conics
parabola
calculus
function
analytic geometry
graphing lines
slope

Find an equation of the tangent line to the parabola $y=x^2$ at point $P(1,1)$.

[hide="hint"]Simple limits are used. I'm assuming this is okay, since limits are in AoPS Volume II[/hide]

this is a calculus problem...
$f'(x) = 2x$
$y = 2(x-1)+1$
more generally, if a line is tangent to a function at a point $x_0$, then the line is
$y = f'(x_0)(x-x_0) + f(x_0)$
but the idea that you are getting at is that a person would see the idea of the derivitve (2 points on a line and finding the limit of the slope of where they are the same point) and not just use the derivitive rule, good for you if you thought of this without knowing the calculus

Um doesn't this involve calculus? Or is there a clever approach without it?[hide]$\frac{d}{dx}x^2=2x$, so the line has slope $2(1)=2$  and contains the point $(1,1)$.  So putting in point-slope form:
$y-1=2(x-1)$, so $\boxed{y=2x-1}$.[/hide]

Here is the solution without using calculus.

Let $y=ax+b$ be the desired tangent line, since the quadratic function $y=x^2$ is only contact with the line at $x=1$,
we can write $x^2-(ax+b)=(x-1)^2\Longleftrightarrow ax+b=2x-1$,yielding the result derived by some guys.

Consider  for $y=x^3$ at $(1,?)$ without using calculus!

Umm, oops. I used limits. Would that be considered calculus?

I thought of a nearby point $Q(x,x^2)$. on the parabola and how the slope of the tangent line would be found by $\lim_{x\to1} \frac{x^2-1}{x-1}$

Then I solved that, got 2 as the slope, and did the rest.

[quote="4everwise"]Kunny, that's a nice solution.[/quote]

 :)

Not a calculus problem, maybe it is a simple geometry problem.

The focus of $y = x^2$ parabola is the point $F(0,\frac14)$, the directrix is the straight-line $y = -\frac14$. using geometric construction we know that the tangent by the point $(1,1)$ is the line bissector of segment $\overline{FP}$, such that the point $P$ is $(1, -\frac14)$.

$m_{FP} = \frac{y_f - y_p}{x_f - x_p} = \frac{\frac14 - (-\frac14)}{0 - 1} = -\frac12$

$t$ means the tangent line by $(1,1)$.

$m_t = -\frac{1}{m_{FP}} = 2$

Therefore, the equation of $t$ is:

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

$\boxed{y = 2x - 1}$

Sorry, but why $x^2-(ax+b)=(x-1)^2\Longleftrightarrow ax+b=2x-1$??? I understand the first part ($x^2-(ax+b)$) that is from y=x² and y=ax+b --> x²=ax+b, but I didn't understand why x²=ax+b are equal to (x-1)²

Because the quadratic equation $x^2-(ax+b)=0$ has only double root $x=1$, we can factorize $x^2-(ax+b)=(x-1)^2$

i suppose that there is a more clear way to show this, if you are finding the intersection of a line and a parabola, you will do
$x^2 = ax + b$
and you want this to be a perfect square since if it is tangent, it only intersects the parabola at one place
$x^2 - ax - b = 0$
$x = \frac{a\pm \sqrt{a^{2} + 4b}}{2}$
for there to be a double root, $a^{2} -4b = 0$ and if it is tangent to the parabola at 1, then we would solve for the intersection at $x = 1$
so $1 = \frac{a}{2}$
$a=2$
$2^{2} = -4b$
$b=-1$
giving $y = 2x-1$
quite clever, thats the bad thing about calculus, once you learn it, you don't see these clever solutions, eventhough the calculus solution is somewhat easier

ahhhhh i got it.

Thx kunny and Altheman.  :)


Let $n, k, p$ be positive integers with $2 \leq k\leq\frac{n}{p+1}$.Let $n$ distinct points on a circle be given. These points are colored blue and red so that exactly $k$ points are blue and, on each arc determined by two consecutive blue points in clockwise direction, there are at least $p$ red points. How many such colorings are there?

The original problem was wrong...I think the points must me labeled. Then we fix a blue point and we have to multiply the answer with $\frac{n}{k}$. We count in how many ways we can put n-kp-k objects in k labeled boxex. This is $\binom{n-kp-k+k-1}{k-1}$. Therefore, the final answer is $\frac{n}{k}\cdot\binom{n-kp-1}{k-1}$


For positive reals $ a,b,c$, prove that
\[ \displaystyle3\sum_{cyc}\frac{a(b\plus{}c)}{b^2\plus{}bc\plus{}c^2}\le 5\plus{}\sum_{cyc}\frac{a^2}{b^2\plus{}bc\plus{}c^2}\]
Hope it is fun. In fact I really wonder whether it is strong (sharp?) or not... just feel that it is not really so easy to see the mathematical structure about what is strong  :D

Anyone interested in this problem or would like to share the solution?

The first step of my solution is expanding, so I don't want to spoil the fun for some people... though I personally think that expanding is the best way to look into the mathematical structure of such problem.  :maybe:

This is a particular case of the inequality

$ \sum \frac {a(b\plus{}c\minus{}ma)}{b^2\plus{}bc\plus{}c^2} \le 2\minus{}m$,

which holds for $ m \ge \frac 4{\sqrt 3}\minus{}2$.


geometry
rotation
inequalities
geometric transformation
trigonometry
triangle inequality

In the figure shown, point $P$ is located inside square $ABCD$. If $PA = 10$,
$PB = 6$ and $PC = 14$, find the area of the square.

[img]6722[/img]

good sombody post the solution? I cant quite grasp what to do with the two solutions here. thanks :)

Can somebody elaborate more the solutions (or hints) presented? Thanks.

[hide="observation"]From the British Flag theorem,

$\overline{PD}^{2}+6^{2}=10^{2}+14^{2}$

$\overline{PD}=2\sqrt{65}$

From the Triangle inequality, 3<AB<16, 8<BC<20. Since AB=BC, 8<AB=BC<16.[/hide]

Was posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=125919]here[/url], solution is second to last post.

[color=brown]Without rotation[/color][hide]
I will solve the problem with $AP=5,BP=3,CP=7$ (dilated by 2).
Let $AB=x,\angle ABP=\delta.$
By cosine theoreme $\cos \delta= \frac{x^{2}+3^{2}-5^{2}}{2.x.3}$ and $\cos (90-\delta)=\frac{x^{2}+3^{2}-7^{2}}{2.x.3}=\sin \delta$.
So  $(\frac{x^{2}+3^{2}-5^{2}}{2.x.3})^{2}+(\frac{x^{2}+3^{2}-7^{2}}{2.x.3})^{2}=1$. Substituting $x^{2}=z$ ($z$ is the required area) and simplifying we get $z^{2}-74z+928=0$ with roots $z_{1}=16$ and $z_{2}=58$.
When $z_{2}=58, x=\sqrt{58}>7$ P is inside ABCD, so this is the solution, but we see that when $z_{1}=16, x=4<7$ and this is another solution where [b]P is outside [/b] the square.[/hide]

[quote="scorpius119"][hide]
Suppose $P$ is rotated 90 degrees counterclockwise around $B$ to point $Q$. Then $BQ=6$, $CQ=10$, and $\angle PBQ=90^\circ$. This gives you enough information about quadrilateral $PBQC$ to find the length of diagonal $\overline{BC}$.[/hide][/quote]

How do we know that $CQ=10$?

[quote="computer_nuke"]How do we know that $CQ=10$?[/quote]

When rotating about B by 90 degrees, the point that was on A lies on C, so $\overline{AP}=\overline{CQ}=10$


[quote="themonster"]Does this mean that you have to be registered in that party? Most of us here are probably under 18 and so that wouldn't be possible. Or does it mean the party whose platform best aligns with your views?[/quote]

The second one.

Just for the record, there are two major political parties in America, the Republican and the Democrat[i][b]ic[/b][/i] parties -- there is no "Democrat Party."

[quote="furious"]alligning yourself with a political party is being closed minded. sure you will like some of their ideas, and maybe even alot of them. but then the party plans to run someone not capable of being president or another office and not capable doing a good job at it (not that this has even happened). but you're alligned with that party so you vote for a not-so-good president or any other office position for that matter.[/quote]

This is a silly argument.  By registering for a party, do I somehow lose my ability to make rational voting decisions?  The fact that I identify myself with a certain set of beliefs doesn't suddenly render my unable to discern if a particular candidate is incompetent.  Should I also not have a political philosophy in order to avoid being closed minded?
(In fact, in New York City, not registering would be to my detriment since we have closed primaries and frequently it is the party primary that determines who the officeholder will be.)

[quote="ankur87"]None. Associating with a party is an influence I don't need. I just roll with whatever I think is right.[/quote]

Exactly- although I tend to the left.

I haven't figured out which of right and left is liberal and which is conservative...help?

Left is said to be liberal and right is said to be conservative.
 :D

I am not devoted to a specific political party.  I just pick whoever seems to satisfy what I beleive is right.

There's a lot of other (4 votes at this time).  What party are the people who voted this category? Green, libertarian ...?

It seems like a lot of you don't want to belong to a specific party. But the country couldn't function without parties.  If there were not at least two, one group of people would rule, and it will be totaliterinism (im horrible at spelling).

[quote="dts"][quote="Farazy"]Left is said to be liberal and right is said to be conservative.[/quote]

In the U.S., yes, but in foreign countries, I think the opposite is sometimes true (left is said to be conservative, right is said to be liberal).[/quote]

You know what really caused a problem in terminology was the fall of the USSR -- as it was falling apart, the communists (= hard left) were also the people trying to preserve some status-quo (= conservatives).

ln(dx/dy), could you perhaps elaborate?  I'm slightly confused how a system without two major political parties would result immediately in a totalitarianism.

or in countries like belgium, there has been a 'left' governement for such a long while, that the 'left' parties (socialists) only try to block any change, and any move to the 'right' is a big change. So progressive/conservative are pretty much relative and country-dependent.

i'm 15, and most of my friends, classmates, etc. do support a certain political party but thats only due to parents, and the majority of the people of the state. do you think that most people are like that, or they actually firmly believe in the views of the political parties???

my political party is a little outdated, but it is still evident in today's society. i'm a democratic-republican :D

[quote="Peter VDD"][quote="ln(dx/dy)"]For the people who said other, what are you?[/quote]what we are? non-americans perhaps? :roll:[/quote]
indeed  :lol:
it's not like in the USA everywhere, you know? we don't have 'democrats' and 'republicans' like you do...  :)

Well, I agree that it's good to be open-minded.  However, in our country, there are two major parties, and independents have virtually no chance in winning an election.  This means that it's probably a good idea to be behind one of them most of the time.  And I'm proud to be a Democrat (yes, they're not perfect either, but anyone's better than Bush).

[quote="bubala"]
And I'm proud to be a Democrat (yes, they're not perfect either, but anyone's better than Bush).
[/quote]
I agree, although I voted for other. I am a "gamer" :P


This is a poem machine.
*points to this topic* :P

The object of the game is to make a poem, with help from people around the world that have intrests in math(a.k.a. members of AoPS). One person posts a line or verse(their choice) that rhymes and still makes a poem. To make the poem snazzy, I would like to make everyone follow a certain rhythm per poem, like in "The Raven" by Edgar Allen Poe. Also, please bold your lines.

Here is he first line:

[b]In an intimidating castle down the bay,[/b]

[b]there was a crazy knight screaming "yay"[/b]

:rotfl:  :rotfl:

That's not in bold.

[b]The Captain of the Guard said "I'll take away"[/b]

[b]But the crazy knight kept screaming "yay"[/b]

please edit that so the poem makes sense.

How should I edit that :(
what does the captain mean by "I'll take away"?

You are supposed to continue the sentence.

[b]"...your favorite horse for sixty-five days"[/b]

[quote="nutz_for2.718281828"][b]"Captain, please, let him stay!"[/b]
[b]"You can't disagree to what I say,[/b]
[b]I'm the Captain, now I slay[/b]
[b]Thee for rising away[/b]
[b]From the rank of knight!" He went away[/b]

try to build up from there. That's a good verse in progress.

P.S. the captain should go for something.(a hint)[/quote]

Line 3 doesn't really flow...

So let's just have the poem this far:

[b]
In an intimidating castle down the bay,
there was a crazy knight screaming "yay"
The Captain of the Guard said "I'll take away
your favorite horse for sixty-five days"
"Captain, please, let him stay!
I'll give anything, let me pay!"[/b]

[b]so the captain puts a pin in a stack of hay,
and said "If you find it out then I may..."[/b]

Do you people remember chocoboom's "continueing poem..."?

[b]Change my mind and let him stay."[/b]

I think it would be better if each person posted a couplet. Normally poems just have 2 lines rhyming, not a ton having the same ending.

[b]And so saying he bowed and trudged away.[/b]

Till now it makes sense. The next person start a new stanza witha a new rhyme, and then on everybody posts two lines each.

ok. Four lines per ending.

[b]The knight hung his head and kicked a stone,[/b]
[b]Just because he was all alone,[/b]

[b]What terrible luck! Not even a phone
To send out an order for an ice cream cone[/b]
(does that make any sense? I'll edit it if you guys don't like it...)

But out in the distance, something was stirring,
It was coming closer, whistling and whirring.

(is whirring a word?)

we can change the rhyme scheme, it changes in many poems.  just make it the same for a couple of stanzas so its clear where it changes.

guys, do it in [b]bold[/b]
[b]
knights came along
with their arms streched long
holding a big bowl of honey
yelling,"give me ur money"[/b]

[b]So the knight went back to the stack of hay
And searched for the little pin all through the day[/b]

Will knights be the only characters here?
Anyway tspa's line was great. and whirring [i]is[/i] a word, star99.

[b]He snuck away, it was easy, no doubt,[/b]
[b]But Sam I Am came up to shout[/b]

 :D  :D

[b]"what is up with all this fuss???
all your horse are belong to us!!!"[/b]

[b]then george bush arrived
with the CIA by his side[/b]

[b]"Georgey dear, don't yell,"[/b]
[b]said his wife, who looked well.[/b]

[b]George went to Iraq and fell,
and rightfully was sent to h*ll[/b]

EDIT THAT!!!!!!!(I'm a VERY strong supporter of Bush.

