modular arithmetic
number theory
Diophantine equation

I got this problem in my Algebra 1 class yesterday:

My lucky number is 34,117. It is a 2N+1 digit number. The first N digits are double the last N digits. The first N+1 digits are three times the last N+1 digits minus 10. Find another number with the same properties.

I just got it after working with it for 4 hours or more. I won't blow the answer, but it was too large for a standard calculator. I reduced it to an equation, but had to modify for every change in the number of digits, and had to divide by 17 around 30 times before I found my answer. does anyone know an easier/faster way to do this? (Perhaps a 17 divisibility rule or an easier equation to work with? )

Here's a 17 divisibility rule:
Test for divisibility by 17. Subtract five times the last digit from the remaining leading truncated number. If the result is divisible by 17, then so was the first number. Apply this rule over and over again as necessary.
Example: 3978-->397-5*8=357-->35-5*7=0. So 3978 is divisible by 17.

The one thing I don't understand is that it states that the first $ N$ digits are double the last $ N$ digits.  When $ N\equal{}1$, this is not the case.

Also, is the $ N$ referred to for a "$ 2N\plus{}1$" digit number different from the first $ N$ digits?  I'm a bit confused.  If you could clarify I could probably help you.

Oh, I see.

I can't really help you much more, but you could assign variables to digits and solve a Diophantine equation?  Good thing it's a relatively simple problem, as all you have to do is find another number with the same properties.  If the problem was more complicated, like for example, find a general formula for a lucky number with proof.  It would be a lot harder.

What is the significance of 17 in all this?  What was your reasoning?

I'm sorry I can't provide a solution, I have to leave right now, but when I get back or tomorrow I'll come and try this problem out.

I couldn't find any other 5 digit ones or any 7 digit ones, I'll try more tomorrow.

Call the first $ N$ digits an $ N$ digit number $ a$, let $ c$ be the $ N\plus{}1$th digit (single digit) and call the last $ N$ digits $ b$. By the two conditions, we get the following.

$ 2b\equal{}a$
$ 10a\plus{}c\equal{}3(10^{N}\cdot c\plus{}b)\minus{}10$

Note that in the second equation, the $ LHS$ is the first $ N\plus{}1$ digits as a number while $ 10^{N} \cdot c \plus{} b$ is the last $ N\plus{}1$ digits. Substituting and rearranging gives the following.

$ 17b\plus{}10\equal{}(3\cdot10^{N}\minus{}1)c$

However the restriction also applies that $ 5\cdot10^{N\minus{}2}\le b <5\cdot 10^{N\minus{}1}$ in order for $ a$ to have $ N$ digits.

Sorry, I don't know if my algebra is right. But from here you can come up with numbers by plugging in $ N$ and testing out some $ b$ and $ c$.

[quote="SnowEverywhere"]$ 10a \plus{} c \equal{} 3(10^{N}\cdot c \plus{} b) \minus{} 10$

[/quote]

This is the equation I was using- but the answer is very large, and for every $ N$, I had to divide by 17 for c=1 and c=2 (These seemed to be the most likely values, although 3 was possible). I tried using a modular system  (by dividing into $ 3(10^N)\minus{}1$)to check for divisibility, but it didn't help with time, since I was only dividing into two numbers anyway. Anyone who knows how to make a better mod system, please tell me! :)

Simplify it the way I did and it becomes a lot easier. From there, you can just plug in $ N$ and solve using the chinese remainder theorem. This theorem pretty much states that for some $ N$ you should just make the following modular equation.

$ 17b \equiv \minus{}10 \pmod {3(10^{N})\minus{}1}$

Now plug in the desired value for $ N$. Solve for congruence in this mod. Then determine congruence for $ c$ mod $ 17$ (same method). Then see if it works for the restrictions I placed in the last post.

[quote="Silent Soliloquy"][quote="TrigMan"]Here's a more clarified version:

My lucky number, 34117, is a 2N+1 digit number, where N = 2.The first N digits are twice the last N digits (34 = 2*17). The first N+1 digits are three times the last N digits minus ten (341 = 117*3 - 10). Find another number with these properties.

Note: The number can be of any odd (2N + 1) digit length, such as 1234567.

The 17 rule really helps with the method I was using, thanks for the help :).[/quote]

How does that make sense,
"The first N+1 digits are 3 times the last N digits minus ten."

You're told N = 2.

N+1 = 3.

N+1 = 3, N = 2. You wrong (First 3 digits) = 3*(Last 3 digits) - 10. You wrote the question wrong?[/quote]

Sorry- It was supposed to say "The first N+1 digits are three times the last N+1 digts minus 10" Sorry for the confusion  :oops:


trigonometry
geometry
trig identities
Law of Sines

Quadrilateral ABC has AB = BC = CD, angle ABC = 70 degrees and ange BCD = 170 degrees. What is the degree measure of angle BAD?

Draw the diagonals of the quadrilateral and label thier intersection E.

[img]http://img139.imageshack.us/img139/7425/quadrilateraltw6.png[/img]

By knowing about isosceles triangles and that the sum of the angles in a triangle is 180, one could fill in the above angles easily. By the law of sines, sin60/AB = sin65/AE, or AE = ABsin65/sin60. Similarly, sin60/CD = sin115/ED, or ED = CDsin115/sin60. It is given that AB = CD, we know that sin60 = sin60, and sin115 = sin 65 because they have the same reference angle and they are both positive. Therefore, AE = ED, and triangle AED is isosceles. That means angle CAD = 30. Adding 55 and 30 together, we get the measure of angle BAD which is 85.


The real numbers $ a,b$ fulfil the conditions
(i) $ 0<a<a\plus{}\frac12\le b$;
(ii) $ a^{40}\plus{}b^{40}\equal{}1$.
Prove that $ b$ has the first 12 digits after the decimal point equal to 9.

[i]Mircea Fianu[/i]


now jee has come real close and its givin me some sleepless nites

i need u ppls xpert advice

i am able to consistently score around 300 to 350 in past years iit papers.

chem and phy give gud returns but maths sux big time.

i got admission in the best class in pune, m prakash academy, but wasted this chance :(

i didn't join any test series like fitjee/brilliants and relied only on the mock test our class conducts and solving past year papers.

i get around 70 in each chem paper, 60 in phy and only 40 in maths.

that gives net total of 340 marks

i have tried very hard to improve maths but all's vain

is there some way to improve maths score in 10 days ?

otherwise with my horrible score i may not feature even in EML

i am not concerned abt chem and phy coz incho prep has helped me a lot and my phy is also pretty gud

[hide]
i know my question is stupid like the ones on goiit, sorry.
but all u guys on aops are imo quality so u can surely help me
[/hide]

Correct me if I'm wrong, but from what I know 340 is an EXCELLENT score? Cutoffs are under 50% or something?

usually expected marks at jee are 60% to 80% of what one scores while solving past year papers or while giving mock tests  :maybe:

this is told by my teachers and it is most likely true coz they have been teaching since past 5 to 10 yrs so it is not just some guess, but verified statistically

even if difficulty level is exact same, other factors come into picture while giving the real exam

and

[hide]
einstein (vinit) has been scoring 400+ in these same tests   :o
[/hide]

Lol someone reply. I am in the same (infact even worse) boat as sagar. I can get good score in math only if i get 2 hours. :(

well if sm1 gets 430 in JEE,how can he get just 60% of wat he gets in practice papers?? :rotfl:  :rotfl:

sum ppl hv minds of steel

others like me crack at time of exam  :P

sagar dude ask your friend vinit how he gets this many marks. And post it here too. I know there is more than just IQ involved for scoring good marks.

@conjurer

lol given at einstein's score at olympiads, i thought everone on aops must have realised that he is supernormal
[hide]
perhaps abnormal  :rotfl:
[/hide]

[hide]
he also scores around 380 to 400 in fitjee tests  :o
[/hide]

[quote="Conjurer"]I know there is more than just IQ involved for scoring good marks.[/quote]
actually no!!!
if u can get ur guesses rite and u hav the iq, u kno nthing more than basic learning ,even  4 absolute mugging questions.... :)

I don't really have the IQ to comprehend what was just said! :O


[quote="kubus"]Well, in the first case I think the weight must be the same, because the total gravity force remains and therefore the box must push the scale (or ground or whatever) with the same force = same weight. This is possible because the air pushed down by the bird pushes on the bottom of the box.

In the second case, the weight can change, it will be probably something between the weight of the cage and the weight of the cage+bird, depending on the specifics of the construction and bird flight.[/quote]
For the 1st one I agree with Kubus and for the 2nd one I agree because the air is not trapped it is free to move in and out of the cage so the force exerted will be a bit less.

yup, i agree with you 2, riddler and kubus..... the only thing is ,in the 1st one the air would be pushed down , ok , but it can be that instead of the air applying a force downwards on the cage , it just gets $compressed$ sort of you know , so the molecules come closer to each other rather than pushing each other dowenwards, in that case it wouldnt be the full weight of the bird that would be acting  on the floor of the box.....dunno if this is true......... :oops:  :oops:

in the second one i totally agree with you 2 ;)

In the first condition,i agree with the thory but just a couple of dumb doubts came up in my mind...

When the bird is hovering,it means it's flapping its wings and pushing the air down with each push.....so wont the weight on the scale change due the momentary pressure of the air hitting the base and then come back to normal and then change.....keep repeating??

And if the bird is hovering,then how is its weight being counted???? :huh:  :huh:

$Yohoho,thisismy300thpost$ all : omighty: me : agent:  : w00t:

[quote="warut_suk"]We put a closed glass-case with air and a small bird inside on a scale. It is obvious that the scale indicates the weight of the bird too, when it is sitting on the bottom of the case. What is the combined weight if the bird is hovering in the mid-air inside the case? What is the weight if we use a cage instead of the closed glass-case?[/quote]Assume the box (cage) weighs nothing, then all you have is the weight of the bird on the scale.  If the bird is flying, does the scale still weigh it?  The scale registers the weight of the bird resting on its pan. If the bird is flying, its wings generate lift. Initially this lift counteracts the bird's weight, so that the bird appears to weigh nothing according to the scale. After the bird lifts off from the scale, the scale has truly nothing to weigh.

[quote="Dr. No"]After the bird lifts off from the scale, the scale has truly nothing to weigh.[/quote]

Well, I think it does. It weighs the increased rebouncing of air molecules (pressure) on the bottom of the cage, due to the bird's flight.

Let's take the whole cage&bird as an object of the system. There is one external force acting on it, namely the gravity (weight) force, even though only the bird has non-zero mass. (Actually there must be some air or something keeping the bird up but let's not talk about this.) In order for this object to remain with zero acceleration on the scales, there MUST be another force opposite to the gravity. That is (besides buyoancy of air - neglect for now) the force coming from the scales! Which must be equal to the weight of the cage&bird. That practically means that the scales SHOW the actual weight of the bird&cage, regardless of where the bird is (if it doesn't move with acceleration)...

[quote="shadysaysurspammed"]In the first condition,i agree with the thory but just a couple of dumb doubts came up in my mind...

When the bird is hovering,it means it's flapping its wings and pushing the air down with each push.....so wont the weight on the scale change due the momentary pressure of the air hitting the base and then come back to normal and then change.....keep repeating??

And if the bird is hovering,then how is its weight being counted???? :huh:  :huh:[/quote]
The avg weight of the bird remains the same.
And the air is applying the pressure on the scale, do you get me?

[quote="bus"]I totally agree with kubus but there's one question coming to my mind, perhaps you'll be able to answer kubus.... if you try to explain this with the air pressure approach (and not as an isolated system), there must be some velocity with which the air pressure propagates down to the bottom of the case - the pressure won't get down there instantly but only after a while (the pressure change will be kind of a wavefront). However, the weightlessness of the bird will get down to the scale much faster than the pressure change - so what happens in between?

[hide="Here is what I think but think about it by yourself first."]Well - I think that the scale will indeed show less weight for a while - because the center of gravity of the case moves downwards. Downwards? Yes, downwards - part of the gravitational force is now not being exerted on the scale but rather being transformed into acceleration.
So, that's just one thing I've realized when reading your last post....[/hide][/quote]

This is what i thought and said in my earlier post that the scale will show a bit less due to the pressure thing....riddler,that concept is fine but read bus's post,that is what i've been trying to say.

