geometry
trapezoid
ratio
geometric transformation
homothety

In the trapezium ABCD, AB is parallel to CD.

AD and BC meet at P.

The two circles whose diameters are AD and BC  meet at E and F.

Prove that PA.PC = PB.PD = PE.PF

This problem seems to require  the Ssa proposition: If 2 triangles have 2 sides and the angle against the larger of the 2 sides congruent, then they are congruent. Likewise, if 2 triangles have 2 sides in the same ratio and the angles against the larger one congruent, then they are similar.

[b]Proof:[/b] (of congruence using the SAS proposition). Let the triangles $\triangle ABC, \triangle A'B'C'$ have the sides AB = A'B', BC = B'C' congruent, $BC \ge AB$, and the same angle $\angle CAB = \angle C'A'B'$. Assume that $AC \neq A'C'$, for example, AC > A'C'. Then there exist a point D on the ray AC different from the point C, such that AD = A'C'. Since AD = A'C', AB = A'B' and the angles $\angle DAB \equiv \angle CAB = \angle C'A'B'$ are equal, the triangles $\triangle ABD \cong \triangle A'B'C'$ are congruent by SAS. Hence, their remaining sides are also congruent, BD = B'C'. But since BC = B'C' as well, the triangles $\triangle BCD$ is isosceles with $\angle BCD = \angle BDC < 90^\circ$. Since AD = A'C' < AC, the angle $\angle BDA = 180^\circ - \angle BDC > 90^\circ$. Therefore, this is the largest angle of the triangle $\triangle ABD$, which means that the side AB opposite to this angle is the largest side of this triangle, AB > BD = BC. This contradicts the assumption that $AB \le BC$. Therefore, the isosceles triangle $\triangle BCD$ cannot exist, which means that the points $D \equiv C$ are identical and the triangles $\triangle ABC \cong \triangle A'B'C'$ are congruent.

If AB > BC, nothing prevents the existence of the isosceles triangle $\triangle BCD$ and it actually does exist. It is clear that the points C, D are the 2 intersections of a circle (B) with radius BC with the ray AC. One of the triangles $\triangle ABC, \triangle ABD$ has the angle $\angle BCA, \angle BDA$ obtuse and the other one acute. If we have 2 pairs of different triangles $\triangle ABC, \triangle ABD$ and $\triangle A'B'C', \triangle A'B'D'$ satisfying the conditions of the Ssa proposition except $BC \ge AB$, we can still deduce congruence of the triangles with the obtuse angles against AB, A'B' and congruence of the triangles with the acute angles against AB, A'B'.

[b]Problem solution:[/b] Since the lines $AB \parallel DC$ are parallel, the triangles $\triangle PAB \sim \triangle PDC$ are similar. Hence, $\frac{PC}{PB} = \frac{PD}{PA}$ and $PA \cdot PC = PB \cdot PD$. This is trivial. Let $O_1, O_2$ be the centers and $r_1 = \frac{AD}{2},\ r_2 = \frac{BC}{2}$ radii of the circles with the diameters AD, BC. The centers $O_1, O_2$ are the midpoints of the diameters AD, BC, hence the center line $O_1O_2 \parallel AB \parallel DC$ and

$\frac{AP}{BP} = \frac{DP}{CP} = \frac{O_1P}{O_2P} = \frac{AD}{BC} = \frac{r_1}{r_2} = \frac{O_1E}{O_2E} = \frac{O_1F}{O_2F}$

The points P, E, F have the same ratio of distances $\displaystyle \frac{r_1}{r_2}$ from the points $O_1, O_2$. The locus of points with this ratio of distances from the points $O_1, O_2$ is a circle (Z) centered on the line $O_1O_2$ and passing through the internal and external homothety centers $H_I, H_E$ of the circles $(O_1), (O_2)$. The points P, E, F all lie on this circle. The line $PH_I$ is the internal bisector of the angle $\angle O_1PO_2$, because it cuts the segment $O_1O_2$ in the ratio $\frac{O_1H_I}{O_2H_I} = \frac{PO_1}{PO_2}$. ($PH_E$ is the external bisector of this angle.) The angles $\angle EPH_I = \angle FPH_I$ spanning the equal arcs $EH_I = FH_I$ of the circle (Z) are equal. Consequently, the angles $\angle EPB = \angle FPA$ are equal and the angles $\angle EPA = \angle FPB$ are also equal. Let K, L be the other intersections of the rays PE, PF (other that E, F) with the circles $(O_1), (O_2)$, respectively. (We could use the other intersections of the rays PE, PF with the circles $(O_2), (O_1)$, respectively, for the same purpose.) Consider the triangle pairs $\triangle PO_1F, \triangle PO_1L$ and $\triangle PO_2E, \triangle PO_2K$. The angles $\angle O_1PF \equiv \angle O_1PL = \angle O_2PE \equiv \angle O_1PK$ are equal. The sides

$\frac{O_1P}{O_2P} = \frac{O_1F}{O_2E} = \frac{O_1L}{O_2K} = \frac{r_1}{r_2}$

or

$\frac{O_1P}{O_2P} = \frac{O_1F}{O_2K} = \frac{O_1L}{O_2E} = \frac{r_1}{r_2}$

are in the same ratio. Clearly, we can use the Ssa proposition without the condition for the congruent angles being against the larger sides. The triangles $\triangle O_1FL, \triangle O_2EK$ are the isosceles triangles that make it impossible to use the Ssa proposition without the side size condition for 2 triangles. But since we have both triangle pairs, we deduce that the triangles $\triangle PO_1L \sim \triangle PO_2E$ with the obtuse angles against $PO_1, PO_2$ are similar and the triangles $\triangle PO_1F \sim \triangle PO_2K$ with the acute angles against $PO_1, PO_2$ are also similar. Therefore, $\frac{PF}{PL} = \frac{PK}{PE} = \frac{r_1}{r_2}$ and $PE \cdot PF = PK \cdot PL$. From similarity of the triangles $\triangle PO_1L \sim \triangle PO_2E$, we know that the central angles $\angle DO_1L \equiv \angle PO_1L = \angle PO_2E \equiv \angle CO_2E$ of the major arcs DL, EC of the circles $(O_1), (O_2)$ are equal. Therefore, the angles $\angle DAL = \angle EKC$ spanning these arcs are equal. Thus the triangles $\triangle PKC \sim \triangle PLA$ are similar, having 2 internal agles equal: $\angle LPA \equiv \angle FPA = \angle EPC \equiv \angle KPC$ and $\angle PAL \equiv \angle DAL = \angle EKC \equiv \angle PKC$. Consequently, $\frac{PA}{PK} = \frac{PL}{PC}$ and $PA \cdot PC = PK \cdot PL = PE \cdot PF$.

Q.E.D.

I think that this problem does not belong to the pre-olympiad section.


function
limit
calculus
calculus computations

1. Given a function f defined on an interval I=(c-p,c+p). Suppose the function g is also defined on I and g(x)=f(x) except, possibly, at a finite set of points $x_{1}, x_{2}.....x_{n}$ in I.
    (a)Show that if $lim_{x\rightarrow c}$ f(x)=L, then $lim_{x\rightarrow c}$ g(x)=L.
    (b)Show that if  $lim_{x\rightarrow c}$f(x) does not exist, then $lim_{x\rightarrow c}$ does not exist.

2. Suppose f is a function with the following property: $lim_{x\rightarrow c}$[f(x)+g(x)] does not exist whenever $lim_{x\rightarrow c}$g(x) does not exist. Prove that $lim_{x\rightarrow c}$f(x)does exist.

3. Let f and g be continuous at c. Prove that if
(a) f(c)>0, then there exists $\delta$>0 such that f(x)>0 for all x$\in (c-\delta, c+\delta)$
(b) f(c<0, then there exists $\delta$>0 such that f(x)<0 for all x$\in (c-\delta, c+\delta)$
(c) f(c)<g(c), then there exists $\delta$>0 such that f(x)<g(x) for all x$\in (c-\delta, c+\delta)$

4. $lim_{x}\rightarrow\infty \sqrt{x^{3}}(\sqrt{x+2}-2\sqrt{x+1}+\sqrt{x})$

5. The function f(x) is defined by \[f(x)=\left\{\begin{array}{cc}x[\frac{1}{x}]+x[x],&\mbox{ if }x\ is\ not\ 0\\ 0, & \mbox{ if }x=0 \end{array}\right.\] where [x] denotes the largest interger not greater than x. Determine the points where f(x) is continuous. Determine whether a point of discontinuity is a removalbe, or a jump, or an infinite discontinuity.

Please, help me. I am a new comer of this forum and also a new learner of Calculus. Thank you very much.

[b]Solution to (4)[/b]: We make the substition $y = \frac1{x}$, where $y > 0.$

Then, $\lim_{x \rightarrow\infty}\sqrt{x^{3}}(\sqrt{x+2}-2\sqrt{x+1}+\sqrt{x})$

$= \lim_{y \to 0}\frac{\sqrt{1+2y}-2\sqrt{1+y}+1}{y^{2}}$ ... $(0/0$ form)

$= \lim_{y \to 0}\frac{\frac{2}{2\sqrt{1+2y}}-\frac{2}{2\sqrt{1+y}}}{2y}$ ... (using L'Hospital's rule)

$= \lim_{y \to 0}\frac{\sqrt{1+y}-\sqrt{1+2y}}{y}\cdot \frac{1}{2(\sqrt{1+2y}\sqrt{1+y})}$

$= \frac12 \cdot \lim_{y \to 0}\frac{(1+y)-(1+2y)}{y(\sqrt{1+y}+\sqrt{1+2y})}$ ... (rationalizing the numerator)

$= \frac12 \cdot \lim_{y \to 0}\frac{-1}{\sqrt{1+y}+\sqrt{1+2y}}$

$= \boxed{-\frac14}.$

-------------------------------
[b]Solution to (3)[/b]:
(a) We are given that $f(c) > 0$ and $f$ is continuous at $c,$ which means that for any $\epsilon > 0,$ there exists a $\delta > 0$ such that $|x-c| < \delta$ implies $|f(x)-f(c)| < \epsilon .$

Let us assume, for the sake of contradiction, that the converse of the given statement is true, i.e. for all $\delta > 0,$ there exists some  $x \in (c-\delta, c+\delta)$ such that $f(x) < 0.$ This means that for any $\epsilon > 0,$ there exist a $\delta_{n}$ such that $|x-c| < \delta_{n}$ implies $|f(x)-f(c)| < \epsilon$ and also a point $x_{n}\in (c-\delta_{n}, c+\delta_{n})$ such that $f(x_{n}) < 0.$ Now, if we consider a sequence of nested closed intervals $[c-\delta_{n}, c+\delta_{n}]$ and take their intersection, then it will contain one and only one point $c_{1}$ (say) such that $f(c_{1}) < 0.$ But, then we must have $c=c_{1},$ which implies $f(c) < 0,$ a contradiction!

And, we are done.

(3a) and (3b) can be done similarly, I guess.

I would be glad if someone would point out any error in the above argument of mine.

5.

When $0<x<1$ the function will not be continuos since there is not division by 0, so we will have a big gap in this segment.
now i will analyse the limits when it approachs a natural number by the left and by the right to see if there is continuity.

$\lim_{x\to+n}F(x)=n/n+n^{2}=n^{2}+1$
in the other case:
$\lim_{x\to-n}F(x)=n/(n-1)+n^{2}-n =n^{2}-n+n/(n-1)$
so for not so large numbers, the functions has a discontinuity when it approaches the natural numbers but is continuous for $\lim_{x\to \infty}$ what can be easily checked analysing the quotient between the two limits. In the negative part happens the same thing....

[b]Solution to (1)[/b]:
[b](a)[/b] Here $I = (c-p, c+p).$ We have $\lim_{x \to c}f(x) = L$ iff for any $\epsilon > 0, \exists\ \delta_{1}$ such that $|x-c| < \delta_{1}$ implies $|f(x)-L| < \epsilon$ for some $\delta_{1}< p.$

Now $f(x) = g(x)$ except possibly at a finite set of points $x_{1}, x_{2}, \ldots , x_{n}$ in $I.$ Let $\delta_{2}= \min (|x_{1}-c|, |x_{2}-c|, \ldots, |x_{n}-c|).$ Then, we must have $f(x) = g(x)$ for all $x$ such that $|x-c| < \delta_{2}.$

Choose $\delta = \min (\delta_{1},\ \delta_{2}).$ Then for any $\epsilon > 0,\ |x-c| < \delta$ implies $|g(x)-L| = |g(x)-f(x)+f(x)-L| \le |g(x)-f(x)|+|f(x)-L| < 0+\epsilon = \epsilon.$

Hence, $\lim_{x \to c}g(x) = L.$ And, we are done.

[b](b) [/b]This is just a consequence of part [b](a)[/b] above. Let us assume, for the sake of contradiction, that $\lim_{x \to c}f(x)$ does not exist and that $\lim_{x \to c}g(x)$ exists. Then the latter implies (using part (a) above) that $\lim_{x \to c}f(x)$ also exists, which is clearly a contradiction! Hence, if $\lim_{x \to c}f(x)$ does not exist, then $\lim_{x \to c}g(x)$ also doesn't exist.

Can anybody solve (2)?
Thanks for your helping hands.

2 is quite easy, break your head with this problem dude!


topology
invariant
function
advanced fields
advanced fields unsolved

this problem i find really hard to tackle:
prove that $(X,d)$ is a compact metric space iff $(X,\varrho)$ is complete, for every $\varrho$ equivalent to $d$.
i would be greatful for any hints, but please don't post the solution.

You can use the 3 following little propositions to prove the statement.

1) $X$ is a compact metric space iff each sequence in $X$ has convergent subsequence.

2) If a Cauchy sequence $(x_n)$ in metric space has a convergent sunsequence then $(x_n)$ converges as well.