[b]in h*ll he heard the babbitt kid
and swooned on the spot, oh yes he did![/b]

@nutz: Why?

[b]God said "I made a mistake"[/b]
[b]I'll put you in heaven to rake.[/b]

Bush was to stupid to realize
This god was someone in disguise

So real God knew quite well
He should send Bush down to h*ll

(to nutz for e)
[b]for george led this country to rot,
and edit my post i shall not!
i have a right to post my view,
it won't be changed by the likes of you![/b]

lol not necessary.  it's a high school thing :lol:

anyways, true story:
[b]so treething sent me math to do,
but it made me feel quite blue,
for i had not on earth a clue. . .[/b]

. . .a long time later. . .

[b]he told me there were typos two![/b]


AMC
AIME
algebra
polynomial
function
quadratics
quadratic formula

13.  Let $P(x)$ be a polynomial with integer coefficients that satisfies $P(17)=10$ and $P(24)=17$. Given that $P(n)=n+3$ has two distinct integer solutions $n_1$ and $n_2$, find the product $n_1*n_2$.

Uggggggggggghhhhhhhhh, I can't see a real nice solution to this problem.  Here's what I did.

[hide]Consider the polynomial $\ Q(x)=P(x) - n - 3$. After that I pounded out the horrible system of equations.  [/hide]
Does anyone have a more elegant method?

[hide]write it as $P(x)=(x-17)(x-24)Q(x)+R(x)$
since there are only 2 roots, $Q(x)=a$, and since the divisor is degree 2, $R(x)$ has degree 1 or 0
$P(x)=a(x-17)(x-24)+bx+c$, then use the 2 function values to find b and c, we get it to be: $P(x)=a(x-17)(x-24)+x-7$

then $a(n-17)(n-24)-10=0$, use the quadratic formula, and you find that it is only integer solutions for $a=-1$, we find those roots to be $19$ and $22$, so our answer is $\boxed{418}$[/hide]


LaTeX
trigonometry
quadratics
algebra
quadratic formula

Try using $\LaTeX$.

[hide="Click me!"]Factoring the equation,
$(2\cos(x)-1)(\cos(x)-5) = 0$.
$2\cos(x)-1=0$ $\textrm{AND}$ $\cos(x)-5 = 0$.
$\cos(x)=\frac{1}{2}$ $\textrm{AND}$ $\cos(x) = 5$.

Obviously, the latter equation is never true.
The first equation is true at $x=\frac{\pi}{3}$ $\textrm{and}$ $x=\frac{5\pi}{3}$. [/hide]

[quote="krnboy32"]Find all solutions in the interval [0 , 2pi] of 2(cos(x))^2-11cos(x)+5=0 thanks.[/quote]

[hide]$2\cos^{2}x-11\cos x+5 = 0$

Letting $t=\cos x$, $2t^{2}-11t+5=0$

So by the Quadratic Formula, $t=5$ or $t=\frac{1}{2}$.

Thus, $\cos x=5$ or $\cos x=\frac{1}{2}$.

Obviously the first case is extraneous, so $\cos x=\frac{1}{2}$ and $x$ is $60^{\circ}$ or $300^{\circ}$.[/hide]

[hide="solution"]Let y=cos(x).

$2y^{2}-11y+5=0$

$\frac{11\pm9}{4}=y$

For x to be a real number, y=1/2. Therefore, $x=\frac{\pi}{3}$ and $x=\frac{5\pi}{3}$[/hide]


Find all the ploynominals $p(x)$ satisfying :
$(\forall x \in \mathbb R) : (x-8)p(2x)=8(x-1)p(x)$

Observe that the only constant $p$ that works is the zero polynomial. Otherwise suppose $p$ is non-constant. Setting $x = 1$ and $x = 8$ gives $p(2) = 0$ and $p(8) = 0$ respectively. Now, setting $x = 4$ gives $p(4) = 0$. Suppose the degree of $p$ is $k$. If we write $\frac{p(2x)}{p(x)} = \frac{8(x-1)}{x-8}$, then for suitably large $x$ we have $2^k = 8$, implying that $k = 3$. It is then easy algebra to show that $p(x) = a(x-2)(x-4)(x-8)$ is a solution for all real $a$.


What is the distance from the vertex to the focus of the parabola defined by the following equation:

$ x^2 \minus{} 8x \minus{} 12y \plus{} 16 \equal{} 0$

Please explain.

You can use the link below to figure out the the focus and the vertex.  Then just use the distance formula.

http://en.wikipedia.org/wiki/Parabola

The given equation factors into: \[ (x\minus{}4)^2\equal{}12y\] Using the standard conic form of a parabola \[ (x\minus{}h)^2\equal{}4p(y\minus{}k)\], we get $ h\equal{}4, p\equal{}3, k\equal{}0$. Also note that the axis of symmetry is parallel to the y-axis. Thus the vertex is at $ (4,0)$ and the directrix is the line $ y\equal{}\minus{}3$. Now the vertex is equidistant from the focus and the directrix (definition of parabola), and since the distance from the vertex and the directrix is 3, the distance from the vertex and the focus is also $ \boxed 3$.

In general, the distance from the focus to the vertex (and vertex to directrix) of $ y\equal{}a(x\minus{}h)^2\plus{}k$ is $ \frac1{4a}$, which is how Yongyi got his second equation.

[hide="Proof"]
Let the quadratic be $ y\equal{}a(x\minus{}h)^2\plus{}k$, so that we have vertex $ (h,k)$. By the definition of a parabola, all points are equidistant from the focus and the directrix. Thus, our focus and directrix are $ (h,k\plus{}p)$ and $ y\equal{}k\minus{}p$ for some $ p$.
To ease up our algebra, we first transform this parabola to $ y\equal{}ax^2$ (since this preserves $ p$). This has focus $ (0,p)$ and directrix $ y\equal{}\minus{}p$. To make the distance calculations easier, we utilize the fact that the point $ \left(\sqrt{\frac pa},p\right)$ is on the parabola. The distance from this to the focus $ (0,p)$ is $ \sqrt{\frac pa}$, and the distance from this to the directrix is $ 2p$. We set these equal to obtain $ \sqrt{\frac pa}\equal{}2p\implies\frac pa\equal{}4p^2\implies p\equal{}\frac1{4a}$, as desired.
[/hide]


1. calculate the volume of a 3.5 mol samlpe of carbon dioxide at 30 degree celsius and 0.90 atm.

I tried tio work it in every possible way but not getting anywhere with it

2. what is the normality of a solution prepared by dissolving 37.5 g citric acid, a triprotic acid with molar mass of 192.14 g in water to make 250. mL solution.

3. how many grams of C will be consumed when 5.00 grams of Na2SO4 react according to the balanced reaction shown?
NaSO4 + 2C --->Na2S + 2 CO2

I need help knowing how to work these can you please show explanation

Thanks in advance

1. Just use the ideal gas law PV = nRT.

[hide="Answer"]$V = \frac{3.5mol \times 0.082LatmK^{-1}mol^{-1}\times (273.15+30)K}{0.90atm}= 96.7L$[/hide].

2. Oneness, what is the definition of normality? Once you know that, you are done.

[hide="Answer"]In this case, normality is the triple of molarity, so

$N = 3 \times \frac{\frac{37.5g}{192.14g/mol}}{0.250L}= 2.34 N$
[/hide].

3. The reaction is

$Na_{2}SO_{4}+2C \longrightarrow Na_{2}S+2CO_{2}$.

[hide="Answer"]The molar mass of sodium sulphate is 142.0 g/mol, and so 5.00 g correspond to 0.0352 mol. Now, according to the reaction stoichiometry, one mole of Na2SO4 reacts with two of C, so 0.0352 mol of Na2SO4 will have to react with 2x0.0352 = 0.0704 mol of C, which corresponds to 0.846 g of carbon.[/hide]


geometry
circumcircle
angle bisector
perpendicular bisector
geometry proposed

Let $ ABC$ be a triangle and $ H$ be its orthocentre. Let $ D$ and $ E$ be the midpoints of the segments $ BC$ and $ AH$, respectively. Denote by $ F$ the orthogonal projection of $ H$ on the internal angle bisector of angle $ BAC$. Prove that $ D$, $ E$ and $ F$ are collinear.

Already posted by Virgil Nicula.Why unknown? http://www.mathlinks.ro/Forum/viewtopic.php?t=169444

I mean I don't know the exact author of this problem. And in the solution of this problem, we have to consider two cases when triangle $ ABC$ is acute and when it's obtuse. Now, I'll present my solution (in case triangle $ ABC$ is acute)

Denote by $ B_1$, $ C_1$ the orthogonal projections of $ B$ and $ C$ on $ AC$, $ AB$, respectively.
We have $ EB_1\equal{}EC_1$ and $ DB_1\equal{}DC_1$, therefore $ DE$ is the perpendicular bisector of the segment $ B_1C_1$.

On the other hand, $ F$ is the projection of $ H$ on the angle bisector of angle $ BAC$, which implies that $ F$ is the middle point of arc $ B_1C_1$ of the circumcircle of quadrilateral $ AB_1HC_1$. It means $ F$ is also lies on the angle bisector of the segment $ B_1C_1$.

Hence, we conclude that $ D$, $ E$ and $ F$ are collinear as desired.

B.t.w. Thank you for your link.

Dear,April,Where did you find this problem?This is very strange that i and you found this problem in one week? :huh:
P.S:Nice solution for a nice problem!

Thank you for your interest. But I think it's not very strange. Because as [b]Kostas Vittas[/b] always quotes: "Nothing is new, under the geometry sun" - [b]Francois Rideau[/b]

And I found this problem in my old notebook.

And to add another link: http://www.mathlinks.ro/Forum/viewtopic.php?t=23523 , though it is not the complete problem.

  darij


abstract algebra
invariant
linear algebra
linear algebra unsolved

Let $ a_0, a_1, \ldots, a_m \in \mathbb{C}$, with $ a_0 \neq 0$ and $ X^p\minus{}1$ does not divide $ a_mX^m \plus{} \cdots \plus{} a_1 X \plus{}a_0$ for any prime $ p$.
$ G$ is a multiplicative group of $ n\times n$ complex matrices s.t. for all $ A,B \in G$, $ \sum_{k\equal{}0}^{m}a_kA^{k}BA^{m\minus{}k}$ is nilpotent.
Show that $ G$ is abelian.

Here a counterexample: $ G$ is the group of regular upper triagonal matrices with $ n > 1$ and  $ f(X) \equal{} (X \minus{} 1)(X \minus{} \alpha)^m$ for $ \alpha\neq \minus{} 1$, $ m\ge 0$.

Note that if $ n>0$ we necessarily have $ f(1)\equal{}0$ such that $ X\minus{}1$ is a divisor of $ f(X)$.

PS: Idea developed in private communication with loup le blanc. Merci beaucoup pour la coopération!

You are right.
I forgot an important detail: $ G$ is finite.
It should work now, I have a proof;)

Hmm, this [b]is[/b] an important piece of information. It might be good to double check problems which seem to be not too easy. Anyway, if you have a proof why not posting it? ;)

[quote="Third Edition"]Hmm, this [b]is[/b] an important piece of information. It might be good to double check problems which seem to be not too easy. [/quote]
Everyone can make a mistake now and then, right?

[quote="Third Edition"]Anyway, if you have a proof why not posting it? ;)[/quote]
I didn't think the purpose of the site was to ask question and answer them yourself, but I can post an answer in a couple of weeks if there is none.

[quote="JC_math"]... but I can post an answer in a couple of weeks if there is none.[/quote]Why waiting to post your solution? I must admit that I am curious to see the trick(s) to tackle this problem.

Sorry for the long wait Third Edition.
I was busy and I moved on to other topics since I posted this qestion.
I managed to find the outline of a solution (although I think when I initially posted that question, I had a shorter solution).

Suppose $ G$ is a minimal non abelian counterexample.
Then we proceed in several steps but the most crucial one (and non trivial) is the following:

[u][b]Step 1:[/b][/u] Show that there exist two primes $ p$, $ q$, a subspace $ V\subset \mathbb{C}^n$ of dimension $ p$, invariant under $ G$ and the restriction $ G_{|V}$ is similar to the group generated by $ A,D$, where $ a_{i,j}\equal{}a$ if $ i\equal{}j\plus{}1$ or $ i\equal{}1$ and$ j\equal{}n$, with $ a^{p^k}\equal{}1$ for some integer $ k$; and  $ D$ is diagonal (non scalar), with $ d_{ii}$ is equal to the product of a $ q$-root of unity $ \delta_i$ and a $ q^{\ell}$-root of unity for some integer $ \ell$.
Then you can assume that $ V\equal{}\mathbb{C}$ and $ a\equal{}q^{\ell}\equal{}1$.

HTH
If not, let me know. A full detailed solution (using that path) would probably take 2 to 3 pages

Thanks for this technical intermediate result. Highly appreciated if you could provide a reference such that there's no need for you to tex things. Thanks again.


vector
analytic geometry
geometry theorems
geometry

Can someone tell me how to interpretate the complex numbers as vectors?

Thanks in advance.

Oh, come on! After this much time, no one can give a simple and straight answer?
Just please help a beginner for God's sake !

Just consider com;ex number $ z\equal{}x\plus{}iy$ as vector $ \overline {OZ}$ with coordinates $ (x,y)$. :wink:

Don't take this the wrong way, but http://en.wikipedia.org/wiki/Complex_plane


calculus
integration
trigonometry
function
logarithms
calculus computations

Hello
$ \displaystyle \int\limits_0^{\frac{\pi }{2}} {\frac{{t\cos t}}{{1 \plus{} {{\sin }^2}t}}dt}$

Can be calculated with residue theory or closed form functions?

hello, the result of your integral contains the so-called dilog function and the value of your integral is approximate $0.38840969994$.
Sonnhard.

