here. The answer given was 5/12.

Case 1: $ 0$ is used

The possible triplets of the digits used is $ \{ 0,2,4 \}$ and $ \{ 0,4,8 \}$.  For each of these there are $ 2\cdot 2\cdot 1\equal{}4$ numbers, so this case contributes $ 2\cdot 4\equal{}8$.

Case 2: $ 0$ is not used.

Then the possible triplets are $ \{ 2,4,6 \}$ and $ \{ 4,6,8 \}$.  There are $ 2\cdot 3!\equal{}12$ numbers in this case.

So the probability is $ \frac{8\plus{}12}{4\cdot 4\cdot 3}\equal{}\boxed{\frac{5}{12}}$


find the value of the sum of n/(n+1)! as n uns from 1 to infinity.

hello, do you mean $ \sum_{n\equal{}1}^\infty\frac{n}{(n\plus{}1)!}$?
Sonnhard.

hello, we have $ \sum_{n\equal{}1}^\infty\frac{n}{(n\plus{}1)!}\equal{}1$.
Sonnhard.

hello, i think it would help if we know that
$ \sum_{n\equal{}1}^\infty\frac{1}{n!}\equal{}e\minus{}1$
Sonnhard.

[b]@ Poincare:[/b] Why don't I write out the first few terms in this form.

$ \left(\frac {1}{1!} \minus{} \frac {1}{2!} \right) \plus{} \left(\frac {1}{2!} \minus{} \frac {1}{3!} \right) \plus{} \left(\frac {1}{3!} \minus{} \frac {1}{4!} \right) \plus{} \left(\frac {1}{4!} \minus{} \frac {1}{5!} \right) \plus{} \ldots$

Can you find an effective way of adding these? There's a [url=http://en.wikipedia.org/wiki/Telescoping_sum]name for this technique[/url] of summing a series that you should familiarize yourself with.

hello, that's ok, but this sum above in my post is well-known.
Sonnhard.

Telescoping is nice and all, but Dr Sonnhard's observation combined with your first remark generalizes to otherwise strange-looking summations of the form $ \sum_{k \equal{} 0}^{\infty} \frac {P(k)}{k!}$ where $ P(k)$ is an arbitrary polynomial.

Consider the series $ S \equal{} \sum_{n \equal{} 0}^{ \plus{} \infty}\frac {P(n)}{(n \plus{} k)!}$, with $ k \in \mathbb{N}_0$, where $ P(n)$ is a polynomial in n. If $ \alpha$ is the degree of P(n), then let's find the constants $ a_i, \,\,i \equal{} 0, ..., \alpha$, such that

$ \frac {P(n)}{(n \plus{} k)!} \equal{} \frac {a_0}{(n \plus{} k)!} \plus{} \frac {a_1}{(n \plus{} k \minus{} 1)!} \plus{} ... \plus{} \frac {a_{\alpha}}{(n \plus{} k \minus{} \alpha)!}$

and each of these terms can be related with $ \sum_{n \equal{} 0}^{ \plus{} \infty}\frac {1}{n!} \equal{} e$.


function
induction
real analysis
real analysis unsolved

Let two bijective and continuous functions$f,g: \mathbb{R}\to\mathbb{R}$ such that : $\left(f\circ g^{-1}\right)(x)+\left(g\circ f^{-1}\right)(x)=2x$ for any real $x$. Show that If we have a value $x_{0}\in\mathbb{R}$ such that $f(x_{0})=g(x_{0})$, then $f=g$.

If $h=f \circ g^{-1}$, then $h^{-1}=g \circ f^{-1}$, so $h(x)+h^{-1}(x)=2x$. Let $d_{x}=h(x)-x$. It follows that
\[ h^n(x)=x+nd_{x}, \forall n \in {\mathbb {Z}}.  \]
For $n<0$, $h^n$ means $(h^{-1})^{-n}$. $h$ is continuous and bijective, hence it is strictly increasing (since it cannot be decreasing). Let $x<y$. Because $h$ is strictly increasing,
\[ h^n(x)<h^n(y) \Rightarrow x+nd_{x}<y+nd_{y} \Rightarrow (y-x)+n(d_{y}-d_{x}) > 0, \forall n \in {\mathbb {Z}},  \]
hence $d_{x}=d_{y}$, or $d_{x}$ constant, so $h(x)=x+c$. Since $h(x_{0})=x_{0}$, $c=0$ and $h=1_{\mathbb {R}}$, i.e, $f=g$.

Thank you! The book had a long and akward solution.  :o .  To clarify one thing. You took $x_{0}=f(x_{0})=g(x_{0})$ to obtain $h(x_{0})=x_{0}$, right?

a mistake...it should say $h(g(x_0)) = g(x_0)$ rather than $h(x_0) = x_0$. the remainder of the proof isn't affected, of course

Ohh..  :oops:  I didn't understood either why $d_{x}=d_{y}$ Can you please explain a bit, if you don't mind?

My guessing: $(y-x)$ is positive and to have LHS positive then $m=d_{y}-d_{x}< 0$.. or $\leq 0$? but $m$ is always positive?   :?

$(y-x)+n(d_{y}-d_{x}) > 0, \forall n \in {\mathbb {Z}}$

if $d_{y}-d_{x} < 0$, then as $n$ goes to $\infty$, $n(d_{y}-d_{x})$ goes to $-\infty$,

and if $d_{y}-d_{x} > 0$, then as $n$ goes to $-\infty$, $n(d_{y}-d_{x})$ goes to $-\infty$.

both of these contradict the statement above.

p.s. esti roman? si eu

Another question :) (after some time) Why $h^n(x)=x+nd_{x}, \forall n \in {\mathbb {Z}}.$ Could this problem be generalised?  :huh:

$h(x) + h^{-1}(x) = 2x$ (**)

and $d_x = h(x) - x$ (by definition), so

$h(x) = x + d_x$ and $h^{-1}(x) = 2x - h(x) = 2x - (h_x + x) = x - d_x$.

then induce: assume $h^n (x) = x+nd_x$ is true for all $n \in [-k, k]$, $k\in \mathbb N$. from (**) we get (by replacing $x$ with $h^k (x)$)

$h^{k+1} (x) + h^{k-1} (x) = 2h^k (x) = 2x+2kd_x$

and since $k-1$ is in the range $[-k, k]$, this becomes

$h^{k+1}(x) + x+(k-1)d_x = 2x+2kd_x$; and then $h^{k+1}(x) = x + (k+1)d_x$.

a similar argument (replacing $x$ with $h^{-k} (x)$ in (**)) gives that $h^{-k-1}(x) = x + (-k-1)d_x$. so the relation holds on $[-k-1, k+1]$--induction complete.


quadratics
algebra
polynomial
function
quadratic formula
number theory
prime factorization

We have

$ x^{3}-x^{2}-x = y^{3}-y^{2}-y = z^{3}-z^{2}-z = 3$

I'm pretty sure we could conclude that x = y = z, so we can eliminate 2 of our 3 variables.

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

Using the quadratic formula,

$ x(x-\left(\frac{1+\sqrt{5}}{2}\right))(x-\left(\frac{1-\sqrt{5}}{2}\right) ) = 3$

That's as much as I know.

Sorry prime factorization,
I think that $ x^{3}-x^{2}-x=3$ is not true... :maybe:

notice that $ x=y=z=1$ is a solution (maybe the only real one) of the original system
and does not verify your equations...

$ (1,1,1)$ is the unique real solution.
Proof 1 ( with Maple): using the method of the Gröbner's basis we obtain:
$ (z-1)(z^{2}+2z+3)P(z)=0$ where $ P$ is a polynomial of degree $ 24$. Obviously $ z^{2}+2z+3>0$ and, with the help of the Sturm's method, we prove that $ P$ hasn't any real root. QED.
Proof 2 (without Maple but with a graphic calculator): we obtain easily: $ p=-x^{6}+(9-3y^{3})x^{4}-3(3-y^{3})^{2}x^{2}+x+y^{2}+(3-y^{3})^{3}-3=0,q=x^{4}+x^{3}+(-6+2y^{3})x^{2}+y+(3-y^{3})^{2}-3=0$.
a) Assume that $ y>1.62$; then $ \alpha=2y^{3}-6>0$ and $ \beta=(3-y^{3})^{2}+y-3>0.11$; $ q=x^{4}+x^{3}+\alpha{x^{2}}+\beta\geq{x^{4}+x^{3}+0.11}>0$ and there doesn't exist any solution in $ x$. Thus $ y\leq1.62$ and by the same proof: $ x,z\leq1.62$.
b) Assume for example that $ z\leq{y}\leq{x}$. The function $ \phi(u)=u+u^{2}+u^{3}$ is strictly increasing; $ \phi(x)+\phi(y)+\phi(z)=9$ thus $ \phi(x)\geq3$ and $ x\geq1$.
c) Assume that $ z\leq{0},y\leq{0}$; then $ x^{2}=3-z-y^{3}\geq3$ and $ x\geq\sqrt{3}$, a contradiction.
Conclusion: $ 0<y\leq{x}$ and $ x\in[1,1.62]$. ($ x,y$ are bounded).
d) With a graphic calculator we have a look on the 2 curves: $ p(x,y)=0,q(x,y)=0$ with the c)-conditions; the intersection is reduced to one point which is necessarily $ (1,1)$. QED.
Remark: certainly there exists a solution which doesn't use any graphic argument.

Claim. The only solution is $ (x,y,z) \equal{} (1,1,1)$.

Proof. Let $ f_1 (x, y, z) \equal{} x \plus{} y^2 \plus{} z^3 \minus{} 3$, $ f_2 (x,y,z) \equal{} y\plus{}z^2\plus{}x^3 \minus{} 3$, and $ f_3 (x,y,z) \equal{} z\plus{}x^2\plus{}y^3 \minus{}3$. Then, let $ f(x,y,z) \equal{} (f_1 \plus{} f_2 \plus{} f_3)(x, y, z) \equal{} \sum x \plus{} \sum x^2 \plus{} \sum x^3 \minus{} 9$. Note that $ f_3 (x,y,z) \equal{} f_2 (y,z,x) \equal{} f_1 (z,x,y)$. Further, note that we cannot have all $ x,y,z \equal{} 0$ because $ f_1 (0,0,0) \equal{} \minus{}3 \neq 0$. Hence, there are there cases to consider: cases where (i) none, (ii) one, or (iii) two of $ x,y,z$ is(are) $ 0$.