3) For equivalent metrics $d$ and $d'$ metric spaces $(X,d)$ and $(X,d')$ have the same Cauchy sequences.

thank you for these hints. i managed to prove the implication from compactness to completness, but i'm starting to believe, that the opposite is false...
if we take two equivalent metrics $d$ and $d'$, metric spaces $(X,d)$ and $(X,d')$ are both complete or both incomplete. hence it suffices to prove the statement for one equivalent metric to $d$. but if we take $d'=d$ the thesis is not correct, $\mathbb{R}$ with natural metric is a simple counterexample...

Completeness is not invariant to homeomorphism. In other words, Vladimir's third "little proposition" is false. Equivalent metrics are just required to induce the same topology; their Cauchy sequences need not be the same.

Consider for instance the usual metric on $\mathbb R$. It's a complete metric. Now consider the homeomorphism $\phi$ from $\mathbb R$ to $(-1,1)$ given by $\phi(x)=\frac 2\pi\arctan x$. The metric $d(x,y)=|\phi(x)-\phi(y)|$ will turn $\mathbb R$ into an incomplete metric space, but $d$ and the usual metric on $\mathbb R$ are equivalent (they both induce the usual topology on $\mathbb R$).

Yes, grobber is right. I had to put

3') Let $d$ and $d'$ be two equivalent metrics on $X$. If $(X,d)$ and $(X,d')$ are both complete then they have the same Cauchy sequences.

instead of 3).

Our metric space $(X,d)$ is compact hence so is $(X,d')$ for $d'$ equivalent to $d$. It remained only to note that each compact metric space is complete.

ok, thanks again. but these hints are only proving the "$\Rightarrow$" implication, or am i missing something? if not, i would be thankful for any more help.

Your problem is not as easy as I thought at the beginning. Frankly, I have difficulties proving (or disproving) the inverse statement. The problem seems to me very interesting.

Ok, so the problem of interest is this:

Let $(X,d)$ be a non-compact metric space. Then we can remetrize $X$ so that it becomes incomplete.

First, notice that we can assume $d$ to be unbounded. In order to see this, proceed as follows. Let $(x_{n})_{n\ge 1}$ be a sequence in $X$ which has no convergent subsequence (it exists, for sequential compactness is equivalent to compactness in metric spaces). Define $f: \{x_{n},\ n\ge 1\}\to\mathbb R$ by $f(x_{n})=n$. $f$ is a continuous function from the closed subset $\{x_{n}\}$ of $X$ into $\mathbb R$. Any continuous function from a closed subset of a metric space into $\mathbb R$ can be extended to a continuous real-valued function on the whole space (I believe this has even been discussed on the forum), so extend $f$ in such a manner, and call the extension $f$ as well. $f: X\to\mathbb R$ is now continuous and unbounded. Finally, replace $d$ by the metric $\tilde d$ defined by $\tilde d(x,y)=d(x,y)+|f(x)-f(y)|$. Checking that $\tilde d$ is an unbounded metric equivalent to $d$ is straightforward. In order to keep the notation simple, revert to $d$ and simply assume $d$ to be unbounded.

My aim is to add a "point at infinity" $p$ to $X$ and to make a metric space out of the new set $\widetilde X=X\cup\{p\}$ in such a way that the relative topology of $X$ coincides with the old one, and $p$ lies in the closure of $X$ in $\widetilde X$. The metric induced on $X$ by that of $\widetilde X$ will then be incomplete: simply consider a sequence $(x_{n})_{n}\subset X$ which converges to $p$ in $\widetilde X$. This will be a Cauchy sequence in the new metric, but it will not be convergent in $X$.

The construction of the topological space $\widetilde X$ immitates that of the Alexandrov compactification. We define the open susbets of $\widetilde X$ to be those sets which are either (a) open subsets of $X$, or (b) complements in $\widetilde X$ of bounded closed subsets of $X$ (a subset $F$ of $X$ is said to be bounded if there is a point $x\in X$ and a positive real $r$ such that $F$ is contained in the ball of radius $r$ centered at $x$, all distances being measured with respect to $d$).

That the above is indeed a (Hausdorff) topology on $\widetilde X$ such that the relative topology it induces on $X$ coincides with the old one should be more or less clear. Another thing that needs checking is that $p$ is not an isolated point. Indeed, every neighborhood of $p$ contains a set of the form $X\setminus F$, where $F$ is a bounded closed subset of $X$. Since $d$ was assumed to be unbounded, $X\setminus F$ is non-empty, so every neighborhood of $p$ in $\widetilde X$ contains points of $X$. Finally, all that's left in order to finish the proof as outlined two paragraphs above is to check that this topology on $\widetilde X$ is metrizable. I'll do this by employing the Nagata-Smirnov Metrization Theorem: a Hausdorff topological space is metrizable iff it's regular and has a basis that is countably locally finite. The regularity should be clear: we already know that $X$ is regular, so it suffices to show that for each closed subset of $\widetilde X$ which does not contain $p$ we can find disjoint open sets $U,V$ around $F$ and $p$ respectively. $F$ is bounded, so just take $U$ to be the set of points having distance $<1$ from $F$ and $V$ to be the complement of the closure of $U$. As for the other condition, let $\mathcal B$ be basis for the topology of $X$ which is countably locally finite. Next, fix a point $x\in X$, and let $V_{n},\ n\ge 1$ be the complement in $\widetilde X$ of the set of points having distance $\le n$ from $x$. The collection $\mathcal B\cup\{V_{n},\ n\ge 1\}$ is a countably locally finite basis for the topology of $\widetilde X$.


please provide the mechanism for  :maybe:  :maybe:

This is just a cationic rearrangement. How about first a mechanistic proposal from yours?

Now the rest of the mechanism should be obvious. Take into account the structure of the final product.


Find all natural number $ x,y,z$ satisfying $ x!\equal{}y!z!$

we cannot find all of them because non countble  but ı want to write an isteresting one
10!=6!.7!


[quote="iamagenius"]why is the following true:

$\frac{\text{amount increase}}{\text{initial amount}}=\text{percent increase}$?[/quote]That's how percent increase is defined, no? Say percent increase a few times, then check what does the word 'percent' means, then check what the word 'increase' means and you will eventually understand.

Well, technically, $\text{percent increase}= 100(\frac{\text{amount increase}}{\text{initial amount}})$
But that's just how it's defined; a percent increase is just seeing how many parts of 100 the amount increases, hence we have:
$\frac{\text{percent increase}}{100}=\frac{\text{amount increase}}{\text{initial amount}}$


What is the least positive integer with exactly six factors.

[hide]using guess and check i get 12[/hide]

[hide] $ 3 \cdot 2 $ $\rightarrow$ $2^2 \cdot 3 = \boxed{12}$[/hide]


The world's largest ball of string is about 2 m in radius. To the nearest order of magnitude, what is the total length L of the string in the ball?

The Solution: Let us assume the ball is spherical with Radiues $ R \equal{} 2 m$. The string in the ball is not closely packed (there are uncountable gaps between nearby sections of string). To allow for these gaps, let us somewhat overestimate the cross-sectional area of the string by assuming the cross section is square, with an edge length $ d \equal{} 4 mm$. Then, with a cross-sectional area of $ d^2$ and a length L, the string occupies a total volume of $ V \equal{} (cross \minus{} sectional area)(length) \equal{} d^2*L$. This is approximately equal to the volume of the ball, given by $ \frac {4}{3}pi*R^3$, which is about $ 4*R^3$ because $ pi$ is about 3. Thus we have $ d^2*L \equal{} 4*R^3 \rightarrow L \equal{} 2 * 10^6 m \equal{} 10^3 km$

Question: How can they assume that the cross-section of the string is a square with edge length 4 mm? I don't get how you get 4 mm from?

Is there a picture?  If not, try to visualize string, or yarn.  Yarn definitely isn't more than .5cm across, and that's probably a really big overestimate.  I would agree and say that 4 mm is probably a fairly conservative estimate.

So you can assume facts like that in physics?

No you cannot. You might have missed the information in the question about the material.

Hmmmm......

But I copied down the exact same problem and solution from the book! :huh:

Well, it's more of a Fermi question than a physics problem. That's why the estimates aren't a problem, they have to be made to show logical and practical reasoning.

A quick, rough estimate of some extremely large quantity that is essentially impossible to measure.  Examples might be:
-How many drops of water are in Lake Erie?
-How many grains of sand are on the beaches of the east coast of the US?

Fermi questions are quite popular during interviews :)

One of my favourite fermi questions is find the rate at which grass grows :P

Like how many grams of tea powder is used in a particular city per day :rotfl:

Yes, that would be an example.  It's similar to one of Fermi's original questions, which was "How many piano tuners are there in Chicago?" or something along those lines.

[quote="JRav"]I wouldn't say that's much of a Fermi question-or not a particularly difficult one anyway.  It can't grow more than a few inches over the course of a few weeks so I could probably get the order of magnitude within 1.  Fermi questions are typically interested in order of magnitude, not an exact number.[/quote]

not if you're looking for m/s or something of that sort

Fermi questions tend not to concern themselves with actual units of measurements.  Like "drop" of water has a pretty clear image to most people.

And besides, if I had a reasonable estimate of the rate that grass grows in cm/s or inches/year, I could just do a unit conversion, which is nothing more than multiplication or division.  It doesn't affect anything.


limit
group theory
abstract algebra
real analysis
real analysis unsolved

Given $\lambda_{n},\epsilon_{n}$ two sequences of real numbers such that $\lim \lambda_{n}=+\infty$, show that the measure of the set of the real numbers $x$ such that $sin(\lambda_{n}x+\epsilon_{n})$ converges has measure $0$.

Does it solve the problem completely ? (i.e, can we really assume that WLG the sequence converges to $0$, considering that its limits can take uncountably many values ?)

In the case $\epsilon_{n}$=0, I think (even though I didn't check it rigorously) that we can prove it using the fact that the required measure is less than the measure of the set of those $x$ such that a subsequence of $(sin(\lambda_{n}x))$ converges. But I believe this set is a measurable subgroup of $\mathbb{R}$, and hence has measure $0$. Not sure I can use this method (if correct) in the general case.


topology
function
advanced fields
advanced fields unsolved

Let $f: C\to C$ be an identity map(Is it the right word? I mean $f(x)=x$), where $C$ is the unit circle on the plane.
Extend $f$ to a continuous map from $\mathbb{R}^2\setminus\{0\}$ to $C$. Can we expect the extension of $f$ on the whole plane?

To answer the secon question, observe that an extension of the function to one from the entire plane to the circle would make it a retract on the circle. In general, one can prove that a retraction induces an epimorphism between fundamental groups, and this would mean that we have an epimorphism from the fundamental group of the plane, which is trivial, to the fundamental group of the circle, which is $\mathbb Z$. Of course, this is absurd.


topology
advanced fields
advanced fields unsolved

$X$ is a path-connected space. $p, q$ are two points in $X$.
$\alpha$ and $\beta$ are two paths from $p$ to $q$.
If $\alpha^{-1}\cdot\gamma\cdot\alpha$ and $\beta^{-1}\cdot\gamma\cdot\beta$ are homotopic for loop $\gamma$ with basepoint $p$, are $\alpha$ and $\beta$ homotopic?
(where $\alpha\cdot\beta$ means the joint of path $\alpha$ and $\beta$)

Of course not. If $\gamma$ is nontrivial, let $\alpha$ be the null loop and $\beta$ be $\gamma$. If $\gamma$ is trivial and the space is not simply connected, let $\alpha$ be the null loop and $\beta$ be any nontrivial loop.

The centralizer of any single element in a nontrivial group is nontrivial.


superior algebra
superior algebra unsolved

Is there any bound on the number of generators of a monomial ideal in $ \mathbb{C}[x,y]$?
Prove your answer!

There is no bound on the number of generators of a monomial ideal in $ \mathbb{C}[x,y]$.

Suppose $ k \ge$ # of generators of any monomial ideal in $ \mathbb{C}[x,y]$.
$ <x^k,x^{k\minus{}1}y,.......,xy^{k\minus{}1},y^k>$ is a monomial ideal in $ \mathbb{C}[x,y]$ with $ k\plus{}1$ generators.
Cotradiction!!


function
vector
limit
calculus
derivative
real analysis
real analysis unsolved

$ f$ is differentiable on $ (a,\infty)$.

$ \frac{f(x)}{x}\rightarrow A$ as $ x\rightarrow \infty$.

I am trying to prove that there exists a sequence $ \{x_n\}, x_n\rightarrow \infty$, such that $ f'(x_n)\rightarrow A$.

Any help would be appreciated.

Use the mean value theorem applied to the interval $ [b,c]$ with fixed $ b$ and $ c$ tending to $ \plus{}\infty$.

First thing to note: WLOG, $ A \equal{} 0.$ If not, replace $ f(x)$ by $ f(x) \minus{} Ax.$

Next: this would be false if you allowed $ f(x)$ to be complex valued (or vector valued). Example: let $ f(x) \equal{} e^{ix}.$ We do have $ \lim_{x\to\infty}\frac {e^{ix}}x \equal{} 0,$ but $ |f'(x)| \equal{} 1$ for all $ x,$ so no sequence of derivatives tends to zero.

So now assume $ f$ is real valued and that $ A \equal{} 0.$

Suppose the conclusion is false. That is, there is no sequence $ x_n$ such that $ f'(x_n)\to0.$ That means that $ f'(x)$ is bounded away from zero for sufficiently large $ x.$ But a derivative satisfies the intermediate value (Darboux) property; hence either $ f'(x)$ is eventually positive or eventually negative.