Let

\begin{eqnarray*} I & := & \int_{0}^{\pi} \frac{x \cos x}{1 + \sin^2 x} \, dx \\
J & := & \int_{0}^{\frac{\pi}{2}} \frac{x \cos x}{1 + \sin^2 x} \, dx.
\end{eqnarray*}

Then

\begin{eqnarray*} I
& = & \int_{0}^{\pi} \frac{x \cos x}{1 + \sin^2 x} \, dx \\
& = & \int_{0}^{\frac{\pi}{2}} \frac{x \cos x}{1 + \sin^2 x} \, dx + \int_{\frac{\pi}{2}}^{\pi} \frac{x \cos x}{1 + \sin^2 x} \, dx \\
& = & \int_{0}^{\frac{\pi}{2}} \frac{x \cos x}{1 + \sin^2 x} \, dx + \int_{0}^{\frac{\pi}{2}} \frac{(x - \pi) \cos x}{1 + \sin^2 x} \, dx \\
& = & 2J - \pi \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{1 + \sin^2 x} \, dx \\
& = & 2J - \pi [ \arctan(\sin x) ]_{0}^{\frac{\pi}{2}} \\
& = & 2J - \frac{\pi^2}{4}.
\end{eqnarray*}

So it remains to calculate the value of $I$. In fact, one can show that $I = \log^2 (1+\sqrt{2}) - \frac{\pi^2}{4}$. You can refer to [url=http://sos440.tistory.com/83]here[/url] for the evaluation of $I$. It is written in Korean, but one would have little difficulty recognizing each calculation in it. Therefore we have

$\int_{0}^{\frac{\pi}{2}} \frac{x \cos x}{1 + \sin^2 x} \, dx \ = \ \frac{1}{2}\log^2(1+\sqrt{2}) \ \approx \ 0.38840969994784799075\cdots .$

By Integral parts, I got $\int_0^{\frac{\pi}{2}} \frac{t\cos t}{1+\sin ^ 2 t}dt=\frac{\pi ^ 2}{8}-\int_0^{\frac{\pi}{2}} \tan ^ {-1}(\sin t)dt.$

How could we calculate $\int_0^{\frac{\pi}{2}} \tan ^ {-1}(\sin t)dt?$

Now we can give reply to http://www.artofproblemsolving.com/Forum/viewtopic.php?f=296&t=287561, thanx to sos440


a)PMI; b)if i<j then (a_i,a_j)=(a_i,c), but (a_i,c)=1 again by PMI.  :huh:


[quote="sharkman"]The ratio of boys to girls in a group is 2:3. If 4 more girls join the group the ratio becomes 2:5. How many boys are in the group?[/quote]

[hide="Solution"]
$ \frac {x}{y} \equal{} \frac {2}{3}$ so $ 3x \equal{} 2y$.
$ \frac {x}{y \plus{} 4} \equal{} \frac {2}{5}$ so $ 5x \equal{} 2y \plus{} 4$.
Substituting, $ 5x \equal{} 3x \plus{} 4$, so $ x \equal{} 2$.
[/hide]

Shouldn't x = 4?

2(y + 4) = 2y + 8, right?


search
number theory unsolved
number theory

Find all natural numbers "m" and "n" such that $ 3^{m}\equal{}26^{n}\plus{}53$. The Problem is from Bulgarian Spring Mathematical Contest  2006 for 8th grade.

If we take congruences mod 3 and 4 we get that m and n are of the form n=2a  m=2b with a,b naturals  then we can rewrite the equation as
(3^a-26^m)(3^a+26^m)=53  since 53 is a prime the only posibility is  3^a-26^m=1   3^a+26^m=53   summing up these two expressions we have 2(3^a)=54  then a=3 ( and replacing in the equation b=1)  then the olny solutions are
n=6  m=2 :)
Daniel

Good work [b]lambruscokid[/b]. Mine solution is a bit different. First of all I obtain that $ 3^{m}$ should end in 9. Therefore $ m\equal{}4k\plus{}2$ $ (k\ge 0)$. Then I transform the equation into $ (3^{2k\plus{}1}\plus{}1)(3^{2k\plus{}1}\minus{}1)\equal{}52(26^{n\minus{}2}13\plus{}1)$. Now I obtain $ 8 | (3^{2k\plus{}1}\plus{}1)(3^{2k\plus{}1}\minus{}1)$ because this is product of two consecutive even numbers. But $ 8 | 52(26^{n\minus{}2}13\plus{}1)$ only when $ 26^{n\minus{}2}\equal{}1$ i.e. $ n\equal{}2$. Now we can easily find that $ m\equal{}6$. Finally the solutions are $ (m,n)\equal{}(6,2)$.


algebra
polynomial
search
invariant
function
algebra unsolved

I am sure I saw it. The most probable place is ML, of course :?

Some incredible facts about this problem: it was discovered by Johannes Faulhaber of Ulm and it was by Paul Mills to prove an old conjecture of Erdos:
  The equation $1^n+2^n+...+(m-1)^n=m^n $ does not have other solutions than those given by $ 1+2=3$. The proof of the conjecture after using the result of this problem is just a formality.

Yes, I know, I almost had a heart attack when seeing that. By the way, I am not sure that what you saw is this strong result, but rather a weaker one, which guaranties the divisibility of the polynomials $P_n$ by the polynomials $P_1$ and $P_2$. This is much stronger.

Indeed, but I don't know how to search such statement on forum. Too few keywords...

[quote="harazi"]I really do not know how to prove the following result:
  Define the polynomials $ P_n$ such that $ P_n(m)=1^n+2^n+...+m^n$ for all m. Prove that there exists polynomials with rational coefficients $Q_n$ such that $ P_{2n}(x)=x(x+1)(2x+1)/6 Q_{2n}(x(x+1)/2)$ and $ P_{2n+1}=x^2(x+1)^2/4 Q_{2n+1}(x(x+1)/2)$[/quote]

I think there is some missing statement about the coefficients being integers (or 1/2 integers, etc).   This is the main point of Faulhaber's polynomials, otherwise it is just a trivial fact that a polynomial $f(y)$ invariant under $y \to -y$ is a function of $y^2$, and a polynomial with $f(-y) = -f(y)$ is $y g(y^2)$ for some $g$.  Here $y = x+1/2$.

This is all that was asked. Anyway, could you please be more explicit in your explanations?

Consider the polynomial $P(x)$ with

$P(0) = 0$  (the summation starts from 0)   and
$P(n) - P(n-1) = n^k$.

Suppose $k = 2m+1$  is odd.

For positive $n$,  $P(n) = 1^k + ... + n^k$.
For negative $n$,
$P(0) - P(-1) = 0$
$P(-1) - P(-2) = -1^k$
....
$P(-n) - P(-n-1) = -n^k$

adding, we find that

$P(-n-1) = P(n) $.

As this is true for infinitely many $n$,   $P(x) = P(-1-x)$ for all $x$

i.e.   $P(y) = P(-y)$  where  $x = - 1/2 + y$

So $P$ is a function of $y^2 = (x+1/2)^2 = x^2 + x + 1/4$; which is a function of $x(x+1)$.

Similar argument for $k=2m$ even.

Again, I think the main point is some integrality statement about the coefficients of Faulhaber's polynomials, not their existence.


Prove that $ \cos 1^\circ$ is an irrational number.

First find (or look up) a formula for $ \cos{5x}$, which shouldn't be hard to find.  Now, we know that $ \cos{45\minus{}30}\equal{}\cos{45}\cos{30}\plus{}\sin{45}\sin{30}\equal{}\frac{\sqrt{3}\plus{}1}{2\sqrt{2}}$, and this is clearly irrational.  Therefore, $ \cos{15}\equal{}16\cos{3}\minus{}20\cos{3}\plus{}5\cos{3}\equal{}\frac{\sqrt{3}\plus{}1}{2\sqrt{2}}$.  If $ \cos{3}$ was rational, then $ 16\cos{3}\minus{}20\cos{3}\plus{}5\cos{3}$ would be rational, which is impossible, so $ \cos{3}$ is irrational.  Finally, using $ \cos{3x}\equal{}4\cos^{3}{x}\minus{}3\cos{x}$, $ \cos{3}\equal{}4\cos^{3}{1}\minus{}3\cos{1}$. And by the same reasoning as above, we conclude that $ \cos{1}$ is irrational.

In fact, $ \cos{1}$ is transcendental, but this is more difficult to prove.


calculus
integration
logarithms
trigonometry
real analysis
real analysis unsolved

i hope i have not asked this one before...  :maybe:

show :D

$ \int_{0}^{1}\;\int_{0}^{1}\;\;\frac{x}{(1\plus{}x^{2}y^{2})\cdot\ln\,(xy)}\;\;\textbf dx\;\textbf dy\;\;\equal{}\;\;\boxed{\ln\left(\frac{\sqrt{\;2\pi^{3}\;}}{\Gamma^{2}\left(\frac{1}{4}\right)}\right)}$

Substitute $ u \equal{} xy \quad \int_0^1 \int_0^y \frac {\frac {u}{y}}{(1 \plus{} u^2) \, \ln u} \frac {du}{y} dy$ and reverse integration order

$ \int_0^1 \underbrace{\int_u^1 \frac {dy}{y^2}}_{(1 \minus{} u)/u} \frac {u}{(1 \plus{} u^2) \, \ln u} \, du \equal{} \int_0^1 \frac {1 \minus{} u}{\ln u} \underbrace{\frac {1}{1 \plus{} u^2}}_{1 \minus{} u^2 \plus{} u^4 \minus{} ...} \, du \equal{} \sum_{k \equal{} 0}^\infty ( \minus{} 1)^k \ln \frac {2k \plus{} 1}{2k \plus{} 2}$

$ \equal{} \ln \prod_{k \equal{} 0}^\infty \frac {(4k \plus{} 1)(4k \plus{} 4)}{(4k \plus{} 2)(4k \plus{} 3)} \equal{} \ln \prod_{k \equal{} 0}^\infty \frac {\left(k \plus{} \frac14\right)(k \plus{} \frac44)}{(k \plus{} \frac24)(k \plus{} \frac34)} \equal{} \ln \frac {\Gamma\left(\frac24\right) \, \Gamma\left(\frac34\right)}{\Gamma\left(\frac14\right) \, \Gamma\left(\frac44\right)} \equal{} \ln \frac {\sqrt {\pi} \;\;\, \frac {\pi}{\sin \frac {\pi}{4}} \, \frac {1}{\Gamma\left(\frac14\right)} }{\Gamma\left(\frac14\right)\cdot 1}$

[quote="Dr. Doe"]
$ \int_0^1 \frac {1 \minus{} u}{\ln u} \underbrace{\frac {1}{1 \plus{} u^2}}_{1 \minus{} u^2 \plus{} u^4 \minus{} ...} \, du \;\equal{}\; \sum_{k \equal{} 0}^\infty ( \minus{} 1)^k \ln \frac {2k \plus{} 1}{2k \plus{} 2}$
[/quote]
doe, how did you do the above step ?   :|

$ \int_0^1 \frac{u^b\minus{}u^a}{\ln u}\, du\equal{}\int_0^1 \int_a^b u^x \, dx\, du
\equal{}\int_a^b \int_0^1 u^x \, du\, dx$ $ \equal{}\int_a^b \frac{1}{x\plus{}1}\, dx\equal{}\ln(1\plus{}x)\Big|_a^b\equal{}\ln \frac{b\plus{}1}{a\plus{}1}$

$ \frac{1}{1\plus{}u^2}\equal{}1\minus{}u^2\plus{}u^4\minus{}u^6\plus{}...$ $ \Rightarrow (1\minus{}u)\, \frac{1}{1\plus{}u^2}\equal{}(1\minus{}u)\minus{}(u^2\minus{}u^3)\plus{}(u^4\minus{}u^5)\minus{}(u^6\minus{}u^7)\plus{}...$

$ \int_0^1 \frac{1\minus{}u}{\ln u}\, \frac{1}{1\plus{}u^2}\, du$$ \equal{}\int_0^1 \frac{1\minus{}u}{\ln u}\, du\minus{}\int_0^1 \frac{u^2\minus{}u^3}{\ln u}\, du\plus{}\int_0^1 \frac{u^4\minus{}u^5}{\ln u}\, du\minus{}\int_0^1 \frac{u^6\minus{}u^7}{\ln u}\, du\plus{}...$

$ \equal{}\ln\frac12\minus{}\ln \frac34\plus{}\ln \frac56\minus{}\ln \frac78\plus{}... \equal{}\sum_{k\equal{}0}^\infty (\minus{}1)^k \ln \frac{2k\plus{}1}{2k\plus{}2}$


Find all positive integer solutions to the equation:

$\frac{1}{x}+\frac{1}{y }= \frac{1}{2000}$

[hide]
The original equation
\[\frac{x+y}{xy}= \frac{1}{2000}, \]
can be rewritten as
\[xy-2000(x+y) = 0. \]
Adding $2000^{2}$ to both sides,
\[(x-2000)(y-2000) = 2000^{2}= 2^{8}\times 5^{6}\]
For each factor of the LHS, there is a corresponding solution $(x,y)$, so there will be $(8+1)(6+1) = 63$ solutions.
[/hide]


advanced fields
advanced fields unsolved

$ (2a\minus{}3b\plus{}5c\plus{}6d\minus{}11e\plus{}3f\plus{}2g)^{10}$

if I open it out into a multinomial,what's  sum of each item's modulus(constant)?

example $ (a\plus{}b\plus{}c)^{4}$ ,its constant is $ 3^{4}$

the answer is $ 4^{10}$,why?

(I'm an AoPs'  rookie.)


calculus
integration
trigonometry
calculus computations

For the first problem, note that $ y^{2}$ is just a constant so get it outside the radical.  Then you're left with $ 1\plus{}\left(\frac{x}{y}\right)^{2}$ underneath the radical.  Make the substitution $ \tan\theta\equal{}\frac{x}{y}$, $ y\sec^{2}\theta\equal{}dx$, and go from there.


