geometry
parallelogram
combinatorics unsolved
combinatorics

Find c (0<c<1) such that cn is the smallest number such that that in any set of n points in the plane one can increase A , where A is the minimum area determined by triangles, by the deletion of at most cn elements.

I have an example where you must eliminate $n/2$ points, however. so we have a lower bound for $c$. (Manuel, I am asuming A>0, right?)

Yes Pascual A>0 . is your bound best  possible?
Can you post your eaxmple with details?

[quote="Pascual2005"]I have an example where you must eliminate $n/2$ points[/quote]

A set of unit squares far from each other?

Pierre.

well there are many examples, and this one is in particular, one of them.

Pierre, i want to do the following thing: I want to color the set of points in $2$ colors in such a way that there is no  monocromatic triangle of area $A$. Is it possible? I think it is

[quote="Pascual2005"]
i want to do the following thing: I want to color the set of points in $2$ colors in such a way that there is no  monocromatic triangle of area $A$. Is it possible? I think it is[/quote]

You mean all the points in the plane? Because that's clearly impossible. :? It's possible to do this for large enough finite numbers of colors, I think, but not $2$.

No, i mean the points in the given set (n points) ;)

Pascual I think that we should prove by geometry that the graph can't have a K6 because by ramsey theory we will always have a monochromatic triangle.

Manuel, I thing we should better prove waht i said, becuase the other thing dont neccesary implies the coloration, but the coloration implies it.

Yes I already realized my silly mistake!

And how many triangles of the minimal area are possible?

Seems I have found a horrible terrible awful dreadful monstrous example that shows that $c>\frac12$. I'll try to describe it later (if I find words; I still don't know how to post pictures :? ) This destroys Pascual's nice 2-coloring conjecture... :(

OK, I simplified my example a lot but it is still quite monstrous. Start with  $n>6$ of points in the generic position (no 3 on one line, no 2 lines parallel, etc.) and call this set a core. Consider all lines through the pairs of core points and call the corresponding set $\mathcal L$. Let $\ell\in\mathcal L$ be a line through two core points $a$ and $b$. Choose some small generic $A$ and construct a parallelogram with base $[a,b]$ and area $2A$ whose other 2 vertices $a_1$ and $b_1$ are far away. Starting with $[a_1,b_1]$ create a stack of parallelograms $a_1b_1b_2a_2$,  $a_2b_2b_3a_3$, ... ,$a_{2k-1}b_{2k-1}b_{2k}a_{2k}$ of area $2A$ such that for every line $\ell'\in\mathcal L$ through the core points $c,d$ (one of which may coincide with $a$ or $b$), there exists a point $a_{2i}$ such that the area of the triangle $cda_{2i}$ is exactly $A$ and such that no triangle of area less than $A$ or a line parallel to one of the lines in $\mathcal L$ is created. If you try to draw a picture, you'll see that it is possible but it is somewhat boring to prove it with words only. Then take another pair of core points and create another such stack of parallelograms and so on. Now assume that we have $m$ core points not thrown away and $m>2$. Then for each pair $a,b$ of these core points, the corresponding stack of parallelograms cannot contain one of the points $a_{2i}$, from where it is clear that it can contain at most half of its points minus 1. This applies to $\frac {m(m-1)}2$ stacks, so the total number of points is not greater than $m$ plus half of the points in stacks minus $\frac{m(m-1)}2$. Thus, for $m>3$ we lose more points in the stacks than get in the core. And for $m\leq 3$ we have less than one half of the points in the core and not more than one half in each stack. Thus, we have to delete strictly more than one half points in this example.

I wonder now if anybody can prove that $c<1$...


I have two questions -

What is the difference between courses A and B?

In the description of the course it says, "Enrollment includes access to one subject class of the students' choice,"  Please clarify.

Here's some information about independent studies, if you haven't seen it already:
http://www.artofproblemsolving.com/Classes/AoPS_C_ClassesI.php

"one subject class of the students' choice" refers to enrollment in classes such as Intermediate Algebra, Introduction to Counting and Probability, or Olympiad Geometry.

I have seen the description.  When I went to register my son for both Intermediate Independent study B and Intermediate Algebra, the total price was for two courses.  Hence the question.

I'm not sure (since this based upon what I've heard from others), but I believe you should just register for the independent study class. Later, you'll receive an e-mail asking which subject class you wish to enroll in.

You just register for the independent study.  Then after you are enrolled in that you can email an instructor to tell them which class you would like to take along with it...or, if you like, you don't have to take the class immediately but can wait and take a class in the future.

If you have accidently signed up for both it shouldn't be a problem.  Just let the instructors know and they will probably work it out.

Exactly - just sign up for the Independent Study - we'll take care of the Subject class separately.


algebra
polynomial
inequalities
vector
inequalities proposed

Let $ P(X) \equal{} a_{n}X^{n} \plus{} a_{n \minus{} 1}X^{n \minus{} 1} \plus{} \ldots \plus{} a_{1}X \plus{} a_{0}$ be a real polynomial of degree $ n$. Suppose $ n$ is an even number and:

a) $ a_{0} > 0$, $ a_{n} > 0$;

b) $ a_{1}^{2} \plus{} a_{2}^{2} \plus{} \ldots \plus{} a_{n \minus{} 1}^{2}\leq\frac {4\min(a_{0}^{2} , a_{n}^{2})}{n \minus{} 1}$.

Prove that $ P(x)\geq 0$ for all real values $ x$.

[i]Laurentiu Panaitopol[/i]

The second property should be:

b) $ a_{1}^{2}\plus{}a_{2}^{2}\plus{}...\plus{}a_{n\minus{}1}^{2}\leq\frac{4\cdot min (a_{0}^{2},a_{n}^{2})}{n\minus{}1}$ (otherwise there's no meaning to the min)

this one already was posted here :
http://www.mathlinks.ro/viewtopic.php?t=133243
question was:
for which $ a_{i}$ next inequality holds:
$ \sum_{i \equal{} 0}^{n}a_{i}x^{i} \geq 0$ for all $ x$

Please help me with this problem,I did not find any hints from the posts above.Thanks.

Sorry to revive this one.But pretty interesting problem.Hope my solution is correct.
The second condition tells us to apply Cauchy-Schwartz.And so do we.
$|a_{n-1}x^{n-1}+...+a_1x|\le \sqrt{a_{n-1}^2+...+a_1}\cdot \sqrt{x^{2n-2}+...+x^2}....(*)$.Now see that $P(x)=a_nx^n+a_0+(a_{n-1}x^{n-1}+...+a_1x)\ge \text{min}(a_0,a_n)(x^n+1)-|a_{n-1}x^{n-1}+...+a_1x|\ge \text{min}(a_0,a_n)(x^n+1)-\sqrt{a_{n-1}^2+...+a_1}\cdot \sqrt{x^{2n-2}+...+x^2}\ge \text{min}(a_0,a_n)\left(x^n+1-\frac{2\sqrt{x^{2n-2}+...+x^2}}{\sqrt{n-1}}\right)$.
Here we used $(*)$ and the second condition.Now we will show that $(n-1)(x^n+1)^2\ge 4\sum_{i=1}^{n-1}x^{2i}$ we can put $n=2m$ and rewrite the equivalent inequality using $x^2=t$.
$(2m-1)(t^{2m}+2t^m+1)\ge 4\sum_{i=1}^{2m-1}t^i$ which follows easily from Karamata .Since $f(u)=t^u$ is convex and the vectors
${a}=(2m-1,2m-1,2m-1,2m-1,2m-2,2m-2,2m-2,2m-2,....,1,1,1,1)$ where each of $2m-1,2m-2,....,1$ occurs $4$ times and ${b}=(2m,2m,....,2m,m,m,...,m,0,0,...,0)$ where each of $2m,m,0$ occur $2m-1,4m-2,2m-1$ times respectively; among ${a}$ and ${b}$ see that $b \succ a$ now Karamata gives us exactly the inequality we are asking for.


integration limit function calculus vector geometry circumcircle

$1-1+1-1+1-1+1-1... = %Error. "FRAC" is a bad command.
{1}{2}$

You had two irreversible steps in the second one, squaring it negates your -1 solution and dividing while numerator was 0 gives you the 1 solution.