WLOG, suppose $ \exists\,\delta > 0$ and $ M$ such that $ f'(x)\ge\delta$ for $ x\ge M.$ But then an application of the Mean Value Theorem tells us that for $ x > M,\ f(x)\ge f(M) \plus{} \delta(x \minus{} M).$

That in turn implies that $ \frac {f(x)}{x}\ge \frac {f(M) \minus{} M}{x} \plus{} \delta.$

If that is true, then $ \lim_{x\to\infty}\frac {f(x)}{x}\ge \delta > 0$ if it exists. This contradicts our hypothesis.


find all answer of $2^{x}-5=11^{y}$ in $\mathbb{N}$

Obviously,we have $n \geq 3$ and easy to show that $x$ is even and $y$ is odd.
Now use $mod 7$ and $mod 11$  ,easy to show that $x=60u+4,y=12t+1$ ($u,t $are nonnegative integer)
Use the quadric residue easy to show $2^{60} \equiv 1(mod 121)$ because $121=8*15+1$
So $2^{60u+4}-5 \equiv 11 (mod 121)$ hence y=1,x=4 is the only the solution of your problem :D  :D


limit
real analysis
real analysis unsolved

Show that $ \lim_{n \to \infty} (1 \plus{} \frac {x}{n} )^x \equal{} e^x, \ \ \forall x > 0$

Maybe $ \lim_{n \to \infty} \left(1 \plus{} \frac {x}{n}\right)^n \equal{} e^x\ ,\ \left(\forall\right)\ x > 0$ .

[quote="§outh§tar"]What does e mean?  :?:[/quote]

If $ e \equal{} \lim_{n \to \infty} \left(1\plus{}\frac1n\right)^n$,
then $ e^x \equal{} \lim_{n \to \infty} \left(1\plus{}\frac1n\right)^{nx} \equal{} \lim_{n \to \infty} \left(1\plus{}\frac{x}{nx}\right)^{nx} \equal{} \lim_{n \to \infty} \left(1\plus{}\frac{x}{n}\right)^{n}$

[quote="PeterP"][quote="§outh§tar"]What does e mean?  :?:[/quote]

If $ e \equal{} \lim_{n \to \infty} \left(1 \plus{} \frac1n\right)^n$,
then $ e^x \equal{} \lim_{n \to \infty} \left(1 \plus{} \frac1n\right)^{nx} \equal{} \lim_{n \to \infty} \left(1 \plus{} \frac {x}{nx}\right)^{nx} \equal{} \lim_{n \to \infty} \left(1 \plus{} \frac {x}{n}\right)^{n}$[/quote]

There's the dirty work of justifying that last step


When a dice is thrown four times, let $ x_1,\ x_2,\ x_3,\ x_4$ be the casts in this order.
For $ P(x_1,\ x_2), O(0,\ 0),\ Q(x_3, \minus{}x_4),$ find the probability such that $ \angle{POQ}$ is acute.

[hide="Answer"]$ 605/1296$, though I'm much more confident in my method than this number.[/hide]

[hide="Solution"]Let $ \angle POQ \equal{} \theta$. If the angle is acute, then $ \cos \theta > 0$. Thus the angle being acute is equivalent to $ 0 < \vec{OP} \cdot \vec{OQ}$. Since $ \vec{OP} \equal{} \langle x_1, x_2 \rangle, \vec{OQ} \equal{} \langle x_3, \minus{}x_4 \rangle$, we have $ \vec{OP} \cdot \vec{OQ} \equal{} x_1x_3 \minus{} x_2x_4$. So we require $ x_1x_3 \minus{} x_2x_4 > 0$, or $ x_1x_3 > x_2x_4$.

Let $ A$ be the number of 4-tuples $ (x_1,x_2,x_3,x_4)$ such that $ x_1x_3 > x_2x_4$, $ B$ the number of 4-tuples such that $ x_1x_3 \equal{} x_2x_4$ and $ C$ the number of 4-tuples such that $ x_1x_3 < x_2x_4$. We want $ A$. By flipping $ x_1$ and $ x_2$, $ x_3$ and $ x_4$, we observe that there is a one-to-one correspondence between the 4-tuples in which $ x_1x_3 > x_2x_4$ and the 4-tuples in which $ x_1x_3 < x_2x_4$. So $ A \equal{} C$. Furthermore, the total number of 4-tuples is $ 6^4$, so $ A \plus{} B \plus{} C \equal{} 2A \plus{} B \equal{} 6^4$. So we can get $ A$ if we solve for $ B$. Thus we need to find the number of 4-tuples such that $ x_1x_3 \equal{} x_2x_4$.

Unfortunately, I don't see an easy way to do this, so I'm going to list how many ways we can obtain a positive integer $ n$ as a product of two not necessarily distinct elements of $ \{1,2,3,4,5,6\}$, where order matters.
$ n$: number of ways
1: 1
2: 2
3: 2
4: 3
5: 2
6: 4
8: 2
9: 1
10: 2
12: 4
15: 2
16: 1
18: 2
20: 2
24: 2
25: 1
30: 2
36: 1

So we want the sums of the squares of all of those. Unless my arithmetic sucks that's going to be $ 86$. So $ B \equal{} 86$. So $ A \equal{} (1296\minus{}86)/2 \equal{} 605$. Then the requested probability is $ 605/1296$.[/hide]


ARML HMMT calculus AMC AIME geometry email

Hey, y'all. Just wanted to know if I'd meet any of ya at ARML this year! I'm going to Penn State.

[b]For Anyone who has voted, please post! We are trying to set up an AoPS gathering at PSU :D. You don't have to leave your name or anything, but maybe we can start a PM chain. [/b]

Penn State for me. Yay!

By the way, someone should move this to the other contests & programs section.

It should be okay. If you guys start a major discussion about ARML, I'll move it.

lol. Actually, moving it may not be a bad idea...I put this up in the wrong post. Not moving it is fine too though.

Anyway, Fierytycoon, I'm sure I saw you at HMMT. I must have at least caught a glimpse of you, because I sat next to the Albany team during HMMT award ceremonies. Too bad I wasn't signed up to AoPS before then...

Anyway, if you can remember me, I was that tall asian guy on Radnor-Marple. Dunno if that rings a bell.

[quote="masamune2387"]
Anyway, Fierytycoon, I'm sure I saw you at HMMT. I must have at least caught a glimpse of you, because I sat next to the Albany team during HMMT award ceremonies. Too bad I wasn't signed up to AoPS before then...

Anyway, if you can remember me, I was that tall asian guy on Radnor-Marple. Dunno if that rings a bell.[/quote]

..except I wasn't on the Albany team. ;) Hmm..I don't know whether I saw you or not..."tall asian guy" fits the profile of about half the students there, lol.

Hmm...I just checked the results again. Were you the guy from Radnor-Marple who placed in top 10 on the Algebra test?

Lol, no that wasn't me. Unfortunately, I screwed up the algebra exam pretty badly (lol i think i got a 16), and came a few points short of placing in the Calculus (a twenty some). What a terrible day...must have been the stress of meeting my long-dist gf for the first time in half a year.

The kid you saw on the list was Huaou, and you should meet him at ARML.

And placing in the top 10...well. Huaou's pretty weird sometimes (and lucky...you can probably sense some jealousy here. Gah, my streak of wins against him is broken). Too bad he butchered up his AMC 12 (105) and AIME (6 or lower)...he's pretty smart though.

Lol you weren't on the Albany team? Well, Alison Miller was, but I dunno anyone else. (lol and why did I think you were on Albany? it clearly says "Massachusetts" next to your name)

What team were you on?

[quote]the stress of meeting my long-dist gf for the first time in half a year. [/quote]

Yep, something like that would do it, heh.

Yeah, I barely missed placing in Algebra (by 2 points) and missed placing in Geometry in a few points. It was weird though how Tiankai got 47's on both the exams he took; I mean, he missed #1 or #2 on both of them???

Oh, and see your PM box.

reminder, this is NOT a private messenger... :-)

[quote="masamune2387"]Lol. Maybe Tiankai worked backwards.

That is quite amusing though.[/quote]

Mm...that's acutally not a bad plan, since the problems are weighted. #1 and #2 are worth nothing in the entire scheme of things anyways - t'would be much better to spend more time and get #10 even if you have to miss #1 and #2.

yea...i worked backwards at Rice MT this yr...it worked fairly well...

Florida doesn't have an ARML team. anyone know why some states don't have them?

It just depends on ur state. I'm pretty sure you could go as a Florida team if you could find a list of all the best pplz in Florida and make a team out of them. I think some regions/schools/districts have their own teams.

according to syntax, oklahoma doesnt have a team either.

Well, I suppose a PM chain would work like this:

I'd be responsible for PMing fierytycoon. Then fierytycoon passes on to some other person...until the chain gets back to me. Or we could PM each other our AIM sn's and start a chat room somewhere.

[quote]WarpedKlown, that sounds like a fine idea, except that I'm not sure how competitive the northern Florida team would be.[/quote]

North florida would probably end up being me shaanan keenan and sean. maybe some other people.

for anyone going to the iowa arml, ill be there.. ill be on the indian team

[quote="JBL"]Football coaches get paid when their teams go to the state championships.  It's not unreasonable to suppose that somewhere, teachers are getting paid to spend their time on an academic endeavor.  But yes, you're right, one would expect the coaches share of expenses to be covered.[/quote]

um.. I highly highly doubt they are getting paid.

so I am just going to say what it costs for Baltimore to send a team.... it costs us 1500 to get the bus (remember that we are A LOT closer to Penn State than Mass),  1000 dollars to house and feed the kids (I think 60 dollars a person), 200 dollars to ARML for sending a team, around 200 dollars for food on Friday, which amounts to nearly 3000 dollars.  (Not to mention the ~200 dollars of t-shirts that CTY usually defray for us) Now with your 200 dollars a person.. that's not even enough to defray the operation costs, let alone pay the coaches.... (We have JHU legal counsel to clear our insurance stuff, and NYC has the NY board of Ed.. I don't know how states would handle their liabilitiy issues)

To run this kind of stuff, one rarely ever goes back to black, hell the cost of running baltimore's math team has so far put me... (~300 in the red) I think.

I know the head coach of NYC Mr. Linker routinely goes several hundred into the red as well.

The baltimore team trains every week (though ... a lot of kids usually don't show up) and we have an operating budget of around 6k (contributed by the Howard Hughes Medical Institute, Dean's office,  Dept of Mathematics and Dept of Applied Mathematics & Statistics) and I STILL go into the red...  It's a shame that we have the \[ but no kids.. but various other places have kids and no \].  Perhaps setsofallsets (if he ever checks his email or this board) and I can combine the two fragments and send one team ;)

It's a shame that academic events aren't nearly as compensated for as athletic events.

[quote="zabelman"][quote]we're staying 2 nights at U of Iowa and we get a T-shirt that says "Texas ARML '04" on it( so that we can show off to ppl who didnt make it  ) and that's bout it...[/quote]

Ya, but the shirts also have an armadillo on them... they're pretty sweet.[/quote]

oh yea...the texas armadillo...sweet...

uh...regarding ur question about iowa sirernest, TEXAS will be there :-D ...just look for 45 kids wearing armadillo T-shirts :-D

[quote]Who are the ppl going to PennState?[/quote]

I'll hopefully be there. If a PM chain has been started on this, can I be added to it? I'd love to meet some of you.

Iowa, anybody?

Hey does anybody know where I can get past ARML problems? obviously got 99 and 00 off the official site, but I'm looking specifically for the 01 problems

I might be able to tell you the problems from memory if you tell me which ones you need (i.e. I might be able to give you most of the individual ones).

yeah, the individual problems would be great. I sware they're not on the internet anywhere.

7. Find the number of ordered pairs of positive numbers of solutions [tex]\displaystyle{(x,y)}[/tex] of solutions to the equations
[tex]\displaystyle{\lfloor x\rfloor+y=5.7, \; x+\lfloor y\rfloor=5.3.}[/tex]

8. Let [tex]\displaystyle{f(x)=(1+x)(1+x^4)(1+x^{16})\cdots}[/tex]. Compute [tex]\displaystyle{\begin{equation*} f^{-1}\Biggl(\frac{8}{5f\big(\frac{3}{8}\big)}\Biggr).\tag{}\end{equation*}}[/tex]

I'll try to get the others in workable form if my memory feels like working.


find the best contanst $a$ such that the inequality:
                          $a(x^2+y^2+z^2)+(2-a)(xy+yz+zx) \geq \sqrt[3]{4}(\sum_{cyclic} \sqrt[3] {x^3y^3+x^3z^3})$ holds for all $x,y,z >0$  :)  :)

[quote="arqady"]$a\leq2,$ since for $a=2$ this is true.[/quote]
I think $a \ge 2$.

[quote="Lovasz"]
I think $a \ge 2$.[/quote]
I think the best contanst $a\leq2.$ ;)

I don't know what happened to variable $a$. Forget it.

[b]Problem:[/b]
Find the best constant k such that the following inequality holds for all positive real numbers $a,b,c$:
\[ \sum_{cyc}x\sqrt[3]{4(y^3+z^3)} \le k(x^2+y^2+z^2)+(2-k)(xy+yz+zx).  \]

[b]Solution:[/b]
We prove it for $k=2$:
\[ \sum_{cyc}x\sqrt[3]{4(y^3+z^3)} \le 2(x^2+y^2+z^2).  \]
For positive reals $y,z$, we have:
\begin{eqnarray*} \sqrt[3]{4(y^3+z^3)} &\le& 2\cdot\frac{y^2+z^2}{y+z} \\ \iff (y-z)^4(y^2+yz+z^2) &\ge& 0 \\ \implies x\sqrt[3]{4(y^3+z^3)} &\le& 2\frac{x}{y+z}(y^2+z^2). \end{eqnarray*}
Summing three inequalities, we must prove:
\begin{eqnarray*} 2(x^2+y^2+z^2) &\ge& 2\sum_{cyc}\frac{x}{y+z}(y^2+z^2) \\ \iff \sum_{cyc}(y^2+z^2)(1-2\frac{x}{y+z}) &\ge& 0. \end{eqnarray*}
And:
\begin{eqnarray*} LHS \ge \sum_{cyc}(y+z)^2(1-2\frac{x}{y+z}) &=& \sum_{cyc}(y+z)^2-\sum_{cyc}2x(y+z) \\ &=& 2(x^2+y^2+z^2)-2(xy+yz+zx) \\ &=& \sum_{cyc}(y-z)^2 \\ &\ge& 0. \end{eqnarray*}
Done.