[quote="vittalar"]hei everybody! does anyone know the result date of rmo tamilnadu[/quote]
It has already come...See This link
http://www.mathlinks.ro/viewtopic.php?t=314669&start=240


Let $a, b, c >0.$ Prove the following inequality:

  $\sqrt [3] {3(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}\geq\sqrt [3] {\frac{a}{b}}+\sqrt [3] {\frac{b}{c}}+\sqrt [3] {\frac{c}{a}}.$ ;)

$3=1+1+1$, rearrange the term $a+b+c$ suitably and apply Holder.

nice solution galeos... mine is not as nice as yours, but let me have a try...

cubing both sides and cancelling some terms gives that we'll have to prove that
$\displaystyle 3+\bigl(\sum_{cyc}\frac{2a}{b}+\frac{3a}{c}\bigl)\geq 3\bigl(\sum_{cyc}\sqrt[3]{\frac{a^2}{bc}}+\sqrt[3]{\frac{bc}{a^2}}\bigl)$
but by am-gm we have that the $LHS$ is greater than $6+\bigl(\sum_{cyc}\frac{2a}{b}+\frac{2a}{c}\bigl)$
then by am-gm, $1+\frac{a}{b}+\frac{a}{c}\geq 3\sqrt[3]{\frac{a^2}{bc}}$ and $1+\frac{b}{a}+\frac{c}{a}\geq 3\sqrt[3]{\frac{bc}{a^2}}$, taking the cyclic sums gives the result. :D


If you have it post your name and highest lvl, i'm Griff336, lvl 12 but i'm a lot better i play a lot of custom games, and probably as good as a lvl 14 or 15 when i'm "on".

[edit]

[i][moderator note: sometimes it takes more than a half hour to respond to a thread, let's try to be a little more patient in the future][/i]

Stating which game you are talking about would help a little here...but I can infer it is Halo 2.  Yeah, I play but I haven't for a long time.  I think I'm level 14 for team slayer.  My gamertag is cyclotomic.


I think it's [hide]d

i^4=1
i^2=-1

subsitution gives 0/i=a, which is 0[/hide]

[hide]Or, $x^4+x^2$ factors to $x^2(x+i)(x-i)$, so $i,-i,0$ are roots.

so the answer is 0 (D)[/hide]

[hide="solution"]$i^4+i^2=0$ so 0 divided by anything is 0[/hide]

[quote="AntonioMainenti"][quote="jli"][hide="solution"]$i^4-i^2=0$ so 0 divided by anything is 0[/hide][/quote]$i^4-i^2=1-(-1)=2$

Something that confuses me about these forums is why in almost every thread people post multiple solutions that are the same (e.g. $i^4+i^2=0$). I think once is enough.. (Note that my post had a slightly different way of looking at the problem.)[/quote]

oops...i meant [b]+[/b]  :blush:

[quote="TripleM"]Thats what the secret text is for - so that everyone can post solutions without knowing what other people have done first. The problems are there for everyone to solve independently, not for someone to find an answer and then everyone else ignore it..[/quote]Right, I got it now. Sorry.


Hexagon ABCDEF has the following properties:

1) diagonals AC, CE and EA are all the same length
2) angles ABC and CDE are both 90 degrees
3) all the sides of the hexagon have lengths which are different integers

a) what is the minimun perimeter of ABCDEF if AC = squareroot of 85?
b) what is the smallest length of AC for which ABCDEF has all these properties?
c) what is the minimun perimeter in this case?

Well, (a) is easy.
First, we get $AC=CE=AE=\sqrt{85}$. Triangles $ABC$ and $CDE$ are right, sides of heagon are integers, which are different hence the sides are 9,2,6,7.(maybe not in this order). We have no restriction on angle $AFE$ hence all we need is $AF$ and $FE$ to be integers so that $AF+FE>AE=\sqrt{85}$. $AF$ and $FE$ can't be 9,2,6,7, or both 5, so the smallest value of $FE+AF$ is achieved when the two side lenghts are 1,10 or 3,8. We can't have their sum 10, because the side lenghts can't be both 5 and can't be 2,6,9,or 7. Hence $min(FE+AF)=11$. The minimum perimeter is $35$.

ok... thanks, i understand now. but can anyone solve parts (b) and (c) ?


geometry
trigonometry
modular arithmetic
area of a triangle
Heron\u0027s formula

This is a collection of formulas. Post them here. You can only make up formulas. Here is mine:

area of n sided polygon with side length s:

$\frac{ns^{2}\tan(90-\frac{180}{n})}{4}$

Or, equivalently, if you are dealing with the radius then the area would be \[\textrm{A}=\frac{nr^{2}\sin\frac{2\pi}{n}}{2}.\]

Can I post problems that involve proving formulas?

Example: Prove that the area of a triangle with sides $a$, $b$,

and $c$, and whose radius is $R$ is $\frac{abc}{4R}$.

[quote="lotrgreengrapes7926"]How do you type "not equivalent to?"  :maybe:[/quote]

$\not \equiv$

quote me. :D

Another:

$[\triangle{ABC}]=\sqrt{p(p-a)(p-b)(p-c)}$, for a triangle with sides $a, b$ and $c$, and $p=\frac{a+b+c}{2}$ (Heron's formula)

[quote="lotrgreengrapes7926"][quote="nutz_for2.718281828"]area of n sided polygon with side length s:

$\frac{ns^{2}\tan(90-\frac{180}{n})}{4}$[/quote]

$\frac{ns^{2}\cot\left(\frac{180}{n}\right)}{4}$ is more...concise. :P

Fermat's Theorem:
$a^{p-1}\equiv1\pmod{p}$ for prime $p$, $a\in\mathbf{N}$ and $a\ne0\pmod{p}$.

How do you type "not equivalent to?"  :maybe:[/quote]

In general, if you ever want to make something "not", then you will either add a "n" to the prefix, or use \not before the command.

Example:
$\neq$ is \neq
$\not\equiv$ is \not\equive
$\not<$ is \not<
$\not\in$ is \not\in
$\not=$ is \not=

Hehe, I have a great formula from my figurate numbers thing I did a while back.

The $n$th figurate number for a figure with $s$ sides is given by: \[a_{n}=\frac{(s-2)(n-1)n}{2}+n=\frac{(s-2)n^{2}-(s-4)n}{2}=\frac{n[(s-2)n-(s-4)]}{2}\] I wrote it in three different forms, but it's just simple fraction rearrangement so you have it in whatever form you like best...


This is just to announce that a few annoying things in the Olympiad package has been fixed by now and the AoPS uses a new version of it. Below are the most important changes that have been made:

1) It is no longer necessary to execute the [b]unitsize()[/b] command before using [b]anglemark[/b] and a few other routines. The size of anglemarks, ticks, etc. is now controlled by the variable [b]markscalefactor[/b]. If you were a fan of the unitsize command and want to get exactly the same pictures as before, just add the line
[b]markscalefactor=1/currentpicture.xunitsize;[/b]
right after your unitsize command. Otherwise, if you like using the defaults in the anglemark, etc. but you want all the sizes to become a bit larger or smaller, merely adjust the value of markscalefactor from its default value $ 0.03$ to something that suits your taste better. Example:
[b]
size(150);
pair A=(0,0), B=(3,1), C=(1,3);
draw(B--A--C);
draw(anglemark(B,A,C));
markscalefactor=0.1;
draw(anglemark(B,A,C),red);
markscalefactor=0.015;
draw(anglemark(B,A,C),green);
[/b]
results in
[asy]size(150); pair A=(0,0), B=(3,1), C=(1,3); draw(B--A--C); draw(anglemark(B,A,C)); markscalefactor=0.1; draw(anglemark(B,A,C),red); markscalefactor=0.015; draw(anglemark(B,A,C),green);[/asy]
with the green mark twice smaller and the red mark about 3 times larger than the default black mark.

2) The possibility to use pens in [b]pathticks[/b] was added. Also, the ticks now follow the curve, not its tangent line as before. Example:
[b]
size(150);
pair A=(0,0), B=(3,1), C=(1,3), D=(2,2);
path P=A..B..C..D;
draw(P,blue);
add(pathticks(P,5,red));
[/b]
results in
[asy]size(150); pair A=(0,0), B=(3,1), C=(1,3), D=(2,2); path P=A..B..C..D; draw(P,blue); add(pathticks(P,5,red));[/asy]
Note that the pen argument is a part of the [b]pathticks[/b] command, not of the [b]add[/b] command.

3) A few outright bugs have been fixed. In particular, it is now possible to make anglemarks with one arc and to find the orthocenters of triangles with obtuse angles.

So, at this point the Olympiad package seems to work as it was supposed to. If you find any more bugs or have any suggestions about what other routines might be useful, post your comments in this thread. Since Maria hasn't put the new version on her web page yet, I'm attaching the file to this post. (The asy extension is not allowed in attachments, so I renamed it olympiad.txt. If you want to use it, don't forget to rename it back :)).

Upgraded geometry_dev package to the last version.

Thanks, but how can i use it? because i'm not good in these things.

[quote=Supernova283]How do I install it? I clicked on the download link, but it just took me to a page with the code in it.
Do I put it in a new Tampermonkey project, like a userscript?[/quote]

in most compilers, you would click upload, from external URL, and then type in the link with all the code in it, then for the file name in this project, just say olympiad.asy

hope this helps!!

Uh, why does [font=monospace]pathticks[/font] return a [font=monospace]pic[/font] anyway?


function
real analysis
real analysis unsolved

Find all real derivable functions f such that if a,b,c, are reals such that $ ab+bc+ca=0$ then $ f(a-b)+f(b-c)+f(c-a)=2f(a+b+c)$.

Seems it is an extension of  IMO problem 2


Find all triplets of positive rational numbers $(m,n,p)$ such that the numbers $m+\frac 1{np}$, $n+\frac 1{pm}$, $p+\frac 1{mn}$ are integers.

[i]Valentin Vornicu, Romania[/i]

Author's Solution:

Let $m=\frac ax$, $n=\frac by$, $p=\frac cz$, with $a,b,c,x,y,z$ positive integers, and $\gcd(a,x) = \gcd(b,y) = \gcd(c,z)=1$. The property in the statement can be rewritten as $\frac ax + \frac{yz}{bc}$, $\frac by + \frac{zx}{ca}$ and $\frac cz + \frac{xy}{ab}$ are integers, that is
\begin{eqnarray} xbc &\mid& abc + xyz \\ ayc &\mid& abc+ xyz \\ abz &\mid& abc+xyz. \end{eqnarray}
From (1) we deduce $x \mid abc$, and as $\gcd(a,x)=1$ we have that $x \mid bc$. From (2) and (3) we deduce similar relations $y \mid ca$, $z\mid ab$. Multiplying these 3 relations we obtain $xyz \mid (abc)^2$. (4)

Again from (1) we have $bc \mid xyz$, and from (2) and (3) we have $ca \mid xyz$ and $ab\mid xyz$, thus, after multiplication, $(abc)^2 \mid (xyz)^3$.  (5)

From (4) and (5) we conclude that $abc$ and $xyz$ have the same prime factors or they are both equal with 1. If they are both equal with 1, then a) is solved, and for b) we obtain the solution $(m,n,p)=(1,1,1)$. So, let us suppose that $abc \neq 1 \neq xyz$. Let us denote the prime factors of $abc$ and $xyz$ by $p_1$, $p_2$, $\ldots$, $p_k$, $k\geq 1$.

Let $x=p_1^{x_1} p_2^{x_2} \cdots p_k^{x_k}$, $y=p_1^{y_1}p_2^{y_2} \cdots p_k^{y_k}$, $z=p_1^{z_1}p_2^{z_2} \cdots p_k^{z_k}$, $a=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$, $b=p_1^{b_1}p_2^{b_2} \cdots p_k^{z_k}$, $c=p_1^{c_1} p_2^{c_2} \cdots p_k^{c_k}$, where $x_i,y_i,z_i,a_i,b_i,c_i$ are non-negative integers such that

\[ x_i+y_i+z_i \geq 1 , \quad a_i+b_i+c_i \geq 1 , \quad \\ x_ia_i = y_ib_i = z_ic_i = 0 \ \textrm{ for all } i \in \{1,2,\ldots, k\} = \overline{1,k}. \]

Let us suppose that for some $i \in \overline{1,k}$ we have $a_i+b_i+c_i < x_i+y_i+z_i$. Then the exponent of $p_i$ in $abc+xyz$ is $a_i+b_i+c_i$. So from (1), (2) and (3) we obtain (and after adding them side by side)
\begin{eqnarray*} x_i + b_i + c_i & \leq & a_i + b_i + c_i \\ a_i + y_i + c_i & \leq & a_i + b_i + c_i \\ a_i + b_i + z_i & \leq & a_i + b_i + c_i \\ ------ & \fbox{+} & ------ \\ x_i + y_i + z_i & \leq & a_i + b_i +c_i , \end{eqnarray*}
which is a contradiction.

Similarly, if for some $i \in \overline{1,k}$ we have $a_i+b_i+c_i > x_i+y_i+z_i$. Then the exponent of $p_i$ in $abc+xyz$ is $x_i+y_i+z_i$.
So from (1), (2) and (3) we obtain (and after adding them side by side)
\begin{eqnarray*} x_i + b_i + c_i & \leq & x_i+y_i+z_i \\ a_i + y_i + c_i & \leq & x_i+y_i+z_i \\ a_i + b_i + z_i & \leq & x_i+y_i+z_i \\ ------ & \fbox{+} & ------ \\ 2(a_i+b_i+c_i ) & \leq & 2(x_i+y_i+z_i) \end{eqnarray*}
which is again a contradiction.

Thus for all $i\in \overline{1,k}$ we have $a_i+b_i + c_i = x_i+y_i+z_i$, which means that $abc=xyz$, so $mnp=\frac{abc}{xyz}=1$.