[quote="jli"][quote="Ihatepie"]0/0=1

1(0)=0

0/0=0

0(0)=0

0=1[/quote]

how does that make any sense?[/quote]
I think what he tried to say is: 0/0 can be anything, so let it be 1 here and 0 there, and therefore 1=0. :rotfl:

It's defined as nullity as of late...look it up.

A "high-tech" approach: $\int_{0}^{\infty}ce^{-cx}dx=1$ for any $c>0$. Letting $c\to 0$, we obtain $1=\lim_{c\to 0}\int_{0}^{\infty}ce^{-cx}dx=\int_{0}^{\infty}\lim_{c\to 0}ce^{-cx}dx=\int_{0}^{\infty}0\,dx=0$

[blinders on] Hey! that sequence of functions even converges [i]uniformly[/i] to zero! What could possibly be wrong with that? [/blinders]

A neat proof of why they do not converge uniformy to zero.

By the way, many of you are intent on proving 1 = -1, which is true in GF(2) anyway.

Oh, the convergence [i]is[/i] uniform on $[0,\infty).$ Even with my blinders on, I'm not wrong about that.

You mean given any $\epsilon > 0$, you can give a $c$ such that that function lies within an $\epsilon$ band around $0$ for [i]all[/i] $x$?

I don't see that. :maybe:

$0\le ce^{-cx}\le c$ for all $x\in [0,\infty)$. So, $c=\epsilon$ works.

Here's one written by Yakov Smirnoff:

$\frac{1}{-1}=\frac{-1}{1}$
$\sqrt{\frac{1}{-1}}=\sqrt{\frac{-1}{1}}$
$\frac{\sqrt{1}}{\sqrt{-1}}=\frac{\sqrt{-1}}{\sqrt{1}}$
$\frac{1}{i}=\frac{i}{1}$
$-i=i$
$-1=1$
But in Soviet Russia, proof writes YOU!

i think the best false proof is the one of all triangles being isosceles (and hence equilateral)

the flaw is quit hard to spot too, unless you know some geo properties :D

(it begins with taking the intersection of angle bisector and perpendicular bisector)

I saw that one:

[hide="from Advanced Euclidean Geometry by Posamentier"]

Take a scalene triangle ABC.
Draw the bisector of C and the perpendicular bisector of AB.
From their intersection G, draw perpendiculars to AC and CB.
Those perpendiculars have their feet at D and F respectively.
Because angle ACG= angleBCG and right angle CDG=right angle CFG,
CDG=CFG (SAA congruence)
Then DG=FG and CD=CF.
Because AG=BG and ADG and BFG are right angles, DAG=FBG.
Therefore DA=FB.
It then follows that AC=BC.
[/hide]

[quote="CircleSquared"]I saw that one:

[hide="from Advanced Euclidean Geometry by Posamentier"]

Take a scalene triangle ABC.
Draw the bisector of C and the perpendicular bisector of A.
From their intersection G, draw perpendiculars to AC and CB.
Those perpendiculars have their feet at D and F respectively.
Because angle ACG= angleBCG and right angle CDG=right angle CFG,
CDG=CFG (SAA congruence)
Then DG=FG and CD=CF.
Because AG=BG and ADG and BFG are right angles, DAG=FBG.
Therefore DA=FB.
It then follows that AC=BC.
[/hide][/quote]I found the flaw!!!   :) [hide]Angles don't have perpendicular bisectors[/hide]

:roll:
I meant angle bisector?

EDIT: Okaaay, never mind.
I don't know what happened, so I'm going back to fix it.

[hide]g is outside triangle abc so the last step is kinda bs?[/hide]

[annoyed]
Ahh snap junggi found the flaw!
 :mad:  :mad:
[/annoyed]

Ok, I'm done.  Junggi wins.
I'm amazed, that's probably the fastest a proof on this thread has been destroyed.

dv=dx implies v=x?

and for the geo one, i think they actually intersect on the circumcircle

[quote="lingomaniac88"][quote="lingomaniac88"][hide="WARNING!  Calculus content ahead."]Evaluate $\int\frac{dx}{x}$.
Use integration by parts to solve.
Let $u=\frac{1}{x}$
Therefore $du=\frac{-dx}{x^{2}}$
Let $dv=dx$
Therefore $v=x$
Substituting in $\int udv=uv-\int vdu$
You obtain $\int\frac{dx}{x}=\left(\frac{1}{x}\right)(x)-\int(x)\left(\frac{-dx}{x^{2}}\right)$
Which yields $\int\frac{dx}{x}=1+\int\frac{dx}{x}$.
Subtract the integral from both sides and that gives: $1=0$.[/hide][/quote]

Nobody's found the flaw in this proof yet, or so it seems...[/quote]

Well, since they are indefinite integrals, I don't think you are allowed to do that...
I think that merely says that $C_{1}= C_{2}+1$, where the c's are the constants of the integrals.

[quote="lingomaniac88"]
Let $dv=dx$
Therefore $v=x$
[/quote]

I don't know anything about calculus, but that step still seems suspicious.

Well, it's easy to prove if you consider all constants to be equivalent.

Edit: so how does integration by parts work for definite integrals?

this is the easiest proof

Find a time machine and go back in time to make the prehistoric people think that the character 1 is the same as the character 0 and viola

Assume, for sake of contradiction that $1 \ne 0$.

Lemma: mathematics is consistent.

Proof: Assume, for sake of contradiction, that mathematics is inconsistent. In that case, Chewbacca lives on Endor. But Chewbacca is a wookie - only ewoks live on Endor! Since wookies are not ewoks, this is a contradiction. Thus, mathematics is consistent.

Now, since mathematics can do arithemtic and is consistent, Godel's Second Incompleteness Theorem implies that it is impossible to prove that mathematics is consistent. However, by the proof of the lemma, it is possible to prove that mathematics is consistent. This is a contradiction.

Thus, $1 = 0$.

Edit: I just realized that the above proof might be invalid. So I proved it's definitely valid:

Assume that the above proof is invalid. But then Chewbacca lives on Endor. This is a contradiction. Thus, the above proof is valid.

Boy, I haven't been this confused since that time I was insulted by William Shakespeare. *start Family Guy flashback*

Look at the monkey! Look at the cute little monkey!


limit
real analysis
real analysis unsolved

For large $ n$, $ \sqrt [n^{2}]{7^{n} \plus{} ( \minus{} 1)^{n}n3^{n} \plus{} n^{5}}\approx 7^{\frac {1}{n}}$, so we can disregard that term.  Expanding $ \sqrt [3]{n^{6} \plus{} 1}$ about the point at infinity we find $ \sqrt [3]{n^{6} \plus{} 1}\approx n^2 \plus{} \frac {1}{3n^4} \plus{} \mathcal{O}(n^{ \minus{} 10})$, so the limit in question is $ \frac {1}{3}$.


[size=150]Let ABC is a triangle,prove that:
$ \frac {3\sqrt {3}}{16}\frac {(b \plus{} c)^2}{s}\geq w_a$.
<=>
$ x,y,z>0$,then:
$ 27(z \plus{} 2x \plus{} y)^6 \geq 1024(z \plus{} x)(x \plus{} y)(y \plus{} z \plus{} x)^3x$
BQ[/size]

[size=150]I do not know how to ask questions,
But so 1F show,
Background problem is a good  problem.
BQ[/size]

[quote="xzlbq"][size=150]Let ABC is a triangle,prove that:
$ \frac {3\sqrt {3}}{16}\frac {(b \plus{} c)^2}{s}\geq w_a$.
<=>
$ x,y,z > 0$,then:
$ 27(z \plus{} 2x \plus{} y)^6 \geq 1024(z \plus{} x)(x \plus{} y)(y \plus{} z \plus{} x)^3x$
BQ[/size][/quote]

[size=150]$ k_a \equal{} m_a\frac {2bc}{b^2 \plus{} c^2}$,(what is k_a name? in english)

$ \frac {3\sqrt {3}}{16}\frac {(b \plus{} c)^2}{s}\geq k_a$.
<=>
$ x,y,z > 0$,then:
$ 27(2x \plus{} z \plus{} y)^4((z \plus{} x)^2 \plus{} (x \plus{} y)^2)^2 \geq 256(2(z \plus{} x)^2 \plus{} 2(x \plus{} y)^2 \minus{} (y \plus{} z)^2)(z \plus{} x)^2(x \plus{} y)^2(x \plus{} y \plus{} z)^2$[/size]


I just remembered that I forgot to sign the aime.  Is that REALLY bad?  :(

Yes.  You'll have to contact the AMC.  They'll (I am guessing) will fax you a form to sign or something.