[quote="Lovasz"]
For positive reals $y,z$, we have:
\[ \begin{eqnarray*} \sqrt[3]{4(y^3+z^3)} &\le& 2\cdot\frac{y^2+z^2}{y+z} \\ \iff (y-z)^4(y^2+yz+z^2) &\ge& 0  \]
[/quote]
Yes! This is a Mildorf's inequality.

[quote="Lovasz"]Let $a \to 0$ and $b=c$, that implies $2a+2-a \ge 4$, means $a \ge 2$, arqady?!?[/quote]
I think you are wrong.
Let $a \to 0$ and $b=c$, that implies $2a+2-a \ge 2\sqrt[3]4.$ ;)
Hence, $2\sqrt[3]4-2\leq a\leq2$ for the best $a.$


Hi,

I want to take the SAT II Physics in December. We will not cover all the areas in the SAT Physics in our AP class by December. Is there any good review book for SAT II? I have Cracking the AP Physics Exam from Princeton Review. If I go over that book, will it be enough for the SAT II?

Thanks.

The Barron's review book got me a 780 with my honors physics class covering barely half of the material.

i got an 800 in physics with the barron's review book and without covering anything about optics in my (AP) class.... there was alot of optics questions on that test!

Why are you taking SAT Physics in December?  :huh:  Assuming you are a junior, I would say to wait until June...but if you're a senior, I think you should get Barron's, since many people claim that Barron's overworks you. (I actually didn't use a review book but still got an 800  :lol: )

Good luck.

Hi,

I took the Physics SAT II at the end of junior year... and all I had taken was mechanics, which is probably only about 25% of the exam if that... I don't even remember what book I bought to help me study. It doesn't matter. I thought I did awful when I came out of the exam, then I learned that the curve is RIDICULOUS. I agree that it is better to wait if you have the opportunity till the end of the year if possible since teaching yourself is annoying... but go for it if you want to.

I took the physics sat II before ap physics even got underway.  It's not that tough, really.  There's this site called hyperphysics that tells you everything you need to know.  Because there's so many problems, you can be sure that they'll all be quite trivial(ie. one-step, plug+chug)

It depends purely upon how well one understands conceptual physics.

I studied like hell my freshman year of high school, but obtained a laughable 700.

My junior year, I got an 800 after finishing the test in thirty minutes.

I would actually recommend a textbook with a textual, conceptual discussion of physics, followed by an actual prep book with silly, simple formulas.

And I think the Princeton Review is more than enough; that's the only book that myself or anyone I know that scored an 800 on the exam ever used.

The Barrons book sucks and is riddled with errors.


Do you think it would help when one listens to music during problem solving?
Does it make more creative?

I think it's disturbing and keeps your mind out of focus. Enjoying music before to put yourself in a relaxed mood is another story ...

no. It breaks your concentration, but if there aer very obnoxious noises, ok.

daaa, frontlaiiiii!!! hooaa, huestas vra devraw, huuhaa! Daas tleee....

I luv asian music when I'm workin'!

I don't really listen to music when I do math, or schoolwork, etc, but I do find it a relaxing accompinant while lying in my bed staring up at the ceiling when I'm bored.

listen to the radio

or some of my music on the computer (indian film music, rock and hip-hop)

Let me see...

Artists:

Achillea
Alcyone
Alquimia
Alquimia & Roedelius
Althea W
Althea W.
Andrew York
Bebel Gilberto
Bill Mize
Billy McLaughlin-Dirk Freymuth
Bobby Cochran
Dan Schwartz
David & Steve Gordon
David Gordon
David Gordon & Steve Gordon
DeLago
Don Ross
Drum N' Space
Ed Gerhard
Eric Johnson
Eric Tingstad
Gary Stadler & Stephannie
Ginkgo Garden
Ginko Garden
Gleisberg
Hands Upon Black Earth
Jaya Lakshmi
Joe Weineck
Joel Andrews
Karunesh
Lawrence Juber-Preston Reed
Leo Kottke
Marcator
The Moontrain Conductors
Nasser Kilada
One At Last
Opera to Relax
Oscar Lopez
Pat Kirtley
Peter Mergener
Peter Roller
Pia
Potosch Potschka
Potsch Potschka
Quango (Thievery Corporation)
Quanjo (Black Star Liner)
Quanjo (Christopher Goze)
Quanjo (Nitin Sawhney)
Quanjo (Stoppa & Nobby)
Roedelius
Roykey
Rudy K
Stella Maris
Stephn Bennett
Steve Gordon
Svensson
TAU
Toires
Tya
Youssou N'Dour
Zingaia

I think that's all of them.

Listening to music only helps me when my little brother is watching Barney in the next room...

Usually the radio.

[quote="alex__ib"]classic music is good, but  that is all!!!....Music disturbes you if you are doing problems,period.[/quote]

BULLSHEEEEETTTTTTTT!!!!!!!!!!

BLLLARGGGGGHHHHHHHH - BBBBBBBLLLARRRRRRRRRGGGGGGG BLAR!!!!!!

yeah, it's so damn situational with me.  Sometimes some good tunes help get the problem solvin' groove goin'  sometimes it just skatters yer focus over the ground like so many sunflower seeds..... AND SUDDENLY YOU'RE A VAMPIRE.  [b]CRAP![/b]

uhh... could you tranlate that into legible typing?

Music is okay, but only things that I have heard several thousand times before so I don't really care about actually paying attention to the music.

[quote="Iversonfan2007"]For the last month or so I've usually been listening to music while doing problems. I dunno, usually loud music (like rap or rock) helps me to focus better by keeping me awake! :lol:[/quote]
That's totally the same with me, but it doesn't really keep me awake.  For me, the louder and rockier the music the harder and better I work.

i listen to thrash metal while doing math

it gets the job done or it gets more (noun)

fear factory, avenged sevenfold, children of bodom, xasther, slayer, rammstein, trivium, stuff like that.


calculus
integration
logarithms
quadratics
function
calculus computations

Pro 1 :
Find : $ \int_{0}^{1}\frac{1}{xlnx}dx$
Pro 2:
Find :
$ \int_{a}^{b}\frac{1}{\sqrt{(x\minus{}a)(b\minus{}x)}}dx;(a<b)$

Pro 3 :
Find $ lim_{x\minus{}>0\plus{}}x^{x^x\minus{}1}$

Pro 4 :
Does $ \int_{0}^{1}\frac{ln(x\plus{}1)}{(x\plus{}1)^2}dx$ converge and why?

Thanks so much :)

1. Try the substitution $ u\equal{}\ln x.$ [hide="Unfortunately"]The integral diverges. Its value is $ \minus{}\infty.$[/hide]

2. Rewrite the quadratic expression inside the square root. Complete the square, so you can recognize this as an arcsine. Alternatively, make a "small" linear substitution to bring it to an integral from $ 0$ to $ 1$ and recognize that as a Beta function.

3. $ x^x\minus{}1\equal{}e^{x\ln x}\minus{}1$ looks like $ x\ln x$ for $ x$ near zero. That means $ x^{x^x\minus{}1}$ looks like $ x^{x\ln x}\equal{}e^{x(\ln x)^2}.$

Now, what can you say about the behavior of $ x(\ln x)^2$ as $ x\to0?$

4. The function being integrated is continuous on $ [0,1].$ This is not an improper integral and there was never any question about its convergence.

It can also be computed quite readily - just do the most natural integration by parts.

Or did you mean to write something else for #4?

[hide]
1/ $ \int \frac {1}{x\ln(x)}dx \equal{} \int\frac {\frac {1}{x}}{\ln(x)} dx \equal{} \ln(\ln(x))$[/hide]

so it diverge to $ \minus{}\infty$  :)

[quote="mehdi cherif"][hide]
1/ $ \int \frac {1}{x\ln(x)}dx \equal{} \int\frac {\frac {1}{x}}{\ln(x)} dx \equal{} \ln(\ln(x))$[/hide]

so it diverge to $ \minus{} \infty$  :)[/quote]
Thanks all :)/ I forgot the last question :D
Could you help me to find the best way to find $ \int x^2\sqrt{x^2\plus{}4}dx$

$ P \equal{} \int x^2 \sqrt{x^2\plus{}4}\,dx \equal{} \frac12 \int 2x \sqrt{x^2\plus{}4} \cdot x \,dx$

$ \equal{} \frac12 \left(\frac{(x^2\plus{}4)^{3/2}}{3/2} x \minus{} \frac23 \int (x^2\plus{}4)^{3/2} \,dx\right)$

$ \equal{} \frac13 x (x^2\plus{}4)^{3/2} \minus{} \frac{16}{3} \int \sec^5t\,dt$

Can you proceed from here?

I wouldnt do what Carcul did, it just makes the problem more difficult.

In this case its always best to use hyperbolic functions, since, note that:
\[ \cosh^2{u} \minus{} \sinh^2{u} \equal{} 1
\]
so put: $ x \equal{} 2\sinh{u}$,

Then our integral will become:
\[ \int x^2\sqrt {x^2 \plus{} 4}\; dx \equal{} 4\int \sinh^2{(2u)}\; du
\]
Which is quite easy...


calculus
integration
function
symmetry
analytic geometry
trigonometry
calculus computations

Draw the picture - this is a butterfly shape.

Since you're explicitly saying that $r=\sqrt{x^{2}+y^{2}},$ you are saying that $r\ge 0.$ Hence you should not have that negative limit of integration in your inner integral.

The integral over the "right wing" of the butterfly should be:

$\int_{-\pi/4}^{\pi/4}\int_{0}^{\sec\theta}f(r)\,r\,dr\,d\theta$

Now we need to add the integral over the "left wing". We could do that by letting $\theta$ run from $\frac{3\pi}4$ to $\frac{5\pi}4,$ but we could also note that with the radial function $f$ and the symmetry of the shape, then the integral over the left wing will be the same as the integral over the right wing - so we don't need to calculate it twice. Just find the integral over the right wing and double it. And while we're at it, everything is also symmetric about the horizontal axis - so we really only need to calculate the integral over the upper half of the right wing. We have arrived at this:

$I=4\int_{0}^{\pi/4}\int_{0}^{\sec\theta}f(r) \,r\,dr\,d\theta$

--

For part (a) you can just do the $d\theta$ integral, getting

$I=2\pi\int_{0}^{1}f(r)\,dr$

[quote="Kent Merryfield"]Draw the picture - this is a butterfly shape.

Since you're explicitly saying that $r=\sqrt{x^{2}+y^{2}},$ you are saying that $r\ge 0.$ Hence you should not have that negative limit of integration in your inner integral.
[/quote]

sort of, because arnt we saying that $r^{2}= x^{2}+y^{2}$ and so then $r = \pm \sqrt{x^{2}+y^{2}}$ ?

[quote="Kent Merryfield"]It doesn't work that way. $\sqrt{w}$ is defined (by convention) to be a positive number. The expression involves $f(\sqrt{x^{2}+y^{2}}).$ That only requires $f$ to be defined for nonnegative arguments.

Put another way, $2=\sqrt{4}.$ You can't square that, and then proclaim that $2=\pm2.$

Now, as a matter of taste and custom: should you allow negative values of $r$ at all in polar coordinates? There are a few curves, such as $r=\cos(5\theta),$ that are more aesthetically pleasing if we allow negative $r,$ so that possibility has conventionally been allowed. But when it comes to integration, allowing negative $r$ has the potential to be terribly confusing. (Are we counting everything once or twice? Does that particular patch of integral count positively or negatively?) It's just safer when doing double integrals to reject negative $r.$

--

On part (c), the answer should be

$I=\int_{-\pi/2}^{\pi/2}\int_{0}^{\cos\theta}f(\tan\theta)\,r\, dr\,d\theta =\frac12\int_{-\pi/2}^{\pi/2}(\cos^{2}\theta)f(\tan\theta) \,d\theta$

Note that $x^{2}+y^{2}\le x$ implies that $x>0$ and hence $\cos\theta>0.$ We should therefore confine the integral to the fourth and first quadrants.[/quote]

hold on, your losing me...

here is my working for part (c), could you please explain to me where i have gone wrong? cheers :)

[b] (c) [/b]
[img]http://img215.imageshack.us/img215/903/picture6ax5.png[/img][/QUOTE]

You went wrong by going all the way around, by allowing $\theta$ to take on all values.

You said that $r^{2}\le r\cos\theta\implies r\le\cos\theta$

Notice that this implication only makes sense if $r\ge0,$ so the conclusion must be $0\le r\le\cos\theta$ - and that's empty unless  $0\le \cos\theta.$ That's what confines us to the fourth and first quadrants.

I don't like arguing entirely algebraically in this instance - far better to have a picture. In this case, the set on which you are integrating is a circular disk. In rectangular coordinates, it is centered at $(1/2,0)$ and has radius $1/2.$ This set lies only in the fourth and first quadrants of the plane.

It was also posible to do one more step, which is to actually perform the $dr$ integral.

[Not directly related to the original problem.] I think $r=\cos 5\theta$ looks the same whether or not negative $r$ are allowed. As $\theta$ varies between $0$ and $2\pi$, it is traced either once or twice, depending on the convention.

Allowing negative $r$ for "polar roses" $r=\cos n\theta$ creates a problem: the rose will have $n$ petals when $n$ is odd and $2n$ petals when $n$ is even. You can't get a 6-petalled flower at all. Rejecting negative $r$ corrects the situation.

Does anyone want to allow negative values of $\rho$ in spherical coordinates $(\rho,\theta,\phi)$?

Yeah, I wasn't thinking about the polar roses when I mentioned the example, but it doesn't really change my point.