Now, using that $abc=xyz$, we transform (1), (2) and (3) into $xbc \mid 2xyz$ and the others, and $xbc \mid 2abc$. From the latter we have $x\mid 2a$, and as $\gcd(x,a)=1$ we have $x\mid 2$. Similarly we obtain $y \mid 2$ and $z\mid 2$. So $xyz\mid 8$. Thus $abc \mid 8$. But not all of $a,b,c$ and $x,y,z$ can be even, so in fact $abc=xyz \mid 4$. A simple computation now reveals that the possible solutions up to permutations (when $(a,b,c)$ permutes $(x,y,z)$ permutes also in the same way) are $(a,b,c,x,y,z)=(4,1,1,1,2,2)$, $(a,b,c,x,y,z)=(2,1,1,1,2,1)$ and $(a,b,c,x,y,z)=(1,1,1,1,1,1)$. So the solutions for the initial problem are
\[ (m,n,p) \in \left \{ (1,1,1),\ \left( 4, \frac 12, \frac 12 \right),\ \left( 2, \frac 12 , 1 \right) \right\} , \ \textrm{ and their permutations}. \]

[quote="harazi"]My solution: put $x=mnp$. By multiplication, $\frac{(x+1)^3}{x^2}$ is integer, thus $x$ is a rational algebraic integer, thus an integer and since it is positive, it must be 1 (guess why). Thus $mnp=1$ and $\frac{m}{2}$ and so on are integers with product 8. From here the problem is solved: one finds the solutions above presented. A little bit easier, isn't it?  ;)[/quote]Haha, I expected a one line solution from either you, or iura/sasha :D Super nice!

[quote="harazi"]. , thus $x$ is a rational algebraic integer, thus an integer [/quote]

Could someone explain this please ??Thank you

Say $\frac{(x+1)^3}{x^2} = k$, then $x^3+(3-k)x^2+3x+1=0$ and if you set $x=\frac ab$ then you have $a \mid 1$ and $b \mid 1$, and $x=\pm 1$

[quote="harazi"]My solution: put $x=mnp$. By multiplication, $\frac{(x+1)^3}{x^2}$ is integer, thus $x$ is a rational algebraic integer, thus an integer and since it is positive, it must be 1 (guess why). Thus $mnp=1$ and $\frac{m}{2}$ and so on are integers with product 8. From here the problem is solved: one finds the solutions above presented. A little bit easier, isn't it?  ;)[/quote]

Very nice solution  :thumbup:

[quote="harazi"]My solution: put $x=mnp$. By multiplication, $\frac{(x+1)^3}{x^2}$ is integer, thus $x$ is a rational algebraic integer, thus an integer and since it is positive, it must be 1 (guess why). Thus $mnp=1$ and $\frac{m}{2}$ and so on are integers with product 8. From here the problem is solved: one finds the solutions above presented. A little bit easier, isn't it?  ;)[/quote]

that's the same with my solution in the olympiad :D

Similar to [b]harazi[/b]'s solution, but posting for storage (and also because I need to write this up anyway).

Set $t := mnp$.  Observe that multiplying all three equations together yields
\[
\mathbb Z\ni\left(m + \frac{1}{np}\right)\left(n + \frac{1}{mp}\right)\left(p + \frac1{mn}\right) = mnp\left(1 + \frac{1}{mnp}\right)^3 = \frac{(t+1)^3}{t^2}.
\]
This means that $t$ is a root of the polynomial $(x+1)^3 - kx^2$ for some (positive) integer $k$, implying $t$ is an algebraic integer.  But $t$ is already known to be rational; thus $t$ is an integer.  The fact that $\gcd(t,t+1) = 1$ further implies that $t = 1$ (else some prime divides the denominator but not the numerator).

Thus $m + \tfrac{1}{np} = 2m$, and analogously the other two integers in the problem statement are $2n$ and $2p$.  Their product is $8mnp = 8$, and so
\[
\{2m,2n,2p\}\in\{\{1,1,8\},\,\,\{1,2,4\},\,\,\{2,2,2\}\}.
\]
It follows that $(m,n,p) = \boxed{(\tfrac12,\tfrac12,4),(\tfrac12,1,2),(1,1,1)}$ and permutations.

Hello from the future! Here's a solution involving just simple $v_p$'s.
Let $m=a$;$n=b$;$p=c$, to avoid having to many $p$'s in this problem.
Assume there exists some prime number $p$ such that $v_p(abc)$>0. (We will contradict this assumption).
We know $v_p(ab)+v_p(bc)+v_p(ca)=v_p((abc)^2)=2v_p(abc)>0$, meaning one of $v_p(ab)$;$v_p(bc)$;$v_p(ca)$ is strictly positive. WLOG, let $v_p(ab)>0$. We have that $\frac{abc+1}{ab} \in \mathbb N$, thus $v_p(abc+1) \ge v_p(ab)$, meaning $p \mid abc+1$, and as by the assumption $v_p(abc) \ge 1$, we get $p \mid 1$, a big big contradiction.
Now the fact that $v_q(abc) \le 0$ for any prime $q$, means that there exists $N \in \mathbb N$ such that $abc=\frac{1}{N}$. Also multiplying the three relations we get $\frac{(\frac{1}{N}+1)^3}{\frac{1}{N}^2} \in \mathbb N$, meaning $\frac{1}{N} \in \mathbb N$ , thus $N=1$. Now substituting back we get $2a \in \mathbb N$;$2b \in \mathbb N$;$2c \in \mathbb N$. And it is easy to check that the only solutions are: $(a,b,c) = {(\tfrac12,\tfrac12,4),(\tfrac12,1,2),(1,1,1)}$ and their permutations.


IQ

ok lets goo

http://www.goodforum.dev.juga.ru/bibu.JPG
http://www.goodforum.dev.juga.ru/untitled.JPG

%(1234)=3421
%(6572)=7265
@(abcd)=badc
@(klmn)=lknm
%(@/.,\)=?

http://www.cybermaths.8m.com/terazi.gif

86243
23830
32434
53071
27834

BARIL
AHPAP
SEHPA
HESAP
HASAR

brothers
show that u are men with IQ
ureee:))))

i don't get any of this :?
what are we supposed to do?

TAKVIM
TAKSIM
MAKSIM
KALBIM
TAHSIL

eti slova oboznachayut sifri

120758
123456
623756
328956
123756
every number is one word

i will help u

1-t
find the others

the other problem i will tell then

[quote="DON FAVOR"]TAKVIM
eti slova oboznachayut sifri
[/quote]
what does that mean?

gunes mya blood brother or sister

great you did it very good

can u do the others?

ok,friend
[quote]
86243
23830
32434
53071
27834

BARIL
AHPAP
SEHPA
HESAP
HASAR
[/quote]
[hide]H:2,A:3,P:4,S:8,E:6,R:0,L:1,I:7,B:5
but you did a wrong at 27834,it must be 26834[/hide]
and the other
[quote]
%(1234)=3421
%(6572)=7265
[/quote]

the first one must be 3412 or the second one must be 7256 or anot.but in this case these aren't true i think.

when we got to the "terazi." question
[hide]we can put 4 cirle, 1 square into the balance[/hide]

[quote]http://www.goodforum.dev.juga.ru/untitled.JPG[/quote]
[hide](3,1),(5,2)[/hide]

[quote]http://www.goodforum.dev.juga.ru/untitled.JPG [/quote]

the cube is so

if i was you i would stop triple posting.  YOU can just edit your first post and add what you have to say.  And all of these posts are within 10 minutes of eachoter. :)

turklere halal olsun!

i know that one of exercieses was wrong:)
i did it

but other 100% true

u did it veri well


[hide] if: "taking 300% means *by 3 then the answer is:

x-2

3x-6=x

3x= x +6

2x=6

x=3  The number is [b] 3 [/b] [/hide]

[hide="And the answer is..."]Let the number be $x$:
When you subtract $2$ from the number, you get $x-2$. When you take $300\%$ of that, you get $3x-6$. Therefore,

$3x-6=x$
$2x=6$
$x=3$[/hide]

[quote="math92"]I am thinking of a number. When I subtract 2 from the number, then take 300% of the result, the number I get is my original number. What is my original number?[/quote]

[hide="and the answer to this problem is..."]
3 :juggle: [/hide]

[hide]Let x = that original number.

300%(x-2) = x
3(x-2) = x
3x - 6 = x
2x=6
[b]x=3[/b][/hide]


Either the problem very easy or there is a typo.

T.P.T.

\[ 4bc \le (a + b + c)^2 + 4a^2 \\ \Rightarrow 4bc \le a^2 + b^2 + c^2 + 4a^2 + 2ab + 2bc + 2ac \\ \Rightarrow 2bc \le a^2 + b^2 + c^2 + 4a^2 + 2ab + 2ac \\ But 2bc \le b^2 + c^2 \\ \]

but all you have proved is that 4bc/(a+b+c)^2+4a^2)<=1 summing up the similars you get that the desired inequality is smaller then 3 and this is not what we want

[quote="SuperMath"]Let $\large a;b;c \geq 0$. Prove that:
$\large \sum \frac{4bc}{(a+b+c)^2+4a^2} \leq 1$[/quote]

Dehomogenizing by setting a + b + c = 2, we transform this into $\sum\frac{4bc}{2^2+4a^2}\leq 1$, what simplifies to $\sum\frac{bc}{1+a^2}\leq 1$. So we have to prove that $\sum\frac{bc}{1+a^2}\leq 1$ for any three nonnegative reals a, b, c satisfying a + b + c = 2. This is a very hard inequality and was discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=57558 .

  Darij


Hi,
Find, according to the values of the natural integer $n$, the remainder in the division of $3^n$ by 7.

:)

You want more? See : http://www.mathlinks.ro/Forum/viewtopic.php?t=80465 and http://www.mathlinks.ro/Forum/viewtopic.php?t=80466 .

We get the cycle of remainders by taking mods:

$3,2,6,4,5,1,3,...$ for $n \equiv 1,2,3,4,5,0$(mod 6)  respectively.


inequalities
calculus
calculus computations

a) Maximize

[b]Sigma([i]i[/i]=1 to [i]i[/i]=n) X[i]i[/i]Y[i]i[/i][/b]   subject to the constraints

[b]Sigma([i]i[/i]=1 to [i]i[/i]=n) (X[i]i[/i])^2 = 1[/b] and [b]Sigma([i]i[/i]=1 to [i]i[/i]=n) (Y[i]i[/i])^2 = 1[/b]

Easier to read in $\text{\LaTeX}.$ Click on this or hold your mouse over it to see what the code looks like.

Maximize  $\sum_{i=1}^nx_iy_i$

subject to $\sum_{i=1}^nx_i^2=1$ and $\sum_{i=1}^ny_i^2=1$

Oh - and I wouldn't bother with Lagrange multipliers with this one. Cauchy's inequality tells you what you need to know.

You are so right...but part b) in the question is asking me to use this to prove the Cauchy-Schwarz Inequality...
so I need to do that by the Lagrange Multipliers way.....


Find non-negative integers a and b where  the expression a<x<b exists with $\sqrt{x}>x$.

It should be 0 and 1, since that is the only region that the square root would work and could be smaller than the original number.

Hmm.. all values of real rational numbers greater than 0 and less than 1 can be expressed as x/y, where y>x. Thus, by Multiplication Property for Equality, (x/y)*y>(x/y)*x, and thus x/y > (x/y)(x/y), x/y > (x/y)^2. Thus, all real rational numbers between 0 and 1 are greater than their square roots! DING DING. However, to prove this for irrational numbers, I am way too irrational to figure that out today.

[quote="Mikey"][quote="pakagawa"]that is the only region that the square root would work and could be smaller than the original number[/quote]

can you prove this?
that only in this region the square root is larger than the original number?[/quote]

$
\sqrt{x}>x\\
x>x^2\\
1>x
$
if x is not zero.  If x is zero then 0>0 which is false. Thus 0<x<1 .


Hello every body.
Davron which of my question do you mean I could not understand  ;)
Abdurashid


articles trigonometry geometry function calculus ratio

I read the following article published in The Australian newspaper on the weekend:

New angle on trigonometry
Brendan O'Keefe
17sep05

MATHS students of Australia, down calculators.

A University of NSW mathematician has devised what he says is a simpler, faster and more accurate method of studying the measurement of triangles.

And it can be done by hand, without calculators or trigonometric tables.

Your saviour: Associate professor Norman Wildberger.

Professor Wildberger, who will launch his new method in a book, Divine Proportions: Rational Trigonometry to Universal Geometry at UNSW next Tuesday, said 2000 years of trigonometry had turned students off mathematics because it relied on complex circular functions to describe the relationships between the sides and corners of triangles.

"If you want to understand circular trigonometric functions properly, you need calculus, you need to understand power series, the inverse trigonometric functions," he said.

"There's a lot of tangential material students have to learn before they get down to actually studying the various relationships in a triangle."

His new method, rational trigonometry, employed two concepts: quadrance and spread.

"Quadrance is a way of measuring the separation between two points: it's the distance squared," he said. "The spread is the ratio of two quadrances obtained by dropping a perpendicular line from one point on one of the lines to another line."

Professor Wildberger said his method "probably" represented a paradigm shift. "It could help surveyors, engineers, navigators ... and teachers and students. It also provides a new framework for Euclydian geometry."

Practitioners in the field cautiously welcomed the idea.

Murdoch University mathematician Ken Harrison, who created the high school maths curriculum for Western Australia, said the system was "very interesting". "He makes a pretty strong argument that his way of looking at trigonometry is simpler but it's one thing to say that and another to get people to change their minds from old habits," Associate Professor Harrison said.

"It simplifies the mathematics dramatically and would save a lot of time."

Australian Association of Mathematics Teachers professional officer Chris Stocks said the new method was "a nice reminder that mathematics is an evolving thing", but there were reservations. "It requires us to replace the idea of distance (which has units and can be measured) with distance squared, which may be mathematically more appealing, but which would be a problem for me as a high school teacher trying to find more concrete approaches for many of my students," Ms Stocks said.

There is also another interesting publication at PhysOrg.com.

I attended a lecture by this prof recently on this subject, and have been following his work (his website is wildegg.com)...so I'm happy to see some press to his theories.

Any comments?

May b if someone can give some examples with basic trigonometric problems where this has advantages it will more intesting and helpful.

http://en.wikipedia.org/wiki/Rational_trigonometry

Kind of interesting, if I should say so myself.

it is  kind of interesting...go wikipedia!


conics
hyperbola
floor function
absolute value

How many ordered pairs of integers $ (x, y)$ are there such that
\[ 0 < \left\vert xy \right\vert < 36?\]

How do you evaluate $ \sum_{k\equal{}1}^{35}d(k)$ without summing all of them up?

