Consider $ 2$ circles that intersect in points $ C$ and $ D$. Let $ AB$ be the common diameter of both circles. Let the verical segment joining the two endpoints of the intersection be $ E$ and $ F$. Let $ G$ Be a point on $ EF$. If $ GC\equal{}1$, $ GD\equal{}2$, and $ AC$ is 6. Find the length of $ DB$.

[hide] Correct answer with unsure solution [/hide]

Well I thought to myself "What if the circles are similar? That means $ GC: GD$ = $ 6: x$. Cross multiplying yields $ x\equal{}12$ which is the correct answer. Can someone prove that this works for all scenarios or likewise, it does not work and was simply a lucky guess.   :ninja:

I don't really understand where exactly a lot of the points go, could you draw a diagram?

As for one of the things you said, all circles are similar.

Yeah but with the diagram, I dont think you can use similar circles. Atleast I dont have a good reason as to how. And I'll try to provide a diagram

All circles are similar, as Ihatepie said.

Also, am I screwing this up, or does "common diameter" imply that the circles are both congruent and coincident? Do you mean common radius? Also, your third sentence makes no sense because while a segment is defined by 2 points, you can't "let it be" two points. In any case, the line segment is already defined: it's CD.

Looking back, if AB is the common radius, it doesn't make sense because CD would not be equal to 3.

Or am I completely failing here?

[quote="randomguy64"]
Or am I completely failing here?[/quote]I thought the same thing; it seems the question is just a little unclear.


For those of you who have Super Smash Bros. Brawl we should have a tournament!!! SIgn up here.

How are you supposed to have a tournament online... with a video game? Lol I feel like I'm missing something really obvious.
  I would sign up (once I figure out how you're supposed to play each other online...) but I've only played regular Smash Bros.

No, this idea fails, whether you like brawl or not. Period.

1. Brawl is not fun (although that's my opinion). Period.
2. Lag. Period.
3. People insisting on items/smashballs. Period.

[quote="dragon96"]Do you really have to start this again?  :mad:  Brawl is way better with items. If you dont like Brawl, just choose Melee stages and characters and turn off items geez.[/quote]
How many times have I told you to not humiliate yourself? Really, could you explain how items are better? Also, basically the only good Melee stage they give is pokemon stadium 1 (pokemon stadium 2 kinda sucks, the conveyor built is really screwed, and the wind allows you to stall really long). Also, just because they are the same characters doesn't mean that they are the same.

1. The physics engine is totally different
2. Each of the characters are redesigned
3. Air dodge
4. Tripping
5. No hitstun
Etc., etc.

I mean, could a person who is pro-brawl support their reasoning for once? I haven't seen a single educated Smash post on AoPS yet (ok, maybe lifeisacircle, but he has nothing better to do than brawl  :P ). Also, could people here please extensively play both brawl and melee before commenting on either one?

Don't get me wrong, i like brawl. In my last post, what I meant to say was "Brawl is not fun relative to melee." Yeah, worded it badly, but now you know.

But whatever you say, you can't argue with reason #2.  :P

dude, this is a thread for people wanting to play brawl

maybe you should make your own thread called "BRAWL SUCKS"?

[quote="dragon96"]He already did:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=217960[/url][/quote]
Notice it's called the smash discussion thread, not the "Brawl sucks, you must listen to me thread." So yeah, if you have a point to make, please just post your reasoning.

I'm not saying brawl isn't fun, but it certainly isnt fun when playing online. It's ridiculous (ly laggy; lag is unaffordable in a game where you sometimes need frame precise timing). And also, you guys need to determine your rules; hopefully you guys know about stage legality and stuff but yeah.

[quote="Klebian"][quote="Nerd_of_the_Ages"]
How many times have I told you to not humiliate yourself? Really, could you explain how items are better? Also, basically the only good Melee stage they give is pokemon stadium 1 (pokemon stadium 2 kinda sucks, the conveyor built is really screwed, and the wind allows you to stall really long).[/quote]
Don't insult others when you're being dumb yourself. Realize that there are 2 pokemon stadiums in brawl, the new one [i]and[/i] the same one from melee, which doesn't have conveyor belt, etc.
Anyway I'm interested in playing but I'd want standard tournament rules: 1v1, 3 or 4 stock, 8 minute time limit, no items, only stages: Final Destination, Battlefield, Smashville, Lylat cruise, and Yoshi's island.

Lag sucks but I've played without lag and that's fun because playing computers sucks too. Lvl 9 = still not a challenge... NotA, what's your FC? Mine is 3866-7847-6621[/quote]

Sorry if I was being offensive. Just tel me whenever I'm not behaving appropriately.

I know there 2, why else do you think i mentioned a pokemon stadium 1 and a pokemon stadium 2? dragon96 suggested me play melee characters in melee courses, but i said basically the only good one is pokemon stadium 1.

About my friend code, I believe I posted it a LONG time ago in that thread about brawl friend codes. Actually, I already added you to my friend list a while ago (maybe 3 or 4 months ago). I'm too lazy to go search out my FC again.  :P

Anyway, even if I do join this for whatever reason, I'll have to work getting better and brawl again, since I've spent all my time improving my melee game.

[quote="Klebian"]lol for what? i think for the most part competitive melee has been replaced by competitive brawl[/quote]
For fun, of course.  :P
And people still play melee, you know. Just not as many.

I dont know how to get my Wii to connect to wireless :(

Although I don't have brawl, I must say.

JIGGLYPUFF ALL THE WAY!

Go with:

No Smashballs, minimum amount of items, only Final Destination.

That's for real skill.

Then build your own map, which is basically all dropblox, and then at the bottom of the map are covered with spikes.

Like

_ _ _ _ _ _ _ _ _ _ _ _

_ _ _ _ _ _ _ _ _ _ _ _

^^^^^^^^^^^^^^

_ = dropblock, ^ = spike. That's fun.

what was the point of reviving a 43 months old thread?

[quote="bluecarneal"][quote="Blackhawk"]this turned in to another "brawl sucks" thread I guess. Where do we even play this mewtwo?[/quote]

The thread is almost 4 years old.[/quote]

thewat? Necro-bump. Please lox.


[quote="A1lqdSchool"]Let ABC be triangle ,AC=2AB. AD,AM respectively are the isector ,median
.The line passing through B ,perpendicular to A,respectively cuts AD,AM at H,K.Calculate S(DHKM)/S(ABC)[/quote]

So, I guess AD is the bisector, not the isector.
But I have no clue what does "the line passing through B, perpendicular to A" mean. Maybe perpendicular to AX for some point X, but I have to many possibilities for X.

oh,Sorry about my mistake :D  :D
yes,AD is bisector and the line passing through B, perpendicular to AD! :)

Let N be the midpoint of AC. Then ABN is isosceles and H is the midpoint of BN.
K is the centroid of ABC.
From Menelaus in BNC we get BD=BC/3.
DM/DB=KM/KA=1/2 so [DKM]/[ABM]=1/9, so [DKM]/[ABC]=1/18.
DK/AB=1/3, so [DHK]/[ABH]=1/9, so [DHK]/[ABN]=1/18, so [DHK]/[ABC]=1/36.
We get the ratio 1/18+1/36=1/12.


Segment $ AC$ has three semicircles with radii $ 4$ cm, $ 6$ cm, and $ 10$ cm. In terms of $ \pi$, how many square centimeters are in the shaded area?

[asy]filldraw(arc((0,0),5,0,180)--(-5,0)--(5,0)--cycle,black);
filldraw(arc((2,0),3,0,180)--(-1,0)--(5,0)--cycle,white);
filldraw(arc((-3,0),2,0,180)--(-5,0)--(-1,0)--cycle,white);
label("$A$",(-5,0),SW);
label("$C$",(5,0),SE);[/asy]

The formula for the area of a circle is pi*radius^2, and the area of a semicircle is the area of a circle divided by 2.  The large semicircle has area pi*10^2/2 or 50 pi.  The two small semicircles are pi*4^2/2 and pi*6^2/2 or 8 pi and 18 pi respectively.  The shaded area is the area of the large semicircle minus the areas of the small semicircles, so its 50pi-8pi-18pi=[b]24pi[/b]

lol forgot to divide everything by 2 and got 48 oof


geometry
trigonometry
number theory proposed
number theory

Find all triangles whose area, side lengths and angle measures(in degrees) are all rational numbers.

triangles with this property must have angles all of whose sines are rational. but the only rational (in degrees) angles with rational sines are 30, 90, 150. (see http://www.mathpages.com/HOME/kmath460/kmath460.htm). but there is no triangle with every angle equal to 30, 90, or 150.

isnt there any elementary solution its from turkey national olympiad
your theorem involves more general condition for example in given triangle we have sinuses and cosines rational so maybe it is simpler to prove there is no triangle like that

see the proof given in the pdf file "linked" by  t0rajir0u [url=http://www.mathlinks.ro/viewtopic.php?t=150658]here[/url]
the proof is given at the beginning of it.
hope that it will help u... :)
i think the proof is elementary and uses nothing "special"....


In this new game: Guess, I will describe an object, and the first one to pm me what the object is correctly, gains points varying on how hard it was.

I will post new hints if someone asks me to.

There may not be more than 5 hints. But you can only ask for a hint one time in a round. When we continue onto the next round, you can ask another hint, but only one.

When you ask for a hint, I will post it so that everyone knows about it.

I will give out one hint at first. This will not count to the 5 that you guys have to count. (Not going to right now.)

This game will end after 50 rounds.

[hide="Sign Ups"]
1.
2.
3.
4.
5.
6.
7.
8.

[/hide]

P.S. (Tell me if I can improve this game somehow.)

Erm 50 rounds is too much, and this belongs in FF.

Hm.. Really? Can someone move it? Let's change it to 20 rounds.

[hide="Sign Up"]
1. bluecarneal
2. auroraak
3. yaofan
4. eggylv999
[/hide]

[quote="AIME15"]You guys do realize that nobody has posted in this thread since May 1, right?[/quote][hide= ][/hide]

If you read you'd realize that IT IS DEAD.  If the mod hasn't posted in a thread for 2 months, it's dead :wink:

Oh..... I haven't seen Jonathan post [b]anywhere[/b] in two months...


geometry
conics
hyperbola
trigonometry
function
calculus
calculus computations

A tangent at any point through a curve meets the co-ordinate axis in $ A$ and $ B$. Such that area of $ \Delta OAB \equal{} 2 \text{ sq.units.}$

Determine equation of the curve.

[hide="I am stuck here:"]I get:

$ \left(x\frac{dy}{dx} \minus{} y\right)^2 \equal{} 4\left(\frac{dy}{dx}\right)$
[/hide]

It's certainly $ y\equal{}\frac{1}{x}$, but I'm not sure how it prove.

Really?  I was pretty sure the answer was $ y \equal{} x \plus{} 2$ ;)  (There are obviously a lot of linear solutions, and you can also do stuff piecewise ($ y \equal{} 2 \minus{} x$ for $ x \leq 1$, $ y \equal{} \frac{1}{x}$ for $ x \geq 1$ works, for example).  The envelope of the linear solutions is the curve $ y \equal{} \pm \frac {1}{x}$.)

[quote="Rajiv"]A tangent at any point through a curve meets the co-ordinate axis in $ A$ and $ B$. Such that area of $ \Delta OAB \equal{} 2 \text{ sq.units.}$

Determine equation of the curve.