[quote="mlok"]Does anyone want to allow negative values of $\rho$ in spherical coordinates $(\rho,\theta,\phi)?$[/quote]
I was already considering saying exactly this. I certainly don't want to deal with $\rho<0.$ And considering the four-dimensional analogue of spherical coordinates (one distance, three angles) would only underline this preference even more.

ok i think i almost have it now....

for part (b) i get:

[url=http://imageshack.us][img]http://img201.imageshack.us/img201/1121/picture6yb5.png[/img][/url]

for (c) i get this answer:

$\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{\frac{1}{cos\theta}}f(tan\theta)rdrd\theta$

On (c), the upper limit of the $dr$ integral is $\cos\theta,$ not $\sec\theta.$ That, and you can go ahead and compute the $dr$ intgral, since $f$ depends only on $\theta.$

(b) is correct. (In my post above, I also took advantage of symmetry about the $x$-axis.)

yep, oops silly mistake, thanks for that.

and if i havent said it before, thank you very much for all your help :)

-sarah


Let $ P,Q$ be polynomials with coefficients in the set $ \{1,2009\}$.Prove that if $ P$ divides Q then $ degP\plus{}1$ divides $ degQ\plus{}1$.

$ x$ divides $ x^2$ but $ 2$ doesn't divide $ 3$?

[quote="Xantos C. Guin"]I presume that "missing" terms are treated as a coefficient of $ 0$.

$ x \equal{} x \plus{} 0$, $ x^2 \equal{} x^2 \plus{} 0x \plus{} 0$,  and $ 0 \not \in \{ 1 , 2009 \}$.[/quote]
Yes,That's right.


We have, for any $n$:
$6^n \equiv 6 \mod 10$
$76^n \equiv 6 \mod 100$
$376^n \equiv 376 \mod 100$

I've done some investigating, and found that there is a unique number which leaves the same remainder $\mod 10,100,1000$ when raised to any power. Those three mods are the only ones I've looked at so far, though.

Now, some questions:
1. Is there a unique number $m$, ($m \neq 0,1$) corresponding to a particular $n$ in $10^n$ which leaves the same residue $\mod 10^n$ when raised to any power? Prove it.
2. If so, why does it keep adding a digit? Sorry, badly worded, but I mean like 6, 76, 376.........

Your assertion is incorrect.

$5^n \equiv 5 \bmod 10$
$25^n \equiv 25 \bmod 100$
$625^n \equiv 625 \bmod 1000$

Yeah, never mind.

I just realized that. :)


That's divergent unless $x=y$; it diverges to zero if $x<y$ and to $\infty$ if $x>y$. The terms for large negative $n$ don't go to 1.


geometry
incenter
geometric transformation
reflection
angle bisector
geometry solved

Let triangle $ABC$ with the incircle $(I)$ and internal bisectors $AD, BE, CF$ . $A_{0}, B_{0}, C_{0}$ be the midpoints of the sides $[BC], [CA], [AB]$, respectively. $A_{1}$ is the intersection of $(I)$ and the line $Dx$ that is tangent to $(I)$. We determine $B_{1}, C_{1}$ similarly.
Prove that the lines $A_{0}A_{1}, B_{0}B_{1}, C_{0}C_{1}$ are concurently.

There are two lines through D tangent to (I)...

well, one of those lines is a side of $ABC$ ... assuming a bit of simmetry, you'll choose either the three sides or the three other tangents ... but trying to make the three sides concur it's not so easy ...

Let the incircle (I) of triangle ABC touch its sides BC, CA, AB at the points X, Y, Z, respectively. Since the lines DX and $DA_{1}$ are the two tangents from the point D to the circle (I) with center I and touch this circle at the points X and $A_{1}$, respectively, these points X and $A_{1}$ must be symmetric to each other with respect to the line DI. The line DI is the angle bisector of the angle CAB (since the point D lies on the angle bisector of the angle CAB by its construction, and the point I, being the incenter of triangle ABC, also lies on this angle bisector). Hence, the points X and $A_{1}$ are symmetric to each other with respect to the angle bisector of the angle CAB. Thus, the point $A_{1}$ is the reflection of the point X in the angle bisector of the angle CAB. Similarly, the points $B_{1}$ and $C_{1}$ are the reflections of the points Y and Z in the angle bisectors of the angles ABC and BCA, respectively. Thus, the problem follows from [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=1307]IMO 1982 problem 2[/url].

  darij


Chú chien than,
Tại sao chú post bài THTT chưa hết hạn lên đây?

http://www.mathlinks.ro/Forum/viewtopic.php?t=136613

Lần sau rút kinh nghiệm em nhé. Chú ở Hà Tây à? Anh cũng có bạn ở đó đấy! :D

Lại có thêm thành viên Việt Nam làm như vậy, lần này là toan hoc , và đây là link

http://www.mathlinks.ro/Forum/viewtopic.php?t=139550

Vẫn là cái cậu toan hoc này http://www.mathlinks.ro/Forum/viewtopic.php?t=145499&sid=d6a57385d736a9bca0808be038224d4c . Tại sao cậu phải làm như vậy? :D Mình đã khoá topic đó rồi.

ko nên hoãn cái sự sung sướng đó lại thày Tuân ạ :P Nói vậy thôi chứ ai mà post lên đây có lẽ ta nên khủng bố bằng ... del bớt bài  :P

Không , ở đây phải hạn chế xoá bài, lock thôi thì được, tôn trọng thành viên hết mức :) Chú cố gắng tham gia bên này cho vui nhá :P

anh cho em 1 chân trợ lí thư kí của mod bên này thì em tham gia liền  :P Mấy hôm nay dạo Mathlinks thấy topic có sự tham gia của thày Tuân là ko lock cũng bị cảnh cáo  :D Thày làm việc hiệu quả thiệt  :P

[quote="zaizai-hoang"]anh cho em 1 chân trợ lí thư kí của mod bên này thì em tham gia liền  [/quote]
Anh còn khoẻ, không cần trợ lý :D

[quote] Mấy hôm nay dạo Mathlinks thấy topic có sự tham gia của thày Tuân là ko lock cũng bị cảnh cáo  :D Thày làm việc hiệu quả thiệt  :P[/quote]
Cảm ơn lời khen tặng.

Anh N.T.TUAN la ai vay? Sao co nguoi goi 'anh', lai co nguoi goi 'thay'?
Em to mo mot chut thoi! :lol:

Chào anh Tuân,anh khoẻ chứ? :p

Anh khoá giùm em cái topic này nhé,bài này trên THTT chưa hết hạn gửi,nói bao nhiêu lần mà cậu every này vẫn vi phạm!

http://www.mathlinks.ro/viewtopic.php?p=1107325#1107325

PS: Ko ngờ trên ML lại có 1 box thế này cho anh em spam thoải mái :D

Nhung khong tinh bai dau!  :rotfl: Tim cho gia su cho anh chua? Anh sap len HN roi day!


1. For any set of positive integers a and b such that a>b, prove that a!/b! must be an integer.

2. Prove that 0!=1.

0!=1 is a definition. You can't really prove it. All attempts are circular reasonings.


If a>b>c and ab+bc+ca=abc prove that

 4/c^2  + 1/(a-b)b   +1/(b-c)c   \geq 4/3

CRUX PROBLEM N1830

Note that the ab+bc+ca= abc is equivalent to 1/a +1/b +1/c = 1.
Using AM/GM, we have :
a \geq 2 \sqrt (b(a-b)) and b \geq 2 \sqrt (c(b-c)).
It follows that :
4/c 2  + 1/(b(a-b)) + 1/(c(b-c))  \geq 4(1/a 2  + 1/b 2  + 1/c 2 )  \geq 4(1/a + 1/b  + 1/c) 2 /3 = 4/3.

Pierre.


function
inequalities
inequalities proposed

1) Let $ f(x,y,z)$ be a function of three variables $ x,y,z$ given by the formula:
\[ f(x,y,z) \equal{} (x \plus{} y)(y \plus{} z)(z \plus{} x) \plus{} \frac {1}{x^3} \plus{} \frac {1}{y^3} \plus{} \frac {1}{z^3}
\]
where  $ x, y, z>0$ and $ x^2 \plus{} y^2 \plus{} z^2 \leq \frac {3}{4}$.
Find the minimum value of the function f(x,y,z)?
2) Let $ f(x,y,z)$ be a function of three variables $ x,y,z$ given by the formula:
\[ f(x,y,z) \equal{} \frac {x^2}{y^2 \plus{} z^2 \plus{} yz} \plus{} \frac {y^2}{x^2 \plus{} z^2 \plus{} xz} \plus{} \frac {z^2}{x^2 \plus{} y^2 \plus{} xy}
\]
where   $ x, y, z>0$ and $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 3$.
Find the minimum value of the function f(x,y,z)?

A very interesting inequality is:
For $ x,y,z>0$ prove that:
$ \frac{x^2}{x^2\plus{}xy\plus{}y^2}\plus{}\frac{y^2}{y^2\plus{}yz\plus{}z^2}\plus{}\frac{z^2}{z^2\plus{}zx\plus{}x^2} \ge 1$
 :)

[quote="vuthailuan"]1) Let $ f(x,y,z)$ be a function of three variables $ x,y,z$ given by the formula:
\[ f(x,y,z) \equal{} (x \plus{} y)(y \plus{} z)(z \plus{} x) \plus{} \frac {1}{x^3} \plus{} \frac {1}{y^3} \plus{} \frac {1}{z^3}\]
where  $ x, y, z > 0$ and $ x^2 \plus{} y^2 \plus{} z^2 \leq \frac {3}{4}$.
Find the minimum value of the function f(x,y,z)?
2) Let $ f(x,y,z)$ be a function of three variables $ x,y,z$ given by the formula:
\[ f(x,y,z) \equal{} \frac {x^2}{y^2 \plus{} z^2 \plus{} yz} \plus{} \frac {y^2}{x^2 \plus{} z^2 \plus{} xz} \plus{} \frac {z^2}{x^2 \plus{} y^2 \plus{} xy}\]
where   $ x, y, z > 0$ and $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 3$.
Find the minimum value of the function f(x,y,z)?[/quote]
Two problem is very easy.For (b),dear my friend, we don't need $ x^2\plus{}y^2\plus{}z^2\equal{}3$ beucause this inequality is homogenus form.You can learn inequality in "Sáng tạo bất đẳng thức" or "Bất đẳng thức suy luận và khám phá".

[quote="Dimitris X"]A very interesting inequality is:
For $ x,y,z > 0$ prove that:
$ \frac {x^2}{x^2 \plus{} xy \plus{} y^2} \plus{} \frac {y^2}{y^2 \plus{} yz \plus{} z^2} \plus{} \frac {z^2}{z^2 \plus{} zx \plus{} x^2} \ge 1$
 :)[/quote]

Normalize to $ xyz \equal{} 1$. Then there exists $ a,b,c$ such that $ x \equal{} \frac {b}{a}$  $ y \equal{} \frac {c}{b}$ $ z \equal{} \frac {a}{c}$.

$ \frac {x^2}{x^2 \plus{} xy \plus{} y^2} \plus{} \frac {y^2}{y^2 \plus{} yz \plus{} z^2} \plus{} \frac {z^2}{z^2 \plus{} zx \plus{} x^2}$

$ \Leftrightarrow \frac {1}{1 \plus{} y/x \plus{} y^2/x^2} \plus{} \frac {1}{1 \plus{} z/y \plus{} z^2/y^2} \plus{} \frac {1}{1 \plus{} x/z \plus{} x^2/z^2}$

$ \Leftrightarrow \frac {b^4}{b^4 \plus{} ab^2c \plus{} a^2c^2} \plus{} \frac {a^4}{a^4 \plus{} a^2bc \plus{} b^2c^2} \plus{} \frac {c^4}{c^4 \plus{} abc^2 \plus{} a^2b^2} \ge \frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{a^4 \plus{} b^4 \plus{} c^4 \plus{} a^2bc \plus{} ab^2c \plus{} abc^2 \plus{} a^2b^2 \plus{} b^2c^2 \plus{} a^2c^2}$ by Cauchy

so it suffices to prove $ a^4 \plus{} b^4 \plus{} c^4 \plus{} 2a^2b^2 \plus{} 2b^2c^2 \plus{} 2c^2a^2 \ge a^4 \plus{} b^4 \plus{} c^4 \plus{} a^bc \plus{} ab^2c \plus{} abc^2 \plus{} a^2b^2 \plus{} b^2c^2 \plus{} a^2c^2$
$ \Leftrightarrow a^2b^2 \plus{} b^2c^2 \plus{} a^2c^2 \ge a^2bc \plus{} ab^2c \plus{} abc^2$ Which is obvious by AMGM or Muirhead

Infact, we don't need $ x,y,z$ to be positive since the denominators are always positive, and having any of $ x,y,z$ negative just decreases the $ xy/yz/zx$ terms without changing anything else


Find all $ n\in N^*$ such that $ n\equal{}\phi(n)\plus{}402$
 :maybe:  :maybe:


modular arithmetic
number theory
greatest common divisor
number theory unsolved

Let n be the positive integer such that 24|(n+1).
Prove that the sum of all the divisors of n is divisible by 24.

$ n\equiv 3\pmod{4}$,so $ n$ cannot be a perfect square,hence $ n$ can be wrriten as
$ n \equal{} p_{1}^{e_{1}}\times p_{2}^{e_{2}}\times \cdots \times p_{k}^{e_{k}}$ with primes $ p_1,p_2,...,p_{k}$ and $ e_1,e_2,...,e_{k}$ are all odd.

$ 1)k \equal{} 1$
the statement is obvious.