The answer above is not quite right.  There is also a solution with a lot fewer cases.

[hide="Solution"]Note that for any positive $ x$, there are $ \left\lfloor\frac{35}x\right\rfloor$ positive values of $ y$ that work.
[list][*]$ x\equal{}1\implies35$
[*]$ x\equal{}2\implies17$
[*]$ x\equal{}3\implies11$
[*]$ x\equal{}4\implies8$
[*]$ x\equal{}5\implies7$
[*]$ x\equal{}6,7\implies5$
[*]$ x\equal{}8\implies4$
[*]$ 9\le x\le11\implies3$
[*]$ 12\le x\le17\implies2$
[*]$ 18\le x\le35\implies1$[/list]
Because of the absolute value, there are four symmetric cases based on the signs of $ x$ and $ y$, so our answer is
\[ 4(35\plus{}17\plus{}11\plus{}8\plus{}7\plus{}2\cdot5\plus{}4\plus{}3\cdot3\plus{}6\cdot2\plus{}18\cdot1)\equal{}\boxed{524}.\][/hide]


Prove that $ (\nexists)$ any function $ f: \mathbb{N}\rightarrow \mathbb{N}$ such that $ f(3n+1)\le f(3n+4)-1$ and $ f(5n+2)\ge f(5n+7)-1\ ,\ (\forall)n\in \mathbb{N}$ .

[hide]If such a function $ f$ exists, from the first property, we have that
$ f(7) \le f(10) \minus{} 1 \le f(13)\minus{}2 \le f(16)\minus{}3 \le f(19)\minus{}4 \le f(22)\minus{}5$,
and from the second,
$ f(7) \ge f(12) \minus{}1 \ge f(17) \minus{} 2 \ge f(22) \minus{} 3$.
But this means
$ f(22)\minus{}3 \le f(7) \le f(22)\minus{}5 \implies \minus{}3 \le \minus{}5$, a contradiction.
Therefore, no such function $ f$ exists.[/hide]

Here's a silly variation: what if we replace both $ \minus{}$ signs in the question with $ \plus{}$ signs?


Here are some of the dumbest IIT practice problems I came across today:-

1)In the sequence $1,2,2,3,3,3,4,4,4,4 .... \text{Find the 150th term(or even generalise)}$

2)How many  $5 \text {digit numbers}$ divisible by $30$ can be formed by the digits  $1,2,3,4,5,0$ wiithout repeating.

3)The mean of the coefficients of the terms of the expansion if $(1+x)^{10}$ is:-

$a)\frac{2^9}{11}$

$b)\frac{2^n}{n+1}$  when $n=10$

$c)\frac{2^{13}}{10}$

$d)\text{None of these}$

[hide="1"]
$\frac{n(n+1)}{2} = 150$
Rounding n up gives 17.
[/hide]

[hide="2"]
It's clear the 5-digit number must be divisible by 10 (end with 0) and must be divisible by 3 (all digits add up to 3). We have to choose 4 out of the 5 remaining digits so that their sum is divisible by 3. The only possibility is if we choose 1, 2, 4 and 5. There are 4*3*2 = 24 permutations of those digits. Therefore there are 24 5-digit numbers.
[/hide]

[hide="3"]
Sum of coefficients are sum of 10th row of pascals triangle which is $2^{10}$ and there are $11$ terms. The arithmetic mean is therefore, $\frac{2^{10}}{11}$ $\boxed{B}$.
[/hide]

[quote="shadysaysurspammed"]Javaman, your first one is wrong.

And for the third one,you actually calculated the sum(or you just know it?)?Yes that is a way  but i found an eaier method in which i am getting either  $b$ or $d$(by taking sum of unit's digit which comes $4$,so the only choices are $b$ or $d$).[/quote]

I thought this is an "easier" method without actually calculating the sum:

Substitute $x=1$ and we get the sum of all coefficients $(1+1)^{10}=2^{10}$, so the average of the 11's terms is $\frac{2^{10}}{11}$.

Besides, I don't see why javaman's first answer is wrong? :?

Yes that's easier,I can't believe blanked out on that method  :wallbash:

I am still not sure of javaman's solution to the 1st one though.....

I like the $(1+1)^{10}$ method. This is linked with the identity that $\binom{a}{0} + \binom{a}{1} + \cdots + \binom{a}{a} = 2^a$.

[hide="Here is the general term of the first sequence and a similar exercise."]
[color=darkblue][b]Answer.[/b] For any $n\in N^*$, $a_n=-[\frac{1-\sqrt{1+8n}}{2}]$, where $[x]=z$ $\Longleftrightarrow z\in Z$ and $z\le x<z+1$ (the integer part $z$ of thr real number $x$).
In the particular case $n: =150$ we obtain $a_{150}=-[\frac{1-\sqrt{1201}}{2}]=$
$-[1-\frac{34,\ldots}{2}]=-[-\frac{33,\ldots}{2}]=-(-17)=17$, i.e. $a_{150}=17\ .$[/color]

[color=darkred][b]Exercise.[/b]

$1.\blacktriangleright$ Ascertain the general term of the sequence:

$a_n\ \vdash\ 1,\underline {3,3,3},\underline {5,5,5,5,5},\underline{7,7,7,7,7,7,7},\ldots$

$2.\blacktriangleright$ Denote the set $A\equiv 2N+1$ of the odd natural numbers. Prove that :

$p\in A\Longrightarrow$ the equation ${[-\sqrt x}]+p+1=0$ has exactly $p$ solutions in $A$,

where $[r]$ is the integer part of the real number $r$.

$3.\blacktriangleright$ Try a generalization of this exercise ![/color][/hide]

[hide="1"]
The 1st $1$ in the sequence comes after 0 numbers have passed.
The 1st $2$ comes along after 1
3 comes after 3
4 comes after 6
5 -10

0  1  3  6  10
  1  2  3  4
    1  1  1
$n_{n}=1/2x^2+bx+c$
$0=1/2+b+c$  or  $b+c=-1/2$
$1=2+2b+c$   or  $2b+c=-1$
$b=-1/2$
$c=0$

So,
$n_{n}=1/2{n^2}-1/2{n}$
Now this is not the generalised sequence but will help us get 150 term.
We want to find $n$ when $n_n=150$
$300=n^2-n$
$n^2-n-300=0$
$18^2-18=306$
So $n=17$
[/hide]

[quote="shadysaysurspammed"]Javaman, your first one is wrong.

And for the third one,you actually calculated the sum(or you just know it?)?Yes that is a way  but i found an eaier method in which i am getting either  $b$ or $d$(by taking sum of unit's digit which comes $4$,so the only choices are $b$ or $d$).[/quote]

I know the nth row of pascals triangle adds up to $2^n$ (starting from the zeroth row). I think my first ones right too?  :?

you know for number 3, all you have to do is find the sum of the coefficients (plug in $1$) and divide it by the number of coefficients, which can easily be found (just $n+1$ where $n$ is the power for the binomial) ;)

EDIT: nvm, i see that someone already solved 3  :)

[quote="shadysaysurspammed"]1)In the sequence $1,2,2,3,3,3,4,4,4,4 .... \text{Find the 150th term(or even generalise)}$[/quote]

After the sequence of $n$'s finish, there have been $\frac{1}{2}n(n+1)$ terms. For example, there are $\frac{1}{2}\cdot3\cdot(3+1)=6$ terms in 1,2,2,3,3,3,3. So $\frac{1}{2}n(n+1)\ge 150 \implies n\ge 17$. Thus the 150th term is $\boxed {17}$.

[quote="Virgil Nicula"][color=darkred][b]Exercise.[/b]

$1.\blacktriangleright$ Ascertain the general term of the sequence:

$a_n\ \vdash\ 1,\underline {3,3,3},\underline {5,5,5,5,5},\underline{7,7,7,7,7,7,7},\ldots$

$2.\blacktriangleright$ Denote the set $A\equiv 2N+1$ of the odd natural numbers. Prove that :

$p\in A\Longrightarrow$ the equation ${[-\sqrt x}]+p+1=0$ has exactly $p$ solutions in $A$,

where $[r]$ is the integer part of the real number $r$.

$3.\blacktriangleright$ Try a generalization of this exercise ![/color][/quote]

$1.\blacktriangleright$

$k^2\ge n\implies k=\lfloor\sqrt{n}\rfloor+1\implies a_n = 2k-1=\boxed {2\lfloor\sqrt{n}\rfloor+1}$.

So the answer is $17$ for the 1st one.I as saying it was wrong because it wasn't there in the four choices given(from where i read it from) :D   You were right javaman

[quote="Virgil Nicula"][color=darkred]
$3.\blacktriangleright$ Try a generalization of this exercise ![/color][/quote]

$3^\circ\blacktriangleright\qquad\text{Define a function } f: \mathbb{R}\to\mathbb{R} \text{ and } g(n)=f(1)+f(2)+\cdots+f(n)$.
$\text{Consider the sequence}\quad \underbrace{f(1),\ldots,f(1)}_{f(1)\text{ times}}, \underbrace{f(2),\ldots,f(2)}_{f(2)\text{ times}}, \ldots$
$\text{If } g^{-1} \text { is the inverse function of } g, \text{ then: }$
$g(k)\ge n\implies k=\lfloor g^{-1}(n)\rfloor+1\implies a_n=f(k)=\boxed {f\left(\lfloor g^{-1}(n)\rfloor+1\right)}$. $\boxed {}$

Is this too general? :P

[quote="shadysaysurspammed"]...$b)\frac{2^n}{n+1}$  when $n=10$...

Why didn't they just say $\frac{2^{10}}{11}$? :huh:


trigonometry
inequalities
inequalities unsolved

$ ABC$ is a trinagle

$ \tan\bigg(\frac{A}{2}\bigg)\plus{}\tan\bigg(\frac{B}{2}\bigg)\plus{}\tan\bigg(\frac{C}{2}\bigg) \geq \sqrt{3}$

no, it isn't true - what is the tangens of a half vertex?

[hide]
this is true!

just use the theorem of jensen, the second derivation of $ \tan{\frac {x}{2}}$ is $ \frac {\tan{\frac {x}{2}}}{2 \cos^{2}{\frac{x}{2}}$ and positive in the intervall, so you can "replace" the $ A,B,C$ by $ 60^{\circ}$ and you get equality in this minimum case.
[/hide]

Naphthalin

Yes, $ f: (0,\frac {\pi}{2})\rightarrow\mathbb{R}$ ,   $ f(x) \equal{} \tan (x)$      is convex functions.

thank's

$ f(x)\equal{}\tan(x)$

$ f"(x)\equal{}?$

i want details please


geometry
geometric transformation
reflection
complex numbers

I've been playing [url=http://www.nationstates.net/cgi-bin/index.cgi]NationSites[/url] for a little while and I started thinking. If the use it for educational purposes then why not here?

So I decided to make the community of AOPS. We first need to make initial decisions to reflect our community. After our nation is established, issues will come in, and we then have many more choices.

There are 4 intial decisions (or 5 if anyone wants to make a flag)

1. Goverment Style - [i][b]Sensible [/b][/i]
2. National Animal - Current Poll (If a mod could kindly attach a poll for National Animal that would be greatly appreciated.)

National Animal Nominations:
Aarvark
Beaver
Goldfish
Sheep

3. Currency
4. National Motto

We need to nominate choices for national animal, Currency, and a National Motto.

This first poll will run for 3 days then we'll get a new poll started for our other initial decisions.

Government style should be a monarchy, with somone like Valentin or Russczyk as our leader, then the leadership hierarchy goes through administrators and then down to the nobles, the administrators, to the peasants, us regular users. I believe this system is implemented already though.

The national animal should definitely be the aardvark. Or the beaver.

Our currency shall be known as Otes, and shall come in three types, the small coins, the big ones, and the pieces of paper that say "AoPS Nation". The smallest unit of money shall be one septillionth of an ote. The largest shall be 7 billion otes, and there must be one note for every amount in between.

The national motto should be something funny. Like "Democracy is too good to share with just anybody." I think it fits.

Anyone agree?  :rotfl:

I basically agree with the monarchy thing. But perhaps we should have an oligarchy, with the admins sharing power, instead of just one?

Why? Goldfish are small and prone to be eaten, and dont taste very good. Besides, they are very slow to react to light (change), so why would we want to be percieved as a slow people?

I agree, an oligarchy would be better.

The flag should look somewhat like this, with a different slogan (not 1+1=2...) see attachment

I say our government should be along these lines: "One man, one vote. The Patrician is the man, and he's got the vote." :rotfl:

Kudos to anyone who recognizes the source of that quote.

I say we have hecatontarchy: rule by 100 people, namely the mods and admins. :)

National animal: sheep

monarchy - leader = Mr. Rusczyk
rest = peasantry
'nuff said

[quote="pkothari13"]Sounds like Pratchett to me! =O! that weirdo.  yeah, i agree with whole oligarchy deal. sounds good to me.[/quote]
Correct! :D

I think the currency should allow complex numbers as legal denominations of money.

why don't we all live in peaceful city-states and allow cows to roam since they'd be the national animal

and the slogan can be: "WACKY, WAVY, INFLATABLE ARMS FLAILING TUBE MAN!!!"

[quote="hello"]I didnt everything in detail but I want to say that a monarchy might not be ideal under any ruler for the simple reason that the ruler will advance his own philosophy. For example, Mr Ruczyk will promote scientific advancement (I think he would) but the majority of average people are not too interested in that, so in that respect, it might be a failure, and a democratic institution is better.[/quote]

You think the majority of AoPSers are against scientific advancement?  Interesting...

No, sorry I didnt read all the posts, my fault. I meant if that is instituted throught the whole world.

I still think the goldfish best represents us. It shows our originality.  :thumbup:

Theres some rescrictions on the flag but I guess I can convert yours.
You can create an image on your computer and upload it as your nation's flag.