[hide="I am stuck here:"]I get:

$ \left(x\frac {dy}{dx} \minus{} y\right)^2 \equal{} 4\left(\frac {dy}{dx}\right)$
[/hide][/quote]
Differentate the equation and you get
$ 2\frac {d^2y}{dx^2}(x^2\frac {dy}{dx} \minus{} xy \minus{} 2) \equal{} 0$.
I think you have missed out an information given in the question that the curve passes through the point (1,1).From here on it is quite obvious :)

I believe only a rectangular hyperbola possesses that property. (wherein the area of the triangle formed by the tangent and the co-ordinate axes is constant)

[quote="pankajsinha"]
Differentate the equation and you get
$ 2\frac {d^2y}{dx^2}(x^2\frac {dy}{dx} \minus{} xy \minus{} 2) \equal{} 0$.
I think you have missed out an information given in the question that the curve passes through the point (1,1).From here on it is quite obvious

[/quote]

Yeah!  :oops:
[hide="What I did was:"]

$ y\equal{}mx\plus{}c \text{  is the equation of the tangent.}
\\
\\
x\minus{} \text{subtangent} \equal{} OA
\\
\\
A(\Delta AOB) \equal{} 2 \equal{} \frac{1}{2}c(OA), c\equal{}OB
\\
\\
\left(x\minus{}\frac{y}{y'}\right)c \equal{} 4
\\
\\
\\
m\equal{}\tan\theta\equal{}\frac{c}{x\minus{}\frac{y}{y'}}
\\
\\
\\
\therefore \left(x\frac{dy}{dx} \minus{} y \right)^2 \equal{} 4\frac{dy}{dx}$[/hide]

The issue is that your setup forces the function to be increasing -- that is, your original differential equation has $ y'$ equal to a square. This is likely because of the way you are calculating area.


probability
geometry
geometric transformation
reflection

A) Points A, B, and C are placed independently and randomly on the boundary of a circle. What is the probability that they line in some semicircle of the circle?

B) Real numbers X and Y are randomly selected from the interval [0,10]. What is the probability that |X-Y| $ \le$ 1/2?

The solution to the first one has been posted all over the place  :wink:

Here's a thread in which it comes in to play: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=171298]Click here[/url]

However, here's an "unconventional" approach to that problem (as in I've never seen anyone use it), it would be interesting if someone did it:

[hide="Yet Another Approach for 1st Problem"]Seeing if they lie on a semicircle is akin to seeing if the triangle they form contains the center.

Let $ A'$, $ B'$ and $ C'$ be the reflections about the center of $ A$, $ B$, and $ C$, respectively.

Let the four triangles $ \triangle ABC, \; \triangle A'BC, \; \triangle AB'C, \; \triangle ABC'$ be the [i]family [/i]of $ \triangle ABC$. Note that $ \triangle ABC$ is in the families of $ \triangle A'BC, \; \triangle AB'C, \; \triangle ABC'$.

Now, prove that in any family, exactly one triangle of the four contains the center. This should lead to the result.[/hide]

[hide="Answer to second"]
We have the lines $ x\minus{}y\equal{}1/2$ and $ y\minus{}x\equal{}1/2$ in the x-y plane, and are bounded to the box $ (x,y) \in [0,10] \times [0,10]$. Find the area within the box contained between those two lines, and then divide by 100. This area is really 100 minus the sum of two right triangles with legs 9.5. So, the probability is $ \frac{100\minus{}9.5 ^2}{100} \equal{} \frac{39}{400}$[/hide]

Problem A can be extended to finding the probability that $ n$ randomly chosen points on a circle lie on a semicircle, or to get even more general, that they lie on an arc of angle $ 2\pi x$ for $ x \in [0,1]$.

[hide="Hint"]Can you geometrically represent all possible cases and the subset in which the condition is satisfied?[/hide]
[hide="Answer"]$ \min(1,nx^{n \minus{} 1})$, or $ \frac {n}{2^{n \minus{} 1}}$ for the semicircle case.[/hide]

EDIT: [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?p=1081010#1081010]Posted yet another extension here[/url]

Please excuse me if I'm incorrect, but I think Tenoreoz meant $ |X\minus{}Y| \leq \frac{1}{2}$, not $ \equal{}$; if it's equal, I think the probability is 0.

[quote="Tenoreoz"]Approach inspired by Zuton Force:

[hide]
[b]Lemma 1[/b]: If any angle of the triangle is obtuse, then the triangle does not contain the center.

[b]Proof[/b]: WLOG, we will let $ \angle C$ be our obtuse angle. Take diameter $ AB$. We have that $ \angle C \equal{} 90$ and the center lies on $ AB$. In order for $ \angle C$ to be obtuse, $ A$ and/or $ B$ must be some other point on $ \widehat{AC}$ or $ \widehat{BC}$, respectively, say $ A'$ and $ B'$. However, clearly $ \triangle A'BC$ or $ \triangle AB'C$ will not contain the center.

Now, WLOG, construct $ \triangle ABC$ such that it contains the center. By Lemma 1, $ \angle A, \angle B, \angle C < 90$. Reflect $ A$ about the center to get $ A'$. Since quadrilateral $ ABA'C$ is cyclic, $ \angle A' \equal{} 180 \minus{} \angle A$, which implies that $ \triangle A'BC$ is obtuse. Repeating this for $ B$ and $ C$ gives the same result. Therefore, exactly one triangle of the four contains the center.

Then applying the result to the first problem shows that the probability is indeed $ \frac {3}{4}$.

[/hide][/quote]
What is WLOG stand for?

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=192385&start=100

for problem A, go down to firecricket's solution, its pretty easy to see by logic (and this was just posted like yesterday  :lol: )

WLOG-without loss of generality.


geometry
symmetry
geometric transformation
reflection
complex numbers
geometry theorems

Inversion is symmetry with respect to a circle or reflection in a circle. Consider symmetry with respect to (or reflection in) a line, which can be considered as a circle of infinite radius. Given a line and a point A not on this line, then the reflection A' of the point A in this line can be constructed by drawing any circle centered on the line and passing through A and then another circle also centered on the line and passing through A. The reflection A'  of A in the given line is the other intersection of the 2 circles. Other than being centered on the given line and passing through A, the 2 circles are arbitrary. Hence, the reflection of A in the given line can be defined as follows:

[color=blue]The point A' is a reflection of the point A in a given line if any circle centered on this line and passing through A also passes through A'.[/color]

A circle centered on a line is perpendicular to this line. So the definition can be modified as follows:

[color=blue]The point A' is a reflection of the point A in a given line if any circle perpendicular to this line and passing through A also passes through A'.[/color]

This allows to generalize the definition for symmetry with respect to (or reflection in or inversion in) a circle:

[color=red]The point A' is a reflection (or inversion) of the point A in a given circle if any circle perpendicular to this circle and passing through A also passes through A'.[/color]

Apart from the basic properties of inversion, such as $OA \cdot OA' = r^2$, where O, r are the center and radius of the inversion circle, and the fact that inversion preserves angles between any curves, including lines and circles, there is the following basic theorem:

[color=red]Any 2 circles (or circle and line) can be transformed by inversion into 2 intersecting lines (if the 2 circles intersect), 2 parallel lines (if the 2 circles touch) or into 2 concentric circles (if the 2 circles do not have a common point).[/color]

Lines, especially parallel lines, or concentric circles are easier to handle than general circles. If the 2 circles intersect at points P, Q, these are the busy points and choosing the inversion center at either of them, say P, transforms them into 2 intersecting lines. Since the points on the inversion circle stay in place, choosing the radius of the inversion circle r = PQ, the inverted circles are tangents to the original circles at Q.

If the 2 circles touch at a point P, this is the busy point and choosing the inversion center at P transforms them into 2 parallel lines perpendicular to the center line of the original circles. If any calculations are to be performed, it helps to have simple expressions for the distance of these 2 parallel lines from each other and for their distances from the inversion center. Let the 2 circles intersect their center line at A, B different from the tangency point P, so $PA = 2r_1, PB = 2r_2$. Choosing the inversion radius $r = 2r_1$ keeps A in place but moves B into B' such that $PB' = \frac{r^2}{PB} = \frac{4r_1^2}{2r_2} = \frac{2r_1^2}{r_2}$, not a convenient position. But even though choosing the inversion radius $r = 2\sqrt{r_1r_2}$ leaves neither A nor B in place, $PA' = \frac{r^2}{2r_1} = \frac{4r_1r_2}{2r_1} = 2r_2$ and $PB' = \frac{r^2}{2r_2} = \frac{4r_1r_2}{2r_2} = 2r_1$, so that A' is a reflection of B in a normal to the center line at the tangency point and B' is a reflection of A in this normal, very simple and convenient.

If the 2 circles do not have a common point, then there is no busy point to put the inversion center in. It is necessary to find a 3rd circle perpendicular to both given circles and centered on their center line and then pick an inversion circle transfoming the 3rd circle into a line perpendicular to the center line. The inversion center is a busy point - one intersection of the 3rd circle with the center line. The center line is inverted into itself and the 2 given circles into some other circles both perpendicular the center line (i.e., centered on it) and perpendicular to the 3rd inverted circle, i.e., to a line perpendicular to the center line, i.e., centered on this perpendicular line. This is possible only of both the inverted circles are centered at the intersection of these 2 perpendicular lines, i.e., if they are concentric.

The remaining question is how to find the circle $c_3$ perpendicular to both the circles $c_1, c_2$. To do this, we use a busy point, one intersection of one of the given circles, say $c_1$, with the center line $l$. Choosing the inversion center here transforms the center line into itself, the circle $c_1$ into a line $c_1' \perp l$ and the circle $c_2$ into a circle $c_2' \perp l$, i.e., centered on $l$. A circle $c_3'$ centered on the intersection X of the perpendicular lines $c_1' \perp l$ and with radius equal to the tangent length from X to $c_2'$  is perpendicular to both $c_1', c_2'$. Inverting $c_1', c_2'$ back to $c_1, c_2$ carries $c_3'$ into $c_3 \perp c_1, c_2$, because inversion preserves angles between curves.

How do we prove that inversion sends a circle into another circle? Or that it sends an angle into an opposite angle?

I manage to answer by myself to the first question, but I used complex numbers, and that's not really a nice solution.

[quote="Philip_Leszczynski"]In his packet of geometry, Kiran Kedlaya mentioned several times that when solving an inversion problem, it is best to look for a "busy point."
Unless I am wildly mistaken, inversions take place with respect to a circle.
So is this "busy point" we are looking for the center of our circle, or are we actually looking for a busy circle?
And if so, given the busy point, do we also need to look to find a convenient radius for our circle of inversion?

Philip Leszczynski[/quote]

dear, it depends on the diagram...

maybe by doing inversion at a busy point, the problem changes to a hard problem.


I was just wondering if anyone knows any good introductory physics books to read in preparation for AP Physics (Mechanics). I plan to take the class next year and wanted to get a jump start.

Are you looking for a textbook or review book?  For textbooks, any of the "University Physics" books should be fine.  I say "any of..." because many authors have books called University Physics, but they pretty much all cover the same material.  Halliday and Resnick is also good (the one also with Walker; the one with Krane is really good but a bit harder).  Kleppner is also good, but pretty difficult.

For review books, I think Princeton Review is pretty good.


Let $ a$ and $ b$ be positive integers such that $ (1+ab)| (a^{2}+b^{2})$. Show that the integer $ \frac{a^{2}+b^{2}}{1+ab}$ must be a perfect square.

It's Imo 1988
See here : http://www.mathlinks.ro/viewtopic.php?t=150371


MATHCOUNTS
percent
geometry
geometric transformation
reflection

1. What percent of the natural numbers from 1 to 100 have at least one digit which is a 6?

2. What about the percent of the natural numbers from 1 to 1000?

Can you generalize?

you forgot number 2

these questions bring back goode memories. ah jes *reflects to when he was a wee lad*

The question asks for the [b]percentage[/b], don't forget.

So, out of the first 100 numbers, [hide]19[/hide] contained a digit six, which makes for exactly [hide]19[/hide]%.  So she was right.  Go do the 1000 case, now :).

tare was prbly reminding bout the 1000 one

Yup, I'm pretty sure she got that one...nice save Syntax, thanks :)

[hide]For 1000:

___ ___ ___

For each spot we must fill in a number. Then, if we subtract the number of numbers we can make without using the digit 6, what remains is the number of numbers we can make using at least one 6.

We can make 9 * 9 * 9 = 729 numbers without using 6. Therefore, there are 1000 - 729 = 271 numbers with at least one 6. Thus, the percentage is 27.1%.

Generally, for numbers from 1 to 10k where k is a positive integer, (10k - 9k) / 10k is the fraction of the total number of numbers containing a given digit from 1 to 9.[/hide]

Very good.

...How old are you? You might be too advanced for this...

Yowza, that was elegant. I woulda done it the other way, and woulda had a harder time generalizing. Nice solution.


vector
linear algebra
matrix
linear algebra unsolved

Prove that there exist a column vector $ a$ with real entries and dimension $ n$ such that $ \sum_{i\equal{}1}^{n}a_i\equal{}1$ and $ a^THa\equal{}\frac{1}{n^2}$, where $ H$ is the Hilbert matrix of dimension $ n$.