[quote="bus"]I totally agree with kubus but there's one question coming to my mind, perhaps you'll be able to answer kubus.... if you try to explain this with the air pressure approach (and not as an isolated system), there must be some velocity with which the air pressure propagates down to the bottom of the case - the pressure won't get down there instantly but only after a while (the pressure change will be kind of a wavefront). However, the weightlessness of the bird will get down to the scale much faster than the pressure change - so what happens in between?

[hide="Here is what I think but think about it by yourself first."]Well - I think that the scale will indeed show less weight for a while - because the center of gravity of the case moves downwards. Downwards? Yes, downwards - part of the gravitational force is now not being exerted on the scale but rather being transformed into acceleration.
So, that's just one thing I've realized when reading your last post....[/hide][/quote]

Well, I'd answer this in a very similar way. Neglecting the bird's take-off (which makes a real mess on the scale I guess;), supposing it is held in the mid-air and someone just turns the gravity on, it will start to push air downwards. Which in total will yield acceleration g of the centor of mass of the box&contents. But this pressure increase will reach the bottom very quickly and there are many effect that make this detailed discussion quite void - the vortices, wing flapping, movement of the bird which is not just hovering there, etc.

[quote="bus"]Well, I was just wondering what would happen in the first few nanoseconds :) after the bird starts hovering - the pressure explanation clearly shows that the scale would [i]have to[/i] show zero (if we consider the cage weightless) at least for a while.
In the light of this finding, think about what happens if the bird instead of starting to hover starts to fly upwards - the center of gravity of the case will move upwards and the scale should show more weight that it did before; however, it also has to show zero (at least at the beginning). So where's the truth?[/quote]

When the bird flaps up,all the air goes down,so all the mass is down,so wont the center of mass go down? :huh:

[quote="bus"]Well, I was just wondering what would happen in the first few nanoseconds :) after the bird starts hovering - the pressure explanation clearly shows that the scale would [i]have to[/i] show zero (if we consider the cage weightless) at least for a while.
In the light of this finding, think about what happens if the bird instead of starting to hover starts to fly upwards - the center of gravity of the case will move upwards and the scale should show more weight that it did before; however, it also has to show zero (at least at the beginning). So where's the truth?[/quote]

Well, you are talking about the first nanoseconds because you want to disregard the effect of the bird on the box. If you really want to do that, everything looks much simpler to me - you consider just the bird and the surrounding molecules of air, not thinking about whether they're in a cage or in a freezer or where. The Newton's Xth law must apply to them also (considering them as a 'closed' system) - their center of mass must accelerate downwards with g. If you want the bird to fly, it must send the molecules down with higher acceleration than g. If you want it to go upwards, it must kick the molecules even harder. Therefore I think the center of mass of the box does not really move upwards, and the scale shows zero. Of course, this is until the molecules sent down by the bird (well, effectively it's just the impulse, not the physical molecules) start to push against the scale.

Is this what you were asking or did I misunderstand?

look, is it true that  is in  the air then the  force by which the bird is pushing the air is $exactly$ equal to the force the air pushes the scale down ???

if its true then, then i would say that after an initial lessening of the weight ( this is due to when the impulse is being transferred by the air to the scale , as bus said) the scale WOULD show the combined weight of the whole system......

Yes, shadysaysurspammed and kubus, you are both right, I thought I'd catch you on this one :).


the secondary structure is the coiling or folding of its chains.  coiling occurs, froming ahelix due to the formation of hydrogen bonds between the nitrogen of one amino acid and the oxygen of another location in another part of the polypeptide chain.
Folding occurs due to hydrogen bonding between different polypeptide chanis lying side by side.

the Teritery Structure is caused by weak interactions among R groups. and between R groups and the fluid Enviroment.

[quote="Azat123"]Is it possible to predict the secondary, tertiary and quaternary structure of a polypeptide sequence based solely on the sequence, and environmental information?
  :?  ;)[/quote]

I believe IBM's Blue Gene supercomputer is designed just for this function in mind.

The quick answer is yes.

The better answer is we can't quickly (or slowly, in most instances) make predictions that are accurate in the real world.  The structures are highly dependent on the history of the material in many instances.

The example I know is that the change in the heme molecule that causes the folding of the red blood cell resulting in early death is a change of a carboyl COOH for the defective OH  group.  Many of the new drugs would be more effective if we could predict with ease where the reactive groups would be to bind with the body's receptors.

Maybe one of you will lead the way to do this better.


function
algebra
domain
polynomial
geometry
3D geometry

Find all the functions from $N$ to $R$ that satisfy this two proprieties:

$f(x + 22) = f(x) \forall x \in N$
$f(x^2y) = (f(x))^2f(y) \forall x, y \in N$

I'm very bad at these, but let's give it a try

[hide]The first thing to is to look at constant functions. Let $f(x)=c$ for all $x$. Then we get, from the second condition, $c=c^3$ or $c=0,1,-1$. Obviously, all three of these solutions work. Let's now plug in $y=0$ into the second condition.

We end up getting $f(0)=f(x)^2f(0)$. Case 1: if $f(0) \neq 0$, then $f(x)^2=1$ for all x, or $f(x)=1, f(x)=-1$ for all $x$. From the first condition and the fact that the domain is the integers, we need something like $f(x)=1$, $f(x)=-1$, $f(x) = 1$ for $x$ odd, $-1$ for $x$ even or vice versa. We've covered the first two, and the last two don't work with the second condition.

Case 2: $f(0)=0.$ But then, $f(22k)=0$ for all $k$. So $f$ can't be a polynomial function... And here's where I realized that we're looking for functions, not polynomials. Oops. Some help?
[/hide]

Well, this function can only spit out 22 values.  So I guess we could do this on a case-by-case basis, though that'd take more time than I care to spend on this problem.

That observation should help, but I don't know how to use it for now. Maybe Andreas has a solution?

Is $x=0$ included in the domain of the function? I'm not sure what $N$ is, and even if it is the set of all natural numbers, $0$ is sometimes included anyways...

It is not mine  :blush:
The function is periodical of period 22 so we can consider it as a function on the integers $mod 22$. $f$ can take at most 22 different values.
If $x = y$ in the second we have $f(x^3) = [f(x)]^3$.
Since the values in the imagine are at most 22 we must have $f(x) \in \{-1, 0, 1\} \forall x$.
So we have $f(x^3) = f(x)$
Values of the cubes $mod 22$:
$1^3 = 1
2^3 = 8
3^3 = 5
4^3 = 20
......$
Now we can write $f(2) = f(2^3) = f(8)$. In this way we can divide the values of $f$ in these groups:
$1 = f(1)
a = f(3) = f(5) = f(-7) = f(9)
b = f(7) = f(-9) = f(-3) = f(-5)
c = f(2) = f(8) = f(6) = f(-4)
d = f(4) = f(-2) = f(-8) = f(-6)
e = f(10)
f = f(11)
g = f(-10)
h = f(-1)
i = f(0)$
From the second equation we can write these:
$1 = f(1) = f(45) = f(9^2 \cdot 5) = f(3)^2f(5) = a^3 \Rightarrow a = 1
h = f(-1) = f(9^2 \cdot (-5)) = f(3)^2f(-5) = f(-5) \Rightarrow b=h
1 = f(45) = .........
b = \pm 1
c = e
d = g
f \in \{0, 1\}
d = c^2
i = cf
c = c^2b
f = bf$
In this way we have four indipendent values:
$1 = f(1) = f(3) = f(5) = f(-7) = f(9)
\pm 1 = b = f(7) = f(-9) = f(-3) = f(-5) = f(-1)
c = f(2) = f(8) = f(6) = f(-4) = f(10)
c^2 = f(4) = f(-2) = f(-8) = f(-6) = f(-10)
f^2 = f = f(11)
cf = f(0)$
with $c = bc^2$ we have $c = 0$ or $b = c$, and $f = bf$ so $f = 0$ or $b = 1$
The value in the first line are the odd squares $(mod 22)$, in the second there are the odd non-squares, then the even non-squares and the even squares.
The divisibility for $11$ and $2$ looks interesting.
The values that $b, c, f$ can have are:
1.$b = 1,c = 1,f = 1$
2.$b = 1,c = 0,f = 1$
3.$b = 1,c = 1,f = 0$
4.$b = -1,c = -1,f = 0$
5.$b = 1,c = 0,f = 0$
6.$b = -1,c = 0,f = 0$
We must prove if these functions make true the equations.
These functions can espressed as:
$\chi_{2}(n) = \{\stackrel{0 if 2 | n}{1 otherwise}$
$\chi_{11}(n) = \{\stackrel{0 if 11 | n}{1 otherwise}$
$\chi_{q}(n) = (0, if 11 | n), (1 if n is a perfect square not multiple of 11), (-1 otherwise)$
In this way the functions are written:
1.$f(n) = 1$
2.$f(n) = \chi_{2}(n)$
3.$f(n) = \chi_{11}(n)$
4.$f(n) = \chi_{q}(n)$
5.$f(n) = \chi_{2}(n)\chi_{11}(n)$
6.$f(n) = \chi_{2}(n)\chi_{q}(n)$
Now it is necessary to verify that $\chi_{2}, \chi_{11}, \chi_{q}$ satisfy the equations of the problem, and we must prove that if $f(n)$ and $g(n)$ are two solutions, then $f(n)g(n)$ it is too.

So finally the solutions of the equation are:

	these six functions
	these six functions with change sign
	the function equal to zero
for a total of $13$ solutions.

I hope it is clear :)


Let $ a,b,c$ such that :$ \frac {1}{a^2 \plus{} b^2 \plus{} 1} \plus{}\frac {1}{b^2 \plus{} c^2 \plus{} 1} \plus{} \frac {1}{c^2 \plus{} a^2 \plus{} 1}\ge 1$.

Prove that $ ab \plus{} bc \plus{} ca\le3$

[quote="thuyanh158"]Let $ a,b,c$ such that :$ \frac {1}{a^2 \plus{} b^2 \plus{} 1} \frac {1}{b^2 \plus{} c^2 \plus{} 1} \plus{} \frac {1}{c^22 \plus{} a^2 \plus{} 1}\ge 1$.

Prove that $ ab \plus{} bc \plus{} ca\le3$[/quote]

It's easy :)
By cauchy-swarchz:
$ 1 \le \sum \frac {1}{a^2 \plus{} b^2 \plus{} 1} \le \sum \frac{c^2\plus{}2}{(a\plus{}b\plus{}c)^2}$
So:$ ab\plus{}bc\plus{}ca \le 3$

[quote="caubetoanhoc94"][quote="thuyanh158"]$ \sum \frac {1}{a^2 \plus{} b^2 \plus{} 1} \le \sum \frac {c^2 \plus{} 2}{(a \plus{} b \plus{} c)^2}$
So:$ ab \plus{} bc \plus{} ca \le 3$[/quote][/quote]

Why does it hold?

[quote="thuyanh158"][quote="caubetoanhoc94"][quote="thuyanh158"]$ \sum \frac {1}{a^2 \plus{} b^2 \plus{} 1} \le \sum \frac {c^2 \plus{} 2}{(a \plus{} b \plus{} c)^2}$
So:$ ab \plus{} bc \plus{} ca \le 3$[/quote]

Why does it hold?[/quote][/quote]

Because:$ (a^2\plus{}b^2\plus{}1)(1\plus{}1\plus{}c^2) \ge (a\plus{}b\plus{}c)^2$


Let Ia, Ib, and Ic be A-,B-, and C-excenter of triangle ABC, respectively. The bisector of ∠BAC,∠ACB,∠CBA intersects BC,CA,AB at D,E,F, respectively. Draw perpendiculars from vertex D to IbIc, E to IcIa, F to IaIb respectively. prove that the three perpendiculars are concurrent.