[Case 1] $ x,y,z \neq 0$. Note that $ y \equal{} ax$ and $ z \equal{} bx$ for some $ a,b \in \mathbb{R} \minus{} \{0\}$. Moreover, $ f_1 (x,y,z) \equal{} 0$ $ \Rightarrow$ $ x \plus{} y^2 \plus{} z^3 \equal{} 3$ $ \Rightarrow$ $ x \plus{} a^2 x^2 \plus{} b^3 x^3 \equal{} 3$ $ \Rightarrow$ $ x^3 \plus{} \frac{a^2}{b^3} x^2 \plus{} \frac{1}{b^3} x \minus{} \frac{3}{b^3} \equal{} 0$, $ f_2 (x,y,z) \equal{} f_1 (y,z,x) \equal{} 0$ $ \Rightarrow$ $ y \plus{} z^2 \plus{} x^3 \equal{} 3$ $ \Rightarrow$ $ ax \plus{} b^2 x^2 \plus{} x^3 \equal{} 3$ $ \Rightarrow$ $ x^3 \plus{} b^2 x^2 \plus{} ax \minus{} 3 \equal{} 0$, $ f_3 (x,y,z) \equal{} f_1 (z,x,y) \equal{} 0$ $ \Rightarrow$ $ z \plus{} x^2 \plus{} y^3 \equal{} 3$ $ \Rightarrow$ $ bx \plus{} x^2 \plus{} a^3 x^3 \equal{} 3$ $ \Rightarrow$ $ x^3 \plus{} \frac{1}{a^3} x^2 \plus{} \frac{b}{a^3} x \minus{} \frac{3}{a^3} \equal{} 0$. Note that these polynomials in $ x$ should be equivalent, meaning all the corresponding coefficients should be the same. So, by just comparing the constant terms, we deduce that $ a^3\equal{}b^3\equal{}1$; now since $ a^3 \equal{} b^3 \equal{} 1$ iff $ a^3 \minus{} 1 \equal{} (a\minus{}1)(a^2\plus{}a\plus{}1) \equal{} 0 \equal{} (b\minus{}1)(b^2\plus{}b\plus{}1) \equal{} b^3\minus{}1$ iff $ a\equal{}b\equal{}1$ since $ a,b \in \mathbb{R}$, we deduce that $ a\equal{}b\equal{}1$. Hence, $ x\equal{}y\equal{}z$. Now, consider $ f(x) \equal{} f_1 (x,x,x) \equal{} f_2 (x,x,x) \equal{} f_3 (x,x,x) \equal{} x \plus{} x^2 \plus{} x^3 \minus{} 3$. Note that $ x\equal{}1$ is a solution to $ f(x) \equal{} 0$ since $ f(1) \equal{} 0$. Moreover, since $ f'(x) \equal{} 1\plus{}2x\plus{}3x^2 \equal{} 3(x \plus{} \frac{1}{3})^2 \plus{} \frac{2}{3} \geq \frac{2}{3} > 0$ for all $ x \in \mathbb{R}$, we deduce that $ f$ is a strictly increasing function and, thus, it cannot have more than one root.

[Case 2] One of $ x,y,z$ is zero; without loss of generality, $ x\equal{}0$ and $ y,z \neq 0$. Then, $ z \equal{} ay$ for some $ a \in \mathbb{R} \minus{} \{0\}$. Note that we should have $ y^2 \plus{} z^3 \minus{}3 \equal{} y\plus{} z^2 \minus{}3 \equal{} z\plus{} y^3 \minus{}3 \equal{} 0$ $ \Rightarrow$ $ y^2 \plus{} a^3 y^3 \minus{} 3 \equal{} y \plus{} a^2 y^2 \minus{} 3 \equal{} ay \plus{} y^3 \minus{} 3 \equal{} 0$ $ \Rightarrow$ $ y^3 \plus{} \frac{1}{a^3} y^2 \minus{} \frac{3}{a^3} \equal{} 0$, $ y^2 \plus{} \frac{1}{a^2} y \minus{} \frac{3}{a^2} \equal{} 0$, and $ y^3 \plus{} ay \minus{} 3 \equal{} 0$. Note that these polynomials in $ y$ should be equivalent, meaning all the corresponding coefficients should be the same. So, by just comparing the constant terms, we deduce that $ a^3\equal{}a^2\equal{}1$; hence, $ a \equal{} 1$. But, note that if $ a\equal{}1$ then $ y^3 \plus{} y^2 \minus{} 3 \equal{} y \plus{} y^2 \minus{}3 \equal{} y \plus{} y^3 \minus{} 3 \equal{} 0$, or $ y^3 \plus{} y^2 \plus{} y \equal{} \frac{9}{2}$, or $ y^3 \equal{} y^2 \equal{} y \equal{} \frac{3}{2}$, which clearly is a contradiction, as there is no such $ y$ satisfying the condition.

[Case 3] Two of $ x,y,z$ are zero; without loss of generality $ x\equal{}y\equal{}0$ and $ z \neq 0$. Then, we have $ z^3 \equal{} z^2 \equal{} z \equal{} 3$, which clearly is a contradiction. We conclude that this is also an impossible case.

Having considered Case 1-3, we are done. Only possible solution is $ x\equal{}y\equal{}z\equal{}1$, as desired.

One possible generalization of a problem of this kind is the following:

For each $ n \in \mathbb{N}$, find all solutions $ (x_1, ..., x_n)$ in reals for the following system of equations:

$ x_1 \plus{} x_2^2 \plus{} ... \plus{} x_n^n \equal{} x_2 \plus{} x_3^2 \plus{} ... \plus{} x_1^n \equal{} ... \equal{} x_n \plus{} x_1^2 \plus{} ... \plus{} x_{n\minus{}1}^n \equal{} n$

One should find that the method used in $ n\equal{}3$ case works in general to find that there is only one trivial solution to this system: $ (1, ..., 1)$ ($ n$ times)


hi all,

here a rather hard pb:
Let a_1,a_2,..,a_n be distinct squarefree integers.
Let s_i=sqrt(a_i). Prove that if
r1,r2,..,rn are rational so that
   r1*s_1+ r2*s_2+..+rn*s_n=0
then r_i=0 for every i.

bye.

Dear Peter, they are square-free! Anyway, I think this problem was already discussed.

look here :
http://www.mathlinks.ro/Forum/viewtopic.php?t=4496&highlight=rational

Pierre.

Then what does square-free mean? :?

I thought it just meant hte numbers without squares, being 2,3,5,6,7,8,10,11,...  :?

It means numbers which are not divisible by any squares. $2^2|8$, so $8$ is not square-free.


$ (x^{2006}\plus{}1)(1\plus{}x^2\plus{}x^4\plus{}\cdots\plus{}x^{2004})\equal{}2006x^{2005}$
$ \Leftrightarrow 1\plus{}x^2\plus{}x^4\plus{}\cdots\plus{}x^{4010}\equal{}2006x^{2005}$
Applying AM-GM inequality, we have $ LSH\ge RHS$ $ (x>0)$
The equality occurs when $ x\equal{}1$

[quote="t0rajir0u"]That doesn't appear to discount the possibility of a negative root.[/quote]
You mean we have to consider the case $ x<0$? :huh:
This is obvious false because  $ LSH>0$, thus $ x>0$


I noticed that some TeX-made pdf documents appear to be very fuzzy. Just as an example, [url=http://journals.cms.math.ca/cgi-bin/vault/public/view/CRUXv23n1/body/PDF/page3-5.pdf?file=page3-5]this[/url] is a section from the online version of Crux (year 1997, which was given public access to  :) ), by zooming it to the maximum, one can see the pixilation on the words. I remember that with Acrobat versions 5 or under, the fuzziness is shown even without the zooming, and it is really irritating as it makes the document really hard to read on screen.

To show contrast, do the same thing to, say Valentin's documents, they do not show the same effects.

I noticed that with a "fuzzy" pdf file, there is a significant lag as you scroll down to another page, whereas it is quite smooth with a non-fuzzy pdf file.

With some files, the regular text appears to be fuzzy but the math seems fine. And some files are exactly the opposite.

Any idea what causes the fuzziness? And if possible, how they should be fixed?

I got no clue at all, but i noticed it as well :?

dunno how to fix it, I guess it's up to the server to install the right drivers for it...

[quote="billzhao"]To show contrast, do the same thing to, say Valentin's documents, they do not show the same effects.

I noticed that with a "fuzzy" pdf file, there is a significant lag as you scroll down to another page, whereas it is quite smooth with a non-fuzzy pdf file.

With some files, the regular text appears to be fuzzy but the math seems fine. And some files are exactly the opposite.

Any idea what causes the fuzziness? And if possible, how they should be fixed?[/quote]
gosh ... I never thought someone would make me a compliment like "hey you're pdf files are smooth and not fuzzy" :D:D

well I do not know where that comes from, but back in 1997 they had some pretty bad versions of TeX and pdftex so it is possible that the difference of software causes this. I usually use the most updated software.

billzhao! Document you mentioned above uses Type 3 fonts, and Valentine's documents operate with Type 1 fonts. In general, Type 3 must die.


I think the problem is very interesting and I think that there are many infinitely persons solved it
Find $x,y,z$be integers s.t $x^3+y^3=4z^3$

If you look to the residue of $a^3$ mod 4, then you find that $x$ and $y$ are both even or both odd.

If $x$ and $y$ are even, $gcd(x,y)=2$ and $gcd(x^3,y^3)=8$. Then $8|x^3+y^3=4z^3$ and thus $2|z^3$, this implies that $8|z^3$.
You can then put : $x=8x'$, $y=8y'$, $z=8z'$. Remplacing that in the first equation, you obtain : $x'^3+y'^3=4z'^3$. Then, use the infinite descente and conclue.

If $x$ and $y$ are odd, $gcd(x,y)=d$ with $d$ odd. We have thus $d^3|z^3$ ==> $d|z$. You can thus use too the infinite descente.

If d=1?????you are wrong????.I think we canot use menthod which you give

yes, you are right ... and I don't see how solve that....

KO9 are u sure that  infinitely many people have solved this question, anyways do you know its solution set.

Bomb


Don't know how to do these...they're from this year's Triangle Stretch

1. In triangle ABC, points D and E lie on sides AB and AC, respectively, so that AD=5, DB=8, AE=6 and EC=11. What is the ratio of the area of triangle ADE to the area of triangle ABC. Express you answer as a common fraction.

2. In triangle ABC points D and E lie on sides AB and BC, respectively, so that $ \frac{AD}{DB}\equal{}\frac{3}{5}$ and $ \frac{CE}{EB}\equal{}\frac{1}{2}$. Segments CD and AE intersect at point F. What is the value of $ \frac{CF}{FD}$? Express your answer as a common fraction.

Number 1: Draw $ \overline{DE}$ and $ \overline{BE}$. It is now clear that $ \frac {[ADE]}{[ABE]} \equal{} \frac5{5 \plus{} 8} \equal{} \frac5{13}\implies[ABE] \equal{} \frac {13}5[ADE]$, and $ \frac {[ABE]}{[ABC]} \equal{} \frac {\frac {13}5[ADE]}{[ABC]} \equal{} \frac6{17}\implies\frac {[ADE]}{[ABC]} \equal{} \frac5{13}\cdot\frac6{17} \equal{} \boxed{\frac {30}{221}}$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=251854]Number 2: Believe it or not, it actually sometimes helps to search and see if somebody has posted it before.[/url]

Thanks!
 :D  :lol:

The second solution was a bit...long?

It's not that long if you know what you're doing. I had the whole diagram labeled in at most $ 20$ seconds, and the equations weren't that difficult to write.


Fie $ABCD$ un tetraedru regulat si $M$ un punct in interiorul lui. Daca $MA=a$, $MB=b$, $MC=c$, $MD=d$, aflati latura tetraedrului, in functie de $a$,$b$,$c$ si $d$.

Nu am nicio idee la problema, dar am fost curios si in plan da ceva de genul $\sqrt{x(a^{2}+b^{2}+c^{2})+yS}$, unde $x,y$ sunt niste constante si $S$ e aria triunghiului cu laturi de lungimi $a,b,c$.

Dak scriem suma celor 4 volume ca volumul mare cred ca obtinem o ecuatie de gradul 3... :roll: Probabil ca ma insel insa :maybe:

idee: notam latura tetraedrului cu un $x$ si distantele de la $M$ la fetele tetraedrului cu $y_{A}$, $y_{B}$, $y_{C}$, $y_{D}$.
stim ca $y_{A}$+ $y_{B}$+ $y_{C}$+ $y_{D}$= $h$, unde $h$ este inaltimea tetraedrului.
Acum trebuie sa exprimam valorile distantelor de la $M$ la fetele tetraedrului in functie de $x$ si de $a$, $b$, $c$, $d$.
Vom lucra in tetraedrul $MABC$. Daca $P$ este proiectia lui $M$ pe plan, iar $B'$, $C'$, $A'$ proiectiile lui $P$ pe laturi, atunci conform $T3\perp$ si tinanad cont de faptul ca $ABC$ este triunghi echilateral, obtinem o ecuatie in $x$ cu parametri $y_{D}$, $a$, $b$, $c$...