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=146373]here[/url] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=282337]here[/url].

Let $ j$ be the column vector of all ones. Let $ a\equal{}\frac1{n^2}\cdot H^{\minus{}1}j$. $ \sum a_i\equal{}j^Ta\equal{}\frac1{n^2}j^TH^{\minus{}1}j\equal{}1$ and $ a^THa\equal{}\frac1{n^4}j^TH^{\minus{}1}HH^{\minus{}1}j\equal{}\frac1{n^4}j^TH^{\minus{}1}j\equal{}\frac1{n^2}$.


How do you guys find the symbols for writing a problem ive  noticed that you all have the signs for theta and sigma and everything and I was just wondering how I could do it because just writing a problem in words can lead to some confusion.

Click LaTeX on the sidebar of this site.


"The decimal expression of $ \frac{1000}{2008}$ repeats with period $ 50$.  Determine the sum of the digits in the repeating portion of the decimal expansion."

I realize that you can simply the fraction into $ \frac{125}{251}$, but from there on I don't know how to do this.  Assume that you have $ 3$ minutes to solve the problem.

I'm not sure, but maybe you could change 125/251 to (period of 50 digits)/999999...999(50 9's)...just a thought. I doubt that's the way to go, though. You have to do something clever, though. Maybe you don't even need to simplify the fraction.

Using my amazing calculator, I got
0.49800796812479003984063745019920318725099601593625...but the sum is 225. 225 is a more normal number...so yeah.

Is this really MC level?

I admit that I don't know what level this problem should be placed in.  However, I thought that some of you brilliant MathCounts people would be able to do this (think of it as Target #8 on a National MathCounts Competition).

But, yes, 225 is the correct answer.

Can anyone do this?

hmmm.. there doesn' seem to be a fast way of doing it... (maybe with mods, but I don't think so)

well... you could realize that 1000/2008 is 4/2008 or 1/502 less than a half...
that might help if you memorized 50 digits of 1/502 :)

Yeah. I have this amazing program that can go up to 24000 digits :D . But a method...hey--you have to know extensively about 2008, don't you? Memorize all the numbers less than 2008 over 2008.lol

I think that was the OTHER number theory problem. :D


Ross Mathematics Program
parameterization
LaTeX

I'm trying to insert a png to my document (TexShop for mac). Here is the compiler error:

Fatal error occurred, the output PDF file is not finished!

I know it's probably not in the right directory. What is the tex source file called? I've tried inserting the png image in every single directory in the hierarchy to no avail. I have no idea where its looking for it. Is there a way i can change where it looks for images, or can someone please tell me where to put the image? You'd think this info would be easily accessible. thanks.

Ross

Put \usepackage{graphicx} in your document's header. Put the .png file in the same dir as your .tex file and use the command \includegraphics[optional parameters]{name.png} to include your png file.

[optionalparameters] can be [height=x in, width=y in] or just one of them [height=3 cm] (don't forget the unit of measurement).

Compile the document with pdfTeX afterwards.


I was just wondering... for the people that go to HMMT,

Does your school pay for the trip? Or do all of you pay for your individual expenses?
Moreover, does this event have anything to do with your school?

I am planning to go to Cambridge for the competition this year with some of my friends, and my school does not wish to do anything for us: no bus, money, nothing. Luckily, some of the local businesses have shown interest in sponsoring us, so my friends and I can be relieved of having to pay hundreds of dollars for something that we're not likely going to win.

I've never been in the past, but I created a math team this year at my school that will attend in 2009, and my principal is willing to pay for transportation, registration, etc., and even willing to buy the AoPS books for my team, though he may just offer a subsidy. However, this is an academy specialized in math and science, so this may be a special case...

There's just barely enough interested students to make a full team, but we probably won't perform very well, looking at past tests. Preparing for it on the weekends is fun, too, though.


Let $ a,b,c >0$ and $ a^2\plus{}b^2\plus{}c^2\equal{}abc$ Prove that !
$ \sum ab \ge 2 \sum a\plus{}9$

[quote="bigbang195"]Let $ a,b,c > 0$ and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} abc$ Prove that !
$ \sum ab \ge 2 \sum a \plus{} 9$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=320625


limit
integration
calculus
inequalities
geometry
geometric transformation
real analysis

Suppose $ f,g\in L^1(\mathbb{R})$. Show that

$ \lim_{y\to\infty}\int_{\minus{}\infty}^\infty |f(x)\plus{}g(x\minus{}y)|dx\equal{}\int_{\minus{}\infty}^\infty|f(x)|dx\plus{}\int_{\minus{}\infty}^\infty |g(x)|dx$.

Fix $ \varepsilon>0$.  If $ t_1$ is large enough, then $ \int_{\mathbb{R}-B(0,t_1)} |f(x)|dx<\varepsilon$.  Similarly, if $ t_2$ is large enough, then $ \int_{\mathbb{R}-B(0,t_2)} |g(x)|dx<\varepsilon$

Then pick $ y$ so large that $ 0+t_1<y-t_2$ and estimate the integral of $ |f(x)-g(x-y)|$ using the triangle inequality ($ |x+y|\geq |x|-|y|$)

[hide="Still here but..."]After a full day of studying, there is probably something dumb I'm missing, but why doesn't this work.

Let $ \epsilon > 0$, then the goal is to find $ y_0\in\mathbb{R}$ such that for all $ y > y_0$ we have

$ \Big|\int_\mathbb{R}|f(x) - g(x - y)|dx - \int_\mathbb{R}|f|dx - \int_\mathbb{R}|g|dx\Big| < \epsilon$.

But \begin{eqnarray*} \Big|\int_\mathbb{R}|f(x) - g(x - y)|dx - \int_\mathbb{R}|f|dx - \int_\mathbb{R}|g|dx\Big| & \leq & \Big|\int_\mathbb{R}|f(x)|dx + \int_\mathbb{R}|g(x - y)|dx - \int_\mathbb{R}|f|dx - \int_\mathbb{R}|g|dx\Big| \\
& = & \Big| \int_\mathbb{R} |g(x - y)|dx - \int_\mathbb{R}|g(x)|dx\Big| \\
& = & 0 \\
& < & \epsilon \end{eqnarray*}

Where the two $ g$ integrals are equal for any fixed $ y$ by the translation invariance of Lebesgue measure over $ \mathbb{R}$. Essentially this means that you don't need to take the limit to get equality which is definitely not true.[/hide]

Ack!  :blush:  A perfect example of why one should always wait until morning before posting something that one knows to not be true, but can't figure out why.

obviously it is the first step that is wrong.

typo aside, you want to estimate
\[ \int_{\mathbb{R}} |f(x) \minus{} g(x \minus{} y)|dx \minus{} \int_{\mathbb{R}} |f(x)|dx \minus{} \int_{\mathbb{R}} |g(x)|dx
\]
it's clearly nonpositive by the triangle inequality (as you noted), but you seem to want to conclude from $ 0\leq a\leq b$, $ 0\leq c,d$ and $ b \minus{} c \minus{} d\leq 0$ that $ |a \minus{} c \minus{} d|\leq |b \minus{} c \minus{} d|$ when in fact the opposite is always true!


The fraction $ \frac{\sqrt{a^2\plus{}x^2}\minus{}(x^2\minus{}a^2)/\sqrt{a^2\plus{}x^2}}{a^2\plus{}x^2}$ reduces to:
$ \textbf{(A)}\ 0 \qquad\textbf{(B)}\ \frac{2a^2}{a^2\plus{}x^2} \qquad\textbf{(C)}\ \frac{2x^2}{(a^2\plus{}x^2)^{\frac{3}{2}}} \qquad\textbf{(D)}\ \frac{2a^2}{(a^2\plus{}x^2)^{\frac{3}{2}}} \qquad\textbf{(E)}\ \frac{2x^2}{a^2\plus{}x^2}$

[hide]

Writing the numerator with a common denominator, $ \frac{\frac{a^2\plus{}x^2\minus{}x^2\plus{}a^2}{\sqrt{a^2\plus{}x^2}}}{a^2\plus{}x^2}\equal{}\frac{2a^2}{(a^2\plus{}x^2)\sqrt{a^2\plus{}x^2}}$, or $ \textbf{(D)}$

[/hide]


$ x,y,z > 0$
$ \sum_{cyc}\frac {yk}{x \plus{} yk \plus{} zk^2}\le1$,$ \forall k\ge0$

[quote="shfdfzhjj"]$ x,y,z > 0$
$ \sum_{cyc}\frac {yk}{x + yk + zk^2}\le1$,$ \forall k\ge0$[/quote]
It's equivalent to $ (k^3-1)^2xyz\geq0.$

[quote="arqady"][quote="shfdfzhjj"]$ x,y,z > 0$
$ \sum_{cyc}\frac {yk}{x + yk + zk^2}\le1$,$ \forall k\ge0$[/quote]
It's equivalent to $ (k^3 - 1)^2xyz\geq0.$[/quote]
Why?Can you post solution?

I don't understand.
You can write clear???

For you, math10!
$ \sum_{cyc}\frac {yk}{x \plus{} yk \plus{} zk^2}\le1\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}ky(y\plus{}kz\plus{}k^2x)(z\plus{}kx\plus{}k^2y)\leq\prod_{cyc}(x\plus{}ky\plus{}k^2z)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}ky(k^3x^2\plus{}k^2y^2\plus{}kz^2\plus{}(k^4\plus{}k)xy\plus{}2k^2xz\plus{}(k^3\plus{}1)yz)\leq$
$ \leq(k^3x^2\plus{}k^2y^2\plus{}kz^2\plus{}(k^4\plus{}k)xy\plus{}2k^2xz\plus{}(k^3\plus{}1)yz)(x\plus{}ky\plus{}k^2z)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(k^3x^3\plus{}(2k^4\plus{}k)x^2y\plus{}(k^5\plus{}2k^2)x^2z\plus{}2k^3xyz\right)\leq$
$ \leq\sum_{cyc}\left(k^3x^3\plus{}(2k^4\plus{}k)x^2y\plus{}(k^5\plus{}2k^2)x^2z\plus{}\frac{1}{3}(k^6\plus{}4k^3\plus{}1)xyz\right)\Leftrightarrow$
$ \Leftrightarrow(k^3\minus{}1)^2xyz\geq0.$

[quote="arqady"]For you, math10!
$ \sum_{cyc}\frac {yk}{x \plus{} yk \plus{} zk^2}\le1\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}ky(y \plus{} kz \plus{} k^2x)(z \plus{} kx \plus{} k^2y)\leq\prod_{cyc}(x \plus{} ky \plus{} k^2z)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}ky(k^3x^2 \plus{} k^2y^2 \plus{} kz^2 \plus{} (k^4 \plus{} k)xy \plus{} 2k^2xz \plus{} (k^3 \plus{} 1)yz)\leq$
$ \leq(k^3x^2 \plus{} k^2y^2 \plus{} kz^2 \plus{} (k^4 \plus{} k)xy \plus{} 2k^2xz \plus{} (k^3 \plus{} 1)yz)(x \plus{} ky \plus{} k^2z)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(k^3x^3 \plus{} (2k^4 \plus{} k)x^2y \plus{} (k^5 \plus{} 2k^2)x^2z \plus{} 2k^3xyz\right)\leq$
$ \leq\sum_{cyc}\left(k^3x^3 \plus{} (2k^4 \plus{} k)x^2y \plus{} (k^5 \plus{} 2k^2)x^2z \plus{} \frac {1}{3}(k^6 \plus{} 4k^3 \plus{} 1)xyz\right)\Leftrightarrow$
$ \Leftrightarrow(k^3 \minus{} 1)^2xyz\geq0.$[/quote]
Yes,you get the point.But I think we can slove it by another way. :lol:

[quote="math10"][quote="arqady"]For you, math10!
$ \sum_{cyc}\frac {yk}{x \plus{} yk \plus{} zk^2}\le1\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}ky(y \plus{} kz \plus{} k^2x)(z \plus{} kx \plus{} k^2y)\leq\prod_{cyc}(x \plus{} ky \plus{} k^2z)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}ky(k^3x^2 \plus{} k^2y^2 \plus{} kz^2 \plus{} (k^4 \plus{} k)xy \plus{} 2k^2xz \plus{} (k^3 \plus{} 1)yz)\leq$
$ \leq(k^3x^2 \plus{} k^2y^2 \plus{} kz^2 \plus{} (k^4 \plus{} k)xy \plus{} 2k^2xz \plus{} (k^3 \plus{} 1)yz)(x \plus{} ky \plus{} k^2z)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(k^3x^3 \plus{} (2k^4 \plus{} k)x^2y \plus{} (k^5 \plus{} 2k^2)x^2z \plus{} 2k^3xyz\right)\leq$
$ \leq\sum_{cyc}\left(k^3x^3 \plus{} (2k^4 \plus{} k)x^2y \plus{} (k^5 \plus{} 2k^2)x^2z \plus{} \frac {1}{3}(k^6 \plus{} 4k^3 \plus{} 1)xyz\right)\Leftrightarrow$
$ \Leftrightarrow(k^3 \minus{} 1)^2xyz\geq0.$[/quote]
thank you very much.
but I do not think this is a good solution. I wish to have a solution using the Cauchy-Schwarz[/quote]
There is, math10 ;)


Find all pairs of functions $ \ f: R\rightarrow\ R$;$ g: R\rightarrow\ R$ satisfy:
$ \  i.$f is increasing;
$ \ ii.$$ \ f(xy)\equal{}g(y)f(x)\plus{}f(y)$ for all $ \ x,y$

Denote $ P(x,y)$ as the property of our function.
First we have $ P(1,y)\Rightarrow g(y)f(1)\equal{}0$
it implies 2 possibilities:
1)$ g(y)\equiv 0$ (trivial) it implies $ f(xy)\equal{}f(y)$ and putting $ x\equal{}0$ gives us the pair of functions $ g(x)\equal{}0 \ f(x)\equal{}a\equal{}f(0)$
2)$ f(1)\equal{}0$ it implies that $ f(0)<f(1)\equal{}0$, because f is increasing
Now observe that $ P(y,y)$ gives us $ g(y)f(y)\equal{}f(y^{2})\minus{}f(y)$
Try $ P(0,y)$. It gives us $ f(0)(1\minus{}g(y))\equal{}f(y)$, wchich combined with $ f(0)<0$ gives us:
$ f(y)(1\minus{}g(y))<0$ for all y
but $ f(y)(1\minus{}g(y))\equal{}f(y)\minus{}f(y)g(y)\equal{}2f(y)\minus{}f(y^{2})<0$ which gives us a contradiction for $ y\equal{}1$

Q.E.D. :wink:

Put $ y\equal{}1$. This yields $ f(x)\equal{}g(1)f(x)\plus{}f(1)$. Solving for $ f(x)$ yields $ f(x)\equal{}\frac{f(1)}{1\minus{}g(1)}$, that holds for all $ x$, so $ f(x)$ is constant. But $ f(x)$ is suposed to be increasing, so there is no solution.   :?:

1) $ f(x)\equal{}0$, $ g$ arbitrary;

2) $ f(x)\equal{}c\ne0$, $ g(x)\equal{}0$.

3) $ g(x)\equal{}sign(x)*|x|^a$, $ f(x)\equal{}cg(x)\minus{}c$, where $ a>0,c>0$. $ sign(x)$ is the sign function.

[quote="Sashsiam_2"]But $ f(x)$ is suposed to be increasing, so there is no solution.   :?:[/quote]
But $ f$ is not strictly increasing...

[quote="basted"]
But $ f$ is not strictly increasing...[/quote]

I believe that strictly increasing=increasing $ \neq$ nondeacreasing :wink:


Hi,

I found a basic tutorial to Beamer and it told me to input the following into the editor I am using:
[code]\documentclass{beamer}
\usepackage{amsmath}
\usetheme{default}
\begin{document}

\begin{frame}
  \frametitle{A sample slide}
A displayed formula:
\[
  \int_{-\infty}^\infty e^{-x^2} \, dx = \sqrt{\pi}
\]
An itemized list:
\begin{itemize}
  \item itemized item 1
  \item itemized item 2
  \item itemized item 3
\end{itemize}
\begin{theorem}
  In a right triangle, the square of hypotenuse equals
  the sum of squares of two other sides.
\end{theorem}
\end{frame}
\end{document}[/code]
I did that however when compiling I was asked whether or not I would like it to install the package beamer.cls (from some website). I selected yes, but it keeps on not installing! Please can you advise me on how I can get this thing working? (It probably is installing, but I am doing something wrong).

By the way, I have been on Start> Miktext >... and tried installing it from there, but the following error message appears which makes no sense to me.
[code]MiKTeX Problem Report
Message: The remote package repository may be corrupted.
Data: http://mirrors.ircam.fr/pub/CTAN/systems/win32/miktex/tm/packages/
Source: mpm.cpp
Line: 2089
MiKTeX: 2.5
OS: Microsoft Windows XP Home Edition Service Pack 2 (Build 2600)
SystemAdmin: yes
PowerUser: no
SharedSetup: yes
BinDir: C:\Program Files\MiKTeX 2.5\miktex\bin
Root0: C:\Documents and Settings\user\Application Data\MiKTeX\2.5
Root1: C:\Documents and Settings\user\Local Settings\Application Data\MiKTeX\2.5
Root2: C:\Documents and Settings\All Users\Application Data\MiKTeX\2.5
Root3: C:\Program Files\MiKTeX 2.5
Install: C:\Program Files\MiKTeX 2.5
UserConfig: C:\Documents and Settings\user\Application Data\MiKTeX\2.5
UserData: C:\Documents and Settings\user\Local Settings\Application Data\MiKTeX\2.5
CommonConfig: C:\Documents and Settings\All Users\Application Data\MiKTeX\2.5
CommonData: C:\Documents and Settings\All Users\Application Data\MiKTeX\2.5 [/code]
Many thanks for any help.  :)

Well, I'm not the indicated person to help ya, since my LaTeX skills are basics. Here's the PDF attached below, is that what you want? (It gave me 4 warnings.)

Anyway I suggest you to download & install full MikTeX distribution, so that you don't have problems in the future.

[quote="stevem"]The problem may be that you are running version 2.5 which is no longer supported. So try downloading the latest version 2.7. You can have both versions running at the same time but it is simpler to uninstall 2.5 and install 2.7. Then tell the editor the new location of MiKTeX.

To save problems like this, why not just install the complete version of MiKTeX? It includes beamer and almost every other package you could wish for. I know it's a very large download but time spent downloading it can save a lot of frustration later if the site is down, MiKTeX is updated or whatever.[/quote] Hi Steve,

I thought that I had downloaded the complete version, for I had left the computer running overnight when I had done so!

I think I know what the problem might be though--I haven't told TeXnic centre where I installed MikTeX. Thanks for telling me about the new version; I should have it set up by the end of summer.

|-.-|: Thanks--that's what I'm after, but the problem is installing the packages.

EDIT: Yes--I do have the file beamer.cab on my computer, but erm... how do I get TeXnic centre to recognise this? (Or have I got the wrong end of the stick again?)

[quote="stevem"]beamer.cab is just the compressed set of files that MiKTeX uses to create the all the files. Beamer files themselves are  located in C:\Program Files\MiKTeX 2.5 \tex\latex\beamer.

What happens if you go to Start, All Programs, MiKTeX 2.5, Browse Packages? Is beamer reported as installed there? If it not then right-click n beamer and choose install. If you get errors, in the same program click on Repository, Change Package Repository ... and point it to the directory containing the beamer.cab file and try installing again.[/quote] Steve, you're great! Thanks a lot. :) (It worked!)

This will also allow me to install other packages, which I have long ago struggled to install. (What's the worse that could happen if I don't upgrade? Just out of interest of course, because I always seem to have a reluctance to "move forward"!)

EDIT: Did I say thanks? I can't believe my eyes! (Since I don't use Word any more, I was looking for an online package to create minutes. I also couldn't get that to work (same error that I had with beamer) but since I've changed repository, it's installing everything without error messages! Cool.  :D

It sounds like the packages were downloaded originally but never actually installed.
[quote="beans"](What's the worse that could happen if I don't upgrade? Just out of interest of course, because I always seem to have a reluctance to "move forward"!)[/quote]As you have the complete version there's not really an urgent need to upgrade. 2.7 does include Xetex which allows you to easily use Windows fonts and you won't easily be able to download the latest packages as the format MiKTeX uses in Repositories has changed.

[quote="stevem"]It sounds like the packages were downloaded originally but never actually installed.[/quote] I have just gone and installed 1266 odd packages now! (Is it bad that this makes me feel 10 feet high?!)
[quote]As you have the complete version there's not really an urgent need to upgrade. 2.7 does include Xetex which allows you to easily use Windows fonts and you won't easily be able to download the latest packages as the format MiKTeX uses in Repositories has changed.[/quote]
Thanks--I'll stick with 2.5 for as long as I can then.  :)

I have a similar problem with beamer. Using MikTex package manager I have tried to dowload it from internet, but I keep getting the message that version 2.5. is not supported any more (even though I have version 2.6.!!!). I have dowloaded the package manually, and I puted the package in C:\Program Files\MikTex 2.6.\tex\latex\beamer, but I keep getting the message that the selected location is not a "local package repository".

Remark: Sorry for bringing an old topic "back to life".

You can't get MiKTeX to install packages like that.
You could install the complete version of 2.7 then you won't have to install package that aren't there as you'll have all that you are likely to need.

Or you could download beamer.zip from [url=http://tug.ctan.org/tex-archive/macros/latex/contrib/beamer/]CTAN[/url] and extract the files to C:\Program Files\MiKTeX 2.5\tex\latex (asuming that's where your latex files are kept), preserving the folder structure.
Then you must install the pgf and xcolor packages, assuming that you haven't already got them.
Finally you must update MiKTeX's database(Settings, Refresh FNDB).

[url=http://tug.ctan.org/tex-archive/macros/latex/contrib/beamer/doc/beameruserguide.pdf]Beamer documentation[/url] pages 12 -13 has the details.

Thanks Stevem!!! Everthing works fine now.   :)


Prove that 2^(3^n)+1 is divisible by 3^(n+1) for all n greater than or equal to 1.

Also check out the website in my sig for more problems that have been compiled for F^3, and weekly problem solving group at my school that my friend and I have started up.

Letting $3^n=3m$ and $2^{3^n}+1=a_n$ then the question want us to prove $3^{n+1}|a_n$

We see that $a_n=2^{3^n}+1=2^{3m}+1=(2^m+1)(4^m-2^m+1)=(a_{n-1})(4^m-2m+1)$

but $4^m-2^m+1\equiv 0\mod 3$

Hence it can be reduce to $a_n\equiv (a_0)3^n\equiv 3^{n+1}\mod 3$   since $a_0=3$

nice solution shyong.your solution is very clear and nice. :D

That was how I had originally done it but I did like shyong's method.


geometry
3D geometry
tetrahedron
circumcircle
geometry proposed

Let $O$ be a point equally distanced from the vertices of the tetrahedron $ABCD$. If the distances from $O$ to the planes $(BCD)$, $(ACD)$, $(ABD)$ and $(ABC)$ are equal, prove that the sum of the distances from a point $M \in \textrm{int}[ABCD]$, to the four planes, is constant.

Let $O_A,O_D$ be the projections of $O$ on $BDC,ABC$ respectively. We immediately see that $O_A,O_D$ are the circumcenters of $BDC,ABC$ respectively, and $\angle BO_AC=\angle BO_DC$, meaning that $\angle BDC=\angle BAC$. After obtaining all the analogous equalities, we see that the sum of the plane angles in all the vertices of the tetrahedron is $\pi$, and it's well-knwon that this means that all the faces of the tetrahedron have equal areas (actually, the faces are congruent triangles). From the equality of the areas, the conclusion follows.