$ 2)k > 1$
The sum of divisors is $ (1 \plus{} p_{1} \plus{} p_{1}^2 \plus{} \cdots \plus{} p_{1}^{e_{1}})(1 \plus{} p_{2} \plus{} p_{2}^2 \plus{} \cdots \plus{} p_{2}^{e_{2}})\cdots (1 \plus{} p_{k} \plus{} p_{k}^2 \plus{} \cdots \plus{} p_{k}^{e_{k}})$

because $ n\equiv 2\pmod{3}$,there exist a prime $ p_{i}\equiv 2\pmod{3}$ among $ p_1,p_2,...,p_{k}$,and then
$ 1 \plus{} p_{i} \plus{} p_{i}^2 \plus{} \cdots \plus{} p_{i}^{e_{i}}\equiv 0\pmod{3}$ since $ e_i$ is odd.

because $ n\equiv 3\pmod{4}$,there exist a prime $ p_{j}\equiv 3\pmod{4}$ among $ p_1,p_2,...,p_{k}$,and then
$ 1 \plus{} p_{j} \plus{} p_{j}^2 \plus{} \cdots \plus{} p_{j}^{e_{j}}\equiv 0\pmod{4}$ since $ e_j$ is odd

and each $ 1 \plus{} p_{u} \plus{} p_{u}^2 \plus{} \cdots \plus{} p_{u}^{e_{u}}(u \equal{} 1,2,...,k)$ are all even number since $ p_u$ and $ e_u$ are odd,
hence we see that the sum is divisible by $ 8$ and by $ 3$,as desired.

EDIT:I noticed my mistake in my first step "$ e_1,...,e_k$ are all odd"
I try to correct it...or anyone?

Not sure why this is specifically true for $ 24$, but if $ \text{gcd}\,(a,24) \equal{} 1$, then $ a(24\minus{}a) \equiv \minus{}a^2 \equiv \minus{}1 \pmod{24}$ (wlog $ a < 12$; check $ a \equal{} 1,5,7,11$). Then taking inverses, the sum of pairs of divisors of $ n$ are divisible by $ 24$.


calculus
integration
trigonometry
real analysis
real analysis unsolved

Please calculate this integral:
$ \int_{0}^{\infty} \frac{x^{2}}{\left(x^{5}\plus{}1\right)\left(x^{6}\plus{}1\right)}dx$

Let $ I\equal{}\int_{0}^{\infty }\frac{x^{2}}{\left( x^{5}\plus{}1\right) \left( x^{6}\plus{}1\right) }dx$

Set $ x\equal{}\frac{1}{y},$

$ I\equal{}\allowbreak \int_{0}^{\infty }\frac{y^{7}}{\left( 1\plus{}y^{5}\right) \left( 1\plus{}y^{6}\right) }dy$

Hence,

$ 2I\equal{}\int_{0}^{\infty }\frac{x^{2}}{x^{6}\plus{}1}dx\equal{}\frac{1}{3}\int_{0}^{\infty }\frac{3x^{2}}{(x^{3})^{2}\plus{}1}dx\equal{}\allowbreak \frac{1}{3}\tan ^{\minus{}1}x^{3}|_{0}^{\infty }\equal{}\allowbreak \frac{\pi }{6}$

Thus, $ \int_{0}^{\infty }\frac{x^{2}}{\left( x^{5}\plus{}1\right) \left( x^{6}\plus{}1\right) }\equal{}\allowbreak \frac{\pi }{12}$ :lol:


Let k be a positive integer and S(k) be the sum of squares of digits of k..Prove that
S(...(S(k))...) (finite S) equal to 1 or 145.
   (Also,I really like Erdos' theorems and their proofs.If somebody knows where I can get them,please show me. :) )


conics
ellipse
rotation
trigonometry
geometry
geometric transformation

How do you solve a problem like this? :?

Let $A$ be the smallest positive angle measure, in degrees, through which the ellipse $5x^2+6xy+5y^2-8=0$ can be rotated counterclockwise to make the major axis vertical. Let $B$ be the square of the distance between the foci of the ellipse. Find the value of $A+B$.

(Also, how do we find the equation of the ellipse after it is rotated?)

The angle you rotate a conic $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ is $\theta$ when $\tan{2\theta}=\frac{B}{A-C}$. After you find the angle, then you simply plug in $x=x'\cos{\theta}-y'\sin{\theta}$ and $y=x'\sin{\theta}+y'\cos{\theta}$, and the $xy$ term will cancel out and you can put it into the regular form that people usually work with.

Thanks for the fast reply.

For $\tan{2\theta}=\frac{B}{A-C}$, there are two possibilities for $\theta$... In this case, since $\tan{2\theta}$ is undefined, $\theta$ is either $\frac{\pi}{4}$ or $\frac{3\pi}{4}$ (for $0< \theta <\pi$). How do we know which is the correct angle? (I believe the other angle makes the major axis horizontal, not vertical.)

You take the one where $\theta$ is between $0$ and $\frac{\pi}{2}$ since you don't want to rotate the $x$-axis more than $90^{\circ}$, so in this case the angle of rotation is $\frac{\pi}{4}$.


function quadratics inequalities algebra

[quote="Raz"]Find the range of the following function:$f(x)=\frac{2^{x}-1}{2^{2x}+2^{x}}$.[/quote]
Let $2^x=t$ then $t>0$.
Since we have $yt^2+(y-1)t+1=0$, so it becomes a quadratic equation problem.

[quote="Raz"]Excuse me, can you write the full solution please? :)[/quote]

use discriminant
$b^2-4ac=(y-1)^2-4y\geq 0$ and solve for $y$ :D

Ok, but it is not enough, because when $D\ge{0}$  that doesn't mean that the qudratic equation has [b]positive[/b] solution. :D  :D

[quote="shyong"][quote="shobber"]add another inequality $y-1>0$.[/quote]

shouldnt it be $y-1< 0$ ?[/quote]
Yep, I forget that it should be $-b$...


geometry function trigonometry AMC AIME algebra polynomial

$1)$ Find $x$

$2)$ Find $x$ in function of $a$ and $b$

1)

let $\triangle ABC$ have $M$ on $AC$ so $AM=MC$. $\angle BAC=7x$, $\angle ACB=2x$ and $\angle MBC=x$.

By the exterior angle theorem, $\angle AMB=3x$.

It follows that $\angle ABM=180-10x$. Then using the law of sines, we have: \[\frac{AM}{BM}=\frac{\sin 180-10x}{\sin 7x}= \frac{MC}{BM}=\frac{\sin x}{\sin 2x}\]  \[\sin 10x*\sin 2x=\sin 7x\sin x\]  \[\cos8x-\cos12x=\cos 6x-\cos 8x\] Now I am not sure what to do...

Well that's also what I got. So no progress till now  :rotfl:

my calculator gives me bad answers...perhaps the problem is different (i remember an aime problem like this, just different numbers)

$x=15^\circ$ is a solution. I just don't know how to reduce the expression :D .

Well, it is certainly possible "in theory" -- expand your cosines out as polynomials and solve for the roots.  $15^{o}$ is the only root which satisfies $10x < 180$, though.


calculus
integration
trigonometry
calculus computations

Let $ \displaystyle f(x)\equal{}\int_{x}^{x\plus{}1}\sin(t^2)dt$

1) Prove that $ |f(x)|<\frac 1x$ for all $ x>0$
I think I have done this correctly. Basically used integration by parts and got $ |f(x)|\le\frac{1}{x(x\plus{}1)}<\frac 1x$

2) Prove that $ 2xf(x)\equal{}\cos(x^2)\minus{}\cos[(x\plus{}1)^2]\plus{}r(x)$ where $ |r(x)|<\frac cx$ and $ c$ is a constant.

My question is how do I go about doing part 2.

That should also come from the integration by parts. Are you using $ dv\equal{}2t\sin(t^2)\,dt$ and $ u\equal{}\frac1{2t}$ in $ \int u\,dv\equal{}uv\minus{}\int v\,du?$

No I used $ u\equal{}t^2$, but I'll give your suggestion a shot.


trigonometry
geometry
3D geometry
algebra
factorization
sum of cubes
complex numbers

so, i was thinking about the cosine of 20, and managed to sort of solve this equation $8x^3-6x-1=0$, where x is the cos of 20.  (this equation comes from the equation for cos(3x).)

so after simplifying a bunch, i got $x=\frac{\sqrt[3]{\frac{1}{2}+\frac{i\sqrt{3}}{2}}+\sqrt[3]{\frac{1}{2}-\frac{i\sqrt{3}}{2}}}{2}$

but the problem is, is there a way to simplify this more?  because it has i in it, even though it is a real number.  how can you get rid of the i?

$8x^3-6x-1=0\Longleftrightarrow 2(4x^3-3x)-1=0,$ we can set $x=\cos \theta.$

[quote="perfectnumber628"] how can you get rid of the i?[/quote]

You can't.  At least, not without expressing the solution in terms of $\cos 20$.   :D

$cos(3x)=4cos^3x-3cosx$

this comes from the cosine of a sum of angles formula (surely it has a better-sounding name than that  :lol: )

so i said 3x=60 and x=20

$cos60=4cos^3 20-3cos20$
$\frac{1}{2}=4cos^3 20-3cos20$

and then i said x=cos20

$0=8x^3-6x-1$

and then i found a website (might have been wikipedia) that said how to solve cubics.  and i got an answer and simplified it a bunch.

rationalize the numerator using sum of cubes factorization

$x=\frac{\sqrt[3]{\frac{1}{2}+\frac{i\sqrt{3}}{2}}+\sqrt[3]{\frac{1}{2}-\frac{i\sqrt{3}}{2}}}{2}$

the sum of cubes factorization is $a^3+b^3=(a+b)(a^2+ab+b^2)$
try multiplying the numerator and denominator by ${{\sqrt[3]{(\frac{1}{2}+\frac{i\sqrt{3}}{2})^2}+\sqrt[3]{(\frac{1}{2}+\frac{i\sqrt{3}}{2})(\frac{1}{2}-\frac{i\sqrt{3}}{2}})}+\sqrt[3]{(\frac{1}{2}-\frac{i\sqrt{3}}{2}}})^2$

of course, the numerator would just equal the sum of the radicands (stuff in the radical), but the denominator is begging for algebraic manipulation.  I think that's what altheman meant, but i'm not gonna try to simplify it ;)

I guess I was more wondering how to use the sum of two cubes factorization in this particular problem.  Could someone please work it out for this specific problem, all the way through?

[quote="perfectnumber628"]how do you know there's no way to express it without using i or cos20?  because doesn't it seem really wierd that cos20 is a real number but you have to use i to express it?[/quote]

Not really.  Evaluating $\cos 20$ is intimately connected to Euler's formula, $e^{i \theta} = \cos \theta + i \sin \theta$, so it makes sense that you would need to apply this formula to a non-special angle in order to evaluate it properly.

Like Kunny, I think it is impossible - cube roots of complex numbers cannot generally be found except without reference to this formula, which will give the answer $\cos 20$.

that's cool, if it is true that this is a real number that needs i to be expressed.  but is there a way to be sure that you need to have i in it?  like, can you prove it?

Well, it doesn't [i]need[/i] the $i$; after all, $\cos 20^o$ is an expression for it which doesn't rely on $i$.  It may be (seems to be?), however, that it's not possible to express this number in terms of radicals of rational numbers alone without implicitly using $i$.

As a side note, the triple angle identities can be used to derive a "simple looking" solution to the cubic equation, and one which is easy to numerically evaluate provided you can take sines and arcsines.

To add some more - Historically speaking , the case where discriminant of the cubic equation is negative, (and thus all roots are real), could not be solved ,in  a systematic way by Cardon's solution. It is/was known as "irreducible case of Cardon's solution.  IOW There is no general arithmetic way to take a cube-root of an imaginary number.

As said above, "trigonometrical" method was the practical  method. (Unless , of course, if you  trial method gave one of the roots). It assumes that you know  (or have access  to) trigonometric tables..


\[ A \equal{}
\left[
\begin{array}{ccccc}
\minus{}a_1 & \minus{}a_2 & ... & \minus{}a_{n\minus{}1} & \minus{}a_{n} \\
       &         & I_{n\minus{}1} &    &     0 \\
\end{array}
\right]\]
a. Find the characteristical polynom $ det(xI_n \minus{} A)$ .
b. Assume you know the roots of the above polynom. Find the eigen vectors for $ A$ and $ A^{T}$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1093232234&t=131111]This topic,[/url] and some of the links from it, are pertinent.

Key words: Companion matrix.

thank you ! Sorry for the possible inconvenience !  :oops: I didn't know what to search for .


quadratics
integration
algebra
polynomial
calculus
number theory

Could you please help me with this?
Suppose that a is an irrational number.
If there is a real no. b such that both  (a+b) and ab are rational, show that a is a quadratic surd, i.e. of the form (r+s^1/2) or (r-s^1/2), where s is not the square of any rational no.

(Seems evident, but I cant think of a formal solution)

Prove that {[] denote G.I.F, f is continous, x>1}

$ \int_{1}^{x} [u]([u] \plus{} 1)f(u)du\equal{} 2 \sum_{i\equal{}1}^{[x]} i \int_{i}^{x}f(u)du$

my attempt
i've assumed that there is no 0 to spoil the problm

if a is irrational and a+b is rational
a and b must be in the form of x$ \pm$y  (where x is rational part and y is irrational)

at the same time ab is rational,i.e. (x+y)(x-y) is rational
$ \Rightarrow$ $ x^2$ - $ y^2$ is rational

$ \Rightarrow$ thus we have, y is in the form of s^1/2 and x is rational(from known properties of rational nos

i can't think any other way
but i know it is definitely wrong way to proceed

a+b is rational and a is irrational
=>a-b is irrational

s=(a-b)^2 = not perfect square of any rational number...
a,b are solutions of
$ x^2 \minus{} (a \plus{} b)x \plus{} ab \equal{} 0$

So
$ r \equal{} a \plus{} b \equal{} rational$
 $ s \equal{} (a \minus{} b)^2$

I posted this question in the number theory forum too, and some one replied:
"$ x^2 \minus{}(a\plus{}b)x\plus{}ab \equal{} 0$  is the minimal polynomial for a over the rationals. The result follows by the quadratic formula."
http://www.mathlinks.ro/viewtopic.php?t=203034
Soumya, I think,he is going the same way as you, but I dont understand his solution .
(I dont understand yours too, but thats not your fault)
Thanks!

I got the solution  the same way as Soumyashant Nayak had done.
There is nothing so difficult in understanding the problem.
Just simple 8th std. level.