The way the problem is stated, it is wrong (note that D, E, F do not really correspond to BAC, ACB, CBA!). Please check it again and fix the bugs.

  darij

[quote="Fang-jh"]Sorry! thank Darij!
Let Ia,Ib,and Ic be A-,B-, and C-excenter of triangle ABC,respectively. the bisector of∠ BAC,∠CBA,∠ACB intersects BC,CA,AB at D,E,F respectively.from vertex D to IbIc,E to IcIa,F to IaIb respectively. prove that the three perpendiculars are concurrent.[/quote]

Come on...
The lines $I_{b}I_{c}$, $I_{c}I_{a}$, $I_{a}I_{b}$ are the external angle bisectors of the angles $BAC$, $CBA$, $ACB$, respectively, and hence perpendicular to the internal angle bisectors of these respective angles. Thus, the perpendiculars to the lines $I_{b}I_{c}$, $I_{c}I_{a}$, $I_{a}I_{b}$ through the points $D$, $E$, $F$ are the internal angle bisectors of the angles $BAC$, $CBA$, $ACB$, and of course these are concurrent.

  darij


Am rezolvat ceea ce m-a interesat din capitolul limite de functii, culegere Ganga, clasa a xi-a. Ma intereseaza rezovarea de la 3 probleme. Multumesc.

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1) problema 6/pag. 286/culegere Ganga/clasa a XI-a (ideea)

Ma intereseaza cum se demonstreaza ca daca  $f:{\mathbb R}\to (0,\infty )$  este strict descrescatoare, atunci exista  $\mathop{\lim }_{x\to \infty }f(x)$  ?

2) (rectificare) problema 9/pag. 286 . Enuntul este:

Fie  $f:{\mathbb R}\to (0,\infty )$  astfel incat  $f(x+1)-f(x)>0,\; \forall x\in{\mathbb R}.$  Sa se arate ca exista  $\mathop{\lim }_{x\to \infty }f(x)$ .

3) 10/286

Fie  $f: [a,b]\to{\mathbb R}$  pentru care exista  $c\in (a,b)$  astfel incat  $\mathop{\lim }_{\begin{array}{l}{x\to c}\\{x<c}\end{array}}f(x)=-\infty$  si  $\mathop{\lim }_{\begin{array}{l}{x\to c}\\{x>c}\end{array}}f(x)=+\infty$ . Aratati ca f nu este monotona pe [\textit{a}, \textit{b}].

\end{document}

[quote="jeanvaljean"]2) (rectificare) problema 9/pag. 286 . Enuntul este:

Fie  $f : \mathbb R \to \left( 0, \infty \right)$  astfel incat $f(x+1)-f(x) > 0, \, \forall x \in \mathbb R$.  Sa se arate ca exista $\lim_{x \to \infty}f(x)$.[/quote]
Tot e greşită. Fie $g : \mathbb R \to \left( 0, \infty \right)$,
\[g(x) = \begin{cases}\frac{\pi}2+\arctan x, \ \ \ \ \ \ \ \ \ \ \forall x \in \mathbb R \setminus \left\{ 1, 2, 3, \ldots \right\}\\ \pi^{e}-\frac1{x}, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{altfel}\\ \end{cases}\]
$\lim_{x \to \infty}g(x)$ nu există.

la problema 2 este greseala de tipar . in loc de 0 este k, k constanta reala.

Am inteles cum se face daca este k real in loc de 0. Dar problemele 1 si 3?

Multumesc

[quote="jeanvaljean"]
1) problema 6/pag. 286/culegere Ganga/clasa a XI-a (ideea)

Ma intereseaza cum se demonstreaza ca daca  $f : \mathbb R \to \left( 0, \infty \right)$  este strict descrescatoare, atunci exista  $\lim_{x \to \infty}f(x)$?
[/quote]
Sirul $\left\{ f(n) \right\}_{n \geq 1}$ este descrescator si pozitiv, deci are limita finita $L \geq 0$. Acuma incearca sa demonstrezi cu ajutorul definitiei ca $\lim_{x \to \infty}f(x) = L$.

[quote="jeanvaljean"]
3) 10/286

Fie  $f: [a,b] \to \mathbb R$  pentru care exista  $c \in (a,b)$  astfel incat  $\lim_{x \to c^{-}}f(x) =-\infty$ si $\lim_{x \to c^{+}}f(x) =+\infty$. Aratati ca $f$ nu este monotona pe $[a,b]$.
[/quote]
1. Exista $p \in (a,c)$ astfel incat $f(p) < f(a)$.
2. Exista $q \in (c,b)$ astfel incat $f(q) > f(a)$.
3. Avem $a < p < q$ cu $f(a) > f(p) < f(q)$, i.e. $f$ nu e monotona.

Va multumesc pentru ajutor. La problema 1),  limita este inff(x) =  L


so i was just thinking about this today.

Joe retains 60% of what he hears. He is given a pop quiz at the end of the class period.  The pop quiz consists of 60% of what he heard in class today.  Without any extra studying or writing notes, what are the odds that Joe scored 100% on the quiz?

I'm not entirely sure, but I think it depends on the number of questions on the quiz. For example:

If there are only 6 questions (chosen out of 10 randomly), then he would have a $ \frac{1}{\binom{10}{6}}\equal{}\frac{1}{210}$ chance.

However, if there are 60 questions (chosen out of 100), then he would have a $ \frac{1}{\binom{100}{60}}\equal{}\frac{1}{13746234145802811501267369720}$ chance.


geometry
geometric transformation
rotation
reflection
Alcumus

George is planning a dinner party for three other couples, his wife, and himself. He plans to seat the four couples around a circular table for 8, and wants each husband to be seated opposite his wife. How many seating arrangements can he make, if rotations and reflections of each seating arrangement are not considered different?  (Note: In this problem, if one seating is a reflection of another, then the two are considered the same!)

Fix one of the men.  Then there are $ 6 \times 4 \times 2\equal{}\boxed{48}$ ways to arrange the rest of the men (they cannot sit across from another man) and the women each only have one choice.

To account for [i]reflections[/i], you have to divide again by $ 2$ to get $ \boxed{24}$.

Wait, doesn't fixing one of the men account for that already?  :maybe:

After doing this problem, my answer was 48. I used a different method that I think accounts for all symmetries. So, because the table is round, the first couple sit and they become reference points for the rest. The second couple has 3 choices where to situate both of them, the third couple has 2, and the fourth couple has 1. So that is 3*2*1=6.  The second, third and fourth couples can switch seats between themselves (the husband can switch with the wife), so I multiplied 6 by 2*2*2, therefore getting 48. The rotations are counted for since the first couple is set as a reference and does not have "4" choices. And I think all other possible over-counting has also been avoided because only three couples can switch between themselves, not four. When all four switch we start over-counting everything by 1, thus we multiplied $2^3$ instead of $2^4$.

I can't see what the "reflections" are in the solution. What caused the reflections? Thanks! :)

It stated reflection are the same in the problem. If it didn't state it, then the answer would be $48$. Let $A, B, C, D$ as men and $1, 2, 3, 4$ as woman and $A$ as the fixed man and couples are $A1$, $B2$, $C3$, and $D4$. Since the problem stated that reflection are the same, $ABCD1234$ is the same as $A4321DCB$. Thus , the answer is $\boxed{24}$. Can someone post the alcumus solution?

[hide=answer here!]24[/hide=answer here!]
this would have been postable if not for the 8 letter thing.

Alcumus, I sometimes feel that your rating system is quite wrong. This Level 23 problem was easier than some Level 15 problems that I've solved. Can someone explain why?

[hide="Easiest solution"]4!=24
[hide="Explanation"](Since the women's seats don't matter as long as their husbands sit somewhere)[/hide]
 [/hide]

[quote=nan0]Potato, but then why don't you account for rotations (dividing by 4)?[/quote]

well, you see, this is what I did in my sol as well. Because you are just choosing the women, and then we know where the men are from that, we don't need to do rotations.

[quote=GTK]I found a better solution:
[hide=My Solution]4! = 24[/hide][/quote]

LOL i think thaz actually it :clap2: good job

[quote=Possible][quote=kidperfekt]1*2*3*4=12[/quote]
Wait wat[/quote]

lol I suck at math

[hide=Solution]
To account for rotations, we fix George to one of the seats (fixing his wife as well). This divides the table into (imaginary) halves. If the husbands all sit together at the top half, there are $3!$ ways to arrange the couples. If one of the wives sits in the top half, then there are $3!\cdot3$ ways to arrange the couples. So, there are $3!+3!\cdot3=\boxed{24}$ seating arrangements George can make.
[/hide]

[quote=bestzack66]Wait isn't this just $4!$? Cuz 4 possibilities for first seating, 3 for second, etc...[/quote]

yea it's $4!$

I just did how many ways can we seat the husbands in a line. (4 factorial). This is because rotations and reflections do not factor in to this problem, and the wives' seats are determined by their husbands (also you could do it the other way around for the same result.)
Surprised this was a level 23 problem! Lol

Yep, just taking $4!$ is valid for this problem. I do see how it might trick some, though. Another valid calculation would be $\frac{6 \cdot 4 \cdot 2}{2}$.


vector
inequalities
linear algebra
linear algebra unsolved

Source: http://arxiv.org/PS_cache/quant-ph/pdf/9711/9711070v2.pdf (relation 8 there)

Let $ \{u_{1},...,u_{n}\}\in C^{n}$ an orthonormal basis of $ C^{n}$ and $ v,v_{1},...,v_{n}$ vectors of norm 1 from $ C^{n}$. If $ |(v_{i},u_{i})|^{2}\geq p$. Prove that $ 2n-2\sqrt{n}\sqrt{p}-2\sqrt{n}\sqrt{n-1}\sqrt{1-p}\leq \sum_{i=1}^{n}\left\Vert v_{i}-v\right\Vert^{2}.$

There's no way to get that inequality without finding the minimum (because we have equality there). Lagrange multipliers aren't the only way to find the minimum, but they are the best way.

Yes, there are cases when we have equality there, so it's tight. What other ways of finding the minimum do you suggest?


[hide="c"] Let $a = 3^{21}$.  We want to find the remainder when $a^{4}+a^{3}+a^{2}+a$ is divided by $3a+1$.  Well, let

$a^{4}+a^{3}+a^{2}+a \equiv R \bmod (3a+1)$

Then

$81a^{4}+81a^{3}+81a^{2}+81a \equiv 81R \bmod (3a+1) \implies$
$(-1)^{4}+3(-1)^{3}+9(-1)^{2}+27(-1) \equiv-20 \equiv 81R \bmod (3a+1)$

It remains to find the multiplicative inverse of $3^{4}\bmod 3^{22}+1$, but this is clearly $-3^{18}$.  Hence

$R \equiv \boxed{ 20 \cdot 3^{18}}\bmod (3^{22}+1)$ [/hide]

We can use
[hide="hint"]
congruences
[/hide]
as follows:
[hide="answer"]
[hide="a"]
Notice that the $81^{21}$, $27^{21}$, $9^{21}$, and $3^{21}$ are all powers of $3^{21}$, and thus
$81^{21}+27^{21}+9^{21}+3^{21}\equiv(-1)^{4}+(-1)^{3}+(-1)^{2}+(-1)(\mod 3^{21}+1)\equiv(0 \mod 3^{21}+1)$
Therefore, $a|81^{21}+27^{21}+9^{21}+3^{21}$
[/hide]
At this point, I got bored and lazy, because the rest of the problem is just computation, so I just stopped. Sorry. Other ppl have answered the question too.
[/hide]


How do you pronounce Rusczyk (sp?) and Lehoczky? :lol:

Those names seem Polish or Russian.  What language is your last name and his, Mr. R?


topology
linear algebra
matrix
advanced fields
advanced fields unsolved

Why a non degenerated 2-form can only be defined in manifolds of even dimension?

for a 2-form $ \omega$ non degeneracy means that the matrix $ \omega (X_iX_j)$ ($ X_i$ is any base of the tangent space at a point) is non singular.
Notice that if the manifold had odd dimension this matrix would be skew-symmetric with odd number of rows and columns and also have non-zero determinant. Why is that a contradiction?


Factorization depends on what field we are working on.


calculus
derivative
function
parameterization
calculus computations

When we want to find the derivative of an implicit function, such as
\[x^{2}+y^{2}=R^{2}\]
we will derive both side and get
\[\frac d{dx}x^{2}+\frac d{dx}y^{2}=0 \]

\[2x+2y\frac{dy}{dx}=0 \]
Thus
\[\frac{dy}{dx}=-\frac{x}{y}\]
Today, I know that we can use the formula
\[\frac{dy}{dx}=-\frac{\partial F/\partial x}{\partial F/\partial y}\]
to find the derivative of the implicit function $F(x,y)=0$. But, why does it work? In addition, does it work for all implicit functions?