A certain fraction r is represented in base b by 0.111111.....while in base 2b it takes the simpler form 0.2b.what is r?

Find the largest integer walue of n so that 3^n divides (25!+26!)

No idea what to do with these..ask for helps//

First Question: Decimals in base b are simple enough: any digit x places to the right of the decimal point is in the $ b^{\minus{}x}$ place. For example, $ .1111..._b\equal{}b^{\minus{}1}\plus{}b^{\minus{}2}\plus{}b^{\minus{}3}\plus{}...\equal{}\frac{b^{\minus{}1}}{1\minus{}b^{\minus{}1}}\equal{}\frac 1{b\minus{}1}$, using the formula for an infinite geometric sum. Do the same thing for $ .2b_{2b}$, which is easier since there's only two digits, and set the 2 fractions equal. b comes out nicely (it should be an integer) and the fraction r follows.

Second Question: Factor gcd(25!, 26!)=25! out of the expression 25!+26! to get 25!(1+26)=25!(27). So now it simply becomes finding the exponent of the largest power of 3 that divides 25!, and adding 3 to it.


Answer the following questions.

(1) Find the maximum value of $ f(x)\equal{}x\plus{}\frac{1}{x}\ (x<0)$.

(2) Find the minimu value of $ g(x)\equal{}x\plus{}\frac{5}{x\minus{}2}\ (x>2)$.

(3) Find the minimum value of $ h(x)\equal{}2x\plus{}\frac{1}{x^2}\ (x>0)$.

(4) Find the maximum value of $ i(x)\equal{}\frac{x}{x^2\plus{}1}$.

How did you solve? and what's the values of $ x$ which give the maximum and minimum values?

(1) $ x \equal{} \minus{}a ; a > 0 \rightarrow x\plus{}\frac{1}{x} \equal{} \minus{}(a\plus{}\frac{1}{a}) \leq \minus{}2$
with equality case : $ x \equal{} \minus{}1$
(2) $ g(x) \equal{} (x\minus{}2)\plus{}\frac{5}{x\minus{}2}\plus{}2 ; x\minus{}2 > 0$
$ \therefore g(x) \geq 2\sqrt{5}\plus{}2$
with equality case : $ x \equal{} 2\plus{}\sqrt{5}$
(3) $ h(x) \equal{} x\plus{}x\plus{}\frac{1}{x^2} \geq 3$
with equality case : $ x \equal{} 1$
(4) $ i(x) \equal{} \frac{x}{x^2\plus{}1} \leq \frac{1}{2} \leftrightarrow \frac{\minus{}(x\minus{}1)^2}{2(x^2\plus{}1)} \leq 0$ which is true
$ \therefore$ maximum value of $ i(x)$ is $ \frac{1}{2}$
with equality case : $ x \equal{} 1$

That's right. :)

For (4) It is suffucient to think the case of $ x>0,\ i(x)\equal{}\frac{x}{x^2\plus{}1}\equal{}\frac{1}{x\plus{}\frac{1}{x}}\leq \frac{1}{2\sqrt{x\cdot \frac{1}{x}}}$.


geometry
parallelogram
analytic geometry

Parallelogram $ ABCD$ has vertices $ A(0,0), B(2,1), C(3,1)$. State the ordered pair which locates vertex $ D$.

hello, $ \overline{BC}$ is parallel to the x-axes, so must be $ \overline{AD}$ also parallel to $ \overline{BC}$ and $ |BC|\equal{}|AD|\equal{}1$, from here we get that the point $ D$ must have the coordinates $ D(1,0)$.
Sonnhard.


this problem is similary arosisi's problem


$ x(x\minus{}3)\plus{}1\equal{}\dfrac{\sqrt{5\plus{}4x}\plus{}3}{2}$

$ x\equal{}?$

good luck!

$ x\in \left(\minus{}\frac{5}{4},0 \right)\cup (3,\plus{}\propto )$

$ \\ x(x\minus{}3)\plus{}1\equal{}\frac{\sqrt{5\plus{}4x}\plus{}3}{2}  \Leftrightarrow ... \\ 2x^2\minus{}6x\minus{}1 \equal{} \sqrt{5\plus{}4x} \Leftrightarrow \\ x^4\minus{}6x^3\plus{}8x^2\plus{}2x\minus{}1\equal{}0  \Leftrightarrow \\ (x^2\minus{}2x\minus{}1)(x^2\minus{}4x\plus{}1)\equal{}0$

$ \\ x \equal{} 1\plus{}\sqrt{2}$ or $ x \equal{} 1\minus{}\sqrt{2}$ or $ x \equal{} 2\plus{}\sqrt{3}$ or $ x \equal{} 2\minus{}\sqrt{3}$

BUT $ 0 <2\minus{}\sqrt{3}<1\plus{}\sqrt{2}<3$

SO $ x \equal{} 1\minus{}\sqrt{2}$ or $ x \equal{} 2\plus{}\sqrt{3}$


probability
function
complementary counting

Three cards are chosen at random from a standard 52-card deck.  What is the probability that they are not all the same color?

I don't understand this question as well. I think the main problem that I didn't understand what this mean $ \binom{x}{y}$ I never face such a fraction like.

$ \binom{x}{y}$ is NOT a fraction.  It denotes the number of ways to pick $ y$ things from a group of $ x$.  Its definition is:

$ \binom{x}{y}\equal{}\frac{x!}{y!(x\minus{}y)!}$.  Usually we use $ \binom{n}{r}$, not $ \binom{x}{y}$.

I think I should read about $ !$ function and $ \binom{n}{r}$ thing also.
OK do I need a calculator to solve such a thing $ \binom{n}{r}$?

No.  Unless it's a really big one such as $ \binom{93}{42}$.

We can cancel out terms with $ x!$ and $ y!$ usually, so the division is usually easier.

This is impossible because there are only two colors in a standard deck.  How can you have 3 different colors if there are only 2?

We want the probability that they are not [b]ALL[/b] the same color...

I tried to make this problem harder than it is by thinking of facecards as neither red nor black, since they are usually multicolored.

Couldn't you do complementary counting? $1-(2*26*25*24/52*51*50) $ by choosing one of the two colors to have all 3 cards of that color, and then there are $26*25*24/52*51*50$ ways to choose 3 cards from that color.

Could someone tell me why the answer isn't simply $1-\frac{1*25*24}{2*51*50}=\frac{15}{17}$? Thanks!

I did $\dfrac{52}{52}\cdot \dfrac{25}{51}\cdot \dfrac{24}{50}=\dfrac{4}{17}$, then subtracted from one to get $\boxed{\dfrac{13}{17}}$

@vacation9, that doesn't work because you have to multiply by $2$ (there are two colors).

[hide=A different solution]There are only two ways this can happen:

1) Two blacks cards, one red card
2) Two red cards, one black card

These two cases are just similar to each other, just switched the colors. It suffices to only solve one of them, and just multiply by 2.

The number of ways of getting two black cards is $\binom{26}2$, and the number of ways of getting one red card is $26$, so the number of ways of getting two black cards and one red card is $\binom{26}2\cdot26$.

Since the number of ways to get any three cards is $\binom{52}3$, the answer is:
$$\frac{\binom{26}2\cdot26\cdot2}{\binom{52}3}=\boxed{\frac{13}{17}}$$


a,b are the legs of one right triangle and c is the hypotenuse. prove that c3>a3+b3(c cuber is greater than a cube + b cube)

[hide]
This can also be solved by the fact that $ sin^3\alpha \plus{} cos^3\alpha <\equal{} sin^2\alpha \plus{} cos^2\alpha \equal{} 1$
Think How??  :)  :D [/hide]

[hide]Let A be the angle between b and c. (a^3+b^3)/c^3=(a/c)^3+(b/c)^3=sin(A)^3+cos(A)^3 and proceed as madness said.[/hide]


quadratics
algebra
difference of squares
special factorizations

(r + 1) (r + 2) (r + 3) (r + 4) -24

Can somebody tell me how to factor this expression completely without expanding it?