Sorry I didn't post the complete solution, but there are other, nicer problems on the form right now :D.


calculus
integration
function
algebra unsolved
algebra

If there are two equations in $x$ of the same degree $n$, and $M$ is a rational integral function of the coefficients of these equations such that

\[Mx, Mx^2, \ldots, Mx^{n-1}\]

are each also rational functions of the same, find the most general form of $M.$

I can't understand it  :blush:  :blush:  :blush:


Monica reduced $ \frac{154}{253}$ to a common fraction by canceling the middle 5s to get $ \frac{14}{23}$. This answer is correct, but the procedure is incorrect. Kia's example worked with Monica's incorrect procedure, too; the fraction $ \frac{242}{x}$ also reduced to its equivalent common fraction when the identical middle digits of the two three-digit integers were canceled out. What is the greatest possible value of $ x$?

http://www.mnspe.org/mathcounts/2007tests/510-07_S_Solutions.pdf

Go to number 25.

Wow, that link has solutions to many of the problems on the FTW.

That's because that particular link leads to one of the Sprint or Target Rounds of MC.


[quote="i_like_pie"][hide]$3x+4(5-x(3x-7)-6)$
$3x+4(5-3x^{2}+7x-6)$
$3x+20-12x^{2}+28x-24$
$-12x^{2}+31x-4$[/hide][/quote]
Your solution looks right...
[hide="a common mistake that i like pie did not make"]Distribute from the inside out first... It makes errors less likely...[/hide]

[quote="i_like_pie"]How can it not be right? It's MY solution, after all. :D[/quote]
sure, tommy... :D :kidding:

I also solved and sure that i_like_pie's answer is true/

[hide="another common mistake that i like pie didnt make"]Watch the negative sign with the first distribution.[/hide]

Looks like I'm good at not making mistakes. :D

[quote="anirudh"][quote="linzk3"]how do you simplify 3x+4[5-x(3x-7)-6]??
please and thanks[/quote]

[hide]
$\\ 3x+4[5-x(3x-7)-6]\\ 3x+4(5-3x^{2}+7x-6)\\ 3x+4(-3x^{2}+7x-1)\\ \therefore3x+20-12x^{2}+28x-24$[/hide][/quote]
That's not simplified.

oops. i forgot the last step...  :blush:

$-12x^{2}+31x-4$ is the anwer


$\sum_{cyc}\frac{x}{\frac{y^{3}+z^{2}}{2}}\leq \sum_{cyc}\frac{x}{yz\sqrt{y}}=\sum_{cyc}\frac{x^{2}}{\sqrt{y}}$
But $(\sum_{cyc}\frac{x^{2}}{\sqrt{y}})^{2}\leq(\sum_{cyc}x^{4})(\sum_{cyc}\frac{1}{x})=(\sum_{cyc}x^{4})(\sum_{cyc}yz)$
So it's enough to prove that \[(\sum_{cyc}x^{4})(\sum_{cyc}yz)\leq 3(\sum_{cyc}x^{6})\] $\leftrightarrow$ \[\sum_{sym}x^{5}y+\sum_{cyc}x^{4}yz\leq 3\sum_{cyc}x^{6}\].
But $\sum_{sym}x^{5}y\leq 2\sum_{cyc}x^{6}$ ,by Muirhead
and $\sum_{cyc}x^{4}yz\leq \sum_{cyc}x^{6}$ ,by rearrangements,and we are done

Thanks.By the way , do you have another solution for this one ?

Yes it's an interesting one ;) .But i think the ideea it's a liitle bit the same.It was a nice inequality.Well done!


projective geometry
geometry
geometric transformation
reflection
LaTeX
geometry proposed

Let $ ABC$ be a triangle. Let $ A_1$, $ B_1$, $ C_1$ be the midpoints of its sides $ BC$, $ CA$, $ AB$, and $ A_2$, $ B_2$, $ C_2$ the midpoints of the altitudes from $ A$, $ B$, $ C$. Show that the lines $ A_1A_2$, $ B_1B_2$, and $ C_1C_2$ meet at one point.

Let AD,BE,CF be the altitudes. $ B_1,C_1,A_2$ are collinear since they are the midpoints of $ AC,AB,AD$, and thus, we get $ \frac {C_1A_2}{A_2B_1} \equal{} \frac {BD}{DC}$ by similar triangles. Simiarly, $ \frac {A_1B_2}{B_2C_1} \equal{} \frac {CE}{EA}$, $ \frac {B_1C_2}{C_2A_1} \equal{} \frac {AF}{FC}$. The result follows by Ceva.

Dear Fang-jl and Mathlinkers,
let A', B', C' be the feet of the A, B, C-altitude of ABC, K the Lemoine's point of ABC, B1, C1 the midpoints of CA, AB, and U, V the meet points of A'B' with A1C1,  B1C1.
According to "Another unlikely concurrence" (http://perso.orange.fr/jl.ayme  vol.1), AU and AV go through K.
According to Pappus's theorem, Pappus, KIM) applied to the hexagone AUC1VBA'A,  A1, K and A2 are collinear.
….
Sincerely
Jean-Louis

Dear Darij Grinberg,

Let A1A3, B1B3, C1C3 be the altitudes of triangle A1B1C1.
A1A3 = AA2 and A1A3//AA2 therefore AA2A1A3 is parallelogram. Otherwise AC1A1B1 is also parallelogram. It means A2 is reflection of A3 in midpoint of B1C1. Similarly with B2, C2.
A1A3, B1B3, C1C3 are concurrent therefore A1A2, B1B2, C1C2 are concurrent at isotomic conjugate of orthocenter of A1B1C1.

Best regards,
Bui Quang Tuan

[quote="jayme"]AU and AV go through K.
[/quote]
Sorry but I think it is wrong since $ U,V$ are on $ A'B'$ and $ A'B'$ doesn't pass $ A$.
Anyway, can you please use Latex?

Sorry, I made a typos: read BV and not AU
Sincerely
Jean-Louis

It is also true for any arbitrary point $ P$ inwardly (or outwardly) to $ \bigtriangleup ABC,$ instead of the orthocenter and its cevian triangle $ \bigtriangleup A'B'C',$ as a direct application of the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=6579]Cevian Nests Theorem[/url], because of the midpoints $ A_{2},\ B_{2},\ C_{2}$ of $ AA',\ BB',\ CC'$ respectively, lie on $ B_{1}C_{1},\ A_{1}C_{1},\ A_{1}B_{1}$ and $ AA_{2}\cap BB_{2}\cap CC_{2}\equiv P.$

Kostas Vittas.

[b][size=100][color=DarkBlue]GENERAL PROBLEM. – A triangle $ \bigtriangleup ABC$ is given and let $ \bigtriangleup A'B'C'$ be, its cevian triangle, with respect to an arbitrary point $ P$ inwardly to it. Let $ A_{1},\ B_{1},\ C_{1}$ be, the midpoints of the side-segments $ BC,\ AC,\ AB$ respectively and also let $ A_{2},\ B_{2},\ C_{2}$ be, the midpoints of the segments $ AA',\ BB',\ CC'$ respectively. Prove that the line segments $ A_{1}A_{2},\ B_{1}B_{2},\ C_{1}C_{2},$ are concurrent at one point.[/color][/size][/b]

[b][size=100]PROOF.[/size][/b] – It is clear that the points $ A_{2},\ B_{2},\ C_{2},$ lie on $ B_{1}C_{1},\ A_{1}C_{1},\ A_{1}B_{1}$ respectively and so, because of $ B_{1}C_{1}\parallel BC,$

based on the [b][size=100]Thales theorem[/size][/b], we have $ \frac{A_{2}B_{1}}{A_{2}C_{1}} \equal{} \frac{A'C}{A'B}$ $ ,(1)$

Similarly we have $ \frac{B_{2}C_{1}}{B_{2}A_{1}} \equal{} \frac{B'A}{B'C}$ $ ,(2)$ and $ \frac{C_{2}A_{1}}{C_{2}B_{1}} \equal{} \frac{C'B}{C'A}$ $ ,(3)$

From $ (1),$ $ (2),$ $ (3)$ $ \Longrightarrow$ $ \frac{A_{2}B_{1}}{A_{2}C_{1}}\cdot \frac{B_{2}C_{1}}{B_{2}A_{1}}\cdot \frac{C_{2}A_{1}}{C_{2}B_{1}} \equal{} \frac{A'C}{A'B}\cdot \frac{C'B}{C'A}\cdot \frac{B'A}{B'C} \equal{} 1$ $ ,(4)$

From $ (4),$ based on the [b][size=100]Ceva theorem[/size][/b], we conclude that the line segments $ A_{1}A_{2},\ B_{1}B_{2},\ C_{1}C_{2},$ are concurrent at one point and the proof is completed.

Kostas Vittas.

[quote="vittasko"][b][size=100][color=DarkBlue]GENERAL PROBLEM. – A triangle $ \bigtriangleup ABC$ is given and let $ \bigtriangleup A'B'C'$ be, its cevian triangle, with respect to an arbitrary point $ P$ inwardly to it. Let $ A_{1},\ B_{1},\ C_{1}$ be, the midpoints of the side-segments $ BC,\ AC,\ AB$ respectively and also let $ A_{2},\ B_{2},\ C_{2}$ be, the midpoints of the segments $ AA',\ BB',\ CC'$ respectively. Prove that the line segments $ A_{1}A_{2},\ B_{1}B_{2},\ C_{1}C_{2},$ are concurrent at one point.[/color][/size][/b]
[/quote]

Some results of me, more general problems

[b]Problem 1.[/b] Given triangle $ ABC$, $ A_1B_1C_1$ and $ A_2B_2C_2$ are cevian triangles of $ P_1$ and $ P_2$ let line $ AA_1\cap B_2C_2 \equal{} \{A_3\}$ similarly have $ B_3,C_3$, let $ AA_2\cap B_1C_1 \equal{} \{A_4\}$ similarly have $ B_4,C_4$ prove that $ A_2A_3,B_2B_3,C_2C_3,A_1A_4,B_1B_4,C_1C_4$ concurrent at a point.

If $ P_2\equiv G$ we get the above problem, from problem 1 in triangle $ ABC$ we denote concurrent point by $ H(P_1,P_2,ABC)$,

[b]Problem 2.[/b]Let $ ABC$ and $ A'B'C'$ be two perspective triangles, let $ A'' \equal{} H(B',C',ABC),B'' \equal{} H(C',A',ABC),C'' \equal{} H(A',B',ABC)$ prove that $ ABC$ and $ A''B''C''$ are perspective.

[b]Problem 3.[/b]Let $ ABC$ and $ A'B'C'$ be two perspective triangles, let $ A_1 \equal{} H(B',C',ABC),B_1 \equal{} H(C',A',ABC),C_1 \equal{} H(A',B',ABC), A_2 \equal{} H(B,C,A'B'C'),B_2 \equal{} H(C,A,A'B'C'),C_2 \equal{} H(A,B,A'B'C')$ prove that $ A_1A_2,B_1B_2,C_1C_2$ are concurrent at perspector of $ ABC$ and $ A'B'C'$.


Find a linear map: A: V->V which is not injective, but is surjective.

I figured out that V would better be R[X], R^R or R^inf.

[quote="steven128"]$K[X] \to K[X], f \mapsto f'$, $\mathrm{char}\,K = 0$[/quote]

The example I was thinking of is just  L(a,b,c,d,...) = (b,c,d,....).   So any L(a,0,0,0,...) = 0  so it's not injective,
but it's clearly surjective.

Other solutions are always welcome :)


algebra
polynomial
calculus
integration
function
real analysis
derivative

let $1<p<+\infty$ be a fixed real number. For $n \geq 1$, one defines $S_{n}: L^{p}(\mathbb{T}) \to L^{p}(\mathbb{T})$ by
$S_{n}(f) = \sum_{-n}^{n}\hat f(k) e^{ikx}$
where $\hat f (k)$ is the $k$-th Fourier coefficient of $f$. Show that there is a constant $C_{p}>0$, independant of $n$, such that:
$\| S_{n}f \|_{p}\leq \| f \|_{p}$

PS: from here, it is very easy to conclude that $S_{n}f$ tends towards $f$ in $L^{p}(\mathbb T)$

[quote="alekk"]Source: Riesz ?[/quote]
Marcel Riesz, to be specific.

The approaches I know involve the Hilbert transform (or conjugate function).