Here's what you can get at the elementary level.

If there exists a differentiable function $Y(x)$ satisfying $x^{2}+Y(x)^{2}=R^{2}$ (for example), then $2x+2Y(x)Y'(x)=0$ by differentiation, and $Y'(x)=-\frac{x}{Y(x)}$. That's just the chain rule, and we can do this for any implicitly defined function. If there is more than one such function, the equation we get holds for all such functions- in this case, the top and bottom halves of the circle.
The hole in all of this is that there is no elementary proof that differentiable functions exist satisfying the equation. For that, we need the Implicit Function Theorem of multivariable calculus, which gives us local existence under some reasonable conditions.

[quote="ifai"]Today, I know that we can use the formula
\[\frac{dy}{dx}=-\frac{\partial F/\partial x}{\partial F/\partial y}\]
to find the derivative of the implicit function $F(x,y)$. But, why does it work? In addition, does it work for all implicit functions?[/quote]

Suppose we define $y$ as a function of $x$ implicitly according to the equation

\[F(x,y)=k\]

In order to derive a formula for $\frac{dy}{dx}$ consider $x$ and $y$ as functions of a parameter $t$ so that our equation is now

\[F(x(t),y(t))=k\]

now differentiate this w.r.t. $t$ to get

\[\frac{dF}{dt}=\frac{\partial F}{\partial x}\frac{dx}{dt}+\frac{\partial F}{\partial y}\frac{dy}{dt}=0\Rightarrow \frac{dy/ dt}{dx/ dt}=-\frac{\partial F/\partial x}{\partial F/\partial y}\]

but $\frac{dy}{dx}=\frac{dy/ dt}{dx/ dt}$ for a parametric curve, hence

\[\frac{dy}{dx}=-\frac{\partial F/\partial x}{\partial F/\partial y}\]

Note that $\frac{dF}{dt}$ is [url=http://mathworld.wolfram.com/TotalDerivative.html]total derivative[/url] of $F(x(t),y(t))$.

[quote="benorin"]
...
now differentiate this w.r.t. $t$ to get
\[\frac{dF}{dt}=\frac{\partial F}{\partial x}\frac{dx}{dt}+\frac{\partial F}{\partial y}\frac{dy}{dt}=0\Rightarrow \frac{dy/ dt}{dx/ dt}=-\frac{\partial F/\partial x}{\partial F/\partial y}\]
but $\frac{dy}{dx}=\frac{dy/ dt}{dx/ dt}$ for a parametric curve, hence

\[\frac{dy}{dx}=-\frac{\partial F/\partial x}{\partial F/\partial y}\]

Note that $\frac{dF}{dt}$ is [url=http://mathworld.wolfram.com/TotalDerivative.html]total derivative[/url] of $F(x(t),y(t))$.[/quote]
Does the total derivative mean
\[\frac{dF}{dt}=\frac{\frac{\partial F}{\partial x}dx+\frac{\partial F}{\partial y}dy}{dt}=\frac{\partial F}{\partial x}\frac{dx}{dt}+\frac{\partial F}{\partial y}\frac{dy}{dt}?\]

Besides, what are the geometric meanings of $\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}$ and $\frac{\partial\Phi}{\partial x}, \frac{\partial\Phi}{\partial y}, \frac{\partial\Phi}{\partial z}$, where $F(x,y)=0$ and $\Phi(x,y,z)=0$.


[quote="orl"]The non-negative numbers $ x_1, x_2, \ldots, x_5$ satisfy $ \sum_{i \equal{} 1}^5 \frac {1}{1 \plus{} x_i} \equal{} 1$.  Prove that $ \sum_{i \equal{} 1}^5 \frac {x_i}{4 \plus{} x_i^2} \leq 1$.[/quote]
$ \sum_{i \equal{} 1}^5 \frac {x_i}{4 \plus{} x_i^2} \leq 1\Leftrightarrow\sum_{i \equal{} 1}^5 \left(\frac{1}{5}\minus{}\frac {x_i}{4 \plus{} x_i^2}\plus{}\frac{3}{4}\left(\frac{1}{1\plus{}x_{i}}\minus{}\frac{1}{5}\right)\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{i\equal{}1}^5\frac{(x_i\minus{}4)^2(x_i\plus{}4)}{(4\plus{}x_i^2)(1\plus{}x_i)}\geq0.$

Nice problem:

Set:
$ a_i \equal{} \frac {1}{1 \plus{} x_i}$
then
$ x_i \equal{} \frac {1 \minus{} a_i}{a_i} \equal{} \frac {\sum_{j \neq i}{a_j}}{a_i}$
Now the i-th fraction of the LHS becomes:
$ \frac {{a_i}\sum_{j \neq i}{a_j}}{4{a_i}^2 \plus{} (\sum_{j \neq i}{a_j})^2} \equal{} \frac {{a_i}\sum_{j \neq i}{a_j}}{\sum_{j \neq i}{({a_i}^2 \plus{} {a_j}^2)} \plus{} 2\sum_{k,j \neq i, j < k}{a_ka_j}}$
Now because $ {a_i}^2 \plus{} {a_j}^2 \geq 2a_ia_j$, the i-th fraction is less or equal to:
$ \frac {{a_i}\sum_{j \neq i}{a_j}}{\sum_{j \neq k}{a_ka_j}}$
Adding the above for $ i \equal{} 1,2,3,4,5$, we get the desired!

Let $y_i = \frac 1{1+x_i}$ for simplicity. Then I claim $$\sum_{i=1}^5 \frac{x_i}{4+x_i^2} = \sum_{i=1}^5 \frac{y_i - y_i^2}{5y_i^2 - 2y_i+1} \geq \sum_{i=1}^5 \frac 34 y_i + \frac 1{20} = \frac 34 + \frac 1{20} \cdot 5 = 1.$$ Indeed, the second inequality simplifies to $$(1-5y_i)^2(3y_i+1) \geq 0,$$ which is obvious as the $y_i$ are positive, so we're done.


geometry
ratio
AMC
AIME
quadratics
algebra
power of a point

Find the area of the large square divided by the area of the small square.

Here's what I've done, but I get the wrong answer:

[hide]Call the side of the large square 2x.  Then the diagonal of the large square is 2xsqrt2 and the radius is xsqrt2.  Then, since half the side of the large square is x, and the radius is xsqrt2, if we draw the radius going straight up through the small square, we find the side of the small square is (xsqrt2 - x).  Squaring this gives 3x^2 - 2x^2sqrt2.  The area of the large square is 4x^2, so 4x^2/(3x^2 - 2x^2sqrt2) is what we need to find.  Factoring out a x^2 gives (x^2(4))/(x^2(3-2sqrt2).  x^2 cancels out and we are left with

4/(3 - 2sqrt2).  Multiplying top and bottom by (3 + 2 sqrt 2 ) gives 12 + 8 sqrt 2, but this is the wrong answer.

[/hide]

Can anybody help me and point out what I did wrong?  Thanks.

I got the ratio of the sides to be 5 and thus the ratio of areas to be 25.  I think what you are forgetting is that ther is a little portion of the radius of the larger circle that extends above the side of the smaller square.  If you want to know how I did it just ask and I ll gladly put it but I ll let you play with it some more now.

Yeah, I bet a bunch of people who took the AIME that year missed it...

I can't seem to figure out how to solve it though

My solution, which i think is cool since you dont actually need to know the radius of the circle.

[hide]Let the side of the big square be 2. Let x be half the side of the little square. Then apply power of a point  -  we get (1+x)(1-x) = 2x(2+2x) (can you see why?). This results in the quadratic equation 5x^2 + 4x -1 = 0, with positive root x = 1/5, so the side of the little square is 2/5, and the ratio of the sides is 5, the area, 25.[hide][/hide][/hide]

I don't get how you applied Power of a Point.

The chords I applied it to intersect in the bottom left vertex of the little square. Can you see how it works now?

Ah, power of a point.  Here's the way I did it:

[hide]

It's easier to just assign values but I just used variables.  Call the side of the big square 2s, the side of the small square 2x, and that little part of the radius that extends past the small square y.

We are looking for the ratio of the areas or 4s^2/4x^2=(s/x)^2.  So we are basically looking for the ratio of the sides.

Using pythagorus we can find that the radius is .

We can also see that .

Next, draw the radius to a vertex of the small square.  We can use pythagorus and get:

Plugging in s rt(2)-s-2x for y we get:

Expand and then collect like terms:

so 5x=s or x=-s.

So s/x=5 or s/x=-1 but -1 is impossible so s/x=5 and (s/x)^2=25.[/hide]

Tetrahedr0n wrote:

Quote:[hide]Then apply power of a point - we get (1+x)(1-x) = 2x(1+2x) (can you see why?).[/hide]

It should be (2+2x) on the Power of a Point.

Thats correct - just a mistype on my part.

Hi Nightflare..
what is the correct ans ?? I'm getting it as 25

it's 25.

btw, how do use big bold italic text?

$asdf$


Probably posted before, but a good exercise in polynomials and inequalities.

Let $P$ be a polynomial witih positive coefficients. Prove that if

$P\left(\frac{1}{x}\right) \ge \frac{1}{P(x)}$

holds for $x=1$, then it holds for every $x>0$.

[hide="hint"][url=http://wangsblog.com/jeffrey/]See blog post[/url][/hide]

[quote="seamusoboyle"]As a matter of interest, does your blog really solve all these problems???

I solved this using cauchy-swarch.[/quote]

Umm, yes... I either solve it or propose it as a practice problem for the subject of the problem of the day. You could go to the link to see it... I also used Cauchy.


Hi. I am studdying for a huge math midturm and I am stuck on this problem:

Find the value for x by first finding the equation of the line passing through all three points:

(-12,-3)

(X,-24)

(8,-73)

Thanks for the help!

This doesn't really belong in the Intermediate forum...

[hide="Solution"]First, find the slope of the line:

$m=\frac{-3+73}{-12-8}=-\frac{70}{20}=-\frac72$

Then, using a given point and the slope, find the $y$-intercept:

$-3=\left(-\frac72\right)\cdot(-12)+b$

$b=-45$

The equation of the line in slope-intercept form is $y=-\frac72x-45$.

Finally, plug in the given $y$ value to find $x$:

$-24=-\frac72x-45$

$21=-\frac72x$

$x=\boxed{-6}$[/hide]


For how many integers $ x$ does a triangle with side lengths $ 10, 24$ and $ x$ have all its angles acute?

$ \textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ \text{more than } 7$

[hide="My Solution"]
First of all, we can narrow our search by finding all side lengths $ x$ such that $ 10, 24, x$ form a triangle.

$ 10 \plus{} 24 > x \implies x < 34\\
10 \plus{} x > 24 \implies x > 14$

From all that, we obtain that $ 15 \le x \le 33$. Now we use the fact that the triangle formed is acute. In order for that to be true, the following must be true.

$ 10^2 \plus{} 24^2 > x^2 \implies x < \sqrt{676}\\
10^2 \plus{} x^2 > 24^2 \implies x > \sqrt{476}$

That implies that $ 22 \le x \le 25$ so there are $ 25 \minus{} 22 \plus{} 1 \equal{} \boxed{(A) 4}$.
[/hide]


MATHCOUNTS
videos
AMC
AMC 10
AIME
AoPS Books

Well, Im in middle school, and procrastinate alot. It makes me feel horrible afterwards and ends up ruining my life, but i cant stop it. Over the sumemr, iv made a ton of goals for next year with math competitions, and have gave myself preperations for what to do to achieve those goals. But whenever i try to do math and such, i end up saying to myself "screw it, im gonna watch south park, i can do that math tommorow". or somthing like that.Now I get almost nothing accomplished becasue of this. I feel like crap whenever i think of what i did during this first half of summer. And then i feel like crying whenever i think of what will happen when Mathcounts states come. Iv made nationals in 6th grade, but only to get all freaking cocky and waste away my 6th grade summer playing video games. In 7th grade i didnt make top 20 at states....now in my 8th grade summer im doing the same thing. Iv tried telling myself that going to nats is one of the best things ever for me and it will be all worth it doing all that math stuff to prepare, but i still end up slacking off. Iv even tried cutting myself to stop procrastinating but i still cant stop. I know theres no silver bullet or w/e to stop this problem, but I have got to know how to (help) stop procrastinating. My next grade year depends on it because mathcounts is the whole point of my middle school years (i mean it) Please can someone offer advice for me?PLEASE!!!!! ;_;

Wake up earlier.