Remark $ 1\plus{}4\equal{}2\plus{}3$, and you are to rearrange the expression as the product of two groups of some quadratic expressions.

thank you for the idea, i seem to kinda got it, wait i'll try to solve it hehehe  :D

Oh, I noticed that $ 4! \equal{} 24$, yielding the result : the given expression has factor of some quadratic expression.

Im sorry but I don't seem to get it. what's with 4!??
Im sorry but Im not good at math  :(

O.K. :lol:

Let $ f(r)\equal{}(r\plus{}1)(r\plus{}2)(r\plus{}3)(r\plus{}4)\minus{}24$, if we solve the equation: $ f(r)\equal{}0$, since $ f(0)\equal{}1\cdot 2\cdot 3\cdot 4\minus{}24\equal{}0$ and $ f(\minus{}5)\equal{}(\minus{}4)(\minus{}3)(\minus{}2)(\minus{}1)\minus{}24\equal{}0$ to get $ f(r)\equal{}r(r\plus{}5)Q(r)$ where $ Q(r)$ is a quadratic quotient.

hehe thank you for explaining it to me because i somehow understand what are you trying to say hehe what a clever trick of getting its factors in your own unique manner without expanding.. truly im impressed  :lol:

Your original idea was better kunny.

\[ (r\plus{}1)(r\plus{}4)(r\plus{}2)(r\plus{}3) \minus{} 24 \equal{}\]
\[ (r^2 \plus{} 5r \plus{} 4)(r^2 \plus{} 5r \plus{} 6) \minus{} 24 \equal{}\]
\[ (r^2\plus{}5r\plus{}5)^2 \minus{} 1 \minus{} 24 \equal{}\]
\[ (r^2\plus{}5r)(r^2\plus{}5r\plus{}10).\]

Difference of squares in the 2nd and 3rd lines.

Hmm, this reminds me of the homework question that we had in the AMC 12 class.

Question: \[ \text{Compute: } \sqrt{(31)(30)(29)(28)\plus{}1}\]

Hey, I was looking at that earlier. Great minds think alike  :lol:

[b]Kunny's [/b]Solution on post # 6 is pretty quick. [b]MellowMelon's [/b]solution is also good, but in his line 2 and 3, expansion did occur but it was not shown. The factoring of a biquadratic (through the difference of 2 squares) emerged beautifully. Hey, just an opinion though. . . .

[quote="gfour84"]the homework question[/quote]

Followup question (semi-giveaway) Prove that $ r(r\plus{}1)(r\plus{}2)(r\plus{}3)$ is never a perfect square for all positive integers $ r$.

Hmmmm, well mine is in that form, but, it is a perfect square just by adding one.  :wink:

[quote="1=2"][quote="gfour84"]the homework question[/quote]

Followup question (semi-giveaway) Prove that $ r(r \plus{} 1)(r \plus{} 2)(r \plus{} 3)$ is never a perfect square for all positive integers $ r$.[/quote]

It's always one less than perfect square, follows from MellowMelon's solution.

[quote="gfour84"]Yea, and from my question  :lol:[/quote]

Your question was just one example. and it's nowhere said it's integer

My point was from all that was already said its kind of obvious already (and, I know, it does not have to be a integer).


function
combinatorics proposed
combinatorics

Let $a_1<a_2<\cdots<a_n$ be positive integers. Some colouring of $\mathbb{Z}$ is periodic with period $t$ such that for each $x\in \mathbb{Z}$ exactly one of $x+a_1,x+a_2,\dots,x+a_n$ is coloured. Prove that $n\mid t$.

[i]Andrei Radulescu-Banu[/i]

We may assume we're working in $\mathbb Z_t$, with the statement of the problem slightly modified to say that for all $x\in\mathbb Z_t$, exactly one of $x+a_i$ is colored. Assume the residues $u_1,\ldots,u_p$ are colored. The number of groups of the form $x+a_1,\ldots,x+a_n$ which contain $u_j$ is the same for all $j$, and each such group (there are $t$ of them, depending only on $x$) contains some $u_j$, which means that we have $t=np$ and we're done.


topology
real analysis
real analysis unsolved

Let $ A \subset \mathbb{R}^d$ and $ s \ge 0$ such that $ \mathcal H^{s*}(A) < \infty$.  Prove that $ \mathcal H^{t*}(A) \equal{} 0$ for every $ t > s$.

I read the other link, and I understand the connection between lebesgue measure and hausdorff outer measure...but I cannot reproduce the proof that you are hitting at.  do you think that you can be slightly more explicit?

thanks!


[quote="shadysaysurspammed"]If the area of a right triangle is equal to its perimeter,find the sides.[/quote]

[hide][size=150]6,8,10[/size]
i tried solving algebraically except it didnt really work

 :? [/hide]

[quote="shadysaysurspammed"]If the area of a right triangle is equal to its perimeter,find the sides.[/quote]

too complicated to do wi/ algebra...

5, 12, 13

As we can see in the above posts, there is not a unique solution

$a^2 = b^2+c^2$

$\mathbb{S} = sr = 2s \Rightarrow r=2$

There are infinite right triangles with $r=2$

I found that, given the length of $b$ we can find $c$ from $c = 4 \cdot \frac{b-2}{b-4}$

And then $a = \sqrt{b^2+c^2}$

Here are some almost-integer examples:

$\\ 4,1 - 84 - 84,1 \\ 4,2 - 44 - 44,2 \\ 4,4 - 24 - 24,4 \\ 4,5 - 20 - 20,5 \\ 4,8 - 14 - 14,8 \\ 5 - 12 - 13 \\ 5,6 - 9 - 10,6 \\ 6 - 8 - 10 \\ 6,5 - 7,2 - 9,7 \\$

Yes there is not just one answer ,the first 2 posts combined hgive the right answers  but method is more important and there is a straight forward algaebric method.

This problem was discussed several times (all in Intermediate). Hence, it's locked. Please use search button before posting.  :)


Suppose that $p$ and $q$ are positive numbers for which $\log_9(p) = \log_{12}(q) = \log_{16}(p+q)$.  What is the value of $q/p$?

[hide]
Let $a$ be a real number so that $9^a=p$, $12^a=q$, and $16^a=p+q$
Note that $(p)(p+q)=q^2$, so $p^2+pq=q^2$.
Dividing by $p^2$, we have $1+\frac{q}{p}=\frac{q^2}{p^2}$.
If we let $x=\frac{q}{p}$, then we have $x^2-x-1=0$, so $x=\frac{1\pm\sqrt{1^2+4}}{2}$.
Since $p$ and $q$ are both positive, reject the negative root, so $\frac{q}{p}=\frac{1+\sqrt{5}}{2}$.
[/hide]

Amazing how that golden ratio just pops up everywhere isnt it.

Cool, that is the Golden Ration. I didn't notice that.


on page thirty it introduces Stewart's theorem and then they show this thing with the angle bisector theorem at just before the example. Could someone explain why that is done and how the equation is derived?

Are you asking about the length of the angle bisector? If so...(not sure if this is the proof in AoPS 2)

So we have $ man \plus{} dad \equal{} bmb \plus{} cnc$.

From the Angle Bisector Theorem, $ \frac bn \equal{} \frac cm\implies bm \equal{} cn$, so substituting this, we have

$ man \plus{} dad \equal{} (bm)b \plus{} (cn)c \equal{} (cn)b \plus{} (bm)c \equal{} bc(m \plus{} n) \equal{} abc$, so dividing both sides by $ a$, we have $ d^2 \plus{} mn \equal{} bc$.