[quote="tittu"]Prove that {[] denote G.I.F, f is continous, x>1}

$ \int_{1}^{x} [u]([u] \plus{} 1)f(u)du \equal{} 2 \sum_{i \equal{} 1}^{[x]} i \int_{i}^{x}f(u)du$[/quote]

If  $ i \leq u < i\plus{}1 \text{ where } i \in \mathbb{Z}, [u]([u] \plus{} 1) \equal{} 2i$. So break the integral into pieces of length 1, and you are through :D


Consider a plane-parallel transparent plate, where the refractive index, $n=n(z)$, varies with distance, $z$, from the lower surface (see figure).

Show that $n_{A}\sin{\alpha}=n_{B}\sin{\beta}$.

The notation is that of the figure.  occurred

Is this problem considered too easy?

Please post a reply.

Slice the plate into $N \rightarrow \infty$ very thin layers, parallel to its sides. Considering that the index of refraction is constant in each thin layer, we can write:
$n_{A}\sin\alpha = n_{1}\sin\phi_{1}$, for the light entering the first layer;
$n_{1}\sin\phi_{1}= n_{2}\sin\phi_{2}$, for refraction between the first and the second layer;
(... and so on for each two adjacent layers, until...)
$n_{N}\sin\phi_{N}= n_{B}\sin\beta$.
From these $N$ equations we get $n_{A}\sin \alpha = n_{B}\sin\beta$.


[quote="1=2"]An equilateral triangle has a side length of 3 inches. Squares are drawn on each of the sides of the equilateral triangle, such that no point in the equilateral triangle is also in the square. The vertices of the squares that are not also vertices of the triangle are the vertices of a hexagon that contains the triangle and the squares. What is the area of that hexagon?[/quote]

[hide="mega hint"]
that hexagon is just a bit smaller than an equilateral triangle...

[/hide]

[quote="1=2"]An equilateral triangle has a side length of 3 inches. Squares are drawn on each of the sides of the equilateral triangle, such that no point in the equilateral triangle is also in the square. The vertices of the squares that are not also vertices of the triangle are the vertices of a hexagon that contains the triangle and the squares. What is the area of that hexagon?[/quote]

[hide="Super Special Awesome"]Make a good, accurate diagram (not very hard) and its easy to see certain things.[/hide]

[hide]The hexagon consists of an equilateral triangle, three congruent squares, and three congruent obtuse triangles.
The area of the equilateral triangle is $ \frac{3^2\cdot\sqrt{3}}{4}\equal{}\frac{9\sqrt{3}}{4}$.
The area of the three squares is $ 3\cdot 3^2\equal{}27$
The area of the three obtuse triangles is $ 3\cdot \frac{3\cdot 3\cdot \sin 120^\circ}{2}\equal{}\frac{27\sqrt{3}}{4}$
The total area is $ \frac{9\sqrt{3}}{4}\plus{}27\plus{}\frac{27\sqrt{3}}{4}\equal{}\boxed{27\plus{}9\sqrt{3}}$[/hide]

[quote="bluelobster"]Whats with the click to reveal hidden content?[/quote]To hide the solution if people want to solve the problem on their own.


Maybe your operating system got deleted somehow?

I have no idea. But one time my 98 just completely disappeared and my dad got mad because he thought i did it. (after 20 mintues of installation on a really slow computer it was back again)

[quote="physics_man"]I'm having trouble starting up my laptop at home (windows 2000 I think).

I press the power button, and the blue Dell page shows up. but then it turns to a black page with a blinking cursor. I can't type anything or do anything really.

does anyone know what's wrong? thanks for any help.[/quote]

[quote="haha_2"]how would that have happened?

and also is there anything I can do?[/quote]

What's going on?  :huh: who is haha_2? who has the problem? or have you created a second profile and are cheeky enough to reveal it plainly at the first chance? :mad:

Anyway the problem could be with your hard disk if its got old. I'm not sure in dell laptops you can change hard disks, but if you can, it will be worth trying.

I don't have experience with the Windows side, but that definitely sounds like a hard drive problem- either the disk is going bad or some system files are corrupted. How much do you have backed up?
You will likely have to replace the drive, and you will almost certainly lose your data. You can try a reformat/reinstall first- if that doesn't work, you'll know it's hardware.

Laptop hard drives can be replaced- I know because I had to have mine replaced last year. Check the warranty policy- if it hasn't been too long since you got the computer, this replacement should be covered.

it turns out that after turning on and off my laptop 5 or 6 times, it will finally start up.

any more ideas what's going on?  :)

I'm not sure whether it is possible for internal hardware of dell laptops to go loose, but i know that similar problems are caused in the PC by RAMs going loose and Hard Disks growing old. If possible, reload your OS and see if the problem persists. Otherwise contact hands on experts.

I've experienced both hard drive and RAM issues on my machine- this sounds more like a hard drive problem. If it were RAM, the boot sequence would fail earlier- the error you're getting seems to be "I can't find the OS". When my RAM went bad (actually, it was the slot), I got nothing but a black screen and three beeps on attempted startup. That was a Mac laptop- the exact signals may be different for you.

If your computer is working right now, back up as much as you can. The problem is not fixed, and it could recur at any time.

Yes when my RAM had been suspected loose my PC would simply beep for some time on switching on and won't go further. After about three days, it would correct itself, only to relapse after about a week (during which I took copious backups, and again started downloading, leaving me on the onset of the next problem with more data to be backed up). The HD problem was more subtle. After the Windows logo, the Welcome screen failed to appear. Instead, either it would reboot or just hang there staring blankly at me. This problem disappeared after I made my 3-year old 40 GB the slave, and the less-than-a-year old 120 GB the master, but mysteriously, so did the other problem, which may lead me to suspect that the RAM wasn't loose at all. But it could have been - its possibilities that count. So if you are getting a startup screen of some kind, your RAM is okay. Get the HD checked.

There are more problems than just RAM or hard drive. It could be many other things including the OS being skrewed up, viruses,or the processor overheating. A good test would be to boot up something like Ubuntu off of a disc, see if it works, and see if it finds your hardware working. Thats a better way than just assuming it is the hard drive or the RAM.

Well, yes. A corrupted system file would have similar symptoms- except that it would be a lot less likely for it to mysteriously start working again.
Booting off a CD successfully does rule out most non-HD hardware problems- and failing indicates that some hardware has gone bad.

As for your other suggestions- a virus couldn't do this, except by breaking critical system files. An overheating processor would have radically different symptoms.
The symptoms reported are a very good match for a hard drive failure. Recommended course of action:
1. Back up everything you can.
2. Attempt to use repair utilities.
3. Take it to someone who knows what they're doing, and get a more certain diagnosis.
4. Wipe it out and reinstall everything.
5. If you're still having problems, replace the drive.

That's what I did- except that "everything I can" was nothing.


logarithms
number theory proposed
number theory

prove or disprove:

every odd number greater than 1 is of the form $ 2^n\plus{}p$ where $ n$ natural, $ p$ prime or 1.

There aren't enough primes for this to have a chance, given how thin powers of $ 2$ are.

The smallest counterexample is $ 127$, as $ 125,123,119,111,95,$ and $ 63$ are all composite. $ 149$ is the next, with $ 147,145,141,133,117,85,$ and $ 21$ all composite.

OK, more detail. In the long run, the average number of ways a number can be represented as $ 2^n\plus{}p$ is $ \frac{2}{\ln 2}\approx 2.885$, since primes have a density of about $ \frac1{\ln N}$ near $ N$. The distribution of primes is modeled fairly well by randomness, so we might guess that whether $ m\minus{}2^{n_1}$ and $ m\minus{}2^{n_2}$ are prime is approximately independent. If this were true, the distribution of values would be approximately Poisson, with about $ 5.6\%$ of all odd numbers not being representable. Empirically, it's a little more spread out than the Poisson distribution: through 30031, I get a mean of $ 2.815$ and a variance of $ 3.084$, with $ 6.2\%$ not representable.
The increased variance comes from numbers $ 2^k\minus{}1$ having prime factors larger than $ k$, so that several of the $ m\minus{}2^n$ values have a common prime factor. We can even exploit this to find infinitely many nonrepresentable numbers: if $ m$ is congruent to $ 3934332335837992627$ mod $ 2^{64}\minus{}1$, then $ m\minus{}2^n$ is divisible by $ 3$ when $ n\equiv 1\mod 2$, by $ 5$ when $ n\equiv 2\mod 4$, by $ 17$ when $ n\equiv 4\mod 8$, by $ 257$ when $ n\equiv 8\mod 16$, by $ 65537$ when $ n\equiv 16\mod 32$, by $ 641$ when $ n\equiv 32\mod 64$, and by $ 6700417$ when $ n\equiv 0\mod 64$.
(Note: $ 641\cdot 6700417\equal{}2^{32}\plus{}1$, and the other primes used are Fermat primes. The product $ 3\cdot 5\cdot 17\cdot 257\cdot 65537\cdot 641\cdot 6700417$ is $ 2^{64}\minus{}1$, and that monstrous 19-digit number is the result of solving the congruences $ m\equiv \minus{}1\mod (2^{32}\minus{}1)\cdot 641,m\equiv 1\mod 6700417$.)

Thank you very much :)

If you have time, could you please take a look at these two problems
http://www.mathlinks.ro/viewtopic.php?p=1255464#1255464
http://www.mathlinks.ro/viewtopic.php?p=1370564#1370564
the first one is unsolved for a long time ;)

Have a nice day!

I like this question, because of the interesting structure if we look beyond the initial question. I don't like those you just posted- they're easy and boring. That sort of request doesn't work- I don't think I've ever responded to one.
If you want to bring attention to an old question, use its own thread.


Solve the equation:$ \sqrt{2x^{2}\plus{}8x\plus{}6}\plus{}\sqrt{x^{2}\minus{}1}\equal{}x\plus{}3$
p/s:could any body show me how to hide a solution?thanks!

To hide stuff use:
[code][hide]text here[/hide][/code]
or to have text for the link:
[code][hide=click text]text[/hide][/code]

No squarings required (almost). The domain for $ \sqrt{2x^2 \plus{} 8x \plus{} 6}$ is $ x \in (\minus{}\infty, \minus{}3] \cup [\minus{}1, \infty)$. The domain for $ \sqrt{x^2 \minus{} 1}$ is $ x \in (\minus{}\infty, \minus{}1] \cup [1, \infty)$. But we need $ x \plus{} 3 \geq 0$, so $ x \in [\minus{}3, \infty)$. Combining the three yields $ x \in \{\minus{}3\} \cup [1, \infty)$.

$ x \equal{} \minus{}3$ doesn't work, while $ x \equal{} 1$ works. For $ x > 1$ we have $ 2x^2 \plus{} 8x \plus{} 6 > (x\plus{}3)^2$, equivalent to $ (x\plus{}3)(x\minus{}1) > 0$. Therefore the only solution is $ x\equal{}1$.

I think that solution needs much knowledge.Please find an another solution

The solution it's not difficult as all the squareroots are positive whenever the first squareroot is greater than the right part of the equation those values for x won't match that's why $ 2x^{2}\plus{}8x\plus{}6 > (x\plus{}3)^{2}$


algebra
function
domain
complex analysis

Show that if $f(z)=\frac{Log z}{(z-1)}$ when $z\neq 1$
                                  $f(z)=1$ when $z=1$

then $f$ is analytic throughout the domain $0<|z|<\infty$, $\pi<Arg z<\pi$

I know that $Log z= \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} (z-1)^n$. then what should I be working on?

Note that the power series you wrote down does not converge on all of the indicated domain. It converges on the disk $\{z: |z-1|<1\}$ (and on part of the boundary of that disk, not that it matters). However, that power series nonetheless holds the key. On that disk, we can cleanly divide the original power series by $z-1$ and obtain another power series in $z-1$ with the same radius of convergence. Since the constant term of this new power series is $1,$ the value of the function at $1$ is $1.$

Away from the special point and over the rest of the indicated domain, we have the quotient of two analytic functions with a never-zero denominator. Such a function is analytic by the quotient rule for derivatives.


algebra
polynomial
real analysis
real analysis unsolved

Find a polynomial ε-solution for:

y''+2y'+y=1,  y(0)=0, y'(0)=0

Note that the auxiliary equation is $ r^{2}\plus{}2r\plus{}1\equal{}0$, which has a double root $ r\equal{}\minus{}1$.  Then the solution is $ y(x)\equal{}Ce^{\minus{}x}$.  You can go from here.

hello, the complete solution is
$ y(x)\equal{}e^{\minus{}x}\plus{}e^{\minus{}x}x\minus{}1$
Sonnhard.

Well, I don't need the solution of the equation, but a polynomial ε-solution for it... :)

What is a polynomial $ \epsilon$ solution?  I'll take a guess that you want an approximation to the solution with some specific error term?  Fine, just compute the Taylor series, it's not hard.


function
integration
inequalities
IMC
college contests

Here's a little something I don't understand:

Problem A6 from IMC 1998

f:[0,1]->R is a continuous function s.t. xf(y)+yf(x)<=1 for all x and y in the domain. Show that \int{0->1}f(x)dx<=\pi/4. Give an example for which equality is reached. You can downolad the solution from here:

[url]http://www.ucl.ac.uk/~ucahjej/imc/imc1998/problems.html[/url]

They give a perfectly valid solution there, but here's what I did:

Let I=\int{0->1}f(x)dx. Integrate xf(y)+yf(x) with respect to x from 0 to 1 to get f(y)/2+y\int{0->1}f(x)dx<=1 => f(y)/2+Iy<=1 for all y in [0,1]. Now integrate this with respect to y from 0 to 1 to get I2+I/2<=1=>I2+I/2-1<=0 for, so I is in the interval [x1, x2], where xi are the roots of the equation x2+x/2-1=0, but the greatest of these roots is approximately 0.780776<0.785398 which is approximately \pi/4 (those numbers were actually copied and pasted from a notebook in Mathematica, so the calculations are quite accurate).