Some people I've talked to have mentioned that if you wake up late, then you get to the afternoon quickly, and you are like, well, today's almost over, it'll be night soon, so I'll just chill and do it tomorrow. Repeat. Instead, when they wake up early, it feels like they have way too many hours head of them, and they feel more compelled to be productive (at least at first). And then, even if you decide to screw off when afternoon comes, you would've accomplished a lot. Also:

- There's less entertaining TV programming on
- No friends on AIM or whatever
- A lot of kids I know find it more natural to play video games at night

Of course, there are other things, like will power and discipline and motivation, but those come from within you. And waking up earlier can serve as a first step for developing these traits, leading to even greater long-term effects.

First and foremost, remember your interests You may not like math at all, and only did it because your parents made you. If you don't really like math, this may be a reason why you haven't been doing it lately. That's why, for people who like doing math, not doing math isn't "procrastination" because math isn't work for us, it's play. If you're being pressured to do math, then you've got to realize that it's not worth doing at all if you don't like it. This is a surprisingly common case.

If you do like math, then putting away the video games is easy. What I would do is hide one of my game disks in a special place (like on top of a really tall bookcase) or in a backpack in a closet in the shed, i.e. someplace random that you will be wayy too lazy to get. If internet is distracting you, unplug it from the wall. This is just a self discipline issue.

Also, while this may seem heartless, cutting yourself to push yourself to be more productive is quite possibly the dumbest thing I have ever heard of. Productivity comes from positive, not negative. Again, if you are being pressured or pressuring yourself into math when you don't really like it, and you are depressed about it, then you should really re-evaluate your self. You don't need math to live a happy life, and you should only do math if you really love it. See a psychiatrist if you are still depressed...

EDIT: Why is MathCounts the whole point of your middle school years? Explain.

That's a lot of negative posts.... so.... HAPPY HAPPY HAPPY POSITIVE HAPPY HAPPY YAY!!!!!!  :D

I have the same problem pretty much... sorry I don't have much good advice... the only thing I can say is to distance math from whatever distracts you... which is pretty obvious...

[quote="Anaxerzia"]
EDIT: Why is MathCounts the whole point of your middle school years? Explain.[/quote]

Is not math the center of your life? There's hardly any other good middle school math competitions.

The Zuton Force- thanks greatly for your help, i guess ill wake up earilier now

jmerry-_- um im pretty sure im not depressed......and me feeling like crap is more of self gulit, and yes.... cutting myself was stupid like im doing that again, but thanks for your care

Anaxerzia- ty, i think people get a little(maybe alot) bored of math, im gonna try to play with it a little and make it more exciting i guess. But i know sure as heck im not quitting math -_- i would feel like a hollow shell if i did. yes... cutting myself is stupid....yea...lol...
Um, for your question, i think its pretty resonable for people to do that. I have freinds and stuff;I just feel that accomplishing things gives people more meaning and somthing more to live for, and im good at mathcounts. Also, iv quit cello and piano becasue of time restraints from my swimming, but over the summer i quit swimming because of AoPS classes. I was acually really good at swimming (4th in state for 2 events) but now all i have is math ;_; But mathcounts is what iv been preparing for all year round since i was around 9 (like most AoPS users) so why shouldnt it be the whole point?

leoxnlin-thanks for your happiness ^^

Why don't you try looking at those other things that you like?

Also there is more to life than contests, and much more to math than math contests. I guess one day you will realize that own your own, though...

OMG. I have this same problem, but I have to say, waking up early, or waking up late and doing it in the dead of night is what works for me.

I'm an AIM addict, so during those 2 periods, there's usually not anyone on, so I hav no excuses, as I'm also toolazy to go downstairs to play video games. So I'll sit down and do my AOPS books or physics or whatever.

EDIT: It doesn't really work for me though, because I'm really sleepy, but it might work for you, idk.

math isnt that important...

if u've already been to nats, do u really need to go again? don't focus on math too much; just do w/e u want to and you should be fine

and by that, i mean do other productive stuff u enjoy doing. maybe the real reason u are procrastinating so much is that u find math boring deep down inside

^^ Coming from a Mathcounts Champion, thats saying a lot.

Also, mathcounts math is very boring: Why don't you look at different math, like a Olympiad stuff. You may find it far less boring and find yourself not procrastinating.

this site keeps tellign me to quit doing math...=(

Well procrastination is worse when you're actually involved with school work. With recreational math it's perfectly fine. The purpose of a summer vacation is to relax and not stress over work - rec math included. (In fact, I haven't done any math in over a month, and I won't even get to since I'm leaving the country for a month in a week)

If I were you, I'd relax, do math every other day, or every other other day.

Here is what works like a miracle for me (and for others that have tried it). I must admit that this isn't really a procrastination problem that I solve, but a lack of focus type of thing. Seems close enough, though. After a few hours of studying I start to lose focus, and so I go for a run.

It is like magic. It clears your head. It wakes you up. There is actually scientific basis for this, too. If you find some sort of cardio exercise you enjoy, you should do it for about 20 minutes a day. There is strong evidence that you grow up to 3 times as many neurons related to learning and memory (so you'll learn things faster and remember them longer). Also, endorphins are released which usually increase happiness and motivation to work.

I cannot stress how much my work habits and mood have improved over the past year since I began running on a regular basis. (But as with anything, if you don't build it into your schedule, it could fall prey itself to procrastination).

Well you said you were setting goals so you should tell your parents these goals and so that you will feel more pressured to do them. Also, If you really don't want to do it, take breaks until you are feeling like doing more math. Watch TV for a while then do some math. You dont have to spend 10 hours a day doing math but 1 or 2 can also help yourself improve as long as you make the most of the time you are spending.

i procrastinate a lot too (especially right before major tests) and for some reason i dont suffer from guilt or stress or whatever. what i do is right before the test, the night before, actually, i read over the textbook like 50 times and study for like 6 hours. and it works for me... u could say i'm used to it. think about it, though: either way, as long as u get the same amount of studying done, u'r good to go. i like to do it at the last minute so then at the test, everything's still fresh in my head. there's just more stress this way.

i didnt really give any usefull tips, did i?

Public service announcement: chat-speak, including the substitution of "u" for "you," "r" for "are," etc., [b]is not welcome[/b] in this forum.  Don't use it.  Thank you.

[quote="dreambold22"]i didnt really give any usefull tips, did i?[/quote]  No, you didn't.  :wink:

Okay, then. I am very sincerely sorry, Mr. Moderator. In the future I will try to watch my grammar and post more useful posts in this forum. Once again, I am very sorry.

[color=green]Moderator says: Thanks![/color]

Back to the original topic: eat ice cream every once and a while. it calms the nerves. I'm not kidding. Or tea. Green tea also calms the nerves. Plus they both taste good.

[quote="dreambold22"] think about it, though: either way, as long as u get the same amount of studying done, u'r good to go. i like to do it at the last minute so then at the test, everything's still fresh in my head. [/quote]

Actually, it isn't the same either way. Study time spread over a long period of time is far more effective (for retention anyway) than the same amount of time crammed the night before.

Studies show that people that cram study do do better on tests that come right after, but they forget really fast as well. For something like Olympiad studying, you are not in it for a grade over a small amount of material, so I suggest the long term approach.

[url=http://www3.interscience.wiley.com/journal/118000121/abstract?CRETRY=1&SRETRY=0]This[/url] isn't the study I learned it from, but they are all over the place now. It is pretty established.

[quote="jmerry"][quote="yoyomattnerdt2"]Now I get almost nothing accomplished becasue of this. I feel like crap whenever i think of what i did during this first half of summer.[/quote]
[quote="yoyomattnerdt2"]Iv even tried cutting myself to stop procrastinating but i still cant stop.[/quote]
I'm not an expert, but this really looks like depression. You should see a psychiatrist. The sooner you stop hurting yourself mentally and physically, the better.[/quote]
CUTTING YOURSELF?!?! you really really REALLY need to stop. or see a specialist, your doctor, etc.
it's bad for you.

I also procrastinate a lot, but I have nothing better to do all day than actually do math all day, so you may want to try removing some distractions, or go somewhere where everyone is studying. For some reason, you work better when the enviorment around you is quiet and everyone is working.

I have a problem when its school: I am a 6th grader (elementary school, not middle school) going to 7th.  Usually in school, before tests, I want to study because I always think I am going to fail, but I can never force myself to actually study. I am worried that junior high will be much harder (because of a sibling), so I want to be able to actually force myself to study. Even though the distractions are all locked up or something, I always rather do something else. Cna I force myself to study for tests other than math ones (I don't need to for these)?

[quote="myyellowducky82"]CUTTING YOURSELF?!?! you really really REALLY need to stop. or see a specialist, your doctor, etc.
it's bad for you.[/quote]

This bothers me for a number of reasons:

$ \blacktriangleright [1]$ If you don't want to read the thread, at least read the OP's posts! He tried it once...

$ \blacktriangleright [2]$ Your tone is quite rude. You're basically saying "wow, you have issues." Freaking out and USING CAPS just makes it worse.

$ \blacktriangleright [3]$ You haven't offered a single brain-cycle's worth of constructive thought.

i think u should tell ur parents (therapist, etc.) about ur problem
that person could help

if u were forced to do math, that's why
if not, see a doctor or something

i have that same exact problem you do, so i know what youre going through. i always put things off until theres no other day to put it off to. what i can suggest is this:

when you get the urge to watch tv, unplug it

when you get the urge to go online, unplug the router and shut down the comp

when you get the urge to play video games, run out the battery (a little fun) and dont recharge

because what it looks liek is youre addicted to procrastination. you think time is endless, so you dont hafta work as hard. its even worsened by the fact that you made nats back in 6th grade. now youre probably thinking you dont hafta work as hard, because you made it at such a young age, and you think youll make it again no matter what. having a mindset like that and having a great brain often leads to laxity and procrastination. i know, because thats what im doing right now. which is why i failed the states MC, the AMC10, the AIME, and multiple other contests. But you should talk to yoiur parents. have them enforce stricter rules around the house, with harsher consequences. although rough at first, the rewards will be grand.

I also procrastinate a lot and I've tried lots of things to stop it. The better results were obtained when I decided to study late in the night, with all the silence. :ninja:

I like doing math, but I can't say that I really love it. I want to get better at solving problems not only to get better results in the contests but simply to have more and more times that worderful sensation "wow, I've solved a difficult problem". So I have to study, and I feel bad when I don't do math for some days. I'm trying to leave this things out and concentrate myself only in the fun, but it's difficult to modify a behavior that you have maintained for years and years. I hope I will have success and finally admire the math in all its beauty and really fall in love with it.  :)

I procastinate all the time in university.  It's bad for you =P

What I found really helps, is studying with others.  Though you may not have anyone close around you who does math, but maybe find someone on the forum to discuss math on a REGULAR basis.  You can talk about problems, new theorems, techniques, new puzzles you found, etc.  Regular talks like this compell you to not come empty handed for a week!

yeahh...

I have a procrastination problem too, and I really need to get myself focused before september because I'm going to Exeter, where schoolwork kills you.

Time management is EVERYTHING.
As soon as I wake up and shower and eat breakfast and stuff like that, my parents make me have a schedule for the day.
like this:

10-11: computer time!
11-12: AoPS Volume 1
12-1: lunch and play!
1-3: violin
3-4: play!
4-6: classic reading

and from 6 i can do whatever i want for the night.

but the schedule only works if you follow it strictly.
I find the schedule really useful.

I need to have something laid out for myself or I end up saying "whatever, i'll do that after 30 more minutes of computer", or something like that, but if I do this, I get everything I need done, and my parents are proud of me.

and contrary to others, I don't recommend doing work early in the morning or at night.
I get tired and I can't focus at those times and I just end up dozing off.

so I hope you get your procrastination problem fixed as I get mine fixed by the end of August!!