Now, notice that $ m \equal{} \frac {ac}{b \plus{} c}$ and $ n \equal{} \frac {ab}{b \plus{} c}$ (which can be obtained from setting $ \frac bn \equal{} \frac c{a \minus{} n}$ and solving for $ n$, and then subtracting from $ a$ for the value of $ m$), so substituting, we have

$ d^2 \plus{} \frac {a^2bc}{(b \plus{} c)^2} \equal{} bc\implies d^2 \equal{} bc \minus{} bc\cdot\frac {a^2}{(b \plus{} c)^2} \equal{} bc\left(1 \minus{} \frac {a^2}{(b \plus{} c)^2}\right)$ (the form $ d^2 \equal{} bc \minus{} mn$ is also important; it's easier to remember as well).

Also, the book says $ \frac{m}{m\plus{}n}\equal{}\frac{c}{b\plus{}c}$. Why is this true(because that makes the solution shorter) ?

Since $ \frac{a_1}{b_1}\equal{}\frac{a_2}{b_2}\equal{}\cdots\equal{}\frac{a_n}{b_n}\equal{}\frac{a_1\plus{}a_2\plus{}\cdots\plus{}a_n}{b_1\plus{}b_2\plus{}\cdots\plus{}b_n}$ (discussed in Volume 1 I think? If not, just set the original equality chain to $ r$, add all of them, then isolate the $ r$ to get the last expression), $ \frac bn\equal{}\frac cm\equal{}\frac{b\plus{}c}{m\plus{}n}\implies\frac c{b\plus{}c}\equal{}\frac m{m\plus{}n}$.


There are $n$ coin in the pile. Two players play the game by alternately performing a move. A move consists of taking $5,7$ or $11$ coins away frome the pile. The unable to perform a move loses the game. Which player – the one playing first or second – has the winning strategy if:
a) $n=2001$
b) $n=5000$

Second player wins if n=2001, and first player wins if n=5000.
The key is to think mod 16. Together after each player has had one turn, the second player can always guarantee the removal of 16 coins unless the first player removed 7 chips first, in which case the second player can remove 7 (to make 14 in total) or 11 (to make 18). But whenever the first player removes 7 first twice, the second player can ensure 32 coins being removed in 4 turns. So since 2001==1 (mod 16). The second player does the following:

If first player removes 5, second player removes 11;
if first player removes 11, second player removes 5;
if first player removes 7 (for the (2n+1)th time), second player removes 7;
if first player removes 7 (for the (2n)th time), second player removes 11;

When n=5000==8(mod 16), first player first removes 7, then it's the same to part 1), which guarantees the victory for the first player. :)

Ηι!

With this strategy we remove in total 1984 coins, that is there are 17 remaining coins

What if player-1 take 7 coins? There are now 10 coins, I can get just 5, and the player-1 finally wins because he'll take the remaining 5 coins

[quote]With this strategy we remove in total 1984 coins, that is there are 17 remaining coins

What if player-1 take 7 coins? There are now 10 coins, I can get just 5, and the player-1 finally wins because he'll take the remaining 5 coins[/quote]

I can't seem to understand your question. If player 1 now takes 7 from the remaining coins, since 17==1 (mod 16), player 2 will take away 7 as I argued above. Now there are 3 coins left. So player 1 loses.

Stupid of me  :oops:

Of course, I can take 7!


[geogebra]eb6f3e86bf50f693d77cefe8ddc7fb28d5e1c4a7[/geogebra]
We have triangle $ ABC$. The intersection points of the angle bisectors and the circumcircle are $ E$, $ D$ and $ F$. Prove that the angle bisectors of $ ABC$ are the altitudes of $ EFD\triangle$.

Cool applet!

[hide]Since AD bisects angle BAC, we know m<BAD=m<DAC.  A, B, C, D, E, and F are concyclic, so we know that the measure of arc BD = measure of arc DC.  Thus arc BD = arc DC = 1/2 arc BC.  Similarly, arc CE = arc EA = 1/2 arc CA and arc AF = arc FB = 1/2 arc AB.  Let H be the intersection of AD and EF.  Note that m<ADF + m<DFC + m<CFE + m<FHD = 180.  Substituting with inscribed angles, 1/2(arc AF) + 1/2(arc DC) + 1/2(arc CE) + m<FHD = 180.  Substituting from our arc equations, 1/4(arc AB) + 1/4(arc BC) + 1/4(arc CA) + m<FHD = 180.  Factoring, 1/4(arc AB + arc BC + arc CA) + m<FHD = 180.  Since arc AB + arc BC + arc CA = 360, we know m<FHD is 90.  Thus HD an altitude of DEF.  By similar argument, the other angle bisectors contain the other altitudes of DEF.[/hide]


Well done guys, Fiachra your score of 34 is brilliant ( another Irish record )Ireland's first silver and so close to Gold and Seamus almost got his HM. Can we claim Martin Orr's score as well ?   Galin will have a few challenges to break the records over the next few IMO competitions, but I am sure he will.
Glad to see the eye of the hurricane stayed NE of Merida and that you are all safe and well.

 :)

Congrats to Fiachra also! It is an impresionable score, even if it's not a gold medal  :( Anyway the hurricane was fun  :roll:  :lol:

Thanks guys! If only id multiplied across by 6 to clear the modular inverses, id have got number 4, but i thought i was on completely wrong track!


topology
advanced fields
advanced fields unsolved

Let $X$ be a metric space, $S\subset X$ is a compact set and $f: S\rightarrow S$ is an isometry.
Prove that $f$ is surjective  :)

Assume the contrary. Let $x\in S\setminus f(S)$. Since $f$ is an isometry $f(S)$ is compact. Therefore,
\[r = d(x, f(S)) \equiv \inf\{d(x,y) | y\in f(S)\}= \min\{d(x,y) | y\in f(S)\}> 0. \]
Consider sequence $x_{k}$ defined as follows: $x_{0}= x$, $x_{k+1}= f(x_{k})$. We get
\[d(x_{0}, x_{k}) \geq d(x, f(S)) \geq r. \]
Since $f$ is an isometry (for $k>2$)
\[d(x_{1}, x_{k}) = d(f(x_{0}), f(x_{k-1})) = d(x_{0}, x_{k-1}) \geq r. \]
Similarly, for every $k\neq l$, $d(x_{k}, x_{l}) \geq r$. Therefore, the sequence $x_{k}$ does not have a limit point. So $S$ is not compact.


Ok,

What happened to all the spam?

[color=#F8F8F8] spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam 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spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam 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spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam 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spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam 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spam spam spam spam spam spam spam spam spam spam spam [/color]

[quote="buzzer11"]poor missouri....[/quote]

you're one to talk

i konw, but i cant help feeling bad... maybe

REVIVAL.........

[url]http://www.youtube.com/watch?v=tbt_PuVAVTU[/url]

It's themed to the "IN 2007"  JibJab song.

I'm not really sure what happened to all the spam, but it seems to be gone now, especially with the infusion of plenty of people who are legitimately good at math and are willing to share their knowledge with less experienced folk.

On a sidenote, there is something wrong with the OP's post, and it is very large. I would advise someone, perhaps a moderator or admin, to look into this when possible. It could be a SQL injection of some sort.

[quote="hrithikguy"]
On a sidenote, there is something wrong with the OP's post, and it is very large. I would advise someone, perhaps a moderator or admin, to look into this when possible. It could be a SQL injection of some sort.[/quote]

I lol'd.

[quote="hrithikguy"]people who are legitimately good at math[/quote]

oxymoron

Hmmm so people [b]by definition[/b] cannot be good at math?

Any predictions as to who will:

a. Make mc nats from missouri next year
b. Probably make USA(J)MO winners/hm from missouri next year
c. Revive this forum

[quote="AwesomeToad"][quote="MCL"]Any predictions as to who'll be the next Victor Wang?[/quote]

In all seriousness, I don't think anyone else will be exactly like Victor Wang for a while. What an odd comparison.[/quote]
Quite odd indeed. Let's make an even comparison instead then: Any predictions as to who'll be the next Ince twins?

Why is this sudden revival wanted by you?