@mychrom: Si nu se ajunge tot la ce ziceam inainte? :maybe:

Da, ciprian, dar am vrut sa prezint in detaliu pasii solutiei.

[quote="dzeta"]Eu am incercat sa scriu suma ariilor pentru problema din plan (la volume nu ma prea pricep) si daca stateam sa scot toti radicalii de pe acolo dadea o ecuatie de grad 16. Sper ca in spatiu da mai bine :)[/quote]

habar n-am, nu am incercat sa scot radicalii, dar cred ca ne-ar fi dat cam la fel.

apropo de volume, stie cineva cum se calculeaza volumul unui tetraedru in functie de distanta dintre doua drepte (opuse) si unghiul dintre ele?

[quote="maky"]daca mai tin eu minte bine era
$V=\frac{AB\cdot CD\cdot d(AB,CD)\cdot \sin\angle(AB,CD)}{6}$
dar e posibil sa fie alt numitor, se deduce cu niste exemple[/quote]

Multumesc mult!

Daca A1, A2, A3, A4, A5 sunt 5 puncte arbitrare din spatiu,
si notam cu dij distanta dintre punctele Ai si Aj, atunci are loc relatia...
Imi este greu sa o scriu aici aceasta relatie, fiindca e vorba de o relatie
in care in membrul I avem un determinant de ordinul 6 asa ca vezi
in ATTACHMENT!

Problema naloaga din g. plana este urmatoarea:

Fie ABC un triunghi echilateral şi M un punct din interiorul lui.
Dacă |MA|=a, |MB|=b şi |MC|=c, aflaţi latura x=|AB|=|BC|=|AC| în funcţie de a, b şi c.

O solutie completa a acestei probleme (si analoaga cu solutia pe care am sugerat-o pentru
problema initiala, din g. tetraedrului intr-o interventie anterioara pe FORUM), o gasiti in
attachment!!!


number theory
relatively prime
number theory unsolved

Nice problem!

1.case - suppose that (y,4) are relatively prime then you will have there a contradiction so there are no solutions in this case.
2.case - suppose that 4 divides y^2 then you see there that 4 doest not divide x^2 then you will have a second contradiction again no solutions. Therefore there are no SOLUTIONS for a x and y element of Integer.

I hope that you will understand if something is not understandable you may say ok i will be very happy to help you .

Thanks

Davron

factorise LHS and see that

$(x-1)^2(x+1)(x^2+x+1)=y^2$

when $x=0$ , $y=\pm 1$

When $x=\pm 1$ , $y=0$

If otherwise we can see that $(x+1)(x^2+x+1)$ must be perfect square .

But $(x+1,x^2+x+1)=1$  ( easily seen since $x^2+x+1=(x+1)^2-x$ and $(x+1,x)=1$ )

So it means $x+1$ and $x^2+x+1$ must be perfect square .

Letting $x+1=k^2$ for some positive integer $k>1$ , then $x^2+x+1=(k^2-1)^2+k^2=k^4-k^2+1$

But $(k^2-1)^2<k^4-k^2+1<k^4$ . So it cant be a perfect square .


geometry
perimeter
geometric transformation
reflection

In triangle ABC, angle B is 90 degrees.  AB=BC=6, and there is a triangle QRS inscribed in ABC such that Q is the midpoint of BC and R is on AB and S is on AC.  What is the minimum perimeter of triangle QRS?

EDITED typo

Perhaps if angle B=90 degrees, AB=BC=6? The figure as stated is not possible (on a euclidean plane, at least).

Given that edit, Solution:

[hide]
We have right triangle ABC with right angle at B and $ AB=BC=6$. Reflect C over segment AB to form point D such that $ \bigtriangleup{ABD}\cong\bigtriangleup{ABC}$. Then, reflect $ \bigtriangleup{ABD}$ over hypotnuse $ AD$ to form another congruent right triangle AED with right angle at E. We now have $ AE\cong{ED}\cong{DB}\cong{BC}\cong{AB}$, all those segments have length 6, C is the farthest point from E, and $ EABD$ is a square of side 6.

Now consider all possible points S on hypotnuse AC. S can be reflected over segment AB to form S' on AD. Then consider Q on BC as given; reflect Q over AB and then that image over AD to form Q' on ED. Clearly $ RS\cong{RS'}$ and $ SQ\cong{S'Q'}$. Thus for all possible points R and S, the perimeter P of $ \bigtriangleup{QRS}$ is such that $ P=QR+RS+SQ=QR+RS'+S'Q'$.

The quantity $ P=QR+RS'+S'Q'$ is minimized when Q, R, S', and Q' are all collinear. Thus the minimized perimeter P occurs when $ QR+RS'+S'Q'\cong{QQ'}$. The segment QQ' is the hypotnuse of right triangle QDQ' with right angle at D, QD=9, DQ'=3, thus the minimum perimeter is $ \boxed{3\sqrt{10}}$.
[/hide]

What if Q is anywhere on BC (not just midpoint).  Using your way, wouldnt the answre be 6?  But thatseems unlikely.  So extension:

In triangle ABC, angle B is 90 degrees. AB=BC=6, and there is a triangle QRS inscribed in ABC such that Q is anywhere of BC and R is anywereh AB and S is anywehre AC. What is the minimum perimeter of triangle QRS?


let $ \triangle ABC$ be isosceles and $ CA \equal{} CB$, $ A_1, B_1$ are two points on $ AB$ and $ \angle A_1 C B_1 \equal{} \angle ACB$, $ \odot O_1$ is the circumcircle of $ \triangle A_1 B_1 C$ and $ \odot O_2$ touches $ \odot O_1$ on the outside , and is tangent to $ CA$, and $ CB$ at $ A_2$ and $ B_2$ respectively. prove that $ A_2 B_2 \equal{} 2AB$

now it's not missing anything

Missing condition $ \angle B_1CA_1 \equal{} \angle CAB$  :?:

does anybody have any ideas? thanks a lot

Hi yetti. Why we must have condition $ \angle B_{1}CA_{1}$ $ \equal{}$ $ \angle CAB$ $ ?$

[quote="tmy1018"]... $ \angle A_1 C B_1 \equal{} \angle ACB$... [/quote]

Sorry, but that's wrong, the correct condition is $ \angle B_1 C A_1 \equal{} \angle CAB \equal{} \angle ABC,$ i.e., the angle of the lines $ CA_1, CB_1$ has to be equal to the [color=red]base[/color] angles of the isosceles $ \triangle ABC.$

Regardless of the angle condition, the problem claims that when the angle $ \angle B_1CA_1$ is [color=red]constant[/color], the circumcircle $ \odot (O_1)$ of the $ \triangle A_1B_1C$ is tangent to a [color=red]fixed[/color] circle $ \odot (O_2),$ given by the fixed tangents $ CA, CB$ from $ C$ and fixed tangency points $ A_2, B_2,$ such that $ A_2B_2 \equal{} 2\ AB.$

Let $ F \in AB$ be foot of the perpenndicular from $ C.$ Inversion with center $ C$ and power $ r^2 \equal{} CF^2$ takes the lines $ CA, CB$ passing through the inversion center $ C$ into themselves, the line $ AB$ tangent to the inversion circle at $ F$ into a circle tangent to the inversion circle at $ F$ and passing through the inversion center $ C,$ i.e., into a circle $ \odot (M, \frac {_r}{^2})$ with diameter $ CF,$ and the circumcircle $ \odot (O_1)$ of the $ \triangle A_1B_1C$ passing through inversion center $ C$ into a line $ A_1^iB_1^i,$ where $ A_1^i, B_1^i$ are images of $ A_1, B_1,$ i.e., intersections of the rays $ (CA_1, (CB_1$ with $ \odot (M, \frac {_r}{^2}).$ Since the angle $ \angle B_1^iCA_1^i \equal{} \angle B_1CA_1$ is fixed, the chord length $ A_1^iB_1^i$ of $ \odot (M, \frac {_r}{^2})$ is also fixed, and the line $ A_1^iB_1^i$ is tangent to a fixed circle $ \odot (M, \varrho)$ concentric with the circle $ \odot (M, \frac {_r}{^2}).$ The fixed circle $ \odot(M, \varrho)$ tangent to the line $ A_1^iB_1^i$ is the inversion image of a fixed circle $ \odot (O_2)$ tangent to $ \odot(O_1).$ Let $ A_1^* \in AB$ be at infinity and $ B_1^* \in AB,$ such that $ \angle B_1^*CA_1^* \equal{} \angle B_1CA_1.$ Then $ CA_1^* \parallel AB$ and $ \angle CB_1^*F \equal{} \angle B_1^*CA_1^* \equal{} \angle B_1CA_1.$ Therefore, tangents to $ \odot (M, \varrho)$ from $ C,$ same as tangents to $ \odot (O_2)$ form $ C,$ and the line $ AB$ form an isosceles triangle with the [color=red]base[/color] angles equal to $ \angle B_1CA_1.$ This establishes the correct angle condition $ \angle B_1CA_1 \equal{} \angle CAB \equal{} \angle ABC.$

The tangency points $ A_2^i, B_2^i$ of $ \odot (M, \varrho)$ with the rays $ (CA, (CB$ are inversion images of $ A_2, B_2$ and $ CA_2 \cdot CA_2^i \equal{} r^2.$ From the similar right triangles $ \triangle MA_2^iC \sim \triangle AFC,$ $ \frac {CA_2^i}{CM} \equal{} \frac {CF}{CA},$ hence $ CA \cdot CA_2^i \equal{} CF \cdot CM \equal{} \frac {r^2}{2},$ leading to $ CA_2 \equal{} 2\ CA$ and $ A_2B_2 \equal{} 2\ AB.$

BTW, the problem (with the correct condition) was posted before at [url]http://www.mathlinks.ro/viewtopic.php?t=23118[/url] and the present solution is based on the problem [url]http://www.mathlinks.ro/viewtopic.php?t=198482[/url].


geometry
geometric transformation
reflection
circumcircle
cyclic quadrilateral
perpendicular bisector
geometry proposed

Let $ \omega$ be a circle with center $ O$. Let $ AB$ and $ CD$ be parpendicular chords of $ \omega$, which intersects at $ Q$. Lines $ AC$ and $ BD$ intersects at $ P$. Let  $ S_1$ be an intersection of $ AB$ and $ PO$, and let $ S_2$ be an intersection of $ CD$ and $ PO$. Prove that line $ PQ$ is tangent to circumsribed circle on triangle $ S_1S_2Q$.