We're in it together!


function
number theory unsolved
number theory

[quote="buzzer11"]I am currently starting a cryptology project, and I was wondering what interesting problems there are in cryptology. Kind of like the famous "unsolved problems" of mathematics but related to cryptology? Thanks in advance![/quote]

I think two major topics are elliptic curves and composite numbers factorization ... .

I don't really understand ECC, can you explain the basics?  :(
I have looked at factorization, and one-way functions. I'm not sure how "easy" and "hard" are defined for one-way functions, either...


A solid sphere of ice with radius 4 is placed on a hollow cone
(vertex pointed down) of radius 4 and height 5. When the ice
melts, how much of the water will overflow the cone? Express
your answer as a common fraction in terms of $ \pi$.

There is ice with volume of $ \frac43 \cdot \pi \cdot 64\equal{}\frac{256\pi}{3}$

We a cone with volume $ \frac13 \cdot \pi \cdot 5 \cdot 16\equal{}\frac{80\pi}{3}$

When we fill the cone all the way with ice, there will be $ \frac{256\pi}{3}\minus{}\frac{80\pi}{3}\equal{}\frac{176\pi}{3}$ of overflow.

Of course, water expands when it freezes, so if this was in real life the number will be a bit less than that.


For integers $r > 0, \ s > 1$ let $A_{rs}= r+s\mathbb{Z}$. We say that $B \in A(r,s)$ iff $B=A_{rs}\cup C$,
where $C$ is either $\emptyset$ or a union of $A_{js}, \ r < j <r+s.$ Is it possible to find integers $n > 0,$
$1 < d_{1}<d_{2}<...$, and sets $B_{i}\in A(n, d_{i})$ such that $B_{1}\subset B_{2}\subset ...$?

Here $\subset$ means subset but not equal and the set $\{d_{i}, \ i \geq 1\}$ is assumed to be an infinite set.

$B\in A(r,s)\ \Longleftrightarrow \ B=r+s\mathbb{Z}\ or\ B=\bigcup_{h\in \{r,r+s-1\}}h+s\mathbb{Z}=[|r,+\infty[$
we suppose that $(B_{i})_{i\in \mathbb{N}^{*}}$ exist.
$B_{i}=n+d_{i}\mathbb{Z}\ or\ B_{i}=\bigcup_{h\in \{r,r+s-1\}}h+s\mathbb{Z}=[|n,+\infty[$
if $\exists i>0,\ such \ that\ B_{i}=[|n,+\infty[\Longrightarrow \ B_{i+1}=B_{i}=[|n,+\infty[$
then $\forall i>0,\ B_{i}=n+d_{i}\mathbb{Z}$
we have,
$B_{i}\subset B_{i+1}\ \Longrightarrow \ n+d_{i}\mathbb{Z}\subset n+d_{i+1}\mathbb{Z}\\ \longrightarrow \ n+d_{i}\in n+d_{i+1}\mathbb{Z}\\ \longrightarrow \ \exists j\in \mathbb{N}^{*},\ n+d_{i}=n+jd_{i+1}\\ \longrightarrow \ d_{i}\ge d_{i+1}\ (contradiction)$

[quote="abc67"]No [b]aviateurpilot,[/b] an element of $A(r,s)$ is not as you said. They are a finite union of arithmetic progressions.
Notice that in the problem $C$ is [b]a,[/b] and not [b]the[/b], union of $A_{js}.$ So $B$ is either $A_{rs}$ or it's in this form:
$B=A_{rs}\cup A_{j_{1}s}\cup ... \cup A_{j_{k}s},$ where $\{j_{1},...,j_{k}\}\subseteq \{r+1,...,r+s-1\}.$[/quote]
sorry, i am very bad in english :blush:

That's ok [b]aviateurpilot[/b]! :)

I forgot to mention that we also want to have $\cup_{i=1}^{\infty}B_{i}\neq \mathbb{Z}$. Here is the problem again:

For integers $r > 0, \ s > 1$ let $A_{rs}= r+s\mathbb{Z}$. We say that $B \in A(r,s)$ iff $B=A_{rs}\cup C$,
where $C$ is either $\emptyset$ or a union of $A_{js}, \ r < j <r+s.$ Is it possible to find integers $n > 0,$
$1 < d_{1}<d_{2}<...$, and sets $B_{i}\in A(n, d_{i})$ such that $\cup_{i=1}^{\infty}B_{i}\neq \mathbb{Z}$ and $B_{1}\subset B_{2}\subset ... ?$
Here $\subset$ means subset but not equal and the set $\{d_{i}, \ i \geq 1\}$ is assumed to be an infinite set.

[quote="abc67"]That's ok [b]aviateurpilot[/b]I forgot to mention that we also want to have $\cup_{i=1}^{\infty}B_{i}\neq \mathbb{Z}$. Here is the problem again:
 [/quote]
but you mention that $n>0$.
then $\bigcup_{i=1}^{\infty}B_{i}\subseteq [|n,+\infty[ \subseteq \mathbb{Z}-\{0\}\neq \mathbb{Z}$

[quote="abc67"][quote="aviateurpilot"]
but you mention that $n>0$.
then $\bigcup_{i=1}^{\infty}B_{i}\subseteq [|n,+\infty[ \subseteq \mathbb{Z}-\{0\}\neq \mathbb{Z}$[/quote]
That's not true!
for example for $n=4, \ d_{1}=2,$ and $B_{1}=A_{42}$, we have $0 \in B_{1}.$
Note that $A_{42}=4+2\mathbb{Z}=2\mathbb{Z}=\{...,-2,0,2,...\}.$
Besides $\bigcup_{i=1}^{\infty}B_{i}\subseteq [|n,+\infty[$ is not true either! :)[/quote]
I forgot $\mathbb{Z}^{-}$  :rotfl:


conics
ellipse
geometry
ratio
similar triangles
geometry proposed

Find the maximum value of the area of an ellipse which is inscribed in the right triangle ABC with AB=AC and BC=2,and an axis of which is pararell to the edge BC.

if you explain 'Circumscribe' I think I can do it

Hi! Thank you for your reply!

Cicumscribe means GAISETUSASERU in Japanese.

kunny

GAISETUSASERU
がいせつさせる

(In Japanese HILAGANA)

I'm thinking..

To:paganinio

I have just edited.

kunny

If we take the figure and stretch it in a direction perpendicular to BC, then the ratio of the area of the ellipse to that of the triangle doesn't change. If we stretch it by the appropriate factor, then the ellipse becomes the incircle of the new triangle. So we need to know the dimentions of the stretched triangle A'BC so that the ratio of the incircle's area to the area of triangle A'BC is maximal.

Let the radius of the incircle of A'BC be r. Using similar triangles we may easily find that the altitude from A' to BC is $\frac{2r}{1-r^2}$, and so we wish to maximize the ratio $\frac{\text{area of circle}}{\text{area of A'BC}}=\frac{\pi r^2}{\frac{2r}{1-r^2}}=\frac{\pi}{2}\cdot r(1-r^2).$

By AM-GM, we have $2=(1-r^2)+(1-r^2)+(2r^2)\ge3\sqrt[3]{(1-r^2)^2(2r^2)}$, which rearranges to $r(1-r^2)\le\frac{2}{3\sqrt3}$. Equality occurs only when $1-r^2=2r^2$, i.e. $r=\frac{1}{\sqrt{3}}$. So $\frac{\text{area of circle}}{\text{area of A'BC}} = \frac{\pi}{2}\cdot r(1-r^2) \le \frac{\pi}{3\sqrt3}$.

So, taking these optimal dimentions and squishing back down to make ABC a right triangle, the area of the ellipse is $\frac{\pi}{3\sqrt3}\cdot [ABC]=\frac{\pi}{3\sqrt3}$.

kunny
forget the "pow"
my interest has turned to this problem.

Is this done with an affine tranformation from the incircle of the given triangle to the incircle of an arbitrary new triangle of which in this case was 'stretched' along the y axis? (by distance A'-A)

this is the best way to stretch the given triangle right?

 :maybe:

Hi, mathzoid. Generally speaking, Japanese high school students don't learn Affin transformation.

This problem seemed to be difficult for Tokyo University entrance applicants.

My solution is very simple.

Enjoy solving the problem!

kunny

this one already done from habelman

thats the only way i think to do that one

 :wink:

Yes, but we can solve it without using Affin Transform.

Just try it!


[color=violet]\[ \begin{array}{l}
 let\_a_n be\_the\_last\_nonzero\_digit\_in\_the\_decimal\_repesentation\_of\_n! \\
 does\_the\_sequence\_a_1 ,a_2 ,a_3 ....a_n \_eventually\_become\_periodic\_after\_a\_finite\_no.\_of\_times? \\
 \end{array}
\][/color]

[b]@ anorithdialga:[/b] Unless you want to center a formula, please enclose it with the dollar-sign key instead of with the slash-bracket notation. That way, you can easily incorporate normal text into your posts. Quote me to take a look at how the dollar-sign notation for LaTeX is used on this forum. Thanks!

[hide="Tip-In"]They repeat with period $ 25$. Can we prove it?[/hide]


geometry
analytic geometry
rotation
trigonometry

Can someone help me with #1, #5, and #6 on the pretest for the olympiad geometry class ([url]http://www.artofproblemsolving.com/Classes/GeomOlympiad/PreTest.pdf[/url])?
I see an algebraic and bashing method for #1, but I can't find an elegant solution.
I almost have #5 done, but I can't think of a complete way to solve it.
As for #6, I'm just not good with constructions.  I saw the answer for it on the third page of that same exact pdf, but of course, it is not a solution (it doesn't explain why the construction solves the problem).

i dont think it would be a good idea for people to post solutions here, maybe pm to you but not post here...

[hide="#1"]For #1, we can easily see $ \angle AOD \equal{} \angle BOC \plus{} 120$. Also, after an easy computation, you find $ BC \equal{} \sqrt {73}$ and $ AD \equal{} 2\sqrt {73}$. Thus we have
$ sin(\frac {AOD}{2}) \equal{} 2sin(\frac {BOC}{2})$
$ sin(\frac {BOC}{2} \plus{} 60) \equal{} 2sin(\frac {BOC}{2})$
We can just guess that $ \frac {BOC}{2} \equal{} 30$, then $ \angle BOC \equal{} 60$, $ \angle AOD \equal{} 180$ (AD is a diameter), and the result follows.[/hide]
[hide="#6"] Let's use complex coordinates :D .
Let A=0, B=a+bi, and C=c+di. We are given a and c. From the given,
$ (a \plus{} bi)cis60 \equal{} c \plus{} di$
$ (a \plus{} bi)(\frac {1}{2} \plus{} \frac {\sqrt {3}i}{2}) \equal{} c \plus{} di$
Setting the real and imaginary parts equal,
$ c \equal{} \frac {a}{2} \minus{} \frac {b\sqrt {3}}{2}$
$ d \equal{} \frac {a\sqrt {3}}{2} \plus{} \frac {b}{2}$
Multiplying the second by $ \sqrt {3}$ and adding, we have
$ 2a \minus{} c \equal{} d\sqrt {3}$
But we were given a and c and it is easy to divide a segment by $ \sqrt {3}$, and rotate a segment by 60, so we are done.
[/hide]

[hide="1"]Using power of point on $ P$ gives $ BP\equal{}8$. Connect $ BD$. Note that $ \triangle PBD$ forms a $ 1\minus{}\sqrt{3}\minus{}2$ triangle so $ \angle CDB \equal{} 30^\circ$ and $ \angle BOC\equal{}60^\circ$. Therefore the radius is equal to $ BC \equal{} \sqrt{73}$ by cosine law. Hence the area is $ 73\pi$.[/hide]

[hide="5"](refering to the attached diagram)
On plane $ AA'DD$, draw a line past $ Y$ parallel to $ XB$. Let its intersection with $ D'A'$ be $ K$. Then $ m$ is line $ B'K$.
From here we can calculate $ A'K \equal{} \frac{1}{6}$, $ B'K'\equal{}\frac{\sqrt{37}}{6}$, and $ A'Z\equal{}\frac{1}{\sqrt{37}}$. Then $ \tan A'ZY \equal{} \frac{A'Y}{A'Z} \equal{} \frac{\sqrt{37}}{3}$.
[/hide]


Imagine that one has 1000 bags

with 1000 coins in each bag. In

one of the bags, all of the coins

are counterfiet and weigh 11g

while in all the other bags, all the

 coins are normal and weight 10g.