[quote="MCL"]Honestly, some people need places to spam.[/quote]

*redirects viewers to my blog*


inequalities
geometry
circumcircle
inradius
trigonometry
probability
inequalities unsolved

Let $ABC$ be a triangle with sides $a,b,c$,circumradius $R$ and inradius $r$. Prove that $\frac{R}{2r} \geq (\frac{64a^2b^2c^2}{(4a^2-(b-c)^2)(4b^2-(c-a)^2)(4c^2-(a-b)^2)})^2$

im very sorry in case this prob is already posted. tried searchin 4 it. in case it is , pls send the link.
thanks

Problem proposed by M.Tetiva in AMM journal.

hey minh can,
could u pleees post the solution to this problem, or mayb a link ot its solution?
pleees do so.
in dire need of help.
thanks

hi venteka:
i saw the problem couple months ago and was not immensely interested in solving it. my interest intensifies recently but the deadline for submitting a solution to editors of The Monthly has passed. i suggest that you talk directly to Marian Tetiva and ask him to give you a hint or a full solution to this new problem. in the meantime i am going to work it out myself. but i think that you need to use a lot of substitution to simplify the right hand side or translate the ineq into all sine or cosine ineq where you can use jensen or am-gm ineq. but i promise not giving it up and a solution should show up on mathlinks screen pretty soon ! (cheers) :wink:  :P

hey zhaobin,
thanks a lot. :omighty:
i hope i got ur solution.
btw, i guess in ur solution, at few places, the squares of expressions have been missed, though the overall seems to fall in place. :maybe:
theres a high probability im wrong( as in most cases  :wink: )
once again thanx.


geometry parallelogram vector trigonometry

In the following parallelogram with screwed up letter labelings, let $\overrightarrow{OA}$ and $\overrightarrow{OB}$ be $\overrightarrow{a}$ and $\overrightarrow{b}$ respectively. Then $\overrightarrow{a}\cdot\overrightarrow{b}=OB(OX)$ using definition of a dot product and simplification.

Then the AoPS book says that this dot product is the length of $\overrightarrow{b}$ times the [i]length of the projection of OA onto OB[/i]. What does that mean? (The projection is segment OX).

If you resolve vector OA into two perpendicular components, one parallel to OB (OX) and one perpendicular to OB (XA) then the one parallel to OB is the perpendicular projection of OA onto OB.

Another way to think of it:
lf x is the measure of the angle AOB, then the magnitude of the dot product of vector OA and OB is:
OB * OA * cosx

Thanks to your explanation, I realized that I interpreted the diagram totally wrong.  :blush:

And also, the fact that the product is OB(OX) comes from the fact that

$AO(OB)(\cos{\angle{AOX}})=(AO)(OB)\frac{OX}{AO}=(OB)(OX)$.


Given a positive real number $a$, prove that

$\frac{1}{\sqrt[4]{a}+\sqrt[8]{a}+1}+\frac{1}{\sqrt[4]{a}-\sqrt[8]{a}+1}-\frac{2(\sqrt[4]{a}-1)}{\sqrt{a}-\sqrt[4]{a}+1}=\frac{4}{a+\sqrt{a}+1}$


Find all $ \ f: R\to R$ such that:
 $ \ f(x)\equal{}f(y) \Leftrightarrow x\equal{}y;$
$ \ f(\frac{x\plus{}y}{x\minus{}y})\equal{}\frac {f(x)\plus{}f(y)}{f(x)\minus{}f(y)}$ for all $ \ x\neq y$.

Let $ x\ne0$ and $ y\equal{}0$.
$ f(x)(f(1)\minus{}1)\equal{}f(0)(1\plus{}f(1))\Rightarrow f(1)\equal{}1,f(0)\equal{}0$.
Obviously $ f$ is odd.

$ \frac{f(xy)\plus{}f(y)}{f(xy)\minus{}f(y)}\equal{}f(\frac{x\plus{}1}{x\minus{}1})\equal{}\frac{f(x)\plus{}1}{f(x)\minus{}1}\Rightarrow f(xy)\equal{}f(x)f(y)$. (*)

Since $ f(x^2)\equal{}f(x)^2$, $ f(x)>0$ when $ x>0$.
Then for $ x>1$,
$ f(\frac{x\plus{}1}{x\minus{}1})\equal{}\frac{f(x)\plus{}1}{f(x)\minus{}1}>0\Rightarrow f(x)>1$

By (*), $ f$ is increasing and $ f(x)\equal{}x^r$ when $ x>0$, where $ r$ is real. It is not hard to get $ r\equal{}1$.

Hence, $ f(x)\equal{}x$.


limit
inequalities
number theory
real analysis
real analysis unsolved

Hope you'll enjoy it.
Let $f : \mathbb{N} \longrightarrow \mathbb{R}$ such that $f ( m + n ) \leq f ( m ) + f ( n )$ for all $m, n \in \mathbb{N}$
Prove that
\[ \ell = \lim_{n \longrightarrow \infty} \frac{f ( n )}{n} = \inf_{n = 1, 2, \ldots} \frac{f ( n )}{n}  \]
Do you see any application to number theory ?  ;)

Fix two coprime positive integers $p,q\in\mathbb N$. If $n$ is large enough, it will be possible to write it as $ap+bq$ for positive integers $p,q$. Let $n=a_np+b_nq$ be the expression of $n$ which achieves the largest coefficient for $p$. In other words, if $n=a_np+b_nq=up+vq$ for positive integers $u,v$, then $a_n\ge u$. It's easy to see that when $n\to\infty,\ \frac{a_n}{b_n}\to\infty$, which means that $n\to\infty$ implies $\frac{a_nf(p)+b_nf(q)}n=\frac{a_np}{a_np+b_nq}\cdot\frac{f(p)}p+\frac{b_nq}{a_np+b_nq}\cdot\frac{f(q)}q\to\frac{f(p)}p\ (*)$. Since $\frac{f(n)}n\le\frac{a_nf(p)+b_nf(q)}n,\ (*)$ shows that $\limsup_{n\to\infty}\frac{f(n)}n\le\frac{f(p)}p$, and since $p$ was arbitrary, we're done: we get precisely $\limsup_{n\to\infty}\frac{f(n)}n=\inf_p\frac{f(p)}p$ (actually, we get $\le$ instead of equality, but the reverse inequality always holds, so we do get equality).

I was rather thinking of applications to number theory (actually of Schnirelmann theorem) than of a number theory flavoured proof  :)
Anyway, your proof is very nice (and quite unexpected). Mine is different, do you want it ?

Let $\ell = \lim_{n \rightarrow \infty} \frac{f ( n )}{n}$
Obviously (excluding the case $\ell = - \infty$, but then the result is clear),
\[ \ell = \lim_{n \rightarrow \infty} \frac{f ( n )}{n} \geq \inf_{n = 1, 2, \ldots} \frac{f ( n )}{n}  \]
Moreover for all positive integer $k, n$ we have

\begin{eqnarray*} \frac{f ( kn )}{kn} & \leq & \frac{kf ( n )}{kn}\\ \frac{f ( kn )}{kn} & \leq & \frac{f ( n )}{n} \end{eqnarray*}

Hence

\begin{eqnarray*} \lim_{k \rightarrow \infty} \frac{f ( kn )}{kn} & \leq & \frac{f ( n )}{n}\\ \ell & \leq & \frac{f ( n )}{n} \end{eqnarray*}
Since $n$ was arbitrary we have $\ell \leq \inf_{n = 1, 2, \ldots} \frac{f ( n )}{n}$.
Therefore
\[ \ell = \lim_{n \rightarrow \infty} \frac{f ( n )}{n} = \inf_{n = 1, 2, \ldots} \frac{f ( n )}{n}  \]


Suppose that the tower is 100 feet east of the road, that you are going 60 feet per second (roughly 40 miles per hour), and that at time t = 0, you are 1000 feet south of the point on the road that is due west of the tower.

1) Develop an equation that gives your distance d from the tower as a function of t. (Express d in feet and t in seconds.)

2) Equations of the form ay^2-bx^2=1 where a and b are postive numbers, give graphs that have a shape called a hyperbola.

Explain, using ideas about transformations of functions, why the graph of your equation in question 1 is a branch of a hyperbola.

I am so lost. Please help!!!