Let ACBD be any cyclic quadrilateral, not necessary with perpendicular diagonals AB, CD. The opposite sides AC, BD still meet at P. Let U, V be midpoints of AC, BD and G the midpoint of UV, the quadrilateral centroid. Reflect O in AC, BD into $ O_1,\ O_2$ and let Q be the midpoint of $ O_1O_2.$ $ AC \equiv PU \perp OO_1,\ BD \equiv PV \perp OO_2$ are perpendicular bisectors of  $ OO_1,\ OO_2,$ P is circumcenter of the $ \triangle OO_1O_2,$ $ UV \parallel O_1O_2$ its midline, OQ its O-median, G is the midpoint of OQ, Q is the quadrilateral anticenter (intersection of maltitudes). $ PQ \perp O_1O_2$ is the perpendicular bisector of $ O_1O_2.$ Then $ \angle APQ \equal{} \angle OO_1O_2 \equal{} \frac {\angle OPO_2}{2} \equal{} \angle VPO \equiv \angle DPO,$ PQ, PO are isogonals WRT the $ \angle APD \equiv \angle CPB.$

If the cyclic quadrilateral ACBD has perpendicular diagonals $ AB \perp CD,$ the anticenter Q is identical with its diagonal intersection, $ Q \equiv AB \cap CD$ (Brahmagupta's theorem) and $ \measuredangle CPS_2 \equiv \measuredangle CPO \equal{} \measuredangle QPB$ (all $ \mod \pi$). From the cyclic ACBD, $ \measuredangle PCS_2 \equiv \measuredangle ACQ \equal{} \measuredangle QBD \equiv \measuredangle QBP.$ Then the $ \triangle CPS_2 \sim \triangle BPQ$ are (oppositely) similar and $ \measuredangle S_1S_2Q \equiv \measuredangle PS_2Q \equal{} \measuredangle S_1QP \Longrightarrow PQ$ is tangent to the circumcircle of the $ \triangle S_1S_2Q.$

Great problem!:P
How did you find this???
It took me half day, but it is very nice :)

[color=darkblue][size=117][b]Very nice problem ![/b][/size] I modified the notations from the initial proposed problem.

[/color][quote="tomek314"][color=darkred]Let $ \omega (O)$ be the circumcircle of the quadrilateral $ ABCD$, where $ AC\perp BD$.

Denote $ \|\begin{array}{c}
R\in AC\cap BD\ ,\ P\in AD\cap BC\\\\
U\in OP\cap BD\ ,\ V\in OP\cap AC\end{array}$ .

Prove that the line $ PR$ is tangent to the circumcircle of $ \triangle\ URV$. [/color] [/quote]
[color=darkblue][b][u]Proof.[/u][/b] From the [b]Hire's theorem[/b] obtain that the point $ R$ belongs to the polar $ p$ of the point $ P$ w.r.t. the circle $ w$, i.e. $ R\in p$ and $ p\perp OP$. Denote $ S\in p\cap BC$. Thus, the points $ P$, $ S$ are harmonical conjugate w.r.t. the pair of points $ \{B,C\}$. From an well-known property results that $ RB\perp RC$ $ \implies$ $ \widehat {PRB}\equiv\widehat {SRB}$, i.e. $ \widehat {PRU}\equiv\widehat {RVU}$ because $ RS\perp UV$ $ \implies$ $ \widehat {SRB}\equiv \widehat {RVU}$. In conclusion, the line $ PR$ is tangent to the circumcircle of the triangle $ URV$.[/color]


Let $x;y;z>0$.Prove that:
 $4(xy+yz+xz)\leq \sqrt{(x+y)(y+z)(x+z)}(\sqrt{x+y}+\sqrt{y+z}+\sqrt{x+z})$

Put $a=\sqrt{x+y},b=\sqrt{y+z},c=\sqrt{z+x}$, then $a,b,c$ are three sides of an acute triangle and we must prove $2a^{2}b^{2}+2b^{2}c^{2}+2c^{2}a^{2}-a^{4}-b^{4}-c^{4}\leq abc(a+b+c)$, or $16S^{2}\leq 2abcp$ or $2r\leq R$. Done! :D


4 balls are to be put in 5 boxes.In how many ways can this be done if
a)Balls are similar and boxes are different
b)balls are different and boxes are similar
c)both boxes and balls are similar

could you explain part a) a little better?  Thanks.

[quote="pilot"]a)Balls are similar and boxes are different
[hide="My Solution"]
If we add one ball to each box the problem becomes putting 8 balls in 5 boxes where each box has at least one ball. The number of ways to do this is ${8 \choose 4}=70$
[/hide]

b)balls are different and boxes are similar
[hide="My Solution"]
Since the boxes are indistinguishable, we consider the numbers of ways the 4 balls can be divided into 1,2,3, or 4 boxes.
1 box $\Rightarrow 1$
2 boxes $\Rightarrow {4 \choose 1} + \frac{{4 \choose 2}}{2}=7$
3 boxes $\Rightarrow {4 \choose 2}=6$
4 boxes $\Rightarrow 1$
Summing these up we get 15.
[/hide]

c)both boxes and balls are similar
[hide="My Solution"]
We do this in the same way as part b, , but here we don't care about which balls are being placed in the boxes, only the number of balls in each box when they are split up.
1 box $\Rightarrow 1$
2 boxes $\Rightarrow 2$
3 boxes $\Rightarrow 1$
4 boxes $\Rightarrow 1$
So there are 5 different ways to do this
[/hide][/quote]

Hi Pilot, ur solutions to part a> and part c> are absolutely correct..

But u need to recheck part( b)..The ans. should be 18, i think

[quote="pilot"]Imagine the 8 balls in a row, you can place 4 dividers to separate them into 5 groups, each group corresponds to the balls in a box so there are C(8,4) ways.[/quote]

I believe you meant "Imagine [i]9 balls[/i] in a row, you can place 4 dividers [i]in the 8 spaces between the 9 balls [/i]to separate them into 5 groups, each group corresponds to the balls in a box so there are C(8,4) ways."

Oops - Yea I meant 9 balls. If you added one ball to each box there would be 4+5=9 balls and 8 spaces between them.

[quote="pilot"]Can anyone figure out what's wrong with my solution to part b?[/quote]

Well,let's approach part b> the following way.

There are FIVE distinct cases

1> All balls are put in a single box.
 No. of ways this can be done is obviously 1

2> 3 balls in one box and the remaining one ball in another box.

No. of ways this can be done = 4C3 = 4

3> 2 balls in one box,remaining two in another box.

 No. of ways this can be done = 4C2 = 6
( this is because once the first set of two balls are selected, the remaining two balls can be placed in only ONE way,
as the boxes are indistinguishable)

4> 2 balls in one box, 1 ball in a second box, and the remaining one in a third box.

  This can again be achieved in 4C2 i.e 6 ways.

5> 4 of the five boxes having one ball each.

  This can again be done in only ONE way.

So , in all we have 1 + 4+ 6+6+1 = 18 ways.

Guys please share ur comments on the above

[quote="pilot"][quote="santosh_blr"]
3> 2 balls in one box,remaining two in another box.

 No. of ways this can be done = 4C2 = 6
( this is because once the first set of two balls are selected, the remaining two balls can be placed in only ONE way,
as the boxes are indistinguishable)
[/quote]

4C2 counts every such arrangement twice. Say the balls were A,B,C,D then choosing A,B for one box and leaving C,D for the other is the same as choosing C,D for the first box and leaving A,B for the second.[/quote]

Ahh..

Thanks Pilot..I didn't observe that..so the no. of ways would be now 15( same as the ans. u obtained)

= 1+ 4 + 4C2/2 + 4C2 + 1


How many times does the digit 8 appear in the list of all integers from 1 to 1000?

Solution:
The easiest approach is to consider how many times 8 can appear in the units place, how many times in the tens place, and how many times in the hundreds place. If we put a 8 in the units place, there are 10 choices for the tens place and 10 choices for the hundreds place (including having no hundreds digit) for a total of 10\times10=100 options, which means that 8 will appear in the ones place 100 times. (If we choose the hundreds place to be equal to 0, we can just think of that as either a two digit or a one digit number.) Likewise, if we put a 8 in the tens place, there are 10 choices for the units place and 10 choices for the hundreds place for a total of 100 options and 100 appearances of 8 in the tens place. Finally, if we put a 8 in the hundreds place, we have 10 options for the units place and 10 options for the tens place for another 100 options and 100 appearances of 8. Since 100 + 100+100=300, there will be a total of \boxed{300} appearances of 8.

can't u see that in the first case one's place digit has been fixed nd ten's nd hundred's place is supplied with 10 nd 10 digits so that i can form a no.888 in this case
now again consider the second case where ten's place digit is fixed by 8 nd similarly one's nd hundred's place r being changedwith 10 nd 10 digits so again i can form a no. 888
it is same in third case
thr r other complication with the solution too my answer for the question is about 271

First, write out your words completely.

You can form 888 three ways. It is counted 3 times in the solution. Unless I misunderstood you, which is highly likely.
what other complications do you see?

Yes, you definitely need to write out your explanation clearly. Without slang and abbreviations, and with capitalization and possibly even $ \LaTeX$.  :)  Because I'm quite certain there are no further complications.

:wink: OK! guys, I will use proper english for you.
Same as I have produced the condition of the number 888 which is counted three times, both of you can see that numbers like 288,388,....,881, 882,.... are also repeating or being counted more than once.

It should be counted twice or thrice... since it has more 8's than others like 811 and that the problem's asking for the total number of 8's, not how many numbers have 8's in them.

First, we find the sum of the number of 8s in the 1s place, the 10s place, and the 100s place. There are 1 from every 10 with an 8 in the units place, giving 100, for the 100 tens. Similarly, there are 10 from each 100 with an 8 in the 10s place, giving 10*10=100, and 100 with one in the hundreds place, for a total of 300. However, numbers with 2 eights are counted twice, so we have to subtract the number of these. For these, we have 3 choices for the position of the other digit, times 9 choices for its value, for a total of 27. Also, 888 is counted 3 times, so it must be subtracted twice. This gives 300-27-2, which is equal to $ \boxed{271}$ eights.

rationalist and gomath888, your answer is the answer to a different problem.  Namely, if the question were "how many integers from 1 to 1000 contain the digit 8", then your answer of 271 would be correct.  (A quicker way to get that answer would be to count how many integers from 1 to 1000 do not contain an 8.)

But the question that was posed is "how many times does 8 appear in the list from 1 to 1000".  Do you see how that's a different question?  In the question that was posed, we need to count the three 8's in 888 three times.  That's why the correct answer is 300.


I also noticed, that when you click to view who is online, it shows all of the members and their location, but it only shows one guest, when there are at least 20-30 online.

EDIT:  I just checked that out.  That is weird!

[quote="fedja"]I looked at the statistics page recently and saw the following:

[i]User statistics
There are 2 registered users, of which 15 (or 750.00%) are administrators (see Project:Administrators). [/i]

Sounds quite weird. Is it a bug in the counting system?[/quote]

Yeah, that's an artifact of tying the wiki and phpbb together - when someone registers now, they go through a different process than the wiki software usually uses, so the # of registered users doesn't get updated.


AMC USA(J)MO USAMO induction geometry inequalities function

number 1...although i did it a lame way where i used induction

I did something like the sum through k terms was divisible by xk, where x was less than k

#6, as i sat there solving it for almost the entire 4.5 hours.

and it made me realize how much i actually loved analytic geometry

which might be why you got #4 wrong :rotfl:

haha probably, cause i think i had three sentences :roll:

#4:

The first time, dinosaurs was not that funny.  However, by about the third time I read it, I couldn't help but laugh.   :D

I liked 1 cuz it was my neatest solution....... 5 seemed pretty cool too but that's just because I like number theory

Whichever one I get the highest score on, which'll probably be 4, or maybe 5, or maybe 1.

#1 was too easy in comparison to the others to be really nice.

#3, on the other hand, was really rewarding.  I was sitting there, trying everything I could think of, when suddenly the signifance of $n^{2}+n-1$ as being $(n+1)(n-1)+n$ hit me...and then the solution followed.

#4...pah, i dislike dinosaurs, wasn't all that great.

#2 I couldn't find a totally rigorous way to show my argument was right, but I gave a good argument, so I might get partial...yet I didn't really like the problem, too analytic.

#6 I took one look and crossed it out

#5, eh, I didn't fully solve it

As you can tell, I like discrete stuff more than continuous stuff...hence the attraction to #3, whose solution was quite pretty.

when the math teacher gave us our test today, she decided that she needed to wait until we had read 4 to see our reactions before going back to her class. the proctor thought it was pretty funny too.