That is, for trigonometric polynomials, define an operater $H$ by this:

If $f(x)=\sum_{k}\widehat{f}(k)e^{ikx}$ then $Hf(x)=\sum_{k}(-i\text{sgn}(k))\widehat{f}(k)e^{ikx},$

where $\text{sgn}(k)=\begin{cases}1&k>0\\0&k=0\\-1&k<0\end{cases}.$

Clearly, $\|Hf\|_{2}\le\|f\|_{2}.$ In fact, $\|Hf\|_{2}^{2}=\|f\|_{2}^{2}-|\widehat{f}(0)|^{2}.$ Hence, the operator $H$ extends by continuity to a bounded operator on $L^{2}.$

The theorem of M. Reisz is that the Hilbert transform is a bounded operator from $L^{p}$ to $L^{p}$ for $1<p<\infty.$

That is, there exists $C_{p}$ such that $\|Hf\|_{p}\le\|f\|_{p}.$

Since $H$ is self-adjoint, we can take $C_{p}=C_{p'}$ where $\frac1p+\frac1{p'}=1.$ $C_{p}$ behaves like $\frac1{p-1}$ as $p\to1$ or (what is the same thing) like $p$ as $p\to\infty.$

Riesz himself proved this using a complex variables argument for even integer values of $p,$ then interpolated between them using a complex method of interpolation (the Riesz-Thorin convexity theorem.) The first proof I learned was essentially a prototype for studying Calderón-Zygmund singular integral operators: first prove that $H$ is weak-type (1,1), then use a real method of interpolation (the Marcinkiewicz interpolation theorem) between that and the strong-type (2,2) result we already have. There are other proofs as well.

Two possible sources, both for this proof and for how the boundedness of the Hilbert transform implies the theorem you asked for, the uniform $L^{p}$ boundedness of the partial sum operators, for $p>1,$ would be the Katznelson book on harmonic analysis and the ultimate original source, Zygmund's [i]Trignometric Series[/i].

A quick overview of some Fourier series convergence results. Let $S_{N}(f)$ be the partial sums of the Fourier series for $f$ and let $\sigma_{N}(f)$ be the Cesaro means of the partial sums - that is, $\sigma_{N}(f)=\frac1{N+1}\sum_{n=0}^{N}S_{n}(f).$

1. If $f$ is of bounded variation, then $S_{N}(f)(x)$ converges everywhere, to the average of the left and right hand limits of $f.$ (This builds on work using more restrictive conditions by Dirichlet and Dini in the mid-19th century, but reaching the final form quoted above wouldn't have happened until the development of the concept of bounded variation, along with the Stieltjes integral, late in the 19th century.)

2. There exists a continuous function whose Fourier series diverges (meaning the sequence $S_{N}(f)(x)$ diverges) at some point. (I think this is from the 1890's.)

3. If $f$ is continuous, then the Cesaro means $\sigma_{N}(f)$ converge to $f$ uniformly. (Fejer, sometime between 1895 and 1905).

4. If $f\in L^{p},\ 1\le p<\infty,$ then $\sigma_{N}(f)\to f$ in the $L^{p}$ sense. (This bootstraps on Fejer's work.)

5. If $f\in L^{p}, 1\le p\le \infty,$ then $\sigma_{N}(f)(x)\to f(x)$ pointwise almost everywhere. (This depends on Lebesgue theorem on the differentiation of the integral.)

6. If $f\in L^{2},$ then $S_{N}(f)\to f$ in the $L^{2}$ sense. In some sense, this is a result that was "already there" (least squares best approximations, etc.) but needed the formalization of $L^{2}$ and Hilbert spaces in general by Lebesgue, F. Riesz, and others between 1900 and 1910 to make complete sense.

7. There exists a function $f\in L^{1}$ such that $S_{N}(f)$ does not tend to $f$ in the $L^{1}$ sense. (Kolmogorov? 1920's?)

8. There exists a function $f\in L^{1}$ such that $S_{N}(f)(x)$ diverges almost everywhere. In fact, you can make that everywhere. (I'm pretty sure this is Kolmogorov.)

9. If $f\in L^{p},\ 1<p<\infty,$ then $S_{N}(f)\to f$ in the $L^{p}$ sense. (M. Riesz, 1920's, and the problem that alekk posted.)

10. If $f\in L^{2},$ then $S_{N}(f)(x)\to f(x)$ pointwise almost everywhere. (Conjectured by Lusin around 1915. Spent nearly 50 years on the list of "famous unsolved problems in mathematics." Proved by Lennart Carleson in the early 1960's.)

11. If $f\in L^{p},p>1$ or if $f\in L\log^{2}L,$ then $S_{N}(f)(x)\to f(x)$ pointwise almost everywhere. (Proved by Hunt using Carleson's techniques, within two years of Carleson's result.)

Note that the best available result concerning the pointwise convergence of the Fourier series of a continuous function is Carleson's theorem, #10.

Thank you very much Kent for all these explanations (what is $L \log L$ ?)
The approach I know shows directly, but it's quite awkward, that the Hilbert transform is bounded from $L^{2m}$ to $L^{2m}$; where $m>1$ is an integer. But I don't like it since it is quite artificial. This is the same conclusion with real interpolation. I will try to show that the Hilbert transform is of weak type (1;1).

$L\log L$ consists of all functions $f$ such that $|f|\log^{+}|f|$ is integrable. Think of it as lying "in between" $L^{1}$ and $L^{p},p>1.$

One thing that will prove useful in dealing with the weak (1,1) boundedness result for the Hilbert transform. Suppose $a(x)$ is supported on an interval $I$ and $\int_{I}a=0.$ Then for $x$ that lie outside of an interval twice the length of $I$ and with the same center (call that center $x_{0}$),

$|H(a)(x)|\le C\frac{\|a\|_{1}}{(x-x_{0})^{2}}$

Is $f\in L\log L$ already known to be sufficient? According to [url=http://www.arxiv.org/abs/math.CA/0307008]Lacey's survey[/url], convergence a.e. holds in $L\log L\log\log\log L$ and fails anywhere below $L\sqrt{\log L/\log\log L}$.

Hmm ... what exactly did Hunt prove? Maybe $L\log^{2}L?$ I know he did $L^{p},p>1.$

Yes, Hunt did $L\log^{2}L$, according to MR review of his 1968 paper. Interesting contrast here: Carleson's 1966 paper was published in [url=http://www.actamathematica.org/]one of the top journals[/url] (no surprise), but Hunt's paper is in the proceedings of a conference held in Southern Illinois University, Edwardsville. Such papers tend to be discounted at annual reviews. SIUE is a fine university, though - have been there and would like to go back.

I'll add one piece of attitude for my part: insisting on partial sums is a narrow-minded way to approach series, and a more flexible attitude towards how to extract a sum from a series is ultimately more useful. The Carleson-Hunt results are a nice historical curiosity and a triumph of analytic technique. However, I see the analytical muscle - the results that say that this whole business of Fourier analysis really works, means something and can be used - as the Fejer-type results (#3, 4, and 5 on my list above) and related results for Abel means, non-tangential limits of harmonic functions, and so on.


A graph has 1991 points and every point has degree at least 1593. Show that there are six points, each of which is joined to the others. Is 1593 the best possible?


Intr-un triunghi $\ ABC$ avem echivalenta
$\cos\frac{A}{2}=2\sin\frac{\pi-A}{4}\cos\frac{B}{2}\Leftrightarrow\ B=C$

Dupa cum stim cu totii  :D exista formula $sin\ x=2sin\frac{B}{2}cos\frac{B}{2}\ (1)$
Reciproca e destul de simpla.
Scriem $A=\pi-2B$ si iese:
$cos(\frac{\pi}{2}-B)=2sin\frac{B}{2}cos\frac{B}{2}$
$sin\ B=2sin\frac{B}{2}cos\frac{B}{2}$, adevarat conform relatiei $(1)$.
Reciproca:
Fie $A=\pi-2B+x$
De aici inlocuim pe $A$ in relatia de demonstrat, inmultind cu $sin\frac{B}{2}$. Deci:
$sin\frac{B}{2}\cdot\ cos(90-B+\frac{x}{2})=sin\ B\cdot\ sin\frac{2B-x}{4}$
$sin\frac{B}{2}\cdot\ sin(B-\frac{x}{2})=sin\ B\cdot\ sin\frac{2B-x}{4}$
De unde, cu $(1)$:
$sin\frac{B}{2}\cdot\ sin\frac{2B-x}{4}cos\frac{2B-x}{4}=sin\frac{B}{2}\cdot\ cos\frac{B}{2}\cdot\ sin\frac{2B-x}{4}$
De aici:
$sin\frac{B}{2}\cdot\ sin\frac{2B-x}{4}(cos\frac{2B-x}{4}-cos\frac{B}{2})=0$
De aici avem mai multe cazuri:
I. $B=0$ imposibil
II.$2B=x$ nu convine
III.$x=0\Rightarrow\ A=\pi-2B\Rightarrow\ B=C$ q.e.d.

Am mai multe comentarii...
1)  Tudor, pentru functiile trigonometrice foloseste \sin,\cos,\tan,\cot, ca sa-ti apara $\sin,\cos,\tan,\cot$ :)
2) [hide="solutie"]
$\cos\frac{A}{2}=2\sin\frac{\pi-A}{4}\cos\frac{B}{2}\Leftrightarrow\sin\frac{B+C}{2}=2\sin\frac{B+C}{4}\cos\frac{B}{2}\Leftrightarrow\ 2\sin\frac{B+C}{4}\cos\frac{B+C}{4}=2\sin\frac{B+C}{4}\cos\frac{B}{2}\Leftrightarrow\frac{B+C}{4}=\frac{B}{2}\Leftrightarrow\ B=C$, cctd[/hide]


In a convex pentagon $ ABCDE$ side $ AB$ is perpendicular to $ CD$ and side $ BC$ is perpendicular to $ DE$. Prove that if $ AB \equal{} AE \equal{} ED \equal{} 1$, then
$ BC \plus{} CD < 1$.


Use Stewart's theorem to deduce the formula of a median.

well Stewarts theorem says:
In triangle $\triangle ABC$, if $D$ lies on the line $BC$ and the segments have lengths
$|AB|=c, |BC|=a, |CA|=b, |AD|=d, |BD|=m, |DC|=n,$ then
 if $D$ lies between $B$ and $C$, then $a(d^2+mn) = mb^2+nc^2,$
 if $D$ lies beyond $C$, then $a(d^2-mn) = mb2-nc^2,$
 if $D$ lies beyond $B$, then $a(d^2-mn) =-mb^2+nc^2.$
(the proof of this is based on the Cosine law)

in the case where $AD$ is the median we have:

If $AD$ is a median of $\triangle ABC$, then $|AD|^2 = \frac{(2|AB|^2+2|AC|^2-|BC|^2)}{4}$.
If the other medians are$ BE $and$ CF$, then
$|AD|^2+|BE|^2+|CF|^2 = 3\frac{(|AB|^2+|BC|^2+|CA|^2)}{4}$.

Let d be the median to side BC (length a) in triangle ABC, splitting BC into lengths m and n (m closer to AC).
Then we have $bmb+cnc=dad+man$.
$b*\frac{a}{2}*b+c*\frac{a}{2}*c=d*a*d+\frac{a}{2}*a*\frac{a}{2}$
$2b*b+2c*c=4d*d+a*a$
$d=\sqrt{\frac{2b^2+2c^2-a^2}{4}}$


[quote="sax"]The real numbers a, b satisfy the equations
a^3 - 3a^2 + 5a-17 = 0
b^3 - 3b^2 + 5b + 11 = 0:
Find a + b.[/quote]
VERY NICE problem :)
[hide="my solution"]
$ (a \minus{} 1)^3 \plus{} 2a \minus{} 16 \equal{} 0$
$ (b \minus{} 1)^3 \plus{} 2b \plus{} 12 \equal{} 0$
Adding them we got:
$ (a \minus{} 1)^3 \plus{} (b \minus{} 1)^3 \plus{} 2(a \plus{} b \minus{} 2) \equal{} 0 \Longleftrightarrow (a \plus{} b \minus{} 2)[(a \minus{} 1)^2 \minus{} (a \minus{} 1)(b \minus{} 1) \plus{} (b \minus{} 1)^2] \plus{} 2(a \plus{} b \minus{} 2) \equal{} 0 \Longleftrightarrow (a \plus{} b \minus{} 2) \equal{} 0$
or
$ (a \minus{} 1)^2 \minus{} (a \minus{} 1)(b \minus{} 1) \plus{} (b \minus{} 1)^2 \plus{} 2 \equal{} 0 \Longleftrightarrow a^2 \minus{} 2a \plus{} 1 \minus{} ab \plus{} a \plus{} b \minus{} 1 \plus{} b^2 \minus{} 2b \plus{} 1 \plus{} 2 \equal{} 0 \Longleftrightarrow a^2 \plus{} b^2 \minus{} a \minus{} b \minus{} ab \plus{} 3 \equal{} 0 \Longleftrightarrow (a \minus{} b)^2 \plus{} (a \minus{} 1)^2 \plus{} (b \minus{} 1)^2 \plus{} 4 \equal{} 0$ which is contradiction.
So $ a \plus{} b \equal{} 2$
 :) [/hide]

ευχαριστώ. ωραία λύση

[quote="sax"]βασικά όμως θα έπρεπε να ήταν (α-1)^3-3α^2+[u][b]10[/b][/u]=0
α+β=3[/u][/b][/quote]
Για κοίτα το καλήτερα.... :P
Που το βρήκες το πρόβλημα???