Using only one weighing on a

scale one must find what bag has

 the counterfiet coins. This is not

a trick question so you can't just

keep adding coins to the same

weighing or something like that.

[hide]Take 1 coin from the first bag, 2 from the second, 3 from the third, etc. or some other assigned number.  This way, we count the number excess of the expected, and the amount corresponds to the number taken from the bag.  Thus, we can track down which particular bag the problem coins came from.[/hide]


Let $ ABC$ be a triangle with side lenghts $ a,b,c$ and let $ S$ be its area .

Prove that: $ S\le\frac{\sqrt{3}}{8}\max(a^2\plus{}b^2,b^2\plus{}c^2,c^2\plus{}a^2)$ .

For the sake of contradiction, let us assume that $ S>\frac{\sqrt{3}}{8}(a^2\plus{}b^2), S>\frac{\sqrt{3}}{8}(b^2\plus{}c^2), S>\frac{\sqrt{3}}{8}(c^2\plus{}a^2)$.

Adding them all up, we have $ 3S>\frac{\sqrt{3}}{4}(a^2\plus{}b^2\plus{}c^2)$.

By [url=http://en.wikipedia.org/wiki/Weitzenb%C3%B6ck's_inequality]Weitzenböck's Inequality[/url], we have $ a^2\plus{}b^2\plus{}c^2\ge 4\sqrt{3}S$, hence $ 3S>\frac{\sqrt{3}}{4}(a^2\plus{}b^2\plus{}c^2)\ge 3S$, a contradiction. $ \blacksquare$


linear algebra
matrix
algebra
polynomial
algebra unsolved

having the matrices:

$A=\left( \begin{array}{ccc}
0 & 0 & 1 \\
0 & 2 & 0 \\
3 & 0 & 0 \end{array} \right);
O_3=\left( \begin{array}{ccc}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \end{array} \right);
I_3=\left( \begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \end{array} \right)
$

Given the set $S(A)=\{X\in M_3(C) | XA=AX\}$

It is known that
if $U,V \in S(A) $then $U\cdot V \in S(A)$
if $Y\in S(A) $and $Y^2=O_3$ then $Y=O_3$

show that if $Z\in S(A)$ and $Z^{2005}=0_3$ then $Z=0_3$.

Notice that since z belongs to s(A)
so does z^n for any n. (condition i)
this implies that z^43 belongs to S(A)
but we also know that z^2005 belongs to S.
so O=z^2048 belongs to S(A).
but then 2048=2^11, using condition ii 11 times we arrive at: z=O.

or because the characteristic polynomial of $Z$ is $-X^3$ (because the eigenvalues of $Z$ are all zero) from Cayley Hamilton we have that $Z^3=O_3$ and from this $Z^4=O_3\Rightarrow Z^2=O_3\Rightarrow Z=O_3$.


conics
hyperbola
ellipse
calculus
calculus computations

I don't know where to go as far as graphing.  I know the the graph touches the x-axis at X= -1 and -5.

y= +/-(5i/2)((x+1)(x+5))^(1/2)

I believe it looks something like a hyperbola but not.  It should start have vertices (-1,0) and (-5,0).  Beyond that, we haven't been given a problem like this before.

It's  an ellipse with center $ C \Leftrightarrow (-3,0) $ and semiaxes which measures are  2  and 5. You can write the equation so:
\[{
(5x+15)^2 +4y^2 =100 \to 25 (x+3)^2 +4y^2 =100 \to \frac {(x+3)^2} {4}}+ \frac {y^2} {25}=1
\]


I am high school teacher / coach and just took on the Mathcounts team this year.  We have our first meeting this coming Thursday.
Two of my high school students agreed to come to the Middle School and help.
Has anyone done this?  Any suggestions for how they might be used?

One of the HS kids is an energetic girl, so I'm hoping that she might  help me attract and retain talented middle school girls.  Maybe she can work with a small group of girls while I work with other students.

Well, at my school we had a high schooler help us 2 years ago. He was a mathcounts alumni and had gotten first at county a few years back though... He helped us a lot... If you can get someone who has done mc before that will help a lot.

I'm a high schooler helping out with my former middle school's MATHCOUNTS club. Each week the teacher gives them a warmup or two to do for the first half hour, and for the second half hour I give them a handout covering a specific topic (see [url=http://mathideas.org/mathcounts/schedule.pdf]here[/url]) with a short reading and several problems to solve during class and more to solve for homework. I keep track of points in a scoreboard on a website I made for the club.

We actually haven't started yet (we're going to start this week) but that's what we've planned.

Experience is one of the most important parts of MATHCOUNTS.  It would be greatly beneficial to your team if you had somebody that has been involved in MC before.

I think that using previous tests (found at mathcounts.org) are the best way to practice.

Or just give them chapters of AoPS vol. 1.

Goodlymaths also has some mathcounts material

Careless errors are basically the biggest downfall.

Make sure they read the problem, especially things like "Express your answer in ________" and "not" as MathCounts problems aren't actually particularly difficult most of the time, (some are) they're just tricky word problems.


calculus
integration
trigonometry
logarithms
calculus computations

The value of:
$ \int_0 ^ \frac {\pi}{4}$$ \frac {\sin x \plus{} \cos x}{\cos^2 x \plus{} sin^4 x}$ $ \cdot dx \equal{} ?$

[hide="Options"]
$ A)$ $ \frac {1}{\sqrt 3} \log({\frac {\sqrt {3} \minus{} 1}{\sqrt {2}}})$
$ B)$ $ \frac {\pi}{4} \minus{} \frac {1}{\sqrt 3} \log({\frac {\sqrt {3} \minus{} 1}{\sqrt {2}}})$
$ C)$ $ \frac {\pi}{4} \plus{} \frac {1}{\sqrt 3} \log({\frac {\sqrt {3} \minus{} 1}{\sqrt {2}}})$
$ D)$ $ \frac {\pi}{4} \plus{} \log({\frac {\sqrt {3} \minus{} 1}{\sqrt {2}}})$
[/hide]

hello, i have $ \int_{0}^{\frac{\pi}{4}}\frac{\sin(x)\plus{}\cos(x)}{\cos^2(x)\plus{}\sin^4(x)}\,dx\equal{}\frac{\pi}{4}\plus{}\frac{\sqrt{3}}{6}\log(2\plus{}\sqrt{3})$.
Sonnhard.

Use substitution $ u\equal{}\tan \left(\frac{x}{2}\right)$.

$ I \equal{} \sqrt{2} \int_0^{1/\sqrt{2}} \frac{1}{\left(u \minus{} \sqrt{\frac32}\right)\left(u \plus{} \sqrt{\frac32}\right)\left(u^2 \plus{} \frac12\right)}\,du$

[quote="Dr Sonnhard Graubner"]hello, i have $ \int_{0}^{\frac {\pi}{4}}\frac {\sin(x) \plus{} \cos(x)}{\cos^2(x) \plus{} \sin^4(x)}\,dx \equal{} \frac {\pi}{4} \plus{} \frac {\sqrt {3}}{6}\log(2 \plus{} \sqrt {3})$.
Sonnhard.[/quote]
Sorry Sonnhard,
U only have the options amongst which the correct answer lies. Thanks for any help,
-Rajiv.

hello, hmm, or is a typo in your problem? In the other case i will show my calculation here.
Sonnhard.

Note that $ \sin^4 x \plus{} \cos^2 x \equal{} \cos^4 x \plus{} \sin^2 x$

The given integral can be written as $ \int_0^{\frac {\pi}{4}} \frac {\cos x}{\sin^4 x \minus{} \sin^2x \plus{} 1} dx \plus{} \int_0^{\frac {\pi}{4}} \frac {\sin x}{\cos^4 x \minus{} \cos^2x \plus{} 1} dx$

The substitutions to make are now obvious

We now have to evaluate $ \int_0^1 \frac {1}{t^4 \minus{} t^2 \plus{} 1} dt$ which upon dividing numerator and denominator by $ t^2$ becomes

 $ \frac {1}{2} \left[ \int_0^1 \frac {d \left(t \minus{} \frac {1}{t}\right)}{\left(t \minus{} \frac {1}{t}\right)^2 \plus{} 1} dt \plus{} \int_0^1 \frac {d \left(t \plus{} \frac {1}{t}\right)}{\left(t \plus{} \frac {1}{t}\right)^2 \minus{} 3} dt \right]$

Both are improper integrals. The first one evaluates to $ \frac {\pi}{2}$ while the second to $ \frac {1}{\sqrt 3} \ln (2 \minus{} \sqrt3)$

Hence the answer is $ \frac {\pi}{4} \plus{} \frac {1}{2\sqrt 3} \ln (2 \minus{} \sqrt3)$

This is equivalent to Option C

Agreed! :)
Thanks a lot.

-$ Rajiv$ $ Patki$


A triangle has sides measuring  3  ,  5  and  n . What value of  n  would make the area of the triangle as big as possible. Express your answer to the nearest hunderdeth?

I think that area is biggest for biggest $ n$. $ 7.99$ 'cause $ 3\plus{}5>n$?

Just because the last side is the biggest possible doesn't mean the area is.
[hide]Let the angle between the 3 and 5 be $ \theta$. The area of the triangle is $ \frac {ab\sin{\theta}}{2}$. This is the maximum when theta is 90 degrees, otherwise $ \sin{\theta} < 1$. Our answer is then $ \sqrt {34}$, so it should be 5.83.[/hide]

EDIT: Forgot this is Mathcounts. Well, you should know that the maximum area occurs in a right triangle, and the proof is basically inside the hide tags.

[geogebra]8fc46f814f5b0ce925dfdc988620a8588a0e46dc[/geogebra]

By shearing, we find that to maximization of height occurs when h=5. (right triangle)

(a nontrig explanation... but essentially xpmath's solution, as the y-axis projection of a path around the circle is the sine of the angle.)

Wow, I used Heron's, that's pretty sad.
By Heron's we want to maximize $ (4\plus{}\frac{n}{2})(4\minus{}\frac{n}{2})(\frac{n}{2}\plus{}1)(\frac{n}{2}\minus{}1)$
Let $ \frac{n^2}{4}\equal{}k$. We want to maximize $ (k\minus{}1)(16\minus{}k)\equal{}\minus{}k^2\plus{}17k\minus{}16$. The vertex is when $ k\equal{}\frac{17}{2}$, or $ n\equal{} \sqrt{34}$.


Let $ \triangle{ABC}$ satisfies these conditions:
1-$ \angle{A} \equal{} 120$ 2-$ AB \equal{} AC$. Let $ P,Q,R,S$ be points on the [u]segments[/u] $ AB,BC,BC,AC$ respectively such that $ Q$ is between $ B$ and $ R$ and :
1-$ \angle{QAR} \equal{} 60$ 2-$ \angle{PQB} \equal{} \angle{AQR}$ 3-$ \angle{SRC} \equal{} \angle{ARQ}$
Prove that the area of triangle $ ARQ$ equals to the sum of  areas of triangles $ PQB$ and $ RSC$.

[quote="shoki"][color=darkred]Let $ ABC$ be an $ A$-isosceles triangle for which $ A = 120^{\circ}$ . Consider the points $ P\in (AB)$ , $ S\in (AC)$ and $ \{Q,R\}\subset (BC)$ so that

$ Q\in (BR)$  . Suppose that $ m(\angle{QAR}) = 60^{\circ}$ , $ \widehat {PQB}\equiv\widehat{AQR}$ and $ \widehat {SRC}\equiv\widehat{ARQ}$ .  Prove that $ [ARQ] = [PQB] + [SRC]$[/color] .[/quote]
[color=darkblue][b][u]Proof 1 (trigonometric)[/u][/b]. Denote $ \left\|\begin{array}{c} D\in BC\ ,\ AD\perp BC \\
 \\
m(\widehat {BAQ}) = x\ ,\ m(\widehat{CAR}) = y \\
 \\
AD = 1\ \ \ \left(x + y = 60^{\circ}\right)\end{array}\right\|\ \Longrightarrow\ \left\|\begin{array}{c} m(\widehat {DAQ}) = y\ ;\ AQ = \frac {1}{\cos y} \\
 \\
m(\widehat {DAR}) = x\ ;\ AR = \frac {1}{\cos x}\end{array}\right\|$ .