Though I thought #1 was a complete giveaway, I have to vote for it because it's the only problem I expect full credit on. I had fun playing with #2 and #4 as well, and enjoyed handwaving them to death, though the graders probably won't be as happy with my "solutions".

#1, only one i'm absolutely sure that i got correct

#1 was just like...look for a pattern...also the problem had a very weak statement...

#4 was reasonably difficult...like you had to really develop a rigirous argument...i bet many people will loose points for assuming certain things...i also liked this problem because it is the first time i have applied one of those minimality-contradiction proofs

i guess the last day was more technical...i was suprised that more people did not get #6...like it was fairly straightforward to calculate the distance using trig...i found the part of the inversion solution..but they added an extra altitude...like looking at their solution it looked very difficult, but it was not really...i also think if the question were to develop a construction, that would have been cool too

My favorite problem is #3 . I allway like combinatoric problem on sets :)

[quote="themonster"][quote="Ubemaya"]What do you guys think of this years' #1 and #4 compared with last years #1 and #2? Easier?[/quote]

I think this year's #1 was about the same difficulty as last year's #1, but maybe a little nicer and more elegant.  This year's #4 definitely seemed harder than last year's #2, though.  And this year's #2 was orders of magnitude harder than last year's #2.[/quote]

Well, I thought last year's #2 was more similar to this year's #1 in terms of difficulty. Also, I do think the amount of people who solved #4 was a lot, but the number of clear,rigorous proofs were minimal.

[quote="Valentin Vornicu"]I have to admit that I liked problem 2 the most out of these problems.[/quote]

Yay finally someone shares my viewpoint! Despite the analytical-ness, I thought, when I saw the solution, wow, this is pretty cool in essence. #3 was kinda ok but there was no chance I could have gotten that so yeah. #4 seemed interesting but I was too lazy to develop a rigorous argument so I gave a lame proof that 8025 was correct. The notation wasn't that funny... #5 was a little weird, what was the motivation for that weird factorization? hmmm...
Then #6 seemed cool but my geometry is blah so obviously didn't get anywhere

I STRONGLY DISLIKE #1 BECAUSE I SCREWED UP ON THE TEST
USED A WEIRD FORM OF CIRCULAR REASONING
GAH

Yes, #2 really was an awesome problem! It was too easy to dismiss it as a rigor demon before seeing the nice solution given.

(Many problems require you to prove rigorously something that seems intuitive, but it doesn't become a [i]rigor demon[/i] until the best solution requires a few pages of technical detail (as if you were required to state associativity every time you used it :P ))

#5 and #6... they're too much based on computation, I think. (well, #5 isn't that bad if you think it's "natural" to consider the difference $(x+1)^{7}-(x^{7}+1)$. I don't.)

[quote="Ubemaya"]I have the feeling that the word "cell" instead of "square" or "block" on #2 wasn't coincidental...[/quote]

Do you mean number 4?

As for #5 there was nothing "natrual" about that problem

[quote="Ubemaya"]I have the feeling that the word "cell" instead of "square" or "block" on #2 wasn't coincidental...[/quote]

Yes, this is supposed to be #4.  I'm pretty sure that that bit wasn't supposed to be a secret, though.  What else would you call the basic unit of structure (and function) in an animal?

It is of Zuming's opinion that #3 was the best USAMO problem in the past 5 years.

For those of us lacking the insight to solve it, though, #4 was the best problem.  The optimal configuration was easy to see, but writing up a succinct proof took considerably ingenuity.  Hand-waving "proofs" were so ubiquitous that 2 was the most frequently awarded score on this problem.

[quote="gighiuhui"]For those of us lacking the insight to solve it, though, #4 was the best problem.  The optimal configuration was easy to see, but writing up a succinct proof took considerably ingenuity.  Hand-waving "proofs" were so ubiquitous that 2 was the most frequently awarded score on this problem.[/quote]
When I logged on here immediately after day 2 was done, I remember feeling incredibly insecure since so many people had confidence in their solutions to that problem and had taken very little time to do it. I personally had ascertained to myself 8025 was the answer and why in about 30 minutes. It took about 2.5 hours more (with "breaks" to make an attempt at number 5) to put it in proof form. The time spent was worth it, I guess. :D

And my proof was definitely not succinct, unless by succinct you mean something that can be written in the USAMO time limit.

how can u guys not think that #1 was the easiest?
definitely, my favorite problem!!!!

We're not talking about the easiest problem, we're talking about our favorite. I, personally, had #1 solved with a written proof on the test in a considerable amount of time under an hour, and I haven't really thought about it sense. That is definitely not true for some of the other problems.

Going back and reading the solution on the AMC website for number 3, i really like it

Oh! I'm sorry, I misunderstood the question. Number 1 is still my favorite problem, but as I look at the all of the problems and solutions again, #3 has been very intriguing. Nice solution!


well can anyone tell me what do u mean by infinite descent,and where do u use it,
pls explain with an example.

Infinite descent is a form of proof by contradiction, in which you assume that there exists a "smallest" solution to something and then show that there's a smaller one, leading to a contradiction.

Here's an example.

[b]Problem.[/b] Prove that $ \sqrt {2}$ cannot be expressed as a rational number.

[i]Proof.[/i] Suppose $ \sqrt {2} \equal{} \frac {p}{q}$, where $ p,q\in\mathbb{N}$. Suppose also that $ p$ and $ q$ are the smallest numbers that satisfy this relation (and are therefore relatively prime).

We find that $ q\sqrt {2} \equal{} p\implies 2q^2 \equal{} p^2$, and so $ p^2$ is even, which means that $ p$ is even. Let $ p \equal{} 2k$.

Now $ 2q^2 \equal{} (2k)^2 \equal{} 4k^2\implies q^2 \equal{} 2k^2$. Therefore, $ q^2$ is even, which means $ q$ is even. Let $ q \equal{} 2j$.

Now $ 4j^2 \equal{} 2k^2\implies 2j^2 \equal{} k^2\implies \frac {k}{j} \equal{} \sqrt {2}$, where $ k \equal{} \frac {p}{2}$ and $ j \equal{} \frac {q}{2}$. Clearly, $ k < p$ and $ j < q$, so we have reached a contradiction in that $ \frac {k}{j}$ is a smaller solution than $ \frac {p}{q}$, which we claimed was the smallest (furthermore, we showed that $ (p,q) \equal{} 2\ne 1$).

Therefore, by infinite descent, the statement is proven. $ \Box$

well thanks grn_trtle,nice example and proof. :lol:


The diagram below shows a $ 4\times4$ rectangular array of points, each of which is $ 1$ unit away from its nearest neighbors.
[asy]unitsize(0.25inch);
defaultpen(linewidth(0.7));

int i, j;
for(i = 0; i < 4; ++i)
	for(j = 0; j < 4; ++j)
		dot(((real)i, (real)j));[/asy]Define a [i]growing path[/i] to be a sequence of distinct points of the array with the property that the distance between consecutive points of the sequence is strictly increasing. Let $ m$ be the maximum possible number of points in a growing path, and let $ r$ be the number of growing paths consisting of exactly $ m$ points. Find $ mr$.

[hide]
(it might be helpful to draw the picture and follow along with the solution)

Label the points 1-16, from left to right, top to bottom, so that 1 and 16 are opposite corners, and 4 and 13 are the other two opposite corners.

The possible lengths of any line segments in the grid (arranged in increasing order) are

$ 1, \sqrt{2}, 2, \sqrt{5}, 2\sqrt{2}, 3, \sqrt{10}, \sqrt{13}, 3\sqrt{2}$

The maximum number of points in any growing path is therefore 10 (9 segments connecting them). We will assume that it is possible to make a growing path with 10 points for now.

First, note that we can make a "shrinking" path and just reverse it to get a growing path. We do this because it is easier to work with large distances first (leaves less possibilities for every point to travel to).

To make a segment of length $ 3\sqrt{2}$, we need to connect two opposite corners. There are 4 possible corners to start from, WLOG assume it is point 1. The next point has to be 16. From here, there are two points that are $ \sqrt{13}$ away from it (2 or 5), and we can assume WLOG that it goes to point 2 because any paths that go to point 5 can be reflected across the diagonal.

From point 2, the only points that are $ \sqrt{10}$ away from it are 13 and 15. However, if we go to point 13, there is no way to move $ 3$ units from that point to another point that hasn't been taken yet, so we have to move to point 15.

Then, the only point $ 3$ units away from 15 is point 3.

From 3, we can only go to point 9 ($ 2\sqrt{2}$ units away), and then we are forced to go to points 7 ($ \sqrt{5}$), 5 ($ 2$), and 10 ($ \sqrt{2}$). From here, there are 3 possible points that are $ 1$ unit away.

We have successfully constructed a "shrinking" path with 10 points, which means there are also growing paths with 10 points and $ m\equal{}10$

The number of shrinking paths is equal to 4 (number of corners to start from) times 2 (number of points that are a distance of $ \sqrt{13}$ from the second point in the path) times 3 (number of points that we could go to for the last step) which is equal to 24.

Therefore, the answer is $ 10\cdot24\equal{}\boxed{240}$

[/hide]

I misread and let m be the number of edges instead of vertices...sigh...

[hide="Answer"]
We place the set of points in the coordinate plane, the first point being $ (0,0)$, and the farthest being $ (3,3)$. Now, we draw the largest possible growing sequence. The sizes possible are (in increasing order) $ 1\to \sqrt{2}\to 2\to \sqrt{5} \to 2\sqrt{2}\to 3\to \sqrt{10} \to \sqrt{13} \to 3\sqrt{2}$. We can see that if we start from the beginning, we will lose a lot of the possibilities. Therefore, we start from the end.

In order for a distance of $ 3\sqrt{2}$, it has to be a diagonal. There are 4 possibilities to choose from. Since it would not matter in the final result, we can start from $ (0,0)$. There is one possibility - $ (3,3)$. We continue this until we reach our final. Multiplying everything together results in 24 of these paths.

There are 10 points in this largest [i]growing path[/i]. Therefore, the answer is $ 10(24)\equal{}\boxed{240}$.[/hide]

[hide=One of the nicer aime problems i've done]
Suppose that the bottom left dot is $(0,0)$ in the coordinate plane. I claim firstly that $m = 10$.

There are nine distinct distances between points in the array: $1, \sqrt{2}, 2, \sqrt{5}, 2\sqrt{2}, 3, \sqrt{10}, \sqrt{13},$ and $3\sqrt{2}$. Observe that these are the distances from $(0,0)$ to all the points lying on or below the line $x = y$; if we take the "starting point" of a segment to be any point other than a corner, we can always translate the segment such that one of the points coincides with a corner and the other lies in the grid, so these are all the distances. Thus $10$ is the maximum number of points. A working construction is $(2,3), (2,2), (3,1), (1,1), (3,2), (1,0), (1,3), (2,0), (0,3), (3,0)$.

Therefore we need to find the number of growing paths consisting of exactly $10$ points. We will instead find the number of "shrinking paths"; that is, paths with the property such that the distance between consecutive points is strictly decreasing.

Consider the first two points of a shrinking path; the distance between them must be $3\sqrt{2}$, so they must be two corners. Without loss of generality, assume the first two points are $(3,0)$ and $(0,3)$ in that order; we can multiply by $4$ at the end to account for all cases.

The next point must be $\sqrt{13}$ away; again, without loss of generality, assume we choose to go to $(2,0)$ from $(0,3)$. We will multiply through by $2$ at the end.

Here we must now go to $(1,3)$; if we go to $(3,3)$, there will be no empty points of a distance $3$ away. After this, our next five steps are, yet again, determined: we proceed to $(1,0), (3,2), (1,1), (3,1),$ and $(2,2)$.