If $ 4^{27}+4^{100}+4^x , x\in \math {Z}$ is a perferct square , what is the

greatest value of $ x$?

:roll:

[quote="PKMathew"]If $ 4^{27} + 4^{100} + 4^x , x\in \math {Z}$ is a perferct square , what is the

greatest value of $ x$?

:roll:[/quote]

Let $ y=x-27$, then the expression $ =(2^{27})^2 (2^{2y}+2(2)^{145}+1)$

$ (2^{2y}+2(2)^{145}+1)>(2^y)^2$

So $ (2^{2y}+2(2)^{145}+1) \ge (2^y+1)^2=2^{2y}+2(2^y)+1$
$ \Leftrightarrow 145 \ge y$, it's easy to see that $ y=145 \Leftrightarrow x=172$ satisfies the condition.

So the greatest value of $ x$ is $ 172$


Given the invertible NxN matrices A, B, C, and D, show that $ (ABCD)^{\minus{}1}\equal{} D^{\minus{}1}C^{\minus{}1}B^{\minus{}1}A^{\minus{}1}$

Just left-multiply both sides by ABCD. LHS is I by definition of inverse. By associativity, the RHS becomes I. Similarly for right-multiplication.


factorial
function
algebra
domain
limit
floor function
number theory

Define a function $f_{n}(x)$ with a domain of the set of prime numbers to be the exponent of $x$ in the prime factorization of $n!$.  For example, $f_{5}(3)=1$ since $5!=2^{3}\times3^\mathbf{1}\times5^{1}$.  Compute $\lim_{n\to\infty}\frac{f_{n}(2)}{f_{n}(5)}$.

[hide="Hint"]As $n$ goes up by one, what is the expected value that $f_{n}(2)$ increases by?  Consider that.[/hide]

If mods think this is too hard, please move it.

[hide="Well..."] $f_{n}(k) = \lfloor \frac{n}{k}\rfloor+\lfloor \frac{n}{k^{2}}\rfloor+...$

Asymptotically (although this is not totally rigorous) we have

$f_{n}(k) \approx \frac{n}{k-1}$

Hence

$\lim_{n \to \infty}\frac{f_{n}(2)}{f_{n}(5)}= \boxed{ 4 }$. [/hide]

I figure..

[hide]

If  and we use the identity  so we add this across the first  partial sum or whatever and we get

 but the left and right sides are asymptotically the same so we put the SQUEEZE on  to get that it's asymptotically to the things, so it's asymptotic to the thing on the right and that's  but as  goes to infinity we have  as our leftovers and that's awesome.

[/hide]


[quote]

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=24555

Now, to prove this inequality, set $ a \equal{} y\plus{}z, b \equal{} z\plus{}x, c \equal{} x\plus{}y$. Then, some calculation yields that $ 2\left(x^{2}\plus{}yz\right)\equal{}2bc\minus{}ca\minus{}ab\plus{}a^{2}$

[/quote]

I cannot see how you come to $ 2bc\minus{}ca\minus{}ab\plus{}a^{2}$. Can anybody help me?

$ 2bc\minus{}ca\minus{}ab\plus{}a^{2}$
$ \equal{}2\left(z\plus{}x\right)\left(x\plus{}y\right)\minus{}\left(x\plus{}y\right)\left(y\plus{}z\right)\minus{}\left(y\plus{}z\right)\left(z\plus{}x\right)\plus{}\left(y\plus{}z\right)^{2}$
$ \equal{}2\left(x^{2}\plus{}yz\right)$.

  Darij

Notice that $ x \equal{}\frac{1}2 (b\plus{}c\minus{}a)$, $ y \equal{}\frac{1}2 (c\plus{}a\minus{}b)$ and $ z \equal{}\frac{1}2 (a\plus{}b\minus{}c)$.

[quote="tmrfea"][quote="darij grinberg"]$ 2bc\minus{}ca\minus{}ab\plus{}a^{2}$
$ \equal{} 2\left(z\plus{}x\right)\left(x\plus{}y\right)\minus{}\left(x\plus{}y\right)\left(y\plus{}z\right)\minus{}\left(y\plus{}z\right)\left(z\plus{}x\right)\plus{}\left(y\plus{}z\right)^{2}$
$ \equal{} 2\left(x^{2}\plus{}yz\right)$.

  Darij[/quote]

First thank you for the express answer.

I see that, but what if you have to start from $ 2\left(x^{2}\plus{}yz\right)$. How do you most elegantly obtain $ 2bc\minus{}ca\minus{}ab\plus{}a^{2}$?[/quote]

It's not a question of elegancy. Just write $ x$, $ y$, $ z$ in terms of $ a$, $ b$, $ c$ (this amounts to solving the linear equations $ a \equal{} y\plus{}z$, $ b \equal{} z\plus{}x$, $ c \equal{} x\plus{}y$ for $ x$, $ y$, $ z$ as unknowns, and the result is $ x \equal{}\frac{b\plus{}c\minus{}a}{2}$, $ y \equal{}\frac{c\plus{}a\minus{}b}{2}$, $ z \equal{}\frac{a\plus{}b\minus{}c}{2}$), and use the resulting expressions to rewrite $ 2\left(x^{2}\plus{}yz\right)$.

[PS. I'm going to move this thread into Pre-Olympiad.]

  Darij


Compute the celullar homology of $ P^2 \times P^2$ and $ T^2 \times T^2$, where $ T^2$ is the $ 2$-torus and $ P^2$ the two dimensional real projective space.

[quote="ZetaX"]I think you can do that calculations yourself  :wink:
You may know that $ H_k(A \times B) \equal{} H_k(A) \times H_k(B)$ applies for you.[/quote]

why is that true?   In general, that is false.

[quote="Hanno"]The cellular complex of the product of two CW-complexes is isomorphic to the tensor product of the cellular complexes of its factors.[/quote]

That does not mean that it's homology factors in that nice way, though. (if this is a response to the above post).

For products of spaces, kunneth theorems are useful:

http://en.wikipedia.org/wiki/K%C3%BCnneth_theorem

In the above cases, you can simply calculate the cellular chain complexes. You do not need Kuenneth.

Of course, but Kunneth can save a lot of time.


find all 10-digit numbers so that:

- the first digit is the number of 1's
- the second digit is the number of 2's
- ...
- the ninth number is the number of 9's
- the tenth number is the number of 0's

cheers :)

i found one *very proud*  :P
[hide]
$2100010006$[/hide]

I think we can establish the following in general for a self-describing number with $n+1$ digits in base $b$ where $n+1\leq b$.  A slight modification to the number makes it
\[a_0a_1a_2...a_n\]
where $a_k$ is the number of $k's$.  Anyway, here is what we know
$a_0$ is nonzero
$a_n$ is zero
$\displaystyle \sum_{i=0}^n a_i=\sum_{i=0}^n ia_i=n+1$

Of course there are more restrictions/equivalences and anybody is welcome to add more.  This gives a basic start

How do you prove 2100010006 is the only number, peeta? ;)

It is indeed possible to construct the general relation for a M-digit integer base M, but it is completely analogue to the the 10-digit case, so I would stick on this one. :)

Nobody knows a source or anything?

[quote="Arne"]I thought the source was "Peter VDD"? :D[/quote]Sorry, I'm not [i]that[/i] old. ;)

I meant: is it already on the board somewhere?

i think i found the important point of the question:

[hide]
the sum of the digits is excactly 10. dont have time anymore to write the entire solution, will do later if nobody else did already
[/hide]

as i said ill give the solution:

[hide]
well the solution is not very beautiful and bruteforce than i would like but anyway, wecan reduce the bf- factor by using that the sum of digits must be $10$.

first the sum of digist is $10$ because each digit stands for the number of digits equal to its position. since there are 10 digits the sum of all those numbers must be 10 as well.

second we go through all possible values of the last digit:
9: since there must be at least a 1 on the ninth digit, and therefore another number somewhere the sum is greater 10 $\implies$ not possible
8: there must at least be a 1 on the 8th digit. hence there must be at least a 2 on the 1st digit, because if there would be a 1 it be a contradiction. so the sum is greater than 10.
7: same argument as for 8, but in addition there must be at least be a 1 on the 2nd digit or later, so the sum is larger 10.
6: is special and we look at it later
5: there must be a 0 on the 6th-9th digit otherwise the sum would be too large. also because there are only 5 zeros there must be 4 digit greater zero. since one of them is the 5th digit either the 3rd or the 4th digit must be greater than 0. hence there would be a an additional 3 or 4 somewhere. and another two digits greater zero. but 5+1+3+1+1=11>10 is to big.
4: the the 6th-9th digits must be zero because otherwise the sum would be too large. hence all the other digits are larger 0. which leads to that there must be a 5 and a 3 which with 4 is 12, which is too large.
3-1: the are 3-1 zeros so there are so there are 6-8 digits greater zero and the sums are in any case way too large.
0: there cant be zero zeros because than there would be a zero ...

6: the 7th-9th digit must be 0 out of reasons explained above. the 6th digit must be 1 because there is at least one, but if there would be more than one sum sum be greater than 12. if the 3rd-5th digit be 1 or larger than 1+3+1+6=11>10. so they are 0. now for the 1st and second digit, since its sum must be 10-6-1=3, that the only possiblity is 1 and 2, i.e. the first digit is 2 and the second is 1. this is therefore the only working solution.

[/hide]

m-m.

try this: $\sum a_i = 10$ and $\sum ia_i = 10$, with a_i the i-th digit :) (and 0 = 10)

I have a shorter, but still brute-force solution.
[hide]

First digit = 3:
3 _ _ _ _ _ _ _ _ _
1 2 3 4 5 6 7 8 9 0
That means  we at least have a 3 and three ones; that makes a sum of 6.
However, the three ones means we have a one in one of these spots: 2,3,4...0. That means thee different elements of the set {2,3,4,5,6...0} are also digits of our number.  Consider the minimum: we have a 1 in the 2 spot, 3 spot, and 0 spot.  That would mean we would have a zero, 2, and 3 in our number, making the sum 3+1+1+1+2+3+0 = 11.  So the digit sum of any such number would be greater than or equal to 11.  However, the digit sum must be 10.  So the first digit can't be 3.  Similar logic eliminates all cases with first digit > 3.  That means we only have to check first digit = 1 or 2, which is pretty easy.[/hide]

that's not really brute force :)

throw in $\sum ia_i = 10$ and you got my solution :)


Hello

Anyone who can prove this;
 $ \prod_{i \equal{} 1 }^{n } (1 \minus{} {{a}_{i }}^{k}) / {{a}_{i } }^{k} \geq {({n}^{k} \minus{} 1)}^{n}$
 if
 $ 0 < {a}_{i}$
and
  $ \sum_{i \equal{} 1}^{n} {a}_{i} \equal{} 1$

I am especially interested in if there are any approaches with chebyshevs inequality.


function
real analysis
real analysis unsolved

Find the minimum value of the function [tex]f(x)=\frac{(x+1)^2(x^{2(n-2)}+2x^{2(n-3)}+\cdots+n-1)}{x(x^{n-2}+2x^{n-3}+\cdots+n-1)}[/tex]. I would like the value of x for which the minimum value occurs, in terms of n.