Apply [u][b]SINUS[/b]' theorem[/u] in $ \left\|\begin{array}{cccc} \triangle ABQ\ : & \frac {BQ}{AQ} = \frac {\sin x}{\sin 30^{\circ}} & \implies & BQ = \frac {2\sin x}{\cos y} \\
 \\
\triangle ACR\ : & \frac {CR}{AR} = \frac {\sin y}{\sin 30^{\circ}} & \implies & CR = \frac { 2\sin y}{\cos x}\end{array}\right\|$ Thus, $ [PQB] + [SRC] = [ARQ]\ \Longleftrightarrow$

$ \frac {BQ^2\cdot\sin 30^{\circ}\cdot\cos y}{\sin\left(x + 60^{\circ} \right)} +$ ${ \frac {CR^2\cdot\sin 30^{\circ}\cdot\cos x}{\sin\left(y + 60^{\circ} \right)} } = AQ\cdot AR\cdot\sin 60^{\circ}\ \Longleftrightarrow\ \frac {4\sin^2x}{\cos y} +$ $ \frac {4\sin^2y}{\cos x} =$ $ \frac {\sqrt 3\sin\left(x + 60^{\circ} \right)}{\cos x\cos y}\ \Longleftrightarrow$

$ 4\sin^2x\cdot\cos x +$ $ 4\cdot\sin^2y\cdot\cos y =$ $ \sqrt 3\cdot\cos\frac {y - x}{2}\ \Longleftrightarrow\ 2\sin x\sin 2x + 2\sin y\sin 2y =$ $ \sqrt 3\cos\frac {y - x}{2}\ \Longleftrightarrow$

$ \cos x - \cos 3x + \cos y - \cos 3y = \sqrt 3\cos\frac {y - x}{2}\ \Longleftrightarrow\ \cos x +$ $ \cos y = \sqrt 3\cos\frac {y - x}{2}$ what is truly because $ x + y = 60^{\circ}$ .[/color]


Does shape affect the rate of diffusion across a membrane? If yes, why?

Yes.  For example, glucose has a faster rate of diffusion than other sugars because of facilitated diffusion.  There is a carrier protein for glucose (but not other sugars), which combines with it on the outer surface, closes around it, and then opens to the inside of the cell where the glucose molecule is released.

partly

it depends much more on the chemical nature than on shape... :)


Let T be a set of natural numbers such that for any a,b numbers of T, (a^2 - ab + b^2) divides a^2.b^2. Prove that the set T is finite.

[quote]
[b]Eigenray wrote:[/b]
If $ (a^2\minus{}ab\plus{}b^2)|a^2b^2$ then $ b\leq \frac2{\sqrt3}a^2$.
[/quote]

Why do you say that?

[quote="Nemion"][quote]
[b]Eigenray wrote:[/b]
If $ (a^2 \minus{} ab \plus{} b^2)|a^2b^2$ then $ b\leq \frac2{\sqrt3}a^2$.
[/quote]Why do you say that?[/quote]
Because it is true.  Write $ a\equal{}dm$, $ b\equal{}dn$ where $ (m,n)\equal{}1$, and see what you get.

[quote="fallinlovewithmaths"]Eigenray you said but if we can find infinitely many a's ,it doesn't work. :|  And i guess there is a small mistake. I quoted this problem, but i think a and b must be distinct numbers because if they are not distinct so b^2 divides b^4 for all b positive integers. Could anybody can solve this problem :?:[/quote]
But we must have $ b < \frac2{\sqrt3}a^2$ for [i]all[/i] $ a,b\in T$.  So for [i]any[/i] $ a\in T$, $ T\subset [1,\frac2{\sqrt3}a^2)$, and is therefore finite.


For $ A,\ B$ be complex matrices $ n \times n$, prove that :

$ det\begin{pmatrix} A & B \\
\minus{} B & A \end{pmatrix} \equal{} det(A^2 \plus{} B^2 \plus{} \sqrt { \minus{} 1}(AB \minus{} BA)).$

[hide]
We will prove that $ det(M)\equal{}det(A^2\plus{}B^2\plus{}i(AB\minus{}BA)\equal{}det(A\plus{}iB).det(A\minus{}iB)$
Supposed that $ x_1,x_2,x_3,\cdots,x_n$ are eigenvalues of $ A\plus{}Bi$ with the corresponding linearry independent (Without loss of generality) eigenvector $ a_1,a_2,a_3,\cdots,a_n$, we will prove that $ x_k$ is also the eigenvalue of $ M$ with the corresponding eigenvector $ [a_k,ia_k]$. $ (A\plus{}Bi)a_k\equal{}x_k.a_k$ then $ M.[a_k,ia_k]\equal{}[(A\plus{}iB)a_k,i(A\plus{}iB)a_k]\equal{}x_k[a_k,ia_k]$
Supposed that $ y_1,y_2,y_3,\cdots,y_n$ are eigenvalues of $ A\minus{}Bi$ with the corresponding linearry independent (Without loss of generality) eigenvector $ b_1,b_2,b_3,\cdots,b_n$, we will prove that $ y_k$ is also the eigenvalue of $ M$ with the corresponding eigenvector $ [i.b_k,b_k]$. $ (A\minus{}Bi)b_k\equal{}y_k.b_k$ then $ M.[i.b_k,b_k]\equal{}[(A\minus{}iB)i.b_k,(A\minus{}iB)b_k]\equal{}y_k[i.b_k,ib_k]$
It is also easy to see that {$ [a_k,ia_k]$} $ \cup$ {$ [ib_k,b_k]$} are linearry independent vectors.
Then the conclusion follows.
[/hide]


[color=darkblue]Given are the triangle $ABC$ for which $A=60^{\circ}$, the points $F_{1}\in (AB)$, $F_{2}\in (AC)$ so that $AF_{1}=AF_{2}=\frac{bc}{2p}$,
the point $S\in (BC)$ so that $\frac{SB}{SC}=\frac{a+b}{a+c}$ and the points $\{P_{1},P_{2}\}\subset (BC)$ so that $BP_{1}=P_{1}P_{2}=P_{2}B$ and $P_{2}\in (P_{1}C)$.
Prove that the lines $F_{1}P_{1}$, $F_{2}P_{2}$ and $AS$ are parallelly and $m(\widehat{BAS})=\frac{B}{2}$, $m(\widehat{CAS})=\frac{C}{2}\ .$

[b]Remark.[/b] See :
http://www.mathlinks.ro/Forum/viewtopic.php?t=105533
http://www.mathlinks.ro/Forum/viewtopic.php?t=105421

I think it is only Thales theorem:
$\frac{SB}{a+b}=\frac{SC}{a+c}=\frac{a}{2a+b+c}$ then
$SB=\frac{a(a+b)}{2a+b+c},BP_{1}=\frac{a}{3}\rightarrow \frac{BP_{1}}{BS}=\frac{2a+b+c}{3(a+b)}$
$BF_{1}=c-\frac{bc}{a+b+c}=\frac{c(a+c)}{a+b+c}$ then
$\frac{BF_{1}}{BA}=\frac{a+c}{a+b+c}$ then we always have
$\frac{a+c}{a+b+c}=\frac{2a+b+c}{3(a+b)}\leftrightarrow a^{2}=b^{2}+c^{2}-bc\leftrightarrow \widehat{A}=60^{0}$


inequalities
search
function
inequalities unsolved

[quote="duc 95"]i have tried some ways but can't solve it!![/quote]

Search for Poland 1992

[quote="duc 95"]Can you tell me the solution?[/quote]

I told you to use the search function because this inequality has been posted many, many times, and you should do the same when you want to post another famous inequality.

For example see http://www.mathlinks.ro/viewtopic.php?t=3889&search_id=898104839


Can we have a button for those two please? I often times forget a </sup> when I'm writing long equations and the whole post screws up (mainly because if I forget to close these two everything after it becomes superscripted/subscripted, including the signature)

Thanks.

mm...yes, i agree that a button would be very convient...


Alot of people own observing instruments, either for just the sake of having one or for actually use and observing.
If you own one post what it is here, including accessories, and some objects you have seen with it.

The scope I use currently is optical tube of a Meade 114-AS and an Orion AstroView EQ Mount.
It has a focal length of 1000mm and an aperture of 114mm, Newtonian Reflector.

Accessories, a Meade Broadband Nebula filter and color filters, bunch of eyepieces including a 5mm Plossl.
Objects, M42, M13 and a bunch of other clusters, M57, M31, Moon, All of the inner planets, Saturn, Jupiter, a couple double stars including Albiero.

I own a pair of strong pair of binoculars. But I think it would be better if I had a telescope though.

Cool. :)
Are they a pair of binoculars made for observing or just regular pair of binoculars?

Made for observing. They have these dials that strengthen or weaken the mirror.

i have a really bad telescope. does anyone know where i can get a decent telescope at a decent price?

Ebay.  :lol:
Or you can search some of the popular telescope brands' websites like [url=http://www.meade.com]Meade[/url] and [url=http://www.telescope.com]Orion[/url], they make alot of scopes that are pretty decent, Orion a little more decent than Meade actually.

Making your own telescope isnt such a bad project, but the quality may be
poor.

I have recently got a Galileo 720 mm refractor..or thats what it says it's called... :D
Need a bit of help...
there's something on it called Mars Eye Finderscope that is supposed to be fixed but it doesn't seem to fit on..
It's extremely hard to fit on and no-one I know can help me out...so I just wanted to know how necessary it's presence is..
Will my telescope work fine without it?
Also, if anyone knows how exactly to fit it can you please tell me?
All the manual says is "Fix Mars Eye Finderscope."...
Thanks in advance!

It is not as popular as Meade or Orion and the telescopes it sells are a little bit more cheaply priced.  The Galileo I have is not really one of its top-of-the-line scopes, so I definitely preferred my Meade when I bought it.  I did not really like the Alt-Az mount and the laser finderscope of that scope had either.
A thing to consider when buying a scope it what you will use it to look for.  Of course, personal preference plays a big role also.

[quote="AstroPhys"]Yeah, I had the same problem with my Galileo reflector, that laser finderscope-thing was kind of a "snug" fit.  The finderscope itself should have thumbscrew at its base that should loosen so it goes on easier.
Without a finderscope like that it would become quite hard to aim the telescope at something in the sky since that is what it is suppose to be used for.  Otherwise it should not effect have the telescope works.[/quote]

But HOW does it fit?? I've been trying and trying...end result- :wallbash_red:
 Anyway, I won't use it till I get some pro to fit it in for me...anyway it's so cloudy that u can't see a thing right now.
Oh by the way, what's the "Brightest star in the sky which you always see overhead and it's near to the moon?" -my bro's text book says it's Mars... :rotfl:  :rotfl:

[quote="ajai"]it depends on where you are. judging by your flag, i believe you're in india in the middle of the monsoon season....i suggest that you visit http://www.skyandtelescope.com ....then get check out the interactive sky map and almanac to determine what is around at whatever time. i use it a lot :D

about your telescope problem: there should be a screw at the base of the scope. look carefully....then twist to the left a lot to loosen the screw. slide on the scope and tighten it by turning it right...[/quote]

Thanx, but I bought 10 star maps today!  :lol:
There's this dealer who gets them from ISRO scientists and stuff so I'm ok...
And I can use the telescope..I JUST CAN'T FIT THE MARS EYE-FINDERSCOPE(tears hair in frustration)...
THAT's my problem..
but does it make a GREAT lot of difference?(this finderscope thing with a red light...)

If you want to aim the telecope at something in particular the finderscope thing helps at getting it in the rid direction.

the finderscope helps in focusing in on certain objects in the celestial sky. if you use a straight device parallel to the telescope and align it where the finder WOULD be, then  finding objects MAY be easier (note the uncertainty). i'm not absolutely sure if that would work, but it won't hurt to try. :D

:) Cool.  What are the brand, model, or specifications?