We have three choices for our final step. Therefore we may multiply $3 \cdot 4 \cdot 2$, giving us $24$ total working paths of length $9$.

Then we multiply $24 \cdot 10 = \boxed{240}$, so $240$ is our answer.


algebra
binomial theorem
geometry unsolved
geometry

can anyone provide geometric proof of binomial theorem?

What exactly do you mean by "geometric"?  The closest thing I can think of is a path-counting argument, which is still essentially combinatorial in nature.

for example ,for $ (a\plus{}b)^2$ and $ (a\plus{}b)^3$ we have geometric proof cosidering a square and a cube.so if we try it for $ n$dimensions then how can we proceed?

Those "proofs" are only useful because they're intuitive:  in higher dimensions, the exact same pattern holds, but determining what it is is no easier than using the binomial theorem itself because of the lack of physical intuition.


As our regulars have noticed, we have overhauled our site.  Please post any bugs you find here & we will fix them (such as broken links, messed up images, or whatever).

The links to the Forum on the Forum page (when you cleck "Forum" while you're in the forum) is directed to the old forum, but that's about it.

I like the moving advertisement :)

er...all the words in the post are messed up...?

also, the page is really big, and thus requires massive scrolling

All the older posts seem to have their HTML + IBBCode disabled.

Not that big of a problem, I suppose.

Osiris - can you point me to a thread?  All the old threads I've looked at render fine.

Mystic - do you mean left-right scrolling?  On my screen, the fonts come up slightly larger than the old site.  They are scalable now, by the way (you can make them bigger or smaller in the standard way).  Also, what do you mean by 'all the words in the posts are messed up?'

I mean that the letters are printed on top of each other. => moderately hard to read. If I double space

like this

then the letters appear

as if they were printed

normally (single-spaced).

Mystic - do you have your browser set to larger-than-normal type?  That would cause the problem you mention (particularly in IE, which has fewer size steps than Netscape).

not sure if this is the same thing as tare was talking about, but when you click 'forum' button, it doesnt work.  And im not sure if this is what osiris was talking about, but the code doesnt seem to work here at triple m's post  http://65.36.154.75/Forum/viewtopic.php?t=2449&start=15  .  I dont know the code, so it could be that he made typos, though.

The forum thing is fixed - next time you refresh your page (or fire up your browser from scratch), that will be fixed for you.

The HTML may be a result of TripleM not enabling html in his post (I 'edited' it and enabled HTML) - earlier posts in that same thread (ComplexZeta's, for example) render fine.

I can't access any class transcripts. It all says the same text:

"You have reached this page in error. If you seek a transcript for a course in which you are enrolled, please log in, then access the class from the My Classes section of the Community."

This is the case for all three classes so I'm guessing that it's something with enrollment scripts (or however you do those) rather than a bug on a particular class.

I just tested the problem sets and they seem to be working fine so it's probably just something with the transcripts instead of enrollment in the course.

TripleM accidentally clicked on back or something so the last code ([/hide]) got typed twice, that's all.

Ctrl+scroll makes the font larger or smaller instantly fyi.

The top "Forum" got fixed but the menu to the left is still linking to the old site.

Sidebar -> fixed.

I'm also working on having three font size settings you can choose from on the index page (as opposed to relying on brower resizing, which doesn't seem to work as well as I'd like).

Huh, it works fine for me, perhaps you're going to another website then coming back...

For those of you who have this problem, do you have the forum set to automatically log you in or not?

I do...but when I logged on today, the new posts thingy showed up so I don't know what's up, but as of now...it works...

It's stopped working for me. It displays the new posts when I come to the site, but after I read a thread, all threads and forums are suddenly marked as read, which makes things difficult.

oh...it's worked for me ever since I pointed it out...

confuted how do you go back to the main screen of the forum with all the boards showing?

I click on the little blue link artofproblemsolving.com Forum Index usually.

Then you should have no problem...maybe you should turn on that "Automatic Log In" thing anyways and hope it works.

This happens on every computer iv been to.  After i post a new message, it says"click here to view your message" and "click here to view return to the forum".  It automatically links me back to the message.

It's supposed to do that after like 5 seconds...

Yes, it's supposed to be automatic after a few seconds.  The site has a JavaScript program that does that.

no.  As soon as it shows up, it starts opening another web page. It waits for less than a second.

What browser (Internet Explorer, Netscape Navigator, other) do you use, and what version?


My school's team got 1st though I wasn't part of it. Only encouraging them.

we did that on 3 questions last year....

[quote="dingzhou"][quote="unimpossible"]Getting along is a huge part of the team round.  Its everybody's job, especially the captain's, to sort it out.[/quote]

no wonder we failed... :maybe:[/quote]

We got 1st. However, I was not part of the team.

The last time I was head of a team was at the Chapter competition--and I had two members constantly questioning my authority and stuff by rewriting every answer, claiming it was "illegible", and asking questions that I would answer over and over again. For example, "Why can't we ask the other team [from our school] for help?"

I grew frustrated, seeing as I was pretty much the only person doing the problems.

...I did [i]try[/i] to sort it out...but I just didn't have the team for it. :mad:

[quote="shentang"]My school's team got 1st though I wasn't part of it. Only [b]encouraging[/b] them.[/quote]
-COUGH-
Yeah we did get first.
:o

[quote="unimpossible"]Getting along is a huge part of the team round.  Its everybody's job, especially the captain's, to sort it out.[/quote]

well the only problem with that was the captain was the one who got the final say, even though we thought her answer was wrong. So she wrote down the wrong answer, and had she written down ours, we would've gotten 3rd in the state.

Our team got ninth, but I got third in the individuals. Don't ask me what we got in the team round or what anybody else on my team got... :maybe:

I wasn't even on a team at states. But if we were, we'd be REALLY bad. :mad:

[quote="Math Geek"]Our team got first.  Our individual rankings were like:

1
2 (me :( )
8
20[/quote]

I honestly don't get why you are sad about getting second at states, but anyways-

Our team ranks were

2(me)
11
40-something
60-something

5th Place Team

Oh, I thought this was asking individual place. But it's still the same. 3rd...

What is wierd:

Last Year: places 13, 21(me?), 25, 27. team 3rd

This year: places 2, 3 (me!), 12, 14(individual), 16. team still 3rd......

Team rankings.
2
4
5
23
I won't say which one I was...
:wink:

phail
9th T.T
1st, 42nd, other two didn't place in top 25%
blargh

[quote="ZhangPeijin"]Dang.
:(
Anyways, I'll do better next year, I hope.[/quote]

Your "mistakes" aren't really mistakes are they...

1st...

Not telling rankings, because I'm not supposed to.

My team got first  :D (i'm on unimpossible's team)

Individual ranks were

2
3
7 (me)
10

Everybody on my team is beast.   :gleam:

[quote="Yongyi781"][quote="mathcrazed"]
Not telling rankings, because I'm not supposed to.[/quote]
Erm... you can.[/quote]

He means his coach doesn't want him to, so let him be :lol:


Five cards are lying on a table as shown.
\[ \begin{matrix} &\qquad&\boxed{\tt{P}}&\qquad&\boxed{\tt{Q}} \\ \\ \boxed{\tt{3}}&\qquad& \boxed{\tt{4}}& \qquad &\boxed{\tt{6}} \end{matrix} \]
Each card has a letter on one side and a whole number on the other side.  Jane said, "If a vowel is on one side of any card, then an even number is on the other side."  Mary showed Jane was wrong by turning over one card.  Which card did Mary turn over?

$A)\quad 3 \qquad B)\quad 4 \qquad C)\quad 6 \qquad D)\quad P \qquad E)\quad Q$

[hide="Im not good at logic"]
Jane said "If a vowel is on one side of any card, then an even number is on the other side." not "If a even number is on one side of any card, then a vowel is on the other side."  P and Q are not vowels, so clearly, the only possible card is 3, which is choice $\boxed{A}$.
[/hide]

[hide="Two Ways"]There are 2 consonants, 2 evens, and 1 odd. The odds are the only ones that are made up of only one card, so
>>A<<

-OR-

For the consonants: can't be them because the statement has nothing to do with consonants
For the evens: don't matter because the statement allows for a consonant also to be on the other side of an even card.
For the odds: Must be them because if they have a vowel on the other side it would invalidate the statement, so
>>A<<[/hide]

hold on, but that would be assuming that each card can't have something that isn't shown, like a Z or an 8, or something... The problem doesn't say that.

No, their answers are right. The question says, "[b]If[/b] a vowel is on one side of any card, [b]then[/b] an even number is on the other side." It wouldn't help to turn over either of the lettered cards, because there's no rule that if you have a consonant on one side, there'll be an odd number on the other. It also wouldn't help to turn over an even-numbered card. The condition is [b]if[/b] there is a vowel, [b]then[/b] there is an even number, not [b]if[/b] there is an even number, [b]then[/b] there is a vowel. Therefore, a card with an even number on one side could have a consonant on the other. The only card left to turn over is $3$.

[hide]Letters dont count, they aren't vowels
The even numbers dont count, it is a one way if, then
therefore, it is the last one, 3[/hide]

oh...I thought you had to prove one of them wrong.

[hide]it can't be the even numbers cause there is 2, and p and q are not vowels so it is A[/hide]


function
limit
calculus
calculus computations

Hi everyone, I have trouble with the question below, could someone help me to solve it please? thanks

The question is: Find the values of a, b, c, and d. In other words find a,b,c and d such that the function f(x) is differentiable.
(this is a piecewise function)

where:  f(x)= (   x+2                    ,     x ≤ 0     )
                      ax^3+bx^2+cx+d   ,   0≤ x ≤2
                      2-x                      ,    x ≥ 2

The function $ f$ is differentiable if and only if the following four conditions are met:
1) at $ x\equal{}0$ the first and second formulas give the same value of $ f$
2) at $ x\equal{}0$ the first and second formulas give the same value of $ f'$
3) at $ x\equal{}2$ the second and third formulas give the same value of $ f$
4) at $ x\equal{}2$ the second and third formulas give the same value of $ f'$

Just a tiny issue, the approach is basically same as that of [b]mlok's[/b] :)

I think the question should be stated something like,
$ f(x) \equal{} x \plus{} 2$ $ \forall$ $ x < 0$
$ f(x) \equal{} ax^3 \plus{} bx^2 \plus{} cx \plus{} d$ $ \forall$ $ 0\leq x < 2$
$ f(x) \equal{} 2 \minus{} x$ $ \forall$ $ 2\leq x$

In that case you cannot calculate $ f(0)$, $ f'(0)$ for the first formula.

To find $ f(0)$ assume $ f$ is continous at $ x\equal{}0$; $ \lim_{x\rightarrow 0}x \plus{} 2 \equal{} c \equal{} f_2(0)$
and you get $ f(0)\equal{}2\equal{}c$ unambiguosly.

Similarly assume that $ f$ is differentiable at $ x \equal{} 0$
Then $ \lim_{x\rightarrow 0^ \plus{} }\frac {f(x) \minus{} f(0)}{x \minus{} 0} \equal{} \lim_{x\rightarrow 0^ \minus{} }\frac {f(x) \minus{} f(0)}{x \minus{} 0}$
That is, $ \lim_{x\rightarrow 0}\frac {(ax^3 \plus{} bx^2 \plus{} cx \plus{} d) \minus{} d}{x} \equal{} \lim_{x\rightarrow 0}\frac {(x \plus{} 2) \minus{} 2}{x}$

similarly for $ x \equal{} 2$


Did you try [url=http://www.google.com/]Google[/url]? In particular [url=http://books.google.com/bkshp?tab=wp]Google Books[/url].