To simplify, let  let the tower be at $(100,0)$ and place you start (t=0) is at $(0,-1000)$ , at the time $t$ you are at $y=-1000+60t$  (x cordinate is zero, of course.
If $z$ is the distance from the tower, at t, then (apply pythogorian relationship)

$z^{2}= (-1000+60t)^{2}+100^{2}$
This of course is like a hyperbola: (

(The general form like $\frac{z^{2}}{a^{2}}-\frac{(t-k)^{2}}{b^{2}}= 1$
[hide](for example a, here is 100  . b would be  (5/3) ...etc   [/hide]


Find the number of subsets $ B$ of the set $ \{1,2,...,2005\}$ such that the sum of the elements of $ B$ is congruent to $ 2006$ modulo $ 2048$.

$ 2^{2005 - 11}$ since there is a unique subset of $ S = \{ 2^i | 0 \leq i < 11 \}$ that must be taken together with any arbitrary subset of $ \{ i | 1 \leq i \leq 2005 \} \backslash S$ for the sum to be congruent to some constant mod $ 2048 = 2^{11}$


conics
parabola
calculus
function
derivative
analytic geometry
graphing lines

Hey all, this is my first post here.  I'm working on some math homework for my calculus class at a community college (I'm in dual-enrollment), and I'm stumped.  That's what has brought me to your online community.

I'll give you the problem as it is stated in my homework:

A car is traveling at night along a highway shaped like a parabola with its vertex at the origin.  The car starts at a point 100 m west and 100 m north of the origin and travels in an easterly direction.  There is a statue located 100 m east and 50 m north of the origin.  At what point on the highway will the car's headlights illuminate the statue?

So, to solve it, I obviously have to find a tangent to the curve at a point such that the tangent also goes through the point where the statue as.

If the parabola has its vertex at the origin and goes through the point (-100, 100), then the equation for the function must be y=(x^2)/100.  Right?

Since this is actually a homework problem and I should be doing it myself, let's not work on it but a variation.  Let's say the equation is y=x^2 and the point where the statue is is (1, 1/2).

I've had some guidance with this problem so far, and this is far as I've gotten.  I should express the point on the curve at which this tangent is as (x, x^2).  Then, I should plug that along with the point (1, 1/2) with the derivative (y'=2x) as the slope into the point slope formula.  Then I should be able to solve for x.  That is the x-coordinate of where this tangent line is tangent to the curve y=x^2.  Is that right?

I've worked through it several times and can't seem to get a right answer.  On my latest try I ended up with x=1/2 and x=3/2.  Neither of the lines tangent to the curve at those points go through the point (1, 1/2).

So what am I doing wrong?  As I said, the first time I introduced this problem above is how it is stated in the book.  So, that's what I'll be working on, but so that y'all don't just tell me the answer to my homework problem, let's work on the hypothetical situation I gave with y=x^2, (1, 1/2), etc.

I'd really appreciate any help you could give me!!

[quote="carmike692000"]I've had some guidance with this problem so far, and this is far as I've gotten.  I should express the point on the curve at which this tangent is as (x, x^2).  Then, I should plug that along with the point (1, 1/2) with the derivative (y'=2x) as the slope into the point slope formula.  Then I should be able to solve for x.  That is the x-coordinate of where this tangent line is tangent to the curve y=x^2.  Is that right?[/quote]

That's how u do the problem, but i get different x-coordinates:

[hide]Using the point-slope form, $x^2-\frac{1}{2}=2x(x-1)$ and moving everything to the right get $x^2-2x+\frac{1}{2}=0$. From quadratic formula get $x=1\pm \frac{\sqrt{2}}{2}$.
Of course, the greater $x$ would mean the headlights would have to be pointing backwards, which probably won't happen. The reasonable value for $x$ would be $1-\frac{\sqrt{2}}{2}$.[/hide]

Be careful naming your variables- make sure you're referring to an arbitrary point on the parabola when you write down a tangent line. Algebra is usually the hard part of this sort of problem for most students- if you don't set it up right, you can't solve it.

$y=\frac{x^2}{100}$ is not the only possible parabola, but it's probably the one intended- and you've just posted a question which is equivalent after scaling.

There's generally no rule against getting help on homework; the real intent is that you understand the solution you hand in. If one of my students asked me, I would help with a problem like this.

Thanks for the advice, y'all.  (No question where I'm from, huh?  :D)

I actually have it figured out now.  It seems I was making two mistakes.  First, when I got to (x^2)/100 - 2x = -50 I factored out an x and said x[(x/100) - 2] = -50.  It seems I'm supposed to use the quadratic formula.  I must admit I'm not sure why, but it works out right when I do.  Second, I forgot to plug my x-values back into y' to find the slope of the tangent line.

So, after correcting those two things and trying again, I found that, for the actual problem (y=(x^2)/100), one of the tangent lines is y=0.585786x - 8.578644.  And so, the point at which the cars headlights illuminate the statue would be x=100 - 50sqrt(2), or x=29.28932188.

I appreciate the help, folks.  And I must say this looks to be an interesting forum.  I kind of registered and posted in a hurry, but what is the purpose of this board?  It looks like I'll have to hang around a bit and check out some of the math on here; looks like I could learn alot!

Factoring isn't a bad idea, but if you want to factor to find roots of a polynomial, you need something like $(x-a)(x-b)=0$; $(x-a)(x-b)=50$ doesn't help, since we don't have such a simple characterization of when a product of numbers is 50.

The main purpose of these boards is for discussion of math contest problems at all levels. Many of the site's members are or were among the top competitors at these contests, and we are here because we like challenging problems. These contest problems ignore the boundaries of the normal curriculum to ask interesting questions, and the solution method is not always obvious from the form of the question.
This "Calculus Computations and Tutorials" section was created to handle questions like yours, and not overwhelm the more advanced or tricky material in the other Calculus/Analysis subforums.

As for where you're from? Take a look at all the other flags people fly- another student from somewhere in the US isn't anything special.

jmerry -  Thanks for the clarification.  I think I see now.  My comment about where I'm from was more about me being from the South with my " y'all ".  I guess that doesn't stand out as much in a world-wide community.

I am getting a totally different answer

the equation to the parabola may be y^2 = -100x or x^2 =100y

As you will see that with y^2 = -100x we get the point to be
say A  (-25(1+sqrt(5))^2 , 50(1+sqrt(5))) and
B  (-25(1-sqrt(5))^2 , 50(1-sqrt(5)))
Out of which point A should be neglected since the coordinates are to the west of starting point.
For point B we can say that the statue will be illuminated if the car has bulbs on its rear portion too.

with x^2 = 100y we get the points to be
say C (50(2+sqrt(2) , 25(2+sqrt(2))^2) and
D  (50(2-sqrt(2) , 25(2-sqrt(2))^2)
But with this equation it means as if the body is going towards south. Isin't it.

I have solved them by the steps that follow underneath :-

1. Finding y'. Let the tangent be at point say E(x1 , y1)
    the equation of tangent at E is (Y-y1)=y'(X-x1)

2. Then substituting Y=50, X=100 in above equation.

3. Replacing x1 or y1 as required to get a quadratic equation
    either in x1 or y1.

4. At last solving the quadratic equation to the values of x1&y1.

Comment if I am wrong.

Waiting eagerly for your reply.


[quote="spinos"]Δίνεται ακολουθία συναρτήσεων $ f_n$ και μια συνεχής συνάρτηση $ f(x)$ στο διάστημα [α,β].
Υποθέτουμε ότι αν $ x_n$ που ανήκει στο [α,β] και $ x_n\to\ x$
Tότε $ f_n(x_n)\to\ f(x)$
Δείξτε ότι η $ f_n$ σύγκλίνει ομοιόμορφα στην $ f$[/quote]

Σκέψου τί γίνεται όταν π.χ. η $ x_n$ είναι μία απαρίθμηση των ρητών του [α,β].