I did smth and I think that it's not very hard to find the minimum now
We can restrict those sums as follows
$\displaystyle x^{n-2}+2x^{n-3}+...+(n-2)x+n-1 = \frac{x^n-1-n(x-1)}{(x-1)^2}$
and the other one is
$\displaystyle x^{2(n-2)}+2x^{2(n-3)}+...+(n-2)x^2+n-1 = \frac{x^{2n}-1-n(x^2-1)}{(x^2-1)^2}$
and now the function is $f(x)=\displaystyle\frac{x^{2n}-1-n(x^2-1)}{x^n-x-nx(x-1)}$

No my friend. Now I need the minimum, in terms of n, and it's not simple at all.

[quote="pcalin"]Find the minimum value of the function $\displaystyle f(x)=\frac{(x+1)^2(x^{2(n-2)}+2x^{2(n-3)}+\cdots+n-1)}{x(x^{n-2}+2x^{n-3}+\cdots+n-1)}$. I would like the value of $x$ for which the minimum value occurs, in terms of $n$.[/quote]

 :)


function
inequalities
algebra solved
algebra

Find the function:
\[ f:R_+ \rightarrow R \]
\[ f(x+y)=f(x^2+y^2) \]
[hide]i think this function is fixed (i mean f(x)=k)[/hide]

Take y = -x and you get f(2x^2) = f(0), whence the result.

i think tou made a mistake beacause x,y are positive numbers so we can't put y=-x and your notation is incorrect. :!:

[hide]

imo , this function should be a constant function

We see that for any positive real $t$ , we can represent it in $\frac{t}{p}+\frac{t}{q}$ where $p,q\in \mathbb{R^+}$ and $\frac 1p+\frac 1q = 1$

So $f(t)=f\left(\frac{t}{p}+\frac{t}{q}\right)=f(rt^2)$ for $r\in \mathbb{R^+}$

form the identity , we know $r=\frac{1}{p^2}+\frac{1}{q^2}=1-2\cdot\frac{1}{pq}\geq \frac{1}{2}$   (*from AM-GM)

So this means for every big number , we can always reduce it using $f(rt^2)=f(t)$ and this implies the function can only be a constant .  :roll: [/hide]

your solutions are very nice and i realy enjoy it.very nice.excellent!... :D

[quote="Arne"]We have \[f(a) = f\left(\frac{a + b}{2} + \frac{a - b}{2}\right) =  f\left(\left(\frac{a + b}{2}\right)^2 + \left(\frac{a - b}{2}\right)^2\right) \\ =  f\left(\left(\frac{a + b}{2}\right)^2 + \left(\frac{b - a}{2}\right)^2\right) = f\left(\frac{b + a}{2} + \frac{b - a}{2}\right) = f(b)\] for any reals $a, b$. Hence result.[/quote]

your idea it's nice :!: but if a-b/2>=0 then b-a/2<=0  and your solution are wrong

Well, I don't know, $\frac{b - a}{2}$ is negative, but $\frac{b - a}{2} + \frac{a + b}{2} = b$ is still positive, and Jensen didn't mention (in the problem statement, I mean) whether $x$ and $y$ must be positive reals, or whether $x + y$ must be positive.

yes i think you are right Arne,and your solution  still nice and cool. :D

hi jensen...
as I told you in class it was a really nice problem and also it's more beautiful solution!!!
but i think arne has solved it some where else too...
 :P  :D


limit
integration
algebra proposed
algebra

[quote="langthang123"]$ S_n \equal{} \sum _k \frac {1}{\sqrt {16n^2 \plus{} 9k}} , k \equal{} 1, 2, 3, ........., n$
$ Lim S_n \equal{} ?$ :?:[/quote]

We have $ S_n\equal{}\frac{1}{4n}\sum_{k\equal{}1}^n\frac{1}{\sqrt{1\plus{}\frac{9k}{16n^2}}}$

We also have $ 1\minus{}\frac{x}{2}\leq\frac{1}{\sqrt{1\plus{}x}}\leq 1$ $ \forall x\geq 0$

And so : $ \frac{1}{4}\minus{}\frac{9}{128n^3}\sum_{k\equal{}1}^nk\leq S_n\leq \frac{1}{4}$

And so $ \lim_{n\rightarrow\plus{}\infty}S_n\equal{}\frac{1}{4}$

[quote="langthang123"][quote]
And so : $ \frac {1}{4} \minus{} \frac {9}{128n^3}\sum_{k \equal{} 1}^nk\leq S_n$

[/quote]
Are you sure ?[/quote]
$ \frac{1}{\sqrt{1\plus{}x}}\geq 1\minus{}\frac{x}{2}$

$ \frac{1}{\sqrt{1\plus{}\frac{9k}{16n^2}}}\geq 1\minus{}\frac{9k}{32n^2}$

$ \sum_{k\equal{}1}^n\frac{1}{\sqrt{1\plus{}\frac{9k}{16n^2}}}\geq \sum_{k\equal{}1}^n1\minus{}\sum_{k\equal{}1}^n\frac{9k}{32n^2}$

$ 4nS_n\geq n\minus{}\frac{9}{32n^2}\sum_{k\equal{}1}^nk$

$ S_n\geq \frac{1}{4}\minus{}\frac{9}{128n^3}\sum_{k\equal{}1}^nk$

I am sure.

What happen if I change "16n^{2}" ---->"5n^{2}" or "20n^{2}"....

and I wonder why you could choose "$ \frac {1}{4} \minus{} \frac {9}{128n^3}\sum_{k \equal{} 1}^nk\leq S_n$"

Can You explain to me?

[quote="wallhieu"]I think Lim S_n = 2/3[/quote]

1) Theory point of view : I think my demo is OK. Show me any fault you discovered.

2) real life point of view :
$ S_1\equal{}0.2$
$ S_2\equal{}0.2274\ldots$
$ S_5\equal{}0.2420\ldots$
$ S_{10}\equal{}0.2462\ldots$
$ S_{50}\equal{}0.2492\ldots$
$ S_{100}\equal{}0.2496\ldots$
$ S_{200}\equal{}0.2498\ldots$

And I think the limit is really $ \frac{1}{4}$  and not $ \frac{2}{3}$

[quote="langthang123"]What happen if I change "16n^{2}" ---->"5n^{2}" or "20n^{2}"....
[/quote]
If $ S_n(a)\equal{}\sum_{k\equal{}1}^n\frac{1}{\sqrt{an^2\plus{}9k}}$ then we have $ S_n(a)\equal{}\frac{1}{\sqrt{a}n}\sum_{k\equal{}1}^n\frac{1}{\sqrt{1\plus{}\frac{9k}{an^2}}}$

We then have $ 1\minus{}\frac{9k}{2an^2}\leq\frac{1}{\sqrt{1\plus{}\frac{9k}{an^2}}}\leq 1$
And so $ n\minus{}\frac{9}{2an^2}\sum_{k\equal{}1}^nk\leq\sum_{k\equal{}1}^n\frac{1}{\sqrt{1\plus{}\frac{9k}{an^2}}}\leq n$
And so $ \frac{1}{\sqrt{a}}\minus{}\frac{9}{2a\sqrt{a}n^3}\sum_{k\equal{}1}^nk\leq S_n(a)\leq \frac{1}{\sqrt{a}}$

And so $ \lim_{n\rightarrow \plus{}\infty}S_n(a)\equal{}\frac{1}{\sqrt{a}}$

[quote="langthang123"]I wonder why you could choose "$ \frac {1}{4} \minus{} \frac {9}{128n^3}\sum_{k \equal{} 1}^nk\leq S_n$"

Can You explain to me?[/quote]

What point is not clear in my post explaining this ?

oh, I understood. Now, can you solve this problem?
$ S_n \equal{} \sum _k \frac{1}{\sqrt{16n^2 \plus{} 9nk}}$

[quote="langthang123"]oh, I understood. Now, can you solve this problem?
$ S_n \equal{} \sum _k \frac {1}{\sqrt {16n^2 \plus{} 9nk}}$[/quote]

This one is simpler :

$ S_n \equal{}\frac{1}{n}( \sum _{k\equal{}1}^n \frac {1}{\sqrt {16 \plus{} 9\frac{k}{n}}})$

And it is easy to show that $ \lim_{n\rightarrow\plus{}\infty}\frac{1}{n}( \sum _{k\equal{}1}^n \frac {1}{\sqrt {16 \plus{} 9\frac{k}{n}}})$ $ \equal{}\int_0^1\frac{dx}{\sqrt{9x\plus{}16}}$ $ \equal{}\frac{2}{9}$


Let be $ a,b,c>0$ . Then prove that :

$ \frac{a\plus{}b\plus{}2c}{(a\plus{}2b\plus{}3c)^2}\plus{}\frac{b\plus{}c\plus{}2a}{(b\plus{}2c\plus{}3a)^2}\plus{}\frac{c\plus{}a\plus{}2b}{(c\plus{}2a\plus{}3b)^2}\le \frac{1}{8}\left(\frac{1}{\sqrt{ab}}\plus{}\frac{1}{\sqrt{bc}}\plus{}\frac{1}{\sqrt{ca}}\right)$

[hide]First make the substitutions $ x\equal{}a\plus{}c$, $ y\equal{}b\plus{}c$, and $ z\equal{}a\plus{}b$.  Then we have $ \frac{x\plus{}y}{(x\plus{}2y)^2} \plus{} \frac{z\plus{}x}{(z\plus{}2x)^2} \plus{} \frac{y\plus{}z}{(y\plus{}2z)^2} \le \frac{1}{8}\left(\frac{1}{\sqrt{ab}} \plus{} \frac{1}{\sqrt{ac}} \plus{}\frac{1}{\sqrt{bc}}\right)$.  By AM-GM we have $ x\ge2\sqrt{ac}$, so $ \frac{2}{x}\le\frac{1}{\sqrt{ac}}.$  Therefore, we have our result if we can prove $ \frac{x\plus{}y}{(x\plus{}2y)^2} \plus{} \frac{z\plus{}x}{(z\plus{}2x)^2} \plus{} \frac{y\plus{}z}{(y\plus{}2z)^2} \le \frac{1}{4} \left(\frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z}\right)$ for $ x,y,z > 0$.  This follows from $ \frac{x\plus{}y}{(x\plus{}2y)^2}\le\frac{1}{4y}$ which can be easily shown to be true because $ x^2>0$.  I probably made a mistake somewhere because this proof works for a strict inequality.[/hide]


I have the following task in my book:

Determine the order of amino acids if the corresponding part of the molecule of the gene consists of the following row of nucleotides A-C-G-C-C-C-A-T-G-G-C-C-G-G-T, where A- adenine, C- cytosine, G- guanine, T- thymine.

And there is a rerference to THE TABLE OF GENETIC CODE. How to use it?

Assuming this is 5' - 3' downstream direction on a DNA strand, then you should analyze each codon (base triplets) at a time on the chart. Match up each codon with its amino acid. The peptide chain will be in the series matching the DNA sequence's codons.

If your book does not have a DNA codon based Amino Acid conversion chart, you must convert the DNA strand into its complementary RNA strand.

If this is already an RNA strand, just consult your chart.

It was a mistake in the book so it was not possible to solve the task. But anyway, thanks for answering! :)


[quote]...First, make sure that the bar near the top of the TeXnicCenter window reads "LaTeX => PDF"; if it doesn't, change it so that it does...[/quote]

I don't understand what it means by "the top of TeXnicCenter window." Does it have to be in PDF form before you can compile? (I tried compiling, but nothing showed up.)

[quote="dragon96"]I tried compiling, but nothing showed up..[/quote]Did you view the file? See green oval in picture below or press F5. If that doesn't work you need to say in detail what you did, including the file you compiled and any errors you got.
[quote="dragon96"]I don't understand what it means by "the top of TeXnicCenter window."[/quote]See red oval below.

When I compiled it, there was no 'No error' message. So I suppose I can't view it if its not compiled.

TexnicCenter always gives information in the bottom window, whatever it does. What does it say? Also post the full tex file and the log file. Are any other files are created?

Well you need to set the bar to PDF before you compile. I think its set to DVI at default.

Yeah, click the dropdown menu and click PDF.