[quote="arthur eddington"]When i first learned that '[/b] Fermat[b]'[/b]s[b] Last[/b] Theorem[b]'[/b] was finally solved by Mr[b] Andrew[/b] Wiles[b],[b] my interest in mathematics was aroused.(I learned it only 2005!) Can anyone of A.O.P.S. Members help me if i there is an available[b] free[/b] copy of '[u]Carl[/u] Friederich[u] Gauss[/u]'[u] classic work entitle '[/u] Disquisitiones[u] Arithemetica[b]?[/b][/u]'[u] Is there any free available copy in the internet[b]?

                                                            Respectfully,
                                                           A.Eddington

[/b][/u][/quote]

not in google books...i think...

[url]http://www.tcs.uni-luebeck.de/pages/balbach/div/disqarith.pdf[/url]

was the only thing i found...


Hey people,
I'm an 11th grader in Bangalore with eyes set on IIT... I've joined BASE but I just wanted your views on whether it's really necessary to be in a coaching class environment to do well there...
'Coz I'm really wondering whether the lousy problem solving and the stress from the teachers about the monthly tests is really worth the amazing peer group you get in these classes. What do you think?
Also I wanted to know the success rate of people who got into IIT with only correspondence courses
and no actual tuition. Can you please advise?
Thanx.

I don't know about others,but I feel coaching matters (for me it matters more than schooling)

But it all depends on you and your way of preparation.

btw,number of 11th graders is increasing :)

well ask Da Vinci
he is also in BASE :P

coaching is necessary to develop a feel

and if u r already good to clear without it, it helps u rock

our mod didn't take any(mind u ANY--not even correspondence) but still got AIR 920 or sth in JEE :oops:
u shud just hav few good teachers in school or surroundings or friends...... :)  :D

hey it means that now i am accompanied by some friends who dont do any coaching classes !

you said you joined BASE! then why this question? plans of discontinuing? :huh:

@rt
read carefully :rotfl:  :rotfl:  :rotfl:

cmon yaar!
\they might have joined some coaching but are $ unsatisfied$ with it!
 :rotfl:  :rotfl:  :rotfl:\

Erm.. yes, I am $ DISsatisfied$ with BASE :D but honestly the only reason I'm there i because my school teachers are [b]brilliantly bad[/b] and my peer group at school has no clue about careers, forget JEE...  :P
I was thinking of discontinuing and working for IIT on my own coz it really doesn't seem difficult, but I kinda agree with all of you that it is necessary if only to give me the feel of the competition..I think I will stick on at least for the 11th.. and then see for the 12th..

BTW it's nice to see 11th graders here..where r u all from?

@skand- Where have you joined for coaching?

EDIT: Da Vinci- One thing I've really been wanting to know- Howz BASE 12th coaching?

And yeah, thanx for the inputs everyone!!

I am in FIITJEE
I also think my school teachers are not so good if you consider standard of IITJEE.

but my teachers in FIITJEE rocks
esp.physics(Parvez Alam Sir)
He is a genius and also the author of FIITJEE package of Rotational mechanics
he is awesome,truly awesome :omighty: admirable,words are not enough to describe him

hey i live in city of nawabs and am doing private coaching
 :) but keeping in touch with the study material of fiitjee(with the help of my friends.............)

[quote="Padmavathi"]Erm.. yes, I am $ DISsatisfied$ with BASE :D but honestly the only reason[b] I'm there i because my school teachers are [b]brilliantly bad[/b][/b] and my peer group at school has no clue about careers, forget JEE...  :P
I was thinking of discontinuing and working for IIT on my own coz it really doesn't seem difficult, but I kinda agree with all of you that it is necessary if only to give me the feel of the competition..I think I will stick on at least for the 11th.. and then see for the 12th..

BTW it's nice to see 11th graders here..where r u all from?

@skand- Where have you joined for coaching?

EDIT: Da Vinci- One thing I've really been wanting to know- Howz BASE 12th coaching?

And yeah, thanx for the inputs everyone!![/quote]

if you discontinue, you'll miss some of the legendary teachers(who have worked in the JEE paper framing committee) :D (they teach for second yr IIT)

and btw, you said you joined BASE because your school teachers are not good! You need coaching for school stuff?  :huh: ......
.......and you also say that you can manage the JEE studies yourself! The statements seem contradictory!  :huh:  :P

hey one correction ritu i had fiitjee materials but equivalently i didnt have them because i never read them  :P
see i will tell what my experiences have been and will let u conclude whetehr u need a coaching or not
for me it was not solving problem ,mugging formulaes that was(and still very much is) important what mattered me is the very  funda understanding of the subject....having not attended coaching i got sufficient time to read books which otehrs didnt purely out of interest.
See my basic interest being amths and physics and having read sufficent analysis before 10th (due to teh excellent peer group i had in my skool....) i felt quite comfortable spendin much time on physics and chem..
See the most imp problem/distraction u have while u self study is that u get loooooooots of time and u mite wanna waste it..this mite not only affect ur concentration but also affect ur rank/performance...but if u can overcoem this and study hard with all the self-cnfidence and not caring about ur peer group who say they ahve finsihed x chaps fropm thsi book and y chaps from other book...
read for teh love of teh subject that's what i wud say which u cant do in coaching
as for coaching they have one great plus point they make ur study welll planned but in that process do destroy ur originality but as skand said u mite have really inspiring teachers.....

correct funda da !
thats the same reason y i hav joined just a maths coaching and which i will be leavin' very soon!

whatever be the matter,you got to say,par rocks :) ,that good rank without coaching leaves me flabbergasted :)
You rock man!!!

but I won't leave coaching despite this,i am not that good as well as I have got some really wonderful teachers here.

nobody is telling u to leave

just keep the passion with u :D  :P

along with the passion to study
 keep alive the passion for competition  :rotfl:  :rotfl:  :rotfl:

be sure of one thing
no coaching let you remain in peace and let die passion for competetion :diablo:

ya thats the only thing i thoroughly like about coaching institutions. they keep  a fire igniting in u!
 :ninja:

First of all I'm really sorry for the late reply..was really caught up...
Phew! That's a lot of advice and I totally agree with all of you that our mod's achievement was AMAZZING!!
I've also joined FIITJEE postal and it seemed quite good so that's y i said I could handle JEE on my own...But in case of doubts or anything, our school teachers are awful..so you can't even expect them to be nice to you for having joined coaching, forget motivating you.. That was my main reason for going into BASE.. and of course like you all mentioned, the FIRE OF COMPETITION remains ignited thanks to the peer group  :)
Maybe I shall take stock at the end of eleventh, and continue till then...
The thing is, for me, I'm more keen on getting in-depth into the subjects and actually enjoying learning them, rather than having a bunch of formulae and results thrust at you to mug up and then you getting tested on these formulae each month..I sometimes get the feeling that some of them are killing the subjects..but science will NEVER die  :D lol..
(No, I'm just saying that in reaction to the vigour with which my friend in the commerce class solves simple addition and subtraction problems(with a calculator!!) in a whole subject called accounts  :rotfl: )

I many a times have felt what we do in coaching classes is adulterated science. Apart from the 'tricks', many concepts are not explained as they should be, as an example, limits and probability. As a result, many students are not able to cope up after the join iits or do research elsewhere...

Well....for my part.....I didn't take part very eagerly in IIT coaching...
I enrolled for FIITJEE RSM.....
still I got a placement....though at the very end.....my special "[b]BATA SHOE PRICE[/b]" roll no. suggests so!! :D

Anyway...I must admit that the Study Material is very good...and useful...
But if you have your conceptions clear....I think no problem will occur....you can crack anything...

[i][color=red]Conception is like pure water....any colour tablet you can dip inside it....it will turn to that colour...may that be IIT tablet, may it be ISI tablet.....may it be something else....[/color][/i]

maybe terrorist group al-qaida tablet :rotfl:  :rotfl:  :rotfl:

What do u mean????????
I am serious......do u think I'm joking????????
:furious: :furious:


calculus
vector
AMC
AMC 12
MIT
college
AoPS Books

Hey guys!

I'm basically a senior in HS, in a course of Physics C (wow, biggest joke ever), Calc BC (decently hard, but still, nothing comparerd to AIME...).

Background: 1.5 yrs of physics, bit of calculus, vector calculus (self study), ODE (self study).

I'm kind of a late starter.  I didn't get interested in math/science olympiads until 5-6 months ago, after taking my 1st AMC12 (and scoring a dismal of 105).  And now, I find all those olympiads, including math, physics and chemistry (the easiest one for me) fascinating.  They are fascinating to me because I mostly can't solve any problems, except for the chemistry one. :blush:

I want to become much more solid in physics, since I am going to a pretty tough college next yr (think MIT), I'd like to get sufficient preparation.  So I have a few questions.

1. Are olympiad problems good preparation if I plan to become a physics major?

2. If I can't handle things like the semifinalist exam for USA physics olympiad, does that mean I'm not cut it to be a successful physics major at a top University?  I love physics, but honestly, competing against guys on this Advanced forum gives me chills down the spine.

3. For guys who are physics majors, did you find the tough physics course have finals that are as difficult as what I would find on a physics olympiad semifinal exam, or IphO exam (those are ridiculous)?

4. Would you advise me to get the Introduction to Mechanics by David Morin, if I plan to become a physics major?  Do you find that kind of tough problems on your intro physics class?

I am capable of doing most Halliday resnick, Tipler's problems, some olympiad level, but that's it.  I would really enjoy some of your experience, as you recall how you did it when you started to become more advanced (I guess you guys were not born physics geniuses...) :D

Thanks.

No one?? :(

I'm thinking about buying David's Morin Intro to classical mechanics, since Terminator made a comparison between that book and the AoPS books, which I like a lot.  Will sb like me benefit from that book, or should I better stick to sth like Resnick Halliday?  I find Resnick Halliday pretty easy, except for the Thermo and optics parts because I didn't study those sections yet. I am also planning to be a physics major.

Morin's book is great.  If you want a comprehensive introduction to elementary (not to say easy) physics, consider reading the Feynman Lectures as well.  I agree that Halliday and Resnick is pretty easy - for better problems, get Irodov's physics problem book.

As for questions 1 and 2, don't worry about not being a beast at physics competitions.  Just keep doing problems, and you will naturally develop your analytical abilities.  Doing well in physics competitions surely doesn't mean you won't be successful as a physics major, because solving problems under time pressure isn't what physics research is about.  You will need strong problem-solving skills, but competitions aren't the only measure of that.

So don't worry about that, and congrats on getting into MIT or whatever college you got into.


I am reading [i]Combinatorics[/i], 2nd edition, by Russell Merris. In section 1.1:

[quote][hide]The Hawaiian alphabet consists of 12 letters, the vowels a, e, i, o, u and the consonants h, k, l, m, n, p, w. Show that 20,736 different 4-letter ‘‘words’’ could be constructed using the 12-letter Hawaiian alphabet.[/hide][/quote]

I simply got an answer to be $ \text{nPr}(12,4)\equal{}11880$. What's wrong with my solution, if any?

How do they define a word? Any string of letters, or must there be one vowel, etc.?

[quote="isabella2296"]I think he means $ 12 \times 11 \times 10 \times 9$.[/quote]
But how does that correspond with the 4?
This notation is confusing.

[quote="Urc"]What is nPr?
National Public Radio?[/quote]

nPr is notation for permutations.

$ \text{nPr } \equal{} \frac{n!}{(n\minus{}r)!}$

This means that $ \text{nPr} (12,4) \equal{} 12 \times 11 \times 10 \times 9$.