Where is my mistake? It's killing me  :? . I realize that those are approximations, so a mistake could easily have been made, but the difference is far too great (0.004...), so it can't be all an error. Maybe there was some mistake in those inequalities involving I?

Please help..

Never mind, I found my flaw (it's a pretty stupid one too :D): I took \int{0->1}y[\int{0->1}f(x)dx]dy to be I2, which is obviously false!

If nobody is interested in this poblem anymore, I guess the topic could be deleted..


Is there a polynomial P(x) with integer coefficients such that P(10) = 400, P(14) = 440 and P(18) = 520 ?

I think there is no such polynomial :

p(x) can be written  400 + (x-10) * (10 + (x-14) * r(x))  with r with integer coefficients. But then r(18) = 5/4 impossible.

yeah moubi you are right. this is a nice and classical question :)

[quote="Valentin Vornicu"]yeah moubi you are right. [/quote]

Are you sure you answered to the right post  :D  :D ??

ups. I saw the french flag, the hidden simbol, and I imediatly thought about moubi. :D sorry for that ...

Don't worry, I'm so [b]proud[/b] I was taken for Moubi  :D

Perhaps I should withdraw the flag ?? (private joke)

Well that does say something about the number of Frenchmen on this site... :D  Belenos, it's probably because you didn't display your flag for so long...

It just that Valentin and I had a short discussion one week ago about the display or not of the National Flag. That's all ....


An integer is called square-full if each of its prime fractors occurs to at least the second power. Prove that there exist infinitely many pairs of consecutive square-full integers.

We only have to investigate numbers to a prime power since x^(mn)=(x^m)^n.  One of those numbers must be divisble by 4 since consecutive numbers must have one even number.  And since anything divisible by 2 but not 4 is not square-full, it has to be divisible by at least 4 (in terms of powers of 2).  Also, two consecutive integers have a gcf of 1.  That's all I can do.  ANy ideas?

Well you have pair of  numbers like  (0,1)  (8,9),  (288,289),  (9800,9801), (352800,352801) ....

The pattern is:

[hide]Let  $k = 17+12\sqrt 2 = 33.970562748477....$
All you have to do is to find $k^n/4$  and Take  one integer by rounding it down, one rounding it up)
when n = 0 , k^0/4 = .25   so numbers are 0,1
        n =1;    k^1/4   = 8. 5 (appox)  so numbers are  8,9
        n =2;    k^2/4    = 288.5(appox)  so the numbers are 288,289
        and so on ... :) :) [/hide]

[hide]In fact if you take these   numbers:  for any  non-negative integer value of $n$

$\frac{(17+12\sqrt 2)^n +(17-12\sqrt 2)^n - 2} 4$
and
$\frac{(17+12\sqrt 2)^n +(17-12\sqrt 2)^n + 2} 4$

You will find, that each one  is a   natural number (easy to prove) , the numbers are consecutive (easy to prove too) and
both are square-full (This is easy to prove too!)
Just plug it in ..n= 0,1,2,3... and you get as many numbers as you wish :)

(In fact all you have to do is to "round up and down"  $\frac{ (17+12\sqrt 12)^n } 4$   or  $(33.97056274...^n)/4$  so it is rater easy to get all these numbers if you have a calculator) ...:)
[/hide]

BTW,  if one asks  of three consecutive  square-full numbers .. well these are not infinite  :)  Can any one find an example of   of such a case.


analytic geometry
graphing lines
slope
limit
calculus
calculus computations

The figure attached shows a right triangle with its hypotenuse OB along the y-axis and its vertex A (x,x^2) on the graph of y = x^2.  Let h represent the length of the hypotenuse, which depends on the
x-coordinate of the point A. Find the value of the limit as x approaches 0 of h.

I'm going to denote the point A as $A(t,t^{2})$ to avoid confusion later. The line along the cord OA has slope $m=\frac{t^{2}-0}{t-0}=t$, hence the line perpendicular to OA that passes through point $A(t,t^{2})$ has slope $-\frac{1}{t}$ and so has equation: $y-t^{2}=-\frac{1}{t}(x-t)$ which has point B as its y-intercept, namely put $x=0$ into the above equation to get $y-t^{2}=-\frac{1}{t}(0-t)\Rightarrow y=t^{2}+1=: h(t)$ now let $t\rightarrow 0$ to get $\lim_{t\rightarrow 0}h(t)=\lim_{t\rightarrow 0}(t^{2}+1)=1$.


Would it be possible for other members of this site to host math jams so we
could have a greater number of math jams? I don't get a chance to attend math jams that often (the ones I have attended have been very good). I don't know if any members would be interested in this but I thought it might be worth a try...

Well, we could have different members host math jams on Instant Messengers, like MSN.

Fierytycoon


Let a,b,c are non-negative numbers such that $ a,b,c\in [0,1]$ and $ ab\plus{}bc\plus{}ca\equal{}1$. Prove that
$ a\plus{}b\plus{}c\plus{}3abc\le \sqrt 2\plus{}1$

[quote="thegod277"]Let a,b,c are non-negative numbers such that $ a,b,c\in [0,1]$ and $ ab + bc + ca = 1$. Prove that
$ a + b + c + 3abc\le \sqrt 2 + 1$[/quote]
Let $ a=max\{a,b,c\}$ and $ f(a,b,c)=a + b + c + 3abc.$ Hence, ${ a\geq\frac{1}{\sqrt3}}$ and
$ f(a,b,c)-f\left(a,\sqrt{(a+b)(a+c)}-a,\sqrt{(a+b)(a+c)}-a\right)=$
$ =\left(\sqrt{a+b}-\sqrt{a+c}\right)^2(1-3a^2)\leq0.$
Thus, it remains to prove that
$ f\left(a,\sqrt{a^2+1}-a,\sqrt{a^2+1}-a\right)\leq1+\sqrt2$ is true for all $ \frac{1}{\sqrt3}\leq a\leq1,$
which is easy to check.

[quote="karis"]when have equality?[/list][/quote]
I think it does not hold.

oh, yes. I think I have a mistake, it is $ a\plus{}b\plus{}c\plus{}3abc\le 8\minus{}4\sqrt 2$, and equality occurs when $ a\equal{}1, b\equal{}c\equal{}\sqrt 2\minus{}1$, but a proof is same as argady.


algebra
polynomial
Vieta
algebra unsolved

Let $ r, s,$ and $ t$ be the roots of the cubic polynomial
$ p(x)\equal{}x^3 \minus{}2007x \plus{}2002$
Determine the value of

$ \frac {r \minus{} 1}{r \plus{} 1} \plus{} \frac {s \minus{} 1}{s \plus{} 1} \plus{} \frac {t \minus{} 1}{t \plus{} 1}$

[hide="Question"]
Is
$ r \plus{} s \plus{} t \equal{} \minus{} 2002$
$ rs \plus{} st \plus{} rt \equal{} 2002$
$ rst \equal{} \minus{} 2002$?[/hide]

$ r \plus{} s \plus{} t \equal{} 0$ and $ rs \plus{} st \plus{} tr \equal{} \minus{}2007$.

I'm not sure of the equations relating sums and products of roots for cubics but if you check wikipedia, you can get a formula for the roots.

Just expand $ (x \minus{} r)(x \minus{} s)(x \minus{} t)$.  Vieta's formulas are not something worth memorizing directly.


There are almost one million subscribers to a national magazine. Of those subscribers, 65% are women and 35% are men. The magazine sponsors a contest and selects 5 winners randomly from their list of subscribers. Find each probability.
    a) P(3 of the winners are men)
    b) P(all 5 winners are men)

I need some help with the procedure, thanks  :lol:

a) $ 35%
$ of 1 million = $ 350,000$, so it's $ \frac{350,000}{1,000,000} * \frac{349,999}{999,999} * \frac{349,998}{999,998} =$ Ill let you simplify that..

b) $ \frac{350,000}{1,000,000} * \frac{349,999}{999,999} * \frac{349,998}{999,998} * \frac{349,997}{999,997} * \frac{349,996}{999,996} =$ You can  finish that too.

If you haven't noticed the pattern, it's probability of choosing one guy times the same minus 1 times (numerator and denominator), and so on. It's because each time you are not counting the person before. Maybe someone else can explain it better though...

Is that just the fundamental counting principle, because on the worksheet, it has something to do with the binomial probability formula - http://www.mathwords.com/b/binomial_probability_formula.htm

Hmm.. I dunno what that is (probability and combinations aint my thing) but see if some1 else can answer that...

[hide="Solution (a)"]
For both of these solutions, I am assuming that, since the amount of subscribers is approximate and very large, that the probability that a random winner is a man is always $ 0.35$ ($ 0.65$ for a woman), no matter how many winners have already been picked.

There are $ \binom{5}{3} \equal{} \frac{5!}{3!2!} \equal{} 10$ combinations of 5 people such that exactly 3 of them are men. For each of these combinations, the probability that it will occur is $ 0.35 ^ 3 \cdot 0.65 ^ 2$, since each combination has three men and two women. (For example the probability that the combination Man-Man-Man-Woman-Woman will occur is $ 0.35 ^ 3 \cdot 0.65 ^ 2$, since we multiply the probabilites that each individual event will occur to find the probability that all of the events will occur.) The probability that any of those combinations will occur is $ \binom{5}{3} \cdot \left( 0.35 ^ 3 \cdot 0.65 ^ 2 \right) \equal{} 10 \cdot 0.35 ^ 3 \cdot 0.65 ^ 2 \equal{} 0.181146875.$
[/hide]

[hide="Solution (b)"]
The probability that the first winner is a man is $ 0.35$, the probability that the second winner is a man is $ 0.35$, and so on. We find the probability that all of these events will occur by multiplying them, so the probability is $ 0.35 ^ 5 \equal{} 0.0052521875$
[/hide]


geometry
circumcircle
similar triangles
angle bisector
geometry proposed

Let point I inside triangle ABC such that ∠ICB =∠IAC,∠IBC = ∠IAB. Prove that the bisectors of angle BAC,BIC meets on BC.

It is enough to prove that $ \frac {AB}{AC} \equal{} \frac {IB}{IC}$ $ ,(1)$

Because of $ \angle IBC \equal{} \angle IAB$ $ ,(2)$ we conclude that the circumcircle $ (O_{1})$ of the triangle $ \bigtriangleup AIB,$ tangents to $ BC$ at point $ B.$

Similarly, because of $ \angle ICB \equal{} \angle IAC$ $ ,(3)$ we conclude that the circumcircle $ (O_{2})$ of the triangle $ \bigtriangleup AIC,$ tangents to $ BC$ at point $ C.$

So, we have $ (MB)^{2} \equal{} (MI)\cdot (MA) \equal{} (MC)^{2}$ $ \Longrightarrow$ $ MB \equal{} MC,$ where $ M\equiv BC\cap AI$ and hence, $ AI$ is the $ A$-median of the given triangle $ \bigtriangleup ABC.$

$ \bullet$ From the similar triangles $ \bigtriangleup AMB\approx \bigtriangleup BMI,$ we have that $ \frac {AB}{IB} \equal{} \frac {AM}{BM}$ $ ,(4)$

From the similar triangles $ \bigtriangleup AMC\approx \bigtriangleup CMI,$ we have that $ \frac {AC}{IC} \equal{} \frac {AM}{CM}$ $ ,(5)$

From $ (4),$ $ (5)$ $ \Longrightarrow$ $ (1)$ and the proof is completed.

Kostas Vittas.

[hide][quote="vittasko"] $ \frac {AB}{IB} \equal{} \frac {AM}{BM}$
$ \frac {AC}{IC} \equal{} \frac {AM}{CM}$ $ ,(5)$
[/quote]
It implies the bisectors of angles ABI,ACI meet D on AI. The figure can help to detect right-angle BDC.[/hide]

[b]New problem:[/b] AB is the common chord oftwo circles (1),(2). The external common tangent of (1),(2) touches (1),(2) at C,D. Prove that the bisectors of angles ACB,ADB are perpendicular to each other.

For the [b][size=100]New problem[/size][/b], it is enough only the argument of the given equalities $ \angle IBC \equal{} \angle IAB$ and $ \angle ICB \equal{} \angle IAC.$

So, let $ E$ be, the intersection point of $ AM$ ( complete the schema [b][size=100]t=212238[/size][/b] ), from the angle bisector of $ \angle ABI.$

It is easy to show that $ ME \equal{} MB \equal{} MC,$ because of $ \angle MEB \equal{} \angle IAB \plus{} \angle ABE \equal{} \angle IBC \plus{} \angle IBE \equal{} \angle MBE.$

Similarly, the angle bisector of $ \angle ACI,$ intersects $ AM$ at the same point $ E$ and we are done $ ($ $ BE\perp EC$ $ ).$

Kostas Vittas.


Let $a,b,c$ be sides of a triangle. Show that

$\frac{b^{2}+c^{2}}{2}+\frac{a^{2}bc}{(b+c)^{2}}\geq \frac{a^{2}}{4}+bc$

[quote="n^4+4"]We can rewrite this as

$\frac{b^{2}-2bc+c^{2}}{2}+\frac{a^{2}bc}{(b+c)^{2}}\ge \frac{a^{2}}{4}$

We then divide by $\frac{a^{2}}{2}$ to obtain

$(\frac{b-c}{a})^{2}+\frac{2bc}{(b+c)^{2}}\ge \frac{1}{2}$

It is then clear by the triangle inequality that we may replace $a$ with $b+c$ and the LHS will decrease while the RHS remains constant.

After doing this, the inequality simplifies to $b^{2}+c^{2}\ge 2bc$[/quote]wait... i think if $b+c>a$ then $\left(\frac{b-c}{a}\right)^{2}+\frac{2bc}{(b+c)^{2}}<\left(\frac{b-c}{a}\right)^{2}+\frac{2bc}{a^{2}}$

He's replacing $a$ with $b+c$, not $b+c$ with $a$  :D