Your flag must be:

1. in JPEG (.jpg) format
2. about 107x71 pixels in size
3. no larger than 10KB
4. appropriate. This means conforming to site etiquette and being suitably flag-like. Images likely to cause widespread offense (e.g. nudity, swastikas) are prohibited, and your nation may be deleted if you post them.

Make some flags people!

Hurry people! Vote! Do you really want our goverment to be sensible?

National Animal Nominations:
Aarvark
Beaver
Goldfish
Sheep

Currency
Complex Numbers or $ e^{ik \pi} $'s

Motto
Democracy is too good to share with just anybody.
1+1=2
One man, one vote. The Patrician is the man, and he's got the vote.
WACKY, WAVY, INFLATABLE ARMS FLAILING TUBE MAN!!!

We need more suggestions for currency!!

We need to choose a base too. I vote for 12; after all, it's already on the T-shirt.

How about Gaussian integers for currency?

Annnnnnnnnnndddddd its decided! Our goverement will be sensible. If a mod would please attach a poll for Animal Nominations, that would be greatly appreciated. If not, we can always do this the old-fashioned way.

Animal Nominations
Aarvark
Beaver
Goldfish
Sheep

Currency
Gausian Integers
Complex Numbers or $e^{ik \pi}$'s
Dollars (Base 12)

Motto
Democracy is too good to share with just anybody.
1+1=2
One man, one vote. The Patrician is the man, and he's got the vote.
WACKY, WAVY, INFLATABLE ARMS FLAILING TUBE MAN!!!

I vote for the sheep and nominate the czyk as the currency, not dollars.


[quote="Aravind Srinivas L"]I just have a doubt regarding the minimum value of exoressions can be ontained from AM GM directly.
This is not true always . could u please correct me if i am wrong.[/quote]
Of course... You cannot solve all inequalities with AM-GM ;)


Given that $ a,b,c$ are sides of triangle, prove that:

\[ a^2\plus{}b^2\plus{}c^2\plus{}a\plus{}b\plus{}c\geq 3  \sqrt[3] {abc(a^2\plus{}b^2\plus{}c^2\plus{}abc\plus{}1)\plus{}a^2b^2\plus{}b^2c^2\plus{}c^2a^2}\]

There is a solution by AM-GM  :)  :cool:

hello,your inequality is equivalent with
$ \frac{1}{27}(a^2\plus{}b^2\plus{}c^2\plus{}a\plus{}b\plus{}c)^3\minus{}abc(a^2\plus{}b^2\plus{}c^2\plus{}abc\plus{}1)\minus{}a^2b^2\minus{}a^2c^2\minus{}b^2c^2\geq0$
setting
$ a\equal{}y\plus{}z,b\equal{}x\plus{}z,c\equal{}x\plus{}y$ and factoring we get
$ 1/27 (x \minus{} z)^2 (8 x \minus{} 3 x^2 \plus{} 24 x^3 \plus{} 8 x^4 \plus{} 9 y \plus{} 12 x y \plus{}
   78 x^2 y \plus{} 48 x^3 y \plus{} 18 y^2 \plus{} 96 x y^2 \plus{} 105 x^2 y^2 \plus{} 45 y^3 \plus{}
   100 x y^3 \plus{} 36 y^4 \plus{} z \plus{} 18 x z \plus{} 6 x^2 z \plus{} 16 x^3 z \plus{} 24 y z \plus{}
   36 x y z \plus{} 66 x^2 y z \plus{} 39 y^2 z \plus{} 90 x y^2 z \plus{} 44 y^3 z \plus{} 3 z^2 \plus{}
   12 x z^2 \plus{} 9 x^2 z^2 \plus{} 21 y z^2 \plus{} 24 x y z^2 \plus{} 21 y^2 z^2 \plus{}
   3 z^3 \plus{} 2 x z^3 \plus{} 6 y z^3 \plus{} z^4)$
this is nonnegative because the sum $ 8x\minus{}3x^2\plus{}24x^3\plus{}8x^4\geq0$ and all other summands are nonnegative.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello,your inequality is equivalent with
$ \frac {1}{27}(a^2 \plus{} b^2 \plus{} c^2 \plus{} a \plus{} b \plus{} c)^3 \minus{} abc(a^2 \plus{} b^2 \plus{} c^2 \plus{} abc \plus{} 1) \minus{} a^2b^2 \minus{} a^2c^2 \minus{} b^2c^2\geq0$
setting
$ a \equal{} y \plus{} z,b \equal{} x \plus{} z,c \equal{} x \plus{} y$ and factoring we get
$ 1/27 (x \minus{} z)^2 (8 x \minus{} 3 x^2 \plus{} 24 x^3 \plus{} 8 x^4 \plus{} 9 y \plus{} 12 x y \plus{} 78 x^2 y \plus{} 48 x^3 y \plus{} 18 y^2 \plus{} 96 x y^2 \plus{} 105 x^2 y^2 \plus{} 45 y^3 \plus{} 100 x y^3 \plus{} 36 y^4 \plus{} z \plus{} 18 x z \plus{} 6 x^2 z \plus{} 16 x^3 z \plus{} 24 y z \plus{} 36 x y z \plus{} 66 x^2 y z \plus{} 39 y^2 z \plus{} 90 x y^2 z \plus{} 44 y^3 z \plus{} 3 z^2 \plus{} 12 x z^2 \plus{} 9 x^2 z^2 \plus{} 21 y z^2 \plus{} 24 x y z^2 \plus{} 21 y^2 z^2 \plus{} 3 z^3 \plus{} 2 x z^3 \plus{} 6 y z^3 \plus{} z^4)$
this is nonnegative because the sum $ 8x \minus{} 3x^2 \plus{} 24x^3 \plus{} 8x^4\geq0$ and all other summands are nonnegative.
Sonnhard.[/quote]

Thanks for your proof. But I think, it is almost impossible to factorize that by hand.  :oops:

The following is my [u]proof[/u]:

We will use two inequalities that are true for $ a,b,c$ being [u]sides of triangle[/u]:

$ ab^3\plus{}bc^3\plus{}ca^3\geq a^2b^2\plus{}b^2c^2\plus{}c^2a^2$

$ a^3b^2\plus{}b^3c^2\plus{}c^3a^2\geq abc(a^2\plus{}b^2\plus{}c^2)$

So by using two inequalities above, and AM-GM inequality, we have:

$ abc(a^2\plus{}b^2\plus{}c^2\plus{}abc\plus{}1)\plus{}a^2b^2\plus{}b^2c^2\plus{}c^2a^2\leq a^3b^2\plus{}b^3c^2\plus{}c^3a^2\plus{}a^2b^2c^2\plus{}abc\plus{}ab^3\plus{}bc^3\plus{}ca^3\equal{}(a^2\plus{}b)(b^2\plus{}c)(c^2\plus{}a)\leq\frac{1}{27}(a^2\plus{}b^2\plus{}c^2\plus{}a\plus{}b\plus{}c)^3$

Therefore, we have proved that:

\[ a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}a\plus{}b\plus{}c\geq 3\sqrt[3]{abc(a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}abc\plus{}1)\plus{}a^{2}b^{2}\plus{}b^{2}c^{2}\plus{}c^{2}a^{2}}\]

 :)  I hope you enjoyed it.


Hey everyone.  Here are a few guidelines for use of this forum.  Feel free to add more.

1) No spamming.  No flaming.  No exceptions.

2) Please keep posts to an ARML level.  Harder problems will be moved to another section of the Forum.

3) Keep answers hidden by hightlighting the answer part of your post and clicking "Hide" above the text box.  This gives others a chance to try the problem for themselves.

4) Don't post any current-year ARML problems until they show up on the ARML website (they already have for 2005).

5) Please use proper forum ettiquette when posting.  It would be nice to have solutions posted, rather than just the answer.

Yeah, ARML isn't like Mathcounts, where state is taken at different times

[quote="MysticTerminator"]just out of curiosity, what is the reason for #4? I wasn't aware that people took ARML late or anything like that[/quote]The San Jose ARML site has some of its events on Friday.

Besides, this rule is not even applicable until about 51 weeks from now.

It is for copyright reasons.  We can't distribute problems until they are "released."

until at least sometime in the afternoon on Saturday -- similar to the AMC/AIME/USAMO locking.

The most critical time, of course is from Friday evening to Saturday morning, when the asynchronicity in which sites have seen which
problems  is greatest.

Maybe you're already planning to do this.

Anyway, thanks for all your work on this wonderful site!


induction
function
inequalities
calculus
inequalities proposed

Hint
For n=2 it is true then by induction. But I have no time to make the computations.
That is only an idea

[quote="silouan"]As rules say I can post an idea[/quote]Where do you see rules, silouan? I've been looking for them, but I can't find... :)

Anyway, your claim for induction surprises me... I don't see it.  :? can you work it out please?

My solution: consider the function $f(x) = \sqrt[\displaystyle x-1]{\left(\frac{1+x^x}{1+x}\right)}$. This function is ascending for positive x, so $f(n+1) > f(n)$.

For $n>1$, we can put both sides of $f(n+1) > f(n)$ to the $n(n-1)^{th}$ power, which keeps the inequality for $n>1$, resulting in the demanded inequality. So, I think it is even true for every real $x>1$. :P

[quote="Peter VDD"] This function is ascending for positive x, so $f(n+1) > f(n)$.[/quote]

You mean increasing, right? How would you prove that it's an increasing function?

Well, I have a very short proof for $n\ge3$. I did not bother to prove for x in [2,3] because f(3)>f(2) is trivial. [the general is true though]

$f(x) = \frac{\sqrt[x-1]{1+x^x}}{\sqrt[x-1]{1+x}}$, the numerator is clearly increasing for $n\ge 3$, the denominator is clearly decreasing, so the total function is increasing as well. [basic calculus] :)


calculus
integration
trigonometry
logarithms
calculus computations

$ \int \frac{x\plus{}\sin x}{1\plus{}\cos x}\ dx$.

$ \int\frac{x+\sin{x}}{1+\cos{x}}dx=\int\frac{x}{1+\cos{x}}dx+\int\frac{\sin{x}}{1+\cos{x}}dx$.

$ \int\frac{x}{1+\cos{x}}dx=\int x\cdot\frac{1}{2}\sec^2{\frac{x}{2}}dx=x\tan{\frac{x}{2}}-\int\tan{\frac{x}{2}}dx$
$ =x\tan{\frac{x}{2}}-2\ln\left|\cos\frac{x}{2}\right|+C$.

So, the answer is $ \boxed{x\tan{\frac{x}{2}}-2\ln\left|\cos{\frac{x}{2}\right|-\ln\left|1+\cos{x}\right|+C}}$.

hello, the answer is
$ \int\frac{x\plus{}\sin(x)}{1\plus{}\cos(x)}\,dx\equal{}x\tan\left(\frac{x}{2}\right)\plus{}C$.
Sonnhard.

Sorry, I had a mistake. $ \int\tan{\frac {x}{2}}dx = - 2\ln\left|\cos{\frac {x}{2}}\right|dx$.

So, the answer is $ x\tan{\frac {x}{2}} + C$. ($ \ln\left|1+\cos{x}\right|=2\ln\left|\cos{\frac{x}{2}\right|+\ln{2}}$)


My Anti-Index (AMC - 10*AIME) is the square root of my real index.

My Anti-Index, or whatever it's called is 42  :D

my AMC 12 score is 137
which is about as close to 1337 as you can get

My Index is the only sequence of three consecutive increasing integers that is likely to make the usamo! And my anti-index is 14, the almost-age of my little brother. (low whistles of amazement)

My index is the only possible eighth power.

[quote]
My AMC 12 index is 6^3.

My AIME is 2^3
[/quote]

Your AIME: 1000
Your AMC : 10001000

My anti-index is one of the answers on the AIME I.

Hmmm...I don't want people to know my AMC 12 score, but let's do it for AIME score...let's see...it's a digit...it's a multiple of three...and it's the first odd square of a prime number. Okay, i guess i really can't do much with it. Wait, if you take the test date, 3/7, subract the three from the seven twice, multiply it by two, add one, and square it...


geometry trigonometry inradius perimeter

A triangle has an area of $16 \text{ in.}^{2}$, and two of the sides of teh traingle have lengths $5 \text{ in.}$ and $7 \text{ in.}$ Find the angle included by these two sides.

So some easy trig allows one to see that
$16=\frac{1}{2}\cdot 5 \cdot 7 \cdot \sin \theta$
$\frac{32}{35}=\sin \theta$
$\sin^{-1}{\frac{32}{35}}\approx 66$

Is there ambiguity in this problem? Couldn't $\sin^{-1}$ also equal $180-66$?

thanks.

[size=75][color=red]preview your posts  :)
b-flat[/color][/size]

[quote="surge"] Couldn't $\sin^{-1}$ also equal $180-66$?[/quote] Yes, I believe it could. :D

[quote="papasfritas"]how do you know it's a right triangle?[/quote]

It does not need to be a right triangle, For all triangles (1/2)a*b*sin(A) is the area of the triangle where angle A is the included angle of the sides with length a and b. This is easily proven by drawing altitude to the side with a or b. And also, yes it could be either 66 or 180-66=114 degrees. Bot hwill create a triangle with area 16 and two sides of length 5 and 7.

The different triangle formulas:

$A=\frac{abSINc}{2}$
where a and b are side lengths and c is the angle between the sides.

$A=\frac{ah_{a}}{2}$
where a is a side and $h_{a}$ is the altitute to that side.

$A=rs$
where when the triangle is inscribed in a circle, r=the inradius of the triangle, s=semiperimeter.

semiperimeter=$\frac{\text{sum of all sides}}{2}$

EDIT: Hero/Heron's formula:

$A=\sqrt{s(s-a)(s-b)(s-c)}$
Where A is the area, s is the semiperemeter, a, b, and c are side lengths.

check out AoPS volume 1, page 144

Don't forget this one:

$A=\sqrt{s(s-a)(s-b)(s-c)}$

Again, $s$ is the semiperimeter, or the perimeter divided by 2.
The side lengths of the triangle are $a$, $b$, and $c$.