Cheerio,

Durandal 1707

Εdit: όχι ότι χρειάζεται να το σκεφτείς έτσι, το παραπάνω ήταν απλώς εποπτικό σχόλειο  :) Η λύση είναι ως εξής:
[hide]Έστω $ \epsilon > 0$ κ΄ έστω ότι η $ f_n$ δε συγκλίνει ομοιόμορφα στην $ f$. Συνεπώς, για κάθε $ n$ θα υπάρχει $ x_n\in [a,b]$ ώστε $ |f_n(x_n) \minus{} f(x_n)| > \epsilon$.
Χωρίς βλάβη της γενικότητας, μπορούμε να υποθέσουμε ότι $ x_n\to x \in [a,b]$ (περνώντας π.χ. σε μία συγκλίνουσα υπακολουθία της $ x_n$). Τότε όμως, $ |f_n(x_n) \minus{} f(x_n)|\leq |f_n(x_n) \minus{} f(x)| \plus{} |f(x) \minus{} f(x_n)|$, οπότε η υπόθεση και η συνέχεια της $ f$ για $ n$ αρκετά μεγάλο ώστε τα παραπάνω να φράσσονται από το $ \epsilon/2$ σε οδηγούν σε άτοπο.
[/hide]


function
MATHCOUNTS
least common multiple

Today in math club, we had a question that was somewhat like this:

"You have 500 stamps of each with values 1 cent through 9 cents. How many ways can you have stamps so the overall value is 10 cents?"

At school, they just taught the boring-old-way of listing it all out. If the problem were extended to 50 cents, or 100 cents, it couldn't be done like that easily. Does anyone have a good way to use generating functions for it?

Thanks.

ido remember a certain conversation about a similar problem. it required a really long and complicated generating function "thing", but i don't remember what it was.
anyone?

[hide="With generating functions..."]The coefficient of $ x^{10}$ in the expansion of $ (1\plus{}x\plus{}x^2\plus{}\cdots\plus{}x^{500})(1\plus{}x^2\plus{}x^4\plus{}\cdots\plus{}x^{1000})(1\plus{}x^3\plus{}x^6\plus{}\cdots\plus{}x^{1500})\cdots(1\plus{}x^9\plus{}x^{18}\plus{}\cdots\plus{}x^{4500})$ or $ \prod_{i\equal{}1}^{9}\left(\sum_{j\equal{}0}^{500}x^{ij}\right)$.[/hide]

That function alone probably wouldn't help, since only a computer could figure that. Syncing it with the Multinomial Theorem might help.

That is what I got, but is there a way to do that without a computer?

Could you eliminate anything > x^10 since its going to give a coefficient higher than the deisred one.

@mihail: that one was simpler in a way actually; i did that manually, but this one is too hard

[quote="#H34N1"]Could you eliminate anything > x^10 since its going to give a coefficient higher than the deisred one.[/quote]

Not a larger coefficient, but a larger exponent.  And yes, of course you can (although you should be a little careful about it).  On the other hand, the other thing you can do is to extend the sum to infinity.  That will let you express the generating function in a nice form, and might facilitate the attempt to find some of its coefficients.

So it would then become $ (1\plus{}x\plus{}x^2\plus{}x^3\plus{}\dots\plus{}x^{10})(1\plus{}x^2\plus{}x^4\plus{}\dots\plus{}x^{10})(1\plus{}x^3\plus{}x^6\plus{}x^9)\dots$, finding the coefficient of $ x^{10}$?

JBL, yeah that's what I meant

In the following link, I showed how to solve a general case of a related problem:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=161837[/url].

The proof didn't use generating functions, although I could have expressed it using generating functions if desired.

For your problem, if you were patient enough, you could develop a closed-form formula for the case of $ n$ cents in a similar way.  The formula would depend on the value of $ n$ mod 2520, the least common multiple of 1, 2, ..., 9.

Considering it was actually a Mathcounts target round, you are under I think 3 minutes? time constraint. I used the function that I posted earlier to solve it, but my friend listed it all out and got 56, wheras the coefficient is somewhere in the 40s. :huh:


Show that for all $x,y$, $2xy\le{x}^2+y^2$. Use this and \[x=\frac{x_i}{\sqrt{x_1^2+x_2^2}},y=\frac{y_i}{\sqrt{y_1^2+y_2^2}}\]with $i=1,2$ to prove the basic schwarz inequality: $x_1y_1+x_2y_2=\sqrt{x_1^2+x_2^2}\sqrt{y_1^2+y_2^2}$. Then prove that equality holds if and only if there exists a $\lambda$ such that $x_1=\lambda{y}_1$ and $x_2=\lambda{y}_2$. (edit: There are many ways to prove the inequality, but use this method to prove the second part too.)

Yes, this is a homework problem, but I already did it. :D

I tried solving it right when you posted it, but I can only show the first step.  :blush:

We can set that y = x + a for some real a.

Then 2xy = 2x(x + a) = 2x^2 + 2xa <= 2x^2 + 2xa + a^2 = x^2 + (x + a)^2 = x^2 + y^2

edit: fixed it

Can you clarify a few things? It says to use $x=...$ and $y=...$ with $i=1,2$; does this mean you sum the two terms (i.e., $x=\dfrac{x_1+x_2}{\sqrt{x_1^2+x_2^2}}$), or do you multiply them, or just use them completely separately? Also, you say to prove the basic inequality, but then you only give an equals sign. Is it supposed to be $\le$ or $\ge$?


Represent the number $989 \cdot 1001 \cdot 1007 +320$ as a product of primes.

Wow... how did you come up with that? Did you use computer calculations? Nice solution anyway.

Well I will admit I calculated N and then looked for some of its prime divisors.  I had a hunch that since it had to be proven that it was composite that its primes were somehow related to the the other numbers.  So I tried a general form of
$N = (p_1+k_1)(p_2+k_2)(p_3+k_3)$ and looked for possible integer values of $k_i$ that would satisfy it.   What I posted then was the resulting solution.  I couldn't find a general method of that type for solving the k's analytically such that it would work for any given number of a certain kind.

[quote="Anonymous"]I had a hunch that since it had to be proven that it was composite that its primes were somehow related to the the other numbers.[/quote]
We can strengthen the above into a "criteria" to generate problems similar to this.
[quote="Anonymous' Criterion"]
The resulting divisors to be found each have constant distance from one of the original three "large numbers"
[/quote]
If the problem was designed to be solved without a calculator, the above would be useful to at least check for a solution.
Take $ p_i$ as before. Generalize the constant term $ 320$ to $ k$. Define $ r_i$ as reasonable numbers such that\[ p_1p_2p_3\plus{}k\equal{}r_1r_2r_3\]
Then, to find a linear relation we subtract the quantity $ p_1$ from each of the subscripted variables and suppose equality still holds.
Then, $ k\equal{}(r_1\minus{}p_1)(r_2\minus{}p_1)(r_3\minus{}p_1)$
But $ p_1$ was arbitrarily chosen; thus, similar equations holds for $ p_2,p_3$.

Specifically in this problem, we can find a quicker solution by looking at divisors of $ 320$:
\[ 320\equal{}2*8*20\equal{}\minus{}10*\minus{}4*8\equal{}\minus{}16*\minus{}10*2\]
Where for each $ i$, $ a_i,b_i,c_i$ are in arithmetic progression.

We could formalize these observations (via interpolation) into a proof of the identity:
\[ a(a\plus{}12)(a\plus{}18)\plus{}320\equal{}(a\plus{}2)(a\plus{}8)(a\plus{}20)\]simply by noting that we have found three values of $ a$ for which it holds, and the coefficients of $ a^3$ on either side are equal
[hide="proof"]
Suppose $ L(a)\equal{}LHS\not\equal{}R(a)\equal{}RHS$. Then, let $ f(a)\equal{}L(a)\minus{}R(a)$. Since the leading coefficients of $ L,R$ are equal, $ deg(f)\equal{}2$. Also, $ f(A_1)\equal{}f(A_2)\equal{}f(A_3)\equal{}0$ for three (distinct) constants that can be found from our three triples ( specifically, 0,-12,-18 ).
Thus \[ |\{r_i|f(r_i)\equal{}0\}|>deg(f)\implies f\equiv 0\]
So $ LHS\equal{}RHS$
[/hide]

If we denote $x=991$ we get:

$(x-2)(x+10)(x+16)+320$

$=x^3+24x^2+108x$

$=x(x+6)(x+18)$

$=991\cdot 997\cdot 1009$

Since all three numbers are prime, we are done.