This is because $ 12!$ is $ 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 10 \times 11 \times 12$ and $ 8!$ is $ 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8$, thus, the number $ 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8$ cancels out.

its simple. it has not been given that repetition is not allowed.so, permutations are 12*12*12*12=20736

i agree wth nick. when something is not given in question, it is because that thing is obvious or just a usual thing and is expected of the reader to interpret it normally and not use combinatorics there,. :D  :D

[quote="isabella2296"]This problem wasn't very clear though. It never defined what a word was.[/quote]

It was probably in the context of the book. AoPS volume 1 (probability section) never defined what a "word" was in some of their problems; they already had it defined a few paragraphs back and there was no need to define it again.

And ThetaPi's original mistake was that he misunderstood the question. Had the question ask how many words could be formed such that no letter was used more than once, he'd be corect.


Same rules except:
You get 1 point for posting a rule.
You get 5 points if nobody posts a new rule in less than 48 hours.
You can challenge a rule if you think that no new rules can be made. If someone posts a new rule in 48 hours, you lose 3 points. But if your challenge is successful, the other person loses 10 and you gain 2.

1. All rules must include an emoticon like  :D.

[hide="score"]
aleph0: 1
[/hide]

1. All rules must include an emoticon like  :D .
2. No rule after this one may contain the letter "t".  :P

[hide="Scores"]cf249: 1
aleph0: 1[/hide]

4 invalidates 1,2,3.

Because, 1,2,3 breaks 4.

1. All rules must include an emoticon like :D.
2. No rule after this one may contain the letter "t". :P
3. All rules will have a minimum of seven words. :)
4. You should say "AIME15 pwns" somewhere in your rule.  :D
5. Lying [e.g. "AIME15 pwns"] :P is bad. One word should be [b]bold[/b] in all rules.
[u]6. "AIME15 pwns" is a [b]false[/b] phrase.  Curiously, all rules should now be underlined.  :) [/u]

@pythag:  Rules made before a given rule don't have to follow that rule.

By the way, post score with your rule.

[hide="score"]
aleph0: 2
cf249: 2
AIME15: 1
pythag011: 1
[/hide]

1. All rules must include an emoticon like :D.
2. No rule after this one may contain the letter "t". :P
3. All rules will have a minimum of seven words. :)
4. You should say "AIME15 pwns" somewhere in your rule.  :D
5. Lying [e.g. "AIME15 pwns"] :P is bad. One word should be [b]bold[/b] in all rules.
[u]6. "AIME15 pwns" is a [b]false[/b] phrase.  Curiously, all rules should now be underlined.  :) [/u]
[u]7. All rules shall have an [b]even[/b] number of words  :o . "AIME15 pwns" is obviously false.[/u]
[u]8. No rule following rule "seven plus one" can have a "p", [b]unless[/b] you are saying "AIME15 pwns". Okay?  :mad:[/u]

[quote="AIME15"]You guys are mean to me :([/quote]

Ahem!!! Post number 153.  :)

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=242496&start=140

1. All rules must include an emoticon like :D.
2. No rule after this one may contain the letter "t". :P
3. All rules will have a minimum of seven words. :)
4. You should say "AIME15 pwns" somewhere in your rule.  :D
5. Lying [e.g. "AIME15 pwns"] :P is bad. One word should be [b]bold[/b] in all rules.
[u]6. "AIME15 pwns" is a [b]false[/b] phrase.  Curiously, all rules should now be underlined.  :) [/u]
[u]7. All rules shall have an [b]even[/b] number of words  :o . "AIME15 pwns" is obviously false.[/u]
[u]8. No rule following rule "seven plus one" can have a "p", [b]unless[/b] you are saying "AIME15 pwns". [/u]
[u]9. No rule may have [b]any[/b] odd number of symbols from now on.  AIME15 no pwns.  :rotfl:[/u]

What do you mean by symbols?
Did you mean to say characters but couldn't use the letter t?

"Rules can't have exactly seven words" in a very longwinded fashion.

EDIT: Or is it "rules with exactly seven words are no longer valid"? In any case, it's a rule.

I'll assume the former, else that breaks a global rule.

7. By won, word htgnel larger 10 = dennab creator, yltelpmoc!

Ok, that was really awkward, and is an easy rule to follow. [Word length < 11]

[quote="james4l"]I'll assume the former, else that breaks a global rule. [/quote]
You assumed correctly.

As for your latest rule, james4l, I don't know if it's acceptable because 10 has three letters when spelled out.

10 here is a number and not a word.

although I originally placed "twenty" instead of 10, that was far too unrestrictive.

[size=200]Rules must have a font size that is greater than or equal to Huge.[/size]

[size=200] Font size must follow an arithmetic sequence.[/size]

I had a good rule, but apparently, the font size can't be larger than 29.

New set of rules:

Rule 1:  All rules must have an odd number of words.

[quote="james4l"]Rule 1:  All rules must have an odd number of words.
Rule 2: A caveat: it must always contain a composite number of words, like this rule here.[/quote]

I'm not that smart... wots a composite #?

A natural number that has more than 2 distinct factors. You can also say a number, not equal to one, that is not prime.

Rule 1: All rules must have an odd number of words.
Rule 2: A caveat: it must always contain a composite number of words, like this rule here.
Rule 3: No rule may have the same number of words as the rule above it, making rules successively longer and even longer.
Rule 4: For guidance purposes, you must place the number of words at the end of the rule. It should be clear as to which is the number, because it will be delineated with square brackets, with the numeral placed inside them, OK people? Something like this: [45]
5: Constraints getting longer and longer make a need for restraint; therefore the 21st and 25th letters of the English alphabet cannot be written in constraints from now on. These letters, for information, are the ones between "t" and "v" and between "x" and "z". These words complete the total. [49]
6: All constraints after this one can't have the word constraint in it. For those that don't know, constraint is a ten letter word. It begins with the letter "c" and ends with the letter "t". The second letter is "o", the third letter is "n", the next letter is "s", the fifth letter is "t", the sixth letter is "r", and there are a few more letters that follow. It is important that all aops members know this word and like it. [81]
7: The ten letter word beginning with "c", "o", "n", "s", and "t" and ending with "r", "a", "i", "n", and "t" was forbidden in the last posted restriction. However, the word "constraints" was not. From now on all constraints have to have the letters "x", "z", and "q" somewhere in them, and those letters have to be in words, like "xenon", "zebra", and "Iraq", that cannot be taken from above restrictions. This task is made harder, since the 21st letter of the English alphabet, which follows most "q"s, is not allowed. [91]


algebra
polynomial
vector
quadratics
linear algebra
matrix
superior algebra

Find the primitive elements for the extensions
Q(sqrt 5, sqrt 10)/Q
Q(sqrt 3, i)
Q(4th root of 3, i)/Q

In each case, find the minimal polynomial over Q.
[Q denotes the rationals]

For the first extension, I guess (sqrt 2)+(sqrt5), since these make up sqrt 5 and sqrt 10.

For the second extension I guess (sqrt 3) + i

I don't know about the third one.

You were doing so well, why not guess $ \sqrt[4]{3}\plus{}i$ for the third one? :P

The way to prove your guesses is to find the degrees of each field extension (not extremely difficult) and then show that the minimal polynomial for your guess has that degree.

For example: if $ \alpha\equal{}\sqrt{3} \plus{} i$, we have $ \alpha^2 \equal{} 2 \plus{} 2\sqrt{3}$, so $ (\alpha^2 \minus{} 2)^2 \equal{} 12$, or $ \alpha^4 \minus{} 4\alpha^2 \minus{} 8 \equal{} 0$.  So if $ x^4\minus{}4x^2 \minus{} 8$ is irreducible, you're done...

Thanks!

I'm also having trouble finding the degree of each extension. For the first extension I know we can split up Q(sqrt 5, sqrt 10)/Q into Q(sqrt 5)/Q, Q(sqrt 10)/Q.  I was told the latter extensions each have degree 2...how is that so?

OK, I'll try to include some details. :)

It's true that $ E \equal{} \mathbb{Q}(\sqrt {5},\sqrt {10})$ is the smallest field containing both $ K \equal{} \mathbb{Q}(\sqrt {5})$ and $ L \equal{} \mathbb{Q}(\sqrt {10})$.  We'll use this in a moment.

First, you want to show that the degrees of $ K/\mathbb{Q}$ and $ L/\mathbb{Q}$ are both 2.  In fact, if $ d$ is not a square, the degree of $ \mathbb{Q}(\sqrt {d})/\mathbb{Q}$ is always $ 2$, because the extension is spanned as a $ \mathbb{Q}$-vector space by $ 1$ and $ \sqrt {d}$. (and these are linearly independent)  More generally, the degree of $ \mathbb{Q}(\alpha)/\mathbb{Q}$ is the degree of the minimal polynomial of $ \alpha$. (if you don't know how to prove this, read about it in an algebra text, or start a new thread)

Anyway, with not too much work, you can show that $ E$ is spanned by the vectors $ 1, \sqrt {5}, \sqrt {10}, \sqrt {50}$, so the degree is at least $ 4$.  Since $ E$ contains $ K$, and $ K/\mathbb{Q}$ has degree 2, the degree of $ E/\mathbb{Q}$ must be 2 or 4.  But if it were 2, we would have $ E \equal{} K$.  Similarly, $ E \equal{} L$, so $ K \equal{} L$, which is false. (why?)  So $ E/\mathbb{Q}$ has degree 4.

[quote="LydianRain"]You were doing so well, why not guess $ \sqrt [4]{3} \plus{} i$ for the third one? :P

The way to prove your guesses is to find the degrees of each field extension (not extremely difficult) and then show that the minimal polynomial for your guess has that degree.

For example: if $ \alpha \equal{} \sqrt {3} \plus{} i$, we have $ \alpha^2 \equal{} 2 \plus{} 2\sqrt {3}$, so $ (\alpha^2 \minus{} 2)^2 \equal{} 12$, or $ \alpha^4 \minus{} 4\alpha^2 \minus{} 8 \equal{} 0$.  So if $ x^4 \minus{} 4x^2 \minus{} 8$ is irreducible, you're done...[/quote]

What is your process for finding the minimal polynomial?

I already showed you my process in as much detail as I know how.  :)

Well, I showed you how to find some polynomial that $ \alpha$ satisfies, now all you need to do is show that it is irreducible.  In this case, it is not so hard to try factoring it into quadratics, but it might be trickier when the degree is 6--perhaps you could get Eisenstein to work.

How to find the minimal polynomial of some $ \alpha$ which is a sum or product of other algebraic numbers $ a$ and $ b$? Interpret multiplication by $ \alpha$ as a linear operator on $ \mathbb{Q}(a,b)$, find a matrix for it, and calculate that matrix's characteristic polynomial; it will be some power of the minimal polynomial of $ \alpha$. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=202382]here[/url] for more.


You know the question with the farmer who ONLY had jellyfish, cows, and octopi on his farm?  Did anyone else's team burst out laughing on it?   :rotfl:   I thought it was kind of funny, but the kid next to me laughed for about half the test.

No one at my chapter actually laughed out loud, but I found it humorous that he was farming jellyfish.

Posts here are kinda supposed to be about questions answer challenges or saying how well you did @ a competition


The Poincare- Bendixson : A noempty compact limit set of a $ C^1$ planar dynamical system which contains no equilibrium is a closed orbit ([url]http://en.wikipedia.org/wiki/Poincar%C3%A9%E2%80%93Bendixson_theorem[/url])

My question is : apply the Poincare- Bendixson to prove that the equation :   $ \,\,\"{x} +\epsilon(x^2-1)\.{x} +x=0$ has at least one closed orbit  ( where $ \epsilon >0$

The hint is : First find two nested closed curves $ C_1, C_2$ such that the flows crosses $ C_1$ inward and crosses $ C_2$ outward

I don't know how use the hint to solve the problem, could you please help me ? thanks