Also $A=\frac{abc}{4R}$ where $a$, $b$, and $c$ are the sides and $R$ is the circumradius.


function limit integration trigonometry vector algebra domain

I've got a query on the definition of "differentiability".
For any real continuous function $ f: {\mathbb R}^n\mapsto {\mathbb R}$, $ \lim_{x\to 0}f(x)$ always exists if infinite value is allowed.
Hence the limit $ d_{\plus{}}(f)(x)\equal{}\lim_{h\to 0\plus{}}{{f(x\plus{}h)\minus{}f(x)}\over h}$ should always exists if we allow it to choose infinite value. Does it means that if infinite value is allowed, all functions are "differentiable on the right"?
Similarly, "

[quote="bchui"]
For any real continuous function $ f: {\mathbb R}^n\mapsto {\mathbb R}$, $ \lim_{x\to 0}f(x)$ always exists if infinite value is allowed.[/quote]If $ f$ is continuous on all of $ \mathbb{R}^ n$, by definition that limit exists and is finite.  It has nothing to do with whether or not infinite values are allowed.
[quote="bchui"]Hence the limit $ d_{ \plus{} }(f)(x) \equal{} \lim_{h\to 0 \plus{} }{{f(x \plus{} h) \minus{} f(x)}\over h}$ should always exists if we allow it to choose infinite value.[/quote]Not true.  Consider the function
\[ f(x) \equal{} \int_0^{|x|}\sin \frac {1}{s}ds
\]
Note that $ f$ is continuous everywhere and infinitely differentiable on $ \mathbb{R}\backslash\{0\}$, and furthermore $ |f'(x)|\leq 1$ for all $ x\in \mathbb{R}\backslash\{0\}$.  However, $ \lim_{h\to 0^ \plus{} }\frac {f(h) \minus{} f(0)}{h}$ does not exist.
Edit: Typo.

[quote="bchui"]... $ d_{ \plus{} }(f)(x) \equal{} \lim_{h\to 0 \plus{} }{{f(x \plus{} h) \minus{} f(x)}\over h}$[/quote]What does $ h\rightarrow 0^\plus{}$ mean while $ h\in \mathbb R^n$? ... and how is the division by the vector $ h$ defined? ;)

[quote="Third Edition"][quote="bchui"]... $ d_{ \plus{} }(f)(x) \equal{} \lim_{h\to 0 \plus{} }{{f(x \plus{} h) \minus{} f(x)}\over h}$[/quote]What does $ h\rightarrow 0^ \plus{}$ mean while $ h\in \mathbb R^n$? ... and how is the division by the vector $ h$ defined? ;)[/quote]
I think the domain should be $ \mathbb R$. In case the domain is a normed vector space, we have to use [url=http://en.wikipedia.org/wiki/Fréchet_derivative]Fréchet derivative[/url] or [url=http://en.wikipedia.org/wiki/Gâteaux_derivative]Gâteaux derivative.[/url]


AMC USA(J)MO USAMO AIME geometry inequalities Euler

Should I go through old MOP notes or redo the AoPS Olympiad Independent Study (by myself obviously). Also, over the course of time I have acquired notes on specific subjects. I'm not sure which approach to take when practicing for this year's USAMO. Any advice?

Well in general, it's a bit too late to start prepping for the USAMO if you are just starting now. I don't know how much of a background you have in mathematics, but even if you did well enough on the AIME to qualify for USAMO, you might still be taken back by the USAMO, namely because the former is problem solving and the latter is proofs. That said, the best way to prep for it, in my opinion, is to go over all the past USAMOs (if u have the time, of course) and try to understand how they proved the things they wanted u to prove. Like problem solving, proofs have their own tricks and method of reasoning, and like problem solving, the best way to learn is to look at solutions and understand them.

um...doing proofs is certainly problem solving.

I think everything'll be fine. Just go over the more recent USAMOs on this website to get a hang of what it's like.  :)

Well, this will (hopefully) be my 2nd time qualifying for USAMO and I did the AoPS IS last summer so I'm not a stranger to proofs. However, I don't know what would be most beneficial at this point so I was just asking for advice.

I'll say try to play around with olympiad questions and see if you can get some progresses in one hour.  Also you can read books like Arts and Crafts in Problem Solving that actually teach you to think and attack problems, and then try to practice thinking and solving a bunch.  I can't really say much because I never went to USAMO (and probably not this year... :( ), but that's what I'll do :)

[quote="darkquantum"][quote="pieterminate"][quote="darkquantum"]Well in general, it's a bit too late to start prepping for the USAMO if you are just starting now. [/quote]
Oh no! :o  :|  :([/quote]

I meant that as like, if you started now, from scratch, with no math background.[/quote]

math background as in 8th algebra? :)

[quote="pieterminate"][/quote][quote="darkquantum"][quote="pieterminate"][quote="darkquantum"]Well in general, it's a bit too late to start prepping for the USAMO if you are just starting now. [/quote]
Oh no! :o  :|  :([/quote]

I meant that as like, if you started now, from scratch, with no math background.[/quote][quote="pieterminate"]

math background as in 8th algebra? :)[/quote]

not to be discouraging, but if that's all you know (no advance number theory, geometry, inequalities, combinatorics, etc.), then it's kinda hopeless unless you're Einstein, Euler, Newton, and Gauss combined together and derive all the advance formulas from scratch in 4.5 hours :rotfl:

[quote="pkerichang"][/quote][quote="pieterminate"][/quote][quote="darkquantum"][quote="pieterminate"][quote="darkquantum"]Well in general, it's a bit too late to start prepping for the USAMO if you are just starting now. [/quote]
Oh no! :o  :|  :([/quote]

I meant that as like, if you started now, from scratch, with no math background.[/quote][quote="pieterminate"]

math background as in 8th algebra? :)[/quote][quote="pkerichang"]

not to be discouraging, but if that's all you know (no advance number theory, geometry, inequalities, combinatorics, etc.), then it's kinda hopeless unless you're Einstein, Euler, Newton, and Gauss combined together and derive all the advance formulas from scratch in 4.5 hours :rotfl:[/quote]

And manage to write proofs for all the questions.

At this point it may be too late to work really hard on a specific subject that you are weak at and expect to tune it to a point where you don't have to worry about it anymore. (severe weaknesses are meant to be dealt with over the summer and fall months)

So maybe the best idea is to work on general problems, don't only work on your strengths but don't go ballistic on your weakness problems either, and also do some mock contests where you take the full time and don't give yourself brownie points for "oh I just made a silly mistake there". It's very hard to acquire entire new sets of skills in a month and a half, but refining skills to a noticeably better point is quite doable.

[quote="hockeyadi23"][/quote][quote="pkerichang"][/quote][quote="pieterminate"][/quote][quote="darkquantum"][quote="pieterminate"][quote="darkquantum"]Well in general, it's a bit too late to start prepping for the USAMO if you are just starting now. [/quote]
Oh no! :o  :|  :([/quote]

I meant that as like, if you started now, from scratch, with no math background.[/quote][quote="pieterminate"]

math background as in 8th algebra? :)[/quote][quote="pkerichang"]

not to be discouraging, but if that's all you know (no advance number theory, geometry, inequalities, combinatorics, etc.), then it's kinda hopeless unless you're Einstein, Euler, Newton, and Gauss combined together and derive all the advance formulas from scratch in 4.5 hours :rotfl:[/quote][quote="hockeyadi23"]

And manage to write proofs for all the questions.[/quote]

Hey, I didn't actually say that I was in 8th grade algebra ;)  or that I was going tu USAMO. :(              But do you actually need a bunch of advanced formulas for USAMO (and how advanced)?
as in Pythagorean..........or..........?

Euler's totient theorem, Fermat's little thm (special case of first :lol:), Fermat's last thm (ok maybe not but whatever), binomial thm (ooh pretty tricky huh), Ptolemy's thm, Euler's thm about polyhedrons (V-E+F=2), lots of other geometry ones, and lots of other number theory ones too. Don't worry, they will accumulate over time :D

oh yeah you need to know things like Ceva's Theorem or Muirhead/Jensen/Chebyshev's Inequality, Fermat's Little/Euler's Generalization that are not taught in regular high school.

[quote="chess64"]Oh I almost forgot!

1. Sturm Method
2. AM-GM-HM
4. Cauchy-Buniakovski-Schwartz
5. Muirhead
6. Rearrangement
7. Cebyshev
8. Minkovski
9. Huygens, Young...
10. Jensen.
11. Power Mean

And some more too. :lol: (I'm not even joking!)[/quote]

you forgot Schur, Bernoulli.

[quote="probability1.01"]Ehh... I don't even know some of those... you don't really need those theorems. The most important ones that aren't either obvious or taught in school:

1. Euler phi function and various properties
2. Cauchy-Schwarz
3. RMS-AM-GM-HM (esp. AM-GM)
4. Ceva/Menelaus
5. Properties of mods (but this is pretty easy)

Everything else is unlikely to appear.[/quote]

Umm, an explanation of these 5? :?  :?:

[quote="probability1.01"]Ehh... I don't even know some of those... you don't really need those theorems. The most important ones that aren't either obvious or taught in school:

1. Euler phi function and various properties
2. Cauchy-Schwarz
3. RMS-AM-GM-HM (esp. AM-GM)
4. Ceva/Menelaus
5. Properties of mods (but this is pretty easy)

Everything else is unlikely to appear.[/quote]

Hmmm, this is actually true. Though rarely a few of the harder problems can be done more easily with some high powered stuff (though someone who does not already know it is unlikely to recognize this solution). As far as inequalities and number theory go, there are already some excellent resources on AoPS. As for obtaining material on the other topics, surfing the web is bound yield something.

[quote="probability1.01"]Ehh... I don't even know some of those... you don't really need those theorems. The most important ones that aren't either obvious or taught in school:

1. Euler phi function and various properties
2. Cauchy-Schwarz
3. RMS-AM-GM-HM (esp. AM-GM)
4. Ceva/Menelaus
5. Properties of mods (but this is pretty easy)

Everything else is unlikely to appear.[/quote]

Yeah... about that.... are all these in the AoPS Vol 2 book?

[quote="chess64"]Oh I almost forgot!

1. Sturm Method
2. AM-GM-HM
4. Cauchy-Buniakovski-Schwartz
5. Muirhead
6. Rearrangement
7. Cebyshev
8. Minkovski
9. Huygens, Young...
10. Jensen.
11. Power Mean

And some more too. :lol: (I'm not even joking!)[/quote]

Sturm method?

sort of vaguely aware what Huygens and Young are but eh...

don't intimidate people for no particular reason.

[quote="chess64"]Oh I almost forgot!

1. Sturm Method
2. AM-GM-HM
4. Cauchy-Buniakovski-Schwartz
5. Muirhead
6. Rearrangement
7. Cebyshev
8. Minkovski
9. Huygens, Young...
10. Jensen.
11. Power Mean

And some more too. :lol: (I'm not even joking!)[/quote]

If this is true, then I know what I'm doing if I make USAMO for the 9 hours...  first: :read:  then:  :wallbash: and  :wallbash_red: then :sleep2: and finally :sleeping:.  :D Seriously that stuff is WAAAY out of my league right now....

ok fine but yeah those are the main olympiad level inequalities, but you don't have to know all of them to do well :)

[quote="chess64"][quote="beta"][quote="hockeyadi23"]

If this is true, then I know what I'm doing if I make USAMO for the 9 hours...  first: :read:  then:  :wallbash: and  :wallbash_red: then :sleep2: and finally :sleeping:.  :D Seriously that stuff is WAAAY out of my league right now....[/quote]

Lol don't worry chess64 is just trying to intimidate you. You only need to know like AM-GM-HM, power mean, cauchy, jensen, holder, and that is enough to solve almost all the olympiad inequalities.[/quote]

well I think those would be intimidating too... :lol:[/quote]
well less intimidating at least (I listed 5 and you listed 11, btw what's #1 and #9?  ).

Haha, yeah, knowing them is one thing, applying them is another story... ;)

http://en.wikipedia.org/wiki/Young's_inequality

Sturm method is actually more difficult, now that I look at it. My friend (olympiad winner in a European country) once gave me a list of inequalities to study, this is that list, but it's actually too hard for olympiad... :blush:

You can also try looking into the topics in the [url=http://www.mathlinks.ro/Forum/index.php?f=55]Inequalities Theorems Forum[/url] ;)

[quote="chess64"]http://en.wikipedia.org/wiki/Young's_inequality[/quote]

Whaa isn't that just weighted AM-GM for 2 variables?

Yeah, generally just don't worry too much about learning the necessary "theorems."  In my opinion at least, you learn all you need to know by doing problems, looking stuff up, etc.  That's the only way you learn how to apply them anyways.  :)

Another great way to learn a theorem is  trying to prove the theorem itself.  You learn a great deal of techniques from proving theorems. Once you know how to prove a theorem, there is a good chance you would remember it and be able to apply it to problems.

For example, try to prove the following:

Prove Jensen from the definition of convex function

Prove weighted AM-GM from Jensen

Prove Holder from weighted AM-GM

Prove Minkowski from Holder.

Prove Power Mean from Holder.

Prove Chebychev from Rearrangement

Prove AM-GM from Rearrangement or other methods

Prove Cauchy-Schwarz using several methods

Prove Schur

etc.

[hide="Caution"]
By the way, never try to prove Muirhead :-) or apply Muirhead. Instead, use it as a guideline since anything can be done with Muirhead can be done with weighted AM-GM.
[/hide]

Cauchy-Schwarz is so obvious in vector form it hardly requires proof.  :D

By the way.  How would you go about proving Karamata?

[quote="pieterminate"][quote="Phelpedo"]If you get the USAMO problem book, there is a glossary of theorems and definitions in the back.[/quote]

What book, and where can we get it? :starwars:[/quote]

Mathematical Olympiads 2000-2001 Problems and Solutions from Around the World
Titu Andreescu, Zuming Feng, and George Lee, Jr.
For example.
you can buy that at
https://enterprise.maa.org/ecomtpro/timssnet/common/tnt_frontpage.cfm
and in the bottom, search "2000-2001"

the book by Titu Andreescu, "Mathematical Olympiad Challenges" is the one i'm using...
its a very very good book

by the way, this is my 2006th post :)